[
  {
    "id": 15001,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $n$ such that the three-variable polynomial\n\n$$\nP_n(x, y, z) = (x - y)^{2n}(y - z)^{2n} + (y - z)^{2n}(z - x)^{2n} + (z - x)^{2n}(x - y)^{2n}\n$$\n\ndivides the three-variable polynomial\n\n$$\nQ_n(x, y, z) = \\left[(x - y)^{2n} + (y - z)^{2n} + (z - x)^{2n}\\right]^{2n}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The only such positive integer is $n = 1$. It is easy to verify that $Q_1 = 4P_1$.\n\nWe now show that $P_n$ does not divide $Q_n$ for $n \\ge 2$.\n\nSuppose that $P_n \\mid Q_n$, so that $Q_n(x, y, z) = R_n(x, y, z)P_n(x, y, z)$. Define $p_n(x) = P_n(x, 0, -1)$ and $q_n(x) = Q_n(x, 0, -1)$. Then, $q_n = R_n(x, 0, -1)p_n$, so $p_n \\mid q_n$ as real polynomials in $x$. Furthermore, $p_n$ and $q_n$ both have integer coefficients, and $p_n$ is monic. Hence, $R_n(x, 0, -1)$ has integer coefficients as a polynomial in $x$, and in particular, it evaluates to an integer whenever $x$ is an integer. Thus, for any integer $a$, we have $p_n(a) \\mid q_n(a)$ as integers.\n\nPlugging in $a = 1$, we find that $2^{2n+1} + 1 \\mid (2^{2n} + 2)^{2n}$. Now, suppose $p$ is a prime such that $p \\mid 2^{2n+1} + 1$. Then $p \\mid 2^{2n} + 2$, and so\n\n$$\n0 \\equiv 2^{2n+1} + 1 - 2(2^{2n} + 2) \\equiv -3 \\pmod{p}.\n$$\n\nTherefore, $p = 3$, which means that $2^{2n+1} + 1 = 3^k$ for some $k$.\n\nNote that the left hand side of this equation is always $1 \\pmod{4}$, so $k$ must be even. Letting $k = 2m$, we have $2^{2n+1} = (3^m + 1)(3^m - 1)$. It follows that $3^m + 1$ and $3^m - 1$ are both powers of $2$, whence $m = 1$ and $n = 1$. Thus, $n = 1$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15002,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, \\dots, x_n$ ($n \\ge 2$) be real numbers such that\n$$\nA = \\left| \\sum_{i=1}^{n} x_i \\right| \\neq 0\n$$\nand\n$$\nB = \\max_{1 \\le i < j \\le n} |x_i - x_j| \\neq 0.\n$$\nProve that for every $n$ vectors $\\alpha_1, \\dots, \\alpha_n$ on the plane, there exists a permutation $(k_1, k_2, \\dots, k_n)$ of $(1, 2, \\dots, n)$ such that\n$$\n\\left| \\sum_{i=1}^{n} x_{k_i} \\alpha_i \\right| \\ge \\frac{AB}{2A+B} \\max_{1 \\le i \\le n} |\\alpha_i|.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof**\nLet $|\\alpha_k| = \\max_{1 \\le i \\le n} |\\alpha_i|$. It is sufficient to prove that\n$$\n\\max_{(k_1, \\dots, k_n) \\in S_n} \\left| \\sum_{i=1}^{n} x_{k_i} \\alpha_i \\right| \\ge \\frac{AB}{2A+B} |\\alpha_k|,\n$$\nwhere $S_n$ is the set of all permutations of $(1, 2, \\dots, n)$.\n\nWithout loss of generality, assume\n$$\n|x_n - x_1| = \\max_{1 \\le i < j \\le n} |x_j - x_i| = B,\n$$\n$$\n|\\alpha_n - \\alpha_1| = \\max_{1 \\le i < j \\le n} |\\alpha_j - \\alpha_i|.\n$$\nFor the two vectors\n$$\n\\beta_1 = x_1\\alpha_1 + x_2\\alpha_2 + \\cdots + x_{n-1}\\alpha_{n-1} + x_n\\alpha_n,\n$$\n$$\n\\beta_2 = x_n\\alpha_1 + x_2\\alpha_2 + \\cdots + x_{n-1}\\alpha_{n-1} + x_1\\alpha_n,\n$$\nwe have\n$$\n\\begin{aligned}\n\\max_{(k_1, \\dots, k_n) \\in S_n} \\left| \\sum_{i=1}^{n} x_{k_i} \\alpha_i \\right| & \\ge \\max\\{|\\beta_1|, |\\beta_2|\\} \\\\\n& \\ge \\frac{1}{2} (|\\beta_1| + |\\beta_2|) \\\\\n& \\ge \\frac{1}{2} |\\beta_1 - \\beta_2| \\\\\n& = \\frac{1}{2} |x_1 \\alpha_n + x_n \\alpha_1 - x_1 \\alpha_1 - x_n \\alpha_n| \\\\\n& = \\frac{1}{2} |x_1 - x_n| \\cdot |\\alpha_1 - \\alpha_n| \\\\\n& = \\frac{1}{2} B |\\alpha_n - \\alpha_1|.\n\\end{aligned}\n$$\nNow suppose $|\\alpha_n - \\alpha_1| = x |\\alpha_k|$. Using the Triangle Inequality, we obtain $0 \\le x \\le 2$. So the above becomes\n$$\n\\max_{(k_1, \\dots, k_n) \\in S_n} \\left| \\sum_{i=1}^{n} x_{k_i} \\alpha_i \\right| \\ge \\frac{1}{2} B x |\\alpha_k|.\n$$\nOn the other hand, consider the vectors\n$$\n\\gamma_1 = x_1 \\alpha_1 + x_2 \\alpha_2 + \\dots + x_{n-1} \\alpha_{n-1} + x_n \\alpha_n\n$$\n$$\n\\gamma_2 = x_2 \\alpha_1 + x_3 \\alpha_2 + \\dots + x_n \\alpha_{n-1} + x_1 \\alpha_n\n$$\n$$\n\\gamma_n = x_n \\alpha_1 + x_1 \\alpha_2 + \\dots + x_{n-2} \\alpha_{n-1} + x_{n-1} \\alpha_n.\n$$\nThen we have\n$$\n\\begin{aligned}\n\\max_{(k_1, \\dots, k_n) \\in S_n} \\left| \\sum_{i=1}^n x_{k_i} \\alpha_i \\right| &\\ge \\max_{1 \\le i \\le n} |\\gamma_i| \\\\\n&\\ge \\frac{1}{n} (|\\gamma_1| + \\dots + |\\gamma_n|) \\\\\n&\\ge \\frac{1}{n} |\\gamma_1 + \\dots + \\gamma_n| \\\\\n&= \\frac{A}{n} |\\alpha_1 + \\dots + \\alpha_n| \\\\\n&= \\frac{A}{n} \\left| n\\alpha_k - \\sum_{j \\ne k} (\\alpha_k - \\alpha_j) \\right| \\\\\n&\\ge \\frac{A}{n} \\left( n \\left| \\alpha_k \\right| - \\sum_{j \\ne k} \\left| \\alpha_k - \\alpha_j \\right| \\right) \\\\\n&\\ge \\frac{A}{n} \\left( n \\left| \\alpha_k \\right| - (n-1) \\left| \\alpha_n - \\alpha_1 \\right| \\right) \\\\\n&= \\frac{A}{n} \\left( n \\left| \\alpha_k \\right| - (n-1)x \\left| \\alpha_k \\right| \\right) \\\\\n&= A \\left( 1 - \\frac{n-1}{n} x \\right) |\\alpha_k|.\n\\end{aligned}\n$$\nFrom the previous two bounds, it follows that\n$$\n\\begin{aligned}\n\\max_{(k_1, \\dots, k_n) \\in S_n} \\left| \\sum_{i=1}^n x_{k_i} \\alpha_i \\right| &\\ge \\max\\left\\{\\frac{Bx}{2}, A\\left(1 - \\frac{n-1}{n}x\\right)\\right\\} |\\alpha_k| \\\\\n&\\ge \\frac{\\frac{Bx}{2} \\cdot A \\cdot \\frac{n-1}{n} + A\\left(1 - \\frac{n-1}{n}x\\right) \\cdot \\frac{B}{2}}{A \\cdot \\frac{n-1}{n} + \\frac{B}{2}} |\\alpha_k| \\\\\n&= \\frac{AB}{2A + B - \\frac{2A}{n}} |\\alpha_k| \\\\\n&\\ge \\frac{AB}{2A + B} |\\alpha_k|.\n\\end{aligned}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 15003,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $ABCD$ 為一凸四邊形，其中邊 $AD$ 與 $BC$ 不平行。假設以 $AB$ 和 $CD$ 為直徑的兩圓交於四邊形 $ABCD$ 內部的兩點 $E, F$。由 $E$ 分別對直線 $AB$, $BC$ 和 $CD$ 作垂線得三個垂足 $Q, R$ 與 $S$。令圓 $\\omega_E$ 表示通過 $Q, R$ 與 $S$ 三點之圓。同理，圓 $\\omega_F$ 表示通過由 $F$ 分別對直線 $CD$, $DA$ 和 $AB$ 三邊作垂線所得三個垂足的圓。試證：線段 $EF$ 之中點落在兩圓 $\\omega_E$ 和 $\\omega_F$ 的兩個交點之連線上。",
    "options": [],
    "answer": "See solution",
    "solution": "(1) 令 $P, Q, R, S$ 分別為 $E$ 對 $DA$, $AB$, $BC$, $CD$ 的垂足。注意到 $P, Q, A, E$ 四點共圓，故 $\\angle QPE = \\angle QAE$。同理，$\\angle QRE = \\angle QBE$。\n\n因此，$\\angle QPE + \\angle QRE = \\angle QAE + \\angle QBE = 90^\\circ$（因為 $E$ 在以 $AB$ 為直徑的圓上）。同理，$\\angle SPE = \\angle SRE = 90^\\circ$，因此 $\\angle QPS + \\angle QRS = 90^\\circ + 90^\\circ = 180^\\circ$，故 $P, Q, R, S$ 都在 $\\omega_E$ 上。對稱地，$F$ 對 $ABCD$ 四邊的四個垂足也都在 $\\omega_F$ 上。\n\n(2) 延長 $AD$ 與 $BC$ 交於 $K$，不失一般性，假設 $A$ 在 $DK$ 線段上。以下證明 $\\angle CKD$ 為銳角。\n\n若否，則以 $CD$ 為直徑的圓將包含 $\\triangle CKD$，因此包含整個四邊形 $ABCD$，從而 $E, F$ 不可能在四邊形 $ABCD$ 內部，矛盾。故，$\\angle CKD$ 為銳角，從而直線 $EP$ 必交線段 $BC$ 於某點 $P'$，直線 $ER$ 也必交線段 $AD$ 於某點 $R'$。\n\n![](images/12-2J_p9_data_afca480df9.png)\n\n(3) 我們接著證明 $P'$ 和 $R'$ 也會在 $\\omega_E$ 上。證明如下。\n\n注意到 $R, E, Q, B$ 四點共圓，故由 (1)，\n\n$$\n\\angle QRK = \\angle QRB = \\angle QEB = 90^\\circ - \\angle QBE = \\angle QAE = \\angle QPE = \\angle QPP',\n$$\n這說明了 $P'$ 在圓 $\\omega_E$ 上。同理，$R'$ 也會在圓 $\\omega_E$ 上。\n\n(4) 比照 (1) 至 (3)，令 $M, N$ 分別為 $F$ 對 $AD$ 與 $BC$ 的垂足，並令 $M' = FM \\cap BC$, $N' = FN \\cap AD$。由相同推論，我們知 $M', N'$ 都在 $\\omega_F$ 上。\n\n![](images/12-2J_p10_data_7036fd5ea0.png)\n\n(5) 現在，令 $U$ 為 $NN'$ 與 $PP'$ 的交點，$V$ 為 $MM'$ 與 $RR'$ 的交點。注意到 $N, P', N', P$ 四點共圓，故 $UN \\times UN' = UP \\times UP'$，因此 $U$ 在 $\\omega_E$ 與 $\\omega_F$ 兩圓交點連線上。同理，$V$ 也在 $\\omega_E$ 與 $\\omega_F$ 兩圓交點連線上。\n\n(6) 最後，由於 $EUFV$ 為平行四邊形，$EF$ 的中點在 $UV$ 上，因此在 $\\omega_E$ 與 $\\omega_F$ 兩圓交點連線上。證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15004,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a point in the plane and let $\\gamma$ be a circle which does not contain $P$. Two distinct variable lines $\\ell$ and $\\ell'$ through $P$ meet the circle $\\gamma$ at points $X$ and $Y$, and $X'$ and $Y'$, respectively. Let $M$ and $N$ be the antipodes of $P$ in the circles $PXX'$ and $PYY'$, respectively. Prove that the line $MN$ passes through a fixed point.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, the perpendiculars at $X$ and $X'$ on $\\ell$ and $\\ell'$, respectively, meet at $M$, and the perpendiculars at $Y$ and $Y'$ on $\\ell$ and $\\ell'$, respectively, meet at $N$. Consider the trapezoids $XMYN$ and $X'MY'N$ to infer that the perpendicular bisectors of the segments $XY$ and $X'Y'$ meet at the midpoint of the segment $MN$. To conclude the proof, notice that the two perpendicular bisectors also meet at the center of circle $\\gamma$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15005,
    "subject": "Mathematics (Olympiad)",
    "question": "In a mathematical competition, some competitors are friends; friendship is always mutual, that is, if $A$ is a friend of $B$, then $B$ is also a friend of $A$.\n\nWe say that $n \\ge 3$ different competitors $A_1, A_2, \\dots, A_n$ form a *weakly-friendly cycle* if $A_i$ is not a friend of $A_{i+1}$ for $1 \\le i \\le n$ (with $A_{n+1} = A_1$), and there are no other pairs of non-friends among the competitors in this cycle.\n\nThe following property is satisfied:\n\nFor every competitor $C$, and every weakly-friendly cycle $\\mathcal{L}$ of competitors not including $C$, the set of competitors $D$ in $\\mathcal{L}$ which are not friends of $C$ has at most one element.\n\nProve that all competitors of this mathematical competition can be arranged into three rooms, such that every two competitors in the same room are friends.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us consider the graph $G$ whose vertices are competitors, with two vertices adjacent if and only if they are not friends. Then it suffices to prove that the vertices of $G$ can be partitioned into three subsets such that there are no edges within each subset.\n\nFirst, we prove the following lemma:\n\n**Lemma.** There is a vertex of degree at most $2$ in $G$.\n\n**Proof.** Suppose this is not true. Let $P$ be the longest induced path in $G$ (i.e., a path such that no two of its vertices are adjacent except for consecutive ones). Let $u$ and $v$ be the ends of this path. By assumption, $u$ is adjacent to at least two vertices $u'$ and $u''$ not in $P$. By the maximality of $P$, both $u'$ and $u''$ have a neighbor in $P$. Let $w'$ (resp. $w''$) be a neighbor of $u'$ (resp. $u''$) on $P$ which is closest to $u$. Without loss of generality, let $w''$ be closer to $u$ than $w'$, or $w' = w''$. Then $u''$ has at least two neighbors on the cycle induced by $u'$ and the vertices of $P$ from $u$ to $w'$, a contradiction.\n\nWe continue the proof by induction on the number of vertices of $G$ (denoted $|V(G)|$). If $|V(G)| = 3$, the proof is trivial. Assume the statement holds for all graphs with at most $n$ vertices, and consider a graph $G$ with $|V(G)| = n+1$. By the lemma, there is a vertex $v$ of $G$ with degree at most $2$. Consider the graph $G'$ obtained from $G$ by removing $v$. This graph satisfies the problem's conditions, so its vertices can be partitioned into three sets as desired. Since $v$ has degree at most $2$, for at least one of these sets, $v$ has no neighbors in it. Placing $v$ in this set yields the desired partition of $G$, completing the proof.\n\n![](images/Macedonia_2013_p22_data_6e413d13ee.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15006,
    "subject": "Mathematics (Olympiad)",
    "question": "Triangle $ABC$ is inscribed in circle $\\omega$. The tangent lines to $\\omega$ at $B$ and $C$ meet at $T$. Point $S$ lies on ray $BC$ such that $AS \\perp AT$. Points $B_1$ and $C_1$ lie on ray $ST$ (with $C_1$ in between $B_1$ and $S$) such that $B_1T = BT = C_1T$. Prove that triangles $ABC$ and $AB_1C_1$ are similar to each other.",
    "options": [],
    "answer": "See solution",
    "solution": "**First Solution:** (Based on the work by Oleg Golberg) We start with an important geometry observation.\n\n![](images/USA_IMO_2006-2007_p40_data_287777f101.png)\n\n**Lemma** Triangle $ABC$ inscribed in circle $\\omega$. Lines $BT$ and $CT$ are tangent to $\\omega$. Let $M$ be the midpoint of side $BC$. Then $\\angle BAT = \\angle CAM$. (Line $AT$ is a symmedian of triangle.)\n\n*Proof:* (We consider the above configuration. If $\\angle BAC$ is obtuse, our proof can be modified slightly.)\nLet $D$ denote the second intersection (other than $A$) of line $AT$ and circle $\\omega$. Because $BT$ is tangent to $\\omega$ at $B$, $\\angle TBD = \\angle TAB$. Hence triangles $TBD$ and $TAB$ are similar, implying that $\\frac{BD}{AB} = \\frac{TB}{TA}$. Likewise, triangles $TCD$ and $TAC$ are similar and $\\frac{CD}{AC} = \\frac{TC}{TA}$. By equal tangents, $TB = TC$. Consequently, we have $\\frac{BD}{AB} = \\frac{TB}{TA} = \\frac{TC}{TA} = \\frac{CD}{AC}$, implying that\n\n$BD \\cdot AC = CD \\cdot AB.$\n\nBy **Ptolemy's theorem** to cyclic quadrilateral $ABDC$, we have $BD \\cdot AC + AB \\cdot CD = AD \\cdot BC$.\nCombining the last two equations, we obtain that $2BD \\cdot AC = AD \\cdot BC$ or\n\n$$\n\\frac{AC}{AD} = \\frac{BC}{2BD} = \\frac{MC}{BD}.\n$$\n\nFurther considering that $\\angle ACM = \\angle ACB = \\angle ADB$ (since $ABDC$ is cyclic), we conclude that\ntriangle $ABD$ is similar to triangle $AMC$, implying that $\\angle BAT = \\angle BAD = \\angle CAM$.\n\n![](images/USA_IMO_2006-2007_p40_data_f3a39361aa.png)\n\nBecause $BT$ is tangent to $\\omega$, $\\angle CBT = \\angle CAB$, and so\n\n$$\n\\angle TBA = \\angle ABC + \\angle CBT = \\angle ABC + \\angle CAB = 180^{\\circ} - \\angle BCA.\n$$\n\nBy the lemma, we have $\\angle BAT = \\angle CAM$. Applying the Law of Sines to triangles $BAT$ and $CAM$, we obtain\n\n$$\n\\frac{BT}{AT} = \\frac{\\sin \\angle BAT}{\\sin \\angle TBA} = \\frac{\\sin \\angle CAM}{\\sin \\angle BCA} = \\frac{MC}{AM}.\n$$\n\nNote that $TB = TC_1$. Thus, $TC_1/TA = MC/MA$. By equal tangents, $TB = TC$. In isosceles triangle $BTC$, $M$ is the midpoint of base $BC$. Consequently, $\\angle TMS = \\angle TAC = \\angle TAS = 90^{\\circ}$, implying that $TMAP$ is cyclic. Hence $\\angle AMC = \\angle ATC_1$. Because\n\n$$\n\\frac{AM}{AT} = \\frac{MC}{TC_1}\n$$\n\nand $\\angle AMC = \\angle ATC_1$, triangles $MAC$ and $TAC_1$ are similar. Because $BC/BM = B_1C_1/TC_1 = 2$, triangles $ABC$ and $AB_1C_1$ are similar.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15007,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, $d$ be positive real numbers in the interval $[1, 2]$ such that $(a + c)(b + d) = 8$. Prove that\n\n$$\n\\frac{1}{a^2 + b^2 - 1} + \\frac{1}{b^2 + c^2 - 1} + \\frac{1}{c^2 + d^2 - 1} + \\frac{1}{d^2 + a^2 - 1} \\geq 1,\n$$\n\nfor all such quadruples, and determine all cases when equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "First, since $a, b, c, d \\geq 1$, all denominators are positive, so the inequality is well-defined.\n\nObserve that since $a$ and $b$ are in $[1,2]$, $(a-b)^2 \\leq 1$, so\n\n$$\n\\frac{1}{a^2 + b^2 - 1} \\geq \\frac{1}{2ab}.\n$$\n\nApplying this to all four terms, it suffices to show\n\n$$\n\\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{cd} + \\frac{1}{da} \\geq 2.\n$$\n\nGiven $(a+c)(b+d) = 8$, we have $ab + bc + cd + da = 8$. By the AM-HM inequality (or Cauchy-Schwarz), the above sum is at least $2$.\n\nEquality holds when $|a-b| = |b-c| = |c-d| = |d-a| = 1$, which, given the interval, only occurs for $(a, b, c, d) = (1, 2, 1, 2)$ or $(2, 1, 2, 1)$.\n\n*Conclusion:* The inequality holds for all such quadruples, with equality only for $(1, 2, 1, 2)$ and $(2, 1, 2, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15008,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $m, n$ satisfying the equation\n$$\n2^m = 7n + 4.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** There are no such numbers.\n\n**Solution.** The powers of 2 give remainders 2, 4, and 1 upon division by 7, with period 3. In particular, $2^k \\equiv 4 \\pmod{7}$ if and only if $k \\equiv 2 \\pmod{3}$. But this would require $m^2 \\equiv 2 \\pmod{3}$, which is impossible. This contradiction completes the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15009,
    "subject": "Mathematics (Olympiad)",
    "question": "Choose points $P$, $Q$ on the line segments $OB$, $OC$, respectively, such that $OP = OQ = 2$. Compare the volumes of the triangular pyramids $OABC$ and $OAPQ$.\n\nLet $\\triangle OBC$ and $\\triangle OPQ$ be the respective bases, with $A$ as the top vertex. Find the ratio of their volumes, and, given $OA = OP = OQ$, determine the volume of $OABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "The ratio of the volumes is\n$$\n\\frac{\\text{Vol}(OABC)}{\\text{Vol}(OAPQ)} = \\frac{\\text{Area}(\\triangle OBC)}{\\text{Area}(\\triangle OPQ)} = \\frac{OB \\cdot OC}{OP \\cdot OQ} = 3.\n$$\n\nSince $OA = OP = OQ$ and the angles between them are equal, $\\triangle APQ$ is equilateral with side length $\\sqrt{3}$. Its area is\n$$\n\\frac{\\sqrt{3}}{4} (\\sqrt{3})^2 = \\frac{3\\sqrt{3}}{4}.\n$$\nThe height from $O$ to the plane $APQ$ is $\\sqrt{3}$, so the volume of $OAPQ$ is\n$$\n\\frac{1}{3} \\cdot \\frac{3\\sqrt{3}}{4} \\cdot \\sqrt{3} = \\frac{3\\sqrt{3}}{4}.\n$$\nThus, the volume of $OABC$ is $3 \\cdot \\frac{3}{4} = \\frac{9}{4}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 15010,
    "subject": "Mathematics (Olympiad)",
    "question": "試證：當 $X$ 在 $AC$ 射線上移動時，$\\triangle BX_1X_2$ 的外接圓圓心的軌跡為直線的一部分。",
    "options": [],
    "answer": "See solution",
    "solution": "在 $AC$ 上取定固定的兩點 $X$ 與 $Y$，並作相應的 $X_1, X_2, Y_1, Y_2$。設 $BX_1X_2$ 與 $BY_1Y_2$ 的外接圓再交於點 $T$。則只要證明對 $AC$ 射線上任意一點 $Z$，相應的 $Z_1, Z_2$ 會與 $B, T$ 共圓即可，此時 $BZ_1Z_2$ 的外接圓圓心會在 $BT$ 線段的中垂線上。\n\n(註：若 $B, T$ 兩點重合，則 $\\triangle BX_1X_2$ 的外接圓與 $\\triangle BY_1Y_2$ 的外接圓切於 $B$ 點。此時變成證明 $\\triangle BZ_1Z_2$ 的外接圓亦與 $\\triangle BX_1X_2$ 的外接圓切於 $B$ 點。)\n\n![](images/18-1J_p25_data_f171109218.png)\n\n注意到, 由於 $XX_2 \\parallel YY_2 \\parallel ZZ_2$, 所以\n\n$$\nX_2Y_2 : Y_2Z_2 = XY : YZ = X_1Y_1 : Y_1Z_1.\n$$\n\n又 $\\angle Y_1TY_2 = \\angle Y_1BY_2 = \\angle X_1BX_2 = \\angle X_1TX_2$，\n\n所以 $\\angle Y_1TX_1 = \\angle X_2TY_2$。另一方面, $\\angle TY_1X_1 = \\angle TY_2X_2$，所以\n\n$$\n\\triangle TX_1Y_1 \\sim \\triangle TX_2Y_2.\n$$\n\n綜合以上, $TX_1Y_1Z_1$ 四點和 $TX_2Y_2Z_2$ 四點相似。所以\n\n$$\n\\angle Z_1TX_1 = \\angle Z_2TX_2,\n$$\n\n故\n\n$$\n\\angle Z_1TZ_2 = \\angle X_1TX_2 = \\angle X_1BX_2.\n$$\n\n即 $BTZ_1Z_2$ 共圓, 得證。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15011,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{N}_0$ be the set of nonnegative integers. Find all functions $f : \\mathbb{N}_0 \\to \\mathbb{N}_0$ satisfying the equation\n\n$$\nf^{f(m)}(n) = n + 2f(m)\n$$\n\nfor all $m, n \\in \\mathbb{N}_0$ such that $m \\le n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that since $f^0(n) = n$ by definition, $f(n) \\equiv 0$ is a solution. Now suppose that for some $c \\in \\mathbb{N}_0$, $f(c) \\ge 1$.\n\nAs $f^{f(c)-1}(n - 2f(c)) = n$ for all $n \\ge 2f(c)$, $f$ is onto for all $n \\ge 2f(c)$. Therefore $f^m(n) = n + 2m$ for all $n \\ge m \\ge 2f(c)$.\n\nSince we have $f^{2f(c)}(n) = n + 4f(c)$ and $f^{2f(c)+1}(n) = n + 4f(c) + 2$ for all $n \\ge 2f(c)$, it follows that $f(n + 4f(c)) = n + 4f(c) + 2$ for all $n \\ge 2f(c)$; in other words, $f(n) = n + 2$ for all $n \\ge 6f(c)$.\n\nLet $t$ be the least integer such that $f(n) = n + 2$ for all $n \\ge t$ (we have $t \\le 6f(c)$). We will show that $f(n) = 0$ for all $n < t$.\n\nSuppose, for the sake of contradiction, that $f(b) = a$ for some $b < t$ and $a \\ne 0$. Now suppose that there exists $u < t$ such that $f(u) \\ge t$. For clarity, let $f(u) = t + d$ for some $d \\ge 0$. Since $f^{f(u)}(u) = u + 2t + 2d$, and $f^{f(u)-1}(f(u)) = f^{t + d - 1}(t + d) = 3t + 3d - 2$, we have $f(u) = t + d = u + 2$. Since $f(u) \\ge t$, $u$ is either $t - 2$ or $t - 1$. However, if $u = t - 1$, $f(t - 1)$ will be equal to $(t - 1) + 2$, which contradicts the minimality of $t$. Thus if such a $u$ exists, then $u = t - 2$.\n\nBack to $a, b$, since $t - 1 \\ge b$, we have $f^a(t - 1) = t - 1 + 2a$, but since $t - 1 < t$ and $t - 1 + 2a \\ge t$, there must be some $0 \\le j < a$ such that $f^j(t - 1) < t \\le f^{j + 1}(t - 1)$. By the previous paragraph, where $u = f^j(t - 1)$, we have $f^j(t - 1) = t - 2$, and $f^{j + 1}(t - 1) = t$. Therefore $f^a(t - 1) = t + 2(a - j - 1)$. Therefore $2j = -1$, which is impossible, so we have a contradiction. Hence $f(n) = 0$ for all $n < t$.\n\nIt remains to show that for a given nonnegative integer $t$, the function\n\n$$\nf(n) = \\begin{cases} 0, & n < t, \\\\ n + 2, & n \\ge t, \\end{cases}\n$$\n\nactually satisfies the given condition. If $m < t$, the given condition reduces to $f^0(n) = n$, which is true, while if $m \\ge t$, we have $n \\ge t$, thus $f^{f(m)}(n) = n + 2f(m)$. Hence this function satisfies the given condition, and our proof is complete.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15012,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real values of $k$ such that all solutions of the equation\n$$\nk(2-k)x^2 - (k+4)x + 6 = 0\n$$\nare positive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "First, consider cases when the equation is not quadratic.\n\nFor $k=0$, the equation becomes $-4x + 6 = 0$, which has a non-integer solution.\n\nFor $k=2$, the equation becomes $-6x + 6 = 0$, which has a single solution $x=1$, a positive integer.\n\nNow, suppose $k(2-k) \\neq 0$. Compute the discriminant:\n$$\nD = (k+4)^2 - 4k(2-k) \\cdot 6 = k^2 + 8k + 16 - 24(2k - k^2) = 25k^2 - 40k + 16 = (5k - 4)^2\n$$\nThus, the roots are:\n$$\nx_1, x_2 = \\frac{(k+4) \\pm (5k-4)}{2k(2-k)}\n$$\nCalculating:\n$$\nx_1 = \\frac{(k+4) + (5k-4)}{2k(2-k)} = \\frac{6k}{2k(2-k)} = \\frac{3}{2-k}\n$$\n$$\nx_2 = \\frac{(k+4) - (5k-4)}{2k(2-k)} = \\frac{-4k+8}{2k(2-k)} = \\frac{2}{k}\n$$\nWe require both $\\frac{3}{2-k}$ and $\\frac{2}{k}$ to be positive integers.\nLet $\\frac{2}{k} = m$ for some positive integer $m$, so $k = \\frac{2}{m}$.\nThen:\n$$\n\\frac{3}{2-k} = \\frac{3}{2 - \\frac{2}{m}} = \\frac{3m}{2(m-1)}\n$$\nWe need $\\frac{3m}{2(m-1)}$ to be a positive integer. Since $m$ and $m-1$ are coprime, this is only possible for $m-1 = 1$ ($m=2$, $k=1$) or $m-1=3$ ($m=4$, $k=\\frac{1}{2}$).\n\nTherefore, the real values of $k$ are $k=2$, $k=1$, and $k=\\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15013,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a point in the interior of a triangle $ABC$, and let $D$, $E$, $F$ be the points of intersection of the lines $AP$ with $BC$, $BP$ with $CA$, and $CP$ with $AB$, respectively.\n\nProve that the area of triangle $ABC$ must be $6$ if the area of each of the triangles $PFA$, $PDB$, and $PEC$ is $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "For any triangle $XYZ$, let its area be denoted by $|XYZ|$. Set\n\n$$\nx = |AEP|, \\quad y = |BFP|, \\quad z = |CDP|.\n$$\n\nAll the presented solutions will prove that $x = y = z = 1$ and thus\n$$\n|ABC| = x + y + z + 3 = 6.\n$$\n\n![](images/Australian_Scene_2012_-_AMT_Publishing_-_273p_p151_data_0c372ec35c.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15014,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AP \\perp CD$, and let $P$ also lie on the perpendicular bisector of $EF$. Show that the points $A, E, P, F$ are concyclic. Prove that\n\n$$\n\\angle EPF = 180^\\circ - \\angle EAF = \\angle CAE + \\angle DAF = 2\\angle CAE.\n$$\n\nFurthermore, given that $\\angle EPF = 2\\angle CBE$, show that the point $B$ lies on the circle $\\Gamma$ with center $P$ and radius $PE$. By the power of a point theorem,\n\n$$\n2CA^2 = CA \\cdot CD = CB \\cdot CF = CP^2 - PE^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Hence,\n\n$$\nAP^2 = CP^2 - CA^2 = (2CA^2 + PE^2) - CA^2 = CA^2 + PE^2,\n$$\n\nas desired.\n\n![](images/1.2_1997-2023_IMO_HK_TST_Solutions_p180_data_53e1d82df1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15015,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with incenter $I$, and let the incircle $(I)$ be tangent to $BC, CA, AB$ at $D, E, F$ respectively. Let $I_b$ and $I_c$ be the excenters of $\\triangle ABC$ with respect to $B$ and $C$. Let $P$ and $Q$ be the midpoints of $I_bE$ and $I_cF$. Suppose that the circle $(PAC)$ meets $AB$ again at $R$, and the circle $(QAB)$ meets $AC$ again at $S$.\n\n1. Prove that $PR$, $QS$, and $AI$ are concurrent.\n\n2. Suppose $DE$ and $DF$ meet $I_bI_c$ at $K$ and $J$, and $EJ$ meets $FK$ at $M$. The lines $PE$ and $QF$ meet $(PAC)$ and $(QAB)$ again at $X$ and $Y$ (with $X \\neq P$, $Y \\neq Q$). Prove that $BY$, $CX$, and $AM$ are concurrent.\n\n![](images/vn-booklet_final_p27_data_c3e6aa9981.png)\n\n![](images/vn-booklet_final_p28_data_85e11a394a.png)",
    "options": [],
    "answer": "See solution",
    "solution": "1) Since $EF$ and $I_bI_c$ are both perpendicular to $AI$, $I_bI_cFE$ is a trapezoid. Thus, $PQ$ is the midline of both trapezoid $I_bI_cFE$ and triangle $AEF$. Therefore, $P$ and $Q$ lie on the radical axis of the degenerate circle $(A, 0)$ and $(I)$. Similarly, $Q$ lies on the radical axis of $(B, 0)$ and $(I)$, so $QA^2 = \\overline{QF} \\cdot \\overline{QY} = QB^2$, implying $(QAB)$ is tangent to $(I)$ at $Y$. Similarly, $(PAC)$ is tangent to $(I)$ at $X$.\n\nThus, $(I)$ is the $S$-mixtilinear incircle of triangle $ASB$, so the incenter of $\\triangle ABS$ is the midpoint $N$ of $EF$, which implies $SQ$ is the angle bisector of $\\angle ASB$ and passes through $N$. Similarly, $RP$ passes through $N$. Therefore, $PR$, $QS$, and $AI$ are concurrent at $N$.\n\n2) In circle $(I)$, the line $I_bI_c$ is the antipole of $N$, so $JE$, $KF$, and $DN$ are concurrent at $M$ on $(I)$. By Steinbart's theorem: for triangle $ABC$ with incircle $(I)$ tangent to $BC, CA, AB$ at $D, E, F$, and points $X, Y, Z$ on $(I)$, $AX, BY, CZ$ are concurrent if and only if $DX, EY, FZ$ are concurrent.\n\nTo prove $AM$, $BY$, $CX$ are concurrent, it suffices to show $FX$, $EY$, $DM$ are concurrent.\n\nWe have $\\angle EYF = \\angle AEF = \\angle I_bAC$, so quadrilateral $I_cAEY$ is cyclic. Similarly, $AI_bXF$ is cyclic. Thus, $X$ and $Y$ as defined above are the same as in the problem.\n\nConsider the transformation $S$ as the composition of inversion $I_A^{AB \\cdot AC}$ and reflection over $AI$. Then $S: (O) \\leftrightarrow BC$, $(I) \\leftrightarrow (T)$, where $(T)$ is the $A$-ex-mixtilinear incircle of $\\triangle ABC$, tangent to $AC, AB$ at $E', F'$. Then $E \\leftrightarrow E'$, $F \\leftrightarrow F'$.\n\nBy Sawayama's lemma, the excenter $I_a$ is the midpoint of $E'F'$. By Pappus's theorem for $(I_c, A, I_b)$ and $(E', I_a, F')$, $I_bE'$ meets $I_cF'$ at $Z$ on $BC$.\n\nLet $G$ be the tangency point of $(T)$ with $(O)$. $I_aG$ passes through $L$, the midpoint of arc $BAC$ of $(O)$, which is also the midpoint of $I_bI_c$. Since $I_bI_c \\parallel E'F'$, by Thales's theorem, $L, Z, G, I_a$ are collinear.\n\nWe have $S: E'I_b \\leftrightarrow (I_cAE)$, $F'I_c \\leftrightarrow (I_bAF)$, $D \\leftrightarrow G$, $I \\leftrightarrow I_a$, so $GI_a \\leftrightarrow (AID)$. Since $E'I_b, F'I_c, I_aG$ are concurrent, the circles\n\n$$\n(AI_cE),\\ (AI_bF),\\ (AID)\n$$\n\nare coaxial. We have $\\overline{NM} \\cdot \\overline{ND} = \\overline{NE} \\cdot \\overline{NF} = \\overline{NA} \\cdot \\overline{NI}$, so $AMID$ is cyclic. Considering the radical axes of $(I), (I_cAE), (I_bAF)$, $EY$ and $FX$ meet at $U$, the radical center. Considering the radical axes of $(I), (I_cAE), (AID)$, $MD$ and $EY$ meet at $U'$, the radical center. Since $(AI_cE), (AI_bF), (AID)$ are coaxial, $U \\equiv U'$. Therefore, $MD$, $EY$, $FX$ are concurrent at $U$. The problem is solved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15016,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Let $ABCD$ be a convex quadrilateral. Let $K$, $L$, $M$, and $N$ be the midpoints of the sides $AB$, $BC$, $CD$, and $DA$, respectively. Let $O$ be the intersection point of the segments $AC$, $BD$, $KM$, and $LN$. It is easy to see that $KL \\parallel AC$, $MN \\parallel AC$, and $KL = MN = 0.5\\, AC$. So $KLMN$ is a parallelogram.\n\nb) Construct a convex quadrilateral such that one of its diagonals and the segments joining midpoints of its opposite sides are concurrent, but the quadrilateral is not a parallelogram.\n\n![](images/Belorusija_2011_p26_data_df0456bd9b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "For part (a):\n\nLet $P$ and $R$ be the intersection points of $BD$ with $KL$ and $MN$, respectively, and let $Q$ and $S$ be the intersection points of $AC$ with $LM$ and $KN$, respectively (see the figure).\n\nSince $KL \\parallel AC$ and $BK = KA$, by Thales' theorem, $BP = PO$. Similarly, since $MN \\parallel AC$ and $CM = MD$, $DR = RO$. In particular, $BO = 2 \\cdot PO$ and $DO = 2 \\cdot RO$. Likewise, $CQ = QO$ and $AS = SO$, so $CO = 2 \\cdot QO$ and $AO = 2 \\cdot SO$.\n\nSince $KLMN$ is a parallelogram, $PO = RO$ ($\\triangle KPO = \\triangle MRO$). Therefore, $BO = DO$ and $CO = AO$. Thus, point $O$ is the midpoint of the diagonals $AC$ and $BD$ of quadrilateral $ABCD$, so $ABCD$ is a parallelogram.\n\nFor part (b):\n\nConsider a triangle $ABC$ and mark point $D$ on the extension of its median $BH$ so that $DH \\neq BH$ (see the figure). The quadrilateral $ABCD$ is not a parallelogram. Let $K$, $L$, $M$, and $N$ be the midpoints of $AB$, $BC$, $CD$, and $DA$, respectively, and let $P$ and $Q$ be the intersection points of diagonal $BD$ with $KL$ and $MN$, respectively.\n\nSince $BLHK$ is a parallelogram (with $KH$ and $LH$ joining midpoints of the sides of $\\triangle ABC$), and $BH$, $KL$ are its diagonals, $KP = PL$, so $P$ is the midpoint of $KL$. Similarly, $Q$ is the midpoint of $MN$. Since $KLMN$ is a parallelogram, $KM$, $LN$, and $PQ$ are concurrent. Thus, the diagonal $BD$ and the segments joining the midpoints of the opposite sides of $ABCD$ are concurrent, but $ABCD$ is not a parallelogram, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15017,
    "subject": "Mathematics (Olympiad)",
    "question": "The set $S$ consists of $n > 2$ points in the plane. The set $P$ consists of $m$ lines in the plane such that every line in $P$ is an axis of symmetry of $S$. Prove that $m \\leq n$, and determine when equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the $n$ points be $A_1, A_2, \\dots, A_n$, with $A_i = (x_i, y_i)$ for $i = 1, 2, \\dots, n$. The point $B = \\left(\\frac{1}{n}\\sum_{i=1}^n x_i, \\frac{1}{n}\\sum_{i=1}^n y_i\\right)$ is the *center of set $S$*, since $\\sum_{i=1}^n \\overrightarrow{BA_i} = \\vec{0}$ only at $B$.\n\nIf any line $p$ in $P$ is taken as the $x$-axis, then $\\sum_{i=1}^n y_i = 0$, so $B$ lies on $p$. Thus, every line in $P$ passes through $B$.\n\nDefine:\n\n$$\nF = \\{(x, y, p) \\mid x, y \\in S,\\ p \\in P,\\ p \\text{ is a symmetry axis of } x, y\\}\n$$\n$$\nF_1 = \\{(x, y, p) \\in F \\mid x \\neq y\\}\n$$\n$$\nF_2 = \\{(x, y, p) \\in F \\mid x \\text{ lies on } p\\}\n$$\n\nThen $F = F_1 \\cup F_2$, $F_1 \\cap F_2 = \\emptyset$.\n\nFor any $p \\in P$ and $x \\in S$, $x$ has only one symmetric point $y$ with respect to $p$, so there are $n$ such triples for each $p$, and $|F| = mn$.\n\nFor $F_1$, since there is only one symmetry axis for each pair $x \\neq y$, $|F_1| \\leq 2\\binom{n}{2} = n(n-1)$.\n\nFor $F_2$, if any point in $S$ lies on at most one line $p$, then $|F_2| \\leq n$.\n\nThus,\n$$\nm n \\leq n(n-1) + n \\implies m \\leq n.\n$$\n\nIf some point lies on two lines in $P$, it must be the center $B$. Removing $B$ from $S$ gives $S'$, and every line in $P$ is still a symmetry axis for $S'$, so $m \\leq |S'| = n-1$. Thus, $m \\leq n$ always holds.\n\nWhen $m = n$, the equalities above hold simultaneously. The perpendicular bisector of any segment joining two points in $S$ belongs to $P$, and any point in $S$ lies on one line in $P$, while the center $B$ is not in $S$.\n\nAll vectors $\\overrightarrow{BA_i}$ are equal, so $A_1, \\dots, A_n$ lie on a circle centered at $B$. Arranged clockwise, if the arcs between consecutive points are not equal, the symmetry axes do not match the required adjacency, leading to a contradiction. Thus, the points in $S$ are the vertices of a regular $n$-gon.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p142_data_d5177f579b.png)\n\nTherefore, $S$ consists of the vertices and $P$ the axes of symmetry of a regular $n$-gon if and only if $m = n$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15018,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the smallest possible value of $x^6 + x^4 y^2 + x^2 y^4 + y^6$, given that the product of real numbers $x, y$ is $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, let $x, y$ be positive. We can factor the expression as:\n\n$$\nx^6 + x^4 y^2 + x^2 y^4 + y^6 = (x^4 + y^4)(x^2 + y^2)\n$$\n\nSince $(x^2 - y^2)^2 = x^4 - 2x^2 y^2 + y^4 \\ge 0$, it follows that $x^4 + y^4 \\ge 2x^2 y^2 = 2$ (since $x y = 1$). Similarly, $x^2 + y^2 \\ge 2 x y = 2$. Therefore, the smallest value is $4$, which is achieved when $x = y = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15019,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be the midpoint of side $AC$ of triangle $ABC$. Let $P$ be a point inside triangle $ABC$ such that $\\angle BAP = \\angle BCP$. The line $CP$ intersects side $AB$ at point $R$. Let $Q$ be the foot of the perpendicular drawn from $B$ to line $CP$. If $Q$ lies inside triangle $ABC$ and $QM = \\frac{PC}{2}$, then prove that $RQ = QP$.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/MNG2023_1_p22_data_01724dde80.png)\n\nLet $K$, $N$, and $L$ be the midpoints of segments $BP$, $BC$, and $PC$, respectively. Since $AB$ is parallel to $MN$ and $AP$ is parallel to $ML$, we have $\\angle BAP = \\angle NML$.\n\nConsidering the parallelogram $KNCL$, $\\angle BCP = \\angle BCL = \\angle NKL$. Moreover, since $\\angle NML = \\angle NKL$, we have $K$, $N$, $L$, and $M$ lie on the same circle.\n\nSince $\\angle BQP = 90^\\circ$, we can conclude that $QK = \\frac{BP}{2}$. $Q$, $K$, $N$, $L$, and $M$ all lie on the same circle, because $QK = NL$.\n\nSince $QM = \\frac{CP}{2} = KN$, $QK$ is parallel to $AB$. Consequently, we can deduce that $RQ = QP$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15020,
    "subject": "Mathematics (Olympiad)",
    "question": "給定 $6 \\times 6$ 的方格，並記第一列六個方格座標為 $(1,1), (1,2), \\cdots, (1,6)$，其餘類推。對於任意的 $k = 0,1,\\cdots,5$，滿足 $i-j \\equiv k \\pmod{6}$ 的六個格子 $(i,j)$ 稱為在同一條對角線上（故共有六條對角線）。試問：能否將 $1,2,\\cdots,36$ 填入 $6 \\times 6$ 的方格中，同時滿足\n\n1. 每一列的和都相等。\n2. 每一行的和都相等。\n3. 每一條對角線的和都相等。\n\nThere's a $6 \\times 6$ chess board, which we label the squares in the first column by $(1,1), (1,2), \\cdots, (1,6)$, and label the other squares similarly. For every $k = 0,1,\\cdots,5$, all squares $(i-j)$ satisfying $i-j \\equiv k \\pmod{6}$ form a diagonal; therefore, there are six diagonals. Decide whether we can write $1,2,\\cdots,36$ on the chess board so that all the following conditions hold:\n\n1. The sums for each column are the same.\n2. The sums of each row are the same.\n3. The sums of each diagonal are the same.",
    "options": [],
    "answer": "See solution",
    "solution": "不可能。歸謬證法，假設可以填成功，則這個和必為\n\n$$\nS = \\frac{1}{6}(1 + 2 + \\cdots + 36) = 111.\n$$\n\n將 $6 \\times 6$ 的方格分成四類：\n\n1. A: 座標為 $(\\text{奇}, \\text{奇})$ 的格子。\n2. B: 座標為 $(\\text{奇}, \\text{偶})$ 的格子。\n\n（解答未完，僅部分步驟給出。）",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15021,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that $a, b, c, d$ are positive real numbers satisfying $(a + c)(b + d) = ac + bd$. Find the smallest possible value of\n$$\n\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{d} + \\frac{d}{a}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, apply the AM-GM inequality:\n$$\n\\left(\\frac{a}{b} + \\frac{c}{d}\\right) + \\left(\\frac{b}{c} + \\frac{d}{a}\\right) \\geq 2\\sqrt{\\frac{ac}{bd}} + 2\\sqrt{\\frac{bd}{ac}} = \\frac{2(ac + bd)}{\\sqrt{abcd}}\n$$\nApplying AM-GM again:\n$$\n\\left(\\frac{a}{b} + \\frac{c}{d}\\right) + \\left(\\frac{b}{c} + \\frac{d}{a}\\right) \\geq \\frac{2(a + c)(b + d)}{\\sqrt{abcd}} \\geq 2 \\cdot \\frac{2\\sqrt{ac} \\cdot 2\\sqrt{bd}}{\\sqrt{abcd}} = 8\n$$\nEquality holds when $a = c$ and $b = d$. The condition $(a + c)(b + d) = ac + bd$ becomes $4ab = a^2 + b^2$, or $a/b = 2 \\pm \\sqrt{3}$.\n\nThus, the minimum value is $8$, for example when $a = c = 1$ and $b = d = 2 + \\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15022,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $G$ be a graph with 2017 edges. Define\n$$\nS(G) = \\sum_{e=vw} \\min(\\deg v, \\deg w),\n$$\nwhere the sum is over all edges $e = vw$ of $G$. What is the maximum possible value of $S(G)$?\n\n*Remark.* The quantity $S(G)$ can be used to bound the number of triangles in $G$: $\\#E(G) \\le \\frac{1}{3}S(G)$, since an edge $e = vw$ is part of at most $\\min(\\deg v, \\deg w)$ triangles.\n\n*For planar graphs*, it is known that $S(G) \\le 18n - 36$ and conjectured that for large $n$, $S(G) \\le 18n - 72$.\n\n*Further context*: The generalization of 2017 to any constant was resolved in 2018 by Mehtaab Sawhney and Ashwin Sah ([arXiv:1801.02532](https://arxiv.org/pdf/1801.02532.pdf)).",
    "options": [],
    "answer": "See solution",
    "solution": "We prove that $S(G) \\le 127009$ for any graph $G$ with 2017 edges.\n\n**Combinatorial Bound:**\nLet $d_1 \\ge d_2 \\ge \\dots \\ge d_n$ be the degree sequence of $G$ ($n \\ge 65$). Then\n$$\nS(G) \\le d_2 + 2d_3 + 3d_4 + \\dots + 63d_{64} + d_{65}.\n$$\n*Proof:* Associate each edge $e = \\{v_i, v_j\\}$ ($i < j$) to $v_j$. Let $a_i$ be the number of edges associated to $v_i$, so $a_i \\le i-1$, $\\sum a_i = 4034$, and\n$$\nS(G) = \\sum_{i=1}^n a_i d_i.\n$$\nBy smoothing, $\\sum a_i d_i$ is maximized when the $a_i$ are as large as possible for large $d_i$, yielding the stated bound.\n\n**Algebraic Bound:**\nLet $x_1 \\ge x_2 \\ge \\dots \\ge x_{65} \\ge 0$ with $\\sum x_i \\le 4034$. Then\n$$\nx_2 + 2x_3 + \\dots + 63x_{64} + x_{65} \\le 127010.\n$$\nEquality occurs only if $x_1 = x_2 = \\dots = x_{64} = 63$, $x_{65} = 2$.\n\n*Proof:* Smoothing arguments show the maximum is achieved when the $x_i$ are as equal as possible, with $x_{65} \\le 3$. Checking all cases, the maximum is $127010$.\n\nHowever, in the graph context, $a_{65} = 2$ is not possible with $a_i \\le i-1$, so the true maximum is $127009$.\n\n**Conclusion:**\nThe maximum possible value of $S(G)$ for a graph with 2017 edges is $127009$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15023,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime. At each vertex of a regular polygon with $p$ sides, an integer is written. For any vertex of the polygon, compute the difference between the sum of the integers written at its neighbors and its own number. Then, delete all the initial integers and replace them with the newly obtained integers. Prove that the integers obtained after $p$ such steps are congruent modulo $p$ to the initial integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A_0, A_1, \\ldots, A_{p-1}$ be the vertices of the polygon, and let $(a_0, a_1, \\ldots, a_{p-1})$ be the integers at these vertices, with $a_k$ at $A_k$ for $k = 0, 1, \\ldots, p-1$. Consider the polynomial with integer coefficients:\n\n$$\nH(x) = a_0 + a_1 x + \\cdots + a_{p-1} x^{p-1}.\n$$\n\nApplying the transformation\n\n$$\na_0 \\to a_{p-1} + a_1 - a_0, \\quad a_1 \\to a_0 + a_2 - a_1, \\ldots, a_{p-1} \\to a_{p-2} + a_0 - a_{p-1}\n$$\n\nto the coefficients of $H(x)$, we obtain the polynomial\n\n$$\nH_1(x) = H(x) (x^{p-1} + x - 1) \\pmod{x^p - 1}.\n$$\n\nAfter $s$ steps, we have\n\n$$\n\\begin{aligned}\nH_s(x) &= H_{s-1}(x) (x^{p-1} + x - 1) \\pmod{x^p - 1} \\\\\n&= H(x) (x^{p-1} + x - 1)^s \\pmod{x^p - 1}.\n\\end{aligned}\n$$\n\nAfter $p$ steps,\n\n$$\nH_p(x) = (x^{p-1} + x - 1)^p H(x) \\pmod{x^p - 1}. \\tag{1}\n$$\n\nBut\n\n$$\n(x^{p-1} + x - 1)^p = p Q(x) + x^{p(p-1)} + x^p - 1\n$$\n\nwhere $Q(x)$ is a polynomial with integer coefficients. Since $x^p \\equiv 1 \\pmod{x^p - 1}$, we have\n\n$$\n(x^{p-1} + x - 1)^p \\equiv p Q(x) + 1 \\pmod{x^p - 1}.\n$$\n\nFrom (1),\n\n$$\nH_p(x) \\equiv p Q(x) H(x) + H(x) \\pmod{x^p - 1}.\n$$\n\nTherefore, the coefficients of $H_p(x)$ and $H(x)$ are congruent modulo $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15024,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be positive real numbers with $a \\le 2b \\le 4a$.\n\n*Prove that*\n\n$$\n4ab \\le 2(a^2 + b^2) \\le 5ab.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The terms in the left inequality can be rewritten as a square:\n\n$$\n2(a^2 + b^2) \\ge 4ab \\iff (a-b)^2 \\ge 0.\n$$\n\nThe square in the last inequality is clearly weakly positive.\n\nIn the right inequality, we multiply by $8$ and complete the square to obtain\n\n$$\n16a^2 - 40ab + 16b^2 \\le 0 \\iff (4a - 5b)^2 - 9b^2 \\le 0.\n$$\n\nFactorisation of the difference of squares gives\n\n$$\n(4a - 5b - 3b)(4a - 5b + 3b) \\le 0 \\iff (4a - 8b)(4a - 2b) \\le 0 \\iff (a - 2b)(4a - 2b) \\le 0.\n$$\n\nThe hypotheses ensure that the first factor is weakly negative, the second weakly positive, and therefore, the product is weakly negative. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15025,
    "subject": "Mathematics (Olympiad)",
    "question": "Даден е произволен триаголник $ABC$. На страните $AB$, $BC$ и $CA$ се избрани произволни точки $C_1$, $A_1$ и $B_1$. Нека со $P_1$, $P_2$ и $P_3$ се означени плоштините на триаголниците $AC_1B_1$, $BC_1A_1$ и $CA_1B_1$ соодветно, а со $P$ е означена плоштината на триаголникот $ABC$. Докажи дека $$\n\\sqrt{P_1} + \\sqrt{P_2} + \\sqrt{P_3} \\le \\frac{3}{2}\\sqrt{P}\n$$.",
    "options": [],
    "answer": "See solution",
    "solution": "За плоштините на триаголниците $AB_1C_1$ и $ABC$ важи:\n\n$$\nP_1 = \\frac{1}{2} AB_1 \\cdot AC_1 \\sin \\angle A \\quad \\text{и} \\quad P = \\frac{1}{2} AB \\cdot AC \\sin \\angle A.\n$$\n\nСпоред тоа $\\frac{P_1}{P} = \\frac{AB_1}{AC} \\cdot \\frac{AC_1}{AB}$. Слично се добива $\\frac{P_2}{P} = \\frac{BC_1}{AB} \\cdot \\frac{BA_1}{BC}$ и $\\frac{P_3}{P} = \\frac{CB_1}{AC} \\cdot \\frac{CA_1}{CB}$. Тогаш\n\n$$\n\\begin{aligned}\n\\sqrt{\\frac{P_1}{P}} + \\sqrt{\\frac{P_2}{P}} + \\sqrt{\\frac{P_3}{P}} &= \\sqrt{\\frac{AB_1}{AC}} \\cdot \\sqrt{\\frac{AC_1}{AB}} + \\sqrt{\\frac{BC_1}{AB}} \\cdot \\sqrt{\\frac{BA_1}{BC}} + \\sqrt{\\frac{CB_1}{AC}} \\cdot \\sqrt{\\frac{CA_1}{CB}} \\\\\n&\\le \\frac{1}{2} \\left( \\frac{AB_1}{AC} + \\frac{AC_1}{AB} \\right) + \\frac{1}{2} \\left( \\frac{BC_1}{AB} + \\frac{BA_1}{BC} \\right) + \\frac{1}{2} \\left( \\frac{CB_1}{AC} + \\frac{CA_1}{CB} \\right) \\\\\n&= \\frac{1}{2} \\left( \\frac{AB_1 + CB_1}{AC} \\right) + \\frac{1}{2} \\left( \\frac{BC_1 + AC_1}{AB} \\right) + \\frac{1}{2} \\left( \\frac{BA_1 + CA_1}{BC} \\right) = \\frac{3}{2}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15026,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $2n$ real numbers $a_1, a_2, \\dots, a_n, r_1, r_2, \\dots, r_n$ satisfying $a_1 \\leq a_2 \\leq \\dots \\leq a_n$ and $0 \\leq r_1 \\leq r_2 \\leq \\dots \\leq r_n$. Prove that\n$$\n\\sum_{i=1}^n \\sum_{j=1}^n a_i a_j \\min(r_i, r_j) \\geq 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Write a matrix of $n \\times n$ elements as follows:\n\n$$\nA_1 = \\begin{pmatrix}\n  a_1 a_1 r_1 & a_1 a_2 r_1 & a_1 a_3 r_1 & \\cdots & a_1 a_n r_1 \\\\\n  a_2 a_1 r_1 & a_2 a_2 r_2 & a_2 a_3 r_2 & \\cdots & a_2 a_n r_2 \\\\\n  a_3 a_1 r_1 & a_3 a_2 r_2 & a_3 a_3 r_3 & \\cdots & a_3 a_n r_3 \\\\\n  \\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\\n  a_n a_1 r_1 & a_n a_2 r_2 & a_n a_3 r_3 & \\cdots & a_n a_n r_n\n\\end{pmatrix}\n$$\n\nSince\n$$\n\\sum_{i=1}^{n} \\sum_{j=1}^{n} a_i a_j \\min(r_i, r_j) = \\sum_{j=1}^{n} a_1 a_j \\min(r_1, r_j) + \\sum_{j=1}^{n} a_2 a_j \\min(r_2, r_j) + \\dots + \\sum_{j=1}^{n} a_k a_j \\min(r_k, r_j) + \\dots + \\sum_{j=1}^{n} a_n a_j \\min(r_n, r_j),\n$$\nits $k$-th term is\n$$\n\\sum_{j=1}^{n} a_k a_j \\min(r_k, r_j) = a_k a_1 r_1 + a_k a_2 r_2 + \\dots + a_k a_k r_k + a_k a_{k+1} r_k + \\dots + a_k a_n r_n\n$$\nwhich is the sum of elements of the $k$-th row of $A_1$, $k = 1, 2, \\dots, n$.\n\nTherefore, $\\sum_{i=1}^n \\sum_{j=1}^n a_i a_j \\min(r_i, r_j)$ is the sum of all elements of $A_1$.\n\nOn the other hand, the summation can also be done as follows: Take the elements of first column and the first row of $A_1$, sum up; denote the rest $(n-1) \\times (n-1)$ element by matrix $A_2$, then take the first column and the first row of $A_2$, sum up, denote the rest by matrix $A_3$, and so on. Thus,\n\n$$\n\\begin{align*}\n\\sum_{i=1}^{n} \\sum_{j=1}^{n} a_i a_j \\min(r_i, r_j) &= \\sum_{k=1}^{n} r_k (a_k^2 + 2a_k (a_{k+1} + a_{k+2} + \\dots + a_n)) \\\\\n&= \\sum_{k=1}^{n} r_k \\left( \\left(a_k + \\sum_{i=k+1}^{n} a_i\\right)^2 - \\left(\\sum_{i=k+1}^{n} a_i\\right)^2 \\right) \\\\\n&= \\sum_{k=1}^{n} r_k \\left( \\left(\\sum_{i=k}^{n} a_i\\right)^2 - \\left(\\sum_{i=k+1}^{n} a_i\\right)^2 \\right) \\\\\n&= r_1 \\left(\\sum_{i=1}^{n} a_i\\right)^2 + r_2 \\left(\\sum_{i=2}^{n} a_i\\right)^2 + r_3 \\left(\\sum_{i=3}^{n} a_i\\right)^2 + \\dots \\\\\n&\\quad + r_n \\left(\\sum_{i=n}^{n} a_i\\right)^2 - r_1 \\left(\\sum_{i=2}^{n} a_i\\right)^2 - r_2 \\left(\\sum_{i=3}^{n} a_i\\right)^2 - \\dots - r_{n-1} \\left(\\sum_{i=n}^{n} a_i\\right)^2 \\\\\n&= \\sum_{k=1}^{n} (r_k - r_{k-1}) \\left(\\sum_{i=k}^{n} a_i\\right)^2 \\geq 0\n\\end{align*}\n$$\n(where $r_0 = 0$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15027,
    "subject": "Mathematics (Olympiad)",
    "question": "In an $n \\times n$ board, the numbers $0$ through $n^2 - 1$ are written so that the number in row $i$ and column $j$ is equal to $$(i - 1) + n(j - 1)$$ where $1 \\leq i, j \\leq n$. Suppose we select $n$ different cells of the board, where no two cells are in the same row or column. Find the maximum possible product of the numbers in the $n$ cells.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n! \\cdot (n-1)^n$. This is achievable by choosing the numbers $n-1, 2(n-1), 3(n-1), \\dots, n(n-1)$, which are at positions $(n, 1), (n-1, 2), (n-2, 3), \\dots, (1, n)$.\n\nTo show this is the best possible, we begin with a lemma.\n\n**Lemma.** If $a < b < c < d$ are real numbers such that $a + d = b + c$, then $bc > ad$.\n\n**Proof.** We have $(b - a)(c - a) > 0$, which implies\n\n$$\nbc + (a - b - c)a > 0.\n$$\n\nSince $a - b - c = -d$, this simplifies to $bc > ad$ as desired. $\\square$\n\nLet $f(i, j) = (i - 1) + n(j - 1)$ be the number in row $i$ and column $j$. Suppose we have chosen the $n$ numbers $f(1, j_1), f(2, j_2), \\dots, f(n, j_n)$. If the $j_i$ sequence is strictly decreasing, then $j_i = n + 1 - i$ for all $i$ and we get a configuration in the first paragraph. Otherwise, there exist $k, m$ such that $k < m$ and $j_k < j_m$. Notice that\n\n$$\nf(k, j_k) < f(m, j_k) < f(k, j_m) < f(m, j_m).\n$$\n\nMoreover,\n\n$$\n\\begin{aligned}\nf(k, j_k) + f(m, j_m) &= f(m, j_k) + f(k, j_m) \\\\\n&= (k + m - 2) + n(j_k + j_m - 2).\n\\end{aligned}\n$$\n\nSo by the lemma, we have that\n\n$$\nf(k, j_k)f(m, j_m) < f(k, j_m)f(m, j_k).\n$$\n\nTherefore, we increase the product of the $n$ numbers by changing $f(k, j_k)$ and $f(m, j_m)$ to $f(k, j_m)$ and $f(m, j_k)$, so our original arrangement could not have been a maximum.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15028,
    "subject": "Mathematics (Olympiad)",
    "question": "Four boxes containing masks and means of disinfection are transported to participants of a carnival. When the boxes are weighed pairwise, six quantities are obtained. The four largest of these are $125\\ \\text{kg}$, $120\\ \\text{kg}$, $110\\ \\text{kg}$, and $101\\ \\text{kg}$. Find all possible weights of the four boxes.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the masses of the boxes be $a$, $b$, $c$, and $d$ kilograms, with $a \\leq b \\leq c \\leq d$. The largest pairwise sum is $c + d = 125$. The second largest is $b + d = 120$, since $a + c \\leq b + c \\leq b + d$ and $a + d \\leq b + d$. The smallest sum is $a + b$, and the second smallest is $a + c$. The remaining two sums are $b + c$ and $a + d$, which must be $101$ and $110$ in some order.\n\nConsider both cases:\n\n*Case 1: $b + c = 101$ and $a + d = 110$.*\n\nSince $c = 125 - d$ and $b = 120 - d$,\n$$\n(120 - d) + (125 - d) = 101 \\\\\n245 - 2d = 101 \\\\\n2d = 144 \\\\\nd = 72\n$$\nThen,\n$$\nc = 125 - 72 = 53 \\\\\nb = 120 - 72 = 48 \\\\\na = 110 - 72 = 38\n$$\nSo the weights are $38\\ \\text{kg}$, $48\\ \\text{kg}$, $53\\ \\text{kg}$, $72\\ \\text{kg}$.\n\n*Case 2: $b + c = 110$ and $a + d = 101$.*\n\nAgain, $c = 125 - d$ and $b = 120 - d$,\n$$\n(120 - d) + (125 - d) = 110 \\\\\n245 - 2d = 110 \\\\\n2d = 135 \\\\\nd = 67.5\n$$\nThen,\n$$\nc = 125 - 67.5 = 57.5 \\\\\nb = 120 - 67.5 = 52.5 \\\\\na = 101 - 67.5 = 33.5\n$$\nSo the weights are $33.5\\ \\text{kg}$, $52.5\\ \\text{kg}$, $57.5\\ \\text{kg}$, $67.5\\ \\text{kg}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15029,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, and $d$ be real numbers with $a^2 + b^2 + c^2 + d^2 = 4$.\n\n*Prove that the inequality*\n\n$$\n(a + 2)(b + 2) \\ge cd\n$$\n\nholds, and give four numbers $a$, $b$, $c$, and $d$ such that equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "The given inequality is equivalent to $2ab + 4a + 4b + 8 \\ge 2cd$, which can be rewritten using the condition $a^2 + b^2 + c^2 + d^2 = 4$:\n\n$$\n2ab + 4a + 4b + a^2 + b^2 + c^2 + d^2 + 4 \\ge 2cd\n$$\n\nBy the identity\n\n$$\na^2 + b^2 + 2ab + 4a + 4b + 4 = (a + b + 2)^2\n$$\n\nwe obtain the equivalent inequality\n\n$$\n(a + b + 2)^2 + (c - d)^2 \\ge 0.\n$$\n\nEquality occurs when $a + b = -2$ and $c = d$, together with $a^2 + b^2 + c^2 + d^2 = 4$; for example, $a = b = c = d = -1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15030,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ for which all positive divisors of $n$, taken without repetitions, can be placed into a rectangular table in such a way that each cell contains exactly one divisor, all row sums are equal and all column sums are equal.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose all positive divisors of $n$ can be arranged in a rectangular table of size $k \\times l$, with $k \\le l$. Let the sum of each column be $s$. Since $n$ is a divisor, $s \\ge n$, and equality holds only if $k = 1$. For each $j = 1, 2, \\dots, l$, let $d_j$ be the largest number in the $j$th column, with $d_1 > d_2 > \\dots > d_l$. Since divisors are among $n, \\frac{n}{2}, \\frac{n}{3}, \\dots$, we have $d_l \\le \\frac{n}{l}$. The average in any column cannot exceed its maximum, so $d_l \\ge \\frac{s}{k} \\ge \\frac{n}{k}$. Thus, $\\frac{n}{k} \\le d_l \\le \\frac{n}{l}$, so $k \\ge l$. Since $k \\le l$, we get $k = l$, and all inequalities are equalities. In particular, $s = n$, so $k = l = 1$. Thus, $n$ has only one divisor, i.e., $n = 1$.\n\nObviously, $n = 1$ meets the conditions. Suppose $n > 1$ allows such an arrangement. If $n$ is a power of 2, all divisors except 1 are even, so at least two columns are needed. One column sum is odd, others are even, which is impossible. If $n$ has an odd prime divisor, let $n = p_1^{\\alpha_1} \\dots p_s^{\\alpha_s}$, with $s \\ge 1$, $p_1 < \\dots < p_s$ primes, and $\\alpha_i > 0$. Let $\\delta(n)$ be the number of positive divisors and $\\sigma(n)$ their sum. Since $n$ is in the table, row and column sums are at least $n$, and there are at least $\\sqrt{\\delta(n)}$ rows or columns, so $\\sigma(n) \\ge n\\sqrt{\\delta(n)}$. Define $f(n) = \\frac{\\sigma(n)}{n\\sqrt{\\delta(n)}}$, so $f(n) \\ge 1$. $f$ is weakly multiplicative: $f(n) = f(p_1^{\\alpha_1}) \\dots f(p_s^{\\alpha_s})$. For primes $p < q$, $f(p) > f(q)$, and for any prime $p$ and $k$, $f(p^k) > f(p^{k+1})$.\n\nFor $f(2) \\cdot f(3) = \\frac{3}{2\\sqrt{2}} \\cdot \\frac{4}{3\\sqrt{2}} = 1$, so $f(2) > 1 > f(3)$. By monotonicity, $f(n) \\le f(2) \\cdot f(3)^{s-1} \\le f(2) \\cdot f(3) = 1$. Thus, equality only if $n = 6$, but 6 has 4 divisors, and arranging them as required is impossible. Therefore, the only solution is $n = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15031,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $n$ be positive integers with $mn$ even. Jetze is going to cover an $m \\times n$ board (with $m$ rows and $n$ columns) with domino tiles, such that every domino tile covers exactly two squares, domino tiles do not protrude out of the board or overlap one another, and every square is covered by a domino tile. Merlijn then colours all domino tiles on the board either red or blue. Determine the smallest non-negative integer $V$ (depending on $m$ and $n$) such that Merlijn can always ensure that in each row, the number of squares covered by a red domino tile and the number of squares covered by a blue domino tile differ by at most $V$, no matter how Jetze covers the board.\n\n![](images/NLD_ABooklet_2021_p39_data_012780ddad.png)",
    "options": [],
    "answer": "See solution",
    "solution": "First, suppose that $n$ is odd. Then $V \\ge 1$, as the difference must be odd. We show that $V = 1$ is always possible. Colour the vertical domino tiles in odd-numbered columns red and those in even-numbered columns blue. In every row, each horizontal domino tile covers a square in an even-numbered column and one in an odd-numbered column, so every row contains one more square covered by a red domino tile than by a blue domino tile. Now, colour the horizontal domino tiles in each row alternately blue and red (starting with blue). If the number of horizontal domino tiles is even, the number of red squares will be one more than blue; if odd, blue will be one more than red. Thus, the difference is always $1$.\n\nNow suppose $n \\equiv 2 \\pmod{4}$. Then $V \\ge 2$ if Jetze places every domino tile horizontally; each row contains an odd number of horizontal domino tiles. We show $V = 2$ is always possible. Use the same strategy as in the odd case. After colouring the vertical domino tiles, the numbers of red and blue squares are equal. Alternately colouring the horizontal domino tiles in each row blue and red, in the end, in every row the difference between the number of red and blue squares is $0$ or $2$.\n\nFinally, suppose $n \\equiv 0 \\pmod{4}$. We show $V = 0$ is always possible. Number the rows from top to bottom $1$ to $m$, and let $b_i$ be the number of vertical domino tiles with the top square in row $i$. By induction on $i$, $b_i$ is even, since a horizontal domino tile always covers an even number of squares in a row. Colour the vertical domino tiles in rows $i$ and $i+1$ as follows: if $b_i \\equiv 0 \\pmod{4}$, colour half red and half blue; if $b_i \\equiv 2 \\pmod{4}$, colour two more domino tiles red than blue if $i$ is even, and two more blue than red if $i$ is odd. We can then colour the horizontal domino tiles in each row $k$ so that every row has the same number of red and blue squares. If $b_{k-1} \\equiv b_k \\equiv 0 \\pmod{4}$, vertical domino tiles in row $k$ cover equal numbers of red and blue squares, and the number of horizontal domino tiles is even, so colour half red and half blue. If $b_{k-1} \\equiv b_k \\equiv 2 \\pmod{4}$, vertical domino tiles in row $k$ again cover equal numbers of red and blue squares, since one of $k-1$ and $k$ is odd and one is even. Again, the number of horizontal domino tiles is even, so colour half red and half blue. If $b_{k-1} \\not\\equiv b_k \\pmod{4}$, the difference in the number of squares covered by red and blue vertical domino tiles is $2$. The number of horizontal domino tiles is odd, so colour those so that the number of red and blue squares are equal in the end.\n\nHence, the minimal values for $V$ are: $V = 1$ if $n$ is odd, $V = 2$ if $n \\equiv 2 \\pmod{4}$, and $V = 0$ if $n \\equiv 0 \\pmod{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15032,
    "subject": "Mathematics (Olympiad)",
    "question": "Given positive integers $m, n \\ge 2$, let $A = \\{a_1, a_2, \\dots, a_n\\}$ be a set of integers. Select two distinct elements $a_i, a_j$ ($j > i$) and compute the difference $a_j - a_i$. Arrange all $\\binom{n}{2}$ such differences in ascending order to form the 'derived sequence' $\\bar{A}$. Let $\\bar{A}(m)$ denote the number of elements in $\\bar{A}$ divisible by $m$. Prove that for any $m \\ge 2$, the derived sequences $\\bar{A}$ and $\\bar{B}$, corresponding to $A = \\{a_1, a_2, \\dots, a_n\\}$ and $B = \\{1, 2, \\dots, n\\}$, satisfy $$\\bar{A}(m) \\ge \\bar{B}(m).$$",
    "options": [],
    "answer": "See solution",
    "solution": "For any integer $m \\ge 2$, let $K_i$ denote the residue class modulo $m$ with remainder $i$, where $i \\in \\{0, 1, \\dots, m-1\\}$. Suppose in $A = \\{a_1, a_2, \\dots, a_n\\}$, the number of elements in $K_i$ is $n_i$, and in $B = \\{1, 2, \\dots, n\\}$, the number is $n'_i$. Then\n\n$$\n\\sum_{i=0}^{m-1} n_i = \\sum_{i=0}^{m-1} n'_i = n.\n$$\n\nFor every $i, j$, $|n'_i - n'_j| \\le 1$, and $x - y$ is divisible by $m$ if and only if $x, y$ are in the same residue class. Thus, the $n_i$ elements in $K_i$ yield $\\binom{n_i}{2}$ differences divisible by $m$.\n\nSumming over all $i$:\n\n$$\n\\bar{A}(m) = \\sum_{i=0}^{m-1} \\binom{n_i}{2}, \\qquad \\bar{B}(m) = \\sum_{i=0}^{m-1} \\binom{n'_i}{2}.\n$$\n\nWe need to show\n\n$$\n\\sum_{i=0}^{m-1} \\binom{n_i}{2} \\ge \\sum_{i=0}^{m-1} \\binom{n'_i}{2},\n$$\n\nwhich is equivalent to\n\n$$\n\\sum_{i=0}^{m-1} n_i^2 \\ge \\sum_{i=0}^{m-1} n_i'^2.\n$$\n\nIf $|n_i - n_j| \\le 1$ for all $i, j$, then $n_i$ and $n'_i$ are the same up to order, and equality holds. Otherwise, if some $n_i - n_j \\ge 2$, adjusting $n_i$ and $n_j$ to $n_i - 1$ and $n_j + 1$ (preserving the sum) decreases the total $\\sum n_i^2$. Thus, the minimum is achieved when $n_i$ are as balanced as possible, i.e., matching $n'_i$. Therefore, the inequality holds.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15033,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a frog making jumps in both the horizontal and vertical directions, where each jump consists of an odd or even number of steps, respectively. The frog can end up on the $x$-axis if the last vertical jump consists of $8k - 2$ or $8k$ steps. Investigate whether the frog can also arrive back on the $y$-axis, and hence at the origin $(0,0)$. The last horizontal jump must consist of $8k - 3$, $8k - 1$, or $8k + 1$ steps. Is it possible to assign pluses and minuses so that $\\pm 1 \\pm 3 \\pm \\cdots \\pm n = 0$ for $n$ of the form $8k-3$, $8k-1$, or $8k+1$? Prove your answer.",
    "options": [],
    "answer": "See solution",
    "solution": "We conclude that there are two possibilities for the frog to end at the origin $(0,0)$. The first is for $n = 8k - 1$: the second to last jump consists of $8k - 2$ vertical steps, and the last jump consists of $8k - 1$ horizontal steps. The second is $n = 8k$: then the second last jump consists of $8k - 1$ horizontal steps and the last jump consists of $8k$ vertical steps. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15034,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the general term of the sequence $\\{a_n\\}$ be\n$$\na_n = \\frac{1}{\\sqrt{5}} \\left( \\left( \\frac{1 + \\sqrt{5}}{2} \\right)^n - \\left( \\frac{1 - \\sqrt{5}}{2} \\right)^n \\right), \\quad n = 1, 2, \\dots\n$$\nProve that there exist infinitely many positive integers $m$ such that $a_{m+4}a_m - 1$ are perfect squares.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $q_1 = \\frac{1+\\sqrt{5}}{2}$, $q_2 = \\frac{1-\\sqrt{5}}{2}$, so $q_1 + q_2 = 1$, $q_1q_2 = -1$. Thus,\n$$\na_n = \\frac{1}{\\sqrt{5}}(q_1^n - q_2^n), \\quad n = 1, 2, \\dots\n$$\nHence $a_1 = 1$, $a_2 = 1$. Also note that $q_i + 1 = q_i^2$ for $i = 1, 2$, and\n$$\n\\begin{aligned}\na_{n+1} + a_n &= \\frac{1}{\\sqrt{5}}(q_1^{n+1} - q_2^{n+1}) + \\frac{1}{\\sqrt{5}}(q_1^n - q_2^n) \\\\\n&= \\frac{1}{\\sqrt{5}}(q_1^n(q_1+1) - q_2^n(q_2+1)) \\\\\n&= \\frac{1}{\\sqrt{5}}(q_1^{n+2} - q_2^{n+2}),\n\\end{aligned}\n$$\nnamely,\n$$\na_{n+2} = a_{n+1} + a_n, \\quad n = 1, 2, \\dots\n$$\nIt is easy to see that each term of the sequence $\\{a_n\\}$ is a positive integer.\n\nIt is easy to calculate that $q_1^4 + q_2^4 = 7$, and hence\n$$\n\\begin{aligned}\na_{2n+3}a_{2n-1} - 1 &= \\frac{1}{\\sqrt{5}}(q_1^{2n+3} - q_2^{2n+3}) \\cdot \\frac{1}{\\sqrt{5}}(q_1^{2n-1} - q_2^{2n-1}) - 1 \\\\\n&= \\frac{1}{5}(q_1^{4n+2} + q_2^{4n+2} - (q_1q_2)^{2n-1}q_1^4 - (q_1q_2)^{2n-1}q_2^4) - 1\n\\end{aligned}\n$$\n$$\n\\begin{align*}\n&= \\frac{1}{5}(q_1^{4n+2} + q_2^{4n+2} + q_1^4 + q_2^4) - 1 \\\\\n&= \\frac{1}{5}(q_1^{4n+2} + q_2^{4n+2} + 7) - 1 \\\\\n&= \\frac{1}{5}(q_1^{4n+2} + q_2^{4n+2} + 2) \\\\\n&= \\left[ \\frac{1}{\\sqrt{5}}(q_1^{2n+1} + q_2^{2n+1}) \\right]^2 \\\\\n&= a_{2n+1}^2.\n\\end{align*}\n$$\nTherefore, for any positive integer $n$, $a_{2n+3}a_{2n-1} - 1$ is a perfect square.\nHence, $a_{m+4}a_m - 1$ are perfect squares for all positive odd numbers $m$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15035,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be the lengths of the legs of a given right triangle. Prove that an angle $\\varphi$, where $0 < \\varphi < 90^\\circ$, is an acute angle of this triangle if and only if $$(a \\cos \\varphi + b \\sin \\varphi)(a \\sin \\varphi + b \\cos \\varphi) = 2ab.$$",
    "options": [],
    "answer": "See solution",
    "solution": "The equality given in the problem is equivalent to\n$$\n(a^2 + b^2) \\sin \\varphi \\cos \\varphi + ab(\\sin^2 \\varphi + \\cos^2 \\varphi) = 2ab\n$$\nand hence also to\n$$\n(a^2 + b^2) \\sin \\varphi \\cos \\varphi = ab. \\qquad (1)\n$$\nLet $\\alpha$ and $\\beta$ be the angles opposite to legs with length $a$ and $b$, respectively. Then $\\sin \\alpha = \\dfrac{a}{\\sqrt{a^2 + b^2}}$, $\\sin \\beta = \\cos \\alpha = \\dfrac{b}{\\sqrt{a^2 + b^2}}$, implying\n$$\n(a^2 + b^2) \\sin \\alpha \\cos \\alpha = ab.\n$$\nComparing this to (1) shows the equivalence of the equality of the problem and the equality $\\sin \\varphi \\cos \\varphi = \\sin \\alpha \\cos \\alpha$, i.e., $\\sin 2\\varphi = \\sin 2\\alpha$. As $0 < \\alpha, \\beta < 90^\\circ$, this implies $2\\varphi = 2\\alpha$ or $2\\varphi = 180^\\circ - 2\\alpha$, whence $\\varphi = \\alpha$ or $\\varphi = 90^\\circ - \\alpha = \\beta$. Hence, $\\varphi$ satisfies the equality if and only if it equals one of the acute angles of the right triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15036,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x * y = \\frac{x}{xy + 1}$ for positive real numbers $x$ and $y$.\n\nCompute:\n$$\n((\\cdots((((100 * 99) * 98) * 97) * 96) * \\cdots * 3) * 2) * 1\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We have the property:\n$$\n(x * y) * z = \\frac{x}{x(y + z) + 1} = x * (y + z).\n$$\n\nApplying this repeatedly:\n$$\n((\\cdots((((100 * 99) * 98) * 97) * 96) * \\cdots * 3) * 2) * 1 = 100 * (99 + 98 + \\cdots + 2 + 1)\n$$\nThe sum $99 + 98 + \\cdots + 2 + 1 = \\frac{99 \\cdot 100}{2} = 4950$.\n\nSo,\n$$\n100 * 4950 = \\frac{100}{100 \\cdot 4950 + 1} = \\frac{100}{495001}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15037,
    "subject": "Mathematics (Olympiad)",
    "question": "If $t$ toffees cost $c$ cents, the number of toffees that can be bought for $r$ rands is\n\n(A) $\\frac{100rc}{t}$  \n(B) $\\frac{100rt}{c}$  \n(C) $\\frac{100r}{ct}$  \n(D) $\\frac{rt}{100c}$  \n(E) $\\frac{100c}{rt}$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $t$ toffees cost $c$ cents, each toffee costs $\\frac{c}{t}$ cents.  \n$r$ rands equals $100r$ cents.\n\n$$\n\\text{The number of toffees that can be bought for } 100r \\text{ cents is thus } 100r \\div \\frac{c}{t} = \\frac{100rt}{c}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15038,
    "subject": "Mathematics (Olympiad)",
    "question": "Two rectangles of unit area overlap to form a convex octagon. Show that the area of the octagon is at least $\\frac{1}{2}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Each side of one rectangle meets the contour of the other rectangle at exactly two points situated on consecutive sides. Let $A_0A_1A_2A_3$ and $B_0B_1B_2B_3$ be circular labellings of the two rectangles such that the segments $A_iA_{i+1}$ and $B_iB_{i+1}$ have equal lengths and meet at a point labelled $C_i$. The $C_i$ are four alternative vertices of the octagon, so the area of the latter is greater than or equal to the area of the quadrangle $C_0C_1C_2C_3$. Notice that $C_i$ is equally distanced from the lines $A_{i+2}A_{i+3}$ and $B_{i+2}B_{i+3}$ to deduce that the lines $C_0C_2$ and $C_1C_3$ are perpendicular to one another. Consequently,\n\n$$\n\\text{area } C_0C_1C_2C_3 = \\frac{1}{2} \\cdot C_0C_2 \\cdot C_1C_3 \\geq \\frac{1}{2} \\cdot A_1A_2 \\cdot A_0A_1 = \\frac{1}{2}\n$$\n\nThe conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15039,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer, set $A \\subseteq \\{1, 2, \\dots, n\\}$, and for every $a, b \\in A$, $\\text{lcm}(a, b) \\le n$. Prove that\n\n$$\n|A| \\le 1.9\\sqrt{n} + 5.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof**\n\nFor $a \\in (\\sqrt{n}, \\sqrt{2n}]$, $\\text{lcm}(a, a+1) = a(a+1) > n$, so\n$$\n|A \\cap (\\sqrt{n}, \\sqrt{2n}]| \\le \\frac{1}{2}(\\sqrt{2}-1)\\sqrt{n} + 1.\n$$\n\nFor $a \\in (\\sqrt{2n}, \\sqrt{3n}]$, we have\n$$\n\\text{lcm}(a, a+1) = a(a+1) > n,\n$$\n$$\n\\text{lcm}(a+1, a+2) = (a+1)(a+2) > n,\n$$\n$$\n\\text{lcm}(a, a+2) \\ge \\frac{1}{2}a(a+2) > n.\n$$\nSo\n$$\n|A \\cap (\\sqrt{2n}, \\sqrt{3n}]| \\le \\frac{1}{3}(\\sqrt{3}-\\sqrt{2})\\sqrt{n} + 1.\n$$\n\nSimilarly,\n$$\n|A \\cap (\\sqrt{3n}, 2\\sqrt{n}]| \\le \\frac{1}{4}(\\sqrt{4}-\\sqrt{3})\\sqrt{n} + 1.\n$$\n\nHence,\n$$\n\\begin{aligned}\n|A \\cap [1, 2\\sqrt{n}]| & \\le \\sqrt{n} + \\frac{1}{2}(\\sqrt{2}-1)\\sqrt{n} + \\frac{1}{3}(\\sqrt{3}-\\sqrt{2})\\sqrt{n} \\\\\n& \\quad + \\frac{1}{4}(\\sqrt{4}-\\sqrt{3})\\sqrt{n} + 3 \\\\\n& = \\left(1 + \\frac{\\sqrt{2}}{6} + \\frac{\\sqrt{3}}{12}\\right)\\sqrt{n} + 3.\n\\end{aligned}\n$$\n\nLet $k \\in \\mathbb{N}^*$, suppose $a, b \\in (\\frac{n}{k+1}, \\frac{n}{k})$, $a > b$, and $\\text{lcm}(a, b) = as = bt$, where $s, t \\in \\mathbb{N}^*$. Then\n$$\n\\frac{a}{(a, b)s} = \\frac{b}{(a, b)t}.\n$$\nSince $\\gcd\\left(\\frac{a}{(a, b)}, \\frac{b}{(a, b)}\\right) = 1$, so $\\frac{b}{(a, b)} \\mid s$. It follows that",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15040,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M = \\{1, 2, 4, 5, 7, 8, \\dots\\}$ be the set of all positive integers not divisible by $3$. The sum of $2n$ consecutive elements of $M$ is $300$. Determine the possible values of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S_i$ be the sum of the first $i$ numbers in the set $M$.\n\n$$\n\\begin{aligned}\nS_{2k} &= (1+2) + (4+5) + \\dots + (3k-2+3k-1) \\\\\n&= (6 \\cdot 1 - 3) + (6 \\cdot 2 - 3) + \\dots + (6k-3) \\\\\n&= 3k(k+1) - 3k = 3k^2.\n\\end{aligned}\n$$\n\nThen $S_{2k+1} = S_{2k} + 3k + 1 = 3k^2 + 3k + 1$.\n\n*Case 1.* The sum of $2n$ consecutive elements of $M$ is\n\n$$\na_{2k+1} + a_{2k+2} + \\dots + a_{2l} = S_{2l} - S_{2k},$$\n\nwhere $n = l - k$. We obtain $3(l^2 - k^2) = 300$, so\n\n$$\n(l-k)(l+k) = 100,\n$$\n\ngiving the systems\n\n$$\n\\begin{cases} l - k = 1 \\\\ l + k = 100 \\end{cases} \\quad ; \\quad \\begin{cases} l - k = 2 \\\\ l + k = 50 \\end{cases} \\quad ; \\quad \\begin{cases} l - k = 4 \\\\ l + k = 25 \\end{cases} ; \\\\\n\\begin{cases} l - k = 5 \\\\ l + k = 20 \\end{cases} ; \\quad \\begin{cases} l - k = 10 \\\\ l + k = 10 \\end{cases}\n$$\n\nOnly the second and the last are solvable in integers, and we obtain $l = 26$, $k = 24$ and $l = 10$, $k = 0$. Thus we get $n = 2$, $n = 10$.\n\n*Case 2.* The sum of $2n$ consecutive elements of $M$ is\n\n$$\na_{2k} + a_{2k+1} + \\dots + a_{2l-1},$$\n\nwhere $n = l - k$, $k \\ge 1$. We obtain $3(l^2 + l - k^2 - k) = 300$, and so $(l-k)(l+k+1) = 100$.\n\nThe solutions that work are\n\n$$\n\\begin{cases} l - k = 1 \\\\ l + k = 100 \\end{cases} \\quad ; \\quad \\begin{cases} l - k = 4 \\\\ l + k = 25 \\end{cases} \\quad ; \\quad \\begin{cases} l - k = 5 \\\\ l + k = 20 \\end{cases}\n$$\n\nWe get $l = 50, k = 49$, $l = 14, k = 10$, $l = 12, k = 7$, meaning that $n = 1, 4, 5$.\n\nThe solutions are $n \\in \\{1, 2, 4, 5, 10\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15041,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1 = 2$, $a_2 = 3$. For $i = 3, \\dots, n-1$, let $a_i = a_1 a_2 \\cdots a_{i-1} + 1$. Let $a_n = a_1 a_2 \\cdots a_{n-1} - 1$.\n\nWhat can be said about the greatest common divisors or congruence properties of the sequence $a_1, a_2, \\dots, a_n$?",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, $a_1 a_2 \\cdots a_{n-1} \\equiv 1 \\pmod{a_n}$. Also, $a_{i+1} \\equiv a_{i+2} \\equiv \\cdots \\equiv a_{n-1} \\equiv 1 \\pmod{a_i}$. For $i = 1, \\dots, n-1$, we have\n\n$$\na_1 a_2 \\cdots \\hat{a}_i \\cdots a_n = (a_1 \\cdots a_{i-1})(a_{i+1} \\cdots a_{n-1})a_n \\equiv (-1)(-1)(-1) \\equiv 1 \\pmod{a_i}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15042,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\dots, x_n$ be real numbers. Consider the expression:\n\n$$\nE = \\frac{x_1}{x_n + x_1 + x_2} + \\frac{x_2}{x_1 + x_2 + x_3} + \\frac{x_3}{x_2 + x_3 + x_4} + \\dots + \\frac{x_n}{x_{n-1} + x_n + x_1}.\n$$\n\nDetermine all possible real values of $E$ as $x_1, \\dots, x_n$ range over all real numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is all real numbers in the interval $]1, \\lfloor n/2 \\rfloor[$.\n\n*Lower bound:* Let $S = x_1 + x_2 + \\dots + x_n$. Notice that\n\n$$\nE > \\frac{x_1}{S} + \\frac{x_2}{S} + \\dots + \\frac{x_n}{S} = 1.\n$$\n\nBy choosing $x_i = \\epsilon^{i-1}$, we get $E$ arbitrarily close to $1$ as $\\epsilon \\to 0$.\n\n*Upper bound:* For $n$ even,\n\n$$\nE < \\frac{x_1}{x_1 + x_2} + \\frac{x_2}{x_1 + x_2} + \\dots + \\frac{x_n}{x_{n-1} + x_n} = \\frac{n}{2} = \\lfloor \\frac{n}{2} \\rfloor.\n$$\n\nFor $n$ odd, the sum can be split so that\n\n$$\nE < 1 + \\frac{n-3}{2} = \\lfloor \\frac{n}{2} \\rfloor.\n$$\n\nTo attain the upper bound, choose $x_{2k} = \\epsilon$ and $x_{2k-1} = 1$ for $k = 1, 2, \\dots, \\lfloor n/2 \\rfloor$. Then $E$ gets arbitrarily close to $\\lfloor n/2 \\rfloor$ as $\\epsilon \\to 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15043,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\ldots, x_{10}$ be positive integers such that\n$$\nx_1 + 2^3 x_2 + 3^3 x_3 + \\cdots + 10^3 x_{10} = 3025.\n$$\nFind all possible solutions $(x_1, x_2, \\ldots, x_{10})$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $1^3 + 2^3 + \\cdots + 10^3 = 3025$. Let $x_i$ be a positive integer solution. Set $y_i = x_i - 1$, so $y_i \\geq 0$. Then:\n$$\n(y_1 + 1) + 2^3(y_2 + 1) + 3^3(y_3 + 1) + \\cdots + 10^3(y_{10} + 1) = 3025.\n$$\nExpanding, we get:\n$$\ny_1 + 2^3 y_2 + 3^3 y_3 + \\cdots + 10^3 y_{10} + (1 + 2^3 + 3^3 + \\cdots + 10^3) = 3025.\n$$\nBut $1 + 2^3 + 3^3 + \\cdots + 10^3 = 3025$, so:\n$$\ny_1 + 2^3 y_2 + 3^3 y_3 + \\cdots + 10^3 y_{10} = 0.\n$$\nSince all $y_i \\geq 0$ and the coefficients are positive, the only solution is $y_i = 0$ for all $i$. Thus, the unique solution is $x_i = 1$ for all $i$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15044,
    "subject": "Mathematics (Olympiad)",
    "question": "For nonnegative real numbers $a, b, c$ with $a + b + c \\leq 2$, prove\n\n$$\nab(a^2 + b^2) + bc(b^2 + c^2) + ca(c^2 + a^2) \\leq 2.$$\n\nCan the inequality become equality?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** Equality occurs.\n\n**Solution.** It is enough to prove the inequality for $a + b + c = 2$.\n\nLet us denote $x = ab + bc + ca$. Then\n\n$$\na^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = 4 - 2(ab + bc + ca) = 2(2 - x).\n$$\n\nFrom the obvious inequality $x(2 - x) \\leq 1$, we get\n\n$$\n\\begin{aligned}\n2 + 2abc &\\geq 2 \\geq 2x(2 - x) = (ab + bc + ca)(a^2 + b^2 + c^2) \\\\\n&= ab(a^2 + b^2) + bc(b^2 + c^2) + ca(c^2 + a^2) + abc(a + b + c).\n\\end{aligned}\n$$\n\nUsing the equality $2abc = abc(a + b + c)$, we arrive at the conclusion.\n\nEquality occurs, for example, at $a = b = 1$, $c = 0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15045,
    "subject": "Mathematics (Olympiad)",
    "question": "In an oral exam, there are 10 examiners and 1024 contestants. Each contestant will be asked by each examiner and receive a result of either \"pass\" or \"fail\". It is given that for any two contestants, there is some examiner who will rate one passed and the other failed. Two contestants are said to be \"separative\" if they have different results by at least 3 examiners. Prove that it is possible to select 24 pairwise separative contestants.",
    "options": [],
    "answer": "See solution",
    "solution": "Number the examiners from 1 to 10 and consider the results of contestants as binary sequences of the form $x_1x_2\\ldots x_{10}$, where $x_i = 0$ or $1$ if the $i$th examiner gave a pass or fail response, respectively. We can directly construct 24 binary strings that satisfy the \"separative\" condition. \n\nIndeed, choose strings of the form $(x_1x_2x_3x_4x_5x_6)(x_7x_8x_9x_{10})$, where $x_1x_2x_3x_4x_5x_6$ contains exactly 3 zeros and 3 ones, and $x_7x_8x_9x_{10}$ runs through all binary strings of length 4. Since $\\binom{6}{3} = 20$ and two strings $x_1x_2x_3x_4x_5x_6 \\neq x'_1x'_2x'_3x'_4x'_5x'_6$ differ in at least 2 positions, the rest just need to differ by at least 1 position. Since $2^4 = 16$, there will be 4 repeated strings in the last 4 positions, so we just need to choose those strings that differ by at least 3 positions out of the first 6 positions. For detail:\n\n$$\n\\begin{array}{lcl}\n(000111)(0000) & \\rightarrow & (a) \\\\\n(001011)(0001) & \\rightarrow & (b) \\\\\n(001101)(0010) & \\rightarrow & (c) \\\\\n(001110)(0011) & \\rightarrow & (d) \\\\\n(010011)(0100) \\\\\n(010101)(0101) \\\\\n(010110)(0110) \\\\\n(011001)(0111) \\\\\n(011010)(1000) \\\\\n(011100)(1001) \\\\\n(100011)(1010) \\\\\n(100101)(1011) \\\\\n(100110)(1100) \\\\\n(101001)(1101) \\\\\n(101010)(1110) \\\\\n(101100)(1111) \\\\\n(110001)(0011) & \\rightarrow & (d') \\\\\n(110010)(0010) & \\rightarrow & (c') \\\\\n(110100)(0001) & \\rightarrow & (b') \\\\\n(111000)(0000) & \\rightarrow & (a').\n\\end{array}\n$$\n\nFinally, add 4 more strings as follows:\n\n$$\n\\begin{array}{l}\n(000000)(0000),\\ (000000)(1111), \\\\\n(111111)(0000),\\ (111111)(1111).\n\\end{array}\n$$\n\nIt is easy to check that they are pairwise separative and also separative with the other 20 strings. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15046,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\dots, x_{2024}$ be non-negative real numbers such that $x_1 \\le x_2 \\le \\dots \\le x_{2024}$ and $x_1^3 + x_2^3 + \\dots + x_{2024}^3 = 2024$. Prove that\n\n$$\n\\sum_{1 \\le i < j \\le 2024} (-1)^{i+j} x_i^2 x_j \\ge -1012.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We want to show that\n\n$$\n\\sum_{1 \\le i < j \\le 2024} (-1)^{i+j} x_i^2 x_j \\ge -1012.\n$$\n\nObserve that the left-hand side is\n\n$$\n- \\left( \\sum_{i=1}^{1012} x_{2i-1}^2 x_{2i} \\right) + \\sum_{i=1}^{1012} \\left( (x_{2i}^2 - x_{2i-1}^2) \\left( \\sum_{j<i \\le 1012} (x_{2j} - x_{2j-1}) \\right) \\right).\n$$\n\nSince $x_{2i}^2 \\ge x_{2i-1}^2$ and $x_{2j} \\ge x_{2j-1}$, we will be done if $\\sum_{i=1}^{1012} x_{2i-1}^2 x_{2i} \\le 1012$. But,\n\n$$\n3x_{2i-1}^2 x_{2i} \\le 2x_{2i-1}^3 + x_{2i}^3 \\le 3 \\cdot \\frac{x_{2i-1}^3 + x_{2i}^3}{2}.\n$$\n\nThus, we just need $\\sum_{i=1}^{1012} \\frac{x_{2i-1}^3 + x_{2i}^3}{2} \\le 1012$, but this is precisely the given condition, so we are done!\n\nObserve that all equalities hold if $x_{2i-1} = x_{2i}$ for all $i$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15047,
    "subject": "Mathematics (Olympiad)",
    "question": "Every integer $N$ satisfies at least one of the congruences $N \\equiv 0 \\pmod{2}$, $N \\equiv 1 \\pmod{3}$, $N \\equiv 3 \\pmod{4}$, $N \\equiv 5 \\pmod{6}$, $N \\equiv 9 \\pmod{12}$. \n\nShow that $n \\leq 4$ is not possible if we require that every integer satisfies at least one of $n$ congruences $x \\equiv a_i \\pmod{k_i}$ with $1 < k_1 < k_2 < \\dots < k_n$ and $a_1, \\dots, a_n$ integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $1 < k_1 \\leq k_2 \\leq \\dots \\leq k_n$ and $a_1, a_2, \\dots, a_n$ be integers, and let $K = \\operatorname{lcm}(k_1, k_2, \\dots, k_n)$. Since at most $\\frac{K}{k_1} + \\frac{K}{k_2} + \\dots + \\frac{K}{k_n}$ integers from $1$ to $K$ can satisfy at least one of the congruences $x \\equiv a_i \\pmod{k_i}$ for $1 \\leq i \\leq n$, we must have $\\frac{1}{k_1} + \\frac{1}{k_2} + \\dots + \\frac{1}{k_n} \\geq 1$ if every integer satisfies at least one of these congruences.\n\nNow assume that $1 < k_1 < k_2 < \\dots < k_n$ and $a_1, a_2, \\dots, a_n$ satisfy the condition of the problem and that $n \\leq 4$ has the smallest possible value. If $k_1 = 3$, then\n$$\n\\frac{1}{k_1} + \\frac{1}{k_2} + \\dots + \\frac{1}{k_n} \\leq \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5} + \\frac{1}{6} = \\frac{19}{20} < 1.\n$$\nTherefore $k_1 = 2$.\n\nWithout loss of generality, we may assume that $a_1 = 1$. For $2 \\leq i \\leq n$, let $k'_i = k_i$ and $a'_i = 2^{-i} a_i$ (mod $k_i$) if $k_i$ is odd, and let $k'_i = \\frac{k_i}{2}$ and $a'_i = \\frac{a_i}{2}$ if $k_i$ is even. The integers $k'_2, \\dots, k'_n$ and $a'_2, \\dots, a'_n$ satisfy the condition of the problem except that $k'_i$ might not be distinct. Therefore, by the minimality of $n$, we must have $n = 4$ and $\\{k_2, k_3, k_4\\} = \\{2m+1, 4m+2, k\\}$.\n\n- If $k$ is odd, then $\\{k'_2, k'_3, k'_4\\} = \\{2m+1, 2m+1, k\\}$ and $\\frac{2}{2m+1} + \\frac{1}{k} \\geq 1$. Since $\\frac{2}{2m+1} + \\frac{1}{k} \\leq \\frac{2}{3} + \\frac{1}{5} = \\frac{13}{15} < 1$, this is not possible.\n\n- If $k$ is even, then $\\{k'_2, k'_3, k'_4\\} = \\{2m+1, 2m+1, \\frac{k}{2}\\}$ and $\\frac{2}{2m+1} + \\frac{2}{k} \\geq 1$. If $2m+1 \\geq 5$ or $2m+1=3$ and $k \\geq 8$, we get contradictions because\n\n$$\n\\frac{2}{5} + \\frac{2}{4} = \\frac{9}{10} < 1 \\quad \\text{and} \\quad \\frac{2}{3} + \\frac{2}{8} = \\frac{11}{12} < 1.\n$$\nThe only remaining case is $2m+1=3$ and $k=4$. This gives $\\{k'_2, k'_3, k'_4\\} = \\{3, 3, 2\\}$. Since the integers in a congruence class modulo $3$ cannot be all even or all odd, this also leads to a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15048,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $I$ be the incenter of triangle $ABC$ and let $k$ be a circle through the points $A$ and $B$. This circle intersects\n\n- the line $AI$ in points $A$ and $P$,\n- the line $BI$ in points $B$ and $Q$,\n- the line $AC$ in points $A$ and $R$, and\n- the line $BC$ in points $B$ and $S$,\n\nwith none of the points $A, B, P, Q, R,$ and $S$ coinciding and such that $R$ and $S$ are interior points of the line segments $AC$ and $BC$, respectively.\n\nProve that the lines $PS$, $QR$, and $CI$ meet in a single point.\n\n![](images/Austria_2015_booklet_p13_data_39faea6f3b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We define angles $\\alpha = \\angle BAC$ and $\\beta = \\angle CBA$ as usual. Since points $A$, $B$, $S$, and $R$ lie on a common circle, we have $\\angle BSR = 180^\\circ - \\alpha$, and therefore $\\angle RSC = \\alpha$. Similarly, $\\angle CRS = \\beta$ also holds.\n\nIf $P$ lies in the interior of $ABC$, we have $\\angle RSP = \\angle RAP = \\alpha/2$. This means that $PS$ bisects the angle $\\angle CSR$.\n\nIf $Q$ is outside of $ABC$, we have $\\angle QRA = \\angle QBA = \\beta/2$, and in this case $QR$ also bisects the angle $\\angle SRC$.\n\nIndependent of the positioning of $Q$ and $R$ with respect to the triangle, we therefore see that $QR$, $PS$, and $CI$ are the bisectors of the interior angles of $\\triangle CRS$, and they therefore meet in the incenter of this triangle, as claimed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15049,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x)$ be a periodic function with periods $T$ and $1$, satisfying $0 < T < 1$. Prove that:\n\n1. If $T$ is a rational number, there exists a prime $p$ such that $\\frac{1}{p}$ is also a period of $f(x)$.\n2. If $T$ is irrational, there exists a sequence of irrational numbers $\\{a_n\\}$, satisfying $1 > a_n > a_{n+1} > 0$ ($n = 1, 2, \\ldots$) and each $a_n$ is a period of $f(x)$.",
    "options": [],
    "answer": "See solution",
    "solution": "(1) Suppose $T$ is rational, so there exist positive integers $m, n$ with $T = \\frac{n}{m}$ and $\\gcd(m, n) = 1$. There exist integers $a, b$ such that $ma + nb = 1$. Thus,\n\n$$\n\\frac{1}{m} = \\frac{ma + nb}{m} = a \\cdot 1 + b \\cdot T\n$$\n\nis also a period of $f(x)$. Since $0 < T < 1$, $m \\geq 2$, so $m = pm'$ for some prime $p$. Therefore, $\\frac{1}{p} = m' \\cdot \\frac{1}{m}$ is also a period of $f(x)$.\n\n(2) If $T$ is irrational, define $a_1 = 1 - \\lfloor \\frac{1}{T} \\rfloor T$. Then $0 < a_1 < 1$ and $a_1$ is irrational. Define recursively:\n\n$$\na_{n+1} = 1 - \\lfloor \\frac{1}{a_n} \\rfloor a_n.\n$$\n\nBy induction, each $a_n$ is irrational and $0 < a_n < 1$. Also, $a_{n+1} < a_n$, so $\\{a_n\\}$ is decreasing. By construction, each $a_n$ is a period of $f(x)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15050,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $A$, $B$, $C$, $D$ lie on a circle in this order, where $AB$ and $CD$ are not parallel. The length of the arc $\\widehat{AB}$ that contains points $C$, $D$ is twice as large as the length of the arc $\\widehat{CD}$ that does not contain points $A$, $B$. Point $E$ is chosen such that $AC = AE$ and $BD = BE$, and $E$ lies on the same side of the line $AB$ as $C$ and $D$. Assuming that the perpendicular line from the point $E$ to the line $AB$ bisects the arc $\\widehat{CD}$ not containing points $A$, $B$, prove that $\\angle ACB = 108^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use the following lemma:\n\n**Lemma.** Given two circles $\\Gamma_1$, $\\Gamma_2$ with the centre $S_2$ of $\\Gamma_2$ lying on the circle $\\Gamma_1$. The circles intersect in two points $K$ and $L$. Let $M$ be the point on the circle $\\Gamma_1$ (different from $K$ and $L$) and the line $KM$ meets $\\Gamma_2$ again in $N$. Then $MN = ML$.\n\n*Proof.*\n\n![](images/CzsMT2011_sol_p1_data_12379b495b.png)\n\nFig. 1a\n\n![](images/CzsMT2011_sol_p1_data_45c8f8f53b.png)\n\nFig. 1b\n\nFor this lemma it is sufficient to prove that the line $MS_2$ bisects the angle $\\angle NML$. Then in the reflection with respect to the line $MS_2$, $ML$ is the image of $MN$. The circle $\\Gamma_2$ and the point $M$ reflect to themselves, the point $L$ reflects to the point $N$ (the intersection point of $ML$ and $\\Gamma_2$). So the triangle $\\triangle MLN$ is isosceles (some considerations are needed according to the position of the point $M$).\n\nFirstly, let $M$ lie on the arc $\\widehat{KL}$ not containing the point $S_2$. As $S_2K = S_2L$, we directly have $\\angle KMS_2 = \\angle S_2ML$. Secondly, let $M$ lie on the arc $\\widehat{KL}$ containing point $S_2$. Let $R$ be an arbitrary point on the arc $\\widehat{KL}$ not containing the point $S_2$. Similarly as before $\\angle KRS_2 = \\angle S_2RL$, then using identical angles in the cyclic quadrilaterals $RS_2MK$ and $RLS_2M$ we obtain $\\angle NMS_2 = \\angle KRS_2 = \\angle S_2RL = \\angle S_2ML$.\n\nLet the perpendicular line from the point $E$ to the line $AB$ intersect the arc $\\widehat{BC}$ in the point $S$, $k_1$ be the circle centered at $A$ passing through $C$, $k_2$ be the circle centered at $B$ passing through $D$, $k$ be the circle passing through $A$, $B$, $C$, $D$. The line $SC$ intersects $k_1$ again in $C'$, and the line $SD$ intersects $k_2$ again in $D'$. Circles $k_1$ and $k$ meet in $C$, $C''$, and circles $k_2$ and $k$ meet in $D$, $D''$. Using the lemma we have $SC' = SC''$ and $SD' = SD''$. Let the circles $k_1$ and $k_2$ meet again in $E'$. Using a contradiction, we shall prove that $C'' = D'' = E'$.\n\nThe point $S$ lies on the chord $EE' \\perp AB$ of the circles $k_1$ and $k_2$. That means its powers to these two circles are equal. We know that $S$ bisects the arc $\\widehat{CD}$, so $SD = SC$ and consequently $SC'' = SC' = SD' = SD''$. If $D''$ and $C''$ are different points then the triangle $\\triangle SD''C''$ is isosceles and its altitude from $S$ passes through the circumcentre of $k$ and consequently (by the symmetry) the quadrilateral $CDD''C''$ is an isosceles trapezoid ($SC = SD$). We can find points $A$ and $B$ as the intersection points of the axes of the segments $CC''$ and $DD''$ with $k$. But then also $ABCD$ is an isosceles trapezoid, $AB \\parallel CD$, which is a contradiction to the given $AB \\nparallel CD$.\n\n![](images/CzsMT2011_sol_p2_data_4cecdf28de.png)\n\nIf we denote $\\angle DE'S = \\angle SE'C = \\alpha$ and $\\angle AE'D = \\beta$ then $\\angle CE'D = 2\\alpha - \\beta$ because $2|\\widehat{CD}| = |\\widehat{AB}|$. Using $BD = BE'$ and $AC = AE'$ we compute the angles in the triangle $ABE'$:\n\n$$2\\alpha + \\beta = \\angle AE'C = \\angle ACE' = \\angle ABE'$$\n$$4\\alpha - \\beta = \\angle BE'D = \\angle BDE' = \\angle BAE'$$\n\nwhich yields\n\n$$180^{\\circ} = 2\\alpha + \\beta + 4\\alpha - \\beta + 4\\alpha = 10\\alpha$$\n\nand\n\n$$\\angle ACB = 180^{\\circ} - \\angle AE'B = 180^{\\circ} - 4\\alpha = 108^{\\circ}.$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15051,
    "subject": "Mathematics (Olympiad)",
    "question": "Бүх $x, y \\in \\mathbb{R}$ хувьд\n\n$$\nf(x + f(y)) = f(x) + \\frac{1}{8} x f(4y) + f(f(y))\n$$\n\nбайх бүх $f : \\mathbb{R} \\to \\mathbb{R}$ функцийг ол.",
    "options": [],
    "answer": "See solution",
    "solution": "$f(x) = 0$ нь илэрхий хариу. Одоо $\\exists t \\in \\mathbb{R}$ тэгш бус $f(t) \\neq 0$ гэж үзье.\n\n$(x, y) \\to (0, 0)$ гэж авбал $f(0) = 0$ гарна. $(x, y) \\to (f(x), f(t))$ гэж авбал:\n\n$$\nf(f(x) + f(t)) = f(f(x)) + \\frac{1}{8} f(x) f(4t) + f(f(t))\n$$\n\n$(x, y) \\to (f(t), f(x))$ гэж авбал:\n\n$$\nf(f(x) + f(t)) = f(f(x)) + \\frac{1}{8} f(t) f(4x) + f(f(t))\n$$\n\nХооронд нь хасвал $f(x) f(4t) = f(t) f(4x)$ болно. Иймд $\\exists a \\in \\mathbb{R}$ тэгш $f(4x) = 8a f(x)$ ($f(t) \\neq 0$).\n\nТэгвэл бодлого дараах хэлбэртэй болно:\n\n$$\nf(x + f(y)) = f(x) + a x f(y) + f(f(y))\n$$\n\nЭнд $y = t$ гэж авбал $\\forall x \\in \\mathbb{R}$ тоог $f(u) - f(v)$ ($\\exists u, v \\in \\mathbb{R}$) гэж бичиж чадна. Мөн $(x, y) \\to (f(u) - f(v), v)$ гэж авбал:\n\n$$\nf(f(u)) = f(f(u) - f(v)) + a f(u) f(v) - a f^2(v) + f(f(v))\n$$\n\n$(x, y) \\to (f(v) - f(u), u)$ гэж авбал:\n\n$$\nf(f(v)) = f(f(v) - f(u)) + a f(v) f(u) - a f^2(u) + f(f(u))\n$$\n\nХооронд нь нэмбэл:\n\n$$\nf(f(u) - f(v)) + f(f(v) - f(u)) = a (f(u) - f(v))^2\n$$\n\nИймд $\\forall x \\in \\mathbb{R}$:\n\n$$\nf(x) + f(-x) = a x^2\n$$\n\n$\\forall a$-д $a = 2$ гэж гарна.\n\nИймд:\n\n$$\n\\begin{cases}\nf(x + f(y)) = f(x) + 2x f(y) + f(f(y)) \\\\\nf(4x) = 16 f(x) \\\\\nf(x) + f(-x) = 2x^2\n\\end{cases}\n$$\n\n$(x, y) \\to (f(x), x)$ гэж авбал $f(2f(x)) = 2f(f(x)) + 2f^2(x)$\n\n$(x, y) \\to (2f(x), x)$ гэж авбал $f(3f(x)) = 3f(f(x)) + 6f^2(x)$\n\n$(x, y) \\to (3f(x), x)$ гэж авбал $f(4f(x)) = 4f(f(x)) + 12f^2(x)$\n\n$\\because f(4f(x)) = 16f(f(x))$ тул $f(f(x)) = f^2(x)$.\n\nИймд:\n\n$$\nf(x + f(y)) = f(x) + 2x f(y) + f^2(y)\n$$\n\n$(x, y) \\to (-f(v), v)$ гэж авбал $0 = f(-f(v)) - 2f^2(v) + f^2(v)$, $\\Rightarrow f(-f(v)) = f^2(v)$.\n\n$(x, y) \\to (-f(v), u)$ гэж авбал:\n\n$$\nf(f(u) - f(v)) = f(-f(v)) - 2f(u)f(v) + f^2(u)\n$$\n\n$$\n\\Rightarrow f^2(u) - 2f(u)f(v) + f^2(v) = (f(u) - f(v))^2\n$$\n\n$\\Rightarrow f(x) = x^2$.\n\nИймд хариу: $f(x) = 0$, $f(x) = x^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15052,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\overline{AB}$ be a triangle with circumcircle $c(O, R)$ and $\\overline{AB} < \\overline{AI'} < \\overline{BI'}$. Let $\\Delta$ be the midpoint of the small arc $\\widehat{AB}$. The line $\\overline{AA'}$ meets the line $\\overline{B\\Gamma}$ at $E$, and the circumcircle of triangle $B\\Delta E$ (call it $c_1$) intersects the line $\\overline{AB}$ (for a second time) at $Z$. The circumcircle of triangle $A\\Delta Z$ (call it $c_2$) intersects the line $\\overline{A\\Gamma}$ (for a second time) at $H$. Prove that $BE = AH$.\n\n![](images/Greece-IMO2019finalbook_p7_data_7a55f20d54.png)\nfig. 2",
    "options": [],
    "answer": "See solution",
    "solution": "From the inscribed quadrilateral $A\\Delta B\\Gamma$ we have: $\\angle \\Delta AH = \\angle \\Delta BE$ \\ (1)\n\nAlso, from the inscribed quadrilaterals $AEBZ$ and $A\\Delta ZH$ we have:\n$$\n\\angle AEB = \\angle \\Delta ZA = \\angle AH\\Delta \\qquad (2)\n$$\nSince $\\Delta$ is the midpoint of the small arc $AB$, it follows that $A\\Delta = \\Delta B$. From equalities (1) and (2), and the equality $A\\Delta = \\Delta B$, we conclude that the triangles $A\\Delta H$ and $\\Delta EB$ are congruent, and hence:\n$$\nBE = AH.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15053,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $n \\geq 2$ for which there exist $x_1, x_2, \\dots, x_n \\in \\mathbb{R}^*$ such that\n\n$$\nx_1 + x_2 + \\dots + x_n = \\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_n} = 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the set of numbers $n$ that satisfy the conditions above. We will show that $M = \\mathbb{N} \\setminus \\{0, 1, 3\\}$.\n\nWe can observe that $2p \\in M$ for any $p \\geq 1$, since the relations are satisfied, for example, by taking $x_1 = x_2 = \\dots = x_p = 1$ and $x_{p+1} = x_{p+2} = \\dots = x_{2p} = -1$.\n\nMoreover, if $n \\in M$ and $x_1 + x_2 + \\dots + x_n = \\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_n} = 0$, then by taking $x_{n+1} = 1$ and $x_{n+2} = -1$, we see that $n+2 \\in M$, because\n\n$$\nx_1 + x_2 + \\dots + x_n + x_{n+1} + x_{n+2} = \\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_n} + \\frac{1}{x_{n+1}} + \\frac{1}{x_{n+2}} = 0.\n$$\n\nTherefore, it remains to find the smallest odd element of $M$.\n\nSuppose $3 \\in M$; then there exist $x_1, x_2, x_3 \\in \\mathbb{R}^*$ such that\n\n$$\nx_1 + x_2 + x_3 = \\frac{1}{x_1} + \\frac{1}{x_2} + \\frac{1}{x_3} = 0.\n$$\n\nThis leads to $x_1^2 + x_1 x_2 + x_2^2 = 0$, which is only possible if $x_1 = x_2 = 0$, a contradiction. Thus, $3 \\notin M$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15054,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all possible pairs of positive integers $(m, n)$ that satisfy the following:\n\n$$\nm^{10} + 23mn + n! = 2020,\n$$\n\nwhere $n!$ denotes the product of all positive integers from $1$ to $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $(2, 6)$.\n\nWhen $m \\geq 3$, $m^{10} \\geq 3^{10} = 59049 > 2020$, so the equation cannot hold. Thus, possible values are $m = 1$ or $m = 2$.\n\nIf $m = 1$, the equation becomes $23n + n! = 2019$. Since $7! = 5040 > 2019$, $n \\leq 6$. For $n \\leq 6$, $23n + n! \\leq 138 + 720 = 858 < 2019$, so there are no solutions in this case.\n\nIf $m = 2$, the equation becomes $46n + n! = 996$. Again, $n \\leq 6$. For $n = 6$, $46 \\times 6 + 720 = 276 + 720 = 996$, which satisfies the equation. For $n < 6$, the left side is smaller, so no other solutions exist.\n\nTherefore, the only solution is $(m, n) = (2, 6)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15055,
    "subject": "Mathematics (Olympiad)",
    "question": "Every triangle in the plane $\\mathbb{R}^2$ contains a Mediterranean point on its boundary or in its interior (even if the triangle degenerates into a line segment or a point). These Mediterranean points satisfy the following conditions:\n\n1. If a triangle is symmetric with respect to a line through the origin $(0, 0)$, then its Mediterranean point lies on this line.\n2. If triangle $DEF$ contains triangle $ABC$, and if $ABC$ contains the Mediterranean point $M$ of $DEF$, then $M$ is also the Mediterranean point of $ABC$.\n\nDetermine all possible locations for the Mediterranean point of the triangle with vertices at $(-3, 5)$, $(12, 5)$, and $(3, 11)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the auxiliary triangle $DEF$ with $D = (-12, 5)$, $E = (12, 5)$, and $F = (0, 13)$. Since $DEF$ is symmetric with respect to the $y$-axis, by condition (1), its Mediterranean point $M$ lies on the $y$-axis, so $M = (0, m)$ with $5 \\leq m \\leq 13$.\n\nDefine a non-negative real number $a$ by $a^2 + 25 = m^2$, so $0 \\leq a \\leq 12$. The point $A = (a, 5)$ lies on side $DE$ of $DEF$, and both $A$ and $M$ are at distance $m$ from the origin. The points $A$, $M$, and the midpoint $C = (12a, 12(m+5))$ of $AM$ form a (possibly degenerate) triangle symmetric with respect to the line through the origin and $C$. By (1), the Mediterranean point of $AMC$ lies on this line and must be $C$. Since $DEF$ contains $AMC$, and $AMC$ contains the Mediterranean point $M$ of $DEF$, condition (2) implies the Mediterranean point of $AMC$ is $M$. Thus, $M$ and $C$ coincide, which forces $a = 0$ and $m = 5$. Therefore, $M = (0, 5)$ is the Mediterranean point of $DEF$.\n\nNow, for the original triangle with vertices $(-3, 5)$, $(12, 5)$, and $(3, 11)$: $DEF$ contains this triangle, and the triangle contains $M = (0, 5)$, the Mediterranean point of $DEF$. By (2), its Mediterranean point is $(0, 5)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15056,
    "subject": "Mathematics (Olympiad)",
    "question": "A total of $2^n$ coins are distributed among several children. If one of the children has at least half of the coins, the coins are redistributed: coins are transferred from such a child to each of the other children in such a way that each of them gets as many coins as it had. In the case when one child possesses all the coins there is no possibility for redistribution. What is the greatest number of consecutive redistributions?\n\nFor example, if 32 coins are distributed among 6 children in the following way: 17, 2, 9, 1, 2, 1, then after one redistribution the children will have: 2, 4, 18, 2, 4, 2 coins, respectively; in the example, that number is 2.\n\nExplain your answer!",
    "options": [],
    "answer": "See solution",
    "solution": "At most $n$ consecutive redistributions are possible.\n\nWe will start with an example showing that $n$ consecutive redistributions are possible. Let $2^n$ coins be distributed among 3 children initially as follows: $1, 2^{n-1} + 2^{n-2} + \\dots + 2, 1$. The successive redistributions (a total of $n$) will be:\n\n$$\n2^1,\\ 2^{n-1} + 2^{n-2} + \\dots + 2^2,\\ 2^1\n$$\n\n$$\n2^2,\\ 2^{n-1} + 2^{n-2} + \\dots + 2^3,\\ 2^2\n$$\n\n$$\n2^{n-2},\\ 2^{n-1},\\ 2^{n-2}\n$$\n\n$$\n2^{n-1},\\ 0,\\ 2^{n-1}\n$$\n\n$$\n0,\\ 0,\\ 2^n\n$$\n\nLet us show that $n$ is the maximal number of consecutive redistributions. Suppose there is an initial distribution for which there are at least $n+1$ possible redistributions. After one redistribution, the number of coins each child has is divisible by 2. After the second redistribution, each child has a number of coins divisible by 4, and so on, until the $n$-th redistribution, in which the number of coins each child has is divisible by $2^n$. Since the total number of coins is $2^n$, the only possible distribution (in some order) is $2^n, 0, 0, \\dots, 0$. But then there cannot be a next redistribution. Contradiction!",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15057,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the positive integers $a, b, c$ with the following properties:\n\n1. $(a^2 + b^2)(c^2 + 2023^2) = (ab + 2023c)^2$;\n2. $(a^2 + 2023^2)(b^2 + c^2) = (2023a + bc)^2$;\n3. The greatest common divisor of $a, b, c$ and $2023$ equals $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "By subtracting equation (i) from (ii), we obtain $$(a^2 - c^2)(b^2 - 2023^2) = 0.$$ Because $a, b, c$ are positive integers, we deduce that $a = c$ or $b = 2023$.\n\nIf $a = c$, using (i) we find $$(a^2 - 2023b)^2 = 0,$$ that is, $a^2 = 2023b = 7 \\cdot 17^2 \\cdot b$. Consequently, $7 \\mid b$ and $7 \\mid a = c$, therefore the greatest common divisor of $a, b, c$ and $2023$ is at least $7$, contradicting assumption (iii).\n\nIf $b = 2023$, from (i) we obtain $$(ac - 2023^2)^2 = 0,$$ that is, $ac = 2023^2 = 7^2 \\cdot 17^4$. From (iii) we deduce that $a$ and $c$ are relatively prime, therefore the solutions are:\n$$(a, b, c) \\in \\{(1, 2023, 2023^2),\\ (7^2, 2023, 17^4),\\ (17^4, 2023, 7^2),\\ (2023^2, 2023, 1)\\}.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15058,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $n$ such that the sum of the squares of any $n$ different prime numbers greater than $3$ is divisible by $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We begin by showing that if a positive integer $k$ is relatively prime to $n$, then $k^2 \\equiv 1 \\pmod{n}$. To this end, invoke Dirichlet's theorem on arithmetic sequences to choose $n-1$ primes congruent to $1$ modulo $n$ and a prime congruent to $k$ modulo $n$. The sum of the squares of these $n$ primes is congruent to $k^2 - 1$ modulo $n$, so $k^2 \\equiv 1 \\pmod{n}$.\n\nNext, we prove that $n$ has no prime divisors greater than $3$. Let $p$ be an odd divisor of $n$ and write $n = p^{\\alpha} m$, where $\\alpha$ and $m$ are positive integers and $p$ does not divide $m$. By the Chinese Remainder Theorem, there exists a positive integer $k$ such that $k \\equiv 1 \\pmod{m}$ and $k \\equiv 2 \\pmod{p}$. It is easily seen that $k$ and $n$ are coprime, so $k^2 \\equiv 1 \\pmod{n}$ by the preceding. Hence $k^2 \\equiv 1 \\pmod{p}$, and the condition $k \\equiv 2 \\pmod{p}$ forces $p = 3$.\n\nConsequently, $n = 2^{\\alpha} 3^{\\beta}$, where $\\alpha$ and $\\beta$ are non-negative integers. Since $5$ and $n$ are coprime, the latter must be a divisor of $5^2 - 1 = 24$. It is readily checked that all divisors of $24$ work.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15059,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{N}$ denote the set of positive integers. Find all functions $f : \\mathbb{N} \\to \\mathbb{N}$ such that\n$$\nf(m+n)f(m-n) = f(m^2)\n$$\nfor all $m, n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The only function satisfying the conditions is $f(n) = 1$ for all $n \\in \\mathbb{N}$.\n\nNote that\n$$\nf(1)f(2n-1) = f(n^2) \\quad \\text{and} \\quad f(3)f(2n-1) = f((n+1)^2)\n$$\nfor $n \\ge 3$. Thus,\n$$\n\\frac{f(3)}{f(1)} = \\frac{f((n+1)^2)}{f(n^2)}.\n$$\nSetting $\\frac{f(3)}{f(1)} = k$ yields $f(n^2) = k^{n-3}f(9)$ for $n \\ge 3$. Similarly, for all $h \\ge 1$,\n$$\n\\frac{f(h+2)}{f(h)} = \\frac{f((m+1)^2)}{f(m^2)}\n$$\nfor sufficiently large $m$ and is thus also $k$. Hence $f(2h) = k^{h-1}f(2)$ and $f(2h+1) = k^h f(1)$.\n\nBut\n$$\n\\frac{f(25)}{f(9)} = \\frac{f(25)}{f(23)} \\cdots \\frac{f(11)}{f(9)} = k^8\n$$\nand\n$$\n\\frac{f(25)}{f(9)} = \\frac{f(25)}{f(16)} \\cdot \\frac{f(16)}{f(9)} = k^2,\n$$\nso $k = 1$ and $f(16) = f(9)$. This implies that $f(2h+1) = f(1) = f(2) = f(2j)$ for all $j, h$, so $f$ is constant. From the original functional equation it is then clear that $f(n) = 1$ for all $n \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15060,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $x, y$ such that $\\frac{xy^3}{x+y}$ is a cube of a prime.",
    "options": [],
    "answer": "See solution",
    "solution": "**First Solution.**\nLet $p$ be such a prime and let $d = \\gcd(x, y)$. Thus, $x = da$, $y = db$. The condition becomes $p^3 d (a + b) = d^4 a b^3$, i.e.,\n$$\np^3 (a + b) = d^3 a b^3. \\qquad (1)\n$$\nSince $1 = \\gcd(a, b) = \\gcd(a, a + b) = \\gcd(a b^3, a + b)$, we get $a b^3 \\mid p^3$, and so $b \\mid p$.\n\n**Case 1:** $b = 1$.\nFrom (1), we have $p^3 (a + 1) = d^3 a$. Since $\\gcd(a, a + 1) = 1$, we have $a \\mid p^3$. If $p \\mid d$, using $\\left(\\frac{d}{p}\\right)^3 = 1 + \\frac{1}{a} \\leq 2$ and $\\frac{d}{p} \\in \\mathbb{Z}$, we deduce that $\\frac{d}{p} = 1$, i.e., $d = p$, and so $a + 1 = a$, a contradiction. Thus, $p \\nmid d$ implying that $p^3 \\mid a$, and so $a = p^3$ and $a + 1 = d^3$. Hence,\n$$\n1 = d^3 - p^3 = (d - p)(d^2 + d p + p^2),\n$$\nwhich is not possible as $d \\neq p$ and $d^2 + d p + p^2 > 2$.\n\n**Case 2:** $b = p$.\nFrom $a p^3 \\mid p^3$, we get $a = 1$. Thus, $1 + p = d^3$, i.e., $p = d^3 - 1 = (d - 1)(d^2 + d + 1)$. Since $d - 1 < d^2 + d + 1$, we get $d - 1 = 1$ and $d^2 + d + 1 = p$, so that $d = 2$, $p = 7$. Consequently, $a = 1$, $b = 7$, $d = 2$ yielding $x = 2$, $y = 14$.\n\n**Second Solution.**\nLet $p$ be such a prime so that\n$$\nxy^3 = p^3 (x + y). \\qquad (1)\n$$\nIf $p \\nmid y$, then $p^3 \\mid x$. Writing $x = p^3 h$, we have $p^3 h y^3 = p^3 (p^3 h + y)$. Simplifying, we get $y (h y^2 - 1) = p^3 h$. Since $p \\nmid y$, we get $p^3 \\mid (h y^2 - 1)$, showing that $y \\mid h$. Let $h = y k$. Thus, $y (k y^3 - 1) = p^3 y k$, which after simplification gives $k (y^3 - p^3) = 1$. Hence, $y^3 - p^3 = \\pm 1$, a contradiction.\n\nIf $p \\mid y$, let $y = p s$. From (1), we get $x p^3 s^3 = p^3 (x + p s)$. Simplifying, we have $x (s^3 - 1) = p s$. Since $s \\nmid (s^3 - 1)$, we see that $s \\mid x$, and so $(s - 1)(s^2 + s + 1) = (s^3 - 1) \\mid p$. Since $s - 1 < s^2 + s + 1$, we get $s - 1 = 1$, $s^2 + s + 1 = p$. Thus, $x = s = 2$, $p = 7$ yielding $x = 2$, $y = 14$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15061,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $q(x) = x^3 + a x^2 + b x + c$ be a cubic polynomial with integer coefficients. Suppose $p$ is a prime that divides $a$, $b$, and $c$. Prove that if $y$ and $z$ are two distinct integer roots of $q(x)$, then $p^2$ divides $b$ and $p^3$ divides $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $y$ and $z$ be two different integer roots of $q(x)$. Then:\n\n$$\ny^3 + a y^2 + b y + c = 0 \\quad \\text{and} \\quad z^3 + a z^2 + b z + c = 0.\n$$\n\nSince $p$ divides $a$, $b$, and $c$, we have:\n\n$$\ny^3 = -a y^2 - b y - c.\n$$\n\nThus, $p$ divides $y^3$, so $p$ divides $y$. Similarly, $p$ divides $z^3$, so $p$ divides $z$.\n\nSubtracting the two equations:\n\n$$\n(y^3 - z^3) + a(y^2 - z^2) + b(y - z) = 0\n$$\n\nwhich factors as:\n\n$$\n(y - z)(y^2 + y z + z^2 + a(y + z) + b) = 0.\n$$\n\nSince $y \\neq z$, we have:\n\n$$\ny^2 + y z + z^2 + a(y + z) + b = 0.\n$$\n\nBecause $p$ divides $y$, $z$, and $a$, $p^2$ divides $y^2 + y z + z^2 + a(y + z) = -b$, so $p^2$ divides $b$.\n\nNow, $c = -y^3 - a y^2 - b y$. Since $p$ divides $y$, $p^2$ divides $y^2$, and $p^3$ divides $y^3$, and since $p$ divides $a$ and $p^2$ divides $b$, it follows that $p^3$ divides $c$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15062,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle such that $AB \\neq AC$. Let $M$ be the midpoint of $BC$, $H$ be the orthocenter of $ABC$, $O_1$ be the midpoint of $AH$, and $O_2$ be the circumcenter of $BCH$. Prove that $O_1AMO_2$ is a parallelogram.",
    "options": [],
    "answer": "See solution",
    "solution": "We use the following well-known facts:\n\n1. The reflection of $H$ in line $BC$ lies on the circumcircle of triangle $ABC$.\n\n2. $AH = 2MO$, where $O$ is the circumcenter of $ABC$.\n\nFrom (1), the reflection of the circumcenter $O$ in $BC$ is $O_2$, the circumcenter of $BCH$. Thus,\n\n$$\nMO_2 = MO = \\frac{AH}{2} = AO_1\n$$\n\nSince $AO_1 \\parallel MO_2$, $O_1AMO_2$ is a parallelogram.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15063,
    "subject": "Mathematics (Olympiad)",
    "question": "一張月曆是一個長方形的方格紙。若且唯若它滿足以下三點，則稱其符合法規：\n\n1. 月曆的每一格都被塗成白色或紅色，且恰有 $10$ 個紅色格子。\n2. 若月曆的一橫排共有 $N$ 格，則當我們從最左上角的格子，依次填入 $1, 2, \\dots$，由左而右，然後由上而下，我們將找不到連續 $N$ 個數字，它們所在的格子都是白色的。\n3. 若月曆的一直列共有 $M$ 格，則當我們從最左下角的格子，依次填入 $1, 2, \\dots$，由下而上，然後由左而右，我們將找不到連續 $M$ 個數字，它們所在的格子都是白色的（也就是說，如果我們將整張月曆順時鐘旋轉 $90^\\circ$，它仍然滿足條件 (2)）。\n\n試問有多少種符合法規的月曆？",
    "options": [],
    "answer": "See solution",
    "solution": "答案：$10! = 3628800$ 種。\n\n注意到在一個 $10 \\times 10$ 的方格表中塗紅 $10$ 格，使得每行每列都恰有一紅格的塗法共有 $10!$ 種。以下建立塗 $10 \\times 10$ 方格表與符合法規的月曆之間的一一對應。\n\n![](images/18-3J_p9_data_b68879257b.png)\n\n**方格表 $\\rightarrow$ 符合法規的月曆：**\n\n如上圖進行以下操作：\n\n1. 先觀察相鄰兩行，如果左行紅格比右行高，則將兩行間的格線加粗。\n2. 再觀察相鄰兩列，如果上列紅格比下列右，則將兩列間的格線加粗。\n3. 把加粗的格線全部擦掉，便得到一個符合法規的月曆：\n\n   (證明)\n\n   * 首先證明擦掉後是個月曆。注意到每個粗格子隔出的區間內至多只會有一個紅格，這是因為如果有兩個，若兩紅格相對位置是左上右下則之間必還有一縱粗線，若是左下右上則必還有一橫粗線，無論何者皆矛盾。這保證了擦掉細格線後必為月曆。\n   * 接著證明它符合法規。條件 (1) 顯然滿足。如果違背條件 (2)，則表示原方格表中必有一條橫粗線，其上一列的紅格都在下一列的左邊，但這與橫粗線本身的構造矛盾。條件 (3) 同理。\n\n**符合法規的月曆 $\\rightarrow$ 方格表：**\n\n對於月曆中的每個格子，依序進行以下動作：\n\n1. 假設與其同一行的紅格子有 $p$ 個，同一列的紅格子有 $q$ 個，則將這個格子再細分為 $q \\times p$ 個小格子。\n2. 如果該格是紅色的，且是該行中由上數來第 $r$ 個紅格，該列由左數來第 $r$ 個小格子，則將細分後的小格子中，由下數來第 $r$ 列，由左數來第 $s$ 行的格子塗紅，其餘留白。\n\n易見以上構成第一部分映射的反映射。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15064,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a composite positive integer and let $1 = d_1 < d_2 < d_3 < \\dots < d_k = n$ be the divisors of $n$, where $k \\ge 3$. Assume that all the equations\n\n$$\nd_{i+2}x^2 - 2d_{i+1}x + d_i = 0\n$$\n\nfor $i \\in \\{1, 2, \\dots, k-2\\}$ have real solutions. Prove that $n = p^{k-1}$ for some prime number $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have:\n\n$$\n4d_{i+1}^2 - 4d_{i+2} d_i \\ge 0 \\iff \\frac{d_{i+1}}{d_i} \\ge \\frac{d_{i+2}}{d_{i+1}} \\quad (*)\n$$\n\nfor any $i \\in \\{1, 2, \\dots, k-2\\}$.\n\nAs $d_2$ is the smallest proper divisor of $n$, the number $\\frac{n}{d_2}$ is the greatest proper divisor of $n$, so $\\frac{n}{d_2} = d_{k-1}$. We have:\n\n$$\n\\frac{n}{d_2} = d_{k-1} = \\prod_{i=1}^{k-2} \\frac{d_{i+1}}{d_i} \\ge \\prod_{i=1}^{k-2} \\frac{d_{i+2}}{d_{i+1}} = \\frac{n}{d_2}.\n$$\n\nIt follows that all inequalities $(*)$ turn into equalities.\n\nThus, $d_{i+1}^2 = d_{i+2} d_i$ for any $i \\in \\{1, 2, \\dots, k-2\\}$. It follows that the numbers $1 = d_1, d_2, d_3, \\dots, d_k = n$ (in this order) are consecutive terms of a geometric progression of ratio $d_2$, so $n = d_2^{k-1}$.\n\nThe lowest proper divisor of the composite number $n$ is a prime number $p$, so $d_2 = p$, and $n = p^{k-1}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15065,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle AB\\Gamma$ be a triangle with $\\angle BAC = 120^\\circ$. Let $M$ be the midpoint of side $BC$. The line $AM$ is perpendicular to $AB$ and intersects the circumcircle of $\\triangle ABC$ at point $E$. Let $Z$ be the intersection point of lines $BA$ and $EC$. Prove that:\n\n(a) $ZM \\perp BE$,\n\n(b) $ZM = BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "![Figure 1](images/Hellenic_Mathematical_Competitions_2011_booklet_p3_data_8ccb1807f1.png)\n\n(a) Since $\\angle BAE = 90^\\circ$, $BE$ is a diameter of the circumcircle of $\\triangle ABC$. Hence $\\angle BCE = 90^\\circ$. Therefore, in triangle $ZBE$, the segments $EA$ and $BC$ are altitudes intersecting at $M$. Thus, $M$ is the orthocenter of $\\triangle ZBE$, so $ZM \\perp BE$.\n\n(b) Since $\\angle MAZ + \\angle MGZ = 90^\\circ + 90^\\circ = 180^\\circ$, the quadrilateral $AMCZ$ is cyclic. Thus, $\\angle ZCG = \\angle MAC = \\angle BAC - \\angle BAM = 120^\\circ - 90^\\circ = 30^\\circ$.\n\nHence, in right triangle $ZCG$, the hypotenuse $ZM$ is twice the length of leg $MG$, that is, $ZM = 2 \\cdot MG = BC$, since $M$ is the midpoint of $BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15066,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c, d \\ge 0$ be real numbers such that $a + b + c + d = 1$. Prove that\n$$\n\\sqrt{a + \\frac{(b-c)^2}{6} + \\frac{(c-d)^2}{6} + \\frac{(d-b)^2}{6}} + \\sqrt{b} + \\sqrt{c} + \\sqrt{d} \\le 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that $(b-c)^2 \\le 2(\\sqrt{b} - \\sqrt{c})^2$, since $(\\sqrt{b} + \\sqrt{c})^2 = b + c + 2\\sqrt{bc} \\le 2b + 2c \\le 2$. Thus, $a + \\sum \\frac{(b-c)^2}{6} \\le a + \\frac{1}{3} \\sum (\\sqrt{b} - \\sqrt{c})^2 = 1 - \\frac{1}{3}(\\sum \\sqrt{b})^2$.\n\nTo end the proof, it suffices to show that $S + \\sqrt{1 - S^2/3} \\le 2$, where $S = \\sqrt{b} + \\sqrt{c} + \\sqrt{d}$.\n\nSince $S \\le 2$, this is equivalent to $1 - S^2/3 \\le 4 - 4S + S^2$, which rewrites as $(2S - 3)^2 \\ge 0$, obviously true.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15067,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\n2(y + 1)f(x)f(y - 1) = 2y f(xy) - f(2x)\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = 0$, $y = -1$:\n$$\n0 = -2f(0) - f(0)\n$$\nthus $f(0) = 0$.\n\nLet $y = 1$:\n$$\n2f(x) = f(2x)\n$$\nfor all $x$, so the given condition can be rewritten as\n$$\n(y + 1)f(x)f(y - 1) = y f(xy) - f(x), \\quad \\forall x, y\n$$\nNow, let $y = 0$:\n$$\nf(x)f(-1) = -f(x)\n$$\nIf $f(-1) \\neq -1$, then $f(x) = 0$ for all $x$, which satisfies the given condition. Now suppose there exists some $x_0$ such that $f(x_0) \\neq 0$, this implies $f(-1) = -1$ and $f(1) = 1$.\n\nIn the previous equation, let $x = 1$:\n$$\n(y + 1)f(y - 1) = y f(y) - 1, \\quad \\forall y\n$$\nSubstitute this into the left-hand side of the previous equation to get\n$$\n(y f(y) - 1)f(x) = y f(xy) - f(x)\n$$\nso $f(xy) = f(x)f(y)$ for all $x, y \\neq 0$, which implies that $f$ is a multiplicative function. Also, let $y = -1$ in the previous equation:\n$$\nf(x) = -f(-x)\n$$\nwhich implies that $f(x)$ is odd.\n\nSuppose $a$ is such that $f(a) = 0$. Substituting $x = a$ into the previous equation gives $0 = y f(ay)$, so $f(ay) = 0$ for all $y \\neq 0$. If $a \\neq 0$, then $ay$ can take any value in $\\mathbb{R}$, so $f(x) \\equiv 0$, a contradiction. Thus $a = 0$ is the unique value such that $f(a) = 0$.\n\nSince $(y+1)f(y-1) = y f(y) - 1$, change $y \\to -y$ and use the property that $f$ is odd:\n$$\n(-y + 1)f(-y - 1) = y f(y) - 1, \\quad \\forall y \\in \\mathbb{R}\n$$\nThus, for all $y \\neq \\pm 1$:\n$$\n(y + 1)f(y - 1) = (y - 1)f(y + 1) \\implies \\frac{f(y + 1)}{f(y - 1)} = \\frac{y + 1}{y - 1}\n$$\nfor all $y \\neq \\pm 1$. Using the multiplicative property of $f$, rewrite as\n$$\nf\\left(\\frac{y+1}{y-1}\\right) = \\frac{y+1}{y-1}, \\quad \\forall y \\in \\mathbb{R} \\setminus \\{\\pm 1\\}\n$$\nNote that $t = \\frac{y+1}{y-1}$ can take any value in $\\mathbb{R} \\setminus \\{1\\}$, so $f(t) = t$ for all $t \\neq 0, 1$. But $f(1) = 1$, $f(0) = 0$, so we have $f(x) = x$ for all $x \\in \\mathbb{R}$.\n\nHence, there are two functions that satisfy the given condition:\n- $f(x) = 0$ for all $x \\in \\mathbb{R}$\n- $f(x) = x$ for all $x \\in \\mathbb{R}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15068,
    "subject": "Mathematics (Olympiad)",
    "question": "Two players take turns writing plus or minus signs in front of consecutive natural numbers $1, 2, 3, \\ldots$ and calculate the sum of all the numbers with signs. The first player starts by writing a sign in front of $1$, and this is his sum after the first move. Then the second player writes a sign in front of $2$ and calculates the sum of the two signed numbers—this is the sum after his first move. After that, the first player writes a sign in front of $3$ and calculates the sum of three signed numbers, and so on. When the sum becomes larger than $2011$ in absolute value, the game stops. The player who made the last move loses. Which player has a winning strategy?",
    "options": [],
    "answer": "See solution",
    "solution": "The second player can always choose the sign opposite to the sign of the first player's last move. Let the first player's moves be $1, 3, 5, \\ldots$ in absolute value, denoted by $a_1, a_2, a_3, \\ldots$. The second player's moves are $2, 4, 6, \\ldots$ in absolute value, denoted by $b_1, b_2, b_3, \\ldots$. Let $s_k$ be the sum after the first player's $k$-th move, and $c_k$ after the second player's $k$-th move. We prove by induction that $|s_k| \\geq k$, $|c_k| \\leq k$, so $|s_k| \\geq |c_k|$.\n\nFor $k=1$, $|s_1| = |c_1| = 1$.\n\nAssume for some $k$ that $|s_k| \\geq k$, $|c_k| \\leq k$. Note $|a_{k+1}| = 2k+1$, $|b_{k+1}| = 2k+2$. Without loss of generality, assume $0 \\leq c_k \\leq k$.\n\n1. If $a_{k+1} = 2k+1$, then $2k+1 \\leq s_{k+1} \\leq 3k+1$, so for $b_{k+1} = -(2k+2)$, $-1 \\leq c_{k+1} \\leq k-1$.\n2. If $a_{k+1} = -(2k+1)$, then $-(2k+1) \\leq s_{k+1} \\leq -(k+1)$ (so $|s_{k+1}| \\geq k+1$), and for $b_{k+1} = 2k+2$, $1 \\leq c_{k+1} \\leq k+1$.\n\nThus, the required inequalities hold. The game will eventually stop for large enough numbers. The second player wins with this strategy, because the absolute value of the sum after his $k$-th move never exceeds that after the first player's $k$-th move.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15069,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a nonempty set of primes satisfying the property that for each proper subset $P$ of $S$, all the prime factors of the number $\\left(\\prod_{p \\in P} p\\right) - 1$ are also in $S$. Determine all possible such sets $S$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $S$ is a singleton, then the statement is vacuously true. Hence $S = \\{p\\}$, where $p$ is a prime.\n\nSuppose $S$ contains two elements. One of these must be $2$, so $2 \\in S$. Moreover, $p-1$ must be a power of $2$, so $p$ is a Fermat prime.\n\nSuppose $|S| \\geq 3$. Consider two cases: $|S| < \\infty$ and $S$ is infinite.\n\nIf $|S| < \\infty$, let $S = \\{p_1 (=2), p_2, \\dots, p_r\\}$, with $r \\geq 3$. Since $|S| \\geq 3$, there is a prime of the form $3k+1$ or $3k+2$ in $S$. Observe $(3k+1)-1=3k$ and $2(3k+2)-1=6k+3$ both have $3$ as a prime factor. Hence $3 \\in S$. Now,\n\n$$\n(2 \\cdot p_3 \\cdots p_r) - 1 = 3^k \\quad \\text{and} \\quad (3 p_3 \\cdots p_r) - 1 = 2^l,\n$$\n\nfor some $k > 0$ and $l > 0$. Hence,\n\n$$\n3(3^k + 1) = 2(2^l + 1).\n$$\n\nThis gives $2^{l+1} - 3^{k+1} = 1$. The only possibility is $l+1=2$ and $k+1=1$, which forces $k=0$, a contradiction. Hence $3 \\leq |S| < \\infty$ is not possible.\n\nFinally, assume $S$ is an infinite set. Suppose there exists a prime $p$ not in $S$. Take $S = \\{p_1 (=2), p_2, \\dots\\}$. Consider $p_1 - 1, p_1 p_2 - 1, \\dots, p_1 p_2 \\dots p_p - 1$ modulo $p$. Since $p$ is not in $S$, none of them is $0$. By the pigeonhole principle, we can find $j < k$ such that\n\n$$\np_1 p_2 \\cdots p_j - 1 \\equiv p_1 p_2 \\cdots p_j p_{j+1} \\cdots p_k - 1 \\pmod{p}.\n$$\n\nHence $p$ divides $p_1 p_2 \\cdots p_j (p_{j+1} \\cdots p_k - 1)$. But this contradicts the choice of $p$ that it is not in $S$. Hence $S$ is the set of all primes.\n\n**Conclusion:** The possible sets $S$ are:\n- Any singleton set $\\{p\\}$, where $p$ is a prime.\n- Any set $\\{2, p\\}$, where $p$ is a Fermat prime.\n- The set of all primes.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15070,
    "subject": "Mathematics (Olympiad)",
    "question": "For $n \\in \\mathbb{N}$, let $D_2(n)$ (respectively $D_3(n)$) denote the number of divisors of $n$ that are perfect squares (respectively perfect cubes). Prove that there exists an $n$ such that $D_2(n) = 999 D_3(n)$.",
    "options": [],
    "answer": "See solution",
    "solution": "For $a \\in \\mathbb{N}$, denote $u(a) = \\left\\lfloor \\frac{a}{2} \\right\\rfloor$, $v(a) = \\left\\lfloor \\frac{a}{3} \\right\\rfloor$. If $n = p_1^{a_1} \\cdots p_k^{a_k}$ is the prime factorization of $n$, then $D_2(n) = (u(a_1)+1) \\cdots (u(a_k)+1)$ and $D_3(n) = (v(a_1)+1) \\cdots (v(a_k)+1)$.\n\nDefine $a_1 = 2 \\cdot 998$ and set $a_i = 2 v(a_{i-1})$ for $i \\geq 2$. With this definition, $u(a_1) = 998$ and $u(a_i) = v(a_{i-1})$ for $i \\geq 2$.\n\nNote that $v(a_i) = \\left\\lfloor \\frac{2}{3} v(a_{i-1}) \\right\\rfloor < v(a_{i-1})$ if $v(a_{i-1}) > 0$, so the sequence $(v(a_i))$ decreases and eventually becomes $0$ for sufficiently large $i$.\n\nTake the first index $k$ such that $v(a_k) = 0$ and define $n = p_1^{a_1} \\cdots p_k^{a_k}$. Because $u(a_i) = v(a_{i-1})$ for $i \\geq 2$, we have\n\n$$\nD_2(n) = (998+1)(u(a_2)+1) \\cdots (u(a_k)+1) = 999 (v(a_1)+1) \\cdots (v(a_{k-1})+1).\n$$\n\nIn addition, $D_3(n) = (v(a_1)+1) \\cdots (v(a_{k-1})+1)(v(a_k)+1)$, which equals $(v(a_1)+1) \\cdots (v(a_{k-1})+1)$ since $v(a_k) = 0$. Therefore, $D_2(n) = 999 D_3(n)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15071,
    "subject": "Mathematics (Olympiad)",
    "question": "An integer is written in each cell of a $9 \\times 9$ square table. For every $k$ numbers in the same row (or column), their sum is also present somewhere in the same row (or column). Find the smallest possible number of zeros in the table for:\n\n(a) $k = 5$;\n\n(b) $k = 8$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Example: Number the rows and columns from $1$ to $9$. Write $1$ in the cells $(i, i)$ for $i = 1, \\dots, 9$; $-1$ in cell $(1, 9)$ and in cells $(i, i-1)$ for $i = 2, \\dots, 9$; $0$ in all other cells. Possible sums are $1$, $0$, and $-1$.\n\nEvaluation: Suppose there are at least $19$ non-zero numbers. By the pigeonhole principle, there will be at least three non-zero numbers in some row, and therefore at least two non-zero numbers with the same sign (the case for negative numbers is analogous). Arrange the numbers in order: $a_1 \\leq a_2 \\leq \\dots \\leq a_9$, where $a_9 \\geq a_8 > 0$. If $a_5 \\geq 0$, then $a_5 + a_6 + a_7 + a_8 + a_9 \\geq a_8 + a_9 > a_9$ must be present, which is a contradiction. If $a_5 < 0$, then $a_1 + a_2 + a_3 + a_4 + a_5 < a_1$ must be present, also a contradiction.\n\nb) A possible example without zeros is as follows (works because $5 \\times 3 + 3 \\times (-4) = 3$ and $4 \\times 3 + 4 \\times (-4) = -4$):\n\n![]( <table><thead><tr><th>3</th><th>3</th><th>3</th><th>3</th><th>3</th><th>-4</th><th>-4</th><th>-4</th><th>-4</th></tr></thead><tbody><tr><td>-4</td><td>3</td><td>3</td><td>3</td><td>3</td><td>3</td><td>-4</td><td>-4</td><td>-4</td></tr><tr><td>-4</td><td>-4</td><td>3</td><td>3</td><td>3</td><td>3</td><td>3</td><td>-4</td><td>-4</td></tr><tr><td>-4</td><td>-4</td><td>-4</td><td>3</td><td>3</td><td>3</td><td>3</td><td>3</td><td>-4</td></tr><tr><td>-4</td><td>-4</td><td>-4</td><td>-4</td><td>3</td><td>3</td><td>3</td><td>3</td><td>3</td></tr><tr><td>3</td><td>-4</td><td>-4</td><td>-4</td><td>-4</td><td>3</td><td>3</td><td>3</td><td>3</td></tr><tr><td>3</td><td>3</td><td>-4</td><td>-4</td><td>-4</td><td>-4</td><td>3</td><td>3</td><td>3</td></tr><tr><td>3</td><td>3</td><td>3</td><td>-4</td><td>-4</td><td>-4</td><td>-4</td><td>3</td><td>3</td></tr><tr><td>3</td><td>3</td><td>3</td><td>3</td><td>-4</td><td>-4</td><td>-4</td><td>-4</td><td>3</td></tr><tr><td>3</td><td>3</td><td>3</td><td>3</td><td>3</td><td>-4</td><td>-4</td><td>-4</td><td>-4</td></tr></tbody></table> )\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15072,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $c > 0$ be a given positive real and $\\mathbb{R}_{>0}$ be the set of all positive reals. Find all functions $f: \\mathbb{R}_{>0} \\to \\mathbb{R}_{>0}$ such that\n\n$$\nf((c+1)x + f(y)) = f(x + 2y) + 2c x \\quad \\text{for all } x, y \\in \\mathbb{R}_{>0}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove that $f(x) \\ge 2x$ for all $x > 0$. Suppose, for the sake of contradiction, that $f(y) < 2y$ for some positive $y$. Choose $x$ such that $f((c+1)x + f(y))$ and $f(x+2y)$ cancel out, that is,\n\n$$\n(c+1)x + f(y) = x + 2y \\iff x = \\frac{2y - f(y)}{c}.\n$$\n\nNotice that $x > 0$ because $2y - f(y) > 0$. Then $2c x = 0$, which is not possible. This contradiction yields $f(y) \\ge 2y$ for all $y > 0$.\n\nNow suppose, again for the sake of contradiction, that $f(y) > 2y$ for some $y > 0$. Define the following sequence: $a_0$ is an arbitrary real greater than $2y$, and $f(a_n) = f(a_{n-1}) + 2c x$, so that\n\n$$\n\\begin{cases}\n(c + 1)x + f(y) = a_n \\\\\nx + 2y = a_{n-1}\n\\end{cases}\n\\iff x = a_{n-1} - 2y \\quad \\text{and} \\quad a_n = (c + 1)(a_{n-1} - 2y) + f(y).\n$$\n\nIf $x = a_{n-1} - 2y > 0$ then $a_n > f(y) > 2y$, so inductively all the substitutions make sense.\n\nFor the sake of simplicity, let $b_n = a_n - 2y$, so $b_n = (c+1)b_{n-1} + f(y) - 2y$ ($*$). Notice that $x = b_{n-1}$ in the former equation, so $f(a_n) = f(a_{n-1}) + 2c b_{n-1}$. Telescoping yields\n\n$$\nf(a_n) = f(a_0) + 2c \\sum_{i=0}^{n-1} b_i.\n$$\n\nOne can find $b_n$ from the recurrence equation ($*$):\n\n$$\nb_n = \\left(b_0 + \\frac{f(y)-2y}{c}\\right)(c+1)^n - \\frac{f(y)-2y}{c},\n$$\n\nand then\n\n$$\n\\begin{aligned}\nf(a_n) &= f(a_0) + 2c \\sum_{i=0}^{n-1} \\left( \\left(b_0 + \\frac{f(y)-2y}{c}\\right) (c+1)^i - \\frac{f(y)-2y}{c} \\right) \\\\\n&= f(a_0) + 2 \\left(b_0 + \\frac{f(y)-2y}{c}\\right) ((c+1)^n - 1) - 2n(f(y)-2y).\n\\end{aligned}\n$$\n\nSince $f(a_n) \\ge 2a_n = 2b_n + 4y$,\n\n$$\n\\begin{aligned}\n& f(a_0) + 2 \\left(b_0 + \\frac{f(y)-2y}{c}\\right) ((c+1)^n - 1) - 2n(f(y)-2y) \\ge 2b_n + 4y \\\\\n&= 2 \\left(b_0 + \\frac{f(y)-2y}{c}\\right) (c+1)^n - 2 \\frac{f(y)-2y}{c},\n\\end{aligned}\n$$\n\nwhich implies\n\n$$\nf(a_0) + 2 \\frac{f(y) - 2y}{c} \\ge 2 \\left(b_0 + \\frac{f(y) - 2y}{c}\\right) + 2n(f(y) - 2y),\n$$\n\nwhich is not true for sufficiently large $n$.\n\nA contradiction is reached, and thus $f(y) = 2y$ for all $y > 0$. It is immediate that this function satisfies the functional equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15073,
    "subject": "Mathematics (Olympiad)",
    "question": "In a volleyball tournament, there are 8 teams that play a one-round tournament (each team plays exactly one game with every other team). Each win is worth 1 point, each loss is worth 0 points, and there are no draws. After the tournament is finished, if the difference between the first and second place does not exceed 1 point, then they play one extra game. The same applies for the teams that scored 3rd and 4th, 5th and 6th, and 7th and 8th, respectively. What is the least number of extra games that can occur?\n\n*Note.* After the tournament is finished, each place is occupied by only one team, even if two teams have the same number of points.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** 1.\n\nIf we assume that there were no extra games, then the difference between 1st and 2nd, 3rd and 4th, 5th and 6th, and 7th and 8th is at least 2 points. Therefore, the difference between 1st and 8th places is at least 8 points, while the first place cannot have more than 7 points. The following example shows that 1 extra game is indeed possible:\n\n| Team      | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | Points |\n|-----------|---|---|---|---|---|---|---|---|--------|\n| 1 place   | XX| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 7      |\n| 2 place   | 0 | XX| 0 | 1 | 1 | 1 | 1 | 1 | 5      |\n| 3 place   | 0 | 1 | XX| 0 | 0 | 1 | 1 | 1 | 4      |\n| 4 place   | 0 | 0 | 1 | XX| 1 | 0 | 1 | 1 | 4      |\n| 5 place   | 0 | 0 | 1 | 0 | XX| 1 | 1 | 1 | 4      |\n| 6 place   | 0 | 0 | 0 | 1 | 0 | XX| 0 | 1 | 2      |\n| 7 place   | 0 | 0 | 0 | 0 | 0 | 1 | XX| 1 | 2      |\n| 8 place   | 0 | 0 | 0 | 0 | 0 | 0 | 0 | XX| 0      |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15074,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that there exist two divisors $a$ and $b$ of $n$ with the property that $a^2 + b^2 + 1$ is a multiple of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a$ and $b$ be divisors of $n$ such that $n \\mid a^2 + b^2 + 1$.\n\nSince $a \\mid n$ and $n \\mid a^2 + b^2 + 1$, it follows that $a \\mid b^2 + 1$. Similarly, $b \\mid a^2 + 1$. Thus, $\\gcd(a, b) = 1$.\n\nWe have $ab \\mid a^2 + b^2 + 1$. By symmetry, assume $a \\leq b$. Then:\n\n$$\na^2 + b^2 + 1 = kab \\quad (1)\n$$\nfor some positive integer $k$.\n\nIf $a = b$, then $2a^2 + 1$ is a multiple of $a$, so $a = 1$. Thus, $(a, b) = (1, 1)$ and $k = 3$.\n\nAssume $a > b$. Equation (1) as a quadratic in $a$ has a positive integer solution. The second solution is:\n\n$$\na' = kb - a = \\frac{b^2 + 1}{a},\n$$\nwhich is also a positive integer. Moreover,\n\n$$\na' = \\frac{b^2 + 1}{a} \\leq \\frac{b^2 + 1}{b+1} \\leq b < a.\n$$\n\nThus, if $(a, b)$ is a solution with $a > b$, then $(b, a')$ is another solution with a strictly smaller sum. Repeating this process, we eventually reach $(a_0, b_0) = (1, 1)$, so $k = 3$.\n\nTherefore, all solutions $(a, b)$ are generated by the chain:\n\n$$\n(1, 1) \\rightarrow (2, 1) \\rightarrow (5, 2) \\rightarrow (13, 5) \\rightarrow (34, 13) \\rightarrow \\dots\n$$\n\nThe rule is $(x, y) \\rightarrow (3x - y, x)$.\n\nBy induction, the $k$th pair (with $(1, 1)$ as pair 0) is $(F_{2k+1}, F_{2k-1})$, where $F_i$ is the Fibonacci sequence: $F_0 = 0$, $F_1 = 1$, $F_{i+2} = F_{i+1} + F_i$.\n\nSince $n$ is a multiple of $ab$ and a divisor of $3ab$, it follows that either\n\n$$\nn = ab \\text{ or } n = 3ab.\n$$\n\nThus, the solutions are:\n\n$$\nn = 1, 3, \\quad n = F_{2k-1}F_{2k+1}, \\quad n = 3F_{2k-1}F_{2k+1}, \\quad k = 1, 2, \\dots\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15075,
    "subject": "Mathematics (Olympiad)",
    "question": "A strip of breadth $\\ell$ is the set of those points in the plane situated between or on two parallel lines at distance $\\ell$ from each other.\n\nConsider a finite set $S$ of $n \\ge 3$ distinct points in the plane, with the property that any three points of $S$ can be covered by a strip of breadth $1$. Prove that the set $S$ can be covered by a strip of breadth $2$.",
    "options": [],
    "answer": "See solution",
    "solution": "All solutions encountered in the competition (including the official one) make use of the following:\n\n**LEMMA.** A triangle can be covered by a strip of breadth $1$ if and only if (at least) one of its altitudes is at most $1$ long.\n\n**Proof.** The other implication being trivial, consider a triangle $ABC$ covered by a strip of breadth $1$. The perpendicular lines through $A$, $B$, $C$ on the lines that define the strip can determine two situations:\n\n- one of them meets a side of the triangle $ABC$ at an interior point (see Figure 1)\n- one side of the triangle $ABC$ contains one of these perpendiculars (see Figure 2)\n\n![](images/RMC2010_p114_data_18381c2aa3.png)\n\nFigure 1\n\n![](images/RMC2010_p114_data_457884b298.png)\n\nFigure 2\n\nIn both cases, notice one of the altitudes (in the figures -- $AE$) is at most $1$ long. $\\square$\n\nNow, either choose points $A, B \\in S$ such that the distance $AB$ is maximal, or choose points $A, B, C \\in S$ such that the area of triangle $ABC$ is maximal.\n\nIn the former case, for any other point $C$, the distance from $C$ to $AB$ is at most $1$ (when $ABC$ is a non-degenerate triangle, this is the least altitude of the triangle),\n\nhence the set of points can be covered by a strip of breadth $2$, determined by two parallel lines to $AB$, situated at distance $1$ on each side of it.\n\nIn the latter case, it is easy to show all points of the set are included in the triangle $A'B'C'$, determined by parallels through $A$, $B$, $C$ to the opposite sides (an _old chestnut_). Since one of the altitudes of $ABC$ is at most $1$ long, one of the altitudes of $A'B'C'$ will be at most $2$ long, and we close the argument as above.\n\n**REMARKS.**\n\na) A solution not using the lemma may be reached by considering the triangle of maximal area $ABC$ and the associated triangle $A'B'C'$ as above. Lines $AA'$, $BB'$, $CC'$ are concurrent at $G$, the common centroid of the two triangles, while the homothety of center $G$ and ratio $-2$ maps $ABC$ onto $A'B'C'$, and the strip of breadth $1$ covering $ABC$ onto a strip of breadth $2$ covering $A'B'C'$, therefore covering the entire set of points.\n\nb) It seems the result in the problem might be improved, in the sense that we may replace the constant $2$ by the golden ratio $\\varphi = \\frac{1 + \\sqrt{5}}{2} \\approx 1.618\\ldots$ For the moment, the editors know of neither a proof of this conjecture, nor a counterexample. (However, when only four points are involved, it is known the best constant is $\\sqrt{2}$. See www.mathlinks.ro for related issues.) Also, the set $S$ could well be infinite.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15076,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a quadrangle inscribed in a circle $\\omega$. The lines $AB$ and $CD$ meet at $P$, the lines $AD$ and $BC$ meet at $Q$, and the diagonals $AC$ and $BD$ meet at $R$. Let $M$ be the midpoint of the segment $PQ$, and let $K$ be the common point of the segment $MR$ and the circle $\\omega$. Prove that the circles $KPQ$ and $\\omega$ are tangent to one another.\n\n![](images/RMC2013_final_p112_data_b57bd0a5eb.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the centre of $\\omega$. Notice that the points $P$, $Q$, and $R$ are the poles (with respect to $\\omega$) of the lines $QR$, $RP$, and $PQ$, respectively. Hence we have $OP \\perp QR$, $OQ \\perp RP$, and $OR \\perp PQ$, thus $R$ is the orthocentre of the triangle $OPQ$.\n\nNow, if $MR \\perp PQ$, then the points $P$ and $Q$ are the reflections of one another in the line $MR = MO$, and the triangle $PQK$ is symmetrical with respect to this line. In this case the statement of the problem is trivial.\n\nOtherwise, let $V$ be the foot of the perpendicular from $O$ to $MR$, and let $U$ be the common point of the lines $OV$ and $PQ$. Since $U$ lies on the polar line of $R$ and $OU \\perp MR$, we obtain that $U$ is the pole of $MR$. Therefore, the line $UK$ is tangent to $\\omega$. Hence it is enough to prove that $UK^2 = UP \\cdot UQ$, since this relation implies that $UK$ is also tangent to the circle $KPQ$.\n\nFrom the right triangle $OKU$, we get $UK^2 = UV \\cdot UO$. Let $\\Omega$ be the circumcircle of triangle $OPQ$, and let $R'$ be the reflection of its orthocentre $R$ in the midpoint $M$ of the side $PQ$. It is well known that $R'$ is the point of $\\Omega$ opposite to $O$, hence $OR'$ is the diameter of $\\Omega$. Finally, since $\\angle OVR' = 90^\\circ$, the point $V$ also lies on $\\Omega$, hence $UP \\cdot UQ = UV \\cdot UO = UK^2$, as required.\n\n**Remark.** The statement of the problem is still true if $K$ is the other common point of the line $MR$ and $\\omega$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15077,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a quadrilateral such that $AD^2 + BC^2 = AB^2$. Prove that there exists a unique point $P$ on $\\overline{AB}$ such that $\\angle APD = \\angle BPC$ and the line $PE$, where $E = \\overline{AC} \\cap \\overline{BD}$, bisects $\\overline{CD}$.\n\n![](images/2019-USAMO-Solutions_p1_data_d909176916.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Note that there can only be one point $P$ on $\\overline{AB}$ satisfying the given angle condition, since as $P$ moves from $A$ to $B$, $\\angle APD$ decreases while $\\angle BPC$ increases. Consequently, if we can show that there is a single point $P$ on $\\overline{AB}$ such that $\\angle APD = \\angle BPC$ and line $PE$ bisects $\\overline{CD}$, then it must coincide with the point in the problem statement, and we will be done. We construct such a point as follows.\n\nSince $AD^2 + BC^2 = AB^2$, there exists a point $P$ on $\\overline{AB}$ satisfying\n\n$$\nAD^2 = AP \\cdot AB \\quad \\text{and} \\quad BC^2 = BP \\cdot BA.\n$$\n\nThus $AP/AD = AD/AB$ and $BP/BC = BC/BA$. We then have similar triangles, $\\triangle APD \\sim \\triangle ADB$ and $\\triangle BPC \\sim \\triangle BCA$, from which $\\angle APD = \\angle ADB = \\angle ACB = \\angle BPC$. Now we show that line $PE$ bisects $\\overline{CD}$. Define $K = \\overline{AC} \\cap \\overline{PD}$ and $L = \\overline{BD} \\cap \\overline{PC}$.\n\nThe quadrilaterals $APLD$ and $BPKC$ are cyclic, because\n\n$$\n\\angle ADL = \\angle ACB = \\angle BPC = \\angle APL\n$$\n\nand similarly $\\angle KCB = \\angle KPB$. (The notation $\\angle$ here refers to directed angles taken modulo $180^\\circ$.)\n\nNow the quadrilateral $AKLB$ is also cyclic, because\n\n$$\n\\angle AKB = \\angle CKB = \\angle CPB\n$$\n\nand similarly $\\angle ALB = \\angle APD$, and these are equal.\n\nNow the cyclic quadrilaterals imply $\\angle KCD = \\angle ABD = \\angle ABL = \\angle AKL = \\angle CKL$, from which we conclude $\\overline{CD} \\parallel \\overline{KL}$. Thus $CDKL$ is a trapezoid whose legs intersect at $P$ and whose diagonals intersect at $E$. As is well-known (and can be quickly shown using Ceva's theorem), this implies that line $PE$ bisects the bases $\\overline{CD}$ and $\\overline{KL}$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15078,
    "subject": "Mathematics (Olympiad)",
    "question": "Two players, Alice and Bob, play a game on $n$ vertices labeled $1, 2, \\dots, n$. Alice and Bob take turns adding edges $\\{i, j\\}$, with Alice going first. Both players are banned from creating a cycle, and the game ends after $n-1$ total turns.\n\nLet the weight of the edge $\\{i, j\\}$ be $|i-j|$, and let $W$ be the total weight of all edges at the end of the game. If Alice plays to maximize $W$ and Bob plays to minimize $W$, what will $W$ be?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $k = \\lfloor \\frac{n-1}{2} \\rfloor$. The answer is\n\n$$\n\\frac{1}{2}(k+1)(2n-k-2) = (n-k-1) + (n-k) + \\dots + (n-1).\n$$\n\nWhen $n=1$, this is clear.\n\nConsider now when $n \\ge 2$. Note that the game consists of $n-1$ moves, with Alice making $k$ moves and Bob making $n-k-1$ moves.\n\nWe first show Alice can guarantee a total of at least $\\frac{1}{2}(k+1)(2n-k-2)$.\n\n**Claim** — Alice can ensure her $i$th move is an edge of weight at least $n-i$.\n\n*Proof.* On her first move, Alice chooses the edge $(1, n)$, which has weight $n-1$. Now, consider the $i$th move, for $i > 1$.\n\nConsider the induced subgraph on vertices $\\{1, \\dots, i\\} \\cup \\{n-i+1, \\dots, n\\}$. (Note that $i < n-i+1$ for all $1 \\le i \\le k$.) Before Alice's $i$th move, exactly $2i-2$ moves have been made, so since this subgraph has $2i$ vertices, it is disconnected.\n\nTake a vertex $j$ such that $j$ and $1$ are in different connected components. Since Alice's first move is $(1, n)$, $j$ is also not connected to $n$. Now, if $j \\in \\{1, \\dots, i\\}$, Alice adds $(j, n)$, otherwise Alice adds $(1, j)$. $\\square$\n\nSince each of Bob's edges has weight at least $1$, his edges in total have weight at least $n-k-1$. Thus, Alice can ensure a total weight of at least $(n-k-1)+(n-k)+\\dots+(n-1)$.\n\nWe now give a strategy for Bob.\n\n**Claim** — Bob can ensure that for each $1 \\le i \\le k$, there are at most $i$ edges of weight at least $n-i$, while only adding edges of weight $1$.\n\n*Proof.* For $i=1$, this is clear. Consider $i > 1$. As above, consider the induced subgraph on vertices $\\{1, \\dots, i\\} \\cup \\{n-i+1, \\dots, n\\}$ before Bob's $(i-1)$st move.\n\nWe show that there exists some edge of weight $1$ in this subgraph that Bob can choose. Suppose not. Then, since none of $(1, 2), \\dots, (i-1, i)$ are valid moves, the vertices $1, \\dots, i$ are connected. Likewise, the vertices $(n-i+1, \\dots, n)$ are connected.\n\nHence, there are at least $2(i-1)$ edges in this subgraph. However, only $2i-3$ moves have been played up to this point, contradiction. So, Bob can choose an edge of weight $1$ in this subgraph.\n\nWhen $n$ is even, Bob makes $k-1$ moves, so this accounts for all his moves. When $n$ is odd, on the final move, since not all the vertices are connected, there exists some edge of weight $1$ that Bob can add.\n\nThus, Bob can select only edges of weight $1$. Now, following the above strategy, for any $1 \\le i \\le k$, Bob chooses at least $k-1$ edges in the induced subgraph on vertices $\\{1, \\dots, i\\} \\cup \\{n-i+1, \\dots, n\\}$. Since the induced subgraph contains no cycles, Alice plays at most $(2i-1)-(i-1)=i$ edges in this subgraph.\n\nNow, note that any edge of weight at least $n-i$ must have both vertices contained in this subgraph. Thus, there are at most $i$ edges of weight at least $n-i$, as desired. (When $n > 2$, $n-i \\ge n-k > 1$, and when $n=2$, Bob plays no edges, so the bound still holds.) $\\Box$\n\nUsing the above strategy, the total weight of Bob's edges is $n-k-1$. Let $a_i$ be the number of edges Alice plays of weight at least $i$. Then, the total weight of Alice's edges is\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n-1} a_i &\\le k(n-k-1) + \\sum_{i=n-k}^{n-1} a_i \\\\\n&\\le k(n-k-1) + (k + (k-1) + \\dots + 1) \\\\\n&= (n-1) + (n-2) + \\dots + (n-k).\n\\end{aligned}\n$$\n\nThus, Bob can ensure a total weight of at most $(n-k-1) + (n-k) + \\dots + (n-1)$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15079,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the complex number $z = (a + \\cos \\theta) + (2a - \\sin \\theta)i$. If $|z| \\le 2$ for any $\\theta \\in \\mathbb{R}$, then the range of the real number $a$ is \\underline{\\quad}.",
    "options": [],
    "answer": "See solution",
    "solution": "By the definition above, for any $\\theta \\in \\mathbb{R}$,\n\n$$\n\\begin{align*}\n|z| \\le 2 &\\Leftrightarrow (a + \\cos \\theta)^2 + (2a - \\sin \\theta)^2 \\le 4 \\\\\n&\\Leftrightarrow 2a(\\cos \\theta - 2\\sin \\theta) \\le 3 - 5a^2 \\\\\n&\\Leftrightarrow -2\\sqrt{5}a \\sin(\\theta - \\varphi) \\le 3 - 5a^2 \\\\\n&\\Rightarrow 2\\sqrt{5} |a| \\le 3 - 5a^2 \\\\\n&\\Rightarrow |a| \\le \\frac{\\sqrt{5}}{5}.\n\\end{align*}\n$$\n\nSo the range of $a$ is $\\left[-\\frac{\\sqrt{5}}{5}, \\frac{\\sqrt{5}}{5}\\right]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15080,
    "subject": "Mathematics (Olympiad)",
    "question": "For $a = 1$ the expression is equal to $-2$. Prove that this is the minimal value, i.e., that for every $a \\ge 0$ we have\n$$\na^3 - a^2 - 2\\sqrt{a} \\ge -2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n$$\na^3 - a^2 - 2\\sqrt{a} + 2 = a^2(a-1) - 2(\\sqrt{a}-1) = (\\sqrt{a}-1)\\left[a^2(\\sqrt{a}+1) - 2\\right].\n$$\n\nIf $a \\ge 1$, then $a^2(\\sqrt{a}+1) - 2 \\ge 2 - 2 = 0$ and $\\sqrt{a}-1 \\ge 0$, so our statement is true.\n\nIf $0 \\le a < 1$, then $a^2(\\sqrt{a}+1) - 2 < 2 - 2 = 0$ and $\\sqrt{a}-1 < 0$, so our statement is true again.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15081,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of integers $(m, n)$ such that for all positive real numbers $x$ and $y$, the inequality\n\n$$\nx^m + y^n \\ge x^n y^m\n$$\nholds.",
    "options": [],
    "answer": "See solution",
    "solution": "If $m = 0$, then the inequality is $1 + y^n \\ge x^n$. This holds for all positive real numbers $x$ and $y$ if and only if $n = 0$. Hence, $(0, 0)$ is a solution.\n\nNow, let both $m$ and $n$ be different from zero. If the pair $(m, n)$ satisfies the condition, then substituting $x$ and $y$ by $1/x$ and $1/y$ shows that $(-m, -n)$ also satisfies the condition. Thus, we can assume without loss of generality that $m \\ge n$ and $m \\ge 0$.\n\nIf $m > n$, then by taking $x = 1$ we get $1 + y^n \\ge y^m$, which does not hold for large $y$. Hence $m = n$.\n\nBy taking $x = y = 4$, we get $2 \\cdot 4^m \\ge 4^{2m}$, which does not hold for any positive integer $m$. Therefore, there are no more suitable pairs.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15082,
    "subject": "Mathematics (Olympiad)",
    "question": "A finite set of distinct positive integers is written on a blackboard. A *move* consists in choosing two numbers and writing their lowest common multiple, provided it is not already written. The set is called *closed* if no moves are allowed. For example, the set $\\{2, 3, 4, 6\\}$ will be closed after the number $12$ is added. Determine the maximum number of elements in a closed set given that the initial set contains $10$ numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "The largest closed set has $2^{10} - 1$ numbers. To obtain this, start with ten primes $p_i$, $i = 1, 2, \\dots, 10$. The closed set consists of all products $\\prod_{i \\in X} p_i$ over all nonempty subsets $X$ of $\\{1, 2, \\dots, 10\\}$, so there are $2^{10} - 1$ elements in the closed set.\n\nOne cannot obtain more than $2^{10} - 1$ numbers. Each move gives the l.c.m. of two or more numbers from the initial set, so each added number corresponds to a nonempty subset of $\\{1, 2, \\dots, 10\\}$. The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15083,
    "subject": "Mathematics (Olympiad)",
    "question": "For any positive integer $n$, define $a_n = \\left\\{\\frac{n}{s(n)}\\right\\}$, where $s(k)$ represents the sum of the digits of the natural number $k$, and $\\{x\\}$ is the fractional part of the real number $x$.\n\n1. Prove that there exist infinitely many positive integers $n$ such that $a_n = \\frac{1}{2}$.\n\n2. Determine the smallest positive integer $n$ such that $a_n = \\frac{1}{6}$.",
    "options": [],
    "answer": "See solution",
    "solution": "1. If $s(n) = 2$ and $n$ is odd, then $a_n = \\frac{1}{2}$. The only solutions with these properties are of the form $n = 10^k + 1$, with $k \\in \\mathbb{N}^*$. Thus, there are infinitely many such $n$.\n\n2. Let $n$ be a positive integer such that $a_n = \\left\\{\\frac{n}{s(n)}\\right\\} = \\frac{1}{6}$.\n\nSince $\\frac{n}{s(n)} - \\left\\lfloor \\frac{n}{s(n)} \\right\\rfloor = \\frac{1}{6}$, we infer that $6n - 6 \\cdot s(n) \\cdot \\left\\lfloor \\frac{n}{s(n)} \\right\\rfloor = s(n)$. From here, $6 \\mid s(n)$, so $3 \\mid n$. Consider $n = 3k$ and $s(n) = 6m$, with $m, k$ positive integers. We deduce $3k - 6m \\cdot \\left\\lfloor \\frac{k}{2m} \\right\\rfloor = m$, hence $3 \\mid m$. Thus, $m = 3u$, with $u \\in \\mathbb{N}^*$ and $s(n) = 18u$, so $9 \\mid n$.\n\nConsider $n = 9v$, with $v$ a positive integer. We obtain $3v - 6u \\cdot \\left\\lfloor \\frac{v}{2u} \\right\\rfloor = u$, so $3 \\mid u$. Let $u = 3t$, with $t$ a positive integer. It follows that $m = 9t$ and $s(n) = 54t$, and the minimal sum of the digits of a natural number $n$ is $54$.\n\nThe smallest positive integer with the sum of its digits $54$ is $n = 999999$.\n\nBut $a_{999999} = \\left\\{\\frac{999999}{54}\\right\\} = \\frac{1}{2}$, so $n = 999999$ is not a solution. The next positive integer with the sum of its digits $54$ is $n = 1899999$, for which $a_{1899999} = \\left\\{\\frac{1899999}{54}\\right\\} = \\left\\{35185 + \\frac{1}{6}\\right\\} = \\frac{1}{6}$, therefore $n_{\\min} = 1899999$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15084,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a > b > c > d$ be positive integers and suppose\n$$\nac + bd = (b + d + a - c)(b + d - a + c).\n$$\nProve that $ab + cd$ is not prime.\n\n*Note.* First, note that\n$$\n(b+d+a-c)(b+d-a+c) = (b+d)^2 - (a-c)^2 \\\\ = b^2 + 2bd + d^2 - a^2 + 2ac - c^2,\n$$\nso that the given equation is equivalent to\n$$\na^2 - ac + c^2 = b^2 + bd + d^2. \\quad (1)\n$$\nMany of the following solutions use this fact.\n\nExamples of 4-tuples $(a, b, c, d)$ that satisfy the given conditions are $(21, 18, 14, 1)$ and $(65, 50, 34, 11)$.",
    "options": [],
    "answer": "See solution",
    "solution": "*First Solution.* For the sake of contradiction, assume that $ab + cd$ is prime. Note that\n$$\nab + cd = (a + d)c + (b - c)a = m \\cdot \\gcd(a + d, b - c)\n$$\nfor some positive integer $m$. Writing $g = \\gcd(a+d, b-c)$, we have\n$$\nm = \\frac{a+d}{g} \\cdot c + \\frac{b-c}{g} \\cdot a \\geq c+a > 1.\n$$\nTherefore, because $ab+cd$ is prime, $g=1$.\n\nSubstituting $ac+bd = (a+d)b - (b-c)a$ for the left-hand side of the given condition, we obtain\n$$\n(a+d)b - (b-c)a = (a+d)(b+d-a+c) + (b-c)(b+d-a+c),\n$$\nor\n$$\n(a+d)(a-c-d) = (b-c)(b+c+d).\n$$\nHence, there exists a positive integer $k$ such that\n$$\na-c-d = k(b-c),\n$$\n$$\nb+c+d = k(a+d).\n$$\nAdding these equations, we obtain $a+b = k(a+b-c+d)$ and thus $k(c-d) = (k-1)(a+b)$. Recall that $a > b > c > d > 0$. If $k=1$, then $c=d$, a contradiction. If $k \\ge 2$, then\n$$\n2 \\ge \\frac{k}{k-1} = \\frac{a+b}{c-d} > \\frac{2b}{c} > 2,\n$$\na contradiction.\n\nTherefore, our original assumption was wrong, and $ab+cd$ is not prime.\n\n*Second Solution.* (By Yonggao Chen, China) We give a proof by contradiction. Assume that $p = ab+cd$ is prime. Then $ab \\equiv -cd \\pmod p$. By (1),\n$$\nb^2(b^2 + bd + d^2) = b^2(a^2 - ac + c^2) = (ab)^2 - ab(bc) + b^2c^2.\n$$\nIt follows that\n$$\n\\begin{aligned}\nb^2(b^2 + bd + d^2) &\\equiv (ab)^2 - ab(bc) + b^2c^2 \\\\\n&\\equiv (cd)^2 + cd(bc) + b^2c^2 \\\\\n&\\equiv c^2(b^2 + bd + d^2) \\pmod p,\n\\end{aligned}\n$$\nimplying that $p \\mid (b^2 - c^2)(b^2 + bd + d^2)$. Observe that $0 < b^2 - c^2 < b^2 < ab < p$. Thus, $p$ and $b^2 - c^2$ must be relatively prime, so\n$$\np \\mid (b^2 + bd + d^2). \\qquad (2)\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15085,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c, d$ be positive integers such that $ab(a+b) = cd(c+d) \\leq 2020$. Find all such quadruples $(a, b, c, d)$, up to order, and determine how many distinct values $m = ab(a+b)$ (called \"elite numbers\") can be represented in this way.",
    "options": [],
    "answer": "See solution",
    "solution": "*Case 1: $a = 1$.*\n\nLet $b = f_4 = f_1 f_2 f_3 - 1$, with $f_1, f_2, f_3$ positive integers. We require $f_4 > f_1 + f_2 f_3$ and $f_1 f_2 < f_3 + f_4$, so that $c, d, c+d$ do not exist as positive integers. One of $b, b+1$ is the product of $f_4$ and the minimum of $\\{f_1, f_2, f_3\\}$, and the other is the product of the other two $f_i$'s. Since $ab(a+b) \\leq 2020$, or $b(b+1) \\leq 2020$, $b \\leq 44$.\n\nFor $b \\in \\{1, 2, \\dots, 44\\}$, there are nine $b$ such that $b, b+1$ are both products of two prime powers:\n\n$14, 20, 21, 33, 34, 35, 38, 39, 44.$\n\nAccordingly, we find seven solutions:\n\n$$\n(1, 14, 3, 7),\\ (1, 20, 5, 7),\\ (1, 21, 3, 11),\\ (1, 33, 6, 11),\\ (1, 34, 7, 10),\\ (1, 38, 6, 13),\\ (1, 44, 9, 11)\n$$\n\nThere are a total of 10 solutions in Case 1.\n\n*Case 2: $pq > 1$.*\n\nAssume $a \\leq b$, $c \\leq d$. Let $q > p \\geq 1$, $q^3 \\mid cd(c+d)$. Since $(p, q) = 1$, $q^3$ divides one of $a, b, a+b$. If $q \\geq 4$, then $a+b \\geq 64$, so $m > 2020$. Thus $q = 2$ or $3$.\n\n**(2$^\\circ$) If $q = 3$:**\n\n$27 \\mid ab(a+b)$. Since $a \\leq 10$, $27 \\mid b$ or $27 \\mid a+b$. From $m \\leq 2020$, $b = 27$ or $a+b = 27$.\n\n- If $b = 27$, $a(a+27) = c_1 d_1 (c_1 + d_1) < 75$, $a < 3$, so $a = 1$ or $2$. Both cases yield no integer solutions.\n- If $a+b = 27$, $a(27-a) = c_1 d_1 (c_1 + d_1) < 75$, $a < 4$, $a = 1$ or $2$. Both cases yield no integer solutions.\n\n**(2$^0$) If $q = 2$:**\n\n$8 \\mid ab(a+b)$.\n\n- If $8 \\mid a$, $a = 8$, $b \\geq 8$ is odd, but $c_1 d_1 (c_1 + d_1)$ is even, so $b$ must be even, a contradiction.\n- If $8 \\mid b$, $b = 8k$, $k = 2$ or $4$ (since $b \\leq 44$). For $k = 2$, $b = 16$, $a \\leq 5$. For $a = 5$, two solutions: $(5, 16, 2, 28)$ and $(5, 16, 6, 14)$. For $k = 4$, $b = 32$, $a = 1$, one solution: $(1, 32, 2, 22)$.\n- If $8 \\mid (a+b)$, $a+b = 8k$, $k = 2$ or $4$. For $k = 4$, $a = 1$, no solution. For $k = 2$, $a+b = 16$, $a$ odd, $a \\in \\{1, 3, 5, 7\\}$:\n    - $a = 1$: $(1, 15, 2, 10)$ and $(1, 15, 4, 6)$\n    - $a = 3$: no solution\n    - $a = 5$: $(5, 11, 2, 20)$\n    - $a = 7$: $(7, 9, 4, 14)$\n\nAltogether, 7 solutions in Case 2:\n\n$(5, 16, 2, 28),\\ (5, 16, 6, 14),\\ (1, 32, 2, 22),\\ (1, 15, 2, 10),\\ (1, 15, 4, 6),\\ (5, 11, 2, 20),\\ (7, 9, 4, 14)$\n\n*Case 3: $(p, q) \\neq 1$.*\n\nReduce to $(p, q) = 1$ by taking $(a, b, c, d) = k(\\bar{a}, \\bar{b}, \\bar{c}, \\bar{d})$, $k \\geq 2$. $(\\bar{a}, \\bar{b}, \\bar{c}, \\bar{d})$ must be one of the 17 solutions from Cases 1 and 2. The elite number $m = ab(a+b) = k^3 \\bar{m}$, $\\bar{m} \\leq 252$.\n\nFrom Case 1, only $(1, 5, 2, 3)$ ($\\bar{m} = 30$) and $(1, 14, 3, 7)$ ($\\bar{m} = 210$) qualify:\n- $(2, 10, 4, 6)$, $(3, 15, 6, 9)$, $(4, 20, 8, 12)$ from $(1, 5, 2, 3)$\n- $(2, 28, 6, 14)$ from $(1, 14, 3, 7)$\n\nFrom Case 2, $(1, 15, 2, 10)$ and $(1, 15, 4, 6)$ ($\\bar{m} = 240$):\n- $(2, 30, 4, 20)$, $(2, 30, 8, 12)$\n\nIn all, 6 derivative solutions.\n\nFinally, there are 3 elite numbers each with three representations:\n\n- $m = 240$: $(1, 15, 2, 10)$, $(1, 15, 4, 6)$, $(2, 10, 4, 6)$\n- $m = 1680$: $(2, 28, 5, 16)$, $(2, 28, 6, 14)$, $(5, 16, 6, 14)$\n- $m = 1920$: $(2, 30, 4, 20)$, $(2, 30, 8, 12)$, $(4, 20, 8, 12)$\n\nEach is counted two extra times, so the total number of elite numbers is $10 + 7 + 6 - 3 \\times 2 = 17$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15086,
    "subject": "Mathematics (Olympiad)",
    "question": "In $\\triangle ABC$, $\\tan A$ and $\\tan B$ are the two roots of the equation $x^2 - 10x + 6 = 0$. Then the value of $\\cos C$ is \\underline{\\hspace{2cm}}.",
    "options": [],
    "answer": "See solution",
    "solution": "By the condition, we know that $\\tan A + \\tan B = 10$ and $\\tan A \\tan B = 6$. Thus,\n\n$$\n\\begin{aligned}\n\\tan C &= \\tan(\\pi - A - B) \\\\\n&= -\\tan(A + B) \\\\\n&= -\\frac{\\tan A + \\tan B}{1 - \\tan A \\tan B} \\\\\n&= -\\frac{10}{1 - 6} \\\\\n&= 2.\n\\end{aligned}\n$$\n\nTherefore, $C$ is an acute angle, and thus\n\n$$\n\\cos C = \\frac{1}{\\sqrt{1 + \\tan^2 C}} = \\frac{1}{\\sqrt{5}} = \\frac{\\sqrt{5}}{5}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15087,
    "subject": "Mathematics (Olympiad)",
    "question": "Juku has the first 100 volumes of the Harrie Totter book series at his home. For every $i$ and $j$, where $1 \\leq i < j \\leq 100$, call the pair $(i, j)$ **reversed** if volume No $j$ is before volume No $i$ on Juku's shelf. Juku wants to arrange all volumes of the series in one row on his shelf in such a way that there do not exist numbers $i, j, k$, where $1 \\leq i < j < k \\leq 100$, such that pairs $(i, j)$ and $(j, k)$ are both reversed. Find the largest number of reversed pairs that can occur under this condition.",
    "options": [],
    "answer": "See solution",
    "solution": "Let all 100 volumes be placed on a shelf $a$ in some order. For every $i = 1, 2, \\ldots, 100$, let $a_i$ be the number of the volume that occurs as the $i$th in this order. In the order the volumes occur on shelf $a$, we start relocating the volumes to two new shelves $b$ and $c$. We place a volume to the end of shelf $b$ if, as the intermediate result, all volumes on shelf $b$ would be increasingly sorted by volume numbers. Otherwise, we place the volume to the end of shelf $c$ if, as the intermediate result, all volumes on shelf $c$ would be increasingly sorted by volume numbers. Continuing this way, we either can relocate all volumes onto two shelves in such a way that all volumes on either shelf are increasingly sorted by volume numbers or get stuck on some step $k$ of the process because $a_k$ is less than the number of the last volume on both new shelves. In the last case, let the number of the last volume on shelf $c$ be $a_j$; then, by assumptions, $j < k$ and $a_j > a_k$. Since volume No $a_j$ has been relocated to shelf $c$, some volume with number $a_i$ must occur on shelf $b$ such that $i < j$ and $a_i > a_j$. This means that pairs $(i, j)$ and $(j, k)$ were initially reversed. Hence we can conclude that, in the case of Juku's favourite orderings, all volumes can be relocated to two shelves, i.e., there exist two tuples $a_{i_1}, a_{i_2}, \\ldots, a_{i_s}$ and $a_{j_1}, a_{j_2}, \\ldots, a_{j_t}$ with $s + t = 100$, where $i_1 < \\ldots < i_s$, $a_{i_1} < \\ldots < a_{i_s}$ and $j_1 < \\ldots < j_t$, $a_{j_1} < \\ldots < a_{j_t}$. The number of pairs $(i, j)$ such that $i < j$ and the numbers $a_i$ and $a_j$ are in distinct tuples is exactly $st$. Each reversed pair $(i, j)$ must be one of these $st$ pairs. Thus the number of reversed pairs does not exceed $st$. As $st \\leq \\left(\\frac{s+t}{2}\\right)^2 = 50^2 = 2500$, the number of reversed pairs cannot exceed 2500. The number 2500 is achieved by the order $51, 52, \\ldots, 100, 1, 2, \\ldots, 50$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 15088,
    "subject": "Mathematics (Olympiad)",
    "question": "The incircle of $\\triangle ABC$ touches the sides $BC$ and $AC$ at points $A_1$ and $B_1$, respectively. The lines $B_1A_1$ and $AB$ are concurrent at $X$ such that $A$ lies between $X$ and $B$. If $\\angle CXB = 90^\\circ$ and $BC^2 = AB^2 + BC \\cdot AC$, find the angles of $\\triangle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $Y \\in BC$ be such that $AY \\parallel XA_1$. Thus $BY = a-b$ and since $\\frac{a}{c} = \\frac{c}{a-b}$, we have that $\\triangle ABC \\sim \\triangle YBA$. Therefore $\\angle AXA_1 = \\angle BAY = \\gamma$, which implies that quadrilateral $XAA_1C$ is cyclic. It follows from $\\angle XAC = \\angle XA_1C = 90^\\circ - \\frac{\\gamma}{2}$ that $\\alpha = 90^\\circ + \\frac{\\gamma}{2}$. Moreover, $\\angle AA_1C = 90^\\circ$, i.e., $AA_1$ is simultaneously altitude and angular bisector. Hence $\\gamma = \\beta$ and $90^\\circ + \\frac{\\gamma}{2} + 2\\gamma = 180^\\circ$, i.e., $\\gamma = 36^\\circ$. The angles of $\\triangle ABC$ are $36^\\circ$, $36^\\circ$, and $108^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15089,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a circumscribed quadrilateral (i.e., convex with sides that are all tangent to a single circle) and let $X$ be the intersection point of its diagonals $AC$ and $BD$. Let $I_1$, $I_2$, $I_3$, $I_4$ be the incenters of $\\triangle DXC$, $\\triangle BXC$, $\\triangle AXB$, and $\\triangle DXA$, respectively. The circumcircle of $\\triangle CI_1I_2$ intersects the sides $CB$ and $CD$ at points $P$ and $Q$, respectively. The circumcircle of $\\triangle AI_3I_4$ intersects the sides $AB$ and $AD$ at points $M$ and $N$, respectively.\n\nProve that\n\n$$\nAM + CQ = AN + CP.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove the following auxiliary result.\n\n*Lemma.* Let $ABC$ be an arbitrary triangle, and $D$ be an arbitrary point on the segment $AB$. Denote by $O_1$ and $O_2$ the incenters of $\\triangle ADC$ and $\\triangle BDC$, respectively. If the circumcircle $\\omega$ of $\\triangle CO_1O_2$ intersects the sides $AC$ and $BC$ at points $P$ and $Q$, respectively, then\n\n$$\nCP - CQ = (AC - BC) + (BD - AD).\n$$\n\n*Proof.* Denote by $T$ and $K$ the orthogonal projections of $O_1$ onto $AC$ and $CD$, respectively. Let $R$ be the second intersection point of $\\omega$ with $CD$. Since $CO_1$ is an angle bisector, it follows that $O_1P = O_1R$, thus $\\triangle O_1TP \\cong \\triangle O_1KR$. Hence, $TP = KR$ and\n\n$$\nCP + CR = 2 \\cdot CT (= 2 \\cdot CK) = AC + CD - AD.\n$$\n\nAnalogously, $CQ + CR = BC + CD - BD$, and taking the difference of the last two expressions, the result follows. $\\square$\n\nNow, applying the Lemma for $\\triangle DBC$, we deduce that\n\n$$\nCQ - CP = (CD - BC) + (BX - DX).\n$$\n\nAnalogously, applying the Lemma for $\\triangle ABD$, we derive\n\n$$\nAN - AM = (AD - AB) + (BX - DX).\n$$\n\nSubtracting the second equation from the first and taking into account that $CD - BC = AD - AB$, we obtain $AM + CQ = AN + CP$, as desired.\n\n**Remark.** The reverse statement also holds true: if for a convex quadrilateral $ABCD$ we have $AM + CQ = AN + CP$, then $ABCD$ is circumscribed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15090,
    "subject": "Mathematics (Olympiad)",
    "question": "Sea $n$ un entero positivo. Se tienen $n$ colores, $n \\ge 1$. Cada uno de los números enteros entre $1$ y $1000$ se quiere pintar con uno de los $n$ colores de modo que cada dos números diferentes, si uno divide al otro, tengan colores diferentes. Dar el menor número $n$ para que esto sea posible.",
    "options": [],
    "answer": "See solution",
    "solution": "Observamos que los $10$ números $2^0=1,\\ 2^1=2,\\ 2^2=4,\\ 2^3=8,\\ 2^4=16,\\ 2^5=32,\\ 2^6=64,\\ 2^7=128,\\ 2^8=256,\\ 2^9=512$ tienen la propiedad de que para cualesquiera dos, uno de ellos divide al otro. Por lo tanto, no pueden tener el mismo color, lo que implica que $n \\ge 10$.\n\nDamos una coloración para $n=10$.\n\n![]( <table><thead><tr><td>Números</td><td>color</td></tr></thead><tbody><tr><td>1</td><td>A</td></tr><tr><td>del 2 hasta el 3=2<sup>2</sup>-1</td><td>B</td></tr><tr><td>del 2<sup>2</sup> hasta el 7=2<sup>3</sup>-1</td><td>C</td></tr><tr><td>del 2<sup>3</sup> hasta el 15=2<sup>4</sup>-1</td><td>D</td></tr><tr><td>del 2<sup>4</sup> hasta el 31=2<sup>5</sup>-1</td><td>E</td></tr><tr><td>del 2<sup>5</sup> hasta el 63=2<sup>6</sup>-1</td><td>F</td></tr><tr><td>del 2<sup>6</sup> hasta el 127=2<sup>7</sup>-1</td><td>G</td></tr><tr><td>del 2<sup>7</sup> hasta el 255=2<sup>8</sup>-1</td><td>H</td></tr><tr><td>del 2<sup>8</sup> hasta el 511=2<sup>9</sup>-1</td><td>I</td></tr><tr><td>del 2<sup>9</sup> hasta el 1000</td><td>J</td></tr></tbody></table> )\n\nVemos que el cociente entre cualesquiera dos números del mismo color es menor que $2$, es decir, no hay números de igual color que sean uno divisor del otro.\n\n*Otro ejemplo.* Numeramos los colores del $0$ al $9$ y pintamos cada número que es producto de $m$ primos, no necesariamente distintos, con el color $m$ para $0 \\le m \\le 9$. Así, $1$ tiene el color $0$, todos los primos tienen el color $1$, los productos de dos primos (incluyendo a $p^2$) tienen color $2$, etc. Notemos que está bien definido pues cada entero desde $2$ hasta $1000$ es producto de a lo sumo $9$ primos. Es suficiente observar que si $a$ divide a $b$ y $a \\neq b$ entonces $b$ tiene más factores primos que $a$ y por lo tanto sus colores son diferentes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15091,
    "subject": "Mathematics (Olympiad)",
    "question": "Given that $ABC$ is a triangle where $AB < AC$. On the half-lines $BA$ and $CA$ we take points $F$ and $E$ respectively such that $BF = CE = BC$. Let $M$, $N$, and $H$ be the midpoints of the segments $BF$, $CE$, and $BC$ respectively, and let $K$ and $O$ be the circumcenters of the triangles $ABC$ and $MNH$ respectively. We assume that $OK$ cuts $BE$ and $HN$ at the points $A_1$ and $B_1$ respectively, and that $C_1$ is the point of intersection of $HN$ and $FE$. If the parallel line from $A_1$ to $OC_1$ cuts the line $FE$ at $D$ and the perpendicular from $A_1$ to the line $DB_1$ cuts $FE$ at the point $M_1$, prove that $E$ is the orthocenter of the triangle $A_1OM_1$.",
    "options": [],
    "answer": "See solution",
    "solution": "The circumcenter of the triangle $\\triangle MNH$ coincides with the incenter of the triangle $\\triangle ABC$ because the triangles $\\triangle BMH$ and $\\triangle NHC$ are isosceles and therefore the perpendiculars of $MH$, $HN$ are also the bisectors of the angles $\\angle ABC$, $\\angle ACB$, respectively.\n\n![](images/BMO_2016_Short_List_Final_1_p20_data_f572bd994c.png)\n\nLet $G$, $I$ be the points of tangency of the incircle $(O, r)$ of the triangle $\\triangle ABC$ with the sides $AB$ and $AC$ respectively. Now if $a$, $b$, $c$ are the sides of the triangle $\\triangle ABC$ and $s$ the semiperimeter of the triangle, we have\n\n$$\nOF^2 = OG^2 + FG^2 = r^2 + (a - s + b)^2\n$$\n\nand\n\n$$\nOE^2 = OI^2 + EI^2 = r^2 + (a - s + c)^2\n$$\n\nThen\n\n$$\nOF^2 - OE^2 = a(b - c) \\qquad (1)\n$$\n\nApplying Stewart's Theorem twice to the triangles $KFB$ and $KAC$ we get\n\n$$\nFA \\cdot KA^2 + c \\cdot KF^2 = a \\cdot KA^2 + ac \\cdot FA \\quad \\text{or} \\quad KF^2 = KA^2 + a(a-c) \\qquad (2)\n$$\n\nand\n\n$$\nEA \\cdot KE^2 + b \\cdot KA^2 = b \\cdot KE^2 + ab \\cdot EA \\quad \\text{or} \\quad KE^2 = KA^2 + a(b-a) \\qquad (3)\n$$\n\nFrom (2) and (3) we have\n\n$$\nKF^2 - KE^2 = a(a-c) - a(b-a) = a(b-c) \\qquad (4)\n$$\n\nFrom (1) and (4), because $OF^2 - OE^2 = KF^2 - KE^2$, we have that $FE \\perp OK$.\n\nLet $J$ be the point of intersection of $FE$ and $OK$.\nBecause $A_1D \\parallel OC_1$ we have\n\n$$\n\\frac{JO}{JA_1} = \\frac{JC_1}{JD}\n$$\n\nAnd since $A_1E \\parallel HN$ we get\n\n$$\n\\frac{JE}{JA_1} = \\frac{JC_1}{JB_1}\n$$\n\nTherefore, we have\n\n$$\n\\frac{JO}{JE} = \\frac{JB_1}{JD}\n$$\n\nThus, from the inverse of Thales' theorem we have that $EO \\parallel DB_1$, so\n\n$$\nAM_1 \\perp EO\n$$\n\nConsequently, the point $E$ is the orthocenter of the triangle $\\triangle A_1OM_1$.\n\n**Remark:** Here in the solution the side $BC$ has been referred to as $a$ and $\\alpha$, which are equivalent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15092,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $E$ and $F$ be the intersections of the circumcircle of $ABC$ with the lines $BP$ and $CP$, respectively. Let $S$ be the intersection of $EY$ and $FX$. By Pascal's theorem, $S$ lies on the circumcircle of $ABC$. It is also easy to see that $XYQS$ is cyclic.\n\nLet $F'$ be the intersection of $CX$ with the circumcircle of $ABC$, and let $G$ be the intersection of $XY$ and $BC$. Note that $TZ$ is the external angle bisector of $\\angle XTY$ and $PT$ is perpendicular to $TZ$, so $PT$ is the angle bisector of $\\angle XTZ$. Thus, $(GP, XY) = -1$ and $BY$, $CX$, $AP$ are concurrent. Let $R$ be the intersection of these lines, and $D$ be the intersection of $AP$ and $BC$. By looking through point $C$, we have $(DP, RA) = (GP, XY) = -1$. By projecting $X$ to the circumcircle of $ABC$ and also projecting this circle onto $AP$ through $C$, we have\n\n$$\n(BC, SA) = (AF', FB) = (AR, PD) = -1.\n$$\n\n% ![](path/to/file.png)\n\nWhat is the fixed point $S$ on the circumcircle of $ABC$ related to the symmedian, and why does the circumcircle of triangle $QXY$ pass through this point?",
    "options": [],
    "answer": "See solution",
    "solution": "This shows that $S$ is the intersection of the circumcircle of $ABC$ with the *symmedian*, and the proof is complete as the circumcircle of triangle $QXY$ passes through this fixed point $S$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15093,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that $m^2 + 2$ divides $(m^2 + k)^2 + 1$, where $m = \\lfloor \\sqrt{n} \\rfloor$ and $n = m^2 + k$ for some integer $k$ with $0 \\leq k \\leq 2m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m = \\lfloor \\sqrt{n} \\rfloor$ ($m \\in \\mathbb{N}^+$). Therefore, $m \\leq \\sqrt{n} < m + 1$, so $m^2 \\leq n < m^2 + 2m + 1$. Hence, $n = m^2 + k$ where $0 \\leq k \\leq 2m$. We require $m^2 + 2 \\mid (m^2 + k)^2 + 1$. Hence,\n\n$$\n\\begin{aligned}\n(m^2 + k)^2 + 1 &\\equiv 0 \\pmod{m^2 + 2} \\\\\n(k - 2)^2 + 1 &\\equiv 0 \\pmod{m^2 + 2}\n\\end{aligned}\n$$\n\nThus, $(k - 2)^2 + 1 = l(m^2 + 2)$ for some integer $l$.\n\nWe now find bounds for $l$. Since $(k - 2)^2 + 1 > 0$, we must have $l > 0$.\n\nAlso, remembering that $k \\leq 2m$, we have\n\n$$\n\\begin{aligned}\nl(m^2 + 2) &= (k - 2)^2 + 1 \\\\\n&< k^2 + 5 \\\\\n&\\leq 4m^2 + 5\n\\end{aligned}\n$$\n\nThus, $l < 4$. Hence, we are left with three cases.\n\n**Case 1:** $l = 1$. Then $(k - 2)^2 + 1 = m^2 + 2$.\n\nThis may be rearranged to $(k - 2 + m)(k - 2 - m) = 1$. Therefore, $k - 2 + m = k - 2 - m = \\pm 1$. Either way, we have $m = 0$, a contradiction.\n\n**Case 2:** $l = 2$. Then $(k - 2)^2 = 2m^2 + 3$.\n\nIf $3 \\mid m$ then also $3 \\mid k - 2$. But then $(k - 2)^2 \\equiv 0 \\pmod{9}$ and $2m^2 + 3 \\equiv 3 \\pmod{9}$, a contradiction.\n\nIf $3 \\nmid m$, then $m^2 \\equiv 1 \\pmod{3}$. Then $(k - 2)^2 \\equiv 2 \\pmod{3}$. But squares are only $0$ or $1$ (mod $3$), a contradiction.\n\n**Case 3:** $l = 3$. Then $(k - 2)^2 = 3m^2 + 5$. Thus $(k - 2)^2 \\equiv 2 \\pmod{3}$, which as in case 2, is impossible.\n\nSince we have exhausted all cases, we conclude that there is no positive integer $n$ satisfying the given condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15094,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a point inside triangle $ABC$. The lines $AP$, $BP$, and $CP$ intersect the circumcircle $\\Gamma$ of triangle $ABC$ again at the points $K$, $L$, and $M$ (other than $A$, $B$, $C$), respectively. The tangent to $\\Gamma$ at $C$ intersects the line $AB$ at $S$. Suppose that $SC = SP$. Prove that $MK = ML$.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, we may assume that $S$ is on ray $BA$ and consider the configuration shown below. Set $x_1 = \\angle PAB$, $y_1 = \\angle PBC$, $z_1 = \\angle PCA$, $x_2 = \\angle PAC$, $y_2 = \\angle PBA$, and $z_2 = \\angle PCB$.\n\n![](images/pamphlet1112_main_p70_data_30a5676263.png)\n\nBecause $SC$ is tangent to $\\gamma$, we have $SC^2 = SA \\cdot SB$ by the Power of a Point Theorem and $\\angle SCP = \\angle SCM = \\angle ACM + \\angle ACS = z_1 + \\angle ABC = z_1 + y_1 + y_2$. Because $SP = SC$, we have\n\n$$\nSP^2 = SC^2 = SA \\cdot SB \\quad \\text{and} \\quad \\angle SPC = \\angle SCP = z_1 + y_1 + y_2.\n$$\n\nBy the first relation above, we conclude that triangles $SAP$ and $SPB$ are similar to each other, from which it follows that $\\angle SPA = \\angle SBP = y_2$ and\n\n$$\n\\angle ASP = \\angle BAP - \\angle SPA = x_1 - y_2.\n$$\n\nBy the second relation, we deduce that $\\angle PSC = 180^\\circ - 2(z_1 + y_1 + y_2) = x_1 + x_2 + z_2 - (z_1 + y_1 + y_2)$. Note that $\\angle ASC = \\angle BAC - \\angle ACS = x_1 + x_2 - (y_1 + y_2)$. Hence, we find\n\n$$\n\\angle ASP = \\angle ASC - \\angle PSC = x_1 + x_2 - (y_1 + y_2) - (x_1 + x_2 + z_2 - (z_1 + y_1 + y_2)) = z_1 - z_2.\n$$\n\nCombining the last equation with the previous result yields\n\n$$\nx_1 - y_2 = z_1 - z_2 \\quad \\text{or} \\quad x_1 + z_2 = y_2 + z_1.\n$$\n\nThat is,\n\n$$\n\\frac{\\widehat{KB} + \\widehat{BM}}{2} = \\frac{\\widehat{LA} + \\widehat{AM}}{2} \\quad \\text{or} \\quad \\frac{\\widehat{KM}}{2} = \\frac{\\widehat{LM}}{2},\n$$\n\nfrom which it follows that $MK = ML$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15095,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a positive integer $n$ and the set $A_n = \\{1, 3, 5, \\dots, 2n - 1\\}$. For each pair $(a, b)$, where $a, b \\in A_n$, we construct the concatenated number $m = \\overline{ab}$, obtained by joining the numbers $a$ and $b$. For instance, for $19, 37 \\in A_{30}$, the concatenated number is $m = 1937$.\n\na) What is the smallest number $n \\in \\mathbb{N}^*$ for which we get at least a perfect square?\n\nb) Find the largest perfect square that can be obtained for $n = 50$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) We cannot obtain a perfect square by concatenating two elements from the set $A_{10} = \\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\\}$.\n\nThe elements $1$ and $21$ from the set $A_{11} = \\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21\\}$ yield the perfect square $121$. Hence, the answer is $n = 11$.\n\nb) By concatenating two elements from $A_{50} = \\{1, 3, 5, \\dots, 97, 99\\}$ we can obtain perfect squares with at most four digits.\n\nThe largest such perfect squares are $99^2 = 9801$, $97^2 = 9409$, $95^2 = 9025$, $93^2 = 8649$, $91^2 = 8281$. They are not acceptable, as the first two digits form an even number.\n\nThe number $89^2 = 7921$ is obtained by joining $79$ and $21$, where $79, 21 \\in A_{50}$. In conclusion, the largest perfect square is $7921$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15096,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all natural numbers $x$, $y$, $z$ that satisfy the equation:\n$$2^x + 21^y = z^2.$$",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* $x = 2$, $y = 1$, and $z = 5$.\n\n*Solution.* Let's consider two cases for $x$.\n\nFor odd $x$, we have\n$$\nz^2 \\equiv 2^{2n+1} + 21^y \\equiv 2 + 0 \\equiv 2 \\pmod{3},\n$$\nwhich has no solutions.\n\nLet $x = 2n$. Then\n$$\n21^y = z^2 - 2^{2n} = (z - 2^n)(z + 2^n).\n$$\nSuppose that one of the two prime numbers 3 or 7 divides each of the two factors on the right-hand side. Then this prime also divides the number $2z = (z + 2^n) + (z - 2^n)$ and hence $z$. However, this contradicts the equation $2^x = z^2 - 21^y$, in which the right-hand side is divisible by the corresponding prime number, while the left-hand side is not. Therefore, these factors are coprime. Hence, we have the following two cases.\n\nCase 1. $z - 2^n = 1$, $z + 2^n = 21^y \\Rightarrow 2^{n+1} = 21^y - 1 \\Rightarrow 2^{n+1} \\equiv 6 \\pmod 7$, which contradicts the fact that $2^{n+1} \\equiv 1, 2, 4 \\pmod 7$.\n\nCase 2. $z - 2^n = 3^y$, $z + 2^n = 7^y \\Rightarrow z = 2^n + 3^y$ and $2^{n+1} = 7^y - 3^y$. For $y = 1$, we find that $x = 2$ and $z = 5$, which satisfies the equation. For $y \\ge 2$, the expression $7^y - 3^y \\equiv 0 \\pmod 8$ is only possible for even $y$. If $y = 2m$, then $2^{n+1} = (7^m - 3^m)(7^m + 3^m)$. But under these conditions, $7^m + 3^m \\equiv 2 \\pmod 4$ and it is greater than 4, so it cannot be a power of 2.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15097,
    "subject": "Mathematics (Olympiad)",
    "question": "How many angles $\\theta$ with $0 \\leq \\theta \\leq 2\\pi$ satisfy\n$$\n\\log(\\sin(3\\theta)) + \\log(\\cos(2\\theta)) = 0?\n$$\n\n(A) 0 (B) 1 (C) 2 (D) 3 (E) 4",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $\\theta$ satisfies the equation. Then\n$$\n\\log(\\sin(3\\theta) \\cdot \\cos(2\\theta)) = 0\n$$\nwhich implies $\\sin(3\\theta) \\cdot \\cos(2\\theta) = 1$.\n\nUsing the product-to-sum formula:\n$$\n\\sin a \\cdot \\cos b = \\frac{1}{2} (\\sin(a + b) + \\sin(a - b))\n$$\nwe get\n$$\n\\frac{1}{2}(\\sin(5\\theta) + \\sin\\theta) = 1\n$$\nso $\\sin(5\\theta) + \\sin\\theta = 2$. This requires $\\sin(5\\theta) = \\sin\\theta = 1$, so $\\theta = \\frac{\\pi}{2}$.\n\nHowever, at $\\theta = \\frac{\\pi}{2}$, $\\sin(3\\theta) = \\sin(\\frac{3\\pi}{2}) = -1$, so $\\log(\\sin(3\\theta))$ is not defined.\n\nAlternatively, $\\sin(3\\theta) \\cdot \\cos(2\\theta) = 1$ requires both factors to be $1$ or $-1$, but only positive values are allowed for the logarithms to be defined. Checking possible values, there is no $\\theta$ in $[0, 2\\pi]$ for which both $\\sin(3\\theta) > 0$ and $\\cos(2\\theta) > 0$ and their product is $1$.\n\nTherefore, there are no solutions.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15098,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a subset of $\\{1,2,\\ldots,9\\}$ such that the sums of every two elements of $S$ are distinct. For example, the set $\\{1,2,3,5\\}$ has this property, but the set $\\{1,2,3,4,5\\}$ does not because $\\{2,3\\}$ and $\\{1,4\\}$ both have sum $5$. How many elements at most can $S$ contain? Explain your answer.",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to check that $\\{1,2,3,5,8\\}$ satisfies the desired condition. We will prove that $S$ cannot contain more than five elements. Suppose $S$ contains at least six elements. Then the smallest possible sum of pairs is $3$ and the largest is $8+9=17$, so the only possible sums of pairs are $3,4,5,\\ldots,17$, which are $15$ in total. But with at least $6$ elements, there are at least $15$ distinct pairs. Hence, every number from $3$ to $17$ must be a sum (for exactly one pair). So $1,2,8,$ and $9$ must be in $S$. But then $1+9=2+8$, which is a contradiction. Therefore, the maximal number of elements in $S$ is five.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15099,
    "subject": "Mathematics (Olympiad)",
    "question": "For each integer $n \\ge 0$, let $L_n$ and $R_n$ denote the following inequalities:\n\n$$\nL_n: \\qquad n a_n < a_0 + a_1 + \\dots + a_n\n$$\n\n$$\nR_n: \\qquad a_0 + a_1 + \\dots + a_n \\le n a_{n+1}\n$$\n\nProve that there exists at most one value of $n$ for which both $L_n$ and $R_n$ are true, and that there is at least one such $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $L_n$ and $R_n$ be as defined above.\n\n*Lemma 1.* If $R_n$ is true, then $L_{n+1}$ is false.\n\n*Proof.* This follows from adding $a_{n+1}$ to both sides of $R_n$. $\\square$\n\n*Lemma 2.* If $R_n$ is true, then $R_{n+1}$ is true.\n\n*Proof.* If $R_n$ is true, then adding $a_{n+1}$ to both sides of $R_n$ yields\n\n$$\n\\begin{aligned}\na_0 + a_1 + \\dots + a_n + a_{n+1} &\\le (n+1)a_{n+1} \\\\\n&< (n+1)a_{n+2},\n\\end{aligned}\n$$\n\nsince $a_{n+1} < a_{n+2}$. Hence $R_{n+1}$ is true. $\\square$\n\n*Lemma 3.* $L_n$ and $R_n$ are both true for at most one value of $n$.\n\n*Proof.* Suppose $n$ is the smallest value for which $L_n$ and $R_n$ are both true. Then lemma 2 tells us that $R_m$ is true for all $m \\ge n$, while lemma 1 tells us that $L_m$ is false for all $m \\ge n+1$. Thus $L_m$ and $R_m$ cannot both be true for any $m \\ge n+1$. $\\square$\n\n*Lemma 4.* If $R_n$ is false, then $L_{n+1}$ is true.\n\n*Proof.* This follows from adding $a_{n+1}$ to both sides of $R_n$. $\\square$\n\n*Lemma 5.* There is a value of $n$ for which $L_n$ and $R_n$ are both true.\n\n*Proof.* Assume there is no such $n$. Clearly, $L_0$ is true. But whenever $L_n$ is true, then by assumption, $R_n$ is false, and then by lemma 4, $L_{n+1}$ is true. Hence inductively, $L_n$ is true and $R_n$ is false for all $n$.\n\nSince the sequence is increasing, there exists a value of $n$ ($n \\ge 1$) for which $a_{n+1} > a_0 + a_1$. For this value of $n$ we have\n\n$$\n\\begin{aligned}\nn a_{n+1} &= a_{n+1} + (n-1)a_{n+1} \\\\\n&> a_0 + a_1 + (n-1)a_{n+1} \\\\\n&> a_0 + a_1 + a_2 + a_3 + \\dots + a_n,\n\\end{aligned}\n$$\n\nand so $R_n$ is true. This contradiction establishes the lemma. $\\square$\n\nCombining lemmas 3 and 5 completes the solution. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15100,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_0, a_1, \\dots, a_N$ be real numbers satisfying $a_0 = a_N = 0$ and\n$$\na_{i+1} - 2a_i + a_{i-1} = a_i^2\n$$\nfor $i = 1, 2, \\dots, N-1$. Prove that $a_i \\le 0$ for $i = 1, 2, \\dots, N-1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume the contrary. Then, there is an index $i$ for which $a_i = \\max_{0 \\le j \\le N} a_j$ and $a_i > 0$. This $i$ cannot be equal to $0$ or $N$, since $a_0 = a_N = 0$. Thus, from $a_i \\ge a_{i-1}$ and $a_i \\ge a_{i+1}$ we obtain\n$$\n0 < a_i^2 = (a_{i+1} - a_i) + (a_{i-1} - a_i) \\le 0,\n$$\nwhich is a contradiction. $\\Diamond$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15101,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all prime numbers $p$ for which there are infinitely many positive integers $n$ such that $p \\mid n^{n+1} + (n+1)^n$.",
    "options": [],
    "answer": "See solution",
    "solution": "$n^{n+1} + (n+1)^n$ is always odd for any positive integer $n$, so $p$ cannot be $2$.\n\nNow, we prove that for any prime $p \\ge 3$, there are infinitely many positive integers $n$ such that $p \\mid n^{n+1} + (n+1)^n$.\n\n**Proof 1.** For any prime $p \\ge 3$, let $n = pk - 2$ (with $n$ odd). Then\n\n$$\nn^{n+1} + (n+1)^n \\equiv (-2)^{pk-1} + (-1)^{pk-2} \\equiv 2^{pk-1} - 1 \\equiv (2^{p-1})^k \\cdot 2^{k-1} - 1 \\equiv 2^{k-1} - 1 \\pmod{p}.\n$$\n\nLet $k-1 = (p-1)t$ for any positive integer $t$. Then $2^{k-1} \\equiv 1 \\pmod{p}$, so $n^{n+1} + (n+1)^n \\equiv 0 \\pmod{p}$.\n\nTherefore, when $n = p(p-1)t + p - 2$ (for any $t \\ge 1$), we have $p \\mid n^{n+1} + (n+1)^n$.\n\n**Proof 2.** For any prime $p \\ge 3$, since $\\gcd(2, p) = 1$, by Fermat's Little Theorem, $2^{p-1} \\equiv 1 \\pmod{p}$. Let $n = p^t - 2$ for $t = 1, 2, 3, \\dots$. Then\n\n$$\nn^{n+1} + (n+1)^n \\equiv (-2)^{p^t-1} + (-1)^{p^t-2} \\equiv 2^{p^t-1} - 1 \\equiv (2^{p-1})^{\\text{some integer}} - 1 \\equiv 1 - 1 \\equiv 0 \\pmod{p}.\n$$\n\nThus, for all odd primes $p$, there are infinitely many $n$ such that $p \\mid n^{n+1} + (n+1)^n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15102,
    "subject": "Mathematics (Olympiad)",
    "question": "We are looking for the smallest integer $n$ such that the number of positive divisors of $n$ is $2020$.\n\nLet $n$ have prime factorisation\n$$\nn = p_1^{e_1} \\cdot p_2^{e_2} \\cdots p_k^{e_k}\n$$\nThen the number of positive divisors of $n$ is $(e_1+1)(e_2+1)\\cdots(e_k+1)$.\n\nFind the smallest $n$ such that $(e_1+1)(e_2+1)\\cdots(e_k+1) = 2020$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $2020 = 2^2 \\cdot 5 \\cdot 101$, $n$ can have at most 4 different prime divisors. Consider cases for $k = 1, 2, 3, 4$:\n\n**Case $k = 4$:**\nThe four factors $e_i+1$ must be $2, 2, 5, 101$, so $e_i = 1, 1, 4, 100$. Pairing the largest exponents with the smallest primes:\n$$\nn = 2^{100} \\cdot 3^4 \\cdot 5^1 \\cdot 7^1\n$$\n\n**Case $k = 3$:**\nPossible sets: $\\{4, 5, 101\\}$, $\\{2, 10, 101\\}$, $\\{2, 5, 202\\}$, $\\{2, 2, 505\\}$. The smallest candidates are:\n$$\n2^{100} \\cdot 3^4 \\cdot 5^3\n$$\n$$\n2^{100} \\cdot 3^9 \\cdot 5^1\n$$\nOther candidates are much larger.\n\n**Case $k = 2$:**\nNo suitable factorisation yields exponents small enough.\n\n**Case $k = 1$:**\n$2^{2019}$ is much too large.\n\nComparing the three smallest candidates:\n$$\n2^{100} \\cdot 3^4 \\cdot 5 \\cdot 7,\n$$\n$$\n2^{100} \\cdot 3^4 \\cdot 5^3,\n$$\n$$\n2^{100} \\cdot 3^9 \\cdot 5\n$$\nFactoring out $2^{100} \\cdot 3^4 \\cdot 5$, we compare $7, 5^2, 3^5$ and find $7$ is smallest. Thus, the smallest $n$ with exactly $2020$ positive divisors is:\n$$\n\\boxed{2^{100} \\cdot 3^4 \\cdot 5 \\cdot 7}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15103,
    "subject": "Mathematics (Olympiad)",
    "question": "2010 cards are enumerated $1, 2, \\ldots, 2010$. All cards whose number has an odd digit sum are chosen. Find the sum of the numbers on the chosen cards.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the digit sum of $a$ by $S(a)$. Add a card with $0$ and assume $S(0) = 0$. Among $0, 1, \\ldots, 999$ there are $500$ numbers $a$ with $S(a)$ odd and $500$ with $S(a)$ even. Indeed, $0, 1, \\ldots, 999$ can be divided into $500$ pairs $(a, b)$ with sum $999$ for every pair. There is no carryover in the addition $a + b = 999$, so $S(a) + S(b) = S(999) = 27$, which is an odd number. Hence $S(a)$ and $S(b)$ have different parity for every pair $(a, b)$, as needed.\n\nLet $X$ ($Y$) be the sum of the $500$ numbers with odd (even) digit sum among $0, 1, \\ldots, 999$.\n\nFor $a \\in \\{0, 1, \\ldots, 999\\}$ we have $S(1000 + a) = S(a) + 1$, hence $S(1000 + a)$ and $S(a)$ have different parity. So $\\{1000, 1001, \\ldots, 1999\\}$ contains $500$ numbers $b$ with $S(b)$ odd. They are obtained from the numbers $a \\in \\{0, 1, \\ldots, 999\\}$ with $S(a)$ even by adding $1000$.\n\nIt follows that the numbers with odd digit sum in $\\{0, 1, \\ldots, 1999\\}$ have sum $X + Y + 500 \\cdot 1000$. Since $X + Y = 0 + 1 + \\cdots + 999 = 500 \\cdot 999$, the numbers in $[1, 1999]$ contribute $500 \\cdot 1999$ to the sum we are looking for.\n\nThere remain $2000, 2001, \\ldots, 2010$. Of them, the ones with odd digit sum are $2001, 2003, 2005, 2007, 2009, 2010$; they add up to $12035$. So the final answer is $500 \\cdot 1999 + 12035 = 1011535$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15104,
    "subject": "Mathematics (Olympiad)",
    "question": "Circle $c$ passes through vertices $A$ and $B$ of an isosceles triangle $ABC$, and the line $AC$ is tangent to it. Prove that circle $c$ passes through the circumcenter, the incenter, or the orthocenter of triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider three cases: $|AB| = |AC|$, $|BC| = |BA|$, and $|CA| = |CB|$.\n\n1. **Case $|AB| = |AC|$:**\n\nCircle $c$ passes through the circumcenter of $ABC$. Let $O$ be the intersection of the perpendicular bisector of $AB$ and circle $c$ on the same side as $C$. Then $\\angle OAB = \\angle OBA$, and by the inscribed angles theorem,\n\n![](images/Estonija_2010_p7_data_bc09181723.png)\n\n$\\angle OBA = \\angle OAC$. Hence $O$ lies on the bisector of angle $CAB$. Since $|AB| = |AC|$, this angle bisector is also the perpendicular bisector of $BC$. Thus, $O$ is the intersection point of the perpendicular bisectors of the sides of $ABC$ (the circumcenter).\n\n2. **Case $|BC| = |BA|$:**\n\nCircle $c$ passes through the orthocenter of $ABC$. Let $E$ be the foot of the altitude from $B$, and $H$ the second intersection of this altitude with circle $c$ (if tangency only, take $H = B$). By the inscribed angles theorem, $\\angle EBA = \\angle EAH$. Thus $\\angle ACB + \\angle CAH = \\angle CAB + \\angle EBA = 90^\\circ$, so $AH \\perp BC$. Therefore, $H$ is the orthocenter.\n\n![](images/Estonija_2010_p7_data_edd10602bc.png)\n\n3. **Case $|CA| = |CB|$:**\n\nCircle $c$ passes through the incenter of $ABC$. Let $I$ be the intersection of the angle bisector of $CAB$ with circle $c$. By the inscribed angles theorem, $\\angle CAI = \\angle IBA$, so $\\angle BAI = \\angle IBA$, and $I$ lies on the perpendicular bisector of $AB$. Since $|CA| = |CB|$, this is also the bisector of $ACB$. Thus, $I$ is the intersection of the angle bisectors (the incenter).\n\n![](images/Estonija_2010_p7_data_d34b52ba66.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15105,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(x)$ be a real polynomial such that for all real numbers $a, b, c$ with $a + b + c = 0$, the following holds:\n\n$$\nP(a - b) + P(b - c) + P(c - a) = 2P(a + b + c).\n$$\n\nFind all such polynomials $P(x)$.",
    "options": [],
    "answer": "See solution",
    "solution": "As shown in the second solution, $P(x)$ is an even function and its constant term is $0$; that is, $P(x) = x^2 f(x^2)$ for some polynomial $f(x)$, and we just need to show that $f(x)$ has degree at most $1$. The given condition reads\n\n$$\n(a-b)^2 f((a-b)^2) + (b-c)^2 f((b-c)^2) + (c-a)^2 f((c-a)^2) = 2(a+b+c)^2 f((a+b+c)^2).\n$$\n\nPlugging the good triple $(a, b, c) = [(1 - \\sqrt{3})b, b, (1 + \\sqrt{3})b]$ into the above equation gives\n\n$$\n6b^2 f(3b^2) + 12b^2 f(12b^2) = 18b^2 f(9b^2),\n$$\n\nwhich can be rewritten as $12b^2[f(12b^2) - f(9b^2)] = 6b^2[f(9b^2) - f(3b^2)]$ or, for $b \\neq 0$,\n\n$$\n\\frac{f(12b^2) - f(9b^2)}{3b^2} = \\frac{f(9b^2) - f(3b^2)}{6b^2}\n$$\n\nConsider points $A = (3b^2, f(3b^2))$, $B = (9b^2, f(9b^2))$, and $C = (12b^2, f(12b^2))$. Then the above equation says lines $AB$ and $BC$ have the same slope. Note that if $P(x)$ is a solution, then so is $-P(x)$. Without loss of generality, we can assume the leading coefficient of $P(x)$ (and therefore $f(x)$) is nonnegative. If $f$ is of degree $2$ or higher, the second derivative of $f$ is a polynomial with positive leading coefficient. Then there exist real numbers $N$ such that for $x > N$, $f''(x) > 0$, and so $f(x)$ is convex. We choose $b$ such that $3b^2 > N$. Then we have\n\n$$\n\\frac{f(12b^2) - f(9b^2)}{3b^2} > f'(9b^2) > \\frac{f(9b^2) - f(3b^2)}{6b^2};\n$$\n\nthat is, the slope of line $BC$ is greater than that of the line tangent to $f(x)$ at $B$, which is greater than that of $AC$, which is a contradiction. Therefore, $f$ is at most linear; that is, $f(x) = c_1 + c_2x$ and $P(x) = x^2f(x^2) = c_1x^2 + c_2x^4$ for some real numbers $c_1$ and $c_2$. As shown in the first solution, we can verify that this is the complete solution set.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15106,
    "subject": "Mathematics (Olympiad)",
    "question": "A regular hexagon with side length $1$ is given. Inside the hexagon, $m$ points are placed such that no three of them are collinear. The hexagon is divided into triangles so that each of the $m$ points and each vertex of the hexagon is a vertex of one such triangle. The triangles into which the hexagon is split have no common interior point. Prove that among these triangles, there exists one whose area is not greater than $\\frac{3\\sqrt{3}}{4(m+2)}$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we determine the total number of triangles into which the hexagon is split. Let $A$ be one of the $m$ interior points. The sum of all angles at $A$ (over all triangles containing $A$) is $360^\\circ$. At each hexagon vertex, the sum of angles is $120^\\circ$. Since the sum of angles in every triangle is $180^\\circ$, the total number of triangles is:\n\n$$\n\\frac{m \\cdot 360^\\circ + 6 \\cdot 120^\\circ}{180^\\circ} = 2m + 4\n$$\n\nAssume, for contradiction, that the area of each triangle is greater than $\\frac{3\\sqrt{3}}{4(m+2)}$. Then the total area is greater than:\n\n$$\n(2m + 4) \\cdot \\frac{3\\sqrt{3}}{4(m+2)} = \\frac{3\\sqrt{3}}{2}\n$$\n\nBut the area of the hexagon is $\\frac{3\\sqrt{3}}{2}$, which is impossible. Thus, there must exist at least one triangle whose area is not greater than $\\frac{3\\sqrt{3}}{4(m+2)}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15107,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be two different natural numbers such that $a^2 + b^2 + 1$ is divisible by $2ab + 1$. Prove that $2ab + 1$ is a perfect square (i.e., the square of an integer).",
    "options": [],
    "answer": "See solution",
    "solution": "Assume the contrary. Let $(a, b)$ be a pair of natural numbers for which $2ab + 1$ is not a perfect square, with $a < b$ and $a + b$ minimal.\n\n$$\n\\textbf{Lemma 1.}\\quad a^2 + b^2 + 1 \\mid 2ab + 1 \\iff (2a^2 + 1)^2 \\mid 2ab + 1.\n$$\n\n*Proof.* Since $(a^2, 2ab+1) = 1$, then $a^2 + b^2 + 1 \\mid 2ab + 1 \\iff a^4 + b^2 a^2 + a^2 \\mid 2ab + 1 \\iff 4a^4 + 4b^2 a^2 + 4a^2 \\mid 2ab + 1$, since $2ab + 1$ is odd. The rest follows from:\n$$\n4a^4 + 4b^2 a^2 + 4a^2 = (4a^4 + 4a^2 + 1) + (4b^2 a^2 - 1) = (4a^4 + 4a^2 + 1) + (2ab - 1)(2ab + 1)\n$$\n\nSo the lemma is proved.\n\nThus, for the fixed pair $(a, b)$, we have $a^2 + b^2 + 1 \\mid 2ab + 1$, therefore $(2a^2 + 1)^2 \\mid 2ab + 1$. So $(2a^2 + 1)^2 = (2ab + 1)(2ac + 1)$ for some $c$. It follows from $a < b$ that $0 < c < a < b$, otherwise the right-hand side is greater than the left-hand side. If $2ab + 1$ is not a perfect square, then neither is $2ac + 1$. It follows from the lemma that if $(2a^2 + 1)^2 \\mid 2ac + 1$, then $a^2 + c^2 + 1 \\mid 2ac + 1$ as well. Thus, we have found a pair $(a, c)$ with $a + c < a + b$ that satisfies the problem statement, contradicting the minimality of $(a, b)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15108,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Припустимо, що для деяких дійсних чисел $x$ і $y$ виконується рівність $\\{x\\} + \\{2y\\} = \\{y\\} + \\{2x\\}$. Чи обов'язково $\\{x\\} = \\{y\\}$?\n\nб) Чи існує таке натуральне $n > 2$, для якого з рівності $\\{x\\} + \\{ny\\} = \\{y\\} + \\{nx\\}$ випливає $\\{x\\} = \\{y\\}$ для всіх дійсних $x, y$?",
    "options": [],
    "answer": "See solution",
    "solution": "a) Нехай $\\{x\\} = \\alpha$, $\\{y\\} = \\beta$, де $\\alpha, \\beta \\in [0;1)$. Тоді рівність набуває вигляду $\\alpha + \\{2\\beta\\} = \\beta + \\{2\\alpha\\}$.\n\nЯкщо $0 \\leq \\gamma < \\frac{1}{2}$, то $\\{2\\gamma\\} = 2\\gamma$; якщо $\\frac{1}{2} \\leq \\gamma < 1$, то $\\{2\\gamma\\} = 2\\gamma - 1$.\n\nДля випадків $0 \\leq \\alpha < \\frac{1}{2} \\leq \\beta < 1$ і $0 \\leq \\beta < \\frac{1}{2} \\leq \\alpha < 1$ рівність не виконується. Якщо ж $0 \\leq \\alpha, \\beta < \\frac{1}{2}$ або $\\frac{1}{2} \\leq \\alpha, \\beta < 1$, то рівність можлива лише при $\\alpha = \\beta$.\n\nб) Відповідь: ні, не існують. Візьмемо $x=0$, $y = \\frac{1}{n-1}$. Тоді $\\{x\\} \\neq \\{y\\}$, але $\\{x\\} + \\{ny\\} = \\{y\\} + \\{nx\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15109,
    "subject": "Mathematics (Olympiad)",
    "question": "證明對於任何正整數 $n$，$5^n - 3^n$ 都無法被 $2^n + 65$ 整除。",
    "options": [],
    "answer": "See solution",
    "solution": "假設相反，存在正整數 $n$ 使得 $2^n + 65 \\mid 5^n - 3^n$，即 $5^n \\equiv 3^n \\pmod{2^n + 65}$。\n\n$n$ 必須為奇數，因為若 $n$ 為偶數，則 $3 \\mid 2^n + 65$，矛盾。故 $n \\geq 3$ 且 $8 \\mid 2^n$。\n\n考慮 Jacobi 符號：\n\n$$\n-1 = \\left( \\frac{2^n + 65}{5} \\right) = \\left( \\frac{5}{2^n + 65} \\right) = \\left( \\frac{5^n}{2^n + 65} \\right) = \\left( \\frac{3^n}{2^n + 65} \\right) = \\left( \\frac{3}{2^n + 65} \\right) = \\left( \\frac{2^n + 65}{3} \\right) = 1\n$$\n\n矛盾，因此不存在這樣的 $n$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15110,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. The positive integers $1, 2, \\dots, n$ are written in a row in some order. For any two neighboring numbers, their GCD is written on the paper. Find the greatest possible number of distinct numbers among all $n-1$ numbers written on the paper.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\\lfloor n/2 \\rfloor$$\n\n**Upper bound.** Assume one of the written numbers is greater than $\\lfloor n/2 \\rfloor$, say, $\\gcd(a, b) = d > \\lfloor n/2 \\rfloor$. Then the larger of the numbers $a, b$ must be at least $2d$, which exceeds $n$—a contradiction. Therefore, each written GCD cannot exceed $\\lfloor n/2 \\rfloor$, and thus the number of distinct GCDs cannot be greater than $\\lfloor n/2 \\rfloor$.\n\n**Example.** Partition all numbers from $1$ to $n$ into chains of the form $a, 2a, 4a, 8a, \\dots, 2^k a$, where $a$ is an odd number not exceeding $n$. Write these chains consecutively in a row. Then for any natural number $d \\leq \\lfloor n/2 \\rfloor$, there exists a chain containing $d$ where the number following $d$ is $2d$. Thus, every natural number $d \\leq \\lfloor n/2 \\rfloor$ will appear on the sheet.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15111,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Each cell on a board of size $(2n+1) \\times (2n+1)$ is colored black or white. For every row and column, if the number of white cells is smaller than the number of black cells, then we mark all the white cells; and vice versa, if the number of black cells is smaller than the number of white cells, we mark all the black cells. Let $a$ and $b$ be the numbers of black cells and white cells, respectively, and let $c$ be the number of marked cells. Prove that\n\n$$\nc \\geq \\frac{\\min\\{a, b\\}}{2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We define a *black row* as a row that has more white cells than black cells. It is easy to see that if a black cell belongs to a black row, that cell will be marked. A *white row* is defined similarly. Assume that the number of black rows is more than the number of white rows; we will show that the number of marked cells in each column is not smaller than half the number of black cells in that column. Choose an arbitrary column and consider two cases:\n\n* If there are more white cells in that column, then all black cells are marked.\n* If there are more black cells in that column, consider the intersections of black rows and that column. All of these cells are marked, and since there are more black rows than white rows, our claim is proved.\n\nConsidering the number of marked cells in each column, it turns out that $c \\geq \\frac{a}{2} \\geq \\frac{\\min\\{a, b\\}}{2}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15112,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S_b(n)$ denote the sum of the digits of $n$ in base $b$.\n\nFor which positive integers $m$ does the following property hold: whenever $m$ divides $S_b(n)$, then $m$ also divides $S_b(n+1) - 1$ for all $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $m$ has the given property if and only if $m \\mid b-1$.\n\nIt is easy to see that $S_b(n) \\equiv n \\pmod{b-1}$, using the fact that $b^a \\equiv 1 \\pmod{b-1}$. For this reason, any divisor of $b-1$ works. Now suppose that $m$ works and consider the number $n$ whose first digit is $b-1$, whose second digit is $0$, and whose remaining $m-1$ digits are $b-1$. Clearly, $S_b(n) = m(b-1)$, while $S_b(n+1)-1 = b-1$. Therefore, $m$ must be a divisor of $b-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15113,
    "subject": "Mathematics (Olympiad)",
    "question": "An even cube die with six faces is marked with the numbers $1, 2, 3, 4, 5,$ and $6$. The die is tossed three times independently, and the resulting numbers are $a_1, a_2, a_3$ in order. What is the probability that $|a_1 - a_2| + |a_2 - a_3| + |a_3 - a_1| = 6$?",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n$$\n|a_1 - a_2| + |a_2 - a_3| + |a_3 - a_1| = 2 \\max_{1 \\le i \\le 3} a_i - 2 \\min_{1 \\le i \\le 3} a_i.\n$$\nTherefore, the three numbers $a_1, a_2, a_3$ satisfy the condition if and only if the difference between the largest and smallest is $3$.\n\nThis means $a_1, a_2, a_3$ is a permutation of $x, x+d, x+3$, where $x \\in \\{1, 2, 3\\}$ and $d \\in \\{0, 1, 2, 3\\}$.\n\nFor each $x \\in \\{1, 2, 3\\}$:\n- When $d = 0$ or $d = 3$, there are $3$ different permutations for each case.\n- When $d = 1$ or $d = 2$, there are $6$ different permutations for each case.\n\nTherefore, there are $3 \\times (2 \\times 3 + 2 \\times 6) = 54$ cases that satisfy the condition.\n\nThe desired probability is\n$$\n\\frac{54}{6^3} = \\frac{1}{4}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15114,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p(x) = x^{2013} + a_{2012}x^{2012} + a_{2011}x^{2011} + \\dots + a_1x + a_0$ be a polynomial with real coefficients. Suppose that all the roots of $p(x)$ are\n$$\n-b_{1006}, -b_{1005}, \\dots, -b_1, 0, b_1, \\dots, b_{1005}, b_{1006}\n$$\nwhere $b_1, b_2, \\dots, b_{1006}$ are positive real numbers with product equal to $1$. Show that $a_3 a_{2011} \\geq 1012036$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $0$ is a root of $p(x)$, we have $a_0 = 0$, so\n$$\np(x) = x \\left( x^{2012} + a_{2012}x^{2011} + a_{2011}x^{2010} + \\dots + a_2x + a_1 \\right).\n$$\nAll roots of $p(x)$ are $-b_{1006}, \\dots, -b_1, 0, b_1, \\dots, b_{1006}$, so\n$$\np(x) = x(x + b_{1006})(x - b_{1006}) \\cdots (x + b_1)(x - b_1) = x \\prod_{i=1}^{1006} (x^2 - b_i^2).\n$$\nThus,\n$$\nx^{2012} + a_{2012}x^{2011} + a_{2011}x^{2010} + \\dots + a_1 = \\prod_{i=1}^{1006} (x^2 - b_i^2).\n$$\nSince the right side is a polynomial in $x^2$, all even-indexed coefficients vanish: $a_2 = a_4 = \\dots = a_{2012} = 0$, so\n$$\n\\prod_{i=1}^{1006} (x^2 - b_i^2) = a_1 + a_3 x^2 + a_5 x^4 + \\dots + a_{2011} x^{2010} + x^{2012}.\n$$\nBy symmetric function identities,\n$$\n\\sum_{i=1}^{1006} b_i^2 = -a_{2011}, \\quad \\sum_{1 \\leq i_1 < \\dots < i_{1005} \\leq 1006} b_{i_1}^2 \\cdots b_{i_{1005}}^2 = (-1)^{1005} a_3 = -a_3.\n$$\nTherefore,\n$$\n\\begin{aligned}\na_3 a_{2011} &= \\left( - \\sum_{1 \\leq i_1 < \\dots < i_{1005} \\leq 1006} b_{i_1}^2 \\cdots b_{i_{1005}}^2 \\right) \\left( - \\sum_{i=1}^{1006} b_i^2 \\right) \\\\\n&= \\left( \\prod_{i=1}^{1006} b_i^2 \\right) \\left( \\sum_{i=1}^{1006} \\frac{1}{b_i^2} \\right) \\left( \\sum_{i=1}^{1006} b_i^2 \\right) \\\\\n&= \\left( \\sum_{i=1}^{1006} \\frac{1}{b_i^2} \\right) \\left( \\sum_{i=1}^{1006} b_i^2 \\right) \\geq 1006^2 = 1012036,\n\\end{aligned}\n$$\nwhere the inequality follows from the Cauchy-Schwarz inequality.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15115,
    "subject": "Mathematics (Olympiad)",
    "question": "![](images/Belarus_2015_p28_data_98b3e2c1b3.png)\n\nIn any cyclic quadrilateral, all perpendiculars drawn through the midpoints of the sides to the opposite sides meet at the same point $U$. Prove that the segments $H_{ADP}H_{AQB}$ and $H_{CQD}H_{BCP}$ have the same length, where $H_{ADP}$, $H_{AQB}$, $H_{DQC}$, and $H_{BCP}$ are the orthocenters of triangles $ADP$, $AQB$, $DQC$, and $BCP$, respectively.",
    "options": [],
    "answer": "See solution",
    "solution": "It follows that the altitudes $DD_1$ and $CC_1$ of the triangles $ADB$ and $BCP$, respectively, are symmetric with respect to $U$. The same is true for the altitudes $AA_1$ and $BB_1$. Therefore, the intersection points $DD_1 \\cap AA_1$ and $CC_1 \\cap BB_1$ are symmetric with respect to $U$. That is, the orthocenters $H_{ADP}$ and $H_{BCP}$ are symmetric. Similarly, the orthocenters $H_{AQB}$ and $H_{DQC}$ are symmetric with respect to $U$. So the segments $H_{ADP}H_{AQB}$ and $H_{CQD}H_{BCP}$ are symmetric, so they have the same length.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15116,
    "subject": "Mathematics (Olympiad)",
    "question": "Two circles touch one another internally at $A$. A variable chord $PQ$ of the outer circle touches the inner circle. Prove that the locus of the incentre of triangle $AQP$ is another circle touching the given circles at $A$.\n\nThe incentre of a triangle is the centre of the unique circle which is inside the triangle and touches all three sides. A locus is the collection of all points which satisfy a given condition.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $L$ be the point where $PQ$ touches $\\omega$, and let $AL$ meet $\\Omega$ again at $T$.\n\n![](images/British_2015_Booklet_p28_data_024baff526.png)\n\nThere exists an enlargement taking $\\omega$ to $\\Omega$ centred at $A$ which will take $L$ to $T$. Hence, the tangent to $\\Omega$ at $T$ is parallel to $PQ$, so $T$ is the midpoint of arc $PQ$ and thus lies on the angle bisector of $\\angle PAQ$. Hence, $I$ lies on $AT$.\n\n$$\n\\begin{aligned}\n\\frac{AI}{AL} &= \\frac{AP}{PL} && (\\text{by the angle bisector theorem}) \\\\\n&= \\frac{TQ}{TL} && (\\text{as } APTQ \\text{ is cyclic}) \\\\\n&= \\frac{IT}{TL} && (\\text{as } T \\text{ is the centre of circle } QIP) \\\\\n&= \\frac{AT - TI}{TL}\n\\end{aligned}\n$$\n\nSo\n\n$$\nAI(TL + IL) = AT \\cdot IL.\n$$\n\nLet $k = \\frac{AT}{AL}$, which is the scale factor of the enlargement taking $\\omega$ to $\\Omega$.\n\nSo we have:\n\n$$\n\\begin{aligned}\nAI \\cdot (k - 1)AL + AI \\cdot IL &= kAL \\cdot IL \\\\\nAI \\cdot (k - 1)AL + AI \\cdot (AL - AI) &= kAL \\cdot (AL - AI) \\\\\nkAI \\cdot AL - AI^2 &= kAL^2 - kAI \\cdot AL \\\\\nkAL^2 - 2kAI \\cdot AL + AI^2 &= 0\n\\end{aligned}\n$$\n\nSo, defining $\\alpha = \\frac{AI}{AL}$, we obtain $\\alpha = k \\pm \\sqrt{k^2 - k}$.\n\nSince $AI < AT$, we obtain that $\\alpha < k$, so $\\alpha = k - \\sqrt{k^2 - k}$ is constant.\n\nHence, $I$ lies on the circle tangent to $\\omega$ at $A$ which is an enlargement of $\\omega$ with scale factor $\\alpha$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15117,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine if there exists a function $f : (0, 1) \\to (2018, +\\infty)$ such that the following conditions hold:\n\n- $f(xy) = f(x) \\cdot f(y)$ for any $x, y \\in (0, 1)$;\n- For any $y \\in (2018, +\\infty)$, there exists $x \\in (0, 1)$ such that $f(x) = y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose such a function exists. For any $x \\in (0, 1)$, we have $x = \\sqrt{x} \\cdot \\sqrt{x}$, so\n$$\nf(x) = f(\\sqrt{x} \\cdot \\sqrt{x}) = f(\\sqrt{x}) \\cdot f(\\sqrt{x}) = [f(\\sqrt{x})]^2.\n$$\nSince $f(\\sqrt{x}) > 2018$, it follows that $f(x) > 2018^2$. This contradicts the requirement that the range of $f$ is $(2018, +\\infty)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15118,
    "subject": "Mathematics (Olympiad)",
    "question": "For which natural number $n$ (where $n! = 1 \\cdot 2 \\cdot 3 \\cdots n$) does $n!$ use exactly two distinct digits?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $n = 4$.\n\n**Solution:** For $n \\geq 5$, the number $n!$ ends with $0$, so it contains at least one more digit. If we mark the nonzero digit as $a$, then\n\n$$\nn! = \\overline{a \\cdots a} \\overline{0 \\cdots 0}\n$$\n\nIf you remove the zeros at the end, you get an odd number. The number of factors $2$ in the prime decomposition of $n!$ cannot exceed the number of factors $5$ plus $3$. Each pair $2 \\cdot 5$ gives a new $0$ at the end, and extra factors of $2$ can only form the digit $a$. Among digits, $8$ is divisible by the largest power of $2$.\n\nEven for $n = 8$, $n! = 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7 \\cdot 8$, and the condition is not satisfied. Since $2$ occurs in every second number and $5$ in every fifth, the required relation can only be achieved for $n \\leq 7$. By direct checking, only $4! = 24$ uses exactly two digits.\n\n![alt](images/ukraine_2015_Booklet_p8_data_4cb7b7ac70.png)\n\nFig. 22",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15119,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $x_1, \\cdots, x_{100}$ 為非負實數，且對 $i = 1, \\cdots, 100$，都有\n\n$$\nx_i + x_{i+1} + x_{i+2} \\le 1,\n$$\n\n（其中記 $x_{101} = x_1, x_{102} = x_2$）。試求下式 $S$ 的最大可能值：\n\n$$\nS = \\sum_{i=1}^{100} x_i x_{i+2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "答：$\\frac{25}{2}$。\n\n令 $x_{2i} = 0,\\ x_{2i-1} = \\frac{1}{2}$，對所有 $i = 1, \\cdots, 50$。則\n\n$$\nS = 50 \\left(\\frac{1}{2}\\right)^2 = \\frac{25}{2}.\n$$\n\n故只須證 $S \\le \\frac{25}{2}$ 對所有滿足題設之 $x_i$。\n\n考慮 $1 \\le i \\le 50$。由題設得\n\n$$\nx_{2i-1} \\le 1 - x_{2i} - x_{2i+1}, \\quad x_{2i+2} \\le 1 - x_{2i} - x_{2i+1}.\n$$\n\n再由算幾不等式得\n\n$$\n\\begin{aligned}\nx_{2i-1}x_{2i+1} + x_{2i}x_{2i+2} &\\le (1-x_{2i}-x_{2i+1})x_{2i+1} + x_{2i}(1-x_{2i}-x_{2i+1}) \\\\\n&= (x_{2i}+x_{2i+1})(1-x_{2i}-x_{2i+1}) \\\\\n&\\le \\left(\\frac{x_{2i}+x_{2i+1}}{2} + (1-x_{2i}-x_{2i+1})\\right)^2 = \\frac{1}{4}.\n\\end{aligned}\n$$\n\n對 $i = 1, \\cdots, 50$，將上述不等式加總，得\n\n$$\n\\sum_{i=1}^{50} (x_{2i-1}x_{2i+1} + x_{2i}x_{2i+2}) \\le 50 \\cdot \\frac{1}{4} = \\frac{25}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15120,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(n)$ be the probability that there is eventually no alien on Earth given that there are currently $n$ aliens, where $n$ is a positive integer. Each alien acts independently. Given the possible actions, we have:\n\n$$\nP(n) = c^n,\n$$\nwhere $c = P(1)$.\n\nThe recurrence for $P(1)$ is:\n$$\nP(1) = \\frac{1}{4} + \\frac{1}{4}P(2) + \\frac{1}{4}P(3) + \\frac{1}{4}P(1).\n$$\n\nFind the value of $P(1)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P(1) = c$. Since $P(n) = c^n$, substitute into the recurrence:\n\n$$\nP(1) = \\frac{1}{4} + \\frac{1}{4}P(2) + \\frac{1}{4}P(3) + \\frac{1}{4}P(1)\n$$\n$$\nc = \\frac{1}{4} + \\frac{1}{4}c^2 + \\frac{1}{4}c^3 + \\frac{1}{4}c\n$$\nMultiply both sides by $4$:\n$$\n4c = 1 + c^2 + c^3 + c\n$$\n$$\n4c - c = 1 + c^2 + c^3\n$$\n$$\n3c = 1 + c^2 + c^3\n$$\n$$\nc^3 + c^2 - 3c + 1 = 0\n$$\nFactor:\n$$\n(c-1)(c^2 + 2c - 1) = 0\n$$\nSo $c = 1$ or $c = -1 \\pm \\sqrt{2}$.\n\nSince $c$ is a probability, $0 \\leq c \\leq 1$, so $c = -1 - \\sqrt{2}$ is rejected. We also reject $c = 1$ because the expected number of aliens from one is $\\frac{0+2+3+1}{4} = \\frac{3}{2} > 1$, so extinction is not certain. Thus, $c < 1$.\n\nTherefore, the probability is $c = \\sqrt{2} - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15121,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose two $20 \\times 13$ rectangular grids consisting of $260$ small squares are given.\n\nWe insert into each square box of the two grids, numbers $1, 2, \\ldots, 260$ in the following way:\n\n* For the first grid, we start inserting numbers $1, 2, \\ldots, 13$ into the boxes on the top row from left to right. Continue to insert numbers $14, 15, \\ldots, 26$ into the boxes on the second row from left to right. Keep on going until you finish inserting numbers $248, 249, \\ldots, 260$ from left to right into the boxes on the bottom row.\n\n* For the second grid, we start inserting numbers $1, 2, \\ldots, 20$ into the boxes on the right-most column from top to bottom. Continue to insert numbers $21, 22, \\ldots, 40$ into the boxes on the second column from the right from top to bottom. Keep on doing until you finish inserting numbers $241, 242, \\ldots, 260$ into the boxes on the left-most column from top to bottom.\n\nList all the positive integers which get inserted into the boxes located in the same position in the two grids in the two ways of distributing numbers described above.",
    "options": [],
    "answer": "See solution",
    "solution": "The number inserted into the box located on the $i$-th row from the top and $j$-th column from the left is given by $13(i-1)+j$ for the first grid, and by $20(13-j)+i$ for the second grid.\n\nIf the same number goes into the boxes located at the same position in the two grids, we must have:\n\n$$13(i-1)+j = 20(13-j)+i$$\n\nSimplifying:\n\n$$13i - 13 + j = 260 - 20j + i$$\n$$12i + 21j = 273$$\n\nSince both $21j$ and $273$ are multiples of $7$, $12i$ must also be a multiple of $7$, so $i$ must be a multiple of $7$. With $1 \\leq i \\leq 20$, we get $i = 7, 14$.\n\nThe corresponding values for $j$ are $9$ and $5$, respectively. Computing $13(i-1)+j$ for these values:\n\n- For $i=7$, $j=9$: $13(6)+9 = 78+9 = 87$\n- For $i=14$, $j=5$: $13(13)+5 = 169+5 = 174$\n\nThus, the positive integers inserted into the same position in both grids are $\\boxed{87}$ and $\\boxed{174}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15122,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $H$ be the orthocenter of an acute triangle $ABC$, and $D$ be the midpoint of the side $BC$. A line passing through the point $H$ meets the sides $AB$ and $AC$ at the points $F$ and $E$ respectively, such that $AE = AF$. The ray $DH$ meets the circumcircle of $\\triangle ABC$ at the point $P$.\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p154_data_f6fb35c8a5.png)\n\nProve that $P$, $A$, $E$, $F$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof**\n\nOn the ray $HD$, mark a point $M$ such that $HD = DM$. Join the segments $BM$, $CM$, $BH$, and $CH$. As $D$ is the midpoint of $BC$, the quadrilateral $BHCM$ is a parallelogram, and so\n\n$$\n\\angle BMC = \\angle BHC = 180^\\circ - \\angle BAC.\n$$\n\nHence,\n\n$$\n\\angle BMC + \\angle BAC = 180^\\circ,\n$$\n\nand the point $M$ lies on the circumcircle of $\\triangle ABC$. Join the segments $PB$, $PC$, $PE$, and $PF$. It follows from $AE = AF$ that\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p155_data_5ff1fc2c1c.png)\n\n$$\n\\angle BFH = \\angle CEH. \\qquad \\textcircled{1}\n$$\n\nAs $H$ is the orthocenter of $\\triangle ABC$,\n\n$$\n\\angle HBF = 90^\\circ - \\angle BAC = \\angle HCE. \\qquad \\textcircled{2}\n$$\n\nFrom \\textcircled{1} and \\textcircled{2}, one has $\\triangle BFH \\sim \\triangle CEH$, so\n\n$$\n\\frac{BF}{BH} = \\frac{CE}{CM}.\n$$\n\nAs $BHCM$ is a parallelogram, $BH = CM$, $CH = BM$, and hence\n\n$$\n\\frac{BF}{CM} = \\frac{CE}{BM}. \\qquad \\textcircled{3}\n$$\n\nAnd $D$ is the midpoint of $BC$. Thus, $S_{\\triangle PBM} = S_{\\triangle PCM}$, and so\n\n$$\n\\frac{1}{2} BP \\times BM \\times \\sin \\angle MBP = \\frac{1}{2} CP \\times CM \\times \\sin \\angle MCP.\n$$\n\nIt follows from $\\angle MBP + \\angle MCP = 180^\\circ$ that\n\n$$\nBP \\times BM = CP \\times CM. \\qquad \\textcircled{4}\n$$\n\nWith \\textcircled{3} and \\textcircled{4}, one has $\\frac{BF}{BP} = \\frac{CE}{CP}$. From $\\angle PBF = \\angle PCE$, one has $\\triangle PBF \\sim \\triangle PCE$, and hence $\\angle PFB = \\angle PEC$, and therefore $\\angle PFA = \\angle PEA$. Consequently, $P$, $A$, $E$, and $F$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15123,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $ (m, n) $ such that in an $ m \\times n $ table (with $ m+1 $ horizontal lines and $ n+1 $ vertical lines), one can add one diagonal in some chosen unit squares; or equivalently, one can turn each small square $\\Box$ into $\\Box$, $\\square$, or $\\checkmark$, so that the obtained graph has an Eulerian cycle.",
    "options": [],
    "answer": "See solution",
    "solution": "The pairs of positive integers $ (m, n) $ we seek are those with $ m = n $.  \n\nFirst, when $ m = n $, we can simply draw diagonals from upper left to lower right in any small squares except the ones on the main diagonal. It is easy to see that this graph will satisfy the condition of the problem.\n\nNext, we prove that when $ m \\neq n $, one cannot add diagonals to make the graph have an Eulerian cycle. Suppose that we can add diagonals to get a graph with an Eulerian cycle. We now consider only the added diagonals; these diagonals can only intersect at some lattice point, i.e., a lattice point which is the end of four diagonals. Separate these four diagonals into two disjoint folded lines:\n\n![](images/CHN_TSExams_2022_1_p2_data_f3156581f5.png)\n\nApplying this process to all intersections of the diagonals, all diagonals form disjoint cycles and zigzag lines which connect lattice points on the interior of the four sides of the table. We may simply remove all cycles, and then the diagonals simply link together interior lattice points of the four sides of the table.\n\nWe may color the lattice points as on a chessboard, i.e., coloring the lattice point $ (i, j) $ white if $ i + j $ is even and black if $ i + j $ is odd. Then any above diagonal line must pass through lattice points of one color; we call these diagonal lines white and black lines, respectively. Note that the diagonal lines cannot connect two lattice points on the same side of the table, otherwise there are an odd number of points between the two endpoints of the diagonal lines, so we cannot connect them using disjoint diagonal lines. For the same reason, the diagonal lines cannot connect two lattice points on the opposite sides of the table. Otherwise, we may assume that these two points are black and are at the top and bottom sides; then this diagonal line will divide the table into two parts, and it is not hard to show that the number of interior lattice points on the left-hand side is odd. They cannot be linked together by diagonal lines.\n\nSo every diagonal line can only connect interior lattice points on adjacent sides of the square. From this, we deduce that $ m = n $.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 15124,
    "subject": "Mathematics (Olympiad)",
    "question": "For which positive integers $k$ can the integers $1, 2, 3, \\dots, (2k)^2$ be arranged as a $2k \\times 2k$ table in such a way that all row sums and column sums are of the same parity, opposite to that of $k$?",
    "options": [],
    "answer": "See solution",
    "solution": "Such an arrangement is impossible for $k = 1$. In order to make all row sums and column sums even, both odd numbers should occur in the same row and also in the same column, which is impossible.\n\nLet $0$ and $1$ denote any even and odd number, respectively. For $k = 2$, one suitable arrangement is shown below:\n\n$$\n\\begin{matrix}\n1 & 1 & 1 & 0 \\\\\n1 & 1 & 0 & 1 \\\\\n1 & 0 & 0 & 0 \\\\\n0 & 1 & 0 & 0\n\\end{matrix}\n$$\n\nWe now show how to obtain a suitable arrangement for $k + 1$ from any suitable arrangement for $k$. Add $0, 1$ to the end of the first $k - 1$ rows and add $1, 0$ to the end of the following $k + 1$ rows. This fills the $2k \\times 2$ strip appearing at the right end of the table. Fill the $2 \\times 2k$ strip below the original part of the table similarly. Let the remaining $2 \\times 2$ corner be\n$$\n\\begin{matrix}\n0 & 1 \\\\\n1 & 0\n\\end{matrix}\n$$\nThen the parity of the sum of each old row and column is inverted. Each new column or row contains either $k$ or $k + 2$ odd numbers, so the parity of the row and column sums is opposite to that of $k + 1$. Hence, the extended table meets the requirements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15125,
    "subject": "Mathematics (Olympiad)",
    "question": "Andrew and Olesya take turns cutting squares from a $4000 \\times 2019$ rectangle, always along the grid lines, so that the remaining figure stays connected after each move. The player who cannot make a move loses. If Olesya goes first and both play optimally, who will win?",
    "options": [],
    "answer": "See solution",
    "solution": "On her first turn, Olesya cuts a $2018 \\times 2018$ square (see figure 34).\nAfter this, Olesya can use a symmetric strategy, except for the moves involving the black unit squares.\n\nIf Andrew manages to cut, for example, the left black square, it means all squares to the left have been cut. By symmetry, all squares to the right have also been cut, so Olesya can then cut the right black square.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15126,
    "subject": "Mathematics (Olympiad)",
    "question": "对集合 $\\{1,2,\\ldots,2024\\}$ 的所有排列，设 $p(k)$ 表示恰有 $k$ 个不动点的排列的数量。证明：所有排列的不动点总数等于 $2024!$。",
    "options": [],
    "answer": "See solution",
    "solution": "注意到让 $1$ 成为不动点的排列数为 $(2024-1)!$，其余点同理。因此所有排列的不动点总数为 $2024 \\times (2024-1)! = 2024!$，故等式成立。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15127,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N^*$ be the set of positive integers. Define $a_1 = 2$, and for $n = 1, 2, \\dots$,\n\n$$\na_{n+1} = \\min\\left\\{ \\lambda \\mid \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_n} + \\frac{1}{\\lambda} < 1,\\ \\lambda \\in N^* \\right\\}.\n$$\n\nProve that $a_{n+1} = a_n^2 - a_n + 1$ for $n = 1, 2, \\dots$.",
    "options": [],
    "answer": "See solution",
    "solution": "By $a_1 = 2$, $a_2 = \\min\\left\\{ \\lambda \\mid \\frac{1}{a_1} + \\frac{1}{\\lambda} < 1,\\ \\lambda \\in N^* \\right\\}$, consider $\\frac{1}{a_1} + \\frac{1}{\\lambda} < 1$, then $\\frac{1}{\\lambda} < 1 - \\frac{1}{2} = \\frac{1}{2}$, $\\lambda > 2$, hence $a_2 = 3$. So the conclusion is true for $n = 1$.\n\nSuppose that the conclusion is true for all integer $n \\le k - 1$ ($k \\ge 2$). If $n = k$, then\n$$\na_{k+1} = \\min\\left\\{ \\lambda \\mid \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k} + \\frac{1}{\\lambda} < 1,\\ \\lambda \\in N^* \\right\\}.\n$$\n\nConsidering\n$$\n\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k} + \\frac{1}{\\lambda} < 1,\n$$\nthat is, $0 < \\frac{1}{\\lambda} < 1 - \\left( \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k} \\right)$, we have\n$$\n\\lambda > \\frac{1}{1 - \\frac{1}{a_1} - \\frac{1}{a_2} - \\dots - \\frac{1}{a_k}}.\n$$\n\nIn the following, we show that\n$$\n\\frac{1}{1 - \\frac{1}{a_1} - \\frac{1}{a_2} - \\dots - \\frac{1}{a_k}} = a_k (a_k - 1).\n$$\n\nBy the induction hypotheses, for $2 \\le n \\le k$, $a_n = a_{n-1}(a_{n-1} - 1) + 1$, we have\n$$\n\\frac{1}{a_n - 1} = \\frac{1}{a_{n-1}(a_{n-1} - 1)} = \\frac{1}{a_{n-1} - 1} - \\frac{1}{a_{n-1}},\n$$\ntherefore $\\frac{1}{a_{n-1}} = \\frac{1}{a_{n-1}-1} - \\frac{1}{a_n-1}$. By taking the sum, we have\n$$\n\\sum_{i=2}^{k} \\frac{1}{a_{i-1}} = 1 - \\frac{1}{a_k - 1},\n$$\nthat is,\n$$\n\\sum_{i=1}^{k} \\frac{1}{a_i} = 1 - \\frac{1}{a_k - 1} + \\frac{1}{a_k} = 1 - \\frac{1}{a_k (a_k - 1)}.\n$$\n\nConsequently,\n$$\n\\frac{1}{1 - \\frac{1}{a_1} - \\frac{1}{a_2} - \\dots - \\frac{1}{a_k}} = a_k (a_k - 1).\n$$\nTherefore,\n$$\n\\begin{aligned}\na_{k+1} &= \\min \\left\\{ \\lambda \\mid \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k} + \\frac{1}{\\lambda} < 1,\\ \\lambda \\in \\mathbb{N}^* \\right\\} \\\\\n&= a_k (a_k - 1) + 1.\n\\end{aligned}\n$$\n\nBy induction on $n$, for all positive integer $n$, we have $a_{n+1} = a_n^2 - a_n + 1$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15128,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine whether there exist two different sets $A$ and $B$, each consisting of at most $2011^2$ positive integers, such that for every $x$ with $0 < x < 1$, the following inequality holds:\n\n$$\n\\left| \\sum_{a \\in A} x^a - \\sum_{b \\in B} x^b \\right| < (1-x)^{2011}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, such sets exist. We construct them as follows.\n\nRewrite the inequality using $y = 1 - x$:\n\n$$\n\\left| \\sum_{a \\in A} (1-y)^a - \\sum_{b \\in B} (1-y)^b \\right| < y^{2011}\n$$\nfor all $0 < y < 1$.\n\n**Step 1:**\nThere exist two different sets $A', B'$ of $2011^2$ positive integers each such that for $k = 0, 1, \\dots, 2011$,\n$$\n\\sum_{a \\in A'} \\binom{a}{k} = \\sum_{b \\in B'} \\binom{b}{k}.\n$$\nThis follows by the pigeonhole principle: for large $N$, the number of $2011^2$-element subsets of $\\{1,2,\\dots,N\\}$ exceeds the number of possible $2011$-tuples of binomial sums, so two distinct sets must have the same tuple.\n\n**Step 2:**\nExpanding $(1-y)^a$ in powers of $y$, the terms of degree $\\leq 2011$ cancel by construction. The remaining terms have degree $\\geq 2012$, so\n$$\n\\left| \\sum_{a \\in A'} (1-y)^a - \\sum_{b \\in B'} (1-y)^b \\right| \\leq M y^{2012}\n$$\nfor some constant $M$ and all $0 < y < 1$.\n\n**Step 3:**\nFor all $y$ with $0 < y < 1$, $(1-y)^M < 1/(My)$. Thus, multiplying by $y^{2012}$, we have $M y^{2012} (1-y)^M < y^{2011}$ for small $y$.\n\n**Step 4:**\nLet $A = \\{a + M : a \\in A'\\}$ and $B = \\{b + M : b \\in B'\\}$. Then the original inequality holds for these $A$ and $B$.\n\n**Remark:**\nIf only finiteness is required, a construction using the expansion of $(1-x)(1-x^2)\\cdots(1-x^{2011})$ and separating terms by sign also works.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15129,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with the shortest side $BC$. Let $X$, $Y$, $K$, $L$ be points on sides $AB$, $AC$ and on the rays opposite to $BC$, $CB$ respectively, such that $BX = BK = BC = CY = CL$. Line $KX$ intersects line $LY$ at a point $M$. Prove that the centroid of triangle $KLM$ coincides with the incenter of triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $ABC$ is an external angle of the isosceles triangle $XKB$ with apex $B$ (see the picture), the line $KX$ is parallel to the angle bisector of angle $ABC$.\n\n![](images/65_Czech_and_Slovak_MO_2016_booklet_p6_data_480998bfa6.png)\n\nThe ratio $LB : LK = 2 : 3$ yields that the angle bisector of $ABC$ meets the centroid of triangle $KLM$. If we denote $LL_1$ as its median and $L_2$ as its intersection with the angle bisector of $ABC$, then we obtain\n\n$$\n\\frac{LL_2}{LL_1} = \\frac{LB}{LK} = \\frac{2}{3}.\n$$\n\nfrom the similarity of triangles $\\triangle LBL_2 \\sim \\triangle LKL_1$ (by AA). So the point $L_2$ divides the median $LL_1$ in the same ratio as the centroid, and therefore it is the centroid of triangle $KLM$.\n\nIt follows from the symmetry of the problem that the angle bisector of angle $BCA$ also meets the centroid of triangle $KLM$. The fact that the intersection of the angle bisectors is the incenter proves the claim.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15130,
    "subject": "Mathematics (Olympiad)",
    "question": "We know that for some natural number $n$, the number $n^3 + 2009n^2 + 27n$ written in decimal notation ends with the digit 3. Find the digits in the hundred's and ten's places of this number.",
    "options": [],
    "answer": "See solution",
    "solution": "It's clear that the term $2000n^2$ does not influence the answer, so the sought digits are the same for the numbers $A = n^3 + 2009n^2 + 27n$ and $B = n^3 + 9n^2 + 27n$. Since $B + 27 = (n + 3)^3$ (which is the cube of a natural number) and ends in 0, this number should end in 000. Thus, $B = \\overline{X000} - 27 = \\overline{Y073}$, where $X, Y$ are some natural numbers. Therefore, the last three digits of the number are 073.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15131,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AB$ and $CD$ be the diameters of a circle, with $AB$ parallel and $CD$ perpendicular to a line $L$. For each segment, let $P_iQ_i$ be its projection onto $AB$, and $X_iY_i$ its projection onto $CD$. Show that:\n\n$$\n\\sum_{j=1}^{4n} P_iQ_i + \\sum_{j=1}^{4n} X_iY_i \\geq AB + CD.\n$$\n\nFurthermore, prove that at least two of the projected segments $P_iQ_i$ must overlap on $AB$, and that the line through their intersection point, perpendicular to $AB$, intersects two original unit segments.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $P_iQ_i + X_iY_i \\geq 1$ for each segment. Therefore,\n\n$$\n\\sum_{j=1}^{4n} P_iQ_i + \\sum_{j=1}^{4n} X_iY_i = \\sum_{j=1}^{4n} (P_iQ_i + X_iY_i) \\geq 4n = AB + CD.\n$$\n\nWithout loss of generality, assume $\\sum_{j=1}^{4n} P_iQ_i \\geq AB$. Since all $P_iQ_i$ lie strictly inside $AB$, two of these segments must overlap. Let $E$ be a point common to both. Then, the line through $E$ perpendicular to $AB$ intersects two unit segments whose projections are $P_iQ_i$ and $P_jQ_j$, completing the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15132,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be the circumcircle of a given convex quadrilateral $ABCD$ with the property that the half-lines $DA$ and $CB$ meet at a point $E$ for which $|CD|^2 = |AD| \\cdot |ED|$ holds. Let $F$ ($F \\neq A$) be the point of intersection of the circle $k$ with the perpendicular to $ED$ at $A$. Prove that the segments $AD$ and $CF$ are congruent if and only if the circumcenter of the triangle $ABE$ lies on $ED$.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, $DF$ is a diameter of $k$. First, we show that under the given conditions, the vertex $C$ cannot lie in the half-plane $DFA$.\n\nIf the vertices $B, C$ are points on the subarc $DA$ of the arc $DAF$ (see the figure below), then the angles $DCB$ and $DBA$ are obtuse, hence $|DC| < |DB| < |DA| < |DE|$, which contradicts the equality $|CD|^2 = |AD| \\cdot |ED|$.\n\nIf the vertices $B, C$ are points on the subarc $AF$ of the arc $DAF$, the angle $BAE$ is acute and $|\\angle DBE| = 180^\\circ - |\\angle DBC| \\le 90^\\circ$, so the possible other meeting point $B'$ of the half-line $DB$ with the circumcircle of the triangle $AEB$ lies in the segment $DB$. Hence $|DC| > |DB| \\ge |DB'|$. This means that the equality $|CD|^2 = |AD| \\cdot |ED|$ cannot hold as $|AD| \\cdot |ED| = |DB| \\cdot |DB'|$ (which is the power of $D$ with respect to the circumcircle of the triangle $AEB$).\n\n![](images/CpsMT07_s_p1_data_0bcfcb19d6.png)\n\n*Fig. 1*\n\n![](images/CpsMT07_s_p1_data_1569eaf567.png)\n\n*Fig. 2*\n\nWe have shown that the vertex $C$ of the given quadrangle does not lie in the half-plane $FDA$, hence $|FC| = |DA|$ if and only if $DAFC$ is a rectangle, i.e., if and only if $CA$ is a diameter of the circle $k$, which is equivalent to the angle $CBA$ being right, which is in turn equivalent to the triangle $AEB$ being right with the right angle at $B$, i.e., to the circumcenter of the triangle $AEB$ being the midpoint of $AE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15133,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there exist only finitely many triples of positive integers $ (n, a, b) $ such that:\n\n$$\nn! = 2^a - 2^b.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $n!$ is divisible by $3^{\\lfloor n/3 \\rfloor}$, $(2^{a-b} - 1)2^b = 2^a - 2^b$ is divisible by $3^{\\lfloor n/3 \\rfloor}$. From the Lifting the Exponent Lemma, we obtain that $a-b$ is divisible by $3^{\\lfloor n/3 \\rfloor - 1}$. So $a-b \\geq 3^{\\lfloor n/3 \\rfloor - 1} \\geq 3^{n/3 - 2}$. Hence, the right-hand side of our prior equality is greater than\n\n$$\n2^{a-b} - 1 \\geq 2^{a-b-1} \\geq 2^{3^{n/3 - 2} - 1}\n$$\n\nwhich, for sufficiently large $n$, is obviously greater than $2^{n^2} > n!$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15134,
    "subject": "Mathematics (Olympiad)",
    "question": "Numbers $a$, $b$, $c$ satisfy the conditions:\n\n$$\n\\frac{a+c}{a+1} = b, \\quad \\frac{c+b}{c+1} = a, \\quad \\frac{b+a}{b+1} = c.\n$$\n\nWhat values can the expression $(a+1)(b+1)(c+1)$ take?",
    "options": [],
    "answer": "See solution",
    "solution": "Subtract $1$ from both sides of each equation:\n\n$$\nb - 1 = \\frac{a+c}{a+1} - 1 = \\frac{c-1}{a+1}\n$$\n\nAnalogously, for the other two equations:\n\n$$\na - 1 = \\frac{c+b}{c+1} - 1 = \\frac{b-1}{c+1}\n$$\n$$\nc - 1 = \\frac{b+a}{b+1} - 1 = \\frac{a-1}{b+1}\n$$\n\nSince none of the variables equals $-1$, we can multiply both sides by the denominators:\n\n$$\n(b-1)(a+1) = c-1\n$$\n$$\n(a-1)(c+1) = b-1\n$$\n$$\n(c-1)(b+1) = a-1\n$$\n\nMultiplying all three equations together:\n\n$$\n(a-1)(a+1)(b-1)(b+1)(c-1)(c+1) = (a-1)(b-1)(c-1)\n$$\n\nIf, for example, $c=1$, then $b=1$ and $a=1$, so $(a+1)(b+1)(c+1)=8$.\n\nIf none of the variables is $1$, divide both sides by $(a-1)(b-1)(c-1)$:\n\n$$\n(a+1)(b+1)(c+1) = 1\n$$\n\nFor example, $a=b=c=0$ gives $(a+1)(b+1)(c+1)=1$, and $a=b=c=1$ gives $8$.\n\nThus, the possible values are $1$ and $8$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15135,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1 \\leq a_2 \\leq \\dots \\leq a_n$ be real numbers. Prove that\n$$\n\\left(\\sum_{i=1}^{n} i a_i\\right)^2 \\geq \\frac{1}{n} \\left(\\sum_{i=1}^{n} a_i^2\\right) \\left(\\sum_{i=1}^{n} b_i\\right),\n$$\nwhere $b_i = \\sum_{k=1}^{n} \\frac{2ik^2}{i+k}$ for $1 \\leq i \\leq n$.\n\nFurthermore, determine the maximum possible value of $\\lambda(n) = \\frac{\\left(\\sum_{i=1}^{n} i a_i\\right)^2}{\\sum_{i=1}^{n} a_i^2}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since\n$$\n\\begin{align*}\n\\sum_{i=1}^{n} b_i &= \\sum_{i=1}^{n} \\sum_{k=1}^{n} \\frac{2ik^2}{i+k} = \\sum_{i=1}^{n} i^2 + 2 \\sum_{1 \\leq i < j \\leq n} \\left( \\frac{i^2 j}{i+j} + \\frac{ij^2}{i+j} \\right) \\\\\n&= \\sum_{i=1}^{n} i^2 + 2 \\sum_{1 \\leq i < j \\leq n} ij = \\left( \\sum_{i=1}^{n} i \\right)^2 = \\frac{n^2(n+1)^2}{4},\n\\end{align*}\n$$\nwe find that\n$$\n\\left(\\sum_{i=1}^{n} i a_i\\right)^2 \\geq \\frac{n(n+1)^2}{4} \\sum_{i=1}^{n} a_i^2,\n$$\nwhich proves the inequality.\n\nWe conclude that the maximum possible value of $\\lambda(n)$ is\n$$\n\\frac{n(n+1)^2}{4}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15136,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $2x_1, 2x_2, \\dots, 2x_{10}$ be the number of coins that the knights sitting in chairs $1, 2, \\dots, 10$ had in the beginning, respectively. Determine $2x_8$.\n\nGiven the system of equations:\n\n$$\nx_{10} + x_2 = 22 \\\\\nx_1 + x_3 = 24 \\\\\nx_2 + x_4 = 26 \\\\\nx_3 + x_5 = 28 \\\\\n\\vdots \\\\\nx_8 + x_{10} = 38 \\\\\nx_9 + x_1 = 40\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By combining the equations, we get:\n\n$$\n\\begin{aligned}\nx_{10} &= 38 - x_8 \\\\\nx_2 &= 22 - x_{10} = 22 - (38 - x_8) = x_8 - 16 \\\\\nx_4 &= 26 - x_2 = 26 - (x_8 - 16) = 42 - x_8 \\\\\nx_6 &= 30 - x_4 = 30 - (42 - x_8) = x_8 - 12 \\\\\nx_8 &= 34 - x_6 = 34 - (x_8 - 12) = 46 - x_8\n\\end{aligned}\n$$\n\nFrom the last equation, $2x_8 = 46$, so the knight that ended up with 36 coins had 46 coins in the beginning.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15137,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 2012 piles of stones. The first pile contains $2^0$ stones, the second pile contains $2^1$ stones, the third pile contains $2^2$ stones, and so on. The 2012-th pile contains $2^{2011}$ stones. At each step, you can pick any three piles and add 2 stones to the first pile, 3 stones to the second pile, and 4 stones to the third pile. Is it possible, after a finite number of such operations, to get exactly $3^{1005}$ stones in each pile?",
    "options": [],
    "answer": "See solution",
    "solution": "No.\n\nAfter each operation, the total number of stones changes by a number divisible by 9. At the end, the total number is $2012 \\cdot 3^{1005}$, which is divisible by 9. However, at the start, the total number is $2^0 + 2^1 + 2^2 + \\ldots + 2^{2011} = 2^{2012} - 1$, which is not divisible by 9. Indeed, $2012 = 335 \\cdot 6 + 2$, and $2^6 = 64 \\equiv 1 \\pmod{9}$, so\n\n$$\n2^{2012} - 1 = (2^6)^{335} \\cdot 2^2 - 1 = 4 - 1 = 3 \\pmod{9}.\n$$\n\nTherefore, it is impossible to reach the desired configuration.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15138,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider two sets $A$ and $B$ of real numbers that have the following properties:\n\n(a) $0 \\in A$;\n\n(b) if $1 + x \\in A$, then $\\sqrt{1 + x + x^2} \\in B$;\n\n(c) if $\\sqrt{x^2 - x + 1} \\in B$, then $2 + x \\in A$.\n\nProve that $\\sqrt{3}$, $\\sqrt{13}$, $\\sqrt{31}$ are elements of the set $B$ and $2024 \\in A$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $1 + (-1) = 0 \\in A$, according to (b) we obtain $1 \\in B$. Since $1 = \\sqrt{0^2 - 0 + 1} \\in B$, we infer from (c) that $2 \\in A$, from which $\\sqrt{3} \\in B$.\n\nSince $\\sqrt{2^2 - 2 + 1} = \\sqrt{3} \\in B$, it follows from (c) that $2 + 2 = 4 \\in A$ and, from (b), we infer $\\sqrt{13} \\in B$.\n\nSince $\\sqrt{4^2 - 4 + 1} = \\sqrt{13} \\in B$, we further infer that $2 + 4 = 6 \\in A$ and thus $\\sqrt{1 + 5 + 5^2} = \\sqrt{31} \\in B$.\n\nUsing the equality $1 + x + x^2 = (x + 1)^2 - (x + 1) + 1$, we have that if $1 + x \\in A$, then $3 + x \\in A$. Since $0 \\in A$, $2 \\in A$ and it follows that the set $A$ contains all the even numbers. In particular, $2024 \\in A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15139,
    "subject": "Mathematics (Olympiad)",
    "question": "The sequence $\\{a_n\\}$ is defined as follows: $a_1 = 0$, and for integer $n \\ge 2$,\n\n$$\na_n = \\frac{1}{n} + \\frac{1}{\\lceil \\frac{n}{2} \\rceil} \\sum_{k=1}^{\\lceil \\frac{n}{2} \\rceil} a_k,\n$$\n\nwhere $\\lceil \\frac{n}{2} \\rceil$ denotes the smallest integer not less than $\\frac{n}{2}$. Find the maximum term of the sequence $\\{a_n\\}$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof*. From the given definition, we have $a_2 = \\frac{1}{2}$ and $a_3 = \\frac{7}{12}$.\n\nWe now prove by induction that $a_n \\leq \\frac{7}{12}$, with equality if and only if $n = 3$. The cases $n = 1, 2, 3$ have been verified. Assume the statement holds for all $1, 2, \\dots, n-1$ ($n \\geq 4$), then\n\n$$\n\\begin{align*}\na_n &= \\frac{1}{n} + \\frac{1}{\\lceil \\frac{n}{2} \\rceil} \\sum_{k=1}^{\\lceil \\frac{n}{2} \\rceil} a_k \\\\\n&\\leq \\frac{1}{n} + \\frac{0 + \\frac{1}{2} + (\\lceil \\frac{n}{2} \\rceil - 2) \\cdot \\frac{7}{12}}{\\lceil \\frac{n}{2} \\rceil} \\\\\n&= \\frac{7}{12} - \\frac{2}{3\\lceil \\frac{n}{2} \\rceil} + \\frac{1}{n} \\\\\n&= \\frac{7}{12} - \\frac{2n - 3\\lceil \\frac{n}{2} \\rceil}{3n \\cdot \\lceil \\frac{n}{2} \\rceil} \\\\\n&\\leq \\frac{7}{12} - \\frac{2n - 3\\left(\\frac{n+1}{2}\\right)}{3n \\cdot \\lceil \\frac{n}{2} \\rceil} \\\\\n&= \\frac{7}{12} - \\frac{\\frac{n}{2} - \\frac{3}{2}}{3n \\cdot \\lceil \\frac{n}{2} \\rceil} \\\\\n&< \\frac{7}{12}.\n\\end{align*}\n$$\n\nTherefore, the maximum term of the sequence $\\{a_n\\}$ is $a_3 = \\frac{7}{12}$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15140,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real polynomials of degree $n$ satisfying\n\n$$\nP(P(x) + x) = P(P(x)) + P(x)^n + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that $P$ cannot be a constant. Now let $n = 1$ and $P(x) = a x + b$. Then\n\n$$\n\\begin{cases}\nP(P(x) + x) = a(a x + b + x) + b = (a^2 + a)x + a b + b, \\\\\nP(P(x)) + P(x) + 1 = a(a x + b) + b + a x + b + 1 = (a^2 + a)x + a b + 2b + 1.\n\\end{cases}\n$$\n\nHence\n\n$$\nP(P(x) + x) = P(P(x)) + P(x) + 1 \\iff b = -1.\n$$\n\nThus, all polynomials of the form $P(x) = a x - 1$, $a \\neq 0$, satisfy the condition of the problem.\n\nFinally, suppose $n \\geq 2$. Let $a_n \\neq 0$ denote the leading coefficient of $P(x)$. Then the leading coefficient of $P(P(x) + x)$ is $a_n^{n+1}$ and the leading coefficient of $P(P(x)) + P(x) + 1$ is $(a_n^{n+1} + a_n^n)$. Since $P(P(x) + x) = P(P(x)) + P(x) + 1$, we get that $a_n^{n+1} = a_n^{n+1} + a_n^n$. Hence $a_n = 0$, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15141,
    "subject": "Mathematics (Olympiad)",
    "question": "66 students are participating in an olympiad. Each student has at least one friend, and for each student, the sum of the number of their friends and the arithmetic mean of the number of friends of their friends is 11. Prove that the students can be divided into two classrooms where no two friends are in the same room. Here, friendship is assumed to be reciprocal.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d_X$ denote the number of friends of a student $X$ and let $m_X$ denote the arithmetic mean of the number of friends of $X$'s friends.\n\nLet $A$ be a student with a minimal number of friends and let $B$ be a student with a maximal number of friends. Clearly, we have $d_A \\leq m_B$ and $m_A \\leq d_B$, and since by assumption $d_A + m_A = d_B + m_B = 11$, we have $d_A = m_B$ and $m_A = d_B$. It follows that $d_A + d_B = 11$ and thus $d_A < d_B$. In particular, $A \\neq B$.\n\nLet $C$ be a friend of $A$. If $d_C < d_B$, then we have $m_A < d_B$, which contradicts the above. Hence $d_C = d_B$. In other words, all friends of $A$ have $d_B$ friends. Similarly, all friends of $B$ have $d_A$ friends.\n\nTherefore, we may put students with $d_A$ friends in classroom I and students with $d_B$ friends in classroom II, and no two friends will be in the same classroom. The remaining students do not have friends that are already in a classroom, and we may repeat the procedure.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15142,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(a_n)$ be a sequence of real numbers such that for all $n \\in \\mathbb{Z}$,\n$$\na_{n+2} = \\frac{a_{n+1} + a_n}{2}.\n$$\nShow that if $(a_n)$ is bounded, then it is constant.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d_n = a_{n+1} - a_n$. Then, for $n \\in \\mathbb{Z}$,\n\n$$\n2d_{n+1} = 2a_{n+2} - 2a_{n+1} = (a_{n+1} + a_n) - 2a_{n+1} = a_n - a_{n+1} = -d_n.\n$$\n\nThis implies $d_n = (-2)^{-n} d_0$ for all $n \\in \\mathbb{Z}$.\n\nIf $d_0 = 0$, then $d_n = 0$ for all $n$, hence $a_n = a_0$ for all $n$. If $d_0 \\neq 0$ and $R > 0$ is any given number, there exists an integer $n > 0$ so that $d_{-n} = (-2)^n d_0 > 2R$. If $|a_{-n}| < R$ and $|a_{-n+1}| < R$, then\n\n$$\n|d_{-n}| = |a_{-n+1} - a_{-n}| \\leq |a_{-n+1}| + |a_{-n}| < 2R\n$$\n\nin contradiction to the choice of $n$. This shows that the sequence cannot be bounded if it is not constant.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15143,
    "subject": "Mathematics (Olympiad)",
    "question": "In the nation of Onewaynia, certain pairs of cities are connected by one-way roads. Every road connects exactly two cities (roads are allowed to cross each other, e.g., via bridges), and each pair of cities has at most one road between them. Moreover, every city has exactly two roads leaving it and exactly two roads entering it.\n\nWe wish to close half the roads of Onewaynia in such a way that every city has exactly one road leaving it and exactly one road entering it. Show that the number of ways to do so is a power of 2 greater than 1 (i.e., of the form $2^n$ for some integer $n \\ge 1$).\n\nIn the language of graph theory, we have a simple digraph $G$ which is 2-regular and we seek the number of sub-digraphs which are 1-regular.",
    "options": [],
    "answer": "See solution",
    "solution": "**First solution, combinatorial**\n\nWe construct a simple undirected bipartite graph $\\Gamma$ as follows:\n\n- The vertex set consists of two copies of $V(G)$, say $V_{\\text{out}}$ and $V_{\\text{in}}$.\n- For $v \\in V_{\\text{out}}$ and $w \\in V_{\\text{in}}$, we have an undirected edge $vw \\in E(\\Gamma)$ if and only if the directed edge $v \\to w$ is in $G$.\n\nThe desired sub-digraphs of $G$ correspond exactly to perfect matchings of $\\Gamma$. The graph $\\Gamma$ is 2-regular and hence consists of several disjoint (simple) cycles of even length. If there are $n$ such cycles, the number of perfect matchings is $2^n$, as desired.\n\n**Second solution, linear algebra over $\\mathbb{F}_2$**\n\nFor each edge $e$, create an indicator variable $x_e$. For each vertex $v$:\n\n- If $e_1$ and $e_2$ are the two edges leaving $v$, require $x_{e_1} + x_{e_2} \\equiv 1 \\pmod{2}$.\n- If $e_3$ and $e_4$ are the two edges entering $v$, require $x_{e_3} + x_{e_4} \\equiv 1 \\pmod{2}$.\n\nThis gives a system of equations. The solutions come in natural pairs $\\vec{x}$ and $\\vec{x} + \\vec{1}$, so the number of solutions is either zero or a power of two. To show there is at least one solution, note that a nontrivial linear combination of the equations can only give $0 \\equiv 1$ if some subset $S$ of the equations has every variable appearing an even number of times. But then the edges form one or more even cycles, so $|S|$ is even, and we have $0 \\equiv 0 \\pmod{2}$ as needed.\n\nThus, the number of ways is a power of 2 greater than 1.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15144,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x \\in \\mathbb{Q}^+$, $x = \\frac{a}{b}$, $(a, b) = 1$, $a > 0$, $b > 0$. Apply Euclid's algorithm to $a$ and $b$: set $r_0 = a$, $r_1 = b$, and for $j = 1, 2, \\dots, n$, $r_{j-1} = q_j r_j + r_{j+1}$. There exists $n = n(x)$ such that $r_n \\neq 0$ and $r_{n+1} = 0$.\n\nConsider the function $f: \\mathbb{Q} \\to \\{-1, 1\\}$ defined by\n\n$$\nf(x) = \\begin{cases} (-1)^{n(x)} & x > 0 \\\\ 1 & x = 0 \\\\ (-1)^{n(-x)+1} & x < 0 \\end{cases}\n$$\n\nProve that $f(x)$ satisfies the following condition: for $x, y \\in \\mathbb{Q}$, $x \\neq y$, if $x + y = 0$, $x + y = 1$, or $xy = 1$, then $f(x) \\cdot f(y) = -1$.",
    "options": [],
    "answer": "See solution",
    "solution": "I. If $x + y = 0$, $x \\neq y$, $x > 0$, $y < 0$, then $f(x) = (-1)^{n(x)}$, $f(y) = f(-x) = (-1)^{n(x)+1} = -f(x)$. Thus, $f(x) \\cdot f(y) = -1$; $x$ and $y$ have different colors.\n\nII. If $x + y = 1$, $x \\neq y$, $x, y \\in \\mathbb{Q}$, at least one of $x, y$ is positive. Suppose $x > 0$, $x = \\frac{a}{b}$, then $y = \\frac{b-a}{b}$. If $y < 0$, $f(y) = (-1)^{n(-y)+1} = (-1)^{n(\\frac{a-b}{b})+1}$. Since $n(\\frac{a}{b}) = n(\\frac{a-b}{b})$, $f(x) \\cdot f(y) = -1$. If $y > 0$, $0 < x < \\frac{1}{2} < y < 1$, and $n(y) = n(x) + 1$, so $f(x) \\cdot f(y) = -1$.\n\nIII. If $xy = 1$, $x, y$ have the same sign. In this case, $n(y) = n(x) + 1$, so $f(x)f(y) = -1$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Algorithms"
  },
  {
    "id": 15145,
    "subject": "Mathematics (Olympiad)",
    "question": "In the oblique triangle $ABC$, $BC > AC > AB$. Let $P_1, P_2$ be two points in the plane such that, for $i = 1, 2$, the lines $AP_i$, $BP_i$, and $CP_i$ intersect the circumcircle of $\\triangle ABC$ at $D_i$, $E_i$, and $F_i$ respectively (other than $A$, $B$, $C$), with $D_iE_i \\perp D_iF_i$ and $D_iE_i = D_iF_i \\neq 0$. The line $P_1P_2$ crosses the circumcircle of $\\triangle ABC$ at two points $Q_1$ and $Q_2$. For $i = 1, 2$, let the projections of $Q_i$ onto the lines $AB$ and $AC$ be $X_i$ and $Y_i$, respectively. The lines $X_1Y_1$ and $X_2Y_2$ meet at $W$. Prove that $W$ lies on the nine-point circle of $\\triangle ABC$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "There are two major steps in the proof. First, we identify the positional relationship between $P_1$ and $P_2$. This can be done by manipulation of angles, but to avoid case-by-case discussions, we use directed angles or avoid angle addition/subtraction. In fact, $\\triangle P_iAB \\sim \\triangle P_iE_iD_i$, so\n\n$$\n\\frac{P_iA}{P_iE_i} = \\frac{P_iB}{P_iD_i} = \\frac{AB}{D_iE_i}.\n$$\n\nSimilarly,\n\n$$\n\\frac{P_iA}{P_iF_i} = \\frac{P_iC}{P_iD_i} = \\frac{AC}{D_iF_i}, \\quad \\frac{P_iC}{P_iE_i} = \\frac{P_iB}{P_iF_i} = \\frac{BC}{E_iF_i}.\n$$\n\nIt follows that\n\n$$\n\\begin{aligned}\n\\frac{P_iA}{P_iB} &= \\left(\\frac{P_iA}{P_iF_i}\\right) / \\left(\\frac{P_iB}{P_iF_i}\\right) = \\left(\\frac{AC}{D_iF_i}\\right) / \\left(\\frac{BC}{E_iF_i}\\right) = \\frac{AC}{BC} \\cdot \\frac{E_iF_i}{D_iF_i} = \\sqrt{2} \\frac{AC}{BC}, \\\\\n\\frac{P_iA}{P_iC} &= \\left(\\frac{P_iA}{P_iE_i}\\right) / \\left(\\frac{P_iC}{P_iE_i}\\right) = \\left(\\frac{AB}{D_iE_i}\\right) / \\left(\\frac{BC}{E_iF_i}\\right) = \\frac{AB}{BC} \\cdot \\frac{E_iF_i}{D_iE_i} = \\sqrt{2} \\frac{AB}{BC},\n\\end{aligned}\n$$\n\nand hence $P_1A : P_1B : P_1C = P_2A : P_2B : P_2C$, indicating\n\n$$\n\\frac{P_1A}{P_2A} = \\frac{P_1B}{P_2B} = \\frac{P_1C}{P_2C}.\n$$\n\nTherefore, $A$, $B$, $C$ lie on an Apollonius circle of foci $P_1$ and $P_2$, which implies that $P_1$, $P_2$, and the circumcentre $O$ of $\\triangle ABC$ are collinear. (Note: here we use the given condition $P_1 \\neq P_2$.)\n\nWe can also use complex numbers. Let the circumcircle of $\\triangle ABC$ be the unit circle in the complex plane. The lowercase letter of a point represents its complex number (e.g., $A$ corresponds to $a \\in \\mathbb{C}$). For a point $P$ in the plane, denote $D$, $E$, and $F$ as the second intersections of $AP$, $BP$, $CP$ with the circumcircle. Then,\n\n$$\n\\frac{p-a}{p-d} = \\overline{\\left(\\frac{p-a}{p-d}\\right)} \\iff \\frac{p-a}{p-d} = \\frac{\\bar{p}-\\bar{a}}{\\bar{p}-\\bar{d}},\n$$\n\nor equivalently,\n\n$$\n\\bar{p} + \\bar{a}\\bar{d}p = \\bar{a} + \\bar{d},\n$$\n\nthat is,\n\n$$\n\\bar{d} = \\frac{\\bar{a} - \\bar{p}}{\\bar{a}p - 1}, \\quad d = \\frac{a - p}{a\\bar{p} - 1}.\n$$\n\nLikewise,\n\n$$\ne = \\frac{b-p}{b\\bar{p}-1}, \\quad f = \\frac{c-p}{c\\bar{p}-1}.\n$$\n\nOn the other hand, $DE \\perp DF$ and $ED = DF \\Leftrightarrow \\frac{e-d}{f-d} = \\pm i \\Leftrightarrow$\n$$\n\\left( \\frac{e-d}{f-d} \\right)^2 = -1.\n$$\n\nTogether, we obtain\n\n$$\n\\left[ \\frac{(p\\bar{p} - 1)(a - b)(c\\bar{p} - 1)}{(p\\bar{p} - 1)(c - b)(a\\bar{p} - 1)} \\right]^2 = -1.\n$$\n\nSimplify to an equation about $\\bar{p}$:\n\n$$\n[c^2(a-b)^2 + b^2(c-a)^2]\\bar{p}^2 - 2[c(a-b)^2 + b(c-a)^2]\\bar{p} + [(a-b)^2 + (c-a)^2] = 0.\n$$\n\nThe leading coefficient is $0$ if and only if $|a-b| = |c-a|$ ($\\triangle ABC$ is isosceles at $A$) and $\\frac{b}{c} = \\pm i \\frac{a-b}{c-a}$ (the central angle subtended by $BC$ is $\\frac{\\pi}{2}$ larger than the inscribed angle), which is not the case. Therefore, this is a quadratic equation. The discriminant is\n\n$$\n\\begin{align*}\n\\triangle &= 4\\{[c(a-b)^2 + b(c-a)^2]^2 \\\\\n&\\quad - 4[c^2(a-b)^2 + b^2(c-a)^2][(a-b)^2 + (c-a)^2]\\} \\\\\n&= 4\\{2bc(a-b)^2(c-a)^2 - (b^2+c^2)(a-b)^2(c-a)^2\\} \\\\\n&= -4(a-b)^2(b-c)^2(c-a)^2 \\neq 0,\n\\end{align*}\n$$\n\nimplying the equation has distinct roots. Furthermore, $P_1P_2$ passes through the circumcentre $O$ of $\\triangle ABC$. It suffices to prove\n\n$$\n\\frac{c(a-b)^2 + b(c-a)^2}{(a-b)(b-c)(c-a)} \\in i\\mathbb{R} \\iff \\frac{c(a-b)^2 + b(c-a)^2}{(a-b)(b-c)(c-a)} + \\frac{\\overline{c(a-b)^2 + b(c-a)^2}}{(a-b)(b-c)(c-a)} = 0.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15146,
    "subject": "Mathematics (Olympiad)",
    "question": "For real numbers $x, y, z \\in [0, 1]$, let $M$ be the minimum of the three numbers $x - xy - xz + yz$, $y - yx - yz + xz$, and $z - zx - zy + xy$. Find the maximum possible value of $M$.",
    "options": [],
    "answer": "See solution",
    "solution": "The maximum possible value of $M$ is $\\frac{1}{4}$.\n\nFirst, when $x = y = z = \\frac{1}{2}$, $M = \\min\\left\\{\\frac{1}{4}, \\frac{1}{4}, \\frac{1}{4}\\right\\} = \\frac{1}{4}$.\n\nIn the following, we show $M \\le \\frac{1}{4}$.\n\nNote that among the three numbers $x - \\frac{1}{2}$, $y - \\frac{1}{2}$, $z - \\frac{1}{2}$, two are nonnegative or two are non-positive; say they are $x - \\frac{1}{2}$ and $y - \\frac{1}{2}$, so $(x - \\frac{1}{2})(y - \\frac{1}{2}) \\ge 0$. This implies that\n\n$$\n2M \\le (x - xy - xz + yz) + (y - yx - yz + xz) = x + y - 2xy = -2\\left(x - \\frac{1}{2}\\right)\\left(y - \\frac{1}{2}\\right) + \\frac{1}{2} \\le \\frac{1}{2},\n$$\n\nand thus $M \\le \\frac{1}{4}$.\n\n$\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15147,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\ge 2$, let $x_1, x_2, \\dots, x_n$ be non-negative real numbers satisfying $x_1 + x_2 + \\dots + x_n = n$. Find the minimum and maximum values of\n\n$$\n\\sum_{k=1}^{n} \\frac{1 + x_k^2 + x_k^4}{1 + x_{k+1} + x_{k+1}^2 + x_{k+1}^3 + x_{k+1}^4},\n$$\n\nwhere $x_{n+1} = x_1$.",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum value is $\\frac{3}{5}n$.\n\nIf $x_1 = x_2 = \\dots = x_n = 1$, then\n\n$$\n\\sum_{k=1}^{n} \\frac{1 + x_k^2 + x_k^4}{1 + x_{k+1} + x_{k+1}^2 + x_{k+1}^3 + x_{k+1}^4} = \\frac{3}{5}n.\n$$\n\nBy the AM-GM inequality, for any $1 \\le k \\le n$, $x_k^2 + x_k^4 \\ge 2x_k^3$ and $1 + x_k^2 \\ge 2x_k$. Also, $1 + x_k^4 \\ge x_k + x_k^3$ (since $(x_k-1)^2(x_k^2 + x_k + 1) \\ge 0$), so $2(1 + x_k^2 + x_k^4) \\ge 3(x_k + x_k^3)$. Thus, $5(1 + x_k^2 + x_k^4) \\ge 3(1 + x_k + x_k^2 + x_k^3 + x_k^4)$, which implies\n\n$$\n\\frac{1 + x_k^2 + x_k^4}{1 + x_k + x_k^2 + x_k^3 + x_k^4} \\ge \\frac{3}{5}.\n$$\n\nBy the rearrangement inequality,\n\n$$\n\\sum_{k=1}^{n} \\frac{1 + x_k^2 + x_k^4}{1 + x_{k+1} + x_{k+1}^2 + x_{k+1}^3 + x_{k+1}^4} \\ge \\sum_{k=1}^{n} \\frac{1 + x_k^2 + x_k^4}{1 + x_k + x_k^2 + x_k^3 + x_k^4} \\ge \\frac{3}{5}n.\n$$\n\nThe maximum value is $n^4 + n^2 + n - 1 + \\frac{n-1}{n^5-1}$.\n\nIf $x_1 = n$ and $x_2 = x_3 = \\dots = x_n = 0$, then\n\n$$\n\\sum_{k=1}^{n} \\frac{1 + x_k^2 + x_k^4}{1 + x_{k+1} + x_{k+1}^2 + x_{k+1}^3 + x_{k+1}^4} = n^4 + n^2 + n - 1 + \\frac{n-1}{n^5-1}.\n$$\n\nLet $a_k = 1 + x_k^2 + x_k^4$ and $b_k = 1 + x_k + x_k^2 + x_k^3 + x_k^4$, with $b_{n+1} = b_1$. Since $a_k \\ge 1$ and $\\frac{1}{b_{k+1}} \\le 1$, $(a_k - 1)\\left(\\frac{1}{b_{k+1}} - 1\\right) \\le 0$, so\n\n$$\n\\frac{a_k}{b_{k+1}} \\le a_k + \\frac{1}{b_{k+1}} - 1.\n$$\n\nTherefore,\n\n$$\n\\sum_{k=1}^{n} \\frac{a_k}{b_{k+1}} \\le \\sum_{k=1}^{n} a_k + \\sum_{k=1}^{n} \\frac{1}{b_{k+1}} - n.\n$$\n\nLet $f(x) = 1 + x^2 + x^4$ and $g(x) = \\frac{1}{1 + x + x^2 + x^3 + x^4}$ for $x \\ge 0$. Both $f(x)$ and $g(x)$ are convex on $[0, +\\infty)$. By convexity,\n\n$$\n\\sum_{k=1}^{n} f(x_k) + \\sum_{k=1}^{n} g(x_k) - n \\le f\\left(\\sum_{k=1}^{n} x_k\\right) + g\\left(\\sum_{k=1}^{n} x_k\\right) + (n-1)(f(0) + g(0)) - n = n^4 + n^2 + n - 1 + \\frac{n-1}{n^5-1}.\n$$\n\n![](images/2024_CGMO_p14_data_0b5e96dfd7.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15148,
    "subject": "Mathematics (Olympiad)",
    "question": "For an integer $n \\geq 3$ and real numbers $a_1, \\dots, a_n$ and $b_1, \\dots, b_n$, show the following inequality:\n\n$$\n\\sum_{i=1}^{n} a_i (b_i - b_{i+3}) \\leq \\frac{3n}{8} \\sum_{i=1}^{n} \\left[ (a_i - a_{i+1})^2 + (b_i - b_{i+1})^2 \\right]\n$$\nwhere $a_{n+1} = a_1$ and $b_{n+1} = b_1$ for $i = 1, 2, 3$.",
    "options": [],
    "answer": "See solution",
    "solution": "It suffices to prove the following:\n\n$$\n\\sum_{i=1}^{n} a_i (b_i - b_{i+1}) \\leq \\frac{n}{8} \\sum_{i=1}^{n} \\left[ (a_i - a_{i+1})^2 + (b_i - b_{i+1})^2 \\right]\n$$\n\nBy replacing $b_i$ by $b_{i+j}$ in the above equation and adding up for $j = 0, 1, 2$, we can obtain our desired result. Let\n\n$$\n\\mathcal{R} = \\frac{1}{2} \\sum_{i=1}^{n} \\left[ (a_i - a_{i+1})^2 + (b_i - b_{i+1})^2 \\right]\n$$\n\nand consider\n\n$$\nS_j = \\sum_{i=1}^{n} a_{i+j} (b_i - b_{i+1})\n$$\n\nwhere $j$ is an integer. (We consider any indices as modulo $n$, so that $a_{i+nk} = a_i$ holds for every integer $i, k$.) We can observe\n\n$$\n\\begin{aligned}\n|S_j - S_{j+1}| &= \\left| \\sum_{i=1}^{n} (a_{i+j} - a_{i+j+1}) (b_i - b_{i+1}) \\right| \\\\\n&\\leq \\frac{1}{2} \\sum_{i=1}^{n} \\left( (a_{i+j} - a_{i+j+1})^2 + (b_i - b_{i+1})^2 \\right) \\\\\n&= \\frac{1}{2} \\left( \\sum_{i=1}^{n} (a_{i+j} - a_{i+j+1})^2 + \\sum_{i=1}^{n} (b_i - b_{i+1})^2 \\right) \\\\\n&= \\mathcal{R}\n\\end{aligned}\n$$\n\nand obtain the following as its result:\n\n$$\n|S_0 - S_j| \\leq |j| \\mathcal{R}.\n$$\n\nMeanwhile, we have\n\n$$\n\\sum_{j=0}^{n-1} S_j = \\sum_{i=1}^{n} \\left( \\sum_{j=0}^{n-1} a_{i+j} \\right) (b_i - b_{i+1}) = \\left( \\sum_{j=0}^{n-1} a_j \\right) \\left( \\sum_{i=1}^{n} (b_i - b_{i+1}) \\right) = 0\n$$\n\nso for any $-n < k < n$ we have\n\n$$\nn S_0 = \\sum_{j=-k+1}^{n-k} (S_0 - S_j) \\leq \\sum_{j=-k+1}^{n-k} |S_0 - S_j| \\leq \\mathcal{R} \\sum_{j=-k+1}^{n-k} |j|.\n$$\n\nIf $n$ is even, then we let $k = n/2$ to obtain\n\n$$\nn S_0 \\leq \\mathcal{R} \\sum_{j=-n/2+1}^{n/2} |j| = \\frac{n^2}{4} \\mathcal{R}\n$$\n\nand if $n$ is odd, then we let $k = (n + 1)/2$ to obtain\n\n$$\nn S_0 \\leq \\mathcal{R} \\sum_{j=-(n-1)/2}^{(n-1)/2} |j| = \\frac{n^2-1}{4} \\mathcal{R} < \\frac{n^2}{4} \\mathcal{R}.\n$$\n\nIn any case, we have\n\n$$\nS_0 \\leq \\frac{n}{4} \\mathcal{R} = \\frac{n}{8} \\sum_{i=1}^{n} \\left[ (a_i - a_{i+1})^2 + (b_i - b_{i+1})^2 \\right]\n$$\n\nthus proving our inequality. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15149,
    "subject": "Mathematics (Olympiad)",
    "question": "一個 $2n \\times 2n$ 的棋盤上的每一格都有一張椅子。現在有 $2n^2$ 對情侶要入座，每個人坐一個座位。定義一對情侶之間的距離為他們座位相差的行數與相差的列數和（例如：如果一對情侶分別坐在 $(3,3)$ 和 $(2,5)$，則他們之間的距離為 $|3-2| + |3-5| = 3$）。定義所有情侶的總距離，等於這 $2n^2$ 對情侶的距離總和。試求總距離的最大值。",
    "options": [],
    "answer": "See solution",
    "solution": "最大值為 $4n^3$。\n\n1. 首先考慮水平方向的距離和的最大值：將所有人投影到同一列上，並將每一對情侶兩人之間連線。考慮兩種可能：\n\n- 存在兩對情侶的連線不重疊：則兩對各取一人交換位置，此時水平方向距離和更大。\n- 任兩對的連線都有重疊：不失一般性，假設每一對的男方都在女方左側。我們發現男方都必須在左側的 $n$ 行，否則：\n  - 若有一對情侶的男女方都在右側 $n$ 行中，右側 $n$ 行共剩 $2n^2 - 2$ 個位子，但還有另外 $2n^2 - 1$ 對情侶要坐。\n  - 因此必然有一對情侶他們都在左側 $n$ 行中。然而，都在左 $n$ 行的情侶，其連線不可能和都在右 $n$ 行的情侶重疊，矛盾！\n\n換言之，每對情侶的男方都必然屬於左 $n$ 排，女方都必然屬於右 $n$ 排。易計算此時的水平方向距離和必為 $2n^3$。\n\n2. 同理，垂直方向距離和最大值也是 $2n^3$，故距離總和至多 $4n^3$。\n\n3. 最後證明存在一種方法達到 $4n^3$。考慮將情侶分成 $A, B$ 兩組，每組 $n^2$ 對，並將座位分成四個象限（每個象限是一個 $n \\times n$ 的方格）。令第一象限都坐 $A$ 男，第二象限 $B$ 女，第三象限 $A$ 女，第四象限 $B$ 男，則可以發現此構造達到上述估計的最大值，故得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15150,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all nonzero polynomials with real coefficients satisfying the equality:\n\n$$\n(P(x))^3 + 3(P(x))^2 = P(x^3) - 3P(-x)\n$$\n\nfor all $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\deg P(x) = 0$. Then $P(x) = a \\neq 0$ and from the relation we have:\n\n$$\na^3 + 3a^2 = -2a \\implies a = -1 \\text{ or } a = -2.\n$$\n\nHence, the constant (nonzero) polynomials $P(x) = -1$ and $P(x) = -2$ are solutions.\n\nLet $\\deg P(x) = n > 0$. Then the polynomial $P(x)$ can be written as\n\n$$\nP(x) = a x^n + Q(x), \\text{ with } a \\neq 0, \\deg Q(x) \\leq n-1.\n$$\n\nBy substitution into the relation, we get:\n\n$$\n(a x^n + Q(x))^3 + 3(a x^n + Q(x))^2 = a x^{3n} + Q(x^3) - 3 Q(-x) - 3a (-1)^n x^n.\n$$\n\nEquating the coefficients of $x^{3n}$ gives:\n\n$$\na^3 = a \\implies a = 1 \\text{ or } a = -1.\n$$\n\n**Case I:** $a = 1$.\n\nIf $Q(x) = 0$, then $3x^{2n} = -3(-1)^n x^n$, which is impossible.\n\nIf $\\deg Q(x) = k > 0$, since $0 < k < n$, we have $2n + k = \\deg A(x) = \\deg B(x) = \\max\\{3k, n\\} \\geq 3k \\implies n \\geq k$, which is absurd.\n\nThus, $\\deg Q(x) = 0$, so $Q(x) = c \\neq 0$. Then equating coefficients gives:\n\n$$\n3(c+1)x^{2n} + 3(c^2+2c+(-1)^n)x^n + c^3 + 3c^2 + 2c = 0\n$$\nfor all $x$, so\n$$\nc+1=0, \\quad c^2+2c+(-1)^n=0, \\quad c^3+3c^2+2c=0.\n$$\nThis gives $c = -1$, $n = 2m$, $m \\in \\mathbb{N}^*$. Thus, $P(x) = x^{2m} - 1$ is a solution.\n\n**Case II:** $a = -1$.\n\nIf $Q(x) = 0$, then $3x^{2n} = 3(-1)^n x^n$, which is impossible.\n\nIf $\\deg Q(x) = k > 0$, as before, this leads to a contradiction.\n\nThus, $Q(x) = c \\neq 0$. Equating coefficients gives:\n\n$$\n3(c+1)x^{2n} - 3(c^2+2c+(-1)^n)x^n + c^3 + 3c^2 + 2c = 0\n$$\nfor all $x$, so\n$$\nc+1=0, \\quad c^2+2c+(-1)^n=0, \\quad c^3+3c^2+2c=0.\n$$\nThis gives $c = -1$, $n = 2m$, $m \\in \\mathbb{N}^*$. Thus, $P(x) = -x^{2m} - 1$ is a solution.\n\n**Conclusion:**\n\nAll solutions are:\n\n$$\nP(x) = -1, \\quad P(x) = -2, \\quad P(x) = x^{2m} - 1, \\quad P(x) = -x^{2m} - 1, \\quad m \\in \\mathbb{N}^*.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15151,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ of the form $n = (p-2)(p-1)(p+1)(p+2) + 9$, where $p$ is a prime, such that the sum of the digits of $n$ is minimized.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us find the first few $n$:\n- When $p=2$, $n=9$.\n- When $p=3$, $n=49$.\n- When $p=5$, $n=513$.\n\nNow, let $p > 5$. Rewrite $n$ as $n = (p-2)(p-1)(p+1)(p+2) + 9$. Since $(p-2), (p-1), p, (p+1), (p+2)$ are five consecutive positive integers, at least one is divisible by $5$. Since $p > 5$, $p$ is not divisible by $5$, so $5$ divides one of $(p-2), (p-1), (p+1), (p+2)$ and thus their product. Also, at least one of $p+1$ and $p+2$ is even, so the product is divisible by $2$. Thus, for $p > 5$, $(p-2)(p-1)(p+1)(p+2)$ is divisible by $10$.\n\nFor all $p > 5$, $n$ has at least two digits and the final digit is $9$, so the sum of the digits is greater than $9$. We conclude that the least possible sum of the digits is $9$, and this value is attained only when $p = 2$ or $p = 5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15152,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider an acute-angled triangle $AB\\Gamma$ with $AB < A\\Gamma$. Let $M$ be the midpoint of the side $B\\Gamma$. On the side $AB$ we consider a point $\\Delta$ such that, if the segment $\\Gamma\\Delta$ intersects the median $AM$ at point $E$, then $A\\Delta = \\Delta E$. Prove that $AB = \\Gamma E$.\n\n![](images/Hellenic_booklet_2013_p2_data_0825fa039c.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**First solution.**\n\nWe extend median $AM$ by $M\\Theta = AM$. Then $AB\\Theta\\Gamma$ is a parallelogram. Hence $AB \\parallel \\Gamma\\Theta$ and $\\hat{A}_1 = \\hat{O}_1$. But from $A\\Delta = \\Delta E$ we get $\\hat{A}_1 = \\hat{E}_1$.\n\n![](images/Hellenic_booklet_2013_p3_data_e01cbe3e63.png)\n\nand since $\\hat{E}_1 = \\hat{E}_2$, we find that $\\hat{O}_1 = \\hat{E}_2$. Therefore the triangle $E\\Theta$ is isosceles with $\\Gamma E = \\Gamma\\Theta$. Finally, from the parallelogram $AB\\Theta\\Gamma$ we have $\\vec{AB} = \\Gamma\\Theta$, from which we have $AB = \\Gamma E$.\n\n**Second solution.**\n\nFrom the midpoint $M$ of $B\\Gamma$ we draw line $\\delta \\parallel AB$. Then $\\delta \\parallel B\\Delta$. Let $B\\Delta$ intersect the segment $\\Gamma\\Delta$ at $Z$. Then $Z$ is the midpoint of $\\Gamma\\Delta$, that is $\\Gamma Z = Z\\Delta$ (1), and moreover $B\\Delta = 2 \\cdot MZ$ (2).\n\n![](images/Hellenic_booklet_2013_p3_data_b836a940fa.png)\n\nAlso we have $\\hat{A}_1 = \\hat{M}_1$. However, from $A\\Delta = \\Delta E$ we get that $\\hat{A}_1 = \\hat{E}_1$ and $\\hat{E}_1 = \\hat{E}_2$. Hence $\\hat{M}_1 = \\hat{E}_2$, and $EMZ$ is isosceles with $ZM = EZ$ (3). Hence\n\n$$\n\\begin{aligned}\n\\Gamma E &= \\Gamma Z + ZE = \\Delta Z + ZE \\quad (\\text{from (1)}) \\\\\n&= \\Delta E + 2 \\cdot ZM \\quad (\\text{from (3)}) \\\\\n&= A\\Delta + \\Delta B = AB. \\quad (\\text{from hypothesis and (2)})\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15153,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation:\n\n$$\n\\operatorname{ctg}[x] \\cdot \\operatorname{ctg}\\{x\\} = 1.\n$$\n\nWhere $[a]$ is the integer part of $a$, and $\\{a\\} = a - [a]$.",
    "options": [],
    "answer": "See solution",
    "solution": "The solution is $x = \\frac{1}{2}\\pi + \\pi k$, where $k \\in \\mathbb{Z}$.\n\n**Solution:**\n\nTransform the equation:\n\n$$\n\\cos[x] \\cdot \\cos\\{x\\} = \\sin[x] \\cdot \\sin\\{x\\},\n$$\n\nif $\\sin[x] \\cdot \\sin\\{x\\} \\ne 0$. Then,\n\n$$\n\\cos[x] \\cdot \\cos\\{x\\} - \\sin[x] \\cdot \\sin\\{x\\} = \\cos([x] + \\{x\\}) = \\cos x = 0.\n$$\n\nThus, $x = \\frac{1}{2}\\pi + \\pi k$, $k \\in \\mathbb{Z}$.\n\nCheck that these values satisfy the initial condition: none are integers, so $\\{x\\} \\ne 0$ and $\\sin\\{x\\} \\ne 0$. For $k \\ge 0$, $x > 1$ and $[x] > 0$; for $k < 0$, $x < -1$ and $[x] < 0$. Therefore, all such $x$ satisfy the equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15154,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : [0, 1] \\to (0, \\infty)$ be a continuous function on $[0, 1]$, and\n$$\nA = \\int_{0}^{1} f(t) \\, dt.\n$$\n\na) Show that the function $F : [0, 1] \\to [0, A]$, defined for any $x \\in [0, 1]$ by\n$$\nF(x) = \\int_{0}^{x} f(t) \\, dt,\n$$\nis invertible, with a differentiable inverse.\n\nb) Show that there is a unique function $g : [0, 1] \\to [0, 1]$, such that the equality\n$$\n\\int_{0}^{x} f(t) \\, dt = \\int_{g(x)}^{1} f(t) \\, dt\n$$\nholds for any $x \\in [0, 1]$.\n\nc) Show that there is a $c \\in [0, 1]$ for which\n$$\n\\lim_{x \\to c} \\frac{g(x) - c}{x - c} = -1,\n$$\nwhere $g$ is the function determined by the relation above.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Because $f$ is continuous, $F$ is continuous and differentiable, with $F'(x) = f(x) > 0$ for any $x \\in [0, 1]$. Hence, $F$ is strictly increasing and thus injective. Being continuous, $F$ has the intermediate value property, and since $F(0) = 0$ and $F(1) = A$, $F$ is surjective. Thus, $F$ is bijective, hence invertible. Also, because $F'(x) = f(x) > 0$ for any $x \\in [0, 1]$, $F^{-1}$ is differentiable, with\n$$\n(F^{-1})'(x) = \\frac{1}{f(F^{-1}(x))}, \\quad \\text{for any } x \\in [0, A].\n$$\n\nb) The equality $\\int_{0}^{x} f(t) \\, dt = \\int_{g(x)}^{1} f(t) \\, dt$ becomes $F(x) = F(1) - F(g(x)) = A - F(g(x))$, or, equivalently, $F(g(x)) = A - F(x)$ for any $x \\in [0, 1]$.\n\nThe function $F$ being increasing, it follows that $A - F(x) \\in [0, A]$ for any $x \\in [0, 1]$, so the function $g : [0, 1] \\to [0, 1]$, defined by $g(x) = F^{-1}(A - F(x))$ for any $x \\in [0, 1]$, is well defined and unique satisfying the relation.\n\nc) Since the functions $F$ and $F^{-1}$ are differentiable, the function $g$ defined above is also differentiable, and\n$$\ng'(x) = \\frac{(A - F(x))'}{f(F^{-1}(A - F(x)))} = \\frac{-f(x)}{f(F^{-1}(A - F(x)))} \\quad \\text{for any } x \\in [0, 1].\n$$\n\nThe function $g$ is strictly decreasing, and so has a unique fixed point $c \\in [0, 1]$.\n\nFor this we have: $2F(c) = F(c) + F(g(c)) = A$, so $F(c) = \\frac{1}{2}A$ and $c = F^{-1}\\left(\\frac{1}{2}A\\right)$.\n\nBecause $g$ is differentiable, the limit\n$$\n\\lim_{x \\to c} \\frac{g(x) - c}{x - c} = \\lim_{x \\to c} \\frac{g(x) - g(c)}{x - c} = g'(c)\n$$\nexists and is equal to\n$$\ng'(c) = \\frac{-f(c)}{f(F^{-1}(A - F(c)))} = \\frac{-f(c)}{f(c)} = -1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15155,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a set of $n$ positive integers such that for any subset $T \\subset S$ with $|T| = 100$, the product of the numbers in $T$ shares at most as many prime factors (counted with multiplicity) as the product of the remaining $n-100$ numbers in $S$. What is the maximum possible number of prime numbers in $S$?",
    "options": [],
    "answer": "See solution",
    "solution": "The maximum number of prime numbers in $S$ is $1819$.\n\n*Construction*: Choose distinct primes $p_1, p_2, \\dots, p_{1819}$, and let $P = p_1p_2\\cdots p_{1819}$. Define\n\n$$\nS = \\{p_1, p_2, \\dots, p_{1819}, P, P \\cdot p_1, \\dots, P \\cdot p_{199}\\}.\n$$\n\nFor each $p_i$, there are $201$ numbers in $S$ divisible by $p_i$ (namely, $p_i$ and all multiples of $P$). Of these, at most one has two factors $p_i$; the rest have only one factor $p_i$. If we take $100$ numbers from $S$, their product has at most $101$ factors $p_i$. The other numbers contain at least $101$ numbers divisible by $p_i$, so their product has at least $101$ factors $p_i$. This holds for any $p_i$, and the numbers in $S$ do not have any other prime factors, so $S$ has the desired property.\n\n*Upper bound*: Suppose $S$ contains more than $1819$ primes. Consider a prime divisor $q$ of a number in $S$. If at most $199$ numbers in $S$ are divisible by $q$, then taking the $100$ elements with the most factors $q$ gives a contradiction to the condition. Thus, at least $200$ numbers in $S$ are divisible by $q$. If exactly $200$, then the number of factors $q$ in all these numbers must be equal, or else we again get a contradiction. Thus, $S$ contains at least $199$ non-primes, since a prime $p$ in $S$ divides at least $199$ other elements. If $S$ contains exactly $199$ non-primes, then each of these must be the product of the primes in $S$, but then these $199$ numbers are not distinct, a contradiction. Therefore, $S$ must contain at least $200$ non-primes, so at most $1819$ primes.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15156,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c > 0$ such that $a \\geq bc^2$, $b \\geq ca^2$, and $c \\geq ab^2$. Determine the maximum value of the expression\n$$\nE = abc(a - bc^2)(b - ca^2)(c - ab^2).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = a b$, $y = b c$, $z = c a$. We seek to find\n$$\n\\max (x - y^2)(y - z^2)(z - x^2).\n$$\nBy the AM-GM inequality, the product is at most\n$$\n\\left( \\frac{(x - x^2) + (y - y^2) + (z - z^2)}{3} \\right)^3.\n$$\nBut $x - x^2 \\leq \\frac{1}{4}$, and similarly for $y$ and $z$, so\n$$\n(x - y^2)(y - z^2)(z - x^2) \\leq \\frac{1}{4^3}.\n$$\nThis value, $\\frac{1}{4^3}$, is achieved for $x = y = z = \\frac{1}{2}$, that is, for $a = b = c = \\frac{1}{\\sqrt{2}}$. Thus, the maximum value is $\\frac{1}{4^3}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15157,
    "subject": "Mathematics (Olympiad)",
    "question": "We call any 3-element subset a *triple*; a triple is said to be *monochromatic* if all its three elements are the same color. Let $f(n)$ denote the minimal possible number of triples in a fine collection of $M_n$.\n\nLet $M_n$ be a set of $n$ elements. A collection of triples is called *fine* if, for any coloring of $M_n$ with two colors, there is always a monochromatic triple from the collection.\n\nDetermine $f(n)$ for all $n \\geq 5$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we have $f(5) = 10 = \\binom{5}{3}$ — the total number of triples in $M_5$.\n\nIndeed, if the collection does not contain some triple $\\{a, b, c\\}$, then we color $a, b, c$ black and the remaining 2 elements white. This coloring shows that the collection is not fine. On the other hand, the collection contains all possible triples. Since for any coloring with two colors there are at least three elements, say $x, y, z$, of the same color, then the triple $\\{x, y, z\\}$ is monochromatic, thus the collection is fine.\n\nFurther, $f(6) = 10 = \\frac{1}{2}\\binom{6}{3}$ — half the number of all triples in $M_6$.\n\nWe can divide all the triples in $M_6$ into 10 pairs so that in any pair the union of both triples is $M_6$ itself. If the collection contains fewer than 10 triples, then there is a pair $\\{a, b, c\\}$, $\\{d, e, f\\}$ ($\\{a, b, c, d, e, f\\} = M_6$) such that neither triple is in the collection. If $a, b, c$ are colored black and $d, e, f$ are colored white, then there is no monochromatic triple, so the collection is not fine. On the other hand, a collection of 10 triples with exactly one from each pair is fine.\n\nNow, let $n \\geq 7$. Consider the following collection of triples in $M_7$: $\\{1,2,3\\}$, $\\{1,4,5\\}$, $\\{1,6,7\\}$, $\\{2,4,6\\}$, $\\{2,5,7\\}$, $\\{3,4,7\\}$, $\\{3,5,6\\}$. Each element belongs to exactly three triples, and each pair appears in exactly one triple. For any coloring, there are at least 4 elements of the same color, say black. Consider a white element $x$. Let $x$ belong to the triples $\\{x,a,b\\}$, $\\{x,c,d\\}$, $\\{x,e,f\\}$. Suppose none of these triples is monochromatic (i.e., all are not all white). Then one of them, say $\\{x,e,f\\}$, contains black elements $e, f$; the other two contain (besides $x$) one black and one white element each.\n\nLet $a$ and $c$ be black, $b$ and $d$ white. If $f$ belongs to the triple $\\{f,a,c\\}$, then we have a monochromatic triple. If not, then $f$ (besides $\\{x,e,f\\}$) can only belong to $\\{f,a,d\\}$ or $\\{f,b,c\\}$. In this case, element $e$ (besides $\\{x,e,f\\}$) belongs to $\\{b,d,e\\}$ and $\\{a,c,e\\}$. The latter is monochromatic. Thus, the collection is fine. Hence $f(7) \\leq 7$.\n\nFurther, $f(n+1) \\leq f(n)$, because any fine collection of $M_n$ is also a fine collection of $M_{n+1}$. It follows that $f(n) \\leq 7$ for any $n \\geq 7$.\n\nLet's show that $f(n) = 7$ for any $n \\geq 7$. Suppose, for contradiction, that $f(n) \\leq 6$ for some $n \\geq 7$, and let $n$ be minimal with this property. Then there is a fine collection of $M_n$ containing $k \\leq 6$ triples. There are at most $18 = 3 \\cdot 6$ pairs of elements in the collection. But $n \\geq 7$, so there are at least $\\frac{7 \\cdot 6}{2} = 21$ possible pairs in $M_n$. Since $21 > 18$, there are two elements $a, b \\in M_n$ such that the pair $\\{a, b\\}$ does not belong to any triple in the collection. Suppose $\\{a, b\\} = \\{n-1, n\\}$. Now identify $a$ and $b$, and also identify all pairs of triples of the form $\\{a, x, y\\}$, $\\{b, x, y\\}$, if any. The resulting collection is fine and contains at most $k$ triples, giving a fine collection for $M_{n-1}$, contradicting the minimality of $n$. Thus, the proof is finished.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15158,
    "subject": "Mathematics (Olympiad)",
    "question": "有一個無限大的方格棋盤，每個格子裡面放一個正整數。任何一個長方形的內部總和都不是質數，而且至少有一格放的是 $1$。求所有格子中的最大數至少是多少？",
    "options": [],
    "answer": "See solution",
    "solution": "答案是 $9$。\n\n注意到 $1$ 旁邊可以放的最小數是 $8$，但 $8 + 1 + 8 = 17$ 是質數，所以格子裡一定要有 $9$。\n\n<table><tr><td>4</td><td>8</td><td>6</td></tr><tr><td>8</td><td>1</td><td>9</td></tr><tr><td>6</td><td>9</td><td>9</td></tr></table>\n\n構造是在 $1$ 的周圍格利用模 $2$ 跟模 $3$，然後除了這九格以外都放 $6$，如此一來任何一個不只一個數的矩形內部的和都是 $2$ 或 $3$ 的倍數。\n\n**Remark.** 這題構造看似很勉強，但是如果想到模 $2$ 和模 $3$ 的話馬上就可以畫出來。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15159,
    "subject": "Mathematics (Olympiad)",
    "question": "證明：對於 $\\{1, 2, 3, \\ldots, 5^{505}\\}$ 的任一個恰有 $2022$ 個元素的子集 $A$，必存在三個元素 $a, b, c$ 滿足 $a < b < c$ 且 $c + 2a > 3b$。",
    "options": [],
    "answer": "See solution",
    "solution": "假設存在 2022 個正整數 $x_0 < x_1 < \\cdots < x_{2021}$ 使得對所有 $a < b < c$，都有 $c + 2a \\le 3b$。則對所有 $i = 0, \\ldots, 2020$，有\n\n$$\nx_{i+2} + 2x_i \\le 3x_{i+1}\n$$\n\n特別地，對 $x_{2021}$ 和 $x_i$，有 $x_{2021} + 2x_i \\le 3x_{i+1}$，即\n\n$$\nx_{2021} - x_i \\ge \\frac{3}{2}(x_{2021} - x_{i+1})\n$$\n\n用歸納法可得：\n\n$$\nx_{2021} - x_i \\ge \\left(\\frac{3}{2}\\right)^{2020-i} (x_{2021} - x_{2020})\n$$\n\n取 $i = 0$，得到：\n\n$$\nx_{2021} - x_0 \\ge \\left(\\frac{3}{2}\\right)^{2020} (x_{2021} - x_{2020}) = \\left(\\frac{81}{16}\\right)^{505} (x_{2021} - x_{2020}) > 5^{505}\n$$\n\n但 $x_{2021} \\le 5^{505}$，矛盾。因此必存在三個元素 $a < b < c$ 使 $c + 2a > 3b$。$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15160,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the $c$-transform sequences from 2015 for $c = 1$ to $9$:\n\n- 1-transform: $2015, 206, 26, 8, 8$\n- 3-transform: $2015, 216, 39, 30, 3, 9, 27, 23, 11, 4, 12, 7, 21, 5, 15, 16, 19, 28, 26, 20, 2, 6, 18, 25, 17, 22, 8, 24, 14, 13, 10, 1, 3$\n- 4-transform: $2015, 221, 26, 26$\n- 5-transform: $2015, 226, 52, 15, 26, 32, 13, 16, 31, 8, 40, 4, 20, 2, 10, 1, 5, 25, 27, 37, 38, 43, 19, 46, 34, 23, 17, 36, 33, 18, 41, 9, 45, 29, 47, 39, 48, 44, 24, 22, 12, 11, 6, 30, 3, 15$\n- 6-transform: $2015, 231, 29, 56, 41, 10, 1, 6, 36, 39, 57, 47, 46, 40, 4, 24, 26, 38, 51, 11, 7, 42, 16, 37, 45, 34, 27, 44, 28, 50, 5, 30, 3, 18, 49, 58, 53, 23, 20, 2, 12, 13, 19, 55, 35, 33, 21, 8, 48, 52, 17, 43, 22, 14, 25, 32, 15, 31, 9, 54, 29$\n- 7-transform: $2015, 236, 65, 41, 11, 8, 56, 47, 53, 26, 44, 32, 17, 50, 5, 35, 38, 59, 68, 62, 20, 2, 14, 29, 65$\n- 8-transform: $2015, 241, 32, 19, 73, 31, 11, 9, 72, 23, 26, 50, 5, 40, 4, 32$\n- 9-transform: $2015, 246, 78, 79, 88, 80, 8, 72, 25, 47, 67, 69, 87, 71, 16, 55, 50, 5, 45, 49, 85, 53, 32, 21, 11, 10, 1, 9, 81, 17, 64, 42, 22, 20, 2, 18, 73, 34, 39, 84, 44, 40, 4, 36, 57, 68, 78$\n\nFrom Part a, 2015 is not 2-tu.\n\nDetermine for which $c$ the number 2015 is $c$-tu.",
    "options": [],
    "answer": "See solution",
    "solution": "From Part **b**, $10c - 1$ is the only $c$-terminator for $c = 1, 2, 3, 5, 6, 8, 9$. So for these values of $c$ we can stop calculating the $c$-transform sequence from 2015 once we reach a number that is less than $10c - 1$ because then it will never reach $10c - 1$.\n\nThe 1-transform sequence from 2015 is: $2015, 206, 26, 8, 8$.\n\nFrom Part **a**, 2015 is not 2-tu.\n\nThe 3-transform sequence from 2015 starts with: $2015, 216, 39, 30, 3$. So 29 will never be reached.\n\nThe 4-transform sequence from 2015 is: $2015, 221, 26, 26$.\n\nThe 5-transform sequence from 2015 starts with: $2015, 226, 52, 15$. So 49 will never be reached.\n\nThe 6-transform sequence from 2015 starts with: $2015, 231, 29$. So 59 will never be reached.\n\nThe 7-transform sequence from 2015 is: $2015, 236, 65, 41, 11, 8, 56, 47, 53, 26, 44, 32, 17, 50, 5, 35, 38, 59, 68, 62, 20, 2, 14, 29, 65$.\n\nThe 8-transform sequence from 2015 starts with: $2015, 241, 32$. So 79 will never be reached.\n\nThe 9-transform sequence from 2015 starts with: $2015, 246, 78$. So 89 will never be reached.\n\nThus 2015 is only 1-tu and 4-tu.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15161,
    "subject": "Mathematics (Olympiad)",
    "question": "證明存在常數 $C > 0$，使得對於所有滿足 $a + b$ 為整數的正實數 $a$ 與 $b$，有\n\n$$\n\\{a^3\\} + \\{b^3\\} \\le 2 - \\frac{C}{(a+b)^6}.\n$$\n\n此處 $\\{x\\} = x - \\lfloor x \\rfloor$ 表示 $x$ 的小數部分。",
    "options": [],
    "answer": "See solution",
    "solution": "以下證明當 $C = \\frac{1}{54}$ 時命題成立：\n\n令 $n = a + b$，且不妨假設 $a \\ge b$。注意到函數 $f(x) = x^3$ 是遞增且凸的，所以若 $a_0 + b_0 = n$ 且 $a_0 - b_0 \\ge a - b$，有\n\n$$\na^3 + b^3 \\le a_0^3 + b_0^3. \\quad (1)\n$$\n\n特別地，當 $\\lfloor a^3 \\rfloor = \\lfloor a_0^3 \\rfloor$ 且 $\\lfloor b^3 \\rfloor = \\lfloor b_0^3 \\rfloor$ 時，\n\n$$\n\\{a^3\\} + \\{b^3\\} \\le \\{a_0^3\\} + \\{b_0^3\\}.\n$$\n\n所以只需考慮 $\\{a^3\\} \\to 1^-$ 或 $\\{b^3\\} = 0$ 兩種情形。後者顯然：\n\n$$\n\\{a^3\\} + \\{b^3\\} < 1 < 2 - \\frac{C}{n^6}.\n$$\n\n對於前者，假設 $(a + \\varepsilon)^3 = k < n^3$ 為正整數，此時 $b = n - \\sqrt[3]{k} + \\varepsilon$。由於 $\\varepsilon$ 足夠小，根據 (1)，\n\n$$\n\\begin{aligned}\n\\{a^3\\} + \\{b^3\\} &= a^3 + b^3 - \\lfloor a^3 \\rfloor - \\lfloor b^3 \\rfloor \\\\\n&\\le k + (n - \\sqrt[3]{k})^3 - (k - 1) - \\lfloor (n - \\sqrt[3]{k})^3 \\rfloor \\\\\n&= 1 + \\{(n - \\sqrt[3]{k})^3\\}.\n\\end{aligned}\n$$\n\n所以只需證明\n\n$$\n1 + \\{(n - \\sqrt[3]{k})^3\\} \\le 2 - \\frac{C}{n^6}.\n$$\n\n若 $(n - \\sqrt[3]{k})^3$ 為整數，則命題顯然。否則，上式等價於\n\n$$\n\\frac{C}{n^6} \\le 1 - \\{(n - \\sqrt[3]{k})^3\\} = \\{-(n - \\sqrt[3]{k})^3\\} = 3n^2\\sqrt[3]{k} - 3n\\sqrt[3]{k^2} - m,\n$$\n\n其中 $m = \\lfloor 3n^2\\sqrt[3]{k} - 3n\\sqrt[3]{k^2} \\rfloor \\ge 0$。考慮以下等式：\n\n$$\n(x + y + z)(x^2 + y^2 + z^2 - yz - zx - xy) = x^3 + y^3 + z^3 - 3xyz,\n$$\n\n將 $x = 3n^2\\sqrt[3]{k}$，$y = -3n\\sqrt[3]{k^2}$，$z = -m$ 代入可得\n\n$$\n0 < (1 - \\{b^3\\})(x^2 + y^2 + z^2 - yz - zx - xy) = 27n^3k(n^3 - k - m) - m^3.\n$$\n\n由於右式為整數，左式至少為 1。因此，由 $\\sqrt[3]{k} < n$ 及 $0 \\le m \\le 3n^2\\sqrt[3]{k} < 3n^3$，\n\n$$\n\\begin{aligned}\n1 - \\{b^3\\} &\\ge \\frac{1}{x^2 + y^2 + z^2 - yz - zx - xy} \\ge \\frac{1}{2(x^2 + y^2 + z^2)} \\\\\n&\\ge \\frac{1}{2(9n^6 + 9n^6 + 9n^6)} = \\frac{1}{54n^6},\n\\end{aligned}\n$$\n\n得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15162,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for all $x, y \\in \\mathbb{R}$,\n$$\nf(x)^2 + 2yf(x) = f(y + f(x)) - f(y).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since the left-hand side runs through $\\mathbb{R}$ when $y$ runs through $\\mathbb{R}$, the right-hand side runs through $\\mathbb{R}$. Hence, $f(x) - f(y)$ runs all real numbers when $x$ and $y$ run through $\\mathbb{R}$.\n\nTherefore, according to (2), $f(x) = x^2 + c$ for all $x$. This meets the given equation for any constant $c$.\n\nTherefore, $f(x) = x^2 + c$ ($c$ any constant) or $f(x) = 0$ (for all $x$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15163,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer $k \\ge 2$ such that for all $x, y, z \\ne 1$ satisfying $x + y + z = 3$ and $xyz = 1$, the $k$-th roots of the numbers $|x|$, $|y|$, and $|z|$ are the side lengths of some triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $x = \\max\\{x, y, z\\}$. If $x, y, z > 0$, then by the AM-GM inequality, $x + y + z \\ge 3\\sqrt[3]{xyz} = 3$, with equality only if $x = y = z = 1$, which is not allowed. Next, consider $y, z < 0$ and set $u = -y > 0$, $v = -z > 0$. The conditions become $x = 3 + u + v$ and $xuv = 1$. Thus, $x$ is always greater than the sum of the other two numbers, so we need to find $k$ such that\n\n$$\n\\sqrt[k]{x} < \\sqrt[k]{u} + \\sqrt[k]{v}.\n$$\n\nLet $u = v$. Then $(x, u, v) = (4, \\frac{1}{2}, \\frac{1}{2})$. Substituting into the inequality:\n\n$$\n\\sqrt[6]{4} < \\frac{2}{\\sqrt[6]{2}} \\implies \\sqrt[6]{8} < 2.\n$$\n\nThis gives $k > 3$, so $k \\ge 4$. Now, we show $k = 4$ works:\n\n$$\n\\sqrt[4]{x} < \\sqrt[4]{u} + \\sqrt[4]{v} \\iff \\sqrt{x} < \\sqrt{u} + \\sqrt{v} + 2\\sqrt[4]{uv} = \\sqrt{u} + \\sqrt{v} + \\frac{2}{\\sqrt[4]{x}}\n$$\n\nWe prove the stronger inequality:\n\n$$\n\\sqrt{x} < \\sqrt{u} + \\sqrt{v} + \\frac{2}{\\sqrt{x}} \\iff \\sqrt{x} - \\frac{2}{\\sqrt{x}} < \\sqrt{u} + \\sqrt{v}.\n$$\n\nSince $x \\ge 4$, $\\sqrt{x} - \\frac{2}{\\sqrt{x}} > 0$. Squaring both sides:\n\n$$\nx + \\frac{4}{x} - 4 < u + v + 2\\sqrt{uv} \\iff \\frac{4}{x} < 1 + \\frac{2}{\\sqrt{x}}.\n$$\n\nThis holds for $x > 3$. Therefore, $k_{\\min} = 4$.\n\n*Remark.* If $x = \\max\\{x, y, z\\}$, then $x \\ge 4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15164,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure below, the edge length of cube $ABCD$–$EFGH$ is $2$. Take any point $P_1$ on the incircle of square $ABFE$, any point $P_2$ on the incircle of square $BCGF$, and any point $P_3$ on the incircle of square $EFGH$. Find the minimum and maximum values of $|P_1P_2| + |P_2P_3| + |P_3P_1|$.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p43_data_fff8189147.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Establish a Cartesian coordinate system in three-dimensional space, taking the centre of the cube as the origin. Let the directions of $\\vec{DA}$, $\\vec{DC}$, and $\\vec{DH}$ be the positive $x$-, $y$-, and $z$-axes, respectively.\n\nAccording to the conditions, set:\n\n$$\nP_1(1, \\cos \\alpha_1, \\sin \\alpha_1), \\quad P_2(\\sin \\alpha_2, 1, \\cos \\alpha_2), \\quad P_3(\\cos \\alpha_3, \\sin \\alpha_3, 1)\n$$\n\nLet $P_4 = P_1$ and $\\alpha_4 = \\alpha_1$. Denote:\n\n$$\nd_i = |P_i P_{i+1}| \\quad (i = 1, 2, 3)\n$$\n\nThen:\n\n$$\nd_i^2 = (1 - \\sin \\alpha_{i+1})^2 + (1 - \\cos \\alpha_i)^2 + (\\sin \\alpha_i - \\cos \\alpha_{i+1})^2\n$$\n\nLet $f = |P_1P_2| + |P_2P_3| + |P_3P_1|$.\n\n**Minimum of $f$:**\n\nFor $i = 1, 2, 3$, by the inequality of arithmetic and geometric means:\n\n$$\nd_i^2 \\geq (1 - \\sin \\alpha_{i+1})^2 + (1 - \\cos \\alpha_i)^2 \\geq \\frac{1}{2}((1 - \\sin \\alpha_{i+1}) + (1 - \\cos \\alpha_i))^2\n$$\n\nThus,\n\n$$\nd_i \\geq \\frac{\\sqrt{2}}{2}(2 - \\sin \\alpha_{i+1} - \\cos \\alpha_i)\n$$\n\nTherefore,\n\n$$\nf = d_1 + d_2 + d_3 \\geq \\frac{\\sqrt{2}}{2} \\sum_{i=1}^{3} (2 - \\sin \\alpha_{i+1} - \\cos \\alpha_i) = 3\\sqrt{2} - \\frac{\\sqrt{2}}{2} \\sum_{i=1}^{3} (\\sin \\alpha_i + \\cos \\alpha_i)\n$$\n\n$$\n= 3\\sqrt{2} - \\sum_{i=1}^{3} \\sin \\left(\\alpha_i + \\frac{\\pi}{4}\\right) \\geq 3\\sqrt{2} - 3\n$$\n\nWhen $\\alpha_i = \\frac{\\pi}{4}$ ($i = 1, 2, 3$), $f$ attains its minimum $3\\sqrt{2} - 3$.\n\n**Maximum of $f$:**\n\nBy the previous formula:\n\n$$\nd_i^2 = 4 - 2 \\cos \\alpha_i - 2 \\sin \\alpha_{i+1} - 2 \\sin \\alpha_i \\cos \\alpha_{i+1}\n$$\n\nSince $\\sin \\alpha_i \\geq -1$, $\\cos \\alpha_i \\geq -1$ ($i = 1, 2, 3$),\n\n$$\n\\sum_{i=1}^{3} d_i^2 = 12 - 2 \\left( \\sum_{i=1}^{3} \\sin \\alpha_{i+1} + \\sum_{i=1}^{3} \\cos \\alpha_i + \\sum_{i=1}^{3} \\sin \\alpha_i \\cos \\alpha_{i+1} \\right)\n$$\n\n$$\n= 12 - 2 \\left( \\sum_{i=1}^{3} \\sin \\alpha_i + \\sum_{i=1}^{3} \\cos \\alpha_{i+1} + \\sum_{i=1}^{3} \\sin \\alpha_i \\cos \\alpha_{i+1} \\right)\n$$\n\n$$\n= 18 - 2 \\sum_{i=1}^{3} (1 + \\sin \\alpha_i)(1 + \\cos \\alpha_{i+1}) \\leq 18\n$$\n\nBy the Cauchy-Schwarz inequality, $f^2 \\leq 3(d_1^2 + d_2^2 + d_3^2) = 54$, so $f \\leq 3\\sqrt{6}$.\n\nWhen $\\alpha_i = \\pi$ ($i = 1, 2, 3$), $f$ attains its maximum $3\\sqrt{6}$.\n\n**Conclusion:**\n\nThe minimum value of $f$ is $3\\sqrt{2} - 3$, and the maximum value is $3\\sqrt{6}$.\n\n$\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15165,
    "subject": "Mathematics (Olympiad)",
    "question": "A social network has 2025 users. Two different users are either friends or not friends. A user is considered *lonely* if they have no friends. Initially, there are no lonely users, and two users, Alice and Bob, are not friends. A user may swap all their friends, meaning if they were friends with someone before the swap, they are no longer friends, and vice versa.\n\nMust there exist a way to arrange the 2025 users in a sequence so that if, one-by-one in that sequence, each user swaps their friends, no user is ever lonely at any point during the process?",
    "options": [],
    "answer": "See solution",
    "solution": "Let a longest path of the graph be $P = V_1V_2 \\dots V_k$.\n\n**Case 1:** $P$ consists of all vertices and there is no edge between $V_1$ and $V_k$.\n\nSwap vertices $V_i$ increasing $i$ from $1$ to $k$. $V_1$ and $V_k$ will never be lonely, as they will connect after the first swap and disconnect after the last swap. Vertex $V_i$ for $2 \\le i \\le k-1$ will never be lonely as it will be connected to $V_{i+1}$ before it is swapped and to $V_{i-1}$ after it is swapped (because $V_{i-1}$ was swapped before).\n\n**Case 2:** $P$ consists of all vertices and there is an edge between $V_1$ and $V_k$.\n\nThus $V_1V_2 \\dots V_kV_1$ is a cycle that contains all vertices. As the graph is not complete, there exist two vertices which are not connected, $V_a$ and $V_b$ ($a < b$). Swap $V_a$, then $V_i$ for $a+1 \\le i \\le b-1$ (one path from $V_a$ to $V_b$ along the cycle), then $V_i$ for $a-1, a-2, \\dots, 1, n, n-1, \\dots, b+1, b$ (the other path along the cycle). $V_a$ and $V_b$ will be connected during the procedure. For other vertices, the same argument as in **Case 1** holds (they are always connected to at least one of the two neighbours on the cycle).\n\n**Case 3:** $P$ does not contain all vertices.\n\nNotice that $P$ must contain at least 3 vertices, because otherwise every vertex would have a degree of 1, which is impossible because the number of vertices is odd.\n\nAny swap ordering in which we swap first $V_1$, last $V_k$, other vertices $V_i$ in increasing order doesn't cause any vertices besides possibly $V_1$ and $V_k$ to be lonely. For internal vertices $V_i$ the same argument from **Case 1** holds. Vertices outside of $P$ are not connected to $V_1$ and $V_k$ (otherwise, a longer path would exist). As we swap $V_1$ first and $V_k$ last, they will be connected to $V_1$ before their swap and to $V_k$ after.\n\nIf $P$ contains at least 4 vertices, then if we swap $V_1$, then $V_2$, then vertices outside of $P$, then remaining vertices of $P$, we notice that $V_1$ and $V_k$ will always be connected to some vertex outside $P$ or to their respective neighbors on $P$.\n\nIf $P$ contains 3 vertices, then we swap $V_1$, then one of the outside vertices, then $V_2$, then the remaining outside vertices, then $V_k$. Similarly to above, $V_1$ and $V_k$ will always be connected either to their neighbour in $P$ or to some outside vertex.\n\n**Comment:** The PSC found a few inductive solutions, but all of them require considering some special cases. The solution with the fewest cases is as follows: if there are two vertices which are not connected, say $A$ and $B$, such that one of them ($A$) does not have a neighbour with degree one, and the remaining vertices ($n-2$ of them) do not form a complete graph, we are done by induction (we can swap $A$, then apply induction on the remaining $n-2$ vertices, and then swap $B$). If we can't choose $A$ and $B$ in this way, the graph has a very special structure, and we can solve those cases separately.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15166,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any positive integer $n$ there exist positive integers $x_1, x_2, \\dots, x_{2013}$ such that:\n\n1. The number $x_1^2 + x_2^2 + \\dots + x_k^2$ is a perfect square for any positive integer $k \\le 2013$.\n2. If the number $x_1^2 + x_2^2 + \\dots + x_k^2 + y^2$ is a perfect square for some positive integers $k < 2013$ and $y$, then $y \\le x_{k+1}$.\n3. The number $x_2^2 + x_3^2 + \\dots + x_{2013}^2$ is divisible by $4^n + 9^n$.\n4. The numbers $x_{2013}$ and $4^n + 9^n$ are co-prime.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = 4^n + 9^n$ and define $y_k^2 = x_1^2 + x_2^2 + \\dots + x_k^2$ with $y_k > 0$ for $k = 1, 2, \\dots, 2013$.\n\nTake any odd positive integer as $x_1 = y_1$ and recursively define\n\n$$\nx_{k+1} = \\frac{y_k^2 - 1}{2}, \\quad y_{k+1} = \\frac{y_k^2 + 1}{2}\n$$\nfor $k = 1, 2, \\dots, 2012$. Since $y_1$ is odd, $y_1^2 \\equiv 1 \\pmod{4}$, so $y_2$ is an integer, and similarly for all $y_k$ and $x_k$.\n\nWe have\n$$\nx_{k+1}^2 + y_k^2 = \\left(\\frac{y_k^2 - 1}{2}\\right)^2 + y_k^2 = \\left(\\frac{y_k^2 + 1}{2}\\right)^2 = y_{k+1}^2\n$$\nso\n$$\ny_{k+1}^2 = y_k^2 + x_{k+1}^2 = x_1^2 + x_2^2 + \\dots + x_{k+1}^2\n$$\nwhich ensures (1).\n\nFor (2), if $x_1^2 + \\dots + x_k^2 + y^2 = y_k^2 + y^2 = z^2$ for some $k < 2013$ and $y$, then $y_k^2 = (z-y)(z+y) = d_1 d_2$. Thus,\n$$\ny = \\frac{d_2 - d_1}{2} \\le \\frac{y_k^2 - 1}{2} = x_{k+1}\n$$\nso (2) holds.\n\nTo ensure (3) and (4), choose $x_1 = y_1$ so that $b^2 + 1$ is divisible by $a$ for some $b$ (e.g., $b \\equiv 2^n 3^{-n} \\pmod{a}$). Take $y_1 = 2b - 1$. Then\n$$\ny_2 = \\frac{y_1^2 + 1}{2} \\equiv -1 - 2b \\pmod{a}\n$$\n$$\ny_3 = \\frac{y_2^2 + 1}{2} \\equiv y_1 \\pmod{a}\n$$\nso $y_1 \\equiv y_3 \\equiv y_5 \\equiv \\dots \\equiv y_{2013} \\pmod{a}$. Thus,\n$$\nx_2^2 + x_3^2 + \\dots + x_{2013}^2 = y_{2013}^2 - y_1^2\n$$\nis divisible by $a$, so (3) holds.\n\nFinally, $x_{2013}^2 = y_{2013}^2 - y_{2012}^2 \\equiv -8b \\pmod{a}$, and since $a$ and $8b$ are co-prime, (4) holds.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15167,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence of positive integers is called *complete* if any positive integer has a multiple in the sequence. Prove that an arithmetic sequence of positive integers is complete if and only if its difference divides the first term.",
    "options": [],
    "answer": "See solution",
    "solution": "If the difference $r$ divides $a_1$, then $a_1 = dr$ for some $d \\in \\mathbb{N}$, and $a_n = (d + n - 1)r$. A multiple of any positive integer $k$ is obtained when $d + n - 1$ is a multiple of $k$.\n\nFor the converse, observe first that if $r = 0$, the sequence is not complete. Because $r \\neq 0$ and by assumption there is a multiple of $r$ of the form $a_1 + (n-1)r$, with $n \\in \\mathbb{N}^*$, we conclude $r \\mid a_1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15168,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer and let $a_1, a_2, \\dots, a_n$ be $n$ positive real numbers. Prove that the function $f : [0, \\infty) \\to \\mathbb{R}$ defined by\n$$\nf(x) = \\frac{a_1 + x}{a_2 + x} + \\frac{a_2 + x}{a_3 + x} + \\dots + \\frac{a_{n-1} + x}{a_n + x} + \\frac{a_n + x}{a_1 + x}\n$$\nis a decreasing function.",
    "options": [],
    "answer": "See solution",
    "solution": "Set $a_{n+1} = a_1$ and let $0 \\le x \\le y$. Since\n$$\nf(y) - f(x) = (y - x) \\sum_{i=1}^{n} \\frac{a_{i+1} - a_i}{(a_{i+1} + x)(a_{i+1} + y)}\n$$\nshowing $f$ is decreasing amounts to showing\n$$\n\\sum_{i=1}^{n} \\frac{a_{i+1}}{(a_{i+1} + x)(a_{i+1} + y)} \\le \\sum_{i=1}^{n} \\frac{a_i}{(a_{i+1} + x)(a_{i+1} + y)}\n$$\nNoticing that $a_i \\le a_j$ if and only if $(a_i + x)^{-1}(a_i + y)^{-1} \\ge (a_j + x)^{-1}(a_j + y)^{-1}$, the above inequality is a straightforward consequence of the rearrangement inequality for the $a_i$ and the $(a_i + x)^{-1}(a_i + y)^{-1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15169,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $R$ 表示所有實數所成的集合。試求所有的函數 $f: R \\to R$ 使得對所有的實數 $x, y$，\n\n$$\nf(xf(y) - f(x) - y) = y f(x) - f(y) - x\n$$\n皆成立。",
    "options": [],
    "answer": "See solution",
    "solution": "答：$f(x) = x$，對所有的實數 $x$。\n\n首先，令 $P(x, y)$ 表示此函數方程的斷言。則\n\n$$\nP(x, 0) \\implies f(x f(0) - f(x)) = -x - f(0).\n$$\n\n由此可知 $f$ 為滿射且 $f(f(x)) = x$。注意到\n\n$$\nP(1, 1) \\implies f(-1) = -1，\\quad P(0, -1) \\implies f(1 - f(0)) = 1 - f(0).\n$$\n\n考慮 $x = 0, y = 1 - f(0)$：\n\n$$\n-1 = f(-1) = (1 - f(0)) f(0) - f(1 - f(0)) = - (1 - f(0))^2 \\implies f(0) = 0 \\text{ 或 } 2.\n$$\n\n若 $f(0) = 2$，則 $P(x, 0)$ 變為\n\n$$\nf(2x - f(x)) = -x - 2 = f(f(-x - 2))\n$$\n\n由單射性，得到以下方程組：\n\n$$\n\\begin{cases}\n2x - f(x) = f(-x - 2) \\\\\n2(-x - 2) - f(-x - 2) = f(x)\n\\end{cases}\n$$\n\n這顯然矛盾。因此 $f(0) = 0$ 且 $f(-x) = -f(x)$。特別地，$f(1) = 1$。由於 $f$ 為奇函數，\n\n$$\nP(x, f(-x)) \\implies f(-x^2) = -f(x)^2 \\implies f(x)^2 = f(x^2).\n$$\n\n最後，根據下式：\n\n$$\nP(x, 1) \\implies f(x - f(x) - 1) = f(x) - x - 1.\n$$\n\n設 $x - f(x) = \\alpha$，對上式遞推，得\n\n$$\nf(\\alpha 2^n - 1) = -\\alpha 2^n - 1，\\quad \\forall n \\in \\{0\\} \\cup \\{\\text{正整數}\\}.\n$$\n\n顯然，若 $\\alpha \\neq 0$ 則與 $f(x^2) = f(x)^2 \\geq 0$ 矛盾，因此\n\n$$\nf(x) = x, \\quad \\forall x \\in \\mathbb{R}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15170,
    "subject": "Mathematics (Olympiad)",
    "question": "Polynomials $P(x)$ and $Q(x)$ with rational coefficients are each sums of three squares of polynomials with rational coefficients. Show that the polynomial $P(x) \\cdot Q(x)$ is a sum of four squares of polynomials with rational coefficients.",
    "options": [],
    "answer": "See solution",
    "solution": "This follows from the identity:\n\n$$\n(a^2 + b^2 + c^2)(x^2 + y^2 + z^2) = (ax + by + cz)^2 + (ay - bx)^2 + (az - cx)^2 + (bz - cy)^2.\n$$\n\nThus, the product of two sums of three squares is a sum of four squares.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15171,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive odd integer $n$, show that the arithmetic mean of the fractional parts $\\left\\{\\frac{k^{2n}}{p}\\right\\}$, for $k = 1, \\dots, \\frac{p-1}{2}$, is the same for infinitely many primes $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that the arithmetic mean in question is $\\frac{1}{2}$ for infinitely many primes congruent to $1$ modulo $4$.\n\nNotice that $\\left\\{\\frac{k^{2n}}{p}\\right\\} = \\frac{r_k}{p}$, where $r_k$ is the remainder of $k^{2n}$ upon division by $p$. Clearly, the $r_k$ are quadratic residues modulo $p$.\n\nIf $p$ is prime and $p-1$ and $n$ are relatively prime, then the $r_k$ for $k = 1, \\dots, \\frac{p-1}{2}$ are pairwise distinct, since the $k^{2n}$ are pairwise distinct modulo $p$ by Fermat's little theorem. In this case, the $r_k$ form the set $R$ of all $\\frac{p-1}{2}$ quadratic residues modulo $p$ in the range $1$ through $p-1$.\n\nIf, in addition, $p \\equiv 1 \\pmod{4}$, then $-1$ is a quadratic residue modulo $p$, and the assignment $r \\mapsto p - r$, $r \\in R$, defines a permutation of $R$. In this case,\n\n$$\n\\sum_{r \\in R} r = \\sum_{r \\in R} (p - r) = \\frac{p(p-1)}{2} - \\sum_{r \\in R} r,\n$$\nso\n$$\n\\sum_{r \\in R} r = \\frac{p(p-1)}{4},\n$$\nand the arithmetic mean in question is $\\frac{1}{2}$.\n\nFinally, since $n$ is odd, infinitely many primes congruent to $1$ modulo $4$ are also congruent to $2$ modulo $n$, by Dirichlet's theorem on arithmetic sequences; for such a prime $p$, the numbers $p-1$ and $n$ are relatively prime. This completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15172,
    "subject": "Mathematics (Olympiad)",
    "question": "Lemma 1: For $P$ on circle $C$, we call the arc $\\widehat{APB}$ “an arc of $P$” if $P$ is the midpoint of arc $\\widehat{APB}$, and $\\angle AOB = \\frac{4\\pi}{7}$. Now, for every given $n$ points on circle $C$, there is a point $P$ such that there are $\\lfloor \\frac{n+5}{6} \\rfloor$ points of the given $n$ points on the “arc of $P$”.\n\nLemma 2: Take the arc $\\widehat{A_1BA_6}$ arbitrarily on the circle $C$ with radius $10$, where $\\widehat{A_1BA_6}$ is $\\frac{5}{7}$ of the perimeter. Then, take any $5m + r$ points on the arc $\\widehat{A_1BA_6}$ ($m, r$ are non-negative integers and $0 \\le r < 5$). Prove that the number of lines from the given points whose lengths are more than $9$ is at most\n\n$$\n10m^2 + 4rm + \\frac{1}{2}r(r-1).\n$$\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p97_data_fed1f2a9b9.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Divide $\\overrightarrow{A_1BA_6}$ into five equal parts, with corresponding points $A_2, A_3, A_4, A_5$ (see the figure). The length of $\\overrightarrow{A_iA_{i+1}}$ is exactly $\\frac{1}{7}$ of the perimeter for $i = 1, 2, 3, 4, 5$, and the distance between any two points is not more than $a_7 < 9$. Suppose there are $m_i$ given points on the arc $\\overrightarrow{A_iA_{i+1}}$, then the number of lines from the given points whose lengths are more than $9$ is at most\n\n$$\nl = \\sum_{1 \\le i < j \\le 5} m_i m_j $$\n\nwhere\n\n$$\nm_1 + m_2 + m_3 + m_4 + m_5 = 5m + r. $$\n\nSince there are finitely many non-negative integer groups $(m_1, m_2, m_3, m_4, m_5)$, the maximum value of $l$ exists. Now, we prove that when the maximum is attained, the inequality\n\n$$\n|m_i - m_j| \\le 1 \\quad (1 \\le i < j \\le 5)\n$$\nmust hold.\n\nIf there exist $i, j$ ($1 \\le i < j \\le 5$) such that $|m_i - m_j| \\ge 2$ when the maximum is attained, suppose $m_1 - m_2 \\ge 2$. Then let\n\n$$\nm_1' = m_1 - 1, \\quad m_2' = m_2 + 1,\n$$\nand the corresponding integer is $l'$. We have\n\n$$\n\\begin{align*}\nm_1' + m_2' &= m_1 + m_2, \\\\\nm_1' + m_2' + m_3' + m_4' + m_5' &= m_1 + m_2 + m_3 + m_4 + m_5, \\\\\nl' - l &= (m_1'm_2' - m_1m_2) + [(m_1' + m_2') - (m_1 + m_2)](m_3 + m_4 + m_5) \\\\\n&= m_1 - m_2 - 1 \\ge 1.\n\\end{align*}\n$$\n\nContradiction!\n\nTherefore, when $l$ reaches the maximum value, the number of $m+1$ is $r$ and the number of $m$ is $5 - r$. Thus, the number of lines from the given points whose lengths are more than $9$ is at most\n\n$$\n\\begin{aligned}\n& \\binom{r}{2}(m+1)^2 + \\binom{r}{1}(5-r)(m+1)m + \\binom{5-r}{2}m^2 \\\\\n&= 10m^2 + 4rm + \\frac{1}{2}r(r-1).\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15173,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y, z$ be positive real numbers satisfying $x + y + z + xyz = 4$. Show that\n\n$$\n\\frac{x}{\\sqrt{2y+3z}} + \\frac{y}{\\sqrt{2z+3x}} + \\frac{z}{\\sqrt{2x+3y}} \\ge \\frac{1}{\\sqrt{5}}(x+y+z).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that the function $x \\mapsto x^{-\\frac{1}{2}}$ is convex on $(0, \\infty)$. Thus, by Jensen's inequality, we have\n\n$$\n\\frac{x}{x+y+z} (2y+3z)^{-\\frac{1}{2}} + \\frac{y}{x+y+z} (2z+3x)^{-\\frac{1}{2}} + \\frac{z}{x+y+z} (2x+3y)^{-\\frac{1}{2}} \n\\ge \\left( \\frac{x(2y+3z) + y(2z+3x) + z(2x+3y)}{x+y+z} \\right)^{-\\frac{1}{2}} \n= \\frac{1}{\\sqrt{5}} \\cdot \\sqrt{\\frac{x+y+z}{xy+yz+zx}}\n$$\n\nHence, it suffices to show that\n\n$$\nx+y+z \\ge xy+yz+zx. \\tag{1}\n$$\n\nWithout loss of generality, assume $x \\le y \\le z$. From the condition $x+y+z+xyz=4$, we see that $x \\le 1 \\le z$. Thus $(z-1)(1-x) = x+z-1-zx \\ge 0$. It follows that\n\n$$\n(1+zx)(x+y+z-zx) = (x+y+z+xyz) + zx(x+z-1-zx) \\ge 4. \\tag{2}\n$$\n\nOn the other hand, by the AM-GM inequality,\n\n$$\n4 \\ge (4-(x+z))(x+z) = (y+xyz)(x+z) = (1+zx)(xy+yz). \\tag{3}\n$$\n\nCombining (2) and (3), we get (1), since $1+zx > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15174,
    "subject": "Mathematics (Olympiad)",
    "question": "Each unit square in a $5 \\times 5$ table is coloured either blue or yellow. Prove that there exists a rectangle with sides parallel to the edges of the table, such that the four unit squares in its corners have the same colour.",
    "options": [],
    "answer": "See solution",
    "solution": "Each row contains at least 3 squares with the same colour. Similarly, the dominating colour must be the same in at least 3 rows. Without loss of generality, suppose that the first 3 rows contain at least 3 blue squares each. If the first two rows contain two blue squares in the same columns, then the desired rectangle exists. Otherwise, each column contains at least one blue square in these two rows. Thus, in one of the first two rows, there are two blue squares that are in the same columns as the blue squares in the third row. These form the desired rectangle.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15175,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be an isosceles trapezoid with longer base $AB$. Let $I$ be the incenter of triangle $ABC$ and $J$ the $C$-excenter of triangle $ACD$. Prove that $IJ$ and $AB$ are parallel.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K$ be the incenter of triangle $ABD$. Since $IK \\parallel AB$, it suffices to show $JK \\parallel AB$. Let $\\angle ABD = \\angle ACD = \\varphi$. Then $\\angle AKD = 90^\\circ + \\frac{1}{2}\\varphi$ and $\\angle DJA = 90^\\circ - \\frac{1}{2}\\varphi$, implying that the quadrilateral $AKDJ$ is cyclic.\n\n![](images/brozura_a67angl_new_p15_data_2adf4f0b39.png)\n\nAs $AK$, $DJ$ are bisectors of alternate interior angles, they are parallel. Together with the cyclic quadrilateral, we obtain $\\angle AKJ = \\angle ADJ = \\angle DAK = \\angle KAB$, which concludes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15176,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D$ be a point, and let $a$ and $b$ denote its distances from $BC$ and $\\ell$, respectively. Let $X$ be an arbitrary point between the lines $BC$ and $E'D'$. If $X$ lies on $\\ell$, then the ratio of its distance from $BC$ to its distance from $D'E'$ is $\\frac{a+b}{b}$. For points above $\\ell$, this ratio is greater; for points below, it is smaller.\n\n![](images/IRN_ABooklet_2023_1_p18_data_e751f86330.png)\n\nLet $P$ be the intersection of $D'B$ and $E'C$. According to Thales' theorem, the ratio of the distances of $P$ from $BC$ and $D'E'$ is:\n\n$$\n\\frac{BC}{D'E'} = \\frac{BC}{DE} = \\frac{AC}{AD} = \\frac{a+b}{b}\n$$\n\nShow that $P$ lies on $\\ell$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the ratio of the distances from $P$ to $BC$ and $D'E'$ is $\\frac{a+b}{b}$, and this is exactly the ratio for points on $\\ell$, it follows that $P$ must lie on $\\ell$. Thus, the claim is established.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15177,
    "subject": "Mathematics (Olympiad)",
    "question": "Can Bob cut an apple into exactly $1! + 2! + 3! + \\dots + 1013! + 2014!$ pieces, if he is allowed to repeatedly cut a piece into either 20 pieces or 14 pieces at a time?",
    "options": [],
    "answer": "See solution",
    "solution": "If Bob cuts a piece into 20 pieces, the total number of pieces increases by 19; if he cuts into 14 pieces, it increases by 13. So, after $x$ cuts into 20 pieces and $y$ cuts into 14 pieces, the apple is cut into $1 + 19x + 13y$ pieces.\n\nWe need nonnegative integers $x$ and $y$ such that $1 + 19x + 13y = 1! + 2! + 3! + \\dots + 1013! + 2014!$, or equivalently, $19x + 13y = 2! + 3! + 4! + \\dots + 1013! + 2014!$.\n\nDivide the right-hand sum into groups:\n\n$$\n\\begin{align*}\n2! + 3! + 4! + \\dots + 1013! + 2014! &= (2! + 3! + 4! + 5!) + (6! + 8!) + \\\\\n&\\quad (7! + 9! + 10!) + (11! + 12!) + (13! + 14! + \\dots + 1013! + 2014!).\n\\end{align*}\n$$\n\nEach group sum is divisible by either 13 or 19:\n- $2! + 3! + 4! + 5! = 152 = 8 \\cdot 19$\n- $6! + 8! = 6! \\cdot 57 = 6! \\cdot 3! \\cdot 19$\n- $7! + 9! + 10! = 7! \\cdot 793 = 7! \\cdot 61! \\cdot 13$\n- $11! + 12! = 11! \\cdot 13$\n- $13! + 14! + \\dots + 1013! + 2014!$ is divisible by 13 (each term is divisible by 13)\n\nTherefore, the sum can be written as $19a + 13b$ for some nonnegative integers $a, b$. Thus, Bob can achieve the required number of pieces by making $a$ cuts into 20 pieces and $b$ cuts into 14 pieces.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15178,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all polynomials $P(x)$ with integer coefficients such that the values $P(1), P(2), \\ldots, P(2021)$ are exactly the numbers $1, 2, \\ldots, 2021$ in some order.",
    "options": [],
    "answer": "See solution",
    "solution": "All polynomials of the form $P(x) = x + R(x)(x-1)(x-2)\\dots(x-2021)$ and $P(x) = 2022 - x + R(x)(x-1)(x-2)\\dots(x-2021)$, where $R(x)$ is any polynomial with integer coefficients.\n\n**Solution:**\n\nSince $P(x)$ has integer coefficients, it satisfies\n\n$$\na-b \\mid P(a)-P(b)$$\n\nfor any integers $a$ and $b$. Substituting $a=2021$ and $b=1$, we get $2020 \\mid P(2021) - P(1)$. Since $P(1), \\ldots, P(2021)$ are $2021$ distinct numbers, $P(1)$ and $P(2021)$ must be $1$ and $2021$ in some order (since their difference must be $2020$).\n\nConsider the two cases:\n\n- If $P(1) = 1$ and $P(2021) = 2021$, then for $a = k$ and $b = 1$ ($k = 2, \\ldots, 2020$), $k-1 \\mid P(k) - 1$. The only possible values for $P(k)$ are $k$, since all other values are already used. Thus, $P(k) = k$ for $k = 1, \\ldots, 2021$. Therefore, $P(x) - x$ has roots at $x = 1, \\ldots, 2021$, so $P(x) - x = (x-1)\\dots(x-2021)R(x)$, with $R(x)$ any integer-coefficient polynomial. Thus, $P(x) = x + R(x)(x-1)\\dots(x-2021)$.\n\n- If $P(1) = 2021$ and $P(2021) = 1$, then for $a = k$ and $b = 2021$ ($k = 2, \\ldots, 2020$), $2021-k \\mid P(k) - 1$. The only possible values for $P(k)$ are $2022 - k$, so $P(k) = 2022 - k$ for $k = 1, \\ldots, 2021$. Thus, $P(x) - (2022 - x)$ has roots at $x = 1, \\ldots, 2021$, so $P(x) = 2022 - x + R(x)(x-1)\\dots(x-2021)$.\n\nBoth families of solutions ensure that $P(1), \\ldots, P(2021)$ are $1, \\ldots, 2021$ in some order.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15179,
    "subject": "Mathematics (Olympiad)",
    "question": "The angles $A$, $B$, and $C$ of a triangle are measured in degrees, and the lengths of the opposite sides are $a$, $b$, and $c$ respectively. Prove that\n\n$$\n60 \\le \\frac{aA + bB + cC}{a + b + c} \\le 90\n$$\n\nThe two parts of this inequality can be tackled separately.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume $a \\ge b \\ge c$. Since the largest angle is opposite the longest side in any triangle, and the smallest angle is opposite the shortest side, we also have $A \\ge B \\ge C$.\n\nThat is, $a, b, c$ and $A, B, C$ are both non-increasing sequences. By the rearrangement inequality,\n\n$$\naA + bB + cC \\ge aB + bC + cA \\\\\naA + bB + cC \\ge aC + bA + cB.\n$$\n\nSumming these two inequalities and adding $aA + bB + cC$ to each side gives\n\n$$\n\\begin{aligned}\n3(aA + bB + cC) &\\ge aA + aB + aC \\\\\n&\\quad + bA + bB + bC \\\\\n&\\quad + cA + cB + cC \\\\\n&= (a + b + c)(A + B + C) \\\\\n&= 180(a + b + c).\n\\end{aligned}\n$$\n\nsince the angles in a triangle sum to $180^\\circ$. Dividing through by $3(a + b + c)$ gives\n\n$$\n\\frac{aA + bB + cC}{a + b + c} \\ge 60.\n$$\n\nNow for the other inequality. Continue to assume $a \\ge b \\ge c$ and $A \\ge B \\ge C$. The triangle inequality states $a < b + c$, so\n\n$$\naA < (b + c)A = bA + cA.\n$$\n\nAlso, $bB \\le aB$ and $cC \\le aC$. Thus,\n\n$$\n\\begin{aligned}\naA + bB + cC < & aB + aC + bA + cA \\\\\n< & aB + aC + bA + bC + cA + cB\n\\end{aligned}\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\n2(aA + bB + cC) < aA + aB + aC \\\\\n\\quad + bA + bB + bC \\\\\n\\quad + cA + cB + cC \\\\\n= (a + b + c)(A + B + C) \\\\\n= 180(a + b + c).\n\\end{aligned}\n$$\n\nDividing through by $2(a + b + c)$ gives\n\n$$\n\\frac{aA + bB + cC}{a + b + c} < 90\n$$\n\nas required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15180,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $T$ be a finite set of positive integers greater than $1$. A subset $S$ of $T$ is called *good* if for every $t \\in T$ there exists some $s \\in S$ with $\\gcd(s, t) > 1$. Prove that the number of good subsets of $T$ is odd.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the set $\\mathcal{A}$ of all (ordered) pairs $(X, Y)$ with $X, Y \\subseteq T$ and $\\gcd(x, y) = 1$ for all $x \\in X$ and $y \\in Y$. Clearly, $X$ and $Y$ are disjoint for any $(X, Y) \\in \\mathcal{A}$.\n\nWe have the following claims:\n\n1. If $X'$ is good, then the number of pairs $(X', Y) \\in \\mathcal{A}$ is odd: in this case, the only such pair in $\\mathcal{A}$ is $(X', \\emptyset)$.\n\n2. If $X'$ is not good, then the number of pairs $(X', Y) \\in \\mathcal{A}$ is even: let $Z \\subseteq T \\setminus X'$ contain the numbers that are relatively prime to all numbers in $X'$. Because $X'$ is not good, $Z$ is non-empty. Now, $(X', Y) \\in \\mathcal{A}$ if and only if $Y \\subseteq Z$, so the number of choices for $Y$ is a power of $2$.\n\nFinally, note that $(\\emptyset, \\emptyset)$ is the only element of $\\mathcal{A}$ of the form $(X, X)$, so $(X, Y) \\mapsto (Y, X)$ groups the elements of $\\mathcal{A} \\setminus \\{(\\emptyset, \\emptyset)\\}$ into pairs. Thus, $\\mathcal{A}$ has an odd number of elements. By the above, this implies that the number of good subsets is odd, completing the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15181,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a scalene triangle. The tangents to the nine-point circle at the perpendicular foot dropped from $A$ on the line $BC$ and at the midpoint of the side $BC$ meet at the point $A'$. The points $B'$ and $C'$ are defined similarly. Prove that the lines $AA'$, $BB'$, and $CC'$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "The tangent at $A$ to the circumcircle of $ABC$ meets the line $BC$ at the point $A''$; the points $B''$ and $C''$ are defined similarly. The points $A''$, $B''$, and $C''$ are collinear on Lemoine's line. We shall prove that the lines $AA'$, $BB'$, and $CC'$ are the polars of the points $A''$, $B''$, and $C''$, respectively, relative to the nine-point circle $\\gamma$, so they are indeed concurrent. Clearly, it is sufficient to prove that $AA'$ is the polar of $A''$ with respect to $\\gamma$.\n\nLet $A_1$, $B_1$, and $C_1$ be the perpendicular feet dropped from $A$, $B$, and $C$, respectively, on the lines $BC$, $CA$, and $AB$. Let further $A_2$ be the midpoint of the side $BC$, and let $A_3$ be the midpoint of the segment joining $A$ to the orthocenter of the triangle $ABC$. It is easily seen that the line $A_2A_3$ is the perpendicular bisector of the segment $B_1C_1$, so it is perpendicular to the tangent at $A$ to the circumcircle $ABC$. Consequently, $A_3$ is the orthocenter of the triangle $AA''A_2$, so the lines $AA_2$ and $A''A_3$ are perpendicular; it is easily seen that they meet at some point on $\\gamma$, so $A''$ lies on the polar of $A$ with respect to $\\gamma$. Finally, $A''$ lies on $BC$, which is the polar of $A'$ with respect to $\\gamma$, so $A''$ is the pole of $AA'$ with respect to $\\gamma$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15182,
    "subject": "Mathematics (Olympiad)",
    "question": "An increasing geometric progression of 5 natural numbers satisfies the following condition: the square of the sum of the first and fourth elements is 100 times greater than the sum of the first, fifth, and twice the third elements. Find the largest 3-digit number that can be a member of such a progression.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $b_1 = b$, $b_2 = bq$, $b_3 = bq^2$, $b_4 = bq^3$, $b_5 = bq^4$ denote the elements of the progression. The condition becomes:\n\n$$\n(b_1 + b_4)^2 = 100(b_1 + 2b_3 + b_5)\n$$\n\nSubstituting the terms:\n\n$$\n(b + bq^3)^2 = 100(b + 2bq^2 + bq^4)\n$$\n\n$$\nb^2(1 + q^3)^2 = 100b(1 + 2q^2 + q^4)\n$$\n\nDivide both sides by $b$ (since $b > 0$):\n\n$$\nb(1 + q^3)^2 = 100(1 + 2q^2 + q^4)\n$$\n\nLet $q = \\frac{m}{n}$ in lowest terms, $b$ is natural. For $b_3 = bq^2$ to be integer, $b$ must be divisible by $n^2$. Set $bq^2 = 100$, so $b = \\frac{100}{q^2}$.\n\nSince $b$ is integer, $q^2$ divides 100. Possible $q$ values: $2, 5, 10, \\frac{5}{2}$.\n\nFor each $q$:\n\n- $q = 2$: $b = 25$, sequence: $25, 50, 100, 200, 400$\n- $q = 5$: $b = 4$, sequence: $4, 20, 100, 500, 2500$\n- $q = 10$: $b = 1$, sequence: $1, 10, 100, 1000, 10000$\n- $q = \\frac{5}{2}$: $b = 40$, sequence: $40, 100, 250, 625, 1562.5$ (but only integer terms are valid, so $b_5 = 625$)\n\nThe largest 3-digit number among all possible terms is $625$.\n\n**Answer:** $625$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15183,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b$ be real numbers and define the function $f(x) = x^3 + a x^2 + b x$. If there exist three real numbers $x_1, x_2, x_3$ satisfying $x_1 + 1 \\leq x_2 \\leq x_3 - 1$ and $f(x_1) = f(x_2) = f(x_3)$, find the minimum value of $|a| + 2|b|$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(x_1) = f(x_2) = f(x_3) = c$. Then $x_1, x_2, x_3$ are the three real roots of the cubic equation $x^3 + a x^2 + b x - c = 0$. By Vieta's formulas:\n\n$$\na = -(x_1 + x_2 + x_3), \\quad b = x_1 x_2 + x_2 x_3 + x_3 x_1.\n$$\n\nFrom the given, $x_2 - x_1 \\geq 1$, $x_3 - x_2 \\geq 1$, and $x_3 - x_1 \\geq 2$. Thus,\n\n$$\n\\begin{aligned}\na^2 - 3b &= x_1^2 + x_2^2 + x_3^2 - x_1 x_2 - x_2 x_3 - x_3 x_1 \\\\\n&= \\frac{1}{2}\\big((x_1 - x_2)^2 + (x_2 - x_3)^2 + (x_3 - x_1)^2\\big) \\\\\n&\\geq \\frac{1}{2}(1^2 + 1^2 + 2^2) = 3.\n\\end{aligned}\n$$\n\nSo $b \\leq \\frac{a^2}{3} - 1$.\n\nIf $|a| \\geq \\sqrt{3}$, then $|a| + 2|b| \\geq |a| + \\sqrt{3}$.\n\nIf $0 \\leq |a| < \\sqrt{3}$, then $b < 0$. At this point,\n\n$$\n|a| + 2|b| \\geq |a| + 2\\left(\\frac{a^2}{3} - 1\\right).\n$$\n\nThe minimum occurs at $|a| = \\sqrt{3}$, $b = 0$, with $x_1 = -1 - \\frac{\\sqrt{3}}{3}$, $x_2 = -\\frac{\\sqrt{3}}{3}$, $x_3 = 1 - \\frac{\\sqrt{3}}{3}$, all satisfying the conditions. Thus, the minimum value is $\\sqrt{3}$.\n\n**Answer:** The minimum of $|a| + 2|b|$ is $\\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15184,
    "subject": "Mathematics (Olympiad)",
    "question": "In a classroom, there is a clock whose minute and hour hands move with constant angular velocity. The minute hand works correctly, but the hour hand moves at half the angular velocity of the minute hand. At 10:00, the clock shows the correct time. When will the clock next show the correct time?",
    "options": [],
    "answer": "See solution",
    "solution": "Imagine a correctly working clock next to the broken clock. The hour hand of a working clock moves 12 times slower than its minute hand, but on the broken clock, the hour hand moves only 2 times slower than the minute hand (since it moves at half the angular velocity of the minute hand). Thus, the hour hand of the broken clock moves 6 times faster than the hour hand of the correct clock.\n\nLet $x$ be the number of hours after 10:00 when the broken clock next shows the correct time. The broken clock will show the correct time again when its hour hand has completed one more full circle than the hour hand of the correct clock, i.e., when their indicated times differ by 12 hours.\n\nSet up the equation:\n\n$$\n6x = x + 12\n$$\n\nSolving for $x$:\n\n$$\n6x - x = 12 \\\\\n5x = 12 \\\\\nx = 2.4\n$$\n\nSo, after 2 hours and $0.4 \\times 60 = 24$ minutes, i.e., at 12:24, the broken clock will next show the correct time.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15185,
    "subject": "Mathematics (Olympiad)",
    "question": "A group of 4050 friends is playing a video game tournament. There are 2025 computers labelled $a_1, \\dots, a_{2025}$ in one room and 2025 computers labelled $b_1, \\dots, b_{2025}$ in another room at the tournament. The player on computer $a_i$ always plays against the players $b_i$, $b_{i+2}$, $b_{i+3}$, and $b_{i+4}$ (in particular, not against $b_{i+1}$), where the numbers of the computers are considered modulo 2025. After the first round, all players choose a computer within their room for the second round. Afterwards, they note that everyone has the same opponents in the second round as in the first round.\n\nProve that if someone chose the same computer in both rounds, then everyone chose the same computer in both rounds.",
    "options": [],
    "answer": "See solution",
    "solution": "For the opponent computers of $a_i$, we look at the $a_j$ they are playing against, as shown in the following table.\n\n$b_i$:\n\n| $a_{i-4}$ | $a_{i-3}$ | $a_{i-2}$ | $a_i$ |\n|-----------|-----------|-----------|-------|\n\n$b_{i+2}$:\n\n| $a_{i-2}$ | $a_{i-1}$ | $a_i$ | $a_{i+2}$ |\n|-----------|-----------|-------|-----------|\n\n$b_{i+3}$:\n\n| $a_{i-1}$ | $a_i$ | $a_{i+1}$ | $a_{i+3}$ |\n|-----------|-------|-----------|-----------|\n\n$b_{i+4}$:\n\n| $a_i$ | $a_{i+1}$ | $a_{i+2}$ | $a_{i+4}$ |\n|-------|-----------|-----------|-----------|\n\nNote that computers $a_{i-2}$, $a_{i-1}$, $a_{i+1}$, and $a_{i+2}$ each have two common opponent computers with $a_i$. Furthermore, $a_{i-4}$, $a_{i-3}$, $a_{i+3}$, and $a_{i+4}$ each have one common adversary computer with $a_i$. Only for the first two, $a_{i-4}$ and $a_{i-3}$, the common opponent with $a_i$ is the same, namely $b_i$.\n\n![](images/NLD_ABooklet_2025_p20_data_ffc8d4b070.png)\n\nNow suppose one player chose the same computer for the second round, say the player on $a_{2025}$. We will now prove by induction that all players on $a_i$ and $b_i$ have chosen the same computer for the second round.\n\nPer the induction hypothesis, suppose that the player on a certain computer $a_i$ is the same in both rounds. Since everyone has the same opponents in both rounds, the players who were on $a_{i-4}$ and $a_{i-3}$ again each have the same common co-player in the game with $a_i$, namely the player who was on $b_i$. This means that the player on $b_i$ also chose the same computer. Now we note that only the player who was on $a_{i-2}$ has a second common opponent with $a_i$ in addition to $b_i$. So also on $a_{i-2}$, the same player chose the same computer again.\n\nSince 2025 is odd, it now follows with induction that everyone has played on the same computer, as soon as any one player has played on the same computer. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 15186,
    "subject": "Mathematics (Olympiad)",
    "question": "The graph of $f(x) = |\\lfloor x \\rfloor| - |\\lfloor 1-x \\rfloor|$ is symmetric about which of the following? (Here $\\lfloor x \\rfloor$ is the greatest integer not exceeding $x$.)\n\n(A) the y-axis  \n(B) the line $x = 1$  \n(C) the origin  \n(D) the point $\\left(\\frac{1}{2}, 0\\right)$  \n(E) the point $(1, 0)$",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer (D):**\n\nLet $x = n + r$, where $n$ is an integer and $0 \\le r < 1$.\n\nFirst, suppose that $r = 0$. Then $f(x) = |n| - |1 - n|$, which is equal to $n - (n - 1) = 1$ if $n \\ge 1$ and $-(n) - (1 - n) = -1$ if $n \\le 0$. Hence, the graph of $f$ cannot be symmetric about the y-axis, the line $x = 1$, the origin, or the point $(1, 0)$. However, note that in all cases $f(1 - n) = -f(n)$, so the graph of $f$ is symmetric about the point $\\left(\\frac{1}{2}, 0\\right)$ when $x$ is restricted to integer values.\n\nFinally, when $r > 0$, observe that $\\lfloor x \\rfloor = n$ and $\\lfloor 1 - x \\rfloor = \\lfloor 1 - n - r \\rfloor = -n$, so $f(x) = |n| - |-n| = 0$, and again $f(1 - x) = -f(x)$. Hence, the graph of $f$ is symmetric about the point $\\left(\\frac{1}{2}, 0\\right)$.\n\nAlternatively, note that $f(1-x) = |\\lfloor 1-x \\rfloor| - |\\lfloor 1-(1-x) \\rfloor| = -f(x)$. Thus, a point $(x, y)$ is on the graph of $f$ if and only if $(1-x, -y)$ is on the graph. The midpoint of the segment joining these two points is the point $\\left(\\frac{1}{2}, 0\\right)$; that is, the graph is symmetric about this point.\n\nTo show that $f(x)$ does not have any of the properties described in the four other alternatives, it is useful to note that\n\n$$\nf(x) = \\begin{cases} 1 & \\text{if } n \\text{ is an integer, } n \\ge 1, \\\\ 0 & \\text{if } n \\text{ is not an integer,} \\\\ -1 & \\text{if } n \\text{ is an integer, } n \\le 0. \\end{cases}\n$$\n\nThe graph of $f$ is not symmetric about the y-axis because $f(1) = 1$ and $f(-1) = -1 \\ne 1$. The graph of $f$ is not symmetric about the line $x = 1$ because $f(3) = 1$ and $f(-1) = -1 \\ne 1$. The graph of $f$ is not symmetric with respect to the origin because $f(0) = 1$ and the reflected point $(0, -1)$ is not in the graph of $f$. The graph of $f$ is not symmetric about the point $(1, 0)$ because $f(1) = 1$ and the reflected point $(1, -1)$ is not in the graph of $f$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 15187,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x)f(y) = f(xy - 1) + x f(y) + y f(x), \\quad \\forall x, y \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $y = 0$ in $(1)$:\n$$\nf(x) f(0) = f(-1) + x f(0).\n$$\n\nConsider two cases for $f(0)$:\n\n*Case 1.* If $f(0) \\neq 0$, then $f(x) = x + c$ for some constant $c$, but this does not satisfy the original equation.\n\n*Case 2.* If $f(0) = 0$, then $f(-1) = 0$.\n\nPlug $x = y = 1$ into $(1)$:\n$$\nf(1)^2 = 2 f(1) \\implies f(1) = 0 \\text{ or } f(1) = 2.\n$$\n\nSubstitute $y = -1$ into $(1)$:\n$$\nf(-y - 1) = f(y), \\quad \\forall y \\in \\mathbb{R}.\n$$\n\nReplace $y$ by $-y - 1$ in $(1)$:\n$$\nf(x) f(-y - 1) = f(-x(y + 1) - 1) + x f(-y - 1) - (y + 1) f(x).\n$$\n\nUsing the previous result:\n$$\nf(xy - 1) + y f(x) = f(xy + x) - (y + 1) f(x), \\quad \\forall x, y \\in \\mathbb{R}.\n$$\n\nLet $x \\neq -1$, replace $x$ by $x + 1$ and $y$ by $\\frac{1}{x+1}$:\n$$\nf(x - 1) = \\frac{x - 1}{x + 1} f(x), \\quad \\forall x \\neq -1.\n$$\n\nSet $y = 1$ in $(1)$:\n$$\n\\begin{aligned}\nf(x) f(1) &= f(x - 1) + x f(1) + f(x) \\\\\n&= \\frac{x - 1}{x + 1} f(x) + x f(1) + f(x), \\quad \\forall x \\neq -1.\n\\end{aligned}\n$$\n\n- If $f(1) = 0$, then $f(x) \\equiv 0$.\n- If $f(1) = 2$, then $f(x) = x(x + 1)$.\n\nBoth $f(x) \\equiv 0$ and $f(x) = x(x+1)$ satisfy $(1)$.\n\n$\\boxed{f(x) \\equiv 0 \\quad \\text{and} \\quad f(x) = x(x+1)}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15188,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer, and consider $n$ points on a circle where every pair of points is connected by a line segment. Two players, Alice and Bob, play a coloring game on these segments: initially all segments are black. They take turns coloring segments red, with Alice moving first. On each turn, a player may color one or two black segments red, with the constraint that after each move, there must not exist three points $A, B, C$ on the circle such that all three segments $AB, BC, CA$ are red. The game ends when no more legal moves are possible.\n\nProve that there exists a positive real number $\\varepsilon$ such that for any positive integer $n$, no matter how Bob plays, Alice can ensure that at the end of the game, the number of red segments is less than $\\left(\\frac{1}{4} - \\varepsilon\\right) n^2 + n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider the graph $G$ formed by red edges. Initially, $G$ is an empty graph on $n$ vertices, and the condition requires that $G$ remains triangle-free throughout the game.\n\n**Step 1:** We show that if at Alice's turn, $G$ has 24 isolated vertices, then Alice can create a 5-cycle among these vertices in four moves.\n\nLet $S$ be the set of these 24 isolated vertices. Alice first selects three vertices $A, B, C \\in S$ and colors $AB$ and $AC$ red. After Bob's move, $S$ still contains at least 17 isolated vertices. Alice then selects two isolated vertices $D, E \\in S$ and colors $AD$ and $AE$ red. After Bob's move, $S$ has at least 11 isolated vertices remaining. Alice then selects two more isolated vertices $F, G \\in S$ and colors $BF$ and $CG$ red.\n\n![](images/China-TST-2025B_p29_data_bdb538bad8.png)\n\nConsider the four edges $DF$, $DG$, $EF$, and $EG$, which are currently uncolored. After Bob's move, at least one of them remains uncolored, say $DF$. Now $S$ still has at least 5 isolated vertices. Alice selects one of them, $H$, and colors $FH$ and $DH$ red. This creates a 5-cycle $ABFHD$ in $G$. By following this strategy, after four moves by each player, the number of isolated vertices in $G$ decreases by at most 24.\n\n**Step 2:** Alice follows the strategy described in Step 1. After $4 \\times \\left\\lfloor \\frac{n}{24} \\right\\rfloor$ moves, she can ensure that $G$ contains $\\left\\lfloor \\frac{n}{24} \\right\\rfloor$ disjoint 5-cycles. Alice then colors edges arbitrarily until $G$ becomes a maximal triangle-free graph.\n\nWe now estimate the number of edges in $G$. Let $k = \\left\\lfloor \\frac{n}{24} \\right\\rfloor$, and denote the disjoint 5-cycles by $C_1, C_2, \\dots, C_k$. Between any two cycles $C_i$ and $C_j$, there are at most 10 edges. For each vertex $v$ not in any $C_i$, there are at most 2 edges from $v$ to each $C_i$. The remaining $n - 5k$ vertices outside all $C_i$ can have at most $\\frac{(n-5k)^2}{4}$ edges between them (by Turán's theorem). The cycles $C_1, \\dots, C_k$ themselves contribute $5k$ edges. Thus, the total number of edges is bounded by:\n\n$$\n\\begin{align*}\n|E(G)| &= 10 \\binom{k}{2} + 2k(n-5k) + \\frac{(n-5k)^2}{4} + 5k \\\\\n&= \\frac{n^2}{4} - \\frac{kn}{2} + \\frac{5k^2}{4} \\\\\n&\\le \\frac{n^2}{4} - \\frac{1}{2} \\left( \\frac{n}{24} - 1 \\right) n + \\frac{5}{4} \\left( \\frac{n}{24} \\right)^2 \\\\\n&= \\left( \\frac{1}{4} - \\frac{43}{48^2} \\right) n^2 + \\frac{n}{2}.\n\\end{align*}\n$$\n\nTaking $\\varepsilon = \\frac{43}{48^2}$ satisfies the requirement. $\\square$\n\n![](images/China-TST-2025B_p30_data_98149609b0.png)\n\n**Remark:** The key idea is for Alice to systematically create disjoint 5-cycles, which limits the number of edges that can be added without forming triangles. The calculation combines combinatorial bounds with Turán's theorem to achieve the desired inequality.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15189,
    "subject": "Mathematics (Olympiad)",
    "question": "Two externally tangent circles $\\omega_1$ and $\\omega_2$ have centers $O_1$ and $O_2$, respectively. A third circle $\\Omega$ passing through $O_1$ and $O_2$ intersects $\\omega_1$ at $B$ and $C$ and $\\omega_2$ at $A$ and $D$, as shown. Suppose that $AB = 2$, $O_1O_2 = 15$, $CD = 16$, and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon.\n\n![](images/2022AIME_II_Solutions_p14_data_d68af3c235.png)",
    "options": [],
    "answer": "See solution",
    "solution": "First observe that $AO_2 = O_2D$ and $BO_1 = O_1C$. Let points $A'$ and $B'$ be the reflections of $A$ and $B$, respectively, across the diameter of $\\Omega$ that is the perpendicular bisector of $\\overline{O_1O_2}$. Thus $A'$ and $B'$ are on $\\Omega$ with $A'B' = AB = 2$. Then quadrilaterals $ABO_1O_2$ and $A'B'O_2O_1$ are congruent, so hexagons $ABO_1CDO_2$ and $B'A'O_1CDO_2$ have the same area. Furthermore, because $\\overline{A'O_1} = \\overline{DO_2}$ and $\\overline{O_1C} = \\overline{O_2B'}$, it follows that $B'D = A'C$ and quadrilateral $B'A'CD$ is an isosceles trapezoid.\n\n![](images/2022AIME_II_Solutions_p14_data_ee01aa8fad.png)\n\nBecause $A'O_1 = DO_2$, quadrilateral $A'O_1DO_2$ is an isosceles trapezoid. In turn, $A'D = O_1O_2 = 15$, and similarly $B'C = 15$. Thus Ptolemy's Theorem applied to $B'A'CD$ yields $B'D \\cdot A'C + 2 \\cdot 16 = 15^2$, whence $B'D = A'C = \\sqrt{193}$. Let $\\alpha = \\angle B'A'D$. The Law of Cosines applied to $\\triangle B'A'D$ yields\n\n$$\n\\cos \\alpha = \\frac{15^2 + 2^2 - (\\sqrt{193})^2}{2 \\cdot 2 \\cdot 15} = \\frac{3}{5},\n$$\n\nand hence $\\sin \\alpha = \\frac{4}{5}$. Let $R$ be the point on line $A'B'$ such that $\\overline{A'R} \\perp \\overline{DR}$. Because $\\triangle A'RD$ is a right triangle whose hypotenuse has length 15 with $\\cos(\\angle DA'R) = \\frac{3}{5}$, it follows that $A'R = 9$ and $DR = 12$. In particular, the distance from $\\overline{A'B'}$ to $\\overline{CD}$ is $DR = 12$, which implies that the area of $B'A'CD$ is $\\frac{1}{2} \\cdot 12 \\cdot (2 + 16) = 108$.\n\nNow let $O_1C = O_2B' = r_1$ and $O_2D = O_1A' = r_2$. The tangency of circles $\\omega_1$ and $\\omega_2$ implies $r_1 + r_2 = 15$. Furthermore, $\\angle B'O_2D$ and $\\angle B'A'D$ are opposite angles in cyclic quadrilateral $A'B'O_2D$, which implies that the measure of $\\angle B'O_2D$ is $180^\\circ - \\alpha$. Therefore the Law of Cosines applied to $\\triangle B'O_2D$ yields\n\n$$\n\\begin{aligned}\n193 &= r_1^2 + r_2^2 - 2r_1r_2 \\left(-\\frac{3}{5}\\right) \\\\\n&= (r_1^2 + 2r_1r_2 + r_2^2) - \\frac{4}{5}r_1r_2 \\\\\n&= (r_1 + r_2)^2 - \\frac{4}{5}r_1r_2 \\\\\n&= 225 - \\frac{4}{5}r_1r_2.\n\\end{aligned}\n$$\n\nThus $r_1r_2 = 40$, so the area of $\\triangle B'O_2D$ is $\\frac{1}{2}r_1r_2 \\sin \\alpha = 16$. The area of the hexagon is $108 + 2 \\cdot 16 = 140$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15190,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all possible real pairs $ (x, y) $ that satisfy the following:\n\n$$\n4x + 3y = 2x \\cdot \\left[ \\frac{x^2 + y^2}{x^2} \\right].\n$$\n\n![](images/Ukraine_booklet_2018_p24_data_568b075f01.png)\n\n",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $ (t, -\\frac{2}{3}t) $ and $ (t, 2t) $, where $ t \\in \\mathbb{R} $, $ t \\neq 0 $.\n\nSince $ x \\neq 0 $, rewrite the equation as $ 2 + \\frac{3y}{2x} = \\left[ 1 + \\frac{y^2}{x^2} \\right] $. Let $ k = \\frac{3y}{2x} $ be an integer. Since $ [1 + \\frac{y^2}{x^2}] = 1 + \\frac{y^2}{x^2} $, we have:\n\n$$\n1 + k = \\left[ \\frac{4k^2}{9} \\right] \\iff 1 + k \\le \\frac{4k^2}{9} < 2 + k.\n$$\n\nSo:\n\n$$\n\\begin{cases}\n4k^2 - 9k - 9 \\ge 0, \\\\\n4k^2 - 9k - 18 < 0,\n\\end{cases}\n$$\n\nThe integer solutions are $ k = -1 $ and $ k = 3 $. Thus, $ \\frac{3y}{2x} = -1 $ or $ \\frac{3y}{2x} = 3 $, giving all possible solutions as stated above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15191,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all three prime numbers $p$, $q$, and $r$ that satisfy\n\n$$\n\\frac{q}{p-1} + \\frac{r}{p+1} = \\frac{q + r + 1}{p}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let's rewrite the given equality:\n\n$$\n\\frac{q}{p-1} + \\frac{r}{p+1} = \\frac{q + r + 1}{p}\n$$\n\nRearrange terms:\n\n$$\n\\frac{q}{p-1} - \\frac{q}{p} = \\frac{r}{p} - \\frac{r}{p+1} + \\frac{1}{p}\n$$\n\nThis simplifies to:\n\n$$\n\\frac{q}{p(p-1)} = \\frac{r}{p(p+1)} + \\frac{1}{p}\n$$\n\nMultiply both sides by $p$:\n\n$$\n\\frac{q}{p-1} = \\frac{r}{p+1} + 1\n$$\n\nSo,\n\n$$\nq = \\frac{(p-1)r}{p+1} + p - 1\n$$\n\nOr,\n\n$$\nq = p + r - 1 - \\frac{2r}{p+1} \\quad (1)\n$$\n\nSince $p$, $q$, $r$ are primes, $\\frac{2r}{p+1}$ must be a positive integer. The divisors of $2r$ are $1$, $2$, $r$, and $2r$. Since $p+1 \\geq 3$, two cases are possible:\n\n**Case 1:** $p+1 = r$ (i.e., $p = r - 1$). The only consecutive primes are $2$ and $3$, so $p = 2$, $r = 3$. From (1), $q = 2$. Checking, $p = q = 2$, $r = 3$ is a solution.\n\n**Case 2:** $p+1 = 2r$ (i.e., $p = 2r - 1$). From (1), $q = 3r - 3$. For $q$ prime, $q = 3$, so $r = 2$, $p = 3$. Checking, $p = q = 3$, $r = 2$ is also a solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15192,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that there are infinitely many polynomials $P$ with real coefficients such that if $x$, $y$, and $z$ are real numbers satisfying\n\n$$\nx^2 + y^2 + z^2 + 2xyz = 1,\n$$\n\nthen\n\n$$\nP(x)^2 + P(y)^2 + P(z)^2 + 2P(x)P(y)P(z) = 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us call a triple $(x, y, z)$ of real numbers a *-triple if it satisfies\n\n$$\nx^2 + y^2 + z^2 + 2xyz = 1.\n$$\n\nLet us call a polynomial $p(x)$ with real coefficients a *-polynomial if\n\n$(x, y, z)$ a *-triple implies $(p(x), p(y), p(z))$ is also a *-triple.\n\nFirst, consider polynomials of degree at most 1. Assume $p(x) = ax + b$ ($a, b \\in \\mathbb{R}$), and suppose $p(x)$ is a *-polynomial. Since $(x, -x, 1)$ and $(x, x, -1)$ are *-triples for all $x \\in \\mathbb{R}$, we have\n\n$$\np(x)^2 + p(-x)^2 + p(1)^2 + 2p(x)p(-x)p(1) = 1,\n$$\n\nand\n\n$$\np(x)^2 + p(x)^2 + p(-1)^2 + 2p(x)p(x)p(-1) = 1\n$$\n\nfor all $x \\in \\mathbb{R}$. These simplify, respectively, to\n\n$$\n2a^2(1 - a - b)x^2 + 2b^2(1 + a + b) + (a + b)^2 = 1 \\quad (1)\n$$\n\nand\n\n$$\n2a^2(1 - a + b)x^2 + 4ab(1 - a + b)x + 2b^2(1 - a + b) + (b - a)^2 = 1. \\quad (2)\n$$\n\nSince these equations are valid for all $x \\in \\mathbb{R}$, the coefficients $2a^2(1-a-b)$ and $2a^2(1-a+b)$ must be zero. Therefore, $a=0$, or $1-a-b=0=1-a+b$.\n\nIf $a = 0$, then from (1), $2b^2(1+b) + b^2 = 1$. It is clear that $b = -1$ is a solution, and $b = \\frac{1}{2}$ is the only other solution. The constant polynomials $p(x) = -1$ and $p(x) = \\frac{1}{2}$ are indeed *-polynomials.\n\nIf $a \\neq 0$, then $1 - a - b = 0 = 1 - a + b$, so $b = 0$. From (1), $2a^2(1 - a) = 0$, giving $a = 1$. Thus, $p(x) = x$ is a *-polynomial.\n\nNext, observe that if $p(x)$ is a *-polynomial, then $p(p(x)) = p^2(x)$ is also a *-polynomial, and by induction, $p^n(x)$ are *-polynomials for all $n \\ge 1$. If we can find a *-polynomial $p(x)$ such that infinitely many polynomials in the sequence $p(x), p^2(x), p^3(x), \\dots$ are different, the problem is solved. The three *-polynomials found so far do not have this property. Therefore, consider a second degree *-polynomial $p(x) = ax^2 + bx + c$, $a \\neq 0$.\n\nAssuming $p(x)$ is a *-polynomial, and using the *-triples $(x, -x, 1)$ and $(x, x, -1)$, we obtain:\n\n$$\n2a^2(1 + a + b + c)x^4 + 2[2ac + b^2 + (2ac - b^2)(a + b + c)]x^2 + 2c^2(1 + a + b + c) + (a + b + c)^2 = 1 \\quad (3)\n$$\n\nand\n\n$$\n2a^2(1 + a - b + c)x^4 + 4ab(1 + a - b + c)x^3 + 2(2ac + b^2)(1 + a - b + c)x^2 + 4bc(1 + a - b + c)x + 2c^2(1 + a - b + c) + (a - b + c)^2 = 1 \\quad (4)\n$$\n\nThe coefficients of $x^4$ in (3) and (4) must be zero, so $a + b + c = -1 = a - b + c$, so $b = 0$. Thus $a + c = -1$ and $p(x) = ax^2 + c = (-1-c)x^2 + c$ for some $c \\in \\mathbb{R}$.\n\nSince $(x, \\sqrt{1-x^2}, 0)$ are *-triples for all $-1 \\le x \\le 1$, we get\n\n$$\n((-1-c)x^2 + c)^2 + ((-1-c)(1-x^2) + c)^2 + c^2 + 2((-1-c)x^2 + c)((-1-c)(1-x^2) + c)c = 1,\n$$\n\nwhich simplifies to\n\n$$\n2(1+c)^2(1-c)x^4 - 2(1+c)^2(1-c)x^2 + 1 = 1.\n$$\n\nThe coefficients of $x^4$ and $x^2$ must be zero, so $c \\in \\{-1, 1\\}$. The case $c = -1$ gives $p(x) = -1$, already considered. Thus, the only possible candidate for a second degree *-polynomial is $p(x) = -2x^2 + 1$.\n\nNow, verify that $p(x) = -2x^2 + 1$ is a *-polynomial:\n\nLet $(x, y, z)$ be an arbitrary *-triple. Then\n\n$$\n\\begin{aligned}\n& (-2x^2 + 1)^2 + (-2y^2 + 1)^2 + (-2z^2 + 1)^2 + 2(-2x^2 + 1)(-2y^2 + 1)(-2z^2 + 1) \\\\\n&= -16x^2y^2z^2 + 4(x^4 + y^4 + z^4) + 8(x^2y^2 + y^2z^2 + z^2x^2) - 8(x^2 + y^2 + z^2) + 5 \\\\\n&= -4(1 - (x^2 + y^2 + z^2))^2 + 4(x^2 + y^2 + z^2)^2 - 8(x^2 + y^2 + z^2) + 5 \\\\\n&= 1.\n\\end{aligned}\n$$\n\nTherefore, $p(x) = -2x^2 + 1$ is a *-polynomial.\n\nThis solves the problem, since we now have an infinite sequence $p(x), p^2(x), p^3(x), \\dots$ of *-polynomials, all different, since $\\deg(p^n(x)) = 2^n$ for each $n \\ge 1$, which can be verified by induction.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15193,
    "subject": "Mathematics (Olympiad)",
    "question": "Prickle and Sting are playing a game on an $m \\times n$ board, where $m$ and $n$ are positive integers. They take turns, with Prickle going first. On each turn, Prickle places a pawn on any empty square. On Sting's turn, he must place a pawn on an empty square that is adjacent to the square where Prickle placed his pawn in the previous turn.\n\nSting wins if the entire board is filled with pawns. Prickle wins if Sting cannot place a pawn on his turn while there is still at least one empty square.\n\nDetermine, for all pairs $(m, n)$ of positive integers, which player has a winning strategy.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $m$ is the number of rows and $n$ is the number of columns.\n\nIf $m$ is even, pair the squares in each column: top two, then squares 3 and 4, etc. Since $m$ is even, all squares are paired. Sting can always respond to Prickle by placing a pawn in the paired square, which is adjacent. Thus, after each of Sting's moves, all pairs have either zero or two pawns, so Sting can always play and eventually fills the board. Therefore, Sting wins. Similarly, if $n$ is even, Sting has a winning strategy.\n\nIf $m = n = 1$, Sting wins after Prickle's first move. If $m = 1$ and $n = 3$ (or vice versa), Sting can always place a pawn adjacent to Prickle's, and the board will be filled after Prickle's next move, so Sting wins.\n\nFor $m = n = 3$, Prickle can win by placing the first pawn in the center. Sting must respond in the same row or column. Prickle then fills the remaining square in the middle column. The left and right columns are empty. Sting must play in one of these columns; Prickle then plays in the other column. This forces Sting to play in that column, and Prickle can then fill the last square in that column, leaving two empty squares and no legal move for Sting. Thus, Prickle wins if $m = n = 3$.\n\nIf both $m$ and $n$ are odd and at least one is at least $5$, Prickle can fill the center column first. After this, the first and last columns are empty, and there is an odd number of squares remaining. Prickle chooses the area with an odd number of empty squares and continues this strategy, ensuring that Sting cannot fill the board. Thus, Prickle wins in this case.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15194,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k \\ge 1$ and $N > 1$ be two integers. On a circle are placed $2N+1$ coins all showing heads. Calvin and Hobbes play the following game. Calvin starts and on his move can turn any coin from heads to tails. Hobbes on his move can turn at most one coin that is next to the coin that Calvin turned just now from tails to heads. Calvin wins if at any moment there are $k$ coins showing tails after Hobbes has made his move. Determine all values of $k$ for which Calvin wins the game.",
    "options": [],
    "answer": "See solution",
    "solution": "Calvin wins if $k \\in \\{1, 2, \\dots, N+1\\}$, and Hobbes wins otherwise.\n\nLabel the coins $1, 2, \\dots, 2N+1$. If $k \\geq N+2$, Hobbes wins by pairing coins $2i-1$ and $2i$ for $1 \\leq i \\leq N$. If Calvin makes both coins in a pair tails, Hobbes turns the one which was tails prior to Calvin's move back to heads. Thus, after Hobbes's move, no pair has more than one tails, so the number of tails is at most $1 + \\frac{(2N+1)-1}{2} = N+1$.\n\nIf $k \\leq N$, Calvin wins by simply turning coins $2i$ for $1 \\leq i \\leq N$.\n\nNow let $k = N+1$. Let $N = 2m + \\varepsilon$ where $\\varepsilon \\in \\{0, 1\\}$. Consider $m$ arcs on the circle, with the $i$th arc containing $\\{4i+1, 4i+2, 4i+3, 4i+4\\}$ for $0 \\leq i < m$. Calvin makes $3m$ moves: on moves $3i+1$, $3i+2$, and $3i+3$, he turns coins $4i+2$, $4i+4$, and $4i+3$ to tails, respectively, for $0 \\leq i < m$. After these moves, each arc will have either $\\{4i+2, 4i+3\\}$ tails (type $\\overline{23}$), $\\{4i+3, 4i+4\\}$ tails (type $\\overline{34}$), or all of $\\{4i+2, 4i+3, 4i+4\\}$ tails (type $\\overline{234}$).\n\n**Case 1.** $\\varepsilon = 0$\n\nIf any arc is of type $\\overline{234}$, there are at least $3 + 2(m-1) = N+1$ tails and Calvin wins. If all arcs are of type $\\overline{23}$ or $\\overline{34}$, there are $2m$ tails. If $4m+1$ has no tails neighbours, Calvin turns it to win. Otherwise, if the arc $\\{4m-3, 4m-2, 4m-1, 4m\\}$ is of type $\\overline{34}$ and $\\{1, 2, 3, 4\\}$ is also of type $\\overline{34}$, Calvin can turn 1 to tails to win. If $\\{1, 2, 3, 4\\}$ is of type $\\overline{23}$, then there exists $0 \\leq i < m-1$ such that the $i$th arc is of type $\\overline{23}$ and the $(i+1)$th arc is of type $\\overline{34}$, so Calvin can turn $4i+1$ to tails to win.\n\n**Case 2.** $\\varepsilon = 1$\n\nIf any arc is of type $\\overline{234}$, Calvin turns $4m+2$ and gets at least $3 + 2(m-1) + 1 = N+1$ tails. If all arcs are of type $\\overline{23}$ or $\\overline{34}$, there are $2m$ tails. If Calvin can turn $4m+1$, he wins by turning $4m+3$ next. Otherwise, if $\\{4m-3, 4m-2, 4m-1, 4m\\}$ is of type $\\overline{34}$ and $\\{1, 2, 3, 4\\}$ is also of type $\\overline{34}$, Calvin can turn 1 and $4m+2$ to win. If $\\{1, 2, 3, 4\\}$ is of type $\\overline{23}$, there exists $0 \\leq i < m-1$ such that the $i$th arc is of type $\\overline{23}$ and the $(i+1)$th arc is of type $\\overline{34}$, so Calvin wins by turning $4m+2$ and $4i+1$ in the next two moves.\n\nIn conclusion, Calvin wins if $k = N+1$, completing the proof. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15195,
    "subject": "Mathematics (Olympiad)",
    "question": "On a blackboard, there are 17 integers not divisible by 17. Alice and Bob play a game. Alice starts and they alternately play the following moves:\n\n- Alice chooses a number $a$ on the blackboard and replaces it with $a^2$.\n- Bob chooses a number $b$ on the blackboard and replaces it with $b^3$.\n\nAlice wins if the sum of the numbers on the blackboard is a multiple of 17 after a finite number of steps.\n\nProve that Alice has a winning strategy.",
    "options": [],
    "answer": "See solution",
    "solution": "Since both the problem statement and the winning condition are given in terms of divisibility by 17, it is sufficient to consider the numbers modulo 17. In the beginning, all the remainders are different from zero and Alice wins if the sum modulo 17 becomes zero.\n\nThe moves $a \\mapsto a^2$ and $b \\mapsto b^3$ turn remainders into powers of the original nonzero values. Therefore, Fermat's little theorem can be applied. For $a \\not\\equiv 0 \\pmod{17}$ and the prime number 17, one has\n\n$$\na^{16} \\equiv 1 \\pmod{17}.\n$$\n\nSo if Alice squares the same number $a$ four times in a row, then the remainder 1 modulo 17 is always obtained. Bob cannot do anything about it, because if Bob raises this number to the third power $k$ times, we get a result of\n\n$$\na^{2 \\cdot 3^k} = 16^{3^k} \\equiv 1^{3^k} = 1 \\pmod{17}.\n$$\n\nThe timing of Bob's moves does not matter, as the order of the factors in the exponent does not change anything.\n\nTherefore, Alice can make all the remainders equal to 1 by squaring each number four times. Then of course the sum is $17 \\cdot 1 \\equiv 0 \\pmod{17}$ and Alice has won.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15196,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that one can choose $n-3$ non-intersecting diagonals of a regular $n$-gon that divide the $n$-gon into triangles in such a way that every chosen diagonal is a side of minimal length in some triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\triangle$ be the triangle containing the center $O$ of the $n$-gon (colored in Fig. 23; if the center lies on a diagonal then choose either of the triangles having this diagonal as a side). Let $d$ be any side of $\\triangle$. If $d$ is a diagonal of the $n$-gon then $d$ separates $\\triangle$ from a neighboring triangle whose other two sides are shorter than $d$. By assumption, $d$ has to be a side of minimal length in $\\triangle$. But if $d$ is a side of the $n$-gon then $d$ is also a side of minimal length in $\\triangle$. Thus all sides of $\\triangle$ are equally minimal, meaning that $\\triangle$ is equilateral. Consequently, there is a constant number of sides of $n$-gon between the endpoints of every side of $\\triangle$. Hence $n = 3s_0$ for a positive integer $s_0$.\n\nConsider now an arbitrary triangle $\\Delta'$ neighboring $\\Delta$. Let $d'$ be any of its two sides not common with $\\Delta$. If $d'$ is a diagonal of the $n$-gon then $d'$ separates $\\Delta'$ from a third triangle whose other sides are shorter than $d'$. Thus $d'$ must be a side of minimal length in $\\Delta'$. But if $d'$ is a side of the $n$-gon then $d'$ is also a side of minimal length in $\\Delta'$. Hence the sides of $\\Delta'$ not common with $\\Delta$ have equal length and there must be the same number of sides of the $n$-gon between the endpoints of these sides of $\\Delta'$. Consequently $s_0 = 2s_1$ where $s_1$ is a positive integer.\n\nIf the sides of equal length of $\\Delta'$ are sides of the $n$-gon then $s_1 = 1$ and $n = 3 \\cdot 2$. Otherwise we can continue similarly to get $s_1 = 2s_2$ where either $s_2 = 1$ or $s_2 = 2s_3$, etc. Thus $n = 3 \\cdot 2^k$ for a natural number $k$. A construction for every $n$ of the form $3 \\cdot 2^k$ follows from the argumentation.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15197,
    "subject": "Mathematics (Olympiad)",
    "question": "For each $i = 1, 2, \\dots, N$, let $a_i$, $b_i$, $c_i$ be integers such that at least one of them is odd. Show that one can find integers $x$, $y$, $z$ such that $xa_i + yb_i + zc_i$ is odd for at least $\\frac{4N}{7}$ different values of $i$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider all the 7 triples $(x, y, z)$, where $x$, $y$, $z$ are either $0$ or $1$, but not all $0$. For each $i$, at least one of the numbers $a_i$, $b_i$, $c_i$ is odd. Thus, among the 7 sums $xa_i + yb_i + zc_i$, 3 are even and 4 are odd. Thus, there are altogether $4N$ odd sums. Therefore, there is a choice of $(x, y, z)$ for which at least $\\frac{4N}{7}$ of the corresponding sums are odd. \n\nYou can think of a table where the rows are numbered $1, 2, \\dots, N$ and the columns correspond to the 7 choices of the triples $(x, y, z)$. The 7 entries in row $i$ are the 7 sums $xa_i + yb_i + zc_i$. Thus, there are 4 odd numbers in each row, making a total of $4N$ odd sums in the table. Since there are 7 columns, one of the columns must contain at least $\\frac{4N}{7}$ odd sums.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15198,
    "subject": "Mathematics (Olympiad)",
    "question": "The bisector of the exterior angle at vertex $C$ of triangle $ABC$ intersects the bisector of the interior angle at vertex $B$ at point $K$. Consider the diameter of the circumcircle of triangle $BCK$ whose one endpoint is $K$. Prove that $A$ lies on this diameter.\n\n![](images/prob1617_p12_data_c4c35b08a5.png)\n\n![](images/prob1617_p12_data_9e76a8dfb2.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $L$ be the second endpoint of the diameter through $K$ of the circumcircle of triangle $BCK$. Then $\\angle KCL = 90^\\circ$, thus $CL$ is the bisector of the interior angle at vertex $C$ of triangle $ABC$, and $\\angle KBL = 90^\\circ$, so $BL$ is the bisector of the exterior angle at vertex $B$ of $ABC$.\n\nThe bisectors of the exterior angles at two vertices of a triangle and the bisector of the interior angle at the third vertex intersect at a common point. Thus, the bisector of the exterior angle at vertex $A$ of triangle $ABC$ passes through points $K$ and $L$. Therefore, point $A$ lies on $KL$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15199,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: (0, \\infty) \\to (0, \\infty)$ be a function such that for all $x, y > 0$,\n$$\nf(x)f(y) = f(xf(y))y = f(f(x)y).\n$$\nProve that $f(xy) = f(x)f(y)$ for all $x, y > 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Putting $x = 1$, we get $f(1)y = f(f(y))$. Now, putting $y = 1$ yields $f(1) = f(f(1))$, while putting $y = f(1)$ leads to $(f(1))^2 = f(f(f(1)))$.\n\nBut $f(f(f(1))) = f(f(1)) = f(1)$, so $f(1)^2 = f(1)$, and as $f(1) > 0$, we have $f(1) = 1$.\n\nHence $f(f(y)) = f(1)y = y$ for all $y > 0$.\n\nNow replace, in the given equation, $y$ by $f(y)$.\n\nWe then obtain $f(x)f(y) = f(xf(f(y))) = f(xy)$ for all $x > 0$ and all $y > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15200,
    "subject": "Mathematics (Olympiad)",
    "question": "Find $A_{2024}$, where\n\n$$\nA_n = 1 \\cdot 2 + 3 \\cdot 4 + 5 \\cdot 8 + \\dots + (2n - 1) \\cdot 2^n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since\n\n$$\n2A_n = 1 \\cdot 4 + 3 \\cdot 8 + \\dots + (2n - 3) \\cdot 2^n + (2n - 1) \\cdot 2^{n+1},\n$$\n\nit follows\n\n$$\n\\begin{align*}\nA_n = 2A_n - A_n &= (2n - 1) \\cdot 2^{n+1} - (1 \\cdot 2 + 2 \\cdot 4 + 2 \\cdot 8 + \\dots + 2 \\cdot 2^n) \\\\\n&= (2n - 1) \\cdot 2^{n+1} - 2 (2 + 4 + 8 + \\dots + 2^n) + 2 \\\\\n&= (2n - 1) \\cdot 2^{n+1} - 4 \\frac{2^n - 1}{2 - 1} + 2 \\\\\n&= (2n - 1) \\cdot 2^{n+1} - 4(2^n - 1) + 2 \\\\\n&= (2n - 1) \\cdot 2^{n+1} - 4 \\cdot 2^n + 4 + 2 \\\\\n&= (2n - 1) \\cdot 2^{n+1} - 4 \\cdot 2^n + 6 \\\\\n&= (2n - 3) \\cdot 2^{n+1} + 6.\n\\end{align*}\n$$\n\nTherefore,\n\n$$\nA_{2024} = (2 \\times 2024 - 3) \\cdot 2^{2025} + 6 = 4045 \\cdot 2^{2025} + 6.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15201,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with circumcenter $O$. The points $P$ and $Q$ are interior points of the sides $CA$ and $AB$, respectively. Let $K$, $L$, and $M$ be the midpoints of the segments $BP$, $CQ$, and $PQ$, respectively, and let $\\Gamma$ be the circle passing through $K$, $L$, and $M$. Suppose that the line $PQ$ is tangent to the circle $\\Gamma$. Prove that $OP = OQ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, the line $PQ$ touches the circle $\\Gamma$ at the point $M$. By the tangent-chord angle theorem, we have $\\angle QMK = \\angle MLK$. Since the points $K$ and $M$ are the midpoints of the segments $BP$ and $PQ$ respectively, $KM \\parallel BQ$, so $\\angle QMK = \\angle AQP$. Thus, $\\angle MLK = \\angle AQP$, and similarly $\\angle MKL = \\angle APQ$. As a result, $\\triangle MKL$ and $\\triangle APQ$ are similar, and thus\n$$\n\\frac{MK}{ML} = \\frac{AP}{AQ}.\n$$\nSince $K$, $L$, and $M$ are the midpoints of $BP$, $CQ$, and $PQ$ respectively, we have\n$$\nKM = \\frac{1}{2}BQ, \\quad LM = \\frac{1}{2}CP.\n$$\nPlugging this into the previous equation, we get $\\frac{BQ}{CP} = \\frac{AP}{AQ}$, i.e.\n$$\nAP \\times CP = AQ \\times BQ.\n$$\nBy the power of a point, we have\n$$\nOP^2 = OA^2 - AP \\times CP = OA^2 - AQ \\times BQ = OQ^2.\n$$\nTherefore $OP = OQ$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15202,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a positive integer. Find all sequences $(a_n)_{n \\ge 1}$ of positive integers such that\n\n$$\na_{n+2}(a_{n+1} - k) = a_n(a_{n+1} + k)$$\n\nfor all $n \\ge 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the given equation by (1). Note that if $a_n \\le k$ for $n \\ge 2$, we get a contradiction, so $a_n > k$ for all $n \\ge 2$. Then (1) becomes\n\n$$a_{n+2} = \\frac{a_n(a_{n+1} + k)}{a_{n+1} - k}.$$ \n\nWe deduce that $a_{n+2} \\ge a_n + 1$.\n\nNow $a_{n+2} = \\frac{a_n(a_{n+1}+k)}{a_{n+1}-k}$ so $a_{n+1} - k \\mid a_n(a_{n+1} + k)$. However, $a_{n+1} - k \\mid a_n(a_{n+1} - k)$ and subtracting gives $a_{n+1} - k \\mid 2ka_n$. Let $r \\ge 1$ such that $r(a_{n+1} - k) = 2ka_n$. Applying (1), $a_{n+2} = a_n + r$. We apply (1) again to get\n\n$$\n\\begin{aligned}\na_{n+3} &= \\frac{a_{n+1}(a_{n+2} + k)}{a_{n+2} - k} = \\frac{a_{n+1}(a_n + r + k)}{a_n + r - k} = a_{n+1} + \\frac{2ka_{n+1}}{a_n + r - k} \\\\\n&= a_{n+1} + \\frac{2ka_{n+1}}{\\frac{r(a_{n+1}-k)}{2k} + r - k} = a_{n+1} + \\frac{4k^2a_{n+1}}{ra_{n+1} + rk - 2k^2}.\n\\end{aligned}\n$$\n\nThus, $ra_{n+1} + rk - 2k^2 \\mid 4k^2a_{n+1}$ so $ra_{n+1} + rk - 2k^2 \\le 4k^2a_{n+1}$. If $r \\ge 4k^2$ we get a contradiction, so $r < 4k^2$. Now we use the fact that if $a, b, c \\in \\mathbb{Z}$, $a, b, c \\ne 0$ then there is a finite number of natural numbers $x$ for which $ax + b \\mid c$. So if $rk \\ne 2k^2$ (whence $r \\ne 2k$), we get a finite number of possibilities for $a_{n+1}$ (which depends on $k$ but not on $r$!). Let $S$ be this finite set. On the other hand, if $r = 2k$, it leads us to $a_{n+1} = a_n + k$.\n\nIn short, for all $n \\ge 2$ we either have $a_n \\in S$ or $a_n = a_{n-1} + k$. But $a_{n+2} \\ge a_n + 1$ for all $n \\ge 2$ so for $n$ sufficiently large, we can't have $a_n \\in S$. So there exists $N \\ge 2$ such that $a_n = a_{n-1} + k$ for all $n \\ge N$. Using $a_{N+1} = a_N + k$ we use (backward) induction to get $a_n = a_{n-1} + k$ for all $n$ so $(a_n)_{n \\ge 1}$ is an arithmetic progression of common difference $k$ and, conversely, these satisfy (1).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15203,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the following inequality:\n\n$$\n\\max\\{|x|-1,\\ x^2-1\\} \\leq \\min\\{x^2-1,\\ 1-x\\},\n$$\n\nwhere $\\max\\{a,b\\} = \\begin{cases} a, & \\text{if } a \\geq b \\\\ b, & \\text{if } a < b \\end{cases}$, and $\\min\\{a,b\\} = \\begin{cases} b, & \\text{if } a \\geq b \\\\ a, & \\text{if } a < b \\end{cases}$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is clear that\n\n$$\n\\max\\{|x|-1,\\ x^2-1\\} \\geq x^2-1 \\geq \\min\\{x^2-1,\\ 1-x\\},\n$$\n\nso the inequality from the problem condition can be satisfied only if\n\n$$\n\\max\\{|x|-1,\\ x^2-1\\} = x^2-1 = \\min\\{x^2-1,\\ 1-x\\},\n$$\n\nand this, in turn, implies that\n\n$$\n|x|-1 \\leq x^2-1 \\leq 1-|x|.\n$$\n\nTherefore, $1-|x| \\geq |x|-1$, which means that $|x| \\leq 1$. Also, we should have that $|x| \\leq |x|^2 = |x| \\cdot |x|$, and since we already know that $|x| \\leq 1$, this can happen only if $|x|=1$ or $|x|=0$, so $x=1$, $x=0$, or $x=-1$.\n\nA simple verification shows that all these\n\n![](images/Ukrajina_2011_p33_data_99473ca3ca.png)\n\nvalues of $x$ satisfy the original problem.\n\nIt is also easy to solve this problem using graphs (fig. 35). Here the graph of the left-hand side is drawn with dash-and-dash line, and the graph of the right-hand side – with dash-and-dot line. They intersect exactly at the specified values of $x$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15204,
    "subject": "Mathematics (Olympiad)",
    "question": "For a quadrilateral $ABCD$, $\\angle ABD = \\angle DBC$ and $AD = CD$. Let $DH$ be the height of $\\triangle ABD$. Prove that $|BC - BH| = HA$.\n\n![](images/UkraineMO2019_booklet_p20_data_b5bf1502f9.png)",
    "options": [],
    "answer": "See solution",
    "solution": "On the ray $BA$, place a segment $BE = BC$. If point $E$ lies on $AB$ (see figure 19), then $\\triangle BCD \\cong \\triangle BED$ due to two pairs of equal sides and the included angle. Thus, $AD = CD = ED$, so $\\triangle ADE$ is isosceles. Here, $HD$ is the height and also the median. Therefore,\n\n$$\nAH = HE = HB - BE = HB - BC \\implies BC = BH - HA.\n$$\n\nIf point $A$ lies on the segment $BE$ (see figure 20), then similarly $\\triangle BCD \\cong \\triangle BED$ and\n\n$$\nAH = HE = BE - HB = BC - HB \\implies BC = BH + HA.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15205,
    "subject": "Mathematics (Olympiad)",
    "question": "For positive integers $x$, $y$, and $z$, prove the inequality:\n\n$$\n\\frac{y}{2x + y} + \\frac{z}{2y + z} + \\frac{x}{2z + x} \\geq 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We prove it using the Cauchy–Schwarz inequality:\n\n$$\n\\left(\\frac{y}{2x+y} + \\frac{z}{2y+z} + \\frac{x}{2z+x}\\right) (x+y+z)^2 = \\left(\\frac{y}{2x+y} + \\frac{z}{2y+z} + \\frac{x}{2z+x}\\right) \\left(y(2x+y) + z(2y+z) + x(2z+x)\\right) \n$$\n\nBy Cauchy–Schwarz,\n\n$$\n\\geq \\left(\\sqrt{\\frac{y}{2x+y}} \\cdot \\sqrt{y(2x+y)} + \\sqrt{\\frac{z}{2y+z}} \\cdot \\sqrt{z(2y+z)} + \\sqrt{\\frac{x}{2z+x}} \\cdot \\sqrt{x(2z+x)}\\right)^2 = (x+y+z)^2.\n$$\n\nTherefore,\n\n$$\n\\frac{y}{2x + y} + \\frac{z}{2y + z} + \\frac{x}{2z + x} \\geq 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15206,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all primes $p$ and all positive integers $a$ and $m$ such that $a \\leq 5p^2$ and $(p-1)! + a = p^m$.",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are:\n\n$(p, a, m) = (2, 1, 1),\\ (2, 3, 2),\\ (2, 7, 3),\\ (3, 1, 1),\\ (3, 7, 2),\\ (3, 25, 3),\\ (5, 1, 2),\\ (2, 15, 4),\\ (5, 101, 3)$.\n\nDirect checks for $p = 2, 3, 5$ yield these solutions.\n\nFor $p \\geq 7$, Wilson's theorem implies $a \\equiv 1 \\pmod{p}$. Also, $p-1 \\mid a-1$, so $a = k p (p-1) + 1$ for some integer $k \\geq 0$. If $k \\geq 6$, then $a \\geq 6p^2 - 6p + 1 > 5p^2$, which is impossible. Thus, $0 \\leq k \\leq 5$.\n\nDividing by $p-1$ gives:\n\n$$\n(p-2)! + k p = p^{m-1} + p^{m-2} + \\dots + 1.\n$$\n\nSince $p \\geq 7$, $2$ and $\\frac{p-1}{2}$ are distinct and appear in $(p-2)!$, so $p-1 \\mid (p-2)!$. This gives $k \\equiv m \\pmod{p-1}$.\n\nIf $m \\geq p$, then\n\n$$\n(p-2)! + k p = p^{m-1} + p^{m-2} + \\dots + 1 > p^{p-1} > (p-1)!,\n$$\n\nwhich is a contradiction. Thus $m \\leq p-1$.\n\nIf $k=0$, then $m = p-1$ and\n\n$$\n(p-2)! = p^m - 1 \\geq p^{p-1} - 1 > (p-1)^{p-1} > (p-1)!,\n$$\n\na contradiction. If $k=1$ or $2$, then $m=1$ or $2$, which is impossible. If $k=3$, then $m=3$, i.e. $(p-2)! = (p-1)^2$, which is impossible.\n\nFor $k=4$, $m=4$ and $(p-2)! = (p-1)(p^2 + 2p - 1)$, so\n\n$$\np-2 \\mid p^2 + 2p - 1 = p^2 - 4 + 2(p-2) + 7,\n$$\n\nwhich only works for $p=7$, but does not yield solutions. For $k=5$, $m=5$ and $(p-2)! = (p-1)(p^3 + 2p^2 + 3p - 1)$, so\n\n$$\np-2 \\mid p^3 + 2p^2 + 3p - 1 = p^3 - 8 + 2(p^2 - 4) + 3(p-2) + 21,\n$$\n\nagain only for $p=7$, which does not yield solutions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15207,
    "subject": "Mathematics (Olympiad)",
    "question": "A mail carrier delivers mail to the 19 houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible?",
    "options": [],
    "answer": "See solution",
    "solution": "Consider 19 consecutive squares and color them yellow (for houses that receive mail) and white (for houses that do not). According to the conditions, no two consecutive yellow cells and no three consecutive white cells are allowed. Let $S_n$ be the number of such colorings for $n$ squares.\n\nDefine:\n- $a_n$: number of ways where the last two cells are $YW$\n- $b_n$: number of ways where the last two cells are $WY$\n- $c_n$: number of ways where the last two cells are $WW$\n\nRecurrence relations:\n- $a_{n+2} = a_n + c_n$\n- $b_{n+2} = a_n + b_n$\n- $c_{n+2} = b_n$\n\nAlso, $S_n = a_n + b_n + c_n$.\n\nFrom the relations, we find:\n$$\nS_{n+2} = 2S_n - S_{n-2} + S_{n-4}, \\quad \\forall n \\geq 5.\n$$\nWith initial values $S_1 = 2$, $S_2 = 3$, $S_3 = 4$, $S_4 = 5$, we compute $S_{19} = 351$.\n\nThus, there are $351$ possible mail delivery patterns.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15208,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence of integers $a_0, a_1, a_2, \\dots$ is called \"kawaii\" if $a_0 = 0$, $a_1 = 1$, and for any positive integer $n$, we have\n\n$$\n(a_{n+1} - 3a_n + 2a_{n-1}) (a_{n+1} - 4a_n + 3a_{n-1}) = 0\n$$\n\nAn integer is called kawaii if it belongs to a kawaii sequence.\n\nSuppose that two consecutive positive integers $m$ and $m+1$ are both kawaii (not necessarily belonging to the same kawaii sequence). Prove that $3$ divides $m$ and that $\\frac{m}{3}$ is kawaii.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\na_{n+1} = \\begin{cases} 3a_n - 2a_{n-1} \\\\ 4a_n - 3a_{n-1} \\end{cases}\n$$\n\nLet $x_n = a_n - a_{n-1}$, with $x_1 = 1 - 0 = 1$.\n\nThe recurrence gives $a_{n+1} - a_n = (3a_n - 2a_{n-1}) - a_n = 2a_n - 2a_{n-1}$ or $(4a_n - 3a_{n-1}) - a_n = 3a_n - 3a_{n-1}$, so $x_{n+1} = 2x_n$ or $3x_n$.\n\nThus, the sequence increments by powers of $2$ or $3$ at each step.\n\nLet $a_n = x_1 + x_2 + \\dots + x_n$.\n\nSuppose $m$ and $m+1$ are both kawaii. Then, for some $n$, $a_n = m$ and for some $k$, $b_k = m+1$ (possibly in different sequences).\n\nConsider the possible increments: since $a_{n+1} = a_n + x_{n+1}$, and $x_{n+1}$ is a multiple of $x_n$, the only way for two consecutive integers to both be kawaii is if $m$ is divisible by $3$ and $\\frac{m}{3}$ is also kawaii (since the recurrence allows scaling by $3$).\n\nTherefore, $3$ divides $m$ and $\\frac{m}{3}$ is kawaii.\n\n![](images/Saudi_Arabia_booklet_2024_p48_data_f8324fc428.png)\n\nFurther, considering the other case for the recurrence, similar reasoning applies.\n\n![](images/Saudi_Arabia_booklet_2024_p48_data_fead7ba086.png)\n\nThus, the result holds.\n\n![](images/Saudi_Arabia_booklet_2024_p48_data_4ad25f7b94.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15209,
    "subject": "Mathematics (Olympiad)",
    "question": "On a piece of $8 \\times 8$ graph paper, what is the minimum number of grids that must be removed so that it is impossible to cut out a \"T\" shape consisting of five grids as shown below?\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p172_data_292ad4c5ba.png)",
    "options": [],
    "answer": "See solution",
    "solution": "At least $14$ grids must be removed. An example is shown below:\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p172_data_6763042fa9.png)\n\nTo ensure no \"T\" shape of five grids remains, divide the $8 \\times 8$ grid into five regions (see below):\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p172_data_92da9728b4.png)\n\nFrom the central region, at least $2$ grids must be removed; otherwise, a \"T\" can be formed (see below):\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p172_data_f23de7b3d6.png)\n\nFor each of the four corner regions, at least $3$ grids must be removed to prevent forming a \"T\". For example, in the upper right corner:\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p172_data_1490f6b0cd.png)\n\nIf only $2$ grids are removed from a corner, a \"T\" can still be formed:\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p173_data_fefaadc5d0.png)\n\nTherefore, the minimum is $2 + 3 \\times 4 = 14$ grids.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 15210,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A\\Beta\\Gamma$ be a triangle and $M$, $N$ the midpoints of $AB$, $A\\Gamma$, respectively. The points $\\Delta$ and $E$ lie on the segment $BN$, such that $\\Gamma\\Delta \\parallel ME$ and $B\\Delta < BE$. Prove that:\n\n$$\nB\\Delta = 2 \\cdot EN.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $ME \\parallel \\Gamma\\Delta$, it follows that $\\angle \\Gamma\\Delta E = \\angle M E \\Delta$, and so\n\n$$\n180^\\circ - \\angle \\Gamma\\Delta E = 180^\\circ - \\angle M E \\Delta \\implies \\angle B\\Delta\\Gamma = \\angle M E N \\quad (1)\n$$\n\nSince the points $M$, $N$ are the midpoints of the sides $AB$, $A\\Gamma$, respectively, we conclude that\n\n$$\nMN \\parallel B\\Gamma, \\quad MN = \\frac{B\\Gamma}{2}. \\quad (2)\n$$\n\nand hence\n\n$$\n\\angle \\Delta B \\Gamma = \\angle M N E, \\quad (3)\n$$\n\nFrom (1) and (3) we get that the triangles $B\\Delta\\Gamma$ and $MEN$ are similar, and so:\n\n$$\n\\frac{B\\Delta}{EN} = \\frac{B\\Gamma}{MN} = 2 \\implies B\\Delta = 2 \\cdot EN.\n$$\n\n![](images/Greek2023_p6_data_ad9e18db57.png \"Figure 1\")\n\nSecond solution. From $\\Gamma$ we draw the parallel to the line $BN$, which meets the line $AE$ at $Z$. Since $N$ is the midpoint of $B\\Gamma$ and $NE \\parallel \\Gamma Z$, it follows that $E$ is the midpoint of $AZ$ and $\\Gamma Z = 2 \\cdot EN$.\n\nSince $M$, $E$ are the midpoints of $AB$, $AZ$, respectively, we have: $ME \\parallel B\\Gamma$.\n\nTherefore the quadrilateral $BZ\\Gamma\\Delta$ is a parallelogram, as it has two pairs of opposite sides parallel. Hence $B\\Delta = \\Gamma Z = 2 \\cdot EN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15211,
    "subject": "Mathematics (Olympiad)",
    "question": "Real numbers $x, y, z$ satisfy\n\n$$\n\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} + x + y + z = 0\n$$\n\nand none of them lies in the open interval $(-1, 1)$. Find the maximum value of $x + y + z$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "By changing $(x, y, z)$ to $(-x, -y, -z)$, the condition remains valid and $x + y + z$ changes sign. Since the ordering of $x, y, z$ is irrelevant and they cannot all have the same sign due to the equation, we may assume $x > 0$, $y > 0$, $z < 0$, and seek the maximum of $V = |x + y + z|$.\n\nTransform the equation to:\n\n$$\nf(x) + f(y) = f(t), \\quad \\text{where} \\quad t := -z > 0, \\quad f(x) = x + \\frac{1}{x},\n$$\n\nwith $x, y, t \\in (1, \\infty)$. The function $f$ is an increasing bijection from $(1, \\infty)$ to $(2, \\infty)$. Thus,\n\n$$\nf(t) = f(x) + f(y) > f(x + y)\n$$\n\nso $t > x + y$, and $x + y + z = x + y - t < 0$. Therefore, maximize $V = t - x - y$.\n\nA valid triple is $(1, 1, t_0)$, where $f(t_0) = 4$, so $t_0 = 2 + \\sqrt{3}$, giving $V = \\sqrt{3}$. We show this is maximal.\n\n**Lemma:** For valid triples $(x, y, t)$ and $(x, y', t')$ with $y' < y$, $t - x - y < t' - x - y'$. Thus, maximizing $V$ leads to $(1, 1, 2 + \\sqrt{3})$.\n\n*Proof of lemma:* Since $f(y') < f(y)$, $f(x) < f(t') < f(t)$, so $1 < t' < t$. Then $1 < t'y' < ty$. From\n\n$$\nf(x) = f(t) - f(y) = (t - y) \\left(1 - \\frac{1}{ty}\\right) = f(t') - f(y') = (t' - y') \\left(1 - \\frac{1}{t'y'}\\right),\n$$\n\nand $0 < 1 - \\frac{1}{t'y'} < 1 - \\frac{1}{ty}$, we get $t - y < t' - y'$, i.e., $t - x - y < t' - x - y'$.\n\nTherefore, the maximum value of $x + y + z$ is $\\boxed{\\sqrt{3}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15212,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$, $D$, and $K$ belong to the sides $AB$, $BC$, and $CA$ of isosceles triangle $ABC$ with apex $B$ such that $AM = 2DC$ and $\\angle AMD = \\angle KDC$. Show that $MD = KD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $FD \\parallel AC$. Then $AF = FM = DC$, hence $\\triangle FMD \\cong \\triangle KDC$ by the side and adjacent angles. Thus, $MD = KD$.\n\n![](images/Ukraine_booklet_2018_p23_data_99cf4ddde0.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15213,
    "subject": "Mathematics (Olympiad)",
    "question": "A directed graph $G$ has 2021 vertices located on a circle. From each vertex, there are 101 directed edges to the next 101 vertices in the counter-clockwise direction. We wish to colour the edges so that for any two vertices $u$ and $v$, one can choose a directed path from $u$ to $v$ in which no two edges have the same colour. What is the minimum number of colours needed?",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum number of colours needed is $\\lceil \\frac{2021}{101} \\rceil = 21$.\n\nLabel the vertices along the cycle as $v_0, v_1, \\dots, v_{2020}$. The distance from $v_0$ to any of $v_1, \\dots, v_{101}$ is 1; from $v_0$ to $v_{102}, \\dots, v_{202}$ is 2, and so on. The longest distance from $v_0$ to $v_{2020}$ is 20.\n\nFor each $i$, $1 \\leq i \\leq 20$, let\n$$\nV_i = \\{v_{101(i-1)+1}, v_{101(i-1)+2}, \\dots, v_{101i}\\}\n$$\nColour the edges that go from vertices of $V_i$ with the $i$-th colour. The edges starting from $v_0$ are coloured with the 21st colour. For any two vertices $v_i$ and $v_j$, we can choose a directed path from $v_i$ to $v_j$ that intersects each $V_\\ell$ in at most one vertex (except possibly $v_i$ and $v_j$ belonging to the same $V_\\ell$). The edges of this path have pairwise different colours.\n\nSuppose we try to colour the edges with only 20 colours. Consider a path that, at each step, jumps from $v_i$ to $v_{i+101}$ (addition modulo 2021). Since $\\gcd(101, 2021) = 1$, this path forms a Hamiltonian cycle $C$.\n\nFor each $v_i$, the sub-path from $v_i$ to $v_{i+2020}$ in $C$ consists of 20 edges, which is the shortest path from $v_i$ to $v_{i+2020}$. With only 20 colours, the edges of this path must all have different colours. Thus, the edge between $v_i$ and $v_{i+101}$ and the edge from $v_{i+2020}$ to $v_{i+2121}$ must be the same colour. Since $\\gcd(2020, 2021) = 1$, this implies all edges in $C$ have the same colour, which is a contradiction.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 15214,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with circumcircle $\\Gamma$ and circumcenter $O$. Denote by $M$ the midpoint of $BC$. Point $D$ is the reflection of $A$ over $BC$, and $E$ is the intersection of $\\Gamma$ and ray $MD$. Let $S$ be the circumcenter of triangle $ADE$. Prove that $A$, $E$, $M$, $O$, and $S$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "First we prove that $A$, $M$, $E$, $S$ are concyclic. Note that $BC$ is the perpendicular bisector of $AD$, so $S$ lies on $BC$. Let $X$ be the intersection of $AD$ and $BC$ as in the figure below.\n\n$$\n\\begin{aligned}\n\\angle EMS &= \\angle DMX \\\\\n&= 90^\\circ - \\angle XDM \\\\\n&= 90^\\circ - \\angle ADE \\\\\n&= \\angle SEA \\\\\n&= \\angle EAS\n\\end{aligned}\n$$\n\nso $AMES$ is cyclic as claimed.\n\nNow we prove that $A$, $M$, $E$, $O$ are concyclic. Let $F$ be the reflection of $D$ over $M$. Then $F$ lies on the same side of $BC$ as $A$ and satisfies $\\triangle FCB \\cong \\triangle DBC \\cong \\triangle ABC$, so $F$ must be the point such that $AFCB$ is an isosceles trapezoid. In particular, $F$ lies on $\\Gamma$. Consequently,\n\n$$\n\\angle OAE = 90^\\circ - \\angle EFA = 90^\\circ - \\angle EMB = \\angle OMB + \\angle BME = \\angle OME\n$$\n\nso $AMEO$ is also cyclic as claimed.\n\n![](images/BW2021_Shortlist_p35_data_8576431068.png)\n\nThus $A$, $E$, $M$, $O$, $S$ are concyclic, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15215,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure, points $C$ and $D$ are on the semicircle with $O$ as its center and $AB$ as its diameter. The tangent line to the semicircle at point $B$ meets the line $CD$ at point $P$. Line $PO$ intersects $CA$ and $AD$ at points $E$ and $F$ respectively. Prove that $OE = OF$.",
    "options": [],
    "answer": "See solution",
    "solution": "Draw line segments $OM$ and $MN$, such that $OM \\perp CD$ and $MN \\parallel AD$. Let $MN \\cap BA = N$, $CN \\cap DA = K$, and connect $BC$ and $BM$. Then we get\n\n$$\n\\angle NBC = \\angle ADC = \\angle NMC,\n$$\n\nwhich means points $N$, $B$, $M$, $C$ are concyclic. Since $O$, $B$, $P$, $M$ are also concyclic points, we obtain\n\n$$\n\\angle OPM = \\angle OBM = 180^\\circ - \\angle MCN,\n$$\n\nthus $CN \\parallel OP$ and\n\n$$\n\\frac{CN}{OE} = \\frac{AN}{AO} = \\frac{NK}{OF}.\n$$\n\nSince $M$ is the midpoint of $CD$, $MN \\parallel DK$, then we get that $N$ is the midpoint of $CK$. Hence, according to the previous result, we have $OE = OF$ as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15216,
    "subject": "Mathematics (Olympiad)",
    "question": "The angle bisector of the acute angle formed at the origin by the graphs of the lines $y = x$ and $y = 3x$ has equation $y = kx$. What is $k$?\n\n(A) $\\frac{1+\\sqrt{5}}{2}$  \n(B) $\\frac{1+\\sqrt{7}}{2}$  \n(C) $\\frac{2+\\sqrt{3}}{2}$  \n(D) $2$  \n(E) $\\frac{2+\\sqrt{5}}{2}$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the line $x = 1$, which intersects the lines $y = x$, $y = 3x$, and $y = kx$ at $(1, 1)$, $(1, 3)$, and $(1, k)$, respectively, as shown below.\n\n![](images/2021_AMC12A_Solutions_Fall_p4_data_f5022e6154.png)\n\nApplying the Angle Bisector Theorem gives\n\n$$\n\\frac{3-k}{\\sqrt{10}} = \\frac{k-1}{\\sqrt{2}},\n$$\n\nfrom which $k = \\frac{1+\\sqrt{5}}{2}$.\n\n**OR**\n\nLet $\\alpha$ and $\\beta$ be the acute angles from the positive x-axis to the lines $y = x$ and $y = 3x$, respectively. Then the bisector has angle $\\gamma = \\frac{1}{2}(\\alpha + \\beta)$ from the positive x-axis. Because $\\tan \\alpha = 1$ and $\\tan \\beta = 3$, it follows that $\\sin \\alpha = \\cos \\alpha = \\frac{1}{\\sqrt{2}}$, $\\sin \\beta = \\frac{3}{\\sqrt{10}}$, and $\\cos \\beta = \\frac{1}{\\sqrt{10}}$. Using a half-angle identity for tangent and the sum identities for sine and cosine gives\n\n$$\n\\begin{aligned}\nk &= \\tan \\gamma = \\frac{\\sin(\\alpha + \\beta)}{1 + \\cos(\\alpha + \\beta)} \\\\\n&= \\frac{\\sin \\alpha \\cos \\beta + \\sin \\beta \\cos \\alpha}{1 + \\cos \\alpha \\cos \\beta - \\sin \\alpha \\sin \\beta} \\\\\n&= \\frac{\\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{10}} + \\frac{3}{\\sqrt{10}} \\cdot \\frac{1}{\\sqrt{2}}}{1 + \\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{10}} - \\frac{1}{\\sqrt{2}} \\cdot \\frac{3}{\\sqrt{10}}} \\\\\n&= \\frac{2}{\\sqrt{5} - 1} \\\\\n&= \\frac{1 + \\sqrt{5}}{2}.\n\\end{aligned}\n$$\n\n**OR**\n\nThe lines $y = x$ and $y = 3x$ form an acute angle at the origin. The bisector of that angle passes through the origin and the midpoint of a line segment between two points on those lines in the first quadrant that are equally distant from the origin. The point $(1, 1)$ is on the line $y = x$ and is $\\sqrt{2}$ units from the origin. Let $(p, 3p)$ be the point on the line $y = 3x$ in the first quadrant that is $\\sqrt{2}$ units from the origin. Then $p^2 + (3p)^2 = 2$, from which $p = \\frac{1}{\\sqrt{5}} = \\frac{1}{5}\\sqrt{5}$. The midpoint between $(1, 1)$ and $(\\frac{1}{5}\\sqrt{5}, \\frac{3}{5}\\sqrt{5})$ is\n\n$$\n\\left( \\frac{1}{2} \\left( 1 + \\frac{1}{5}\\sqrt{5} \\right), \\frac{1}{2} \\left( 1 + \\frac{3}{5}\\sqrt{5} \\right) \\right) = \\left( \\frac{\\sqrt{5}+5}{10}, \\frac{3\\sqrt{5}+5}{10} \\right).\n$$\n\nThe slope of the line through this midpoint and the origin is\n\n$$\nk = \\frac{3\\sqrt{5} + 5}{\\sqrt{5} + 5} = \\frac{1 + \\sqrt{5}}{2}.\n$$\n\n**Note:** The value $\\frac{1+\\sqrt{5}}{2}$ is the golden ratio.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 15217,
    "subject": "Mathematics (Olympiad)",
    "question": "Given positive integers $m, n$ and non-negative real numbers $a_0, a_1, \\dots, a_m, b_0, b_1, \\dots, b_n$. Define\n\n$$\nc_k = \\max_{i+j=k} a_i b_j \\quad (k = 0, 1, \\dots, m+n).\n$$\n\nProve that:\n\n$$\n\\frac{c_0 + c_1 + \\dots + c_{m+n}}{m+n+1} \\ge \\frac{a_0 + a_1 + \\dots + a_m}{m+1} \\cdot \\frac{b_0 + b_1 + \\dots + b_n}{n+1}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the set of points $\\mathcal{P} = \\{(i_0, j_0), (i_1, j_1), \\dots, (i_{m+n}, j_{m+n})\\}$ as a \"monotone path\" from $(0,0)$ to $(m,n)$ if $(i_0, j_0) = (0,0)$, $(i_{m+n}, j_{m+n}) = (m,n)$, and for every index $0 \\le k < m+n$, we have\n\n$$\n(i_{k+1}, j_{k+1}) \\in \\{(i_k + 1, j_k), (i_k, j_k + 1)\\}.\n$$\n\nNoting that $i_k + j_k = k$, we get\n\n$$\n\\sum_{k=0}^{m+n} a_{i_k} b_{j_k} \\le \\sum_{k=0}^{m+n} c_{i_k + j_k} = \\sum_{k=0}^{m+n} c_k.\n$$\n\nWe call the left-hand side the weight of the monotone path $\\mathcal{P}$, denoted $W(\\mathcal{P})$. To prove the conclusion, it suffices to show: there exists a monotone path $\\mathcal{P}$ such that\n\n$$\nW(\\mathcal{P}) \\ge \\frac{m+n+1}{(m+1)(n+1)} \\sum_{i=0}^{m} a_i \\sum_{j=0}^{n} b_j.\n$$\n\n**Lemma:** Let $0 \\le x, y \\le 1$, then\n\n$$\nxy + \\max \\left\\{ \\frac{m+n}{m(n+1)}(1-x), \\frac{m+n}{(m+1)n}(1-y) \\right\\} \\ge \\frac{m+n+1}{(m+1)(n+1)}.\n$$\n\n*Proof of the lemma:* Let the left-hand side be $L$. By symmetry, assume $\\frac{m+n}{m(n+1)}(1-x) \\ge \\frac{m+n}{(m+1)n}(1-y)$, then\n\n$$\ny \\ge 1 - \\frac{(m+1)n}{m(n+1)}(1-x).\n$$\n\nCombining with $x \\ge 0$,\n\n$$\n\\begin{aligned}\nL &\\ge x \\left(1 - \\frac{(m+1)n}{m(n+1)}(1-x)\\right) + \\frac{m+n}{m(n+1)}(1-x) \\\\\n  &= \\frac{m+n+1}{(m+1)(n+1)} + \\frac{n}{m(m+1)(n+1)}((m+1)x - 1)^2 \\\\\n  &\\ge \\frac{m+n+1}{(m+1)(n+1)}.\n\\end{aligned}\n$$\n\nThis completes the lemma.\n\nReturning to the main proof, use induction on $m+n$. When $m=0$ or $n=0$, the proposition is clear. Assume it holds for $m+n < \\ell$, consider $m+n = \\ell$ with $m, n > 0$. By homogeneity, assume $\\sum_{i=0}^m a_i = \\sum_{j=0}^n b_j = 1$, so $0 \\le a_0, b_0 \\le 1$. Let\n\n$$\nS = \\max_{\\text{monotone path } \\mathcal{P}} W(\\mathcal{P}).\n$$\n\nBy induction, there exists a monotone path $P_0$ from $(1,0)$ to $(m,n)$ such that\n\n$$\nW(P_0) \\ge \\frac{m+n}{m(n+1)} \\sum_{i=1}^{m} a_i \\sum_{j=0}^{n} b_j = \\frac{m+n}{m(n+1)}(1-a_0).\n$$\n\nLet $\\mathcal{P}$ be the path from $(0,0)$ to $(1,0)$ then $P_0$, so\n\n$$\nS \\ge W(\\mathcal{P}) = a_0 b_0 + \\frac{m+n}{m(n+1)}(1-a_0).\n$$\n\nSimilarly, going from $(0,0)$ to $(0,1)$ and then using induction,\n\n$$\nS \\ge a_0 b_0 + \\frac{m+n}{(m+1)n}(1-b_0).\n$$\n\nCombining and using the lemma,\n\n$$\nS \\ge a_0 b_0 + \\max \\left\\{ \\frac{m+n}{m(n+1)}(1-a_0), \\frac{m+n}{(m+1)n}(1-b_0) \\right\\} \\ge \\frac{m+n+1}{(m+1)(n+1)}.\n$$\n\nThis completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15218,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f$ defined on the positive real numbers and taking real values such that\n\n$$\nf(x) + f(y) \\leq \\frac{f(x + y)}{2}, \\quad \\frac{f(x)}{x} + \\frac{f(y)}{y} \\geq \\frac{f(x + y)}{x + y}\n$$\n\nfor all positive real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Substitute $x = y = t$ (with $t > 0$) into the inequalities:\n\n- $4f(t) \\leq f(2t)$\n- $4f(t) \\geq f(2t)$\n\nThus, $f(2t) = 4f(t)$ for $t > 0$. Repeating this $m$ times gives $f(2^m t) = 2^{2m} f(t)$.\n\nLet $g(x) = \\frac{f(x)}{x}$. We show $g(nt) = n g(t)$ for any positive integer $n$ and $t > 0$. For $n = 2^m$, this is already shown. Using $g(x) + g(y) \\geq g(x + y)$, for positive integers $n$ and $t > 0$:\n\n$$\ng(nt) \\leq n g(t)\n$$\n\nChoose $m$ so $2^m > n$:\n\n$$\ng(2^m t) \\leq g(nt) + g((2^m - n)t) \\leq n g(t) + (2^m - n) g(t) = 2^m g(t)\n$$\n\nBut $g(2^m t) = 2^m g(t)$, so $g(nt) + g((2^m - n)t) = n g(t) + (2^m - n) g(t)$. Thus, $g(nt) = n g(t)$.\n\nNext, we show $g$ is monotone decreasing. For $t > 0$:\n\n- $f(t) + f(2t) \\leq \\frac{f(3t)}{2}$\n- $f(t) = t g(t)$, $f(2t) = 2t g(2t) = 4t g(t)$, $f(3t) = 3t g(3t) = 9t g(t)$\n\nSo $5 g(t) \\leq \\frac{9}{2} g(t)$, hence $g(t) \\leq 0$ for $t > 0$. For $0 < x \\leq y$, $g(x) \\geq g(x) + g(y - x) \\geq g(y)$, so $g$ is monotone decreasing.\n\nLet $g(1) = a \\leq 0$. We show $g(t) = a t$ for $t > 0$. Suppose $g(t) < a t$ for some $t > 0$. There exists a rational $\\frac{p}{q} > t$ with $g(t) < \\frac{p a}{q}$. But $g(\\frac{p}{q}) = \\frac{1}{q} g(p) = \\frac{p}{q} g(1) = \\frac{p a}{q}$, so $g(t) < g(\\frac{p}{q})$ with $\\frac{p}{q} > t$, contradicting monotonicity. Similarly, $g(t) > a t$ leads to a contradiction. Thus, $g(t) = a t$ for $t > 0$.\n\nTherefore, $f(x) = x g(x) = a x^2$ with $a \\leq 0$. For such $f$:\n\n- $f(x) + f(y) - \\frac{f(x + y)}{2} = \\frac{a}{2} (x - y)^2 \\leq 0$\n- $\\frac{f(x)}{x} + \\frac{f(y)}{y} - \\frac{f(x + y)}{x + y} = 0$\n\nThus, all such $f$ satisfy the problem's conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15219,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle with $AB < AC$, inscribed in the circle $c(O, R)$. The extension of the altitude $AD$ intersects the circumcircle $c(O, R)$ at $E$, and the perpendicular bisector ($\\mu$) of the side $AB$ meets $AD$ at $L$. The line $BL$ meets $AC$ at $M$ and the circumcircle $c(O, R)$ at $N$. Finally, $EN$ meets ($\\mu$) at $Z$. Prove that:\n\n$$\nMZ \\perp BC \\Leftrightarrow (CA = CB \\text{ or } Z \\equiv O).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Direct.** Let $MZ \\perp BC$. Then $MZ \\parallel AD$, since $AD \\perp BC$. Therefore we have: $Z\\hat{M}C = E\\hat{A}C$. From the quadrilateral $EANC$ we have: $E\\hat{A}C = E\\hat{N}C$ and hence we get\n\n$$\nZ\\hat{M}C = E\\hat{A}C = E\\hat{N}C = Z\\hat{N}C.\n$$\n\nTherefore the quadrilateral $MNCZ$ is cyclic and then it follows\n\n$$\nA\\hat{C}Z = M\\hat{C}Z = M\\hat{N}Z = B\\hat{N}E.\n$$\n\n![](images/Hellenic_Mathematical_Competitions_2011_booklet_p13_data_60fa9b72f0.png \"Figure 3\")\n\nAlso we have $B\\hat{N}E = B\\hat{A}E$, (from the cyclic quadrilateral $ABEN$). Therefore we have\n\n$$\nA\\hat{C}Z = B\\hat{A}E = B\\hat{A}D. \\qquad (1)\n$$\n\nFrom the right angled triangle $ABD$ we have\n\n$$\nB\\hat{A}D = 90^\\circ - A\\hat{B}D = 90^\\circ - A\\hat{B}C, \\qquad (2)\n$$\n\nand then\n\n$$\nA\\hat{C}Z = 90^\\circ - A\\hat{B}C. \\qquad (3)\n$$\n\nSince $O$ is the circumcentre of the acute angled triangle $ABC$ we get\n\n$$\nA\\hat{C}O = 90^\\circ - A\\hat{B}C. \\qquad (4)\n$$\n\nFrom (3) and (4) we obtain $A\\hat{C}Z = 90^\\circ - A\\hat{B}C = A\\hat{C}O$ and since the points $Z$ and $O$ lie on the same half plane with respect to $AC$, it follows that the points $C$, $O$ and $Z$ are collinear.\n\nTherefore we distinguish the cases:\n\n* $O \\equiv Z$, and then we have our result.\n* $O \\not\\equiv Z$, and then the straight line $OZ$ is the perpendicular bisector of the side $AB$. Since $C$ belongs to $OZ$, we conclude that $CA = CB$.\n\n### Converse\n\nIf $CA = CB$ or $O \\equiv Z$, then the points $C$, $O$ and $Z$ are collinear, because they belong to the perpendicular bisector of the side $AB$. Hence\n\n$$\nA\\hat{C}Z = A\\hat{C}O = 90^\\circ - A\\hat{B}C = B\\hat{A}D = B\\hat{A}E = B\\hat{N}E = M\\hat{N}Z \\Rightarrow M\\hat{C}Z = M\\hat{N}Z,\n$$\n\nand so the quadrilateral $MNCZ$ is cyclic. It gives that\n\n$$\nZ\\hat{M}C = Z\\hat{N}C = E\\hat{N}C = E\\hat{A}C.\n$$\n\nTherefore $ZM \\parallel AE$ and since $AE \\perp BC$, it follows that $ZM \\perp BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15220,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Find the value of $\\sigma_1 = a + b + c$.\n\nb) Is the following system of equations consistent?\n\nLet $\\sigma_1 = a + b + c$, $\\sigma_2 = ab + bc + ca$, $\\sigma_3 = abc$. The given equalities are:\n\n$$\n\\sigma_1 = \\sigma_1^2 - 2\\sigma_2 \\quad (1), \\quad \\sigma_1^2 = \\sigma_1^3 - 3\\sigma_1\\sigma_2 + 3\\sigma_3. \\quad (2)\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "a) $3$;\n\nb) Yes.\n\n(Solution of M.Mankevich, E.Dovgialo.)\n\nFrom the given equalities, we have the system:\n\n$$\n\\sigma_1 = \\sigma_1^2 - 2\\sigma_2 \\quad (1), \\quad \\sigma_1^2 = \\sigma_1^3 - 3\\sigma_1\\sigma_2 + 3\\sigma_3. \\quad (2)\n$$\n\nFrom (1) we have $\\sigma_1^2 = \\sigma_1^3 - 2\\sigma_1\\sigma_2$. Then from (2) it follows that\n\n$$\n-2\\sigma_1\\sigma_2 = -\\sigma_1\\sigma_2 + 3\\sigma_3. \\tag{3}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15221,
    "subject": "Mathematics (Olympiad)",
    "question": "The square is partitioned into 4 equal rectangles and 1 square (see the figure below). The area of the small square is $16$, and the area of each rectangle is $96$. Find the side lengths of the rectangles.\n\n![](images/Ukrajina_2011_p7_data_b47aa3981b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The side of the small square is $4$, since $\\sqrt{16} = 4$. The difference between the two sides of the rectangle is $4$. Let the sides of the rectangle be $x$ and $x+4$.\n\nThe area of the large square is $16 + 4 \\times 96 = 400$, so its side length is $20$.\n\nThe side of the large square consists of two different sides of the rectangle, so $2x + 4 = 20$. Solving for $x$:\n\n$$\n2x + 4 = 20 \\\\\n2x = 16 \\\\\nx = 8\n$$\n\nTherefore, the sides of the rectangle are $8$ and $12$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15222,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, let $H$ be the orthocenter and $AK$ an altitude. Circle $w$ passes through points $A$ and $K$ and intersects sides $AB$ and $AC$ at points $M$ and $N$, respectively. The line through $A$ parallel to $BC$ intersects the circumscribed circles of triangles $AHM$ and $AHN$ a second time at points $X$ and $Y$, respectively. Prove that $XY = BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $Z$ be the point where circle $w$ meets line $BC$; then $AZ$ is a diameter of $w$. If $K = Z$, $w$ is tangent to $BC$, and since $AK \\perp BC$, the center of $w$ lies on $AK$. If $K \\neq Z$, then $\\angle AKZ = 90^\\circ$, so $AZ$ is a diameter of $w$. Thus, $\\angle AMZ = 90^\\circ$. Let line $MZ$ meet the circumscribed circle of $\\triangle XAH$ again at $S$; then $\\angle AMS = 90^\\circ$. Also, $\\angle SHA = \\angle SMA$, so $SH \\perp AH$. Therefore, $AX \\parallel SH \\parallel BC$, and $XAHS$ is a rectangle, so $AX = SH$. Note that $CH \\perp AB$ and $ZM \\perp AB$, so $HC \\parallel SZ$. Thus, $SHCZ$ is a parallelogram, so $SH = ZC$. Therefore, $AX = SZ$. Similarly, $AY = BZ$, so $XY = BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15223,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that\n\n$$\n\\sum_{n=1}^{2022} \\frac{n^2}{(n+1)!} - \\sum_{n=1}^{2022} \\frac{1}{n!} < 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n\n$$\n\\frac{n^2}{(n+1)!} - \\frac{1}{n!} = \\frac{n(n+1) - 2(n+1) + 1}{(n+1)!} = \\frac{1}{(n-1)!} - \\frac{2}{n!} + \\frac{1}{(n+1)!}\n$$\n\nWith this identity, we have:\n\n$$\n\\begin{aligned}\n& \\sum_{n=1}^{2022} \\frac{n^2}{(n+1)!} - \\sum_{n=1}^{2022} \\frac{1}{n!} = \\sum_{n=0}^{2021} \\frac{1}{n!} - 2 \\sum_{n=1}^{2022} \\frac{1}{n!} + \\sum_{n=2}^{2023} \\frac{1}{n!} \\\\\n& = \\frac{1}{0!} - \\frac{1}{1!} - \\frac{1}{2022!} + \\frac{1}{2023!} = \\frac{1}{2023!} - \\frac{1}{2022!} < 0.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15224,
    "subject": "Mathematics (Olympiad)",
    "question": "Sea el punto $A$ dado por:\n\n$$\nA \\equiv \\left( \\frac{2r^2\\delta}{r^2 + \\delta^2 - \\Delta^2}, \\frac{2r(\\delta^2 - \\Delta^2)}{r^2 + \\delta^2 - \\Delta^2} \\right).\n$$\n\nDetermina la pendiente de la recta *AE* y analiza los casos en los que los denominadores de las expresiones pueden anularse. Explica el significado geométrico de estos casos.",
    "options": [],
    "answer": "See solution",
    "solution": "La pendiente de la recta *AE* viene dada por la diferencia entre sus coordenadas $y$:\n\n$$\n\\frac{2r^3\\Delta^2}{(r^2 + \\delta^2)(r^2 + \\delta^2 - \\Delta^2)}\n$$\n\ndividida entre la diferencia de sus coordenadas $x$:\n\n$$\n\\frac{2r^2\\delta\\Delta^2}{(r^2 + \\delta^2)(r^2 + \\delta^2 - \\Delta^2)}\n$$\n\nEsto es igual a $-\\frac{r}{\\delta}$, que coincide con la pendiente de *DI*, por lo tanto, hemos terminado.\n\n*Nota:* Puede haber denominadores que se anulen, casos que deben tratarse aparte. Los denominadores que podrían anularse son $r^2 - (\\delta - \\Delta)^2$, $r^2 - (\\delta + \\Delta)^2$ y $r^2 + \\delta^2 - \\Delta^2$. Si $r^2 = (\\delta - \\Delta)^2$, la recta *AB* sería vertical y la coordenada $x$ de $A$ sería\n\n$$\n\\frac{2(\\delta - \\Delta)^2\\delta}{(\\delta - \\Delta)^2 + \\delta^2 - \\Delta^2}\n$$\n\nidéntica a la de $B$, quedando la coordenada $y$ de $A$ determinada por la recta *AC*.\n\nDe forma análoga, si $r^2 = (\\delta + \\Delta)^2$, la recta *AC* sería vertical, siendo la coordenada $x$ de $A$ igual a $\\delta + \\Delta$, y su coordenada $y$ determinada por la recta *AB*.\n\nFinalmente, como $\\Delta = \\frac{a}{2}$ y $\\delta = \\left|\\frac{a}{2} - \\frac{c+a-b}{2}\\right| = \\frac{|b-c|}{2}$, tenemos que $r^2 = \\Delta^2 - \\delta^2$ es equivalente a\n\n$$\nr^2 = \\frac{a^2 - (b-c)^2}{4} = \\frac{a^2 - b^2 - c^2 + 2bc}{4} = bc \\sin^2 \\frac{A}{2} = r^2 \\cot \\frac{B}{2} \\cot \\frac{C}{2},\n$$\n\ndonde se ha usado el resultado conocido $r = 4R \\sin \\frac{A}{2} \\sin \\frac{B}{2} \\sin \\frac{C}{2}$. Esto sería equivalente a $\\cos \\frac{B+C}{2} = 0$, lo cual es absurdo pues $B+C < 90^\\circ$. La condición para que ese tercer denominador se anule es $B+C = 180^\\circ$, lo que significa que el punto $A$ está \"en el infinito\", es decir, que las rectas *AB* y *AC* son paralelas.\n\nEn definitiva, los denominadores que pueden anularse no suponen un problema en la solución, sino que indican que dos de las rectas definidas son verticales.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15225,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$ and two points $D \\in AC$, $E \\in BD$ such that $\\angle DAE = \\angle AED = \\angle ABC$. Show that $BE = 2CD$ if and only if $\\angle ACB = 90^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\alpha = \\angle A$, $\\beta = \\angle B$, and $\\gamma = \\angle C$. Then\n\n$$\n\\frac{BE}{\\sin(\\alpha - \\beta)} = \\frac{AB}{\\sin(\\pi - \\beta)}, \\quad \\frac{CD}{\\sin(\\alpha - \\beta)} = \\frac{BC}{\\sin 2\\beta}, \\quad \\frac{AB}{\\sin \\gamma} = \\frac{BC}{\\sin \\alpha}\n$$\n\nand so\n\n$$\n\\frac{BE}{CD} = \\frac{\\sin 2\\beta \\sin \\gamma}{\\sin \\beta \\sin \\alpha} = \\frac{2 \\cos \\beta \\sin \\gamma}{\\sin \\alpha} = \\frac{2 \\cos \\beta \\sin \\gamma}{\\sin \\beta \\cos \\gamma + \\cos \\beta \\sin \\gamma}\n$$\n\nTherefore, $BE = 2CD$ if and only if $\\cos \\gamma = 0$, i.e., $\\gamma = 90^\\circ$. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15226,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x)f(y) = f(xy - 1) + x f(y) + y f(x), \\quad \\forall x, y \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $y = 0$ in (1):\n$$\nf(x) f(0) = f(-1) + x f(0).\n$$\n\nConsider two cases for $f(0)$:\n\n**Case 1:** $f(0) \\neq 0$ implies $f(x) = x + c$ for all $x \\in \\mathbb{R}$ with $c$ constant, which is not a solution.\n\n**Case 2:** $f(0) = 0$ implies $f(-1) = 0$.\n\nPlug $x = y = 1$ into (1):\n$$\nf(1)^2 = 2 f(1) \\implies f(1) = 0 \\text{ or } f(1) = 2.\n$$\nFrom (1), substitute $y = -1$:\n$$\nf(-y - 1) = f(y), \\quad \\forall y \\in \\mathbb{R}. \\tag{2}\n$$\nReplace $y$ by $-y - 1$ in (1):\n$$\nf(x) f(-y-1) = f(-x(y+1)-1) + x f(-y-1) - (y+1) f(x), \\quad \\forall x, y \\in \\mathbb{R}.\n$$\nUsing (2):\n$$\nf(xy - 1) + y f(x) = f(xy + x) - (y + 1) f(x), \\quad \\forall x, y \\in \\mathbb{R}. \\tag{3}\n$$\nLet $x \\neq -1$, replace $x$ by $x + 1$ and $y$ by $\\frac{1}{x+1}$ in (3):\n$$\nf(x-1) = \\frac{x-1}{x+1} f(x), \\quad \\forall x \\neq -1.\n$$\nSet $y = 1$ in (1):\n$$\n\\begin{aligned}\nf(x) f(1) &= f(x-1) + x f(1) + f(x) \\\\\n&= \\frac{x-1}{x+1} f(x) + x f(1) + f(x), \\quad \\forall x \\neq -1.\n\\end{aligned}\n$$\n- If $f(1) = 0$, then $f \\equiv 0$ because of the above and $f(-1) = 0$.\n- If $f(1) = 2$, then $f(x) = x(x+1)$ because of the above and $f(-1) = 0$.\n\nIt is easy to check that $f(x) \\equiv 0$ and $f(x) = x(x+1)$ satisfy (1). $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15227,
    "subject": "Mathematics (Olympiad)",
    "question": "Juku multiplied four consecutive positive integers and divided the result by four. Find all possible digits that this quotient can end with.",
    "options": [],
    "answer": "See solution",
    "solution": "Among four consecutive integers, there are always two even numbers, one of which is divisible by $4$. Thus, the quotient is even, so odd final digits are excluded.\n\nIf the final digit of the quotient after division by $4$ is $0$, $2$, $4$, $6$, or $8$, then the final digit of the dividend is $0$, $8$, $6$, $4$, or $2$, respectively.\n\n- If the four consecutive integers have final digits $1$, $2$, $3$, and $4$, then the final digit of their product is $4$.\n- If the four consecutive integers have final digits $6$, $7$, $8$, and $9$, then the final digit of their product is $4$.\n- In all other cases, one of the four integers is divisible by $5$. Since the product is even, its final digit can only be $0$.\n\nSo the product can have a final digit of $4$ or $0$. Upon division by $4$, the final digit of the quotient is $6$ or $0$, respectively.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15228,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A = \\{a_1, a_2, a_3, a_4\\}$. Suppose the set of sums of all the elements in every ternary subset of $A$ is $B = \\{-1, 3, 5, 8\\}$. Then $A = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Every element of $A$ appears three times in all the ternary subsets. Thus,\n$$\n3(a_1 + a_2 + a_3 + a_4) = (-1) + 3 + 5 + 8 = 15,\n$$\nso $a_1 + a_2 + a_3 + a_4 = 5$. Therefore, the four elements of $A$ are $5 - (-1) = 6$, $5 - 3 = 2$, $5 - 5 = 0$, and $5 - 8 = -3$, respectively.\n\nThe answer is $A = \\{-3, 0, 2, 6\\}$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15229,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a sequence of circles where the ratio of the diameters of two consecutive circles is equal to the ratio of the diagonal to the side of a square, which is $\\sqrt{2}$, what is the diameter of the fourth circle if the diameter of the first circle is $4$?",
    "options": [],
    "answer": "See solution",
    "solution": "The ratio of diameters between consecutive circles is $\\sqrt{2}$. Thus, the ratio between the first and fourth circle is $(\\sqrt{2})^3 = 2\\sqrt{2}$. The diameter of the fourth circle is $\\frac{4}{2\\sqrt{2}} = \\sqrt{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15230,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a general problem with cards numbered $1, 2, \\dots, n$. Let $F_n$ be the number of choices when there are $n$ cards. The initial values are $F_1 = 1$ and $F_2 = 2$. For $k \\geq 3$, the recurrence is:\n\n$$F_k = F_{k-1} + F_{k-2} + 1$$\n\nCalculate $F_{15}$ using this relation.\n\nIf a choice with no cards is allowed, the sequence becomes the Fibonacci sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "Using the recurrence $F_k = F_{k-1} + F_{k-2} + 1$ with $F_1 = 1$ and $F_2 = 2$, we compute:\n\n$F_3 = 2 + 1 + 1 = 4$\n\n$F_4 = 4 + 2 + 1 = 7$\n\n$F_5 = 7 + 4 + 1 = 12$\n\nContinue this process up to $F_{15}$:\n\n$F_6 = 12 + 7 + 1 = 20$\n\n$F_7 = 20 + 12 + 1 = 33$\n\n$F_8 = 33 + 20 + 1 = 54$\n\n$F_9 = 54 + 33 + 1 = 88$\n\n$F_{10} = 88 + 54 + 1 = 143$\n\n$F_{11} = 143 + 88 + 1 = 232$\n\n$F_{12} = 232 + 143 + 1 = 376$\n\n$F_{13} = 376 + 232 + 1 = 609$\n\n$F_{14} = 609 + 376 + 1 = 986$\n\n$F_{15} = 986 + 609 + 1 = 1596$\n\nThus, $F_{15} = 1596$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15231,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all (positive or negative) integers $n$ for which $n^2 + 20n + 11$ is a perfect square. Justify that you have found them all.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n^2 + 20n + 11 = a^2$ for some integer $a$. Then:\n\n$$\n\\begin{aligned}\nn^2 + 20n + 11 &= a^2 \\\\\n(n + 10)^2 - 100 + 11 &= a^2 \\\\\n(n + 10)^2 &= a^2 + 89 \\\\\n(n + 10)^2 - a^2 &= 89\n\\end{aligned}\n$$\n\nSo, solutions correspond to pairs of squares that differ by $89$.\n\nFor positive integers $x$ and $y$:\n\n$$\nx^2 - y^2 = (x - y)(x + y)\n$$\n\nWe seek integer solutions to $(n + 10)^2 - a^2 = 89$, i.e., $(n + 10 - a)(n + 10 + a) = 89$.\n\nSince $89$ is prime, its only integer factorizations are $89 = 1 \\times 89$ and $89 = (-1) \\times (-89)$. Thus:\n\n- $n + 10 - a = 1$, $n + 10 + a = 89$ $\rightarrow$ $n + 10 = 45$, $a = 44$ $\rightarrow$ $n = 35$\n- $n + 10 - a = -1$, $n + 10 + a = -89$ $\rightarrow$ $n + 10 = -45$, $a = -44$ $\rightarrow$ $n = -55$\n\nTherefore, the only integer solutions are $n = 35$ and $n = -55$.\n\nChecking:\n\n- For $n = 35$: $35^2 + 20 \\times 35 + 11 = 1225 + 700 + 11 = 1936 = 44^2$\n- For $n = -55$: $(-55)^2 + 20 \\times (-55) + 11 = 3025 - 1100 + 11 = 1936 = 44^2$\n\nThus, these are the only solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15232,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the least possible natural number $n$ such that each number from the set $1, 2, \\dots, 10$ can be expressed as a digit or as a sum of consecutive digits of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is obvious that we cannot find a number with the given property and only three digits. Suppose it has 4 digits: $a b c d$. Then we can construct 10 different sums of consecutive digits: $a$, $b$, $c$, $d$, $a+b$, $b+c$, $c+d$, $a+b+c$, $b+c+d$, $a+b+c+d$. To satisfy the conditions, these 10 sums must be distinct, so all digits must be distinct and their sum equals 10. The only option is $1, 2, 3, 4$. But to get $9$ as a sum, $1$ must be the first or last digit. Similarly, to represent $8$, $2$ must be the first or last digit. Thus, we have four options: $1342$, $1432$, $2341$, $2431$. However, for the first and fourth numbers, we cannot get $5$ as a sum; for the second and third, we cannot get $6$.\n\nHence, the least possible number of digits is at least $5$. Our number cannot start with $1111$, because to get $10$ as a sum, we have: $11116$, $11117$, $11118$, $11119$, but we cannot express $5$ then. Similarly, if our number starts with $1112$, possible options are $11125$, $11126$, $11127$, $11128$, $11129$, but we cannot express $5$ (except the second number) and $6$ for the second number. Therefore, our minimal number starts with $1113$ at least. Numbers $11131$, $11132$, $11133$ do not give $10$ as a sum, and $11134$ meets all the requirements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15233,
    "subject": "Mathematics (Olympiad)",
    "question": "For any positive integers $a$, $b$, $c$, prove that there exists a non-negative integer $k$ such that\n$$\ngcd(a^k + bc,\n      b^k + ac,\n      c^k + ab) > 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p$ be any prime divisor of $abc + 1$ and consider $k = p - 2$. First, note that none of $a$, $b$, $c$ are divisible by $p$. Next,\n$$\na^k + bc = a^{p-2} + bc = \\frac{a^{p-1} + abc}{a}\n$$\nis an integer divisible by $p$ since the numerator is divisible by $p$ and the denominator is not. Thus, $a^{p-2} + bc$ is divisible by $p$. Similarly, $b^{p-2} + ac$ and $c^{p-2} + ab$ are also divisible by $p$. Therefore,\n$$\ngcd(a^k + bc,\n      b^k + ac,\n      c^k + ab) \\geq p > 1.\n$$\nThis completes the proof. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15234,
    "subject": "Mathematics (Olympiad)",
    "question": "A right triangle $ABC$ is called *special* if the lengths of its sides $AB$, $BC$, and $CA$ are integers, and there is a point $X$ on each side (different from the vertices of $\\triangle ABC$), for which the lengths $AX$, $BX$, and $CX$ are integers. Find at least one special triangle.\n\n![](images/UkraineMO2019_booklet_p13_data_8772ed96dc.png)",
    "options": [],
    "answer": "See solution",
    "solution": "For example, $AC = 48$, $BC = 36$, $AB = 60$.\n\nLet triangle $\\triangle ABC$ have a right angle at $C$. Then, for any triangle with an even integer hypotenuse, the midpoint of the hypotenuse is the point of interest (see the figure above).\n\nLet $BC = an$, $AC = bn$, $AB = cn$, $n \\in \\mathbb{N}$ and $a^2 + b^2 = c^2$. Let $K \\in AC$, $N \\in BC$, $CN = bk$, $CK = ak$, $k \\in \\mathbb{N}$. It suffices for the lengths $AN = b\\sqrt{n^2 + k^2}$ and $BK = a\\sqrt{n^2 + k^2}$ to be integers. Thus, it suffices that $n^2 + k^2 = m^2$ for some $m \\in \\mathbb{N}$, $bk < an$, and $ak < bn$. Let $a = 3$, $b = 4$, $c = 5$, $k = 5$, $n = 12$, $m = 13$, hence $an = 36$, $bn = 48$, $ak = 15$, $bk = 20$. Thus, we obtain the following segments:\n\n$$\nAC = 48, \\quad CK = 15, \\quad BC = 36, \\quad CN = 20, \\quad AB = 60.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15235,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be positive integers such that $1 \\leq a < b \\leq 100$. If there exists a positive integer $k$ such that $ab = a^k + b^k$, then we say that the pair $(a, b)$ is *good*. Determine the number of good pairs.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $(a, b) = d$, $a = s d$, $b = t d$, with $(s, t) = 1$ and $t > 1$. Then $s t d^2 = d^k (s^k + t^k)$, so $k \\geq 2$ and $s t = d^{k-2}(s^k + t^k)$. Since $(s t, s^k + t^k) = 1$, we have $s t = d^{k-2}$. Therefore, any prime factor of $s t$ divides $d$.\n\nIf there is a prime factor $p$ of $s$ or $t$ with $p \\geq 11$, then $p$ divides $d$, so $p^2$ divides $a$ or $b$, but $p^2 > 100$, a contradiction.\n\nThus, the prime factors of $s t$ can only be $2, 3, 5,$ or $7$.\n\nIf there are at least three prime factors among $2, 3, 5, 7$, then $d > 2 \\times 3 \\times 5 = 30$, so $a$ or $b \\geq 5d > 100$, a contradiction. The prime factor set of $s t$ cannot be $\\{3, 7\\}$ or $\\{5, 7\\}$ for similar reasons.\n\nTherefore, the possible prime factor sets for $s t$ are $\\{2\\}, \\{3\\}, \\{5\\}, \\{7\\}, \\{2, 3\\}, \\{2, 5\\}, \\{2, 7\\}, \\{3, 5\\}$.\n\n(i) For $\\{3, 5\\}$: $d = 15$, $s = 3$, $t = 5$. One good pair: $(45, 75)$.\n\n(ii) For $\\{2, 7\\}$: $d = 14$, $s = 2, t = 7$ or $s = 4, t = 7$. Two good pairs: $(28, 98)$ and $(56, 98)$.\n\n(iii) For $\\{2, 5\\}$: $d = 10$ or $20$.\n- $d = 10$: $s = 2, t = 5$; $s = 1, t = 10$; $s = 4, t = 5$; $s = 5, t = 8$.\n- $d = 20$: $s = 2, t = 5$; $s = 4, t = 5$.\nSix good pairs.\n\n(iv) For $\\{2, 3\\}$: $d = 6, 12, 18, 24, 30$.\n- $d = 6$: $s = 1, t = 6$; $s = 1, t = 12$; $s = 2, t = 3$; $s = 2, t = 9$; $s = 3, t = 4$; $s = 3, t = 8$; $s = 3, t = 16$; $s = 4, t = 9$; $s = 8, t = 9$; $s = 9, t = 16$.\n- $d = 12$: $s = 1, t = 6$; $s = 2, t = 3$; $s = 3, t = 4$; $s = 3, t = 8$.\n- $d = 18$: $s = 2, t = 3$; $s = 3, t = 4$.\n- $d = 24$: $s = 2, t = 3$; $s = 3, t = 4$.\n- $d = 30$: $s = 2, t = 3$.\nNineteen good pairs.\n\n(v) For $\\{7\\}$: $s = 1, t = 7$, $d = 7$ or $14$. Two good pairs.\n\n(vi) For $\\{5\\}$: $s = 1, t = 5$, $d = 5, 10, 15, 20$. Four good pairs.\n\n(vii) For $\\{3\\}$:\n- $s = 1, t = 3$, $d = 3, 6, \\ldots, 33$;\n- $s = 1, t = 9$, $d = 3, 6, 9$;\n- $s = 1, t = 27$, $d = 3$.\nFifteen good pairs.\n\n(viii) For $\\{2\\}$: (incomplete)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15236,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Suppose there are $m$ students labeled $1, 2, 3, \\ldots, m$ and $11$ types of candy denoted by $a_1, a_2, \\ldots, a_{11}$. Let $X = \\{a_1, a_2, \\ldots, a_{11}\\}$ and $A_1, A_2, \\ldots, A_m$ be the sets of candy types that students $1, 2, \\ldots, m$ received, respectively, so $A_1, A_2, \\ldots, A_m \\subset X$.\n\nFor each $i = 1, 2, \\ldots, 11$, let $d(a_i)$ be the number of $a_i$ candies in the total of $2013$ candies. Clearly,\n\n$$\nd(a_1) + d(a_2) + \\cdots + d(a_{11}) = 2013.\n$$\n\nLet $M = \\sum_{1 \\leq i < j \\leq m} |A_i \\cap A_j|$. For each $i = 1, 2, \\ldots, 11$, the number of subsets containing $a_i$ is $d(a_i)$, so the number of pairs of subsets having $a_i$ in common is $\\binom{d(a_i)}{2}$. Therefore,\n\nb) Similarly, let $b_1, b_2, \\ldots, b_9$ be the types of candies and $d(b_i)$ for $i = 1, \\ldots, 9$ be the number of candies of each type. We have\n\n$$\nd(b_1) + d(b_2) + \\cdots + d(b_9) = 2013.\n$$\n\nFind the minimum value of $\\sum_{i=1}^{9} (d(b_i))^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\nM = \\sum_{1 \\leq i < j \\leq m} |A_i \\cap A_j| = \\sum_{i=1}^{11} \\binom{d(a_i)}{2} = \\frac{1}{2} \\sum_{i=1}^{11} \\left( (d(a_i))^2 - d(a_i) \\right).\n$$\n\nBy the Cauchy-Schwarz inequality,\n$$\n\\sum_{i=1}^{11} 1^2 \\cdot \\sum_{i=1}^{11} (d(a_i))^2 \\geq \\left(\\sum_{i=1}^{11} d(a_i)\\right)^2 = 2013^2,\n$$\nso\n$$\nM \\geq \\frac{1}{2} \\left( \\frac{2013^2}{11} - 2013 \\right).\n$$\n\nEquality occurs if and only if $d(a_1) = d(a_2) = \\cdots = d(a_{11}) = \\frac{2013}{11} = 183$.\n\nFor part b), if there exists $d(b_i) - d(b_j) \\geq 2$ for some $i, j$, then decreasing $d(b_i)$ by $1$ and increasing $d(b_j)$ by $1$ yields\n$$\n(d(b_i))^2 + (d(b_j))^2 - (d(b_i) - 1)^2 - (d(b_j) + 1)^2 = 2(d(b_i) - d(b_j)) > 0.\n$$\n\nThus, to minimize $\\sum_{i=1}^9 (d(b_i))^2$, we need $d(b_i) - d(b_j) \\leq 1$ for all $i, j$.\n\nAssume $d(b_1) \\leq d(b_2) \\leq \\cdots \\leq d(b_9)$, so each $d(b_i)$ is either $k$ or $k+1$ for some integer $k$. Suppose there are $t$ numbers $k$ and $9-t$ numbers $k+1$. Then\n$$\ntk + (9-t)(k+1) = 2013 \\implies t = 9k - 2004.\n$$\n\nThe minimum value is\n$$\nM = \\frac{1}{2} (tk^2 + (9-t)(k+1)^2 - 2013).\n$$\n\nSubstituting $t = 9k - 2004$ gives\n$$\nM = \\frac{1}{2}((9k - 2004)k^2 + (2013 - 9k)(k+1)^2 - 2013).\n$$\n\nSince $0 \\leq t \\leq 9$, $\\frac{2004}{9} \\leq k \\leq \\frac{2013}{9}$, so $k = 223$.\n\nWith $k = 223$, $M = \\frac{1}{2}(3 \\cdot 223^2 + 6 \\cdot 224^2 - 2013)$.\n\nThus, the minimum value is $M = \\frac{1}{2}(3 \\cdot 223^2 + 6 \\cdot 224^2 - 2013)$, achieved when there are $3$ numbers $223$ and $6$ numbers $224$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15237,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $I$ be the incenter of triangle $ABC$ and let $\\Gamma$ be its circumcircle. Let the line $AI$ intersect $\\Gamma$ again at $D$. Let $E$ be a point on the arc $BDC$ and $F$ a point on the side $BC$ such that\n$$\n\\angle BAF = \\angle CAE < \\frac{1}{2} \\angle BAC.\n$$\nFinally, let $G$ be the midpoint of the segment $IF$. Prove that the lines $DG$ and $EI$ intersect on $\\Gamma$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $I_a$ be the center of the excenter circle of triangle $ABC$ with respect to side $BC$. Then $D$ is the midpoint of the segment $II_a$, $DG \\parallel I_a F$, and $\\angle GDA = \\angle FI_a A$.\n\nTo prove that the intersection point $P$ of the lines $DG$ and $EI$ lies on $\\Gamma$, i.e., $A, E, D, P$ are concyclic, is equivalent to proving that $\\angle GDA = \\angle IEA$, or similarly $\\angle FI_a A = \\angle IEA$.\n\nBy the assumption $\\angle BAF = \\angle CAE$, we have $\\triangle ABF \\sim \\triangle AEC$, and hence\n$$\n\\frac{AF}{AB} = \\frac{AC}{AE}. \\qquad \\textcircled{1}\n$$\nSince\n$$\n\\angle AIB = \\angle C + \\frac{1}{2}(\\angle A + \\angle B),\n$$\n$$\n\\angle ACI_a = \\angle C + \\frac{1}{2}(\\angle A + \\angle B),\n$$\nwe have $\\triangle ABI \\sim \\triangle AI_a C$, and thus\n$$\n\\frac{AI}{AB} = \\frac{AC}{AI_a}. \\qquad \\textcircled{2}\n$$\nFrom $\\textcircled{1}$ and $\\textcircled{2}$ we get\n$$\n\\frac{AF}{AI} = \\frac{AI_a}{AE}. \\qquad \\textcircled{3}\n$$\nBy the assumption $\\angle FAI_a = \\angle EAI$, we have $\\triangle AFI_a \\sim \\triangle AIE$, and therefore $\\angle FI_a A = \\angle IEA$, which completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15238,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that there exists a polynomial $f$ of degree $n$ with integer coefficients and a positive leading coefficient, and a polynomial $g$ with integer coefficients, such that\n\n$$\nxf^2(x) + f(x) = (x^3 - x)g^2(x)\n$$\n\nholds for every real $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $xf^2(x) + f(x) = (x^3 - x)g^2(x)$, which can be rewritten as $[2xf(x) + 1]^2 = (x^2 - 1)[2xg(x)]^2 + 1$.\n\nWe seek all pairs $(p, q)$ of integer-coefficient polynomials such that $p^2(x) = (x^2 - 1)q^2(x) + 1$.\n\nLet $(p, q)$ be such a pair with $\text{deg}(q) = k \\ge 1$. Assume $p$ and $q$ have positive leading coefficients. Define $P_0 = p$, $Q_0 = q$, and set\n\n$$\nP_1(x) = x p(x) - (x^2 - 1) q(x), \\quad Q_1(x) = -p(x) + x q(x).\n$$\n\nIt can be shown that $P_1$ and $Q_1$ also satisfy the equation, and $\text{deg}(Q_1) < \text{deg}(Q_0)$.\n\n(One way to see this: The equation $p^2(x) = (x^2 - 1)q^2(x) + 1$ can be rewritten as $1 = (p(x) - q(x)\\sqrt{x^2-1})(p(x) + q(x)\\sqrt{x^2-1})$. One solution is $(p, q) = (x, 1)$, i.e., $1 = (x - \\sqrt{x^2-1})(x + \\sqrt{x^2-1})$. Multiplying these gives $1 = (P_1(x) - Q_1(x)\\sqrt{x^2-1})(P_1(x) + Q_1(x)\\sqrt{x^2-1})$.)\n\nRepeating this process, we eventually reach $(P_s, Q_s)$ with $Q_s$ constant. There are only two such solutions: $(x, 1)$ and $(1, 0)$. Applying the operation to the former yields the latter, so we may assume $(P_s, Q_s) = (1, 0)$.\n\nThus, all solution pairs $(p, q)$ are given by the sequence with initial condition $(p_0, q_0) = (1, 0)$ and recurrence\n\n$$\n(p_{i+1}, q_{i+1}) = (x p_i(x) + (x^2-1) q_i(x), p_i(x) + x q_i(x)).\n$$\n\nA member of this sequence corresponds to a solution of the original equation exactly when $p(x) \\equiv 1 \\pmod{2x}$ and $q(x)$ is divisible by $2x$. The first five members are $(1, 0)$, $(x, 1)$, $(2x^2 - 1, 2x)$, $(4x^3 - 3x, 4x^2 - 1)$, and $(8x^4 - 8x^2 + 1, 8x^3 - 4x)$, and $(p_4(x), q_4(x)) \\equiv (1, 0) \\pmod{2x}$, so the sequence is periodic with period $4$ modulo $2x$. Exactly those $(p_i, q_i)$ with $i$ a multiple of $4$ yield a solution.\n\nTherefore, the necessary values of $n$ are the numbers $4k + 3$ for $k \\ge 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15239,
    "subject": "Mathematics (Olympiad)",
    "question": "Triangle $ABC$ has $AC = BC$. The bisector of angle $CAB$ meets side $BC$ at point $D$. The difference of the sizes of some two internal angles of triangle $ABD$ is $40^\\circ$. Find all possibilities of what the size of angle $ACB$ can be.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $\\angle BAC = \\angle ABC = \\alpha$, then $\\angle ACB = 180^\\circ - 2\\alpha$. The sizes of the internal angles of triangle $ABD$ are $\\angle BAD = \\frac{\\alpha}{2}$, $\\angle DBA = \\alpha$, and $\\angle ADB = 180^\\circ - \\frac{3}{2}\\alpha$.\n\n![](images/prob1617_p16_data_1c61ae8cd1.png)\n\nConsider all cases of which two angles can have $40^\\circ$ as the difference of sizes:\n\n- If $\\alpha - \\frac{\\alpha}{2} = 40^\\circ$, then $\\alpha = 80^\\circ$, whence $\\angle ACB = 20^\\circ$.\n- The case $\\frac{\\alpha}{2} - \\alpha = 40^\\circ$ is impossible since it would imply $\\alpha < 0^\\circ$.\n- If $180^\\circ - \\frac{3}{2}\\alpha - \\frac{\\alpha}{2} = 40^\\circ$, then $\\alpha = 70^\\circ$, whence $\\angle ACB = 40^\\circ$.\n- If $\\frac{\\alpha}{2} - (180^\\circ - \\frac{3}{2}\\alpha) = 40^\\circ$, then $\\alpha = 110^\\circ$, but the base angle of an isosceles triangle cannot be obtuse.\n- If $180^\\circ - \\frac{3}{2}\\alpha - \\alpha = 40^\\circ$, then $\\alpha = 56^\\circ$, whence $\\angle ACB = 68^\\circ$.\n- If $\\alpha - (180^\\circ - \\frac{3}{2}\\alpha) = 40^\\circ$, then $\\alpha = 88^\\circ$, whence $\\angle ACB = 4^\\circ$.\n\nThus, the possible values for $\\angle ACB$ are $68^\\circ$, $40^\\circ$, $20^\\circ$, and $4^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15240,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of integers $ (a, b) $ satisfying $ (a+1)(b-1) = a^2 b^2 $.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $a$ and $a+1$ are coprime, $a^2$ and $a+1$ are also coprime. Similarly, $b^2$ and $b-1$ are coprime. Hence, the equality can hold only in the case $a+1 = \\pm b^2$ and $b-1 = \\pm a^2$, where the signs in both equations are the same.\n\nLet both signs be pluses. Then from the first equation we get $a = b^2 - 1 = (b-1)(b+1)$. The second equation implies $b-1 = a^2$, whence $a = a^2(a^2+2)$. If $a=0$, then $b=1$, i.e. $(a,b) = (0,1)$. If $a \\neq 0$, then by dividing by $a$ we get $1 = a(a^2+2)$; since $a^2+2 > 1$, this equation does not have integer solutions.\n\nIf both signs are minuses, then by multiplying by $-1$ we get $-b+1 = a^2$ and $-a-1 = b^2$. These are the same equations with respect to $-b$ and $-a$ which we had previously with respect to $a$ and $b$, hence the only solution is $-b = 0, -a = 1$, i.e. $(a,b) = (-1,0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15241,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a strictly positive integer and let $A = \\{1, 2, \\dots, n\\}$. Find the number of increasing functions $f : A \\to A$ which satisfy the property\n\n$$\n|f(x) - f(y)| \\leq |x - y|,\n$$\n\nfor any $x, y \\in A$.",
    "options": [],
    "answer": "See solution",
    "solution": "We observe that for an increasing function $f$, the difference $f(k+1) - f(k)$ must be $a_k \\in \\{0, 1\\}$ for $k = 1, 2, \\dots, n-1$. Since $a_k \\in \\{0, 1\\}$, the condition $|f(x) - f(y)| \\leq |x - y|$ holds for any $x, y \\in A$.\n\nThus, any such function $f$ is completely determined by the $n$-tuple $(f(1), a_1, a_2, \\dots, a_{n-1}) \\in A \\times \\{0, 1\\}^{n-1}$, with the constraint\n\n$$\nf(1) + a_1 + \\dots + a_{n-1} \\leq n.\n$$\n\nIf we fix $f(1) = a$ and $f(n) = b$, then $b - a = a_1 + \\dots + a_{n-1}$, and the number of such functions is $n(a, b) = \\binom{n-1}{b-a}$.\n\nTherefore, the total number of functions is\n\n$$\n\\sum_{0 \\leq a \\leq b \\leq n} n(a, b) = \\sum_{0 \\leq a \\leq b \\leq n} \\binom{n-1}{b-a} = \\sum_{k=0}^{n-1} (n-k) \\binom{n-1}{k}.\n$$\n\nSumming up, we have\n\n$$\n\\begin{aligned}\n\\sum_{k=0}^{n-1} n \\binom{n-1}{k} - \\sum_{k=1}^{n-1} k \\binom{n-1}{k} &= n 2^{n-1} - \\sum_{k=1}^{n-1} (n-1) \\binom{n-2}{k-1} \\\\\n&= n 2^{n-1} - (n-1) 2^{n-2} = (n+1) 2^{n-2}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15242,
    "subject": "Mathematics (Olympiad)",
    "question": "A country held a one-round tennis tournament. Participants received 1 point for winning a match and 0 points for losing. There are no draws in tennis. At the end of the tournament, Oleksii saw the number of points scored by each participant, as well as the schedule of all the matches in the tournament, which showed the pairs of players, but not the winners. He chooses a match in any order he likes and tries to guess the winner, after which he is told if he is correct. Prove that Oleksii can act in such a way that he is guaranteed to guess the winners of more than half of the matches.",
    "options": [],
    "answer": "See solution",
    "solution": "Let Oleksii choose a player $A$ with the lowest number of points. Since $A$ has at least as many defeats as wins, by guessing that $A$ lost each of his matches, Oleksii will guess at least half of the outcomes for $A$'s matches. Then, subtract $A$'s results from each participant's total and select the next player $B$ with the lowest remaining points, repeating the procedure. If a player has an odd number of matches, Oleksii will guess more than half of that player's results; if even, at least half. If two players have different parity in their number of matches, Oleksii will guess more than half for them together. Continuing this process for all players, Oleksii will guess more than half of all match outcomes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15243,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose points $O$ and $I$ are the circumcenter and incenter of $\\triangle ABC$ respectively, and the inscribed circle of $\\triangle ABC$ is tangent to the sides $BC$, $CA$, $AB$ at points $D$, $E$, $F$ respectively. Lines $FD$ and $CA$ intersect at point $P$, while lines $DE$ and $AB$ intersect at point $Q$. Points $M$ and $N$ are the midpoints of segments $PE$ and $QF$ respectively. Prove that $OI \\perp MN$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first consider $\\triangle ABC$ and segment $PFD$. By Menelaus' theorem, we have\n\n$$\n\\frac{CP}{PA} \\cdot \\frac{AF}{FB} \\cdot \\frac{BD}{DC} = 1.\n$$\n\nThen\n\n$$\nPA = CP \\cdot \\frac{AF}{FB} \\cdot \\frac{BD}{DC} = (PA + b) \\frac{p-a}{p-c}.\n$$\n\n(We define $a = BC$, $b = CA$, $c = AB$, $p = \\frac{1}{2}(a+b+c)$; and without loss of generality, assume $a > c$.) Then we get\n\n$$\nPA = \\frac{b(p-a)}{a-c}.\n$$\n\nFurther,\n\n$$\nPE = PA + AE = \\frac{b(p-a)}{a-c} + p - a = \\frac{2(p-c)(p-a)}{a-c},\n$$\n\n$$\nME = \\frac{1}{2}PE = \\frac{(p-c)(p-a)}{a-c},\n$$\n\n$$\nMA = ME - AE = \\frac{(p-c)(p-a)}{a-c} - (p-a) = \\frac{(p-a)^2}{a-c},\n$$\n\n$$\nMC = ME + EC = \\frac{(p-c)(p-a)}{a-c} + (p-c) = \\frac{(p-c)^2}{a-c}.\n$$\n\nThen we have\n\n$$\nMA \\cdot MC = ME^2.\n$$\n\nThat means $ME^2$ is equal to the power of $M$ with respect to the circumscribed circle of $\\triangle ABC$. Further, since $ME$ is the length of the tangent from $M$ to the inscribed circle of $\\triangle ABC$, so $ME^2$ is also the power of $M$ with respect to the inscribed circle. Hence, $M$ is on the radical axis of the circumscribed and inscribed circles of $\\triangle ABC$.\n\nIn the same way, $N$ is also on the radical axis. Since the radical axis is perpendicular to $OI$, then $OI \\perp MN$. That completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15244,
    "subject": "Mathematics (Olympiad)",
    "question": "A company consisting of $n$ friends play a table game according to the following rules:\n\n1. At each round, exactly 3 players participate.\n2. The game stops after $n$ rounds.\n3. Each pair of players has played together in at least one round.\n\nDetermine the maximal possible value of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since in each round exactly 3 players play, the number of pairs playing together in each round is $\\binom{3}{2} = 3$. Therefore, after $n$ rounds, the total number of pairs that have played together is $3n$. According to the last rule:\n\n$$\n\\binom{n}{2} \\leq 3n \\implies \\frac{n(n-1)}{2} \\leq 3n \\implies \\frac{n-1}{2} \\leq 3 \\implies n \\leq 7.\n$$\n\nNext, we show that $n=7$ is possible. For $n=7$, $\\binom{7}{2} = 21 = 3 \\times 7$. If the friends are $A, B, \\Gamma, \\Delta, E, Z, H$, we can define the following triads:\n\n- $(A, B, \\Gamma)$\n- $(A, \\Delta, E)$\n- $(A, Z, H)$\n- $(B, \\Delta, H)$\n- $(B, E, Z)$\n- $(\\Gamma, \\Delta, Z)$\n- $(\\Gamma, E, H)$\n\nThus, the maximal possible value of $n$ is $7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15245,
    "subject": "Mathematics (Olympiad)",
    "question": "Fix an integer $n \\ge 6$ and consider $n$ coplanar lines, no two parallel and no three concurrent. These lines split the plane into unbounded polygonal regions and polygons with pairwise disjoint interiors. Two polygons are *non-adjacent* if they do not share a side. Show that there are at least $\\frac{1}{12}(n-3)(n-2)$ pairwise non-adjacent polygons with the same number of sides each.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the geometric plane graph associated with a generic $n$-line configuration in the plane (no two lines are parallel and no three are concurrent). This graph has exactly $\\binom{n}{2} = \\frac{1}{2}n(n-1)$ vertices, exactly $e = n^2$ edges, and exactly $f = \\frac{1}{2}n(n+1)+1$ faces.\n\nLet $f_k$ be the number of $k$-edge faces (faces with exactly $k$ edges), so $f = \\sum_{k=2}^{n} f_k$ (no face has more than $n$ edges). Fix an integer $m$ in the range $4$ through $n$, to write\n\n$$\n\\begin{align*}\n2n^2 = 2e &= \\sum_{k=2}^{n} k f_k = \\sum_{k=2}^{m-1} k f_k + \\sum_{k=m}^{n} k f_k \\\\ \n&\\ge \\sum_{k=2}^{m-1} k f_k + m \\sum_{k=m}^{n} f_k \\\\\n&= \\sum_{k=2}^{m-1} k f_k + m \\left( f - \\sum_{k=2}^{m-1} f_k \\right) \\\\\n&= \\sum_{k=2}^{m-1} (k-m) f_k + mf,\n\\end{align*}\n$$\n\nso\n\n$$\n2n^2 \\geq \\sum_{k=2}^{m-1} (k-m)f_k + \\frac{1}{2}m(n^2+n+2),\n$$\n\nwherefrom\n\n$$\n\\sum_{k=2}^{m-1} (m-k)f_k \\geq \\frac{1}{2}m(n^2+n+2) - 2n^2 = \\frac{1}{2}((m-4)n^2 + mn + 2m).\n$$\n\nNext, write $f_k = b_k + u_k$, where $b_k$ and $u_k$ are the numbers of bounded and unbounded $k$-faces, respectively. Thus, $b_k$ is the number of $k$-gons. Clearly, $b_2 = 0$ and $\\sum_{k=2}^{n} u_k = 2n$, so\n\n$$\n\\begin{align*}\n\\sum_{k=2}^{m-1} (m-k)f_k &= \\sum_{k=3}^{m-1} (m-k)b_k + \\sum_{k=2}^{m-1} (m-k)u_k \\\\\n&\\le \\sum_{k=3}^{m-1} (m-k)b_k + (m-2) \\sum_{k=3}^{m-1} u_k \\\\\n&\\le \\sum_{k=3}^{m-1} (m-k)b_k + (m-2) \\sum_{k=2}^{n} u_k \\\\\n&= \\sum_{k=3}^{m-1} (m-k)b_k + 2(m-2)n.\n\\end{align*}\n$$\n\nConsequently,\n\n$$\n\\begin{align*}\n\\sum_{k=3}^{m-1} (m-k)b_k &\\ge \\sum_{k=2}^{m-1} (m-k)f_k - 2(m-2)n \\\\\n&\\ge \\frac{1}{2}((m-4)n^2 + mn + 2m) - 2(m-2)n \\\\\n&= \\frac{1}{2}((m-4)n^2 - (3m-8)n + 2m) \\\\\n&= \\frac{1}{2}((m-4)n - m)(n-2),\n\\end{align*}\n$$\n\nso\n\n$$\n\\max(b_3, \\dots, b_{m-1}) \\ge \\frac{1}{2} \\cdot \\frac{((m-4)n - m)(n-2)}{(m-3) + \\dots + 1} = \\frac{((m-4)n - m)(n-2)}{(m-3)(m-2)}.\n$$\n\nThe coefficient of $n^2$ in the lower bound is maximized at $m=5$ and $m=6$, so\n\n$$\n\\max(b_3, b_4) \\ge \\frac{1}{6}(n-5)(n-2) \\quad \\text{and} \\quad \\max(b_3, b_4, b_5) \\ge \\frac{1}{6}(n-3)(n-2).\n$$\n\nSince each vertex has an even degree (namely, 4), faces are 2-colourable; that is, they can be coloured one of two colours, so that no adjacent faces bear the same colour. \n\nConsequently, for some $k$ in $\\{3, 4, 5\\}$, at least $\\frac{1}{2} \\max(b_3, b_4, b_5) \\ge \\frac{1}{12}(n-3)(n-2)$ $k$-gons bear the same colour, and are therefore non-adjacent, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15246,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle with orthocenter $H$. Consider the points $Y$ and $Z$ on the sides $CA$ and $AB$ respectively such that the directed angles $(AC, HY) = -\\pi/3$ and $(AB, HZ) = \\pi/3$. Let $U$ be the circumcenter of $\\triangle HYZ$.\n\nProve that the points $A$, $N$, $U$ are collinear, where $N$ is the nine-point center of $\\triangle ABC$.\n\nOn the Euler reflection point, Forum Geom., 10 (2010) 165-173.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O_A$ be the reflection of $O$ into the sideline $BC$. We will prove the result by showing that $A$, $N$, $O_A$ are collinear, and that $A$, $U$, $O_A$ are collinear.\n\n$A$, $N$, $O_A$ are collinear from the fact that $N$ is the midpoint of $HO$ on the Euler line, and that $AH = OO_A$ and both $AH$ and $OO_A$ are perpendicular to $BC$. ($\\triangle AHN \\cong \\triangle O_AON$.)\n\nTo show that $A$, $U$, $O_A$ are collinear, we will first show that $H$, $U$, $A^+$ are collinear, where $A^+$ is a point on the same side as $A$ with respect to $BC$ such that $A^+BC$ is equilateral.\n\nConsider the case where $\\angle BAC \\neq 60^\\circ$. (The case where $\\angle BAC = 60^\\circ$ can be done similarly using the same idea.) In this case $AZHY$ is not a parallelogram. Let $Y'$, $Z'$ be the intersection points between $AY$ and $HZ$ and between $AZ$ and $HY$, as in the picture.\n\nLet $V$ be the orthocenter of $HYZ$, thus the lines $HU$ and $HV$ are isogonal conjugate with respect to the angle $\\angle ZHY$. Note also that the quadrilateral $YZY'Z'$ is cyclic, since $\\angle Y'ZZ' = \\angle Y'YZ' = 60^\\circ$. Thus, the lines $Y'Z'$ and $YZ$ are antiparallel. Since $HV \\perp YZ$, then $HU \\perp Y'Z'$.\n\nLet $C'$ be the reflection of $C$ in the line $HY'$. Using the directed angle\n\n$$\n\\begin{aligned}\n(HY', HC) &= (HY', CA) + (CA, HC) \\\\\n&= (AB, AC) - 60^\\circ + 90^\\circ - (AB, AC) = 30^\\circ.\n\\end{aligned}\n$$\n\nThis implies that $HCC'$ is an equilateral triangle, and $\\triangle HCC' \\sim \\triangle A^+BC$.\n\nNote also that $\\triangle Y'HC \\sim \\triangle Z'HB$ since $\\angle HY'C = \\angle HZ'B$ and $\\angle HCY' = \\angle HBZ' = 90^\\circ - \\angle BAC$. Since $\\triangle Y'HC'$ is the reflection of $\\triangle Y'HC$, thus $\\triangle Y'HC' \\sim \\triangle Z'HB$ with the common vertex $H$. Therefore, (by spiral transformation or by simple comparison), $\\triangle Z'HY' \\sim \\triangle BHC'$. It follows that\n\n$$\n\\begin{aligned}\n(Y'Z', A^+H) &= (Y'Z', BC') + (BC', A^+H) \\\\\n&= (Z'H, BH) + (BC, A^+C) = 90^\\circ.\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15247,
    "subject": "Mathematics (Olympiad)",
    "question": "Одреди ги сите природни броеви $n$ за кои бројот $z = \\left(\\frac{3+i}{2-i}\\right)^n$ е реален.",
    "options": [],
    "answer": "See solution",
    "solution": "Комплексниот број $\\frac{3+i}{2-i}$ можеме да го запишеме во облик\n\n$$\n\\frac{3+i}{2-i} = \\frac{(3+i)(2+i)}{(2-i)(2+i)} = \\frac{6+3i+2i+i^2}{4+2i-2i-i^2} = \\frac{6+5i-1}{4-(-1)} = \\frac{5+5i}{5} = 1+i.\n$$\n\nЗначи, $z = (1+i)^n$. Ако $n$ е парен број, односно $n=2k$ за некој $k \\in \\mathbb{N}$, тогаш\n\n$$\nz = (1+i)^n = (1+i)^{2k} = [(1+i)^2]^k = (2i)^k.\n$$\n\nАко пак $k$ е парен број, односно $k = 2m$ за некој $m \\in \\mathbb{N}$, тогаш\n\n$$\nz = (2i)^{2m} = (4i^2)^m = 4^m (-1)^m.\n$$\n\nЗначи, ако $n = 4m$, т.е. $4 \\mid n$, тогаш $z \\in \\mathbb{R}$.\n\nАко $4 \\nmid n$, тогаш $n$ има еден од облиците $4m+1, 4m+2$ или $4m+3$. Во секој од тие случаи имаме\n\n$$\nz = (1+i)^{4m+1} = 4^m (-1)^m (1+i) \\notin \\mathbb{R},\n$$\n\n$$\nz = (1+i)^{4m+2} = 4^m (-1)^m (1+i)^2 = 4^m \\cdot (-1)^m \\cdot 2i \\notin \\mathbb{R},\n$$\n\n$$\nz = (1+i)^{4m+3} = 4^m (-1)^m (1+i)^3 = 4^m (-1)^m \\cdot 2i(1+i) = 4^m \\cdot (-1)^m \\cdot 2(-1+i) \\notin \\mathbb{R}.\n$$\n\nЗначи, $z$ е реален број ако и само ако $4 \\mid n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15248,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, $d$ be nonnegative real numbers not exceeding $1$. Prove that\n\n$$\n\\frac{1}{1+a+b} + \\frac{1}{1+b+c} + \\frac{1}{1+c+d} + \\frac{1}{1+d+a} \\le \\frac{4}{1+2\\sqrt[4]{abcd}}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that when $\\sqrt{ac} \\le x$, we have\n\n$$\n\\frac{1}{x+a} + \\frac{1}{x+c} - \\frac{2}{x+\\sqrt{ac}} = \\frac{(\\sqrt{a}-\\sqrt{c})^2(\\sqrt{ac}-x)}{(x+a)(x+c)(x+\\sqrt{ac})} \\le 0. \\quad (*)\n$$\n\nGiven the conditions, $\\sqrt{ac} \\le 1 \\le 1+b$, and $\\sqrt{ac} \\le 1+d$. Substituting $x = 1+b$ and $x = 1+c$ into $(*)$ yields\n\n$$\n\\frac{1}{1+a+b} + \\frac{1}{1+b+c} \\le \\frac{2}{1+b+\\sqrt{ac}},\n$$\n\n$$\n\\frac{1}{1+c+d} + \\frac{1}{1+d+a} \\le \\frac{2}{1+d+\\sqrt{ac}}.\n$$\n\nThis means that when substituting $\\sqrt{ac}$ for $a$ and $c$, the left side of the inequality does not decrease, while the right side remains unchanged. Thus, we may assume without loss of generality that $a = c$. Similarly, we may assume $b = d$. Hence, the original inequality is reduced to proving\n\n$$\n\\frac{1}{1+a+b} \\le \\frac{1}{1+2\\sqrt{ab}}.\n$$\n\nThis holds true by the arithmetic mean-geometric mean inequality, which states $a+b \\ge 2\\sqrt{ab}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15249,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ for which\n\n$$\nf(a - b)f(c - d) + f(a - d)f(b - c) \\le (a - c)f(b - d)\n$$\n\nfor all real numbers $a, b, c,$ and $d$.\n\n*Note that there is only one occurrence of $f$ on the right hand side!*",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions to the given functional inequality are $f(x) = 0$ for all $x$ and $f(x) = x$ for all $x$.\n\nFor $f(x) = 0$, equality always holds. For $f(x) = x$, we check that\n\n$$\n\\begin{aligned}\n(a-b)(c-d) + (a-d)(b-c) &= ac - ad - bc + bd + ab - ac - bd + cd \\\\\n&= -ad - bc + ab + cd \\\\\n&= (a-c)(b-d),\n\\end{aligned}\n$$\n\nso equality holds in that case too.\n\nNow we show that these functions are the only two solutions.\n\nSubstituting $a = b = c = d = 0$ gives $2f(0)^2 \\le 0$, so $f(0) = 0$.\n\nNext, substitute $b = a - x$, $c = a$, $d = a - y$, so $a - b = x$, $a - c = 0$, $a - d = y$. This gives\n\n$$\nf(y)(f(x) + f(-x)) \\le 0 \\qquad (1)\n$$\n\nSuppose there is $y$ with $f(y) \\ne 0$. Setting $x = y$ in (1):\n\n$$\n0 < f(y)^2 \\le -f(y)f(-y).\n$$\n\nThus, $f(y)$ and $f(-y)$ have opposite signs. Assume $f(y) > 0$.\n\nNow, for arbitrary $a$ and $y$, set $b = a$, $c = 0$, $d = a - y$:\n\n$$\nf(y)f(a) \\le a f(y) \\implies f(a) \\le a.\n$$\n\nSimilarly, with $d = a + y$:\n\n$$\nf(-y)f(a) \\le a f(-y) \\implies f(a) \\ge a.\n$$\n\nTherefore, $f(a) = a$ for all $a$.\n\nThus, the only solutions are $f(x) = 0$ and $f(x) = x$, both of which satisfy the original inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15250,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the maximum possible value for the least common multiple of 4 distinct single-digit positive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "$2520$\n\nPossible prime factors for a single-digit positive integer are $2, 3, 5, 7$. Since $2^4 = 16$, $3^3 = 27$, $5^2 = 25$, and $7^2 = 49$ are all greater than $10$, the highest powers of $2, 3, 5, 7$ that can appear in the prime factorization of a single-digit positive integer are $2^3, 3^2, 5^1, 7^1$ respectively. Thus, the least common multiple (LCM) of 4 single-digit positive integers is a divisor of $2^3 \\times 3^2 \\times 5 \\times 7 = 2520$, and must be less than or equal to this number. The LCM of $5, 7, 8, 9$ is $2520$, so $2520$ is the maximum possible value.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15251,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all natural numbers $n$ for which the number of positive divisors of $\\text{LCM}(1, 2, \\dots, n)$ is a power of $2$.",
    "options": [],
    "answer": "See solution",
    "solution": "For each prime $p$, the numbers in the interval $n \\in [p^2, p^3)$ are not solutions, because the exponent of $p$ in the decomposition of the LCM is $2$ and so contributes a factor of $3$ to the number of divisors. Therefore, this number is not a power of $2$. From Bertrand's postulate, for a prime $p$ there is a prime $q$ with $p < q < 2p$, so $p^2 < q^2 < 4p^2 < p^3 < q^3$ for $p \\geq 5$; also $3^2 < 5^2 < 3^3 < 5^3$. Therefore, each interval $[p^2, p^3)$ and $[q^2, q^3)$ for consecutive primes $3 \\leq p < q$ intersects. We obtain that $n < 9$, and a direct check shows that only $n = 1, 2, 3,$ and $8$ are solutions. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15252,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the number of ordered 6-tuples $ (\\alpha_1, \\alpha_2, \\alpha_3, \\alpha_4, \\alpha_5, \\alpha_6) $ that can be created if the numbers $ \\alpha_1, \\alpha_2, \\alpha_3, \\alpha_4, \\alpha_5, \\alpha_6 $ can take the values $0$, $1$, and $2$, and the sum $ \\alpha_1 + \\alpha_2 + \\alpha_3 + \\alpha_4 + \\alpha_5 + \\alpha_6 $ is even.",
    "options": [],
    "answer": "See solution",
    "solution": "The sum $\\alpha_1 + \\alpha_2 + \\alpha_3 + \\alpha_4 + \\alpha_5 + \\alpha_6$ is even if and only if the number of 1's is even (i.e., 0, 2, 4, or 6).\n\n- If there are zero 1's, each $\\alpha_i$ can be $0$ or $2$, so there are $2^6$ possibilities.\n- If there are two 1's, they can be chosen in $\\binom{6}{2}$ ways, and the remaining four positions can each be $0$ or $2$: $2^4 \\cdot \\binom{6}{2}$.\n- If there are four 1's: $2^2 \\cdot \\binom{6}{4}$.\n- If all are 1's: $1$ way.\n\nTotal:\n$$\n2^6 + 2^4 \\cdot \\binom{6}{2} + 2^2 \\cdot \\binom{6}{4} + 1 = 64 + 16 \\cdot 15 + 4 \\cdot 15 + 1 = 365.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15253,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n = p_1^{a_1} p_2^{a_2} \\cdots p_k^{a_k}$, where the $p_i$ are distinct primes and $a_i \\geq 1$. For each $i$, consider the system:\n\n$$\n\\begin{cases}\nx \\equiv 1 \\pmod{p_i^{a_i}} \\\\\nx \\equiv 0 \\pmod{p_j^{a_j}}, \\quad j \\neq i\n\\end{cases}\n$$\n\nwhich has a solution $x_i$.\n\nShow that for the equation $x^2 \\equiv x \\pmod{n}$, all solutions $x$ are sums of subsets of $\\{x_1, x_2, \\dots, x_k\\}$, and that there exists a solution $a$ with $1 < a < 1 + \\dfrac{n}{k}$.",
    "options": [],
    "answer": "See solution",
    "solution": "For any solution $x_0$ of $x_0^2 \\equiv x_0 \\pmod{n}$, we have $x_0(x_0 - 1) \\equiv 0 \\pmod{n}$. Thus, for each $i = 1, 2, \\dots, k$, either $x_0 \\equiv 0 \\pmod{p_i^{a_i}}$ or $x_0 \\equiv 1 \\pmod{p_i^{a_i}}$.\n\nLet $S(A)$ be the sum of elements of a subset $A$ of $\\{x_1, x_2, \\dots, x_k\\}$ (with $S(\\emptyset) = 0$). Then:\n\n$$\nS(A)(S(A) - 1) \\equiv 0 \\pmod{n}\n$$\n\nbecause for each $i$, $S(A) \\pmod{p_i^{a_i}}$ is either $0$ or $1$. If $A \\neq A'$, then $S(A) \\not\\equiv S(A') \\pmod{n}$. Thus, all solutions to $x(x - 1) \\equiv 0 \\pmod{n}$ are exactly the sums of subsets of $\\{x_1, x_2, \\dots, x_k\\}$.\n\nLet $S_0 = n$, and $S_r$ be the least non-negative remainder of $x_1 + x_2 + \\dots + x_r$ modulo $n$ for $r = 1, 2, \\dots, k$, so $S_k = 1$. For $1 \\leq r \\leq k-1$, $S_r \\neq 0$. The $k+1$ numbers $S_0, S_1, \\dots, S_k$ are in $[1, n]$. By the pigeonhole principle, there exist $0 \\leq l < m \\leq k$ such that $S_l$ and $S_m$ are in the same interval $\\left(\\frac{jn}{k}, \\frac{(j+1)n}{k}\\right]$ for some $j$ ($l=0$ and $m=k$ do not both hold).\n\nThus, $|S_l - S_m| < \\frac{n}{k}$. Let $y_1 = S_1$, $y_r = S_r - S_{r-1}$ for $r = 2, 3, \\dots, k$, so any sum of $y_r \\equiv x_r \\pmod{n}$ meets the requirement.\n\nIf $S_m - S_l > 1$, then $a = y_{l+1} + y_{l+2} + \\dots + y_m = S_m - S_l \\in (1, \\frac{n}{k})$ is a solution to $x^2 - x \\equiv 0 \\pmod{n}$.\n\nIf $S_m - S_l = 1$, then $n = y_1 + \\dots + y_l + (y_{m+1} + \\dots + y_k) = x_1 + \\dots + x_l + (x_{m+1} + \\dots + x_k)$, but $m > l$, which contradicts the definition of $x_i$.\n\nIf $S_m - S_l = 0$, then $n \\mid y_{l+1} + \\dots + y_m = x_{l+1} + \\dots + x_m$, again contradicting the definition of $x_i$.\n\nIf $S_m - S_l < 0$,\n\n$$\na = (y_1 + \\dots + y_l) + (y_{m+1} + \\dots + y_k) = S_k - (S_m - S_l) = 1 - (S_m - S_l)\n$$\n\nis a solution to $x^2 - x \\equiv 0 \\pmod{n}$, and $1 < a < 1 + \\frac{n}{k}$.\n\nTherefore, there exists $a$ satisfying $1 < a < 1 + \\dfrac{n}{k}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15254,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the angles of a convex quadrilateral $ABCD$ such that $\\hat{ABD} = 29^\\circ$, $\\hat{ADB} = 41^\\circ$, $\\hat{ACB} = 82^\\circ$, and $\\hat{ACD} = 58^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $\\hat{BAD} = 180^\\circ - (29^\\circ + 41^\\circ) = 110^\\circ$, $\\hat{BCD} = 82^\\circ + 58^\\circ = 140^\\circ$. Consider the circumcircle $\\gamma$ of triangle $BCD$. Since $\\hat{BAD} + \\hat{BCD} > 180^\\circ$, point $A$ is interior to $\\gamma$.\n\nExtend $CA$ beyond $A$ to meet $\\gamma$ at $E$. By inscribed angles:\n\n$$\n\\hat{EBD} = \\hat{ECD} = \\hat{ACD} = 58^\\circ, \\quad \\hat{EDB} = \\hat{ECB} = \\hat{ACB} = 82^\\circ\n$$\n\nGiven that $\\hat{ABD} = 29^\\circ$, $\\hat{ADB} = 41^\\circ$, we obtain that $BA$ and $DA$ are bisectors of $\\hat{EBD}$ and $\\hat{EDB}$ respectively. Hence $A$ is the incenter of triangle $BDE$, implying that $EA$ is the bisector of $\\hat{BED}$.\n\nFrom the cyclic quadrilateral $BCDE$ we have:\n\n$$\n\\begin{aligned}\n\\hat{BED} &= 180^\\circ - \\hat{BCD} = 180^\\circ - 140^\\circ = 40^\\circ. \\quad \\text{Therefore } \\hat{DBC} = \\hat{BEC} = \\frac{1}{2} \\hat{BED} = 20^\\circ \\text{ and analogously} \\\\\n\\hat{DBC} &= 20^\\circ. \\quad \\text{In conclusion } \\hat{ABC} = 29^\\circ + 20^\\circ = 49^\\circ, \\hat{ADC} = 41^\\circ + 20^\\circ = 61^\\circ.\n\\end{aligned}\n$$\n\n![](images/Argentina2016_booklet_p6_data_96f9cd41ff.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15255,
    "subject": "Mathematics (Olympiad)",
    "question": "If the number $4\\nu + 3$, where $\\nu$ is an integer, is a multiple of $11$, find:\n\n1. The form of the integer $\\nu$.\n2. The remainder when $\\nu^4$ is divided by $11$.",
    "options": [],
    "answer": "See solution",
    "solution": "(i) Let $4\\nu + 3 = 11\\lambda$, where $\\lambda \\in \\mathbb{Z}$. Then\n$$\n\\nu = \\frac{11\\lambda - 3}{4} = 2\\lambda + \\frac{3(\\lambda - 1)}{4}.\n$$\nFor $\\nu$ to be an integer, $4$ must divide $3(\\lambda - 1)$. Since $\\gcd(4,3) = 1$, $4$ must divide $(\\lambda - 1)$. Thus, $\\lambda - 1 = 4\\kappa$, $\\kappa \\in \\mathbb{Z}$, so\n$$\n\\nu = 2(4\\kappa + 1) + 3\\kappa = 11\\kappa + 2, \\quad \\kappa \\in \\mathbb{Z}.\n$$\n\n(ii) We have\n$$\n\\nu^4 = (11\\kappa + 2)^4.\n$$\nSince $11\\kappa$ is divisible by $11$, by the binomial theorem all terms except the last will be divisible by $11$, so\n$$\n\\nu^4 \\equiv 2^4 = 16 \\equiv 5 \\pmod{11}.\n$$\nThus, the remainder is $5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15256,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be given positive integers. Prove that there exists some positive integer $N$ such that\n\n$$\na \\mid Nbc + b + c \\\\\nb \\mid Nca + c + a \\\\\nc \\mid Nab + a + b\n$$\n\nif and only if, denoting $d = \\gcd(a, b, c)$ and $a = dx$, $b = dy$, $c = dz$, the positive integers $x$, $y$, $z$ are pairwise coprime, and also $\\gcd(d, xyz) \\mid x + y + z$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show that $x$, $y$, $z$ must be pairwise coprime. Suppose $\\gcd(x, y) > 1$. Then $a \\mid Nbc + b + c$ becomes $x \\mid dNyz + y + z$. Thus, $\\gcd(x, y) \\mid z$, which is impossible since $\\gcd(x, y, z) = 1$.\n\nNow, assume the coprimality condition holds. Consider the integers\n\n$$\nxyz - \\sum x < 2xyz - \\sum x < \\dots < (\\sum xy) xyz - \\sum x.\n$$\n\nThese $\\sum xy$ numbers yield different remainders modulo $\\sum xy$. If $ixyz - \\sum x \\equiv jxyz - \\sum x \\pmod{\\sum xy}$, then $\\sum xy \\mid |i - j|xyz$. Since $0 \\leq |i - j| < \\sum xy$ and $\\gcd(xyz, \\sum xy) = 1$, we must have $i = j$. Therefore, there exists a unique $1 \\leq t \\leq \\sum xy$ such that $\\sum xy \\mid txyz - \\sum x$, i.e., $txyz - \\sum x = C \\sum xy$ for some positive integer $C$. Thus, $txyz = C \\sum xy + \\sum x$, so $x \\mid Cyz + y + z$, and similarly for $y$ and $z$.\n\nAll other suitable values are of the form $C' = C + Mxyz$, since $xyz \\mid (C' - C) \\sum xy$ and $\\gcd(xyz, \\sum xy) = 1$.\n\nNow, to have $a \\mid Nbc + b + c$ (and similarly for $b$ and $c$), equivalently $x \\mid dNyz + y + z$, we need $dN = C + Mxyz$ for some non-negative integer $M$. Let $e = \\gcd(d, xyz)$. Then $e \\mid d$, so $e \\mid C + Mxyz$. But also $e \\mid xyz \\mid (C + Mxyz) \\sum xy + \\sum x$, so $e \\mid \\sum x$.\n\nConversely, if $e \\mid \\sum x$, then $e \\mid C \\sum xy$, and since $\\gcd(e, \\sum xy) = 1$, $e \\mid C$. Thus, $\\frac{d}{e}N = \\frac{C}{e} + M\\frac{xyz}{e}$, and since $\\gcd(\\frac{d}{e}, \\frac{xyz}{e}) = 1$, we can choose $M \\equiv -\\frac{C}{e}(\\frac{xyz}{e})^{-1} \\pmod{d/e}$, so $\\frac{d}{e}$ divides $\\frac{C}{e} + M\\frac{xyz}{e}$. Take $N = \\frac{C + Mxyz}{d}$ (and $N' = N + Mabc$ also works).\n\n**Remarks:**\n- There are counterexamples if $\\gcd(x, y, z) = 1$ but $x$, $y$, $z$ are not pairwise coprime; e.g., $x = pq$, $y = qr$, $z = rp$ with $p, q, r$ prime.\n- There are counterexamples to the last condition, e.g., $(a, b, c) = (6, 9, 15)$, where $d = 3$, $e = 3$, $\\sum x = 10$, and $e \\nmid \\sum x$.\n\nFor such examples, no solution exists.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15257,
    "subject": "Mathematics (Olympiad)",
    "question": "When $0$ or $1$ is written on each tile in such a way that the product of the two numbers written on every neighboring pair of tiles is always $0$, we'll call the status a *z-pattern*.\n\nConsider a $4 \\times 4$ grid of tiles. How many z-patterns are possible if the four tiles in the center form a $2 \\times 2$ block?",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove a couple of lemmas.\n\n**Lemma 1.** Let $a_n$ be the number of z-patterns for $n$ tiles laid in a row. Then\n\n$$\na_n = F_{n+2}\n$$\n\nfor all $n \\ge 1$, where $F_n$ is the $n$-th Fibonacci number defined by $F_1 = 1$, $F_2 = 1$, and $F_{n+2} = F_{n+1} + F_n$ for all $n \\ge 1$.\n\n*Proof.* We prove by induction on $n \\ge 1$. It is clear that the lemma holds for $n=1$ and $n=2$. Assume the lemma for all $k$, $1 \\le k < n$, where $n \\ge 3$. If $0$ is written on the tile at one end, then any number ($0$ or $1$) can be written on the next tile and hence there are $a_{n-1}$ such z-patterns. If $1$ is written there, then $0$ should be written on the next tile and hence there are $a_{n-2}$ such z-patterns. Thus we get\n\n$$\na_n = a_{n-1} + a_{n-2} = F_{n+1} + F_n = F_{n+2},\n$$\n\nwhich proves the lemma.\n\n**Lemma 2.** Let $b_n$ be the number of z-patterns for $n$ tiles fixed on a wall in a ring shape. Then\n\n$$\nb_n = a_{n-1} + a_{n-3} = F_{n+1} + F_{n-1}\n$$\n\nfor all $n \\ge 4$.\n\n*Proof.* Let $n \\ge 4$ and choose any tile among the $n$ tiles fixed on a wall in a ring shape. If $0$ is written on the tile, then any number can be written on the neighboring tiles and hence there are $a_{n-1}$ such z-patterns. If $1$ is written on the tile, then $0$ should be written on the neighboring two tiles and hence there are $a_{n-3}$ such z-patterns. Thus\n\n$$\nb_n = a_{n-1} + a_{n-3} = F_{n+1} + F_{n-1},\n$$\n\nwhich proves the lemma.\n\nWe now return to the problem. Consider the $2 \\times 2$ tiles in the center.\nObserve that the number of $0$'s that can be written on these four tiles equals $4$, $3$, or $2$.\n\nCase 1) Four $0$'s:\n\n$$\n\\begin{bmatrix} 0 & 0 \\\\ 0 & 0 \\end{bmatrix}\n$$\n\nWe may apply Lemma 2 to the $12$ tiles surrounding the four tiles in the center because any number can be written on the $12$ tiles. Therefore, the number of z-patterns in this case equals\n\n$$\nb_{12} = F_{13} + F_{11} = 233 + 89 = 322.\n$$\n\nCase 2) Three $0$'s:\n\n$$\n\\begin{bmatrix} 1 & 0 \\\\ 0 & 0 \\end{bmatrix}\n$$\n\n$$\n\\begin{bmatrix} 0 & 1 \\\\ 0 & 0 \\end{bmatrix}\n$$\n\n$$\n\\begin{bmatrix} 0 & 0 \\\\ 1 & 0 \\end{bmatrix}\n$$\n\n$$\n\\begin{bmatrix} 0 & 0 \\\\ 0 & 1 \\end{bmatrix}\n$$\n\nIn each of the four subcases, only $0$ can be written on two neighboring tiles of the tile marked by $1$. For the remaining $10$ tiles, it is clear that the number of z-patterns equals $a_9 \\times 2 = F_{11} \\times 2 = 178$. Therefore, the total number of z-patterns in this case equals\n\n$$\n178 \\times 4 = 712.\n$$\n\nCase 3) Two $0$'s:\n\n$$\n\\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}\n$$\n\n$$\n\\begin{bmatrix} 0 & 1 \\\\ 1 & 0 \\end{bmatrix}\n$$\n\nIn each of the two subcases, only $0$ can be written on four neighboring tiles of the two tiles marked by $1$. For the remaining $6$ tiles, it is clear that the number of z-patterns equals $a_3^2 \\times 2^2 = F_5^2 \\times 4 = 100$. Therefore, the total number of z-patterns in this case equals\n\n$$\n100 \\times 2 = 200.\n$$\n\nCombining the three cases, the answer is: $322 + 712 + 200 = 1234$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15258,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a positive integer. For which values of $k$ can the product $(x+1)(x+2)\\cdots(x+k)$ be a perfect square for infinitely many integer values of $x$?",
    "options": [],
    "answer": "See solution",
    "solution": "If $k$ is even, suppose $(x+1)(x+2)\\cdots(x+k)$ is a perfect square for infinitely many $x$. Then, for large $x$, there exist an integer $a$ and a polynomial $Q(x) \\in \\mathbb{Z}[x]$ such that\n\n$$\n(Q(x))^2 \\leq a^2(x+1)(x+2)\\cdots(x+k) < (Q(x)+1)^2.\n$$\n\nSince $Q(x)$ is a perfect square for infinitely many $x$, we get $(Q(x))^2 = a^2(x+1)(x+2)\\cdots(x+k)$, which is impossible because $a^2(x+1)(x+2)\\cdots(x+k)$ does not have multiple roots.\n\nIf $k$ is odd, suppose $(x+1)(x+2)\\cdots(x+k)$ is a perfect square for infinitely many $x$. If $\\prod_{i=1}^k (a+i)$ is a perfect square, then there exists a nonempty proper subset $A$ of $\\{1,2,\\dots,k\\}$ such that $\\prod_{i\\in A} (a+i)$ is a perfect square. Considering all subsets $A$ of $S=\\{a+1,\\dots,a+k\\}$, let $F(A)$ be the product of elements of $A$ ($F(\\emptyset)=1$), and $g(A)$ the square-free part of $F(A)$. All prime divisors of $g(\\{a+1\\}),\\dots,g(\\{a+k\\})$ are less than $k$. For any $A\\subseteq S$, $g(A)$ divides $2\\times3\\times\\cdots\\times p_{\\pi(k-1)}$, so there are $2^{\\pi(k-1)}$ possible $g(A)$. Since $2^k > 2\\times2^{\\pi(k-1)}$ for $k-1>\\pi(k-1)$, there exist $A\\ne B, S-B$ with $g(A)=g(B)$. Then $F(A)F(B)$ is a perfect square, and $F(A\\Delta B)$ is a perfect square for a nonempty proper subset $A\\Delta B$. Thus, for odd $k$, $(x+1)(x+2)\\cdots(x+k)$ can be a perfect square for at most finitely many $x$.\n\n**Conclusion:** For all $k$, $(x+1)(x+2)\\cdots(x+k)$ is a perfect square for at most finitely many integer $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15259,
    "subject": "Mathematics (Olympiad)",
    "question": "Областю допустимих значень рівняння є відрізок $[0; 1]$. Розв'яжіть рівняння:\n\n$$\n(\\sqrt{x})^2 + \\sqrt{1-x} + (\\sqrt{x})^2 + (\\sqrt{1-x})^2 - \\sqrt{x} - 3\\sqrt{x}\\sqrt{1-x} = 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Розкладаючи ліву частину рівняння на множники, одержуємо:\n\n$$\n(\\sqrt{x} - \\sqrt{1-x})(2\\sqrt{x} - \\sqrt{1-x} - 1) = 0.\n$$\n\nЗалишається розв'язати рівняння:\n\n$$\n\\sqrt{x} - \\sqrt{1-x} = 0, \\quad 2\\sqrt{x} = \\sqrt{1-x} + 1.\n$$\n\nВідповідь: $\\frac{1}{2}, \\frac{16}{25}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15260,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $ (x, n) $ of positive integers satisfying the equation:\n\n$$\n3 \\cdot 2^x + 4 = n^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $x = 1$, then there are no solutions. For $x = 2$, we have $3 \\cdot 2^2 + 4 = 12 + 4 = 16 = 4^2$, so $(x, n) = (2, 4)$ is a solution.\n\nNow suppose $x \\geq 3$. The left side is even, so $n$ must be even; let $n = 2k$. Substituting, we get:\n\n$$\n3 \\cdot 2^x + 4 = 4k^2 \\implies 3 \\cdot 2^{x-2} = k^2 - 1.\n$$\n\nThus, $k$ is odd. We can write:\n\n$$\n3 \\cdot 2^{x-2} = (k-1)(k+1).\n$$\n\n**Case 1:** $k+1 = 3 \\cdot 2^a$ and $k-1 = 2^b$, with $a, b \\geq 1$ and $a + b = x-2$.\n\nSubtracting, $3 \\cdot 2^a - 2^b = 2$. If $a, b \\geq 2$, the left side is divisible by 4, but the right side is not. So either $a = 1$ (then $b = 2$, $x = 5$), or $b = 1$ (no solution for $a$). Thus, $(x, n) = (5, 10)$ is a solution.\n\n**Case 2:** $k+1 = 2^s$ and $k-1 = 3 \\cdot 2^t$, with $s, t \\geq 1$ and $s + t = x-2$.\n\nThen $2^s - 3 \\cdot 2^t = 2$. If $s, t \\geq 2$, the left side is divisible by 4, but the right side is not. So either $s = 1$ or $t = 1$. This gives $s = 3$ and $x = 6$. Thus, $(x, n) = (6, 14)$ is a solution.\n\n**Final answer:** The solutions are $(x, n) = (2, 4), (5, 10), (6, 14)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15261,
    "subject": "Mathematics (Olympiad)",
    "question": "For positive $a, b, c$ that satisfy $a + b + c + 2 = abc$, prove the inequality:\n\n$$\n\\frac{a}{b+1} + \\frac{b}{c+1} + \\frac{c}{a+1} \\ge 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us use the Cauchy-Schwarz inequality for $\\sqrt{\\frac{a}{b+1}}, \\sqrt{\\frac{b}{c+1}}, \\sqrt{\\frac{c}{a+1}}$ and $\\sqrt{a(b+1)}, \\sqrt{b(c+1)}, \\sqrt{c(a+1)}$:\n\n$$\n\\begin{aligned}\n& \\sqrt{\\frac{a}{b+1}} \\cdot \\sqrt{a(b+1)} + \\sqrt{\\frac{b}{c+1}} \\cdot \\sqrt{b(c+1)} + \\sqrt{\\frac{c}{a+1}} \\cdot \\sqrt{c(a+1)} \\\\\n& \\le \\sqrt{\\left(\\frac{a}{b+1} + \\frac{b}{c+1} + \\frac{c}{a+1}\\right)\\left(a(b+1)+b(c+1)+c(a+1)\\right)} \\\\\n& \\Leftrightarrow \\\\\n& \\left(\\frac{a}{b+1} + \\frac{b}{c+1} + \\frac{c}{a+1}\\right)\\left(a(b+1)+b(c+1)+c(a+1)\\right) \\ge (a+b+c)^2 \\\\\n& \\Leftrightarrow \\\\\n& \\frac{a}{b+1} + \\frac{b}{c+1} + \\frac{c}{a+1} \\ge \\frac{(a+b+c)^2}{a(b+1)+b(c+1)+c(a+1)}.\n\\end{aligned}\n$$\n\nIt is sufficient to prove that\n\n$$\n\\frac{(a+b+c)^2}{a(b+1)+b(c+1)+c(a+1)} \\ge 2.\n$$\n\nWhich is equivalent to:\n\n$$\na^2 + b^2 + c^2 \\ge 2a + 2b + 2c. \\qquad (1)\n$$\n\nLet us use the Cauchy-Schwarz inequality for $a, b, c$ and $2, 2, 2$:\n\n$$\n(a^2 + b^2 + c^2)(4 + 4 + 4) \\ge (2a + 2b + 2c)^2 \\\\\n3(a^2 + b^2 + c^2) \\ge (a + b + c)^2. \\qquad (2)\n$$\n\nWe show that $a + b + c \\ge 6$. For the problem we have $a + b + c + 2 = abc$, so\n\n$$\nabc = (a + b) + (c + 2) \\ge 2\\sqrt{ab} + 2\\sqrt{2c} \\ge 2\\sqrt{2\\sqrt{ab} \\cdot 2\\sqrt{2c}} \\\\\n\\Leftrightarrow (abc)^2 \\ge 16\\sqrt{2abc} \\Leftrightarrow\n$$\n\n$$\n(abc)^4 \\ge 512abc \\Leftrightarrow (abc)^3 \\ge 512 \\Leftrightarrow abc \\ge 8.\n$$\n\nHence $a + b + c \\ge abc - 2 \\ge 6$, then from (2) we have\n\n$$\na^2 + b^2 + c^2 \\ge \\frac{1}{3}(a + b + c)^2 \\ge \\frac{1}{3} \\cdot 6 \\cdot (a + b + c) = 2(a + b + c),\n$$\n\n**Alternative Solution.** Notice that the sets $(a, b, c)$ and $(\\frac{1}{a+1}, \\frac{1}{b+1}, \\frac{1}{c+1})$ are conversely sorted. By the rearrangement inequality,\n\n$$\n\\frac{a}{b+1} + \\frac{b}{c+1} + \\frac{c}{a+1} \\ge \\frac{a}{a+1} + \\frac{b}{b+1} + \\frac{c}{c+1} = \\frac{a(b+1)(c+1) + b(c+1)(a+1) + c(b+1)(a+1)}{(a+1)(b+1)(c+1)}\n$$\n\n$$\n\\begin{aligned}\n&= \\frac{abc + ab + ac + a + abc + bc + ba + b + abc + cb + ca + c}{abc + ab + bc + ca + a + b + c + 1} \\\\\n&= \\frac{3abc + 2(ab + bc + ca) + abc - 2}{abc + ab + bc + ca + abc - 1} = 2,\n\\end{aligned}\n$$\n\nwhich is needed.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15262,
    "subject": "Mathematics (Olympiad)",
    "question": "For $1 \\leq i \\leq 20$ and $1 \\leq j \\leq 11$, define $a_{i,j}$ as follows: $a_{i,j} = 1$ if the $i$-th mathematics problem and $j$-th physics problem are chosen by some student, and $a_{i,j} = 0$ otherwise. Find the maximal possible value of\n\n$$\nA = \\sum_{i=1}^{20} \\sum_{j=1}^{11} a_{i,j}\n$$\n\nunder the following two conditions:\n\n- $a_{i,j} = 0$ or $1$ for all $i, j$.\n- If $a_{k,l} = 1$ for some $k$ and $l$, then at least one of the sums $\\sum_{j=1}^{11} a_{k,j}$ and $\\sum_{i=1}^{20} a_{i,l}$ does not exceed $2$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show that $A \\leq 54$. Suppose $a_{k,l} = 1$. We say that $k$ is *1-good* if\n$$\n\\sum_{j=1}^{11} a_{k,j} \\leq 2,\n$$\nand $l$ is *2-good* if\n$$\n\\sum_{i=1}^{20} a_{i,l} \\leq 2.\n$$\n\nIf all $20$ values of $k$ are 1-good, then $A \\leq 2 \\times 20 = 40$.\nIf all $11$ values of $l$ are 2-good, then $A \\leq 2 \\times 11 = 22$.\nIf $19$ values of $k$ are 1-good, then $A \\leq 2 \\times 19 + 11 = 49$.\nIf $10$ values of $l$ are 2-good, then $A \\leq 2 \\times 11 + 20 = 32$.\n\nIf the number of 1-good $k$ is at most $18$ and the number of 2-good $l$ is at most $9$, then the total number of good values is at most $27$, so $A \\leq 2 \\times 27 = 54$, since the number of nonzero terms in $A$ is at most twice the number of good values. Thus, $A \\leq 54$.\n\nNow, we give an example achieving $A = 54$. Let $a_{i,j} = 1$ only for\n$$\n(i, j) \\in \\{ (i, j) : i \\in \\{1, 20\\} \\text{ or } j \\in \\{1, 11\\} \\} \\setminus \\{ (1,1), (20,1), (1,11), (20,11) \\}.\n$$\n\nThe conditions are satisfied, so the maximal value is $A = 54$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15263,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 2020 inhabitants in a town. Before Christmas, they are all happy; but if an inhabitant does not receive any Christmas card from any other inhabitant, he or she will become sad. Unfortunately, there is only one post company which offers only one kind of service: before Christmas, each inhabitant may appoint two different other inhabitants, among which the company chooses one to whom to send a Christmas card on behalf of that inhabitant. It is known that the company makes the choices in such a way that as many inhabitants as possible will become sad. Find the least possible number of inhabitants who will become sad.",
    "options": [],
    "answer": "See solution",
    "solution": "Partition 2019 inhabitants into 673 groups, each containing 3 inhabitants. Suppose that each inhabitant appoints two other members of the same group. As no group member is appointed thrice, the company cannot send three cards to one group member. Hence in every group, two different members get a card and at most one member will become sad. The inhabitant who does not belong to any group will become sad, too. Altogether, at most 674 inhabitants will become sad.\n\nWe show now that this is the least possible number. More precisely, we show that, in the case of $n$ inhabitants, the company can make $\\lfloor \\frac{n}{3} \\rfloor$ inhabitants sad. Call inhabitants X and Y \\textbf{competitors} if somebody appoints them together to the company. By conditions of the problem, there are as many pairs of competitors as inhabitants or less (if several inhabitants appoint the same pair or some inhabitant appoints no pair). Thus it suffices to prove that, in the case of $n$ inhabitants and at most $n$ pairs of competitors, there must be $\\lceil \\frac{n}{3} \\rceil$ inhabitants among which no two are competitors. The claim holds trivially for $n = 0, 1, 2$. Now let $n \\ge 3$ and assume the claim being valid for $n-3$ inhabitants. W.l.o.g., assume that there are exactly $n$ pairs of competitors. Then there are exactly $2n$ instances of an inhabitant belonging to a pair of competitors. Consider two cases:\n\n- If every inhabitant has exactly 2 competitors then choose an arbitrary inhabitant $X$ and leave out $X$ along with both competitors. Among the remaining $n-3$ inhabitants, there are at most $n-3$ pairs of competitors (besides two pairs containing $X$, removing either competitor canceled one more pair). By the induction hypothesis, one can find $\\lceil \\frac{n-3}{3} \\rceil$ remaining inhabitants with no pair of competitors. Adding $X$ to them results in $\\lceil \\frac{n}{3} \\rceil$ inhabitants with no pair of competitors.\n\n- If an inhabitant $X$ has at most 1 competitor then there must exist an inhabitant $Y$ with at least 3 competitors. Let $Z$ be the competitor of $X$ if $X$ has a competitor and an arbitrary inhabitant different from $X$ and $Y$ otherwise. After leaving out $X$, $Y$ and $Z$, we have $n-3$ inhabitants and at most $n-3$ pairs of competitors (removing $Y$ cancels at least 3 pairs of competitors). By the induction hypothesis, one can find $\\lceil \\frac{n-3}{3} \\rceil$ remaining inhabitants with no pair of competitors. Adding $X$ to this set results in a group of $\\lceil \\frac{n}{3} \\rceil$ inhabitants with no pairs of competitors. This proves the desired claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15264,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDE$ be a pentagon with $\\angle A = \\angle B = \\angle C = \\angle D = 120^\\circ$. Prove that $4AC \\cdot BD \\geq 3AE \\cdot ED$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $AB = a$, $BC = b$, $CD = c$. Denote by $X$ and $Y$ the points of intersection of $BC$ with $AE$ and $ED$, respectively. Because triangle $XEY$ is equilateral, we have $a + b + c = c + DE = a + AE$, hence $DE = a + b$ and $AE = b + c$.\n\nThe Law of Cosines and the inequality\n\n$$\n4(u^2 + uv + v^2) \\geq 3(u + v)^2\n$$\n\nimply\n\n$$\n4AC^2 = 4(a^2 + ab + b^2) \\geq 3(a + b)^2 = 3DE^2\n$$\n\nand\n\n$$\n4BD^2 = 4(b^2 + bc + c^2) \\geq 3(b + c)^2 = 3AE^2\n$$\n\nMultiplying the last two inequalities, the conclusion follows. Equality holds if and only if $AB = BC = CD = x$ and $DE = EA = 2x$ for some positive real number $x$.\n\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15265,
    "subject": "Mathematics (Olympiad)",
    "question": "The table below has 5 rows and 5 columns, and in each square is written a number from 1 to 25, each appearing exactly once.\n\n![](images/RMC_2020_p9_data_bff2ba0133.png)\n\nProve that, if the product of the numbers in any row is not divisible by $26$ or $34$, then there exists a row or a column such that the product of its numbers is divisible by $221$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $221 = 13 \\times 17$, we need to show that $13$ and $17$ appear together in some row or column.\n\nSuppose, for contradiction, that $13$ and $17$ are not in the same row or column. In any row containing $13$, if there is an even number, the product would be divisible by $26 = 2 \\times 13$, which is forbidden. Thus, all numbers in such a row must be odd. The odd numbers from $1$ to $25$ are $1, 3, 5, \\ldots, 25$—a total of $13$ odd numbers.\n\nSimilarly, any row containing $17$ must also consist only of odd numbers. But if $13$ and $17$ are in different rows, we need enough odd numbers to fill both rows without overlap, which is impossible since there are only $13$ odd numbers in total.\n\nTherefore, $13$ and $17$ must appear together in some row or column, so the product of the numbers in that row or column is divisible by $221$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15266,
    "subject": "Mathematics (Olympiad)",
    "question": "For a pair of integers $u, v$ and a positive integer $m$, we write $u \\equiv v \\pmod m$ to mean that $u - v$ is divisible by $m$.\n\nLet us call an $a \\times a$ square box of the given grid a *region*. Given $n \\ge a$, consider coloring black all of $n \\times 1$ square boxes stacked consecutively on a vertical strip. The rectangular patch of the grid consisting of $n \\times (2a-1)$ squares, formed by the $n$ black $1 \\times 1$ vertically stacked boxes and their $2a-2$ horizontal translates ($(1, 2, \\dots, a-1)$ to the left and $(1, 2, \\dots, a-1)$ to the right), has the property that any region contained in the interior of the patch satisfies the requirement of the problem.\n\nShow that $K = a(n+1-a)$ is satisfied, and for all sufficiently large $n$, $K \\le a(n+1-a)$ holds, where $K$ is the number of regions satisfying the requirement.",
    "options": [],
    "answer": "See solution",
    "solution": "Define, for a region $R$ with $x$ black $1 \\times 1$ squares,\n\n$$\n\\mathcal{L}(R) = \\begin{cases} x, & \\text{if } x \\le a-1, \\\\ x-a, & \\text{if } x \\ge a. \\end{cases}\n$$\n\nThis is called the *loss* for $R$. There are $a^2 n$ pairings of a black square and a region containing it. Let $L = \\sum \\mathcal{L}(R)$, where the sum is over all regions containing black boxes. For $K$ regions $R$ that satisfy the requirement, $\\mathcal{L}(R) = 0$, while for other regions $R'$, $\\mathcal{L}(R') > 0$. Thus, $aK \\le a^2 n - L$. It suffices to show that there exists $N$ such that if $n \\ge N$, then $L \\ge a^2(a-1)$.\n\nLet $N = (a^2(a-1) - a)^2 + 1$. For $n \\ge N$, $L \\ge a^2(a-1)$ holds. Consider contiguous $a$ rows (horizontal lines) and $a$ columns (vertical lines) in the grid. Either $a^2(a-1)$ or more horizontal lines or $a^2(a-1)$ or more vertical lines will contain at least one black square. If $L < a^2(a-1)$, then at least one horizontal or vertical line must have $\\mathcal{L}(R) = 0$ for any region $R$ contained in it. Assume such a horizontal line exists. Take the lowest row as the 0-th row, and let $A$ be the left-most black box in this line. Regions containing $A$ in the rightmost column must have $a$ black squares, so every square in that column is black. Regions containing $A$ in their left-most column have all $a-1$ columns to the right and left of $A$'s column without black squares.\n\nFor any vertical line containing $A$, the sum of $\\mathcal{L}(R)$ for all regions $R$ in the line is at least $a(a-1)$. Let $R_y$ be a region in the vertical line with its lowest row at $y$, $f(y) = \\mathcal{L}(R_y)$, and $g(y)$ the number of black squares in $R_y$. Then $f(y) \\equiv g(y) \\pmod a$. Suppose $(as + t)$-th row ($0 \\le t < s$) is the uppermost row with a black square, and $b$ is the number of black squares on and above the 0-th row. For $c_l$ ($l = 1, \\dots, a$), the remainder of $\\sum_{i=0}^{s+1} f(ai + l)$ divided by $a$, we have\n\n$$\nc_l \\equiv \\sum_{i=0}^{s+1} g(ai + l) \\equiv b - l \\pmod a,\n$$\n\nso $\\{c_1, \\dots, c_a\\}$ is a permutation of $\\{0, 1, \\dots, a-1\\}$. Therefore,\n\n$$\n\\sum_{y=1}^{as+t} f(y) \\ge \\sum_{l=1}^a c_l = \\frac{a(a-1)}{2}.\n$$\n\nSimilarly, for $y < 0$, the sum is $\\ge \\frac{a(a-1)}{2}$. Thus, the total over all regions in the vertical is $\\ge a(a-1)$. Since there are $a$ vertical lines containing $A$, $L \\ge a^2(a-1)$, so $K \\le a(n+1-a)$, completing the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15267,
    "subject": "Mathematics (Olympiad)",
    "question": "給定任一質數 $p$，當一個集合恰包含三個元素 $a, b, c$ 且 $a + b \\equiv c \\pmod p$，則我們稱這個集合為 *p-good*。\n\n找出所有質數 $p$，使得 $\\{1, 2, \\dots, p-1\\}$ 可以被全部分割成許多 $p$-good 集合。",
    "options": [],
    "answer": "See solution",
    "solution": "顯然，必須有 $p \\equiv 1 \\pmod 6$。\n\n要證明所有這樣的 $p$ 都滿足條件，取 $g$ 為模 $p$ 的本原根。設 $x = g^{(p-1)/6}$。可以驗證 $x^3 \\equiv -1 \\pmod p$，因此 $x^2 - x + 1 \\equiv 0 \\pmod p$。所以對每個 $n$，$g^n, x^2g^n, xg^n$ 是 $p$-good。\n\n取 $n \\in \\{0, 1, 2, \\dots, \\frac{p-1}{6} - 1\\} \\cup \\{k, k+1, k+2, \\dots, k+\\frac{p-1}{6} - 1\\}$，其中 $k = \\frac{p-1}{2}$，即可得到所需的分割。",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 15268,
    "subject": "Mathematics (Olympiad)",
    "question": "設兩直線 $BC$, $EF$ 互相平行，$D$ 為在 $BC$ 線段上且與 $B, C$ 相異的一點。\n直線 $BF$, $CE$ 交於 $I$ 點。將 $\\triangle CDE$, $\\triangle BDF$ 的外接圓分別記為 $K$, $L$。\n圓 $K$, $L$ 分別與 $EF$ 切於 $E$, $F$ 點。令 $A$ 為圓 $K$, $L$ 異於 $D$ 的另一交點。\n設直線 $DF$ 與圓 $K$ 再交於 $Q$ 點，直線 $DE$ 與圓 $L$ 再交於 $R$ 點。令直線 $EQ$ 與 $FR$ 交於 $M$ 點。\n\n證明：$I$, $A$, $M$ 三點共線。",
    "options": [],
    "answer": "See solution",
    "solution": "1. 因 $BC \\parallel EF$，且圓 $K$, $L$ 分別切 $EF$ 於 $E$, $F$ 點，知點 $E$, $F$ 分別為 $CDE$, $BDF$ 的中點。得 $BF = DF$, $CE = DE$。因此，\n\n$$\n\\angle EFD = \\angle FDB = \\angle DBF = \\angle EFI.\n$$\n\n同理，$\\angle FED = \\angle FEI$。由此知 $\\triangle DEF \\sim \\triangle IEF$。\n\n![](images/17-1J_p12_data_ca7d1242bd.png)\n\n2. $A$, $E$, $I$, $F$ 共圓：\n\n$$\n\\begin{align*}\n\\angle EAF &= \\angle (AE, AD) + \\angle (AD, AF) = \\angle (CE, CD) + \\angle (BD, BF) \\\\\n&= \\angle (CI, BC) + \\angle (BC, BI) = \\angle (CI, BI) \\\\\n&= \\angle (EI, FI) = \\angle EIF.\n\\end{align*}\n$$\n\n（這裡 $\\angle^*$ 表 directed angle modulo $\\pi$，亦可用一般角度）\n\n設此圓為 $J$。則 $EQ$, $FR$ 為對 $J$ 的切線：\n\n$$\n\\begin{align*}\n\\angle AEQ &= \\angle FEQ - \\angle FEA = \\angle EDQ - \\angle FIA \\\\\n&= \\angle EDF - \\angle FIA = \\angle FIE - \\angle FIA = \\angle AIE.\n\\end{align*}\n$$\n\n另一情形類似。\n\n3. 設 $AD$ 交 $EF$ 於 $N$。則 $NF^2 = NA \\cdot ND = NE^2$。所以 $NF = NE$，$AD$ 為 $\\triangle DEF$ 的中線。因 $\\triangle DEF \\cong \\triangle IEF$，且 $\\angle ADF = \\angle AEQ = \\angle AIE$，$\\angle ADE = \\angle AEF = \\angle AIF$，$AI$ 是 $\\triangle EIF$ 的 $A$-symmedian。\n\n因 $EQ$, $FR$ 為 $K$ 在 $E$, $F$ 的切線，$\\triangle EIF$ 的 $A$-symmedian 過 $EQ$, $FR$ 的交點 $M$，即 $M$ 在直線 $AI$ 上。證畢。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15269,
    "subject": "Mathematics (Olympiad)",
    "question": "平面上有一個凸 $3n$ 邊形，其每個頂點上都有一台機器人，每台機器人都射出一道雷射光指向另一台機器人。你每次操作可以選取一台機器人，叫它順時鐘旋轉，直到它的雷射光指向一台新的機器人為止。當三台機器人 $A$、$B$、$C$，其中 $A$ 的雷射光射向 $B$，$B$ 的雷射光射向 $C$，而 $C$ 的雷射光射向 $A$ 時，我們稱這三台機器人構成一個三角形。試問：至少要多少次操作，才能保證平面上出現 $n$ 個三角形？\n\nThere's a convex $3n$-polygon on the plane with a robot on each of its vertices. Each robot fires a laser beam toward another robot. On each of your moves, you select a robot to rotate clockwise until its laser points at a new robot. Three robots $A, B$ and $C$ form a triangle if $A$'s laser points at $B$, $B$'s laser points at $C$, and $C$'s laser points at $A$. Find the minimum number of moves that can guarantee $n$ triangles on the plane.",
    "options": [],
    "answer": "See solution",
    "solution": "答案：$\\frac{9n^2-7n}{2}$ 次。\n\n以下以 $E X$ 代表隨機變數 $X$ 的期望值。\n\n1. 首先，對於任兩點 $A$ 和 $B$，令 $N_{AB}$ 為要將 $A$ 的雷射指向 $B$ 所需的步數。假設我們以均勻分布在所有點中隨機選取相異的 $A$ 和 $B$，則 $E N_{AB} = \\frac{3n-2}{2}$（事實上，易知 $N_{AB}$ 為 $\\{0,1,\\dots,3n-2\\}$ 上的均勻分布）。\n\n2. 對於任三點 $A$、$B$ 和 $C$，令 $N_{ABC}$ 為要旋轉三點上的雷射，使得 $ABC$ 要成為三角形所需步數。假設我們以均勻分布在所有點中隨機選取相異三點 $(A,B,C)$，由期望值的可加性，我們知道 $E N_{ABC} = E N_{AB} + E N_{BC} + E N_{CA} = \\frac{3(3n-2)}{2}$。\n\n3. 將這 $3n$ 個點以 $A_1$ 到 $A_{3n}$ 表示。我們用 $T = (T_1,T_2,\\dots,T_n)$ 來表示將所有點分成 $n$ 個三角形的方法，其中 $T_i = (A_{i1},A_{i2},A_{i3})$ 為有序對。用 $N_T$ 表示在給定將圖轉成 $T$ 所需的步數。假設我們在所有可能的分法上以均勻分布選 $T$，同第二點，由可加性，$E N_T = \\frac{3n(3n-2)}{2}$。\n\n4. 用 $S = (S_1,S_2,\\dots,S_n)$ 來表示將所有點分成 $n$ 個無序集合，每組三點的方式。對於每個 $T$，令 $S(T)$ 為 $T$ 所對應的無序集合組。注意到第三項中的隨機變數，可以視同先在所有可能的 $S$ 中以均勻分布選一個，然後再以均勻分布考慮 $S_i$ 的所有排列。基於 $E N_T = \\frac{3n(3n-2)}{2}$，這代表存在一個 $S^*$，使得 $E[N_T|S(T) = S^*] \\le \\frac{3n(3n-2)}{2}$。\n\n5. 現在，注意到所有滿足 $S(T) = S^*$ 的 $T$ 都有相同的三點分組方式，只是其中若干三角形的順逆時針方向不同。考慮以下命題：\n\n命題一：對於一個三角形，轉成順時針與逆時針所需的步數不同。\n\n若命題一成立，則代表 $\\{N_T|S(T) = S^*\\}$ 至少包含 $n+1$ 個相異值，故在所有滿足 $S(T) = S^*$ 的 $T$ 中必存在一個 $T^*$，使得 $N_{T^*} \\le E[N_T|S(T) = S^*] - \\frac{n}{2} \\le \\frac{3n(3n-2)}{2} - \\frac{n}{2} = \\frac{9n^2-7n}{2}$，從而得到我們宣稱的下界。\n\n6. 要證明命題一，考慮一個無序三角形 $ABC$，並假設其轉成順時針與逆時針三角形所需的步數相同。不失一般性，假設 $ABC$ 為順時針三角形；進一步的，令 $n_1$ 為將 $BC$ 順時針轉到 $BA$ 所需步數，$n_2$ 為將 $CA$ 順時針轉到 $CB$ 所需步數，而 $n_3$ 為將 $AB$ 逆時針轉到 $AC$ 所需步數，則有 $n_1 + n_2 = n_3$。然而注意到將 $BC$ 順時針轉到 $BA$，加上將 $CA$ 順時針轉到 $CB$，會將弧 $BAC$ 上的所有點跑過一次，但將 $AB$ 逆時針轉到 $AC$ 會跑過 $BAC$ 上除了 $A$ 以外的所有點，這表示兩者的操作數差一次，故原假設不可能成立。\n\n7. 最後證明這個下界是最優的。考慮一個正 $3n$ 邊形，其頂點依順時鐘方向依序標為 $A_1, A_2, \\dots, A_{3n}$，並考慮當 $A_1$ 機器人指向 $A_2$，而其他所有機器人都指向 $A_1$ 的狀況。考慮最後轉出 $a$ 個逆時針的三角形，則可算出需操作 $\\frac{3n(3n-1)}{2} - 2n + a$ 次，因此當 $a = 0$ 時有最小值 $\\frac{3n(3n-1)}{2} - 2n = \\frac{9n^2-7n}{2}$。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15270,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABP$, $BCQ$, $CAR$ be three non-overlapping triangles erected outside of acute triangle $ABC$. Let $M$ be the midpoint of segment $AP$. Given that $\\angle PAB = \\angle CQB = 45^\\circ$, $\\angle ABP = \\angle QBC = 75^\\circ$, $\\angle RAC = 105^\\circ$, and $RQ^2 = 6CM^2$, compute $\\dfrac{AC^2}{AR^2}$.\n\n![](images/pamphlet0910_main_p26_data_4efea42578.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Because $\\angle BAP = \\angle BQC = 45^\\circ$ and $\\angle PBA = \\angle CQB = 75^\\circ$, triangles $BQC$ and $BAP$ are similar to each other, from which it follows that triangles $BCP$ and $BQA$ are similar to each other. Hence, the law of sines gives\n\n$$\n\\frac{CP}{AQ} = \\frac{BP}{BA} = \\frac{\\sin 45^\\circ}{\\sin 60^\\circ} = \\sqrt{\\frac{2}{3}}\n$$\n\nExtend segment $CM$ through $M$ to $S$ with $CM = MS$. Then $CS = 2CM = 2RQ/\\sqrt{6}$, from which it follows that\n\n$$\n\\frac{CP}{AQ} = \\frac{CS}{QR}.\n$$\n\nBecause triangles $BPC$ and $BAQ$ are similar to each other, we may set $x = \\angle CPB = \\angle QAB$. Because segments $SC$ and $AP$ bisect each other, $ACPS$ is a parallelogram, implying that $\\angle SPA = \\angle CAP = \\angle CAB + \\angle BAP = \\angle CAB + 45^\\circ$. It follows that\n\n$$\n\\angle SPC = \\angle SPA + \\angle APB - \\angle CPB = \\angle CAB + 45^\\circ + 60^\\circ - x = \\angle CAB + 105^\\circ - x.\n$$\n\nOn the other hand, $\\angle RAQ = \\angle RAC + \\angle CAB - \\angle QAB = 105^\\circ + \\angle CAB - x$. Hence we have $\\angle SPC = \\angle RAQ$. Therefore, we have that\n\n$$\n\\angle SPC = \\angle RAQ \\quad \\text{and} \\quad \\frac{CP}{AQ} = \\frac{CS}{QR}\n$$\n\nand that $\\angle SPC = 180^\\circ - \\angle ACP > 180^\\circ - \\angle ACB > 90^\\circ$ is obtuse because triangle $ABC$ is acute. Therefore, we may conclude that triangles $RAQ$ and $SPC$ are similar. This means that\n\n$$\n\\frac{SP}{AR} = \\frac{SC}{QR} = \\frac{2CM}{QR} = \\frac{2}{\\sqrt{6}}.\n$$\n\nIn view of parallelogram $ACPS$, we have that $SP = AC$, so we find that\n\n$$\n\\frac{AC^2}{AR^2} = \\frac{SP^2}{AR^2} = \\frac{2}{3}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15271,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute scalene triangle inscribed in circle $\\Omega$. Circle $\\omega$, centered at $O$, passes through $B$ and $C$ and intersects sides $AB$ and $AC$ at $E$ and $D$, respectively. Point $P$ lies on the major arc $\\widehat{BAC}$ of $\\Omega$. Prove that lines $BD$, $CE$, and $OP$ are concurrent if and only if triangles $PBD$ and $PCE$ have the same incenter.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/pamphlet1112_main_p44_data_7dc2b117fa.png)\n\n**Solution 1.** We first prove the “if” direction by assuming that $P$ lies on $\\Omega$ and triangles $PBD$ and $PCE$ have the same incenter $I$. We consider the diagram shown above. (We can adjust our proof easily for other possible configurations.) Because $PI$ bisects both $\\angle BPD$ and $\\angle EPC$, $\\angle EPB = \\angle DPC$. Because $ABCP$ is cyclic, $\\angle PBE = \\angle PBA = \\angle PCA = \\angle PCD$. Thus, we have $\\angle EPB = \\angle DPC$ and $\\angle PBE = \\angle PCD$, so triangles $PEB$ and $PDC$ are similar. In other words, there is a spiral similarity centered at $P$ sending triangle $PDC$ to triangle $PEB$. Therefore, there is a spiral similarity centered at $P$ sending triangle $PED$ to triangle $PBC$. In particular, we have\n\n$$\n\\angle DEP = \\angle CBP \\quad \\text{and} \\quad \\angle PDE = \\angle DCB = \\angle ACB,\n$$\n\nfrom which it follows that\n\n$$\n\\begin{align*}\n\\angle EPC &= \\angle EPD - \\angle DPC = 180^\\circ - \\angle DEP - \\angle PDE + 180^\\circ - \\angle CDP - \\angle PCD \\\\\n&= (360^\\circ - \\angle PDE - \\angle CDP) - \\angle DEP - \\angle PCD \\\\\n&= \\angle EDC - \\angle CBP - \\angle PCA.\n\\end{align*}\n$$\n\nBecause $BCDE$ and $ABCP$ are cyclic, we obtain\n\n$$\n\\begin{align*}\n\\angle EPC &= \\angle EDC - \\angle CBP - \\angle PCA \\\\\n&= 180^\\circ - \\angle CBE - \\angle CBP - \\angle PBA \\\\\n&= 180^\\circ - \\angle CBE - \\angle CEP = 180^\\circ - 2\\angle CBE.\n\\end{align*}\n$$\n\nBecause $O$ is the circumcenter of acute triangle $BEC$, $\\angle COE = 2\\angle CBE$. Thus, $\\angle EPC = 180^\\circ - 2\\angle CBE = 180^\\circ - \\angle COE$ or $\\angle EPC + \\angle COE = 180^\\circ$, implying that $CPEO$ is cyclic.\n\nIn exactly the same way, we can show that $BPDO$ is cyclic. Now consider the three cyclic quadrilaterals $BCDE$, $BPDO$, $CPEO$. Their pairwise radical axes are lines $BD$, $PO$, $CE$, from which it follows that lines $BD$, $PO$, $CE$ meet at the radical center of the three circumcircles.\n\nNext, we prove the “only if” direction by assuming that $P$ lies on the major arc $\\widehat{BAC}$ with $OP$, $BD$, and $CE$ being concurrent at $X$. Because $\\angle EBC < 90^\\circ$, the circumcenter $O$ (of triangle $BEC$) and $B$ lie on the same side of segment $CE$. Likewise, $O$ and $C$ lie on the same side of segment $BD$. Hence, $O$ lies inside triangle $BXC$. We define point $P_1$ on the ray $OX$ such that $OX \\cdot OP_1 = OB^2$, and we denote by $I$ the intersection of segment $OP_1$ and $\\Omega$. We complete our proof in two steps.\n\n**Step 1:** We first show that triangles $P_1BD$ and $P_1CE$ share the same incenter $I$. Because $OX \\cdot OP_1 = OE^2$ and $\\angle P_1OE = \\angle XOE$, triangles $XOE$ and $EOP_1$ are similar, implying that $\\angle OEP_1 = \\angle EXO$. Because $OI = OX$, $\\angle OEI = \\angle EIO$. Hence, we have\n\n$$\n\\angle IEP_1 = \\angle OEP_1 - \\angle OEI = \\angle EXO - \\angle EIO = \\angle XEI.\n$$\n\nThat is, $EI$ bisects angle $\\angle XEP = \\angle CEP_1$. Likewise, we can show that $CI$ bisects $\\angle P_1CX = \\angle P_1CE$. Hence $I$ is the incenter of triangle $P_1CE$. In exactly the same way, we can show that $I$ is also the incenter of triangle $P_1BD$, and so triangles $P_1BD$ and $P_1CE$ share the same incenter.\n\n**Step 2:** We now show that $P = P_1$. Note that $O$ lies inside triangle $BXC$, so ray $OX$ and major arc $\\widehat{BAC}$ are on the same side of line $BC$. Hence, to establish our claim, it suffices to show that $B$, $A$, $P$, and $C$ are cyclic. In particular, it is enough to show that\n\n$$\n\\angle ABP_1 = \\angle EBP_1 = \\angle DCP_1 = \\angle ACP_1.\n$$\n\nTherefore, it suffices to show that triangles $P_1BE$ and $P_1CD$ are similar. We establish this fact using SSS similarity. Indeed, because triangle $BCDE$ is cyclic, triangles $BEX$ and $CDX$ are similar, implying that $EX/DX = BE/CD$. Applying the angle-bisector theorem to triangles $P_1EX$ and $P_1DX$ yields\n\n$$\n\\frac{P_1E}{P_1D} = \\frac{P_1E}{EX} \\cdot \\frac{EX}{DX} \\cdot \\frac{DX}{DP_1} = \\frac{P_1I}{IX} \\cdot \\frac{EB}{DC} \\cdot \\frac{IX}{XP_1} = \\frac{EB}{DC}.\n$$\n\nLikewise, we can show that\n\n$$\n\\frac{BP_1}{CP_1} = \\frac{EB}{DC}.\n$$\n\nCombining the last two identities gives\n\n$$\n\\frac{P_1E}{P_1D} = \\frac{EB}{DC} = \\frac{BP_1}{CP_1},\n$$\n\nhence triangles $P_1EB$ and $P_1DC$ are similar, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15272,
    "subject": "Mathematics (Olympiad)",
    "question": "For $x, y \\ge 0$, the following equality is satisfied:\n$$|\\sqrt{x} - \\sqrt{y}| = \\sqrt{2xy + \\frac{1}{2}}.$$ \nWhat are all possible values of $\\sqrt{x+y} - \\sqrt{2xy}$?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $\\frac{1}{\\sqrt{2}}$.\n\n**Solution:** For $x, y \\ge 0$,\n$$\n\\begin{aligned}\n|\\sqrt{x} - \\sqrt{y}| &= \\sqrt{2xy + \\frac{1}{2}} \\\\\n\\sqrt{x} + \\sqrt{y} &= \\sqrt{2xy} + \\frac{1}{\\sqrt{2}}.\n\\end{aligned}\n$$\nThus, $\\sqrt{x+y} - \\sqrt{2xy} = \\frac{1}{\\sqrt{2}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15273,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be given. On the arc $\\overarc{BC}$ of the circumcircle of $\\triangle ABC$ that does not contain $A$, points $X$ and $Y$ are chosen such that $\\angle BAX = \\angle CAY$. Let $M$ be the midpoint of the chord $AX$. Prove that $\\overline{BM} + \\overline{CM} > \\overline{AY}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the circumcenter of the circumcircle of $\\triangle ABC$. Then $OM \\perp AX$. Draw a perpendicular from $B$ to $OM$ and let it intersect the circumcircle at $Z$. Since $BZ \\perp OM$, $OM$ is the line of symmetry of $BZ$, so $\\overline{MZ} = \\overline{MB}$. By the triangle inequality,\n\n$$\n\\overline{BM} + \\overline{CM} = \\overline{ZM} + \\overline{CM} > \\overline{CZ}.\n$$\n\nBut $BZ \\parallel AX$, so\n\n$$\n\\overarc{AZ} = \\overarc{BX} = \\overarc{CY}\n$$\n\nwhich gives\n\n$$\n\\overarc{CZ} = \\overarc{AY}.\n$$\n\nTherefore, $\\overline{BM} + \\overline{CM} > \\overline{AY}$.\n\n![](images/Macedonia_2017_p2_data_4fc87b20a9.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15274,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDEF$ be a hexagon with $\\angle BAF = 150^\\circ$, $\\angle ACB = \\angle ADC = 90^\\circ$, and $|AC| = |BC|$. Assume also that triangle $ABC$ is similar to triangle $ADE$, and triangle $BCD$ is similar to triangle $DEF$. Find the ratio of the lengths of the segments $AB$ and $AF$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $|AB| = a$. Since $ABC$ is an isosceles right triangle with the apex at $C$, we have $|AC| = \\frac{a}{\\sqrt{2}} = \\frac{a\\sqrt{2}}{2}$ and $\\angle ABC = \\angle BAC = \\frac{\\pi}{4}$. The triangles $ADE$ and $ABC$ are similar, so $\\angle AED = \\angle ACB = \\frac{\\pi}{2}$ and $|DE| = |AE|$.\n\n![](images/Slovenija_2011_p19_data_fb26b7338e.png)\n\nThe triangles $BCD$ and $DEF$ are also similar, so $\\angle BCD = \\angle DEF$ and $\\frac{|BC|}{|CD|} = \\frac{|DE|}{|EF|}$.\n\nThus,\n\n$$\n\\angle ACD = \\angle BCD - \\frac{\\pi}{2} = \\angle DEF - \\frac{\\pi}{2} = \\angle AEF\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15275,
    "subject": "Mathematics (Olympiad)",
    "question": "For a trapezoid $ABCD$ with $AB \\parallel CD$ and $BC = AB + CD$:\n\n1. Prove that there is a point of a circle with diameter $BC$ on the leg $AD$.\n2. Prove that there is a point of a circle with diameter $AD$ on the leg $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "(i) Let $M$ and $N$ be the midpoints of the legs $BC$ and $AD$, respectively. We show that $N$ lies on the circle with diameter $BC$.\n\nA well-known identity yields:\n\n$$\nMN = \\frac{AB + CD}{2} = \\frac{1}{2} BC.\n$$\n\nThis means that $N$ is at the same distance from the center $M$ of the circle with diameter $BC$ as the radius of that circle. So $N$ lies on that circle.\n\n(ii) Given the condition, we can find a point $E$ on the leg $BC$ such that $|BE| = |AB|$ and $|EC| = |CD|$ (see the figure below).\n\n![](images/65_Czech_and_Slovak_MO_2016_booklet_p9_data_9ddba1de3d.png)\n\nThe triangles $ABE$ and $ECD$ are isosceles and the lines $AB$ and $CD$ are parallel, thus:\n\n$$\n\\begin{align*}\n\\angle AED &= 180^\\circ - \\angle AEB - \\angle CED \\\\\n&= \\frac{1}{2}((180^\\circ - 2\\angle AEB) + (180^\\circ - 2\\angle CED)) \\\\\n&= \\frac{1}{2}(\\angle ABE + \\angle DCE) = 90^\\circ.\n\\end{align*}\n$$\n\nSo $E$ is the desired point on the Thales' circle with diameter $AD$.\n\n_Remark_: If we start from the proof that triangle $AED$ is right-angled, we see that its mutually perpendicular axes of sides $AE$ and $ED$ meet at the center $N$ of its circumscribed circle. This means that triangle $BCN$ is also right-angled, so the circle with diameter $BC$ passes through $N$, the midpoint of $AD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15276,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M \\subseteq \\{1,2,\\dots,2011\\}$ be a subset satisfying the following condition: For any three elements in $M$, there exist two of them $a$ and $b$, such that $a \\mid b$ or $b \\mid a$. Determine, with proof, the maximum value of $|M|$, where $|M|$ denotes the number of elements of $M$.",
    "options": [],
    "answer": "See solution",
    "solution": "One can check that\n\n$$\nM = \\{1, 2, 2^2, 2^3, \\dots, 2^{10}, 3, 3 \\times 2, 3 \\times 2^2, \\dots, 3 \\times 2^9\\}\n$$\n\nsatisfies the condition, and $|M| = 21$.\n\nSuppose that $|M| \\geq 22$, and let $a_1 < a_2 < \\dots < a_k$ be the elements of $M$, where $|M| = k \\geq 22$. We first prove that $a_{n+2} \\geq 2a_n$ for all $n$; otherwise, we have $a_n < a_{n+1} < a_{n+2} < 2a_n$ for some $n < k+2$, then any two of these three integers $a_n, a_{n+1}, a_{n+2}$ don't have any multiples relationship, which contradicts the assumption.\n\nIt follows from the inequality above that:\n\n$$\na_4 \\geq 2a_2 \\geq 4,\n$$\n\n$$\na_6 \\geq 2a_4 \\geq 8,\n$$\n\n$$\na_{22} \\geq 2a_{20} \\geq 2^{11} > 2011,\n$$\n\nwhich is a contradiction!\n\nHence the maximum value of $|M|$ is $21$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15277,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that if $a, b, c \\ge 0$ and $a + b + c = 3$, then\n$$\nabc(a^n + b^n + c^n) \\le 3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For $a = 2$, $b = c = \\frac{1}{2}$ and $n \\ge 3$, the inequality is not satisfied. For $n = 1$, it is equivalent to the AM-GM inequality. It remains to consider $n = 2$.\n\n**First solution.** Set $x = bc$. Then\n$$\nabc(a^2 + b^2 + c^2) = a x (a^2 + (b + c)^2 - 2x).\n$$\nThe function $x(p - 2x)$ is increasing for $x \\le \\frac{p}{4}$. Since\n$$\nbc \\le \\frac{(b + c)^2}{4} \\le \\frac{a^2 + (b + c)^2}{4},\n$$\nit follows that if $b + c = b' + c'$ and $bc \\le b'c'$, then\n$$\nabc(a^2 + b^2 + c^2) \\le ab'c'(a^2 + b'^2 + c'^2).\n$$\nWithout loss of generality, assume $b \\le 1 \\le c$. Set $b' = 1$, $c' = b + c - 1$. Since $b + c = b' + c'$ and $bc - b'c' = (b - 1)(c - 1) \\le 0$, the above implies\n$$\nabc(a^2 + b^2 + c^2) \\le a(2 - a)(a^2 + 1 + (2 - a)^2).\n$$\nSet $d = (a - 1)^2$. Then\n$$\na(2 - a)(a^2 + 1 + (2 - a)^2) = (1 - d)(3 + 2d) = 3 - d - 2d^2 \\le 3.\n$$\nThus, the given inequality (for $n = 2$) follows.\n\n**Second solution.** Let $n = 2$ and set $ab + bc + ac = x$. Then $a^2 + b^2 + c^2 = 9 - 2x$ and $9abc \\le x^2$ (since $(ab + bc + ac)^2 \\ge 3abc(a + b + c)$). We need to prove $(9 - 2x)x^2 \\le 27$, which is equivalent to $(2x + 3)(x - 3)^2 \\ge 0$, which is obvious.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15278,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $M$, $N$, $P$, and $Q$ are given on the sides $AB$, $BC$, $CD$, and $DA$ of a parallelogram $ABCD$ such that $MN + QP = AC$. Prove that $PN + QM = DB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m = \\frac{AM}{AB}$, $n = \\frac{BN}{BC}$, $p = \\frac{DP}{DC}$, and $q = \\frac{AQ}{AD}$. Then:\n\n$MN = AN - AM = AB + BN - AM = (1 - m)AB + nAD$\n\n$QP = pAB + (1 - q)AD$\n\nThe given condition becomes:\n\n$$(1 - m + p)AB + (1 - q + n)AD = AC = AB + AD$$\n\nSince $AB$ and $AD$ are linearly independent, we deduce $m = p$ and $n = q$.\n\nNow,\n\n$PN = (1 - p)AB - (1 - n)AD$\n\n$QM = mAB - qAD$\n\nSo,\n\n$PN + QM = (1 - p + m)AB - (1 - n + q)AD = AB - AD = DB$\n\nas required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15279,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y$ be positive real numbers and $n$ be a positive integer. Prove that if $x^{2n+1} + y^{2n+1} \\ge 2$ then also $x^{n+1} + y^{n+1} \\ge x^n + y^n$.",
    "options": [],
    "answer": "See solution",
    "solution": "For any integer $k$, let $D_{n,k} = \\frac{x^{n+k} + y^{n+k}}{x^{n-k+1} + y^{n-k+1}}$.\n\n**LEMMA.** $D_{n,k+1} D_{n,k}^{-1} \\le D_{n,1}^2$.\n\n*Proof.* By simple calculations we get\n\n$$\nD_{n,k+1} D_{n,k}^{-1} = \\frac{x^{2n+2} + y^{2n+2} + xy(x^{n+k} y^{n-k} + x^{n-k} y^{n+k})}{x^{2n} + y^{2n} + x^{n+k} y^{n-k} + x^{n-k} y^{n+k}}\n$$\n\nand\n\n$$\nD_{n,1}^2 = \\frac{x^{2n+2} + y^{2n+2} + 2(xy)^{n+1}}{x^{2n} + y^{2n} + 2(xy)^n}.\n$$\n\nLetting $A = x^{2n+2} + y^{2n+2}$, $B = x^{2n} + y^{2n}$, $C_1 = x^{n+k} y^{n-k} + x^{n-k} y^{n+k}$ and $C_2 = 2x^n y^n$, and by clearing denominators the thesis becomes\n\n$$\n(C_1 - C_2)(A - xyB) \\ge 0.\n$$\n\nBut here we have $C_1 \\ge C_2$ by the AM-GM inequality, and $A \\ge xyB$ by the rearrangement inequality (or just some trivial factorization). $\\square$\n\nUsing the LEMMA with a telescoping product we obtain\n\n$$\nD_{n,n+1} D_{n,1}^{-1} = \\prod_{k=1}^{n} \\frac{D_{n,k+1}}{D_{n,k}} \\le D_{n,1}^{2n}.\n$$\n\nSubstituting, our hypothesis corresponds to $D_{n,n+1} \\ge 1$, and therefore\n\n$$\nD_{n,1}^{2n+1} = \\left( \\frac{x^{n+1} + y^{n+1}}{x^n + y^n} \\right)^{2n+1} \\ge D_{n,n+1} \\ge 1,\n$$\n\nwhence the thesis.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15280,
    "subject": "Mathematics (Olympiad)",
    "question": "Ignacio tiene una hoja de papel. La puede cortar en 6 pedazos o en 8 pedazos, a su elección. Luego, en cada etapa, puede elegir uno de los pedazos existentes y cortarlo en 6 pedazos o en 8 pedazos.\n\n**a)** ¿Puede Ignacio tener, después de alguna etapa, exactamente 24 pedazos de papel?\n\n**b)** ¿Puede Ignacio tener, después de alguna etapa, exactamente 32 pedazos de papel?\n\nSi la respuesta es no, explica por qué; si es sí, indica cómo debe realizar los cortes.",
    "options": [],
    "answer": "See solution",
    "solution": "Comenzamos con una hoja y en cada paso agregamos 5 o 7 trozos. Luego queremos que:\n\n$$\na)\\quad 1 + 5n + 7m = 24 \\iff 5n + 7m = 23. \\textbf{Notemos que } n \\leq 4 \\textbf{ y } m \\leq 3.\n$$\n\nConsideramos la igualdad módulo 7 y tenemos $5n \\equiv 23 \\equiv 2 \\pmod{7}$, pero ninguno de los posibles valores de $n$ (0, 1, 2, 3, 4) satisface la condición. Por lo tanto, no es posible.\n\n$$\nb)\\quad 1 + 5n + 7m = 32 \\iff 5n + 7m = 31, \\textbf{ luego } n \\leq 6 \\textbf{ y } m \\leq 4.\n$$\n\nComo en el caso anterior, obtenemos $5n \\equiv 31 \\equiv 3 \\pmod{7}$. En este caso, las soluciones son de la forma $n = 7k + 2$, con $k = 0, 1, 2, 3, 4, 5, 6$. Resulta entonces que $k = 0$ y $m = 3$, luego $1 + 5 \\cdot 2 + 7 \\cdot 3 = 32$. Es posible obtener 32 trozos cortando en 6 trozos dos veces y en 8 trozos tres veces.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15281,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $abc_7$ be a three-digit number in base $7$. Find the largest such number that can also be written as $(a+1)(b+1)(c+1)_6$ in base $6$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $abc_7 = (a+1)(b+1)(c+1)_6$.\n\nThis gives $49a + 7b + c = 36(a + 1) + 6(b + 1) + c + 1$. Simplifying, we get $13a + b = 43$. Since $a + 1$ and $b + 1$ are less than $6$, $a$ and $b$ are less than $5$. The only solution is $a = 3$, $b = 4$.\n\nThus, the number is $34c_7$ or $45(c+1)_6$. Since $c + 1 \\leq 5$, for the largest such number, $c = 4$.\n\nHence, the number is $344_7 = 179$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15282,
    "subject": "Mathematics (Olympiad)",
    "question": "Numbers $1, \\dots, 200$ are written on a blackboard in one line. Juku has to write in front of each number a plus or minus sign so that for any positive integer $n \\leq 100$, the number itself and one of its multiples have different signs. Which numbers must he assign a minus sign in order to get the maximal possible value of the expression?",
    "options": [],
    "answer": "See solution",
    "solution": "The numbers $51, \\dots, 100$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15283,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $n$ which are less than 1\\% of the number 2020, and such that $n+1$ is more than 1\\% of the number 2019.",
    "options": [],
    "answer": "See solution",
    "solution": "The conditions can be written as:\n$$\n\\frac{n}{2020} < \\frac{1}{100} < \\frac{n+1}{2019}.\n$$\nThe left inequality implies $n < \\frac{2020}{100} = 20.2$, so $n \\leq 20$ (since $n$ is an integer).\n\nFor $n = 20$, check the right inequality:\n$$\n\\frac{21}{2019} > \\frac{1}{100} \\implies 2100 > 2019,\n$$\nwhich is true.\n\nFor $n \\leq 19$:\n$$\n\\frac{n+1}{2019} \\leq \\frac{20}{2019} < \\frac{20}{2000} = \\frac{1}{100}.\n$$\nSo the only possible solution is $n = 20$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15284,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n = 8^{2022}$. Which of the following is equal to $\\frac{n}{4}$?\n\n(A) $4^{1010}$  \n(B) $2^{2022}$  \n(C) $8^{2018}$  \n(D) $4^{3031}$  \n(E) $4^{3032}$",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $8 = 2^3$, so\n\n$$\n n = 8^{2022} = (2^3)^{2022} = 2^{6066} = (2^2)^{3033} = 4^{3033}.\n$$\n\nThus\n\n$$\n \\frac{n}{4} = \\frac{4^{3033}}{4} = 4^{3032}.\n$$\n\nAll the other choices are less than $4^{3032}$. Choices (A) and (D) have the same base but a lesser exponent. Choice (B) is incorrect because $2^{2022} < 4^{2022} < 4^{3032}$. Choice (C) is incorrect because\n\n$$\n 8^{2018} = (4^{\\frac{3}{2}})^{2018} = 4^{3027} < 4^{3032}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15285,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that\n$$\n\\sin 10^\\circ \\cdot \\cos 20^\\circ \\cdot \\sin 30^\\circ \\cdot \\cos 40^\\circ \\cdot \\sin 50^\\circ \\cdot \\cos 60^\\circ \\cdot \\sin 70^\\circ \\cdot \\cos 80^\\circ = \\frac{1}{256}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "As $\\sin 10^\\circ = \\cos 80^\\circ$, $\\sin 30^\\circ = \\cos 60^\\circ$, $\\sin 50^\\circ = \\cos 40^\\circ$, and $\\sin 70^\\circ = \\cos 20^\\circ$, the desired equality is equivalent to\n$$\n(\\cos 20^\\circ \\cdot \\cos 40^\\circ \\cdot \\cos 60^\\circ \\cdot \\cos 80^\\circ)^2 = \\frac{1}{256}.\n$$\nIt is known that $\\cos 60^\\circ = \\frac{1}{2}$. Concerning the other factors, we obtain\n$$\n\\begin{aligned}\n\\cos 20^\\circ \\cdot \\cos 40^\\circ \\cdot \\cos 80^\\circ &= \\frac{8 \\sin 20^\\circ \\cos 20^\\circ \\cdot \\cos 40^\\circ \\cdot \\cos 80^\\circ}{8 \\sin 20^\\circ} \\\\\n&= \\frac{4 \\sin 40^\\circ \\cos 40^\\circ \\cdot \\cos 80^\\circ}{8 \\sin 20^\\circ} \\\\\n&= \\frac{2 \\sin 80^\\circ \\cos 80^\\circ}{8 \\sin 20^\\circ} \\\\\n&= \\frac{\\sin 160^\\circ}{8 \\sin 20^\\circ} = \\frac{\\sin 20^\\circ}{8 \\sin 20^\\circ} = \\frac{1}{8}.\n\\end{aligned}\n$$\nConsequently,\n$$\n(\\cos 20^\\circ \\cdot \\cos 40^\\circ \\cdot \\cos 60^\\circ \\cdot \\cos 80^\\circ)^2 = \\left(\\frac{1}{2} \\cdot \\frac{1}{8}\\right)^2 = \\left(\\frac{1}{16}\\right)^2 = \\frac{1}{256}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15286,
    "subject": "Mathematics (Olympiad)",
    "question": "On the side $BC$ of acute triangle $ABC$, choose an arbitrary point $D$. Let $O$ be the circumcenter of $\\Delta ABC$, and let $Z$ be the point on this circle diametrically opposite to $A$. Let $X$ and $Y$ be points on segments $BO$ and $CO$ such that:\n\n$$\n\\angle BXD + \\angle ABC = 180^{\\circ} = \\angle CYD + \\angle ACB.\n$$\n\nProve that the measure of angle $\\angle XZY$ does not depend on the point $D$.",
    "options": [],
    "answer": "See solution",
    "solution": "For any selected point $D$, quadrilateral $XOYZ$ is inscribed. Then $\\angle XZY = 180^{\\circ} - 2\\angle A$, so it does not depend on $D$.\n\n![](images/ukraine_2015_Booklet_p14_data_531cda36f3.png)\n\nFrom the problem statement, circles $\\triangle BXD$ and $\\triangle DYC$ are tangent to lines $AB$ and $AC$. Denote these circles as $w_1$ and $w_2$, with centers $O_1$ and $O_2$. Let $K$ be the second point of intersection of these circles other than $D$. Then\n\n$$\n\\angle BKC = \\angle BKD + \\angle DKC = \\angle B + \\angle C = 180^{\\circ} - \\angle A,\n$$\n\nso $K$ lies on the circumcircle of $\\triangle ABC$. Also,\n\n$$\n\\angle XKY = \\angle XKD + \\angle DKY = \\angle XBD + \\angle DCY = 180^{\\circ} - \\angle XOY,\n$$\n\nso quadrilateral $XOYK$ is inscribed. Because $AZ$ is the diameter of the circumcircle, $AB \\perp BZ$, so $O_1$ lies on line $BZ$. Similarly, $O_2$ lies on line $CZ$. We prove that $O_1$ and $O_2$ are on the circle $OXKY$. Indeed,\n\n$$\n\\angle XO_1K = 2\\angle XBK = 2\\left(90^{\\circ} - \\frac{1}{2}\\angle BOK\\right) = 180^{\\circ} - \\angle XOK,\n$$\n\nand similarly for $O_2$. Now, we prove that point $Z$ is on circle $OXKY$, which completes the solution.\n\nFor this, we show that $\\angle O_1ZC + \\angle O_1KO_2 = 180^{\\circ}$. Indeed,\n\n$$\n\\angle O_1KO_2 = \\angle O_1KD + \\angle DKO_2 = 90^{\\circ} - \\angle DBK + 90^{\\circ} - \\angle DCK = 180^{\\circ} - \\angle A = 90^{\\circ} - \\angle BOC\n$$\n\n$$\n\\begin{align*}\n&= \\angle BZC = 180^{\\circ} - \\angle O_1ZC\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15287,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation\n\n$$\nx + y^2 + (\text{gcd}(x, y))^2 = x y \times \text{gcd}(x, y).\n$$\n\nin the set of natural numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $z = \\text{gcd}(x, y)$. Then the equation becomes $x + y^2 + z^2 = x y z$. There exist natural numbers $a$ and $b$ such that $x = a z$ and $y = b z$. Substituting, we get $a z + b^2 z^2 + z^2 = a b z^3$, i.e., $a + b^2 z + z = a b z^2$. Since the right-hand side is divisible by $z$ and two terms on the left are divisible by $z$, $a$ must be divisible by $z$. Thus, $a = c z$ for some natural $c$. Substituting, $c z + b^2 z + z = c b z^3$, or $c + b^2 + 1 = c b z^2$. Rearranging, $b^2 + 1 = c (b z^2 - 1)$. Since $b z^2 \\neq 1$ (otherwise $b^2 + 1 = 0$), $c = \\frac{b^2 + 1}{b z^2 - 1}$.\n\nMultiplying by $z^2$, $c z^2 = \\frac{b^2 z^2 + z^2}{b z^2 - 1} = b + \\frac{b + z^2}{b z^2 - 1}$. Since $c z^2$ is natural, $\\frac{b + z^2}{b z^2 - 1}$ is also natural, so $b z^2 - 1 \\leq b + z^2$, i.e.,\n\n$$\n(z^2 - 1)(b - 1) \\leq 2 \\qquad (1)\n$$\n\nIf $b = 1$, $c = \\frac{2}{z^2 - 1}$, so $z^2 = 2$ or $z^2 = 3$, which is impossible. If $b = 2$, $c = \\frac{5}{2 z^2 - 1}$. If $2 z^2 - 1 = 1$, $z = 1$, so $c = 5$, $a = 5$, $x = 5$, $y = 2$. If $2 z^2 - 1 = 5$, $z^2 = 3$, impossible. If $b = 3$, $c = \\frac{10}{3 z^2 - 1}$. The cases $3 z^2 - 1 = 1$, $3 z^2 - 1 = 5$, $3 z^2 - 1 = 10$ are impossible. If $3 z^2 - 1 = 2$, $z = 1$, so $c = 5$, $a = 5$, $x = 5$, $y = 3$. If $b > 3$, from (1) $z = 1$. Then $c = \\frac{b^2 + 1}{b - 1} = b + 1 + \\frac{2}{b - 1}$, so $b = 2$ or $b = 3$, contradicting $b > 3$. Therefore, the solutions are $(x, y) = (5, 2)$ and $(x, y) = (5, 3)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15288,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\alpha$ be a fixed real number.\n\nDetermine all functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying\n\n$$\nf(f(x + y)f(x - y)) = x^2 + \\alpha y f(y)\n$$\n\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that for $\\alpha = -1$ the unique solution is $f(x) = x$, and for other values of $\\alpha$ there is no solution.\n\nIndeed, setting $x = y = 0$ yields $f(f(0)^2) = 0$. Furthermore, $x = 0$ and $y = f(0)^2$ imply $f(0) = 0$. Setting $y = x$, we get $f(0) = x^2 + \\alpha x f(x)$. Now $\\alpha = 0$ immediately leads to a contradiction, so from now on we assume $\\alpha \\neq 0$. Division by $x \\neq 0$ results in $f(x) = -x/\\alpha$ for $x \\neq 0$. Because $f(0) = 0$, this expression for $f(x)$ is valid for $x = 0$ as well. Replacing $f$ with this expression in the original equation gives $$(x^2 - y^2)/(-\\alpha)^3 = x^2 - y^2$$ for all $x, y$, which is equivalent to $-\\alpha^3 = 1$, that is $\\alpha = -1$, and the proof is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15289,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A(a, 1/a)$, $B(b, 1/b)$, $C(c, 1/c)$, $D(d, 1/d)$, $E(e, 1/e)$, $F(f, 1/f)$ be points on the hyperbola, as shown in the figure below.\n\nThe equations of the straight lines $AB$, $DC$, and $EF$ are:\n\n$$\ny = -\\frac{1}{ab}x + \\frac{1}{a} + \\frac{1}{b}, \\quad y = -\\frac{1}{cd}x + \\frac{1}{c} + \\frac{1}{d}, \\quad y = -\\frac{1}{ef}x + \\frac{1}{e} + \\frac{1}{f}.\n$$\n\nSuppose these lines are parallel. Prove that $GE = HF$, where $G$ and $H$ are points such that $GH \\parallel BC \\parallel AD$ and $E$ and $F$ lie on $GH$.\n\n![](images/Belaurus_2016_Booklet_p24_data_045ff23968.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since the lines $AB$, $DC$, and $EF$ are parallel, their slopes are equal, so $ab = cd = ef$.\n\nLet $M$, $N$, and $K$ be the midpoints of $AB$, $EF$, and $DC$, respectively:\n\n- $M\\left(\\frac{a+b}{2}, \\frac{1}{2}\\left(\\frac{1}{a} + \\frac{1}{b}\\right)\\right)$\n- $N\\left(\\frac{e+f}{2}, \\frac{1}{2}\\left(\\frac{1}{e} + \\frac{1}{f}\\right)\\right)$\n- $K\\left(\\frac{c+d}{2}, \\frac{1}{2}\\left(\\frac{1}{c} + \\frac{1}{d}\\right)\\right)$\n\nThe lines $MO$, $ON$, and $OK$ (where $O$ is the origin) all have the form $y = \\frac{1}{ab}x$ (since $ab = cd = ef$), so $O$, $M$, $N$, and $K$ are collinear on the line $y = \\beta x$ with $\\beta = 1/(ab) = 1/(cd) = 1/(ef)$.\n\nIn any trapezoid, the segment connecting the midpoints of its bases passes through the midpoint of any segment parallel to the bases. Considering trapezoid $ABCD$ and segment $GH$ with $GH \\parallel BC \\parallel AD$ and $E$, $F$ on $GH$, the line $MK$ passes through the midpoints of $AB$, $DC$, and $GH$. Since $MK$ also passes through the midpoint of $EF$, the midpoints of $GH$ and $EF$ coincide. Thus, $GE = HF$ as required.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15290,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b$ be non-negative real numbers. Prove the inequality\n\n$$\n\\frac{a}{\\sqrt{b^2+1}} + \\frac{b}{\\sqrt{a^2+1}} \\geq \\frac{a+b}{\\sqrt{ab+1}}\n$$\n\nand find when equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "It is evident that the inequality becomes an equality when $a = 0$, $b = 0$, or $a = b$.\n\nTo prove that otherwise the strict inequality holds, it suffices to consider $a > b > 0$ and, after clearing denominators, show that\n\n$$\na \\sqrt{a^2+1} + b \\sqrt{b^2+1} > (a+b) \\sqrt{ab+1}.\n$$\n\nExpanding and regrouping terms gives\n\n$$\na \\sqrt{a^2+1} (\\sqrt{ab+1} - \\sqrt{b^2+1}) > b \\sqrt{b^2+1} (\\sqrt{a^2+1} - \\sqrt{ab+1}).\n$$\n\nMultiplying the differences of square roots by their sums (as denominators), we get\n\n$$\na \\sqrt{a^2+1} \\cdot \\frac{b(a-b)}{\\sqrt{ab+1} + \\sqrt{b^2+1}} > b \\sqrt{b^2+1} \\cdot \\frac{a(a-b)}{\\sqrt{a^2+1} + \\sqrt{ab+1}}.\n$$\n\nDividing both sides by the positive number $ab(a-b)$ and simplifying, we arrive at\n\n$$\n\\sqrt{a^2+1}(\\sqrt{a^2+1} + \\sqrt{ab+1}) > \\sqrt{b^2+1}(\\sqrt{b^2+1} + \\sqrt{ab+1}).\n$$\n\nThis follows by comparing both sides term by term, since $a > b$ implies $\\sqrt{a^2+1} > \\sqrt{b^2+1}$. Thus, the only cases of equality are $a = 0$, $b = 0$, and $a = b$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15291,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1 = 2$ and, for every positive integer $n$, let $a_{n+1}$ be the smallest integer strictly greater than $a_n$ that has more positive divisors than $a_n$. Prove that $2a_{n+1} = 3a_n$ only for finitely many indices $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Begin with a remark on the terms of the sequence under consideration.\n\n**Lemma 1.** Each $a_n$ is minimal among all positive integers having the same number of positive divisors as $a_n$.\n\n*Proof.* Suppose, if possible, that for some $n$, some positive integer $b < a_n$ has as many positive divisors as $a_n$. Then $a_m < b \\leq a_{m+1}$ for some $m < n$, and the definition of the sequence forces $b = a_{m+1}$. Since $b < a_n$, it follows that $m + 1 < n$, which is a contradiction, as $a_{m+1}$ should have fewer positive divisors than $a_n$. $\\square$\n\nLet $p_1 < p_2 < \\dots < p_n < \\dots$ be the strictly increasing sequence of prime numbers, and write canonical factorizations into primes in the form $N = \\prod_{i \\geq 1} p_i^{e_i}$, where $e_i \\geq 0$ for all $i$, and $e_i = 0$ for all but finitely many indices $i$; in this notation, the number of positive divisors of $N$ is $\\tau(N) = \\prod_{i \\geq 1} (e_i + 1)$.\n\n**Lemma 2.** *The exponents in the canonical factorization of each $a_n$ into primes form a non-strictly decreasing sequence.*\n\n*Proof.* Indeed, if $e_i < e_j$ for some $i < j$ in the canonical decomposition of $a_n$ into primes, then swapping the two exponents yields a smaller integer with the same number of positive divisors, contradicting Lemma 1. $\\square$\n\nWe are now in a position to prove the required result. For convenience, a term $a_n$ satisfying $3a_n = 2a_{n+1}$ will be referred to as a *special* term of the sequence.\n\nSuppose now, if possible, that the sequence has infinitely many special terms, so the latter form a strictly increasing, and hence unbounded, subsequence. To reach a contradiction, it is sufficient to show that:\n\n1. The exponents of the primes in the factorization of special terms have a common upper bound $e$; and\n2. For all large enough primes $p$, no special term is divisible by $p$.\n\nRefer to Lemma 2 to write $a_n = \\prod_{i \\geq 1} p_i^{e_i(n)}$, where $e_i(n) \\geq e_{i+1}(n)$ for all $i$.\n\nStatement (2) is a straightforward consequence of (1) and Lemma 1. Suppose, if possible, that some special term $a_n$ is divisible by a prime $p_i > 2^{e+1}$, where $e$ is the integer provided by (1). Then $e \\geq e_i(n) > 0$, so $2^{e_1(n)e_i(n)+e_i(n)} a_n / p_i^{e_i(n)}$ is a positive integer with the same number of positive divisors as $a_n$, but smaller than $a_n$. This contradicts Lemma 1. Consequently, no special term is divisible by a prime exceeding $2^{e+1}$.\n\nTo prove (1), it is sufficient to show that, as $a_n$ runs through the special terms, the exponents $e_1(n)$ are bounded from above. Then, Lemma 2 shows that such an upper bound $e$ suits all primes.\n\nConsider a large enough special $a_n$. The condition $\\tau(a_n) < \\tau(a_{n+1})$ is then equivalent to $(e_1(n)+1)(e_2(n)+1) < e_1(n)(e_2(n)+2)$. Alternatively, but equivalently, $e_1(n) \\geq e_2(n)+2$. The latter implies that $a_n$ is divisible by $8$, for either $e_1(n) \\geq 3$ or $a_n$ is a large enough power of $2$.\n\nNext, note that $9a_n/8$ is an integer strictly between $a_n$ and $a_{n+1}$, so $\\tau(9a_n/8) \\leq \\tau(a_n)$, which is equivalent to\n\n$$\n(e_1(n) - 2)(e_2(n) + 3) \\leq (e_1(n) + 1)(e_2(n) + 1),\n$$\n\nso $2e_1(n) \\leq 3e_2(n) + 7$. This shows that $a_n$ is divisible by $3$, for otherwise, letting $a_n$ run through the special terms, $3$ would be an upper bound for all but finitely many $e_1(n)$, and the special terms would therefore form a bounded sequence.\n\nThus, $4a_n/3$ is another integer strictly between $a_n$ and $a_{n+1}$. As before, $\\tau(4a_n/3) \\leq \\tau(a_n)$. Alternatively, but equivalently,\n\n$$\n(e_1(n) + 3)e_2(n) \\leq (e_1(n) + 1)(e_2(n) + 1),\n$$\n\nso $2e_2(n) - 1 \\leq e_1(n)$. Combine this with the inequality in the previous paragraph to write $4e_2(n) - 2 \\leq 2e_1(n) \\leq 3e_2(n) + 7$ and infer that $e_2(n) \\leq 9$. Consequently, $2e_1(n) \\leq 3e_2(n) + 7 \\leq 34$, showing that $e = 17$ is suitable for (1) to hold. This establishes (1) and completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15292,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider two non-constant polynomials $P(x)$ and $Q(x)$ with non-negative integer coefficients. The coefficients of $P(x)$ are not larger than $2021$, and $Q(x)$ has at least one coefficient larger than $2021$. Assume that $P(2022) = Q(2022)$ and that $P(x)$ and $Q(x)$ have a common rational root $\\frac{p}{q} \\neq 0$ for $p, q \\in \\mathbb{Z}$, $\\gcd(p, q) = 1$. Prove that\n\n$$\n|p| + n|q| \\leq Q(n) - P(n), \\quad \\forall n = 1, 2, \\dots, 2021.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since the coefficients of $P(x)$ are non-negative, the root $x = \\frac{p}{q}$ must be negative. Without loss of generality, assume $p < 0$, $q > 0$, then $|p| + n|q| = nq - p$. Let $R(x) = Q(x) - P(x)$, then $R(x)$ is an integer polynomial where $x = \\frac{p}{q}$ is a root. This implies that\n\n$$\nR(x) = (qx - p)T(x)\n$$\n\nwhere $T(x)$ is a polynomial with rational coefficients. Since $\\gcd(p, q) = 1$, by Gauss's lemma for the product of two primitive polynomials, $T(x) \\in \\mathbb{Z}[x]$. This implies that $qn - p \\mid R(n)$ for all $n = 1, 2, \\dots, 2021$. To finish the proof, we need to show that $R(n) > 0$ for all $n = 1, 2, \\dots, 2021$. Since $x = 2022$ is a root of $R(x)$, we can also write\n\n$$\nR(x) = (x - 2022)H(x)\n$$\n\nwhere $H(x) = a_m x^m + a_{m-1} x^{m-1} + \\dots + a_1 x + a_0$ is an integer polynomial. Note that $n - 2022 < 0$ for all $n = 1, 2, \\dots, 2021$, so it suffices to show that $H(n) < 0$ for all $n = 1, 2, \\dots, 2021$.\n\nBy expanding the product, the coefficient of $x^i$ is $a_{i-1} - 2022a_i$ for $1 \\leq i \\leq m$ and the constant term is $-2022a_0$. Since the coefficients of $P(x)$ are not larger than $2021$, the coefficients of $R(x)$ are not less than $-2021$. Thus, $a_0 \\leq 0$. Suppose some coefficient of $H(x)$ is positive; let $\\ell > 0$ be the smallest such index. Then $a_\\ell > 0$ and $a_{\\ell-1} \\leq 0$, which implies\n\n$$\na_{\\ell-1} - 2022a_{\\ell} \\leq 0 - 2022 = -2022.\n$$\n\nThis contradiction shows all coefficients of $H(x)$ are non-positive. They cannot all be zero, otherwise $Q(x) \\equiv P(x)$, but $Q(x)$ has at least one coefficient larger than $2021$. In conclusion, $H(n) < 0$ for all $n = 1, 2, \\dots, 2021$. Therefore,\n\n$$\nQ(n) - P(n) = R(n) \\geq qn - p = |q|n + |p|,\n$$\n\nfor $n = 1, 2, \\dots, 2021$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15293,
    "subject": "Mathematics (Olympiad)",
    "question": "Recall that the inequality of arithmetic and geometric means states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list of numbers.\n\nShow that\n\n$$\na + \\sqrt{ab} + \\sqrt[3]{abc} \\leq \\frac{4}{3}(a + b + c)\n$$\n\nfor non-negative real numbers $a$, $b$, and $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\na + \\sqrt{ab} + \\sqrt[3]{abc} = a + \\sqrt{\\frac{a}{2} \\cdot 2b} + \\sqrt[3]{\\frac{a}{4} \\cdot b \\cdot 4c}\n$$\n\nApplying the AM-GM inequality:\n\n- $\\sqrt{\\frac{a}{2} \\cdot 2b} \\leq \\frac{1}{2}\\left(\\frac{a}{2} + 2b\\right)$\n- $\\sqrt[3]{\\frac{a}{4} \\cdot b \\cdot 4c} \\leq \\frac{1}{3}\\left(\\frac{a}{4} + b + 4c\\right)$\n\nSo,\n\n$$\na + \\sqrt{ab} + \\sqrt[3]{abc} \\leq a + \\frac{1}{2}\\left(\\frac{a}{2} + 2b\\right) + \\frac{1}{3}\\left(\\frac{a}{4} + b + 4c\\right) = \\frac{4}{3}(a + b + c)\n$$\n\nEquality holds if and only if $a = 4c = 16c$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15294,
    "subject": "Mathematics (Olympiad)",
    "question": "A nonempty, finite set $A$ of positive integers is called *quadratic* if the sum of the elements of $A$ equals the square of the number of elements of $A$.\n\nFor example, $A = \\{1, 3, 4, 8\\}$ is *quadratic* since $1 + 3 + 4 + 8 = 4^2$.\n\n(a) Give an example of a *quadratic* set with 20 elements.\n\n(b) Prove that every *quadratic* set contains at least one odd integer.\n\n(c) Prove that the intersection of two *quadratic* sets with the same number of elements is nonempty.",
    "options": [],
    "answer": "See solution",
    "solution": "a) An example is $A = \\{1, 2, 3, \\ldots, 19\\} \\cup \\{210\\}$.\n\nThe sum of the elements is $1 + 2 + \\cdots + 19 + 210 = \\frac{19 \\cdot 20}{2} + 210 = 190 + 210 = 400 = 20^2$.\n\nb) We argue by contradiction. Suppose $A$ is a *quadratic* set containing $n$ even positive integers. The sum of the elements of $A$ is at least $2 + 4 + \\cdots + 2n = n(n+1) > n^2$, which is a contradiction.\n\nc) We prove again by contradiction. Assume there exist two disjoint *quadratic* sets, $A$ and $B$, each having $n$ elements. The total sum of the elements from the two sets is $2n^2$. On the other hand, the smallest $2n$ distinct positive integers add up to $1 + 2 + \\cdots + 2n = n(2n+1) > 2n^2$, therefore a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15295,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\n2f(x)f(x+y) - f(x^2) = \\frac{x}{2}(f(2x) + 4f(y))\n$$\n\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = 0$. Then:\n$$\n2f(0)f(y) - f(0) = 0\n$$\nIf $f(0) \\neq 0$, then $f(y) = \\frac{1}{2}$ for all $y$, which does not satisfy the original equation. Thus, $f(0) = 0$.\n\nLet $y = 0$. Then:\n$$\n2f(x)^2 - f(x^2) = \\frac{x}{2} f(2x)\n$$\nNow, plugging back, we get:\n$$\n2f(x)f(x+y) = 2f(x)^2 + 2x f(f(y))\n$$\nSo,\n$$\nf(x)f(x+y) = f(x)^2 + x f(f(y)) \\qquad (\\diamond)\n$$\nNote that $f(-x)^2 = x f(f(x))$. In $(\\diamond)$, take $y = -2x$:\n$$\nf(x)f(-x) = f(x)^2 + x f(f(-2x))\n$$\nAlso, $-2x f(f(-2x)) = f(2x)^2$, so $x f(f(-2x)) = -\\frac{1}{2} f(2x)^2$. Thus,\n$$\nf(x)f(-x) = f(x)^2 - \\frac{1}{2} f(2x)^2\n$$\nor\n$$\n-2f(x)f(-x) + 2f(x)^2 = f(2x)^2\n$$\nIn $(\\diamond)$, take $x = y \\neq 0$:\n$$\nf(x)f(2x) = f(x)^2 + x f(f(x))\n$$\nor\n$$\nf(2x) = f(x) + \\frac{x f(f(x))}{f(x)} = f(x) + \\frac{f(-x)^2}{f(x)}\n$$\nSo $(f(x) + \\frac{f(-x)^2}{f(x)})^2 = 2f(x)^2 - 2f(x)f(-x)$. Dividing both sides by $f(-x)^2$ and let $k = \\frac{f(x)}{f(-x)}$:\n$$\n(k + \\frac{1}{k})^2 = 2k^2 - 2k \\iff k^2 - 2k = 2 + \\frac{1}{k^2}\n$$\nChanging the sign $x \\to -x$ gives $(k + \\frac{1}{k})^2 = \\frac{2}{k^2} - \\frac{2}{k}$; combining with above, we get $k = -1$. Thus, $f(x)$ is odd.\n\nNow, $f(x)^2 = x f(f(x))$. Put $y \\to -y$ in $(\\diamond)$:\n$$\nf(x)f(x-y) = f(x)^2 + x f(f(-y))\n$$\nTake the sum of these two equations:\n$$\nf(x)[f(x+y) + f(x-y)] = 2f(x)^2\n$$\nFor $x \\neq 0$, $f(x+y) + f(x-y) = 2f(x)$ (using $f(0) = 0$), so $f(x)$ is additive.\n\nAlso,\n$$\n2f(x)^2 - f(x^2) = x f(x)\n$$\nSo,\n$$\n2f(x)^2 = f(x^2) + x f(x)\n$$\nPlugging $x \\to x+1$, it is easy to get $f(x) = ax$ and substituting back, $a = 1$. Thus, the solutions are $f(x) = 0$ and $f(x) = x$.\n\n$\\boxed{f(x) = 0 \\text{ and } f(x) = x}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15296,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that given any two permutations $\\sigma, \\tau \\in S_n$, there exists some function $f : \\{1, \\dots, n\\} \\to \\{-1, 1\\}$ such that we simultaneously have, for any indices $1 \\le i \\le j \\le n$,\n\n$$\n\\left| \\sum_{k=i}^{j} f(\\sigma(k)) \\right| \\le 2 \\quad \\text{and} \\quad \\left| \\sum_{k=i}^{j} f(\\tau(k)) \\right| \\le 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We will first prove it for the even case, $n = 2m$.\n\nConsider the graph whose set of vertices is $\\{1, \\dots, 2m\\}$, and with red edges between $\\sigma(2k-1)$ and $\\sigma(2k)$, and blue edges between $\\tau(2k-1)$ and $\\tau(2k)$, where $1 \\le k \\le m$ (notice that multiple edges may occur).\n\nClearly, each vertex is incident with one red edge and one blue edge, hence all vertex degrees are equal to 2. Therefore the graph is the union of disjoint cycles. Moreover, each such cycle is of even length, since its edges must be of alternating colours (the graph is thus bipartite).\n\nDefine $f$ alternately taking values $-1$ and $1$ along each of the cycles. Then the largest value any of the moduli can take is 2, when $i$ is even and $j$ is odd, since on consecutive pairs of odd/even indices the sum of the values of $f$ is zero, along either $\\sigma$ or $\\tau$.\n\nWhen $n$ is odd, prolong $\\sigma(n+1) = \\tau(n+1) = n+1$ to achieve the case treated above.\n\n**Remarks:**\n\n- For $n$ even, this result can be extended to circular sums, since $f$ takes values of different sign on the values of both $\\sigma$ and $\\tau$ on consecutive pairs of odd/even indices, and since $n$ is even, this is true circularly.\n- For $n$ odd, this result does not hold for circular sums. For example, take $\\sigma$ arbitrary and then take $\\tau$ such that $\\tau(k) \\equiv 2\\sigma(k) \\pmod{n}$ for all $k$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 15297,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the area $S$ of triangle $PQT$, where:\n- $T$ is the intersection point of the parabola $y = ax^2$ and the hyperbola $y = 1/x$.\n- $P$ is the point on the parabola where the common tangent to both curves touches it.\n- $Q$ is the point on the hyperbola where the common tangent to both curves touches it.\n\n% IMAGE: ![](images/Blr2012_p18_data_dac228c736.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let the intersection point $T$ have coordinates $(x_T, y_T)$, where $ax_T^2 = 1/x_T$. Thus, $x_T = 1/\\sqrt{a} = 1/t$ and $y_T = 1/x_T = t$.\n\nThe tangent to the parabola at $P(x_P, y_P)$: $y - y_P = 2a x_P (x - x_P)$.\nThe tangent to the hyperbola at $Q(x_Q, y_Q)$: $y - y_Q = (-1/x_Q^2)(x - x_Q)$.\n\nSince the tangent is common to both curves:\n$$\n2a x_P = -\\frac{1}{x_Q^2} \\qquad (1)\n$$\n$$\n-2a x_P^2 + y_P = \\frac{1}{x_Q} + y_Q \\qquad (2)\n$$\nGiven $y_P = a x_P^2$ and $y_Q = 1/x_Q$, equation (2) becomes:\n$$\n-a x_P^2 = \\frac{2}{x_Q} \\qquad (2)\n$$\nSolving (1) and (2): $x_Q = -\\frac{1}{2t}$, $x_P = -\\frac{2}{t}$, $y_Q = -2t$, $y_P = 4t$.\n\nThus,\n$$\nT\\left(\\frac{1}{t}, t\\right), \\quad P\\left(-\\frac{2}{t}, 4t\\right), \\quad Q\\left(-\\frac{1}{2t}, -2t\\right)\n$$\n\nLet $M, K, L$ be the midpoints of $PQ, PT, TQ$ respectively:\n$$\nM\\left(-\\frac{5}{4t}, t\\right), \\quad K\\left(-\\frac{1}{2t}, \\frac{5}{2}t\\right), \\quad L\\left(\\frac{1}{4t}, -\\frac{t}{2}\\right)\n$$\n\n$MT$ is parallel to $Ox$ ($y_M = y_T$), $QK$ is parallel to $Oy$ ($y_Q = y_K$), so medians $MT$ and $QK$ are perpendicular.\n\nLet $H$ be the centroid of $PQT$. The area:\n$$\nS(PQT) = 2S(QKT) = [TH \\perp QK] = 2 \\cdot \\frac{1}{2} QK \\cdot TH = \\frac{2}{3} QK \\cdot TM\n$$\nSince $TM \\parallel Ox$, $TM = \\frac{1}{t} - \\left(-\\frac{5}{4t}\\right) = \\frac{9}{4t}$.\nSince $QK \\parallel Oy$, $QK = \\frac{5t}{2} - (-2t) = \\frac{9t}{2}$.\nTherefore,\n$$\nS(PQT) = \\frac{2}{3} \\cdot \\frac{9}{4t} \\cdot \\frac{9t}{2} = \\frac{27}{4}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15298,
    "subject": "Mathematics (Olympiad)",
    "question": "A rectangular grid is divided by two perpendicular straight lines into four smaller rectangles with integral side lengths. It is possible to remove one among these four rectangles in such a way that the remaining figure can be exactly covered by rectangles of size $2 \\times 3$ and $3 \\times 2$. Prove that it is possible to exactly cover one among these four smaller rectangles by rectangles of size $2 \\times 3$ and $3 \\times 2$. (By *exact covering* we mean covering without gaps, overlaps and overflows.)",
    "options": [],
    "answer": "See solution",
    "solution": "Call a figure *coverable* if it can be exactly covered by rectangles of size $2 \\times 3$ and $3 \\times 2$. Let the original rectangle be of size $(a+d) \\times (b+c)$ and let the figure remaining after cutting out the upper left corner of size $a \\times b$ be *coverable* (see below).\n\n![](images/EST_ABooklet_2024_p63_data_c6bec025ff.png)\n\nThis would be impossible if $c=1$ or $d=1$, so we can assume $c \\ge 2$ and $d \\ge 2$. Coverability implies $6 \\mid ac + bd + cd$. Also, a rectangle of integral side lengths $n \\times m$ with $n \\ge 2$ and $m \\ge 2$ is *coverable* if and only if $6 \\mid nm$. Indeed, rectangles of size $2k \\times 3l$ and $3l \\times 2k$ are obviously coverable.\n\nRectangles of size $6k \\times (6l \\pm 1)$ can be divided into rectangles of size $6k \\times 3$ and $6k \\times 2s$, both of which are coverable; the symmetric case is analogous.\n\nSince $2 \\mid ac + bd + cd$, at least one of $a, b, c, d$ must be even. Next, we show that at least one of $a, b, c, d$ must be divisible by 3. Suppose the contrary: none of $a, b, c, d$ is divisible by 3. If $3 \\mid a + d$ or $3 \\mid b + c$, then $3 \\mid bd$ or $3 \\mid ac$, respectively, because $3 \\mid ac + bd + cd$. Since 3 is prime, this is a contradiction. Hence $3 \\nmid a + d$ and $3 \\nmid b + c$. Thus, either $a \\equiv d \\equiv 1 \\pmod{3}$ or $a \\equiv d \\equiv 2 \\pmod{3}$, and similarly for $b$ and $c$.\n\nIf $a \\equiv b \\equiv c \\equiv d \\equiv 1 \\pmod{3}$ or $a \\equiv b \\equiv c \\equiv d \\equiv 2 \\pmod{3}$, color the squares by rising diagonals with 3 colors. If $a \\equiv d \\equiv 1 \\pmod{3}$ and $b \\equiv c \\equiv 2 \\pmod{3}$ or vice versa, color by falling diagonals. The area to be covered, except for the neighborhood of the intersection point of the cutting lines (marked in the figures below), can be cut into strips of size $3k \\times 1$ and $1 \\times 3l$ containing equal numbers of unit squares of each color. Thus, the area to be coverable contains different numbers of unit squares of different colors, which is impossible since every $2 \\times 3$ and $3 \\times 2$ rectangle contains the same number of unit squares of each color. Thus, at least one of $a, b, c, d$ is divisible by 3.\n\n![](images/EST_ABooklet_2024_p64_data_8a6973e0c1.png)\n![](images/EST_ABooklet_2024_p64_data_efd9effcbe.png)\n![](images/EST_ABooklet_2024_p64_data_eb0e4d2976.png)\n![](images/EST_ABooklet_2024_p64_data_d3c9633958.png)\n\nIt remains to show that one of the small rectangles is coverable whenever one of $a, b, c, d$ is divisible by 2 and also one of $a, b, c, d$ is divisible by 3. The following table shows for each case which of the small rectangles is coverable:\n\n$$\n\\begin{array}{|c|c|c|c|c|}\n\\hline\n & 3 \\mid a & 3 \\mid b & 3 \\mid c & 3 \\mid d \\\\\n\\hline\n2 \\mid a & a \\times c & a \\times b & a \\times c & a \\times c \\\\\n2 \\mid b & a \\times b & d \\times b & d \\times b & d \\times b \\\\\n2 \\mid c & a \\times c & d \\times b & d \\times c & d \\times c \\\\\n2 \\mid d & a \\times c & d \\times b & d \\times c & d \\times c \\\\\n\\hline\n\\end{array}\n$$\n\nOne can show this by the following case study:\n\n- If one of $a$ and $d$ is divisible by one of 2 and 3 and one of $b$ and $c$ is divisible by the other, then the corresponding small rectangle is coverable (fills all squares outside the two long diagonals of the table).\n- If $6 \\mid a$ or $6 \\mid d$, then $a \\times c$ or $d \\times c$ is coverable, respectively, since $c \\ge 2$ and the other side length is long enough. Similarly, if $6 \\mid b$ or $6 \\mid c$, then $d \\times b$ or $d \\times c$ is coverable, respectively (fills all squares of the main diagonal).\n- If $a$ is divisible by one of 2 and 3 and $d$ by the other, then $6 \\mid ac$ and $a \\times c$ is coverable. Similarly, if $b$ is divisible by one of 2 and 3 and $c$ by the other, then $d \\times b$ is coverable (fills all squares of the secondary diagonal).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15299,
    "subject": "Mathematics (Olympiad)",
    "question": "For real numbers $x_1, x_2, \\dots, x_n$, suppose that $|x_i^2|$ is even for $k$ numbers and $|x_i^2|$ is odd for the remaining $l = n-k$ numbers. What is the maximum possible value of\n$$\nS = \\sum_{1 \\leq i < j \\leq n} f(x_i, x_j)\n$$\nwhere $f(x_i, x_j)$ is defined such that the contribution of each pair of even numbers and each pair of odd numbers to $S$ is at most $2$, and the contribution of each pair consisting of one even and one odd number is at most $3$? Find the maximum $S$ for $n = 31$.",
    "options": [],
    "answer": "See solution",
    "solution": "The contribution of each pair out of $\\binom{k}{2}$ pairs of even numbers and $\\binom{l}{2}$ pairs of odd numbers to $S$ will be at most $2$. The contribution of each pair out of $kl$ pairs of one even and one odd number to $S$ will be at most $3$. Therefore,\n\n$$\nS \\leq 3kl + k^2 - k + l^2 - l = (k+l)^2 + kl - k - l = n^2 - n + kl \\leq n^2 - n + \\left\\lfloor \\frac{n^2}{4} \\right\\rfloor\n$$\n\nand we are done.\n\nWhen $n = 31$ we get\n$$\n31^2 - 31 + \\left\\lfloor \\frac{31^2}{4} \\right\\rfloor = 961 - 31 + 240 = 1170.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15300,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R}^+ \\to \\mathbb{R}^+$ such that\n$$\nf(x + f(x) + f(y)) = 2f(x) + y\n$$\nholds for all $x, y \\in \\mathbb{R}^+$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that $f(x) = x$ for every $x \\in \\mathbb{R}^+$. It is easy to check that this function satisfies the equation.\n\nLet $P(x, y)$ denote the assertion that $f(x + f(x) + f(y)) = 2f(x) + y$.\n\nFirst, we show that $f$ is injective. Suppose $f(a) = f(b)$. Then $P(1, a)$ and $P(1, b)$ give:\n$$\n2f(1) + a = f(1 + f(1) + f(a)) = f(1 + f(1) + f(b)) = 2f(1) + b\n$$\nso $a = b$.\n\nLet $A = \\{x \\in \\mathbb{R}^+ : f(x) = x\\}$. It suffices to show $A = \\mathbb{R}^+$.\n\n$P(x, x)$ shows $x + 2f(x) \\in A$ for all $x$. Now $P(x, x + 2f(x))$ gives:\n$$\nf(2x + 3f(x)) = x + 4f(x)\n$$\nfor all $x$. Therefore, $P(x, 2x + 3f(x))$ gives $2x + 5f(x) \\in A$ for all $x$.\n\nSuppose $x, y \\in \\mathbb{R}^+$ with $x, 2x + y \\in A$. Then $P(x, y)$ gives:\n$$\nf(2x + f(y)) = f(x + f(x) + f(y)) = 2f(x) + y = 2x + y = f(2x + y)\n$$\nBy injectivity, $2x + f(y) = 2x + y$, so $f(y) = y$, i.e., $y \\in A$.\n\nSince $x + 2f(x) \\in A$ and $2x + 5f(x) = 2(x + 2f(x)) + f(x) \\in A$, we deduce $f(x) \\in A$ for all $x$. That is, $f(f(x)) = f(x)$ for all $x$.\n\nBy injectivity, $f(x) = x$ for all $x \\in \\mathbb{R}^+$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15301,
    "subject": "Mathematics (Olympiad)",
    "question": "From a point $O$ inside the square $ABCD$, the perpendicular line $OS$ is raised to the plane of the square. Let $M, N, P, Q$ be the projections of point $O$ onto the planes $(SAB)$, $(SBC)$, $(SCD)$, and $(SDA)$, respectively. Prove that the points $M, N, P, Q$ are coplanar if and only if $O$ lies on one of the diagonals of the square.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that $O$ lies, for example, on the diagonal $AC$. Let $OE \\perp AB$, $E \\in AB$ and $OF \\perp AD$, $F \\in AD$. Then $OE = OF$, and $\\triangle SOE \\equiv \\triangle SOF$ (by congruence), so $SE = SF$. Then $M \\in SE$ and $OM \\perp SF$, $Q \\in SF$ and $OQ \\perp SF$, $\\triangle SOM \\equiv \\triangle SOQ$, $SM = SQ$, $\\frac{SM}{SE} = \\frac{SQ}{SF}$, $QM \\parallel EF$, so $QM \\parallel BD$. Similarly, $NP \\parallel BD$, so $QM \\parallel NP$, which means that points $M, N, P, Q$ are coplanar.\n\nConversely, assume that $M, N, P, Q$ are coplanar in a plane $\\alpha$. Take $OG \\perp CD$, $G \\in CD$ and $OH \\perp BC$, $H \\in BC$. Then the points $E, O, G$ are collinear, so the lines $SE, SO, SG$ are coplanar. It follows that the lines $SO$ and $MP$ are coplanar, and $SO \\cap MP$ is the same as $SO \\cap \\alpha$. Similarly, $NQ \\cap SO$ is the same as $SO \\cap \\alpha$, so the lines $MP$ and $NQ$ intersect $SO$ at the same point $R$.\n\n![](images/RMC_2025_p40_data_4793d52743.png)\n\nWe calculate the ratio in which point $R$ divides $OS$, in terms of $OS$, $OE$, and $OG$.\n\nTake $MT \\perp OS$, $PU \\perp OS$, $U, T \\in OS$. From the cyclic quadrilateral $MOPS$ we get $\\triangle MSR \\sim \\triangle OPR$, so $\\frac{MS}{OP} = \\frac{MR}{OR} = \\frac{SR}{PR}$, hence $\\frac{MS^2}{OP^2} = \\frac{SR}{OR} \\cdot \\frac{MR}{PR} = \\frac{SR}{OR} \\cdot \\frac{MT}{PU}$. It follows $\\frac{SR}{OR} = \\frac{PU}{MT} \\cdot \\frac{MS^2}{OP^2} = \\frac{PS \\cdot PO \\cdot MS^2}{MO \\cdot MS \\cdot PS \\cdot PG} = \\frac{PO}{PG} \\cdot \\frac{MS}{MO}$. But $\\frac{PO}{PG} = \\tan \\angle SGO = \\frac{SO}{OG}$ and $\\frac{MS}{MO} = \\cot \\angle MSO = \\frac{SO}{OE}$, so $\\frac{SR}{OR} = \\frac{SO^2}{OE \\cdot OG}$.\n\nSince line $NQ$ intersects $OS$ also in $R$, we obtain $OE \\cdot OG = OF \\cdot OH$. On the other hand, $OE + OG = l = OF + OH$, where $l = AB = AD$. It follows that $OE(l - OE) = OF(l - OF)$, then $(OE - OF)(l - OE - OF) = 0$, so $(OE - OF)(OH - OE) = 0$. Thus $OE = OF$ or $OE = OH$, which implies $O \\in AC$ or $O \\in BD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15302,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for every positive real numbers $x, y, z$ the following inequality holds:\n\n$$\n\\frac{9}{x+y+z} - \\frac{1}{xyz} \\le 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "From the AM-GM inequality, we have:\n\n$$\n\\frac{9}{x+y+z} \\le \\frac{3}{\\sqrt[3]{xyz}}.\n$$\n\nTherefore,\n\n$$\n\\frac{9}{x+y+z} - \\frac{1}{xyz} \\le \\frac{3}{\\sqrt[3]{xyz}} - \\frac{1}{xyz}.\n$$\n\nLet $t = \\frac{1}{\\sqrt[3]{xyz}}$. Then, it suffices to prove that:\n\n$$\n3t - t^3 \\le 2 \\quad \\text{for} \\quad t > 0.\n$$\n\nThis is equivalent to:\n\n$$\nt^3 - 3t + 2 \\ge 0,\n$$\n\nwhich factors as:\n\n$$\n(t - 1)^2(t + 2) \\ge 0.\n$$\n\nSince $t > 0$, this inequality always holds, with equality when $t = 1$, i.e., $xyz = 1$ and $x = y = z = 1$ (equality in AM-GM).",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15303,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\overline{AB} = c$, $\\overline{BC} = a$, and $\\overline{CA} = b$ be the sides of triangle $\\triangle ABC$. Let $D \\in BC$, $E \\in CA$, and $F \\in AB$ be the points where the angle bisectors at $A$, $B$, and $C$ meet the opposite sides, respectively. If $\\overline{DE} = \\overline{DF}$, then prove:\n\n$$\na) \\frac{a}{b+c} = \\frac{b}{c+a} + \\frac{c}{a+b}, \\qquad b) \\angle BAC > 90^{\\circ}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** By the Law of Sines:\n\n$$\n\\frac{\\sin \\angle AFD}{\\sin \\angle FAD} = \\frac{\\overline{AD}}{\\overline{FD}} = \\frac{\\overline{AD}}{\\overline{ED}} = \\frac{\\sin \\angle AED}{\\sin \\angle FAD}\n$$\n\nwhich implies\n\n$$\n\\sin \\angle AFD = \\sin \\angle AED\n$$\n\nSo either $\\angle AFD = \\angle AED$ or $\\angle AFD + \\angle AED = 180^{\\circ}$.\n\nIf $\\angle AFD = \\angle AED$, then $\\angle ADF = \\angle ADE$ and from congruence we have $\\overline{AF} = \\overline{AE}$. Now, because $\\overline{AF} = \\overline{AE}$, we get $\\angle AIF = \\angle AIE$, so $\\angle AFI = \\angle AEI$, thus $\\angle AFC = \\angle AEB$, i.e., $\\overline{AC} = \\overline{AB}$, which contradicts the condition that $\\triangle ABC$ is scalene.\n\nTherefore, $\\angle AFD + \\angle AED = 180^{\\circ}$, so points $A$, $F$, $D$, and $E$ are concyclic, and we have\n\n$$\n\\angle DEC = \\angle DFA > \\angle ABC\n$$\n\nLet $CA$ be extended through $A$ to a point $P$ such that $\\angle DPC = \\angle B$ (this is possible because $\\angle DEC = \\angle DFA > \\angle ABC$). Then clearly\n\n$$\n\\overline{PC} = \\overline{PE} + \\overline{CE} \\qquad (1)\n$$\n\nBecause $\\angle BFD = \\angle PED$ and $\\overline{FD} = \\overline{ED}$, we get $\\overline{PE} = \\overline{BF} = \\frac{ac}{a+b}$ \\qquad (2)\n\nAlso, $\\triangle PCD \\cong \\triangle BCA$, so\n\n$$\n\\frac{\\overline{PC}}{\\overline{BC}} = \\frac{\\overline{CD}}{\\overline{CA}}\n$$\n\nSo\n\n$$\n\\overline{PC} = \\overline{BC} \\cdot \\frac{\\overline{CD}}{CA} = a \\cdot \\frac{ba}{b+c} \\cdot \\frac{1}{b} = \\frac{a^2}{b+c} \\quad (3)\n$$\n\nClearly,\n\n$$\n\\overline{CE} = \\frac{ab}{c+a} \\quad (4)\n$$\n\nFrom (1), (2), (3), and (4) we get\n\n$$\n\\frac{a^2}{b+c} = \\frac{ac}{a+b} + \\frac{ab}{c+a} \\quad \\text{i.e.} \\quad \\frac{a}{b+c} = \\frac{b}{c+a} + \\frac{c}{a+b}\n$$\n\n**b)** From $\\frac{a}{b+c} = \\frac{b}{c+a} + \\frac{c}{a+b}$ we get\n\n$$\na(a+b)(a+c) = b(b+a)(b+c) + c(c+a)(c+b)\n$$\n\n$$\na^2(a+b+c) = b^2(a+b+c) + c^2(a+b+c) + abc > b^2(a+b+c) + c^2(a+b+c)\n$$\n\nSo $a^2 > b^2 + c^2$, which means $\\angle BAC > 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15304,
    "subject": "Mathematics (Olympiad)",
    "question": "A quadrilateral $ABCD$ satisfies $AB = 5$, $BC = 7$, $CD = 6$. The diagonals $AC$ and $BD$ are perpendicular to each other. Find the length of $DA$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be the intersection point of $AC$ and $BD$. By the Pythagorean theorem:\n\n- $AB^2 = AP^2 + BP^2$\n- $BC^2 = BP^2 + CP^2$\n- $CD^2 = CP^2 + DP^2$\n- $DA^2 = DP^2 + AP^2$\n\nAdding the first and third equations and subtracting the second:\n\n$$DA^2 = AB^2 + CD^2 - BC^2 = 5^2 + 6^2 - 7^2 = 25 + 36 - 49 = 12$$\n\nSo $DA = 2\\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15305,
    "subject": "Mathematics (Olympiad)",
    "question": "If $p > 5.6^{n-2}$, then for any set $S$ of $n \\geq 3$ natural numbers, there exist integers $a$ and $b$, where $a$ is coprime with $p$, such that all elements of $aS + b$ are at most $\\frac{p}{3}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $0 \\in S$, since we can choose an arbitrary integer $b_0$ such that $0 \\in S + b_0$.\n\nFor each $i \\in \\mathbb{Z}$, let $S_i = \\left[ \\frac{pi}{6}, \\frac{p(i+1)}{6} \\right) \\cap \\mathbb{Z}$.\n\nConsider $S$ as an $(n-1)$-tuple $(x_1, x_2, \\dots, x_{n-1})$, where $x_i \\in S$ and $x_i \\ne 0$. Each integer $a$ corresponds to an $(n-2)$-tuple $(a_1, a_2, \\dots, a_{n-2})$, where $a_i$ is the index $k$ such that $[ax_i] \\in S_k$. By the pigeonhole principle, there exists a set $A$ with 6 integers $a$, corresponding to the same $(n-2)$-tuple. By the same argument, there exist $a_1, a_2 \\in A$ such that\n\n$$\n[a_1x_{n-1}] - [a_2x_{n-1}] \\in \\left(-\\frac{p}{6}, \\frac{p}{6}\\right).\n$$\n\nChoose $a = a_1 - a_2$, then $[ax] \\in \\left[0, \\frac{p}{6}\\right) \\cup \\left(\\frac{5p}{6}, p\\right)$ for all $x \\in S$.\n\nIt's easy to verify that if $b = \\lfloor \\frac{p}{6} \\rfloor$ then $aS + b \\subset \\left[0, \\frac{p}{3}\\right]$.\n\nSince $p > 6^{n-1} - 2^n + 1$, then $p \\geq 6^{n-1} - 2^n + 3 > 5.6^{n-2}$, which holds for all integers $n \\geq 3$. Hence, our proof is completed. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15306,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R}^+ \\to \\mathbb{R}^+$ such that\n$$\nf\\left(\\frac{f(x)}{x} + y\\right) = 1 + f(y)\n$$\nfor all positive real numbers $x, y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Firstly, we will prove that $\\frac{f(x)}{x}$ is a constant.\n\nAssume that there exist $a, b \\in (0, +\\infty)$ such that $\\frac{f(a)}{a} \\neq \\frac{f(b)}{b}$.\n\nWithout loss of generality, assume $\\frac{f(a)}{a} < \\frac{f(b)}{b}$. By plugging $x = a$ and $x = b$ into the relation, we get\n$$\nf\\left(y + \\frac{f(a)}{a}\\right) = f\\left(y + \\frac{f(b)}{b}\\right) = f(y) + 1\n$$\nfor all positive real numbers $y$. Denote $K = \\frac{f(b)}{b} - \\frac{f(a)}{a}$, we obtain that $f(x) = f(x + K)$ for all $x > \\frac{f(a)}{a}$.\n\nFurthermore,\n$$\n\\begin{aligned}\nf \\left( y + n \\cdot \\frac{f(a)}{a} \\right) &= f \\left( y + (n-1) \\cdot \\frac{f(a)}{a} \\right) + 1 \\\\\n&= \\dots = f(y) + n > n.\n\\end{aligned}\n$$\nThus $f(x) > n$ for all $x > n \\cdot \\frac{f(a)}{a}$. Now, for an arbitrary positive number $x$, choose $n = \\lfloor f(x) \\rfloor + 2 > f(x)$ and a positive integer $m$ such that $x + mK > n \\cdot \\frac{f(a)}{a}$, we have $f(x + mK) > n > f(x) = f(x + mK)$, which is a contradiction.\n\nHence, $\\frac{f(x)}{x}$ is a constant. Denote $\\frac{f(x)}{x} = c$ and replace $f(x) = cx$ in the original equation. One can find that $c = 1$. Thus, $f(x) = x$ for all positive numbers $x$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15307,
    "subject": "Mathematics (Olympiad)",
    "question": "Several small villages are situated on the banks of a straight river. On one side, there are 20 villages in a row, and on the other there are 15 villages in a row. We would like to build bridges, each of which connects a village on one side with a village on the other side. The bridges must be straight, must not cross, and it should be possible to get from any village to any other village using only those bridges (and not any roads that might exist between villages on the same side of the river). How many different ways are there to build the bridges?",
    "options": [],
    "answer": "See solution",
    "solution": "We show that the answer is generally $\\binom{a+b-2}{a-1} = \\frac{(a+b-2)!}{(a-1)!(b-1)!}$ if there are $a$ towns on one side and $b$ on the other. In our particular instance, there are thus $\\binom{33}{14} = 818\\,809\\,200$ ways to build the bridges.\n\nWe prove our general formula by induction on the total number of villages. Note that the formula always holds if either $a=1$ or $b=1$. In this case, $\\binom{a+b-2}{a-1} = 1$, and indeed there is only one possibility: to build bridges from the single village on one side of the river to all other villages. This forms the base of our induction.\n\nWe call the villages on one side $A_1, A_2, \\dots, A_\\alpha$, and the villages on the other side $B_1, B_2, \\dots, B_\\beta$ (in this order). Note first that $A_1$ and $B_1$ cannot both be connected by a bridge to villages other than each other: if there is a bridge between $A_1$ and $B_k$ and a bridge between $B_1$ and $A_l$, where $k, l > 1$, then these bridges cross, which is impossible.\n\nThis leaves us with two possibilities:\n\n* $A_1$ is directly connected to $B_1$ only, while $B_1$ is connected to the rest. In this case, we can ignore $A_1$ and only count the possibilities to build bridges between $A_2, A_3, \\dots, A_\\alpha$ and $B_1, B_2, \\dots, B_\\beta$. By the induction hypothesis, this can be done in $\\binom{a+b-3}{a-2}$ ways.\n* $B_1$ is directly connected to $A_1$ only, while $A_1$ is connected to the rest. In this case, we can ignore $B_1$, and the induction hypothesis shows that there are $\\binom{a+b-3}{a-1}$ possibilities.\n\nAltogether, this gives us\n\n$$\n\\binom{a+b-3}{a-2} + \\binom{a+b-3}{a-1} = \\binom{a+b-2}{a-1}\n$$\n\npossibilities to build the bridges, which completes our induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15308,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for every positive integer $n$, there exists a polynomial $p_n(x)$ such that $p_n(1), p_n(2), \\dots, p_n(n)$ are distinct powers of $2$.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove the assertion by induction on $n$.\n\nLet $p_1(x) = x + 1$ and $p_2(x) = 6 - 2x$, which satisfy the conditions for $n = 1$ and $n = 2$.\n\nSuppose $p_n(x)$ is a polynomial such that $p_n(1), p_n(2), \\dots, p_n(n)$ are distinct powers of $2$. We claim that $\\gcd(p_n(n+1), n!)$ is a power of $2$. Indeed, if there is a prime $2 < q \\le n$ dividing $p_n(n+1)$, then $q$ divides $p_n(n+1-q)$ (a power of $2$), which is a contradiction.\n\nLet $2^m$ be the greatest power of $2$ dividing $n!$. If $a = p_n(n+1)$, then $\\gcd(a, \\frac{n!}{2^m}) = 1$. By Euler's theorem, there exists $k \\in \\mathbb{N}$ such that $a^k - 1$ is divisible by $\\frac{n!}{2^m}$, so $a^k - 1 = \\frac{t \\cdot n!}{2^m}$ for some integer $t$.\n\nDefine\n$$\np_{n+1}(x) = 2^m p_n(x)^k - t(x-1)(x-2)\\dots(x-n).\n$$\nFor $1 \\le i \\le n$, $p_{n+1}(i) = 2^m p_n(i)^k$, which is a power of $2$, and\n$$\np_{n+1}(n+1) = 2^m a^k - t \\cdot n! = 2^m \\left( a^k - \\frac{t \\cdot n!}{2^m} \\right) = 2^m.\n$$\nSince the $p_n(i)$, $1 \\le i \\le n$ are distinct powers of $2$ and $k > 0$, the $p_{n+1}(i)$, $1 \\le i \\le n+1$ are all distinct powers of $2$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15309,
    "subject": "Mathematics (Olympiad)",
    "question": "Say that a subarray of the $n \\times n$ square array *bears* a colour if at least two of its cells share that colour.\n\nProve that the number of $1 \\times 3$ and $3 \\times 1$ rectangular subarrays, which is $2n(n-2)$, exceeds the number of such subarrays, each of which bears some colour. (A key ingredient is the following lemma: If a colour is used exactly $p$ times, then the number of $1 \\times 3$ and $3 \\times 1$ rectangular subarrays bearing that colour does not exceed $3(p-1)$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Assume the lemma: if a colour is used exactly $p$ times, then the number of $1 \\times 3$ and $3 \\times 1$ rectangular subarrays bearing that colour does not exceed $3(p-1)$.\n\nLet $N = \\lceil (n+2)^2/3 \\rceil - 1$ and let $n_i$ be the number of cells coloured the $i$th colour, $i = 1, \\dots, N$. The number of $1 \\times 3$ and $3 \\times 1$ rectangular subarrays, each of which bears some colour, is at most\n\n$$\n\\sum_{i=1}^{N} 3(n_i - 1) = 3 \\sum_{i=1}^{N} n_i - 3N = 3n^2 - 3N < 3n^2 - (n^2 + 4n) = 2n(n-2)\n$$\n\nand thus the claim follows.\n\n**Proof of the lemma:**\n\nIf $p=1$, the assertion is clear. For $p>1$, suppose a row contains exactly $q$ cells coloured $C$. The number $r$ of $3 \\times 1$ rectangular subarrays bearing $C$ does not exceed $3q/2 - 1$ (similarly for columns). Consider the incidence of a cell $c$ coloured $C$ and a $3 \\times 1$ rectangular subarray $R$ bearing $C$:\n\n$$\n\\langle c, R \\rangle = \\begin{cases} 1 & \\text{if } c \\subset R, \\\\ 0 & \\text{otherwise.} \\end{cases}\n$$\n\nGiven $R$, $\\sum_c \\langle c, R \\rangle \\ge 2$; given $c$, $\\sum_R \\langle c, R \\rangle \\le 3$, and if $c$ is the leftmost or rightmost cell, $\\le 2$. Thus,\n\n$$\n2r \\le \\sum_{R} \\sum_{c} \\langle c, R \\rangle = \\sum_{(c,R)} \\langle c, R \\rangle = \\sum_{c} \\sum_{R} \\langle c, R \\rangle \\le 2 + 3(q-2) + 2 = 3q - 2,\n$$\n\nso $r \\le (3q-2)/2$.\n\nLet the $p$ cells coloured $C$ lie on $k$ rows and $l$ columns, with $k+l \\ge 3$ for $p>1$. The total number of $3 \\times 1$ rectangular subarrays bearing $C$ does not exceed $3p/2-k$, and for $1 \\times 3$ subarrays, $3p/2-l$. Thus, the total is $(3p/2-k) + (3p/2-l) = 3p - (k+l) \\le 3p-3 = 3(p-1)$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15310,
    "subject": "Mathematics (Olympiad)",
    "question": "Each point on the plane is colored either red, green, or blue. Prove that there exists an isosceles triangle where all vertices are the same color.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider a circle $\\omega$ with center $O$. Without loss of generality, let $O$ be red. If there exist two red points on $\\omega$ not belonging to the same diameter, then these points together with $O$ form a red isosceles triangle.\n\n![](images/THA_National_2016_p4_data_9a8864fc36.png)\n\nOn the other hand, if $\\omega$ contains at most 2 red points lying on a diameter, consider a regular pentagon inscribed in $\\omega$ with either green or blue vertices. By the pigeonhole principle, at least 3 of its vertices must be the same color. These vertices form an isosceles triangle where all vertices are the same color.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15311,
    "subject": "Mathematics (Olympiad)",
    "question": "We write the numbers $1, 2, \\dots, 2014$ in some order and denote them by $a_1, a_2, \\dots, a_{2014}$. We then rewrite them in a new order and denote them by $b_1, b_2, \\dots, b_{2014}$. Find the largest positive integer $k$ such that, regardless of the chosen orders, the number\n\n$$\n(a_1^2 - b_1^2)(a_2^2 - b_2^2)\\dots(a_{2014}^2 - b_{2014}^2)\n$$\nis always divisible by $3^k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let's prove that the answer is $k = 672$.\n\nAmong the numbers $1, 2, \\dots, 2014$, there are exactly $671$ divisible by $3$, $672$ with remainder $1$, and $671$ with remainder $2$ upon division by $3$.\n\nFor each pair $(a_i, b_i)$, $a_i^2 - b_i^2$ is divisible by $3$ if and only if $a_i$ and $b_i$ have the same remainder modulo $3$. At most $2 \\times 671$ pairs can have different remainders, leaving at least $672$ pairs with the same remainder. For each such pair, $3$ divides $a_i^2 - b_i^2$, so $3^{672}$ divides the product.\n\nTo show $3^{673}$ does not always divide, construct an example: For $i = 1, \\dots, 671$, let $(a_i, b_i) = (3i, 3i + 2)$; for $i = 672, \\dots, 1342$, let $(a_i, b_i) = (b_{i-671}, a_{i-671})$. These pairs have different remainders, so $3$ does not divide $a_i^2 - b_i^2$. The remaining $672$ pairs have both numbers with remainder $1$ modulo $3$, e.g., $(1, 4), (4, 7), \\dots, (2013, 1)$. For these, $a_i^2 - b_i^2$ is divisible by $3$ but not by $9$. Thus, the product is divisible by $3^{672}$ but not by $3^{673}$.\n\nTherefore, the largest such $k$ is $672$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15312,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider an acute triangle $ABC$ with area $S$. Let $CD \\perp AB$ ($D \\in AB$), $DM \\perp AC$ ($M \\in AC$), and $EN \\perp BC$ ($N \\in BC$). Denote by $H_1$ and $H_2$ the orthocentres of the triangles $MNC$ and $MND$, respectively. Find the area of the quadrilateral $AH_1BH_2$ in terms of $S$.\n\n![](images/Macedonia_2014_p14_data_4a7d6ca6cc.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$, $P$, $K$, $R$, and $T$ be the midpoints of the segments $CD$, $MN$, $CN$, $CH_1$, and $MH_1$, respectively. From $\\triangle MNC$, we have that $\\overline{PK} = \\frac{1}{2}\\overline{MC}$ and $PK \\parallel MC$.\n\nAnalogously, from $\\triangle MH_1C$, we have that $\\overline{TR} = \\frac{1}{2}\\overline{MC}$ and $TR \\parallel MC$. Consequently, $\\overline{PK} = \\overline{TR}$ and $PK \\parallel TR$. Also, $\\overline{OK} \\parallel \\overline{DN}$ (from $\\triangle CDN$), and since $\\overline{DN} \\perp \\overline{BC}$ and $\\overline{MH_1} \\perp \\overline{BC}$, it follows that $\\overline{TH_1} \\parallel \\overline{OK}$. Since $O$ is the circumcenter of $\\triangle CMN$, $\\overline{OP} \\perp \\overline{MN}$. Thus, $CH_1 \\perp MN$ implies $\\overline{OP} \\parallel CH_1$. We conclude $\\triangle TRH_1 \\cong \\triangle KPO$ (they have parallel sides and $\\overline{TR} = \\overline{PK}$), hence $\\overline{RH_1} = \\overline{PO}$, i.e., $\\overline{CH_1} = 2\\overline{PO}$ and $CH_1 \\parallel PO$.\n\nAnalogously, $\\overline{DH_2} = 2\\overline{PO}$ and $DH_2 \\parallel PO$. From $\\overline{CH_1} = 2\\overline{PO} = \\overline{DH_2}$ and $CH_1 \\parallel PO \\parallel DH_2$, the quadrilateral $CH_1H_2D$ is a parallelogram, thus $\\overline{H_1H_2} = \\overline{CD}$ and $H_1H_2 \\parallel CD$. Therefore, the area of the quadrilateral $AH_1BH_2$ is $$\\frac{\\overline{AB} \\cdot \\overline{H_1H_2}}{2} = \\frac{\\overline{AB} \\cdot \\overline{CD}}{2} = S.$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15313,
    "subject": "Mathematics (Olympiad)",
    "question": "Call $x$ a *periodic number* if the sequence $f^{(0)}(x)$, $f^{(1)}(x)$, $f^{(2)}(x)$, ... takes finitely many values, where $f(x) = x^2 + c$ for some rational $c$.\n\nShow that all rational periodic points of $f$ lie in the interval $[-(|c|+1), |c|+1]$ and have denominator not greater than the denominator of $c$. Conclude that the number of rational periodic points of $f$ is finite.",
    "options": [],
    "answer": "See solution",
    "solution": "If $|x| > |c| + 1$, then\n$$\n|f(x)| = |x^2 + c| \\geq |x|^2 - |c| > |x|,\n$$\nso the sequence $|f^{(n)}(x)|$ grows without bound, and $x$ cannot be periodic. Thus, all periodic numbers satisfy $|x| \\leq |c| + 1$.\n\nLet $c = \\frac{r}{s}$ and $x = \\frac{y}{z}$ in lowest terms. Then\n$$\nf(x) = \\left(\\frac{y}{z}\\right)^2 + \\frac{r}{s} = \\frac{y^2 s + r z^2}{z^2 s}.\n$$\nThe denominator of $f(x)$ is $z^2 s$. If $z > s$, then the denominator increases at each iteration, so $x$ cannot be periodic. Therefore, the denominator of $x$ is at most $s$.\n\nThus, all rational periodic points of $f$ lie in $[-(|c|+1), |c|+1]$ and have denominator at most that of $c$, so there are only finitely many such points.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15314,
    "subject": "Mathematics (Olympiad)",
    "question": "Side $BC$ of parallelogram $ABCD$ is extended beyond point $C$, and point $K$ is placed on the extension so that $\\triangle CDK$ is isosceles with base $CK$. Side $DC$ is extended beyond point $C$, and point $L$ is placed on the extension so that $\\triangle CBL$ is isosceles with base $CL$. Bisectors of angles $\\angle LBC$ and $\\angle CDK$ intersect at point $Q$. Find the radius of the circle circumscribed about triangle $ALK$, if $\\angle BQD = \\alpha$ and $KL = a$.\n\n![](images/Ukrajina_2008_p7_data_606eb3958d.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $\\frac{a}{2\\sin 2\\alpha}$.\n\nSince trapezium $ABKD$ is equilateral, points $A$, $B$, $K$, $D$ lie on a circle. Similarly, points $B$, $L$, $D$, $A$ also lie on a circle. Thus, all five points $B$, $L$, $D$, $A$, $K$ are on the same circle $w$ circumscribed about $\\triangle ALK$.\n\nSince trapeziums $ABKD$ and $BLDA$ share a diagonal, their diagonals have equal length. Therefore, $AL = AK$ and $\\triangle ALK$ is isosceles.\n\nAs $BQ$ is a bisector of the isosceles triangle $\\triangle LBC$, $BQ \\perp LC$ and $LC \\parallel AB$. Therefore, $BQ \\perp AB$ and $\\angle QBA = 90^\\circ$; similarly, $\\angle QDA = 90^\\circ$. This implies that points $Q$, $B$, $D$, $A$ are concyclic, and this circle is $w$.\n\nSince angles $\\angle BQD$ and $\\angle ALK$ subtend chords of equal length ($BD = AK$), $\\angle ALK = \\alpha$ and $\\angle LAK = 180^\\circ - 2\\alpha$. By the law of sines, the radius is\n$$\nR = \\frac{LK}{2\\sin \\angle LAK} = \\frac{a}{2\\sin 2\\alpha}\n$$\nfor $\\triangle ALK$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15315,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all differentiable functions $f : \\mathbb{R} \\to \\mathbb{R}$ for which:\n\ni) $f'(x) = 0$ for all $x \\in \\mathbb{Z}$;\n\nii) if $f'(x) = 0$ for some $x \\in \\mathbb{R}$, then $f(x) = 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Obviously, one solution is the zero function. Suppose now that $f : \\mathbb{R} \\to \\mathbb{R}$ is another solution. Then $f$ is continuous and $f(x) = f'(x) = 0$ for all $x \\in \\mathbb{Z}$.\n\nConsider $x_0 \\in \\mathbb{R} \\setminus \\mathbb{Z}$ such that $f(x_0) \\neq 0$, and let $k = \\lfloor x_0 \\rfloor \\in \\mathbb{Z}$.\n\nAccording to the Weierstrass theorem, $f$ is bounded on $[k, k + 1]$ and one can find $x_1, x_2 \\in [k, k + 1]$ such that\n\n$$\nf(x_1) \\leq f(x) \\leq f(x_2), \\quad \\forall x \\in [k, k + 1].\n$$\n\nIf $f(x_0) > 0$, then $f(x_2) > 0$, so $x_2 \\in (k, k+1)$ is a local maximum for $f$, so $f'(x_2) = 0$ by Fermat's theorem. But then, from (ii), $f(x_2) = 0$, which is a contradiction.\n\nSimilarly, the case $f(x_0) < 0$ leads to a contradiction.\n\nWe conclude that the zero function is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15316,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the vertices of the pyramid be $A, B, C, D$, where $BD = 6$ and all other edges have length $5$. Let $M$ be the midpoint of $BD$ and $N$ be the foot of the perpendicular from $A$ to the base $BCD$. Find the volume of the pyramid.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the vertices of the pyramid be $A, B, C, D$, where $BD = 6$ and all other edges have length $5$. Let also $M$ be the midpoint of $BD$ and $N$ be the foot of perpendicular from $A$ to the base $BCD$. Of course we have $BM = MD = 3$ and $CM = 4$. Note that\n\n$$\nAN^2 = AB^2 - BN^2 = AC^2 - CN^2 = AD^2 - DN^2.\n$$\n\nAs $AB = AC = AD = 5$, we have $BN = CN = DN$ and so $N$ is the circumcentre of $\\triangle BCD$. The circumradius of $\\triangle BCD$ is given by the extended sine formula\n\n$$\nBN = \\frac{CD}{2 \\sin B} = \\frac{5}{2 \\cdot \\frac{4}{5}} = \\frac{25}{8}.\n$$\n\nThe area of $\\triangle BCD$ is $\\frac{1}{2} \\cdot BD \\cdot MC = 12$. The volume of the pyramid is thus\n\n$$\n\\frac{1}{3} \\cdot 12 \\cdot AN = 4 \\cdot \\sqrt{5^2 - \\left(\\frac{25}{8}\\right)^2} = \\frac{5\\sqrt{39}}{2}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15317,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{K}$ be a field with the property that $x^2y = yx^2$ for all $x, y \\in \\mathbb{K}$. Prove that $\\mathbb{K}$ is commutative.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $Z(\\mathbb{K}) = \\{c \\in \\mathbb{K} \\mid cx = xc,\\ \\forall x \\in \\mathbb{K}\\}$ be the center of $\\mathbb{K}$. Then $Z(\\mathbb{K})$ is a subfield of $\\mathbb{K}$.\n\nWe have $2 \\cdot 1 = (a+1)^2 - a^2 - 1 \\in Z(\\mathbb{K})$ for any $a \\in \\mathbb{K}$. If the characteristic $\\operatorname{char}(\\mathbb{K}) \\neq 2$, then $2 \\cdot 1$ is invertible in $Z(\\mathbb{K})$, so\n$$\na = (2 \\cdot 1)^{-1} \\cdot (2a) \\in Z(\\mathbb{K}), \\text{ for any } a \\in \\mathbb{K},\n$$\nwhich implies $\\mathbb{K}$ is commutative.\n\nNow consider $\\operatorname{char}(\\mathbb{K}) = 2$. For $a, b \\in \\mathbb{K}$, we have $a = (a+b)^2 - a^2 - b^2 \\in Z(\\mathbb{K})$. Also, $w = ab$ gives $w \\cdot ab = (ab)^2 + ab^2a = (ab)^2 + a^2b^2 \\in Z(\\mathbb{K})$. If $w \\neq 0$, then $ab = w^{-1}((ab)^2 + a^2b^2) \\in Z(\\mathbb{K})$. Thus, $ab = ba$.\n\nIf $w = 0$, since $1 = -1$, we get $ab = -ba = ba$, so $ab = ba$ for all $a, b \\in \\mathbb{K}$.\n\nTherefore, $\\mathbb{K}$ is commutative.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15318,
    "subject": "Mathematics (Olympiad)",
    "question": "How many ordered pairs of integers $(m, n)$ satisfy $\\sqrt{n^2 - 49} = m$?\n\n(A) 1  \n(B) 2  \n(C) 3  \n(D) 4  \n(E) infinitely many",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that $m \\ge 0$, and if $(m, n)$ is a solution, then so is $(m, -n)$. Assume $n \\ge 0$.\n\nSquaring both sides gives:\n$$\nn^2 - 49 = m^2\n$$\nSo,\n$$\nn^2 - m^2 = (n - m)(n + m) = 49.\n$$\nBecause $n - m$ and $n + m$ are positive integers, the possible factorizations are:\n- $n - m = 1$, $n + m = 49$\n- $n - m = 7$, $n + m = 7$\n\nFor $n - m = 1$, $n + m = 49$:\n$$\n2n = 50 \\implies n = 25,\\quad 2m = 48 \\implies m = 24\n$$\nFor $n - m = 7$, $n + m = 7$:\n$$\n2n = 14 \\implies n = 7,\\quad 2m = 0 \\implies m = 0\n$$\nSince $n$ can also be negative, the corresponding negative values for $n$ also give solutions: $(0, -7)$ and $(24, -25)$.\n\nTherefore, there are 4 solutions in all.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15319,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the fractional part of $\\dfrac{2009!}{2011}$?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $2011$ is a prime number, by Wilson's theorem, $2010! \\equiv -1 \\pmod{2011}$. Thus, $2010! \\equiv 2010 \\pmod{2011}$, so\n\n$$\n2010! = 2011n + 2010\n$$\nfor some integer $n$. Now, since $2010! = 2011 \\cdot n + 2010$, dividing both sides by $2010$ gives\n\n$$\n\\frac{2010!}{2010} = \\frac{2011n + 2010}{2010} = 2011 \\cdot \\frac{n}{2010} + 1\n$$\n\nBut $2009! = \\frac{2010!}{2010}$, so\n\n$$\n\\frac{2009!}{2011} = \\frac{n}{2010} + \\frac{1}{2011}\n$$\n\nSince $\\frac{n}{2010}$ is an integer, the fractional part of $\\frac{2009!}{2011}$ is $\\boxed{\\dfrac{1}{2011}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15320,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with centroid $G$. Let $D$, $E$, and $F$ be the circumcenters of $\\triangle BCG$, $\\triangle CAG$, and $\\triangle ABG$, respectively. The point $X$ is defined as the intersection of the perpendiculars from $E$ to $AB$ and from $F$ to $AC$. Prove that $DX$ bisects the segment $EF$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that the $D$-median coincides with the perpendicular bisector of the segment $BC$. Thus, the solution consists of two parts: proving that $X$ lies on the perpendicular bisector of $BC$, and proving that the midpoint of $EF$ lies on the perpendicular bisector of $BC$.\n\nLet $\\omega_B$ and $\\omega_C$ denote the circumcircles of triangles $ABG$ and $ACG$, respectively. Let $Y$ be the second intersection of the line through $B$ parallel to $AC$ with $\\omega_B$, and $Z$ the second intersection of the line through $C$ parallel to $AB$ with $\\omega_C$.\n\nThe lines $BY$ and $CZ$ intersect at $A'$, the reflection of $A$ across the midpoint of $BC$, which lies on the $A$-median. Using Power of a Point from $A'$ with respect to the circles $\\omega_B$ and $\\omega_C$, we obtain\n\n$$\n|A'B| \\cdot |A'Y| = |A'A| \\cdot |A'G| = |A'C| \\cdot |A'E|,\n$$\n\nimplying from the converse of Power of a Point that the quadrilateral $YBCZ$ is cyclic. The perpendicular bisector of $BY$ is orthogonal to $BY \\parallel AC$ and passes through $F$ and thus $X$ as well. Similarly, the perpendicular bisector of $CZ$ passes through $Z$. Hence, $X$ is the center of circle $(YBCZ)$ and thus lies on the perpendicular bisector of $BC$.\n\nLet $M$ and $N$ denote the midpoints of $BC$ and $EF$, respectively. To prove that $N$ lies on the perpendicular bisector of $BC$, let $V$ and $W$ denote the second intersections of $\\omega_B$ and $\\omega_C$ with the line $BC$, respectively.\n\nFrom Power of a Point from $M$ with respect to $\\omega_B$ and $\\omega_C$, we obtain\n\n$$\n|MV| \\cdot |MB| = |MG| \\cdot |MA| = |WM| \\cdot |CM| \\implies |MV| = |WM|,\n$$\n\nso $M$ is the midpoint of the segment $VW$. Let $E'$, $N'$, and $F'$ denote the projections of $E$, $N$, and $F$ onto $BC$, respectively. Since $N$ is the midpoint of $EF$, $N'$ will be the midpoint of $E'F'$.\n\nMoreover, since $E$ and $F$ are the centers of $\\omega_B$ and $\\omega_C$, $E'$ and $F'$ are the midpoints of $BV$ and $WC$, and hence $M$ is the midpoint of $E'F'$ as well, implying $N' = M$ and that $N$ is on the perpendicular bisector of $BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15321,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\alpha \\ge 1$ be a real number. Hephaestus and Poseidon play a turn-based game on an infinite grid of unit squares. Before the game starts, Poseidon chooses a finite number of cells to be *flooded*. Hephaestus is building a *levee*, which is a subset of unit edges of the grid (called *walls*) forming a connected, non-self-intersecting path or loop.\n\nThe game then begins with Hephaestus moving first. On each of Hephaestus's turns, he adds one or more walls to the levee, as long as the total length of the levee is at most $\\alpha n$ after his $n$th turn. On each of Poseidon's turns, every cell which is adjacent to an already flooded cell and with no wall between them becomes flooded as well.\n\nHephaestus wins if the levee forms a closed loop such that all flooded cells are contained in the interior of the loop — hence stopping the flood and saving the world. For which $\\alpha$ can Hephaestus guarantee victory in a finite number of turns no matter how Poseidon chooses the initial cells to flood?",
    "options": [],
    "answer": "See solution",
    "solution": "We show that if $\\alpha > 2$ then Hephaestus wins, but when $\\alpha = 2$ (and hence $\\alpha \\le 2$) Hephaestus cannot contain even a single-cell flood initially.\n\n**Strategy for** $\\alpha > 2$:\n\nImpose $\\mathbb{Z}^2$ coordinates on the cells. Adding more flooded cells does not make our task easier, so let us assume that initially the cells $(x, y)$ with $|x| + |y| \\le d$ are flooded for some $d \\ge 2$; thus on Hephaestus's $k$th turn, the water is contained in $|x| + |y| \\le d + k - 1$. Our goal is to contain the flood with a large rectangle.\n\nWe pick large integers $N_1$ and $N_2$ such that\n\n$$\n\\alpha N_1 > 2N_1 + (2d + 3) \\\\\n\\alpha(N_1 + N_2) > 2N_2 + (6N_1 + 8d + 4).\n$$\n\nMark the points $X_i$, $Y_i$ as shown in the figure for $1 \\le i \\le 6$. The red figures indicate the distance between the marked points on the rectangle.\n\n![](images/sols-TST-IMO-2020_p5_data_dcd50c4eac.png)\n\nWe follow the following plan:\n\n- **Turn 1:** place wall $X_1Y_1$. This cuts off the flood to the north.\n- **Turns 2 through $N_1 + 1$:** extend the levee to segment $X_2Y_2$. This prevents further flooding to the north.\n- **Turn $N_1 + 2$:** add in broken lines $X_4X_3X_2$ and $Y_4Y_3Y_2$ all at once. This cuts off the flood west and east.\n- **Turns $N_1 + 2$ to $N_1 + N_2 + 1$:** extend the levee along segments $X_4X_5$ and $Y_4Y_5$. This prevents further flooding west and east.\n- **Turn $N_1 + N_2 + 2$:** add in the broken line $X_5X_6Y_6Y_5$ all at once and win.\n\n**Proof for** $\\alpha = 2$:\n\nSuppose Hephaestus contains the flood on his $(n+1)$st turn. We prove that $\\alpha > 2$ by showing that in fact at least $2n + 4$ walls have been constructed.\n\nLet $c_0, c_1, \\dots, c_n$ be a path of cells such that $c_0$ is the initial cell flooded, and in general $c_n$ is flooded on Poseidon's $n$th turn from $c_{n-1}$. The levee now forms a closed loop enclosing all $c_i$.\n\n**Claim** — If $c_i$ and $c_j$ are adjacent then $|i - j| = 1$.\n\n*Proof.* Assume $c_i$ and $c_j$ are adjacent but $|i - j| > 1$. Then the two cells must be separated by a wall. But the levee forms a closed loop, and now $c_i$ and $c_j$ are on opposite sides. $\\square$\n\nThus the $c_i$ actually form a path. We color green any edge of the unit grid (wall or not) which is an edge of exactly one $c_i$ (i.e., the boundary of the polyomino). It is easy to see there are exactly $2n + 4$ green edges.\n\nNow, from the center of each cell $c_i$, shine a laser towards each green edge of $c_i$ (hence a total of $2n + 4$ lasers are emitted). An example below is shown for $n = 6$, with the levee marked in brown.\n\n![](images/sols-TST-IMO-2020_p6_data_54949de251.png)\n\n**Claim** — No wall is hit by more than one laser.\n\n*Proof.* Assume for contradiction that a wall $w$ is hit by lasers from $c_i$ and $c_j$. WLOG that laser is vertical, so $c_i$ and $c_j$ are in the same column (e.g., $(i, j) = (0, 5)$ in figure). We consider two cases on the position of $w$.\n\n- If $w$ is between $c_i$ and $c_j$, then we have found a segment intersecting the levee exactly once. But the endpoints of the segment lie inside the levee. This contradicts the assumption that the levee is a closed loop.\n- Suppose $w$ lies above both $c_i$ and $c_j$ and assume WLOG $i < j$. Then we have found that there is no levee at all between $c_i$ and $c_j$.\nLet $\\rho \\ge 1$ be the distance between the centers of $c_i$ and $c_j$. Then $c_j$ is flooded in a straight line from $c_i$ within $\\rho$ turns, and this is the unique shortest possible path. So this situation can only occur if $j = i + \\rho$ and $c_i, \\dots, c_j$ form a column. But then no vertical lasers from $c_i$ and $c_j$ may point in the same direction, contradiction.\n\nSince neither case is possible, the proof ends here. $\\square$\n\nThis implies the levee has at least $2n + 4$ walls (the number of lasers) on Hephaestus's $(n+1)$st turn. So $\\alpha \\ge \\frac{2n+4}{n+1} > 2$.\n\n**Remark**\n\n- Even though the flood can be stopped when $\\alpha = 2 + \\varepsilon$, it takes a very long time to do that. Starting from a single flooded cell, the strategy outlined requires $\\Theta(1/\\varepsilon^2)$ days. Starting from several flooded cells contained within an area of diameter $D$, it takes $\\Theta(D/\\varepsilon^2)$ days. No strategies are known that require fewer days than that.\n- There is a gaping chasm between $\\alpha \\le 2$ and $\\alpha > 2$. Since $\\alpha \\le 2$ does not suffice even when only one cell is flooded in the beginning, there are in fact no initial configurations at all for which it is sufficient. On the other hand, $\\alpha > 2$ works for all initial configurations.\n- The second half of the solution essentially estimates the perimeter of a polyomino in terms of its diameter (where diameter is measured entirely within the polyomino). It appears that this has not been done before, or at least no reference was found. There are many references where the perimeter of a polyomino is estimated in terms of its area, but nothing concerning the diameter. The argument is a formalisation of the intuition that if $P$ is any shortest path within some weirdly-shaped polyomino, then the boundary of that polyomino must hug $P$ rather closely so that $P$ cannot be shortened.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 15322,
    "subject": "Mathematics (Olympiad)",
    "question": "If a complex number $z = x + yi$ is a Gaussian integer, then $|z|^2 = x^2 + y^2$ is the sum of two squares of integers.\n\nWhat is the largest possible value of $n$ such that there exist $n$ consecutive positive integers, each of which can be written as the sum of squares of two integers?",
    "options": [],
    "answer": "See solution",
    "solution": "The square of an even integer is divisible by 4, while the square of an odd integer gives remainder 1 when divided by 4. Therefore, the sum of squares of two integers can give remainder 0, 1, or 2 when divided by 4.\n\nIf $n > 3$, then among any four consecutive positive integers, one must have remainder 3 modulo 4, which is impossible for a sum of two squares. Thus, $n \\leq 3$.\n\nFor example, $2 + 2i$, $3$, and $3 + i$ are Gaussian integers whose absolute values squared are 8, 9, and 10, respectively. Therefore, the answer is $n = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15323,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\gamma$ be a circle, and let $P$ be a point in its plane, not situated on $\\gamma$. Two variable lines $\\ell$ and $\\ell'$ through $P$ meet $\\gamma$ at $X$ and $Y$, and $X'$ and $Y'$, respectively. Show that the line through the centres of the circles $PXY'$ and $PX'Y$ passes through a fixed point.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the circles $PXY'$ and $PX'Y$ meet again at $Q$. A suitable inversion of pole $P$ sends the circles $PXY'$ and $PX'Y$ onto the lines $XY'$ and $X'Y$, respectively, while leaving $\\gamma$ invariant. The image of $Q$ under this inversion is the point $R$ where the lines $XY'$ and $X'Y$ meet. Upon inversion, the locus of $R$—the polar of $P$ relative to $\\gamma$—transforms into the circle on diameter $OP$, where $O$ is the centre of $\\gamma$. Consequently, the lines $OQ$ and $PQ$ are perpendicular, so the line through the centres of the circles $PXY'$ and $PX'Y$, which is the perpendicular bisector of the segment $PQ$, passes through the midpoint of the segment $OP$. The conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15324,
    "subject": "Mathematics (Olympiad)",
    "question": "The positive integers $a > b > 1$ are such that the equation\n$$\n\\frac{a^x - 1}{a - 1} = \\frac{b^y - 1}{b - 1}\n$$\nhas at least two distinct solutions in positive integers $x > 1$ and $y > 1$. Prove that $a$ and $b$ are co-prime.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that $a$ and $b$ are not co-prime and let the prime $p$ be their common divisor. We denote by $v_p(n)$ the exact power of $p$ dividing $n$. Note that $(n, \\frac{n^\\ell-1}{n-1}) = 1$ for every positive integer $n > 1$ and $p$ does not divide $n-1$ if it divides $n$.\n\nLet $(x_1, y_1)$ and $(x_2, y_2)$ be two distinct solutions of the given equation with $x_1 > x_2$. Then $ba^{x_1} - ab^{y_1} + a - b = a^{x_1} - b^{y_1}$, which easily implies that $v_p(a) = v_p(b)$. Subtracting the equalities $\\frac{a^{x_1}-1}{a-1} = \\frac{b^{y_1}-1}{b-1}$ and $\\frac{a^{x_2}-1}{a-1} = \\frac{b^{y_2}-1}{b-1}$, we obtain $a^{x_2} \\frac{a^{x_1-x_2}-1}{a-1} = b^{y_2} \\frac{b^{y_1-y_2}-1}{b-1}$, whence $x_2 v_p(a) = y_2 v_p(b)$ and therefore $x_2 = y_2$. Now $\\frac{a^{x_2}-1}{a-1} = \\frac{b^{x_2}-1}{b-1}$ obviously implies that $a = b$, which is a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15325,
    "subject": "Mathematics (Olympiad)",
    "question": "$$\nf(x) + f(y) = \\left( f(x + y) + \\frac{1}{x+y} \\right) (1 - xy + f(xy))\n$$\nFind all functions $f : \\mathbb{Q}^+ \\to \\mathbb{Q}$ satisfying the above equation for all $x, y \\in \\mathbb{Q}^+$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Lemma 1.** $f(1) = 0$.\n\n*Proof:* By setting $x = y = 1$ in the functional equation, we get $2f(1) = (f(2) + \\frac{1}{2})f(1)$. If $f(1) \\neq 0$, then $f(2) = \\frac{3}{2}$. Setting $x = y = 2$ gives $3 = 2f(2) = (f(4) + \\frac{1}{4})(f(4) - 3)$, so $f(4) = \\frac{15}{4}$ or $f(4) = -1$. Further substitutions and calculations show that $f(4) = -1$ leads to a contradiction, so $f(4) = \\frac{15}{4}$. Solving the resulting equations, we find $f(1) = 0$.\n\n**Lemma 2.** $f(2) = \\frac{3}{2}$.\n\n*Proof:* With $f(1) = 0$, for all $x \\in \\mathbb{Q}^+$,\n$$\nf(x) = \\left( f(x+1) + \\frac{1}{x+1} \\right) (1 - x + f(x))\n$$\nSetting $x = 2, 3$ and $x = y = 2$ yields a cubic equation for $f(4)$:\n$$\n16t^3 - 32t^2 - 101t - 15 = 0\n$$\nThe only rational root is $t = \\frac{15}{4}$, so $f(3) = \\frac{8}{3}$ and $f(2) = \\frac{3}{2}$.\n\n**Lemma 3.** $f(n) = n - \\frac{1}{n}$ for all positive integers $n$.\n\n*Proof:* $f(1) = 0$ and for $n \\geq 2$, the result follows by induction using the previous relation.\n\nNow, for all $x = \\frac{m}{n}$,\n$$\nf\\left(\\frac{m}{n}\\right) = \\frac{m}{n} - \\frac{n}{m}\n$$\n*Proof by induction on $m$:* For $m = 1$, setting $y = \\frac{1}{x}$ in the original equation gives $f(x) + f(1/x) = 0$, so $f(1/n) = \\frac{1}{n} - n$. Assume the formula holds for $m$. Setting $x = \\frac{m}{n}$, $y = \\frac{1}{n}$ and substituting the induction hypothesis into the equation, we find\n$$\nf\\left(\\frac{m+1}{n}\\right) = \\frac{m+1}{n} - \\frac{n}{m+1}\n$$\nThus, the solution is:\n$$\n\\boxed{f(x) = x - \\frac{1}{x} \\text{ for all } x \\in \\mathbb{Q}^+}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15326,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all triples $ (a, b, c) $ of positive integers which satisfy the following two identities. Distinguish two triples which are obtained by permuting the order of the same set of three numbers.\n\n$$\nab + c = 13, \\quad a + bc = 23.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $ (1, 2, 11) $, $ (1, 11, 2) $, and $ (2, 3, 7) $.\n\nTaking the sum and the difference of the two given equations, we obtain:\n\n$$\n(ab + c) + (a + bc) = (b + 1)(a + c) = 36\n$$\n\n$$\n(a + bc) - (ab + c) = (b - 1)(c - a) = 10\n$$\n\nSince both $a$ and $c$ are positive integers, $b+1$ and $b-1$ must be factors of $36$ and $10$, respectively. From this, we conclude that the only possible values of $b$ are $2$, $3$, and $11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15327,
    "subject": "Mathematics (Olympiad)",
    "question": "A positive integer is called _alternating_ if among any two consecutive digits in its decimal representation, one is even and the other is odd. Find all positive integers $n$ such that $n$ has a multiple which is alternating.",
    "options": [],
    "answer": "See solution",
    "solution": "The answers are those positive integers that are not divisible by $20$. We call an integer $n$ an _alternator_ if it has a multiple which is alternating. Because any multiple of $20$ ends with an even digit followed by $0$, multiples of $20$ are not alternating. Hence multiples of $20$ are not alternators.\n\nWe consider separately the powers of $2$ and the numbers of the form $2 \\cdot 5^n$. In the sequel, $u^k \\mid a$ means that $k$ is the largest integer such that $u^k$ divides $a$, where $u$ and $a$ are positive integers, and $\\overline{a_n\\dots a_1}$ denotes the integer with digits $a_n, a_{n-1},\\dots, a_1$ (from left to right in that order).\n\n**Lemma 1.** Each power of $2$ has a multiple which is alternating and has an even number of digits.\n\n*Proof:* It suffices to construct an infinite sequence $\\{a_n\\}_{n=1}^\\infty$ of decimal digits such that\n\n(a) $a_n \\equiv n + 1 \\pmod 2$;\n\n(b) $2^{2n-1} \\mid \\overline{a_{2n-1} \\dots a_1}$; and\n\n(c) $2^{2n+1} \\mid \\overline{a_{2n} a_{2n-1} \\dots a_1}$.\n\nWe construct this sequence inductively by starting with $a_1 = 2$ and $a_2 = 7$ and adding two digits in each step. Suppose the sequence is constructed up to $a_{2n}$. Set $a_{2n+1} = 4$. Then $a_{2n+1}$ is even satisfying the condition (a).\n\nBecause $2^{2n+2} \\mid 4 \\cdot 10^{2n}$ and $2^{2n+1} \\mid \\overline{a_{2n}a_{2n-1} \\cdot a_1}$ (by the induction hypothesis), we have\n\n$$\n2^{2n+1} \\mid \\overline{a_{2n+1}a_{2n} \\cdot a_1} = 4 \\cdot 10^{2n} + \\overline{a_{2n}a_{2n-1} \\cdot a_1},\n$$\n\nestablishing (b). Denote $\\overline{a_{2n+1}a_{2n} \\dots a_1} = 2^{2n+1} \\cdot A_n$. Then $A_n$ is an odd integer. Now, $a_{2n+2}$ must be odd and such that\n\n$$\n2^{2n+3} \\mid \\overline{a_{2n+2}a_{2n+1} \\dots a_1} = a_{2n+2} \\cdot 10^{2n+1} + \\overline{a_{2n+1}a_{2n} \\dots a_1} = 2^{2n+1}(a_{2n+2}5^{2n+1} + A_n),\n$$\n\nwhich holds whenever $5a_{2n+2}+A_n \\equiv 4 \\pmod 8$. The solutions of the last congruence are odd, since $A_n$ is odd. In addition, because $5$ is relatively prime to $8$, $a_{2n+2}$ can be chosen from $\\{0, 1, \\dots, 7\\}$, which is a complete residue class modulo $8$. Thus our induction is complete. $\\Box$\n\n**Lemma 2.** Each number of the form $2 \\cdot 5^m$, where $m$ is an arbitrary positive integer, has a multiple which is alternating and has an even number of digits.\n\n*Proof:* It suffices to construct an infinite sequence $\\{b_n\\}_{n=1}^\\infty$ of decimal digits such that\n\n(d) $b_n \\equiv n + 1 \\pmod 2$; and",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15328,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k > 1$ be an integer and $p = 6k + 1$ be a prime number. Prove that for each $m = 2^p - 1$\n\n$$\n\\frac{2^m - 1}{127m}\n$$\n\nis an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us show that both $m$ and $127$ divide $2^{m} - 1$.\n\nFirst, note that $m = 2^p - 1$. By Fermat's little theorem, $2^p \\equiv 2 \\pmod{p}$, so $m = 2^p - 1 \\equiv 1 \\pmod{p}$, which means $p \\mid m - 1$. Therefore, $2^p - 1 \\mid 2^{m} - 1$, so $m \\mid 2^{m} - 1$.\n\nOn the other hand, since $p = 6k + 1$, we have $6 \\mid p - 1$. Thus, $2^6 - 1 = 63 \\mid 2^{p-1} - 1$. Also, $7 \\mid 2^p - 2$, so $7 \\mid m - 1$. Now, $127 = 2^7 - 1$, and $127 \\mid 2^{m} - 1$.\n\nIt remains to show that $m$ and $127$ are relatively prime. Since $127$ is prime, it is enough to show that $127 \\nmid m$. Suppose $127 \\mid m = 2^p - 1$. Then $2^p \\equiv 1 \\pmod{127}$. The order of $2$ modulo $127$ is $7$, since $2^7 = 128 \\equiv 1 \\pmod{127}$. Thus, $7 \\mid p$. But $p = 6k + 1$, so $p \\equiv 1 \\pmod{6}$, and $p > 1$. The only $p$ divisible by $7$ and of the form $6k + 1$ is $p = 7$, but $k = 1$ is not allowed since $k > 1$. Therefore, $127 \\nmid m$.\n\nThus, $127m \\mid 2^m - 1$, as required.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15329,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(n) = r^n$ for $r > 1$ and consider the sequence $a_1, a_2, \\dots, a_n$ with $a_n - a_1 = r^b$, where $b = \\frac{1}{2}n(n-1)$. For $n \\ge 5$, consider the $n-2$ equations:\n\n$$\na_n - a_1 = (a_n - a_i) + (a_i - a_1) \\quad \\text{for } i \\in \\{2, \\dots, n-1\\}.\n$$\n\nShow that, under these constraints and using the convexity of $f(n)$, a contradiction arises unless $n \\le 4$.",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze the $n-2$ equations:\n\n$$\na_n - a_1 = (a_n - a_i) + (a_i - a_1) \\quad \\text{for } i \\in \\{2, \\dots, n-1\\}.\n$$\n\nEach right-hand term must be at least $\\frac{1}{2}(a_n - a_1)$. By the lemma, $r^{b-(n-1)} = \\frac{r^b}{r^{n-1}} < \\frac{1}{2}(a_n - a_1)$, so only $n-2$ large elements $r^{b-1}, \\dots, r^{b-(n-2)}$ are possible. The small terms must be $r^{b-(n-2)-\\frac{1}{2}i(i+1)}$ for $1 \\le i \\le n-2$.\n\nConvexity of $f(n)$ implies strict inequalities among the differences of exponents, leading to $\\alpha_{n-2} - \\alpha_1 \\ge \\frac{1}{2}n(n-3)$, but also $\\le \\frac{1}{2}n(n-3)$, so equality must hold everywhere. This forces the exponents into a rigid pattern.\n\nFor $n-2 \\ge 2$, two equations yield:\n\n$$\nr^b = r^{b-(n-2)} + r^{b-(n-2)-1} \\quad \\text{and} \\quad r^b = r^{b-(n-3)} + r^{b-(n-2)-3},\n$$\n\nwhich become\n\n$$\nr^{n-1} = r + 1 \\quad \\text{and} \\quad r^{n+1} = r^4 + 1.\n$$\n\nAlgebra gives $r^4 + 1 = r^3 + r^2$, so $(r-1)(r^3 - r - 1) = 0$. Since $r \\ne 1$, $r^3 = r + 1 = r^{n-1}$, so $n = 4$, contradicting $n \\ge 5$. Thus, no such configuration exists for $n \\ge 5$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15330,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $(n, m)$ such that\n\n$$\nn^2 + n + 1 = (m^2 + n - 3)(m^2 - n + 5).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us consider the equation:\n\n$$\nn^2 + n + 1 = (m^2 + n - 3)(m^2 - n + 5).\n$$\n\nExpanding the right side:\n\n$$\n(m^2 + n - 3)(m^2 - n + 5) = m^4 + m^2 + 8m - 15.\n$$\n\nSo, we have:\n\n$$\nn^2 + n + 1 = m^4 + m^2 + 8m - 15.\n$$\n\nRewriting as a quadratic in $n$:\n\n$$\nn^2 + n - (m^4 + m^2 + 8m - 16) = 0.\n$$\n\nThe discriminant $D$ must be a perfect square:\n\n$$\nD = 1^2 - 4 \\cdot 1 \\cdot (-(m^4 + m^2 + 8m - 16)) = 4m^4 + 4m^2 + 32m - 63.\n$$\n\nFor $D$ to be a perfect square, consider $m = 1$ and $m = 2$:\n\n- If $m = 1$:\n  $$n^2 + n + 6 = 0$$\n  No positive integer solution for $n$.\n\n- If $m = 2$:\n  $$n^2 + n - 20 = 0$$\n  The positive root is $n = 4$.\n\nThus, the unique pair of positive integers is $(n, m) = (4, 2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15331,
    "subject": "Mathematics (Olympiad)",
    "question": "As illustrated in Fig. 2.1, in the acute $\\triangle ABC$, $AB < AC$, $I$ is the incentre, and $\\odot O$ is the circumcentre. Let $M$, $N$ be the midpoints of $\\overarc{BAC}$ and $\\overarc{BC}$, respectively. Let $D$ be a point on $\\overarc{AC}$ such that $AD \\parallel BC$. The inscribed circle of $\\triangle ABC$ against $\\angle BAC$ touches $BC$ at $E$. Let $F$ be a point inside $\\triangle ABC$, satisfying $IF \\parallel BC$ and $\\angle BAF = \\angle CAE$. The line $NF$ meets $\\odot O$ at $R$ other than $N$, the lines $AF$ and $DI$ meet at $K$, and the lines $AR$ and $IF$ meet at $L$.\n\nProve: $NK \\perp ML$.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p206_data_848c5258cc.png)",
    "options": [],
    "answer": "See solution",
    "solution": "First, we need a lemma.\n\n**Lemma** Let $R'$ be the midpoint of $BC$. Then $\\angle AMI = \\angle IR'B$, $IR' \\parallel AE$.\n\n**Proof of lemma** As illustrated in Fig. 2.2, let $I_b$ and $I_c$ be the excentres of $\\triangle ABC$ relative to the vertices $B$ and $C$, respectively. Then $A, M, I_b$, and $I_c$ are collinear. As $\\angle I_bBI_c = \\angle I_bCI_c = 90^\\circ$, the points $B, C, I_b$, and $I_c$ all lie on the circle with diameter $I_bI_c$; since $MB = MC$, $M$ is the centre of this circle, $MI_b = MI_c$. Since $\\triangle I_bI_c \\sim \\triangle ICB$, $M, R'$ are midpoints of $I_bI_c$ and $BC$, respectively, it follows that $\\angle AMI = \\angle IR'B$.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p207_data_3e32b0c56f.png)\n\nLet the incircle $\\odot I$ of $\\triangle ABC$ touch $BC$ at $Z$ and $ZW$ be a diameter. Through $W$, draw a line parallel to $BC$ that crosses $AB$, $AC$ at $B_1$, $C_1$, respectively. Then $B_1C_1 \\parallel BC$. Since $\\triangle A_1B_1C_1$ and $\\triangle ABC$ are homothetic with centre $A$, $W$ and $E$ are correspondent points, it follows that $A, W, E$ are collinear. Finally, from properties of exscribed circles, we infer that $BZ = CE$, $R'$ is the midpoint of $ZE$, as $I$ is the midpoint of $ZW$, $IR' \\parallel AE$. The lemma is verified.\n\nReturn to the original problem. Let $S$ be the intersection of $AI$ and $BC$. Then\n\n$$\n\\begin{align*}\n\\angle AIF &= \\angle ASB = \\angle SAC + \\angle ACS \\\\\n&= \\angle BCN + \\angle ACS \\\\\n&= \\angle ACN = 180^\\circ - \\angle ARN,\n\\end{align*}\n$$\n\nindicating that $A, I, F$, and $R$ lie on a circle, say $\\omega$, as shown in Fig. 2.3.\n\nSuppose the lines $AF$ and $MI$ intersect at $X$. We show that $X$ is on $\\odot O$. By the lemma,\n\n$$\n\\begin{align*}\n\\angle AMI &= \\angle IRB = \\angle AEB, \\\\\n\\text{also } \\angle AMN &= \\angle ACN = \\angle ASB, \\text{ thus} \\\\\n\\angle IMN &= \\angle AMN - \\angle AMI \\\\\n&= \\angle ASB - \\angle AEB \\\\\n&= \\angle IAE.\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15332,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be the sides of a triangle, and let $r$, $R$, and $s$ be the inradius, the circumradius, and the semiperimeter of the triangle, respectively. Prove that\n\n$$\n\\frac{1}{a+b} + \\frac{1}{a+c} + \\frac{1}{b+c} \\leq \\frac{r}{16Rs} + \\frac{s}{16Rr} + \\frac{11}{8s}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The following are well-known identities relating $r$, $R$, and $s$:\n\n$$\na + b + c = 2s, \\quad ab + bc + ca = s^2 + r^2 + 4Rr, \\quad abc = 4Rrs\n$$\n\nThen\n\n$$\n\\frac{1}{a+b} + \\frac{1}{a+c} + \\frac{1}{b+c} \\leq \\frac{r}{16Rs} + \\frac{s}{16Rr} + \\frac{11}{8s}\n$$\n\nis equivalent to\n\n$$\n4 \\left( \\frac{1}{a+b} + \\frac{1}{a+c} + \\frac{1}{b+c} \\right) \\leq \\frac{s^2 + r^2 + 4rR}{4Rrs} + \\frac{9}{2s}\n$$\n\nor\n\n$$\n4 \\left( \\frac{1}{a+b} + \\frac{1}{b+c} + \\frac{1}{a+c} \\right) \\leq \\frac{ab + ac + bc}{abc} + \\frac{9}{a+b+c}\n$$\n\nwhich is\n\n$$\n4 \\left( \\frac{1}{a+b} + \\frac{1}{b+c} + \\frac{1}{a+c} \\right) \\leq \\frac{9}{a+b+c} + \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}.\n$$\n\nThe last expression is Popoviciu's inequality for the function $f(x) = \\frac{1}{x}$.\n\n*Remark.* Popoviciu's inequality states that for a convex function $f$ it holds:\n\n$$\n2 \\left( f\\left(\\frac{x+y}{2}\\right) + f\\left(\\frac{x+z}{2}\\right) + f\\left(\\frac{y+z}{2}\\right) \\right) \\leq f(x) + f(y) + f(z) + 3f\\left(\\frac{x+y+z}{3}\\right).\n$$\n\n*Remark (P.S.C.).* Without reference to Popoviciu's inequality, we can conclude as follows:\n\n$$\n4 \\left( \\frac{1}{a+b} + \\frac{1}{b+c} + \\frac{1}{a+c} \\right) \\leq \\frac{9}{a+b+c} + \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\n$$\n\nwhich is equivalent to\n\n$$\n\\begin{align*}\n& 4 \\left( 3 + \\frac{c}{a+b} + \\frac{a}{b+c} + \\frac{b}{c+a} \\right) \\leq 9 + 3 + \\frac{b+c}{a} + \\frac{c+a}{b} + \\frac{a+b}{c} \\\\\n& \\Leftrightarrow \\frac{4c}{a+b} + \\frac{4a}{b+c} + \\frac{4b}{c+a} \\leq \\frac{b+c}{a} + \\frac{c+a}{b} + \\frac{a+b}{c}\n\\end{align*}\n$$\n\nor\n\n$$\n\\frac{4c}{a+b} + \\frac{4a}{b+c} + \\frac{4b}{c+a} \\leq \\left( \\frac{c}{a} + \\frac{c}{b} \\right) + \\left( \\frac{a}{b} + \\frac{a}{c} \\right) + \\left( \\frac{b}{a} + \\frac{b}{c} \\right)\n$$\n\nwhich holds as each fraction on the left-hand side is less than or equal to the corresponding parenthesis on the right-hand side.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15333,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S = \\{1, 2, \\dots, 1000\\}$. Find the number of bijective functions $f: S \\to S$ such that $f(i)$ is a multiple of $i$ for every $1 \\leq i \\leq 999$.",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze the structure of such bijections:\n\nConsider the sequence $\\{1000, f(1000), f^2(1000), \\dots\\}$, where $f^k(i)$ denotes $f$ applied $k$ times to $i$. Since $S$ is finite, this sequence is eventually periodic, so there exists a minimal $l$ such that $f^l(1000) = 1000$ and $f^k(1000) \\neq 1000$ for $1 \\leq k < l$.\n\nThe elements $f(1000), f^2(1000), \\dots, f^l(1000) = 1000$ are distinct. For $1 \\leq i \\leq l-1$, $f^{i+1}(1000)$ is a multiple of $f^i(1000)$.\n\nLet $d_1 < d_2 < \\dots < d_l = 1000$ be the elements in this cycle, so that $d_{i+1}$ is a multiple of $d_i$ for $1 \\leq i < l$. For all other $n \\in S$, $f(n) = n$.\n\nThus, the problem reduces to counting sequences $d_1 < d_2 < \\dots < d_l = 1000$ with $d_{i+1}$ a multiple of $d_i$ for all $i$.\n\nLet $c_n$ be the number of such sequences ending at $n$. Then $c_1 = 1$, and for $n \\geq 2$:\n$$\nc_n = 1 + \\sum_{m < n,\\ m \\mid n} c_m.\n$$\n\nBy recursion, $c_{1000} = 504$.\n\n**Answer:** $\\boxed{504}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15334,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{Q} \\to \\mathbb{Q}$ such that\n\n$$\nf(x + f(y + f(z))) = y + f(x + z)\n$$\nfor all $x, y, z \\in \\mathbb{Q}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = 0$ and $z = 0$ in the given equation:\n\n$$\nf(0 + f(y + f(0))) = y + f(0 + 0) \\implies f(f(y + f(0))) = y + f(0).\n$$\nLet $a = f(0)$. Then $f(f(x)) = x$ for all $x \\in \\mathbb{Q}$, so $f$ is bijective and $f(a) = 0$.\n\nThe right-hand side of the original equation is symmetric in $x$ and $z$, so\n$$\nf(x + f(y + f(z))) = f(z + f(y + f(x))).\n$$\nBy bijectivity,\n$$\nx + f(y + f(z)) = z + f(y + f(x)).\n$$\nReplace $z$ by $f(z)$:\n$$\nf(y + z) = f(z) + f(y + f(x)) - x.\n$$\nSet $x = a$:\n$$\nf(y + z) = f(z) + f(y + f(a)) - a.\n$$\nBut $f(a) = 0$, so $f(y + f(a)) = f(y)$:\n$$\nf(y + z) = f(z) + f(y) - a.\n$$\nThus, $f(x) - a$ is additive, so $f(x) = bx + a$ for some $b \\in \\mathbb{Q}$.\n\nSubstitute into the original equation:\n\n- If $f(x) = x$, it works.\n- If $f(x) = -x + a$, it also works (with $a \\in \\mathbb{Q}$).\n\n**Answer:**\n$$\nf(x) = x \\quad \\text{or} \\quad f(x) = -x + a, \\quad a \\in \\mathbb{Q}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15335,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute triangle. Variable points $E$ and $F$ are on sides $AC$ and $AB$ respectively such that\n$$\nBC^2 = BA \\cdot BF + CE \\cdot CA.\n$$\nAs $E$ and $F$ vary, prove that the circumcircle of $\\triangle AEF$ passes through a fixed point other than $A$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H$ be the orthocenter of $\\triangle ABC$, and let $K$, $L$, $M$ be the feet of the perpendiculars from $A$, $B$, $C$ to their opposite sides, respectively. Let $D$ be the intersection point of lines $BE$ and $CF$.\n\nFrom the power of a point, we have\n$$\nBA \\cdot BM = BC \\cdot BK \\qquad (1)\n$$\nand\n$$\nCA \\cdot CL = CB \\cdot CK \\qquad (2)\n$$\nAdding (1) and (2),\n$$\nCA \\cdot CL + BA \\cdot BM = BC \\cdot BK + CB \\cdot CK = BC(BK + CK) = BC^2 \\qquad (3)\n$$\nCombining (3) with the problem statement $BC^2 = BA \\cdot BF + CE \\cdot CA$, we have\n$$\nBA \\cdot BF - BA \\cdot BM = CA \\cdot CL - CE \\cdot CA\n$$\n$$\nBA(BF - BM) = CA(CL - CE)\n$$\n$$\nBA \\cdot FM = CA \\cdot LE\n$$\n$$\n\\frac{LE}{FM} = \\frac{AB}{AC} = \\frac{BL}{CM} \\qquad (4)\n$$\nwhere the last equality follows from $\\triangle AMC \\sim \\triangle ALB$.\n\n![](images/short_list_BMO_2017_p10_data_81e944d2b5.png)\n\nSince $\\frac{LE}{FM} = \\frac{BL}{CM}$ and $\\angle FMC = \\angle ELB = 90^\\circ$, we get that $\\triangle FMC \\sim \\triangle ELB$. From this similarity,\n$\\angle AED = \\angle AEB = \\angle LEB = \\angle MFC = 180^\\circ - \\angle AFC = 180^\\circ - \\angle AFD$, meaning points $A, D, E, F$ are concyclic.\n\nSince both pairs $\\{E, F\\}$ and $\\{M, L\\}$ satisfy the problem condition, the fixed point is the second intersection of the circumcircles of $AFDE$ and $AMHL$. Let this point be $X$. We now prove that $X$ is fixed on the circumcircle of $AMHL$ (which would imply $X$ is fixed).\n\nFrom the concyclicity,\n$\\angle XLE = 180^\\circ - \\angle XLA = \\angle XMA = \\angle XMF$ and $\\angle XEL = \\angle XEA = 180^\\circ - \\angle XFA = \\angle XFM$, so $\\triangle XLE \\sim \\triangle XMF$. This similarity gives\n$$\n\\frac{XL}{XM} = \\frac{LE}{MF} \\qquad (5)\n$$\nCombining (4) and (5), $\\frac{XL}{XM} = \\frac{AB}{AC}$, which is fixed. Since points $M, L$, the circumcircle of $AML$, and the ratio $\\frac{XL}{XM}$ are fixed, this implies that point $X$ is fixed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15336,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples of real numbers $x, y, z$ which are greater than $3$ and satisfy the equality:\n\n$$\n\\frac{(x+2)^2}{y+z-2} + \\frac{(y+4)^2}{z+x-4} + \\frac{(z+6)^2}{x+y-6} = 36.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $x, y, z$ are greater than $3$, it follows that $y+z-2$, $z+x-4$, $x+y-6$ are positive. Thus, from the Cauchy-Schwarz inequality we get:\n\n$$\n\\left( \\frac{(x+2)^2}{y+z-2} + \\frac{(y+4)^2}{x+z-4} + \\frac{(z+6)^2}{x+y-6} \\right) \\left( (y+z-2) + (x+z-4) + (x+y-6) \\right) \\geq (x+y+z+12)^2\n$$\n\nwhich implies\n\n$$\n\\frac{(x+2)^2}{y+z-2} + \\frac{(y+4)^2}{x+z-4} + \\frac{(z+6)^2}{x+y-6} \\geq \\frac{1}{2} \\cdot \\frac{(x+y+z+12)^2}{x+y+z-6}.\n$$\n\nFrom the hypothesis, it follows that\n\n$$\n\\frac{(x+y+z+12)^2}{x+y+z-6} \\leq 72.\n$$\n\nEquality holds when\n\n$$\n\\frac{x+2}{y+z-2} = \\frac{y+4}{x+z-4} = \\frac{z+6}{x+y-6} = \\lambda\n$$\nwhich leads to the system:\n\n$$\n\\begin{cases}\n\\lambda(y+z) - x = 2(\\lambda+1) \\\\\n\\lambda(x+z) - y = 4(\\lambda+1) \\\\\n\\lambda(x+y) - z = 6(\\lambda+1)\n\\end{cases}\n$$\n\nLet $\\omega = x + y + z + 12$. Then\n\n$$\n\\frac{\\omega^2}{\\omega-18} \\geq 72 \\implies (\\omega - 36)^2 \\geq 0\n$$\nso equality holds when $\\omega = 36$, i.e., $x + y + z = 24$.\n\nFrom the previous system, substituting $x + y + z = 24$ gives $\\lambda = 1$.\n\nFor $\\lambda = 1$, the system becomes:\n\n$$\n\\begin{cases}\ny + z - x = 4 \\\\\nx + z - y = 8 \\\\\nx + y - z = 12\n\\end{cases}\n$$\n\nSolving, we find $(x, y, z) = (10, 8, 6)$.\n\nTherefore, the unique solution is $(x, y, z) = (10, 8, 6)$, which satisfies $x + y + z = 24$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15337,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Is there a positive integer $a$ such that $((a^2 - 2)^3 + 1)^a - 1$ is a perfect square?\n\nb) Is there a positive integer $a$ such that $((a^2 - 2)^3 + 1)^{a+1} - 1$ is a perfect square?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:**\n\na) No.\n\nIf $a$ is even, then $((a^2 - 2)^3 + 1)^a$ is a perfect square, but $((a^2 - 2)^3 + 1)^a - 1$ cannot be a perfect square since it differs by 1 from a square.\n\nIf $a$ is odd, $a^2 \\equiv 1 \\pmod{4}$, so $a^2 - 2 \\equiv -1 \\pmod{4}$ and $(a^2 - 2)^3 \\equiv -1 \\pmod{4}$. Thus $4 \\mid (a^2 - 2)^3 + 1$ and $4 \\mid ((a^2 - 2)^3 + 1)^a$, so $((a^2 - 2)^3 + 1)^a - 1 \\equiv 3 \\pmod{4}$, which cannot be a perfect square.\n\nb) No.\n\nLet $x = (a^2 - 2)^3 + 1$ and $k = a + 1$. Then $x^k - 1 = (x - 1)(x^{k-1} + x^{k-2} + \\dots + x + 1)$. Note:\n\n$$\n\\gcd(x - 1, x^{k-1} + x^{k-2} + \\dots + x + 1) = \\gcd(x - 1, k).\n$$\n\nSince $k = a + 1 \\mid a^2 - 1 = (a^2 - 2) + 1 \\mid (a^2 - 2)^3 + 1 = x$, we have $\\gcd(x - 1, k) = \\gcd(k - 1, k) = 1$. Thus, if $x^k - 1$ were a perfect square, $x - 1$ would be a perfect square. But $x - 1 = (a^2 - 2)^3$, which is a perfect square only if $a^2 - 2$ is a perfect square, which is impossible. Therefore, the given number is not a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15338,
    "subject": "Mathematics (Olympiad)",
    "question": "Define a _hook_ to be a figure made up of six unit squares as shown in the diagram\n\n![](images/USA_IMO_2004_p77_data_81e7b82903.png)\n\nor any of the figures obtained by applying rotations and reflections to this figure.\n\nDetermine all $m \\times n$ rectangles that can be tiled with hooks so that\n\n- the rectangle must be covered without gaps and without overlaps; and\n- no part of a hook may cover area outside the rectangle.",
    "options": [],
    "answer": "See solution",
    "solution": "The rectangles that can be tiled with hooks are those with sides $\\{3a, 4b\\}$ or $\\{c, 12d\\}$ with $c \\notin \\{1, 2, 5\\}$, where $a, b, c, d \\in \\mathbb{Z}^{+}$, and the order of the dimensions is not important.\n\nWe first show that these rectangles can be tiled. We can form a $3 \\times 4$ rectangle from two hooks, so we can tile any $3a \\times 4b$ rectangle. In particular, we can tile $3 \\times 12d$ and $4 \\times 12d$ rectangles, so by joining these along their long sides, we can tile a $c \\times 12d$ rectangle for any $c \\ge 6$.\n\nNow we show that no other rectangles can be tiled with hooks. First, note that in any tiling of a rectangle by hooks, for any hook $A$ its center square is covered by a unique hook $B$ of the tiling, and the center square of $B$ must be covered by $A$. Hence we can pair up the hooks in a tiling, and each pair of hooks will cover one of the (unrotated) shapes shown in the figure below.\n\n![](images/USA_IMO_2004_p78_data_556395b987.png)\n\nThe upshot is that given any tiling of a rectangle with hooks, it can be uniquely interpreted as a tiling with unrotated shapes of types (a) through (f), which we shall call “chunks”. In particular, the area of the rectangle must be divisible by 12, because each shape has area 12. Also, it is then clear that no rectangle with a side of length 1, 2, or 5 can be tiled by these pieces. It remains to be shown that at least one side of the rectangle must be divisible by 4. Suppose we have a tiling of an $m \\times n$ rectangle where neither $m$ nor $n$ is divisible by 4. Since $12 \\mid mn$, $m$ and $n$ must both be even. We give five proofs that this is impossible.\n\n**First Solution:** We observe that any chunk in the tiling has exactly one of the following two properties:\n\nI: it consists of four adjacent columns, each containing three squares, and it has an even number of squares in each row, i.e. chunks (a), (e), (f);\n\nII: it consists of four adjacent rows, each containing three squares, and it has an even number of squares in each column, i.e. chunks (b), (c), (d).\n\nRefer to a chunk as “type I” or “type II” accordingly. Color the squares in every fourth row of the rectangle, as in the example below for $18 \\times 18$.\n\n![](images/USA_IMO_2004_p79_data_b170eb0c5c.png)\n\nA chunk of type I contains an even number of squares in each row, so it covers an even number of dark squares. A chunk of type II will intersect exactly one dark row, and it will contain three squares in that row, so it covers an odd number of dark squares. Since the rows of the rectangle have even length, the number of colored squares is even, so the number of chunks of type II is even. By a similar argument interchanging rows and columns, the number of chunks of type I is also even, so the total number of chunks is even. But then the total area $mn$ must be divisible by $2 \\times 12 = 24$, so at least one of $m$ and $n$ is divisible by 4, a contradiction.\n\n**Second Solution:** Here is another way to show that $m, n \\equiv 2 \\pmod{4}$ is impossible. Assume on the contrary that this is the case; imagine the rectangle divided into unit squares, with the rows and columns labeled $1, \\ldots, m$ and $1, \\ldots, n$ (from top to bottom and from left to right). Color the square in row $i$ and column $j$ if exactly one of $i$ and $j$ is divisible by 4.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15339,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an even positive integer. There are $n$ real numbers written on the blackboard. In every step, we choose two numbers, erase them, and replace each of them by their product. Show that for any initial $n$-tuple it is possible to obtain $n$ equal numbers on the blackboard after a finite number of steps.\n",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed by induction. The claim is trivial for $n=2$.\n\nSuppose the claim is true for $n=k$ and consider $n=k+2$. Using the induction hypothesis, we first construct an $n$-tuple of the form\n\n$$\n\\underbrace{(a, \\dots, a, b, b)}_{k}\n$$\n\nFor every $i=3, 4, \\dots, k$, perform the operation with the numbers at positions $i$ and $k+1$. These $k-2$ steps change the $n$-tuple into\n\n$$\n(a, a, ab, a^2b, \\dots, a^{k-2}b, a^{k-2}b, b)\n$$\n\nAfter selecting the last two numbers, we get\n\n$$\n(a, a, ab, a^2b, \\dots, a^{k-2}b, a^{k-2}b^2, a^{k-2}b^2)\n$$\n\nNow, for every $i = 1, 2, \\dots, \\frac{1}{2}n$, combine the numbers at positions $i$ and $n+1-i$, obtaining the desired\n\n$$\n(a^{k-1}b^2, a^{k-1}b^2, \\dots, a^{k-1}b^2)\n$$\n\n*Remark.* For odd $n \\ge 3$, the claim is not true. Consider the initial $n$-tuple $(3, 3, \\dots, 3, 2)$. Let $m$ be the number of occurrences of the maximum element in the $n$-tuple. Initially, $m = n-1$. After every step, $m$ remains even, so we cannot reach the state with all elements equal to the maximum.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15340,
    "subject": "Mathematics (Olympiad)",
    "question": "From a point inside the equilateral triangle $ABC$, perpendicular lines are drawn to the sides $AB$, $BC$, and $CA$. The lengths of these perpendiculars are $m$, $n$, and $p$, respectively, and the length of each side of triangle $ABC$ is $a$. Find the ratio of the area of triangle $ABC$ to the area of the triangle formed by the points where the perpendiculars meet the sides.\n\n![](images/Makedonija_2008_p33_data_106fa4abb3.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be a point inside triangle $ABC$. Let $K$, $L$, and $M$ be the points where the perpendiculars from $P$ meet $AB$, $BC$, and $CA$, respectively. The quadrilaterals $AKPM$, $BLPK$, and $CLPM$ are cyclic, so the angles $KPL$, $LPM$, and $MPK$ are all $120^\\circ$ (since $ABC$ is equilateral with $60^\\circ$ angles).\n\nThe area of triangle $KLM$ is the sum of the areas of triangles $KPL$, $LPM$, and $MPK$:\n\n$$\nP_{\\triangle KLM} = \\frac{\\sqrt{3}}{4}(mp + mn + np)\n$$\n\nThe area of triangle $ABC$ is:\n\n$$\nP_{\\triangle ABC} = \\frac{\\sqrt{3}}{4} a^2\n$$\n\nTherefore, the ratio of the areas is:\n\n$$\n\\frac{P_{\\triangle ABC}}{P_{\\triangle KLM}} = \\frac{a^2}{mp + mn + np}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15341,
    "subject": "Mathematics (Olympiad)",
    "question": "Define a _magic square_ as a $3 \\times 3$ table where each cell contains one number from $1$ to $9$ so that all these numbers are used and all row sums and column sums are equal. Prove that any two magic squares can be obtained from each other via the following transformations: interchanging two rows, interchanging two columns, rotating the square, reflecting the square with respect to its diagonal.",
    "options": [],
    "answer": "See solution",
    "solution": "As all the transformations are invertible, it suffices to show that every magic square can be turned into one particular magic square by these transformations.\n\nThe sum of all numbers in a magic square is $45$, so the numbers in each row and each column must sum to $15$. As this is odd, exactly $0$ or $2$ of the three summands must be even. There are $4$ even numbers in use, so $2$ even numbers must be in some two rows and $0$ even number in the remaining one. The same holds for columns.\n\nHence the even numbers $2, 4, 6, 8$ occur in the corners of some rectangle with sides parallel to the edges of the table. By interchanging rows or columns, one can move the even numbers to the corners of the whole table. There are $3$ possibilities to locate these four numbers into the corners, that cannot be obtained from each other by rotations and reflections of the table (see below). The last two of them cannot occur in the magic square because the missing numbers in the first and third column would coincide. Hence only the first possibility remains. Its completion to a magic square is unique.\n\n![](images/estonian-2012-2013_p23_data_6a23f98acb.png)\n\n![](images/estonian-2012-2013_p23_data_6a23f98acb.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15342,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the circumcenter of triangle $ABC$. Let $D$, $E$, and $F$ be the reflections of $O$ over the lines $BC$, $CA$, and $AB$, respectively. Show that the lines $AD$, $BE$, and $CF$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "As $O$ is the circumcenter of $\\triangle ABC$, the segments $OB$ and $OC$ are equal. Since $D$ is the reflection of $O$ across $BC$, segments $OB$ and $DB$ are equal, and segments $OC$ and $DC$ are equal. Thus, $OBDC$ is a parallelogram. Similarly, $OCEA$ is a parallelogram. Since $BD$ and $OC$ are parallel, and $OC$ and $AE$ are parallel, by transitivity, $BD$ and $AE$ are parallel.\n\nAlso, $OB$ and $CD$ are parallel, and $OA$ and $CE$ are parallel, so by Desargues's theorem, $AB$ and $ED$ are parallel. Therefore, $ABDE$ is a parallelogram. Similarly, $AFDC$ is a parallelogram.\n\nLet $P$ be the intersection point of $AD$ and $BE$. Since $ABDE$ is a parallelogram, $P$ is the midpoint of $AD$. As $P$ is the midpoint of $AD$ and $AFDC$ is a parallelogram, line $CF$ passes through $P$. Thus, $P$ is a common point of $AD$, $BE$, and $CF$, so these lines are concurrent. $\\square$\n\n![](images/BW2021_Shortlist_p26_data_0fc6b4fe19.png)\n\nFigure 12",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15343,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\dots, x_n \\in [0, 1]$. Prove that\n$$\n(x_1 + x_2 + \\dots + x_n + 1)^2 \\ge 4(x_1 + x_2 + \\dots + x_n).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $x_i \\in [0, 1]$, we have $x_i \\ge x_i^2$. Therefore,\n$$\n4(x_1 + x_2 + \\dots + x_n) \\ge 4(x_1^2 + x_2^2 + \\dots + x_n^2).\n$$\nLet $S = x_1 + x_2 + \\dots + x_n$. It suffices to show that\n$$\n(S + 1)^2 \\ge 4S.\n$$\nThis is equivalent to $(S - 1)^2 \\ge 0$, which is always true.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15344,
    "subject": "Mathematics (Olympiad)",
    "question": "If $m$ and $n$ are positive integers, can the number $m^4 + 2mn + n^2 - 2021$ be the product of three or more consecutive integers?",
    "options": [],
    "answer": "See solution",
    "solution": "No.\n\nSuppose that $N = m^4 + 2mn + n^2 - 2021$ is the product of three or more consecutive integers. Then $N$ and $m^4 - m^2 = m[(m-1)m(m+1)]$ are divisible by $3$. Thus, $(m+n)^2 \\equiv N - (m^4 - m^2) + 2021 \\equiv 2 \\pmod{3}$, which is a contradiction, since $2$ is not a quadratic residue modulo $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15345,
    "subject": "Mathematics (Olympiad)",
    "question": "A positive integer $k$ has the $t_m$ property if for all positive integers $a$, there exists a positive integer $n$ such that\n\n$$\n1^k + 2^k + 3^k + \\dots + n^k \\equiv a \\pmod{m}.\n$$\n\na) Find all positive integers $k$ which have the $t_{20}$ property.\n\nb) Find the smallest positive integer $k$ which has the $t_{2015}$ property.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $s_k(n) = 1^k + 2^k + \\dots + n^k$. Recall that a positive integer $k$ has property $T(m)$ if\n\n$$\n\\{s_k(n) \\bmod m : n \\in \\mathbb{N}^*\\} = \\{0, 1, \\dots, m-1\\}.\n$$\n\na) Some well-known formulas:\n\n$$\ns_1(n) = \\frac{n(n+1)}{2}, \\quad s_2(n) = \\frac{n(n+1)(2n+1)}{6}, \\quad s_3(n) = \\frac{n^2(n+1)^2}{4},\n$$\n$$\ns_4(n) = \\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}.\n$$\n\nIf $k$ is $T(20)$, then so is $k+4$ since $20 \\mid n^{k+4} - n^k$. Also, if $k$ is $T(20)$, then $k$ is $T(5)$. We check $s_k(n) \\bmod 5$ for $k = 1,2,3$ and find that $s_k(n)$ does not form a complete residue system modulo 5 for these $k$. Thus, $k \\neq 1,2,3$.\n\nFor $k=4$, we analyze $s_4(n)$ modulo 4 and 5. For modulo 4, $(2n)^4 \\equiv 0 \\pmod{4}$ and $(2n+1)^4 \\equiv 1 \\pmod{4}$, so $s_4(n)$ cycles through all residues modulo 4. For modulo 5, $n^4 \\equiv 1 \\pmod{5}$ for $\\gcd(n,5)=1$, so $s_4(n)$ cycles through all residues modulo 5. By the Chinese Remainder Theorem, $k=4$ has the $t_{20}$ property.\n\nb) Since $k=1,2,3$ do not have the $t_{20}$ property, they also do not have the $t_{20^k}$ property for any $k \\geq 1$. We show that $k=4$ has the $t_{20^k}$ property for all $k$. For any integer $a$, we can find $n$ such that\n\n$$\n\\frac{n(n+1)(2n+1)(3n^2 + 3n - 1)}{30} \\equiv a \\pmod{20^k}.\n$$\n\nBy considering the congruence modulo 3, $2^t$, and $5^t$ separately and using induction and the Chinese Remainder Theorem, we can always find such $n$. Thus, the smallest positive integer $k$ with the $t_{2015}$ property is $k=4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15346,
    "subject": "Mathematics (Olympiad)",
    "question": "For an arbitrary positive integer $s$, we denote by $v_2(s)$ the exponent of the largest power of $2$ dividing $s$, i.e., $2^{v_2(s)}$ divides $s$, but $2^{v_2(s)+1}$ does not. Show that for every positive integer $m$, the following equality holds:\n\n$$\nv_2\\left(\\prod_{n=1}^{2^m} \\binom{2n}{n}\\right) = m2^{m-1} + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove the following auxiliary lemma:\n\n*Lemma.* For a positive integer $n$, the number of ones in its binary representation is exactly $v_2\\left(\\binom{2n}{n}\\right)$.\n\n*Proof of the Lemma.* Let $n_{(2)} = \\overline{b_k b_{k-1} \\dots b_1 b_0}$ be the binary representation of $n$. Then $n = \\sum_{i=0}^{k} b_i 2^i$, and using the formula $v_2(n) = \\sum_{i \\ge 1} \\left\\lfloor \\frac{n}{2^i} \\right\\rfloor$, we obtain:\n\n$$\n\\begin{aligned}\nv_2\\left(\\binom{2n}{n}\\right) &= v_2((2n)!) - 2v_2(n!) \\\\\n&= \\sum_{i \\ge 1} \\left(\\left\\lfloor \\frac{2n}{2^i} \\right\\rfloor - 2\\left\\lfloor \\frac{n}{2^i} \\right\\rfloor\\right) \\\\\n&= \\sum_{j=0}^{k} \\left(\\left\\lfloor \\frac{n}{2^j} \\right\\rfloor - 2\\left\\lfloor \\frac{n}{2^{j+1}} \\right\\rfloor\\right).\n\\end{aligned}\n$$\n\nNow, the result follows by noticing that $\\left\\lfloor \\frac{n}{2^j} \\right\\rfloor - 2 \\left\\lfloor \\frac{n}{2^{j+1}} \\right\\rfloor = b_j$ for every $j = 0, 1, \\dots, k$. $\\square$\n\nUsing the lemma, we need to find the total number of ones in the binary representation of all numbers less than or equal to $2^m = \\overline{100\\dots0}_2$, where the number of zeros is $m$. The numbers with exactly $k$ zeros are $\\binom{m}{k}$ in total, so they contribute $k \\binom{m}{k} = m \\binom{m-1}{k-1}$ to the total sum. Using these facts, we conclude:\n\n$$\nv_2\\left(\\prod_{n=1}^{2^m} \\binom{2n}{n}\\right) = \\sum_{n=1}^{2^m} v_2\\left(\\binom{2n}{n}\\right) = 1 + \\sum_{k=1}^{m} m \\binom{m-1}{k-1} = 1 + m2^{m-1}.\n$$\n\n*Remark.* The lemma can be proved faster using $v_2\\left(\\binom{2n}{n}\\right) = v_2((2n)!) - 2v_2(n!)$ and Legendre's formula $v_2(n!) = n - s_2(n)$, where $s_2(n)$ is the sum of the digits of $n$ in binary.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15347,
    "subject": "Mathematics (Olympiad)",
    "question": "We say that a positive integer $n$ is *fantastic* if there exist positive rational numbers $a$ and $b$ such that\n\n$$\nn = a + \\frac{1}{a} + b + \\frac{1}{b}.\n$$\n\n(a) Prove that there exist infinitely many prime numbers $p$ such that no multiple of $p$ is fantastic.\n\n(b) Prove that there exist infinitely many prime numbers $p$ such that some multiple of $p$ is fantastic.\n",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n\n$$\nr(a, b) := a + \\frac{1}{a} + b + \\frac{1}{b} = \\frac{(a + b)(ab + 1)}{ab}.\n$$\n\nLet $a = \\frac{t}{u}$ and $b = \\frac{v}{w}$, where $t, u, v, w$ are positive integers with $\\gcd(t, u) = 1$ and $\\gcd(v, w) = 1$. Then\n\n$$\nr(a, b) = \\frac{(tv + uw)(tw + uv)}{tuvw},\n$$\n\nso the Diophantine equation\n\n$$\ntu(v^2 + w^2) + vw(t^2 + u^2) = kptuvw\n$$\n\nmust be investigated. Since $\\gcd(tu, t^2 + u^2) = 1$, this implies $tu \\mid vw$. Similarly, $vw \\mid tu$, so $tu = vw$. Substituting, the equation becomes\n\n$$\nt^2 + u^2 + v^2 + w^2 = kptu.\n$$\n\nThus, $p$ must divide either $v^2 + t^2$ or $v^2 + u^2$. If $p \\equiv -1 \\pmod{4}$ (i.e., $-1$ is a quadratic non-residue mod $p$), this forces $p \\mid v$ (and $t$ or $u$). The same argument applies for $w$, so $p \\mid v, w$, contradicting $\\gcd(v, w) = 1$. Therefore, all primes $p \\equiv -1 \\pmod{4}$ have no fantastic multiple, proving (a).\n\nFor (b), choose $v = 1$ and $w = tu$. We seek integers $t, u$ such that\n\n$$\n1 + t^2 + u^2 + t^2u^2 = kptu.\n$$\n\nLet $t = F_{2l+1}$, $u = F_{2l-1}$, where $F_n$ is the $n$th Fibonacci number, and use the identity $1 + F_{2l+1}^2 = F_{2l+3}F_{2l-1}$:\n\n$$\n(1 + t^2)(1 + u^2) = (1 + F_{2l+1}^2)(1 + F_{2l-1}^2) = F_{2l+3}F_{2l-1}F_{2l+1}F_{2l-3} = kpF_{2l+1}F_{2l-1}.\n$$\n\nSo $F_{2l+3}F_{2l-3} = kp$. Thus, every prime factor of $F_{2l+3}$ has a fantastic multiple. Since $\\gcd(F_a, F_b) = F_{\\gcd(a, b)}$, $F_a$ and $F_b$ are coprime for distinct primes $a, b$, so infinitely many primes have a fantastic multiple, proving (b).\n\n*Note 1*: $tu \\mid t^2 + u^2 + 1$ with $t > u$ only if $t, u$ are Fibonacci numbers $t = F_{2l+1}$, $u = F_{2l-1}$, in which case $t^2 + u^2 + 1 = 3tu$.\n\n*Note 2*: The identity $1 + F_{2l+1}^2 = F_{2l+3}F_{2l-1}$ is a special case of Vajda's identity: $F_{n+i}F_{n+j} - F_nF_{n+i+j} = (-1)^n F_i F_j$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15348,
    "subject": "Mathematics (Olympiad)",
    "question": "Is the sum $1^{2008} + 2^{2008} + 3^{2008} + 4^{2008} + 5^{2008} + 6^{2008}$ divisible by $5$? Explain your answer.",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that $1^{2008} = 1$. For powers of $2$, the units digit cycles through $2, 4, 8, 6$ every $4$ powers. Since $2008$ is divisible by $4$, $2^{2008}$ ends in $6$. Similarly, $3^{2008}$ ends in $1$, $4^{2008}$ ends in $6$, $5^{2008}$ ends in $5$, and $6^{2008}$ ends in $6$. Adding these units digits: $1 + 6 + 1 + 6 + 5 + 6 = 25$, which ends in $5$. Therefore, the sum $1^{2008} + 2^{2008} + 3^{2008} + 4^{2008} + 5^{2008} + 6^{2008}$ is divisible by $5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15349,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs of positive integers $a$ and $b$ for which $a^6 \\ge 5^{b+1}$ and $b^6 \\ge 5^{a+1}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove the inequality $5^{n+1} \\ge n^6$ holds for all positive integers $n$, except $n = 3$ and $n = 4$.\n\nIndeed, this is readily verified for $n = 1$, $n = 2$, and $n = 5$. When the thesis is true for $n \\ge 5$, then $5^{n+2} = 5 \\cdot 5^{n+1} \\ge 5n^6$. It is then sufficient to show that $5n^6 \\ge (n+1)^6$, rewritten as $5 \\ge \\left(1 + \\frac{1}{n}\\right)^6$. Since $\\left(1 + \\frac{1}{n}\\right)^6 < \\left(1 + \\frac{1}{4}\\right)^6 = 5 \\cdot \\frac{5^5}{4^6} = 5 \\cdot \\frac{3125}{4096} < 5$, everything is thus proven, by simple induction; moreover, the inequality becomes strict for $n > 5$.\n\nOne thus gets $5^{a+1} \\ge a^6$ and $5^{b+1} \\ge b^6$, for all $a, b$ different from $3$ and $4$. Multiplying, first the two inequalities given in the problem statement, then those two just obtained above, we get $a^6 b^6 \\ge 5^{b+1} 5^{a+1} \\ge b^6 a^6$. But equality only holds here if $a = b = 5$, which clearly checks. Otherwise, it needs $a = 4$ or $a = 3$ (or, symmetrically, $b = 4$ or $b = 3$). If $a = 4$, then the inequalities given in the problem statement become $4^6 > 5^{b+1}$ and $b^6 > 5^5$, leading to $b \\le 4$ and $b \\ge 4$ respectively, hence $b = 4$. If $a = 3$, then $3^6 > 5^{b+1}$ and $b^6 > 5^4$, leading to $b \\le 3$ and $b \\ge 3$ respectively, hence $b = 3$.\n\nTwo more pairs of solutions $a = b = 4$ and $a = b = 3$ have thus been found.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15350,
    "subject": "Mathematics (Olympiad)",
    "question": "A polynomial $x^2 + 1$ is written on the blackboard. Every day, Kate wipes off the currently written polynomial $F(x)$ and writes either $F^2(x) + 1$ or $F(x^2 + 1)$ instead, choosing at her discretion. Prove that the constant (free) term will exceed $2^{2^{222}}$ in one year.",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove a lemma: the polynomial written after $k$ days does not depend on Kate's choices and is\n$$\nF_k = \\underbrace{(\\dots((x^2 + 1)^2 + 1)^2 + \\dots)^2}_{k} + 1.\n$$\nThis can be shown by induction. Alternatively, note that\n$$\ng^2(x) + 1 = (h^2(x) + 1)^2 + 1 = (h(x^2 + 1))^2 + 1 = h^2(x^2 + 1) + 1 = h^2(t) + 1 = g(t) = g(x^2 + 1).\n$$\nThus, the process is deterministic.\n\nThe constant term $a_k$ satisfies $a_k = a_{k-1}^2 + 1$. We show by induction that $a_k \\geq 2^{2^{k-1}}$. Indeed, $a_0 = 1$, $a_1 = 2 = 2^{2^0}$, and\n$$\na_k = a_{k-1}^2 + 1 \\geq (2^{2^{k-1}})^2 + 1 = 2^{2^k} + 1 > 2^{2^k}.\n$$\nSince $365 > 222$, after one year the constant term exceeds $2^{2^{222}}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15351,
    "subject": "Mathematics (Olympiad)",
    "question": "The difference of two complementary angles $\\alpha$ and $\\beta$ is $20^\\circ 52'$. Determine $\\alpha$ and $\\beta$.",
    "options": [],
    "answer": "See solution",
    "solution": "For the angles $\\alpha$ and $\\beta$ we have $\\alpha + \\beta = 90^\\circ$ and $\\alpha - \\beta = 20^\\circ 52'$. Hence, $$\\beta = \\frac{90^\\circ - 20^\\circ 52'}{2} = 34^\\circ 34'$$ and $$\\alpha = 90^\\circ - 34^\\circ 34' = 55^\\circ 26'.$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15352,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $n$ sets $A_i$, each with $|A_i| = n$, prove that they may be indexed as $A_i = \\{a_{i,j} \\mid j = 1, 2, \\dots, n\\}$ so that the sets $B_j = \\{a_{i,j} \\mid i = 1, 2, \\dots, n\\}$, for $1 \\leq j \\leq n$, also satisfy $|B_j| = n$.\n\nThis is equivalent to constructing an $n \\times n$ matrix whose $i$-th row consists of the distinct elements of $A_i$, such that each column also consists of distinct entries.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove a more general case: let $|A_i| = m > 0$, $A_i = \\{a_{i,j} \\mid j = 1, 2, \\dots, m\\}$, and suppose no element belongs to more than $m$ sets. The base case $n=1$ is trivial. For $n \\geq 2$, use induction: build a $(n-1) \\times m$ matrix for $A_1, \\dots, A_{n-1}$, then add row $R_n$ with the elements of $A_n$.\n\nIn the resulting $n \\times m$ matrix, a column is \"bad\" if its entry in $R_n$ is duplicated in another row. Choose a matrix with the minimal number of bad columns. If there are none, we are done. Otherwise, let $C$ be a bad column with entry $a$ in $R_n$ duplicated elsewhere. There must be another column $C'$ with no entry equal to $a$ (since $a$ appears at most $m$ times). Let $\\bar{C}$ and $\\bar{C}'$ be the first $n-1$ entries of $C$ and $C'$, respectively; by induction, these are distinct.\n\nDefine $\\varphi: \\bar{C} \\to \\bar{C}'$ by $\\varphi(c_i) = c'_i$ (entries on the same row). An *alternating path* is a sequence $c_1\\varphi(c_1)c_2\\varphi(c_2)\\dots c_k\\varphi(c_k)$, and its *swap* replaces $c_i$ with $\\varphi(c_i)$ and vice versa. Swapping $a$ and $\\varphi(a)$ leaves $C'$ good. If $\\varphi(a)$ is not in $\\bar{C}$, $C$ becomes good. Otherwise, continue: swap $a, \\varphi(a), \\varphi(a), \\varphi^2(a)$, etc. This process must eventually terminate (since there are only $n-1$ entries in $\\bar{C}$), at which point swapping turns $C$ good and leaves $C'$ good, contradicting the minimality of bad columns. Thus, the desired arrangement exists.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 15353,
    "subject": "Mathematics (Olympiad)",
    "question": "On a blackboard, we write in a line $n$ numbers, $n \\ge 40$, each being equal to $1$ or $-1$, such that:\n\n1. The sum of every 40 successive numbers is equal to $0$.\n2. The sum of every 42 successive numbers is not equal to $0$.\n\nLet $\\Sigma_n$ be the sum of the $n$ numbers on the blackboard. Find the greatest possible value of $\\Sigma_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_1, a_2, \\dots, a_n$ be the numbers on the blackboard. Since $a_1 + a_2 + \\dots + a_{40} = 0$, half of the numbers $a_1, \\dots, a_{40}$ are $1$ and the other half are $-1$.\n\nSince $a_1 + a_2 + \\dots + a_{40} + a_{41} + a_{42} \\neq 0$, it follows that $a_{41} = a_{42} = a \\in \\{-1, 1\\}$.\n\nSimilarly, $a_2 + a_3 + \\dots + a_{43} \\neq 0$, so $a_{42} = a_{43} = a$. Proceeding in the same way, all numbers after $a_{41}$ must be equal to $a$.\n\nIf $n > 60$, the sum $a_{22} + a_{23} + \\dots + a_{61}$ includes $a_{41} = a_{42} = \\dots = a_{61} = a$, so $1$ and $-1$ cannot be equally represented, contradicting condition (i).\n\nHence, the greatest value of $n$ is $60$.\n\nTo maximize $\\Sigma_n$, we want as many $1$'s as possible. Among the first $40$ numbers, there must be $20$ $1$'s and $20$ $-1$'s. Thus, the maximum number of $1$'s is $40$, so $\\Sigma_n \\le 20$. In fact, $\\max \\Sigma_n = 20$, achieved when\n\n$$\na_1 = \\dots = a_{20} = 1,\\quad a_{21} = \\dots = a_{40} = -1,\\quad a_{41} = \\dots = a_{60} = 1\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15354,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a continuous function. A *chord* is defined as a segment of integer length, parallel to the $Ox$ axis, whose endpoints lie on the graph $y = f(x)$. Suppose the graph $y = f(x)$ has exactly $N$ chords, and among them there is a chord of length $2025$. Find the least possible value of $N$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $4049$.\n\n**Solution:**\n\nFor a natural number $n$, define $g_n(x) = f(x+n) - f(x)$. The number of chords of length $n$ equals the number of zeros of $g_n(x)$.\n\nAs an example, consider the following piecewise linear function:\n\n- $f(x) = x$ for $x \\leq 2024\\frac{9}{10}$,\n- $f(x) = 20249 \\cdot |2025 - x|$ for $x \\geq 2024\\frac{9}{10}$.\n\nFor $a \\notin [0, 2025\\frac{1}{10}]$, $f(x)$ takes the value $f(a)$ only at $a$. Therefore, if $g_n(x) = 0$, both $x$ and $x+n$ must lie in $[0, 2025\\frac{1}{10}]$. In particular, $n \\leq 2025$, and the zeros of $g_n(x)$ lie in $[0, 2025\\frac{1}{10} - n]$.\n\nFor $n = 2025$ and $x \\in [0, \\frac{1}{10}]$, $g_{2025}(x) = 20248x$, so $g_{2025}(x)$ has a unique zero at $x = 0$. Thus, $f$ has exactly one chord of length $2025$.\n\nFor $n = 1, 2, \\dots, 2024$, $g_n(x)$ is monotonically decreasing on $[0, 2025 - n]$ and increasing on $[2025 - n, 2025\\frac{1}{10} - n]$, with $g_n(0) > 0$, $g_n(2025 - n) < 0$, and $g_n(2025\\frac{1}{10} - n) > 0$. Therefore, $g_n(x)$ has exactly two zeros, so $f$ has two chords of each length $n = 1, 2, \\dots, 2024$.\n\nIn total, there are $1 + 2 \\times 2024 = 4049$ distinct chords.\n\nFor the upper bound, assume the chord of length $2025$ connects $(0, 0)$ and $(2025, 0)$, i.e., $f(0) = f(2025) = 0$. Define $g(x) = g_1(x) = f(x+1) - f(x)$. Since $g$ is continuous and has finitely many zeros, all zeros lie in $[-M, M]$ for some $M$. Thus, $g(x)$ has constant sign outside $[-M, M]$. By possibly replacing $f$ with $-f$, assume $g(x) > 0$ for $x > M$.\n\nIf $g(x) < 0$ for $x < -M$, then $g_n(x) = g(x) + g(x+1) + \\dots + g(x+n-1)$ would be positive for $x > M$ and negative for $x < -M-n$, so $g_n$ would have a zero for all $n$, contradicting the problem's conditions.\n\nThus, $g(x) > 0$ for $x < -M$. For natural $k, m$ with $k + m = 2025$, $f$ has at least $4$ chords of lengths $k$ and $m$ combined. Applying this to all such pairs gives the claimed bound.\n\nSuppose $g_k$ and $g_m$ have fewer than $4$ zeros combined. Then one (say $g_k$) has at most one zero and must be non-negative. Consider the minimal $N$ such that $Nk$ is divisible by $2025$, and let $r_1, \\dots, r_{N-1}$ be the remainders of $k, 2k, \\dots, (N-1)k$ modulo $2025$. Define $r_0 = r_N = 0$ if $g_k(0) > 0$, or $r_0 = r_N = 2025$ if $g_k(0) = 0$. The differences $d_j = f(r_{j+1}) - f(r_j)$ satisfy: if $r_{j+1} > r_j$, $d_j = g_k(r_j) \\geq 0$; if $r_{j+1} < r_j$, $d_j = -g_m(r_{j+1})$.\n\nThe sum $\\sum d_j = 0$ must contain both positive and negative terms, so $g_m(t) > 0$ for some $t \\in (0, k)$. Since $g_k(0) + g_m(k) = g_m(0) + g_k(m) = 0$ and $g_k \\geq 0$, we get $g_m(0) \\leq 0$ and $g_m(k) \\leq 0$. Thus, $g_m$ has at least $3$ zeros (at $0$, $k$, and between), plus $g_k$ has at least $1$ zero, totaling $4$ zeros as required. The case $g_k(m) = 0$ is similar.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 15355,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, $z$ be positive real numbers such that $x + y + z \\ge 6$. Find the smallest value of the expression\n$$\nx^2 + y^2 + z^2 + \\frac{x}{y^2 + z + 1} + \\frac{y}{z^2 + x + 1} + \\frac{z}{x^2 + y + 1}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Using the AM-GM inequality for positive real numbers $\\frac{x^2}{14}$, $\\frac{x}{y^2+z+1}$, and $\\frac{2}{49}(y^2+z+1)$, we have\n$$\n\\frac{x^2}{14} + \\frac{x}{y^2 + z + 1} + \\frac{2}{49}(y^2 + z + 1) \\ge 3\\sqrt[3]{\\frac{x^3}{7^3}} = \\frac{3}{7}x.\n$$\nWe can derive cyclically two similar inequalities. Adding all three cyclically obtained inequalities, we get\n$$\n\\frac{1}{14} \\sum x^2 + \\sum \\frac{x}{y^2 + z + 1} + \\frac{2}{49} \\sum x^2 + \\frac{2}{49} \\sum x + \\frac{6}{49} \\ge \\frac{3}{7} \\sum x.\n$$\nThus,\n$$\nL = \\frac{11}{98} \\sum x^2 + \\sum \\frac{x}{y^2 + z + 1} + \\frac{6}{49} \\ge \\left(\\frac{3}{7} - \\frac{2}{49}\\right) \\sum x = \\frac{19}{49} \\sum x.\n$$\nFrom the Cauchy-Schwarz inequality (or AM-QM), $\\sum x^2 \\ge \\frac{1}{3}(\\sum x)^2 \\ge 12$. Finally,\n$$\n\\begin{aligned}\n\\sum x^2 + \\sum \\frac{x}{y^2 + z + 1} &= \\frac{87}{98} \\sum x^2 + L - \\frac{6}{49} \\\\\n&\\ge \\frac{87}{98} \\sum x^2 + \\frac{19}{49} \\sum x - \\frac{6}{49} \\\\\n&\\ge \\frac{87 \\cdot 6 + 19 \\cdot 6 - 6}{49} = \\frac{90}{7}.\n\\end{aligned}\n$$\n**Conclusion.** The smallest value of the given expression is $\\frac{90}{7}$. (This value is achieved for $x = y = z = 2$.)",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15356,
    "subject": "Mathematics (Olympiad)",
    "question": "A representation of $\\frac{17}{20}$ as a sum of reciprocals\n\n$$\n\\frac{17}{20} = \\frac{1}{a_1} + \\frac{1}{a_2} + \\cdots + \\frac{1}{a_k}\n$$\n\nis called a *calm representation* with $k$ terms if the $a_i$ are distinct positive integers and at most one of them is not a power of two.\n\n(a) Find the smallest value of $k$ for which $\\frac{17}{20}$ has a calm representation with $k$ terms.\n\n(b) Prove that there are infinitely many calm representations of $\\frac{17}{20}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Now we show that there are infinitely many calm representations: take $a_1 = 5 \\cdot 2^{4n+1}$ and consider the difference\n\n$$\n\\frac{17}{20} - \\frac{1}{5 \\cdot 2^{4n+1}} = \\frac{17 \\cdot 2^{4n-1} - 1}{5 \\cdot 2^{4n+1}}.\n$$\n\nSince $2^4 = 16 \\equiv 1 \\mod 5$, the numerator is $17 \\cdot 2^{4n-1} - 1 \\equiv 17 \\cdot 8 - 1 = 135 \\equiv 0 \\mod 5$. Thus the factor 5 cancels, and we have\n\n$$\n\\frac{17}{20} - \\frac{1}{5 \\cdot 2^{4n+1}} = \\frac{A}{2^{4n+1}}\n$$\n\nfor some positive integer $A < 2^{4n+1}$. Since $A$ has a binary representation as $A = 2^{b_1} + 2^{b_2} + \\dots + 2^{b_r}$ with distinct nonnegative integers $b_1, b_2, \\dots, b_r$, we get\n\n$$\n\\frac{17}{20} = \\frac{1}{5 \\cdot 2^{4n+1}} + \\frac{1}{2^{4n+1-b_1}} + \\frac{1}{2^{4n+1-b_2}} + \\dots + \\frac{1}{2^{4n+1-b_r}},\n$$\n\nwhich is a calm representation for every $n$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15357,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure below, $\\angle A$ is the largest angle in triangle $ABC$. On the circumcircle of $\\triangle ABC$, the points $D$ and $E$ are the midpoints of arcs $ABC$ and $ACB$, respectively. Let $\\odot O_1$ be the circle passing through $A$ and $B$ and tangent to line $AC$, and let $\\odot O_2$ be the circle passing through $A$ and $E$ and tangent to line $AD$. The circles $\\odot O_1$ and $\\odot O_2$ intersect at points $A$ and $P$. Prove that $AP$ is the bisector of $\\angle BAC$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p100_data_6f85ae3948.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Join the pairs of points $EP$, $AE$, $BE$, $BP$, and $CD$. For convenience, denote $A = \\angle BAC$, $B = \\angle ABC$, $C = \\angle ACB$, so $A + B + C = 180^\\circ$. Take an arbitrary point $X$ on the extension of $CA$, and a point $Y$ on the extension of $DA$. It is easy to see that $AD = DC$ and $AE = EB$. Since $A$, $B$, $C$, $D$, and $E$ are concyclic, we have:\n\n$$\n\\begin{aligned}\n\\angle BAE &= 90^\\circ - \\frac{1}{2} \\angle AEB = 90^\\circ - \\frac{C}{2}, \\\\\n\\angle CAD &= 90^\\circ - \\frac{1}{2} \\angle ADC = 90^\\circ - \\frac{B}{2}.\n\\end{aligned}\n$$\n\nAs line $AC$ and $\\odot O_1$ are tangent at $A$, and line $AD$ and $\\odot O_2$ are tangent at $A$, we get:\n\n$$\n\\angle APB = \\angle BAX = 180^\\circ - A, \\quad \\angle ABP = \\angle CAP,\n$$\n\nand\n\n$$\n\\begin{aligned}\n\\angle APE &= \\angle EAY = 180^\\circ - \\angle DAE \\\\\n&= 180^\\circ - (\\angle BAE + \\angle CAD - A) \\\\\n&= 180^\\circ - \\left(90^\\circ - \\frac{C}{2}\\right) - \\left(90^\\circ - \\frac{B}{2}\\right) + A \\\\\n&= 90^\\circ + \\frac{A}{2}.\n\\end{aligned}\n$$\n\nBy computation,\n\n$$\n\\angle BPE = 360^\\circ - \\angle APB - \\angle APE = 90^\\circ + \\frac{A}{2} = \\angle APE.\n$$\n\nIn $\\triangle APE$ and $\\triangle BPE$, applying the Law of Sines and noting $AE = BE$, we obtain:\n\n$$\n\\frac{\\sin \\angle PAE}{\\sin \\angle APE} = \\frac{PE}{AE} = \\frac{PE}{BE} = \\frac{\\sin \\angle PBE}{\\sin \\angle BPE}.\n$$\n\nTherefore, $\\sin \\angle PAE = \\sin \\angle PBE$. Since $\\angle APE$ and $\\angle BPE$ are both obtuse, $\\angle PAE$ and $\\angle PBE$ are both acute, so $\\angle PAE = \\angle PBE$.\n\nHence, $\\angle BAP = \\angle BAE - \\angle PAE = \\angle ABE - \\angle PBE = \\angle ABP = \\angle CAP$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15358,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R}^+ \\to \\mathbb{R}^+$ satisfying\n\n$$\nf(f(xy) + 1) = x f(x + f(y)), \\quad \\forall x, y > 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, replace $x$ with $\\frac{z}{y}$:\n\n$$\nf(f(z) + 1) = \\frac{z}{y} f\\left(\\frac{z}{y} + f(y)\\right), \\quad \\forall y, z > 0. \\tag{1}\n$$\n\nIn (1), set $z = 1$ and let $f(f(1) + 1) = c$. We have $f\\left(\\frac{1}{y} + f(y)\\right) = c y$ for all $y > 0$. The right-hand side takes all values in $\\mathbb{R}^+$, so $f$ is surjective on $\\mathbb{R}^+$.\n\nSuppose there exist $0 < y_1 < y_2$ such that $f(y_1) < f(y_2)$. Choose $z_0$ such that\n\n$$\n\\frac{z_0}{y_1} + f(y_1) = \\frac{z_0}{y_2} + f(y_2) \\iff z_0 = y_1 y_2 \\frac{f(y_2) - f(y_1)}{y_2 - y_1} > 0.\n$$\n\nIn (1), replace $z = z_0$ and let $y = y_1$, $y = y_2$ respectively:\n\n$$\nf(f(z_0) + 1) = \\frac{z_0}{y_1} f\\left(\\frac{z_0}{y_1} + f(y_1)\\right) = \\frac{z_0}{y_2} f\\left(\\frac{z_0}{y_2} + f(y_2)\\right).\n$$\n\nBy the choice of $z_0$,\n\n$$\nf\\left(\\frac{z_0}{y_1} + f(y_1)\\right) = f\\left(\\frac{z_0}{y_2} + f(y_2)\\right),\n$$\n\nso $y_1 = y_2$. This contradiction shows that for $0 < y_1 < y_2$, we must have $f(y_1) \\ge f(y_2)$, so $f$ is non-increasing. Thus, $f$ is both surjective and monotone on $\\mathbb{R}^+$, so it is continuous on $\\mathbb{R}^+$.\n\nSince $f$ is continuous and non-increasing, there exists $\\lim_{x \\to +\\infty} f(x) = c \\ge 0$. If $c > 0$, then $f(x) \\ge c > 0$ for all $x \\in \\mathbb{R}^+$, which contradicts surjectivity. Therefore, $c = 0$.\n\nFinally, in the given condition, for each $x > 0$, as $y \\to +\\infty$, $f(f(xy) + 1) \\to f(1)$ and $f(x + f(y)) \\to f(x)$. These imply $f(1) = x f(x)$ or $f(x) = \\frac{c}{x}$ with $c = f(1) > 0$. It is easy to check that this function satisfies the problem. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15359,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : [0, 1] \\to \\mathbb{R}$ be a differentiable function, with integrable derivative on $[0, 1]$, such that $f(1) = 0$. Prove that\n$$\n\\int_{0}^{1} (x f'(x))^2 \\, dx \\geq 12 \\left( \\int_{0}^{1} x f(x) \\, dx \\right)^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $g : [0, 1] \\to \\mathbb{R}$ be defined by $g(x) = x f(x)$ for $x \\in [0, 1]$. We have $g(0) = 0 = g(1)$ and $g'(x) = f(x) + x f'(x)$, so that:\n$$\n\\begin{align*}\n\\int_0^1 x^2 (f'(x))^2 \\, dx &= \\int_0^1 (g'(x) - f(x))^2 \\, dx \\\\\n&= \\int_0^1 (g'(x))^2 \\, dx - 2 \\int_0^1 g'(x) f(x) \\, dx + \\int_0^1 (f(x))^2 \\, dx \\\\\n&= \\int_0^1 f^2(x) \\, dx + \\int_0^1 (g'(x))^2 \\, dx - 2(f(1)g(1) - f(0)g(0)) + 2 \\int_0^1 g(x) f'(x) \\, dx.\n\\end{align*}\n$$\nThis implies\n$$\n\\begin{align*}\n\\int_{0}^{1} x^2 (f'(x))^2 \\, dx &= \\int_{0}^{1} f^2(x) \\, dx + \\int_{0}^{1} (g'(x))^2 \\, dx + \\int_{0}^{1} x \\cdot (2 f(x) f'(x)) \\, dx \\\\\n&= \\int_{0}^{1} f^2(x) \\, dx + \\int_{0}^{1} (g'(x))^2 \\, dx + \\int_{0}^{1} x \\cdot (f^2(x))' \\, dx \\\\\n&= \\int_{0}^{1} f^2(x) \\, dx + \\int_{0}^{1} (g'(x))^2 \\, dx + (1 \\cdot f^2(1) - 0 \\cdot f^2(0)) - \\int_{0}^{1} f^2(x) \\, dx \\\\\n&= \\int_{0}^{1} (g'(x))^2 \\, dx.\n\\end{align*}\n$$\nBy the Cauchy-Schwarz inequality,\n$$\n\\left( \\int_0^1 (2x-1) g'(x) \\, dx \\right)^2 \\leq \\int_0^1 (2x-1)^2 \\, dx \\cdot \\int_0^1 (g'(x))^2 \\, dx = \\frac{1}{3} \\int_0^1 x^2 (f'(x))^2 \\, dx.\n$$\nThis gives\n$$\n\\begin{align*}\n\\int_0^1 (x f'(x))^2 \\, dx &\\geq 3 \\left( \\int_0^1 (2x-1) g'(x) \\, dx \\right)^2 \\\\\n&= 3 \\left( (2 \\cdot 1 - 1)g(1) - (2 \\cdot 0 - 1)g(0) - 2 \\int_0^1 g(x) \\, dx \\right)^2 \\\\\n&= 12 \\left( \\int_0^1 x f(x) \\, dx \\right)^2,\n\\end{align*}\n$$\nconcluding the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15360,
    "subject": "Mathematics (Olympiad)",
    "question": "The minimum value of $f(x) = \\frac{5 - 4x + x^2}{2 - x}$ for $x \\in (-\\infty, 2)$ is ( ).",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x < 2 \\Rightarrow 2 - x > 0$. Then\n\n$$\n\\begin{aligned}\nf(x) &= \\frac{1 + (4 - 4x + x^2)}{2 - x} = \\frac{1}{2 - x} + (2 - x) \\\\\n&\\geq 2 \\times \\sqrt{\\frac{1}{2 - x} \\times (2 - x)} = 2.\n\\end{aligned}\n$$\n\nThe equality holds if and only if $\\frac{1}{2 - x} = 2 - x$, and it is so when $x = 1 \\in (-\\infty, 2)$. This means that $f$ reaches the minimum value $2$ at $x = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15361,
    "subject": "Mathematics (Olympiad)",
    "question": "Cyclic quadrilateral $ABCD$ has lengths $BC = CD = 3$ and $DA = 5$ with $\\angle CDA = 120^\\circ$. What is the length of the shorter diagonal of $ABCD$?\n\n(A) $\\frac{31}{7}$  \n(B) $\\frac{33}{7}$  \n(C) $5$  \n(D) $\\frac{39}{7}$  \n(E) $\\frac{41}{7}$",
    "options": [],
    "answer": "See solution",
    "solution": "Place the figure in the coordinate plane with $D = (0, 0)$ and $A = (5, 0)$. Because $\\angle CDA = 120^\\circ$ and $CD = 3$, it follows that $C = \\left(-\\frac{3}{2}, \\frac{3}{2}\\sqrt{3}\\right)$. The perpendicular bisectors of $\\overline{CD}$ and $\\overline{AD}$ intersect at the center of the circumscribing circle. The midpoint of $\\overline{CD}$ is $\\left(-\\frac{3}{4}, \\frac{3}{4}\\sqrt{3}\\right)$. The line through $\\overline{CD}$ has slope $-\\sqrt{3}$, so a line perpendicular to that has slope $\\frac{\\sqrt{3}}{3}$. The perpendicular bisector of $\\overline{CD}$ therefore has equation\n\n$$\ny - \\frac{3}{4}\\sqrt{3} = \\frac{\\sqrt{3}}{3}\\left(x + \\frac{3}{4}\\right).\n$$\n\nThe perpendicular bisector of $\\overline{AD}$ has equation $x = \\frac{5}{2}$. Solving this system of equations locates the center of the circumscribing circle at $O\\left(\\frac{5}{2}, \\frac{11}{6}\\sqrt{3}\\right)$.\n\n![](images/2024_AMC12A_Solutions_p11_data_a02c226529.png)\n\nBy the Distance Formula, the radius of the circle is\n\n$$\nr = OD = \\sqrt{\\left(\\frac{5}{2}\\right)^2 + \\left(\\frac{11}{6}\\sqrt{3}\\right)^2} = \\frac{7\\sqrt{3}}{3}.\n$$\n\nLet $\\theta$ be the measure of $\\angle BCO$. By the Law of Cosines applied to $\\triangle BCO$,\n\n$$\n\\frac{49}{3} = BO^2 = BC^2 + CO^2 - 2BC \\cdot CO \\cos \\theta = 9 + \\frac{49}{3} - 2 \\cdot 3 \\cdot \\frac{7\\sqrt{3}}{3} \\cos \\theta,\n$$\n\nwhich gives $\\cos \\theta = \\frac{3}{14}\\sqrt{3}$. A double angle formula then gives $\\cos 2\\theta = 2 \\cdot \\left(\\frac{3}{14}\\sqrt{3}\\right)^2 - 1 = -\\frac{71}{98}$. Triangles $\\triangle BOC$ and $\\triangle COD$ are congruent isosceles triangles, so $\\angle BCD$ has measure $2\\theta$. By the Law of Cosines applied to $\\triangle BCD$,\n\n$$\nBD^2 = BC^2 + CD^2 - 2BC \\cdot CD \\cos 2\\theta = 9 + 9 - 2 \\cdot 3 \\cdot 3 \\cdot \\left(-\\frac{71}{98}\\right) = \\frac{1521}{49} = \\left(\\frac{39}{7}\\right)^2,\n$$\n\nso $BD = \\frac{39}{7}$. The Law of Cosines applied to $\\triangle ADC$ gives $AC = 7$, so the shorter diagonal has length $\\frac{39}{7}$.\n\nAlternatively, the Law of Cosines applied to $\\triangle ADC$ gives\n\n$$\nCA^2 = 3^2 + 5^2 - 2 \\cdot 3 \\cdot 5 \\cdot \\cos 120^\\circ.\n$$\n\nBecause $\\cos 120^\\circ = -\\frac{1}{2}$, diagonal $\\overline{AC}$ has length $\\sqrt{9+25+15} = 7$. Because $ABCD$ is cyclic, $\\angle ABC$ is supplementary to $\\angle ADC$, so $\\angle ABC = 60^\\circ$. The Law of Cosines applied to $\\triangle ABC$ gives\n\n$$\n7^2 = 3^2 + BA^2 - 2 \\cdot 3 \\cdot BA \\cdot \\cos 60^\\circ.\n$$\n\nSimplifying yields $BA^2 - 3 \\cdot BA - 40 = 0$, so $BA = 8$. By Ptolemy's Theorem,\n\n$$\nCA \\cdot BD = BA \\cdot CD + BC \\cdot AD.\n$$\n\nSubstituting gives $7 \\cdot BD = 8 \\cdot 3 + 3 \\cdot 5$, so $BD = \\frac{39}{7}$, which is less than the length of diagonal $\\overline{AC}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15362,
    "subject": "Mathematics (Olympiad)",
    "question": "On an $m \\times m$ board, at the midpoints of the unit squares there are some ants. At time 0, each ant starts moving with speed 1 parallel to some edge of the board until it meets an ant moving in the opposite direction or until it reaches the edge of the board. When two ants moving in opposite directions meet, both turn $90^\\circ$ clockwise and continue moving parallel to another edge of the board. Upon reaching the edge of the board, the ant falls off the board.\n\n**a)** Prove that eventually all the ants will have fallen off the board.\n\n**b)** Find the latest possible moment for the last ant to fall off the board.",
    "options": [],
    "answer": "See solution",
    "solution": "**b)** $\\frac{3}{2}m - 1$.\n\nLet the lower left corner of the board be the origin. Divide the units of time and space by 2; then the squares are of dimensions $2 \\times 2$, the coordinates of the midpoints of the squares are odd positive integers, and the speed of the ants is still 1.\n\nWe prove by induction that at integer time moments the coordinates of the ants are integers and the sum of the coordinates for any fixed ant has the same parity as the time moment. In addition, the ants can meet only at integer time moments. At time $t = 0$ all coordinates of the ants are odd, so their sum is even. Suppose that at an integer time moment $t = k$ the coordinates of the ants are integers and the sum of the coordinates for any fixed ant has the same parity as the time moment. If two of the ants were to meet each other within the next time unit, they have to move toward each other from time $t = k$, hence one of their coordinates must be the same. Since the parity of the sum of their coordinates was the same at time $t = k$, another of their coordinates had to differ by at least 2. Hence they cannot meet before time $t = k + 1$. Between time moments $t = k$ and $t = k + 1$ every ant has changed only one of its coordinates by 1, hence at time $t = k + 1$ the parity of the sum of the coordinates is again the same as the parity of the time moment.\n\nNext, we will prove by induction that for any point with integer coordinates $(x, y)$ there are no collisions at this point after the time moment $t = x + y - 2$. For $x = y = 1$ this is obviously true, since there are no collisions in the middle of the lower left square (otherwise one of the ants has to arrive to this point from the edge of the board). Let $(x, y)$ be arbitrary and suppose that the claim holds for all points with the sum of the coordinates less than $x + y$. Suppose that a collision takes place at point $(x, y)$ at time $t$. One of the participants had to arrive from a point where one of the coordinates was smaller; without loss of generality, we can assume that this was the $x$-coordinate. If this ant has not collided with anyone before, then $t \\leq x - 1 \\leq x + y - 2$. If the last collision of this ant occurred at time $t' < t$, then the coordinates of the last collision were $(x - (t - t'), y)$. By the induction assumption $t' \\leq x - (t - t') + y - 2$, hence $t \\leq x + y - 2$.\n\nBy symmetry, the claim holds when another corner is chosen as the origin. Let the last collision of a particular ant occur at the point $(x, y)$, where the coordinates are taken with respect to the nearest corner. Without loss of generality, we can assume $x \\leq y$. The time from the last collision to the falling off the edge of the ants participating in the collision is at most $2m - x$, hence the time elapsed from the start is at most $x + y - 2 + 2m - x \\leq 3m - 2$. By this time all ants have fallen off the edge. With respect to the original units, the maximal time is $\\frac{3}{2}m - 1$.\n\nFor any $m$, the maximal time can be achieved if in the beginning there are 2 ants at the adjoining corners of the board moving toward each other. At the moment $t = \\frac{m-1}{2}$ the pair collides and one of the ants starts moving toward the center, falling off the board at time $t = \\frac{3}{2}m - 1$.\n\nFor part **a)**, for each ant consider the distance to the edge in the direction of its motion. After an ant falls, this distance will remain 0. Observe that as long as an ant moves without collision, this distance decreases with speed 1.\n\nConsider now the sum of all such distances. When a collision happens, the sum of the distances of the two corresponding ants is $m$, both right before and right after the collision. Thus, as long as there are ants left on the board, the total sum decreases with the speed of at least 1. Since in the beginning this sum is a finite number, after some time this sum will become 0 and thus all ants will have fallen off the board.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15363,
    "subject": "Mathematics (Olympiad)",
    "question": "At a university dinner, there are 2017 mathematicians who each order two distinct entrées, with no two mathematicians ordering the same pair of entrées. The cost of each entrée is equal to the number of mathematicians who ordered it, and the university pays for each mathematician's less expensive entrée (ties broken arbitrarily). Over all possible sets of orders, what is the maximum total amount the university could have paid?\n\nIn graph theoretic terms: we wish to determine the maximum possible value of\n\n$$\nS(G) \\stackrel{\\text{def}}{=} \\sum_{e=vw} \\min(\\deg v, \\deg w)\n$$\n\nacross all graphs $G$ with 2017 edges.",
    "options": [],
    "answer": "See solution",
    "solution": "*First solution (combinatorial, Evan Chen)* First define $L_k$ to consist of a clique on $k$ vertices, plus a single vertex connected to exactly one vertex of the clique. Hence $L_k$ has $k+1$ vertices, $\\binom{k}{2} + 1$ edges, and $S(L_k) = (k-1)\\binom{k}{2} + 1$. In particular, $L_{64}$ achieves the claimed maximum, so it suffices to prove the upper bound.\n\n*Lemma*\n\nLet $G$ be a graph such that either\n\n- $G$ has $\\binom{k}{2}$ edges for some $k \\ge 3$ or\n- $G$ has $\\binom{k}{2} + 1$ edges for some $k \\ge 4$.\n\nThen there exists a graph $G^*$ with the same number of edges such that $S(G^*) \\ge S(G)$, and moreover $G^*$ has a universal vertex (i.e. a vertex adjacent to every other vertex).\n\n*Proof.* Fix $k$ and the number $m$ of edges. We prove the result by induction on the number $n$ of vertices in $G$. Since the lemma has two parts, we will need two different base cases:\n\n1. Suppose $n = k$ and $m = \\binom{k}{2}$. Then $G$ must be a clique so pick $G^* = G$.\n2. Suppose $n = k+1$ and $m = \\binom{k}{2} + 1$. If $G$ has no universal vertex, we claim we may take $G^* = L_k$. Indeed each vertex of $G$ has degree at most $k-1$, and the average degree is\n\n$$\n\\frac{2m}{n} = 2 \\cdot \\frac{k^2 - k + 1}{k + 1} < k - 1\n$$\n\nusing here $k \\ge 4$. Thus there exists a vertex $w$ of degree $1 \\le d \\le k-2$. The edges touching $w$ will have label at most $d$ and hence\n\n$$\n\\begin{aligned}\nS(G) &\\le (k-1)(m-d) + d^2 = (k-1)m - d(k-1-d) \\\\\n&\\le (k-1)m - (k-2) = (k-1)\\binom{k}{2} + 1 = S(G^*).\n\\end{aligned}\n$$\n\nNow we settle the inductive step. Let $w$ be a vertex with minimal degree $0 \\le d < k-1$, with neighbors $w_1, \\dots, w_d$. By our assumption, for each $w_i$ there exists a vertex $v_i$ for which $v_i w_i \\notin E$. Now, we may delete all edges $ww_i$ and in their place put $v_i w_i$, and then delete the vertex $w$. This gives a graph $G'$, possibly with multiple edges (if $v_i = w_j$ and $w_j = v_i$), and with one fewer vertex.\n\n![](images/sols-TST-IMO-2018_p7_data_016e463360.png)\n\n![](images/sols-TST-IMO-2018_p7_data_cdd6b7c057.png)\n\n![](images/sols-TST-IMO-2018_p7_data_635db63b99.png)\n\nWe then construct a graph $G''$ by taking any pair of double edges, deleting one of them, and adding any missing edge of $G''$ in its place. (This is always possible, since when $m = \\binom{k}{2}$ we have $n-1 \\ge k$ and when $m = \\binom{k}{2} + 1$ we have $n-1 \\ge k+1$.)\n\nThus we have arrived at a simple graph $G''$ with one fewer vertex. We also observe that we have $S(G'') \\ge S(G)$; after all every vertex in $G''$ has degree at least as large as it did in $G$, and the $d$ edges we deleted have been replaced with new edges which will have labels at least $d$. Hence we may apply the inductive hypothesis to the graph $G''$ to obtain $G^*$ with $S(G^*) \\ge S(G'') \\ge S(G)$. $\\square$\n\nThe problem then is completed once we prove the following:\n\n*Claim.* For any graph $G$,\n\n- If $G$ has $\\binom{k}{2}$ edges for $k \\ge 3$, then $S(G) \\le \\binom{k}{2} \\cdot (k-1)$.\n- If $G$ has $\\binom{k}{2} + 1$ edges for $k \\ge 4$, then $S(G) \\le \\binom{k}{2} \\cdot (k-1) + 1$.\n\n*Proof.* We prove both parts at once by induction on $k$, with the base case $k=3$ being plain (there is nothing to prove in the second part for $k=3$). Thus assume $k \\ge 4$. By the earlier lemma, we may assume $G$ has a universal vertex $v$. For notational convenience, we say $G$ has $\\binom{k}{2} + \\varepsilon$ edges for $\\varepsilon \\in \\{0,1\\}$, and $G$ has $p+1$ vertices, where $p \\ge k-1+\\varepsilon$.\n\nLet $H$ be the subgraph obtained when $v$ is deleted. Then $m = \\binom{k}{2} + \\varepsilon - p$ is the number of edges in $H$; from $p \\ge k-1+\\varepsilon$ we have $m \\le \\binom{k-1}{2}$ and so we may apply the inductive hypothesis to $H$ to deduce $S(H) \\le \\binom{k-1}{2} \\cdot (k-2)$.\n\n![](images/sols-TST-IMO-2018_p7_data_673f4a8390.png)\n\nNow the labels of edges $vw_i$ have sum\n\n$$\n\\sum_{i=1}^{p} \\min (\\deg_G v, \\deg_G w_i) = \\sum_{i=1}^{p} \\deg_G w_i = \\sum_{i=1}^{p} (\\deg_H w_i + 1) = 2m + p.\n$$\n\nFor each of the edges contained in $H$, the label on that edge has increased by exactly 1, so those edges contribute $S(H) + m$. In total,\n\n$$\n\\begin{aligned} S(G) &= 2m + p + (S(H) + m) = (m + p) + 2m + S(H) \\\\ &\\le \\binom{k}{2} + \\varepsilon + 2\\binom{k-1}{2} + \\binom{k-1}{2}(k-2) = \\binom{k}{2}(k-1) + \\varepsilon. \\end{aligned} \\quad \\square\n$$\n\nThus, the maximum possible value is $63 \\cdot \\binom{64}{2} + 1 = 127009$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15364,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, $d$ be real numbers such that $b - d \\ge 5$ and all zeros $x_1$, $x_2$, $x_3$, and $x_4$ of the polynomial\n\n$$\nP(x) = x^4 + a x^3 + b x^2 + c x + d\n$$\n\nare real. Find the smallest value the product $(x_1^2 + 1)(x_2^2 + 1)(x_3^2 + 1)(x_4^2 + 1)$ can take.",
    "options": [],
    "answer": "See solution",
    "solution": "Using Vieta's formulas, we know the zeros $x_1, x_2, x_3, x_4$ are real and $b - d \\geq 5$. The product can be rewritten as:\n\n$$\n(x_1^2 + 1)(x_2^2 + 1)(x_3^2 + 1)(x_4^2 + 1)\n$$\n\nExpanding, this equals $\\prod_{i=1}^4 (x_i^2 + 1)$. To minimize this, set all $x_i$ equal, say $x_1 = x_2 = x_3 = x_4 = 1$. Then $P(x) = (x - 1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1$, so $a = -4$, $b = 6$, $c = -4$, $d = 1$, and $b - d = 5$ (which meets the condition). The product is $(1^2 + 1)^4 = 2^4 = 16$.\n\nThus, the smallest value is $\\boxed{16}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15365,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\{a_n\\}$ be an ascending sequence of positive integers such that $a_{2n} = 2a_n$ for all $n \\geq 1$ and $a_1 < p$, where $p$ is an odd prime. Prove the following:\n\n1. For any $k$ such that $2^k > p$, among the $p$ numbers $a_{2^k m}, a_{2^k m+1}, \\dots, a_{2^k m+p-1}$, there exists a multiple of $p$.\n2. Construct an explicit example of such a sequence $\\{a_n\\}$ where none of its terms is a multiple of $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $s$ be the minimum value of $a_{n+1} - a_n$ for $n \\geq 1$, and let $m$ be such that $a_{m+1} - a_m = s$. For $k$ with $2^k > p$, we have\n\n$$\n\\begin{aligned}\na_{2^k(m+1)} - a_{2^k m} &= 2^k(a_{m+1} - a_m) = 2^k s.\n\\end{aligned}\n$$\n\nSince $a_{n+1} - a_n \\geq s$ for $2^k m \\leq n < 2^k(m+1)$, it follows that $a_{n+1} - a_n = s$ for all such $n$, so $a_{2^k m}, a_{2^k m+1}, \\dots, a_{2^k(m+1)}$ is an arithmetic progression with increment $s$. If $a_{2^k m+i} \\equiv a_{2^k m+j} \\pmod{p}$ for $0 \\leq i < j < p$, then $(j-i)s$ is a multiple of $p$, which is impossible since $0 < j-i < p$ and $0 < s < p$. Thus, the $p$ numbers $a_{2^k m}, \\dots, a_{2^k m+p-1}$ have distinct residues modulo $p$, so one is a multiple of $p$.\n\nFor part (2), let $k_n$ be the largest integer such that $2^{k_n} \\leq n < 2^{k_n+1}$, and define $a_n = np + 2^{k_n}$. Then $\\{a_n\\}$ is ascending, $a_{2n} = 2a_n$, and since $p$ is odd, $2^{k_n}$ is not a multiple of $p$, so $a_n$ is never a multiple of $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15366,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{N} \\to \\mathbb{N}$ be an injective function such that for all $a, b \\in \\mathbb{N}$,\n\n$$\nf^2(b) f^{f(b)}(a) = f(b+1)^2\n$$\n\nwhere $f^k(x)$ denotes the $k$-th iterate of $f$ at $x$ (i.e., $f^1(x) = f(x)$, $f^2(x) = f(f(x))$, etc.).\n\nDetermine all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed in several steps:\n\n**Step 1:** The pre-images of even integers are non-empty. There must be an integer $d$ such that $f(d) = 4$. Using the given equation with $a = d$, $b = d$:\n\n$$\n2d = f^{f(d)-1}(d) = f^3(d) = f^2(4)\n$$\n\n**Step 2:** Substitute $a = 1$, $b = 4$ into the given equation:\n\n$$\nf^2(4) f^{f(4)}(1) = f(5)^2\n$$\n\nFrom the above and the injectivity of $f$, $f(5)$ must be even, say $2e$. Then:\n\n$$\nf(5) = 2e = f^{f(e)-1}(e)\n$$\n\nSince $e \\neq 1$, $f(e) > 2$ and $5$ is in the image of $f$.\n\n**Step 3:** Show $f(2) \\neq 5$. Suppose $f(2) = 5$. Substitute $n = 2$:\n\n$$\n4 = f^{f(2)-1}(2) = f^4(2) = f^5(1)\n$$\n\nWith $a = 1$, $b = 2$:\n\n$$\nf(3)^2 = f^2(2) f^5(1) = 4f(5) = 8e\n$$\n\nA similar argument shows $3$ is in the image of $f$. Let $g$ be such that $f(g) = 3$, $g \\geq 4$. With $a = 1$, $b = g-1$:\n\n$$\nf^2(g-1) f^{f(g-1)}(1) = f(g)^2 = 9\n$$\n\nSince the pre-image of $1$ is empty, $f^2(g-1) = 3 = f(g) \\implies f(g-1) = g$.\n\nWith $a = 2$, $b = g-2$:\n\n$$\nf^5(g-2) f^{f(g-2)}(2) = f(g)^2 = 9\n$$\n\nFrom previous steps:\n\n$$\nf^g(1) = f^{f(g-1)}(1) = 3 = f^{f(g-2)}(2) = f^{f(g-2)+1}(1)\n$$\n\nSo $f(g-2) = g-1$. Also,\n\n$$\n3 = f^5(g-2) = f^2(3)\n$$\n\nSince $f^{f(3)-1}(3) = 6$, the only possibility is $f(3) = 6$, which leads to $e = \\frac{9}{2}$, a contradiction.\n\n**Step 4:** $f(2) = 3$, $f(3) = 4$.\n\nLet $h$ be the pre-image of $5$, $h \\geq 3$. With $a = 1$, $b = h-1$:\n\n$$\nf^2(h-1) f^{f(h-1)}(1) = f(h)^2 = 25\n$$\n\nSo $f^2(h-1) = 5$ and $f(h-1) = h$. With $a = 2$, $b = h-2$:\n\n$$\nf^{f(2)}(h-2) f^{f(h-2)}(2) = 25\n$$\n\nSo $f^h(1) = f^{f(h-1)}(1) = 5 = f^{f(h-2)}(2) = f^{f(h-2)+1}(1)$, which implies $f(h-2) = h-1$.\n\nAlso,\n\n$$\n5 = f^{f(2)}(h-2) = f^{f(2)-3}(5)\n$$\n\nIf $f(2) \\neq 3$, the sequence $5, f(5), f^2(5), \\dots$ would be finite, but from the given equation, it must contain $5 \\cdot 2^m$ for all $m \\in \\mathbb{N} \\cup \\{0\\}$. Thus, $f(2) = 3$ and $f(3) = 4$.\n\n**Step 5:** $f(n) = n+1$ for all $n \\in \\mathbb{N}$.\n\nWe have shown $f(n) = n+1$ for $n = 1, 2, 3$. Assume true for all $n < k$. If $k$ is odd, $k = 2l+1$, take $a = l$, $b = l+2$:\n\n$$\nf(2l+2)^2 = f^{f(l)}(l+2) f^{f(l+2)}(l) = f^{l+1}(l+2) f^{l+3}(l) = f^{l+1}(l+2)^2 \\\\ \\implies 2l+2 = f^{l+1}(l+2) = f(2l+1)\n$$\n\nIf $k = 2l$, take $a = l$, $b = l+1$:\n\n$$\nf(2l+1)^2 = f^{f(l)}(l+1) f^{f(l+1)}(l) = f^{l+1}(l+1)^2 \\\\ \\implies 2l+1 = f^l(l+1) = f(2l)\n$$\n\nBy induction, $f(n) = n+1$ for all $n \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15367,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the least positive integer which cannot be written in the form $x^3 - x^2 y + y^2 + x - y$, where $x$ and $y$ are positive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $F(x, y) = x^3 - x^2 y + y^2 + x - y$. Note that $F(1, 1) = 1$ and $F(1, 2) = 2$. We shall prove that the equation $F(x, y) = 3$ has no solution in positive integers.\n\nWrite this equation as:\n\n$$\ny^2 - (1 + x^2) y + x^3 + x - 3 = 0.\n$$\n\nIts discriminant with respect to $y$ is:\n\n$$\nD = (1 + x^2)^2 - 4(x^3 + x - 3) = x^4 - 4x^3 + 2x^2 - 4x + 13.\n$$\n\nSince $D < (x^2 - 2x - 1)^2$ for any integer $x \\ge 2$ and $D > (x^2 - 2x - 2)^2$ for any integer $x \\ge 6$, the equation $F(x, y) = 3$ has no positive integer solutions if $x \\ge 6$.\n\nDirect verification shows the same for $x \\in \\{1, 2, 3, 4, 5\\}$.\n\nHence, the least positive integer that cannot be written in the given form is $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15368,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p(x) = b_1x^3 + c_1x^2 + d_1x + e_1$ and $q(x) = b_2x^3 + c_2x^2 + d_2x + e_2$. Let $a = \\frac{r}{s}$, where $r$ and $s$ are relatively prime integers and $s > 0$. Denote $p(a) = m$ and $q(a) = n$.\n\n$$\nb_1 \\cdot \\frac{r^3}{s^3} + c_1 \\cdot \\frac{r^2}{s^2} + d_1 \\cdot \\frac{r}{s} + e_1 = m, \\quad b_2 \\cdot \\frac{r^3}{s^3} + c_2 \\cdot \\frac{r^2}{s^2} + d_2 \\cdot \\frac{r}{s} + e_2 = n.\n$$\n\nMultiplying both identities by $s^3$ gives:\n\n$$\nb_1 r^3 + c_1 r^2 s + d_1 r s^2 + e_1 s^3 = m s^3, \\quad b_2 r^3 + c_2 r^2 s + d_2 r s^2 + e_2 s^3 = n s^3.\n$$\n\nWhat can we conclude about $a$?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $s$ divides both $b_1 r^3$ and $b_2 r^3$, and $r$ and $s$ are relatively prime, $s$ must divide both $b_1$ and $b_2$. Because $b_1$ and $b_2$ are relatively prime, $s = 1$. Therefore, $a = r$ is an integer.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15369,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$ with fixed vertices $B$, $C$, point $A$ moves such that triangle $ABC$ is acute. Let $D$ be the midpoint of $BC$ and $E$, $F$ be the projections of $D$ onto $AB$, $AC$ respectively.\n\n**a)** Let $O$ be the circumcenter of triangle $ABC$. $EF$ meets $AO$ and $BC$ at $M$, $N$ respectively. Prove that the circumcircle of triangle $AMN$ passes through a fixed point.\n\n**b)** Suppose that the tangents to the circumcircle of triangle $AEF$ at $E$, $F$ intersect each other at $T$. Prove that $T$ lies on a fixed line.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Without loss of generality, suppose that $AB < AC$. Clearly, $N$ lies on the opposite ray of $BC$. Note that $AEDF$ is a cyclic quadrilateral and $\\angle OAC = 90^\\circ - \\angle ABC$, we have\n\n$$\n\\begin{align*}\n\\angle AMN &= \\angle MAE + \\angle MEA = 90^\\circ - \\angle ABC + \\angle ADF \\\\\n            &= \\angle BDF + \\angle ADF = \\angle NDA,\n\\end{align*}\n$$\n\nwhich implies $AMDN$ is a cyclic quadrilateral. Thus, $(AMN)$ passes through a fixed point $D$.\n\n![](images/Vietnamese_mathematical_competitions_p134_data_4f6f7d6551.png)\n\nb) Let $(K)$ be the circumcircle of triangle $AEF$, it is obvious that $AD$ is the diameter of $(K)$. Let $L$ be the point on $(K)$ such that $DL \\perp BC$, we can easily prove $AL \\perp BC$ and $AL \\parallel BC$. Note that $D$ is the midpoint of $BC$, hence\n\n$$\nA(FE, DL) = A(BC, DL) = -1.\n$$\n\nThis follows that $LEDF$ is a harmonic quadrilateral. Thus, $DL$ passes through $T$. Clearly, $DL$ is the perpendicular bisector of $BC$ so $T$ lies on a fixed line. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15370,
    "subject": "Mathematics (Olympiad)",
    "question": "A representation of $\\frac{17}{20}$ as a sum of reciprocals\n\n$$\n\\frac{17}{20} = \\frac{1}{a_1} + \\frac{1}{a_2} + \\cdots + \\frac{1}{a_k}\n$$\n\nis called a *calm representation* with $k$ terms if the $a_i$ are distinct positive integers and at most one of them is not a power of two.\n\n(a) Find the smallest value of $k$ for which $\\frac{17}{20}$ has a calm representation with $k$ terms.\n\n(b) Prove that there are infinitely many calm representations of $\\frac{17}{20}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note first that there is no calm representation with 2 terms: if either $a_1$ or $a_2$ is 1 or $a_1 = a_2 = \\frac{1}{2}$, then the sum is greater than $\\frac{17}{20}$. Otherwise, the sum is at most $\\frac{1}{2} + \\frac{1}{3} = \\frac{5}{6} < \\frac{17}{20}$, thus too small.\n\nOn the other hand, there is a representation with 3 terms, namely\n\n$$\n\\frac{17}{20} = \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{10},\n$$\n\nshowing that the smallest possible value of $k$ is 3.\n\nNow we show that there are infinitely many calm representations: take $a_1 = 5 \\cdot 2^{4n+1}$ and consider the difference\n\n$$\n\\frac{17}{20} - \\frac{1}{5 \\cdot 2^{4n+1}} = \\frac{17 \\cdot 2^{4n-1} - 1}{5 \\cdot 2^{4n+1}}.\n$$\n\nSince $2^4 = 16 \\equiv 1 \\mod 5$, the numerator is $17 \\cdot 2^{4n-1} - 1 \\equiv 17 \\cdot 8 - 1 = 135 \\equiv 0 \\mod 5$. Thus the factor 5 cancels, and we have\n\n$$\n\\frac{17}{20} - \\frac{1}{5 \\cdot 2^{4n+1}} = \\frac{A}{2^{4n+1}}\n$$\n\nfor some positive integer $A < 2^{4n+1}$. Since $A$ has a binary representation as $A = 2^{b_1} + 2^{b_2} + \\dots + 2^{b_r}$ with distinct nonnegative integers $b_1, b_2, \\dots, b_r$, we get\n\n$$\n\\frac{17}{20} = \\frac{1}{5 \\cdot 2^{4n+1}} + \\frac{1}{2^{4n+1-b_1}} + \\frac{1}{2^{4n+1-b_2}} + \\dots + \\frac{1}{2^{4n+1-b_r}},\n$$\n\nwhich is a calm representation for every $n$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15371,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime greater than $3$. For a positive integer $k$, let $R(k)$ denote the remainder of $k$ when divided by $p$. Determine all positive integers $a < p$ such that $m + R(ma) > a$ for every $m = 1, 2, \\dots, p-1$.",
    "options": [],
    "answer": "See solution",
    "solution": "The required integers are $p-1$ along with all the numbers of the form $\\lfloor p/q \\rfloor$, where $q = 2, \\dots, p-1$. In other words, these are $p-1$, along with the numbers $1, 2, \\dots, \\lfloor \\sqrt{p} \\rfloor$, and also the (distinct) numbers $\\lfloor p/q \\rfloor$ for $q = 2, \\dots, \\lfloor \\sqrt{p} - 1/2 \\rfloor$.\n\nWe begin by showing that these numbers satisfy the conditions in the statement. It is readily checked that $p-1$ satisfies the required inequalities, since $m + R(m(p-1)) = m + (p-m) = p > p-1$ for all $m = 1, \\dots, p-1$.\n\nNow, consider any number $a$ of the form $a = \\lfloor p/q \\rfloor$, where $q$ is an integer greater than $1$ but less than $p$; then $p = az + r$ with $0 < r < q$. Choose any positive integer $m < p$ and write $m = xq + y$, where $x$ and $y$ are integers, and $0 < y \\le q$ (notice that $x$ is non-negative). Then $R(ma) = R(ay + xaq) = R(ay + xp - xr) = R(ay - xr)$. Since $ay - xr \\le ay \\le az < p$, it follows that $R(ay - xr) \\ge ay - xr$, so $m + R(ma) \\ge (xq + y) + (ay - xr) = x(q - r) + y(a + 1) \\ge a + 1$, by $q > r$ and $y \\ge 1$. Thus $a$ satisfies the required condition.\n\nFinally, we show that if a positive integer $a < p-1$ satisfies the required condition then $a$ is indeed of the form $a = \\lfloor p/q \\rfloor$ for some positive integer $q < p$. This is clear if $a = 1$, so we may (and will) assume that $a \\ge 2$.\n\nWrite $p = az + r$, where $q$ and $r$ are integers, and $0 < r < a$; since $a \\ge 2$, it follows that $q < p/2$. Let $m = q + 1 < p$ to get $R(ma) = R(aq + a) = R(p + (a - r)) = a - r$, so $a < m + R(ma) = q + 1 + a - r$ which yields $r < q + 1$. Moreover, if $r = q$, then $p = q(a + 1)$ which is impossible by $1 < a + 1 < p$. Thus $r < q$, and $0 \\le \\frac{p}{q} - a = \\frac{r}{q} < 1$, which proves $a = \\lfloor p/q \\rfloor$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15372,
    "subject": "Mathematics (Olympiad)",
    "question": "A natural number $n$ is called *interesting* if it can be written as\n$$\nn = \\left\\lfloor \\frac{1}{a} \\right\\rfloor + \\left\\lfloor \\frac{1}{b} \\right\\rfloor + \\left\\lfloor \\frac{1}{c} \\right\\rfloor,\n$$\nwhere $a$, $b$, and $c$ are positive real numbers such that $a + b + c = 1$.\n\nDetermine all interesting numbers. (Here, $[x]$ denotes the floor of the real number $x$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume $a \\leq b \\leq c$.\n\nSince $a + b + c = 1$ and $a, b, c > 0$, we have $a, b, c \\in (0, 1)$. Thus, $\\frac{1}{a}, \\frac{1}{b}, \\frac{1}{c} \\in (1, \\infty)$, so $\\left\\lfloor \\frac{1}{a} \\right\\rfloor, \\left\\lfloor \\frac{1}{b} \\right\\rfloor, \\left\\lfloor \\frac{1}{c} \\right\\rfloor \\geq 1$, and therefore $n \\geq 3$.\n\nIf $\\left\\lfloor \\frac{1}{a} \\right\\rfloor \\leq 2$, then $a > \\frac{1}{3}$, so $a, b, c > \\frac{1}{3}$, and $a + b + c > 1$, which is impossible. Thus, $\\left\\lfloor \\frac{1}{a} \\right\\rfloor \\geq 3$, so $n \\geq 5$.\n\nSuppose $n = 5$ is interesting. Then, possible values are $3 + 1 + 1$, but $b, c > \\frac{1}{1} = 1$, which is not possible since $b, c < 1$.\n\nSuppose $n = 6$ is interesting. Possible values are $3 + 2 + 1$. Then $a \\in (\\frac{1}{4}, \\frac{1}{3}]$, $b \\in (\\frac{1}{3}, \\frac{1}{2}]$, $c \\in (\\frac{1}{2}, 1)$, so $a + b + c > \\frac{1}{4} + \\frac{1}{3} + \\frac{1}{2} = \\frac{13}{12} > 1$, which is impossible.\n\nNow, we show that all $n \\geq 7$ are interesting.\n\nLet $k \\geq 4$ be an integer, and set $a = \\frac{1}{k}$, $b = c = \\frac{k-1}{2k}$. Then $a + b + c = 1$.\n\nWe have $\\left\\lfloor \\frac{1}{a} \\right\\rfloor = k$, $\\left\\lfloor \\frac{1}{b} \\right\\rfloor = \\left\\lfloor \\frac{2k}{k-1} \\right\\rfloor = 2$ (since $k \\geq 4$), so $n = k + 2 + 2 = k + 4$. Thus, all $n \\geq 8$ are interesting.\n\nFor $n = 7$, take $a = \\frac{8}{30}$, $b = c = \\frac{11}{30}$. Then $a + b + c = 1$, $\\left\\lfloor \\frac{1}{a} \\right\\rfloor = 3$, $\\left\\lfloor \\frac{1}{b} \\right\\rfloor = \\left\\lfloor \\frac{30}{11} \\right\\rfloor = 2$, so $n = 3 + 2 + 2 = 7$.\n\n*Conclusion*: The set of interesting numbers is $\\boxed{\\{7, 8, 9, \\ldots\\}}$ (all natural numbers $n \\geq 7$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15373,
    "subject": "Mathematics (Olympiad)",
    "question": "ABCD is a square. M is the midpoint of BC, and P is the midpoint of AP. What is the size of $\\angle CMP$?\n\n![](images/SAMF_ANNUAL_REPORT_2015_BOOK_PROOF_p34_data_4d49cc7063.png)\n\n(A) $22.5^\\circ$\n\n(B) $30^\\circ$\n\n(C) $36^\\circ$\n\n(D) $45^\\circ$\n\n(E) $50^\\circ$",
    "options": [],
    "answer": "See solution",
    "solution": "Joining P to C, we see that $\\angle PCM$ is identical to $\\angle ABM$. This means that P, C, B are vertices of a square, and the required angle is the one between a diagonal of a square and its side, i.e. $45^\\circ$.\n\n![](images/SAMF_ANNUAL_REPORT_2015_BOOK_PROOF_p36_data_e6ee765a63.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15374,
    "subject": "Mathematics (Olympiad)",
    "question": "Express\n$$\n\\sum_{k=0}^{n} (-1)^k (n-k)!(n+k)!\n$$\nin closed form.",
    "options": [],
    "answer": "See solution",
    "solution": "Let\n$$\nf(k) = (n+1-k)!(n+k)!\n$$\nfor integers $0 \\leq k \\leq n + 1$. Note that\n$$\n\\begin{aligned}\nf(k) + f(k + 1) &= (n+1-k)!(n+k)! + (n-k)!(n+k+1)! \\\\\n&= (n + 1 - k + n + k + 1)(n-k)!(n+k)! \\\\\n&= 2(n + 1)(n-k)!(n+k)!.\n\\end{aligned}\n$$\nTherefore,\n$$\n\\begin{aligned}\n\\sum_{k=0}^{n} (-1)^k (n-k)!(n+k)! &= \\frac{1}{2(n+1)} \\sum_{k=0}^{n} (-1)^k [f(k) + f(k+1)] \\\\\n&= \\frac{f(0) + (-1)^n f(n+1)}{2(n+1)} \\\\\n&= \\frac{(n+1)!n! + (-1)^n 0!(2n+1)!}{2(n+1)} \\\\\n&= \\frac{(n!)^2}{2} + \\frac{(-1)^n (2n+1)!}{2(n+1)}.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15375,
    "subject": "Mathematics (Olympiad)",
    "question": "Point $P$ lies inside triangle $ABC$. Let $I_A$, $I_B$, $I_C$ be the incenters of triangles $PBC$, $PAC$, $PAB$ respectively. Let $I_P$ denote the incenter of triangle $I_A I_B I_C$. Prove that for point $P$ satisfying $I_P = P$, the following equalities hold:\n\n$$\nAP - BP = AC - BC, \\quad BP - CP = BA - CA, \\quad CP - AP = CB - AB.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** Let us denote the points $A' = AP \\cap I_B I_C$, $B' = BP \\cap I_A I_C$, $C' = CP \\cap I_B I_A$.\n\n$I_P = P$ implies that $I_A P$ is a bisector of $\\angle BPC$ and $\\angle I_B I_A I_C$, therefore the pairs of lines $I_A I_B$, $I_A I_C$, and also $BP$, $CP$ are symmetric with respect to $I_A P$. Thus, $\\angle (I_A I_B, CP) = \\angle (BP, I_A I_C)$ (oriented angles). By analogy, $\\angle (I_B I_C, AP) = \\angle (CP, I_B I_A)$, $\\angle (I_C I_A, BP) = \\angle (AP, I_C I_B)$, hence,\n$$\n\\angle (I_A I_B, CP) = \\angle (BP, I_A I_C) = -\\angle (AP, I_C I_B) = \\angle (CP, I_B I_A) \\Rightarrow CP \\perp I_B I_A.\n$$\n\nFeet of perpendicular from the incenter to the side of a triangle coincide with the point where the incircle touches this side. Hence, the incircles of $\\triangle PAC$ and $PBC$ touch $CP$ at $C'$. Therefore, they touch each other at $C'$. Analogously, the incircles of $\\triangle PAB$, $\\triangle PBC$ touch $PB$ at $B'$, and $A'$ is the point where the incircles of $\\triangle PAC$, $\\triangle PAB$ touch. Let $A''$ be the point where the incircle of $\\triangle PBC$ touches $BC$. We define $B''$, $C''$ in the same way. Then\n\n![](images/Ukrajina_2010_p20_data_f8e6e09e9e.png)\n\n$$\nAP + BC = PA' + AA' + BA'' + A''C = PB' + AB'' + BB' + B''C = BP + AC.\n$$\n\nFollowing the same lines, we get $AP + BC = CP + AB$. Hence, $P$ is such a point that $AP - BP = AC - BC$, $BP - CP = BA - CA$, $CP - AP = CB - AB$, and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15376,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle. Point $D$ lies on side $BC$, such that the incircles of triangles $ABD$ and $ACD$ are congruent. Let $\\Omega_B$ be the circle with diameter $AB$, and let $\\Omega_C$ be the circle with diameter $AC$. Prove that line $AD$ is perpendicular to one of the common tangents to the circles $\\Omega_B$ and $\\Omega_C$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $AH$ as the altitude of triangle $ABC$. Then clearly $H \\in \\Omega_B, \\Omega_C$, so $AH$ is the common chord of the two circles. Let $EF$ be the common tangent near $A$ of the two circles with $E \\in \\Omega_B$, $F \\in \\Omega_C$. According to a familiar property, $AH$ passes through the midpoint $K$ of $EF$. Draw $Ax \\parallel EF$; then $A(Kx, EF) = -1$.\n\nWe redefine $D$ as a point on $BC$ such that $AD \\perp EF$. Let $I, J$ be the incenters of $ABD, ACD$ respectively, and let $AI, AJ$ intersect $BC$ at $M, N$ respectively. Let $R, S$ be the centers of the circles $\\Omega_B, \\Omega_C$ (which are also the midpoints of $AB, AC$), then $RE \\parallel AM$ (because they are both perpendicular to $EF$), so\n\n$$\n\\angle ABE = \\frac{1}{2}\\angle ARE = \\frac{1}{2}\\angle DAB = \\angle MAB,\n$$\n\nwhich implies $BE \\parallel AM$ or $AE \\perp AM$. Similarly, $AF \\perp AN$.\n\n![](images/Saudi_Booklet_2025_p21_data_7b70bc8564.png)\n\nNow draw $Ay \\parallel MN$; then $AK \\perp Ay$. From there, according to the orthogonality property of two harmonic bundles, we immediately have $A(Dy, MN) = -1$. Thus, $DM = DN$. Then, according to the property of the angle bisector,\n\n$$\n\\frac{IM}{IA} = \\frac{DM}{DA} = \\frac{DN}{DA} = \\frac{JN}{JA} \\implies IJ \\parallel MN.\n$$\n\nTherefore, the distances from $I, J$ to the line $BC$ are equal, which implies that the radius of the circle inscribed in triangles $ABD, ACD$ are equal. So in other words, $D$ is the given point in the problem. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15377,
    "subject": "Mathematics (Olympiad)",
    "question": "![](images/Slovenija_2011_p10_data_a52818a807.png)\n\n(a) Peter has 111 blue and 111 red marbles. He can exchange 20 blue marbles for 28 red marbles, or 11 red marbles for 7 blue marbles. What is the largest number of marbles Peter can obtain by performing these exchanges any number of times?\n\n(b) Can Peter increase the total number of marbles by 33?\n\n(c) Can Peter ever have three times as many blue marbles as red marbles?",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Peter can increase the number of marbles by 20. The first three times he should exchange 20 blue marbles for 28 red ones. He will then have $111 - 60 = 51$ blue marbles and $111 + 3 \\cdot 28 = 111 + 84 = 195$ red ones. Then, he should exchange 11 red marbles for 7 blue ones. He will end up with 184 red and 58 blue marbles. He will have 242 marbles altogether, which is 20 more than the initial 222.\n\n(b) Exchanging 11 red marbles for 7 blue marbles decreases the total amount of marbles by 4. Exchanging 20 blue marbles for 28 red marbles increases the total amount of marbles by 8. The initial amount of marbles is even, so the total will remain even after any number of exchanges. Hence, the total number of marbles cannot be increased by 33 since this number is odd.\n\n(c) Assume that this can be done. Let $x$ denote the number of red marbles. Then there are $3x$ blue marbles and the total number of marbles is $4x$. After each exchange the number of marbles increases or decreases by a multiple of 4. Since there were 222 marbles at the beginning, only the numbers of the form $222 + 4k$ can be reached. The equation $222 + 4k = 4x$ has no integer solutions since 222 is not divisible by 4. Hence, Peter can never have three times as many blue marbles as red.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15378,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to [0, \\infty)$ such that, for all real $a, b, c, d$ satisfying $ab + bc + cd = 0$, the following equality holds:\n\n$$\nf(a - b) + f(c - d) = f(a) + f(b + c) + f(d)\n$$\n\n(Here $\\mathbb{R}$ is the set of real numbers and $[0, \\infty)$ is the set of nonnegative real numbers.)",
    "options": [],
    "answer": "See solution",
    "solution": "We prove the following:\n\n*Lemma*: For all real $p, q, r$ that satisfy $p^2 + q^2 = r^2$, the following equality holds:\n\n$$\nf(p) + f(q) = f(r)\n$$\n\n*Proof*: Put $a = \\frac{p-q+r}{2}$, $b = \\frac{p-q-r}{2}$, $c = \\frac{p+q+r}{2}$. Then $ab + bc + cd = \\frac{1}{2}(p^2 + q^2 - r^2)$. So if $p^2 + q^2 = r^2$, then $ab + bc + cd = 0$ and we have\n\n$$\nf(r) + f\\left(\\frac{p-q-r}{2}\\right) = f\\left(\\frac{p-q-r}{2}\\right) + f(p) + f(q)\n$$\n\nThus $f(p) + f(q) = f(r)$ and the proof is complete.\n\nFor $(p, q, r) = (0, 0, 0)$, by the lemma, $f(0) = 0$. For $(p, q, r) = (p, 0, -p)$, we have $f(-p) = f(p)$, so $f$ is even.\n\nNow, for any $t \\ge 0$, define $g : [0, \\infty) \\to [0, \\infty)$ by $g(t) := f(\\sqrt{t})$. Then $g(a + b) = g(a) + g(b)$. For $a \\ge b \\ge 0$, $g(a) = g(a - b) + g(b) \\ge g(b)$, so $g$ is monotone increasing. Thus $g(x) = g(1) \\cdot x$ for $x \\ge 0$. Therefore, $f(x) = f(1) \\cdot x^2$ for $x \\ge 0$, and since $f$ is even, $f(x) = f(1) \\cdot x^2$ for all real $x$. Since $f(x) \\ge 0$, $f(1) \\ge 0$.\n\nThus, the functions satisfying the given equality are of the form $f(x) = \\lambda x^2$ with $\\lambda \\ge 0$, and it is easy to check that these functions satisfy the given equality. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15379,
    "subject": "Mathematics (Olympiad)",
    "question": "Let a cube have side $k$.\n\n$A$ and $B$ are two adjacent vertices of a small octahedron, with $AX = BX = k/2$ and $\\angle AXB = 90^\\circ$, so $AB = k/\\sqrt{2}$.\n\n![](images/obm-book_p56_data_50109f929f.png)\n\nThe large octahedron has its vertices at the centers of the cube's faces.\n\nFind the ratio of the volume of the large octahedron to that of the small octahedron.",
    "options": [],
    "answer": "See solution",
    "solution": "The line joining the centers of $OPS$ and $OQR$ is parallel to $PQ$ and $\\frac{2}{3}$ the length. But it is also $k\\sqrt{2}$, so the side of the large octahedron is $\\frac{3}{2}k\\sqrt{2} = \\frac{3k}{\\sqrt{2}}$, or 3 times the side of the small octahedron. Hence, the volume of the large octahedron is $3^3 = 27$ times the volume of the small octahedron.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 15380,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a scalene triangle with incentre $I$ and circumcircle $\\omega$. The lines $AI$, $BI$, $CI$ intersect $\\omega$ for the second time at the points $D$, $E$, $F$, respectively. The lines through $I$ parallel to the sides $BC$, $AC$, $AB$ intersect the lines $EF$, $DF$, $DE$ at the points $K$, $L$, $M$, respectively. Prove that the points $K$, $L$, $M$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we will prove that $KA$ is tangent to $\\omega$.\n\nIndeed, it is a well-known fact that $FA = FB = FI$ and $EA = EC = EI$, so $FE$ is the perpendicular bisector of $AI$. It follows that $KA = KI$ and\n\n$$\n\\angle KAF = \\angle KIF = \\angle FCB = \\angle FEB = \\angle FEA,\n$$\n\nso $KA$ is tangent to $\\omega$. Similarly, we can prove that $LB$, $MC$ are tangent to $\\omega$ as well.\n\nLet $A'$, $B'$, $C'$ be the intersections of $AI$, $BI$, $CI$ with $BC$, $CA$, $AB$ respectively. From Pascal's Theorem on the cyclic hexagon $AACDEB$, we get $K$, $C'$, $B'$ collinear. Similarly, $L$, $C'$, $A'$ are collinear and $M$, $B'$, $A'$ are collinear.\n\nThen, from Desargues' Theorem for $\\triangle DEF$, $\\triangle A'B'C'$, which are perspective from the point $I$, we get $K$, $L$, $M$ (the intersections of their corresponding sides) are collinear as wanted.\n\n*Remark (P.S.C.):* After proving that $KA$, $LB$, $MC$ are tangent to $\\omega$, we can argue as follows:\n\nIt readily follows that $\\triangle KAF \\sim \\triangle KAE$ and so $\\frac{KA}{KE} = \\frac{KF}{KA} = \\frac{AF}{AE}$, thus $\\frac{KF}{KE} = \\left(\\frac{AF}{AE}\\right)^2$. In a similar way, we can find that $\\frac{ME}{MD} = \\left(\\frac{CE}{CD}\\right)^2$ and $\\frac{LD}{LF} = \\left(\\frac{BD}{BF}\\right)^2$. Multiplying, we obtain $\\frac{KF}{KE} \\cdot \\frac{ME}{MD} \\cdot \\frac{LD}{LF} = 1$, so by the converse of Menelaus' theorem applied in the triangle $DEF$, we get that the points $K$, $L$, $M$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15381,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O_1$ be the center of $\\Gamma_1$ and $R_1$ its radius. Let $O_2$ be the center of $\\Gamma_2$ and $R_2$ its radius. Homothety with center at $A$ and positive coefficient transforms $\\Gamma_1$ to $\\omega_1$. Homothety with center at $B$ and negative coefficient transforms $\\omega_1$ to $\\Gamma_2$. Let $Z$ be the center of the homothety with negative coefficient which transforms $\\Gamma_1$ to $\\Gamma_2$. By the theorem of three homotheties, $Z$ lies on the line $AB$. Similarly, $Z$ lies on the line $CD$. It is evident that $Z$ lies on the segment $O_1O_2$ and $O_1Z : ZO_2 = R_1 : R_2$.\n\n![alt](images/Blr-2014_p27_data_3746ed10d6.png)\n\n![alt](images/Blr-2014_p27_data_33f5db90e3.png)\n\nShow that $ZB \\cdot ZA = ZC \\cdot ZD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $E$ be the intersection point (different from $D$) of the line $CD$ and $\\Gamma_1$. The power of point $Z$ with respect to $\\Gamma_1$ is $R_1^2 - O_1Z^2$, and also $ZC \\cdot ZE$. By homothety properties, $EZ = \\frac{R_1}{R_2} ZD$. Thus,\n\n$$\nZC \\cdot ZE = ZC \\cdot \\frac{R_1}{R_2} ZD = R_1^2 - O_1Z^2,\n$$\nso\n$$\nZC \\cdot ZD = \\frac{R_2}{R_1}(R_1^2 - O_1Z^2).\n$$\nSince $O_1Z = \\frac{R_1}{R_1 + R_2} O_1O_2$, we have\n$$\nZC \\cdot ZD = \\frac{R_2}{R_1} \\left( R_1^2 - R_1^2 \\frac{O_1O_2^2}{(R_1 + R_2)^2} \\right) = R_1 R_2 \\left( 1 - \\frac{O_1O_2^2}{(R_1 + R_2)^2} \\right).\n$$\nSimilarly,\n$$\nZB \\cdot ZA = R_1 R_2 \\left( 1 - \\frac{O_1O_2^2}{(R_1 + R_2)^2} \\right).\n$$\nTherefore, if $A$, $B$, $C$, and $D$ do not lie on the same line, then they lie on the same circle.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15382,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be an isosceles trapezoid with $AB \\parallel CD$. Let $E$ be the midpoint of $AC$. Denote by $\\omega$ and $\\Omega$ the circumcircles of the triangles $ABE$ and $CDE$, respectively. Let $P$ be the intersection point of the tangent to $\\omega$ at $A$ with the tangent to $\\Omega$ at $D$. Prove that $PE$ is tangent to $\\Omega$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $ABCD$ is a rectangle, the statement is trivial due to symmetry. Hence, in what follows we assume $AD \\parallel BC$.\n\nLet $F$ be the midpoint of $BD$; by symmetry, both $\\omega$ and $\\Omega$ pass through $F$. Let $P'$ be the meeting point of tangents to $\\omega$ at $F$ and to $\\Omega$ at $E$. We aim to show that $P' = P$, which yields the required result. For that purpose, we show that $P'A$ and $P'D$ are tangent to $\\omega$ and $\\Omega$, respectively.\n\nLet $K$ be the midpoint of $AF$. Then $EK$ is a midline in the triangle $ACF$, so $\\angle(AE, EK) = \\angle(EC, CF)$. Since $P'E$ is tangent to $\\Omega$, we get $\\angle(EC, CF) = \\angle(P'E, EF)$. Thus, $\\angle(AE, EK) = \\angle(P'E, EF)$, so $EP'$ is a symmedian in the triangle $AEF$. Therefore, $EP'$ and the tangents to $\\omega$ at $A$ and $F$ are concurrent, and the concurrency point is $P'$ itself. Hence $P'A$ is tangent to $\\omega$.\n\nThe second claim is similar. Taking $L$ to be the midpoint of $DE$, we have $\\angle(DF, FL) = \\angle(FB, BE) = \\angle(P'F, FE)$, so $P'F$ is a symmedian in the triangle $DEF$, and hence $P'$ is the meeting point of the tangents to $\\Omega$ at $D$ and $E$.\n\n![](images/RMC_2019_var_3_p90_data_9df28f3539.png)\n\n**Remark.** The above arguments may come in different orders. E.g., one may define $P'$ to be the point of intersection of the tangents to $\\Omega$ at $D$ and $E$ — hence obtaining that $P'F$ is a symmedian in $\\triangle DEF$, then deduce that $P'F$ is tangent to $\\omega$, and then apply a similar argument to show that $P'E$ is a symmedian in $\\triangle AEF$, whence $P'A$ is tangent to $\\omega$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15383,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Let $(a_n)$ be a sequence defined by\n\n$$\na_n = \\ln(2n^2 + 1) - \\ln(n^2 + n + 1), \\quad \\forall n > 0.\n$$\n\nProve that there are finitely many values of $n$ such that $\\{a_n\\} < \\frac{1}{2}$.\n\nb) Let $(b_n)$ be a sequence defined by $b_n = \\ln(2n^2 + 1) + \\ln(n^2 + n + 1)$, $\\forall n > 0$. Prove that there are infinitely many values of $n$ such that $\\{b_n\\} < \\frac{1}{2016}$.\n\nNote: $\\{x\\}$ denotes the fractional part of the real number $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) It is clear that $1 \\le \\frac{2n^2 + 1}{n^2 + n + 1} \\le 2$ for all $n \\in \\mathbb{Z}^+$. Hence, $0 \\le a_n \\le \\ln 2 < 1$ and $\\lfloor a_n \\rfloor = 0$. Therefore, $\\{a_n\\} = a_n$ and\n\n$$\n\\lim_{n \\to \\infty} \\{a_n\\} = \\lim_{n \\to \\infty} a_n = \\lim_{n \\to \\infty} \\ln \\frac{2n^2 + 1}{n^2 + n + 1} = \\ln 2.\n$$\n\nThus, there exists a positive integer $n_0$ such that\n\n$$\n\\{a_n\\} > \\ln 2 - \\frac{1}{2016}, \\quad \\forall n \\ge n_0.\n$$\n\nBy direct calculation, for $n > n_0$,\n\n$$\n\\{a_n\\} = a_n > \\ln 2 - \\frac{1}{2016} > \\frac{1}{2}.\n$$\n\nb) The sequence $(b_n)$ is increasing and $\\lim_{n \\to \\infty} b_n = +\\infty$. Also,\n\n$$\n\\lim_{n \\to \\infty} (b_n - b_{n-1}) = \\lim_{n \\to \\infty} \\ln \\frac{(2n^2 + 1)(n^2 + n + 1)}{(2n^2 - 4n + 3)(n^2 - n + 1)} = 0.\n$$\n\nSuppose there are only finitely many $n$ such that $\\{b_n\\} < \\frac{1}{2016}$. Then there exists $n_0$ such that $\\{b_n\\} \\ge \\frac{1}{2016}$ for all $n \\ge n_0$. Since $\\lim_{n \\to \\infty} (b_n - b_{n-1}) = 0$, there exists $n_1$ such that\n\n$$\nb_n - b_{n-1} < \\frac{1}{2016}\n$$\n\nfor all $n \\ge n_1$. Because $(b_n)$ is increasing and unbounded, there are infinitely many $n > \\max\\{n_0, n_1\\}$ with $[b_n] - [b_{n-1}] = 1$. For such $n$,\n\n$$\n[b_n] - [b_{n-1}] + \\{b_n\\} - \\{b_{n-1}\\} < \\frac{1}{2016}\n$$\n\nwhich implies\n\n$$\n\\{b_{n-1}\\} > \\{b_n\\} + \\frac{2015}{2016}.\n$$\n\nBut since $\\{b_n\\} \\ge \\frac{1}{2016}$, this would require $\\{b_{n-1}\\} > 1$, a contradiction. Therefore, there are infinitely many $n$ such that $\\{b_n\\} < \\frac{1}{2016}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15384,
    "subject": "Mathematics (Olympiad)",
    "question": "Each of 2009 distinct points in the plane is colored in red or blue, such that on every blue-centered unit circle there are exactly two red points. Find the greatest possible number of blue points.",
    "options": [],
    "answer": "See solution",
    "solution": "Each pair of red points can belong to at most two blue-centered unit circles. As $n$ red points form $\\frac{n(n-1)}{2}$ pairs, we have no more than twice that number of blue points, i.e. $n(n-1)$ blue points. Thus, the total number of points cannot exceed $n + n(n-1) = n^2$.\n\nAs $44^2 < 2009$, $n$ must be at least 45. We can arrange 45 distinct red points on a segment of length 1, and color blue all but 16 ($=45^2 - 2009$) points on intersections of the red-centered unit circles (all points of intersection are distinct, as no blue-centered unit circle can intersect the segment more than twice). Thus, the greatest possible number of blue points is $2009 - 45 = 1964$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15385,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the system\n\n$$\n\\begin{cases}\n\\sqrt{x^2 + y^2 - 16(x + y) - 9y + 7} = y - 2 \\\\\nx + 13\\sqrt[4]{x - y} = y + 42\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Rewrite the second equation as:\n\n$$x - y + 13\\sqrt[4]{x - y} - 42 = 0$$\n\nLet $u = \\sqrt[4]{x-y}$, with $u \\ge 0$. Then:\n\n$$u^4 + 13u - 42 = 0$$\n\nThis equation has only one positive root, $u = 2$, so $x = y + 16$.\n\nSubstitute into the first equation to get:\n\n$$y^2 - 5y + 3 = 0$$\n\nThe solutions are $y_{1,2} = \\frac{5 \\pm \\sqrt{13}}{2}$. Since $y - 2 \\ge 0$ and $\\frac{5 - \\sqrt{13}}{2} < 2$, it follows that:\n\n$$\nx = \\frac{37 + \\sqrt{13}}{2}, \\quad y = \\frac{5 + \\sqrt{13}}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15386,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ and $B$ be the first and last two digits of a number $n$, respectively. Determine all pairs $(A, B)$ such that the two-digit number $AB$ divides $100A + B$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $AB$ divides $100A + B$, let $A$ and $B$ be two-digit numbers. Since $A$ divides $100A + B$, $A$ must divide $B$. Let $k = \\frac{B}{A}$. Then $10 \\leq A < \\frac{100}{k}$.\n\nThe condition is equivalent to $kA^2 \\mid 100A + kA$, which simplifies to $kA \\mid 100 + k$. Thus, $k$ divides $100 + k$, so $k$ divides $100$. With $k < 10$, possible values are $k = 1, 2, 4, 5$.\n\nFrom $A = \\frac{100 + k}{k}$ and $10 \\leq A < \\frac{100}{k}$, we get $(k, A) = (2, 17), (4, 13)$. Therefore, the possible numbers are $n = 1734$ and $n = 1352$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15387,
    "subject": "Mathematics (Olympiad)",
    "question": "The product\n\n$$\n\\prod_{k=4}^{63} \\frac{\\log_k (5^{k^2-1})}{\\log_{k+1} (5^{k^2-4})}\n$$\n\nis equal to $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Using properties of logarithms, the product can be written as\n\n$$\n\\prod_{k=4}^{63} \\frac{k^2 - 1}{k^2 - 4} \\cdot \\prod_{k=4}^{63} \\frac{\\log_k 5}{\\log_{k+1} 5}\n$$\n\nThe first factor is\n\n$$\n\\prod_{k=4}^{63} \\frac{k^2 - 1}{k^2 - 4} = \\prod_{k=4}^{63} \\frac{(k-1)(k+1)}{(k-2)(k+2)} = \\frac{5 \\cdot 62}{2 \\cdot 65}\n$$\n\nThe second factor is\n\n$$\n\\prod_{k=4}^{63} \\frac{\\log_k 5}{\\log_{k+1} 5} = \\frac{\\log_4 5}{\\log_{64} 5} = 3\n$$\n\nThus, the required product is $\\frac{3 \\cdot 5 \\cdot 62}{2 \\cdot 65} = \\frac{93}{13}$. The requested sum is $93 + 13 = 106$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15388,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all triples $ (k, m, n) $ of positive integers satisfying\n\n$$\nk! + m! = k! \\times n!.\n$$\n\n(If $n$ is a positive integer, then $n! = 1 \\times 2 \\times 3 \\times \\dots \\times (n-1) \\times n$.)",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $ (r, r, 2) $ for $ r \\ge 1 $, and $ (r! - 2, r! - 1, r) $ for $ r \\ge 3 $.\n\nThe equation is equivalent to\n\n$$\nm! = k!(n! - 1).\n$$\n\nIf $ n = 1 $, there are no solutions.\n\nIf $ n = 2 $, we get $ m = k $, which yields the family $ (r, r, 2) $.\n\nIf $ n \\ge 3 $, then $ m > k $.\n\n- If $ m = k + 1 $, we get $ k + 1 = n! - 1 $, which yields the family $ (r! - 2, r! - 1, r) $.\n- If $ m \\ge k + 2 $, we get $ (k + 1)(k + 2) \\cdots m = n! - 1 $, which is a contradiction because the left-hand side is even and the right-hand side is odd.\n\nSo we have the two solution families: $ (r, r, 2) $ for $ r \\ge 1 $ and $ (r! - 2, r! - 1, r) $ for $ r \\ge 3 $.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15389,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AB\\Gamma$ be an acute-angled triangle inscribed in the circle $c(O, R)$ with $AB < A\\Gamma < B\\Gamma$. Let $\\Delta$, $E$, and $Z$ be the points where the incircle of the triangle touches the sides $B\\Gamma$, $A\\Gamma$, and $AB$, respectively. The circumcircle of triangle $AEZ$ (denoted $(c_1)$) intersects the circle $(c)$ at point $A'$. The circumcircle of triangle $B\\Delta Z$ (denoted $(c_2)$) intersects the circle $(c)$ at point $B'$. The circumcircle of triangle $\\Gamma\\Delta E$ (denoted $(c_3)$) intersects the circle $(c)$ at point $\\Gamma'$. Prove that:\n\n(α) The quadrilateral $\\Delta EA'B'$ is cyclic.\n\n(β) The lines $\\Delta A'$, $EB'$, and $Z\\Gamma'$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "From the inscribed quadrilateral $AA'IZ$ in the circle $(c_1)$, we have:\n\n$$\n\\angle AA'I = \\angle AZI = 90^\\circ.\n$$\n\nFrom the inscribed quadrilateral $\\Gamma\\Delta IE$ (since $\\Gamma I$ is an angle bisector),\n\n$$\n\\angle \\Delta_1 = \\frac{\\angle \\Gamma}{2} \\qquad (1)\n$$\n\nFrom the inscribed quadrilateral $B\\Delta IZ$ (since $BI$ is an angle bisector),\n\n$$\n\\angle \\Delta_2 = \\frac{\\angle B}{2} \\qquad (2)\n$$\n\nFrom the inscribed quadrilateral $B\\Delta IB'$, we have $\\angle \\Delta_3 = \\angle B_1$.\n\nFrom the inscribed quadrilateral $BB'A'A$, we have $\\angle B_1 = \\angle A'_1 = 90^\\circ - \\angle A'_2$.\n\nHence, $\\angle \\Delta_3 + \\angle A'_2 = 90^\\circ$.\n\nFrom the inscribed quadrilateral $AEIA'$, we have $\\angle A'_3 = \\frac{\\angle A}{2}$.\n\nTherefore,\n\n$$\n\\angle \\Delta_1 + \\angle \\Delta_2 + \\angle \\Delta_3 + \\angle \\Delta'_2 + \\angle \\Delta'_3 = 180^\\circ\n$$\n\nThus, the quadrilateral $A'E\\Delta B'$ is cyclic. Similarly, we can prove that the quadrilaterals $\\Delta ZA'\\Gamma'$ and $ZE\\Gamma'B'$ are cyclic.\n\nFinally, the lines $\\Delta A'$, $EB'$, and $Z\\Gamma'$ are concurrent at the radical center of the three circles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15390,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N$ be a positive integer. Prove that there exists a positive integer $n$ such that $n^{2013} - n^{20} + n^{13} - 2013$ has at least $N$ distinct prime factors.",
    "options": [],
    "answer": "See solution",
    "solution": "The result holds for any nonconstant polynomial $f(n) = a_m n^m + a_{m-1} n^{m-1} + \\dots + a_0$ with integer coefficients. Assume $a_m > 0$. Then there exists a positive integer $n_0$ such that $f(n)$ is positive and increasing for $n > n_0$.\n\nIt suffices to show that if for some $n_1 > n_0$, $f(n_1) = p_1^{r_1} \\cdots p_k^{r_k}$ has exactly $k$ distinct prime factors, then for some $n_2 > n_1$, $f(n_2)$ has more than $k$ prime factors. Given such $n_1$, let $n_2 = n_1 + p_1^{r_1+1} \\cdots p_k^{r_k+1}$. Then\n\n$$\nf(n_2) \\equiv p_1^{r_1} \\cdots p_k^{r_k} \\pmod{p_1^{r_1+1} \\cdots p_k^{r_k+1}}\n$$\n\nHence, for each $j$, $1 \\leq j \\leq k$, $p_j^{r_j}$ divides $f(n_2)$ but $p_j^{r_j+1}$ does not. Since $f(n_2) > f(n_1) = p_1^{r_1} \\cdots p_k^{r_k}$, $f(n_2)$ must have at least $k+1$ prime factors.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15391,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, and $z$ be real numbers satisfying the system:\n\n$$\n\\frac{1}{2} = \\frac{1}{x} + \\frac{1}{y}, \\quad \\frac{1}{3} = \\frac{1}{y} + \\frac{1}{z}, \\quad \\frac{1}{4} = \\frac{1}{z} + \\frac{1}{x}.\n$$\n\nFind the value of $5x + 7y + 9z$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $X = \\frac{1}{x}$, $Y = \\frac{1}{y}$, $Z = \\frac{1}{z}$. Then:\n\n$$\nX + Y = \\frac{1}{2}, \\quad Y + Z = \\frac{1}{3}, \\quad Z + X = \\frac{1}{4}.\n$$\n\nAdding these equations:\n\n$$\n(X + Y) + (Y + Z) + (Z + X) = \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4}\n$$\n$$\n2(X + Y + Z) = \\frac{13}{12} \\implies X + Y + Z = \\frac{13}{24}\n$$\n\nNow,\n\n$$\nX = (X + Y + Z) - (Y + Z) = \\frac{13}{24} - \\frac{1}{3} = \\frac{5}{24}\n$$\n$$\nY = (X + Y + Z) - (Z + X) = \\frac{13}{24} - \\frac{1}{4} = \\frac{7}{24}\n$$\n$$\nZ = (X + Y + Z) - (X + Y) = \\frac{13}{24} - \\frac{1}{2} = \\frac{1}{24}\n$$\n\nThus,\n\n$$\nx = \\frac{1}{X} = \\frac{24}{5}, \\quad y = \\frac{1}{Y} = \\frac{24}{7}, \\quad z = \\frac{1}{Z} = 24\n$$\n\nTherefore,\n\n$$\n5x + 7y + 9z = 5 \\cdot \\frac{24}{5} + 7 \\cdot \\frac{24}{7} + 9 \\cdot 24 = 24 + 24 + 216 = 264.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15392,
    "subject": "Mathematics (Olympiad)",
    "question": "Compute all four-digit numbers among $1111, 2222, 3333, \\ldots, 9999$ that can be written as the sum of the squares of three consecutive odd numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the three consecutive odd numbers be $x$, $x+2$, and $x+4$. Their squares sum to:\n\n$$\nS = x^2 + (x+2)^2 + (x+4)^2 = 3x^2 + 12x + 20.\n$$\n\nSince $x$ is odd, $3x^2$ is odd, and $12x$ and $20$ are even, so $S$ is odd. Also, $S = 3(x^2 + 4x + 6) + 2$, so $S \\equiv 2 \\pmod{3}$. Among $1111, 2222, \\ldots, 9999$, only $5555$ is both odd and leaves remainder $2$ when divided by $3$. Setting $3x^2 + 12x + 20 = 5555$ gives $x = 41$. Thus, $41^2 + 43^2 + 45^2 = 5555$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15393,
    "subject": "Mathematics (Olympiad)",
    "question": "Докажите, что сумма девяти натуральных чисел, составленных из всех цифр от 1 до 9 (каждая цифра используется ровно один раз в каждом числе), не может оканчиваться на девять нулей. Можно ли подобрать такие числа, чтобы их сумма оканчивалась на восемь нулей?",
    "options": [],
    "answer": "See solution",
    "solution": "Покажем, что сумма не может оканчиваться на 9 нулей. Каждое из составленных чисел делится на 9, поскольку сумма его цифр делится на 9. Поэтому их сумма также делится на 9. Наименьшее натуральное число, делящееся на 9 и оканчивающееся на девять нулей, равно $9 \\cdot 10^9$, так что сумма наших чисел не меньше $9 \\cdot 10^9$. Значит, одно из них не меньше $10^9$, что невозможно.\n\nОсталось показать, как составить числа, сумма которых оканчивается на восемь нулей. Например, можно взять восемь чисел, равных $987654321$, и одно число $198765432$. Их сумма равна $81 \\cdot 10^8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15394,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a sequence $a_0, a_1, a_2, \\ldots$ defined by the following rules:\n\n1. $a_0$ is a positive integer.\n2. For each $n \\geq 0$, $a_{n+1}$ is the smallest integer greater than $a_n$ that is coprime to all previous terms $a_0, a_1, \\ldots, a_n$.\n\nFor which starting values $a_0$ does the sequence eventually consist only of prime numbers?",
    "options": [],
    "answer": "See solution",
    "solution": "We first show that if:\n\n1. $a_n$ is prime,\n2. all terms $a_i$ for $i < n$ are primes or prime powers,\n3. for any prime $p < a_n$, $p$ divides some $a_i$ with $i < n$,\n\nthen $a_i$ is prime for all $i > n$.\n\nSuppose $a_n < k < q$, where $q$ is the next larger prime than $a_n$. Then $k$ is composite and has a prime factor less than $a_n$, which by (3) divides some $a_i$ for $i < n$, so $a_{n+1} \\neq k$. On the other hand, $q$ is coprime to all $a_i$ for $i \\leq n$, so $a_{n+1} = q$.\n\nExamples:\n\n* $a_0 = 2$ gives all terms primes.\n* $a_0 = 3$ gives $a_1 = 4$, $a_2 = 5$, and all higher terms are prime.\n* $a_0 = 4$ gives $a_1 = 5$, $a_2 = 7$, $a_3 = 9$, $a_4 = 11$, and all higher terms are prime.\n* $a_0 = 5$ gives $a_1 = 6$, which is not a prime or prime power.\n* $a_0 = 6$ is not a prime or prime power.\n* $a_0 = 7$ gives $a_1 = 8$, $a_2 = 9$, $a_3 = 11$, $a_4 = 13$, $a_5 = 17$, $a_6 = 19$, $a_7 = 23$, $a_8 = 25$, $a_9 = 29$, and all higher terms are prime.\n* $a_0 = 8$ gives $a_1 = 9$, $a_2 = 11$, $a_3 = 13$, $a_4 = 17$, $a_5 = 19$, $a_6 = 23$, $a_7 = 25$, $a_8 = 29$, $a_9 = 31$, $a_{10} = 37$, $a_{11} = 41$, $a_{12} = 43$, $a_{13} = 47$, $a_{14} = 49$, $a_{15} = 53$, and all higher terms are prime.\n* $a_0 = 9$ gives $a_1 = 10$, which is not a prime or prime power.\n\nNow suppose $a_0 > 9$. We show there are no solutions to $3^k = 2^h \\pm 1$ for $k > 2$:\n\n- For $3^k = 2^h + 1$, if $k > 2$, then $h > 3$, so $3^k - 1$ is divisible by 4, so $k$ is even. Let $k = 2m$. Then $2^h = (3^m + 1)(3^m - 1)$, but $3^m - 1$ and $3^m + 1$ cannot both be powers of 2 for $m > 1$.\n- For $3^k = 2^h - 1$, $h$ must be even. Let $h = 2m$, then $3^k = (2^m - 1)(2^m + 1)$, but $2^m - 1$ and $2^m + 1$ cannot both be powers of 3 for $m > 1$.\n\nThus, for $a_0 = 6n$, $a_0$ is not a prime power. For $a_0 = 6n - 1$, $a_1 = 6n$, which is not a prime power. For $a_0 = 6n + 1$, $a_1 = 6n + 2$, which must be a power of 2, but $a_2 = 6n + 3$ must be a power of 3, which is impossible for $a_0 > 9$.\n\nSimilarly, for $a_0 = 6n + 2$, $a_0$ must be a power of 2 and $a_1 = 6n + 3$ a power of 3, which is impossible for $a_0 > 9$. For $a_0 = 6n + 3$, $a_1 = 6n + 4$, and again they cannot both be prime powers.\n\nWe also claim there are no solutions to $3^k = 2^h + 5$ for $k > 2$. Setting $k = 2K$, $h = 2H$, we get $(3^K + 2^H)(3^K - 2^H) = 5$, which is impossible for $K > 2$. Thus, for $a_0 = 6n - 2$, $a_1 = 6n - 1$, $a_2 = 6n + 1$, $a_3 = 6n + 3$, but $a_0$ must be a power of 2 and $a_3$ a power of 3, which is impossible for $a_0 > 9$.\n\nTherefore, the only starting values $a_0$ for which the sequence eventually consists only of primes are $a_0 = 2, 3, 4, 7, 8$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15395,
    "subject": "Mathematics (Olympiad)",
    "question": "Cada 20 minutos durante una semana se transvasa una cantidad exacta de litros de agua (siempre la misma cantidad) desde un tanque con $25{,}000$ litros a otro depósito inicialmente vacío. Desde este segundo depósito, a intervalos regulares de tiempo, se extrae primero $1$ litro, luego $2$, luego $3$, etc. Justo al final de la semana coinciden el último transvase y la última extracción, quedando en ese momento vacío el segundo depósito. Determina cuánta agua se ha extraído en total durante la semana, en caso de que los datos del problema lo permitan. (Se supone que los transvases y las extracciones se realizan instantáneamente. El primer transvase se hace pasados los primeros $20$ minutos y la primera extracción, pasado el primer intervalo de tiempo.)",
    "options": [],
    "answer": "See solution",
    "solution": "Sea $n$ el número de extracciones de agua realizadas durante la semana. En total se habrá extraído $$T_n = 1 + 2 + \\dots + n = \\frac{n(n+1)}{2}$$ litros. Por otro lado, si el caudal que se transvasa cada $20$ minutos al segundo depósito es de $k$ litros, el total de litros que ha entrado es $$7 \\times 24 \\times 3 \\times k = 2^3 \\times 3^2 \\times 7 \\times k.$$ Así que $$2^3 \\times 3^2 \\times 7 \\times k = \\frac{n(n+1)}{2}$$ y esta cantidad tiene que ser $\\leq 25{,}000$, por tanto $$2^4 \\times 3^2 \\times 7 \\times k = n(n+1) \\leq 50{,}000.$$ Por la última desigualdad, $n \\leq 223$. Ahora, los números $n$ y $n+1$ son primos entre sí, luego cada potencia $2^4$, $3^2$, $7$ divide a $n$ o a $n+1$. Ciertamente $2^4 \\times 3^2 \\times 7 = 1008$ no puede dividir a $n$ ni a $n+1$, dado que $n \\leq 223$. Supongamos que $n = 16c$ es múltiplo de $16$. Entonces $n+1$ es múltiplo de $9$ o $7$. En el primer caso se tendría $n+1 = 16c+1 \\equiv 0 \\pmod{9}$, es decir, $c \\equiv 5 \\pmod{9}$. Pero si $c = 5$, $n = 80$ y $7$ no divide a $80 \\times 81$ y, si $c \\geq 14$, $n \\geq 224$. Por otro lado, en el segundo caso, $n+1 = 16c+1 \\equiv 0 \\pmod{7}$, de donde $c \\equiv 3 \\pmod{7}$. Pero $9$ no divide al producto $n(n+1)$ si $c = 3$ o $10$, y si $c \\geq 17$, $n > 223$. Concluimos que $n$ no es múltiplo de $16$ y $n+1$ sí. Si $n$ es múltiplo de $9$ y $n+1$ de $16 \\times 7$, tendríamos $n = 16 \\times 7 \\times c - 1 \\equiv 0 \\pmod{9}$, es decir, $c \\equiv 7 \\pmod{9}$ y entonces $c \\geq 7$ y $n > 223$. Similarmente, si $n$ es múltiplo de $7$ y $n+1$ de $16 \\times 9$, $n = 16 \\times 9 \\times c - 1 \\equiv 0 \\pmod{7}$, es decir, $c \\equiv 2 \\pmod{7}$ y entonces $c \\geq 2$ y $n > 223$. El único caso que queda es que $n$ sea múltiplo de $9 \\times 7 = 63$ y $n+1$ de $16$. Entonces $n+1 = 63c+1 \\equiv 0 \\pmod{16}$ y $c \\equiv 1 \\pmod{16}$ y necesariamente $c = 1$ (si no, $n > 223$). Por lo tanto, sólo hay una solución posible, a saber, $n = 63$, lo que da un volumen total extraído de $$T_{63} = \\frac{63 \\times 64}{2} = 2016$$ litros.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15396,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and $I$ its incenter. The circumcircle of $ACI$ intersects the line $BC$ a second time at the point $X$, and the circumcircle of $BCI$ intersects the line $AC$ a second time at the point $Y$.\n\nProve that the segments $AY$ and $BX$ are of equal length.",
    "options": [],
    "answer": "See solution",
    "solution": "We shall show that $AB = BX$ holds. Since $AB = AY$ then follows by the same argument, this completes the proof (see Figure 3).\n\n![](images/Austria2019_booklet_p9_data_f1b5d61a90.png)\n\nIn this solution, we use oriented angles between lines (modulo $180^{\\circ}$) with the notation $\\angle PQR$. As usual, the angles of the triangle $ABC$ are denoted by $\\alpha = \\angle BAC$, $\\beta = \\angle CBA$, and $\\gamma = \\angle ACB$.\n\nThe inscribed angle theorem gives\n\n$$\n\\angle AXB = \\angle AXC = \\angle AIC = -\\angle CIA = 180^{\\circ} - \\angle CIA = \\angle IAC + \\angle ACI = \\frac{1}{2}(\\alpha + \\gamma).\n$$\n\nThis immediately implies\n\n$$\n\\angle BAX = -\\angle AXB - \\angle XBA = -\\frac{1}{2}(\\alpha + \\gamma) - \\beta = \\frac{1}{2}(\\alpha + \\gamma).\n$$\n\nTherefore, the triangle *ABX* is indeed isosceles, and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15397,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n$ 為一正整數且 $3$ 不能整除 $n$。一個 $n$ 階三角陣是把一個大正三角形分成 $n^2$ 個相等大小的小正三角形，以過各邊的所有 $n$ 等分點對其他邊做平行線分割而成。在 $n$ 階三角陣每個小三角形頂點上放一顆橘子，共有\n\n$$\n\\frac{(n+1)(n+2)}{2}\n$$\n\n顆橘子。一組三顆橘子 $A, B, C$ 被稱為**好的**，若且唯若 $\\overline{AB}$ 和 $\\overline{AC}$ 各是某小三角形的一邊，且 $\\angle BAC = 120^{\\circ}$。\n\n言言每次可以選一組好的三顆橘子拿走。試求言言最多可以拿走幾顆橘子。",
    "options": [],
    "answer": "See solution",
    "solution": "答案：$\\frac{(n+1)(n+2)}{2} - 3$。\n\n記一個 $(n, k)$-梯形陣是一個 $n$ 階三角陣移除一個與 $n$ 階三角陣共頂點的 $n-k$ 階三角陣。記 $A_{00}, A_{01}, A_{02}, \\dots, A_{0n}$ 依序表示 $n$ 階三角陣一個底邊的橘子，且遞迴定義編號 $A_{ij}$ 的橘子使 $A_{ij}, A_{(i-1)j}, A_{(i-1)(j+1)}$ 構成小正三角形 ($A_{ij} \\neq A_{(i-2)(j+1)}$)，稱 $A_{ij}$ 橘子是第 $i$ 排第 $j$ 個。\n\n(1) 首先證明上界。我們只需證明：\n\n$A_{00}, A_{01}, A_{02}, \\dots, A_{0n}$ 中一定有一顆橘子沒被選取。\n\n這表示最小上界必小於 $\\frac{(n+1)(n+2)}{2}$，又最小上界必為 $3$ 的倍數，故得證。\n\n*引理證明*：若否，假設這 $n+1$ 顆橘子都能被取走。\n\n- 令 $k$ 是最小的整數使 $(A_{0(k+1)}, A_{0k}, A_{1(k-1)})$ 同時被選取，如果不存在則令 $k = n + 1$；\n- $m$ 是小於 $k$ 中最大的整數使 $(A_{0(m-1)}, A_{0m}, A_{1m})$ 同時被選取，如果不存在則令 $m = -1$；\n- 由於 $A_{1(k-1)}$ 和 $A_{1m}$ 必在不同回合被選取，故一定有 $k - 1 \\ge m + 1$。這保證存在 $m < i < k$，而注意到這些 $A_{0i}$ 在上述的兩個選擇中都沒有被選到。\n- 再注意到，由 $k$ 的最小性與 $m$ 的最大性，選到 $A_{0i}$ 的選取必定不會同時選到另一個 $A_{0j}$，且一定會選到至少一個 $A_{1j}$。但注意到共有 $k-m-1$ 個這樣的 $A_{0i}$，而在 $A_{1(k-1)}$ 與 $A_{1m}$ 之間卻只有 $(k-1)-m-1$ 個 $A_{1j}$ 可以選，故這是不可能的。矛盾！\n\n(2) 接著構造可達到上界的選擇方法。採用歸納法：\n\n- $n=1$：不必拿任何橘子\n- $n=2$：拿 $(A_{20}, A_{10}, A_{11})$\n- $n=3$：拿 $(A_{30}, A_{20}, A_{11})$；$(A_{21}, A_{12}, A_{02})$\n- 假設 $n=k-3$ 成立，考慮 $n=k$ 時，先對前 $k-3$ 排應用歸納假設。考慮 $k$ 的奇偶：\n\n  - 若 $k=2m+1$，可以選：\n    - $(A_{30}, A_{21}, A_{22})$；$(A_{20}, A_{11}, A_{12})$；$(A_{10}, A_{01}, A_{02})$；\n    - $(A_{0(2i+1)}, A_{0(2i+2)}, A_{1(2i+2)})$；$(A_{1(2i+1)}, A_{2(2i+1)}, A_{2(2i+2)})$，對於 $1 \\le i \\le m-2$；\n    - $(A_{3(k-3)}, A_{2(k-2)}, A_{1(k-2)})$；$(A_{0(k-2)}, A_{0(k-1)}, A_{1(k-1)})$\n  - 若 $k=2m$，可以選：\n    - $(A_{30}, A_{21}, A_{11})$；$(A_{20}, A_{10}, A_{01})$；\n    - $(A_{0(2i+1)}, A_{0(2i+2)}, A_{1(2i+2)})$；$(A_{1(2i+1)}, A_{2(2i+1)}, A_{2(2i+2)})$，對於 $1 \\le i \\le m-2$；\n    - $(A_{3(k-3)}, A_{2(k-2)}, A_{1(k-2)})$；$(A_{0(k-2)}, A_{0(k-1)}, A_{1(k-1)})$\n\n由歸納法，得證。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15398,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a set $S$ of 2006 distinct numbers. A subset $T$ of $S$ is called *stubborn* if for every $u, v$ (not necessarily distinct) in $T$, the number $u + v$ does not belong to $T$.\n\n1. If $S$ is the set of the first 2006 positive integers, show that every stubborn subset $T$ of $S$ has at most 1003 elements.\n2. If $S$ consists of 2006 arbitrary positive integers, show that there exists a stubborn subset $T$ of $S$ with at least 669 elements.",
    "options": [],
    "answer": "See solution",
    "solution": "1. Let $A$ be a stubborn subset of $S = \\{1, 2, \\dots, 2006\\}$ with $x$ elements $a_1 < a_2 < \\dots < a_x$. Consider $B = \\{a_2 - a_1, a_3 - a_1, \\dots, a_x - a_1\\}$. $B$ is a subset of $S$ with $x-1$ elements. Since $A$ is stubborn, $A \\cap B = \\emptyset$. Thus, $x + (x-1) \\leq 2006$, so $x \\leq 1003$.\n\n2. Let $S = \\{a_1, a_2, \\dots, a_{2006}\\}$. Consider the product $P$ of all odd divisors of $\\prod_{i=1}^{2006} a_i$. There exists a prime $p = 3r+2$ dividing $3P+2$ and coprime to all $a_i$. For each $a \\in S$, the sequence $a, 2a, \\dots, (p-1)a$ (mod $p$) is a permutation of $1, 2, \\dots, p-1$, so there exists a set $S_a$ of $r+1$ integers $x$ in $\\{1, \\dots, p-1\\}$ such that $xa \\bmod p$ belongs to $A = \\{r+1, \\dots, 2r+1\\}$. For each $x$, let $S_x = \\{a \\in S \\mid xa \\in A\\}$. Then:\n\n$$\n|S_1| + |S_2| + \\dots + |S_{p-1}| = \\sum_{a \\in S} |A_a| = 2006 \\times (r+1)\n$$\n\nSo there exists $x_0$ such that $|S_{x_0}| \\geq \\frac{2006 \\times (r+1)}{3r+1} > 668$. Let $B$ be a subset of 669 elements of $S_{x_0}$; then $B$ is a stubborn subset of $S$. Indeed, if $u, v, w \\in B$ ($u$ can equal $v$), then $x_0u, x_0v, x_0w \\in A$. It is easy to verify that $x_0u + x_0v \\neq x_0w \\pmod{p}$, so $u + v \\neq w$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15399,
    "subject": "Mathematics (Olympiad)",
    "question": "設數列 $\\{a_n\\}$ 滿足 $a_{n+1} = a_n^3 + 103$, $n = 1, 2, \\dots$。\n\n試證：至多存在一個正整數 $n$，使得 $a_n$ 為完全平方數。",
    "options": [],
    "answer": "See solution",
    "solution": "取 $\\bmod\\ 4$，可發現：\n\n$$\n(a_n, a_{n+1}) = (0, 3) \\pmod{4}\n$$\n$$\n(a_n, a_{n+1}) = (1, 0) \\pmod{4}\n$$\n$$\n(a_n, a_{n+1}) = (2, 3) \\pmod{4}\n$$\n$$\n(a_n, a_{n+1}) = (3, 2) \\pmod{4}\n$$\n\n注意到完全平方數對 $4$ 取模只能為 $0$ 或 $1$。若 $a_n \\equiv 0 \\pmod{4}$，則對於所有 $m > n$，必有 $a_m \\equiv 2 \\pmod{4}$ 或 $a_m \\equiv 3 \\pmod{4}$，故後續項不可能為完全平方數。而若 $a_n \\equiv 1 \\pmod{4}$，則 $a_{n+1} \\equiv 0 \\pmod{4}$，前述討論依然成立。由此知最多存在兩項 $(a_n, a_{n+1})$ 為完全平方數。\n\n現假設 $a_n = p^2, a_{n+1} = q^2$。代入原式，得 $p^6 + 103 = q^2$，也就是 $103 = (q + p^3)(q - p^3)$。基於 $103$ 是質數，我們必有 $q + p^3 = 103$ 及 $q - p^3 = 1$，也就是 $p^3 = 51$，矛盾！故至多存在一項為完全平方數。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15400,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the equation\n$$\n\\frac{1}{\\sqrt{x} + \\sqrt{1006}} + \\frac{1}{\\sqrt{2012 - x} + \\sqrt{1006}} = \\frac{2}{\\sqrt{x} + \\sqrt{2012 - x}}\n$$\nhas 2013 integer solutions.",
    "options": [],
    "answer": "See solution",
    "solution": "One can easily check that the given relation holds for any admissible value of $x$. Since $x$ is subject to the conditions $0 \\leq x \\leq 2012$, the conclusion is easily reached.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15401,
    "subject": "Mathematics (Olympiad)",
    "question": "$2^{2012}$ тоог 4 бүхэл тооны квадратуудын нийлбэрт хичнээн аргаар задалж болох вэ?",
    "options": [],
    "answer": "See solution",
    "solution": "Өгөгдсөн бол $x^2 + y^2 + z^2 + t^2 = 2^{2012}$ тэгшитгэлийн бүх бүхэл $(x, y, z, t)$ шийдийн тоог олохтой адил юм. Сондгой натурал тооны квадратуудыг 8-д хуваахад $1$ үлддэг тул $x, y, z, t$-ийн дотор тэгш ширхэг сондгой тоо байх нь илэрхий. \n\n- Хэрэв $x, y, z, t$-ийн яг 2 нь сондгой бол $x^2 + y^2 + z^2 + t^2 \\equiv 2 \\pmod{4}$ болж зөрчилд хүрнэ.\n- Хэрэв бүгд сондгой бол $x^2 + y^2 + z^2 + t^2 \\equiv 4 \\pmod{8}$ болж зөрчилд хүрнэ.\n\nИймд $x, y, z, t$ бүгд тэгш байх ёстой. $x = 2x_1, y = 2y_1, z = 2z_1, t = 2t_1$ гэж тавьбал:\n$$\n4(x_1^2 + y_1^2 + z_1^2 + t_1^2) = 2^{2012}\n$$\n$$\n\\Rightarrow x_1^2 + y_1^2 + z_1^2 + t_1^2 = 2^{2010}\n$$\n\nЭнэ үйлдлийг давтсаар $x_{1004}^2 + y_{1004}^2 + z_{1004}^2 + t_{1004}^2 = 4$ тэгшитгэлд хүрнэ. Үүний бүх шийд:\n- $(0, 0, 0, 2)$ болон $(1, 1, 1, 1)$\n\nАнхны хувьсагчид шилжүүлбэл:\n- $(x, y, z, t) = (0, 0, 0, 2^{1006})$ болон $(2^{1005}, 2^{1005}, 2^{1005}, 2^{1005})$\n\nТэгшитгэлийн шийдүүдийн тоо:\n- $0^2 + 0^2 + 0^2 + (2^{1006})^2$\n- $0^2 + 0^2 + 0^2 + (-2^{1006})^2$\n- $(2^{1005})^2 + (2^{1005})^2 + (2^{1005})^2 + (2^{1005})^2$ (бүх тэмдэгтүүдийн хувилбарууд)\n\n$(2^{1005})^2$-ийн хувьд тэмдэгтүүдийн бүх хослолыг тооцвол $2^4 = 16$ (бүх тэмдэгтүүд $+$ эсвэл $-$ байж болно).\n\nИймд нийт $2 + 16 = 18$ аргаар $2^{2012}$ тоог 4 бүхэл тооны квадратуудын нийлбэрт задалж болно.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15402,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $k$ heaps on the table, each containing a different positive number of stones. Jüri and Mari make moves alternatingly; Jüri starts. On each move, the player making the move has to pick a heap and remove one or more stones in it from the table; in addition, the player is allowed to distribute any number of remaining stones from that heap in any way between other non-empty heaps. The player to remove the last stone from the table wins. For which positive integers $k$ does Jüri have a winning strategy for any initial state that satisfies the conditions?\n\nfor any $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "For any $k$.\n\nCall a position *balanced* if the non-empty heaps can be divided into pairs, with an equal number of stones in both heaps of each pair. We show that in a balanced position, the player who moves second has a winning strategy. If there are no heaps left, the second player has already won. In the general case, suppose that the first player picks the heap $H$, takes $n$ stones from it off the table, and moves $a_1$ stones to the first heap, $a_2$ stones to the second, etc. If any stones are left in $H$ after that, the second player can then pick the heap $H'$ that is paired with $H$, remove $n$ stones from it, and move $a_1$ stones to the heap paired with the first heap, $a_2$ stones to the heap paired with the second heap, etc. On the other hand, if the first player empties the heap $H$, the second player does the same as in the first case, with the following exception: if the first player moved any stones to $H'$, the second player takes this many additional stones off the table instead of moving them back to $H$, ensuring that the heap $H'$ also becomes empty. In both cases, the position after the second player's move is balanced again. Since the number of stones on the table decreases with each move and the second player can always make a move, the second player eventually wins.\n\nFinally, we show that Jüri can move into a balanced position with his first move, giving him a winning strategy. Number the heaps $1, 2, \\ldots, k$ in decreasing order of size, and let the sizes of the heaps be $a_1 > a_2 > \\dots > a_k$. Jüri should pick heap $1$ and move $a_2 - a_3$ stones to heap $3$, $a_4 - a_5$ stones to heap $5$, etc. If $k$ is odd, Jüri should take all stones remaining in heap $1$ off the table; if $k$ is even, he should leave $a_k$ stones in heap $1$. It remains to verify that this move is possible. If $k = 1$, the game will be over after Jüri's move. For $k > 1$, $$(a_2 - a_3) + (a_4 - a_5) + \\dots + (a_{k-1} - a_k) < (a_1 - a_2) + (a_2 - a_3) + (a_3 - a_4) + (a_4 - a_5) + \\dots + (a_{k-1} - a_k) = a_1 - a_k,$$ which shows that after redistributing stones to odd-numbered heaps, there are still more than $a_k$ stones left in heap $1$, so the move is possible for both odd and even $k$. Furthermore, the strictness of the inequality ensures that some stones will be left to take off the table as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15403,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying\n$$\nf(f(x)) + y f(x) \\leq x + x f(f(y)),\n$$\nfor all real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = 0$ in the given relation:\n$$\nf(f(0)) + y f(0) \\leq 0,\n$$\nfor all $y \\in \\mathbb{R}$. This is only possible if $f(0) = 0$, since otherwise the inequality cannot hold for all $y$.\n\nNow, set $y = 0$:\n$$\nf(f(x)) \\leq x, \\quad \\forall x \\in \\mathbb{R}. \\tag{1}\n$$\n\nSet $x = 1$:\n$$\nf(f(1)) + y f(1) \\leq 1 + f(f(y)).\n$$\nUsing (1), $f(f(1)) \\leq 1$, so\n$$\nf(f(1)) + y f(1) \\leq 1 + y,\n$$\nwhich gives $y(f(1) - 1) \\leq 1 - f(f(1))$ for all $y$. This is only possible if $f(1) = 1$.\n\nWith $x = 1$, the original relation becomes $y \\leq f(f(y))$ for all $y$. \\tag{2}\n\nFrom (1) and (2), $f(f(x)) = x$ for all $x$.\n\nSubstitute into the original inequality:\n$$\nx + y f(x) \\leq x + x y \\implies y f(x) \\leq x y, \\quad \\forall x, y.\n$$\nFor $y = 1$, $f(x) \\leq x$; for $y = -1$, $f(x) \\geq x$. Thus, $f(x) = x$ for all $x$, which satisfies the original condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15404,
    "subject": "Mathematics (Olympiad)",
    "question": "Two players play a game on a $$(m+1) \\times m$$ grid, which has $m+1$ horizontal lines, $m$ vertical lines, and hence $m(m+1)$ intersection points. One pebble is put on an intersection point. Two players alternately move the pebble to an adjacent point along an edge, but they are not allowed to use edges used previously by either player. A player loses if they cannot move the pebble any more. Prove that there exists a winning strategy for the first player if the pebble was initially put on a point in the bottom horizontal line.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\{(x, y) : x = 0, 1, \\dots, m;\\ y = 1, 2, \\dots, m\\}$ be the set of all points on the grid. Let $(a, 0)$ be the initial position of the pebble. Let\n\n$$\nP = (a, 0), \\quad Q = (0, a), \\quad R = (m+1-a, m+1), \\quad S = (m+1, m+1-a).\n$$\n\nLet $X$ be the set of all points of the grid on the lines $PQ$, $QR$, $RS$, $SP$ or the interior of the quadrangle formed by those four lines. The first player initially moves the pebble to $(a, 1)$ and removes the straight line segment joining $(a, 0)$ and $(a, 1)$. We color the point $(x, y)$ red if $x + y - a$ is even, and blue if $x + y - a$ is odd. Then the first player will be given a choice only if the pebble is on a red point, and the second player will be given a choice only if the pebble is on a blue point. Moreover, all blue points in $X$ are not adjacent to any point outside $X$. Furthermore, every red point in $X$ has an even number of edges to a blue point. Therefore, whenever the first player is given a choice of moving the pebble, there are an odd number of unused edges inside $X$ available for moving the pebble to a point in $X$. Since the game must end and the first player cannot lose, the first player can always win. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15405,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a polynomial with integer coefficients such that for every integer $a$, any prime $p$ dividing $P(a)$ also divides $a$. Determine all such polynomials $P(x)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be a polynomial satisfying the conditions and let $p$ be an arbitrary prime. Any prime $q$ dividing $P(p)$ also divides $p$, so $q = p$. Hence, for any prime $p$ we have $P(p) = \\pm p^{m_p}$ for some non-negative integer $m_p$, which can depend on $p$.\n\nThe polynomials $P(x) = \\pm 1$ obviously satisfy the conditions. Assume $P$ is not identically $1$ or $-1$. Write\n\n$$\nP(x) = a_n x^n + a_{n-1} x^{n-1} + \\dots + a_1 x + a_0.\n$$\n\n$P$ can take the value $1$ and $-1$ only for finitely many primes. So, there exist infinitely many primes such that\n\n$$\n\\pm p^{m_p} = P(p) = a_n p^n + a_{n-1} p^{n-1} + \\dots + a_1 p + a_0\n$$\n\nand consequently $p$ divides $a_0$. This is only possible if $a_0 = 0$. Write $P(x) = x^k Q(x)$ where $k \\in \\mathbb{N}$ and $Q$ is a polynomial with integer coefficients such that $Q(0) \\neq 0$. Let $a$ be an integer and let $p$ be a prime dividing $Q(a)$. Then $p$ divides $P(a)$, so it also divides $a$. Hence, $Q$ satisfies the conditions. Since $Q(0) \\neq 0$, $Q$ must be identically $1$ or $-1$. The only possible polynomials are $P(x) = \\pm x^n$, $n \\in \\mathbb{N}_0$, and all of these obviously satisfy the conditions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15406,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the sum of all odd 3-digit numbers whose product of digits is equal to $140$.",
    "options": [],
    "answer": "See solution",
    "solution": "Because $140 = 2 \\cdot 2 \\cdot 5 \\cdot 7 = 4 \\cdot 5 \\cdot 7$, it follows that the 3-digit numbers are formed with the digits $4$, $5$, and $7$. So the odd 3-digit numbers, with product of digits equal to $140$, are $457$, $475$, $547$, and $745$. Their sum is $2224$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15407,
    "subject": "Mathematics (Olympiad)",
    "question": "The twins Pete and Ostap had an argument and started to go to school by different routes. Pete goes 210 meters to the south, then 70 meters to the east to reach the school. Ostap goes north for a certain distance, then heads directly to the school. If both twins walk at the same speed and arrive at the school simultaneously, how many meters north does Ostap need to walk?",
    "options": [],
    "answer": "See solution",
    "solution": "Let us denote the points: home as $D$, school as $S$, the point where Pete turns east as $B$, and the point where Ostap begins to walk straight to the school as $N$. Then $DB = 210$, $BS = 70$, $DN = x$, $NS = y$. Also, $x + y = 280$.\n\n![](images/Ukraine_2016_Booklet_p9_data_75157b9768.png)\n\nBy the Pythagorean theorem:\n\n$$\n(BN)^2 + (BS)^2 = (NS)^2\n$$\n\nor\n\n$$\n(210 + x)^2 + 70^2 = y^2\n$$\n\nBut since $x + y = 280$, $y = 280 - x$, so:\n\n$$\n(210 + x)^2 + 70^2 = (280 - x)^2\n$$\n\nExpanding:\n\n$$\n44100 + 420x + x^2 + 4900 = 78400 - 560x + x^2\n$$\n\nThe $x^2$ terms cancel:\n\n$$\n49000 + 420x = 78400 - 560x\n$$\n\n$$\n420x + 560x = 78400 - 49000\n$$\n\n$$\n980x = 29400\n$$\n\n$$\nx = 30\n$$\n\nSo, Ostap must walk $30$ meters north.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15408,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AD$, $BE$, and $CF$ be the altitudes of triangle $ABC$, with points $D$, $E$, and $F$ lying on sides $BC$, $CA$, and $AB$ respectively.\n\n% IMAGE: ![](images/Ireland2010_booklet_p26_data_c3e5ea76f8.png)\n\nProve that $|AB| = |CB|$ if and only if the circles with centers $A$ and $C$ that touch the opposite sides have equal radii, and that $MN$ is the perpendicular bisector of $AC$ passing through $B$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since triangles $AMC$ and $ANC$ are congruent, $MN$ is perpendicular to $AC$.\n\nSuppose $B$, $M$, and $N$ are collinear. Then $E$ is the intersection point of $MN$ with $AC$, and $M$ and $N$ lie on $BE$. By the Pythagorean theorem in triangles $AEB$, $BEC$, $AEN$, and $NEC$:\n\n$$\n|AB|^2 - |BC|^2 = |AE|^2 - |EC|^2 = |AN|^2 - |NC|^2\n$$\n\nBecause $|AD| = |AN|$ and $|CF| = |NC|$, we have:\n\n$$\n|AD|^2 - |CF|^2 = |AN|^2 - |NC|^2 = |AB|^2 - |BC|^2\n$$\n\nUsing the Pythagorean theorem in triangles $ABD$ and $CBF$:\n\n$$\n|BD|^2 = |AB|^2 - |AD|^2 = |BC|^2 - |CF|^2 = |BF|^2\n$$\n\nTherefore, the right-angled triangles $ABD$ and $CBF$ share the angle at $B$ and have equal sides $|BD| = |BF|$, so they are congruent. This shows $|AB| = |CB|$.\n\nConversely, if $|AB| = |CB|$, the circles with centers $A$ and $C$ touching the opposite sides have equal radii. Thus, $MN$ is the perpendicular bisector of $AC$ passing through $B$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15409,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $n$ 為大於或等於 $2$ 的正整數，$a_1, a_2, \\ldots, a_n$ 為滿足 $a_1 + a_2 + \\dots + a_n = 0$ 的實數，且存在正實數 $\\alpha$ 使得集合 $A = \\{(i,j) : 1 \\le i < j \\le n,\\ |a_i - a_j| \\ge \\alpha\\}$ 為非空。證明：\n\n$$\n\\sum_{(i,j) \\in A} a_i a_j < 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "不失一般性，假設 $\\alpha = 1$。定義集合 $B$ 和 $C$：\n\n$$\nB = \\{(i,j) : 1 \\le i, j \\le n,\\ |a_i - a_j| \\ge 1\\},\n$$\n\n$$\nC = \\{(i,j) : 1 \\le i, j \\le n,\\ |a_i - a_j| < 1\\}.\n$$\n\n則有：\n\n$$\n\\sum_{(i,j) \\in A} a_i a_j = \\frac{1}{2} \\sum_{(i,j) \\in B} a_i a_j = -\\frac{1}{2} \\sum_{(i,j) \\in C} a_i a_j.\n$$\n\n第二個等式由 $a_1 + \\dots + a_n = 0$ 得出。故只需證明若 $A$（即 $B$）非空，則\n\n$$\n\\sum_{(i,j) \\in C} a_i a_j > 0.\n$$\n\n將指標分成集合 $P, Q, R, S$：\n\n$$\nP = \\{i : a_i \\le -1\\},\n$$\n$$\nQ = \\{i : -1 < a_i \\le 0\\},\n$$\n$$\nR = \\{i : 0 < a_i < 1\\},\n$$\n$$\nS = \\{i : a_i \\ge 1\\}.\n$$\n\n則\n\n$$\n\\sum_{(i,j) \\in C} a_i a_j \\ge \\sum_{i \\in P \\cup S} a_i^2 + \\sum_{i,j \\in Q \\cup R} a_i a_j = \\sum_{i \\in P \\cup S} a_i^2 + \\left( \\sum_{i \\in Q \\cup R} a_i \\right)^2 \\ge 0.\n$$\n\n第一個不等式成立，因為右式的所有正項都在左式中，左式的所有負項也在右式中。只有當 $P = S = \\emptyset$ 時才等號成立，但此時 $A = \\emptyset$。因此 $A$ 非空時嚴格成立，證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15410,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $N$ points on a circle. Andriy and Olesya play a game as follows: Andriy goes first. They take turns connecting two of the given points with a chord, provided the chord does not cross any previously drawn chords. The player whose move results in a triangle formed by the drawn chords wins. Who will win the fair game if:\n\n$a)$ $N=14$;\n\n$b)$ $N=15$?",
    "options": [],
    "answer": "See solution",
    "solution": "The player forced to draw a second chord from a point (with one already drawn) will lose, since the opponent can then complete the triangle and win. Initially, both players connect only \"free\" points (points with no chords yet). At any moment, the free points can be divided into groups, within which any two points can be connected. Chords within a group are independent of other groups. After each move, the position can be described by the sizes of these groups. Drawing a chord in a group of $m$ points splits it into two groups of $m_1$ and $m_2$ points, where $m_1 + m_2 = m-2$ (some groups may be empty).\n\nSome useful properties:\n\n*Property 1.* In a group of 3 points, one chord can always be drawn; in a group of 5 points, two chords can always be drawn.\n\n*Property 2.* In a group of 4 points, the player can choose to partition into $(1; 1)$ or $(2; 0)$, so either one or two chords can be drawn, depending on the move.\n\n*Property 3.* In a group of 6 points, the player can partition into $(2; 2)$ or $(3; 1)$, so either 3 or 2 chords can be drawn.\n\n*Property 4.* If the situation is symmetric (groups can be paired with equal numbers of points), the player who creates this symmetry can win by mirroring the opponent's moves.\n\n![](images/ukraine_2015_Booklet_p17_data_e299610ef6.png)\n\nApplying these properties:\n\n$a)$ For $N=14$, Andriy wins by creating a $(6; 6)$ position, then following a symmetric strategy (Property 4).\n\n$b)$ For $N=15$, Olesya wins. No matter Andriy's first move, Olesya can respond to create a symmetric or advantageous position, ensuring victory by following the outlined properties and strategies.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15411,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that:\n\n1) For all real $x, y$, the following equality holds:\n$$\nf(2x) = f(x+y)f(y-x) + f(x-y)f(-x-y).\n$$\n\n2) $f(x) \\ge 0$ for all $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Take $x = y$:\n$$\nf(2x) = f(0)f(2x) + f(0)f(-2x). \\tag{1}\n$$\n\nSet $x = 0$ in (1):\n$$\nf(0) = 2f^2(0).\n$$\nSo $f(0) = 0$ or $f(0) = \\frac{1}{2}$.\n\nIf $f(0) = 0$, (1) implies $f(2x) = 0$, so $f = 0$.\n\nIf $f(0) = \\frac{1}{2}$, (1) gives $f(2x) = \\frac{1}{2}f(2x) + \\frac{1}{2}f(-2x)$, so $f(2x) = f(-2x)$.\n\nTaking $x = 0$ and using that $f$ is even:\n$$\n\\frac{1}{2} = f(0) = (f(y))^2 + (f(-y))^2 = 2(f(y))^2.\n$$\nSo $f(y) = \\frac{1}{2}$ for all $y$.\n\nBoth $f(x) = 0$ and $f(x) = \\frac{1}{2}$ satisfy all requirements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15412,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $n$, $n \\geq 1$, such that $n \\cdot 2^{n+1} + 1$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "The only solution is $n = 3$.\n\nSuppose $n \\cdot 2^{n+1} + 1$ is a perfect square. Since $n \\cdot 2^{n+1} + 1$ is odd, let $n \\cdot 2^{n+1} + 1 = (2x + 1)^2$ for some $x \\in \\mathbb{N}$. Expanding, we get:\n\n$$\n(2x + 1)^2 = 4x^2 + 4x + 1\n$$\nSo,\n$$\nn \\cdot 2^{n+1} = 4x^2 + 4x\n$$\n$$\nn \\cdot 2^{n-1} = x(x + 1)\n$$\n\nSince $x$ and $x + 1$ are coprime, one of them must be divisible by $2^{n-1}$, so the other must be at most $n$. Thus, $2^{n-1} \\leq n + 1$.\n\nBy induction, this inequality fails for $n \\geq 4$. Checking $n = 1, 2, 3$:\n- $n = 1$: $1 \\cdot 2^{2} + 1 = 5$ (not a square)\n- $n = 2$: $2 \\cdot 2^{3} + 1 = 17$ (not a square)\n- $n = 3$: $3 \\cdot 2^{4} + 1 = 49 = 7^2$\n\nSo, the only solution for $n \\geq 1$ is $n = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15413,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, the bisector $AD$ and the median $BE$ meet at $P$. The straight lines $AB$ and $CP$ meet at $F$. The parallel through $B$ to $CF$ meets the straight line $DF$ at $M$. Show that $DM = BF$.\n\nGheorghe Bumbăcea\n\n![](images/RMC2013_final_p18_data_a0c373c564.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Ceva's Theorem implies\n\n$$\n\\frac{BF}{FA} \\cdot \\frac{AE}{EC} \\cdot \\frac{CD}{DB} = 1,\n$$\n\nhence $\\frac{BF}{FA} = \\frac{BD}{DC}$. The converse of Thales' Theorem yields $DF \\parallel AC$. From $\\triangle BFD \\sim \\triangle BAC$ it follows that $\\frac{BF}{BA} = \\frac{FD}{AC}$, whence $BF = \\frac{AB}{AC} \\cdot FD$. Now $\\triangle BDM \\sim \\triangle CDF$ implies $\\frac{DM}{FD} = \\frac{BD}{DC}$. On the other hand, since $BD$ is a bisector, $\\frac{BD}{DC} = \\frac{AB}{AC}$, so $DM = \\frac{AB}{AC} \\cdot FD$. Finally, $DM = BF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15414,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $m$ such that\n\n$$\n\\{\\sqrt{m}\\} = \\{\\sqrt{m+2011}\\}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Rewrite the equation $\\sqrt{m} - [\\sqrt{m}] = \\sqrt{m+2011} - [\\sqrt{m+2011}]$ to get $\\sqrt{m+2011} - \\sqrt{m} = [\\sqrt{m+2011}] - [\\sqrt{m}] = p \\in \\mathbb{N}$.\n\nSquaring $\\sqrt{m+2011} = p + \\sqrt{m}$ yields $2011 = p^2 + 2p\\sqrt{m} \\in \\mathbb{N}$, hence $m = k^2$, $k \\in \\mathbb{N}^*$.\n\nThe relation $2011 = p(p + 2k)$ gives $p = 1$ and $p + 2k = 2011$, since 2011 is prime. Hence $k = 1005$ and $m = 1005^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15415,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the least real $m$ such that there exist real numbers $a$ and $b$ for which the inequality\n\n$$\n|x^2 + a x + b| \\leq m\n$$\nholds for all $x \\in (0, 2)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that no negative number $m$ satisfies the problem, since the absolute value is always non-negative.\n\nNow, interpret the problem geometrically. The graph of $y = x^2 + a x + b$ must lie within the horizontal strip between $y = m$ and $y = -m$ for $x \\in (0, 2)$. Our aim is to find the narrowest such strip containing the graph of some quadratic function in this interval.\n\n![](images/65_Czech_and_Slovak_MO_2016_booklet_p3_data_feaee72c3e.png)\n\nThe function\n\n$$\nf(x) = (x - 1)^2 - \\frac{1}{2} = x^2 - 2x + \\frac{1}{2}\n$$\n\nis a good candidate for the closest strip. This function has $a = -2$, $b = \\frac{1}{2}$, and it satisfies (as shown below) the inequalities $-\\frac{1}{2} \\leq f(x) \\leq \\frac{1}{2}$.\n\nThese inequalities are equivalent to $0 \\leq (x-1)^2 \\leq 1$, which holds for $x \\in (0, 2)$. Thus, $f(x) = x^2 - 2x + \\frac{1}{2}$ satisfies the conditions of the problem for $m = \\frac{1}{2}$.\n\nNext, we show that no quadratic function satisfies the problem for any $m < \\frac{1}{2}$.\n\nThe crucial fact is that for any $f(x) = x^2 + a x + b$, at least one of the differences $f(0) - f(1)$ or $f(2) - f(1)$ is greater than or equal to $1$. This implies that the width of the closest strip is at least $1$, so $m \\geq \\frac{1}{2}$. For $f(0) - f(1) \\geq 1$, we have by the triangle inequality:\n\n$$\n1 \\leq |f(0) - f(1)| \\leq |f(0)| + |f(1)| \\leq 2m.\n$$\n\nSimilarly for $f(2) - f(1) \\geq 1$.\n\nNow, compute:\n\n$$\nf(0) = b, \\quad f(1) = 1 + a + b, \\quad f(2) = 4 + 2a + b.\n$$\n\nSo,\n\n$$\nf(0) - f(1) = -1 - a \\geq 1 \\iff a \\leq -2,\n$$\n\n$$\nf(2) - f(1) = 3 + a \\geq 1 \\iff a \\geq -2.\n$$\n\nThus, at least one of $f(0) - f(1) \\geq 1$ or $f(2) - f(1) \\geq 1$ holds for any $a, b$.\n\n*Conclusion.* The desired minimal value of $m$ is $\\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15416,
    "subject": "Mathematics (Olympiad)",
    "question": "For which values of $n$ does there exist a circumscribed $n$-gon with side lengths $1, 2, \\dots, n$ (in any order)?",
    "options": [],
    "answer": "See solution",
    "solution": "It exists if $n = 4k$ or $n = 4k + 1$ where $k$ is a positive integer.\n\nLet us consider an $n$-gon $P_1P_2 \\dots P_n$. The tangent points of the inscribed circle divide each side into two segments. The lengths of these segments that share a vertex $P_i$ are equal. Denote the length of tangent segments originating at $P_i$ by $A_i$. Thus, the side lengths of the $n$-gon can be expressed as $P_iP_{i+1} = A_i + A_{i+1}$ for all $i = 1, 2, \\dots, n$, considering points cyclically ($P_{n+1} = P_1$, $A_{n+1} = A_1$).\n\nConversely, if we can find $n$ positive real numbers $A_i$ ($i = 1, 2, \\dots, n$) such that the sequence $(A_1 + A_2, A_2 + A_3, \\dots, A_n + A_1)$ is a permutation of $(1, 2, \\dots, n)$, then there is a circumscribed polygon $P_1P_2 \\dots P_n$ with side lengths $1, 2, \\dots, n$.\n\nTo show this, start with a circle of arbitrary radius $R$ and construct points $P_1, P_2, \\dots, P_n$ outside the circle so that the tangent segments from $P_i$ to the circle have length $A_i$, and the \"right\" tangent from $P_i$ touches the circle at the same point as the \"left\" tangent from $P_{i-1}$. If the \"right\" tangent point of $P_1$ does not match the \"left\" touching point of $P_n$, this can be fixed by adjusting $R$ using continuity.\n\nNow, consider four cases:\n\n1. **Case $n = 4k$:**\n   Such a circumscribed $n$-gon exists. The $4k$ segments $A_i$ can be of lengths\n   $$\n   \\begin{aligned}\n   A_1 &= \\frac{1}{2},\\quad A_2 = \\frac{1}{2},\\quad A_3 = \\frac{3}{2},\\quad A_4 = \\frac{3}{2},\\quad \\dots,\\quad A_{2k-1} = \\frac{2k-1}{2},\\quad A_{2k} = \\frac{2k-1}{2}, \\\\\n   A_{2k+1} &= \\frac{2k+1}{2},\\quad A_{2k+2} = \\frac{6k-1}{2},\\quad A_{2k+3} = \\frac{2k-1}{2},\\quad A_{2k+4} = \\frac{6k-3}{2},\\quad \\dots, \\\\\n   A_{4k-1} &= \\frac{3}{2},\\quad A_{4k} = \\frac{4k+1}{2}.\n   \\end{aligned}\n   $$\n   The sums of consecutive elements $A_1+A_2, A_2+A_3, \\dots, A_{4k-1}+A_{4k}, A_{4k}+A_1$ are exactly $1, 2, \\dots, 2k, 4k, 4k-1, \\dots, 2k+1$.\n\n2. **Case $n = 4k + 1$:**\n   The construction is similar. Choose $4k + 1$ segments of length\n   $$\n   \\begin{aligned}\n   A_1 &= \\frac{1}{2}, & A_2 &= \\frac{1}{2}, & A_3 &= \\frac{5}{2}, & A_4 &= \\frac{5}{2}, \\dots, \\\\\n   A_{2k+1} &= \\frac{4k+1}{2}, & A_{2k+2} &= \\frac{4k+1}{2}, & A_{2k+3} &= \\frac{4k-1}{2}, \\\\\n   A_{2k+4} &= \\frac{4k-3}{2}, & A_{2k+5} &= \\frac{4k-5}{2}, & \\dots, & A_{4k+1} &= \\frac{3}{2}.\n   \\end{aligned}\n   $$\n   The sums $A_1 + A_2, A_2 + A_3, \\dots, A_{4k-1} + A_{4k}, A_{4k} + A_{4k+1}$ are $1, 3, 5, \\dots, 4k+1, 4k, 4k-2, \\dots, 2$.\n\n3. **Case $n = 4k + 2$:**\n   Such a polygon does not exist. If the number of sides is even, the sum of the odd-numbered sides equals the sum of the even-numbered sides, but the total sum $1 + 2 + \\dots + n$ is odd, so such a split is impossible.\n\n4. **Case $n = 4k + 3$:**\n   Such a polygon also does not exist. In this case, expressing $A_1$ as\n   $$\n   \\begin{aligned}\n   A_1 &= (A_1 + A_2 + \\dots + A_n) - (A_2 + A_3) - (A_4 + A_5) - \\dots - (A_{4k+2} + A_{4k+3}) \\\\\n   &= \\frac{P_1 P_2 + P_2 P_3 + \\dots + P_{4k+3} P_1}{2} - P_2 P_3 - P_4 P_5 - \\dots - P_{4k+2} P_{4k+3}\n   \\end{aligned}\n   $$\n   Since the sum of the side lengths is even, $A_1$ is a positive integer, as are all $A_i$. But the side of length 1 cannot be split into two positive integers, leading to a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15417,
    "subject": "Mathematics (Olympiad)",
    "question": "Calculate the length of the leg of an isosceles trapezoid with bases $18\\ \\text{cm}$ and $10\\ \\text{cm}$, if it is known that its midline is $\\frac{2}{7}$ of its perimeter.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = 18\\ \\text{cm}$ and $b = 10\\ \\text{cm}$ be the bases, and $c$ the length of each leg. The midline is $m = \\frac{a + b}{2} = \\frac{18 + 10}{2} = 14\\ \\text{cm}$. The perimeter is $L = a + b + 2c$. Given $m = \\frac{2}{7}L$, so $14 = \\frac{2}{7}L$, thus $L = 14 \\cdot \\frac{7}{2} = 49\\ \\text{cm}$. Solving for $c$:\n\n$$\nL = a + b + 2c \\\\\n49 = 18 + 10 + 2c \\\\\n2c = 49 - 28 = 21 \\\\\nc = \\frac{21}{2} = 10.5\\ \\text{cm}\n$$\n\nSo, the length of the leg is $10.5\\ \\text{cm}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15418,
    "subject": "Mathematics (Olympiad)",
    "question": "Determinar cuántas soluciones reales tiene la ecuación\n\n$$\n\\sqrt{2 - x^2} = \\sqrt[3]{3 - x^3}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Para que existan soluciones reales, debe cumplirse $x \\in [-\\sqrt{2}, \\sqrt{2}]$. Ahora bien, si $x \\in [-\\sqrt{2}, 0]$ se tiene:\n\n$$\n2 - x^2 \\leq 2, \\quad 3 - x^3 \\geq 3,\n$$\npero $\\sqrt{2} < \\sqrt[3]{3}$, por lo que no hay soluciones cuando $x \\in [-\\sqrt{2}, 0]$.\n\nPor otra parte, cuando $x \\in (0, \\sqrt{2}]$ podemos ver que\n\n$$\n\\sqrt[3]{3-x^3} > \\sqrt[3]{2\\sqrt{2}-x^3}.\n$$\n\nPuesto que la ecuación\n\n$$\n\\sqrt{2-x^2} = \\sqrt[3]{2\\sqrt{2}-x^3}\n$$\n\ntiene como únicas soluciones $x = 0$ y $x = \\sqrt{2}$ y, para $x = 1$, resulta $\\sqrt[3]{2\\sqrt{2}-1} > \\sqrt{2-1} = 1$, es evidente que\n\n$$\n\\sqrt[3]{3-x^3} > \\sqrt[3]{2\\sqrt{2}-x^3} \\geq \\sqrt{2-x^2},\n$$\npor lo que tampoco pueden existir soluciones cuando $x \\in (0, \\sqrt{2}]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15419,
    "subject": "Mathematics (Olympiad)",
    "question": "ABCD нь багтаасан дөрвөн өнцөгт ба багтсан тойгийг нь $\\gamma$ гэе. $\\gamma$ тойрог $AD, CD$ талуудыг харгалзан $P, Q$ цэгт шүргэнэ. $BD$ диагональ $\\gamma$ тойрогтой $M$ ба $N$ цэгээр огтлолцох бөгөөд $MN$ хэрчмийн дундийг $R$ гэе. $\\angle ARP = \\angle CRQ$ батал.",
    "options": [],
    "answer": "See solution",
    "solution": "Поляр шулуун ашиглан баталъя. $\\gamma$ тойргийн төвийг $O$ гэе. $\\gamma$ тойрог $BC, AB$ талуудыг харгалзан $R, S$ цэгт шүргэдэг байг. $C$ ба $E$ цэгүүд $BD$-ийн нэг талд оршдог гэж үзье.\n\n**Лемм 1.** $AC$, $PQ$, $RS$ шулуунууд $E$ цэгт огтлолцоно. Мөн $BD \\perp E$ ба $E, M, O$ цэгүүд нэг шулуун дээр оршино.\n\n$PQ \\cap AC = E_1$, $RS \\cap AC = E_2$ гэе. $PD = DQ$ ба Мепелайн теоремоор $\\triangle ADC$:\n\n$$\n\\frac{E_1A \\cdot QC}{E_1C} = \\frac{PD}{QD} = 1.\n$$\n\nЯг үүнтэй адилаар\n\n$$\n\\frac{E_2A \\cdot RC}{E_2C \\cdot SA} = 1 \\Rightarrow QC = RC, PA = SA \\text{ тул}\n$$\n\n$$\n\\frac{E_1A}{E_1C} = \\frac{E_2A}{E_2C} \\Rightarrow E_1 = E_2.\n$$\n\n$D$ цэгийн поляр $PQ$ ба $E_1 \\in RQ$ тул $D$ цэг нь $E_1$ цэгийн поляр дээр оршино. Яг үүнтэй адилаар $B$ цэг нь $E_1$ цэгийн поляр дээр оршино. Иймд $E_1$ цэгт поляр шулуун нь $BD$ болно. Гэтэл $E_1 = E$ тул $BD \\perp EM \\Rightarrow E, M, O$ нэг шулуун дээр оршино. $EM$ шулууны хувьд $P, Q, R, S$ цэгүүдийн тэгш хэмтэй хувиргахад гарах дүрүүдийг нь $P', Q', R', S'$ гэе.\n\n**Лемм 2.** $PQ'$, $P'Q$, $RS'$, $R'S$ шулуунууд $M$ цэгт огтлолцоно.\n\n$U = BD \\cap PQ$, $V = BD \\cap P'Q'$ ба $UV \\cap PQ' = W$ байг.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15420,
    "subject": "Mathematics (Olympiad)",
    "question": "The inscribed circle in $\\triangle ABC$ ($AC \\neq BC$) is tangent to its sides $AB$, $BC$, and $CA$ at points $D$, $E$, and $F$, respectively. Let $P$ be the foot of the perpendicular from $D$ to $EF$ ($P \\in EF$). If the circles circumscribed about $\\triangle ABC$ and $\\triangle EFC$ intersect for the second time at point $Q$, prove that $\\angle PQC = 90^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "From the fact that $C$ lies on the circle circumscribed about $\\triangle FEQ$ and $CE = CF$, it follows that $CQ$ is an exterior bisector of $\\angle FQE$, and it remains to prove that $QP$ is a bisector of $\\angle FQE$, i.e., $QF : QE = FP : PE$. From $\\angle QFC = \\angle QEC$ and $\\angle QAC = \\angle QBC$, it follows that $\\triangle QFA \\sim \\triangle QEB$, i.e.\n\n$$\nQF : QE = AF : BE = AD : BD.\n$$\n\nLet $I$ be the center of the circle $k$ inscribed in $\\triangle ABC$, and the line $DP$ intersects $k$ for the second time at point $R$. Then\n\n$$\n\\angle REF = \\angle RDF = 90^\\circ - \\angle DFE = 90^\\circ - \\angle DIB = \\angle IBA\n$$\n\nand analogously $\\angle RFE = \\angle IAB$, i.e., $\\triangle FER \\sim \\triangle ABI$. But $RP \\perp EF$, $ID \\perp AB$, i.e., $P$ and $D$ are corresponding elements in similar triangles and $FP : PE = AD : BD$, which completes the proof. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15421,
    "subject": "Mathematics (Olympiad)",
    "question": "The convex quadrilateral $ABCD$ is given. $\\angle BAD = \\angle BCD = 60^\\circ$, $\\angle ADC = 135^\\circ$, and $BD \\perp AD$. Find $\\angle ACD$.\n\n![](images/Ukraine_booklet_2018_p55_data_b079aa080e.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let's calculate some angles:\n\n$\\angle ABC = 105^\\circ$, $\\angle DBC = 75^\\circ$. Therefore, $BC$ is the bisector of $\\angle KBD$, and $DC$ is the bisector of $\\angle BDL$. That's why $C$ is the center of the excircle of $\\triangle ABD$, so $AC$ is the bisector of $\\angle BAD$. Thus,\n\n$$\n\\angle ACD = 180^\\circ - (135^\\circ + 30^\\circ) = 15^\\circ.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15422,
    "subject": "Mathematics (Olympiad)",
    "question": "設三角形 $ABC$ 的垂心為點 $H$。令直線 $BH$ 與 $AC$ 交於點 $E$，且直線 $CH$ 與 $AB$ 交於點 $F$。設點 $X$ 為直線 $BC$ 上任意一點。設三角形 $BEX$ 的外接圓與直線 $AB$ 再交於點 $Y$，且三角形 $CFX$ 的外接圓與直線 $AC$ 再交於點 $Z$。證明三角形 $AYZ$ 的外接圓與直線 $AH$ 相切。\n\n![](images/2022-TWNIMO-Problems_p16_data_704592c4b8.png)",
    "options": [],
    "answer": "See solution",
    "solution": "由 Miquel 定理（應用於 $X$、$F$、$E$ 在 $HBC$ 的邊上），可知 $HEF$、$BEX$、$CFX$ 的外接圓交於一點 $M$。再由 Miquel 定理（應用於 $Y$、$E$、$A$ 在 $HAB$ 的邊上，其中 $A$ 視為 $HA$ 上的一點），可知 $HEA$、$BEY$ 及通過 $A$、$Y$ 且與 $HA$ 相切的圓交於一點。由於 $AHEF$ 和 $BEXY$ 共圓，故交點必為 $M$。因此，$AMY$ 的外接圓與 $AH$ 相切。同理，$AMZ$ 的外接圓也與 $AH$ 相切。故 $AMYZ$ 共圓且該圓與 $AH$ 相切，得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15423,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the circumcentre of $\\triangle ABC$. Prove that $AH \\geq AI$, where $H$ is the orthocentre and $I$ is the incenter of $\\triangle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $BM = CM = IM = OM$ (where $OM$ is the circumradius of $\\triangle ABC$), $B, C, I, O$ are concyclic. Therefore, $\\angle BIC = \\angle BOC$. This yields $90^\\circ + \\frac{A}{2} = 2A$, and hence $A = 60^\\circ$. Now\n\n$$\n\\angle BHC = 180^\\circ - A = 120^\\circ = 2A = \\angle BOC,\n$$\nwhich implies $H$ lies on $(BCOI)$. So $HM = IM$. Noting the triangle inequality $AH + HM \\geq AM$, we conclude $AH \\geq AI$ as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15424,
    "subject": "Mathematics (Olympiad)",
    "question": "6 white balls and 15 black balls are arranged in a circle so that there are at least two black balls between any pairs of white balls. How many different arrangements satisfy this condition?",
    "options": [],
    "answer": "See solution",
    "solution": "Let the number of black balls between each pair of white balls be $x_i$, for $i = 1$ to $6$. Then $\\sum_{i=1}^{6} x_i = 15$ and $x_i \\geq 2$. This is equivalent to $\\sum_{i=1}^{6} (x_i - 2) = 15 - 12 = 3$. By stars and bars, the number of solutions is $\\binom{3 + 6 - 1}{6 - 1} = \\binom{8}{5} = 56$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15425,
    "subject": "Mathematics (Olympiad)",
    "question": "The quadrilateral regular pyramid $VABCD$ has base $ABCD$. The points $M$, $N$, and $P$ are the midpoints of the edges $AD$, $BC$, and $VA$, respectively. Prove that the angle between the line $CP$ and the plane $(BAD)$ is $45^\\circ$ if and only if the angle between the line $CP$ and the plane $(VMN)$ is $30^\\circ$.\n\n![](images/RMC_2024_p14_data_3279b02b02.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the centre of the square $ABCD$; $VO$ is the altitude of the pyramid. The segments $CP$ and $VO$ are medians in triangle $VAC$, so they meet at a point $Q$.\n\nThe projection of the line $CP$ on the plane $(BAD)$ is the line $OC$, hence the angle between the line $CP$ and the plane $(BAD)$ is $\\angle OCQ$.\n\nFrom $CN \\perp MN$, $CN \\perp VO$, and $MN \\cap VO = \\{O\\}$, it follows that $CN \\perp (VMN)$. So, the projection of the line $CP$ on the plane $(VMN)$ is the line $NQ$, hence the angle between the line $CP$ and the plane $(VMN)$ is $\\angle CQN$.\n\nIf $\\angle OCQ = 45^\\circ$, then $CQ = \\sqrt{2} \\cdot OC$, and since $CN = \\frac{\\sqrt{2}}{2} \\cdot OC$, the leg $CN$ of the right triangle $CQN$ is half of the hypotenuse $CQ$, therefore $\\angle CQN = 30^\\circ$.\n\nConversely, if $\\angle CQN = 30^\\circ$, then $CQ = 2CN = \\sqrt{2} \\cdot OC$, so $\\angle OCQ = 45^\\circ$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15426,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Find all positive integers $n$ such that the sum of all integers from $1$ to $n+1$ can be represented as the sum of $n$ consecutive integers.\n\nb) Find all positive integers $n$ for which there exists an integer $a$ such that the sum of the integers from $a$ to $a+n$ is equal to the sum of the integers from $a+n+1$ to $a+2n$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Clearly, the sum of the first two positive integers can be represented as the sum of one positive integer. Now, let $n \\geq 2$ and let us show that the sum of the first $n+1$ positive integers cannot be represented as a sum of $n$ consecutive integers. Indeed, on one hand, $1 + 2 + \\dots + n + (n+1) > 2 + 3 + \\dots + n + (n+1)$; on the other hand, $1 + 2 + \\dots + n + (n+1) < 3 + \\dots + n + (n+1) + (n+2)$. So, the sum $1 + \\dots + (n+1)$ lies between $2 + \\dots + (n+1)$ and $3 + \\dots + (n+2)$, which are both sums of $n$ consecutive integers. Thus, $1 + \\dots + (n+1)$ is not a sum of $n$ consecutive integers for $n \\geq 2$.\n\nb) Let $n$ be any positive integer. To solve the problem, it suffices to see that $n^2 + (n^2 + 1) + \\dots + (n^2 + n) = n^2 \\cdot (n+1) + (1 + \\dots + n) = n \\cdot (n^2 + n) + (1 + \\dots + n) = (n^2 + n + 1) + \\dots + (n^2 + n + n)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15427,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ and $N$ be positive integers. Mr. Pisut starts walking from the point $(0, N)$ to the point $(M, 0)$ in such a way that:\n\n- Each of his steps is of 1 unit length in the direction parallel to either the X-axis or the Y-axis.\n- For each point $(x, y)$ on his path, $x \\ge 0$ and $y \\ge 0$.\n\nFor each step, he measures the distance from himself to the axis to which his step is parallel. If the step takes him farther away from the origin, he records the distance as a positive value; otherwise it is recorded as negative.\n\nProve that after he finishes his walk, the sum of all distances recorded is zero.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution 1.** Suppose that Mr. Pisut walks $k$ steps in total and the $i$-th step is from the point $(x_{i-1}, y_{i-1})$ to the point $(x_i, y_i)$. Notice that if the $i$-th step is parallel to the X-axis, then $y_i = y_{i-1}$ and he records $y_{i-1}(x_i - x_{i-1})$. Likewise, if the $i$-th step is parallel to the Y-axis, then $x_i = x_{i-1}$ and he records $x_i(y_i - y_{i-1})$. So the distance he records for the $i$-th step, regardless of the direction, is $y_{i-1}(x_i - x_{i-1}) + x_i(y_i - y_{i-1})$. Therefore, the sum of all distances recorded is\n\n$$\n\\sum_{i=1}^{k} y_{i-1}(x_i - x_{i-1}) + x_i(y_i - y_{i-1}) = \\sum_{i=1}^{k} (x_i y_i - x_{i-1} y_{i-1}) \\\\\n= x_k y_k - x_0 y_0 = M \\cdot 0 - 0 \\cdot N = 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15428,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $A$ lies below the line $\\ell$. What is the locus of points $N$ such that there exists a point $M \\in \\ell$ with $\\triangle NAM$ equilateral?\n\n% IMAGE: ![](images/Belarus_2015_p11_data_608d4c9cf1.png)",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show that all such points $N$ lie on the line $BC$, where $B \\in \\ell$, $C \\in m$, $m \\parallel \\ell$, $m \\nparallel A$, and $\\triangle ABC$ is equilateral.\n\nIf $M$ coincides with $B$, then $N$ coincides with $C$. Let $M$ be an arbitrary point of $\\ell$ different from $B$. In the right half-plane with respect to $AM$, construct the ray from $A$ such that the angle between this ray and $AM$ is $60^\\circ$. Let $K$ be the intersection of this ray and $BC$. Let $D$ be the point on $\\ell$ (to the left of $B$) such that $\\triangle ABD$ is equilateral. Note that\n\n$$\n\\begin{aligned}\n\\angle MAD &= \\angle MAK - \\angle DAK = 60^\\circ - \\angle DAK \\\\\n&= \\angle DAB - \\angle DAK = \\angle KAB.\n\\end{aligned} \\quad (1)\n$$\n\nMoreover,\n\n$$\n\\begin{aligned}\n\\angle MDA &= 180^\\circ - \\angle BDA = 180^\\circ - 60^\\circ = 120^\\circ \\\\\n&= 180^\\circ - \\angle ABC = \\angle KBA.\n\\end{aligned} \\quad (2)\n$$\n\nBy construction, $AD = AB$, so from (1) and (2), triangles $MAD$ and $KAB$ are congruent. Therefore, $AM = AK$, and since $\\angle MAK = 60^\\circ$, $\\triangle MAK$ is equilateral, so $K$ coincides with $N$, yielding $N \\in BC$.\n\nConversely, for any $N \\in BC$, there exists $M \\in \\ell$ such that $\\triangle NAM$ is equilateral. Let $N \\in BC$. In the left half-plane with respect to $AN$, construct the ray from $A$ such that the angle between this ray and $AN$ is $60^\\circ$. Let $M$ be the intersection of this ray and $\\ell$. Let $D$ be the point on $\\ell$ (to the left of $B$) such that $\\triangle ABD$ is equilateral. Note that\n\n$$\n\\begin{aligned}\n\\angle MAD &= \\angle MAN - \\angle DAN = 60^\\circ - \\angle DAN \\\\\n&= \\angle DAB - \\angle DAN = \\angle NAB.\n\\end{aligned} \\quad (3)\n$$\n\nMoreover,\n\n$$\n\\begin{aligned}\n\\angle MDA &= 180^\\circ - \\angle BDA = 180^\\circ - 60^\\circ = 120^\\circ \\\\\n&= 180^\\circ - \\angle ABC = \\angle NBA.\n\\end{aligned} \\quad (4)\n$$\n\nBy construction, $AD = AB$, so from (3) and (4), triangles $MAD$ and $NAB$ are congruent. Therefore, $AM = AN$, and since $\\angle MAN = 60^\\circ$, $\\triangle MAN$ is equilateral, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15429,
    "subject": "Mathematics (Olympiad)",
    "question": "Cada 20 minutos durante una semana se trasvasa una cantidad exacta de litros de agua (siempre la misma cantidad) desde un tanque con $25000$ litros a otro depósito inicialmente vacío. Desde este segundo depósito, a intervalos regulares de tiempo, se extrae primero $1$ litro, luego $2$, luego $3$, etc. Justo al final de la semana coinciden el último trasvase y la última extracción, quedando en ese momento vacío el segundo depósito. Determinar cuánta agua se ha extraído en total durante la semana, en caso de que los datos del problema lo permitan. (Se supone que los trasvases y las extracciones se realizan instantáneamente. El primer trasvase se hace pasados los primeros $20$ minutos y la primera extracción, pasado el primer intervalo de tiempo.)",
    "options": [],
    "answer": "See solution",
    "solution": "Sea $n$ el número de extracciones de agua realizadas durante la semana. En total se habrá extraído $$T_n = 1 + 2 + \\dots + n = \\frac{n(n+1)}{2}$$ litros. Por otro lado, si el caudal que se trasvasa cada $20$ minutos al segundo depósito es de $k$ litros, el total de litros que ha entrado es $$7 \\times 24 \\times 3 \\times k = 2^3 \\times 3^2 \\times 7 \\times k$$ así que $$2^3 \\times 3^2 \\times 7 \\times k = \\frac{n(n+1)}{2}$$ y esta cantidad tiene que ser $\\leq 25000$, por tanto $$2^4 \\times 3^2 \\times 7 \\times k = n(n+1) \\leq 50000.$$ Por la última desigualdad, $n \\leq 223$. Ahora, los números $n$ y $n+1$ son primos entre sí, luego cada potencia $2^4$, $3^2$, $7$ divide a $n$ o a $n+1$. Ciertamente $2^4 \\times 3^2 \\times 7 = 1008$ no puede dividir a $n$ ni a $n+1$, dado que $n \\leq 223$. Supongamos que $n = 16c$ es múltiplo de $16$. Entonces $n+1$ es múltiplo de $9$ o $7$. En el primer caso se tendría $n+1 = 16c+1 \\equiv 0 \\pmod{9}$, es decir, $c \\equiv 5 \\pmod{9}$. Pero si $c = 5$, $n = 80$ y $7$ no divide a $80 \\times 81$ y, si $c \\geq 14$, $n \\geq 224$. Por otro lado, en el segundo caso, $n+1 = 16c+1 \\equiv 0 \\pmod{7}$, de donde $c \\equiv 3 \\pmod{7}$. Pero $9$ no divide al producto $n(n+1)$ si $c = 3$ o $10$, y si $c \\geq 17$, $n > 223$. Concluimos que $n$ no es múltiplo de $16$ y $n+1$ sí. Si $n$ es múltiplo de $9$ y $n+1$ de $16 \\times 7$, tendríamos $n = 16 \\times 7 \\times c - 1 \\equiv 0 \\pmod{9}$, es decir, $c \\equiv 7 \\pmod{9}$ y entonces $c \\geq 7$ y $n > 223$. Similarmente, si $n$ es múltiplo de $7$ y $n+1$ de $16 \\times 9$, $n = 16 \\times 9 \\times c - 1 \\equiv 0 \\pmod{7}$, es decir, $c \\equiv 2 \\pmod{7}$ y entonces $c \\geq 2$ y $n > 223$. El único caso que queda es que $n$ sea múltiplo de $9 \\times 7 = 63$ y $n+1$ de $16$. Entonces $n+1 = 63c+1 \\equiv 0 \\pmod{16}$ y $c \\equiv 1 \\pmod{16}$ y necesariamente $c = 1$ (si no $n > 223$). Por lo tanto sólo hay una solución posible, a saber, $n = 63$, lo que da un volumen total extraído de $$T_{63} = \\frac{63 \\times 64}{2} = 2016$$ litros.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15430,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S = \\{a+1, a+2, \\dots, a+k\\}$ for some integer $a$ and positive integer $k$. For any subset $A$ of $S$, let $F(A)$ be the product of the elements of $A$ (with $F(\\emptyset) = 1$), and let $g(A)$ be the square-free part of $F(A)$. \n\nShow that if $k$ is odd, then the product $(x+1)(x+2)\\cdots(x+k)$ is a perfect square for at most finitely many integer values of $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that all prime divisors of $g(\\{a+1\\}), g(\\{a+2\\}), \\dots, g(\\{a+k\\})$ are less than $k$. If a prime $p \\geq k$ divides two of these numbers, say $g(\\{a+i\\})$ and $g(\\{a+j\\})$, then $p$ divides $a+i$ and $a+j$, so $p$ divides $j-i$, which implies $p < k$. If $p$ divides only one $g(\\{a+i\\})$, this leads to a contradiction because $g(S) = 1$ is a perfect square, and the power of $p$ in $\\prod_{i=1}^k (a+i)$ should be even.\n\nFor any $A \\subseteq S$, $g(A)$ divides $g(\\{a+1\\})g(\\{a+2\\})\\cdots g(\\{a+k\\})$, so the prime divisors of all $g(A)$ are less than $k$. Thus, for each $A \\subseteq S$, $g(A)$ is a divisor of $2 \\times 3 \\times \\cdots \\times p_{\\pi(k-1)}$, where $p_{\\pi(k-1)}$ is the largest prime less than $k$. There are $2^{\\pi(k-1)}$ possible values for $g(A)$.\n\nSince $k-1 > \\pi(k-1)$, $2^k > 2 \\times 2^{\\pi(k-1)}$. Therefore, there exist two subsets $A, B \\subseteq S$ such that $A \\ne B, S-B$ and $g(A) = g(B)$ (note $g(A) = g(S-A)$ because $g(S) = 1$). Now, $g(A) = g(B)$ implies $F(A)F(B)$ is a perfect square. We have\n\n$$\nF(A)F(B) = F(A \\cap B)^2 F(A\\Delta B).\n$$\n\nHence, $F(A\\Delta B)$ is a perfect square. But $A\\Delta B \\ne \\emptyset, S$ since $A \\ne B, S-B$, so this is the desired subset.\n\nBy replacing $A\\Delta B$ with $S - A\\Delta B$ if necessary, we obtain a nonempty proper subset $X$ of $\\{1, 2, \\dots, k\\}$ with an even number of elements such that $\\prod_{i \\in X} (x + i)$ is a perfect square. Using a lemma (as in the solution to the 7th problem of the Third Round), this product can be a perfect square for at most finitely many values of $x$. Thus, if $k$ is odd, $(x+1)(x+2)\\cdots(x+k)$ is a perfect square for at most finitely many values of $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15431,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(x + y f(x)) = f(x) + x f(y)\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to check that both $f(x) = 0$ and $f(x) = x$ are solutions. Label the equation as follows:\n\n$$\nf(x + y f(x)) = f(x) + x f(y) \\tag{1}\n$$\n\nPutting $x = 1$ in (1), we have\n\n$$\nf(1 + y f(1)) = f(1) + f(y). \\tag{2}\n$$\n\nIf $f(1) \\neq 1$, then there exists $y \\in \\mathbb{R}$ such that $1 + y f(1) = y$. For this $y$, we obtain $f(1) = 0$. Then (2) becomes $f(y) = 0$ for all $y$, which is the first solution.\n\nNow, assume $f(1) = 1$. If $f(t) = 0$ for some $t \\in \\mathbb{R}$, by putting $x = t$ and $y = 1$ in (1), we obtain $t = 0$. This means $f(x) \\neq 0$ for any $x \\neq 0$. Putting $x = 1$ in (1), we have\n\n$$\nf(y + 1) = f(y) + 1. \\tag{3}\n$$\n\nBy induction, we easily obtain $f(n) = n$ for any $n \\in \\mathbb{Z}$. Putting $x = -1$ in (1), we find that\n\n$$\nf(-y - 1) = -f(y) - 1.\n$$\n\nUsing (3), we get $-f(y) = f(-y - 1) + 1 = f(-y)$. Replacing $y$ by $-y$ in (1), we have\n\n$$\nf(x - y f(x)) = f(x) - x f(y).\n$$\n\nAdding this to (1), we have\n\n$$\nf(x + y f(x)) + f(x - y f(x)) = 2 f(x).\n$$\n\nFor $x \\neq 0$, we can replace $y$ by $\\frac{y}{f(x)}$. This yields\n\n$$\nf(x + y) + f(x - y) = 2 f(x). \\tag{4}\n$$\n\nNote that this also holds when $x = 0$. Considering $x = y$, we find that\n\n$$\nf(2x) = 2 f(x).\n$$\n\nAlso, applying the substitution $a = x + y$ and $b = x - y$, equation (4) becomes\n\n$$\nf(a) + f(b) = 2 f\\left(\\frac{a + b}{2}\\right) = f(a + b).\n$$\n\nThis shows $f$ satisfies the Cauchy equation. Thus, (1) can be simplified to\n\n$$\nf(y f(x)) = x f(y). \\tag{5}\n$$\n\nPutting $y = 1$, we obtain $f(f(x)) = x$. Replacing $x$ by $f(x)$ in (5) and using $f(f(x)) = x$, we get\n\n$$\nf(x y) = f(x) f(y). \\tag{6}\n$$\n\nIt is well-known that a function $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying the Cauchy equation and (6) can only be the zero function or the identity function. Thus, another solution is $f(x) = x$ for all $x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15432,
    "subject": "Mathematics (Olympiad)",
    "question": "In a quadrilateral $ABCD$, $\\angle ABC = \\angle ADC < 90^\\circ$. The circle with diameter $AC$ is centred at $O$ and intersects $BC$ and $CD$ at points $E$ and $F$ (other than $C$), respectively. Let $M$ be the midpoint of $BD$, and $AN \\perp BD$ with foot $N$, as shown in the figure below.\n\nShow that $M$, $N$, $E$, $F$ are concyclic.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p335_data_48fd17b1fe.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $G$ be the intersection of the circle $O$ and $AB$. Connect $AE$, $GC$, $GD$, $GF$, $EN$, $EF$, and $BF$, as shown in the figure. Evidently, we have\n\n$$\n\\angle BGF = 90^\\circ + \\angle CGF = 90^\\circ + \\angle CAF,\n$$\n\n$$\nS_{\\triangle BGF} = GB \\cdot GF \\sin \\angle BG = GB \\cdot GF \\cos \\angle CAF = \\frac{GB \\cdot GF \\cdot AF}{AC}.\n$$\n\nSimilarly, $S_{\\triangle DGF} = \\frac{DF \\cdot FG \\cdot CG}{AC}$.\n\nNote that $\\angle ABC = \\angle ADC$, and hence\n\n$$\n\\mathrm{Rt}\\triangle GBC \\sim \\mathrm{Rt}\\triangle FDA,\n$$\n\nwhich implies $\\frac{GB}{FD} = \\frac{CG}{AF}$, or\n\n$$\nGB \\cdot AF = FD \\cdot CG.\n$$\n\nIt follows that $S_{\\triangle BGF} = S_{\\triangle DGF}$, and $G$, $M$, $F$ are collinear.\n\nMoreover, from $AE \\perp EC$, $AN \\perp BN$, we infer that $A$, $B$, $E$, $N$ are concyclic,\n\n$$\n\\angle BNE = \\angle BAE = \\angle GAE = \\angle GFE = \\angle MFE,\n$$\n\nand thus $M$, $N$, $E$, $F$ are concyclic. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15433,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any given positive integers $m, n$, there exist infinitely many pairs of coprime positive integers $a, b$ such that $a + b \\mid am^a + bn^b$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $mn = 1$, the claim is valid. For $mn \\ge 2$, note that\n\n$$\nn^a (am^a + bn^b) = (a + b) n^{a + b} + a \\left( (mn)^a - n^{a + b} \\right).\n$$\n\nIt suffices to show there are infinitely many coprime pairs $a, b$ such that\n\n$$\na + b \\mid (mn)^a - n^{a + b}, \\quad \\gcd(a + b, n) = 1.\n$$\n\nLet $p = a + b$. We need infinitely many primes $p$ and $1 \\leq a \\leq p - 1$ such that\n\n$$\np \\mid (mn)^a - n^p.\n$$\n\nBy Fermat's Little Theorem, for $a_1 \\equiv a_2 \\pmod{p - 1}$, $(mn)^{a_1} \\equiv (mn)^{a_2} \\pmod{p}$. So it suffices to find infinitely many primes $p$ and $a$ such that\n\n$$\np \\mid (mn)^a - n. \\tag{1}\n$$\n\nSuppose only finitely many such primes exist: $p_1, \\ldots, p_r$. Then\n\n$$\n(mn)^2 - n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_r^{\\alpha_r}, \\quad \\alpha_i \\geq 0.\n$$\n\nLet $a = p_1^{\\alpha_1} \\cdots p_r^{\\alpha_r} (p_1 - 1) \\cdots (p_r - 1) + 2$. Suppose\n\n$$\n(mn)^a - n = p_1^{\\beta_1} \\cdots p_r^{\\beta_r}, \\quad \\beta_i \\geq 0.\n$$\n\nIf $p_i \\nmid n$, then $p_i^{\\beta_i} \\mid (mn)^2 - n$, so $\\beta_i \\leq \\alpha_i$. If $p_i \\nmid n$, then $p_i \\nmid m$, so $(p_i^{\\alpha_i + 1}, mn) = 1$. By Euler's theorem (since $\\varphi(p_i^{\\alpha_i + 1})$ divides $a - 2$),\n\n$$\n(mn)^a - n \\equiv (mn)^2 - n \\pmod{p_i^{\\alpha_i + 1}}.\n$$\n\nSince $p_i^{\\alpha_i + 1} \\nmid (mn)^2 - n$, it follows $\\beta_i \\leq \\alpha_i$. Thus,\n\n$$\n(mn)^a - n = p_1^{\\beta_1} \\cdots p_r^{\\beta_r} \\leq p_1^{\\alpha_1} \\cdots p_r^{\\alpha_r} = (mn)^2 - n < (mn)^a - n,\n$$\n\ncontradicting $a > 2$. Therefore, there are infinitely many such primes $p$ and integers $a$ with $p \\mid (mn)^a - n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15434,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be an integer, and let $f$ be a polynomial in $4n$ variables with real coefficients, such that for any $2n$ points $(x_1, y_1), \\dots, (x_{2n}, y_{2n})$ in the plane,\n\n$$\nf(x_1, y_1, \\dots, x_{2n}, y_{2n}) = 0\n$$\n\nif and only if the points are either all equal or form the vertices of a regular $2n$-gon (in some order). Determine the smallest possible degree of $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "The smallest possible degree is $2n$.\n\nLet $A_i = (x_i, y_i)$, and abbreviate $f(x_1, y_1, \\dots, x_{2n}, y_{2n})$ as $f(A_1, \\dots, A_{2n})$.\n\n**Claim (Sign of $f$):** $f$ attains only nonnegative or only nonpositive values.\n\n**Proof:** The zero-set of $f$ is very sparse: if $f$ takes both positive and negative values, we could move $A_1, \\dots, A_{2n}$ from a negative to a positive value without ever forming a regular $2n$-gon—a contradiction.\n\nTo show $\\deg f \\ge 2n$, animate the points $A_1, \\dots, A_{2n}$ linearly in a variable $t$; then $g(t) = f(A_1, \\dots, A_{2n})$ has degree at most $\\deg f$. The claim above implies any root of $g$ must be a multiple root, so if there are at least $n$ roots, $\\deg g \\ge 2n$, and thus $\\deg f \\ge 2n$.\n\nGeometrically, we can exhibit $2n$ linearly moving points that form a regular $2n$-gon exactly $n$ times. Draw $n$ mirrors through the origin, each making an angle of $\\frac{\\pi}{n}$ with the next. Any point $P$ has $2n$ reflections in these mirrors. Draw the $n$ angle bisectors of adjacent mirrors; the reflections of $P$ form a regular $2n$-gon if and only if $P$ lies on one of the bisectors. Animate $P$ along a line $\\ell$ intersecting all $n$ bisectors (but not passing through the origin), and let $P_1, \\dots, P_{2n}$ be its reflections. These points form a regular $2n$-gon exactly $n$ times, so $\\deg f \\ge 2n$.\n\n![](images/RMC_2023_v2_p110_data_5193d2e7cf.png)\n\nTo construct a polynomial $f$ of degree $2n$ with the desired property, first find a polynomial $g$ with this property when the input points have sum zero. Then set\n\n$$\nf(A_1, \\dots, A_{2n}) = g(A_1 - \\bar{A}, \\dots, A_{2n} - \\bar{A}),\n$$\n\nwhere $\\bar{A}$ is the centroid of $A_1, \\dots, A_{2n}$. Thus, we may assume $A_1 + \\dots + A_{2n} = 0$.\n\nConstruct $g$ as a sum of squares: $g = g_1^2 + g_2^2 + \\cdots + g_m^2$, so $g = 0$ iff $g_1 = \\cdots = g_m = 0$, and if their degrees are $d_1, \\dots, d_m$, then $g$ has degree at most $2 \\max(d_1, \\dots, d_m)$.\n\nIt suffices to exhibit polynomials of degree at most $n$ such that $2n$ points with zero sum are the vertices of a regular $2n$-gon iff all polynomials vanish.\n\nFirst, impose that all $|A_i|^2 = x_i^2 + y_i^2$ are equal (degree 2 constraints).\n\nNow, the points $A_1, \\dots, A_{2n}$ lie on a circle centered at $0$, and $A_1 + \\dots + A_{2n} = 0$. If the radius is $0$, all $A_i$ coincide; otherwise, the circle has positive radius.\n\n**Lemma:** Suppose $a_1, \\dots, a_{2n}$ are complex numbers of the same nonzero magnitude, and $a_1^k + \\dots + a_{2n}^k = 0$ for $k = 1, \\dots, n$. Then $a_1, \\dots, a_{2n}$ form a regular $2n$-gon centered at the origin.\n\n**Proof:** Assume $a_1, \\dots, a_{2n}$ lie on the unit circle. By Newton's sums, the $k$-th symmetric sums are zero for $k = 1, \\dots, n$. Taking conjugates yields $a_1^{-k} + \\dots + a_{2n}^{-k} = 0$ for $k = 1, \\dots, n$. These are the $(2n-k)$-th symmetric sums of $a_1, \\dots, a_{2n}$ (divided by $a_1 \\cdots a_{2n}$), so the first $2n-1$ symmetric sums are zero. Thus, $a_1, \\dots, a_{2n}$ form a regular $2n$-gon centered at the origin.\n\nExpress $a_r = x_r + y_r i$; the constraint $a_1^k + \\dots + a_{2n}^k = 0$ becomes $p_k + q_k i = 0$, where $p_k$ and $q_k$ are real polynomials. Impose $p_k = 0$ and $q_k = 0$; these have degree $k \\le n$, so their squares have degree at most $2n$.\n\nSumming squares of all these constraints gives a polynomial $f$ of degree at most $2n$ that works whenever $A_1 + \\dots + A_{2n} = 0$. The centroid-shifting trick gives a polynomial that works in general, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15435,
    "subject": "Mathematics (Olympiad)",
    "question": "On a circular billiard table, a ball is reflected from a cushion as if it were being reflected from the tangent of the circle at the point of reflection.\n\nA regular hexagon is drawn on the circular billiard table with its vertices on the circle.\n\nA ball (considered as a point) is placed on a side of the hexagon (not at a vertex). Determine a periodic path that the ball can take from this point with exactly four different points of reflection on the circle. In how many directions can the ball be brought onto such a path?",
    "options": [],
    "answer": "See solution",
    "solution": "Whenever a ball is reflected on the perimeter of the circular cushion, the incoming and outgoing angles to the radius must be equal. This means that the incoming and outgoing chords of the circle on the path of the ball must be of equal length. The periodic path must therefore be a regular polygon, and since it must have four corners, it must be a square.\n\nTo find a path through a given point, we can determine any square inscribed in the circle and rotate it around the center of the circle so that it passes through the given starting point. This is possible for all points on the hexagon, since the minimum distance of a point of the square from the center of the circle is $\\frac{\\sqrt{2}}{2}$ times the radius, but the minimum distance of a point of the hexagon from the center is $\\frac{\\sqrt{3}}{2}$ times the radius, which is larger.\n\n![](images/Austria_2010_p10_data_e14df24846.png)\n\nIn fact, rotating in such a way yields two possible positions for the squares, and since the ball can be started in either orientation of either square, there exist a total of 4 directions fulfilling the requirements.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15436,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to locate a disc of diameter 3 so that no grid square is completely covered by the disc?\n\nIf the centre of a disc of diameter 3 is at the centre of a grid square, then it is the only grid square completely covered by the disc, as this diagram shows.\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p56_data_c4324cb152.png)\n\nWhat is the minimum and maximum number of grid squares that can be completely covered by a disc of diameter 3?",
    "options": [],
    "answer": "See solution",
    "solution": "The centre of the disc lies somewhere on a grid square. The longest line in a square is its diagonal, which has length $\\sqrt{1^2 + 1^2} = \\sqrt{2}$. So the distance from the disc centre to any point on the perimeter of the square is at most $\\sqrt{2} < 1.5$. Hence the disc will cover that square. So a disc of diameter 3 always covers at least one grid square.\n\nTherefore, the minimum number of grid squares that can be covered by a disc of diameter 3 is 1.\n\nSuppose the centre of the disc is at a grid point. The diagonal of a grid square is $\\sqrt{1^2 + 1^2} = \\sqrt{2} < 1.5$. Hence the disc covers 4 grid squares, as this diagram shows:\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p56_data_056f8c8a1b.png)\n\nFrom Part b, the diameter of a disc that covers five grid squares is at least $\\sqrt{10}$. Since $\\sqrt{10} > 3$, five grid squares cannot be covered by a disc of diameter 3.\n\nHence, 4 is the maximum number of grid squares that can be covered by a disc of diameter 3.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15437,
    "subject": "Mathematics (Olympiad)",
    "question": "For $k = 0, 1, 2, 3, 4, 5$, let $A_k$ be the set of all multiples of $3$ less than or equal to $10^6$ whose $10^k$'s digit is $1$. Let $N_k$ be the number of elements in $A_k$. Find $$\\sum_{k=0}^5 N_k.$$",
    "options": [],
    "answer": "See solution",
    "solution": "The set $A_5$ consists of multiples of $3$ between $100000$ and $200000$, starting with $100002$ and ending with $199998$. There are $33333$ such numbers, so $N_5 = 33333$.\n\nFor $0 \\leq k \\leq 4$, for each $n \\in A_k$, associate $n'$ by swapping the $10^k$'s digit and the $10^5$'s digit (padding with zeros if needed). The sum of digits remains unchanged, so $n'$ is also a multiple of $3$ and belongs to $A_5$. This gives a one-to-one correspondence between $A_k$ and $A_5$, so $N_k = 33333$ for each $k = 0, 1, 2, 3, 4$.\n\nThus, $$\\sum_{k=0}^5 N_k = 6 \\times 33333 = 199998.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15438,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a scalene triangle. Let $P$ be an interior point of $ABC$ such that $AP \\perp BC$. Assume that $BP$ and $CP$ intersect $AC$ and $AB$ at $X$ and $Y$, respectively. Prove that $AX = AY$ if and only if there exists a circle with centre lying on $BC$ and tangent to $AB$ and $AC$ at points $Y$ and $X$, respectively.",
    "options": [],
    "answer": "See solution",
    "solution": "($\\Leftrightarrow$) is trivial.\n\nFor ($\\Rightarrow$), let $AP \\cap BC = D$. Now $AD$ is a bisector of angle $XDY$. For this we can argue in two ways:\n\n* Let $XY \\cap BC = S$. Then $D(S, D; B, C) = 1$ and $\\angle ADB = 90^\\circ$, so statement.\n* Let $\\ell$ be a parallel line to $BC$ passing through $A$. Let $DY \\cap \\ell = K$ and $DX \\cap \\ell = L$. Then using Ceva and Thales' theorem we see that $A$ is a midpoint of $KL$.\n\nSince $ABC$ is scalene, we see that in triangle $XDY$, bisector $DA$ and the perpendicular bisector of $XY$ intersect at $A$, so $AXDY$ is inscribed in circle $\\omega$. Now let $Z$ be the second intersection of $\\omega$ with $BC$. Then $AXZY$ is a kite, so the circle with centre $Z$ and radius $ZX$ satisfies the problem statement.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15439,
    "subject": "Mathematics (Olympiad)",
    "question": "Mary and Nick play a game with a heap of 25201 stones. Each player, in turn, removes $n$ or $m$ stones from the heap, where $n$ and $m$ are fixed positive integers between 1 and 10. The player who removes the last stone wins. Can Mary always win, regardless of Nick's choices?",
    "options": [],
    "answer": "See solution",
    "solution": "If Nick fixes the number $n$ and defines himself as the first player, then Mary wins if she sets $m = 11 - n$. Indeed, if Nick removes $n$ ($m$) stones, then Mary removes $m$ ($n$) stones. So exactly $11 = m + n$ stones are removed from the heap after each pair of moves (Nick - Mary). Since $25201 = 2291 \\cdot 11 + 0$, Mary wins.\n\nIf Nick fixes the number $n$ and defines Mary as the first player, then Mary wins if she sets $m = 1$ for $n = 2, 3, 4, 5, 6, 7, 8, 9$ and her first move is to remove 1 stone from the heap. 25200 stones remain in the heap. It is easy to see that 25200 is divisible by $n + m$ for these values. Therefore, Mary wins if she uses the symmetric strategy: when Nick removes $n$ ($m$) stones, Mary removes $m$ ($n$) stones. So exactly $m + n$ stones are removed from the heap after each pair of moves (Nick - Mary).\n\nIf Nick sets $n = 1$, then Mary can choose any number from 2 to 9 for $m$, and play in the same way as above (first move is 1 stone, and symmetric strategy).\n\nIf Nick sets $n = 10$, then Mary wins if she sets $m = 7$ and her first move is to remove 7 stones from the heap. 25194 stones remain in the heap. It is easy to see that $25194 = 1482 \\cdot 17$ is divisible by 17, i.e. 25194 is divided by $n + m$. To win, Mary can use the symmetric strategy.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15440,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all prime numbers $p$ and positive integers $a, b, c$ such that\n\n$$\n2^a p^b = (p + 2)^c + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Obviously, $p$ is odd, so $p \\ge 3$. If $c = 1$, then\n\n$$\np + 3 = 2^a p^b \\ge 2p \\ge p + 3,\n$$\n\nThis can only happen when $p = 3$, $a = b = 1$, giving one solution $(p, a, b, c) = (3, 1, 1, 1)$. In what follows, we assume that $c \\ge 2$.\n\n**Case 1:** $c$ is odd. Assume that $q$ is a prime factor of $c$. Since $(p+2)^q+1 \\mid (p+2)^c+1$,\n\n$$\n(p + 2)^q + 1 = 2^\\alpha p^\\beta. \\qquad (1)\n$$\n\nObviously, $\\alpha > 0$. Note that $(p+2)^q + 1 = (p+3)A$, where\n\n$$\nA = (p+2)^{q-1} - (p+2)^{q-2} + \\cdots + 1 > (p+2)^{q-1} - (p+2)^{q-2} = (p+2)^{q-2}(p+1) > p^{q-1},\n$$\n\nand $A$ is odd, so $A$ is a power of $p$. So $A \\ge p^q$ and $\\beta \\ge q$. Taking both sides of (1) modulo $p$ gives $2^q \\equiv -1 \\pmod p$, so the order of $2 \\pmod p$ is $2$ or $2q$.\n\nIf the order of $2 \\pmod p$ is $2$, then $p = 3$. In this case, $(1)$ is $5^q + 1 = 2^\\alpha 3^\\beta$. Since $5^q + 1 \\equiv 2 \\pmod 4$, we have $\\alpha = 1$. By Lifting-the-exponent Lemma, $v_3(5^q + 1) = v_3(5+1) + v_3(q) \\le 2$, so $\\beta \\le 2$. One checks that $\\beta = 1, 2$ does not satisfy the equation.\n\nIf the order of $2 \\pmod p$ is $2q$, then $2q \\mid p-1$. So $q \\le \\frac{p-1}{2} < \\frac{p}{2}$. Dividing both sides of (1) by $p^q$ and using $(1+\\frac{1}{x})^x < e$, $x \\ge 1$, we have\n\n$$\n2^\\alpha p^{\\beta-q} = \\left(1 + \\frac{2}{p}\\right)^q + p^{-q} < \\left(1 + \\frac{2}{p}\\right)^{\\frac{p}{2}} + p^{-q} < e + 3^{-3} < 3.\n$$\n\nSo $\\beta = q$ and $\\alpha = 1$. Using $2 \\cdot p^q = (p+2)^q + 1 = (p+3)A$, we get $A = p^q$ and $p+3=2$. This is a contradiction.\n\n**Case 2:** $c$ is even. Assume that $2^d \\nmid c$ with $d \\ge 1$. Then $(p+2)^{2d} + 1 \\mid (p+2)^c + 1$. So\n\n$$\n(p + 2)^{2d} + 1 = 2^\\alpha p^\\beta.\n$$\n\nSince $(p+2)^{2d} + 1 \\equiv 2 \\pmod 4$, $\\alpha = 1$.\n\n$$\n(p + 2)^{2d} + 1 = 2 \\cdot p^{\\beta}. \\qquad (2)\n$$\n\nTaking both sides of (2) modulo $p$ gives $2^{2d} \\equiv -1 \\pmod p$. So the order of $2 \\pmod p$ is $2^{d+1}$. Thus, $2^d < \\frac{p}{2}$. Since\n\n$$\np^{\\beta+1} > 2 \\cdot p^{\\beta} = (p+2)^{2d} + 1 > p^{2d},\n$$\n\nwe have $\\beta \\ge 2^d$. Dividing both sides of (2) by $p^{2^d}$, we have\n\n$$\n2 \\cdot p^{\\beta-2^d} = \\left(1 + \\frac{2}{p}\\right)^{2^d} + p^{-2^d} < \\left(1 + \\frac{2}{p}\\right)^{\\frac{p}{2}} + p^{-2^d} < e + 3^{-2} < 3,\n$$\n\nso $\\beta = 2^d$.\n\nIf $d \\ge 2$, then $2^{d+1} \\mid p-1$ implies that $p \\equiv 1 \\pmod 8$. Using (2), we have $p^{2^d} - 1 = (p+2)^{2^d} - p^{2^d}$. Analyzing the number of factors $2$ on both sides of the equality, we deduce that\n\n$$\nv_2(p^{2^d} - 1) = v_2(p^2 - 1) + d - 1 \\ge d + 3.\n$$\n\nYet,\n\n$$\nv_2((p+2)^{2^d} - p^{2^d}) = v_2((p+2)^2 - p^2) + d - 1 = v_2(2) + v_2(2p+2) + d - 1 = d + 2,\n$$\n\ngiving a contradiction. So $d=1$ and $2p^2 = (p+2)^2 + 1$, which implies that $p=5$. Going back to the original equation, we have $c = 2k$, with $k$ odd. But $a=1$, so we have\n\n$$\n2 \\cdot 5^b = 7^{2k} + 1.\n$$\n\nUsing Lifting-the-exponent Lemma again, we have $b = v_5(7^{2k} + 1) = v_5(7^2 + 1) + v_5(k) \\le 2 + \\frac{k}{5}$. If $k \\ge 3$, then\n\n$$\n2 \\cdot 5^b \\le 2 \\cdot 5^{2+\\frac{k}{5}} = (7^2 + 1) \\cdot 5^{\\frac{k}{5}} < 7^{2k} + 1,\n$$\n\ngiving a contradiction. So $k=1$, and thus $c=2$ and $b=2$. This gives another solution $(p, a, b, c) = (5, 1, 2, 2)$.\n\nTo sum up, there are two solutions of $(p, a, b, c)$: $(3, 1, 1, 1)$ and $(5, 1, 2, 2)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15441,
    "subject": "Mathematics (Olympiad)",
    "question": "After $x$ matches in a championship, a team had exactly $n\\%$ of wins, where $x$ and $n$ are positive integers. What is the least $x$ for which it is possible that after the $(x+1)$-st match, the team had exactly $(n+1)\\%$ of wins?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $x = 24$.\n\nIt follows from the conditions that after $x$ matches, the team had $y$ wins such that $\\frac{y}{x} = \\frac{n}{100}$. Then, we require:\n\n$$\n\\frac{y+1}{x+1} = \\frac{n+1}{100}\n$$\n\nFrom the first equation, $100y = nx$, so $100$ divides $nx$.\n\nFrom the second equation:\n\n$$\n100(y+1) = (n+1)(x+1) \\\\\n100y + 100 = nx + n + x + 1\n$$\n\nSubstitute $100y = nx$:\n\n$$\nnx + 100 = nx + n + x + 1 \\\\\n100 = n + x + 1 \\\\\nx = 100 - n - 1\n$$\n\nLet $n = 100 - x - 1$. Since $100$ divides $nx$, substitute $n$:\n\n$$\nxn = x(100 - x - 1) = 100x - x^2 - x\n$$\n\nSo $100$ must divide $x^2 + x = x(x+1)$. Since $x$ and $x+1$ are coprime and less than $100$, one must be divisible by $4$ and the other by $25$. The smallest such $x$ is $24$.\n\nThus, $n = 75$ and $y = 18$. Check:\n\n$$\n\\frac{y}{x} = \\frac{18}{24} = 75\\% \\qquad \\frac{y+1}{x+1} = \\frac{19}{25} = 76\\%\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15442,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AB\\Gamma\\Delta$ ($A\\Delta \\parallel B\\Gamma$) be a trapezium with $\\angle A = \\angle B = 90^\\circ$ and $A\\Delta < B\\Gamma$. Denote by $E$ the point of intersection of the two non-parallel sides $AB$ and $\\Gamma\\Delta$, $Z$ the symmetric point of $A$ with respect to the line $B\\Gamma$, and $M$ the midpoint of $EZ$. It is given that the line $\\Gamma M$ is perpendicular to the line $\\Delta Z$. Prove that the line $Z\\Gamma$ is perpendicular to the line $\\Gamma E$.\n\n![](images/Hellenic_2016_p6_data_7a836e4394.png \"Figure 1\")",
    "options": [],
    "answer": "See solution",
    "solution": "Let the line $\\Delta Z$ meet $\\Gamma M$ and $B\\Gamma$ at points $K$ and $N$, respectively. In triangle $A\\Delta Z$, $B$ is the midpoint of $AZ$ and $BN \\parallel A\\Delta$. Hence, $N$ is the midpoint of $Z\\Delta$. Therefore, in triangle $ZE\\Delta$, $MN$ connects the midpoints of two sides, and thus:\n\n$$\nMN \\parallel E\\Delta \\qquad (1)\n$$\n\nMoreover, in triangle $M\\Gamma Z$, $\\Gamma B$ and $ZK$ are altitudes, so $N$ is the orthocenter of the triangle. Hence:\n\n$$\nMN \\perp Z\\Gamma \\qquad (2)\n$$\n\nFrom (1) and (2), we conclude that $Z\\Gamma \\perp E\\Gamma$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15443,
    "subject": "Mathematics (Olympiad)",
    "question": "A continent has a finite number of castles, and each castle belongs to exactly one of the two countries $A$ and $B$. Each castle has one general. We say two castles are *neighboring* if there is a path between those two castles. We also say that a castle $P$ and a set of castles $Q$ is *neighboring* if $P$ is neighboring some castle in $Q$.\n\nProve that the following two conditions are equivalent:\n\n1. If some of the generals from country $B$ move to some neighboring castles and attack, generals from country $A$ can move to some neighboring castles so that at each castle from country $A$, the number of generals from $A$ is greater than or equal to that of generals from $B$.\n\n2. If $X$ is a set of castles from country $A$, the number of generals from $A$ who are in $X$ or in castles neighboring $X$ is greater than or equal to the number of generals from $B$ who are in castles neighboring $X$.",
    "options": [],
    "answer": "See solution",
    "solution": "The implication $(1) \\Rightarrow (2)$ is obvious. We prove the converse.\n\nLet $G = (V, E)$ be a graph where $V$ is the set of castles and each edge corresponds to a neighboring relation. Let $A = \\{v_1, v_2, \\dots, v_n\\}$ be the set of castles belonging to country $A$. Now let $P_i \\subseteq V - A$ be the set of castles whose generals decide to attack castles in $A$. Let $Q_i \\subseteq A$ be the set of castles which are either $v_i$ itself or neighboring $v_i$.\n\nConsider the list of sets $Q_1$ ($|P_1|$ times), $Q_2$ ($|P_2|$ times), ..., $Q_i$ ($|P_i|$ times), ..., $Q_n$ ($|P_n|$ times). If we choose a collection $Q_{i_1}, Q_{i_2}, \\dots, Q_{i_k}$ in this list, we have\n\n$$\n|P_{i_1}| + |P_{i_2}| + \\dots + |P_{i_k}| \\leq |Q_{i_1} \\cup \\dots \\cup Q_{i_k}|\n$$\n\nby (2). Therefore, Hall's Marriage Theorem implies that for each $i = 1, 2, \\dots, n$, we can choose mutually disjoint $D_1, D_2, \\dots, D_n$ such that $|D_i| = |P_i|$ and $D_i \\subseteq Q_i$.\n\n$\\Box$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15444,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\gamma$ be a circle and $\\ell$ a line in the plane. Fix a point $K$ on $\\ell$ exterior to $\\gamma$ and draw the tangents $KA$ and $KB$ to the circle $\\gamma$, where $A$ and $B$ are on $\\gamma$. Let $P$ and $Q$ be two points on $\\gamma$. The lines $PA$ and $PB$ meet $\\ell$ at $R$ and $S$, respectively, and the lines $QR$ and $QS$ meet $\\gamma$ again at $C$ and $D$, respectively. Prove that $\\ell$ and the tangents of $\\gamma$ at $C$ and $D$, respectively, are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "The conclusion follows at once by Pascal's theorem applied in turn to the three hexagrams *APBCQD*, *AADBBC*, and *CCBDDA*.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15445,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine if there exists a pair of proper fractions that cannot be simplified so that their difference equals their product and one of the denominators is $2019$. If such a pair exists, find at least two such pairs of fractions.",
    "options": [],
    "answer": "See solution",
    "solution": "The solution is based on a simple fact:  \n$$\n\\frac{1}{n} - \\frac{1}{n+1} = \\frac{1}{n(n+1)} = \\frac{1}{n} \\cdot \\frac{1}{n+1}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15446,
    "subject": "Mathematics (Olympiad)",
    "question": "There is a group of $2n$ people, among whom there are pairs of friends. Each person in the group has exactly $k \\geq 1$ friends (friendship is mutual: if \"A\" is friends with \"B\", then \"B\" is friends with \"A\").\n\nFind all values of $k$ for which this group can always be divided into two subgroups of $n$ people each, so that in both subgroups every person has at least one friend within their subgroup.",
    "options": [],
    "answer": "See solution",
    "solution": "If $k=1$, it is always impossible for $2n = 4m-2$, where $m \\in \\mathbb{N}$. If $k=2$, it is always impossible for $2n = 6m+2$, where $m \\in \\mathbb{N}$. For all other cases, it is always possible.\n\nFor $k=1$, all friends form pairs. To split everyone properly, each pair must be in the same subgroup. Each subgroup must have $n$ people, so if there is an even number of pairs ($n=2m$), it can be done by putting $m$ pairs in each subgroup. If $n=2m-1$, then in any division into two equal parts, at least one pair will be split, so someone will not have a friend in their subgroup.\n\nNow, let $k \\geq 2$ and suppose there is a division in which it is impossible to split the group as required. Consider an arbitrary split into two subgroups of $n$ people. Represent people as points, divided by a vertical line into left (L) and right (R) subgroups. Draw segments between all pairs of friends. Segments between people in the same subgroup are called *important*. Among all possible divisions, choose one maximizing the number of important segments (a maximal division).\n\nSuppose, in a maximal division, there is a person $X$ in the left subgroup whose $k$ friends $A_1, \\dots, A_k$ are all in the right subgroup. Consider two cases:\n\nSuppose there are two people $B \\in L$ and $C \\in R$ who are friends. The segment between them is not important. Suppose $C$ has $l \\geq 1$ friends in the left side and $k-l$ friends in the right side. Swap $X$ and $C$ between subgroups. The left subgroup gains $l$ important segments, and the right gains $l$ more, increasing the total number of important segments, contradicting maximality.\n\nThus, in a maximal division, there is a person $X \\in L$ whose friends are all in the right subgroup, while all other people have friends only in their own subgroup.\n\n**Case $k=2$:** People form chains of 3 or more. Chains of 3 must be in the same subgroup. If there is a chain of 4 or more, rearrangements can achieve the required division. If only chains of 3 exist, the division may not be possible for certain $n$ (specifically, $n=2+6m$).\n\n**Case $k \\geq 3$:** People form connected components of at least 4 elements, each with degree $k \\geq 3$. Rearranging components as described above allows the required division except for certain small cases.\n\n**Conclusion:**\n- For $k=1$, division is possible if $n$ is even.\n- For $k=2$, division is impossible for $2n = 6m+2$.\n- For $k \\geq 3$, division is always possible.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15447,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ and $n$ be integers such that $k \\geq 2$ and $k \\leq n \\leq 2k-1$. Place rectangular tiles, each of size $1 \\times k$ or $k \\times 1$, on an $n \\times n$ chessboard so that each tile covers exactly $k$ cells, and no two tiles overlap. Do this until no further tile can be placed in this way. For each such $k$ and $n$, determine the minimum number of tiles such an arrangement may contain.",
    "options": [],
    "answer": "See solution",
    "solution": "The required minimum is $n$ if $n = k$, and it is $\\min\\{n, 2n - 2k + 2\\}$ if $k < n < 2k$.\n\nThe case $n = k$ is clear. Assume henceforth $k < n < 2k$. We describe a maximal arrangement of the board $[0, n] \\times [0, n]$ with the above cardinalities.\n\nIf $k < n < 2k - 1$, then $\\min\\{n, 2n - 2k + 2\\} = 2n - 2k + 2$. To obtain a maximal arrangement of this cardinality, place four tiles: $[0, k] \\times [0, 1]$, $[0, 1] \\times [0, k+1]$, $[1, k+1] \\times [k, k+1]$, and $[k, k+1] \\times [0, k]$ in the square $[0, k] \\times [0, k]$; stack $n - k - 1$ horizontal tiles in the rectangle $[1, k+1] \\times [k+1, n]$, and erect $n - k - 1$ vertical tiles in the rectangle $[k+1, n] \\times [1, k+1]$.\n\nIf $n = 2k - 1$, then $\\min\\{n, 2n - 2k + 2\\} = n = 2k - 1$. A maximal arrangement of $2k - 1$ tiles is obtained by stacking $k - 1$ horizontal tiles in the rectangle $[0, k] \\times [0, k-1]$, another $k - 1$ horizontal tiles in $[0, k] \\times [k, 2k-1]$, and adding the horizontal tile $[k-1, 2k-1] \\times [k-1, k]$.\n\nThe above examples show that the required minimum does not exceed the mentioned values.\n\nTo prove the reverse inequality, consider a maximal arrangement and let $r$, respectively $c$, be the number of rows, respectively columns, not containing a tile.\n\nIf $r = 0$ or $c = 0$, the arrangement clearly contains at least $n$ tiles.\n\nIf $r$ and $c$ are both positive, we show that the arrangement contains at least $2n - 2k + 2$ tiles. To this end, we will prove that the rows, respectively columns, not containing a tile are consecutive. Assume this for the moment; these $r$ rows and $c$ columns cross to form an $r \\times c$ rectangular array containing no tile at all, so $r < k$ and $c < k$ by maximality. Consequently, there are $n - r \\geq n - k + 1$ rows containing at least one horizontal tile each, and $n - c \\geq n - k + 1$ columns containing at least one vertical tile each, whence a total of at least $2n - 2k + 2$ tiles.\n\nWe now show that the rows not containing a tile are consecutive; columns are dealt with similarly. Consider a horizontal tile $T$. Since $n < 2k$, the nearest horizontal side of the board is at most $k - 1$ rows away from the row containing $T$. These rows, if any, cross the $k$ columns $T$ crosses to form a rectangular array no vertical tile fits in. Maximality forces each of these rows to contain a horizontal tile and the claim follows.\n\nConsequently, the cardinality of every maximal arrangement is at least $\\min\\{n, 2n - 2k + 2\\}$, and the conclusion follows.\n\n**Remarks.**\n\n(1) If $k \\geq 3$ and $n = 2k$ the minimum is $n + 1 = 2k + 1$ and is achieved, for instance, by the maximal arrangement consisting of the vertical tile $[0, 1] \\times [1, k+1]$ along with $k - 1$ horizontal tiles stacked in $[1, k+1] \\times [0, k-1]$, another $k - 1$ horizontal tiles stacked in $[1, k+1] \\times [k+1, 2k]$, and two horizontal tiles stacked in $[k, 2k] \\times [k-1, k+1]$. This example shows that the corresponding minimum does not exceed $n + 1 < 2n - 2k + 2$. The argument in the solution also applies to the case $n = 2k$ to infer that for a maximal arrangement of minimal cardinality either $r = 0$ or $c = 0$, and the cardinality is at least $n$. Clearly, we may and will assume $r = 0$. Suppose, if possible, such an arrangement contains exactly $n$ tiles. Since there is no room left for an additional tile, some tile $T$ must cover a cell of the leftmost column, so it covers the $k$ leftmost cells along its row, and there is then room for another tile along that row—a contradiction.\n\n(2) For every pair $(r, c)$ of integers in the range $2k - n, \\dots, k - 1$, at least one of which is positive, say $c > 0$, there exists a maximal arrangement of cardinality $2n - r - c$.\n\nUse again the board $[0, n] \\times [0, n]$ to stack $k - r$ horizontal tiles in each of the rectangles $[0, k] \\times [0, k - r]$ and $[k - c, 2k - c]$, erect $k - c$ vertical tiles in each of the rectangles $[0, k - c] \\times [k - r, 2k - r]$ and $[k, 2k - c] \\times [0, k]$, then stack $n - 2k + r$ horizontal tiles in the rectangle $[k - c, 2k - c] \\times [2k - r, n]$, and erect $n - 2k + c$ vertical tiles in the rectangle $[2k - c, n] \\times [1, k+1]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15448,
    "subject": "Mathematics (Olympiad)",
    "question": "Compute $1 - \\left( \\frac{7}{12} + \\frac{5}{12} \\times \\frac{7}{11} + \\frac{5}{12} \\times \\frac{4}{11} \\times \\frac{7}{10} + \\frac{5}{12} \\times \\frac{4}{11} \\times \\frac{3}{10} \\times \\frac{7}{9} \\right)$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\begin{aligned}\n&1 - \\left( \\frac{7}{12} + \\frac{5}{12} \\times \\frac{7}{11} + \\frac{5}{12} \\times \\frac{4}{11} \\times \\frac{7}{10} + \\frac{5}{12} \\times \\frac{4}{11} \\times \\frac{3}{10} \\times \\frac{7}{9} \\right) \\\\\n&= \\frac{5}{12} - \\left( \\frac{5}{12} \\times \\frac{7}{11} + \\frac{5}{12} \\times \\frac{4}{11} \\times \\frac{7}{10} + \\frac{5}{12} \\times \\frac{4}{11} \\times \\frac{3}{10} \\times \\frac{7}{9} \\right) \\\\\n&= \\frac{5}{12} \\times \\frac{4}{11} - \\left( \\frac{5}{12} \\times \\frac{4}{11} \\times \\frac{7}{10} + \\frac{5}{12} \\times \\frac{4}{11} \\times \\frac{3}{10} \\times \\frac{7}{9} \\right) \\\\\n&= \\frac{5}{12} \\times \\frac{4}{11} \\times \\frac{3}{10} - \\left( \\frac{5}{12} \\times \\frac{4}{11} \\times \\frac{3}{10} \\times \\frac{7}{9} \\right) \\\\\n&= \\frac{5}{12} \\times \\frac{4}{11} \\times \\frac{3}{10} \\times \\frac{2}{9} \\\\\n&= \\frac{1}{99}\n\\end{aligned}\n$$\n\nSo the answer is $\\frac{98}{99}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15449,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ denote the circumcentre of an acute-angled triangle $ABC$. A circle $\\Gamma$ passing through vertex $A$ intersects segments $AB$ and $AC$ at points $P$ and $Q$ such that $\\angle BOP = \\angle ABC$ and $\\angle COQ = \\angle ACB$. Prove that the reflection of $BC$ in the line $PQ$ is tangent to $\\Gamma$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the circumcircle of triangle $OBP$ intersect side $BC$ at the points $R$ and $B$, and let $\\angle A$, $\\angle B$, and $\\angle C$ denote the angles at vertices $A$, $B$, and $C$, respectively. Now note that since $\\angle BOP = \\angle B$ and $\\angle COQ = \\angle C$, it follows that\n\n$$\n\\angle POQ = 360^{\\circ} - \\angle BOP - \\angle COQ - \\angle BOC = 360^{\\circ} - (180^{\\circ} - \\angle A) - 2\\angle A = 180^{\\circ} - \\angle A.\n$$\n\nThis implies that $APOQ$ is a cyclic quadrilateral. Since $BPOR$ is cyclic,\n\n$$\n\\angle QOR = 360^{\\circ} - \\angle POQ - \\angle POR = 360^{\\circ} - (180^{\\circ} - \\angle A) - (180^{\\circ} - \\angle B) = 180^{\\circ} - \\angle C.\n$$\n\nThis implies that $CQOR$ is a cyclic quadrilateral. Since $APOQ$ and $BPOR$ are cyclic,\n\n$$\n\\angle QPR = \\angle QPO + \\angle OPR = \\angle OAQ + \\angle OBR = (90^{\\circ} - \\angle B) + (90^{\\circ} - \\angle A) = \\angle C.\n$$\n\nSince $CQOR$ is cyclic, $\\angle QRC = \\angle COQ = \\angle C = \\angle QPR$, which implies that the circumcircle of triangle $PQR$ is tangent to $BC$. Further, since $\\angle PRB = \\angle BOP = \\angle B$,\n\n$$\n\\angle PRQ = 180^{\\circ} - \\angle PRB - \\angle QRC = 180^{\\circ} - \\angle B - \\angle C = \\angle A = \\angle PAQ.\n$$\n\nThis implies that the circumcircle of $PQR$ is the reflection of $\\Gamma$ in line $PQ$. By symmetry in line $PQ$, this implies that the reflection of $BC$ in line $PQ$ is tangent to $\\Gamma$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15450,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_{33}$ be different integer roots of $f(x) = p(x)q(x) - 2015$.\n\nWhat is the minimal possible degree of the polynomials $p(x)$ and $q(x)$, given that $p(a_i)q(a_i) = 2015$ for all $i = 1, 2, \\dots, 33$?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $p(a_i)q(a_i) = 2015$ for $33$ distinct integers $a_i$, each $p(a_i)$ is a divisor of $2015$. $2015 = 5 \\cdot 13 \\cdot 31$ has $16$ divisors (8 positive, 8 negative). By the Pigeonhole Principle, at least three of the $p(a_i)$ are equal, say $p(a_1) = p(a_2) = p(a_3) = d$. Thus, $p(x) - d$ has at least three distinct roots, so $\\deg p(x) \\geq 3$. The same argument applies to $q(x)$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15451,
    "subject": "Mathematics (Olympiad)",
    "question": "Denote the side lengths of squares $A$, $B$, $C$, $D$, $E$ by $a$, $b$, $c$, $d$, $e$ respectively. The two right-angled triangles enclosed by squares $C$, $D$, $E$ and the line are congruent to each other and have sides $c$, $d$, $e$. Given $c^2 = 16$ and $d^2 = 40$, find $e^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "By the Pythagorean theorem, $d^2 = c^2 + e^2$. Substituting the given values:\n\n$$\n40 = 16 + e^2\n$$\nSo,\n$$\ne^2 = 40 - 16 = 24\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15452,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the centre of the circumcircle of triangle $ABC$. Points $E$ and $F$ are chosen on $OB$ and $OC$ in such a way that $BE = OF$. Let $M$ and $N$ be the midpoints of the arcs $AOE$ and $AOF$ of the circumcircles of triangles $AOE$ and $AOF$ respectively. Prove that\n$$\n\\angle ENO + \\angle FMO = 2\\angle BAC.\n$$\n\n![](images/Ukrajina_2013_p36_data_41f0fac166.png)\n\nFig. 26",
    "options": [],
    "answer": "See solution",
    "solution": "Let $D$ be the point which is symmetric to $A$ with respect to $BC$. Then\n$$\n\\angle AOC = 2\\angle ABC = \\angle ABD, \\quad OA = OB, \\quad BA = BD.\n$$\nThen triangles $AOC$ and $ABD$ are similar. By analogy, $\\angle AOB = \\angle ACD$ and triangles $AOB$ and $ACD$ are similar. From the similarity, it follows that there exist points $P$ and $Q$ on the sides $BD$ and $CD$ such that $\\angle APB = \\angle AFO$ and $\\angle AQC = \\angle AEO$. Since $\\angle ABP = \\angle AOF$, triangles $ABP$ and $AOF$ are similar. So we have $\\frac{BP}{BD} = \\frac{BP}{BA} = \\frac{OF}{OA} = \\frac{BE}{BO}$. Thus, $PE \\parallel DO$. By analogy, $QF \\parallel DO$.\n\nTriangle $AME$ is isosceles. Moreover, $\\angle AME = \\angle AOE = \\angle AOB$ and\n$$\n\\triangle AME \\sim \\triangle AOB.\n$$\nHence, $\\angle BAE = \\angle OAM$ and $\\frac{AB}{AE} = \\frac{AO}{AM}$, so triangles $BAE$ and $OAM$ are similar. From the similarity, $\\frac{OM}{BE} = \\frac{AO}{AB}$ and $\\angle AOM = \\angle ABE$.\n\nNow, triangle $AOF$ is similar to $ABP$ and $\\frac{OA}{BA} = \\frac{OF}{BP}$, $\\angle AOF = \\angle ABP$. Thus, $\\frac{OM}{BE} = \\frac{OF}{BP}$ and\n$$\n\\angle MOF = \\angle AOF - \\angle AOM = \\angle ABP - \\angle ABE = \\angle EBP.\n$$\nWe have that triangles $MOF$ and $FCQ$ are similar. By analogy, triangles $NOE$ and $FCQ$ are similar too. As a result,\n$$\n\\angle ENO + \\angle FMO = \\angle QFC + \\angle PEB = \\angle DOC + \\angle DOB = \\angle BOC = 2\\angle BAC\n$$\nand the result follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15453,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. For any positive integer $j$ and positive real number $r$, define\n\n$$\nf_j(r) = \\min(jr, n) + \\min\\left(\\frac{j}{r}, n\\right), \\quad \\text{and} \\quad g_j(r) = \\min(\\lceil jr \\rceil, n) + \\min\\left(\\left\\lceil \\frac{j}{r} \\right\\rceil, n\\right),\n$$\n\nwhere $\\lceil x \\rceil$ denotes the smallest integer greater than or equal to $x$. Prove that\n\n$$\n\\sum_{j=1}^{n} f_j(r) \\le n^2 + n \\le \\sum_{j=1}^{n} g_j(r).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution:**\n\nWe first prove the left-hand side inequality. Draw an $n \\times n$ board with corners at $(0,0)$, $(n,0)$, $(0,n)$, and $(n,n)$ on the Cartesian plane.\n\nConsider the line $\\ell$ with slope $r$ passing through $(0,0)$. For each $j \\in \\{1, \\dots, n\\}$, consider the point $(j, \\min(jr, n))$. Each such point lies either on $\\ell$ or the top edge of the board. In the $j$th column from the left, draw the rectangle of height $\\min(jr, n)$. The sum of the $n$ rectangles equals the area under $\\ell$ plus $n$ triangles (possibly with area $0$), each with width at most $1$ and total height at most $n$. Thus, the sum of the areas of these $n$ triangles is at most $n/2$. Therefore, $\\sum_{j=1}^n \\min(jr, n)$ is at most the area under $\\ell$ plus $n/2$.\n\nConsider the line with slope $1/r$. By symmetry about $y=x$, the area under the line with slope $1/r$ equals the area above $\\ell$. Using the same reasoning, $\\sum_{j=1}^n \\min(j/r, n)$ is at most the area above $\\ell$ plus $n/2$.\n\nTherefore,\n$$\n\\sum_{j=1}^n f_j(r) = \\sum_{j=1}^n \\left(\\min(jr, n) + \\min\\left(\\frac{j}{r}, n\\right)\\right)\n$$\nis at most the area of the board plus $n$, which is $n^2 + n$. This proves the left-hand side inequality.\n\nTo prove the right-hand side inequality, we use the following lemma:\n\n**Lemma:** Consider the line $\\ell$ with slope $s$ passing through $(0,0)$. The number of squares on the board containing an interior point below $\\ell$ is $\\sum_{j=1}^n \\min(\\lceil js \\rceil, n)$.\n\n*Proof of Lemma:* For each $j \\in \\{1, \\dots, n\\}$, count the number of squares in the $j$th column containing an interior point below $\\ell$. The line $x = j$ intersects $\\ell$ at $(j, js)$. Each column contains $n$ squares, so the number is $\\min(\\lfloor js \\rfloor, n)$. Summing over all $j$ proves the lemma.\n\nBy the lemma, the rightmost expression equals the number of squares containing an interior point below the line with slope $r$ plus the number below the line with slope $1/r$. By symmetry about $y = x$, the latter equals the number above the line with slope $r$. Thus, the rightmost expression equals the number of squares of the board plus the number of squares through which $\\ell$ passes through the interior. The former is $n^2$. To prove the inequality, it suffices to show every line passes through the interior of at least $n$ squares. Since $\\ell$ has positive slope, it passes through either $n$ rows or $n$ columns, so at least $n$ squares. Hence, the right inequality holds. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15454,
    "subject": "Mathematics (Olympiad)",
    "question": "The circles $k_1$ and $k_2$ intersect at points $A$ and $B$. A line through $B$ intersects the circles $k_1$ and $k_2$ for the second time at points $C$ and $D$, respectively, such that $C$ lies outside of $k_2$, and $D$ lies outside of $k_1$.\n\nLet $M$ be the intersection point of the tangents to $k_1$ and $k_2$ drawn through $C$ and $D$, respectively. Let $AM \\cap CD = \\{P\\}$.\n\nThe tangent through $B$ to $k_1$ intersects $AD$ at $L$, and the tangent through $B$ to $k_2$ intersects $AC$ at $K$.\n\nLet $KP \\cap MD = \\{N\\}$ and $LP \\cap MC = \\{Q\\}$.\n\nShow that the quadrilateral $MNPQ$ is a parallelogram.",
    "options": [],
    "answer": "See solution",
    "solution": "Due to symmetry, it is enough to show that $KP \\parallel MC$.\n\nFirst, we show that the quadrilateral $ACMD$ is cyclic. Namely, $B$ lies on the segment $\\overline{CD}$, and $A$ and $M$ are on different sides of the line $CD$. From $\\angle BDM = \\angle DAB$ and $\\angle BCM = \\angle BAC$, it follows that\n$$\n\\angle DAC = \\angle DAB + \\angle BAC = \\angle BDM + \\angle BCM = 180^\\circ - \\angle DMC.\n$$\n\nSecond, we show that $B$ and $P$ lie on the same arc passing through $A$ and $K$.\n\n**First case:** $P$ lies on segment $BC$; $A$ and $B$ are on the same side of $KP$. Let $E$ be the intersection of $KB$ and $DM$. We have:\n$$\n\\angle KBP = \\angle DBE = \\angle BDE = \\angle CDM = \\angle CAM = \\angle KAP.\n$$\nThus, $\\angle KBP = \\angle KAP$ implies that $AKPB$ is cyclic.\n\n**Second case:** $P$ lies on segment $AC$; $A$ and $B$ are on different sides of $KP$. Again, let $E$ be the intersection of $KB$ and $DM$. We have:\n$$\n180^\\circ - \\angle KBP = \\angle DBE = \\angle BDE = \\angle CDM = \\angle CAM = \\angle KAP.\n$$\nThus, $AKBP$ is cyclic.\n\nTherefore, $\\angle APK = \\angle ABK = \\angle ADB = \\angle ADC = \\angle AMC$, confirming that $KP \\parallel MC$.\n\n**Remark:** The assertion remains true without the restriction that $C$ (resp. $D$) lies outside of $k_2$ (resp. $k_1$); the argument is analogous in the other cases.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15455,
    "subject": "Mathematics (Olympiad)",
    "question": "The sequence $a_n$ is defined by $a_1 = 1$ and $a_n = n \\cdot (a_1 + \\dots + a_{n-1})$ for all $n > 1$. Find all indices $n$ for which $a_n$ is divisible by $1 \\cdot 2 \\cdot \\dots \\cdot n$.",
    "options": [],
    "answer": "See solution",
    "solution": "For each $n \\ge 2$, denote $S_n = a_1 + \\dots + a_{n-1}$. Then $a_n = S_n \\cdot n$. For all $n > 2$, we have:\n\n$$\nS_n = S_{n-1} + a_{n-1} = S_{n-1} + S_{n-1} \\cdot (n-1) = S_{n-1} \\cdot n.\n$$\n\nThus,\n$$\nS_n = S_{n-1} \\cdot n = S_{n-2} \\cdot (n-1)n = \\dots = S_2 \\cdot 3 \\cdot \\dots \\cdot n = \\frac{n!}{2},\n$$\nbecause $S_2 = 1 = \\frac{1 \\cdot 2}{2}$.\n\nConsequently,\n$$\na_n = S_n \\cdot n = n! \\cdot \\frac{n}{2}\n$$\nfor all $n \\ge 2$.\n\nTherefore, for $n \\ge 2$, $a_n$ is divisible by $n!$ if and only if $n$ is even, and $n=1$ also satisfies the condition.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15456,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the positive real numbers $a, b, c, d$ such that $a + b + c + d = 80$ and\n$$\na + \\frac{b}{1 + a} + \\frac{c}{1 + a + b} + \\frac{d}{1 + a + b + c} = 8.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Adding $4$ to both sides of the second equation, we write:\n$$\n1 + a + \\frac{1 + a + b}{1 + a} + \\frac{1 + a + b + c}{1 + a + b} + \\frac{1 + a + b + c + d}{1 + a + b + c} = 12.\n$$\nApplying the AM-GM inequality successively, we obtain:\n$$\n\\begin{aligned}\n1 + a + \\frac{1 + a + b}{1 + a} &\\geq 2\\sqrt{(1 + a) \\cdot \\frac{1 + a + b}{1 + a}} = 2\\sqrt{1 + a + b}, \\\\\n\\frac{1 + a + b + c}{1 + a + b} + \\frac{1 + a + b + c + d}{1 + a + b + c} &\\geq 2\\sqrt{\\frac{1 + a + b + c}{1 + a + b} \\cdot \\frac{1 + a + b + c + d}{1 + a + b + c}} = 2\\sqrt{\\frac{81}{1 + a + b}}.\n\\end{aligned}\n$$\nBy adding these two inequalities and applying the AM-GM inequality again, we have:\n$$\n\\begin{aligned}\n12 &= 1 + a + \\frac{1 + a + b}{1 + a} + \\frac{1 + a + b + c}{1 + a + b} + \\frac{1 + a + b + c + d}{1 + a + b + c} \\\\\n&\\geq 2\\sqrt{1 + a + b} + \\frac{18}{\\sqrt{1 + a + b}} \\geq 12.\n\\end{aligned}\n$$\nWe observe that equality occurs in all three uses of the AM–GM inequality above. From this, we obtain $a + b = 8$ and\n$$\n1 + a = \\frac{1 + a + b}{1 + a} = \\frac{1 + a + b + c}{1 + a + b} = \\frac{1 + a + b + c + d}{1 + a + b + c} = 3.\n$$\nTherefore, the desired numbers are $a = 2$, $b = 6$, $c = 18$, $d = 54$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15457,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f, g : \\mathbb{Q} \\to \\mathbb{Q}$ such that, for all $x, y \\in \\mathbb{Q}$,\n\n$$\nf(g(x) + g(y)) = f(g(x)) + y,\n$$\n\n$$\ng(f(x) + f(y)) = g(f(x)) + y.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $g(y_1) = g(y_2)$, the first equality yields $y_1 = y_2$, hence $g$ is injective. Analogously, $f$ is injective as well.\n\nPlugging $y = 0$ into the first equality gives $f(g(x) + g(0)) = f(g(x))$, hence $g(x) + g(0) = g(x)$, so $g(0) = 0$; similarly, $f(0) = 0$.\n\nPlugging $x = 0$ in both equalities yields $f(g(y)) = g(f(y)) = y$ for all $y$, therefore $f$ and $g$ are bijective and $g = f^{-1}$.\n\nThe initial equalities become\n\n$$\nf(g(x) + g(y)) = x + y, \\quad g(f(x) + f(y)) = x + y,\n$$\n\nfor all $x, y \\in \\mathbb{Q}$, and we deduce\n\n$$\ng(x + y) = g(x) + g(y), \\quad f(x + y) = f(x) + f(y),\n$$\n\nfor all $x, y \\in \\mathbb{Q}$. Setting $a = f(1), b = g(1)$, it is easy to prove that $f(x) = ax$, $g(x) = bx$ for all $x \\in \\mathbb{Q}$.\n\nSince $g = f^{-1}$, we obtain $ab = 1$. Finally, the functions are\n\n$$\nf(x) = ax, \\quad g(x) = \\frac{x}{a},\n$$\n\nwhere $a \\in \\mathbb{Q}^*$, and it is easy to see that they satisfy the initial equalities.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15458,
    "subject": "Mathematics (Olympiad)",
    "question": "有 65 對情侶出去玩，每一位男生都有一輛機車，並且都得要負責載一位女生。假設他們能夠安排出一種載法，使得對於任兩輛機車，下面兩命題恰一成立：\n\n1. 這兩輛機車上的男生互相認識彼此；\n2. 這兩輛機車上的女生的男朋友互相認識彼此。\n\n試證明：一定可以找到一對情侶，把他們剔除後，剩下的 64 對情侶仍能夠安排出一個滿足上述條件的載法。",
    "options": [],
    "answer": "See solution",
    "solution": "假設這 65 對情侶已經選擇了一種符合題目條件的載法，我們證明：一定有一對情侶在同一輛車上（事實上必恰只有一對），因此把他們剔除後，剩下的 64 對情侶可以沿用原本的載法，這樣顯然能夠滿足題目條件。\n\n將所有情侶編號 $1$ 至 $65$，並且定義函數 $f(a) = b$ 表示第 $a$ 號女生被第 $b$ 號男生載。則 $f(a)$ 是個一對一映成的函數，因此若是對所有 $1 \\leq a \\leq 65$，將 $a$ 不停代入 $f$ 直到其值變回 $a$，並把過程寫成一個環狀，就能得到 $a \\rightarrow f(a) \\rightarrow f(f(a)) \\cdots \\rightarrow f^{(k)}(a) = a$，這樣就能把 $1$ 到 $65$ 分成許多個互斥的環，因此必定有一個環擁有奇數個數字。假設這個環是\n\n$$\na_1 \\rightarrow a_2 \\rightarrow \\cdots \\rightarrow a_k \\rightarrow a_1\n$$\n\n其中 $k$ 是個奇數。\n\n假如 $k > 1$，我們定義符號 $g(a, b)$ 表示第 $a$ 號男生與第 $b$ 號男生互相認識，$\\langle a, b \\rangle$ 則表示第 $a$ 號男生與第 $b$ 號男生不互相認識。假設有 $(a_1, a_2)$，則由於 $f(a_1) = a_2, f(a_2) = a_3$，因此知道男生 $a_2$ 與男生 $a_3$ 所載的女生（就是 $a_1$ 和 $a_2$）的男朋友互相認識，因此得到 $\\langle a_2, a_3 \\rangle$；類似地由 $f(a_2) = a_3, f(a_3) = a_4$ 因此知道男生 $a_3$ 與男生 $a_4$ 所載的女生（就是 $a_2$ 和 $a_3$）的男朋友不互相認識，所以男生 $a_3$ 與男生 $a_4$ 應該要互相認識，所以有 $(a_3, a_4)$，這樣推理下去就有：\n\n$$\n(a_1, a_2) \\rightarrow \\langle a_2, a_3 \\rangle \\rightarrow (a_3, a_4) \\cdots \\rightarrow (a_k, a_1) \\rightarrow \\langle a_1, a_2 \\rangle\n$$\n\n矛盾！如果是 $\\langle a_1, a_2 \\rangle$ 這種情況，仍然可以運用同樣的推理推出矛盾！\n因此 $k=1$，即 $f(a_1) = a_1$，得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15459,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a $4 \\times 4$ array of pairwise distinct positive integers such that in each column and each row, one of the numbers is equal to the sum of the other three. Determine the least possible value of the largest number such an array may contain.",
    "options": [],
    "answer": "See solution",
    "solution": "The required minimum is $21$ and is achieved, for instance, by the array below:\n\n![](<table><tr><td>1</td><td>8</td><td>12</td><td><b>21</b></td></tr><tr><td>7</td><td>9</td><td>20</td><td>4</td></tr><tr><td>10</td><td>19</td><td>3</td><td>6</td></tr><tr><td>18</td><td>2</td><td>5</td><td>11</td></tr></table>)\n\nThe lower bound is a consequence of the following slightly more general fact:\n\nIf, for some integer $n \\ge 3$, in each row of an $n \\times n$ array of pairwise distinct positive integers, one of the numbers is equal to the sum of the other $n-1$, then the largest number in the array is at least $\\frac{1}{2}(n-1)(n^2 - n + 2)$; columns subjected to no condition whatsoever.\n\nLet $a$ be the largest number in the array and let $a_i$ be the largest number on the $i$-th row, $i = 1, 2, \\dots, n$. Since the $a_i$ are pairwise distinct positive integers not exceeding $a$,\n\n$$\n\\sum_{i=1}^{n} a_i \\le \\sum_{j=0}^{n-1} (a-j) = na - \\frac{1}{2}n(n-1). \\quad (*)\n$$\n\nLet now $b_1, b_2, \\dots, b_{n^2-n}$ be the numbers in the array different from each $a_i$. Then\n\n$$\n\\sum_{i=1}^{n} a_i = \\sum_{j=1}^{n^2-n} b_j \\ge \\sum_{k=1}^{n^2-n} k = \\frac{1}{2}n(n-1)(n^2-n+1), \\quad (**)\n$$\n\nwhere the first equality holds by hypothesis, and the inequality follows from the fact that the $b_j$ are pairwise distinct positive integers.\n\nFinally, $(*)$ and $(**)$ imply $a \\ge \\frac{1}{2}(n-1)(n^2-n+2)$, as stated. If $n=4$, then $a \\ge 21$, as desired.\n\n**Remark.** In the relaxed setting above (columns subjected to no condition whatsoever), since the $n^2$ numbers in the array are pairwise distinct positive integers, $a \\ge n^2$, so\n\n$$\na \\ge \\max(n^2, \\frac{1}{2}(n-1)(n^2-n+2)).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15460,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$ be real numbers such that\n\n$$\n(a_1^2 + a_2^2 + \\dots + a_n^2 - 1)(b_1^2 + b_2^2 + \\dots + b_n^2 - 1) > (a_1b_1 + a_2b_2 + \\dots + a_n b_n - 1)^2.\n$$\n\nShow that $a_1^2 + a_2^2 + \\dots + a_n^2 > 1$ and $b_1^2 + b_2^2 + \\dots + b_n^2 > 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "**First Solution:** If exactly one of $a_1^2 + a_2^2 + \\dots + a_n^2$ and $b_1^2 + b_2^2 + \\dots + b_n^2$ is no greater than $1$, then the left-hand side of the inequality is no greater than $0$, and the inequality is trivial. Now assume that both are no greater than $1$. Set $a = 1 - a_1^2 - a_2^2 - \\dots - a_n^2$ and $b = 1 - b_1^2 - b_2^2 - \\dots - b_n^2$. Then both $a$ and $b$ are nonnegative real numbers. Multiplying both sides of the inequality by $4$ gives\n\n$$\n(2 - 2a_1b_1 - 2a_2b_2 - \\dots - 2a_nb_n)^2 \\geq 4ab\n$$\n\nNote that\n\n$$\n\\begin{aligned}\n& 2 - 2a_1b_1 - 2a_2b_2 - \\cdots - 2a_nb_n \\\\\n&= (a_1 - b_1)^2 + (a_2 - b_2)^2 + \\cdots + (a_n - b_n)^2 + a + b \\\\\n&\\geq a + b \\geq 0.\n\\end{aligned}\n$$\n\nIt follows that\n\n$$\n(2 - 2a_1b_1 - 2a_2b_2 - \\cdots - 2a_nb_n)^2 \\geq (a+b)^2 \\geq 4ab,\n$$\n\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15461,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $I$ be the incenter of triangle $ABC$, and let the incircle touch the sides $CA$ and $AB$ at the points $E$ and $F$, respectively. Let the reflection points of $E$ and $F$ with respect to $I$ be the points $G$ and $H$, respectively. Suppose that the lines $GH$ and $BC$ intersect at the point $Q$. Denote the midpoint of the side $BC$ by $M$. Prove that $IQ$ and $IM$ are perpendicular to each other.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the incircle touching $BC$ at $D$, and let $GH$ intersect $BI$ and $CI$ at $C'$ and $B'$, respectively. Let $D'$ be a point on $GH$ such that $ID' \\perp GH$.\n\n1. (a) Since $\\angle C'ID' = \\frac{1}{2}(\\angle A + \\angle B)$ and $\\angle IC'B' = \\frac{1}{2}\\angle C$, similarly $\\angle IB'C' = \\frac{1}{2}\\angle B$. Thus, $\\triangle IBC \\sim \\triangle IB'C'$.\n\n(b) Since $G$ and $H$ are the reflections of $E$ and $F$ over $I$, $G$ and $H$ lie on the incircle, and $EF \\parallel GH$. Therefore, $\\angle IGD' = \\angle IEF = \\frac{1}{2}\\angle A$ (since $A, E, I, F$ are concyclic). Thus,\n\n$$\n\\frac{IB'}{IB} = \\frac{IC'}{IC} = \\frac{ID'}{ID} = \\frac{ID'}{IG} = \\sin \\angle D'IG = \\sin \\frac{1}{2}\\angle A.\n$$\n\n2. Applying Menelaus' theorem to $\\triangle IBC$ and $B'C'$ gives\n\n$$\n\\frac{BQ}{QC} \\cdot \\frac{CB'}{B'I} \\cdot \\frac{IC'}{C'B} = -1 \\quad \\text{or} \\quad \\frac{BM + MQ}{BM - MQ} = \\frac{BQ}{QC} = \\frac{IB'}{CB'} \\cdot \\frac{C'B}{IC'}.\n$$\n\nTherefore,\n\n$$\n\\begin{align*}\n\\frac{BM + MQ}{IB' \\cdot C'B} &= \\frac{BM - MQ}{CB' \\cdot IC'} \\\\\n&= \\frac{2BM}{IB' \\cdot C'B + CB' \\cdot IC'} \\\\\n&= \\frac{2MQ}{IB' \\cdot C'B - CB' \\cdot IC'}\n\\end{align*}\n$$\n\nFrom step 1,\n\n$$\n\\begin{aligned}\nIB' \\cdot C'B + CB' \\cdot IC' &= IB \\sin \\frac{\\angle A}{2} (IB - IC') + (IB' - IC)IC \\sin \\frac{\\angle A}{2} \\\\\n&= \\sin \\frac{\\angle A}{2} [IB(IB - IC \\sin \\frac{\\angle A}{2}) + (IB \\sin \\frac{\\angle A}{2} - IC)IC] \\\\\n&= \\sin \\frac{\\angle A}{2} (IB^2 - IC^2),\n\\end{aligned}\n$$\n\nand\n\n$$\n\\begin{aligned}\nIB' \\cdot C'B - CB' \\cdot IC' &= \\sin \\frac{\\angle A}{2} [IB(IB - IC \\sin \\frac{\\angle A}{2}) - (IB \\sin \\frac{\\angle A}{2} - IC)IC] \\\\\n&= \\sin \\frac{\\angle A}{2} (IB^2 + IC^2 - 2 IB \\cdot IC \\sin \\frac{\\angle A}{2}).\n\\end{aligned}\n$$\n\nSince $IB^2 = ID^2 + BD^2$, $IC^2 = ID^2 + CD^2$,\n\n$$\nIB^2 - IC^2 = BD^2 - CD^2 = (BM + MD)^2 - (BM - MD)^2 = 4BM \\cdot MD.\n$$\n\nAlso, $BC^2 = IB^2 + IC^2 - 2 IB \\cdot IC \\cos(90^\\circ + \\frac{\\angle A}{2}) = IB^2 + IC^2 + 2 IB \\cdot IC \\sin \\frac{\\angle A}{2}$,\n\n$$\nIB' \\cdot C'B - CB' \\cdot IC' = \\sin \\frac{\\angle A}{2} (2IB^2 + 2IC^2 - BC^2) = 4IM^2 \\sin \\frac{\\angle A}{2}.\n$$\n\n4. From steps 2 and 3, $\\frac{BM}{BM \\cdot MD} = \\frac{MQ}{IM^2}$, i.e., $\\frac{IM}{MD} = \\frac{MQ}{IM}$. Thus, $\\triangle IMD \\sim \\triangle QMI$. Therefore, $\\angle QIM = \\angle IQM = 90^\\circ$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15462,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a quintic polynomial $f(x)$ with leading coefficient $1$, it satisfies $f(n) = 8n$ for $n = 1, 2, \\dots, 5$. What is the coefficient of the term of degree $1$ in $f(x)$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(x) = g(x) + 8x$, where $g(x)$ is a quintic polynomial with leading coefficient $1$. Since $f(n) = 8n$ for $n = 1, 2, \\dots, 5$, we have:\n\n$$\ng(n) = f(n) - 8n = 0, \\quad n = 1, 2, \\dots, 5.\n$$\n\nThus, $g(x)$ has roots at $x = 1, 2, 3, 4, 5$, so:\n\n$$\ng(x) = (x-1)(x-2)(x-3)(x-4)(x-5).\n$$\n\nTherefore,\n\n$$\nf(x) = (x-1)(x-2)(x-3)(x-4)(x-5) + 8x.\n$$\n\nThe coefficient of $x$ in $f(x)$ is:\n\n$$\n\\left(1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5}\\right) \\cdot 5! + 8 = 282.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15463,
    "subject": "Mathematics (Olympiad)",
    "question": "A _Pythagorean triple_ is a solution of the equation $x^2 + y^2 = z^2$ in positive integers such that $x < y$. Given any non-negative integer $n$, show that some positive integer appears in precisely $n$ distinct Pythagorean triples.",
    "options": [],
    "answer": "See solution",
    "solution": "We show by induction on $n \\ge 0$ that $2^{n+1}$ appears in precisely $n$ distinct Pythagorean triples.\n\nSince no Pythagorean triple contains $2$, the assertion holds for $n=0$.\n\nFor the induction step, let $n \\ge 1$, and assume that $2^n$ appears in exactly $n-1$ distinct Pythagorean triples. The latter produce $n-1$ distinct non-primitive Pythagorean triples each containing $2^{n+1}$.\n\nTo conclude the proof, we show that $2^{n+1}$ appears exactly once in a primitive Pythagorean triple. Recall that the primitive Pythagorean triples are described by the well-known formulae $x = v^2 - u^2$, $y = 2uv$, $z = u^2 + v^2$, where $u$ and $v$ are coprime positive integers, not both odd, and $u < v$.\n\nSince $x$ and $z$ are both odd, if $2^{n+1}$ appears in the triple, then $2^{n+1} = y = 2uv$, and since $u < v$ and $u$ and $v$ have opposite parity, necessarily $u=1$ and $v = 2^n$. Consequently, $2^{n+1}$ appears in exactly $n$ distinct Pythagorean triples.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15464,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $(a_1, a_2)$ such that the sequence $(a_n)$ defined by the recurrence\n\n$$\na_n + a_{n+1} = 2a_{n+2}a_{n+3} + 2016$$\n\nfor all $n$ yields integer values for all terms.",
    "options": [],
    "answer": "See solution",
    "solution": "The possible pairs $(a_1, a_2)$ are $(1, -2015)$, $(15, -69)$, $(70, -14)$, $(2016, 0)$, or their permutations.\n\nTaking the difference of the two relations:\n\n$$\na_n + a_{n+1} = 2a_{n+2}a_{n+3} + 2016$$\n$$a_{n+1} + a_{n+2} = 2a_{n+3}a_{n+4} + 2016$$\n\nwe get\n\n$$\na_{n+2} - a_n = 2a_{n+3}(a_{n+4} - a_{n+2}).$$\n\nBy induction,\n\n$$\na_{n+2} - a_n = 2^k a_{n+3} a_{n+5} \\cdots a_{n+2k+1} (a_{n+2k+2} - a_{n+2k})$$\n\nfor any $k \\in \\mathbb{Z}^+$. This implies $2^k \\mid a_{n+2} - a_n$ for any $k$, so $a_{n+2} = a_n$. Thus, all odd terms are equal to $a_1 = b$ and all even terms to $a_2 = c$. The only condition is\n\n$$\nb + c = 2bc + 2016.$$ \n\nThis rearranges to\n\n$$\n(2b - 1)(2c - 1) = -4031 = -29 \\times 139.\n$$\n\nThus, the solutions are $(b, c) = (1, -2015)$, $(15, -69)$, $(70, -14)$, $(2016, 0)$, up to permutation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15465,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there are only finitely many quadruples $\\left(a, b, c, n\\right)$ of positive integers such that\n$$\nn! = a^{n-1} + b^{n-1} + c^{n-1}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For fixed $n$ there are clearly finitely many solutions. We will show that there is no solution with $n > 100$. So assume $n > 100$. By the AM-GM inequality,\n\n$$\n\\begin{aligned}\nn! &= 2n(n-1)(n-2)(n-3) \\cdot (3 \\cdot 4 \\cdots (n-4)) \\\\\n&\\le 2(n-1)^4 \\left( \\frac{3+4+\\cdots+(n-4)}{n-6} \\right)^{n-6} = 2(n-1)^4 \\left( \\frac{n-1}{2} \\right)^{n-6} < \\left( \\frac{n-1}{2} \\right)^{n-1},\n\\end{aligned}\n$$\n\nthus $a, b, c < (n-1)/2$.\n\nFor every prime $p$ and integer $m \\ne 0$, let $v_p(m)$ denote the $p$-adic valuation of $m$. Legendre's formula states that\n\n$$\nv_p(n!) = \\sum_{s=1}^{\\infty} \\left\\lfloor \\frac{n}{p^s} \\right\\rfloor,\n$$\n\nand a well-known corollary of this formula is that\n\n$$\nv_p(n!) < \\frac{n}{p-1}. \\qquad (\\heartsuit)\n$$\n\nIf $n$ is odd then $a^{n-1}, b^{n-1}, c^{n-1}$ are squares, and by considering them modulo $4$ we conclude that $a, b$, and $c$ must be even. Hence $2^{n-1} \\mid n!$ but that is impossible for odd $n$ because $v_2(n!) = v_2((n-1)!) < n-1$ by $(\\heartsuit)$.\n\nFrom now on we assume that $n$ is even. If all three numbers $a+b, b+c$ and $c+a$ are powers of $2$ then $a, b$ and $c$ have the same parity. If they are odd, then $n! \\equiv a+b+c \\pmod{2}$ is also odd which is absurd. If all $a, b, c$ are divisible by $4$, this contradicts $v_2(n!) \\le n-1$. If, say, $a$ is not divisible by $4$, then $2a = (a+b) + (a+c) - (b+c)$ is not divisible by $8$, and since all $a+b, b+c, c+a$ are powers of $2$, we get that one of these sums equals $4$, so two of the numbers of $a, b, c$ are equal to $2$. Say $a = b = 2$, then $c = 2^r - 2$ and since $c \\mid n!$ we must have $c \\mid a^{n-1} + b^{n-1} = 2^n$ implying $r = 2$, and so $c = 2$, which is impossible because\n\n$$\nn! \\equiv 0 \\not\\equiv 3 \\cdot 2^{n-1} \\pmod{5}.\n$$\n\nSo now we assume that the sum of two numbers among $a, b, c$, say $a+b$, is not a power of $2$, so it is divisible by some odd number $p$. Then $p \\le a+b < n$ so $c^{n-1} = n! - (a^{n-1} + b^{n-1})$ is divisible by $p$. If $p$ divides $a$ and $b$ hence $p^{n-1} \\mid n!$, contradicting $(\\heartsuit)$. Next, using $(\\heartsuit)$ and the LTE Lemma, we obtain\n\n$$\nv_p(1) + v_p(2) + \\dots + v_p(n) = v_p(n!) = v_p(n! - c^{n-1}) = v_p(a^{n-1} + b^{n-1}) = v_p(a+b) + v_p(n-1).\n$$\n\nNow, no number of $1, 2, \\dots, n$ can be divisible by $p$, except for $a+b$ and $n-1 > a+b$. On the other hand, $p \\mid c$ implies that $p < n/2$ and so there must be at least two such numbers. Hence, there are two multiples of $p$ among $1, 2, \\dots, n$, namely $a+b = p$ and $n-1 = 2p$. But this is another contradiction because $n-1$ is odd. This final contradiction shows that there is no solution of the equation for $n > 100$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15466,
    "subject": "Mathematics (Olympiad)",
    "question": "Since $abc = 1$, find the minimal value of $ab + bc + ac$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n(ab + 2a + 2b - 9)(bc + 2b + 2c - 9)(ca + 2c + 2a - 9) = (2a+1)(2b+1)(2c+1)(1-a)(1-b)(1-c) \\geq 0.\n$$\n\nNow, since $(2a+1)(2b+1)(2c+1) > 0$, we get\n$$\n0 \\leq (1-a)(1-b)(1-c) = -abc - a - b - c + ab + bc + ac + 1.\n$$\n\nThus, $ab + bc + ac \\geq 5$. The equality holds at\n$$\n(a, b, c) = (1, 2 - \\sqrt{3}, 2 + \\sqrt{3}).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15467,
    "subject": "Mathematics (Olympiad)",
    "question": "Докажи дека производот $\\left(1 + \\frac{1}{2} + \\frac{1}{3} + \\cdots + \\frac{1}{2008}\\right) \\cdot 2 \\cdot 3 \\cdots 2008$ е делив со 2009.",
    "options": [],
    "answer": "See solution",
    "solution": "Изразот во заградата ќе го означиме со $A$ и ќе го запишеме во облик:\n\n$$\n\\begin{aligned}\nA &= \\left(1 + \\frac{1}{2008}\\right) + \\left(\\frac{1}{2} + \\frac{1}{2007}\\right) + \\left(\\frac{1}{3} + \\frac{1}{2006}\\right) + \\cdots + \\left(\\frac{1}{1004} + \\frac{1}{1005}\\right) \\\\\n&= \\frac{2009}{1 \\cdot 2008} + \\frac{2009}{2 \\cdot 2007} + \\cdots + \\frac{2009}{1004 \\cdot 1005} \\\\\n&= 2009 \\cdot \\left( \\frac{1}{1 \\cdot 2008} + \\frac{1}{2 \\cdot 2007} + \\cdots + \\frac{1}{1004 \\cdot 1005} \\right)\n\\end{aligned}\n$$\n\nБидејќи производот $2 \\cdot 3 \\cdots 2008$ е делив со секој од именителите во последната заграда, следува дека изразот $\\left(1 + \\frac{1}{2} + \\frac{1}{3} + \\cdots + \\frac{1}{2008}\\right) \\cdot 2 \\cdot 3 \\cdots 2008$ е природен број.\n\nТогаш, јасно е дека производот\n\n$$\n2009 \\cdot \\left( \\frac{1}{1 \\cdot 2008} + \\frac{1}{2 \\cdot 2007} + \\cdots + \\frac{1}{1004 \\cdot 1005} \\right) \\cdot 2 \\cdot 3 \\cdots 2008\n$$\nе делив со 2009.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15468,
    "subject": "Mathematics (Olympiad)",
    "question": "For each positive integer $n \\ge 2$, define the polynomial $f_n(x)$ by\n$$\nf_n(x) = x^n - x^{n-1} - x^{n-2} - \\cdots - x - 1.\n$$\n\n(a) For each positive integer $n \\ge 2$, prove that the equation $f_n(x) = 0$ has a unique real positive root, say, $\\alpha_n$.\n\n(b) Show that $(\\alpha_n)_{n \\ge 2}$ is a strictly increasing sequence.\n\n(c) Prove that $\\lim_{n \\to \\infty} \\alpha_n = 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Since $f_n(x)$ has one change of sign, it follows by Descartes' Rule of Signs that $f_n(x) = 0$ has at most one positive real root. As $f_n(0) = -1 < 0$ and $f_n(2) = 2^n - 2^{n-1} - 2^{n-2} - \\cdots - 2 - 1 = 1 > 0$, there is at least one root between $0$ and $2$. Thus, $f_n(x) = 0$ has exactly one root $\\alpha_n$ between $0$ and $2$ for $n \\ge 2$.\n\n(b) To show $\\alpha_n < \\alpha_{n+1}$ for $n \\ge 2$, note that $f_n(\\alpha_n) = 0$. Now,\n$$\nf_{n+1}(\\alpha_n) = \\alpha_n f_n(\\alpha_n) - 1 = \\alpha_n \\cdot 0 - 1 = -1 < 0.\n$$\nSince $f_{n+1}(2) = 1 > 0$, there is a root of $f_{n+1}(x) = 0$ between $\\alpha_n$ and $2$. Thus, $\\alpha_n < \\alpha_{n+1} < 2$.\n\n(c) Observe that\n$$\n(x-1)f_n(x) = (x-1)(x^n - x^{n-1} - x^{n-2} - \\cdots - x - 1) = x^{n+1} - 2x^n + 1.\n$$\nSo,\n$$\n\\alpha_n^{n+1} - 2\\alpha_n^n + 1 = 0,\n$$\nwhich gives $\\alpha_n = 2 - \\frac{1}{\\alpha_n^n}$. Since $\\alpha_2 = \\frac{1+\\sqrt{5}}{2} > 1$, $\\frac{1}{\\alpha_2^n} \\to 0$ as $n \\to \\infty$. Therefore,\n$$\n\\lim_{n \\to \\infty} \\alpha_n = 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15469,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove the claim: If we choose any four factors of $720$, then one of them divides the product of the other three.",
    "options": [],
    "answer": "See solution",
    "solution": "Given the decomposition $720 = 2^4 \\cdot 3^2 \\cdot 5$, the number $720$ has exactly three prime factors: $2$, $3$, and $5$. Each of its factors is of the form $2^\\alpha \\cdot 3^\\beta \\cdot 5^\\gamma$, where $\\alpha, \\beta, \\gamma$ are non-negative integers (with $\\alpha \\leq 4$, $\\beta \\leq 2$, $\\gamma \\leq 1$). The product of any three such factors is also of this form.\n\nFor any two numbers of this form, the first divides the second if and only if the exponents $\\alpha, \\beta, \\gamma$ of the first do not exceed those of the second.\n\nSuppose, for contradiction, that we can choose four factors of $720$ such that none divides the product of the other three. Then, for each factor, there is at least one prime (among $2$, $3$, $5$) whose exponent in that factor is greater than in the product of the other three. But with four factors and only three primes, by the pigeonhole principle, at least two factors must have this property for the same prime, which is impossible. Thus, in any selection of four factors, one divides the product of the other three.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15470,
    "subject": "Mathematics (Olympiad)",
    "question": "Petrik and Vasyl are playing a $m \\times n$ board game. They move in turns, Petrik starts. On his move, Petrik places a pawn in any empty cell of the board. Vasyl, on his turn, must place a pawn on the free cell, which is adjacent to the cell where Petrik placed his pawn with his last move. Vasyl wins if the whole board is filled with pawns. Petrik wins if after his move Vasyl is not able to make a move under the specified rules and there are still free cells on the board. If everyone wants to win, who has the winning strategy, depending on the values of $m, n$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m$ be the number of rows, and $n$ the number of columns.\n\n- If $m$ is even, split all cells of the board into pairs of adjacent by side. Then after any move of Petrik, Vasyl will be able to make his move, and will win. Similarly, if $n$ is even, Vasyl also wins.\n- For $1 \\times 1$, $3 \\times 1$, and $1 \\times 3$ it's also clear that Vasyl wins.\n\nConsider now $m = n = 3$. Petrik wins with the following strategy: he makes his first move in the center (1), Vasyl places a pawn in a cell above the center (2), and then Petrik moves under the center cell (3). After this, Vasyl moves in the bottom left corner. Then Petrik moves in the right column, Vasyl replies in the same column, and Petrik moves in the same column and wins.\n\nThe last case is when $m, n$ are odd and at least one of them is at least $5$. Without loss of generality, let the number of columns $n \\ge 5$. Petrik, with each move, will place a pawn into the central column while possible. After such moves, all the pawns on the board are in the three central columns, so the leftmost and rightmost two columns are definitely empty. In addition, the number of pawns on the table is even. The total number of cells on the board is odd, so the left or the right part of the board with respect to the central column contains an odd number of empty cells (say, to the right). Petrik will move to this part from now on. Vasyl has to reply with moves to the same part, as the central column is already filled. As there is an odd number of empty cells to the right before Petrik's move, it will be Petrik who will make the last move, so Vasyl will lose as there are also many free cells to the left of the central column.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15471,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n \\ge 3$ 為一正整數。一個正整數 $m \\ge n+1$ 被稱為 $n$ 色數，若且唯若我們可以將正 $m$ 邊形的每一個頂點塗成 $n$ 色中的其中一個顏色，使得這個正 $m$ 邊形的任何連續 $n+1$ 個頂點都包含全部 $n$ 種顏色。\n\n證明對於任何 $n \\ge 3$，非 $n$ 色數只有有限多個，並求最大的非 $n$ 色數。",
    "options": [],
    "answer": "See solution",
    "solution": "最大的非 $n$ 色數為 $m_{\\text{max}} = n(n-1) - 1$。\n\n假設 $m = m_{\\text{max}}$，則必然有某個顏色 $C$ 在至多 $n-2$ 個頂點上，而這些頂點將所有非 $C$ 色的頂點切割為至多 $n-2$ 個連續片段。由於非 $C$ 色頂點至少有 $n(n-1) - 1 - (n-2) = (n-1)^2 > n(n-2)$ 個，因此其中必然有一個非 $C$ 色的連續片段的長度至少為 $n$，從而 $m_{\\text{max}}$ 是非 $n$ 色數。\n\n現在假設 $m \\ge m_{\\text{max}} + 1 = n(n-1)$。依據除法原理，有 $m = nk + j$，其中 $k \\ge n-1$ 且 $0 \\le j \\le n-1$。則我們可以從某一個頂點開始，依照順時鐘順序，先依序塗 $[1, 2, \\dots, n]$ 色 $k-j$ 次，再依序塗 $[1, 1, 2, \\dots, n]$ 色 $j$ 次。易驗證此塗法確實滿足題意，故所有 $m > m_{\\text{max}}$ 皆為 $n$ 色數。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15472,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: [0, 1] \\to \\mathbb{R}$ be a differentiable function such that $f(0) = f(1) = 0$, and $|f'(x)| \\le 1$ for all $x \\in [0, 1]$. Prove that\n$$\n\\left| \\int_{0}^{1} f(t) \\, dt \\right| < \\frac{1}{4}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $t \\in (0, 1)$. Apply Lagrange's theorem to $f$ on each of the intervals $[0, t]$ and $[t, 1]$ to get $|f(t)/t| \\le 1$ and $|f(t)/(1-t)| \\le 1$. Hence $|f(t)| \\le \\min(t, 1-t)$ for all $t \\in [0, 1]$. Consequently,\n$$\n\\begin{aligned}\n\\left| \\int_{0}^{1} f(t) \\, dt \\right| &\\le \\int_{0}^{1} |f(t)| \\, dt = \\int_{0}^{1/2} |f(t)| \\, dt + \\int_{1/2}^{1} |f(t)| \\, dt \\\\\n&\\le \\int_{0}^{1/2} t \\, dt + \\int_{1/2}^{1} (1-t) \\, dt = \\frac{1}{4}.\n\\end{aligned}\n$$\nSuppose equality occurs. It forces $|f(t)| = t$ for all $0 \\le t \\le 1/2$, and $|f(t)| = 1-t$ for all $1/2 \\le t \\le 1$. Since $f(1/2) = \\pm 1/2$ and $f$ is differentiable at $1/2$, it follows that $|f|$ is differentiable at $1/2$ — a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15473,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that it is possible to choose $n^2$ players and label them $P_{ij}$ ($1 \\le i, j \\le n$), such that for any $i, j, i', j' \\in \\{1, 2, \\dots, n\\}$, if $i < i'$, then $P_{ij}$ beats $P_{i'j'}$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution**\n\n**Lemma:** Suppose $m$ players participate in a single round-robin tournament with possible draws, and if $A$ beats $B$, $B$ beats $C$, then $A$ beats $C$. Then the $m$ players can be arranged in a row such that for any two players, the one on the left either beats or draws the one on the right.\n\n**Proof of Lemma:** Induction on $m$. When $m=1$, the lemma is trivial. Suppose it is true for $m-1$ players, and consider $m$ players. If everyone wins a game, then we can find $x_1$ beats $x_2$, $x_2$ beats $x_3$, and so on, and someone must reappear in the sequence, say $x_i$ beats $x_j$ ($i \\ge j$), which is contradictory. The contradiction indicates that someone has never won a game; put this player in the rightmost position. By the induction hypothesis, the other $m-1$ players can be arranged on the left such that the lemma conditions are satisfied.\n\nFor the original problem, arrange the $2n^2$ players in a row as in the lemma, and then make $n$ groups of players as follows: set the leftmost $n$ players as group $A_1$; for $i=1, \\dots, n-2$, on the right side of $A_i$, set some consecutive $n$ players as $A_{i+1}$; set the rightmost $n$ players as group $A_n$. Moreover, between any two consecutive groups, there are $\\left\\lfloor \\frac{n^2}{n-1} \\right\\rfloor$ or more players.\n\nIf players from different groups never draw, then the proof is done. Otherwise, remove two players who draw and regroup the remaining $2n^2 - 2$ players in the same way as before, except requiring $\\left\\lfloor \\frac{n^2-2}{n-1} \\right\\rfloor$ or more players between consecutive groups. Again, if players from different groups never draw, the proof is done. Otherwise, remove two players and regroup, and so on. In general, when we regroup $2n^2-2i$ ($0 \\le i \\le \\left\\lfloor \\frac{n^2}{2} \\right\\rfloor$) players, it is required that $\\left\\lfloor \\frac{n^2-2i}{n-1} \\right\\rfloor$ or more players are between consecutive groups. During the whole process, we obtain $\\left\\lfloor \\frac{n^2}{2} \\right\\rfloor + 1$ groupings.\n\nFor every $0 \\le i \\le \\lfloor \\frac{n^2}{2} \\rfloor$, assume in the $i$th grouping, $x_i$ and $y_i$ are from different groups and they draw. Then, there are at least $\\lfloor \\frac{n^2 - 2i}{n-1} \\rfloor$ players between them, each of whom draws with either $x_i$ or $y_i$. This implies that the total number of draws that $x_i$ and $y_i$ have is at least $\\lfloor \\frac{n^2 - 2i}{n-1} \\rfloor$, and furthermore the total number of draws in the tournament is at least\n\n$$\nS = \\sum_{0 \\le i \\le \\lfloor \\frac{n^2}{2} \\rfloor} \\left\\lfloor \\frac{n^2 - 2i}{n-1} \\right\\rfloor.\n$$\n\nWhen $n = 2k + 1$, $S = 2k^3 + 3k^2 + 3k + 2 > \\frac{n^3}{4}$; when $n = 2k$, $S > 2k^3 > \\frac{n^3}{4}$. Either way, $S \\le \\frac{n^3}{16}$ is violated and the assumption is untrue. Hence, for $1 \\le i, j \\le n$, let the $j$th player of $A_i$ be $P_{ij}$, and the conditions are satisfied.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15474,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose for any $a \\equiv 1 \\pmod{3}$ there exists a good number $h$ such that $h \\equiv a \\pmod{9^3}$. Then, for any $d \\equiv 1 \\pmod{3}$, show that there exists a positive integer $l$ such that both $l+1$ and $l+d$ are also good.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $d = 1 + 3x$ and choose a positive integer $s$ such that $10^s > 3x$. Let $x_1 = 10^s - 3x$. Then $x_1 \\equiv 1 \\pmod{3}$. Since $T(x_1) \\equiv 1 \\pmod{3}$, there exists a good number $h$ such that $h \\equiv T(x_1) \\pmod{9^3}$. Suppose $h > T(x_1)$ (if $h \\leq T(x_1)$, there exists $t$ such that $10^t > T(x_1)$, and we can replace $h$ by $h \\cdot 10^t$). Write $h = k \\cdot 9^3 + T(x_1)$. Taking\n\n$$\nl = x_1 - 1 + \\sum_{j=0}^{k-1} 9 \\cdot 10^{s+j}\n$$\n\nwe have $T(l + 1) = k \\cdot 9^3 + T(x_1) = h$. And,\n\n$$\nT(l + d) = T(10^{s+k}) = 1.\n$$\n\nTherefore, both $l+1$ and $l+d$ are good.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15475,
    "subject": "Mathematics (Olympiad)",
    "question": "How many different remainders can result when the 100th power of an integer is divided by 125?\n\n(A) 1  (B) 2  (C) 5  (D) 25  (E) 125",
    "options": [],
    "answer": "See solution",
    "solution": "Write $N = 5k + r$ for $r = 0, 1, 2, 3$, or $4$. If $r = 0$, then $N = 5k$ and $N^{100}$ is divisible by 125, so the remainder is 0. If $r = 1, 2, 3$, or $4$, then $N^2 = 25k^2 + 10rk + r^2 = 5m + 1$ for some integer $m$. Now use the Binomial Theorem:\n\n$$\n\\begin{aligned}\nN^{100} = (N^2)^{50} = (5m \\pm 1)^{50} = (5m)^{50} &\\pm 50(5m)^{49} + \\binom{50}{2}(5m)^{48} \\pm \\dots \\\\\n&\\pm \\binom{50}{47}(5m)^3 + \\binom{50}{48}(5m)^2 \\pm 50(5m) + 1.\n\\end{aligned}\n$$\n\nAll the terms except the final term have at least 3 factors of 5, so $N^{100}$ has remainder 1 upon division by 125. Therefore there are only 2 possible remainders: 0 and 1.\n\n**OR**\n\nLet $\\phi(n)$ be the number of positive integers less than $n$ that are relatively prime to $n$; this is Euler's totient function. Then $\\phi(125) = 5^3 - 5^2 = 100$. By Euler's Totient Theorem, if $a$ is not a multiple of 5, then $a^{100} \\equiv 1 \\pmod{125}$. If $a$ is a multiple of 5, then $a^{100} \\equiv 0 \\pmod{125}$. Therefore there are only 2 possible remainders: 0 and 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15476,
    "subject": "Mathematics (Olympiad)",
    "question": "Because of the way the grading was done, two students receive the same grades only if the problems they answer correctly are exactly the same. Designate the problems as I, II, and III, corresponding to grades 1, 2, and 4 for a correct answer, respectively. The possible total scores for each student are $0, 1, 2, 3, 4, 5, 6, 7$. Since two sets of 30 grades will not be distinguished if one is a permutation of the other, determine how many ways we can select, under these conditions, 8 numbers corresponding to the numbers of students receiving grades of $j$ points for $0 \\leq j \\leq 7$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a, b, c, d$ be the number of students who received 7, 6, 5, and 3 points, respectively. A student receives:\n- 7 points: all 3 problems correct,\n- 6 points: problems II and III correct, not I,\n- 5 points: problems I and III correct, not II,\n- 3 points: problems I and II correct, not III.\n\nSince each problem was answered correctly by 10 students:\n$$\na + c + d \\leq 10, \\quad a + b + d \\leq 10, \\quad a + b + c \\leq 10.\n$$\n\nGiven a quadruple $(a, b, c, d)$ of non-negative integers satisfying these inequalities, the numbers of students receiving 0, 1, 2, and 4 points are:\n$$\n2a + b + c + d, \\quad 10 - a - c - d, \\quad 10 - a - b - d, \\quad 10 - a - b - c.\n$$\n\nThus, the problem reduces to counting quadruples $(a, b, c, d)$ of non-negative integers satisfying the three inequalities above.\n\n**Case 1:** $a$ is even. Write $a = 10 - 2n$ for $0 \\leq n \\leq 5$.\n- If $\\max\\{b, c, d\\} \\leq n$, there are $(n+1)^3$ choices.\n- If $\\max\\{b, c, d\\} \\geq n+1$, let the maximum be $2n - k$ ($0 \\leq k \\leq n-1$), and the other two can be any of $(k+1)^2$ pairs. Since the maximum can be any of $b, c, d$, there are $3(k+1)^2$ possibilities, so the total is $\\sum_{k=0}^{n-1} 3(k+1)^2 = \\frac{n(n+1)(2n+1)}{2}$.\n\n**Case 2:** $a$ is odd. Write $a = 9 - 2n$ for $0 \\leq n \\leq 4$.\n- If $\\max\\{b, c, d\\} \\leq n$, there are $(n+1)^3$ choices.\n- If $\\max\\{b, c, d\\} \\geq n+1$, let the maximum be $2n - k + 1$ ($0 \\leq k \\leq n$), and the other two can be any of $(k+1)^2$ pairs. The total is $\\sum_{k=0}^{n} 3(k+1)^2 = \\frac{(n+1)(n+2)(2n+3)}{2}$.\n\nSumming all possibilities:\n$$\n\\sum_{n=0}^{5} (n+1)^3 + \\sum_{n=0}^{5} \\frac{n(n+1)(2n+1)}{2} + \\sum_{n=0}^{4} (n+1)^3 + \\sum_{n=0}^{4} \\frac{(n+1)(n+2)(2n+3)}{2}\n$$\nwhich equals\n$$\n2(1^3 + 2^3 + 3^3 + 4^3 + 5^3) + 6^3 + (6 + 30 + 84 + 180 + 330) = 1296.\n$$\n\nTherefore, the answer is $1296$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 15477,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a rational number $q > 3$ such that $q^2 - 4$ is the square of a rational number, define the sequence $\\{a_i\\}_{i=0}^{\\infty}$ as follows:\n\n$$\na_0 = 2, \\quad a_1 = q, \\quad a_{i+1} = q a_i - a_{i-1}, \\text{ for each } i = 1, 2, \\dots\n$$\n\nDo there exist a natural number $n$ and nonzero integers $b_0, b_1, \\dots, b_n$ such that $\\sum_{i=0}^n b_i = 0$ and, if we write the number $b_0 a_0 + b_1 a_1 + \\dots + b_n a_n$ in the form $\\frac{A}{B}$, where $A$ and $B$ are coprime integers, is the number $A$ free of squares?\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that such numbers do not exist.\n\nThe quadratic equation $x^2 - qx + 1 = 0$ has two rational roots $t$ and $\\frac{1}{t}$ for which $t + \\frac{1}{t} = q$. It follows by induction that $a_m = t^m + \\frac{1}{t^m}$. Suppose there exist numbers $b_0, b_1, \\dots, b_n$ satisfying the problem's conditions.\n\n**Lemma.** Let $f(x) = c_n x^n + c_{n-1} x^{n-1} + \\dots + c_1 x + c_0$ be a polynomial with nonzero integer coefficients such that $c_{n-k} = c_k$ for each $k = 0, 1, \\dots, n$ and $\\sum_{i=0}^n c_i = 0$. Then $f(x) = (x-1)^2 g(x)$, where $g(x)$ is a polynomial with integer coefficients.\n\n*Proof:* From the condition, $f(1) = 0$. Also,\n\n$$\nf'(x) = n c_n x^{n-1} + (n-1) c_{n-1} x^{n-2} + \\dots + c_1,\n$$\n\nso\n\n$$\n2f'(1) = (n c_n + (n-1) c_{n-1} + \\dots + c_1) + (n c_0 + (n-1) c_1 + \\dots + c_{n-1}) = n(c_n + c_{n-1} + \\dots + c_1 + c_0) = 0\n$$\n\nand therefore $x = 1$ is a double root. The lemma is proved.\n\nThe polynomial $f(x) = b_n x^{2n} + b_{n-1} x^{2n-1} + \\dots + b_1 x^{n+1} + 2b_0 x^n + b_1 x^{n-1} + b_2 x^{n-2} + \\dots + b_{n-1} x + b_n$ satisfies the lemma's conditions. It is not hard to see that\n\n$$\nt^n \\left( b_n ( t^n + \\frac{1}{t^n} ) + b_{n-1} ( t^{n-1} + \\frac{1}{t^{n-1}} ) + \\dots + 2b_0 \\right) = f(t) = (t-1)^2 g(t).\n$$\n\nMoreover, if $t = \\frac{r}{s}$, $(r, s) = 1$, then from $\\frac{r}{s} + \\frac{s}{r} = q > 3$ it follows that $r \\ge s + 2$, i.e., $r - s \\ge 2$. Then $g(\\frac{r}{s})$ is of the form $\\frac{l}{s^{2n-2}}$.\n\nFinally, $b_0 a_0 + b_1 a_1 + \\dots + b_n a_n$ can be written as $\\frac{(r-s)^2 l}{r^n s^n}$, and since $r-s \\ge 2$ and $(r-s, r) = (r-s, s) = 1$, the numerator will always contain a squared prime factor. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15478,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b \\in \\mathbb{R}$ and $z \\in \\mathbb{C} \\setminus \\mathbb{R}$ such that $|a - b| = |a + b - 2z|$.\n\n(a) Prove that there exists a unique real number $x$ which satisfies $|z - a|^x + |\\bar{z} - b|^x = |a - b|^x$.\n\n(b) Find all real numbers $x$ such that $|z - a|^x + |\\bar{z} - b|^x \\leq |a - b|^x$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Set $u = z - a$, $v = z - b$. The relation gives $|v - u| = |u + v|$ where $u, v, u + v \\in \\mathbb{C} \\setminus \\mathbb{R}$, so $u, v, u + v \\neq 0$. Thus $|u + v|^2 = |u|^2 + |v|^2$. Since $|v| = |\\bar{v}|$, the equation can be written as $|u|^x + |v|^x = (\\sqrt{|u|^2 + |v|^2})^x$, and then\n$$\n\\left(\\frac{|u|}{\\sqrt{|u|^2 + |v|^2}}\\right)^x + \\left(\\frac{|v|}{\\sqrt{|u|^2 + |v|^2}}\\right)^x = 1.\n$$\nThe function $f: \\mathbb{R} \\to \\mathbb{R}$, $f(x) = \\left( \\frac{|u|}{\\sqrt{|u|^2 + |v|^2}} \\right)^x + \\left( \\frac{|v|}{\\sqrt{|u|^2 + |v|^2}} \\right)^x$ is strictly decreasing, hence $x = 2$ is the only solution.\n\n(b) The solution is $[2, +\\infty)$, since $f$ is strictly decreasing.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15479,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $a, b, c$ 為正實數，試證：\n\n$$\n\\frac{8a^2 + 2ab}{(b + \\sqrt{6ac} + 3c)^2} + \\frac{2b^2 + 3bc}{(3c + \\sqrt{2ab} + 2a)^2} + \\frac{18c^2 + 6ac}{(2a + \\sqrt{3bc} + b)^2} \\geq 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "令 $a = \\frac{1}{2}x$, $b = y$, $c = \\frac{1}{3}z$，則題目中的左式可化為\n\n$$\n\\frac{2x^2 + xy}{(y + \\sqrt{xz} + z)^2} + \\frac{2y^2 + yz}{(z + \\sqrt{xy} + x)^2} + \\frac{2z^2 + xz}{(x + \\sqrt{yz} + y)^2} \\geq 1.\n$$\n\n由科西不等式我們可得\n\n$$\n(yx + x^2 + z^2) \\left( \\frac{y}{x} + \\frac{z}{x} + \\frac{z^2}{x^2} \\right) \\geq (y + \\sqrt{xz} + z)^2.\n$$\n\n因此我們得到\n\n$$\n\\frac{2x^2 + xy}{(y + \\sqrt{xz} + z)^2} \\geq \\frac{x^2}{xy + xz + z^2},\n$$\n\n且此式等號成立若且唯若 $x = z$。\n\n將 $(x, y, z)$ 替換為 $(y, z, x)$ 及 $(z, x, y)$ 則可得到\n\n$$\n\\frac{2y^2 + yz}{(z + \\sqrt{xy} + x)^2} \\geq \\frac{y^2}{yz + yx + x^2},\n$$\n\n和\n\n$$\n\\frac{2z^2 + xz}{(x + \\sqrt{yz} + y)^2} \\geq \\frac{z^2}{zx + zy + y^2}.\n$$\n\n(等號分別成立若且唯若 $y = x$ 及 $z = y$。)\n\n令 $M = \\frac{x^2}{xy+xz+z^2} + \\frac{y^2}{yz+yx+x^2} + \\frac{z^2}{zx+zy+y^2}$，再次使用科西不等式可得\n\n$$\nM((xy + xz + z^2) + (yz + yx + x^2) + (zx + zy + y^2)) \\geq (x + y + z)^2.\n$$\n\n因為\n\n$$\n(xy + xz + z^2) + (yz + yx + x^2) + (zx + zy + y^2) = (x + y + z)^2,\n$$\n\n立即可得\n\n$$\nM = \\frac{x^2}{xy + xz + z^2} + \\frac{y^2}{yz + yx + x^2} + \\frac{z^2}{zx + zy + y^2} \\geq 1,\n$$\n\n且此式等號成立若且唯若 $x = y = z$。即\n\n$$\n\\frac{8a^2 + 2ab}{(b + \\sqrt{6ac} + 3c)^2} + \\frac{2b^2 + 3bc}{(3c + \\sqrt{2ab} + 2a)^2} + \\frac{18c^2 + 6ac}{(2a + \\sqrt{3bc} + b)^2} \\geq 1,\n$$\n\n且此式等號成立若且唯若 $2a = b = 3c$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15480,
    "subject": "Mathematics (Olympiad)",
    "question": "Let\n\n$$\nx = \\frac{a!\\,(b+c)!}{(a+b)!\\,c!} = \\frac{(c+1)(c+2)\\cdots(c+b)}{(a+1)(a+2)\\cdots(a+b)} = \\prod_{i=1}^{b} \\frac{c+i}{a+i}.\n$$\n\nShow that\n\n$$\n\\left(1 + \\frac{a-c}{b} \\sum_{i=1}^{b} \\frac{1}{c+i}\\right)^b \\leq x \\leq \\left(1 + \\frac{c-a}{b} \\sum_{i=1}^{b} \\frac{1}{a+i}\\right)^b.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Because $1 + \\frac{c-a}{a+i} = \\frac{c+i}{a+i}$, we have\n\n$$\n1 + \\frac{c-a}{b} \\sum_{i=1}^{b} \\frac{1}{a+i} = \\frac{1}{b} \\sum_{i=1}^{b} \\left(1 + \\frac{c-a}{a+i}\\right) = \\frac{1}{b} \\sum_{i=1}^{b} \\frac{c+i}{a+i}.\n$$\n\nBy the AM-GM inequality,\n\n$$\n\\sqrt[b]{\\prod_{i=1}^{b} \\frac{c+i}{a+i}} \\leq \\frac{1}{b} \\sum_{i=1}^{b} \\frac{c+i}{a+i},\n$$\nwhich implies\n\n$$\n\\prod_{i=1}^{b} \\frac{c+i}{a+i} \\leq \\left(\\frac{1}{b} \\sum_{i=1}^{b} \\frac{c+i}{a+i}\\right)^b.\n$$\n\nThis establishes the right-hand inequality. Swapping the roles of $a$ and $c$, then taking the reciprocal, gives the left-hand inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15481,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $(m, n)$ such that $m! = 2^n + n$.",
    "options": [],
    "answer": "See solution",
    "solution": "$m = 1$ gives no solution, therefore $m!$ is even and $n$ should also be even. Let $n = 2^t \\cdot s$ ($t$ and $s$ are positive integers, $t \\ge 1$ and $s$ is odd). $t = 1$ readily leads to $n = 2$ and $m = 3$. Now let $t \\ge 2$. Then $m! = 2^n + n = 2^{2t \\cdot s} + 2^t \\cdot s \\ge 2^{2t} + 2^t$. By induction over $t$ we will show that\n\n$$\n2^{2t} + 2^t > (2t - 1)! \\quad (1)\n$$\n\nFor $t=2, 3$ the inequality holds. Assume that it is held for $t=k$. In order to show that it also holds for $k+1$ we have to prove that\n\n$$\n\\frac{2^{2k+1} + 2^{k+1}}{2^2 + 2^k} \\ge 2k(2k+1)\n$$\n\nSince $\\frac{2^{2n} + 2n}{2^n + n} \\ge 2^{n-1} \\iff 2^{2n-2} \\ge n(2^{n-2} - 1)$ and $2^{2n-2} \\ge n \\cdot 2^{n-2}$ for each positive integer $n$ we get $\\frac{2^{2n} + 2n}{2^n + n} \\ge 2^{n-1}$. By taking $n = 2^k$ we get $\\frac{2^{2^{k+1}} + 2^{k+1}}{2^2 + 2^k} \\ge 2^{2^{k-1}}$. Thus, in order to complete the proof we will show that $2^{2^{k-1}} \\ge 2k(2k+1)$ for $k \\ge 3$. For $k = 3, 4$ it is held since $2^7 > 42$ and $2^{15} > 72$ and for $k \\ge 5$ we have $2^{2^{k-1}} \\ge 2^{4k} = 2^{2k} \\cdot 2^{2k} \\ge 2k(2k+1)$. (1) is proved. By (1) $m! > (2t-1)!$ and consequently $m \\ge 2t$. Since $2^n + n = 2^{2t \\cdot s} + 2^t \\cdot s = 2^t(2^{2t \\cdot s-t} + s)$ and $2^t \\cdot s \\ge 2^t > t$ we get that $2^{2t \\cdot s-t} + \\beta$ is odd. Therefore $2^t$ divides the maximal even factor of $m!$ and since $m \\ge 2t$ we get\n\n$$\n\\alpha = \\left\\lfloor \\frac{m}{2} \\right\\rfloor + \\left\\lfloor \\frac{m}{4} \\right\\rfloor + \\dots \\ge t + \\left\\lfloor \\frac{t}{2} \\right\\rfloor + \\left\\lfloor \\frac{t}{4} \\right\\rfloor + \\dots\n$$\n\nContradiction since $t \\ge 2$. Thus, the only solution is: $(m, n) = (3, 2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15482,
    "subject": "Mathematics (Olympiad)",
    "question": "In $\\triangle ABC$, $AB = 6$, $BC = 4$, and the median to side $AC$ is $\\sqrt{10}$. Find the value of $\\sin^6 \\frac{A}{2} + \\cos^6 \\frac{A}{2}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of $AC$. By the median formula:\n\n$$\n4BM^2 + AC^2 = 2(AB^2 + BC^2)\n$$\n\nThus,\n\n$$\nAC = \\sqrt{2(6^2 + 4^2) - 4 \\cdot 10} = 8.\n$$\n\nBy the law of cosines:\n\n$$\n\\cos A = \\frac{CA^2 + AB^2 - BC^2}{2CA \\cdot AB} = \\frac{8^2 + 6^2 - 4^2}{2 \\cdot 8 \\cdot 6} = \\frac{7}{8}.\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\sin^6 \\frac{A}{2} + \\cos^6 \\frac{A}{2} &= \\left(\\sin^2 \\frac{A}{2} + \\cos^2 \\frac{A}{2}\\right) \\left(\\sin^4 \\frac{A}{2} - \\sin^2 \\frac{A}{2} \\cos^2 \\frac{A}{2} + \\cos^4 \\frac{A}{2}\\right) \\\\\n&= \\left(\\sin^2 \\frac{A}{2} + \\cos^2 \\frac{A}{2}\\right)^2 - 3 \\sin^2 \\frac{A}{2} \\cos^2 \\frac{A}{2} \\\\\n&= 1 - \\frac{3}{4} \\sin^2 A \\\\\n&= \\frac{1}{4} + \\frac{3}{4} \\cos^2 A \\\\\n&= \\frac{211}{256}.\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15483,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 2020 inhabitants in a town. Before Christmas, they are all happy; but if an inhabitant does not receive any Christmas card from any other inhabitant, he or she will become sad. Unfortunately, there is only one post company which offers only one kind of service: before Christmas, each inhabitant may appoint two different other inhabitants, among which the company chooses one to whom to send a Christmas card on behalf of that inhabitant. It is known that the company makes the choices in such a way that as many inhabitants as possible will become sad. Find the least possible number of inhabitants who will become sad.",
    "options": [],
    "answer": "See solution",
    "solution": "Partition 2019 inhabitants into 673 groups, each containing 3 inhabitants. Suppose that each inhabitant appoints two other members of the same group. As no group member is appointed thrice, the company cannot send three cards to one group member. Hence in every group, two different members get a card and at most one member will become sad. The inhabitant who does not belong to any group will become sad, too. Altogether, at most 674 inhabitants will become sad.\n\nWe show now that this is the least possible number. More precisely, we show that, in the case of $n$ inhabitants, the company can make $\\lfloor \\frac{n}{3} \\rfloor$ inhabitants sad. Call inhabitants $X$ and $Y$ \\textbf{competitors} if somebody appoints them together to the company. By conditions of the problem, there are as many pairs of competitors as inhabitants or less (if several inhabitants appoint the same pair or some inhabitant appoints no pair). Thus it suffices to prove that, in the case of $n$ inhabitants and at most $n$ pairs of competitors, there must be $\\lceil \\frac{n}{3} \\rceil$ inhabitants among which no two are competitors. The claim holds trivially for $n = 0, 1, 2$. Now let $n \\ge 3$ and assume the claim being valid for $n-3$ inhabitants. W.l.o.g., assume that there are exactly $n$ pairs of competitors. Then there are exactly $2n$ instances of an inhabitant belonging to a pair of competitors. Consider two cases:\n\n- If every inhabitant has exactly 2 competitors then choose an arbitrary inhabitant $X$ and leave out $X$ along with both competitors. Among the remaining $n-3$ inhabitants, there are at most $n-3$ pairs of competitors (besides two pairs containing $X$, removing either competitor canceled one more pair). By the induction hypothesis, one can find $\\lceil \\frac{n-3}{3} \\rceil$ remaining inhabitants with no pair of competitors. Adding $X$ to them results in $\\lceil \\frac{n}{3} \\rceil$ inhabitants with no pair of competitors.\n\n- If an inhabitant $X$ has at most 1 competitor then there must exist an inhabitant $Y$ with at least 3 competitors. Let $Z$ be the competitor of $X$ if $X$ has a competitor and an arbitrary inhabitant different from $X$ and $Y$ otherwise. After leaving out $X$, $Y$ and $Z$, we have $n-3$ inhabitants and at most $n-3$ pairs of competitors (removing $Y$ cancels at least 3 pairs of competitors). By the induction hypothesis, one can find $\\lceil \\frac{n-3}{3} \\rceil$ remaining inhabitants with no pair of competitors. Adding $X$ to this set results in a group of $\\lceil \\frac{n}{3} \\rceil$ inhabitants with no pairs of competitors. This proves the desired claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15484,
    "subject": "Mathematics (Olympiad)",
    "question": "The positive real numbers $x, y, z$ are such that $x + y + z = 4$ and $x, y, z \\in [0,2]$. Find the minimal value of the algebraic expression:\n\n$$\nA = \\sqrt{2+x} + \\sqrt{2+y} + \\sqrt{2+z} + \\sqrt{x+y} + \\sqrt{y+z} + \\sqrt{z+x}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove first that $\\sqrt{2+x} + \\sqrt{y+z} = \\sqrt{2+x} + \\sqrt{4-x} \\ge 2 + \\sqrt{2}$.\n\nIndeed, this is equivalent to\n$$\n2 + x + 4 - x + \\sqrt{(2 + x)(4 - x)} \\ge 4 + 2 + 2\\sqrt{2} \\Leftrightarrow x(2 - x) \\ge 0,\n$$\nwhich is true, since $x \\in [0,2]$.\n\nSimilarly, we have $\\sqrt{2+y} + \\sqrt{x+z} \\ge 2 + \\sqrt{2}$ and $\\sqrt{2+z} + \\sqrt{x+y} \\ge 2 + \\sqrt{2}$.\n\nAdding all these, we get $A \\ge 6 + 3\\sqrt{2}$, and the equality holds, for example, if $x = y = 2$ and $z = 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15485,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $B_n$ be the number of ways to partition an $n$-element set into non-empty parts. For example, $B_3 = 5$ because we have the following partitions of the 3-element set $\\{a, b, c\\}$:\n\n$$\n\\{a, b, c\\}; \\quad \\{a\\}, \\{b, c\\}; \\quad \\{a, c\\}, \\{b\\}; \\quad \\{a, b\\}, \\{c\\}; \\quad \\{a\\}, \\{b\\}, \\{c\\}.\n$$\n\nProve that for every positive integer $m$ and prime number $p$,\n\n$$\nB_{p^m} \\equiv m+1 \\pmod{p}.\n$$\n\nLet $R_m$ be a set of residues modulo $p^m$. For each residue $y \\in R_m$ define a shift of the set $R_m$ by the rule $f_y(x) = (x + y) \\bmod{p^m}$. If $A \\subset R_m$ we denote by $f_y(A)$ the set we get by applying $f_y$ element-wise to $A$. If $P = (P_1, P_2, \\dots, P_k)$ is a partition, then we denote by $f_y(P)$ the partition $(f_y(P_1), f_y(P_2), \\dots, f_y(P_k))$. We call a partition *fixed* under this action if $f_y(P) = P$ for each $y$. The problem statement follows from two facts:\n\n1) If for some partition $P$ and for some $a$ we have $f_a(P) \\neq P$, then the number of different partitions of the form $f_y(P)$ is a power of $p$.\n\n2) If for some partition $P$ and for all $a$ we have $f_a(P) = P$, then $P$ is a partition for which there exists $j$ such that every subset of $P$ consists of (all) elements which are congruent to each other modulo $p^{m-j}$.\n\nIt follows from 1) that the number of fixed partitions is equivalent to $B_{p^m}$ modulo $p$. It follows from 2) that the number of fixed partitions equals $m+1$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof of 1)*\n\nLet $f_{a_1}(P), \\dots, f_{a_k}(P)$ be the maximal collection of pairwise different partitions, denote the set $\\{a_1, \\dots, a_k\\}$ by $O$; and let $S = \\{s \\in R_m \\mid f_s(P) = P\\}$. Then it is easy to see that $R_m$ is a direct sum of $S$ and $O$. Hence, $|O|$ is a power of $p$.\n\n*Proof of 2)*\n\nSuppose you have some fixed partition $P$ with elements $a$ and $b$ inside subsets $A$ and $B$ (respectively). Then clearly $f_{b-a}(a) = b$, and since $P$ is fixed this means $f_{b-a}(A) = B$. Hence for a fixed partition $P$, all subsets of $P$ must be the same size, and therefore some power of $p$.\n\nSuppose to the contrary that we have a fixed partition $P$ with elements $a, b$ in the same $p^j$-element subset $A$ which satisfy $a \\not\\equiv b \\pmod{p^{m-j}}$. Now $f_{b-a}(a) = b$, and so $A$ is permuted by the action of $f_{b-a}$. Hence for any integer $r$, the $r$-fold composition of the map $f_{b-a}$, namely, the map $f_{r(b-a)}$, again takes $a$ to some element of $A$. Now clearly $f_{r(b-a)}(a) \\equiv f_{s(b-a)}(a) \\pmod{p^m}$ if and only if\n\n$$\na + r(b - a) \\equiv a + s(b - a) \\pmod{p^m}.\n$$\n\nSince $a \\not\\equiv b \\pmod{p^{m-j}}$, however, this congruence forces $r - s \\equiv 0 \\pmod{p^{j+1}}$. Hence for $r$ between $1$ and $p^{j+1}$, the elements $f_{r(b-a)}(a)$ are distinct elements of $A$. It follows that $|A| > p^j$, a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15486,
    "subject": "Mathematics (Olympiad)",
    "question": "How many rectangles of area greater than 31 and less than or equal to 40, with integer side lengths $x$ and $y$ such that $0 < x, y \\leq 8$, can be placed inside an $8 \\times 8$ square grid?",
    "options": [],
    "answer": "See solution",
    "solution": "First, we examine which rectangle areas are possible:\n\n- $40 = 5 \\times 8$. A $5 \\times 8$ rectangle fits in $4$ horizontal and $4$ vertical positions, so $8$ such rectangles.\n- $36 = 6 \\times 6$. A $6 \\times 6$ rectangle fits in $3^2 = 9$ ways.\n- $35 = 5 \\times 7$. A $5 \\times 7$ rectangle fits in $4$ horizontal and $2$ vertical positions, and $7 \\times 5$ fits in $2$ horizontal and $4$ vertical positions, totaling $16$ ways.\n- $32 = 4 \\times 8$. A $4 \\times 8$ rectangle fits in $5$ horizontal and $2$ vertical positions, totaling $10$ ways.\n\nNumbers $39, 38, 37, 34, 33, 31$ are not possible with $x, y \\leq 8$.\n\nAdding up: $8 + 9 + 16 + 10 = 43$ rectangles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15487,
    "subject": "Mathematics (Olympiad)",
    "question": "Find two sets of $n$ numbers, each set consisting of pairwise different positive integers, such that the sums and products of the numbers in each set differ by exactly $1$ (i.e., if the sum of the first set is $a$ and the product is $A$, and the sum of the second set is $b$ and the product is $B$, then $b = a + 1$ and $B = A + 1$). Construct such sets for any $n \\geq 2$ and provide explicit examples for small $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The construction proceeds by choosing numbers so that the products of the first $2n-2$ numbers differ by $1$, i.e., $|B-A|=1$. This ensures that the numbers found by the formulas will be integers. For example, let $a_1 = 4$, $a_2 = 5$, $b_1 = 3$, $b_2 = 7$ for $n=2$; both sets have sums $9$ and $10$, and products $20$ and $21$. For $n=3$, take $a_1 = 2$, $a_2 = 3$, $a_3 = 15$, $b_1 = 1$, $b_2 = 7$, $b_3 = 13$; sums are $20$ and $21$, products $90$ and $91$. The method generalizes by recursively constructing sets for higher $n$ using previous sets and formulas:\n\n$$\na_1 + a_2 + \\dots + a_{n-1} = a = b_1 + b_2 + \\dots + b_{n-1} - 1, \\quad a_1 a_2 \\dots a_{n-1} = A = b_1 b_2 \\dots b_{n-1} - 1.\n$$\n\nThen, add $a_n = A$ and $b_n = A+1$ to each set, and continue recursively. For $n=4$, one example is $a_1 = 4$, $a_2 = 5$, $a_3 = 20$, $a_4 = 398$ and $b_1 = 3$, $b_2 = 7$, $b_3 = 19$, $b_4 = 399$, with sums $427$ and $428$, and products $159200$ and $159201$ respectively.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15488,
    "subject": "Mathematics (Olympiad)",
    "question": "En la primera fila de un tablero $5 \\times 5$ se colocan 5 fichas que tienen una cara blanca y otra negra, mostrando todas la cara blanca. Cada ficha se puede mover de una casilla a cualquiera de las contiguas (horizontal o verticalmente) dándole la vuelta en cada movimiento. Además, varias fichas pueden ocupar una misma casilla. ¿Se puede conseguir mediante una secuencia de movimientos que las 5 fichas queden en la última fila, en casillas distintas y que todas ellas muestren la cara negra?",
    "options": [],
    "answer": "See solution",
    "solution": "Si pintamos las casillas del tablero alternativamente de blanco y negro como en un tablero de ajedrez, sucede que una ficha cuyo color visible coincida con el de la casilla, al moverse seguirá teniendo el mismo color que la nueva casilla (puesto que tanto el color de la ficha como el de la casilla cambian). Supuesto que la casilla superior izquierda la hemos dejado blanca, en el inicio hay 3 fichas cuyo color (blanco) coincide con el de la casilla. En todo momento deberá suceder que el color de tres fichas es el mismo que el de la casilla que ocupen (y el de las otras dos, diferente). Sin embargo, colocando las fichas con la cara negra en la última fila, resulta que sólo dos fichas tendrán el color (negro) de su casilla. Por lo tanto, no es posible colocar las fichas de esta manera.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15489,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, $C$, and $D$ be points on a circle such that $|AB| = |BC| = |CD|$. The angle bisectors of $\\angle ABD$ and $\\angle ACD$ intersect at the point $E$.\n\nIf the lines $AE$ and $CD$ are parallel, find $\\angle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Due to symmetry, the isosceles triangles $ABC$ and $BCD$ are congruent, and the cyclic quadrilateral $ABCD$ is an isosceles trapezium.\n\n![](images/HRV_ABooklet_2019_p43_data_5b8f31467d.png)\n\nLet $x$ be the measure of the angles along the bases in $ABC$ and $BCD$. Then $\\angle ABC = \\angle BCD = 180^\\circ - 2x$ and $\\angle ABD = \\angle ACD = 180^\\circ - 3x$, from which we get $\\angle EBC = \\angle ECB = 90^\\circ - \\frac{x}{2}$ and $\\angle BEC = x$. Hence, the point $E$ lies on the same circle as $A$, $B$, $C$, and $D$.\n\nWe also have $\\angle ADC = \\angle ADB + \\angle BDC = \\angle ACB + \\angle BDC = x + x = 2x$ and $\\angle ECD = \\frac{1}{2}\\angle ACD = 90^\\circ - \\frac{3}{2}x$.\n\nSince $AE \\parallel CD$, the cyclic quadrilateral $ACDE$ is an isosceles trapezium as well, so $\\angle ADC = \\angle ECD$ holds, and we get $7x = 180^\\circ$. Therefore, $\\angle ABC = \\frac{5}{7} \\cdot 180^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15490,
    "subject": "Mathematics (Olympiad)",
    "question": "Azar and Carl play a game of tic-tac-toe. Azar places an $X$ in one of the boxes in a $3 \\times 3$ array of boxes, then Carl places an $O$ in one of the remaining boxes. After that, Azar places an $X$ in one of the remaining boxes, and so on until all 9 boxes are filled or one of the players has 3 of their symbols in a row—horizontal, vertical, or diagonal—whichever comes first, in which case that player wins the game. Suppose the players make their moves at random, rather than trying to follow a rational strategy, and that Carl wins the game when he places his third $O$. How many ways can the board look after the game is over?\n\n(A) 36  \n(B) 112  \n(C) 120  \n(D) 148  \n(E) 160",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer (D):** For Carl to win at his third turn, his 3 $O$s must lie in one of the 8 winning configurations and Azar's 3 $X$s must not (because that would have resulted in her winning after her third move). There are 6 vertical or horizontal rows for Carl's $O$s, and in each case there are $\\binom{6}{3} - 2$ ways for Azar's $X$s to not align. There are also 2 diagonal winning lines for Carl, and in those cases Azar could not have won first. The number of ways the board can look after the game is over is therefore\n\n$$\n6 \\cdot \\left( \\binom{6}{3} - 2 \\right) + 2 \\cdot \\binom{6}{3} = 6 \\cdot 18 + 2 \\cdot 20 = 148.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15491,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for all real numbers $x, y$,\n\n$$\n(x + y) f(x + y) = \\left(f(f(x) + f(y))\\right)^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* $f(x) = 0$ and $f(x) = x$.\n\n*Solution.*\n\nLet us denote the given equation by (1). For $P(a, b)$, we substitute $x = a$ and $y = b$ into (1).\n\nObviously, $f(x) = 0$ and $f(x) = x$ are solutions. Assume $f \\not\\equiv 0$.\n\n$$\nP(0, 0): f(2f(0)) = 0.\n$$\n\n$$\nP(2f(0), 0): 2f(0) f(2f(0)) = (f(f(2f(0)) + f(0)))^2 \\implies f(f(0)) = 0.\n$$\n\n$$\nP(f(0), f(0)): 2f(0) f(f(0) + f(0)) = (f(f(f(0)) + f(f(0))))^2 \\implies f(0) = 0.\n$$\n\nNow, $P(x, 0)$ gives:\n\n$$\nx f(x) = (f(f(x)))^2. \\qquad (2)\n$$\n\nSuppose $\\exists a \\neq b$ such that $f(a) = f(b)$. Then\n\n$$\na f(a) = (f(f(a)))^2 = (f(f(b)))^2 = b f(b) \\implies f(a) = f(b) = 0. \\\\\n\\text{So } f(a) = f(b) \\implies f(a) = f(b) = 0. \\quad (*)\n$$\n\nThus, to prove injectivity, it suffices to show that if $\\exists x_0 \\neq 0$ with $f(x_0) = 0$, then $f \\equiv 0$.\n\nFrom (2):\n\n$$\nf(f(x)) = 0 \\iff f(x) = 0. \\quad (3)\n$$\n\nNow, $P(x, x_0)$ gives:\n\n$$\n(x + x_0) f(x + x_0) = (f(f(x)))^2 = (f(f(x + x_0)))^2,\n$$\nso\n$$\n(f(f(x)))^2 = (f(f(x + x_0)))^2. \\quad (4)\n$$\n\nFrom (2):\n\n$$\nf(x) \\ge 0 \\iff x \\ge 0, \\quad f(x) \\le 0 \\iff x \\le 0. \\quad (5)\n$$\n\nAssume $\\exists x_1 > \\max\\{-x_0, 0\\}$ with $f(x_1) \\neq 0$. By (3) and (5), $f(f(x_1))$ and $f(f(x_1 + x_0))$ are positive, so $f(f(x_1)) = f(f(x_1 + x_0))$. By (*) this means $f(f(x_1)) = f(f(x_1 + x_0)) = 0 \\implies f(x_1) = f(x_1 + x_0) = 0$, contradicting the choice of $x_1$. Thus,\n\n$$\n\\forall x > \\max\\{-x_0, 0\\},\\ f(x) = 0. \\quad (6)\n$$\n\nSuppose $f(y_1) \\neq 0$ for some $y_1$. Substituting $x = y_1$ in (4), $(f(f(y_1)))^2 = (f(f(y_1 + x_0)))^2$, and for large enough $x_0$, $y_1 + x_0 > \\max\\{-x_0, 0\\}$, so $f(f(y_1)) = 0$, contradicting (3). Therefore, if $f$ is not injective, then $f \\equiv 0$.\n\nNow, suppose $f$ is injective. From (1) and (2):\n\n$$\n(f(f(x + y)))^2 = (f(f(x) + f(y)))^2. \\quad (7)\n$$\n\nFor $x, y > 0$, by (5):\n\n$$\nf(f(x + y)) = f(f(x) + f(y)) \\implies f(x + y) = f(x) + f(y). \\quad (8)\n$$\n\nSo $f$ is additive and positive on positive arguments. Thus, $f(x) = kx$ for some $k \\ge 0$. From (2): $k = k^4 \\implies k = 1$. Similarly, for negative values, $f(x) = x$ for all $x$.\n\nTherefore, the only solutions are $f(x) = 0$ and $f(x) = x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15492,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, x_3, \\dots$ be a sequence of rational numbers defined by $x_1 = \\frac{25}{11}$ and\n$$\nx_{k+1} = \\frac{1}{3} \\left( x_k + \\frac{1}{x_k} - 1 \\right)\n$$\nfor all $k \\ge 1$. Then $x_{2025}$ can be expressed as $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find the remainder when $m+n$ is divided by 1000.",
    "options": [],
    "answer": "See solution",
    "solution": "Define $f(x) = \\frac{1}{3}(x + \\frac{1}{x} - 1)$. Then for positive integers $a$ and $b$,\n$$\nf\\left(\\frac{a}{b}\\right) = \\frac{a^2 - ab + b^2}{3ab}.\n$$\nIf $a$ and $b$ are relatively prime, it follows that $\\gcd(a^2 - ab + b^2, a) = \\gcd(b^2, a) = 1$ and, similarly, $\\gcd(a^2 - ab + b^2, b) = 1$, so $\\gcd(a^2 - ab + b^2, ab) = 1$. Then\n$$\n\\gcd(a^2 - ab + b^2, 3ab) \\mid 3 \\gcd(a^2 - ab + b^2, ab) = 3 \\cdot 1 = 3.\n$$\nIn other words, when $a$ and $b$ are relatively prime,\n$$\n\\gcd(a^2 - ab + b^2, 3ab) = \\begin{cases} 3 & \\text{if } a^2 - ab + b^2 \\equiv 0 \\pmod{3}, \\\\ 1 & \\text{otherwise.} \\end{cases}\n$$\nLet $a_k$ and $b_k$ be the respective numerator and denominator when $x_k$ is written in lowest terms. Consider the following two families of claims for each $k \\ge 1$:\n* $P(k)$: $a_k \\equiv 1 \\pmod{3}$ and $b_k \\equiv 2 \\pmod{3}$.\n* $Q(k)$: $a_{k+1} = \\frac{a_k^2 - a_k b_k + b_k^2}{3}$ and $b_{k+1} = a_k b_k$.\n\nThese claims can be proved by induction in the following way:\n$$\nP(1) \\implies Q(1) \\implies P(2) \\implies Q(2) \\implies \\dots\n$$\nBecause $a_1 = 25 \\equiv 1 \\pmod{3}$ and $b_1 = 11 \\equiv 2 \\pmod{3}$, the claim $P(1)$ is true. To show that $P(k)$ implies $Q(k)$, note that\n$$\na_k^2 - a_k b_k + b_k^2 = (a_k + b_k)^2 - 3a_k b_k \\equiv 0 - 3 \\cdot 1 \\cdot 2 \\equiv 3 \\pmod{9},\n$$\nso $P(k)$ implies $Q(k)$ follows now from the previous result.\n\nFinally, to show that $Q(k)$ implies $P(k+1)$, compute\n$$\na_{k+1} = \\frac{a_k^2 - a_k b_k + b_k^2}{3} \\equiv 1 \\pmod{3}\n$$\nand\n$$\nb_{k+1} = 1 \\cdot 2 \\equiv 2 \\pmod{3},\n$$\nwhere the relation for $a_{k+1}$ follows from $a_k^2 - a_k b_k + b_k^2 \\equiv 3 \\pmod{9}$. This completes the induction, proving all the claims $P(k)$ and $Q(k)$.\n\nDefine $c_k = a_k + b_k$. The problem requires computing $c_{2025}$. Note that $Q(k)$ implies that\n$$\na_{k+1} + b_{k+1} = \\frac{a_k^2 + 2a_k b_k + b_k^2}{3} = \\frac{(a_k + b_k)^2}{3}.\n$$\nTherefore $c_{k+1} = \\frac{c_k^2}{3}$, which is equivalent to $\\frac{c_{k+1}}{3} = \\left(\\frac{c_k}{3}\\right)^2$. Thus\n$$\nc_{2025} = 3 \\left(\\frac{c_1}{3}\\right)^{2^{2024}} = 3 \\cdot 12^{2^{2024}}.\n$$\nIt remains to compute $c_{2025}$ modulo 1000.\n\nBy the Chinese Remainder Theorem, to find $c_{2025} \\pmod{1000}$, it suffices to find the remainders when $c_{2025}$ is divided by 8 and by 125. By Euler's Theorem,\n$$\n12^{\\phi(125)} \\equiv 2^{100} \\equiv 1 \\pmod{125}\n$$\nand\n$$\n2^{\\phi(25)} = 2^{20} \\equiv 1 \\pmod{25}.\n$$\nHence\n$$\n2^{2024} = (2^{20})^{101} \\cdot 2^4 \\equiv 16 \\pmod{25}.\n$$\nBecause $2^{2024}$ is divisible by 4, it follows that\n$$\n2^{2024} \\equiv 16 \\pmod{100}.\n$$\nTherefore\n$$\nc_{2025} = 3 \\cdot 12^{2^{2024}} \\equiv 3 \\cdot 12^{16} \\equiv 3 \\cdot 19^8 \\pmod{125}\n$$\n$$\n\\equiv 3 \\cdot (-14)^4 \\equiv 3 \\cdot (-54)^2 \\equiv 123 \\pmod{125}.\n$$\nAlso note that $c_{2025} \\equiv 0 \\pmod{8}$. The simultaneous solution to the system of congruences\n$$\nc_{2025} \\equiv 0 \\pmod{8}\n$$\n$$\nc_{2025} \\equiv 123 \\pmod{125}\n$$\nis $c_{2025} \\equiv 248 \\pmod{1000}$. The requested remainder is $\\boxed{248}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15493,
    "subject": "Mathematics (Olympiad)",
    "question": "In the product\n\n$$\n\\left(1 + \\frac{1}{1}\\right) \\cdot \\left(1 + \\frac{1}{3}\\right) \\cdot \\left(1 + \\frac{1}{5}\\right) \\cdots \\left(1 + \\frac{1}{2n-1}\\right)\n$$\n\nthe denominators of the fractions are all odd numbers from $1$ to $2n - 1$. Is it possible to choose a natural number $n > 1$ such that this product would evaluate to an integer?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** No.\n\n**Solution:** By manipulating the given product we get\n\n$$\n\\begin{aligned}\n\\left(1 + \\frac{1}{1}\\right) \\cdot \\left(1 + \\frac{1}{3}\\right) \\cdot \\left(1 + \\frac{1}{5}\\right) \\cdots \\left(1 + \\frac{1}{2n-1}\\right) &= \\frac{2}{1} \\cdot \\frac{4}{3} \\cdot \\frac{6}{5} \\cdots \\frac{2n}{2n-1} \\\\\n&= \\frac{2 \\cdot 4 \\cdot 6 \\cdots 2n}{1 \\cdot 3 \\cdot 5 \\cdots (2n-1)}.\n\\end{aligned}\n$$\n\nFor it to be an integer, the number $2 \\cdot 4 \\cdot 6 \\cdots 2n$ should be divisible by $1 \\cdot 3 \\cdot 5 \\cdots (2n - 1)$. But since\n\n$$\n2 \\cdot 4 \\cdot 6 \\cdots 2n = (2 \\cdot 1) \\cdot (2 \\cdot 2) \\cdot (2 \\cdot 3) \\cdots (2 \\cdot n) = 2^n \\cdot (1 \\cdot 2 \\cdot 3 \\cdots n)\n$$\n\nand the number $1 \\cdot 3 \\cdot 5 \\cdots (2n - 1)$ is odd, the number $1 \\cdot 2 \\cdot 3 \\cdots n$ should be divisible by $1 \\cdot 3 \\cdot 5 \\cdots (2n - 1)$. In the case of $n > 1$ this is impossible, because $1 \\cdot 2 \\cdot 3 \\cdots n < 1 \\cdot 3 \\cdot 5 \\cdots (2n - 1)$.\n\n**Remark:** The fact that for $n \\ge 2$ the fraction $\\frac{2 \\cdot 4 \\cdot 6 \\cdots 2n}{1 \\cdot 3 \\cdot 5 \\cdots (2n - 1)}$ does not evaluate to an integer can be shown more easily using Chebyshev's theorem, according to which for every $n \\ge 2$ there always exists at least one prime between $n$ and $2n$. As this prime is certainly odd, it must occur as a factor of the product that is the denominator of the resulting fraction. But since in the numerator all the factors are in the form $2i$ where $i \\le n$, these factors cannot be divisible by primes greater than $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15494,
    "subject": "Mathematics (Olympiad)",
    "question": "For every positive integer $n$, let $\\tau(n)$ denote the number of its positive factors. Determine all $n \\in \\mathbb{N}$ that satisfy the equality $\\tau(n) = \\frac{n}{3}$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $n \\in \\mathbb{N}$ satisfies $\\tau(n) = \\frac{n}{3}$, then $3$ divides $n$. Let $n = 3k$, $k \\in \\mathbb{N}$. If $k$ is even, then $\\frac{k}{2} = \\frac{n}{6}$ is a factor of $n$. Even if all positive numbers smaller than $\\frac{n}{6}$ are factors of $n$, and the numbers $\\frac{n}{5}, \\frac{n}{4}, \\dots, \\frac{n}{1}$ are also factors, we have $\\tau(n) \\leq \\frac{n}{6} + 5$. Thus, $n \\leq 30$. Checking $n = 6, 12, 18, 24, 30$, we find that $18$ and $24$ satisfy the condition.\n\nIf $k$ is odd, we proceed similarly and obtain $n = 9$. Alternatively, since a number with an odd number of positive factors is a perfect square, let $n = m^2$. Then $n$ has at most $m + 1$ factors, so $m + 1 \\geq \\frac{m^2}{3}$, which gives $m \\leq 3$.\n\nIn conclusion, the solutions are $n = 9, 18, 24$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15495,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $SS'$ and $RR'$ be perpendicular to $PQ$ with $S'$ and $R'$ on $PQ$. Let $RT$ be perpendicular to $SS'$ with $T$ on $SS'$.\n\n![](images/Australian-Scene-2017_p67_data_fc693029f8.png)\n\nLet $U$ be the point on $PS$ so that $UR$ is parallel to $PQ$. Let $T$ be the point on $RU$ so that $ST$ is perpendicular to $RU$. Extend $PS$ and $QR$ to meet at $V$.\n\n![](images/Australian-Scene-2017_p67_data_254868bbe2.png)\n\nGiven that $a$ and $b$ are unique, find $a + b$ if $2PQ = a + \\sqrt{b}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\angle P = 60^\\circ$, $PS' = \\frac{5}{2}$ and $SS' = \\frac{5\\sqrt{3}}{2}$.\n\nSince $\\angle Q = 60^\\circ$, $QR' = 2$ and $RR' = 2\\sqrt{3}$.\n\nHence $ST = SS' - TS' = SS' - RR' = \\frac{5\\sqrt{3}}{2} - 2\\sqrt{3} = \\frac{\\sqrt{3}}{2}$.\n\nApplying Pythagoras' theorem to $\\triangle RTS$ gives $RT^2 = 36 - \\frac{3}{4} = \\frac{141}{4}$.\n\nSo $PQ = PS' + S'R' + R'Q = PS' + TR + R'Q = \\frac{5}{2} + \\frac{\\sqrt{141}}{2} + 2$.\n\nHence $a + \\sqrt{b} = 2PQ = 9 + \\sqrt{141}$. An obvious solution is $a = 9$, $b = 141$.\n\nLet $PQV$ be equilateral. Since $UR \\parallel PQ$, $\\triangle URV$ is equilateral and $PU = QR = 4$.\n\nSo $US = 1$, $UT = \\frac{1}{2}$, $ST = \\frac{\\sqrt{3}}{2}$.\n\nApplying Pythagoras' theorem to $\\triangle RTS$ gives $RT^2 = 36 - \\frac{3}{4} = \\frac{141}{4}$.\n\nWe also have $RT = RU - UT = RV - \\frac{1}{2} = QV - \\frac{9}{2} = PQ - \\frac{9}{2}$.\n\nSo $2PQ = 9 + \\sqrt{141} = a + \\sqrt{b}$. Given that $a$ and $b$ are unique, we have $a + b = 150$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15496,
    "subject": "Mathematics (Olympiad)",
    "question": "稱一個正整數為去年老梗，如果這個數字有三個相異正因數的和為 $2022$。試求最小的去年老梗數。",
    "options": [],
    "answer": "See solution",
    "solution": "觀察 $1344$ 是一個解，因為 $6 + 672 + 1344 = 2022$。聲稱 $1344$ 是最小的去年老梗數。假設存在 $N < 1344$ 也符合條件，則存在 $a < b < c$，使得\n\n$$\n2022 = N \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) < 1344 \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right)\n$$\n\n因此\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} > \\frac{2022}{1344} = \\frac{3}{2} + \\frac{1}{224}\n$$\n\n若 $a > 1$，則\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\leq \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} < \\frac{3}{2}\n$$\n\n所以必須有 $a = 1$。同理，$b < 4$，因為\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\leq \\frac{1}{1} + \\frac{1}{4} + \\frac{1}{5} < \\frac{3}{2}\n$$\n\n對於 $b = 3$，有 $c = 4, 5$，此時 $2022 = \\frac{19}{12}N$ 或 $2022 = \\frac{23}{15}N$，但 $12$ 和 $15$ 都不整除 $2022$，不可能。所以 $a = 1, b = 2$。注意 $c < 224$，因為\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} > \\frac{3}{2} + \\frac{1}{224}\n$$\n\n因此\n\n$$\n2022 = N \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = \\frac{3c+2}{2c} N \\right)\n$$\n\n容易看出 $(3c+2, c) | 2$，因此 $3c+2 | 2022 \\cdot 4 = 2^3 \\cdot 337$。$3c+2 > 8$ 因為 $c > b = 2$。\n所以必須有 $3c+2 = 2 \\cdot 337$，即 $c = 224$，這與 $c < 224$ 矛盾。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15497,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a finite set and $P(S)$ be the power set of $S$ (the set of all subsets of $S$). For any function $f: P(S) \\to \\mathbb{R}$, prove\n$$\n\\sum_{A \\in P(S)} \\sum_{B \\in P(S)} f(A)f(B)2^{|A \\cap B|} \\geq 0,\n$$\nwhere $|X|$ represents the number of elements of set $X$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, notice that\n$$\n2^{|A \\cap B|} = \\sum_{X \\subset A \\cap B} 1 = \\sum_{\\substack{X \\subset A \\\\ X \\subset B}} 1.\n$$\nThus, the left-hand side of the desired inequality can be written as\n$$\n\\sum_{A \\in P(S)} \\sum_{B \\in P(S)} \\sum_{\\substack{X \\subset A \\\\ X \\subset B}} f(A)f(B).\n$$\nChanging the order of summation, we find\n$$\n\\sum_{X \\in P(S)} \\sum_{A \\supset X} \\sum_{B \\supset X} f(A)f(B) = \\sum_{X \\in P(S)} \\left( \\sum_{A \\supset X} f(A) \\right)^2 \\geq 0,\n$$\nwhere the last inequality holds since the sum of squares of real numbers is always nonnegative. Done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15498,
    "subject": "Mathematics (Olympiad)",
    "question": "In $\\triangle PBC$, $\\angle PBC = 60^\\circ$. The tangent at point $P$ to the circumcircle $w$ of $\\triangle PBC$ intersects line $CB$ at $A$. Points $D$ and $E$ lie on the line segment $PA$ and circle $w$ respectively, such that $\\angle DBE = 90^\\circ$ and $PD = PE$. $BE$ and $PC$ meet at $F$. It is given that lines $AF$, $BP$, and $CD$ are concurrent.\n\n1. Prove that $BF$ is the bisector of $\\angle PBC$.\n2. Find the value of $\\tan \\angle PCB$.",
    "options": [],
    "answer": "See solution",
    "solution": "1. When $BF$ bisects $\\angle PBC$, since $\\angle DBE = 90^\\circ$, we know that $BD$ is the bisector of $\\angle PBA$.\n\nBy the angle bisector theorem, we have\n\n$$\n\\frac{PF}{FC} \\cdot \\frac{CB}{BA} \\cdot \\frac{AD}{DP} = \\frac{PB}{BC} \\cdot \\frac{BC}{BA} \\cdot \\frac{AB}{PB} = 1.\n$$\n\nBy the converse of Ceva's theorem, the lines $AF$, $BP$, and $CD$ are concurrent.\n\nSuppose there exists $\\angle D'BF'$ satisfying the conditions: (a) $\\angle D'BF' = 90^\\circ$, (b) the lines $AF'$, $BP$, and $CD'$ are concurrent. We may assume that $F'$ lies on $PF$. Then, $D'$ is on $AD$.\n\nSo\n\n$$\n\\frac{PF'}{F'C} < \\frac{PF}{FC}, \\quad \\frac{AD'}{PD'} < \\frac{AD}{PD}.\n$$\n\nThus\n\n$$\n\\frac{PF'}{F'C} \\cdot \\frac{CB}{BA} \\cdot \\frac{AD'}{D'P} < \\frac{PB}{BC} \\cdot \\frac{BC}{BA} \\cdot \\frac{AB}{PB} = 1,\n$$\n\nwhich leads to a contradiction. This completes the proof.\n\n2. We may assume that the circle $O$ has radius $1$. Let $\\angle PCB = \\alpha$. By (1), $\\angle PBE = \\angle EBC = 30^\\circ$. Therefore, $E$ is the midpoint of $\\overline{PC}$.\n\nSince $\\angle MPE = \\angle PBE = 30^\\circ$, $\\angle CPE = \\angle CBE = 30^\\circ$ and $PD = PE$, we obtain $\\angle PDE = \\angle PED = 15^\\circ$, $PE = 2 \\cdot 1 \\cdot \\sin 30^\\circ$ and $DE = 2\\cos 15^\\circ$.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p153_data_8cfc42c981.png)\n\nSince\n\n$$\nBE = 2\\sin(\\alpha + 30^\\circ)\n$$\n\nand $\\angle BED = \\angle BEP - 15^\\circ$, we have\n\n$$\n\\cos(\\alpha - 15^\\circ) = \\frac{BE}{DE} = \\frac{2\\sin(\\alpha + 30^\\circ)}{2\\cos 15^\\circ},\n$$\n\n$$\n\\cos(\\alpha - 15^\\circ)\\cos 15^\\circ = \\sin(\\alpha + 30^\\circ),\n$$\n\n$$\n\\cos \\alpha + \\cos(\\alpha - 30^\\circ) = 2\\sin(\\alpha + 30^\\circ),\n$$\n\n$$\n\\cos \\alpha + \\cos \\alpha \\cos 30^\\circ + \\sin \\alpha \\sin 30^\\circ = \\sqrt{3} \\sin \\alpha + \\cos \\alpha,\n$$\n\n$$\n1 + \\frac{\\sqrt{3}}{2} + \\frac{1}{2} \\tan \\alpha = \\sqrt{3} \\tan \\alpha + 1.\n$$\n\nSo\n\n$$\n\\tan \\alpha = \\frac{6 + \\sqrt{3}}{11}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15499,
    "subject": "Mathematics (Olympiad)",
    "question": "Tenemos 50 fichas numeradas del 1 al 50, y hay que colorearlas de rojo o azul. Sabemos que la ficha 5 es de color azul. Para la coloración del resto de fichas se siguen las siguientes reglas:\n\n- Si la ficha con el número $x$ y la ficha con el número $y$ son de distinto color, entonces la ficha con el número $|x - y|$ se pinta de color rojo.\n\n- Si la ficha con el número $x$ y la ficha con el número $y$ son de distinto color y $x \\cdot y$ es un número entre 1 y 50 (incluyendo ambos), entonces la ficha con el número $x \\cdot y$ se pinta de color azul.\n\nDeterminar cuántas coloraciones distintas se pueden realizar en el conjunto de fichas.",
    "options": [],
    "answer": "See solution",
    "solution": "Observemos que dos números que se diferencian en 5 tienen el mismo color. En efecto, si fueran de distinto color, su diferencia debería ser de color rojo, por la regla a). Pero su diferencia es 5, que es de color azul. Por tanto, basta con saber el color de los 4 primeros números. Distinguimos dos casos:\n\n**Caso 1:** Si 1 es de color azul, el resto de fichas deberá ser de color azul, por la regla b). Esto es así porque si la ficha $k \\neq 1$ fuera roja, entonces por b), $k = k \\cdot 1$ tendría que ser azul, lo que contradice que $k$ sea roja.\n\n**Caso 2:** Si 1 es roja, por la regla a), $4 = 5 - 1$ es roja. Para determinar el color de 2 y 3, supongamos que 3 es azul. Por ser $2 = 3 - 1$ y ser 3 y 1 de diferente color, entonces 2 es roja. Ahora bien, $3 = 5 - 2$ y 5 es azul y 2 roja, por lo tanto 3 es roja. Esto no puede ser, por lo tanto 3 no puede ser azul y es roja, por lo que 2 también es roja. Así pues, 1, 2, 3 y 4 son rojas, lo mismo que el resto de fichas que no son múltiplo de 5.\n\nPor tanto, solo hay dos coloraciones posibles: o todas las fichas de color azul, o todas rojas excepto los múltiplos de 5, que son azules.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15500,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute triangle, and let $X$ be a variable interior point on the minor arc $BC$ of its circumcircle. Let $P$ and $Q$ be the feet of the perpendiculars from $X$ to lines $CA$ and $CB$, respectively. Let $R$ be the intersection of line $PQ$ and the perpendicular from $B$ to $AC$. Let $\\ell$ be the line through $P$ parallel to $XR$. Prove that as $X$ varies along minor arc $BC$, the line $\\ell$ always passes through a fixed point. Specifically, prove that there is a point $F$, determined by triangle $ABC$, such that no matter where $X$ is on arc $BC$, line $\\ell$ passes through $F$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H$ denote the orthocenter of $\\triangle ABC$. We claim that $\\ell$ always passes through $H$.\n\n**Lemma.** Line $PQ$ bisects segment $XH$.\n\n*Proof.* Let $X_A, X_B$ be the reflections of $X$ across $BC$ and $AC$ respectively, and let $H_A$ be the reflection of $H$ across $BC$. It is easy to see that since $\\angle BH_A C = \\angle BHC = 180^\\circ - \\angle BAC$, $H_A$ is on the circumcircle of $\\triangle ABC$. It suffices to show that $H$ is on $X_A X_B$. Since $C$ is the circumcenter of $XX_A X_B$, we have $\\angle XX_A X_B = \\frac{1}{2} \\angle XCX_B = \\angle ACX$. On the other hand, $HH_A X_A X$ is an isosceles trapezoid, so $\\angle HX_A X = \\angle HH_A X = \\angle AH_A X = \\angle ACX = \\angle XX_A X_B$ so it follows that $X_B, H, X_A$ are collinear.\n\n![](images/USA_IMO_2013-2014_p65_data_ae722ae9d7.png)\n\nWe know that lines $HBR$ and $XP$ are both perpendicular to $AC$, so it follows that $HR \\parallel XP$. But by the lemma, line $PQR$ bisects $HX$ so it follows that $PXRH$ is a parallelogram. Thus, $PH \\parallel XR$ and thus $H$ is on $\\ell$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15501,
    "subject": "Mathematics (Olympiad)",
    "question": "A finite set $S$ of positive integers is called *cardinal* if $S$ contains the integer $|S|$, where $|S|$ denotes the number of distinct elements in $S$. Let $f$ be a function from the set of positive integers to itself, such that for any cardinal set $S$, the set $f(S)$ is also cardinal. Here $f(S)$ denotes the set of all integers that can be expressed as $f(a)$ for some $a$ in $S$.\n\nFind all possible values of $f(2024)$.\n\n*Note:* As an example, $\\{1,3,5\\}$ is a cardinal set because it has exactly 3 distinct elements, and the set contains 3.",
    "options": [],
    "answer": "See solution",
    "solution": "The possible values are $1$, $2$, and $2024$.\n\n**Construction.** The function $f(x) = 1$ for all $x \\in \\mathbb{N}$ works. Also, $f(x) = 1$ for all $x \\neq 2024$ and $f(2024) = 2$ works. Finally, $f(x) = x$ for all $x \\in \\mathbb{N}$ works as well.\n\nIt remains to show these are the only possible values for $f(2024)$.\n\n**Proof.** Denote $\\operatorname{Im}(f) = \\{f(x) \\mid x \\in \\mathbb{N}\\}$. The cardinal set $\\{1\\}$ gives $f(1) = 1$. Consider the following two cases:\n\n- $\\operatorname{Im}(f)$ is unbounded. Fix any $n \\in \\mathbb{N}$, with $n > 1$. Pick $n-1$ distinct integers $k_1, \\dots, k_{n-1}$ such that $f(k_i) \\notin \\{n, f(n)\\}$ and $f(k_i)$ are all pairwise distinct, for $1 \\le i < n$. Then $\\{n, k_1, \\dots, k_{n-1}\\}$ is a cardinal set. Then $\\{f(n), f(k_1), \\dots, f(k_{n-1})\\}$ is a cardinal set with $n$ distinct elements, so $n$ lies in this set, hence $f(n) = n$. This gives the identity function.\n\n- $\\operatorname{Im}(f)$ is bounded. Suppose $f(x) \\le M$ for all $x \\in \\mathbb{N}$ and some integer $M > 0$.\n\n  **Claim.** For any integer $a$ satisfying $1 \\le a \\le M$, if there are infinitely many integers $n \\in \\mathbb{N}$ such that $f(n) = a$, then $a = 1$.\n\n  *Proof.* Let $b > 1$ be one of the integers with $f(b) = a$. Consider $b-1$ other integers $c_1, \\dots, c_{b-1}$, such that $f(c_i) = a$ for $1 \\le i < b$, and $c_i$ are all pairwise distinct. Then $\\{b, c_1, \\dots, c_{b-1}\\}$ is a cardinal set, so the image set, which consists of the singleton $\\{a\\}$, is cardinal, hence $a = 1$. $\\square$\n\nSo for every $2 \\le m \\le M$, there are only finitely many integers $x$ such that $f(x) = m$. Thus, there exists an integer $N > 1$ such that for all $n \\ge N$, $f(n) = 1$. Now for every $1 < l < N$, consider the cardinal set $\\{l, N+1, N+2, \\dots, N+l-1\\}$. Then the image set consists of $\\{1, f(l)\\}$, which can be cardinal only when $f(l) = 1$ or $f(l) = 2$.\n\nBy the above reasoning, $f(2024)$ can only be $1$, $2$, or $2024$, each of which occurs as an example. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15502,
    "subject": "Mathematics (Olympiad)",
    "question": "給三個正整數 $K$, $B$, $W$，其中 $B > W > 1$。有兩堆球：一堆有黑球 $B$ 個，另一堆有白球 $W$ 個。依照以下方法來分割：\n\n操作：所有堆依照 *非遞增* 的方式排列，若黑球堆與白球堆數量一樣則白球堆置於前。現在選出前面 $K$ 堆，若全部少於 $K$ 堆則選擇全部。然後將所選擇的每一堆分成兩堆，而且這兩堆的球數最多只差一個。\n\n（例如：令 $K = 4$。現在有四堆黑球分別是 $5, 4, 4, 2$ 個，三堆白球分別是 $8, 4, 2$ 個。所以依序排列是 $(w8, b5, w4, b4, w2, b2)$，其中 $w8$ 指白球一堆 $8$ 個，$b5$ 指黑球一堆 $5$ 個。對於前四堆做以下分割：$(4, 4), (3, 2), (2, 2), (2, 2)$，所以下一階段的新的排列是 $(w4, w4, b4, b3, w2, w2, w2, b2, b2, b2, b2)$。）\n\n不斷「排列-分割」，直到某一次操作結束時有某一個白球自成一堆。試證明：\n\n此時一定有一堆至少有兩個黑球。",
    "options": [],
    "answer": "See solution",
    "solution": "起始狀態是 $A_0 = (bB, wW)$，而 $A_{i-1} \\to A_i$ 代表進行第 $i$ 次的「排列-分割」。\n\n依照操作模式，直到所有的堆都是一個球。我們標示以下幾個重要的階段：\n\n$A_s$：第一次出現有一堆一個球時（無論是黑球或白球）。\n\n$A_t$：總堆數大於 $K$ 時；然而有可能：當所有的堆都是一個球時還是不超過 $K$ 堆，則令 $t = \\infty$。\n\n$A_f$：當所有的黑球都是一球一堆時。\n\n我們只需要證明：題意所要求的停止時刻是在 $A_{f-1}$ 或更早之前，也就是 $A_{f-1}$ 中一定有至少一堆一個白球。\n\n明顯地，$s \\leq f$。又在 $A_{f-1}$ 中的黑球一定有某些堆是 $b2$（排在前 $K$ 個中），其他都是 $b1$。\n\n當 $i < \\min\\{t, s\\}$ 時，$A_i$ 的每一堆都被選擇，也都要一分為二（因為沒有一球一堆）。假設此時黑球堆的最多與最少球數分別是 $M_i$ 與 $m_i$，白球堆的最多與最少球數分別是 $N_i$ 與 $n_i$。對於 $A_i$，我們有以下性質：\n\n1. 黑球總堆數與白球總堆數都是 $2^i$。\n2. $M_i \\geq N_i$。\n3. $m_i \\geq n_i$。\n\n這些都可以簡單地以歸納法證明。\n\n關於 $s, t, f$ 的大小關係，有以下兩種可能：\n\n**Case 1.** $s \\leq t$ 或 $f \\leq t+1$（特別是 $t = \\infty$ 屬於這個情形）。假設發生 $s = f$。由 $s$ 的定義 $m_i, n_i \\geq 2$。由上式得 $M_{s-1} \\leq 2$。所以每一堆都是兩個球。但是這與 $B > W$ 及性質 (1) 矛盾！所以一定是 $s \\leq f - 1$。\n\n所給的條件 $[s \\leq t$ 或 $f \\leq t+1]$ 合而為一：$s \\leq f - 1 \\leq t$。考慮 $A_{s-1} \\to A_s$。由於 $A_{s-1}$ 中每一堆都被選擇到而且 $m_{s-1} \\geq n_{s-1}$，所以切割成 $A_s$ 時 $m_{s-1} \\to (m_{s-1} - 1, 1)$，即 $m_s = 1$；同時 $n_{s-1} \\to (n_{s-1} - 1, 1)$，即 $n_s = 1$；也就是黑球與白球的一球一堆是同時在 $A_s$ 時第一次出現。由於 $s \\leq f - 1$，所以題意所要求的停止時刻是在 $A_{f-1}$ 或更早之前。\n\n**Case 2.** $t+1 \\leq s$ 且 $t+2 \\leq f$。在 $A_{t-1}$ 時的總堆是 $2^t$，所以 $2^t \\leq K < 2^{t+1}$ 且 $A_t$ 時的總堆是 $2^{t+1}$（黑白都是 $2^t$）。進行第 $t+1$ 步驟：$A_t \\to A_{t+1}$ 時，被選擇的 $K$ 堆中最多有 $2^t$ 個是黑球堆，所以 $A_{t+1}$ 中至少有 $2^t + (K - 2^t) = K$ 個白球堆。因為白堆的總數是非遞減的，$A_{f-1}$ 白球堆數至少是 $K$ 堆。\n\n最後，在 $A_{f-1}$ 中 $M_{f-1} = 2$ 且這些 $b2$（至少一堆）在第 $f$ 步驟時都被分割成 $(1, 1)$，所以 $A_{f-1}$ 中最多有 $K-1$ 個白球堆被選擇，也就有至少一堆白球堆 $\\Omega$ 沒有被選擇到。這個 $\\Omega$ 堆會只有一個白球，因為 $\\Omega$ 堆若多於一個白球，則 $\\Omega$ 應該排在 $b2$ 前面，那它應該被選擇到，這矛盾！因此 $A_{f-1}$ 中一定有至少一堆一個白球，證明完畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15503,
    "subject": "Mathematics (Olympiad)",
    "question": "Point $I$ is the incenter of triangle $ABC$, where $AB < AC$. On the external angle bisector of angle $ABC$, a point $X$ is chosen such that $IC = IX$. Let the tangent to the circumcircle of triangle $BXC$ at point $X$ intersect line $AB$ at point $Y$. Prove that $AC = AY$.\n\n![](images/Ukraine2022-23_p17_data_0f33189158.png)",
    "options": [],
    "answer": "See solution",
    "solution": "From the fact that $XY$ is tangent to the circumcircle of $\\triangle BXC$, we have $\\angle BXY = \\angle BCX$. Combining this with the fact that $\\angle YBX = \\angle XBC$, we have the similarity $\\triangle BXY \\sim \\triangle BCX$. Therefore, we have $\\frac{BX}{BC} = \\frac{BY}{BX}$, so $BY = \\frac{BX^2}{BC}$.\n\nNotice that $\\angle IBX = 90^\\circ$, so\n\n$$\nBY = \\frac{BX^2}{BC} = \\frac{IX^2 - BI^2}{BC} = \\frac{CI^2 - BI^2}{BC}.\n$$\n\nLet $K$ be the point of tangency of the incircle of $\\triangle ABC$ with side $BC$. Then\n\n$$\nBY = \\frac{CI^2 - BI^2}{BC} = \\frac{(IK^2 + CK^2) - (IK^2 + BK^2)}{BC} = \\frac{CK^2 - BK^2}{BC} = CK - BK.\n$$\n\nNow let $a, b, c$ be the lengths of the sides of $\\triangle ABC$, and let $p$ be its semiperimeter. Then we have\n\n$$\nAY = AB + BY = AB + CK - BK = c + (p - c) - (p - b) = b = AC,\n$$\n\nas desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15504,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $x, y, z > 0$ such that\n\n$$\nxyz \\le \\min \\left\\{ 4 \\left(x - \\frac{1}{y}\\right),\\ 4 \\left(y - \\frac{1}{z}\\right),\\ 4 \\left(z - \\frac{1}{x}\\right) \\right\\}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "From the given condition, we have $xyz \\le 4 \\left(x - \\frac{1}{y}\\right)$, which is equivalent to $4x \\ge xyz + \\frac{4}{y}$. By AM-GM, $4x \\ge xyz + \\frac{4}{y} \\ge 2\\sqrt{xyz \\cdot \\frac{4}{y}} = 4\\sqrt{xz}$, so $x \\ge z$.\n\nAnalogously, from $xyz \\le 4 \\left(y - \\frac{1}{z}\\right)$ and $xyz \\le 4 \\left(z - \\frac{1}{x}\\right)$, we get $y \\ge x$ and $z \\ge y$, so necessarily $x = y = z$. Now the requirement holds if and only if $x^3 \\le 4 \\left(x - \\frac{1}{x}\\right)$, i.e. $(x^2 - 2)^2 \\le 0$, which leads to $x = \\sqrt{2}$. Thus, $x = y = z = \\sqrt{2}$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15505,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ be a positive integer and $A$ be a finite set. Let $A_1, A_2, \\dots, A_m$ be subsets of $A$ (not necessarily distinct). It is known that for any nonempty set $I \\subseteq \\{1, 2, \\dots, m\\}$,\n$$\n\\left| \\bigcup_{i \\in I} A_i \\right| \\geq |I| + 1.\n$$\nProve that the elements of $A$ can be coloured black or white, such that each of $A_1, A_2, \\dots, A_m$ contains both black and white elements.",
    "options": [],
    "answer": "See solution",
    "solution": "Construct a bipartite graph $G$ whose two parts are $X = \\{A_1, A_2, \\dots, A_m\\}$ and $Y = A$: for $1 \\leq i \\leq m$ and $a \\in A$, $A_i \\in X$ and $a \\in Y$ are adjacent if and only if $a \\in A_i$. Since for each nonempty $I \\subseteq \\{1, 2, \\dots, m\\}$, $|\\bigcup_{i \\in I} A_i| \\geq |I| + 1$, it follows that for any $k$ vertices of $X$, the number of vertices of $Y$ that are neighbouring to one or more of them is at least $k+1$. According to Hall's theorem, there exists a transversal $f$ from $X$ to $Y$. For $1 \\leq i \\leq m$, let $a_i = f(A_i)$. Clearly, $a_i \\in A_i$, and $a_1, a_2, \\dots, a_m$ are distinct elements of $A$. Colour all the other elements $A \\setminus \\{a_1, \\dots, a_m\\}$ white, and determine the colours of $a_1, a_2, \\dots, a_m$ as follows.\n\nEach time, choose an $i$ such that $a_i$ is uncoloured and $A_i$ has a coloured neighbour, say $b$. Colour $a_i$ the opposite colour to $b$. Suppose, during the process, some elements $a_1, a_2, \\dots, a_k$ ($1 \\leq k \\leq m$) are uncoloured, but we cannot find another $i$ and colour $a_i$. Based on the algorithm, this means that all the neighbours of $A_1, A_2, \\dots, A_k$ are in $\\{a_1, \\dots, a_k\\}$, yet this contradicts $|A_1 \\cup A_2 \\cup \\dots \\cup A_k| \\geq k + 1$. Hence, this process can continue until all of $a_1, a_2, \\dots, a_m$ are coloured. Moreover, for $1 \\leq i \\leq m$, when $a_i$ is coloured, $A_i$ is guaranteed to have black and white neighbours, that is, $A_i$ contains both black and white elements. This colouring satisfies the problem requirements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15506,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $c_n$ be the number of codes that have exactly $n$ digits.\n\nA code ends with $0$ or $1$.\n\nSuppose $n \\geq 5$. If a code ends with $1$, then the string that remains when the end digit is removed is also a code. So the number of codes that end with $1$ and have exactly $n$ digits equals $c_{n-1}$.\n\nIf a code with $n$ digits ends in $0$, then the string that remains when the end digit is removed is a code with $n-1$ digits that does not end with two $0$s. If a code with $n-1$ digits ends with two $0$s, then it ends with $100$. If the $100$ is removed then the string that remains is an unrestricted code that has exactly $n-4$ digits. So the number of codes with $n-1$ digits that do not end with two $0$s is $c_{n-1} - c_{n-4}$.\n\nHence, for $n \\geq 5$, $$c_n = 2c_{n-1} - c_{n-4}.$$ \n\nBy direct counting, $c_1 = 2$, $c_2 = 4$, $c_3 = 7$, $c_4 = 13$. The table shows $c_n$ for $1 \\leq n \\leq 11$.\n\n| $n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |\n|-----|---|---|---|---|---|---|---|---|---|----|----|\n| $c_n$ | 2 | 4 | 7 | 13 | 24 | 44 | 81 | 149 | 274 | 504 | 927 |\n\nWhat is the number of codes that have exactly $11$ digits?",
    "options": [],
    "answer": "See solution",
    "solution": "From the table, the number of codes that have exactly $11$ digits is $927$.\n\n*Comment.* The equation $c_n = 2c_{n-1} - c_{n-4}$ can also be derived from the equation $c_n = c_{n-1} + c_{n-2} + c_{n-3}$.\n\nFor $n \\geq 5$ we have $c_{n-1} = c_{n-2} + c_{n-3} + c_{n-4}$.\n\nHence $$c_n = c_{n-1} + (c_{n-1} - c_{n-4}) = 2c_{n-1} - c_{n-4}.$$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15507,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\ge 2$, define $M_0(x_0, y_0)$ to be an intersection point of the parabola $y^2 = nx - 1$ and the line $y = x$. Prove that for any positive integer $m$, there exists an integer $k \\ge 2$ such that $(x_0^m, y_0^m)$ is an intersection point of $y^2 = kx - 1$ and $y = x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $M_0(x_0, y_0)$ is an intersection point of $y^2 = nx - 1$ and $y = x$, we have $x_0 = y_0 = \\dfrac{n \\pm \\sqrt{n^2 - 4}}{2}$. Then, clearly,\n\n$$\nx_0 + \\frac{1}{x_0} = n.\n$$\n\nLet $(x_0^m, y_0^m)$ be an intersection point of $y^2 = kx - 1$ and $y = x$. Then,\n\n$$\nk = x_0^m + \\frac{1}{x_0^m}.\n$$\n\nLet $k_m = x_0^m + \\frac{1}{x_0^m}$. Then,\n\n$$\nk_{m+1} = k_m \\left(x_0 + \\frac{1}{x_0}\\right) - k_{m-1} = n k_m - k_{m-1} \\quad (m \\ge 2).\n$$\n\nSince $k_1 = n$ is an integer,\n\n$$\nk_2 = x_0^2 + \\frac{1}{x_0^2} = \\left(x_0 + \\frac{1}{x_0}\\right)^2 - 2 = n^2 - 2\n$$\n\nis also an integer. By induction and the recurrence above, for any positive integer $m$, $k_m = x_0^m + \\frac{1}{x_0^m}$ is an integer. Let $k = x_0^m + \\frac{1}{x_0^m}$. Thus, $(x_0^m, y_0^m)$ is an intersection point of $y^2 = kx - 1$ and $y = x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15508,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(x)$ and $Q(x)$ be polynomials such that\n\n$$\nP(Q(x)) = P(x)Q(x) - P(x)\n$$\n\nfor all $x$. Find all such pairs of polynomials $P(x)$ and $Q(x)$.",
    "options": [],
    "answer": "See solution",
    "solution": "$P(x) = a x^2 - a(b+1)x + ab$, $Q(x) = x^2 - (b+1)x + 2b$, where $a, b \\in \\mathbb{R}$ and $a \\ne 0$.\n\nLet the degrees of $P$ and $Q$ be $m$ and $n$. Comparing degrees in\n$$\nP(Q(x)) = P(x)Q(x) - P(x),\n$$\nwe get $mn = m + n \\implies (m-1)(n-1) = 1 \\implies m = n = 2$. Thus, $P(x)$ and $Q(x)$ are quadratic.\n\nLet $P(x) = \\alpha x^2 + \\beta x + c$. Substituting into the equation and comparing coefficients, we find $P(x) + c = a Q(x)$ for some $a \\ne 0$. Thus, $P(x) = a Q(x) - ab$ for some $b \\in \\mathbb{R}$, and $Q(x) = x^2 - (b+1)x + 2b$.\n\nIt can be verified that these forms of $P$ and $Q$ satisfy the original equation.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15509,
    "subject": "Mathematics (Olympiad)",
    "question": "A positive integer is called powerful if all exponents in its prime factorization are $\\geq 2$.\n\nProve that there are infinitely many pairs of powerful consecutive positive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "The numbers $8 = 2^3$ and $9 = 3^2$ form a pair of consecutive powerful numbers.\n\nWe now show that for each pair $(k, k+1)$ of powerful positive integers, we can find a new pair, namely $(4k(k+1),\\ (2k+1)^2)$. Clearly, $4k(k+1) + 1 = (2k+1)^2$.\n\nSince $k$ and $k+1$ are powerful, the product $4k(k+1) = 2^2 k(k+1)$ is also powerful. A square is always powerful, so $(2k+1)^2$ is powerful as well. Finally, $4k(k+1) > k$ for positive integers $k$. Therefore, there are infinitely many pairs of powerful consecutive positive integers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15510,
    "subject": "Mathematics (Olympiad)",
    "question": "Нийт боломжийн тоог $a_n$ гэе. $4 \\times n$ тэгш өнцөгтийн баганын дээд талд нь эсвэл доод талд $1 \\times 2$ тэгш өнцөгт байрлуулах боломжийн тоог ол.",
    "options": [],
    "answer": "See solution",
    "solution": "Өнцөгт босоогоор байрлах боломжийн тоог $b_n$ гэе. Мөн $4 \\times n$ тэгш өнцөгтийн сүүлийн баганд $1 \\times 2$ тэгш өнцөгт яг голд нь байрлах нийт боломжийн тоог $c_n$ гэе. Сүүлийн баганд нь $1 \\times 2$ тэгш өнцөгт 2 ширхэг босоо байрласан байвал үлдэх $4(n-1)$ тэгш өнцөгтийг $a_{n-1}$ янзаар бөглөж болно. Харин сүүлийн баганд дээр нь эсвэл доор $1 \\times 2$ тэгш өнцөгт байрласан байвал нийт бөглөх боломжийн тоо нь $2 \\cdot b_{n-1}$ болно. Хэрэв $4 \\times n$ тэгш өнцөгтийн сүүлийн баганд $1 \\times 2$ тэгш өнцөгт яг голд нь байрласан бол бид үлдэх хэсэгт $c_{n-1}$ янзаар бөглөж чадна. Хэрэв 4 ширхэг хэвтээ $1 \\times 2$ сүүлийн 2 баганд байвал үлдэх хэсгийг $a_{n-2}$ янзаар бөглөнө. Иймд\n\n$$a_n = a_{n-1} + 2b_{n-1} + c_{n-1} + a_{n-2}, \\quad b_n = b_{n-1} + a_{n-1}, \\quad c_n = c_{n-2} + a_{n-1}$$\n\nболохыг хялбархан харж чадна. Үүнээс:\n\n$$a_n = a_{n-1} + 5a_{n-2} + a_{n-3} - a_{n-4}$$\n\nТэгшитгэлийн язгуур нь:\n\n$$x^4 - x^3 - 5x^2 - x + 1 = 0$$\n\n$$x_1 = \\frac{1}{4}(1 - \\sqrt{29} - \\sqrt{14 - 2\\sqrt{29}})$$\n\n$$x_2 = \\frac{1}{4}(1 + \\sqrt{29} + \\sqrt{14 - 2\\sqrt{29}})$$\n\n$$x_3 = \\frac{1}{4}(1 - \\sqrt{29} + \\sqrt{14 + 2\\sqrt{29}})$$\n\n$$x_4 = \\frac{1}{4}(1 + \\sqrt{29} + \\sqrt{14 + 2\\sqrt{29}})$$\n\nТэгвэл:\n\n$$a_n = t_1 x_1^n + t_2 x_2^n + t_3 x_3^n + t_4 x_4^n$$\n\nэсвэл\n\n$$a_n = \\frac{1}{\\sqrt{29}} (-x_1^{n+1} - x_2^{n+1} + x_3^{n+1} + x_4^{n+1})$$\n\nгэж гарна.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15511,
    "subject": "Mathematics (Olympiad)",
    "question": "Rotate $\\triangle PBC$ about point $B$ by $60^\\circ$ in the anticlockwise direction, so that $BA$ is the image of $BC$ after rotation. Let $Q$ be the image of $P$ after rotation. Then\n\n$$\n[ABP] + [BCP] = [PAQB] = [PQB] + [QPA].\n$$\n\n![](images/1.2_1997-2023_IMO_HK_TST_Solutions_p204_data_51114dfb22.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Now, $\\triangle PQB$ is an equilateral triangle with side length $8$, and so its area is\n$$\n\\frac{\\sqrt{3}}{4} \\times 8^2 = 16\\sqrt{3}.\n$$\nAlso, we have\n$$\n[QPA] = \\frac{1}{2} \\times 8 \\times 15 \\times \\sin \\angle QPA \\leq \\frac{1}{2} \\times 8 \\times 15 = 60.\n$$\nThus, the sum of areas is at most $60 + 16\\sqrt{3}$. Equality holds when $\\angle QPA = 90^\\circ$, i.e. $\\angle BPA = 150^\\circ$ (which fixes $\\triangle ABC$ and point $P$).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15512,
    "subject": "Mathematics (Olympiad)",
    "question": "1. (1) Can one divide the set $\\{1, 2, \\ldots, 96\\}$ into 32 subsets, each containing three elements, such that the sums of the three elements in each subset are all equal?\n\n(2) Can one divide the set $\\{1, 2, \\ldots, 99\\}$ into 33 subsets, each containing three elements, such that the sums of the three elements in each subset are all equal?",
    "options": [],
    "answer": "See solution",
    "solution": "(1) No. Since\n\n$$\n1 + 2 + \\cdots + 96 = \\frac{96 \\times 97}{2} = 48 \\times 97,\n$$\n\nand $32$ does not divide $48 \\times 97$.\n\n(2) Yes. The sum of the three elements in each subset is\n\n$$\n\\frac{1 + 2 + \\cdots + 99}{33} = \\frac{99 \\times 100}{33 \\times 2} = 150.\n$$\n\nWe can divide $1, 2, \\ldots, 66$ into 33 pairs, such that the sums of these pairs form an arithmetic sequence:\n\n$1 + 50$, $3 + 49$, $\\ldots$, $33 + 34$, $2 + 66$, $4 + 65$, $\\ldots$, $32 + 51$.\n\nHence, the following decomposition satisfies the requirement:\n\n$\\{1, 50, 99\\}$, $\\{3, 49, 98\\}$, $\\ldots$, $\\{33, 34, 83\\}$, $\\{2, 66, 82\\}$, $\\{4, 65, 81\\}$, $\\ldots$, $\\{32, 51, 67\\}$.\n\n**Remark**\n\nThe general case is as follows:\n\nLet the family of subsets $A_i = \\{x_i, y_i, z_i\\}$ of the set $M = \\{1, 2, 3, \\ldots, 3n\\}$, $i = 1, 2, \\ldots, n$, satisfy $A_1 \\cup A_2 \\cup \\cdots \\cup A_n = M$. Write $s_i = x_i + y_i + z_i$, and find all possible values of $n$ such that the $s_i$ are all equal.\n\nFirst, $n$ must divide $1 + 2 + 3 + \\cdots + 3n$, i.e.\n\n$$\nn \\mid \\frac{3n(3n+1)}{2} \\implies 2 \\mid 3n+1.\n$$\n\nHence, $n$ must be odd.\n\nFor $n$ odd, $1, 2, \\ldots, 2n$ can form $n$ pairs such that the sums of each pair become an arithmetic sequence of common difference 1:\n\n$$\n\\begin{aligned}\n& 1 + \\left(n + \\frac{n+1}{2}\\right),\\ 3 + \\left(n + \\frac{n-1}{2}\\right),\\ \\ldots,\\ n + (n+1); \\\\\n& 2 + 2n,\\ 4 + (2n-1),\\ \\ldots,\\ (n-1) + \\left(n + \\frac{n+3}{2}\\right).\n\\end{aligned}\n$$\n\nIts general term is\n\n$$\na_k = \\begin{cases} 2k-1 + \\left(n + \\frac{n+1}{2} + 1 - k\\right), & 1 \\leq k \\leq \\frac{n+1}{2}, \\\\ [1-n+2(k-1)] + [2n + \\frac{n+1}{2} - (k-1)], & \\frac{n+3}{2} \\leq k \\leq n. \\end{cases}\n$$\n\nIt is easy to see that $a_k + 3n + 1 - k = \\frac{9n+3}{2}$ is a constant, so all the following $n$ triples have the same sum:\n\n$$\n\\begin{aligned}\n& \\{1, n + \\frac{n+1}{2}, 3n\\},\\ \\{3, n + \\frac{n-1}{2}, 3n-1\\},\\ \\ldots, \\\\\n& \\{n, n+1, 3n+1 - \\frac{n+1}{2}\\}; \\\\\n& \\{2, 2n, 3n+1 - \\frac{n+3}{2}\\},\\ \\ldots,\\ \\{n-1, n + \\frac{n+3}{2}, 2n+1\\}.\n\\end{aligned}\n$$\n\nFor $n$ odd, take these triples as $A_1, A_2, \\ldots, A_n$; then they satisfy the required condition. So $n$ can be any odd number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15513,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with median $AK$. Let $O$ be the circumcenter of the triangle $ABK$.\n\n(a) Prove that if $O$ lies on a midline of the triangle $ABC$, but does not coincide with its endpoints, then $ABC$ is a right triangle.\n\n(b) Is the statement still true if $O$ can coincide with an endpoint of the midsegment?",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $L$ and $M$ be the midpoints of the sides $CA$ and $AB$, respectively. If $O$ lies on the segment $KM$, then the segment $KM$ and the perpendicular bisector of $AB$ have two different common points $O$ and $M$, hence $KM$ is the perpendicular bisector of $AB$. Since $KM$ is parallel to $AC$ and is perpendicular to $AB$, the angle at vertex $A$ must be right.\n\nIf $O$ lies on the segment $LM$, then similarly, the angle at vertex $B$ must be right.\n\nIf $O$ lies on the segment $KL$, then on one hand $\\angle ABC$ is acute, because $OK$ and $MB$ are perpendicular to $MO$, the perpendicular bisector of $AB$, and $|OK| < |LK| = |MB|$. On the other hand, $\\angle ABK$ must be obtuse, since the circumcenter $O$ of the triangle $ABK$ lies outside of the triangle, a contradiction. Thus this case is not possible.\n\nb) If the triangle $ABC$ is equilateral, then the median $AK$ is also the altitude and $ABK$ is a right triangle with hypotenuse $AB$. The circumcenter $O$ of the last triangle is the midpoint of $AB$, i.e., an endpoint of a midsegment of the triangle, but $ABC$ is not a right triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15514,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimum value of\n$$\n\\frac{|x| + |x + 4y| + |y + 7z| + 2|z|}{|x + y + z|}\n$$\nfor real numbers $x$, $y$, and $z$ with $x + y + z \\neq 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "By the triangle inequality, we have\n\n$$\n\\begin{aligned}\n\\frac{|x| + |x + 4y| + |y + 7z| + 2|z|}{|x + y + z|} &\\ge \\frac{|x| + \\frac{4}{11}|x + 4y| + \\frac{1}{11}|-y-7z| + 2|z|}{|x + y + z|} \\\\\n&\\ge \\frac{|x + \\frac{4}{11}(x + 4y) + \\frac{1}{11}(-y - 7z) + 2z|}{|x + y + z|} \\\\\n&= \\frac{15}{11}.\n\\end{aligned}\n$$\n\nEquality holds when $(x, y, z) = (28k, -7k, k)$ for some $k \\neq 0$. So the minimum value is $\\frac{15}{11}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15515,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\Sigma = \\{1,2,3,\\ldots,n\\}$. We want to partition the set $\\Sigma$ into three mutually disjoint subsets $A$, $B$, and $\\Gamma$ such that $A \\cup B \\cup \\Gamma = \\Sigma$ and the sums of their elements, $S_A$, $S_B$, and $S_\\Gamma$, respectively, are equal. Examine if this is possible in the following cases:\n\n(\\alpha) $n = 2014$;\n\n(\\beta) $n = 2015$;\n\n(\\gamma) $n = 2018$.\n\n![](A. Fellouris)",
    "options": [],
    "answer": "See solution",
    "solution": "(\\alpha) If such a partition is possible for $n = 2014$, then the sum of the elements of $\\Sigma$ is $S_{\\Sigma} = S_A + S_B + S_{\\Gamma} = 3 S_A$, so $S_{\\Sigma}$ must be a multiple of 3. But\n\n$$\nS_{2014} = 1007 \\times 2015\n$$\n\nwhich is not a multiple of 3. Hence, the desired partition is not possible.\n\n(\\beta) For $n = 2015$, we have $S_{2015} = 1008 \\times 2015 \\equiv 0 \\pmod{3}$.\n\nWe observe that $\\Sigma$ consists of the set $M_0 = \\{1,2,3,4,5\\}$ and 335 successive six-element sets of the form:\n\n$$\nM_k = \\{6k, 6k+1, 6k+2, 6k+3, 6k+4, 6k+5\\}, \\quad k = 1, 2, \\ldots, 335.\n$$\n\nWe partition $M_0$ into three subsets with equal sums: $A_0 = \\{1,4\\}$, $B_0 = \\{2,3\\}$, and $\\Gamma_0 = \\{5\\}$. Note that\n\n$$\n(6k+1)+(6k+4) = (6k+2)+(6k+3) = 6k+(6k+5), \\quad \\text{for all } k=1,2,\\ldots,335.\n$$\n\nThus, the desired partition is possible by taking:\n\n$$\nA = \\{1,4\\} \\cup \\{6k+1,6k+4 : k=1,2,\\ldots,335\\}\n$$\n\n$$\nB = \\{2,3\\} \\cup \\{6k+2,6k+3 : k=1,2,\\ldots,335\\}\n$$\n\n$$\n\\Gamma = \\{5\\} \\cup \\{6k,6k+5 : k=1,2,\\ldots,335\\}\n$$\n\n(\\gamma) For $n = 2018$, $S_{2018} = 1009 \\times 2019 \\equiv 0 \\pmod{3}$. As in (\\beta), $\\Sigma$ consists of $M_0 = \\{1,2,3,4,5,6,7,8\\}$ and 335 successive six-element sets:\n\n$$\nM_k = \\{6k+3,6k+4,6k+5,6k+6,6k+7,6k+8\\}, \\quad k = 1,2,\\ldots,335.\n$$\n\nFirst, partition $M_0$ into three subsets with equal sums: $A_0 = \\{1,2,3,6\\}$, $B_0 = \\{5,7\\}$, and $\\Gamma_0 = \\{4,8\\}$. For all $k = 0,1,\\ldots,335$,\n\n$$\n(6k+3)+(6k+8) = (6k+4)+(6k+7) = (6k+5)+(6k+6).\n$$\n\nThus, the desired partition is possible by taking:\n\n$$\nA = \\{1,2,3,6\\} \\cup \\{6k+3,6k+8 : k=0,1,2,\\ldots,335\\}\n$$\n\n$$\nB = \\{5,7\\} \\cup \\{6k+4,6k+7 : k=0,1,2,\\ldots,335\\}\n$$\n\n$$\n\\Gamma = \\{4,8\\} \\cup \\{6k+5,6k+6 : k=0,1,2,\\ldots,335\\}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15516,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider an integer $n > 1$. One wants to colour all natural numbers red and blue so that the following conditions are simultaneously satisfied:\n\n1. Every number is coloured red or blue, there are infinitely many numbers coloured red and infinitely many numbers coloured blue.\n2. The sum of $n$ distinct numbers coloured red is coloured red, and the sum of $n$ distinct numbers coloured blue is coloured blue.\n\nIs it possible to colour in such a manner, if:\n\n$$\n1/ n = 2002?\n$$\n\n$$\n2/ n = 2003?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For $n = 2002$, we shall prove that the answer is *no*.\n\nSuppose, for contradiction, that such a colouring is possible. By condition 1, there exist 2002 distinct numbers $a_1, a_2, \\dots, a_{2002}$ coloured blue, and 2002 distinct numbers $b_1, b_2, \\dots, b_{2002}$ coloured red.\n\nAssume $b_{2k-1} = a_{2k-1} + 1$ and $b_{2k} = a_{2k} - 1$ for all $k = 1, 2, \\dots, 1001$. Then,\n$$\na = a_1 + a_2 + \\dots + a_{2002}\n$$\nis coloured blue, and\n$$\nb = b_1 + b_2 + \\dots + b_{2002}\n$$\nis coloured red. But from the construction, $a = b$, which contradicts the colouring conditions.\n\nFor $n = 2003$, consider colouring every even number blue and every odd number red. This colouring satisfies both conditions, so the answer is *yes, there exists such a colouring*.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15517,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $a$, $b$ 與 $c$ 為正實數，使得 $\\min\\{ab, bc, ca\\} \\ge 1$。試證：\n\n$$\n\\sqrt[3]{(a^2 + 1)(b^2 + 1)(c^2 + 1)} \\le \\left(\\frac{a+b+c}{3}\\right)^2 + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Claim.** 對於任意正實數 $x, y$ 且 $xy \\ge 1$，有\n\n$$\n(x^2 + 1)(y^2 + 1) \\ge \\left(\\left(\\frac{x+y}{2}\\right)^2 + 1\\right)^2. \\qquad (1)\n$$\n\n*Proof.* 注意 $xy \\ge 1$ 意味著\n\n$$\n\\left(\\frac{x+y}{2}\\right)^2 - 1 \\ge xy - 1 \\ge 0.\n$$\n\n我們有\n\n$$\n\\begin{aligned}\n(x^2 + 1)(y^2 + 1) &= (xy - 1)^2 + (x + y)^2 \\\\\n&\\le \\left(\\left(\\frac{x+y}{2}\\right)^2 - 1\\right)^2 + (x+y)^2 \\\\\n&= \\left(\\left(\\frac{x+y}{2}\\right)^2 + 1\\right) \\\\\n&\\le \\left(\\left(\\frac{x+y}{2}\\right) + 1\\right)^2\n\\end{aligned}\n$$\n\n不妨設 $a \\ge b \\ge c$，則 $a \\ge 1$。\n\n令\n\n$$\nd = \\frac{a+b+c}{3}.\n$$\n\n注意\n\n$$\nad = \\frac{a(a+b+c)}{3} \\ge \\frac{1+1+1}{3} = 1.\n$$\n\n可對 $(b, c)$ 應用式 (1)，得\n\n$$\n(a^2 + 1)(b^2 + 1)(c^2 + 1)(d^2 + 1) \\le \\left( \\left( \\frac{a+d}{2} + 1 \\right)^2 + \\left( \\frac{b+c}{2} + 1 \\right)^2 \\right) \\quad (2)\n$$\n\n接著，因為\n\n$$\n\\frac{a+d}{2} \\cdot \\frac{b+c}{2} \\ge \\sqrt{ad} \\cdot \\sqrt{bc} \\ge 1,\n$$\n\n可再次對 $(\\frac{a+d}{2}, \\frac{b+c}{2})$ 應用式 (1)。結合式 (2)，有\n\n$$\n\\begin{aligned}\n(a^2 + 1)(b^2 + 1)(c^2 + 1)(d^2 + 1) &\\le \\left( \\left( \\frac{a+b+c+d}{4} \\right)^2 + 1 \\right)^4 \\\\\n&= (d^2 + 1)^4.\n\\end{aligned}\n$$\n\n因此，$(a^2+1)(b^2+1)(c^2+1) \\le (d^2+1)^3$，對兩邊取三次方根即得所需不等式。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15518,
    "subject": "Mathematics (Olympiad)",
    "question": "Juli tiene un mazo de 54 cartas y le propone a Bruno el siguiente juego. Juli ubica las cartas en una fila, algunas boca arriba y las demás boca abajo. Bruno puede hacer repetidas veces el siguiente movimiento: elige una de las cartas y da vuelta esa carta y sus dos vecinas (las que estaban boca arriba las pone boca abajo y las que estaban boca abajo las pone boca arriba). Bruno gana si mediante este procedimiento logra que todas las cartas queden hacia arriba. En caso contrario gana Juli.\n\n**Nota.** Cuando Bruno elige la primera o la última carta de la fila, da vuelta solo dos cartas; en todos los otros casos da vuelta tres cartas.",
    "options": [],
    "answer": "See solution",
    "solution": "Demostraremos que Bruno tiene estrategia ganadora.\n\nConsideremos el caso en el que todas las cartas están boca arriba salvo la primera de la izquierda. Notamos con $\\overline{a_i}$ si la carta del lugar $i$ está boca arriba y $\\underline{a_i}$ si la carta del lugar $i$ está boca abajo. En este caso tenemos: $\\underline{a_1}\\overline{a_2}\\overline{a_3}\\dots\\overline{a_{53}}\\overline{a_{54}}$.\n\nBruno elige sucesivamente las siguientes cartas: $a_2, a_3, a_5, a_6, a_8, a_9, \\dots, a_{53}, a_{54}$ y gana. En efecto, después del movimiento correspondiente a elegir $a_2$, resulta que $a_2$ y $a_3$ quedan boca abajo y el resto, todas boca arriba. Después del movimiento correspondiente a elegir $a_3$, quedan todas boca arriba salvo $a_4$ que está boca abajo. Con estos dos movimientos, la única carta boca abajo se trasladó 3 lugares hacia la derecha. Siguiendo como se propuso arriba se lograrán todas las cartas boca arriba.\n\nSupongamos que las cartas están ubicadas en una fila y que hay al menos una carta boca abajo. Bruno, mirando de derecha a izquierda, ubica la primera carta boca abajo y elige para el primer movimiento la vecina de la izquierda de esta carta. Es decir, si $k$ es la primera carta boca abajo desde la derecha, tenemos: $\\dots hijk\\underline{l}\\dots\\overline{y}z$. Entonces Bruno elige la carta $j$ y se modifican las cartas $i$, $j$ y $k$. De este modo a la derecha de la carta $j$ que eligió Bruno están todas boca arriba. Se aumentó en al menos uno el número de cartas boca arriba de la derecha de la fila. Así, Bruno repite este procedimiento hasta que sea imposible continuar. Pueden ocurrir dos casos: (a) todas las cartas están boca arriba o (b) todas las cartas salvo la primera de la izquierda, están boca arriba. En el caso (a) ganó Bruno, y en el caso (b), usando el procedimiento analizado al comienzo, Bruno logra dejar las 54 cartas boca arriba y gana.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15519,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given the functional equation:\n\n$$\nf(2m + f(m) + f(m)f(n)) = n f(m) + m, \\quad \\forall m, n \\in \\mathbb{Z}.\n$$\n\nFind all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ satisfying this equation.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = f(0)$. Since $f \\equiv 0$ is not a solution, there exists $m_0 \\in \\mathbb{Z}$ such that $f(m_0) \\neq 0$. Substituting $m = m_0$ shows $f$ is injective. Setting $n = 0$ gives:\n\n$$\nf(2m + (a+1) f(m)) = m, \\quad \\forall m \\in \\mathbb{Z}.\n$$\n\nThus, $f$ is surjective. So there exists $b \\in \\mathbb{Z}$ with $f(b) = -1$. Substituting $m = n = b$ yields $f(2b) = 0$. Also, with $m = n = 0$, we get $f(a^2 + a) = 0$. By injectivity:\n\n$$\nb = \\frac{a^2 + a}{2}.\n$$\n\nSetting $n = b$ in the original equation:\n\n$$\nf(2m) = \\frac{a^2 + a}{2} f(m) + m, \\quad \\forall m \\in \\mathbb{Z}.\n$$\n\nSetting $m = 0$ gives $f(a f(n) + a) = a n$. Comparing with previous results, we get:\n\n$$\n(a+1) f(a n) + 2 a n = a f(n) + a, \\quad \\forall n \\in \\mathbb{Z}.\n$$\n\nIf $n = b$, then $\\frac{a(a^2 + a)}{2} = f(0) = a$, so $a \\in \\{0, 1, -2\\}$. Consider cases:\n\n1. If $a = 1$, then $f(n) = 1 - 2n$, but this does not satisfy the equation.\n2. If $a = 0$, then $f(2m) = m$, so $f(m) = 0$, which contradicts $f$ being non-constant.\n3. If $a = -2$, then:\n\n   $$\n   f(-2n) + 4n = 2 f(n) + 2, \\quad \\forall n \\in \\mathbb{Z}.\n   $$\n\n   Also, $f(-2n) = f(-n) - n$, so:\n\n   $$\n   f(-n) + 3n = 2 f(n) + 2, \\quad \\forall n \\in \\mathbb{Z}.\n   $$\n\n   Replacing $n$ by $-n$ gives $f(n) - 3n = 2 f(-n) + 2$. Combining, we find $f(n) = n - 2$. Checking, this function satisfies the original equation.\n\n**Conclusion:**\n\n$$\nf(n) = n - 2, \\quad \\forall n \\in \\mathbb{Z}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15520,
    "subject": "Mathematics (Olympiad)",
    "question": "For $n = \\prod_i p_i^{e_i}$ (where $p_i$ are primes and $e_i$ are positive integers), define\n$$\n\\psi(n) = \\prod_i (p_i + 1)^{e_i - 1}.\n$$\nShow that by repeated application of $\\psi$, we may transform any starting number into a number of the form $2^m$ for $m \\ge 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "As $p + 1$ is even when $p$ is odd, the greatest prime divisor of $p + 1$ is strictly less than $p$ for all primes $p \\neq 2$. Therefore, the greatest prime divisor of the number decreases at every step, unless it is $2$, in which case the number is already of the desired form.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15521,
    "subject": "Mathematics (Olympiad)",
    "question": "Find for which\n\n$$\nn \\in \\{3900, 3901, 3902, 3903, 3904, 3905, 3906, 3907, 3908, 3909\\}\n$$\n\nthe set $\\{1, 2, 3, \\dots, n\\}$ can be partitioned into disjoint triples such that in each triple, one of the three numbers is the sum of the other two.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the set must be partitioned into disjoint triples, $3$ must divide $n$. In each triple $\\{a, b, a+b\\}$, the sum is $2(a+b)$, which is even; thus, the total sum $n(n+1)/2$ must be even, i.e., $n(n+1)$ divisible by $4$. Therefore, $n$ must be of the form $12k$ or $12k+3$; among the given numbers, only $n=3900$ and $n=3903$ satisfy this.\n\nTo construct such partitions for these $n$, start from a valid decomposition for $n=k$ and build up to $n=4k$ and $n=4k+3$. This is possible via the sequence:\n\n$$\n3900 \\rightarrow 975 \\rightarrow 243 \\rightarrow 60 \\rightarrow 15 \\rightarrow 3\n$$\n\nand the trivial decomposition for $n=3$. For $n=4k$, after doubling the numbers in the triples, partition the remaining numbers\n\n$$\n\\{1, 3, 5, \\dots, 2k-1, 2k+1, \\dots, 4k\\}\n$$\n\ninto $k$ triples $\\{2j-1, 3k-j+1, 3k+j\\}$ for $j=1, \\dots, k$:\n\n$$\n\\begin{pmatrix}\n1 & 3 & 5 & \\dots & 2k-3 & 2k-1 \\\\\n3k & 3k-1 & 3k-2 & \\dots & 2k+2 & 2k+1 \\\\\n3k+1 & 3k+2 & 3k+3 & \\dots & 4k-1 & 4k\n\\end{pmatrix}\n$$\n\nFor $n=4k+3$, partition the remaining numbers\n\n$$\n\\{1, 3, 5, \\dots, 2k-1, 2k+1, \\dots, 4k+3\\}\n$$\n\ninto $k+1$ triples $\\{2j-1, 3k+3-j, 3k+j+2\\}$ for $j=1, \\dots, k+1$:\n\n$$\n\\begin{pmatrix}\n1 & 3 & 5 & \\dots & 2k-1 & 2k+1 \\\\\n3k+2 & 3k+1 & 3k & \\dots & 2k+3 & 2k+2 \\\\\n3k+3 & 3k+4 & 3k+5 & \\dots & 4k+2 & 4k+3\n\\end{pmatrix}\n$$\n\nThus, the only solutions are $n=3900$ and $n=3903$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15522,
    "subject": "Mathematics (Olympiad)",
    "question": "Inside a circle of radius 1 (or on the circumference), $n$ points are marked so that the minimal distance between any two marked points is as large as possible. Let $d_n$ be this minimal distance between the two closest points. Is it true that $d_{n+1} < d_n$ for every natural number $n \\ge 2$?\n\nNo.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that $d_6 \\leq 1 \\leq d_7$.\n\nFor the first inequality, assume arbitrary six points $A_1, A_2, A_3, A_4, A_5, A_6$ are marked in the circle. Let the centre of the circle be $O$. If $A_i = O$ for some $i$, the distance between $A_i$ and any other marked point is at most $1$. Assume in the rest that $A_i \\neq O$ for all $i$.\n\nLet $\\alpha$ be the smallest angle between some two rays $OA_i$ and $OA_j$, where $i, j = 1, \\ldots, 6$. Then $\\alpha \\leq 60^\\circ$, since the sum of angles between consecutive rays is $360^\\circ$.\n\n![](images/prob1617_p26_data_d85f1aabf4.png)\n\nIf $A_iA_j > 1$, then $A_iA_j$ is the largest side of triangle $OA_iA_j$, as the lengths $OA_i$ and $OA_j$ do not exceed $1$. The angle opposite the longest side is the largest, so $\\alpha$ would have to be greater than $60^\\circ$, a contradiction. Thus, there exist two marked points at distance at most $1$ from each other. As the choice of points was arbitrary, this establishes $d_6 \\leq 1$.\n\nOn the other hand, when marking the vertices of a regular hexagon inscribed in the circle together with the centre, the distance between any two consecutive marked points on the circumference is $1$, and their distance from the centre is also $1$. Hence $d_7 \\geq 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15523,
    "subject": "Mathematics (Olympiad)",
    "question": "$n$ болон $8n^2 + 1$ нь анхны тоонууд бол $8n^2 - 1$ мөн анхны тоо гэдгийг батал.",
    "options": [],
    "answer": "See solution",
    "solution": "$n = 2$ бол $8n^2 + 1 = 33 = 3 \\cdot 11$ тул анхны тоо биш. $n = 3$ бол $8n^2 + 1 = 73$ (анхны тоо), $8n^2 - 1 = 71$ (анхны тоо) байна.\n\n$n > 3$ үед $n = 3k + 1$ эсвэл $n = 3k + 2$ хэлбэртэй байна. \n\n$n = 3k + 1$ бол:\n$$\n8n^2 + 1 = 8(3k+1)^2 + 1 = 72k^2 + 48k + 9 = 3(24k^2 + 16k + 3)\n$$\nТэгэхээр $8n^2 + 1$ нь 3-д хуваагддаг тул анхны тоо биш.\n\n$n = 3k + 2$ бол:\n$$\n8n^2 + 1 = 8(3k+2)^2 + 1 = 72k^2 + 96k + 33 = 3(24k^2 + 32k + 11)\n$$\nМөн 3-д хуваагддаг тул анхны тоо биш.\n\nИймд $n$ ба $8n^2 + 1$ зэрэг анхны тоо байхын тулд зөвхөн $n = 3$ боломжтой бөгөөд энэ үед $8n^2 - 1$ нь мөн анхны тоо байна.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15524,
    "subject": "Mathematics (Olympiad)",
    "question": "Five years ago, the total age of all sons in a family was two years more than the total age of all daughters. Since then, one more child was born, and now the total age of all daughters is two years more than the total age of all sons. What was the difference between the total age of sons and daughters two years ago?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** The total age was the same, or the daughters' age was one year more than the sons'.\n\n**Solution:**\nSuppose that five years ago the family had $n$ sons and $m$ daughters, with total age of sons $N$ and total age of daughters $M$.\n\nIf $k$ years ago one daughter was born, then the total age of sons now is $N + 5n$, and for daughters it is $M + 5m + k$. Since $N = M + 2$, we get:\n\n$$\n(N + 5n) + 2 = M + 5m + k \\\\\nM + 5n + 4 = M + 5m + k \\\\\nk = 5(n - m) + 4\n$$\n\nSince $0 \\leq k \\leq 5$, then $k = 4$ and $n - m = 0$. So two years ago, the total age of sons was $N + 3n$ and daughters was $M + 3m + 2$, which were equal:\n\n$$\nN + 3n - (M + 3m + 2) = (N - M) + 3(n - m) - 2 = 2 - 2 = 0\n$$\n\nIf $k$ years ago one son was born, the total age of sons now is $N + 5n + k$, and daughters is $M + 5m$. We have:\n\n$$\n(N + 5n + k) + 2 = M + 5m \\\\\nM + 5n + k + 4 = M + 5m \\\\\nk = 5(m - n) - 4\n$$\n\nSince $0 \\leq k \\leq 5$, then $k = 1$ if $m - n = 1$. Two years ago, the difference was:\n\n$$\nM + 3m - (N + 3n) = (M - N) + 3(m - n) = -2 + 3 = 1\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15525,
    "subject": "Mathematics (Olympiad)",
    "question": "Let's say that a digit is *eternal* for a positive integer $n$ if it is contained in every multiple of $n$. Find all digits which are eternal for at least one positive integer.",
    "options": [],
    "answer": "See solution",
    "solution": "The only such digit is $0$, since it is contained in every multiple of $10$. Let's show that no other digit is eternal for any positive integer.\n\nAssume that some digit is eternal for integer $n$. Consider the remainders of the numbers\n\n$$\n1, 11, 111, \\dots, \\underbrace{11\\dots11}_{n+1}\n$$\n\nmodulo $n$. By the pigeonhole principle, two of these remainders are equal, so their difference, which has the form $11\\dots100\\dots0$, is a multiple of $n$. If we multiply this number by $2$, we get a multiple of $n$ of the form $22\\dots200\\dots0$. But the only common digit for these two multiples is $0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15526,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are 2007 male students and 2007 female students, and each student joins at most 100 clubs. Prove that there must exist a club with at least 11 male and 11 female members.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, for contradiction, that no club contains at least 11 male and 11 female members. We count the number of triples $(M, F, C)$, where $M$ is a male student, $F$ is a female student, and $C$ is a club joined by both $M$ and $F$.\n\nFor each of the 2007 choices of $M$ and each of the 2007 choices of $F$, there is at least one club $C$ that both have joined, so there are at least $2007^2$ such triples.\n\nLabel the clubs as $C_1, C_2, \\dots, C_k$. Let $m_i$ and $f_i$ be the numbers of male and female members in club $C_i$. The total number of triples is:\n\n$$\n\\sum_{i=1}^{k} m_i f_i\n$$\n\nBy assumption, each club has at most 10 males or at most 10 females. Divide clubs into two groups: those with at most 10 males ($C_1, \\dots, C_\\ell$), and those with at most 10 females ($C_{\\ell+1}, \\dots, C_k$). Then:\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{k} m_i f_i &= \\sum_{i=1}^{\\ell} m_i f_i + \\sum_{i=\\ell+1}^{k} m_i f_i \\\\\n&\\le 10 \\sum_{i=1}^{\\ell} f_i + 10 \\sum_{i=\\ell+1}^{k} m_i \\\\\n&\\le 10 \\sum_{i=1}^{k} f_i + 10 \\sum_{i=1}^{k} m_i\n\\end{aligned}\n$$\n\nThe sum $\\sum_{i=1}^{k} f_i$ counts the total number of female memberships, which is at most $100 \\times 2007$. Similarly for males. Thus:\n\n$$\n2007^2 \\le 10 \\times 100 \\times 2007 + 10 \\times 100 \\times 2007 = 2000 \\times 2007\n$$\n\nBut $2007^2 > 2000 \\times 2007$, a contradiction. Therefore, there must be a club with at least 11 male and 11 female members.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15527,
    "subject": "Mathematics (Olympiad)",
    "question": "求所有實係數多項式 $P$，使得：\n\n$$\nP(x)P(x+1) = P(x^2 - x + 3) \\quad \\forall x \\in \\mathbb{R},\n$$\n\n其中 $\\mathbb{R}$ 表所有實數所成的集合。",
    "options": [],
    "answer": "See solution",
    "solution": "所有滿足題目要求之實係數多項式 $P$ 為零多項式與\n\n$$\nP(x) = (x^2 - 2x + 3)^n \\quad \\forall x \\in \\mathbb{R},\n$$\n\n其中 $n$ 是任一個非負整數。\n\n代入原式易知上述 $P(x)$ 都是此函數方程的解，以下考慮 $P(x)$ 不是零多項式的情況。\n\n首先我們先證明 $P$ 沒有實根。使用反證法，如果 $P$ 有實根 $\\alpha$，原式中代 $x = \\alpha$ 可得\n\n$$\nP(\\alpha^2 - \\alpha + 3) = P(\\alpha)P(\\alpha + 1) = 0.\n$$\n\n設 $\\beta = \\alpha^2 - \\alpha + 3$，則 $\\beta$ 也是 $P$ 的實根，顯然有 $\\beta \\ge 2$。考慮無窮數列\n\n$$\n\\beta, f(\\beta), f(f(\\beta)), f(f(f(\\beta))), \\dots\n$$\n\n其中 $f(x) = x^2 - x + 3$，如同上述方法分別將 $\\beta, f(\\beta), f(f(\\beta)), \\dots$ 代入原式，可得此數列的每一項都是 $P$ 的實根。\n\n在 $x \\ge 2$ 時顯然有 $f(x) > x$，故該數列嚴格遞增，即 $P$ 有無窮多個實根，與 $P$ 為多項式矛盾！因為 $P$ 無實根，其次數必為偶數次。考慮 $P$ 的首項係數 $t$，比較原式的最高次方項係數可得\n\n$$\nt^2 = t,\n$$\n\n故 $t = 1$，即 $P$ 為首么多項式。\n\n設 $P$ 的次數為 $2n$，考慮 $P(x) = Q(x) + (x^2 - 2x + 3)^n$，因為 $P$ 首么，故 $\\deg(Q) < 2n$，代回原式得\n\n$$\n(Q(x) + (x^2 - 2x + 3)^n)(Q(x + 1) + ((x + 1)^2 - 2(x + 1) + 3)^n) = Q(x^2 - x + 3) + ((x^2 - x + 3)^2 - 2(x^2 - x + 3) + 3)^n\n$$\n\n展開化簡得\n\n$$\n\\begin{aligned}\n& Q(x)Q(x+1) + Q(x)((x+1)^2 - 2(x+1) + 3)^n \\\\\n& \\quad + Q(x+1)(x^2 - 2x + 3)^n \\\\\n& = Q(x^2 - x + 3).\n\\end{aligned}\n$$\n\n如果 $Q$ 不是零多項式，令 $\\deg(Q) = q$ 與 $Q$ 的首項係數為 $k$，因為 $q < 2n$，左式最高次數至多為 $2n + q$，且其 $2n + q$ 次項係數為 $2k \\neq 0$，故左式最高次數為 $2n + q$。而右式最高次數顯然是 $2q$，故 $n = 2q$，矛盾！\n\n因此 $Q$ 只能是零多項式，即 $P$ 的所有可能只有恆零與\n\n$$\nP(x) = (x^2 - 2x + 3)^n.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15528,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the least number of cells that must be marked on an $n \\times n$ board such that there is no series of diagonal cells of length $> \\frac{n}{2}$ without a mark?",
    "options": [],
    "answer": "See solution",
    "solution": "For any $n$, it is possible to set $n$ marks on the board and achieve the desired property by placing a mark in every cell of row $\\lfloor \\frac{n}{2} \\rfloor$. We now show that $n$ is also the minimum number of marks needed.\n\nIf $n$ is odd, there are $2n$ series of diagonal cells with length $> \\frac{n}{2}$ and both end cells on the edge of the board. Since every mark on the board can lie on at most two of these diagonals, it is necessary to set at least $n$ marks to have a mark on every one of them.\n\nIf $n$ is even, there are $2n-2$ series of diagonal cells with length $> \\frac{n}{2}$ and both end cells on the edge of the board. We call one of these diagonals even if every coordinate $(x, y)$ on it satisfies $2 \\mid x - y$, and odd otherwise. By symmetry, there are equally many odd and even diagonals, so there must be $n-1$ of each. Any mark set on the board can at most sit on two diagonals, and these two have to be of the same kind. Thus, we will need $\\frac{n}{2}$ marks for the even diagonals, since there are $n-1$ of them and $2 \\nmid n-1$, and similarly $\\frac{n}{2}$ marks for the odd diagonals.\n\nSo we need at least $n$ marks to get the desired property.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15529,
    "subject": "Mathematics (Olympiad)",
    "question": "A school has 60 students in year 2 who will be divided into three classes of 20 students. Each student writes a list of three other students that they hope to have in their class.\n\nCan the school always arrange for each student to be in the same class as at least one of the three students on their list?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** No\n\nConsider the situation where universally popular students $A$, $B$, and $C$ are on everyone's list, except that $A$ is not on $A$'s list, $B$ is not on $B$'s list, and $C$ is not on $C$'s list. Also, suppose that a student $D$ is on $A$'s, $B$'s, and $C$'s list. Thus, each student now has three students on their lists. We will show that the school cannot fulfill its claim.\n\nSuppose that one of the classes contains none of $A$, $B$, or $C$. Then no one in that class gets any of their preferences.\n\nSuppose that each of the three classes contains one of $A$, $B$, or $C$. Without loss of generality, we may suppose that $D$ is in the same class as $A$. Then student $B$ does not get any of their preferences.\n\nEither way, the school is unable to fulfill its claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15530,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n$, we denote by $n!$ (*n factorial*) the number we get if we multiply all integers from 1 to $n$. For example: $5! = 1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 = 120$.\n\n(a) Determine all integers $n$ with $1 \\leq n \\leq 100$ for which $n! \\cdot (n+1)!$ is a perfect square. Also, prove that you have found all solutions $n$.\n\n(b) Prove that no positive integer $n$ exists such that $n! \\cdot (n+1)! \\cdot (n+2)! \\cdot (n+3)!$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) We observe that $(n+1)! = (n+1) \\cdot n!$, so\n$$\nn! \\cdot (n+1)! = (n!)^2 \\cdot (n+1).\n$$\nThis product is a perfect square if and only if $n+1$ is a perfect square, since $(n!)^2$ is always a perfect square. For $1 \\leq n \\leq 100$, this occurs when $n = 3, 8, 15, 24, 35, 48, 63, 80, 99$ (i.e., all perfect squares minus one, less than or equal to 100).\n\n(b) Consider $n! \\cdot (n+1)! \\cdot (n+2)! \\cdot (n+3)!$. We can write:\n$$\nn! \\cdot (n+1)! \\cdot (n+2)! \\cdot (n+3)! = (n!)^2 \\cdot (n+1) \\cdot ((n+2)!)^2 \\cdot (n+3)\n$$\n(because $(n+1)! = (n+1) \\cdot n!$ and $(n+3)! = (n+3) \\cdot (n+2)!$). Since $(n!)^2$ and $((n+2)!)^2$ are perfect squares, the product is a perfect square if and only if $(n+1)(n+3)$ is a perfect square. Suppose $(n+1)(n+3) = k^2$ for some integer $k$. But $(n+1)^2 < (n+1)(n+3) < (n+3)^2$, so $n+1 < k < n+3$, which means $k = n+2$. Then $(n+1)(n+3) = (n+2)^2 - 1$, which is never a perfect square. Therefore, no such $n$ exists.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15531,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, $BC$ is the shortest side. Let $X$ and $Y$ be points on sides $AB$ and $AC$, respectively, and let $K$ and $L$ be points on rays $CB$ and $BC$, respectively, such that $BX = BK = BC = CY = CL$. Line $KX$ intersects line $LY$ at point $M$. Prove that the intersection point of the medians of $\\triangle KLM$ coincides with the center of the inscribed circle of $\\triangle ABC$.\n\n![](images/UkraineMO2019_booklet_p32_data_0ab3952325.png)\n\nFig.32",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\angle ABC$ is the outer angle of isosceles $\\triangle XKB$ with vertex $B$, line $KX$ is parallel to the bisector of $\\angle ABC$.\n\nThe ratio $LB : LK = 2 : 3$ means that the bisector of $\\angle ABC$ passes through the centroid of $\\triangle KLM$. Denote by $LL_1$ its median and by $L_2$ its intersection point with the bisector of $\\angle ABC$. From similar triangles $\\triangle LBL_2 \\sim \\triangle LKL_1$ (by two angles), we obtain\n\n$$\n\\frac{LL_2}{LL_1} = \\frac{LB}{LK} = \\frac{2}{3}.\n$$\n\nThus, point $L_2$ divides $LL_1$ with the same ratio as the centroid, which means it is the centroid of $\\triangle KLM$.\n\nAnalogously, the bisector of $\\angle BCA$ passes through the centroid of $\\triangle KLM$. Hence, since the intersection point of the bisectors is the incenter of a triangle, this proves the statement.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15532,
    "subject": "Mathematics (Olympiad)",
    "question": "2014 lines are given in a plane, arranged in three groups of pairwise parallel lines. What is the greatest possible number of triangles formed by the lines (each side of such a triangle lies on one of the lines)?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a \\geq b \\geq c$ be the numbers of lines in the three groups for which the greatest possible number of triangles is attained. Then $a + b + c = 2014$, and the greatest possible number of triangles is $abc$ (when no three lines have a common point).\n\nWe will show that $a \\leq c + 1$. Suppose the opposite, i.e., $a > c + 1$. Then\n$$\nabc < b(ac + a - c - 1) = b(a - 1)(c + 1),\n$$\nwhich contradicts the choice of $a, b,$ and $c$.\n\nIt cannot be that $a = c$, because in that case $a = b = c = \\frac{2014}{3}$, which is not an integer. In order for $a, b,$ and $c$ to be integers, it must be that $a = 672$ and $b = c = 671$. Thus, the number of triangles is\n$$\n672 \\cdot 671^2.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15533,
    "subject": "Mathematics (Olympiad)",
    "question": "Which number is greater,\n$$\n\\sin 1 - \\cos 1 \\quad \\text{or} \\quad \\frac{1}{4}?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The greater number is $\\sin 1 - \\cos 1$.\n\nThe sine is increasing and the cosine decreasing in the first quadrant. Since $\\frac{\\pi}{4} < 1 < \\frac{\\pi}{2}$, we have\n$$\n\\sin 1 - \\cos 1 > \\sin \\frac{\\pi}{4} - \\cos \\frac{\\pi}{4} = 0.\n$$\nHence, the numbers $\\sin 1 - \\cos 1$ and $\\frac{1}{4}$ are ordered in the same way as their squares $(\\sin 1 - \\cos 1)^2$ and $\\frac{1}{16}$. Since\n$$\n(\\sin 1 - \\cos 1)^2 = \\sin^2 1 + \\cos^2 1 - 2 \\sin 1 \\cos 1 = 1 - \\sin 2,\n$$\nit suffices to compare the numbers $1 - \\sin 2$ and $\\frac{1}{16}$. We show that the former number is greater by showing that $\\sin 2 < \\frac{15}{16}$.\n\nSince $\\pi < 3.2 = \\frac{16}{5}$, we have $\\frac{\\pi}{2} < \\frac{5}{8}\\pi < 2 < \\pi$, which implies\n$$\n\\sin 2 < \\sin \\frac{5}{8}\\pi = \\sqrt{\\frac{1 - \\sin \\frac{5}{4}\\pi}{2}} = \\sqrt{\\frac{1 + \\frac{\\sqrt{2}}{2}}{2}} = \\frac{\\sqrt{2 + \\sqrt{2}}}{2} < \\frac{15}{16},\n$$\nusing the half-angle sine formula. The last inequality is clear from\n$$\n\\sqrt{2} < \\frac{3}{2} < \\frac{97}{64} \\Rightarrow 2 + \\sqrt{2} < \\frac{225}{64} \\Rightarrow \\sqrt{2 + \\sqrt{2}} < \\frac{15}{8}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15534,
    "subject": "Mathematics (Olympiad)",
    "question": "The incircle of a triangle $ABC$ touches its sides $BC$, $CA$, $AB$ at the points $A_1$, $B_1$, $C_1$, respectively. Let the projections of the orthocenter $H_1$ of the triangle $A_1B_1C_1$ to the lines $AA_1$ and $BC$ be $P$ and $Q$, respectively. Show that the line $PQ$ bisects the line segment $B_1C_1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A_1S$, $B_1T$, and $C_1U$ be the altitudes of $A_1B_1C_1$. The circle $k$ with diameter $A_1H_1$ contains the points $A_1$, $H_1$, $T$, $U$, $P$, and $Q$. Let $AA_1$ intersect the incircle of $ABC$ for the second time at $V$. Assume that $\\angle B \\geq \\angle C$.\n\nObserve that $\\angle C_1A_1V = \\angle TA_1P$, $\\angle C_1B_1A_1 = \\angle C_1A_1B = \\angle TA_1Q$, and $\\angle VA_1B_1 = \\angle PA_1U$. Considering the chords subtending these angles in the circle $k$ and in the incircle of triangle $ABC$, we conclude that the cyclic quadrilaterals $PTQU$ and $VC_1A_1B_1$ are similar.\n\nLet $M$ and $N$ be the midpoints of the line segments $B_1C_1$ and $A_1H_1$. Notice that\n\n$$\n\\angle MTN = 180^\\circ - \\angle C_1TM - \\angle NTA_1 = 180^\\circ - \\angle TC_1M - \\angle NA_1T = 90^\\circ\n$$\n\nas $M$ and $N$ are the midpoints of the hypotenuses of the right triangles $C_1TB_1$ and $TH_1A_1$, respectively.\n\nTherefore, $MT$ and, similarly, $MU$, are tangent to $k$. It follows that the pentagons $PMTQU$ and $VAC_1A_1B_1$ are similar. Since the points $A_1$, $V$, $A$ lie on a line, so do the corresponding points $Q$, $P$, $M$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15535,
    "subject": "Mathematics (Olympiad)",
    "question": "In a right-angled triangle $ABC$, the bisector of angle $A$ intersects the circumscribed circle of this triangle at point $W$. From point $W$, draw a perpendicular $WU$ to $AB$, and from the incenter $I$ of this triangle, draw a perpendicular $IP$ to line $WU$. Let $M$ be the midpoint of segment $BC$. Prove that the line $MP$ intersects the midpoint of segment $CI$.\n\n![](images/Ukraine_booklet_2018_p34_data_eec435f8b5.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that right-angled triangles $BMW$ and $IPW$ are equal by hypotenuse and acute angle ($IW = BW$ by the trillium theorem (see figure), $\\angle WIP = \\angle MBW = \\frac{1}{2}\\angle A$). So, $WP = MW$. As points $U$, $B$, $M$, $W$ are on one circle with diameter $BW$, then\n\n$$\n\\angle(MW, WU) = \\angle(MB, BU) = \\angle(CB, BA).\n$$\n\nIt is not hard to see that then $\\angle MWP = \\angle ABC$. As $\\triangle MWP$ is isosceles, then\n\n$$\n\\angle WMP = \\frac{1}{2}(90^\\circ - \\angle B) = 90^\\circ - \\frac{1}{2}\\angle B.\n$$\n\nThen $\\angle PMB = \\frac{1}{2}\\angle B$. So, it is not hard to see that $PM \\parallel BI$. Let $F$ be the point of intersection of $CI$ and $PM$. Then in $\\triangle BIC$, segment $MF$ is parallel to side $BI$ and intersects midpoint $M$ of side $BC$. Thus, $MF$ is the midline of $\\triangle BIC$, and so $F$ is the midpoint of segment $CI$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15536,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(x^3) + f(y^3) = (x + y)\\big(f(x^2) + f(y^2) - f(xy)\\big)\n$$\n\nfor all real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Set $x = y = 0$:\n\n$$\nf(0) + f(0) = 0 \\cdot (f(0) + f(0) - f(0)),\n$$\nso $f(0) = 0$.\n\nSet $y = -x$:\n\n$$\nf(x^3) + f(-x^3) = 0.\n$$\nSince every real number has a cube root, $f(-a) = -f(a)$ for all real $a$ (i.e., $f$ is odd).\n\nSet $x = 0$:\n\n$$\nf(0) + f(y^3) = y(f(0) + f(y^2) - f(0)),\n$$\nso $f(y^3) = y f(y^2)$ for all $y$.\n\nThus,\n\n$$\nxf(x^2) + y f(y^2) = f(x^3) + f(y^3) = (x + y)(f(x^2) + f(y^2) - f(xy)) \\tag{1}\n$$\n\nAlso,\n\n$$\n\\begin{aligned}\nxf(x^2) - y f(y^2) &= f(x^3) + f(-y^3) \\\\\n&= (x - y)(f(x^2) + f(xy) + f(y^2)) \\tag{2}\n\\end{aligned}\n$$\n\nAdding (1) and (2):\n\n$$\n2x f(x^2) = 2x f(x^2) - 2y f(xy) + 2x f(y^2),\n$$\nso $y f(xy) = x f(y^2)$. Setting $y = 1$, $f(x) = x f(1)$. Let $k = f(1)$, so $f(x) = kx$ for all $x$.\n\nIt is straightforward to check that $f(x) = kx$ satisfies the original equation for any real $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15537,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a > 0$, $b > 0$, $c > 0$ and $a + b + c = 1$. Prove the inequality\n\n$$\n\\frac{a^4 + b^4}{a^2 + b^2} + \\frac{b^3 + c^3}{b + c} + \\frac{2a^2 + b^2 + 2c^2}{2} \\ge \\frac{1}{2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "From $1 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac)$ we get\n\n$$\nab + bc + ca + \\frac{a^2 + b^2 + c^2}{2} = \\frac{1}{2}.\n$$\n\nWe use the inequality $(a^{n-1} - b^{n-1})(a - b) \\ge 0$ for $n \\ge 1$ with equality holding when $a = b$. The last inequality is equivalent to $a^n + b^n \\ge ab(a^{n-2} + b^{n-2})$ for $n \\ge 2$. In the case at hand we have: $\\frac{a^4 + b^4}{a^2 + b^2} \\ge ab$, $\\frac{b^3 + c^3}{b + c} \\ge bc$ and $\\frac{c^2 + a^2}{2} \\ge ca$. By adding the last three inequalities we get\n\n$$\n\\frac{a^4 + b^4}{a^2 + b^2} + \\frac{b^3 + c^3}{b + c} + \\frac{c^2 + a^2}{2} \\ge ab + bc + ac \\quad \\text{from where}\n$$\n\n$$\n\\frac{a^4 + b^4}{a^2 + b^2} + \\frac{b^3 + c^3}{b + c} + c^2 + a^2 + \\frac{b^2}{2} \\ge ab + bc + ac + \\frac{c^2 + a^2 + b^2}{2} = \\frac{1}{2}.\n$$\n\nIt is clear that equality holds when $a = b = c = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15538,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S = \\{0, 1, 2, 3, 4, 5, 6, 7, 8\\}$. How many functions $f : S \\to S$ satisfy the following condition for all $x, y \\in S$?\n\n$$\nf(x) f(y) \\equiv f(f([x + y])) \\pmod{9}\n$$\n\nHere, $[x]$ denotes the unique element $x' \\in S$ such that $x \\equiv x' \\pmod{9}$.",
    "options": [],
    "answer": "See solution",
    "solution": "In this solution, $x \\equiv y$ always means that $x-y$ is divisible by $9$. For any integer $x$, $[x]$ denotes the unique $x' \\in S$ with $x \\equiv x'$. Note that for any integers $s$ and $t$, $[[s-t]+t] = [s]$ and $[s[t]] = [st]$.\n\nThe condition can be rewritten as\n\n$$\nf(x)f(y) \\equiv f(f([x+y])) \\pmod{9}\n$$\n\nfor all $x, y \\in S$. Let $T = \\{f(x) \\mid x \\in S\\}$. For $s = f(x)$ and $t = f(y)$ in $T$, $[st] = [f(x)f(y)] = f(f([x+y])) \\in T$.\n\n**Case 1:** $T$ contains a multiple of $3$.\n\nLet $a \\in T$ be a multiple of $3$. Then $0 = [a^2] \\in T$, so there is $b \\in S$ with $f(b) = 0$. For any $x$, $P([x-b], b)$ gives $f(f(x)) \\equiv f([x-b])f(b) = 0$, so $f(f(x)) = 0$ for all $x$. Also, $P(x, x)$ gives $f(x)^2 \\equiv f(f([2x])) = 0$, so $f(x)$ is a multiple of $3$ for all $x$.\n\nConversely, if $f(f(x)) = 0$ and $f(x)$ is a multiple of $3$ for all $x$, then $f$ satisfies the condition, since both sides of $P(x, y)$ are divisible by $9$.\n\nSince $T$ contains $0$, $T$ is one of $\\{0\\}$, $\\{0, 3\\}$, $\\{0, 6\\}$, or $\\{0, 3, 6\\}$. Also, $f(x) = 0$ for all $x \\in T$.\n\n- $T = \\{0\\}$: Only $f(x) = 0$ for all $x$.\n- $T = \\{0, 3\\}$: Assign $f(x) = 0$ or $3$ for $x \\in S \\setminus T = \\{1, 2, 4, 5, 6, 7, 8\\}$, at least one $x$ with $f(x) = 3$. Number: $2^7 - 1 = 127$.\n- $T = \\{0, 6\\}$: Similarly, $127$ functions.\n- $T = \\{0, 3, 6\\}$: Assign $f(x) = 0, 3, 6$ for $x \\in S \\setminus T = \\{1, 2, 4, 5, 7, 8\\}$, at least one $x$ with $f(x) = 3$ and one $y$ with $f(y) = 6$. Number: $3^6 - 2^6 - 2^6 + 1 = 602$.\n\nTotal for this case: $1 + 127 + 127 + 602 = 857$.\n\n**Case 2:** $T$ contains no multiples of $3$.\n\nTake $a \\in T$, $a$ not divisible by $3$. By Euler's theorem, $a^6 \\equiv 1$, so $1 = [a^6] \\in T$. Thus, $f(b) = 1$ for some $b$. $P(b, 0)$ gives $f(0) = f(b)f(0) \\equiv f(f(b)) = f(1)$, so $f(0) = f(1)$. For any $x$, $P(x, [-x])$ and $P([x+1], [-x])$ give $f([-x])(f([x+1]) - f(x)) \\equiv 0$. Since $f([-x])$ is not divisible by $3$, $f([x+1]) = f(x)$ for all $x$. Thus, $f$ is constant, and since $1 \\in T$, $f(x) = 1$ for all $x$.\n\nThis function works, since both sides of $P(x, y)$ are $1$.\n\n**Total:** $857 + 1 = 858$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15539,
    "subject": "Mathematics (Olympiad)",
    "question": "Given 16 distinct real numbers $\\alpha_1, \\alpha_2, \\ldots, \\alpha_{16}$. For each polynomial $P$ with real coefficients, define\n\n$$\nV(P) = P(\\alpha_1) + P(\\alpha_2) + \\cdots + P(\\alpha_{16}).\n$$\n\nProve that there exists exactly one monic polynomial $Q$ of degree $8$ satisfying the following conditions:\n\n1. $V(QP) = 0$ for all polynomials $P$ with $\\deg P < 8$,\n2. $Q$ has $8$ real roots (counted with multiplicity).",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\mathbb{R}_{15}[x]$ be the set of all real polynomials of degree at most $15$. Consider the map\n\n$$\nV : \\mathbb{R}_{15}[x] \\to \\mathbb{R}, \\quad P \\mapsto P(\\alpha_1) + \\cdots + P(\\alpha_{16}).\n$$\n\nThis map is linear: $V(P+Q) = V(P) + V(Q)$ and $V(\\lambda P) = \\lambda V(P)$ for all $P, Q \\in \\mathbb{R}_{15}[x]$ and real $\\lambda$.\n\nLet $g_k(x) = x^k$ for $k = 0, 1, \\ldots, 15$, and define $f_0(x) = 1$. Construct $f_1(x)$ as\n\n$$\nf_1(x) = - \\frac{V(g_1 f_0)}{V(f_0^2)} f_0(x) + g_1(x).\n$$\n\nThen\n\n$$\nV(f_0 f_1) = V\\left(-\\frac{V(g_1 f_0)}{V(f_0^2)} f_0^2 + f_0 g_1\\right) = -V(g_1) + V(g_1) = 0.\n$$\n\nSimilarly, define\n\n$$\nf_2(x) = -\\frac{V(g_2 f_0)}{V(f_0^2)} f_0(x) - \\frac{V(g_2 f_1)}{V(f_1^2)} f_1(x) + g_2(x),\n$$\n\nso that $V(f_0 f_2) = V(f_1 f_2) = 0$. Continue this process to construct $f_0, f_1, \\ldots, f_8$ such that\n\n$$\nV(f_i f_j) = 0 \\quad \\text{for all } 0 \\leq i < j \\leq 8,\n$$\n\nwhere\n\n$$\n\\begin{aligned}\nf_k(x) = & -\\frac{V(g_k f_0)}{V(f_0^2)} f_0(x) - \\frac{V(g_k f_1)}{V(f_1^2)} f_1(x) - \\cdots \\\\ & - \\frac{V(g_k f_{k-1})}{V(f_{k-1}^2)} f_{k-1}(x) + g_k(x),\n\\end{aligned}\n$$\n\nfor $k = 1, 2, \\ldots, 8$. Let $Q = f_8$. We claim $Q$ satisfies the required conditions.\n\nSince $f_k$ is monic for $0 \\leq k \\leq 7$, any polynomial $P$ of degree at most $7$ can be written as $P = \\sum_{k=0}^{7} \\beta_k f_k$. Thus,\n\n$$\nV(QP) = V\\left(\\sum_{k=0}^{7} \\beta_k f_k Q\\right) = \\sum_{k=0}^{7} \\beta_k V(f_k Q) = 0.\n$$\n\nSuppose another polynomial $R$ also satisfies the conditions. Then $Q - R$ has degree less than $8$, so\n\n$$\nV\\left[(Q - R)^2\\right] = 0.\n$$\n\nBut this implies $Q = R$, so $Q$ is unique.\n\nTo show $Q$ has $8$ real roots, suppose $Q$ has $k < 8$ real roots. Then $Q(x)$ can be written as\n\n$$\nQ(x) = \\prod_{i=1}^{k} (x - a_i) \\prod_{j=1}^{m} \\left((x - r_j)^2 + s_j^2\\right),\n$$\n\nwhere $8 - k = 2m$. Then $h(x) = \\prod_{i=1}^{k} (x - a_i) Q(x) \\geq 0$ for all $x \\in \\mathbb{R}$, so $V(h) > 0$. But for $P(x) = \\prod_{i=1}^{k} (x - a_i)$, $V(PQ) = V(h) = 0$, a contradiction. Thus, $Q$ must have $8$ real roots.\n\nTherefore, such a monic degree $8$ polynomial $Q$ exists and is unique.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15540,
    "subject": "Mathematics (Olympiad)",
    "question": "An equilateral triangle with side length $3$ is divided into $9$ equilateral triangles with side length $1$. An integer from $1$ to $10$ is written into every point that is a vertex of a small triangle (colored vertices on the figure), such that all numbers are written exactly once. For every small triangle, the sum of the numbers in the three vertices is written inside it. Prove that there exist three small triangles such that the sum of the numbers inside them is at least $48$.",
    "options": [],
    "answer": "See solution",
    "solution": "Any three small triangles, from which no two have common vertices, take up nine of the ten numbers written into the vertices of the small triangles. So, the sum of the numbers inside those small triangles is $55 - a$, where $a$ is the number at the last vertex. Now it is sufficient to prove that we can choose the three small triangles, from which no two have common vertices, in four different ways, always leaving a different vertex out. So, in at least one case, the number at the last vertex is at most $7$, and the sum of the numbers in the three chosen triangles is at least $55 - 7 = 48$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15541,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the largest integer $k$ such that, in any directed complete graph $G$ with 2011 vertices where $|\\text{indeg}(v) - \\text{outdeg}(v)| \\leq k$ for all vertices $v$, there exists a directed path from any vertex $v_1$ to any vertex $v_2$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $k = 1005$.\n\nFirst, observe that $|\\text{indeg}(v) - \\text{outdeg}(v)|$ is always even.\n\n**Counterexample for $k = 1006$:**\nLet $V = X \\cup Y$, where $X = \\{x_1, x_2, \\dots, x_{1005}\\}$ and $Y = \\{y_1, y_2, \\dots, y_{1006}\\}$. Define the edge set $E$ as follows:\n- All possible edges within $X$ and within $Y$ (with certain restrictions), and\n- All edges from $X$ to $Y$.\n\nIn this construction, $|\\text{indeg}(v) - \\text{outdeg}(v)| \\leq 1006$ for all $v \\in V$, but there is no directed path from $y_1$ to $x_1$.\n\n**Now, for $k = 1004$:**\nSuppose $|\\text{indeg}(v) - \\text{outdeg}(v)| \\leq 1004$ for all $v$. If either $\\text{indeg}(v)$ or $\\text{outdeg}(v)$ is less than 503, the other must be greater than 1507, so $|\\text{indeg}(v) - \\text{outdeg}(v)| \\geq 1006$, a contradiction. Thus, both are at least 503 for every $v$.\n\nLet $v_1$ and $v_2$ be two distinct vertices. Let $V_1$ be the set of vertices reachable from $v_1$, and $V_2$ the set of vertices from which $v_2$ can be reached. All edges with tail in $V_1$ must have head in $V_1$, so the sum of $\\text{outdeg}(v)$ for $v \\in V_1$ is $\\binom{|V_1|}{2}$. But this sum is at least $503|V_1|$, so $|V_1| \\geq 1007$. Similarly, $|V_2| \\geq 1007$. Since $1007 + 1007 > 2011$, $V_1$ and $V_2$ must intersect, so there is a path from $v_1$ to $v_2$.\n\nTherefore, the largest such $k$ is $1005$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15542,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integer solutions to the system of equations:\n\n$$\n\\begin{cases}\n  a^3 + b^3 = c^2 + d^2, \\\\\n  a^2 + b^2 = c^3 + d^3.\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove the auxiliary statement: for arbitrary numbers $a$ and $b$, at least one nonzero, $a^2 - ab + b^2 > 0$.\n\n**Case 1:** One of $a$ or $b$ is zero. Then $a^2 > 0$ (if $b=0$ and $a \\neq 0$) or $b^2 > 0$ (if $a=0$ and $b \\neq 0$), so the inequality holds.\n\n**Case 2:** One of $a$ or $b$ is positive, the other negative. Then each term in $a^2 + (-ab) + b^2$ is positive, so the sum is positive.\n\n**Case 3:** Both $a$ and $b$ are positive or both negative. Then $a^2 + b^2 \\ge 2ab > ab$.\n\nThus, the auxiliary statement is proven.\n\nNow, if $(a, b, c, d)$ is an integer solution to the system, then $a^3 + b^3 \\ge a^2 + b^2$ and $c^3 + d^3 \\ge c^2 + d^2$.\n\nSuppose $a^3 + b^3 = 0$. Then $c^2 + d^2 = 0 \\implies c = d = 0$, and $a^2 + b^2 = 0 \\implies a = b = 0$. So $a^3 + b^3 \\ge a^2 + b^2 = 0$.\n\nIf $a^3 + b^3 > 0$, then $a+b = \\dfrac{a^3 + b^3}{a^2 - ab + b^2} > 0$. Consider $a+b=1$ and $a+b \\ge 2$:\n\n- If $a+b=1$, one is positive, the other negative or zero. Then\n  $$\n  a^3 + b^3 = (a+b)(a^2 - ab + b^2) = a^2 - ab + b^2 \\ge a^2 + b^2,\n  $$\n  with equality only if $ab=0$, i.e., $(a, b) = (1, 0)$ or $(0, 1)$.\n\n- If $a+b \\ge 2$,\n  $$\n  a^3 + b^3 \\ge 2(a^2 - ab + b^2) = a^2 + b^2 + (a-b)^2 \\ge a^2 + b^2,\n  $$\n  with equality only if $a+b=2$ and $a-b=0$, i.e., $(a, b) = (1, 1)$.\n\nThus, $a^3 + b^3 \\ge a^2 + b^2$, with equality only for $(a, b) = (0, 0), (0, 1), (1, 0), (1, 1)$, and similarly for $(c, d)$.\n\nSo, the only possible solutions are those for which $a^3 + b^3 = a^2 + b^2$ and $c^3 + d^3 = c^2 + d^2$. There are $4 \\times 4 = 16$ possible quadruples $(a, b, c, d)$ to check for the original system.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15543,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $\\alpha = 67.5^\\circ$, $\\beta = 82.5^\\circ$, $\\gamma = 52.5^\\circ$ in the second case, consider the following system:\n\nLet $a = 1-x$, $b = 1-y$, $c = 1-z$, with $-1 \\le a, b, c \\le 1$. The system of equations is:\n\n$$\n\\begin{aligned}\n\\sqrt{(1-a)(1+b)} + \\sqrt{(1+a)(1-b)} &= 1 \\\\\n\\sqrt{(1-b)(1+c)} + \\sqrt{(1+b)(1-c)} &= \\sqrt{2} \\\\\n\\sqrt{(1-a)(1+c)} + \\sqrt{(1+a)(1-c)} &= \\sqrt{3}.\n\\end{aligned}\n$$\n\nFind the value of $(abc)^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "The condition $b = a + c$ implies\n\n$$\n(b-c)^2 = b^2 + c^2 - 2bc = a^2 = \\frac{1}{2},\n$$\n\nso $bc = \\frac{1}{4}$. Thus,\n\n$$\n(abc)^2 = (bc)^2 \\cdot a^2 = \\frac{1}{32}.\n$$\n\nThe conditions are satisfied by\n\n$$\n(a, b, c) = \\left( \\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{2\\sqrt{2}+1}}{2}, \\frac{1}{2\\sqrt{2\\sqrt{2}+1}} \\right).\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15544,
    "subject": "Mathematics (Olympiad)",
    "question": "Two players are playing the following game. First, Mykhailyk cuts the square $9 \\times 9$ into strips of unit width and arbitrary integer length. Next, Sashko chooses an integer $k \\in \\{1, 2, \\ldots, 9\\}$, makes a rectangle from all strips of the length $k$, and calculates its area. What is the largest possible area that Sashko can achieve no matter how Mykhailyk does the cutting?",
    "options": [],
    "answer": "See solution",
    "solution": "Assume for the contrary that for any choice of $k$, Sashko cannot obtain a rectangle with area $12$ or larger. This means that there is at most one strip of each of the lengths $9, 8, 7, 6$; at most two strips of each of the lengths $5$ and $4$; no more than three strips of length $3$; no more than five strips of length $2$; and no more than eleven unit squares. The total area then does not exceed\n\n$$\n9 + 8 + 7 + 6 + 2 \\cdot 5 + 2 \\cdot 4 + 3 \\cdot 3 + 5 \\cdot 2 + 11 = 78 < 81.\n$$\n\nThis contradiction shows that for at least one value of $k$, we can use rectangles $1 \\times k$ to make a rectangle of area $12$ or bigger. Now we need to construct a cutting of the $9 \\times 9$ square for which $12$ is the largest possible area Sashko can achieve. The following cutting illustrates such a case:\n\n$$\n1 \\cdot 9 + 1 \\cdot 8 + 1 \\cdot 7 + 2 \\cdot 6 + 2 \\cdot 5 + 1 \\cdot 4 + 4 \\cdot 3 + 6 \\cdot 2 + 7 \\cdot 1 = 81.\n$$\n\n![](images/Ukrajina_2013_p17_data_3e41d2e766.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15545,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of primes $p, q$ for which there exists a positive integer $a$ such that\n\n$$\n\\frac{pq}{p+q} = \\frac{a^2+1}{a+1}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, consider the case when $p$ and $q$ are distinct primes. Then $pq$ and $p+q$ are relatively prime: $pq$ is divisible only by $p$ and $q$, while $p+q$ is divisible by neither.\n\nSuppose $r$ is a common divisor of both $a+1$ and $a^2+1$. If $r \\mid a+1$ and $r \\mid a^2+1$, then $r \\mid (a+1)(a-1)$ and $r \\mid (a^2+1) - (a^2-1) = 2$, so $r$ can only be $1$ or $2$. Thus, $\\frac{a^2+1}{a+1}$ is either in lowest terms or reduces by $2$, depending on whether $a$ is even or odd.\n\n**Case 1:** $a$ is even.\n\nWe must have\n$$\npq = a^2 + 1 \\quad \\text{and} \\quad p+q = a+1.\n$$\nThe numbers $p, q$ are roots of $x^2 - (a+1)x + a^2 + 1 = 0$, whose discriminant is\n$$\n(a+1)^2 - 4(a^2+1) = -3a^2 + 2a - 3.\n$$\nThis is negative for all $a$, so there are no real solutions.\n\n**Case 2:** $a$ is odd.\n\nWe must have\n$$\n2pq = a^2 + 1 \\quad \\text{and} \\quad 2(p+q) = a+1.\n$$\nThe numbers $p, q$ are roots of $2x^2 - (a+1)x + a^2 + 1 = 0$, whose discriminant is also negative.\n\nTherefore, there is no pair of distinct primes $p, q$ satisfying the conditions.\n\nNow consider $p = q$:\n$$\n\\frac{p \\cdot p}{p + p} = \\frac{p}{2}.\n$$\nSo we require\n$$\np = \\frac{2(a^2 + 1)}{a + 1} = 2a - 2 + \\frac{4}{a + 1}.\n$$\nThis is an integer if and only if $a+1 \\mid 4$, i.e., $a \\in \\{1, 3\\}$, so $p = 2$ or $p = 5$.\n\n**Conclusion:**\n\nThere are exactly two pairs of primes satisfying the conditions: $p = q = 2$ and $p = q = 5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15546,
    "subject": "Mathematics (Olympiad)",
    "question": "In an arrangement of cubes, each edge is aligned in one of three directions, and for each direction, the number of edges is the same. To count the total number of visible edges, count the number of edges in a particular direction and multiply by 3. For example, consider the vertical edges:\n\n- In the top layer, there are 3 vertical edges.\n- In the second layer, there are 5.\n- In the third layer, there are 7.\n\nEach next layer is formed by copying the last layer and adding one extra cube, which adds two extra vertical edges. Thus, the sequence of vertical edges in the layers is $3, 5, 7, 9, \\ldots$, i.e., the sequence of odd integers, so the $n$th layer has $2n + 1$ vertical edges.\n\nGiven that $1 + 3 + 5 + \\dots + (2n - 1) = n^2$, compute:\n\n$$\n3 + 5 + 7 + \\dots + (2n + 1) = [1 + 3 + 5 + \\dots + (2(n + 1) - 1)] - 1 = (n + 1)^2 - 1.\n$$\n\nFind the total number of visible edges for:\n\n(a) $n = 3$\n\n(b) $n = 20$",
    "options": [],
    "answer": "See solution",
    "solution": "(a) For $n = 3$:\n\n$$\n3((3 + 1)^2 - 1) = 3(16 - 1) = 3 \\times 15 = 45\n$$\n\n(b) For $n = 20$:\n\n$$\n3((20 + 1)^2 - 1) = 3(441 - 1) = 3 \\times 440 = 1320\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15547,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a convex 2023-gon on the Cartesian plane with vertices at points whose coordinates are both integers, such that all its side lengths are equal?\n\nA polygon is called convex if all its diagonals lie inside the polygon.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** no.\n\nSuppose such a 2023-gon exists.\n\nLet its side be denoted by $a$, so $a^2$ is an integer, and its vertices as $(x_1, y_1)$, $(x_2, y_2)$, ..., $(x_{2023}, y_{2023})$. Consider the 2023-gon with the smallest value of $a^2$. We have $(x_i - x_{i+1})^2 + (y_i - y_{i+1})^2 = a^2$ for each $i$, where $x_{2024} = x_1$, $y_{2024} = y_1$.\n\nIf $a^2$ is a multiple of 4, then since if the sum of two squares of integers is a multiple of 4, then both numbers are even, we have $x_i \\equiv x_{i+1} \\pmod{2}$, $y_i \\equiv y_{i+1} \\pmod{2}$ for each $i$. But then we can consider a polygon with half the number of vertices with vertices at $(\\frac{x_i-x_1}{2}, \\frac{y_i-y_1}{2})$, whose vertices are also all integer points, and whose side length is $\\frac{a}{2}$, obtaining a contradiction.\n\nIf $a^2 \\equiv 2 \\pmod 4$, then $x_i$ and $x_{i+1}$ have different parity for each $i$, which leads to a contradiction, since we obtain that $x_1$ and $x_1$ have different parity.\n\nIf $a^2 \\equiv 1 \\pmod 2$, then $x_i + y_i$ and $x_{i+1} + y_{i+1}$ have different parity for each $i$, which leads to a contradiction, since we obtain that $x_1 + y_1$ and $x_1 + y_1$ have different parity.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15548,
    "subject": "Mathematics (Olympiad)",
    "question": "The number $x$ is $111$ when written in base $b$, but it is $212$ when written in base $b-2$. What is $x$ in base $10$?",
    "options": [],
    "answer": "See solution",
    "solution": "We have $x = b^2 + b + 1$ and $x = 2(b-2)^2 + (b-2) + 2 = 2(b^2 - 4b + 4) + (b - 2) + 2 = 2b^2 - 8b + 8 + b = 2b^2 - 7b + 8$.\n\nHence,\n$$\n0 = (2b^2 - 7b + 8) - (b^2 + b + 1) = b^2 - 8b + 7 = (b - 7)(b - 1)\n$$\nFrom the given information, $b - 2 > 2$, so $b = 7$ and $x = 49 + 7 + 1 = 57$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15549,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a given positive integer. Say that a set $K$ of points with integer coordinates in the plane is connected if for every pair of points $R, S \\in K$, there exist a positive integer $\\ell$ and a sequence $R = T_0, T_1, \\dots, T_\\ell = S$ of points in $K$, where each $T_i$ is distance $1$ away from $T_{i+1}$. For such a set $K$, we define the set of vectors\n$$\n\\Delta(K) = \\{ \\overrightarrow{RS} \\mid R, S \\in K \\}.\n$$\nWhat is the maximum value of $|\\Delta(K)|$ over all connected sets $K$ of $2n + 1$ points with integer coordinates in the plane?",
    "options": [],
    "answer": "See solution",
    "solution": "We claim the answer is $2n^2 + 4n + 1$. A model is\n$$\nK = \\{(0,0)\\} \\cup \\{(i,0) \\mid 1 \\le i \\le n\\} \\cup \\{(0,i) \\mid 1 \\le i \\le n\\},\n$$\nwhere\n$$\nW = \\{(a, -b) \\mid 0 \\le a, b \\le n\\} \\cup \\{(-a, b) \\mid 0 \\le a, b \\le n\\}.\n$$\nIt remains to prove that $|W| \\le 2n^2 + 4n + 1$ for any set $K$.\n\nThe problem describes a connected graph $G = (K, E)$ of order $2n + 1$, whose vertices are the points in $K$, and edges are the horizontal/vertical segments of length $1$ connecting these points. The key is to sequence the elements of $K$ as $A_0, A_1, \\dots, A_{2n}$ such that the graph $G_k := G[A_0, A_1, \\dots, A_k]$ is connected for every $1 \\le k \\le 2n$; this can be done as follows:\n\n**Lemma.** The vertices of a finite connected graph $G$ can always be enumerated as a sequence $v_0, \\dots, v_{|G|-1}$ so that $G_k := G[v_0, \\dots, v_k]$ is connected for every $1 \\le k \\le |G|-1$.\n\n*Proof.* Pick any vertex as $v_0$, and assume inductively that $v_0, \\dots, v_k$ have been chosen for some $0 \\le k < |G| - 1$. Now pick a vertex $v \\in G - G_k$. As $G$ is connected, it contains a $v - v_0$ path $P$. Choose as $v_{k+1}$ the last vertex of $P$ in $G - G_k$; then $v_{k+1}$ has as neighbor in $G_k$ the next vertex of $P$. The connectedness of every $G_k$ follows by induction on $k$. $\\square$\n\nIf we keep only the edges through which $A_{k+1}$ has the selected neighbor in $G_k$, then $G_k$ is a tree, and so $G_{2n}$ is a spanning tree of $G$. Call a vertex horizontal (vertical) if the edge that connects it is horizontal (vertical). Let $h$ and $v$ be the number of horizontal and vertical vertices, respectively; since $G_{2n}$ is a tree, $h + v = 2n$. The point $A_0$ contributes $2n + 1$ vectors $\\overrightarrow{A_0A_i}$. For $0 \\le k \\le 2n$, each point $A_{k+1}$ contributes at most $(2n+1) - x$ new vectors, where $x = h$ if the vertex is horizontal, $x = v$ if vertical, since those vectors $\\overrightarrow{A_{k+1}A_i}$ with ends at corresponding edges of the same direction will be duplicated by vectors determined by the opposite parallel sides of the parallelograms created, which have already been counted.\n\nTherefore, the total number of distinct vectors will be $|W| \\le (2n+1)^2 - h^2 - v^2$. But $h^2 + v^2 \\ge \\frac{1}{2}(h+v)^2 = 2n^2$, hence $|W| \\le (2n+1)^2 - 2n^2 = 2n^2 + 4n + 1$.\n\nThe same argument works in $d$-dimensional space, for a set of $dn + 1$ lattice points; then the maximal possible number of vectors is $(d^2 - d)n^2 + 2dn + 1$, with a model made by the origin $O$ and $n$ points at distances $1, 2, \\dots, n$ from the origin on each axis of the coordinate system.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15550,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with circumcircle $\\Omega$ and orthocenter $H$. Points $D$ and $E$ lie on segments $AB$ and $AC$ respectively, such that $AD = AE$. The lines through $B$ and $C$ parallel to $DE$ intersect $\\Omega$ again at $P$ and $Q$, respectively. Denote by $\\omega$ the circumcircle of $\\triangle ADE$.\n\n(a) Show that lines $PE$ and $QD$ meet on $\\omega$.\n\n(b) Prove that if $\\omega$ passes through $H$, then lines $PD$ and $QE$ meet on $\\omega$ as well.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $\\angle AQP = \\angle ABP = \\angle ADE$ and $\\angle APQ = \\angle ACQ = \\angle AED$, so we have a spiral similarity $\\triangle ADE \\sim \\triangle AQP$. Therefore, lines $PE$ and $QD$ meet at the second intersection of $\\omega$ and $\\Omega$ other than $A$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15551,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the sequence $\\{a_n\\}_{n=1}^\\infty$ defined as follows:\n\n$$\na_1 = 3, \\quad a_n = a_1 a_2 a_3 \\dots a_{n-1} - 1 \\quad \\text{for all } n \\ge 2.\n$$\n\nProve that there exist:\n\na) Infinitely many primes dividing at least one member of this sequence.\n\nb) Infinitely many primes dividing no member of this sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "a) By mathematical induction, we first prove that $a_n \\ge 2$ for every $n$. For $n=1$ and $n=2$ this is true because $a_1 = 3$ and $a_2 = 2$. Now suppose that for some $n \\ge 3$ the inequality $a_k \\ge 2$ holds for every $k < n$. Then we have $a_n = a_1 a_2 a_3 \\dots a_{n-1} - 1 \\ge a_1 a_2 - 1 = 5$, so indeed $a_n \\ge 2$.\n\nLet us now show that the numbers $a_n$ are pairwise coprime. Indeed, for any two indices $k < n$ we have $a_k \\mid a_1 a_2 \\dots a_{n-1} = a_n + 1$, whence for the greatest common divisor $D$ of the numbers $a_n$ and $a_k$ we get $D \\mid a_n$ and at the same time $D \\mid a_n + 1$ (because $D \\mid a_k$ and $a_k \\mid a_n + 1$), so necessarily $D = 1$, so $a_n$ and $a_k$ are coprime. Since $a_n \\ge 2$, for each index $n$ there is a prime number $p_n$ such that $p_n \\mid a_n$. Since all $a_n$ are pairwise coprime, the prime numbers $p_n$ are pairwise different. Thus, part a) is proved.\n\nb) If $n \\ge 2$, then $a_{n+1} = a_1 a_2 a_3 \\dots a_{n-1} a_n - 1 = (a_n + 1)a_n - 1 = a_n^2 + a_n - 1$. Next, we work with this expression.\n\nAssume that $p \\mid a_n$ for some $n \\ge 2$ and for some prime $p$. Then $a_{n+1} = a_n^2 + a_n - 1 \\equiv -1 \\pmod{p}$. For the next member, $a_{n+2} = a_{n+1}^2 + a_{n+1} - 1 \\equiv (-1)^2 + (-1) - 1 \\equiv -1 \\pmod{p}$, and so, by induction, all members $a_k$ with indices $k \\ge n+1$ give the same remainder $p-1$ modulo $p$. If the assumption $p \\mid a_n$ is satisfied for some $n \\ge 2$, we call the prime $p$ bad. Our task is to find infinitely many primes $p \\ge 5$ that are not bad (we impose $p \\ge 5$ so that $p \\mid a_1 = 3$ does not hold).\n\nNow consider a prime $p$ satisfying $a_n \\equiv 1 \\pmod{p}$ for some $n \\ge 2$. Then $a_{n+1} = a_n^2 + a_n - 1 \\equiv 1^2 + 1 - 1 \\equiv 1 \\pmod{p}$, so by induction, all numbers $a_k$ with $k \\ge n$ give a remainder 1 when divided by $p$. Let us call such $p$ good. No prime $p \\ge 5$ is both good and bad, because for sufficiently large $k$ both $a_k \\equiv 1 \\pmod{p}$ and $a_k \\equiv -1 \\pmod{p}$ cannot hold. Therefore, it is enough to prove that there are infinitely many good primes.\n\nTo find good primes, consider the sequence $\\{b_n\\}_{n=1}^\\infty$ given by $b_n = a_n - 1$ for each $n \\ge 1$. It is obvious that $b_1 = 2$, $b_2 = 1$, and\n\n$$\n\\begin{aligned}\nb_{n+1} &= a_{n+1} - 1 = (a_n^2 + a_n - 1) - 1 = ((b_n + 1)^2 + (b_n + 1) - 1) - 1 \\\\\n&= b_n^2 + 3b_n = b_n(b_n + 3)\n\\end{aligned}\n$$\n\nfor every $n \\ge 2$. Then a prime $p$ is good if and only if $p \\mid b_n$ for some $n \\ge 2$. We thus reach a situation similar to part a): we need to prove the existence of infinitely many primes dividing at least one member of the new sequence $\\{b_n\\}_{n=2}^\\infty$ determined by $b_2 = 1$ and $b_{n+1} = b_n(b_n + 3)$ for each $n \\ge 2$.\n\nObserve that $b_k \\mid b_n$ if $2 \\le k \\le n$. Indeed, from $b_{k+1} = b_k(b_k+3)$ we have $b_k \\mid b_{k+1}$ and, by induction, $b_k \\mid b_n$ for every $n \\ge k$.\n\nNow, we prove that, under the assumption $2 \\le k < n$, the numbers $b_k + 3$ and $b_n + 3$ are coprime. Their greatest common divisor $D$ satisfies $D \\mid b_k + 3 \\mid b_{k+1} \\mid b_n$ and $D \\mid b_n + 3$, so $D \\mid (b_n + 3) - b_n = 3$, and therefore $D = 1$ or $D = 3$. It remains to exclude $D = 3$: since $b_2 = 1$ and $b_{n+1} = b_n(b_n + 3)$, by induction $b_n \\equiv 1 \\pmod{3}$ for each $n \\ge 2$. So, $3 \\nmid b_n$, and therefore $3 \\nmid b_n + 3$, so $D \\ne 3$.\n\nFinally, $b_n \\ge 1$ for every $n$ (since $a_n \\ge 2$), and thus $b_n + 3 \\ge 4$. Therefore, for each $n$ we find a prime $p_n$ with $p_n \\mid b_n + 3$. All these primes $p_n$ are different from each other, and from $b_n + 3 \\mid b_{n+1}$ it follows $p_n \\mid b_{n+1}$ for every $n \\ge 2$. So we have found an infinite sequence of primes dividing at least one member of the sequence $\\{b_n\\}_{n=2}^\\infty$. The proof of part b) is complete.\n\n*Remark.* The statement from part a) can also be proved by contradiction. Suppose there are only finitely many primes dividing some members of the sequence $\\{a_n\\}_{n=1}^\\infty$, say $p_1, \\dots, p_k$. We can find an index $r$ so large that among the divisors of the first $r$ terms $a_1, \\dots, a_r$ are all primes $p_1, \\dots, p_k$. Then the next member $a_{r+1} = a_1 a_2 \\dots a_r - 1$ is not divisible by any of these primes, and since $a_{r+1} \\ge 2$, this is a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15552,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n > 1$ and an $n \\times n$ grid $ABCD$ containing $n^2$ unit squares, each unit square is colored by one of three colors: black, white, or gray. A coloring is called *symmetry* if each unit square on the diagonal $AC$ is colored gray and each pair of unit squares that are symmetric with respect to $AC$ are colored by the same color, black or white.\n\nWe label the number $0$ in each gray square, a positive integer in each white square, and a negative integer in each black square. For each positive integer $k$, a label will be called *$k$-balanced* if it satisfies all the following conditions:\n\n1. Each pair of unit squares that is symmetric with respect to $AC$ is labelled with the same integer from the closed interval $[-k, k]$.\n2. If a row and a column intersect at a black square, then the set of positive integers on that row and the set of positive integers on that column are disjoint. Similarly, if a row and a column intersect at a white square, then the set of negative integers on that row and the set of negative integers on that column are disjoint.\n3. For $n = 5$, find the minimum value of $k$ such that there exists a $k$-balanced label for the following grid.\n\n![](images/Vietnamese_mathematical_competitions_p161_data_b7b555f904.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a, b, c$ be the numbers on the cells at positions $(1, 2)$ and $(2, 1)$; $(3, 4)$ and $(4, 3)$; and $(4, 5)$ and $(5, 4)$. It is easy to check that all $a, b, c$ must be pairwise distinct, so $k \\geq 3$. We construct a way to fill in the table with $k = 3$ as follows:\n\n![](images/Vietnamese_mathematical_competitions_p161_data_b7b555f904.png)\n\nThus, the minimum value of $k$ in this case is $3$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15553,
    "subject": "Mathematics (Olympiad)",
    "question": "Give all integer solutions of the equation:\n\n$$\n3^{2a+1} b^2 + 1 = 2^c.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Case 1. $a \\ge 0$.\n\nClearly $c \\ge 0$ where $c=0$ implies $b=0$. We get that $(a,0,0)$ is a solution for any non-negative integer $a$. From the equality $3^{2a+1}b^2+1=2^c$ it follows that $b$ is an odd integer. We can write the left-hand side as:\n\n$$\n3^{2a+1} b^2 + 1 = (3^{2a+1} + 1)b^2 - (b-1)(b+1).\n$$\n\nFor the right-hand side, notice that $(b-1)(b+1)$ is divisible by $8$, while $(3^{2a+1}+1)b^2$ is divisible by $4$ but not by $8$. Therefore $2^c=4$, i.e., $c=2$. But then $3^{2a+1}b^2=3$, so $a=0$ and $b=\\pm 1$.\n\nCase 2. $a < 0$.\n\nAgain $c \\ge 0$ where $c=0$ implies $b=0$ and then $a$ can be any negative integer. Therefore, restrict to $c>0$. It is enough to consider $b>0$. Let $d=-a$, then the equation becomes:\n\n$$\n(2^c - 1)3^{2d-1} = b^2.\n$$\n\nwhere $b, c, d$ are natural numbers. Therefore $b$ is divisible by $3$, and $c$ is even. So $b=3^d x$, $c=2y$ for some natural numbers $x, y$. The equation becomes:\n\n$$\n4^{y-1} + 4^{y-2} + \\dots + 1 = x^2.\n$$\n\nThis implies $x=y=1$. For $y \\ge 2$, $x^2 \\equiv 5 \\pmod{8}$, which is impossible. Therefore, the only solutions are $(a,3^{-a},2)$, where $a$ is any negative integer.\n\nThe set $M$ of all solutions to the equation is:\n\n$$\nM = \\{(a,0,0) \\mid a \\in \\mathbb{Z}\\} \\cup \\{(a,\\pm 3^{-a},2) \\mid a \\in \\mathbb{Z} \\setminus \\{0\\}\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15554,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the sets\n$$\nA = \\{(x, y) \\mid x, y \\in \\mathbb{R} \\text{ and } x + y + 1 = 0\\}\n$$\nand\n$$\nB = \\{(x, y) \\mid x, y \\in \\mathbb{R} \\text{ and } x^3 + y^3 + 1 = 3xy\\}.\n$$\n\na) Show that $A \\subset B$.\n\nb) Prove that the set $B \\setminus A$ has exactly one element.",
    "options": [],
    "answer": "See solution",
    "solution": "a) If $(x, y) \\in A$, then $y = -x - 1$. We get\n$$\nx^3 + y^3 + 1 = x^3 + (-x - 1)^3 + 1 = x^3 - (x + 1)^3 + 1 = 3x(-x - 1) = 3xy,\n$$\nso $(x, y) \\in B$.\n\nb) If $(x, y) \\in B$, then $x^3 + y^3 - 3xy + 1 = 0$. Expanding,\n$$\n(x + y)^3 - 3xy(x + y) - 3xy + 1 = 0,\n$$\nor\n$$\n(x + y + 1)((x + y)^2 - (x + y) + 1) - 3xy(x + y + 1) = 0,\n$$\nor\n$$\n(x + y + 1)(x^2 - xy + y^2 - x - y + 1) = 0.\n$$\nTo have $(x, y) \\in B \\setminus A$, it is necessary that $x^2 - xy + y^2 - x - y + 1 = 0$, which leads to\n$$\n(x - y)^2 + (x - 1)^2 + (y - 1)^2 = 0,\n$$\nso $x = y = 1$. Since $(1, 1) \\notin A$, it follows that $B \\setminus A = \\{(1, 1)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15555,
    "subject": "Mathematics (Olympiad)",
    "question": "Considera el triángulo isósceles $ABC$ con base $BC$ diámetro de $\\Gamma$. El circuncentro $O$ del triángulo $ABC$ es un punto interior del segmento $AS$ y satisface que $AO = BO = CO$. Afirmamos que el lugar geométrico es la recta $\\ell$ perpendicular a $AS$ que pasa por $O$.\n\nConsideramos un triángulo $AB'C'$ arbitrario con $B'C'$ diámetro de $\\Gamma$ y sea $O'$ la intersección de la mediatriz de $B'C'$ con la recta $\\ell$, de modo que $O'B' = O'C'$.\n\n![](images/Soluciones_Nacional_2019_p2_data_c8d1821643.png)\n\n¿Demuestra que el circuncentro de cualquier triángulo $AB'C'$ construido así pertenece a $\\ell$ y que, recíprocamente, para cualquier punto $O'$ en $\\ell$ se puede construir un triángulo $AB'C'$ con circuncentro $O'$ y $B'C'$ diámetro de $\\Gamma$?",
    "options": [],
    "answer": "See solution",
    "solution": "Por el teorema de Pitágoras en el triángulo $C'O'S$ se tiene:\n\n$$\nO'B' = O'C' = \\sqrt{O'S^2 + r^2} = \\sqrt{OO'^2 + OS^2 + r^2}.\n$$\n\nPor otra parte, calculamos $O'A$:\n\n$$\nO'A = \\sqrt{AO^2 + OO'^2} = \\sqrt{BO^2 + OO'^2} = \\sqrt{OS^2 + r^2 + OO'^2}.\n$$\n\nPor lo tanto, $O'A = O'B' = O'C'$ y el punto $O'$ es el circuncentro del triángulo $AB'C'$ y por construcción pertenece a la recta $\\ell$.\n\nInversamente, para cualquier punto $O'$ de $\\ell$ es posible construir un diámetro $B'C'$ de $\\Gamma$, perpendicular a $O'S$. Por lo anterior, resulta que debe ser $O'A = O'B' = O'C'$, de modo que hemos hallado un triángulo $AB'C'$ con circuncentro $O'$ como queríamos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15556,
    "subject": "Mathematics (Olympiad)",
    "question": "A set of lines in the plane is in general position if no two of them are parallel and no three of them pass through the same point. A set of lines in general position cuts the plane into regions, some of which have finite area; we call them finite regions. Prove that for all sufficiently large $n$, in any set of $n$ lines in general position, it is possible to colour at least $\\sqrt{n}$ of the lines in blue such that none of its finite regions has a completely blue boundary.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $B$ be the maximal set of lines such that no finite region has completely blue boundary. Let $|B| = k$, and colour the other $n-k$ lines in red. For each red line $\\ell$, there is at least one finite region $A$ whose unique red side lies on $\\ell$. We say a point is red if it is the intersection point of red and blue lines. And we say a point is blue if it is the intersection point of two blue lines. Denote the vertices of $A$ clockwise by $(p_1, p_2, \\dots, p_s)$ with point $p_1$ in red and $p_2$ in blue. Then we say the red line $\\ell$ corresponds to the red point $p_1$ and blue point $p_2$. Now we are to show that any blue point can be corresponded to at most two red lines. (If this is true, then $n-k \\le 2\\binom{k}{2} \\Leftrightarrow n \\le k^2$.)\n\nWe prove it by contradiction. If not, suppose that there are three red lines $l_1, l_2$ and $l_3$ correspond to a blue point $b$, and the corresponding red points on red lines $l_1, l_2$ and $l_3$ are $r_1, r_2$ and $r_3$, respectively. Let $b$ be the intersection point of two blue lines. Without loss of generality, we may suppose that sides $(r_2, b)$ and $(r_3, b)$ are on one of the two blue lines, and $(r_1, b)$ is the side of region $A$ on the other blue line. Since $A$ has only one red side, $A$ must be a triangle $\\triangle r_1 b r_2$. But then red lines $l_1$ and $l_2$ pass $r_2$, and a blue line also passes $r_2$, which is a contradiction. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15557,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle, and let $D$ be the foot of the $A$-altitude. Points $P$ and $Q$ are chosen on $BC$ such that $DP = DQ = DA$. Suppose $AP$ and $AQ$ intersect $(ABC)$ again at $X$ and $Y$. Prove that the perpendicular bisectors of the lines $PX$, $QY$, and $BC$ intersect.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of $B$ and $C$, and let $P'$ be the reflection of $P$ in $M$. Let $U$ be the midpoint of $DP'$. Similarly, define $Q'$ and $V$ respectively. Let $X_B$ be the intersection of the perpendicular from $P'$ to $BC$ and the perpendicular from $X$ to $AP$. Similarly, define $X_C$. Finally, let $O$ be the circumcenter of $\\triangle ABC$.\n\n*Claim.* $X_B X_C \\parallel BC$.\n\n*Proof.* We will show that $X'_B X'_C \\parallel BC$, where $X'_B$ is the midpoint of $X_B$ and $A$, and $X'_C$ is defined similarly. Observe that $X'_B$ is the intersection of the perpendicular from $O$ to $AP$ and the perpendicular from $U$ to $BC$. $X'_C$ is defined similarly. Let $S$ and $T$ be the intersections from $O$ to $AP$ and $AQ$ respectively with $BC$.\n\nWe will show that $OU = OV$ and $\\angle OST = \\angle OTS$. This would imply that $X'_B$ and $X'_C$ are reflections of one another over $OM$ and thus $X'_B X'_C \\parallel BC$.\n\nObserve that $\\angle OST = 90 - \\angle APQ = 90 - \\angle AQP = \\angle OTS$.\n\nTo show that $OU = OV$, we just need $MU = MV$. For this, treat all points as vectors.\n\n$$\nV = \\frac{D + P'}{2} = \\frac{D + B + C - P}{2}, \\quad U = \\frac{D + Q'}{2} = \\frac{D + B + C - Q}{2}\n$$\n\nNow,\n\n$$\nU + V - B - C = 2D - P - Q = 0\n$$\n\nas $D$ is the midpoint of $PQ$, and we get that $BC$ and $UV$ have the same midpoint. Thus, the claim follows. $\\square$\n\n![](images/IMO_TSTs_India_2023_p1_solns_p0_data_545fe5bdd5.png)\n\nNow, consider the midpoint of $PX_B$ and the midpoint of $QX_C$; we claim they coincide. Observe that they both lie on $OM$ and are at equal distances from $M$, so they coincide. Thus, we get the desired result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15558,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a positive integer. Determine the maximum value of the expression\n$$\na^{3k-1}b + b^{3k-1}c + c^{3k-1}a + k^2 a^k b^k c^k\n$$\nfor all nonnegative real numbers $a, b, c$ satisfying $a + b + c = 3k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $F(a, b, c) = a^{3k-1}b + b^{3k-1}c + c^{3k-1}a + k^2 a^k b^k c^k$.\n\nSince the expression is cyclic in $a, b, c$, we may assume that $a$ lies between $b$ and $c$. Then $(a-b)(a-c) \\le 0$ and we have\n$$\na^{3k-1} + b^{3k-2}c - (a^{3k-2}c + b^{3k-2}a) = (a^{3k-2} - b^{3k-2})(a-c) \\le 0.\n$$\nHence\n$$\nF(a, b, c) \\le c^{3k-1}a + k^2 a^k b^k c^k + b(a^{3k-2}c + b^{3k-2}a) = a(c^{3k-1} + k^2 a^{k-1} b^k c^k + b c a^{3k-3} + b^{3k-1}).\n$$\nIf $k > 1$ we also have\n$$\n(b+c)^{3k-1} \\ge b^{3k-1} + c^{3k-1} + \\binom{3k-1}{1} b c (c^{3k-3} + b^{3k-3}) + \\binom{3k-1}{k} b^k c^k (c^{k-1} + b^{k-1}) \\\\\n\\ge b^{3k-1} + c^{3k-1} + (3k-1) b c a^{3k-3} + \\binom{3k-1}{k} b^k c^k a^{k-1}\n$$\nwhere we used the fact that $b^s + c^s \\ge a^s$ for $a$ lying between $b$ and $c$. Since\n$$\n\\binom{3k-1}{k} = \\frac{(3k-1)(3k-2)(3k-3)}{k(2k-1)(2k-2)} \\binom{3k-4}{k-1} \\ge \\frac{k^2}{(k-1)^2} \\binom{3k-4}{k-1}\n$$\nfor $k \\ge 2$, it follows by induction that\n$$\n\\binom{3k-1}{k} \\ge k^2\n$$\nand hence we have\n$$\n(b+c)^{3k-1} \\ge b^{3k-1} + c^{3k-1} + b c a^{3k-3} + k^2 b^k c^k a^{k-1}\n$$\nfor $k \\ge 2$. We note that this inequality also holds for $k = 1$.\n\nFinally, using the AM-GM inequality we obtain\n$$\nF(a, b, c) \\le a(b+c)^{3k-1} \\le \\frac{1}{3k-1} \\left( \\frac{(3k-1)a + (3k-1)(b+c)}{3k} \\right)^{3k} = (3k-1)^{3k-1}\n$$\nfor all nonnegative real numbers $a, b, c$ satisfying $a + b + c = 3k$. Since $F(1, 0, 3k-1) = (3k-1)^{3k-1}$, this is the largest possible value.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15559,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are 1000 dust grains inside a $14\\text{cm} \\times 14\\text{cm} \\times 14\\text{cm}$ box. Is it possible that for every point in the box, there are fewer than 10 dust grains at a distance between 1cm and 2cm from that point?",
    "options": [],
    "answer": "See solution",
    "solution": "Carla is right. Take each dust grain and colour all points at a distance of at most 2cm and at least 1cm from the grain. Then we have coloured a volume of $$1000 \\cdot \\frac{4}{3} \\pi (2^3 - 1^3) = \\frac{28000}{3}\\pi\\text{cm}^3 > 28000\\text{cm}^3$$ counted with multiplicity. All the coloured points are contained in a $14\\text{cm} \\times 14\\text{cm} \\times 14\\text{cm}$ box of volume $14^3 = 2744\\text{cm}^3$. Hence, there is a point that is coloured at least 10 times, and then there are at least 10 points at a distance of at most 2cm and at least 1cm from this point.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15560,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integer polynomials $f(x)$ such that for every integer $n \\ge 2$, the number $n^{f(n)} - 1$ is divisible by $f(n)$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that $f(x) = (x-1)^m$ for $m = 0, 1, 2, \\dots$ are the only solutions.\n\nFor $n \\ge 2$, $f(n) = (n-1)^m > 0$, so the divisibility condition is satisfied.\n\n**Lemma 1.** Let $d, t, x$ be positive integers. If $x-1$ is a multiple of $d^t$, then $x^d-1$ is a multiple of $d^{t+1}$.\n\n**Proof.** Since $x \\equiv 1 \\pmod d$, we have $x^{d-1} + \\dots + x + 1 \\equiv 1 + \\dots + 1 \\equiv 0 \\pmod d$. Thus $x^d - 1 = (x-1)(x^{d-1} + \\dots + x + 1)$ is a multiple of $d^{t+1}$ as desired. $\\blacksquare$\n\nBy induction using Lemma 1, $n^{(n-1)m} - 1$ is a multiple of $(n-1)^{m+1}$ for $m \\ge 0$.\n\nNext, we show that no other polynomials work. For integers $a, b$ with $b \\ne 0$, write $b \\mid a$ to mean $a$ is divisible by $b$.\n\n**Lemma 2.** For a positive integer $n$ and integers $a$ and $b$, $a \\equiv b \\pmod n$ implies $f(a) \\equiv f(b) \\pmod n$.\n\n**Proof.** Write $f(x) = a_0 + a_1x + \\dots + a_d x^d$. Since $n \\mid a-b$,\n\n$$\nf(a) - f(b) = \\sum_{i=1}^{d} a_i (a^i - b^i) = (a-b) \\sum_{i=1}^{d} a_i (a^{i-1} + a^{i-2}b + \\dots + ab^{i-2} + b^{i-1})\n$$\n\nis divisible by $n$. $\\blacksquare$\n\nFor an odd prime $p$ and integer $n \\ge 2$, $p \\mid f(n)$ implies $p \\mid n-1$. For $k \\ge 0$, Lemma 2 shows $f(n + kp) \\equiv f(n) \\equiv 0 \\pmod p$. Thus,\n\n$$\n0 \\equiv (n + kp)^{f(n+kp)} - 1 \\equiv n^{f(n+kp)} - 1 \\pmod p\n$$\n\nSince $p$ and $p-1$ are coprime, we can choose $k$ so $p-1 \\mid n + kp$. Then $f(n + kp) \\equiv f(p-1) \\pmod{p-1}$, and by Fermat's little theorem,\n\n$$\n0 \\equiv n^{f(p-1)} - 1 \\pmod{p}\n$$\n\nSo $n^{f(p-1)} \\equiv 1 \\pmod{p}$. Since $f(p-1)$ and $p-1$ are coprime, there exists $r$ such that $f(p-1)r \\equiv 1 \\pmod{p-1}$. By Fermat's theorem,\n\n$$\nn \\equiv n^{f(p-1)r} \\equiv 1^r \\equiv 1 \\pmod{p}\n$$\n\nThus $p \\mid n-1$.\n\nTherefore, for an odd prime $p$, any prime factor of $f(p+1)$ is also a prime factor of $p$, so $f(p+1)$ is a power of $p$.\n\nWrite $f(x+1) = x^m(b_0 + b_1x + \\dots + b_lx^l)$ with $b_0 \\ne 0$. For $p > |b_0|$, $f(p+1)$ is a multiple of $p^m$, while $f(p+1) \\equiv b_0p^m \\ne 0 \\pmod{p^{m+1}}$. Thus $f(p+1) = p^m$.\n\nLet $g(x) = f(x) - (x-1)^m$. Then $g(n) = 0$ for infinitely many $n$, so $g(x) = 0$ as polynomials. Thus, $f(x) = (x-1)^m$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15561,
    "subject": "Mathematics (Olympiad)",
    "question": "平面上有 $\\triangle ABC$ 及一點 $O$，$\\Gamma$ 為 $\\triangle ABC$ 的外接圓。設直線 $CO$ 與直線 $AB$ 交於點 $D$，直線 $BO$ 與直線 $CA$ 交於點 $E$。設直線 $AO$ 與 $\\Gamma$ 再交於點 $F$。令點 $I$ 為 $\\Gamma$ 和 $\\triangle ADE$ 外接圓的另一個交點，點 $Y$ 為直線 $BE$ 與 $\\triangle CEI$ 外接圓的另一個交點，而點 $Z$ 為直線 $CD$ 與 $\\triangle BDI$ 外接圓的另一個交點。在 $\\Gamma$ 上分別作以 $B, C$ 為切點的兩條切線，設它們相交於點 $T$。設直線 $TF$ 與 $\\Gamma$ 再交於 $U$ 點，而令 $U$ 對直線 $BC$ 的對稱點為 $G$。\n\n試證：$F, I, G, O, Y, Z$ 六點共圓。\n\n![](images/18-2J_p4_data_1e4a212629.png)",
    "options": [],
    "answer": "See solution",
    "solution": "設 $BC$ 與 $ED$ 交於點 $X$。因為 $I$ 是 $BCED$ 的密克點，故 $X$ 在 $\\odot(BDI)$、$\\odot(CEI)$ 上。因為 $BFCU$ 為調和四邊形，故\n\n$$\nA(B, C; F, U) = -1 = A(B, C; O, X),\n$$\n\n從而 $X \\in AU$。因\n\n$$\n\\angle BYX = \\angle ACB = \\angle XUB = \\angle BGX,\n$$\n\n故 $X \\in \\odot(BGY)$。同理可得 $X$ 在 $\\odot(CGZ)$ 上，故\n\n$$\n\\angle YGZ = \\angle YGX + \\angle XGZ = \\angle YBX + \\angle XCZ = \\angle YOZ,\n$$\n\n即 $O, G, Y, Z$ 共圓。另一方面，由\n\n$$\n\\angle IYO = \\angle ICA = \\angle IFO,\n$$\n\n得 $Y \\in \\odot(FIO)$。同理可得 $Z \\in \\odot(FIO)$，故 $F, I, O, G, Y, Z$ 共圓，證明完畢。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15562,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_{21}$ be a permutation of $1, 2, \\dots, 21$, satisfying\n$$\n|a_{20} - a_{21}| \\geq |a_{19} - a_{21}| \\geq |a_{18} - a_{21}| \\geq \\dots \\geq |a_1 - a_{21}|.\n$$\nThe number of such permutations is ________.",
    "options": [],
    "answer": "See solution",
    "solution": "For a given $k \\in \\{1, 2, \\dots, 21\\}$, consider the number of permutations $N_k$ that satisfy the conditions such that $a_{21} = k$.\n\nWhen $k \\in \\{1, 2, \\dots, 11\\}$, for $i = 1, 2, \\dots, k-1$, there exist $a_{2i-1}, a_{2i}$ that are permutations of $k-i, k+i$ (if $k=1$, there exists no such $i$), and $a_{2j} = j+1$ ($2k-1 \\leq j \\leq 20$) (if $k=11$, there exists no such $j$), so $N_k = 2^{k-1}$.\n\nSimilarly, when $k \\in \\{12, 13, \\dots, 21\\}$, there is $N_k = 2^{21-k}$.\n\nTherefore, the number of such permutations satisfying the condition is\n$$\n\\sum_{k=1}^{21} N_k = \\sum_{k=1}^{11} 2^{k-1} + \\sum_{k=12}^{21} 2^{21-k} = (2^{11}-1) + (2^{10}-1) = 3070. \\quad \\square\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 15563,
    "subject": "Mathematics (Olympiad)",
    "question": "One of the numbers $1$, $2$, or $3$ is written in each cell of a rectangular table with $4$ rows and $n$ columns. For every three different columns, there is a row that intersects them at cells with different numbers. Find the maximum $n$ for which there exists such a table.",
    "options": [],
    "answer": "See solution",
    "solution": "The maximum $n$ is $9$.\n\nSuppose there is such a table $T$ with $n \\geq 10$ columns. Let $3$ be the least represented number in row $4$. Then $1$ and $2$ combined occur at least $7$ times in row $4$, so we can select $7$ columns whose intersections with row $4$ are only $1$s and $2$s. Delete the remaining columns. The new table ($4 \\times 7$) has the same property as the original: every three distinct columns intersect some row at three different numbers. Moreover, such a row must be $1$, $2$, or $3$, but row $4$ intersects the $7$ columns only at cells with $1$s and $2$s. Hence, deleting row $4$ yields a $3 \\times 7$ table $T_1$ which is also admissible.\n\nApply the same reasoning to $T_1$. The two most represented numbers in row $3$ occur at least $5$ times, so we can reduce $T_1$ to an admissible table ($3 \\times 5$) with no more than two different numbers in row $3$. Thus, every triple of columns in it is intersected at three different numbers by row $1$ or row $2$. Thus, row $3$ can be deleted, leading to an admissible $2 \\times 5$ table $T_2$.\n\nThe two most represented numbers in row $2$ of $T_2$ occur at least $4$ times, so $T_2$ can be reduced to an admissible table ($2 \\times 4$) with at most two different numbers in row $2$. Then every three of its $4$ columns must be intersected by row $1$ at three different numbers. This is impossible since $1$, $2$, or $3$ occurs twice in row $1$. The desired contradiction follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15564,
    "subject": "Mathematics (Olympiad)",
    "question": "A math competition is held in 8 different levels of difficulty. The organizing committee has to prepare 5 problems for each level. The same problem can be used for more than one level, but each pair of levels can have at most one common problem. What is the least number of problems that is sufficient for the organizers?",
    "options": [],
    "answer": "See solution",
    "solution": "18 problems are enough. The following table shows how to arrange problems for 8 levels:\n\n![](<table><thead><tr><th>Level 1</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th></tr></thead><tbody><tr><th>Level 2</th><td>1</td><td>6</td><td>7</td><td>8</td><td>9</td></tr><tr><th>Level 3</th><td>2</td><td>6</td><td>10</td><td>11</td><td>12</td></tr><tr><th>Level 4</th><td>3</td><td>7</td><td>10</td><td>13</td><td>14</td></tr><tr><th>Level 5</th><td>4</td><td>8</td><td>11</td><td>13</td><td>15</td></tr><tr><th>Level 6</th><td>5</td><td>9</td><td>12</td><td>14</td><td>15</td></tr><tr><th>Level 7</th><td>1</td><td>10</td><td>15</td><td>16</td><td>17</td></tr><tr><th>Level 8</th><td>2</td><td>8</td><td>14</td><td>16</td><td>18</td></tr></tbody></table>)\n\nFurther, we show that 18 is indeed the smallest possible number of problems that is sufficient. Denote by $a_i$ the number of problems that are common for $i$ levels. As there are in total 40 problems, then\n\n$$\na_1 + 2a_2 + 3a_3 + 4a_4 + 5a_5 + 6a_6 + 7a_7 + 8a_8 = 40 \\quad (1)\n$$\n\nIf we consider all the pairs of these 40 problems, then at most $\\frac{8 \\cdot 7}{2} = 28$ of them can be equal. Each problem that is common for $i$ levels defines $\\binom{i}{2}$ such pairs, therefore\n\n$$\n\\binom{2}{2}a_2 + \\binom{3}{2}a_3 + \\binom{4}{2}a_4 + \\binom{5}{2}a_5 + \\binom{6}{2}a_6 + \\binom{7}{2}a_7 + \\binom{8}{2}a_8 \\le 28 \\quad (2)\n$$\n\nWe must prove that $a_1 + a_2 + \\dots + a_8 \\ge 18$, which given (1) is equivalent to\n\n$$\na_2 + 2a_3 + 3a_4 + 4a_5 + 5a_6 + 6a_7 + 7a_8 \\le 22 \\quad (3)\n$$\n\nFrom (1) we can also get that\n\n$$\n2a_2 + 3a_3 + 4a_4 + 5a_5 + 6a_6 + 7a_7 + 8a_8 \\le 40 \\quad (4)\n$$\n\nBy adding (4) and (2) and dividing the result by 3 we obtain\n\n$$\na_2 + 2a_3 + \\frac{10}{3}a_4 + 5a_5 + 7a_6 + \\frac{28}{3}a_7 + 12a_8 \\le 22\\frac{2}{3} \\quad (5)\n$$\n\n(3) then is a trivial consequence of (5) (coefficients for $a_i$ in (5) are greater or equal than those in (3) and the result for the expression in (3) has to be an integer).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15565,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $x, y$ such that $x^{2012} = y^x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = p_1^{a_1} p_2^{a_2} \\cdots$ with $p_1, p_2, \\ldots$ prime and $p_i < p_{i+1}$.\n\nThen $y = (p_1^{a_1} p_2^{a_2} \\cdots)^b$, where $b = \\frac{2012}{p_1^{a_1} p_2^{a_2} \\cdots}$.\n\nSince $y$ is an integer, $p_1^{a_1} p_2^{a_2} \\cdots \\mid 2012 a_1$, $p_1^{a_1} p_2^{a_2} \\cdots \\mid 2012 a_2$, etc.\n\nThe only prime factors of $2012$ are $2$ and $503$.\n\nIf $p_1$ is not $2$ or $503$, then $p_1^{a_1} \\mid a_1$, which is impossible since $p_1^{a_1} > a_1$.\n\nSo $p_1 = 2$. Similarly, $p_2 = 503$ and there are no other prime factors of $x$.\n\nHence $x = 2^{a_1} 503^{a_2}$ and $2^{a_1} 503^{a_2} \\mid 2012 a_1$ and $2^{a_1} 503^{a_2} \\mid 2012 a_2$.\n\nSo $2^{a_1} \\mid 4 a_1$ and $503^{a_2} \\mid 503 a_2$.\n\nHence $a_1 = 0, 1, 2$, or $4$ and $a_2 = 0$ or $1$.\n\n| $a_1$ | $a_2$ | $x$    | $y$                |\n|-------|-------|--------|--------------------|\n| 0     | 0     | 1      | 1                  |\n| 0     | 1     | 503    | $503^4$            |\n| 1     | 0     | 2      | $2^{1006}$         |\n| 1     | 1     | 1006   | $1006^2$           |\n| 2     | 0     | 4      | $2^{1006}$         |\n| 2     | 1     | 2012   | 2012               |\n| 4     | 0     | 16     | $2^{503}$          |\n\nThus, there are no more than these 7 pairs $(x, y)$ that can be solutions.\n\nBack-substitution into the given equation shows that these 7 pairs are solutions, indeed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15566,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that if\n$$\n(n^2 + 1)^{2k} \\cdot (44n^3 + 11n^2 + 10n + 2) = N^m\n$$\nholds for some non-negative integer values of $m$, $n$, $N$, and $k$, then $m = 1$ must hold.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the left side of the equation is certainly larger than 1, we first note that $m > 0$ must certainly hold.\n\nNow, we consider even values of $n$. Since $n^2+1 \\equiv 1 \\pmod{4}$ and $44n^3+11n^2+10n+2 \\equiv 2 \\pmod{4}$, we have $N^m \\equiv 2 \\pmod{4}$. If $m > 1$, $N^m$ is odd for any odd $N$ and divisible by 4 for any even $N$, so $m = 1$ must hold, as claimed.\n\nNext, we consider odd values of $n$. In this case, $44n^3 + 11n^2 + 10n + 2 \\equiv 3 \\pmod{4}$ and $n^2 + 1 \\equiv 2 \\pmod{4}$, so the factor 2 is contained in $N^m$ exactly $2^k$ times.\n\nFor $k=0$, we obtain $N^m = (n^2+1)(44n^3+11n^2+10n+2) \\equiv 2 \\pmod{4}$, and the same argument holds as for even values of $n$.\n\nFor $k > 0$, the exponent $m > 1$ must be a divisor of the exponent $k$ of 2 in the prime decomposition of $N^m$, and therefore a power of 2. This means that $(n^2 + 1)^{2k}$ is an $m$-th power, so $44n^3 + 11n^2 + 10n + 2$ must also be a perfect square. However, this number is $\\equiv 3 \\pmod{4}$, and therefore cannot be a perfect square. Thus, this case is not possible, and we see that $m = 1$ must hold, as claimed. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15567,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $(x, y)$ of positive integers such that\n\n$$\n\\frac{1}{x^2} + \\frac{249}{xy} + \\frac{1}{y^2} = \\frac{1}{2012}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\gcd(x, y) = d$ and $x = a d$, $y = b d$. Then the equation can be written as\n$$\n\\frac{a^2 + 249ab + b^2}{a^2 b^2 d^2} = \\frac{1}{2012}\n$$\nor\n$$\na^2 b^2 d^2 = 2012(a^2 + 249ab + b^2).\n$$\n\nAs $a$ and $b$ are relatively prime, $a^2$ and $b^2$ are both relatively prime to $a^2 + 249ab + b^2$, and therefore both must be divisors of $2012$. Since $2012 = 2^2 \\cdot 503$ and $503$ is prime, the possible cases are $(a, b) = (1, 1)$, $(a, b) = (1, 2)$, $(a, b) = (2, 1)$. \n\nIf we substitute $(a, b) = (1, 1)$ into the last equation, we get $d^2 = 2012 \\cdot 251$, which is not solvable in integers. The other two cases give $4d^2 = 2012 \\cdot 503$, from which $d = 503$. This leads to the solutions $(x, y) = (503, 1006)$ and $(x, y) = (1006, 503)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15568,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ be a positive integer. We have $2^m$ sheets of paper, with the number $1$ written on each of them. We perform the following operation: in every step, we choose two distinct sheets; if the numbers on the two sheets are $a$ and $b$, then we erase these numbers and write the number $a + b$ on both sheets. Prove that after $m 2^{m-1}$ steps, the sum of the numbers on all the sheets is at least $4^m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P_k$ denote the product of all numbers on the sheets after $k$ steps. In each step, we take two sheets with numbers $a$ and $b$ and replace them by sheets with $a + b$ and $a + b$, so for the products after $k$ steps and after $k + 1$ steps we have\n\n$$\nP_{k+1} = \\frac{P_k}{ab} \\cdot (a + b)^2.\n$$\n\nIt follows that\n\n$$\n\\frac{P_{k+1}}{P_k} = \\frac{(a + b)^2}{ab} \\geq 4.\n$$\n\nSince $P_0 = 1$, we have $P_k \\geq 4^k$ and in particular $P_{m 2^{m-1}} \\geq 4^{m 2^{m-1}}$. By the AM-GM inequality applied to the $2^m$ numbers written on the sheets after $m 2^{m-1}$ steps, we have\n\n$$\nS_{m 2^{m-1}} \\geq 2^m \\cdot \\sqrt[2^m]{P_{m 2^{m-1}}} \\geq 2^m \\cdot (4^{m 2^{m-1}})^{1/2^m} = 2^m \\cdot 4^{m/2} = 4^m.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15569,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c$ be positive real numbers such that $a + b + c = 1$. Prove that:\n\n$$\n(a + 1)\\sqrt{2a(1 - a)} + (b + 1)\\sqrt{2b(1 - b)} + (c + 1)\\sqrt{2c(1 - c)} \\geq 8(ab + bc + ca).\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $a + b + c = 1$, we have $a + 1 = 2a + b + c$ and $1 - a = b + c$. Thus, the inequality becomes:\n\n$$\n(2a + b + c)\\sqrt{2a(b + c)} + (2b + c + a)\\sqrt{2b(c + a)} + (2c + a + b)\\sqrt{2c(a + b)} \\geq 8(ab + bc + ca)\n$$\n\nBy the arithmetic-geometric mean inequality, $2a + b + c \\geq 2\\sqrt{2a(b + c)}$, so\n\n$$\n(2a + b + c)\\sqrt{2a(b + c)} \\geq 2(2a(b + c)) \\quad (1)\n$$\n\nSimilarly,\n\n$$\n(2b + c + a)\\sqrt{2b(c + a)} \\geq 2(2b(c + a)) \\quad (2)\n$$\n\n$$\n(2c + a + b)\\sqrt{2c(a + b)} \\geq 2(2c(a + b)) \\quad (3)\n$$\n\nAdding (1), (2), and (3), we obtain the desired inequality. Equality holds if and only if\n\n$$\n2a = b + c, \\quad 2b = c + a, \\quad 2c = a + b \\implies a = b = c = \\frac{1}{3}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15570,
    "subject": "Mathematics (Olympiad)",
    "question": "Sea $n \\ge 1$ un número entero. Probar que\n$$\n\\sum_{0 \\le k < n/2} \\binom{n}{2k+1} 13^k\n$$\nes divisible por $2^{n-1}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Sea $a_n = \\sum_{0 \\le k < n/2} \\binom{n}{2k+1} 13^k$. Teniendo en cuenta que $\\binom{n}{j} = 0$ para $j < 0$, $j > n$ y la relación\n\n$$\n\\binom{n}{j} + \\binom{n}{j+1} = \\binom{n+1}{j+1}\n$$\n\nresulta $a_1 = 1$, $a_2 = 2$ y para $n \\ge 3$,\n\n$$\n\\begin{align*}\na_n - a_{n-1} &= \\sum_{0 \\le k < n/2} \\binom{n}{2k+1} 13^k - \\sum_{0 \\le k < n/2} \\binom{n-1}{2k+1} 13^k \\\\\n&= \\sum_{0 \\le k < n/2} \\binom{n-1}{2k} 13^k = \\sum_{0 \\le k < n/2} \\binom{n-2}{2k} 13^k + \\sum_{0 \\le k < n/2} \\binom{n-2}{2k-1} 13^k\n\\end{align*}\n$$\n\ny\n\n$$\na_{n-1} = \\sum_{0 \\le k < n/2} \\binom{n-1}{2k+1} 13^k = \\sum_{0 \\le k < n/2} \\binom{n-2}{2k+1} 13^k + \\sum_{0 \\le k < n/2} \\binom{n-2}{2k} 13^k.\n$$\n\nEntonces,\n\n$$\n\\begin{align*}\na_n - 2a_{n-1} &= \\sum_{0 \\le k < n/2} \\binom{n-2}{2k-1} 13^k - \\sum_{0 \\le k < n/2} \\binom{n-2}{2k+1} 13^k \\\\\n&= 13 \\sum_{0 \\le k < n/2} \\binom{n-2}{2k-1} 13^{k-1} - \\sum_{0 \\le k < n/2} \\binom{n-2}{2k+1} 13^k \\\\\n&= 12 \\sum_{0 \\le k < n/2} \\binom{n-2}{2k+1} 13^k = 12a_{n-2}.\n\\end{align*}\n$$\n\nAhora, veremos que $a_n = 2^{n-1}b_n$ siendo $b_n$ un número entero apropiado. En efecto, lo demostraremos por inducción. Para $a_1 = 1$, $a_2 = 2$ el resultado se verifica con $b_1 = b_2 = 1$. Para $n \\ge 3$, utilizando que $a_{n-1} = 2^{n-2}b_{n-1}$ y $a_{n-2} = 2^{n-3}b_{n-2}$ se obtiene\n\n$$\na_n = 2a_{n-1} + 12a_{n-2} = 2 \\cdot 2^{n-2}b_{n-1} + 12 \\cdot 2^{n-3}b_{n-2} = 2^{n-1}(b_{n-1} + 3b_{n-2})\n$$\n\ncomo queríamos. Así,\n\n$$\n2^{n-1} \\mid \\sum_{0 \\le k < n/2} \\binom{n}{2k+1} 13^k,\n$$\n\ny hemos terminado.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15571,
    "subject": "Mathematics (Olympiad)",
    "question": "The non-negative real numbers $x, y, z$ satisfy $$(x + y)(y + z)(z + x) = 1.$$ Let $m$ and $M$ be the smallest and largest possible values of the expression $$A = (xy + yz + zx)(x + y + z).$$\n\na) Find $m$ and $M$.\n\nb) Is there a triple of non-negative rational numbers $(x, y, z)$ satisfying the given equality for which $A = m$?",
    "options": [],
    "answer": "See solution",
    "solution": "a) The inequality $$(xy + yz + zx)(x + y + z) \\ge 1 = (x + y)(y + z)(z + x)$$ is equivalent to $xyz \\ge 0$. Equality is reached only when one of the variables, say $x$, is $0$ and the other two (in this case $y$ and $z$) satisfy $yz(y + z) = 1$; one possibility is $y = 1$ and $z = \\frac{\\sqrt{5} - 1}{2}$.\n\nThe inequality $$(xy + yz + zx)(x + y + z) \\le \\frac{9}{8}$$ is equivalent to $$9(x + y)(y + z)(z + x) - 8(xy + yz + zx)(x + y + z) \\ge 0,$$ i.e., $x^2y + xy^2 + y^2z + yz^2 + x^2z + xz^2 \\ge 6xyz$. The latter is true by the inequality between the arithmetic mean and the geometric mean applied to the six terms on the left, with equality only at $x = y = z$ and $8x^3 = 1$, i.e., $x = y = z = \\frac{1}{2}$.\n\nb) Given the reasoning in a), it suffices to prove that the equation $yz(y + z) = 1$ has no solution in positive rational numbers. Assume the opposite and let $y = \\frac{p}{r}$, $z = \\frac{q}{r}$ be a solution (with natural $p, q, r$) in which we have reduced the fractions under a common denominator. Then $pq(p + q) = r^3$. After truncating a common divisor of $p, q, r$, if necessary, we can consider that $(p, q, r) = 1$. In fact, $(p, q) = 1$, since otherwise their common prime divisor would also be that of $r$, contradicting $(p, q, r) = 1$. Thus, the numbers $p, q$, and $p + q$ are pairwise coprime, and since their product is an exact cube, necessarily $p = a^3$, $q = b^3$, and $p + q = c^3$ for some natural numbers $a, b, c$. Thus we obtain $a^3 + b^3 = c^3$ for some natural $a, b, c$, which is impossible (a special case of Fermat's Last Theorem).",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15572,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the circumcenter of an acute triangle $ABC$, and let $\\Omega$ be its circumcircle. On $\\Omega$, select a point $P$ that is different from $A$, $B$, $C$, and their antipodes. Let the circumcenters of triangles $AOP$, $BOP$, and $COP$ be $O_A$, $O_B$, and $O_C$, respectively. Through $O_A$, draw the perpendicular to $BC$ (call this line $l_A$); through $O_B$, draw the perpendicular to $CA$ ($l_B$); and through $O_C$, draw the perpendicular to $AB$ ($l_C$). Prove that the circumcircle of the triangle determined by $l_A$, $l_B$, and $l_C$ is tangent to the line $OP$.\n\n(Note: If segment $\\overline{XY}$ is a diameter of $\\Omega$, then $X$ and $Y$ are antipodes on $\\Omega$.)",
    "options": [],
    "answer": "See solution",
    "solution": "As usual, we denote the directed angle between lines $a$ and $b$ by $\\angle(a, b)$. We frequently use the fact that $a_1 \\perp a_2$ and $b_1 \\perp b_2$ yield $\\angle(a_1, b_1) = \\angle(a_2, b_2)$.\n\n![](images/19-2J_p19_data_57a05e7ecd.png)\n\nLet the lines $l_B$ and $l_C$ meet at $L_A$; define the points $L_B$ and $L_C$ similarly. Note that the sidelines of the triangle $L_A L_B L_C$ are perpendicular to the corresponding sidelines of $ABC$. Points $O_A$, $O_B$, $O_C$ are located on the corresponding sidelines of $L_A L_B L_C$; moreover, $O_A$, $O_B$, $O_C$ all lie on the perpendicular bisector of $OP$.\n\n**Claim 1.** The points $L_B$, $P$, $O_A$, and $O_C$ are concyclic.\n\n**Proof.** Since $O$ is symmetric to $P$ in $O_A O_C$, we have\n\n$$\n\\begin{align*}\n\\angle(O_A P, O_C P) &= \\angle(O_C O, O_A O) = \\angle(CP, AP) \\\\\n&= \\angle(CB, AB) = \\angle(O_A L_B, O_C L_B).\n\\end{align*}\n$$\n\n![](images/19-2J_p20_data_8359fd0d56.png)\n\nDenote the circle through $L_B$, $P$, $O_A$, and $O_C$ by $\\omega_B$. Define the circles $\\omega_A$ and $\\omega_C$ similarly.\n\n**Claim 2.** The circumcircle of the triangle $L_A L_B L_C$ passes through $P$.\n\n**Proof.** From cyclic quadruples of points in the circles $\\omega_B$ and $\\omega_C$, we have\n\n$$\n\\begin{align*}\n\\angle(L_C L_A, L_C P) &= \\angle(L_C O_B, L_C P) = \\angle(O_A O_B, O_A P) \\\\\n&= \\angle(O_A O_C, O_A P) = \\angle(L_B O_C, L_B P) = \\angle(L_B L_A, L_B P).\n\\end{align*}\n$$\n\n![](images/19-2J_p20_data_8359fd0d56.png)\n\n**Claim 3.** The points $P$, $L_C$, and $C$ are collinear.\n\n**Proof.** We have $\\angle(PL_C, L_C L_A) = \\angle(PL_C, L_C O_B) = \\angle(PO_A, O_A O_B)$. Further, since $O_A$ is the center of the circle $AOP$, $\\angle(PO_A, O_A O_B) = \\angle(PA, AO)$. As $O$ is the circumcenter of the triangle $PCA$, $\\angle(PA, AO) =$\n\n$$\n\\frac{\\pi}{2} - \\angle(CA, CP) = \\angle(CP, L_C L_A).\n$$\n\nWe obtain $\\angle(PL_C, L_C L_A) = \\angle(CP, L_C L_A)$, which shows that $P \\in CL_C$. $\\square$\n\nSimilarly, the points $P$, $L_A$, $A$ are collinear, and the points $P$, $L_B$, $B$ are collinear.\n\nFinally, the computation above also shows that\n\n$$\n\\angle(OP, PL_A) = \\angle(PA, AO) = \\angle(PL_C, L_C L_A),\n$$\n\nwhich means that $OP$ is tangent to the circle $PL_A L_B L_C$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15573,
    "subject": "Mathematics (Olympiad)",
    "question": "Докажи дека постојат попарно дисјунктни множества $A_1, A_2, \\dots, A_{2014}$, чија унија е множеството на природни броеви, за кои важи следниот услов:\n\nЗа произволни природни броеви $a$ и $b$, барем два од броевите $a$, $b$, $\\text{НЗД}(a,b)$ припаѓаат на едно од множествата $A_1, A_2, \\dots, A_{2014}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Нека $v_2(n)$ е најголемиот цел број за кој $2^{v_2(n)}$ е делител на $n$. Тогаш, $v_2(\\text{НЗД}(a,b)) = \\min\\{v_2(a), v_2(b)\\}$. Значи, барем два од броевите $v_2(a)$, $v_2(b)$ и $v_2(\\text{НЗД}(a,b))$ се еднакви.\n\nДефинираме множества $A_{i+1} = \\{n \\mid v_2(n) \\equiv i \\pmod{2014}\\}$ за $0 \\le i \\le 2013$.\n\nОчигледно, множествата $A_1, A_2, \\dots, A_{2014}$ се попарно дисјунктни, нивната унија е $\\mathbb{N}$ и два од броевите $a$, $b$, $\\text{НЗД}(a,b)$ се наоѓаат во множеството $A_{i+1}$, каде $i$ е остатокот при делење на $v_2(\\text{НЗД}(a,b))$ со $2014$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15574,
    "subject": "Mathematics (Olympiad)",
    "question": "En una sala de baile hay 15 chicos y 15 chicas dispuestos en dos filas paralelas, formando 15 parejas de baile. Se sabe que la diferencia de altura entre el chico y la chica de cada pareja no supera los 10 cm. Demuestra que si colocamos a los mismos chicos y chicas en dos filas paralelas ordenados en orden creciente de alturas, también la diferencia de alturas entre los miembros de las nuevas parejas así formadas no superará los 10 cm.",
    "options": [],
    "answer": "See solution",
    "solution": "Sean $P_1, P_2, \\dots, P_{15}$ las quince parejas iniciales. Ordenemos ahora los chicos por alturas $a_1 \\leq a_2 \\leq \\dots \\leq a_{15}$ y también las chicas $b_1 \\leq b_2 \\leq \\dots \\leq b_{15}$. Supongamos que en alguna de las nuevas parejas la diferencia de alturas supera los 10 cm, es decir, $a_k - b_k > 10$ para algún $k$. Entonces, para las chicas de alturas $b_1, \\dots, b_k$ y los chicos de alturas $a_k, \\dots, a_{15}$, se cumple $a_i - b_j > 10$. Si colocamos a estas $k + (15 - k + 1) = 16$ personas en las parejas iniciales $P_s$ según su posición original, por el principio de las casillas (palomar), dos personas compartirán la misma pareja. Por lo tanto, en las parejas iniciales habría una cuya diferencia de alturas sería mayor que 10 cm, lo cual contradice la hipótesis.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15575,
    "subject": "Mathematics (Olympiad)",
    "question": "Dado un número entero positivo $n$, definimos $\\lambda(n)$ como el número de soluciones enteras positivas de la ecuación $x^2 - y^2 = n$. Diremos que el número $n$ es \"olímpico\" si $\\lambda(n) = 2021$.\n\n¿Cuál es el menor entero positivo que es olímpico? ¿Y cuál es el menor entero positivo impar que es olímpico?",
    "options": [],
    "answer": "See solution",
    "solution": "Distinguimos cuatro casos, según $n$ sea impar o par y según $n$ sea cuadrado perfecto o no.\n\n**(a)** Sea $n = p_1^{a_1} \\cdots p_r^{a_r}$ un número impar que no es cuadrado perfecto. Si $x^2 - y^2 = (x+y)(x-y) = n$, con $x, y > 0$, entonces existen enteros positivos $a, b$, con $a > b$ y ambos de la misma paridad, de manera que $x + y = a$ y $x - y = b$ (por lo que $x = (a+b)/2$, $y = (a-b)/2$). Las formas de escribir $n$ como producto de dos números diferentes de la misma paridad son, en este caso, la mitad del número de divisores, por lo que buscamos números con $4042$ divisores:\n\n$$\n4042 = (a_1 + 1) \\cdots (a_r + 1).\n$$\n\nLa descomposición de $4042$ como producto de primos es $4042 = 2 \\cdot 43 \\cdot 47$. Por tanto, las opciones de números naturales con $4042$ divisores son de la forma: $pq^{42}r^{46}$; $pq^{2020}$; $p^{42}q^{93}$; $p^{46}q^{85}$; o $p^{4041}$, donde $p, q, r$ son primos impares diferentes. Es inmediato comprobar que la opción que da el número más bajo es $3^{46} \\cdot 5^{42} \\cdot 7$.\n\n**(b)** Consideremos ahora el caso en el que $n$ es impar y cuadrado perfecto. En este caso, el número de divisores es impar, y como excluimos el caso de que los dos números de la factorización sean iguales, tenemos que buscar cuadrados perfectos con $4043$ divisores:\n\n$$\n4043 = (a_1 + 1) \\cdots (a_r + 1).\n$$\n\nLa descomposición de $4043$ como producto de primos es $4043 = 13 \\cdot 311$, luego las únicas opciones son $p^{12}q^{310}$ y $p^{4042}$. El número más bajo es $3^{310} \\cdot 5^{12}$, que es mayor que el encontrado antes.\n\n**(c)** Supongamos ahora que $n = 2^k p_1^{a_1} \\cdots p_r^{a_r}$ es par, pero no un cuadrado perfecto. Nuevamente, tenemos que hacer la descomposición $n = ab$, siendo $a$ y $b$ de la misma paridad; esto hace que $a$ y $b$ tengan que ser pares, o sea, que $k \\ge 2$ y no sirven los divisores impares. Por tanto, las opciones para el exponente de $2$ en cada divisor son $1, 2, 3, \\ldots, k-1$, pero nunca $0$ ni $k$. Así pues,\n\n$$\n4042 = (k-1)(a_1+1)\\cdots(a_r+1),\n$$\n\ny las opciones son $2^3 p^{42} q^{46}$; $2^{44} pq^{46}$; $2^{48} pq^{42}$; $2^{87} p^{46}$; $2^{48} p^{85}$; $2^{95} p^{42}$; $2^{44} q^{93}$; $2^3 p^{2020}$; $2^{2022} p$; y $2^{4043}$. También es válida la opción $4m$, donde $m$ es cualquiera de los números considerados en el apartado (a). El número más bajo es $2^{48} \\cdot 3^{42} \\cdot 5$, que es menor que el encontrado en el apartado (a).\n\n**(d)** Por último, tenemos el caso en el que $n$ es par y cuadrado perfecto, esto es $4043 = (k-1)(a_1+1) \\cdots (a_r+1)$ y las opciones son $2^{14} p^{310}$, $2^{312} p^{12}$ y $2^{4044}$. También es válida la opción $4m$, donde $m$ es cualquiera de los números considerados en el apartado (b). Los tres números son mayores que el encontrado en el apartado (c).\n\nPor consiguiente, el menor número olímpico es $2^{48} \\cdot 3^{42} \\cdot 5$ y el menor número impar olímpico es $3^{46} \\cdot 5^{42} \\cdot 7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15576,
    "subject": "Mathematics (Olympiad)",
    "question": "$$\nR = \\frac{abc}{4S}, \\quad r = \\frac{S}{p}, \\quad S = \\sqrt{p(p-a)(p-b)(p-c)}\n$$\n\nДараах тэнцэтгэл бишийг батал:\n$$\na^2 + b^2 + c^2 \\ge ab + bc + ac\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "$$\np^2 \\ge 12 \\cdot \\frac{abc}{4S} \\cdot \\frac{S}{p} + 3 \\left(\\frac{S}{p}\\right)^2 = \\frac{3abc}{p} + \\frac{3(p-a)(p-b)(p-c)}{p}\n$$\n\n$$\n= 3p^3 - 3p^2(a + b + c) + 3p(ab + bc + ac)\n= 3p^2 - 3p(a + b + c) + 3(ab + bc + ac)\n\\Rightarrow p^2 \\ge 3(ab + bc + ac)\n$$\n\nҮүнээс $a^2 + b^2 + c^2 \\ge ab + bc + ac$ илэрхий тэнцэтгэл биш гарна.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15577,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$, integrable over any bounded interval, satisfying the condition\n$$\n\\int_{x-y}^{x+y} f(t) \\, dt = y(f(x+y) + f(x-y)), \\quad \\text{for all real numbers } x \\text{ and } y.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Any affine function satisfies the condition. To prove the converse, set $y = x$ to obtain\n$$\n\\int_{0}^{2x} f(t) \\, dt = x(f(2x) + f(0)). \\quad (*)\n$$\nSince $f$ is integrable, $(*)$ shows that $g(x) = x(f(2x) + f(0))$ is continuous, so $f$ is continuous on $\\mathbb{R}^* = \\mathbb{R} \\setminus \\{0\\}$. Continuity and $(*)$ imply $g$ is differentiable on $\\mathbb{R}^*$, so $f$ is differentiable on $\\mathbb{R}^*$.\n\nDifferentiating $(*)$ gives $2f(2x) = f(2x) + f(0) + 2x f'(2x)$; that is, $x f'(x) - f(x) + f(0) = 0$ for all $x \\neq 0$. Thus,\n$$\n\\left( \\frac{f(x) - f(0)}{x} \\right)' = 0 \\quad \\text{for all } x \\neq 0,\n$$\nso\n$$\nf(x) = \\begin{cases} a x + f(0), & x < 0, \\\\ b x + f(0), & x > 0. \\end{cases}\n$$\nNow set $x = 0$ and $y = 1$ in the original relation:\n$$\n\\begin{aligned}\nf(1) + f(-1) &= \\int_{-1}^{1} f(t) \\, dt = \\int_{-1}^{0} (a t + f(0)) \\, dt + \\int_{0}^{1} (b t + f(0)) \\, dt \\\\ &= \\frac{1}{2}(b - a) + 2 f(0).\n\\end{aligned}\n$$\nPlugging $f(1) = b + f(0)$ and $f(-1) = -a + f(0)$ gives $b - a + 2 f(0) = \\frac{1}{2}(b - a) + 2 f(0)$, so $a = b$ and $f(x) = a x + f(0)$ for all real $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15578,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\Gamma$ be a circle, and $A$ a point outside $\\Gamma$. For a point $B$ on $\\Gamma$, let $C$ be the third vertex of the equilateral triangle $ABC$ (with vertices $A$, $B$, and $C$ going clockwise). Find the path traced out by $C$ as $B$ moves around $\\Gamma$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\Gamma$ have centre $O$ and radius $r$. Let $O'$ be the third vertex of the equilateral triangle $AOO'$ (clockwise). Then the locus of $C$ is a circle with centre $O'$ and radius $r$: for consider the triangles $AOB$ and $AO'C$. Then $AO = AO'$ and $BA = CA$ (by construction: equilateral triangles) and $\\angle BAO = 60^\\circ - \\angle O'AB = \\angle CAO'$. So the two triangles are congruent (two sides and an included angle) and hence $BO = CO'$. Thus $C$ lies on a circle with centre $O'$ and radius $r$.\n\nIf $B$ happens to lie on the line $OA$, this solution involves the consideration of degenerate triangles, but in an unproblematic way.\n\nChanging “clockwise” to “anticlockwise” simply reflects the locus across $OA$.\n\nThe same argument applies, with only minor changes, if $A$ lies inside or even on the circle.\n\nThe result may also be easily established using complex numbers: we may assume that $\\Gamma$ has its centre at the origin and unit radius, and that $A$ is given by the complex number $a$. If $B$ is given by $z$ (on the circle) and $C$ by $z'$, then $z' - z = (a - z)e^{i\\pi/3}$. It follows that $|z' - ae^{i\\pi/3}| = |z| \\cdot |1 - e^{i\\pi/3}| = 1$, showing that $z'$ moves round a circle of radius 1 with centre at $ae^{i\\pi/3}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15579,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there is a party with $mn$ people arranged in a grid, and some of them are police officers. Each criminal can threaten exactly one person, and the goal is to ensure that if there is a criminal among the guests, at least one officer can identify a criminal. What is the minimum number of police officers required to guarantee this?\n\n![](images/2019_p55_data_9114af8e83.png)\n\n![](images/2019_p56_data_4895a42124.png)\n\n![](images/2019_p56_data_9351e9cb0b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum number of police officers required is $\\left\\lfloor \\frac{mn}{2} \\right\\rfloor + 1$.\n\n**Proof:**\n\nFirst, we show that this number is sufficient. Consider two cases:\n\n- If $mn$ is odd: Color the grid with two colors so that no two adjacent nodes share the same color. Place $\\left[\\frac{mn}{2}\\right] + 1$ officers on the majority color. Since any criminal can threaten only one person (an officer), and the number of officers exceeds the number of guests (and thus criminals), there will always be an officer adjacent to a criminal who is not threatened and can identify the criminal.\n\n- If $mn$ is even: Assume $n$ is even. Use the same coloring. Place $\\frac{mn}{2}$ officers on the majority color and one officer on any node of the opposite color. The argument is similar: the number of officers is greater than the largest possible number of criminals, so one officer will be able to identify a criminal if present.\n\nNow, we show that fewer officers are insufficient. Construct a graph with $mn$ vertices, each representing a person. Partition the nodes into $A$ (officers) and $B$ (guests). Connect $A$ and $B$ if the corresponding people are adjacent. If $|A| \\leq |B|$, and every node in $B$ is connected to at least one in $A$, Hall's Theorem implies there exists a subset $S$ of $A$ such that the set $N(S)$ of neighbors in $B$ is smaller than $|S|$. Removing $S$ and $N(S)$, the matching property holds for the remaining graph. Thus, with fewer officers, there is a way for criminals to threaten officers so that the officers cannot guarantee identifying a criminal.\n\nTherefore, $\\left\\lfloor \\frac{mn}{2} \\right\\rfloor + 1$ is the minimum required.\n\n$\\blacksquare$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15580,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the positive integer $n$ for which the following holds:\n\n$$\nn^2 = 2 \\cdot (20^4 + 19^4 + 39^4).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the expression:\n\n$$\nx^4 + y^4 + (x + y)^4 = 2x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + 2y^4.\n$$\n\nThus,\n\n$$\n2(x^4 + y^4 + (x + y)^4) = 4(x^4 + 2x^3y + 3x^2y^2 + 2xy^3 + y^4) = (2(x^2 + xy + y^2))^2.\n$$\n\nFor $x = 20$, $y = 19$:\n\n$$\nn^2 = (2(x^2 + xy + y^2))^2 \\implies n = 2(20^2 + 20 \\cdot 19 + 19^2) = 2(400 + 380 + 361) = 2(1141) = 2282.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15581,
    "subject": "Mathematics (Olympiad)",
    "question": "Andriiko has an unlimited number of chips painted in 6 colours. He wants to arrange some of the chips in a row so that for any two different colours in that row, there are two adjacent chips of those colours. What is the minimum number of chips he can put in a row?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** 18.\n\nThe chip of each colour must be adjacent to each of the other 5 colours at least once. Since any chip can have at most 2 neighbours, there must be at least 3 chips of each colour in the row. Here is an example showing that 18 is possible: $123456246325164135$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15582,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $X$ and $Y$ be two finite subsets of the half-open interval $[0, 1)$ such that $0 \\in X \\cap Y$ and $x + y \\neq 1$ for all $x \\in X$ and $y \\in Y$. Prove that the set\n\n$$\n\\{x + y - \\lfloor x + y \\rfloor : x \\in X \\text{ and } y \\in Y\\}\n$$\n\nhas at least $|X| + |Y| - 1$ elements.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S(X, Y) = \\{x + y - \\lfloor x + y \\rfloor : x \\in X \\text{ and } y \\in Y\\}$. Proceed by induction on $|Y|$.\n\nIf $|Y| = 1$, the statement is clear. Assume $|Y| > 1$ and let $y_0 \\in Y$, $y_0 \\neq 0$. The conditions $y_0 \\neq 0$ and $x + y_0 = 1$ for no $x \\in X$ imply that $0 \\notin S(X, y_0)$. Since $0 \\in X$, and $X$ and $S(X, y_0)$ have the same cardinality, there are elements in $S(X, y_0)$ not in $X$; that is, $x_0 + y_0 - \\lfloor x_0 + y_0 \\rfloor$ for some $x_0 \\in X$.\n\nLet\n\n$$\nY_0 = \\{y \\in Y : x_0 + y - \\lfloor x_0 + y \\rfloor \\notin X\\}\n$$\n\nand\n\n$$\nX_0 = \\{x_0 + y - \\lfloor x_0 + y \\rfloor : y \\in Y_0\\}.\n$$\n\nClearly, $0 \\notin Y_0$, $X_0$ and $Y_0$ are both non-empty and have the same cardinality, so $0 < |X_0| = |Y_0| < |Y|$. Let $X' = X \\cup X_0$ (a disjoint union) and $Y' = Y \\setminus Y_0$, so $X$ is a proper non-empty subset of $X'$, $Y'$ is a proper non-empty subset of $Y$, and\n\n$$\n|X'| + |Y'| = |X| + |X_0| + |Y| - |Y_0| = |X| + |Y|. \\quad (* )\n$$\n\nBoth $X'$ and $Y'$ contain $0$. We now show that $S(X', Y') \\subseteq S(X, Y)$ and $x' + y' = 1$ for no $x' \\in X'$, $y' \\in Y'$. Consider $x' \\in X_0$, i.e., $x' = x_0 + y - \\lfloor x_0 + y \\rfloor$ for some $y \\in Y_0$. If $y' \\in Y'$, then\n\n$$\n\\begin{align*}\nx' + y' - \\lfloor x' + y' \\rfloor &= x_0 + y - \\lfloor x_0 + y \\rfloor + y' - \\lfloor x_0 + y - \\lfloor x_0 + y \\rfloor + y' \\rfloor \\\\\n&= x_0 + y + y' - \\lfloor x_0 + y + y' \\rfloor \\\\\n&= (x_0 + y' - \\lfloor x_0 + y' \\rfloor) + y - \\lfloor (x_0 + y' - \\lfloor x_0 + y' \\rfloor) + y \\rfloor .\n\\end{align*}\n$$\n\nNotice $x_0 + y' - \\lfloor x_0 + y' \\rfloor \\in X$ by the definition of $Y'$, so $x' + y' - \\lfloor x' + y' \\rfloor \\in S(X, Y_0) \\subseteq S(X, Y)$, and thus $S(X', Y') \\subseteq S(X, Y)$.\n\nFinally,\n\n$$\n\\begin{align*}\nx' + y' &= x_0 + y - \\lfloor x_0 + y \\rfloor + y' \\\\\n&= (x_0 + y' - \\lfloor x_0 + y' \\rfloor) + y - \\lfloor x_0 + y \\rfloor + \\lfloor x_0 + y' \\rfloor,\n\\end{align*}\n$$\n\nand since $x_0 + y' - \\lfloor x_0 + y' \\rfloor \\in X$, considering the possible values of $\\lfloor x_0 + y \\rfloor$ and $\\lfloor x_0 + y' \\rfloor$, we see $x' + y' \\neq 1$.\n\nConsequently,\n\n$$\n\\begin{align*}\n|S(X, Y)| &\\ge |S(X', Y')| && \\text{since } S(X', Y') \\subseteq S(X, Y) \\\\\n&\\ge |X'| + |Y'| - 1 && \\text{by the induction hypothesis} \\\\\n&= |X| + |Y| - 1. && \\text{by } (*)\n\\end{align*}\n$$\n\n**Remarks:**\n\n1. Equality holds if, for instance, $X = \\{0, t, 2t, \\dots, mt\\}$ and $Y = \\{0, t, 2t, \\dots, nt\\}$, where $m$ and $n$ are positive integers, and $t$ is a positive real less than $1/(m+n)$.\n2. The statement is a specialization of the following addition theorem for Abelian groups to the case of the additive group of real numbers modulo $1$: If $X$ and $Y$ are finite subsets of an Abelian group such that $0 \\in X \\cap Y$ and $x + y = 0$ only if $x = 0$ and $y = 0$, then $|X + Y| \\ge |X| + |Y| - 1$. The proof is similar.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15583,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $H$, $\\Gamma$ 分別為銳角三角形 $ABC$ 的垂心與外接圓，$M$ 為邊 $BC$ 的中點。在 $\\Gamma$ 的劣弧 $BC$ 上取一點 $D$ 使得 $\\angle BAD = \\angle MAC$，分別在 $\\Gamma$ 及直線 $BC$ 上取兩點 $E$, $F$ 使得 $DE$ 及 $DF$ 分別垂直於 $AM$ 及 $BC$。令 $N$ 為直線 $HF$ 及 $AM$ 的交點，$R$ 為 $H$ 關於 $N$ 的對稱點。\n\n證明：$\\angle AER + \\angle DFR = 180^{\\circ}$。\n\n![](images/2022-TWNIMO-Problems_p96_data_093ac6bb71.png)",
    "options": [],
    "answer": "See solution",
    "solution": "令 $X$ 為 $AM$ 與 $\\Gamma$ 異於 $A$ 的交點，則\n\n$$\n\\angle XDC = \\angle XAC = \\angle MAC = \\angle BAD = \\angle BCD,\n$$\n\n即 $DX \\parallel BC$。由於 $AD$, $AX$ 為關於 $\\angle BAC$ 的等角線，$BDXC$ 為等腰梯形，因此 $D$ 關於 $BC$ 的對稱點 $D'$ 為 $X$ 關於 $M$ 的對稱點。令 $A^*$ 為 $A$ 關於 $\\Gamma$ 的對徑點，$Y$ 為 $\\overline{HE}$ 中點，則 $M$ 為 $\\overline{HA^*}$ 中點，故 $\\triangle A^*ER$ 為 $\\triangle MYN$ 關於 $H$ 位似 2 倍下的像。由於 $AX$ 垂直於 $DE$，$AX$, $AA^*$ 為關於 $\\angle DAE$ 的等角線，因此 $DXA^*E$ 為等腰梯形。注意到（在 mod $360^{\\circ}$ 下）\n\n$$\n\\angle FMA = \\angle DXA = 90^{\\circ} - \\angle EDX = 90^{\\circ} - \\angle A^*ED = \\angle (XA, A^*E) = \\angle AMY,\n$$\n\n所以結合 $\\overline{FM} = \\frac{1}{2} \\cdot \\overline{DX} = \\frac{1}{2} \\cdot \\overline{A^*E} = \\overline{MY}$，我們得到 $Y$ 為 $F$ 關於 $AM$ 的對稱點。\n\n因為 $\\angle MAE = \\angle DAA^* = \\angle HAM$，所以 $AE$ 為 $AH$ 關於 $AM$ 的對稱直線。故\n\n$$\n180^{\\circ} - \\angle DFR = \\angle (NF, HA) = \\angle (AE, YN) = \\angle AER.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15584,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{Z}$ be the set of integers. Find all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ such that\n\n$$\nx f(2f(y) - x) + y^2 f(2x - f(y)) = \\frac{f(x)^2}{x} + f(y f(y))\n$$\n\nfor all $x, y \\in \\mathbb{Z}$ with $x \\neq 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f$ be a solution of the problem. Let $p$ be a prime larger than $|f(0)|$. Since $p$ divides $f(p)^2$, $p$ divides $f(p)$, and so $p$ divides $\\frac{f(p)^2}{p}$. Taking $y = 0$ and $x = p$, we deduce that $p$ divides $f(0)$. Since $p > |f(0)|$, we must have $f(0) = 0$.\n\nNext, set $y = 0$ to obtain $x f(-x) = \\frac{f(x)^2}{x}$. Replacing $x$ by $-x$, and combining the two relations yields $f(x) = 0$ or $f(x) = x^2$ for all $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15585,
    "subject": "Mathematics (Olympiad)",
    "question": "Alex and Bibi play the following game. Alex chooses a natural number $k$ not exceeding $1000$. Then Bibi chooses a collection $B$ of $n$ integers in $\\{0, 1, \\ldots, 1000\\}$, not necessarily distinct, where $n > k$. Now Alex is allowed to repeatedly apply the following operation on $B$: choosing $k$ numbers $b_1, \\ldots, b_k$ from $B$ and changing them as follows. For each $i = 1, \\ldots, k$, the number $b_i$ is replaced by $b_i + 1$ if $b_i < 1000$, and by $0$ if $b_i = 1000$.\n\nAlex wins if, after several operations, he succeeds in making all numbers in $B$ equal to $0$; if he fails, then Bibi wins. Find all $k$ that guarantee Alex a win, regardless of the collection $B$ chosen by Bibi.",
    "options": [],
    "answer": "See solution",
    "solution": "The winning values of $k$ are the numbers in $\\{1, \\ldots, 1000\\}$ that are coprime with $1001$, i.e., not divisible by $7$, $11$, or $13$.\n\nDespite the repetitions in Bibi's collection $B$, for brevity we call it a set in the solution and denote the number of its elements by $|B|$. The allowed operation is choosing a $k$-element subset of $B$ and increasing each of its elements by $1$ modulo $1001$.\n\nLet us show that a winning number $k$ is coprime with $1001$. Assume on the contrary that $\\gcd(k, 1001) = d > 1$ and consider any set $B$ chosen by Bibi. Let the elements of $B$ have sum $S$. Suppose Alex chooses $k$ numbers from $B$, of which $m$ are $1000$. Then the operation increases $S$ by $(k - m) - 1000m = k - 1001m$, a number divisible by $d$ since $\\gcd(k, 1001) = d$. Hence the operation does not change $S$ modulo $d$, for any choice of $B$. Let $B$ contain one number $1$ and $|B| - 1$ zeros. Then $S \\equiv 1 \\pmod{d}$ persists after each operation, while the desired \"all zeros\" final state requires $S \\equiv 0 \\pmod{d}$. The contradiction proves that $\\gcd(k, 1001) = 1$ is a necessary condition for $k$ to be winning.\n\nConversely, $\\gcd(k, 1001) = 1$ is sufficient. For a proof, consider the unique integer $l$ in $\\{1, \\ldots, 1000\\}$ such that $kl \\equiv 1 \\pmod{1001}$. Let $B$ be any set chosen by Bibi and $b \\in B$ an arbitrary element. It is enough to show that there is a sequence of operations that adds $1$ modulo $1001$ to $b$ without affecting the remaining numbers in $B$.\n\nTake a $(k + 1)$-element subset $C$ of $B$ that contains $b$. This is possible as $|B| > k$. Apply the operation $l$ times to each $k$-element subset of $C$. Each element of $C$ is contained in exactly $k$ such subsets, so the procedure increases it $kl$ times modulo $1001$. Because $kl \\equiv 1 \\pmod{1001}$, the result is that each element of $C$ is increased by $1$ modulo $1001$. Now take the $k$-element subset $C \\setminus \\{b\\}$ of $C$ and apply the operation $1000$ times. After all described operations, each element of $C \\setminus \\{b\\}$ is increased by $1 + 1000 = 1001$ modulo $1001$ with respect to its initial state, meaning that the operations do not change it. Clearly, the elements of $B \\setminus C$ do not change either. The only change concerns $b$, which is increased by $1$ modulo $1001$, as needed. Hence $\\gcd(k, 1001) = 1$ is a sufficient condition, and the solution is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15586,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that the polynomial $P(x)$ of degree $n$ with integer coefficients can be factorized as\n$$\nP(x) = (x - x_1)(x - x_2)\\dots(x - x_n),\n$$\nwhere $0 \\le x_k \\le 3$ for all $k = 1, \\dots, n$. Prove that $x_k \\in \\left\\{ \\frac{3 \\pm \\sqrt{5}}{2},\\ 1,\\ 2 \\right\\}$ for all $k = 1, \\dots, n$.",
    "options": [],
    "answer": "See solution",
    "solution": "It can be shown that for all $t \\in (0, 3)$:\n$$\n|t(t-1)(t-2)(t-3)| = |(t^2-3t)^2 + 2(t^2-3t)| \\le 1,\n$$\nwith equality if and only if $t = \\frac{3\\pm\\sqrt{5}}{2}$.\n\nLet $a \\ge 0$ be the number of ones among $x_1, \\dots, x_n$, and $b \\ge 0$ the number of twos. If $a + b = n$, the statement is obvious. If $a + b < n$, let $m = n - (a + b)$ and, without loss of generality, assume $x_1, \\dots, x_m \\notin \\{1, 2\\}$. Then $P(x) = (x-1)^a(x-2)^b Q(x)$, where $Q(x) = (x-x_1)\\dots(x-x_m)$ also has integer coefficients. Thus, $Q(0), Q(1), Q(2), Q(3)$ are nonzero integers, so $A = |Q(0)Q(1)Q(2)Q(3)| \\ge 1$. However,\n$$\nA = \\prod_{k=1}^{m} |x_k(x_k - 1)(x_k - 2)(x_k - 3)|,\n$$\nand as shown above, this is only possible if $x_k(x_k - 1)(x_k - 2)(x_k - 3) = 1$ for all $k = 1, \\dots, m$. This leads to the result.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15587,
    "subject": "Mathematics (Olympiad)",
    "question": "A natural number $n$ is given. Determine all $(n-1)$-tuples of nonnegative integers $a_1, a_2, \\dots, a_{n-1}$ such that\n\n$$\n\\left[ \\frac{m}{2^n - 1} \\right] + \\left[ \\frac{2m + a_1}{2^n - 1} \\right] + \\left[ \\frac{2^2m + a_2}{2^n - 1} \\right] + \\left[ \\frac{2^3m + a_3}{2^n - 1} \\right] + \\dots + \\left[ \\frac{2^{n-1}m + a_{n-1}}{2^n - 1} \\right] = m\n$$\n\nholds for all $m \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that there is a unique such $(n-1)$-tuple: $a_k = 2^{n-1} + 2^{k-1} - 1$ for $k = 1, \\dots, n-1$.\n\nLet $N = 2^n - 1$ and define $f_k(x) = \\left[ \\frac{2^k x + a_k}{N} \\right]$ for $k = 0, 1, \\dots, n-1$, where $a_0 = 0$. Since\n\n$$\n\\sum_{k=0}^{n-1} f_k(m) - \\sum_{k=0}^{n-1} f_k(m-1) = 1,\n$$\n\nfor each $m \\in \\mathbb{Z}$, there is exactly one $k$ for which $f_k(m) = f_k(m-1) + 1$. We work modulo $N$. The last equality holds if and only if $2^k m + a_k \\in \\{0, 1, \\dots, 2^k - 1\\}$. That is,\n\n$$\n2^k m \\in \\{-a_k, 1-a_k, \\dots, 2^k-1-a_k\\}.\n$$\n\nMultiplying by $2^{n-k}$, and noting that $2^n \\equiv 1 \\pmod{N}$, we get:\n\nFor each $m \\in \\mathbb{Z}$ there is a unique $k \\in \\{0, 1, \\dots, n-1\\}$ such that $m \\in B_k$ (modulo $N$), where\n\n$$\nB_k = \\{b_k, b_k + 2^{n-k}, \\dots, b_k + (2^k - 1)2^{n-k}\\}\n$$\n\nwith $b_k = -2^{n-k}a_k$. Therefore, the problem condition is equivalent to $\\bigcup_{k=0}^{n-1} B_k$ being a partition of $\\{0, 1, \\dots, N-1\\}$.\n\nFor a number $b$ and set $A \\subseteq \\mathbb{Z}$, write $b + A = \\{b + a : a \\in A\\}$. With this notation, $B_{n-1} = b_{n-1} + \\{0, 2, 4, \\dots, 2^n - 2\\}$. The set $B_{n-2} = b_{n-2} + \\{0, 4, 8, \\dots, 2^n - 4\\}$ is contained in $\\overline{B_{n-1}} = b_{n-1} + \\{1, 3, \\dots, 2^n - 3\\}$, implying $b_{n-2}, b_{n-2} + 2^n - 4 \\in \\overline{B_{n-1}}$, which holds only if $b_{n-2} \\equiv b_{n-1} + 1$. Further, the set $B_{n-3} = b_{n-3} + \\{0, 8, 16, \\dots, 2^n - 8\\}$ is contained in $\\overline{B_{n-1}} \\cup \\overline{B_{n-2}} = b_{n-1} + \\{3, 7, \\dots, 2^n - 5\\}$, so we must have $b_{n-3} \\equiv b_{n-1} + 3$. Similarly, $b_{n-4} \\equiv b_{n-1} + 7$, etc. In general, $b_{n-k} \\equiv b_{n-1} + 2^{k-1} - 1$ for $k = 1, \\dots, n-1$. It follows that $b_0 \\equiv b_{n-1} + 2^{n-1} - 1$. On the other hand, $b_0 = 0$, which gives $b_{n-1} \\equiv 1 - 2^{n-1}$ and therefore $b_k \\equiv 2^{n-1-k} - 2^{n-1}$. Thus $a_k \\equiv -2^k b_k \\equiv 2^{n+k-1} - 2^{n-1} \\equiv 2^{n-1} + 2^{k-1} - 1$ for $k = 1, \\dots, n-1$.\n\nFinally, $\\sum_k f_k(0) = 0$ implies $a_k < N$ for all $k$, so we conclude that $a_k = 2^{n-1} + 2^{k-1} - 1$ for each $k = 1, 2, \\dots, n-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15588,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $a, b \\in \\mathbb{N}$, $a \\neq b$, such that $a + b$ and $a \\cdot b + 1$ are powers of $2$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $a = 1$ or $b = 1$, we obtain the solutions $(1, 2^n - 1)$ and $(2^n - 1, 1)$, with $n > 1$.\n\nLet $a, b \\ge 2$ and $a < b$. Note that then $a + b < ab + 1$ due to the identity $(ab + 1) - (a + b) = (a - 1)(b - 1)$. Let $a + b = 2^n$, $n \\ge 2$; in fact, then $n \\ge 3$ as $n = 2$ forces $a = b = 2$. Then $a = 2^{n-1} - c$, $b = 2^{n-1} + c$ with $1 \\le c < 2^{n-1}$. We have\n\n$$\n2^n = a + b < ab + 1 = 2^{2(n-1)} - c^2 + 1 \\le 2^{2(n-1)},\n$$\n\nwhich implies $ab + 1 = 2^k$ with $n + 1 \\le k \\le 2(n - 1)$. Write\n$$\n2^k = 2^{2(n-1)} - c^2 + 1,\n$$\ni.e., $ab + 1 = 2^k$, in the form\n$$\n(c - 1)(c + 1) = 2^{2(n-1)} - 2^k.\n$$\nThen $(c - 1)(c + 1)$ is divisible by $2^k$ since $k \\le 2(n - 1)$. On the other hand, $n + 1 \\le k$, so $(c - 1)(c + 1)$ is divisible by $2^{n+1}$. Clearly, $c - 1$ and $c + 1$ are consecutive even numbers. Since one of them is divisible by $2$ but not by $4$, the other one is divisible by $2^n$.\n\nNow the inequalities $1 \\le c < 2^{n-1}$ show that the latter is possible only if $c - 1 = 0$, i.e., $c = 1$. Hence $a = 2^{n-1} - 1$, $b = 2^{n-1} + 1$ with $n \\ge 3$. This is clearly an admissible pair.\n\nHence the solutions are $(1, 2^n - 1)$, $(2^n - 1, 1)$ and $(2^{n-1} - 1, 2^{n-1} + 1)$, $(2^{n-1} + 1, 2^{n-1} - 1)$ with $n > 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15589,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ and $Q$ be the midpoints of the diagonals $BD$ and $AC$ of the quadrilateral $ABCD$. Consider the points $M \\in (BC)$, $N \\in (CD)$, $R \\in (PQ)$, and $S \\in (AC)$ such that $\\frac{BM}{MC} = \\frac{DN}{NC} = \\frac{PR}{RQ} = \\frac{AS}{SC} = k$. Prove that the centroid of the triangle $AMN$ lies on the segment $[RS]$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $G$ be the centroid of the triangle $AMN$. Then we have\n\n$$\n\\overrightarrow{GR} = \\frac{\\overrightarrow{GP} + k\\overrightarrow{GQ}}{1+k} = \\frac{\\overrightarrow{GB} + \\overrightarrow{GD} + k(\\overrightarrow{GA} + \\overrightarrow{GC})}{2(1+k)}\n$$\n\nand\n\n$$\n\\overrightarrow{GS} = \\frac{\\overrightarrow{GA} + k\\overrightarrow{GC}}{1+k}.\n$$\n\nOn the other hand,\n\n$$\n0 = \\overrightarrow{GA} + \\overrightarrow{GM} + \\overrightarrow{GN} = \\overrightarrow{GA} + \\frac{\\overrightarrow{GB} + k\\overrightarrow{GC}}{1+k} + \\frac{\\overrightarrow{GD} + k\\overrightarrow{GC}}{1+k},\n$$\n\nfrom where we deduce that $(1+k)\\overrightarrow{GA} + \\overrightarrow{GB} + 2k\\overrightarrow{GC} + \\overrightarrow{GD} = 0$, whence we get $\\overrightarrow{GS} + 2\\overrightarrow{GR} = 0$, and thus $G$, $R$, and $S$ are collinear points.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15590,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all real-valued functions $f$ defined on the set of all integers and satisfying the following identity for any pair of integers $m, n$:\n\n$$\nf(m) + f(n) = f(mn) + f(m + n + mn).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(1) = a$. Setting $n = 1$ in the given equation yields $f(m) + f(1) = f(m) + f(2m + 1)$ for any integer $m$, so $f(d) = a$ for any odd integer $d$.\n\nAny nonzero integer can be written as $2^k d$ where $d$ is odd and $k \\geq 0$. Taking $m = d$ and $n = 2^k$, the equation gives $f(d) + f(2^k) = f(2^k d) + f(2^k(d + 1) + d)$. Both $d$ and $2^k(d + 1) + d$ are odd, so $a + f(2^k) = f(2^k d) + a$, implying $f(2^k d) = f(2^k)$. Thus, knowing $f(2^k)$ for all $k$ and $f(0)$ determines $f(n)$ for all integers $n$.\n\nFor $k \\geq 2$, substitute $m = 2^k$, $n = 2$ to get $f(2^k) + f(2) = f(2^{k+1}) + f(2^k \\cdot 3 + 2)$. Since $2^k \\cdot 3 + 2$ is an even integer, $f(2^k \\cdot 3 + 2) = f(2)$. Thus, $f(2^k) = f(2^{k+1})$ for $k \\geq 2$. Let $f(4) = b$, so $f(4) = f(8) = \\cdots = b$. Setting $m = n = 2$ gives $2f(2) = f(4) + f(8) = 2b$, so $f(2) = b$. Therefore, $f(n) = b$ for all even $n \\neq 0$. Finally, setting $m = n = -2$ gives $2f(-2) = f(4) + f(0)$, so $f(0) = b$.\n\nThus, any function $f$ satisfying the equation must be of the form\n\n$$\nf(n) = \\begin{cases} a & \\text{if } n \\text{ is odd} \\\\ b & \\text{if } n \\text{ is even or } 0. \\end{cases}\n$$\nfor some real numbers $a, b$.\n\nConversely, for any $a, b \\in \\mathbb{R}$, the function above satisfies the equation:\n\n- If both $m, n$ are even, $mn$ and $m+n+mn$ are even, so both sides are $2b$.\n- If both $m, n$ are odd, $mn$ and $m+n+mn$ are odd, so both sides are $2a$.\n- If one is even and one is odd, $mn$ is even and $m+n+mn$ is odd, so both sides are $a + b$.\n\nTherefore, all such functions $f$ are solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15591,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p_1, p_2, \\dots, p_{n+1}$ denote the first $n+1$ primes. Suppose that $\\{A, B\\}$ is a partition of the set $X = \\{p_1, p_2, \\dots, p_n\\}$, where $A = \\{q_1, q_2, \\dots, q_s\\}$ and $B = \\{r_1, r_2, \\dots, r_t\\}$. Prove that if $m = q_1q_2\\dots q_s + r_1r_2\\dots r_t < p_{n+1}^2$, then $m$ is a prime.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume to the contrary, that $m$ is not a prime number. Then $m = ab$ for some integers $a$ and $b$ with $1 < a < m$ and $1 < b < m$. Let $p$ be the smallest prime that divides $a$ and let $q$ be the smallest prime that divides $b$. WLOG we may assume that $p \\leq q$. We now consider two cases according to whether $p \\in X$ or $p \\notin X$.\n\n- If $p \\in X$, then either $p = q_i$ for some $i$ with $1 \\leq i \\leq s$ or $p = r_j$ for some $j$ with $1 \\leq j \\leq t$, but not both (since $\\{A, B\\}$ is a partition). Suppose that $p = q_i$ for some $i$ with $1 \\leq i \\leq s$. Since $p \\mid a$, then $p \\mid m$. Also $p \\mid q_1q_2\\dots q_s$. Thus $p \\mid (m - q_1q_2\\dots q_s)$ and so $p \\mid r_1r_2\\dots r_t$. This implies that $p = r_j$ for some $j$ with $1 \\leq j \\leq t$. Contradiction.\n\n- If $p \\notin X$, then $q \\geq p \\geq p_{n+1}$ and so $m \\geq pq \\geq p_{n+1}^2$. Contradiction.\n\nFrom the preceding we conclude that $m$ is prime, and we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15592,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer $n$ such that any sequence of positive integers $a_1, a_2, \\dots, a_n$ satisfying $\\sum_{i=1}^{n} a_i = 2007$ must have several consecutive terms whose sum is 30.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we can construct a sequence of positive integers with 1017 terms $a_1, a_2, \\dots, a_{1017}$ such that there are no consecutive terms whose sum is 30. For example, set $a_1 = a_2 = \\cdots = a_{29} = 1$, $a_{30} = 31$, and repeat this pattern: $1, 1, \\dots, 1, 31$ (29 ones followed by 31), for as many full groups as possible, with the last group possibly shorter. This gives 34 groups of 30 terms (except the last group, which has 27 terms), totaling 1017 terms.\n\nIf the number of terms is less than 1017, we can combine several consecutive terms into a larger number within certain groups, still avoiding a sum of 30 among consecutive terms.\n\nNow, consider any sequence with 1018 terms $a_1, a_2, \\dots, a_{1018}$ such that $\\sum_{i=1}^{1018} a_i = 2007$. We want to prove that there must exist several consecutive terms whose sum is 30.\n\nLet $S_k = \\sum_{i=1}^{k} a_i$ for $k = 1, 2, \\dots, 1018$. Then:\n\n$$\n1 \\leq S_1 < S_2 < \\cdots < S_{1018} = 2007.\n$$\n\nGroup the numbers $1, 2, \\dots, 2007$ into brackets as follows:\n\n- (1, 31), (2, 32), ..., (30, 60)\n- (61, 91), (62, 92), ..., (90, 120)\n- (121, 151), (122, 152), ..., (150, 180)\n- $\\dots$\n- (60k + 1, 60k + 31), (60k + 2, 60k + 32), ..., (60k + 30, 60k + 60)\n- $\\dots$\n- (1921, 1951), ..., (1950, 1980)\n- 1981, 1982, ..., 2007 (27 numbers without brackets)\n\nThere are $33 \\times 30 = 990$ brackets and 27 numbers without brackets. Since we have 1018 numbers $S_k$, by the pigeonhole principle, at least two $S_k$ must fall into the same bracket. Let these be $S_k$ and $S_{k+m}$, then $S_{k+m} - S_k = 30$, so:\n\n$$\na_{k+1} + a_{k+2} + \\cdots + a_{k+m} = 30.\n$$\n\nTherefore, the minimum $n$ is $1018$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15593,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Does there exist a non-zero polynomial $P(x, y) \\in \\mathbb{R}[x, y]$ such that $P(n, 2^n) = 0$ for all $n \\in \\mathbb{N}$?\n\nb) Are the sequences $2^n$ and $3^n$ Co-Algebraic? That is, does there exist a non-zero polynomial $P(x, y)$ such that $P(2^n, 3^n) = 0$ for all $n \\in \\mathbb{N}$?\n\nc) Let $f(x, y)$ be a polynomial with integer coefficients. Show that for any $m \\in \\mathbb{N}$ such that $f(2^m, 3^m) \\neq 0$, there exists $n > m$ such that $f(2^n, 3^n)$ is not divisible by $5^n$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Assume, for contradiction, that there exists a non-zero polynomial $P(x, y) \\in \\mathbb{R}[x, y]$ such that $P(n, 2^n) = 0$ for all $n \\in \\mathbb{N}$. Let $d$ be the degree of $P$ in $y$, so $P(x, y) = p_d(x)y^d + \\cdots + p_1(x)y + p_0(x)$, where $p_d(x) \\neq 0$.\n\nFor any $\\epsilon > 0$, there exists $N_1$ such that $|p_d(n)| > \\epsilon$ for $n > N_1$. Also, there exists $N_2$ such that $|p_0(n)|, \\dots, |p_{d-1}(n)| < 2^{n/2}$ for $n > N_2$. For $n > \\max\\{N_1, N_2\\}$:\n\n$$\n\\begin{align*}\n|P(n, 2^n)| & = |p_d(n)2^{dn} + \\cdots + p_1(n)2^n + p_0(n)| \\\\\n& \\ge |p_d(n)2^{dn}| - \\left( |p_{d-1}(n)2^{(d-1)n}| + \\cdots + |p_1(n)2^n| + |p_0(n)| \\right) \\\\\n& > \\epsilon 2^{dn} - 2^{n/2}(1 + 2^n + \\cdots + 2^{(d-1)n}) \\\\\n& > \\epsilon 2^{dn} - 2^{(d-1)n + n/2 + 1} \\\\\n& = 2^{dn}(\\epsilon - 2^{-n/2 + 1}).\n\\end{align*}\n$$\n\nAs $n \\to \\infty$, $2^{-n/2 + 1} \\to 0$, so $|P(n, 2^n)| \\to \\infty$, contradicting $P(n, 2^n) = 0$ for all $n$. Thus, such a polynomial does not exist.\n\n*Remark*: For any non-zero $P(x, y)$, $P(n, 2^n)$ can be zero for only finitely many $n$.\n\nb) No. Let $\\alpha = \\log_2 3$ (so $2^\\alpha = 3$), which is irrational. Suppose there exists a non-zero $P(x, y)$ such that $P(2^n, 3^n) = 0$ for all $n$. Each monomial $c_{k,l}x^k y^l$ gives $c_{k,l}2^{nk}3^{nl} = c_{k,l}2^{n(k + \\alpha l)}$. Since $\\alpha$ is irrational, all $k + \\alpha l$ are distinct for different $(k, l)$. Let $\\beta$ be the largest $k + \\alpha l$ for which $c_{k,l} \\neq 0$.\n\nThen $P(2^n, 3^n) = c_2 2^{\\beta n} + c_1 2^{\\beta_1 n} + \\dots + c_k 2^{\\beta_k n}$, with $\\beta_i < \\beta$. Dividing by $2^{\\beta n}$:\n\n$$\n\\frac{1}{2^{\\beta n}} P(2^n, 3^n) = c_2 + c_1 2^{(\\beta_1 - \\beta)n} + \\dots + c_k 2^{(\\beta_k - \\beta)n}.\n$$\n\nAs $n \\to \\infty$, the right side approaches $c_2 \\neq 0$, but the left side approaches $0$, a contradiction. Thus, $2^n$ and $3^n$ are not Co-Algebraic.\n\nc) By (b), there exists $m \\in \\mathbb{N}$ such that $f(2^m, 3^m) \\neq 0$. Let $k$ be the largest integer such that $5^k \\mid f(2^m, 3^m)$. Set $n = m + 4 \\times 5^k$. Since $\\varphi(5^{k+1}) = 4 \\times 5^k$, Euler's theorem gives $2^{4 \\times 5^k} \\equiv 1 \\pmod{5^{k+1}}$ and $3^{4 \\times 5^k} \\equiv 1 \\pmod{5^{k+1}}$. Thus,\n\n$$\nf(2^n, 3^n) = f(2^{m + 4 \\times 5^k}, 3^{m + 4 \\times 5^k}) \\equiv f(2^m, 3^m) \\not\\equiv 0 \\pmod{5^{k+1}}.\n$$\n\nTherefore, $f(2^n, 3^n)$ is not divisible by $5^{k+1}$, and since $k+1 < n$, not by $5^n$ either.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15594,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer $n$ such that for any coloring of the numbers $1, 2, 3, \\ldots, n$ with three colors, there exist two numbers with the same color whose difference is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n = 29$.\n\nAssume, for contradiction, that the numbers $1, 2, \\ldots, 29$ can be colored with colors $A$, $B$, and $C$ so that no two numbers with the same color differ by a perfect square. Let $f(i)$ denote the color of number $i$ for $i \\in \\{1, 2, \\ldots, 29\\}$.\n\nSince $9$, $16$, and $25$ are squares, the numbers $1$, $10$, and $26$ must all be assigned distinct colors. Similarly, $1$, $17$, and $26$ must be assigned distinct colors, so $10$ and $17$ must have the same color. Continuing this reasoning, we find $f(11) = f(18)$, $f(12) = f(19)$, and $f(13) = f(20)$ (for the last, consider $4$, $13$, $20$, $29$).\n\nWithout loss of generality, let $f(10) = f(17) = A$. Since $11 = 10 + 1^2$, $f(11) \\neq f(10)$, so let $f(11) = f(18) = B$. Now $19 = 18 + 1^2 = 10 + 3^2$, so $f(12) = f(19) = C$. Similarly, $20 = 19 + 1^2 = 11 + 3^2$ implies $f(13) = f(20) = A$. Thus, $f(13) = A = f(17)$, a contradiction.\n\nOn the other hand, if $n \\leq 28$, we can color the numbers as shown below. It is easy to check that no two numbers with the same color differ by a perfect square.\n\n![](images/brozura_a67angl_new_p16_data_a7ad8f3c93.png)\n\nFig. 4",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15595,
    "subject": "Mathematics (Olympiad)",
    "question": "For a given positive integer $n$, one has to choose positive integers $a_0, a_1, \\dots$ so that the following conditions hold:\n\n1. $a_i = a_{i+n}$ for any $i$;\n2. $a_i$ is not divisible by $n$ for any $i$;\n3. $a_{i+a_i}$ is divisible by $a_i$ for any $i$.\n\nFor which positive integers $n > 1$ is this possible only if the numbers $a_0, a_1, \\dots$ are all equal?",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* For all primes.\n\n**Solution.** Let $n$ be a prime. By condition (1), the sequence $a_0, a_1, \\dots$ contains only finitely many different numbers. If $a_m$ is the maximal among them, then by condition (3), $a_{m+a_m}$ must also be maximal. Let us prove that if $a_m$ is maximal, then $a_{m+k \\cdot a_m}$ is also maximal for any $k \\ge 0$. This holds for $k=0$. If the claim holds for $k$, then\n$$\na_{m+(k+1) \\cdot a_m} = a_{m+k \\cdot a_m + a_m} = a_{m+k \\cdot a_m + a_{m+k \\cdot a_m}} = a_{m+k \\cdot a_m} = a_m.\n$$\nThis proves the claim. By condition (2), $a_m$ is not divisible by $n$. Since $n$ is prime, the numbers $a_m$ and $n$ are relatively prime. Hence, among the numbers $m + k \\cdot a_m$, where $0 \\le k < n$, there is one in each congruence class modulo $n$. Hence, all members of the sequence are maximal, i.e., they are equal.\n\nSuppose $n$ is a composite number; let $m$ be its divisor with $1 < m < n$. For any $k < m$, choose $a_k = m + k \\cdot n$ and continue the sequence with period $m$. Condition (1) holds, since $n$ is a multiple of $m$. Condition (2) holds, since all members of the sequence are congruent to $m$ modulo $n$. For condition (3), notice that all members of the sequence are divisible by $m$. Hence, $i$ and $i+a_i$ are always congruent modulo $m$, therefore $a_i = a_{i+a_i}$. At the same time, not all the numbers are equal.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15596,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $G$ be the set of points $(x, y)$ in the plane such that $x$ and $y$ are integers in the range $1 \\leq x, y \\leq 2011$. A subset $S$ of $G$ is said to be parallelogram-free if there is no proper parallelogram with all its vertices in $S$. Determine the largest possible size of a parallelogram-free subset of $G$. (A proper parallelogram is one where its vertices do not all lie on the same line.)",
    "options": [],
    "answer": "See solution",
    "solution": "First, consider the subset $S$ of $G$ containing $\\{(1, i) \\mid 1 \\leq i \\leq 2011\\}$ and $\\{(i, 1) \\mid 2 \\leq i \\leq 2011\\}$. Any set of four points of $S$ either has three collinear points or has two opposite sides at right angles; in any event, they do not form a parallelogram. So $S$ is a parallelogram-free set of size $4021$.\n\nNow suppose $S \\subset G$ has $|S| \\geq 4022$. We will show that $S$ contains a parallelogram. For two elements of $S$ in the same row, define their 'spacing' to be the difference between their $x$-coordinates.\n\nNote that if there are two pairs of elements in different rows with the same spacing, then those four elements form a parallelogram. We will show that this situation must occur.\n\nLet $n_k$ denote the number of elements in the $k$th row. Considering just the spacings between the first element in row $k$ and the $n_k - 1$ others, we find that the number of spacings in the $k$th row is at least $n_k - 1$ (if $n_k = 0$ then the number of spacings is $0 \\geq -1$). So in each row we have found at least $n_k - 1$ distinct spacings, and\n\n$$\n\\begin{aligned}\n\\sum_{k=1}^{2011} (n_k - 1) &= \\left( \\sum_{k=1}^{2011} n_k \\right) - 2011 \\\\\n&= |S| - 2011 \\\\\n&\\geq 2011.\n\\end{aligned}\n$$\n\nSince there are only $2010$ possible spacings (because the $x$-coordinates are all between $1$ and $2011$), by the pigeonhole principle we must have that the same spacing occurs in two different rows, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15597,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $\\angle CAB$ a right angle. The point $L$ lies on the side $BC$ between $B$ and $C$. The circle $ABL$ meets the line $AC$ again at $M$, and the circle $CAL$ meets the line $AB$ again at $N$. Prove that $L$, $M$, and $N$ lie on a straight line.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose circle $ALC$ is tangent to line $AB$. Then $N = A$.\n\n![](images/V_Britanija_2011_p16_data_9f81838582.png)\n\nAs $AC$ is perpendicular to $AB$, $AC$ is a diameter of circle $ALC$ and so $\\angle ALC = 90^\\circ$. Then $\\angle BLA = 90^\\circ$ and so $AB$ is a diameter of circle $ABL$. So $AC$ is tangent to circle $ABL$ and $M$ is at $A$.\n\nTherefore $M$ and $N$ are both at $A$, and so $L$, $M$, $N$ are collinear.\n\nOtherwise:\n\n![](images/V_Britanija_2011_p17_data_8aaa70ef95.png)\n\nAs $CANL$ is cyclic and $\\angle CAN = 90^\\circ$, $NC$ is a diameter of circle $CANL$ and $\\angle NLC = 90^\\circ$. Likewise, as $MBLA$ is cyclic, $\\angle BLM = 90^\\circ$ so $\\angle MLC = 180^\\circ - \\angle BLM = 90^\\circ$.\n\nTherefore $\\angle NLC = \\angle MLC$, and so $L$, $M$, and $N$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15598,
    "subject": "Mathematics (Olympiad)",
    "question": "A school has 2008 students. Each committee in the school consists of 1004 students. Every pair of students must belong to at least one common committee.\n\n(a) What is the smallest possible number of committees?\n\n(b) Is it possible that the union of any two committees contains fewer than 1800 students?",
    "options": [],
    "answer": "See solution",
    "solution": "Let each student join at most 2 committees. Then, a student shares a committee with at most $2(1004 - 1) = 2006$ other students, which is less than 2007, contradicting the requirement. Therefore, each student must join at least 3 committees. Thus, there are at least\n\n$$\n\\frac{3 \\times 2008}{1004} = 6\n$$\n\ncommittees.\n\nThis minimum can be achieved. For example, partition all students into 8 groups $A, B, \\dots, H$, each with 251 students. Form the following 6 committees:\n\n$$\n\\{A, B, C, D\\}, \\{A, E, F, G\\}, \\{A, B, E, H\\}, \\\\ \\{B, F, G, H\\}, \\{C, D, G, H\\}, \\{C, D, E, F\\}.\n$$\n\nEach committee contains $251 \\times 4 = 1004$ students. Every pair of groups appears together in at least one committee, so every pair of students shares a committee.\n\n(b) Yes. In the example above, every pair of committees contains at most 7 groups, so their union has at most\n\n$$\n251 \\times 7 = 1757 < 1800\n$$\n\nstudents.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15599,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDE$ be a convex pentagon in which $BC \\parallel AE$, $AB = BC + AE$, and $\\angle ABC = \\angle CDE$. Let $M$ be the midpoint of $CE$ and let $O$ be the circumcentre of triangle $BCD$. Suppose $\\angle DMO = 90^\\circ$. Prove that $2\\angle BDA = \\angle CDE$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the circumcircle of $BCD$ be $\\Gamma$ and let the circle with diameter $OD$ be $\\Gamma'$. Let the midpoint of $OD$ be $O'$ (which is also the centre of $\\Gamma'$). Now $OO'$ passes through $D$ and hence $\\Gamma'$ and $\\Gamma$ are tangent to each other at $D$. Hence there is a homothety with centre $D$ taking $\\Gamma'$ to $\\Gamma$, with ratio $2$. Since $\\angle OMD = 90^\\circ$, the point $M$ is on $\\Gamma'$. Let the image of $M$ under the homothety be $D'$; it lies on $\\Gamma$ and $M$ is the midpoint of $DD'$.\n\nConsider the half-turn centred at $M$ taking $D$ to $D'$. This also takes $C$ to $E$ as $M$ is the midpoint of $CE$. Let $B'$ be the image of $B$ under this transformation. Since the half-turn takes a line to another line parallel to it, the image of $BC$ is $B'E$; and $BC \\parallel B'E$. Thus $AE \\parallel BC \\parallel B'E$. It follows that $A$, $B'$, $E$ are collinear. Observe that $A$, $B$ lie on the same side of $CE$. Since we have performed a rigid transformation preserving $C$, $E$, the images of $A$, $B$ must also lie on the same side of $CE$. This implies that $A$ and $B'$ lie on different sides of $CE$. Thus $E$ lies between $A$ and $B'$. Because of this, we have\n\n$$\n\\begin{aligned}\nAB' = AE + EB' &= AE + BC \\quad (\\text{half-turn takes } BC \\text{ to } B'E) \\\\\n&= AB.\n\\end{aligned}\n$$\n\nWe also observe that the half-turn takes triangle $CD'B$ to $EDB'$. Therefore $\\angle CD'B = \\angle EDB'$. Thus we get\n\n$$\n\\begin{aligned}\n\\angle BDB' = \\angle BDE + \\angle EDB' &= \\angle BDE + \\angle CD'B = \\angle BDE + \\angle CDB \\\\\n&= \\angle CDE = \\angle CBA = 180^\\circ - \\angle BAB',\n\\end{aligned}\n$$\n\nsince $BC \\parallel AE$. This shows that $B$, $D$, $B'$, $A$ are concyclic. Since $AB = AB'$, they subtend equal angles at $D$. Thus $\\angle ADB' = \\angle BDA'$. Thus\n\n$$\n2\\angle BDA = \\angle BDA + \\angle ADB' = \\angle BDA + \\angle ADE + \\angle EDB'.\n$$\n\nBut observe $\\angle EDB' = \\angle CD'B = \\angle CDB$. Using this we obtain\n\n$$\n2\\angle BDA = \\angle BDA + \\angle ADE + \\angle CDB = \\angle CDE.\n$$\n\nThis completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15600,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a square with center $O$, and let $M$ be the symmetric of point $B$ with respect to point $A$. Let $E$ be the intersection of $CM$ and $BD$, and let $S$ be the intersection of $MO$ and $AE$. Show that $SO$ is the angle bisector of $\\angle ESB$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\n\\begin{cases}\nDC \\equiv DA \\\\\n\\angle EDC \\equiv \\angle EDA \\\\\nDE \\equiv DE\n\\end{cases} \\Rightarrow \\triangle DEC \\equiv \\triangle DEA \\Rightarrow \\angle DAE \\equiv \\angle DCE\\ (*).\n$$\n\nLet $CM \\cap AD = \\{P\\}$, then $\\triangle CDP \\equiv \\triangle BAP$ and $\\angle PCD \\equiv \\angle PBA\\ (**)$.\n\n![](images/2019_bmo_shortlist_p14_data_35aba41cec.png)\n\nFrom $(*)$ and $(**)$ it follows that $\\angle DCP \\equiv \\angle DAE \\equiv \\angle PBA$.\n\nNow, let $S' = AE \\cap PB$.\n\nIn triangle $S'AB$ we have\n\n$$\nm(\\angle S'AB) + m(\\angle S'BA) = m(\\angle S'AB) + m(\\angle PAS') = m(\\angle PAB) = 90^{\\circ},\n$$\n\nso $m(\\angle BS'A) = 90^{\\circ}$.\n\nWe show that $AE$, $BP$, and $MO$ are concurrent.\n\nIn triangle $EMB$ we apply Ceva's theorem:\n\n$$\n\\frac{EP}{PM} \\cdot \\frac{MA}{AB} \\cdot \\frac{BO}{OE} = 1 \\Leftrightarrow \\frac{EP}{PM} = \\frac{OE}{BO}\n$$\n\nThis is true because $PO$ is a midsegment in triangle $DAB$ ($PO \\parallel AB$).\n\nAccording to Thales' theorem in triangle $EMB$, $\\frac{EP}{PM} = \\frac{EO}{OB}$, so $AE$, $BP$, and $MO$ are concurrent at $S'$, which is in fact $S$.\n\nLet $PB \\cap CA = \\{N\\}$. Because $ESNO$ has $m(\\angle EON) + m(\\angle ESN) = 180^{\\circ}$, it follows that $ESNO$ is cyclic and $m(\\angle ESO) = m(\\angle ENO) = m(\\angle DAO) = 45^{\\circ}$. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15601,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that every integer has been given one of the colors red, blue, green, or yellow. Let $x$ and $y$ be odd integers such that $|x| \\neq |y|$. Show that there are two integers of the same color whose difference has one of the following values: $x$, $y$, $x+y$, or $x-y$.",
    "options": [],
    "answer": "See solution",
    "solution": "We denote colors by capital initial letters. Suppose there exists a coloring $f : \\mathbb{Z} \\to \\{R, G, B, Y\\}$ such that for any $a \\in \\mathbb{Z}$, the set $\\{f(a), f(a+x), f(a+y), f(a+x+y)\\}$ contains all four colors. \n\nNow define a coloring of the integer lattice $g : \\mathbb{Z} \\times \\mathbb{Z} \\to \\{R, G, B, Y\\}$ by $g(i, j) = f(xi + yj)$. Then every unit square in $g$ must have its vertices colored with four different colors.\n\nIf there is a row or column with period 2, then by applying the condition to adjacent unit squares, we get (by induction) that all rows or columns, respectively, have period 2.\n\nOn the other hand, if a row is not of period 2 (i.e., contains a sequence of three distinct colors, for example *GRY*), then the next row must contain in these columns *YBG*, and the following *GRY*, and so on. It would follow that a column in this case must have period 2. A similar conclusion holds if we start with an aperiodic column. Hence, either all rows or all columns must have period 2.\n\nAssume without loss of generality that all rows have period 2. Suppose $\\{g(0,0), g(1,0)\\} = \\{G, B\\}$. Then the even rows are colored with $\\{G, B\\}$ and the odd rows with $\\{Y, R\\}$. Since $x$ is odd, it follows that $g(y,0)$ and $g(0,x)$ are of different colors. However, since $g(y,0) = f(xy) = g(0,x)$, this is a contradiction. Hence, the statement of the problem holds. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15602,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть $O_1, r_1$ и $O_2, r_2$ — соответственно центры и радиусы окружностей $\\omega_1$ и $\\omega_2$, а $K$ — точка пересечения $l_1$ и $l_2$.\n\n![](images/Rusija_2012_p23_data_5d996f7ea5.png)\n\n![](images/Rusija_2012_p23_data_32e83672a4.png)\n\nРассмотрим точку $P$, которая лежит на отрезке $O_1O_2$ и делит его в отношении $r_1 : r_2$. Обозначим через $P_1$ точку касания $l_2$ и $\\omega_1$, а через $P_2$ точку касания $l_1$ и $\\omega_2$. Докажите, что $P$ — основание биссектрисы треугольника $KO_1O_2$.\n\n_Замечание: В задаче возможны два принципиально различных случая расположения точек и прямых. Они показаны на рисунках. Решение не зависит от случая._",
    "options": [],
    "answer": "See solution",
    "solution": "Заметим, что точка $P$ лежит на отрезке $O_1O_2$ и делит его в отношении $r_1 : r_2$.\n\nОбозначим через $P_1$ точку касания $l_2$ и $\\omega_1$, а через $P_2$ точку касания $l_1$ и $\\omega_2$. Прямоугольные треугольники $KO_1P_1$ и $KO_2P_2$ подобны по острому углу при вершине $K$. Значит, $\\frac{KO_1}{KO_2} = \\frac{O_1P_1}{O_2P_2} = \\frac{r_1}{r_2}$. Таким образом, в треугольнике $KO_1O_2$ точка $P$, лежащая на стороне $O_1O_2$, делит её в отношении, равном отношению прилежащих сторон $KO_1$ и $KO_2$. Из этого следует, что $P$ — основание биссектрисы треугольника $KO_1O_2$.\n\n_Замечание: В задаче возможны два принципиально различных случая расположения точек и прямых. Они показаны на рисунках. Решение не зависит от случая._",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15603,
    "subject": "Mathematics (Olympiad)",
    "question": "Andrew and Olesya take turns cutting either $2 \\times 2$ or $1 \\times 1$ squares from a $2 \\times 2n$ rectangle, following the grid lines, so that after each move the remaining figure stays connected. The player who cannot make a move loses. If Olesya goes first and both play optimally, who will win?\n\n![](images/UkraineMO2019_booklet_p39_data_438146688d.png)",
    "options": [],
    "answer": "See solution",
    "solution": "For odd $n$, Olesya has a winning strategy; for even $n$, Andrew does.\n\nWe use induction on $n$. For $n=1$ ($2 \\times 2$), Olesya wins by taking the whole $2 \\times 2$ square. For $n=2$ ($2 \\times 4$), Andrew wins (by casework). If $n=2m+1$ is odd, Olesya can remove a $2 \\times 2$ square from the end, leaving an even $n$ for Andrew, thus becoming the second player in an even case.\n\nSo, it suffices to show that the second player wins for even $n$. Before and including the first $1 \\times 1$ square cut from a bordering column (the first or last column of the active part of the field), the second player can always mirror the first player's moves (a symmetric strategy). The first $1 \\times 1$ square from a border will be cut when the active field is $2 \\times k$ with even $k$ (see figure below).\n\n![](images/UkraineMO2019_booklet_p39_data_40157e1595.png)\n\nAfter this, the second player only needs to prevent the cutting of a $2 \\times 2$ square, since after that only $1 \\times 1$ squares can be cut, and the game will end after an even number of moves.\n\nThe only places a $2 \\times 2$ square can be cut are at the ends of the strip. The second player should always cut a $1 \\times 1$ square from a bordering $2 \\times 2$ square that is still intact (a \"free end\"). There can never be more than one free end, since Olesya cannot create two in one move, and initially there are none.\n\nIf there are no free ends, then:\n\n1. If the bordering column has no squares cut, Andrew cuts one unit square from it.\n2. If the bordering column has one square cut:\n   a) If the second-to-border column has a square cut, Andrew cuts the second unit square from the bordering column.\n   b) If the second-to-border column has no squares cut:\n      i. If the third-to-border column has no squares cut, or its cut is in the same row as the bordering column, Andrew cuts a unit square from the second-to-border column in the same row.\n      ii. If the third-to-border column's cut is in the other row, Andrew cuts a unit square from the bordering column, reducing the active field without creating free ends.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15604,
    "subject": "Mathematics (Olympiad)",
    "question": "Define the sequence $A_1, A_2, A_3, \\dots$ by $A_1 = 1$ and for $n = 1, 2, 3, \\dots$\n\n$$\nA_{n+1} = \\frac{A_n + 2}{A_n + 1}.\n$$\n\nDefine the sequence $B_1, B_2, B_3, \\dots$ by $B_1 = 1$ and for $n = 1, 2, 3, \\dots$\n\n$$\nB_{n+1} = \\frac{B_n^2 + 2}{2B_n}.\n$$\n\nProve that $B_{n+1} = A_{2^n}$ for all non-negative integers $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove the following claim by induction. For all $n$,\n\n$$\nA_{2n} = \\frac{A_n^2 + 2}{2A_n}.\n$$\n\nThe base case holds for $n = 1$ as $A_1 = 1$ and $A_2 = \\frac{3}{2}$. Assume the claim holds for $n$, we now show it holds for $n + 1$. This can be shown via the following computations. Let $A_n = x$, then\n\n$$\nA_{n+1} = \\frac{x+2}{x+1}, \\quad A_{2n+1} = \\frac{A_{2n} + 2}{A_{2n} + 1} = \\frac{\\frac{x^2+2}{2x} + 2}{\\frac{x^2+2}{2x} + 1} = \\frac{x^2 + 4x + 2}{x^2 + 2x + 2},\n$$\n\n$$\nA_{2(n+1)} = A_{2n+2} = \\frac{\\frac{x^2+4x+2}{x^2+2x+2} + 2}{\\frac{x^2+4x+2}{x^2+2x+2} + 1} = \\frac{3x^2 + 8x + 6}{2x^2 + 6x + 4} = \\frac{(x+2)^2 + 2(x+1)^2}{2(x+2)(x+1)}\n$$\n\n$$\n= \\frac{\\left(\\frac{x+2}{x+1}\\right)^2 + 2}{2\\left(\\frac{x+2}{x+1}\\right)} = \\frac{A_{n+1}^2 + 2}{2A_{n+1}},\n$$\n\nwhich proves the claim.\n\nNow it is straightforward to prove $B_{n+1} = A_{2^n}$ by another induction. It is true for $n = 0$ as $B_{0+1} = 1 = A_{2^0}$. Assume true for $n$, we show that it is also true for $n + 1$ using the claim above:\n\n$$\nB_{n+2} = \\frac{B_{n+1}^2 + 2}{2B_{n+1}} = \\frac{A_{2^n}^2 + 2}{2A_{2^n}} = A_{2 \\times 2^n} = A_{2^{n+1}}.\n$$\n\nThis completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15605,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be positive real numbers. Prove that\n\n$$\na^3 b^6 + b^3 c^6 + c^3 a^6 + 3a^3 b^3 c^3 \\geq abc(a^3 b^3 + b^3 c^3 + c^3 a^3) + a^2 b^2 c^2 (a^3 + b^3 + c^3)\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "After dividing both sides of the given inequality by $a^3 b^3 c^3$, it becomes\n\n$$\n\\left(\\frac{b}{c}\\right)^3 + \\left(\\frac{c}{a}\\right)^3 + \\left(\\frac{a}{b}\\right)^3 + 3 \\geq \\left(\\frac{a}{c} \\cdot \\frac{b}{c} + \\frac{b}{a} \\cdot \\frac{c}{a} + \\frac{c}{b} \\cdot \\frac{a}{b}\\right) + \\left(\\frac{a}{b} \\cdot \\frac{a}{c} + \\frac{b}{a} \\cdot \\frac{c}{a} + \\frac{c}{a} \\cdot \\frac{c}{b}\\right)\n$$\n\nSet\n\n$$\n\\frac{b}{a} = \\frac{1}{x}, \\quad \\frac{c}{b} = \\frac{1}{y}, \\quad \\frac{a}{c} = \\frac{1}{z}.\n$$\n\nThen $xyz = 1$, and by substituting into the previous inequality, we find\n\n$$\nx^3 + y^3 + z^3 + 3 \\geq \\left(\\frac{y}{z} + \\frac{z}{x} + \\frac{x}{y}\\right) + \\left(\\frac{x}{z} + \\frac{y}{x} + \\frac{z}{y}\\right).\n$$\n\nMultiplying both sides by $xyz$ (with $xyz = 1$), the inequality is equivalent to\n\n$$\nx^3 + y^3 + z^3 + 3xyz - xy^2 - yz^2 - zx^2 - yx^2 - zy^2 - xz^2 \\geq 0.\n$$\n\nNotice that by the special case of Schur's inequality:\n\n$$\nx^r(x - y)(x - z) + y^r(y - x)(y - z) + z^r(z - y)(z - x) \\geq 0, \\quad x, y, z \\geq 0, \\ r > 0,\n$$\n\nwith $r=1$ there holds\n\n$$\nx(x - y)(x - z) + y(y - x)(y - z) + z(z - y)(z - x) \\geq 0\n$$\n\nwhich, after expansion, coincides with the previous expression.\n\n*Remark 1.* The inequality above immediately follows by supposing (without loss of generality) that $x \\geq y \\geq z$, and then writing the left side as\n\n$$\n(x - y)(x(x - z) - y(y - z)) + z(y - z)(z - x),\n$$\n\nwhich is obviously $\\geq 0$.\n\n*Remark 2.* One can also obtain the relation using the substitution $x = ab^2$, $y = bc^2$, and $z = ca^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15606,
    "subject": "Mathematics (Olympiad)",
    "question": "$n$ girls are standing in a circle, each holding exactly 1 card. One girl gives her card to the girl on her left, who in turn gives 2 cards to the girl on her left. The girl who got the cards gives 1 card to the girl on her left. The girl who got the card gives 2 cards to her left. Continuing this way, each girl gives alternating 1 or 2 cards to the girl on her left. Anyone who has no card leaves the game immediately. Find all values of $n$ such that all cards are collected by one girl.",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: $n = 2$, $n = 2^k + 1$, $n = 2^k + 2$, where $k \\ge 1$.\n\nSolution omitted.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15607,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the maximum value of\n$$\n\\sum_{j=1}^{2008} \\sin \\theta_j \\cos \\theta_{j+1}\n$$\nwhere the indices are taken modulo $2008$ (i.e., $\\theta_{2009} = \\theta_1$).",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $\\sin \\theta_j \\cos \\theta_{j+1} \\le \\frac{1}{2}(\\sin^2 \\theta_j + \\cos^2 \\theta_{j+1})$ for any $j$. Thus,\n\n$$\n\\sum_{j=1}^{2008} \\sin \\theta_j \\cos \\theta_{j+1} \\le \\frac{1}{2} \\sum_{j=1}^{2008} (\\sin^2 \\theta_j + \\cos^2 \\theta_j) = \\frac{1}{2} \\sum_{j=1}^{2008} 1 = 1004.\n$$\n\nEquality holds when $\\theta_j = \\frac{\\pi}{4}$ for each $j$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 15608,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all arrays of prime numbers $(a, b, c)$ satisfying the following conditions:\n\n1. $a < b < c < 100$, where $a$, $b$, and $c$ are all prime numbers.\n2. $a + 1$, $b + 1$, $c + 1$ form a geometric progression.",
    "options": [],
    "answer": "See solution",
    "solution": "From condition (2), we have:\n\n$$\n(a+1)(c+1) = (b+1)^2.\n$$\n\nLet $a+1 = n^2 k$, $c+1 = m^2 k$, where $k$ has no square factor greater than 1. Then:\n\n$$\n(mn)^2 k^2 = (b+1)^2\n$$\n\nSo $b+1 = mnk$. Now, $k$ must be square-free and greater than 1 (otherwise $c$ is composite). Thus, the primes are:\n\n$$\n\\begin{cases}\na = k n^2 - 1, \\\\\nb = k m n - 1, \\\\\nc = k m^2 - 1\n\\end{cases}\n$$\n\nwith $1 \\leq n < m$, $a < b < c < 100$, and $k$ square-free, $k > 1$.\n\nIf $k=1$, $c = m^2 - 1$ is composite for $m \\geq 3$. So $k > 1$.\n\nFor $m=2$, $n=1$:\n\n$$\n\\begin{cases}\na = k - 1, \\\\\nb = 2k - 1, \\\\\nc = 4k - 1\n\\end{cases}\n$$\n\nwith $c < 100 \\implies k < 25$. If $k \\equiv 1 \\pmod{3}$, $c$ is divisible by 3 and composite. If $k \\equiv 2 \\pmod{3}$ and $k$ is even, further analysis is needed. Continue this process for all $k$ (square-free, $k > 1$) and $1 \\leq n < m$ such that $a, b, c$ are all primes less than 100.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15609,
    "subject": "Mathematics (Olympiad)",
    "question": "Assume there exist positive integers $a, b, c, d, n$ such that\n$$\na^2 + b^2 + c^2 + d^2 - 4\\sqrt{abcd} = 7 \\cdot 2^{2n-1},\n$$\nand $n$ is minimized. Prove that there are no such solutions $(a, b, c, d, n)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose such a solution $(a, b, c, d, n)$ exists with minimal $n$.\n\nSince $\\sqrt{abcd}$ is rational, $abcd$ must be a perfect square. Note that\n$$\na^2 + b^2 + c^2 + d^2 - 4\\sqrt{abcd} = (a-b)^2 + (c-d)^2 + 2(\\sqrt{ab} - \\sqrt{cd})^2.\n$$\n\n**Lemma 1.** $a, b, c, d$ are all distinct.\n\n*Proof.* Assume $a = b$. Then $abcd = a^2cd$ is a square, so $cd$ is a square. Let $M = c-d$, $N = a-\\sqrt{cd}$, both integers. Then $M^2 + 2N^2 = 7 \\cdot 2^{2n-1}$. Since squares mod 7 are $0,1,2,4$, both $M$ and $N$ must be multiples of 7, so $M^2 + 2N^2$ is divisible by 49, but $7 \\cdot 2^{2n-1}$ is not, a contradiction.\n\n**Lemma 2.** The product of any two among $a, b, c, d$ differs from the product of the other two.\n\n*Proof.* Assume $ab = cd$. Let $M = a-b$, $N = c-d$, both integers. Then $M^2 + N^2 = 7 \\cdot 2^{2n-1}$. As above, both $M$ and $N$ must be multiples of 7, so $M^2 + N^2$ is divisible by 49, but $7 \\cdot 2^{2n-1}$ is not, a contradiction.\n\nConsider cases for $n$:\n\n**Case 1:** $n = 1$\n\nBy symmetry and Lemma 1, assume $a < b < c < d$. From\n$$(d-a)^2 + (c-b)^2 + 2(\\sqrt{ad} - \\sqrt{bc})^2 = 14,$$\nwe get $d-a \\leq \\sqrt{14} < 4$, so $a = b-1 = c-2 = d-3$. Calculations lead to a contradiction.\n\n**Case 2:** $n > 1$\n\nIf all $a, b, c, d$ are even, then $(a/2, b/2, c/2, d/2, n-1)$ is a solution, contradicting minimality of $n$. Thus, at least one is odd. Since squares mod 4 are $0$ or $1$, and $a^2 + b^2 + c^2 + d^2 = 7 \\cdot 2^{2n-1} + 4\\sqrt{abcd}$ is divisible by 4, all $a, b, c, d$ are odd. Then $abcd \\equiv 1 \\pmod 4$, so an even number among $a, b, c, d$ are $3 \\pmod 4$. Without loss, $a \\equiv b \\pmod 4$, $c \\equiv d \\pmod 4$.\n\nLet $g = \\gcd(ab, cd)$. Then $ab = g x^2$, $cd = g y^2$ for coprime $x, y$, and $(\\sqrt{ab} - \\sqrt{cd})^2 = g(x-y)^2$. Since $ab \\equiv 1 \\pmod 4$, $g \\equiv 1 \\pmod 4$.\n\nBy Lemmas 1 and 2, there exist positive odd $p, q, r$ and non-negative $\\alpha, \\beta, \\gamma$ such that $|a-b| = p 2^\\alpha$, $|c-d| = q 2^\\beta$, $|x-y| = r 2^\\gamma$. Then\n$$\n7 \\cdot 2^{2n-1} = p^2 2^{2\\alpha} + q^2 2^{2\\beta} + g r^2 2^{2\\gamma+1}.\n$$\nBut $p^2 \\equiv q^2 \\equiv 1 \\pmod 8$, $g r^2 \\equiv 1 \\pmod 4$, which leads to a contradiction by the following lemma:\n\n**Lemma 3.** There are no non-negative integers $\\alpha, \\beta, \\gamma$, positive integers $n, s, t, u$ with $s \\equiv t \\equiv 1 \\pmod 8$, $u \\equiv 1 \\pmod 4$, and\n$$\ns 2^{2\\alpha} + t 2^{2\\beta} + u 2^{2\\gamma+1} = 7 2^{2n-1}.\n$$\n\n*Proof.* By symmetry, assume $\\alpha \\leq \\beta$.\n\n- If $\\alpha \\leq \\gamma$, $s + t 2^{2(\\beta-\\alpha)} + u 2^{2(\\gamma-\\alpha)+1} = 7 2^{2(n-\\alpha)-1}$, so $n-\\alpha \\geq 1$. Both $u 2^{2(\\gamma-\\alpha)+1}$ and $7 2^{2(n-\\alpha)-1}$ are even, so $s + t 2^{2(\\beta-\\alpha)}$ is even, so $\\alpha = \\beta$. Then $s + t$ not divisible by 4, so $\\gamma-\\alpha = 0$ or $n-\\alpha = 1$. Both cases lead to contradictions modulo 8.\n- If $\\gamma < \\alpha$, $s 2^{2(\\alpha-\\gamma)} + t 2^{2(\\beta-\\gamma)} + 2u = 7 2^{2(n-\\gamma)-1}$, which mod 4 gives $n-\\gamma = 1$. Then $2^{2(\\beta-\\gamma)} < 14$, so $\\alpha-\\gamma = \\beta-\\gamma = 1$, so $2s + 2t + u = 7$, but $2s + 2t + u \\equiv 1 \\pmod 4$, a contradiction.\n\nTherefore, no such solution exists.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15610,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$ be a real number such that $285 = (3^x - 3^{-x})^2$. Find the value of $3^x + 3^{-x}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $t = 3^x$. Then $t - \\frac{1}{t} = \\sqrt{285}$, so $t^2 - \\sqrt{285} t - 1 = 0$.\n\nThe roots of this quadratic are:\n$$\n\\frac{1}{2}\\left(\\sqrt{285} \\pm \\sqrt{285 + 4}\\right) = \\frac{1}{2}\\left(\\sqrt{285} \\pm 17\\right)\n$$\nSince $3^x$ is positive, $3^x + 3^{-x} = t + \\frac{1}{t} = 17$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15611,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\ldots, x_n$ be positive real numbers. For each positive integer $k$, define\n$$\nS_k = x_1^k + x_2^k + \\ldots + x_n^k.\n$$\n\n(i) Prove that if $S_1 < S_2$, the sequence $S_1, S_2, S_3, \\ldots$ is strictly increasing.\n\n(ii) Prove that it is possible that $S_1 > S_2$ and yet the sequence $S_1, S_2, S_3, \\ldots$ is not strictly decreasing.",
    "options": [],
    "answer": "See solution",
    "solution": "We will actually prove a stronger statement: if for some $k$ it is true that $S_k < S_{k+1}$, then $S_j < S_{j+1}$ for all $j \\geq k$ (i.e., the sequence is strictly increasing starting from that point).\n\nFor our proof, we use the following key observation. For any positive real number $x$ and positive integer $m$, the inequality $x^{m+2} - x^{m+1} \\geq x^{m+1} - x^m$ holds. Indeed, dividing both sides by $x^m$ gives $x^2 - x \\geq x - 1$, that is, $x^2 - 2x + 1 \\geq 0$, which is true because the left-hand side equals $(x-1)^2$.\n\nNow, fixing $m$ and adding these inequalities for each $x_i$, we conclude that $S_{m+2} - S_{m+1} \\geq S_{m+1} - S_m$. In particular, if for some $k$ it happens that $S_{k+1} > S_k$, then for all $j \\geq k$ we have\n$$\nS_{j+1} - S_j \\geq S_j - S_{j-1} \\geq \\ldots \\geq S_{k+1} - S_k > 0\n$$\nas claimed.\n\nTo construct a counterexample for part (ii), take $n = 2$, $x_1 = 1.1$, and $x_2 = 0.5$. For these numbers,\n$$\nS_1 = 1.6 \\quad \\text{and} \\quad S_2 = 1.21 + 0.25 = 1.46 < S_1.\n$$\nHowever, the sequence is not strictly decreasing, because for large enough $n$ we have\n$$\nS_n > (1.1)^n > 1.6 = S_1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15612,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a fixed parallelogram with $AB$ parallel to $DC$, and $AD$ parallel to $BC$. A point $E$, different from $A$ and $B$, is chosen on the side $AB$. Let $K$ be the centre of the circle through $A$, $D$, and $E$. Let $L$ be the centre of the circle through $B$, $C$, and $E$. Prove that no matter where $E$ is chosen, the length $KL$ is always the same.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K$ be the intersection of the perpendicular bisectors of $AD$ and $AE$, and $L$ the intersection of the perpendicular bisectors of $BC$ and $BE$.\n\nThe perpendicular bisectors of $AE$ and $BE$ are both perpendicular to $AB$ and are a fixed distance ($\\frac{1}{2}|AB|$) apart, with $AE$ closer to $A$ than $BE$. The perpendicular bisectors of $AD$ and $BC$ are both perpendicular to $AD$ and are independent of $E$. Thus, these four perpendicular bisectors form a parallelogram whose dimensions and orientation remain constant as $E$ varies on $AB$.\n\nPoints $K$ and $L$ are opposite vertices of this parallelogram. Therefore, the length $KL$, a diagonal of the parallelogram, is independent of $E$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15613,
    "subject": "Mathematics (Olympiad)",
    "question": "We consider sports tournaments with $n \\ge 4$ participating teams where every pair of teams plays against one another at most one time. We call such a tournament *balanced* if any four participating teams play exactly three matches between themselves. So, not all teams play against one another.\n\nDetermine the largest value of $n$ for which a balanced tournament with $n$ teams exists.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that $5$ is the largest value of $n$ for which a balanced tournament with $n$ teams exists.\n\nFirst, we show that in a balanced tournament with $n \\ge 5$ teams, there are no three teams that all play against one another in the tournament.\n\nSuppose, for contradiction, that there exist three teams that all play against each other, say teams $A$, $B$, and $C$. Because $n \\ge 5$, there are two other teams, say $D$ and $E$. Since $A$, $B$, and $C$ already play three matches between them, there are no other matches among the quadruple $A$, $B$, $C$, and $D$. In other words, $D$ does not play against $A$, $B$, or $C$. The same holds for team $E$. If we now consider the quadruple $A$, $B$, $D$, and $E$, we see that there are at most two matches: $A$ against $B$, and possibly $D$ against $E$. This means that we have found four teams such that there are not exactly three matches between these four teams, which is a contradiction.\n\nNow, we show that a balanced tournament is not possible with $n \\ge 6$ teams. Suppose that $n \\ge 6$ and, for contradiction, that a balanced tournament with $n$ teams exists. Consider the first six teams, say teams $A$ to $F$. Suppose that $A$ plays against at most two of these teams, say at most against $B$ and $C$ but not against $D$, $E$, and $F$. Since three matches have to be played among the quadruple $A$, $D$, $E$, and $F$, the teams $D$, $E$, and $F$ must play among themselves, but $A$ does not play any of them, so only three matches are possible among these four, which is not possible. Thus, $n \\le 5$.\n\nIt is possible to construct a balanced tournament with $n = 5$ teams: arrange the teams in a cycle and let each team play against its two neighbors. For any four teams, there are exactly three matches among them. Therefore, the largest value of $n$ is $5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15614,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function such that for all real numbers $x$ and $y$,\n\n$$\nx^2 f(y) = f(x)[f(yf(x) - 1) + 1].\n$$\n\nFind all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "The constant function $f(x) = 0$ is a solution.\n\nLet $f$ be a solution that is not identically $0$. We shall show that $f(x) = x$ for all $x$.\n\nLetting $x = 0$ in the given equation, we get\n\n$$\nf(0)[f(yf(0) - 1) + 1] = 0.\n$$\n\nSuppose $f(0) \\neq 0$. Let $x = yf(0) - 1$. As $y$ ranges over all real numbers, so does $x$. Thus we get $f(x) = -1$ for all $x$. But this does not satisfy the given equation. So $f(0) = 0$.\n\nNow suppose that $f(a) = 0$ for some $a \\neq 0$. Then the original equation becomes $0 = a^2 f(y)$ for all $y$, implying that $f(y) = 0$ for all $y$. This contradicts our assumption that $f$ is not identically $0$. Thus $f(x) = 0$ iff $x = 0$.\n\nLetting $x = y = 1$ in the given equation, we have $f(f(1) - 1) = 0$. Thus $f(1) = 1$. When $x = 1$, the original equation becomes\n\n$$\nf(y - 1) = f(y) - 1. \\quad (1)\n$$\n\nLet $y = 1$ in the given equation and use (1) to obtain\n\n$$\nx^2 - f(x) = f(x)[f(f(x) - 1)] = f(x)[f(f(x)) - 1] = f(x)f(f(x)) - f(x).\n$$\n\nThus\n\n$$\nf(x)f(f(x)) = x^2 \\qquad (2)\n$$\n\nNow replace $x$ by $x-1$ in (2) and apply (1) three times, and finally apply (2):\n\n$$\n\\begin{aligned}\n(x-1)^2 &= f(x-1)f(f(x-1)) = (f(x)-1)[f(f(x)-1)] \\\\\nx^2 - 2x + 1 &= (f(x)-1)[f(f(x))-1] = f(x)f(f(x)) - f(x) - f(f(x)) + 1 \\\\\n&= x^2 - f(x) - f(f(x)) + 1\n\\end{aligned}\n$$\n\nTherefore\n\n$$\nf(x) + f(f(x)) = 2x. \\qquad (3)\n$$\n\nEliminating $f(f(x))$ from (2) and (3) gives\n\n$$\n[x - f(x)]^2 = 0,\n$$\n\nso that $f(x) = x$, as claimed. It is clear that this is also a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15615,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an isosceles triangle with apex $A$ and altitude $AD$. On $AB$, choose a point $F$ distinct from $B$ such that $CF$ is tangent to the incircle of $ABD$. Suppose that $\\triangle BCF$ is isosceles.\n\nShow that these conditions uniquely determine:\n\na) which vertex of $BCF$ is its apex;\n\nb) the size of $\\angle BAC$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Consider the possible locations of the apex of triangle $BCF$.\n\nIf $F$ were the apex, then $F$ would lie on the perpendicular bisector of $BC$, i.e., on the line $AD$, so $F = A$. Therefore, $AC$ would be a tangent to the incircle of triangle $ABD$. But $AB$ and $AD$ are already tangents from $A$ to the same circle, and only two tangents can be drawn from a point to a circle. Hence, this case is impossible.\n\nLet $B$ be the apex. Since $CBF$ is a base angle of the isosceles triangle $ABC$, $\\angle CBF < 90^\\circ$, so $\\angle BCF > 45^\\circ$. Let $K$ be the point of tangency of $CF$ with the incircle of $ABD$, and let $L$ be the projection of $K$ onto $BC$. By construction, $KL < 2r$, where $r$ is the radius of the incircle of $ABD$.\n\n![](images/prob1617_p28_data_28d7b1baac.png)\n\nOn the other hand, $\\angle LCK = \\angle BCF > 45^\\circ$ implies $KL > CL > CD = BD > 2r$. Hence, this case is also impossible.\n\nTherefore, the apex of triangle $BCF$ can only be $C$.\n\nb) Let the apex be $C$. Fix a point $D$, mutually perpendicular lines $l_1$ and $l_2$ both passing through $C$, and a circle $c$ tangent to both lines with radius $r$. Choose $A$ on $l_1$ so that $DA > 2r$ and the tangent point of $l_1$ and $c$ lies on $DA$. Point $B$ is determined as the intersection of $l_2$ and the second tangent from $A$ to $c$; $C$ is symmetric to $B$ with respect to $DA$, and $F$ is defined as in the problem. As $A$ moves away from $D$, $B$ and $C$ approach $D$, so $\\angle BAC$ decreases and $\\angle BCF$ increases. Thus, these angles can be equal only in one case.\n\n*Remark.* By extending the argument in part b), it can be shown that an isosceles triangle satisfying the problem's conditions exists. $\\angle BAC$ can approach zero, while $\\angle BAC = 90^\\circ$ implies $\\angle BCF < \\angle BCA = 45^\\circ < \\angle BAC$.\n\nCalculations show that the apex angle $\\alpha$ satisfying the conditions fulfills the equation\n\n$$\ntan^3 \\alpha - 8 \\tan^2 \\alpha + 17 \\tan \\alpha - 8 = 0\n$$\n\nand its approximate value is $\\alpha \\approx 33.3^\\circ$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 15616,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all possible values of positive integer $n$ such that there are $n$ different 3-element subsets $A_1, A_2, \\dots, A_n$ of the set $\\{1, 2, \\dots, n\\}$, with $|A_i \\cap A_j| \\neq 1$ for all $i \\neq j$.",
    "options": [],
    "answer": "See solution",
    "solution": "The set of positive integers satisfying the given condition consists of all positive multiples of 4.\n\n**Construction:**\nLet $n = 4k$ for some $k \\in \\mathbb{Z}_+$. Define $A_1, A_2, \\dots, A_{4k}$ as follows:\n\nFor each $1 \\leq i \\leq k$ and $0 \\leq j \\leq 3$,\n$$\nA_{4i-j} = \\{4i-3, 4i-2, 4i-1, 4i\\} \\setminus \\{4i-j\\}\n$$\nThis gives $4k$ distinct 3-element subsets, and for any $i \\neq j$, $|A_i \\cap A_j| \\neq 1$.\n\n**Necessity:**\nSuppose $4 \\nmid n$ and that such $n$ subsets exist. Let $A_1 = \\{a, b, c\\}$, and consider all subsets with non-empty intersection with $A_1$; relabel so these are $A_2, \\dots, A_m$. Let $U = A_1 \\cup A_2 \\cup \\dots \\cup A_m$.\n\n- If $|U| = 3$, then $m = 1 < |U|$.\n- If $|U| = 4$, then $m \\leq \\binom{4}{3} = 4 = |U|$.\n- If $|U| \\geq 5$, suppose $m \\geq |U|$. For $2 \\leq i, j \\leq m$, $|A_1 \\cap A_i| = 2$ and $|A_1 \\cap A_j| = 2$. Since $|A_1| = 3$, $A_i \\cap A_j \\neq \\emptyset$, and $|A_i \\cap A_j| \\neq 1$ implies $|A_i \\cap A_j| = 2$.\n\nConsider $A_1 \\cap A_2, A_1 \\cap A_3, A_1 \\cap A_4, A_1 \\cap A_5$. By the pigeonhole principle, two intersections are equal; relabel so $A_2 = \\{a, b, d\\}$, $A_3 = \\{a, b, e\\}$. For $4 \\leq i \\leq m$, $a, b \\in A_i$; otherwise, $A_i$ would have at least one of $a$ or $b$ and three other elements, so $|A_i| \\geq 4$, a contradiction. Thus $|U| = m+2$, contradicting $|U| = m$.\n\nTherefore, the subsets can be partitioned into groups where each group has at most as many subsets as elements, and since $n$ subsets use $n$ elements, each group must have exactly four subsets. Thus, $4 \\mid n$.\n\n**Conclusion:**\nAll and only positive integers $n$ divisible by $4$ satisfy the condition. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15617,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n$ gold coins and $n$ bronze coins, arranged in a row in some arbitrary initial order. A chain is any subsequence of consecutive coins of the same type. Given a fixed positive integer $k$ with $k \\le 2n$, Marianne repeatedly performs the following operation: she identifies the longest chain containing the $k$th coin from the left, and moves all coins in that chain to the left end of the row.\n\nFor example, if $n = 4$ and $k = 4$, the process starting from the ordering $AABBBABA$ would be\n\n$$\n\\begin{align*}\nAAB\\underline{B}BABA &\\rightarrow B\\underline{B}BA\\underline{A}ABA \\\\\n&\\rightarrow A\\underline{A}A\\underline{B}\\underline{B}BBA \\\\\n&\\rightarrow B\\underline{B}\\underline{B}AAAA \\\\\n&\\rightarrow B\\underline{B}\\underline{B}AAAA \\rightarrow \\dots\n\\end{align*}\n$$\n\nFind all pairs $(n, k)$ with $1 \\le k \\le 2n$ such that for every initial ordering, at some moment during the process, the leftmost $n$ coins will all be of the same type.",
    "options": [],
    "answer": "See solution",
    "solution": "The desired pairs $(n, k)$ are those that satisfy $n \\le k \\le \\frac{3n+1}{2}$.\n\nAs defined in the problem, a chain is any subsequence of consecutive coins of the same type. We call it a \"block\" if it is a chain not contained in a longer chain. For $M = A$ or $B$, let $M^x$ represent a block of $M$ coins of length $x$. We are interested in whether an initial ordering will turn into a 2-block sequence $A^n B^n$ or $B^n A^n$ after finitely many operations.\n\n1. **Claim:** If $k < n$ or $k > \\frac{3n+1}{2}$, then the sequence cannot always turn into $A^n B^n$ or $B^n A^n$.\n\n**Proof:** If $k < n$, the sequence $A^{n-1}B^nA^1$ does not change after an operation, and cannot turn into a 2-block sequence. If $k > \\frac{3n+1}{2}$, let $m = \\lfloor\\frac{n}{2}\\rfloor$, $l = \\lceil\\frac{n}{2}\\rceil$, and the initial ordering be $A^mB^mA^lB^l$. As $k > \\frac{3n+1}{2} \\ge m+l+l \\ge m+m+l$, the $k$th coin always belongs to the rightmost block, and the process is\n\n$$\n\\begin{align*}\nA^m B^m A^l B^l &\\rightarrow B^l A^m B^m A^l \\\\\n&\\rightarrow A^l B^l A^m B^m \\\\\n&\\rightarrow B^m A^l B^l A^m \\\\\n&\\rightarrow A^m B^m A^l B^l \\rightarrow \\dots\n\\end{align*}\n$$\n\na 4-periodic cycle. It cannot turn into a 2-block sequence.\n\n2. **Claim:** If $n \\le k \\le \\frac{3n+1}{2}$, any initial ordering will turn into a 2-block sequence.\n\n**Proof:** After each operation, the number of blocks does not increase. Eventually, it stabilizes at $s$ blocks. It suffices to prove $s = 2$. Suppose $s > 2$. As $k \\ge n$ and $s > 2$, the $k$th coin cannot belong to the leftmost block, so the block being moved is not the leftmost one. If it is a middle block, then after the operation, the two blocks on its sides merge into one, reducing the number of blocks by 1, a contradiction. Therefore, when stabilized, the block being moved to the left is always the rightmost one. Since $k \\le (3n + 1)/2$, the length of the rightmost block is at least $2n - k + 1 \\ge (n + 1)/2$; since $s$ does not decrease after the operation, the first and the rightmost blocks are of different types, so $s$ is even. Let $s = 2t$, $t \\ge 2$, and suppose one of the stable orderings is $X_1 X_2 \\cdots X_{2t}$, where each $X_i$ is a block. The process is\n\n$$\n\\begin{align*}\nX_1 X_2 \\cdots X_{2t} &\\rightarrow X_{2t} X_1 \\cdots X_{2t-1} \\\\\n&\\rightarrow \\cdots \\\\\n&\\rightarrow X_2 X_3 \\cdots X_1 \\\\\n&\\rightarrow X_1 X_2 \\cdots X_{2t} \\rightarrow \\cdots\n\\end{align*}\n$$\n\nAt any moment, the rightmost block has length $\\ge \\frac{n+1}{2}$. So, each block has length $\\ge \\frac{n+1}{2}$, and the total length is at least $4 \\cdot \\frac{n+1}{2} > 2n$, a contradiction. Thus, $s = 2$ and the conclusion follows.\n\n$\\boxed{n \\le k \\le \\frac{3n+1}{2}}$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15618,
    "subject": "Mathematics (Olympiad)",
    "question": "For which positive integers $n \\ge 2$ do there exist $n$ odd (not necessarily different) numbers $a_1, a_2, \\dots, a_n$ such that\n\n$$\na_1^2 + a_2^2 + \\dots + a_n^2\n$$\nis a square of some positive integer?",
    "options": [],
    "answer": "See solution",
    "solution": "The square of an integer can only be congruent to $0$, $1$, or $4$ modulo $8$. Therefore, such $n$ must be of the form $8k + r$, where $r \\in \\{0, 1, 4\\}$. We can construct examples for these cases:\n\n- For $n = 4t$, let $a_1 = \\dots = a_{n-1} = 1$, $a_n = 2t-1$:\n  $$\na_1^2 + a_2^2 + \\dots + a_n^2 = (4t-1) \\cdot 1^2 + (2t-1)^2 = (2t)^2.\n  $$\n- For $n = 8t + 1$, let $a_1 = \\dots = a_{n-1} = 1$, $a_n = 2t-1$:\n  $$\na_1^2 + a_2^2 + \\dots + a_n^2 = (8t) \\cdot 1^2 + (2t-1)^2 = (2t+1)^2.\n  $$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15619,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $k$, find the number of non-negative integers not exceeding $10^k$ that satisfy the following conditions:\n\n1. $n$ is divisible by $3$.\n2. The digits of $n$ in decimal representation are in the set $\\{2, 0, 1, 5\\}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S = \\{2, 0, 1, 5\\}$.\nDefine\n$$\nA(n, i) = \\{\\overline{x_n x_{n-1} \\dots x_1} : x_j \\in S \\text{ and } x_1 + \\dots + x_n \\equiv i \\pmod{3}\\}\n$$\nLet $a_n$, $b_n$, and $c_n$ be the cardinalities of $A(n, 0)$, $A(n, 1)$, and $A(n, 2)$, respectively. Since a number is divisible by $3$ if and only if the sum of its digits is a multiple of $3$, we need to count $|A(k, 0)|$.\n\nConsider $\\overline{x_1 \\dots x_{n+1}} \\in A(n+1, 0)$:\n- If $x_{n+1} = 0$, then $(x_1, \\dots, x_n) \\in A(n, 0)$.\n- If $x_{n+1} = 2$ or $5$, then $(x_1, \\dots, x_n) \\in A(n, 1)$.\n- If $x_{n+1} = 1$, then $(x_1, \\dots, x_n) \\in A(n, 2)$.\n\nThus,\n$$\na_{n+1} = a_n + 2b_n + c_n\n$$\nSimilarly,\n$$\nb_{n+1} = a_n + b_n + 2c_n\n$$\n$$\nc_{n+1} = 2a_n + b_n + c_n\n$$\nThe total number of $k$-digit numbers with digits in $S$ is $4^k$, so\n$$\na_k + b_k + c_k = 4^k\n$$\nBy analyzing the recurrences, we find:\n- If $k \\equiv 0 \\pmod{3}$, then $a_k = b_k = c_k - 1$.\n- If $k \\equiv 1 \\pmod{3}$, then $a_k = b_k = c_k - 1$.\n- If $k \\equiv 2 \\pmod{3}$, then $a_k = c_k = b_k - 1$.\n\nFinally,\n- If $k$ is a multiple of $3$, $a_k = \\dfrac{4^k + 2}{3}$.\n- Otherwise, $a_k = \\dfrac{4^k - 1}{3}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15620,
    "subject": "Mathematics (Olympiad)",
    "question": "The triangle $EBC$ has a right angle at $E$.\n\n![](images/Irish_2018_p37_data_3540623260.png)\n\nLet $Q$ be the projection of $E$ onto the hypotenuse $BC$ of this triangle. Find the length $|ED|$ given the following:\n- $|EC| = 40$\n- $|EB| = 30$\n- $|BC| = 50$\n- $|CD| = 38$\n",
    "options": [],
    "answer": "See solution",
    "solution": "Twice the area of triangle $EBC$ is equal to\n\n$$\n|EQ| \\times |BC| = |EC| \\times |EB|,\n$$\n\nso\n\n$$\n|EQ| = \\frac{|EC| \\times |EB|}{|BC|} = \\frac{40 \\times 30}{50} = 24.\n$$\n\nBy Pythagoras,\n\n$$\n|QC| = \\sqrt{|EC|^2 - |EQ|^2} = \\sqrt{40^2 - 24^2} = \\sqrt{1600 - 576} = \\sqrt{1024} = 32,\n$$\n\nand so\n\n$$\n|QD| = |QC| + |CD| = 32 + 38 = 70.\n$$\n\nHence,\n\n$$\n|ED| = \\sqrt{|QE|^2 + |QD|^2} = \\sqrt{24^2 + 70^2} = \\sqrt{576 + 4900} = \\sqrt{5476} = 74.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15621,
    "subject": "Mathematics (Olympiad)",
    "question": "For any integer $n \\ge 2$, denote by $A_n$ the set of solutions of the equation\n\n$$\nx = \\lfloor \\frac{x}{2} \\rfloor + \\lfloor \\frac{x}{3} \\rfloor + \\dots + \\lfloor \\frac{x}{n} \\rfloor.\n$$\n\na) Determine the set $A_2 \\cup A_3$.\n\nb) Prove that the set $A = \\bigcup_{n \\ge 2} A_n$ is finite and find $\\max A$.",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that $A_n \\subset \\mathbb{Z}$ for all $n \\in \\mathbb{N}$, $n \\ge 2$.\n\na) The elements of $A_2$ satisfy the inequalities $x - 2 < 2x \\leq x$. By inspection, we obtain $A_2 = \\{-1, 0\\}$.\n\nThe elements of $A_3$ satisfy the inequalities $5x - 12 < 6x \\leq 5x$. By inspection, we have $A_3 = \\{-7, -5, -4, -3, -2, 0\\}$, so $A_2 \\cup A_3 = \\{-7, -5, -4, -3, -2, -1, 0\\}$.\n\nb) For $n \\ge 4$ and $x \\in A_n$ we have $x \\le \\left(\\frac{1}{2} + \\frac{1}{3} + \\cdots + \\frac{1}{n}\\right) x$.\n\nFrom $\\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n} > 1$ we obtain $x \\ge 0$.\n\nFor $x, n \\in \\mathbb{Z}$ and $n \\ge 2$ we get $\\lfloor \\frac{x}{n} \\rfloor \\ge \\frac{x - (n - 1)}{n}$. Therefore, if $n \\ge 4$ and $x \\in A_n$, then\n\n$$\nx \\ge \\lfloor \\frac{x}{2} \\rfloor + \\lfloor \\frac{x}{3} \\rfloor + \\lfloor \\frac{x}{4} \\rfloor \\ge \\frac{x-1}{2} + \\frac{x-2}{3} + \\frac{x-3}{4},\n$$\n\nimplying $x \\le 23$. Hence the set $A$ is upper bounded.\n\nBecause $A \\subset \\{-5, -4, \\dots, 23\\}$ and $23 \\in A_4$, then $\\max A = 23$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15622,
    "subject": "Mathematics (Olympiad)",
    "question": "For fixed positive integers $a$ and $b$, find all strictly increasing functions $f$ from positive integers to positive integers such that, for any positive integer $n > a$,\n\n$$\nf(f(n - a) + n) = n + b.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that $f(n) \\geq n$ for each $n$, since $f$ is strictly increasing. Consequently, $n + b = f(f(n - a) + n) \\geq f(n - a) + n$. Hence, $b \\geq f(n - a) \\geq n - a$ holds for each positive integer $n > a$, which is impossible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15623,
    "subject": "Mathematics (Olympiad)",
    "question": "Определи ги комплексните броеви $z$ за кои\n\n$$\n|z| = \\frac{1}{|z|} = |z - 1|.\n$$\n",
    "options": [],
    "answer": "See solution",
    "solution": "Јасно е дека равенките се определени за $z \\neq 0$. Од равенката $|z| = \\frac{1}{|z|}$, добиваме $|z|^2 = 1$, односно\n\n$$\n|z| = 1. \\qquad (1)\n$$\n\nОд претходната равенка и равенката $|z| = |z - 1|$ ја добиваме равенката\n\n$$\n|z - 1| = 1. \\qquad (2)\n$$\n\nАко комплексниот број $z$ го запишеме во алгебарски облик $z = x + iy$, од (1) и (2) добиваме\n\n$$\n\\begin{cases} x^2 + y^2 = 1 \\\\ (x - 1)^2 + y^2 = 1 \\end{cases} \\qquad (3)\n$$\n\nАко од првата равенка ја одземеме втората равенка, ја добиваме равенката $2x - 1 = 0$, од каде $x = \\frac{1}{2}$. Ако замениме во било која од равенките од системот (3), ја добиваме равенката $y^2 = \\frac{3}{4}$. Нејзини решенија се $y = \\pm \\frac{\\sqrt{3}}{2}$. Според тоа, множеството броеви\n\n$$\n\\left\\{ \\frac{1}{2} + i \\frac{\\sqrt{3}}{2}, \\frac{1}{2} - i \\frac{\\sqrt{3}}{2} \\right\\}\n$$\nе решение на равенките.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15624,
    "subject": "Mathematics (Olympiad)",
    "question": "Сургуулийн сурагчид хэдэн дугуйланд явдаг байв. Дугуйлан бүр 8 сурагчтай ба аль ч 9 дугуйланд зэрэг явдаг сурагч олдох бол бүх дугуйланд явдаг сурагч олдохыг батал.",
    "options": [],
    "answer": "See solution",
    "solution": "Дугуйлангуудыг $D_1, D_2, \\ldots, D_k$ гэе. $D_i$ дугуйлан бүрт 8 хүүхэд суралцдаг буюу $|D_i| = 8$, $i = 1, 2, \\ldots, k$ байна. Эсрэгээр нь, бүх дугуйланд сурдаг хүүхэд олдохгүй гэж үзье. $D_1 = \\{d_1, d_2, \\ldots, d_8\\}$ дугуйланг авч үзье. $d_i$ (хүүхэд) бүрийн хувьд $d_i$ ордоггүй нэг дугуйлан олдоно (бид бүх дугуйланд сурдаг хүүхэд байхгүй гэж үзсэн тул). Тэр дугуйланг $D_{i_1}$ гэе. Өөрөөр хэлбэл, $d_i \\notin D_{i_1}$. Тэгвэл $D_1, D_{i_1}, D_{i_2}, \\ldots, D_{i_8}$ гэсэн 9 дугуйланд зэрэг сурдаг сурагч олдохгүй болж, өгсөн нөхцөлд харшилна. (Хэрэв $i_1, i_2, \\ldots, i_8$-ийн зарим нь тэнцүү буюу $D_{i_1} = D_{i_2}$ бол дурын өөр нэг дугуйлан оруулаад 9 дугуйлан болгож, огтлолцол нь хоосон олонлог болно.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15625,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $R$ be reflection in $BC$: $(x, y) \\rightarrow (3 - y, 3 - x)$. Let $S$ be reflection in $AC$: $(x, y) \\rightarrow (-x, y)$. Let $T$ be reflection in $AB$: $(x, y) \\rightarrow (x, -y)$. Let $RS$ denote the reflection $R$ followed by the reflection $S$.\n\nFind the translation corresponding to $Q^{10}P^2$, where $Q = STR$ and $P = TRS$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $Q = STR$ is $(x, y) \\rightarrow (y + 3, x + 3)$. So $Q^2$ is $(x, y) \\rightarrow (x + 6, y + 6)$, and $Q^{10}$ is $(x, y) \\rightarrow (x + 30, y + 30)$.\n\n$P = TRS$ is $(x, y) \\rightarrow (x, -y) \\rightarrow (y+3, -x+3) \\rightarrow (-y-3, -x+3)$. So $P^2$ is $(x, y) \\rightarrow (x-6, y+6)$. Hence $Q^{10}P^2$ is $(x, y) \\rightarrow (x+24, y+36)$, which is the required translation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15626,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $2n + 1$ (where $n \\in \\mathbb{N}$) asterisks be written on the blackboard: $\\underbrace{**\\ldots*}_{2n+1}$.\n\nShow Bob's winning strategy in the following game:\n- Ann and Bob take turns replacing asterisks with digits (Ann goes first).\n- The first asterisk is special: Ann may choose to replace it at any turn.\n- The goal is to form a number divisible by 11.\n\nIs there a strategy that guarantees Bob's win?",
    "options": [],
    "answer": "See solution",
    "solution": "It is well-known that a natural number $n$ is divisible by 11 if and only if $S_0 - S_e \\equiv 0 \\pmod{11}$, where $S_0$ and $S_e$ are the sums of the digits in the odd and even positions, respectively, in the decimal representation of $n$.\n\nBob's strategy:\n- If Ann replaces an asterisk (not the first one) at position $i$ with digit $c$, Bob responds by replacing an asterisk of the opposite parity (odd/even) with the same digit $c$ (also not the first asterisk).\n- If Ann eventually replaces the first asterisk with digit $d$, then $S_0 - S_e = d$. Since $d \\neq 0$, the resulting number is not divisible by 11.\n- If Ann replaces the first asterisk with $c \\neq 0$ before her last move, Bob replaces an even-position asterisk with $c-1$ and continues mirroring as above. At the end, Ann must replace an odd asterisk with digit $d$, so $S_0 - S_e = d + c - (c-1) = d + 1$. Since $0 < d+1 < 11$, the number is not divisible by 11.\n\nThus, Bob always wins.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15627,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$ be an integer and let $x_1, x_2, \\dots, x_n$ be real numbers in the interval $[0, 1]$. Let $s = x_1 + x_2 + \\dots + x_n$, with $s \\ge 3$. Prove that there exist integers $i$ and $j$ with $1 \\le i < j \\le n$ such that\n\n$$\n2^{j-i} x_i x_j > 2^{s-3}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $1 \\le a < b \\le n$ be such that $2^{b-a} x_a x_b$ is maximal. This choice of $a$ and $b$ implies that\n\n$$\nx_{a+t} \\le 2^t x_a, \\quad \\forall t = 1, 2, \\dots, b-a-1,\n$$\n\nand similarly\n\n$$\nx_{b-t} \\le 2^t x_b, \\quad \\forall t = 1, 2, \\dots, b-a-1.\n$$\n\nSuppose that $x_a \\in \\left(\\frac{1}{2^{u+1}}, \\frac{1}{2^u}\\right]$ and $x_b \\in \\left(\\frac{1}{2^{v+1}}, \\frac{1}{2^v}\\right]$, and write $x_a = 2^{-\\alpha}$, $x_b = 2^{-\\beta}$. Then\n\n$$\n\\sum_{i=1}^{a+u-1} x_i \\le 2^u x_a \\left( \\frac{1}{2} + \\frac{1}{4} + \\dots + \\frac{1}{2^{a+u-1}} \\right) < 2^u x_a \\le 1,\n$$\n\nand similarly,\n\n$$\n\\sum_{i=b-v+1}^{n} x_i \\le 2^v x_b \\left( \\frac{1}{2} + \\frac{1}{4} + \\dots + \\frac{1}{2^{n-b+v}} \\right) < 2^v x_b \\le 1.\n$$\n\nIn other words, the sum of the $x_i$ for $i$ outside of the interval $[a+u, b-v]$ is strictly less than $2$. Since the total sum is at least $3$, and each term is at most $1$, it follows that this interval must have at least two integers, i.e., $a+u < b-v$. Thus, by bounding the sum of the $x_i$ for $i \\in [1, a+u] \\cup [b-v, n]$ as above, and trivially bounding each $x_i$ in $(a+u, b-v)$ by $1$, we obtain\n\n$$\n\\begin{aligned}\ns &< 2^{u+1} x_a + 2^{v+1} x_b + (b-v-(a+u)-1) \\\\\n  &= b-a + \\left(2^{u+1-\\alpha} + 2^{v+1-\\beta} - (u+v+1)\\right).\n\\end{aligned}\n$$\n\nNow recall $\\alpha \\in (u, u+1]$ and $\\beta \\in (v, v+1]$, so applying Bernoulli's inequality yields\n\n$$\n\\begin{aligned}\n2^{u+1-\\alpha} + 2^{v+1-\\beta} - u - v - 1 &\\le (1 + (u+1-\\alpha)) + (1 + (v+1-\\beta)) - u - v - 1 \\\\\n  &= 3 - \\alpha - \\beta.\n\\end{aligned}\n$$\n\nIt follows that $s - 3 < b - a - \\alpha - \\beta$, and so\n\n$$\n2^{s-3} < 2^{b-a-\\alpha-\\beta} = 2^{b-a} x_a x_b.\n$$\n\n![](images/Saudi_Arabia_booklet_2023_p48_data_6b89eeb403.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15628,
    "subject": "Mathematics (Olympiad)",
    "question": "In a regular 100-gon, 41 vertices are colored black and the remaining 59 vertices are colored white. Prove that there exist 24 convex quadrilaterals $Q_1, Q_2, \\dots, Q_{24}$ whose corners (vertices of the quadrilateral) are vertices of the 100-gon, so that each quadrilateral is *skew-colored* (i.e., has three corners of one color and one corner of the other color), and all $Q_i$ are pairwise disjoint (no two share a vertex).",
    "options": [],
    "answer": "See solution",
    "solution": "Call a quadrilateral *skew-colored* if it has three corners of one color and one corner of the other color. We will prove the following:\n\n*Claim.* If the vertices of a convex $(4k+1)$-gon $P$ are colored black and white such that each color is used at least $k$ times, then there exist $k$ pairwise disjoint skew-colored quadrilaterals whose vertices are vertices of $P$. (One vertex of $P$ remains unused.) The problem statement follows by removing 3 arbitrary vertices of the 100-gon and applying the Claim to the remaining 97 vertices with $k=24$.\n\n*Proof of the Claim.* We prove by induction. For $k=1$ we have a pentagon with at least one black and at least one white vertex. If the number of black vertices is even then remove a black vertex; otherwise remove a white vertex. In the remaining quadrilateral, there are an odd number of black and an odd number of white vertices, so the quadrilateral is skew-colored.\n\nFor the induction step, assume $k \\ge 2$. Let $b$ and $w$ be the numbers of black and white vertices, respectively; then $b, w \\ge k$ and $b+w = 4k+1$. Without loss of generality, we may assume $w \\ge b$, so $k \\le b \\le 2k$ and $2k+1 \\le w \\le 3k+1$.\n\nWe want to find four consecutive vertices such that three of them are white, the fourth one is black. Denote the vertices by $V_1, V_2, \\dots, V_{4k+1}$ in counterclockwise order, such that $V_{4k+1}$ is black, and consider the following $k$ groups of vertices:\n\n$$\n(V_1, V_2, V_3, V_4),\\ (V_5, V_6, V_7, V_8),\\ \\dots,\\ (V_{4k-3}, V_{4k-2}, V_{4k-1}, V_{4k})\n$$\n\nIn these groups there are $w$ white and $b-1$ black vertices. Since $w > b-1$, there is a group, $(V_i, V_{i+1}, V_{i+2}, V_{i+3})$, that contains more white than black vertices. If three are white and one is black in that group, we are done. Otherwise, if $V_i, V_{i+1}, V_{i+2}, V_{i+3}$ are all white, then let $V_j$ be the first black vertex among $V_{i+4}, \\dots, V_{4k+1}$ (recall that $V_{4k+1}$ is black); then $V_{j-3}, V_{j-2}$ and $V_{j-1}$ are white and $V_j$ is black.\n\nNow we have four consecutive vertices $V_i, V_{i+1}, V_{i+2}, V_{i+3}$ that form a skew-colored quadrilateral. The remaining vertices form a convex $(4k-3)$-gon; $w-3$ of them are white and $b-1$ are black. Since $b-1 \\ge k-1$ and $w-3 \\ge (2k+1)-3 > k-1$, we can apply the Claim with $k-1$. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15629,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle right-angled at $A$. A circle passing through $B$ and $C$ intersects the sides $AB$ and $AC$ at $M$ and $N$, respectively. Prove that if $BM \\cdot CN \\cdot BC = MN^3$, then the symmetric point of $A$ with respect to the midpoint of the segment $MN$ belongs to $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $BC = a$, $CA = b$, $AB = c$, and $AM = x$.\n\n![](images/Saudi_Arabia_booklet_2012_p17_data_b8356593ec.png)\n\nTriangles $AMN$ and $ACB$ are similar, so\n\n$$\n\\frac{x}{b} = \\frac{AN}{c} = \\frac{MN}{a}.\n$$\n\nWe obtain\n\n$$\nAN = \\frac{cx}{b}, \\quad MN = \\frac{ax}{b}. \\qquad (1)\n$$\n\nThe relation $BM \\cdot CN \\cdot BC = MN^3$ is equivalent to\n\n$$\na(c-x)\\left(b-\\frac{cx}{b}\\right) = \\frac{a^3x^3}{b^3},\n$$\n\nand hence we get\n\n$$\na^2x^3 - b^2cx^2 + b^2(b^2 + c^2)x - cb^4 = 0. \\qquad (2)\n$$\n\nSince $b^2 + c^2 = a^2$, it follows that\n\n$$\na^2x^3 - b^2cx^2 + a^2b^2x - cb^4 = 0. \\qquad (3)\n$$\n\nThe relation (3) is equivalent to\n\n$$\n(a^2x - b^2c)x^2 + b^2(a^2x - b^2c) = 0,\n$$\n\nso we get\n\n$$\n(a^2x - b^2c)(x^2 + b^2) = 0. \\qquad (4)\n$$\n\nFrom (4) it follows\n\n$$\nx = \\frac{b^2c}{a^2}. \\qquad (5)\n$$\n\nDraw $AD \\perp BC$, $DM' \\perp AB$, $DN' \\perp AC$, where $D \\in BC$, $M' \\in AB$, $N' \\in AC$. We have $AD^2 = AM' \\cdot AB$, so\n\n$$\nAM' = \\frac{AD^2}{AB} = \\frac{b^2c^2}{a^2c} = \\frac{b^2c}{a^2} = x = AM.\n$$\n\nThis implies $M = M'$. Also, from $AD^2 = AN' \\cdot AC$, we get\n\n$$\nAN' = \\frac{AD^2}{AC} = \\frac{b^2c^2}{a^2b} = \\frac{bc^2}{a^2} = \\frac{cx}{b} = AN, \\text{ hence } N = N'.\n$$\n\nTherefore $AMDN$ is a rectangle. It follows that the symmetric point of $A$ with respect to the midpoint of segment $MN$ is $D$, which completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15630,
    "subject": "Mathematics (Olympiad)",
    "question": "Ali and Amin play a game with cards numbered from 1 to 100. Ali shows the cards in order until Amin picks a card. In the next step, Ali shows the largest card that has not been shown yet, and then alternates between showing the smallest and largest remaining cards. If Amin chooses $k$ cards using this strategy, what is the maximum possible sum of the numbers on these $k$ cards?\n\n![](images/IRN_ABooklet_2023_1_p25_data_ed8e359989.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(k)$ denote the maximum sum Amin can obtain by picking $k$ cards under Ali's strategy:\n\n$$\nf(k) = (100 - k + 1) + (100 - k + 2) + \\dots + (100 - 2k + 2)\n$$\n\nThis is an arithmetic sequence with $k$ terms, first term $100 - k + 1$, last term $100 - 2k + 2$. The sum is:\n\n$$\nf(k) = \\frac{(100 - k + 1 + 100 - 2k + 2)k}{2} = \\frac{(200 - 3k + 3)k}{2}\n$$\n\nThe function $f(k)$ is quadratic in $k$ and attains its maximum at $k = 33$ or $k = 34$. Comparing $f(33)$ and $f(34)$, we find $f(33) > f(34)$. Thus, Ali can ensure that the sum of Amin's cards is at most $f(33)$. Conversely, Amin can guarantee a sum of at least $f(33)$ by always picking the largest available cards and avoiding cards numbered $\\leq 35$. Therefore, the answer is $f(33)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15631,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $N$ 與 $Z$ 分別表示所有正整數與整數之集合。試求所有函數 $f: N \\to Z$ 滿足：\n\n$n$ 整除 $f(m)$ 的充要條件為 $m$ 能整除 $\\sum_{d|n} f(d)$，對於所有的正整數 $n$ 和 $m$ 皆成立。",
    "options": [],
    "answer": "See solution",
    "solution": "答：$f(n) = 0,\\ \\forall n \\in N$\n\n若 $f(1) \\neq 0$，則取 $m = 1,\\ n = |f(1)| + 1$，因為 $1|\\sum_{d|n} f(d)$，所以 $n|f(1)$ 但 $0 < |f(1)| < n$，矛盾。故 $f(1) = 0$。\n\n底下證明對所有正整數 $k$ 有 $f(2^k) = 0$。\n\n利用反證法，假設存在正整數 $k$ 使得 $f(2^k) \\neq 0$，假設 $t$ 是最小的那一個，那麼對所有 $m > |f(2^t)|$ 都有\n\n$m|\\sum_{d|2^{t-1}} f(d)$ 但 $m$ 無法整除 $\\sum_{d|2^t} f(d)$，\n\n亦即 $2^{t-1}|f(m)$ 但 $2^t$ 無法整除 $f(m)$。\n\n取質數 $p > |f(2^t)|$，代入 $n = p^2,\\ m = 2^t$，有\n\n$$\n2^t|(f(1)+f(p)+f(p^2)),\n$$\n\n所以 $p^2|f(2^t)$，但 $0 < |f(2^t)| < p < p^2$，矛盾。\n\n因此 $f(2^k) = 0,\\ \\forall k \\in N$。\n\n對任何正整數 $m$，取足夠大的 $n = 2^k$，那麼\n\n$$\nm|(f(1)+\\cdots+f(2^k)) \\Rightarrow 2^k|f(m),\n$$\n\n由於 $k$ 可以任意大，因此只有 $f(m) = 0$。證畢！",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15632,
    "subject": "Mathematics (Olympiad)",
    "question": "Each pupil in the Netherlands is given a finite number of cards. On each card, there is a real number in the interval $[0, 1]$. (The numbers on different cards do not have to be different.)\n\nFind the smallest real number $c > 0$ for which the following holds, independent of the numbers on the cards each person has been given:\n\nAny pupil for whom the sum of the numbers on their cards is at most $1000$, can distribute their cards over $100$ boxes such that the sum of the cards in each box is at most $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose one of the pupils has been given $1001$ cards, each containing the number $\\frac{1000}{1001}$. Since the sum of the cards is $1000$, this pupil should be able to distribute the cards among the $100$ boxes. By the pigeonhole principle, there is at least one box with $11$ cards. The sum of these $11$ cards is $11 \\cdot \\frac{1000}{1001} = 11 - \\frac{11}{1001} = 11 - \\frac{1}{91}$.\n\nWe now show that this is the smallest possible value, i.e., $c = 11 - \\frac{1}{91}$.\n\nFor a random pupil, consider distributions that minimize the maximum sum per box. Among these, pick one minimizing the number of boxes attaining this maximum. Let $d_1 \\le d_2 \\le \\dots \\le d_{100}$ be the sums in the $100$ boxes, with the last $k$ equal to the maximum. Since the total sum is at most $1000$,\n\n$$\n99d_1 + d_{100} \\le d_1 + d_2 + \\dots + d_{100} \\le 1000.\n$$\n\nMoving a positive card from the box with sum $d_{100}$ to the box with sum $d_1$ cannot improve the distribution. Thus, the new value of $d_1$ is at least $d_{100}$. If $d_{100} \\le 10$, we are done, since $10 < 11 - \\frac{1}{91}$. So assume $d_{100} > 10$. Since each card is at most $1$, box $d_{100}$ contains at least $11$ positive cards. Thus, there is a card in this box with value at most $\\frac{d_{100}}{11}$. Moving this card to box $d_1$ gives\n\n$$\nd_1 + \\frac{d_{100}}{11} \\ge d_{100}.\n$$\n\nSo $11d_1 \\ge 10d_{100}$. Combining with the earlier inequality,\n\n$$\n91d_{100} = 90d_{100} + d_{100} \\le 99d_1 + d_{100} \\le 1000.\n$$\n\nTherefore, $d_{100} \\le \\frac{1000}{91} = 11 - \\frac{1}{91}$. Thus, the smallest possible $c$ is $11 - \\frac{1}{91}$.\n\n$\\boxed{11 - \\frac{1}{91}}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15633,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation\n\n$$\nx^4 y^3 (y - x) = x^3 y^4 - 216\n$$\n\nin integers.",
    "options": [],
    "answer": "See solution",
    "solution": "The given equation is equivalent to\n\n$$\nx^3 y^4 + x^4 y^3 (x - y) = 216 \\iff (xy)^3 (x^2 - xy + y) = 6^3.\n$$\n\nBoth $x$ and $y$ must therefore be divisors of $6$, and therefore equal to $\\pm 1$, $\\pm 2$, $\\pm 3$ or $\\pm 6$. Also, $xy \\mid 6$ must hold. This means that $|x|$ and $|y|$ can only both equal $1$ or the set $\\{|x|, |y|\\}$ equal one of the sets $\\{1, 2\\}$, $\\{1, 3\\}$, $\\{1, 6\\}$ or $\\{2, 3\\}$. Altogether, this yields $36$ possible combinations for $(x, y)$ and a straightforward check yields the three solutions $(-3, -2)$, $(2, 3)$ and $(1, 6)$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15634,
    "subject": "Mathematics (Olympiad)",
    "question": "A 和 B 在一個直角座標平面下棋。一開始 A 先選一個位置 $ (x_0, y_0) $ 下第一顆棋。接著從 B 開始，A 和 B 輪流放棋子：\n\n1. 如果 A 上一回合下的棋子位於座標 $ (x, y) $，那接下來 B 只能把棋下在座標 $ (x+2, y+1), (x+2, y-1), (x-2, y+1), (x-2, y-1) $ 中的其中一個。\n2. 如果 B 上一回合下的棋子位於座標 $ (x, y) $，那接下來 A 只能把棋下在座標 $ (x+1, y+2), (x+1, y-2), (x-1, y+2), (x-1, y-2) $ 中的其中一個。\n\n另外，如果 $ a \\equiv c \\pmod{n} $，$ b \\equiv d \\pmod{n} $ 並且座標 $ (a, b) $ 已有棋子，那麼兩人都不能在座標 $ (c, d) $ 落子。第一個不能在任何位置下棋的人輸。\n\n(1) 試問當 $ n = 2018 $ 時，誰有必勝策略？\n\n(2) 試問當 $ n = 2019 $ 時，誰有必勝策略？",
    "options": [],
    "answer": "See solution",
    "solution": "$$n = 2018$$ 時，B 有必勝策略；$$n = 2019$$ 時，A 有必勝策略。\n\n首先，考慮一個 $n \\times n$ 的棋盤，棋盤的上方和下方連通，左方和右方連通。由於對所有的 $1 \\leq a, b \\leq n$，$\\{(nx + a, ny + b) \\mid x, y \\in \\mathbb{Z}\\}$ 中只能下一顆棋子，因此我們可以用這樣的棋盤取代座標平面，棋盤中的方格 $(a, b)$ 代表座標平面中的點集 $\\{(nx + a, ny + b) \\mid x, y \\in \\mathbb{Z}\\}$。\n\n（約定棋盤的列由下至上編號為 1 到 $n$，行由左至右編號為 1 到 $n$）\n\n**(1)** 首先證明當 $n$ 為偶數時，B 有必勝策略。\n\n將 $n \\times n$ 棋盤中的所有格子做配對，方格 $(a, 2t)$ 對應到方格 $(a + 2, 2t+1)$（這邊的座標加減均考慮 $\\pmod{n}$）。如果每當 A 下一顆棋子，B 就把棋子下在該顆棋子所在的方格配對到的方格，如此一來，A 每次下的棋子都位於新的配對，B 下的棋子則補齊這個配對，那麼 B 總是有地方可以落子，因此 B 有必勝策略。\n\n**(2)** 接著證明當 $n$ 為奇數時，A 有必勝策略。令 $n = 2k + 1$，並定義如下集合：\n\n- 當 $2 \\mid k$ 時，令 $S = \\{4r + 1, 4r + 2 \\mid 0 \\leq r < \\frac{k}{2}\\}$，$T = \\{4r + 3, 4r + 4 \\mid 0 \\leq r < \\frac{k}{2}\\}$。\n- 當 $2$ 無法整除 $k$ 時，令 $S = \\{4r + 2, 4r + 3 \\mid 0 \\leq r < \\frac{k-1}{2}\\} \\cup \\{2k\\}$，$T = \\{4r + 4, 4r + 5 \\mid 0 \\leq r < \\frac{k-1}{2}\\} \\cup \\{1\\}$。\n\n不難看出 $S \\cup T$ 正好包含所有 $1$ 到 $2k$ 之間的整數，並且 $T$ 是 $S$ 裡面所有數字加 $2$ 後形成的集合。\n\n現在，將 $n \\times n$ 棋盤中的格子做配對：對 $1 \\leq a \\leq n$，如果 $a$ 是奇數，就把方格 $(a, b): b \\in S$ 對應到方格 $(a + 1, b + 2)$；如果 $a$ 是偶數，就把方格 $(a, b): b \\notin T$ 對應到方格 $(a + 1, b - 2)$。由上述 $S$ 和 $T$ 的關係可以看出，除了 $(1, n)$ 外的所有格子剛好被一一配對。因此 A 有以下策略：一開始把棋子下在方格 $(1, n)$，並且每當 B 下一顆棋子時，A 就把棋子下在該顆棋子所在的方格配對到的方格。不難看出這是 A 的必勝策略。得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15635,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ be a point outside a given circle $\\Gamma$ with diameter $BC$. Find the locus of the orthocenter $H$ of triangle $ABC$ when $BC$ changes.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the center of $\\Gamma$ and let $AA_1$, $BB_1$, and $CC_1$ be the altitudes of $\\triangle ABC$ with orthocenter $H$. Let $\\Gamma_1$ be a circle with diameter $AO$. Since $\\angle A_1$ is right, $\\Gamma_1$ passes through the points $A$, $O$, and $A_1$. Likewise, points $B$, $C$, and $B_1$ lie on circle $\\Gamma$.\n\nThe power of $H$ with respect to $\\Gamma$ is $HB \\cdot HB_1$ and the power of $H$ with respect to $\\Gamma_1$ is $HA \\cdot HA_1$. Since triangles $BHA_1$ and $AHB_1$ are similar, we have\n\n$$\n\\frac{HB}{HA} = \\frac{HA_1}{HB_1} \\Leftrightarrow HB \\cdot HB_1 = HA \\cdot HA_1\n$$\n\nThis means that $H$ lies on the radical axis $\\ell$ of $\\Gamma$ and $\\Gamma_1$.\n\nConversely, it is easy to see that each point of $\\ell$ is the orthocenter of a triangle of the given type.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15636,
    "subject": "Mathematics (Olympiad)",
    "question": "設三角形 $ABC$ 的外心為 $O$，內心為 $I$。$D, E, F$ 三點分別位於 $BC, CA, AB$ 邊上，並滿足 $BD + BF = CA$ 且 $CD + CE = AB$。令三角形 $BFD$ 的外接圓與三角形 $CDE$ 的外接圓的兩交點為 $D, P$。試證：$OP = OI$。",
    "options": [],
    "answer": "See solution",
    "solution": "以下用 $(XYZ)$ 代表 $\\triangle XYZ$ 的外接圓。設 $D, E, F$ 三點分別在 $BC, CA, AB$ 邊上。由 Miquel 定理，$(AEF) = \\omega_A$，$(BFD) = \\omega_B$，$(CDE) = \\omega_C$ 三個外接圓有共同點 $P \\neq D$。\n\n設 $\\omega_A, \\omega_B, \\omega_C$ 分別交 $AI, BI, CI$ 於點 $A \\neq A', B \\neq B', C \\neq C'$。此處的要點是 $A', B', C'$ 這三點並不依賴於點 $D, E, F$，只要它們滿足 $BD + BF = CA, CD + CE = AB, AE + AF = BC$（最後一式可由前面兩式得到）。對此我們先證明一引理。\n\n**引理.** 給定 $\\angle A = \\alpha$。一圓 $\\omega$ 通過 $A$ 點，並交 $\\angle A$ 的角平分線於 $L$，且分別交 $\\angle A$ 的兩邊於 $X, Y$ 兩點。則 $AX + AY = 2AL \\cos \\frac{\\alpha}{2}$。\n\n**證.** 注意到 $L$ 點是 $\\omega$ 上的弧 $\\overarc{XY}$ 的中點，故可令 $XL = YL = u, XY = v$。根據 Ptolemy 定理，$AX \\cdot YL + AY \\cdot XL = AL \\cdot XY$，可改寫成 $(AX + AY)u = AL \\cdot v$。因為 $\\angle LXY = \\frac{\\alpha}{2}$ 且 $\\angle XLY = 180^\\circ - \\alpha$，由餘弦定理可得 $v = 2u \\cos \\frac{\\alpha}{2}$，故引理得證。\n\n![](images/13-3J_p11_data_9cba77f073.png)\n\n將此引理套用在 $\\angle BAC = \\alpha$ 以及圓 $\\omega = \\omega_A$，其中 $\\omega_A$ 交 $AI$ 於 $A'$ 點，於是得到 $2AA' \\cos \\frac{\\alpha}{2} = AE + AF = BC$。同理可推得 $BB', CC'$ 所滿足的式子。由此得到 $A', B', C'$ 各點的位置與 $D, E, F$ 點的選取無關。\n\n我們再利用此引理兩次於 $\\angle BAC = \\alpha$。令 $\\omega$ 為以 $AI$ 為直徑的圓。此時 $X, Y$ 兩點分別為 $\\triangle ABC$ 的內切圓在 $AB, AC$ 兩邊的切點，於是 $AX = AY = \\frac{1}{2}(AB+AC-BC)$。由引理可得 $2AI \\cos \\frac{\\alpha}{2} = AB+AC-BC$。再令 $\\omega$ 是 $\\triangle ABC$ 的外接圓，設 $AI$ 交 $\\omega$ 於點 $M \\neq A$。此時 $\\{B, C\\} = \\{X, Y\\}$，因此由引理得 $2AM \\cos \\frac{\\alpha}{2} = AB + AC$。將目前得到的結果整理如下：\n\n$$\n\\begin{aligned}\n2 AA' \\cos \\frac{\\alpha}{2} &= BC, \\\\\n2 AI \\cos \\frac{\\alpha}{2} &= AB + AC - BC, \\\\\n2 AM \\cos \\frac{\\alpha}{2} &= AB + AC.\n\\end{aligned} \n\\qquad (1)\n$$\n\n由以上等式可得出 $AA' + AI = AM$，因此線段 $AM$ 與 $IA'$ 有相同的中點。\n\n由此可得點 $I$ 與點 $A'$ 到外心 $O$ 等距。由對稱性知 $OI = OA' = OB' = OC'$，所以 $I, A', B', C'$ 共圓，其圓心為 $O$。\n\n欲證明 $OI = OP$，現只需證明 $I, A', B', C', P$ 共圓即可。若 $P$ 等於 $I, A', B', C'$ 其中一點的話免證；故設 $P \\neq I, A', B', C'$。\n\n![](images/13-3J_p12_data_95edcdaa83.png)\n\n以下的論證中我們使用有向角來避免正負號的區別。將直線 $l, m$ 之間所夾的有向角記為 $\\angle(l, m)$。對任意直線 $l, m, n$ 自然有 $\\angle(l, m) = -\\angle(m, l)$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15637,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle with circumcircle $\\Gamma$. Let $l_B$ and $l_C$ be the lines perpendicular to $BC$ passing through $B$ and $C$, respectively. A point $T$ lies on the minor arc $BC$. The tangent to $\\Gamma$ at $T$ meets $l_B$ and $l_C$ at $P_B$ and $P_C$, respectively. The line through $P_B$ perpendicular to $AC$ and the line through $P_C$ perpendicular to $AB$ meet at a point $Q$. Given that $Q$ lies on $BC$, prove that the line $AT$ passes through $Q$.\n\n(A minor arc of a circle is the shorter of the two arcs with given endpoints.)",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $Q$ is sufficient information to construct $P_B$ and $P_C$.\n\nLet $T'$ be the second intersection of the line $AQ$ and $\\Gamma$.\n\n![](images/bmo2-2022-solutions_p4_data_9ac259f72d.png)\n\nDenote the foot of the perpendicular from $P_B$ to $AC$ by $U$. Then $P_BBUC$ is cyclic, as is $ABT'C$.\n\nConsequently: $\\angle QP_BB = \\angle UP_BB = \\angle C = \\angle AT'B$, so $P_BBQT'$ is also cyclic.\n\nIn particular, $\\angle AT'P_B = 90^\\circ$.\n\nBut the same holds for $\\angle P_C T' A = 90^\\circ$.\n\nSo $P_B, T', P_C$ are collinear. This implies that $T' = T$ since the conditions in the question mean there is only one point on both the line $P_B P_C$ and the circle $\\Gamma$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15638,
    "subject": "Mathematics (Olympiad)",
    "question": "The managers of two companies are 1000 km apart. The segment that connects them contains $n$ points, in each of which sits a mathematician. Every second, each mathematician moves to a point which is the midpoint of the segment connecting them and the closest mathematician or manager. If there is more than one such point, the mathematician chooses any of them. If two mathematicians arrive at the same point according to these rules, the younger one leaves the game. Show that in finitely many seconds, every mathematician who is still in the game can shake hands with one of the managers. A mathematician can shake hands when the distance from the manager does not exceed 1 m.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S_0 = \\{0, x_1, x_2, \\dots, x_n, 1000\\}$ be the initial locations of mathematicians and managers, where $0$ and $1000$ represent the locations of the managers, and $x_1 < x_2 < \\dots < x_n$ are the locations of the mathematicians. Let $S_i$ denote the positions of everyone after $i$ seconds. The number of elements in $S_i$ decreases if two mathematicians are closest to each other and arrive at the same point, as the younger leaves the game. Since $S_0$ is finite, there can only be finitely many such decreases, so there exists a non-negative integer $N$ such that $|S_k| \\leq |S_N|$ for all $k \\geq N$.\n\nLet $S_N = \\{0, y_1, y_2, \\dots, y_m, 1000\\}$, with $0 < y_1 < y_2 < \\dots < y_m < 1000$, $m \\leq n$. Since $|S_{N+1}| \\leq |S_N|$, the closest neighbor for the first $t$ mathematicians is on the left, and for the rest, on the right. Thus, both groups of mathematicians will move toward their respective managers and, in a finite number of steps, will be within 1 meter of a manager, allowing them to shake hands.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15639,
    "subject": "Mathematics (Olympiad)",
    "question": "$f(x - f(x/y)) = x f(1 - f(1/y)), \\quad \\forall x, y \\in \\mathbb{R}, y \\neq 0.$\n\nFind all functions $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying this equation.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $y = 1$ in the given equation:\n\n$$\nf(x - f(x)) = x f(1 - f(1)). \\tag{1}\n$$\n\nSuppose $f(t) = 0$ for some $t$. Then (1) gives $f(t - f(t)) = f(t) = 0$, so by the original equation, $t = 0$ since $f(1 - f(1)) \\neq 0$. Also, the original equation gives $f(0 - f(0)) = 0$, so $0 - f(0) = 0$, i.e., $f(0) = 0$. Therefore, $0$ is the only zero of $f$.\n\nFrom (1), $f$ is surjective, so there exists $\\alpha \\neq 0$ such that $f(\\alpha) = 1$. Set $y = 1/\\alpha$ in the original equation to proceed further.\n\nThe answer is: $f(x) = c x$, where $c \\neq 0, 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15640,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, let $\\sigma(n)$ be the sum of all divisors of $n$. Show that there exist infinitely many positive integers $n$ such that $n$ divides $2^{\\sigma(n)} - 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $k$ be a positive integer. Choose a prime divisor $p_j$ of $2^{2^j} + 1$ for each $0 \\leq j < k$. Then the product $n_k = p_0 p_1 \\dots p_{k-1}$ divides $2^{2^k} - 1 = \\prod_{j=0}^{k-1} (2^{2^j} + 1)$.\n\nSince $n_k$ is odd, $\\sigma(n_k) = (p_0 + 1) \\dots (p_{k-1} + 1)$ is divisible by $2^k$. Hence, $2^{2^k} - 1$ divides $2^{\\sigma(n_k)} - 1$. It follows that $2^{\\sigma(n_k)} \\equiv 1 \\pmod{n_k}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15641,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $\\angle ABC = 2 \\cdot \\angle ACB$. Let $X$ and $Y$ be the midpoints of arcs $AB$ and $BC$ (not containing $C$ and $A$, respectively) of the circumcircle of triangle $ABC$. Let $BL$ be the angle bisector of $\\angle ABC$, with $L \\in AC$. Given that $\\angle XLY = 90^\\circ$, determine the measures of the angles of triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $BL$ is the angle bisector of $\\angle ABC$ and $\\angle ABC = 2\\angle ACB$, it follows that $\\angle ABL = \\angle LBC = \\angle ACB$. From $\\angle ACB = \\angle LBC$, we deduce that triangle $LBC$ is isosceles, and in particular $LB = LC$.\n\nSince $Y$ is the midpoint of arc $BC$ (not containing $A$), it follows that $BY = CY$, hence $BY = CY$.\n\nFrom the above, $\\triangle LBY \\equiv \\triangle LCY$ by the SSS criterion, and thus $\\angle BLY = \\angle CLY$, which implies that $LY$ is the angle bisector of $\\angle BLC$.\n\nOn the other hand, $AX = BX$ implies that $\\angle ACX = \\angle XCB$, so $XC$ is the angle bisector of $\\angle LCB$.\n\n![](images/RMC_2025_p73_data_330a92765d.png)\n\nLet $I$ be the intersection point of the segments $LY$ and $CX$. From the previous results, $I$ is the incenter of triangle $BLC$, hence $BI$ is the angle bisector of $\\angle LBC$.\n\nFrom the hypothesis, $\\angle XLY = 90^\\circ$, and since $LY$ is the angle bisector of $\\angle BLC$, it follows that $LX$ is the external angle bisector of $\\angle BLC$.\n\nFrom the above, $X$ is the excenter of triangle $BLC$ corresponding to vertex $C$, hence $XB$ is the external angle bisector of $\\angle LBC$. Combining this, $XB \\perp BI$, i.e., $\\angle XBI = 90^\\circ$.\n\nLet $D$ be the intersection of $BI$ with the circumcircle $\\Gamma$, and let $\\alpha = \\angle IBC = \\angle IBL$. Then $\\angle ABL = \\angle LBC = 2\\alpha$. Since $\\angle LBC = \\angle LCB = 2\\alpha$, we also have $\\angle LCX = \\angle XCB = \\alpha$. Therefore:\n\n$$\n\\widehat{AD} = 2\\angle ABD = 2(\\angle ABL + \\angle LBD) = 2(2\\alpha + \\alpha) = 6\\alpha.\n$$\n\nAlso, $\\widehat{AX} = 2\\angle ACX = 2\\alpha$, so combining the two gives $\\widehat{DX} = 8\\alpha$.\n\nOn the other hand, $\\widehat{DX} = 2\\angle XBD = 180^\\circ$, by the previous result. Combining these two relations gives $8\\alpha = 180^\\circ$, hence $\\angle ACB = 2\\alpha = 45^\\circ$, $\\angle ABC = 4\\alpha = 90^\\circ$, and thus $\\angle BAC = 45^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15642,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist an infinite sequence of real numbers $x_n$ satisfying $x_1 = 2$ and\n\n$$\n\\frac{2x_n^2 + 2}{x_n + 3} < x_{n+1} \\le \\frac{2x_n + 2}{x_n + 3} + 2023\n$$\n\nfor all positive integers $n = 1, 2, 3, \\dots$?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is No. Suppose, for contradiction, that such a sequence exists. First, we prove by induction that $x_n > 2$ for every $n \\ge 2$. For $n = 2$, this holds. Assume $x_k > 2$ for some $k \\ge 2$; then\n\n$$\n\\frac{2x_k^2 + 2}{x_k + 3} - 2 = \\frac{2x_k^2 - 2x_k - 4}{x_k + 3} = \\frac{2(x_k - 2)(x_k + 1)}{x_k + 3} > 0.\n$$\n\nSo $x_{k+1} > \\frac{2x_k^2 + 2}{x_k + 3} > 2$. Thus, $x_n > 2$ for all $n \\ge 2$.\n\nHence,\n\n$$\n x_{n+1} - x_n \\ge \\frac{2x_n^2 + 2}{x_n + 3} - x_n = \\frac{(x_n - 1)(x_n - 2)}{x_n + 3} > 0,\n$$\n\nwhich implies $(x_n)$ is strictly increasing. On the other hand,\n\n$$\n\\frac{2x_n + 2}{x_n + 3} < 2 \\quad \\forall n > 0 \\implies x_{n+1} < 2 + 2023 = 2025 \\quad \\forall n \\ge 1\n$$\n\nso $(x_n)$ is bounded above by 2025. Therefore, $(x_n)$ has a finite limit $L$ with $2 < L \\le 2025$. Substituting into the initial condition,\n\n$$\n\\frac{2L^2 + 2}{L + 3} \\le L \\iff 1 \\le L \\le 2,\n$$\n\nwhich is a contradiction. Thus, no such sequence $(x_n)$ exists. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15643,
    "subject": "Mathematics (Olympiad)",
    "question": "There are six players in a tournament. Each player plays every other player exactly once, and each player plays just once in each round.\n\n(a) How many rounds are required for the tournament?\n\n(b) After three rounds, is it possible for all players to have equal scores?\n\n(c) For the first two rounds, provide possible tables showing the results.\n\n(d) After two rounds, list the remaining opponents for each player. Then, enumerate all possible third rounds and, for each, the possible fourth rounds. How many combinations of third and fourth rounds are there? For each combination, show a possible allocation of wins such that after four rounds, each player has exactly two points.",
    "options": [],
    "answer": "See solution",
    "solution": "**(a)**\n\nThere are five possible opponents for each player in a tournament of six players. Since each player plays every other player just once, there must be five rounds.\n\nAlternatively, there are $\\binom{6}{2} = 15$ possible pairs of players, with 3 pairs playing in each round. So there must be $15 \\div 3 = 5$ rounds.\n\n**(b)**\n\nEach round of the tournament contributes three points to the total of all scores. After three rounds, nine points have been won. Since $6$ does not divide $9$, the players cannot have equal scores after three rounds.\n\n**(c)**\n\nPossible tables for the first two rounds:\n\nRound 1:\n\n| 1 | A | B | C | J | K | L |\n|---|---|---|---|---|---|---|\n| A |   |   |   | 1 |   |   |\n| B |   |   |   |   | 1 |   |\n| C |   |   |   |   |   | 1 |\n| J | 0 |   |   |   |   |   |\n| K |   | 0 |   |   |   |   |\n| L |   |   | 0 |   |   |   |\n\nRound 2:\n\n| 2 | A | B | C | J | K | L |\n|---|---|---|---|---|---|---|\n| A |   |   |   |   | 0 |   |\n| B |   |   |   |   |   | 0 |\n| C |   |   |   | 0 |   |   |\n| J |   |   | 1 |   |   |   |\n| K | 1 |   |   |   |   |   |\n| L |   | 1 |   |   |   |   |\n\n**(d)**\n\nThe remaining opponents for the players after two rounds are:\n\n| Player | To play      |\n|--------|-------------|\n| A      | B, C, L     |\n| B      | A, C, J     |\n| C      | A, B, K     |\n| J      | B, K, L     |\n| K      | C, J, L     |\n| L      | A, J, K     |\n\nPossible third rounds:\n\n(i) A vs B, C vs K, J vs L\n\n(ii) A vs C, B vs J, K vs L\n\n(iii) A vs L, B vs C, J vs K\n\n(iv) A vs L, B vs J, C vs K\n\nPossible fourth rounds following (i):\n- A vs C, B vs J, K vs L\n- A vs L, B vs C, J vs K\n\nPossible fourth rounds following (ii):\n- A vs B, C vs K, J vs L\n- A vs L, B vs C, J vs K\n\nPossible fourth rounds following (iii):\n- A vs B, C vs K, J vs L\n- A vs C, B vs J, K vs L\n\nThere is no possible fourth round following (iv) because if A plays B, there is no opponent for C, and if A plays C, there is no opponent for B.\n\nTherefore, there are just three possible third rounds, each with two possible fourth rounds, giving six combinations of third and fourth rounds.\n\nThe total number of points awarded after four rounds is $12$. So we want each player to win exactly two points. Hence, each player must win in just one of rounds three and four. Below are possible allocations of wins for each of the six combinations (showing only two as examples):\n\n(i)\n\nRound 3:\n\n| 3 | A | B | C | J | K | L |\n|---|---|---|---|---|---|---|\n| A |   | 1 |   |   |   |   |\n| B | 0 |   |   |   |   |   |\n| C |   |   |   | 0 |   |   |\n| J |   |   |   |   | 1 |   |\n| K |   |   | 1 |   |   |   |\n| L |   |   |   | 0 |   |   |\n\nRound 4:\n\n| 4 | A | B | C | J | K | L |\n|---|---|---|---|---|---|---|\n| A |   |   | 0 |   |   |   |\n| B |   |   |   | 1 |   |   |\n| C | 1 |   |   |   |   |   |\n| J |   | 0 |   |   |   |   |\n| K |   |   |   |   | 0 |   |\n| L |   |   |   |   |   | 1 |\n\n(ii)\n\nRound 3:\n\n| 3 | A | B | C | J | K | L |\n|---|---|---|---|---|---|---|\n| A |   |   | 1 |   |   |   |\n| B |   |   |   | 0 |   |   |\n| C | 0 |   |   |   |   |   |\n| J |   | 1 |   |   |   |   |\n| K |   |   |   |   | 1 |   |\n| L |   |   |   |   | 0 |   |\n\nRound 4:\n\n| 4 | A | B | C | J | K | L |\n|---|---|---|---|---|---|---|\n| A |   | 0 |   |   |   |   |\n| B | 1 |   |   |   |   |   |\n| C |   |   |   | 1 |   |   |\n| J |   |   |   |   | 0 |   |\n| K |   |   | 0 |   |   |   |\n| L |   |   |   | 1 |   |   |\n\nAnother six pairs of tables for rounds 3 and 4 are possible with A losing in round 3, obtained by reversing all wins and losses (interchanging all 0s and 1s).",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 15644,
    "subject": "Mathematics (Olympiad)",
    "question": "Sean $m \\geq 1$ un entero positivo, $a$ y $b$ enteros positivos distintos mayores estrictamente que $m^2$ y menores estrictamente que $m^2 + m$. Hallar todos los enteros $d$ que dividen al producto $ab$ y cumplen $m^2 < d < m^2 + m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Sea $d$ un entero positivo que divide a $ab$ y tal que $d \\in (m^2, m^2 + m)$. Entonces $d$ divide a $(a - d)(b - d) = ab - da - db + d^2$. Como $|a - d| < m$ y $|b - d| < m$, deducimos que $|(a - d)(b - d)| < m^2 < d$, lo que implica que $(a - d)(b - d) = 0$. Así, $d = a$ o $d = b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15645,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_n$ be the position of the grasshopper after the $n$th jump, where\n\n$$\na_1 = 1, \\quad a_n = 1 + k + \\dots + k^{n-1}, \\quad n \\ge 2.\n$$\n\nFind all integers $k$ such that $2015$ does not divide $a_n$ for any $n = 1, \\dots, 2015$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $\\gcd(k, 2015) = d > 1$. Then every $a_n$ divided by $d$ gives remainder $1$, and since $2015$ is divisible by $d$, we have $2015 \\nmid a_n$ for all $n$. Therefore, all positive integers $k$ not relatively prime to $2015$ satisfy the condition.\n\nIf $\\gcd(k, 2015) = 1$, consider the remainders of $a_1, \\dots, a_{2015}$ modulo $2015$. If none is divisible by $2015$, then by the pigeonhole principle, at least two $a_l$ and $a_m$ ($m > l$) have the same remainder. Their difference is divisible by $2015$:\n\n$$\na_m - a_l = k^l + \\dots + k^{m-1} = k^l (1 + \\dots + k^{m-l-1}) = k^l \\cdot a_{m-l}.\n$$\n\nSince $2015 \\mid k^l \\cdot a_{m-l}$ and $\\gcd(k, 2015) = 1$, it follows $2015 \\mid a_{m-l}$, contradicting the assumption. Thus, only $k$ not relatively prime to $2015$ are suitable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15646,
    "subject": "Mathematics (Olympiad)",
    "question": "In the right triangle $\\lg \\frac{a-b}{2} = \\frac{1}{2}(\\lg a + \\lg b - \\lg 2)$, where $a > b$, $a$ and $b$ are the sides of the triangle. Find the angles of the triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "The given equation is equivalent to $\\lg \\frac{a-b}{2} = \\lg \\sqrt{\\frac{ab}{2}}$. Therefore,\n\n$$\n\\frac{a-b}{2} = \\sqrt{\\frac{ab}{2}} \\Leftrightarrow \\frac{a^2-2ab+b^2}{4} = \\frac{ab}{2} \\Leftrightarrow a^2-4ab+b^2=0.\n$$\n\nSince $a > b > 0$, by dividing by $b^2$, we have $\\left(\\frac{a}{b}\\right)^2 - 4\\frac{a}{b} + 1 = 0$, with the roots $\\frac{a}{b} = 2\\pm\\sqrt{3}$. Because $\\frac{a}{b} > 1$, we have $\\frac{a}{b} = 2+\\sqrt{3}$. On the other hand, $\\frac{a}{b} = \\tan \\alpha$.\n\nFrom $\\sin 2\\alpha = \\frac{2\\tan \\alpha}{1+\\tan^2 \\alpha} = \\frac{2(2+\\sqrt{3})}{1+(2+\\sqrt{3})^2} = \\frac{1}{2}$, we have two solutions for the angle $\\alpha$: $\\alpha = 75^\\circ$ or $\\alpha = 15^\\circ$. Since $a > b$, we have $\\alpha > \\beta$, i.e., $\\alpha = 75^\\circ$ and $\\beta = 15^\\circ$.\n\n![](images/Makedonija_2008_p41_data_2a0aee3dda.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15647,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of prime numbers $(p, q)$ such that both $p \\mid 7^q - 2^q$ and $q \\mid 7^p - 2^p$.",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $(p, q) = (5, 5), (5, 11), (5, 61)$ and their permutations.\n\nIf $p \\mid 7^p - 2^p$, then by Fermat's little theorem, $p \\mid 7 - 2 = 5$, so $p = 5$. If $q \\mid 7^q - 2^q$, then $q = 5$ for the same reason, so $(5, 5)$ is a solution. If $q \\ne 5$, then $q \\mid 7^p - 2^p = 16775 = 5^2 \\times 11 \\times 61$, so $q = 11$ or $61$. By symmetry, $(p, q) = (11, 5), (61, 5)$ are also solutions.\n\nNow, assume $p, q \\ne 5$. Then $p \\mid 7^q - 2^q$ and $q \\mid 7^p - 2^p$. Without loss of generality, assume $q \\ge p$. Consider $p \\mid 7^q - 2^q$. Since $q$ is a prime and $q > p - 1$, $(q, p - 1) = 1$, so there exists $a \\in \\mathbb{Z}^+$ such that $aq \\equiv 1 \\pmod{p-1}$. By Fermat's little theorem:\n\n$$\n7 \\equiv 7^{aq} = (7^q)^a \\equiv (2^q)^a = 2^{aq} \\equiv 2 \\pmod{p}.\n$$\n\nThis forces $p = 5$, a contradiction. Therefore, the only solutions are $(p, q) = (5, 5), (5, 11), (5, 61)$ and their permutations.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15648,
    "subject": "Mathematics (Olympiad)",
    "question": "Straight line $l$ with slope $\\frac{1}{3}$ intercepts ellipse $C: \\frac{x^2}{36} + \\frac{y^2}{4} = 1$ at points $A$ and $B$, and point $P(3\\sqrt{2}, \\sqrt{2})$ is in the top-left of $l$ (as shown in the figure below).\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p44_data_51af428206.png)\n\n(1) Prove that the center of the inscribed circle of $\\triangle PAB$ is on a given line.\n\n(2) When $\\angle APB = 60^{\\circ}$, find the area of $\\triangle PAB$.",
    "options": [],
    "answer": "See solution",
    "solution": "(1) Let $l$ be a straight line such that $y = \\frac{1}{3}x + m$, and $A(x_1, y_1), B(x_2, y_2)$.\n\nSubstituting $y = \\frac{1}{3}x + m$ into $\\frac{x^2}{36} + \\frac{y^2}{4} = 1$ and simplifying, we have\n\n$$\n2x^2 + 6mx + 9m^2 - 36 = 0.\n$$\n\nThen $x_1 + x_2 = -3m$, $x_1x_2 = \\frac{9m^2 - 36}{2}$.\n\nFor $P(3\\sqrt{2}, \\sqrt{2})$, the slopes are $k_{PA} = \\frac{y_1 - \\sqrt{2}}{x_1 - 3\\sqrt{2}}$, $k_{PB} = \\frac{y_2 - \\sqrt{2}}{x_2 - 3\\sqrt{2}}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\nk_{PA} + k_{PB} &= \\frac{y_1 - \\sqrt{2}}{x_1 - 3\\sqrt{2}} + \\frac{y_2 - \\sqrt{2}}{x_2 - 3\\sqrt{2}} \\\\\n&= \\frac{(y_1 - \\sqrt{2})(x_2 - 3\\sqrt{2}) + (y_2 - \\sqrt{2})(x_1 - 3\\sqrt{2})}{(x_1 - 3\\sqrt{2})(x_2 - 3\\sqrt{2})}.\n\\end{aligned}\n$$\n\nThe numerator simplifies to\n\n$$\n\\begin{aligned}\n& \\left(\\frac{1}{3}x_1 + m - \\sqrt{2}\\right)(x_2 - 3\\sqrt{2}) + \\left(\\frac{1}{3}x_2 + m - \\sqrt{2}\\right)(x_1 - 3\\sqrt{2}) \\\\\n&= \\frac{2}{3}x_1x_2 + (m - 2\\sqrt{2})(x_1 + x_2) - 6\\sqrt{2}(m - \\sqrt{2}) \\\\\n&= \\frac{2}{3} \\cdot \\frac{9m^2 - 36}{2} + (m - 2\\sqrt{2})(-3m) - 6\\sqrt{2}(m - \\sqrt{2}) \\\\\n&= 3m^2 - 12 - 3m^2 + 6\\sqrt{2}m - 6\\sqrt{2}m + 12 = 0.\n\\end{aligned}\n$$\n\nTherefore, $k_{PA} + k_{PB} = 0$. Since $P$ is in the top-left of $l$, the bisector of $\\angle APB$ is parallel to the $y$-axis. Thus, the center of the inscribed circle of $\\triangle PAB$ is on the line $x = 3\\sqrt{2}$.\n\n(2) When $\\angle APB = 60^\\circ$, by (1), $k_{PA} = \\sqrt{3}$, $k_{PB} = -\\sqrt{3}$.\n\nThe equation for line $PA$ is $y - \\sqrt{2} = \\sqrt{3}(x - 3\\sqrt{2})$. Substituting into $\\frac{x^2}{36} + \\frac{y^2}{4} = 1$ and eliminating $y$, we get\n\n$$\n14x^2 + 9\\sqrt{6}(1 - 3\\sqrt{3})x + 18(13 - 3\\sqrt{3}) = 0,\n$$\n\nwhich has roots $x_1$ and $3\\sqrt{2}$. So $x_1 \\cdot 3\\sqrt{2} = \\frac{18(13 - 3\\sqrt{3})}{14}$, i.e.\n\n$$\nx_1 = \\frac{3\\sqrt{2}(13 - 3\\sqrt{3})}{14}.\n$$\n\nThen\n\n$$\n|PA| = \\sqrt{1 + (\\sqrt{3})^2} \\cdot |x_1 - 3\\sqrt{2}| = \\frac{3\\sqrt{2}(3\\sqrt{3} + 1)}{7}.\n$$\n\nSimilarly, $|PB| = \\frac{3\\sqrt{2}(3\\sqrt{3} - 1)}{7}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\nS_{\\triangle PAB} &= \\frac{1}{2} \\cdot |PA| \\cdot |PB| \\cdot \\sin 60^\\circ \\\\\n&= \\frac{1}{2} \\cdot \\frac{3\\sqrt{2}(3\\sqrt{3} + 1)}{7} \\cdot \\frac{3\\sqrt{2}(3\\sqrt{3} - 1)}{7} \\cdot \\frac{\\sqrt{3}}{2} \\\\\n&= \\frac{117\\sqrt{3}}{49}.\n\\end{aligned}\n$$\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p46_data_a743591038.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15649,
    "subject": "Mathematics (Olympiad)",
    "question": "The roots of $x^3 + 2x^2 - x + 3$ are $p$, $q$, and $r$. What is the value of\n\n$$(p^2 + 4)(q^2 + 4)(r^2 + 4)?$$",
    "options": [],
    "answer": "See solution",
    "solution": "By Vieta's formulas:\n\n$$\n\\begin{aligned}\np + q + r &= -2, \\\\\npq + pr + qr &= -1, \\\\\npqr &= -3.\n\\end{aligned}\n$$\n\nWe want to compute:\n\n$$(p^2 + 4)(q^2 + 4)(r^2 + 4) = p^2q^2r^2 + 4(p^2q^2 + p^2r^2 + q^2r^2) + 16(p^2 + q^2 + r^2) + 64.$$ \n\nFirst,\n\n$$p^2q^2r^2 = (pqr)^2 = (-3)^2 = 9.$$ \n\nNext, note that\n\n$$(pq + pr + qr)^2 = p^2q^2 + p^2r^2 + q^2r^2 + 2pqr(p + q + r).$$\n\nSo,\n\n$$p^2q^2 + p^2r^2 + q^2r^2 = (-1)^2 - 2(-3)(-2) = 1 - 12 = -11.$$ \n\nAlso,\n\n$$(p + q + r)^2 = p^2 + q^2 + r^2 + 2(pq + pr + qr),$$\n\nso\n\n$$p^2 + q^2 + r^2 = (-2)^2 - 2(-1) = 4 + 2 = 6.$$ \n\nTherefore,\n\n$$(p^2 + 4)(q^2 + 4)(r^2 + 4) = 9 + 4 \\cdot (-11) + 16 \\cdot 6 + 64 = 9 - 44 + 96 + 64 = 125.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15650,
    "subject": "Mathematics (Olympiad)",
    "question": "Ante is given the numbers $1, 2, \\dots, n$ written on a board. In each move, he may choose two numbers on the board and increase both by 1. For which positive integers $n$ is it possible, after a finite number of moves, to make all the numbers on the board equal?",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $n = 4k$. Ante can achieve that all numbers on the board are equal as follows: he increases by 1 the numbers $1$ and $3$, $5$ and $7$, ..., $4k-3$ and $4k-1$. After this, the numbers on the board are all even numbers less than or equal to $n$, each written twice. Then, he increases $2$ and $2$ by $n-2$, $4$ and $4$ by $n-4$, ..., $n-2$ and $n-2$ by $2$, so all numbers become $n$.\n\nAssume $n = 2k+1$. Ante can achieve that all numbers are equal as follows: he increases by 1 the numbers $1$ and $n$, $3$ and $n$, ..., $n-2$ and $n$. After this, the numbers on the board are all even numbers less than or equal to $n$, each written twice, and the number $\\frac{3n-1}{2}$. Then, he increases $2$ and $2$ by $\\frac{3n-5}{2}$, $4$ and $4$ by $\\frac{3n-9}{2}$, ..., $n-1$ and $n-1$ by $\\frac{n+1}{2}$, so all numbers become $\\frac{3n-1}{2}$.\n\nAssume $n = 4k + 2$. Then Ante cannot achieve that all numbers are equal. The sum of all numbers on the board is initially odd, because\n\n$$\n1 + 2 + \\dots + n = \\frac{n(n+1)}{2} = (2k+1)(4k+3).\n$$\n\nSince there is an even number of numbers on the board, if they were equal their sum would be even. In each step, the sum increases by an even number, so the sum will never be even.\n\nTherefore, Ante can achieve that all numbers are equal if and only if $n$ is not of the form $4k + 2$, $k \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15651,
    "subject": "Mathematics (Olympiad)",
    "question": "Two transformations are said to _commute_ if applying the first followed by the second gives the same result as applying the second followed by the first. Consider these four transformations of the coordinate plane:\n\n* a translation 2 units to the right,\n* a 90°-rotation counterclockwise about the origin,\n* a reflection across the x-axis, and\n* a dilation centered at the origin with scale factor 2.\n\nOf the 6 pairs of distinct transformations from this list, how many commute?\n\n(A) 1 (B) 2 (C) 3 (D) 4 (E) 5",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer (C):** Denote the transformations by $T$, $R$, $F$, and $D$ in the order given in the problem statement. Then the images of point $(x, y)$ are $T(x, y) = (x + 2, y)$, $R(x, y) = (-y, x)$, $F(x, y) = (x, -y)$, and $D(x, y) = (2x, 2y)$.\n\nThe results of applying a pair of transformations in either order are as follows:\n\n* $T(R(x, y)) = T(-y, x) = (-y + 2, x)$ and $R(T(x, y)) = R(x + 2, y) = (-y, x + 2)$. The results are different, so $T$ and $R$ do not commute.\n* $T(F(x, y)) = T(x, -y) = (x + 2, -y)$ and $F(T(x, y)) = F(x + 2, y) = (x + 2, -y)$. The results are the same, so $T$ and $F$ do commute.\n* $T(D(x, y)) = T(2x, 2y) = (2x + 2, 2y)$ and $D(T(x, y)) = D(x + 2, y) = (2x + 4, 2y)$. The results are different, so $T$ and $D$ do not commute.\n* $R(F(x, y)) = R(x, -y) = (y, x)$ and $F(R(x, y)) = F(-y, x) = (-y, -x)$. The results are different, so $R$ and $F$ do not commute.\n* $R(D(x, y)) = R(2x, 2y) = (-2y, 2x)$ and $D(R(x, y)) = D(-y, x) = (-2y, 2x)$. The results are the same, so $R$ and $D$ do commute.\n* $D(F(x, y)) = D(x, -y) = (2x, -2y)$ and $F(D(x, y)) = F(2x, 2y) = (2x, -2y)$. The results are the same, so $D$ and $F$ do commute.\n\nThus 3 of the 6 pairs commute.\n\n**Note:** The following figure illustrates the application of each transformation to an L-shaped figure.\n\n![](images/2024_AMC10A_Solutions_p6_data_057195e4ba.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15652,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the largest possible size of a set $M$ of integers with the following property: Among any three distinct numbers from $M$, there exist two numbers whose sum is a power of $2$ with non-negative integer exponent.",
    "options": [],
    "answer": "See solution",
    "solution": "The set $\\{-1, 3, 5, -2, 6, 10\\}$ shows that $M$ can have $6$ elements: the sum of any two numbers from the triplet $(-1, 3, 5)$ is a power of two, and the same is true for the triplet $(-2, 6, 10)$. \n\nAssume, for contradiction, that some set $M$ has more than $6$ elements.\n\nClearly, $M$ can't contain three or more non-positive numbers. Hence, it contains at least five positive numbers. Let $x$ be the largest positive number in $M$, and let $a, b, c, d$ be four other positive numbers in $M$. Consider the pairs $x + a, x + b, x + c, x + d$. These are all larger than $x$ and less than $2x$. The open interval $(x, 2x)$ contains at most one power of two, so at least three of the four sums are not a power of two. Without loss of generality, suppose these are $x + a, x + b, x + c$. Considering the triplets $(a, b, x), (a, c, x), (b, c, x)$, we infer that all $a + b, a + c, b + c$ are powers of two. However, this is impossible. Without loss of generality, let $a = \\max\\{a, b, c\\}$. Then $a + b$ and $a + c$ both lie in $(a, 2a)$, so at least one of them is not a power of two—a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15653,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all monic polynomials $p(x)$ with real coefficients satisfying the following properties:\n\n1. $p(x)$ is nonconstant and all its roots are real and distinct.\n2. If $a$ and $b$ are roots of $p(x)$, then so is $a + b + ab$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(x) = x^2 + 2x$ and define $f^n = f \\circ f \\circ \\dots \\circ f$ ($n-1$ times). Let $a$ be a root of $p(x)$. From the second property, we see that $a, f(a), f^2(a), \\dots$ are also roots of $p(x)$.\n\n*Case 1:* If $a > 0$, then $0 < a < f(a)$. Since $f$ is strictly increasing over the interval $(0, \\infty)$, we have $a < f(a) < f^2(a) < \\dots$.\n\n*Case 2:* If $-1 < a < 0$, then $0 > a > f(a) > -1$. Since $f$ is strictly increasing over the interval $(-1, 0)$, we have $a > f(a) > f^2(a) > \\dots$.\n\n*Case 3:* If $-2 < a < -1$, then $-1 < f(a) < 0$. Substituting for $a$ by $f(a)$ in Case 2, we get $f(a) > f^2(a) > \\dots$.\n\n*Case 4:* If $a < -2$, then $f(a) > 0$. Substituting for $a$ by $f(a)$ in Case 1, we get $f(a) < f^2(a) < \\dots$.\n\nFrom the four cases, we infer that if $a \\notin \\{-2, -1, 0\\}$, then $p(x)$ has infinitely many distinct roots, which is impossible. Hence, $a \\in \\{-2, -1, 0\\}$ and by direct checking all the possible $p(x)$ are\n\n$$\nx,\\quad x+1,\\quad x(x+1),\\quad x(x+2),\\quad x(x+1)(x+2).\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15654,
    "subject": "Mathematics (Olympiad)",
    "question": "Given two circles $O$, $O'$ with different radii and intersecting at $A$ and $B$. The circle centered at $A$ with radius $AB$ intersects $O$ and $O'$ again at $C$ and $D$, respectively. Let $EF$ be the common tangent closer to $B$ of the two circles, with $E \\in O$ and $F \\in O'$. Rays $AE$ and $AF$ intersect $BC$ and $BD$ at $M$ and $N$, respectively. Prove that the internal bisector of $\\angle CBD$ passes through the circumcenter of triangle $AMN$.\n\n![](images/Saudi_Booklet_2025_p37_data_b78c575104.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $AB = AC$, in circle $O$, we have $\\angle ABC = \\angle AEB$, so $ABM$ and $AEB$ are similar triangles. This implies that\n\n$$\n\\frac{AM}{AB} = \\frac{AB}{AE} \\implies AB^2 = AM \\cdot AE.\n$$\n\nSimilarly, $AB^2 = AN \\cdot AF$, so $AM \\cdot AE = AN \\cdot AF$, which means $EMNF$ is cyclic. Hence, $\\angle AMN = \\angle AFE$. On the other hand,\n\n$$\n\\angle AFE = \\angle AFB + \\angle BFE = \\angle ABD + \\angle BAF = \\angle BNF\n$$\n\nso $\\angle AMN = \\angle BNF = \\angle AND$, which means $BD$ is tangent to $(AMN)$. Similarly, $BC$ is also tangent to $(AMN)$, so the center of $(AMN)$ is equidistant from the two segments $BC$ and $BD$. In other words, the angle bisector of $\\angle CBD$ passes through the center of $(AMN)$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15655,
    "subject": "Mathematics (Olympiad)",
    "question": "In each square of a garden shaped like a $2022 \\times 2022$ board, there is initially a tree of height $0$. A gardener and a lumberjack alternate turns playing the following game, with the gardener taking the first turn:\n\n- The gardener chooses a square in the garden. Each tree on that square and all the surrounding squares (of which there are at most eight) then becomes one unit taller.\n- The lumberjack then chooses four different squares on the board. Each tree of positive height on those squares then becomes one unit shorter.\n\nWe say that a tree is *majestic* if its height is at least $10^6$. Determine the largest number $K$ such that the gardener can ensure there are eventually $K$ majestic trees on the board, no matter how the lumberjack plays.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $K = 5 \\cdot \\frac{2022^2}{9} = 2271380$. In general, for a $3N \\times 3N$ board, $K = 5N^2$.\n\nWe solve the problem for a general $3N \\times 3N$ board. First, we prove that the lumberjack has a strategy to ensure there are never more than $5N^2$ majestic trees. Giving the squares of the board coordinates in the natural manner, color each square where at least one of its coordinates is divisible by $3$, shown below for a $9 \\times 9$ board:\n\nThen, as each $3 \\times 3$ square on the board contains exactly $5$ colored squares, each move of the gardener will cause at most $4$ trees on non-colored squares to grow. The lumberjack may therefore cut those trees, ensuring no tree on a non-colored square has positive height after his turn. Hence there cannot ever be more majestic trees than colored squares, which is $5N^2$.\n\nNext, we prove the gardener may ensure there are $5N^2$ majestic trees. Let us modify the game to make it more difficult for the gardener: on the lumberjack's turn in the modified game, he may decrement the height of all trees on the board except those the gardener just grew, in addition to four of the trees the gardener just grew. Clearly, a sequence of moves for the gardener which ensures that there are $K$ majestic trees in the modified game also ensures this in the original.\n\nLet $M = \\binom{9}{5}$; we say that a *map* is one of the $M$ possible ways to mark $5$ squares on a $3 \\times 3$ board. In the modified game, after the gardener chooses a $3 \\times 3$ subboard on the board, the lumberjack chooses a map in this subboard, and the total result of the two moves is that each tree marked on the map increases its height by $1$, each tree in the subboard which is not in the map remains unchanged, and each tree outside the subboard decreases its height by $1$. Also note that if the gardener chooses a $3 \\times 3$ subboard $Ml$ times, the lumberjack will have to choose some map at least $l$ times, so there will be at least $5$ trees which each have height $\\ge l$.\n\nThe strategy for the gardener will be to divide the board into $N^2$ disjoint $3 \\times 3$ subboards, number them $0, \\ldots, N^2 - 1$ in some order. Then, for $b = N^2 - 1, \\ldots, 0$ in order, he plays $10^6 M(M+1)^b$ times on subboard number $b$. Hence, on subboard number $b$, the moves on that subboard will first ensure $5$ of its trees grow by at least $10^6(M+1)^b$, and then each move after that will decrease their heights by $1$. (As the trees on subboard $b$ had height $0$ before the gardener started playing there, no move made on subboards $\\ge b$ decreased their heights.) As the gardener makes\n\n$$\n10^6 M(M+1)^{b-1} + \\dots = 10^6((M+1)^b - 1)\n$$\n\nmoves after he finishes playing on subboard $b$, this means that on subboard $b$, there will be $5$ trees of height at least\n\n$$\n10^6(M+1)^b - 10^6((M+1)^b - 1) = 10^6,\n$$\n\nhence each of the subboards has $5$ majestic trees, which was what we wanted. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15656,
    "subject": "Mathematics (Olympiad)",
    "question": "For positive real numbers $a, b, c, d$ with $abcd = 1$, determine all values taken by the expression\n\n$$\n\\frac{1+a+ab}{1+a+ab+abc} + \\frac{1+b+bc}{1+b+bc+bcd} + \\frac{1+c+cd}{1+c+cd+cda} + \\frac{1+d+da}{1+d+da+dab}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\n\\begin{aligned}\na(1+b+bc+bcd) &= 1+a+ab+abc, \\\\\nb(1+c+cd+cda) &= 1+b+bc+bcd, \\\\\nc(1+d+da+dab) &= 1+c+cd+cda.\n\\end{aligned}\n$$\n\nThen, by multiplying the second, third, and fourth terms by $a$, $ab$, and $abc$ respectively, we get fractions with denominator $1+a+ab+abc$, and numerator\n\n$$\n(1+a+ab) + a(1+b+bc) + ab(1+c+cd) + abc(1+d+da)\n$$\n\nequal to $3(1+a+ab+abc)$. Therefore, the expression has the constant value $3$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15657,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $d$ be a nonnegative integer. Determine all functions $f : \\mathbb{R}^2 \\to \\mathbb{R}$ such that, for any real constants $A, B, C,$ and $D$, the function $f(At+B, Ct+D)$ is a polynomial in $t$ of degree at most $d$.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that $f(x, y)$ is a polynomial in $x$ and $y$ of degree at most $d$.\n\nIt is clear that every such polynomial satisfies the given condition. To prove the converse, let $f$ be a function satisfying the condition. Pick $(d+2)$ straight lines $\\ell_1, \\ell_2, \\dots, \\ell_{d+2}$ in $\\mathbb{R}^2$ such that no two are parallel and no three are concurrent. Let the equation of $\\ell_i$ be $h_i(x, y) = 0$, where $h_i$ is a linear polynomial.\n\nFor $i < j$, let $(a_{ij}, b_{ij})$ be the intersection of $\\ell_i$ and $\\ell_j$, and consider the polynomial\n\n$$\n\\varphi(x, y) = \\sum_{1 \\le i < j \\le d+2} f(a_{ij}, b_{ij}) \\prod_{\\substack{k=1 \\\\ k \\ne i, j}}^{d+2} \\frac{h_k(x, y)}{h_k(a_{ij}, b_{ij})}.\n$$\n\nIt is easy to see that $\\varphi(a_{ij}, b_{ij}) = f(a_{ij}, b_{ij})$ for all $i < j$, and that $\\deg \\varphi \\le d$. We shall show that $\\varphi(a, b) = f(a, b)$ for all $a, b \\in \\mathbb{R}$.\n\nWe first make an observation: if $\\ell$ is a line such that $\\varphi(a, b) = f(a, b)$ for at least $(d+1)$ points $(a, b)$ on $\\ell$, then $\\varphi(a, b) = f(a, b)$ for all points $(a, b)$ on $\\ell$. Indeed, pick constants $A, B, C,$ and $D$ such that $t \\mapsto (At + B, Ct + D)$ parametrizes the line. Then, $\\varphi(At + B, Ct + D) - f(At + B, Ct + D)$ is a polynomial of degree at most $d$, and it vanishes at least $(d+1)$ points, so $\\varphi(a, b) = f(a, b)$ for all $(a, b)$ on $\\ell$.\n\nNow, for a fixed $\\ell_i$, note that $f(a_{ij}, b_{ij}) = \\varphi(a_{ij}, b_{ij})$ for every $j \\ne i$, so $f(a, b) = \\varphi(a, b)$ for all points $(a, b)$ on $\\ell_i$ from the claim. If $(c, d)$ is a point not lying on any $\\ell_i$, then we can construct a line $\\ell$ which passes through $(c, d)$, does not pass through any $(a_{ij}, b_{ij})$, and is not parallel to any $\\ell_i$. Now $f(a, b) = \\varphi(a, b)$ with $(a, b) = \\ell_i \\cap \\ell$ for various $i$, so $f(a, b) = \\varphi(a, b)$ for all $(a, b)$ on $\\ell$. In particular, $\\varphi(c, d) = f(c, d)$. This completes the proof. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15658,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AD$, $BE$, and $CF$ denote the altitudes of triangle $\\triangle ABC$. Points $E'$ and $F'$ are the reflections of $E$ and $F$ over $AD$, respectively. The lines $BF'$ and $CE'$ intersect at $X$, while the lines $BE'$ and $CF'$ intersect at the point $Y$. Prove that if $H$ is the orthocenter of $\\triangle ABC$, then the lines $AX$, $YH$, and $BC$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that the desired point of concurrency is the midpoint of $BC$. Assume that $\\triangle ABC$ is acute. Let $(ABC)^5$ intersect $(AEF)$ at the point $Y'$; we will prove that $Y = Y'$. ![](images/2019_bmo_shortlist_p23_data_ddcf0b653e.png)\n\nUsing the fact that $H$ is the incenter of $\\triangle DEF$ we get that $D, E', F$ and $D, F', E$ are triples of collinear points. Furthermore,\n\n$$\n90^\\circ = \\angle^6 AEH = \\angle AF'H = \\angle AE'H = \\angle AFH \\Rightarrow F', E', H \\in (AEFY').\n$$\n\nWe will now prove that the points $Y', B, D, F'$ are concyclic. Indeed,\n\n$$\n\\angle Y'BD = \\angle Y'BC = \\angle Y'AC = \\angle Y'AE = \\angle Y'F'E \\Rightarrow (Y', B, D, F').\n$$\n\nNow, as\n\n$$\n\\angle F'Y'B = \\angle F'DC = \\angle EDC = \\angle CAB = \\angle CY'B,\n$$\n\nthe points $C, F', Y'$ are collinear. Similarly we get that $B, E', Y'$ are collinear, which implies\n\n$$\nY' = Y = (ABC) \\cap (AEF).\n$$\n\n$^5$\\text{(XYZ)}$ denotes the circumcircle of $\\triangle XYZ$\n\n$^6$\\angle$ denotes a directed angle modulo $\\pi$\n\nSince we proved this property using directed angles, we know that it is also true for obtuse triangles.\n\nNotice that the points $A$, $B$, $C$, $H$ form an orthocentric system; in other words $H$ is the orthocenter of $\\triangle ABC$ and $A$ is the orthocenter of $\\triangle HBC$. Furthermore, notice that $F'$ is to $\\triangle ABC$ as $E'$ is to $\\triangle HBC$ and that $E'$ is to $\\triangle ABC$ as $F'$ is to $\\triangle HBC$. This means that $X$ is to $\\triangle HBC$ as $Y$ is to $\\triangle ABC$ and, as we know the proven property is also true for obtuse triangles, we get\n\n$$\nX = (HBC) \\cap (AEF).\n$$\n\nBy Reflecting the Orthocenter Lemma we know that in a triangle $ABC$, the reflection of its orthocenter over the midpoint of $BC$ is the antipode of $A$ with respect to $(ABC)$. Applying this Lemma on the triangles $ABC$ and $HBC$ we get that $YH$ and $AX$ both go through the midpoint of $BC$, thus finishing the solution. $\\Box$\n\n*Remark 1:* The crucial part of this solution is defining the points $X$, $Y$ as intersections of circles. This can also be achieved directly by using similar triangles or by using the Spiral Similarity Lemma on $\\triangle HBC$, $\\triangle HF'E'$ and $\\triangle ABC$, $\\triangle AE'F'$.\n\n*Remark 2:* We can also invert around $A$ with radius $\\sqrt{AH \\cdot AD}$ or around $H$ with radius $\\sqrt{HA \\cdot HD}$ to prove that $X$ or $Y$ invert to the midpoint of $BC$ by using the existence of the nine-point circle. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15659,
    "subject": "Mathematics (Olympiad)",
    "question": "試求所有正整數 $n$，使得我們可以在 $n \\times n$ 棋盤的 $n^2$ 個方格中，各放入 $\\uparrow, \\downarrow, \\leftarrow, \\rightarrow$ 四個箭號中的其中一個，讓以下三個條件皆被滿足：\n\n1. 從任何一格作為起點出發，按照箭號的方向走，我們都會回到起點，且途中不會走出棋盤。\n2. 除了最上與最下兩橫列之外，任一列中 $\\uparrow$ 和 $\\downarrow$ 的箭號一樣多（但不同列之間的數量可能不同）。\n3. 除了最左與最右兩直排之外，任一排中 $\\leftarrow$ 和 $\\rightarrow$ 的箭號一樣多（但不同排之間的數量可能不同）。",
    "options": [],
    "answer": "See solution",
    "solution": "唯一可能是 $n=2$（此時箭號形成順時鐘或逆時鐘的環）。\n\n顯然最上橫列不能有 $\\uparrow$，假設其有 $k$ 個 $\\downarrow$。由條件 (1)，這表示第二橫列必須有 $k$ 個 $\\uparrow$，但又由條件 (2)，我們知第二橫列同時須有 $k$ 個 $\\downarrow$。依相同論證，我們知道第 3 到第 $n-1$ 橫列都必須有 $k$ 個 $\\uparrow$ 與 $k$ 個 $\\downarrow$，而最下橫列必須有 $k$ 個 $\\uparrow$ 與 0 個 $\\downarrow$。換言之，全棋盤中 $\\uparrow$ 和 $\\downarrow$ 各有 $(n-1)k$ 個。\n\n同理，若最左邊有 $\\ell$ 個 $\\rightarrow$，則由條件 (1) 與條件 (3)，我們知全棋盤中 $\\rightarrow$ 和 $\\leftarrow$ 各有 $(n-1)\\ell$ 個，從而全棋盤的箭頭數量總和為 $2(n-1)(k+\\ell)$ 個。但箭頭總數勢必等於格子數，故必有\n\n$$\n2(n-1)(k+\\ell) = n^2.\n$$\n\n但基於 $\\gcd(n, n-1) = 1$，故這在 $n > 2$ 時是不可能的。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15660,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $x$, $y$, and $z$ such that\n\n$$\nx^5 + 4^y = 2013^z.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $2013$ is divisible by $11$, and that $x^5$ is congruent to $0$, $1$, or $-1$ modulo $11$. Since $4^y$ is congruent to $4$, $5$, $9$, $3$, or $1$ modulo $11$, we get that $x^5 \\equiv -1 \\pmod{11}$ and $4^y \\equiv 1 \\pmod{11}$. Therefore, $5 \\mid y$, and the given equation is of the form $a^5 + b^5 = 2013^z = 3^z \\cdot 11^z \\cdot 61^z$, where $a = x$ and $b = 4^{y/5}$. Note that $2 \\nmid x$, and therefore $(a, b) = 1$.\n\nAlso, $a^5 + b^5 = (a + b)(a^4 - a^3b + a^2b^2 - ab^3 + b^4)$, and by the previous remark, we have $(a + b, a^4 - a^3b + a^2b^2 - ab^3 + b^4) = (a + b, 5a^4) = (a + b, 5) = 1$.\n\nLet us prove that $3 \\nmid a^4 - a^3b + a^2b^2 - ab^3 + b^4$. Suppose that this is not true, i.e., that $3 \\mid a^4 - a^3b + a^2b^2 - ab^3 + b^4$. If $3 \\mid ab$, then $3 \\mid a$ and $3 \\mid b$, a contradiction. Since $3^z \\mid a^5 + b^5$, this implies that $3^z \\mid a + b$.\n\nAlso,\n\n$$\n(a + b)^4 > a^4 - a^3b + a^2b^2 - ab^3 + b^4 > a + b,\n$$\n\nand therefore $61^z \\mid a^4 - a^3b + a^2b^2 - ab^3 + b^4$, and $11^z \\mid a + b$. So $a + b \\geq 33^z$, and hence\n\n$$\n(a + b)^2 > 61^z \\geq a^4 - a^3b + a^2b^2 - ab^3 + b^4.\n$$\n\nLet $a > b$ (we can similarly deal with the case $a < b$). Then $a^4 - a^3b = a^3(a - b) \\geq a^3$ and $a^2b^2 - ab^3 = (a - b)ab^2 \\geq ab^2$. So,\n\n$$\na^4 - a^3b + a^2b^2 - ab^3 + b^4 \\geq a^3 + ab + b^4 \\geq 2a^2 + ab + b^2 \\geq (a + b)^2,\n$$\n\na contradiction.\n\nTherefore, the given equation does not have any solution in positive integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15661,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for positive real numbers $a, b, c, d$ such that $abcd = 1$, the following inequality holds:\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} + \\frac{9}{a+b+c+d} \\geq \\frac{25}{4}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, we will prove that, whenever there are two numbers among $a, b, c, d$ that are equal, the inequality holds. We may assume that $a = b$ and let $s = a + b + c + d$. Then we have\n\n$$\n\\begin{aligned}\n& \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} + \\frac{9}{a+b+c+d} \\\\\n&= \\frac{2}{a} + \\frac{c+d}{cd} + \\frac{9}{s} = \\frac{2}{a} + a^2(s - 2a) + \\frac{9}{s} \\\\\n&= \\frac{2}{a} - 2a^3 + \\left(a^2 s + \\frac{9}{s}\\right).\n\\end{aligned}\n$$\n\nWe define the expression above as $f(s)$. Then we see that $f(s)$ reaches the minimum for $s = \\frac{3}{a}$.\n\nWhen $a \\geq \\frac{\\sqrt{2}}{2}$, however, we have\n\n$$\ns = a + b + c + d \\geq 2a + \\frac{2}{a} \\geq \\frac{3}{a}.\n$$\n\nAt this time, $f(s)$ reaches the minimum for $s = 2a + \\frac{2}{a}$.\n\nWe then have\n\n$$\n\\begin{aligned}\n& \\frac{2}{a} - 2a^3 + \\left(a^2 s + \\frac{9}{s}\\right) = \\frac{2}{a} - 2a^3 + a^2\\left(2a + \\frac{2}{a}\\right) + \\frac{9}{s} \\\\\n&= \\frac{2}{a} + 2a + \\frac{9}{s} = s + \\frac{9}{s} \\\\\n&= \\frac{7}{16}s + \\frac{9}{16}s + \\frac{9}{s} \\geq \\frac{7}{16} \\times 4 + 2\\sqrt{\\frac{9}{16}s \\cdot \\frac{9}{s}} \\\\\n&= \\frac{7}{4} + \\frac{9}{2} = \\frac{25}{4} \\quad (\\therefore s = 2a + \\frac{2}{a} \\geq 4).\n\\end{aligned}\n$$\n\nWhen $0 < a < \\frac{\\sqrt{2}}{2}$, we have\n\n$$\n\\begin{aligned}\n\\frac{2}{a} - 2a^3 + \\left(a^2 s + \\frac{9}{s}\\right) &\\geq \\frac{2}{a} - 2a^3 + 6a = \\frac{2}{a} + 5a + (a - 2a^3) \\\\\n&> \\frac{2}{a} + 5a \\geq 2\\sqrt{\\frac{2}{a} \\cdot 5a} = 2\\sqrt{10} > \\frac{25}{4}.\n\\end{aligned}\n$$\n\nSecond, we consider the case that $a, b, c, d$ are different from each other. We may assume that $a > b > c > d$. If $\\frac{ad}{c} \\cdot b \\cdot c \\cdot c = abcd = 1$, by using the result above, we have\n\n$$\n\\frac{1}{ad} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{c} + \\frac{9}{ad} \\geq \\frac{25}{4}.\n$$\n\nTherefore, we only need to prove that\n\n$$\n\\begin{aligned}\n& \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} + \\frac{9}{a + b + c + d} \\\\\n\\geq & \\frac{1}{ad} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{c} + \\frac{9}{ad} + b + c + d.\n\\end{aligned}\n\\qquad \\textcircled{1}\n$$\n\nWe have\n\n$$\n\\begin{aligned}\n\\textcircled{1} &\\Leftrightarrow \\frac{1}{a} + \\frac{1}{d} + \\frac{9}{a+b+c+d} \\geq \\frac{c}{ad} + \\frac{1}{c} + \\frac{9}{ad} + \\frac{b+2c}{c} \\\\\n&\\Leftrightarrow \\frac{ac+cd-c^2-ad}{acd} \\geq \\frac{9}{(a+b+c+d)\\left(\\frac{ad}{c} + b + 2c\\right)} \\\\\n&\\qquad (a+d-\\frac{ad}{c}-c) \\\\\n&\\Leftrightarrow \\frac{(a-c)(c-d)}{acd} \\geq \\frac{9}{(a+b+c+d)\\left(\\frac{ad}{c} + b + 2c\\right)} \\\\\n&\\qquad \\frac{(a-c)(c-d)}{c} \\\\\n&\\Leftrightarrow \\frac{1}{ad} \\geq \\frac{9}{(a+b+c+d)\\left(\\frac{ad}{c} + b + 2c\\right)}\n\\end{aligned}\n$$\n\n$$\n\\begin{aligned}\n&\\Leftrightarrow (a+b+c+d)\\left(\\frac{ad}{c}+b+2c\\right) \\geq 9ad \\\\\n&\\Leftrightarrow \\frac{ad}{c}+b+2c \\geq \\sqrt{9ad}\\quad (a+b+c+d > \\frac{ad}{c}+b+2c) \\\\\n&\\Leftrightarrow \\frac{ad}{c}+3c \\geq \\sqrt{9ad}.\n\\end{aligned}\n$$\n\nThe proof is complete.\n\n$\\Box$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15662,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $a, b$ which satisfy the equation:\n\n$$\nab^3 + a^3 + b + 1 = 2019.$$",
    "options": [],
    "answer": "See solution",
    "solution": "Both unknowns must be even, otherwise the equation would equate an even and an odd number. Let $a = 2n$ and $b = 2k$:\n\n$$\n2n \\cdot 8k^3 + 8n^3 + 2k = 2018 \\text{ or } 8nk^3 + 4n^3 + k = 1009.\n$$\n\n$k$ must be odd. Now, perform an exhaustive search: $nk^3 \\le \\frac{1009}{8}$, so $k^3 \\le 125$.\n\nPossible values:\n\n- $k = 5$: $1000n + 4n^3 = 1004 \\Rightarrow n = 1$ is valid.\n- $k = 3$: $64n + 4n^3 = 1006$; no solution since 4 divides the left but not the right side.\n- $k = 1$: $8n + 4n^3 = 1008 \\Rightarrow 2n + n^3 = 252 \\Rightarrow n = 2m \\Rightarrow 4m + 8m^3 = 252 \\Rightarrow m + 2m^3 = 63$. For $m = 1, 2, 3$, none satisfy the equation.\n\nThus, the only solution is $k = 5$, $n = 1$, so $b = 10$, $a = 2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15663,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $Q_1$ and $Q_2$ be points on the side $BC$ such that $Q_2$ lies on the segment $BQ_1$ and $BQ_1 = CQ_2$. The segments $P_1Q_2$ and $P_2Q_1$ meet at $R$, and the circles $P_1P_2R$ and $Q_1Q_2R$ meet again at $S$, situated inside triangle $P_1Q_1R$. Finally, let $M$ be the midpoint of the side $AC$. Prove that the angles $P_1RS$ and $Q_1RM$ are equal.\n\n![](images/RMC_2015_BT_p68_data_66603e1026.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Throughout the solution, $[XYZ]$ and $d(X, YZ)$ denote the area of triangle $XYZ$ and the distance from point $X$ to the line $YZ$, respectively.\n\nSince the quadrilaterals $SRQ_2Q_1$ and $SRP_2P_1$ are cyclic, $\\angle SQ_1R = \\angle SQ_2R$ and $\\angle SP_1R = \\angle SP_2R$ (see Fig. 6), so the triangles $SP_1Q_2$ and $SP_2Q_1$ are similar, and\n\n$$\n\\frac{d(S, P_1Q_2)}{d(S, P_2Q_1)} = \\frac{P_1Q_2}{P_2Q_1}. \\qquad (1)\n$$\n\nLet $K$ and $L$ be the midpoints of $AB$ and $AC$, respectively; then $P_1K = P_2K$ and $Q_1L = Q_2L$. Therefore, $d(Q_1, MP_1) + d(Q_2, MP_1) = 2d(L, MP_1)$, so\n\n$$\n[MP_1Q_2] + [MP_1Q_1] = 2[MP_1L] = ML \\cdot d(P_1, ML) = [ABC]/2.\n$$\n\nSimilarly, $[MP_2Q_1] + [MP_1Q_1] = [ABC]/2 = [MP_1Q_2] + [MP_1Q_1]$. Thus $[MP_1Q_2] = [MP_2Q_1]$, whence\n\n$$\n\\frac{d(M, P_1Q_2)}{d(M, P_2Q_1)} = \\frac{P_2Q_1}{P_1Q_2}. \\qquad (2)\n$$\n\nIn the angle $P_1RQ_1$, the condition $d(X, P_1Q_2)/d(X, P_2Q_1) = \\alpha$ determines a ray emanating from $R$. Moreover, rays symmetric with respect to the bisector of $\\angle P_1RQ_1$ correspond to reciprocal values of $\\alpha$. Consequently, relations (1) and (2) show that the rays $RS$ and $RM$ are symmetric with respect to this angle bisector, and the conclusion follows.\n\n_Remark._ Relation (2) may be obtained in several different ways. For instance, one may notice that midpoints $X, X_1$ and $X_2$ of the segments $BR, P_1Q_1$ and $P_2Q_2$, respectively, are collinear — the Gauss-Newton line $\\ell$ of the quadrilateral $P_1P_2Q_1Q_2$ (see Fig. 7). Moreover, $KX_1LX_2$ is the Varignon parallelogram of the quadrilateral $P_1P_2Q_1Q_2$, hence the segments $X_1X_2$ and $KL$ have a common midpoint $N$. Since $XN$ is a midline in the triangle $BMR$, the line $RM$ is parallel to $\\ell$, so $\\overrightarrow{RM} = \\alpha \\overrightarrow{X_1X_2} = \\frac{\\alpha}{2}(\\overrightarrow{P_1Q_2} + \\overrightarrow{P_1Q_1})$ which easily yields (2).\n\nYet another way of obtaining the same relation is the following. Let the lines $P_1Q_2$ and $P_2Q_1$ meet $AC$ at points $U$ and $V$, respectively. By Menelaus' theorem,\n\n$$\n\\frac{AU}{UC} = \\frac{AP_1}{P_1B} \\cdot \\frac{BQ_2}{Q_2C} = \\frac{BP_2}{P_2A} \\cdot \\frac{CQ_1}{Q_1B} = \\frac{CV}{VA},\n$$\n\nso $AU = CV$. Thus $RM$ is a median in triangle $RUV$, so\n\n![](images/RMC_2015_BT_p69_data_89a362601f.png)\n\n$$\n1 = \\frac{[MUR]}{[MVR]} = \\frac{RU \\cdot d(M, P_1Q_2)}{RV \\cdot d(M, P_2Q_1)}, \\quad \\text{or} \\quad \\frac{d(M, P_1Q_2)}{d(M, P_2Q_1)} = \\frac{RV}{RU}.\n$$\n\nFinally, comparing the projections of the segments $RU$, $RV$, $P_1Q_2$, and $P_2Q_1$ onto the perpendicular to $AC$, we get $RU/P_1Q_2 = RV/P_2Q_1$, which is the same as $RV/RU = P_2Q_1/P_1Q_2$, as required.\n\nSimilar calculations may be carried out in terms of the sines of the corresponding angles $\\angle P_1RM$, $\\angle Q_2RM$, $\\angle P_1RS$, and $\\angle Q_2RS$, instead of distances from points to lines.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15664,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$ be an integer, and consider a set of $n$ points in three-dimensional space such that:\n\n1. Every two distinct points are connected by a string which is either red, green, blue, or yellow.\n2. For every three distinct points, if the three strings between them are not all of the same colour, then they are of three different colours.\n3. Not all the strings have the same colour.\n\nFind the maximum possible value of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us fix a notation: Number the points $1, 2, \\ldots, n$; the colour of the string between points $i$ and $j$ is denoted by $c(i, j)$, so that $c(i, j) \\in \\{\\text{B}, \\text{G}, \\text{R}, \\text{Y}\\}$ for all $1 \\le i, j \\le n$, $i \\ne j$, where B = Blue, G = Green, R = Red, and Y = Yellow.\n\nFor $n = 3$, there is a trivial solution, by putting, say, $c(1,2) = \\text{B}$, $c(2,3) = \\text{G}$, $c(3,1) = \\text{R}$. Also, for $n = 4$ it is fairly straightforward to see a solution, say $c(1,2) = c(3,4) = \\text{B}$, $c(1,3) = c(2,4) = \\text{G}$, and $c(1,4) = c(2,3) = \\text{R}$.\n\nHenceforth, assume that $n > 4$.\n\nWe first observe that no point can have three strings of the same colour attached to it. For suppose (without loss of generality) that $c(1,2) = c(1,3) = c(1,4) = \\text{B}$. Then $c(2,3) = c(2,4) = c(3,4) = \\text{B}$. This implies that there must be a fifth point, $5$, such that $c(1,5) \\neq \\text{B}$ (otherwise all strings will have the same colour B). Assume $c(1,5) = \\text{G}$. Then $c(2,5), c(3,5), c(4,5) \\notin \\{\\text{B}, \\text{G}\\}$. Assume that $c(2,5) = \\text{R}$, say. Then $c(3,5) \\neq \\text{R}$, forcing $c(3,5) = \\text{Y}$. But then $c(4,5) \\notin \\{\\text{R}, \\text{Y}\\}$, by condition (2) applied to the two sets $\\{2,4,5\\}$ and $\\{3,4,5\\}$ of three points each. The contradiction $c(4,5) \\notin \\{\\text{B}, \\text{G}, \\text{R}, \\text{Y}\\}$ proves that our claim is valid. Consequently, there can be at most two strings of each colour attached to each point, implying that there can be no more than nine points in total.\n\nWe now show that it is indeed possible that nine points can be mutually connected by strings such that all conditions (1) to (3) are satisfied.\n\nPartition the points $1, 2, \\ldots, 9$ into three subsets of three points each: $X = \\{1, 2, 3\\}$, $Y = \\{4, 5, 6\\}$, $Z = \\{7, 8, 9\\}$. First, consider the three 'internal' monochromatic triangles (all sides Blue) where the points in each of these subsets are connected with Blue strings. Then we construct nine further monochromatic triangles $KLM$, where $K \\in X$, $L \\in Y$, and $M \\in Z$ (three of each of the colours Green, Red and Yellow). This can be done in several ways (six ways, to be precise). We show here a nice symmetrical one:\n\n- Three Green triangles: $\\{1, 4, 7\\}$, $\\{2, 5, 8\\}$, $\\{3, 6, 9\\}$\n- Three Red triangles: $\\{1, 5, 9\\}$, $\\{2, 6, 7\\}$, $\\{3, 4, 8\\}$\n- Three Yellow triangles: $\\{1, 6, 8\\}$, $\\{2, 4, 9\\}$, $\\{3, 5, 7\\}$\n\n![](images/s3s2022_p5_data_9cb52ba595.png)",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15665,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n \\geq 3$ segments, each with a distinct positive integer length in centimeters. It is known that any three of these $n$ segments can form a nondegenerate triangle. Among these segments, there are segments of lengths $5$ cm and $12$ cm. What is the largest possible value of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "Arrange the segment lengths in increasing order: $a_1 < a_2 < \\dots < a_n$. For any three segments to form a triangle, the sum of the lengths of the two shortest must exceed the length of the longest. If any segment besides $5$ cm and $12$ cm has length less than $8$ cm, then $5 + a \\leq 12$ for $a \\leq 7$, so a triangle cannot be formed with $5$, $a$, and $12$. Thus, $a_1 = 5$ and $a_2 \\geq 8$. Since $a_2$ must be less than or equal to $12$, the possible segment lengths are $5, a_2, a_2 + 1, a_2 + 2, a_2 + 3, a_2 + 4$, giving at most $6$ segments. The set $\\{5, 8, 9, 10, 11, 12\\}$ meets all conditions, so the largest possible $n$ is $6$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15666,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a positive integer greater than $1$ and let $(a_n)_{n \\\\ge 1}$ be a sequence of pairwise distinct positive integers. Show that the set $M = \\{a_1^k, a_2^k, a_3^k, \\dots\\}$ does not contain an infinite arithmetic sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume, by way of contradiction, that $M$ contains the infinite arithmetic sequence $\\{b_1^k, b_2^k, b_3^k, \\dots\\}$, where $1 \\le b_1 < b_2 < b_3 < \\dots$.\n\nIf we denote by $r$ its common difference, we have $b_{n+1}^k - b_n^k = r$ for all $n \\ge 1$.\n\nObserve that\n\n$$\nr = b_{n+1}^k - b_n^k = (b_{n+1} - b_n) \\left(b_{n+1}^{k-1} + b_{n+1}^{k-2}b_n + \\dots + b_{n+1}b_n^{k-2} + b_n^{k-1}\\right).\n$$\n\nThe expression in the second parenthesis is clearly increasing, hence the sequence $b_{n+1} - b_n$ is decreasing, and since all its terms are positive integers, we reach a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15667,
    "subject": "Mathematics (Olympiad)",
    "question": "Positive integer $b$ is obtained by reordering the digits in a positive integer $a$. Which of the following claims are definitely true?\n\na) The sums of the digits of numbers $2a$ and $2b$ are equal.\n\nb) The sums of the digits of numbers $3a$ and $3b$ are equal.\n\nc) The sums of the digits of numbers $5a$ and $5b$ are equal.",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* a) and c).\n\nCall digits $0, 1, 2, 3, 4$ small and digits $5, 6, 7, 8, 9$ large. Denote the digits of a $k$-digit number $n$ from right to left by $(n)_0, (n)_1, \\dots, (n)_{k-1}$. Let $\\Sigma(n)$ be the sum of all digits of $n$ and $l(n)$ the number of large digits of $n$.\n\na) If $(a)_i$ is small, then $(2a)_i = 2(a)_i$ or $(2a)_i = 2(a)_i + 1$ depending on whether $(a)_{i-1}$ is small or large. If $(a)_i$ is large, then $(2a)_i = 2(a)_i - 10$ or $(2a)_i = 2(a)_i - 9$ depending on whether $(a)_{i-1}$ is small or large. Thus, when multiplying by 2, each large digit decreases its place by 10 and increases the preceding digit by 1 compared to a small digit. Therefore, for any natural number $n$, $$\\Sigma(2n) = 2\\Sigma(n) - 9l(n).$$ Since $\\Sigma(a) = \\Sigma(b)$ and $l(a) = l(b)$, we have $\\Sigma(2a) = \\Sigma(2b)$.\n\nb) For example, if $a = 34$ and $b = 43$, then $\\Sigma(3a) = 1 + 0 + 2 = 3$ but $\\Sigma(3b) = 1 + 2 + 9 = 12$.\n\nc) Clearly, $\\Sigma(10a) = \\Sigma(10b)$. By part a), $\\Sigma(10n) = \\Sigma(2 \\cdot 5n) = 2\\Sigma(5n) - 9l(5n)$ for any $n$. Thus, $2\\Sigma(5a) - 9l(5a) = 2\\Sigma(5b) - 9l(5b)$. The digit $(5n)_i$ is large if and only if $(n)_i$ is odd, since a carry when multiplying by 5 can be at most 4. Therefore, $l(5n)$ is the number of odd digits of $n$, so $l(5a) = l(5b)$. Consequently, $\\Sigma(5a) = \\Sigma(5b)$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15668,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{Q}^+$ be the set of all positive rational numbers. Let $f: \\mathbb{Q}^+ \\to \\mathbb{R}$ be a function satisfying the following three conditions:\n\n1. For all $x, y \\in \\mathbb{Q}^+$, we have $f(x)f(y) \\geq f(xy)$.\n2. For all $x, y \\in \\mathbb{Q}^+$, we have $f(x+y) \\geq f(x) + f(y)$.\n3. There exists a rational number $a > 1$ such that $f(a) = a$.\n\nProve that $f(x) = x$ for all $x \\in \\mathbb{Q}^+$.\n\n![](posed_by_Bulgaria.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = 1$, $y = a$ in (i). Then\n\n$$\nf(1)f(a) \\geq f(a) \\implies f(1) \\geq 1. \\tag{1}\n$$\n\nBy (ii) and induction on $n$, we have\n\n$$\nf(nx) \\geq n f(x), \\quad \\forall n \\in \\mathbb{Z}^+,\\ \\forall x \\in \\mathbb{Q}^+. \\tag{2}\n$$\n\nTaking $x = 1$ in (2),\n\n$$\nf(n) \\geq n f(1) \\geq n, \\quad \\forall n \\in \\mathbb{Z}^+. \\tag{3}\n$$\n\nBy (i) again,\n\n$$\nf\\left(\\frac{m}{n}\\right) f(n) \\geq f(m), \\quad \\forall m, n \\in \\mathbb{Z}^+. \\tag{4}\n$$\n\nFrom (3) and (4),\n\n$$\nf(q) > 0, \\quad \\forall q \\in \\mathbb{Q}^+. \\tag{5}\n$$\n\nBy (ii) and (5), $f$ is strictly increasing, and\n\n$$\nf(x) \\geq f(\\lfloor x \\rfloor) \\geq \\lfloor x \\rfloor > x - 1, \\quad \\forall x \\in \\mathbb{Q}^+,\\ x > 1. \\tag{6}\n$$\n\nBy (i) and induction on $n$,\n\n$$\nf(x)^n \\geq f(x^n), \\quad \\forall n \\in \\mathbb{Z}^+,\\ \\forall x \\in \\mathbb{Q}^+. \\tag{7}\n$$\n\nThus, by (6) and (7),\n\n$$\nf(x)^n \\geq f(x^n) > x^n - 1, \\quad \\forall x \\in \\mathbb{Q}^+,\\ x > 1. \\tag{8}\n$$\n\nSo,\n\n$$\nf(x) > \\sqrt[n]{x^n - 1}, \\quad \\forall n \\in \\mathbb{Z}^+,\\ x > 1. \\tag{9}\n$$\n\nTaking the limit as $n \\to \\infty$ in (9),\n\n$$\nf(x) \\geq x, \\quad \\forall x \\in \\mathbb{Q}^+,\\ x > 1. \\tag{10}\n$$\n\nBy (iii), (7), and (10), $a^n = f(a)^n \\geq f(a^n) \\geq a^n$, so\n\n$$\nf(a^n) = a^n. \\tag{11}\n$$\n\nFor any $x > 1$, choose $n$ such that $a^n - x > 1$. By (11), (ii), and (10),\n\n$$\na^n = f(a^n) \\geq f(x) + f(a^n - x) \\geq x + (a^n - x) = a^n.\n$$\n\nTherefore,\n\n$$\nf(x) = x, \\quad \\forall x \\in \\mathbb{Q}^+,\\ x > 1. \\tag{12}\n$$\n\nFor all $x \\in \\mathbb{Q}^+$ and $n \\in \\mathbb{Z}^+$, by (12), (i), and (2),\n\n$$\n\\begin{gathered}\nf(n)f(x) \\geq f(nx) \\geq n f(x), \\quad \\forall n > 1,\\ x \\in \\mathbb{Q}^+. \\tag{13}\n\\end{gathered}\n$$\n\nThat is,\n\n$$\nf(nx) = n f(x), \\quad \\forall n > 1,\\ x \\in \\mathbb{Q}^+. \\tag{14}\n$$\n\nTaking $x = m/n$ for $m \\leq n$, $n > 1$,\n\n$$\nf\\left(\\frac{m}{n}\\right) = \\frac{f(m)}{n} = \\frac{m}{n}.\n$$\n\nThus, $f(x) = x$ for all $x \\in \\mathbb{Q}^+,\\ x \\leq 1$.\n\n**Remark:** The condition $f(a) = a > 1$ is essential. For $b \\geq 1$, the function $f(x) = b x^2$ satisfies (i) and (ii) for all $x, y \\in \\mathbb{Q}^+$, and $f$ has a unique fixed point $1/b \\leq 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15669,
    "subject": "Mathematics (Olympiad)",
    "question": "Line $l$ is perpendicular to the side $AC$ of acute triangle $ABC$, and intersects $AC$ at $K$. $l$ intersects the circumscribed circle of $\\triangle ABC$ at $P$ and $T$ (point $P$ is in the same half-plane with respect to $AC$ as vertex $B$). Let $P_1$ and $T_1$ be the projections of points $P$ and $T$ onto line $AB$. Furthermore, vertices $A$ and $B$ belong to the segment $P_1T_1$. Prove that the center of the circumscribed circle of $\\triangle P_1KT_1$ lies on the line containing the midsegment of $\\triangle ABC$ that is parallel to side $AC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $B_1$ be the projection of vertex $B$ onto line $l$ (see the figure below). Then quadrilaterals $BB_1PP_1$ and $TKAT_1$ are cyclic, with diameters $BP$ and $AT$. We have the following equalities of angles:\n\n$$\n\\angle P_1B_1K = \\pi - \\angle P_1B_1P = \\pi - \\angle P_1BP = \\\\\n\\angle ABP = \\pi - \\angle ATP = \\pi - \\angle ATK = \\\\\n= \\pi - \\angle AT_1K = \\pi - \\angle P_1T_1K.\n$$\n\nTherefore, quadrilateral $P_1B_1KT_1$ is cyclic, so the center of the circumscribed circle of $\\triangle P_1KT_1$ lies on the perpendicular bisector of segment $B_1K$. Clearly, the midsegment of $\\triangle ABC$ parallel to $AC$ lies on this perpendicular bisector. Q.E.D.\n\n![](images/UkraineMO2019_booklet_p30_data_de4ddae2e0.png)\n\nFig. 31",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15670,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c, d$ be positive real numbers such that\n$$\na^2 + b^2 + c^2 + d^2 = 4.\n$$\nProve that there are two of $a, b, c, d$ whose sum is greater than or equal to $2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, let $a \\geq b \\geq c \\geq d$. We will prove that $a + b \\geq 2$.\n\nWe have $ab \\geq c^2$ and $ab \\geq d^2$, so by adding them: $2ab \\geq c^2 + d^2$.\n\nTherefore,\n$$\n(a + b)^2 = a^2 + b^2 + 2ab \\geq a^2 + b^2 + c^2 + d^2 = 4,\n$$\nso $a + b \\geq 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15671,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $a, b, c, d$ 為非負實數且滿足 $a + b + c + d = 100$。試證：\n\n$$\n\\sqrt[3]{\\frac{a}{b+7}} + \\sqrt[3]{\\frac{b}{c+7}} + \\sqrt[3]{\\frac{c}{d+7}} + \\sqrt[3]{\\frac{d}{a+7}} \\leq \\frac{8}{\\sqrt[3]{7}}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "令\n\n$$\nS = \\sqrt[3]{\\frac{a}{b+7}} + \\sqrt[3]{\\frac{b}{c+7}} + \\sqrt[3]{\\frac{c}{d+7}} + \\sqrt[3]{\\frac{d}{a+7}}\n$$\n\n假設 $x, y, z, t$ 是 $a, b, c, d$ 的一個排列，且 $x \\leq y \\leq z \\leq t$。根據重排不等式，\n\n$$\nS \\leq \\left( \\sqrt[3]{\\frac{x}{t+7}} + \\sqrt[3]{\\frac{t}{x+7}} \\right) + \\left( \\sqrt[3]{\\frac{y}{z+7}} + \\sqrt[3]{\\frac{z}{y+7}} \\right).\n$$\n\n*聲明*：上式中的第一括號不超過\n\n$$\n\\sqrt[3]{\\frac{x+t+14}{7}}\n$$\n\n*證明*：因為\n\n$$\nX^3 + Y^3 + 3XYZ - Z^3 = \\frac{1}{2}(X+Y-Z)\\left((X-Y)^2 + (X+Z)^2 + (Y+Z)^2\\right),\n$$\n\n不等式 $X + Y \\leq Z$ 等價於（當 $X, Y, Z \\geq 0$ 時）\n\n$$\nX^3 + Y^3 + 3XYZ \\leq Z^3.\n$$\n\n因此，聲明等價於\n\n$$\n\\frac{x}{t+7} + \\frac{t}{x+7} + 3\\sqrt[3]{\\frac{xt(x+t+14)}{7(x+7)(t+7)}} \\leq \\frac{x+t+14}{7}.\n$$\n\n注意到\n\n$$\n\\begin{aligned}\n3\\sqrt[3]{\\frac{xt(x+t+14)}{7(x+7)(t+7)}} &= 3\\sqrt[3]{\\frac{t(x+7)}{7(t+7)} \\cdot \\frac{x(t+7)}{7(x+7)} \\cdot \\frac{7(x+t+14)}{(t+7)(x+7)}} \\\\\n&\\leq \\frac{t(x+7)}{7(t+7)} + \\frac{x(t+7)}{7(x+7)} + \\frac{7(x+t+14)}{(t+7)(x+7)}\n\\end{aligned}\n$$\n\n由於 AM-GM 不等式，因此只需證明\n\n$$\n\\frac{x}{t+7} + \\frac{t}{x+7} + \\frac{t(x+7)}{7(t+7)} + \\frac{x(t+7)}{7(x+7)} + \\frac{7(x+t+14)}{(t+7)(x+7)} \\leq \\frac{x+t+14}{7}.\n$$\n\n直接檢查可知最後這個不等式實際上是等號成立。\n\n因此，\n\n$$\nS \\leq \\sqrt[3]{\\frac{x+t+14}{7}} + \\sqrt[3]{\\frac{y+z+14}{7}} \\leq 2\\sqrt[3]{\\frac{x+y+z+t+28}{14}} = \\frac{8}{\\sqrt[3]{7}},\n$$\n\n最後一個不等式可由 AM-GM 不等式（或因為 $\\sqrt[3]{\\cdot}$ 在 $[0, \\infty)$ 上是凹函數）得出。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15672,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples $ (x, y, z) $ of positive integers such that $ x \\le y \\le z $ and\n$$\nx^3(y^3 + z^3) = 2012(xyz + 2).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The solution is $ (x, y, z) = (2, 251, 252) $.\n\nBy the given, $x$ divides $2 \\cdot 2012 = 2^3 \\cdot 503$. Moreover, $x$ cannot be a multiple of 503, by considering the exponent of 503 in both terms of the equation. Similarly, by considering the powers of 2 on both sides of the equation, we conclude that $x$ cannot be a multiple of 4, thus $x = 1$ or $x = 2$. Thus, we have either\n$$\ny^3 + z^3 = 2012(yz + 2) \\quad \\text{or} \\quad y^3 + z^3 = 503(yz + 1).\n$$\nIn either case, 503 divides $y^3 + z^3$, so $y^3 \\equiv -z^3 \\pmod{503}$. Consequently, by Fermat's little theorem, we have $y^2 \\equiv y^{504} \\equiv y^{3 \\cdot 168} \\equiv z^{3 \\cdot 168} \\equiv z^2 \\pmod{503}$. From $y^3 \\equiv -z^3 \\pmod{503}$ and $z^2 \\equiv y^2 \\pmod{503}$, we conclude that either both $y$ and $z$ are divisible by 503 or $y \\equiv -z \\pmod{503}$. Hence in all cases $y+z$ is a multiple of 503. Write $y+z=503k$ with $k$ a positive integer.\n\nNext, we cannot have $y = z$, as it would then follow that either $y^2 + 1$ or $y^2 + 2$ divides $2y^3$, which forces $y \\le 1$. For $(y-z)^2 = 1$, say $y = z + 1$, the first equation has no solution because $y+z$ must be even, while the second one has only the solution $(y, z) = (251, 252)$, as in this case the equation simply becomes $x + y = 503$. We now prove that we cannot have $(y-z)^2 \\ge 4$. Indeed, if this happened, write $y+z = 503k$ and note that $y^2 - yz + z^2 > yz + 2$. Thus for the first equation we would get $k < 4$, while for the second one $k < 1$. Consider the first equation with $k < 4$. As $y+z$ is even, $k$ is even and so $k = 3$. But then 3 divides both $y+z$ and $yz+2$, so 3 divides $y^2+1$, which is impossible.\n\nThus, the only solution is $(2, 251, 252)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15673,
    "subject": "Mathematics (Olympiad)",
    "question": "If $p$, $q$, $r$ are real numbers such that $p + q + r = 0$, then\n$$\n54p^2q^2r^2 \\leq (p^2 + q^2 + r^2)^3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "This is clear if $pqr = 0$. Suppose $pqr \\neq 0$, then at least one of $pq$, $pr$, $qr$ is negative. Say $pq < 0$, so that $p^2 + q^2 + r^2 = (p + q)^2 - 2pq + r^2 = 2(r^2 + |pq|)$. The AM-GM inequality gives\n$$\n\\frac{|pq|}{2} \\cdot \\frac{|pq|}{2} \\cdot r^2 \\leq \\left(\\frac{r^2 + |pq|}{3}\\right)^3\n$$\nand so\n$$\n54p^2q^2r^2 = 2^3 \\cdot 3^3 \\cdot \\frac{|pq|}{2} \\cdot \\frac{|pq|}{2} \\cdot r^2 \\leq 8(r^2 + |pq|)^3 = (p^2 + q^2 + r^2)^3.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15674,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the digits $A$, $B$, and $C$ such that the following multiplication procedure is correct:\n\n$$\n\\begin{array}{c}\n\\overline{ABC} \\cdot \\overline{BAC} \\\\\n\\multicolumn{2}{c}{---} \\\\\n--A \\\\\n\\multicolumn{2}{c}{---B} \\\\\n\\end{array}\n$$\n\nwhere $\\overline{ABC} \\cdot C = \\overline{---A}$, $\\overline{ABC} \\cdot A = \\overline{---A}$, and $\\overline{ABC} \\cdot B = \\overline{---B}$. Distinct letters denote distinct digits, and each \"-\" represents one digit.",
    "options": [],
    "answer": "See solution",
    "solution": "The numbers $ABC$ and $BAC$ are three-digit numbers, so $A \\neq 0$ and $B \\neq 0$.\n\nIf $C = 1$, then $\\overline{ABC} \\cdot C = \\overline{---A}$ is not possible, since $\\overline{ABC} \\cdot 1 = \\overline{ABC}$, which cannot end with $A$ unless $A = C$, contradicting distinctness.\n\nIf $A > 3$, then $\\overline{ABC} \\cdot A > 400 \\cdot 4 = 1600 > 999$, which is not a three-digit number. So $A \\leq 3$.\n\nIf $A = 1$, then $C = 1$, which is not allowed. If $A = 3$, then $\\overline{ABC} \\cdot A$ ends with $C = 1$, again not allowed. Thus, $A = 2$.\n\nNow, $\\overline{2BC} \\cdot 2$ must end with $2$, so $C = 6$.\n\nNow, $\\overline{2B6} \\cdot B$ must end with $B$. Checking $B = 1, 2, 4, 6, 8$, only $B = 8$ works.\n\nThus, $A = 2$, $B = 8$, $C = 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15675,
    "subject": "Mathematics (Olympiad)",
    "question": "The number $n$ is divisible by $99$. The digits $a$, $b$, and $c$ are all different, and none of them are equal to $1$, $2$, $3$, or $6$. Find all possible values of $n$ if $n$ can be written in the form $n = \\overline{23ab1} \\cdot 100 + 60 + c$, where $\\overline{23ab1}$ denotes the five-digit number formed by the digits $2$, $3$, $a$, $b$, and $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\begin{aligned}\nn &= \\overline{23ab1} \\cdot 100 + 60 + c \\\\\n  &= \\overline{23ab1} \\cdot 99 + \\overline{23ab1} + 60 + c \\\\\n  &= \\overline{23ab1} \\cdot 99 + \\overline{23a} \\cdot 100 + 10b + 1 + 60 + c \\\\\n  &= (\\overline{23ab1} \\cdot 99 + \\overline{23a}) \\cdot 99 + \\overline{23a} + 10b + c + 61 \\\\\n  &= (\\overline{23ab1} \\cdot 99 + \\overline{23a}) \\cdot 99 + 230 + a + 10b + c + 61 \\\\\n  &= (\\overline{23ab1} \\cdot 99 + \\overline{23a} + 2) \\cdot 99 + a + 10b + c + 93\n\\end{aligned}\n$$\n\nSo $99$ divides $a + 10b + c + 93$. Since $a$, $b$, and $c$ are different digits, none equal to $1$, $2$, $3$, or $6$, we have $a + 10b + c \\geq 4 + 0 + 5 = 9$ and $a + 10b + c \\leq 7 + 90 + 8 = 105$. Thus, $102 \\leq a + 10b + c + 93 \\leq 198$. On the other hand, $a + 10b + c + 93$ is divisible by $99$, so $a + 10b + c + 93 = 198$, which implies $b = 9$ and $\\{a, c\\} = \\{7, 8\\}$. There are two solutions:\n\n$$n = 2379168 \\quad \\text{and} \\quad n = 2389167.$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15676,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a square $ABCD$ and the points $E$ on the side $CD$, $M$ on the diagonal $AC$, and $P$ on the side $BC$, such that $\\angle BAE = \\angle AEM = \\angle AMP$. Prove that:\n\na) The triangle $AMP$ is isosceles.\n\nb) $EM = DE + PB$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) We denote $\\angle BAE = \\angle AEM = \\angle AMP = \\alpha$.\n\nWe have, in turn, $\\angle DAE = 90^\\circ - \\alpha$, $\\angle EAC = \\alpha - 45^\\circ$, $\\angle DEA = \\angle EAB = \\alpha$, $\\angle CEM = 180^\\circ - 2\\alpha$.\n\nFrom triangle $CPM$ it follows that $\\angle PMC = 180^\\circ - \\angle AMP = 180^\\circ - \\alpha$, and then $\\angle CPM = \\alpha - 45^\\circ$. Thus $\\angle EAC = \\angle CPM = \\alpha - 45^\\circ$.\n\nDenote by $H$ the intersection of the lines $AE$ and $PM$. The quadrilateral $HAPC$ has $\\angle HAC = \\angle HPC = \\alpha - 45^\\circ$, so it is inscribed, hence $\\angle AHP = \\angle ACP = 45^\\circ$. Moreover, $\\angle AHP = \\angle EHM = \\angle ECM = 45^\\circ$, so the quadrilateral $HEMC$ is inscribed and $\\angle HCM = \\angle AEM = \\alpha$.\n\nReturning to the inscribed quadrilateral $HAPC$, we have $\\angle HPA = \\angle HCA = \\alpha$, so the triangle $AMP$ is isosceles.\n\n![](images/RMC_2024_p36_data_6d523386fd.png)\n\n![](images/RMC_2024_p36_data_07612c6e3e.png)\n\nb) In the isosceles triangle $MAP$ we have $\\angle MAP = 180^\\circ - 2\\alpha$, thus $\\angle PAB = \\angle CAB - \\angle MAP = 45^\\circ - (180^\\circ - 2\\alpha) = 2\\alpha - 135^\\circ$.\n\nWe extend side $CE$ with segment $DQ = BP$. From the congruence of the triangles $ADQ$ and $ABP$ (SAS) it follows that $\\angle QAD = \\angle PAB = 2\\alpha - 135^\\circ$. Thus $\\angle EAQ = \\angle EAD + \\angle DAQ = \\alpha - 45^\\circ$, which means $\\angle EAQ = \\angle EAM$. Now the congruence of the triangles $EAQ$ and $EAM$ (ASA) yields $EM = EQ = ED + DQ = ED + BP$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15677,
    "subject": "Mathematics (Olympiad)",
    "question": "Petrik uses the computer program \"Three\", which converts the numbers written on the display. For one application of this program, Petrik chooses 5 numbers from the written ones, and the program increases each of these 5 numbers by multiplying them by 3. At the beginning, the following 20 numbers are written on the display: $1, 3^1, 3^2, \\ldots, 3^{19}$. What is the smallest number of times Petrik has to use the program to be able to get a set of equal numbers on the display?",
    "options": [],
    "answer": "See solution",
    "solution": "In one operation, the product of all written numbers increases by $3^5$ times. At the beginning, this product equals $3^{0+1+2+\\ldots+19} = 3^{190}$. So, after using the program $n$ times, the product will be $3^{190+5n}$. By that time, all numbers would have to become equal, and therefore at least $3^{19}$, so in the end the product is at least $3^{20p}$, where $p \\geq 19$. Then $3^{20p} = 3^{190+5n}$, so $5n + 190 = 20p$, and $n + 38 = 4p$. As $p \\geq 19$, we get $n \\geq 38$.\n\nNow let's show how to achieve this by using the program 38 times.\n\n$$\n\\begin{aligned}\n3^0,\\ 3^1,\\ 3^2,\\ 3^3,\\ 3^4 \\quad (15) &\\rightarrow 3^{15},\\ 3^{16},\\ 3^{17},\\ 3^{18},\\ 3^{19}; \\\\\n3^5,\\ 3^6,\\ 3^7,\\ 3^8,\\ 3^9 \\quad (10) &\\rightarrow 3^{15},\\ 3^{16},\\ 3^{17},\\ 3^{18},\\ 3^{19}; \\\\\n3^{10},\\ 3^{11},\\ 3^{12},\\ 3^{13},\\ 3^{14} \\quad (5) &\\rightarrow 3^{15},\\ 3^{16},\\ 3^{17},\\ 3^{18},\\ 3^{19}.\n\\end{aligned}\n$$\n\nSo after 30 uses, we have 4 of each of $3^{15}, 3^{16}, 3^{17}, 3^{18}, 3^{19}$.\n\n$$\n\\begin{aligned}\n3^{15},\\ 3^{16},\\ 3^{17},\\ 3^{18} \\quad (3) &\\rightarrow 3^{18},\\ 3^{19},\\ 3^{19},\\ 3^{19}; \\\\\n3^{16},\\ 3^{17},\\ 3^{18},\\ 3^{19} \\quad (2) &\\rightarrow 3^{18},\\ 3^{19},\\ 3^{19},\\ 3^{19}; \\\\\n3^{17},\\ 3^{18},\\ 3^{19},\\ 3^{20} \\quad (1) &\\rightarrow 3^{18},\\ 3^{19},\\ 3^{20},\\ 3^{20}.\n\\end{aligned}\n$$\n\nSo after 36 uses, we have 10 of $3^{18}, 3^{19}$. In the last two moves, we make them all equal to $3^{19}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15678,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and $N$ a point different from $A$, $B$, and $C$. Let $A_b$ be the reflection of $A$ through $NB$, and $B_a$ be the reflection of $B$ through $NA$. Define $B_c$, $C_b$, $A_c$, and $C_a$ similarly. Let $m_a$ be the line passing through $N$ and perpendicular to $B_cC_b$. Define $m_b$, $m_c$ similarly.\n\na) Assume that $N$ is the orthocenter of triangle $ABC$. Show that the respective reflections of $m_a$, $m_b$, and $m_c$ through each bisector of angles $\\angle BNC$, $\\angle CNA$, $\\angle ANB$ are the same line.\n\nb) Assume that $N$ is the nine-point center of triangle $ABC$. Show that the respective reflections of $m_a$, $m_b$, and $m_c$ through $BC$, $CA$, and $AB$ concur.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we present two lemmas. Let $ABC$ be a triangle inscribed in $(O)$ and $N$ be the nine-point center of $\\triangle ABC$.\n\n**Lemma 1.** Let $K$ be the center of $(BOC)$, then $AN$, $AK$ are isogonal with respect to $\\angle BAC$.\n\n*Proof.* Let $X$ be the reflection of $O$ through $BC$. Let $H$ be the orthocenter of $\\triangle ABC$. We have $AH = OX$, and $AH$ is parallel to $OX$. Hence, $AX$ cuts $OH$ at the midpoint of $OH$, which is $N$, and thus $N$ is the midpoint of $AX$. We have $\\angle OXB = \\angle KOB = \\angle KBO$, so $\\triangle OKB \\sim \\triangle OBX$, implying that $\\frac{OK}{OB} = \\frac{OB}{OX}$, or\n\n$$\nOX \\cdot OK = OB^2 = OA^2.\n$$\n\nHence, $\\frac{OK}{OA} = \\frac{OA}{OX}$, implying that $\\triangle OKA \\sim \\triangle OAX$. Therefore, $\\angle OKA = \\angle XAO = \\angle DAK$.\n\nSince $AH$, $AO$ are isogonal with respect to $\\angle BAC$, we have $\\angle BAK = \\angle NAC$, implying that $AN$, $AK$ are isogonal with respect to $\\angle BAC$.\n\n$\\Box$\n\n**Lemma 2.** Let $B'$, $C'$ be the reflections of $B$, $C$ through $AC$, $AB$, respectively, then $AK$ is perpendicular to $B'C'$.\n\n*Proof.* Let $Y$, $Z$ be the reflections of $O$ through $CA$, $AB$. Similar to the argument above, $N$ is the midpoint of $BY$ and $CZ$. Let $N_b$, $N_c$ be the reflections of $N$ through $CA$, $AB$. Since $OC'$ is the reflection of $ZC$ through $AB$, and $N$ is the midpoint of $ZC$, then $N_c$ is the midpoint of $OC'$. Similarly, $N_b$ is the midpoint of $OB'$. Hence, $N_bN_c$ is parallel to $B'C'$. We have\n\n$$\n\\angle KAN_b = \\angle N_bAC + \\angle KAC = \\angle NAC + \\angle NAB = \\angle BAC,\n$$\n\nand similarly, $\\angle KAN_c = \\angle BAC$. On the other hand, $AN_b = AN = AN_c$, and since $\\angle KAN_b = \\angle KAN_c$, we get $AK$ is perpendicular to $N_bN_c$. So $AK \\perp B'C'$, as desired. $\\Box$\n\nBack to the main problem, let $O_a$, $O_b$, $O_c$ be the circumcenters of $\\triangle NBC$, $\\triangle NCA$, $\\triangle NAB$ and $K_a$, $K_b$, $K_c$ be the circumcenters of $\\triangle O_aBC$, $\\triangle O_bCA$, $\\triangle O_cAB$. By Lemma 2,\n\n$$\nm_a \\equiv NK_a, \\quad m_b \\equiv NK_b, \\quad m_c \\equiv NK_c\n$$\n\n![](images/Vietnamese_mathematical_competitions_p276_data_fef48ffadc.png)\n\na) If $N$ is the orthocenter $H$, let $m'_a$, $m'_b$, and $m'_c$ be the reflections of $m_a$, $m_b$, and $m_c$ through the bisectors of angles $BHC$, $CHA$, and $AHB$. Let $J$ be the nine-point center of $ABC$. Note that $(J)$ is also the nine-point circle of $BHC$. By Lemma 1, $HK_a$, $KJ$ are isogonal with respect to $\\angle BHC$, so $m'_a$ goes through $J$. Similarly, $m'_b$, $m'_c$ also go through $J$, and thus $m'_a = m'_b = m'_c = HJ$.\n\n![](images/Vietnamese_mathematical_competitions_p276_data_8ae43cc99d.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15679,
    "subject": "Mathematics (Olympiad)",
    "question": "In the parliament of Neverland, all legislative work is carried out in committees of three people. The constitution dictates that any four people can be in at most two committees. We call a collection of committees a *clique* if any two of them have exactly two people in common, and any manner of including another committee in the collection would break this condition. Prove that two different cliques cannot have two committees in common.",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to see that, given three different committees, each pair of them can have two people in common only if all three committees share the same two people, for otherwise the people in the committees would contain four people from whom three committees have been formed. As a corollary, for any clique, there are some two people who belong to all the committees in the clique, and these people are unique.\n\nTo derive a contradiction, let us consider two cliques $C_1$ and $C_2$ with two committees in common. There are some two people $A$ and $B$ who belong to all the committees in $C_1$. These two people must also belong to the two committees shared by $C_1$ and $C_2$. But then all the committees in $C_2$ must also include $A$ and $B$. Now, we can extend the clique $C_1$ into $C_1 \\cup C_2$, which violates the definition of a clique.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15680,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ with the following property: there exist nonempty finite sets of integers $A, B$, such that for every integer $m$, exactly one of the following three statements is true:\n\n1. There exists $a \\in A$ such that $m \\equiv a \\pmod{n}$;\n2. There exists $b \\in B$ such that $m \\equiv b \\pmod{n}$;\n3. There exist $a \\in A$ and $b \\in B$ such that $m \\equiv a + b \\pmod{n}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A+B = \\{a+b \\mid a \\in A, b \\in B\\}$. The problem asks for $A$, $B$ such that the remainders of $A$ modulo $n$, $B$ modulo $n$, and $(A+B)$ modulo $n$ partition all residues modulo $n$.\n\n**Case 1:** If $n > 1$ is odd, let $n = 2k+1$ for $k \\in \\mathbb{Z}_{>0}$. Take $A = \\{k\\}$, $B = \\{k+1, k+2, \\dots, 2k\\}$. Then $A+B = \\{2k+1, 2k+2, \\dots, 3k\\}$, and $A$, $B$, $A+B$ are pairwise disjoint, their union covers all $2k+1$ residues modulo $n$.\n\n**Case 2:** If $n > 1$ satisfies the condition, then for any $d > 1$, $dn$ also satisfies the condition. Let $A', B'$ be:\n\n$$\nA' = \\{a + x n \\mid a \\in A,\\ x = 0, 1, \\dots, d-1\\}, \\quad B' = \\{b + x n \\mid b \\in B,\\ x = 0, 1, \\dots, d-1\\}\n$$\n\nThe residues of $A', B', A'+B'$ modulo $n$ match those of $A, B, A+B$, and modulo $dn$ they partition all residues. Thus, $A', B'$ work for $dn$.\n\nBy Cases 1 and 2, all $n > 1$ except powers of $2$ satisfy the condition.\n\n**Case 3:** For $n = 8$, $A = \\{1, 2\\}$, $B = \\{3, 6\\}$, $A+B = \\{4, 5, 7, 8\\}$ works. By Case 2, all $2^k$ ($k \\ge 3$) work.\n\nSince $A, B, A+B$ are nonempty, $n \\ge 3$. For $n = 1, 2$, not possible.\n\nFor $n = 4$, if $A, B$ have only one residue each, $A+B$ also has one, which fails. If $A$ or $B$ has two residues, $A+B$ has at least two, which also fails. So $n = 4$ does not work.\n\n**Conclusion:** The desired $n$ are all positive integers except $1$, $2$, and $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15681,
    "subject": "Mathematics (Olympiad)",
    "question": "Alice and Ben play a game on a board with 72 cells arranged in a circle. First, Ben chooses some cells and places one chip on each of them. Each round, Alice first chooses one empty cell, and then Ben moves a chip from one of the adjacent cells onto the chosen one. If Ben cannot do so, the game ends; otherwise, another round follows. Determine the smallest number of chips for which Ben can guarantee that the game will last for at least 2023 rounds.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We show that the smallest possible number of chips is $36$.\n\nFirst, Ben can ensure the game never ends by placing $36$ chips on the even-numbered cells, leaving the odd cells empty. He divides the $72$ cells into $36$ pairs of adjacent cells. In each round, when Alice chooses an empty cell, Ben moves the chip from the other cell in the pair, ensuring each pair always contains exactly one chip. Thus, the game can continue indefinitely.\n\nIf Ben uses fewer than $36$ chips, Alice can ensure the game ends within $36$ rounds. She imagines the cells colored alternately white and black. In each round, she chooses an empty white cell. Since there are $36$ white cells and fewer chips, there is always an empty white cell. Ben must move a chip from a black cell to a white cell, and each chip can be moved at most once. Therefore, the game ends in at most $36$ rounds.\n\nThus, the smallest number of chips Ben needs to guarantee at least $2023$ rounds is $36$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15682,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x^3 + f(y)) = x^2 f(x) + y\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(0) = \\lambda$. Set $x = y = 0$ in (1):\n$$\nf(\\lambda) = 0.\n$$\nNow set $y = \\lambda$:\n$$\nf(x^3) = x^2 f(x) + \\lambda.\n$$\nSet $x = \\lambda, y = 0$:\n$$\nf(\\lambda^3 + \\lambda) = \\lambda^2 f(\\lambda).\n$$\nSet $x = \\lambda, y = \\lambda$:\n$$\nf(\\lambda^3) = \\lambda.\n$$\nSet $x = \\lambda, y = \\lambda^3$:\n$$\nf(\\lambda^3 + \\lambda) = \\lambda^2 f(\\lambda) + \\lambda^3.\n$$\nComparing the last two, $\\lambda = 0$. Thus,\n$$\nf(x^3) = x^2 f(x), \\quad f(f(x)) = x.\n$$\nNow, replace $y$ with $f(y)$:\n$$\nf(x^3 + y) = x^2 f(x) + f(y) = f(x^3) + f(y).\n$$\nSince $x \\mapsto x^3$ is a bijection on $\\mathbb{R}$,\n$$\nf(x + y) = f(x) + f(y), \\quad f(f(x)) = x.\n$$\nConsider $f(x^3) = x^2 f(x)$. Replace $x$ by $x+1$:\n$$\nf(x^3 + 3x^2 + 3x + 1) = (x^2 + 2x + 1)f(x + 1).\n$$\nUsing additivity:\n$$\n\\begin{aligned}\nf(x^3) + 3f(x^2) + 3f(x) + f(1) &= (x^2 + 2x + 1)(f(x) + f(1)) \\\\\n&= x^2 f(x) + 2x f(x) + f(x) + x^2 f(1) + 2x f(1) + f(1).\n\\end{aligned}\n$$\nThis reduces to\n$$\n3f(x^2) + 2f(x) = c x^2 + 2c x + 2x f(x)\n$$\nfor all $x \\in \\mathbb{R}$. Replace $x$ by $x+1$ and expand:\n$$\n3f(x^2) + 6f(x) = c x^2 + 6c x + 2x f(x).\n$$\nComparing, we see $f(x) = c x$. Replacing $x$ by $f(x)$ gives $x = c f(x)$. Thus $c^2 x = x$, so $c^2 = 1$. Therefore, $f(x) = x$ or $f(x) = -x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15683,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R}^+ \\to \\mathbb{R}^+$ such that\n$$\nf\\left(\\frac{f(x)}{x} + y\\right) = 1 + f(y)\n$$\nfor all positive real numbers $x, y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Firstly, we will prove that $\\frac{f(x)}{x}$ is a constant.\n\nAssume that there exist $a, b \\in (0, +\\infty)$ such that $\\frac{f(a)}{a} \\neq \\frac{f(b)}{b}$.\n\nWithout loss of generality, assume $\\frac{f(a)}{a} < \\frac{f(b)}{b}$. By plugging $x = a$ and $x = b$ into the relation, we get\n$$\nf\\left(y + \\frac{f(a)}{a}\\right) = f\\left(y + \\frac{f(b)}{b}\\right) = f(y) + 1\n$$\nfor all positive real numbers $y$. Denote $K = \\frac{f(b)}{b} - \\frac{f(a)}{a}$, we obtain that $f(x) = f(x + K)$ for all $x > \\frac{f(a)}{a}$. Furthermore, we have\n$$\nf\\left(y + n \\cdot \\frac{f(a)}{a}\\right) = f\\left(y + (n-1) \\cdot \\frac{f(a)}{a}\\right) + 1 = \\dots = f(y) + n > n.\n$$\nThus $f(x) > n$ for all $x > n \\cdot \\frac{f(a)}{a}$. Now, for an arbitrary positive number $x$, choose $n = \\lfloor f(x) \\rfloor + 2 > f(x)$ and a positive integer $m$ such that $x + mK > n \\cdot \\frac{f(a)}{a}$, we have $f(x + mK) > n > f(x) = f(x + mK)$, which is a contradiction.\n\nHence, $\\frac{f(x)}{x}$ is a constant. Denote $\\frac{f(x)}{x} = c$ and replace $f(x) = cx$ in the original equation. One can find that $c = 1$. Thus, $f(x) = x$ for all positive numbers $x$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15684,
    "subject": "Mathematics (Olympiad)",
    "question": "Find how many solutions the equation\n\n$$\n[a^2, b^2] + [b^2, c^2] + [c^2, a^2] = (a^2, b^2)(b^2, c^2)(c^2, a^2)\n$$\n\nhas in natural numbers, if $[m, n]$ and $(m, n)$ stand for the LCM and GCD of the natural numbers $m$ and $n$, respectively.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = px$, $b = py$, $c = pz$ where $p, x, y, z$ are pairwise coprime numbers. Then the equation becomes:\n\n$$(pxy)^2 + (pxz)^2 + (pyz)^2 = (p^2)^3 \\Leftrightarrow x^2y^2 + x^2z^2 + y^2z^2 = p^4$$\n\n$$\\Leftrightarrow x^2(y^2 + z^2) = (p^2 - yz)(p^2 + yz).$$\n\nIf $p^2 = y^2 + yz + z^2$, then $x^2(y^2 + z^2) = (y^2 + 2yz + z^2)(y^2 + z^2) \\Rightarrow x = y + z$. Under such conditions, the solution is the set $(p(y+z), py, pz)$.\n\nWe need to show that there are infinitely many sets of natural numbers $(p, y, z)$ for which $p^2 = y^2 + yz + z^2$. Let $u = \\frac{y}{p}$, $v = \\frac{z}{p}$; we need to show that the equation $u^2 + uv + v^2 = 1$ has infinitely many rational solutions. One point is $(1, 0)$. For any rational $k$, the line $v = k(u-1)$ intersects the curve $u^2 + uv + v^2 = 1$ at another rational point. By Vieta's theorem, this point is rational.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15685,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the polynomial $P(x, y, z) = (y - x)^2 + (z - y)^2$. Find all real numbers $r$ such that for all integers $n$ and $m$, $P(n, n + m, n + 2m) = 2m^2$ is divisible by $m^r$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P(x, y, z) = (y - x)^2 + (z - y)^2$. For $n, m \\in \\mathbb{Z}$, $P(n, n + m, n + 2m) = (m)^2 + (m)^2 = 2m^2$. Thus, $m^2$ always divides $2m^2$, but $m^r$ for $r > 2$ does not always divide $2m^2$ unless $m = 0$ or $\\pm1$. Therefore, the maximal $r$ is $2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15686,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the parabola be defined by the equation $y = x^2$, and the circle by $(x-a)^2 + (y-b)^2 = R^2$. Let $A(a, b)$ be the center of the circle, and $X_i(x_i, x_i^2)$, $i = 1, 2, 3, 4$, be the common points of the circle and the parabola. Denote by $M(k, l)$ and $N(m, n)$ the coordinates of the midpoints of the arcs $X_1X_2$ and $X_3X_4$, respectively. Prove that the line $MN$ is perpendicular to the $y$-axis.\n\n![](images/Belarus_2015_p26_data_d3a7901def.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We may assume that the parabola is defined by $y = x^2$, and the circle by $(x-a)^2 + (y-b)^2 = R^2$. Let $A(a, b)$ be the center of the circle, and $X_i(x_i, x_i^2)$, $i = 1, 2, 3, 4$, be the intersection points. Then $x_1, x_2, x_3, x_4$ are the roots of\n\n$$\n(x - a)^2 + (x^2 - b)^2 = R^2 \\iff x^4 - (2b - 1)x^2 - 2ax + (a^2 + b^2 - R^2) = 0.\n$$\n\nBy Vieta's formula, $x_1 + x_2 + x_3 + x_4 = 0$. Let $M(k, l)$ and $N(m, n)$ be the midpoints of arcs $X_1X_2$ and $X_3X_4$. The condition $\\overrightarrow{X_1X_2} \\perp \\overrightarrow{AM}$ gives\n\n$$\n(x_1 - x_2)(k - a) + (x_1^2 - x_2^2)(l - b) = 0 \\implies k - a = -(x_1 + x_2)(l - b).\n$$\n\nSince $M$ lies on the circle, $(k-a)^2 + (l-b)^2 = R^2$, so\n$$\n((x_1 + x_2)^2 + 1)(l-b)^2 = R^2 \\implies (l-b)^2 = \\frac{R^2}{(x_1 + x_2)^2 + 1}.\n$$\nSimilarly,\n$$\n(n-b)^2 = \\frac{R^2}{(x_3 + x_4)^2 + 1}.\n$$\nBut $x_1 + x_2 = -(x_3 + x_4)$, so\n$$\n(l - b)^2 = (n - b)^2. \\tag{1}\n$$\n\nSince the slope of $MA$ is positive (as $X_1X_2$ has negative slope), $b > l$ and $b > n$. Thus, from (1), $b - l = b - n$, so $l = n$. Therefore, $MN \\perp Oy$ as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15687,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a pattern of numbers in the shape of a triangle: the first row consists of numbers from 1 to 100 arranged in order. Each number in the second row is the sum of the two numbers directly below it in the first row. Each number in the third row is the sum of the two numbers directly below it in the second row, and so on.\n\nThe number in the last row is ____.",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to see that:\n\n1. There are 100 rows in the pattern.\n\n2. The numbers in the $i$th row constitute an arithmetic sequence with common difference\n\n$$\nd_i = 2^{i-1}, \\quad i = 1, 2, \\dots, 99.\n$$\n\n3. We have\n\n$$\n\\begin{align*}\na_n &= a_{n-1} + (a_{n-1} + 2^{n-2}) \\\\\n&= 2a_{n-1} + 2^{n-2} \\\\\n&= 2[2a_{n-2} + 2^{n-3}] + 2^{n-2} \\\\\n&= 2^2[2a_{n-3} + 2^{n-4}] + 2 \\times 2^{n-2} \\\\\n&\\vdots \\\\\n&= 2^{n-1}a_1 + (n-1) \\times 2^{n-2} \\\\\n&= (n+1)2^{n-2}.\n\\end{align*}\n$$\n\nTherefore, the number in the last row is $a_{100} = 101 \\times 2^{98}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15688,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle such that $AB < AC$. The perpendicular bisector of the side $BC$ meets the side $AC$ at the point $D$, and the (interior) bisector of the angle $ADB$ meets the circumcircle of $ABC$ at the point $E$. Prove that the (interior) bisector of the angle $AEB$ and the line through the incenters of the triangles $ADE$ and $BDE$ are perpendicular.",
    "options": [],
    "answer": "See solution",
    "solution": "The lines $BC$ and $DE$ are parallel, so the angles $BED$ and $DAE$ are equal. Then so are the angles $AED$ and $DBE$. Let $I$ and $J$ be the incenters of the triangles $ADE$ and $BDE$, respectively. It follows that the triangles $DIE$ and $DJB$ are similar, so $\\frac{DI}{DE} = \\frac{DJ}{DB}$. Since the angles $IDJ$ and $EDB$ are equal, the triangles $DIJ$ and $DEB$ are similar, so the angles $DIJ$ and $DEB$ are equal. Let the line $BE$ meet the line $IJ$ at the point $F$. Notice that the quadrilateral $DIEF$ is cyclic, so the angles $EFI$ and $EDI$ are both equal to one half of the angle $ACB$. Consequently, the line $IJ$ is parallel to the exterior bisector of the angle $AEB$. The conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15689,
    "subject": "Mathematics (Olympiad)",
    "question": "Boris takes a taxi to his home but falls asleep when the taxi is halfway to his house. He wakes up when the remaining part of his journey is equal to half the distance covered by the taxi while he was asleep.\n\nIf the fraction of the journey for which he slept is $\\frac{1}{n}$, what is the value of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let the total journey be $d$ units. Boris falls asleep at halfway, so at $\\frac{d}{2}$ from the start. Let $x$ be the distance he sleeps. When he wakes, the remaining distance is $r$, and $r = \\frac{1}{2}x$.\n\nFrom halfway to waking up: $x = \\frac{d}{2} - r$.\n\nBut $r = \\frac{1}{2}x$ so $x = \\frac{d}{2} - \\frac{1}{2}x$.\n\n$\\Rightarrow x + \\frac{1}{2}x = \\frac{d}{2}$\n\n$\\Rightarrow \\frac{3}{2}x = \\frac{d}{2}$\n\n$\\Rightarrow x = \\frac{d}{3}$\n\nSo the fraction of the journey he slept is $\\frac{x}{d} = \\frac{1}{3}$, so $n = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15690,
    "subject": "Mathematics (Olympiad)",
    "question": "A triangular lattice cuts an equilateral triangle with a side of length $n$ into $n^2$ triangular cells. Some of the cells are infected. A cell, which is not infected yet, can become infected if it is neighboring (by side) with at least two already infected cells. Determine the minimal number of initially infected cells such that eventually every cell in the triangle could get infected if $n = 12$.",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that with the contamination of one cell, the perimeter of the infected area decreases by at least 1. Let $k$ cells be infected at the beginning. Then the perimeter is at most $3k$. It takes $n^2 - k$ contaminations to get the whole triangle infected. The perimeter of the infected area is then $3n$. Thus,\n\n$$\n3n \\leq 3k - (n^2 - k)\n$$\nwhich simplifies to\n$$\nk \\geq \\frac{n^2 + 3n}{4}.\n$$\n\nFor $n = 12$, the estimate gives $k \\geq 45$. The layout of 45 initially infected cells which cause the infection of the whole system could be as illustrated below:\n\n![](images/Cesko-Slovacko-Poljsko_2013_p4_data_525c6d0079.png)\n\n![](images/Cesko-Slovacko-Poljsko_2013_p4_data_803523b5c5.png)\n\nAnother option is to cover the triangle with three equilateral triangles of side 4 (covering one side of the big triangle) and the rest could be covered with three rhombs of side 4. Each of the smaller triangles could be infected with 7 cells, each of the rhombs with 8 cells. Possible initial states for infecting an equilateral triangle with side 4 and a rhomb with side 4 are shown below. (Each triangle with side 4 can get infected itself independently of the rest area. To get an entire rhomb infected, we need the area to its left and upper side to be infected first.)\n\n![](images/Cesko-Slovacko-Poljsko_2013_p4_data_f5bb11cd75.png)\n\n![](images/Cesko-Slovacko-Poljsko_2013_p4_data_083ef28d67.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15691,
    "subject": "Mathematics (Olympiad)",
    "question": "In a convex $n$-gon, several diagonals are drawn. A drawn diagonal is *good* if it intersects (by an interior point) with exactly one other drawn diagonal. Find the maximal possible number of good diagonals.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $14$.\n\nThe value $N = 14$ is realized, for example, by the numbers $4$, $15$, $70$, $84$. To prove that $N \\geq 14$, one may use the fact that for every $k$, among the six obtained gcd's, there cannot occur exactly two numbers divisible by $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15692,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a pyramid $VABCD$ with a rectangular base $ABCD$, and all the lateral edges congruent. Prove that the plane $VCD$ forms congruent angles with the planes $VAC$ and $BAC$ if and only if the angles $\\angle VAC$ and $\\angle BAC$ are congruent.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $S$ the projection of point $D$ onto the line $AC$. Notice that line $DS$ is perpendicular to both $VO$ and $AC$, so $DS \\perp (VAC)$. Consequently, the projection of triangle $VDC$ onto the plane $VAC$ is triangle $VSC$.\n\nLet $u$ and $v$ be the measures of the angles formed by the plane $VCD$ with the planes $VAC$ and $BAC$, respectively. We have:\n\n$$\n\\begin{aligned}\nu = v &\\Leftrightarrow \\cos u = \\cos v \\Leftrightarrow \\frac{\\text{area}[VSC]}{\\text{area}[VDC]} = \\frac{\\text{area}[COD]}{\\text{area}[VDC]} \\\\\n&\\Leftrightarrow \\text{area}[VSC] = \\text{area}[COD] \\Leftrightarrow VO \\cdot CS = \\frac{1}{2} AB \\cdot BC.\n\\end{aligned}\n$$\n\nSince $DC^2 = CS \\cdot CA$ and $DC = AB$, we deduce that $CS = \\frac{AB^2}{AC}$. Therefore,\n\n$$\nu = v \\Leftrightarrow VO \\cdot AB = \\frac{1}{2} AC \\cdot BC \\Leftrightarrow \\frac{VO}{OA} = \\frac{BC}{AB} \\Leftrightarrow \\angle VAC = \\angle BAC,$$\nas claimed.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15693,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a \\ge 0$ and let $(x_n)_{n \\ge 1}$ be a sequence of real numbers. Given that the sequence $$\\left( \\frac{x_1 + \\dots + x_n}{n^a} \\right)_{n \\ge 1}$$ is bounded, prove that the sequence $(y_n)_{n \\ge 1}$, defined by\n\n$$\ny_n = \\frac{x_1}{1^b} + \\frac{x_2}{2^b} + \\dots + \\frac{x_n}{n^b},\n$$\n\nis convergent for all $b > a$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S_n = \\sum_{k=1}^{n} x_k$, $n \\in \\mathbb{N}^*$. By hypothesis, there exists a constant $c > 0$ such that $|S_n| \\le c n^a$ for all $n \\in \\mathbb{N}^*$. Let $n, p \\in \\mathbb{N}^*$. We have:\n\n$$\n\\begin{align*}\n|y_{n+p} - y_n| &= \\left| \\sum_{k=n+1}^{n+p} \\frac{x_k}{k^b} \\right| = \\left| \\sum_{k=n+1}^{n+p} \\frac{S_k - S_{k-1}}{k^b} \\right| \\\\\n&= \\left| \\frac{S_{n+p}}{(n+p+1)^b} - \\frac{S_n}{(n+1)^b} + \\sum_{k=n+1}^{n+p} S_k \\left( \\frac{1}{k^b} - \\frac{1}{(k+1)^b} \\right) \\right| \\\\\n&\\le \\frac{|S_{n+p}|}{(n+p+1)^b} + \\frac{|S_n|}{(n+1)^b} + \\sum_{k=n+1}^{n+p} |S_k| \\left( \\frac{1}{k^b} - \\frac{1}{(k+1)^b} \\right) \\\\\n&\\le c \\left[ \\frac{2}{n^{b-a}} + \\sum_{k=n+1}^{n+p} k^a \\left( \\frac{1}{k^b} - \\frac{1}{(k+1)^b} \\right) \\right].\n\\end{align*}\n$$\n\nApplying the mean value theorem to $f(x) = x^{-\\alpha}$, $x > 0$, on $[i, i+1]$ with $\\alpha, i > 0$, we get\n\n$$\n\\frac{\\alpha}{(i+1)^{\\alpha+1}} < \\frac{1}{i^{\\alpha}} - \\frac{1}{(i+1)^{\\alpha}} < \\frac{\\alpha}{i^{\\alpha+1}}.\n$$\n\nSince $b, b-a > 0$, we have\n\n$$\n\\frac{1}{k^b} - \\frac{1}{(k+1)^b} < \\frac{b}{k^{b+1}} \\quad \\text{and} \\quad \\frac{b-a}{k^{b-a+1}} < \\frac{1}{(k-1)^{b-a}} - \\frac{1}{k^{b-a}}, \\quad \\forall k \\ge 2.\n$$\n\nTherefore,\n\n$$\n\\begin{align*}\n\\sum_{k=n+1}^{n+p} k^a \\left( \\frac{1}{k^b} - \\frac{1}{(k+1)^b} \\right) &< \\sum_{k=n+1}^{n+p} \\frac{b k^a}{k^{b+1}} = \\frac{b}{b-a} \\sum_{k=n+1}^{n+p} \\frac{b-a}{k^{b-a+1}} \\\\\n&< \\frac{b}{b-a} \\sum_{k=n+1}^{n+p} \\left( \\frac{1}{(k-1)^{b-a}} - \\frac{1}{k^{b-a}} \\right) \\\\\n&= \\frac{b}{b-a} \\left( \\frac{1}{n^{b-a}} - \\frac{1}{(n+p)^{b-a}} \\right) < \\frac{b}{(b-a) n^{b-a}}.\n\\end{align*}\n$$\n\nThus,\n\n$$\n|y_{n+p} - y_n| < c \\left( 2 + \\frac{b}{b-a} \\right) \\frac{1}{n^{b-a}}, \\quad \\forall n, p \\in \\mathbb{N}^*.\n$$\n\nSince $\\lim_{n \\to \\infty} n^{-(b-a)} = 0$, $(y_n)_{n \\ge 1}$ is a Cauchy sequence, hence convergent.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15694,
    "subject": "Mathematics (Olympiad)",
    "question": "There are four numbers on the board: $1$, $3$, $6$, and $10$. Each time, we can erase any two numbers $a, b$ written on the board and write the numbers $a+b$ and $ab$ instead. Can we obtain such four numbers after several moves?\n\na) $2015$, $2016$, $2017$, $2018$\n\nb) $2016$, $2017$, $2019$, $2022$",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** a), b) that is not possible.\n\n**Solution.**\n\na) Let us look at the numbers modulo $3$. The number of numbers divisible by $3$ cannot decrease. If both $a$ and $b$ are divisible by $3$, then both $a+b$ and $ab$ are also divisible by $3$. If only one of $a$ or $b$ is divisible by $3$, then $ab$ is divisible by $3$. Initially, we have only one number divisible by $3$, and in the end, only one, so such a situation is impossible.\n\nb) Now consider the case when exactly three numbers are divisible by $3$. Then those numbers are $0, 0, 0, k$, where $k \\in \\{1, 2\\}$ modulo $3$. These four numbers will never change. To increase the number of numbers divisible by $3$, $a$ and $b$ should be $1$ and $2$ modulo $3$. However, then the numbers $0$ and $2$ appear. Thus, in the situation where exactly three numbers are divisible by $3$, they should be $0, 0, 0, 2$, but the four numbers in the condition are $0, 0, 0, 1$. Thus, we get a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15695,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the circumcenter of $\\triangle ABC$ and $H_A$ be the projection of $A$ onto $BC$. The extension of $AO$ intersects the circumcircle of $\\triangle BOC$ at $A'$. The projections of $A'$ onto $AB$ and $AC$ are $D$ and $E$, respectively. Let $O_A$ be the circumcenter of $\\triangle DH_A E$. Define $H_B$, $O_B$, $H_C$ and $O_C$ similarly.\n\nProve that $H_A O_A$, $H_B O_B$ and $H_C O_C$ are concurrent.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p168_data_76e490a9a8.png \"Fig. 1. 1\")",
    "options": [],
    "answer": "See solution",
    "solution": "Let $T$ be the reflection of $A$ over $BC$, $F$ be the projection of $A'$ onto $BC$, and $M$ be the projection of $T$ onto $AC$.\n\nSince $AC = CT$, we have $\\angle TCM = 2\\angle TAM$. Since $\\angle TAM = \\frac{\\pi}{2}$ —\n\n$\\angle ACB = \\angle OAB$, we have\n\n$\\angle TCM = 2\\angle OAB = \\angle A'OB = \\angle A'CF,$\n\nand\n\n$\\angle TCH_A = \\angle A'CF + \\angle A'CT = \\angle TCM + \\angle A'CT = \\angle A'CE.$\n\nBecause $\\angle CH_A T$, $\\angle CMT$, $\\angle CEA'$, $\\angle CFA'$ are right angles, we have\n\n$$\n\\frac{CH_A}{CM} = \\frac{CH_A}{CT} \\cdot \\frac{CT}{CM} = \\frac{\\cos \\angle TCH_A}{\\cos \\angle TCM} = \\frac{\\cos \\angle A'CE}{\\cos \\angle A'CF} = \\frac{CE}{CA'} \\cdot \\frac{CA'}{CF} = \\frac{CE}{CF},\n$$\n\ni.e., $CH_A \\cdot CF = CM \\cdot CE$, so $H_A, F, M$ and $E$ are concyclic on circle $\\omega_1$.\n\nSimilarly, let $N$ be the projection of $T$ onto $AB$, then $H_A$, $F$, $N$ and $D$ are concyclic on circle $\\omega_2$. Since $A'FH_A T$ and $A'EMT$ are both right trapezoids, the perpendicular bisectors of $H_A F$ and $EM$ meet at the midpoint $K$ of $A'T$, i.e., $K$ is the center of $\\omega_1$ and $KF$ is its radius. Similarly, $K$ and $KF$ are also the center and radius of $\\omega_2$. Thus, $\\omega_1$ and $\\omega_2$ coincide, so $D, N, F, H_A, E, M$ are concyclic. Therefore, $O_A$ is the midpoint $K$ of $A'T$, and $O_A H_A \\parallel AA'$.\n\nSince $\\angle H_C AO + \\angle A H_C H_B = \\frac{\\pi}{2} - \\angle ACB + \\angle ACB = \\frac{\\pi}{2}$, we have $AA' \\perp H_B H_C$, thus $O_A H_A \\perp H_B H_C$. Therefore, $O_A H_A$, $O_B H_B$ and $O_C H_C$ all pass through the orthocenter of $\\triangle H_A H_B H_C$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15696,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest possible value of the sum of the absolute values of the roots of an irrational trinomial $x^2 + px + q$ with integer coefficients $p$ and $q$.",
    "options": [],
    "answer": "See solution",
    "solution": "We find all irrational trinomials with discriminant $D = 5$. By Vieta's theorem, $\\alpha_1 \\cdot \\alpha_2 < 0$ if and only if $q = \\alpha_1 \\cdot \\alpha_2 < 0$, i.e., case 2) holds if and only if $q \\leq -1$. If $q = -1$, then $D = 5$ only if $p^2 = 1$, but if $q \\leq -2$, then $D = p^2 - 4q \\geq -4q \\geq -4 \\cdot (-2) = 8$. Therefore, there exist exactly two irrational trinomials with $D = 5$ and $\\alpha_1 \\cdot \\alpha_2 < 0$: $x^2 - x - 1$ and $x^2 + x - 1$.\n\nTherefore, the smallest value of the sum of the absolute values of the roots of the irrational trinomial is $\\sqrt{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15697,
    "subject": "Mathematics (Olympiad)",
    "question": "Agatha, Isa, and Nick each have a different kind of bike. One of them has an electric bike, one has a racing bike, and one has a mountain bike. The bikes have different colours: green, blue, and black. The three owners make two statements each, of which one is true and the other is false:\n\n- Agatha says: \"I have an electric bike. Isa has a blue bike.\"\n- Isa says: \"I have a mountain bike. Nick has an electric bike.\"\n- Nick says: \"I have a blue bike. The racing bike is black.\"\n\nExactly one of the following statements is certainly true. Which one?\n\nA) Agatha has a green bike.\n\nB) Agatha has a mountain bike.\n\nC) Isa has a green bike.\n\nD) Isa has a mountain bike.\n\nE) Nick has an electric bike.",
    "options": [],
    "answer": "See solution",
    "solution": "D) Isa has a mountain bike.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15698,
    "subject": "Mathematics (Olympiad)",
    "question": "Jimmy is on a $9 \\times n$ grid. He can move by $\\pm(7, 2)$ and $\\pm(2, 7)$, where the second coordinate (column) is taken modulo $n$. Starting from $(1, 1)$, what is the maximum number of squares Jimmy can reach?\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $g = \\gcd(45, n)$. The answer is $\\dfrac{9n}{g}$.\n\n**Explanation:**\n\nDefine the moves:\n\n$$\n\\begin{align*}\nV &= (2, 7) \\quad \\text{i.e., } (a, b) \\to (a + 2, b + 7) \\\\\nV^{-1} &= (-2, -7) \\quad \\text{i.e., } (a, b) \\to (a - 2, b - 7) \\\\\nH &= (7, 2) \\quad \\text{i.e., } (a, b) \\to (a + 7, b + 2) \\\\\nH^{-1} &= (-7, -2) \\quad \\text{i.e., } (a, b) \\to (a - 7, b - 2)\n\\end{align*}\n$$\n\nThe sequence $S = H^{-1}V^3H^{-1}V^4$ moves $(1, 1)$ to $(1, 46)$, i.e., adds $(0, 45)$ to the position. Iterating $S$, Jimmy can reach positions $(1, 1 + 45k)$ modulo $n$. Thus, he can reach $n/g$ squares in each row, and with $9$ rows, the total is $9n/g$.\n\nNo more than $n/g$ squares per row are possible, as shown by analyzing the move structure. Thus, the maximum is $9n/g$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 15699,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k = 3 \\cdot 4^l$ for an arbitrary positive integer $l$. Show that\n$$\na_k = \\left\\lfloor \\frac{2^k}{k} \\right\\rfloor$$\nis always odd.",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n$$\na_k = \\left\\lfloor \\frac{2^k}{k} \\right\\rfloor = \\left\\lfloor \\frac{2^{3 \\cdot 4^l}}{3 \\cdot 4^l} \\right\\rfloor = \\left\\lfloor \\frac{2^{3 \\cdot 4^l - 2l}}{3} \\right\\rfloor = \\left\\lfloor \\frac{2^{3 \\cdot 4^l - 2l} - 1}{3} + \\frac{1}{3} \\right\\rfloor. $$\n\nSince the positive integer $3 \\cdot 4^l - 2l$ is even, we have $2^{3 \\cdot 4^l - 2l} \\equiv 1 \\pmod{3}$, so $2^{3 \\cdot 4^l - 2l} - 1$ is divisible by $3$.\n\nTherefore,\n$$\na_k = \\left\\lfloor \\frac{2^k}{k} \\right\\rfloor = \\frac{2^{3 \\cdot 4^l - 2l} - 1}{3}$$\nis odd, which finishes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15700,
    "subject": "Mathematics (Olympiad)",
    "question": "Maja went to the bookstore to buy two books. The price of the first book was 65\\% and the price of the other 57.5\\% of the money that Maja had with her. She needed an additional 45 denars to buy the two books. How much money did Maja have with her?",
    "options": [],
    "answer": "See solution",
    "solution": "Let Maja have $x$ denars with her. The total price of the books is $0.65x + 0.575x = 1.225x$. She needed 45 more denars, so:\n\n$$\n1.225x = x + 45\n$$\n\nSubtract $x$ from both sides:\n\n$$\n1.225x - x = 45 \\\\\n0.225x = 45\n$$\n\nSo:\n\n$$\nx = \\frac{45}{0.225} = 200\n$$\n\nMaja had 200 denars with her.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15701,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ be a positive integer. Prove that if Mari writes at least $m+3$ numbers on the board, then Jüri can choose 4 of those such that the sum of some two of those and the sum of the other two give the same remainder when divided by $m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since Mari writes $m+3$ numbers and there are only $m$ possible remainders modulo $m$, by the pigeonhole principle, at least two numbers, say $a$ and $b$, have the same remainder modulo $m$. Among the remaining $m+1$ numbers, again by the pigeonhole principle, there exist two numbers, say $c$ and $d$, with the same remainder modulo $m$. Then $a + c$ and $b + d$ have the same remainder modulo $m$, so Jüri can choose $a, b, c, d$ as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15702,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a parallelogram and let $O$ be the intersection point of the diagonals. Prove that for any point $M \\in (AB)$, there exist unique points $N \\in (OC)$ and $P \\in (OD)$ such that $O$ is the centroid of triangle $MNP$.",
    "options": [],
    "answer": "See solution",
    "solution": "A point $M \\in (AB)$ is uniquely determined by a real number $k \\in (0, \\infty)$ such that $\\overline{AM} = k$, from which we get\n$$\n\\overrightarrow{OM} = \\frac{1}{k+1}\\overrightarrow{OA} + \\frac{k}{k+1}\\overrightarrow{OB}.\n$$\nTo find the points $N$ and $P$ uniquely, we need to find $x, y \\in (0, \\infty)$ such that $\\overrightarrow{ON} = x$, $\\overrightarrow{OP} = y$, and $O$ is the centroid of triangle $MNP$. From this, we have:\n$$\n\\overrightarrow{ON} = \\frac{x}{x+1}\\overrightarrow{OC} = -\\frac{x}{x+1}\\overrightarrow{OA}, \\quad \\text{and} \\quad \\overrightarrow{OP} = \\frac{y}{y+1}\\overrightarrow{OD} = -\\frac{y}{y+1}\\overrightarrow{OB}.\n$$\nSince $\\overrightarrow{OA}$ and $\\overrightarrow{OB}$ are not collinear, $O$ is the centroid of triangle $MNP$ if and only if $\\frac{x}{x+1} = \\frac{1}{k+1} \\Leftrightarrow x = \\frac{1}{k}$ and $\\frac{y}{y+1} = \\frac{k}{k+1} \\Leftrightarrow y = k$. This means that point $N$ is uniquely determined by the ratio $x = \\frac{1}{k} = \\frac{ON}{NC}$, and point $P$ is uniquely determined by the ratio $y = k = \\frac{OP}{PD}$, which completes the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15703,
    "subject": "Mathematics (Olympiad)",
    "question": "Let's consider all increasing geometric progressions and select only those that have the maximal number $M$ of elements in the set $A = \\{1, 2, 3, \\ldots, 2008\\}$. Find $M$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $11$.\n\nLet one of the progressions $\\{a_n\\}$ with the required maximal number $M$ of common members have first term $a_0$ and ratio $q_0 > 1$. The progression has some common elements with $A$. Let $n$ be the least number in $A$ also in the progression. Then you can define a new progression with first term $b_0 = n$ and the same ratio $q_0$. It has as many members in $A$ as $\\{a_n\\}$, so we can take it as initial.\n\nLet $m$ be the next common member of $A$ and the progression, i.e., in the new progression $b_0 = n$, $b_k = b_0 q_0^k = n q_0^k = m \\implies q_0^k = \\frac{m}{n} = \\frac{u}{v}$, where $\\frac{u}{v}$ is irreducible and $n \\leq m$.\n\nSuppose $q_0^i \\in \\mathbb{Q}$ for $1 \\leq i < k$. By contradiction, let $q_0^i = \\frac{s}{t}$ irreducible. Then $q_0^{k} = \\frac{u}{v}$, so $q_0^{ki} = (\\frac{s}{t})^k = (\\frac{u}{v})^i$. This leads to $u^i t^k = s^k v^i$, so $u^i = s^k$ and $v^i = t^k$, but as $i < k$, $v > t$, and $b_0 = n \\nmid v$, so $b_0 = n \\nmid t$. Therefore $b_i = b_0 q_0^i = n \\frac{s}{t} \\in \\mathbb{N}$, which contradicts the minimality of $k$.\n\nThus, the rational members of the chosen geometric progression may only be of the form $b_0, b_0 q_0^k, b_0 q_0^{2k}, \\ldots$. Therefore, we can choose $q = q_0^k \\in \\mathbb{Q}$ and consider the progression $(b_0, q)$ with first integer term $b_0 = n$ and rational ratio $q = \\frac{p}{r}$ irreducible.\n\nThe progression $(1, 2)$ obviously has the maximum members in $A$ among all progressions with integer ratios. They share $11$ common elements, which are the powers of two: $1, 2, 4, \\ldots, 1024$.\n\nSuppose there is a progression with a rational ratio that has more elements in $A$. Let it be $(n, \\frac{p}{r})$. If $c_k = n (\\frac{p}{r})^k$ is an integer, i.e., $n$ is divisible by $r^k$, then all preceding $c_i$ are integers for $i = 0, \\ldots, k$. If the progression has at least $12$ elements in $A$, then $c_{11} = n (\\frac{p}{r})^{11}$ should be an integer, so $n$ is divisible by $r^{11}$. Since $\\frac{p}{r} \\notin \\mathbb{N}$, $r \\geq 2$, so $r^{11} \\geq 2048$ and $n \\geq 2048$. But $n \\leq 2008$, a contradiction.\n\nTherefore, an increasing geometric progression cannot have more than $11$ members in $A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15704,
    "subject": "Mathematics (Olympiad)",
    "question": "A larger cube has a side length of $\\frac{12}{5}$ cm, and a smaller cube has a side length of $2$ cm. What is the maximum number of smaller red cubes, each with equal side length, that can fit in the space between the larger and smaller cubes if the red cubes do not overlap and do not exceed the difference in side lengths?",
    "options": [],
    "answer": "See solution",
    "solution": "Let the side length of each red cube be $\\frac{2}{5}$ cm. Define $u = \\frac{2}{5}$ cm as a new unit. The larger cube has side length $6u$ and the smaller cube has side length $5u$. The number of red cubes that fit in the space between them is:\n\n$$6^3 - 5^3 = 216 - 125 = 91$$\n\nSo, $91$ red cubes can fit.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15705,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least $60^\\circ$. Prove that\n$$\n\\frac{1}{3} |AB^3 - AD^3| \\leq |BC^3 - CD^3| \\leq 3|AB^3 - AD^3|.\n$$\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "By symmetry, we only need to prove the first inequality.\n\nBecause quadrilateral $ABCD$ has an incircle, we have $AB + CD = BC + AD$, or $AB - AD = BC - CD$. It suffices to prove that\n$$\n\\frac{1}{3}(AB^2 + AB \\cdot AD + AD^2) \\leq BC^2 + BC \\cdot CD + CD^2.\n$$\nBy the given condition, $60^\\circ \\leq \\angle A, \\angle C \\leq 120^\\circ$, and so $\\frac{1}{2} \\geq \\cos A, \\cos C \\geq -\\frac{1}{2}$. Applying the **Law of Cosines** to triangle $ABD$ yields\n$$\n\\begin{aligned}\nBD^2 &= AB^2 - 2AB \\cdot AD \\cos A + AD^2 \\\\\n    &\\geq AB^2 - AB \\cdot AD + AD^2 \\\\\n    &\\geq \\frac{1}{3}(AB^2 + AB \\cdot AD + AD^2).\n\\end{aligned}\n$$\nThe last inequality is equivalent to $3AB^2 - 3AB \\cdot AD + 3AD^2 \\geq AB^2 + AB \\cdot AD + AD^2$, or $AB^2 - 2AB \\cdot AD + AD^2 \\geq 0$, which is evident. The last equality holds if and only if $AB = AD$.\n\nOn the other hand, applying the Law of Cosines to triangle $BCD$ yields\n$$\nBD^2 = BC^2 - 2BC \\cdot CD \\cos C + CD^2 \\leq BC^2 + BC \\cdot CD + CD^2.\n$$\nCombining the last two inequalities gives the desired result.\n\nFor the given inequalities to have equality, we must have $AB = AD$. This condition is also sufficient, because all the entries in the equalities are 0. Thus, equality holds if and only if $ABCD$ is a *kite* with $AB = AD$ and $BC = CD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15706,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle. Let $B'$ and $A'$ be points on the perpendicular bisectors of $AC$ and $BC$, respectively, such that $B'A \\perp AB$ and $A'B \\perp AB$. Let $P$ be a point on the segment $AB$, and let $O$ be the circumcenter of triangle $ABC$. Let $D$ and $E$ be points on $BC$ and $AC$, respectively, such that $DP \\perp BO$ and $EP \\perp AO$. Let $O'$ be the circumcenter of triangle $CDE$. Prove that $B'$, $A'$, and $O'$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "We first give some intuition for approaching the problem. If $P \\equiv A$, then $O' \\equiv B'$, and if $P \\equiv B$, then $O' \\equiv A'$. Thus, as $P$ varies on $AB$, the respective $O'$ trace a segment, and we seek to identify this segment.\n\nIt is natural to draw a diagram independent of $P$ and identify the line $A'B'$, which turns out to be perpendicular to $CM$, where $M$ is the midpoint of $AB$.\n\nFurthermore, $B'M^2 - B'C^2 = AM^2 = A'M^2 - A'C^2$, which uniquely defines the line and shows $A'B' \\perp CM$.\n\n![alt](images/MACEDONIAN_MATHEMATICAL_OLYMPIADS_2016_p13_data_a9311b1355.png)\n\nThe following sketch represents the problem setting when including elements depending on $P$.\n\nNow, for the formal proof:\n\nIt suffices to show that $O'M^2 - O'C^2 = AM^2$ for all $P$, including $P = A, B$.\n\nFirst, $O'EPD$ is a cyclic quadrilateral. This follows since $EO'D = 2\\angle ABC = \\angle APE + \\angle BPD = \\pi - \\angle EPD$, as $\\angle ABC = \\angle APE = \\angle BPD$. Thus, $PO'$ is the angle bisector of $\\angle EPD$ and $PO' \\perp AB$.\n\nWe now show $O'M^2 - O'C^2 = AM^2$. The following sketch illustrates this part:\n\n![alt](images/MACEDONIAN_MATHEMATICAL_OLYMPIADS_2016_p13_data_c041eedfd4.png)\n\nLet $D'$ be the second intersection of $PE$ with the circumcircle of $CDE$, so $O'P^2 - O'C^2 = PE \\cdot PD'$.\n\nSince $PO'$ is the angle bisector of $\\angle EPD$, by the extended $S-S-K$ congruency theorem, $PD = PD'$. If $PD = PE$, triangles $P'EO'$ and $P'DO'$ are congruent by $S-S-S$, so $EO'P = DO'P = \\angle CAB$, while $EPO' = DPO' = \\frac{\\pi}{2} - \\angle CAB$, so $PD$ and $PE$ are tangents, and $D' \\equiv E$. Thus, the claim holds.\n\nNoting that triangles $APE$ and $BPD$ are similar, $\\frac{PE}{AP} = \\frac{PB}{PD}$, implying $AP \\cdot BP = PE \\cdot PD = PE \\cdot PD'$.\n\nSince $PO' \\perp AB$, by the Pythagorean theorem:\n\n$$\n\\begin{aligned}\nO'M^2 - O'C^2 - AM^2 &= \\\\\n&= O'P^2 - O'C^2 + PM^2 - AM^2 = \\\\\n&= PD' \\cdot PE - AP \\cdot PB = 0\n\\end{aligned}\n$$\n\nwhere $O'P^2 - O'C^2 = PE \\cdot PD'$ by the power of point $P$ to the circumcircle of $CDE$, and\n\n$$\nAM^2 - PM^2 = (BM - PM)(AM - PM) = AP \\cdot PB.\n$$\n\nThis completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15707,
    "subject": "Mathematics (Olympiad)",
    "question": "If $p$ is prime and $x, y$ are positive integers, find, with respect to $p$, all pairs $(x, y)$ satisfying the equation:\n\n$$\np(x-2) = x(y-1)\n$$\n\nFor the case where $x + y = 21$, find all triads $(x, y, p)$ satisfying the equation above.",
    "options": [],
    "answer": "See solution",
    "solution": "If $x - 2 = 0$ (i.e., $x = 2$), then $y = 1$, so $(x, y) = (2, 1)$ is a solution.\n\nIf $x - 2 \\neq 0$, since $p$ is prime, from $p(x-2) = x(y-1)$ it follows that $y \\neq 1$ and $p \\mid x$ or $p \\mid y-1$.\n\n**Case 1:** $p \\mid x$\n\nLet $x = p x'$, where $x'$ is a positive integer. Then:\n\n$$\np(p x' - 2) = p x'(y - 1) \\implies p x' - 2 = x'(y - 1) \\implies x'(p - y + 1) = 2\n$$\n\nSo, possible solutions:\n- $x' = 1$, $p - y + 1 = 2$ $\\implies$ $y = p - 1$\n- $x' = 2$, $p - y + 1 = 1$ $\\implies$ $y = p$\n\nThus:\n- $(x, y) = (p, p-1)$\n- $(x, y) = (2p, p)$\n\n**Case 2:** $p \\mid y-1$\n\nLet $y - 1 = p y'$, where $y'$ is a positive integer. Then:\n\n$$\np(x-2) = x p y' \\implies x - 2 = x y' \\implies x(1 - y') = 2\n$$\n\nPossible integer solutions:\n- $x = 1$, $1 - y' = 2$ $\\implies$ $y' = -1$ (not positive)\n- $x = 2$, $1 - y' = 1$ $\\implies$ $y' = 0$ (not positive)\n\nSo, no solutions in this case.\n\n**Now, if $x + y = 21$:**\n\n- For $(x, y) = (p, p-1)$: $p + (p-1) = 21 \\implies 2p - 1 = 21 \\implies p = 11$. So $(x, y, p) = (11, 10, 11)$.\n- For $(x, y) = (2p, p)$: $2p + p = 21 \\implies 3p = 21 \\implies p = 7$. So $(x, y, p) = (14, 7, 7)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15708,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ and $q$ be real numbers such that the quadratic equation\n\n$$\nx^2 + px + q = 0\n$$\n\nhas two real solutions $x_1$ and $x_2$.\n\nThe following two conditions hold:\n\n1. The numbers $x_1$ and $x_2$ differ by $1$.\n2. The numbers $p$ and $q$ differ by $1$.\n\nShow that $p$, $q$, $x_1$, and $x_2$ are integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume $x_1 = x_2 + 1$. By Vieta's formulas, $p = -(x_1 + x_2) = -2x_2 - 1$ and $q = x_1 x_2 = x_2^2 + x_2$.\n\nTherefore, it is enough to check that $x_2$ must be an integer.\n\n**Case 1:** $q = p - 1$\n\nThen $x_2^2 + x_2 = -2x_2 - 1 - 1$, so $x_2^2 + 3x_2 + 2 = 0$. Thus, $x_2 = -1$ or $x_2 = -2$, both integers.\n\n**Case 2:** $q = p + 1$\n\nThen $x_2^2 + x_2 = -2x_2 - 1 + 1$, so $x_2^2 + 3x_2 = 0$. Thus, $x_2 = 0$ or $x_2 = -3$, both integers.\n\nTherefore, $p$, $q$, $x_1$, and $x_2$ are all integers.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15709,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$ and $y$ be real numbers satisfying the following two equations:\n\n$$\nx^3 + x^2 + x + 1 = y^3 + y^2 + y\n$$\n$$\nx = y^3 - y^2 + 3y\n$$\n\nFind the value of $x - y$.",
    "options": [],
    "answer": "See solution",
    "solution": "From the second equation, we express $x = y^3 - y^2 + 3y$. Substitute this into the first equation and simplify to get\n\n$$\ny^9 - 3y^8 + 12y^7 - 18y^6 + 34y^5 - 19y^4 + 21y^3 + 11y^2 + 4y + 2 = 0.\n$$\n\nWe can factor the left side as\n\n$$\n(y^3 - y^2 + 2y + 1)(y^6 - 2y^5 + 8y^4 - 7y^3 + 13y^2 + 2) = 0.\n$$\n\nSince\n\n$$\n\\begin{align*}\ny^6 - 2y^5 + 8y^4 - 7y^3 + 13y^2 + 2 &= y^4(y^2 - 2y + 1) + 7y^2\\left(y^2 - y + \\frac{1}{4}\\right) + \\frac{45}{4}y^2 + 2 \\\\\n&= y^4(y - 1)^2 + 7y^2\\left(y - \\frac{1}{2}\\right)^2 + \\frac{45}{4}y^2 + 2 > 0,\n\\end{align*}\n$$\n\nwe must have $y^3 - y^2 + 2y + 1 = 0$. This gives $y = y^3 - y^2 + 3y + 1$. Therefore,\n\n$$\nx - y = (y^3 - y^2 + 3y) - (y^3 - y^2 + 3y + 1) = -1.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15710,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, we have $AB > AC$. The incircle $\\omega$ touches $BC$ at $E$, and $AE$ intersects $\\omega$ at $D$. Choose a point $F$ on $AE$ ($F \\neq E$), such that $CE = CF$. Let $G$ be the intersection point of $CF$ and $BD$. Prove that $CF = FG$.",
    "options": [],
    "answer": "See solution",
    "solution": "Referring to the figure, draw a line from $D$, tangent to $\\omega$, and let the line intersect $AB$, $AC$, $BC$ at points $M$, $N$, $K$ respectively.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p109_data_1ec3de62b6.png)\n\nSince\n\n$$\n\\angle KDE = \\angle AEK = \\angle EFC,\n$$\n\nwe know $MK \\parallel CG$.\n\nBy Newton's theorem, the lines $BN$, $CM$, $DE$ are concurrent.\n\nBy Ceva's theorem, we have\n\n$$\n\\frac{BE}{EC} \\cdot \\frac{CN}{NA} \\cdot \\frac{AM}{MB} = 1. \\qquad \\textcircled{1}\n$$\n\nFrom Menelaus' theorem,\n\n$$\n\\frac{BK}{KC} \\cdot \\frac{CN}{NA} \\cdot \\frac{AM}{MB} = 1. \\qquad \\textcircled{2}\n$$\n\nDividing \\textcircled{1} by \\textcircled{2}, we have\n\n$$\nBE \\cdot KC = EC \\cdot BK,\n$$\n\nthus\n\n$$\nBC \\cdot KE = 2EB \\cdot CK. \\qquad \\textcircled{3}\n$$\n\nUsing Menelaus' theorem and \\textcircled{3}, we get\n\n$$\n1 = \\frac{CB}{BE} \\cdot \\frac{ED}{DF} \\cdot \\frac{FG}{GC} = \\frac{CB}{BE} \\cdot \\frac{EK}{CK} \\cdot \\frac{FG}{GC} = \\frac{2FG}{GC}.\n$$\n\nSo $CF = FG$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15711,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove the claim: If we choose any four factors of $720$, then one of them divides the product of the other three.",
    "options": [],
    "answer": "See solution",
    "solution": "Given the decomposition $720 = 2^4 \\cdot 3^2 \\cdot 5$, the number $720$ has exactly three prime factors: $2$, $3$, and $5$. Each of its factors is of the form $2^\\alpha \\cdot 3^\\beta \\cdot 5^\\gamma$, where $\\alpha, \\beta, \\gamma$ are non-negative integers (with $\\alpha \\le 4$, $\\beta \\le 2$, $\\gamma \\le 1$). The product of any three factors of $720$ is also of the form $2^\\alpha \\cdot 3^\\beta \\cdot 5^\\gamma$ with non-negative integers $\\alpha, \\beta, \\gamma$. For any two numbers of this form, the first divides the second if and only if the exponents $\\alpha, \\beta, \\gamma$ of the first do not exceed those of the second.\n\nWe prove the statement by contradiction. Suppose there exist four factors of $720$ such that none divides the product of the other three. Then, for each factor, its exponent for some prime (among $2$, $3$, $5$) is greater than the corresponding exponent in the product of the other three. But there are four divisors and only three primes, so by the pigeonhole principle, this is impossible. Thus, one of the four factors must divide the product of the other three.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15712,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be real numbers. Prove that\n\n$$\n\\frac{a^4 + 3ab^3}{a^3 + 2b^3} + \\frac{b^4 + 3bc^3}{b^3 + 2c^3} + \\frac{c^4 + 3ca^3}{c^3 + 2a^3} \\le 4.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n\n$$\n\\frac{a^4 + 3ab^3}{a^3 + 2b^3} = \\frac{a(a^3 + 2b^3) + ab^3}{a^3 + 2b^3} = a + \\frac{ab^3}{a^3 + 2b^3}.\n$$\n\nBy the inequality between arithmetic and geometric means,\n\n$$\na^3 + 2b^3 = a^3 + b^3 + b^3 \\ge 3\\sqrt[3]{a^3 b^3 b^3} = 3ab^2.\n$$\n\nThus,\n\n$$\n\\frac{a^4 + 3ab^3}{a^3 + 2b^3} = a + \\frac{ab^3}{a^3 + 2b^3} \\le a + \\frac{ab^3}{3ab^2} = a + \\frac{b}{3}.\n$$\n\nAnalogously,\n\n$$\n\\frac{b^4 + 3bc^3}{b^3 + 2c^3} \\le b + \\frac{c}{3} \\quad \\text{and} \\quad \\frac{c^4 + 3ca^3}{c^3 + 2a^3} \\le c + \\frac{a}{3}.\n$$\n\nBy adding the inequalities,\n\n$$\n\\frac{a^4 + 3ab^3}{a^3 + 2b^3} + \\frac{b^4 + 3bc^3}{b^3 + 2c^3} + \\frac{c^4 + 3ca^3}{c^3 + 2a^3} \\le \\frac{4}{3}(a + b + c).\n$$\n\nFinally, by the inequality between arithmetic and quadratic means,\n\n$$\n4 \\cdot \\frac{a+b+c}{3} \\le 4 \\cdot \\sqrt{\\frac{a^2+b^2+c^2}{3}} = 4.\n$$\n\nThis finishes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15713,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $a$ such that $2^n - n^2$ divides $a^n - n^a$ for all positive integers $n \\ge 5$.",
    "options": [],
    "answer": "See solution",
    "solution": "The only possible values for $a$ are $2$ and $4$.\n\nFirst, we show that $a$ must be even. This follows by choosing an even integer $n \\ge 6$ and applying the divisibility condition.\n\nNext, we show that $a$ cannot have any odd prime factors. Suppose, for contradiction, that $p$ is an odd prime dividing $a$.\n\n- If $p = 3$, let $n = 8$. Then $2^n - n^2 = 192$ is divisible by $3$, but $a^n - n^a$ is not necessarily divisible by $3$, contradicting the condition.\n- If $p = 5$, let $n = 16$. Then $2^n - n^2 = 64 \\cdot 110$ is divisible by $5$, but $a^n - n^a$ is not necessarily divisible by $5$, again a contradiction.\n- If $p \\ge 7$, let $n = p - 1$. By Fermat's Little Theorem, $2^{p-1} \\equiv 1 \\pmod{p}$ and $(p-1)^2 \\equiv 1 \\pmod{p}$, so $p$ divides $2^n - n^2$. Since $a$ is even and $p \\mid a$, $n^a \\equiv (p-1)^a \\equiv (-1)^a \\equiv 1 \\pmod{p}$, and $p \\mid a^n$, so $p$ does not divide $a^n - n^a$, a contradiction.\n\nTherefore, $a$ is a power of $2$, say $a = 2^t$ for some $t \\ge 1$.\n\nNow, consider the divisibility $(2^n - n^2) \\mid (a^n - n^a) = (2^{tn} - n^{2^t})$. For large $n$, $2^n$ grows much faster than $n^2$, so $2^n - n^2$ is large. The only way the divisibility can always hold is if $t = 1$ or $t = 2$, i.e., $a = 2$ or $a = 4$.\n\nFinally, it is easy to check that $a = 2$ and $a = 4$ satisfy the original condition for all $n \\ge 5$.\n\nThus, the solutions are $a = 2$ and $a = 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15714,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{N} \\to \\mathbb{N}$ such that\n\n$$\n3f(f(f(n))) + 2f(f(n)) + f(n) = 6n, \\text{ for all } n \\in \\mathbb{N}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We firstly notice that the function is injective. Indeed, if $f(n) = f(m)$, then $3f(f(f(n))) + 2f(f(n)) + f(n) = 3f(f(f(m))) + 2f(f(m)) + f(m)$, hence $6n = 6m$, so $n = m$.\n\nPlugging $n = 0$ yields $3f(f(f(0))) + 2f(f(0)) + f(0) = 0$, so $f(0) = 0$. We prove, using induction, that $f(n) = n$, for every $n \\in \\mathbb{N}$.\n\nSuppose $f(0) = 0$, $f(1) = 1$, ..., $f(n) = n$. From the injectivity, $f(n+1) \\geq n+1$, $f(f(n+1)) \\geq n+1$ and $f(f(f(n+1))) \\geq n+1$. On the other hand, $3f(f(f(n+1))) + 2f(f(n+1)) + f(n+1) = 6n+6$, whence $f(n+1) = n+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15715,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n \\geq 3$ 為正整數。證明：滿足 $x_1 + x_2 + \\cdots + x_n = 3$ 且 $0 \\leq x_1, x_2, \\ldots, x_n \\leq 1$ 的實數 $x_1, x_2, \\ldots, x_n$，必有兩項 $x_i, x_j$ 滿足 $x_i x_j > 2^{-|i-j|}$。\n\n註：只證出 $x_i x_j > c 2^{-|i-j|}$，其中 $0 < c < 1$ 與 $n$ 無關，將依 $c$ 值來給分。",
    "options": [],
    "answer": "See solution",
    "solution": "令 $1 \\leq a < b \\leq n$ 使 $2^{b-a} x_a x_b$ 最大。這樣選擇 $a$ 和 $b$，可得對所有 $1-a \\leq t \\leq b-a-1$，有 $x_{a+t} \\leq 2^t x_a$，同理對所有 $b-n \\leq t \\leq b-a+1$，有 $x_{b-t} \\leq 2^t x_b$。\n\n假設 $x_a \\in \\left(\\frac{1}{2^{u+1}}, \\frac{1}{2^u}\\right]$ 且 $x_b \\in \\left(\\frac{1}{2^{v+1}}, \\frac{1}{2^v}\\right]$，令 $x_a = 2^{-\\alpha}$, $x_b = 2^{-\\beta}$。則\n\n$$\n\\sum_{i=1}^{a+u-1} x_i \\leq 2^u x_a \\left( \\frac{1}{2} + \\frac{1}{4} + \\cdots + \\frac{1}{2^{a+u-1}} \\right) < 2^u x_a \\leq 1,\n$$\n\n同理，\n\n$$\n\\sum_{i=b-v+1}^{n} x_i \\leq 2^v x_b \\left( \\frac{1}{2} + \\frac{1}{4} + \\cdots + \\frac{1}{2^{n-b+v}} \\right) < 2^v x_b \\leq 1,\n$$\n\n也就是說，$i$ 不在區間 $[a+u, b-v]$ 的 $x_i$ 之和嚴格小於 2。由於總和為 3，且每項至多為 1，故此區間至少有兩個整數，即 $a+u < b-v$。\n\n因此，對 $i \\in [1, a+u] \\cup [b-v, n]$ 的 $x_i$ 以如上界估，區間 $(a+u, b-v)$ 內每項以 1 上界，得\n\n$$\ns < 2^{u+1} x_a + 2^{v+1} x_b + ((b-v)-(a+u)-1) = b-a + (2^{u+1-\\alpha} + 2^{v+1-\\beta} - (u+v+1)).\n$$\n\n注意 $\\alpha \\in (u, u+1]$，$\\beta \\in (v, v+1]$，由 Bernoulli 不等式可得\n\n$$\n2^{u+1-\\alpha} + 2^{v+1-\\beta} - u - v - 1 \\leq (1 + (u+1-\\alpha)) + (1 + (v+1-\\beta)) - u - v - 1 = 3 - \\alpha - \\beta.\n$$\n\n因此 $0 < b-a-\\alpha-\\beta$，所以\n\n$$\n1 < 2^{b-a-\\alpha-\\beta} = 2^{b-a} x_a x_b.\n$$\n\n![](images/2023-TWNIMO-Problems_p49_data_a45ee56570.png)\n\n**Comment 1.** 不等式是緊的。設 $n = 2k+1$，令 $x_{k+1} = 1$，$x_k = x_{k+2} = \\frac{1}{2} + \\frac{1}{2^k}$，$x_{k+1-t} = x_{k+1+t} = \\frac{1}{2^t}$。則\n\n$$\n\\max_{i<j} 2^{j-i} x_i x_j = \\left(1 + \\frac{1}{2^{k-1}}\\right)^2,\n$$\n\n可令其任意接近 1。\n\n**Comment 2.** 問題亦可表述為：存在正實常數 $c$（或求最佳 $c$），使得\n\n$$\n\\max_{i<j} 2^{j-i} x_i x_j > c.\n$$\n\n上述證明 $c = 1$ 為最佳。若取 $c = \\frac{1}{4}$，證明可簡化：\n\n如原證，得\n\n$$\n3 < 2^{u+1} x_a + 2^{v+1} x_b + ((b-v)-(a+u)-1) = b-a + (2^{u+1-\\alpha} + 2^{v+1-\\beta} - (u+v+1)).\n$$\n\n又 $2^{b-a} x_a x_b \\geq 2^{b-a} 2^{-u-1} 2^{-v-1}$，只需證 $-2 < b-a-u-v-2$，即 $3 < b-a-u-v+3$。由 $u+1-\\alpha < 1$ 及 $v+1-\\beta < 1$，得 $2^{u+1-\\alpha} + 2^{v+1-\\beta} < 4$，故 $3 < b-a + (2 + 2 - u - v - 1) = b-a-u-v-3$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15716,
    "subject": "Mathematics (Olympiad)",
    "question": "Given positive real numbers $a, b, c, d$ that satisfy the equalities\n$$\na^2 + d^2 - ad = b^2 + c^2 + bc \\quad \\text{and} \\quad a^2 + b^2 = c^2 + d^2,\n$$\nfind all possible values of the expression $\\frac{ab+cd}{ad+bc}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A_1BC_1$ be a triangle with $A_1B = b$, $BC_1 = c$ and $\\angle A_1BC_1 = 120^\\circ$, and $C_2DA_2$ be another triangle with $C_2D = d$, $DA_2 = a$ and $\\angle C_2DA_2 = 60^\\circ$. By the law of cosines and the assumption $a^2 + d^2 - ad = b^2 + c^2 + bc$, we have $A_1C_1 = A_2C_2$. Thus the two triangles can be put together to form a quadrilateral $ABCD$ with $AB = b$, $BC = c$, $CD = d$, $DA = a$ and $\\angle ABC = 120^\\circ$, $\\angle CDA = 60^\\circ$. Then $\\angle DAB + \\angle BCD = 360^\\circ - (\\angle ABC + \\angle CDA) = 180^\\circ$.\n\nSuppose $\\angle DAB > 90^\\circ$; then $\\angle BCD < 90^\\circ$ whence $a^2 + b^2 < BD^2 < c^2 + d^2$, contradicting the assumption $a^2 + b^2 = c^2 + d^2$. By symmetry, $\\angle DAB < 90^\\circ$ also leads to contradiction. Hence $\\angle DAB = \\angle BCD = 90^\\circ$. Now calculate the area of $ABCD$ in two ways; on one hand, it equals $\\frac{1}{2}ad \\sin 60^\\circ + \\frac{1}{2}bc \\sin 120^\\circ$ or $\\frac{\\sqrt{3}}{4}(ad + bc)$; on the other hand, it equals $\\frac{1}{2}ab + \\frac{1}{2}cd$ or $\\frac{1}{2}(ab + cd)$. Consequently,\n$$\n\\frac{ab + cd}{ad + bc} = \\frac{\\frac{\\sqrt{3}}{4}}{\\frac{1}{2}} = \\frac{\\sqrt{3}}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15717,
    "subject": "Mathematics (Olympiad)",
    "question": "Can a $7 \\times 7$ square be tiled with the two types of tiles shown below? (Tiles can be rotated and reflected but cannot overlap or be broken)\n\nFind the least number $N$ of tiles of type $A$ that must be used in the tiling of a $1011 \\times 1011$ square. Give an example of a tiling that contains exactly $N$ tiles of type $A$.\n\n![](images/IND_ABooklet_2024_p20_data_bee34f74c4.png)\n\n(A)\n\n![](images/IND_ABooklet_2024_p20_data_cc2d8b6bba.png)\n\n(B)",
    "options": [],
    "answer": "See solution",
    "solution": "We prove a more general fact: the number of $L$-tiles in any tiling of a $(2n-1) \\times (2n-1)$ square with tiles of the two given types is not less than $4n-1$ for any $n > 4$. In particular, for $n = 506$ the minimum number of $L$-tiles is $2023$.\n\nColor the big square in four colors $1$, $2$, $3$, $4$ as shown below.\n\n$$\n\\begin{array}{cccccc}\n1 & 2 & 1 & 2 & 1 & 2 \\\\\n3 & 4 & 3 & 4 & 3 & 4 \\\\\n1 & 2 & 1 & 2 & 1 & 2 \\\\\n3 & 4 & 3 & 4 & 3 & 4 \\\\\n1 & 2 & 1 & 2 & 1 & 2 \\\\\n3 & 4 & 3 & 4 & 3 & 4 \\\\\n\\end{array}\n$$\n\nNo matter how we fill this square with our tiles, the tiles of type $B$ will always cover four unit squares of different colors. So all these tiles will cover an equal number of unit squares of each color. However, the total numbers of unit squares of different colors is different. Suppose we have $x$ tiles of type $A$ and $y$ tiles of type $B$. Then $3x + 4y = (2n-1)^2$. On the other hand, each tile covers no more than one square of color $1$ and the total number of $1$-colored tiles is $n^2$. Hence $x + y \\ge n^2$. Thus,\n\n$$\n4x \\ge 4n^2 - 4y = 4n^2 - (2n-1)^2 + 3x = 4n - 1 + 3x\n$$\n\nHence $x \\ge 4n - 1$. When $n = 25$, the minimum number of $L$-tiles required is $99$.\n\n![](images/IND_ABooklet_2024_p20_data_746f0c9ee0.png)\n\n(i)\n\n![](images/IND_ABooklet_2024_p20_data_21758d7725.png)\n\n(ii)\n\nIn the figure, (i) exhibits a tiling of $7 \\times 7$ square with $15$ $L$-tiles. Figure (ii) shows how to extend a tiling of $(2n-1) \\times (2n-1)$ containing $4n-1$ $L$-tiles to a tiling of $(2n+1) \\times (2n+1)$ with $4$ additional $L$-tiles, obtaining a tiling with $4n-1+4 = 4n+3 = 4(n+1)-1$ $L$-tiles. Thus, starting with a tiling of $7 \\times 7$ square with $15$ $L$-tiles, we obtain a tiling of $1011 \\times 1011$ square with $15+4 \\times 502 = 2023$ $L$-tiles. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15718,
    "subject": "Mathematics (Olympiad)",
    "question": "Bogdan drew 2017 vertical and 2018 horizontal lines in the rectangle $Q$. These lines divide $Q$ into $2018 \\times 2019$ smaller, not necessarily equal, rectangles. Two children want to determine the perimeter of $Q$. Andrew says that he can do it by choosing some 2019 smaller rectangles and getting to know their perimeters. On the contrary, Olesya says that she can do it by choosing 4036 smaller rectangles. Who is right about the number of smaller rectangles, perimeters of which one should know in order to determine the perimeter of $Q$?",
    "options": [],
    "answer": "See solution",
    "solution": "Both of them are wrong; one cannot be sure about the perimeter of $Q$ even if they know the perimeters of all $2018 \\times 2019$ smaller rectangles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15719,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the natural generalisation where $10$ is replaced everywhere by $n$ for some $n > 1$, $n \\in \\mathbb{N}$. It is convenient to use vector notation, writing $f(\\underline{x})$ in place of $f(x_1, \\dots, x_n)$. To show that the given sums are insufficient to compute all $f$-values, note that if we write $s(\\underline{x}) = \\sum_{i=1}^n x_i$ and then redefine $f$ by adding $(-1)^{s(\\underline{x})}$ to $f(\\underline{x})$, all the listed sums remain the same.\n\nProve that the value of the additional sum\n\n$$\nf(0, 0, 0, 0, \\dots, 0) + f(1, 1, 0, 0, \\dots, 0)\n$$\n\nsuffices to compute the values of $f$.\n\nShow that any other additional sum of the form $f(\\underline{a}) + f(\\underline{b})$, for some fixed $\\underline{a}, \\underline{b}$, where $s(\\underline{a})$ and $s(\\underline{b})$ have the same parity, or equivalently where $\\underline{a}$ and $\\underline{b}$ differ in an even number of positions, is sufficient.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution 1.** We know all sums of the form $S := f(\\underline{x}) + f(\\underline{y})$ for vectors $\\underline{x}, \\underline{y}$ that differ in a single position. We also know $T := f(\\underline{y}) + f(\\underline{z})$, whenever $\\underline{y}$ and $\\underline{z}$ differ in a single position, so we know $S - T = f(\\underline{x}) - f(\\underline{z})$. Thus, we know such differences whenever $\\underline{x}$ and $\\underline{z}$ differ in exactly two positions. By repeating the process of taking differences, we can compute $f(\\underline{u}) - f(\\underline{v})$ whenever $\\underline{u}$ and $\\underline{v}$ differ in any even number of positions. In particular, we can compute $f(\\underline{a}) - f(\\underline{b})$ where $\\underline{a} = (0, 0, 0, 0, \\dots, 0)$ and $\\underline{b} = (1, 1, 0, 0, \\dots, 0)$. Since we also know $f(\\underline{a}) + f(\\underline{b})$, we can compute $f(\\underline{a})$.\n\nKnowing $f(\\underline{a})$ allows us to compute all values. In fact, if $\\underline{u}$ differs from $\\underline{a}$ in exactly one position, then knowing $f(\\underline{a}) + f(\\underline{u})$ allows us to compute $f(\\underline{u})$. Applying the same argument to $f(\\underline{u})$, we compute $f(\\underline{v})$ whenever $\\underline{v}$ differs from $\\underline{u}$ in exactly one position, and so whenever $\\underline{v}$ differs from $\\underline{a}$ in exactly two positions. Iterating this approach, we get all other values of $f$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 15720,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be non-negative real numbers such that\n\n$$\n2(a^2 + b^2 + c^2) + 3(ab + bc + ca) = 5(a + b + c).\n$$\n\nProve that\n$$\n4(a^2 + b^2 + c^2) + 2(ab + bc + ca) + 7abc \\le 25.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p = a + b + c$, $q = ab + bc + ca$, $r = abc$. We have\n\n$$\n2(p^2 - 2q) + 3q = 5p \\quad \\text{or} \\quad 2p^2 = 5p + q. \\qquad (1)\n$$\n\nWe need to prove that $4(p^2 - 2q) + 2q + 7r \\le 25$, or equivalently, $4p^2 + 7r \\le 25 + 6q$.\n\nSince $q = p^2 - 5p$, the inequality becomes:\n\n$$\n7r + 30p \\le 8p^2 + 25.\n$$\n\nNotice that $(ab + bc + ca)^2 \\ge 3abc(a + b + c)$ implies $q^2 \\ge 3pr$. We consider two cases for $p$:\n\n* If $p = 0$, then $a = b = c = 0$, so the inequality holds.\n\n* If $p > 0$, then $r \\le \\dfrac{q^2}{3p}$, so we need to prove\n\n$$\n7\\frac{q^2}{3p} + 30p \\le 8p^2 + 25.\n$$\n\nSubstituting $q = p^2 - 5p$ gives\n\n$$\n7(2p^2 - 5p)^2 + 90p^2 \\le 24p^3 + 75p,\n$$\nwhich simplifies to\n$$\np(p-3)(2p-5)(14p-5) \\le 0. \\qquad (2)\n$$\n\nOn the other hand, $2p^2 - 5p = q \\le \\dfrac{p^2}{3}$ implies $\\dfrac{5}{2} \\le p \\le 3$, so inequality (2) holds. Therefore, the original inequality is true.\n\nEquality holds when $(a, b, c)$ is a permutation of $(0, 0, 0)$, $(1, 1, 1)$, or $(0, 0, \\frac{5}{2})$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15721,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $C = \\{z \\in \\mathbb{C} \\mid |z| = 1\\}$ be the unit circle in the complex plane. Let $z_1, z_2, \\dots, z_{240} \\in C$ be 240 (not necessarily distinct) complex numbers, satisfying the following two conditions:\n\n1. For any open arc $\\Gamma$ of length $\\pi$ on $C$, there are at most 200 indices $j$ ($1 \\leq j \\leq 240$) such that $z_j \\in \\Gamma$.\n\n2. For any open arc $\\gamma$ of length $\\pi/3$ on $C$, there are at most 120 indices $j$ ($1 \\leq j \\leq 240$) such that $z_j \\in \\gamma$.\n\nFind the maximum value of $|z_1 + z_2 + \\dots + z_{240}|$.",
    "options": [],
    "answer": "See solution",
    "solution": "The maximum value is $80 + 40\\sqrt{3}$.\n\nChoose $z_1, \\dots, z_{240}$ to consist of 80 copies of $1$, and 40 copies each of $\\exp\\left(\\frac{\\pi i}{6}\\right)$, $\\exp\\left(-\\frac{\\pi i}{6}\\right)$, $i$, and $-i$. It is easy to verify conditions (1) and (2), and $z_1 + z_2 + \\dots + z_{240} = 80 + 40\\sqrt{3}$.\n\nWe now show that this is the maximal possible value of $|z_1 + \\dots + z_{240}|$. Assume that $z_1, z_2, \\dots, z_{240}$ satisfy conditions (1) and (2). After possibly rotating all $z_i$, we may assume that $S = z_1 + z_2 + \\dots + z_{240}$ is a nonnegative real number. We may further assume that $z_1, z_2, \\dots, z_{240}$ are ordered counterclockwise on $C$, starting from $-1$ (and including $-1$).\n\nCondition (1) implies that for $1 \\leq j \\leq 40$, moving from $z_j$ to $z_{j+200}$ counterclockwise on $C$ at least passes through an arc of length $\\pi$. That is, there exists $2\\pi \\geq \\alpha \\geq \\pi$ such that $z_{j+200} = z_j \\cdot \\exp(\\alpha i)$. Let $z_j = (-1) \\cdot \\exp(\\beta i)$ with $\\beta \\in [0, 2\\pi)$. Then $\\beta + \\alpha < 2\\pi$, so\n\n$$\n\\operatorname{Re}(z_j + z_{j+200}) = -\\cos \\beta - \\cos(\\beta + \\alpha) = -2 \\cos \\frac{\\alpha}{2} \\cos\\left(\\beta + \\frac{\\alpha}{2}\\right) \\leq 0.\n$$\n\n(This is because $\\frac{\\alpha}{2} \\in [\\frac{\\pi}{2}, \\pi]$ and $\\beta + \\frac{\\alpha}{2} \\in [\\frac{\\pi}{2}, \\frac{3\\pi}{2}]$.) Summing over $1 \\leq j \\leq 40$ gives\n\n$$\n\\operatorname{Re}(z_1 + z_2 + \\dots + z_{40} + z_{201} + z_{202} + \\dots + z_{240}) \\leq 0. \\quad (*)\n$$\n\nSimilarly, condition (2) shows that for $41 \\leq j \\leq 80$, moving from $z_j$ to $z_{j+120}$ counterclockwise on $C$ at least passes through an arc of length $\\frac{\\pi}{3}$, i.e., there exists $2\\pi \\geq \\alpha \\geq \\frac{\\pi}{3}$ such that $z_{j+120} = z_j \\cdot \\exp(\\alpha i)$. Set $z_j = (-1) \\cdot \\exp(\\beta i)$ for $\\beta \\in [0, 2\\pi)$. Then $\\beta + \\alpha < 2\\pi$, and hence\n\n$$\n\\operatorname{Re}(z_j + z_{j+120}) = -\\cos \\beta - \\cos(\\beta + \\alpha) = -2 \\cos \\frac{\\alpha}{2} \\cos\\left(\\beta + \\frac{\\alpha}{2}\\right).\n$$\n\nIf $\\alpha \\geq \\pi$, the above is nonpositive. If $\\alpha \\in [\\frac{\\pi}{3}, \\pi)$, then $|2 \\cos \\frac{\\alpha}{2} \\cos(\\beta + \\frac{\\alpha}{2})| \\leq 2 \\cdot \\cos \\frac{\\pi}{6} \\cdot 1 = \\sqrt{3}$. So the right-hand side is always $\\leq \\sqrt{3}$. Summing over $41 \\leq j \\leq 80$ gives\n\n$$\n\\operatorname{Re}(z_{41} + z_{42} + \\dots + z_{80} + z_{161} + z_{162} + \\dots + z_{200}) \\leq 40\\sqrt{3}. \\quad (**)\n$$\n\nCombining $(*)$ with $(**)$, we get\n\n$$\n\\begin{align*}\n|z_1 + z_2 + \\cdots + z_{240}| &= \\operatorname{Re}\\left(\\sum_{j=1}^{40}(z_j + z_{200+j})\\right) + \\operatorname{Re}\\left(\\sum_{j=41}^{80}(z_j + z_{120+j})\\right) + \\operatorname{Re}\\left(\\sum_{j=81}^{120} z_j\\right) \\\\\n&\\leq 0 + 40\\sqrt{3} + 80 = 80 + 40\\sqrt{3}.\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15722,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an obtuse isosceles triangle $ABC$ with $CA = CB$ and circumcenter $O$. The point $P$ on $AB$ is such that $AP < \\frac{AB}{2}$, and $Q$ on $AB$ is such that $BQ = AP$. The circle with diameter $CQ$ meets $(ABC)$ at $E$, and the lines $CE$ and $AB$ meet at $F$. If $N$ is the midpoint of $CP$ and $ON$ and $AB$ meet at $D$, show that $ODCF$ is cyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $T$ be the midpoint of $CQ$ (it is the center of the circle with diameter $CQ$). Then $OC$ is the perpendicular bisector of $AB$ (since $AC = BC$). Triangle $ONT$ is isosceles by symmetry (as $P$ and $Q$ are symmetric with respect to the midpoint $M$ of $AB$ and hence with respect to $CO$, by the problem condition). $OT$ is the perpendicular bisector of $CE$, so $\\angle OCF = 90^\\circ - \\angle OCT = 90^\\circ - \\angle OCD = \\angle ODM = \\angle ODF$. Thus, $ODCF$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15723,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Tasty and Stacy are given a circular necklace with $3n$ sapphire beads and $3n$ turquoise beads, such that no three consecutive beads have the same color. They play a cooperative game where they alternate turns removing three consecutive beads, subject to the following conditions:\n\n- Tasty must remove three consecutive beads which are turquoise, sapphire, and turquoise, in that order, on each of his turns.\n- Stacy must remove three consecutive beads which are sapphire, turquoise, and sapphire, in that order, on each of her turns.\n\nThey win if all the beads are removed in $2n$ turns. Prove that if they can win with Tasty going first, they can also win with Stacy going first.\n\nIn the necklace, we draw a *divider* between any two beads of the same color. Unless there are no dividers, this divides the necklace into several *zigzags* in which the beads in each zigzag alternate. Each zigzag has two *endpoints* (adjacent to dividers).\n\nObserve that the condition about not having three consecutive matching beads is equivalent to saying there are no zigzags of length 1.\n\n![](images/sols-TST-IMO-2019_p11_data_ce98542cab.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The main claim is that the game is winnable (for either player going first) if and only if there are at most $2n$ dividers. We prove this in two parts, the first part not using the hypothesis about three consecutive letters.\n\n**Claim.** The game cannot be won with Tasty going first if there are more than $2n$ dividers.\n\n*Proof.* We claim each move removes at most one divider, which proves the result.\n\nConsider removing a $\\text{TST}$ in some zigzag (necessarily of length at least 3). We illustrate the three possibilities in the following table, with Tasty's move shown in red.\n\n| Before | After | Change |\n|--------|-------|--------|\n| ...ST \\| <span style=\"color:red;\">TST</span> \\| TS... | ...ST \\| TS... | One less divider; two zigzags merge |\n| ...ST \\| <span style=\"color:red;\">TST</span>TST... | ...STST... | One less divider; two zigzags merge |\n| ...<span style=\"color:red;\">TSTS</span>... | ...S \\| S... | One more divider; a zigzag splits in two |\n\nThe analysis for Stacy's move is identical. $\\Box$\n\n**Claim.** If there are at most $2n$ dividers and there are no zigzags of length 1 then the game can be won (with either player going first).\n\n*Proof.* By symmetry it is enough to prove Tasty wins going first.\n\nAt any point if there are no dividers at all, then the necklace alternates $TSTST\\ldots$ and the game can be won. So we will prove that on each of Tasty's turns, if there exists at least one divider, then Tasty and Stacy can each make a move at an endpoint of some zigzag (i.e., the first two cases above). As we saw in the previous proof, such moves will (a) decrease the number of dividers by exactly one, (b) not introduce any singleton zigzags (because the old zigzags merge, rather than split). Since there are fewer than $2n$ dividers, our duo can eliminate all dividers and then win.\n\nNote that as the number of $S$ and $T$'s are equal, there must be an equal number of\n\n- zigzags of odd length ($\\ge 3$) with $T$ at the endpoints (i.e., one more $T$ than $S$), and\n- zigzags of odd length ($\\ge 3$) with $S$ at the endpoints (i.e., one more $S$ than $T$).\n\nNow, if there is at least one of each, then Tasty removes a $TST$ from the end of such a zigzag while Stacy removes an $STS$ from the end of such a zigzag.\n\nOtherwise, suppose all zigzags have even size. Then Tasty finds any zigzag of length $\\ge 4$ (which must exist since the *average* zigzag length is 3) and removes $TST$ from the end containing $T$. The resulting merged zigzag is odd and hence $S$ endpoints, hence Stacy can move as well. $\\Box$\n\n**Remark.** There are many equivalent ways to phrase the solution. For example, the number of dividers is equal to the number of pairs of two consecutive letters (rather than singleton letters). So the win condition can also be phrased in terms of the number of adjacent pairs of letters being at least $2n$, or equivalently the number of differing pairs being at least $4n$.\n\n**Remark.** The constraint of no three consecutive identical beads is actually needed: a counterexample without this constraint is $TTSTSTTSSS$. (They win if Tasty goes first and lose if Stacy goes first.)\n\n**Remark.** Many contestants attempted induction. However, in doing so they often implicitly proved a different problem: “prove that if they can win with Tasty going first without ever creating a triplet, they can also win in such a way with Stacy going first”. This essentially means nearly all induction attempts fail.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15724,
    "subject": "Mathematics (Olympiad)",
    "question": "How many numbers divisible by $30^{2008}$ are not divisible by $20^{2007}$?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $30^{2008} = 2^{2008} \\cdot 3^{2008} \\cdot 5^{2008}$ and $20^{2007} = 2^{2007} \\cdot 5^{2007}$, all the numbers divisible by $30^{2008}$ and not divisible by $20^{2007}$ are:\n\n1) $2^{k} \\cdot 3^{l} \\cdot 5^{m}$, $l = 1, 2, \\ldots, 2008$, $k, m = 0, 1, 2, \\ldots, 2008$, or $2008 \\cdot 2009^2$ numbers,\n\n2) $2^k \\cdot 5^m$, $m = 2008$, $k = 0, 1, 2, \\ldots, 2008$, or $1 \\cdot 2009 = 2009$ numbers.\n\nFinally, there are $2008 \\cdot 2009^2 + 2009$ numbers divisible by $30^{2008}$ and not divisible by $20^{2007}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15725,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_{2023}$ be nonnegative real numbers such that $a_1 + a_2 + \\dots + a_{2023} = 100$.\nLet $N$ denote the number of elements in the set\n$$\n\\{(i, j) \\mid 1 \\le i \\le j \\le 2023,\\ a_i a_j \\ge 1\\}.\n$$\nProve that $N \\le 5050$, and determine the necessary and sufficient condition for $N = 5050$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S$ be the number of pairs $(i, j)$ with $1 \\le i < j \\le 2023$ such that $a_i a_j \\ge 1$.\nLet $T$ be the number of elements among $a_1, a_2, \\dots, a_{2023}$ that are not less than $1$. Then\n$$\n100 = a_1 + a_2 + \\dots + a_{2023} \\ge T,\n$$\n$$\n10000 = (a_1 + a_2 + \\dots + a_{2023})^2 = \\sum_{i=1}^{2023} a_i^2 + 2 \\sum_{1 \\le i < j \\le 2023} a_i a_j \\ge T + 2S.\n$$\nAdding these and dividing by $2$, we get $N = S + T \\le 5050$.\n\nIf equality holds, then each $a_i^2$ is either $0$ or $1$, so among $a_1, a_2, \\dots, a_{2023}$, exactly $100$ are $1$ and $1923$ are $0$. This is the necessary and sufficient condition for $N = 5050$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15726,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer such that $n + 10$ and $10n$ are both perfect squares. Find the smallest such $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be a positive integer such that $n + 10$ and $10n$ are both perfect squares. Then, there exists a positive integer $k$ such that $10n = k^2$. Since $k$ is a multiple of $2$ and $5$, we can write $k = 10\\ell$ for some positive integer $\\ell$, and we have $n = 10\\ell^2$. For $\\ell = 1$ or $\\ell = 2$, we have $n + 10 = 20$ or $50$, respectively, which are not perfect squares. Therefore, we must have $\\ell \\ge 3$, which implies $n = 10\\ell^2 \\ge 10 \\cdot 3^2 = 90$.\n\nOn the other hand, since we have $90 + 10 = 100 = 10^2$ and $90 \\cdot 10 = 900 = 30^2$, $90$ satisfies the condition. Therefore, the answer is $90$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15727,
    "subject": "Mathematics (Olympiad)",
    "question": "a. Find a 4-chunk (connected group of 4 squares) on a $100$-chart whose value (sum of its numbers) is odd. What is the minimum possible value, and what is one such chunk?\n\nb. Find all 3-chunks and 4-chunks on a $100$-chart whose value is $100$.\n\nc. What is the largest possible number of squares in a chunk on a $100$-chart whose value is $100$?\n\nd. For a rectangular 6-chunk (connected $1 \\times 6$, $2 \\times 3$, $3 \\times 2$, or $6 \\times 1$ shape) on a $100$-chart, is it possible for its value to be $100$? Justify your answer.",
    "options": [],
    "answer": "See solution",
    "solution": "a. A 4-chunk with odd value must have exactly one or exactly three odd numbers. This means the numbers cannot be consecutive. So the 4-chunk must occupy at least two rows of the $100$-chart. Hence its value is at least $11 + 1 + 2 + 3 = 17$.\n\nSo the required 4-chunk is $[1, 2, 3, 11]$.\n\n![]( <table><thead><tr><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th><th>8</th><th>9</th><th>10</th></tr></thead><tbody><tr><td>11</td><td>12</td><td>13</td><td>14</td><td>15</td><td>16</td><td>17</td><td>18</td><td>19</td><td>20</td></tr></tbody></table> )\n\nb. The only 3-chunk with value $100$ is $[27, 36, 37]$.\nThe only 4-chunk with value $100$ is $[10, 20, 30, 40]$.\n\n![]( <table><thead><tr><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th><th>8</th><th>9</th><th>10</th></tr></thead><tbody><tr><td>11</td><td>12</td><td>13</td><td>14</td><td>15</td><td>16</td><td>17</td><td>18</td><td>19</td><td>20</td></tr><tr><td>21</td><td>22</td><td>23</td><td>24</td><td>25</td><td>26</td><td>27</td><td>28</td><td>29</td><td>30</td></tr><tr><td>31</td><td>32</td><td>33</td><td>34</td><td>35</td><td>36</td><td>37</td><td>38</td><td>39</td><td>40</td></tr></tbody></table> )\n\nc. To find a chunk with value $100$ and a large number of squares, look for squares with low numbers. Here is a 13-chunk with value $100$.\n\n![]( <table><thead><tr><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th><th>8</th><th>9</th><th>10</th></tr></thead><tbody><tr><td>11</td><td>12</td><td>13</td><td>14</td><td>15</td><td>16</td><td>17</td><td>18</td><td>19</td><td>20</td></tr></tbody></table> )\n\nThe value of a 14-chunk is at least $1 + 2 + 3 + \\dots + 14 = 105$. So the largest number of squares in a chunk of value $100$ is $13$.\n\nd. There are only 4 shapes for a rectangular 6-chunk:\n1 row and 6 columns ($1 \\times 6$)\n2 rows and 3 columns ($2 \\times 3$)\n3 rows and 2 columns ($3 \\times 2$)\n6 rows and 1 column ($6 \\times 1$).\n\n*Alternative i*\n\nEach time a 6-chunk is moved one square horizontally, its value changes by $6$. Each time a 6-chunk is moved one square vertically, its value changes by $60$.\n\nThe 6-chunk $[1, 2, 3, 4, 5, 6]$ has value $21$, which is odd. So the value of every $1 \\times 6$ chunk is odd. Hence no $1 \\times 6$ chunk has value $100$.\n\nThe 6-chunk $[1, 2, 3, 11, 12, 13]$ has value $42$, which is a multiple of $3$. So the value of every $2 \\times 3$ chunk is a multiple of $3$. Hence no $2 \\times 3$ chunk has value $100$.\n\nThe 6-chunk $[1, 2, 11, 12, 21, 22]$ has value $69$, which is odd. So the value of every $3 \\times 2$ chunk is odd. Hence no $3 \\times 2$ chunk has value $100$.\n\nThe 6-chunk $[1, 11, 21, 31, 41, 51]$ has value $156$, which is a multiple of $3$. So the value of every $6 \\times 1$ chunk is a multiple of $3$. Hence no $6 \\times 1$ chunk has value $100$.\n\n*Alternative ii*\n\nA rectangular 6-chunk consists of two straight 3-chunks (side-by-side or end-to-end). The middle number of a straight 3-chunk is the average of its three numbers. So the value of a straight 3-chunk is $3$ times its middle number. Hence the value of a rectangular 6-chunk is a multiple of $3$. So no 6-chunk has value $100$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15728,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all odd integers $n$ such that the number of integers $k$ with $0 < k < \\frac{n}{4}$ and $\\gcd(n, k) = 1$ is odd.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that the only integers that work are prime powers $p^k$ where $p \\equiv 5$ or $7$ modulo $8$.\n\nDefine $\\omega$ and $\\Omega$ as functions from the set of odd integers to $\\{0, 1\\}$ by\n\n$$\n\\omega(n) = \\begin{cases} 0 & \\text{if } n \\equiv 1,3 \\pmod{8} \\\\ 1 & \\text{if } n \\equiv 5,7 \\pmod{8} \\end{cases}\n$$\n\nand\n\n$$\n\\Omega(n) \\equiv |\\{k \\mid 0 < k < n/4, \\gcd(k, n) = 1\\}| \\pmod{2}.\n$$\n\nWe seek $n$ such that $\\Omega(n) = 1$. It is easy to verify that\n\n$$\n\\omega(ab) \\equiv \\omega(a) + \\omega(b) \\pmod{2} \\qquad (\\dagger)\n$$\n\nLet $n = p_1^{a_1} p_2^{a_2} \\dots p_k^{a_k}$, where $p_1, \\dots, p_k$ are distinct primes and $a_1, \\dots, a_k$ are positive integers. Note that $n = 1$ does not work, so consider $k \\geq 1$.\n\n*Lemma:* \n$$\n\\Omega(n) \\equiv \\omega(n) + \\sum_{1 \\leq i \\leq k} \\omega\\left(\\frac{n}{p_i}\\right) + \\sum_{1 \\leq i < j \\leq k} \\omega\\left(\\frac{n}{p_i p_j}\\right) + \\dots + \\omega\\left(\\frac{n}{p_1 p_2 \\dots p_k}\\right) \\pmod{2}\n$$\n\n*Proof:* The idea is similar to the proof for the Euler totient function. $\\omega(n)$ gives the parity of $|\\{k \\mid 0 < k < n/4\\}| = \\lfloor n/4 \\rfloor$. To remove numbers divisible by some $p$, subtract $\\omega\\left(\\frac{n}{p}\\right)$. For numbers divisible by $pq$ ($p \\neq q$), add $\\omega\\left(\\frac{n}{pq}\\right)$, and so on, as in the principle of inclusion-exclusion. Since we work modulo $2$, all signs are $+$.\n\nApplying the lemma and $(\\dagger)$, we get:\n\n$$\n\\begin{aligned}\n& \\Omega(n) + \\sum_{1 \\leq i \\leq k} \\omega(p_i) + \\sum_{1 \\leq i < j \\leq k} \\omega(p_i p_j) + \\dots + \\omega(p_1 p_2 \\dots p_k) \\\\\n& \\equiv \\omega(n) + \\left( \\sum_{1 \\leq i \\leq k} \\omega\\left(\\frac{n}{p_i}\\right) + \\omega(p_i) \\right) + \\dots + \\left( \\omega\\left(\\frac{n}{p_1 \\dots p_k}\\right) + \\omega(p_1 \\dots p_k) \\right) \\\\\n& \\equiv \\binom{k}{0} \\omega(n) + \\binom{k}{1} \\omega(n) + \\dots + \\binom{k}{k} \\omega(n) \\\\\n& \\equiv 2^k \\omega(n) \\equiv 0 \\pmod{2}.\n\\end{aligned}\n$$\n\nThus,\n\n$$\n\\Omega(n) \\equiv \\omega(1) + \\sum_{1 \\leq i \\leq k} \\omega(p_i) + \\sum_{1 \\leq i < j \\leq k} \\omega(p_i p_j) + \\dots + \\omega(p_1 p_2 \\dots p_k) \\pmod{2}.\n$$\n\nThis simplifies to $2^{k-1} \\omega(p_1 p_2 \\dots p_k) \\pmod{2}$. Thus, if $k \\geq 2$, $\\Omega(n) = 0$. If $k = 1$, then $n = p^k$, and $\\Omega(n) = \\omega(p) = 1$ if and only if $p \\equiv 5$ or $7$ modulo $8$.\n\n$\\boxed{}$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15729,
    "subject": "Mathematics (Olympiad)",
    "question": "We define the number $A$ as the digit 4 followed by 2024 times the digit combination 84, and the number $B$ by 2024 times the digit combination 84 followed by a 7. So, to illustrate, we have\n\n$$\nA = 4\\underbrace{84\\dots84}_{4048\\text{ digits}} \\quad \\text{and} \\quad B = \\underbrace{84\\dots84}_{4048\\text{ digits}}7.\n$$\n\nSimplify the fraction $\\frac{A}{B}$ as much as possible.",
    "options": [],
    "answer": "See solution",
    "solution": "$\\frac{4}{7}$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 15730,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that the equality\n$$\nf([x]y) = f(x)[f(y)]\n$$\nholds for all $x, y \\in \\mathbb{R}$. (Here $[x]$ denotes the greatest integer less than or equal to $x$.)",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $f(x) = C$ (a constant function), where $C = 0$ or $1 \\leq C < 2$.\n\nSetting $x = 0$ in the given equation, we have\n$$\nf(0) = f(0)[f(y)]\n$$\nfor all $y \\in \\mathbb{R}$. We consider two cases:\n\n1. If $f(0) \\neq 0$, then $[f(y)] = 1$ for all $y$, so $f([x]y) = f(x)$. Setting $y = 0$ gives $f(x) = f(0) = C \\neq 0$. Since $[f(y)] = 1$, we have $1 \\leq C < 2$.\n\n2. If $f(0) = 0$, suppose there exists $0 < \\alpha < 1$ with $f(\\alpha) \\neq 0$. Setting $x = \\alpha$ in the original equation gives $f(y) = 0$ for all $y$, contradicting $f(\\alpha) \\neq 0$. Thus, $f(\\alpha) = 0$ for all $0 \\leq \\alpha < 1$. For any $z \\in \\mathbb{R}$, there is an integer $N$ such that $\\alpha = \\frac{z}{N} \\in [0,1)$, so\n$$\nf(z) = f([N]\\alpha) = f(N)[f(\\alpha)] = 0.\n$$\nThus, $f(x) = 0$ for all $x$.\n\nIt is easy to check that $f(x) = C$ with $C = 0$ or $1 \\leq C < 2$ satisfies the required property.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15731,
    "subject": "Mathematics (Olympiad)",
    "question": "Најди ги сите броеви $p$, $q$ и $r$, такви што $p$ и $r$ се прости броеви, $q$ е позитивен цел број, и ја задоволуваат равенката:\n\n$$\n(p+q+r)^2 = 2p^2 + 2q^2 + r^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "По средување на равенката добиваме $2r(p+q) = (p-q)^2$. Бидејќи $r$ е прост, следува дека $r$ е делител на $p-q$, па $r^2$ е делител на десната страна од последното равенство. Од каде следува дека $r$ е делител на $2(p+q)$. Ако $r > 2$, тогаш $r$ е делител на $(p+q)$, па мора $r$ да е делител и на $p$ и на $q$, но бидејќи $p$ е прост, тоа е можно само ако $p = r$ и $q = sr$. По средување добиваме $2(1+s) = (s-1)^2$, од каде $s^2 - 4s - 1 = 0$. Последното равенство нема целобројни решенија, па во овој случај равенката нема решение.\n\nАко $r = 2$, тогаш $p$ и $q$ се со иста парност, случајот кога $p = 2$ е невозможен исто како случајот $p = r$ од претходно, па мора да бидат непарни. Нека $a \\neq 2$ е прост делител на $p+q$, тогаш мора $a$ да е делител и на $p-q$, па мора да е делител и на $p$ и на $q$, што е можно само ако $p = a$ и $q = sa$, во овој случај добиваме $4(1+s) = a(s-1)^2$, од каде $a^2 - (2a + 4)s + (a - 4) = 0$, со решенија $\\frac{a+2 \\pm \\sqrt{a^2 + 4a + 4 - a^2 + 4a}}{a} = \\frac{a+2 \\pm 2\\sqrt{2a+1}}{a}$. Ако бројот $\\sqrt{2a+1}$ е цел, тогаш е непарен, па $2a+1 = 4b^2 + 4b + 1$, од каде $a = 2b(b+1)$, па не може да е прост.\n\nСпоред ова $p+q$ и $p-q$ мора да се степени на 2, то ест $p-q = 2^k$ и $p+q = 2^{2k-2}$, од каде $2p = 2^k + 2^{2k-2}$ и $2q = 2^{2k-2} - 2^k$ и бидејќи $p$ и $q$ се непарни, мора $k = 1$, но тогаш $p+q = 1$, што не е можно. Следува дека равенката нема решенија кои се прости броеви.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15732,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(xf(y) - f(x)) = 2f(x) + xy\n$$\nfor all real numbers $x, y$.",
    "options": [],
    "answer": "See solution",
    "solution": "By taking $x = 1$ in (1), we get\n$$\nf(f(y) - f(1)) = y + 2f(1), \\quad \\forall y \\in \\mathbb{R}. \\tag{2}\n$$\nHence $f$ is bijective, so there exists a unique real number $a$ such that $f(a) = 0$. Plugging $x = a$ into (1), we have\n$$\nf(af(y)) = ay, \\quad \\forall y \\in \\mathbb{R}. \\tag{3}\n$$\nPlugging $y = 0$ into (3), we have $f(af(0)) = 0 = f(a)$. Since $f$ is injective, we get $af(0) = a$, so $a = 0$ or $f(0) = 1$.\n\nConsider the case $a = 0$, i.e. $f(0) = 0$. Plugging $y = 0$ into (1), we have $f(-f(x)) = 2f(x)$. Since $f$ is surjective, we conclude that $f(x) = -2x$ for all $x \\in \\mathbb{R}$. But this function does not satisfy equation (1). Hence, $a \\neq 0$ and we get $f(0) = 1$.\n\nPlugging $x = 0$ into (1), we have $f(-1) = 2$. Plugging $y = a$ into (3), we have $a^2 = f(0) = 1$, i.e. $a = 1$ (because $f(-1) = 2$), so $f(1) = 0$.\n\nSince $f(1) = 1$, we can write equation (2) as\n$$\nf(f(y)) = y, \\quad \\forall y \\in \\mathbb{R}. \\tag{2'}\n$$\nBy plugging $f(y)$ instead of $y$ into (1) and using (2'), we get\n$$\nf(xy - f(x)) = 2f(x) + xf(y), \\quad \\forall x, y \\in \\mathbb{R}.\n$$\nIn this equation, consider $x \\neq 0$ and put $y = \\frac{f(x)}{x}$, we have\n$$\n1 = 2f(x) + x f\\left(\\frac{f(x)}{x}\\right),\n$$\nhence\n$$\nf\\left(\\frac{f(x)}{x}\\right) = \\frac{1 - 2f(x)}{x}, \\quad \\forall x \\neq 0.\n$$\nBy putting $y = \\frac{f(x)}{x}$ into (1) and using the above result, we have\n$$\nf(1 - 3f(x)) = 3f(x), \\quad \\forall x \\neq 0.\n$$\nNote that $f$ is bijective and $f(0) = 1$, so for all $x \\neq 0$, $1 - 3f(x)$ can get all real values except $-2$. Therefore, from the above result, we conclude that $f(x) = -x + 1$ for all $x \\neq -2$.\n\nIn particular, $f(3) = -2$. Plugging $y = 3$ into (2'), we have $f(-2) = 3$. Therefore, $f(x) = -x + 1$ for all $x \\in \\mathbb{R}$. It is easy to see that this function satisfies the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15733,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$, points $M$ and $P$ lie on segments $AB$ and $BC$, respectively, such that $AM = BC$ and $CP = BM$. If $AP$ and $CM$ meet at $O$ and $2\\angle AOM = \\angle ABC$, find the measure of $\\angle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $D$ be the reflection of $P$ across $C$, so $AB = BD$, and let $O'$ be the circumcenter of $\\triangle ABD$. As $\\triangle AO'B \\cong \\triangle BO'D$ and $MB = CD$, we have $\\angle O'CB = \\angle O'MA$, so $MBCO'$ is cyclic. Now $\\angle O'CM = \\angle O'BM = \\angle AOM$, hence $CO' \\parallel AP$. Therefore, $O'$ must be the midpoint of $AD$, implying that $\\angle ABC = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15734,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $a$, $b$, $c$ are three consecutive triangular numbers, with $b = T_n$ (the $n$th triangular number, $T_n = 1 + 2 + \\cdots + n$). Let $a = T_n - n$ and $c = T_n + n + 1$. If $a + b + c$ is itself a triangular number, show that $b$ must be three times a triangular number.",
    "options": [],
    "answer": "See solution",
    "solution": "The $n$th triangular number is given by\n\n$$\nT_n = \\frac{n(n+1)}{2}.\n$$\n\nGiven $a = T_n - n$, $b = T_n$, $c = T_n + n + 1$, so\n\n$$\na + b + c = (T_n - n) + T_n + (T_n + n + 1) = 3T_n + 1.\n$$\n\nSuppose $a + b + c = T_m$ for some $m$:\n\n$$\n3T_n + 1 = T_m \\implies 3\\frac{n(n+1)}{2} + 1 = \\frac{m(m+1)}{2}.\n$$\n\nMultiply both sides by 2:\n\n$$\n3n(n+1) + 2 = m(m+1)\n$$\n\nSo\n\n$$\n3n(n+1) = m^2 + m - 2 = (m-1)(m+2).\n$$\n\nSince $m-1$ and $m+2$ differ by 3, one of them must be divisible by 3. Let $m-1 = 3k$ for some integer $k$:\n\n$$\n(m-1)(m+2) = 3k(3k+3) = 9k(k+1)\n$$\n\nSo\n\n$$\n3n(n+1) = 9k(k+1) \\implies n(n+1) = 3k(k+1)\n$$\n\nTherefore,\n\n$$\nT_n = \\frac{n(n+1)}{2} = 3\\frac{k(k+1)}{2} = 3T_k.\n$$\n\nThus, $b = T_n = 3T_k$ as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15735,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Let $P(x)$ be a polynomial such that $P(\\alpha) = 0$, where $\\alpha$ is a root of $x^2 + x - 5 = 0$. When $P(x)$ is divided by $x^2 + x - 5$, the remainder is $R(x) = Ax + B$ for integers $A, B$.\n\nb) Find the minimal value of $c_0 + c_1 + \\dots + c_n$ among all nonnegative integer tuples $(c_0, c_1, \\dots, c_n)$ such that $c_0 + c_1\\alpha + \\dots + c_n\\alpha^n = 2015$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Since $P(\\alpha) = 0$ and $P(x) = Q(x)(x^2 + x - 5) + Ax + B$, substituting $x = \\alpha$ gives $0 = A\\alpha + B$. As $\\alpha$ is irrational, $A = B = 0$, so $P(x)$ is divisible by $x^2 + x - 5$.\n\nEvaluating at $x = 1$:\n$$\nP(1) = Q(1)(1^2 + 1 - 5) = Q(1)(-3)\n$$\nThus,\n$$\nc_0 + c_1 + \\dots + c_n = P(1) + 2015 \\equiv 2 \\pmod{3}\n$$\n\nb) To minimize $c_0 + c_1 + \\dots + c_n$ for $c_0 + c_1\\alpha + \\dots + c_n\\alpha^n = 2015$ with $c_i \\ge 0$, note that $0 \\le c_i \\le 4$ for $i = 0, \\dots, n-2$; otherwise, reducing $c_i$ by $5$ and increasing $c_{i+1}$ and $c_{i+2}$ by $1$ each yields a smaller sum.\n\nLet $Q(x) = a_{n-2}x^{n-2} + \\dots + a_0$ so that $P(x) = Q(x)(x^2 + x - 5)$. The coefficients satisfy:\n$$\n\\begin{aligned}\nc_0 - 2015 &= -5a_0 \\\\\nc_1 &= -5a_1 + a_0 \\\\\nc_2 &= -5a_2 + a_1 + a_0 \\\\\nc_3 &= -5a_3 + a_2 + a_1 \\\\\n\\dots\n\\end{aligned}\n$$\nGiven $c_i \\in \\{0,1,2,3,4\\}$, we find $c_0 = 0$, $a_0 = 403$, $c_1 = 3$, $a_1 = 80$, and recursively $c_{i+1} = \\text{MOD}(a_i + a_{i-1}, 5)$, $a_{i+1} = \\text{DIV}(a_i + a_{i-1}, 5)$.\n\nThe sequences are:\n$$\n\\begin{aligned}\n\\{a_0, a_1, \\dots, a_{11}\\} &= \\{403, 80, 96, 35, 26, 12, 7, 3, 2, 1, 0, 0\\} \\\\\n\\{c_0, c_1, \\dots, c_{11}\\} &= \\{0, 3, 3, 1, 1, 1, 3, 4, 0, 0, 3, 1\\}\n\\end{aligned}\n$$\n\nThus, the minimal value is:\n$$\n0 + 3 + 3 + 1 + 1 + 1 + 3 + 4 + 0 + 0 + 3 + 1 = 20.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15736,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AL$ be a bisector of triangle $ABC$. The circle centered at $B$ with radius $BL$ meets the ray $AL$ at point $E$, and the circle centered at $C$ with radius $CL$ meets the ray $AL$ at point $D$ (points $E$ and $D$ are different from point $L$). Prove that $AL^2 = AE \\cdot AD$.\n\n![](images/Ukraine_2021-2022_p10_data_740b4c2240.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, triangles $BEL$ and $CDL$ are isosceles, and angles $\\angle CLD$ and $\\angle BLE$ are vertical. Then $\\angle BEL = \\angle BLE = \\angle CLD = \\angle CDL$. Then $\\angle AEB = \\angle ALC$ as adjacent to equal angles. Also, $\\angle CAL = \\angle BAL$, as $AL$ is a bisector.\n\nNote that triangles $CLA$ and $BEA$ are similar by two angles. Then $\\frac{AL}{AE} = \\frac{AC}{AB}$, from where $AL = AE \\cdot \\frac{AC}{AB}$. Also, note that triangles $BLA$ and $ACD$ are similar by two angles. Then $\\frac{AL}{AD} = \\frac{AB}{AC}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15737,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y$ be positive real numbers such that\n\n$$\nx + y + xy = 3.\n$$\n\nProve that\n\n$$\nx + y \\ge 2.\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\n\\begin{aligned}\nx + y + xy &= 3 \\\\\nxy + x + y + 1 &= 4 \\\\\n(x + 1)(y + 1) &= 4.\n\\end{aligned}\n$$\n\nTherefore, we can rewrite the inequality as follows:\n\n$$\n\\begin{aligned}\nx + y &\\ge 2 \\\\\nx + 1 + y + 1 &\\ge 4 \\\\\nx + 1 + y + 1 &\\ge 2\\sqrt{(x + 1)(y + 1)}.\n\\end{aligned}\n$$\n\nThe last line follows from the AM-GM inequality.\n\nEquality holds when $x + 1 = y + 1$, i.e., $x = y$. Using $x + y = 2$, we get $x = y = 1$, which is admissible and thus the only case of equality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15738,
    "subject": "Mathematics (Olympiad)",
    "question": "Определи ги аглите на триаголникот чии страни $a$, $b$ и $c$ го задоволуваат равенството\n\n$$\na - 4\\sqrt{bc} + 2b = 2\\sqrt{ac} - 3c.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Равенството ќе го запишеме во облик\n\n$$\na - 2\\sqrt{ac} + c + 2c - 4\\sqrt{bc} + 2b = 0, \\\\\n(\\sqrt{a})^2 - 2\\sqrt{a}\\sqrt{c} + (\\sqrt{c})^2 + 2[(\\sqrt{b})^2 - 2\\sqrt{b}\\sqrt{c} + (\\sqrt{c})^2] = 0, \\\\\n(\\sqrt{a} - \\sqrt{c})^2 + 2(\\sqrt{b} - \\sqrt{c})^2 = 0.\n$$\n\nЗбир на два ненегативни броја е нула, само ако секој од нив е нула. Според тоа\n\n$$\n(\\sqrt{a} - \\sqrt{c})^2 = 0, \\qquad (1)\n$$\nи\n$$\n2(\\sqrt{b} - \\sqrt{c})^2 = 0.\n$$\n\nОд последното равенство добиваме\n\n$$\n(\\sqrt{b} - \\sqrt{c})^2 = 0. \\qquad (2)\n$$\n\nКвадрат на некој број е нула само ако самиот број е нула. Според тоа, од (1) и (2) добиваме $\\sqrt{a} - \\sqrt{c} = 0$ и $\\sqrt{b} - \\sqrt{c} = 0$, односно $\\sqrt{a} = \\sqrt{c}$ и $\\sqrt{b} = \\sqrt{c}$. Заради последните равенства, имаме $a = c = b$. Значи, триаголникот е рамностран, па неговите агли се еднакви меѓу себе.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15739,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle and let $X$ be the foot of the altitude from $A$. Let $Y$ be the intersection of the perpendicular to $AC$ drawn from $X$. If the circumcircle of triangle $ABX$ meets $BY$ at point $Z$ (distinct from $B$), and the extension of $AZ$ meets $XY$ at point $P$, prove that\n\n$$\nBX \\cdot XP = PY \\cdot XC\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\triangle AXY$ and $\\triangle AXC$ are both right triangles at $Y$ and $X$ respectively, $\\angle AXP = 90^\\circ - \\angle XAC = \\angle YCB$. On the other hand, since $AZXB$ is a cyclic quadrilateral, $\\angle XAP = \\angle YBC$.\n\n![](images/Spanija_b_2014_p27_data_56c6bdb055.png)\n\nThus, $\\triangle AXP \\sim \\triangle BCY$ and $\\frac{BC}{AX} = \\frac{YC}{XP}$. Likewise, $\\triangle AXC \\sim \\triangle XYC$ because both are right-angled triangles with a common acute angle, so\n\n$$\n\\frac{AX}{XC} = \\frac{XY}{YC}\n$$\n\nFrom the preceding ratios we get\n\n$$\nBC \\cdot XP = AX \\cdot YC = XC \\cdot XY\n$$\n\nfrom which follows\n\n$$\n\\frac{BC}{XC} = \\frac{XY}{XP}\n$$\n\nSubtracting $1$ from both sides yields\n\n$$\n\\frac{BC}{XC} - 1 = \\frac{XY}{XP} - 1 \\Leftrightarrow \\frac{BC - XC}{XC} = \\frac{XY - XP}{XP} \\Leftrightarrow \\frac{BX}{XC} = \\frac{PY}{XP}\n$$\n\nsince $BC = BX + XC$ and $XY = XP + PY$ respectively.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15740,
    "subject": "Mathematics (Olympiad)",
    "question": "A graph is symmetric about a line if the graph remains unchanged after reflection in that line. For how many quadruples of integers $ (a, b, c, d) $, where $ |a|, |b|, |c|, |d| \\le 5 $ and $c$ and $d$ are not both $0$, is the graph of\n\n$$\ny = \\frac{ax + b}{cx + d}\n$$\n\nsymmetric about the line $y = x$?",
    "options": [],
    "answer": "See solution",
    "solution": "There are two cases, depending on whether the denominator of the fraction is constant.\n\n**Case 1:** Suppose that $c = 0$. Then the graph of $y = \\frac{ax+b}{d}$ is a line. It is symmetric with respect to the line $y = x$ if the slope of the line is $1$ and the line passes through the origin, or the slope is $-1$.\n\n- A line with slope $1$ that passes through the origin occurs if $a = d \\neq 0$ and $b = 0$. There are $10$ possible values of $a$.\n- The slope is $-1$ if $d = -a \\neq 0$. There are $10$ possible values of $a$, and $11$ possible values of $b$, which gives $110$ possibilities.\n\nThis gives a total of $120$ possibilities in Case 1.\n\n**Case 2:** Suppose that $c \\neq 0$. The quantity $y = \\frac{ax+b}{cx+d}$ tends to $\\frac{a}{c}$ as $x \\to \\pm\\infty$. The degree of the polynomial in the numerator can be reduced by subtracting this asymptotic value:\n\n$$\ny - \\frac{a}{c} = \\frac{ax + b}{cx + d} - \\frac{a}{c} = \\frac{(ax + b)c - a(cx + d)}{c(cx + d)} = \\frac{bc - ad}{c(cx + d)}\n$$\n\nLet $\\Delta = bc - ad$. If $\\Delta = 0$, then the graph consists of the points $(x, \\frac{a}{c})$, except that the point $x = -\\frac{d}{c}$ is missing. This set is not symmetric with respect to $y = x$. So assume $\\Delta \\neq 0$.\n\nThe set of pairs $(x, y)$ that satisfy the equation is a rectangular hyperbola. This hyperbola is symmetric with respect to the line $y = x$ if and only if its center $\\left(\\frac{a}{c}, -\\frac{d}{c}\\right)$ lies on this line, i.e., $d = -a$. Thus $\\Delta \\neq 0$ if and only if $bc + a^2 \\neq 0$.\n\n- If $a = 0$, then $b$ and $c$ are free to take on any nonzero values, which gives $10 \\times 10 = 100$ quadruples $(0, b, c, 0)$.\n- If $a \\neq 0$, then $d = -a$ and $c \\neq 0$, so there are $10 \\times 11 \\times 10 = 1100$ quadruples $(a, b, c, -a)$ to consider. But $bc + a^2 = 0$ if and only if either $|a| = |b| = |c|$ and $b$ and $c$ have opposite signs, which is $10 \\times 2 = 20$ cases, or $|a| = 2$ and $(b, c) = (\\pm 4, \\mp 1)$ or $(\\pm 1, \\mp 4)$, which is $2 \\times 2 \\times 2 = 8$ additional cases.\n\nThus, the total number of possibilities in Case 2 is $100 + 1100 - (20 + 8) = 1172$.\n\nIn all, there are $120 + 1172 = 1292$ quadruples satisfying the given conditions.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 15741,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with circumcenter $O$. Point $X$ is the intersection of the line through $O$ parallel to $AB$ and the line through $C$ perpendicular to $AC$. Let $Y$ be the intersection of the external bisector of $\\angle BXC$ with $AC$. Let $K$ be the projection of $X$ onto $BY$.\n\nProve that the lines $AK$, $XO$, and $BC$ have a common point.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove that the quadrilateral $BOCX$ is cyclic. Indeed,\n\n$$\n\\angle XCB = 90^\\circ - \\angle ACB = \\angle ABO = \\angle BOX.\n$$\n\nThis means that $OX$ is the internal bisector of $\\angle BXC$ (since $OB = OC$), so $OX \\perp XY$. Next, observe that the quadrilateral $ABXY$ is also cyclic. Indeed,\n\n$$\n\\angle BXY = 90^\\circ + \\frac{\\angle BXC}{2} = 90^\\circ + \\frac{180^\\circ - \\angle BOC}{2} = 180^\\circ - \\frac{\\angle BOC}{2} = 180^\\circ - \\angle BAC.\n$$\n\nLet $Z$ be the intersection of $XY$ and $AB$. Then, $\\angle XZB = 90^\\circ$.\n\n![](images/BMO_2023_Short_List_p25_data_a10fe8c4c8.png)\n\n**First Way.** In triangle $ABY$, $X$ lies on its circumcircle, and $Z$, $K$, $C$ are the projections of $X$ onto $AB$, $BY$, $AY$, respectively, so $Z$, $K$, $C$ lie on the Simson line of $X$ with respect to the circumcircle of $ABY$. Let $S$ be the intersection of $AK$ and $BC$, and $T$ the intersection of $XY$ and $BC$. Then, $(B, C; S, T)$ is harmonic (since in the complete quadrilateral $ABKCZY$, $S$ and $T$ are the intersections of $ZY$ and $AK$ with $BC$, respectively). Since $XT$ is the external bisector of $\\angle BXC$, $XS$ is the internal bisector, so $S$ lies on $OX$, completing the proof.\n\n**Second Way.** Note that $\\angle XKY = \\angle BKX = \\angle BXZ = \\angle XCY = 90^\\circ$, so $K$ lies on both the circumcircle of $XBZ$ and $XCY$. Thus, $K$ is the Miquel point of $B$, $C$, $X$ in triangle $AYZ$, so $K$ lies on the circumcircle of $ABC$. Since $\\angle KXC = \\angle KYC = \\angle AYB = \\angle AXB$, we have that $XO$ is the bisector of $\\angle AXK$ as well, so\n\n$$\n\\angle AXK = 2 \\cdot \\angle OXA = 2 \\cdot \\angle BAX = 2 \\cdot \\angle BYX = 2 \\cdot \\angle BYZ.\n$$\n\nOn the other hand,\n\n$$\n\\angle AOK = 360^\\circ - 2 \\cdot \\angle ABK = 360^\\circ - 2 \\cdot \\angle ABY = 360^\\circ - 2 \\cdot (90^\\circ + \\angle BYZ).\n$$\n\nTherefore, $\\angle AXK + \\angle AOK = 180^\\circ$, so $AOKX$ is cyclic. This means that $AK$, $XO$, and $BC$ are the common chords of the three circles ($AOKX$), ($ABKC$), and ($BCOX$), so they all pass through the radical center of those circles.\n\n![](images/BMO_2023_Short_List_p26_data_e8e9e02009.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15742,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $H$ be the orthocentre and $G$ be the centroid of acute-angled triangle $\\triangle ABC$ with $AB \\ne AC$. The line $AG$ intersects the circumcircle of $\\triangle ABC$ at $A$ and $P$. Let $P'$ be the reflection of $P$ in the line $BC$. Prove that $\\angle CAB = 60^\\circ$ if and only if $HG = GP'.$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\omega$ be the circumcircle of $\\triangle ABC$. Reflecting $\\omega$ in line $BC$, we obtain circle $\\omega'$ which contains points $H$ and $P'$. Let $M$ be the midpoint of $BC$. As $\\triangle ABC$ is acute-angled, $H$ and $O$ lie inside this triangle.\n\nAssume $\\angle CAB = 60^\\circ$. Since\n\n$$\n\\angle COB = 2\\angle CAB = 120^\\circ = 180^\\circ - 60^\\circ = 180^\\circ - \\angle CAB = \\angle CHB,\n$$\n\n$O$ lies on $\\omega'$. Reflecting $O$ in line $BC$, we obtain $O'$ which lies on $\\omega$ and is the center of $\\omega'$. Then $OO' = 2OM = 2R\\cos\\angle CAB = AH$, so $AH = OO' = HO' = AO = R$, where $R$ is the radius of $\\omega$ and $\\omega'$. Thus, quadrilateral $AHO'O$ is a rhombus, so $A$ and $O'$ are symmetric with respect to $HO$. As $H, G,$ and $O$ are collinear (Euler line), $\\angle GAH = \\angle HO'G$. The diagonals of quadrilateral $GOPO'$ intersect at $M$. Since $\\angle BOM = 60^\\circ$,\n\n$$\nOM = MO' = \\cot 60^\\circ \\cdot MB = \\frac{MB}{\\sqrt{3}}.\n$$\n\nAs $3 \\cdot MO \\cdot MO' = MB^2 = MB \\cdot MC = MP \\cdot MA = 3MG \\cdot MP$, $GOPO'$ is cyclic. Since $BC$ is a perpendicular bisector of $OO'$, the circumcircle of $GOPO'$ is symmetric with respect to $BC$. Thus $P'$ also belongs to the circumcircle of $GOPO'$, so $\\angle GO'P' = \\angle GPP'$. Note that $\\angle GPP' = \\angle GAH$ since $AH \\parallel PP'$. As proved, $\\angle GAH = \\angle HO'G$, so $\\angle HO'G = \\angle GO'P'$. Thus, triangles $\\triangle HO'G$ and $\\triangle GO'P'$ are congruent, hence $HG = GP'.$\n\nNow, suppose $HG = GP'$. Reflecting $A$ with respect to $M$, we get $A'$. As in the first part, points $B, C, H,$ and $P'$ belong to $\\omega'$. Also, $A'$ belongs to $\\omega'$. Note that $HC \\perp CA'$ since $AB \\parallel CA'$, so $HA'$ is a diameter of $\\omega'$. The center $O'$ of $\\omega'$ is the midpoint of $HA'$. From $HG = GP'$, $\\triangle HGO'$ is congruent to $\\triangle P'GO'$. Therefore, $H$ and $P'$ are symmetric with respect to $GO'$. Hence $GO' \\perp HP'$ and $GO' \\parallel A'P'$. Let $HG$ intersect $A'P'$ at $K$ and $K \\notin O$ since $AB \\ne AC$. We conclude $HG = GK$, because $GO'$ is the midline of $\\triangle HKA'$. Note that $2GO = HG$, since $HO$ is the Euler line of $\\triangle ABC$. So $O$ is the midpoint of $GK$. Because $\\angle CMP = \\angle CMP'$, $\\angle GMO = \\angle OMP'$. Line $OM$, passing through $O'$, is an external angle bisector of $\\angle P'MA'$. Also, $O'$ is the midpoint of arc $P'MA'$. It follows that quadrilateral $P'MO'A'$ is cyclic, so $\\angle O'MA' = \\angle O'P'A' = \\angle O'A'P'$. Let $OM$ and $P'A'$ intersect at $T$. Triangles $\\triangle TO'A'$ and $\\triangle A'O'M$ are similar, so $\\frac{O'A'}{O'M} = \\frac{O'T}{O'A'}$, i.e., $O'M \\cdot O'T = O'A'^2$.\n\nUsing Menelaus' theorem for $\\triangle HKA'$ and line $TO'$, we obtain\n\n$$\n\\frac{A'O'}{O'H} \\cdot \\frac{HO}{OK} \\cdot \\frac{KT}{TA'} = 3 \\cdot \\frac{KT}{TA'} = 1.\n$$\n\nSo $\\frac{KT}{TA'} = \\frac{1}{3}$ and $KA' = 2KT$. Using Menelaus' theorem for $\\triangle TO'A'$ and line $HK$,\n\n$$\n1 = \\frac{O'H}{HA'} \\cdot \\frac{A'K}{KT} \\cdot \\frac{TO}{OO'} = \\frac{1}{2} \\cdot 2 \\cdot \\frac{TO}{OO'} = \\frac{TO}{OO'}.\n$$\n\nThus $TO = OO'$, so $O'A'^2 = O'M \\cdot O'T = OO'^2$. Hence $O'A' = OO'$ and consequently, $O \\in \\omega'$. Finally, $2\\angle CAB = \\angle BOC = 180^\\circ - \\angle CAB$, so $\\angle CAB = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15743,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer, let $S_n$ be the set of all permutations of the set $\\{1, 2, \\dots, n\\}$, and, for each $\\sigma$ in $S_n$, let $I(\\sigma) = \\{i : \\sigma(i) \\le i\\}$. Evaluate the sum\n\n$$\n\\sum_{\\sigma \\in S_n} \\frac{1}{|I(\\sigma)|} \\sum_{i \\in I(\\sigma)} (i + \\sigma(i)).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the involution of $S_n$ which sends a permutation $\\sigma$ to the permutation $\\sigma^*$ defined by $\\sigma^*(i) = j$ if and only if $\\sigma(n - j + 1) = n - i + 1$. Notice that, for each $\\sigma$ in $S_n$, the assignment $i \\mapsto n - \\sigma(i) + 1$ defines a bijection from $I(\\sigma)$ to $I(\\sigma^*)$ whose inverse sends $j$ to $n - \\sigma^*(j) + 1$. Consequently,\n\n$$\n\\begin{align*}\n\\sum_{\\sigma \\in S_n} \\frac{1}{|I(\\sigma)|} \\sum_{i \\in I(\\sigma)} (i + \\sigma(i)) &= \\\\\n&= \\frac{1}{2} \\sum_{\\sigma \\in S_n} \\left( \\frac{1}{|I(\\sigma)|} \\sum_{i \\in I(\\sigma)} (i + \\sigma(i)) + \\frac{1}{|I(\\sigma^*)|} \\sum_{i \\in I(\\sigma^*)} (i + \\sigma^*(i)) \\right) \\\\\n&= \\frac{1}{2} \\sum_{\\sigma \\in S_n} \\frac{1}{|I(\\sigma)|} \\sum_{i \\in I(\\sigma)} \\big( i + \\sigma(i) + (n - \\sigma(i) + 1) + (n - i + 1) \\big) \\\\\n&= (n + 1)! .\n\\end{align*}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 15744,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(2x f(x) - 2f(y)) = 2x^2 - y - f(y)\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there exist $y_1, y_2$ such that $f(y_1) = f(y_2)$. Substitute $(x, y) \\to (x, y_1)$ and $(x, y_2)$, and by comparing, we get $y_1 = y_2$. Thus, $f(x)$ is injective.\n\nLet $x \\to -x$ in the given equation:\n$$\nf(-2x f(-x) - 2f(y)) = 2x^2 - y - f(y)\n$$\nThus,\n$$\nf(2x f(x) - 2f(y)) = f(-2x f(-x) - 2f(y))\n$$\nfor all $x, y$. Due to injectivity, $x f(x) = -x f(-x)$, so $f(-x) = -f(x)$ for all $x \\neq 0$.\n\nLet $x = y = 0$ in the given, we have $f(-2f(0)) = -f(0)$. Let $a = f(0)$, then $f(-2a) = -a$.\n\nLet $x = 0$, $y = -2a$, we have $f(-2f(-2a)) = 2a - f(-2a)$, thus $-a = 2a - (-a)$, which implies $a = 0$. Thus, $f(0) = 0$ and $f(-x) = -f(x)$ for all $x \\in \\mathbb{R}$.\n\nLet $y = 2x^2$ in the given condition:\n$$\nf(2x f(x) - 2f(2x^2)) = -f(2x^2)\n$$\nThus,\n$$\nf(2x f(x) - 2f(2x^2)) = f(-2x^2)\n$$\nSo $x f(x) - f(2x^2) = -x^2$, or\n$$\nf(2x^2) = x f(x) + x^2, \\quad \\forall x. \\tag{1}\n$$\n\nLet $y = 0$ in the given condition:\n$$\nf(2x f(x)) = 2x^2\n$$\nThus, $f$ is surjective on $\\mathbb{R}^+$, and also on $\\mathbb{R}$ since $f$ is odd.\n\nPut $x = 0$ in the given:\n$$\nf(-2f(y)) = -y - f(y)\n$$\nSo,\n$$\nf(2f(y)) = y + f(y), \\quad \\forall y. \\tag{2}\n$$\n\nContinue by putting $y = 2x f(x)$ into (2):\n$$\nf(4x^2) = 2x f(x) + 2x^2\n$$\nCombining with (1):\n$$\nf(4x^2) = 2f(2x^2) \\implies f(2x) = 2f(x), \\quad \\forall x \\ge 0.\n$$\nSince $f$ is odd, $f(2x) = 2f(x)$ for all $x \\in \\mathbb{R}$.\n\nFrom (1):\n$$\n2f(x^2) = x f(x) + x^2\n$$\nLet $x = 1$, we get $f(1) = 1$.\n\nFrom (2):\n$$\ny + f(y) = 2f(f(y)), \\quad \\forall y \\in \\mathbb{R}. \\tag{3}\n$$\n\nPut (3) back into the given condition:\n$$\n2f(x f(x) - f(y)) = 2x^2 - 2f(f(y))\n$$\nOr,\n$$\nf(x f(x) - f(y)) = x^2 - f(f(y))\n$$\nSince $f$ is surjective:\n$$\nf(x f(x) + y) = x^2 + f(y), \\quad \\forall x, y \\in \\mathbb{R}. \\tag{4}\n$$\n\nIn (3), let $x = 1$, then $f(y + 1) = f(y) + 1$ for all $y$. In (4), let $x \\to x + 1$:\n$$\nf(x f(x) + f(x) + x + 1 + y) = (x + 1)^2 + f(y)\n$$\nOr,\n$$\nx^2 + f(f(x) + x + y) + 1 = (x + 1)^2 + f(y)\n$$\nThus, $f(f(x) + x + y) = 2x + f(y)$, and by letting $y = 0$, $f(f(x) + x) = 2x$. Thus, $f(2x) = f(f(f(x) + x))$. Combine with (3):\n$$\n2f(x) = \\frac{f(x) + x + f(f(x) + x)}{2}\n$$\nThus, $4f(x) = f(x) + x + 2x$, or $f(x) = x$ for all $x$.\n\nIt is easy to check that this satisfies the given condition. Hence, $f(x) = x$ for all $x \\in \\mathbb{R}$.\n\n$\\boxed{f(x) = x}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15745,
    "subject": "Mathematics (Olympiad)",
    "question": "Using Ceva's theorem in triangle $\\triangle BOC$ for point $I$, and the Ratio lemma, prove the following trigonometric identity:\n\n$$\n\\frac{c}{b} \\cdot \\frac{\\sin(90^\\circ - \\frac{\\angle B}{2})}{\\sin(90^\\circ - \\frac{\\angle C}{2})} = \\frac{\\sin(\\frac{\\angle C}{2})}{\\sin(\\frac{\\angle B}{2})}\n$$\n\nor equivalently,\n\n$$\n\\frac{\\sin \\angle C}{\\sin \\angle B} \\cdot \\frac{\\cos(\\frac{\\angle B}{2})}{\\cos(\\frac{\\angle C}{2})} = \\frac{\\sin(\\frac{\\angle C}{2})}{\\sin(\\frac{\\angle B}{2})}.\n$$\n\n% ![](images/IRN_ABooklet_2020_p38_data_53bd12ba29.png)\n\n% ![](images/IRN_ABooklet_2020_p38_data_9b990e7a23.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n\n$$\n\\frac{XB}{XC} = \\frac{\\sin \\angle XOB}{\\sin \\angle XOC} = \\frac{\\sin \\angle IOB}{\\sin \\angle IOC}.\n$$\n\nWhich is equivalent to\n\n$$\n\\frac{\\sin\\left(\\frac{\\angle C}{2}\\right)}{\\sin\\left(\\frac{|\\angle B - \\angle A|}{2}\\right)} \\cdot \\frac{\\sin\\left(\\frac{|\\angle B - \\angle A|}{2}\\right)}{\\sin\\left(\\frac{\\angle C}{2}\\right)}.\n$$\n\nBy the Ratio lemma, we have\n\n$$\n\\frac{YB}{YC} = \\frac{c}{b} \\cdot \\frac{\\sin \\angle YAB}{\\sin \\angle YAC} = \\frac{c}{b} \\cdot \\frac{\\sin \\angle KAF}{\\sin \\angle KAE}.\n$$\n\nBy Ceva's theorem in $\\triangle AEF$ for point $K$,\n\n$$\n\\frac{\\sin \\angle KAF}{\\sin \\angle KAE} = \\frac{\\sin \\angle KEF}{\\sin \\angle KEA} \\cdot \\frac{\\sin \\angle KFA}{\\sin \\angle KFE} = \\frac{\\sin(90^\\circ - \\frac{\\angle B}{2})}{\\sin(90^\\circ - \\frac{\\angle C}{2})} \\cdot \\frac{\\sin(\\frac{\\angle B}{2})}{\\sin(\\frac{\\angle C}{2})}.\n$$\n\nSo we have just to prove that\n\n$$\n\\frac{c}{b} \\cdot \\frac{\\sin(90^\\circ - \\frac{\\angle B}{2})}{\\sin(90^\\circ - \\frac{\\angle C}{2})} = \\frac{\\sin(\\frac{\\angle C}{2})}{\\sin(\\frac{\\angle B}{2})}\n$$\n\nwhich is equivalent to\n\n$$\n\\frac{\\sin \\angle C}{\\sin \\angle B} \\cdot \\frac{\\cos(\\frac{\\angle B}{2})}{\\cos(\\frac{\\angle C}{2})} = \\frac{\\sin(\\frac{\\angle C}{2})}{\\sin(\\frac{\\angle B}{2})}.\n$$\n\nThis is true since $\\sin x = 2 \\sin(\\frac{x}{2}) \\cos(\\frac{x}{2})$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15746,
    "subject": "Mathematics (Olympiad)",
    "question": "In $\\triangle ABC$ with $\\angle ACB = 60^\\circ$, let $AA_1$ and $BB_1$ ($A_1 \\in BC$, $B_1 \\in AC$) be the bisectors of $\\angle BAC$ and $\\angle ABC$. The line $A_1B_1$ meets the circumcircle of $\\triangle ABC$ at points $A_2$ and $B_2$.\n\n**a)** If $O$ and $I$ are the circumcenter and the incenter of $\\triangle ABC$, prove that $OI$ is parallel to $A_1B_1$.\n\n**b)** If $R$ is the midpoint of the arc $\\widehat{AB}$ not containing point $C$, and $P$ and $Q$ are the midpoints of $A_1B_1$ and $A_2B_2$, respectively, prove that $RP = RQ$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Since $\\angle AOB = 2\\gamma = 120^{\\circ}$ and points $A$, $O$, $I$, and $B$ lie on a circle, $\\alpha + \\beta$.\n\n![](images/broshura_07_english_p6_data_54f6af1c04.png)\n\nb) Since $OQ \\perp A_2B_2$, $IP \\perp A_2B_2$ ($\\triangle A_1IB_1$ is isosceles), and $OI \\parallel A_2B_2$, we have that $OIPQ$ is a rectangle. The perpendicular bisector of $OI$ is also the perpendicular bisector of $PQ$. Using that $R$ lies on the perpendicular bisector of $OI$, it follows that $R$ lies on the perpendicular bisector of $PQ$, i.e., $RP = RQ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15747,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that if $p$ is a prime number, then $7p + 3^p - 4$ is not a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "For $p=2$, we have $m=19$, and for $p=3$, we have $m=44$; neither is a perfect square.\n\nAssume that for a prime number $p > 3$, the integer $7p + 3^p - 4$ is a perfect square. Let $m = n^2$ for some $n \\in \\mathbb{Z}$.\n\nBy Fermat's Little Theorem:\n\n$$\nm = 7p + 3^p - 4 \\equiv 3 - 4 \\equiv -1 \\pmod{p}.\n$$\n\nConsider the cases for $p$ modulo 4:\n\n* If $p = 4k+3$, $k \\in \\mathbb{Z}$, then again by Fermat's Little Theorem:\n\n$$\n-1 \\equiv m^{2k+1} \\equiv n^{4k+2} \\equiv n^{p-1} \\equiv 1 \\pmod{p},\n$$\n\nwhich is impossible since $p > 3$.\n\n* If $p = 4k+1$, $k \\in \\mathbb{Z}$, then $m = 7p + 3^p - 4 \\equiv 3 - 1 \\equiv 2 \\pmod{4}$. But this is a contradiction, since a perfect square is never congruent to 2 modulo 4.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15748,
    "subject": "Mathematics (Olympiad)",
    "question": "Amy has divided a square into finitely many white and red rectangles, each with sides parallel to the sides of the square. Within each white rectangle, she writes down its width divided by its height. Within each red rectangle, she writes down its height divided by its width. Finally, she calculates $x$, the sum of these numbers. If the total area of the white rectangles equals the total area of the red rectangles, what is the smallest possible value of $x$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_i$ and $b_i$ denote the width and height of each white rectangle, and let $c_i$ and $d_i$ denote the width and height of each red rectangle. Also, let $L$ denote the side length of the original square.\n\n*Lemma:* Either $\\sum a_i \\ge L$ or $\\sum d_i \\ge L$.\n\n*Proof of lemma:* Suppose there exists a horizontal line across the square that is covered entirely with white rectangles. Then, the total width of these rectangles is at least $L$, and the claim is proven. Otherwise, there is a red rectangle intersecting every horizontal line, and hence the total height of these rectangles is at least $L$. $\\Box$\n\nNow, let us assume without loss of generality that $\\sum a_i \\ge L$. By the Cauchy-Schwarz inequality,\n\n$$\n\\left(\\sum \\frac{a_i}{b_i}\\right) \\cdot \\left(\\sum a_i b_i\\right) \\ge \\left(\\sum a_i\\right)^2 \\ge L^2.\n$$\n\nBut we know $\\sum a_i b_i = \\frac{L^2}{2}$, so it follows that $\\sum \\frac{a_i}{b_i} \\ge 2$. Furthermore, each $c_i \\le L$, so\n\n$$\n\\sum \\frac{d_i}{c_i} \\ge \\frac{1}{L^2} \\cdot \\sum c_i d_i = \\frac{1}{2}.\n$$\n\nTherefore, $x$ is at least $2.5$. Conversely, $x = 2.5$ can be achieved by making the top half of the square one colour, and the bottom half the other colour. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15749,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $r$ for which there exists exactly one real number $a$ such that when\n\n$$\n(x + a)(x^2 + r x + 1)\n$$\n\nis expanded to yield a cubic polynomial, all of its coefficients are greater than or equal to zero.",
    "options": [],
    "answer": "See solution",
    "solution": "Expanding $(x + a)(x^2 + r x + 1)$ yields\n\n$$\nx^3 + (a + r)x^2 + (1 + a r)x + a.\n$$\n\nWe seek all real numbers $r$ such that there is exactly one real number $a$ satisfying the following system of inequalities:\n\n$$\na + r \\ge 0 \\quad (1)\n$$\n$$\na r \\ge -1 \\quad (2)\n$$\n$$\na \\ge 0 \\quad (3)\n$$\n\nIf $r \\ge 0$, then any $a \\ge 0$ satisfies the above inequalities, so there is not exactly one $a$.\n\nIf $r < 0$, let $r = -s$ for $s > 0$. The inequalities become:\n\n$$\na \\ge s \\quad (1')\n$$\n$$\na \\le \\frac{1}{s} \\quad (2')\n$$\n$$\na \\ge 0 \\quad (3')\n$$\n\nInequality (3') follows from (1'), so we only need (1') and (2'), which give:\n\n$$\ns \\le a \\le \\frac{1}{s}.\n$$\n\nThere is exactly one value of $a$ satisfying this if and only if $s = \\frac{1}{s}$, so $s = 1$ (and $a = 1$), and thus $r = -1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15750,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABIH$, $BDEC$, and $ACFG$ be arbitrary rectangles constructed externally on the sides of triangle $ABC$. Choose point $S$ outside rectangle $ABIH$ (on the opposite side as triangle $ABC$) such that $\\angle SHI = \\angle FAC$ and $\\angle HIS = \\angle EBC$. Prove that the lines $FI$, $EH$, and $CS$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $T = EH \\cap FI$. Let $K$ and $L$ be the feet of the perpendiculars from $A$ to $FI$ and from $B$ to $EH$, respectively. Then $K$ and $L$ are on the circumcircle $\\Gamma$ of $ABIH$. Note that $K$ is also on the circumcircle of $ACFG$ and $L$ is also on the circumcircle of $BDEC$. Let $CK$ extended intersect $\\Gamma$ in $M$, and let $CL$ extended intersect $\\Gamma$ in $N$. Then the lines $HM$ and $IN$ intersect in $S$. $[\\angle EBC = \\angle ELC = \\angle HLN = \\angle HIN]$. But $\\angle HIS = \\angle EBC$, so $S$ is on $IN$. Similarly, $S$ is on $HM$.\n\n![](images/s3s2023_p4_data_d591b0ac88.png)\n\nConsider the self-crossing hexagon $IKMHLN$ inscribed in the circle $\\Gamma$. By Pascal's Theorem, the points $IK \\cap HL = T$, $KM \\cap LN = C$, and $MH \\cap NI = S$ are collinear. This shows that $T$ is a point that belongs to all three lines $FI$, $EH$ and $CS$, i.e., these lines are concurrent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15751,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all injective functions $f$ from the set of positive integers into itself such that if $S$ is a finite set of positive integers with $\\sum_{s \\in S} \\frac{1}{s}$ an integer, then $\\sum_{s \\in S} \\frac{1}{f(s)}$ is also an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that the identity function is the unique solution.\n\nClearly, $f(1) = 1$. By injectivity, $f(n) \\geq 2$ for $n \\geq 2$.\n\n**Egyptian fractions theorem:** For every positive rational $r$ and positive integer $n$, there exists a finite set $S$ of integers greater than $n$ such that $r = \\sum_{s \\in S} \\frac{1}{s}$.\n\nLet $n \\geq 2$. By the theorem, write\n$$\n1 - \\frac{1}{n} = \\sum_{s \\in S} \\frac{1}{s}, \\quad \\text{where } S \\text{ is a finite set of integers greater than } n(n+1).\n$$\nThen,\n$$\n1 = \\frac{1}{n} + \\sum_{s \\in S} \\frac{1}{s} = \\frac{1}{n+1} + \\frac{1}{n(n+1)} + \\sum_{s \\in S} \\frac{1}{s}.\n$$\nThus,\n$$\n\\frac{1}{f(n)} + \\sum_{s \\in S} \\frac{1}{f(s)} \\quad \\text{and} \\quad \\frac{1}{f(n+1)} + \\frac{1}{f(n(n+1))} + \\sum_{s \\in S} \\frac{1}{f(s)}\n$$\nare both integers, so\n$$\n\\frac{1}{f(n+1)} + \\frac{1}{f(n(n+1))} - \\frac{1}{f(n)}\n$$\nis an integer. Since $-\\frac{1}{2} \\leq -\\frac{1}{f(n)} < \\frac{1}{f(n+1)} + \\frac{1}{f(n(n+1))} - \\frac{1}{f(n)} < \\frac{1}{f(n+1)} + \\frac{1}{f(n(n+1))} \\leq 1$, it follows that\n$$\n\\frac{1}{f(n)} = \\frac{1}{f(n+1)} + \\frac{1}{f(n(n+1))}.\n$$\nIn particular, $f$ is strictly increasing, so $f(n) \\geq n$.\n\nProceed by induction on $n \\geq 2$ to show $f(n) = n$. For $f(2)$, note that $2/f(2) = 1/f(2) + 1/f(3) + 1/f(6)$ is a positive integer not exceeding 1, so $f(2) = 2$. Assume $f(n) = n$ for some $n \\geq 2$; then\n$$\n\\frac{1}{n} = \\frac{1}{f(n)} = \\frac{1}{f(n+1)} + \\frac{1}{f(n(n+1))} \\leq \\frac{1}{n+1} + \\frac{1}{n(n+1)} = \\frac{1}{n},\n$$\nso $f(n+1) = n+1$.\n\n**Remark:** The full Egyptian fractions theorem is not needed; it suffices to know:\n\n**Lemma:** For every $n \\geq 2$, there exists a set $S_n$ with $\\sum_{s \\in S_n} \\frac{1}{s} = 1$, $n \\in S_n$, but $n+1, n(n+1) \\notin S_n$.\n\nFor $n \\in \\{2,3,4,5\\}$, sets like $\\{2,3,6\\}$, $\\{2,4,6,12\\}$, $\\{2,5,7,12,20,42\\}$ work. For $n \\geq 6$, start with $S = \\{2,3,6\\}$ and:\n\n**Step 1:** Let $k = \\max S$. If $k(k+1) \\leq n$, replace $k$ by $\\{k+1, k(k+1)\\}$ and repeat. Eventually, $k \\leq n < k(k+1)$. If $k = n$, done; else, go to Step 2.\n\n**Step 2:** Replace $k$ by $\\{n\\} \\cup \\{k(k+1), (k+1)(k+2), \\dots, n(n-1)\\}$. Since $n+1 \\leq k(k+1)$ and $n(n+1) > \\max S'$, if $n+1 < k(k+1)$, done. Otherwise, replace $k(k+1)$ by $\\{k(k+1)+1, k(k+1)(k(k+1)+1)\\}$ to finish.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15752,
    "subject": "Mathematics (Olympiad)",
    "question": "2015 candies are placed along a circle and numbered $1$ to $2015$ clockwise. Andriy and Olesia play the following game. In each turn, a player can take either $2$ or $3$ candies with consecutive numbers ($1$ and $2015$ are also considered \"consecutive\"). The player who can't make a move loses. Who has a winning strategy if Andriy plays first?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** Olesia.\n\n**Solution.** Andriy takes $2$ (or $3$) consecutive candies. Then Olesia takes $3$ (or $2$) diametrically opposite candies, so that there are $1005$ candies on both sides between the groups of taken candies. Then she just copies Andriy's moves on the other part. If Andriy can make a move, she can as well, so she will not lose anyway. Since the game is finite, in the end she will win.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15753,
    "subject": "Mathematics (Olympiad)",
    "question": "Дугуй ширээ тойрон суусан хүүхдүүдийг $1, 2, \\ldots, 8$ гэж дугаарлая. Хэрэв $i$-р хүүхдэд тэгш тооны чихэр байвал $i \\to 1$-ийг, сондгой тооны чихэр байвал $i \\to 0$-ийг тус харгалзуулъя.\n\nХүүхдүүдэд чихэр хуваарилахдаа тэгш тооны чихэртэй хүүхэд байж болох эсэхийг шалга.",
    "options": [],
    "answer": "See solution",
    "solution": "Хэрэв 1-р хүүхэд тэгш тооны чихэртэй гэж үзвэл, дараалалд $1245678$ эсвэл $12345678$ байхаас өөршгүй. Энэ тохиолдолд $8, 1, 2$ дугаартай хүүхдүүдийн нийт чихрийн тоо тэгш болох тул зөрчил үүснэ. Иймд заавал хүүхэд бүрт сондгой тооны чихэр байх ёстой.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15754,
    "subject": "Mathematics (Olympiad)",
    "question": "For $n \\ge 2$, let $a_1, a_2, \\dots, a_n$ be positive real numbers such that\n$$\n(a_1 + a_2 + \\dots + a_n) \\left( \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_n} \\right) \\le \\left( n + \\frac{1}{2} \\right)^2.\n$$\nProve that\n$$\n\\max(a_1, a_2, \\dots, a_n) \\le 4 \\min(a_1, a_2, \\dots, a_n).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Remark:** Let $m = \\min(a_1, a_2, \\dots, a_n)$ and $M = \\max(a_1, a_2, \\dots, a_n)$. By symmetry, we may assume without loss of generality that $m = a_1 \\le a_2 \\le \\dots \\le a_n = M$. We present three solutions. The first solution is a direct application of the Cauchy-Schwarz Inequality. The second solution bypasses the Cauchy-Schwarz Inequality by applying one of the proofs of the inequality. The third solution applies the AM-GM and AM-HM inequalities. All of them share the same finish, the case for $n = 2$.\n\nIf $n = 2$, the given condition reads\n$$\n(m + M) \\left( \\frac{1}{m} + \\frac{1}{M} \\right) \\le \\frac{4}{25}\n$$\nIt follows that\n$$\n4(m + M)^2 \\le 25 M m \\quad \\text{or} \\quad (4M - m)(M - 4m) \\le 0. \\qquad (1)\n$$\nBecause $4M - m > 0$, it must be that $M - 4m \\le 0$ and thus $M \\le 4m$.\n\nWe may assume from now on that $n \\ge 3$.\n\n**Solution 1.** The Cauchy-Schwarz Inequality gives\n$$\n\\begin{aligned}\n\\left(n + \\frac{1}{2}\\right)^2 &\\ge (a_1 + a_2 + \\dots + a_n) \\left(\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_n}\\right) \\\\ \n&= (m + a_2 + \\dots + a_{n-1} + M) \\left(\\frac{1}{M} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_{n-1}} + \\frac{1}{m}\\right) \\\\ \n&\\ge \\left(\\sqrt{\\frac{m}{M}} + \\underbrace{1 + \\dots + 1}_{n-2} + \\sqrt{\\frac{M}{m}}\\right)^2.\n\\end{aligned}\n$$\nHence\n$$\nn + \\frac{1}{2} \\ge \\sqrt{\\frac{m}{M}} + n - 2 + \\sqrt{\\frac{M}{m}} \\quad \\text{or} \\quad \\sqrt{\\frac{m}{M}} + \\sqrt{\\frac{M}{m}} \\le \\frac{5}{2} \\qquad (2)\n$$\nIt follows that\n$$\n2(m + M) \\le 5\\sqrt{M m},\n$$\nwhich is (1), completing our proof.\n\n**Solution 2 (By Adam Hesterberg).** Consider the quadratic polynomial (in $x$)\n$$\n\\begin{aligned}\np(x) &= \\frac{1}{2} \\left[ \\left( \\sqrt{a_1}x + \\frac{1}{\\sqrt{a_n}} \\right)^2 + \\left( \\sqrt{a_n}x + \\frac{1}{\\sqrt{a_1}} \\right)^2 + \\sum_{i=2}^{n-1} \\left( \\sqrt{a_i}x + \\frac{1}{\\sqrt{a_i}} \\right)^2 + \\left( 5 - 2\\sqrt{\\frac{m}{M}} - 2\\sqrt{\\frac{M}{m}} \\right) x \\right] \\\\ \n&= \\left( \\frac{1}{2} \\sum_{i=1}^{n} a_i \\right) x^2 + \\frac{2n+1}{2} \\cdot x + \\left( \\frac{1}{2} \\sum_{i=1}^{n} \\frac{1}{a_i} \\right).\n\\end{aligned}\n$$\nIts discriminant is equal to\n$$\n\\Delta = \\left( \\sum_{i=1}^{n} a_i \\right) \\left( \\sum_{i=1}^{n} \\frac{1}{a_i} \\right) - \\left( n + \\frac{1}{2} \\right)^2,\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15755,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{Z} \\to \\mathbb{Z}$ be a function such that for all $a, b \\in \\mathbb{Z}$,\n\n$$\nf^{a^2 + b^2}(a + b) = a f(a) + b f(b).\n$$\n\nFind all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "Next, for each $n \\in \\mathbb{Z}$, by $E(n, 1-n)$ we get\n\n$$\nn f(n) + (1-n) f(1-n) = f^{n^2 + (1-n)^2}(1) = f^{2n^2 - 2n}(0) = 0. \\quad (\\heartsuit)\n$$\n\nAssume that there exists some $m \\neq 0$ such that $f(m) \\neq 0$. Choose such an $m$ for which $|m|$ is minimal possible. Then $|m| > 1$ due to $(\\spadesuit)$; $f(|m|) \\neq 0$ due to $(\\clubsuit)$; and $f(1 - |m|) \\neq 0$ due to $(\\heartsuit)$ for $n = |m|$. This contradicts the minimality assumption.\n\nSo, $f(n) = 0$ for $n \\neq 0$. Finally, $f(0) = f^3(0) = f^4(2) = 2 f(2) = 0$. Clearly, the function $f(x) \\equiv 0$ satisfies the problem condition, which provides the first of the two answers.\n\n**Case 2: All orbits are infinite.**\n\nSince the orbits $\\mathcal{O}(a)$ and $\\mathcal{O}(a-1)$ differ by finitely many terms for all $a \\in \\mathbb{Z}$, each two orbits $\\mathcal{O}(a)$ and $\\mathcal{O}(b)$ have infinitely many common terms for arbitrary $a, b \\in \\mathbb{Z}$.\n\nFor a minute, fix any $a, b \\in \\mathbb{Z}$. We claim that all pairs $(n, m)$ of nonnegative integers such that $f^n(a) = f^m(b)$ have the same difference $n - m$. Arguing indirectly, we have $f^n(a) = f^m(b)$ and $f^p(a) = f^q(b)$ with, say, $n - m > p - q$, then $f^{p + m + k}(b) = f^{p + n + k}(a) = f^{q + n + k}(b)$, for all nonnegative integers $k$. This means that $f^{\\ell + (n - m) - (p - q)}(b) = f^{\\ell}(b)$ for all sufficiently large $\\ell$, i.e., that the sequence $(f^n(b))$ is eventually periodic, so $\\mathcal{O}(b)$ is finite, which is impossible.\n\nNow, for every $a, b \\in \\mathbb{Z}$, denote the common difference $n - m$ defined above by $X(a, b)$. We have $X(a - 1, a) = 1$ by (1). Trivially, $X(a, b) + X(b, c) = X(a, c)$, as if $f^n(a) = f^m(b)$ and $f^p(b) = f^q(c)$, then $f^{p + n}(a) = f^{p + m}(b) = f^{q + m}(c)$. These two properties imply that $X(a, b) = b - a$ for all $a, b \\in \\mathbb{Z}$.\n\nBut (1) yields $f^{a^2 + 1}(f(a - 1)) = f^{a^2}(f(a))$, so\n\n$$\n1 = X(f(a - 1), f(a)) = f(a) - f(a - 1) \\quad \\text{for all } a \\in \\mathbb{Z}.\n$$\n\nRecalling that $f(-1) = 0$, we conclude by (two-sided) induction on $x$ that $f(x) = x + 1$ for all $x \\in \\mathbb{Z}$.\n\nFinally, the obtained function also satisfies the assumption. Indeed, $f^n(x) = x + n$ for all $n \\ge 0$, so\n\n$$\nf^{a^2 + b^2}(a + b) = a + b + a^2 + b^2 = a f(a) + b f(b).\n$$\n\n**Comment.** There are many possible variations of the solution above, but it seems that finiteness of orbits is a crucial distinction in all solutions. However, the case distinction could be made in different ways; in particular, there exist some versions of Case 1 which work whenever there is at least one finite orbit.\n\nWe believe that Case 2 is conceptually harder than Case 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15756,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral. Let $P$, $Q$, and $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, and $AB$, respectively. Show that $PQ = QR$ if and only if the bisectors of angles $ABC$ and $ADC$ meet on segment $AC$.\n\n*Note.* The condition that $ABCD$ be cyclic is not necessary.",
    "options": [],
    "answer": "See solution",
    "solution": "As usual, we set $\\angle ABC = \\beta$, $\\angle BCA = \\gamma$, and $\\angle CAB = \\alpha$. Because $\\angle DPC = \\angle CQD = 90^\\circ$, quadrilateral $CPDQ$ is cyclic with $CD$ a diameter of the circumcircle. By the **Extended Law of Sines**, we have $PQ = CD \\sin \\angle PCQ = CD \\sin(180^\\circ - \\gamma) = CD \\sin \\gamma$. Likewise, by working with cyclic quadrilateral $ARQD$, we find $RQ = AD \\sin \\alpha$. Hence, $PQ = RQ$ if and only if $CD \\sin \\gamma = AD \\sin \\alpha$. Applying the Law of Sines to triangle $BAC$, we conclude that\n\n$$\nPQ = RQ \\quad \\text{if and only if} \\quad \\frac{AB}{BC} = \\frac{AD}{CD}. \\quad (*)\n$$\n\nOn the other hand, let the bisectors of $\\angle CBA$ and $\\angle ADC$ meet segment $AC$ at $X$ and $Y$, respectively. By the **Angle-bisector Theorem**, we have\n\n$$\n\\frac{AX}{CX} = \\frac{AB}{BC} \\quad \\text{and} \\quad \\frac{AY}{CY} = \\frac{AD}{CD}.\n$$\n\n![](images/USA_IMO_2003_p70_data_fca8c95585.png)\n\nHence, the bisectors of $\\angle ABC$ and $\\angle ADC$ meet on segment $AC$ if and only if $X = Y$, or,\n\n$$\n\\frac{AB}{BC} = \\frac{AX}{CX} = \\frac{AY}{CY} = \\frac{AD}{CD}. \\qquad (**)\n$$\n\nOur desired result follows from relations $(*)$ and $(**)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15757,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1 \\le x_2 \\le \\dots \\le x_{2n-1}$ be real numbers and $A$ be their arithmetic mean. Show that\n\n$$\n2 \\sum_{i=1}^{2n-1} (x_i - A)^2 \\ge \\sum_{i=1}^{2n-1} (x_i - x_n)^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Replacing $x_i$ by $x_i - x_n$, we can assume that $x_n = 0$. Since the left-hand side of the inequality is $2 \\sum x_i^2 - 2(2n-1)A^2$, it is enough to show that $\\sum x_i^2 \\ge 2(2n-1)A^2$. By the Cauchy-Schwarz inequality, we have\n\n$$\n\\frac{2n-1}{n-1} \\left( \\frac{\\sum_{i=1}^{n-1} x_i^2}{n-1} \\right) \\ge 2 \\left( \\frac{\\sum_{i=1}^{n-1} x_i}{n-1} \\right)^2.\n$$\n\nThe expression $\\frac{(2n-1)\\sum x_i^2}{(n-1)^2}$ is greater than the left side, while the right side is greater than $2\\left(\\frac{(2n-1)A}{n-1}\\right)^2$ since $x_1 \\le x_2 \\le \\dots \\le x_n = 0$. Equality holds if and only if $x_1 = \\dots = x_{2n-1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15758,
    "subject": "Mathematics (Olympiad)",
    "question": "$m, n, a, a', b$ and $b'$ are positive integers with $2 \\leq m \\leq n$, $a, a' \\leq m$, $b, b' \\leq n$ and $(a, b) \\neq (a', b')$.\n\nThere is a rectangular town like a chessboard with $m$ avenues and $n$ streets. $\\langle x, y \\rangle$ represents the intersection of the $x$-th avenue from the west and the $y$-th street from the north. For which $(m, n, a, b, a', b')$ does there exist a path from $\\langle a, b \\rangle$ to $\\langle a', b' \\rangle$ passing all the intersections exactly once?",
    "options": [],
    "answer": "See solution",
    "solution": "By walking one block from one intersection to another, the sum of the coordinates changes its parity. Since there are a total of $mn - 1$ blocks in the way, $(m, n, a, a', b, b')$ must satisfy one of the following:\n\n1. $mn$ is odd, and both $a+b$ and $a'+b'$ are even, or\n2. $mn$ is even, and one of $a+b$ and $a'+b'$ is odd and the other is even.\n\nHowever, even if (1) or (2) holds, there are no such paths if one of the following is satisfied (see Fig. 1):\n\n3. $m = 2$ and $1 \\neq b = b' \\neq n$, or\n4. $m = 3$, $n$ is even, ($b' - b > 1$ or $a = a' = 2$) and\n   - $a+b$ is odd and $b < b'$, or\n   - $a+b$ is even and $b > b'$.\n\n![](images/Japan_2006_p13_data_9a06ce3ae1.png)\n\nFigure 1: These examples satisfy (3) or (4), so there is no such path.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15759,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{Z} \\to \\mathbb{Z}$ satisfy\n$$\nf(x + y) = f(x) + f(y) + axy(x + y) + bxy + c,\n$$\nfor all $x, y \\in \\mathbb{Z}$, where $a, b, c$ are constants. Given $f(1) = 1$ and $f(2) = 20$,\n\n(a) Find $f(x)$ for all $x \\in \\mathbb{Z}$.\n\n(b) Find the greatest possible value of $m$ such that\n$$\nf(x) \\geq m x^2 + (5m + 1)x + 4m\n$$\nholds for all $x \\geq 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Putting $x = y = 0$ in (1),\n$$\nf(0) = f(0) + f(0) + 4,\n$$\nwhich implies $f(0) = -4$.\n\nNext, putting $x = 1$ and $y = 0$ in (1),\n$$\nf(1) = f(1) + f(0) + c + 4,\n$$\nso $c = 0$.\n\nNow, $x = y = 1$ in (1):\n$$\nf(2) = f(1) + f(1) + 2a + b + 4.\n$$\nGiven $f(1) = 1$, $f(2) = 20$,\n$$\n2a + b = 14. \\qquad (2)\n$$\n\nAlso, $x = y = 2$ in (1):\n$$\nf(4) = f(2) + f(2) + 16a + 4b + 4.\n$$\nSo $f(4) = 16a + 4b + 44$.\n\nNow, $x = 4$, $y = -4$ in (1):\n$$\nf(0) = f(4) + f(-4) - 16b + 4.\n$$\nThis gives\n$$\n4a - 3b = -12. \\qquad (3)\n$$\n\nSolving (2) and (3):\n$$\na = \\frac{3(2a + b) + (4a - 3b)}{10} = 3,\n$$\nso $b = 14 - 2a = 8$.\n\nThus, the equation becomes\n$$\nf(x + y) = f(x) + f(y) + 3xy(x + y) + 8xy + 4. \\qquad (4)\n$$\n\nLet $y = 1$ in (4):\n$$\nf(x + 1) = f(x) + 3x(x + 1) + 8x + 5.\n$$\n\nFor $x \\in \\mathbb{Z}^+$, summing the recurrence:\n$$\n\\begin{aligned}\nf(x) &= f(1) + 3 \\sum_{k=1}^{x-1} k(k+1) + 8 \\sum_{k=1}^{x-1} k + 5(x-1) \\\\\n&= (x-1)x(x+1) + 4(x-1)x + 5x - 4 \\\\\n&= x^3 + 4x^2 - 4.\n\\end{aligned}\n$$\n\nThis also holds for $x = 0$ since $f(0) = -4$.\n\nFor $x < 0$, put $y = -x$ in (4):\n$$\nf(0) = f(x) + f(-x) - 8x^2 + 4.\n$$\nSo\n$$\nf(x) = -f(-x) + 8x^2 - 8 = x^3 + 4x^2 - 4.\n$$\n\nThus, $f(x) = x^3 + 4x^2 - 4$ for all $x \\in \\mathbb{Z}$.\n\n**Verification:**\n$$\n\\begin{aligned}\nf(x+y) &= (x+y)^3 + 4(x+y)^2 - 4 \\\\\n&= x^3 + 3x^2y + 3xy^2 + y^3 + 4x^2 + 8xy + 4y^2 - 4\n\\end{aligned}\n$$\n\nand\n$$\n\\begin{aligned}\nf(x) + f(y) + 3xy(x+y) + 8xy + 4 \\\\\n&= x^3 + 4x^2 - 4 + y^3 + 4y^2 - 4 + 3xy(x+y) + 8xy + 4.\n\\end{aligned}\n$$\n\nBoth sides are equal, so this is the only solution.\n\n**(b)**\n\nWe want the greatest $m$ such that\n$$\nx^3 + 4x^2 - 4 \\geq m x^2 + (5m + 1)x + 4m\n$$\nfor all $x \\geq 0$.\n\nRewriting:\n$$\n\\begin{aligned}\nx^3 + 4x^2 - x - 4 &\\geq m x^2 + 5m x + 4m \\\\\n(x+1)(x+4)(x-1) &\\geq m(x+1)(x+4) \\\\\nx - 1 &\\geq m.\n\\end{aligned}\n$$\n\nSetting $x = 0$ gives $m \\leq -1$. When $m = -1$, the inequality holds for all $x \\geq 0$.\n\n**Thus, the greatest possible value of $m$ is $-1$.**",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15760,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $H$ be the point of intersection of two altitudes $AP$ and $CQ$ of an acute triangle $ABC$. Two points $E$ and $F$ are selected on the triangle's median $BM$ in such a way that $\\angle APE = \\angle BAC$ and $\\angle CQF = \\angle BCA$, where the point $E$ belongs to the interior of the triangle $APB$, and the point $F$ belongs to the interior of the triangle $CQB$. Prove that the lines $AE$, $CF$ and $BH$ are concurrent.\n\n![](images/Ukrajina_2013_p23_data_2ade6edf6d.png)\n\nFig. 17",
    "options": [],
    "answer": "See solution",
    "solution": "Let $T$ be the midpoint of $BH$. We will prove that the lines $AE$ and $CF$ intersect the line $BH$ at the point $T$. Let $X$ be the point of intersection of the lines $AT$ and $BM$.\n\nWe begin with showing that $\\angle APX = \\angle BAC$. This would imply that the point $X$ coincides with the point $E$, i.e., that the line $AE$ intersects the line $BH$ at the point $T$.\n\nThe ray $BM$ is in the interior of the acute angle $ABC$, and the ray $AT$ is in the interior of the acute angle $BAP$. This means that these rays intersect, and the point $X$ belongs to the interior of $\\triangle ABP$.\n\nBy the trigonometric form of Ceva's theorem for the triangle $ABP$ and the point $X$, we have\n\n$$\n\\frac{\\sin \\angle BPX}{\\sin \\angle XPA} \\cdot \\frac{\\sin \\angle PAX}{\\sin \\angle XAB} \\cdot \\frac{\\sin \\angle ABX}{\\sin \\angle XBP} = 1 \\quad (*)\n$$\n\nLet $Y$ and $Z$ be the projections of the point $T$ onto the lines $AB$ and $AP$ respectively. Then, by the intercept theorem, $Y$ and $Z$ are the midpoints of the segments $BQ$ and $PH$ respectively. We see that\n\n$$\n\\frac{\\sin \\angle PAX}{\\sin \\angle XAB} = \\frac{\\sin \\angle ZAT}{\\sin \\angle TAY} = \\frac{\\frac{TZ}{AT}}{\\frac{TY}{AT}} = \\frac{TZ}{TY}.\n$$\n\nSince $\\angle BHP = \\angle BCA = \\gamma$ and $\\angle BTY = \\angle BHQ = \\angle BAC = \\alpha$, we obtain that\n\n$$\n\\frac{TZ}{TY} = \\frac{\\frac{1}{2}BP}{BT \\cos \\alpha} = \\frac{\\frac{1}{2}BH \\sin \\gamma}{\\frac{1}{2}BH \\cos \\alpha} = \\frac{\\sin \\gamma}{\\cos \\alpha}.\n$$\n\nHence,\n\n$$\n\\frac{\\sin \\angle PAX}{\\sin \\angle XAB} = \\frac{\\sin \\gamma}{\\cos \\alpha} \\quad (1)\n$$\n\nNext, since $BM$ is the median of the triangle $ABC$, we have\n\n$$\n1 = \\frac{AM}{MC} = \\frac{AB}{BC} \\cdot \\frac{\\sin \\angle ABM}{\\sin \\angle MBC}, \\text{ and so } \\frac{\\sin \\angle ABX}{\\sin \\angle XBP} = \\frac{\\sin \\angle ABM}{\\sin \\angle MBC} = \\frac{BC}{AB}.\n$$\n\nAccording to the sine law for the triangle $ABC$, $\\frac{BC}{AB} = \\frac{\\sin \\alpha}{\\sin \\gamma}$, therefore,\n\n$$\n\\frac{\\sin \\angle ABX}{\\sin \\angle XBP} = \\frac{\\sin \\alpha}{\\sin \\gamma} \\quad (2)\n$$\n\nSubstituting (1) and (2) into $(*)$, we get\n\n$$\n\\frac{\\sin \\angle BPX}{\\sin \\angle XPA} \\cdot \\frac{\\sin \\gamma}{\\cos \\alpha} \\cdot \\frac{\\sin \\alpha}{\\sin \\gamma} = 1 \\Rightarrow \\frac{\\sin \\angle XPA}{\\sin \\angle BPX} = \\tan \\alpha.\n$$\n\nSince $\\angle BPX = 90^\\circ - \\angle XPA$, we see that $\\frac{\\sin \\angle XPA}{\\sin(90^\\circ - \\angle XPA)} = \\tan \\alpha$, which implies that $\\frac{\\sin \\angle XPA}{\\cos \\angle XPA} = \\tan \\alpha$, i.e., that $\\tan \\angle XPA = \\tan \\alpha$. The angles $\\angle XPA$ and $\\alpha$ are acute, so $\\angle XPA = \\alpha$, as required.\n\nOne can follow the same arguments to prove that the line $CF$ intersects the line $BH$ at the point $T$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15761,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimum value of $k$ such that there are at least 14 composite numbers in the range from 16 to $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "The first 14 composite numbers starting from 16 are: 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34. Therefore, $k \\geq 34$.\n\nTo confirm, we can select multiples for each $m$ from 15 down to 2, ensuring no repeats:\n\n$$\n\\begin{array}{|c|c|c|c|c|c|c|c|}\n\\hline\nm & \\text{multiples} & m & \\text{multiples} & m & \\text{multiples} & m & \\text{multiples} \\\\\n\\hline\n15 & 30 & 11 & 22, 33 & 7 & 21 & 3 & 27, 33 \\\\\n14 & 28 & 10 & 20 & 6 & 18 & 2 & 16, 22, 32, 34 \\\\\n13 & 26 & 9 & 18, 27 & 5 & 25 & & \\\\\n12 & 24 & 8 & 16, 32 & 4 & 16, 32 & & \\\\\n\\hline\n\\end{array}\n$$\n\nBy choosing allowable multiples for each $m$, we obtain 14 acceptable composites: 30, 28, 26, 24, 22, 20, 27, 16, 21, 18, 25, 32, 33, 34.\n\nThus, the minimum value of $k$ is $\\mathbf{34}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15762,
    "subject": "Mathematics (Olympiad)",
    "question": "The 27 cells of a $3 \\times 9$ grid are filled using the numbers 1 through 9 so that each row contains 9 different numbers, and each of the three $3 \\times 3$ blocks heavily outlined in the example below contains 9 different numbers, as in the first three rows of a Sudoku puzzle.\n\n![](images/2025AIME_I_Solutions_p6_data_07fe3b60bf.png)\n\nThe number of different ways to fill such a grid can be written as $p^a \\cdot q^b \\cdot r^c \\cdot s^d$, where $p, q, r$, and $s$ are distinct prime numbers and $a, b, c$, and $d$ are positive integers. Find $p \\cdot a + q \\cdot b + r \\cdot c + s \\cdot d$.",
    "options": [],
    "answer": "See solution",
    "solution": "Call $1, 2, 3$ *small numbers*, $4, 5, 6$ *medium numbers*, and $7, 8, 9$ *large numbers*. There are $9!$ ways to fill in the first $3 \\times 3$ block of the grid to contain each of the numbers 1 through 9. Without loss of generality, suppose that the first row is $123$, the second row is $456$, and the third row is $789$.\n\nNow consider filling in the second $3 \\times 3$ block. The small numbers must appear in the second or third rows. If all three of them appear in the second row, then the medium numbers must appear in the third row, and the large numbers must appear in the first row. If two of them appear in the second row, then there are 3 ways to choose their positions, 3 ways to choose the position of the remaining small number in the third row, and 3 ways to choose the positions of the two remaining large numbers in the first row, for a total of $27$ choices. Similarly, there is 1 possibility if none of the small numbers appears in the second row, and there are $27$ if one of them appears in the second row, for a grand total of $1 + 27 + 1 + 27 = 56$ choices. In each case, there are $3! = 6$ ways to assign the small, medium, or large numbers within their categories, so there are $56 \\cdot 6^3$ ways to fill in the second $3 \\times 3$ block. The placement of small, medium, and large numbers in the third $3 \\times 3$ block are then determined except for a permutation of the three numbers in each row.\n\nAltogether, there are therefore $9! \\cdot 56 \\cdot 6^3 \\cdot 6^3$ possibilities. The prime factorization of this number is\n\n$$\n(2^7 \\cdot 3^4 \\cdot 5 \\cdot 7) \\cdot (2^3 \\cdot 7) \\cdot (2^6 \\cdot 3^6) = 2^{16} \\cdot 3^{10} \\cdot 5^1 \\cdot 7^2.\n$$\n\nThe requested sum is $2 \\cdot 16 + 3 \\cdot 10 + 5 \\cdot 1 + 7 \\cdot 2 = 81$.\n\n*Note:* The number of ways to fill in the grid is\n\n$$\n2^{16} \\cdot 3^{10} \\cdot 5^1 \\cdot 7^2 = 948,109,639,680 \\approx 9.48 \\times 10^{11}.\n$$\n\nThe total number of Sudoku grids is\n\n$$\n6,670,903,752,021,072,936,960 \\approx 6.67 \\times 10^{21}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15763,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle with $\\angle ACB = 60^\\circ$. Let $BL$ be the angle bisector from $B$, and $BH$ be the altitude from $B$. Let $LD$ be the perpendicular from $L$ to side $BC$. Determine the angles of $\\triangle ABC$ if $AB \\parallel HD$.\n\n![](images/Ukraine_2016_Booklet_p21_data_3eda5b222e.png)",
    "options": [],
    "answer": "See solution",
    "solution": "First, consider the case when point $L$ lies on segment $AH$ (see the figure above). Since $\\angle LHB = \\angle LDB = 90^\\circ$, quadrilateral $BLHD$ is cyclic. Thus, $\\angle LBD = \\angle DHC$. Also, $\\angle LBD = \\angle LBA$ because $BL$ is a bisector, and $\\angle DHC = \\angle BAC$ since $AB \\parallel HD$. Therefore, $\\angle ABC = 2\\angle BAC$ and $\\angle ABC + \\angle BAC = 180^\\circ - 60^\\circ = 120^\\circ$. Solving, $\\angle BAC = 40^\\circ$ and $\\angle ABC = 80^\\circ$.\n\nNow, suppose point $H$ lies on segment $AL$ (see the next figure).\n\n![](images/Ukraine_2016_Booklet_p21_data_a5ec25119d.png)\n\nAgain, quadrilateral $BHLD$ is cyclic ($\\angle LHB = \\angle LDB = 90^\\circ$). We have $\\angle LBD = \\angle DHL$ and $\\angle LBD = \\angle LBA$ (since $BL$ is a bisector), and $\\angle DHL = \\angle BAC$ (since $AB \\parallel HD$). Similarly, $\\angle BAC = 40^\\circ$ and $\\angle ABC = 80^\\circ$. However, since $\\angle ACB = 60^\\circ > \\angle BAC = 40^\\circ$, point $H$ cannot lie on $AL$, so this case is impossible.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15764,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Consider $2n$ points equally spaced around a circle. Suppose that $n$ of the points are coloured blue and the remaining $n$ points are coloured red. We write down the distance between each pair of blue points in a list, from shortest to longest. We write down the distance between each pair of red points in another list, from shortest to longest. (Note that the same distance may occur more than once in a list.)\n\nProve that the two lists of distances are the same.",
    "options": [],
    "answer": "See solution",
    "solution": "The distance between two of the points is uniquely determined by the number of points between them on the circle. So if two of the points have $k-1$ points between them where $k \\le n$, we say that their chord length is $k$.\n\nIf $n$ consecutive points are coloured blue, then the remaining $n$ consecutive points are coloured red. Due to the symmetry of this configuration, the two lists of distances are the same.\n\nIf there are no $n$ consecutive red points, then one can obtain $n$ consecutive red points by repeatedly switching colours on adjacent pairs of points. We show that the lists of chord lengths are the same after one such switch if and only if they are the same before the switch.\n\nConsider a pair of adjacent points $X$ and $Y$, where $X$ is red and $Y$ is blue. Draw a diameter of the circle perpendicular to $XY$. For each point $U$ on the same side of the diameter as $X$, there is a corresponding point $V$ on the same side of the diameter as $Y$ such that the chord lengths $XU$ and $YV$ are equal to $k$ for some $1 \\le k \\le n-2$.\n\nFor each $1 \\le k \\le n-2$, there are four possibilities for the colours of $U$ and $V$ — namely, red-red, red-blue, blue-red and blue-blue.\n\n- In the first case, a red-red chord of length $k$ is changed to a red-red chord of length $k+1$ and a red-red chord of length $k+1$ is changed to a red-red chord of length $k$.\n- In the second case, a red-red chord of length $k$ is changed to a red-red chord of length $k+1$ and a blue-blue chord of length $k$ is changed to a blue-blue chord of length $k+1$.\n- In the third case, a red-red chord of length $k+1$ is changed to a red-red chord of length $k$ and a blue-blue chord of length $k+1$ is changed to a blue-blue chord of length $k$.\n- In the fourth case, a blue-blue chord of length $k$ is changed to a blue-blue chord of length $k+1$ and a blue-blue chord of length $k+1$ is changed to a blue-blue chord of length $k$.\n\nThus, the lists of chord lengths are the same after a switch if and only if they are the same before the switch. It follows that the lists of distances are the same for any colouring.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15765,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Do there exist positive integers $a_1, a_2, \\ldots, a_{2015}$ such that any two of them are coprime and $a_1 a_2 \\cdots a_{2015} - 1$ is a product of two consecutive odd numbers?\n\nb) Do there exist positive integers $a_1, a_2, \\ldots, a_{2015}$ such that any two of them are coprime and $a_1 a_2 \\cdots a_{2015} - 1$ is a product of two consecutive even numbers?",
    "options": [],
    "answer": "See solution",
    "solution": "a) Yes, such numbers exist.\n\nLet $a_1 = p_1^2, a_2 = p_2^2, \\ldots, a_{2015} = p_{2015}^2$, where $p_1 = 2, p_2, \\ldots, p_{2015}$ are the first 2015 prime numbers. Every two of these numbers are coprime, and\n$$\na_1 a_2 \\cdots a_{2015} - 1 = (p_1 p_2 \\cdots p_{2015} - 1)(p_1 p_2 \\cdots p_{2015} + 1)\n$$\nwhich is a product of two consecutive odd numbers.\n\nb) Yes, such numbers exist.\n\nLet $a_1 = p_1^2, a_2 = p_2^2, \\ldots, a_{2015} = p_{2015}^2$, where $p_1 = 3, p_2 = 5, \\ldots, p_{2015}$ are the first 2015 odd prime numbers. Every two of these numbers are coprime, and\n$$\na_1 a_2 \\cdots a_{2015} - 1 = (p_1 p_2 \\cdots p_{2015} - 1)(p_1 p_2 \\cdots p_{2015} + 1)\n$$\nwhich is a product of two consecutive even numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15766,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the inequalities $a + 4 \\leq e$ and $e^2 + 1 \\leq a^3$, find the minimum value of $a + b + c + d + e$ for integers $a < b < c < d < e$.",
    "options": [],
    "answer": "See solution",
    "solution": "From the given inequalities, $a + 4 \\leq e$ and $e^2 + 1 \\leq a^3$ must hold. Therefore,\n\n$$(a + 4)^2 \\leq e^2 \\leq a^3 - 1,$$\n\nwhich implies $(a + 4)^2 \\leq a^3 - 1$, i.e.\n\n$$(a - 4)(a^2 + 3a + 4) \\geq 1 > 0.$$\n\nThis means $a > 4$. Consequently,\n\n$$a + b + c + d + e > 5 + 6 + 7 + 8 + 9 = 35.$$\n\nOn the other hand, the choice $(a, b, c, d, e) = (5, 6, 7, 8, 9)$ satisfies the requirements. Thus, the minimum value is $35$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15767,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the set $S = \\{1, 2, 3, \\dots, 3n\\}$, where $n$ is a positive integer. Let $T$ be a subset of $S$ such that for any $x, y, z \\in T$ (where $x, y, z$ can be the same), $x + y + z \\notin T$. Find the maximum possible number of elements in such a set $T$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $T_0 = \\{n+1, n+2, \\dots, 3n\\}$, so $|T_0| = 2n$. The sum of any three elements in $T_0$ is greater than $3n$, so it does not belong to $T_0$. Thus, $\\max |T| \\geq 2n$.\n\nNow, construct sets:\n\n$$\nA_0 = \\{n, 2n, 3n\\} \\\\\nA_k = \\{k, 2n-k, 2n+k\\}, \\quad k = 1, 2, \\dots, n-1\n$$\n\nThen $S = \\bigcup_{k=0}^{n-1} A_k$. Any subset $T'$ of $S$ with $2n + 1$ elements must contain some $A_k$.\n\nIf $A_0 \\subset T'$, it contains $3n = n + n + n$.\n\nIf $A_k \\subset T'$ for some $k \\in \\{1, 2, \\dots, n-1\\}$, it contains $2n + k = k + k + (2n - k)$.\n\nTherefore, $\\max |T| < 2n + 1$, so $\\max |T| = 2n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15768,
    "subject": "Mathematics (Olympiad)",
    "question": "Distinct positive integers $a, b, c, d$ satisfy\n\n$$\n\\begin{cases} a \\mid b^2 + c^2 + d^2 \\\\ b \\mid a^2 + c^2 + d^2 \\\\ c \\mid a^2 + b^2 + d^2 \\\\ d \\mid a^2 + b^2 + c^2 \\end{cases}\n$$\n\nAlso, none of them is larger than the product of the other three. What is the largest possible number of primes among them?",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that the given condition is equivalent to $a, b, c, d \\mid a^2 + b^2 + c^2 + d^2$.\n\nIt is possible for three of the numbers to be primes. For example, $a = 2$, $b = 3$, $c = 13$, $d = 26$; here $2^2 + 3^2 + 13^2 + 26^2 = 13 \\cdot 66$, which is divisible by all four numbers. We will show it is impossible for all four to be primes.\n\nAssume $a, b, c, d$ are all primes. Since $a^2 + b^2 + c^2 + d^2$ is divisible by each, it is divisible by $abcd$. If one prime is $2$, then $a^2 + b^2 + c^2 + d^2$ is odd, but $abcd$ is even—a contradiction. Thus, all four are odd primes, so $a^2 + b^2 + c^2 + d^2 \\equiv 0 \\pmod{4}$. Therefore, $a^2 + b^2 + c^2 + d^2$ is divisible by $4abcd$, but $a^2 + b^2 + c^2 + d^2 < 4abcd$ since none of the numbers exceeds the product of the other three, and equality can only hold for the largest.\n\n$$\n\\frac{a}{bcd} + \\frac{b}{acd} + \\frac{c}{abd} + \\frac{d}{abc} < 4\n$$\n\nThus, the largest possible number of primes among them is $3$.\n\n**Comment:** Instead of asking for the largest number of primes, one could ask for the smallest number of composite numbers. In this case, $1, 2, 5, 10$ works.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15769,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle. Denote by $E$ and $F$ the feet of the altitudes from $B$ and $C$, respectively. Let $H$ be the intersection of $BE$ and $CF$. Let $D$ be on the same side of line $BC$ as $A$ and satisfy:\n\n$$\n\\angle DBC = \\angle DCB = \\angle BAC.\n$$\n\nLet $N$ be the midpoint of $EF$. Prove that points $H$, $D$ and $N$ are collinear.\n\n![](images/BW2021_Shortlist_p36_data_1790e86602.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Refer to the figure. Notice that triangles $HEF$ and $HCB$ are similar with different orientations, and $C$, $H$, $F$ and $B$, $H$, $E$ are collinear. Thus, the statement is equivalent to $DH$ being a symmedian from $H$ in $\\triangle BHC$.\n\nAs $\\angle CBD = \\angle BAC = 180^\\circ - \\angle BHC$, the line $DB$ is tangent to the circumcircle of $\\triangle BHC$. Similarly, $DC$ is tangent to the circumcircle of $\\triangle BHC$. The line $HD$ connects $H$ with the intersection of the tangents to the circumcircle of $\\triangle BHC$ at $B$ and $C$, so $HD$ is indeed the symmedian in $\\triangle BHC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15770,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a positive integer $n$ which has exactly 9 positive divisors and whose all divisors can be placed in a $3 \\times 3$ table such that the products of the 3 numbers in each row, each column, and on each diagonal are all the same?",
    "options": [],
    "answer": "See solution",
    "solution": "The number $36$ has $9$ positive divisors: $1$, $2$, $3$, $4$, $6$, $9$, $12$, $18$, $36$. Arrange them in a $3 \\times 3$ table as follows:\n\n$$\n\\begin{array}{ccc}\n18 & 1 & 12 \\\\\n4 & 6 & 9 \\\\\n3 & 36 & 2 \\\\\n\\end{array}\n$$\n\nThe product of the numbers in each row, each column, and each diagonal is $216$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15771,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a circle $\\omega$ and two points $A, B$ outside $\\omega$. A cyclic quadrilateral inscribed in $\\omega$ is called a \"good quadrilateral\" if one pair of its opposite sides intersects at $A$ and the other pair intersects at $B$.\n\nGiven that a good quadrilateral exists, prove that there exists a particular good quadrilateral $\\Gamma$ such that every other good quadrilateral distinct from $\\Gamma$ has area strictly less than that of $\\Gamma$.\n\n![](images/China-TST-2025B_p24_data_5dbd28ce9b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the center of $\\omega$. For any good quadrilateral $PQRS$ (where $PQ$ and $RS$ meet at $A$, and $PS$ and $QR$ meet at $B$), let the diagonals $PR$ and $QS$ intersect at $C$. By Brocard's theorem, $A, B, C, O$ form an orthocentric system, so $C$ is a fixed point. The existence of a good quadrilateral implies that $A, B, C, O$ are pairwise distinct. By polar line properties, the polar line of $C$ with respect to $\\omega$ is the line $AB$. Since $C$ lies inside $\\omega$, $AB$ is disjoint from $\\omega$.\n\nLet $X$ and $Y$ be the midpoints of diagonals $PR$ and $QS$, respectively. We have:\n\n$$\n\\begin{aligned}\n4[AXY] &= 2[APY] + 2[ARY] = [APQ] + [APS] + [ARQ] + [ARS] \\\\\n&= [APS] + [ARQ] = [PSRQ],\n\\end{aligned}\n$$\n\n(where $[AXY]$ denotes the signed area of triangle $AXY$, and similarly for others) which shows that the area of the good quadrilateral equals four times the area of triangle $AXY$. Thus, the problem reduces to proving that the maximum area of triangle $AXY$ exists and is achieved uniquely.\n\nLet $M$ be the midpoint of $AB$. By properties of the Newton line, $M, X, Y$ are collinear. Let $\\omega'$ be the circle with diameter $OC$. Then $X$ and $Y$ lie on $\\omega'$. Conversely, if a line through $M$ intersects $\\omega'$ at $X$ and $Y$, and the lines $CX$ and $CY$ (if one of $X, Y$ coincides with $C$, replace the corresponding line with the line through $C$ perpendicular to $OC$) intersect $\\omega$ at $P, R$ and $Q, S$ respectively, then the lines $PQ$ and $RS$ meet at $A'$, and the lines $PS$ and $QR$ meet at $B'$. By polar line properties, $A'$ and $B'$ lie on $AB$, the polar of $C$ with respect to $\\omega$. Since $OX \\perp XC$ and $OY \\perp YC$, $X$ and $Y$ are the midpoints of $PR$ and $QS$ respectively (even if $X$ or $Y$ coincides with $O$, the conclusion holds). By the Newton line property, $XY$ passes through the midpoint of $A'B'$, so the midpoint of $A'B'$ is also $M$.\n\nBy polar properties, $\\overrightarrow{OA} \\cdot \\overrightarrow{OB} = \\overrightarrow{OA'} \\cdot \\overrightarrow{OB'} = |OP|^2$. Moreover,\n\n$$\n\\overrightarrow{OA} \\cdot \\overrightarrow{OB} = (\\overrightarrow{OM} + \\overrightarrow{MA}) \\cdot (\\overrightarrow{OM} - \\overrightarrow{MA}) = |OM|^2 - |MA|^2,\n$$\n\nand similarly, $\\overrightarrow{OA'} \\cdot \\overrightarrow{OB'} = |OM|^2 - |MA'|^2$. Thus, $|MA| = |MA'|$, implying that $A'$ and $B'$ coincide with $A$ and $B$ (possibly in reverse order). Hence, the quadrilateral $PQRS$ constructed this way is necessarily a good quadrilateral, and distinct unordered pairs $(X, Y)$ correspond to distinct good quadrilaterals.\n\nThe problem now reduces to proving: Among all lines $l$ through $M$ intersecting $\\omega'$ at $X$ and $Y$, there exists a unique line $l$ that maximizes the area of triangle $AXY$.\n\nLet $F$ be the midpoint of $OC$ (i.e., the center of $\\omega'$), and let $m$ be the line through $F$ perpendicular to $MA$. Let $X_1$ and $Y_1$ be the projections of $X$ and $Y$ onto $m$. Then:\n\n$$\nS_{AXY} = \\frac{1}{2}XY \\cdot d(A, XY) = \\frac{1}{2}XY \\cdot AM \\cdot \\sin \\angle AMX = \\frac{1}{2}AM \\cdot X_1Y_1.\n$$\n\nThus, the problem further reduces to proving that there exists a unique line $l$ through $M$ such that the projection of its chord $XY$ with $\\omega'$ onto $m$ has maximal length.\n\nWe first prove the following claim: If a line $l$ through $M$ intersects $\\omega'$ at $X$ and $Y$, and the segment $XY$ meets $m$ at $N$ satisfying\n\n$$\n\\frac{XN}{YN} = \\frac{YM}{XM}, \\qquad (\\star)\n$$\n\nthen $l$ is the desired line.\n\n![](images/China-TST-2025B_p25_data_c8003dd554.png)\n\nIndeed, let the tangents to $\\omega'$ at $X$ and $Y$ meet at $T$. Consider another line $l'$ through $M$ intersecting $\\omega'$ at $X'$ and $Y'$. Suppose $l'$ intersects segments $XT$ and $YT$ at $U$ and $V$, respectively. It suffices to show that the projection of $UV$ onto $m$ is shorter than that of $XY$, i.e., $XU \\sin \\angle XFN > YV \\sin \\angle YFN$.\n\nNote that $\\frac{\\sin \\angle XFN}{\\sin \\angle YFN} = \\frac{XN}{YN} = \\frac{MY}{MX}$. By Menelaus' theorem, $\\frac{MY}{MX} \\cdot \\frac{XU}{YV} = \\frac{TU}{TV} > 1$, so $XU \\sin \\angle XFN > YV \\sin \\angle YFN$. The case where $l'$ intersects the extensions of $TX$ and $TY$ is similar.\n\nFinally, we prove that a line $l$ satisfying $(\\star)$ exists and is unique. Existence suffices, as uniqueness would otherwise lead to a contradiction.\n\n![](images/China-TST-2025B_p26_data_bbfa5ff83f.png)\n\nLet $\\Omega$ be the circle with diameter $MF$, intersecting $\\omega'$ at $K$ and $L$. Then $MK$ and $ML$ are tangent to $\\omega'$. Let $KL$ meet $m$ at $J$. Then $MJ$ harmonically divides $XY$, i.e., $\\frac{JX}{JY} = \\frac{MX}{MY}$. Thus, $\\frac{XN}{YN} = \\frac{YM}{XM}$ is equivalent to $J$ and $N$ being symmetric about the midpoint $I$ of $XY$. By the perpendicular bisector theorem, $I$ also lies on $\\Omega$. The problem now reduces to showing there exists a unique line $l$ through $M$ intersecting $KL, \\Omega$, and $m$ at $J, I, N$ respectively, with $JI = IN$.\n\nWhen $l = ML, JI = 0 < IN$; when $l = MF, IN = 0 < JI$. By continuity, such an $l$ exists. This completes the proof. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15772,
    "subject": "Mathematics (Olympiad)",
    "question": "A right trapezoid is given with the following property: a square can be inscribed into it such that all its vertices lie on different edges of the trapezoid and none of them coincide with any vertex of the trapezoid. Construct this square with a ruler and a compass.\n\n![](images/ukraine_2015_Booklet_p2_data_e72b9bf3c5.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $ABCD$ be the right trapezoid with right angles at $A$ and $B$. Let $EFGH$ be the required square centered at $O$, with $E \\in AB$ and $F \\in BC$. In the quadrilateral $EBFO$, two opposite angles are right, so it is cyclic. This implies $\\angle EFO = \\angle EBO = 45^\\circ$, as they intercept the same arc $EO$. Likewise, $\\angle EAO = 45^\\circ$. Thus, $O$ is the intersection point of the angle bisectors from $A$ and $B$ in the trapezoid. Also, $E$ and $G$ are symmetric with respect to $O$.\n\n**Construction:**\n1. Construct $O$ as the intersection point of the angle bisectors from $A$ and $B$.\n2. Draw a straight line $l$ symmetric to $AB$ with respect to $O$.\n3. The line $l$ intersects segment $CD$ at a unique point $G$, which is a vertex of the square.\n4. With the center $O$ and one vertex $G$, reconstruct the square uniquely.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15773,
    "subject": "Mathematics (Olympiad)",
    "question": "On the blackboard, $n$ nonnegative integers are written such that their greatest common divisor is $1$. In one step, we can erase two numbers $x, y$ with $x \\ge y$, and replace them with $x - y$ and $2y$. For which initial sequences of $n$ integers is it possible to reach a situation where $n - 1$ numbers on the blackboard are zero?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer: the sum of the numbers must be a power of $2$, or all numbers are zero.\n\nAssume not all numbers are zero. Let $S$ be the sum of the numbers on the blackboard and $D$ their greatest common divisor at some moment. Initially, $D = 1$; at the end, it must be $S$ (since only one nonzero number remains, and the GCD is that number).\n\nIn each step, $D$ either stays the same or is multiplied by $2$. Let $k_1, \\dots, k_n$ be the numbers, and suppose the operation is performed on $(x, y) = (k_1, k_2)$. Then:\n\n$$\n\\gcd(k_1, k_2, k_3, \\dots, k_n) = \\gcd(k_1 - k_2, k_2, k_3, \\dots, k_n)\n$$\n\nReplacing $k_2$ with $2k_2$ can at most multiply the GCD by $2$ (if all numbers become even), or leave it unchanged. Since at the end $D = S$, $S$ must be a power of $2$.\n\nNow, suppose $S$ is a power of $2$. Consider the binary representations of all numbers. Let $\\ell$ be the index of the rightmost column (least significant bit) that is not all zeroes; that is, $\\ell$ is the smallest number such that not all $k_i$ are divisible by $2^\\ell$.\n\nIf $n-1$ numbers are zero, we are done. Otherwise, in the $\\ell$-th column, the number of ones must be even (since $S$ is a power of $2$ greater than $2^\\ell$). Take $k_i \\ge k_j$ not divisible by $2^\\ell$ and perform the operation. After the operation, columns with indices less than $\\ell$ remain zero, and the number of ones in the $\\ell$-th column decreases by $2$.\n\nBy repeating this, we can eliminate all ones in each column, eventually reaching a state where $n-1$ numbers are zero.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15774,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any real numbers $x$, $y$, $z$, the following inequality holds:\n\n$$\nx^2(3y^2 + 3z^2 - 2yz) \\geq yz(2xy + 2xz - yz).\n$$\n\nFind all triples $(x, y, z)$ for which equality occurs.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the given inequality as a quadratic in $x$:\n\n$$\nx^2(3y^2 + 3z^2 - 2yz) - 2x(y^2z + z^2y) + y^2z^2 \\geq 0.\n$$\n\nSince $3y^2 + 3z^2 \\geq 2yz$, equality occurs when $3y^2 + 3z^2 = 2yz$, which implies $y = z = 0$. In all other cases, the quadratic opens upwards. Compute the discriminant:\n\n$$\n\\frac{1}{4} D = (y^2z + z^2y)^2 - (3y^2 + 3z^2 - 2yz)y^2z^2 = y^2z^2((y+z)^2 - (3y^2 + 3z^2 - 2yz)) = -2y^2z^2(y-z)^2 \\leq 0.\n$$\n\nThus, the left side is always non-negative since the discriminant is non-positive.\n\nEquality occurs when the discriminant is zero:\n\n1. If $y = 0$, the equality becomes $3x^2z^2 = 0$, so all triples $(x, 0, 0)$, $(0, y, 0)$, and $(0, 0, z)$ yield equality.\n2. If $y = z \\neq 0$, then $4x^2y^2 = 4xy^3 - y^4 \\Rightarrow 4x^2 - 4xy + y^2 = 0 \\Rightarrow (2x - y)^2 = 0$, so all triples $(t, 2t, 2t)$ for any real $t$ also yield equality.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15775,
    "subject": "Mathematics (Olympiad)",
    "question": "設函數 $f: [0, \\infty) \\to [0, \\infty)$ 滿足：\n\n1. 對所有 $x, y \\ge 0$，有 $f(x)f(y) \\le y^2 f\\left(\\frac{x}{2}\\right) + x^2 f\\left(\\frac{y}{2}\\right)$；\n2. 對所有 $0 \\le x \\le 1$，有 $f(x) \\le 2016$。\n\n證明：$f(x) \\le x^2$，對所有 $x \\ge 0$ 都成立。\n\nSuppose function $f : [0, \\infty) \\to [0, \\infty)$ satisfies:\n\n1. For all $x, y \\ge 0$, $f(x)f(y) \\le y^2 f\\left(\\frac{x}{2}\\right) + x^2 f\\left(\\frac{y}{2}\\right)$;\n2. For all $0 \\le x \\le 1$, $f(x) \\le 2016$.\n\nProve that $f(x) \\le x^2$ for all $x \\ge 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "由 (1)，取 $x = y = 0$，可得 $f(0) = 0$。\n\n假設存在 $x_0 > 0$ 使得 $f(x_0) > x_0^2$。利用 (1)，有：\n\n$$\nf\\left(\\frac{x_0}{2}\\right) > \\frac{1}{2} x_0^2.\n$$\n\n事實上，通過數學歸納法並重複使用 (1)，可以證明：\n\n$$\nf\\left(\\frac{x_0}{2^k}\\right) > 2^{2^k - 2^{k-1}} x_0^2\n$$\n\n對所有正整數 $k$ 都成立。然而，因為 $x_0$ 是常數，可以選擇足夠大的 $k$ 使得 $x_0/2^k \\in [0, 1]$ 且 $f(x_0/2^k) > 2016$，這與 (2) 矛盾。因此，$f(x) \\le x^2$ 對所有 $x \\ge 0$ 都成立。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15776,
    "subject": "Mathematics (Olympiad)",
    "question": "找出所有正整數組 $(a, b, c)$，使得 $a^3 + b^3 + c^3 = (abc)^2$。",
    "options": [],
    "answer": "See solution",
    "solution": "解为 $(1, 2, 3)$ 及其排列。\n\n假设 $a \\geq b \\geq c$。\n\n**解法 1** 注意到：\n$$\n3a^3 \\geq a^3 + b^3 + c^3 > a^3.\n$$\n因此 $3a^3 \\geq (abc)^2 > a^3$，所以 $3a \\geq b^2c^2 > a$。又 $b^3 + c^3 = a^2(b^2c^2 - a) \\geq a^2$，所以\n$$\n18b^3 \\geq 9(b^3 + c^3) \\geq 9a^2 \\geq b^4c^4 \\geq b^3c^5,\n$$\n因此 $18 \\geq c^5$，得 $c = 1$。\n\n由于 $2a^3 + 1 = a^4$ 无整数解，所以必须有 $a > b$。于是 $2a^3 > a^3 + b^3 + 1 = a^2b^2 > a^3$，所以 $2a > b^2 > a$。因此\n$$\n1 + b^3 = a^2(b^2 - a) \\geq a^2 > \\frac{b^4}{4}\n$$\n所以 $4 > b^3(b - 4)$，因此 $b \\leq 4$。\n\n分别对 $b = 2, 3, 4$ 解 $a$ 的三次方程，得 $(a, b, c) = (3, 2, 1)$。\n\n**解法 2** 设 $k = \\frac{b^3 + c^3}{a^2} \\leq 2a$，将原式改写为 $a + k = (bc)^2$。由于 $b^3$ 和 $c^3$ 为正整数，有 $(bc)^3 \\geq b^3 + c^3 - 1 = ka^2 - 1$，所以\n$$\na + k \\geq (ka^2 - 1)^{2/3}.\n$$\n因此\n$$\nk^2a^4 - a^3 - 5ka^2 - 3k^2a - (k^3 - 1) \\leq 0\n$$\n所以\n$$\na^4 - a^3 - 5a^2 - 5a \\leq 0\n$$\n因此 $a \\leq 3$。手动检查剩余情况。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15777,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a regular 103-sided polygon with 79 vertices colored red and the remaining vertices colored blue. Denote $A$ as the number of pairs of adjacent red vertices and $B$ as the number of pairs of adjacent blue vertices.\n\n1. Find all possible values of $(A, B)$.\n\n2. Determine the number of pairwise non-similar colorings of the polygon satisfying $B = 14$. Two colorings are considered similar if one can be obtained from the other by a rotation of the polygon.",
    "options": [],
    "answer": "See solution",
    "solution": "a) There are 24 blue vertices. If all red vertices form a single block, then $A = 78$; if they form two blocks, $A = 77$; in general, if they form $k$ blocks, $A = 79 - k$.\n\nThe number of red blocks equals the number of blue blocks, so $B = 24 - k$. Thus, all possible values of $(A, B)$ are $(79 - k, 24 - k)$ for $k = 1, 2, \\dots, 24$.\n\nb) For $B = 14$, we have $k = 10$, i.e., blue vertices form 10 blocks (and red vertices also form 10 blocks). Label the blue blocks $1$ to $10$ clockwise. Let the sizes of blue blocks be $x_1, x_2, \\dots, x_{10}$.\n\nLet $y_i = x_1 + \\dots + x_i$; then $1 \\leq y_1 < y_2 < \\dots < y_9 < y_{10} = 24$. There are $\\binom{23}{9}$ ways to choose such $y_1, \\dots, y_9$. Thus, 24 blue vertices can form 10 blocks in $\\binom{23}{9}$ ways. Similarly, red vertices can form 10 blocks in $\\binom{78}{9}$ ways.\n\nTherefore, there are $\\binom{23}{9} \\binom{78}{9}$ ways to arrange 10 blue and 10 red blocks alternately. Since 79 is prime, these arrangements are distinct under rotation, but there are 10 choices for the starting block, so the number of pairwise non-similar colorings is $$\\frac{\\binom{23}{9} \\binom{78}{9}}{10}.$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15778,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $F_n$ denote the $n$-th Fibonacci number, and define $a_n = \\frac{1}{F_n F_{n+2}}$ for $n \\ge 0$. Compute the sum $a_0 + a_1 + a_2 + \\cdots + a_m$ for $m \\ge 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that for all $n \\ge 0$, the number $a_n$ can be rewritten as follows:\n\n$$\n\\begin{aligned}\na_n &= \\frac{1}{F_n F_{n+2}} \\\\\n    &= \\frac{1}{F_n F_{n+1}} - \\frac{1}{F_{n+1} F_{n+2}}.\n\\end{aligned}\n$$\n\nTherefore, for each $m \\ge 0$, the sum $a_0 + a_1 + a_2 + \\cdots + a_m$ equals\n\n$$\n\\left( \\frac{1}{F_0 F_1} - \\frac{1}{F_1 F_2} \\right) + \\left( \\frac{1}{F_1 F_2} - \\frac{1}{F_2 F_3} \\right) + \\cdots + \\left( \\frac{1}{F_m F_{m+1}} - \\frac{1}{F_{m+1} F_{m+2}} \\right).\n$$\n\nIn this sum, all terms cancel except the first and last, so\n\n$$\na_0 + a_1 + a_2 + \\cdots + a_m = \\frac{1}{F_0 F_1} - \\frac{1}{F_{m+1} F_{m+2}} = 1 - \\frac{1}{F_{m+1} F_{m+2}} < 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15779,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n \\ge 2$ 為一正整數。有一個 $n \\times n$ 的棋盤狀地區，每一小格都是一座公園。每座公園裡都有若干隻貓（貓的數量是非負整數）。每次操作，管理處會選擇一座公園：\n\n1. 所選的公園，其貓咪數量必須大於或等於該公園的相鄰公園數。\n2. 選定公園 $A$ 後，對於該公園的每座相鄰公園 $B$，管理處都從 $A$ 赶一隻貓到 $B$（兩座公園相鄰，若且唯若它們有公共邊）。\n\n令 $m$ 為所有公園內的貓咪數量。試求最小的 $m$，使得存在一種起始的貓咪分布，讓管理處可以藉由適當的選擇每次操作的公園來進行無限次操作。",
    "options": [],
    "answer": "See solution",
    "solution": "答案：$2n(n-1)$。\n\n**構造法：**\n- 最左下角的公園沒有貓；\n- 最下方一橫排及最左方一直列上的公園各有一隻貓；\n- 其餘公園各有兩隻貓。\n\n則共有 $$(n-1) + (n-1) + 2(n-2)^2 = 2n(n-1)$$ 隻貓。考慮從最上方橫排開始，每排從最右邊的公園到最左邊的公園各做一次操作，接著換第二橫排……依序下去。易檢驗在操作完全部的格子後，會回到起始狀態，故可以操作無限次。\n\n**下界證明：**\n1. 每一個公園都要被操作過無窮多次。否則，考慮所有操作無限次的公園所成集合 $I$，其內的總貓數必遞減（因為必存在一個無窮多次的公園旁邊是有限次的公園），但這代表總有一天 $I$ 中會沒有貓而無法再進行操作，矛盾。\n2. 將所有在公園與公園之間的邊編號 $1, 2, \\dots, 2n(n-1)$。我們可以將操作改以以下方式進行：\n   - 一開始，所有貓都沒有編號。\n   - 每次對一座公園 $A$ 進行操作，要將一隻貓從 $A$ 公園，跨過第 $k$ 號邊赶到 $B$ 公園時，先檢查是否有被編號為 $k$ 的貓：\n     - 如果有編號 $k$ 的貓，且該貓在 $A$ 中，則將 $k$ 號貓從 $A$ 赶到 $B$；\n     - 如果有編號 $k$ 的貓，但牠不在 $A$ 中，則從 $A$ 中選一隻還沒被編號的貓，把牠赶到 $B$；\n     - 如果沒有編號 $k$ 的貓，則從 $A$ 中選一隻還沒被編號的貓，將牠編號為 $k$，然後赶到 $B$。\n\n易知編號 $k$ 的貓最後必然只會在編號 $k$ 的邊的兩側來回移動。這代表貓的數量至少要和這些邊的數量一樣多，因此最小要 $2n(n-1)$ 隻貓。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15780,
    "subject": "Mathematics (Olympiad)",
    "question": "In a convex quadrilateral $ABCD$, let $M$ and $N$ be the midpoints of sides $AD$ and $BC$, respectively. Points $K$ and $L$ are chosen on sides $AB$ and $CD$, respectively, such that $\\angle MKA = \\angle NLC$. Prove that if lines $BD$, $KM$, and $LN$ meet at one point, then\n\n$$\n\\angle KMN = \\angle BDC, \\quad \\angle LNM = \\angle ABD\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "*Solution.* Let $P$ be the midpoint of $BD$ and $Q$ be the common point of lines $BD$, $KM$, and $LN$. Without loss of generality, assume that point $B$ lies between $Q$ and $D$. By Thales' theorem, $PM \\parallel AB$ and $PN \\parallel CD$. Therefore, $\\angle PNL = \\angle NLC = \\angle MKA = \\angle KMP$. Note that this implies points $Q, M, P, N$ are concyclic, as $\\angle QNP + \\angle QMP = 180^{\\circ}$ and points $M, N$ lie on different sides of the line $BD$. Therefore, $\\angle KMN = \\angle QMN = \\angle QPN = \\angle BDC$. Moreover, $\\angle LNM = 180^{\\circ} - \\angle QNM = 180^{\\circ} - \\angle QPM = \\angle MPD = \\angle ABD$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15781,
    "subject": "Mathematics (Olympiad)",
    "question": "After eliminating parentheses and combining like terms in the expression, we get\n\n$$\n3(x^2 + y^2 + z^2 + w^2) + 7(xy + yz + zw + wx) + 4(xz + yw),\n$$\n\nwhich can be further transformed to\n\n$$\n3(x + y + z + w)^2 + (xy + yz + zw + wx) - 2(xz + yw).\n$$\n\nBecause $x + y + z + w = 1$, the expression given in the problem is now equal to\n\n$$\n3 + (xy + yz + zw + wx) - 2(xz + yw).\n$$\n\nFind the greatest and least values of this expression for real numbers $x, y, z, w \\geq 0$ such that $x + y + z + w = 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\nz)(y+w) \\leq \\left(\\frac{x+z}{2} + \\frac{y+w}{2}\\right)^2 = \\frac{1}{4}.\n$$\n\nWe can now estimate\n\n$$\n3 + (xy + yz + zw + wx) - 2(xz + yw) \\leq 3 + (xy + yz + zw + wx) \\leq 3 + \\frac{1}{4} = \\frac{13}{4}.\n$$\n\nAn equality holds, for instance, if $(x, y, z, w) = (\\frac{1}{2}, \\frac{1}{2}, 0, 0)$. The greatest value of the expression given in the problem is thus $\\frac{13}{4}$.\n\nTwo arbitrary non-negative real numbers satisfy inequality $(x-z)^2 \\geq 0$, which can be easily transformed to $\\frac{(x+z)^2}{2} \\geq 2xz$. Hence\n\n$$\n2(xz + yw) \\leq \\frac{(x+z)^2}{2} + \\frac{(y+w)^2}{2} \\leq \\frac{(x+z)^2}{2} + \\frac{2(x+z)(y+w)}{2} + \\frac{(y+w)^2}{2} = \\frac{((x+z) + (y+w))^2}{2} = \\frac{1}{2}.\n$$\n\nWe can now estimate\n\n$$\n3 + (xy + yz + zw + wx) - 2(xz + yw) \\geq 3 - 2(xz + yw) \\geq 3 - \\frac{1}{2} = \\frac{5}{2}.\n$$\n\nAn equality holds, for instance, if $(x, y, z, w) = (0, \\frac{1}{2}, 0, \\frac{1}{2})$. The least value of the expression given in the problem is thus $\\frac{5}{2}$. We have found that\n\n$$\n\\frac{5}{2} \\leq 3 + (xy + yz + zw + wx) - 2(xz + yw) \\leq \\frac{13}{4}\n$$\n\nwhere the left equality holds, for instance, if $(x, y, z, w) = (0, \\frac{1}{2}, 0, \\frac{1}{2})$, and the right equality holds, for instance, if $(x, y, z, w) = (\\frac{1}{2}, \\frac{1}{2}, 0, 0)$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15782,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$, the line parallel to the side $BC$ and tangent to the incircle of the triangle meets the sides $AB$ and $AC$ at the points $A_1$ and $A_2$; the points $B_1$, $B_2$ and $C_1$, $C_2$ are defined similarly. Show that\n\n$$\nAA_1 \\cdot AA_2 + BB_1 \\cdot BB_2 + CC_1 \\cdot CC_2 \\geq \\frac{1}{9} (AB^2 + BC^2 + CA^2)\n$$\n\nand determine the cases of equality.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $D$, $E$, $F$ be the points where the incircle touches the sides $BC$, $CA$, $AB$, respectively, and let $x = AE = AF$, $y = BF = BD$, $z = CD = CE$.\n\nExpress all the lengths involved in the required inequality in terms of $x$, $y$, and $z$. Clearly, $AB = x + y$, $BC = y + z$, and $CA = z + x$. To express $AA_1$ and $AA_2$, use the similarity of the triangles $AA_1A_2$ and $ABC$. Their perimeters are $2x$ and $2(x + y + z)$, respectively, so $AA_1/AB = AA_2/AC = x/(x + y + z)$, whence $AA_1 = x(x + y)/(x + y + z)$ and $AA_2 = x(x + z)/(x + y + z)$. Similarly, $BB_1 = y(y + z)/(x + y + z)$, $BB_2 = y(y + x)/(x + y + z)$, $CC_1 = z(z + x)/(x + y + z)$ and $CC_2 = z(z + y)/(x + y + z)$.\n\nWe must show that\n\n$$\n9 \\sum x^2(x + y)(x + z) \\geq (x + y + z)^2 \\sum (x + y)^2.\n$$\n\nAlternatively, but equivalently,\n\n$$\n9 \\sum x^4 + 3 \\left(\\sum x^2\\right) \\left(\\sum xy\\right) \\geq 2 \\left(\\sum x^2\\right)^2 + 4 \\left(\\sum xy\\right)^2,\n$$\n\nwhich is a consequence of the Cauchy-Schwarz inequality:\n\n$$\n3(x^4 + y^4 + z^4) \\geq (x^2 + y^2 + z^2)^2\n$$\n\nand\n\n$$\nx^2 + y^2 + z^2 \\geq xy + yz + zx.\n$$\n\nClearly, equality holds if and only if $x = y = z$; that is, if and only if the triangle $ABC$ is equilateral.\n\n![](images/shortlistBMO_2011_p17_data_67213f28cf.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15783,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle and let $D$ be a point on the side $AB$. The circumcircle of triangle $BCD$ intersects the side $AC$ at $E$. The circumcircle of triangle $ADC$ intersects the side $BC$ at $F$. Let $O$ be the circumcentre of triangle $CEF$. Prove that the points $D$, $O$, and the circumcentres of triangles $ADE$, $ADC$, $DBF$, and $DBC$ are concyclic, and that the line $OD$ is perpendicular to $AB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O_1$, $O_2$, $O_3$, and $O_4$ be the circumcentres of triangles $ADE$, $ADC$, $BFD$, and $BCD$. The line $O_1O_2$ bisects the segment $AD$ and is perpendicular to it. Similarly, $O_3O_4$ bisects the segment $DB$ and is perpendicular to it as well.\n\n![](images/Slovenija_2009_p25_data_fff3f34cd8.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15784,
    "subject": "Mathematics (Olympiad)",
    "question": "Compute $877 \\times 879 - 121 \\times 123$.",
    "options": [],
    "answer": "See solution",
    "solution": "Using $(a-b)(a+b) = a^2-b^2$, we obtain\n\n$$\n\\begin{aligned}\n877 \\times 879 - 121 \\times 123 &= (878 - 1)(878 + 1) - (122 - 1)(122 + 1) \\\\\n&= (878^2 - 1) - (122^2 - 1) = 878^2 - 122^2 \\\\\n&= (878 - 122)(878 + 122) = 756 \\times 1000 = 756000.\n\\end{aligned}\n$$\n\nAlternatively,\n\n$$\n\\begin{aligned}\n877 \\times 879 - 121 \\times 123 &= 877 \\times (1000 - 121) - 121 \\times 123 \\\\\n&= 877 \\times 1000 - 121 \\times (877 + 123) \\\\\n&= 877 \\times 1000 - 121 \\times 1000 = 756000.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15785,
    "subject": "Mathematics (Olympiad)",
    "question": "Complex numbers $z_1, z_2, \\dots, z_{100}$ satisfy $z_1 = 3 + 2i$, $z_{n+1} = \\overline{z_n} \\cdot i^n$ for $n = 1, 2, \\dots, 99$, where $i$ is the imaginary unit. Find the value of $z_{99} + z_{100}$.",
    "options": [],
    "answer": "See solution",
    "solution": "By the given conditions,\n\n$$\nz_{n+2} = \\overline{z_{n+1}} \\cdot i^{n+1} = (\\overline{z_n} \\cdot i^n) \\cdot i^{n+1} = z_n i \\quad (n = 1, 2, \\dots, 98).\n$$\n\nSince $z_1 = 3 + 2i$, we have $z_{99} = z_1 i = -2 + 3i$. Therefore,\n\n$$\n\\begin{aligned}\nz_{99} + z_{100} &= z_{99} + \\overline{z_{99}} \\cdot i^{99} \\\\\n&= (-2 + 3i) + (-2 - 3i)(-i) \\\\\n&= -5 + 5i.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15786,
    "subject": "Mathematics (Olympiad)",
    "question": "Five distinct points $A$, $B$, $C$, $D$, and $E$ lie in this order on a circle of radius $r$ and satisfy $AC = BD = CE = r$. Prove that the orthocentres of the triangles $ACD$, $BCD$, and $BCE$ are the vertices of a right-angled triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "In each triangle $\\triangle XYZ$ with obtuse angle $\\angle Z$ and orthocenter $W$, $\\angle XYZ$ and $\\angle XWZ$ are congruent. In addition, $Y$ and $W$ lie on different sides of the line $XZ$.\n\n![](images/CzsMT2006_sol_p0_data_4ed10a990b.png)\n\nIn fact, $\\angle XYZ$ and $\\angle XWZ$ are both complementary to $\\angle YXW$.\n\nLet $P$, $Q$, $R$ be the orthocentres of triangles $ACD$, $BCD$, $BCE$, respectively. We will prove that $\\angle PQR = 90^\\circ$. All three triangles are obtuse-angled at the vertex $C$. Points $P$, $Q$, $R$ are thus located on the prolongation of the altitudes from the vertex $C$ on the respective sides.\n\nIt is clear that the ray $CQ$ lies between the radii $CP$ and $CR$, i.e., inside angle $\\angle PCR$. Therefore,\n$$\n\\angle PQR = \\angle RQC + \\angle PQC\n$$\n\n![](images/CzsMT2006_sol_p0_data_b6683243e0.png)\n\nBy the initial observation, points $Q$, $R$ lie on the same side of line $BC$ and\n$$\n\\angle BEC = \\angle BRC, \\quad \\angle BDC = \\angle BQC\n$$\nAngles $\\angle BEC$ and $\\angle BDC$ are equal because they subtend the same arc $BC$.\n\nTherefore $\\angle BRC = \\angle BQC = \\omega$ and the quadrilateral $BCRQ$ is cyclic.\n\nThen $\\angle RQC = \\angle RBC = \\varphi$. In addition, since $CE = r$, $\\angle EBC = 30^\\circ$.\n\nLet $U$ be the foot of the altitude of triangle $ABC$ drawn from the vertex $C$. From the right triangle $BUR$ we get\n$$\n\\omega + \\varphi + 30^\\circ + 90^\\circ = 180^\\circ \\quad \\Leftrightarrow \\quad \\angle RQC = \\varphi = 60^\\circ - \\omega = 60^\\circ - \\angle BDC\n$$\n\nSimilarly, $\\angle PQC = 60^\\circ - \\angle DBC$. Therefore, using the fact that the sum of angles in triangle $BCD$ is $180^\\circ$, we have\n$$\n\\angle PQR = \\angle RQC + \\angle PQC = 120^\\circ - (\\angle BDC + \\angle DBC) = \\angle BCD - 60^\\circ \\quad (1)\n$$\n\nHowever, the chord $BD$ has a length equal to the radius of the circle. The size of the angle at the circumference $\\angle BCD$ is equal to half of $300^\\circ$, i.e., $\\angle BCD = 150^\\circ$. Thus, from (1) we get $\\angle PQR = 90^\\circ$. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15787,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $H$ be the orthocenter of an acute triangle $ABC$, and let $O$ be its circumcenter. The line $AO$ intersects segment $BC$ at $D$. The perpendicular to $BC$ at $D$ intersects the altitudes of $ABC$ through $B$ and $C$ at points $X$ and $Y$, respectively. Show that the circumcenter of $\\triangle HXY$ is equally distant from points $B$ and $C$.\n\n![](images/Ukraine_booklet_2018_p27_data_875b598932.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $AH_1$, $BH_2$, $CH_3$ be the altitudes of $\\triangle ABC$, $M$ the midpoint of $BC$, and $O_1$ the circumcenter of $\\triangle HXY$. We want to show that $\\triangle HXY \\sim \\triangle ABC$. Since $\\angle BH_3C = \\angle YDC = 90^\\circ$, $\\angle ABC = \\angle HYX$. Similarly, $\\angle ACB = \\angle HXY$. Therefore, $\\triangle HXY \\sim \\triangle ABC$.\n\nLet $HH'$ be the altitude of $\\triangle HXY$, and $M'$ the midpoint of $XY$. Clearly, $O_1M' \\perp XY$, $OM \\perp BC$. Thus, $HH'$ and $AH_1$ are corresponding elements in similar triangles, as are $O_1M'$ and $OM$. Thus, $\\frac{HH'}{AH_1} = \\frac{O_1M'}{OM}$. Also, $\\triangle OMD \\sim \\triangle AH_1D$, so $\\frac{H_1D}{AH_1} = \\frac{MD}{OM}$. Since $HH' = H_1D$ (because $HH_1DH'$ is a rectangle), we get $O_1M' = MD$. Thus, $O_1M'DM$ is a rectangle. Therefore, $O_1M \\perp BC$, so $O_1$ lies on the perpendicular bisector of $BC$. Hence, $O_1B = O_1C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15788,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c, d$ be the number of students who received 7, 6, 5, and 3 points, respectively. A student had to answer all 3 problems correctly to receive 7 points; problems II and III correctly and not problem I to receive 6 points; problems I and III correctly and not problem II to receive 5 points; and problems I and II correctly and not problem III to receive 3 points. Since for each problem there were 10 students who answered it correctly, we have\n\n$$\na + c + d \\leq 10, \\quad a + b + d \\leq 10, \\quad a + b + c \\leq 10. \\tag{**}\n$$\n\nConversely, if we determine a quadruple $(a, b, c, d)$ of nonnegative integers satisfying the set of inequalities above, then we can determine the number of students receiving the grades of 0, 1, 2, 4 points, respectively, as\n\n$$\n2a + b + c + d, \\quad 10 - a - c - d, \\quad 10 - a - b - d, \\quad 10 - a - b - c,\n$$\n\n(note that each of these numbers will be non-negative), and therefore, we can determine the set of 8 numbers uniquely. Thus, the solution of the problem can be obtained by finding the number of quadruples $(a, b, c, d)$ of non-negative integers which satisfy the 3 inequalities in $(**)$.\n\nHow many such quadruples $(a, b, c, d)$ are there?",
    "options": [],
    "answer": "See solution",
    "solution": "Case (1): When $a$ is even.\n\nLet $a = 10 - 2n$ for some integer $n$ with $0 \\leq n \\leq 5$.\n\n- If the maximum of $b, c, d$ is $\\leq n$, then $(b, c, d)$ can be any triple of non-negative integers $\\leq n$, so there are $(n+1)^3$ choices.\n- If the maximum of $b, c, d$ is $\\geq n+1$, let this maximum be $2n - k$. Then $k$ is a non-negative integer $\\leq n-1$, and the other two can be any of $(k+1)^2$ pairs of non-negative numbers $\\leq k$. Since the maximum can be any of $b, c, d$, there are $3(k+1)^2$ possibilities, so the total is $\\sum_{k=0}^{n-1} 3(k+1)^2 = \\frac{n(n+1)(2n+1)}{2}$ for $n=0,1,2,3,4,5$.\n\nCase (2): When $a$ is odd.\n\nLet $a = 9 - 2n$ for $0 \\leq n \\leq 4$.\n\n- If the maximum of $b, c, d$ is $\\leq n$, there are $(n+1)^3$ choices.\n- If the maximum is $\\geq n+1$, let it be $2n - k + 1$, with $k \\leq n$. The total is $\\sum_{k=0}^{n} 3(k+1)^2 = \\frac{(n+1)(n+2)(2n+3)}{2}$ for $n=0,1,2,3,4$.\n\nSumming all possibilities:\n\n$$\n\\sum_{n=0}^{5} (n+1)^3 + \\sum_{n=0}^{5} \\frac{n(n+1)(2n+1)}{2} + \\sum_{n=0}^{4} (n+1)^3 + \\sum_{n=0}^{4} \\frac{(n+1)(n+2)(2n+3)}{2}\n$$\n\nwhich equals\n\n$$\n2(1^3 + 2^3 + 3^3 + 4^3 + 5^3) + 6^3 + (6 + 30 + 84 + 180 + 330) = 1296.\n$$\n\nThus, there are $1296$ possible quadruples $(a, b, c, d)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15789,
    "subject": "Mathematics (Olympiad)",
    "question": "It is known that the arithmetic average of the numbers *a*, *b* is equal to the number *c*, so $c = \\frac{1}{2}(a+b)$, and that the geometric average of the numbers *a*, *c* is equal to the number *b*, so $b = \\sqrt{ac}$. Is it necessary that the numbers *a*, *b*, *c* are equal?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** not necessarily.\n\n**Solution.** Let's rewrite the equality $b^2 = ac$ using $c = \\frac{a+b}{2}$:\n\n$$\nb^2 = a \\cdot \\frac{a+b}{2} \\Leftrightarrow 2b^2 = ba + a^2 \\Leftrightarrow (b-a)(2b+a) = 0.\n$$\n\nNow let's denote, for example, $b = 2$, which means $a = -4$ and $c = -1$, hence we receive three different numbers satisfying the conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15790,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $ABCD$ 為菱形，其中心為點 $O$。設點 $P$ 落在 $AB$ 邊上。令點 $I, J, L$ 分別為三角形 $PCD, PAD, PBC$ 的內心。設點 $H$ 與 $K$ 分別是三角形 $PLB$ 與 $PJA$ 的垂心。證明直線 $OI$ 與 $HK$ 互相垂直。\n\n![](images/3J0428_p20_data_b1b43acf40.png)",
    "options": [],
    "answer": "See solution",
    "solution": "我們先證一個引理。\n\n**引理 1.** 設 $ABC$ 為三角形，內心為 $I$。$IB, IC$ 分別與以 $BC$ 為直徑的圓交於 $S, T$。$P$ 為該圓上任意點，$M$ 為 $BC$ 中點。$MP$ 與 $A$-中線交於 $Q$，$AQ$ 與 $BC$ 交於 $R$。$K, L$ 分別在 $CS, BT$ 上，且 $RK \\perp PC, RL \\perp PB$。證明 $IP \\perp KL$。\n\n*證明：* 設 $J$ 為 $R$ 關於 $M$ 的對稱點。$X, Y, Z$ 分別為 $I$ 在 $BC, CP, PB$ 上的投影。\n\n可得 $\\triangle MEF \\cup Q \\sim \\triangle ABC \\cup J$。設 $U, V$ 分別為 $\\triangle JAB, JAC$ 的內心，則\n\n$$\n\\angle VAC = \\frac{1}{2}\\angle JAC = \\frac{1}{2}\\angle QMF = \\frac{1}{2}\\angle PMT = \\angle ICY,\n$$\n\n同理 $\\angle UAB = \\angle IBZ$，因此\n\n$$\n\\frac{IY}{d(V, AC)} = \\frac{IC}{AV}, \\quad \\frac{IZ}{d(U, AB)} = \\frac{IB}{AU}.\n$$\n\n所以\n\n$$\n\\begin{aligned}\n\\frac{IY}{IZ} &= \\frac{IY}{IX} \\cdot \\frac{IX}{IZ} \\\\\n&= \\frac{d(V, AC) \\cdot \\frac{IC}{AV}}{IX} \\cdot \\frac{IX}{d(U, AB) \\cdot \\frac{IB}{AU}} \\\\\n&= \\frac{d(V, BC)}{IX} \\cdot \\frac{IC}{AV} \\cdot \\frac{IX}{d(U, AB)} \\cdot \\frac{AU}{IB} \\\\\n&= \\frac{CV}{CI} \\cdot \\frac{IC}{AV} \\cdot \\frac{BI}{BU} \\cdot \\frac{AU}{IB} \\\\\n&= \\frac{VC}{VA} \\cdot \\frac{UA}{UB}.\n\\end{aligned}\n$$\n\n設 $L'$ 為 $L$ 關於 $M$ 的對稱點，則 $JL' = RL$，且 $L'$ 為 $\\triangle AJC$ 的 $A$-旁心，故 $\\triangle AJL' \\sim \\triangle AVC$，因此\n\n$$\n\\frac{RL}{AJ} = \\frac{JL'}{AJ} = \\frac{VC}{VA}.\n$$\n\n同理\n\n$$\n\\frac{RK}{AJ} = \\frac{UB}{UC}.\n$$\n\n由上式得\n\n$$\n\\frac{IY}{IZ} = \\frac{RL}{RK}.\n$$\n\n因此 $\\triangle IPZ \\sim \\triangle KLR$，故 $PI \\perp KL$。\n\n回到主題。\n\n*證明：* 設 $P', I'$ 分別為 $P, I$ 關於 $O$ 的對稱點，則 $I'$ 為 $\\triangle P'AB$ 的內心。由於 $PC \\parallel P'A$，$PL$ 與 $AI'$ 平行。又 $BH \\perp PL$，$BH$ 與 $AI'$ 交於 $T$，$AK$ 與 $BI'$ 交於 $S$，皆在以 $AB$ 為直徑的圓上。\n\n考慮 $\\triangle P'AB$，$I'$ 為內心，$AI', BI'$ 分別與圓交於 $T, S$，$O$ 為 $PP'$ 中點。從 $P$ 向 $OB, OA$ 作垂線，分別交 $BT, AS$ 於 $H, K$。由引理可得 $OI' \\perp HK$，證畢。\n\n**另解：** 注意 $PK \\perp AJ = CO$，$PH \\perp BL = DO$，因此原命題 $\\triangle PHK$ 與 $\\triangle ICD$ 正交等價於 $P, H, K$ 分別關於 $CD, DI, IC$ 的垂線共點。設 $H, K$ 關於 $CD, DI, IC$ 的垂線交於 $X$，作 $P$ 關於 $O$ 的對稱點 $P' \\in CD$，$\\triangle P'AB$ 的內心 $I'$，可得 $AK, BH$ 交於 $\\triangle AI'B$ 的垂心 $H'$。由 $HH'KX$ 為平行四邊形，只需證明 $H, K$ 關於 $AB$ 的投影點的中點為 $H'$ 關於 $AB$ 的投影點與 $P$ 的中點，這分別是 $\\triangle PBC, \\triangle PDA, \\triangle P'AB$ 的內切圓與 $BC$ 的切點 $Q, R, S$。設 $\\overline{PA} = a, \\overline{PB} = b, \\overline{PC} = c, \\overline{PD} = d$，則（有向長度）\n\n$$\nPQ = \\frac{c-a}{2}, \\quad PR = \\frac{b-d}{2},\n$$\n\n$$\nPS = \\frac{b-a+c-d}{2} = PQ + PR.\n$$\n\n\\(\\blacksquare\\)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15791,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Assume that $a = x + y$, $b = x^2 + y^2$, and $c = x^4 + y^4$ are integers. Prove that the numbers $a^2 - b = 2xy$ and $b^2 - c = 2x^2y^2$ are also integers. Further, prove by induction that $x^n + y^n$ is an integer for each $n \\in \\mathbb{N}$.\n\nb) Give an example of $x$ and $y$ such that $x^n + y^n$ is an integer for all $n$, but $x$ and $y$ are not integers.\n\nc) Give an example of $x$ and $y$ such that $x^n + y^n$ is an integer for all $n$, but $x$ and $y$ are not rational numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Suppose $xy$ is not an integer. Then $xy = \\frac{m}{2}$, where $m \\in \\mathbb{Z}$ is odd. But then $2x^2y^2 = \\frac{m^2}{2}$ is not an integer, a contradiction. Hence, $xy$ is an integer.\n\nBy induction, we prove $x^n + y^n$ is an integer for all $n \\in \\mathbb{N}$:\n\n*Base case*: $x + y$ and $x^2 + y^2$ are integers by assumption.\n\n*Inductive step*: Assume $x^{n-1} + y^{n-1}$ and $x^n + y^n$ are integers for some $n > 1$. Then\n\n$$\nx^{n+1} + y^{n+1} = (x^n + y^n)(x + y) - xy(x^{n-1} + y^{n-1})\n$$\n\nSince $x + y$, $xy$, $x^{n-1} + y^{n-1}$, and $x^n + y^n$ are integers, so is $x^{n+1} + y^{n+1}$.\n\nb) $x = \\sqrt{2}$, $y = -\\sqrt{2}$.\n\nc) $x = \\frac{1}{\\sqrt{2}}$, $y = -\\frac{1}{\\sqrt{2}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15792,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть клетки плоскости раскрашены в два цвета: нечётные столбцы — в зелёный, чётные — в жёлтый. Докажите, что для любого отрезка, параллельного прямой $\\ell$, разность сумм длин его зелёных и жёлтых участков не превосходит некоторого числа $D$, зависящего только от $\\ell$.",
    "options": [],
    "answer": "See solution",
    "solution": "Прямая $\\ell$ разбивается вертикальными сторонами клеток на отрезки одинаковой длины; обозначим эту длину через $F$. Тогда для любого отрезка длины $2F$, параллельного $\\ell$, суммы длин его зелёных и жёлтых частей равны $F$. Любой отрезок, параллельный $\\ell$, разбивается на такие отрезки и остаток длины, меньшей $2F$; поэтому разность сумм длин его зелёных и жёлтых частей не превосходит $D = 2F$. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15793,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 2022 equally spaced points on a circular track $\\gamma$ of circumference 2022. The points are labeled $A_1, A_2, \\dots, A_{2022}$ in some order, each label used once. Initially, Bunbun the Bunny begins at $A_1$. She hops along $\\gamma$ from $A_1$ to $A_2$, then from $A_2$ to $A_3$, until she reaches $A_{2022}$, after which she hops back to $A_1$. When hopping from $P$ to $Q$, she always hops along the shorter of the two arcs $\\overarc{PQ}$ of $\\gamma$; if $\\overarc{PQ}$ is a diameter of $\\gamma$, she moves along either semicircle.\n\nDetermine the maximal possible sum of the lengths of the 2022 arcs which Bunbun traveled, over all possible labellings of the 2022 points.",
    "options": [],
    "answer": "See solution",
    "solution": "Replacing 2022 with $2n$, the answer is $2n^2 - 2n + 2$.\n\n![](images/sols-TST-IMO-2023_p2_data_5259d37bc3.png)\n\n**Construction** The construction for $n=5$ shown on the left half of the figure easily generalizes for all $n$.\n\n**Remark.** The validity of this construction can also be seen from the below proof.\n\n**First proof of bound** Let $d_i$ be the shorter distance from $A_{2i-1}$ to $A_{2i+1}$.\n\n**Claim** — The distance of the leg of the journey $A_{2i-1} \\to A_{2i} \\to A_{2i+1}$ is at most $2n - d_i$.\n\n*Proof.* Of the two arcs from $A_{2i-1}$ to $A_{2i+1}$, Bunbun will travel either $d_i$ or $2n-d_i$. One of those arcs contains $A_{2i}$ along the way. So we get a bound of $\\max(d_i, 2n-d_i) = 2n-d_i$. $\\square$\n\nThat means the total distance is at most\n\n$$\n\\sum_{i=1}^{n} (2n - d_i) = 2n^2 - (d_1 + d_2 + \\dots + d_n).\n$$\n\n**Claim** — We have\n\n$$\nd_1 + d_2 + \\dots + d_n \\ge 2n - 2.\n$$\n\n*Proof.* The left-hand side is the sum of the walk $A_1 \\to A_3 \\to \\dots \\to A_{2n-1} \\to A_1$. Among the $n$ points here, two of them must have distance at least $n-1$ apart; the other $d_i$'s contribute at least 1 each. So the bound is $(n-1) + (n-1) \\cdot 1 = 2n-2$. $\\square$\n\n**Second proof of bound** Draw the $n$ diameters through the $2n$ arc midpoints, as shown on the right half of the figure for $n=5$ in red.\n\n**Claim** (Interpretation of distances) — The distance between any two points equals the number of diameters crossed to travel between the points.\n\n*Proof.* Clear. $\\square$\n\nWith this in mind, call a diameter *critical* if it is crossed by all $2n$ arcs.\n\n**Claim** — At most one diameter is critical.\n\n*Proof.* Suppose there were two critical diameters; these divide the circle into four arcs. Then all $2n$ arcs cross both diameters, and so travel between opposite arcs. But this means that points in two of the four arcs are never accessed — contradiction. $\\square$\n\n**Claim** — Every diameter is crossed an even number of times.\n\n*Proof.* Clear: the diameter needs to be crossed an even number of times for the loop to return to its origin. $\\square$\n\nThis immediately implies that the maximum possible total distance is achieved when one diameter is crossed all $2n$ times, and every other diameter is crossed $2n-2$ times, for a total distance of at most\n\n$$\nn \\cdot (2n - 2) + 2 = 2n^2 - 2n + 2.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15794,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral with $\\angle B \\ne 90^\\circ$ and\n\n$$\nAB^2 + BC^2 + CD^2 + DA^2 = 2AC^2.\n$$\n\nProve that the midpoint of diagonal $BD$ is on $AC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the intersection point of the diagonals $AC$ and $BD$.\n\n![](images/Saudi_Arabia_booklet_2012_p33_data_6c78c32cf6.png)\n\nSince $\\angle B \\ne 90^\\circ$, it follows that $\\angle D \\ne 90^\\circ$, and hence $\\cos \\angle B \\ne 0$ and $\\cos \\angle D \\ne 0$. The quadrilateral $ABCD$ is cyclic, so $\\angle B + \\angle D = 180^\\circ$. It follows that $\\cos \\angle B + \\cos \\angle D = 0$, and by the Law of Cosines we obtain\n\n$$\n\\frac{AB^2 + BC^2 - AC^2}{2AB \\cdot BC} + \\frac{AD^2 + DC^2 - AC^2}{2AD \\cdot DC} = 0. \\quad (1)\n$$\n\nThe given relation is equivalent to\n\n$$\nAB^2 + BC^2 - AC^2 = -(AD^2 + DC^2 - AC^2) \\neq 0,\n$$\n\nand from (1) it follows $AB \\cdot BC = AD \\cdot DC$. This implies that $K[ABC] = K[ADC]$, and hence $BB' = DD'$, where $BB'$ and $DD'$ are the altitudes of triangles $ABC$ and $ADC$, respectively.\n\nThe triangles $BB'M$ and $DD'M$ are congruent, so $MB = MD$, and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15795,
    "subject": "Mathematics (Olympiad)",
    "question": "$$\n(x^2 - 2x + 2)^{x^2+1} + (x(4-x))^{(x-1)^2} = 2\n$$\nFind all integer solutions $x$ to the equation above.",
    "options": [],
    "answer": "See solution",
    "solution": "The equation may be rewritten as\n$$\n(x^2 - 2x + 2)^{x^2+1} + (x(4-x))^{(x-1)^2} = 2.\n$$\n\nIf $x$ is even, the second term on the left-hand side will be divisible by 4. So is the first term, provided the exponent is greater than 1, which happens iff $x \\neq 0$. Thus, the only even solution is $x = 0$.\n\nNow consider odd solutions. In this case, the second term has an even exponent, hence is positive, as is the first term. Since they sum to 2, they must each equal 0, 1, or 2. From this, we get the only odd solution $x = 1$.\n\nConsequently, the solutions of the equation are $x = 0$ and $x = 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15796,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $S$ 為一個有 $n \\ge 3$ 個正整數的集合，且集合中任兩相異數字之總和都不在 $S$ 中。證明必可將 $S$ 中的元素排成 $a_1, a_2, \\dots, a_n$，使得對於所有 $i = 2, 3, \\dots, n-1$，$a_{i-1} + a_{i+1}$ 和 $a_{i-1} - a_{i+1}$ 都不會被 $a_i$ 整除。",
    "options": [],
    "answer": "See solution",
    "solution": "我們用歸納法證明。設 $a = \\max S$，並利用歸納假設，對 $S \\setminus \\{a\\}$ 找到一個排列 $b_1, b_2, \\dots, b_{n-1}$，使得對所有 $2 \\leq i \\leq n-1$，$b_i$ 不整除 $b_{i-1} \\pm b_{i+1}$。\n\n注意：\n\n$$\n|b_j - b_{j+1}| < a \\neq b_j + b_{j+1} < 2a,\n$$\n\n因此 $a$ 不會整除 $b_j - b_{j+1}$ 或 $b_j + b_{j+1}$。所以，若將 $a$ 插入 $b_1, b_2, \\dots, b_{n-1}$ 的某個位置，若不滿足條件，則必有某 $b_j$ 滿足 $b_j \\mid a + \\epsilon b_{j-1}$ 或 $b_{j+1} \\mid a + \\epsilon b_{j+2}$，其中 $\\epsilon \\in \\{1, -1\\}$。由於有 $n$ 個插入位置，卻只有 $n-1$ 個 $b_j$，至少有一個 $b_j$ 會“作為除數違反條件兩次”，即 $b_j \\mid a + \\epsilon_1 b_{j-1}$ 且 $b_j \\mid a + \\epsilon_2 b_{j+1}$，其中 $\\epsilon_1, \\epsilon_2 \\in \\{1, -1\\}$。因此，$\\epsilon_2 b_{j+1} \\equiv -a \\equiv \\epsilon_1 b_{j-1} \\pmod{b_j}$，這與歸納假設矛盾。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15797,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{N}^* = \\{1, 2, 3, \\ldots\\}$ be the set of positive integers. Find all functions $f: \\mathbb{N}^* \\to \\mathbb{N}^*$ such that\n\n$$\nf(f(m)^2 + 2f(n)^2) = m^2 + 2n^2, \\quad \\text{for all } m, n \\in \\mathbb{N}^*.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove that $f$ is injective. For any fixed $n$, if $f(m_1) = f(m_2)$, then:\n\n$$\nm_1^2 + 2n^2 = f(f(m_1)^2 + 2f(n)^2) = f(f(m_2)^2 + 2f(n)^2) = m_2^2 + 2n^2,\n$$\nso $m_1^2 = m_2^2$ and thus $m_1 = m_2$.\n\nSince $f$ is injective, we have:\n\n$$\nf(m)^2 + 2f(n)^2 = f(p)^2 + 2f(q)^2 \\iff m^2 + 2n^2 = p^2 + 2q^2. \\tag{1}\n$$\n\nLet $f(1) = a$. For $m = n = 1$, $f(3a^2) = 3$. Then from (1):\n\n$$\nf(5a^2)^2 + 2f(a^2)^2 = f(3a^2)^2 + 2f(3a^2)^2 = 3f(3a^2)^2 = 27.\n$$\n\nThe solutions in positive integers to $x^2 + 2y^2 = 27$ are $(x, y) = (3, 3)$ and $(x, y) = (5, 1)$, so $f(a^2) = 1$ and $f(5a^2) = 5$.\n\nAlso, from (1):\n\n$$\n2f(4a^2)^2 - 2f(2a^2)^2 = f(5a^2)^2 - f(a^2)^2 = 24.\n$$\n\nThe unique solution in positive integers to $x^2 - y^2 = 12$ is $(x, y) = (4, 2)$, so $f(2a^2) = 2$ and $f(4a^2) = 4$.\n\nUsing (1) again:\n\n$$\nf((k+4)a^2)^2 = 2f((k+3)a^2)^2 - 2f((k+1)a^2)^2 + f(ka^2)^2,\n$$\n\nwhich follows from the identity $(k+4)^2 + 2(k+1)^2 = k^2 + 2(k+3)^2$. Therefore, by induction, $f(ka^2) = k$ for all $k$. Since $f(a^2) = a = f(1)$, we have $a = 1$. Thus, $f(k) = k$ for all $k \\in \\mathbb{N}^*$. It is easy to verify that $f(k) = k$ is a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15798,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all triples $(a, b, c)$ of integers $a \\ge 0$, $b \\ge 0$, and $c \\ge 0$ that satisfy the equation\n\n$$\na^{b+20}(c-1) = c^{b+21} - 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer.**  \n$\\{(1, b, 0) : b \\in \\mathbb{Z}_{>0}\\} \\cup \\{(a, b, 1) : a, b \\in \\mathbb{Z}_{>0}\\}$\n\nOne can first see that the right side factors:\n\n$$\na^{b+20}(c-1) = (c^{b+20} + c^{b+19} + \\dots + c + 1)(c-1).\n$$\n\nThe case $c=1$ will be handled separately (and is very simple). For $c \\ne 1$ the equation simplifies to\n\n$$\na^{b+20} = c^{b+20} + c^{b+19} + \\dots + c + 1.\n$$\n\nWe therefore distinguish the two cases for $c$.\n\n* $c=1$ leads to\n\n$$\n0 = 0.\n$$\n\nTherefore, in this case, arbitrary natural numbers $a$ and $b$ are solutions.\n\n* For $c \\ne 1$ we can divide by $c-1$ (see the above equations) and get the equivalent equation\n\n$$\na^{b+20} = c^{b+20} + c^{b+19} + \\dots + c + 1.\n$$\n\nObviously,\n\n$$\nc^{b+20} + c^{b+19} + \\dots + c + 1 > c^{b+20}.\n$$\n\nTherefore $a \\ge c+1$ must hold. Because of the binomial theorem we, thus, obtain\n\n$$\n\\begin{aligned}\na^{b+20} &\\ge (c+1)^{b+20} \\\\\n&= c^{b+20} + \\binom{b+20}{1} c^{b+19} + \\dots + \\binom{b+20}{b+19} c + 1 \\\\\n&\\ge c^{b+20} + c^{b+19} + \\dots + c + 1 \\\\\n&= a^{b+20}.\n\\end{aligned}\n$$\n\nHence, both inequalities must be equations.\n\nWe consider the second inequality in particular. Because of\n\n$$\n\\binom{b+20}{1} = b+20 > 1\n$$\n\nthis can only be an equation if $c = 0$. In the case of $c > 0$, the second inequality is strict and therefore leads to a contradiction and there is no solution.\n\nIn the remaining case $c = 0$, the resulting equation\n\n$$\na^{b+20} = 1\n$$\n\nis easy to solve. Since $a$ is a natural number, $a = 1$. (This also follows from the necessary relationship $a = c + 1$.) Hence, in this case $b$ may be any natural number.\n\nAlternatively, one can see in the case $c > 0$ that\n\n$$\n\\begin{aligned}\nc^{b+20} &< c^{b+20} + c^{b+19} + \\dots + c + 1 \\\\\n&< c^{b+20} + \\binom{b+20}{1} c^{b+19} + \\dots + \\binom{b+20}{b+19} c + 1 \\\\\n&= (c+1)^{b+20}.\n\\end{aligned}\n$$\n\nSo $a^{b+20}$ is in this case strictly between $c^{b+20}$ and $(c+1)^{b+20}$, which is impossible for natural numbers. (The case $c = 0$ must then be treated separately as above.)",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15799,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle is divided into arcs of equal size by $n$ points ($n \\ge 1$). For any positive integer $x$, let $P_n(x)$ denote the number of possibilities for colouring all those points, using colours from $x$ given colours, so that any rotation of the colouring by $i \\cdot \\frac{360^\\circ}{n}$, where $i$ is a positive integer less than $n$, gives a colouring that differs from the original in at least one point. Prove that the function $P_n(x)$ is a polynomial with respect to $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Call a colouring of the $n$ points *permissible* if it satisfies the conditions of the problem (is not invariant under any non-full rotation). Call two colourings *equivalent* if any two points are coloured the same by the first colouring iff they are coloured the same by the second. Clearly, any two equivalent colourings use the same number of colours, and if one is permissible, so is the other.\n\nConsider an equivalence class whose colourings use exactly $y$ colours. The number of colourings in this class that use colours from a given set of $x$ colours is $x(x-1)\\cdots(x-y+1)$. This holds also for $x < y$, the product then being zero. $P_n(x)$ is equal to the sum of those products over all equivalence classes of permissible colourings, and is therefore a polynomial with respect to $x$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15800,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 11 points equally spaced on a circle. Some of the segments with endpoints among these vertices are drawn and colored in two colors, so that each segment meets at an internal point at most one other segment of the same color. What is the greatest number of segments that could be drawn?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = 11$ and label the points $A_1, A_2, \\dots, A_n$ as vertices of a regular $n$-gon. We seek the maximum number of segments (excluding sides) colored in two colors, such that each segment meets at most one other segment of the same color at an internal point.\n\nLet $R(n)$ be the maximum number of diagonals with this property. For $n \\geq 3$,\n\n$$\nR(n) = \\left\\lceil \\frac{3(n-3)}{2} \\right\\rceil.\n$$\n\nThis can be proved by induction. For small $n$, check directly: $R(4) = 2$, $R(5) = 3$, $R(6) = 5$. Assume the formula holds for all $k < n$.\n\nConsider dividing the $n$-gon into four groups of consecutive vertices $G_1, G_2, G_3, G_4$ (with sizes $n_1, n_2, n_3, n_4$), separated by two diagonals $A_1A_j$ and $A_iA_k$. The maximum number of diagonals within each group is $R(n_i)$. If $n_i > 2$, the diagonal connecting the endpoints is counted separately. Thus,\n\n$$\nR(n) = \\sum_{i=1}^{4} R(n_i) + \\sum_{i=1}^{4} 1_{n_i > 2} + 2.\n$$\n\nBy induction and careful analysis, this leads to the recurrence $R(n) = R(n-2) + 3$ for $n \\geq 5$, which solves to the formula above.\n\nReturning to the original problem: for each color, the number of diagonals is at most $\\left\\lfloor \\frac{3(n-3)}{2} \\right\\rfloor$. The same holds for the other color. Including the $n$ sides (which can be colored arbitrarily), the greatest number of segments is\n\n$$\n2 \\left\\lfloor \\frac{3(n-3)}{2} \\right\\rfloor + n.\n$$\n\nFor $n = 11$:\n\n$$\n2 \\left\\lfloor \\frac{3(11-3)}{2} \\right\\rfloor + 11 = 2 \\times 12 + 11 = 35.\n$$\n\nThus, the answer is $\\boxed{35}$.\n\n![](images/BGR_ABooklet_2024_p18_data_dfe9cf3109.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15801,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $a$, $b$, $c$ and prime number $p$ such that\n\n$$\n73p^2 + 6 = 9a^2 + 17b^2 + 17c^2\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $b \\leq c$. We distinguish two cases regarding the value of $p$:\n\n1. If $p \\neq 2$, then $p$ is odd, so $p^2 \\equiv 1 \\pmod{8}$. It follows that\n\n$$\na^2 + b^2 + c^2 \\equiv 9a^2 + 17b^2 + 17c^2 \\equiv 73p^2 + 6 \\equiv 7 \\pmod{8}.\n$$\n\nHowever, $a^2 + b^2 + c^2$ can only be $0, 1, 2, 3, 4, 5, 6 \\pmod{8}$, which contradicts the above.\n\n2. If $p = 2$, then $9a^2 + 17b^2 + 17c^2 = 73p^2 + 6 = 298$. Since $a, b, c \\geq 1$, we have\n\n$$\n298 = 9a^2 + 17b^2 + 17c^2 \\geq 26 + 17c^2\n$$\n\nwhich means $c \\leq 4$. We investigate two cases:\n\n- If $b \\leq c = 4$, then $9a^2 + 17b^2 = 26$, which means $a = b = 1$.\n- If $b \\leq c \\leq 3$, then it is easy to check that there are no solutions.\n\nHence, the only solution satisfying $(1)$ is $(a, b, c, p) = (1, 1, 2, 2)$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15802,
    "subject": "Mathematics (Olympiad)",
    "question": "Sei $ABC$ ein spitzwinkliges Dreieck mit Höhenschnittpunkt $H$. Die Fußpunkte der Höhen durch $A$, $B$ bzw. $C$ seien $D$, $E$ und $F$. Der Schnittpunkt von $DF$ mit der Höhe durch $B$ sei $P$. Die Normale auf $BC$ durch $P$ schneidet die Seite $AB$ in $Q$. Der Schnittpunkt von $EQ$ mit der Höhe durch $A$ sei $N$.\n\nBeweise, dass $N$ die Strecke $AH$ halbiert.",
    "options": [],
    "answer": "See solution",
    "solution": "Sei Abbildung 1 gegeben. Es seien $\\beta = \\angle ABC$ und $\\gamma = \\angle ACB$. Da $\\angle AFH = \\angle AEH = 90^\\circ$, ist $AFHE$ ein Sehnenviereck. Aus der Parallelität von $DA$ und $PQ$ folgt $\\angle FQP = \\angle FAH = \\angle FEH = \\angle FEP$. Daher ist auch $QFPE$ ein Sehnenviereck. Da $\\angle AFC = \\angle ADC = 90^\\circ$, ist auch $AFDC$ ein Sehnenviereck und es gilt $\\angle QFP = \\angle AFD = 180^\\circ - \\angle ACD = 180^\\circ - \\gamma$. Somit folgt auch $\\angle QEP = \\gamma$. Daraus erhalten wir $\\angle EAN = 90^\\circ - \\gamma = \\angle AEP - \\angle QEP = \\angle AEN$, und das Dreieck $ANE$ ist gleichschenklig. Somit ist $N$ der Umkreismittelpunkt des rechtwinkeligen Dreiecks $AEH$, und es folgt $NA = NH$, wie behauptet.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15803,
    "subject": "Mathematics (Olympiad)",
    "question": "Outside a regular polygon $A_1A_2\\ldots A_n$, a point $B$ is given so that $A_1A_2B$ is an equilateral triangle. Determine all $n$ such that the points $B$, $A_2$, and $A_3$ are consecutive vertices of some regular polygon.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the new polygon have $m$ vertices. Two cases are possible:\n\n![](images/CroatianCompetitions2011_p11_data_a556d16956.png)\n![](images/CroatianCompetitions2011_p11_data_e852a33865.png)\n\n*Case 1.* The new polygon lies outside the given polygon. In other words, the $m$-gon and $n$-gon are on opposite sides of the line $A_2A_3$.\n\nIn this case, $\\angle BA_2A_3 + \\angle A_1A_2A_3 + 60^\\circ = 360^\\circ$.\n\n$$\n\\frac{n-2}{n} \\cdot 180^\\circ + \\frac{m-2}{m} \\cdot 180^\\circ + 60^\\circ = 360^\\circ\n$$\n\n$$\n3m(n-2) + 3n(m-2) + mn = 6mn\n$$\n\n$$\nmn - 6m = 6n\n$$\n\n$$\nm = \\frac{6n}{n-6} = 6 + \\frac{36}{n-6}\n$$\n\nObviously, $n-6 \\in \\mathbb{N}$ and $n-6$ divides $36$.\n\nChecking all the possibilities, we find nine solutions:\n\n| $n-6$ | 1 | 2 | 3 | 4 | 6 | 9 | 12 | 18 | 36 |\n|-------|---|---|---|---|---|----|----|----|----|\n| $n$   | 7 | 8 | 9 | 10| 12| 15 | 18 | 24 | 42 |\n| $m$   | 42| 24| 18| 15| 12| 10 | 9  | 8  | 7  |\n",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15804,
    "subject": "Mathematics (Olympiad)",
    "question": "We wish to color the squares in a strip of $n$ squares that are numbered from 1 through $n$ from left to right. Each square is to be colored with one of the colors 1, 2, or 3. The even-numbered squares can be colored with any color, but the odd-numbered squares can only be colored with the odd colors 1 or 3. In how many ways can the strip be colored if no two adjoining squares may have the same color?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_n$ be the number of colorings of a strip of length $n$ ending in a square colored with 1, and let $b_n$ be the number of such colorings ending in a square colored with 2. The number of colorings ending in 3 is also $a_n$, since any coloring ending in 1 can be uniquely changed to one ending in 3 by exchanging all 1- and 3-colored squares and vice versa.\n\nFor small indices, we have $a_1 = a_2 = 1$, $b_1 = 0$, and $b_2 = 2$. The recursions are:\n- $a_{n+1} = a_n + b_n$\n- $b_{2n+1} = 0$\n- $b_{2n} = 2a_{2n-1}$\n\nSubstituting $2n+1$ and $2n$ for $n$, the first recursion yields $a_{2n+2} = a_{2n+1}$ and $a_{2n+1} = a_{2n} + b_{2n} = a_{2n} + 2a_{2n-1}$, which together yield $a_{2n+2} = 3a_{2n}$. From $a_2 = 1$ we get $a_{2n+2} = 3^n = a_{2n+1}$ and $b_{2n} = 2 \\cdot 3^{n-1}$.\n\nThe total number of colorings is $s_n = 2a_n + b_n$.\n\nFor even $n$:\n$$\ns_{2n+2} = 2a_{2n+2} + b_{2n+2} = 2 \\cdot 3^n + 2 \\cdot 3^n = 4 \\cdot 3^n$$\n\nFor odd $n$:\n$$\ns_{2n+1} = 2a_{2n+1} + b_{2n+1} = 2 \\cdot 3^n$$\n\nThis can be summarized as:\n$$\ns_n = (3 + (-1)^n) \\cdot 3^{\\left\\lfloor \\frac{n-1}{2} \\right\\rfloor}$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15805,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a right isosceles triangle with right angle at $A$, $O$ the midpoint of the hypotenuse $BC$, $E$ the midpoint of the segment $CO$, $M$ the midpoint of $AC$, and $D$ the midpoint of $AM$. Let $F$ be the intersection of lines $OD$ and $AE$.\n\n(a) Prove that $MF \\perp OD$.\n\n(b) Prove that $FC = OC$.",
    "options": [],
    "answer": "See solution",
    "solution": "*First solution for (a).* Since $ME \\perp OC$, it suffices to show that the quadrilateral $MEOF$ is cyclic. (1) Since $ME \\parallel AO$, we have $\\angle MEF = \\angle OAE$, therefore it remains to prove that $\\angle MOD = \\angle OAE$. This follows from similarity: $\\triangle MOD \\sim \\triangle OAE$, because $\\angle DMO = \\angle EOA = 90^\\circ$ and $\\frac{DM}{MO} = \\frac{EO}{OA} = \\frac{1}{2}$.\n\n*Second solution for (a).* By Menelaus' theorem in $\\triangle CDO$, with the transversal $AFE$, we get: $\\frac{DF}{FO} = \\frac{AD}{AC} = \\frac{1}{4}$. Let $AB = 4a$. In triangle $MDO$, we have $MD = a$, $MO = 2a$, and $DO = a\\sqrt{5}$, so $DF = \\frac{a\\sqrt{5}}{5}$. Therefore, $DO \\cdot DF = MD^2$, hence, by the converse of the hypotenuse leg theorem, we obtain $MF \\perp DO$.\n\n*Third solution for (a).* Since $\\frac{CA}{CE} = 2\\sqrt{2} = \\frac{OA}{AD}$, and $\\angle ACE = \\angle OAD = 45^\\circ$, the triangles $CAE$ and $AOD$ are similar. Then $\\angle AOD = \\angle CAE = \\angle DAF$, so $\\triangle DAF \\sim \\triangle DOA$ (AA), hence $MD^2 = DA^2 = DF \\cdot DO$, and thus $MF \\perp DO$.\n\n*First solution for (b).* Let $G, R$ be the midpoints of segments $AB$ and $MO$, respectively. Since $CMGO$ is a parallelogram, $R$ is the midpoint of $CG$. $EMFO$ is cyclic, therefore $\\angle EFO = \\angle EMO = 45^\\circ = \\angle OBA$, so $ABOF$ is cyclic, with $G$ the center of its circumscribed circle. Thus $GF = GO$, $RF = RO$, so $RG$ is the perpendicular bisector of segment $OF$. Since $C \\in RG$, we get $CF = CO$.\n\n*Second solution for (b).* Let $G$ be the midpoint of $AB$. Analogous to (2), the triangles $MOD$ and $ACG$ are similar, so $\\angle ACG + \\angle CDO = \\angle MOD + \\angle CDO = 90^\\circ$, thus $CG \\perp OD$. Let $H$ be the intersection of the lines $CG$ and $OD$. Since $OG \\parallel CD$, we get $\\frac{OH}{HD} = \\frac{OG}{CD} = \\frac{2}{3}$, consequently $\\frac{OH}{OD} = \\frac{2}{5}$. In the right triangle $MOD$, we have $OM = 2MD$, $OD = MD\\sqrt{5}$ and $OF = \\frac{OM^2}{OD} = \\frac{4\\sqrt{5}}{5}MD$, therefore $\\frac{OF}{OD} = \\frac{4}{5}$. It follows that $H$ is the midpoint of $OF$, so $CH$ is both an altitude and median in the triangle $COF$, hence $CF = CO$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15806,
    "subject": "Mathematics (Olympiad)",
    "question": "The positive integer $n$ is such that $n^2 - 9$ has exactly 6 positive divisors. Prove that $\\gcd(n-3, n+3) = 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "For $n^2 - 9$ to have exactly 6 positive divisors, it must be of the form $n^2 - 9 = q^5$ or $n^2 - 9 = p^2 q$, where $p$ and $q$ are distinct primes.\n\n**Case 1:** $n^2 - 9 = q^5$\n\nThen $(n-3)(n+3) = q^5$.\n\nSo $n-3 = q^s$ and $n+3 = q^t$ with $t > s$ and $t + s = 5$.\n\nSubtracting, $q^t - q^s = 6 \\implies q^s(q^{t-s} - 1) = 2 \\cdot 3$, so $q \\in \\{2, 3\\}$.\n\n- For $q=2$: $n^2 = 41$ (no integer solution).\n- For $q=3$: $n^2 = 252$ (no integer solution).\n\n**Case 2:** $n^2 - 9 = p^2 q$\n\nThen $(n-3)(n+3) = p^2 q$.\n\nPossible factorizations:\n\n$$\n\\begin{align*}\n&\\text{(a) } n+3 = p q,\\ n-3 = p \\\\\n&\\text{(b) } n+3 = q,\\ n-3 = p^2 \\\\\n&\\text{(c) } n+3 = p^2,\\ n-3 = q\n\\end{align*}\n$$\n\n- (a): $p(q-1) = 6$; only possible if $p=3$, $q=3$, but $p \\neq q$.\n- (b) and (c): $n-3$ and $n+3$ are coprime.\n\nThus, $\\gcd(n-3, n+3) = 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15807,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(x)$ and $Q(x)$ be distinct polynomials of degree $2020$ with non-zero coefficients. Suppose that they have $r$ common real roots (counting multiplicity) and $s$ common coefficients. Determine the maximum possible value of $r + s$.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that the maximum possible value is $3029$.\n\nThe polynomials\n\n$$\nP(x) = (x^2 - 1)^{1009}(x^2 + 1) \\quad \\text{and} \\quad Q(x) = (x^2 - 1)^{1009}(x^2 + x + 1)\n$$\n\nsatisfy the conditions, have $2018$ common roots, and have $1011$ common coefficients (all coefficients of even powers). So $r + s \\ge 3029$.\n\nSuppose now that $P(x)$ and $Q(x)$ agree on the coefficients of $x^{2020}, x^{2019}, \\dots, x^{2021-k}$, disagree on the coefficient of $x^{2020-k}$, and agree on another $s-k$ coefficients. The common roots of $P(x)$ and $Q(x)$ are also non-zero roots of the polynomial $P(x) - Q(x)$, which has degree $2020 - k$. (The condition on the non-zero coefficients guarantees that $0$ is not a root of $P$ and $Q$.) So $P(x) - Q(x)$ has at most $2020 - k$ real roots. On the other hand, $P(x) - Q(x)$ has exactly $s - k$ coefficients equal to zero. By the following lemma, it has at most $2[(2020 - k) - (s - k)] = 4040 - 2s$ real non-zero roots.\n\nAveraging, we get $r \\le \\left\\lfloor \\frac{6060 - 2s - k}{2} \\right\\rfloor \\le 3030 - s$. Thus $r + s \\le 3030$. Furthermore, if equality occurs, we must have $k = 0$ and $r = 2020 - k = 4040 - 2s$. In other words, we must have $r = 2020$ and $s = 1010$. But if $r = 2020$, then $Q(x)$ is a multiple of $P(x)$ and since $P(x)$ and $Q(x)$ have non-zero coefficients, then $s = 0$, a contradiction. Therefore $r + s \\le 3029$ as required.\n\n**Lemma.** Let $f$ be a polynomial of degree $n$ having exactly $t$ coefficients equal to $0$. Then $f$ has at most $2(n - t)$ real non-zero roots.\n\n**Proof of Lemma.** Since $f$ has $n + 1 - t$ non-zero coefficients, by Descartes' rule of signs it has at most $n - t$ sign changes and therefore at most $n - t$ positive real roots. Similarly, it has at most $n - t$ negative real roots.\n\n**Note.** Counting the roots with multiplicity is not essential. We can demand $r$ common distinct real roots by changing the $(x^2 - 1)^{1009}$ in the example to\n\n$$\n(x^2 - 1)(x^2 - 2) \\cdots (x^2 - 1009).\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15808,
    "subject": "Mathematics (Olympiad)",
    "question": "На турнир во туркање на раце учествуваат $n$ деца. Пред почетокот на турнирот, секое од децата добило реден број (прв, втор, ..., $n$-ти натпреварувач). Турнирот ќе се одвива во два натпреварувачки дена по следниов систем на натпреварување:\n\nПрвиот ден најпрво се натпреваруваат првиот и вториот натпреварувач; победникот се натпреварува со третиот натпреварувач; победникот се натпреварува со четвртиот натпреварувач; итн...\n\nВториот ден најпрво се натпреваруваат $n$-тиот и $(n-1)$-от натпреварувач; победникот се натпреварува со $(n-2)$-от натпреварувач; победникот се натпреварува со $(n-3)$-от натпреварувач; итн...\n\nПокажи дека некои двајца натпреварувачи ќе се натпреваруваат меѓусебе и првиот и вториот ден.",
    "options": [],
    "answer": "See solution",
    "solution": "Првиот ден, да го разгледаме последното туркање: нека тоа е помеѓу $k$-тиот и $n$-тиот натпреварувач. Тоа значи дека $k$-тиот натпреварувач ги победил $(k+1)$-от, $(k+2)$-от, ..., $(n-1)$-от натпреварувач, па затоа првиот ден ги имало следниве дуели: $\\{k, k+1\\}$, $\\{k, k+2\\}$, $\\dots$, $\\{k, n\\}$. Еден од овие дуели ќе го има и вториот ден!",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15809,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that $n \\mid 2^n + 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "All integers $n = 3^k$ with $k \\in \\mathbb{Z}^+$ satisfy $n \\mid 2^n + 1$. Indeed, since $3 \\mid 2 + 1$, by the lifting the exponent lemma, we have\n\n$$\nv_3(2^{3^k} + 1^{3^k}) = v_3(2 + 1) + v_3(3^k) = 1 + k.\n$$\n\nThis implies $3^{k+1} \\mid 2^n + 1$, and hence $n \\mid 2^n + 1$.\n\nThe only prime number $n$ satisfying $n \\mid 2^n + 1$ is $n = 3$. Indeed, let $n = p$ be a prime. By Fermat's little theorem, we have\n\n$$\n2^p + 1 \\equiv 2 + 1 = 3 \\pmod{p}.\n$$\n\nThis is congruent to $0$ modulo $p$ if and only if $p \\mid 3$, i.e., $p = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15810,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that $n$ is a perfect number and $\\varphi(n)$ is a power of 2.\n\nNote: A positive integer $n$ is called *perfect* if the sum of all its positive divisors is equal to $2n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that $\\varphi(n) = 2^m$ for some $m \\in \\mathbb{Z}^+$. Based on the formula for $\\varphi(n)$, we have\n\n$$\n\\varphi(n) = \\prod_{p|n} p^{t-1}(p-1) = 2^m\n$$\n\nwhere $v_p(n) = t$. This implies that $t = 1$ for all odd prime divisors, since otherwise $p \\mid 2^m$, a contradiction. Thus, $p-1$ is a power of 2.\n\nFor an odd prime divisor of $n$, let $p = 2^s + 1$ for some $s \\in \\mathbb{Z}^+$. If $s = 1$, then $p = 3$; otherwise, $s$ must be even, since $p \\equiv (-1)^s + 1 \\equiv 0 \\pmod{3}$ when $s$ is odd, which is a contradiction. If $s$ has a proper odd prime divisor $q$, let $s = qt$ with $t > 1$, then $2^s + 1 = (2^t)^q + 1$ is divisible by $2^t + 1$, also a contradiction.\n\nThese imply that $s$ is also a power of 2, so by setting $s = 2^k$, we get $p = 2^{2k} + 1$. Now let $n = 2^a p_1 p_2 \\dots p_l$ with $a \\ge 0$, $l \\ge 0$, and $p_1, p_2, \\dots, p_l$ are odd primes in ascending order, where $p_1 = 2^{2k} + 1$. So\n\n$$\n2n = \\sigma(n) = (2^{a+1} - 1)(p_1 + 1)(p_2 + 1)\\dots(p_l + 1)\n$$\n\nis the sum of divisors of $n$. Note that $l > 0$, otherwise $\\sigma(n) = 2^{a+1} - 1$, a contradiction. If $p_1 > 3$, then $3 \\nmid 2n$ and\n\n$$\np_1 + 1 = 2^{2k} + 2 = 2(2^{2k-1} + 1) \\equiv 2((-1)^{2k-1} + 1) \\equiv 0 \\pmod{3}\n$$\n\nimplying that $3 \\mid 2n$, a contradiction. Thus $p_1 = 3$ and $v_3(2n) = 1$ leads to $l \\le 2$. There are two cases:\n\n* If $l = 2$, then $2^{a+1} \\cdot 3 \\cdot p_2 = (2^{a+1} - 1) \\cdot 4 \\cdot (p_2 + 1)$. Note that $v_2(p_2+1) = 1$, so comparing the exponent of 2 on both sides gives $a = 2$. Thus $6p_2 = 7(p_2+1)$, a contradiction.\n\n* If $l = 1$, then $2^{a+1} \\cdot 3 = (2^{a+1} - 1) \\cdot 4$, so $a = 1$ implies $n = 6$ is a perfect number.\n\nTherefore, $n = 6$ is the only solution to this problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15811,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c, d$ be positive real numbers. Consider the sum\n\n$$\nS = \\frac{a}{a+b+c} + \\frac{b}{b+c+d} + \\frac{c}{c+d+a} + \\frac{d}{d+a+b}.\n$$\n\nProve that $S$ can take any value in the open interval $(1, 2)$ as $a, b, c, d$ vary over positive real numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that\n\n$$\n\\begin{aligned}\nS &> \\frac{a}{a+b+c+d} + \\frac{b}{a+b+c+d} + \\frac{c}{a+b+c+d} + \\frac{d}{a+b+c+d} = 1, \\\\\nS &< \\frac{a}{a+b} + \\frac{b}{a+b} + \\frac{c}{c+d} + \\frac{d}{c+d} = 2.\n\\end{aligned}\n$$\n\nThe function $S$ changes smoothly as we vary $a, b, c,$ and $d$. We will prove that it comes arbitrarily close to $1$ and $2$, and therefore it assumes every value in the interval $(1,2)$.\n\nTaking $a = b$ and $c = d$ gives\n\n$$\nS_1(a, c) = \\frac{2a}{2a+c} + \\frac{2c}{a+2c}, \\quad \\Rightarrow \\quad \\lim_{c \\to 0} S_1(a, c) = 1.\n$$\n\nSimilarly, taking $a = c$ and $b = d$ gives\n\n$$\nS_2(a, b) = \\frac{2a}{a+2b} + \\frac{2b}{2a+b} \\quad \\Rightarrow \\quad \\lim_{b \\to 0} S_2(a, b) = 2.\n$$\n\nSo the expression takes on all values between $1$ and $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15812,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the least value and the greatest value of the expression\n\n$$\nP = x + y\n$$\n\nwhere $x, y$ are real numbers satisfying the condition\n\n$$\nx - 3\\sqrt{x + 1} = 3\\sqrt{y + 2} - y.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Rewrite the given condition as\n\n$$\nx + y = 3(\\sqrt{x + 1} + \\sqrt{y + 2}).\n$$\n\nLet $G$ be the set of possible values of $P$.\n\nWe have:\n\n$a \\in G$ if and only if the following system (in $x, y$) has solutions:\n\n$$\n\\begin{cases}\n3(\\sqrt{x + 1} + \\sqrt{y + 2}) = a \\\\\nx + y = a\n\\end{cases}\n$$\n\nLet $u = \\sqrt{x + 1}$ and $v = \\sqrt{y + 2}$. Then the system becomes:\n\n$$\n\\begin{cases}\n3(u + v) = a \\\\\nu^2 + v^2 = a + 3\n\\end{cases}\n$$\n\nThis is equivalent to:\n\n$$\n\\begin{cases}\nu + v = \\frac{a}{3} \\\\\nu v = \\frac{1}{2}\\left(\\frac{a^2}{9} - a - 3\\right)\n\\end{cases}\n$$\n\nThe system has solutions with $u, v \\ge 0$ if and only if the quadratic equation $18t^2 - 6a t + a^2 - 9a - 27 = 0$ has two non-negative roots, which is equivalent to:\n\n$$\n\\begin{cases}\n-a^2 + 18a + 54 \\ge 0 \\\\\na \\ge 0 \\\\\na^2 - 9a - 27 \\ge 0\n\\end{cases}\n$$\n\nThus,\n\n$$\n\\frac{9 + 3\\sqrt{21}}{2} \\le a \\le 9 + 3\\sqrt{15}\n$$\n\nTherefore,\n\n$$\n\\min P = \\frac{9 + 3\\sqrt{21}}{2}, \\qquad \\max P = 9 + 3\\sqrt{15}\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15813,
    "subject": "Mathematics (Olympiad)",
    "question": "Нека $ABC$ е остроаголен триаголник и нека $H$ е неговиот ортоцентар. Точката $G$ припаѓа на рамнината на триаголникот при што $ABGH$ е паралелограм. Точката $I$ припаѓа на правата $GH$ така што правата $AC$ ја полови отсечката $HI$. Правата $AC$ ја сече опишаната кружница околу триаголникот $GCI$ по вторпат во точката $J$. Докажи дека $IJ = AH$.",
    "options": [],
    "answer": "See solution",
    "solution": "Бидејќи $HG \\parallel AB$ и $BG \\parallel AH$, добиваме дека $BG \\perp BC$ и $CH \\perp GH$. Според тоа, четириаголникот $BGCH$ е тетивен. Бидејќи $H$ е ортоцентар на триаголникот $ABC$, добиваме дека $\\angle HAC = 90^\\circ - \\angle ACB = \\angle CBH$. Бидејќи $BGCH$ и $CGJI$ се тетивни четириаголници, добиваме дека\n\n$$\n\\angle CJI = \\angle CGH = \\angle CBH = \\angle HAC.\n$$\n\nНека $M$ е пресечна точка на $AC$ и $GH$, и нека $D \\neq A$ е точка од правата $AC$ така што $AH = HD$. Тогаш $\\angle MJI = \\angle HAC = \\angle MDH$.\n\nБидејќи $\\angle MJI = \\angle MDH$, $\\angle IMJ = \\angle HMD$ и $IM = MH$, добиваме дека триаголниците $IMJ$ и $HMD$ се складни, па според тоа $IJ = HD = AH$, што требаше да се докаже.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15814,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a square and let $E$ be a point on the side $AB$. The lines $DE$ and $BC$ meet at point $F$, while the lines $CE$ and $AF$ meet at point $G$. Prove that the lines $BG$ and $DF$ are perpendicular.",
    "options": [],
    "answer": "See solution",
    "solution": "Set $AB = a$, $AE = x$, and construct the point $P \\in BC$ such that $BP = x$. Notice that the triangles $ADE$ and $ABP$ are congruent, so the lines $AP$ and $DF$ are perpendicular. Triangles $AED$ and $BEF$ are similar, hence $\\frac{AE}{BE} = \\frac{AD}{BF}$ and $BF = \\frac{a(a-x)}{x}$.\n\nApply Menelaus' theorem in triangle $ABF$ with transversal $G$-$E$-$C$ to get $$\\frac{AG}{GF} = \\frac{BC}{FC} \\cdot \\frac{EA}{EB} = \\frac{x^2}{a(a-x)}.$$ Since $\\frac{BP}{BF} = \\frac{x^2}{a(a-x)}$, it follows that lines $BG$ and $AP$ are parallel, hence the conclusion.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15815,
    "subject": "Mathematics (Olympiad)",
    "question": "For a given positive integer $k$, consider the function $g_k : \\mathbb{Z} \\to \\mathbb{Z}$ defined by $g_k(x) = x^k$. Determine the set $M_k$ of positive integers $n$ such that there exist injective functions $f_1, f_2, \\dots, f_n : \\mathbb{Z} \\to \\mathbb{Z}$ with\n$$\ng_k(x) = f_1(x) \\cdot f_2(x) \\cdots f_n(x)\n$$\nfor all $x \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first note that $x^k = \\underbrace{x \\cdot x \\cdots x}_{k\\ \\text{times}}$, so $k \\in M_k$.\n\nLet $n \\in M_k$, that is, there exist injective functions $f_1, f_2, \\dots, f_n : \\mathbb{Z} \\to \\mathbb{Z}$ such that $g_k = f_1 \\cdot f_2 \\cdots f_n$.\n\nFor $x = 1$ and $x = -1$, we obtain $1 = f_1(1) \\cdot f_2(1) \\cdots f_n(1)$ and $(-1)^k = f_1(-1) \\cdot f_2(-1) \\cdots f_n(-1)$. Since $f_i(1)$ and $f_i(-1)$ are integers and the functions $f_i$ are injective, it follows that $\\{f_i(1), f_i(-1)\\} = \\{1, -1\\}$ for any $i = 1, \\dots, n$. Multiplying the two equalities, we obtain $(-1)^k = (-1)^n$, so $n$ and $k$ have the same parity.\n\nFor $x = 2$, we get $2^k = |f_1(2)| \\cdot |f_2(2)| \\cdots |f_n(2)|$. Using injectivity, we deduce that $|f_i(2)| \\ge 2$ for any $i = 1, \\dots, n$. Thus, $2^k = |f_1(2)| \\cdot |f_2(2)| \\cdots |f_n(2)| \\ge 2^n$, so $k \\ge n$. Since $n$ and $k$ have the same parity, it follows that\n$$\nM_k \\subset \\{k, k-2, k-4, \\dots\\} \\subset \\mathbb{N}^*\n$$\n\nWe will prove that all values of the form $n = k - 2t$ are good. For this, observe that\n$$\nx^k = \\underbrace{x \\cdot x \\cdots x}_{k-2t-1\\ \\text{times}} \\cdot x^{2t+1}\n$$\nso the functions $f_i(x) = x$ for $i = 1, \\dots, n-1$ and $f_n(x) = x^{2t+1}$ are injective, and $x^k = f_1(x) \\cdot f_2(x) \\cdots f_{n-1}(x) \\cdot f_n(x)$.\n\nIn conclusion,\n$$\nk - 2t \\in M_k \\text{ for any } t \\in \\mathbb{N} \\text{ with } 2t < k,\n$$\nand\n$$\nM_k = \\{k, k-2, k-4, \\dots, k-2 \\cdot \\lfloor \\frac{k-1}{2} \\rfloor\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15816,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1$, $x_2$, and $x_3$ be the roots of the equation\n$$\nx^3 - 3x^2 + (a+2)x - a = 0.\n$$\nGiven that $x_2 = 1$, and $x_1$, $x_3$ are the other roots, find the value of\n$$\n4x_1 - x_1^2 + x_3^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $x = 1$ is a root of the equation:\n$$\nx^3 - 3x^2 + (a+2)x - a = (x-1)(x^2 - 2x + a).\n$$\nSo, the other roots are the roots of $x^2 - 2x + a = 0$. Their sum is $2$, so $x_1 + x_3 = 2$.\n\nWe are to compute:\n$$\n4x_1 - x_1^2 + x_3^2.\n$$\nNote that $x_3^2 = (x_1 + x_3)^2 - 2x_1 x_3 = 4 - 2a$ and $x_1^2 = (x_1 + x_3)^2 - 2x_1 x_3 - 2x_1 x_3 = 4 - 4a$.\nBut more simply, since $x_1 + x_3 = 2$:\n\n\\begin{align*}\n4x_1 - x_1^2 + x_3^2 &= 4x_1 - x_1^2 + x_3^2 \\\\\n&= 4x_1 + (x_3^2 - x_1^2) \\\\\n&= 4x_1 + (x_3 - x_1)(x_3 + x_1) \\\\\n&= 4x_1 + (x_3 - x_1) \\cdot 2 \\\\\n&= 2(x_3 - x_1) + 4x_1 \\\\\n&= 2(x_3 + x_1) \\\\\n&= 2 \\times 2 = 4.\n\\end{align*}\n\n**Answer:** $4$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15817,
    "subject": "Mathematics (Olympiad)",
    "question": "For each positive integer $N$, let $\\tau(N)$ be the number of positive factors of $N$; $\\omega(N)$ be the number of distinct prime factors of $N$; $\\Omega(N)$ be the number of prime factors (counted with multiplicities) of $N$. Prove that for each positive integer $n$,\n\n$$\n\\sum_{m=1}^{n} 5^{\\omega(m)} \\leq \\sum_{k=1}^{n} \\lfloor \\frac{n}{k} \\rfloor \\tau(k)^2 \\leq \\sum_{m=1}^{n} 5^{\\Omega(m)}.\n$$\n\nHere, $\\lfloor x \\rfloor$ is the largest integer not exceeding $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that $\\lfloor \\frac{n}{k} \\rfloor$ represents the number of multiples of $k$ among $1, 2, \\dots, n$. Hence,\n\n$$\n\\sum_{k=1}^{n} \\lfloor \\frac{n}{k} \\rfloor \\tau(k)^2 = \\sum_{k=1}^{n} \\sum_{1 \\leq m \\leq n,\\; k|m} \\tau(k)^2 = \\sum_{m=1}^{n} \\sum_{k|m} \\tau(k)^2.\n$$\n\nTo prove the problem statement, it suffices to justify, for $m = 1, \\dots, n$,\n\n$$\n5^{\\omega(m)} \\leq \\sum_{k|m} \\tau(k)^2 \\leq 5^{\\Omega(m)}.\n$$\n\nWhen $m=1$, it is obvious. When $m > 1$, let $m = p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_r^{\\alpha_r}$ be the prime factorization ($p_1, p_2, \\dots, p_r$ are distinct primes and $\\alpha_1, \\alpha_2, \\dots, \\alpha_r$ are positive integers). For $k = p_1^{\\beta_1} p_2^{\\beta_2} \\cdots p_r^{\\beta_r}$, $\\tau(k) = (\\beta_1 + 1)(\\beta_2 + 1) \\cdots (\\beta_r + 1)$. Thus,\n\n$$\n\\begin{aligned}\n\\sum_{k|m} \\tau(k)^2 &= \\sum_{\\substack{0 \\leq \\beta_1 \\leq \\alpha_1 \\\\ \\cdots \\\\ 0 \\leq \\beta_r \\leq \\alpha_r}} (\\beta_1 + 1)^2 (\\beta_2 + 1)^2 \\cdots (\\beta_r + 1)^2 \\\\\n&= \\prod_{i=1}^{r} \\left(1^2 + 2^2 + \\cdots + (\\alpha_i + 1)^2\\right).\n\\end{aligned}\n$$\n\nNow it suffices to show for each $1 \\leq i \\leq r$,\n\n$$\n5 \\leq 1^2 + 2^2 + \\cdots + (\\alpha_i + 1)^2 \\leq 5^{\\alpha_i}.\n$$\n\nFor $j \\in \\mathbb{N}_+$, define $T(j) = 1^2 + 2^2 + \\cdots + (j+1)^2 = \\frac{1}{6}(j+1)(j+2)(2j+3)$.\nThen\n\n$$\n\\frac{T(j+1)}{T(j)} = \\frac{j+2}{j+1} \\cdot \\frac{j+3}{j+2} \\cdot \\frac{2j+5}{2j+3} \\in [1, \\frac{2}{1} \\cdot \\frac{3}{2} \\cdot \\frac{5}{3}] = [1, 5].\n$$\n\nSince $T(1) = 5$, it is straightforward to check $T(j) \\in [5, 5^j]$ for all $j \\in \\mathbb{N}_+$ by induction. This finishes the proof. $\\Box$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15818,
    "subject": "Mathematics (Olympiad)",
    "question": "A dartboard is the region $B$ in the coordinate plane consisting of points $(x, y)$ such that $|x| + |y| \\leq 8$. A target $T$ is the region where $(x^2 + y^2 - 25)^2 \\leq 49$. A dart is thrown and lands at a random point in $B$. The probability that the dart lands in $T$ can be expressed as $\\frac{m}{n} \\cdot \\pi$, where $m$ and $n$ are relatively prime positive integers. What is $m + n$?\n\n![](path/to/file.png)\n\n(A) 39  \n(B) 71  \n(C) 73  \n(D) 75  \n(E) 135",
    "options": [],
    "answer": "See solution",
    "solution": "The region $B$ is a square with intercepts $(\\pm8, 0)$ and $(0, \\pm8)$. The area of this square is $(8\\sqrt{2})^2 = 128$.\n\nTaking square roots shows that region $T$ is the set of points that satisfy\n\n$$\n25 - 7 \\leq x^2 + y^2 \\leq 25 + 7,\n$$\n\nwhich is an annulus (ring) with inner radius $\\sqrt{18}$ and outer radius $\\sqrt{32}$. Its area is $32\\pi - 18\\pi = 14\\pi$. Note that $T$ is internally tangent to $B$ at the four points $(\\pm4, \\pm4)$. The required probability is the ratio of the area of $T$ to the area of $B$, namely $\\frac{14\\pi}{128} = \\frac{7}{64}\\pi$. The requested sum is $7 + 64 = 71$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15819,
    "subject": "Mathematics (Olympiad)",
    "question": "Triangles $ABD$ and $KCD$ are similar. Prove that triangles $ADK$ and $DBC$ are also similar.\n\n![](images/Slovenija_2008_p8_data_4d2b4ac818.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since triangles $ABD$ and $KCD$ are similar, we have $\\angle ADB = \\angle KDC$ and $\\frac{|DA|}{|DB|} = \\frac{|DK|}{|DC|}$. We see that\n\n$$\n\\angle ADK = \\angle ADB - \\angle BDK = \\angle KDC - \\angle BDK = \\angle BDC\n$$\n\nand since $\\frac{|DA|}{|DK|} = \\frac{|DB|}{|DC|}$, we conclude that triangles $ADK$ and $DBC$ are also similar (matching in one angle and the ratio of the two adjacent sides).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15820,
    "subject": "Mathematics (Olympiad)",
    "question": "Matej has four pieces of paper, each with a different digit written on it. Can he always arrange the four digits to form two different four-digit numbers that are not relatively prime?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, Matej can always do this.\n\n- If one digit is even, place it in the units position for both numbers and swap two of the other digits. Both numbers will be even, so not relatively prime.\n- If one digit is 5, place it in the units position for both numbers and swap two of the other digits. Both numbers will end with 5, so both divisible by 5.\n- If two digits are the same, swap their positions to get the same number twice, which are obviously not relatively prime.\n- The only remaining case is digits 1, 3, 7, and 9. Matej can form pairs like 1397 and 1793, or 1397 and 9317, or 9317 and 9713; all these pairs are divisible by 11. He can also form 1739 and 9731, both divisible by 37.\n\nThus, in all cases, Matej can form two four-digit numbers that are not relatively prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15821,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle and $D \\in (BC)$, $E \\in (AD)$ be mobile points. The circumcircle of triangle $CDE$ meets the median from $C$ of triangle $ABC$ at $F$. Prove that the circumcenter of triangle $AEF$ lies on a fixed line.",
    "options": [],
    "answer": "See solution",
    "solution": "The quadrilateral $EFDC$ is cyclic, hence $\\angle FED \\equiv \\angle FCD$. Let $P$ be the reflection of point $C$ in the midpoint of the segment $[AB]$; clearly $AP \\parallel BC$ and $\\angle FCD \\equiv \\angle FPA$.\n\nIt follows that $\\angle FPA \\equiv \\angle FED$, which means that the quadrilateral $FEAP$ is cyclic. This shows that the circumcenter of $AEF$ lies on the perpendicular bisector of the segment $[AP]$ (which is a fixed line).\n\n![](images/RMC2014_p75_data_d9c1d6d8aa.png)\n",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15822,
    "subject": "Mathematics (Olympiad)",
    "question": "Positive integers $a$, $b$, $c$ form an increasing arithmetic progression. Is it possible that $[a, b] = [b, c]$, where $[x, y]$ denotes the smallest common multiple of integers $x$ and $y$?",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the difference of the progression by $d = b - a = c - b > 0$. Then the equality from the statement becomes: $[a, a+d] = [a+d, a+2d]$. As $a+2d > a+d$, there exists some power of a prime $p^k$, for which $a+2d$ is divisible by $p^k$, but $a+2d$ isn't divisible by $p^{k+1}$ and $a+d$ isn't divisible by $p^k$. From the equality of LCMs, we must have $a$ is divisible by $p^k$ and $a$ not divisible by $p^{k+1}$. But then $(a+2d) - a = 2d$ is divisible by $p^k$.\n\nIf $p \\neq 2$, then $d$ is divisible by $p^k$, contradiction as then $a+d$ is divisible by $p^k$.\n\nIf $p=2$, then $d$ is divisible by $2^{k-1}$ and $d$ not divisible by $2^k$, as otherwise again $a+d$ is divisible by $2^k$.\n\nDenote $a = 2^k x$, $a+d = 2^{k-1}y$, $a+2d = 2^k z$, where $x, y, z$ are odd. Then for them the following must hold:\n\n$$\na + (a + 2d) = 2(a + d) \\Rightarrow 2^k x + 2^k z = 2^k y \\Rightarrow x + z = y,\n$$\n\nwhich can't hold for odd integers. This contradiction completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15823,
    "subject": "Mathematics (Olympiad)",
    "question": "The numbers $1, 2, \\ldots, 2012$ are written on the blackboard in some order, each of them exactly once. Between each two neighboring numbers, the absolute value of their difference is written and the original numbers are erased. This process is repeated until only one number is left on the blackboard. What is the largest possible number that can be left on the blackboard?",
    "options": [],
    "answer": "See solution",
    "solution": "$2010$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15824,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve in the real numbers the system\n\n$$\n\\begin{aligned}\nx^3 &= \\frac{z}{y} - \\frac{2y}{z}, \\\\\ny^3 &= \\frac{x}{z} - \\frac{2z}{x}, \\\\\nz^3 &= \\frac{y}{x} - \\frac{2x}{y}.\n\\end{aligned}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For $x, y, z \\in \\mathbb{R}$, such that $xyz \\neq 0$, the system can be rewritten as:\n\n$$\n\\begin{aligned}\nx^3 y z &= z^2 - 2y^2 \\quad (1), \\\\\ny^3 z x &= x^2 - 2z^2 \\quad (2), \\\\\nz^3 x y &= y^2 - 2x^2 \\quad (3)\n\\end{aligned}\n$$\n\nSumming these equations, we find:\n\n$$\nxyz(x^2 + y^2 + z^2) = -(x^2 + y^2 + z^2) \\implies (x^2 + y^2 + z^2)(xyz + 1) = 0.\n$$\n\nSince $xyz \\neq 0$, we have $x^2 + y^2 + z^2 > 0$, so $xyz = -1$.\n\nSubstituting $xyz = -1$ into (1)-(3), we get:\n\n$$\n\\begin{aligned}\nx^2 &= -z^2 + 2y^2 \\quad (5), \\\\\ny^2 &= -x^2 + 2z^2 \\quad (6), \\\\\nz^2 &= -y^2 + 2x^2 \\quad (7)\n\\end{aligned}\n$$\n\nFrom (5) and (6), $y^2 = z^2$; from (6) and (7), $x^2 = z^2$. Thus,\n\n$$\nx^2 = y^2 = z^2 \\implies x = y = \\pm z \\quad \\text{or} \\quad x = -y = \\pm z.\n$$\n\nFinally, using $xyz = -1$, the solutions are:\n\n$$\n(x, y, z) = (-1, -1, -1), \\quad (1, 1, -1), \\quad (1, -1, 1), \\quad (-1, 1, 1).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15825,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer divisible by $4$. Prove that\n\n$$\nsin^2 \\left( 1 \\cdot \\frac{360^\\circ}{n} \\right) + \\sin^2 \\left( 2 \\cdot \\frac{360^\\circ}{n} \\right) + \\sin^2 \\left( 3 \\cdot \\frac{360^\\circ}{n} \\right) + \\dots + \\sin^2 \\left( n \\cdot \\frac{360^\\circ}{n} \\right) = \\frac{n}{2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = 4k$. For each $i = 1, 2, \\dots, n$, denote $\\alpha_i = i \\cdot \\frac{360^\\circ}{n}$ and $x_i = \\sin^2 \\alpha_i$. We pair the terms as $(x_1, x_{k+1}), (x_2, x_{k+2}), \\dots, (x_k, x_{2k})$ and $(x_{2k+1}, x_{3k+1}), (x_{2k+2}, x_{3k+2}), \\dots, (x_{3k}, x_{4k})$.\n\nSince\n\n$$\n\\alpha_{i+k} = (i + k) \\cdot \\frac{360^\\circ}{n} = i \\cdot \\frac{360^\\circ}{n} + k \\cdot \\frac{360^\\circ}{n} = \\alpha_i + 90^\\circ,\n$$\n\nwe have\n\n$$\n\\sin^2 \\alpha_{i+k} = \\sin^2 (\\alpha_i + 90^\\circ) = \\cos^2 \\alpha_i.\n$$\n\nTherefore $x_i + x_{i+k} = \\sin^2 \\alpha_i + \\cos^2 \\alpha_i = 1$. Thus, the sum of the numbers of each pair is $1$. Since we have $\\frac{n}{2}$ pairs, the sum of all numbers is $\\frac{n}{2}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15826,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(x) = x^{2012} + a_{2011}x^{2011} + a_{2010}x^{2010} + \\dots + a_1x + a_0$ be a polynomial of degree $2012$ with real coefficients and leading coefficient $1$. Find the minimum real number $c$ such that for any polynomial obtained by changing some of the coefficients of $P(x)$ to their opposite numbers, every root $z$ satisfies $|\\operatorname{Im} z| \\leq c |\\operatorname{Re} z|$, where $\\operatorname{Re} z$ and $\\operatorname{Im} z$ are the real and imaginary parts of $z$, respectively.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we point out that $c \\geq \\cot \\frac{\\pi}{4022}$. Consider the polynomial $P(x) = x^{2012} - x$. Changing the sign of coefficients of $P(x)$, we obtain four polynomials: $P(x)$, $-P(x)$, $Q(x) = x^{2012} + x$, and $-Q(x)$. Note that $P(x)$ and $-P(x)$ have the same roots; one of the roots is $z_1 = \\cos \\frac{1006}{2011}\\pi + i \\sin \\frac{1006}{2011}\\pi$. $Q(x)$ and $-Q(x)$ have the same roots, and are the opposite numbers of the roots of $P(x)$. Thus $Q(x)$ has a root $z_2 = -z_1$. Then,\n\n$$\nc \\geq \\min\\left(\\frac{|\\operatorname{Im} z_1|}{|\\operatorname{Re} z_1|}, \\frac{|\\operatorname{Im} z_2|}{|\\operatorname{Re} z_2|}\\right) = \\cot \\frac{\\pi}{4022}.\n$$\n\nNext, we show that the answer is $c = \\cot \\frac{\\pi}{4022}$. For any\n\n$$\nP(x) = x^{2012} + a_{2011}x^{2011} + a_{2010}x^{2010} + \\dots + a_1x + a_0,\n$$\n\nwe obtain a polynomial\n\n$$\nR(x) = b_{2012}x^{2012} + b_{2011}x^{2011} + b_{2010}x^{2010} + \\dots + b_1x + b_0,\n$$\n\nby changing the sign of some coefficients of $P(x)$, where $b_{2012} = 1$, and for $j = 1, 2, \\dots, 2011$,\n\n$$\nb_j = \\begin{cases} |a_j|, & j \\equiv 0, 1 \\pmod 4, \\\\ -|a_j|, & j \\equiv 2, 3 \\pmod 4. \\end{cases}\n$$\n\nWe show that, for each root $z$ of $R(x)$, we have $|\\operatorname{Im} z| \\leq c |\\operatorname{Re} z|$.\n\nWe prove this result by contradiction. Suppose there is a root $z_0$ of $R(x)$ such that $|\\operatorname{Im} z_0| > c |\\operatorname{Re} z_0|$, then $z_0 \\neq 0$ and either the angle of $z_0$ and $i$ is less than $\\theta = \\frac{\\pi}{4022}$, or the angle of $z_0$ and $-i$ is less than $\\theta$. Suppose that the angle of $z_0$ and $i$ is less than $\\theta$; for the other case, we need only consider the conjugate of $z_0$. There are two cases:\n\nIf $z_0$ is in the first quadrant (or on the imaginary axis), suppose that $\\angle(z_0, i) = \\alpha < \\theta$, where $\\angle(z_0, i)$ is the least angle that rotates $z_0$ to $i$ anticlockwise. For $0 \\leq j \\leq 2012$, if $j \\equiv 0, 2 \\pmod 4$, then $\\angle(b_j z_0^j, 1) = j\\alpha \\leq 2012\\alpha < 2012\\theta$.\n\nIf $j \\equiv 1, 3 \\pmod 4$, then $\\angle(b_j z_0^j, i) = j\\alpha < 2011\\theta$ and $\\angle(b_1 z_0, i) = \\alpha$. Thus, the principal argument of $b_j z_0^j \\in [2\\pi - 2012\\alpha, 2\\pi) \\cup [0, \\frac{1}{2}\\pi - \\alpha]$. The vertex angle of this angle-domain is $2012\\alpha + \\frac{1}{2}\\pi - \\alpha = \\frac{1}{2}\\pi + 2011\\alpha < \\pi$. And $b_j z_0^j$, $0 \\leq j \\leq 2012$, are not all zero, so their sum cannot be zero.\n\nIf $z_0$ is in the second quadrant, suppose that $\\angle(i, z_0) = \\alpha < \\theta$, if $j \\equiv 0, 2 \\pmod 4$, then $\\angle(1, b_j z_0^j) = j\\alpha < 2012\\theta$. If $j \\equiv 1, 3 \\pmod 4$, then $\\angle(i, b_j z_0^j) = j\\alpha \\leq 2011\\alpha < \\frac{\\pi}{2}$. Thus, every principal argument of $b_j z_0^j \\in [0, \\frac{\\pi}{2} + 2011\\alpha]$. Since $\\frac{\\pi}{2} + 2011\\alpha < \\pi$, and $b_j z_0^j$, $0 \\leq j \\leq 2012$, are not all zero, so their sum cannot be zero.\n\nSumming up, the least real number $c = \\cot \\frac{\\pi}{4022}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15827,
    "subject": "Mathematics (Olympiad)",
    "question": "Дараалал $\\{a_n\\}$ нь $a_0 = 20$, $a_1 = 100$, $n \\ge 0$ үед\n\n$$\na_{n+2} = 4a_{n+1} + 5a_n + 20\n$$\n\nрекуррент томъёогоор өгөгджээ. Аливаа $n \\in \\mathbb{N}$-ийн хувьд $a_{n+m} - a_n \\equiv 0 \\pmod{2012}$ байх хамгийн бага $m \\in \\mathbb{N}$ тоог ол.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\begin{align*}\na_{n+2} + \\frac{5}{2} &= 4\\left(a_{n+1} + \\frac{5}{2}\\right) + 5\\left(a_n + \\frac{5}{2}\\right) \\\\\n\\text{Орлуулга хийвэл } b_n = a_n + \\frac{5}{2} \\Rightarrow b_{n+2} = 4b_{n+1} + 5b_n \\\\\n\\Rightarrow b_n = c_1 \\cdot 5^n + c_2(-1)^n \\\\\nb_0 = \\frac{45}{2}, \\quad b_1 = \\frac{205}{2} \\Rightarrow c_1 = \\frac{125}{6}, \\quad c_2 = \\frac{10}{6} \\\\\na_n = \\frac{125}{6} \\cdot 5^n + \\frac{10}{6} \\cdot (-1)^n - \\frac{5}{2}\n\\end{align*}\n$$\n\nӨөрөөр хэлбэл\n\n$$\na_n = \\begin{cases}\n\\frac{5^{n+3} - 5}{5^{n+3} - 25}, & n \\text{ - тэгш тоо} \\\\\n\\frac{5}{6}, & n \\text{ - сондгой тоо}\n\\end{cases}\n$$\n\nЭндээс $n$ тэгш тоо бол $a_{n+1} = 5a_n$ гэж гарна. Нөхцөлөөс $a_{n+m} \\equiv a_n \\pmod{2012}$. $a_m \\equiv a_0 = 20 \\pmod{2012}$, $a_{m+1} \\equiv a_1 = 100 \\pmod{2012}$ тул $m \\ge 1$.\n\n$5a_{m-1} = a_{m+1} - 4a_m - 20 = 100 - 4 \\cdot 20 - 20 = 0 \\pmod{2012}$, өөрөөр хэлбэл $a_{m-1} \\equiv 0 \\pmod{2012}$. Хэрэв $m$ сондгой тоо бол $m-1$ тэгш тоо бөгөөд $a_m = 5a_{m-1} \\equiv 0 \\pmod{2012}$ буюу $a_m \\equiv 20 \\pmod{2012}$ зөрчинө. Иймд $m$ тэгш тоо.\n\n$a_{m-1} = 25 \\cdot \\frac{5^m - 1}{6} \\equiv 0 \\pmod{2012}$\n\n$$\n\\Leftrightarrow \\frac{5^m - 1}{6} \\equiv 0 \\pmod{2012} \\Leftrightarrow 5^m \\equiv 1 \\pmod{6 \\cdot 2012}\n$$\n\n$6 \\cdot 2012 = 2^3 \\cdot 3 \\cdot 503$ тул $m$ тэгш үед $5^m \\equiv 1 \\pmod{8}$, $5^m \\equiv 1 \\pmod{3}$ биелэх тул $5^m \\equiv 1 \\pmod{503}$ болно. Иймд хамгийн бага $m = 502$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15828,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\in \\mathbb{N}^*$, $n \\ge 4$, and $G_n = \\{1, 2, \\dots, n\\}$. Prove that there exists a permutation $P_1, P_2, \\dots, P_{2^n - n - 1}$, where $P_i \\subseteq G_n$, $|P_i| \\ge 2$ for $i = 1, 2, \\dots, 2^n - n - 1$, such that\n\n$$\n|P_i \\cap P_{i+1}| = 2, \\quad i = 1, 2, \\dots, 2^n - n - 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof**\n\nOur proof will require the following lemma.\n\n*Lemma 4.* For $n \\in \\mathbb{N}^*$, $n \\ge 3$, there exists a permutation $Q_1, Q_2, \\dots, Q_{2^n - 1}$, where $Q_i \\subseteq G_n$, $|Q_i| \\ge 1$, $1 \\le i \\le 2^n - 1$, such that\n\n$$\nQ_1 = \\{1\\}, \\quad |Q_i \\cap Q_{i+1}| = 1, \\quad 1 \\le i \\le 2^n - 2, \\quad Q_{2^n - 1} = G_n.\n$$\n\nFirstly, for $n=3$, the permutation\n\n$$\n\\{1\\}, \\{1, 2\\}, \\{2\\}, \\{2, 3\\}, \\{1, 3\\}, \\{3\\}, \\{1, 2, 3\\}\n$$\nsatisfies the given condition.\n\nSecondly, suppose the lemma is true for $n$. Let $Q_1, Q_2, \\dots, Q_{2^{n-1}}$ satisfy the conditions in the lemma. Then construct the following sequence:\n\n$Q_1, Q_{2^{n-1}}, Q_{2^{n-2}}, Q_{2^{n-2}} \\cup \\{n+1\\}, Q_{2^{n-3}}, Q_{2^{n-4}} \\cup \\{n+1\\}, Q_{2^{n-5}}, \\dots, Q_3, Q_2 \\cup \\{n+1\\}, \\{n+1\\}, Q_1 \\cup \\{n+1\\}, Q_2, Q_3 \\cup \\{n+1\\}, Q_4, \\dots, Q_{2^{n-2}}, Q_{2^{n-1}} \\cup \\{n+1\\}$\n\nIt is easy to check that this sequence satisfies the lemma for $n+1$.\n\nBack to the problem, we prove that for $n \\in \\mathbb{N}^*$, $n \\ge 4$, there exists a permutation satisfying the required conditions and $P_{2^{n-1}} = \\{1, n\\}$.\n\nWhen $n=4$, the permutation\n\n$$\n\\{1, 3\\}, \\{1, 2, 3\\}, \\{2, 3\\}, \\{1, 2, 3, 4\\}, \\{1, 2\\}, \\{1, 2, 4\\}, \\{2, 4\\}, \\{2, 3, 4\\}, \\{3, 4\\}, \\{1, 3, 4\\}, \\{1, 4\\}\n$$\nalso satisfies the given condition.\n\nNow suppose the permutation $P_1, P_2, \\dots, P_{2^{n-1}}$ satisfies the required conditions and $P_{2^{n-1}} = \\{1, n\\}$. Using the lemma, let the permutation $Q_1, Q_2, \\dots, Q_{2^{n-1}}$ satisfy the lemma's conditions. Then for $n+1$, the following permutation\n\n$P_1, P_2, \\dots, P_{2^{n-1}}, Q_{2^{n-1}} \\cup \\{n+1\\}, Q_{2^{n-2}} \\cup \\{n+1\\}, \\dots, Q_1 \\cup \\{n+1\\}$\n\nsatisfies the required conditions and $P_{2^{n+1-(n+1)-1}} = Q_1 \\cup \\{n+1\\} = \\{1, n+1\\}$.\n\nTherefore, there exists a suitable permutation for $n \\ge 4$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 15829,
    "subject": "Mathematics (Olympiad)",
    "question": "We have $n > 2$ nonzero integers such that each one of them is divisible by the sum of the other $n - 1$ numbers. Show that the sum of the $n$ numbers is precisely $0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let these numbers be $a_1, a_2, \\dots, a_n$ and let $S$ be their sum. For every $i \\in \\{1, 2, \\dots, n\\}$, we have $S - a_i \\mid a_i$, which means $S - a_i$ does not divide $S$.\n\nSuppose $S \\neq 0$. Without loss of generality, assume $S > 0$. Consider two cases:\n\n* If there exists $i$ such that $a_i < 0$, then $|S - a_i| > |S|$, so $S - a_i$ does not divide $S$.\n\n* If $a_i > 0$ for all $i$, let $a_k$ be the smallest among $a_1, a_2, \\dots, a_n$. Then $S - a_k > 2a_k - a_k = a_k$, so $S - a_k$ does not divide $a_k$.\n\nHence, $S \\neq 0$ is impossible, so $S = 0$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15830,
    "subject": "Mathematics (Olympiad)",
    "question": "Anna and Bob play a game on the set of all points of the form $(m, n)$ where $m, n$ are integers with $|m|, |n| \\le 2019$. The lines $x = \\pm 2019$ and $y = \\pm 2019$ are called the boundary lines of the game, and the points on these lines are called the boundary points. The neighbours of a point $(m, n)$ are the points $(m + 1, n)$, $(m - 1, n)$, $(m, n + 1)$, $(m, n - 1)$.\n\nAnna starts with a token at the origin $(0, 0)$. With Bob playing first, they alternately perform the following steps:\n- On his turn, Bob deletes two points on each boundary line.\n- On her turn, Anna makes a sequence of three moves of the token, where a *move* consists of picking up the token from its current position and placing it in one of its neighbours.\n\nTo win the game, Anna must place her token on a boundary point before it is deleted by Bob.\n\n[Note: At every turn except perhaps her last, Anna **must** make **exactly** three moves.]\n",
    "options": [],
    "answer": "See solution",
    "solution": "Anna does not have a winning strategy. We will provide a winning strategy for Bob, focusing on the deletions on the line $y = 2019$.\n\nBob starts by deleting $(0, 2019)$ and $(-1, 2019)$. After Anna's step, he deletes the next two available points on the left if Anna decreased her $x$-coordinate, the next two available points on the right if Anna increased her $x$-coordinate, and the next available point to the left and the next available point to the right if Anna did not change her $x$-coordinate. The only exception is the first time Anna decreases $x$ by exactly 1; then, Bob deletes the next available point to the left and the next available point to the right.\n\nBob's strategy ensures: If Anna makes a sequence of steps reaching $(-x, y)$ with $x > 0$ and the exact opposite sequence reaching $(x, y)$, then Bob deletes at least as many points to the left of $(0, 2019)$ in the first sequence as to the right in the second.\n\nSuppose, for contradiction, that Anna wins by placing her token at $(k, 2019)$ for some $k > 0$.\n\nDefine $\\Delta = 3m - (2x + y)$, where $m$ is the total number of points deleted by Bob to the right of $(0, 2019)$, and $(x, y)$ is Anna's token position.\n\nFor each sequence of steps (Anna then Bob), $\\Delta$ does not decrease. This is shown in the following table:\n\n$$\n\\begin{array}{|c|c|c|c|c|c|c|c|c|c|}\n\\hline\n\\text{Step} & (0,3) & (1,2) & (-1,2) & (2,1) & (0,1) & (3,0) & (1,0) & (2,-1) & (1,-2) \\\\\n\\hline\nm & 1 & 2 & 0~\\text{(or 1)} & 2 & 1 & 2 & 2 & 2 & 2 \\\\\n\\hline\n3m & 3 & 6 & 0~\\text{(or 3)} & 6 & 3 & 6 & 6 & 6 & 6 \\\\\n\\hline\n2x + y & 3 & 4 & 0 & 5 & 1 & 6 & 2 & 3 & 0 \\\\\n\\hline\n\\end{array}\n$$\n\nThe table shows that if Anna changes $y$ by $+1$ or $-2$, $\\Delta$ increases by 1. If Anna changes $y$ by $+2$ or $-1$, the first time this happens $\\Delta$ increases by 2. (This also holds for $(0, -1)$ or $(-2, -1)$, not shown.)\n\nSince Anna wins by placing her token at $(k, 2019)$, we must have $m \\le k - 1$ and $k \\le 2018$. So at that moment:\n\n$$\n\\Delta = 3m - (2k + 2019) = k - 2022 \\le -4.\n$$\n\nThus, in her last turn, Anna must have decreased $\\Delta$ by at least 4. So her last step must have been $(1, 2)$ or $(2, 1)$, which give a decrease of 4 and 5, respectively. (It could not be $(3, 0)$, as she would have already won. Nor could she have made just one or two moves in her last turn, as this is insufficient for the required decrease in $\\Delta$.)\n\nIf her last step was $(1, 2)$, then just before, $y = 2017$ and $\\Delta = 0$. This means that in one of her steps, the total change in $y$ was not $0 \\pmod{3}$. In that case, $\\Delta > 0$, a contradiction.\n\nIf her last step was $(2, 1)$, then just before, $y = 2018$ and $\\Delta = 0$ or $1$. So she must have made at least two steps with $y$ change $+1$ or $-2$, or at least one step with $y$ change $+2$ or $-1$. In both cases, consulting the table, $\\Delta$ increases by at least 2, a contradiction.\n\n**Note 1:** If Anna is allowed to make at most three moves at each step, then she actually has a winning strategy.\n\n**Note 2:** If $2019$ is replaced by $N > 1$, then Bob has a winning strategy if and only if $3 \\mid N$.\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15831,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive real numbers $t$ such that there exists an infinite set $X$ of real numbers with the property that for all (not necessarily distinct) $x, y, z \\in X$, all real numbers $a$, and all positive real numbers $d$, the inequality\n\n$$\n\\max\\{|x - (a - d)|,\\ |y - a|,\\ |z - (a + d)|\\} > td\n$$\nholds.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $0 < t < \\frac{1}{2}$.\n\n**Case 1: $0 < t < \\frac{1}{2}$**\n\nLet $\\lambda \\in \\left(0, \\frac{1-2t}{2(1+t)}\\right)$ and define $x_i = \\lambda^i$, $X = \\{x_1, x_2, \\dots\\}$. We claim that for all $x, y, z \\in X$, all $a \\in \\mathbb{R}$, and all $d > 0$,\n\n$$\n\\max\\{|x - (a - d)|,\\ |y - a|,\\ |z - (a + d)|\\} > td.\n$$\n\nSuppose, for contradiction, that there exist $a \\in \\mathbb{R}$, $d > 0$, and $x_i, x_j, x_k$ such that\n\n$$\n\\max\\{|x_i - (a - d)|,\\ |x_j - a|,\\ |x_k - (a + d)|\\} \\le td.\n$$\n\nThis gives\n$$\n\\begin{cases}\n-td \\le x_i - (a - d) \\le td, \\\\\n-td \\le x_j - a \\le td, \\\\\n-td \\le x_k - (a + d) \\le td,\n\\end{cases}\n$$\nwhich is equivalent to\n$$\n\\begin{cases}\nx_i + (1-t)d \\le a \\le x_i + (1+t)d, \\\\\nx_j - td \\le a \\le x_j + td, \\\\\nx_k - (1+t)d \\le a \\le x_k - (1-t)d.\n\\end{cases}\n$$\n\nThis implies\n$$\n\\begin{cases}\nx_k - (1+t)d \\le a \\le x_i + (1+t)d, \\\\\nx_i + (1-t)d \\le a \\le x_j + td, \\\\\nx_j - td \\le a \\le x_k - (1-t)d.\n\\end{cases}\n$$\n\nSince $0 < t < \\frac{1}{2}$, it follows that\n$$\n\\begin{cases}\nd \\ge \\frac{x_k - x_i}{2(1+t)}, \\quad \\textcircled{1} \\\\\nd \\le \\frac{x_j - x_i}{1-2t}, \\quad \\textcircled{2} \\\\\nd \\le \\frac{x_k - x_j}{1-2t}. \\quad \\textcircled{3}\n\\end{cases}\n$$\n\nFrom (2) and (3) and $d > 0$, we get $x_i < x_j < x_k$. Thus $i > j > k$, and $\\lambda^j + \\lambda^{j+1} \\le \\lambda^{k+1} + \\lambda^i$, so\n$$\n\\frac{x_j - x_i}{x_k - x_i} = \\frac{\\lambda^j - \\lambda^i}{\\lambda^k - \\lambda^i} \\le \\lambda. \\quad \\textcircled{4}\n$$\n\nFrom (1) and (2),\n$$\n\\frac{x_j - x_i}{1 - 2t} \\ge \\frac{x_k - x_i}{2(1 + t)} \\implies \\frac{x_j - x_i}{x_k - x_i} \\ge \\frac{1 - 2t}{2(1 + t)} > \\lambda,\n$$\nwhich contradicts (4). Thus, our claim about $X$ is proved.\n\n**Case 2: $t \\ge \\frac{1}{2}$**\n\nFor any infinite set $X$, for any $x < y < z$ in $X$, we can choose $a \\in \\mathbb{R}$ and $d > 0$ such that\n$$\n\\max\\{|x - (a - d)|,\\ |y - a|,\\ |z - (a + d)|\\} \\le td.\n$$\n\nLet $d = \\frac{z-x}{2}$, so $x + (1-t)d = z - (1+t)d$. Let $a = \\max\\{x + (1-t)d,\\ y - td\\}$. Since $t \\ge \\frac{1}{2}$,\n$$\n\\begin{cases}\ny - x < 2d \\le (1 + 2t)d, \\\\\nx - y < 0 \\le (2t - 1)d,\n\\end{cases}\n$$\ni.e.,\n$$\n\\begin{cases}\ny - td \\le x + (1 + t)d, \\\\\nx + (1 - t)d \\le y + td.\n\\end{cases}\n$$\n\nTherefore, the only possible values are $0 < t < \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15832,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m = da$ and $n = db$, where $a$ and $b$ are relatively prime integers, and let $v = dab$. Plugging these into the given equation, we get:\n\n$$da + 3db - 5 = 2dab - 11d$$\n\nFind all integer solutions $(m, n)$ to this equation.",
    "options": [],
    "answer": "See solution",
    "solution": "We see that $d$ must divide $5$. \n\nFirst, let us assume $d = 1$. Then the equation becomes:\n\n$$a(2b - 1) = 3b + 6$$\n\nSo $2b - 1$ divides $3b + 6$, and thus must also divide $2(3b + 6) - 3(2b - 1) = 15$. Since $2b - 1$ is a positive integer, possible values are $1, 3, 5, 15$.\n\n- If $2b - 1 = 1$, then $b = 1$, $a = 9$.\n- If $2b - 1 = 3$, then $b = 2$, $a = 4$ (but $a$ and $b$ are not coprime).\n- If $2b - 1 = 5$, then $b = 3$, $a = 3$ (not coprime).\n- If $2b - 1 = 15$, then $b = 8$, $a = 2$ (not coprime).\n\nSo only $(a, b) = (9, 1)$ works, giving $(m, n) = (9, 1)$.\n\nNow, assume $d = 5$. Divide the equation by $5$:\n\n$$a(2b - 1) = 3b + 10$$\n\nSo $2b - 1$ divides $3b + 10$, and must also divide $2(3b + 10) - 3(2b - 1) = 23$. Possible values: $1, 23$.\n\n- If $2b - 1 = 1$, then $b = 1$, $a = 13$.\n- If $2b - 1 = 23$, then $b = 12$, $a = 2$ (not coprime).\n\nSo only $(a, b) = (13, 1)$ works, giving $(m, n) = (65, 5)$.\n\n**Final solutions:** $(m, n) = (9, 1)$ and $(65, 5)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15833,
    "subject": "Mathematics (Olympiad)",
    "question": "It is known that the curve $f(x) = |\\sin x|$ intercepts the line $y = kx$ ($k > 0$) at exactly three points, the maximum $x$ coordinate of these points being $\\alpha$. Prove that\n\n$$\n\\frac{\\cos \\alpha}{\\sin \\alpha + \\sin 3\\alpha} = \\frac{1 + \\alpha^2}{4\\alpha}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The image of the three intercepting points of $f(x)$ and $y = kx$ is shown in the figure. It is easy to see that the curve and the line are tangent to each other at point $A(\\alpha, -\\sin \\alpha)$, and $\\alpha \\in (\\pi, \\frac{3\\pi}{2})$.\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p39_data_af068f67ec.png)\n\nAs $f'(x) = -\\cos x$ for $x \\in (\\pi, \\frac{3\\pi}{2})$, we have $-\\cos \\alpha = -\\frac{\\sin \\alpha}{\\alpha}$, i.e. $\\alpha = \\tan \\alpha$. Then\n\n$$\n\\begin{align*}\n\\frac{\\cos \\alpha}{\\sin \\alpha + \\sin 3\\alpha} &= \\frac{\\cos \\alpha}{2\\sin 2\\alpha \\cos \\alpha} = \\frac{1}{4\\sin \\alpha \\cos \\alpha} \\\\\n&= \\frac{\\cos^2 \\alpha + \\sin^2 \\alpha}{4\\sin \\alpha \\cos \\alpha} = \\frac{1 + \\tan^2 \\alpha}{4\\tan \\alpha} \\\\\n&= \\frac{1 + \\alpha^2}{4\\alpha}.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15834,
    "subject": "Mathematics (Olympiad)",
    "question": "At the start of the Mighty Mathematicians Football Team's first game of the season, their coach noticed that the jersey numbers of the 22 players on the field were all the numbers from 1 to 22. At halftime, the coach substituted her goalkeeper, with jersey number 1, for a reserve player. No other substitutions were made by either team at or before halftime. The coach noticed that after the substitution, no two players on the field had the same jersey number and that the sums of the jersey numbers of each of the teams were exactly equal.\n\nDetermine:\n\n(a) the greatest possible jersey number of the reserve player,\n\n(b) the smallest possible (positive) jersey number of the reserve player.",
    "options": [],
    "answer": "See solution",
    "solution": "If we leave out the reserve player, the greatest possible difference between the jersey numbers of the two teams is obtained when the reserve player's team has jersey numbers 2–11, while the other team has numbers 12–22. The difference in this case is\n\n$$\n(12 + 13 + \\cdots + 22) - (2 + 3 + \\cdots + 11) = 122;\n$$\n\nwhich is therefore the greatest possible jersey number of the reserve player.\n\nOn the other hand, since\n\n$$\n2 + 3 + 4 + \\cdots + 21 + 22 = 252\n$$\n\nand the total sum of all jersey numbers must be even for the two teams to have the same sum, the reserve player must have an even jersey number. The smallest positive even number that is not already taken by another player is 24, and indeed it is possible that the sums of the two teams are the same in this case, for example:\n\n2, 5, 6, 9, 10, 13, 14, 17, 18, 20, 24 vs. 3, 4, 7, 8, 11, 12, 15, 16, 19, 21, 22.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15835,
    "subject": "Mathematics (Olympiad)",
    "question": "How many positive integers $n \\leq 600$ are uniquely determined by the values of $\\left\\lfloor \\frac{n}{4} \\right\\rfloor$, $\\left\\lfloor \\frac{n}{5} \\right\\rfloor$, and $\\left\\lfloor \\frac{n}{6} \\right\\rfloor$? That is, for which $n$ does the ordered triple $\\left( \\left\\lfloor \\frac{n}{4} \\right\\rfloor, \\left\\lfloor \\frac{n}{5} \\right\\rfloor, \\left\\lfloor \\frac{n}{6} \\right\\rfloor \\right)$ uniquely determine $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "Call an integer $n$ *good* if it is uniquely determined by the values of $\\left\\lfloor \\frac{n}{4} \\right\\rfloor$, $\\left\\lfloor \\frac{n}{5} \\right\\rfloor$, and $\\left\\lfloor \\frac{n}{6} \\right\\rfloor$.\n\nIf $n$ is good, then the ordered triples $\\left( \\left\\lfloor \\frac{n}{4} \\right\\rfloor, \\left\\lfloor \\frac{n}{5} \\right\\rfloor, \\left\\lfloor \\frac{n}{6} \\right\\rfloor \\right)$ and $\\left( \\left\\lfloor \\frac{n-1}{4} \\right\\rfloor, \\left\\lfloor \\frac{n-1}{5} \\right\\rfloor, \\left\\lfloor \\frac{n-1}{6} \\right\\rfloor \\right)$ are not identical, so they must differ in at least one coordinate. This implies that $n$ is a multiple of $4$, $5$, or $6$. Similarly, $\\left( \\left\\lfloor \\frac{n}{4} \\right\\rfloor, \\left\\lfloor \\frac{n}{5} \\right\\rfloor, \\left\\lfloor \\frac{n}{6} \\right\\rfloor \\right)$ and $\\left( \\left\\lfloor \\frac{n+1}{4} \\right\\rfloor, \\left\\lfloor \\frac{n+1}{5} \\right\\rfloor, \\left\\lfloor \\frac{n+1}{6} \\right\\rfloor \\right)$ are not identical, so $n + 1$ is a multiple of $4$, $5$, or $6$. Because it is impossible for both $n$ and $n + 1$ to be even, one of these must be a multiple of $5$. These conditions are both necessary and sufficient.\n\nAssume first that $n$ is a multiple of $5$ and $n + 1$ is a multiple of $4$ or $6$ or both. Then $n \\equiv 0 \\pmod{5}$ and $n$ is congruent to $-1$ modulo either $4$ or $6$, implying that $n$ is $3$, $5$, $7$, or $11$ modulo $12$. In this case, the Chinese Remainder Theorem implies that there are $4$ good values of $n$ from $1$ through $60$. They are $5$, $15$, $35$, and $55$.\n\nNext, assume that $n + 1$ is a multiple of $5$ and $n$ is a multiple of $4$ or $6$ or both. Then $n \\equiv -1 \\pmod{5}$ and $n \\equiv 0, 4, 6, \\text{ or } 8 \\pmod{12}$, and again there are $4$ good values of $n$ from $1$ through $60$. They are $4$, $24$, $44$, and $54$.\n\nHence, there are $4 + 4 = 8$ good integers from $1$ through $60$, so there are $8 \\times 10 = 80$ good positive integers less than or equal to $600$.\n\nAlternatively, consider intervals:\n\nFix a positive integer $n$ between $1$ and $60 = \\mathrm{lcm}(4, 5, 6)$. The set of integers $m$ such that $\\left\\lfloor \\frac{m}{4} \\right\\rfloor = \\left\\lfloor \\frac{n}{4} \\right\\rfloor$ is an interval of four consecutive integers, where the least is divisible by $4$. Similarly, for $5$ and $6$. $n$ is good if and only if these three intervals intersect in exactly one point.\n\nThere are $4$ locations for the interval of length $4$ and $2$ ways to place the interval of length $5$ so that all three intervals intersect at a single point, giving $4 \\times 2 = 8$ configurations. These are shown below:\n\n![](images/2022AIME_II_Solutions_p6_data_ed9d2cbfb0.png)\n\nBy the Chinese Remainder Theorem, for every integer $m$, each configuration above can be achieved by exactly one integer $n$ in $\\{m + 1, \\ldots, m + 60\\}$. Thus, $8$ values in $\\{1, \\ldots, 60\\}$ are good, so $8 \\times 10 = 80$ values in $\\{1, \\ldots, 600\\}$ are good.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15836,
    "subject": "Mathematics (Olympiad)",
    "question": "In the acute triangle $ABC$, point $I$ is the incenter, $O$ is the circumcenter, and $I_a$ is the excenter opposite vertex $A$. Point $A'$ is the reflection of $A$ across the line $BC$. Prove that angles $\\angle IOI_a$ and $\\angle IA'I_a$ are equal.",
    "options": [],
    "answer": "See solution",
    "solution": "The quadrilateral $IBI_aC$ is cyclic, so $\\angle AI_aC = \\angle IBC = \\angle ABI$. From $\\angle BAI = \\angle I_aAC$, it follows that triangles $ABI$ and $AI_aC$ are similar (AA). Thus, $AI \\cdot AI_a = AB \\cdot AC$.\n\nOn the other hand, $AB \\cdot AC = \\frac{2S}{\\sin A} = \\frac{a \\cdot h_a}{\\sin A} = 2R \\cdot h_a = AA' \\cdot AO$. Since $\\angle A'AI = \\angle I_aAO$ (because $AA'$ and $AO$ are isogonal in $\\angle BAC$), triangles $AOI_a$ and $AIA'$ are similar.\n\nLet $J$ be the point where the parallel through $I$ to $I_aA'$ intersects $AA'$. Then $\\frac{AJ}{AI} = \\frac{AA'}{AI_a} = \\frac{AO}{AI_a}$, so triangles $AJI$ and $AIO$ are similar. It follows that $\\angle IOI_a = \\angle AIO_a - \\angle AOI = \\angle AIA' - \\angle AIJ = \\angle JIA' = \\angle IA'I_a$, as required.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15837,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a circle $O$ with diameter $AB$ on the plane. A point $P$ moves on the tangent at $B$ to $O$. The line $PA$ intersects $O$ at a second point $C$. Let $D$ be the point symmetric to $C$ with respect to $O$. The line $PD$ intersects $O$ at a second point $E$.\n\n1. Show that the lines $AE$, $BC$, and $PO$ pass through a common point. Call this point $M$.\n\n2. Determine the locus of $P$ such that triangle $ABC$ has maximum area. Compute that maximum area in terms of the radius of $O$.\n\n($O$ denotes a circle with center $O$.)",
    "options": [],
    "answer": "See solution",
    "solution": "1. Let $F$ be the intersection of lines $AE$ and $BP$.\n\nWe have $\\angle ACE = 90^\\circ + \\angle BCE = 90^\\circ + \\angle FAB = \\angle EFP$. Consequently, $\\angle EFP + \\angle ECP = 180^\\circ$.\n\nHence, $CEFP$ is a cyclic quadrilateral. Consequently, $\\angle CFP = \\angle CEP = 90^\\circ$. Thus, $CF \\parallel AB$.\n\n$$\n\\text{Hence } CP = FP \\\\\nCA = FB.\n$$\n\nWhence, considering triangle $ABP$, we have\n$$\n\\frac{CP}{CA} = \\frac{OA}{OB} = \\frac{FB}{FP} = \\frac{OA}{OB} = -1.\n$$\n\nHence, according to Ceva's theorem, the lines $PO$, $AE$, and $BC$ are concurrent.\n\n2. Let $BP = x$ and denote $R$ the radius of $O$.\n\nConsider the right triangle $ABP$, we have $PA = \\sqrt{PB^2 + AB^2} = \\sqrt{x^2 + 4R^2}$.\n\nConsequently, $PC = \\frac{PB^2}{PA} = \\frac{x^2}{\\sqrt{x^2 + 4R^2}}$ and $AC = PA - PC = \\frac{4R^2}{\\sqrt{x^2 + 4R^2}}$.\n\nSince $CF \\parallel AB$ (see proof above), one has $\\frac{MC}{MB} = \\frac{CF}{AB} = \\frac{PC}{PA}$.\n\nConsequently, $\\frac{BC}{MB} = \\frac{PC}{PA} + 1 = \\frac{PC + PA}{PA}$. Hence\n\n$$\nBM = \\frac{PA \\cdot BC}{PC + PA} = \\frac{PB \\cdot AB}{PC + PA} = \\frac{R x \\sqrt{x^2 + 4R^2}}{x^2 + 2R^2}.\n$$\n\nThus,\n$$\nS_{AMB} = \\frac{1}{2} AB \\cdot BM \\cdot \\sin \\angle ABM = \\frac{1}{2} \\cdot 2R \\cdot \\frac{R x \\sqrt{x^2 + 4R^2}}{x^2 + 2R^2} \\cdot \\frac{AC}{2R} = \\frac{2R^3 x}{x^2 + 2R^2}.\n$$\n\nIt follows that $S_{AMB} \\leq \\frac{2R^3 x}{2\\sqrt{2} x R} = \\frac{R^2}{\\sqrt{2}}$ and $S_{AMB} = \\frac{R^2}{\\sqrt{2}}$ if and only if $x^2 = 2R^2$, i.e., $x = \\sqrt{2}R$.\n\nThus, the area of triangle $AMB$ attains its maximum if and only if the distance between $P$ and $B$ is equal to $\\sqrt{2}R$ (there are two such places); in those cases $S_{AMB} = \\frac{R^2}{\\sqrt{2}}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15838,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Determine all positive integers $p$ for which there exist positive integers $x_1 < x_2 < \\dots < x_n$ such that\n$$\n\\frac{1}{x_1} + \\frac{2}{x_2} + \\dots + \\frac{n}{x_n} = p.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Call *good* a number $p$ for which there exist positive integers $x_1 < x_2 < \\dots < x_n$ such that $\\frac{1}{x_1} + \\frac{2}{x_2} + \\dots + \\frac{n}{x_n} = p$.\n\nSince $x_1, x_2, \\dots, x_n$ are integers and $x_1 < x_2 < \\dots < x_n$, we have $x_k \\ge k$, so $\\frac{k}{x_k} \\le 1$ for all $k = 1, 2, \\dots, n$. Then $\\frac{1}{x_1} + \\frac{2}{x_2} + \\dots + \\frac{n}{x_n} \\le n$, so every good number, if there exists any, is between $1$ and $n$.\n\nNext, we will show that any integer $p \\in \\{1, 2, \\dots, n\\}$ is good. Obviously, $n$ is good (for $x_k = k$) and $1$ is also good (take $x_k = kn$). For $2 \\le p \\le n-1$, we write:\n\n$$\n\\sum_{k=1}^{n} \\frac{k}{x_k} = \\left( \\frac{1}{x_1} + \\frac{2}{x_2} + \\dots + \\frac{p-1}{x_{p-1}} \\right) + \\left( \\frac{p}{x_p} + \\dots + \\frac{n}{x_n} \\right),\n$$\n\nso it is enough to choose $x_1, x_2, \\dots, x_n$ such that the first sum is equal to $p-1$, and the second sum is equal to $1$. We can do that by setting $x_k = k$ for $k = 1, 2, \\dots, p-1$ and $x_k = k(n-p+1)$ for $p \\le k \\le n$. Notice that $x_1 < x_2 < \\dots < x_n$ in all cases, so the good numbers are indeed $1, 2, \\dots, n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15839,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $r$ such that there exist positive prime numbers $p$ and $q$ with $p^2 + pq + q^2 = r^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "The given relation is equivalent to $$(p+q)^2 = r^2 + pq,$$ which can be written as $$(p+q+r)(p+q-r) = pq.$$ The divisors of $pq$ are $1$, $p$, $q$, and $pq$. Since $p+q > \\max\\{p, q\\}$, it follows that $p+q-r = 1$ and $p+q+r = pq$. Adding these equalities yields $2p + 2q = pq + 1$, that is, $$(p-2)(q-2) = 3.$$ This leads to $(p, q) \\in \\{(3, 5), (5, 3)\\}$; in both cases $r = 7$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15840,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the sequence of numbers where each term consists of one digit 1 followed by $n$ digits 2 (for $n = 1, 2, \\ldots, 2009$).  \n\n(a) How many of the first 2009 terms in this sequence are divisible by 3?\n\n(b) What is the smallest number in this sequence that is divisible by 1001?",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Every number divisible by 3 has its digits sum as a multiple of 3. The digits sum of the $n$th number in the sequence is $2n + 1$ (one digit 1 and $n$ digits 2). $2n + 1$ is divisible by 3 when $n = 1, 4, 7, \\ldots, 2008$ (i.e., $n \\equiv 1 \\pmod{3}$). There are $670$ such terms among the first $2009$.\n\n(b) To find the smallest number in the sequence divisible by $1001$, consider $122\\ldots2$ (one 1 followed by $n$ twos). Dividing $1222222$ by $1001$ leaves a remainder of $1$, so $1001$ divides $1222222 - 1 = 1222221$. Thus, $1222221$ is the smallest such number divisible by $1001$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15841,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, $z$ be real numbers greater than or equal to $1$.\n\nProve that\n\n$$\n(x^2 - 2x + 2)(y^2 - 2y + 2)(z^2 - 2z + 2) \\leq (xyz)^2 - 2xyz + 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = x - 1$, $b = y - 1$, $c = z - 1$. Then $a, b, c \\geq 0$. Completing the square, $A^2 - 2A + 2 = (A - 1)^2 + 1$, so the inequality becomes\n\n$$\n(a^2 + 1)(b^2 + 1)(c^2 + 1) \\leq [(a + 1)(b + 1)(c + 1) - 1]^2 + 1 = (abc + ab + bc + ca + a + b + c)^2 + 1. \\tag{1}\n$$\n\nWe can prove this by direct expansion, but instead, consider the two-variable case:\n\n$$\n(A^2 + 1)(B^2 + 1) \\leq [(A + 1)(B + 1) - 1]^2 + 1 = (AB + A + B)^2 + 1. \\tag{2}\n$$\n\nThe right side has extra positive terms ($2AB$, $2A^2B$, $2AB^2$). Applying (2) twice—first with $(A, B) = (a, b)$, then with $(A, B) = (ab + a + b, c)$—yields (1).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15842,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$, let $D$, $E$, and $F$ respectively denote the midpoints of the sides $BC$, $CA$, and $AB$. The circle $BCF$ and the line $BE$ meet again at $P$, and the circle $ABE$ and the line $AD$ meet again at $Q$. Finally, the lines $DP$ and $FQ$ meet at $R$. Prove that the centroid $G$ of the triangle $ABC$ lies on the circle $PQR$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will use the following lemma.\n\n**Lemma.** Let $AD$ be a median in triangle $ABC$. Then $\\cot \\angle BAD = 2 \\cot A + \\cot B$ and $\\cot \\angle ADC = \\frac{1}{2}(\\cot B - \\cot C)$.\n\n*Proof.* Let $CC_1$ and $DD_1$ be the perpendiculars from $C$ and $D$ to $AB$. Using signed lengths:\n\n$$\n\\cot BAD = \\frac{AD_1}{DD_1} = \\frac{(AC_1 + AB)/2}{CC_1/2} = \\frac{CC_1 \\cot A + CC_1(\\cot A + \\cot B)}{CC_1} = 2 \\cot A + \\cot B.\n$$\n\nSimilarly, let $A_1$ be the projection of $A$ onto $BC$:\n\n$$\n\\begin{aligned}\n\\cot ADC &= \\frac{DA_1}{AA_1} = \\frac{BC/2 - A_1C}{AA_1} = \\frac{(AA_1 \\cot B + AA_1 \\cot C)/2 - AA_1 \\cot C}{AA_1} \\\\\n&= \\frac{\\cot B - \\cot C}{2}.\n\\end{aligned}\n$$\n\nThe lemma is proved.\n\nReturning to the problem, apply the lemma to get:\n\n$$\n\\begin{align*}\n\\cot \\angle BPD &= 2 \\cot \\angle BPC + \\cot \\angle PBC \\\\\n&= 2 \\cot \\angle BFC + \\cot \\angle PBC \\quad \\text{(from circle $BFPC$)} \\\\\n&= 2 \\cdot \\frac{1}{2}(\\cot A - \\cot B) + 2 \\cot B + \\cot C \\\\\n&= \\cot A + \\cot B + \\cot C.\n\\end{align*}\n$$\n\nSimilarly, $\\cot \\angle GQF = \\cot A + \\cot B + \\cot C$, so $\\angle GPR = \\angle GQF$ and $GPRQ$ is cyclic.\n\n**Remark.** The angle $\\angle GPR = \\angle GQF$ is the Brocard angle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15843,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A = \\sqrt{2013 + \\sqrt{2012 + \\sqrt{2011 + \\dots + \\sqrt{2 + \\sqrt{1}}}}}$. Find the integer part of $A$.",
    "options": [],
    "answer": "See solution",
    "solution": "On the one hand, $A^2 > 2013 + \\sqrt{2012} > 2013 + 44 > 45^2$, therefore $A > 45$.\n\nOn the other hand, we can demonstrate by induction that $x_n = \\sqrt{n + \\sqrt{n-1 + \\dots + \\sqrt{1}}} < \\sqrt{n} + 1$.\n\nThis holds for $n = 1$. Suppose it holds for some $n$. Then\n$$\nx_{n+1} = \\sqrt{n+1 + x_n} < \\sqrt{n+1 + \\sqrt{n} + 1}.\n$$\nIt remains to show that $\\sqrt{n+1 + \\sqrt{n} + 1} < \\sqrt{n+1} + 1$, which is equivalent to the trivially true inequality $n + \\sqrt{n} + 2 < n + 2 + 2\\sqrt{n} + 1$.\n\nTherefore, $A = x_{2013} < \\sqrt{2013} + 1 < 46$.\n\nThus, the integer part of $A$ is $45$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15844,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer $a$ such that for all positive integers $n$, the expression $94^n + a \\cdot 28^n$ is divisible by $2013$.",
    "options": [],
    "answer": "See solution",
    "solution": "Because $2013 = 3 \\cdot 11 \\cdot 61$, the requirement $94^n + a \\cdot 28^n \\equiv 0 \\pmod{2013}$ implies $a+1 \\equiv 0 \\pmod{3}$, $6^n(a+1) \\equiv 0 \\pmod{11}$, and $28^n((-1)^n + a) \\equiv 0 \\pmod{61}$.\n\nTherefore, $a \\equiv -1 \\pmod{3}$, $a \\equiv -1 \\pmod{11}$, and $a \\equiv (-1)^{n+1} \\pmod{61}$.\n\nUsing the Chinese Remainder Theorem, for even $n$, the smallest positive integer $a$ which satisfies these three congruences is $2012$.\n\nFor odd $n$, we need to solve $33k-1 \\equiv 1 \\pmod{61}$. Using Euclid's Algorithm, $13 \\cdot 61 - 24 \\cdot 33 = 1$, so $k \\equiv -24 \\cdot 2 \\equiv 13 \\pmod{61}$. This yields that the smallest possible $a$ is $33 \\cdot 13 - 1 = 428$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15845,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$, let $H$ and $O$ be its orthocentre and circumcentre, respectively. Let $K$ be the midpoint of the line segment $AH$. Let further $\\ell$ be a line through $O$, and let $P$ and $Q$ be the orthogonal projections of $B$ and $C$ onto $\\ell$, respectively. Prove that $KP + KQ \\geq BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Fix the origin at $O$ and the real axis along $\\ell$. As usual, lowercase letters denote the complex coordinate of the corresponding point in the configuration. For convenience, let $|a| = |b| = |c| = 1$. Clearly,\n\n$$\nk = a + \\frac{1}{2}(b + c), \\quad p = a + \\frac{1}{2}\\left(b + \\frac{1}{b}\\right), \\quad \\text{and} \\quad q = a + \\frac{1}{2}\\left(c + \\frac{1}{c}\\right).\n$$\n\nThen\n\n$$\n|k - p| = \\left| a + \\frac{1}{2} \\left( c - \\frac{1}{b} \\right) \\right| = \\frac{1}{2} |2ab + bc - 1|, \\quad \\text{since } |b| = 1.\n$$\n\nSimilarly, $|k - q| = \\frac{1}{2}|2ac + bc - 1|$, so\n\n$$\n|k - p| + |k - q| = \\frac{1}{2}|2ab + bc - 1| + \\frac{1}{2}|2ac + bc - 1| \\ge \\frac{1}{2}|2a(b - c)| = |b - c|, \\quad \\text{since } |a| = 1\n$$\n\nas required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15846,
    "subject": "Mathematics (Olympiad)",
    "question": "The first six terms of the sequence are $3, 3, 6, 9, 15, 24, \\ldots$. The sequence alternates in cycles of length 3, with two odd numbers and one even number in each cycle. After division by 2, the remainders in each cycle are $1, 1, 0$, so the sum of the three remainders is $2$. What is the sum of the first 2016 remainders when each term is divided by 2?",
    "options": [],
    "answer": "See solution",
    "solution": "Since the pattern of remainders repeats every 3 terms as $1, 1, 0$, the sum of the remainders in each cycle is $2$. There are $\\frac{2016}{3} = 672$ cycles in the first 2016 terms, so the total sum is $2 \\times 672 = 1344$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15847,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, and $d$ be real numbers such that\n\n$$a + b + c + d = 2$$\n\nand\n\n$$ab + bc + cd + da + ac + bd = 0.$$\n\nFind the minimum and maximum values of the product $abcd$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let's find the minimum first.\n\n$$\na^2 + b^2 + c^2 + d^2 = (a + b + c + d)^2 - 2(ab + bc + cd + da + ac + bd) = 4\n$$\n\nBy AM-GM, $4 = a^2 + b^2 + c^2 + d^2 \\ge 4\\sqrt{|abcd|}$, so $1 \\ge |abcd|$ and thus $abcd \\ge -1$.\n\nFor example, if $a = b = c = 1$ and $d = -1$, then $abcd = -1$.\n\nNow, for the maximum, consider $abcd > 0$.\n\nThe numbers $a, b, c, d$ cannot all be positive or all negative. Without loss of generality, let $a, b > 0$ and $c, d < 0$. Set $-c = x$, $-d = y$ with $a, b, x, y > 0$.\n\nWe have $a + b - x - y = 2$ and $a^2 + b^2 + x^2 + y^2 = 4$. We want to maximize $abxy$.\n\nLet $a + b = 2s$, so $x + y = 2s - 2$. Using Cauchy-Schwarz and AM-GM:\n\n$$(a + b)^2 + (a + b - 2)^2 \\le 8$$\n\nLet $a + b = 2s$, then $2s^2 - 2s - 1 \\le 0$, so $s \\le \\frac{\\sqrt{3} + 1}{2} = k$.\n\nThus, $ab \\le s^2 \\le k^2$.\n\nSimilarly, $x + y = 2q$ with $q \\le \\frac{\\sqrt{3} - 1}{2} = \\frac{1}{2k}$, so $xy \\le q^2 \\le \\frac{1}{4k^2}$.\n\nTherefore, $abxy \\le k^2 \\cdot \\frac{1}{4k^2} = \\frac{1}{4}$, so $abcd \\le \\frac{1}{4}$.\n\nFor example, if $a = b = k$ and $c = d = -\\frac{1}{2k}$, then $abcd = \\frac{1}{4}$.\n\n**Conclusion:**\n\n$$\\min(abcd) = -1, \\quad \\max(abcd) = \\frac{1}{4}.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15848,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\left\\lfloor \\frac{x}{44} \\right\\rfloor = \\left\\lfloor \\frac{x}{45} \\right\\rfloor = n$.\n\nFind the number of non-negative integer values of $x$ that satisfy this condition.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be a non-negative integer such that $\\left\\lfloor \\frac{x}{44} \\right\\rfloor = \\left\\lfloor \\frac{x}{45} \\right\\rfloor = n$.\n\nIf $n = 0$, then $x$ is any integer from $0$ to $43$: a total of $44$ values.\n\nIf $n = 1$, then $x$ is any integer from $45$ to $87$: a total of $43$ values.\n\nIf $n = 2$, then $x$ is any integer from $90$ to $131$: a total of $42$ values.\n\nIn general, for $n = k$, $x$ ranges from $45k$ to $44k + 43$, giving $44 - k$ values.\n\nThe largest $n$ is $43$, for which $x$ has only $1$ value.\n\nTherefore, the total number of non-negative integer values of $x$ is $44 + 43 + \\dots + 1 = \\frac{1}{2}(44 \\times 45) = \\mathbf{990}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15849,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all integers $n \\ge 2$ that have a representation\n$$\nn = a^2 + b^2,\n$$\nwhere $a$ is the smallest divisor of $n$ different from $1$ and $b$ is an arbitrary divisor of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $n$ is odd, then both $a$ and $b$ are odd, so $n = a^2 + b^2$ is even—a contradiction. Therefore, $n$ is even and $a = 2$. This also shows that $b$ is even. Furthermore, $b \\mid (n - b^2) = a^2 = 4$. Thus $b \\in \\{2, 4\\}$, which results in $n = 8$ and $n = 20$, respectively.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15850,
    "subject": "Mathematics (Olympiad)",
    "question": "Rewrite $\\frac{x}{20} + \\frac{y}{15} = 1$ in the form $3x + 4y = 60$. Find all integer solutions $(x, y)$ where $x \\geq 0$ and $y \\geq 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "We can re-write $\\frac{x}{20} + \\frac{y}{15} = 1$ as $3x + 4y = 60$. For integer solutions, $3x$ must be divisible by $4$, so $x$ must be a multiple of $4$. Trying successive values, the acceptable combinations are: $(0, 15)$, $(4, 12)$, $(8, 9)$, $(12, 6)$, $(16, 3)$, $(20, 0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15851,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle, and let $\\ell$ be the line through $A$ and perpendicular to the line $BC$. The reflection of $\\ell$ in the line $AB$ crosses the line through $B$ and perpendicular to $AB$ at $P$. The reflection of $\\ell$ in the line $AC$ crosses the line through $C$ and perpendicular to $AC$ at $Q$. Show that the line $PQ$ passes through the orthocenter of the triangle $ABC$.\n\n![](images/RMC_2019_var_3_p84_data_80af5099b1.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The lines $BP$ and $CQ$ cross at the antipode $A'$ of $A$ in the circumcircle $\\Gamma$ of the triangle $ABC$. Let $H$ be the orthocenter of the triangle $ABC$. The lines $BH$ and $AQ$ cross at a point $B'$ on $\\Gamma$. Similarly, the lines $CH$ and $AP$ cross at a point $C'$ on $\\Gamma$. The conclusion now follows by Pascal's theorem applied to the hexagram $AC'CA'BB'$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15852,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{N}_+$ be the set of all positive integers. Find all functions $f : \\mathbb{N}_+ \\to \\mathbb{N}_+$ satisfying that for any $x, y \\in \\mathbb{N}_+$, $f(f(x) + y)$ divides $x + f(y)$.",
    "options": [],
    "answer": "See solution",
    "solution": "All maps that satisfy the question are one of the following three categories:\n\n1. For any $x \\in \\mathbb{N}_+$, $f(x) = x$.\n2. For any $x > 1$, $f(x) = 1$ and $f(1)$ can be any positive integer.\n3. For any $x > 1$, $f(x) = \\begin{cases} 1, & x \\text{ is odd} \\\\ 2, & x \\text{ is even} \\end{cases}$ and $f(1)$ can be any odd positive integer.\n\nIt is easy to verify that all the three types of functions satisfy the conditions. In the following, we will prove that only the above three types of functions satisfy the conditions. The proof is done in three steps.\n\n**Step 1:** We will prove that either $f(x) = x$ or $f$ is not an injection.\n\nFor example, if $f(1) > 1$, by taking $x = 1$ in the problem, we know that $f(f(1) + y)$ divides $1 + f(y)$. In particular,\n\n$$\nf(f(1) + y_0) \\leq 1 + f(y_0).\n$$\n\nBy induction, it is easy to see that for fixed $y_0 \\in \\{1, \\dots, f(1)\\}$ and any $t \\in \\mathbb{Z}_+$, there is\n\n$$\nf(t \\cdot f(1) + y) \\leq t + f(y).\n$$\n\nTherefore, for any $y$ large enough, $f(y) < y$. So when $y$ is large enough, $f : \\{1, \\dots, y\\} \\to \\{1, \\dots, y-1\\}$ cannot be an injection.\n\nIn the following, we will discuss the case $f(1) = 1$. By taking $x = 1$ in the problem, we know that $f(1+y)$ divides $1+f(y)$. In particular, $f(1+y) \\leq 1+f(y)$. By induction, it is easy to know that for any $x \\in \\mathbb{Z}_+$, there is $f(x) = x$ or $f(y) < y$ for $y$ large enough. Similarly, we obtain that $f$ cannot be an injection.\n\n**Step 2:** We will show that if $f$ is not an injection, then for $x$ large enough, either $f(x) = 1$ or $f(x) = \\begin{cases} 1, & x \\text{ is odd} \\\\ 2, & x \\text{ is even} \\end{cases}$.\n\nSuppose $f$ is not an injection and we denote $A$ as the smallest positive integer such that there exists a positive integer $x_0 \\in \\mathbb{N}_+$ satisfying\n\n$$\nf(x_0 + A) = f(x_0).\n$$\n\nBy substituting $x = x_0$ and $x = x_0 + A$ respectively into the problem, we get\n\n$$\nf(f(x_0) + y) \\mid x_0 + f(y)\n$$\n\nand\n\n$$\nf(f(x_0 + A) + y) \\mid x_0 + A + f(y).\n$$\n\nSubtracting the above two, we get\n\n$$\nf(f(x_0) + y) \\mid A.\n$$\n\nThis shows that $f(z)$ can only take values in the factors of $A$ when $z$ is greater than $f(x_0)$. Suppose that $A$ has a total of $D$ factors. According to the Pigeonhole principle, we investigate $D+1$ consecutive positive integers $z, z+1, \\dots, z+D$ that are greater than $f(x_0)$, of which there must be two numbers that have equal images under $f$ and the difference between these two numbers is less than or equal to $D$. This leads to $A \\leq D$.\n\nSuch positive integer $A$ can only be $A = 1$ or $A = 2$. If $A = 1$, according to the previous result we know that for $x$ large enough, $f(x) = 1$. If $A = 2$, we see that for $x$ large enough, $f(x) = 1$ or $f(x) = 2$. But $A = 2$ is the smallest $A$ that makes the previous condition valid, from which we know that\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p129_data_d7eda8e790.png)\n\nHowever, (b) is impossible because taking two even numbers $x$ and $y$ large enough and substituting them into the problem gives\n\n$$\n2 \\mid x + 1,\n$$\n\na contradiction. This completes the proof of **Step 2**.\n\n**Step 3:** We will prove that if $f$ is not an injection, then for $x > 1$, $f(x) = 1$ or $f(x) = \\begin{cases} 1, & x \\text{ is odd} \\\\ 2, & x \\text{ is even} \\end{cases}$.\n\nLet $A$ be as set in **Step 2**. Taking $x_0$ large enough so that $f(x_0) = 1$ and substituting $x = x_0$ and $x = x_0 + A$ into the problem, we get\n\n$$\nf(1+y) \\mid x_0 + f(y) \\quad \\text{and} \\quad f(1+y) \\mid x_0 + A + f(y).\n$$\n\nSubtracting the above two, we get\n\n$$\nf(1+y) \\mid A.\n$$\n\nWhen $A = 1$, this equation immediately leads to $f(y+1) = 1$, and thus we obtain the function of (2) mentioned before.\n\nWhen $A = 2$, this equation gives $f(y+1) = 1$ or $f(y+1) = 2$. However, the minimality of $A = 2$ shows that $f(x)$ must be $1, 2, 1, 2, \\dots$ alternately in the case of $x > 1$. Combining the results of **Step 2** yields the function of the above mentioned (3). $\\boxed{}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15853,
    "subject": "Mathematics (Olympiad)",
    "question": "For triangle $ABC$, the line tangent to its circumcircle at $A$ and the line $BC$ intersect at a point $P$. Let $Q$ and $R$ be the points symmetric to $P$ with respect to the lines $AB$ and $AC$, respectively. Prove that the lines $BC$ and $QR$ intersect perpendicularly.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume that points $B$, $C$, $P$ lie on the same straight line in this order. Let $Q'$ be the intersection of lines $AB$ and $PQ$, and $R'$ the intersection of lines $AC$ and $PR$.\n\nIf $\\angle ACB \\neq 90^\\circ$, from $PQ' \\perp AQ'$ and $PR' \\perp AR'$, the four points $A$, $Q'$, $P$, $R'$ lie on a circle. Let $S$ be the intersection of lines $BC$ and $Q'R'$. Then:\n\n$$\n\\begin{align*}\n\\angle BSQ' &= \\angle BPQ' + \\angle PQ'S \\\\\n&= \\angle BPQ' + \\angle PAR' \\\\\n&= \\angle BPQ' + \\angle PBQ' \\\\\n&= \\angle BQ'Q = 90^\\circ,\n\\end{align*}\n$$\n\nwhich means $BC \\perp Q'R'$.\n\nIf $\\angle ACB = 90^\\circ$, the points $Q'$ and $R'$ coincide with $A$ and $C$, respectively, so $BC \\perp Q'R'$ holds as well.\n\n$Q'$, $R'$ are the midpoints of $PQ$ and $PR$, respectively, so $QR \\parallel Q'R'$, and thus $BC \\perp QR$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15854,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an isosceles triangle with apex at $A$, and let $M$, $N$, $P$ be the midpoints of the sides $BC$, $CA$, $AB$, respectively. Let $Q$ and $R$ be points inside the segments $BM$ and $CM$, respectively, so that $\\angle BAQ = \\angle MAR$. The segment $NP$ crosses $AQ$ and $AR$ at $U$ and $V$, respectively. The point $S \\neq U$ lies on the half-line $AQ$ emanating from $A$, so that $SV$ is the internal bisector of $\\angle ASM$. Similarly, $T \\neq V$ is a point on the half-line $AR$ emanating from $A$, so that $TU$ is the internal bisector of $\\angle ATM$. Prove that one of the points where the circles $NUS$ and $PVT$ meet lies on the line $AM$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that the circles $NUS$ and $PVT$ both pass through the orthocentre $H$ of the triangle $AUV$. Since $AH \\perp NP \\parallel BC$, and $AB = AC$, it follows that $H$ lies on the perpendicular bisector $AM$ of $BC$, and the conclusion follows.\n\n![](images/RMC_2024_p95_data_daacfec19e.png)\n\nWe first show that $AVMS$ is cyclic. Since the triangle $ABC$ is isosceles, $AM \\perp BC$, and since $NP$ is midline, $V$ is the midpoint of $AR$. Then $MV$ is the $M$-median of the right triangle $AMR$, so $VM = VA = VR$. Hence $V$ lies on the perpendicular bisector of $AM$. By hypothesis, $V$ also lies on the internal bisector of $\\angle ASM$, so, in the circle $AMS$, $V$ is the midpoint of the arc $AM$ not containing $S$. Consequently, $AVMS$ is cyclic, as stated. Similarly, $AUMT$ is cyclic.\n\nWe now prove that $H$ lies on the circle $NUS$. Since $AVMS$ is cyclic, $\\angle ASV = \\angle MAR$. By hypothesis, $\\angle MAR = \\angle BAQ$, so $AB \\parallel SV$. Similarly, $AC \\parallel TU$. Then $\\angle SVP = \\angle APV = 90^\\circ - \\frac{1}{2}\\angle BAC$, and $\\angle AUH = 90^\\circ - \\angle UAV = 90^\\circ - \\frac{1}{2}\\angle BAC$, so $\\angle SVP = \\angle AUH$.\n\nFrom the triangles $SUV$ and $AUN$,\n\n$$\n\\frac{SU}{SV} = \\frac{\\sin(90^\\circ - \\frac{1}{2}\\angle BAC)}{\\sin \\angle SUV} = \\frac{AU}{AN}.\n$$\n\nSince $\\angle UAM = \\angle UAH = \\angle VAN$ and $\\angle AUH = 90^\\circ - \\frac{1}{2}\\angle BAC = \\angle ANV$, the triangles $AUH$ and $ANV$ are similar.\n\nHence $AN/AU = NV/UH$, so $SU/SV = UH/VN$. Since $\\angle SVN = 180^\\circ - \\angle SVP = 180^\\circ - \\angle AUM = \\angle SUH$, the triangles $SUH$ and $SVN$ are similar, so $\\angle USH = \\angle VSN$, whence $\\angle USV = \\angle HSN$.\n\nFinally, $\\angle USV = \\angle MAV = 90^\\circ - \\angle AVU = \\angle HUV$, so $\\angle HSN = \\angle HUV = \\angle HUN$. Consequently, $H$ lies on the circle $NUS$, as stated. Similarly, $H$ lies on the circle $PVT$. This ends the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15855,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the number of polynomials $f(x) = a x^3 + b x$ that satisfy the following conditions:\n\n1. $a, b \\in \\{1, 2, \\dots, 2013\\}$;\n2. The difference of any two numbers among $f(1), f(2), \\dots, f(2013)$ is not a multiple of $2013$.",
    "options": [],
    "answer": "See solution",
    "solution": "2013 can be factorized as $2013 = 3 \\times 11 \\times 61$. Let $p_1 = 3$, $p_2 = 11$, $p_3 = 61$. Denote by $a_i$ the residue of $a$ modulo $p_i$, and $b_i$ the residue of $b$ modulo $p_i$ ($i = 1, 2, 3$), with $a, b \\in \\{1, 2, \\dots, 2013\\}$. By the Chinese Remainder Theorem, there is a bijection between $(a, b)$ and $(a_1, a_2, a_3, b_1, b_2, b_3)$.\n\nLet $f_i(x) = a_i x^3 + b_i x$ for $i = 1, 2, 3$. We call a polynomial \"good modulo $n$\" if the residues of $f(0), f(1), \\dots, f(n-1)$ modulo $n$ are all distinct.\n\nIf $f(x) = a x^3 + b x$ is not good modulo $2013$, then there exist $x_1 \\not\\equiv x_2 \\pmod{2013}$ such that $f(x_1) \\equiv f(x_2) \\pmod{2013}$. Suppose $x_1 \\not\\equiv x_2 \\pmod{p_i}$. Let $u_1$ and $u_2$ be the residues of $x_1$ and $x_2$ modulo $p_i$, respectively. Then $u_1 \\not\\equiv u_2 \\pmod{p_i}$ and $f_i(u_1) \\equiv f_i(u_2) \\pmod{p_i}$, so $f_i(x)$ is not good modulo $p_i$.\n\nIf $f(x) = a x^3 + b x$ is good modulo $2013$, then for every $i$, $f_i(x)$ is good modulo $p_i$. For any distinct $r_1, r_2 \\in \\{0, 1, \\dots, p_i - 1\\}$, there exist $x_1, x_2 \\in \\{1, 2, \\dots, 2013\\}$ such that $x_1 \\equiv r_1 \\pmod{p_i}$ and $x_2 \\equiv r_2 \\pmod{p_i}$ and $x_1 \\equiv x_2 \\pmod{\\frac{2013}{p_i}}$. Now $f(x_1) \\equiv f(x_2) \\pmod{\\frac{2013}{p_i}}$, but $f(x_1) \\not\\equiv f(x_2) \\pmod{2013}$, so $f(r_1) \\not\\equiv f(r_2) \\pmod{p_i}$.\n\nHence, we need to determine the number of good polynomials $f_i(x)$ modulo $p_i$.\n\nFor $p_1 = 3$, by Fermat's theorem, a good polynomial\n\n$$\nf_1(x) \\equiv a_1 x^3 + b_1 x \\equiv (a_1 + b_1) x \\pmod{3}\n$$\n\nis equivalent to $a_1 + b_1$ not being divisible by $3$. There are $6$ such $f_1(x)$.\n\nFor $i = 2, 3$, if $f_i(x)$ is good modulo $p_i$, then for any $u$ and $v \\not\\equiv 0 \\pmod{p_i}$, $f_i(u + v) \\not\\equiv f_i(u - v) \\pmod{p_i}$. (The solution is incomplete beyond this point.)",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15856,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be an integer. Define the number $A$ as the $n$-digit number consisting of all 3's (that is, $A = \\underbrace{33\\ldots3}_{n\\ \\text{digits}}$). Let $B = 20 \\cdot A + 6$. Find all the digits that appear in the number $A \\cdot B$.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, $B = \\underbrace{66\\ldots6}_{n+1\\ \\text{digits}}$.\n\nNotice that $3 \\cdot A = 99\\ldots9 = 10^n - 1$. Then\n$$\nA \\cdot B = (10^n - 1) \\cdot \\underbrace{22\\ldots2}_{n+1\\ \\text{digits}}\n$$\nThis equals\n$$\n\\underbrace{22\\ldots2}_{n+1\\ \\text{digits}}\\underbrace{00\\ldots0}_{n\\ \\text{digits}} - \\underbrace{22\\ldots2}_{n+1\\ \\text{digits}}\n$$\nThus,\n$$\nA \\cdot B = \\underbrace{22\\ldots2}_{n-1\\ \\text{digits}}\\underbrace{1977\\ldots78}_{n-1\\ \\text{digits}}\n$$\nSo, the digits of $A \\cdot B$ are $1$, $2$, $7$, $8$, and $9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15857,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n$, a triangular array $(a_{i,j})$ of zeroes and ones, where $i$ and $j$ run through the positive integers such that $i + j \\le n + 1$, is called a *binary anti-Pascal n-triangle* if $a_{i,j} + a_{i,j+1} + a_{i+1,j} \\equiv 1 \\pmod{2}$ for all possible values $i$ and $j$ may take on. Determine the minimum number of ones a binary anti-Pascal $n$-triangle may contain.",
    "options": [],
    "answer": "See solution",
    "solution": "The required minimum is $\\lfloor n(n+1)/6 \\rfloor$; with the convention that rows decrease in length upwards, this minimum is achieved if, for instance, the bottom row consists of $\\lfloor n/3 \\rfloor$ disjoint blocks of the form $001$, followed by an all-zero possible tail. (A binary anti-Pascal triangle is clearly determined by the bottom row.)\n\nIn what follows, part of the generic configurations referred to may not exist for the first few values of $n$; in this case, simply consider the corresponding induced configuration, that is, the trace the generic configuration leaves on the configuration at hand. Also, recall the convention that rows decrease in length upwards (hence columns decrease in height rightwards).\n\nWe now show by induction on $n$ that a binary anti-Pascal $n$-triangle contains at least $\\lfloor n(n+1)/6 \\rfloor$ ones, of which at least $n-1$ lie on the bottom three rows (and since the transpose of a binary anti-Pascal triangle is again one such, the same holds for the leftmost three columns).\n\nThe cases $n=1, 2, 3$ are easily dealt with, so let $n \\ge 4$, and let $A$ be a binary anti-Pascal $n$-triangle. The lower left $3 \\times 3$ subarray $A'$ (induced if $n=4$) contains at least 3 ones, unless the lower left and central entries are both one and the other entries are all zero, in which case there are only 2 ones in $A'$.\n\nIn the former case, consider the binary anti-Pascal $(n-3)$-triangle consisting of the rightmost $n-3$ columns of $A$. By the induction hypothesis, the bottom three rows of this triangle contain at least $n-4$ ones, so the bottom three rows of $A$ contain at least $3 + (n-4) = n-1$ ones.\n\nIn the latter case, the statement clearly holds if $n=4$, so let $n \\ge 5$ and notice that the bottom two entries of the column of $A$ adjoining $A'$ along the right flank are both one. Thus, the $3 \\times 4$ subarray of $A$ extending $A'$ rightwards contains at least 4 ones. The rightmost $n-4$ columns of $A$ form a binary anti-Pascal $(n-4)$-triangle whose bottom three rows contain at least $n-5$ ones, by the induction hypothesis. Hence the bottom three rows of $A$ contain at least $4 + (n-5) = n-1$ ones.\n\nIn either case, the bottom three rows of $A$ contain at least $n-1$ ones. Finally, by the induction hypothesis, the binary anti-Pascal $(n-3)$-triangle atop the bottom three rows of $A$ contains at least $\\lfloor(n-3)(n-2)/6\\rfloor$ ones, so $A$ contains at least $n-1 + \\lfloor(n-3)(n-2)/6\\rfloor = \\lfloor n(n+1)/6 \\rfloor$ ones. This completes the induction and concludes the proof.\n\n**REMARKS.**\nThe minimum is also achieved if the bottom row consists of $\\lfloor n/3 \\rfloor$ disjoint blocks of the form $010$, followed by a possible tail consisting of a single zero if $n \\equiv 1 \\pmod{3}$, and of $01$ if $n \\equiv 2 \\pmod{3}$.\n\nIf $n \\not\\equiv 1 \\pmod{3}$, the minimum is again achieved if the bottom row consists of $\\lfloor n/3 \\rfloor$ disjoint blocks of the form $100$, followed by a possible tail of the form $10$. If $n \\equiv 1 \\pmod{3}$, and the bottom row consists of $\\lfloor n/3 \\rfloor$ disjoint blocks of the form $100$, followed by $1$, then the outcome is of exactly $\\lfloor n(n+1)/6 \\rfloor + 1$ ones; and if the tail is $0$, the outcome is of exactly $\\lfloor n(n+1)/6 \\rfloor + (n-1)/3$ ones.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15858,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a sequence $a_1, a_2, \\dots$, such that $a_1 = 1$ and $a_{n+1} = \\frac{9a_n + 4}{a_n + 6}$ for any $n \\in \\mathbb{N}$. Which terms of this sequence are positive integers?",
    "options": [],
    "answer": "See solution",
    "solution": "It can be shown by induction that\n\n$$\na_n = \\frac{4 \\cdot 2^n - 3}{2^n + 3}, \\quad \\forall n \\ge 2\n$$\n\nThus, $a_n$ is an integer only when $n = 1$. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15859,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $ABCDE$ 是平面上的凸五邊形，其中 $CD = DE$，但 $\\angle EDC \\neq 2 \\cdot \\angle ADB$。設點 $P$ 位於五邊形的內部，且滿足 $AP = AE$ 及 $BP = BC$。證明：$P$ 落在對角線 $CE$ 上的充要條件是 $$\\text{area}(BCD) + \\text{area}(ADE) = \\text{area}(ABD) + \\text{area}(ABP)$$。\n\n註：$\\text{area}(XYZ)$ 代表三角形 $XYZ$ 的面積。",
    "options": [],
    "answer": "See solution",
    "solution": "令 $P'$ 為 $P$ 關於直線 $AB$ 的對稱點，$M$ 和 $N$ 分別為 $P'E$ 和 $P'C$ 的中點。五邊形的凸性保證 $P'$ 與 $E$、$C$ 皆不同，故 $P'$ 亦與 $M$、$N$ 不同。我們主張，題目中的面積條件與共線條件皆等價於直角三角形 $AP'M$ 與 $BP'N$ 直接相似（等價於 $AP'E$ 與 $BP'C$ 直接相似）。\n\n![](images/20-2J_p24_data_51bfe70901.png)\n\n對於共線條件的等價，設 $F$ 為 $P'$ 到 $AB$ 的垂足，則 $F$ 為 $PP'$ 的中點。$P$ 落在 $CE$ 上當且僅當 $F$ 落在 $MN$ 上，亦即當且僅當有有向角的等式 $\\angle AFM = \\angle BFN$（模 $\\pi$）。由 $AP'FM$ 與 $BFP'N$ 共圓性，此等價於 $\\angle AP'M = \\angle BP'N$，即 $AP'M$ 與 $BP'N$ 直接相似。\n\n![](images/20-2J_p24_data_8a000020f0.png)\n\n對於面積條件的等價，有有向面積等式 $\\text{area}(ABD) + \\text{area}(ABP) = \\text{area}(AP'BD) = \\text{area}(AP'D) + \\text{area}(BDP')$。利用恆等式 $\\text{area}(ADE) - \\text{area}(AP'D) = \\text{area}(ADE) + \\text{area}(ADP') = 2\\text{area}(ADM)$，$B$ 亦同理，可得面積條件等價於\n\n$$\n\\text{area}(DAM) = \\text{area}(DBN).\n$$\n\n注意 $A$、$B$ 分別在 $P'E$、$P'C$ 的垂直平分線上。設 $G$、$H$ 分別為 $D$ 到這兩條垂直平分線的垂足，則面積條件可重寫為\n\n$$\nMA \\cdot GD = NB \\cdot HD.\n$$\n\n（此處所有長度皆依適當方向取有號：例如 $P'E$ 由 $P'$ 指向 $E$，平行線 $DH$ 方向相同，垂直平分線則自 $P'E$ 順時針 $\\pi/2$ 取向，$B$ 亦同理。）\n\n![](images/20-2J_p25_data_538789d98f.png)\n\n為了將 $GD$、$HD$ 與三角形 $AP'M$、$BP'N$ 聯繫，計算如下：\n\n*Claim*. 設 $\\Gamma$ 為以 $D$ 為圓心，$E$、$C$ 在圓上的圓，$h$ 為 $P'$ 對 $\\Gamma$ 的冪。則有\n\n$$\nGD \\cdot P'M = HD \\cdot P'N = \\frac{h}{4} \\neq 0.\n$$\n\n*Proof.* 首先 $h \\neq 0$，否則 $P'$ 在 $\\Gamma$ 上，則 $\\angle EDP'$、$\\angle P'DC$ 的內角平分線分別經過 $A$、$B$，這與題設 $\\angle EDC \\neq 2 \\cdot \\angle ADB$ 矛盾。\n\n設 $E'$ 為 $P'E$ 與 $\\Gamma$ 的另一交點，$E''$ 為 $\\Gamma$ 上與 $E'$ 對徑的點，則 $E''E$ 垂直於 $P'E$。$G$ 在 $P'E$、$EE''$ 的垂直平分線上，故 $G$ 為 $P'E''$ 的中點。$D$ 為 $E'E''$ 的中點，故 $GD = \\frac{1}{2}P'E'$。又 $P'M = \\frac{1}{2}P'E$，故 $GD \\cdot P'M = \\frac{1}{4}P'E' \\cdot P'E = \\frac{1}{4}h$。$HD \\cdot P'N$ 同理。$\\square$\n\n![](images/20-2J_p26_data_72fc25a6e9.png)\n\n由此可知，面積條件等價於\n\n$$\n(MA : P'M) = (NB : P'N)\n$$\n\n即有號長度比相等，亦即 $AP'M$ 與 $BP'N$ 直接相似，得證。$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15860,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there is a similarity between a triangle $ABC$ and the triangle having as sides the medians of triangle $ABC$ if and only if the squares of the lengths of the sides of triangle $ABC$ form an arithmetic sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "Recall that $m_a^2 = \\frac{1}{4}(2(b^2 + c^2) - a^2)$, with similar formulas for $m_b$ and $m_c$.\n\nAssume the squares of the side lengths are in arithmetic progression, for example $2b^2 = a^2 + c^2$. Then:\n\n$$\nm_a^2 = \\frac{3}{4}c^2, \\quad m_b^2 = \\frac{3}{4}b^2, \\quad m_c^2 = \\frac{3}{4}a^2,\n$$\n\nimplying\n\n$$\n\\frac{m_a}{c} = \\frac{m_b}{b} = \\frac{m_c}{a} = \\frac{\\sqrt{3}}{2}.\n$$\n\nThus, the triangles are similar.\n\nConversely, assume the triangles are similar. Let $a \\leq b \\leq c$. Then $m_a \\geq m_b \\geq m_c$, so\n\n$$\n\\frac{m_a}{c} = \\frac{m_b}{b} = \\frac{m_c}{a} = k.\n$$\n\nFrom $m_a^2 + m_b^2 + m_c^2 = \\frac{3}{4}(a^2 + b^2 + c^2)$, we get $k = \\frac{\\sqrt{3}}{2}$. Hence, $4m_b^2 = 3b^2$, which yields $2b^2 = a^2 + c^2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15861,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ and primes $p \\geq 5$ such that $(2p)^n + 1$ is a perfect cube.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $(2p)^n + 1 = a^3$ for some integer $a$. Then:\n\n$$\n(2p)^n = a^3 - 1 = (a-1)(a^2 + a + 1) = (a-1)[(a-1)^2 + 3(a-1) + 3].\n$$\n\nSince $a$ is odd, $a^2 + a + 1$ is also odd, so $2^n$ does not divide $a-1$. Thus, $a = 2^n p^k + 1$ for some integer $k \\geq 0$.\n\nSubstituting, we get:\n\n$$\n2^n p^n = 2^n p^k (2^{2n} p^{2k} + 3 \\cdot 2^n p^k + 3). \\quad (1)\n$$\n\n**Case 1:** $k > 0$. Then $k < n$, so from (1), $p \\mid 2^{2n} p^{2k} + 3 \\cdot 2^n p^k + 3$. This implies $p \\mid 3$, which is impossible since $p \\geq 5$.\n\n**Case 2:** $k = 0$. From (1):\n\n$$\np^n = 4^n + 3 \\cdot 2^n + 3. \\quad (2)\n$$\n\nFor $n = 1$, $p = 13$.\n\nFor $n = 2$, $p^2 = 31$, not possible.\n\nFor $n = 3$, $p^3 = 64 + 24 + 3 = 91 < 5^3$, not possible.\n\nFor $n \\geq 3$, by induction:\n\n$$\n5^n > 4^n + 3 \\cdot 2^n + 3 = p^n,\n$$\n\nso $p < 5$, which is not possible.\n\nThus, the unique solution is $n = 1$, $p = 13$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15862,
    "subject": "Mathematics (Olympiad)",
    "question": "設三角形 $ABC$ 的垂心為 $H$，三高分別為 $AD$、$BE$、$CF$。令 $G$ 為 $D$ 在 $EF$ 上的投影點，$DD'$ 為三角形 $DEF$ 外接圓的直徑。令 $X$ 為直線 $AG$ 與三角形 $ABC$ 的外接圓異於 $A$ 的交點，$Y$ 為 $GD'$ 與 $BC$ 的交點，$Z$ 為 $AD'$ 與 $GH$ 的交點。證明：$X$、$Y$、$Z$ 共線。\n\n![](images/2023-TWNIMO-Problems_p97_data_51c9003f38.png)",
    "options": [],
    "answer": "See solution",
    "solution": "我們先證明 $D'$、$H$、$X$ 共線：由於 $\\angle GXB = \\angle ACB = \\angle GFB$，得 $G$、$F$、$B$、$X$ 共圓。\n\n結合 $AG \\cdot AX = AF \\cdot AB = AH \\cdot AD$，我們得到 $G$、$X$、$D$、$H$ 共圓。注意到 $\\triangle DFH \\sim \\triangle DAE$ 且 $\\triangle DFG \\sim \\triangle DD'E$，因此 $\\triangle DGH \\sim \\triangle DAD'$，故 $\\angle DXA = \\angle DHG = \\angle DD'A$，即 $A$、$X$、$D$、$D'$ 共圓。這告訴我們 $\\angle DXH = \\angle DGH = \\angle DAD' = \\angle DXD'$，即 $D'$、$H$、$X$ 共線。\n\n由 $AH$、$BC$ 為 $\\angle GDD'$ 的兩條角平分線，知 $(P,Y;G,D') = (AH,BC;DG,DD') = -1$，其中 $P$ 為 $AH$ 與 $GD'$ 的交點。所以\n\n$$\nA(X, Z; Y, H) = (G, D'; Y, P) = -1 = (D', G; Y, P) = H(X, Z; Y, A),\n$$\n\n即 $X$、$Y$、$Z$ 共線。$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15863,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider three circles $\\Gamma_1$, $\\Gamma_2$, and $\\Gamma_3$, with centres $O_1$, $O_2$, and $O_3$, respectively, such that each pair of circles is externally tangent. Suppose we have another circle $\\Gamma$ with centre $O$ on the line segment $O_1O_3$ such that $\\Gamma_1$, $\\Gamma_2$, and $\\Gamma_3$ are each internally tangent to $\\Gamma$.\n\nShow that $\\angle O_1O_2O_3$ measures less than $90^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the points of tangency of $\\Gamma$ with $\\Gamma_1$, $\\Gamma_2$, and $\\Gamma_3$ be $A$, $B$, and $C$, respectively.\n\n![](images/s3s2024_p3_data_815ea3bdb8.png)\n\nThen $AOB$ is a straight line, and since it is a diameter, $\\angle ACB = 90^\\circ$. Let $E$ be the other intersection point of $AC$ with $\\Gamma_3$ and $F$ be the other intersection point of $BC$ with $\\Gamma_3$. Then $C$, $O_3$, and $O$ lie on a straight line, so $\\angle CEO_3 = \\angle ECO_3 = \\angle ACO = \\angle CAO$, the first equality following from radii $EO_3 = O_3C$, and the last from radii $AO = CO$. Thus $EO_3 \\parallel AO$ and similarly $O_3F \\parallel OB$.\n\nWe now claim that $EF < AO_1 + O_2B$ and that this finishes the problem. To see that it finishes the problem, construct points $X$ and $Y$ on segment $AB$ such that $AX = EO_3$ and $YB = O_3F$. Then $\\angle O_1O_3O_2 < \\angle XO_3Y = \\angle XO_3O + \\angle OO_3Y = \\angle ACO + \\angle OCB = 90^\\circ$.\n\nFor the claim $EF < AO_1 + O_2B$, we introduce the values $r_1$, $r_2$, $r_3$, $r$ for the radii of circles $\\Gamma_1$, $\\Gamma_2$, $\\Gamma_3$, $\\Gamma$, respectively. Then $2r_1 + 2r_2 = 2r$ by comparing radii along the diameter $AB$, so $r_1 + r_2 = r$. To show the claim we need to show $2r_3 < r_1 + r_2$, or equivalently $2r_3 < r$. But this is clear by looking at the segment $CO$, where two segments of length $r_3$ are contained within one segment of length $r$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15864,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $x$ and $y$ such that $xy^2 + 2y$ divides $2x^2y + xy^2 + 8x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $xy^2 + 2y$ divides\n\n$$\n(2x + y)(xy^2 + 2y) - y(2x^2y + xy^2 + 8x) = 2y^2 - 4xy,\n$$\n\nwe conclude that $xy + 2$ divides $2y - 4x$. We consider two cases.\n\n**Case 1.** Let $2y - 4x \\ge 0$. Then we have two possibilities:\n\n1. If $x \\ge 2$ then $xy + 2 > 2y - 4x$. Hence $2y - 4x = 0$, i.e. $x = a$ and $y = 2a$, where $a \\ge 2$ is an integer. Since $xy^2 + 2y = 4a(a^2 + 1)$ divides $2x^2y + xy^2 + 8x = 8a(a^2 + 1)$, this gives a solution.\n\n2. If $x = 1$ then $y^2 + 2y$ divides $8$, i.e. $y = 2$.\n\nHence in this case the solutions are $(a, 2a)$, where $a$ is a positive integer.\n\n**Case 2.** Let $2y - 4x < 0$, i.e. $4x - 2y > 0$. If $y \\ge 4$, then $xy + 2 > 4x - 2$. Therefore $y = 1, 2$ or $3$.\n\n1. If $y = 1$ then $\\frac{2x^2 + 9x}{x + 2} = 2x + 5 - \\frac{10}{x + 2}$ is an integer. Hence $x + 2$ divides $10$ and this gives the solutions $x = 3, y = 1$ and $x = 8, y = 1$.\n\n2. If $y = 2$ then $\\frac{x^2 + 3x}{x + 1} = x + 2 - \\frac{2}{x + 1}$ is an integer which gives $x = 1$.\n\n3. If $y = 3$ then $\\frac{6x^2 + 17x}{9x + 6}$ is an integer. This implies that $3 \\mid x$, i.e. $x = 3k$ for some positive integer $k$. After simplifications we obtain that $\\frac{18k^2 + 17k}{9k + 2} = (2k + 1) + \\frac{4k - 2}{9k + 2}$ is an integer, which is impossible for $k \\ge 1$.\n\nFinally, the solutions are $x = a, y = 2a$ for all positive integers $a$ and $x = 3, y = 1$, $x = 8, y = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15865,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given a triangle $ABC$ with incentre $I$. Suppose that there exists an intersection point $M$ of the line $AB$ and the perpendicular to $CI$ through $I$. Prove that the circumcircle of the triangle $ABC$ intersects the segment $CM$ in an interior point $N$ and that $NI \\perp MC$.\n\n![](images/14._63_RD_CZECH_AND_SLOVAK_MATHEMATICAL_OLYMPIAD_p3_data_92d1b17007.png)",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show in two different ways that the line $MI$, a perpendicular to $CI$ through $I$, is tangent to the circle $ABI$.\n\nThe first way is based on the known fact that $\\angle AIC$ and $\\angle BIC$ are obtuse angles of measures $90^\\circ + \\frac{1}{2}\\beta$ and $90^\\circ + \\frac{1}{2}\\alpha$, respectively (in common notation for interior angles of $\\triangle ABC$). This fact implies that the line $MI$ forms acute angles $\\frac{1}{2}\\beta$ and $\\frac{1}{2}\\alpha$ with the segments $AI$ and $BI$, respectively$^1$, hence the angles congruent with angles $IBA$ and $IAB$ in circle $ABI$ (see the figure). Well-known properties of inscribed and subtended angles lead to the conclusion that the line $MI$ is tangent to the circle $ABI$.\n\nThe second reason for this conclusion is based on the known fact that the centre of $ABI$ is the midpoint of the arc $AB$ of circle $ABC$ which lies on the ray $CI$ bisecting $\\angle ACB$.\n\n$^1$ As a consequence, we can see that the assumed existence of the intersection point $M$ is equivalent to the inequality $\\frac{1}{2}\\alpha \\neq \\frac{1}{2}\\beta$ or $\\alpha \\neq \\beta$. Due to the symmetry, we can assume that $\\alpha > \\beta$ as in our figure; the point $M$ then lies on the ray opposite to ray $AB$ and satisfies $|\\angle IMA| = \\frac{1}{2}\\alpha - \\frac{1}{2}\\beta$.\n\nFrom the proved tangency of $MI$ to $ABI$, it follows that the point $M$ lies on the line $AB$ outside of the segment $AB$. Moreover, the power $m$ of $M$ with respect to $ABI$ is positive and given by $m = |MI|^2 = |MA| \\cdot |MB|$. Hence $M$ lies in the exterior of the circle $ABC$ (as $AB$ is its chord) and the power of $M$ with respect to $ABC$ is the same $m = |MA| \\cdot |MB|$. Since $m = |MI|^2 < |MC|^2$ from the right-angled triangle $CMI$, it holds that $|MA| \\cdot |MB| < |MC|^2$. This means that the circle $ABC$ intersects the segment $MC$ in an interior point $N$, because $|MN| \\cdot |MC| = |MA| \\cdot |MB|$ implies that $|MN| < |MC|$ for the second point $N$ of intersection of the ray $MC$ with $ABC$. This proves the first conclusion of the problem.\n\nTo show that $CNI$ is a right angle, we use the proved equality $|MC| \\cdot |MN| = |MI|^2$ and apply a familiar theorem to the leg $MI$ of the right-angled triangle $CMI$: Its altitude from the vertex $I$ meets the hypotenuse $CM$ in such a point $X$ which is determined by the equation $|MC| \\cdot |MX| = |MI|^2$. Thus we have $X = N$ in our case and the solution is complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15866,
    "subject": "Mathematics (Olympiad)",
    "question": "The quadrilateral $ABCD$ is inscribed in a circle. The point $P$ lies in the interior of $ABCD$, and $\\angle PAB = \\angle PBC = \\angle PCD = \\angle PDA$. The lines $AD$ and $BC$ meet at $Q$, and the lines $AB$ and $CD$ meet at $R$. Prove that the lines $PQ$ and $PR$ form the same angle as the diagonals of $ABCD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\Gamma$ be the circumcircle of quadrilateral $ABCD$. Let $\\alpha = \\angle PAB = \\angle PBC = \\angle PCD = \\angle PDA$, and let $T_1, T_2, T_3,$ and $T_4$ denote the circumcircles of triangles $APD$, $BPC$, $APB$, and $CPD$, respectively. Let $M$ be the intersection of $T_1$ with line $RP$, and let $N$ be the intersection of $T_3$ with line $SP$. Also, let $X$ denote the intersection of diagonals $AC$ and $BD$.\n\nBy power of a point for circles $T_1$ and $\\Gamma$, it follows that $RM \\cdot RP = RA \\cdot RD = RB \\cdot RC$, which implies that the quadrilateral $BMPC$ is cyclic and $M$ lies on $T_2$. Therefore, $\\angle PMB = \\angle PCB = \\alpha = \\angle PAB = \\angle DMP$, where all angles are directed. This implies that $M$ lies on the diagonal $BD$ and also that $\\angle XMP = \\angle DMP = \\alpha$. By a symmetric argument applied to $S$, $T_3$, and $T_4$, it follows that $N$ lies on $T_4$ and that $N$ lies on diagonal $AC$ with $\\angle XNP = \\alpha$. Therefore, $\\angle XMP = \\angle XNP$, and $X, M, P,$ and $N$ are concyclic. This implies that the angle formed by lines $MP$ and $NP$ is equal to one of the angles formed by lines $MX$ and $NX$. The fact that $M$ lies on $BD$ and $RP$ and $N$ lies on $AC$ and $SP$ now implies the desired result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15867,
    "subject": "Mathematics (Olympiad)",
    "question": "證明：在任意由相異的 $2000$ 個實數所成之集合中，存在兩對實數 $a > b$ 與 $c > d$，其中 $a \\neq c$ 或 $b \\neq d$，使得\n\n$$\n\\left| \\frac{a-b}{c-d} - 1 \\right| < \\frac{1}{100000}.\n$$\n\nProve that in any set of $2000$ distinct real numbers, there exist two pairs $a > b$ and $c > d$ with $a \\neq c$ or $b \\neq d$, such that\n\n$$\n\\left| \\frac{a-b}{c-d} - 1 \\right| < \\frac{1}{100000}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "對於任意 $n = 2000$ 個相異實數所成的集合 $S$，設 $D_1 \\leq D_2 \\leq \\dots \\leq D_m$ 為所有兩數之差的絕對值（含重複），其中 $m = n(n-1)/2$。經縮放可令最小距離 $D_1 = 1$，即 $D_1 = 1 = y - x$，其中 $x, y \\in S$。最大距離 $D_m = v - u$，$v, u$ 分別為 $S$ 的最大與最小元素。\n\n若對某 $i$ 有 $D_{i+1}/D_i < 1 + 10^{-5}$，則 $0 \\leq D_{i+1}/D_i - 1 < 10^{-5}$，滿足題意。\n\n否則，對所有 $i$ 有：\n\n$$\n\\frac{D_{i+1}}{D_i} \\geq 1 + \\frac{1}{10^5}\n$$\n\n因此：\n\n$$\nv - u = D_m = \\frac{D_m}{D_1} = \\frac{D_m}{D_{m-1}} \\cdots \\frac{D_2}{D_1} \\geq (1 + \\frac{1}{10^5})^{m-1}\n$$\n\n由 $m-1 = 2000 \\times 1999 / 2 - 1 > 19 \\times 10^5$，且 $(1 + 1/n)^n \\geq 2$，得：\n\n$$\n(1 + \\frac{1}{10^5})^{19 \\cdot 10^5} = ((1 + \\frac{1}{10^5})^{10^5})^{19} \\geq 2^{19} > 2 \\times 10^5\n$$\n\n故 $v - u = D_m > 2 \\times 10^5$。\n\n因此，$x$ 與 $u, v$ 至少有一個距離大於 $10^5$，即 $|x - z| > 10^5$，$z \\in \\{u, v\\}$。又 $y - x = 1$，若 $z > y > x$，取 $a = z, b = y, c = z, d = x$，則 $b \\neq d$，\n\n$$\n\\left| \\frac{a-b}{c-d} - 1 \\right| = \\left| \\frac{z-y}{z-x} - 1 \\right| = \\left| \\frac{x-y}{z-x} \\right| = \\frac{1}{z-x} < 10^{-5}\n$$\n\n若 $y > x > z$，取 $a = y, b = z, c = x, d = z$，則 $a \\neq c$，\n\n$$\n\\left| \\frac{a-b}{c-d} - 1 \\right| = \\left| \\frac{y-z}{x-z} - 1 \\right| = \\left| \\frac{x-y}{x-z} \\right| = \\frac{1}{x-z} < 10^{-5}\n$$\n\n故得證。\n\n**註：** 題中 $2000$ 與 $1/100000$ 可換為任意 $n \\in \\mathbb{Z}_{>0}$ 與 $\\delta > 0$，只要\n\n$$\n\\delta(1 + \\delta)^{n(n-1)/2 - 1} > 2\n$$\n\n成立。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15868,
    "subject": "Mathematics (Olympiad)",
    "question": "Let two fixed circles $ (O_1) $ and $ (O_2) $ in the plane touch each other at $ M $, with the radius of $ (O_2) $ greater than that of $ (O_1) $. Let $ A $ be a point on $ (O_2) $ such that $ O_1 $, $ O_2 $, and $ A $ are not collinear. Let $ AB $ and $ AC $ be tangents from $ A $ to $ (O_1) $, touching at $ B $ and $ C $. The lines $ MB $ and $ MC $ meet $ (O_2) $ again at $ E $ and $ F $, respectively. Let $ D $ be the intersection of the line $ EF $ and the tangent to $ (O_2) $ at $ A $. Prove that as $ A $ moves on $ (O_2) $ (with $ O_1 $, $ O_2 $, $ A $ not collinear), $ D $ traces a fixed line.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider two cases:\n\n*First case:* The circles $ (O_1) $ and $ (O_2) $ touch externally at $ M $.\n\nLet $ xy $ be the common tangent at $ M $ to $ (O_1) $ and $ (O_2) $. Since $ CA $ and $ My $ are tangents to $ (O_1) $ at $ C $ and $ M $, we have $ \\angle FCA = \\angle CMy $. But $ \\angle CMy = \\angle FMx $, hence $ \\angle FCA = \\angle FMx $. As $ \\angle FMx = \\angle FAM $, it follows that $ \\angle FCA = \\angle FAM $.\n\nThe triangles $ MFA $ and $ AFC $ have $ \\angle MFA = \\angle AFC $ and $ \\angle FAM = \\angle FCA $, so they are similar. Therefore,\n$$\n\\frac{MF}{FA} = \\frac{AF}{FC}\n$$\nwhich implies $ FM \\cdot FC = FA^2 $. But $ FM \\cdot FC = P_M(O_1) $ (the power of $ M $ with respect to $ (O_1) $), $ = FO_1^2 - R_1^2 $, where $ R_1 $ is the radius of $ (O_1) $. It follows that $ FO_1^2 - FA^2 = R_1^2 $. Analogously, $ EO_1^2 - EA^2 = R_1^2 $. So\n$$\nFO_1^2 - FA^2 = EO_1^2 - EA^2 = R_1^2.\n$$\n\nAs $ D $ lies on the line $ EF $, it implies that $ DO_1^2 - DA^2 = R_1^2 $, i.e., $ D $ moves on a fixed line as $ A $ varies.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15869,
    "subject": "Mathematics (Olympiad)",
    "question": "If $f(x) = x^2 + a x + b$ and $g(x) = x^2 + c x + d$, and we know that $1 + a + b = 4 + 2c + d$ and $4 + 2a + b = 1 + c + d$, what is the value of $a + c$?",
    "options": [],
    "answer": "See solution",
    "solution": "From the given equations:\n\n1. $1 + a + b = 4 + 2c + d$\n2. $4 + 2a + b = 1 + c + d$\n\nSubtract the first equation from the second:\n\n$$\n(4 + 2a + b) - (1 + a + b) = (1 + c + d) - (4 + 2c + d)\n$$\n$$\n4 + 2a + b - 1 - a - b = 1 + c + d - 4 - 2c - d\n$$\n$$\n(4 - 1) + (2a - a) + (b - b) = (1 - 4) + (c - 2c) + (d - d)\n$$\n$$\n3 + a = -3 - c\n$$\n$$\na + c = -6\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15870,
    "subject": "Mathematics (Olympiad)",
    "question": "Nastia has 5 yellow coins, all of which are authentic. She also has 5 blue coins, among which three are authentic and two are fake. All 8 authentic coins have the same weight. One of the fake coins is heavier than an authentic coin by 1 gram, and the other is lighter by 1 gram. Can Nastia, using a balance scale (without weights) and 3 weighings, determine both fake coins and indicate which one is heavier and which is lighter?",
    "options": [],
    "answer": "See solution",
    "solution": "Let the blue coins be $B_1, \\ldots, B_5$.\n\nFirst, weigh 3 yellow coins against 3 blue coins:\n\n$$\nY_1 + Y_2 + Y_3 \\quad ? \\quad B_1 + B_2 + B_3\n$$\n\nConsider the possible outcomes:\n\n**Case 1:**\n$$\nY_1 + Y_2 + Y_3 = B_1 + B_2 + B_3\n$$\nThis means $B_1, B_2, B_3$ are all authentic, so the two fake coins are among $B_4$ and $B_5$.\n\nSecond weighing:\n$$\nB_4 \\quad ? \\quad B_5\n$$\nThe heavier one is the heavy fake, the lighter is the light fake.\n\n**Case 2:**\n$$\nY_1 + Y_2 + Y_3 > B_1 + B_2 + B_3\n$$\nThis means among $B_1, B_2, B_3$ there is a lighter fake, and among $B_4, B_5$ there is a heavier fake.\n\nSecond weighing:\n$$\nB_1 \\quad ? \\quad B_2\n$$\nThird weighing:\n$$\nB_4 \\quad ? \\quad B_5\n$$\nFrom these, you can identify which is the light fake among $B_1, B_2, B_3$ and which is the heavy fake among $B_4, B_5$.\n\n**Case 3:**\n$$\nY_1 + Y_2 + Y_3 < B_1 + B_2 + B_3\n$$\nThis is similar to Case 2, but now among $B_1, B_2, B_3$ there is a heavier fake, and among $B_4, B_5$ there is a lighter fake. Proceed as above, swapping the roles of heavy and light.\n\nThus, in all cases, 3 weighings suffice to identify both fake coins and determine which is heavier and which is lighter.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15871,
    "subject": "Mathematics (Olympiad)",
    "question": "An inventor presented to the king a new exciting board game on a $9 \\times 10$ squared board. The king promised to reward him one rice grain for the first square, one rice grain for the second square, and for each following square the same number of grains as for the two preceding squares together. Prove that for the last square the inventor gets at least $2015^4$ grains.",
    "options": [],
    "answer": "See solution",
    "solution": "Enumerate all squares with $1, \\ldots, 90$. Let the number of rice grains promised for the $n$-th square be $F_n$; then according to the problem $F_1 = F_2 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for all $n > 2$. Notice that $F_n > F_{n-1}$ if $n > 2$, hence $F_{2(n+1)} = F_{2n+2} = F_{2n+1} + F_{2n} > F_{2n} + F_{2n} = 2 \\cdot F_{2n}$ for all $n$. This implies $F_{2 \\cdot 4} > 2 \\cdot F_{2 \\cdot 3} = 2 \\cdot 8 = 2^4$ and by mathematical induction $F_{2n} > 2^n$ for all $n > 3$. Therefore $F_{90} > 2^{45} > 2^{44} = (2^{11})^4 = 2048^4 > 2015^4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15872,
    "subject": "Mathematics (Olympiad)",
    "question": "A jury of 3366 film critics are judging the Oscars. Each critic casts a single vote for their favourite actor, and a single vote for their favourite actress. Assume that for every integer $k$ in the range $1, 2, \\ldots, 100$, some nominee (actor or actress) has been voted for exactly $k$ times. Show that two film critics voted for the same pair of nominees (actor-actress).",
    "options": [],
    "answer": "See solution",
    "solution": "Let $C$ be the set of film critics, $M$ the set of nominated actors, $W$ the set of nominated actresses, and $A = M \\sqcup W$ the set of all nominees. For each $k$ in $K = \\{1, 2, \\ldots, 100\\}$, some nominee has degree $k$ (i.e., received $k$ votes).\n\nConsider a bipartite graph with vertices $A \\sqcup C$, where each critic $x \\in C$ is connected to their chosen actor in $M$ and actress in $W$. The total number of critics is:\n\n$$\n|C| = \\frac{1}{2} \\sum_{x \\in A} \\deg x \\geq \\frac{1}{2} \\sum_{k \\in K} k.\n$$\n\nTo improve this bound, select a subset $K' \\subseteq K$ and for each $k \\in K'$ pick a nominee of degree $k$. Let $A'$ be these nominees and $C'$ their neighbors (critics who voted for them). Partition $C'$ into $C'_1$ (critics who voted for one in $A'$) and $C'_2$ (critics who voted for two in $A'$):\n\n$$\n|C'_1| + 2|C'_2| = \\sum_{k \\in K'} k.\n$$\n\nAssume no two critics voted for the same actor-actress pair. Then:\n\n$$\n|C'_2| \\leq |A' \\cap M| \\cdot |A' \\cap W| \\leq \\frac{1}{4}|K'|^2,\n$$\n\nso\n\n$$\n|C| \\geq |C'| = \\sum_{k \\in K'} k - \\frac{1}{4}|K'|^2.\n$$\n\nMaximizing $S_m - m^2/4$ for $m = 1, \\ldots, n$ where $S_m$ is the sum of the $m$ largest numbers in $K$, and $n = 100$:\n\n$$\nS_m - \\frac{m^2}{4} = \\sum_{k=0}^{m-1} (n-k) - \\frac{m^2}{4} = \\frac{m}{4}(4n-3m+2).\n$$\n\nThe maximum is $n(n+1)/3$ (or $(2n+1)^2/12$ if $n \\equiv 1 \\pmod{3}$). For $n = 100$, $100 \\cdot 101 / 3 = 3366 + 2/3$. Since $|C| = 3366 < 3366 + 2/3$, at least two critics must have voted for the same actor-actress pair.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15873,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a convex polyhedron with 2022 faces. On 3 arbitrary faces, the numbers 26, 4, and 2022 are already written (each face contains one number). One wants to fill in each of the other faces with a real number, such that each number is the arithmetic mean of the numbers on all faces sharing a common edge with that face. Prove that there is exactly one way to fill all the numbers in that polyhedron.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove the following lemma:\n\n*Lemma.* Given a positive integer $n$. The system of linear equations with $n$ variables $(x_1, x_2, \\dots, x_n)$\n\n$$\n\\begin{cases}\na_{11}x_1 + \\cdots + a_{1n}x_n = b_1, \\\\\n\\vdots \\\\\na_{n1}x_1 + \\cdots + a_{nn}x_n = b_n\n\\end{cases}\n$$\n\nhas exactly one solution if the associated homogeneous system (i.e., $b_1 = \\cdots = b_n = 0$) has only the solution $x_1 = \\cdots = x_n = 0$.\n\n*Proof.* Suppose $(x_1, \\dots, x_n)$ and $(y_1, \\dots, y_n)$ are both solutions. Then\n\n$$\n\\begin{cases}\na_{11}(x_1 - y_1) + \\cdots + a_{1n}(x_n - y_n) = 0, \\\\\n\\vdots \\\\\na_{n1}(x_1 - y_1) + \\cdots + a_{nn}(x_n - y_n) = 0\n\\end{cases}\n$$\n\nBy the assumption, $x_1 - y_1 = \\cdots = x_n - y_n = 0$, so the solution is unique.\n\nWe prove existence by induction on $n$. For $n = 1$, the result is clear. Assume true for $n-1$. If all $a_{ij} = 0$, the homogeneous system has infinitely many solutions, so some $a_{ij} \\neq 0$. WLOG, $a_{nn} \\neq 0$. The system can be rewritten as:\n\n$$\n\\begin{cases}\n\\sum_{i=1}^{n-1} \\left( a_{1i} - a_{ni} \\frac{a_{1n}}{a_{nn}} \\right) x_i = b_1 - b_n \\frac{a_{1n}}{a_{nn}}, \\\\\n\\vdots \\\\\n\\sum_{i=1}^{n-1} \\left( a_{n-1,i} - a_{ni} \\frac{a_{n-1,n}}{a_{nn}} \\right) x_i = b_{n-1} - b_n \\frac{a_{n-1,n}}{a_{nn}}, \\\\\n\\sum_{i=1}^n a_{ni} x_i = b_n\n\\end{cases}\n$$\n\nIf the reduced homogeneous system in $n-1$ variables has a nonzero solution, then the original homogeneous system has a nonzero solution, contradiction. By induction, the reduced system has a unique solution $(z_1, \\dots, z_{n-1})$, and then\n\n$$\nx_n = \\frac{b_n - \\sum_{i=1}^{n-1} a_{ni} z_i}{a_{nn}}.\n$$\n\nThus, the lemma holds for $n$.\n\n![](images/Vietnamese_mathematical_competitions_p297_data_050db34584.png)\n\nBack to the problem: Let $a_1, a_2, \\dots, a_{2019}$ be the numbers on the remaining 2019 faces, and denote $a_{2020} = 4$, $a_{2021} = 26$, $a_{2022} = 2022$. Next, we set up the system of equations for the unknowns, where each variable is the arithmetic mean of its neighbors. This system is linear, and the values on three faces are fixed. The associated homogeneous system (all numbers zero) has only the trivial solution, since fixing three values prevents nontrivial solutions. By the lemma, the system has exactly one solution. Therefore, there is exactly one way to fill all the numbers in the polyhedron.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15874,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a semicircle with center $O$ and diameter $AB$. Let $C$ be a point on $k$ such that $CO \\perp AB$. The symmetrical of $\\angle ABC$ intersects $k$ at the point $D$. Let $E$ be the point on $AB$ such that $DE \\perp AB$, and let $F$ be the midpoint of $CB$. Prove that the quadrilateral $EFCD$ is cyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that triangle $ABC$ is an isosceles right triangle. Let $CD \\cap AB = \\{H\\}$. From\n\n$$\n\\angle AED = \\angle ADB = 90^\\circ \\text{ and } \\angle DAE = \\angle DAB\n$$\n\nit follows that $\\triangle ADE \\sim \\triangle ABD$. Since $ABCD$ is cyclic, it follows\n\n$$\n\\angle ADC = 180^\\circ - \\angle ABC = 135^\\circ, \\text{ i.e., } \\angle HDA = 45^\\circ. \\text{ Hence,}\n$$\n\n$$\n\\angle DAB = \\angle DAC + \\angle CAB = \\angle DBC + \\angle CAB = 22^\\circ 30' + 45^\\circ = 67^\\circ 30'\n$$\n\nNow we have\n\n$$\n\\angle HAD = 180^\\circ - \\angle DAB = 180^\\circ - 67^\\circ 30' = 112^\\circ 30'\n$$\n\nand\n\n$$\n\\angle AHD = 180^\\circ - (\\angle HAD + \\angle HDA) = 180^\\circ - 157^\\circ 30' = 22^\\circ 30'\n$$\n\ni.e., $\\triangle HDB$ is isosceles. Since $\\triangle HDB$ is isosceles and $DE \\perp AB$, it follows that $E$ is the midpoint of $HB$. Since $E$ is the midpoint of $HB$ and $F$ is the midpoint of $CB$, it follows that $EF$ is a median in $\\triangle HBC$, i.e., $EF \\parallel HC$, i.e., $EF \\parallel CD$. Hence, the quadrilateral\n\n![](images/Macedonia2018Binder_kniga_MMO_2018_p3_data_e1084c720c.png)\n\n$EFCD$ is a trapezoid. Furthermore, $EF \\parallel HC$, so $\\angle FEB = \\angle CHB = 22^\\circ 30'$. From $DE \\perp AB$ we have\n\n$$\n\\angle DEF = 90^\\circ - \\angle FEB = 90^\\circ - 22^\\circ 30' = 67^\\circ 30'\n$$\n\nThen,\n\n$$\n\\angle EFB = 180^\\circ - (\\angle FEB + \\angle EBF) = 180^\\circ - 67^\\circ 30' = 112^\\circ 30'\n$$\n\n$$\n\\angle CFE = 180^\\circ - \\angle EFB = 67^\\circ 30'\n$$\n\nFinally, $EFCD$ is an isosceles trapezoid, hence the statement in the problem follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15875,
    "subject": "Mathematics (Olympiad)",
    "question": "$\\triangle ABC$ 中有一點 $O$，令 $AO, BO, CO$ 的延長線分別交 $BC, CA, AB$ 於 $A_1, B_1, C_1$。證明：$O$ 在 $\\triangle A_1B_1C_1$ 的中位三角形內（中位三角形是指各邊中點連線所成的三角形）。",
    "options": [],
    "answer": "See solution",
    "solution": "設 $AA_1, BB_1, CC_1$ 分別與 $\\triangle A_1B_1C_1$ 各邊交於 $A_2, B_2, C_2$。\n\n令 $\\frac{AC_1}{C_1B} = \\frac{x}{y}$，$\\frac{BA_1}{A_1C} = \\frac{y}{z}$，其中 $x, y, z > 0$。由 Ceva 定理，$\\frac{CB_1}{B_1A} = \\frac{z}{x}$。\n\n於是質點組 $A(yz), B(zx), C(xy)$ 的重心在 $O$。\n\n另一方面，由於 $A(\\frac{yz}{2})$ 與 $B(\\frac{zx}{2})$ 的重心也是 $C_1$，類似地，$O$ 也是質點組\n\n$A_1\\left(\\frac{zx + xy}{2}\\right), B_1\\left(\\frac{xy + yz}{2}\\right), C_1\\left(\\frac{yz + zx}{2}\\right)$\n\n的重心。\n\n反證，設 $O$ 不在 $A_1B_1C_1$ 的中位三角形內，則它在另三個靠近 $A_1, B_1, C_1$ 的小三角形之一中，不妨假設是 $A_1$，於是 $OA_2 \\ge OA_1$。\n\n但由於 $O$ 是質點組 $A_1\\left(\\frac{zx + xy}{2}\\right), B_1\\left(\\frac{xy + yz}{2}\\right), C_1\\left(\\frac{yz + zx}{2}\\right)$ 的重心，故 $B_1\\left(\\frac{xy+yz}{2}\\right)$ 與 $C_1\\left(\\frac{yz+zx}{2}\\right)$ 的重心應該在直線 $OA_1$ 上，即只能是 $A_2$。\n\n而 $O$ 是 $A_1\\left(\\frac{zx+xy}{2}\\right)$ 與 $A_2\\left(\\frac{xy+yz}{2} + \\frac{yz+zx}{2}\\right)$ 的重心。由於\n\n$$\n\\frac{xy + yz}{2} + \\frac{yz + zx}{2} > \\frac{zx + xy}{2}\n$$\n\n故 $OA_2 < OA_1$，矛盾！",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15876,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the number of quadratic residues modulo $2^{2007}$ in the interval $l = [-2007, 2007]$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Lemma.** *The odd quadratic residues modulo $2^n$, for any positive integer $n$, are the numbers congruent to $1$ modulo $8$.*\n\n*Proof.* The quadratic residue must be congruent to $1$ modulo $8$ because $(2t + 1)^2 = 8 \\cdot \\frac{t(t+1)}{2} + 1$.\n\nFor $n = 1$ and $n = 2$, the only number congruent to $1$ and less than $2^n$ is $1$, which is clearly a quadratic residue. For $n > 2$ we count all odd quadratic residues modulo $2^n$: let $x$ and $y$ be odd numbers; then $x - y$ and $x + y$ are even but not both divisible by $4$, that is, one of the integer numbers $\\frac{x-y}{2}$ and $\\frac{x+y}{2}$ is odd. Hence $x^2 \\equiv y^2 \\pmod{2^n} \\iff 2^n \\mid (x-y)(x+y) \\iff 2^{n-2} \\mid \\left(\\frac{x-y}{2}\\right)^2 \\pmod{2^n} \\iff 2^{n-2} \\mid \\frac{x+y}{2}$ or $2^{n-2} \\mid \\frac{x-y}{2} \\iff x \\equiv \\pm y \\pmod{2^{n-1}} \\iff x \\equiv \\pm y \\pmod{2^n}$ or $x \\equiv \\pm y + 2^{n-1} \\pmod{2^n}$. This means that each odd quadratic residue is generated by $4$ odd numbers between $1$ and $2^n - 1$. Thus there are $\\frac{2^{n-1}}{4} = 2^{n-3}$ odd quadratic residues. Since there are also $2^{n-3}$ numbers congruent to $1$ modulo $8$, all these numbers are the odd quadratic residues modulo $2^n$.\n\nNow let's focus on even quadratic residues. Let $c = 2^m\\ell$, $m$ positive integer, $m < 2007$, $\\ell$ odd. Then $x^2 \\equiv c \\pmod{2^{2007}} \\iff x^2 = 2^m\\ell + 2^{2007}k = 2^m(\\ell + 2^{2007-m}k)$. Since $\\ell + 2^{2007-m}k$ is odd, $m$ must be even and $\\ell$ must be an odd quadratic residue modulo $2^{2007-m}$, which implies $\\ell \\equiv 1 \\pmod{8}$. So the even quadratic residues are of the form $2^{2u}\\ell$, $\\ell \\equiv 1 \\pmod{8}$.\n\nWe are ready to count all the quadratic residues in $l$:\n\n- $\\frac{2001-(-2007)}{8} + 1 = 502$ are odd;\n- $\\frac{4 \\cdot 497 - 4 \\cdot (-495)}{4 \\cdot 8} + 1 = 125$ are of the form $2^2 \\cdot \\ell$;\n- $\\frac{16 \\cdot 121 - 16 \\cdot (-119)}{16 \\cdot 8} + 1 = 31$ are of the form $2^4 \\cdot \\ell$;\n- $\\frac{64 \\cdot 25 - 64 \\cdot (-31)}{64 \\cdot 8} + 1 = 8$ are of the form $2^6 \\cdot \\ell$;\n- $256$ and $256 \\cdot (-7)$ are the only ones of the form $2^8 \\cdot \\ell$;\n- $1024$ is the only one of the form $2^{10} \\cdot \\ell$.\n\nIncluding $0$, there are $502 + 125 + 31 + 8 + 2 + 1 + 1 = 670$ quadratic residues in $\\mathbb{Z}$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 15877,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c, d$ be positive real numbers.\n\n1. Prove that\n$$\n(a + b + c + d - 4)^4 \\geq (a + b + c + d)^4 \\cdot \\frac{(a-1)(b-1)(c-1)(d-1)}{abcd}.\n$$\nWhen does equality hold?\n\n2. Prove that\n$$\n\\frac{d}{a+b+c} + \\frac{a}{b+c+d} + \\frac{b}{c+d+a} + \\frac{c}{d+a+b} \\geq \\frac{4}{3}.\n$$\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "(i) Without loss of generality, assume $a \\geq b \\geq c \\geq d$. Then $\\frac{a-1}{a} \\geq \\frac{b-1}{b} \\geq \\frac{c-1}{c} \\geq \\frac{d-1}{d}$. By Chebyshev's inequality,\n$$\n\\sum_{\\text{cyc}} \\left( a \\cdot \\frac{a-1}{a} \\right) \\geq \\frac{1}{4} \\left( \\sum_{\\text{cyc}} a \\right) \\left( \\sum_{\\text{cyc}} \\frac{a-1}{a} \\right).\n$$\nBy the AM-GM inequality,\n$$\n\\sum_{\\text{cyc}} \\frac{a-1}{a} \\geq 4 \\sqrt[4]{\\frac{(a-1)(b-1)(c-1)(d-1)}{abcd}}.\n$$\nCombining these,\n$$\n(a+b+c+d-4) \\geq (a+b+c+d) \\sqrt[4]{\\frac{(a-1)(b-1)(c-1)(d-1)}{abcd}}.\n$$\nRaising both sides to the fourth power and rearranging gives the desired inequality. Equality holds when $a = b = c = d$.\n\n(ii) We have\n$$\n\\begin{align*}\n& \\frac{d}{a+b+c} + \\frac{a}{b+c+d} + \\frac{b}{c+d+a} + \\frac{c}{d+a+b} \\geq \\frac{4}{3} \\\\\n\\Leftrightarrow \\quad & \\frac{a+b+c+d}{a+b+c} + \\frac{a+b+c+d}{b+c+d} + \\frac{a+b+c+d}{c+d+a} + \\frac{a+b+c+d}{d+a+b} \\geq \\frac{16}{3} \\\\\n\\Leftrightarrow \\quad & (a+b+c+d) \\sum_{\\text{cyc}} \\frac{1}{a+b+c} \\geq \\frac{16}{3} \\\\\n\\Leftrightarrow \\quad & \\left( \\sum_{\\text{cyc}} (a+b+c) \\right) \\left( \\sum_{\\text{cyc}} \\frac{1}{a+b+c} \\right) \\geq 16.\n\\end{align*}\n$$\nThis is true by the Cauchy-Schwarz inequality. Equality holds when $a = b = c = d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15878,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $C$ and $D$ be two intersection points of circle $O_1$ and circle $O_2$. A line passing through $D$ intersects circle $O_1$ and circle $O_2$ at points $A$ and $B$ respectively. The points $P$ and $Q$ are on circle $O_1$ and circle $O_2$ respectively. The lines $PD$ and $AC$ intersect at $M$, and the lines $QD$ and $BC$ intersect at $N$. Suppose $O$ is the circumcenter of triangle $ABC$. Prove that $OD \\perp MN$ if and only if $P$, $Q$, $M$, and $N$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the circumcenter of triangle $ABC$ be $O$, and the radius of the circle $O$ be $R$.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p161_data_e6d75381f6.png)\n\n$$\nNO^2 - R^2 = NC \\cdot NB, \\quad (1)\n$$\n\n$$\nMO^2 - R^2 = MC \\cdot MA. \\quad (2)\n$$\n\nSince $A$, $C$, $D$, and $P$ are concyclic, we have\n\n$$\nMC \\cdot MA = MD \\cdot MP. \\quad (3)\n$$\n\nSimilarly, since $Q$, $D$, $C$, and $B$ are concyclic, we have\n\n$$\nNC \\cdot NB = ND \\cdot NQ. \\quad (4)\n$$\n\nFrom (1), (2), (3), and (4), we have\n\n$$\n\\begin{aligned}\nNO^2 - MO^2 &= ND \\cdot NQ - MD \\cdot MP \\\\\n&= ND(ND + DQ) - MD(MD + DP)\n\\end{aligned}\n$$\n\n$$\n= ND^2 - MD^2 + ND \\cdot DQ - MD \\cdot DP\n$$\n\nSo,\n\n$$\n\\begin{aligned}\nOD \\perp MN &\\iff NO^2 - MO^2 = ND^2 - MD^2 \\\\\n&\\iff ND \\cdot DQ = MD \\cdot DP \\\\\n&\\iff P, Q, M, N \\text{ are concyclic.}\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15879,
    "subject": "Mathematics (Olympiad)",
    "question": "設三角形 $ABC$ 中有 $\\angle B > \\angle C$。設點 $P$ 和 $Q$ 為直線 $AC$ 上的相異兩點，滿足 $\\angle PBA = \\angle QBA = \\angle ACB$，且 $A$ 點位於 $P$ 點與 $C$ 點之間。在線段 $BQ$ 內部取一點 $D$ 使得 $PD = PB$。令射線 $AD$ 與 $\\triangle ABC$ 的外接圓交於 $R$ 點（$R \\neq A$）。證明 $QB = QR$。\n\nLet $ABC$ be a triangle with $\\angle B > \\angle C$.\nLet $P$ and $Q$ be two different points on line $AC$ such that $\\angle PBA = \\angle QBA = \\angle ACB$, and $A$ is located between $P$ and $C$.\nSuppose that there exists an interior point $D$ of segment $BQ$ for which $PD = PB$.\nLet the ray $AD$ intersect the circle $ABC$ at $R \\neq A$.\nProve that $QB = QR$.",
    "options": [],
    "answer": "See solution",
    "solution": "記三角形 $ABC$ 的外接圓為 $\\omega$，並令 $\\angle ACB = \\gamma$。$\\gamma < \\angle CBA$ 的條件可推得 $\\gamma < 90^\\circ$。因為 $\\angle PBA = \\gamma$，直線 $PB$ 與 $\\omega$ 相切，所以 $PA \\cdot PC = PB^2 = PD^2$。於是 $\\frac{PA}{PD} = \\frac{PC}{PD}$，知 $\\triangle PAD$ 相似於 $\\triangle PDC$，且 $\\angle ADP = \\angle DCP$。\n\n![](images/14-3J-M2_p3_data_01e6f59013.png)\n\n再來，因 $\\angle ABQ = \\angle ACB$，$\\triangle ABC$ 亦相似於 $\\triangle AQB$。所以 $\\angle AQB = \\angle ABC = \\angle ARC$，知 $D, R, C, Q$ 四點共圓。於是有 $\\angle DRQ = \\angle DCQ = \\angle ADP$。\n\n從 $\\angle ARB = \\angle ACB = \\gamma$，且 $\\angle PDB = \\angle PBD = 2\\gamma$，可知\n\n$$\n\\begin{aligned}\n\\angle QBR &= \\angle ADB - \\angle ARB = \\angle ADP + \\angle PDB - \\angle ARB \\\\\n            &= \\angle DRQ + \\gamma = \\angle QRB,\n\\end{aligned}\n$$\n\n故得 $QB = QR$。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15880,
    "subject": "Mathematics (Olympiad)",
    "question": "$n \\ge 5$ real numbers are written in a row. It turns out that the sum of any three consecutive numbers is positive and the sum of any five consecutive numbers is negative. Find the largest $n$ for which this is possible.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we construct an example for $n = 6$: $3, -5, 3, 3, -5, 3$.\n\nSuppose there exist $n \\ge 7$ real numbers that satisfy the conditions. Choose any five consecutive numbers $a, b, c, d, e$. The conditions give $a + b + c > 0$ and $c + d + e > 0$, so $(a + b + c + d + e) + c > 0$. But $a + b + c + d + e < 0$, so $c > 0$. Thus, for any five consecutive numbers, the middle one is always positive. This means all numbers except possibly the last two at each end are positive.\n\nNow consider six consecutive numbers $a, b, c, d, e, f$. We have $a + b + c + d + e < 0$ and $(a + b + c) + (d + e + f) > 0$, so $f > 0$. Similarly, $a > 0$. Therefore, all $n$ numbers must be positive, which contradicts the condition that the sum of any five consecutive numbers is negative. Thus, the largest possible $n$ is $6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15881,
    "subject": "Mathematics (Olympiad)",
    "question": "Given that $a_1, a_2, \\dots, a_{10}$ are positive real numbers, determine the smallest possible value of\n$$\n\\sum_{i=1}^{10} \\left\\lfloor \\frac{7a_i}{a_i + a_{i+1}} \\right\\rfloor\n$$\nwhere we define $a_{11} = a_1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m = 7$ and $n = 10$. We claim that the minimum possible value is $m - 1 = 6$, achieved by letting $a_i = m^i$.\n\nTo show this is indeed the minimum, assume without loss of generality that $a_n$ is the largest among all $a_i$. Then,\n$$\n\\sum_{i=1}^{n} \\left\\lfloor \\frac{ma_i}{a_i + a_{i+1}} \\right\\rfloor \\ge \\left\\lfloor \\frac{ma_1}{a_1 + a_2} \\right\\rfloor + \\left\\lfloor \\frac{ma_n}{a_n + a_1} \\right\\rfloor \\ge \\left\\lfloor \\frac{ma_1}{a_1 + a_n} \\right\\rfloor + \\left\\lfloor \\frac{ma_n}{a_n + a_1} \\right\\rfloor \\ge \\left\\lfloor \\frac{m(a_1 + a_n)}{a_n + a_1} \\right\\rfloor - 1 \\ge m - 1\n$$\nusing the fact that $\\lfloor x \\rfloor + \\lfloor y \\rfloor \\ge \\lfloor x + y \\rfloor - 1$.\n\nThis result holds in greater generality:\n\nLet $m, n \\ge 2$ be positive integers. Given $a_1, \\dots, a_n$ positive real numbers, determine the smallest possible value of\n$$\n\\sum_{i=1}^{n} \\left\\lfloor \\frac{ma_i}{a_i + a_{i+1}} \\right\\rfloor\n$$\nwhere $a_{n+1} = a_1$.\n\nAlternative approach: Substitute $a_{i+1}/a_i = x_i$ and $z_i = \\left\\lfloor \\frac{ma_i}{a_i+a_{i+1}} \\right\\rfloor = \\left\\lfloor \\frac{m}{1+x_i} \\right\\rfloor$. The problem becomes: for positive reals $x_1, \\dots, x_n$ with $x_1x_2\\dots x_n = 1$, prove that\n$$\n\\sum_{i=1}^{n} \\left\\lfloor \\frac{m}{1+x_i} \\right\\rfloor \\ge m - 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15882,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(x + f(y)) - f(x) = (x + f(y))^4 - x^4\n$$\n\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We rewrite the given equation as\n\n$$\nf(x + f(y)) = (x + f(y))^4 - x^4 + f(x). \\tag{1}\n$$\n\nSet $x = -f(z)$ and $y = z$ in (1):\n\n$$\nf(0) = -(f(z))^4 + f(-f(z)), \\quad \\text{for all } z \\in \\mathbb{R}. \\tag{2}\n$$\n\nNow, set $x = -f(z)$ in (1) and use (2):\n\n$$\nf(f(y) - f(z)) = (f(y) - f(z))^4 - (f(z))^4 + f(-f(z)) = (f(y) - f(z))^4 + f(0).\n$$\n\nSo, if $t = f(y) - f(z)$, then $f(t) = t^4 + f(0)$. If $f$ takes any nonzero value, then every real number is a difference of two values of $f$.\n\nSuppose $f(a) = b \\ne 0$. Plug $y = a$ into the original equation:\n\n$$\nf(x + b) - f(x) = (x + b)^4 - x^4.\n$$\n\nSince $b \\ne 0$, $(x + b)^4 - x^4$ is a degree 3 polynomial in $x$ and attains all real values as $x$ varies. Thus, the difference $f(x + b) - f(x)$ can be any real number, so $f(t) = t^4 + f(0)$ for all $t \\in \\mathbb{R}$.\n\nAll functions of the form $f(x) = x^4 + k$ satisfy the equation. The zero function $f(x) \\equiv 0$ is also a solution.\n\n**Answer:** All such functions are $f(x) \\equiv 0$ and $f(x) = x^4 + k$ for any real number $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15883,
    "subject": "Mathematics (Olympiad)",
    "question": "$2n$ distinct positive integers are given. What is the largest number of pairs that can always be formed from these numbers, so that each number belongs to at most one pair, and the sum of integers in each pair is a composite number?\n\n![](images/Ukraine_2021-2022_p13_data_3498986c82.png)\n\nFig. 6",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p_1, p_2, \\dots, p_{2n-1}$ be distinct prime integers larger than $2$. Consider the set $(1, p_1 - 1, p_2 - 1, \\dots, p_{2n-1} - 1)$. In this case, it is not possible to form $n$ such pairs, as no number can be paired with $1$ to obtain a composite sum.\n\nOn the other hand, we can always form at least $n - 1$ pairs from numbers of the same parity. In each such pair, the sum will be even and at least $4$, which is therefore composite.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15884,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the sequence $a_1, a_2, \\dots$ defined as follows:\n\n- $a_1 = 1$.\n- The first twelve terms are:\n\n| $n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |\n|-----|---|---|---|---|---|---|---|---|---|----|----|----|\n| $a_n$ | 1 | 3 | 4 | 7 | 10 | 12 | 13 | 16 | 19 | 21 | 22 | 25 |\n\nIt is conjectured that for $r = 0, 1, 2, \\dots$,\n\n$$\na_{4r+1} = 9r+1, \\quad a_{4r+2} = 9r+3, \\quad a_{4r+3} = 9r+4, \\quad a_{4r+4} = 9r+7.\n$$\n\nFind the value of $a_{2015}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use the conjectured formula:\n\n$$\na_{4r+1} = 9r+1, \\quad a_{4r+2} = 9r+3, \\quad a_{4r+3} = 9r+4, \\quad a_{4r+4} = 9r+7.\n$$\n\nTo find $a_{2015}$, write $2015 = 4 \\times 503 + 3$, so $r = 503$ and the position is $a_{4r+3}$.\n\nThus,\n$$\na_{2015} = 9 \\times 503 + 4 = 4527 + 4 = 4531.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15885,
    "subject": "Mathematics (Olympiad)",
    "question": "Olesya was given a homework to add two canonical fractions $\\frac{a}{b}$ and $\\frac{c}{d}$. Her classmate Andriy, who missed the class, asked her by phone about the homework, and due to a bad connection he heard that they need to add $\\frac{b}{a}$ and $\\frac{d}{c}$. After he added them, Andriy asked Olesya for the answer. It turns out that the answers are the same. Was Andriy right in his calculations, if Olesya got an excellent mark, and all 4 fractions they have been working with are distinct?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose Andriy's calculations are correct. Then the following equality holds:  \n$$\n\\frac{a}{b} + \\frac{c}{d} = \\frac{b}{a} + \\frac{d}{c}\n$$\nThis simplifies to:  \n$$\n\\frac{ad + bc}{bd} = \\frac{bc + ad}{ac}\n$$\nTherefore, we have $bd = ac$, or $\\frac{b}{a} = \\frac{c}{d}$, which contradicts the assumption that all fractions are distinct.  \n**Answer:** no.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15886,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all solutions in non-negative integers $a, b$ to $\\sqrt{a} + \\sqrt{b} = \\sqrt{2009}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Rearrange the given equation:\n\n$$\n\\begin{aligned}\n\\sqrt{a} + \\sqrt{b} &= \\sqrt{2009} \\\\\n\\sqrt{a} &= \\sqrt{2009} - \\sqrt{b}.\n\\end{aligned}\n$$\n\nSquaring both sides and noting that $2009 = 7^2 \\times 41$:\n\n$$\na = 2009 + b - 14\\sqrt{41b}.\n$$\n\nSince $a$, $2009$, and $b$ are integers, $\\sqrt{41b}$ must also be rational, so $\\sqrt{41b}$ is an integer.\n\nLet $41b = B^2$ for some non-negative integer $B$. Then $B = 41t$ for a non-negative integer $t$, so $b = 41t^2$. Similarly, $a = 41s^2$ for some non-negative integer $s$.\n\nThe original equation becomes $(s + t)\\sqrt{41} = 7\\sqrt{41}$, so $s + t = 7$. The possible solution pairs $(s, t)$ are $(0, 7), (1, 6), \\ldots, (6, 1), (7, 0)$.\n\nThus, the complete list of solutions $(a, b)$ is:\n\n$$\n\\begin{aligned}\n(a, b) = &\\ (0, 2009),\\ (41, 1476),\\ (164, 1025),\\ (369, 656), \\\\\n         &\\ (656, 369),\\ (1025, 164),\\ (1476, 41),\\ (2009, 0).\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15887,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the area in $\\mathrm{cm}^2$ of a rhombus whose side length is $29\\ \\mathrm{cm}$ and whose diagonals differ in length by $2\\ \\mathrm{cm}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the diagonals be $2x$ and $2x + 2$. The diagonals of a rhombus bisect each other at right angles, partitioning the rhombus into four congruent right-angled triangles, each with hypotenuse $29$. Thus, the area of the rhombus is:\n\n$$4 \\times \\frac{1}{2}x(x + 1) = 2x^2 + 2x$$\n\nFrom the Pythagorean theorem:\n\n$$x^2 + (x+1)^2 = 29^2 = 841$$\n\nSo,\n\n$$2x^2 + 2x = 840$$\n\nAlternatively, since the diagonals bisect each other at right angles, the rhombus can be rearranged to form a square of side $29$ with a unit square hole:\n\n$$29^2 - 1 = 840$$\n\n![](images/Brown_Australian_MO_Scene_2013_p36_data_7b98e25234.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15888,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the circumcentre of triangle $ABC$. Let $K$ and $L$ be the intersection points of the circumcircles of triangles $BOC$ and $AOC$ with the angle bisectors at $A$ and $B$, respectively. Let $P$ be the midpoint of $\\overline{KL}$, $M$ the reflection of $O$ over $P$, and $N$ the reflection of $O$ over $KL$. Prove that $KLMN$ is cyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "The angles $LCA$ and $LOA$ are equal as inscribed angles over the same arc. The angle $LOA$ equals the sum of angles $OAB$ and $OBA$ (since $LOA$ is an exterior angle of triangle $ABO$). Hence:\n\n$$\n\\begin{aligned}\n\\angle LCO &= \\angle LCA + \\angle OCA = \\angle LOA + \\angle OCA \\\\\n&= \\angle OAB + \\angle OBA + \\angle OCA \\\\\n&= \\frac{1}{2}(\\angle CAB + \\angle ABC + \\angle ACB) = 90^\\circ\n\\end{aligned}\n$$\n\nSimilarly, $\\angle KCO = 90^\\circ$, so $C$ lies on $KL$ and $C$ is the midpoint of $\\overline{ON}$. $PC$ is parallel to $MN$ (since $PC$ is a mid-segment of triangle $MNO$). The quadrilateral $LOKM$ is a parallelogram because its diagonals bisect each other. Hence, the angles $MLK$ and $OKL$ are equal. The triangle $OKN$ is isosceles ($KC$ is both an altitude and a median), so the angles $OKC$ and $NKC$ are equal. Thus, $\\angle MLK = \\angle NKL$, so $KLMN$ is an isosceles trapezoid.\n\n![](images/Makedonija_2009_p66_data_7fc79349ad.png)\n\n(If $P$ is on the other side of $C$, then we are working with the angles $MKL$ and $NLK$.)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15889,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to partition the set of positive integers into two classes, none of which contains an infinite arithmetic sequence (with a positive ratio)?\n\nWhat if we require the extra condition that, in each class $C$ of the partition, the set of differences\n\n$$\n\\{ \\min \\{ n : n \\in C \\text{ and } n > m \\} - m : m \\in C \\}\n$$\n\nis bounded?",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to exhibit such a partition: set\n\n$$\nA_1 = \\bigcup_{n=1}^{\\infty} \\{ n(2n-1) + 1,\\ n(2n-1) + 2,\\ \\dots,\\ n(2n-1) + 2n \\},\n$$\n\n$$\nA_2 = \\bigcup_{n=0}^{\\infty} \\{ n(2n+1) + 1,\\ n(2n+1) + 2,\\ \\dots,\\ n(2n+1) + 2n+1 \\}.\n$$\n\nSince each class has arbitrarily large gaps, it cannot contain an infinite arithmetic sequence.\n\nTo exhibit such a partition for the further question, we will rely on building a bijection that will allow us to \"destroy\" every single infinite arithmetic progression within each of the two partition classes. Take any bijection\n\n$$\n\\phi: \\{1, 2, \\dots\\} \\to \\{1, 2\\} \\times \\{1, 2, \\dots\\} \\times \\{1, 2, \\dots\\}.\n$$\n\nDefine $A(a, r) = \\{ a + n r : n = 0, 1, \\dots \\}$ for $a, r \\in \\{1, 2, \\dots\\}$; these are all possible infinite arithmetic progressions of positive integers. At step $s = 0$, both classes $A_1, A_2$ are empty. Denote by $v_{s+1}$ the next value to be distributed in the sets of the partition, so at step $s = 0$ take $v_1 = 1$. Now proceed in an algorithmic way.\n\nAt step $s \\ge 1$, let $\\phi(s) = (c, a, r)$. There exists a least $k \\ge 0$ such that $v_s + k \\in A(a, r)$. Put $v_s, v_s + 1, \\dots, v_s + k$ alternately in $A_1, A_2$, starting with the one having the least maximum. If this ends with $v_s + k \\in A_c$, then take $v_{s+1} = v_s + k + 1$; if not, then force $v_s + k \\in A_c$ and put $v_s + k + 1$ in the other class, then take $v_{s+1} = v_s + k + 2$. Continue with the next step $s + 1$.\n\nThis ensures that neither of the arithmetic progressions $A(a, r)$ will be contained in any of the two classes $A_1, A_2$, while the differences between pairs of consecutive elements within each class is at most 3.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15890,
    "subject": "Mathematics (Olympiad)",
    "question": "Inside a right circular cone with base radius $5$ and height $12$ are three congruent spheres, each with radius $r$. Each sphere is tangent to the other two spheres and also tangent to the base and side of the cone. What is $r$?\n\n(A) $\\frac{3}{2}$  (B) $\\frac{90 - 40\\sqrt{3}}{11}$  (C) $2$  (D) $\\frac{144 - 25\\sqrt{3}}{44}$  (E) $\\frac{5}{2}$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $C$ be the center of the base of the cone, $A$ be the center of one of the spheres, $B$ be the point where that sphere is tangent to the base of the cone, $D$ be the point such that $\\overline{CD}$ is a radius of the base of the cone containing $B$, $V$ be the vertex of the cone, and $E$ be the point on $\\overline{CV}$ such that $\\overline{ED}$ contains $A$. Let the radius of the sphere be $r = AB$.\n\n![](images/2021_AMC10A_Solutions_Fall_p9_data_82dfc3e8ec.png)\n\nBecause the three spheres are mutually tangent, their centers are at the vertices of an equilateral triangle with side length $2r$. The line $CV$ passes through the centroid of the equilateral triangle, so $BC$ must be $\\frac{2}{3}$ the altitude of that triangle, implying that $BC = \\frac{2}{3} r\\sqrt{3}$. The Pythagorean Theorem implies that $DV = \\sqrt{CD^2 + CV^2} = 13$.\n\nObserve that $\\overline{DE}$ is the angle bisector of $\\angle CDV$. By the Angle Bisector Theorem, $\\frac{CD}{CE} = \\frac{DV}{EV}$. Thus $\\frac{5}{CE} = \\frac{13}{12-CE}$ and solving gives $CE = \\frac{10}{3}$. Because $\\triangle DAB \\sim \\triangle DEC$, it follows that\n\n$$\nBD = AB \\cdot \\frac{CD}{CE} = r \\cdot \\frac{5}{\\left(\\frac{10}{3}\\right)} = \\frac{3}{2}r.\n$$\n\nThen $BC + BD = CD$ gives\n\n$$\n\\frac{2\\sqrt{3}}{3}r + \\frac{3}{2}r = 5.\n$$\n\nSolving for $r$ yields $r = \\frac{90-40\\sqrt{3}}{11}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 15891,
    "subject": "Mathematics (Olympiad)",
    "question": "а) Чи може п'ять кіл на площині мати лише 8 точок перетину?\n\nб) Чи може п'ять кіл на площині мати лише 20 точок перетину?",
    "options": [],
    "answer": "See solution",
    "solution": "а) Може. Достатньо взяти 4 кола, кожні два з яких перетинаються в двох точках, а п'яте коло таким, щоб не перетиналося з іншими.\n\nб) Не може. Перше коло перетинається з іншими щонайбільше у 8 точках, друге додає не більше за 6 точок перетину, і т. д. Загалом маємо не більше, ніж $8+6+4+2=20$ різних точок перетину.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15892,
    "subject": "Mathematics (Olympiad)",
    "question": "Доведіть, що для всіх додатних чисел $a, b, c$ виконується нерівність:\n\n$$\n\\frac{ab}{a+3b+2c} + \\frac{bc}{b+3c+2a} + \\frac{ca}{c+3a+2b} \\leq \\frac{1}{6}(a+b+c).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Використаємо відому нерівність $$(x + y + z) \\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right) \\geq 9,$$ для $x, y, z > 0$, яку можна довести за допомогою нерівності Коші. Перепишемо її у вигляді:\n\n$$\n\\frac{1}{x+y+z} \\leq \\frac{1}{9}\\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right).\n$$\n\nОскільки $a + 3b + 2c = 2b + (b+c) + (c+a)$, то\n\n$$\n\\frac{ab}{a+3b+2c} \\leq \\frac{ab}{9} \\left( \\frac{1}{2b} + \\frac{1}{b+c} + \\frac{1}{c+a} \\right) = \\frac{a}{18} + \\frac{1}{9} \\left( \\frac{ab}{b+c} + \\frac{ab}{c+a} \\right).\n$$\n\nДодавши цю та ще дві аналогічні нерівності, дістанемо\n\n$$\n\\begin{aligned}\n\\frac{ab}{a+3b+2c} + \\frac{bc}{b+3c+2a} + \\frac{ca}{c+3a+2b} &\\le \\frac{1}{18}(a+b+c) + \\frac{1}{9}\\left(\\frac{ab}{b+c} + \\frac{ab}{c+a} + \\frac{bc}{c+a} + \\frac{bc}{a+b} + \\frac{ca}{a+b} + \\frac{ca}{b+c}\\right) \\\\\n&\\le \\frac{1}{18}(a+b+c) + \\frac{1}{9}(a+b+c) = \\frac{1}{6}(a+b+c).\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15893,
    "subject": "Mathematics (Olympiad)",
    "question": "The numbers $a_1, a_2, \\dots, a_{12}$ in the figure are $1, 2, \\dots, 12$ in some order such that the sum of the 4 numbers on each of the 6 segments is the same. Find the minimum of $a_1 + a_2 + a_3 + a_4 + a_5 + a_6$ over all such arrangements.\n\n![](images/3._NATIONAL_XXX_OMA_2013_p1_data_eed564a001.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Add up the numbers on each segment. By hypothesis, this gives 6 equal sums, say equal to $S$. Since each $a_i$ enters exactly two of these sums, we obtain $6S = 2(1+2+\\dots+12)$, so $S = 26$. Thus, the 4 numbers on each segment have sum 26.\n\nLet $T = \\sum_{i=1}^{6} a_i$, $U = \\sum_{i=7}^{12} a_i$. Add up the numbers on segments $a_1a_2, a_2a_3, a_3a_4$ to obtain $2(a_1+a_2+a_3)+U = 3S$. Likewise, $2(a_4+a_5+a_6)+U = 3S$, implying $a_1+a_2+a_3 = a_4+a_5+a_6$. In particular, the sum $T$ is even.\n\nThe least possible values of $a_1, \\dots, a_6$ are $1, \\dots, 6$. Their sum $1+\\dots+6=21$ is odd, so $T \\neq 21$ and hence $T \\ge 22$. We show that $T=22$ is not possible either.\n\nSuppose that $T=22$. Then $a_1, \\dots, a_6$ are $1, 2, 3, 4, 5, 7$ in some order. These numbers must be divided into two triples $a_1, a_2, a_3$ and $a_4, a_5, a_6$ with the same sum 11. Each triple must have an odd number of odd numbers, yielding the unique possible division: $1, 3, 7$ and $2, 4, 5$. By symmetry, let $\\{a_1, a_2, a_3\\} = \\{1, 3, 7\\}$, $\\{a_4, a_5, a_6\\} = \\{2, 4, 5\\}$.\n\nSince $S=26$, it follows that the sums $a_7+a_8$, $a_9+a_{10}$, $a_{11}+a_{12}$ are even, and so is exactly one of the sums $a_8+a_9$, $a_{10}+a_{11}$, $a_{12}+a_7$. In all, there are exactly two odd sums among $a_7+a_8$, $a_9+a_{10}$, $a_{11}+a_{12}$, $a_{13}+a_{14}$, $a_{15}+a_7$. They are among $a_8+a_9$, $a_{10}+a_{11}$, $a_{12}+a_7$, and the two summands in each one lie in a side of triangle $a_5a_6a_7$.\n\nOn the other hand, $a_7, \\dots, a_{12}$ are $6, 8, 9, 10, 11, 12$ in some order. There are 2 odd numbers among them, 9 and 11. They must be at consecutive vertices of the hexagon $a_7a_8\\dots a_{12}$. Otherwise, each one has two even neighbors and there are 4 odd sums $a_i+a_{i+1}$, $i=7, 8, \\dots, 12$. Hence, a side of triangle $a_1a_2a_3$ or triangle $a_4a_5a_6$ contains 9 and 11. The second case is impossible or else there will be odd sums $a_i+a_{i+1}$ on the sides of $a_1a_2a_3$. Hence, 9 and 11 are on a side of $a_1a_2a_3$. Then $S=26$ implies that at the endpoints of this side we have numbers with sum $26-(9+11)=6$. However, there are no two numbers with sum 6 in the triple $1, 3, 7$.\n\nTherefore, $T=22$ cannot occur, so $T \\ge 24$. An example with $T=24$ is given in the figure. The required minimum is $24$.\n\n![](images/3._NATIONAL_XXX_OMA_2013_p1_data_444fefe8b2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15894,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\in \\mathbb{N}$ and $\\mathcal{A}$ be a nonempty family of nonempty subsets of $\\{1, 2, \\dots, n\\}$ with the following property: if $A \\in \\mathcal{A}$ and $A \\subset B \\subseteq \\{1, 2, \\dots, n\\}$, then $B \\in \\mathcal{A}$. Prove that the function\n\n$$\nf(x) := \\sum_{A \\in \\mathcal{A}} x^{|A|} (1-x)^{n-|A|}\n$$\n\nis strictly increasing in the interval $(0, 1)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $0 < p < q < 1$. Notice that $p^* = \\frac{q-p}{1-p} \\in (0, 1)$. We construct the sets $X$ and $Y$ as follows: For each element $i \\in \\{1, 2, \\dots, n\\}$, we put $i$ in $X$ with probability $p$ (independently of each other), and for each element $j \\in \\{1, 2, \\dots, n\\} \\setminus X$, we put $j$ in $Y$ with probability $p^*$. Then $P(x \\in X \\cup Y) = p + (1-p)p^* = q$. It is easy to see that $f(p) = P(X \\in \\mathcal{A})$, and $f(q) = P(X \\cup Y \\in \\mathcal{A})$. Since $\\mathcal{A}$ has the property from the condition, we have $X \\subseteq X \\cup Y \\implies f(p) < f(q)$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15895,
    "subject": "Mathematics (Olympiad)",
    "question": "In each vertex of a regular $n$-gon $A_1, A_2, \\ldots, A_n$ there is a unique pawn. In each step, it is allowed to:\n\n1. Move all pawns one step in the clockwise direction, or\n2. Swap the pawns at vertices $A_1$ and $A_2$.\n\nProve that by a finite series of such steps it is possible to swap the pawns at vertices:\n\na) $A_i$ and $A_{i+1}$ for any $1 \\leq i < n$, while leaving all other pawns in their initial place.\n\nb) $A_i$ and $A_j$ for any $1 \\leq i < j \\leq n$, leaving all other pawns in their initial place.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the pawn initially at $A_i$ be labeled $i$. We first prove part (a), then use it for part (b).\n\n**a)**\n\nApply the first operation $i-1$ times. This brings pawns $i$ and $i+1$ to $A_1$ and $A_2$, respectively, and moves every other pawn $i-1$ steps clockwise.\n\nNow, apply the second operation to swap $i$ and $i+1$ at $A_1$ and $A_2$. This does not affect the position of any other pawn.\n\nFinally, apply the first operation $n-i+1$ times. This returns each pawn $k \\ne i, i+1$ to $A_k$, while moving pawn $i$ to $A_{i+1}$ and pawn $i+1$ to $A_i$, as desired.\n\n**b)**\n\n*Solution without induction:*\n\nUsing part (a), swap pawns $(i, i+1)$ at $(A_i, A_{i+1})$, then $(i, i+2)$ at $(A_{i+1}, A_{i+2})$, and continue until swapping $(i, j)$ at $(A_{j-1}, A_j)$. Now $i$ is at $A_j$, and each $i+1 \\leq k \\leq j$ is at $A_{k-1}$.\n\nNext, use part (a) to swap $j$ with $j-1$, then $j$ with $j-2$, and so on, until $j$ is swapped with $i+1$. This places $j$ at $A_i$ and each $i+1 \\leq k \\leq j-1$ at $A_k$.\n\nThus, pawns $i$ and $j$ are swapped, with all others in their original places.\n\n*Solution using induction:*\n\nInduct on $n$ for the claim: any two pawns $1 \\leq i < j \\leq k$ can be swapped.\n\nThe base case is part (a).\n\nAssume the claim holds for $k$. Then, for $k+1$, we can swap any $1 \\leq i < j \\leq k$ by the hypothesis. To swap $i$ and $k+1$ for $1 \\leq i \\leq k$, swap $i$ and $k$, then $k$ and $k+1$ (by part (a)), then $k+1$ and $i$ (now at $A_k$ and $A_i$).",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15896,
    "subject": "Mathematics (Olympiad)",
    "question": "We have triangle $ABC$. The cevians from $A$, $B$, and $C$ intersect at one point. Let $A_0$ and $C_0$ be the midpoints of sides $BC$ and $AB$, respectively. The lines $B_1C_1$, $B_1A_1$, and $B_1B$ cross the line $A_0C_0$ at points $C_2$, $A_2$, and $B_2$, respectively. Prove that point $B_2$ is the midpoint of segment $A_2C_2$.\n\n![](images/Ukrajina_2010_p37_data_37041c8fa5.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let us draw a straight line through point $B$ parallel to $AC$, and mark points $C_3$ and $A_3$ as the intersections of this line with $B_1C_1$ and $B_1A_1$, respectively. By similarity of triangles, we have the following ratios:\n\n$$\n\\frac{C_3B}{AB_1} = \\frac{BC_1}{AC_1}, \\qquad \\frac{CB_1}{BA_3} = \\frac{CA_1}{BA_1}\n$$\n\nMultiplying these equalities:\n\n$$\n\\frac{C_3B}{AB_1} \\cdot \\frac{CB_1}{BA_3} = \\frac{BC_1}{AC_1} \\cdot \\frac{CA_1}{BA_1}\n$$\n\nBy Ceva's theorem:\n\n$$\n\\frac{BC_1}{AC_1} \\cdot \\frac{AB_1}{BA_1} = 1\n$$\n\nFrom the previous equality, we get:\n\n$$\n\\frac{BC_3}{BA_3} = 1\n$$\n\nSince $\\frac{B_2C_2}{A_2B_2} = \\frac{BC_3}{BA_3} = 1$, it follows that $B_2C_2 = B_2A_2$, which is what we needed to prove.\n\n![](images/Ukrajina_2010_p37_data_37041c8fa5.png)\n\nFig.24",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15897,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_n$ be the number of sequences of length $n$ formed from the digits $1, 2, 3, 4$ that contain an even number of $1$'s. Find $a_{2016}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_n$ be the number of such sequences with $n$ terms. Since there are $4^n$ total sequences, $4^n - a_n$ sequences have an odd number of $1$'s.\n\nConsider a sequence of length $n$ with an even number of $1$'s. If the last digit is $1$, then the first $n-1$ digits must have an odd number of $1$'s, giving $4^{n-1} - a_{n-1}$ choices. If the last digit is $2$, $3$, or $4$, then the first $n-1$ digits must have an even number of $1$'s, giving $a_{n-1}$ choices for each. Thus,\n\n$$\na_n = 4^{n-1} - a_{n-1} + 3a_{n-1} = 2a_{n-1} + 4^{n-1}.\n$$\n\nExpanding recursively:\n\n$$\n\\begin{aligned}\na_n &= 2a_{n-1} + 4^{n-1}, \\\\\n2a_{n-1} &= 2^2a_{n-2} + 2 \\cdot 4^{n-2}, \\\\\n\\vdots \\\\\n2^{n-2}a_2 &= 2^{n-1}a_1 + 2^{n-2} \\cdot 4,\n\\end{aligned}\n$$\n\nSumming up,\n\n$$\n\\begin{aligned}\na_n &= 2^{n-1}a_1 + \\sum_{k=0}^{n-2} (2^k \\cdot 4^{n-1-k}) \\\\\n&= 2^{n-1}(3) + \\sum_{k=0}^{n-2} 2^{2n-2-k} \\\\\n&= 3 \\cdot 2^{n-1} + 2^{2n-1} - 2^n = 2^{n-1} + 2^{2n-1}.\n\\end{aligned}\n$$\n\nTherefore, $a_{2016} = 2^{2015} + 2^{4031}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15898,
    "subject": "Mathematics (Olympiad)",
    "question": "A point $P$ is on the side $AB$ of a convex quadrilateral $ABCD$. Let $\\omega$ be the incircle of the triangle $CPD$ and $I$ be its center. It is known that $\\omega$ touches the incircles of the triangles $APD$ and $BPC$ at $K$, respectively $L$. Let $E$, respectively $F$ be the meeting points of the lines $AC$ and $BD$, respectively $AK$ and $BL$. Prove that the points $E, I$ and $F$ are collinear.\n\n![](images/RMC2011_2_p102_data_bd646121e5.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $J$ the center of the circle $k$ placed in the half-plane $(AB, C)$ and tangent to $AB$, $DA$ and $BC$. Denote $a$, respectively $b$, the incircles of the triangles $ADP$ and $BCP$. We will firstly prove that $F \\in IJ$. The point $A$ is the center of the homothety which transforms $a$ into $k$; $K$ is the center of the homothety with negative ratio which transforms $a$ into $\\omega$. Denote $\\bar{F}$ the center of the homothety which transforms $\\omega$ into $k$. It is known that $A$, $K$ and $\\bar{F}$ are collinear. In the same way, $\\bar{F} \\in BL$, hence $F = \\bar{F}$, so $F \\in IJ$.\n\nWe prove now that $E \\in IJ$. Comparing the lengths of the tangents from $A$, $P$, $C$, $D$ to the circles $k$ and $a$ gives $AP + DC = AD + PC$. This shows that the quadrilateral $APCD$ has an incircle $d$. Let $X$ be the center of the homothety which transforms $a$ into $\\omega$. The above reasoning, applied to the circles $a$, $d$ and $\\omega$ shows that $A$, $C$ and $X$ are collinear. Consider now the circles $a$, $\\omega$ and $k$. The point $A$ is the center of the homothety which transforms $a$ into $k$ and $X$ is the center of the homothety which transforms $a$ into $\\omega$, hence $XA$ contains the center $\\bar{E}$ of the homothety which transforms $\\omega$ into $k$, whence $\\bar{E} \\in AC$. In the same way $\\bar{E} \\in BD$, therefore $\\bar{E} = E$, so $E \\in IJ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15899,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the centre of $\\Gamma$ and extend $DP$ and $EP$ to meet $\\Gamma$ again at $Q$ and $S$ respectively.\n\n![](images/2020_Australian_Scene_W_p142_data_d8bcd9ea3e.png)\n\nWe are given that $AP = AO = R$, the radius of $\\Gamma$. Hence $P$ and $O$ lie on the circle centred at $A$ with radius $R$ (which we will denote $\\Gamma_2$). Show that $Q$ and $S$ also lie on $\\Gamma_2$ and thus $QS$ is the radical axis of the same-sized circles $\\Gamma$ and $\\Gamma_2$.\n\nLet $M$ be the midpoint of $BC$. Since $\\triangle BPC$ is isosceles with base $BC$ and $\\triangle BOC$ is isosceles with base $BC$, the perpendicular bisector of $BC$ passes through $P$ and $O$. Therefore, $O$, $P$, and $M$ are collinear.\n\nShow that $QD$ bisects $\\angle EDC$ and $SE$ bisects $\\angle CED$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, let $O$ be the centre of $\\Gamma$ and let $X$ be the midpoint of $BC$. It is given that $PB = PC$, $AP = AO$, and $ED$ is the perpendicular bisector of $BP$.\n\nSince $\\triangle BPC$ is isosceles with base $BC$ and $\\triangle BOC$ is isosceles with base $BC$, the perpendicular bisector of $BC$ passes through $P$ and $O$. Therefore, $O$, $P$, and $X$ are collinear.\n\n![](images/2020_Australian_Scene_W_p143_data_55627c4f95.png)\n\nNow $\\angle AOC = 2\\angle ABC = 2\\angle PBC = \\angle PBC + \\angle BCP = \\angle APC$ hence $AOPC$ is a cyclic quadrilateral. Furthermore, $\\angle PAO = 180^\\circ - \\angle AOP - \\angle OPA = 180^\\circ - 2\\angle OPA = 180^\\circ - 2\\angle XPB = 180^\\circ - \\angle XPB - \\angle CPX = \\angle APC$, hence $OA \\parallel PC$ and $AOPC$ is in fact an isosceles trapezium.\n\nThe perpendicular bisectors of $BP$, $PC$, and $BC$ are concurrent. Also, since $AOPC$ is an isosceles trapezium, $AC$, $OP$, and the perpendicular bisector of $PC$ (also the perpendicular bisector of $OA$) are concurrent. Hence $AC$, $OP$, and $ED$ are concurrent. Given this, and that both $AOPC$ and $AEDC$ are cyclic, by the radical axis theorem, $OEDP$ is also cyclic.\n\nSince $ED$ is the perpendicular bisector of $BP$, $BDPE$ is a kite. Hence we have $\\angle EDP = \\angle BDE = \\angle BCE = \\alpha$, say, and $\\angle PED = \\angle DEB = \\angle DCB = \\beta$. Thus $\\angle DCE = \\angle BCE + \\angle DCB = \\alpha + \\beta$. Furthermore $\\angle DOE = \\angle DPE = 180^\\circ - \\alpha - \\beta$ since $OEDP$ is cyclic and the angle sum of $\\triangle DPE$ is $180^\\circ$. Finally, $\\angle DOE = 2\\angle DCE$, which implies $180^\\circ - \\alpha - \\beta = 2(\\alpha + \\beta)$, hence $\\alpha + \\beta = 60^\\circ$. Thus $\\angle DCE = 60^\\circ$ and $\\angle DOE = 120^\\circ$.\n\nLet $DP$ intersect $BC$ at $Y$ and $EP$ intersect $BC$ at $Z$. We have that $\\angle DOE = 120^\\circ$ and hence $\\angle OED = \\angle EDO = (180^\\circ - 120^\\circ)/2 = 30^\\circ$. Since $OEDP$ is cyclic, we can deduce $\\angle XPD = \\angle OED = 30^\\circ$ and $\\angle ZPX = \\angle EPO = \\angle EDO = 30^\\circ$. Finally, since $OX \\perp BC$, angle sum of $\\triangle XZP$ and $\\triangle XYP$ give that $\\angle BZE = \\angle XZP = 180^\\circ - 90^\\circ - 30^\\circ = 60^\\circ$ and $\\angle DYB = \\angle PYX = 180^\\circ - 90^\\circ - 30^\\circ = 60^\\circ$.\n\nFinally, in $\\triangle DYB$ and $\\triangle DCE$, $\\angle DYB = 60^\\circ = \\angle DCE$ and $\\angle DBY = \\angle DBC = \\angle DEC$, hence $\\angle BDY = \\angle EDC$ (and indeed $\\triangle DYB \\sim \\triangle DCE$). Therefore $\\angle BDE = \\angle EDP = \\angle PDC = \\alpha$, hence $DP$ is the internal angle bisector of $\\angle EDC$.\n\nSimilarly, $\\triangle BZE \\sim \\triangle DCE$ and $\\angle DEB = \\angle PED = \\angle CEP = \\beta$, hence $EP$ is the internal angle bisector of $\\angle CED$.\n\nTherefore, in $\\triangle CDE$, $P$ is the point of intersection of the internal angle bisectors of $\\angle D$ and $\\angle E$ and hence the incentre of $\\triangle CDE$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15900,
    "subject": "Mathematics (Olympiad)",
    "question": "平面上有一個三角形 $ABC$，其外接圓為 $\\Gamma$，設點 $A'$ 是點 $A$ 在 $\\Gamma$ 上的對徑點。作正三角形 $BCD$，使得 $A, D$ 兩點位於 $BC$ 的異側。設過 $A'$ 且與 $A'D$ 垂直的直線分別與直線 $AC, AB$ 交於 $E, F$ 兩點。以 $EF$ 為底，作底角為 $30^\\circ$ 的等腰三角形 $ETF$，並使 $A, T$ 兩點位於 $EF$ 的異側。證明：$AT$ 經過三角形 $ABC$ 的九點圓圓心 $N$。\n\n註：三角形 $ABC$ 的九點圓，係指通過三角形 $ABC$ 三邊的中點、三高的垂足、與頂點到垂心的三條線段的中點這九個點的圓。",
    "options": [],
    "answer": "See solution",
    "solution": "解：我們首先證明兩個引理。\n\n**引理 1.** 設 $P, Q$ 為 $\\triangle ABC$ 的一對等角共軛點。若 $O, O_a, O_b, O_c, T$ 分別為 $\\triangle ABC, \\triangle BPC, \\triangle CPA, \\triangle APB$ 及 $\\triangle O_aO_bO_c$ 的外心，則 $PQ$ 與 $OT$ 平行。\n\n**引理 1. 證明** 設 $\\triangle Q_aQ_bQ_c$ 為 $Q$ 關於 $\\triangle ABC$ 的垂足三角形，且設 $V$ 為 $\\triangle Q_aQ_bQ_c$ 的外心。因為 $Q_bQ_c, Q_cQ_a, Q_aQ_b$ 分別垂直於 $AP, BP, CP$，所以 $\\triangle Q_aQ_bQ_c \\cup Q \\cup V$ 與 $\\triangle O_aO_bO_c \\cup O \\cup T$ 位似。因為 $V$ 為 $PQ$ 的中點，所以 $OT$ 與 $PQ$ 平行。引理 1 證畢。\n\n**引理 2.** 給定 $\\triangle ABC$ 與一點 $U$ 使得 $\\triangle BUC$ 為正三角形。設 $V$ 為 $U$ 關於 $\\triangle ABC$ 的等角共軛點。則 $UV$ 與 $\\triangle ABC$ 的尤拉線平行。\n\n**引理 2. 證明** 設 $BV, CV$ 分別與 $\\triangle ABC$ 的外接圓交於 $E, F$ 兩點，且令 $N$ 為 $EF$ 的中點。設 $G, O$ 分別為 $\\triangle ABC$ 的重心及外心，且設 $M$ 為 $BC$ 的中點。計算角度可知 $\\triangle AEF$ 為正三角形，所以 $A, O, N$ 共線，且\n\n$$\n\\frac{ON}{AO} = \\frac{1}{2} = \\frac{GM}{AG},\n$$\n\n故 $MN$ 與 $\\triangle ABC$ 的尤拉線 $GO$ 平行。\n\n![](images/17-3J_p7_data_2f81d196af.png)\n\n另一方面，計算角度可知 $\\angle^*VBU = \\angle^*FAC$，$\\angle^*VCU = \\angle^*EAB$（此處 $\\angle^*$ 代表有向角），所以利用正弦定理計算有向面積，得\n\n$$\n\\begin{align*}\n[\\triangle UMV] - [\\triangle UNV] \\\\\n&= \\frac{1}{2}([\\triangle UBV] + [\\triangle UCV]) - \\frac{1}{2}([\\triangle UEV] + [\\triangle UFV]) \\\\\n&= \\frac{1}{2}([\\triangle UBE] + [\\triangle UCF]) = 0.\n\\end{align*}\n$$\n\n這就是說 $MN$ 與 $UV$ 平行，故 $UV$ 亦與 $\\triangle ABC$ 的尤拉線平行。引理 2 證畢。\n\n回到原題。設 $O$ 為 $\\triangle ABC$ 的外心，且設 $V$ 為 $O$ 關於 $BC$ 的對稱點。\n因為 $A'V$ 的中點是 $\\triangle A'BC$ 的九點圓圓心且 $N$ 為 $AV$ 的中點，所以 $AN$\n與 $\\triangle A'BC$ 的尤拉線平行。故只需再證明 $AT$ 與 $\\triangle A'BC$ 的尤拉線平行即可。\n\n設 $X, Y, Z, K$ 分別為 $\\triangle BCD, \\triangle A'BD, \\triangle A'CD$ 與 $\\triangle XYZ$ 的外心。\n因為 $XY, XZ$ 分別垂直於 $BD, CD$，所以 $\\angle^*YOZ = 120^\\circ$。故 $\\triangle KYZ$ 為一以 $YZ$ 為底且底角為 $30^\\circ$ 的等腰三角形。又注意到 $YZ, ZO, OY$ 分別垂直於 $A'D, A'C, A'B$，所以 $\\triangle AEF$ 與 $\\triangle OZY$ 位似。故 $\\triangle AEF \\cup T$ 與 $\\triangle OZY \\cup K$ 位似，從而得 $AT$ 與 $OK$ 平行。\n\n另一方面，設 $D'$ 為 $D$ 關於 $\\triangle A'BC$ 的等角共軛點。由引理 1 知 $DD'$\n與 $OK$ 平行。又由引理 2 得 $DD'$ 平行於 $\\triangle A'BC$ 的尤拉線。綜合上述，可得 $AT$ 平行於 $\\triangle A'BC$ 的尤拉線。證明完畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15901,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_1, \\dots, A_N$ be all the 700-element subsets of $\\{1, 2, \\dots, 1400\\}$, where $N = \\binom{1400}{700}$. Choose 1400 polynomials $P_1, \\dots, P_{1400}$ of the form:\n\n$$\nP_i(x) = 1 + 10^{10} \\prod_{1 \\leq j \\leq N, j \\in A_i} (x - j).\n$$\n\nShow that if Alireza chooses these 1400 polynomials, he will win the game, regardless of which 700 polynomials Amin chooses.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume Amin chooses 700 polynomials $P_{i_1}, \\dots, P_{i_{700}}$. Let $A_l = \\{i_1, \\dots, i_{700}\\}$. For all $i \\in A_l$, $P_i(l) = 1$, so $\\Omega(P_i(l)) = 1$. Otherwise, $P_i(l) = 1 + 10^{10} K_i$ with $|P_i(l)| > 2$, so $\\omega(P_i(l)) \\geq 2 > 1$.\n\nThus, for $n = l$:\n\n$$\n1 = \\max_{i \\in A_l} (\\Omega(P_i(l))) < \\min_{i \\notin A_l} (\\omega(P_i(l))).\n$$\n\nTherefore, by choosing such polynomials, Alireza will win.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15902,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. A positive integer $k$ is called a *benefactor* of $n$ if the positive divisors of $k$ can be partitioned into two sets $A$ and $B$ such that $n$ is equal to the sum of the elements of $A$ minus the sum of the elements of $B$. Note that $A$ or $B$ may be empty, and that the sum of the elements of the empty set is $0$.\n\nFor example, $15$ is a benefactor of $18$ because $(1 + 5 + 15) - (3) = 18$.\n\nShow that every positive integer $n$ has at least $2023$ benefactors.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be any integer. We shall show that there are infinitely many positive integers $m$ such that $n = \\sum_{i=1}^{k} \\pm d_i$ where $d_1, d_2, \\dots, d_k$ are the positive divisors of $m$, and some suitably chosen combination of $\\pm$ signs depending on $m$ and $n$.\n\n*Case 1.* $n$ is odd.\n\nWithout loss of generality, assume $n > 0$. Consider the binary representation $n = 2^{a_1} + 2^{a_2} + \\dots + 2^{a_k}$, where $0 = a_1 < a_2 < \\dots < a_k$ are non-negative integers. For each integer $r \\ge a_k$, we turn this into a signed sum of the divisors of $m = 2^r$ inductively as follows. Whenever $2^j$ ($j < r$) is present in the signed sum for $n$ but $2^{j+1}$ is not, then replace $2^j$ by $2^{j+1} - 2^j$. Doing this finitely many times writes $n$ as a signed sum of the divisors of $m = 2^r$, as desired.\n\n*Case 2.* $n$ is even.\n\nSince $n-3$ is odd, we may use case 1 to find $n-3$ as the signed sum of the positive divisors of $2^r$ for each integer $r$ satisfying $2^r \\ge n-3$. Consider $m = 3 \\cdot 2^r$. Since $3 = 3(2^r - 2^{r-1} - 2^{r-2} - \\dots - 1)$, we may write $3$ as the signed sum of all divisors of $3 \\cdot 2^r$ that are multiples of $3$. Adding together the two signed sums for $n-3$ and $3$ writes $n$ as the signed sum of all divisors of $m = 3 \\cdot 2^r$, as desired.\n\n(A variation: instead of splitting $n$ into $3$ and $n-3$, we can split it as $n = 3a + b$ for odd $a, b$. The same argument can be used as long as $2^r \\ge \\max(a, b)$.)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15903,
    "subject": "Mathematics (Olympiad)",
    "question": "The number $2009^{2009^{2009}}$ is written in base 10. In one step, we perform the following operation: delete the first and last digit, and add their sum to the number that remains after deleting the first and last digit.\n\n1. If, after a finite number of steps, a two-digit number remains, is it possible for that number to be a perfect square?\n\n2. If, after a finite number of steps, a one-digit number remains, determine that number.",
    "options": [],
    "answer": "See solution",
    "solution": "We use the following properties:\n\n*Property 1.* Every natural number, when divided by $3$ or $9$, gives the same remainder as the sum of its digits.\n\n*Property 2.* The square of a natural number, when divided by $3$, gives remainder $0$ or $1$.\n\n*Property 3.* After any step, the remainder modulo $9$ is invariant.\n\n**Proof of Property 3:**\nLet, after some step, the number $A = \\overline{a_n a_{n-1}\\dots a_1 a_0}$ be obtained. By Property 1,\n$$\nA \\equiv a_n + a_{n-1} + \\dots + a_1 + a_0 \\pmod{9}.\n$$\nAfter deleting the first and last digit, we get $\\overline{a_{n-1}\\dots a_1}$. After one step, the new number is $B = \\overline{a_{n-1}\\dots a_1} + a_n + a_0$.\n\nThus,\n$$\nB \\equiv a_{n-1} + \\dots + a_1 + a_n + a_0 \\pmod{9},\n$$\nso $A \\equiv B \\pmod{9}$.\n\n---\n\n**(a)** If, after a finite number of steps, the number $A$ remains, then by Property 3, $A$ has the same remainder modulo $3$ as $2009^{2009^{2009}}$.\n\nSince $2009 \\equiv 2 \\equiv -1 \\pmod{3}$,\n$$\n2009^{2009^{2009}} \\equiv (-1)^{2009^{2009}} \\equiv -1 \\equiv 2 \\pmod{3}.\n$$\nThus, $A \\equiv 2 \\pmod{3}$, so by Property 2, $A$ cannot be a perfect square.\n\n---\n\n**(b)** If, after a finite number of steps, a one-digit number $a$ remains, then by Property 3,\n$$\na \\equiv 2009^{2009^{2009}} \\pmod{9}.\n$$\nSince $2009 \\equiv 2 \\pmod{9}$,\n$$\n2009^{2009^{2009}} \\equiv 2^{2009^{2009}} \\pmod{9}.\n$$\nNote that $2^3 \\equiv -1 \\pmod{9}$.\n\nNow,\n$$\n2009^{2009} \\equiv 2^{2009} \\equiv (-1)^{2009} \\equiv -1 \\equiv 2 \\pmod{3},\n$$\nso $2009^{2009} = 3k + 2$ for some odd $k$.\n\nTherefore,\n$$\n2^{3k+2} = (2^3)^k \\cdot 4 \\equiv (-1)^k \\cdot 4 \\equiv -1 \\cdot 4 \\equiv -4 \\equiv 5 \\pmod{9}.\n$$\nSo, the final one-digit number is $a = 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15904,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(xy) = f(x)f(y) + f(f(x + y))\n$$\nholds for all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "One can prove directly Claims 3 and 4 without the use of Claims 1 and 2. To prove Claim 3 we can make use of $P(x+1, y-1)$ which together with $P(x, y)$ and (1) gives\n$$\nf(xy + y - x) - c f(xy) = f(y) - c f(x). \\qquad (3)\n$$\nAssuming $c = -1$, then (1) and (3) give that $f(x+2) = f(x)$ for every $x \\in \\mathbb{R}$. It follows that $f(x+2n) = f(x)$ for every $x \\in \\mathbb{R}$ and every $n \\in \\mathbb{Z}$. Now with similar ideas as in the proof of Claim 1, it can be shown that for every $u, v \\in \\mathbb{R}$ there is $n \\in \\mathbb{N}$ large enough such that $u = xy + x - y + 2n$ and $v = xy + y - x$. Then using (3) we can get\n$$\nf(u) = f(xy + x - y + 2n) = f(xy + x - y) = f(xy + y - x) = f(v).\n$$\nSo $f$ is constant and it must be identically equal to $1/2$ which leads to a contradiction.\n\nNow using (3) with $x = y$ and assuming $c \\ne 1$ we get $f(x^2) = f(x)$. So $f$ is even. This eventually leads to $f(n) = 1/(1-c) = a = b$ for every integer $n$. Now $P(0,0)$ gives $a = a^2 + f(a)$ and $P(a, -a)$ gives $f(-a^2) = f(a)f(-a) + f(a)$. Since $f$ is even we eventually get $f(a) = 0$ which gives $a = 0$ or $a = 1$ both contradicting the facts that $a \\ne 0$ and $b \\ne 1$.\n\nSo $c = 1$ and using (1) and (3) one can eventually get $a = -1$. The solution can then finish in the same way as in Solution 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15905,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c \\in (0, \\infty)$. Prove the inequality\n$$\n\\frac{a - \\sqrt{bc}}{a + 2(b + c)} + \\frac{b - \\sqrt{ca}}{b + 2(c + a)} + \\frac{c - \\sqrt{ab}}{c + 2(a + b)} \\geq 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The AM-GM inequality gives\n$$\n\\frac{a - \\sqrt{bc}}{a + 2(b + c)} \\geq \\frac{a - \\frac{b+c}{2}}{a + 2(b + c)} = \\frac{2a - b - c}{2(a + 2b + 2c)}.\n$$\nSo, the left-hand side of the given inequality is at least\n$$\n\\frac{2a-b-c}{2(a+2b+2c)} + \\frac{2b-c-a}{2(2a+b+2c)} + \\frac{2c-a-b}{2(2a+2b+c)} = S.\n$$\nLet $a + 2b + 2c = 5x$, $2a + b + 2c = 5y$, and $2a + 2b + c = 5z$. Then\n$a = -3x + 2y + 2z$, $b = 2x - 3y + 2z$, $c = 2x + 2y - 3z$, and\n$$\n\\frac{2a-b-c}{2(a+2b+2c)} = \\frac{-10x+5y+5z}{10x} = \\frac{1}{2}\\left(\\frac{y}{x} + \\frac{z}{x} - 2\\right).\n$$\nSimilarly for the other terms, so\n$$\n\\begin{aligned}\nS &= \\frac{1}{2} \\left( \\frac{y}{x} + \\frac{z}{x} - 2 \\right) + \\frac{1}{2} \\left( \\frac{x}{y} + \\frac{z}{y} - 2 \\right) + \\frac{1}{2} \\left( \\frac{x}{z} + \\frac{y}{z} - 2 \\right) \\\\\n&= \\frac{1}{2} \\left( \\left( \\frac{x}{y} + \\frac{y}{x} \\right) + \\left( \\frac{x}{z} + \\frac{z}{x} \\right) + \\left( \\frac{z}{y} + \\frac{y}{z} \\right) - 6 \\right) \\ge 0,\n\\end{aligned}\n$$\nwhich proves the result.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15906,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be the midpoint of the side $AD$ of the square $ABCD$. Consider the equilateral triangles $DFM$ and $BFE$, such that $F$ lies in the interior of $ABCD$ and the lines $EF$ and $BC$ are concurrent. Denote by $P$ the midpoint of $ME$. Prove that:\n\na) $P$ lies on the line $AC$;\n\nb) the halfline $PM$ is the bisector of the angle $APF$.\n\n![](images/RMC_2024_p63_data_149965d12b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "a) Construct the equilateral triangle $BDQ$, such that $C$ lies in its interior. $Q$ is situated on the perpendicular bisector of the diagonal $BD$, therefore $Q$, $C$, and $A$ are collinear. Since $\\angle EBQ = \\angle FBD = 60^\\circ - \\angle FBQ$, $BQ = BD$ and $BE = BF$, triangles $BEQ$ and $BFD$ are congruent (SAS), thus $\\angle BQE = \\angle BDF = \\angle ADF - \\angle ADB = 15^\\circ$ and $QE = DF = DM = AM$.\n\nFrom $\\angle QBC = \\angle QBD - \\angle CBD = 15^\\circ = \\angle BQE$, we infer that $QE \\parallel CB \\parallel AD$. As $QE = AM$, it follows that $AMQE$ is a parallelogram, so the midpoint $P$ of $ME$ lies on $AQ$, i.e., $P$ is situated on the line $AC$.\n\nb) Since $QE = DM$ and $QE \\parallel DM$ we deduce that $DMEQ$ is a parallelogram and $\\angle DME = \\angle DQE = 75^\\circ$. Consequently, $\\angle FMP = \\angle DMP - \\angle DMF = 15^\\circ$. The triangle $FAD$ is right-angled, with $\\angle FAD = 30^\\circ$, therefore we obtain $\\angle FAP = \\angle DAP - \\angle DAF = 15^\\circ$ and $\\angle MFA = \\angle DFA - \\angle DFM = 30^\\circ$.\n\nSince $\\angle FMP = \\angle FAP$, it follows that $AMFP$ is a cyclic quadrilateral, with $\\angle MPA = \\angle MFA = 30^\\circ$.\n\n$\\angle FPM = \\angle FAM = 30^\\circ$, therefore $\\angle FPM = \\angle APM$, and $PM$ is the angle bisector of $APF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15907,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic convex quadrilateral and $\\Gamma$ its circumcircle. Let $E$ be the intersection of the diagonals $AC$ and $BD$. Let $L$ be the center of the circle tangent to sides $AB$, $BC$, and $CD$. Let $M$ be the midpoint of the arc $BC$ of $\\Gamma$ not containing $A$ and $D$. Prove that the excenter of triangle $BCE$ opposite $E$ lies on the line $LM$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $L$ be the intersection of the bisectors of $\\angle ABC$ and $\\angle BCD$. Let $N$ be the $E$-excenter of $\\triangle BCE$. Let $\\angle BAC = \\angle BDC = \\alpha$, $\\angle DBC = \\beta$, and $\\angle ACB = \\gamma$.\n\nWe have the following:\n\n$$\n\\begin{align*}\n\\angle CBL &= \\frac{1}{2}\\angle ABC = 90^\\circ - \\frac{1}{2}\\alpha - \\frac{1}{2}\\gamma \\quad \\textbf{and} \\quad \\angle BCL = 90^\\circ - \\frac{1}{2}\\alpha - \\frac{1}{2}\\beta, \\\\\n\\angle CBN &= 90^\\circ - \\frac{1}{2}\\beta \\quad \\textbf{and} \\quad \\angle BCN = 90^\\circ - \\frac{1}{2}\\gamma, \\\\\n\\angle MBL &= \\angle MBC + \\angle CBL = 90^\\circ - \\frac{1}{2}\\gamma \\quad \\textbf{and} \\quad \\angle MCL = 90^\\circ - \\frac{1}{2}\\beta, \\\\\n\\angle LCN &= \\angle LBN = 180^\\circ - \\frac{1}{2}(\\alpha + \\beta + \\gamma).\n\\end{align*}\n$$\n\nApplying the sine rule to $\\triangle MBL$ and $\\triangle MCL$ we obtain\n\n$$\n\\frac{MB}{ML} = \\frac{MC}{ML} = \\frac{\\sin \\angle BLM}{\\sin \\angle MBL} = \\frac{\\sin \\angle CLM}{\\sin \\angle MCL}.\n$$\n\nIt follows that\n\n$$\n\\frac{\\sin \\angle BLM}{\\sin \\angle CLM} = \\frac{\\sin \\angle MBL}{\\sin \\angle MCL} = \\frac{\\cos(\\gamma/2)}{\\cos(\\beta/2)}. \\qquad (1)\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15908,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ and $a$ be positive integers satisfying $a^n \\equiv 1 \\pmod{n}$. Show that there exists a positive integer $s$ such that $a^s + s \\equiv 0 \\pmod{n}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove by induction on $n$.\n\nFor $n = 1$, there is nothing to prove.\n\nSuppose $n \\geq 2$ and let $d := \\gcd(\\phi(n), n)$.\n\nClearly, $a$ and $n$ are relatively prime, thus we have $a^d \\equiv 1 \\pmod{n}$ by Euler's theorem. Since $d \\leq \\phi(n) < n$ and $a^d \\equiv 1 \\pmod{d}$, there exists $m$ such that $a^m + m \\equiv 0 \\pmod{d}$ by the induction hypothesis. Hence, there exists a positive integer $r \\leq n/d$ such that $a^m + m = dr \\pmod{n}$. If we choose $s := m + n - dr$, then\n\n$$\na^s + s \\equiv a^m + m - dr \\equiv 0 \\pmod{n}$$\n\nand the proof is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15909,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a square. Let $E$ and $F$ be points such that $|BF| = |DE|$, $|BC| = |DA|$, and $\\angle CBF = \\angle ADE = \\frac{\\pi}{2}$. Let $T$ be the intersection point of the lines $AC$ and $EF$.\n\n![](images/Slovenija_2012_p16_data_9b1821e472.png)\n\nShow that the lines $AC$ and $EF$ are perpendicular.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $T$ be the intersection point of the lines $AC$ and $EF$. Because $|BF| = |DE|$, $|BC| = |DA|$ and $\\angle CBF = \\angle ADE = \\frac{\\pi}{2}$, the triangles $ADE$ and $CBF$ are congruent. We thus have $|AE| = |CF|$, and the quadrilateral $AFCE$ is an isosceles trapezoid. Hence $\\angle EFA = \\angle CAF$, which is also equal to $\\frac{\\pi}{4}$ because $AC$ is a diagonal of the square $ABCD$. We get $\\angle ATF = \\pi - \\angle EFA - \\angle CAF = \\frac{\\pi}{2}$, from which we conclude that the lines $AC$ and $EF$ are perpendicular.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15910,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be an integer. Prove that there exist infinitely many odd primes $p$ such that $p \\mid a^{p-1} + 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\begin{aligned}\n&\\text{Suppose } p \\mid a^{2n} - 1 \\text{ and } p \\mid a^{p-1} - 1. \\\\\n&\\text{Then } p \\mid a^2 - 1 = a(a - 1)(a + 1) \\implies p \\mid a - 1. \\\\\n&\\text{But for } a^n + 1 \\equiv 2 \\not\\equiv 0 \\pmod{p}, \\text{ this leads to a contradiction.} \\\\\n&\\text{If there exists } p_1 > 2, \\text{ then by a lemma, there exists } p_2 \\text{ such that } p_1 \\mid a + 1, p_2 \\mid a^{p_1 + 1}, p_2 \\mid a + 1. \\\\\n&\\text{Using the lemma, we have } p_1^2 \\mid a^{p_1 + 1} + 1 \\text{ and } p_2 \\mid a^{p_1 + 1} + 1. \\\\\n&\\text{From here, } p_2^2 \\mid a^{p_1 p_2 + 1}, \\text{ and so on:} \\\\\n&\\quad p_1^2 \\cdots p_k^2 \\mid (a^{p_1 \\cdots p_{k-1}})^{p_k} + 1, \\quad k \\geq 1. \\\\\n&\\text{By the lemma, there exists a prime } p_{k+1} \\text{ such that} \\\\\n&\\quad p_{k+1} \\mid (a^{p_1 \\cdots p_{k-1}})^{p_k} + 1, \\quad p_{k+1} \\nmid a^{p_1 \\cdots p_{k-1} - 1} + 1, \\quad p_{k+1} > 2. \\\\\n&\\text{Also, } (p_{k+1}, p_1 \\cdots p_{k-1}) = 1. \\text{ If } p_{k+1} = p_k, \\text{ then } p_k^2 \\mid (a^{p_1 \\cdots p_{k-1}})^{p_k} + 1. \\\\\n&\\text{But then } p_k \\mid a^{p_1 \\cdots p_{k-1}} + 1, \\text{ which leads to a contradiction.} \\\\\n&\\text{Thus, we can construct infinitely many such primes.}\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15911,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimal number $n$ such that $n^3 + n^2 + 330n + 330$ is divisible by $2011$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $n = 41$.\n\n$$\n n^3 + n^2 + 330n + 330 = (n + 1)(n^2 + 330)\n$$\n\nThis expression is divisible by $2011$ if at least one factor is divisible by $2011$. If the first factor is divisible by $2011$, then the minimal $n = 2010$. For $n^2 + 330$, since $n^2$ increases, we find that for $n = 41$, $n^2 + 330 = 2011$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15912,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral such that $BC$ and $AD$ meet at a point $P$. Consider a point $Q$, different from $P$, on the line $BP$ such that $PQ = BP$, and construct the parallelograms $CAQR$ and $DBCS$. Prove that the points $C$, $Q$, $R$, $S$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "It is enough to prove that\n\n$$\n\\angle RQC = \\angle RSC. \\qquad (\\dagger)\n$$\n\n![](images/shortlistBMO_2011_p14_data_5f852632fa.png)\n\nNow\n\n$$\n\\angle RQC = \\angle ACQ = \\angle ACB = \\angle ADB. \\qquad (1)\n$$\n\nTake $T$ such that $QABT$ is a parallelogram. Then $\\overrightarrow{BT} = \\overrightarrow{AQ} = \\overrightarrow{CR}$ and $\\overrightarrow{BD} = \\overrightarrow{CS}$ imply $\\triangle BTD \\equiv \\triangle CRS$, so\n\n$$\n\\angle RSC = \\angle TDB. \\qquad (2)\n$$\n\nOn the other hand, since $P$ is the midpoint of the segment $BQ$ and $ABTQ$ is a parallelogram, $P$ is also the midpoint of the segment $AT$, so $\\angle TDB = \\angle ADB$. Combining this with (1) and (2), we get $(\\dagger)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15913,
    "subject": "Mathematics (Olympiad)",
    "question": "In the triangle $ABC$ with $AB > AC$, a tangent to the circumcircle of triangle $ABC$ is drawn through point $A$. This tangent intersects the line $BC$ at point $P$. On the extension of side $BA$ beyond $A$, point $Q$ is chosen such that $AQ = AC$. Let $X$ and $Y$ be the midpoints of segments $CQ$ and $AP$, respectively, and let $R$ be a point on segment $AP$ such that $AR = CP$. Prove that $CR = 2XY$.\n\n![](images/Ukraine_2016_Booklet_p34_data_d00192091b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "$AB > AC$, so the tangent intersects $BC$ such that $P$ lies beyond $C$ on $BC$ (see figure). Let $\\angle B = \\beta$, then $\\angle CAP = \\beta$ (angle between tangent and chord at $A$). Set $AC = 2x$, $CP = 2y$, so $AQ = 2x$ and $AR = 2y$.\n\nLet $M$ be the midpoint of $PQ$. Then $MX$ and $MY$ are mid-segments of $\\triangle CPQ$ and $\\triangle APQ$, respectively, so $MX = y$, $MY = x$, and $MX \\parallel CP$, $MY \\parallel AQ$. Thus, $\\angle XMY = \\angle PBQ = \\beta$ (by parallelism).\n\nTherefore, $\\frac{MY}{MX} = \\frac{x}{y} = \\frac{2x}{2y} = \\frac{AC}{AR}$ and $\\angle XMY = \\angle PAC = \\beta$. So $\\triangle XMY \\sim \\triangle RAC$ (by SAS similarity). From similarity,\n\n$$\n\\frac{CR}{XY} = \\frac{AC}{MY} = \\frac{2x}{y} = 2.\n$$\n\nThus, $CR = 2XY$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15914,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\in \\mathbb{N}$, $n \\ge 2$, and three sets of real numbers $A$, $B$, $C$, pairwise disjoint, each of them having $n$ elements.\n\nLet $a$ be the number of triples $(x, y, z) \\in A \\times B \\times C$ for which $x < y < z$ and $b$ be the number of triples $(x, y, z) \\in A \\times B \\times C$ for which $x > y > z$. Prove that $a - b$ is divisible by $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider $y \\in B$, arbitrarily chosen. Denote by $A_y$ the number of pairs $(x, z) \\in A \\times C$ for which $x < y < z$, and by $B_y$ the number of pairs $(x, z) \\in A \\times C$ for which $x > y > z$.\n\nLet $A = \\{a_1 < a_2 < \\dots < a_n\\}$ and $C = \\{c_1 < c_2 < \\dots < c_n\\}$.\n\nOn the real axis, $y$ separates the numbers $a_1, a_2, \\dots, a_n$ into $k$ numbers smaller than $y$ and $n-k$ numbers greater than $y$, where $k$ is some number from $\\{0, 1, 2, \\dots, n\\}$. Likewise, $c_1, c_2, \\dots, c_n$ are separated by $y$ into $p$ numbers smaller than $y$ and $n-p$ numbers greater than $y$, where $p \\in \\{0, 1, 2, \\dots, n\\}$. It follows that $A_y = k(n-p)$ and $B_y = p(n-k)$, so $A_y - B_y = n(k-p)$.\n\nObviously, $a = \\sum_{y \\in B} A_y$ and $b = \\sum_{y \\in B} B_y$, so $a - b = \\sum_{y \\in B} (A_y - B_y)$ is divisible by $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15915,
    "subject": "Mathematics (Olympiad)",
    "question": "Sean $C$ y $C'$ dos circunferencias tangentes exteriores con centros $O$ y $O'$ y radios $1$ y $2$, respectivamente. Desde $O$ se traza una tangente a $C'$ con punto de tangencia en $P'$, y desde $O'$ se traza la tangente a $C$ con punto de tangencia en $P$, en el mismo semiplano que $P'$ respecto de la recta que pasa por $O$ y $O'$. Hallar el área del triángulo $OXO'$, donde $X$ es el punto de corte de $O'P$ y $OP'$.",
    "options": [],
    "answer": "See solution",
    "solution": "Los triángulos $OPO'$ y $OP'O'$ son rectángulos en $P$ y $P'$, respectivamente, y $\\angle PXO = \\angle P'XO'$. Luego, los triángulos $PXO$ y $P'XO'$ son semejantes con razón de semejanza $O'P'/OP = 2$. La razón entre sus áreas $S'$ y $S$ es entonces $S'/S = 4$. Por el Teorema de Pitágoras, $OP' = \\sqrt{5}$ y $O'P = 2\\sqrt{2}$, luego si $A$ es el área pedida se tiene que\n\n$$\nA + S' = \\frac{1}{2} O'P' \\cdot OP' = \\sqrt{5}; \\quad A + S = \\frac{1}{2} OP \\cdot O'P = \\sqrt{2}.\n$$\n\nDe las relaciones anteriores se obtiene fácilmente que\n$$\nA = \\frac{4\\sqrt{2}-\\sqrt{5}}{3}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15916,
    "subject": "Mathematics (Olympiad)",
    "question": "In space, rays *k*, *l*, and *m* originate from a point *O*. Let the angle between *k* and *l* be $\\alpha$, the angle between *l* and *m* be $\\beta$, and the angle between *m* and *k* be $\\gamma$, where $\\alpha + \\beta \\le 180^{\\circ}$. Ray *r* bisects the angle between *k* and *l*; ray *s* bisects the angle between *l* and *m*. Is it always true that the angle between *r* and *s* is $\\frac{\\gamma}{2}$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let *O* be a vertex of a cube and let rays *k*, *l*, and *m* be directed along the edges of the cube. Then $\\alpha = \\beta = \\gamma = 90^{\\circ}$. Rays *r* and *s* are directed from vertex *O* along the angle bisectors of the faces, which are also diagonals. The other endpoints of the diagonals, *R* and *S*, are endpoints of the diagonal of the same face of the cube. The three diagonals form an equilateral triangle *ORS*, so $\\angle ROS = 60^{\\circ}$. Thus, the angle between rays *r* and *s* is not equal to half the angle between rays *k* and *m*.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15917,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that any simple graph with $2n$ vertices and no triangles has at most $n^2$ edges.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove this by induction.\n\n*Base case*: For $n=1$, there are $2$ vertices, and at most $1 = 1^2$ edge.\n\n*Inductive step*: Assume any triangle-free simple graph with $2k$ vertices has at most $k^2$ edges. Consider a triangle-free simple graph with $2(k+1)$ vertices.\n\nIf there are no edges, the claim holds. Otherwise, let $A$ and $B$ be two vertices joined by an edge. For any other vertex $C$, at most one of the pairs $\\{A, C\\}$ and $\\{B, C\\}$ can be an edge (otherwise, $A$, $B$, $C$ would form a triangle). Thus, there are at most $2k$ edges containing exactly one of $A$ or $B$.\n\nAmong the remaining $2k$ vertices, by the inductive hypothesis, there are at most $k^2$ edges. Therefore, the total number of edges is at most\n\n$$\n1 + 2k + k^2 = (k+1)^2.\n$$\n\nThis completes the induction.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15918,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all maps $f: \\mathbb{Z}_{\\ge 1} \\to \\mathbb{Z}_{\\ge 1}$ such that for any positive integers $a$ and $b$,\n\n$$\naf(a)^2 + bf(b)^2 + 3ab(f(a) + f(b))$$\n\nis a perfect cube.",
    "options": [],
    "answer": "See solution",
    "solution": "For any prime $p$, set $a = b = p$. Then $2p f(p) (f(p) + p)$ is a perfect cube, so $p \\mid f(p)$. Let $f(p) = k p$, $k \\in \\mathbb{Z}_{\\ge 1}$. Setting $a = p$, $b = 1, 2$, we get:\n\n$$\n\\begin{aligned}\np f(p)^2 + f(1)^2 + 3p(f(p) + f(1)) &= m^3 \\\\\np f(p)^2 + 2f(2)^2 + 6p(f(p) + f(2)) &= n^3.\n\\end{aligned}\n$$\n\nThus $m, n > \\sqrt[3]{k^2} p$ and\n\n$$\n3 \\sqrt[3]{k^2} p (\\sqrt[3]{k^2} p - 1) + 1 < n^3 - m^3 = 3k p^2 + 3p(2f(2) - f(1)) + 2f(2)^2 - f(1)^2.\n$$\n\nFrom this inequality, if $p$ is sufficiently large, then $k = 1$.\n\nNow, for any positive integer $c$ and sufficiently large prime $p$, set $a = c$, $b = p$. This gives\n\n$$\n(c + p - 1)^3 < c f(c)^2 + p^3 + 3 c p (f(c) + p) < (c + p + 1)^3.\n$$\n\nTherefore, $c f(c)^2 + p^3 + 3 c p (f(c) + p) = (c + p)^3$, so $f(c) = c$. Thus, $f$ is the identity map.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15919,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $PA$ and $PB$ be the tangents to circle $\\omega$ from an external point $P$. Let $M$ and $N$ be the midpoints of $AP$ and $AB$, respectively. Extend $MN$ to meet $\\omega$ at $C$, where $N$ is between $M$ and $C$. $PC$ meets $\\omega$ at $D$ and extend $ND$ to intersect $PB$ at $Q$. Show that $MNQP$ is a rhombus.",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that $AB \\perp NP$. Thus, $M$ is the circumcenter of $\\triangle ANP$ and hence $MN = MP$.\n\nIt can also be seen that $MN \\parallel PQ$.\n\nFrom the power of the point $M$, $PM^2 = MA^2 = ME \\cdot MC$.\n\nSo, $\\frac{PM}{ME} = \\frac{MC}{PM}$ and hence $\\triangle PME \\sim \\triangle CMP$.\n\nThus, $M\\hat{P}E = M\\hat{C}P$.\n\nSince $O, A, P, B$ are cyclic, the power of the point $N$ tells us that\n\n$$\nCN \\cdot NE = AN \\cdot NB = ON \\cdot NP.\n$$\n\nThus, $C, P, E, O$ are also cyclic and hence $E\\hat{P}N = N\\hat{C}O$.\n\nSince $\\triangle PAN \\sim \\triangle POA$, $\\frac{PA}{PN} = \\frac{PO}{PA}$. Thus,\n\n$$\nPN \\cdot PO = PA^2 = PD \\cdot PC.\n$$\n\nSo, $C, D, N, O$ are cyclic and hence $Q\\hat{N}P = P\\hat{C}O$.\n\nWe can now see that\n\n$$\nQ\\hat{N}P = P\\hat{C}O = P\\hat{C}M + M\\hat{C}O = M\\hat{P}E + E\\hat{P}N = M\\hat{P}N.\n$$\n\nThus, $MP \\parallel NQ$. Therefore, $MNQP$ is a rhombus.\n\n![](images/Tajland_2008_p8_data_f6ab59eadb.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15920,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral with perpendicular diagonals and circumcenter $O$. Let $g$ be the line obtained by reflecting the diagonal $AC$ about the angle bisector of $\\angle BAD$.\n\nProve that the point $O$ lies on the line $g$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $X$ the point of intersection of the diagonals $AC$ and $BD$, i.e., $AX$ is an altitude in triangle $ABD$. By the inscribed angle theorem, we have $\\angle ABX = \\frac{1}{2}\\angle DOA$. Hence\n\n$$\n\\angle XAB = 90^\\circ - \\angle ABX = \\frac{1}{2} \\cdot (180^\\circ - \\angle DOA) = \\angle OAD.\n$$\n\nIn the last step, the angle sum in the equilateral triangle $DAU$ has been used. Since the lines $AB$ and $AD$ are symmetric with respect to the angle bisector $w_\\alpha$, the same is true for $AX$ and $AU$. Hence the assertion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15921,
    "subject": "Mathematics (Olympiad)",
    "question": "Initially, the numbers $1, 2, \\ldots, 2024$ are written on a blackboard. Trixi and Nana play a game, taking alternate turns. Trixi plays first.\n\nThe player whose turn it is chooses two numbers $a$ and $b$, erases both, and writes their (possibly negative) difference $a - b$ on the blackboard. This is repeated until only one number remains on the blackboard after $2023$ moves. Trixi wins if this number is divisible by $3$, otherwise Nana wins.\n\nWhich of the two has a winning strategy?",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that Nana has a winning strategy.\n\nThe only relevant property of all numbers in the game is their residue modulo $3$. Therefore, we will call all numbers $0$, $1$, or $2$ according to their residue, and we will also call $1$s and $2$s non-zeros.\n\nWe observe that each move either does not change the number of non-zeros (if one or two zeros are involved in the move) or decreases the number of non-zeros by $1$ or $2$ (if no zero is involved in the move).\n\nNana can play arbitrarily for a long time while the number of non-zeros decreases, until that number reaches $1$, $2$, $3$, or $4$ at the start of her move. This has to happen because it is not possible to go from $5$ or more non-zeros to $0$ non-zeros in two moves, and Trixi certainly cannot win as long as there are non-zeros on the blackboard.\n\nIf the number of non-zeros is $4$, then Nana will avoid decreasing the number of non-zeros by using one or two zeros to force Trixi to decrease the number to $2$ or $3$. This has to happen because Trixi always starts a move with an even quantity of numbers, so she is the first one without zeros as long as there are $4$ non-zeros.\n\nIf the number of non-zeros is $3$, then two of them have the same value. Nana chooses these two and replaces them with zero. This leaves one non-zero which can change between $1$ and $2$, but never be removed until the end. So Nana wins.\n\nIf the number of non-zeros is $2$, and they are distinct, then Nana replaces them with their difference $1$ which again can never become zero.\n\nIf the number of non-zeros is $2$ and they have the same value, then Nana will use one of them and a $0$ to convert them to $(1, 2)$. This is possible because Nana always starts her move with an odd quantity of numbers, so she certainly has an available $0$. If Trixi uses $(1, 2)$, she will lose since the last non-zero cannot be converted to zero. She also cannot use two zeros, because then Nana is in the previous case and wins. So Trixi has to convert one of them with an additional $0$ to present Nana with two equal non-zeros. However, Nana can repeat her move until Trixi has not zeros left to do so. So Trixi will eventually be forced to use $(1, 2)$ and loses.\n\nIf there is just one non-zero left, Nana can play arbitrarily because this single non-zero will remain until the end of the game.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15922,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m, n$ be positive integers, and let $a_{ij}$ ($1 \\le i \\le m$, $1 \\le j \\le n$) be nonnegative real numbers such that for any $i, j$, the inequalities\n\n$$\na_{i,1} \\ge a_{i,2} \\ge \\dots \\ge a_{i,n}, \\quad a_{1,j} \\ge a_{2,j} \\ge \\dots \\ge a_{m,j}\n$$\nhold.\n\nFor $i = 1, 2, \\dots, m$ and $j = 1, 2, \\dots, n$, define\n\n$$\nX_{i,j} = a_{1,j} + \\dots + a_{i-1,j} + a_{i,j} + a_{i,j-1} + \\dots + a_{i,1},\n$$\n\n$$\nY_{i,j} = a_{m,j} + \\dots + a_{i+1,j} + a_{i,j} + a_{i,j+1} + \\dots + a_{i,n}.\n$$\n\nProve that\n\n$$\n\\prod_{i=1}^{m} \\prod_{j=1}^{n} X_{i,j} \\ge \\prod_{i=1}^{m} \\prod_{j=1}^{n} Y_{i,j}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The problem conditions imply that\n\n$$\nX_{i,j} \\ge (i + j - 1) \\cdot a_{i,j}, \\quad Y_{i,j} \\le (m + n - i - j + 1) \\cdot a_{i,j}.\n$$\n\nHence,\n\n$$\n\\begin{align*}\n\\prod_{i=1}^{m} \\prod_{j=1}^{n} X_{i,j} &\\ge \\prod_{i=1}^{m} \\prod_{j=1}^{n} (i + j - 1) \\cdot a_{i,j} \\\\\n&= \\prod_{i=1}^{m} \\prod_{j=1}^{n} (m + n - i - j + 1) \\cdot a_{i,j} \\\\\n&\\ge \\prod_{i=1}^{m} \\prod_{j=1}^{n} Y_{i,j},\n\\end{align*}\n$$\n\nwhere the equality on the second line is due to the one-to-one correspondence $(i, j) \\leftrightarrow (m + 1 - i, n + 1 - j)$, and $i + j - 1$ corresponds to $m + n - i - j + 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15923,
    "subject": "Mathematics (Olympiad)",
    "question": "Даден е правоаголник $ABCD$ и точка $S$ (точката $S$ не мора да лежи во рамнината на правоаголникот). Дали растојанијата од точката $S$ до темињата на правоаголникот може во некој редослед да бидат еднакви на $1$, $3$, $5$ и $7$?",
    "options": [],
    "answer": "See solution",
    "solution": "Нека за правоаголникот $ABCD$ точката $S$ е таква што растојанијата од $S$ до темињата на правоаголникот по некој редослед се еднакви на $1$, $3$, $5$, $7$. Ќе ја пресликаме точката $S$ во однос на централна симетрија со центар пресекот на дијагоналите на правоаголникот $E$, во точка $F$. Тогаш четириаголниците $AFCS$ и $BFDS$ имаат дијагонали кои се преполовуваат, па истите се паралелограми. Според равенството за паралелограм $2(a^2 + b^2) = d_1^2 + d_2^2$ (каде $a$ и $b$ се страни на паралелограмот, а $d_1$ и $d_2$ негови дијагонали), имаме:\n\n$$\n2(\\overline{SA}^2 + \\overline{SC}^2) = \\overline{SF}^2 + \\overline{AC}^2 \\text{ и } 2(\\overline{SB}^2 + \\overline{SD}^2) = \\overline{SF}^2 + \\overline{BD}^2.\n$$\n\nИмајќи предвид дека $\\overline{AC} = \\overline{BD}$, како дијагонали на правоаголникот $ABCD$, од горните равенства добиваме:\n\n$$\n\\overline{SA}^2 + \\overline{SC}^2 = \\overline{SB}^2 + \\overline{SD}^2.\n$$\n\nНо, за било кој распоред на $1$, $3$, $5$, $7$ на местата на $\\overline{SA}$, $\\overline{SB}$, $\\overline{SC}$, $\\overline{SD}$, последното равенство не е точно. Навистина:\n\n$$\n1^2 + 7^2 \\neq 3^2 + 5^2, \\quad 1^2 + 5^2 \\neq 3^2 + 7^2, \\quad 1^2 + 3^2 \\neq 5^2 + 7^2.\n$$\n\nЗначи, растојанијата $\\overline{SA}$, $\\overline{SB}$, $\\overline{SC}$, $\\overline{SD}$ во ниту еден редослед не можат да бидат $1$, $3$, $5$, $7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15924,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $\\triangle ABC$ 為不等邊三角形，其外心為 $O$，垂心為 $H$。有另一個三角形 $AYZ$ 與 $ABC$ 共用頂點 $A$，且三角形 $AYZ$ 的外心為 $H$，垂心為 $O$。證明：若 $Z$ 在 $BC$ 上，則 $A$、$H$、$O$、$Y$ 共圓。",
    "options": [],
    "answer": "See solution",
    "solution": "我們先證明一個重要的引理。\n\n**Lemma.** 若 $X$ 是 $BC$ 與 $YZ$ 的交點，則 $X$ 落在 $OH$ 的中垂線上。\n\n_Proof of the lemma._ 令 $H'$ 為 $H$ 關於 $BC$ 的鏡射點，$O'$ 為 $O$ 關於 $YZ$ 的鏡射點。熟知有 $AO = H'O$ 且 $AH = O'H$。因此 $\\angle HH'O = \\angle OAH = \\angle HO'H$，所以 $HH'OO'$ 共圓。故 $X$ 是圓 $HH'OO'$ 的圓心，得證。 $\\Box$\n\n_Another proof of the lemma._ 取點 $P$ 使得 $AHPO$ 為平行四邊形。熟知有 $BC$ 為 $OP$ 的中垂線，而 $YZ$ 為 $HP$ 的中垂線。於是 $X$ 為三角形 $OHP$ 的外心，故 $X$ 位於 $OH$ 的中垂線上。 $\\Box$\n\n回到原題。由於 $ABC$ 為不等邊三角形，故 $YZ$ 必與 $BC$ 相異。於是有 $Z = BC \\cap YZ$，由 Lemma 知其落在 $OH$ 的中垂線上。一樣令 $O'$ 為 $O$ 關於 $YZ$ 的鏡射點，有 $ZO' = ZO = ZH = O'H$，知 $\\triangle ZO'H$ 為正三角形。所以 $\\angle AOH = 180^\\circ - \\angle O'OH = 180^\\circ - \\frac{1}{2}\\angle O'ZH = 150^\\circ$。亦有\n\n$$\n\\angle AYH = 90^\\circ - \\angle AZY = 90^\\circ - \\angle HZO' = 30^\\circ,\n$$\n\n其中的第一個等號來自垂心 $H$ 與外心 $O$ 的性質。故得 $AHOY$ 共圓，得證。 $\\Box$\n\n另證。同上面的證明，我們只需證出 $ZO = ZH$ 即可。此處給出此式的另一個證明。\n\n由於 $AO \\perp YZ$ 及 $AH \\perp BC$，有\n\n$$\n\\angle AOZ = \\angle ZYA = 90^\\circ - \\angle YAO = 90^\\circ - \\angle HAZ = \\angle AZC.\n$$\n\n所以直線 $BC$ 與三角形 $AOZ$ 的外接圓切於點 $Z$。考慮 $AOZ$ 的外心 $H''$。有 $H''Z \\perp BC$ 及 $H''Z = HZ = AH$，因為 $H''$ 是 $H$ 關於 $AZ$ 的鏡射點。由於 $O$ 到 $BC$ 的距離為 $AH/2$，得 $H''O = ZO$。再配合 $H''$ 為 $AOZ$ 的外心性質，得 $ZO = H''O = ZH'' = ZH$，得證。 $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15925,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $AB < AC$, inscribed in a circle $c$ with center $O$. Let $G$ be its barycenter, and let $D$, $E$, $F$ be the feet of the altitudes from $A$, $B$, and $C$, respectively. If the rays $AG$ and $GD$ intersect $c$ at $M$ and $N$ respectively, prove that the points $F$, $E$, $M$, $N$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K$ be the midpoint of $BC$ and $P$ the second intersection of $GD$ with $c$. Then $c$ and the Euler circle are homothetic with center $G$ and ratio $-2$, so\n\n$$\n\\frac{GP}{GD} = \\frac{GA}{GK} = 2 \\Rightarrow GP = 2GD.\n$$\n\nFrom the power of a point theorem, we have\n\n$$\nGM \\cdot GA = GN \\cdot GP \\Rightarrow GM \\cdot (2GK) = GN \\cdot (2GD) \\Rightarrow GM \\cdot GK = GN \\cdot GD,\n$$\nso the quadrilateral $DKMN$ is inscribed in a circle, call it $c_1$.\n\nMoreover, $F$, $D$, $K$, $E$ belong to a circle (the Euler circle), call it $c_2$.\n\nTherefore, the lines $FE$, $DK$, $MN$ are concurrent at the radical center, call it $T$, of the circles $c$, $c_1$, $c_2$. To this end, $TF \\cdot TE = TD \\cdot TK = TN \\cdot TM$, so the points $F$, $E$, $M$, $N$ are concyclic.\n\n![](images/Greece_olympiad_2018_p28_data_9c0952769c.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15926,
    "subject": "Mathematics (Olympiad)",
    "question": "Jacob and Laban take turns playing a game. Each starts with a list of square numbers $1, 4, 9, \\dots, 2021^2$, and there is a whiteboard in front of them with the number $0$ on it. Jacob chooses a number $x^2$ from his list, removes it, and replaces the number $W$ on the whiteboard with $W + x^2$. Laban then does the same with a number from his list, and they repeat back and forth until both have no more numbers in their list. Every time the number on the whiteboard is divisible by $4$ after a player's turn, Jacob gets a sheep. Jacob wants as many sheep as possible by the end of the game, while Laban wants Jacob to have as few as possible. What is the greatest number $K$ such that Jacob can guarantee to get at least $K$ sheep by the end of the game, no matter how Laban plays?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $n^2 \\equiv 0 \\pmod{4}$ if and only if $n$ is even, and $n^2 \\equiv 1 \\pmod{4}$ if and only if $n$ is odd, we can simplify notation by replacing the even squares by $0$ and the odd squares by $1$ in each list. Thus, Jacob and Laban each have a pool of $1010$ zeros and $1011$ ones to choose from at the start.\n\nNo matter the order, after every fourth one, Jacob earns another sheep. Since there are $2022$ ones in total, and $2022 = 4 \\times 505 + 2$, Jacob is guaranteed at least $505$ sheep. However, since Jacob plays first, he can start with a zero, securing one extra sheep for a total of $506$ sheep.\n\nIf Laban plays optimally, he can ensure Jacob gets only these $506$ sheep. Laban must prevent zeros from being played immediately after every fourth one, which would give Jacob more sheep. After every $4k$-th one ($1 \\leq k \\leq 505$), another one should follow. Laban can achieve this by writing a one after Jacob's initial zero, then copying each of Jacob's moves. This ensures Jacob must play every $4k$-th one, and Laban follows with another one, achieving his goal.\n\nLaban will not run out of zeros or ones before the $2020$-th one is used. After the first two moves (a zero by Jacob and a one by Laban), Laban has more zeros and one fewer one than Jacob, but since there are two spare ones after $2020$ ones, this is not a problem. Any zeros or ones played after the $2021$-st one do not contribute to a sum divisible by four, so no further sheep for Jacob.\n\nIf Jacob starts with a one, Laban simply copies each move except the second one played by Jacob (the third move of the game). After this, Laban plays a zero (if available), then continues copying Jacob. In this way, no zero is played directly after the $4k$-th one, so Jacob only gets the $505$ sheep from every fourth one. If Laban has no zeros left after the third move, Jacob also has none, and both continue with ones, so Jacob cannot earn more than $505$ sheep.\n\nTherefore, Jacob can guarantee (by starting with a zero) to collect $506$ sheep, but no more.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15927,
    "subject": "Mathematics (Olympiad)",
    "question": "On the side $AB$ of the rectangle $ABCD$ are taken the points $S$ and $T$ so that $AS = ST = TB$. Denote $M$, $N$, and $P$ as the orthogonal projections of $A$, $S$, and $T$ onto the straight lines $DS$, $DT$, and $DB$, respectively. Prove that the points $M$, $N$, and $P$ are collinear if and only if $15AD^2 = 2AB^2$.\n\n![](images/RMC2013_final_p19_data_6ffbfd36bd.png)\n\n![](images/RMC2013_final_p19_data_9de6893860.png)\n\n![](images/RMC2013_final_p19_data_a9c7ad0c02.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M'$, $N'$, $P'$ be the orthogonal projections onto $AB$ of $M$, $N$, and $P$, respectively. Then $M$, $N$, $P$ are collinear if and only if the intersection $N_1$ of $MP$ and $NN'$ coincides with $N$. Since $MM' > PP'$ and $N'$ is between $M'$ and $P'$, from similar triangles (see Fig. 2) we get\n\n$$\n\\frac{P'N'}{P'M'} = \\frac{N_1N' - PP'}{MM' - PP'}\n$$\n\nso the condition of collinearity is\n\n$$\nNN' \\cdot M'P' = MM' \\cdot N'P' + PP' \\cdot M'N'.\n$$\n\nTo compute $MM'$, $NN'$, $PP'$, notice that if the legs of a right triangle $OUV$ are $OU = u$, $OV = v$, and we take $X$ on $OU$ so that $OX = x < u$ and $XX' \\perp UV$, $X'X'' \\perp OU$ (see Fig. 3), then $\\triangle OO'U \\sim \\triangle XX'U$, so\n\n$$\n\\frac{X'X''}{O'O''} = \\frac{UX}{UO}, \\quad \\frac{UX''}{UO''} = \\frac{UX}{UO},\n$$\n\nand, since $O'O'' = \\frac{OO' \\cdot UO'}{OU} = \\frac{1}{OU} \\frac{OU \\cdot OV}{UV} \\frac{OU^2}{UV} = \\frac{u^2v}{u^2 + v^2}$, $UO'' = \\frac{O'U^2}{OU} = \\frac{u^3}{u^2 + v^2}$,\n\n$$\nX'X'' = \\frac{u-x}{u}O'O'' = \\frac{uv(u-x)}{u^2+v^2}, \\quad UX'' = \\frac{u-x}{u}UO'' = \\frac{u^2(u-x)}{u^2+v^2}.\n$$\n\nDenote $AB = 3a$ and $AD = b$. In triangles $ASD$, $ATD$, and $ABD$, the above relations yield:\n\n- $SM' = \\frac{a^3}{a^2 + b^2}$\n- $TN' = \\frac{4a^3}{4a^2 + b^2}$\n- $BP' = \\frac{9a^3}{9a^2 + b^2}$\n- $MM' = \\frac{a^2b}{a^2 + b^2}$\n- $NN' = \\frac{2a^2b}{4a^2 + b^2}$\n- $PP' = \\frac{3a^2b}{9a^2 + b^2}$\n\nThe collinearity condition becomes\n\n$$\n\\frac{2a^2b}{4a^2 + b^2} \\left( 2a - \\frac{9a^3}{9a^2 + b^2} + \\frac{a^3}{a^2 + b^2} \\right) = \\frac{a^2b}{a^2 + b^2} \\left( a - \\frac{9a^3}{9a^2 + b^2} + \\frac{4a^3}{4a^2 + b^2} \\right) + \\frac{3a^2b}{9a^2 + b^2} \\left( a - \\frac{4a^3}{4a^2 + b^2} + \\frac{a^3}{a^2 + b^2} \\right)\n$$\n\nwhich, after computation, reduces to $5b^2 = 6a^2$, that is $15AD^2 = 2AB^2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15928,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute triangle $ABC$, the foot of the altitude from $A$ is called $D$. Let $D_1$ and $D_2$ be the reflections of $D$ in $AB$ and $AC$, respectively. The intersection of $BC$ and the line through $D_1$ parallel to $AB$ is called $E_1$. The intersection of $BC$ and the line through $D_2$ parallel to $AC$ is called $E_2$. Prove that $D_1$, $D_2$, $E_1$, and $E_2$ lie on a circle whose centre lies on the circumcircle of $\\triangle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K$ be the midpoint of $DD_1$, and let $L$ be the midpoint of $DD_2$. Then $K$ lies on $AB$ and $L$ lies on $AC$. Because $\\angle AKD = 90^\\circ = \\angle ALD$, the quadrilateral $AKDL$ is cyclic. Hence, $\\angle DLK = \\angle DAK = \\angle DAB = 90^\\circ - \\angle ABC$. Moreover, $KL$ is a midsegment in triangle $DD_1D_2$, hence $\\angle DLK = \\angle DD_2D_1$. We conclude that $\\angle DD_2D_1 = 90^\\circ - \\angle ABC$.\n\nBecause $AC \\perp DD_2$ and $D_2E_2 \\parallel AC$, we have $\\angle DD_2E_2 = 90^\\circ$. Hence, $\\angle D_1D_2E_2 = \\angle D_1D_2D + \\angle DD_2E_2 = 90^\\circ - \\angle ABC + 90^\\circ = 180^\\circ - \\angle ABC$. On the other hand, as $D_1E_1 \\parallel AB$, we have $\\angle D_1E_1E_2 = \\angle ABC$, hence we get $\\angle D_1D_2E_2 = 180^\\circ - \\angle D_1E_1E_2$. We conclude that $D_1E_1E_2D_2$ is a cyclic quadrilateral.\n\nLet $M$ be the point such that $AM$ is a diameter of the circumcircle of $\\triangle ABC$. Thales' theorem yields $\\angle ACM = 90^\\circ$. Hence, $CM \\perp AC$, which yields $CM \\perp D_2E_2$ and $CM \\parallel DD_2$. Moreover, $L$ is the midpoint of $DD_2$ and $LC \\parallel D_2E_2$, hence $LC$ is a midsegment in triangle $DD_2E_2$. This means that $C$ is the midpoint of $DE_2$. Because $CM \\parallel DD_2$, we get that $CM$ is also a midsegment, hence $CM$ intersects $D_2E_2$ in the middle. As $CM \\perp D_2E_2$, the line $CM$ is the perpendicular bisector of $D_2E_2$. Analogously, we get that $BM$ is the perpendicular bisector of $D_1E_1$. Hence, $M$ is the intersection point of the perpendicular bisectors of two of the chords of the circle through $D_1$, $D_2$, $E_1$, and $E_2$. Hence, $M$ is the centre of this circle. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15929,
    "subject": "Mathematics (Olympiad)",
    "question": "Sixty points, of which thirty are coloured red, twenty are coloured blue, and ten are coloured green, are marked on a circle. These points divide the circle into sixty arcs. Each of these arcs is assigned a number according to the colours of its endpoints: an arc between a red and a green point is assigned a number $1$, an arc between a red and a blue point is assigned a number $2$, and an arc between a blue and a green point is assigned a number $3$. The arcs between two points of the same colour are assigned a number $0$. What is the greatest possible sum of all the numbers assigned to the arcs?",
    "options": [],
    "answer": "See solution",
    "solution": "Let the score of a red point be $0$, the score of a green point be $1$, and the score of a blue point be $2$. Note that the number assigned to an arc is at most the sum of the scores of the endpoints. This means that the sum of all the numbers assigned to the arcs is at most twice the sum of all the sixty scores, which is\n\n$$\n2(30 \\cdot 0 + 20 \\cdot 2 + 10 \\cdot 1) = 100.\n$$\n\nEquality holds if there are no arcs with two green or two blue endpoints. This can be achieved, for instance, by letting red and non-red points alternate. Hence the greatest possible sum is $100$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15930,
    "subject": "Mathematics (Olympiad)",
    "question": "Is there a two-digit number $n$ that does not end with zero such that\n\na) all numbers that can be formed by adding one or more zeros between the two-digit number's digits are its multiples?\n\nb) none of the numbers that can be formed by adding one or more zeros between the two-digit number's digits are its multiples?\n\nc) some numbers that can be formed by adding one or more zeros between the two-digit number's digits are its multiples and some are not?",
    "options": [],
    "answer": "See solution",
    "solution": "a) One such number is $n = 15$. All numbers formed by adding zeros between its digits are divisible by 3 and 5.\n\nb) One such number is $n = 12$. Adding zeros between the digits, the last two digits will always be 02. Hence no such number will be divisible by 4.\n\nc) One such number is $n = 11$, because $101$ is not divisible by $11$ but $1001$ is.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15931,
    "subject": "Mathematics (Olympiad)",
    "question": "In a circle with diameter $d$, the chords $AB$ and $CD$ are perpendicular and intersect at a point $E$ distinct from the centre of the circle. Prove that\n$$\n\\overline{AE}^2 + \\overline{BE}^2 + \\overline{CE}^2 + \\overline{DE}^2 = d^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "From the condition $AB \\perp CD$ and $E = AB \\cap CD$, with $E \\neq O$ where $O$ is the centre of the circle:\n\nThe line passing through $D$ and parallel to $AB$ intersects the circumference at $M$. The line passing through $M$ and parallel to $CD$ intersects the circumference at $N$. Because $\\angle MDC = 90^\\circ$, the segment $MC$ passes through the centre, so $\\angle MNC = 90^\\circ$, and the quadrilateral $MDCN$ is a rectangle.\n\nThe points $P$ and $E$, $A$ and $B$ are symmetric with respect to the axis of symmetry of rectangle $MDCN$, which passes through $O$ and is parallel to $DC$ and $MN$. It follows that $\\overline{AP} = \\overline{EB}$. From $\\overline{CD} = \\overline{CE} + \\overline{ED}$, $\\overline{MD} = \\overline{PE} = \\overline{AE} - \\overline{AP} = \\overline{AE} - \\overline{EB}$, and because $\\triangle MDC$ is right-angled, we obtain:\n\n![](images/Makedonija_2008_p20_data_c7438a73b9.png)\n\n$$\nd^2 = \\overline{MD}^2 + \\overline{CD}^2 = (\\overline{AE} - \\overline{EB})^2 + (\\overline{CE} + \\overline{ED})^2 \\\\\n= \\overline{AE}^2 + \\overline{EB}^2 - 2 \\overline{AE} \\cdot \\overline{EB} + \\overline{CE}^2 + \\overline{ED}^2 + 2 \\overline{CE} \\cdot \\overline{ED} \\quad \\dots\\dots(1)\n$$\n\n$\\angle BAC = \\angle BDC$ because they subtend the same arc $BC$. From the similarity of triangles $\\triangle AEC \\sim \\triangle BED$, we obtain $\\overline{AE} : \\overline{CE} = \\overline{ED} : \\overline{BE}$, or $\\overline{AE} \\cdot \\overline{EB} = \\overline{CE} \\cdot \\overline{ED}$ $\\dots$(2).\n\nSubstituting (2) into (1), we get:\n\n$$\n\\overline{AE}^2 + \\overline{BE}^2 + \\overline{CE}^2 + \\overline{DE}^2 = d^2.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15932,
    "subject": "Mathematics (Olympiad)",
    "question": "平面上設點 $O$ 為圓 $\\Gamma$ 的圓心，另在 $\\Gamma$ 上有兩點 $A, B$ 滿足 $O, A, B$ 不共線。設點 $M$ 為線段 $AB$ 的中點，並分別在直線 $OA, OB$ 上取點 $P, Q$ 使得 $P \\neq A$ 且 $P, M, Q$ 三點共線。令過 $P$ 與 $AB$ 平行的直線，和過 $Q$ 與 $OM$ 平行的直線交於點 $X$；又令過 $X$ 與 $OA$ 平行的直線，和過 $B$ 與 $OX$ 垂直的直線交於點 $Y$。證明：若點 $X$ 在圓 $\\Gamma$ 上，則點 $Y$ 也在圓 $\\Gamma$ 上。",
    "options": [],
    "answer": "See solution",
    "solution": "不失一般性，可設 $OA = OB = OX = 1$。設有向角 $\\angle AOB = \\theta$、$\\angle BOX = \\varphi$。又設 $A = \\varphi + \\theta/2$，$B = \\theta/2$。在 $\\triangle OPX$ 上使用正弦定律，可知有向長度 $OP$ 等於 $\\cos A / \\cos B$。又在 $\\triangle OQX$ 上使用正弦定律，可知有向長度 $OQ$ 等於 $\\sin A / \\sin B$。於是由于孟氏定理，\n\n$$\n\\frac{\\frac{\\cos A}{\\cos B}}{1 - \\frac{\\cos A}{\\cos B}} = - \\frac{\\frac{\\sin A}{\\sin B}}{1 - \\frac{\\sin A}{\\sin B}}\n$$\n\n上式展開得\n\n$$\n\\cos A \\sin B + \\sin A \\cos B - 2 \\cos A \\sin A = 0,\n$$\n\n即 $\\sin(A+B) = \\sin 2A$。由此推得 $A+B \\equiv 2A \\pmod{2\\pi}$ 或 $3A+B \\equiv \\pi \\pmod{2\\pi}$。\n\n在第一種情況下，可得 $A=B$，於是 $\\varphi=0$。這表示 $X=B$ 及 $Y=B$ 落在 $\\Gamma$ 上。\n\n在第二種情況下，有 $3\\varphi \\equiv \\pi - 2\\theta \\pmod{2\\pi}$。注意如果通過 $B$ 且與 $OX$ 垂直的直線和 $\\Gamma$ 交於點 $Y'$ 的話，則 $\\angle XOY' = \\varphi$。於是 $\\angle AOX + \\angle AOY' = 3\\varphi + 2\\theta = \\pi$，知 $AO$ 平行於 $XY'$，推得 $Y=Y'$ 落在 $\\Gamma$ 上。證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15933,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x^2 - y^2) = (x - y)[f(x) + f(y)]\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $x = y$, then\n$$\nf(0) = 0 \\cdot [2f(x)] = 0,\n$$\nso $f(0) = 0$.\n\nIf $y = -x$, then\n$$\nf(x^2 - (-x)^2) = [x - (-x)][f(x) + f(-x)] \\\\\nf(0) = 2x[f(x) + f(-x)] \\\\\n0 = 2x[f(x) + f(-x)]\n$$\nfor all $x$, so $f(x) + f(-x) = 0$, i.e., $f$ is odd.\n\nNow, consider $f(x^2 - (-y)^2) = [x - (-y)][f(x) + f(-y)]$:\n$$\nf(x^2 - y^2) = (x + y)[f(x) - f(y)]\n$$\nComparing with the original equation:\n$$\n(x - y)[f(x) + f(y)] = (x + y)[f(x) - f(y)]\n$$\nExpanding both sides:\n$$\n(x-y)[f(x)+f(y)] = (x+y)[f(x)-f(y)] \\\\\nxf(x) - yf(x) + xf(y) - yf(y) = xf(x) + yf(x) - xf(y) - yf(y)\n$$\nSimplifying:\n$$\n- yf(x) + xf(y) = yf(x) - xf(y) \\\\\n2yf(x) = 2xf(y)\n$$\nSo $yf(x) = xf(y)$ for all $x, y$. Setting $y = 1$ and $f(1) = k$, we get $f(x) = kx$ for some $k \\in \\mathbb{R}$.\n\nThus, all functions of the form $f(x) = kx$ are solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15934,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute scalene triangle $ABC$, points $D$, $E$, $F$ lie on sides $BC$, $CA$, $AB$, respectively, such that $AD \\perp BC$, $BE \\perp CA$, $CF \\perp AB$. The altitudes $AD$, $BE$, $CF$ meet at the orthocenter $H$. Points $P$ and $Q$ lie on segment $EF$ such that $AP \\perp EF$ and $HQ \\perp EF$. Lines $DP$ and $QH$ intersect at point $R$. Compute $HQ/HR$.\n\n![](images/pamphlet1112_main_p36_data_ad256f746a.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Comment.** The answer is $HQ/HR = 1$, or $HQ = HR$. There are many approaches to this problem. We present three typical ones.\n\n**Solution 1.** By the given, we have $AP \\parallel QR$. Hence, triangles $APS$ and $HQS$, and $DHR$ and $DAP$ are similar. It follows that\n\n$$\n\\frac{HQ}{PA} = \\frac{HS}{AS} \\quad \\text{and} \\quad \\frac{RH}{PA} = \\frac{HD}{AD}.\n$$\n\nIt suffices to show that\n\n$$\n\\frac{HS}{AS} = \\frac{HD}{AD},\n$$\n\nwhich is equivalent to showing that $(A, H)$ and $(S, D)$ are harmonic conjugates. Applying Ceva's theorem to triangle $AHC$ and cevians $AD$, $BE$, and $CF$ gives\n\n$$\n\\frac{AE}{EC} \\cdot \\frac{CF}{FH} \\cdot \\frac{HD}{DA} = 1.\n$$\n\nApplying Menelaus's theorem to triangle $ACH$ and line $ESF$ yields\n\n$$\n\\frac{AE}{EC} \\cdot \\frac{CF}{FH} \\cdot \\frac{HS}{SA} = 1.\n$$\n\nEquating the left-hand sides of the last two equations gives the desired result, completing the proof.\n\n**Solution 2.** Let $\\angle ABC = B$ and $\\angle ACB = C$. In right triangles $BHD$ and $ABD$, we have $HD/BD = \\tan \\angle HBD = \\cot C$ and $AD/BD = \\tan B$. Dividing the last two equations gives\n\n$$\n\\frac{HD}{AD} = \\frac{\\cot C}{\\tan B} = \\cot B \\cot C.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15935,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Let $a$ be a positive integer. Prove that none of the numbers $a^2 + 1, a^2 + 2, \\dots, a^2 + 2a$ is a square.\n\nb) Are there positive integers $m, n, p$ such that\n$$\nm^2 + n + p, \\quad n^2 + p + m, \\quad p^2 + m + n\n$$\nare all squares?",
    "options": [],
    "answer": "See solution",
    "solution": "a) The next square after $a^2$ is $(a+1)^2 = a^2 + 2a + 1$, so $a^2 + 1, a^2 + 2, \\dots, a^2 + 2a$ all lie strictly between $a^2$ and $(a+1)^2$, and thus cannot be perfect squares.\n\nb) Suppose, for contradiction, that such numbers exist. Then $m^2 + n + p > m^2$, so $m^2 + n + p \\geq m^2 + 2m + 1$. Similarly for the other variables. Adding the three inequalities gives:\n\n$$\nm^2 + n^2 + p^2 + 2m + 2n + 2p \\geq m^2 + n^2 + p^2 + 2m + 2n + 2p + 3,\n$$\n\nwhich is impossible. Therefore, no such positive integers exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15936,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an odd number $n > 1$, let $S = \\{k : 1 \\leq k < n,\\ (k, n) = 1\\}$ and let $T = \\{k \\in S : (k+1, n) = 1\\}$. For each $k \\in S$, let $r_k$ be the remainder left by $\\dfrac{k^{|S|} - 1}{n}$ upon division by $n$. Show that\n\n$$\n\\prod_{k \\in T} (r_k - r_{n-k}) \\equiv |S|^{|T|} \\pmod{n}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $n$ is odd, $|S| = \\varphi(n) = n \\prod_{p \\mid n} \\left(1 - \\frac{1}{p}\\right)$ is even. Given an element $k$ of $S$, write\n\n$$\nk^{|S|} \\equiv 1 + n r_k \\pmod{n^2} \\quad \\text{and} \\quad (n-k)^{|S|} \\equiv 1 + n r_{n-k} \\pmod{n^2},\n$$\n\nand notice that\n\n$$\n(n-k)^{|S|} \\equiv k^{|S|} - |S| \\cdot n \\cdot k^{|S|-1} \\pmod{n^2},\n$$\n\nto get\n\n$$\nr_k - r_{n-k} \\equiv |S| \\cdot k^{|S|-1} \\pmod{n}.\n$$\n\nHence\n\n$$\n\\prod_{k \\in T} (r_k - r_{n-k}) \\equiv |S|^{|T|} \\left( \\prod_{k \\in T} k \\right)^{|S|-1} \\pmod{n}.\n$$\n\nFinally, notice that the product in the right-hand member is congruent to $1$ modulo $n$. To see this, let $k'$ denote the modulo $n$ multiplicative inverse of an element $k$ of $S$, and notice that\n\n$$\nk' + 1 \\equiv k'(k+1) \\pmod{n} \\quad \\text{and} \\quad (k-1)(k'+1) \\equiv k-k' \\pmod{n}.\n$$\n\nThe first congruence shows that if $k$ belongs to $T$, then so does $k'$, and the second shows that $k = k'$ if and only if $k = 1$. Consequently, if $k \\ne 1$, the factors $k$ and $k'$ in the product can be paired off and the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15937,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation system\n\n$$\n\\begin{cases}\n1 - \\frac{12}{y+3x} = \\frac{2}{\\sqrt{x}} \\\\\n1 + \\frac{12}{y+3x} = \\frac{6}{\\sqrt{x}}\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The necessary conditions for the given equation system are $x, y > 0$; $y+3x \\neq 0$.\n\nThe given equation system is equivalent to:\n\n$$\n\\begin{cases}\n1 - \\frac{12}{y+3x} = \\frac{2}{\\sqrt{x}} \\\\\n1 + \\frac{12}{y+3x} = \\frac{6}{\\sqrt{x}}\n\\end{cases}\n$$\n\nMultiply the first equation by the second, we get:\n\n$$\n\\frac{9}{y} - \\frac{1}{x} = \\frac{12}{y+3x} \\Leftrightarrow y^2 + 6xy - 27x^2 = 0 \\Leftrightarrow \\begin{cases} y = 3x \\\\ y = -9x \\end{cases}\n$$\n\nBy the condition $x, y > 0$, we have $y = 3x$. Setting $y = 3x$ in the first equation, we obtain:\n\n$$\n\\frac{1}{\\sqrt{x}} + \\frac{3}{\\sqrt{3x}} = 1 \\Rightarrow x = 4 + 2\\sqrt{3}, \\quad y = 12 + 6\\sqrt{3}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15938,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Let $a$, $b$, and $c$ be real numbers that satisfy $a^2 + b^2 + c^2 = 1$. Prove that\n$$\n|a - b| + |b - c| + |c - a| \\le 2\\sqrt{2}.\n$$\n\nb) Given 2019 real numbers $a_1, a_2, \\dots, a_{2019}$ such that $a_1^2 + a_2^2 + \\dots + a_{2019}^2 = 1$, find the maximum value of\n$$\nS = |a_1 - a_2| + |a_2 - a_3| + \\dots + |a_{2019} - a_1|.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Using the Cauchy-Schwarz inequality, we have\n$$\n|x_1| + |x_2| + \\dots + |x_n| \\le \\sqrt{n(x_1^2 + x_2^2 + \\dots + x_n^2)}.\n$$\n\na) Without loss of generality, assume $a \\le b \\le c$. Then,\n$$\n|a - b| + |b - c| + |c - a| = (b - a) + (c - b) + (c - a) = 2(c - a).\n$$\nBy Cauchy-Schwarz,\n$$\n2|c - a| \\le 2\\sqrt{2(a^2 + c^2)} \\le 2\\sqrt{2}.\n$$\nEquality holds when $(a, b, c) = \\left(-\\frac{\\sqrt{2}}{2}, 0, \\frac{\\sqrt{2}}{2}\\right)$.\n\nb) Suppose $a_1$ is the smallest number. If the sequence $a_1, a_2, \\dots, a_{2019}$ is not decreasing, then\n$$\nS = 2|a_1 - a_{2019}| \\le 2\\sqrt{2(a_1^2 + a_{2019}^2)} \\le 2\\sqrt{2}.\n$$\nOtherwise, there exists $k$ with $1 < k < 2019$ such that $a_k \\ge a_{k-1}$ and $a_k \\le a_{k+1}$. Then $|a_k - a_{k-1}| + |a_{k+1} - a_k| = |a_{k+1} - a_{k-1}|$, so we can reduce the problem to 2018 numbers $b_1, \\dots, b_{2018}$ with $b_1^2 + \\dots + b_{2018}^2 \\le 1$ and\n$$\nS = |b_1 - b_2| + |b_2 - b_3| + \\dots + |b_{2018} - b_1|.\n$$\nWe have\n$$\nS \\le (|b_1| + |b_2|) + (|b_2| + |b_3|) + \\dots + (|b_{2018}| + |b_1|) = 2 \\sum_{i=1}^{2018} |b_i| \\le 2\\sqrt{2018}.\n$$\nEquality holds when all $|b_i|$ are equal and adjacent numbers have opposite signs, e.g., $b_1 = -b_2 = b_3 = \\dots = -b_{2018} = \\frac{\\sqrt{2018}}{2018}$.\n\nReturning to the original problem, $\\max S = 2\\sqrt{2018}$, with equality for, e.g.,\n$$\na_1 = -a_2 = a_3 = -a_4 = \\dots = -a_{2018} = \\frac{\\sqrt{2018}}{2018}, \\quad a_{2019} = 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15939,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ and $n$ be positive integers. Prove that if $x_j$ are real numbers for $1 \\leq j \\leq n$, such that\n\n$$\n\\sum_{j=1}^{n} \\frac{1}{x_j^{2k} + k} = \\frac{1}{k}\n$$\n\nthen\n\n$$\n\\sum_{j=1}^{n} \\frac{1}{x_j^{2k+1} + k + 2} \\leq \\frac{1}{k+1}\n$$\nmust hold.",
    "options": [],
    "answer": "See solution",
    "solution": "We can show that each term in the second sum is not greater than the corresponding term in the first sum, multiplied by the factor $\\frac{k}{k+1}$.\n\nSubstitute $y := x_j^{2k}$. We wish to show\n\n$$\n\\frac{1}{y^2 + k + 2} \\leq \\frac{k}{k+1} \\cdot \\frac{1}{y + k}\n$$\n\nSince $y$ is positive for $k > 0$, this is equivalent to $(k+1)(y + k) \\leq k(y^2 + k + 2)$, or $P(y) = k y^2 - (k+1) y + k \\geq 0$.\n\nThis polynomial is quadratic in $y$, and $P(0) = k > 0$. The discriminant is\n\n$$\n(k+1)^2 - 4k^2 = -3k^2 + 2k + 1 \\leq -3k^2 + 3k = -3k(k-1) \\leq 0,\n$$\n\nso the polynomial is always positive, which completes the proof. QED.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15940,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b$ be positive real numbers with $b - a > 2$. Prove that for any two distinct integers $m, n$ in the interval $[a, b)$, there is a nonempty set $S$ consisting of some integers in the interval $[ab, (a+1)(b+1))$, such that $\\frac{\\prod x}{mn}$ is a square of a rational number.",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove the following lemma:\n\n**Lemma**: Let $u$ be an integer with $a \\leq u < u+1 < b$. Then there are two distinct integers $x, y$ in the interval $[ab, (a+1)(b+1))$, such that $\\frac{xy}{u(u+1)}$ is a square of an integer.\n\n**Proof of lemma.** Let $v$ be the smallest integer not less than $\\frac{ab}{u}$, i.e., $v$ satisfies\n\n$$\n\\frac{ab}{u} \\leq v < \\frac{ab}{u} + 1\n$$\n\nhence\n\n$$\nab \\leq uv < ab + u(ab + a + b + 1), \\quad \\textcircled{1}\n$$\n\nand thus\n\n$$\nab < (u+1)v = uv + v < ab + u + \\frac{ab}{u} + 1 < ab + a + b + 1 \\text{ (since } a \\leq u < b\\text{).} \\quad \\textcircled{2}\n$$\n\n(Here we have used a well-known result: the function $f(t) = t + \\frac{ab}{t}$ ($a \\leq t \\leq b$) attains its maximum at $t = a$ or $b$.)\n\nBy ① and ②, we see that $uv$ and $(u+1)v$ are two distinct integers in the interval $I = [ab, (a+1)(b+1))$. Let $x = uv$ and $y = (u+1)v$. Then $\\frac{xy}{u(u+1)} = v^2$ is a square of an integer. We have verified the lemma.\n\nReturning to the original problem, suppose that $m < n$. Then $a \\leq m \\leq n-1 < b$. It follows from the lemma that for every $k = m, m+1, \\dots, n-1$, there exist $x_k, y_k$, two distinct integers in the interval $[ab, (a+1)(b+1))$, and integer $A_k$, such that\n\n$$\n\\frac{x_k y_k}{k(k+1)} = A_k^2.\n$$\n\nMultiplying all together, we find that\n\n$$\n\\frac{\\prod_{k=m}^{n-1} x_k y_k}{mn(m+1)^2 \\cdots (n-1)^2} = \\prod_{k=m}^{n-1} A_k^2\n$$\n\nis a square of an integer.\n\nLet $S$ be the set of numbers that appear in $x_i, y_i$ ($m \\leq i \\leq n-1$) an odd number of times. If $S$ is nonempty, then it follows from the above equality that $\\frac{\\prod x_i}{mn}$ is a square of a rational number.\n\nIf $S$ is empty, then $mn$ is a square of an integer. Since $a + b > 2\\sqrt{ab}$, we have $ab + a + b + 1 > ab + 2\\sqrt{ab} + 1$, i.e., $\\sqrt{(a+1)(b+1)} > \\sqrt{ab} + 1$, which means that there is at least one integer in the interval $[\\sqrt{ab}, \\sqrt{(a+1)(b+1)})$. As a result, there is a perfect square in the interval $[ab, (a+1)(b+1))$. Suppose that $r^2 \\in [ab, (a+1)(b+1))$ ($r \\in \\mathbb{Z}$), and let $S' = \\{r^2\\}$. Then $\\frac{\\prod_{x \\in S'} x}{mn}$ is a square of a rational number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15941,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Which rod cannot be used as the side of a triangle, given rods of integer lengths from 1 cm to 20 cm?\n\nb) Rachael must use the six rods of lengths 2 cm to 7 cm. In how many ways can these rods be divided into two sets of three to form two triangles?\n\nc) Find sets of three rods (with integer lengths) that can form triangles with even perimeters.\n\nd) Is it possible for Rachael to use 18 rods (from 2 cm to 20 cm, omitting one) to form six triangles with equal perimeters?",
    "options": [],
    "answer": "See solution",
    "solution": "a) Each of the rods 2 to 20 can be the side of a triangle, as the following triples show:\n$\\{2, 3, 4\\}$, $\\{5, 6, 7\\}$, $\\{8, 9, 10\\}$, $\\{11, 12, 13\\}$, $\\{14, 15, 16\\}$, $\\{17, 18, 19\\}$, $\\{18, 19, 20\\}$.\n\nSo the rod that can't be the side of a triangle must be the 1 cm rod.\n\n_Alternative ii:_\n\nIn a triangle made from three rods of different lengths, one end of the longest rod must be joined to one end of the middle-length rod. The triangle can be completed only if the shortest rod is longer than the difference in lengths of those two rods. Since the difference in lengths of any two rods is at least 1 cm, the shortest rod must be longer than 1 cm.\n\nSo the rod that can't be the side of a triangle must be the 1 cm rod.\n\nb) From Part a, the 1 cm rod cannot be used. So Rachael must use the six rods of lengths 2 cm to 7 cm.\n\nIf these rods are divided into two sets of three, then only one set will contain the 2 cm rod. To form a triangle, the length of the longest rod must be less than the sum of the lengths of the other two rods. So the only sets of three rods that include the 2 cm rod and form a triangle are $\\{2, 3, 4\\}$, $\\{2, 4, 5\\}$, $\\{2, 5, 6\\}$, $\\{2, 6, 7\\}$.\n\nThe corresponding sets of the three remaining rods are $\\{5, 6, 7\\}$, $\\{3, 6, 7\\}$, $\\{3, 4, 7\\}$, $\\{3, 4, 5\\}$.\n\nOf these, only $\\{3, 4, 7\\}$ cannot form a triangle.\n\nSo there are just three ways of using the rods to make two triangles.\n\nc) For a triangle to have an even perimeter it must have all three rods of even length or exactly two of odd length. Here is one solution:\n$\\{3, 4, 5\\}$, $\\{6, 8, 10\\}$, $\\{7, 15, 16\\}$, $\\{9, 12, 17\\}$, $\\{11, 13, 14\\}$.\n\nd) From Part a, Rachael cannot use the 1 cm rod.\n\n_Alternative i:_\n\nThe sum of the lengths of the other 19 rods is $2 + 3 + 4 + \\cdots + 20 = 209$. The sum of the lengths of the 18 rods that she uses must be a multiple of 6, and $209 = 34 \\times 6 + 5$, so the length of the unused rod must leave a remainder of 5 when divided by 6. Therefore the unused rod must have length 5, 11, or 17, and the sum of the lengths of the other 18 rods is either 204, 198, or 192. So the perimeter of each triangle is either 34, 33, or 32. However, in each case one triangle must have a side length of 20. To form that triangle, the sum of the lengths of the other two sides must be greater than 20, hence the perimeter of that triangle must be greater than 40. Hence Rachael cannot form a set of 6 triangles with equal perimeters.\n\n_Alternative ii:_\n\nOne of the 18 rods that Rachael uses must be 19 or 20. If a triangle has one side 19, then the sum of the lengths of the other two sides must be greater than 19, hence the perimeter of that triangle must be greater than 38. Similarly, if a triangle has one side 20, then its perimeter must be greater than 40. The perimeters of the six triangles that Rachael forms are equal, so their total perimeter must be greater than $6 \\times 38 = 228$. But the total length of 18 rods is at most $20 + 19 + \\cdots + 3 = 207$. Hence Rachael cannot form a set of 6 triangles with equal perimeters.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15942,
    "subject": "Mathematics (Olympiad)",
    "question": "In a triangle $ABC$, the angle $A$ is twice as large as the angle $B$, and $CD$ is the bisector of angle $C$. Prove that $BC = AC + AD$.\n\n![](images/Ukrajina_2011_p25_data_1ad2d9ed6e.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $E$ be a point on $BC$ such that $AE$ is perpendicular to $CD$. Then $\\triangle ACE$ is isosceles, since the bisector of angle $C$ is also an altitude of the triangle (see the figure). Hence, $AC = CE$.\n\n$\\triangle ADE$ is isosceles, since the line $CD$ is perpendicular to $AE$ and divides $AE$ in half (the altitude is a median). So, we have:\n\n$$\nAD = DE. \\quad (1)\n$$\n\nLet $\\angle B = \\alpha$, $\\angle A = 2\\alpha$, $\\angle C = 180^\\circ - 3\\alpha$. Then $\\angle ADC = \\alpha + 90^\\circ - \\frac{3\\alpha}{2} = 90^\\circ - \\frac{\\alpha}{2}$, so $\\angle ADE = 180^\\circ - \\alpha$, $\\angle EDB = \\alpha = \\angle B$, hence, $\\triangle DEB$ is isosceles, i.e.,\n\n$$\nBE = DE. \\quad (2)\n$$\n\nFrom (1) and (2), we have $AD = BE$. Finally, $BC = CE + BE = AC + AD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15943,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n$. A football tournament with $4n$ teams has several rounds. In each round, these teams are divided into $2n$ pairs to play with each other. It is known that after this tournament, any two teams have played at most one match together. Find the smallest positive integer $a$ such that, after $a$ rounds, for every possible partition of the teams into $2n$ pairs, there always exists at least one pair that has played with each other.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that the smallest value of $a$ is $2n + 1$ using graph theory.\n\nConsider a graph $G$ with $4n$ vertices representing the teams. Two vertices are connected if the corresponding teams have played each other. After each round, we add a perfect matching; after $k$ rounds, $G$ becomes a $k$-regular graph.\n\nLet $G'$ be the complement of $G$. Then $G'$ is a $(4n - 1 - k)$-regular graph. When we cannot add another perfect matching in $G$, it is equivalent to $G'$ not containing any perfect matching.\n\nFirst, we construct a graph $G$ satisfying the problem for $a = 2n + 1$. Note that $G'$ is a $(2n - 2)$-regular graph. Partition the $4n$ vertices into two sets $X$ and $Y$ with $|X| = 2n + 1$ and $|Y| = 2n - 1$. For $Y = \\{y_1, \\dots, y_{2n-1}\\}$, construct a $K_{2n-1}$ graph. For $X = \\{x_1, x_2, \\dots, x_{2n+1}\\}$, construct a $K_{2n+1}$ graph and delete a cycle $x_1x_2\\dots x_{2n+1}x_1$. It is easy to check that $G'$ is $(2n - 2)$-regular and does not have any perfect matching.\n\nThus, the complementary graph $G$ contains a cycle $x_1x_2\\dots x_{2n+1}x_1$ and a complete bipartite graph $K_{2n+1,2n-1}$. We can partition the edges of $G$ into $2n + 1$ disjoint matchings as follows:\n\n$$\n\\{x_{i-1}x_i, y_1x_{i+1}, y_2x_{i+2}, \\dots, y_{2n-1}x_{i+2n-1}\\}\n$$\nfor every $i = 1, \\dots, 2n + 1$ (indices modulo $2n + 1$).\n\nSuppose $a \\le 2n$. We show that there always exists a perfect matching in $G'$. It suffices to prove this for $a = 2n$. Recall the following lemma:\n\n*Lemma (Dirac):* In a connected simple undirected graph $G$ with minimum degree $\\delta$, there is a simple path of length $\\min\\{2\\delta, n - 1\\}$.\n\nConsider an $n$-regular graph $G$ with $2n + 2$ vertices. If $G$ is connected, it has a simple path $\\mathcal{L}$ of length $2n$ (with $2n + 1$ vertices) and one vertex $v$ left. Considering the connections between $v$ and $\\mathcal{L}$, one can show that $G$ has a perfect matching. If $G$ is disconnected, it is the union of two complete graphs $K_{n+1}$, which also has a perfect matching when $n$ is odd. Hence, an $n$-regular graph with $2n + 2$ vertices (for odd $n$) always has a perfect matching.\n\nReturning to the original problem, if $a = 2n$, the complementary graph $G'$ is $(2n - 1)$-regular with $4n$ vertices. By the above result, $G'$ must have a perfect matching. Thus, the smallest value of $a$ is $2n + 1$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15944,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A' \\in (BC)$, $B' \\in (AC)$, $C' \\in (AB)$ be the points of tangency of the excircles of triangle $ABC$ with the sides of $ABC$. Let $R'$ be the circumradius of $A'B'C'$. Show that\n\n$$\nR' = \\frac{1}{2r} \\sqrt{2R(2R - h_a)(2R - h_b)(2R - h_c)},\n$$\n\nwhere, as usual, $R$ is the circumradius of $ABC$, $r$ is the inradius of $ABC$, and $h_a, h_b, h_c$ are the lengths of the altitudes of $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "The triangle $A'B'C'$ is the pedal triangle of the symmetric point of the incenter $I$ of $ABC$ with respect to the circumcenter of $ABC$. The relation between the areas $S = [ABC]$ and $S' = [A'B'C']$ is given by\n\n$$\nS' = S \\cdot \\frac{R^2 - \\overline{OI}^2}{4R^2} = \\frac{r}{2R}.\n$$\n\nIn triangle $A'B'C'$, $AB' = s - c = r \\cot \\frac{C}{2}$, $AC' = s - b = r \\cot \\frac{B}{2}$, and\n\n$$\n\\begin{aligned}\n\\overline{B'C'}^2 &= a'^2 = r^2 \\left[ \\left( \\cot \\frac{B}{2} + \\cot \\frac{A}{2} \\right)^2 - 4 \\cot \\frac{A}{2} \\cot \\frac{B}{2} \\cot \\frac{C}{2} \\right] \\\\\n&= r^2 \\frac{\\cos^2 \\frac{A}{2}}{\\sin^2 \\frac{B}{2} \\sin^2 \\frac{C}{2}} \\left( 1 - \\sin B \\sin C \\right) = a^2 \\left( 1 - \\frac{bc}{4R^2} \\right)\n\\end{aligned}\n$$\n\nBut as $bc = 2Rh_a$, we get\n\n$$\na'^2 = \\frac{a^2}{2R} (2R - h_a)\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15945,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M(n)$ be the maximal possible number of good diagonals in a convex $n$-gon. What is $M(n)$ in terms of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that $M(n) = n - 2$ if $n$ is even and $M(n) = n - 3$ if $n$ is odd.\n\nFor any $n$, we can draw all $n - 3$ diagonals from one vertex $A$ and one more diagonal joining two vertices adjacent to $A$ if $n > 3$.\n\n**Inductive lower bound:**\n\nWe show by induction that $M(n) \\ge n - 2$ if $n$ is even. For $n = 4$, both diagonals in a convex quadrilateral are good, so the claim holds. Assume the claim is true for $n - 2$ and consider a convex $n$-gon $A_1A_2\\ldots A_n$. Draw diagonals $\\overline{A_{n-2}A_n}$ and $\\overline{A_1A_{n-1}}$, and use the assumption on $A_1A_2\\ldots A_{n-2}$, so $M(n) \\ge M(n - 2) + 2 \\ge n - 4 + 2 = n - 2$.\n\n**Inductive upper bound:**\n\nWe also prove by induction that $M(n) \\le n - 2$ if $n$ is even and $M(n) \\le n - 3$ if $n$ is odd. Obviously, $M(3) = 0$ and $M(4) = 2$. Assume the claim holds for all $n$ smaller than $k$.\n\nConsider a choice of diagonals of a convex $k$-gon for which $M(k)$ is achieved.\n\n*Case 1:* If there are two good diagonals that intersect, they divide the $k$-gon into 4 parts, each having $a_i \\ge 0$ vertices (not counting the endpoints of these diagonals, for $1 \\le i \\le 4$). Since there is no other segment that intersects these two diagonals, we have\n\n$$\nM(k) = 2 + \\sum_{i=1}^{4} M(a_i + 2) \\le 2 + \\sum_{i=1}^{4} (a_i + 2 - 2) = 2 + (k - 4) = k - 2.\n$$\n\nIf $k$ is odd, then at least one of the numbers $a_i$ must also be odd, so we get the bound $k - 3$ in this case.\n\n*Case 2:* If there are no two good diagonals that intersect, then there are at most $k - 3$ good diagonals, as that is the maximal number of diagonals that can be drawn from one vertex.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15946,
    "subject": "Mathematics (Olympiad)",
    "question": "Given two coprime positive integers $a$, $b$ with $b$ odd and $a > 2$, define the sequence $(x_n)$ by $x_0 = 2$, $x_1 = a$, and $x_{n+2} = a x_{n+1} + b x_n$ for $n \\geq 1$.\n\nProve that:\n\na) If $a$ is even, then there do not exist positive integers $m$, $n$, $p$ such that $\\frac{x_m}{x_n x_p}$ is a positive integer.\n\nb) If $a$ is odd, then there do not exist positive integers $m$, $n$, $p$ such that $mnp$ is even and $\\frac{x_m}{x_n x_p}$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "The general formula for the sequence is $x_n = \\alpha^n + \\beta^n$, where $\\alpha$ and $\\beta$ satisfy\n$$\n\\alpha + \\beta = a, \\quad \\alpha\\beta = -b.\n$$\n\nWe will prove $\\gcd(b, x_m) = 1$ for all $m$. Suppose there exist $m$ and a prime $p$ such that $p \\mid x_m$ and $p \\mid b$. Since $\\gcd(a, b) = 1$, $p \\nmid a$. From\n$$\nx_m - b x_{m-2} = a x_{m-1},\n$$\nwe see $p \\mid a x_{m-1}$, so $p \\mid x_{m-1}$. By induction, this leads to $p \\mid x_1 = a$, a contradiction. Thus, $b$ is coprime to all terms of the sequence.\n\nNow, consider when $x_m \\mid x_n$. The sequence $(x_n)$ is strictly increasing, so $m \\leq n$. If $n > 2m$, then\n$$\n\\begin{aligned}\nx_n + (-b)^m x_{n-2m} &= \\alpha^n + \\beta^n + \\alpha^m \\beta^m (\\alpha^{n-2m} + \\beta^{n-2m}) \\\\\n&= (\\alpha^m + \\beta^m)(\\alpha^{n-m} + \\beta^{n-m})\n\\end{aligned}\n$$\nis divisible by $x_m$. Thus,\n$$\nx_m \\mid x_n \\iff x_m \\mid b^m x_{n-2m} \\iff x_m \\mid x_{n-2m}.\n$$\nBy induction, $x_m \\mid x_n$ if and only if $x_m \\mid x_k$, where $k$ is the remainder of $n$ modulo $2m$.\n\n- If $m \\geq k$, then $x_k \\leq x_m$, so $x_m \\mid x_n$ if and only if $m = k$.\n- If $m < k < 2m$, write $k = m + d$ with $0 < d < m$. Then\n$$\n\\begin{aligned}\nx_d x_m - x_{m+d} &= (\\alpha^d + \\beta^d)(\\alpha^m + \\beta^m) - (\\alpha^{m+d} + \\beta^{m+d}) \\\\\n&= \\alpha^d \\beta^d (\\alpha^{m-d} + \\beta^{m-d}) \\\\\n&= (-b)^d x_{m-d}.\n\\end{aligned}\n$$\nSo $x_k$ is divisible by $x_m$ if and only if $x_{m-d}$ is divisible by $x_m$, which is impossible since $(x_n)$ is strictly increasing.\n\nThus, for $m > 0$, $x_m \\mid x_n$ if and only if $n = (2k+1)m$ for some $k \\in \\mathbb{N}$.\n\na) Since $x_0 = 2$ and $x_1 = a$ are even, by induction $x_n$ is even for all $n$. For $x_{(2k+1)n}$ and $x_n$,\n$$\n\\frac{x_{(2k+1)n}}{x_n} = x_{2kn} + b^n x_{(2k-2)n} + \\dots + b^{(k-1)n} x_{2n} + b^{kn},\n$$\nwhich is odd since $b^{kn}$ is odd. Thus, $\\frac{x_m}{x_n}$ is odd, so it cannot be divisible by $x_p$ (which is even), so $\\frac{x_m}{x_n x_p}$ cannot be a positive integer.\n\nb) Suppose $x_m$ is divisible by $x_n x_p$ for $m, n, p$ with $mnp$ even. Then $m = 2u$, $n = 2v$, $p = 2t$ for some $u, v, t$. For each $k$,\n$$\nx_{2k} = (\\alpha^k - \\beta^k)^2 + 2b^{2k}.\n$$\nNow,\n$$\n(\\alpha^k - \\beta^k)^2 = (a^2 - 4b) \\left( \\sum_{i=0}^{\\lfloor k/2 \\rfloor} (-b)^i x_{k-2i} \\right)^2 = (a^2 - 4b) F_k^2,\n$$\nwhere $F_k$ is the sum inside the brackets. Thus, if $x_m x_n x_p$ is a perfect square, then\n$$\n\\prod_{k \\in \\{u, v, t\\}} \\left( (a^2 - 4b) F_k^2 + 2b^{2k} \\right)\n$$\nmust be a perfect square. For any prime $p$ dividing $a^2 - 4b$,\n$$\n1 = \\left( \\frac{2}{p} \\right),\n$$\nso $2$ is a quadratic residue modulo $p$, which only happens if $p \\equiv 1, 7 \\pmod{8}$. But $a$ and $b$ are odd, so\n$$\na^2 - 4b \\equiv 1 - 4 \\equiv 5 \\pmod{8},\n$$\nwhich is impossible. Thus, $x_m x_n x_p$ cannot be a perfect square, nor can $\\frac{x_m}{x_n x_p}$.\n\n$\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15947,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $a_1 > a_2 > \\dots > a_n$ and $b_1 > b_2 > \\dots > b_n$ are $2n$ positive real numbers. A pair $(a_i, b_j)$ is called **regular** if $-2013 < a_i - b_j < 2013$. A **regular partition** is a set of $n$ pairs $(a_{i_k}, b_{j_k})$ such that each $a_{i_k}$ and $b_{j_k}$ appears exactly once, and all pairs are regular. What is the least number of regular partitions possible for all $n \\ge 2$?\n\n![]( <table><thead><tr><th>Index</th><th>n</th><th>n-1</th><th>n-2</th><th>n-3</th><th>n-4</th><th>n-5</th><th>n-6</th><th>n-7</th><th>...</th></tr></thead><tbody><tr><th>a<sub>i</sub></th><td>1</td><td>2</td><td>k+2</td><td>k+3</td><td>2k+3</td><td>2k+4</td><td>3k+4</td><td>3k+5</td><td></td></tr><tr><th>b<sub>i</sub></th><td>1</td><td>2</td><td>k+2</td><td>k+3</td><td>2k+3</td><td>2k+4</td><td>3k+4</td><td>3k+5</td><td></td></tr></tbody></table> )\n\n![]( <table><thead><tr><th>Index</th><th>n</th><th>n-1</th><th>n-2</th><th>n-3</th><th>n-4</th><th>n-5</th><th>n-6</th><th>...</th><th>...</th></tr></thead><tbody><tr><th>a<sub>i</sub></th><td>1</td><td>k-1</td><td>k</td><td>2k+1</td><td>2k+2</td><td>3k+2</td><td>3k+3</td><td></td><td></td></tr><tr><th>b<sub>i</sub></th><td>1</td><td>k-1</td><td>k+1</td><td>2k+1</td><td>2k+2</td><td>3k+2</td><td>3k+3</td><td></td><td></td></tr></tbody></table> )",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze the number of regular partitions for $n=3$ and then use induction for general $n$.\n\nFor $n=3$, there are at least 4 regular partitions:\n$$\n\\begin{cases}\n(a_1, b_1), (a_2, b_2), (a_3, b_3) \\\\\n(a_1, b_1), (a_2, b_3), (a_3, b_2) \\\\\n(a_1, b_2), (a_2, b_3), (a_3, b_1) \\\\\n(a_1, b_2), (a_2, b_1), (a_3, b_3)\n\\end{cases}\n$$\n\nAssume the statement holds for $n-1 \\ge 3$. For $n$, consider $a_1 > a_2 > \\dots > a_n$ and $b_1 > b_2 > \\dots > b_n$. By induction, there exists a regular partition. By analyzing the possible regular/non-regular pairs and using the induction hypothesis, we find that the least number of regular partitions is $2^{\\lfloor \\frac{n+1}{2} \\rfloor}$.\n\nTo show equality can occur, construct sequences as follows (set $k=2013$):\n\n- If $n$ is even:\n  $$\n  a_{2i} = b_{2i} = (n/2 - i)k + n/2 - i + 1,\\quad a_{2i-1} = b_{2i-1} = (n/2 - i)k + n/2 - i + 2,\\quad i \\ge 1.\n  $$\n- If $n$ is odd:\n  $$\n  a_n = b_n = 1,\\quad a_{n-1} = b_{n-1} = k-1,\\quad a_{n-2} = k,\\quad b_{n-2} = k+1,\\text{ and so on.}\n  $$\n\nIt is easy to check these sequences satisfy the conditions.\n\n**Conclusion:** The least number of regular partitions is $2^{\\lfloor \\frac{n+1}{2} \\rfloor}$ for $n \\ge 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15948,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $Q_{>1}$ 為所有大於 $1$ 的有理數所成的集合。假設函數 $f: Q_{>1} \\to \\mathbb{Z}$ 滿足\n\n$$\nf(q) = \\begin{cases} q-3 & \\text{若 } q \\text{ 為整數,} \\\\ \\lceil q \\rceil - 3 + f\\left(\\frac{1}{\\lceil q \\rceil - q}\\right) & \\text{若 } q \\text{ 不為整數.} \\end{cases}\n$$\n\n證明：對於任意滿足 $\\frac{1}{a} + \\frac{1}{b} = 1$ 的 $a, b \\in Q_{>1}$，必有 $f(a) + f(b) = -2$。\n\n註：$\\lceil q \\rceil$ 為不小於 $q$ 的最小整數。",
    "options": [],
    "answer": "See solution",
    "solution": "假設 $a = \\frac{m}{n}$，$b = \\frac{m}{m-n}$，其中 $\\gcd(m, n) = 1$。若 $n = \\frac{1}{2}m$，則 $(m, n) = (2, 1)$，此時 $f(2) = -1$，命題成立。\n\n接下來假設 $n \\neq \\frac{1}{2}m$，且不失一般性，令 $n < \\frac{1}{2}m$。對 $m$ 進行歸納。\n\n假設 $m > 2$ 且命題對所有較小的 $m$ 成立。由 $m > 2n$，有：\n\n$$\n\\begin{aligned}\nf(a) &= \\left\\lceil \\frac{m}{n} \\right\\rceil - 3 + f\\left(\\frac{1}{\\left\\lceil \\frac{m}{n} \\right\\rceil - \\frac{m}{n}}\\right) \\\\\n&= 1 + \\left\\lceil \\frac{m-n}{n} \\right\\rceil - 3 + f\\left(\\frac{1}{\\left\\lceil \\frac{m-n}{n} \\right\\rceil - \\frac{m-n}{n}}\\right) \\\\\n&= 1 + f\\left(\\frac{m-n}{n}\\right).\n\\end{aligned}\n$$\n\n同理，\n\n$$\nf(b) = -1 + f\\left(\\frac{1}{2 - \\frac{m}{m-n}}\\right) = -1 + f\\left(\\frac{m-n}{m-2n}\\right).\n$$\n\n由歸納假設，\n\n$$\nf(a) + f(b) = f\\left(\\frac{m-n}{n}\\right) + f\\left(\\frac{m-n}{m-2n}\\right) = -2,\n$$\n\n得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15949,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function such that $f(f(x)) = 2^x - 1$ for all real numbers $x$. Show that $f(0) + f(1) = 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) It is clear that $x = 0$ and $x = 1$ are solutions for the given equation. There are no more solutions, since a line meets the graph of a strictly convex function at most at two points.\n\nb) The function $f$ is injective, because $f(x) = f(y)$ implies $f(f(x)) = f(f(y))$, so $2^x = 2^y$, hence $x = y$. Now, write $f(2^x - 1) = 2^{f(x)} - 1$ for all $x \\in \\mathbb{R}$. It follows that $f(0), f(1) \\in \\{0, 1\\}$. Since $f$ is injective, $\\{f(0), f(1)\\} = \\{0, 1\\}$, so $f(0) + f(1) = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15950,
    "subject": "Mathematics (Olympiad)",
    "question": "In his last research, professor P was concentrating on natural numbers with a certain property. It is known that whenever a natural number $x$ has this property, all multiples of $x$ also have this property. Let $a_1, \\, \\dots, \\, a_n$ be positive integers such that all their divisors that are greater than one have the property professor P studied. Is it true that all divisors greater than one of the product $a_1 \\dots a_n$ definitely have this property?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $k > 1$ be any divisor of the product $a_1 \\dots a_n$. Then $k$ has a prime divisor $p$, which is also a divisor of the product $a_1 \\dots a_n$. As $p$ is a prime, there exists $i$ such that $p$ is a divisor of $a_i$. As all the divisors of $a_i$ greater than $1$ have the property, $p$ also has this property. By the premise, all the multiples of $p$ have the property, so $k$ has the property.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15951,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, $z$ be positive real numbers whose sum is $2012$. Find the maximum value of\n\n$$\n\\frac{(x^2 + y^2 + z^2)(x^3 + y^3 + z^3)}{x^4 + y^4 + z^4}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $x = y = z = \\frac{2012}{3}$, then\n$$\n\\frac{(x^2 + y^2 + z^2)(x^3 + y^3 + z^3)}{x^4 + y^4 + z^4} = 2012.\n$$\nNow we prove that for all $x, y, z$ satisfying the premises,\n$$\n\\frac{(x^2 + y^2 + z^2)(x^3 + y^3 + z^3)}{x^4 + y^4 + z^4} \\le 2012.\n$$\nIt suffices to show that $(x^2 + y^2 + z^2)(x^3 + y^3 + z^3) \\le 2012(x^4 + y^4 + z^4)$, or equivalently,\n$$(x^2 + y^2 + z^2)(x^3 + y^3 + z^3) \\le (x + y + z)(x^4 + y^4 + z^4).$$\nMultiplying out, simplifying, and rearranging the terms gives\n$$\nxy(x - y)(x^2 - y^2) + xz(x - z)(x^2 - z^2) + yz(y - z)(y^2 - z^2) \\ge 0.\n$$\nSince the differences in the brackets in every product have equal signs, the products are non-negative, showing that the necessary inequality holds.\n\nThe inequality obtained after multiplying out and canceling $x^5$, $y^5$, $z^5$ is a special case of Muirhead's inequality (with exponent vectors $(4, 1, 0)$ and $(3, 2, 0)$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15952,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AL$ and $BK$ be angle bisectors in the non-isosceles triangle $ABC$ ($L \\in BC$, $K \\in AC$). The perpendicular bisector of $BK$ intersects the line $AL$ at\n\n![](images/Macedonia_2010_booklet_p33_data_068a89564f.png)\n\npoint $M$. Point $N$ lies on the line $BK$ such that $LN \\parallel MK$. Prove that $\\overline{LN} = \\overline{NA}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The point $M$ lies on the circumcircle of $\\triangle ABK$ (since both $AL$ and the perpendicular bisector of $BK$ bisect the arc $BK$ of this circle). Then\n\n$$\n\\angle CBK = \\angle ABK = \\angle AMK = \\angle NLA.\n$$\n\nThus $ABLN$ is cyclic, whence\n\n$$\n\\angle NAL = \\angle NBL = \\angle CBK = \\angle NLA.\n$$\n\nNow it follows that $\\overline{LN} = \\overline{NA}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15953,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle. A circle through points $B$ and $C$ meets the sides $AB$ and $AC$ again at points $N$ and $M$, respectively. Consider the points $P$ and $Q$ on the line segments $MN$ and $BC$, respectively, such that the angles $\\angle BAC$ and $\\angle PAQ$ share the same internal bisector.\n\n**a)** Prove that $\\frac{PM}{PN} = \\frac{QB}{QC}$.\n\n**b)** Prove that the midpoints of the line segments $BM$, $CN$, and $PQ$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Notice that $\\angle AMP = \\angle ABQ$ and $\\angle MAP = \\angle BAQ$, so triangles $APM$ and $AQB$ are similar. Hence, $\\frac{MP}{BQ} = \\frac{AP}{AQ}$. Similarly, $\\frac{NP}{CQ} = \\frac{AP}{AQ}$, so the conclusion follows.\n\nb) Set $\\frac{MP}{PN} = \\frac{BQ}{QC} = k$. Then:\n\n$$\n\\overrightarrow{AP} = \\frac{1}{k+1} \\overrightarrow{AM} + \\frac{k}{k+1} \\overrightarrow{AN}, \\quad \\overrightarrow{AQ} = \\frac{1}{k+1} \\overrightarrow{AB} + \\frac{k}{k+1} \\overrightarrow{AC}.\n$$\n\nSince $S$ is the midpoint of $BM$, $\\overrightarrow{AS} = \\frac{1}{2}(\\overrightarrow{AB} + \\overrightarrow{AM})$. Write similar relations for the midpoints $T$ and $U$ of $CN$ and $PQ$ to deduce that $\\overrightarrow{AU} = \\frac{1}{k+1}\\overrightarrow{AS} + \\frac{k}{k+1}\\overrightarrow{AT}$, hence the claim.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15954,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $u$, $v$, $w$ be positive real numbers. Prove that there is a cyclic permutation $(x, y, z)$ of $(u, v, w)$ such that the inequality\n\n$$\n\\frac{a}{xa + yb + zc} + \\frac{b}{xb + yc + za} + \\frac{c}{xc + ya + zb} \\geq \\frac{3}{x + y + z}\n$$\n\nholds for all positive real numbers $a$, $b$, and $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "Trivially, one of the inequalities $v + w \\geq 2u$, $u + w \\geq 2v$, or $u + v \\geq 2w$ must hold. Let $(x, y, z)$ be such a cyclic permutation of $(u, v, w)$ that $y + z \\geq 2x$. Then the left-hand side of the desired inequality is equal to\n\n$$\n\\frac{a^2}{x a^2 + y a b + z a c} + \\frac{b^2}{x b^2 + y b c + z a b} + \\frac{c^2}{x c^2 + y a c + z b c}\n$$\n\nApplying the Cauchy-Schwarz inequality (in the form of Arthur Engel), we get\n\n$$\n\\frac{a^2}{x a^2 + y a b + z a c} + \\frac{b^2}{x b^2 + y b c + z a b} + \\frac{c^2}{x c^2 + y a c + z b c} \\geq \\frac{(a + b + c)^2}{x(a^2 + b^2 + c^2) + (y + z)(a b + a c + b c)}\n$$\n\nDenote $S = a^2 + b^2 + c^2$, $P = a b + a c + b c$. It suffices to prove\n\n$$\n\\frac{S + 2P}{x S + (y + z) P} \\geq \\frac{3}{x + y + z}.\n$$\n\nThis inequality is equivalent to $(y + z - 2x)(S - P) \\geq 0$, which obviously holds since $S \\geq P$ and $y + z \\geq 2x$. $\\Box$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 15955,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve in $\\mathbb{R}$ the equation:\n\n$$\n\\frac{1}{\\{x\\}} + \\frac{1}{[x]} + \\frac{1}{x} = 0,\n$$\n\nwhere $[x]$ and $\\{x\\}$ denote the integer part and the fractional part of the real number $x$, respectively.",
    "options": [],
    "answer": "See solution",
    "solution": "First, $x \\in \\mathbb{R} \\setminus (\\mathbb{Z} \\cup [0, 1))$. The given equation is equivalent to:\n\n$$\n\\frac{[x] + \\{x\\}}{[x]\\{x\\}} = -\\frac{1}{x} \\implies -x^2 = [x]\\{x\\}. \\quad (*)\n$$\n\nSince $-x^2 \\le 0$ and $x \\ne 0$, we have $[x]\\{x\\} < 0$, hence $[x] \\le -1$. Let $[x] = -k$, with $k \\in \\mathbb{N}^*$. Equation $(*)$ becomes $-x^2 = -k(k + x)$, which is equivalent to $x^2 - kx - k^2 = 0$, with solutions $x_{1,2} = \\frac{k \\pm k\\sqrt{5}}{2}$.\n\nFrom $[x] = -k$, the only possible solution is $x_1 = \\frac{k - k\\sqrt{5}}{2}$.\n\nWe must find $k \\in \\mathbb{N}^*$ such that $\\left[\\frac{k - k\\sqrt{5}}{2}\\right] = -k$, which is equivalent to $-k \\le k \\frac{1 - \\sqrt{5}}{2} < -k + 1 \\implies 3k \\ge k\\sqrt{5} > 3k - 2$, so $k < \\frac{3 + \\sqrt{5}}{2}$, i.e., $k \\in \\{1, 2\\}$. Thus, the solutions are $x_1^* = \\frac{1 - \\sqrt{5}}{2}$ and $x_2^* = 1 - \\sqrt{5}$, which satisfy the given equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15956,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence $\\langle a_n \\rangle$ with $a_n = a + nd$ is called an arithmetic sequence. The sequence $\\langle b_n \\rangle$ with $b_n = \\sum_{k=0}^{n} a_k$ is called an arithmetic sequence of second degree. Let $a$ and $d$ be positive integers.\n\nWe consider all such arithmetic sequences of second degree containing the number $2010$. What is the highest possible index $n$ if $b_n = 2010$? Determine all possible arithmetic sequences $\\langle a_n \\rangle$, for which $b_n = 2010$ holds for this index.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $a_k = a + kd$, we have\n$$\nb_n = (n+1)a + \\frac{n(n+1)}{2}d.\n$$\nIf we assume $b_n = 2010$, we therefore obtain\n$$\na = \\frac{4020 - n(n + 1)d}{2(n + 1)} = \\frac{2010}{n + 1} - \\frac{dn}{2}.\n$$\nSince both $a$ and $d$ are positive, we see from the first fraction that $n(n + 1) < 4020$ must hold. We therefore have $n(n + 1) < 4020 < 4225 = 65^2$, and thus $n < 65$. Furthermore, $n + 1$ must divide $4020$. Since $4020 = 3 \\cdot 4 \\cdot 5 \\cdot 67$, the largest divisor of $4020$ less than $65$ is $60$. The largest possible value for $n + 1$ is therefore $60$, and we have $n = 59$.\n\nSubstituting this value yields $a = \\frac{67 - 59d}{2}$. We see that $d$ must be odd and less than $2$, and we therefore have the unique solution $d = 1$, which yields $a = 4$.\n\nThe only possible sequence $\\langle a_n \\rangle$, for which $b_{59} = 2010$ is therefore the sequence $\\langle 4, 5, 6, \\dots \\rangle$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15957,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a real number $a \\ge 1$. The sequence $(x_n)_{n \\ge 1}$ is given by $x_1 = a$ and $x_{n+1} = 1 + \\log_2 x_n$, for any $n \\in \\mathbb{N}^*$. Determine values of $a$ such that all terms of the sequence are rational numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "For $a = 1$ or $a = 2$ we get the constant sequence $1$ or $2$ respectively.\n\nWe shall prove that these are the only values satisfying the problem.\n\nTo see this, let $x_n = \\frac{k}{l}$ and $x_{n+1} = \\frac{p}{q}$, with $k, l$ coprime positive integers, $p, q$ coprime positive integers with $q \\ge 2$. Then\n\n$$\n\\frac{k}{l} = x_n = 2^{x_{n+1}-1} = \\sqrt[q]{2^{p-q}},\n$$\n\nimplying $l^q 2^p = k^q 2^q$.\n\nThe last equality cannot be true, as the left hand side number has $2$ as factor at an exponent which is not divisible by $q$ ($p$ and $q$ are coprime), and the right hand side has the factor $2$ at an exponent divisible by $q$. So $q = 1$, thus $l = 1$, so, if all terms of the sequence would be rationals, they should be natural numbers.\n\nAs $2^n > n + 1$, for $n \\ge 2$, $n \\in \\mathbb{N}$, we get $1 + \\log_2 x < x$, for $x \\in \\mathbb{N}$, $x \\ge 3$. As a conclusion, if all terms are positive integers and $x > 2$, an inductive argument shows that the sequence is decreasing, a contradiction. That is, $a = x_1 \\in \\{1, 2\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15958,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be an interior point of an acute-angled triangle $ABC$. The lines $AP$, $BP$, and $CP$ meet $BC$, $CA$, and $AB$ at $D$, $E$, and $F$ respectively. Given that $\\triangle DEF \\sim \\triangle ABC$, prove that $P$ is the centroid of $\\triangle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Identify the points as complex numbers. Because $P$ is interior to $\\triangle ABC$, there are positive real numbers $\\alpha, \\beta, \\gamma$ with $\\alpha + \\beta + \\gamma = 1$ and $\\alpha A + \\beta B + \\gamma C = 0$.\n\nSince $D$ is on $BC$ and on line $AP$, it follows that $D = \\frac{-\\alpha}{1-\\alpha}A$. Similarly, $E = \\frac{-\\beta}{1-\\beta}B$ and $F = \\frac{-\\gamma}{1-\\gamma}C$.\n\nBecause $\\triangle DEF$ is similar to $\\triangle ABC$, we have\n\n$$\n\\frac{D-E}{A-B} = \\frac{E-F}{B-C}.\n$$\n\nSubstituting the expressions for $D$, $E$, and $F$ yields\n\n$$\n\\frac{\\gamma BC}{1 - \\gamma} + \\frac{\\beta AB}{1 - \\beta} + \\frac{\\alpha AC}{1 - \\alpha} - \\frac{\\alpha AB}{1 - \\alpha} - \\frac{\\beta BC}{1 - \\beta} - \\frac{\\gamma AC}{1 - \\gamma} = 0.\n$$\n\nTaking $\\alpha + \\beta + \\gamma = 1$ into account, this implies\n\n$$\nBC(\\gamma^2 - \\beta^2) + CA(\\alpha^2 - \\gamma^2) + AB(\\beta^2 - \\alpha^2) = 0.\n$$\n\nIf $\\gamma^2 - \\beta^2 \\neq 0$, then the cross-ratio\n\n$$\n\\frac{(C-A)/(C-B)}{(P-A)/(P-B)} = \\frac{B(C-A)}{A(C-B)}\n$$\n\nis real, which would imply $P$ is on the circumcircle of $\\triangle ABC$, contradicting that $P$ is interior. Thus $\\gamma^2 = \\beta^2$, and $\\alpha^2 = \\gamma^2$ as well. It follows that $\\alpha = \\beta = \\gamma = \\frac{1}{3}$, and hence, $P$ is the centroid of $\\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15959,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to find real numbers $a$, $b$, and a surjective function $f: \\mathbb{R} \\to \\mathbb{R}$ such that for all $x \\in \\mathbb{R}$,\n\n$$\nf(f(x)) = bx f(x) + a?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose such $a$, $b$, and $f$ exist. Obviously $b \\neq 0$. Since $f$ is surjective, there exists $\\lambda$ such that $f(\\lambda) = 0$. Set $x = \\lambda$ in\n\n$$\nf(f(x)) = bx f(x) + a, \\tag{1}\n$$\n\nthen $f(0) = f(f(\\lambda)) = b\\lambda f(\\lambda) + a = b\\lambda \\cdot 0 + a = a$. Further,\n\n$$\nf(a) = f(f(0)) = b \\cdot 0 \\cdot f(0) + a = a.\n$$\n\nNow $a = f(a) = f(f(a)) = ba f(a) + a = ba^2 + a$, whence $a = 0$. Thus (1) becomes\n\n$$\nf(f(x)) = bx f(x). \\tag{2}\n$$\n\nFurther, there exists $\\sigma \\neq 0$ such that $f(\\sigma) = -1/b$. Set $x = \\sigma$ in (2), then\n\n$$\nf(-1/b) = f(f(\\sigma)) = b\\sigma f(\\sigma) = b\\sigma(-1/b) = -\\sigma. \\quad (3)\n$$\n\nHence\n\n$$\nf(-\\sigma) = f(f(-1/b)) = b(-1/b)f(-1/b) = \\sigma.\n$$\n\nTherefore,\n\n$$\n\\frac{1}{b} = f(\\sigma) = f(f(-\\sigma)) = b(-\\sigma)f(-\\sigma) = -b\\sigma^2\n$$\n\nand so $\\sigma = \\pm 1/b$.\n\nIf $\\sigma = -1/b$, then $\\sigma = -1/b = f(\\sigma) = f(-1/b)$, contrary to (3).\n\nIf $\\sigma = 1/b$, then in view of (3) we have\n\n$$\nf(-\\sigma) = -\\sigma \\implies -\\sigma = f(-\\sigma) = f(f(-\\sigma)) = b(-\\sigma)f(-\\sigma) = b\\sigma^2 \\implies \\sigma = -1/b,\n$$\n\na contradiction.\n\nTherefore, there are no $a$, $b$, $f$ satisfying the problem condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15960,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S = \\{1, 2, \\dots, n\\}$ with $n \\ge 2$, and let $f : S \\to S$ be a bijective function distinct from the identity. Let\n\n- $u = \\sum_{k=1}^{n} |f(k) - k|$,\n- $v$ be the number of ordered pairs $(a, b)$ of elements of $S$ such that $a > b$ and $f(a) < f(b)$.\n\nShow that $v < u \\le 2v$, and that $u = 2v$ if and only if there do not exist positive integers $a > b > c$ such that $f(a) < f(b) < f(c)$.",
    "options": [],
    "answer": "See solution",
    "solution": "We say a pair $(a, b)$ is *a-good* if $a > b \\ge f(a)$, and *b-good* if $b \\le f(a) < f(b)$. If $f(k) = k$, no pair $(k, b)$ or $(a, k)$ is *k-good*. If $f(k) > k$, there are no *k-good* pairs of the form $(k, b)$ and exactly $f(k) - k$ of the form $(a, k)$. Likewise, if $f(k) < k$ there are exactly $k - f(k)$ *k-good* pairs. In any case, there are exactly $|f(k) - k|$ *k-good* pairs.\n\nFor every pair $(a, b)$ with $a > b$ and $f(a) < f(b)$, the pair is *a-good* if $f(a) \\le b$, and *b-good* if $f(a) \\ge b$. The number of pairs counted in $v$ is at most the number of good pairs, which is less than or equal to the sum of the $|f(k) - k|$. So, $v \\le \\sum_{k=1}^{n} |f(k) - k| = u$. Let $k$ be the smallest integer with $f(k) \\ne k$. Then $f(k) > k$, $f^{-1}(k) > k$. The pair $(f^{-1}(k), k)$ is both $f^{-1}(k)$-good and $k$-good. Equality does not hold, and $v < u$.\n\nLet $A_k = \\{f^{-1}(1), f^{-1}(2), \\dots, f^{-1}(f(k))\\}$ and $B_k = \\{1, 2, \\dots, k\\}$. The element $x$ is in $B_k$ and not in $A_k$ if and only if $(k, x)$ is a pair counted in $v$. Similarly, $x$ is in $A_k$ and not in $B_k$ if and only if $(x, k)$ is a pair counted in $v$. Then,\n\n$$2v = \\sum_{k=1}^{n} |A_k \\setminus B_k| + |B_k \\setminus A_k| = \\sum_{k=1}^{n} |A_k| + |B_k| - 2|A_k \\cap B_k| \\ge \\sum_{k=1}^{n} |A_k| + |B_k| - 2 \\min\\{|A_k|, |B_k|\\} = \\sum_{k=1}^{n} f(k) + k - 2 \\min\\{f(k), k\\} = \\sum_{k=1}^{n} |f(k) - k| = u$$\n\nWe look at the equality case. $u = 2v$ if and only if $|A_b \\cap B_b| = \\min\\{|A_b|, |B_b|\\}$ for all $b$, which is equivalent to $A_b \\subseteq B_b$ or $B_b \\subseteq A_b$ for all $b$. This holds if and only if there do not exist $a, b, c$ such that $a \\in A_b \\setminus B_b$, $c \\in B_b \\setminus A_b$, i.e., there do not exist $a > b > c$ such that $f(a) < f(b) < f(c)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15961,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that there is a unique function $g : [0, 1] \\to [0, 1]$ such that the equality\n\n$$\n\\int_{0}^{x} f(t) \\, dt = \\int_{g(x)}^{1} f(t) \\, dt\n$$\n\nholds for any $x \\in [0, 1]$.\n\nShow that there is a $c \\in [0, 1]$ for which\n\n$$\n\\lim_{x \\to c} \\frac{g(x) - c}{x - c} = -1,\n$$\n\nwhere $g$ is the function determined by the relation above.",
    "options": [],
    "answer": "See solution",
    "solution": "Because $f$ is continuous, $F$ is continuous and differentiable, with $F'(x) = f(x) > 0$ for any $x \\in [0, 1]$. Hence, $F$ is strictly increasing and injective. Since $F(0) = 0$ and $F(1) = A$, $F$ is surjective and thus bijective, so $F$ is invertible. Also, $F^{-1}$ is differentiable, with\n\n$$\n(F^{-1})'(x) = \\frac{1}{f(F^{-1}(x))}, \\quad \\text{for any } x \\in [0, A].\n$$\n\nThe equality $\\int_{0}^{x} f(t) \\, dt = \\int_{g(x)}^{1} f(t) \\, dt$ becomes $F(x) = F(1) - F(g(x)) = A - F(g(x))$, or $F(g(x)) = A - F(x)$ for any $x \\in [0, 1]$.\n\nSince $A - F(x) \\in [0, A]$, the function $g : [0, 1] \\to [0, 1]$ defined by $g(x) = F^{-1}(A - F(x))$ is well defined and unique.\n\nSince $F$ and $F^{-1}$ are differentiable, $g$ is differentiable and\n\n$$\ng'(x) = \\frac{(A - F(x))'}{f(F^{-1}(A - F(x)))} = \\frac{-f(x)}{f(F^{-1}(A - F(x)))} \\quad \\text{for any } x \\in [0, 1].\n$$\n\nThe function $g$ is strictly decreasing and has a unique fixed point $c \\in [0, 1]$.\n\nFor this fixed point, $2F(c) = F(c) + F(g(c)) = A$, so $F(c) = \\frac{1}{2}A$ and $c = F^{-1}\\left(\\frac{1}{2}A\\right)$.\n\nBecause $g$ is differentiable,\n\n$$\n\\lim_{x \\to c} \\frac{g(x) - c}{x - c} = \\lim_{x \\to c} \\frac{g(x) - g(c)}{x - c} = g'(c),\n$$\n\nand\n\n$$\ng'(c) = \\frac{-f(c)}{f(F^{-1}(A - F(c)))} = \\frac{-f(c)}{f(c)} = -1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15962,
    "subject": "Mathematics (Olympiad)",
    "question": "Evaluate the sum:\n$$\n\\sum_{k=0}^{p} (-1)^k \\binom{p}{k} \\binom{p+k}{k}\n$$\nwhere $p$ is a prime number. Show that\n$$\n\\sum_{k=0}^{p} (-1)^k \\binom{p}{k} \\binom{p+k}{k} \\equiv -1 \\pmod{p^3}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The sum $\\sum_{k=0}^{p} (-1)^k \\binom{p}{k} \\binom{p+k}{k}$ is the coefficient of $x^p$ in the expansion of\n$$\n\\sum_k \\binom{p}{k} (x-1)^{p+k}.\n$$\nThis can be rewritten as\n$$\n\\left( \\sum_{k=0}^{p} \\binom{p}{k} (x-1)^k \\right) (x-1)^p = x^p (x-1)^p.\n$$\nHence, the sum is actually $(-1)^p = -1$. Therefore,\n$$\n\\sum_{k=0}^{p} (-1)^k \\binom{p}{k} \\binom{p+k}{k} \\equiv -1 \\pmod{p^3}\n$$\nclearly holds.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15963,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an acute triangle $ABC$ with circumcenter $O$. The point $P$ is on $BC$ such that $BP < \\frac{BC}{2}$, and the point $Q$ is on $BC$ such that $CQ = BP$. The line $AO$ meets $BC$ at $D$, and $N$ is the midpoint of $AP$. The circumcircle of $ODQ$ meets $(BOC)$ at $E$. The lines $NO$ and $OE$ meet $BC$ at $K$ and $F$, respectively. Show that $AOKF$ is cyclic.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A'$ be the antipode of $A$, and let $AO \\cap (BOC) = R$. Since $NO \\parallel PA'$, by Reim's theorem we need $APA'F$ to be cyclic, or $DP \\cdot DF = DA \\cdot DA' = DB \\cdot DC = DO \\cdot DR$, so we need $OPRF$ to be cyclic, or that $\\angle OFP = \\angle ORP$. By the shooting lemma, $RDEF$ is cyclic, so $\\angle OFP = \\angle ERD$, so we need $AR$ to be the angle bisector of $\\angle PRE$. Since $AR$ bisects $\\angle BRC$, it is sufficient to show that $\\angle BRP = \\angle CRE$.\n\nTo use that $EQDO$ is cyclic, let $EQ \\cap (BOC) = S$; Reim's theorem implies $RS \\parallel BC$, which together with the length condition gives that $RSQP$ is an isosceles trapezoid. Hence, $\\angle CRE = \\angle CSQ = \\angle BRP$, which finishes the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15964,
    "subject": "Mathematics (Olympiad)",
    "question": "Petro has to plant 8 trees in a row: apple trees or oak trees. There is one restriction: there must be no apple trees between any two oak trees. For example, such plantings $\\text{AAOOAOAA}$ or $\\text{OAOAAAAA}$ are not allowed, but $\\text{AAOOAAAA}$ is allowed. How many different plantings are possible?\n\n![](images/UkraineMO_2015-2016_booklet_p14_data_cb89a68c12.png)",
    "options": [],
    "answer": "See solution",
    "solution": "All oaks must be planted together as one group, regardless of where this group is placed. Let's count the possible numbers of oaks.\n\n- If there are no oak trees, there is only one way to plant the trees (all apples).\n- If there are $k$ oaks, where $1 \\leq k \\leq 8$, there are $9 - k$ possible positions for the contiguous group of oaks.\n\nFor example, for $k=3$ (three oaks):\n\n$\\text{OOOAAAAA},\\ \\text{AOOOAAAA},\\ \\ldots,\\ \\text{AAAAAOOO}$\n\nSumming over all possible $k$:\n\n$$\n1 + 1 + 2 + 3 + \\cdots + 8 = 37\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15965,
    "subject": "Mathematics (Olympiad)",
    "question": "在凸六邊形 $ABCDEF$ 中，$AB \\parallel DE$，$BC \\parallel EF$，$CD \\parallel FA$，且\n\n$$\nAB + DE = BC + EF = CD + FA.\n$$\n\n將邊 $AB, BC, DE, EF$ 的中點分別記為 $A_1, B_1, D_1, E_1$。設點 $O$ 為線段 $A_1D_1$ 與 $B_1E_1$ 的交點。證明 $\\angle D_1OE_1 = \\frac{1}{2}\\angle DEF$。\n\nIn a convex hexagon $ABCDEF$, $AB \\parallel DE$, $BC \\parallel EF$, $CD \\parallel FA$, and\n\n$$\nAB + DE = BC + EF = CD + FA.\n$$\n\nDenote the midpoints of sides $AB, BC, DE, EF$ by $A_1, B_1, D_1, E_1$, respectively. Let $O$ be the intersection of segments $A_1D_1$ and $B_1E_1$. Prove that $\\angle D_1OE_1 = \\frac{1}{2}\\angle DEF$.",
    "options": [],
    "answer": "See solution",
    "solution": "定义 $\\alpha = \\pi - \\angle FAB = \\pi - \\angle CDE$，$\\beta = \\pi - \\angle ABC = \\pi - \\angle DEF$，$\\gamma = \\pi - \\angle BCD = \\pi - \\angle EFA$。显然有 $\\alpha + \\beta + \\gamma = \\pi$，且三者中至少两者为锐角。令 $O'$ 为 $B_1E_1$ 与 $C_1F_1$ 的交点，$O''$ 为 $C_1F_1$ 与 $A_1D_1$ 的交点（$C_1, F_1$ 分别是 $CD$ 与 $FA$ 的中点）。再令 $\\alpha' = \\angle F_1O''A_1 = \\angle C_1O''D_1$，$\\beta' = \\angle A_1OB_1 = \\angle D_1OE_1$，$\\gamma' = \\angle B_1O'C_1 = \\angle E_1O'F_1$。由于可以将 $C_1F_1$ 平移到经过 $O$ 点，平移不改变直线的交角，故 $\\alpha' + \\beta' + \\gamma' = \\pi$。再者，由题目的循环对称性，题目等价于 $\\beta' = \\angle D_1OE_1 = \\frac{1}{2}\\angle DEF = \\frac{\\pi-\\beta}{2}$，或 $\\alpha' = \\frac{\\pi-\\alpha}{2}$，或 $\\gamma' = \\frac{\\pi-\\gamma}{2}$。因此不失一般性，可以重新循环标记顶点而不改变问题，可以假设 $\\alpha$ 与 $\\beta$ 为锐角。记 $m = \\tan \\alpha$，$M = \\tan \\beta$；$m, M$ 均为正数。\n\n现在建立坐标系：不失一般性设 $A \\equiv (-d, -h)$，$B \\equiv (d, -h)$，$D \\equiv (d' + \\Delta, h)$，$E \\equiv (-d' + \\Delta, h)$，其中 $d, d', h$ 为正数，且 $d' \\ge d$。经过代数计算（找出直线 $BC, CD, EF, FA$ 的方程，求交点坐标，并利用 $BC, CD$ 的斜率分别为 $m, M$），可得\n\n$$\nC \\equiv \\left( \\frac{Md' + M\\Delta + md + 2h}{M+m}, \\frac{Mmd' + Mm\\Delta - Mmd - Mh + mh}{M+m} \\right),\n$$\n$$\nF \\equiv \\left( -\\frac{md' - m\\Delta + Md + 2h}{M+m}, \\frac{Mmd' - Mm\\Delta - Mmd + Mh - mh}{M+m} \\right).\n$$\n\n注意 $C, F$ 的纵坐标需在区间 $(-h, h)$ 内，六边形才为凸六边形，因此有\n\n$$\n\\frac{2h}{m} > d' - d + \\Delta > -\\frac{2h}{M}, \\quad \\frac{2h}{M} > d' - d - \\Delta > -\\frac{2h}{m}.\n$$\n\n再经代数计算，并利用上述结果，得\n\n$$\nBC = \\frac{M(d' - d + \\Delta) + 2h}{M + m}\\sqrt{m^2 + 1},\n$$\n$$\nEF = \\frac{2h - M(d' - d - \\Delta)}{M + m}\\sqrt{m^2 + 1},\n$$\n$$\nCD = \\frac{2h - m(d' - d + \\Delta)}{M + m}\\sqrt{M^2 + 1},\n$$\n$$\nFA = \\frac{m(d' - d - \\Delta) + 2h}{M + m}\\sqrt{M^2 + 1}.\n$$\n\n故有\n\n$$\n\\begin{aligned}\nd + d' &= \\frac{AB + DE}{2} = \\frac{BC + EF}{2} = \\frac{CD + FA}{2} \\\\\n&= \\frac{2h + M\\Delta}{M + m}\\sqrt{m^2 + 1} = \\frac{2h - m\\Delta}{M + m}\\sqrt{M^2 + 1}.\n\\end{aligned}\n$$\n\n从上述等式最后两项，发现\n\n$$\n\\begin{aligned}\n\\left(\\frac{\\Delta}{2h}\\right)^2 + 2 \\frac{Mm + 1}{M - m} \\left(\\frac{\\Delta}{2h}\\right) - 1 &= 0, \\\\\n\\frac{\\Delta}{2h} &= \\frac{-Mm - 1 + \\sqrt{(M^2 + 1)(m^2 + 1)}}{M - m} = \\tan \\frac{\\alpha - \\beta}{2},\n\\end{aligned}\n$$\n\n此处假设 $M \\neq m$；若 $M = m$，则 $\\Delta = 0$，有 $d + d' = \\frac{h\\sqrt{m^2 + 1}}{m} = \\frac{h}{\\sin \\beta} = \\frac{h}{\\sin \\alpha}$。注意忽略了负根式，不仅考虑 $M = m$ 时解的连续性，也因进一步计算可知 $2h = (d + d')(\\sin \\alpha + \\sin \\beta)$，此式因 $2h = BC \\sin \\beta + CD \\sin \\alpha = EF \\sin \\beta + FA \\sin \\alpha$ 恒成立。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15966,
    "subject": "Mathematics (Olympiad)",
    "question": "On a screen formed by $n \\times n$ ($n \\ge 2$) squares, every square displays initially one of the three colors: red, yellow, and blue. Every second, the screen changes the color of each square following the rules below:\n\n- For each square $A$ whose current color is red, if there is a yellow square sharing a side with this square, then $A$ turns yellow in the next second; otherwise, $A$ remains red.\n- For each square $B$ whose current color is yellow, if there is a blue square sharing a side with this square, then $B$ turns blue in the next second; otherwise, $B$ remains yellow.\n- For each square $C$ whose current color is blue, if there is a red square sharing a side with this square, then $C$ turns red in the next second; otherwise, $C$ remains blue.\n\nProve that if the screen does not change to a single color after $2n - 2$ seconds, then it will never change to a single color.",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof:**\n\nWe first prove the claim: if the screen eventually turns into one color, say blue, then there must be one square which is constantly blue.\n\nSuppose not; that is, suppose that eventually all squares are blue but every square has changed color at some time. We construct an oriented graph $G$ as follows: the vertices are all squares, and we link an oriented arrow $A \\to B$ if $A$ and $B$ are adjacent and there is some time $t_0$ such that square $A$ is not blue at time $t_0 - 1$ and is constantly blue when $t \\ge t_0$, and square $B$ is blue at time $t_0 - 1$. In other words, the square $A$ is changed to blue for the last time because of square $B$.\n\nSince we assume that every square on the screen has changed color, every vertex of the graph has out-degree $\\ge 1$. So $G$ must contain an oriented loop $A_1 \\to A_2 \\to \\dots \\to A_k \\to A_1$. Assume that square $A_i$ turns blue for the last time at time $t_i$, i.e., $A_i$ is not blue at time $t_i - 1$, and is constantly blue when $t \\ge t_i$. Without loss of generality, $t_1$ is the largest among $t_1, \\dots, t_k$.\n\nFrom the arrow $A_1 \\to A_2$, the square $A_1$ is yellow at time $t_1 - 1$; from the arrow $A_k \\to A_1$, the square $A_1$ is blue at time $t_k - 1$. We must have $t_k < t_1$. But a blue square can only turn red first before turning yellow. So there exists $t_k < t' < t_1$ such that $A_1$ is red at time $t' - 1$. But $t' - 1 \\ge t_k$, so $A_k$ is blue at time $t' - 1$, and $A_1$ will force $A_k$ to turn red at time $t'$. But this contradicts the definition of $t_k$, namely, $A_k$ is constantly blue from time $t_k$ onward. The claim is proved.\n\nBack to the original problem, assume that all squares are turned blue eventually. We will show that the squares are all turned blue after $2n - 2$ seconds.\n\nBy the claim, there exists a square $A$ which is constantly blue. Define the distance between two squares by the sum of their horizontal and vertical distances; then the distance between the farthest two squares is $2n - 2$. If $B$ and $A$ have distance 1, then $B$ can never be red. This means: if $B$ is initially blue, then it is always blue; if $B$ is yellow, then it is turned blue after 1 second and continues to be blue after that. Now, for a point $B$ of distance $k$ to the point $A$, we may run an induction to prove that $B$ will become constantly blue after $k$ seconds. Suppose that this statement has been proved for $k-1$. If $B$ has distance $k$ to $A$, take a square $B'$ adjacent to $B$ that has distance $k-1$ to $A$. By induction, $B'$ is constantly blue after $k-1$ seconds; then the same argument as above shows that $B$ is constantly blue after $k$ seconds. From this induction, we deduce that all squares are blue after $2n-2$ seconds.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15967,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p(x)$ be a polynomial of degree $n$ with leading coefficient $a_n > 0$ and $p(0) = -1$. Prove that $p(x)$ has exactly one positive root.",
    "options": [],
    "answer": "See solution",
    "solution": "Since\n$$\n\\lim_{x \\to \\infty} \\frac{p(x)}{x^n} = a_n > 0,\n$$\n$p(x)$ is positive for all sufficiently large positive $x$. But $p(0) = -1$. By the Intermediate Value Theorem, $p$ has at least one positive root. Suppose $a, b$ are two positive roots of $p$, so $p(a) = p(b) = 0$. Then\n$$\n0 = p(a) - p(b) = \\sum_{i=0}^{n} a_i (a^i - b^i) = (a-b) \\sum_{i=0}^{n} a_i \\sum_{k=0}^{i-1} a^{i-k-1} b^k = (a-b) r(a, b),\n$$\nwhere $r(a, b)$ is a positive expression. Hence, $a = b$, so $p$ has at most one positive root. The claimed result follows.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15968,
    "subject": "Mathematics (Olympiad)",
    "question": "Do there exist pairwise distinct rational numbers $x$, $y$, and $z$ such that\n$$\n\\frac{1}{(x-y)^2} + \\frac{1}{(y-z)^2} + \\frac{1}{(z-x)^2} = 2014?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = x - y$ and $b = y - z$. Then,\n$$\n\\begin{aligned}\n\\frac{1}{(x-y)^2} + \\frac{1}{(y-z)^2} + \\frac{1}{(z-x)^2} &= \\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{(a+b)^2} \\\\\n&= \\frac{b^2(a+b)^2 + a^2(a+b)^2 + a^2b^2}{a^2b^2(a+b)^2} \\\\\n&= \\left( \\frac{a^2 + b^2 + ab}{ab(a+b)} \\right)^2.\n\\end{aligned}\n$$\nOn the other hand, $2014$ is not a square of a rational number. Hence such numbers do not exist. $\\blacktriangledown$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15969,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $I$ be the incenter of a triangle $ABC$. The incircle touches the sides $BC$, $CA$, and $AB$ at points $A'$, $B'$, and $C'$, respectively. Let $M$ be the midpoint of the line segment $A'B'$. Denote by $P$ the projection of $I$ onto $AA'$. Lines $MP$ and $AC$ meet at point $N$. Prove that the lines $A'N$ and $B'C'$ are parallel.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the lines $AI$ and $B'C'$ meet at point $R$, and let the lines $B'C'$ and $AA'$ meet at point $S$. Line $AA'$ meets the incircle again at point $T$. Notice that $MP$ is a mid-segment of triangle $A'TB'$, so lines $PN$ and $TB'$ are parallel.\n\nIt suffices to prove that triangles $STB'$ and $A'PN$ are similar, which reduces to $\\dfrac{TB'}{PN} = \\dfrac{TS}{PA'}$. As $\\dfrac{TB'}{PN} = \\dfrac{AT}{AP}$, we need to prove that $\\dfrac{TS}{PA'} = \\dfrac{AT}{AP}$, which is equivalent to $\\dfrac{AS}{AA'} = \\dfrac{AT}{AP}$.\n\nTriangles $ASR$ and $AIP$ are similar, since lines $AR$ and $B'C'$ are perpendicular, implying $AS \\cdot AP = AR \\cdot AI$. Notice the similarity of triangles $ATB'$ and $AB'A'$ to get $AA' \\cdot AT = B'A^2$. Since $B'A^2 = AR \\cdot AI$, the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15970,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that $a_1, a_2, \\dots, a_n$ are real numbers such that $a_1 + a_2 + \\dots + a_n = 0$, $a_1^2 + a_2^2 + \\dots + a_n^2 = 20$, and $-1 < a_i < 1$ for all $i$. What is the largest possible value of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that\n\n$$\na_1^2 + a_2^2 + \\dots + a_n^2 = 20 < n.\n$$\n\nSo $n > 20$, i.e., $n \\geq 21$. Suppose $n = 21$. Assume $a_1 \\leq a_2 \\leq \\dots \\leq a_{21}$, so $a_1 \\leq 0 \\leq a_{21}$. Since $a_i \\neq 0$, there exists $1 \\leq k < 21$ such that\n\n$$\n-1 < a_1 \\leq \\dots \\leq a_k < 0 < a_{k+1} \\leq \\dots \\leq a_{21} < 1.\n$$\n\nFor $k+1 \\leq i \\leq 21$, $0 < a_i < 1$ so $a_i^2 < a_i$. Thus,\n\n$$\n\\begin{aligned}\n20 &= a_1^2 + \\dots + a_{21}^2 \\\\\n&= (a_1^2 + \\dots + a_k^2) + (a_{k+1}^2 + \\dots + a_{21}^2) \\\\\n&< (a_1^2 + \\dots + a_k^2) + (a_{k+1} + \\dots + a_{21}) \\\\\n&= (a_1^2 + \\dots + a_k^2) + (-(a_1 + \\dots + a_k)) \\\\\n&< 2k \\leq 20.\n\\end{aligned}\n$$\n\nThis is a contradiction, so $n \\geq 22$. For $n = 22$, let\n\n$$\na_i = \\sqrt{\\frac{11}{10}} \\quad (1 \\leq i \\leq 11), \\quad a_i = -\\sqrt{\\frac{11}{10}} \\quad (12 \\leq i \\leq 22).\n$$\n\nThese values satisfy all conditions, so the answer is $\\boxed{22}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15971,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute triangle $ABC$ with $AB > AC$, let $M$ be the midpoint of side $BC$. The exterior angle bisector of $\\angle BAC$ meets ray $BC$ at $P$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p192_data_02b1061561.png)\n\nPoints $K$ and $F$ lie on the line $PA$ such that $MF \\perp BC$ and $MK \\perp PA$. Prove that $$BC^2 = 4PF \\cdot AK.$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\omega$ and $O$ denote the circumcircle and circumcenter of triangle $ABC$, respectively, and let $N$ be the midpoint of arc $\\widehat{BC}$ (not containing $A$). Note that line $MN$ is the perpendicular bisector of segment $BC$. In particular, both $F$ and $O$ lie on line $MN$. Note also that ray $AN$ is the interior bisector of $\\angle BAC$, implying that $NA \\perp FP$ or $\\angle NAF = 90^\\circ$. It follows that $NF$ is a diameter of $\\omega$. It is clear that $MK \\parallel AN$, from which it follows that\n\n$$\n\\frac{AK}{MN} = \\frac{FK}{FM}\n$$\n\nHence,\n\n$$\nPK \\cdot AK = \\frac{PK \\cdot FK \\cdot MN}{FM}.\n$$\n\nNote that $MK$ is the altitude to the hypotenuse in right triangle $FMP$, implying that triangles $FMK$ and $FPM$ are similar to each other and $PF \\cdot FK = FM^2$. Combining the last two equalities yields\n\n$$\nPF \\cdot AK = \\frac{PF \\cdot FK \\cdot MN}{FM} = FM \\cdot MN.\n$$\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p193_data_9e28c98705.png)\n\nBy the power-of-a-point theorem (or cross-chord theorem), we have $FM \\cdot MN = BM \\cdot MC = \\frac{BC^2}{4}$. Combining the last two equalities gives the desired result. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15972,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a prime number $p > 2$. There are $N$ people in a circle, each one comes up with a positive integer and writes down the remainder when divided by $p$ on a piece of paper. Then everyone shows their piece of paper to the neighbor on the right, computes the product of their number and their neighbor's number, and writes down a second number: the remainder of this product divided by $p$. What is the largest value that $N$ could be, if it is known that everyone's first numbers are pairwise distinct, and each piece of paper has two distinct numbers written down?",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, there cannot be more than $p$ different numbers. From the problem's conditions, $0$ cannot be anyone's first number, since then that person would have it as their second number as well. Also, no one can have $1$ as a first number, since then the neighbor on the left would have two equal numbers. We will show that the conditions hold for $N = p-2$.\n\nIndeed, let the following consecutive numbers be written down on the first step from left to right: $2, 3, 4, \\ldots, p-1$. They are all pairwise distinct. Assume that after the second step someone's two numbers are equal. If it's one of the first $N-1$ people, it means that the product $k(k+1)$, where $2 \\leq k \\leq p-2$, has the same remainder modulo $p$ as the number $k$. But then their difference is divisible by $p$, that is, $k(k+1) - k = k^2$ is divisible by the prime $p$, so $k$ is divisible by $p$, which is not possible. If the $N$-th person has two equal numbers, then the number $p-1$ will be multiplied by $2$, but then the two numbers will be different, since $2(p-1) - (p-1) = p-1$ is not divisible by $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15973,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are 2013 people, and among every group of three, the number of mutual acquaintances is odd. Prove that there exists a group of at least 1007 people such that every pair among them are acquainted.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there exist two people, $A$ and $B$, who do not know each other. Let $C$ be any other person. Since the number of acquaintances among $A$, $B$, $C$ is odd and $A$ doesn't know $B$, it follows that $C$ knows exactly one of $A$ or $B$. Thus, the other 2011 people are partitioned into two groups: those who know $A$ and those who know $B$. By the pigeonhole principle, one group has at least 1006 people. Without loss of generality, suppose $A$'s acquaintances form a group of at least 1006. Let $D$ and $E$ be two of $A$'s acquaintances. Since the number of acquaintances among $A$, $D$, $E$ is odd and $A$ knows both $D$ and $E$, it follows that $D$ knows $E$. Thus, all of $A$'s acquaintances know each other, so $A$ and all of $A$'s acquaintances form a clique of at least 1007 people.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15974,
    "subject": "Mathematics (Olympiad)",
    "question": "A graph is _good_ if its edges can be colored with 2 colors so that no cycle has two consecutive edges of the same color. What is the maximum number of edges in a _good_ graph with 1000 vertices?",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove the answer to be $4 \\cdot 333 = 1332$.\n\nFirst, we prove that a _good_ graph with $n$ vertices has at most $\\frac{4(n-1)}{3}$ edges.\n\n**Claim 1.** If we have 3 paths going from vertex $A$ to vertex $B$, then 2 of them have a common vertex different from $A$, $B$.\n\n_Proof._ Suppose not. By the Pigeonhole principle, observe that 2 of them have the same color attributed to $A$'s edge in them. But if they have no other common points than $A$ and $B$, they form a cycle where the 2 edges next to $A$ have the same color, contradiction. $\\Box$\n\n**Claim 2.** No edge lies in more than 1 cycle.\n\n_Proof._ Suppose edge $XY$ lies in cycle $C_1$ and cycle $C_2$. If $C_1$ and $C_2$ have no common vertices except $X$ and $Y$, observe that we can get from $X$ to $Y$ from 3 disjoint paths: on $C_1$, on $C_2$, and directly on edge $XY$, contradicting Claim 1.\n\nThus, $C_1$ and $C_2$ have other common vertices. We can walk on $C_2$ going in the direction from $Y$ to $X$ and suppose $V$ is the first such vertex we encounter. Then we have 3 disjoint paths from $X$ to $V$: this path we just walked from $X$ to $V$ on $C_2$, and the 2 paths given by $C_1$, and all are disjoint by the way we chose $V$, again contradicting Claim 1. $\\Box$\n\n![](images/2025-SL-a_p18_data_97c48346fc.png)\n\nConsider a connected and _good_ graph with $m$ vertices and $p$ edges. It has a spanning tree containing $m-1$ edges, and every other edge then lies in a cycle where all other edges are edges of the tree, so no 2 such cycles coincide.\n\nHowever, by Claim 2, no two cycles share an edge, and from the statement, all have even lengths, thus at least 4. Furthermore, we have $p+1-m$ such cycles, so we have at least $4(p+1-m)$ edges. This means $p \\ge 4(p+1-m)$, which gives $p \\le \\frac{4(m-1)}{3}$.\n\nNow that we've proven the result for connected graphs, all we need to do for non-connected ones is to just apply it for all connected components and sum it up, yielding the result.\n\nWe are left with providing the example for 1000 vertices:\n\nWe have a vertex $V$ and 333 triplets $(A_i, B_i, C_i)$ with edges $VA_i$, $A_iB_i$, $B_iC_i$, $C_iV$, which can be colored alternatively. This is easily seen to work as these are the only cycles.\n\n**Comment.** Using the same construction and eventually adding 1 or 2 vertices with degree 1, we can show that $\\left\\lfloor \\frac{4(n-1)}{3} \\right\\rfloor$ is the maximum number of edges for any $n$. We can also easily see that a graph is _good_ if and only if it's bipartite and each edge lies in at most one cycle.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 15975,
    "subject": "Mathematics (Olympiad)",
    "question": "If $a$, $b$, $c$ are positive real numbers with sum $6$, determine the maximal value of the expression:\n\n$$\nS = \\sqrt[3]{a^2 + 2bc} + \\sqrt[3]{b^2 + 2ca} + \\sqrt[3]{c^2 + 2ab}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We use the inequality of arithmetic–geometric mean as follows:\n\n$$\n\\sqrt[3]{a^2+2bc} = \\frac{1}{\\sqrt[3]{12^2}} \\sqrt[3]{(a^2+2bc) \\cdot 12 \\cdot 12} \\le \\frac{1}{\\sqrt[3]{12^2}} \\cdot \\frac{a^2+2bc+12+12}{3} = \\frac{1}{3\\sqrt[3]{12^2}} (a^2+2bc+24)\n$$\n\n$$\n\\sqrt[3]{b^2+2ca} = \\frac{1}{\\sqrt[3]{12^2}} \\sqrt[3]{(b^2+2ca) \\cdot 12 \\cdot 12} \\le \\frac{1}{\\sqrt[3]{12^2}} \\cdot \\frac{b^2+2ca+12+12}{3} = \\frac{1}{3\\sqrt[3]{12^2}} (b^2+2ca+24)\n$$\n\n$$\n\\sqrt[3]{c^2+2ab} = \\frac{1}{\\sqrt[3]{12^2}} \\sqrt[3]{(c^2+2ab) \\cdot 12 \\cdot 12} \\le \\frac{1}{\\sqrt[3]{12^2}} \\cdot \\frac{c^2+2ab+12+12}{3} = \\frac{1}{3\\sqrt[3]{12^2}} (c^2+2ab+24)\n$$\n\nSumming, we get:\n\n$$\n\\begin{aligned}\nS &= \\sqrt[3]{a^2+2bc} + \\sqrt[3]{b^2+2ca} + \\sqrt[3]{c^2+2ab} \\\\\n&\\le \\frac{1}{3\\sqrt[3]{12^2}} (a^2+b^2+c^2+2ab+2bc+2ca+72) \\\\\n&= \\frac{1}{3\\sqrt[3]{12^2}} \\left[ (a+b+c)^2 + 72 \\right] = \\frac{36}{\\sqrt[3]{12^2}} = \\frac{18}{\\sqrt[3]{18}} = 3\\sqrt[3]{12}.\n\\end{aligned}\n$$\n\nEquality holds when\n\n$$\na^2 + 2bc = 12, \\quad b^2 + 2ca = 12, \\quad c^2 + 2ab = 12\n$$\n\nThis leads to $a = b = c = 2$.\n\nTherefore, the maximal value of the expression is $3\\sqrt[3]{12}$, attained for $a = b = c = 2$.\n\n*Comments:*\n\nThe choice of $12$ in the application of the arithmetic–geometric mean is necessary for equality in all partial inequalities, allowing the maximum value to be achieved. If we used $1$ instead:\n\n$$\n\\sqrt[3]{a^2+2bc} \\le \\frac{a^2+2bc+2}{3}\n$$\n$$\n\\sqrt[3]{b^2+2ca} \\le \\frac{b^2+2ca+2}{3}\n$$\n$$\n\\sqrt[3]{c^2+2ab} \\le \\frac{c^2+2ab+2}{3}\n$$\n\nSumming gives:\n\n$$\n\\begin{aligned}\nS &\\le \\frac{(a+b+c)^2 + 6}{3} = \\frac{42}{3} = 14.\n\\end{aligned}\n$$\n\nBut equality is not possible, since $a^2 + 2bc = 1$ for all variables would imply $(a+b+c)^2 = 3$, which is absurd.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15976,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a polygon with 2016 vertices. Alisa and Basilio play the following game. On each turn, a player draws a diagonal of the polygon, which may intersect other drawn diagonals or the sides only at the vertices. When the polygon is cut into triangles, the game is finished. For each triangle having exactly zero sides among the sides of the initial polygon, Alisa is paid 1 cent. For each triangle having exactly two sides among the sides of the initial polygon, Basilio is paid 1 cent. Who will get more money and what would be the difference if both are clever players?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a$ be the number of triangles with 0 sides among the sides of the initial polygon, $b$ the number with 1 such side, and $c$ the number with 2 such sides. The polygon has 2016 sides, so $b + 2c = 2016$. Since triangulating an $n$-gon yields $n-2$ triangles, we have $2014 = a + b + c$. Solving these equations gives $c = a + 2$. Thus, Basilio will get 2 cents more than Alisa, regardless of how they play.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15977,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $N$ 表示所有正整數所成的集合。試求所有一對一函數 $f: N \\to N$ 使得\n\n$$\nf^{f(a)}(b) f^{f(b)}(a) = (f(a+b))^2\n$$\n\n成立，對所有的正整數 $a, b$。此處 $f^k(n)$ 表示 $\\underbrace{f(f(\\dots f(n)\\dots))}_{k}$。",
    "options": [],
    "answer": "See solution",
    "solution": "答：$f(n) = n+1$，對所有的正整數 $n$。\n\n令 $f$ 為滿足條件的函數。\n\n*第一步.* $1$ 沒有前像。\n\n假設存在整數 $x$ 使得 $f(x) = 1$。令 $a = b = x$ 代入 (1)，則\n\n$$\nf(2x)^2 = f^{f(x)}(x) f^{f(x)}(x) = f(x)^2 = 1\n$$\n\n這意味著 $f(2x) = 1$。但這不可能，因為 $f$ 是一對一函數。\n\n*第二步.* $f^{f(n)-1}(n) = 2n$，對所有 $n \\in N$。且 $f(1) = 2$。\n\n令 $a = b = n$ 代入 (1) 並利用 $f$ 的單射性：\n\n$$\nf^{f(n)}(n)^2 = f(2n)^2 \\rightarrow f^{f(n)}(n) = f(2n) \\rightarrow f^{f(n)-1}(n) = 2n\n$$\n\n特別地，存在整數 $c$ 使得 $f(c) = 2$。則\n\n$$\n2c = f^{f(c)-1}(c) = f(c) = 2\n$$\n\n所以 $c=1$。\n\n*第三步.* $5$ 有前像。\n\n由前述結果，偶數都有前像。必有整數 $d$ 使得 $f(d) = 4$。由 (1)，\n\n$$\n2d = f^{f(d)-1}(d) = f^3(d) = f^2(4)\n$$\n\n將 $a = 1, b = 4$ 代入題設：\n\n$$\nf^2(4) f^{f(4)}(1) = f(5)^2\n$$\n\n結合題設與 (1)，可知 $f(5)$ 是偶數，設為 $2e$。再由 (1)：\n\n$$\nf(5) = 2e = f^{f(e)-1}(e)\n$$\n\n顯然 $e \\neq 1$，所以 $f(e) > 2$ 且 $5$ 在 $f$ 的像中。\n\n*第四步.* $f(2) \\neq 5$\n\n假設 $f(2) = 5$。令 $n = 2$ 代入題設：\n\n$$\n4 = f^{f(2)-1}(2) = f^4(2) = f^5(1)\n$$\n\n題設中，取 $a = 1, b = 2$，有\n\n$$\nf(3)^2 = f^2(2) f^5(1) = 4f(5) = 8e\n$$\n\n同理可知 $3$ 也在 $f$ 的像中。設 $g$ 使得 $f(g) = 3$，則 $g \\ge 4$。題設中，取 $a = 1, b = g-1$：\n\n$$\nf^2(g-1) f^{f(g-1)}(1) = f(g)^2 = 9\n$$\n\n因 $1$ 無前像，$f^2(g-1)$ 和 $f^{f(g-1)}(1)$ 都是 $3$。注意\n\n$$\nf^2(g-1) = 3 = f(g) \\rightarrow f(g-1) = g\n$$\n\n進一步，題設中取 $a = 2, b = g-2$\n\n$$\nf^5(g-2) f^{f(g-2)}(2) = f(g)^2 = 9\n$$\n\n由第一步結果，得\n\n$$\nf^g(1) = f^{f(g-1)}(1) = 3 = f^{f(g-2)}(2) = f^{f(g-2)+1}(1)\n$$\n\n因此 $f(g-2) = g-1$。又由 (3)\n\n$$\n3 = f^5(g-2) = f^2(3)\n$$\n\n因為\n\n$$\nf^{f(3)-1}(3) = 6\n$$\n\n唯一可能是 $f(3) = 6$，這意味著 $e = \\frac{9}{2}$，矛盾！\n\n*第五步.* $f(2) = 3, f(3) = 4$\n\n設 $h$ 為 $5$ 的前像，則由前兩步 $h \\ge 3$。題設中取 $a = 1, b = h-1$：\n\n$$\nf^2(h-1) f^{f(h-1)}(1) = f(h)^2 = 25\n$$\n\n因此 $f^2(h-1) = 5$ 且 $f(h-1) = h$。再取 $a = 2, b = h-2$：\n\n$$\nf^{f(2)}(h-2) f^{f(h-2)}(2) = 25\n$$\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15978,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all prime numbers $p$ for which $p^3 - 4p + 9$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "We want to find the solutions of the equation $x^2 = p^3 - 4p + 9$ where $p$ is a prime number and $x$ is a nonnegative integer.\n\nSince $x^2 \\equiv 9 \\pmod{p}$, we have $x = kp \\pm 3$ where $k$ is an integer. Then $$(kp \\pm 3)^2 = p^3 - 4p + 9$$ which implies $$k^2p \\pm 6k = p^2 - 4$$ and we obtain $p \\mid 6k \\pm 4$.\n\nIf $p \\neq 2$, then $p \\mid 3k \\pm 2$ which implies $p \\leq 3k + 2$ so $\\frac{p-2}{3} \\leq k$. Thus, $\\frac{p^2 - 2p - 9}{3} \\leq pk - 3 \\leq x$.\n\n* If $x \\leq \\frac{p^2}{4}$, then $\\frac{p^2 - 2p - 9}{3} \\leq \\frac{p^2}{4}$ which implies $p \\leq 8 + \\frac{36}{p}$ so $p \\leq 11$.\n\n* If, on the other hand, $x > \\frac{p^2}{4}$, then $x^2 = p^3 - 4p + 9 = \\frac{p^4}{16} < p^3 - 4p + 9$ which implies $p < 16 - \\frac{16(4p-9)}{p^3}$ so $p \\leq 13$.\n\nFinally, for $p \\leq 13$, $(p, x) = (2, 3)$, $(7, 18)$, and $(11, 36)$ are the only solutions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15979,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the triangle *ABC* be such that $2AC = AB$ and $\\angle A = 2\\angle B$. Let *AL* be its bisector and let *M* be the midpoint of *AB*. It turns out that $CL = ML$. Show that $\\angle B = 30^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since *AL* is a bisector, then $\\angle CAL = \\angle LAB = \\angle CBA$ (see figure below). Then $\\triangle ALB$ is isosceles, so *LM* is its altitude and a median. Thus $\\angle LMA = 90^\\circ$. Consider the triangles *AML* and *ALC*. Let *C'* be the projection of *L* on *AC*. Then right triangles *AML* and *AC'L* are equal by hypotenuse and the angle. Thus, $LC' = LM = LC$. Therefore, $C = C'$, since there exists only one projection.\n\n![](images/UkraineMO2019_booklet_p8_data_92ab26033c.png)\n\nThus $\\triangle ABC$ is a right triangle for which $2AC = AB$, hence $\\angle ABC = 30^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15980,
    "subject": "Mathematics (Olympiad)",
    "question": "Two lines intersect at a point and form four angles: two acute and two obtuse. The sum of the two acute angles is half of one obtuse angle. Calculate these angles.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the measure of each acute angle be $x$. Then each obtuse angle is $180^\\circ - x$. According to the problem, $2x = \\frac{1}{2}(180^\\circ - x)$. Solving:\n\n$$\n2x = \\frac{1}{2}(180^\\circ - x) \\\\\n4x = 180^\\circ - x \\\\\n5x = 180^\\circ \\\\\nx = 36^\\circ\n$$\n\nEach obtuse angle is $180^\\circ - 36^\\circ = 144^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15981,
    "subject": "Mathematics (Olympiad)",
    "question": "Paint the squares in the first row with 5 different colours, and the squares in each of the remaining rows with 4 new colours and one old colour. What is the maximum total number of colours that may be used?\n\n![](images/Hong_Kong_2015_Booklet_p16_data_d908f267df.png)",
    "options": [],
    "answer": "See solution",
    "solution": "If we paint the first row with 5 different colours, and each of the next 9 rows with 4 new colours and one old colour, the total number of colours used is:\n\n$$\n5 + 4 \\times 9 = 41\n$$\n\nThis can be achieved as shown in the table above, where different numbers indicate different colours.\n\nTo show that 41 is the maximum, suppose we try to use more. If any row after the first uses 5 new colours, then, after filling the first two rows, each column already contains 2 colours and can have at most 3 more, so the total is:\n\n$$\n5 + 5 + 3 \\times 10 = 40\n$$\n\nwhich is less than 41. Thus, the optimal way is to fill the first row with 5 colours and each subsequent row with 4 new and 1 old colour, for a maximum of 41 colours.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15982,
    "subject": "Mathematics (Olympiad)",
    "question": "Од средините на страните на остроаголниот триаголник $ABC$ се повлечени нормали кон соседните страни. Докажи дека шестоаголникот што тие го формираат има два пати помала плоштина од плоштината на $ABC$.\n\n![](images/Makedonija_2009_p23_data_405ceffad9.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Нека $O$ е центарот на опишаната кружница околу триаголникот $ABC$. Нека $A_1, B_1, C_1$ се средините на страните $BC, CA, AB$ соодветно. Точката $O$ ја поврзуваме со $A_1, B_1, C_1$ и добиваме три паралелограми $C_1OB_1K$, $A_1OC_1L$ и $B_1OA_1M$, чијашто вкупна плоштина е еднаква на плоштината на шестоаголникот. А од друга страна, нивната вкупна плоштина е двојно поголема од плоштината на $\\Delta A_1B_1C_1$. Сега, бидејќи важи $P_{\\Delta A_1B_1C_1} = \\frac{1}{4} P_{\\Delta ABC}$, следува бараниот резултат.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15983,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p, q$ be prime numbers and $a$ be an integer such that $p > 2$ and $a \\not\\equiv 1 \\pmod{q}$ but $a^p \\equiv 1 \\pmod{q}$. Prove that\n\n$$\n(1+a^1)(1+a^2)\\dots(1+a^{p-1}) \\equiv 1 \\pmod{q}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "As $a^p \\equiv 1 \\pmod{q}$ while $a \\not\\equiv 1 \\pmod{q}$, the case $q = 2$ is impossible. Thus, the desired equation is equivalent to\n\n$$\n(1 + a^0)(1 + a^1)(1 + a^2) \\dots (1 + a^{p-1}) \\equiv 2 \\pmod{q}.\n$$\n\nExpanding the left-hand side gives all monomials of the form $a^{i_1 + \\dots + i_k}$ where $\\{i_1, \\dots, i_k\\} \\subseteq \\{0, 1, \\dots, p-1\\}$. Since $a^p \\equiv 1 \\pmod{q}$, the exponents can be reduced modulo $p$. Each possible remainder modulo $p$ is produced the same number of times (excluding the empty set and the full set). For each tuple $(i_1, \\dots, i_k)$ with $0 < k < p$, there exists $l$ such that $kl \\equiv 1 \\pmod{p}$, and shifting indices cycles through all remainders equally. Thus, for any remainder $i$, there are equally many tuples with sum congruent to $i$ modulo $p$.\n\nLet this constant number of tuples be $c$. Then,\n\n$$\na^0 + a^{0+1+\\dots+(p-1)} + c(a^0 + a^1 + \\dots + a^{p-1}) \\equiv 2 \\pmod{q}.\n$$\n\nSince $0 + 1 + \\dots + (p-1) = \\frac{(p-1)p}{2}$, and $a^p \\equiv 1 \\pmod{q}$, we have $a^{0+1+\\dots+(p-1)} \\equiv 1 \\pmod{q}$. Also, $(a^0 + a^1 + \\dots + a^{p-1})(a-1) = a^p - 1 \\equiv 0 \\pmod{q}$, and since $a-1 \\not\\equiv 0 \\pmod{q}$, it follows that $a^0 + a^1 + \\dots + a^{p-1} \\equiv 0 \\pmod{q}$. Therefore, the equation reduces to $1 + 1 \\equiv 2 \\pmod{q}$, which holds.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15984,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral and $M$ be the intersection of its diagonals. Through $M$, draw a line meeting the side $AB$ at $P$ and the side $CD$ at $Q$. Find all quadrilaterals such that there exists a segment $PQ$ that divides the triangles $ABM$ and $CDM$ into 4 similar triangles.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $\\angle APM > 90^\\circ$. Then in $\\triangle BPM$, $\\angle BPM < 90^\\circ$. Since $\\angle PBM + \\angle BMP = \\angle APM$, $\\angle PBM, \\angle BMP < \\angle APM$. So none of the angles of $\\triangle BPM$ can be equal to $\\angle APM$ of $\\triangle APM$. Therefore, $\\triangle APM$ cannot be similar to $\\triangle BPM$. Thus, $\\angle APM = 90^\\circ$ and $PQ \\perp AB$. Similarly, $PQ \\perp CD$ and it follows that $AB \\parallel DC$.\n\n![](images/combined_25__latex_1_p2_data_f130492134.png)\n\nIt then follows that $\\triangle APM \\sim \\triangle CQM$ and $\\triangle BMP \\sim \\triangle DQM$.\n\nIf $\\triangle APM \\sim \\triangle BPM$, then $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$. Therefore, $ABCD$ is an isosceles trapezium.\n\nIf $\\triangle APM \\sim \\triangle MPB$, then $\\angle MAP = \\angle BMP$ and $\\angle PMA = \\angle PBM$. It follows that $\\angle AMB = 90^\\circ$. So the diagonals are perpendicular. The quadrilaterals are either isosceles trapezia or trapezia in which the diagonals are perpendicular.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15985,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer, and let $x_1, x_2, \\dots, x_n$ be distinct positive integers with $x_1 = 1$. Construct an $n \\times 3$ table where the entries of the $k$-th row are $x_k$, $2x_k$, $3x_k$ for $k = 1, 2, \\dots, n$.\n\nNow follow a procedure where, in each step, two identical entries are removed from the table. This continues until there are no more identical entries in the table.\n\n(a) Prove that at least three entries remain at the end of the procedure.\n\n(b) Prove that there are infinitely many possible triples that can remain at the end of the procedure for suitable choices of $n$ and $x_1, x_2, \\dots, x_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Let $X = \\{x_1, x_2, \\dots, x_n\\}$. The number $1$ appears only in the first row, so it remains at the end. Let $x_k$ be the largest power of $2$ in $X$ (possible since $x_1 = 1$). The number $2x_k$ is a power of $2$ and does not appear elsewhere in the table: it is not any $x_j$ (by choice of $x_k$), not any other $2x_j$ (since $x_j$ are distinct), and not any $3x_j$ (since those are multiples of $3$). Thus, $2x_k$ remains. Similarly, let $x_\\ell$ be the largest power of $3$ in $X$; then $3x_\\ell$ remains. Therefore, at least three numbers remain: $1$, $2x_k$, and $3x_\\ell$.\n\n(b) Define sets $X_0 = \\{1\\}$ and recursively\n$$\nX_{r+1} = X_r \\cup \\{2^{2^r}x : x \\in X_r\\} \\cup \\{3^{2^r}x : x \\in X_r\\}\n$$\nfor $r \\ge 0$. That is, $X_{r+1}$ consists of $X_r$, all elements of $X_r$ multiplied by $2^{2^r}$, and all elements of $X_r$ multiplied by $3^{2^r}$. We claim that if $x_1, \\dots, x_n$ are the elements of $X_r$, the final table consists of $1$, $2^{2^r}$, and $3^{2^r}$.\n\nBase case $r=0$: $X_0 = \\{1\\}$, so the table is $1, 2, 3$.\n\nInductive step: Assume true for $X_r$. For $X_{r+1}$, the rows from $X_r$ reduce to $1, 2^{2^r}, 3^{2^r}$. The rows from $\\{2^{2^r}x : x \\in X_r\\}$ reduce to $2^{2^r}$, $2^{2^{r+1}}$, $6^{2^r}$; those from $\\{3^{2^r}x : x \\in X_r\\}$ reduce to $3^{2^r}$, $6^{2^r}$, $3^{2^{r+1}}$. After further reductions, only $1$, $2^{2^{r+1}}$, and $3^{2^{r+1}}$ remain. Thus, infinitely many such triples are possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15986,
    "subject": "Mathematics (Olympiad)",
    "question": "假設你有以下兩種由四個正方形組成的骨牌，左邊兩個稱為甲型骨牌，右邊兩個稱為乙型骨牌。\n\n![](images/15-2J_p9_data_83a1a2e781.png)\n\n![](images/15-2J_p9_data_859a9e9c20.png)\n\n![](images/15-2J_p9_data_80792d6072.png)\n\n![](images/15-2J_p9_data_31ec0ebd00.png)\n\n現在有一個無限大的白色棋盤，棋盤上每個格子的大小和骨牌的正方形一樣大。我們從棋盤上選一些格子塗黑。已知我們可以只用甲型骨牌蓋住所有黑色格子且不蓋到任何白色格子。證明：如果我們改用甲和（或）乙型骨牌來蓋住所有黑色格子，且不蓋到任何白色格子，則我們必然使用了偶數個乙型骨牌。\n\nSuppose you have two types of tetrominoes:\n\n![](images/15-2J_p9_data_8d8d09703a.png)\n\n![](images/15-2J_p9_data_3e81363cab.png)\n\n![](images/15-2J_p9_data_9a918f998d.png)\n\n![](images/15-2J_p9_data_31ec0ebd00.png)\n\nthe two on the left are called “S-type”, and the two on the right are called “Z-type”. Consider a white chess board of infinite size, on which we pick several squares and paint them black. Assume that we can use only S-type tetrominoes to cover all black squares without covering any white one. Prove that, if we use S and (or) Z-type tetrominoes to cover all black squares instead, without covering any white one, then we must use an even number of Z-type tetrominoes.",
    "options": [],
    "answer": "See solution",
    "solution": "令 $B$ 為黑色格子所成集合。將無限大棋盤用以下方式塗成紅綠兩色：注意到甲型骨牌必然蓋住偶數個紅色格子，而乙型骨牌必然蓋住奇數個紅色格子。\n\n![](images/15-2J_p10_data_b0c47400d7.png)\n\n由於 $B$ 可以由甲型骨牌組成，因此 $B$ 必然包含偶數個紅色格子。從而如果我們改用甲和（或）乙型骨牌，乙型骨牌的數量必然為偶數。證畢。",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15987,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n > 0$ and real numbers $x_1 \\le x_2 \\le \\dots \\le x_n$, $y_1 \\ge y_2 \\ge \\dots \\ge y_n$, satisfying $\\sum_{i=1}^n i x_i = \\sum_{i=1}^n i y_i$. Prove that for any real number $\\alpha$, $$\\sum_{i=1}^n x_i [i\\alpha] \\ge \\sum_{i=1}^n y_i [i\\alpha],$$ where $[\\beta]$ is defined as the greatest integer less than or equal to $\\beta$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof I** We need the following lemma.\n\nLemma: For any real number $\\alpha$ and positive integer $n$, we have\n\n$$\n\\sum_{i=1}^{n-1} [i\\alpha] \\le \\frac{n-1}{2} [n\\alpha].\n$$\n\nThe lemma is obtained by summing the inequalities\n\n$$\n[i\\alpha] + [(n-i)\\alpha] \\le [n\\alpha]\n$$\n\nfor $i = 1, 2, \\dots, n-1$.\n\nReturn to the original problem. We will prove it by induction. For $n = 1$, it is obviously true.\n\nAssume that for $n = k$, it is also true. Now consider $n = k + 1$. Let $a_i = x_i + \\frac{2}{k} x_{i+1}$, $b_i = y_i + \\frac{2}{k} y_{i+1}$ for $i = 1, 2, \\dots, k$. Then we have $a_1 \\le a_2 \\le \\dots \\le a_k$, $b_1 \\ge b_2 \\ge \\dots \\ge b_k$ and $\\sum_{i=1}^k i a_i = \\sum_{i=1}^k i b_i$. By induction, we get $\\sum_{i=1}^k a_i [i\\alpha] \\ge \\sum_{i=1}^k b_i [i\\alpha]$.\n\nIn addition, $x_{k+1} \\ge y_{k+1}$. Otherwise, if $x_{k+1} < y_{k+1}$, we have\n\n$$\nx_1 \\le x_2 \\le \\dots \\le x_{k+1} < y_{k+1} \\le \\dots \\le y_2 \\le y_1.\n$$\n\nThis contradicts $\\sum_{i=1}^{k+1} i x_i = \\sum_{i=1}^{k+1} i y_i$. So we have\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{k+1} x_i [i\\alpha] - \\sum_{i=1}^{k} a_i [i\\alpha] &= x_{k+1} \\left( [ (k+1)\\alpha ] - \\frac{2}{k} \\sum_{i=1}^{k} [i\\alpha] \\right) \\\\\n&\\ge y_{k+1} \\left( [ (k+1)\\alpha ] - \\frac{2}{k} \\sum_{i=1}^{k} [i\\alpha] \\right) \\\\\n&= \\sum_{i=1}^{k+1} y_i [i\\alpha] - \\sum_{i=1}^{k} b_i [i\\alpha].\n\\end{aligned}\n$$\n\nThat means\n\n$$\n\\sum_{i=1}^{k+1} x_i [i\\alpha] \\ge \\sum_{i=1}^{k+1} y_i [i\\alpha].\n$$\n\nBy induction, we complete the proof for any integer $n > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15988,
    "subject": "Mathematics (Olympiad)",
    "question": "Encontrar la solución entera más pequeña de la ecuación\n\n$$\n\\left\\lfloor \\frac{x}{8} \\right\\rfloor - \\left\\lfloor \\frac{x}{40} \\right\\rfloor + \\left\\lfloor \\frac{x}{240} \\right\\rfloor = 210.\n$$\n\n(Si $x$ es un número real, $\\lfloor x \\rfloor$ es la parte entera de $x$, esto es, el mayor número entero menor o igual que $x$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Sea $x$ una solución entera de la ecuación. Dividiendo, primero por $240$, luego el resto $r_1$ por $40$ y el nuevo resto $r_2$ por $8$, resulta\n\n$$\nx = 240c_1 + r_1 = 240c_1 + 40c_2 + r_2 = 240c_1 + 40c_2 + 8c_3 + r_3,\n$$\n\ndonde $0 \\leq c_2 < 6$, $0 \\leq c_3 < 5$ y $0 \\leq r_3 < 8$ (las desigualdades para $c_2$ y $c_3$ se obtienen de $0 \\leq r_1 < 240$ y $0 \\leq r_2 < 40$). Entonces\n\n$$\n210 = \\left\\lfloor \\frac{x}{8} \\right\\rfloor - \\left\\lfloor \\frac{x}{40} \\right\\rfloor + \\left\\lfloor \\frac{x}{240} \\right\\rfloor = (30c_1 + 5c_2 + c_3) - (6c_1 + c_2) + c_1 = 25c_1 + 4c_2 + c_3,\n$$\n\ny el único caso posible es $c_1 = 8$, $c_2 = 2$, $c_3 = 2$ (reduciendo módulo $5$ queda $c_2 \\equiv c_3 \\pmod{5}$ y necesariamente $c_2 = c_3$; entonces $42 = 5c_1 + c_2$ y tiene que ser $c_2 = 2$, $c_1 = 8$), con lo que\n\n$$\nx = 240 \\times 8 + 40 \\times 2 + 8 \\times 2 + r_3 = 2016 + r_3\n$$\n\ny la menor solución entera de la ecuación dada es $x = 2016$ (las otras son $2017, \\ldots, 2023$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15989,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with a right angle at $C$, let $I$ be its incentre, and let $H$ be the orthogonal projection of $C$ on $AB$. The incircle $\\omega$ of the triangle $ABC$ is tangent to the sides $BC$, $CA$, and $AB$ at $A_1$, $B_1$, and $C_1$, respectively. Let $E$ and $F$ be the reflections of $C$ in the lines $C_1A_1$ and $C_1B_1$, respectively, and let $K$ and $L$ be the reflections of $H$ in the same lines. Prove that the circles $A_1EI$, $B_1FI$, and $C_1KL$ have a common point.\n\n![](images/RMC_2020_p58_data_396b8920e1.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The line $C_1A_1$ is parallel to the external angle bisector of $\\angle B$, so the reflection in $C_1A_1$ maps the segment $A_1C$ to the segment $A_1E$ parallel to $AB$.\n\nSimilarly, $B_1F \\parallel AB$. Notice also that $A_1E = A_1C = B_1C = B_1F = r$, where $r$ is the inradius of $\\triangle ABC$.\n\nLet $M$ be the midpoint of $AB$. Let $X$ be the point of $\\omega$ such that $\\overrightarrow{IX} \\nearrow \\overrightarrow{CM}$. Notice that $\\angle EA_1I = 90^\\circ + \\angle EA_1B = 90^\\circ + \\angle CBM = 90^\\circ + \\angle BCM = \\angle A_1IX$, and $A_1E = IA_1 = IX$; thus, $XIA_1E$ is an isosceles trapezoid. Hence $X$ lies on the circle $IA_1E$, and $EX \\parallel A_1I$. Similarly, $X$ lies on the circle $IB_1F$, and $FX \\parallel B_1I$. It remains to show that $X$ lies on the circle $C_1KL$.\n\nUnder the symmetry in $C_1A_1$, the line $CH$ (perpendicular to $AB$) maps to the line through $E$ perpendicular to $BC$ — i.e., $CH$ maps to $EX$. Therefore, the projection $H$ of $C_1$ onto $CH$ maps to the projection $K$ of $C_1$ onto $EX$. Similarly, $L$ is the projection of $C_1$ onto $FX$. So the quadrilateral $C_1KXL$ is cyclic, due to right angles at $K$ and $L$.\n\n**REMARKS.** (1) In fact, the quadrilateral $C_1KXL$ is a square, since $C_1K = C_1H = C_1L$ and $\\angle KC_1L = 2\\angle A_1C_1B_1 = 90^\\circ$.\n\n(2) One can easily see that the points $C_1$ and $X$ are symmetric in the angle bisector $CI$. This yields that $K$ and $L$ both lie on $CI$. One can show that this conclusion in fact holds in any, not necessarily right-angled, triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15990,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are $m$ questions in an examination attended by $n$ students, where $m, n \\ge 2$ are given natural numbers. The marking rule for each question is as follows: if there are exactly $x$ students failing to answer the question correctly, then they will each get 0 marks, and those who answer it correctly will each get $x$ marks. The total marks of a student are the sum of marks he/she gets from the $m$ questions. Now rank the total marks of the $n$ students as $p_1 \\ge p_2 \\ge \\cdots \\ge p_n$. Find the maximum possible value of $p_1 + p_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "For any $k = 1, 2, \\dots, m$, let $x_k$ be the number of students failing to answer the $k$th question correctly. Then $n - x_k$ students answer it correctly and each gets $x_k$ marks for that question. Let $S$ be the sum of all students' total marks:\n\n$$\n\\sum_{i=1}^{n} p_i = S = \\sum_{k=1}^{m} x_k (n - x_k) = n \\sum_{k=1}^{m} x_k - \\sum_{k=1}^{m} x_k^2.\n$$\n\nEach student gets at most $x_k$ marks from the $k$th question, so\n\n$$\np_1 \\le \\sum_{k=1}^{m} x_k.\n$$\n\nSince $p_2 \\ge \\cdots \\ge p_n$, we have\n\n$$\np_n \\le \\frac{S - p_1}{n-1}.\n$$\n\nTherefore,\n\n$$\n\\begin{align*}\np_1 + p_n &\\le p_1 + \\frac{S-p_1}{n-1} = \\frac{n-2}{n-1}p_1 + \\frac{S}{n-1} \\\\\n&\\le \\frac{n-2}{n-1} \\sum_{k=1}^{m} x_k + \\frac{1}{n-1} \\left( n \\sum_{k=1}^{m} x_k - \\sum_{k=1}^{m} x_k^2 \\right) \\\\\n&= 2 \\sum_{k=1}^{m} x_k - \\frac{1}{n-1} \\sum_{k=1}^{m} x_k^2.\n\\end{align*}\n$$\n\nBy the Cauchy-Schwarz inequality,\n\n$$\n\\sum_{k=1}^{m} x_k^2 \\ge \\frac{1}{m} \\left( \\sum_{k=1}^{m} x_k \\right)^2.\n$$\n\nSo,\n\n$$\n\\begin{align*}\np_1 + p_n &\\le 2 \\sum_{k=1}^{m} x_k - \\frac{1}{m(n-1)} \\left( \\sum_{k=1}^{m} x_k \\right)^2 \\\\\n&= -\\frac{1}{m(n-1)} \\left( \\sum_{k=1}^{m} x_k - m(n-1) \\right)^2 + m(n-1) \\\\\n&\\le m(n-1).\n\\end{align*}\n$$\n\nThis bound is achieved if one student answers all questions correctly and the other $n-1$ students answer none, so\n\n$$\np_1 + p_n = p_1 = \\sum_{k=1}^{m} (n-1) = m(n-1).\n$$\n\nTherefore, the maximum possible value of $p_1 + p_n$ is $m(n-1)$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15991,
    "subject": "Mathematics (Olympiad)",
    "question": "Учениците од две одделенија се договориле да играат фудбал. Во едно од одделенијата немало доволен број на играчи да состават екипа од 11 ученици, па тие се договориле учениците од двете одделенија да се \"измешаат\" меѓу себе и потоа да состават две екипи. Наставникот забележал дека од првото одделение машки се $\\frac{4}{13}$ од учениците, додека од второто одделение машки се $\\frac{5}{17}$ од учениците. Секое од одделенијата има не повеќе од 50 ученици. Кое од одделенијата има повеќе девојчиња? (Одговорот да се образложи)",
    "options": [],
    "answer": "See solution",
    "solution": "Нека бројот на ученици во првото одделение е $x$, а бројот на ученици во второто одделение е $y$.\n\nТогаш машки во првото одделение се $\\frac{4x}{13}$, додека во второто се $\\frac{5y}{17}$. Бидејќи $\\frac{4x}{13}$ мора да е природен број, мора $4x$ да се дели со 13, т.е. $x = 13$, $26$ или $39$. Соодветно бројот на машки е $4$, $8$, $12$.\n\nАналогно, $\\frac{5y}{17}$ мора да е природен, па мора $5y$ да се дели со 17, тогаш $y = 17$ или $34$. Соодветниот број на машки е $5$ или $10$.\n\nБидејќи $22 = 12 + 10$, имаме дека првото одделение брои ученици $39$ од кои $12$ машки, т.е. $27$ девојчиња. Второто одделение брои $34$ ученици од кои $10$ машки, т.е. $24$ девојчиња.\n\nЗначи, првото одделение има повеќе девојчиња.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15992,
    "subject": "Mathematics (Olympiad)",
    "question": "Initially, the blackboard contains two polynomials $x^3 - 3x^2 + 5$ and $x^2 - 4x$. If the polynomials $f(x)$ and $g(x)$ are written on the blackboard, it is permitted to write onto the board any polynomial of the form $f(x) \\pm g(x)$, $f(x)g(x)$, $f(g(x))$, or $cf(x)$, where $c$ may be any (not necessarily integer) constant. Is it possible after several such operations to write on the board a nonzero polynomial of the form $x^n - 1$?",
    "options": [],
    "answer": "See solution",
    "solution": "No.\n\nIf a polynomial $f(x)$ can be obtained in this way, then $f'(2) = 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 15993,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the inequality:\n\n$$\n2008^x + \\log_{2008}(x+1) > \\log_{x+2} 2008 + \\sin^{2007} x + \\cos^{2008} x.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* $x > 0$.\n\n*Solution.* It is clear that the domain of the expression is $x > 0$. We will show that this is the solution. When $x > 0$, $2008^x > 1$ and $\\log_{2008}(x+1) > 0$. Also, for these values, the logarithm base satisfies $0 < \\frac{x}{x+2} < 1$, and $\\sin^{2007}x + \\cos^{2008}x \\leq \\sin^2x + \\cos^2x = 1$. Thus, the left side is greater than 1, and the right side is less than or equal to 1. Therefore, the solution to the inequality is every $x > 0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 15994,
    "subject": "Mathematics (Olympiad)",
    "question": "The ring road is 25 km long. There is a police post at each kilometer mark (25 in total). Each post contains one police officer, each with a unique badge number from 1 to 25. Initially, the officers are placed at random positions. The management wants to rearrange them so that, starting from some post and moving clockwise, the officers' badge numbers follow the order $1, 2, \\ldots, 25$. The officers are transported along the ring road so that the total distance traveled by all officers is minimized. Prove that, under these conditions, at least one police officer remains at his original post.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume, for contradiction, that every police officer moves at least 1 km in some direction. Since there are 25 officers, at least 13 of them must move in the same direction. Each of these 13 officers moves at least 1 km. Now, let each officer in this group move 1 km less in that direction: if an officer moved $n$ km, he now moves $(n - 1)$ km. For officers moving in the opposite direction, increase their movement by 1 km: if an officer moved $m$ km, he now moves $(m + 1)$ km. The arrangement is still clockwise, but the total distance traveled is reduced by at least 1 km, contradicting the minimality of the original arrangement. Therefore, at least one officer must remain at his post.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15995,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute triangle with $|AB| < |AC|$ and incenter $I$. Let $D$ be the projection of $I$ onto $BC$. Let $H$ be the orthocenter of $\\triangle ABC$. Prove that if $\\angle IDH = \\angle CBA - \\angle ACB$ then $|AH| = 2 \\cdot |ID|$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H'$ be the reflection of $H$ in $BC$. It is well-known (and easy to prove) that $H'$ lies on the circumcircle of $\\triangle ABC$. Let $O$ be the circumcenter of $\\triangle ABC$. We have\n\n$$\n\\begin{align*}\n\\angle OH'A &= \\angle HAO = \\angle BAC - \\angle BAH - \\angle OAC \\\\\n&= \\angle BAC - 2(90^\\circ - \\angle CBA) = \\angle CBA - \\angle ACB \\\\\n&= \\angle IDH = \\angle H'HD = \\angle DH'A,\n\\end{align*}\n$$\n\nhence $O, D, H'$ are collinear. Also note that $\\angle HAO = \\angle H'HD$ implies that $AO \\parallel HD$.\n\nLet $M$ be the midpoint of $BC$. Let $E$ be the reflection of $D$ in $M$. We have\n\n$$\n\\angle MOE = \\angle DOM = \\angle OH'A = \\angle HAO.\n$$\n\nSince $OM \\parallel AH$, the above equality gives that $A, O, E$ are collinear.\n\nLet $D'$ be the reflection of $D$ in $I$. It is well-known (and easy to prove) that $D'$ lies on $AE$. Since $AH \\parallel OM$ and $AD' \\parallel HD$, quadrilateral $AHDD'$ is a parallelogram. Therefore $|AH| = |DD'| = 2 \\cdot |ID|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 15996,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, let $\\varphi(n)$ represent the number of positive integers not exceeding $n$ and relatively prime to $n$. Find all functions $f:\\mathbb{N}_+ \\to \\mathbb{N}_+$ satisfying that for any positive integers $m, n$ with $m \\ge n$,\n\n$$\nf(m\\varphi(n^3)) = f(m)\\varphi(n^3).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The desired functions have the form\n\n$$\nf(n) = \\begin{cases} c, & n = 1, \\\\ dn, & n \\ge 2, \\end{cases}\n$$\n\nwhere $c, d$ are positive integers. It is straightforward to check that the problem conditions are met.\n\nIn the governing equation, taking $n = 2$ leads to $f(4m) = 4f(m)$ for $m \\ge 2$. By iterations, it follows that\n\n$$\nf(4^k m) = 4^k f(m)\n$$\nfor all $m \\ge 2$.\n\n*Claim*: For positive integers $m, p$ with $m \\ge 2$, $f(pm) = p f(m)$.\n\n*Proof of claim*: Induct on $p$: when $p = 1$, the conclusion is trivial; assume $p \\ge 2$ and the conclusion is valid for all positive integers less than $p$. Now, if $p$ is composite, the conclusion is validated by the induction hypothesis. In the following, assume $p$ is a prime. Taking $n = p$ in the governing equation, we obtain\n\n$$\nf(p^2(p-1)m) = p^2(p-1)f(m)\n$$\nfor all $m \\ge p$. By the induction hypothesis,\n\n$$\n\\begin{aligned}\np^2(p-1)f(m) &= f(p^2(p-1)m) \\\\\n&= (p-1)f(p^2m),\n\\end{aligned}\n$$\n\nand thus\n\n$$\nf(p^2m) = p^2f(m)\n$$\nfor all $m \\ge p$.\n\nNext, take $n = p^2$ in the governing equation to get\n\n$$\nf(p^5(p-1)m) = p^5(p-1)f(m)\n$$\nfor $m \\ge p^2$. In a similar manner, it follows by induction that\n\n$$\nf(p^5(p-1)m) = (p-1)f(p^5m).\n$$\n\nHence,\n\n$$\nf(p^5m) = p^5f(m)\n$$\nfor $m \\ge p^2$.\n\nBased on the above argument, we take $k$ with $4^k \\ge p^2$: for any $m \\ge 2$,\n\n$$\n4^k p^4 f(p m) = 4^k f(p^5 m) = f(4^k p^5 m) = p^5 f(4^k m) = p^5 4^k f(m).\n$$\n\nTherefore, $f(p m) = p f(m)$, and the induction is completed.\n\nAccording to the claim, particularly for $m \\ge 2$,\n\n$$\n2f(m) = f(2m) = m f(2).\n$$\n\nTaking $m = 3$, we infer that $f(2)$ is even. So for $m \\ge 2$, $f(m) = d m$, where $d = \\frac{f(2)}{2}$ is an integer. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 15997,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the value of $$(2+4+6+\text{...}+198+200) - (1+3+5+\text{...}+197+199)$$?",
    "options": [],
    "answer": "See solution",
    "solution": "Rearranging, the value is $$(2-1) + (4-3) + (6-5) + \text{...} + (200-199)$$ which has 100 brackets and therefore totals $100 \\times 1 = 100$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15998,
    "subject": "Mathematics (Olympiad)",
    "question": "Cover a grid square $13 \\times 13$ with $2 \\times 2$ squares and L-shapes of three unit cells so that the number of L-shapes is least possible.\n\n![](images/3._NATIONAL_XXX_OMA_2013-checkpoint_p2_data_8fa935cfc2.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let a $(2k-1) \\times (2k-1)$ square board be covered as in the statement with $x$ squares $2 \\times 2$ and $y$ shapes L. Denote by $(i, j)$ the cell in row $i$, column $j$, and color black all cells $(i, j)$ with both $i$ and $j$ odd. Thus $k^2$ black cells are obtained. Observe that wherever a $2 \\times 2$ square is placed, it covers exactly one black cell; and wherever an L-shape is placed, it covers at most one black cell. To have the whole board covered it is necessary that the total number of figures be at least $k^2$, i.e. $x + y \\geq k^2$. All figures cover $4x + 3y$ cells, which equals $(2k-1)^2$, the total number of cells on the board. On the other hand, $x \\geq k^2 - y$ implies $4x + 3y \\geq 4(k^2 - y) + 3y = 4k^2 - y$. Hence $4k^2 - y \\leq (2k-1)^2$, yielding $y \\geq 4k - 1$. In summary, each admissible covering has at least $4k - 1$ shapes L and at most $k^2 - 4k + 1$ squares $2 \\times 2$. A $13 \\times 13$ board corresponds to the case $k = 7$, so the number of L-shapes is at least $4 \\cdot 7 - 1 = 27$; the number of $2 \\times 2$ squares is at most $7^2 - 4 \\cdot 7 + 1 = 22$. The example in the figure shows a covering with 22 squares $2 \\times 2$ and 27 shapes L. Hence the minimum number of L-shapes is 27.\n\n![](images/3._NATIONAL_XXX_OMA_2013-checkpoint_p2_data_fd6bbaa1c5.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 15999,
    "subject": "Mathematics (Olympiad)",
    "question": "In the convex pentagon $ABCDE$, $|\\angle CBA| = |\\angle BAE| = |\\angle AED|$ holds. On the sides $AB$ and $AE$, there are points $P$ and $Q$, respectively, such that $|AP| = |BC| = |QE|$ and $|AQ| = |BP| = |DE|$. Prove that $CD \\parallel PQ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $|BC| = |AP| = |EQ|$, $|BP| = |AQ| = |ED|$ and $|\\angle CBP| = |\\angle PAQ| = |\\angle QED|$, the triangles $PBC$, $QAP$ and $DEQ$ are congruent by the condition SAS.\n\nHence $|CP| = |PQ| = |QD|$ and also\n\n$$\n|\\angle CPQ| = 180^{\\circ} - |\\angle BPC| - |\\angle APQ| = 180^{\\circ} - |\\angle PQA| - |\\angle EQD| = |\\angle PQD|.\n$$\n\nThis means that by the condition SAS, the isosceles equilateral triangles $CPQ$ and $DQP$ are also congruent. It follows that their altitudes from $C$ and $D$ to the common opposite side $PQ$ have the same lengths, and hence $CD \\parallel PQ$.\n\n*Remark.* The observation that the triangles $CPQ$ and $DQP$ are isosceles and congruent can also be obtained by reasoning that they are two (not colored above) corresponding parts of congruent quadrilaterals $QABC$ and $DEAP$. The congruence of these quadrilaterals is a consequence of congruences $\\triangle QAB \\cong \\triangle DEA$ and $\\triangle ABC \\cong \\triangle EAP$.\n\nIn the following solution, we specify that the congruence of quadrilaterals $QABC$ and $DEAP$ is a certain rotation. Thanks to this, we also complete the new solution differently (without using the altitudes of the congruent triangles).\n\n*Another solution.* Let $S$ denote the circumcenter of $BAE$. Obviously, $|BA| = |AE|$. Therefore, in the rotation with center $S$ by the oriented angle $BSA$, $B \\to A \\to E$, and therefore $P \\to Q$.\n\n![](images/CZE_ABooklet_2023_p14_data_9082f3258e.png)\n\nAnother consequence of $B \\to A \\to E$ is the congruence of the four angles $SBA$, $SAB$, $SAE$ and $SEA$. It follows that the angle bisectors of congruent angles $CBA$, $BAE$ and $AED$ are respectively the rays $BS$, $AS$ and $ES$. In our rotation, the image of the oriented angle $CBS$ is the oriented angle $BAS$, so with respect to $|BC| = |AP|$, $C \\to P$ holds. The same is true from the oriented angles $SAE$ and $SED$, the equality $|AQ| = |ED|$ then leads to $Q \\to D$. Together we have $C \\to P \\to Q \\to D$, which implies that the line segments *CD* and *PQ* have a common perpendicular bisector—the bisector of *CD* bisects the angle *CSD*, and therefore bisects the angle *PSQ*, and therefore is also the perpendicular bisector of *PQ*. Because of the common bisector, the lines *CD* and *PQ* are parallel.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16000,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real values of $\\lambda$ such that for all $a, b \\ge 0$, the following inequality holds:\n\n$$\n\\frac{a+b}{2} \\ge \\lambda\\sqrt{ab} + (1-\\lambda)\\sqrt{\\frac{a^2+b^2}{2}}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $a = b \\ge 0$, the inequality holds for all $\\lambda \\in \\mathbb{R}$. Assume $a \\ne b$.\n\nWe have:\n\n$$\n\\frac{a+b}{2} \\ge \\lambda\\sqrt{ab} + (1-\\lambda)\\sqrt{\\frac{a^2+b^2}{2}}\n$$\n\nwhich is equivalent to\n\n$$\na+b \\ge 2\\lambda\\sqrt{ab} + (2-2\\lambda)\\sqrt{\\frac{a^2+b^2}{2}}.\n$$\n\nAfter manipulation, this leads to\n\n$$\n2\\lambda \\ge 1 - \\frac{(a-b)^2}{(\\sqrt{2a^2+2b^2}+(a+b))(\\sqrt{2a^2+2b^2}+(a+b)+2\\sqrt{ab})}.\n$$\n\nLet $a = 1$, $b = 1+\\epsilon$ with $\\epsilon > 0$. The denominator is at least $16$, so\n\n$$\n2\\lambda \\ge 1 - \\frac{\\epsilon^2}{16}.\n$$\n\nAs $\\epsilon \\to 0$, $1 - \\frac{\\epsilon^2}{16}$ approaches $1$, so $2\\lambda \\ge 1$, i.e., $\\lambda \\ge \\frac{1}{2}$.\n\nFor $\\lambda = \\frac{1}{2}$, the inequality holds, and for larger $\\lambda$ the right side decreases, so the inequality holds for all $\\lambda \\ge \\frac{1}{2}$.\n\n**Answer:** $\\boxed{\\left[\\frac{1}{2},\\,+\\infty\\right)}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16001,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Show that there are functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying\n$$\nf(f(x) + y f(y)) = x + (f(y))^n\n$$\nfor any $x, y \\in \\mathbb{R}$, if and only if $n = 2$. Determine those functions.",
    "options": [],
    "answer": "See solution",
    "solution": "For $y = 0$, we have $f(f(x)) = x + (f(0))^n$ for any $x \\in \\mathbb{R}$, implying that $f$ is bijective. Let $a$ such that $f(a) = 0$. For $y = a$, we deduce $f(f(x)) = x$, so $f(0) = 0$.\n\nFor $x = 0$, we get $f(y f(y)) = (f(y))^n$ for real $y$. Substituting $y$ with $f(y)$, we get $(f(y))^n = y^n$.\n\n* For odd $n$, we get $f(y) = y$, which gives a contradiction in the given relation.\n* For even $n$, we obtain $f(y) \\in \\{-y, y\\}$ for any $y \\in \\mathbb{R}$. Thus $y^n \\in \\{-y^2, y^2\\}$ for all real $y$, which implies $n = 2$.\n\nThus, $f(f(x) + y f(y)) = x + y^2$ for $x, y \\in \\mathbb{R}$.\n\nIt is easy to check that the identity function and minus the identity function verify the relation. We shall show that they are the only ones. If not, suppose there are $a, b \\neq 0$ such that $f(a) = a$ and $f(b) = -b$. Then for $x = a, y = b$, we should have $f(a - b^2) = a + b^2 \\in \\{a - b^2, -a + b^2\\}$, that is $a = 0$ or $b = 0$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16002,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, on the side $AC$ the points $F$ and $L$ with $AF = LC < \\frac{1}{2}AC$ are chosen. Find the angle $FBL$, if $AB^2 + BC^2 = AL^2 + LC^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $\\angle FBL = 90^\\circ$.\n\nLet $AB = c$, $BC = a$, $AF = LC = x$. According to the known formula for the length of a median (see the figure below):\n\n![](images/Ukrajina_2010_p38_data_8dbe54d9c5.png)\n\n$$\n\\begin{aligned}\nBM^2 = m^2 &= \\frac{1}{4}(2c^2 + 2a^2 - (2b + 2x)^2) = \\frac{1}{4}(2(2b + x)^2 + 2x^2 - (2b + 2x)^2) \\\\\n&= \\frac{1}{4}(8b^2 + 8bx + 4x^2 - 4b^2 - 8bx - 4x^2) = b^2.\n\\end{aligned}\n$$\n\nTherefore, in $\\triangle FBL$, the median is half of the base, so this triangle is right-angled and $\\angle FBL = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16003,
    "subject": "Mathematics (Olympiad)",
    "question": "A rabbit is at some point $ (x, y) $ in the Euclidean plane. There are some (possibly infinitely many) landmines, which are circles with any radius that do not intersect except possibly at one point (tangent). Every move, the rabbit can hop a distance of exactly $1$, but cannot land in the interior of a landmine (but may possibly land on the edge). Suppose that neither the rabbit nor the origin is in a landmine. Find the minimum $\\varepsilon$ so that the rabbit can always reach a point with a distance of at most $\\varepsilon$ from the origin, regardless of the landmine configuration.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim the answer is $\\frac{1}{\\sqrt{2}}$.\n\nFirstly, we prove that there is a configuration of landmines such that the rabbit cannot be closer than $\\frac{1}{\\sqrt{2}}$ to the origin.\n\nConsider a square packing of circles, with centers $\\left(\\frac{2m+1}{\\sqrt{2}}, \\frac{2n+1}{\\sqrt{2}}\\right)$ for all $m, n \\in \\mathbb{Z}$ and radii $\\frac{1}{\\sqrt{2}}$.\n\n![](images/2025-SL-a_p20_data_c2f7639fc6.png)\n\nIf the rabbit begins on one of the tangency points, it can only move to other tangency points and thus will always be at least $\\frac{1}{\\sqrt{2}}$ away from the origin.\n\nNow we prove that $\\frac{1}{\\sqrt{2}}$ can always be achieved. Firstly, we reduce the distance to $\\le 1$. Denote the origin by $O$ and the rabbit's position by $A$. Let $d = OA$ and suppose that $d > 1$. Consider the circles $(O, d)$ and $(A, 1)$ which intersect at two points, $X$ and $Y$.\n\n![](images/2025-SL-a_p20_data_143261cd3b.png)\n\n*Claim.* No landmine can fully contain the red arc *XY*.\n\n*Proof.* If a landmine were to exist, consider the center of the landmine. Since it is closer to *X* and *Y* than *A*, it must be in the blue region bounded by the perpendicular bisectors of *XA* and *YA*. But since it is also closer to *X* and *Y* than *O*, it must be in the green region bounded by the perpendicular bisectors of *XO* and *YO*.\n\n![](images/2025-SL-a_p21_data_c3984ff75d.png)\n\nNotice that the two regions are disjoint, meaning the center of the landmine cannot exist. $\\Box$\n\nHence, the rabbit can hop onto the arc. Notice that the maximum distance from the arc to $O$ is at $XO = YO = \\sqrt{d^2-1}$. Hence the squared distance between the rabbit and $O$ decreases by at least $1$ every move and thus the rabbit can eventually achieve a distance of at most $1$ from $O$.\n\nNow, if the distance $OA$ is still greater than $\\frac{1}{\\sqrt{2}}$, then $\\frac{1}{\\sqrt{2}} < OA \\le 1$. Thus we consider the circles $(O, \\frac{1}{\\sqrt{2}})$ and $(A, 1)$. Since $OP^2 + OA^2 > 1 = AP^2$, $\\angle POA$ is acute, thus $O$ and $A$ lie on opposite sides of the line $PQ$. Therefore, similar to before, the perpendicular bisectors form two disjoint regions, no landmine fully contains the red arc $PQ$.\n\n![](images/2025-SL-a_p21_data_780c8f6537.png)\n\nTherefore, the rabbit can hop onto the red arc and will be of distance at most $\\frac{1}{\\sqrt{2}}$ from $O$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16004,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $Q(x) = x^3 - 12x$. Find the range of $\\Delta(t)$, where $\\Delta(t)$ is the distance between the largest and smallest real roots of the equation $Q(x) = -t$ as $t$ varies over $\\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, $Q'(x) = 3x^2 - 12$ has roots at $x = -2$ and $x = 2$, so $Q$ has a local maximum at $(-2, 16)$ and a local minimum at $(2, -16)$. For $-16 \\le t \\le 16$, the equation $Q(x) = -t$ has three real roots; otherwise, it has only one real root, so $\\Delta(t) = 0$ for $t < -16$ or $t > 16$.\n\nLet $u \\le v \\le w$ be the roots. Since the critical points are at $x = -2$ and $x = 2$, we have $u \\le -2 \\le v \\le 2 \\le w$, so $-2 \\le v \\le 2$. The sum and product of the roots give $u + v + w = 0$ and $uv + vw + uw = -12$. Thus, $u + w = -v$ and $uw = v^2 - 12$.\n\nThe distance between the largest and smallest roots is $w - u$. We have:\n\n$$\n(w - u)^2 = (w + u)^2 - 4uw = (-v)^2 - 4(v^2 - 12) = v^2 - 4v^2 + 48 = 48 - 3v^2\n$$\n\nAs $v$ ranges from $-2$ to $2$, $v^2$ ranges from $0$ to $4$, so $(w - u)^2$ ranges from $48 - 3 \\cdot 4 = 36$ to $48$. Thus, $w - u$ ranges from $6$ to $4\\sqrt{3}$.\n\nTherefore, the range of $\\Delta(t)$ is $\\{0\\} \\cup [6, 4\\sqrt{3}]$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16005,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть $\\alpha = \\frac{-1 + \\sqrt{29}}{2}$ — корень уравнения $\\alpha^2 + \\alpha = 7$. Докажите, что для любого натурального числа $n$ сумму в $n$ рублей можно набрать монетами достоинств $1, \\alpha, \\alpha^2, \\ldots$ так, чтобы ни одна из монет не встречалась более 6 раз. Также покажите, что предъявленное значение $\\alpha$ — единственное возможное.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Пусть $\\alpha = \\frac{-1 + \\sqrt{29}}{2}$, тогда $\\alpha^2 + \\alpha = 7$ и $\\alpha > 2$. Для любого натурального $m$ имеем $(2\\alpha)^m = a_m + b_m\\sqrt{29}$, где $a_m, b_m$ — целые числа, $a_m < 0 < b_m$ при нечётных $m$, $a_m > 0 > b_m$ при чётных $m$, значит $\\alpha^m$ иррационально.\n\nРассмотрим все способы набрать $n$ рублей монетами достоинств $1, \\alpha, \\alpha^2, \\ldots$. Выберем способ с наименьшим числом монет. Если какая-то монета $\\alpha^i$ встречается хотя бы 7 раз, то можно заменить 7 монет $\\alpha^i$ на монеты $\\alpha^{i+1}$ и $\\alpha^{i+2}$, так как $\\alpha^{i+1} + \\alpha^{i+2} = 7\\alpha^i$, уменьшив число монет, что невозможно по выбору. Значит, ни одна монета не встречается более 6 раз.\n\n*Замечание.* Для любого допустимого $\\alpha$ сумма 7 рублей набирается только как $7 = \\alpha + \\alpha^2$, значит, предъявленное $\\alpha$ — единственное.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16006,
    "subject": "Mathematics (Olympiad)",
    "question": "![](images/15-3J_p31_data_549ef384aa.png)\n\nLet $\\mathbb{Q}^+$ be the set of all positive rational numbers. Find all functions $f : \\mathbb{Q}^+ \\to \\mathbb{Q}^+$ satisfying $f(1) = 1$ and\n\n$$\nf(x + n) = f(x) + n f\\left(\\frac{1}{x}\\right)\n$$\nfor all positive integers $n$ and all $x \\in \\mathbb{Q}^+$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f : \\mathbb{Q}^+ \\to \\mathbb{Q}^+$ satisfy $f(1) = 1$ and $f(x + n) = f(x) + n f\\left(\\frac{1}{x}\\right)$ for all positive integers $n$ and all $x \\in \\mathbb{Q}^+$.\n\nDefine $g : \\mathbb{Q}^+ \\to \\mathbb{N}_0$ as follows: for any $\\frac{p}{q}$ (with $p, q$ coprime positive integers), $g\\left(\\frac{p}{q}\\right)$ is the minimal number of steps needed to reduce $(p, q)$ to $(1, k)$ or $(k, 1)$ by repeatedly replacing the larger number with its remainder upon division by the smaller. This $g$ is well-defined and $g(x) = g(x^{-1})$.\n\nWhen $0 < x < 1$ and $n$ is a positive integer,\n$$\ng(x + n) = g(x) + 1. \\tag{1}\n$$\n\nWe prove by induction on $g\\left(\\frac{p}{q}\\right)$ that $f\\left(\\frac{p}{q}\\right) = p$ for all coprime $p, q$.\n\n**Base case:** $g\\left(\\frac{p}{q}\\right) = 0$ means $p = 1$ or $q = 1$.\n- If $q = 1$, substitute $(x, n) = (1, p-1)$:\n  $$\nf(p) = p. \\tag{2}\n  $$\n- If $p = 1$, substitute $(q, 1)$ and use (2) to get $f\\left(\\frac{1}{q}\\right) = 1$.\n\n**Inductive step:** Assume true for $g\\left(\\frac{p}{q}\\right) = i-1$.\n- **Case 1:** $\\frac{p}{q} > 1$. Let $n = \\lfloor \\frac{p}{q} \\rfloor$, so $g\\left(\\frac{p-nq}{q}\\right) = i-1$. Substitute $(\\frac{p-nq}{q}, n)$:\n  $$\nf\\left(\\frac{p}{q}\\right) = (p-nq) + n q = p. \\tag{3}\n  $$\n- **Case 2:** $\\frac{p}{q} < 1$. Then $\\frac{q}{p} > 1$, so $g\\left(\\frac{q}{p}\\right) = i$. From (3), $f\\left(\\frac{p+q}{p}\\right) = p+q$, $f\\left(\\frac{q}{p}\\right) = q$. Substitute $(\\frac{q}{p}, 1)$:\n  $$\nf\\left(\\frac{p}{q}\\right) = (p+q) - q = p.\n  $$\n\nThus, by induction, $f\\left(\\frac{p}{q}\\right) = p$ for all coprime $p, q$.\n\n**Conclusion:**\n$$\nf\\left(\\frac{p}{q}\\right) = p, \\quad \\text{for all coprime positive integers } p, q.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16007,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be two real numbers in the interval $(0, 1)$, so that $a$ is a rational number and\n\n$$\n\\{na\\} \\ge \\{nb\\}, \\text{ for every natural number } n.\n$$\n\nProve that $a = b$.\n\n(We denote by $\\{x\\}$ the fractional part of the real number $x$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = \\frac{p}{q}$, where $p$ and $q$ are nonzero natural numbers, relatively prime, with $p < q$.\n\nThen $0 = \\{qa\\} \\ge \\{qb\\} \\ge 0$, thus $qb$ is a natural number. It follows that $b = \\frac{s}{q}$, where $s$ is a nonzero natural number.\n\nConsidering $n = 1$ in the relation from the hypothesis, we get that $\\{a\\} \\ge \\{b\\}$, wherefrom $p \\ge s$.\n\nConsidering $n = q - 1$ in the relation from the hypothesis, we get that $\\{-\\frac{p}{q}\\} \\ge \\{-\\frac{s}{q}\\}$, that is $1 - \\frac{p}{q} \\ge 1 - \\frac{s}{q} \\iff p \\le s$.\n\nSo $p = s$, hence $a = b$, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16008,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $(A, B)$ such that increasing $A$ by $B\\%$ yields the same result as decreasing $B$ by $A\\%$.",
    "options": [],
    "answer": "See solution",
    "solution": "We write the condition as an equation relating $A$ and $B$:\n\n$$\nA \\left(1 + \\frac{B}{100}\\right) = B \\left(1 - \\frac{A}{100}\\right)\n$$\n\nMultiplying both sides by $100$:\n\n$$\nA(100 + B) = B(100 - A)\n$$\n\nExpanding and rearranging:\n\n$$\nA(100 + B) + B(100 - A) = 0\n$$\n$$\n100A + AB + 100B - AB = 100A + 100B = 0\n$$\n\nBut this simplifies to:\n\n$$\nA(100 + B) = B(100 - A)\n$$\n$$\nA(100 + B) + A(100 - B) = 0\n$$\n\nAlternatively, solving for $A$:\n\n$$\nA(100 + B) = B(100 - A)\n$$\n$$\nA(100 + B) + A^2 = 100B\n$$\n$$\nA(100 + B) + A^2 - 100B = 0\n$$\n\nBut the solution proceeds as:\n\n$$\nA = \\frac{100B}{2B + 100} = \\frac{50B}{B + 50} = 50 - \\frac{2500}{B + 50}\n$$\n\nThus, $B + 50$ must be a divisor of $2500$. The divisors are $1, 2, 4, 5, 10, 20, 25, 50, 100, 125, 250, 500, 625, 1250, 2500$. Only three of these yield three-digit $B$ values: $250, 500, 625$. The corresponding pairs $(A, B)$ are:\n\n- $(40, 200)$\n- $(45, 450)$\n- $(46, 575)$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16009,
    "subject": "Mathematics (Olympiad)",
    "question": "$a_{n+2} + \\frac{5}{2} = 4(a_{n+1} + \\frac{5}{2}) + 5(a_n + \\frac{5}{2})$.\n\nFind the smallest positive even integer $m$ such that $a_{m-1} \\equiv 0 \\pmod{2012}$, given $a_0 = 20$, $a_1 = 100$, and $a_{n+2}$ is defined recursively as above.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $b_n = a_n + \\frac{5}{2}$. Then:\n$$\n\\begin{align*}\nb_{n+2} &= 4b_{n+1} + 5b_n \\\\\nb_n &= c_1 \\cdot 5^n + c_2(-1)^n \\\\\nb_0 &= \\frac{45}{2}, \\quad b_1 = \\frac{205}{2} \\\\\nc_1 = \\frac{125}{6}, \\quad c_2 = \\frac{10}{6} \\\\\na_n = \\frac{125}{6} \\cdot 5^n + \\frac{10}{6} \\cdot (-1)^n - \\frac{5}{2}\n\\end{align*}\n$$\nAlternatively,\n$$\na_n = \\begin{cases} \\frac{5^{n+3} - 5}{5^{n+3} - 25}, & n \\text{ even} \\\\ \\frac{5}{6}, & n \\text{ odd} \\end{cases}\n$$\n\nFor $n$ even, $a_{n+1} = 5a_n$. We seek $m$ even such that $a_{m-1} \\equiv 0 \\pmod{2012}$.\n\nFrom the formula:\n$$\na_{m-1} = 25 \\cdot \\frac{5^m - 1}{6} \\equiv 0 \\pmod{2012}\n$$\nSo,\n$$\n\\frac{5^m - 1}{6} \\equiv 0 \\pmod{2012} \\implies 5^m \\equiv 1 \\pmod{12072}\n$$\nwhere $12072 = 6 \\times 2012 = 2^3 \\times 3 \\times 503$.\n\nSince $m$ is even, $5^m \\equiv 1 \\pmod{8}$ and $5^m \\equiv 1 \\pmod{3}$ always hold. Thus, we need $5^m \\equiv 1 \\pmod{503}$.\n\nThe order of $5$ modulo $503$ is $502$, so the smallest such $m$ is $m = 502$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16010,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be a positive integer, and let $A_1A_2 \\cdots A_{2n}$ be a convex polygon with $2n$ sides inscribed in a circle. It is known that there exists a point $P$ within this polygon such that\n\n$$\n\\angle PA_1A_2 = \\angle PA_2A_3 = \\cdots = \\angle PA_{2n-1}A_{2n} = \\angle PA_{2n}A_1.\n$$\n\nProve that\n\n$$\n\\prod_{i=1}^{n} |A_{2i-1}A_{2i}| = \\prod_{i=1}^{n} |A_{2i}A_{2i+1}|,\n$$\n\nwhere $A_{2n+1} = A_1$.",
    "options": [],
    "answer": "See solution",
    "solution": "All subscripts are understood modulo $2n$. Denote $\\angle PA_1A_2 = \\angle PA_2A_3 = \\cdots = \\alpha$. For $1 \\leq i \\leq 2n$, extend $A_iP$ to intersect the circumference at point $B_i$. Note that\n\n$$\n\\angle A_iB_iB_{i-1} = \\angle A_iA_{i-1}P = \\angle A_{i+1}A_iB_i = \\alpha.\n$$\n\nHence, $A_iA_{i+1}B_{i-1}B_i$ is an isosceles trapezoid. (In fact, $B_1B_2 \\cdots B_{2n}$ is the polygon obtained by rotating $A_2A_3 \\cdots A_{2n}A_1$ clockwise by $2\\alpha$ on the circumference.) In particular, $A_iA_{i+1} = B_{i-1}B_i$. Also, by $\\triangle A_{i-1}PA_i \\sim \\triangle B_iPB_{i-1}$, we get\n\n$$\n\\frac{A_{i-1}P}{B_iP} = \\frac{A_iP}{B_{i-1}P} = \\frac{A_{i-1}A_i}{B_{i-1}B_i} = \\frac{A_{i-1}A_i}{A_iA_{i+1}}.\n$$\n\nTaking the product of the first and last equalities over all odd $i$, we get\n\n$$\n\\prod_{i=1}^{n} \\frac{A_{2i-2}A_{2i-1}}{A_{2i-1}A_{2i}} = \\prod_{i=1}^{n} \\frac{A_{2i-2}P}{B_{2i-1}P}.\n$$\n\nTaking the product of the second and last equalities over all even $i$, we get\n\n$$\n\\prod_{i=1}^{n} \\frac{A_{2i-1}A_{2i}}{A_{2i}A_{2i+1}} = \\prod_{i=1}^{n} \\frac{A_{2i}P}{B_{2i-1}P}.\n$$\n\nNoting that the right-hand sides of the two equations above are equal, we have\n\n$$\n\\prod_{i=1}^{n} \\frac{A_{2i-2}A_{2i-1}}{A_{2i-1}A_{2i}} = \\prod_{i=1}^{n} \\frac{A_{2i-1}A_{2i}}{A_{2i}A_{2i+1}} \\implies \\prod_{i=1}^{n} (A_{2i}A_{2i+1})^2 = \\prod_{i=1}^{n} (A_{2i-1}A_{2i})^2.\n$$\n\nThis completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16011,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{Z} \\to \\mathbb{Z}$ be a surjective function such that for all $x, y, z \\in \\mathbb{Z}$,\n$$\nf(x + y + z) = f(x) + f(y) + f(z) - 1.\n$$\nFind all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze the given functional equation and its consequences.\n\nLet $f : \\mathbb{Z} \\to \\mathbb{Z}$ be surjective and satisfy\n$$\nf(x + y + z) = f(x) + f(y) + f(z) - 1, \\quad \\forall x, y, z \\in \\mathbb{Z}.\n$$\n\n**Case 1:** $f(0) = 1$.\n\nLet $x = b^3 - 3b$, $y = b$, $z = 0$ (with $b \\neq 0$):\n$$\nf(4b - b^3) = 1 \\iff 4b - b^3 = 0 \\iff b = 2, -2.\n$$\n\n**Case 1.1:** $f(-2) = -1$.\n\nThen,\n$$\nf(x) + f(-2 - x) = 0, \\quad \\forall x \\in \\mathbb{Z} \\qquad (1)\n$$\n\nChoose $a \\in \\mathbb{Z}$ with $f(a) = 2$. Set $y = a, z = 0$:\n$$\nf(2x + a) = 2f(x), \\quad \\forall x \\in \\mathbb{Z} \\qquad (2)\n$$\n\nIn particular, $f(a-4) = 2(-2) = -2$. Using (1), $f(2-a) = 2$. Thus,\n$$\nf(2x + (2 - a)) = 2f(x) = f(2x + a).\n$$\n\nIf $2-a \\neq a$, $f$ would take only finitely many values on $2\\mathbb{Z} + a$, contradicting surjectivity. Thus, $a = 1$ and\n$$\nf(2x + 1) = 2f(x), \\quad \\forall x \\in \\mathbb{Z} \\qquad (3)\n$$\n\nWe claim $f(x) = x + 1$ for all $x \\in \\mathbb{N}$. By induction:\n- If $f(r-1) = r$ and $f(s-1) = s$, then $f(rs-1) = rs$ (set $x = r, y = s, z = 0$).\n- For primes $p$, suppose $f(m-1) = m$ for all $m < p$.\n\nChoose $c$ with $f(c) = p$. If $c \\not\\equiv -1 \\pmod{p}$, take $d$ so $0 \\leq pd + c < p-1$. Set $x = d, y = c, z = 0$:\n$$\npd + c + 1 = f(pd + c) = pf(d).\n$$\nThis forces $pd + c + 1$ divisible by $p$, a contradiction. If $c + 1$ is a multiple of $2p$, $c$ is odd, which is impossible by (3). Thus, $c \\equiv p-1 \\pmod{2p}$.\n\nNow,\n$$\nf(2px + 2c + 1) = 2f(px + c) = 2pf(x) = pf(2x + 1) = f(2px + p + c).\n$$\nIf $2c + 1 \\neq p + c$, $f$ would take finitely many values, a contradiction. Thus, $c = p-1$.\n\nBy induction, $f(x) = x + 1$ for all $x \\in \\mathbb{N}$, and by (1),\n$$\nf(x) = x + 1, \\quad \\forall x \\in \\mathbb{Z}.\n$$\n\n**Case 1.2:** $f(2) = -1$.\n\nChoose $a$ with $f(a) = 2$. Then,\n$$\nf(a + 4) = 2f(2) = -2 \\implies f(-2 - a) = 2.\n$$\nThe same argument gives $a = -1$. By induction, $f(1 - x) = x$ for $x \\in \\mathbb{N}$, so $f(x) = 1 - x$ for all $x \\in \\mathbb{Z}$.\n\n**Case 2:** $f(0) = -1$.\n\nChoose $e$ with $f(e) = 1$ and set $y = e, z = 0$:\n$$\nf(x - e) = -f(x) = 0 \\implies f(x - 2e) = -f(x - e) = f(x).\n$$\nThus, $f$ is periodic, contradicting surjectivity.\n\n**Conclusion:**\nThe only solutions are\n$$\nf(x) = x + 1, \\quad \\forall x \\in \\mathbb{Z},\n$$\nand\n$$\nf(x) = 1 - x, \\quad \\forall x \\in \\mathbb{Z}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16012,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p_1, p_2, \\dots, p_{42}$ be 42 pairwise different primes. Prove that the number\n$$\n\\sum_{j=1}^{42} \\frac{1}{p_j^2 + 1}\n$$\ncannot be equal to the reciprocal $\\frac{1}{n^2}$ of a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that the sum can be written as the reciprocal of a perfect square $n^2$. Let $P := \\prod_{j=1}^{42} (p_j^2 + 1)$ be the product of all denominators. Then:\n$$\n\\sum_{j=1}^{42} \\frac{1}{p_j^2 + 1} = \\frac{1}{n^2} \\iff n^2 \\cdot \\sum_{j=1}^{42} \\frac{P}{p_j^2 + 1} = P.\n$$\nConsider both sides modulo 3.\n\n**Case 1:** If none of the $p_j$ equals 3, each $p_j^2 + 1 \\equiv -1 \\pmod{3}$, so $P \\equiv 1 \\pmod{3}$ and each $\\frac{P}{p_j^2+1} \\equiv -1 \\pmod{3}$. The left side is divisible by 3, a contradiction.\n\n**Case 2:** If $p_j = 3$ for some $j$, then $3^2 + 1 \\equiv 1 \\pmod{3}$, so $P \\equiv -1 \\pmod{3}$. In the sum, one term is $\\equiv -1$ and 41 terms are $\\equiv 1$, so the sum is $\\equiv 1$. Since $n^2 \\equiv 0$ or $1 \\pmod{3}$, the left side cannot be $\\equiv -1$, again a contradiction.\n\nThus, the sum cannot be the reciprocal of a perfect square, as claimed.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16013,
    "subject": "Mathematics (Olympiad)",
    "question": "Define the sequence $\\{a_n\\}_{n \\ge 0}$ by $a_0 = 0$, $a_1 = 1$, $a_2 = 2$, $a_3 = 6$, and\n\n$$\na_{n+4} = 2a_{n+3} + a_{n+2} - 2a_{n+1} - a_n, \\quad n \\ge 0.\n$$\n\nProve that $n^2$ divides $a_n$ for infinitely many positive integers $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "From the recursive relation, we compute $a_4 = 12$, $a_5 = 25$, $a_6 = 48$. Thus, $\\frac{a_1}{1} = 1$, $\\frac{a_2}{2} = 1$, $\\frac{a_3}{3} = 2$, $\\frac{a_4}{4} = 3$, $\\frac{a_5}{5} = 5$, $\\frac{a_6}{6} = 8$, that is, $\\frac{a_n}{n} = F_n$ for $n = 1, 2, 3, 4, 5, 6$, where $\\{F_n\\}_{n \\ge 1}$ is the Fibonacci sequence.\n\nWe prove by induction that $a_n = nF_n$ for all $n \\ge 1$. Assume $a_k = kF_k$ for $k = n, n+1, n+2, n+3$. Then:\n\n$$\n\\begin{aligned}\na_{n+4} &= 2(n+3)F_{n+3} + (n+2)F_{n+2} - 2(n+1)F_{n+1} - nF_n \\\\\n&= 2(n+3)F_{n+3} + (n+2)F_{n+2} - 2(n+1)F_{n+1} - n(F_{n+2} - F_{n+1}) \\\\\n&= 2(n+3)F_{n+3} + 2F_{n+2} - (n+2)F_{n+1} \\\\\n&= 2(n+3)F_{n+3} + 2F_{n+2} - (n+2)(F_{n+3} - F_{n+2}) \\\\\n&= (n+4)(F_{n+3} + F_{n+2}) = (n+4)F_{n+4}.\n\\end{aligned}\n$$\n\nThus, $a_n = nF_n$ for all $n \\ge 1$.\n\nNow, it suffices to prove that $n$ divides $F_n$ for infinitely many positive integers $n$. Using the Binet formula for Fibonacci numbers:\n\n$$\nF_n = \\frac{1}{\\sqrt{5}} \\left[ \\left( \\frac{1+\\sqrt{5}}{2} \\right)^n - \\left( \\frac{1-\\sqrt{5}}{2} \\right)^n \\right] = \\frac{1}{2^n \\sqrt{5}} \\sum_{k=0}^{n} \\binom{n}{k} (1-(-1)^k)(\\sqrt{5})^k.\n$$\n\nFrom this, for $n = 5^l$:\n\n$$\nF_{5^l} = \\frac{1}{2^{5^l-1}} \\sum_{k=0}^{\\frac{5^l-1}{2}} \\binom{5^l}{2k+1} 5^k.\n$$\n\nWe show that for $k = 0, \\dots, l-1$, $5^l$ divides $\\binom{5^l}{2k+1} 5^k$. Indeed,\n\n$$\n\\binom{5^l}{2k+1} = \\frac{5^l(5^l-1)\\cdots(5^l-2k)}{1 \\cdot 2 \\cdots (2k+1)}.\n$$\n\nFor every $a < 5^l$, $\\exp_5(5^a - a) = \\exp_5(a)$, so $\\exp_5((2k)!)= \\exp_5(5^l - 1)\\cdots(5^l - 2k)$. Thus,\n\n$$\n\\exp_5\\left(\\binom{5^l}{2k+1}\\right) = l - \\exp_5((2k+1)) \\ge l - \\exp_5(5^k) = l - k,\n$$\n\nsince $2k+1 \\le 5^k$. Therefore, for every positive integer $l$, $5^l$ divides $F_{5^l}$, so $(5^l)^2$ divides $a_{5^l}$. Thus, $n^2$ divides $a_n$ for infinitely many $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16014,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all triples $(k, m, n)$ of positive integers satisfying\n\n$$\nk! + m! = k! \\times n!.\n$$\n\n(If $n$ is a positive integer, then $n! = 1 \\times 2 \\times 3 \\times \\cdots \\times (n-1) \\times n$.)",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $(r, r, 2)$ for $r \\ge 1$, and $(r! - 2, r! - 1, r)$ for $r \\ge 3$.\n\nThe equation is equivalent to\n\n$$\nm! = k!(n! - 1).\n$$\n\nIf $n = 1$, there are no solutions.\n\nIf $n = 2$, we get $m = k$, which yields the family $(r, r, 2)$.\n\nIf $n \\ge 3$, then $m > k$.\n\n- If $m = k + 1$, we get $k + 1 = n! - 1$, which yields the family $(r! - 2, r! - 1, r)$.\n- If $m \\ge k + 2$, we get $(k + 1)(k + 2) \\cdots m = n! - 1$, which is a contradiction because the left-hand side is even and the right-hand side is odd.\n\nSo, the two solution families are $(r, r, 2)$ for $r \\ge 1$ and $(r! - 2, r! - 1, r)$ for $r \\ge 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16015,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ be a positive integer and let $A$, respectively $B$, be an alphabet with $m$, respectively $2m$ letters. Let $n$ be an even integer greater than or equal to $2m$. Let $a_n$ be the number of words of length $n$ made of letters from $A$ such that every letter in $A$ occurs a positive even number of times. Let $b_n$ be the number of words of length $n$ made of letters from $B$ such that every letter in $B$ occurs an odd number of times. Determine the ratio $b_n/a_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The required ratio is $2^{n-m}$.\n\nTo prove this, let $A^n$ and $B^n$ denote the sets of words described in the statement, so $a_n = |A^n|$ and $b_n = |B^n|$, and think of $B$ as an extension of $A = \\{a_1, \\dots, a_m\\}$ by a disjoint copy $\\bar{A} = \\{\\bar{a}_1, \\dots, \\bar{a}_m\\}$.\n\nDeletion of all bars in a word in $B^n$ produces a word in $A^n$. Now let $\\alpha$ be a word in $A^n$ and let $a_i$ occur $k_i$ times in $\\alpha$; the $k_i$ are positive even integers which add up to $n$. Since there are exactly $2^{k_i-1}$ distinct ways to bar $a_i$ an odd number of times in $\\alpha$, the preimage of $\\alpha$ under deletion of all bars has exactly $2^{k_1-1} \\cdots 2^{k_m-1} = 2^{n-m}$ elements. The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16016,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with an obtuse angle at $A$. Let $Q$ be a point (other than $A$, $B$, or $C$) on the circumcircle of the triangle, on the same side of chord $BC$ as $A$, and let $P$ be the other end of the diameter through $Q$. Let $V$ and $W$ be the feet of the perpendiculars from $Q$ onto $CA$ and $AB$ respectively. Prove that the triangles $PBC$ and $AWV$ are similar.\n\n[Note: the circumcircle of the triangle $ABC$ is the circle which passes through the vertices $A$, $B$, and $C$.]\n\n![](images/V_Britanija_2008_p14_data_449f69ca4f.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Because $ABPC$ is cyclic, $\\angle BAC + \\angle BPC = 180^\\circ$, and so\n\n$$\n\\angle BPC = \\angle WAV. \\qquad (1)\n$$\n\nBecause $\\angle QWA + \\angle QVA = 180^\\circ$, $AWQV$ is cyclic.\n\nSince $PQ$ is a diameter of the circumcircle of triangle $ABC$, $\\angle PAQ = 90^\\circ$. Also, $AQ$ is a diameter of the circumcircle of the quadrilateral $AWQV$ so that $PA$ is a tangent to this circle.\n\nBy the alternate segment theorem, $\\angle PAB = \\angle AVW$. By angles in the same segment, $\\angle PAB = \\angle PCB$. Therefore\n\n$$\n\\angle PCB = \\angle AVW. \\qquad (2)\n$$\n\nEquations (1) and (2) imply that $\\triangle AVW$ has two angles equal to the corresponding angles in $\\triangle PCB$.\n\nTherefore $\\triangle AVW$ is similar to $\\triangle PCB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16017,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples $(a, b, c)$ of integers that satisfy the following relations:\n\n$$\na^2 + a = b + c, \\quad b^2 + b = a + c, \\quad c^2 + c = a + b.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Adding the three equations gives:\n$$\na^2 + b^2 + c^2 + a + b + c = 2(a + b + c)\n$$\nwhich simplifies to\n$$\na^2 + b^2 + c^2 = a + b + c. \\tag{*}\n$$\nFor any integer $x$, $x^2 \\geq x$, with equality only if $x = 0$ or $x = 1$.\n\nTherefore, equality in $(*)$ is possible only if $a^2 = a$, $b^2 = b$, and $c^2 = c$.\nThis gives $a = 0$ or $1$, $b = 0$ or $1$, $c = 0$ or $1$.\n\nSubstituting into the original equations, only $(a, b, c) = (0, 0, 0)$ and $(1, 1, 1)$ satisfy all three.\n\nThus, the required triples are $(0, 0, 0)$ and $(1, 1, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16018,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Prove that $D$, $G$, and $K$ are collinear.\n\nb) Let $M$ be a point on $KB$ and $N$ a point on $KC$ such that $IM \\perp AC$ and $IN \\perp AB$. The perpendicular bisector of $IK$ intersects $MN$ at $H$. Suppose that $IH$ meets $AB$ and $AC$ at $P$ and $Q$, respectively. Prove that the circumcircle of triangle $APQ$ intersects $(O)$ again at a point on $AI$.\n\n![](images/Vietnamese_mathematical_competitions_p304_data_d0860d5540.png)",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $T$ be the projection of $A$ on $GD$. It is well known that $T$ is the second intersection of $(AEF)$ and $(O)$. We observe that $\\triangle TEB \\sim \\triangle TFC$, then\n\n$$\n\\frac{TB}{TC} = \\frac{BE}{CF}.\n$$\n\nOn the other hand,\n\n$$\n\\frac{BE}{CF} = \\frac{BE}{IB} \\cdot \\frac{IC}{CF} = \\frac{\\cos ACB}{\\cos ABC} = \\frac{BD}{CD'}\n$$\n\nwhich implies that $TBDC$ is a harmonic quadrilateral, or $TD$ passes through $K$, which is the intersection of the tangents at $B$ and $C$ of $(O)$.\n\nb) Let $X$, $Y$ be the intersections of $KB$, $KC$ with $IN$, $IM$, respectively. The tangents at $B$ and $C$ of $(O)$ meet the tangent at $A$ at $X'$, $Y'$. Because $IX \\parallel OX'$ and $IY \\parallel OY'$, then\n\n$$\n\\triangle KX'Y' \\sim \\triangle KXY\n$$\n\nwith $O$, $I$ corresponding. Note that $O$ is the incenter of $KX'Y'$, so $I$ is the incenter of triangle $KXY$.\n\nLet $(KXY)$ meet $IY$, $IX$ at $R$, $S$ respectively. Clearly, $R$, $S$ are the circumcenters of triangles $KIX$ and $KIY$. Hence, $RS$ is the perpendicular bisector of $IK$, so $H$ lies on $RS$. Applying Pascal's theorem for\n$$\n\\begin{pmatrix} K & X & R \\\\ S & Y & K \\end{pmatrix}\n$$\nwe obtain that the tangent at $K$ of $(KXY)$, $RS$, and $MN$ are concurrent, which means $KH$ is the tangent of $(KXY)$. By angle chasing, we have\n\n$$\n\\begin{aligned}\n\\angle HIK &= \\angle HKI = \\angle HKB + \\angle BKI \\\\\n&= \\angle KYX + 90^\\circ - \\angle BAC = 270^\\circ - 2\\angle ABC - \\angle BAC \\\\\n&= 90^\\circ + \\angle ACB - \\angle ABC\n\\end{aligned}\n$$\n\nAlso, $\\angle(AO, BC) = \\angle OAC + \\angle ACB = 90^\\circ - \\angle ABC + \\angle ACB = \\angle HIK$. Note that $IK \\perp BC$, thus $IH \\perp AO$. Hence, $PBQC$ is cyclic or $I$ has the same power to $(ABC)$ and $(APQ)$, which means $AI$ passes through the second intersection of $(APQ)$ and $(O)$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16019,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle. Point $M$ and $N$ lie on sides $AC$ and $BC$ respectively such that $MN \\parallel AB$. Points $P$ and $Q$ lie on sides $AB$ and $CB$ respectively such that $PQ \\parallel AC$. The incircle of triangle $CMN$ touches segment $AC$ at $E$. The incircle of triangle $BPQ$ touches segment $AB$ at $F$. Line $EN$ and $AB$ meet at $R$, and lines $FQ$ and $AC$ meet at $S$. Given that $AE = AF$, prove that the incenter of triangle $AEF$ lies on the incircle of triangle $ARS$.\n\n![](images/pamphlet0910_main_p49_data_3dabe244d9.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\omega_1, \\omega_C, \\omega_B$, and $\\omega$ denote the incircles of triangles $ABC$, $MNC$, $PBQ$, and $ARS$, respectively. Denote by $I$ and $I_1$ the incenters of triangles $ABC$ and $ARS$, respectively. Let $\\omega_1$ touch sides $AB$ and $AC$ at $R_1$ and $S_1$, respectively.\n\nIt is clear that there is a homothety $\\mathbf{H}_1$ centered at $C$ sending triangle $CMN$ to $CAB$, and that images of $M, E, N$, and line $EN$ under $\\mathbf{H}_1$ are $A, S_1, B$, and line $S_1B$. In particular, $BS_1 \\parallel RE$ with $AB/AR = AS_1/AE$. In exactly the same way, we can prove that $CR_1 \\parallel SF$ with $AC/AS = AR_1/AF$. By equal tangents, we have $AS_1 = AR_1$. By the given condition, $AE = AF$. It follows that\n\n$$\n\\frac{AB}{AR} = \\frac{AS_1}{AE} = \\frac{AR_1}{AF} = \\frac{AC}{AS},\n$$\n\nimplying that $BC \\parallel RS$. Thus, there is a homothety $\\mathbf{H}$ centered at $A$ sending triangle $ABC$ to triangle $ARS$. It is clear that the images of $S_1, R_1, \\omega_1$, and $I_1$ under $\\mathbf{H}$ are $E, F, \\omega$, and $I$, respectively. Thus, $I$ lies on $\\omega$, which is what we wish to show, if and only if $I_1$ lies on $\\omega_1$. But the latter claim holds because the midpoint of minor arc $\\widehat{R_1S_1}$ on $\\omega_1$ is the incenter of triangle $AR_1S_1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16020,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\gamma_1$ and $\\gamma_2$ be two circles tangent at point $T$, and let $\\ell_1$ and $\\ell_2$ be two lines through $T$. The lines $\\ell_1$ and $\\ell_2$ meet $\\gamma_1$ again at points $A$ and $B$, respectively, and $\\gamma_2$ at points $A_1$ and $B_1$, respectively. Let $X$ be a point in the complement of $\\gamma_1 \\cup \\gamma_2 \\cup \\ell_1 \\cup \\ell_2$. The circles $ATX$ and $BTX$ meet $\\gamma_2$ again at points $A_2$ and $B_2$, respectively. Prove that the lines $TX$, $A_1B_2$, and $A_2B_1$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the circle $ATX$ and the line $A_2B_1$ meet again at the point $Y$, and let the circle $BTX$ and the line $A_1B_2$ meet again at the point $Z$. Notice that the lines $AB$, $AY$, and $AZ$ are all parallel to the line $A_1B_1$, so the points $A$, $B$, $Y$, and $Z$ are collinear.\n\nNext, consider the cyclic quadrilaterals $A_1B_1A_2B_2$, $A_1B_1B_2T$, and $BTB_2Z$ to get successively $\\angle B_2A_2Y = \\angle B_1A_1B_2 = \\angle B_1TB_2 = \\angle BZB_2$, and infer that the points $A_2$, $B_2$, $Y$, and $Z$ are concyclic. It then follows that the lines $TX$, $A_1B_2$, and $A_2B_1$ are the radical axes of the pairs of circles $ATX$ and $BTX$, $BTX$ and $A_2B_2YZ$, and $A_2B_2YZ$ and $ATX$, respectively, whence the conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16021,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be a positive real number such that $a^3 = 6(a+1)$. Prove that the equation\n\n$$\nx^2 + a x + a^2 - 6 = 0\n$$\n\nhas no real solution.",
    "options": [],
    "answer": "See solution",
    "solution": "To show that $x^2 + a x + a^2 - 6 = 0$ has no real solution, it suffices to show that its discriminant is negative.\n\nThe discriminant is:\n$$\n\\Delta = a^2 - 4(a^2 - 6) = a^2 - 4a^2 + 24 = -3a^2 + 24 = 3(8 - a^2)\n$$\n\nFor real solutions, we require $\\Delta \\geq 0$, i.e., $a^2 \\leq 8$.\n\nGiven $a^3 = 6(a+1)$, we have:\n$$\na^3 = 6a + 6 \\implies a^3 - 6a - 6 = 0\n$$\nBut $a^2 = 6 + \\frac{6}{a}$, so\n$$\na^2 \\geq 6 + \\frac{6\\sqrt{2}}{4} = 6 + \\frac{3\\sqrt{2}}{2} > 8\n$$\nThis contradicts $a^2 \\leq 8$. Therefore, the equation has no real solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16022,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Determine if there exist positive integer numbers $x, y, z$ such that\n$$\n2016 = x^3 + y^3 + z^3\n$$\n\nb) Determine if there exist positive integer numbers $x, y, z, t$ such that\n$$\n2016 = x^3 + y^3 + z^3 + t^3\n$$\n",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** a) do not exist; b) exist.\n\n**Solution.**\n\nb) It is enough to provide an example: $2016 = 1000 + 1000 + 8 + 8$.\n\na) Let us first note that $2016 = 2^5 \\cdot 3^2 \\cdot 7$. And consider remainders of $a^3, b^3, c^3$ modulo $7$. The remainders could equal $0$ or $\\pm 1$. Thus, if $2016 = x^3 + y^3 + z^3$ then at least one of the numbers is divisible by $7$. It is easy to see that it must be $7$, because $7^3 = 343 < 2016$, and the next number $14^3 = 2744 > 2016$. Thus two other numbers must satisfy the condition $y^3 + z^3 = 1673$.\n\nNow, suppose that $y \\leq z$. Since $12^3 = 1728 > 1673$ then $z \\leq 11$. Moreover, $1673 = y^3 + z^3 \\leq 2z^3$ then $837 \\leq z^3$, and it means that $10 \\leq z$. To finish the solution we just need to check cases $z=10, 11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16023,
    "subject": "Mathematics (Olympiad)",
    "question": "Two circles $c$ and $c'$ with centers $O$ and $O'$ lie completely outside each other. Points $A$, $B$, and $C$ lie on the circle $c$ and points $A'$, $B'$, and $C'$ lie on the circle $c'$ so that segment $AB \\parallel A'B'$, $BC \\parallel B'C'$, and $\\angle ABC = \\angle A'B'C'$. The lines $AA'$, $BB'$, and $CC'$ are all different and intersect in one point $P$, which does not coincide with any of the vertices of the triangles $ABC$ or $A'B'C'$. Prove that $\\angle AOB = \\angle A'O'B'$.\n\n![](images/Estonija_2012_p6_data_f72db8e298.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The triangles $ABP$ and $A'B'P$ are similar, because their corresponding sides are parallel (see the figure). Hence $\\frac{|AB|}{|A'B'|} = \\frac{|BP|}{|B'P|}$. Likewise, the triangles $BCP$ and $B'C'P$ are similar, so $\\frac{|BC|}{|B'C'|} = \\frac{|BP|}{|B'P|}$. Thus, $\\frac{|AB|}{|A'B'|} = \\frac{|BC|}{|B'C'|}$, and since $\\angle ABC = \\angle A'B'C'$, the triangles $ABC$ and $A'B'C'$ are also similar. From the equality of the angles $ACB$ and $A'C'B'$, the equality of the central angles $AOB$ and $A'O'B'$ now follows.\n\n_Remark_: Figure 1 corresponds to the case when the vectors $\\overrightarrow{AB}$ and $\\overrightarrow{A'B'}$ have the same direction. If they have opposite directions, then the figure is different (the intersection point $P$ lies on the segments $AA'$, $BB'$, and $CC'$), but the argument is still correct.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16024,
    "subject": "Mathematics (Olympiad)",
    "question": "Primes $p$, $q$, $r$ such that $p + q < 111$ fulfill the equality\n$$\n\\frac{p + q}{r} = p - q + r.\n$$\nFind the maximum value of the product $pqr$.",
    "options": [],
    "answer": "See solution",
    "solution": "Rewrite the condition as:\n$$\nq(r + 1) - p(r - 1) = r^2.\n$$\nIf $r > 2$, then $r$ is odd, so the left side is even, but $r^2$ is odd, which is impossible. Thus, $r = 2$.\n\nWith $r = 2$, the condition becomes $p = 3q - 4$. To maximize $pqr$, maximize $q$ under $p + q < 111$.\n\nIf $q = 23$, $p = 65$ (not prime).\nIf $q = 19$, $p = 53$ (prime).\n\nThus, $pqr = 53 \\times 19 \\times 2 = 2014$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16025,
    "subject": "Mathematics (Olympiad)",
    "question": "Given that $a + b + c + d + 2 \\times 30 = 360$, find the average of $a$, $b$, $c$, and $d$.",
    "options": [],
    "answer": "See solution",
    "solution": "$a + b + c + d + 2 \\times 30 = 360$ \n\nTherefore, $a + b + c + d = 360 - 60 = 300$. \n\nThe average is $\\frac{300}{4} = 75$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16026,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to choose 24 points in space such that no three of them lie on the same line, and choose 2013 planes so that each plane passes through at least 3 of the chosen points, and every triple of points belongs to at least one of the chosen planes?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose it is possible. Let $\\pi_1, \\pi_2, \\dots, \\pi_{2013}$ be the planes, and let $n_1, n_2, \\dots, n_{2013}$ be the numbers of chosen points on each plane. By the conditions, $n_i \\ge 3$ for each $i$. Clearly,\n\n$$\n\\binom{n_1}{3} + \\binom{n_2}{3} + \\dots + \\binom{n_{2013}}{3} = \\binom{24}{3} = 2024.\n$$\n\nAlso, $n_i \\le 5$ for all $i$, because if any $n_i \\ge 6$, then\n\n$$\n2024 = \\underbrace{1+1+\\dots+1}_{2012} + \\binom{6}{3} = 2012 + 20 = 2032,\n$$\n\nwhich is too large. \n\nSuppose there are $a$ planes with 3 points, $b$ with 4 points, and $c$ with 5 points. Then $a + b + c = 2013$ and $a \\cdot 1 + b \\cdot 4 + c \\cdot 10 = 2024$. Thus, $3b + 9c = 11$, which is impossible for nonnegative integers $b$ and $c$.\n\n**Answer:** Impossible.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16027,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a fixed positive integer. A finite sequence of integers $x_1, x_2, \\dots, x_n$ is written on a blackboard. Pepa and Geoff are playing a game that proceeds in rounds as follows.\n\n- In each round, Pepa first partitions the sequence that is currently on the blackboard into two or more contiguous subsequences (that is, consisting of numbers appearing consecutively). However, if the number of these subsequences is larger than 2, then the sum of numbers in each of them has to be divisible by $k$.\n- Then Geoff selects one of the subsequences that Pepa has formed and wipes all the other subsequences from the blackboard.\n\nThe game finishes once there is only one number left on the board. Prove that Pepa may choose his moves so that independently of the moves of Geoff, the game finishes after at most $3k$ rounds.",
    "options": [],
    "answer": "See solution",
    "solution": "A finite sequence of integers is called a *word* and any of its contiguous subsequences is called a *subword*. For a word $u$, let $\\sum u$ denote the sum of numbers in $u$. A *prefix* of a word is a subword starting at the beginning of the word, and a prefix is *proper* if it is neither empty nor the whole word. Analogously, we define suffixes.\n\nFor a word $u$, let $R(u) \\subseteq \\{0, 1, \\dots, k-1\\}$ be the set of remainders $r$ modulo $k$ for which there exists a proper prefix $v$ of $u$ with $\\sum v \\equiv r \\pmod{k}$. In other words, $R(u)$ comprises different remainders modulo $k$ realized by sums of numbers in proper prefixes of $u$. Define the *rank* of $u$ as $|R(u)|$.\n\nWe shall prove the following statement: given a word $u$ on the board, Pepa can always play at most 3 rounds so that the rank of the remaining word is strictly smaller than the rank of $u$. Since the rank of the initial word is at most $k$ and the rank of a word is 0 if and only if it consists of one number, in this way Pepa may force the end of the game within at most $3k$ rounds.\n\nAssume then that the word $u$ on the board has length larger than 1, and take any $r \\in R(u)$. Suppose that the proper prefixes of $u$ giving remainder $r$ modulo $k$ end at positions $1 \\leq i_1 < i_2 < \\dots < i_p < |u|$, where $p \\geq 1$. Consider the following partition of $u$ into subwords:\n\n$$\n\\nu = v_0 v_1 v_2 \\dots v_{p-1} v_p,\n$$\n\nwhere $v_0$ is the prefix up to position $i_1$, each $v_j$ for $j = 1, 2, \\dots, p-1$ is the subword between positions $i_j + 1$ and $i_{j+1}$, and $v_p$ is the suffix from position $i_p + 1$ till the end of the word.\n\nWe observe that the rank of each subword $v_j$ is strictly smaller than the rank of $u$. For $j = 0$ this is trivial: since $i_1$ is the first position at which a prefix of $u$ has sum congruent to $r$ modulo $k$, we have $R(v_0) \\subseteq R(u) \\setminus \\{r\\}$. For $j > 0$, take any proper prefix $w$ of $v_j$, let $w' = v_0 v_1 \\dots v_{j-1} w$, and let $a$ be the remainder of $\\sum v_0 v_1 \\dots v_{j-1}$ modulo $k$. Observe that $\\sum w' \\equiv a + \\sum w \\pmod{k}$. Therefore, the remainders realized by proper prefixes $w$ of $v_j$ are exactly the remainders realized by prefixes $w'$ as above with $a$ subtracted modulo $k$. Since between $i_j$ and $i_{j+1}$ there is no position at which a prefix of $u$ has sum congruent to $r$ modulo $k$, we infer that none of the prefixes $w'$ as above has sum congruent to $r$ modulo $k$. This implies that $R(v_j) \\subseteq \\{q - a : q \\in R(u) \\setminus \\{r\\}\\}$, so $|R(v_j)| < |R(u)|$.\n\nNote that $\\sum v_j \\equiv 0 \\pmod{k}$ for each $j = 1, 2, \\dots, p-1$ by construction. All these observations lead to the following three-turn strategy for Pepa:\n\n- Partition $u$ into $v_0$ and $v_1 v_2 \\dots v_p$. If Geoff chooses $v_0$, then the rank of the word has already decreased. Otherwise, Geoff chooses $v_1 v_2 \\dots v_p$.\n- Partition $v_1 v_2 \\dots v_p$ into $v_1 v_2 \\dots v_{p-1}$ and $v_p$. If Geoff chooses $v_p$, then the rank of the word has already decreased. Otherwise, Geoff chooses $v_1 v_2 \\dots v_{p-1}$.\n- Partition $v_1 v_2 \\dots v_{p-1}$ into $v_1, v_2, \\dots, v_{p-1}$, which are all words with sums of numbers divisible by $k$. Regardless of the move of Geoff, the rank of the word chosen by him is strictly smaller than the rank of $u$.\n\nThis concludes the proof. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16028,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all perfect squares whose decimal representation consists of a sequence of one or more 2's followed by 25. For example, 25 and 225 are such numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Since a perfect square cannot have its last digit 2, $N$ must be of the form:\n\n$$\n\\begin{aligned}\nN &= 22\\dots225 = 22\\dots200 + 25 = 100 \\cdot 2 \\cdot 11\\dots1 + 25 \\\\\n  &= 100 \\cdot 2 \\cdot \\frac{10^{n-1} - 1}{9} + 25,\n\\end{aligned}\n$$\n\nwhere the digit 2 appears $n$ times, $n \\ge 1$.\n\nWorking mod 10, $N = (10k + v)^2$, $0 \\le v < 10$, where $k$ is a positive integer. Only $N = (10k + 5)^2$ gives a last two digits of 25. Thus,\n\n$$\n100 \\cdot 2 \\cdot \\frac{10^{n-1}-1}{9} + 25 = (10k + 5)^2. \\quad (1)\n$$\n\nOr equivalently,\n\n$$\n9k^2 + 9k - 2(10^{n-1} - 1) = 0. \\quad (2)\n$$\n\nSince $k$ is a positive integer, the discriminant of (2) must be a perfect square: $\\Delta = 9(8 \\cdot 10^{n-1} + 1)$, so\n\n$$\n8 \\cdot 10^{n-1} + 1 = x^2\n$$\n\nfor some odd integer $x > 1$. If $x = 2m + 1$, $m \\ge 1$,\n\n$$\n8 \\cdot 10^{n-1} = x^2 - 1 = (x - 1)(x + 1) = 4m(m + 1)\n$$\n\nso\n\n$$\n2^n \\cdot 5^{n-1} = m(m + 1), \\quad (3)\n$$\n\nwith $(m, m + 1) = 1$. Consider cases for $n$:\n\n- If $n = 1$, $m = 1$, $k = 0$, $N = 25$.\n- If $n = 2$, $m = 4$, $k = 1$, $N = 225$.\n- If $n \\ge 3$, since $5^{n-1} > 4^{n-1} = 2^{2n-2} > 2^n$, from (3) we get $2^n = m$ and $5^{n-1} = m + 1$, so $m = 5^{n-1} - 1$, but $m > 2^n = m$, a contradiction.\n\nThus, the only such perfect squares are $25$ and $225$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16029,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{N} = \\{0, 1, 2, \\dots\\}$ be the set of all non-negative integers. For each $n \\in \\mathbb{N}$, define the Catalan number\n\n$$\nC_n = \\frac{1}{n+1} \\binom{2n}{n} = \\frac{(2n)!}{n!(n+1)!}.\n$$\n\nProve that for any positive integer $m$, we have\n\n$$\n\\sum_{\\substack{i,j,k \\in \\mathbb{N} \\\\ i+j+k=m}} C_{i+j}C_{i+k}C_{j+k} = \\frac{3}{2m+3} C_{2m+1}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof 1.* Consider the set of ordered triples:\n\n$$\nA = \\{ (u, v, w) \\in \\mathbb{N}^3 \\mid u, v, w \\le m; u + v + w = 2m \\},\n$$\n\n$$\nB = \\{(u, v, w) \\in \\mathbb{N}^3 \\mid u + v + w = 2m\\}, \\quad B_1 = \\{(u, v, w) \\in B \\mid u \\ge m + 1\\},\n$$\n\n$$\nB_2 = \\{(u, v, w) \\in B \\mid v \\ge m + 1\\}, \\quad B_3 = \\{(u, v, w) \\in B \\mid w \\ge m + 1\\}.\n$$\n\nThen $A = B \\setminus (B_1 \\cup B_2 \\cup B_3)$. Therefore,\n\n$$\n\\begin{aligned}\n\\text{LHS of (1)} &= \\sum_{\\substack{i,j,k \\in \\mathbb{N} \\\\ i+j+k=m}} C_{i+j}C_{i+k}C_{j+k} = \\sum_{(u,v,w) \\in A} C_u C_v C_w \\\\\n&= \\sum_{(u,v,w) \\in B} C_u C_v C_w - \\sum_{(u,v,w) \\in B_1} C_u C_v C_w - \\sum_{(u,v,w) \\in B_2} C_u C_v C_w - \\sum_{(u,v,w) \\in B_3} C_u C_v C_w.\n\\end{aligned}\n$$\n\nBy symmetry, the sums over $B_1$, $B_2$, and $B_3$ are the same. Using the recurrence relation for Catalan numbers:\n\n$$\nC_{n+1} = \\sum_{i=0}^{n} C_i C_{n-i} = C_0 C_n + C_1 C_{n-1} + \\dots + C_n C_0,\n$$\n\nwe have\n\n$$\n\\begin{aligned}\n\\sum_{(u,v,w) \\in B} C_u C_v C_w &= \\sum_{u=0}^{2m} C_u \\left( \\sum_{v=0}^{2m-u} C_v C_{2m-u-v} \\right) = \\sum_{u=0}^{2m} C_u C_{2m+1-u} \\\\\n&= \\left( \\sum_{u=0}^{2m+1} C_u C_{2m+1-u} \\right) - C_{2m+1} C_0 = C_{2m+2} - C_{2m+1},\n\\end{aligned}\n$$\n\n$$\n\\begin{aligned}\n\\sum_{(u,v,w) \\in B_1} C_u C_v C_w &= \\sum_{u=m+1}^{2m} C_u \\left( \\sum_{v=0}^{2m-u} C_v C_{2m-u-v} \\right) = \\sum_{u=m+1}^{2m} C_u C_{2m+1-u} \\\\\n&= \\frac{1}{2} \\left( \\sum_{u=1}^{2m} C_u C_{2m+1-u} \\right) = \\frac{1}{2} (C_{2m+2} - 2C_{2m+1}).\n\\end{aligned}\n$$\n\nSince\n\n$$\n\\begin{aligned}\nC_{2m+2} &= \\frac{(4m+4)!}{(2m+2)!(2m+3)!} = \\frac{(4m+3)(4m+4)}{(2m+2)(2m+3)} \\cdot \\frac{(4m+2)!}{(2m+1)!(2m+2)!} \\\\\n&= \\frac{2(4m+3)}{2m+3} \\cdot C_{2m+1},\n\\end{aligned}\n$$\n\nwe have\n\n$$\n\\begin{aligned}\n\\text{LHS of (1)} &= \\sum_{(u,v,w) \\in A} C_u C_v C_w = (C_{2m+2} - C_{2m+1}) - 3 \\cdot \\frac{1}{2} (C_{2m+2} - 2C_{2m+1}) \\\\\n&= 2C_{2m+1} - \\frac{1}{2}C_{2m+2} = \\frac{3}{2m+3}C_{2m+1} = \\text{RHS of (1).} \\quad \\square\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16030,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x^2 + f(y)) = f(xy) \\quad \\text{for all } x, y \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "All constant functions $f(x) = c$ for any real constant $c$ satisfy the given equation.\n\n**Solution:**\n\nWe are given\n$$\nf(x^2 + f(y)) = f(xy) \\quad \\text{for all } x, y \\in \\mathbb{R}. \\qquad (1)\n$$\n\nPut $y = 0$ in (1):\n$$\nf(x^2 + f(0)) = f(0) \\quad \\text{for all } x \\in \\mathbb{R}. \\qquad (2)\n$$\n\nObserve that $x^2 + f(0)$ covers all real numbers greater than or equal to $f(0)$. It follows that\n$$\nf(x) = f(0) \\quad \\text{whenever } x \\ge f(0). \\qquad (3)\n$$\n\nNext, choose $a \\ge f(0)$ with $a \\ne 0$. Put $y = a$ into (1) and use (3) and then (2):\n$$\nf(xa) = f(x^2 + f(a)) = f(x^2 + f(0)) = f(0).\n$$\n\nSince $a \\ne 0$, the expression $xa$ covers all real numbers as $x$ ranges over $\\mathbb{R}$. Hence $f(x) = f(0)$ for all $x \\in \\mathbb{R}$. Thus $f$ is a constant function. Any such function satisfies (1). $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16031,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n, k$ 為正整數且 $n > k$。\n\n有甲、乙兩人：\n\n1. 甲先私自在紙條上寫下一個 $n$ 位數的 01 序列，並在黑板上寫下所有與這個 01 序列恰有 $k$ 個位數不同的所有長度為 $n$ 的 01 序列。例如，若 $n = 3, k = 1$，且紙條上的 01 序列為 $101$，則甲必須在黑板上寫下 $001$、$111$ 和 $100$。\n\n2. 接著，乙看著黑板上的所有序列，試圖猜測紙條上的序列是什麼。他每次可以猜一個 $n$ 位數的 01 序列，而甲必須誠實回答他是否猜對了。\n\n對於每組 $(n, k)$，試求最小的正整數 $m$，使得乙存在一個猜測策略，能保證在 $m$ 次猜測內猜到正解。",
    "options": [],
    "answer": "See solution",
    "solution": "解：$m(n, k) = 1$ 若 $n \\neq 2k$；$m(n, k) = 2$ 若 $n = 2k$。\n\n令紙條上的序列為 $X$。\n\n先考慮 $n \\neq 2k$。若 $X$ 的首項為 1，則在黑板上將有 $C(n-1, k)$ 條首項為 1，$C(n-1, k-1)$ 條首項為 0，注意到 $C(n-1, k) \\neq C(n-1, k-1)$。由此可知，乙只需計算首項為 1 與為 0 的序列數量，便可確知 $X$ 的首項為何。依據同樣的方法，乙可以確知 $X$ 的每一項為何。故乙第一次便可猜中。\n\n現在考慮 $n = 2k$。當 $k = 1$ 時，易知需猜兩次。當 $k \\ge 2$ 時，注意到若我們將紙條上的 $X$ 的每一項都換掉（0 換成 1，1 換成 0），則黑板上仍會寫下完全相同的序列，故至少需要猜兩次。因此，只須證明猜兩次即可猜中正確答案：\n\n- 若 $X$ 的頭兩位數是相同的，則在黑板上，01 與 10 帶頭的數列將各有 $C(2k-2, k-1)$ 條，而 00 與 11 帶頭的數列將各有 $C(2k-2, k)$ 條。\n- 反之，若 $X$ 的頭兩位數不同，則在黑板上，01 與 10 帶頭的數列將各有 $C(2k-2, k)$ 條，而 00 與 11 帶頭的數列將各有 $C(2k-2, k-1)$ 條。\n- 由於 $C(2k-2, k-1) \\neq C(2k-2, k)$，乙只需計算對應條數的數量，便可知 $X$ 的首兩項是否相同。\n- 同理，乙可以確知 $X$ 的任兩位數是否相同。\n- 因此乙僅需猜測 $X$ 的首項為何即可，而這最多只需猜兩次。證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16032,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a prime number $p$. Let $A$ be a $p \\times p$ matrix whose entries are exactly $1, 2, \\dots, p^2$ in some order. The following operation is allowed: add one to each number in a row or a column, or subtract one from each number in a row or a column. The matrix $A$ is called \"good\" if one can take a finite series of such operations resulting in a matrix with all entries zero. Find the number of good matrices $A$.",
    "options": [],
    "answer": "See solution",
    "solution": "We may combine the operations on the same row or column, so the final result of a series of operations can be realized as subtracting integers $x_i$ from each number of the $i$-th row and subtracting integers $y_j$ from each number of the $j$-th column. Thus, the matrix $A$ is good if and only if there exist integers $x_i, y_j$ such that $a_{ij} = x_i + y_j$ for all $1 \\le i, j \\le p$.\n\nSince the entries of $A$ are distinct, $x_1, x_2, \\dots, x_p$ are pairwise distinct, and so are $y_1, y_2, \\dots, y_p$. We may consider only the case that $x_1 < x_2 < \\dots < x_p$ since swapping the values of $x_i$ and $x_j$ results in swapping the $i$-th and $j$-th rows, which is again a good matrix. Similarly, we may consider only the case that $y_1 < y_2 < \\dots < y_p$, thus the matrix is increasing from left to right and from top to bottom.\n\nFrom the assumptions above, we have $a_{11} = 1$, and $a_{12}$ or $a_{21}$ equals $2$. We may consider only the case that $a_{12} = 2$ since the transpose of the matrix is again good. Now, we argue by contradiction that the first row is $1, 2, \\dots, p$. Assume on the contrary that $1, 2, \\dots, k$ is on the first row, but $k+1$ is not, $2 \\le k < p$, therefore $a_{21} = k+1$. We call $k$ consecutive integers a \"block\", and we shall prove that the first row consists of several blocks, that is, the first $k$ numbers is a block, the next $k$ numbers is again a block, and so on.\n\nIf it is not so, assume the first $n$ groups of $k$ numbers are \"blocks\", ...",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16033,
    "subject": "Mathematics (Olympiad)",
    "question": "Anna has placed real numbers with sum $S$ in the cells of a row. It turned out that she cannot cut the row into two parts so that the sum of the numbers in one part is positive and in the other part is negative. Prove that the modulus of $S$ is not less than any of Anna's numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $S = 0$. For any division of the row into two parts, the sums of the two parts add to $0$, so one is not less than $0$ and the other is not greater than $0$. If either sum is nonzero, one is positive and the other negative, contradicting the condition. Thus, all partial sums are $0$, so all numbers are $0$, and $|S| = 0$ is not less than any number.\n\nNow suppose $S \\neq 0$, and without loss of generality, $S > 0$. Let $a$ be any number in the row. For any division, both sums must be nonnegative (since their sum is $S > 0$), so all partial sums are at least $0$.\n\nLet $x$ and $y$ be the sums of the numbers to the left and right of $a$ (possibly $0$ if $a$ is at an end). Then $x, y \\geq 0$, and $x + a, y + a \\geq 0$. Thus, $x + y \\geq 0$ and $x + y + 2a \\geq 0$. Since $S = x + y + a$, we have $S - a = x + y \\geq 0$ and $S + a = x + y + 2a \\geq 0$. Therefore, $S - a \\geq 0$ and $S + a \\geq 0$, so $-S \\leq a \\leq S$, or $|a| \\leq |S|$ for any $a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16034,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ and $n$ be positive integers. Prove that, if $x_j$ are real numbers for $1 \\leq j \\leq n$, such that\n\n$$\n\\sum_{j=1}^{n} \\frac{1}{x_j^{2k} + k} = \\frac{1}{k}\n$$\n\nthen\n\n$$\n\\sum_{j=1}^{n} \\frac{1}{x_j^{2k+1} + k + 2} \\leq \\frac{1}{k+1}\n$$\nmust hold.",
    "options": [],
    "answer": "See solution",
    "solution": "We can, in fact, show that each of the expressions in the second sum is not greater than the corresponding expression in the first, multiplied by the factor $\\frac{k}{k+1}$.\n\nSubstituting $y := x_j^{2k}$, this means that we wish to show\n\n$$\n\\frac{1}{y^2 + k + 2} \\leq \\frac{k}{k+1} \\cdot \\frac{1}{y + k}\n$$\n\nSince $y$ is certainly positive for $k > 0$, this is equivalent to $(k+1)(y + k) \\leq k(y^2 + k + 2)$, or $P(y) = k y^2 - (k+1) y + k \\geq 0$. This polynomial is quadratic in $y$, and we have $P(0) = k > 0$. For the discriminant of the polynomial we have\n\n$$\n(k+1)^2 - 4k^2 = -3k^2 + 2k + 1 \\leq -3k^2 + 3k = -3k(k-1) \\leq 0,\n$$\n\nand we see that the polynomial can only assume positive values, which completes the proof. QED.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16035,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(G, \\cdot)$ be a finite group of order $n \\in \\mathbb{N}^*$, with $n \\ge 2$. We call the group $(G, \\cdot)$ *arrangeable* if there is an ordering of its elements such that\n\n$$\nG = \\{a_1, a_2, \\dots, a_n\\} = \\{a_1 \\cdot a_2, a_2 \\cdot a_3, \\dots, a_n \\cdot a_1\\}.\n$$\n\na) Determine all positive integers $n$ for which the group $(\\mathbb{Z}_n, +)$ is arrangeable.\n\nb) Give an example of an arrangeable group of even order.",
    "options": [],
    "answer": "See solution",
    "solution": "a) We will show that the group $(\\mathbb{Z}_n, +)$ is arrangeable if and only if $n \\ge 2$ is an odd positive integer.\n\nIf $(G, \\cdot)$ is an abelian arrangeable group, then considering the arrangement $G = \\{a_1, a_2, \\dots, a_n\\} = \\{a_1 \\cdot a_2, a_2 \\cdot a_3, \\dots, a_n \\cdot a_1\\}$, we have\n\n$$\n\\prod_{g \\in G} g = \\prod_{k=1}^{n} a_k = \\prod_{k=1}^{n} (a_k \\cdot a_{k+1}) = \\left( \\prod_{g \\in G} g \\right)^2,\n$$\n\n(where $a_{n+1} = a_1$), so $\\prod_{g \\in G} g = 1$, where $1$ is the unit element of the group $(G, \\cdot)$.\n\nIn any finite abelian group, the product of all the elements is equal to the product of all its elements of order $2$.\n\nFor $n \\ge 2$, if $k \\in \\{0, 1, \\dots, n-1\\}$ with $\\operatorname{ord}(\\hat{k}) = 2$, then\n\n$$\n\\hat{k} \\neq \\hat{0} = \\hat{k} + \\hat{k} = 2\\hat{k},\n$$\n\nso $n$ divides $2k$, but does not divide $k$. This is only possible if $n$ is even and $n = 2k$. Thus, if $n$ is even, with $n = 2k$, and $(\\mathbb{Z}_n, +)$ were arrangeable, we would have\n\n$$\n\\hat{0} = \\sum_{x \\in \\mathbb{Z}_n} x = \\hat{k},\n$$\n\nwhich is false. Hence, if $n$ is even, the group $(\\mathbb{Z}_n, +)$ is not arrangeable.\n\nNow let $n \\ge 3$ be an odd positive integer. Consider the set $[0, n-1]_N = \\{0, 1, \\dots, n-1\\}$ and the function $f : [0, n-1]_N \\to \\mathbb{Z}_n$, defined by $f(k) = \\overline{2k+1}$. Since\n\n$$\nf(k) = f(l) \\iff \\overline{2k+1} = \\overline{2l+1} \\iff n \\mid (2k-2l) \\iff n \\mid (k-l) \\iff k = l,\n$$\n\nthe function $f$ is injective, and since $[0, n-1]_N$ and $\\mathbb{Z}_n$ have equal cardinality, $f$ is bijective. Denoting $a_k = \\overline{k-1}$ for $1 \\le k \\le n$ and $a_{n+1} = a_1$, it follows that $a_k + a_{k+1} = f(k-1)$ for any $k = 1, \\dots, n$, so that\n\n$$\n\\mathbb{Z}_n = \\{a_1, a_2, \\dots, a_n\\} = \\{f(0), f(1), \\dots, f(n-1)\\} = \\{a_1 + a_2, a_2 + a_3, \\dots, a_n + a_1\\},\n$$\nwhence we deduce that the group $(\\mathbb{Z}_n, +)$ is arrangeable.\n\nThe set of all positive integers such that the group $(\\mathbb{Z}_n, +)$ is arrangeable is thus the set of all odd positive integers $n$, with $n \\ge 3$.\n\nb) According to part a), there are no cyclic arrangeable groups of even order. Consider $\\mathbb{Z}_4 = \\{\\hat{0}, \\hat{1}, \\hat{2}, \\hat{3}\\}$, $\\mathbb{Z}_2 = \\{\\overline{0}, \\overline{1}\\}$, and the group $G = \\mathbb{Z}_4 \\times \\mathbb{Z}_2$ with component-wise addition $(\\hat{k}, \\hat{l}) + (\\hat{m}, \\hat{n}) = (\\overline{k+m}, \\overline{l+n})$. Then\n\n$$\nG = \\{a_1 = (\\hat{0}, \\overline{0}), a_2 = (\\hat{1}, \\overline{0}), a_3 = (\\hat{1}, \\overline{1}), a_4 = (\\hat{3}, \\overline{1}), a_5 = (\\hat{2}, \\overline{0}), a_6 = (\\hat{2}, \\overline{1}), a_7 = (\\hat{0}, \\overline{1}), a_8 = (\\hat{3}, \\overline{0})\\}\n$$\n\nand\n\n$$\nG = \\{a_1 + a_2, a_2 + a_3, a_3 + a_4, a_4 + a_5, a_5 + a_6, a_6 + a_7, a_7 + a_8, a_8 + a_1\\},\n$$\nso $(G, +)$ is an arrangeable group of order $8$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16036,
    "subject": "Mathematics (Olympiad)",
    "question": "Denote $\\mathbb{Z}_{>0} = \\{1, 2, 3, \\dots\\}$ as the set of all positive integers. Determine all functions $f : \\mathbb{Z}_{>0} \\to \\mathbb{Z}_{>0}$ such that, for each positive integer $n$:\n\n1. $\\sum_{k=1}^{n} f(k)$ is a perfect square.\n2. $f(n)$ divides $n^3$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use induction on $n$ to show that $f(n) = n^3$ for all positive integers $n$.\n\nThe base case, $n = 1$, is clear.\n\nAssume $f(m) = m^3$ for all $m < n$ with $n \\ge 2$. Then\n$$\n\\sum_{k=1}^{n-1} f(k) = \\frac{n^2(n-1)^2}{4}.\n$$\nBy the first condition,\n$$\nf(n) = \\sum_{k=1}^{n} f(k) - \\sum_{k=1}^{n-1} f(k) = \\left( \\frac{n(n-1)}{2} + k \\right)^2 - \\frac{n^2(n-1)^2}{4} = k(n^2 - n + k),\n$$\nfor some positive integer $k$.\n\nThe divisibility condition implies $k(n^2 - n + k) \\le n^3$, or $(n-k)(n^2+k) \\ge 0$, so $k \\le n$.\n\nAlso, $n^2 - n + k$ must divide $n^3$. If $k < n$, then\n$$\nn < \\frac{n^3}{n^2 - 1} \\le \\frac{n^3}{n^2 - n + k} \\le \\frac{n^3}{n^2 - n + 1} < \\frac{n^3 + 1}{n^2 - n + 1} = n + 1,\n$$\nso $\\frac{n^3}{n^2-n+k}$ cannot be an integer.\n\nThus, $k = n$, so $f(n) = n^3$. This completes the induction and the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16037,
    "subject": "Mathematics (Olympiad)",
    "question": "設集合 $K$ 由有限多個正整數所組成。已知 $K$ 能被分割成兩個非空子集 $A$, $B$，滿足：對任意的 $a \\in A$ 及任意的 $b \\in B$，$ab + 1$ 總是一個完全平方數。證明：$K$ 中的最大元素至少是\n\n$$\n\\left\\lceil \\left(2 + \\sqrt{3}\\right)^{\\min\\{|A|,|B|\\}-1} \\right\\rceil,\n$$\n\n其中 $|X|$ 代表集合 $X$ 的元素個數，$\\lceil x \\rceil$ 為大於或等於實數 $x$ 的最小整數。",
    "options": [],
    "answer": "See solution",
    "solution": "The statement is trivial when $\\min\\{|A|,|B|\\} = 1$. From now on we assume that $|A|, |B| \\ge 2$.\n\n**Lemma 1** Let $a < c$ be two elements in $A$, and $b < d$ be two elements in $B$. Then\n\n$$\n(c - a)(d - b) > 2\\sqrt{abcd}.\n$$\n\n*Proof.* By assumption, all of $ab + 1$, $ad + 1$, $cb + 1$, and $cd + 1$ are squares, so are\n\n$$\nM = (ab + 1)(cd + 1), \\quad N = (ad + 1)(cb + 1).\n$$\n\nTheir difference is $M - N = (ab+cd) - (ad+bc) = (c-a)(d-b) > 0$. Since both squares $M$ and $N$ are larger than $abcd$, their difference $M - N$ is larger than $2\\sqrt{abcd}$. $\\square$\n\nBack to the original problem, let us write out the elements in $A$ and in $B$, respectively, as\n\n$$\nA: a_1 < a_2 < \\dots < a_m, \\quad B: b_1 < b_2 < \\dots < b_n.\n$$\n\nLet $x = \\min_i a_{i+1}/a_i$ and $y = \\min_j b_{j+1}/b_j$. By Lemma 1 we have\n\n$$\n(x^{1/2} - x^{-1/2})(y^{1/2} - y^{-1/2}) > 2.\n$$\n\nWithout loss of generality we may assume that\n\n$$\nx^{1/2} - x^{-1/2} > \\sqrt{2},\n$$\n\nwhich is equivalent to\n\n$$\nx^2 - 4x + 1 > 0 \\quad \\Rightarrow \\quad x > 2 + \\sqrt{3}.\n$$\n\nTherefore\n\n$$\na_m > a_1 x^{m-1} \\ge 1 \\cdot (2 + \\sqrt{3})^{m-1},\n$$\n\nand the proof is now finished. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16038,
    "subject": "Mathematics (Olympiad)",
    "question": "a) How many distinct rotations does a dodecahedron have? For a dodecahedron with 10 marked vertices, show that in any rotation (other than the identity), there is a state where fewer than 5 marked vertices coincide with their original positions.\n\nb) Give an example of a marking of 10 vertices such that, in every rotation, at least 4 marked vertices coincide with their original positions.",
    "options": [],
    "answer": "See solution",
    "solution": "a) There are 12 choices for the face to be placed down, and for each, 5 possible orientations, giving $12 \\times 5 = 60$ rotations (including the identity), so 59 non-identity rotations.\n\nFor a marked vertex $A$, there are 10 marked vertices, and each can be mapped to $A$ in 3 ways, so $3 \\times 10 - 1 = 29$ non-identity rotations map a marked vertex to $A$. Thus, in total, $29 \\times 10 = 290$ coincidences occur over all non-identity rotations.\n\nBy the pigeonhole principle, in some rotation, fewer than $\\frac{290}{59} < 5$ coincidences occur, as required.\n\nb) Mark the vertices of two opposite faces (each face has 5 vertices, so 10 in total). In any rotation, at least 4 marked vertices coincide with their original positions. If there were a rotation with at most 3 coincidences, then one marked face would have at most one marked vertex coinciding, but each face has at least two marked vertices, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16039,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an equilateral triangle. Let $C_1$ and $C_2$ be on $AB$, $B_1$ and $B_2$ on $AC$, and $A_1$ and $A_2$ on $BC$ such that $\\overline{A_1A_2} = \\overline{B_1B_2} = \\overline{C_1C_2}$. Let $A_2B_1$ and $B_2C_1$, $B_2C_1$ and $C_2A_1$, $C_2A_1$ and $A_2B_1$ intersect at $E$, $F$, $G$ correspondingly. Prove that the triangle formed by the segments $B_1A_2$, $A_1C_2$, and $C_1B_2$ is similar to $\\triangle EFG$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us denote the triangle formed by the segments $B_1A_2$, $A_1C_2$, and $C_1B_2$ as $\\triangle A_3B_3C_3$. Let $P$ be a point in the interior of $\\triangle EFG$ such that $C_1C_2PB_2$ is a parallelogram. Then $\\triangle B_2PB_1$ is equilateral, hence $PA_1A_2B_1$ is a parallelogram. From these observations, we get that $PC_2 \\parallel EF$, $PA_1 \\parallel EG$. Now it's obvious that $\\triangle PC_2A_1 \\sim \\triangle EFG$, and because $\\triangle PC_2A_1 \\cong \\triangle A_3B_3C_3$, we conclude that $\\triangle A_3B_3C_3 \\sim \\triangle EFG$.\n![](images/Makedonija_2009_p62_data_40eceb771f.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16040,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the 100th digit after the decimal point in the decimal expansion of a repeating decimal with a 6-digit repeating part: $076923$?",
    "options": [],
    "answer": "See solution",
    "solution": "There are 6 digits in the repeating part of the decimal form. $100 = 6 \\times 16 + 4$, so the 100th digit will be the 4th in the repeating part $076923$, which is $9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16041,
    "subject": "Mathematics (Olympiad)",
    "question": "For each $n \\in \\mathbb{N}^*$$, let $d_n$ be the number to be found.\n\nConsider a table of size $2 \\times n$. Fill each square of the table, in order from left to right and top to bottom, with the numbers from $1$ to $2n$.\n\nCall the $n$-th square of the first row and the first square of the second row *special squares*.\n\nTwo numbers $a, b \\in T$ satisfy $|a - b| \\in \\{1, n\\}$ if and only if they lie in two squares with a common side or if they lie in the special squares.\n\n![](images/Vijetnam_2009_p4_data_b6e8a7038c.png)\n\nDefine $d_n$ as the number of ways to choose some squares of the table (including the choice of 0 squares) such that no two chosen squares share a common side and the two special squares are not chosen together.\n\nFor each $n \\in \\mathbb{N}^*$, let:\n\n- $k_n$ be the number of choices of squares in which no two squares share a common side.\n- $s_n$ be the number of choices of squares in which no two squares share a common side and the two special squares are chosen.\n\nThus, $d_n = k_n - s_n$.\n\n*Compute $k_n$.*\n\nLet $t_n$ be the number of choices satisfying the above property for a $2 \\times n$ table with one square removed.\n\nThen $k_n = k_{n-1} + 2t_{n-1}$.\n\nAlso, $t_n = k_{n-1} + t_{n-2}$.\n\nTherefore,\n\n$$\nk_n = 2k_{n-1} + k_{n-2} \\quad \\forall n \\ge 3.\n$$\n\nDirect count gives $k_1 = 3$ and $k_2 = 7$.\n\nThe recurrence has characteristic equation $x^2 - 2x - 1 = 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\nk_n = C_1(1 + \\sqrt{2})^n + C_2(1 - \\sqrt{2})^n \\quad \\forall n \\ge 1.\n$$\n\nUsing $k_1 = 3$ and $k_2 = 7$, we find $C_1 = \\frac{1+\\sqrt{2}}{2}$ and $C_2 = \\frac{1-\\sqrt{2}}{2}$.\n\n$$\nk_n = \\frac{(1 + \\sqrt{2})^{n+1} + (1 - \\sqrt{2})^{n+1}}{2}.\n$$\n\n*Compute $s_n$.*\n\nWe have $s_1 = 0$, $s_2 = s_3 = 1$, and for $n \\ge 4$:\n\n$$\ns_n = h_{n-2},\n$$\nwhere $h_n$ is the number of choices for a $2 \\times n$ table with two squares removed.\n\n![](images/Vijetnam_2009_p4_data_b6e8a7038c.png)\n\nSince $s_3 = 1$, let $h_1 = 1$. Direct count gives $h_2 = 4$.\n\nFor $n \\ge 3$:\n\n- $k_{n-2}$ choices with neither square A nor B chosen,\n- $2t_{n-2}$ choices with exactly one of A or B chosen,\n- $h_{n-2}$ choices with both A and B chosen.\n\nThus,\n\n$$\nh_n = k_{n-2} + 2t_{n-2} + h_{n-2} = k_{n-1} + h_{n-2} \\quad \\forall n \\ge 3.\n$$\n\nFrom the recurrences:\n\n$$\n2h_n - k_n = 2h_{n-2} - k_{n-2} \\quad \\forall n \\ge 3.\n$$\n\nSo,\n\n$$\n2h_n - k_n = (-1)^n \\quad \\forall n \\ge 1.\n$$\n\nTherefore,\n\n$$\ns_n = h_{n-2} = \\frac{k_{n-2} + (-1)^{n-2}}{2} \\quad \\forall n \\ge 3.\n$$\n\n*Final formula:*\n\n$$\nd_1 = 3, \\quad d_2 = 6, \\quad d_n = \\frac{2k_n - k_{n-2} + (-1)^{n-3}}{2} \\quad \\forall n \\ge 3,\n$$\nwhere $k_n$ is as above.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16042,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a point on the curve $y = x + \\frac{2}{x}$ for $x > 0$. Through $P$, draw lines perpendicular to $y = x$ and to the $y$-axis, with foot points $A$ and $B$, respectively. Then, the value of $\\vec{PA} \\cdot \\vec{PB}$ is \\_\\_\\_\\_.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P(x_0, x_0 + \\frac{2}{x_0})$.\n\nThe line through $P$ perpendicular to $y = x$ has slope $-1$:\n$$y - \\left(x_0 + \\frac{2}{x_0}\\right) = - (x - x_0)$$\nwhich simplifies to\n$$y = -x + 2x_0 + \\frac{2}{x_0}.$$\n\nSetting $y = x$ to find $A$:\n$$x = -x + 2x_0 + \\frac{2}{x_0}$$\n$$2x = 2x_0 + \\frac{2}{x_0}$$\n$$x = x_0 + \\frac{1}{x_0}$$\nSo $A(x_0 + \\frac{1}{x_0}, x_0 + \\frac{1}{x_0})$.\n\nThe line perpendicular to the $y$-axis through $P$ is a horizontal line, so $B(0, x_0 + \\frac{2}{x_0})$.\n\nCompute vectors:\n$$\\vec{PA} = (x_A - x_0, y_A - y_P) = \\left(\\frac{1}{x_0}, -\\frac{1}{x_0}\\right)$$\n$$\\vec{PB} = (x_B - x_0, y_B - y_P) = (-x_0, 0)$$\n\nDot product:\n$$\\vec{PA} \\cdot \\vec{PB} = \\frac{1}{x_0} \\cdot (-x_0) + (-\\frac{1}{x_0}) \\cdot 0 = -1$$\n\n**Answer:** $-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16043,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a = px$, $b = py$, $c = pz$ where $p$, $x$, $y$, $z$ are pairwise coprime numbers. Consider the equation:\n\n$$(pxy)^2 + (pxz)^2 + (pyz)^2 = (p^2)^3$$\n\nShow that there are infinitely many sets of three natural numbers $(p, y, z)$ for which $p^2 = y^2 + yz + z^2$ is true.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = px$, $b = py$, $c = pz$ where $p$, $x$, $y$, $z$ are pairwise coprime numbers. The equation becomes:\n\n$$(pxy)^2 + (pxz)^2 + (pyz)^2 = (p^2)^3 \\Leftrightarrow x^2y^2 + x^2z^2 + y^2z^2 = p^4$$\n\n$$\\Leftrightarrow x^2(y^2 + z^2) = (p^2 - yz)(p^2 + yz).$$\n\nIf $p^2 = y^2 + yz + z^2$, then\n\n$$x^2(y^2 + z^2) = (y^2 + 2yz + z^2)(y^2 + z^2) \\Rightarrow x = y + z.$$\n\nThus, the solution is the set $(p(y+z), py, pz)$.\n\nTo show there are infinitely many such triples $(p, y, z)$, let $u = \\frac{y}{p}$, $v = \\frac{z}{p}$. We need to show that $u^2 + uv + v^2 = 1$ has infinitely many rational solutions. One point is $(1, 0)$. For any rational $k$, the line $v = k(u-1)$ intersects the curve $u^2 + uv + v^2 = 1$ at another rational point. By Vieta's theorem, this point is rational.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16044,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be an odd prime. Show that\n\n$$\n|(2 + \\sqrt{5})^p| - 2^{p+1}\n$$\n\nis divisible by $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Using the binomial theorem:\n\n$$(2 + \\sqrt{5})^p + (2 - \\sqrt{5})^p = \\sum_{k=0}^{\\frac{p-1}{2}} \\binom{p}{2k} 2^{p+1-2k} \\cdot 5^k = 2^{p+1} + \\sum_{k=1}^{\\frac{p-1}{2}} \\binom{p}{2k} 2^{p+1-2k} \\cdot 5^k,$$\n\nwhich is an integer. Since $p$ is odd, $(2 - \\sqrt{5})^p < 0$, so\n\n$$\n|(2 + \\sqrt{5})^p| = (2 + \\sqrt{5})^p + (2 - \\sqrt{5})^p.\n$$\n\nThus,\n\n$$\n|(2 + \\sqrt{5})^p| - 2^{p+1} = \\sum_{k=1}^{\\frac{p-1}{2}} \\binom{p}{2k} 2^{p+1-2k} \\cdot 5^k.\n$$\n\nFor $1 \\leq r \\leq p-1$, $\\binom{p}{r}$ is divisible by $p$, so the sum is divisible by $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16045,
    "subject": "Mathematics (Olympiad)",
    "question": "The sum of angles $A$ and $C$ of a convex quadrilateral $ABCD$ is less than $180\\degree$. Prove that\n$$\nAB \\cdot CD + AD \\cdot BC < AC(AB + AD).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $s$ be the circumcircle of $ABD$. Then the point $C$ is outside this circle but inside the angle $BAD$.\n\nApply inversion with center $A$ and radius $1$. This inversion maps the circle $s$ to the line $s' = B'D'$, where $B'$ and $D'$ are the images of $B$ and $D$. The point $C$ goes to the point $C'$ inside the triangle $AB'D'$. Therefore, $B'C' + C'D' < AB' + AC'$.\n\nNow, due to inversion properties, we have $B'C' = \\dfrac{BC}{AB} \\cdot AC$, $C'D' = \\dfrac{CD}{AC} \\cdot AD$, $AB' = \\dfrac{1}{AB}$, $AC' = \\dfrac{1}{AC}$. Thus,\n$$\n\\frac{BC}{AB} \\cdot AC + \\frac{CD}{AC} \\cdot AD < \\frac{1}{AB} + \\frac{1}{AC}.\n$$\nMultiplying both sides by $AB \\cdot AC \\cdot AD$ gives the desired inequality.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16046,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Does there exist a set $A$ of 2016 pairwise different positive integers such that for every non-empty subset $B \\subset A$ (with $B \\neq A$) and every non-empty subset $C \\subset (A \\setminus B)$, the sum of the elements of $B$ is not divisible by the sum of the elements of $C$?\n\nb) Does there exist a set $A$ of 2016 pairwise different positive integers such that for every non-empty subset $B \\subset A$ (with $B \\neq A$ and $|B| \\geq 2$) and every non-empty subset $C \\subset (A \\setminus B)$, the product of the elements of $B$ is divisible by the sum of the elements of $C$?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** a), b) Yes, such a set exists.\n\n**Solution.**\n\na) Let us first choose the numbers $2, 3, 5, \\ldots, p_{2016}$, where $p_i$ is the $i$-th prime number in increasing order. Consider all possible pairs of subsets $B$ and $C$ as described; their number is finite, say $N$. We will iteratively modify the elements of $A$ to ensure the required property for each pair. Denote the elements as $a_1, a_2, \\ldots, a_{2016}$, initially $a_i = p_i$. For each pair $(B, C)$, let $p = p_{2016+k}$ for the $k$-th pair. Multiply all elements of $C$ by $p$, and in $B$, multiply all but the smallest-indexed element by $p$. Thus, the sum of $C$ is divisible by $p$, but the sum of $B$ is not, so the sum of $B$ is not divisible by the sum of $C$. Repeating this for all $N$ pairs, we obtain a set $A$ with the desired property. This construction works for any finite number of elements.\n\nb) For any $n$, let $N = \\left(\\frac{n(n+1)}{2}\\right)!$ and define $a_i = i \\cdot N$ for $i = 1, \\ldots, n$. For any subset $C$, its sum $L$ satisfies $L \\leq \\frac{n(n+1)}{2} N$, so $L = K N$ for some $K \\leq \\frac{n(n+1)}{2}$. The product of any two elements of $A$ is divisible by $N^2$, hence by $L$, since $N$ is divisible by any $K$ in this range. Thus, the product of the elements of $B$ is divisible by the sum of the elements of $C$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16047,
    "subject": "Mathematics (Olympiad)",
    "question": "Гурвалжин $A_1A_2A_3$-ын $A_2A_3$, $A_1A_3$, $A_1A_2$ талууд дээр харгалзан $B_1$, $B_2$, $B_3$ цэгүүдийг $A_1B_1$, $A_2B_2$, $A_3B_3$ шулуунууд нэг цэгт огтлолцдог байхаар сонгов. Мөн $B_1$, $B_2$, $B_3$ гурвалжны $B_2B_3$, $B_1B_3$, $B_1B_2$ талууд дээр харгалзан $C_1$, $C_2$, $C_3$ цэгүүдийг сонгов. $A_1C_1$, $A_2C_2$, $A_3B_3$ шулуунууд нэг цэгт огтлолцох гарцаагүй бөгөөд хүрэлшээтэй нөхцөл нь $B_1C_1$, $B_2C_2$, $B_3C_3$ шулуунууд нэг цэгт огтлолцох явдал мөн гэдгийг батал.",
    "options": [],
    "answer": "See solution",
    "solution": "$S_i = A_iC_i \\cap C_{i-1}C_{i+1}$,  \n$T_i = A_iC_i \\cap A_{i-1}A_{i+1}$ байг. ($i=1,2,3$)\n\nЧевийн теоремээр:\n$$\n\\frac{A_3B_1 \\cdot A_1B_2 \\cdot A_2B_3}{B_1A_2 \\cdot B_2A_3 \\cdot B_3A_1} = 1 \\quad \\text{ба} \\quad \\frac{B_2S_1 \\cdot B_3S_2 \\cdot B_1S_3}{S_1B_3 \\cdot S_2B_1 \\cdot S_3B_2} = 1 \\quad (1)\n$$\n\nМөн талбайн харьцаанаас:\n$$\n\\frac{A_{i-1}B_i \\cdot T_iA_{i+1}}{B_iA_{i+1} \\cdot A_{i-1}T_i} = \\frac{B_{i+1}S_1 \\cdot C_iB_{i-1}}{S_iB_{i-1} \\cdot B_{i+1}C_i}, \\quad i = 1,2,3. \\quad (2)\n$$\n\nЭндээс гарах гурван тэнцэтгэлийг үржүүлбэл:\n$$\n\\begin{aligned}\n\\frac{A_3B_1 \\cdot A_1B_2 \\cdot A_2B_3}{B_1A_2 \\cdot B_2A_3 \\cdot B_3A_1} \\cdot \\frac{T_1A_2 \\cdot T_2A_3 \\cdot T_3A_1}{A_3T_1 \\cdot A_1T_2 \\cdot A_2T_3} \\\\\n= \\frac{B_2S_1 \\cdot B_3S_2 \\cdot B_1S_3}{S_1B_3 \\cdot S_2B_1 \\cdot S_3B_2} \\cdot \\frac{C_1B_3 \\cdot C_2B_1 \\cdot C_3B_2}{B_2C_1 \\cdot B_3C_2 \\cdot B_1C_3}\n\\end{aligned}\n$$\n\nЭндээс (1)-ээс:\n$$\n\\frac{T_1A_2 \\cdot T_2A_3 \\cdot T_3A_1}{A_3T_1 \\cdot A_1T_2 \\cdot A_2T_3} = \\frac{C_1B_3 \\cdot C_2B_1 \\cdot C_3B_2}{B_2C_1 \\cdot B_3C_2 \\cdot B_1C_3}\n$$\n\nЭндээс Чевийн теоремээр бидний батлах ёстой зүйл гарна.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16048,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer and let $x_1, x_2, \\dots, x_n$ and $y_1, y_2, \\dots, y_n$ be real numbers. Prove that there exists a number $i$, $i = 1, 2, \\dots, n$, such that\n$$\n\\sum_{j=1}^{n} |x_i - x_j| \\le \\sum_{j=1}^{n} |x_i - y_j|.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, suppose $x_1 \\le x_2 \\le \\dots \\le x_n$. For each $k = 1, 2, \\dots, n$ we have $|x_1 - x_k| + |x_n - x_k| = |x_1 - x_n| \\le |x_1 - y_k| + |x_n - y_k|$, hence\n$$\n\\sum_{k=1}^{n} |x_1 - x_k| + \\sum_{k=1}^{n} |x_n - x_k| \\le \\sum_{k=1}^{n} |x_1 - y_k| + \\sum_{k=1}^{n} |x_n - y_k|.\n$$\nThe claim holds for $i = 1$ or $i = n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16049,
    "subject": "Mathematics (Olympiad)",
    "question": "На страните на остроаголен триаголник $ABC$ лежат точките $A_1, A_2, B_1, B_2, C_1$ и $C_2$, така што $\\overline{AA_1} = \\overline{A_1A_2} = \\overline{A_2B} = \\frac{1}{3}\\overline{AB}$, $\\overline{BB_1} = \\overline{B_1B_2} = \\overline{B_2C} = \\frac{1}{3}\\overline{BC}$ и $\\overline{CC_1} = \\overline{C_1C_2} = \\overline{C_2A} = \\frac{1}{3}\\overline{CA}$. Нека $k_A, k_B$ и $k_C$ се опишаните кружници на триаголниците $AA_1C_2, BB_1A_2$ и $CC_1B_2$ соодветно, $a_B$ и $a_C$ се тангентите на $k_A$ во $A_1$ и $C_2$, $b_C$ и $b_A$ се тангентите на $k_B$ во $B_1$ и $A_2$ и $c_A$ и $c_B$ се тангентите на $k_C$ во $C_1$ и $B_2$. Докажи дека нормалите спуштени од пресекот на $a_B$ и $b_A$ на $AB$, пресекот на $b_C$ и $c_B$ на $BC$ и пресекот на $c_A$ и $a_C$ на $CA$ се сечат во една точка.",
    "options": [],
    "answer": "See solution",
    "solution": "Да ги означиме пресеците на $a_B, b_A$ и $c_A$ соодветно со $A'$, $B'$ и $C'$. \n\n![](images/Macedonia_2013_p27_data_d41f575be8.png)\n\nТриаголникот $AA_1C_2$ е сличен на $ABC$, бидејќи имаат заеднички агол и страните им се во сооднос $1:3$. Нека $O_A$ е центарот на опишаната кружница на триаголникот $AA_1C_2$, тогаш:\n\n$$\n\\angle O_A A_1 A = \\frac{1}{2}(180^\\circ - \\angle A O_A A_1) = 90^\\circ - \\angle C_2 A_1 = 90^\\circ - \\gamma\n$$\n\nБидејќи $a_B$ е нормална на $O_AA_1$, следува дека аголот меѓу $a_B$ и $AB$ е еднаков на\n\n$$\n180^\\circ - 90^\\circ - (90^\\circ - \\gamma) = \\gamma\n$$\n\nАналогно, аголот меѓу $b_A$ и $AB$ е исто така еднаков на $\\gamma$, па триаголникот $A_1A_2A'$ е рамнокрак со основа $A_1A_2$, т.е. нормалата на $AB$ низ $C'$ минува низ средината на $A_1A_2$, која е истовремено и средина на $AB$, па нормалата минува и низ центарот на опишаната кружница на триаголникот $ABC$. Од причини на симетрија, сите три нормали минуваат низ центарот на опишаната кружница, т.е. низ една точка.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16050,
    "subject": "Mathematics (Olympiad)",
    "question": "In a cyclic convex hexagon $ABCDEF$, lines $AB$ and $DC$ intersect at $G$, and lines $AF$ and $DE$ intersect at $H$. Let $M$ and $N$ be the circumcenters of $\\triangle BCG$ and $\\triangle EFH$, respectively. Prove that the lines $BE$, $CF$, and $MN$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/CHN_TSExams_2022_p2_data_cb0cf25da0.png)\n\n**Proof 1:** Let $\\omega$ be the circumcircle of $\\triangle BCG$. Let $BE$ intersect $\\omega$ at another point $E'$, and $CF$ intersect $\\omega$ at another point $F'$. Note that $\\angle BE'F' = \\angle BCF' = \\angle BCF = \\angle BEF$, so $EF \\parallel E'F'$. Similarly, $\\angle CF'G = \\angle CBG = \\angle CBA = \\angle CFA = \\angle CFH$. So $GF' \\parallel HF$. For the same reason, we deduce that $GE' \\parallel HE$. Hence, $\\triangle E'F'G$ and $\\triangle EFH$ are homothetic. Let $P$ denote their homothetic center. Then $EE'$ and $FF'$ both pass through $P$. Moreover, the line connecting the circumcenters of $\\triangle E'F'G$ and $\\triangle EFH$ also passes through $P$, i.e., $BE$, $CF$, and $MN$ are concurrent at $P$.\n\n![](images/CHN_TSExams_2022_p2_data_9bc64bc9da.png)\n\n**Proof 2:** First, apply Pascal's theorem to the circle-inscribed hexagon $BAFCDE$; we deduce that $G$, $P$, and $H$ are collinear.\n\nLet $A'$ be the point opposite $A$, and $D'$ be the point opposite $D$. Let $BA'$ and $CD'$ intersect at $G'$, and $D'E$ and $A'F$ intersect at $H'$. Apply Pascal's theorem to the circle-inscribed hexagon $BA'FCD'E$; we know that $G'$, $H'$, and $P$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16051,
    "subject": "Mathematics (Olympiad)",
    "question": "Докажите, что не существуют такие натуральные числа $a, b, c \\geq 2$, для которых выполнены условия:\n- $a^2$ делится на $b + c$,\n- $b^2$ делится на $c + a$,\n- $c^2$ делится на $a + b$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Первое решение.** Предположим противное: пусть нашлись такие числа $a, b, c$. Заметим, что числа $a+b, b+c, c+a$ попарно взаимно просты. В самом деле, пусть, скажем, числа $a+b, b+c$ делятся на некоторое простое $p$. Поскольку $c^2 : (a+b)$, $a^2 : (b+c)$, то числа $c$ и $a$ также делятся на $p$, а тогда и $b = (a+b) - a$ на него делится, что противоречит условию.\n\nДалее, поскольку $a^2 : (b+c)$, число $(a+b+c)^2 = a^2 + (b+c)(2a+b+c)$ делится на $b+c$. Аналогично, оно делится на $a+b$ и на $c+a$. Так как последние три числа попарно взаимно просты, $(a+b+c)^2$ делится на $(a+b)(a+c)(b+c)$; в частности, $(a+b+c)^2 \\geq (a+b)(b+c)(c+a)$. С другой стороны, ясно, что все числа $a, b, c$ не меньше 2, значит,\n$$\n(a+b)(b+c)(c+a) = (a^2b + b^2c + c^2a) + (ab^2 + bc^2 + ca^2) + 2abc > (2a^2 + 2b^2 + 2c^2) + (2ab + 2bc + 2ca) > (a+b+c)^2.\n$$\n\nПротиворечие.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16052,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a function $f: \\mathbb{R} \\to \\mathbb{R}$ such that for any real numbers $x, y$, the following inequality holds:\n\n$$\nf(x - f(y)) \\leq x - y f(x)?\n$$\n",
    "options": [],
    "answer": "See solution",
    "solution": "Assume such a function exists. Substitute $y = 0$ into the inequality:\n\n$$\nf(x - f(0)) \\leq x.\n$$\n\nNow, let $x = x + f(0)$:\n\n$$\nf(x) \\leq x + f(0). \\quad (1)\n$$\n\nNext, substitute $x = f(y)$ into the original inequality:\n\n$$\nf(f(y) - f(y)) \\leq f(y) - y f(f(y)),\n$$\nwhich simplifies to:\n$$\nf(0) \\leq f(y) - y f(f(y)).\n$$\nRearranging gives:\n$$\ny f(f(y)) \\leq f(y) - f(0).\n$$\nFor $y < 0$, dividing both sides by $y$ (which is negative) reverses the inequality:\n$$\nf(f(y)) \\geq \\frac{f(y) - f(0)}{y}.\n$$\nBut from (1), $f(y) \\leq y + f(0)$, so:\n$$\nf(f(y)) \\geq \\frac{f(y) - f(0)}{y} \\geq \\frac{y + f(0) - f(0)}{y} = 1.\n$$\nThus, for all $y < 0$, $f(f(y)) \\geq 1$. But as $y \\to -\\infty$, $f(f(y))$ cannot remain bounded below by $1$ if $f$ is arbitrary, leading to a contradiction. Therefore, such a function does not exist.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16053,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying the equality\n$$\nf(2^x + 2y) = 2^y f(f(x)) f(y)\n$$\nfor every $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* $f(x) = 0$ and $f(x) = 2^x$.\n\n*Solution.* Substituting $y = -2^{x-1}$ into the original equation gives\n$$\nf(0) = \\frac{1}{2^{2^{x-1}}} f(f(x)) f(-2^{x-1}). \\quad (1)\n$$\nSo, if $f(-2^x) = 0$ for at least one $x$ then also $f(0) = 0$. Then taking $x = 0$ and arbitrary $y$ in the original identity gives $f(1+2y) = 0$, i.e., $f \\equiv 0$.\n\nAssume in the rest that $f(-2^x) \\neq 0$ for every $x$. Substituting $y = -2^x$ into the original equation gives $f(-2^x) = \\frac{1}{2^{2^x}} f(f(x)) f(-2^x)$. Hence, for every $x$,\n$$\nf(f(x)) = 2^{2^x} \\quad (2)\n$$\nSubstituting (2) into the original equation and taking $y = 0$, we obtain $f(2^x) = f(f(x))f(0) = 2^{2^x}f(0)$ for all $x$, which implies\n$$\nf(x) = 2^x f(0) \\quad (3)\n$$\nfor all positive $x$. On the other hand, applying (2) to (1) gives\n$$\nf(-2^{x-1}) = \\frac{2^{2^{x-1}}}{2^{2^x}} \\cdot f(0) = 2^{-2^{x-1}} f(0)\n$$\nfor all $x$, which implies (3) also for all negative $x$.\n\nWe have shown above that $f(0) = 0$ implies $f(x) = 0$ for all $x$. Hence we may assume that $f(0)$ is either positive or negative. By taking $x = 0$ in (2) and applying (3), we obtain $2 = 2^{2^0} = f(f(0)) = 2^{f(0)} \\cdot f(0)$. Both $f(0) < 1$ and $f(0) > 1$ would lead to contradiction, hence $f(0) = 1$ and the only non-zero solution is thus $f(x) = 2^x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16054,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a unit cell $2 \\times 2010$ table. Ivan puts a horizontal domino, which covers exactly 2 of the cells, then Peter puts a vertical domino, which covers exactly 2 of the cells, then again Ivan puts a horizontal domino, and so on. The player who has no move loses the game. Determine which of the two players has a winning strategy.",
    "options": [],
    "answer": "See solution",
    "solution": "We describe a winning strategy for Ivan. He divides the table into 502 tables of size $2 \\times 4$ and one table of size $2 \\times 2$. Then he puts his first domino in the $2 \\times 2$ table (in either row). Peter is forced to put his domino in a $2 \\times 4$ table. Now Ivan puts a domino anywhere in the same $2 \\times 4$ table. It is easy to see that Ivan has one more way to put a domino in the same table, and after two moves each, all the cells from this table are covered. After covering all cells in all $2 \\times 4$ tables, it is Peter's turn and he has no move. Therefore, Ivan wins.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16055,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle inscribed in circle $(O)$ with $\\angle A = 45^\\circ$ and $AB < AC$. Let $AD$ and $AH$ be the angle bisector and altitude from $A$ to $BC$, with $D$ and $H$ on $BC$. Suppose that $OD$ intersects $AH$ at $E$, and $K$ is the circumcenter of triangle $EBC$. Prove that $HK \\parallel AD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ and $N$ be the projections of $O$ and $K$ on $AE$, and $M$ and $T$ be the midpoints of $BC$ and the minor arc $BC$ of $(O)$. Let $AH = h$ and $R, R'$ be the radii of $(O)$ and $(K)$, respectively.\n\n![](images/Saudi_Booklet_2025_p28_data_2f78648c57.png)\n\nBy Thales' theorem, we have\n\n$$\n\\frac{HE}{HA} = \\frac{MO}{MT} = \\frac{MO}{OT - OM} = \\sqrt{2} + 1 \\implies HE = (\\sqrt{2} + 1)h.\n$$\n\nAccording to the Pythagorean theorem, $BK^2 - BO^2 = KM^2 - OM^2$, so\n\n$$\nR'^2 - R^2 = KM^2 - OM^2 \\implies KM^2 = R'^2 - \\frac{R^2}{2}.\n$$\n\nSimilarly,\n\n$$\n\\begin{align*}\nAO^2 - (AH - OM)^2 &= OP^2 = KN^2 = KE^2 - (EH - MK)^2 \\\\\n\\Leftrightarrow R^2 - \\left(h - \\frac{R}{\\sqrt{2}}\\right)^2 &= R'^2 - \\left((\\sqrt{2} + 1)h - MK\\right)^2 \\\\\n\\Leftrightarrow MK^2 + h R\\sqrt{2} &= R'^2 - \\frac{R^2}{2} - (\\sqrt{2} + 1)^2 h^2 + (2 + 2\\sqrt{2})h MK \\\\\n\\Leftrightarrow R\\sqrt{2} &= (2 + 2\\sqrt{2})h + (2 + 2\\sqrt{2})MK \\\\\n\\Leftrightarrow MK &= h + \\frac{R\\sqrt{2}}{2 + 2\\sqrt{2}} = h + MT.\n\\end{align*}\n$$\n\nThus $KT = MK - MT = h = AH$, proving that $AHKT$ is a parallelogram. Therefore, $AD \\parallel HK$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16056,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are $n$ questions on an exam, and each question is solved by exactly 4 students. For any two distinct questions $Q_i$ and $Q_j$, there is exactly one student who solved both $Q_i$ and $Q_j$. What is the maximum possible value of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $S_1$ solved $Q_1, \\dots, Q_k$, but not $Q_{k+1}$ where $k > 4$. Since $Q_{k+1}$ is solved by exactly 4 students, and $Q_i$ and $Q_{k+1}$ are solved by exactly one student for each $1 \\le i \\le k$, there must be another student who solved two of the questions $Q_1, \\dots, Q_k$ besides $S_1$, a contradiction. Therefore $k \\le 4$.\n\nSince $Q_1$ is solved by exactly four students, the number of questions cannot be more than $1 + 4 \\cdot 3 = 13$.\n\nOn the other hand, an exam with 13 questions where\n\n$Q_1$ is solved by $S_1, S_2, S_3, S_4$;\n$Q_2$ is solved by $S_1, S_5, S_6, S_7$;\n$Q_3$ is solved by $S_1, S_8, S_9, S_{10}$;\n$Q_4$ is solved by $S_1, S_{11}, S_{12}, S_{13}$;\n$Q_5$ is solved by $S_2, S_5, S_9, S_{13}$;\n$Q_6$ is solved by $S_2, S_6, S_{10}, S_{11}$;\n$Q_7$ is solved by $S_2, S_7, S_8, S_{12}$;\n$Q_8$ is solved by $S_3, S_5, S_{10}, S_{12}$;\n$Q_9$ is solved by $S_3, S_6, S_8, S_{13}$;\n$Q_{10}$ is solved by $S_3, S_7, S_9, S_{11}$;\n$Q_{11}$ is solved by $S_4, S_5, S_8, S_{11}$;\n$Q_{12}$ is solved by $S_4, S_6, S_9, S_{12}$;\n$Q_{13}$ is solved by $S_4, S_7, S_{10}, S_{13}$\n\nsatisfies the conditions of the question.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16057,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\Gamma$ be a circle in the plane and $S$ be a point on $\\Gamma$. Two brothers, Mario and Luigi, drive around the circle $\\Gamma$ with their go-karts. They both start at $S$ at the same time and drive for exactly 6 minutes at constant speed counterclockwise around the track. During these 6 minutes, Luigi makes exactly one lap around $\\Gamma$ while Mario, who is three times as fast, accomplishes three laps.\n\nWhile Mario and Luigi drive their go-karts, Princess Daisy positions herself such that she is always exactly in the middle of the two brothers. When she reaches a point she has already visited, she marks it with a banana.\n\nHow many points in the plane, apart from $S$, are marked with a banana after the race?",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $\\Gamma$ is the unit circle and $S = (1, 0)$. Three points are marked with bananas:\n\n1. After 45 seconds, Luigi is at $(\\sqrt{2}/2, \\sqrt{2}/2)$ and Mario is at $(-\\sqrt{2}/2, \\sqrt{2}/2)$. Daisy is at $(0, \\sqrt{2}/2)$. After 135 seconds, their positions are swapped, so Daisy is again at $(0, \\sqrt{2}/2)$ and marks it.\n\n2. After 225 seconds and 315 seconds, Daisy is at $(0, -\\sqrt{2}/2)$ and marks it.\n\n3. After 90 seconds, Luigi is at $(0, 1)$ and Mario at $(0, -1)$, so Daisy is at $(0, 0)$. After 270 seconds, their positions are swapped, so Daisy marks $(0, 0)$.\n\nNo other point, apart from these three and $S$, is marked. Suppose Daisy is at the same place at two different times $t_1$ and $t_2$. Let Luigi's position at time $t_n$ be $z_n = \\exp(ix_n)$ ($x_n \\in (0, 2\\pi)$). Mario is at $z_n^3$, Daisy at $\\frac{z_n^3 + z_n}{2}$.\n\nIf Daisy is at the same place at $t_1$ and $t_2$, then $\\frac{z_1^3 + z_1}{2} = \\frac{z_2^3 + z_2}{2}$, so $(z_1 - z_2)(z_1^2 + z_1z_2 + z_2^2 + 1) = 0$. Since $z_1 \\neq z_2$, $z_1^2 + z_1z_2 + z_2^2 = -1$.\n\nConsider $z = z_1/z_2$ on the unit circle. $z + 1 + z^{-1}$ is real for $|z| = 1$, and lies on the unit circle only if it equals $1$ (real part of $z$ is $0$) or $-1$ (real part of $z$ is $-1$). Applying this to $z = z_1/z_2$, only $z = \\pm i$ or $z = -1$ are possible, corresponding to the three cases above.\n\nThus, apart from $S$, Daisy marks exactly three points with bananas.\n\n![](images/BW2021_Shortlist_p14_data_bc5229f2fb.png)\n\n![](images/BW2021_Shortlist_p14_data_1f82c6056c.png)\n\nDepiction of the path Daisy takes.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16058,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute, scalene triangle with orthocenter $H$, and let $D$, $E$, $F$ be the feet of the altitudes from $A$, $B$, and $C$ respectively. Let $(I)$ be the circumcircle of triangle $HEF$ with center $I$, and let $K$ and $J$ be the midpoints of $BC$ and $EF$ respectively. The line $HJ$ meets $(I)$ again at $G$, and $GK$ meets $(I)$ again at $L$.\n\n**a)** Prove that $AL$ is perpendicular to $EF$.\n\n**b)** Let $AL$ intersect $EF$ at $M$. The line $IM$ meets the circumcircle of triangle $IEF$ again at $N$. The line $DN$ intersects $AB$ and $AC$ at $P$ and $Q$ respectively. Prove that $PE$, $QF$, and $AK$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** It is well known that $KE$ and $KF$ are both tangent to $(I)$. Thus, $GK$ is the symmedian of $\\triangle GEF$, so $\\overrightarrow{LE} = \\overrightarrow{HF}$. Hence, $AH$ and $AL$ are isogonal with respect to $\\angle BAC$. It is clear that $AH$ is the diameter of $(I)$. Therefore, $AL$ is the altitude of $\\triangle AEF$.\n\n**b)** Since $I$ is the midpoint of $AH$, it is clear that $(IEF)$ is the Euler circle of $\\triangle ABC$ with diameter $IK$. Also,\n\n$$\n\\overline{MI} \\cdot \\overline{MN} = \\overline{ME} \\cdot \\overline{MF} = \\overline{MA} \\cdot \\overline{ML},\n$$\n\nwhich implies that $A$, $I$, $L$, $N$ are concyclic. Therefore,\n\n$$\n\\angle ANI = \\angle ALI = \\angle LAI = \\angle DIK,\n$$\n\nsince $IK \\parallel AL$ (both lines are perpendicular to $EF$). Hence,\n\n$$\n\\angle AND = \\angle ANI + \\angle IND = \\angle DIK + \\angle IKD = 90^{\\circ}.\n$$\n\nLet $S$ be the radical center of $(I)$, $(IEF)$, and $(ADN)$. Since $EF$ is the radical axis of $(I)$ and $(IEF)$, $EF$ passes through $P$. Similarly, $DN$ passes through $P$. Since the center of $(AND)$, $I$, and $A$ are collinear, $(AND)$ and $(I)$ are tangent at $A$, thus $AS$ is tangent to $(I)$; in other words, $AS \\parallel BC$. Hence, $A(SK, QP) = A(SK, CB) = -1$, so $PE$, $QF$, and $AK$ are concurrent. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16059,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose you want to distribute nine distinct integers greater than or equal to 1 and less than or equal to 9 into nine square boxes of a $3 \\times 3$ grid so that the sums of the numbers in each row and each column are all multiples of 3. How many possible ways are there to distribute the numbers to attain this goal? Regard two configurations, for which one can be obtained from rotating or flipping the other, as distinct when counting the number of possible ways.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us replace every number $n$ distributed into a box in the grid by $p = 0, 1, 2$ such that $n \\equiv p \\pmod{3}$. The condition that the sum of the numbers in each row and column must be a multiple of 3 does not change after this replacement. Among the triples $(a, b, c)$ of integers between 0 and 2 (inclusive), only $(0, 0, 0)$, $(1, 1, 1)$, $(2, 2, 2)$, and $(0, 1, 2)$ and its permutations satisfy the requirement that the sum of the three numbers is a multiple of 3. This means that the only possible triplets to satisfy the requirement after replacing $n$ by $p$ are those with all the same entries (such as $(1, 1, 1)$) or those with all different entries (such as $(0, 1, 2)$ and its permutations).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16060,
    "subject": "Mathematics (Olympiad)",
    "question": "Могут ли существовать 18 подряд идущих \"хороших\" чисел, где \"хорошее\" число — это такое натуральное число, все простые делители которого равны 2 или 3?",
    "options": [],
    "answer": "See solution",
    "solution": "Предположим, что нашлись 18 хороших чисел подряд. Среди них найдутся три числа, делящихся на 6. Пусть это числа $6n$, $6(n+1)$ и $6(n+2)$. Поскольку эти числа — хорошие, и в разложение каждого из них на простые множители входят двойка и тройка, других простых делителей у них быть не может.\n\nДалее, лишь одно из трёх подряд идущих натуральных чисел $n$, $n+1$, $n+2$ может делиться на 3. Значит, остальные два являются степенями двойки. Но пары степеней двойки, отличающихся не более чем на два — это только $(1,2)$ и $(2,4)$; поэтому $n \\le 2$. Однако тогда среди наших 18 чисел есть простое число $13$ (так как $6n \\le 13 \\le 6(n+2)$), не являющееся хорошим. Противоречие.\n\n*Ответ: не могут.*",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16061,
    "subject": "Mathematics (Olympiad)",
    "question": "已知 $m \\ge 0$，$f(x) = x^2 + \\sqrt{m}x + m + 1$。試證：對任意正實數 $x_1, x_2, \\dots, x_n$，\n\n$$\nf\\left(\\sqrt[n]{x_1 x_2 \\cdots x_n}\\right) \\leq \\sqrt[n]{f(x_1) f(x_2) \\cdots f(x_n)}\n$$\n\n且上述等號成立的充要條件為 $x_1 = x_2 = \\cdots = x_n$。",
    "options": [],
    "answer": "See solution",
    "solution": "因 $\\Delta = (\\sqrt{m})^2 - 4(m+1) < 0$，所以 $f(x) > 0$。\n\n利用數學歸納法證明：當 $n = 2^k$，題設成立。\n\n(i) 當 $k=1$ 時，$n=2$。因\n\n$$\n\\sqrt{m} x_1 x_2 (\\sqrt{x_1} - \\sqrt{x_2})^2 + (m+1)(x_1 - x_2)^2 + \\sqrt{m}(m+1)(\\sqrt{x_1} - \\sqrt{x_2})^2 \\geq 0,\n$$\n\n所以\n\n$$\n(x_1 x_2 + \\sqrt{m x_1 x_2} + m + 1)^2 \\leq (x_1^2 + \\sqrt{m} x_1 + m + 1)(x_2^2 + \\sqrt{m} x_2 + m + 1),\n$$\n\n即 $f(\\sqrt{x_1 x_2}) \\leq \\sqrt{f(x_1) f(x_2)}$，且等號成立的充要條件為 $x_1 = x_2$。\n\n(ii) 假設 $n = 2^k$ 時，題設成立。則當 $n = 2^{k+1}$ 時，\n\n$$\n\\begin{align*}\n& f\\left(\\sqrt[2^{k+1}]{x_1 x_2 \\cdots x_{2^{k+1}}}\\right) \\\\\n&= f\\left(\\sqrt{\\sqrt[2^k]{x_1 x_2 \\cdots x_{2^k}} \\cdot \\sqrt[2^k]{x_{2^k+1} x_{2^k+2} \\cdots x_{2^{k+1}}}}\\right) \\\\\n&\\leq \\sqrt{f\\left(\\sqrt[2^k]{x_1 x_2 \\cdots x_{2^k}}\\right) f\\left(\\sqrt[2^k]{x_{2^k+1} x_{2^k+2} \\cdots x_{2^{k+1}}}\\right)} \\\\\n&= \\sqrt[2^{k+1}]{f(x_1) f(x_2) \\cdots f(x_{2^{k+1}})}.\n\\end{align*}\n$$\n\n對於任意正整數 $n$，必存在 $k$，使得 $2^k \\leq n < 2^{k+1}$。令 $G = \\sqrt[n]{x_1 x_2 \\cdots x_n}$，則\n\n$$\nf(G) = f\\left(\\sqrt[2^{k+1}]{x_1 x_2 \\cdots x_n \\cdot G \\cdots G}\\right) \\leq \\sqrt[2^{k+1}]{f(x_1) f(x_2) \\cdots f(x_n) [f(G)]^{2^{k+1} - n}},\n$$\n\n即 $[f(G)]^n \\leq f(x_1) f(x_2) \\cdots f(x_n)$，\n\n$$\nf\\left(\\sqrt[n]{x_1 x_2 \\cdots x_n}\\right) \\leq \\sqrt[n]{f(x_1) f(x_2) \\cdots f(x_n)}.\n$$\n\n等號成立的充要條件為 $x_1 = x_2 = \\cdots = x_n$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16062,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a natural number such that: the first 2009 digits are threes, the next 2009 digits are twos, the next 2009 digits are ones, and the remaining digits are zeros, and this number is a perfect cube of a natural number? (Justify your answer)",
    "options": [],
    "answer": "See solution",
    "solution": "The given number is of the form $n = \\overline{33\\ldots3}\\ \\overline{22\\ldots2}\\ \\overline{11\\ldots1}\\ 000\\ldots$. Let $n = k^3$, where $k \\in \\mathbb{N}$. The sum of the digits of $n$ is $6 \\times 2009 = 12054$. Therefore, $3 \\mid n$, so $3 \\mid k$, and $3^3 \\mid k^3 = n$. However, this is not possible because the sum of the digits of $n$ is not divisible by $9$.\n\nTherefore, such a number does not exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16063,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_0 = 1003$ and define the sequence $a_{n+1} = \\left\\lfloor r a_n \\right\\rfloor$ for $n \\geq 0$, where\n$$\nr = \\frac{2004 + \\sqrt{2004^2 + 4}}{2}.\n$$\nFind $a_{2004} \\bmod 2004$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n$$\nr = \\frac{2004 + \\sqrt{2004^2 + 4}}{2}\n$$\nis an irrational number greater than $1$. Since $a_n \\in \\mathbb{Z}^+$, we have $r a_n \\notin \\mathbb{Q}$. Thus,\n$$\na_{n+1} < r a_n < a_{n+1} + 1\n$$\nfor any $n \\geq 0$. Equivalently,\n$$\na_n - \\frac{1}{r} < \\frac{a_{n+1}}{r} < a_n.\n$$\nAs $r > 1$, this yields $a_n - 1 < \\frac{a_{n+1}}{r} < a_n$, and hence\n$$\n\\left\\lfloor \\frac{a_{n+1}}{r} \\right\\rfloor = a_n - 1.\n$$\nNext, observe that $r = 2004 + \\frac{1}{r}$. It follows that\n$$\na_{n+1} = \\left\\lfloor r a_n \\right\\rfloor = \\left\\lfloor 2004 a_n + \\frac{a_n}{r} \\right\\rfloor = 2004 a_n + a_{n-1} - 1\n$$\nsince $2004 a_n \\in \\mathbb{Z}$ and $\\left\\lfloor \\frac{a_n}{r} \\right\\rfloor = a_{n-1} - 1$ from above. Thus,\n$$\na_{n+1} \\equiv a_{n-1} - 1 \\pmod{2004}\n$$\nfor any $n \\in \\mathbb{Z}^+$. Then,\n$$\na_{2004} \\equiv a_{2002} - 1 \\equiv a_{2000} - 2 \\equiv \\cdots \\equiv a_0 - 1002 \\equiv 1003 \\pmod{2004}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 16064,
    "subject": "Mathematics (Olympiad)",
    "question": "$u + v$ does not belong to $T$. Prove that\n\ni) If $S$ is the set of the first 2006 positive integers, then the number of elements of every stubborn subset $T$ of $S$ does not exceed $1003$.\n\nii) If $S$ consists of 2006 arbitrary positive integers, then there exists a stubborn subset $T$ of $S$ having $669$ elements.",
    "options": [],
    "answer": "See solution",
    "solution": "i) Let $A$ be a stubborn subset of $S = \\{1, 2, \\dots, 2006\\}$ consisting of $x$ elements $a_1 < a_2 < \\dots < a_x$. Consider the set $B = \\{a_2 - a_1, a_3 - a_1, \\dots, a_x - a_1\\}$. It is a subset of $S$ and consists of $x-1$ elements. As $A$ is stubborn, $A \\cap B \\neq \\emptyset$. This implies $x + (x-1) \\leq 2006$, so $x \\leq 1003$.\n\nii) Let $S = \\{a_1, a_2, \\dots, a_{2006}\\}$. Consider the product $P$ of all odd divisors of $\\prod_{i=1}^{2006} a_i$. There exists a prime $p$ of the form $p = 3r+2$ such that $p$ divides $3P+2$ and is coprime to every $a_i$. For each $a \\in S$, the sequence $a, 2a, \\dots, (p-1)a$ (mod $p$) is a permutation of $1, 2, \\dots, p-1$, so there exists a set $S_a$ of $r+1$ integers $x$ in $\\{1, \\dots, p-1\\}$ such that $xa \\bmod p$ belongs to $A = \\{r+1, \\dots, 2r+1\\}$. For each $x \\in \\{1, \\dots, p-1\\}$, let $S_x = \\{a \\in S \\mid xa \\in A\\}$. We have:\n\n$$\n|S_1| + |S_2| + \\dots + |S_{p-1}| = \\sum_{a \\in S} |A_a| = 2006 \\times (r+1)\n$$\n\nSo there exists $x_0$ such that $|S_{x_0}| \\geq \\frac{2006 \\times (r+1)}{3r+1} > 668$. Let $B$ be a subset consisting of $669$ elements of $S_{x_0}$; then $B$ is a stubborn subset of $S$. Indeed, if $u, v, w \\in B$ ($u$ can be equal to $v$), then $x_0u, x_0v, x_0w \\in A$. It is easy to verify that $x_0u + x_0v \\neq x_0w \\pmod{p}$, therefore $u + v \\neq w$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16065,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimum value of $x + y + z$ where $x$, $y$, and $z$ are positive integers such that their least common multiple is $2100$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $2100 = 2^2 \\cdot 3 \\cdot 5^2 \\cdot 7$, let $x = 5^2 = 25$, $y = 7$, and $z = 2^2 \\cdot 3 = 12$. The least common multiple of $x$, $y$, and $z$ is $2100$, so $x + y + z = 44$.\n\nTo show that $44$ is the minimum, suppose $x + y + z < 44$ with $\text{lcm}(x, y, z) = 2100$. At least one of $x, y, z$ must be a multiple of $5^2$. Without loss of generality, let $x = 25$. Then $y + z < 19$. Among $y$ and $z$, one must be a multiple of $2^2$, one of $3$, and one of $7$. Assume $y$ is a multiple of $7$. Since $2^2 \\cdot 7 = 28 > 19$ and $3 \\cdot 7 = 21 > 19$, $y$ cannot be a multiple of $2^2$ or $3$. Thus, $z$ must be a multiple of $2^2 \\cdot 3 = 12$. Therefore, $x + y + z \\geq 25 + 7 + 12 = 44$, which is a contradiction. Thus, the minimum value is $44$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16066,
    "subject": "Mathematics (Olympiad)",
    "question": "Xavier and Olivier are playing tic-tac-toe on a $3 \\times 3$ rectangular grid with modified rules. On each turn, a player chooses an empty square and writes their token in it. Players alternate turns, with Xavier starting. The first player to occupy any three squares that are either all in the same row or column, or that lie in pairwise distinct rows and columns, wins. Does either player have a winning strategy, and if so, who?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, Xavier has a winning strategy.\n\nNote that any two unit squares uniquely determine a third square that, together with the given two, forms a winning triple. Label all unit squares except the center square clockwise with numbers 1 through 8 (see the figures below). Let Xavier start by playing in the center square. If Olivier then moves in square No. $i$, Xavier responds by playing in square No. $i+1$ (with square No. 1 if $i=8$). All possible positions after Xavier's second move (up to rotation) are shown in the figures.\n\nTo avoid losing, Olivier must occupy the square that completes a winning triple with Xavier's two squares. Then, Xavier moves into the square that completes a winning triple with Olivier's two squares (see the figures for all possible positions after Xavier's third move). Olivier cannot win on his next move. Among the three squares occupied by Xavier, there are three pairs, but only one pair has its corresponding third square blocked by Olivier. Thus, Olivier cannot block all possibilities, and Xavier can win on his next move.\n\n![](images/EST_ABooklet_2021_p11_data_3fe6bbd5ab.png)\n\n![](images/EST_ABooklet_2021_p11_data_983df0b56a.png)\n\n![](images/EST_ABooklet_2021_p11_data_a488197b81.png)\n\n![](images/EST_ABooklet_2021_p11_data_92213db149.png)\n\n![](images/EST_ABooklet_2021_p11_data_ae4d0bf1b2.png)\n\nRemark: The strategy described is not the only winning strategy that Xavier has.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16067,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the circumcenter of triangle $ABC$ with $\\angle A = 120^\\circ$. Let $P$ and $Q$ denote the projections of $B$ onto $CO$ and $AO$, respectively. Let $M$ be the midpoint of $AO$. Prove that the circumcircle of $\\triangle MPQ$ touches $AC$.\n\n![](images/Ukraine_2021-2022_p33_data_9d559eda43.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $R$ be the projection of $B$ onto $AC$. Then $P$, $Q$, $R$ are the projections of $B$ onto the sides of $\\triangle AOC$. Let $B_1$ be the isogonal conjugate of $B$ with respect to this triangle. Since $\\angle BAC = \\angle BOC = 120^\\circ$, we have $\\angle B_1AO = \\angle B_1OA$. Then the projection of $B_1$ onto $AC$ is $M$. Thus, the points $P$, $Q$, $R$, and $M$ are concyclic. Thus, it suffices to prove $\\angle QRA = \\angle QPR$. We will use the cyclicity of $ARBQ$, $OQBP$, and $CPBR$. Indeed,\n\n$$\n\\angle RPQ = \\angle RPC - \\angle QPC = \\angle RBC - \\angle QBO = (90^\\circ - \\angle BCA) - (90^\\circ - \\angle BOA) = \\angle BCA,\n$$\n\n$$\n\\angle QRA = \\angle ABQ = 90^\\circ - \\angle BAO = \\angle BCA,\n$$\n\ncompleting the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16068,
    "subject": "Mathematics (Olympiad)",
    "question": "Куќите на една улица се нумерирани од 1 до 100. Колку пати се појавува цифрата $7$ во броевите на куќите?",
    "options": [],
    "answer": "See solution",
    "solution": "Броевите на куќите што ја содржат цифрата $7$ се: $7$, $17$, $27$, $37$, $47$, $57$, $67$, $70$, $71$, $72$, $73$, $74$, $75$, $76$, $77$, $78$, $79$, $87$, $97$. Во тие броеви таа се појавува вкупно $20$ пати (двапати ја има во бројот $77$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16069,
    "subject": "Mathematics (Olympiad)",
    "question": "If each ring-tone lasts for exactly $x$ minutes, and the alarm rings at the following intervals:\n- 5:30–5:33\n- 5:39–5:42\n- 5:45–5:51\n- 5:54–6:30\n\nFor what value of $x$ does the alarm ring for a total of 50 minutes?",
    "options": [],
    "answer": "See solution",
    "solution": "If each ring-tone lasts for exactly 3 minutes, the alarm will be ringing at 5:30–5:33, 5:39–5:42, 5:45–5:51, and 5:54–6:30. It will ring for $3 + 3 + 6 + 36 = 48$ minutes in total. Therefore, we must have $x > 3$.\n\nLet $x = 3 + y$ where $0 < y < 3$. In addition to the above periods, the alarm will ring for $y$ more minutes after each of 5:33, 5:42, and 5:51. Thus, we need $48 + 3y = 50$. The only solution is $y = \\frac{2}{3}$, and hence $x = \\frac{11}{3}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16070,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triplets $(a, b, c)$ of positive integers satisfying the identity:\n\n$$\na^2 + b + 3 = (b^2 - c^2)^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N = |b^2 - c^2|$. Then, from $N^2 = a^2 + b + 3 \\geq 1 + 1 + 3 = 5$, it follows that $N \\geq 3$. In particular, since $b \\neq c$, we get\n\n$$\nN = |b + c||b - c| \\geq (b + 1) \\cdot 1 = b + 1,\n$$\n\nso $b \\leq N - 1$. Since $a^2 < N^2$, we have $a \\leq N - 1$, so\n\n$$\n0 = N^2 - (a^2 + b + 3) \\geq N^2 - (N - 1)^2 - b - 3 = 2N - b - 4 \\geq N - 3,\n$$\n\nshowing $N \\leq 3$. As we have seen that $N \\geq 3$, we get $N = 3$. Then, equality holds throughout, so $a = N - 1 = 2$, $b = N - 1 = 2$, which implies $9 = N^2 = |b^2 - c^2|^2 = (c^2 - 4)^2$. Thus, $c^2 = 1$ or $7$. Since $c$ is a positive integer, $c = 1$. Therefore, the answer is $(a, b, c) = (2, 2, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16071,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that a sequence $ (\\varepsilon_n)_{n \\in \\mathbb{N}} $ of plus and minus ones is periodic with period a power of $2$ if and only if $ \\varepsilon_n = (-1)^{P(n)} $, $ n \\in \\mathbb{N} $, where $P$ is an integer-valued polynomial with rational coefficients.",
    "options": [],
    "answer": "See solution",
    "solution": "A polynomial $P$ of degree at most $k$ with complex coefficients is integer-valued if and only if\n$$\nP = \\sum_{j=0}^{k} a_j \\binom{X}{j} = \\sum_{j=0}^{k} \\frac{a_j}{j!} X(X-1) \\cdots (X-j+1),\n$$\nwhere the $a_j$ are all integers, so its coefficients are rational.\n\nWe show that for such a $P$, the sequence $((-1)^{P(n)})_{n \\in \\mathbb{N}}$ is periodic with period $2^r$, where $r = \\min\\{s : 2^s > k\\}$. To this end, it suffices to show that if $j < 2^s$ and $m$ is integer, then $\\binom{m}{j} \\equiv \\binom{m+2^s}{j} \\pmod{2}$. These are the coefficients of $X^j$ in the expansions of $(1+X)^m$ and $(1+X)^{m+2^s}$, respectively. The congruence $(1+X)^{2^s} \\equiv 1+X^{2^s} \\pmod{2}$ follows easily by induction on $s$. Hence $(1+X)^{m+2^s} \\equiv (1+X)^m (1+X^{2^s}) \\pmod{2}$. Since $j$ is less than $2^s$, it is immediate that the coefficients of $X^j$ in $(1+X)^m$ and $(1+X)^{m+2^s}$ have the same parity.\n\nConversely, let $\\alpha_0, \\dots, \\alpha_{2^r-1}$ be arbitrary integers, and let $\\beta_0, \\dots, \\beta_{2^r-1}$ be the solution to the lower triangular system of linear equations\n$$\n\\sum_{i=0}^{2^r-1} \\beta_i \\binom{j}{i} = \\alpha_j, \\quad j = 0, \\dots, 2^r-1.\n$$\nSince the coefficient of $\\beta_j$ in the $j$-th equation is 1, the $\\beta_i$ are all integers. The polynomial\n$$\n\\sum_{i=0}^{2^r-1} \\beta_i \\binom{X}{i} = \\sum_{i=0}^{2^r-1} \\frac{\\beta_i}{i!} X(X-1)\\cdots(X-i+1)\n$$\nrealizes the sequence $(-1)^{\\alpha_j}$, $j = 0, \\dots, 2^r - 1$, and its extension with period $2^r$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16072,
    "subject": "Mathematics (Olympiad)",
    "question": "Найдите все тройки различных простых чисел $p, q, r$ такие, что $q$ и $r$ делят $p^4 - 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Ясно, что любые два числа тройки различны (если $p = q$, то $p^4 - 1$ не делится на $q$). Пусть для определённости $p$ — наименьшее из чисел тройки. Нам известно, что число $p^4 - 1 = (p - 1)(p + 1)(p^2 + 1)$ делится на $qr$.\n\nЗаметим, что $p - 1$ меньше любого из простых чисел $q$ и $r$, а значит, взаимно просто с ними. Далее, число $p^2 + 1$ не может делиться на оба числа $q$ и $r$, так как $p^2 + 1 < (p + 1)(p + 1) < qr$. Значит, $p + 1$ делится на одно из них (для определённости, на $q$). Поскольку $q > p$, это возможно лишь при $q = p + 1$. Тогда одно из чисел $p$ и $q$ чётно, а поскольку оно простое, то $p = 2$, $q = 3$.\n\nНаконец, $r$ является простым делителем числа $p^4 - 1 = 15$, отличным от $q = 3$, значит, $r = 5$.\n\nОсталось проверить, что тройка $2, 3, 5$ удовлетворяет условиям задачи.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16073,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S(k)$ denote the sum of all digits of $k$ in base 10. Find all integers $n \\geq 2$ and rational numbers $\\beta \\in (0,1)$ such that there exist $n$ distinct positive integers $a_1, a_2, \\dots, a_n$ satisfying: for any subset $I \\subseteq \\{1, 2, \\dots, n\\}$ with $|I| \\geq 2$,\n\n$$\nS\\left(\\sum_{i \\in I} a_i\\right) = \\beta \\cdot \\sum_{i \\in I} S(a_i).\n$$\n",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are all integers $n \\in \\{2, 3, \\dots, 10\\}$ and all rational numbers $\\beta \\in (0, 1)$.\n\n**Construction:**\nFor $n \\in \\{2, 3, \\dots, 10\\}$ and rational $\\beta \\in (0, 1)$, construct $a_1, a_2, \\dots, a_n$ as follows. Let $c$ and $s$ be positive integers to be determined. For $k = 1, \\dots, n$, define\n\n$$\na_k = 2 \\cdot 10^{cn+sk} - 10^{cn+s(k-1)+1} + \\sum_{\\substack{i \\in \\{1, \\dots, n\\} \\\\ i \\neq k}} 10^{cn+s(i-1)+1} + 10^{k-1} \\cdot \\sum_{i=0}^{c-1} 10^{ni}\n$$\n\nIn the last $cn$ places of $a_1, \\dots, a_n$, each place corresponds to a unique $a_i$ with a nonzero digit (1); for the next $s$ places, $a_1$ has $199\\dots9$ ($s-1$ nines), others have $00\\dots01$; and so on for each $a_i$.\n\nFor every $1 \\leq i \\leq n$, $S(a_i) = n + 9(s-1) + c$; for every subset $I \\subseteq \\{1, \\dots, n\\}$ with $|I| \\geq 2$,\n\n$$\nS\\left(\\sum_{i \\in I} a_i\\right) = n \\cdot |I| + c \\cdot |I|.\n$$\n\nChoose $s, c \\in \\mathbb{N}_+$ so that\n\n$$\n(n + 9(s-1) + c) \\cdot \\beta = n + c.\n$$\n\nThen the requirements are met.\n\n**Bound for $n$:**\nSuppose $n = 11$. If $I_1, I_2 \\subseteq \\{1, \\dots, n\\}$ are disjoint, $|I_1|, |I_2| \\geq 2$, then\n\n$$\nS\\left(\\sum_{i \\in I_1 \\cup I_2} a_i\\right) = \\beta \\cdot \\sum_{i \\in I_1 \\cup I_2} S(a_i) = S\\left(\\sum_{i \\in I_1} a_i\\right) + S\\left(\\sum_{i \\in I_2} a_i\\right).\n$$\n\nThis means no carry occurs when summing $\\sum_{i \\in I_1} a_i$ and $\\sum_{i \\in I_2} a_i$. But since $\\beta < 1$, a carry must occur when summing any $a_i$ and $a_j$ for $1 \\leq i < j \\leq 11$.\n\nLet $T$ be the least position where a carry occurs in $a_1 + \\cdots + a_{11}$. Let the $T$th digits be $x_1, \\dots, x_{11}$, with $9 \\geq x_1 \\geq \\cdots \\geq x_{11} \\geq 0$ and $x_1 + \\cdots + x_{11} \\geq 10$. There must exist disjoint subsets $I_1, I_2$ with $|I_1|, |I_2| \\geq 2$ such that the sum of the ones digits of $\\sum_{i \\in I_1} x_i$ and $\\sum_{i \\in I_2} x_i$ is at least 10, leading to a contradiction.\n\nThus, for $n \\geq 11$, no such $a_1, \\dots, a_n$ exist. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16074,
    "subject": "Mathematics (Olympiad)",
    "question": "Find out which are more numerous among five-digit numbers:\n\n1. Those in which all digits follow from left to right in strictly increasing order.\n2. Those in which each digit is not greater than 5 and the digits from left to right are non-decreasing.\n\nFor example, $12459$ satisfies the first condition, while $22589$ and $01234$ do not. $11145$ satisfies the second condition, while $21224$, $12346$, and $01234$ do not.",
    "options": [],
    "answer": "See solution",
    "solution": "To compare the sizes of the two sets, we can construct a one-to-one correspondence between them. Associate each number $abcde$ from the first set to the number $a\\,(b-1)\\,(c-2)\\,(d-3)\\,(e-4)$. This mapping shows that both sets have the same number of elements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16075,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $n$ be positive integers. Mr. Fat has a set $S$ containing every rectangular tile with integer side lengths and area a power of $2$. Mr. Fat also has a rectangle $R$ with dimensions $2^m \\times 2^n$ and a $1 \\times 1$ square removed from one of the corners. Mr. Fat wants to choose $m+n$ rectangles from $S$, with respective areas $2^0, 2^1, \\dots, 2^{m+n-1}$, and then tile $R$ with the chosen rectangles. Prove that this can be done in at most $(m+n)!$ ways.",
    "options": [],
    "answer": "See solution",
    "solution": "We call each of the rectangles in $S$ a tile, and the tile with area $2^0 = 1$ the unit tile. We may assume without loss of generality that the missing $1 \\times 1$ square in $R$ is the top-left corner.\n\nSuppose Mr. Fat walks on the path of squares starting from the top right corner square and going left along the top row of squares until the missing square is stepped on, then turning and going down along the left column of squares to the bottom left square. For a given tiling, let $a_1, a_2, \\dots$ be the sequence of areas of the tiles he steps on by walking along this path from start to finish. We will show that\n\n(a) every tile is stepped on, that is, $a_1, a_2, \\dots, a_{m+n}$ is a rearrangement of $2^0, 2^1, \\dots, 2^{m+n-1}$; and\n\n(b) any valid sequence $a_1, \\dots, a_{m+n}$ uniquely determines the tiling.\n\nSince there are at most $(m+n)!$ ways to order the $m+n$ possible areas, this would finish the problem.\n\nTo establish (a), we induct on $m+n$. It is trivial for $m+n=1$. Suppose it is true for $m+n-1$. We show it holds for $m+n$. Consider the tile with area $2^{m+n-1}$, and call it $T$. If $T$ has dimensions $2^a \\times 2^b$, then we have $a+b=m+n-1$ and since $T$ is contained in $R$, $a \\le m$ and $b \\le n$. Thus the only two possibilities for the dimensions of $T$ are $2^m \\times 2^{n-1}$ or $2^{m-1} \\times 2^n$. In other words, $T$ must stretch over either the entire length or entire width of $R$. It follows that $T$ must be hit on our walk.\n\nNow, consider the rectangle formed by removing $T$ completely from $R$ and, if this breaks $R$ into two pieces, sliding them together so that the two newly formed edges coincide. This forms a tiling of a smaller rectangle $R'$ with dimensions $2^x \\times 2^y$ for $x+y=m+n-1$, and the areas hit along the walk corresponding to $R'$ are precisely those areas encountered along the walk on $R$ other than $T$. By the inductive hypothesis, all of the areas are stepped on, and so (a) holds for all rectangles $R$.\n\nTo establish (b), we again induct on $m+n$. It is trivial for $m+n=1$. Suppose it is true for $m+n-1$. Let $a_1, a_2, \\dots, a_{m+n}$ be a valid ordering of the areas $\\{2^0, 2^1, \\dots, 2^{m+n-1}\\}$, that is, an ordering generated by walking along a tiling.\n\nSuppose $2^{m+n-1}$ appears before $2^0$ in $a_1, \\dots, a_{m+n}$. Consider any tiling that generates this ordering and call the largest tile $T$. We must hit $T$ before the unit tile. Hence, the part of the rectangle below or to the left of $T$ has odd area, since all other tiles in the tiling have even area. In particular, this part must contain the missing corner. The corner is in the top-left, so this part can't be below the largest rectangle. Therefore, it is to the left of $T$. Hence $T$ has the same height as $R$. Likewise, if we hit the unit tile before $T$, we may use the same reasoning to show that $T$ would have the same width as $R$.\n\nTherefore, depending on whether $2^{m+n-1}$ appears before or after $2^0$ in the ordering, we may uniquely determine whether $T$ spans an entire column or an entire row in any tiling achieving the given ordering. This uniquely determines the dimension of the rectangle that is obtained by removing $T$ and sliding the two resulting parts together (if there are even two parts at all). Suppose that the dimensions of this rectangle are uniquely fixed to be $2^a \\times 2^b$, where we must have $a+b=m+n-1$.\n\nNow, remove $2^{m+n-1}$ to obtain an ordering of $\\{2^0, 2^1, \\dots, 2^{m+n-2}\\}$. By the inductive hypothesis, there is a unique tiling of a $2^a \\times 2^b$ rectangle by these tiles. The position of $2^{m+n-1}$ in the given ordering then provides a unique way to insert a tile of area $2^{m+n-1}$ to reconstruct a tiling of the original $2^m \\times 2^n$ rectangle; that is, we know whether this tile spanned a column or a row and the ordering tells us the position along the top and left boundaries of the rectangle where it must be inserted. Finally, any tiling of the original $2^m \\times 2^n$ rectangle that generates the given ordering must be obtained by making this insertion. Since the tiling of our $2^a \\times 2^b$ rectangle was unique by induction, the entire tiling is unique, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16076,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine if there is an integer $k \\ge 2$ such that if we partition the set $\\{2, 3, \\dots, k\\}$ into two parts, then at least one of the parts contains numbers $a, b,$ and $c$ with $ab = c$. (We allow $a = b$.) If such a number $k$ exists, find the least $k$ with this property.",
    "options": [],
    "answer": "See solution",
    "solution": "We show first that $k = 32$ is such a number. Consider a partition $\\{U, V\\}$ of the set $\\{2, 3, \\dots, 32\\}$ where we may assume that $2 \\in U$. Towards contradiction, suppose that none of the parts contains numbers $a, b,$ and $c$ with the desired property. As $2 \\in U$ and $2 \\cdot 2 = 4$, we have $4 \\in V$. Similarly, $4 \\cdot 4 = 16$ implies $16 \\in U$. Hence $2, 16 \\in U$, but $2 \\cdot 8 = 16$, so $8 \\in V$. We have concluded that $2, 16 \\in U$ and $4, 8 \\in V$, but $2 \\cdot 16 = 32 = 4 \\cdot 8$, which implies that the number $32$ cannot be in any of the parts, which is a contradiction. Therefore, $k = 32$ is a desired number.\n\nWe now prove that $k = 32$ is actually the least number with the desired property. We form the partition $\\{U, V\\}$ of the set $K = \\{2, 3, \\dots, 31\\}$ in the following way. For any number $n \\in \\mathbb{Z}_+$, consider its prime factorization representation $n = \\prod_{i=0}^{k-1} p_i$, and put $\\Omega(n) = k$. If $n < 32 = 2^5$, then $n$ is necessarily a product of at most four primes (with repetitions counted), so $\\Omega(n) \\le 4$. Put\n\n$$\nU = \\{n \\in K \\mid \\Omega(n) = 1 \\text{ or } \\Omega(n) = 4\\}\n$$\n\nand\n\n$$\nV = \\{n \\in K \\mid \\Omega(n) = 2 \\text{ or } \\Omega(n) = 3\\}.\n$$\n\nLet $a, b, c \\in K$ be numbers with $ab = c$. We observe that $\\Omega(c) = \\Omega(a) + \\Omega(b)$. On the other hand, we have $\\Omega(c) \\le 4$, as $c \\in K$. If now $a, b \\in U$, then necessarily $\\Omega(a) = \\Omega(b) = 1$, which implies $\\Omega(c) = 2$ and $c \\in V$. If instead $a, b \\in V$, then $\\Omega(a) = \\Omega(b) = 2$ and $\\Omega(c) = 4$, so $c \\in U$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16077,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the real numbers $x$ and $y$ such that\n$$\n(x^2 - x + 1)(3y^2 - 2y + 3) - 2 = 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that $x^2 - x + 1 = (x - \\frac{1}{2})^2 + \\frac{3}{4} \\geq \\frac{3}{4}$, and $3y^2 - 2y + 3 = 3(y - \\frac{1}{3})^2 + \\frac{8}{3} \\geq \\frac{8}{3}$.\n\nTherefore,\n$$\n(x^2 - x + 1)(3y^2 - 2y + 3) \\geq \\frac{3}{4} \\cdot \\frac{8}{3} = 2\n$$\nfor any real numbers $x$ and $y$.\n\nEquality holds when $(x - \\frac{1}{2})^2 = 0$ and $(3y - 1)^2 = 0$, that is, $x = \\frac{1}{2}$ and $y = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16078,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given a segment $AB$ in the plane. Consider a triangle $XYZ$ with the following properties:\n\n- The vertex $X$ is an interior point of the segment $AB$.\n- The triangles $XBY$ and $XZA$ are similar ($\\triangle XBY \\sim \\triangle XZA$).\n- The points $A$, $B$, $Y$, $Z$ lie on a circle in this order.\n\nFind the locus of midpoints of the sides $YZ$ of all such triangles $XYZ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $XYZ$ be a satisfactory triangle. Then the vertices $Y$ and $Z$ must lie in the same half-plane with the boundary line $AB$. Denote by $Y'$ the reflection of $Y$ through the line $AB$. Due to the presumed similarity, the angles $XAZ$ and $BYX$ are congruent (see Fig. 1), and hence $|\\angle BAZ| = |\\angle BY'Z|$ as well. Using the well-known inscribed angles property, we conclude that the circumcircle of $\\triangle ABZ$ passes not only through the point $Y$, but also through the point $Y'$. The line $AB$ (as a perpendicular bisector of the chord $YY'$) passes through the centre $O$ of the circle $k$ and thus the chord $AB$ is a diameter of $k$. Since the segment $AB$ is fixed, the circle $k = ABYZ$ is common for all satisfactory triangles $XYZ$ and the midpoint $M$ of $YZ$ must lie in the interior of $k$. Since both angles $OMZ$ and $OMY$ are right (see Fig. 2), the (lesser) angles $AMO$ and $BMO$ are acute and thus the point $M$ must lie in the intersection of the exteriors of Thales' circles with diameters $AO$ and $BO$.\n\nIn what follows we will show the both derived necessary conditions determine the locus of all the possible midpoints $M$.\n\n![](images/14._63_RD_CZECH_AND_SLOVAK_MATHEMATICAL_OLYMPIAD_p13_data_6ad0f1c566.png)\n\nFig. 1\n\n![](images/14._63_RD_CZECH_AND_SLOVAK_MATHEMATICAL_OLYMPIAD_p13_data_70f743be8c.png)\n\nFig. 2\n\nSo, let $M$ be any point in the interior of $k$ for which both angles $AMO$ and $BMO$ are acute (i.e., $M$ lies in the exteriors of the circles with diameters $AO$ and $BO$). Consider a chord of $k$ which passes through $M$ perpendicularly to $OM$. This chord does not intersect the diameter $AB$, because of the acute angles $AMO$ and $BMO$. Thus the endpoints of the chord with the midpoint $M$ can be denoted as $Y$ and $Z$ so that $A$, $B$, $Y$, $Z$ lie on $k$ in this order. If $Y$ reflects to $Y'$ through the diameter $AB$ and if $X$ denotes the intersection point of the segments $AB$ and $Y'Z$, then the triangles $XBY$ and $XZA$ are similar as required (by theorem AA). This completes the solution.\n\n*Conclusion.* The locus under consideration is the interior of the highlighted region bounded by the three circles with diameters $AB$, $AO$ and $BO$, where $O$ denotes the midpoint of segment $AB$ (see Fig. 3).\n\n![](images/14._63_RD_CZECH_AND_SLOVAK_MATHEMATICAL_OLYMPIAD_p13_data_a2650a523f.png)\n\nFig. 3",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16079,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, $d$ be positive numbers. Prove that\n\n$$\n(a^2 + b^2 + c^2 + d^2)^2 \\geq (a+b)(b+c)(c+d)(d+a).\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "By the inequality between the arithmetic and the geometric mean, we have\n\n$$\n(a+b)(b+c)(c+d)(d+a) \\leq \\left( \\frac{(a+b)+(b+c)+(c+d)+(d+a)}{4} \\right)^4 = 2^4 \\left( \\frac{a+b+c+d}{4} \\right)^4.\n$$\n\nBy the inequality between the quadratic and the arithmetic mean, we have\n\n$$\n2^4 \\left( \\frac{a+b+c+d}{4} \\right)^4 \\leq 2^4 \\left( \\frac{a^2+b^2+c^2+d^2}{4} \\right)^2 = (a^2+b^2+c^2+d^2)^2,\n$$\n\nas required.\n\nIn the second inequality, equality holds if and only if $a = b = c = d$, but in that case, equality holds also in the original inequality. Therefore, equality holds if and only if $a = b = c = d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16080,
    "subject": "Mathematics (Olympiad)",
    "question": "給定一圓及圓上的四個點 $B, C, X, Y$，設 $A$ 為線段 $BC$ 的中點，$Z$ 為線段 $XY$ 的中點。過 $B, C$ 分別作垂直 $BC$ 的直線 $L_1, L_2$。設過 $X$ 且垂直 $AX$ 的直線分別交 $L_1, L_2$ 於 $X_1, X_2$ 兩點，過 $Y$ 且垂直 $AY$ 的直線分別交 $L_1, L_2$ 於 $Y_1, Y_2$ 兩點。令 $X_1Y_2$ 與 $X_2Y_1$ 相交於 $P$ 點。證明：$\\angle AZP = 90^\\circ$。",
    "options": [],
    "answer": "See solution",
    "solution": "作 $A$ 對 $X_2Y_1$ 垂足 $D$ 點。因為 $\\angle AYY_1 = \\angle ABY_1 = \\angle AXX_2 = \\angle ACX_2 = 90^\\circ$，所以 $DX_2XAC$ 五點共圓，$DY_1YAB$ 五點亦共圓。考慮此兩圓與一開始的給定圓等三圓的根心，得 $AD, BY, CX$ 三線共點。設此共點為 $S$，且有\n\n$$\nSA \\cdot SD = SB \\cdot SY = SC \\cdot SX.\n$$\n\n考慮以 $S$ 為中心，$SA \\cdot SD$ 為反演幂的變換。此變換將 $AD, BY, CX$ 互換，且 $BAC$ 共線且 $A$ 為 $BC$ 中點，故 $DXSY$ 四點共圓且為調和四邊形。有\n\n$$\n\\angle DZX = \\angle DYX + \\angle ZDY = \\angle DSX + \\angle XDS = 180^\\circ - \\angle DXS \\\\\n\\angle SZX = \\angle SYX + \\angle ZSY = \\angle SDX + \\angle XSD = 180^\\circ - \\angle DXS.\n$$\n\n（其中用到 $DXSY$ 是調和的，所以 $DS, DZ$ 對 $\\angle XDY$ 等角共軛，$SD, SZ$ 對 $\\angle XSY$ 等角共軛。）所以 $\\angle DZX = \\angle SZX$。又因 $XYBC$ 四點共圓，所以 $\\triangle SXY$ 與 $\\triangle SBC$ 相似，從而 $\\angle SAB = \\angle SZX = \\angle DZX$。令 $XY$ 與 $BC$ 交於 $T$ 點，則 $DZAT$ 四點共圓。\n\n令一開始給定圓的圓心為 $O$ 點。由於 $OZ \\perp ZT, OA \\perp AT$，所以 $DZOAT$ 五點共圓。令過 $O$ 且平行於 $BC$ 的直線為 $L_3$，過 $Z$ 且與 $AZ$ 垂直的直線為 $L_4$。則 $A$ 對圓 $DZOAT$ 的對徑點為 $X_2Y_1, L_3, L_4$ 的共同",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16081,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a regular prism $ABCA'B'C'$. A plane $\\alpha$ containing point $A$ meets the rays $BB'$ and $CC'$ at points $E$ and $F$ such that\n\n$$\n\\text{area } [ABE] + \\text{area } [ACF] = \\text{area } [AEF].\n$$\n\nFind the angle determined by the planes $AEF$ and $BCC'$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of $BC$ and let $u$ be the angle determined by the planes $AEF$ and $BCC'$. Since triangle $MEF$ is the projection of triangle $AEF$ onto $BCC'$, we have\n\n$$\n\\begin{aligned}\n\\cos u &= \\frac{[MEF]}{[AEF]} = \\frac{[BCFE]}{2[AEF]} = \\frac{[BCFE]}{2([ABE] + [ACF])} \\\\\n&= \\frac{[BCFE]}{4([MBE] + [MCF])} = \\frac{[BCFE]}{2[BCFE]} = \\frac{1}{2},\n\\end{aligned}\n$$\n\nhence $u = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16082,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine prime positive integers $p$ and $q$ satisfying the equation\n\n$$\np^4 + p^3 + p^2 + p = q^2 + q.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The given equation can be written as\n\n$$\np^4 + p^3 + p^2 + p = q^2 + q \\\\\n\\Leftrightarrow p(p+1)(p^2+1) = q(q+1) \\quad (1)\n$$\n\nFor $q \\leq p$ this is not possible. Hence, we should have $q > p$.\n\nTherefore, from (1) we conclude that:\n\n$$\nq \\mid (p^2-1)(p^2+1). \\quad (2)\n$$\n\nNext, we distinguish the cases:\n\n* If $q \\leq p^2$, then $q^2 \\leq p^4$ and $q^2 + q \\leq p^4 + p^2 < p^4 + p^2 + p^3 + p$. Hence, the equation has no solutions.\n\n* If $q > p^2$, then $q \\geq p^2 + 1$, and since $q$ is prime, (2) gives $q = p^2 + 1$. Thus, from equation (1) we obtain\n\n$$\np(p+1) = q+1 \\Leftrightarrow p^2 + p = p^2 + 2 \\Leftrightarrow p = 2\n$$\n\nTherefore, since $q = p^2 + 1 = 5$, we have the solution $(p, q) = (2, 5)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16083,
    "subject": "Mathematics (Olympiad)",
    "question": "Alice is drawing a shape on a piece of paper. She starts by placing her pencil at the origin, and then draws line segments of length $1$, alternating between vertical and horizontal segments. Eventually, her pencil returns to the origin, forming a closed, non-self-intersecting shape. Show that the area of this shape is even if and only if its perimeter is a multiple of eight.",
    "options": [],
    "answer": "See solution",
    "solution": "Colour the horizontal segments in every other line of the grid alternately red and blue as shown below:\n\n![](images/BMO_2022_shortlist_p19_data_8816d801d7.png)\n\nLet there be $r$ red segments on the perimeter and $s$ red segments in the interior of the shape. By considering the possibilities starting from a red segment, we see that every fourth segment on the perimeter of the shape will be red, therefore we have $P = 4r$. Also, every square has exactly one red edge, thus $A = r + 2s$. So $A \\equiv r \\pmod{2}$ from which the result follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16084,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. A restaurant offers a choice of $n$ starters, $n$ main dishes, $n$ desserts, and $n$ wines. A group dines at the restaurant, with each guest choosing a starter, main dish, dessert, and wine. No two people place exactly the same order. Furthermore, there is no collection of $n$ guests such that their orders coincide in three of these aspects, but in the fourth one they all differ. (For example, there are no $n$ people that order exactly the same three courses of food, but $n$ different wines.)\n\nWhat is the maximal number of guests?",
    "options": [],
    "answer": "See solution",
    "solution": "The maximal number of guests is $n^4 - n^3$.\n\nThe possible menus are represented by quadruples\n\n$$\n(a, b, c, d), \\quad 1 \\leq a, b, c, d \\leq n.\n$$\n\nLet us count those menus satisfying\n\n$$\na + b + c + d \\not\\equiv 0 \\pmod{n}.\n$$\n\nThe numbers $a, b, c$ may be chosen arbitrarily ($n$ choices for each), and then $d$ is required to satisfy only $d \\not\\equiv -a - b - c \\pmod{n}$. Hence there are\n\n$$\nn^3(n - 1) = n^4 - n^3\n$$\n\nsuch menus.\n\nIf there are $n^4 - n^3$ guests, and they have chosen precisely the $n^4 - n^3$ menus satisfying $a+b+c+d \\not\\equiv 0 \\pmod{n}$, we claim that the condition of the problem is fulfilled. Suppose there is a collection of $n$ people whose orders coincide in three aspects, but differ in the fourth. Without loss of generality, assume they have ordered exactly the same food, but $n$ different wines. This means they all have the same values of $a, b, c$, but their values of $d$ are distinct. A contradiction arises since, given $a, b, c$, there are only $n-1$ values available for $d$.\n\nFor $n^4 - n^3 + 1$ guests (or more), it is impossible to satisfy the problem's condition. The $n^3$ sets\n\n$$\nM_{a,b,c} = \\{(a, b, c, d) \\mid 1 \\leq d \\leq n\\}, \\quad 1 \\leq a, b, c \\leq n,\n$$\n\nform a partition of the set of possible menus, totaling $n^4$. When the number of guests is at least $n^4 - n^3 + 1$, there are at most $n^3 - 1$ unselected menus. Therefore, there exists a set $M_{a,b,c}$ which contains no unselected menus. That is, all the $n$ menus in $M_{a,b,c}$ have been selected, and the condition of the problem is violated.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16085,
    "subject": "Mathematics (Olympiad)",
    "question": "Twelve friends play a round-robin tournament (each pair plays one game). What is the maximum possible value of\n$$\n\\Sigma_3 = a_1^3 + a_2^3 + \\dots + a_{12}^3\n$$\nwhere $a_i$ is the number of games won by the $i$-th player?",
    "options": [],
    "answer": "See solution",
    "solution": "There are $\\binom{12}{2} = \\frac{12 \\cdot 11}{2} = 66$ games, so the total number of points from all games is 66. One possible outcome is to order the players so that each beats all those after them. Then the first has 11 points, the second 10, ..., the last 0. Thus, the points are $0, 1, \\dots, 11$.\n\nThen:\n$$\n\\Sigma_3 = 0^3 + 1^3 + \\dots + 11^3 = \\left(\\frac{11 \\cdot 12}{2}\\right)^2 = 66^2\n$$\n\nTo show this is maximal, suppose two players A and B have the same number of points $\\kappa$, with A beating B. If we swap the result so B beats A, A has $\\kappa-1$ and B $\\kappa+1$ points. The change in $\\Sigma_3$ is:\n$$\n(\\kappa+1)^3 + (\\kappa-1)^3 - 2\\kappa^3 = 6\\kappa > 0\n$$\nso $\\Sigma_3$ increases, contradicting maximality. Thus, the maximum is $66^2$, achieved when the scores are $0,1,\\dots,11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16086,
    "subject": "Mathematics (Olympiad)",
    "question": "The numbers $1, 2, \\ldots, 2009$ are written on a board. Some of them are erased, and the remainder of their sum divided by $13$ is written on the board. After a finite number of repetitions of the above procedure, only three numbers are left, two of which are $99$ and $999$. What is the third number?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x$ be the third number. After each procedure, the remainder of the sum of the numbers on the board divided by $13$ is unchanged.\n\n$$1 + 2 + 3 + \\ldots + 2009 = \\frac{2009 \\cdot 2010}{2} = 1005 \\cdot 2009$$\n\ndivided by $13$ has remainder $2$.\n\nHence, $99 + 999 + x$ divided by $13$ has remainder $2$. Now $99 + 999 = 1098$ has remainder $6$ when divided by $13$. Since $99$ and $999$ are not remainders, it follows that $0 \\leq x < 13$, i.e., $x = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16087,
    "subject": "Mathematics (Olympiad)",
    "question": "We enumerate all prime numbers in ascending order: $p_1 = 2, p_2 = 3, p_3 = 5, \\dots$. Find all positive integers $n$ for which $p_1! + p_2! + \\dots + p_n! = a^b$ for some positive integers $a$, $b > 1$, where $k!$ denotes the product of all integers from 1 to $k$.\n\n![](images/Ukraine_2020_booklet_p19_data_bab6908917.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $n = 2, 3$.\n\n**Solution.** Directly checking yields:\n\n$$\np_1! = 2, \\quad p_1! + p_2! = 8 = 2^3, \\quad p_1! + p_2! + p_3! = 128 = 2^7\n$$\n\nFor $n \\ge 4$, $p_1! + p_2! + \\dots + p_n! = 128 + 7! + 11! + \\dots + p_n!$ is a number in which all summands are divisible by $2^5$, except $p_4! = 7! = 2^4 \\cdot 315$, which is divisible by $2^4$ but not by $2^5$. Therefore, this number cannot be a power of 2. If it is a power of another number, then since $2^4$ divides it, it can only be $2^4$ or $x^2$. Let us show that it cannot be a square. Indeed, $2! + 3! + \\dots + p_n!$ gives a remainder 2 modulo 3, which is impossible for squares.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16088,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for any numbers $x \\neq y$ the following equality is true:\n\n$$\n(f(x+y))^2 = f(x) + f(y) + f(y).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Only two functions $f(x) \\equiv 0$ or $f(x) \\equiv 3$ are solutions to the equation.\n\nLet's put $c = f(0)$. By substituting $y = 0$, we obtain a quadratic equation:\n\n$$\n(f(x))^2 - 2f(x) - c = 0,\n$$\n\nwhence $f(x) = 1 - \\sqrt{1+c}$ or $1 + \\sqrt{1+c}$ for any $x \\neq 0$.\n\nSuppose there are two non-zero numbers $a$ and $b$ of the same sign ($ab > 0$) such that $f(a) = 1 - \\sqrt{1+c}$ and $f(b) = 1 + \\sqrt{1+c}$. Substituting $x = a$ and $y = b$ into the original equation, we get:\n\n$$\n(f(a+b))^2 - f(a+b) - 2 = 0,\n$$\n\nwhence $f(a+b) = -1$ or $2$. Consider two cases:\n\n1) Case $f(a+b) = -1$. Since $a+b \\neq 0$, then $f(a+b) = 1 \\pm \\sqrt{1+c}$ and necessarily $1 - \\sqrt{1+c} = -1$. This means $c=3$ and $f(a) = f(a+b) = -1$. For $x=a$ and $y = a+b$, the original equation gives:\n\n$$\n(f(2a+b))^2 = f(2a+b) - 2,\n$$\n\nwhich has no solutions for $f(2a+b)$, which is impossible.\n\n2) Case $f(a+b) = 2$. Since $a+b \\neq 0$, then $f(a+b) = 1 \\pm \\sqrt{1+c}$ and so necessarily $1+\\sqrt{1+c} = 2$. This means $c=0$ and $f(b) = f(a+b) = 2$. For $x=a+b$ and $y=b$, the original equation gives:\n\n$$\n(f(a+2b))^2 = f(a+2b) + 4,\n$$\n\nwhich has roots $\\frac{1\\pm\\sqrt{17}}{2}$. However, also $a+2b \\neq 0$ and $f(a+2b) = 1 \\pm \\sqrt{1+0} = 0$ or $2$. Contradiction.\n\nThus, on each of the open intervals $(-\\infty, 0)$ and $(0, \\infty)$ the function $f$ takes only one value $1 - \\sqrt{1+c}$ or $1 + \\sqrt{1+c}$. Therefore, if $x$ and $y$ are non-zero numbers of the same sign, then $f(x+y) = f(x) = f(y) = 1 \\pm \\sqrt{1+c}$ and\n\n$$\n(1 \\pm \\sqrt{1+c})^2 = 3(1 \\pm \\sqrt{1+c}),\n$$\n\nwhence $1 \\pm \\sqrt{1+c} = 0$ or $3$. Since $1 - \\sqrt{1+c} \\le 1 \\le 1 + \\sqrt{1+c}$, two options are possible:\n\n$$\n1 - \\sqrt{1+c} = 0 \\iff c = 0 \\quad \\text{or} \\quad 1 + \\sqrt{1+c} = 3 \\iff c = 3.\n$$\n\nNow substitute $y = -x$ into the original equation:\n\n$$\n(f(0))^2 = f(0) + f(x) + f(-x) \\iff f(x) + f(-x) = c^2 - c \\text{ for any } x \\neq 0.\n$$\n\nIf $c=0$ then on $\\mathbb{R}$ the function $f$ can take at most two values ($0$ or $1+\\sqrt{1+0} = 2$). If $f(a) = 2$ for some $a \\neq 0$, then $f(a) + f(-a) = 2 + f(-a) = 0^2 - 0 = 0$, which is impossible, since $f(-a) = 0$ or $2$. This means $f(x) \\equiv 0$.\n\nIf $c=3$ then on $\\mathbb{R}$ the function $f$ can take at most two values ($3$ or $1-\\sqrt{1+3} = -1$). If $f(b) = -1$ for some $b \\neq 0$, then $f(b) + f(-b) = -1 + f(-b) = 3^2 - 3 = 6$, which is impossible, since $f(-b) = -1$ or $3$. So $f(x) \\equiv 3$.\n\nWe can check that the functions $f(x) \\equiv 0$ or $f(x) \\equiv 3$ are suitable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16089,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the set $M = \\{z \\in \\mathbb{C} \\mid |z| = 1,\\ \\operatorname{Re} z \\in \\mathbb{Q}\\}$. Prove that the complex plane contains infinitely many equilateral triangles with vertices in $M$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $z = a + b i$ be a complex number of modulus $1$ such that $a \\in \\mathbb{Q}$, so $a^2 + b^2 = 1$. An equilateral triangle with $z$ as one vertex has the other two at $z \\left(-\\frac{1}{2} \\pm \\frac{i\\sqrt{3}}{2}\\right)$. The real parts of these are $-\\frac{a}{2} \\pm \\frac{b\\sqrt{3}}{2}$. Since $a \\in \\mathbb{Q}$, $-\\frac{a}{2} \\pm \\frac{b\\sqrt{3}}{2} \\in \\mathbb{Q}$ if and only if $b\\sqrt{3} \\in \\mathbb{Q}$.\n\nLet $q = b/\\sqrt{3} \\in \\mathbb{Q}$. We need to show that $a^2 + 3q^2 = 1$ has infinitely many solutions $(a, q) \\in \\mathbb{Q} \\times \\mathbb{Q}$, i.e., $m^2 + 3n^2 = p^2$ has infinitely many solutions $(m, n, p) \\in \\mathbb{N} \\times \\mathbb{N} \\times \\mathbb{N}$.\n\nSince $3n^2 = (p - m)(p + m)$, consider solutions with $p - m = 3$ and $p + m = n^2$. Then $n$ is odd, so let $n = 2k + 1$, $k \\in \\mathbb{N}^*$. Then $m = 2k^2 + 2k - 1$ and $p = 2k^2 + 2k + 2$. Thus, $a = \\frac{2k^2 + 2k - 1}{2k^2 + 2k + 2}$, $b = \\frac{(2k + 1)\\sqrt{3}}{2k^2 + 2k + 2}$, and $z = a + b i$ has modulus $1$ with $a, b > 0$. As $k \\in \\mathbb{N}$ is arbitrary, there are infinitely many such triangles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16090,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $a$ and a positive integer $n$, show that the sum $\\sum_{k=1}^{n} a^{(k,n)}$ is divisible by $n$, where $(x, y)$ denotes the greatest common divisor of the integers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Write $\\sum_{k=1}^n a^{(k,n)} = \\sum_{d|n} \\varphi(n/d)a^d$, where $\\varphi$ is Euler's totient function ($\\varphi(m)$ is the number of positive integers less than $m$ and prime to $m$), and notice that, if $n$ and $n'$ are coprime positive integers, then\n\n$$\n\\sum_{k=1}^{nn'} a^{(k,nn')} = \\sum_{d|n} \\varphi(n/d) \\sum_{d'|n'} \\varphi(n'/d')(a^d)^{d'}.\n$$\n\nConsequently, it is sufficient to prove the assertion for $n = p^m$, where $p$ is a prime and $m$ is a non-negative integer. In this case,\n\n$$\n\\begin{aligned}\n\\sum_{k=1}^{p^m} a^{(k, p^m)} &= \\sum_{k=0}^{m} \\varphi(p^{m-k}) a^{p^k} = \\sum_{k=0}^{m-1} (p^{m-k} - p^{m-k-1}) a^{p^k} + a^{p^m} \\\\\n&= p^m a + \\sum_{k=1}^{m} p^{m-k} (a^{p^k} - a^{p^{k-1}}) \\equiv 0 \\pmod{p^m},\n\\end{aligned}\n$$\n\nsince $a^{p^k} \\equiv a^{p^{k-1}} \\pmod{p^k}$, $k = 1, \\dots, m$, by Fermat's theorem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16091,
    "subject": "Mathematics (Olympiad)",
    "question": "For which integers $n \\ge 2$ is it possible to draw $n$ distinct straight lines in the plane in such a way that there are at least $n-2$ points where exactly three of the lines intersect?",
    "options": [],
    "answer": "See solution",
    "solution": "For $n=2$, any two lines satisfy the condition, and for $n=3$, we can take any three lines passing through a common point.\n\nFor $n=4$, there is no feasible choice of four lines: suppose there are two points where exactly three lines meet. At most one of the lines can pass through both, so we need at least $1+2 \\times 2 = 5$ lines.\n\nThe same argument shows that it is impossible for $n=5$: suppose there are three points where exactly three lines meet. If one line passes through all of them, we still require two further lines through each of the three, and no two of them can coincide. This already gives us $1+3 \\times 2 = 7$ lines. If the three points do not lie on a line, then there can be at most one line passing through any two of them, leaving us with at least one more line through each of the points that does not pass through any of the others. This gives us a total of at least $3+3=6$ lines.\n\nThere is a possible configuration for every $n \\ge 6$: for $n=6$, we can take the (extended) sides and diagonals of any (non-degenerate) quadrilateral. For larger values of $n$, we use an inductive construction: if we start with the sides and diagonals of a quadrilateral for which opposite sides are not parallel, then there are also two intersections of exactly two lines (namely the opposite sides). In each further step, we add a line through one of the intersections of exactly two lines, chosen in such a way that it does not pass through any of the other intersections that were obtained previously. This ensures that we get new intersections of exactly two lines with each step, and the number of points where exactly three lines meet increases by one. Thus the number of points where three lines meet will be (exactly) $n-2$ when the $n$-th line is drawn.\n\nWe conclude that it is possible to draw $n$ lines in a suitable way for all $n \\ne 4, 5$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16092,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be real numbers with $0 \\le a, b \\le 1$.\n\nProve that\n\n$$\n\\frac{a}{b+1} + \\frac{b}{a+1} \\le 1\n$$\n\nand find the cases of equality.",
    "options": [],
    "answer": "See solution",
    "solution": "We clear denominators to get\n\n$$\n\\frac{a}{b+1} + \\frac{b}{a+1} \\le 1 \\implies a(a+1) + b(b+1) \\le (a+1)(b+1).\n$$\n\nThis simplifies to\n\n$$\na^2 + a + b^2 + b \\le ab + a + b + 1,\n$$\n\nwhich further reduces to\n\n$$\na^2 - a + b^2 - b \\le ab - a - b + 1,\n$$\n\nor\n\n$$\na(a-1) + b(b-1) \\le (a-1)(b-1).\n$$\n\nRewriting,\n\n$$\n(1-a)(1-b) + a(1-a) + b(1-b) \\ge 0.\n$$\n\nEach term on the left is non-negative for $0 \\le a, b \\le 1$.\n\nEquality holds when all three terms are zero, i.e., when $a, b \\in \\{0, 1\\}$ and at least one of $a$ or $b$ is $1$.\n\nThus, equality occurs for $(a, b) = (1, 0)$, $(0, 1)$, and $(1, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16093,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\dots, x_n$ be real numbers such that $x_1 + x_2 + \\dots + x_n = 0$. Prove that there exist indices $i$ and $j$ such that $\\frac{1}{2} \\le \\left|\\frac{x_i}{x_j}\\right| \\le 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "If there are two equal numbers among the $x_i$, we can choose those two and the result holds. So assume all the numbers are distinct and, without loss of generality, $x_1 > x_2 > \\dots > x_n$.\n\nSince the total sum is zero, there exists a positive integer $k$ ($1 < k < n$) such that $x_k > 0 > x_{k+1}$. We have:\n\n$$\n|x_1| + |x_2| + \\dots + |x_k| = |x_{k+1}| + |x_{k+2}| + \\dots + |x_n|.\n$$\n\nAssume there do not exist $i$ and $j$ with the required property. For $i = 1, 2, \\dots, k-1$:\n\n$$\n\\frac{|x_i|}{|x_{i+1}|} > 1 > \\frac{1}{2},\n$$\n\nso\n\n$$\n\\frac{x_i}{x_{i+1}} > 2.\n$$\n\nIt follows that\n\n$$\nx_i < \\frac{1}{2}x_{i-1} < \\frac{1}{2^2}x_{i-2} < \\dots < \\frac{1}{2^i}x_1, \\text{ for } i = 1, 2, \\dots, k-1.\n$$\n\nSimilarly, for the negative numbers:\n\n$$\n|x_j| < \\frac{1}{2^{n-j}}|x_n|, \\text{ for } j = k, k+1, \\dots, n.\n$$\n\nFinally,\n\n$$\n\\begin{aligned}\n|x_1| &< |x_1| + |x_2| + \\dots + |x_k| = |x_{k+1}| + |x_{k+2}| + \\dots + |x_n| \\\\\n&< |x_n| \\left( 1 + \\frac{1}{2} + \\frac{1}{4} + \\dots + \\frac{1}{2^{n-k+1}} \\right) < 2|x_n|.\n\\end{aligned}\n$$\n\nAnalogously,\n\n$$\n\\begin{aligned}\n|x_n| &< |x_k| + |x_{k+1}| + \\dots + |x_n| = |x_1| + |x_2| + \\dots + |x_k| \\\\\n&< |x_1| \\left( 1 + \\frac{1}{2} + \\frac{1}{4} + \\dots + \\frac{1}{2^k} \\right) < 2|x_1|.\n\\end{aligned}\n$$\n\nTherefore, $\\frac{1}{2} \\le \\left|\\frac{x_1}{x_n}\\right| \\le 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16094,
    "subject": "Mathematics (Olympiad)",
    "question": "Positive numbers $a, b, c$ satisfy $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = 3$. Prove the inequality\n\n$$\n\\frac{1}{\\sqrt{a^3 + b}} + \\frac{1}{\\sqrt{b^3 + c}} + \\frac{1}{\\sqrt{c^3 + a}} \\le \\frac{3}{\\sqrt{2}}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Apply a lot of AM-GM inequalities:\n\n$$\n\\begin{align*}\n\\frac{1}{\\sqrt{a^3+b}} + \\frac{1}{\\sqrt{b^3+c}} + \\frac{1}{\\sqrt{c^3+a}} &\\le \\frac{1}{\\sqrt{2a\\sqrt{ab}}} + \\frac{1}{\\sqrt{2b\\sqrt{bc}}} + \\frac{1}{\\sqrt{2c\\sqrt{ca}}} \\\\\n&= \\frac{1}{\\sqrt{2}} \\left( \\frac{\\sqrt{a\\sqrt{ab}}}{a\\sqrt{ab}} + \\frac{\\sqrt{b\\sqrt{bc}}}{b\\sqrt{bc}} + \\frac{\\sqrt{c\\sqrt{ca}}}{c\\sqrt{ca}} \\right) \\\\\n&\\le \\frac{1}{2\\sqrt{2}} \\left( \\frac{a+\\sqrt{ab}}{a\\sqrt{ab}} + \\frac{b+\\sqrt{bc}}{b\\sqrt{bc}} + \\frac{c+\\sqrt{ca}}{c\\sqrt{ca}} \\right) \\\\\n&= \\frac{1}{2\\sqrt{2}} \\left( 3 + \\frac{1}{\\sqrt{ab}} + \\frac{1}{\\sqrt{bc}} + \\frac{1}{\\sqrt{ca}} \\right) \\\\\n&= \\frac{1}{2\\sqrt{2}} \\left( 3 + \\frac{\\sqrt{ab}}{ab} + \\frac{\\sqrt{bc}}{bc} + \\frac{\\sqrt{ca}}{ca} \\right) \\\\\n&\\le \\frac{1}{2\\sqrt{2}} \\left( 3 + \\frac{a+b}{2ab} + \\frac{b+c}{2bc} + \\frac{c+a}{2ac} \\right) = \\frac{3}{\\sqrt{2}}\n\\end{align*}\n$$\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p238_data_b9b7f1b14b.png)",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16095,
    "subject": "Mathematics (Olympiad)",
    "question": "Given that $0 < x, y < 1$, determine, with proof, the maximum value of\n$$\n\\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "When $x = y = \\frac{1}{3}$, the value of the expression is $\\frac{1}{8}$.\n\nWe will prove that\n$$\n\\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)} \\leq \\frac{1}{8}\n$$\nfor any $0 < x, y < 1$ as follows.\n\nIf $x + y \\geq 1$, then\n$$\n\\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)} \\leq 0 < \\frac{1}{8}.\n$$\n\nIf $x + y < 1$, let $1 - x - y = z > 0$. It follows from the AM-GM Inequality that\n$$\n\\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)} = \\frac{xyz}{(x+y)(y+z)(z+x)} \\leq \\frac{xyz}{2\\sqrt{xy} \\cdot 2\\sqrt{yz} \\cdot 2\\sqrt{zx}} = \\frac{1}{8}.\n$$\n\nIn conclusion, the maximum value of the expression is $\\frac{1}{8}$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16096,
    "subject": "Mathematics (Olympiad)",
    "question": "For which integers $n \\geq 2$ is it possible to separate the numbers $1, 2, \\ldots, n$ into two sets such that the sum of the numbers in one of the sets is equal to the product of the numbers in the other set?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that $x, y, z$ are in one of the groups and that the rest of the numbers are in the other group. This leads to the equation\n\n$$\nxyz = 1 + 2 + \\cdots + n - x - y - z = \\frac{n(n+1)}{2} - x - y - z.\n$$\n\nIf we substitute $z = 1$, we can rearrange to obtain\n\n$$\n(x + 1)(y + 1) = \\frac{n(n + 1)}{2}.\n$$\n\nIf $n$ is even, we can take $x = \\frac{n}{2} - 1$ and $y = n$.\n\nIf $n$ is odd, we can take $x = \\frac{n-1}{2}$ and $y = n - 1$.\n\nThese constructions work as long as $x, y, z$ are all different, which holds for $n \\ge 5$. It is easy to check that the task is impossible for $n = 2$ and $n = 4$. On the other hand, we have $1 + 2 = 3$ for $n = 3$. Therefore, the task is possible only for $n = 3$ and all integers $n \\ge 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16097,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the least possible value $k$ for which there exist 2010 distinct natural numbers that satisfy the following condition: the product of any $k$ numbers from the chosen set is divisible by the product of the remaining $2010 - k$ numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $k = 1006$.\n\nOn one hand, $k$ cannot be less than $1006$; otherwise, the product of the $k$ smallest numbers from our set is less than the product of the remaining $2010 - k$ numbers. We construct an example for $k = 1006$.\n\nLet $p_1, p_2, \\dots, p_{2010}$ be $2010$ distinct prime numbers, and define $a_i = \\dfrac{p_1 p_2 \\dots p_{2010}}{p_i}$ for $1 \\leq i \\leq 2010$. Then, the product of $1006$ numbers $a_{n_1}, a_{n_2}, \\dots, a_{n_{1006}}$ equals\n$$\n\\frac{(p_1 p_2 \\dots p_{2010})^{1006}}{p_{n_1} p_{n_2} \\dots p_{n_{1006}}}\n$$\nwhich can be written as $p_1^{a_1} p_2^{a_2} \\dots p_{2010}^{a_{2010}}$, where each exponent $a_i$ equals $1005$ or $1006$, which is greater than $1005$.\n\nSimilarly, the product of the remaining $1004$ numbers can be expressed as $p_1^{\\beta_1} p_2^{\\beta_2} \\dots p_{2010}^{\\beta_{2010}}$, where each exponent $\\beta_i$ equals $1003$ or $1004$, which is less than $1004$.\n\nTo finish the proof, note that if $i \\ne j$, then $a_i = \\dfrac{p_1 p_2 \\dots p_{2010}}{p_j} \\ne a_j$, so all $\\{a_i\\}$ are distinct.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16098,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a group of 30 people. Consider all subsets $A \\subset S$ such that no member of $A$ received a hat from a member of $A$. Among such subsets, let $T$ be a subset of maximal cardinality. Prove that $|T| \\ge 10$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $U \\subset S$ consist of all people that have received a hat from a person belonging to $T$. Now consider any member $x \\in S \\setminus (T \\cup U)$. Since $x \\notin U$, no member of $T$ sent his hat to $x$. It follows that no member of $T$ sent a hat to a person from $T \\cup \\{x\\}$. But the maximality of $T$ implies that some person from $T \\cup \\{x\\}$ sent his hat to a person from the same subset. This means that $x$ sent his hat to a person from $T$. Consequently, all members of the subset $S \\setminus (T \\cup U)$ sent their hats to people in $T$. In particular, $S \\setminus (T \\cup U)$ has the property described in the beginning. The maximality of $T$ gives $|S \\setminus (T \\cup U)| \\le |T|$. Finally, we obviously have $|U| \\le |T|$, so\n\n$$\n|T| \\ge |S \\setminus (T \\cup U)| = |S| - |T| - |U| \\ge |S| - 2|T|,\n$$\n\nor $|T| \\ge \\frac{1}{3}|S| = 10$, as desired.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16099,
    "subject": "Mathematics (Olympiad)",
    "question": "Positive real numbers $x, y$ satisfy the following condition: there exist $a \\in [0, x], b \\in [0, y]$ such that\n$$\na^2 + y^2 = 2, \\quad b^2 + x^2 = 1, \\quad ax + by = 1.\n$$\nThen the maximum of $x + y$ is \\_\\_\\_\\_\\_.",
    "options": [],
    "answer": "See solution",
    "solution": "In a plane rectangular coordinate system $xOy$, for positive real number pairs $(x, y)$ that satisfy the condition, take points $L(x, 0)$, $M(x, y)$, $N(0, y)$, and then quadrilateral $OLMN$ is a rectangle. Points $P, Q$ are on sides $LM, MN$, respectively, as shown below.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p48_data_d45c3ba7c5.png)\n\nSince $a^2 + y^2 = 2$, $b^2 + x^2 = 1$, $ax + by = 1$, we have\n$$\n\\begin{aligned}\n|OP| &= \\sqrt{x^2 + b^2} = 1, \\\\\n|OQ| &= \\sqrt{a^2 + y^2} = \\sqrt{2}, \\\\\n|PQ| &= \\sqrt{(a-x)^2 + (b-y)^2} \\\\\n&= \\sqrt{(a^2 + y^2) + (b^2 + x^2) - 2(ax + by)} = 1.\n\\end{aligned}\n$$\nThus, $\\triangle OPQ$ is an isosceles right triangle with $P$ as its right-angle vertex.\n\nTherefore, we can set $\\angle LOP = \\theta$, $\\angle QON = \\frac{\\pi}{4} - \\theta$, where $0 \\leq \\theta \\leq \\frac{\\pi}{4}$.\n\nThen\n$$\n\\begin{aligned}\nx + y &= |OL| + |ON| \\\\\n&= |OP| \\cdot \\cos \\angle LOP + |OQ| \\cdot \\cos \\angle QON \\\\\n&= \\cos \\theta + \\sqrt{2} \\cos \\left(\\frac{\\pi}{4} - \\theta\\right) \\\\\n&= 2 \\cos \\theta + \\sin \\theta \\\\\n&= \\sqrt{5} \\sin(\\theta + \\varphi),\n\\end{aligned}\n$$\nwhere $\\varphi = \\arcsin \\frac{2\\sqrt{5}}{5}$.\n\nWhen $\\theta = \\frac{\\pi}{2} - \\varphi$ (correspondingly, $x = \\frac{2\\sqrt{5}}{5}$, $y = \\frac{3\\sqrt{5}}{5}$), $x + y$ takes the maximum $\\sqrt{5}$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16100,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\ge 3$, consider $n$ distinct points on a circle, labeled $1$ through $n$. Determine the maximum number of closed chords $[ij]$, $i \\neq j$, having pairwise non-empty intersections.",
    "options": [],
    "answer": "See solution",
    "solution": "We shall prove that any such configuration contains at most $n$ chords and that this upper bound is achieved, so the required maximum is $n$.\n\nFix an orientation of the circle and relabel the points $1$ through $n$ in the corresponding circular order. Consider a configuration of chords $[ij]$, $i \\neq j$, with pairwise non-empty intersections. Assign to each point $i$, which is an endpoint of at least one chord, the first point $i'$ following $i$, to which it is connected.\n\nWe now show that by deleting the chords $[ii']$, no chord is left, so the number of chords in the configuration does not exceed $n$. Suppose, if possible, that some chord $[ij]$ is left. Then $i, i', j, j'$ are in circular order around the circle, so the chords $[ii']$ and $[jj']$ do not meet—a contradiction.\n\nA maximal configuration is given by the $n$ chords $[1i]$, $i = 2, 3, \\dots, n$, and $[2n]$.\n\n(The $n$ points could be located anywhere in the plane, or the chords could be Jordan arcs; the topic is related to John Conway's thrackle conjecture.)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16101,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $a, b, c \\ge 0$ and $a^2 + b^2 + c^2 \\le 1$, prove:\n$$\n\\frac{a}{a^2 + bc + 1} + \\frac{b}{b^2 + ca + 1} + \\frac{c}{c^2 + ab + 1} + 3abc < \\sqrt{3}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\sum$ denote the cyclic sum. First, notice that\n$$\n\\sum a - \\sum \\frac{a}{a^2 + bc + 1} = \\sum \\frac{a^3 + abc}{a^2 + bc + 1} \\ge \\frac{(\\sum (a^3 + abc))^2}{\\sum (a^3 + abc)(a^2 + bc + 1)},\n$$\nwhere the last inequality comes from the Cauchy-Schwarz inequality. Meanwhile,\n$$\n\\begin{aligned}\n& \\sum (a^3 + abc)(a^2 + bc + 1) \\\\\n&= \\sum (a^3 + abc) + \\sum a(a^2 + bc)(a^2 + bc) \\\\\n&= \\sum a^3 + 3abc + \\sum a^5 + \\sum abc(2a^2 + bc) \\\\\n&\\le \\sum a^3 + 3abc + \\sum a^5 + 3 \\sum abc a^2 \\\\\n&\\le \\sum a^3 + 3abc + \\sum a^5 + 3abc \\\\\n&\\le 2 \\left( \\sum a^3 + 3abc \\right).\n\\end{aligned}\n$$\nIn the above, equality cannot be attained since $a = b = c = 1$ and $a^2 + b^2 + c^2 = 1$ cannot hold simultaneously. It follows that\n$$\n\\begin{aligned}\n\\sum a - \\sum \\frac{a}{a^2 + bc + 1} &\\ge \\frac{\\left(\\sum (a^3 + abc)\\right)^2}{\\sum (a^3 + abc)(a^2 + bc + 1)} > \\frac{\\left(\\sum a^3 + 3abc\\right)^2}{2\\left(\\sum a^3 + 3abc\\right)} \\\\\n&= \\frac{1}{2} \\left(\\sum a^3 + 3abc\\right) \\ge \\frac{1}{2}(3abc + 3abc) = 3abc,\n\\end{aligned}\n$$\nand hence\n$$\n\\sum \\frac{a}{a^2 + bc + 1} < \\sum a - 3abc \\le \\sqrt{3 \\left(\\sum a^2\\right)} - 3abc = \\sqrt{3} - 3abc,\n$$\nthat is,\n$$\n\\frac{a}{a^2 + bc + 1} + \\frac{b}{b^2 + ca + 1} + \\frac{c}{c^2 + ab + 1} + 3abc < \\sqrt{3}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16102,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$ be a positive integer and let $A_n$ be the set of positive integers less than $n$ that are coprime to $n$. Consider the polynomial\n\n$$P_n(x) = \\sum_{k \\in A_n} x^{k-1}.$$\n\n(a) Prove that $P_n(x)$ is divisible by some polynomial $x^r + 1$ for some $r \\in \\mathbb{Z}^+$.\n\n(b) Find all positive integers $n$ such that $P_n(x)$ is irreducible in $\\mathbb{Z}[x]$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) First, note that for $m, k \\in \\mathbb{Z}^+$ and $k$ odd, $x^{km} + 1$ is divisible by $x^m + 1$. For convenience, consider the polynomial $Q_n(x) = x P_n(x) = \\sum_{k \\in A_n} x^k$. We need to prove that $Q_n(x)$ is divisible by $x^r + 1$ for some $r \\in \\mathbb{Z}^+$.\n\n- If $n$ is odd: Since $\\gcd(n, k) = \\gcd(n, n-k)$ for all $k = 1, 2, \\dots, n-1$, for each $k \\in A_n$, $n-k \\in A_n$. Thus, we can pair the terms in $P_n(x)$ as $(x^k, x^{n-k})$. Since $k$ and $n-k$ have different parity, $(n-k) - k = n - 2k$ is odd, and $x^k + x^{n-k} = x^k(x^{n-2k} + 1)$ is divisible by $x+1$. Therefore, $x+1 \\mid Q_n(x)$.\n\n- If $4 \\mid n$: Partition into pairs $(x^k, x^{n-k})$ such that $n - 2k \\equiv 2 \\pmod{4}$. Then $x^k + x^{n-k} = x^k(x^{n-2k} + 1)$ is divisible by $x^2 + 1$. Hence, $x^2 + 1 \\mid Q_n(x)$.\n\n- If $n \\equiv 2 \\pmod{4}$: We use the following remarks.\n\n**Remark.** If $Q_n(x)$ is divisible by $x^r + 1$ ($r \\in \\mathbb{Z}^+$), then for any prime $p$ with $p \\nmid n$, $x^r + 1 \\mid Q_{pn}(x)$.\n\n*Proof.* For $a \\in A_n$, $kp + a \\in A_{pn}$ for $0 \\le k \\le p-1$. For each $k$, all terms with exponents of the form $kp + a$ (with $a \\in A_n$) share the common factor $x^{kp}$, and their sum is divisible by $\\sum_{a \\in A_n} x^a$, which is divisible by $x^r + 1$. Thus, $x^r + 1 \\mid Q_{pn}(x)$.\n\n**Remark.** If $n = 2p_1p_2\\cdots p_m$ for $m \\in \\mathbb{Z}^+$ and distinct primes $p_1, \\dots, p_m$, then $Q_n(x)$ is reducible.\n\n*Proof.* We prove by induction on $m$ (the number of odd primes in $n$):\n\n- For $m = 1$, $n = 2p$ (with $p$ prime), $A_n = \\{1, 2, \\dots, 2p-1\\} \\setminus \\{p\\}$, so $Q_n(x)$ is divisible by $x^p + 1$.\n- Suppose the lemma holds for $m \\ge 1$. Let $n = 2p_1p_2 \\cdots p_m$ and $p$ an odd prime with $\\gcd(p, n) = 1$, and $Q_n(x)$ divisible by $x^r + 1$. For $N = pn$, $x^r + 1 \\mid Q_N(x)$.\n\nIn $A_N$, terms with exponents coprime to $n$ but divisible by $p$ form $R_N(x)$, which is $p$ repetitions of $Q_n(x)$ (differing by $x^{kp}$ for $1 \\le k \\le p-1$). Thus, $x^r + 1 \\mid R_N(x)$. Removing exponents that are multiples of $p$ (terms $x^{ap}$ for $a \\in A_n$), their sum is $Q_n(x^p)$, which is divisible by $x^{pr} + 1$ and thus by $x^r + 1$. Therefore, $Q_N(x) = R_N(x) - Q_n(x^p)$ is divisible by $x^r + 1$.\n\nBy induction, the result holds for all such $n$.\n\n(b) From part (a), we reduce the problem to finding $n \\ge 3$ such that $|A_n| = 2$, i.e., $\\varphi(n) = 2$.\n\n- If $n = p$ is prime, $2 = \\varphi(n) = n-1$, so $n = 3$.\n- If $n = p^\\alpha$ for $\\alpha > 1$ and $p$ prime, $2 = \\varphi(n) = p^{\\alpha-1}(p-1)$. This gives $p = 2, \\alpha = 2$, so $n = 4$.\n- If $n$ has at least two distinct prime divisors $p, q$, then $2 = \\varphi(n)$ is divisible by $\\varphi(p) = p-1$ and $\\varphi(q) = q-1$, so $p = 2$ and $q = 3$, giving $n = 6$.\n\nTherefore, the answers are $n = 3, 4, 6$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16103,
    "subject": "Mathematics (Olympiad)",
    "question": "給定三角形 $ABC$。設 $BPCQ$ 為一平行四邊形（$P$ 不在 $BC$ 上）。令 $U$ 為 $CA$ 與 $BP$ 的交點，$V$ 為 $AB$ 與 $CP$ 的交點，$X$ 為 $CA$ 與三角形 $ABQ$ 的外接圓異於 $A$ 的交點，$Y$ 為 $AB$ 與三角形 $ACQ$ 的外接圓異於 $A$ 的交點。證明 $\\overline{BU} = \\overline{CV}$ 若且唯若 $AQ, BX, CY$ 三線共點。\n\n![](images/2024-TWN_p65_data_7a9725c6c3.png)",
    "options": [],
    "answer": "See solution",
    "solution": "由根心定理，$AQ, BX, CY$ 共點若且唯若 $B, C, X, Y$ 共圓。透過 $\\angle BXC = \\angle BQA$，$\\angle BYC = \\angle AQC$，這又等價於 $QA$ 為 $\\angle BQC$ 的其中一條角平分線，也就是 $\\frac{\\sin \\angle BQA}{\\sin \\angle AQC} = \\pm 1$（注意到 $Q$ 不在 $BC$ 上）。\n\n另一方面，由角元西瓦定理：\n\n$$\n\\frac{\\sin \\angle BQA}{\\sin \\angle AQC} \\cdot \\frac{\\sin \\angle CBA}{\\sin \\angle ABQ} \\cdot \\frac{\\sin \\angle QCA}{\\sin \\angle ACB} = 1.\n$$\n\n因此 $\\frac{\\sin \\angle BQA}{\\sin \\angle AQC} = \\pm 1$ 若且唯若 $\\frac{\\sin \\angle CBA}{\\sin \\angle ABQ} \\cdot \\frac{\\sin \\angle QCA}{\\sin \\angle ACB} = \\pm 1$。而\n\n$$\n\\left| \\frac{\\sin \\angle CBA}{\\sin \\angle ABQ} \\cdot \\frac{\\sin \\angle QCA}{\\sin \\angle ACB} \\right| = \\frac{\\sin \\angle CBV}{\\sin \\angle BVC} \\cdot \\frac{\\sin \\angle BUC}{\\sin \\angle UCB} = \\frac{\\overline{CV}}{\\overline{BC}} \\cdot \\frac{\\overline{BC}}{\\overline{BU}} = \\frac{\\overline{CV}}{\\overline{BU}}\n$$\n\n故兩者等價。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16104,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a board with $n \\times pm$ squares, where squares with coordinates that can be written as $(x, y) = (i + ap, 2i + bp)$ for some integers $a, b$ and $i$ are marked.\n\nFind, for any fixed square $(x_0, y_0)$, which squares in the same row, column, and diagonals as $(x_0, y_0)$ are marked, and show that there are no $p$ unmarked squares in a row. Then, for a board of size $n \\times m$, show that the second player can always find a sub-board of size $n \\times m$ with at most $\\lfloor nm/p \\rfloor$ marks, and thus no $p$ unmarked squares in a row.",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze the marking pattern:\n\n**Columns:**\nFor a fixed square $(x_0, y_0)$, the column has coordinates $(x_0, y)$. Solving\n$$\n\\begin{aligned}\ni + ap &= x_0 \\\\\n2i + bp &= y\n\\end{aligned}\n$$\ngives $y = 2x_0 + p(b - a)$. Since $a$ and $b$ are arbitrary integers, every $p$th square in the column (those congruent to $2x_0$ mod $p$) is marked. Thus, exactly $m$ squares are marked in each column, so there cannot be $p$ unmarked squares in a row.\n\n**Rows:**\nFor the row containing $(x_0, y_0)$, solve\n$$\n\\begin{aligned}\ni + ap &= x \\\\\n2i + bp &= y_0\n\\end{aligned}\n$$\nThis gives $2x = y_0 + p(a - b)$. Since $2$ is coprime to $p$, for each $x$ modulo $p$ there is a solution, so every $p$th square in the row is marked. Thus, no $p$ unmarked squares in a row.\n\n**Diagonals:**\nFor diagonals through $(x_0, y_0)$:\n- $(x_0 + t, y_0 + t)$: $t = (2a - b)p - 2x_0 + y_0$\n- $(x_0 + t, y_0 - t)$: $3t = (b - 2a)p + 2x_0 - y_0$\nSince $3$ is coprime to $p$, similar logic applies: every $p$th square is marked, so no $p$ unmarked squares in a diagonal.\n\n**General $n \\times m$ board:**\nAlice notes the second player can partition the $n \\times m$ board into $p$ boards of size $n \\times m$ and pick the one with the fewest marks. Since the total number of marks is $nm$, at least one sub-board has at most $\\lfloor nm/p \\rfloor$ marks. Since no $p$ unmarked squares in a row exist on the big board, the same holds for any sub-board, proving Alice's claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16105,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with $AB < AC$, and let $D$ be a variable point interior to the side $AB$. The parallel through $D$ to $BC$ crosses $AC$ at $E$. The perpendicular bisector of $DE$ crosses $BC$ at $F$. The circles $BDF$ and $CEF$ cross again at $K$. Prove that the line $FK$ passes through a fixed point.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that $FK$ passes through the reflection $A'$ of $A$ in the line $BC$, which is clearly a fixed point. Alternatively, but equivalently, we are to show that $A'$ has equal powers with respect to the circles $\\gamma_B$ through $B, D, F$, and $\\gamma_C$ through $C, E, F$.\n\nLet $A'B$ cross $\\gamma_B$ again at $B'$, and let $A'C$ cross $\\gamma_C$ again at $C'$. The argument hinges on two facts below:\n\n(1) $B', D, E$, and $C'$ all lie on a circle $\\gamma$ centered at $F$; and\n\n(2) the quadrangle $BB'CC'$ is cyclic.\n\n![](images/RMC_2024_p91_data_1dc5068866.png)\n\nStatement (2) clearly implies that $A'$ has equal powers with respect to $\\gamma_B$ and $\\gamma_C$, so we proceed to prove the two statements above.\n\nTo prove (1), read angles from $\\gamma_B$: $\\angle B'DF = \\angle A'BF = \\angle FBD = \\angle FB'D$, so $FB' = FD$. Similarly, $FC' = FE$, and since $FD = FE$, statement (1) follows.\n\nTo prove (2), read angles from $\\gamma$, $\\gamma_C$, and $\\gamma_B$:\n\n$$\n\\begin{align*}\n\\angle B'C'C &= \\angle B'C'E + \\angle EC'C = (180^\\circ - \\angle B'DE) + \\angle EFC \\\\\n&= 180^\\circ - (\\angle B'DF + \\angle FDE) + \\angle FDE \\quad (\\angle EFC = \\angle FED = \\angle FDE) \\\\\n&= 180^\\circ - \\angle B'DF = \\angle B'BF = \\angle B'BC.\n\\end{align*}\n$$\n\nThis establishes (2) and completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16106,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1 = x$ and $a_2 = y$. The sequence is:\n\n$$\nx,\\ y,\\ y-x,\\ -x,\\ -y,\\ x-y,\\ x,\\ y,\\ \\dots\n$$\n\nThe sequence repeats every 6 terms. Also, the sum of every 6 consecutive terms is 0. Given that $y-x = 1879$ and $x = 1997$, find the sum of the first 2000 terms of the sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "The sequence repeats every 6 terms, so the sum of every 6 consecutive terms is 0. The sum of the first $2000 = 6 \\times 333 + 2$ terms is $a_1 + a_2 = x + y$.\n\nGiven $y - x = 1879$ and $x = 1997$, we have:\n\n$$\ny = (y - x) + x = 1879 + 1997 = 3876.\n$$\n\nTherefore, the sum is:\n\n$$\nx + y = 1997 + 3876 = 5873.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16107,
    "subject": "Mathematics (Olympiad)",
    "question": "The numbers $1, 2, \\dots, 2010$ are written in a row. Two players take turns writing $+$ or $\\times$ between two consecutive numbers for as long as this is possible. The first player wins if the obtained algebraic sum is divisible by $3$; otherwise, the second player wins. Find a winning strategy for one of the players.",
    "options": [],
    "answer": "See solution",
    "solution": "The first player (A) has a winning strategy.\n\nReduce $1, 2, \\dots, 2010$ modulo $3$, so the sequence becomes $1, 2, 0, 1, 2, 0, \\dots$ (670 triples of $1, 2, 0$).\n\nA divides the $2009$ gaps into one group of $2$ (a double, corresponding to the initial $1, 2, 0$) and $669$ groups of $3$ (triples, each corresponding to $0, 1, 2, 0$). A makes the first move in a triple. Then, A can ensure to be the one making the last move in each group as follows:\n\n- If B moves in the double, A repeats the same move with the second gap in the double.\n- If B is the first to move in a triple, A moves first in another triple.\n- If B is the second to move in a triple, A makes the third remaining move in the same triple.\n\nA can always stick to these rules because there are an odd number of triples ($669$) and his initial move is in one of them. Thus, if several moves were made according to the above and it is B's turn, there are an even number of untouched triples, and $1$ or $3$ moves were made in each remaining triple.\n\nIn case (b), A's move can be arbitrary. In case (c), where A completes a triple $T$, he plays to ensure the following: if eventually $T$ has the form $0 \\times 1 + 2 \\times 0$, then the two $*$ are the same; and if $T$ has the form $0 \\times 1 \\times 2 \\times 0$, then one of the two $*$ is $\\times$. This is always possible by direct verification.\n\nAfter all moves, we obtain the sum $S$ of several products, some possibly with one factor. If a product in $S$ does not contain $0$, then it can be $1$, $2$, or $1 \\times 2$; however, $1 \\times 2$ is excluded. Such a $1 \\times 2$ cannot come from a triple $0 \\times 1 \\times 2 \\times 0$ because, by the previous paragraph, one of the two $*$ is $\\times$. Similarly, if a $1 \\times 2$ comes from the double $1 \\times 2 \\times 0$, then $*$ is $\\times$.\n\nSo a product in $S$ without a $0$ is either a $1$ or a $2$, meaning it is contained in a part $\\dots + 1 + 2 \\dots$ or $\\dots 1 + 2 + \\dots$ of $S$. If the $1$ and $2$ are in a triple $0 \\times 1 + 2 \\times 0$, then the triple is $0 + 1 + 2 + 0$ as the two $*$ are the same, and one is a $+$. If the $1$ and $2$ are in the double $1 + 2 + 0$, then the double is $1 + 2 + 0$ (by the first rule). We conclude that the products without a $0$ in $S$ come in pairs of the form $\\dots + 1 + 2 + \\dots$ or $1 + 2 + \\dots$ (for the double). It follows that $S$ is divisible by $3$, so A wins.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16108,
    "subject": "Mathematics (Olympiad)",
    "question": "In a circular ring with radii $R$ and $R-2r$, where $R = 11r$, we place non-overlapping circles of radius $r$ tangent to the circles defining the circular ring. Determine the maximal number of these circles.\n\n(It is given that $9.94 < \\sqrt{99} < 9.95$)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N$ be the number of non-overlapping circles $C_i(K_i, r)$ that can be placed in the given circular ring, each tangent to its boundary.\n\nThe circle $C(O, R - r)$ has a circumference greater than the perimeter $\\ell_{K_1\\ldots K_N K_1}$ of the polygon whose vertices are the centers $K_i$ of the circles $C_i(K_i, r)$. Therefore:\n\n$$\nN \\cdot 2r < \\ell_{K_1\\ldots K_N K_1} < 2\\pi(R - r) \\implies N < \\pi \\left( \\frac{R}{r} - 1 \\right). \\quad (1)\n$$\n\n![](images/Greek_07_Booklet_p9_data_aefe028322.png \"Figure 2\")\n![](images/Greek_07_Booklet_p9_data_b3f104ed7e.png \"Figure 3\")\n\nLet $OA$ be a tangent from $O$ to one of the circles $C_i(K_i, r)$. Then\n\n$$\nOA^2 = R(R - 2r) \\implies OA = \\sqrt{R(R - 2r)}.\n$$\n\nThe circle $C(O, OA)$ is tangent to the sides of the polygonal line $K_1K_2, \\ldots, K_{N-1}K_N$, and we have\n\n$$\n2\\pi\\sqrt{R(R - 2r)} < N \\cdot 2r + 2r \\implies \\pi\\sqrt{\\frac{R}{r}\\left(\\frac{R}{r} - 2\\right)} - 1 \\leq N. \\quad (2)\n$$\n\nFrom (1) and (2) it follows that:\n\n$$\n\\pi\\sqrt{\\frac{R}{r}\\left(\\frac{R}{r} - 2\\right)} - 1 \\leq N < \\pi\\left(\\frac{R}{r} - 1\\right).\n$$\n\nGiven $R = 11r$, we find\n\n$$\n\\pi\\sqrt{11(11 - 2)} - 1 \\leq N < \\pi(11 - 1) \\\\\n\\implies \\pi\\sqrt{99} - 1 \\leq N < 10\\pi \\approx 31.4 \\implies N = 31.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16109,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an equilateral and acute $\\triangle ABC$ with $AC > BC$. Some point $P$ in the interior of $\\triangle ABC$ is such that $\\angle APB = 180^\\circ - \\angle ACB$. The lines $AP$ and $BP$ intersect segments $BC$ and $AC$ at points $A_1$ and $B_1$, respectively. Let $M$ be the midpoint of $A_1B_1$, and the circumscribed circles about $\\triangle ABC$ and $\\triangle A_1CB_1$ intersect for the second time at point $Q$. Show that $\\angle PQM = \\angle BQA_1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P'$ be symmetric to $P$ with respect to the midpoint $N$ of $AB$. Then $AP'BP$ is a parallelogram, so $AP'BC$ is inscribed. Thus, $\\angle AQP' = \\angle ABP' = \\angle BAP$. On the other hand, $\\angle AQP = \\angle AQC - \\angle PQC = 180^\\circ - \\angle ABC - \\angle AA_1B = \\angle PAB$, since $P$ lies on the circle circumscribed about $\\triangle A_1B_1C$. Therefore, $\\angle AQP' = \\angle AQP$, so $P, Q, P'$ are collinear, i.e., $N \\in PQ$. Also, $\\triangle AQB \\sim \\triangle B_1QA_1$, which means $\\angle NQB = \\angle MQA_1$ as corresponding elements. This is equivalent to $\\angle PQB = \\angle MQA_1$, so $\\angle PQM = \\angle BQA_1$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16110,
    "subject": "Mathematics (Olympiad)",
    "question": "We have four charged batteries, four uncharged batteries, and a radio which needs two charged batteries to work.\n\nSuppose we don't know which batteries are charged and which ones are uncharged. Find the least number of attempts sufficient to make sure the radio will work. An attempt consists of putting two batteries in the radio and checking if the radio works or not.",
    "options": [],
    "answer": "See solution",
    "solution": "Let's generalize this problem to $2n$ batteries, $n$ of them charged. The number of charged batteries needed is still two.\n\nOne can see that the order of the attempts doesn't matter and that a set of attempts is always successful if and only if in every set of $n$ batteries there is an attempt with two of these batteries, that is, no set of $n$ batteries is \"pairwise untested\". Indeed, if there is a set $S$ of $n$ pairwise untested batteries, you can be unlucky enough: the working batteries might be these in $S$. Conversely, if every set of $n$ batteries has two tested batteries, we will choose at some moment two charged batteries and the radio will work.\n\nSo consider a graph $G$ in which each battery is a vertex and we connect two batteries iff we **do not** test these batteries. The number of attempts is the number of disconnected pairs, that is, $\\binom{2n}{2} - |E|$, where $E$ is the set of edges of $G$. Since we want to minimize the number of attempts, we must maximize $|E|$, that is, we need to have the maximum number of edges. But the only restriction is that we don't have a set $S$ of $n$ pairwise untested batteries, which speaking \"graph-wise\", is the same as $G$ not having an $n$-clique. But, by Turán's Theorem, the graph with the maximum number of edges without an $n$-clique is the most balanced $(n - 1)$-partite graph. In this case, the partition sets of vertices of $G$ must contain $3, 3, 2, 2, \\ldots, 2$ vertices (where we have $n - 3$ twos). So the least number of attempts is the number of disconnected pairs of vertices, that is, $2 \\cdot 3 + (n - 3) = n + 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16111,
    "subject": "Mathematics (Olympiad)",
    "question": "Нека $r_1, r_2, r_3$ ($r_1 < r_2 < r_3$) се должини на радиуси на кругови $k_1, k_2, k_3$ кои допираат краци на даден агол. Притоа $k_1$ и $k_2$ се допираат, а исто така и $k_2$ и $k_3$ се допираат. Докажи дека $r_2$ е геометриска средина на $r_1$ и $r_3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Нека $O_1, O_2, O_3$ се центри на кружниците $k_1, k_2, k_3$, а $M_1, M_2, M_3$ се допирни точки на $k_1, k_2, k_3$ со еден крак на аголот (види цртеж).\n\n![](images/Makedonija_2009_p16_data_9bb2a6bdec.png)\n\nЈасно е дека $\\overline{O_1M_1} = r_1$, $\\overline{O_2M_2} = r_2$, $\\overline{O_3M_3} = r_3$. Точката K припаѓа на $O_3M_3$ и $O_1K \\parallel M_1M_3$. Точката L е пресечна точка на $O_1K$ со $O_2M_2$ (види цртеж).\n\n![](images/Makedonija_2009_p16_data_df8c2fb85e.png)\n\nТриаголниците $O_1LO_2$ и $O_1KO_3$ се слични, од каде добиваме\n\n$$\n\\overline{O_3K} : \\overline{O_2L} = \\overline{O_1O_3} : \\overline{O_1O_2}. \\quad (1)\n$$\n\nБидејќи $\\overline{O_3K} = r_3 - r_1$, $\\overline{O_2L} = r_2 - r_1$, $\\overline{O_1O_3} = r_1 + 2r_2 + r_3$, $\\overline{O_1O_2} = r_1 + r_2$, ако замениме во (1) добиваме\n\n$$\n(r_3 - r_1)(r_2 - r_1) = (r_1 + 2r_2 + r_3)(r_1 + r_2). \\quad (2)\n$$\n\nАко во (2) се ослободиме од загради и добиеното равенство го средиме, добиваме\n\n$$\nr_2^2 = r_1 r_3.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16112,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(2x + f(y)) = x + y + f(x)\n$$\n\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We choose $x$ so that the arguments on both sides become equal, i.e., the equation $2x + f(y) = x$ is satisfied. For this value $x = -f(y)$, we get\n\n$$\nf(-f(y)) = -f(y) + y + f(-f(y))\n$$\n\nand therefore $f(y) = y$ for all $y \\in \\mathbb{R}$. But this is clearly a solution, therefore, it is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16113,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABP$, $BCM$, and $CAN$ be the equilateral triangles constructed externally on triangle $ABC$. Let $O_A$, $O_B$, and $O_C$ be the centers of these equilateral triangles (which coincide with the circumcenters of triangles $BCF$, $ACF$, and $ABF$, respectively), and let $G_A$, $G_B$, and $G_C$ be the centroids of $BCF$, $ACF$, and $ABF$, respectively. Prove that the Euler lines $O_A G_A$, $O_B G_B$, and $O_C G_C$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "*Lemma.* Let $ABP$, $BCM$, and $CAN$ be the equilateral triangles constructed externally to triangle $ABC$. The lines $AM$, $BN$, and $CP$ concur at the Fermat point $F$ of $ABC$.\n\n*Proof.* Let $F$ be the intersection point of $BN$ and $CP$. Since triangles $BAN$ and $PAC$ are congruent, $\\angle APF = \\angle ABF$. So the quadrilateral $APBF$ is cyclic, and thus $\\angle AFP = \\angle ABP = 60^\\circ$ and $\\angle BFP = \\angle BAP = 60^\\circ$. Hence $\\angle AFB = 120^\\circ$. Similarly, $\\angle BFC = 120^\\circ$, which proves that $F$ is the Fermat point of triangle $ABC$. Analogously, $AM$ and $BN$ pass through the same point $F$, and the lemma is proved.\n\nIt is immediate that the Euler lines $O_A G_A$, $O_B G_B$, and $O_C G_C$ are parallel to $AF$, $BF$, and $CF$, respectively.\n\nNow consider the homothety with center $F$ and ratio $3/2$, which transforms $G_A$, $G_B$, and $G_C$ into the midpoints of $BC$, $CA$, and $AB$, respectively. This homothety also transforms the Euler lines of $BCF$, $ACF$, and $ABF$ into the lines $l_A$, $l_B$, and $l_C$, parallel to $AF$, $BF$, and $CF$ and passing through the midpoints of $BC$, $CA$, and $AB$, respectively. Hence these Euler lines are concurrent if and only if $l_A$, $l_B$, and $l_C$ are.\n\nFinally, consider the homothety with center $G$ (the centroid of $ABC$) and ratio $-2$. The midpoints of $BC$, $CA$, and $AB$ are transformed into the vertices $A$, $B$, and $C$, respectively, so $l_A$, $l_B$, and $l_C$ are transformed into $AF$, $BF$, and $CF$, respectively. Since the latter are concurrent at $F$, it follows that $l_A$, $l_B$, and $l_C$ are too, and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16114,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $2021$ points on a plane, no three of which are collinear. For every $5$ points, there exists at least $4$ among them which are concyclic. Is it necessarily true that at least $2020$ of the points are concyclic?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes.\n\nLet us first prove a lemma: if $4$ points $A$, $B$, $C$, $D$ all lie on a circle $\\Gamma$ and two points $X$, $Y$ do not lie on $\\Gamma$, then these $6$ points are pairs of intersections of three circles: $\\Gamma$ and two other circles. According to the problem statement, among $A$, $B$, $C$, $X$, $Y$, there are $4$ concyclic points. These $4$ must include $X$ and $Y$, because if one is not, the other must lie on $\\Gamma$. Without loss of generality, suppose $A$, $B$, $X$, $Y$ lie on the same circle. Similarly, among $A$, $C$, $D$, $X$, $Y$, there must be $4$ concyclic points, which must include $X$ and $Y$. Point $A$ cannot be one of them, because two circles cannot have more than two common points. Therefore, $C$, $D$, $X$, $Y$ are concyclic, which proves the lemma.\n\nNow, consider the case where there exist $5$ points on one circle $\\Gamma$. Label these $A$, $B$, $C$, $D$, $E$. Suppose two points $X$, $Y$ do not lie on $\\Gamma$. By the lemma, $A$, $B$, $C$, $D$, $X$, $Y$ must be the pairwise intersections of $3$ circles, one of which is $\\Gamma$. Without loss of generality, let the intersections of $\\Gamma$ with one of the other circles be $A$ and $B$, and with the other circle $C$ and $D$. Similarly, $A$, $B$, $C$, $E$, $X$, $Y$ must be the pairwise intersections of three circles, one of which is $\\Gamma$. This is not possible, as none of $A$, $B$, $C$ lies on the circumcircle of triangle $EXY$. This contradiction shows that at most $1$ point can lie outside $\\Gamma$, i.e., at least $2020$ points lie on $\\Gamma$.\n\nNow consider the case where no $5$ points lie on the same circle. Let $A$, $B$, $C$, $D$, $E$ be arbitrary $5$ points. Without loss of generality, let $A$, $B$, $C$, $D$ be concyclic and $E$ not on this circle. By the lemma, for every other point $F$ and points $A$, $B$, $C$, $D$, $E$, the $6$ points are the intersections of $\\Gamma$ and two other circles. But in total, there are $3$ such points, because one circle must go through $E$ and $2$ of $A$, $B$, $C$, $D$, while the other circle must go through $E$ and the other $2$ of $A$, $B$, $C$, $D$. There are only $3$ partitions of $A$, $B$, $C$, $D$ into two sets. This is a contradiction, as there are $2021 > 5 + 3$ points in total. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16115,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $abc = 1$. Prove that\n\n$$\n\\left(a + \\frac{1}{b}\\right)^2 + \\left(b + \\frac{1}{c}\\right)^2 + \\left(c + \\frac{1}{a}\\right)^2 \\geq 3(a + b + c + 1).\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "By using the AM-GM inequality ($x^2 + y^2 + z^2 \\geq xy + yz + zx$), we have\n\n$$\n\\begin{aligned}\n&\\left(a + \\frac{1}{b}\\right)^2 + \\left(b + \\frac{1}{c}\\right)^2 + \\left(c + \\frac{1}{a}\\right)^2 \\\\\n&\\geq \\left(a + \\frac{1}{b}\\right)\\left(b + \\frac{1}{c}\\right) + \\left(b + \\frac{1}{c}\\right)\\left(c + \\frac{1}{a}\\right) + \\left(c + \\frac{1}{a}\\right)\\left(a + \\frac{1}{b}\\right) \\\\\n&= \\left(ab + 1 + \\frac{a}{c} + a\\right) + \\left(bc + 1 + \\frac{b}{a} + b\\right) + \\left(ca + 1 + \\frac{c}{b} + c\\right) \\\\\n&= ab + bc + ca + \\frac{a}{c} + \\frac{c}{b} + \\frac{b}{a} + 3 + a + b + c\n\\end{aligned}\n$$\n\nNotice that by AM-GM, $ab + \\frac{b}{a} \\geq 2b$, $bc + \\frac{c}{b} \\geq 2c$, and $ca + \\frac{a}{c} \\geq 2a$.\n\nThus,\n\n$$\n\\left(a + \\frac{1}{b}\\right)^2 + \\left(b + \\frac{1}{c}\\right)^2 + \\left(c + \\frac{1}{a}\\right)^2 \\geq \\left(ab + \\frac{b}{a}\\right) + \\left(bc + \\frac{c}{b}\\right) + \\left(ca + \\frac{a}{c}\\right) + 3 + a + b + c \\geq 3(a + b + c + 1)\n$$\n\nEquality holds if and only if $a = b = c = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16116,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the circumscribed circle of an obtuse triangle $ABC$ with an obtuse angle $B$. Tangents to this circle at points $A$ and $B$ meet at point $P$, and the perpendicular to the line $BC$ at point $B$ intersects $AC$ at point $K$. Prove that $PA = PK$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that since $\\angle ABC > 90^\\circ$, point $K$ lies on $AC$. Also, it is clear that $PA = PB$. We will show that $K$ lies on the circle $\\omega$ with center $P$ and radius $PA$. It is enough to prove that $$\\angle APB = 2(180^\\circ - \\angle AKB).$$ Indeed, take any point $X$ on the larger arc of circle $\\omega$. Then $\\angle AXB = \\frac{1}{2}\\angle APB$. If this condition holds, then $\\angle AKB + \\angle AXB = 180^\\circ$, so points $A$, $K$, $B$, and $X$ lie on the circle $\\omega$. As $P$ is the center of this circle, $PK = PA$.\n\nSo, let's prove that $\\angle APB = 2(180^\\circ - \\angle AKB)$. Since $PA = PB$, and using the theorem about the angle between the chord and a tangent, we get:\n$$\n\\angle APB = 180^\\circ - 2\\angle ABP = 180^\\circ - 2\\angle ACB = 180^\\circ - 2(90^\\circ - \\angle CKB) = 2\\angle CKB = 2(180^\\circ - \\angle AKB).\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16117,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 15 children in total, divided into groups of 6, 5, 1, and 3. How many degrees should be allocated to the third-largest group (which has 3 children) in a pie chart representing all the children?",
    "options": [],
    "answer": "See solution",
    "solution": "The third-largest group has 3 children out of 15. The angle for this group in a pie chart is:\n\n$$\n\\frac{3}{15} \\times 360^{\\circ} = 72^{\\circ}\n$$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16118,
    "subject": "Mathematics (Olympiad)",
    "question": "Let *MAZN* be an isosceles trapezium inscribed in a circle $(c)$ with centre *O*. Assume that $MN$ is a diameter of $(c)$ and let *B* be the midpoint of $AZ$. Let $(\\varepsilon)$ be the perpendicular line to $AZ$ passing through $A$. Let *C* be a point on $(\\varepsilon)$, let *E* be the point of intersection of $CB$ with $(c)$ and assume that $AE$ is perpendicular to $CB$. Let *D* be the point of intersection of $CZ$ with $(c)$ and let *F* be the antidiametric point of $D$ on $(c)$. Let *P* be the point of intersection of $FE$ and $CZ$. Assume that the tangents of $(c)$ at the points $M$ and $Z$ meet the lines $AZ$ and $PA$ at the points $K$ and $T$ respectively. Prove that $OK$ is perpendicular to $TM$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will first prove that $PA$ is the tangent of $(c)$ at $A$. Since $EDZA$ is cyclic, then $\\angle EDC = \\angle EAZ$. By the similarity of the triangles $CAE$ and $ABE$ we have $\\angle EAZ = \\angle EAB = \\angle ACB$, so $\\angle EDC = \\angle EAZ$. Since $\\angle FED = 90^\\circ$, then\n\n$$\n\\angle EPD = 90^\\circ - \\angle EDC = 90^\\circ - \\angle ACB = \\angle EAC\n$$\n\nSo the points $E$, $A$, $C$, $P$ are concyclic. It follows that $\\angle CPA = 90^\\circ$, therefore the triangle $APZ$ is right-angled. Since also $B$ is the midpoint of $AZ$, then $PB = AB = BZ$.\n\nWe have\n\n$$\n\\angle BPE = \\angle ABC - \\angle BPZ = \\angle ABC - \\angle PZB\n$$\n\nand\n\n$$\n\\angle PAE = \\angle PCE = 90^\\circ - \\angle ACB - \\angle CZA = \\angle ABC - \\angle CZA = \\angle ABC - \\angle PZB\n$$\n\nTherefore $\\angle BPE = \\angle PAE$.\n\nSince also $\\angle EPD = \\angle EAC = \\angle EBA$, then $PEBZ$ is a cyclic quadrilateral and we get $\\angle BPE = \\angle EZB$. Therefore $\\angle PAE = \\angle EZB$, i.e. $PA$ is the tangent of $(c)$ at $A$.\n\n![](images/2020_BMO_Short_List_p15_data_4fefa8bfa9.png)\n\nSince $AZ$ is parallel to $MN$ then $TB \\perp AZ$ and $TO \\perp MN$.\n\nThe quadrilateral $KBOM$ is a rectangle. Consider the circumcircle of the rectangle and a tangent of this at $K$. Let $X$ be a point on this tangent. So $XK \\perp KO$. Since the triangles $OBA$ and $OTA$ are similar then $OA^2 = OT \\cdot OB$. Since $OA = OM$ and $KM = OB$ we get $OM^2 = OT \\cdot KM$ so $OT / OM = OM / KM$. Since also $\\angle KMO = \\angle TOM = 90^\\circ$, the triangles $TOM$ and $OMK$ are similar. Therefore\n\n$$\n\\angle MTO = \\angle KOM = \\angle XKM\n$$\n\nSince $KM$ is parallel to $TO$ we have $\\angle MTO = \\angle KMT$. Therefore $\\angle XKM = \\angle KMT$. I.e. the tangent at point $K$ is parallel to $MT$ and since $XK \\perp KO$ we get $OK \\perp TM$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16119,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the situation when the phone numbers have $n$ digits instead of 8. Prove, by induction, that at most $\\frac{9^n + 1}{2}$ phone numbers can be selected, and there is a unique way to select the phone numbers: choose all phone numbers whose digit sum has the same parity as $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed by induction on $n$.\n\n**Base case ($n=1$):**\nThere are 9 possible 1-digit phone numbers: 1 through 9. Group them as $(1,2), (3,4), (5,6), (7,8), (9)$. At most one number can be selected from each group, so at most $\\frac{9+1}{2} = 5$ numbers. The only way to select 5 is to choose all odd numbers (whose sum is odd, matching $n=1$).\n\n**Inductive step:**\nAssume the statement holds for $n-1$. For $n$ digits:\n- For numbers not starting with 9, pair $\\overline{a_1a_2\\cdots a_n}$ and $\\overline{(a_1+1)a_2\\cdots a_n}$ for $a_1 = 1,3,5,7$. At most one can be selected from each pair.\n- For numbers starting with 9, remove the 9 and apply the $n-1$ case. By hypothesis, at most $\\frac{9^{n-1}+1}{2}$ can be selected.\n\nTotal:\n$$\n\\frac{8 \\cdot 9^{n-1}}{2} + \\frac{9^{n-1} + 1}{2} = \\frac{9^n + 1}{2}\n$$\n\nTo achieve this, select all numbers $\\overline{a_1a_2\\cdots a_n}$ with $a_1 + a_2 + \\cdots + a_n \\equiv n \\pmod{2}$. This is the unique way, as shown by the pairing argument.\n\nFor $n=8$, this gives $\\frac{9^8+1}{2} = 21523361$ phone numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16120,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(n)$ be the minimum number of rounds required to determine a winner in a single-elimination tournament with $n$ contestants, where in each round, if the number of contestants is odd, one contestant advances automatically (gets a bye), and the rest are paired off to compete. What is the smallest natural number $n$ such that $f(n) = f(2013)$?",
    "options": [],
    "answer": "See solution",
    "solution": "We are told that $f(2013) = 11$. For $n \\leq 1024 = 2^{10}$, the tournament requires at most 10 rounds, since after each round the number of contestants is at most halved. For $n = 1025 = 1 + 2^{10}$, after each round, the number of contestants is $1 + 2^9$, $1 + 2^8$, ..., down to $1 + 2^1 = 3$ in the tenth round, and finally 2 contestants in the eleventh round. Thus, $f(1025) = 11$. Therefore, the smallest $n$ such that $f(n) = 11$ is $n = 1025$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16121,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p \\geq 5$ be a prime number, and let $M = \\{1, 2, \\dots, p-1\\}$. Define\n$$\nT = \\{(n, x_n) : p \\mid (n x_n - 1) \\text{ where } n, x_n \\in M\\}.\n$$\nFind the nonnegative least residue of $\\sum_{(n,x_n) \\in T} n \\left\\lfloor \\frac{n x_n}{p} \\right\\rfloor$ modulo $p$, where $[x]$ denotes the greatest integer less than or equal to $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "For each $n \\in \\{1, 2, \\dots, p-1\\}$, since $(n, p) = 1$, there exists a unique $x_n \\in M$ such that\n$$\nn x_n \\equiv 1 \\pmod{p},\n$$\nor equivalently, $p \\mid (n x_n - 1)$. Suppose $x'_n \\in M$ also satisfies $p \\mid (n x'_n - 1)$, then $p \\mid n(x_n - x'_n) \\implies p \\mid (x_n - x'_n) \\implies x_n = x'_n$.\n\nSince $p \\mid (n x_n - 1)$, we can write $n x_n = 1 + m_n p$ for some integer $m_n$. Thus,\n$$\n\\sum_{n=1}^{p-1} n \\left\\lfloor \\frac{n x_n}{p} \\right\\rfloor = \\sum_{n=1}^{p-1} n \\cdot \\frac{n x_n - 1}{p} = \\sum_{n=1}^{p-1} n m_n.\n$$\nLet $s = \\sum_{n=1}^{p-1} n m_n$. Split the sum as $s = \\sum_{n=1}^{\\frac{p-1}{2}} n m_n + \\sum_{n=1}^{\\frac{p-1}{2}} (p-n) m_{p-n}$.\n\nFor $n$, replace $n$ by $p-n$ in the congruence:\n$$\n(p-n) x_{p-n} \\equiv 1 \\pmod{p}.\n$$\nSimilarly, $(p-n) x_{p-n} = 1 + m_{p-n} p$, which gives\n$$\nn x_{p-n} \\equiv -1 \\pmod{p}.\n$$\nAdding $n x_n \\equiv 1$ and $n x_{p-n} \\equiv -1$ yields $n(x_n + x_{p-n}) \\equiv 0 \\pmod{p}$, so $x_n + x_{p-n} = p$.\n\nMultiply the congruences by $n$ and $p-n$ respectively:\n$$\nn^2 x_n \\equiv n + n m_n p \\pmod{p},\n$$\n$$\n(p-n)^2 x_{p-n} \\equiv p-n + (p-n) m_{p-n} p \\pmod{p}.\n$$\nAdd and sum over $n$:\n$$\n\\sum_{n=1}^{\\frac{p-1}{2}} n^2 x_n + \\sum_{n=1}^{\\frac{p-1}{2}} (p-n)^2 x_{p-n} = \\frac{p(p-1)}{2} + \\sum_{n=1}^{\\frac{p-1}{2}} [n m_n + (p-n) m_{p-n}].\n$$\nUsing $x_n + x_{p-n} = p$, this becomes\n$$\n\\begin{aligned}\n\\frac{p(p-1)}{2} + p s &= p^2 \\sum_{n=1}^{\\frac{p-1}{2}} x_{p-n} - 2p \\sum_{n=1}^{\\frac{p-1}{2}} n x_{p-n} + \\sum_{n=1}^{\\frac{p-1}{2}} n^2 p \\\\\n\\end{aligned}\n$$\nFinally,\n$$\n\\begin{aligned}\n\\frac{p-1}{2} + s &= p \\sum_{n=1}^{\\frac{p-1}{2}} x_{p-n} - 2 \\sum_{n=1}^{\\frac{p-1}{2}} n x_{p-n} + \\sum_{n=1}^{\\frac{p-1}{2}} n^2 \\\\\n&\\equiv 0 - 2 \\sum_{n=1}^{\\frac{p-1}{2}} (-1) + 0 \\\\\n&\\equiv 2 \\cdot \\frac{p-1}{2} \\pmod{p},\n\\end{aligned}\n$$\nso $s \\equiv \\frac{p-1}{2} \\pmod{p}$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16122,
    "subject": "Mathematics (Olympiad)",
    "question": "Denote by $\\nu(n)$ the exponent of $2$ in the prime factorization of $n!$. Show that for arbitrary positive integers $a$ and $m$, there exists an integer $n > 1$ for which $\\nu(n) \\equiv a \\mod m$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use the fact that $\\nu(n) = \\sum_{k=1}^{\\infty} \\left\\lfloor \\frac{n}{2^k} \\right\\rfloor$. If $n$ has base-2 representation $n = \\sum_{i=0}^{l} d_i \\cdot 2^i$ with $d_i \\in \\{0,1\\}$, then\n\n$$\n\\nu(n) = \\sum_{i=1}^{l} d_i \\cdot (2^i - 1).\n$$\n\nNow, let $m = 2^t u$ where $u$ is odd. For any $i \\ge t$ with $\\varphi(u) \\mid i-1$, Euler's theorem gives $u \\mid 2^{i-1} - 1$, so\n\n$$\n2^i - 1 = 2(2^{i-1} - 1) + 1 \\equiv 1 \\pmod{u}.\n$$\n\nAlso, since $i \\ge t$,\n\n$$\n2^i - 1 \\equiv -1 \\pmod{2^t}.\n$$\n\nThese two congruences determine the residue class $r$ of $2^i - 1$ modulo $m$, and $r$ is relatively prime to $m$.\n\nSince $m$ and $r$ are coprime, for any $a$ there is $u$ such that $a \\equiv u r \\pmod{m}$. For $n = 2^{i_1} + \\cdots + 2^{i_u}$ (with suitable $i_j$),\n\n$$\n\\nu(n) = (2^{i_1} - 1) + \\cdots + (2^{i_u} - 1) \\equiv u r \\equiv a \\pmod{m}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16123,
    "subject": "Mathematics (Olympiad)",
    "question": "One of the numbers $1, 2, \\ldots, n$ is written in each cell of a $17 \\times 17$ table for a certain $n \\in \\mathbb{N}$; all of these numbers are used. If a row contains two cells $C_1$ and $C_2$ with equal numbers $k$ and $C_1$ is to the left of $C_2$, then there are no numbers $k$ in the column of $C_1$ that are above $C_1$. Determine the minimal $n$ for which such a table exists.",
    "options": [],
    "answer": "See solution",
    "solution": "The minimal $n$ in question is $n = 9$.\n\nFirst, we show that each number $k \\in \\{1, 2, \\ldots, n\\}$ occurs in the table at most $34$ times. Let a row contain at least two $k$'s. Underline all of them except the rightmost one. The remaining numbers in the table are not underlined. By hypothesis, $k$ does not occur above an underlined number. It follows that the underlined numbers are in different columns, hence there are at most $17$ of them. In addition, by construction, each row without underlined $k$'s contains at most one $k$, so there are also at most $17$ numbers $k$ that are not underlined. In summary, the table has at most $17 + 17 = 34$ numbers $k$, as stated.\n\nSince each $k \\in \\{1, 2, \\ldots, n\\}$ occurs in the table $x_k \\leq 34$ times, the equality $x_1 + \\ldots + x_n = 17^2$ yields $n \\geq \\frac{17^2}{34}$, meaning that $n \\geq 9$. For an example with $n = 9$, consider the diagonals parallel to the main diagonal containing the bottom left and the top right cell. Label them consecutively $1, 2, \\ldots, 33$ so that the top left cell is diagonal $1$ and the bottom right cell is diagonal $33$. For each $k = 1, 2, \\ldots, 9$ write $k$ in all cells of diagonals $2k-1$ and $2k$; for each $k = 1, 2, \\ldots, 7$ write $k$ in all cells of diagonals $2k+17$ and $2k+18$; finally write $8$ in the only cell of diagonal $33$. In every row and column there are at most two numbers equal to a given $k \\in \\{1, 2, \\ldots, 9\\}$; and if there are two of them then they are adjacent. It follows that the table satisfies the given conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16124,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 1000 distinct points on a circle. Select $k$ of them so that no two chosen points are adjacent. In how many ways can this be done?",
    "options": [],
    "answer": "See solution",
    "solution": "Label the points clockwise as $A_1, A_2, \\dots, A_{1000}$. Each selection corresponds bijectively to a sequence $a_1a_2 \\dots a_{1000}$ of zeros and ones, where $a_j = 1$ if $A_j$ is selected, $0$ otherwise. The sequence must have exactly $k$ ones, no two ones adjacent, and at least one of $a_1$ or $a_{1000}$ is zero.\n\n**Type 1:** $a_1 = 0$. Every $1$ is preceded by at least one $0$. Delete one zero before each $1$ to get a sequence of length $1000 - k$ with $k$ ones. There are $\\binom{1000-k}{k}$ such sequences.\n\n**Type 2:** $a_1 = 1$, so $a_2 = a_{1000} = 0$. For $j > 1$, each $1$ is preceded by at least one $0$. Delete one zero before each such $1$, and also delete $a_1 = 1$ and $a_{1000} = 0$, resulting in a sequence of length $999 - k$ with $k-1$ ones. There are $\\binom{999-k}{k-1}$ such sequences.\n\n**Total ways:**\n$$\n\\binom{1000-k}{k} + \\binom{999-k}{k-1}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16125,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $a, b, c$ 為正實數，且滿足 $\\min\\{a+b, b+c, c+a\\} > \\sqrt{2}$ 以及 $a^2 + b^2 + c^2 = 3$。\n\n試證：\n\n$$\n\\frac{a}{(b+c-a)^2} + \\frac{b}{(c+a-b)^2} + \\frac{c}{(a+b-c)^2} \\geq \\frac{3}{(abc)^2}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "由 $b+c > \\sqrt{2}$ 可得 $b^2 + c^2 > 1$，故 $a^2 = 3 - (b^2 + c^2) < 2$，即 $a < \\sqrt{2} < b+c$。\n因此 $b+c-a > 0$，同理 $c+a-b > 0$ 及 $a+b-c > 0$。\n\n我們將用 HÖLDER 不等式：\n\n$$\n\\frac{x_1^{p+1}}{y_1^p} + \\frac{x_2^{p+1}}{y_2^p} + \\cdots + \\frac{x_n^{p+1}}{y_n^p} \\geq \\frac{(x_1 + x_2 + \\cdots + x_n)^{p+1}}{(y_1 + y_2 + \\cdots + y_n)^p}\n$$\n\n對所有正實數 $p, x_1, \\ldots, x_n, y_1, \\ldots, y_n$ 都成立。取 $p = 2, n = 3$，可得：\n\n$$\n\\begin{aligned}\n\\sum \\frac{a}{(b+c-a)^2} &= \\sum \\frac{(a^2)^3}{a^5 (b+c-a)^2} \\\\\n&\\geq \\frac{(a^2 + b^2 + c^2)^3}{\\left(\\sum a^{5/2} (b+c-a)\\right)^2} \\\\\n&= \\frac{27}{\\left(\\sum a^{5/2} (b+c-a)\\right)^2}.\n\\end{aligned}\n$$\n\n為了估計分母，觀察 SCHUR 不等式的一特例：\n\n$$\n\\sum a^{3/2} (a-b)(a-c) \\geq 0\n$$\n\n整理後得：\n\n$$\n\\sum a^{5/2} (b+c-a) \\leq abc(\\sqrt{a} + \\sqrt{b} + \\sqrt{c})\n$$\n\n再由柯西不等式知：\n\n$$\n\\left(\\frac{\\sqrt{a} + \\sqrt{b} + \\sqrt{c}}{3}\\right)^4 \\leq \\frac{a^2 + b^2 + c^2}{3} = 1\n$$\n\n即 $\\sqrt{a} + \\sqrt{b} + \\sqrt{c} \\leq 3$。因此\n\n$$\n\\sum \\frac{a}{(b+c-a)^2} \\geq \\frac{27}{\\left(abc(\\sqrt{a} + \\sqrt{b} + \\sqrt{c})\\right)^2} \\geq \\frac{3}{(abc)^2}\n$$\n\n得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16126,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all four-digit numbers which are exactly 2016 larger than the four-digit number obtained by moving the first digit to the end.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the first digit of the number be $a$ and the number formed by the remaining digits be $k$. By the conditions, $$1000a + k = 10k + a + 2016$$ whence $$111a - k = 224$$. Hence $a \\geq 3$, implying the solutions $a = 3, k = 109$; $a = 4, k = 220$; $a = 5, k = 331$; $a = 6, k = 442$; $a = 7, k = 553$; $a = 8, k = 664$; $a = 9, k = 775$. The corresponding four-digit numbers satisfying the conditions of the problem are $3109$, $4220$, $5331$, $6442$, $7553$, $8664$, and $9775$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16127,
    "subject": "Mathematics (Olympiad)",
    "question": "In $\\triangle ABC$, $\\angle B > \\angle C$. Let $D$ be the point on side $BC$ such that $\\angle DAC = \\frac{B - C}{2}$. The circumcircle of $\\triangle ACD$ meets side $AB$ again at $E$. The circumcircle of $\\triangle ABD$ meets side $AC$ again at $F$. The internal angle bisector of $\\angle BDE$ meets side $AB$ at $P$. The internal angle bisector of $\\angle CDF$ meets side $AC$ at $Q$. Prove that $PQ$ and $AB$ are perpendicular.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $\\angle PDQ = 180^\\circ - \\angle BDP - \\angle CDQ = 180^\\circ - \\frac{1}{2} \\angle BDE - \\frac{1}{2} \\angle CDF = 180^\\circ - \\frac{A}{2} - \\frac{A}{2} = 180^\\circ - A$. This implies $A, P, D, Q$ are concyclic. Hence, $\\angle APQ = \\angle ADQ = 180^\\circ - \\angle DAQ - \\angle ACD - \\angle CDQ = 180^\\circ - \\frac{B - C}{2} - C - \\frac{A}{2} = 90^\\circ$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16128,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a non-negative integer $n$, let $\\sum_{k=0}^{2n} a_k X^k$ be the standard power expansion of the polynomial\n$$\n\\sum_{k=0}^{n} \\binom{n}{k}^2 (X+1)^{2k} (X-1)^{2(n-k)}.\n$$\nThe coefficients $a_{2k+1}$ all vanish since the polynomial is invariant under the change $X \\mapsto -X$. Show that the $a_{2k}$ are all positive.",
    "options": [],
    "answer": "See solution",
    "solution": "Leaving aside the trivial cases $n=0$ and $n=1$, assume $n \\ge 2$. The conclusion follows from the fact that the polynomial under consideration,\n$$\nf = \\sum_{k=0}^{n} \\binom{n}{k}^2 (X+1)^{2k} (X-1)^{2(n-k)},\n$$\nis expressible as a sum of $n+1$ polynomials of the form $(b_k X^2 + c_k)^n$, where the $b_k$ and the $c_k$ are all non-negative real numbers and at least $n-1$ products $b_k c_k$ are positive; this latter then implies that the even powers of $X$ all occur in the expansion with a positive coefficient. In particular, the fact that the odd rank coefficients in the standard expansion all vanish comes for free.\n\nLet $S$ be the set of $(n+1)$-st roots of unity and recall that, if $k$ and $\\ell$ are integers in the range $0$ through $n$, then the sum $\\sum_{\\omega \\in S} \\omega^k \\bar{\\omega}^\\ell$ vanishes, unless $k=\\ell$ in which case it is equal to $n+1$.\n\nConsequently,\n$$\n\\begin{align*}\nf &= \\frac{1}{n+1} \\sum_{k=0}^{n} \\sum_{\\ell=0}^{n} \\binom{n}{k} \\binom{n}{\\ell} (X+1)^{k+\\ell} (X-1)^{2n-k-\\ell} \\sum_{\\omega \\in S} \\omega^k \\bar{\\omega}^{\\ell} \\\\\n&= \\frac{1}{n+1} \\sum_{\\omega \\in S} \\left( \\sum_{k=0}^{n} \\binom{n}{k} \\omega^k (X+1)^k (X-1)^{n-k} \\right) \\\\\n&\\qquad \\cdot \\left( \\sum_{\\ell=0}^{n} \\binom{n}{\\ell} \\bar{\\omega}^{\\ell} (X+1)^{\\ell} (X-1)^{n-\\ell} \\right) \\\\\n&= \\frac{1}{n+1} \\sum_{\\omega \\in S} (\\omega(X+1) + X-1)^n (\\bar{\\omega}(X+1) + X-1)^n \\\\\n&= \\frac{1}{n+1} \\sum_{\\omega \\in S} \\left( (2+\\omega+\\bar{\\omega})X^2 + 2 - \\omega - \\bar{\\omega} \\right)^n.\n\\end{align*}\n$$\nFinally, since each $\\omega + \\bar{\\omega}$ is a real number whose absolute value does not exceed $2$ and $\\omega + \\bar{\\omega} < 2$ for at least $n-1$ roots $\\omega \\neq \\pm 1$, each summand above is a polynomial of the desired form. This completes the proof.\n\nClearly, the $a_k$ are all integral. Carrying out calculations, the $a_{2k}$ can explicitly be expressed in terms of $n$ and $k$ alone:\n$$\na_{2k} = \\frac{1}{n+1} \\binom{n}{k} \\sum_{\\omega \\in S} (2 + \\omega + \\bar{\\omega})^k (2 - \\omega - \\bar{\\omega})^{n-k} = \\binom{2k}{k} \\binom{2n-2k}{n-k}.\n$$\nAs a side fact, since the $a_{2k+1}$ all vanish and the $a_{2k}$ are all positive, the corresponding polynomial function is minimized over the real numbers at the origin, where it achieves a minimum of $\\sum_{k=0}^{n} \\binom{n}{k}^2 = \\binom{2n}{n}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16129,
    "subject": "Mathematics (Olympiad)",
    "question": "Sean $a, b, c, d$ números reales tales que\n\n$$\na + b + c + d = 0 \\quad \\text{y} \\quad a^2 + b^2 + c^2 + d^2 = 12.\n$$\n\nHalla el valor mínimo y el valor máximo que puede tomar el producto $abcd$, y determina para qué valores de $a, b, c, d$ se consiguen ese mínimo y ese máximo.",
    "options": [],
    "answer": "See solution",
    "solution": "De las condiciones del enunciado, tenemos que no todos los números tienen el mismo signo. El producto $abcd$ tomará un valor positivo cuando dos números sean positivos y dos negativos, así que buscaremos el máximo suponiendo que $a, b > 0$ y $c, d < 0$. Notemos que\n\n$$\n2(ab + cd) \\leq a^2 + b^2 + c^2 + d^2 = 12, \\quad (1)\n$$\n\ncon lo que $ab + cd \\leq 6$. Utilizando la desigualdad entre las medias aritmética y geométrica,\n\n$$\n(ab) \\cdot (cd) \\leq \\left(\\frac{ab + cd}{2}\\right)^2 \\leq 9. \\quad (2)\n$$\n\nLa igualdad en (2) se alcanza cuando $ab = cd = 3$, con lo que (1) obliga a que sea $(a-b)^2 = (c-d)^2 = 0$; es decir, $a = b$ y $c = d$. Así pues, $a = b = \\sqrt{3}$, $c = d = -\\sqrt{3}$. El valor máximo de la expresión es por tanto $9$.\n\nPara hallar el valor mínimo supondremos que $a, b, c > 0$ y que $d < 0$. (Si tres de los números son negativos y el otro positivo, considerando sus opuestos tenemos que el valor de $abcd$ permanece invariante.) Por tanto, $d = -(a + b + c)$ y\n\n$$\na^2 + b^2 + c^2 + d^2 = 2(a^2 + b^2 + c^2 + ab + bc + ca) = 12.\n$$\n\nEsto quiere decir que\n\n$$\n(a+b+c)^2 = a^2+b^2+c^2+ab+bc+ca+ab+bc+ca \\leq 6+\\frac{a^2+b^2+c^2+ab+bc+ca}{2} \\leq 9,\n$$\n\ndonde se ha usado que, por la desigualdad de Cauchy, $ab+bc+ca \\leq a^2+b^2+c^2$. Por tanto, $a+b+c \\leq 3$. El problema es equivalente a encontrar el máximo valor posible de $abc(a + b + c)$. Usando nuevamente la desigualdad entre las medias aritmética y geométrica,\n\n$$\nabc \\leq \\frac{(a + b + c)^3}{27} \\quad \\text{y por tanto} \\quad abc(a + b + c) \\leq \\frac{(a + b + c)^4}{27} \\leq 3.\n$$\n\nEn consecuencia, el valor mínimo es $-3$ y se alcanza en los casos $(3, -1, -1, -1)$ y $(1, 1, 1, -3)$ (o permutaciones de estos).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16130,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the maximal natural number, with all distinct digits, such that the difference between any two consecutive digits is at least 2.",
    "options": [],
    "answer": "See solution",
    "solution": "It is clear that our number must have 10 digits.\n\nWe start with $9, 7, 5$ (since $8$ cannot be the second or third digit).\n\n$9, 7, 5, 8, 6, 4$ are the first six digits. We can't have $3$ after, so we use $2$, and finally, the answer is: $9758642031$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16131,
    "subject": "Mathematics (Olympiad)",
    "question": "Each point in the plane is labelled with a real number. For each cyclic quadrilateral $ABCD$ in which the line segments $AC$ and $BD$ intersect, the sum of the labels at $A$ and $C$ equals the sum of the labels at $B$ and $D$.\n\nProve that all points in the plane are labelled with the same number.",
    "options": [],
    "answer": "See solution",
    "solution": "For any point $P$ in the plane, let $f(P)$ denote its label. Consider two points $A$ and $B$, and construct any cyclic pentagon $ABPQR$ whose vertices lie in that order.\n\nThen we have $f(A) + f(Q) = f(P) + f(R) = f(B) + f(Q)$.\n\nThis implies that the arbitrarily chosen points $A$ and $B$ satisfy $f(A) = f(B)$.\n\nSo it is necessarily true that all points in the plane are labelled with the same number.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16132,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose set $X = \\{1, 2, \\dots, 20\\}$. $A$ is a subset of $X$. The number of elements of $A$ is at least 2, and all the elements of $A$ can be arranged as consecutive positive integers. Then, how many such sets $A$ are there?",
    "options": [],
    "answer": "See solution",
    "solution": "Each set $A$ can be uniquely determined by its minimum element $a$ and maximum element $b$, where $a, b \\in X$ and $a < b$. For each pair $(a, b)$, $A = \\{a, a+1, \\dots, b\\}$ is a valid set. The number of such pairs is $\\binom{20}{2} = 190$, so there are $190$ such sets $A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16133,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the sequence $\\{x_n\\}$ be defined by the recurrence relation $x_{n+1} = a x_n^m + b$, where $a$, $b$ are real numbers and $m$ is a positive integer. Determine the necessary and sufficient conditions on $a$, $b$, and $m$ for the sequence $\\{x_n\\}$ to be bounded.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the sequence $\\{x_n\\}$ defined by $x_{n+1} = a x_n^m + b$.\n\n**Case 1:** $b > 0$\n\nEach term $x_n > 0$. The sequence is bounded if and only if the equation $a x^m + b = x$ has positive real roots. If not, the function $p(x) = a x^m + b - x$ attains a minimum $t > 0$ on $(0, +\\infty)$, so $x_{n+1} - x_n \\geq t$ and the sequence is unbounded.\n\nIf $a x^m + b = x$ has a positive real root $x_0$, then by induction, $x_n < x_0$ for all $n$, so the sequence is bounded.\n\nThe equation $a x^m + b = x$ has positive roots if and only if the minimum of $a x^{m-1} + \\frac{b}{x}$ on $(0, +\\infty)$ is at most $1$. By the mean inequality:\n\n$$\na x^{m-1} + \\frac{b}{x} \\geq m \\sqrt[m]{\\frac{a b^{m-1}}{(m-1)^{m-1}}}\n$$\n\nThus, the sequence is bounded if and only if\n$$\nm \\sqrt[m]{\\frac{a b^{m-1}}{(m-1)^{m-1}}} \\leq 1, \\quad \\text{that is,}\\quad a b^{m-1} \\leq \\frac{(m-1)^{m-1}}{m^m}.\n$$\n\n**Case 2:** $b < 0$, $m$ odd\n\nLet $y_n = -x_n$. Then $y_{n+1} = a y_n^m + (-b)$, and $y_1 = -b > 0$. The sequence $\\{x_n\\}$ is bounded if and only if $\\{y_n\\}$ is bounded, so the same condition as above applies.\n\n**Case 3:** $b < 0$, $m$ even\n\nThe analysis is similar, but care must be taken with sign changes.\n\n**Summary:**\n\nThe sequence $\\{x_n\\}$ defined by $x_{n+1} = a x_n^m + b$ is bounded if and only if\n$$\na b^{m-1} \\leq \\frac{(m-1)^{m-1}}{m^m}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16134,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the side length of the base and the height of the regular pyramid $PABCD$ be equal. Point $G$ is the centroid of face $\\triangle PBC$. What is the sine of the angle between line $AG$ and the base $ABCD$?",
    "options": [],
    "answer": "See solution",
    "solution": "Take the midpoint $M$ of $BC$, so $G$ lies on $PM$. Project $P$ and $G$ onto the base $ABCD$; their projections are $O$ (the center of the base square) and $H$ (lying on $OM$), respectively.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p47_data_0d9aa26577.png)\n\nWe have $\\frac{GH}{PO} = \\frac{HM}{OM} = \\frac{GM}{PM} = \\frac{1}{3}$.\n\nLet $AB = PO = 6$. Then $GH = \\frac{PO}{3} = 2$, $OH = \\frac{2}{3}OM = 2$. Since $AO = 3\\sqrt{2}$ and $\\angle AOH = 135^\\circ$, we compute:\n\n$$\nAH^2 = AO^2 + OH^2 - 2AO \\cdot OH \\cdot \\cos 135^\\circ = 34,\n$$\n\nso $AG = \\sqrt{AH^2 + GH^2} = \\sqrt{38}$.\n\nThe angle between $AG$ and the base $ABCD$ is $\\angle GAH$, so\n\n$$\n\\sin \\angle GAH = \\frac{GH}{AG} = \\frac{2}{\\sqrt{38}} = \\frac{\\sqrt{38}}{19}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16135,
    "subject": "Mathematics (Olympiad)",
    "question": "A polynomial $P(x)$ with integer coefficients satisfies $P(n^2) = 0$ for some nonzero integer $n$. Show that $P(a^2) \\neq 1$ must hold for any nonzero rational number $a$.",
    "options": [],
    "answer": "See solution",
    "solution": "The assertion is obvious if $P(x)$ is constant, so assume $P(x)$ has degree $m > 0$. Since $P(x)$ has integer coefficients and $P(n^2) = 0$, there exists a polynomial $Q(x)$ with integer coefficients such that\n\n$$\nP(x) = Q(x)(x - n^2)\n$$\n\nSuppose $P(a^2) = 1$ for some nonzero rational $a$. Write $a = \\frac{q}{p}$ with $p > 0$, $q \\neq 0$, and $\\gcd(p, q) = 1$. Substituting $x = a^2$ gives\n\n$$\nP(a^2) = Q(a^2)(a^2 - n^2) = \\frac{1}{p^{2m}} \\left( p^{2(m-1)} Q\\left(\\left(\\frac{q}{p}\\right)^2\\right) \\right) (q^2 - (pn)^2)\n$$\n\nSince $P(a^2) = 1$, we have\n\n$$\n\\left( p^{2(m-1)} Q\\left(\\left(\\frac{q}{p}\\right)^2\\right) \\right) (q^2 - (pn)^2) = p^{2m}\n$$\n\nBoth $p^{2(m-1)} Q\\left(\\left(\\frac{q}{p}\\right)^2\\right)$ and $q^2 - (pn)^2$ are integers, so $q^2 - (pn)^2$ must divide $p^{2m}$. But $q - pn$ and $q + pn$ are coprime to $p$, and since $n, p, q \\neq 0$, $q - pn$ and $q + pn$ cannot both have absolute value $1$. Thus, $q^2 - (pn)^2$ cannot divide $p^{2m}$, a contradiction. Therefore, $P(a^2) \\neq 1$ for any nonzero rational $a$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16136,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer, and $f$ a function defined on the set $D$ of positive divisors of $n$ with values in $\\mathbb{Z}$. Prove that the following two statements are equivalent:\n\n(A) For any positive divisor $m$ of $n$,\n\n$$\nn \\mid \\sum_{d \\mid m} f(d) \\binom{n/d}{m/d}.\n$$\n\n(B) For any positive divisor $k$ of $n$,\n\n$$\nk \\mid \\sum_{d \\mid k} f(d).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For the given mapping $f : D \\to \\mathbb{Z}$, define $g : D \\to \\mathbb{Z}$ by\n\n$$\ng(k) = \\sum_{d \\mid k} f(d), \\quad \\forall k \\in D.\n$$\n\nBy the M\"obius inversion, $f$ is uniquely determined by $g$:\n\n$$\nf(k) = \\sum_{d \\mid k} \\mu\\left(\\frac{k}{d}\\right) g(d), \\quad \\forall k \\in D.\n$$\n\nThis gives\n\n$$\n\\begin{align*}\n\\sum_{d \\mid m} f(d) \\binom{n/d}{m/d} &= \\sum_{d \\mid m} \\sum_{x \\mid d} \\mu\\left(\\frac{d}{x}\\right) g(x) \\binom{n/d}{m/d} \\\\\n&= \\sum_{x \\mid m} g(x) \\left( \\sum_{x \\mid d,\\ d \\mid m} \\mu\\left(\\frac{d}{x}\\right) \\binom{n/d}{m/d} \\right) \\\\\n&= \\sum_{x \\mid m} g(x) \\left( \\sum_{s \\mid \\frac{m}{x}} \\mu(s) \\binom{n/(xs)}{m/(xs)} \\right). \\tag{1}\n\\end{align*}\n$$\n\n**Lemma:** If $b \\mid a$, then\n$$\n\\sum_{k \\mid b} \\mu(k) \\binom{a/k}{b/k} \\equiv 0 \\pmod{a}.\n$$\n\nAssuming the lemma, we show (A) and (B) are equivalent.\n\n**(B) $\\Rightarrow$ (A):** Suppose $x \\mid g(x)$ for every $x \\in D$. By the lemma,\n$$\n\\frac{n}{x} \\mid \\sum_{s \\mid \\frac{m}{x}} \\mu(s) \\binom{n/(xs)}{m/(xs)},\n$$\nso plugging into (1) gives (A).\n\n**(A) $\\Rightarrow$ (B):** Assume (A). Use induction to show $k \\mid g(k)$ for all $k \\in D$. Suppose $k \\mid g(k)$ for all $k < m$, $k \\in D$, and consider $k = m$. By the lemma and induction,\n$$\n\\begin{align*}\n0 &\\equiv \\sum_{d \\mid m} f(d) \\binom{n/d}{m/d} \\\\\n&\\equiv \\sum_{x \\mid m} g(x) \\left( \\sum_{s \\mid \\frac{m}{x}} \\mu(s) \\binom{n/(xs)}{m/(xs)} \\right) \\\\\n&\\equiv g(m) \\cdot \\frac{n}{m} + \\sum_{x \\mid m,\\ x < m} g(x) \\left( \\sum_{s \\mid \\frac{m}{x}} \\mu(s) \\binom{n/(xs)}{m/(xs)} \\right) \\\\\n&\\equiv g(m) \\cdot \\frac{n}{m} \\pmod{n},\n\\end{align*}\n$$\nso $m \\mid g(m)$. This completes the induction.\n\n**Proof of the lemma:**\n\n$$\n\\sum_{k \\mid b} \\mu(k) \\binom{a/k}{b/k} = \\frac{a}{b} \\sum_{k \\mid b} \\mu(k) \\binom{a/k - 1}{b/k - 1}.\n$$\n\nLet $b = \\prod_{i=1}^t p_i^{\\beta_i}$. It suffices to show for each $1 \\leq i \\leq t$,\n$$\n\\sum_{k \\mid b} \\mu(k) \\binom{a/k - 1}{b/k - 1} \\equiv 0 \\pmod{p_i^{\\beta_i}}.\n$$\n\nLet $p_1^{\\beta_1} = p^{\\beta}$. Write $\\binom{u}{v}$ as $C(u, v)$. Then,\n$$\n\\begin{align*}\n\\sum_{k \\mid b} \\mu(k) C\\left(\\frac{a}{k} - 1, \\frac{b}{k} - 1\\right) &= \\sum_{I \\subset \\{1, \\dots, t\\}} (-1)^{|I|} C\\left(\\frac{a}{\\prod_{i \\in I} p_i} - 1, \\frac{b}{\\prod_{i \\in I} p_i} - 1\\right) \\\\\n&= \\sum_{J \\subset \\{2, \\dots, t\\}} (-1)^{|J|} \\left[ C\\left(\\frac{a}{\\prod_{j \\in J} p_j} - 1, \\frac{b}{\\prod_{j \\in J} p_j} - 1\\right) \n- C\\left(\\frac{a}{p \\prod_{j \\in J} p_j} - 1, \\frac{b}{p \\prod_{j \\in J} p_j} - 1\\right) \\right].\n\\end{align*}\n$$\n\nIt suffices to show: if $u$ and $v$ are multiples of $p^\\beta$, then\n$$\nC(u-1, v-1) - C\\left(\\frac{u}{p} - 1, \\frac{v}{p} - 1\\right) \\equiv 0 \\pmod{p^{\\beta}}. \\tag{2}\n$$\n\nLet $w = u - v$. Then\n$$\n\\begin{aligned}\nC(u-1, v-1) &= \\frac{\\prod_{0 \\leq x < \\frac{v}{p},\\ 1 \\leq r < p} (xp + r + w)}{\\prod_{0 \\leq x < \\frac{v}{p},\\ 1 \\leq r < p} (xp + r)} \\cdot C\\left(\\frac{u}{p} - 1, \\frac{v}{p} - 1\\right).\n\\end{aligned}\n$$\n\nAs $w$ is a multiple of $p^{\\beta}$,\n$$\nC(u-1, v-1) \\prod_{0 \\leq x < \\frac{v}{p},\\ 1 \\leq r < p} (xp + r) \\equiv C\\left(\\frac{u}{p} - 1, \\frac{v}{p} - 1\\right) \\prod_{0 \\leq x < \\frac{v}{p},\\ 1 \\leq r < p} (xp + r) \\pmod{p^{\\beta}},\n$$\nso (2) follows. The lemma is proved.\n\n**Alternate proof of the lemma:**\n\nLet $X$ be a finite set, $h: X \\to X$ a bijection with $h^{(a)} = \\mathrm{Id}_X$. Under $h$, $X$ decomposes into orbits of length $l \\mid a$. For each $k \\mid a$, let\n$$\nm(k) = \\#\\{x \\in X : h^{(k)}(x) = x\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16137,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the minimum value of\n\n$$\n\\frac{(x + y + |x - y|)^2}{xy}\n$$\n\nfor positive $x$, $y$?",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $x \\ge y$. Then:\n\n$$\n\\frac{(x + y + |x - y|)^2}{xy} = \\frac{(x + y + x - y)^2}{xy} = \\frac{(2x)^2}{xy} = \\frac{4x^2}{xy} = \\frac{4x}{y} \\ge 4\n$$\n\nEquality holds when $x = y$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16138,
    "subject": "Mathematics (Olympiad)",
    "question": "定義費氏數：$F_0 = 0, F_1 = 1, F_2 = 1$，且對所有正整數 $n$，有 $F_{n+2} = F_{n+1} + F_n$。\n\n證明存在一個正整數 $N$，使得對於任意非常數、長度為 $N$ 的正整數等差數列，其中必有一項，當其寫成若干個費氏數之和時，至少需要 $2024$ 個費氏數（費氏數可重複）。",
    "options": [],
    "answer": "See solution",
    "solution": "以下約定：將正整數寫成費氏數之和時，使用的費氏數要最大。例如：$18$ 雖然可以寫成 $8+5+5$，但我們寫成 $13+5$。\n\n令 $k$ 為正整數，$\\mathcal{F}$ 為費氏數形成的集合。定義集合加法 $A+B := \\{a+b : a \\in A, b \\in B\\}$。令 $N(k)$ 為集合 $\\underbrace{\\mathcal{F} + \\dots + \\mathcal{F}}_{k}$ 中最長的等差數列的長度。\n\n我們將用數學歸納法證明 $N(k)$ 皆存在且有限，則取 $N = N(2023) + 1$ 即為所求。\n\n**Base case:** $k=1$ 時，$N(1) \\leq 4$，因為 $\\mathcal{F}$ 中最長的等差數列為 $0,1,2,3$，共 $4$ 項。\n\n接下來我們證明兩個引理。\n\n**Lemma 1.** 若 $a > F_n$，且 $a$ 可被寫為 $k$ 個 $F_i$ 的和，其中每個 $F_i \\leq F_n$，那麼 $a$ 亦可被寫為 $k$ 個 $F_i$ 的和，滿足 $F_i \\leq F_n$，且其中有 $F_n$。\n\n*Lemma 1 證明.* 設 $a = \\sum_{j=1}^{k} F_{a(j)}$。假設 $a$ 不能被寫成那樣的和。注意到 $2F_t = F_{t+1} + F_{t-2}$。若\n\n- $F_{a(j)}$ 之中有兩個連續的費氏數 $F_i + F_{i+1}$，我們將其改為 $F_{i+2} + F_0$。\n- $F_{a(j)}$ 之中有兩個重複的費氏數 $F_i + F_i$，且這兩項不是 $0$，我們將其改為 $F_{i+1} + F_{i-2}$。\n\n不斷進行這兩項操作。若在這過程中 $F_n$ 從來沒有出現，則我們可以將 $a$ 寫為 $\\sum_{j=1}^{k} F_{b(j)}$，$F_{b(j)} < F_n$，且除了 $0$ 之外，不存在重複或連續的費氏數。這樣的數的最大值為\n\n$$\nF_{n-1} + F_{n-3} + F_{n-5} \\leq F_n < a,\n$$\n\n矛盾。故 $a$ 可寫為 $k$ 個至多為 $F_n$ 的費氏數之和，且其中有 $F_n$。$\\boxed{}$\n\n特別地，若 $F_n < a < F_{n+1}$，且 $a$ 可被寫為 $k$ 個費氏數之和，則我們可要求其中一個為 $F_n$。\n\n**Lemma 2.** 給定正整數 $M$，則對任意的 $4M$ 項正整數等差數列\n\n$$\nA, A+d, \\dots, A+(4M-1)d,\n$$\n\n都存在整數 $\\ell$，使得在區間 $[F_\\ell, F_{\\ell+1}]$ 中該等差數列至少出現 $M$ 項。\n\n*Lemma 2 證明.* $d=0$ 時不必證，故設 $d>0$。若 $A+(3M)d$ 到 $A+(4M-1)d$ 皆落在某段 $[F_\\ell, F_{\\ell+1}]$ 中，結論成立。否則，知在這之中有一個 $F_\\ell$ 將其切開，可設為 $A+md \\leq F_\\ell < A+(m+1)d$，其中 $m \\geq 3M$。\n\n注意到 $A+(3M)d \\geq F_3 = 1$。且當 $\\ell \\geq 3$ 時，我們有 $1.5F_{\\ell-1} = F_{\\ell-1} + \\frac{1}{2}(F_{\\ell-2} + F_{\\ell-3}) \\leq F_\\ell$。於是由 $1.5F_{\\ell-1} \\leq F_\\ell < A+(m+1)d$，我們可以得到\n\n$$\nF_{\\ell-1} < \\frac{A}{1.5} + \\frac{m+1}{1.5}d \\leq A + (m+1-M)d,\n$$\n\n故第 $m+1-M$ 項至第 $m$ 項皆落於區間 $[F_{\\ell-1}, F_\\ell]$ 中。$\\boxed{}$\n\n註：這段的估計不是最緊的，比如將 $1.5$ 換成任意介於 $1$ 和 $\\frac{1+\\sqrt{5}}{2}$ 的 $r$，而 $4$ 換成任意滿足 $\\frac{L-1}{r} > L-2$ 的數字 $L$ 皆可。\n\n最後我們證明 $N(k) < 4N(k-1)+4$。由數學歸納法假設 $N(k-1)$ 是有限的，那麼對於任意長為 $4N(k-1)+4$ 的等差數列中，存在一段長度為 $N(k-1)+1$ 的子數列 $a_1, a_2, \\dots, a_{N(k-1)+1}$ 落在某段 $[F_\\ell, F_{\\ell+1}]$ 中。由 Lemma 1 我們知道，若 $a_i$ 皆可被寫為 $k$ 項費氏數之和，則可要求這 $k$ 項中出現 $F_\\ell$。如此一來，$\\langle a_i - F_\\ell \\rangle$ 就是一個長度為 $N(k-1)+1$，但每項皆可被寫為 $k-1$ 個費氏數之和的等差數列，和 $N(k-1)$ 的定義矛盾。所以 $N(k)$ 至多為 $4N(k-1)+3$。本題證畢。$\\boxed{}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16139,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be positive real numbers such that\n$$\na^2 + ab + ac - bc = 0.\n$$\n\na) Show that if two of the numbers $a$, $b$, and $c$ are equal, then at least one of the numbers $a$, $b$, and $c$ is irrational.\n\nb) Show that there exist infinitely many triples $(m, n, p)$ of positive integers such that\n$$\nm^2 + mn + mp - np = 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "a) The relation is symmetric in $b$ and $c$, so consider two cases:\n\n- If $a = b$, then $a^2 + a^2 + ac - ac = 2a^2 = 0$, which is impossible for positive $a$.\n- If $b = c$, then $a^2 + 2ab - b^2 = 0$, so $(a + b)^2 = 2b^2$, giving $a + b = b\\sqrt{2}$ and $a = b(\\sqrt{2} - 1)$. If $b$ is rational, $a$ is irrational; if $b$ is irrational, the claim holds.\n\nb) Let $m, n, p$ be positive integers. Set $n = mu$, $p = mv$ for integers $u, v$. The equation becomes $1 + u + v - uv = 0$, or $(u - 1)(v - 1) = 2$. For example, $u = 2$, $v = 3$ gives the triples $(m, 2m, 3m)$ for any $m \\in \\mathbb{N}^*$, and there are infinitely many such $m$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16140,
    "subject": "Mathematics (Olympiad)",
    "question": "Assume there exist positive integers $m$ and $n$ such that\n$$\nk + 4 = m^3 \\quad \\text{and} \\quad k^2 + 5k + 2 = n^3.\n$$\nShow that such $k$ does not exist.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there exist positive integers $m$ and $n$ such that\n$$\nk + 4 = m^3 \\quad \\text{and} \\quad k^2 + 5k + 2 = n^3.\n$$\nThen the product\n$$\n(mn)^3 = (k+4)(k^2+5k+2) = k^3 + 9k^2 + 22k + 8\n$$\nis also a cube of a positive integer.\nSince\n$$\n(k+2)^3 = k^3 + 6k^2 + 12k + 8 < k^3 + 9k^2 + 22k + 8 < k^3 + 9k^2 + 27k + 27 = (k+3)^3,\n$$\nwe have that $(mn)^3$ is between two consecutive cubes, which is impossible. Therefore, such $k$ does not exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16141,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that\n\n$$\nf(x+n) = \\frac{f(x)}{(n^2 + 2nx)f(x) + 1}, \\quad n = 1, 2, \\dots\n$$\n\nFrom this, show that\n\n$$\nf\\left(\\frac{1}{n}\\right) = n^2 = \\frac{1}{\\left(\\frac{1}{n}\\right)^2}, \\quad n = 1, 2, \\dots\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By letting $x = \\frac{1}{n}$ in the previous result, we have\n\n$$\nf\\left(n + \\frac{1}{n}\\right) = \\frac{f\\left(\\frac{1}{n}\\right)}{\\left(n^2 + 2\\right)f\\left(\\frac{1}{n}\\right) + 1}.\n$$\n\nAlso, setting $y = \\frac{1}{x}$ in another relation gives\n\n$$\nf(x) + f\\left(\\frac{1}{x}\\right) + 2 = \\frac{1}{f\\left(x + \\frac{1}{x}\\right)}.\n$$\n\nTherefore,\n\n$$\nf(n) + f\\left(\\frac{1}{n}\\right) + 2 = \\frac{1}{f\\left(n + \\frac{1}{n}\\right)} = n^2 + 2 + \\frac{1}{f\\left(\\frac{1}{n}\\right)}.\n$$\n\nSince $f(n) = \\frac{1}{n^2}$, it follows that $f\\left(\\frac{1}{n}\\right) = n^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16142,
    "subject": "Mathematics (Olympiad)",
    "question": "There is a central train station at point $O$, which is connected to other train stations $A_1, A_2, \\dots, A_8$ with tracks. There is also a track between stations $A_i$ and $A_{i+1}$ for each $1 \\le i \\le 8$ (here $A_1 = A_9$). The length of each track $A_iA_{i+1}$ is equal to $1$, and the length of each track $OA_i$ is equal to $2$, for each $1 \\le i \\le 8$.\n\nThere are also $8$ trains $B_1, B_2, \\dots, B_8$, the speed of the train $B_j$ is $j$. Trains can move only by the tracks above, in both directions. No time is wasted on changing directions. If two or more trains meet at some point, they will move together from now on, with the speed equal to that of the fastest of them.\n\nIs it possible to arrange trains into stations $A_1, A_2, \\dots, A_8$ (each station has to contain one train initially), and to organize their movement in such a way, that all trains arrive at $O$ in time $t < \\frac{1}{2}$?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** yes.\n\nArrange the trains clockwise as follows: $B_8, B_1, B_7, B_2, B_6, B_3, B_5$, and $B_4$. Now let's show how they should move to meet in time.\n\nLet $B_8$ move through the station of $B_1$ (at this time $B_1$ stays at place and is waiting for $B_8$), and they arrive to $O$ in time $t_8 = \\frac{3}{8}$.\n\nLet train $B_7$ move through the station of $B_2$ (at this time $B_2$ stays at place and is waiting for $B_7$), and they arrive to $O$ in time $t_7 = \\frac{3}{7}$.\n\nLet train $B_6$ move through the station of $B_3$ (at this time $B_3$ stays at place and is waiting for $B_6$), and they arrive to $O$ in time $t_6 = \\frac{1}{2}$.\n\n$B_5$ goes to $O$ right away in time $t_5 = \\frac{2}{5}$.\n\n$B_4$ goes to $O$ right away in time $t_4 = \\frac{1}{2}$.\n\nNow we see that two groups — $B_6 + B_3$ and $B_4$ — arrive at the time $\\frac{1}{2}$, but the groups $B_8 + B_1$ and $B_7 + B_2$ arrive faster than in $\\frac{1}{2}$, and their speeds are larger than those of the first groups. So, the group $B_8 + B_1$ upon its arrival to $O$ goes to the group $B_6 + B_3$, unites with it, and goes to the point $O$ with its speed. Similarly, group $B_7 + B_2$ will get $B_4$ to $O$ faster.\n\nIt's possible to find the time of arrival of each group to $O$. For example, for the last group this time is $\\frac{37}{77} < \\frac{1}{2}$. Indeed, group $B_7 + B_2$ arrives at $O$ at $\\frac{3}{7}$, at this time $B_4$ has travelled $\\frac{12}{7}$ in the direction of point $O$. The time before their meet is $\\frac{2}{4+7} = \\frac{2}{77}$, and it's equal to the time spent to arrive back to $O$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16143,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples of positive integers $a, b, c$ such that\n$$\na + [a, b] = b + [b, c] = c + [c, a],\n$$\nwhere $[x, y]$ denotes the least common multiple of $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $a = b = c$.\n\n**Solution:** Note that $b$, $[a, b]$, and $[b, c]$ are divisible by $b$, so $a$ is divisible by $b$. Similarly, $c$ is divisible by $a$, and $b$ is divisible by $c$, so $a = b = c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16144,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $VABCD$ be a regular pyramid with square base $ABCD$. Suppose that on the line $AC$ there is a point $M$ such that $VM = MB$ and the plane $(VMB)$ is perpendicular to the plane $(VAB)$. Prove that $4AM = 3AC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $VM = MB$ and $M$ lies on $AC$, and $(VMB) \\perp (VAB)$, $M$ is determined by these conditions. Let $P$ be the midpoint of $VB$. The angle between the planes $(VAB)$ and $(VBM)$ is $\\angle APM$, so $\\angle APM = 90^\\circ$.\n\nTriangles $MPA$ and $POA$ are similar, so $\\frac{MA}{PA} = \\frac{PA}{OA}$. Therefore, $AM = \\frac{PA^2}{OA} = \\frac{3}{4}AC$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16145,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all integers $n \\geq 2$, having at least four positive divisors, with the property that for any two distinct divisors $d_1$ and $d_2$ of $n$, such that $1 < d_1 < d_2 < n$, the number $d_2 - d_1$ is also a divisor of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "There are three solutions: $n = 6$, $n = 8$, and $n = 12$.\n\nFirst, notice that if $n$ is odd, all its divisors are also odd, so if $d_1$ and $d_2$ are two divisors such that $1 < d_1 < d_2 < n$, then $d_2 - d_1$ is even, so $d_2 - d_1 \\nmid n$. Therefore, $n$ is even.\n\nConsider $n = 2^a \\cdot b$, where $a, b \\in \\mathbb{N}_{\\geq 1}$ and $b$ is odd.\n\nIf $b = 1$, then $n = 2^a$, with $a \\in \\mathbb{N}^*$. For $a \\in \\{1, 2\\}$ there are no solutions, and for $a \\geq 4$, choosing $d_1 = 2^{a-3}$ and $d_2 = 2^{a-1}$, it follows that $d_2 - d_1 = 3 \\cdot 2^{a-3} \\nmid n$, a contradiction. Consequently, $a = 3$, so $n = 8$. It's trivial to see that $n = 8$ is a solution of the problem.\n\nFor $b \\geq 3$ we have two cases.\n\nIf $a = 1$, then $n = 2b$. Choose $d_1 = 2$ and $d_2 = b$; we obtain $b - 2 \\mid 2b$, so $b - 2 \\mid 2b - 2(b - 2) = 4$. Since $b$ is odd, we infer that $b = 3$, so $n = 6$, which is a solution.\n\nIf $a \\geq 2$, choose $d_1 = 2^a$ and $d_2 = 2^{a-1} b$; we obtain $2^{a-1}(b-2) \\mid 2^a b$, so $b - 2 \\mid 2b$. Since $b$ is odd, similarly it follows that $b = 3$. Therefore, $n = 2^a \\cdot 3$.\n\nChoose now $d_1 = 3$ and $d_2 = 2^a$; it results that $2^a - 3 \\mid 2^a \\cdot 3$, so $2^a - 3 \\mid 2^a \\cdot 3 - 3(2^a - 3) = 9$, from which we obtain $a = 2$. Hence, $n = 12$ and we easily verify that it is a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16146,
    "subject": "Mathematics (Olympiad)",
    "question": "Several natural numbers are given on a line. We perform a transformation as follows: for every pair of consecutive integers on the line, write the sum of those two numbers in the middle of them. After 2013 such steps, how many times does the number 2013 appear on the line if:\n\na) The given numbers are $1$ and $1000$?\n\nb) The given numbers are $1, 2, 3, \\ldots, 1000$ in increasing order from left to right?",
    "options": [],
    "answer": "See solution",
    "solution": "a) We first observe that one cannot write the number $2013$ between a pair $(a, b)$ with $a + b > 2013$. Using this observation, it is easy to check that the number $2013$ is written only twice after $2013$ steps: in the $8$th and the $1013$th steps.\n\nb) We add an extra number $1$ after $1000$ on the line. We perform $2013$ such steps on this new line and count how many times the number $2013$ is written. We construct a sequence $A$ of pairs inductively as follows. Initially,\n\n$$\nA = \\{(1, 2), (2, 3), \\dots, (999, 1000), (1000, 1)\\}.\n$$\n\nIn each step of the transformation, whenever we write a number $k = a + b$ between the pair $(a, b)$ (from left to right), we add two pairs $(a, k)$ and $(k, b)$ into the sequence $A$.\n\nWe will prove by induction on $a + b$ that an ordered pair $(a, b)$ appears in the sequence $A$ only if $\\gcd(a, b) = 1$, and if $\\gcd(a, b) = 1$ then the pair $(a, b)$ appears exactly once in the sequence $A$.\n\nSuppose that this statement holds for any $a + b < k$, we show that it also holds when $a + b = k$. Without loss of generality, assume $a < b$ (the case $a > b$ is similar). The pair $(a, b)$ is added to the sequence $A$ whenever we write the number $b$ between the pair $(a, b - a)$. Since $a + (b - a) = b < k$, the pair $(a, b - a)$ appears in the sequence $A$ only if $\\gcd(a, b - a) = 1$, and if $\\gcd(a, b - a) = 1$ then the pair $(a, b - a)$ appears exactly once in the sequence $A$. Since $\\gcd(a, b) = \\gcd(a, b - a)$, the statement holds for $a + b = k$. By induction, the statement holds for any $(a, b)$. This implies that the number of times that $2013$ is written is the number of pairs $(a, 2013 - a)$ with $\\gcd(a, 2013 - a) = 1$. Therefore, the number $2013$ is written $\\varphi(2013) = 1200$ times on the new line.\n\nBy part a), the number $2013$ is written twice between $1000$ and $1$. Hence, the number $2013$ is written $1200 - 2 = 1198$ times on the given line.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16147,
    "subject": "Mathematics (Olympiad)",
    "question": "Three players A, B, and C play the following game. At the beginning, each player has a sheet of paper with their own name written on it. Player A chooses one of the other players and replaces the name on that player's sheet with the name on his own sheet. Then player B makes a similar move, then player C, and after that the turn goes to player A again. The game ends when all the sheets have the same name written on them, and the winner is the player whose name it is. Does any player have a winning strategy (a strategy that allows a player to win no matter what the opponents play)?",
    "options": [],
    "answer": "See solution",
    "solution": "No.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16148,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number and let $m, n$ be integers greater than $1$ such that $n \\mid m^{p(n-1)} - 1$. Prove that $\\gcd(m^{n-1} - 1, n) > 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Set $\\alpha = v_p(n-1)$ and write $n-1 = r \\cdot p^{\\alpha}$. Let $q$ be an arbitrary prime divisor of $n$, and set $d = \\operatorname{ord}_q(m)$. Since\n\n$$\nm^{p(n-1)} = m^{r \\cdot p^{\\alpha+1}} \\equiv 1 \\pmod{q},\n$$\n\nit follows that $d \\mid r \\cdot p^{\\alpha+1}$. If $v_p(d) \\neq \\alpha + 1$, then clearly $d \\mid n-1$, and therefore $q$ is a common divisor of $n$ and $m^{n-1} - 1$. Otherwise, suppose that $v_p(d) = \\alpha + 1$ for all prime divisors of $n$. Together with $m^{q-1} \\equiv 1 \\pmod{q}$ by Fermat's Little Theorem, we have\n\n$$\nd \\mid q-1 \\implies v_p(q-1) \\geq v_p(d) = \\alpha + 1.\n$$\n\nThus, any prime divisor $q$ of $n$ satisfies $q \\equiv 1 \\pmod{p^{\\alpha+1}}$. Because $n$ is a product of these prime divisors, we deduce that $n \\equiv 1 \\pmod{p^{\\alpha+1}}$. However, this contradicts $v_p(n-1) = \\alpha$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16149,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\dots, x_{2008}$ be real numbers such that $x_1^2 = 999^2$ and for $2 \\leq n \\leq 2008$, $x_n^2 = |x_{n-1} + 1|^2 = (x_{n-1} + 1)^2$. Find the minimum value of $S = x_1 + x_2 + \\dots + x_{2008}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S = x_1 + x_2 + \\dots + x_{2008}$. Since $x_1^2 = 999^2$, and $x_n^2 = (x_{n-1} + 1)^2$ for $2 \\leq n \\leq 2008$, we have:\n\n$$\n\\begin{aligned}\nx_1^2 + x_2^2 + \\dots + x_{2008}^2 &= 999^2 + (x_1 + 1)^2 + \\dots + (x_{2007} + 1)^2 \\\\\n&= (x_1^2 + \\dots + x_{2007}^2) + 2(x_1 + \\dots + x_{2007}) + 2007 + 999^2 \\\\\n&= (x_1^2 + \\dots + x_{2007}^2) + 2(S - x_{2008}) + 1000008\n\\end{aligned}\n$$\n\nFrom this, we obtain:\n\n$$\n2S = x_{2008}^2 + 2x_{2008} - 1000008 = (x_{2008} + 1)^2 - 1000009.\n$$\n\nBecause $|x_n| = |x_{n-1} + 1|$, the parity of $x_n$ alternates as $n$ increases. Since $x_1$ is odd, $x_{2008}$ is even, so $(x_{2008} + 1)^2 \\geq 1$. Thus,\n\n$$\nS \\geq \\frac{1 - 1000009}{2} = -500004.\n$$\n\nOn the other hand, if we set\n\n$$\nx_n = n - 1000 \\quad (1 \\leq n \\leq 1000); \\quad x_n = -1 \\quad (1001 \\leq n \\leq 2008,\\ n \\text{ odd}); \\quad x_n = 0 \\quad (1001 \\leq n \\leq 2008,\\ n \\text{ even})\n$$\n\nthen this choice satisfies the conditions, and since $x_{2008} = 0$, we get $S = -500004$. Therefore, the minimum value that $S$ can take is $-500004$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16150,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the function\n\n$$\nf(x) = -x + \\sqrt{(a+x)(b+x)}\n$$\n\nwhere $a$ and $b$ are two given distinct positive real numbers. Prove that for every real number $s$ belonging to the interval $(0,1)$, there exists a unique real number $\\alpha$ such that\n\n$$\nf(\\alpha) = \\left(\\frac{a^s + b^s}{2}\\right)^{\\frac{1}{s}}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "It is easily seen that $f(x)$ is a continuous function on $[0, \\infty)$. We shall prove the following assertions:\n\n1. $f(x)$ is strictly increasing on $[0, \\infty)$.\n2. $f(0) = \\sqrt{ab}$, $\\lim_{x \\to \\infty} f(x) = \\frac{a+b}{2}$.\n3. For every $s$ with $0 < s < 1$, we have:\n\n$$\n\\sqrt{ab} < \\left(\\frac{a^s + b^s}{2}\\right)^{\\frac{1}{s}} \\le \\frac{a+b}{2}.\n$$\n\nWith these assertions, the intermediate value theorem of continuous functions proves the existence of a unique $\\alpha$ such that $f(\\alpha) = \\left(\\frac{a^s + b^s}{2}\\right)^{\\frac{1}{s}}$.\n\n**Proof of 1:**\n\n$$\nf'(x) = -1 + \\frac{2x+a+b}{2\\sqrt{(a+x)(b+x)}} = \\frac{(\\sqrt{a+x}-\\sqrt{b+x})^2}{2\\sqrt{(a+x)(b+x)}} > 0.\n$$\n\n**Proof of 2:**\n\nIt is evident that $f(0) = \\sqrt{ab}$ and\n\n$$\n\\lim_{x \\to \\infty} f(x) = \\lim_{x \\to \\infty} \\frac{-x^2 + (a+x)(b+x)}{x + \\sqrt{(a+x)(b+x)}} = \\lim_{x \\to \\infty} \\frac{a+b+\\frac{ab}{x}}{1+\\sqrt{\\left(\\frac{a}{x}+1\\right)\\left(\\frac{b}{x}+1\\right)}} = \\frac{a+b}{2}.\n$$\n\n**Proof of 3:**\n\nThe first inequality is deduced from the arithmetic-geometric mean inequality.\n\nPut $m = \\left(\\frac{a^s + b^s}{2}\\right)^{\\frac{1}{s}}$, $x = \\frac{a}{m}$, $y = \\frac{b}{m}$. Clearly, $x^s + y^s = 2$ and Bernoulli's inequality implies that\n\n$$\nx = (1 + x^s - 1)^{\\frac{1}{s}} \\ge 1 + \\frac{x^s - 1}{s}.\n$$\n\n![alt](path \"title\")\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16151,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be positive real numbers satisfying $abc = 1$. Prove that\n\n$$\n\\frac{a^2}{b+c} + \\frac{b^2}{c+a} + \\frac{c^2}{a+b} \\geq \\frac{9}{3 + a^2 + b^2 + c^2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that\n\n$$\n\\begin{aligned}\n\\sqrt[3]{(a+b)(b+c)(c+a)} &\\leq \\frac{(a+b)+(b+c)+(c+a)}{3} \\\\\n&= \\frac{2a + 2b + 2c}{3} \\\\\n&\\leq \\frac{(a^2+1) + (b^2+1) + (c^2+1)}{3} \\\\\n&= \\frac{3 + a^2 + b^2 + c^2}{3}.\n\\end{aligned}\n$$\n\nHence, we have\n\n$$\n\\frac{a^2}{b+c} + \\frac{b^2}{c+a} + \\frac{c^2}{a+b} \\geq 3 \\cdot \\sqrt[3]{\\frac{a^2 b^2 c^2}{(a+b)(b+c)(c+a)}} \\geq \\frac{9}{3 + a^2 + b^2 + c^2}.\n$$\n\nIt is easy to check that equality holds if and only if $a = b = c = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16152,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $ (A, +, \\cdot) $ is a ring in which $ x^2 = 0 $ only for $ x = 0 $. Let $ B = \\{ a \\in A \\mid a^2 = 1 \\} $. Show that:\n\n(a) $ ab - ba = bab - a $, for all $ a \\in A $ and $ b \\in B $.\n\n(b) $ (B, \\cdot) $ is a group.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Let $ a \\in A $ and $ b \\in B $. Since $ ((b-1)a(b+1))^2 = (b-1)a(b+1)(b-1)a(b+1) = (b-1)a(b^2-1)a(b+1) = 0 $, we get $ (b-1)a(b+1) = 0 $, hence $ ab - ba = bab - a $.\n\n(b) Let $ a $ and $ b $ be two elements of $ B $. Write\n\n$$\n\\begin{align*}\n(ab - ba)^2 &= abab - ab^2a - ba^2b + baba \\\\\n&= a(bab) + (bab)a - 2 \\\\\n&= a(ab - ba + a) + (ab - ba + a)a - 2 \\quad \\text{(by part (a))} \\\\\n&= a^2b - aba + a^2 + aba - ba^2 + a^2 - 2 = 0,\n\\end{align*}\n$$\nto get $ (ab - ba)^2 = 0 $, hence $ ab = ba $. Consequently, $ (ab)^2 = a^2b^2 = 1 $, i.e., $ ab \\in B $. Moreover, $ a^{-1} = a $. It follows that $ B $ is a subgroup of $ U(A) $.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16153,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integer pairs $(m, n)$ such that $1 \\leq n \\leq 20$, $\\frac{1}{4} < \\frac{n-m}{n} < \\frac{2}{7}$, and $n-m$ and $n$ are relatively prime.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p = n - m$ and $q = n$. Finding all positive integer pairs $(m, n)$ is equivalent to finding all integer pairs $(p, q)$ such that:\n\n- $1 \\leq q \\leq 20$\n- $\\frac{1}{4} < \\frac{p}{q} < \\frac{2}{7}$\n- $\\gcd(p, q) = 1$\n\nThe inequality $\\frac{1}{4} < \\frac{p}{q} < \\frac{2}{7}$ is equivalent to $\\frac{7}{2}p < q < 4p$.\n\nConsider possible values for $p$:\n\n- If $p \\leq 0$, $q$ doesn't exist because $\\frac{p}{q} \\leq 0$.\n- If $p = 1$, then $\\frac{7}{2} < q < 4$, so $q$ doesn't exist.\n- If $p = 2$, then $7 < q < 8$, so $q$ doesn't exist.\n- If $p = 3$, then $\\frac{21}{2} < q < 12$, so $q = 11$.\n- If $p = 4$, then $14 < q < 16$, so $q = 15$.\n- If $p = 5$, then $\\frac{35}{2} < q < 20$, so $q = 18, 19$.\n- If $p \\geq 6$, $q > \\frac{7}{2}p \\geq 21$, which is outside the allowed range for $q$.\n\nThus, the integer pairs $(p, q)$ are $(3, 11)$, $(4, 15)$, $(5, 18)$, $(5, 19)$. Therefore, the positive integer pairs $(m, n)$ are $(8, 11)$, $(11, 15)$, $(13, 18)$, $(14, 19)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16154,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $a \\in [-2, \\infty)$, $r \\in [0, \\infty)$, and let $n$ be a positive integer.\n\nShow that\n\n$$\nr^{2n} + a r^n + 1 \\geq (1 - r)^{2n}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $r = 0$, the relation is obvious.\n\nOtherwise, dividing by $r^{2n}$, one gets the same inequality with $r$ replaced by $\\frac{1}{r}$, so one can assume $r \\in (0, 1]$.\n\nSince $r^{2n} + a r^n + 1 \\geq r^{2n} - 2 r^n + 1 = (1 - r^n)^2$ and $1 - r^n \\geq 0$, it is enough to prove that $1 - r^n \\geq (1 - r)^n$.\n\nThis follows from $r^n + (1 - r)^n \\leq r + (1 - r) = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16155,
    "subject": "Mathematics (Olympiad)",
    "question": "Calculate the value of the sum:\n\n$$\nS = \\cos\\left(\\frac{2\\pi}{2010}\\right) + 2\\cos\\left(\\frac{2\\pi \\cdot 2}{2010}\\right) + 3\\cos\\left(\\frac{2\\pi \\cdot 3}{2010}\\right) + 2\\cos\\left(\\frac{2\\pi \\cdot 4}{2010}\\right) + 5\\cos\\left(\\frac{2\\pi \\cdot 5}{2010}\\right) + \\dots\n$$\nwhere the coefficients correspond to the values of Euler's totient function $\\varphi(n)$ for divisors $n$ of 2010.",
    "options": [],
    "answer": "See solution",
    "solution": "It is well known that\n\n$$\n\\sum_{k=1}^{n} \\cos\\left(\\frac{2\\pi k}{n}\\right) = 0 \\quad \\text{if } n \\ge 2.\n$$\n\nWe can write $S$ as a linear combination of such sums:\n\n$$\nS = \\sum_{d|2010} \\varphi(d) \\sum_{\\ell=1}^{2010/d} \\cos\\left(\\frac{2\\pi d\\ell}{2010}\\right).\n$$\n\nEach term $\\cos\\left(\\frac{2\\pi k}{2010}\\right)$ appears with coefficient $\\sum_{d|\\gcd(k, 2010)} \\varphi(d)$, which equals $\\gcd(k, 2010)$ by Gauss's formula $\\sum_{d|n} \\varphi(d) = n$.\n\nAll inner sums vanish except for $d = 2010$, where $\\sum_{\\ell=1}^1 \\cos\\left(\\frac{2\\pi \\cdot 2010 \\cdot \\ell}{2010}\\right) = \\cos(2\\pi) = 1$. Thus,\n\n$$\nS = \\varphi(2010) = 528.\n$$",
    "topic": "Number Theory",
    "subtopic": "Number-Theoretic Functions"
  },
  {
    "id": 16156,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for all positive integers $n$, the equation $a^n + 2010b^n = c^{n+1}$ has infinitely many natural solutions $a$, $b$, $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider $a = b = 2011k^{n+1}$, where $k$ is a positive integer. Then:\n\n$$\n\\begin{align*}\na^n + 2010b^n &= 2011^n k^{n(n+1)} + 2010 \\cdot 2011^n k^{n(n+1)} \\\\\n&= 2011^n k^{n(n+1)} (1 + 2010) \\\\\n&= 2011^{n+1} k^{n(n+1)}\n\\end{align*}\n$$\n\nThis can be written as $(2011k^n)^{n+1} = c^{n+1}$, so $c = 2011k^n$. Since $k$ can be any positive integer, there are infinitely many natural solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16157,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a scalene triangle, and let $I$ be its incentre. The lines $AI$, $BI$, $CI$ meet the circle $ABC$ again at $D$, $E$, $F$, respectively. The line through $I$ parallel to the line $BC$ meets the line $EF$ at $K$; the points $L$ and $M$ are defined similarly. Prove that the points $K$, $L$, $M$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "The idea is to show the triangles $KFA$ and $KAE$ are similar, whereupon $\\dfrac{KF}{KE} = \\left(\\dfrac{AF}{AE}\\right)^2$. Similarly, $\\dfrac{LD}{LF} = \\left(\\dfrac{BD}{BF}\\right)^2$, $\\dfrac{ME}{MD} = \\left(\\dfrac{CE}{CD}\\right)^2$, and the conclusion follows by the converse of the Menelaus theorem applied to the triangle $DEF$.\n\nTo show the triangles $KFA$ and $KAE$ are similar, it is sufficient to prove that the angles $KAF$ and $KEA$ are congruent.\n\n![](images/RMC_2015_BT_p92_data_0492569dfa.png)\n\nIt is a well-known fact that $EF$ is the perpendicular bisector of the segment $AI$—standard angle chase shows that $(EAI, EIA)$ and $(FAI, FIA)$ are pairs of congruent angles—so the angles $KAI$ and $KIA$ are congruent. Consequently, so are the angles $KAF$ and $KIF$. The latter is in turn congruent to the angle $AEF$, since they are both congruent to half the angle $ACB$. This ends the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16158,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\alpha$, $\\beta$, and $\\gamma$ be the angles of a triangle. Prove that\n\n$$\n\\cos(\\alpha) \\cos(3\\alpha) + \\cos(\\beta) \\cos(3\\beta) + \\cos(\\gamma) \\cos(3\\gamma) + \\frac{7}{4} \\geq 2 \\cos(\\alpha) \\cos(\\beta) \\cos(\\gamma).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, recall that $2 \\cos(\\alpha) \\cos(\\beta) \\cos(\\gamma) = 1 - \\cos^2(\\alpha) - \\cos^2(\\beta) - \\cos^2(\\gamma)$. Now, using the identity $\\cos(3x) = 4 \\cos^3(x) - 3 \\cos(x)$, we see that\n\n$$\n\\begin{aligned}\n4 \\left[ \\cos(\\alpha) \\cos(3\\alpha) + \\cos(\\beta) \\cos(3\\beta) + \\cos(\\gamma) \\cos(3\\gamma) \\right] - 8 \\cos(\\alpha) \\cos(\\beta) \\cos(\\gamma) + 7 &= 16 \\left[ \\cos^4(\\alpha) + \\cos^4(\\beta) + \\cos^4(\\gamma) \\right] \\\\ &\\quad - 8 \\left[ \\cos^2(\\alpha) + \\cos^2(\\beta) + \\cos^2(\\gamma) \\right] + 3 \\\\ &= \\sum \\left(4 \\cos^2(\\theta) - 1\\right)^2 \\geq 0.\n\\end{aligned}\n$$\n\nThe condition for equality is $\\alpha = \\beta = \\gamma = \\pi/3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16159,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of distinct prime numbers $(p, q)$ such that $p^p + q^q + 1$ is divisible by both $p$ and $q$.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, $p \\ne q$. So we may assume that $p < q$ without loss of generality.\n\nAssume that $p = 2$. Then since\n$$\nq^q + 5 \\equiv 5 \\equiv 0 \\pmod{q},\n$$\nthe only possible prime for $q$ is $5$. Furthermore, $(p, q) = (2, 5)$ satisfies the above condition.\n\nNow we assume that both $p$ and $q$ are odd primes. Since $p^p + 1 \\equiv 0 \\pmod{q}$, $p^{p-1} - p^{p-2} + \\dots - p + 1$ is divisible by $q$ and $p^{2p} \\equiv 1 \\pmod{q}$. On the other hand, Fermat's Little Theorem says\n$$\np^{q-1} \\equiv 1 \\pmod{q}.\n$$\nIf $\\gcd(2p, q-1) = 2$, then $p^2 \\equiv 1 \\pmod{q}$. Hence $p \\equiv 1 \\pmod{q}$ or $p \\equiv -1 \\pmod{q}$. Therefore\n$$\n0 \\equiv p^{p-1} - p^{p-2} + \\dots - p + 1 \\equiv 1 \\pmod{q},\n$$\nwhich is a contradiction.\n\nAssume that $\\gcd(2p, q - 1) = 2p$, that is, $q \\equiv 1 \\pmod p$. In this case we have\n$$\n0 \\equiv p^p + q^q + 1 \\equiv p^p + 1 + 1 \\equiv 2 \\pmod{p},\n$$\nwhich is again a contradiction.\n\nTherefore $(p, q) = (2, 5)$ or $(5, 2)$ is the only pair of primes satisfying the above condition.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16160,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer and $1 \\le r \\le n$. Define\n\n$$\nS_r^n := \\sum_{k=1,\\; k \\neq r}^{n} \\cot \\left( \\frac{(k-r)\\pi}{n+1} \\right).\n$$\n\nFind a closed form for $S_r^n$ in terms of $r$ and $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use the identities $\\cot(x + \\pi) = \\cot(x)$ and $\\cot(-x) = -\\cot(x)$ for all $x \\notin \\mathbb{Z}\\pi$.\n\nFirst, note that\n$$\n\\cot\\left(\\frac{(k-r)\\pi}{n+1}\\right) = \\cot\\left(\\frac{(n+1+k-r)\\pi}{n+1}\\right).\n$$\n\nWe rewrite the sum:\n$$\n\\begin{align*}\nS_r^n &= \\sum_{k=1}^{r-1} \\cot \\left( \\frac{(n+1+k-r)\\pi}{n+1} \\right) + \\sum_{k=r+1}^{n} \\cot \\left( \\frac{(k-r)\\pi}{n+1} \\right) \\\\\n&= \\sum_{k=n+2}^{n+r} \\cot \\left( \\frac{(k-r)\\pi}{n+1} \\right) + \\sum_{k=r+1}^{n} \\cot \\left( \\frac{(k-r)\\pi}{n+1} \\right) \\\\\n&= \\sum_{k=r+1}^{n+r} \\cot \\left( \\frac{(k-r)\\pi}{n+1} \\right) - \\cot \\left( \\frac{(n+1-r)\\pi}{n+1} \\right).\n\\end{align*}\n$$\n\nNow, using Gauss's trick of reversing the sum:\n$$\n\\begin{align*}\n2 \\sum_{k=r+1}^{n+r} \\cot \\left( \\frac{(k-r)\\pi}{n+1} \\right) &= 2 \\sum_{j=1}^{n} \\cot \\left( \\frac{j\\pi}{n+1} \\right) \\\\\n&= \\sum_{j=1}^{n} \\left( \\cot \\left( \\frac{j\\pi}{n+1} \\right) + \\cot \\left( \\frac{(n+1-j)\\pi}{n+1} \\right) \\right)\n\\end{align*}\n$$\n\nBut\n$$\n\\cot\\left(\\frac{(n+1-j)\\pi}{n+1}\\right) = \\cot\\left(\\pi - \\frac{j\\pi}{n+1}\\right) = -\\cot\\left(\\frac{j\\pi}{n+1}\\right),\n$$\nso the sum is zero.\n\nTherefore,\n$$\nS_r^n = -\\cot\\left(\\frac{(n+1-r)\\pi}{n+1}\\right) = \\cot\\left(\\frac{r\\pi}{n+1}\\right).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16161,
    "subject": "Mathematics (Olympiad)",
    "question": "$a, b, c > 0$ бол\n\n$$\n\\frac{a+b+3c}{3a+3b+2c} + \\frac{a+3b+c}{3a+2b+3c} + \\frac{3a+b+c}{2a+3b+3c} \\geq \\frac{15}{8}\n$$\n\nтэгшитгэл биш биелэхийг батал.",
    "options": [],
    "answer": "See solution",
    "solution": "Орлуулга хийе: $x = 2a + 3b + 3c$, $y = 3a + 2b + 3c$, $z = 3a + 3b + 2c$. Энд $a, b, c > 0$ тул $x, y, z > 0$.\n\n$a = \\frac{-5x + 3y + 3z}{8}$, $b = \\frac{3x - 5y + 3z}{8}$, $c = \\frac{3x + 3y - 5z}{8}$.\n\nДараах илэрхийлэлүүд гарна:\n\n$a + b + 3c = \\frac{7x + 7y - 9z}{8}$\n\na + 3b + c = \\frac{7x - 9y + 7z}{8}\n\n3a + b + c = \\frac{-9x + 7y + 7z}{8}\n\nТэгшитгэл бишийг $x, y, z > 0$-ээр илэрхийлбэл:\n\n$$\n\\frac{27}{8} + \\frac{7}{8}\\left(\\left(\\frac{x}{z} + \\frac{z}{x}\\right) + \\left(\\frac{y}{z} + \\frac{z}{y}\\right) + \\left(\\frac{x}{y} + \\frac{y}{x}\\right)\\right) \\geq -\\frac{27}{8} + \\frac{42}{8} = \\frac{15}{8}\n$$\n\nЭнд $\\frac{x}{z} + \\frac{z}{x}$, $\\frac{y}{z} + \\frac{z}{y}$, $\\frac{x}{y} + \\frac{y}{x} \\geq 0$ гэсэн илэрхий тэнцэтгэл бишийг ашиглав. $x = y = z \\Leftrightarrow a = b = c$ үед тэнцэл болно.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16162,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all sequences $a_1, a_2, a_3, \\dots$ of nonnegative integers such that $a_1 < a_2 < a_3 < \\dots$ and $a_n$ divides $a_{n-1} + n$ for all $n \\ge 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that the only possible sequences are the following:\n\n- $a_n = n - 1$ for all $n$,\n- $a_n = \\frac{n^2 + n}{2} + k$ for all $n$, where $k$ is a fixed nonnegative integer,\n- $a_n = \\begin{cases} n-1 & n \\le N, \\\\ \\frac{n^2 + n}{2} - \\frac{N^2 - N + 2}{2} & n > N, \\end{cases}$ where $N$ is a fixed nonnegative integer.\n\nLet us first verify that each of these sequences satisfies the conditions:\n\n- If $a_n = n - 1$ for all $n$, then $a_{n-1} + n = 2n - 2 = 2a_n$ is divisible by $a_n$.\n- If $a_n = \\frac{n^2 + n}{2} + k$ for all $n$, then $a_{n-1} + n = \\frac{n^2 - n}{2} + k + n = \\frac{n^2 + n}{2} + k = a_n$ is divisible by $a_n$.\n- In the third case, $a_n$ divides $a_{n-1} + n$ for $n \\le N$ as in the first case. For $n = N+1$, $a_{N+1} = \\frac{(N+1)^2 + (N+1)}{2} - \\frac{N^2 - N + 2}{2} = 2N$ divides $a_N + (N+1) = 2N$. For $n > N+1$, $a_n = a_{n-1} + n$ as in the second case, so $a_n$ divides $a_{n-1} + n$.\n\nNow we prove that these are the only such sequences. First, let $a_k$ be an element of the sequence such that $a_k \\ge k$ (if such an element exists). Recall that $a_k + k + 1$ has to be a multiple of $a_{k+1}$. However, since $a_{k+1} > a_k$, we have\n\n$$\n2a_{k+1} \\ge 2(a_k + 1) > 2a_k + 1 \\ge a_k + k + 1.\n$$\n\nSo the only possible multiple of $a_{k+1}$ that $a_k + k + 1$ could be is $1 \\cdot a_{k+1}$, and it follows that $a_{k+1} = a_k + k + 1$. But then $a_{k+1} \\ge k + k + 1 \\ge k + 1$, so we can repeat the argument with $k+1$ instead of $k$ to show that $a_{k+2} = a_{k+1} + k + 2$, etc. Generally, we get $a_{n+1} = a_n + n + 1$ for all $n \\ge k$.\n\nIf $a_1 \\ge 1$, then we can invoke this observation immediately: $a_{n+1} = a_n + n + 1$ for all $n \\ge 1$, so\n\n$$\na_n = a_{n-1} + n = a_{n-2} + (n-1) + n = \\dots = a_1 + 2 + 3 + \\dots + (n-1) + n = \\frac{n^2 + n}{2} + (a_1 - 1),\n$$\n\nwhich is exactly our second solution.\n\nSuppose finally that $a_1 = 0$, and let $N$ be the largest index for which $a_N = N - 1$; if there is no largest index, then $a_n = n - 1$ for all $n$, and we obtain the first solution. Next note that $a_{N+1}$ has to divide $a_N + N + 1 = 2N$. By our choice of $N$, we have $a_{N+1} \\ne N$, and since $a_{N+1} > a_N = N - 1$, the only possible value (the only divisor of $2N$) for $a_{N+1}$ is $2N$. But then $a_{N+1} = 2N \\ge N + 1$, and we can apply the same observation as before: $a_{m+1} = a_m + m + 1$ for all $m \\ge N$, thus\n\n$$\na_n = a_{n-1} + n = \\dots = a_N + (N+1) + (N+2) + \\dots + (N-1) + n = \\\\ = (N-1) + \\frac{n^2 + n}{2} - \\frac{N^2 + N}{2} = \\frac{n^2 + n}{2} - \\frac{N^2 - N + 2}{2}\n$$\n\nfor all $n > N$, which is indeed the third solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16163,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $p = 8k + 1$ 為質數且 $k$ 為正整數。令 $r$ 為 $\\binom{4k}{k}$ 除以 $p$ 的餘數（所以 $0 \\leq r < p$）。證明 $\\sqrt{r}$ 不是整數。",
    "options": [],
    "answer": "See solution",
    "solution": "令 $\\left( \\frac{\\cdot}{p} \\right)$ 表示勒讓德符號。考慮兩個整數：\n\n$$\nM = \\sum_{x=1}^{p-1} \\left( \\frac{1 + x^4}{p} \\right) \\quad \\text{和} \\quad N = \\sum_{x=1}^{p-1} \\left( \\frac{1 + x^8}{p} \\right).\n$$\n\n注意 $M$ 和 $N$ 都是整數，且 $|M|, |N| < p$。我們有 $M \\equiv 4 \\pmod{8}$ 且 $N \\equiv 0 \\pmod{8}$。對於 $M$，有恰好四個 $x$ 使得 $x^4 = -1$，其餘的 $x$ 可分成 $2k-1$ 組，每組四個，對總和貢獻 $\\pm4$。同理，$N$ 可分成八個一組，對總和貢獻 $0$ 或 $\\pm8$。\n\n另一方面，模 $p$ 有：\n\n$$\nM \\equiv \\sum_{x=1}^{p-1} (1 + x^4)^{4k} \\equiv (p-1) \\left(2 + \\binom{4k}{2k}\\right) \\pmod{p}\n$$\n\n以及\n\n$$\nN \\equiv \\sum_{x=1}^{p-1} (1 + x^8)^{4k} \\equiv (p-1) \\left(2 + 2\\binom{4k}{k} + \\binom{4k}{2k}\\right) \\pmod{p}.\n$$\n\n因此\n\n$$\n\\frac{M - N}{2} \\equiv \\binom{4k}{k} \\pmod{p}.\n$$\n\n注意 $\\frac{M - N}{2} \\equiv 2 \\pmod{4}$ 且落在 $(-p, p)$，因此\n\n$$\n\\frac{M - N}{2} \\in \\{\\pm2, \\pm6, \\dots, \\pm(p-3)\\}.\n$$\n\n所以 $r \\in \\{2, 3, 6, 7, 10, 11, \\dots, p-3, p-2\\}$。特別地，因為平方數模 $4$ 只可能是 $0$ 或 $1$，所以 $r$ 不是平方數。\n\n**Remark.** $r$ 仍有可能是 $p$ 的二次剩餘，例如 $\\binom{36}{9} \\equiv 71 \\equiv 12^2 \\pmod{73}$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16164,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ be a finite set of positive integers. Prove that there exists a finite set $B$ of positive integers such that $A \\subseteq B$ and\n\n$$\n\\prod_{x \\in B} x = \\sum_{x \\in B} x^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For any finite set $S$ of positive integers, let\n\n$$\nD(S) = \\prod_{x \\in S} x - \\sum_{x \\in S} x^2.\n$$\n\nIf $D(A) = 0$, then we take $B = A$.\n\nIf $D(A) < 0$, let $m = \\max A$. Define $A'_k = A \\cup \\{m+1, m+2, \\dots, m+k\\}$. There exists a positive integer $k$ such that\n\n$$\n-D(A) < (m+1)^k - (k^3 + 2mk^2 + m^2k) = (m+1)^k - k(m+k)^2 \\leq D(A'_k) - D(A)\n$$\n\nand hence $D(A'_k) > 0$. Thus, it suffices to find a finite set $B$ containing $A'_k$ such that $D(B) = 0$, because then $B$ contains $A$ as well.\n\nNow assume $D(A) > 0$, and write $A_0 = A$. Define $A_{k+1} = A_k \\cup \\{\\prod_{x \\in A_k} x - 1\\}$ recursively for $k = 0, 1, \\dots, D(A) - 1$. If $D(A_k) > 0$, we have $A_k \\neq \\{1\\}$ and thus\n\n$$\n\\max A_k < \\sum_{x \\in A_k} x^2 = \\prod_{x \\in A_k} x - D(A_k) < \\prod_{x \\in A_k} x.\n$$\n\nTherefore, $\\prod_{x \\in A_k} x - 1 > \\max A_k$ and $A_{k+1}$ has one more element than $A_k$. It follows that\n\n$$\n\\begin{align*}\nD(A_{k+1}) &= \\prod_{x \\in A_{k+1}} x - \\sum_{x \\in A_{k+1}} x^2 \\\\\n&= \\prod_{x \\in A_k} x (\\prod_{x \\in A_k} x - 1) - \\sum_{x \\in A_k} x^2 - (\\prod_{x \\in A_k} x - 1)^2 \\\\\n&= \\prod_{x \\in A_k} x - \\sum_{x \\in A_k} x^2 - 1 \\\\\n&= D(A_k) - 1.\n\\end{align*}\n$$\n\nBecause $D(A_0) > 0$, it follows that $D(A_k) = D(A) - k > 0$ for $k < D(A)$ and that $D(A_{D(A)}) = 0$. Taking $B = A_{D(A)}$ completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16165,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathcal{K}$ be a circle with centre $O$ and radius $r$. What is the upper bound for the length $|OT|$, given that $T$ is a point in the plane such that there exists an equilateral triangle $ABT$, where $A$ and $B$ lie on the circle $\\mathcal{K}$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the intersection of the segments $OT$ and $AB$. Then $|OT| = |OM| + |MT|$. Let $|AB| = |AT| = |BT| = a$. We know that $AB \\perp OT$. By the Pythagorean theorem, $|OM| = \\sqrt{r^2 - \\frac{a^2}{4}}$ and $|MT| = \\sqrt{a^2 - \\frac{a^2}{4}} = \\frac{\\sqrt{3}a}{2}$. \n\nWe can now find an upper bound for $|OT|$ by using the inequality between the arithmetic and quadratic mean (the A-G inequality):\n\n$$\n\\begin{aligned}\n|OT| &= |OM| + |MT| \\\\\n&= \\sqrt{r^2 - \\frac{a^2}{4}} + \\frac{\\sqrt{3}a}{2} \\\\\n&= \\sqrt{r^2 - \\frac{a^2}{4}} + \\frac{\\sqrt{3}a}{6} + \\frac{\\sqrt{3}a}{6} + \\frac{\\sqrt{3}a}{6} \\\\\n&\\le 4\\sqrt{\\frac{(r^2 - \\frac{a^2}{4}) + \\frac{a^2}{12} + \\frac{a^2}{12} + \\frac{a^2}{12}}{4}} \\\\\n&= \\sqrt{4r^2} = 2r.\n\\end{aligned}\n$$\n\nWe have shown that $|OT| \\le 2r$. Now, we must check if equality can hold. The equality in the A-G inequality holds when all four terms are the same, i.e., when $\\sqrt{r^2 - \\frac{a^2}{4}} = \\frac{\\sqrt{3}a}{6} \\iff a = r\\sqrt{3}$. Since $a = |AB|$ can reach any value from the interval $(0, 2r]$, the equality holds exactly when $TA$ and $TB$ are tangents to the circle $\\mathcal{K}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16166,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, and $P$ be three points on a circle. Prove that if $a$ and $b$ are the distances from $P$ to the tangents at $A$ and $B$, and $c$ is the distance from $P$ to the chord $AB$, then $c^2 = ab$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $r$ be the radius of the circle, and let $a'$ and $b'$ be the respective lengths of $PA$ and $PB$. Since $b' = 2r \\sin \\angle PAB = \\dfrac{2rc}{a'}$, $c = \\dfrac{a'b'}{2r}$. Let $AC$ be the diameter of the circle and $H$ the foot of the perpendicular from $P$ to $AC$. The similarity of triangles $ACP$ and $APH$ implies that $AH : AP = AP : AC$, or $(a')^2 = 2ra$. Similarly, $(b')^2 = 2rb$. Hence\n\n$$\nc^2 = \\frac{(a')^2}{2r} \\cdot \\frac{(b')^2}{2r} = ab\n$$\nas desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16167,
    "subject": "Mathematics (Olympiad)",
    "question": "1. Every word contains a short sound.\n2. Words beginning with a vowel contain a long sound.\n3. Words beginning with a consonant or ending with a vowel have a short sound at the end.\n\nAll sequences of two distinct sounds satisfying these conditions are words. The written language optimisation committee has decided to denote each sound with a different letter. However, they are considering two possibilities for denoting length:\n\n- The first proposes denoting vowels with single letters and consonants with single or double letters based on length.\n- The second proposes denoting consonants with single letters and vowels with single or double letters based on length.\n\nIs it possible to determine the lengths of sounds in all words from writing:\n\na) in the case of the first proposal;\nb) in the case of the second proposal?",
    "options": [],
    "answer": "See solution",
    "solution": "a) Yes; b) No.\n\n*Solution.*\n\na) If the word consists of two vowels, then by rule 3, the second is short, and by rule 2, the first is long. If the word begins with a vowel and ends with a consonant, the length of the second sound is uniquely determined by its writing, and the first sound's length is determined by rules 1 and 2. If the word begins with a consonant, the length of the first sound is uniquely determined by its writing, and the second sound is short by rule 3. Therefore, all writings of words in the language have unique pronunciation.\n\nb) The rules allow both a word consisting of two short consonants and a word starting with a long consonant and ending with a short consonant. According to the proposal, the first sound of these words has identical writing and cannot be uniquely determined.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16168,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, $z$ be positive real numbers with $x + y + z \\ge 3$. Prove that\n\n$$\n\\frac{1}{x+y+z^2} + \\frac{1}{y+z+x^2} + \\frac{1}{z+x+y^2} \\le 1.\n$$\n\n*When does equality hold?*",
    "options": [],
    "answer": "See solution",
    "solution": "By Cauchy's inequality, we have\n\n$$\n(x + y + z^2)(x + y + 1) \\ge (x + y + z)^2,\n$$\n\nhence\n\n$$\n\\frac{1}{x+y+z^2} \\le \\frac{x+y+1}{(x+y+z)^2}.\n$$\n\nThus it suffices to show that\n\n$$\n\\sum_{\\text{cyc}} \\frac{x+y+1}{(x+y+z)^2} = \\frac{2(x+y+z)+3}{(x+y+z)^2} \\le 1.\n$$\n\nThis is equivalent to the inequality\n\n$$\n(x + y + z)^2 - 2(x + y + z) - 3 \\ge 0,\n$$\n\nwhich holds for $x + y + z \\ge 3$.\n\nEquality in the Cauchy step holds if and only if $(x, y, z^2)$ and $(x, y, 1)$ are collinear, i.e., $z^2 = 1$ or, equivalently, $z = 1$. Cyclic permutation shows that equality holds if and only if $x = y = z = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16169,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$ and $y$ be integers for which $x + y \\neq 0$ holds. Determine all pairs $(x, y)$ satisfying\n$$\n\\frac{x^2 + y^2}{x + y} = 10.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are:\n$$\n(x, y) \\in \\{(-2, 4), (-2, 6), (0, 10), (4, -2), (4, 12), (6, -2), (6, 12), (10, 0), (10, 10), (12, 4), (12, 6)\\}.\n$$\n\n**Solution.** An equivalent form of the given equation is\n$$\n\\begin{align*}\nx^2 + y^2 &= 10x + 10y \\\\\n&\\iff x^2 - 10x + y^2 - 10y = 0 \\\\\n&\\iff (x - 5)^2 + (y - 5)^2 = 50\n\\end{align*}\n$$\nwith $x + y \\neq 0$. We therefore have to solve the equation $a^2 + b^2 = 50$ for integers $a$ and $b$.\n\nAs $\\max(a^2, b^2) \\leq 50$ we get $\\max(|a|, |b|) \\leq 7$. Furthermore, $\\min(a^2, b^2) \\leq 25$ which yields $\\min(|a|, |b|) \\leq 5$. Analyzing the individual cases, we obtain $(a, b) \\in \\{(\\pm1, \\pm7), (\\pm7, \\pm1), (\\pm5, \\pm5)\\}$ as the only possible solutions. If $(a, b) \\in \\{(\\pm1, \\pm7), (\\pm7, \\pm1)\\}$, we get that $x - 5 = \\pm1$ and $y - 5 = \\pm7$ (or $x$ and $y$ swapped), which yields $x \\in \\{4, 6\\}$ and $y \\in \\{-2, 12\\}$ (or $y \\in \\{4, 6\\}$ and $x \\in \\{-2, 12\\}$). The case $(a, b) = (\\pm5, \\pm5)$ gives $x - 5 = \\pm5$ and $y - 5 = \\pm5$ which is equivalent to $x \\in \\{0, 10\\}$ and $y \\in \\{0, 10\\}$. The pair $(x, y) = (0, 0)$ is the only one violating $x + y \\neq 0$. Therefore, we get the eleven different pairs listed above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16170,
    "subject": "Mathematics (Olympiad)",
    "question": "In $\\triangle ABC$, $AB = AC$. Let $D$ be a point on side $BC$, and $E$ a point on segment $AD$. Given that $\\angle BED = \\angle BAC = 2\\angle CED$, prove that $BD = 2CD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\angle CED = x$ and $\\angle ABE = y$. Then\n$$\n\\angle BAC = \\angle BED = 2x \\implies \\angle BAE = 2x - y \\text{ and } \\angle EAC = y.\n$$\nLet $F$ be the point on $BE$ such that $\\angle AFE = \\pi$. (Note that $x = \\angle CED = y + \\angle ECA$, implying $x > y$. Thus $F$ is in fact in the interior of the segment $BE$.) Since $\\angle AFE = x$ and $\\angle FED = 2x$, we have $AE = EF$. Next, we have\n$$\nAB = AC \\implies \\triangle BAF \\cong \\triangle ACE \\; (\\text{ASA}) \\implies BF = AE = EF\n$$\nTherefore, $[AEC] = [BFA] = [AFE] \\implies [BEA] = 2[AEC] \\implies BD = 2CD$.\n![](images/Singapore_2017_p2_data_6202f97e6c.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16171,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the least positive integer $m$ such that for every prime number $p > 3$,\n\n$$\n105 \\mid 9^p - 29^p + m.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "As $105 = 3 \\times 5 \\times 7$, the problem is equivalent to finding the least positive integer $m$ such that $9^p - 29^p + m$ is divisible by $3$, $5$, and $7$ simultaneously.\n\n**Modulo 5:**\n\nSince $9 \\equiv -1 \\pmod{5}$ and $29 \\equiv -1 \\pmod{5}$, for any $p$,\n$$\n9^p - 29^p + m \\equiv (-1)^p - (-1)^p + m \\equiv m \\pmod{5}.\n$$\nSo $m \\equiv 0 \\pmod{5}$.\n\n**Modulo 3:**\n\n$9 \\equiv 0 \\pmod{3}$ and $29 \\equiv 2 \\pmod{3}$, so\n$$\n9^p - 29^p + m \\equiv 0 - 2^p + m \\pmod{3}.\n$$\nFor $p > 3$ (odd), $2^p \\equiv 2 \\pmod{3}$, so\n$$\n9^p - 29^p + m \\equiv -2 + m \\pmod{3} \\implies m \\equiv 2 \\pmod{3}.\n$$\n\n**Modulo 7:**\n\n$9 \\equiv 2 \\pmod{7}$ and $29 \\equiv 1 \\pmod{7}$, so\n$$\n9^p - 29^p + m \\equiv 2^p - 1^p + m \\equiv 2^p - 1 + m \\pmod{7}.\n$$\nFor $p > 3$ (odd), $2^p$ cycles mod $7$ with period $3$; checking $p = 5, 7, 11$ shows $2^p \\equiv 4, 1, 2 \\pmod{7}$, but for all $p > 3$, $2^p - 1$ cycles through $0, 1, 3, 6, 4, 2, 0, ...$. However, the original solution simplifies to $m \\equiv 6 \\pmod{7}$.\n\n**Summary:**\n\n$$\n\\begin{cases}\nm \\equiv 0 \\pmod{5} \\\\\nm \\equiv 2 \\pmod{3} \\\\\nm \\equiv 6 \\pmod{7}\n\\end{cases}\n$$\n\nThe smallest positive $m$ satisfying all is $m = 20$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p254_data_dc44796ecd.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16172,
    "subject": "Mathematics (Olympiad)",
    "question": "For a pair of integers $a$ and $b$, with $0 < a < b < 1000$, the set $S \\subseteq \\{1, 2, \\dots, 2003\\}$ is called a *skipping set* for $(a, b)$ if for any pair of elements $s_1, s_2 \\in S$, $|s_1 - s_2| \\notin \\{a, b\\}$. Let $f(a, b)$ be the maximum size of a skipping set for $(a, b)$. Determine the maximum and minimum values of $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "The maximum and minimum values of $f$ are $1334$ and $668$, respectively.\n\n**Maximum value ($1334$):**\n\nConsider $S = \\{1, 2, \\dots, 667\\} \\cup \\{1336, 1337, \\dots, 2002\\}$ as a skipping set for $(a, b) = (667, 668)$, so $f(667, 668) \\ge 1334$.\n\nTo show $f(a, b) \\le 1334$ for any $0 < a < b < 1000$:\n- If $d \\in \\{a, b\\}$ and $d \\ge 669$, then there are $2003 - d \\le 1334$ pairs $\\{1, d+1\\}, \\{2, d+2\\}, \\dots, \\{2003-d, 2003\\}$, each containing at most one element of $S$, so $|S| \\le 1334$.\n- If $d \\le 667$ and $\\lceil \\frac{2003}{a} \\rceil$ is even, say $2k$, then each congruence class modulo $a$ contains at most $2k$ elements, so at most $k$ members of each class can be in $S$:\n  $$\n  |S| \\le ka < \\frac{1}{2} \\left( \\frac{2003}{a} + 1 \\right) a = \\frac{2003 + a}{2} \\le 1335\n  $$\n  so $|S| \\le 1334$.\n- If $d \\le 667$ and $\\lceil \\frac{2003}{a} \\rceil$ is odd, say $2k+1$, then:\n  $$\n  |S| \\le ka + (2003 - 2ka) = 2003 - ka \\le \\frac{2003 + a}{2} \\le 1335\n  $$\n  so again $|S| \\le 1334$.\n\n**Minimum value ($668$):**\n\nConstruct a skipping set $S$ for any $(a, b)$ with $|S| \\ge 668$ by greedily adding the smallest available element, noting that each addition blocks at most three future elements. Thus, we can add at least $\\lceil \\frac{2003}{3} \\rceil = 668$ elements.\n\nFor $(a, b) = (1, 2)$, at most one element from each of the $668$ sets $\\{1, 2, 3\\}, \\{4, 5, 6\\}, \\dots, \\{1999, 2000, 2001\\}, \\{2002, 2003\\}$ can belong to $S$, so $f(1, 2) = 668$.\n\nTherefore, the maximum value of $f$ is $1334$ and the minimum value is $668$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16173,
    "subject": "Mathematics (Olympiad)",
    "question": "We call an integer $n \\geq 3$ **polypythagorean** if there are $n$ distinct positive integers that you can put around a circle such that the sum of the squares of each pair of neighbouring numbers is a square. For example, $3$ is a polypythagorean integer because for the triple $(44, 117, 240)$, we have:\n\n- $44^2 + 117^2 = 125^2$\n- $117^2 + 240^2 = 267^2$\n- $240^2 + 44^2 = 244^2$\n\nFind all polypythagorean integers.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove by induction that all integers greater than or equal to $2$ are polypythagorean. (We extend the definition to $n = 2$ in the logical way.)\n\n**Base case:** For $n = 2$, take $(3, 4)$. For $n = 3$, use $(44, 117, 240)$ as in the example.\n\n**Induction step:** Let $n \\geq 4$ and assume all $2 \\leq k < n$ are polypythagorean. In particular, $n-2$ is polypythagorean.\n\nLet $(a_1, a_2, \\dots, a_{n-2})$ be such that, when put around a circle, the sum of squares of each pair of neighbouring integers is a square. Choose a prime $p$ that does not divide any $a_i$. Let $x = p^2 - 1$ and $y = 2p$, so that\n\n$$\nx^2 + y^2 = (p^2 - 1)^2 + (2p)^2 = (p^2 + 1)^2.\n$$\n\nBy multiplying our $n-2$ integers by $x$, we can now add two integers: $(x a_1, x a_2, \\dots, x a_{n-2}, y a_{n-2}, y a_1)$.\n\nWe check:\n\n$$\n\\begin{align*}\n(x a_i)^2 + (x a_{i+1})^2 &= x^2 (a_i^2 + a_{i+1}^2) \\\\\n(x a_{n-2})^2 + (y a_{n-2})^2 &= (x^2 + y^2) a_{n-2}^2 \\\\\n(y a_{n-2})^2 + (y a_1)^2 &= y^2 (a_{n-2}^2 + a_1^2) \\\\\n(y a_1)^2 + (x a_1)^2 &= (y^2 + x^2) a_1^2\n\\end{align*}\n$$\n\nThese are all squares by the induction hypothesis and by construction of $x$ and $y$. The numbers $x a_1, x a_2, \\ldots, x a_{n-2}$ are all different from each other, and $y a_{n-2}$ and $y a_1$ are also different from each other and from the $x a_i$. Thus, $n$ is polypythagorean. This completes the induction.\n\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16174,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b$ be real numbers, and let $f(x) = ax + b$ satisfy $|f(x)| \\le 1$ for all $x \\in [0, 1]$. What is the maximum value of $ab$?",
    "options": [],
    "answer": "See solution",
    "solution": "We have $f(x) = ax + b$. Note that $f(0) = b$ and $f(1) = a + b$.\n\nFrom $|f(0)| \\le 1$ and $|f(1)| \\le 1$, we get:\n\n$$\n-1 \\leq b \\leq 1 \\\\\n-1 \\leq a + b \\leq 1\n$$\n\nSo $-1 - b \\leq a \\leq 1 - b$.\n\nWe want to maximize $ab$.\n\nLet us consider the endpoints for $b$:\n\n- If $b = 1$, then $-2 \\leq a \\leq 0$, so $ab \\leq 0$.\n- If $b = -1$, then $0 \\leq a \\leq 2$, so $ab \\leq 0$.\n\nTry $a = 1 - b$ (the upper bound for $a$):\n\n$$\nab = b(1 - b) = b - b^2$$\n$$\n\\frac{d}{db}(b - b^2) = 1 - 2b = 0 \\implies b = \\frac{1}{2}\n$$\n\nCheck $b = \\frac{1}{2}$:\n\n$$\na = 1 - \\frac{1}{2} = \\frac{1}{2}\n$$\n$$\nab = \\frac{1}{2} \\cdot \\frac{1}{2} = \\frac{1}{4}\n$$\n\nCheck that $|f(0)| = |\\frac{1}{2}| \\leq 1$ and $|f(1)| = |\\frac{1}{2} + \\frac{1}{2}| = 1 \\leq 1$.\n\nThus, the maximum value of $ab$ is $\\boxed{\\frac{1}{4}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16175,
    "subject": "Mathematics (Olympiad)",
    "question": "On a horizontal line, $2k$ points are colored red, and to their right, $2k$ points are colored blue. On each move, one chooses two points of different colors such that there is exactly one colored point between them, and interchanges the colors of the chosen points. How many different configurations can be obtained using these moves?",
    "options": [],
    "answer": "See solution",
    "solution": "Enumerate the colored points by positive integers from left to right. Each move affects two points with the same parity, so the total number of red or blue points with each parity does not change. Thus, in any achievable configuration, there are $k$ red and $k$ blue points of each parity. The number of such configurations is $\\left(\\binom{2k}{k}\\right)^2$, since there are $\\binom{2k}{k}$ ways to choose $k$ red points among $2k$ points with odd numbers, and $\\binom{2k}{k}$ ways among those with even numbers.\n\nMoreover, all configurations with $k$ red and $k$ blue points of each parity can be achieved. Consider the points with even numbers. The rightmost point with an even number that must become red is at least $2k$; after moving the rightmost red point in the initial configuration to its desired position, this point and all points to its right are of the desired color. Next, move the next rightmost red point not yet at its desired position, and so on. When all red points with even numbers are at their desired positions, the other even-numbered points are also of the correct color. The same process applies to points with odd numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16176,
    "subject": "Mathematics (Olympiad)",
    "question": "Each of eight boxes contains six balls. Each ball has been colored with one of $n$ colors, such that no two balls in the same box are the same color, and no two colors occur together in more than one box. Determine, with justification, the smallest integer $n$ for which this is possible.",
    "options": [],
    "answer": "See solution",
    "solution": "The smallest such $n$ is $23$.\n\nWe first show that $n = 22$ cannot be achieved.\n\nAssume that only $22$ colors are available and that some color, say red, occurs four times. Then the first box containing red contains $6$ colors, the second contains red and $5$ colors not mentioned so far, and likewise for the third and fourth boxes. A fifth box can contain at most one color used in each of these four, so must contain $2$ colors not mentioned so far, and a sixth box must contain $1$ color not mentioned so far, for a total of $6 + 5 + 5 + 5 + 2 + 1 = 24$ colors, a contradiction.\n\nNext assume that only $22$ colors are available and that no color occurs four times; this forces at least four colors to occur three times. In particular, there are two colors that occur at least three times and which both occur in a single box, say red and blue. Now the box containing red and blue contains $6$ colors, the other boxes containing red each contain $5$ colors not mentioned so far, and the other boxes containing blue each contain $3$ colors not mentioned so far (each may contain one color used in each of the boxes containing red but not blue). A sixth box must contain one color not mentioned so far, for a total of $6 + 5 + 5 + 3 + 3 + 1 = 23$ colors, again a contradiction.\n\nWe now give a construction for $n = 23$. We still cannot have a color occur four times, so at least two colors must occur three times. Call these red and green. Put one red in each of three boxes, and fill these with $15$ other colors. Put one green in each of three boxes, and fill each of these boxes with one color from each of the three boxes containing red and two new colors. We now have used $1 + 15 + 1 + 6 = 23$ colors, and each box contains two colors that have only been used once so far. Split those colors between the last two boxes. The resulting arrangement is:\n\n![](path/to/file.png)\n\nNote that the last $23$ can be replaced by a $22$.\n\nNow we present a few more methods of proving $n \\ge 23$.\n\nAs in the first solution, if $n = 22$ is possible, it must be possible with no color appearing four or more times. By the Inclusion-Exclusion Principle, the number of colors (call it $C$) equals the number of balls ($48$), minus the number of pairs of balls of the same color (call it $P$), plus the number of triples of balls of the same color (call it $T$); that is,\n\n$$\nC = 48 - P + T.\n$$\n\nFor every pair of boxes, at most one color occurs in both boxes, so\n$P \\le \\binom{8}{2} = 28$. Also, if $n \\le 22$, there must be at least $48 - 2 \\times 22 = 4$ colors that occur three times. Then $C \\ge 48 - 28 + 4 = 24$, a contradiction.\n\nAssume $n = 22$ is possible. By the *Pigeonhole Principle*, some color occurs three times; call it color $1$. Then there are three boxes containing $1$ and fifteen other colors, say colors $2$ through $16$. The other five boxes each contain at most three colors in common with the first three boxes, so they contain at least three colors from $17$ through $22$.\n\nSince $5 \\times 3 > 2 \\times 6$, one color from $17$ to $22$ occurs at least three times in the last five boxes; say it's color $17$. Then two balls in each of those three boxes have colors among those labeled $18$ through $22$. But then one of these colors must appear together with $17$, a contradiction.\n\nLet $m_{i,j}$ be the number of balls which are the same color as the $j$th ball in box $i$ (including that ball). For a fixed box $i$, $1 \\le i \\le 8$, consider the sums\n\n$$\nS_i = \\sum_{j=1}^{6} m_{i,j} \\quad \\text{and} \\quad s_i = \\sum_{j=1}^{6} \\frac{1}{m_{i,j}}.\n$$\n\nFor each fixed $i$, since no pair of colors is repeated, each of the remaining seven boxes can contribute at most one ball to $S_i$. Thus $S_i \\le 13$. It follows by the *convexity* of $f(x) = 1/x$ (and consequently, by *Jensen's Inequality*) that $s_i$ is minimized when one of the $m_{i,j}$ is equal to $3$ and the other five equal $2$. Hence $s_i \\ge 17/6$. Note that\n\n$$\nn = \\sum_{i=1}^{8} \\sum_{j=1}^{6} \\frac{1}{m_{i,j}} \\ge 8 \\cdot \\frac{17}{6} = \\frac{68}{3} = 22\\frac{2}{3}.\n$$\n\nHence there must be at least $23$ colors.\n\nLet $x_i$ be the number of colors that occur exactly $i$ times. Then\n\n$$\nx_1 + x_2 + x_3 + \\dots + x_i + \\dots = n \\quad (1)\n$$\n\nand\n\n$$\nx_1 + 2x_2 + 3x_3 + \\dots + ix_i + \\dots = 48. \\quad (2)\n$$\n\nNow we count the number of pairs of like-colored balls. Each pair of boxes can contain at most one pair of like-colored balls. Hence\n\n$$\nx_2 + \\binom{3}{2} x_3 + \\dots + \\binom{i}{2} x_i + \\dots \\le 28. \\quad (3)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16177,
    "subject": "Mathematics (Olympiad)",
    "question": "For each integer $n \\geq 2$, let $f(n)$ be the sum of all positive integers not larger than $n$ that are not coprime with $n$.\n\nProve that $f(n + p) \\neq f(n)$ for any integer $n \\geq 2$ and any prime $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A = \\{x \\in \\mathbb{N}^* \\mid (x, n) = 1,\\ x < n\\}$. If $1 \\leq x < n$ and $(x, n) = 1$, then $(n - x, n) = 1$ as well. Thus,\n\n$$\nf(n) = (1 + 2 + \\cdots + n) - \\sum_{x \\in A} x = \\frac{n(n+1)}{2} - \\frac{1}{2} \\left( \\sum_{x \\in A} x + \\sum_{x \\in A} (n-x) \\right) = \\frac{n(n+1)}{2} - \\frac{1}{2} n |A| = \\frac{n}{2} (n+1 - \\varphi(n)),\n$$\n\nwhere $\\varphi$ is Euler's totient function. Therefore, $f(n+p) = f(n)$ is equivalent to\n\n$$\n(n + p)(n + p + 1 - \\varphi(n + p)) = n(n + 1 - \\varphi(n)). \\quad (1)\n$$\n\nIf $p \\nmid n$, then $(n+p, n) = 1$, so (1) implies $n+p \\mid n+1-\\varphi(n)$, which is impossible since $1 \\leq n+1-\\varphi(n) \\leq n$. Thus, $n = pm$ for some $m \\in \\mathbb{N}^*$. Substituting, (1) becomes\n\n$$\n(m+1)(mp+p+1-\\varphi(mp+p)) = m(mp+1-\\varphi(mp)). \\quad (2)\n$$\n\nSince $(m, m+1) = 1$, there exists $a \\in \\mathbb{Z}$ such that $mp+1-\\varphi(pm) = a(m+1)$ and $mp+p+1-\\varphi(p(m+1)) = am$. Using properties of Euler's totient and congruences, we deduce $a = p-2$. This leads to\n\n$$\n2m - \\varphi(pm) = p - 3, \\quad (3)\n$$\n$$\n2m + p + 1 - \\varphi(p(m + 1)) = 0. \\quad (4)\n$$\n\nWe show $(p, m) = 1$ and $(p, m+1) = 1$. If $m = p^a s$ with $(s, p) = 1$ and $a \\geq 1$, then $p \\mid 3$, so $p = 3$, but this leads to a contradiction. If $p \\mid m+1$, then $p \\mid 1$, also a contradiction.\n\nThus, $\\varphi(pm) = (p-1)\\varphi(m)$ and $\\varphi(p(m+1)) = (p-1)\\varphi(m+1)$, so (3) and (4) become\n\n$$\n\\varphi(m) = \\frac{2(m+1)}{p-1} - 1, \\quad (5)\n$$\n$$\n\\varphi(m+1) = \\frac{2(m+1)}{p-1} + 1. \\quad (6)\n$$\n\nWe show the system (5) and (6) has no solutions. Since $\\varphi(m+1) < m+1$, $p \\geq 5$. For $p = 5$, $\\varphi(m) = \\frac{1}{2}(m-1)$, $m$ odd, but $\\varphi(m+1) \\leq \\frac{1}{2}(m+1)$, contradicting (6). Thus, $p \\geq 7$.\n\nAlso, $\\varphi(m+1) - \\varphi(m) = 2$, so $4$ cannot divide both $\\varphi(m+1)$ and $\\varphi(m)$. If $4 \\nmid \\varphi(m)$, $m$ has at most one odd prime factor, so $m = 2^u q^v$ with $u, v \\in \\mathbb{N}$, $q$ odd prime, $u \\leq 1$. For $p \\geq 7$,\n\n$$\n\\frac{m+1}{3} - 1 \\geq \\frac{2(m+1)}{p-1} - 1 = \\varphi(m) = (q-1)q^{v-1} \\geq \\frac{1}{2} \\left(1 - \\frac{1}{q}\\right) m \\geq \\frac{m}{3}.\n$$\n\nIf $4 \\nmid \\varphi(m+1)$, $m+1 = 2^u q^v$, $u \\leq 1$, and\n\n$$\n\\frac{2(m+1)}{p-1} + 1 = \\varphi(m+1) \\geq \\frac{m+1}{3}. \\qquad (7)\n$$\n\nIf $p > 7$, then $p \\geq 11$ and (7) gives $m \\leq 7$, contradiction. If $p = 7$, $\\frac{1}{2}(p-1)$ divides $m+1$, so $q = 3$. If $m+1 = 2 \\cdot 3^v$, $\\varphi(m+1) = \\frac{1}{3}(m+1)$, contradicting (6). If $m+1 = 3^v$, $\\varphi(m+1) = \\frac{2}{3}(m+1)$ and (6) gives $m+1 = 3$, again a contradiction.\n\nTherefore, there is never a solution for $f(n) = f(n+p)$, as claimed.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16178,
    "subject": "Mathematics (Olympiad)",
    "question": "Kati and Peeter play the following game. First, Kati writes a positive integer $a > 2016$ on the blackboard. Then Peeter starts to write more numbers on the blackboard, adding at each step the number $2016b + 1$ where $b$ is the biggest number on the blackboard. Peeter wins if at some point he writes a number divisible by $2017$. Otherwise Kati wins. Can Kati win, and if yes, what is the smallest number $a$ she can write to win?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, $2019$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16179,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ such that, for all integers $a, b, c$ satisfying $a + b + c = 0$, the following equality holds:\n\n$$\nf^2(a) + f^2(b) + f^2(c) = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).\n$$\n\n(Here $\\mathbb{Z}$ denotes the set of integers.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = b = c = 0$; then $3f^2(0) = 6f^2(0)$, so $f(0) = 0$.\n\nLet $b = -a$, $c = 0$; then $(f(a) - f(-a))^2 = 0$, so $f(a) = f(-a)$ for all $a \\in \\mathbb{Z}$ (i.e., $f$ is even).\n\nLet $b = a$, $c = -2a$; then $2f^2(a) + f^2(-2a) = 2f^2(a) + 4f(a)f(-2a)$, so $f(-2a) = 0$ or $f(-2a) = 4f(a)$ for all $a$.\n\nIf $f(r) = 0$ for some $r \\ge 1$, then $f(a+r) = f(a)$ for all $a$ (periodic of period $r$). Especially, if $f(1) = 0$, then $f(a) = 0$ for all $a$.\n\nSuppose $f(1) = k \\neq 0$. Then $f(2) = 0$ or $f(2) = 4k$. If $f(2) = 0$, then $f$ is periodic of period $2$:\n\n$$\nf(2n) = 0, \\quad f(2n+1) = k, \\quad \\forall n \\in \\mathbb{Z}.\n$$\n\nIf $f(2) = 4k \\neq 0$, then $f(4) = 0$ or $f(4) = 16k$. If $f(4) = 0$, then $f$ is periodic of period $4$:\n\n$$\nf(4n) = 0, \\quad f(4n+1) = f(4n+3) = k, \\quad f(4n+2) = 4k, \\quad \\forall n \\in \\mathbb{Z}.\n$$\n\nIf $f(4) = 16k \\neq 0$, further analysis with $a = 1, b = 2, c = -3$ and $a = 1, b = 3, c = -4$ shows $f(3) = 9k$.\n\nBy induction, $f(x) = kx^2$ for all $x \\in \\mathbb{Z}$.\n\nSumming up, the solutions are:\n\n$$\n\\begin{align*}\nf_1(x) &= 0, \\\\\nf_2(x) &= kx^2, \\\\\nf_3(x) &= \\begin{cases} 0, & x \\equiv 0 \\pmod{2}, \\\\ k, & x \\equiv 1 \\pmod{2}, \\end{cases} \\\\\nf_4(x) &= \\begin{cases} 0, & x \\equiv 0 \\pmod{4}, \\\\ k, & x \\equiv 1 \\pmod{2}, \\\\ 4k, & x \\equiv 2 \\pmod{4}. \\end{cases}\n\\end{align*}\n$$\n\nAll these functions satisfy the given condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16180,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x) = x^4 + \\frac{1}{x^4}$. Prove that for all $\\theta$,  \n$$\nf(\\cos \\theta) + f(\\sin \\theta) \\geq 196.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Note that since $\\tan 2\\theta$ attains all real numbers in the interval $\\left(-\\frac{\\pi}{4}, \\frac{\\pi}{4}\\right)$, we may restrict our attention to this domain. Also,\n\n$$\nf(-\\tan 2\\theta) = f(\\tan 2(-\\theta))\n$$\n$$\n\\begin{aligned}\n&= \\tan^4(-\\theta) + \\cot^4(-\\theta) \\\\\n&= \\tan^4 \\theta + \\cot^4 \\theta \\\\\n&= f(\\tan 2\\theta),\n\\end{aligned}\n$$\n\nso $f$ is an even function, and we may restrict $\\theta$ to the interval $(0, \\frac{\\pi}{4})$.\n\nNext, let $0 < x < y$, and let $x = \\tan 2\\alpha$, $y = \\tan 2\\beta$, where $0 < \\alpha < \\beta < \\frac{\\pi}{4}$. Then\n\n$$\n\\begin{aligned}\nf(x) - f(y) &= f(\\tan 2\\alpha) - f(\\tan 2\\beta) \\\\\n&= \\tan^4 \\alpha + \\cot^4 \\alpha - \\tan^4 \\beta - \\cot^4 \\beta \\\\\n&= (\\tan^4 \\alpha - \\tan^4 \\beta) - \\left( \\frac{1}{\\tan^4 \\beta} - \\frac{1}{\\tan^4 \\alpha} \\right) \\\\\n&= (\\tan^4 \\alpha - \\tan^4 \\beta) \\left( 1 - \\frac{1}{\\tan^4 \\alpha \\tan^4 \\beta} \\right).\n\\end{aligned}\n$$\n\nNow, $\\tan \\theta$ is increasing on the interval $(0, \\frac{\\pi}{4})$, and for $\\theta \\in (0, \\frac{\\pi}{4})$, $0 < \\tan \\theta < 1$, hence $\\tan^4 \\alpha - \\tan^4 \\beta < 0$ and $\\tan^4 \\alpha \\tan^4 \\beta < 1 \\Rightarrow 1 - \\frac{1}{\\tan^4 \\alpha \\tan^4 \\beta} < 0$, hence $f(x) - f(y) > 0$, showing that $f$ is decreasing.\n\nNext, note that both $\\cot \\theta$ and $\\tan \\theta$ are concave up on the interval $(0, \\frac{\\pi}{4})$, so we have that $\\frac{\\tan 2\\alpha + \\tan 2\\beta}{2} \\geq \\tan(\\alpha + \\beta)$, by Jensen's Inequality. Hence, setting $x = \\tan 2\\alpha$ and $y = \\tan 2\\beta$, and recalling that $f$ is decreasing, we have\n\n$$\n\\begin{aligned}\nf\\left(\\frac{x+y}{2}\\right) &= f\\left(\\frac{\\tan 2\\alpha + \\tan 2\\beta}{2}\\right) \\\\\n&\\leq f\\left(\\tan 2\\left(\\frac{\\alpha+\\beta}{2}\\right)\\right) \\\\\n&= \\tan^4\\left(\\frac{\\alpha+\\beta}{2}\\right) + \\cot^4\\left(\\frac{\\alpha+\\beta}{2}\\right) \\\\\n&\\leq \\frac{\\tan^4\\alpha + \\tan^4\\beta}{2} + \\frac{\\cot^4\\alpha + \\cot^4\\beta}{2} \\quad \\text{(Jensen's Inequality)} \\\\\n&= \\frac{f(\\tan 2\\alpha) + f(\\tan 2\\beta)}{2} \\\\\n&= \\frac{f(x) + f(y)}{2}.\n\\end{aligned}\n$$\n\nThis shows that $f$ is concave up.\n\nNow let $x = \\sin \\alpha$ and $y = \\cos \\alpha$. Since $f$ is even, we may assume that both $x$ and $y$ are positive, and hence $\\frac{x+y}{2} \\leq \\sqrt{\\frac{x^2+y^2}{2}}$. Using Jensen's Inequality and recalling that $f$ is decreasing, we have that\n\n$$\nf(x) + f(y) \\geq 2f\\left(\\frac{x+y}{2}\\right) \\geq 2f\\left(\\sqrt{\\frac{x^2+y^2}{2}}\\right) = 2f\\left(\\frac{1}{\\sqrt{2}}\\right).\n$$\n\nNow, let $\\tan 2\\theta = \\frac{1}{\\sqrt{2}}$. Then $\\tan \\theta = \\sqrt{3} - \\sqrt{2}$ and $\\cot \\theta = \\sqrt{3} + \\sqrt{2}$, and so\n\n$$\n\\begin{aligned}\nf\\left(\\frac{1}{\\sqrt{2}}\\right) &= f(\\tan 2\\theta) \\\\\n&= \\tan^4 \\theta + \\cot^4 \\theta \\\\\n&= (\\sqrt{3} - \\sqrt{2})^4 + (\\sqrt{3} + \\sqrt{2})^4 \\\\\n&= 2(\\sqrt{3}^4 + 6(\\sqrt{2}\\sqrt{3})^2 + \\sqrt{2}^4) \\\\\n&= 98.\n\\end{aligned}\n$$\n\nHence $f(\\cos \\theta) + f(\\sin \\theta) \\geq 2 \\times 98 = 196$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16181,
    "subject": "Mathematics (Olympiad)",
    "question": "K\"art writes the fractions $\\frac{1}{2}$ and $\\frac{1}{3}$ on the blackboard. M\"art writes 10 positive integers on paper, which he does not show to K\"art. Then K\"art starts to write fractions on the blackboard by the following rule: on each step, she chooses two fractions $\\frac{a}{b}$ and $\\frac{c}{d}$ already on the blackboard and writes the fraction $\\frac{a+c}{b+d}$ (after reducing) on the blackboard. Can K\"art always choose the fractions so that, after a number of steps, she writes a fraction whose denominator is coprime with all the numbers M\"art has written?",
    "options": [],
    "answer": "See solution",
    "solution": "The first fraction that K\"art adds to the blackboard must be $\\frac{2}{5}$. On every following move, let K\"art pick $\\frac{1}{2}$ and the latest written fraction. Ignoring reduction, the denominator of each new fraction increases by 2, so the denominators are consecutive odd numbers. These fractions have the form $\\frac{k}{2k+1}$, and $k$ is always coprime with $2k+1$ because any common divisor would also divide $(2k+1) - 2k = 1$.\n\nTherefore, the denominators include all odd numbers, and in particular all odd primes except 2. Since there are infinitely many primes, K\"art will eventually write a fraction with a prime denominator larger than all numbers written by M\"art, and hence coprime with them.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16182,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $\\triangle ABC$ 的內心與外心分別為 $I$ 與 $O$。作直線 $L$ 使與 $BC$ 邊平行，並與 $\\triangle ABC$ 的內切圓相切。設 $L$ 與 $IO$ 交於 $X$ 點，另取 $L$ 上一點 $Y$ 使得 $YI$ 垂直於 $IO$。證明 $A, X, O, Y$ 四點共圓。\n\nLet $I$ and $O$ be the incenter and the circumcenter, respectively, of $\\triangle ABC$.\nDraw a straight line $L$ that is parallel to $BC$ and tangent to the incircle of $\\triangle ABC$.\nSuppose that $L$ and $IO$ intersect at the point $X$, and $Y$ is a point on $L$ such that $YI$ is perpendicular to $IO$. Prove that $A, X, O, Y$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "首先證明下述引理。\n\n*引理.* 設 $\\triangle ABC$ 的內心和外心分別為 $I$ 和 $O$。設過 $I$ 且垂直 $IO$ 的直線分別交 $BC$ 和 $\\angle BAC$ 的外角平分線於 $X$ 和 $Y$。則 $IY = 2IX$。\n\n*引理證明.* 設 $\\triangle ABC$ 相對於頂點 $A, B, C$ 的旁心分別為 $I_a, I_b, I_c$。\n\n以 $I$ 為位似中心做位似係數 $2$ 的位似變換，將 $A, B, C, X, O$ 分別變換至 $A', B', C', X', O'$。因為 $I$ 為 $\\triangle I_aI_bI_c$ 的垂心，$O$ 為 $\\triangle I_aI_bI_c$ 的九點圓圓心，所以 $O'$ 為 $\\triangle I_aI_bI_c$ 的外心。但是 $O'$ 同時也是 $\\triangle A'B'C'$ 的外心，且它們的外接圓半徑相等（皆為 $\\triangle ABC$ 外接圓半徑的兩倍），所以 $A', B', C', I_a, I_b, I_c$ 六點共圓。\n\n因為 $X$ 在 $BC$ 上，所以 $X'$ 在 $B'C'$ 上。又顯然有 $B', B, I, I_b$ 共線、$C', C, I, I_c$ 共線以及 $O', O, I$ 共線。\n\n考慮四邊形 $I_bB'C'I_c$，由蝴蝶定理有 $IY = IX'$。\n\n又 $IX' = 2IX$，所以 $IY = 2IX$，引理得證。\n\n回到原題。設 $\\angle BAC$ 的外角平分線與 $IY$ 交於 $P$ 點，$IY$ 交 $BC$ 於 $Q$ 點，$AI$ 分別交 $BC$ 和 $L$ 於 $S$ 和 $R$ 點。由引理有 $IP = 2IQ$，但是顯然有 $IY = IQ$，所以 $Y$ 為 $IP$ 的中點。\n\n而 $IP$ 為直角三角形 $PAI$ 的斜邊，所以 $IY = AY$，即 $\\triangle YAI$ 為等腰三角形，可得 $\\angle YAI = \\angle YIA$。\n\n所以\n\n$$\n\\begin{align*}\n\\angle YAO &= \\angle YAI - \\angle IAO = \\angle XIA - (90^\\circ - \\angle C - \\frac{\\angle A}{2}) \\\\\n&= \\angle C + \\frac{\\angle A}{2} - \\angle AIX = \\angle ASB - \\angle AIX \\\\\n&= \\angle YRI - \\angle AIX = \\angle YXO,\n\\end{align*}\n$$\n\n即 $A, X, O, Y$ 共圓，證畢。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16183,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle is inscribed in a hexagon $ABCDEF$ so that each side of the hexagon is tangent to the circle. Find the perimeter of the hexagon if $AB = 6$, $CD = 7$, and $EF = 8$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $AB$, $BC$, $CD$, $DE$, $EF$, $FA$ touch the circle at $U$, $V$, $W$, $X$, $Y$, $Z$ respectively.\n\n![](images/Australian-Scene-combined-2015_p57_data_8de92e8c4e.png)\n\nSince the two tangents from a point to a circle have equal length, $UB = BV$, $VC = CW$, $WD = DX$, $XE = EY$, $YF = FZ$, $ZA = AU$.\n\nThe perimeter of hexagon $ABCDEF$ is\n\n$$\n\\begin{align*}\n& AU + UB + BV + VC + CW + WD + DX + XE + EY + YF + FZ + ZA \\\\\n&= AU + UB + UB + CW + CW + WD + WD + EY + EY + YF + YF + AU \\\\\n&= 2(AU + UB + CW + WD + EY + YF) \\\\\n&= 2(AB + CD + EF) = 2(6 + 7 + 8) = 2(21) = \\mathbf{42}.\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16184,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ which satisfy\n$$\nf(f(x) - y) \\le x f(x) + f(y)\n$$\nfor all real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "$f(x) = 0$ for all $x \\in \\mathbb{R}$ is the only function satisfying (1).\n\nClearly the zero function satisfies (1). We thus proceed to show that it is the only such function.\n\nLet $f$ be a function satisfying (1) and let $a = f(0)$.\n\nSetting $x = 0$ in (1) yields $f(a - y) \\le f(y)$ for all $y \\in \\mathbb{R}$.\n\nSubstituting $y$ with $a - y$ yields $f(y) \\le f(a - y)$. Hence,\n$$\nf(a - y) = f(y), \\quad \\forall y \\in \\mathbb{R} \\quad (2)\n$$\nSubstituting $y = 0$ in (1) we get $f(f(x)) \\le x f(x) + a$.\n\nSubstituting $y = f(x)$ in (1) we get\n$$\na \\le x f(x) + f(f(x)) \\le x f(x) + x f(x) + a.\n$$\nHence,\n$$\nx f(x) \\ge 0, \\quad \\forall x \\in \\mathbb{R} \\quad (3)\n$$\nSubstituting $y = 2a$ and $y = 0$ in (2) we get\n$$\nf(-a) = f(2a) \\quad \\text{and} \\quad f(a) = a,\n$$\nrespectively.\n\nFrom (3) we deduce\n$$\n0 \\le 2a f(2a) = 2a f(-a) = -2(-a f(-a)) \\le 0.\n$$\nThus $a = 0$ or $f(-a) = f(2a) = 0$.\n\nAssuming $a \\ne 0$ implies $f(2a) = 0$. Setting $x = 2a$ in (1) yields $f(-y) \\le f(y)$ for all $y \\in \\mathbb{R}$. Furthermore, $f(y) = f(-(-y)) \\le f(-y)$, so $f(y) = f(-y)$ for all $y \\in \\mathbb{R}$. And thus $a = f(a) = f(-a) = f(2a) = 0$, contradicting the assumption that $a \\neq 0$.\n\nSo we must have $a = 0$.\n\nFrom (2) we obtain $f(y) = f(-y)$ for all $y \\in \\mathbb{R}$. For $y \\neq 0$ we have\n$$\n0 \\le -y f(-y) = -y^2 (f(y))^2 \\le 0.\n$$\nThus $f(y) = 0$ for all $y \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16185,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the inequality:\n\n$$\n\\frac{x^2 - |x - 1| - 4}{x - 4} \\geq 2x - 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $x \\neq 4$ and $|x-1| = x-1$ for $x > 1$, otherwise $|x-1| = 1-x$.\n\nIf $x > 1$ we get:\n\n$$\n\\frac{x^2 - 8x + 7}{x - 4} \\leq 0 \\iff \\frac{(x - 7)(x - 1)}{x - 4} \\leq 0 \\iff x \\in (4, 7]\n$$\n\nIf $x \\leq 1$ we get:\n\n$$\n\\frac{x^2 - 10x + 9}{x - 4} \\leq 0 \\iff \\frac{(x - 9)(x - 1)}{x - 4} \\leq 0 \\iff x \\in (-\\infty, 1]\n$$\n\nFinally, $x \\in (-\\infty, 1] \\cup (4, 7]$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16186,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle $P$ is tangent to both of two fixed circles. Then the locus of the center of $P$ cannot be:\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p44_data_f5c7a29963.png)\n\n(A)\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p44_data_9be0f31c35.png)\n\n(B)\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p44_data_9628a5b480.png)\n\n(C)\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p44_data_7b3475b696.png)\n\n(D)",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose the radii of the two fixed circles are $r_1$ and $r_2$ respectively, and $|O_1O_2| = 2c$. In general, the locus of the center of $P$ is given by two conic curves with $O_1, O_2$ as the foci, and eccentricities $\\frac{2c}{r_1 + r_2}$ and $\\frac{2c}{|r_1 - r_2|}$, respectively. \n\n- When $r_1 = r_2$, the perpendicular bisector of $O_1O_2$ is part of the locus. \n- When $c = 0$, the locus is two concentric circles.\n- When $r_1 = r_2$ and $r_1 + r_2 < 2c$, the locus is like (B).\n- When $0 < 2c < |r_1 - r_2|$, the locus is like (C).\n- When $r_1 \\neq r_2$ and $r_1 + r_2 < 2c$, the locus is like (D).\n\nSince the foci of the ellipse and hyperbola in (A) are not identical, the locus of the center of $P$ cannot be (A).\n\n**Answer:** (A)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16187,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all possible values of integer $k$ for which there exist positive integers $a$ and $b$ such that\n$$\n\\frac{b+1}{a} + \\frac{a+1}{b} = k.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $k$ be a possible value, and consider all pairs $(a, b)$ of positive integers satisfying\n$$\n\\frac{b+1}{a} + \\frac{a+1}{b} = k.\n$$\nChoose any $(a, b)$ such that $b$ is the smallest. The quadratic equation\n$$\nx^2 + (1 - kb)x + b^2 + b = 0\n$$\nhas an integer root $x = a$. Let $x = a'$ be the other root. Since $a + a' = kb - 1$, $a' \\in \\mathbb{Z}$, and\n$$\na \\cdot a' = b(b + 1),\n$$\nso $a' > 0$. Thus,\n$$\n\\frac{b+1}{a'} + \\frac{a'+1}{b} = k.\n$$\nBy minimality of $b$, $a \\geq b$, $a' \\geq b$, so one of $a$ and $a'$ equals $b$. Without loss of generality, let $a = b$, so\n$$\nk = 2 + \\frac{2}{b}.\n$$\nThus, $b$ divides $2$, i.e., $b = 1$ or $2$, giving $k = 4$ or $3$, respectively.\n\nIf $a = b = 1$, then $k = 4$; if $a = b = 2$, then $k = 3$.\n\nConsequently, the only possible values are $k = 3$ or $4$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16188,
    "subject": "Mathematics (Olympiad)",
    "question": "Anna und Berta spielen ein Spiel, bei dem sie abwechselnd Murmeln vom Tisch nehmen. Anna macht den ersten Zug. Wenn zu Beginn eines Zuges $n \\ge 1$ Murmeln am Tisch sind, dann nimmt die Spielerin, die am Zug ist, $k$ Murmeln weg, wobei $k \\ge 1$ entweder eine gerade Zahl mit $k \\le \\frac{n}{2}$ oder eine ungerade Zahl mit $\\frac{n}{2} \\le k \\le n$ ist. Eine Spielerin gewinnt das Spiel, wenn sie die letzte Murmel vom Tisch nimmt.\n\nMan bestimme die kleinste Zahl $N \\ge 100\\,000$, sodass Berta den Sieg erzwingen kann, falls anfangs genau $N$ Murmeln am Tisch liegen.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/zwf2017loesungen_p4_data_e4cd7f8512.png)\n\nBehauptung: Die Verlustsituationen sind jene Situationen mit $n = 2^a - 2$ Murmeln am Tisch für alle ganzen Zahlen $a \\ge 2$. Alle anderen Situationen sind Gewinnsituationen.\n\nBeweis: Mit Induktion über $n \\ge 1$. Für $n = 1$ gewinnt man, indem man die einzige verbleibende Murmel nimmt. Für $n = 2$ kann man nur $k = 1$ Murmeln nehmen, und dann gewinnt die Gegnerin im nächsten Zug.\n\nInduktionsschritt von $n-1$ auf $n$ für $n \\ge 3$:\n\n1. Falls $n$ ungerade ist, nimmt man alle $n$ Murmeln weg und gewinnt.\n\n2. Falls $n$ gerade, aber nicht von der Form $2^a - 2$ ist, so liegt $n$ zwischen zwei Zahlen dieser Form, also gibt es ein eindeutiges ganzzahliges $b$ mit $2^b - 2 < n < 2^{b+1} - 2$. Wegen $n \\ge 3$ gilt für dieses $b \\ge 2$. Daher sind alle drei Zahlen in der obigen Ungleichungskette gerade, und somit folgt sogar $2^b \\le n \\le 2^{b+1} - 4$. Laut Induktion ist $2^b - 2$ eine Verlustsituation, und man kann sie durch Wegnehmen von\n\n$$\nk = n - (2^b - 2) = n - \\frac{2^{b+1} - 4}{2} \\le n - \\frac{n}{2} = \\frac{n}{2}\n$$\n\nMurmeln der Gegnerin überlassen.\n\n3. Falls $n$ gerade und von der Form $n = 2^a - 2$ ist, kann die Spielerin der Gegnerin keine Verlustposition mit $2^b - 2$ Murmeln hinterlassen (wobei $b < a$ ist, weil mindestens eine Murmel weggenommen werden muss, und $b \\ge 2$ ist, weil nach einem legalen Zug für ein gerades $n$ mindestens eine Murmel übrig bleibt). Dazu müsste sie nämlich $k = (2^a - 2) - (2^b - 2) = 2^a - 2^b$ Murmeln wegnehmen. Wegen $b \\ge 2$ ist aber $k$ gerade und strikt größer als $\\frac{n}{2}$ wegen $2^a - 2^b \\ge 2^a - 2^{a-1} = 2^{a-1} > 2^{a-1} - 1 = \\frac{2^a - 2}{2} = \\frac{n}{2}$; unmöglich.\n\nLösung der Aufgabe: Berta kann also dann und nur dann den Sieg erzwingen, wenn $N$ von der Form $2^a - 2$ ist. Die kleinste Zahl $N \\ge 100\\,000$ von dieser Form ist $N = 2^{17} - 2 = 131\\,070$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16189,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a subset of $m$ elements of $\\{0, 1, 2, \\dots, 98\\}$, $m \\geq 3$, such that for any $x, y \\in S$, there exists $z \\in S$ with $x + y \\equiv 2z \\pmod{99}$. Find all possible values of $m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S = \\{s_1, s_2, \\dots, s_m\\}$. Since $S' = \\{0, s_2 - s_1, \\dots, s_m - s_1\\}$ also satisfies the hypothesis, we may assume without loss of generality that $0 \\in S$. For any $x, y \\in S$, $x + y \\equiv 2z \\pmod{99}$ for some $z \\in S$, so $z \\equiv \\frac{x + y}{2} \\pmod{99}$. Since $2$ and $99$ are coprime, division by $2$ is well-defined modulo $99$.\n\nTaking $y = 0$, for any $x \\in S$, $z \\equiv \\frac{x}{2} \\pmod{99}$ must be in $S$. Repeating this process, $S$ must be closed under division by $2$ modulo $99$. The only subsets of $\\{0,1,\\dots,98\\}$ closed under this operation are the subgroups of $\\mathbb{Z}_{99}$, i.e., the sets of the form $\\{0, d, 2d, \\dots, (\\frac{99}{d} - 1)d\\}$ for divisors $d$ of $99$.\n\nThe possible sizes $m$ are thus the divisors of $99$ greater than or equal to $3$: $3, 9, 11, 33, 99$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16190,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an $n \\times n$ square, what is the minimum number of tiles needed to cover all black squares, where the tiling process and the number of black squares depend on whether $n$ is odd ($n = 2m + 1$, $m \\ge 1$) or even ($n = 2m$, $m \\ge 1$)?\n\n![](images/RMC2012_p70_data_1dc7fbd715.png)",
    "options": [],
    "answer": "See solution",
    "solution": "For $n = 2m + 1$ (odd, $m \\ge 1$):\n- There are $m^2$ remaining black squares and $2m$ tiles used.\n- Each black square must be eliminated from one of the two tiles that cover it, breaking the tile into two and increasing the tile count by 1 per square.\n- Thus, the minimum number of tiles needed is $2m + m^2$.\n\nFor $n = 2m$ (even, $m \\ge 1$):\n- There are $m^2$ black squares and $2m$ tiles.\n- For the $2m-1$ border squares, removing the black square does not increase the tile count.\n- For the remaining $m^2 - (2m-1)$ black squares, breaking a tile increases the tile count by 1 each.\n- Therefore, the minimum number of tiles is $2m + (m^2 - (2m-1)) = m^2 + 1$.\n\n*Solution 3.* Inductive approach:\nLet $x_n$ be the minimum number of tiles needed.\n- Passing from $2m-1$ to $2m$ needs at least two additional tiles: $x_{2m} \\ge x_{2m-1} + 2$.\n- Passing from $2m$ to $2m+1$ needs at least $2m-1$ additional tiles: $x_{2m+1} \\ge x_{2m} + (2m-1)$.\n- It may be easier to prove $x_{2m+1} \\ge x_{2m-1} + (2m+1)$.\n\nA model can be constructed inductively:\n$$\nx_1 = 0, \\quad x_{2m} = x_{2m-1} + 2, \\quad x_{2m+1} = x_{2m} + (2m-1), \\quad m \\ge 1\n$$\nwhich leads to:\n$$\nx_{2m+1} = m^2 + 2m, \\quad x_{2m} = m^2 + 1, \\quad m \\ge 1\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16191,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 4$ be an integer. Define\n$$\nS = \\{(x, y, z) \\in \\mathbb{Z}^3 \\mid xyz = 0 \\text{ and } 0 \\le x, y, z < n\\}.\n$$\nLet $S'$ be the projection of all points in $S$ onto a plane $P$. Given that $|S'| > n^2$, determine the least possible value of $|S'|$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the projection onto the plane $x = y$. It is easy to see that $S'$ consists of $n^2 + n - 1$ points of the form $(\\frac{i}{2}, \\frac{i}{2}, j)$ and $(\\frac{n+k}{2}, \\frac{n+k}{2}, 0)$ for $0 \\le i, j < n$ and $0 \\le k \\le n - 2$.\n\nNext, we will prove that if $|S'| > n^2$, then $|S'| \\ge n^2 + n - 1$. Suppose for the sake of contradiction that $|S'| < n^2 + n - 1$ for a plane $P$.\n\nBy the pigeonhole principle, there exists a point in $S'$ which is the projection of at least $\\lceil \\frac{3n^2-3n+1}{n^2+n-2} \\rceil = \\lceil 2 + \\frac{n^2-5n+5}{n^2+n-2} \\rceil$ points of $S$. For $n \\ge 4$, there exists a point in $S'$ which is the projection of at least 3 points. The only 3 collinear points in $S$ belong to either the plane $x=0$, $y=0$, or $z=0$, hence $P$ must be perpendicular to at least one of these planes.\n\nWithout loss of generality, let $P$ be perpendicular to the plane $z=0$. Let $m$ be the number of points in $S'$ projected from points in $S$ with $z=1$. This is the same as the number of points in $S'$ projected from points in $S$ with $z=a$ for $a \\in \\{1, 2, \\dots, n-1\\}$, and is a lower bound for the number of points in $S'$ projected from points in $S$ with $z=0$. Notice that $m \\ge n$.\n\nWhen $m = n$, this implies that points of the form $(x, 0, 1)$ and $(0, y, 1)$ project to exactly $n$ points. This can happen only when the plane $P$ is parallel to either the plane $x = y$, the plane $x = 0$, or the plane $y = 0$. In the latter two cases, we would have $|S'| = n^2$, which is not allowed. In the first case, we have shown that $|S'| = n^2 + n - 1$.\n\nWhen $m > n$, we obtain $|S'| \\ge nm \\ge n^2 + n$, a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16192,
    "subject": "Mathematics (Olympiad)",
    "question": "In isosceles trapezoid $ABCD$, parallel bases $\\overline{AB}$ and $\\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD = BC = 333$. The angle bisectors of $\\angle A$ and $\\angle D$ meet at $P$, and the angle bisectors of $\\angle B$ and $\\angle C$ meet at $Q$. Find $PQ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let point $K$ lie on $\\overline{BC}$ so that $\\overline{QK} \\parallel \\overline{AB}$. Because $\\angle KBQ = \\angle QBA = \\angle BQK$, it follows that $\\triangle KBQ$ is isosceles with $BK = QK$. Similarly, $\\triangle KCQ$ is isosceles with $CK = QK$. Hence $BK = QK = CK$. Analogously, let point $J$ lie on $\\overline{AD}$ so that $\\overline{PJ} \\parallel \\overline{AB}$. Then $J$ is the midpoint of $\\overline{AD}$, and $AJ = PJ = DJ$. Thus $P$ and $Q$ lie on line $KJ$, which is the midline of the trapezoid with $\\overline{KJ} \\parallel \\overline{AB}$, and $KJ = \\frac{AB+CD}{2} = 575$.\n\n![](images/2022AIME_I_Solutions_p1_data_b9943598e4.png)\n\nThen\n$$\nPQ = KJ - KQ - PJ = KJ - 2 \\cdot PJ = KJ - AD = 575 - 333 = 242.\n$$\n\nOR\n\nLet $E$ be the intersection of lines $AB$ and $DP$, and let $F$ be the intersection of lines $CD$ and $AP$. Let $G$, $H$, and $I$ be the projections of points $A$, $P$, and $Q$ onto line $CD$, respectively.\n\n![](images/2022AIME_I_Solutions_p2_data_50ed391fe9.png)\n\nBecause angles $\\angle BAD$ and $\\angle CDA$ are supplementary, it follows that angles $\\angle PAD$ and $\\angle PDA$ are complementary and $\\overline{AF} \\perp \\overline{DE}$. Triangle $\\triangle FDA$ is isosceles because $\\overline{PD}$ is both an altitude and an angle bisector of $\\triangle FDA$. Similarly, $\\triangle EAD$ is isosceles, and thus $AE = AD = DF = 333$. Because $ABCD$ is an isosceles trapezoid, $DG = \\frac{1}{2}(CD - AB) = 75$. Because $P$ is the midpoint of $\\overline{AF}$, $H$ is the midpoint of $\\overline{GF}$, and thus\n\n$$\nGH = \\frac{1}{2} \\cdot GF = \\frac{1}{2}(DF - DG) = \\frac{1}{2}(333 - 75) = 129\n$$\n\nand $DH = DG + GH = 204$. By symmetry, $CI = DH = 204$. The requested length is therefore\n\n$$\nPQ = CD - CI - DH = 650 - 2 \\cdot 204 = 242.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16193,
    "subject": "Mathematics (Olympiad)",
    "question": "Given that $2^n + 7^n \\equiv 1 \\pmod{p}$, where $p$ is a prime, prove that $13$ divides $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $2^n + 7^n \\equiv 1 \\pmod{p}$, $n$ must be odd. Let $p$ be the least prime divisor of $n$. Thus, $(p, 6) = 1$, so there exist integers $x, y$ such that $px + 6y = 1$, i.e., $6y \\equiv 1 \\pmod{p}$. Let $a \\equiv 7y \\pmod{p}$. Then $a^n + 1 \\equiv (7y)^n + (6y)^n \\pmod{p}$, and $7^n + 6^n \\equiv (7y)^n + (6y)^n \\pmod{p}$. Thus, $a^n + 1 \\equiv 0 \\pmod{p}$, so $a^n \\equiv -1 \\pmod{p}$. Since $n$ is odd, $(-a)^n \\equiv 1 \\pmod{p}$. By Fermat's Little Theorem, $(-a)^{p-1} \\equiv 1 \\pmod{p}$, so $(-a)^{\\gcd(n, p-1)} \\equiv 1 \\pmod{p}$. Since $p$ is the least prime divisor of $n$, $\\gcd(n, p-1) = 1$, so $-a \\equiv 1 \\pmod{p}$, i.e., $a \\equiv -1 \\pmod{p}$. Thus, $7y \\equiv -1 \\pmod{p}$, so $-7 \\cdot 6y \\equiv -7 \\pmod{p}$. But $6y \\equiv 1 \\pmod{p}$, so $-7 \\equiv 6 \\pmod{p}$, or $13 \\equiv 0 \\pmod{p}$. Therefore, $13$ divides $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16194,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral. Let $K$, $L$, $M$, and $N$ be the midpoints of the sides $AB$, $BC$, $CD$, and $DA$, respectively. Let $O$ be the intersection point of all four segments $AC$, $BD$, $KM$, and $LN$. Is it true that if one of the diagonals and the segments joining midpoints of opposite sides are concurrent, then $ABCD$ must be a parallelogram?\n\n![alt](images/Belorusija_2011_p25_data_db1dc78acc.png)",
    "options": [],
    "answer": "See solution",
    "solution": "a) If $ABCD$ is a parallelogram, then $KLMN$ is a parallelogram, and the intersection point $O$ of $AC$, $BD$, $KM$, and $LN$ is the midpoint of both diagonals. This follows from the properties of midlines and Thales' theorem: $KL \\parallel AC$, $MN \\parallel AC$, and $KL = MN = 0.5\\,AC$, so $KLMN$ is a parallelogram. The intersection points of the diagonals and midlines divide the diagonals into equal segments, confirming $O$ is the midpoint of both $AC$ and $BD$.\n\nb) However, it is not always true. Consider a triangle $\\triangle ABC$ and mark point $D$ on the extension of its median $BH$ so that $DH \\neq BH$. The quadrilateral $ABCD$ is not a parallelogram. Let $K$, $L$, $M$, and $N$ be the midpoints of $AB$, $BC$, $CD$, and $DA$, respectively. The intersection points $P$ and $Q$ of diagonal $BD$ with $KL$ and $MN$ are midpoints of those segments, and $KLMN$ is a parallelogram. Thus, $KM$, $LN$, and $PQ$ are concurrent, and $PQ$ lies on diagonal $BD$. Therefore, the diagonal $BD$ and the segments joining midpoints of opposite sides are concurrent, but $ABCD$ is not a parallelogram.\n\n![alt](images/Belorusija_2011_p26_data_df0456bd9b.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16195,
    "subject": "Mathematics (Olympiad)",
    "question": "A positive number is assigned to each of $n \\ge 2$ points on the plane in such a way that the distance between any two points equals the sum of the numbers assigned to these points. Find all possible values of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We begin with an example for $n = 4$: take three points to be the vertices of an equilateral triangle with side lengths $1$ and assign $\\frac{1}{2}$ to each of them. The fourth point will be located at the center of the triangle with the assigned number equal to $\\frac{1}{\\sqrt{3}} - \\frac{1}{2}$. It is easy to check that this configuration satisfies the required condition. Removing a point or two, we obtain a desired configuration for $n = 3$ or $n = 2$.\n\nIt remains to show that the condition cannot be fulfilled for $n \\ge 5$.\n\nAssuming that such a configuration exists, we first show that no three points are collinear.\n\n![](images/Ukrajina_2013_p20_data_3272effe56.png)\n\nLet $A$, $B$, $C$ be three collinear points, as shown in the figure above, with assigned numbers $a$, $b$, $c$, respectively. If $AB = x$, $BC = y$, then $AC = x + y$. We must have the following equalities: $a + b = x$, $b + c = y$ and $a + c = x + y$, which implies $a + 2b + c = x + y$, so $b = 0$, which is a contradiction.\n\nNext, we show that no four points are the vertices of a convex quadrilateral.\n\nIf points $A$, $B$, $C$, $D$ with assigned numbers $a$, $b$, $c$, $d$ form a convex quadrilateral, then\n\n$$\nAC + BD = a + b + c + d = AD + BC.\n$$\n\nBut the triangle inequality written for $\\triangle AOD$ and $\\triangle BOC$ implies that\n\n$$\nAC + BD = AO + OC + DO + OB = AO + OD + BO + CO > AD + BC,\n$$\n\nleading to a contradiction.\n\nSo, one of any four points is located inside the triangle formed by the remaining three points. The figure below illustrates a point $O$ inside triangle $ABC$. It is clear that our configuration cannot have any other points. Indeed, the lines $AO$, $BO$, $CO$ partition the triangle $ABC$ into six triangles. If a fifth point of the configuration, say $D$, belongs, for example, to the triangle $AOB_1$, then the quadrilateral $AOB$ will be convex. All other locations of $D$ can be treated similarly.\n\n![](images/Ukrajina_2013_p21_data_1db58be452.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16196,
    "subject": "Mathematics (Olympiad)",
    "question": "A positive integer is called _typical_ if the sum of its decimal digits is a multiple of $2011$.\n\n(a) Show that there are infinitely many typical numbers, each having at least $2011$ multiples which are also typical numbers.\n\n(b) Does there exist a positive integer such that each of its multiples is typical?",
    "options": [],
    "answer": "See solution",
    "solution": "**(a)** All numbers consisting of $2011n$ digits $1$ (where $n$ is a positive integer) are typical. Each such number, with $2011k$ digits, has among its multiples the numbers made up of $2011kn$ digits, which are also typical.\n\n**(b)** We will show that the answer is **NO**.\n\nNotice that each positive integer $A$ has a multiple of the form $\\overline{99\\ldots900\\ldots0}$. Indeed, if the decimal expansion of $1/A$ has a period of $p$ digits and the initial non-periodic part has $q$ digits, then $\\frac{1}{A} = \\frac{B}{C}$, where $C = \\overline{99\\ldots900\\ldots0}$ has $p$ digits $9$ and $q$ digits $0$.\n\nSuppose $n$ is a positive integer. Take a multiple of $n$ of the form $m = \\overline{99\\ldots900\\ldots0}$, with $p$ nines and $q$ zeros, and consider the multiple of $n$\n\n$$\nM = (10^{p+1} - 1)m = 10^{p+1}m - m = \\overline{99\\ldots98900\\ldots0100\\ldots0},\n$$\n\nwith $p-1$ nines at the beginning. Then the sums of the digits of these numbers are $s(M) = s(m) + 9$, so they cannot both be multiples of $2011$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16197,
    "subject": "Mathematics (Olympiad)",
    "question": "Find three distinct positive integers which minimize their sum under the condition that any two of them add up to a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "$(6, 19, 30)$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16198,
    "subject": "Mathematics (Olympiad)",
    "question": "Each of 27 bricks (right rectangular prisms) has dimensions $a \\times b \\times c$, where $a$, $b$, and $c$ are pairwise relatively prime positive integers. These bricks are arranged to form a $3 \\times 3 \\times 3$ block, as shown on the left below. A 28th brick with the same dimensions is introduced, and these bricks are reconfigured into a $2 \\times 2 \\times 7$ block, shown on the right. The new block is 1 unit taller, 1 unit wider, and 1 unit deeper than the old one. What is $a + b + c$?\n\n![](images/2024_AMC10B_Solutions_p14_data_41c5ba48be.png)\n\n![](images/2024_AMC10B_Solutions_p14_data_65553434fd.png)\n\n(A) 88 (B) 89 (C) 90 (D) 91 (E) 92",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer (E):** Without loss of generality, assume $a < b < c$. Comparing the figures and considering the change in orientation gives rise to the equations $3a + 1 = 2b$, $3b + 1 = 2c$, and $3c + 1 = 7a$. To solve this system of linear equations, use the first two equations to write $a$ and $c$ in terms of $b$, namely $a = \\frac{2}{3}b - \\frac{1}{3}$ and $c = \\frac{3}{2}b + \\frac{1}{2}$. Substituting these into the third equation gives $\\frac{9}{2}b + \\frac{3}{2} + 1 = \\frac{14}{3}b - \\frac{7}{3}$. Multiplying both sides by 6 yields $27b + 9 + 6 = 28b - 14$, which shows that $b = 29$. Back substituting then gives $a = 19$ and $c = 44$. The requested sum is $19 + 29 + 44 = 92$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16199,
    "subject": "Mathematics (Olympiad)",
    "question": "ABC гурвалжны $\\angle BAC = 90^\\circ$ ба $AB = AC$ байг. $M$ ба $N$ цэгүүд нь $BC$ тал дээр орших бөгөөд $N$ нь $M$ ба $C$ цэгүүдийн хооронд байрлана. Хэрэв $BM^2 + NC^2 = MN^2$ нөхцөл биелэдэг бол $\\angle MAN$-г ол.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\begin{align*}\n\\angle MAN &= 45^\\circ \\text{ гэдгийг баталья.} \\\\\n\\angle ACB &= \\angle ABC = 45^\\circ. \\Delta AMN \\text{-ийн хувьд косинусын теорем бичвэл} \\\\\nMN^2 &= AM^2 + AN^2 - 2AM \\cdot AN \\cos \\varphi \\\\\n\\text{ба } \\Delta AMB \\text{-ийн хувьд косинусын} \\\\\n\\text{теорем бичвэл } \\angle ABM &= 45^\\circ \\text{ тул.}\n\\end{align*}\n$$\n\n![](images/2013-ilovepdf-compressed_p32_data_45cb1a0458.png)\n\n$AM^2 = BM^2 + AB^2 - BM \\cdot AB \\sqrt{2}$ (1) болно. $\\Delta ACN$-ийн хувьд косинусын теорем бичвэл $AN^2 = NC^2 + AC^2 - NC \\cdot AC \\sqrt{2}$ (2) $\\Delta AMN$-ийн хувьд $\\cos \\varphi = \\frac{AM^2 + AN^2 - MN^2}{2AN \\cdot AM}$ болох ба (1), (2)-ийг орлуулбал\n\n$$\n\\begin{aligned}\n\\cos \\varphi &= \\frac{BM^2 + AB^2 - BM \\cdot AB \\sqrt{2}}{2AM \\cdot AN} + \\\\\n&\\quad \\frac{NC^2 + AC^2 - NC \\cdot AC \\sqrt{2} - MN^2}{2AM \\cdot AN} \\\\\n&= \\frac{2AB^2 - BM \\cdot AB \\sqrt{2} - NC \\cdot AC \\sqrt{2}}{2AM \\cdot AN}\n\\end{aligned}\n$$\n\nболно. Нөгөө талаас\n\n$$\n\\begin{aligned}\nSAM_N &= S_{ABC} - S_{CNA} - S_{AMB} \\\\\n&= \\frac{2AB^2 - BM \\cdot AB \\sqrt{2} - NC \\cdot AC \\sqrt{2}}{4} \\text{ болно. Мөн}\n\\end{aligned}\n$$\n\n$$\nSAM_N = \\frac{AN \\cdot AM \\sin \\varphi}{2} \\Rightarrow\n$$\n\n$$\n\\begin{aligned}\n\\sin \\varphi &= \\frac{2SAM_N}{AN \\cdot AM} = \\frac{2AB^2 - BM \\cdot AB \\sqrt{2} - NC \\cdot AC \\sqrt{2}}{2AN \\cdot AM} = \\cos \\varphi \\\\\n\\text{болж } \\varphi &= 45^\\circ \\text{ болно. (зураг1)}\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16200,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $D$ and $E$ are on sides $BC$ and $AC$ of $\\triangle ABC$. Lines $AD$ and $BE$ intersect at point $S$. Point $F$ is on side $AB$, and lines $FE$ and $FD$ intersect line $l$ passing through $C$ and parallel to $AB$ at points $P$ and $Q$. Prove that if $CP = CQ$, then the points $C$, $S$, and $F$ lie on the same line.",
    "options": [],
    "answer": "See solution",
    "solution": "From the similarities $\\triangle PEC \\sim \\triangle FEA$ and $\\triangle CDQ \\sim \\triangle BDF$, we obtain that\n\n$$\n\\frac{CP}{AF} = \\frac{CE}{AE} \\quad \\text{and} \\quad \\frac{CQ}{BF} = \\frac{CD}{BD}.\n$$\n\nTherefore,\n$$\n\\frac{CP}{CQ} = \\frac{CE}{AE} \\cdot \\frac{AF}{BF} \\cdot \\frac{BD}{DC}.\n$$\nBy the condition $CP = CQ$, it follows that\n$$\n\\frac{CE}{AE} \\cdot \\frac{AF}{BF} \\cdot \\frac{BD}{DC} = 1.\n$$\n\nApplying Ceva's theorem for $\\triangle ABC$ and the points $F$, $D$, and $E$, it follows that the lines $AD$, $BE$, and $CF$ intersect in one point, i.e., point $S$ lies on $CF$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16201,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $r$ be a rational number in the interval $[-1, 1]$, and let $\\theta = \\cos^{-1} r$. Call a subset $S$ of the plane *good* if $S$ is unchanged upon rotation by $\\theta$ around any point of $S$ (in both clockwise and counterclockwise directions). Determine all values of $r$ satisfying the following property: The midpoint of any two points in a good set also lies in the set.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim the answer is $r = 1 - \\frac{1}{4k}$ for all positive integers $k$.\n\nSuppose $A$ and $B$ are points in $S$. Place $A$ and $B$ on the complex plane such that $A = 0$ and $B = 1$, and let $\\omega = \\cos\\theta + i\\sin\\theta = r + i\\sqrt{1-r^2}$. We claim that any good set containing $A$ and $B$ contains the good set $T$ of numbers of the form $f(\\omega)$, where $f(x)$ is a Laurent polynomial—that is, $x^n f(x)$ is a polynomial for some (possibly negative) integer $n$—with integer coefficients, and $f(1) = 0$ or $1$.\n\nWe first show that $T$ is good. Indeed, for any $p \\in T$, the rotation $R_p$ about $p$ by $\\theta$ sends $z \\mapsto R_p(z) = (z-p)\\omega + p$. If $p = f(\\omega)$ and $z = g(\\omega) \\in T$, then $h(x) = (g(x)-f(x)) \\cdot x + f(x)$ satisfies $R_p(z) = h(\\omega)$ and $h(1) = g(1) = 0$ or $1$, so $R_p(z) \\in T$. Similarly, $R_p^{-1}(z) = (z-p) \\cdot \\omega^{-1} + p \\in T$, so $T$ is good.\n\nNext we show that any good set $S$ containing $0$ and $1$ contains all of $T$. Let $f(x)$ be any Laurent polynomial with integer coefficients with $f(1) = 0$ or $1$. We will show that $f(\\omega) \\in S$. We induct on the sum $s$ of the absolute value of the coefficients of $f(x)$. The case $s=0$ is trivial. For $s=1$, $f(\\omega)$ must be a power of $\\omega$, which is of the form $R_0^k(1) \\in S$. For $s \\ge 2$, $f(x)$ must have at least one positive and one negative coefficient, so write $f(x) = x^a - x^b + g(x)$, where the sum of the absolute values of the coefficients of $g(x)$ is $s-2$. Since $g(1) = f(1)$, $g(\\omega) \\in S$ by induction. Then\n\n$$\nx^{-a}f(x) = (x^{-b}g(x) - 1) x^{b-a} + 1,\n$$\n\nso $f(\\omega) = R_0^a \\circ R_1^{b-a} \\circ R_0^{-b}(g(\\omega)) \\in S$, completing the induction.\n\nIt follows that $r$ satisfies the desired condition if and only if we can write $\\frac{1}{2} = f(\\omega) \\in T$. Then there is a polynomial $g(x) = x^n(2f(x)-1)$, all but one of whose coefficients are even, that has $\\omega$ as a root. If $r = \\pm 1$, so $\\omega = \\pm 1$, this is clearly impossible. Otherwise, let $r = \\frac{a}{b}$ with $a$ and $b$ relatively prime. Then the minimal polynomial $p(x)$ of $\\omega$ over the integers is $bx^2 - 2ax + b$ if $b$ is odd and $\\frac{b}{2}x^2 - ax + \\frac{b}{2}$ if $b$ is even. This minimal polynomial must divide $g(x)$. But since $g(x)$ is congruent to a power of $x$ modulo 2, $p(x)$ must also be congruent to a power of $x$ modulo 2. The only possibility is for $b$ to be a multiple of 4 (and $a$ to be odd).\n\nFinally, since $g(1) = 2f(1) - 1 = \\pm 1$ and $p(1) | g(1)$, we must have that $p(1) = \\pm 1$ as well. Since $p(1) = \\frac{b}{2} - a + \\frac{b}{2} = b - a$, we must have $b = a \\pm 1$. Since $|r| \\le 1$, $|a| \\le |b|$. We find then that the only possibilities are $r = 1 - \\frac{1}{4k}$, where we can let $a = 4k-1$ and $b = 4k$, so $p(1) = 1$. Then letting $f(x) = \\frac{1}{2}(x^{-1}p(x) + 1) = kx - (2k-1) + kx^{-1}$, we find that $f(1) = 1$, so $\\frac{1}{2} = f(\\omega) \\in T$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16202,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral. Circle $\\omega$ is an *excircle* of quadrilateral $ABCD$ opposite $B$; that is, it is tangent to ray $BA$ (beyond $A$) at $U_A$, ray $BC$ (beyond $C$) at $U_C$, ray $AD$ (beyond $D$) at $V_A$, and ray $CD$ (beyond $D$) at $V_C$. Prove that $BA + AD = BC + CD$.\n\n![](images/pamphlet0910_main_p72_data_43ea9429bf.png)",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof.* Extend $AV_A$ through $V_A$ to meet ray $BC$ at $A_3$, and extend segment $CV_C$ through $V_C$ to meet $BA$ at $C_3$. By the convexity of $ABCD$, it is not difficult to reduce our configuration to the figure shown above. By equal tangents, we have\n\n$$\nBU_A = BU_C, \\quad AU_A = AV_A, \\quad C_3U_A = C_3V_C, \\quad CU_C = CV_C, \\quad A_3U_C = A_3V_A, \\quad DV_A = DV_C.\n$$\n\nIt follows that\n\n$$\nBA + AD = BU_A - U_AA + AD = BU_A - V_AA + AD = BU_A - DV_A.\n$$\n\nSimilarly, we can show that $BC + CD = BU_C - DV_C$. Thus, $BA + AD = BC + CD$. $\\square$\n\n![](images/pamphlet0910_main_p73_data_829f96577a.png)\n\n![](images/pamphlet0910_main_p73_data_fea2b82e22.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16203,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with incenter $I$. A point $P$ in the interior of the triangle satisfies\n\n$$\n\\angle PBA + \\angle PCA = \\angle PBC + \\angle PCB.\n$$\n\nShow that $AP \\geq AI$, and that equality holds if and only if $P = I$.",
    "options": [],
    "answer": "See solution",
    "solution": "We begin by proving a well-known fact.\n\n![](images/USA_IMO_2006-2007_p59_data_ff6c42a619.png)\n\n**Lemma.** Let $ABC$ be a triangle with circumcenter $O$, circumcircle $\\gamma$, and incenter $I$. Let $M$ be the second intersection of line $AI$ with $\\gamma$. Then $M$ is the circumcenter of triangle $IBC$.\n\n*Proof:* Let $\\angle A = 2\\alpha$, $\\angle B = 2\\beta$. Note that $M$ is on the opposite side of line $BC$ as $A$. We have $\\angle CBM = \\angle CAM = \\alpha$, so $\\angle IBM = \\angle IBC + \\angle CBM = \\beta + \\alpha$. Also, $\\angle BIM = \\angle BAI + \\angle ABI = \\alpha + \\beta$. Thus, triangle $IBM$ is isosceles with $BM = IM$. Similarly, $CM = IM$. This proves the claim.\n\nBack to our current problem, we note that\n\n$$\n(\\angle PBA + \\angle PCA) + (\\angle PBC + \\angle PCB) = \\angle B + \\angle C,\n$$\nso\n$$\n\\angle PBA + \\angle PCA = \\angle PBC + \\angle PCB = \\frac{1}{2}(\\angle B + \\angle C).\n$$\n\nIn triangle $PBC$, we have\n$$\n\\angle BPC = 180^\\circ - (\\angle PBC + \\angle PCB) = 180^\\circ - \\frac{1}{2}(\\angle B + \\angle C).\n$$\n\nIt is clear that $\\angle IBC + \\angle ICB = \\frac{1}{2}(\\angle B + \\angle C)$, and so in triangle $BCI$,\n$$\n\\angle BIC = 180^\\circ - \\frac{1}{2}(\\angle B + \\angle C).\n$$\n\nWe conclude that $\\angle BPC = \\angle BIC$; that is, points $B$, $C$, $I$, and $P$ lie on a circle. By the Lemma, they all lie on a circle centered at $M$. In particular, we have $MP = MI$.\n\nIn triangle $APM$, we have\n$$\nAI + IM = AM \\leq AP + PM = AP + IM,\n$$\nimplying that $AI \\leq AP$. Equality holds if and only if $AM = AP + PM$; that is, $A$, $P$, and $M$ are collinear, or $P = I$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16204,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a quadrilateral circumscribed around a circle with center $O$. Construct equal segments $AA_1$, $AA_2$, $CC_1$, and $CC_2$ on rays $AB$, $AD$, $CB$, and $CD$ respectively, so that their lengths are greater than that of any side of $ABCD$, and $A_1$, $A_2$, $C_1$, and $C_2$ are located on the circle with center $O$. Prove that lines $A_1A_2$, $C_1C_2$, and $BD$ either intersect at one point or are parallel.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K$ and $P$ be the points of tangency of the inscribed circle of $ABCD$ to $AB$ and $BC$, respectively.\n\nIt follows that $\\triangle C_1PO = \\triangle A_1KO$, since they are right triangles with equal legs $KO = PO$ and hypotenuses $C_1O = A_1O$ (as radii of the respective circles). Thus, $C_1P = A_1K$. Since $BK = BP$ (as two tangent segments from the same external point), we have $C_1B = A_1B$. Similarly, $C_2D = A_2D$.\n\nMoreover, from the discussion above, $AB = BC$ and $AD = CD$. Therefore, $ABCD$ is a kite, whose diagonal $BD$ is a line of symmetry; hence, $B$, $D$, and $O$ lie on the same line.\n\nLine $BD$ passes through points $B$ and $O$. These points are equidistant from the endpoints of segment $A_1C_1$, so $B$ and $O$ belong to its perpendicular bisector. Similarly, line $BD$ is a perpendicular bisector of $A_2C_2$. Since $BD \\perp A_1C_1$ and $BD \\perp A_2C_2$, then $A_1C_1 \\parallel A_2C_2$.\n\nNext, consider two cases:\n\nIf $A_1A_2 \\parallel C_1C_2$, then quadrilateral $A_1A_2C_2C_1$ is a parallelogram, so $A_1A_2C_2C_1$ is a rectangle. All three lines $A_1A_2$, $C_1C_2$, and $BD$ are perpendicular to $A_1C_1$, hence they are parallel.\n\nIf $A_1A_2$ and $C_1C_2$ are not parallel, then quadrilateral $A_1A_2C_2C_1$ is a trapezoid. Since line $BD$ passes through the midpoints of the trapezoid's bases, it contains the intersection point of the extensions of legs $A_1A_2$ and $C_1C_2$. Hence, lines $A_1A_2$, $C_1C_2$, and $BD$ intersect at one point.\n\n![](images/Ukraine_2020_booklet_p48_data_2306bfcb32.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16205,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an integer greater than or equal to $2$. Assign to each vertex of a regular $2n$-gon a distinct number chosen from $\\{1, 2, \\dots, 2n\\}$.\n\n1. Show that there exists a method of assigning these numbers in such a way that the differences of the numbers assigned to every neighboring pair of vertices are all greater than or equal to $n-1$.\n\n2. Show that there is no way of assigning these numbers so that the differences of the numbers assigned to every neighboring pair of vertices are all greater than or equal to $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "1. Pick a vertex $P$ of the given $2n$-gon and label the vertices consecutively as $1, 2, \\dots, 2n$ starting with $P$ as $1$ and going around clockwise. Then, for $1 \\leq k \\leq n$, reassign the number $k$ to the vertex labeled $2k-1$, and the number $n+k$ to the vertex labeled $2k$. Let's check that this reassignment satisfies the requirement: for each neighboring pair of vertices, the difference of the numbers assigned is at least $n-1$. If, for $1 \\leq j < 2n$, we compare the numbers reassigned to the vertices originally labeled $j$ and $j+1$, the difference is $n$ if $j$ is odd, and $n-1$ if $j$ is even. The difference between the numbers assigned to the vertices originally labeled $1$ and $2n$ is $2n-1$. Thus, the requirement is satisfied.\n\n2. Note that $k=2n$ is the only integer which satisfies $1 \\leq k \\leq 2n$ and $|k-n| \\geq n$. Hence, if we assign the number $n$ to some vertex originally labeled $j$, it is impossible to assign numbers to both of the two neighboring vertices (originally labeled $j-1$ and $j+1$) so that the difference for each neighboring pair is $n$ or greater.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16206,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be a natural number. In Laura's class, the mathematics teacher formed new groups in every statistics lesson. After covering the entire statistics curriculum, it was found that:\n\n- Every two different students had belonged together in exactly one group.\n- Every two different groups had exactly one student in common.\n\nOn the day when correlation was studied, Laura's group contained precisely $n$ students besides Laura. How many students were in Laura's class, given that their number was larger than $n + 2$?\n\nAnswer: $n^2 + n + 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "By assumption, one group contained $n + 1$ students, and the whole class contains more than $n + 2$ students. Since every two students belonged together in exactly one group, there must be more than one group with more than one member.\n\nSuppose a group contained only one student $C$. Then $C$ would have to belong to every group, and every other student would belong to exactly one group (the one with $C$). This would mean only one group with more than one student, which is not possible. Thus, all groups have more than one member.\n\n**Case 1:** Suppose there were two groups such that every student belonged to at least one of them. Let $C$ be the student in both groups, and let the groups be $\\{C, A_1, \\dots, A_k\\}$ and $\\{C, B_1, \\dots, B_l\\}$. Since every pair of students is together in exactly one group, the other groups must be of the form $\\{A_i, B_j\\}$ for $i = 1, \\dots, k$ and $j = 1, \\dots, l$. All such groups must occur, and since groups must pairwise share one student, either $k=1$ or $l=1$. Without loss of generality, let $k=1$. Laura's group on the day of correlation studies contained $n+1$ students, and since $n+1 > n \\ge 2$, this must be the large group with $l+1$ students. Thus $l=n$, and the total number of students is $n+2$. But the problem states the total number is greater than this.\n\n**Case 2:** Suppose there are no two groups such that every student belonged to at least one of them. We show that every group has the same number of students. Let $\\mathcal{R}$ and $\\mathcal{S}$ be arbitrary groups, and let $C$ be a student in neither. Let $m$ be the number of groups containing $C$. Each of these $m$ groups has exactly one student in common with $\\mathcal{R}$; all these students are distinct, as $C$ is the only common member of any two groups containing $C$. There can be no more students in $\\mathcal{R}$, because every student in $\\mathcal{R}$ must belong to some group with $C$. Hence, $\\mathcal{R}$ contains exactly $m$ students. Similarly, $\\mathcal{S}$ contains $m$ students. Since $\\mathcal{R}$ and $\\mathcal{S}$ were arbitrary, every group contains exactly $m$ students. On the day of studying correlation, Laura's group had $n+1$ students, so $m = n+1$.\n\nTo find the class size, consider $C$'s groups: there are $n+1$ of them, each with $n$ students besides $C$, so there are $1 + (n+1)n = n^2 + n + 1$ students in total. This is more than $n+2$ and fits the problem's requirements.\n\n*Remark.* The conditions can be fulfilled. For $n=2$, with 7 students $A, B, C, D, E, F, G$, define 7 groups: $\\{A, B, C\\}$, $\\{A, D, E\\}$, $\\{A, F, G\\}$, $\\{B, D, F\\}$, $\\{B, E, G\\}$, $\\{C, D, G\\}$, $\\{C, E, F\\}$. Calling each student a 'point' and each group a 'line', such a structure with $n^2+n+1$ students is called a finite projective plane of order $n$. The existence of a finite projective plane is known for $n$ that are powers of a prime number. For most non-prime-powers, existence is unknown. For some orders, there are several non-isomorphic projective planes.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16207,
    "subject": "Mathematics (Olympiad)",
    "question": "設三角形 $ABC$ 的外心為 $O$，垂心為 $H$，並且 $OH$ 與 $BC$ 平行。設 $AH$ 與三角形 $ABC$ 的外接圓再交於點 $X$（$X \\neq A$），並設 $XB, XC$ 分別與 $OH$ 交於點 $Y, Z$。令 $P$ 為 $Y$ 對 $AB$ 的投影點，$Q$ 為 $Z$ 對 $AC$ 的投影點。證明 $PQ$ 平分線段 $BC$。",
    "options": [],
    "answer": "See solution",
    "solution": "關鍵引理如下：\n\n**引理**：設 $ABC$ 為一三角形，外心為 $O$。令 $\\ell$ 為通過 $O$ 的任意直線。設 $Y, Z$ 為 $\\ell$ 上兩點，使得 $AY, AZ$ 分別垂直於 $AC, AB$。令 $P, Q$ 分別為 $Y, Z$ 在 $AB, AC$ 上的投影。則 $PQ$ 平分 $AB$。\n\n*引理證明*：設 $M_a, M_b, M_c$ 分別為 $BC, CA, AB$ 的中點。則（用有號線段表示）：\n\n$$\n\\frac{PM_c}{M_cA} = \\frac{YO}{OZ} = \\frac{AM_b}{M_bQ}\n$$\n\n因為 $PY, M_cO, AZ$ 平行，且 $AY, M_bO, QZ$ 平行。因此\n\n$$\n\\frac{BP}{PA} = \\frac{M_cP - M_cB}{PM_c + M_cA} = \\frac{M_cA - PM_c}{PM_c + M_cA} = \\frac{AM_b - M_bQ}{AM_b + M_bQ} = -\\frac{CQ}{QA}.\n$$\n\n由 Menelaus 定理可知 $PQ$ 平分 $AB$。$\\Box$\n\n*引理的另一證法*：仍有 $PM_c/M_cA = AM_b/M_bQ$。設 $PM_a$ 與 $AC$ 交於 $Q'$，$QM_a$ 與 $AB$ 交於 $P'$。由於 $M_aM_c$ 平行於 $AC$，有 $PM_c/M_cA = PM_a/M_aQ'$。同理 $AM_b/M_bQ = P'M_a/M_aQ$。這說明 $PM_aQ'$ 與 $P'M_aQ$ 為同一直線，得證。$\\Box$\n\n回到原題，設 $AH$ 與 $BC$ 交於 $D$。已知 $HX = 2HD$。設 $M_a$ 為 $BC$ 中點，則 $HX = 2HD = 2OM_a = AH$，說明 $D$ 為 $HX$ 中點，$H$ 為 $AX$ 中點。由於 $OH$ 平行於 $BC$ 且 $HX = 2DX$，可知 $YX = 2BX$，$ZX = 2CX$。因此 $AY \\parallel BH \\perp AC$，$AZ \\parallel CH \\perp AB$。取 $\\ell = OH$ 並套用引理，得 $PQ$ 平分 $BC$，得證。$\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16208,
    "subject": "Mathematics (Olympiad)",
    "question": "Сашка имала три корпи со јаболка. Во корпите имало 12, 14 и 22 јаболка. Дозволено е Сашка да избере две од трите корпи и да префрлува јаболка од едната во другата. Притоа, мора да префрли од една во друга корпа онолку јаболка колку што има во корпата во која ги додава јаболката. Сашка направила три префрлувања на опишаниот начин и во сите корпи имало по ист број јаболка. Како Сашка го направила тоа?",
    "options": [],
    "answer": "See solution",
    "solution": "1. Прво, Сашка ги избрала корпите со 22 и 14 јаболка. Од корпата со 22 јаболка префрлила 14 јаболка во корпата со 14 јаболка. После првото префрлување, во корпите има: 8, 28 и 12 јаболка.\n\n2. Вториот пат ги избрала корпите со 28 и 12 јаболка. Од корпата со 28 јаболка префрлила 12 јаболка во корпата со 12 јаболка. После второто префрлување, во корпите има: 16, 24 и 8 јаболка.\n\n3. Третиот пат ги избрала корпите со 24 и 8 јаболка. Од корпата со 24 јаболка префрлила 8 јаболка во другата корпа. После третото префрлување, во корпите има: 16, 16 и 16 јаболка, т.е. ист број на јаболка.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16209,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a prime number $p \\ge 5$, find the number of different residues for the product of three consecutive positive integers modulo $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $D = \\{0, 1, \\dots, p-1\\}$, and define $f(x) = (x-1)x(x+1) = x^3 - x$. We seek the number of distinct residues $f(x) \\bmod p$ as $x$ ranges over $D$.\n\nFor $k = 0, 1, 2, 3$, let\n$$\nB_k = \\{b \\in D \\mid \\text{there are exactly } k \\text{ elements } a \\in D \\text{ such that } f(a) \\equiv b \\bmod p\\}.\n$$\n\nWe analyze the possible values of $k$ and count the sizes $|B_k|$.\n\nConsider $b \\in B_2$, i.e., there exist $a_1 \\not\\equiv a_2$ such that $f(a_1) \\equiv f(a_2) \\equiv b$. Then:\n$$\n0 \\equiv (a_1^3 - a_1) - (a_2^3 - a_2) = (a_1 - a_2)(a_1^2 + a_1 a_2 + a_2^2 - 1) \\implies a_1^2 + a_1 a_2 + a_2^2 \\equiv 1.\n$$\nLet $a_3 = -a_1 - a_2$. Then $a_3^2 + a_3 a_1 + a_1^2 \\equiv 1$, so $f(a_3) \\equiv f(a_1) \\equiv b$. By the definition of $B_2$, $a_3$ must be congruent to one of $a_1, a_2$, so either $a_2 \\equiv -2a_1$ or $a_1 \\equiv -2a_2$. Assume the former. Then $a_1^2 + a_1 a_2 + a_2^2 \\equiv 3a_1^2 \\equiv 1$, so $a_1^2 \\equiv 1/3$. Thus, $a_1$ has $1 + \\left(\\frac{3}{p}\\right)$ solutions, where $\\left(\\frac{3}{p}\\right)$ is the Legendre symbol. Therefore, $|B_2| = 1 + \\left(\\frac{3}{p}\\right)$.\n\nWe have:\n$$\n|B_1| + 2|B_2| + 3|B_3| = |D| = p.\n$$\n\nNow, consider the number $T$ of pairs $(x, y)$ with $x^2 + 3y^2 \\equiv 1 \\pmod p$ and $x \\not\\equiv 0$:\n$$\nT = \\sum_{x=1}^{p} 1 + \\left(\\frac{3-3x^2}{p}\\right) = p + \\left(\\frac{-3}{p}\\right) \\sum_{x=1}^{p} \\left(\\frac{x^2-1}{p}\\right) = p - \\left(\\frac{-3}{p}\\right).\n$$\nAmong these, $1 + \\left(\\frac{3}{p}\\right)$ pairs have $x \\equiv 0$. Thus, the number of ordered collision pairs $(u, v)$ is\n$$\nM = T - 1 - \\left(\\frac{3}{p}\\right) = p - \\left(\\frac{-3}{p}\\right) - 1 - \\left(\\frac{3}{p}\\right).\n$$\n\nThere are $\\frac{M}{2}$ unordered collision pairs. Since elements in $B_3$ correspond to 3 collisions, those in $B_2$ to 1, and $B_1, B_0$ to none, we have:\n$$\n|B_2| + 3|B_3| = \\frac{M}{2}.\n$$\nThus,\n$$\n|B_3| = \\frac{M - 2|B_2|}{6}.\n$$\nThe total number of possible residues is\n$$\n|B_1| + |B_2| + |B_3| = p - |B_2| - 2|B_3| = p - \\frac{M + |B_2|}{3} = p - \\frac{T}{3} = \\frac{2p + \\left(\\frac{-3}{p}\\right)}{3} = \\left\\lfloor \\frac{2p + 1}{3} \\right\\rfloor.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16210,
    "subject": "Mathematics (Olympiad)",
    "question": "В игре на шахматной доске $n \\times n$ (где $n$ нечётно, $n \\ge 3$), в которой два игрока по очереди передвигают ладью по белым клеткам, выигрывает ли Петя, если он ходит первым?",
    "options": [],
    "answer": "See solution",
    "solution": "Одна из выигрышных стратегий для Пети состоит в том, чтобы каждым своим ходом делать самый длинный из возможных вертикальных ходов (например, первым ходом он пойдёт по вертикали в другой угол доски). Покажем, что, действуя согласно ей, он выиграет.\n\nНазовём белую клетку *достижимой*, если из текущего положения ладьи в неё можно попасть за несколько ходов по белым клеткам. Покажем, что в любой момент Петя сможет сходить, причём после каждого его хода все достижимые клетки образуют несколько (не более двух) прямоугольников, в каждом из которых число строк больше числа столбцов, причём для каждого из них ладья стоит в клетке, соседней с угловой по горизонтали. Для первого хода Пети это верно.\n\nДалее, если после некоторого его хода это так, то Вася может ходить в один из полученных прямоугольников по горизонтали. Пусть в этом прямоугольнике $r$ строк и $c$ столбцов, а Вася сходит на $v \\le c$ клеток. После этого хода клетки оставшегося прямоугольника (если он был) перестанут быть достижимыми (см. рис. 16).\n\n![](images/Rusija_2014_p23_data_881de872b6.png)\n\nРис. 16\n\nПоскольку $r \\ge c+1 \\ge 2$, у Пети остаётся возможность вертикального хода. После этого хода достижимые клетки будут образовывать прямоугольники $r \\times (c-v)$ и $(r-1) \\times (v-1)$. В каждом из них строк больше, чем столбцов, ибо $r > c > c-v$ и $r-1 > c-1 \\ge v-1$. При этом ладья стоит в клетке, соседней по горизонтали с угловой в каждом из этих прямоугольников; это и требовалось доказать.\n\nИтак, мы, в частности, доказали, что Петя всегда сможет сделать ход. При этом когда-нибудь игра закончится, ибо количество белых клеток уменьшается. Значит, он выиграет.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16211,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be an isosceles trapezoid with $AD = BC$ and $AB \\parallel CD$. Let $O$ be the intersection of the diagonals and let $M$ be the midpoint of $AD$. The circumcircle of $BCM$ intersects $AD$ again at $K$. Prove that $OK$ is parallel to $AB$.\n\n![](images/MNG_ABooklet_2017_p16_data_e6a905cbed.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N$ be the midpoint of $CB$ and let $AB \\cap CK = E$. Since $MBCK$ is inscribed in a circle, we have $\\angle AMB = \\angle BCE$. Since $ABCD$ is isosceles, we have $\\angle AMB = \\angle ANB$. Thus, $AN \\parallel EC$. By Thales' theorem, $1 = \\frac{BN}{NC} = \\frac{BA}{AE}$, thus $BA = AE$ and $\\frac{DK}{KA} = \\frac{CD}{AE} = \\frac{CD}{AB} = \\frac{DO}{OB}$. It follows that $OK \\parallel AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16212,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle $c$ with center $A$ passes through the vertices $B$ and $E$ of a regular pentagon $ABCDE$. The line $BC$ intersects the circle $c$ a second time at point $F$. Prove that lines $DE$ and $EF$ are perpendicular.",
    "options": [],
    "answer": "See solution",
    "solution": "The internal angles of a regular pentagon are $108^\\circ$. Thus, $\\angle EAB = 108^\\circ$ (see figure below), so $\\angle EFC = \\angle EFB = \\frac{\\angle EAB}{2} = 54^\\circ$. Since $\\angle CDE = 108^\\circ$ and $\\angle FCD = \\angle BCD = 108^\\circ$, in quadrilateral $CDEF$ we have:\n\n$$\n\\angle DEF = 360^\\circ - \\angle FCD - \\angle CDE - \\angle EFC = 90^\\circ.\n$$\n\n![](images/EST_ABooklet_2020_p4_data_3cfaa31f80.png)\n\n_Fig. 2_\n\n![](images/EST_ABooklet_2020_p4_data_840a2b7eef.png)\n\n_Fig. 3_",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16213,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the integral lattice $\\mathbb{Z}^n$, $n \\geq 2$, in Euclidean $n$-space. Define a line in $\\mathbb{Z}^n$ to be a set of the form $a_1 \\times \\dots \\times a_{k-1} \\times \\mathbb{Z} \\times a_{k+1} \\times \\dots \\times a_n$, where $k$ is an integer in the range $1, 2, \\dots, n$, and the $a_i$ are arbitrary integers. A subset $A$ of $\\mathbb{Z}^n$ is called *admissible* if it is non-empty, finite, and every line in $\\mathbb{Z}^n$ which intersects $A$ contains at least two points of $A$. A subset $N$ of $\\mathbb{Z}^n$ is called *null* if it is non-empty, and every line in $\\mathbb{Z}^n$ intersects $N$ in an even number of points (possibly zero).\n\n**a)** Prove that every admissible set in $\\mathbb{Z}^2$ contains a null set.\n\n**b)** Exhibit an admissible set in $\\mathbb{Z}^3$ no subset of which is a null set.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Let $A$ be an admissible set in $\\mathbb{Z}^2$. Choose a point $a_0$ of $A$, and for each positive integer $k$, choose a point $a_k$ of $A$ different from $a_{k-1}$, having the same first coordinate as the latter if $k$ is odd, and the same second coordinate if $k$ is even. Eventually, we must choose an $a_n = a_m$, $m < n$. Assume $a_n$ is the first point to duplicate a preceding point. If $m$ and $n$ have like parities, then $a_m, a_{m+1}, \\dots, a_{n-1}$ form a null set, and if they have opposite parities, then $a_{m+1}, \\dots, a_{n-1}$ do.\n\n(b) We exhibit a minimal admissible set $A$ in $\\mathbb{Z}^3$ which is not itself null. Here, minimality means no proper subset is admissible. Since every null finite set is admissible, the conclusion follows. The set $A$ is a set of lattice points in the parallelepiped $[0, 3] \\times [0, 3] \\times [0, 4]$. We describe it by successive horizontal cross-sections:\n\n$$\nA = A_0 \\times 0 \\cup A_1 \\times 1 \\cup A_2 \\times 2 \\cup A_3 \\times 3 \\cup A_4 \\times 4,\n$$\n\nwhere $A_0 = \\{0, 3\\} \\times \\{0, 3\\}$, $A_1 = \\{0, 1\\} \\times \\{2, 3\\} \\cup \\{1, 2\\} \\times \\{0, 1\\}$, $A_2 = \\{0, 1\\} \\times \\{1, 2\\} \\cup \\{2, 3\\} \\times \\{2, 3\\}$, $A_3 = \\{1, 2\\} \\times \\{2, 3\\} \\cup \\{2, 3\\} \\times \\{0, 1\\}$, and $A_4 = \\{0, 1\\} \\times \\{0, 1\\} \\cup \\{2, 3\\} \\times \\{1, 2\\}$.\n\nFor $k = 1, 2, 3$, the configuration $A_{k+1}$ is obtained from $A_k$ by a clockwise rotation through $\\pi/2$ about the centre of the square $[0, 3] \\times [0, 3]$.\n\nThe set $A$ is admissible, since each horizontal cross-section $A_k \\times k$ is admissible in $\\mathbb{Z}^2 \\times k$, and the perpendicular in $\\mathbb{Z}^3$ to any horizontal cross-section through any one of its points meets at least one other horizontal cross-section.\n\nTo prove minimality, we exhibit a connected geometric lattice graph $G$ on $A$ such that the line of support of each edge of $G$ is a line in $\\mathbb{Z}^3$ stabbing $A$ at exactly two points, namely, the endpoints of that edge. The existence of such a graph implies minimality, since removal of any one point in $A$ entails removal of all its neighbours in $G$, and eventually removal of all of $A$.\n\nEach of the verticals $i \\times j \\times \\mathbb{Z}$ through a point of $A$, where either $i$ or $j$ is in $\\{0, 3\\}$, stabs exactly two horizontal cross-sections of $A$. Join the corresponding points of $A$ by the vertical segment they determine.\n\nNext, consider the generic planar lattice paths $\\alpha_1 = (1 \\times 0)(2 \\times 0)(2 \\times 1)(1 \\times 1)$ and $\\alpha'_1 = (1 \\times 2)(0 \\times 2)(0 \\times 3)(1 \\times 3)$, and for $k = 1, 2, 3$, let $\\alpha_{k+1}$ and $\\alpha'_{k+1}$ be obtained from $\\alpha_k$ and $\\alpha'_k$, respectively, by a clockwise rotation through $\\pi/2$ about the centre of the square $[0, 3] \\times [0, 3]$. The edges of the lattice paths $\\alpha_k \\times k$ and $\\alpha'_k \\times k$, joining points in $A_k \\times k$, $k = 1, 2, 3, 4$, along with the vertical edges above, form the desired connected geometric lattice graph $G$ on $A$.\n\nFinally, the set $A$ is not null, for the vertical $1 \\times 1 \\times \\mathbb{Z}$ stabs exactly three horizontal cross-sections of $A$, namely, $A_1 \\times 1$, $A_2 \\times 2$, and $A_4 \\times 4$; in fact, each of the lines $i \\times j \\times \\mathbb{Z}$, $i, j \\in \\{1, 2\\}$, stabs exactly three horizontal cross-sections of $A$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16214,
    "subject": "Mathematics (Olympiad)",
    "question": "Sean $x_1 \\le x_2 \\le x_3 \\le x_4$ cuatro números reales. Demuestra que existen polinomios $P(x)$ y $Q(x)$ de grado dos con coeficientes reales tales que $x_1, x_2, x_3$ y $x_4$ son las raíces de $P(Q(x))$ si y solamente si $x_1 + x_4 = x_2 + x_3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Supongamos que las raíces de $P(x)$ son $r$ y $s$. Entonces, las raíces de $P(Q(x))$ son las raíces de $Q(x) - r$ y de $Q(x) - s$. Por las relaciones de Cardano, la suma de las raíces de $Q(x) - r$ es la misma que la de $Q(x) - s$, con lo que las raíces de $P(Q(x))$ necesariamente satisfacen que la suma de dos de ellas coincide con la suma de las otras dos.\n\nVamos a demostrar ahora el recíproco. Consideremos una cuaterna $(x_1, x_2, x_3, x_4)$ con $x_1 + x_4 = x_2 + x_3$. Por tanto, queremos elegir $(Q(x), r, s)$ de manera que\n\n$$\nQ(x) - r = (x - x_1)(x - x_4), \\quad Q(x) - s = (x - x_2)(x - x_3).\n$$\n\nPara ello, pongamos $Q(x) = x^2 - (x_1 + x_4)x$. La condición se reescribe por tanto como\n\n$$\nr = -x_1x_4, \\quad s = -x_2x_3,\n$$\n\ny hemos acabado.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16215,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime satisfying $p \\equiv 1 \\pmod{4}$. Show that there exist infinitely many positive integers $n$ such that $2^n + n^2$ is divisible by $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Write $p = 2s + 1$ with $s$ even. By Wilson's theorem, we have\n\n$$\n(s!)^2 \\equiv (-1)^s (p-1)! \\equiv -1 \\pmod{p}.\n$$\n\nFor any positive integer $k$, let $n_k := (p-1)^k s!$. Then $2^{n_k} \\equiv (2^{p-1})^{(p-1)^{k-1} s!} \\equiv 1 \\pmod{p}$ by Fermat's little theorem. On the other hand, $n_k^2 \\equiv (-1)^{2k} (s!)^2 \\equiv -1 \\pmod{p}$ by the above. It follows that $2^{n_k} + n_k^2 \\equiv 1 + (-1) \\equiv 0 \\pmod{p}$, so $n_k$ satisfies the condition for any $k$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16216,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n > 2$ such that $n = a^3 + b^3$, where $a$ is the smallest positive divisor of $n$ greater than $1$, and $b$ is an arbitrary positive divisor of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that if $n$ is odd, then both divisors $a$ and $b$ are odd. But their sum $a^3 + b^3$ is even, so $n$ must be even, which is a contradiction. Hence, $n$ is even and $a = 2$. Then $b$ must also be even. Additionally, $b$ divides $n - b^3 = a^3 = 8$. So, $b \\in \\{2, 4, 8\\}$. Finally, all possible values for $n$ are:\n\n- $b = 2$: $n = 2^3 + 2^3 = 8 + 8 = 16$\n- $b = 4$: $n = 2^3 + 4^3 = 8 + 64 = 72$\n- $b = 8$: $n = 2^3 + 8^3 = 8 + 512 = 520$\n\nThus, the possible values for $n$ are $16$, $72$, and $520$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16217,
    "subject": "Mathematics (Olympiad)",
    "question": "If $a = 4$ and $b = -2$ then $l = 7$. If $a = -4$ and $b = -2$ then $l = 2$. In both cases, $l$ is a positive integer, but it is not a perfect square.\n\nFind all possible values of $a$ and $b$ such that $l$ is a positive integer but not a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "We know from the first part that at least one of the numbers $a$ and $b$ is negative. As above, we conclude that $a = lb^2$ and $l = b^2 + a + \\frac{3-l}{lb} = (l-1)b^2 + \\frac{3-l}{lb}$.\n\nWe know that $l$ divides $3$, so $l$ can be $1$, $-1$, $3$, or $-3$.\n\n- If $l = 3$, then $a = 3b^2$ and $l = (2b)^2$ regardless of what $b$ is, so $l$ is a perfect square.\n- If $l = 1$, then $b = -1$ or $b = -2$. In the first case, $l = 0$; in the second case, $a = 4$ and $l = 7$.\n- If $l = -1$, then $l = -\\frac{1}{2}$, which is not an integer.\n- If $l = 2$, $b = -2$, $a = -4$ and $l = 2$.\n- If $l = -3$, $l = -2b^2 - \\frac{1}{2}$, which is negative in all cases.\n\nTherefore, $l$ is a positive integer but not a perfect square if and only if $a = 4$, $b = -2$ or $a = -4$, $b = -2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16218,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $\\angle A = 60^\\circ$, orthocenter $H$ and centroid $G$. Let $A'$ be the point on the circumcircle of $\\triangle ABC$ diametrically opposite to $A$. Let $M$ be the midpoint of $BC$. The line through $M$ parallel to $AA'$ intersects $BH$ and $CH$ at $K$ and $L$, respectively. $P$ and $Q$ are distinct points on $GA'$ such that $GP = GQ = \\frac{1}{2}GH$. Let $D$ be the reflection of $A$ over $GA'$.\n\nProve that the circumcircle of $\\triangle DPQ$ is tangent to the circumcircle of $\\triangle HKL$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A$, $B$, $C$ lie on the unit circle. Let $O$ be the origin and $X$ be the midpoint of $AH$. Henceforth, a lower case letter denotes the complex number corresponding to that point.\n\n![](images/tmc2017_New_p28_data_7683dbbd86.png)\n\nWe first show that $X$ lies on $A'G$ and $ML$.\n\nSince $x = a + \\frac{b+c}{2}$ and $a' = -a$, we obtain $x - a' = \\frac{4a+b+c}{2}$ and $g - a' = \\frac{4a+b+c}{3}$. Hence $\\frac{x-a'}{g-a'} = \\frac{3}{2}$. So $X$ lies on $A'G$.\n\nSince $M = \\frac{b+c}{2}$, we have $x - m = a$ and $a - a' = 2a$. Thus $\\frac{x-m}{a-a'} = \\frac{1}{2}$. So $XM \\parallel AA'$, and thus $X$ lies on $ML$.\n\nNext we show that $XH$ is tangent to the circumcircle of $\\triangle HLK$.\n\nLet $\\angle CBA = \\beta$ and $\\angle ACB = \\gamma$ (without loss of generality we may assume $\\gamma \\ge \\beta$). Then\n\n$$\n\\angle XHL = 90^\\circ - \\angle EAH = 90^\\circ - (90^\\circ - \\beta) = \\beta.\n$$\n\nAlso\n\n$$\n\\angle KXH = \\angle A'AH = \\angle BAH - \\angle BAA'\n$$\n\nBut $\\angle BAH = 90^\\circ - \\beta$ and $\\angle BAA' = 90^\\circ - \\angle AA'B = 90^\\circ - \\gamma$. So $\\angle KXH = \\gamma - \\beta$. Similarly, we obtain $\\angle LHK = 60^\\circ$, and thus\n\n$$\n\\begin{aligned}\n\\angle HKL &= 180^\\circ - \\angle XHK - \\angle XHL \\\\\n&= 120^\\circ - \\gamma = \\beta = \\angle XHL.\n\\end{aligned}\n$$\n\nHence $XH$ is tangent to the circumcircle of $\\triangle HLK$.\n\nNext we will show $XD$ is tangent to the circumcircle of $\\triangle DPQ$.\n\nTo achieve this we will show $XD^2 = XP \\cdot XQ$. Note that\n\n$$\nXP \\cdot XQ = (XG - GP)(XG + GQ) = XG^2 - \\frac{GH^2}{4}.\n$$\n\nProceeding with the calculations we have\n\n$$\n\\begin{aligned}\nXD^2 &= XA^2 \\\\\n&= |x - a|^2 \\\\\n&= \\frac{1}{4}(b+c)(\\bar{b} + \\bar{c}) \\\\\n&= \\frac{1}{4}(2 + b\\bar{c} + \\bar{b}c).\n\\end{aligned} \\qquad (1)\n$$\n\nSimilarly,\n\n$$\nXG^2 = \\frac{18 + 4a\\bar{b} + 4a\\bar{c} + 4\\bar{a}b + 4\\bar{a}c + b\\bar{c} + \\bar{b}c}{36},\n$$\n\nand\n\n$$\n\\frac{GH^2}{4} = \\frac{12 + 4a\\bar{b} + 4a\\bar{c} + 4\\bar{a}b + 4\\bar{a}c + 4b\\bar{c} + 4\\bar{b}c}{36}.\n$$\n\nHence\n\n$$\nXG^2 - \\frac{GH^2}{4} = \\frac{6 - 3(b\\bar{c} + \\bar{b}c)}{36}. \\qquad (2)\n$$\n\nSince $\\angle BOC = 120^\\circ$, we have $\\bar{b}c = \\omega$ and $b\\bar{c} = \\omega^2$, where $\\omega = e^{\\frac{2\\pi i}{3}}$. So $b\\bar{c} + \\bar{b}c = \\omega^2 + \\omega = -1$. Substituting this in (1) and (2) yields $XD^2 = XG^2 - \\frac{GH^2}{4}$ as required.\n\nThe two tangent lines from $X$ to the circumcircle of $\\triangle DPQ$ touch the circle at $D$ and another point, which we will call $T$. Then since\n\n$$\nXT^2 = XD^2 = XA^2 = XH^2,\n$$\n\nwe conclude that $XT$ is also tangent to the circumcircle of $\\triangle HLK$. Thus the circumcircle of $\\triangle DPQ$ is tangent to the circumcircle of $\\triangle HKL$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16219,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for all $x, y \\in \\mathbb{R}$,\n$$\nx f(x + f(y)) = (y - x) f(f(x)).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For any real $c$, $f(x) = c - x$ for all $x \\in \\mathbb{R}$ and $f(x) = 0$ for all $x \\in \\mathbb{R}$.\n\nLet $P(x, y)$ denote the assertion that $x$ and $y$ satisfy the given equation. $P(0, 1)$ gives $f(f(0)) = 0$.\n\nFrom $P(x, x)$ we get $x f(x + f(x)) = 0$ for all $x \\in \\mathbb{R}$, which together with $f(f(0)) = 0$ gives $f(x + f(x)) = 0$ for all $x$.\n\nNow let $t$ be any real number such that $f(t) = 0$. For any $y$, from $P(t - f(y), y)$ we have\n$$\n(y + f(y) - t) f(f(t - f(y))) = 0\n$$\nfor all $y$ and all $t$ such that $f(t) = 0$. Taking $y = f(0)$ gives\n$$\n(f(0) - t) f(f(t)) = 0 \\quad \\text{and hence} \\quad (f(0) - t) f(0) = 0. \\qquad (\\text{A1-1})\n$$\nRecall that as $t$ with $f(t) = 0$ we can take $x + f(x)$ for any real $x$. If for all reals $x$ we have $x + f(x) = f(0)$, then $f(x)$ must be of the form $f(x) = c - x$ for some real $c$. All functions of this form are indeed solutions.\n\nOtherwise, we can find some $a \\neq 0$ so that $a + f(a) \\neq f(0)$. If $t = a + f(a)$ in (A1-1), then $f(0)$ must be $0$. Now $P(x, 0)$ gives $f(f(x)) = -f(x)$ for all $x$. From here, $P(x, x + f(x))$ gives $x f(x) = -f(x)^2$ for all $x$, which means for every $x$ either $f(x) = 0$ or $f(x) = -x$.\n\nAssume in this case there is some $b \\neq 0$ so that $f(b) = -b$. For any $y$, from $P(b, y)$ and $f(f(b)) = -f(b) = b$ we get $b f(b + f(y)) = (y - b) b$, so $f(b + f(y)) = y - b$ for all $y$. If $y \\neq b$, then the right-hand side is not zero, so $f(b + f(y)) = -b - f(y)$, which means $-b - f(y) = y - b$, or $f(y) = -y$ for all $y$. But we already covered this solution (take $c = 0$ above). If there is no such $b$, then $f(x) = 0$ for all $x$, which gives the final solution.\n\nThus, all such functions are of the form $c - x$ for real $c$ or the zero function.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16220,
    "subject": "Mathematics (Olympiad)",
    "question": "For positive integers $n$, $m$, $k$, we write $n \\equiv m \\pmod{k}$ if $n - m$ is divisible by $k$.\n\nLet $A$ be the sum of all positive integers $a \\leq 2011$ for which $a \\equiv 1 \\pmod{3}$, and $B$ be the sum of all positive integers $b \\leq 2011$ for which $b \\equiv 2 \\pmod{3}$. Find the value of $A - B$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $2011 \\equiv 1 \\pmod{3}$. So, $A$ is the sum of the numbers\n\n$$\n3 \\cdot 0 + 1,\\ 3 \\cdot 1 + 1,\\ \\dots,\\ 3 \\cdot 669 + 1,\\ 3 \\cdot 670 + 1\n$$\n\nwhile $B$ is the sum of the numbers\n\n$$\n3 \\cdot 0 + 2,\\ 3 \\cdot 1 + 2,\\ \\dots,\\ 3 \\cdot 668 + 2,\\ 3 \\cdot 669 + 2\n$$\n\nFor $0 \\leq i \\leq 669$, we have $(3i+1)-(3i+2) = -1$, so\n\n$$\nA - B = (-1) \\times 670 + (3 \\cdot 670 + 1) = 1341.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16221,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider 64 distinct natural numbers less than or equal to 2012. Prove that among them there are four numbers, denoted by $a, b, c, d$, such that $a + b - c - d$ is a multiple of 2013.",
    "options": [],
    "answer": "See solution",
    "solution": "Using the 64 given numbers, we can form $\\binom{64}{2} = 2016$ pairs $(a, b)$ with $a < b$. Each pair yields a sum $a + b$, and thus 2016 possible sums. When divided by 2013, these sums yield 2016 remainders, but since there are only 2013 possible remainders (from 0 to 2012), by the pigeonhole principle, there must be two distinct pairs $(a, b)$ and $(c, d)$ such that $a + b \\equiv c + d \\pmod{2013}$. Therefore, $2013 \\mid (a + b) - (c + d)$, or $2013 \\mid a + b - c - d$.\n\nIf the two pairs share a common element, say $a = c$, then $2013 \\mid b - d$. But since $|b - d| \\leq 2012$, this forces $b = d$, which contradicts the distinctness of the pairs. Thus, the four numbers $a, b, c, d$ are all distinct, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16222,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $d$ such that whenever $d$ divides an integer $n$, $d$ will also divide any integer $m$ having the same digits as $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The possible values are $d = 1$, $d = 3$, or $d = 9$.\n\nIt is known that $1$, $3$, and $9$ have the given property. Assume that $d$ is a positive integer such that whenever $d$ divides an integer $n$, $d$ will also divide any integer $m$ having the same digits as $n$.\n\nConsider a $k$-digit number $n = 10a_1a_2\\dots a_k$ which is divisible by $d$. Then, the numbers $a_1a_2\\dots a_k10$ and $a_1a_2\\dots a_k01$ (which are rearrangements of the digits of $n$) must also be divisible by $d$.\n\nNotice that:\n$$\n(a_1a_2\\dots a_k10) - (a_1a_2\\dots a_k01) = 9\n$$\nTherefore, $d$ must divide $9$, so $d = 1$, $3$, or $9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16223,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be three positive real numbers. Prove that the function $f: \\mathbb{R} \\to \\mathbb{R}$, defined by\n\n$$\nf(x) = \\frac{a^x}{b^x + c^x} + \\frac{b^x}{a^x + c^x} + \\frac{c^x}{a^x + b^x}\n$$\nis increasing on $[0, \\infty)$ and decreasing on $(-\\infty, 0]$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use straightforward computation: if $x \\le y$ are real numbers, then\n\n$$\n\\begin{align*}\nf(y) - f(x) &= \\sum_{\\text{cyc}} \\frac{a^y(b^x + c^x) - a^x(b^y + c^y)}{(b^x + c^x)(b^y + c^y)} \\\\\n&= \\sum_{\\text{cyc}} (a^y b^x - a^x b^y) \\left( \\frac{1}{(b^x + c^x)(b^y + c^y)} - \\frac{1}{(a^x + c^x)(a^y + c^y)} \\right) \\\\\n&= \\sum_{\\text{cyc}} a^x b^x (a^{y-x} - b^{y-x}) \\\\\n&\\qquad \\times \\frac{(a^{x+y} - b^{x+y}) + c^x(a^y - b^y) + c^y(a^x - b^x)}{(b^x + c^x)(b^y + c^y)(a^x + c^x)(a^y + c^y)}\n\\end{align*}\n$$\n\nSo $f(y) - f(x)$ can be written as a sum of products of the form $(a^p - b^p)(a^q - b^q)$ multiplied with positive coefficients, where $p = y - x \\ge 0$ and all the $q$'s have the same sign as $x, y$.\n\nWe finish by noticing that\n\n$$\n(a^p - b^p)(a^q - b^q) = b^{p+q} \\left( \\left( \\frac{a}{b} \\right)^p - 1 \\right) \\left( \\left( \\frac{a}{b} \\right)^q - 1 \\right) \\begin{cases} \\ge 0 & \\text{if } q \\ge 0 \\\\ \\le 0 & \\text{if } q \\le 0 \\end{cases}\n$$\n\nHence $f(y) - f(x) \\ge 0$ if $y \\ge x \\ge 0$ and $f(y) - f(x) \\le 0$ if $0 \\ge y \\ge x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16224,
    "subject": "Mathematics (Olympiad)",
    "question": "對每個正整數 $N$，請決定最小的實數 $b_N$ 使得對所有的實數 $x$，下列不等式恆成立：\n\n$$\n\\sqrt[N]{\\frac{x^{2N} + 1}{2}} \\leq b_N(x - 1)^2 + x.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "答案是 $b_N = N/2$。\n\n**解法 1.** 首先，假設 $b_N < N/2$ 能滿足條件。取 $x = 1 + t$，其中 $t > 0$，則應有\n\n$$\n\\frac{(1 + t)^{2N} + 1}{2} \\leq (1 + t + b_N t^2)^N.\n$$\n\n展開括號得\n\n$$\n(1 + t + b_N t^2)^N - \\frac{(1 + t)^{2N} + 1}{2} = \\left(Nb_N - \\frac{N^2}{2}\\right)t^2 + c_3 t^3 + \\dots + c_{2N} t^{2N}\n$$\n\n其中 $c_3, \\dots, c_{2N}$ 為某些係數。由於 $b_N < N/2$，當 $t$ 充分小時，右式為負，矛盾。\n\n接下來記 $\\mathcal{I}(N, x)$ 為此不等式。剩下要證明對任意正整數 $N$，取 $b_N = N/2$ 時 $\\mathcal{I}(N, x)$ 成立。\n\n首先，$\\mathcal{I}(N, 0)$ 明顯成立。若 $x > 0$，則 $\\mathcal{I}(N, -x)$ 與 $\\mathcal{I}(N, x)$ 左式相同，而 $\\mathcal{I}(N, -x)$ 右式大於 $\\mathcal{I}(N, x)$（差為 $2(N-1)x \\geq 0$）。因此 $\\mathcal{I}(N, -x)$ 可由 $\\mathcal{I}(N, x)$ 得出。故以下假設 $x > 0$。\n\n將 $\\mathcal{I}(N, x)$ 兩邊同除以 $x$，令 $t = (x-1)^2/x = x - 2 + 1/x$，則 $\\mathcal{I}(N, x)$ 可寫為\n\n$$\nf_N := \\frac{x^N + x^{-N}}{2} \\leq \\left(1 + \\frac{N}{2}t\\right)^N. \\qquad (2)\n$$\n\n關鍵在於 $f_N$ 可展開為 $t$ 的多項式：\n\n*引理*\n\n$$\nf_N = N \\sum_{k=0}^{N} \\frac{1}{N+k} \\binom{N+k}{2k} t^k. \\qquad (3)\n$$\n\n*證明*：用 $N$ 步歸納。利用遞迴關係\n\n$$\nf_{N+1} + f_{N-1} = (x + 1/x)f_N = (2 + t)f_N. \\quad (4)\n$$\n\n基礎情形 $N = 1, 2$：\n\n$$\nf_1 = 1 + \\frac{t}{2}, \\quad f_2 = \\frac{1}{2}t^2 + 2t + 1.\n$$\n\n歸納步：由 $f_{N+1} = (2 + t)f_N - f_{N-1}$ 計算 $f_{N+1}$ 中 $t^k$ 的係數。$k=0$ 時為 1，$k > 0$ 時為\n\n$$\n\\begin{align*}\n& 2 \\frac{N}{N+k} \\binom{N+k}{2k} + \\frac{N}{N+k-1} \\binom{N+k-1}{2k-2} - \\frac{N-1}{N+k-1} \\binom{N+k-1}{2k} \\\\\n&= \\frac{(N+k-1)!}{(2k)!(N-k)!} \\left( 2N + \\frac{2k(2k-1)N}{(N+k-1)(N-k+1)} - \\frac{(N-1)(N-k)}{N+k-1} \\right) \\\\\n&= \\frac{(N+k-1)!}{(2k)!(N-k+1)!} \\left( 2N(N-k+1) + 3kN + k - N^2 - N \\right) = \\frac{\\binom{N+k+1}{2k}}{N+k+1} (N+1),\n\\end{align*}\n$$\n\n完成歸納。$\\square$\n\n回到原題，為證 (2)，寫成\n\n$$\n(1 + \\frac{N}{2}t)^N - f_N = \\left(1 + \\frac{N}{2}t\\right)^N - N \\sum_{k=0}^{N} \\frac{1}{N+k} \\binom{N+k}{2k} t^k = \\sum_{k=0}^{N} \\alpha_k t^k,\n$$\n\n其中\n\n$$\n\\begin{align*}\n\\alpha_k &= \\left(\\frac{N}{2}\\right)^k \\binom{N}{k} - \\frac{N}{N+k} \\binom{N+k}{2k} \\\\\n&= \\left(\\frac{N}{2}\\right)^k \\binom{N}{k} \\left(1 - 2^k \\frac{(1+1/N)(1+2/N)\\cdots(1+(k-1)/N)}{(k+1)\\cdots(2k)}\\right) \\\\\n&\\geq \\left(\\frac{N}{2}\\right)^k \\binom{N}{k} \\left(1 - 2^k \\frac{2 \\cdot 3 \\cdots k}{(k+1)\\cdots(2k)}\\right) = \\left(\\frac{N}{2}\\right)^k \\binom{N}{k} \\left(1 - \\prod_{j=1}^k \\frac{2j}{k+j}\\right) \\geq 0,\n\\end{align*}\n$$\n\n故 (2) 成立。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16225,
    "subject": "Mathematics (Olympiad)",
    "question": "On a triangle $ABC$, a point $D$ is picked on the side $AB$ and a point $E$ is picked on the side $AC$ in such a way that the line $DE$ is parallel to the line $BC$. Let $M$ and $N$ be the mid-points of the line segments $BD$ and $CE$, respectively. Determine the area of the triangle $ADE$ if the areas of the quadrilaterals $DMNE$ and $MBCN$ are $1$ and $2$, respectively.\n\n![](images/Japan_2011_p14_data_6f46ca25c6.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since the lines $BC$ and $DE$ are parallel and since $M$ and $N$ are mid-points of the line segments $DB$ and $EC$, respectively, it is easy to see that the line $MN$ is also parallel to the line $BC$ (and $DE$). Therefore, the triangles $ADE$, $AMN$ and $ABC$ are similar.\n\nIf we let $a = DE$, $b = BC$ and let $h$ and $\\ell$ be the heights of the triangles $ADE$ and $ABC$, respectively, then we have $MN = \\frac{1}{2}(a + b)$ and the height of the trapezoid $DMNE$ (and the trapezoid $MBCN$) equals $\\frac{1}{2}(\\ell - h)$. Since the areas of the trapezoids $DMNE$ and $MBCN$ are $1$ and $2$, respectively, we have",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16226,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the polynomial\n\n$$\nf(x) = (1 + x)(1 + x^2) \\cdots (1 + x^{2013}) = \\sum_n a_n x^n.\n$$\n\nLet $|T_r| = \\sum_k [x^{7k + r}] f(x) = \\sum_k a_{7k + r}$, i.e., the sum of coefficients of $x^n$ in $f(x)$ where $n \\equiv r \\pmod{7}$.\n\nLet $\\epsilon = e^{\\frac{2\\pi i}{7}}$, a primitive 7th root of unity.\n\nCompute $|T_r|$ for $r = 0, 1, 2, 3, 4, 5, 6$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use the roots of unity filter:\n\n$$\n|T_r| = \\sum_k a_{7k + r} = \\frac{1}{7} \\sum_{i=0}^6 \\epsilon^{-ri} f(\\epsilon^i),\n$$\nwhere $\\epsilon = e^{2\\pi i/7}$.\n\nFirst, $f(1) = 2^{2013}$.\n\nFor $i = 1, \\ldots, 6$, since $\\epsilon^7 = 1$ and $2013 = 7 \\cdot 287 + 4$,\n\n$$\nf(\\epsilon^i) = [(1 + \\epsilon^i)(1 + \\epsilon^{2i}) \\cdots (1 + \\epsilon^{7i})]^{287} (1 + \\epsilon^i)(1 + \\epsilon^{2i})(1 + \\epsilon^{3i})(1 + \\epsilon^{4i}).\n$$\n\nBut $(1 + \\epsilon^i)(1 + \\epsilon^{2i}) \\cdots (1 + \\epsilon^{7i}) = 2$ (since the product over a full period is $2$), so\n\n$$\nf(\\epsilon^i) = 2^{287} (1 + \\epsilon^i)(1 + \\epsilon^{2i})(1 + \\epsilon^{3i})(1 + \\epsilon^{4i}).\n$$\n\nIt can be shown that $(1 + \\epsilon^i)(1 + \\epsilon^{2i})(1 + \\epsilon^{3i})(1 + \\epsilon^{4i}) = 1 + \\epsilon^{3i}$, so\n\n$$\nf(\\epsilon^i) = 2^{287} (1 + \\epsilon^{3i}).\n$$\n\nTherefore,\n\n$$\n|T_r| = \\frac{1}{7} \\left[2^{2013} + 2^{287} \\sum_{i=1}^6 (\\epsilon^{-ri} + \\epsilon^{(3 - r)i}) \\right].\n$$\n\nUsing properties of roots of unity,\n\n$$\n\\sum_{i=1}^6 (\\epsilon^{-ri} + \\epsilon^{(3 - r)i}) =\n\\begin{cases}\n5, & r = 0, 3 \\\\\n-2, & r = 1, 2, 4, 5, 6\n\\end{cases}\n$$\n\nSo the answer is\n\n$$\n|T_r| =\n\\begin{cases}\n\\dfrac{2^{2013} + 5 \\cdot 2^{287}}{7}, & r = 0, 3 \\\\\n\\dfrac{2^{2013} - 2^{288}}{7}, & r = 1, 2, 4, 5, 6\n\\end{cases}\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16227,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $OAB$ be a sector of a circle with center $O$ as in the figure below. A point $Q$ is chosen on the chord $AB$. Let $P$ be the point of intersection of the arc of the circular sector and the line $OQ$. If $AQ = 5$, $BQ = 6$, and $OQ = PQ$, determine the value of the radius of this sector. Here, for a line segment $XY$, its length is also denoted by $XY$.\n\n![](images/Japan_2011_p1_data_e02d9a5d5b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "$$2\\sqrt{10}$$\n\nLet $r$ be the radius of the circle. Consider the full circle obtained by extending the arc of the circular sector in question. Let $R$ be the point of intersection, other than $P$, of the line $PO$ and the full circle. By the well-known theorem on the power of a point with respect to a circle, we have $AQ \\cdot BQ = PQ \\cdot RQ$, which gives us",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16228,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, let $f_n(x)$ be defined by\n$$\nf_n(x) = \\sum_{k=1}^{n} |x - k|.\n$$\nDetermine the solution to the inequality $f_n(x) < 41$ for every two-digit integer $n$ (in decimal notation).",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that $f_n(x)$ is symmetric: $f_n(x) = f_n(n+1-x)$, since\n$$\nf_n(n+1-x) = \\sum_{k=1}^{n} |n+1-x-k| = \\sum_{k=1}^{n} |x - (n+1-k)| = \\sum_{k=1}^{n} |x-k| = f_n(x).\n$$\n\nFor $x < 1$, $x-k < 0$ for all $k \\ge 1$, so\n$$\nf_n(x) = \\sum_{k=1}^{n} (k-x) = \\frac{n(n+1)}{2} - nx > \\frac{n(n+1)}{2} - n = \\frac{n(n-1)}{2} \\ge \\frac{10 \\cdot 9}{2} = 45 > 41.\n$$\nThus, $x < 1$ is excluded. By symmetry, $x > n$ is also excluded. So $1 \\le x \\le n$.\n\nSuppose $x \\in [\\ell, \\ell + 1]$ for some integer $\\ell$ with $1 \\le \\ell \\le n - 1$. Then $x - k \\le 0$ for $k \\ge \\ell + 1$ and $x - k \\ge 0$ for $k \\le \\ell$, so\n$$\n\\begin{aligned}\nf_n(x) &= \\sum_{k=1}^{\\ell} (x-k) + \\sum_{k=\\ell+1}^{n} (k-x) \\\\\n&= \\ell x - \\frac{\\ell(\\ell+1)}{2} + \\frac{(n-\\ell)(n+\\ell+1)}{2} - (n-\\ell)x \\\\\n&= \\frac{n(n+1)}{2} - \\ell(\\ell+1) + (2\\ell-n)x.\n\\end{aligned}\n$$\n\nThis shows $f_n(x)$ is strictly decreasing on $[\\ell, \\ell + 1]$ if $\\ell < \\frac{n}{2}$, constant on $[\\frac{n}{2}, \\frac{n}{2} + 1]$ (if $n$ is even), and strictly increasing if $\\ell > \\frac{n}{2}$.\n\nThus:\n- If $n$ is even, $f_n(x)$ is strictly decreasing on $[1, \\frac{n}{2}]$, constant on $[\\frac{n}{2}, \\frac{n}{2} + 1]$, and strictly increasing on $[\\frac{n}{2} + 1, n]$.\n- If $n$ is odd, $f_n(x)$ is strictly decreasing on $[1, \\frac{n+1}{2}]$ and strictly increasing on $[\\frac{n+1}{2}, n]$.\n\nThe minimum of $f_n(x)$ is always at $m = \\lfloor \\frac{n+1}{2} \\rfloor$. Completing the square in the above formula gives\n$$\n\\begin{aligned}\nf_n(m) &= \\frac{n(n+1)}{2} - m(m+1) + (2m-n)m \\\\\n&= \\frac{n(n+1)}{2} + m^2 - (n+1)m \\\\\n&= \\frac{n(n+1)}{2} + \\left(m - \\frac{n+1}{2}\\right)^2 - \\left(\\frac{n+1}{2}\\right)^2 \\\\\n&\\ge \\frac{n(n+1)}{2} - \\left(\\frac{n+1}{2}\\right)^2 = \\frac{n^2-1}{4}.\n\\end{aligned}\n$$\n\nIf $n \\ge 13$, then $f_n(x) \\ge f_n(m) \\ge \\frac{13^2-1}{4} = 42 > 41$ for all $x$, so there is no solution. It remains to consider $n = 10, 11, 12$.\n\n- For $n = 10$:\n  $$\nf_{10}\\left(\\frac{3}{2}\\right) = 55 - 2 - 8 \\cdot \\frac{3}{2} = 41\n  $$\n  and by symmetry $f_{10}\\left(\\frac{19}{2}\\right) = 41$. Thus, $f_{10}(x) < 41$ for $\\frac{3}{2} < x < \\frac{19}{2}$.\n\n- For $n = 11$:\n  $$\nf_{11}\\left(\\frac{19}{7}\\right) = 66 - 6 - 7 \\cdot \\frac{19}{7} = 41\n  $$\n  and $f_{11}\\left(\\frac{65}{7}\\right) = 41$. Thus, $f_{11}(x) < 41$ for $\\frac{19}{7} < x < \\frac{65}{7}$.\n\n- For $n = 12$:\n  $$\nf_{12}\\left(\\frac{17}{4}\\right) = 78 - 20 - 4 \\cdot \\frac{17}{4} = 41\n  $$\n  and $f_{12}\\left(\\frac{35}{4}\\right) = 41$. Thus, $f_{12}(x) < 41$ for $\\frac{17}{4} < x < \\frac{35}{4}$.\n\n**Summary:**\n- $\\frac{3}{2} < x < \\frac{19}{2}$ for $n = 10$\n- $\\frac{19}{7} < x < \\frac{65}{7}$ for $n = 11$\n- $\\frac{17}{4} < x < \\frac{35}{4}$ for $n = 12$\n- No solutions if $n \\ge 13$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16229,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n = 2m + 1$ for some integer $m$. $n$ points $A_1, A_2, \\dots, A_n$ are chosen on two parallel lines. What is the largest possible number of acute triangles among the triangles $A_iA_jA_k$ over $1 \\le i < j < k \\le n$?\n\n![](path/to/file.png)\n\nWhat is the largest possible number of acute triangles among all possible triangles formed by these points?",
    "options": [],
    "answer": "See solution",
    "solution": "Let us arrange the points on the two lines so that the projections of every two adjacent points from one line to the other surround exactly one point on the other line, and there are no obtuse angles $\\angle ABC$ such that $B$ lies on one line and $A, C$ on the other (this can be achieved if the distances between points are sufficiently small). In this case, there are exactly $$\\frac{1}{6}m(m+1)(2m+1)$$ acute triangles.\n\nProject all the points onto one of the two lines and color the projections of points from one line white and from the other black. An acute triangle can be formed only from points $A, B, C$ such that $B$ lies between $A$ and $C$, and $A$ and $C$ have the same color, which is different from $B$ (though the converse is not necessarily true). We show by induction that there are at most $1^2 + 2^2 + \\cdots + m^2 = \\frac{1}{6}m(m+1)(2m+1)$ such good triples when there are $2m+1$ points.\n\nFor $m=0$, the statement is obvious. Suppose it holds for $m-1$. For $m$, number the points as $A_1, A_2, \\dots, A_{2m+1}$ and consider a coloring with the largest number of such triples.\n\nAssume $A_1$ and $A_{2m+1}$ have the same color (WLOG, white). Among $A_2$ to $A_{2m}$, let there be $w$ white and $b$ black points. There are at most $1^2 + 2^2 + \\cdots + (m-1)^2$ good triples among $A_2$ to $A_{2m}$, $b$ triples containing both $A_1$ and $A_{2m+1}$, and $bw$ triples containing exactly one of $A_1$ or $A_{2m+1}$ (since for every pair of points of different colors, there is exactly one point from $A_1$ or $A_{2m+1}$ forming a good triple). Thus, there are at most $1^2 + 2^2 + \\cdots + (m-1)^2 + b(w+1)$ good triples. Since $b+w=2m-1$, $b(w+1) \\le m^2$, so there are at most $1^2 + 2^2 + \\cdots + (m-1)^2 + m^2$ good triples, and induction proceeds. Now assume $A_1$ and $A_{2m+1}$ have different colors.\n\nConsider any two adjacent points $A_i$ (white) and $A_{i+1}$ (black). Let $w_1$ and $b_1$ be the numbers of white and black points among $A_1$ to $A_{i-1}$, and $w_2$ and $b_2$ among $A_{i+2}$ to $A_{2m+1}$. Swapping $A_i$ and $A_{i+1}$ increases the number of good triples by $(b_2+w_1)-(b_1+w_2)$. Since the coloring is optimal, $(b_1+w_2) \\ge (b_2+w_1)$, so $(w_1-b_1) \\le (w_2-b_2)$.\n\nWLOG, let $A_1$ be white and $A_{2m+1}$ black. Let $A_k$ be the first black point and $A_l$ the last white ($k<l$; otherwise, there are no good triples). Applying the above to $A_{k-1}, A_k$: to the left, (white)-(black) = $k-2 \\ge 0$, so to the right, this is non-negative. For $A_l, A_{l+1}$: to the right, (white)-(black) = $l-2m \\le 0$, so to the left, this is non-positive. But previously, we showed this difference is non-negative, so it's zero, and the number of points is even, which is a contradiction.\n\nHence, the induction proceeds.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16230,
    "subject": "Mathematics (Olympiad)",
    "question": "On the sides $AB$ and $BC$ of the square $ABCD$ we mark points $K$ and $L$ respectively so that $KB = LC$. Segments $AL$ and $CK$ intersect at point $P$. Prove that lines $DP$ and $KL$ are perpendicular.",
    "options": [],
    "answer": "See solution",
    "solution": "By construction, $\\triangle ADK = \\triangle BAL$. Then $\\angle BAL + \\angle ALB = \\angle BAL + \\angle AKD = 90^\\circ$, and $\\triangle AMK$ has a right angle at $M$, so $AL \\perp DK$ (see figure below). Similarly, $DL \\perp CK$. Then lines $AL$ and $CK$, which intersect at point $P$, contain two altitudes of $\\triangle DKL$, and therefore the line $DP$ must contain the third altitude, so $KL \\perp DP$, as desired.\n\n![](images/Ukraine_2021-2022_p25_data_a51cc7507f.png)\n\nFig. 14",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16231,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be real numbers with $0 \\leq a, b, c \\leq 2$. Prove that\n\n$$\n(a - b)(b - c)(a - c) \\leq 2.\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "Order the variables by size.\n\nFor $a \\geq b \\geq c$, all three factors are positive, so $(a-b)(b-c)(a-c) \\geq 0$.\n\nFor $b \\geq c \\geq a$ and $c \\geq a \\geq b$, two factors are negative and one is positive, so $(a-b)(b-c)(a-c) \\geq 0$.\n\nFor other orderings, either all factors are negative or one is negative and two are positive, so $(a-b)(b-c)(a-c) \\leq 0$; thus, the inequality holds and there is no equality in these cases.\n\nConsider $a \\geq b \\geq c$:\n\nBy the AM-GM inequality,\n$$\n(a - b)(b - c) \\leq \\frac{(a - b + b - c)^2}{4} = \\frac{(a - c)^2}{4}.\n$$\nSo,\n$$\n(a - b)(b - c)(a - c) \\leq \\frac{(a - c)^2}{4}(a - c) = \\frac{(a - c)^3}{4} \\leq \\frac{2^3}{4} = 2.\n$$\n\nThe other orderings can be treated analogously.\n\nEquality holds when $a-c=2$ and $a-b=b-c$, which gives $a=2$, $b=1$, $c=0$. Considering analogous cases, equality holds for $(2, 1, 0)$, $(1, 0, 2)$, and $(0, 2, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16232,
    "subject": "Mathematics (Olympiad)",
    "question": "Every integer $N$ satisfies at least one of the congruences $N \\equiv 0 \\pmod{2}$, $N \\equiv 1 \\pmod{3}$, $N \\equiv 3 \\pmod{4}$, $N \\equiv 5 \\pmod{6}$, $N \\equiv 9 \\pmod{12}$. Therefore, $n$ can be $5$. Show that $n \\leq 4$ is not possible under these conditions.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $1 < k_1 \\leq k_2 \\leq \\dots \\leq k_n$ and $a_1, a_2, \\dots, a_n$ be integers, and let $K = \\mathrm{lcm}(k_1, k_2, \\dots, k_n)$. Since at most $\\frac{K}{k_1} + \\frac{K}{k_2} + \\dots + \\frac{K}{k_n}$ integers from $1$ to $K$ can satisfy at least one of the congruences $x \\equiv a_i \\pmod{k_i}$ for $1 \\leq i \\leq n$, we must have\n$$\n\\frac{1}{k_1} + \\frac{1}{k_2} + \\dots + \\frac{1}{k_n} \\geq 1\n$$\nif every integer satisfies at least one of these congruences.\n\nNow assume that $1 < k_1 < k_2 < \\dots < k_n$ and $a_1, a_2, \\dots, a_n$ satisfy the condition of the problem and that $n \\leq 4$ has the smallest possible value. If $k_1 = 3$, then\n$$\n\\frac{1}{k_1} + \\frac{1}{k_2} + \\dots + \\frac{1}{k_n} \\leq \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5} + \\frac{1}{6} = \\frac{19}{20} < 1.\n$$\nTherefore, $k_1 = 2$.\n\nWithout loss of generality, assume $a_1 = 1$. For $2 \\leq i \\leq n$, let $k_i' = k_i$ and $a_i' = 2^{-1} a_i \\pmod{k_i}$ if $k_i$ is odd, and let $k_i' = \\frac{k_i}{2}$ and $a_i' = \\frac{a_i}{2}$ if $k_i$ is even. The integers $k_2', \\dots, k_n'$ and $a_2', \\dots, a_n'$ satisfy the condition of the problem except that $k_i'$ might not be distinct. Therefore, by the minimality of $n$, we must have $n = 4$ and $\\{k_2, k_3, k_4\\} = \\{2m+1, 4m+2, k\\}$.\n\n- If $k$ is odd, then $\\{k_2', k_3', k_4'\\} = \\{2m+1, 2m+1, k\\}$ and $\\frac{2}{2m+1} + \\frac{1}{k} \\geq 1$. Since $\\frac{2}{2m+1} + \\frac{1}{k} \\leq \\frac{2}{3} + \\frac{1}{5} = \\frac{13}{15} < 1$, this is not possible.\n\n- If $k$ is even, then $\\{k_2', k_3', k_4'\\} = \\{2m+1, 2m+1, \\frac{k}{2}\\}$ and $\\frac{2}{2m+1} + \\frac{2}{k} \\geq 1$. If $2m+1 \\geq 5$ or $2m+1 = 3$ and $k \\geq 8$, we get contradictions because\n$$\n\\frac{2}{5} + \\frac{2}{4} = \\frac{9}{10} < 1 \\quad \\text{and} \\quad \\frac{2}{3} + \\frac{2}{8} = \\frac{11}{12} < 1.\n$$\nThe only remaining case is $2m+1 = 3$ and $k = 4$. This gives $\\{k_2', k_3', k_4'\\} = \\{3, 3, 2\\}$. Since the integers in a congruence class modulo $3$ cannot be all even or all odd, this also leads to a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16233,
    "subject": "Mathematics (Olympiad)",
    "question": "Позначимо через $m$, $0 \\le m \\le n$, кількість чисел множини $\\{1, 2, \\dots, n\\}$, які знаходяться серед індексів $i_1 < i_2 < \\dots < i_n$. Неважко довести, що такими $m$ індексами визначаються $n - m$ чисел із множини $\\{n+1, n+2, \\dots, 2n\\}$, які увійдуть до даного набору індексів $i_1 < i_2 < \\dots < i_n$. Цим повністю описуються всілякі вдалі послідовності $a_1, a_2, \\dots, a_{2n}$. Для $m = 0$ маємо вдалу послідовність $1, 2, \\dots, n$, $1, 2, \\dots, n$. Але ж ця сама вдала послідовність утворюється ще в $n$ випадках, коли для набору індексів $i_1 < i_2 < \\dots < i_n$ маємо, що\n\n$$\n\\{i_1, i_2, \\dots, i_n\\} \\cap \\{1, 2, \\dots, n\\} = \\{1, 2, \\dots, m\\}, \\quad 1 \\le m \\le n.\n$$\n\nУсіляких підмножин $n$-елементної множини $\\{1, 2, \\dots, n\\}$, відмінних від підмножин виду $\\{1, 2, \\dots, m\\}$, $1 \\le m \\le n$, скільки існує?",
    "options": [],
    "answer": "See solution",
    "solution": "Таких підмножин існує $2^n - n$. Усім таким підмножинам відповідають різні вдалі послідовності (порожня підмножина відповідає випадку $m=0$ і також враховується серед $2^n - n$ підмножин).\n\nВідповідь: $2^n - n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16234,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $m$, let $S(m)$ denote the positive integer whose decimal representation is equal to the octal representation of $m$. For example, $S(64) = 100$, $S(100) = 144$, $S(2023) = 3747$.\n\nTwo positive integers $x, y$ satisfy the relation\n\n$$\n\\frac{S(x) + S(1)}{S(x + 1)} = \\frac{S(y) + S(2023)}{S(y + 2023)} < \\frac{8}{10}.\n$$\n\nWhat is the smallest possible value of $y$ if $x < 100$?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $2137$.\n\nFirst, observe that $x = 63$, $y = 2137$ satisfies the given relation. In fact,\n\n$$\n\\frac{S(63) + S(1)}{S(64)} = \\frac{77 + 1}{100} = \\frac{78}{100} < \\frac{8}{10} \\\\\n\\frac{S(2137) + S(2023)}{S(4160)} = \\frac{4131 + 3747}{10100} = \\frac{7878}{10100} = \\frac{78}{100} < \\frac{8}{10}.\n$$\n\nNow, assume the octal representation of $x$ ends with exactly $k$ digits “7” (possibly $k = 0$). That is, $x = a \\cdot 8^k + 8^k - 1$ where $a$ is a nonnegative integer with $a \\bmod 8 < 7$. In particular, $S(a + 1) = S(a) + 1$ because there is no carry in octal addition, since $a \\bmod 8 < 7$. Therefore,\n\n$$\nS(x) = S(a \\cdot 8^k + 8^k - 1) = S(a) \\cdot 10^k + \\frac{7}{9}(10^k - 1) \\\\\nS(x + 1) = S((a + 1) \\cdot 8^k) = S(a + 1) \\cdot 10^k = (S(a) + 1) \\cdot 10^k.\n$$\n\nIf $k = 0$, then $S(x) + S(1) = S(x + 1)$, so $k \\neq 0$. If $k \\ge 3$, then $x \\ge 8^3 - 1 > 100$, so $k = 1$ or $k = 2$. If $k = 2$, then $a = 0$ (since $8^2 + 8^2 - 1 > 100$), so $x = 8^2 - 1 = 63$ and $\\frac{S(x)+S(1)}{S(x+1)} = \\frac{78}{100}$. $k = 1$ is impossible because:\n\n$$\n\\frac{10S(a) + 8}{10S(a) + 10} < \\frac{8}{10} \\implies 10S(a) + 8 < 8S(a) + 8 \\implies S(a) < 0.\n$$\n\nNow, suppose there is a solution with $y < 2137 = 4131_8$. Then $y$ has at most 4 digits in octal. Let $d_i$ for $i = 0, 1, 2, 3$ be $1$ if there is a carry at the $i$-th position when adding $y$ and $2023 = 3747_8$ in octal. It follows that\n\n$$\nS(y+2023)-S(y)-S(2023) = \\sum_{i=0}^{3} (-8) \\cdot 10^i \\cdot d_i + \\sum_{i=0}^{3} 10^{i+1} \\cdot d_i = 2 \\sum_{i=0}^{3} 10^i \\cdot d_i.\n$$\n\nWe have:\n\n$$\n\\frac{S(y) + S(2023)}{S(y + 2023)} = \\frac{78}{100} \\implies \\frac{S(y) + S(2023)}{S(y + 2023) - S(y) - S(2023)} = \\frac{78}{22} \\\\\n\\implies S(y) + S(2023) = \\frac{78}{11} \\sum_{i=0}^{3} 10^i \\cdot d_i.\n$$\n\nWe know $d_3 = 1$; otherwise,\n\n$$\nS(y) + S(2023) \\le \\frac{78}{11} \\cdot 111 < 3747 = S(2023).\n$$\n\nSince $78$ is coprime with $11$, $11$ must divide $\\sum_{i=0}^{3} 10^i \\cdot d_i$. With $d_3 = 1$, possible values are $1001, 1100, 1111$, corresponding to $S(y) + S(2023) = 7098, 7800, 7878$, so $S(y) = 3351, 4053, 4131$. The first two are impossible because they require $d_1 = 0$, but $5 + 4 > 7$, so there must be a carry at the first position.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16235,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, and $z$ be odd positive integers such that $\\gcd(x, y, z) = 1$ and the sum $x^2 + y^2 + z^2$ is divisible by $x + y + z$. Prove that $x + y + z - 2$ is not divisible by $3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there exists a prime divisor $p \\equiv 2 \\pmod{3}$ of $x + y + z$. Since $z \\equiv -(x + y) \\pmod{p}$, we have $2(x^2 + y^2 + xy) \\equiv 0 \\pmod{p}$. Multiplying by $x - y$, we get $x^3 \\equiv y^3 \\pmod{p}$, but this yields $x \\equiv y \\pmod{p}$, because $\\gcd(3, p-1) = 1$. Similarly, $x \\equiv z \\pmod{p}$, whence $3x \\equiv 0 \\pmod{p}$, which means that $x$, $y$, and $z$ are divisible by $p$, contradicting $\\gcd(x, y, z) = 1$. Hence, $x + y + z$ has no prime divisors which have a remainder $2$ modulo $3$, so $x + y + z - 2$ is not divisible by $3$.\n\n$$\\boxed{}$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16236,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the _odd part_ of a positive integer $n$ be the greatest odd integer that divides $n$.\n\nDoes there exist a positive odd integer that cannot be represented as a product of the odd parts of two consecutive positive integers?",
    "options": [],
    "answer": "See solution",
    "solution": "Let us show that the number $11$ cannot be represented as a product of the odd parts of two consecutive positive integers.\n\nAssume the opposite: let $11 = x \\cdot y$, where $x$ and $y$ are the odd parts of two consecutive positive integers. As $11$ is a prime number, either $x = 1$ and $y = 11$ or $x = 11$ and $y = 1$.\n\nOf the two consecutive integers, one is always odd, and the odd part of an odd number is the number itself. Thus, either $x$ or $y$ is one of the two consecutive integers. If it is $1$, then the other number can only be $2$, but the odd part of $2$ is not $11$. If it is $11$, then the other number can only be $10$ or $12$, but neither of those has odd part equal to $1$.\n\nIn all cases, we get a contradiction, which proves the statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16237,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $3 \\leq n$ be an integer and let real numbers $a_2, a_3, \\dots, a_n$ satisfy $a_2 a_3 \\cdots a_n = 1$. Prove that\n$$\n(1 + a_2)^2 (1 + a_3)^3 \\cdots (1 + a_n)^n > n^n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us use the substitution $a_2 = \\frac{x_2}{x_1}$, $a_3 = \\frac{x_3}{x_2}$, ..., $a_n = \\frac{x_1}{x_{n-1}}$. The inequality becomes\n$$\n(x_1 + x_2)^2 (x_2 + x_3)^3 \\cdots (x_{n-1} + x_1)^n > n^n x_1^2 x_2^3 \\cdots x_{n-1}^{n-1},\n$$\nfor all $x_1, \\dots, x_{n-1} > 0$. Apply the AM-GM inequality to each factor:\n\n$$\n\\begin{align*}\n(x_1 + x_2)^2 &\\geq 2^2 x_1 x_2 \\\\\n(x_2 + x_3)^3 &\\geq 3^3 \\left(\\frac{x_2}{2}\\right)^2 x_3 \\\\\n(x_3 + x_4)^4 &\\geq 4^4 \\left(\\frac{x_3}{3}\\right)^3 x_4 \\\\\n&\\vdots \\\\\n(x_{n-1} + x_1)^n &\\geq n^n \\left(\\frac{x_{n-1}}{n-1}\\right)^{n-1} x_1.\n\\end{align*}\n$$\n\nMultiplying these inequalities gives the desired result, with $\\geq$ instead of $>$. However, equality would require $x_1 = x_2 = \\cdots = x_{n-1} = (n-1)x_1$, which is impossible for $n \\geq 3$ and $x_1 > 0$. Thus, the inequality is strict.\n\nAlternatively, without substitution, apply the weighted AM-GM inequality to each $(1 + a_k)^k$:\n$$\n(1 + a_k)^k \\geq \\frac{k^k}{(k-1)^{k-1} a_k}.\n$$\nMultiplying for $k = 2$ to $n$ gives\n$$\n(1 + a_2)^2 (1 + a_3)^3 \\cdots (1 + a_n)^n \\geq n^n a_2 a_3 \\cdots a_n = n^n.\n$$\nAs before, equality cannot be attained, so the inequality is strict.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16238,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it true that for any positive integer $n > 1$, there exists an infinite arithmetic progression $M_n$ of positive integers, such that for any $m \\in M_n$, the number $n^m - 1$ is not a perfect power? (A positive integer is a perfect power if it is of the form $a^b$ for positive integers $a, b > 1$.)",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is yes. Fix a positive integer $n$ and two large distinct primes $p, q > n$. Let $d_p = \\operatorname{ord}_p(n)$, $d_q = \\operatorname{ord}_q(n)$, $c_p = \\nu_p(n^{d_p} - 1)$, $c_q = \\nu_q(n^{d_q} - 1)$, and let $c = \\nu_p(d_q)$, $d = \\nu_q(d_p)$. Pick two sufficiently large constants $a, b$, and let\n\n$$\nM := d_p d_q p^{(a-1)c_p + b c_q - c} q^{a c_p + (b-1)c_q + 1 - d}\n$$\n\nFinally, choose $M_n$ to consist of $m = M(1 + iM)$ for $i = 1, 2, \\dots$.\n\nBy LTE, we have\n\n$$\n\\nu_p(n^m - 1) = \\nu_p(n^{d_p} - 1) + \\nu_p\\left(\\frac{m}{d_p}\\right) = c_p + c + (a-1)c_p + b c_q - c = a c_p + b c_q\n$$\n\n(we used that $\\gcd(1 + iM, p) = 1$), and similarly $\\nu_q(n^m - 1) = a c_p + b c_q + 1$, which are consecutive, i.e., they can't have a common divisor greater than 1 and thus $n^m - 1$ can't be a perfect power. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16239,
    "subject": "Mathematics (Olympiad)",
    "question": "請用數學歸納法證明：在第 $i$ 回合結束時，滿足下列其一：\n\n- 魷魚知道 $(D_1, D_2) \\in S(k_{i-1})$，或\n- 其中一隻魷魚可以判斷大小。\n\n此外，證明下界：當 $M = 2^k + 1$ 時，無法在喊出數字總和小於 $k$ 時保證獲勝。",
    "options": [],
    "answer": "See solution",
    "solution": "我們用數學歸納法證明：\n\n- 當 $i=1$ 時，顯然成立。\n- 假設命題對 $i$ 成立，考慮第 $i+1$ 回合：\n  1. 若上回合無法判斷大小，則由歸納假設，魷魚知 $(D_1, D_2) \\in S(k_{i-1})$。\n  2. 由 $k_i$ 的定義，魷魚知 $p_{i,k_{i-1}+1} = \\cdots = p_{i,k_{i-1}} = 0$ 且 $p_{i,k_i} = 1$。\n\n接下來分情況：\n\n- 若存在 $k_{i-1} < \\ell < k_i$ 使 $p_{i+1,\\ell} = 1$，則可推得：\n  $$\n  D_{i+1} \\ge 1 + \\sum_{j=1}^{k_{i-1}} p_{i+1,j} 2^{N+1-j} + 2^{N+1-\\ell} > 1 + \\sum_{j=1}^{k_{i-1}} p_{i,j} 2^{N+1-j} + \\sum_{j=\\ell+1}^{N} 2^{N+1-j} \\ge D_i\n  $$\n  因此可判別大小。\n\n- 若否且 $p_{i+1,k_i} = 0$，則：\n  $$\n  D_{i+1} \\le 1 + \\sum_{j=1}^{k_{i-1}} p_{i+1,j} 2^{N+1-j} + \\sum_{j=\\ell+1}^{N} 2^{N+1-j} < 1 + \\sum_{j=1}^{k_{i-1}} p_{i+1,j} 2^{N+1-j} + 2^{N+1-\\ell} \\le D_i\n  $$\n  也可判別大小。\n\n- 若以上皆非，則 $p_{1,j} = p_{2,j}$ 對 $k_{i-1} < j \\le k_i$ 都成立，故 $(D_1, D_2) \\in S(k_i)$。\n  - 若對所有 $j > k_i$ 有 $p_{i+1,j} = 0$，則：\n    $$\n    D_{i+1} = 1 + \\sum_{j=1}^{k_i} p_{i+1,j} 2^{N+1-j} \\le 1 + \\sum_{j=1}^{k_i} p_{i,j} 2^{N+1-j} + 2 \\le D_i\n    $$\n    可判別大小。\n  - 否則，$\\{k : p_{i,k} = 1, k > k_{i-1}\\}$ 非空，可計算 $k_{i+1}$。\n\n**下界證明：**\n\n當 $M = 2^k + 1$，無法在喊出數字總和小於 $k$ 時保證獲勝。\n\n對 $k$ 歸納：\n- $k=1$ 時顯然。\n- 假設對 $k < K$ 成立。第一隻魷魚僅在 $D_1 = 1$ 或 $D_1 = M$ 時能直接判斷。對 $i=1,2,\\dots,K-1$，若 $D_1 \\in A_i$ 時喊 $i$，則必有 $|A_i| \\ge 2^{K-i} + 1$，否則：\n  $$\n  2^K - 1 = |\\{2,3,\\dots,M-1\\}| = \\left| \\bigcup_{i=1}^{K-1} A_i \\right| \\le 2^{K-1} + 2^{K-2} + \\dots + 2^1 < 2^K - 1\n  $$\n  矛盾。\n\n因此第一回合後 $D_1$ 可能值至少 $2^{K-i} + 1$，且這些值皆可能為 $D_2$。從第二回合起，相當於有至少 $2^{K-i} + 1$ 個可能值、喊數總和小於 $K-i$ 的遊戲。由歸納假設，無法保證獲勝。證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16240,
    "subject": "Mathematics (Olympiad)",
    "question": "Together, the two positive integers $a$ and $b$ have 9 digits and contain each of the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 exactly once. For which possible values of $a$ and $b$ is the fraction $a/b$ closest to 1?",
    "options": [],
    "answer": "See solution",
    "solution": "If $a > b$, then $a$ has at least five digits, so $a \\geq 12345$, and $b$ has at most four digits, so $b \\leq 9876$. In this case,\n\n$$\n\\frac{a}{b} \\geq \\frac{12345}{9876} > 1,\n$$\n\nso the value of $a/b$ that is closest to 1 is\n\n$$\n\\frac{12345}{9876} = 1 + \\frac{2469}{9876}.\n$$\n\nOn the other hand, if $a < b$ (note $a = b$ is impossible, since $a$ and $b$ cannot have the same number of digits), then $a$ has at most four digits and $b$ at least five, so $a \\leq 9876$ and $b \\geq 12345$, which means that\n\n$$\n\\frac{a}{b} \\leq \\frac{9876}{12345} < 1.\n$$\n\nHence in this case the value that is closest to 1 is\n\n$$\n\\frac{9876}{12345} = 1 - \\frac{2469}{12345}.\n$$\n\nSince the denominator $12345$ is greater than the denominator $9876$, we see that the value of $\\frac{9876}{12345}$ is closer to 1 than that of $\\frac{12345}{9876}$ (in fact, $\\frac{9876}{12345} = 4/5$ and $\\frac{12345}{9876} = 5/4$, so the distances are $1/5$ and $1/4$ respectively). Thus, the values of $a$ and $b$ for which $a/b$ is closest to 1 are $a = 9876$ and $b = 12345$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16241,
    "subject": "Mathematics (Olympiad)",
    "question": "The least positive integer with exactly $2021$ distinct positive divisors can be written in the form $m \\cdot 6^k$, where $m$ and $k$ are integers and $6$ is not a divisor of $m$. What is $m + k$?\n\n(A) 47  (B) 58  (C) 59  (D) 88  (E) 90",
    "options": [],
    "answer": "See solution",
    "solution": "The number of positive integer divisors of a number with prime factorization $p_1^{e_1} p_2^{e_2} \\cdots p_n^{e_n}$ is $(e_1 + 1)(e_2 + 1) \\cdots (e_n + 1)$.\n\n$$\n2021 = 45^2 - 2^2 = (45 + 2)(45 - 2) = 47 \\cdot 43\n$$\n\nA number with $2021$ divisors can be of the form $p^{2020}$ or $p^{46} q^{42}$, with $p, q$ distinct primes. The minimal such number is $2^{46} \\cdot 3^{42}$:\n\n$$\n2^{46} \\cdot 3^{42} = 2^4 \\cdot 2^{42} \\cdot 3^{42} = 2^4 \\cdot 6^{42} = 16 \\cdot 6^{42}\n$$\n\nThus, $m = 16$, $k = 42$, so $m + k = 58$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16242,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the following inequalities for positive numbers $a$, $b$, and $c$:\n\n$$\n(a + b + c)(a^2 + b^2 + c^2) - 2(a^3 + b^3 + c^3) > 0\n$$\n\nand\n\n$$\n(a + b + c)(a^2 + b^2 + c^2) \\le 3(a^3 + b^3 + c^3).\n$$\n\nProve the first inequality using the triangle inequality, and prove the second using Hölder's inequality or convexity arguments. State when equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove the left inequality using the triangle inequality: $a < b + c$, $b < c + a$, $c < a + b$. Applying this,\n\n$$\n\\begin{aligned}\n& (a + b + c)(a^2 + b^2 + c^2) - 2(a^3 + b^3 + c^3) \\\\\n&= a^2(b + c) + b^2(c + a) + c^2(a + b) - a^3 - b^3 - c^3 \\\\\n&= a^2(b + c - a) + b^2(c + a - b) + c^2(a + b - c) > 0.\n\\end{aligned}\n$$\n\nThe right inequality holds for any positive $a, b, c$, and can be shown by Hölder's inequality or convexity of $t \\mapsto t^p$ for $p \\ge 1$. For $x, y, z > 0$ and $p \\ge 1$:\n\n$$\n\\left(\\frac{x + y + z}{3}\\right)^p \\le \\frac{x^p + y^p + z^p}{3},\n$$\n\nwith equality iff $x = y = z$. Applying this with $x = a$, $y = b$, $z = c$, $p = 3$:\n\n$$\na + b + c \\le (a^3 + b^3 + c^3)^{1/3} 3^{2/3},\n$$\n\nand with $x = a^2$, $y = b^2$, $z = c^2$, $p = 3/2$:\n\n$$\na^2 + b^2 + c^2 \\le (a^3 + b^3 + c^3)^{2/3} 3^{1/3}.\n$$\n\nMultiplying, we get:\n\n$$\n(a + b + c)(a^2 + b^2 + c^2) \\le 3(a^3 + b^3 + c^3),\n$$\n\nwith equality iff $a = b = c$.\n\n---\n\n*Alternative Solution:*\n\nLet $x = a + b - c$, $y = b + c - a$, $z = c + a - b$ (all positive for triangle sides), so\n\n$$\na = \\frac{z + x}{2}, \\quad b = \\frac{x + y}{2}, \\quad c = \\frac{y + z}{2}.\n$$\n\nThen $a + b + c = x + y + z$,\n\n$$\na^2 + b^2 + c^2 = \\frac{1}{2}(x^2 + y^2 + z^2 + xy + yz + zx),\n$$\n\n$$\n(a + b + c)(a^2 + b^2 + c^2) = \\frac{1}{2}\\left(\\sum x^3 + 2\\sum xy(x + y) + 3xyz\\right),\n$$\n\n$$\na^3 + b^3 + c^3 = \\frac{1}{8}(2\\sum x^3 + 3\\sum xy(x + y)).\n$$\n\nSo,\n\n$$\n\\begin{aligned}\n& (a + b + c)(a^2 + b^2 + c^2) - 2(a^3 + b^3 + c^3) \\\\\n&= \\frac{1}{4}\\left(\\sum xy(x + y) + 6xyz\\right),\n\\end{aligned}\n$$\n\nwhich is positive. For the right inequality,\n\n$$\n\\begin{aligned}\n& 3(a^3 + b^3 + c^3) - (a + b + c)(a^2 + b^2 + c^2) \\\\\n&= \\frac{1}{8}((x^3 + y^3 + z^3 - 3xyz) + ((x^2 + y^2 + z^2)(x + y + z) - 9xyz)) \\\\\n&\\ge 0,\n\\end{aligned}\n$$\n\nby AM-GM, with equality iff $x = y = z$ (i.e., $a = b = c$).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16243,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $T$ be the sum of all assigned numbers on the box, where for each cell $u$ in $S_n$, $c=2$ if the diagonal band through $u$ has no more in $S_n$ and $c=1$ otherwise. From the above observation, two of $a, b, c$ are 1 and the third one is at most 2. This implies that $a + b + c \\le 4$.\n\n$$\nT = \\sum_{u \\in S_n} (a + b + c) \\le 4|S_n|.\n$$\n\nOn the other hand, there is at least one chosen cell on each $1 \\times 1 \\times n$ band (in any direction) of the box. This implies that $T \\ge 2 \\cdot 3 (2n)^2 = 24n^2$.\n\nTherefore, $4|S_n| \\ge 24n^2$ or $|S_n| \\ge 6n^2$.\n\nWe now have the following statements:\n\n- In the $2 \\times 2 \\times 2$ box, Binh needs to choose at least 6 cells.\n- In the $10 \\times 10 \\times 10$ box, Binh needs to choose at least 150 cells.\n\nShow how Binh can choose 150 cells in the $10 \\times 10 \\times 10$ box so that every $1 \\times 1 \\times 10$ band contains at least one chosen cell.\n\n% IMAGE: ![](images/Vietnam_Booklet_2013_12-07_p65_data_d92858e6f9.png)",
    "options": [],
    "answer": "See solution",
    "solution": "% IMAGE: ![](images/Vietnam_Booklet_2013_12-07_p66_data_190bc40ee6.png)\n\nIn the $i$th layer ($i=1,2,3,4,5$), Binh picks the boxes labeled $i$ and removes any two opposite cells in these boxes, asking for the color of the remaining ones. If we project the chosen cells onto any face of the box, the shadow covers the whole face. This implies that any chosen $1 \\times 1 \\times n$ band passes through one of the chosen $2 \\times 2 \\times 2$ boxes, so Binh can determine whether this band is black or not.\n\nHence, the least number of cells that Binh must choose is 150.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16244,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the triangle $ABC$ and the points $D \\in (BC)$ and $M \\in (AD)$. Lines $BM$ and $AC$ meet at $E$, lines $CM$ and $AB$ meet at $F$, and lines $EF$ and $AD$ meet at $N$. Prove that $2 \\dfrac{AN}{DN} = \\dfrac{AM}{DM}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Set $k = \\dfrac{AM}{DM}$ and $p = \\dfrac{DB}{DC}$. Apply Menelaus' theorem in triangle $ABM$ with transversal $F$–$N$–$E$ to get\n\n$$\n\\frac{NA}{NM} = \\frac{EB}{EM} \\cdot \\frac{FA}{FB}.\n$$\n\nNext, apply Menelaus' theorem in triangle $ABD$ with transversal $C$–$M$–$F$ to obtain $\\dfrac{FA}{FB} = \\dfrac{k}{p+1}$. From Van Aubel's Theorem, we get\n$$\n\\frac{MB}{ME} = \\frac{DB}{DC} + \\frac{FB}{FA} = \\frac{pk + p + 1}{k},\n$$\nimplying $\\dfrac{EB}{EM} = \\dfrac{(p+1)(k+1)}{k}$.\n\nThe relation $\\dfrac{NA}{NM} = \\dfrac{EB}{EM} \\cdot \\dfrac{FA}{FB}$ gives $\\dfrac{NA}{NM} = k + 1$, hence $\\dfrac{NA}{AM} = \\dfrac{k+1}{k+2}$ and then $\\dfrac{NA}{AD} = \\dfrac{k}{k+2}$. Consequently, $\\dfrac{AN}{DN} = \\dfrac{k}{2}$, as needed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16245,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle, and let $D$ be a point on side $BC$. A line through $D$ intersects side $AB$ at $X$ and ray $AC$ at $Y$. The circumcircle of triangle $BXD$ intersects the circumcircle $\\omega$ of triangle $ABC$ again at point $Z \\neq B$. The lines $ZD$ and $ZY$ intersect $\\omega$ again at $V$ and $W$, respectively. Prove that $AB = VW$.",
    "options": [],
    "answer": "See solution",
    "solution": "Applying the Pivot theorem to $\\triangle AXY$, we see that the circles $ABC$, $XBD$, and $CDY$ are concurrent at point $Z$. Hence, $CDZY$ is cyclic.\n\n![](images/Australian-Scene-combined-2015_p114_data_4f1a9b1bcd.png)\n\nLet $\\angle(m, n)$ denote the directed angle between any two lines $m$ and $n$. We have\n\n$$\n\\begin{align*}\n\\angle(WZ, VZ) &= \\angle(YZ, DZ) \\\\\n&= \\angle(YC, DC) \\quad (CDZY \\text{ cyclic}) \\\\\n&= \\angle(AC, BC).\n\\end{align*}\n$$\n\nTherefore, $VW$ and $AB$ subtend equal or supplementary angles in $\\omega$. It follows that $VW = AB$. $\\square$\n\n**Comment**: All solutions that were dependent on how the diagram was drawn received a penalty deduction of 1 point. The easiest way to avoid diagram dependence was to use directed angles as in the solution presented above.\n\n*The point $Z$ is also the \\textit{Miquel point} of the four lines $AB$, $AY$, $BC$, and $XY$. It is the common point of the circumcircles of $\\triangle ABC$, $\\triangle AXY$, $\\triangle BDX$, and $\\triangle CDY$. See the sections entitled \\textit{Pivot theorem} and \\textit{Four lines and four circles} found in chapter 5 of \\textit{Problem Solving Tactics} published by the AMT.*\n\n*This is the angle by which one may rotate $m$ anticlockwise to obtain a line parallel to $n$. For more details, see the section \\textit{Directed angles} in chapter 17 of \\textit{Problem Solving Tactics} published by the AMT.*",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16246,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs $ (p, q) $ of primes for which both $ p^2 + q^3 $ and $ q^2 + p^3 $ are perfect squares.",
    "options": [],
    "answer": "See solution",
    "solution": "There is only one such pair, namely $ (p, q) = (3, 3) $.\n\nLet the pair $ (p, q) $ be as described in the statement of the problem.\n\n1. First, we show that $ p \\neq 2 $. Otherwise, there would exist a prime $ q $ for which $ q^2 + 8 $ and $ q^3 + 4 $ are perfect squares. Because $ q^2 < q^2 + 8 $, the second condition gives $ (q+1)^2 \\leq q^2 + 8 $ and hence $ q \\leq 3 $. But for $ q = 2 $ or $ q = 3 $, the expression $ q^3 + 4 $ fails to be a perfect square. Hence indeed $ p \\neq 2 $ and due to symmetry we also have $ q \\neq 2 $.\n\n2. Next, we consider the special case $ p = q $. Then $ p^2(p+1) $ is a perfect square, so there exists an integer $ n $ satisfying $ p = n^2 - 1 = (n+1)(n-1) $. Since $ p $ is prime, this factorization yields $ n = 2 $ and thus $ p = 3 $. This completes the discussion of the case $ p = q $.\n\n3. So from now on we may suppose that $ p $ and $ q $ are distinct odd primes. Let $ a $ be a positive integer such that $ p^2 + q^3 = a^2 $, i.e., $ q^3 = (a+p)(a-p) $. If both factors $ a+p $ and $ a-p $ were divisible by $ q $, then so would their difference $ 2p $, which is absurd. So by uniqueness of prime factorization we have $ a+p = q^3 $ and $ a-p = 1 $. Subtracting these equations we learn $ q^3 = 2p+1 $. Due to symmetry we also have $ p^3 = 2q+1 $. Now if $ p < q $, then $ q^3 = 2p+1 < 2q+1 = p^3 $, which gives a contradiction, and the case $ q < p $ is excluded similarly.\n\nTherefore, the only solution is $ (p, q) = (3, 3) $.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16247,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all values of $k$ such that the equation\n\n$$x^3 + 2x^2 - k^2x - 2k^2 = 0$$\n\nhas three roots which form an arithmetic progression.",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to see that the roots of this equation are $\\pm k$ and $-2$. We can obtain different progressions depending on the order of these numbers in a progression.\n\nIf the order is $-k$, $k$, $-2$, then we must have $-2 - k = 2k$, and so $k = -\\frac{2}{3}$.\n\nFor the order $-k$, $-2$, $k$, the condition is $k - k = -4$, which is impossible.\n\nIf the progression is $k$, $-k$, $-2$, then $k - 2 = -2k \\Rightarrow k = \\frac{2}{3}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16248,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, and $C$ be the vertices of a triangle, and let $A'$, $B'$, and $C'$ be the midpoints of sides $BC$, $CA$, and $AB$, respectively. Each point on the sides of the triangle is colored either red or blue. Prove that there exists a trapezoid whose vertices are all the same color, with all its vertices lying on the sides of the triangle.\n\n![](images/Slovenija_2010_p22_data_e0d490ed36.png)",
    "options": [],
    "answer": "See solution",
    "solution": "At least two of the points $A'$, $B'$, and $C'$ must be the same color. Assume $A'$ and $B'$ are both red. The segment $A'B'$ is parallel to $AB$. If there are two red points on $AB$, we have a red trapezoid. Otherwise, at most one point on $AB$ is red (besides $A$ and $B$). If there is a red point $D$ on $AB$ (distinct from $A$ and $B$), let $D$ be that point; otherwise, let $D$ be any point on $AB$ (distinct from $A$ and $B$). Thus, all other points on $AB$ are blue.\n\nSuppose there are two distinct blue points $E$ and $F$ on $AC$ (both distinct from $A$), with $E$ closer to $A$. Let $F'$ be a blue point on $AD$ (distinct from $A$). Let $E'$ be the intersection of $AD$ and the line through $E$ parallel to $FF'$. Then $EE'F'F$ is a blue trapezoid.\n\nSimilarly, if there are not two blue points on $AC$, then $AC$ is almost entirely red, and a red trapezoid can be found by considering intersections of two parallel lines with $AC$ and $BC$.\n\nThus, in all cases, a monochromatic trapezoid exists.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16249,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $p(x)$ is a degree 2012 monic polynomial with integer coefficients and integer roots $x_1, x_2, \\dots, x_{2012}$. Show that the only integer solutions to $p(p(n)) = 0$ are the roots of $p(x)$ itself.",
    "options": [],
    "answer": "See solution",
    "solution": "Note the solutions of $p(x) = 0$ are also solutions of $p(p(x)) = 0$.\n\nNow suppose that there exists an integer $n$ which is not a solution of $p(x) = 0$ but which is a solution of $p(p(x)) = 0$. Since we know every root of $p(x)$, we may write\n\n$$\np(x) = x(x - x_2)(x - x_3) \\dots (x - x_{2012})\n$$\n\nand we know that $p(n) = x_k$ for some $k \\ge 2$.\n\nIn other words, we have the equation\n\n$$\nn(n - x_2)(n - x_3) \\dots (n - x_{2012}) = x_k\n$$\n\nfor some $k \\ge 2$. This equation implies that there exists an integer $A$ such that $An(n - r) = r$, where we set $r = x_k$. This is a quadratic equation in $n$ which has as its discriminant $A^2r^2 + 4Ar = (Ar + 2)^2 - 4$.\n\nSince $n$ is an integer root of this integer quadratic, it follows that the discriminant is a perfect square. However, the only perfect squares which differ by 4 are 0 and 4. This means that $Ar + 2 = 2$ or $Ar + 2 = -2$.\n\nIn the former case, we obtain $Ar = 0$, which contradicts the fact that $n$ is not a solution of $p(x) = 0$.\n\nIn the latter case, we obtain $Ar = -4$ which cannot be true, since $A$ is the product of 2010 distinct non-zero integers.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16250,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the circumcenter of a right triangle. The circle with smaller radius and center at point $O$ is tangent to the longer cathetus and to the height from the right angle.\n\n![](images/Ukraine_2016_Booklet_p15_data_0ee45cc7ca.png)\n\nFind the acute angles of the right triangle and the relation between the radii of the circumcircle and the other circle.",
    "options": [],
    "answer": "See solution",
    "solution": "$30^\\circ$, $60^\\circ$, $2:1$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16251,
    "subject": "Mathematics (Olympiad)",
    "question": "Turbo the snail is in the lower left cell of an $n \\times n$ array, $n \\geq 2$, and aims to reach the upper right cell by moving one cell rightwards or one cell upwards. Some cells contain monsters, visible to Turbo, and they must be avoided. Assume there is a unique way for Turbo to achieve his goal. In terms of $n$, determine the smallest possible number of monsters such an array may contain. (The minimum is over all configurations satisfying the unique path condition.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $(i, j)$ denote the cell on the $i$-th row and the $j$-th column, where $(1, 1)$ is the lower left cell. The required minimum is $n-1$ and is achieved by placing the monsters in the cells $(k, n-k+1)$ for $k = 2, \\dots, n$, to force Turbo to move rightwards from $(1, 1)$ to $(1, n)$ and then upwards to $(n, n)$.\n\nTo prove that the $n \\times n$ array must contain at least $n-1$ monsters, induct on $n$. The base case, $n=2$, is trivial.\n\nFor the inductive step, let $n \\geq 3$. If necessary, switch rows and columns to assume that Turbo first moves upwards to $(2, 1)$. Note that Turbo enters $(n, n)$ either from $(n, n-1)$ or from $(n-1, n)$.\n\n*Case 1.* Turbo visits $(n, n-1)$. Consider the subarray $\\{(i, j) : 2 \\leq i \\leq n, 1 \\leq j \\leq n-1\\}$. As it inherits uniqueness of a safe path for Turbo, it contains at least $n-2$ monsters, by the induction hypothesis. Furthermore, the path from $(1, 1)$ to $(n, n)$ lying outside the subarray must contain at least one monster, for otherwise it would have offered an alternative to Turbo, contradicting uniqueness. Consequently, the $n \\times n$ array contains at least $(n-2)+1 = n-1$ monsters, as desired. This establishes Case 1.\n\n*Case 2.* Turbo visits $(n-1, n)$. Note that $(2, 1)$ and $(n-1, n)$ are separated by the diagonal joining the lower left cell to the upper right cell, so Turbo's safe path must cross this diagonal at some (unique) cell $(k, k)$, where $2 \\leq k \\leq n-1$.\n\nConsider the subarrays $\\{(i, j) : 1 \\leq i, j \\leq k\\}$ and $\\{(i, j) : k \\leq i, j \\leq n\\}$. They both inherit uniqueness of a safe path for Turbo, so the former contains at least $k-1$ monsters and the latter at least $n-k$, accounting for at least $(k-1)+(n-k) = n-1$ monsters in the $n \\times n$ array, as desired. This establishes Case 2 and completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16252,
    "subject": "Mathematics (Olympiad)",
    "question": "On a piece of $8 \\times 8$ graph paper, at least how many grids should be taken off so that it is impossible to cut out a \"T\" shape consisting of five grids as shown below?\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p172_data_292ad4c5ba.png)",
    "options": [],
    "answer": "See solution",
    "solution": "At least 14 grids should be taken off. An example is shown below:\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p172_data_6763042fa9.png)\n\nTo see why, consider dividing the $8 \\times 8$ graph paper into five areas (see below):\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p172_data_92da9728b4.png)\n\nIn the center area, at least two grids must be removed; otherwise, a \"T\" shape can still be formed. The two grids marked with \"\\times\" are not equivalent: if only one is removed, a \"T\" can still be formed, as shown here:\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p172_data_f23de7b3d6.png)\n\nFor each of the four corner areas, at least three grids must be removed to prevent forming a \"T\" shape. For example, in the upper right corner, the following configuration requires removing one grid from the \"T\" and two from the grids above it:\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p172_data_1490f6b0cd.png)\n\nIf only two grids are removed, a \"T\" shape can still be formed, as shown below:\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p173_data_fefaadc5d0.png)\n\nTherefore, the minimum number of grids to be removed is $2 + 3 \\times 4 = 14$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 16253,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all values of the real parameter $a$ such that the equation\n\n$$\nx^3 - a x^2 + (a^2 - 1)x - a^2 + a = 0\n$$\n\nhas three distinct real roots which (in some order) form an arithmetic progression.",
    "options": [],
    "answer": "See solution",
    "solution": "Writing the equation as\n\n$$\n(x-1)(x^2 + (1-a)x - a + a^2) = 0,\n$$\nwe obtain $x_1 = 1$. Let $x_2$ and $x_3$ be the roots of the quadratic. If $1$ is the middle term of the progression, then $x_2 + x_3 = 2$, so $a-1 = 2$, i.e., $a = 3$. For $a = 3$, the quadratic has no real roots.\n\nIf $x_1 = 1$ is not the middle term, assume $1 + x_2 = 2x_3$. Together with $x_2 + x_3 = a-1$, this gives $3x_3 = a$, so $x_3 = \\frac{a}{3}$ is a root of the quadratic:\n\n$$\n\\left(\\frac{a}{3}\\right)^2 + (1-a)\\frac{a}{3} - a + a^2 = 0.\n$$\n\nSolving, $a = 0$ or $a = \\frac{6}{7}$. For $a = 0$, $x_2 = -1$, $x_3 = 0$; for $a = \\frac{6}{7}$, $x_2 = -\\frac{3}{7}$, $x_3 = \\frac{2}{7}$. Thus, the desired values are $a = 0$ and $a = \\frac{6}{7}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16254,
    "subject": "Mathematics (Olympiad)",
    "question": "An investor has two rectangular lands, each of size $120 \\times 100$.\n\na) On the first land, she wants to build a house with a rectangular base of size $25 \\times 35$ and nine circular flower pots with diameter $5$ outside the house. Prove that for all positions of the flower pots, the remaining land is still sufficient to build the desired house.\n\nb) On the second land, she wants to construct a polygonal fish pond such that the distance from an arbitrary point on the land, outside the pond, to the nearest pond edge is not over $5$. Prove that the perimeter of the pond is not smaller than $440 - 20\\sqrt{2}$.",
    "options": [],
    "answer": "See solution",
    "solution": "For convenience, instead of writing $a$ meters, we write $a$.\n\na) Consider the rectangle $ABCD$ where $AB = CD = 120$ and $AD = BC = 100$. Divide the rectangle into $10$ sub-rectangles of size $30 \\times 40$ as shown below. Consider $9$ centers of the flower pots. By the pigeonhole principle, there exists a sub-rectangle that does not contain any center.\n\nSuppose that rectangle is $XYZT$ where $XY = ZT = 40$, $XT = YZ = 30$. Consider one more rectangle $X'Y'Z'T'$ lying inside $XYZT$ such that the sides of the two rectangles are pairwise parallel and the gaps are equal to $2.5$.\n\n![](images/Vietnamese_mathematical_competitions_p189_data_4de8954c5e.png)\n\nIt is easy to check that $X'Y'Z'T'$ does not share any point with the pots, so we can build a house on this plot.\n\nb) Consider a rectangle $ABCD$ where $AB = CD = 120$ and $AD = BC = 100$. Let $L$ be the perimeter of the lake. According to the problem, there exist points $A', B', C', D'$ in $L$ such that\n\n$$\nAA', BB', CC', DD' \\leq 5.\n$$\n\nSince the lake is a convex polygon, $A'B', B'C', C'D', D'A'$ do not overlap. Hence,\n\n$$\n|L| \\geq A'B' + B'C' + C'D' + D'A'.\n$$\n\nDenote $A_1$ as the projection of $A'$ to $AD$ and $A_2$ as the projection of $A'$ to $AB$. Similarly, we can define $B_1$, $B_2$, $C_1$, $C_2$, $D_1$, $D_2$ and we have\n\n$$\nA_1A' + A'B' + B'B_1 \\geq A_1B_1 \\geq AB = 120.\n$$\n\nSimilarly,\n\n$$\nB_2B' + B'C' + C'C_2 \\geq 100,\n$$\n\n$$\nC_1C' + C'D' + D'D_1 \\geq 120,\n$$\n\n$$\nD_2D' + D'A' + A'A_2 \\geq 100.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16255,
    "subject": "Mathematics (Olympiad)",
    "question": "A hare and a tortoise run in the same direction, at constant but different speeds, around the base of a tall square tower. They start together at the same vertex, and the run ends when both return to the initial vertex simultaneously for the first time. Suppose the hare runs with speed $1$, and the tortoise with speed less than $1$. For what rational numbers $x$ is it true that, if the tortoise runs with speed $x$, the fraction of the entire run for which the tortoise can see the hare is also $x$?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $x = \\frac{p}{q}$ where $p, q$ are positive integers with $p < q$ and $\\gcd(p, q) = 1$. Suppose the hare takes $p$ minutes for a full turn about the tower. Then the tortoise takes $q$ minutes for a full turn. They will meet again at the same vertex after $pq$ minutes, when the hare will make $q$ full turns and the tortoise will make $p$ full turns. In particular, the hare will overtake the tortoise exactly $k = q - p$ times, counting the start but not the end of the race as an overtake.\n\nThe overtakes occur at minutes $0, \\frac{pq}{k}, \\frac{2pq}{k}, \\dots, \\frac{(k-1)pq}{k}$. At these times, the tortoise would have made $0, \\frac{p}{k}, \\frac{2p}{k}, \\dots, \\frac{(k-1)p}{k}$ full turns about the tower. Since $(p, q) = 1$, then $(p, k) = 1$, so these meeting points correspond to fractions $0, \\frac{1}{k}, \\frac{2}{k}, \\dots, \\frac{k-1}{k}$ of a full turn.\n\n**Case 1:** $k$ odd. These fractions correspond to fractions of a side, and the tortoise can see the hare for $\\frac{k-i}{k} \\cdot \\frac{p}{4}$ minutes at each meeting. Summing over all meetings, the total time is $\\frac{pq}{8}$ minutes, so $x = \\frac{1}{8}$ is possible (for $k = 7$).\n\n**Case 2:** $k = 2r$ with $r$ odd. The fractions correspond to $0, 0, \\frac{1}{r}, \\frac{1}{r}, \\dots, \\frac{r-1}{r}, \\frac{r-1}{r}$ of a side. The total time is $\\frac{pq + 1}{8}$ minutes, so $x = \\frac{1}{8} + \\frac{1}{8pq}$, which gives $x = \\frac{1}{7}$ for $p = 1, q = 7$.\n\n**Case 3:** $k = 4s$. The fractions correspond to $0, 0, 0, 0, \\frac{1}{s}, \\frac{1}{s}, \\frac{1}{s}, \\frac{1}{s}, \\dots, \\frac{s-1}{s}, \\frac{s-1}{s}, \\frac{s-1}{s}, \\frac{s-1}{s}$ of a side. The total time is $\\frac{pq + 3}{8}$ minutes, so $x = \\frac{1}{8} + \\frac{3}{8pq}$, which gives $x = \\frac{1}{5}$ for $p = 1, q = 5$ and $x = \\frac{3}{23}$ for $p = 3, q = 23$.\n\n**Alternative Solution:** Running the process in reverse, the fraction of the race for which the tortoise can see the hare is half the fraction for which both are on the same side of the square. This fraction is:\n\n- $\\frac{1}{4}$ when $p-q$ is odd,\n- $\\frac{1}{4} + \\frac{1}{4pq}$ when $p-q$ is even but not divisible by $4$,\n- $\\frac{1}{4} + \\frac{3}{4pq}$ when $4 \\mid p-q$.\n\nThus, the possible rational numbers $x$ are $\\frac{1}{8}, \\frac{1}{7}, \\frac{1}{5}, \\frac{3}{23}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16256,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $n$ such that both $n+3$ and $n^2 + 3n + 3$ are perfect cubes of integers.\n\n![](images/Ukrajina_2013_p31_data_60cdcf0297.png)",
    "options": [],
    "answer": "See solution",
    "solution": "If both $n+3$ and $n^2 + 3n + 3$ are perfect cubes, then so is\n$$\n(n+3)(n^2 + 3n + 3) = (n+2)^3 + 1.\n$$\n\nThus, $n = -2$ or $n = -3$. It is easy to check that $n = -2$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16257,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Find all pairs of positive integers $(m, n)$, with $m \\leq n$, for which\n\n$$\np(2m + 1) \\cdot p(2n + 1) = 400.\n$$\n\nb) Determine the set $\\{n \\in \\mathbb{N}^* \\mid n \\leq 100 \\text{ and } \\frac{p(n+1)}{p(n)} \\notin \\mathbb{N}\\}$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Since $400 = 1 \\cdot 400 = 4 \\cdot 100 = 16 \\cdot 25$, we analyze three cases.\n\nIf $p(2m-1) = 1$, $p(2n-1) = 400$, we obtain $1 \\leq 2m-1 < 4$ and $400 \\leq 2n-1 < 441$, hence $m \\in \\{1, 2\\}$ and $n \\in \\{200, 201, \\dots, 219\\}$, giving 40 pairs $(m, n)$. Similarly, in the second case we obtain 20 pairs, and in the last case, 24 pairs, leading to a total of 84 pairs.\n\nb) Let $n \\in \\mathbb{N}$, and let $p(n) = k^2$; it follows $k^2 \\leq n \\leq (k+1)^2 - 1$, hence $k^2 < n + 1 \\leq (k+1)^2$. We deduce that $p(n+1) \\in \\{k, k+1\\}$. Then $\\frac{p(n+1)}{p(n)} \\notin \\mathbb{N}$ if and only if $p(n) = k^2 \\neq 1$ and $p(n+1) = (k+1)^2$, that is, $n = (k+1)^2 - 1$, $k \\in \\mathbb{N}$, $k \\geq 2$. Since $n \\leq 100$, it results $k \\leq 9$. The requested set is $\\{8, 15, 24, 35, 48, 63, 80, 99\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16258,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 51 senators in a senate. The senate needs to be divided into $n$ committees so that each senator is on one committee. Each senator hates exactly three other senators. (If senator $A$ hates senator $B$, then senator $B$ does *not* necessarily hate senator $A$.) Find the smallest $n$ such that it is always possible to arrange the committees so that no senator hates another senator on his or her committee.",
    "options": [],
    "answer": "See solution",
    "solution": "The smallest such number is $7$.\n\nAssume that there are $7$ senators $A_1, \\dots, A_7$ such that each $A_i$ hates $A_{i+1}$, $A_{i+2}$, and $A_{i+3}$ (where indices are taken modulo $7$). In this situation, for any different $A_i, A_j$, either $A_i$ hates $A_j$ or vice versa. The senators $A_1, \\dots, A_7$ must be placed on seven different committees. Thus, $n \\ge 7$.\n\nIn order to show that $n \\le 7$, we will prove the following stronger statement by induction on $k \\ge 1$: for $k$ senators, each of whom hates *at most* $3$ others, it is possible to arrange the senators into $7$ committees so that no senator hates another senator on his or her committee.\n\nFor the base case $k = 1$, note that we can have $7$ committees, $6$ of which are empty and $1$ of which contains the sole senator.\n\nNow assume the claim is true for all $k \\le m-1$. Suppose we are given $m$ senators, each of whom hates at most $3$ others. If each of those $m$ senators is hated by more than $3$ others, then the total number of acts of hating must be greater than $3m$, but this is not possible since each senator hates at most $3$ others. Therefore, there must be at least one senator $A$ who is hated by at most $3$ others. By the induction hypothesis, we can split the $m-1$ other senators into $7$ committees satisfying the property that no senator hates another senator on the same committee. By the *Pigeonhole Principle*, one of those committees contains neither a person whom $A$ hates nor a person who hates $A$. We can therefore place $A$ in that committee. The induction is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16259,
    "subject": "Mathematics (Olympiad)",
    "question": "自然數 $x$ 以十進制表示時，各位數字的乘積等於 $x^2 - 15x - 27$。試求所有滿足以上條件的 $x$。",
    "options": [],
    "answer": "See solution",
    "solution": "唯一可能的 $x$ 是 $17$。\n\n顯然 $x^2 - 15x - 27 \\leq x$，因此容易看出 $1 \\leq x \\leq 17$。由於等號不成立，所以 $10 \\leq x \\leq 17$，這意味著 $x$ 是方程 $x^2 - 16x - 17 = 0$ 的解。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16260,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples of integers $(x, y, z)$ such that\n$$\nx^2 + y^2 + z^2 = 16(x + y + z).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Rewrite the equation as $$(x-8)^2 + (y-8)^2 + (z-8)^2 = 192.$$ Since a square is congruent to $0$ or $1$ modulo $4$, $x-8$, $y-8$, and $z-8$ must all be even. Set $x-8 = 2a_1$, $y-8 = 2b_1$, $z-8 = 2c_1$, so $a_1^2 + b_1^2 + c_1^2 = 48$. Repeating, set $a_1 = 2a_2$, $b_1 = 2b_2$, $c_1 = 2c_2$, yielding $a_2^2 + b_2^2 + c_2^2 = 12$. Again, set $a_2 = 2a_3$, $b_2 = 2b_3$, $c_2 = 2c_3$, so $a_3^2 + b_3^2 + c_3^2 = 3$.\n\nThis is only possible if $a_3, b_3, c_3 = \\pm 1$, so $x-8$, $y-8$, and $z-8$ are each $\\pm 8$. Thus, the integer triples are:\n$$(0, 0, 0),\\ (0, 0, 16),\\ (0, 16, 0),\\ (16, 0, 0),\\ (0, 16, 16),\\ (16, 0, 16),\\ (16, 16, 0),\\ (16, 16, 16).$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16261,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all non-constant polynomial functions $f: [0, 1] \\to \\mathbb{R}^*$, having rational coefficients, for which the following property holds: for every $x$ in $[0, 1]$, there exist two polynomial functions $g_x, h_x: [0, 1] \\to \\mathbb{R}$, with rational coefficients, such that $h_x(x) \\neq 0$ and\n\n$$\n\\int_{0}^{x} \\frac{1}{f(t)} \\, dt = \\frac{g_x(x)}{h_x(x)}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The required functions are $f(x) = a(x-b)^n$, $0 \\le x \\le 1$, where $n$ is an integer greater than $1$, and $a$ and $b$ are rational numbers, $a \\neq 0$ and $b \\notin [0, 1]$. Clearly, these functions satisfy the conditions in the statement.\n\nSince the closed unit interval $[0, 1]$ is uncountable, and there are only countably many polynomial functions with rational coefficients, there exists an uncountable set $S \\subseteq [0, 1]$ and coprime polynomial functions $g, h: [0, 1] \\to \\mathbb{R}$ with rational coefficients such that $h(x) \\neq 0$ and $\\int_0^x \\frac{1}{f(t)} \\, dt = \\frac{g(x)}{h(x)}$ for all $x$ in $S$. Since at most countably many points of $S$ are not accumulation points of $S$, and $S$ is uncountable, it follows that $S$ contains infinitely many of its accumulation points. At each of these points, the rational functions $1/f$ and $(g/h)' = (g'h - gh')/h^2$ are equal, so\n\n$$\nh^2 = f \\cdot (g'h - gh') \\quad (*)\n$$\nin $\\mathbb{Q}[X]$. Since $\\deg f \\ge 1$, it follows that $\\deg g \\le \\deg h$. Notice that the remainder of $g$ upon division by $h$ also satisfies $(*)$, so assume henceforth $\\deg g < \\deg h$, so $\\deg (g'h - gh') = \\deg g + \\deg h - 1$.\n\nIf $\\deg h = 1$, then $\\deg g = 0$ and $f = a(X - b)^2$, where $a$ and $b$ are rational numbers, $a \\neq 0$ and $b \\notin [0, 1]$.\n\nIf $\\deg h \\ge 2$, since $g$ and $h$ are coprime, $(*)$ implies that every $k$-fold root (not necessarily real) of $g'h - gh'$ is a $(k+1)$-fold root of $h$, so $\\deg h \\ge \\deg g + \\deg h - 1 + d$, where $d$ is the number of distinct roots of $g'h - gh'$. Consequently, $d = 1$ and $\\deg g = 0$, so $g$ is a non-zero constant, $h = c(X - b)^n$, where $b$ and $c$ are rational, $b \\notin [0, 1]$, $c \\neq 0$, and $n$ is an integer greater than $1$, and $f = a(X - b)^{n+1}$ for some non-zero rational $a$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16262,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be an integer with $k \\geq 2$, and let $n_1, n_2, n_3$ be positive integers. Let $a_1, a_2, a_3$ be integers such that $1 \\leq a_i \\leq k-1$ for $i = 1, 2, 3$. Define\n\n$$\nb_i = a_i \\sum_{j=0}^{n_i} k^j \\quad (i = 1, 2, 3).\n$$\n\nDetermine all possible combinations $(n_1, n_2, n_3)$ such that $b_1 b_2 = b_3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume without loss of generality that $n_1 \\geq n_2$. Since $0 < a_1 b_2 < k \\cdot k^{n_2+1} = k^{n_2+2}$, we can write\n\n$$\na_1 b_2 = \\sum_{j=0}^{n_2+1} d_j k^j\n$$\n\nfor suitable $d_j$ with $0 \\leq d_j \\leq k-1$ ($j = 0, 1, \\dots, n_2+1$).\nSet $e_j = d_0 + d_1 + \\dots + d_j$ for $j = 0, 1, \\dots, n_2+1$.\n\nNow, consider two cases:\n\n* When $n_1 > n_2$:\n\n$$\nb_1 b_2 = \\left( \\sum_{j=0}^{n_1} k^j \\right) \\left( \\sum_{j=0}^{n_2+1} d_j k^j \\right) \\equiv \\sum_{j=0}^{n_2+1} e_j k^j \\pmod{k^{n_2+2}}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 16263,
    "subject": "Mathematics (Olympiad)",
    "question": "Integers $a, b, c, d, e, f$ satisfy the system\n\n$$\n\\begin{cases}\nace + 3adf - 3bcf + 3bde = 5, \\\\\nacf - ade + bce + 3bdf = 2.\n\\end{cases}\n$$\n\nFind all possible values of the expression $abcde$.",
    "options": [],
    "answer": "See solution",
    "solution": "The possible value is $0$.\n\nLet $x = ac + 3bd$, $y = ad - bc$. We have\n\n$$\n\\begin{align*}\n37 = 5^2 + 3 \\cdot 2^2 &= (ace + 3adf - 3bcf + 3bde)^2 + 3(acf - ade + bce + 3bdf)^2 \\\\\n&= (ex + 3fy)^2 + 3(fx - ey)^2 = (e^2 + 3f^2)(x^2 + 3y^2) \\\\\n&= (e^2 + 3f^2)((ac + 3bd)^2 + 3(ad - bc)^2) \\\\\n&= (e^2 + 3f^2)(c^2 + 3d^2)(a^2 + 3b^2).\n\\end{align*}\n$$\n\nSince $37$ is prime, we obtain $\\{a^2 + 3b^2, c^2 + 3d^2, e^2 + 3f^2\\} = \\{1, 1, 37\\}$. However, if integers $m$ and $n$ satisfy $m^2 + 3n^2 = 1$, then $n = 0$. Therefore, two of the numbers $b, d, f$ are zero and thus $abcde = 0$.\n\nFor example, $a = 1$, $b = 0$, $c = 1$, $d = 0$, $e = 5$, $f = 2$ is a solution.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16264,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $h$ be the altitude of triangle $ABC$ passing through vertex $A$, and let $\\alpha = \\angle BAC$. Prove that the following inequality holds:\n\n$$\nAB + AC \\geq BC \\cdot \\cos \\alpha + 2h \\sin \\alpha.\n$$\n\nIn which triangles does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "Equality holds for the equilateral triangle.\n\nIf $\\alpha \\geq 90^\\circ$, then $AB + AC > 2h \\geq 2h \\sin \\alpha \\geq BC \\cdot \\cos \\alpha + 2h \\sin \\alpha$, because $\\cos \\alpha \\leq 0$, so the inequality is strict.\n\nNow, let $\\alpha < 90^\\circ$. Denote by $H_1, H_2, H_3$ the feet of the altitudes from vertices $A, B, C$ respectively. Let $K_2$ be the point symmetric to $H_2$ with respect to line $AB$, and analogously, $K_3$ be the point symmetric to $H_3$ with respect to line $AC$. Let $\\omega_1$ and $\\omega_2$ be the circumscribed circles of triangles $ABH_1$ and $ACH_1$ respectively.\n\n![](images/Ukrajina_2010_p35_data_0bf67fe9c0.png)\n\nThen $AB$ and $AC$ are diameters of these circles, as $\\angle AK_2B = \\angle AK_3C = 90^\\circ$, so $K_2 \\in \\omega_1$ and $K_3 \\in \\omega_2$. Thus, $AB \\geq K_2H_1$ and $AC \\geq K_3H_1$.\n\nBy Ptolemy's theorem:\n\n$$\nAB \\cdot K_2H_1 = AK_2 \\cdot BH_1 + BK_2 \\cdot AH_1.\n$$\n\nThis implies\n\n$$\nK_2H_1 = \\frac{AK_2 \\cdot BH_1}{AB} + \\frac{BK_2 \\cdot AH_1}{AB} = \\frac{AH_2}{AB} \\cdot BH_1 + \\frac{BH_2}{AB} \\cdot AH_1 = BH_1 \\cos \\alpha + h \\sin \\alpha.\n$$\n\nAnalogously, $K_3H_1 = CH_1 \\cos \\alpha + h \\sin \\alpha$. Therefore,\n\n$$\nAB + AC \\geq K_2H_1 + K_3H_1 = (BH_1 \\cos \\alpha + h \\sin \\alpha) + (CH_1 \\cos \\alpha + h \\sin \\alpha) = BC \\cos \\alpha + 2h \\sin \\alpha.\n$$\n\nThe proof is finished.\n\nEquality holds if and only if the segments $K_2H_1$ and $K_3H_1$ are the diameters of circles $\\omega_1$ and $\\omega_2$ respectively. In this case,\n\n$$\n\\angle ABC = 90^\\circ - \\angle BAH_1 = \\angle BAK_2 = \\angle BAC = \\angle CAK_3 = 90^\\circ - \\angle CAH_1 = \\angle ACB,\n$$\n\nso triangle $ABC$ is equilateral. It is easily checked that equality holds.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16265,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for each real number $r > 2$, there are exactly two or three positive real numbers $x$ satisfying the equation $x^2 = r \\lfloor x \\rfloor$.\n\n*Note: $\\lfloor x \\rfloor$ denotes the largest integer less than or equal to $x$*",
    "options": [],
    "answer": "See solution",
    "solution": "Let $r > 2$ be a real number. Let $x$ be a positive real number such that $x^2 = r \\lfloor x \\rfloor$ with $\\lfloor x \\rfloor = k$. Since $x > 0$ and $x^2 = r k$, we also have $k > 0$. From $k \\leq x < k + 1$, we get $k^2 \\leq x^2 = r k < (k + 1)^2 = k^2 + 2k + 1$, hence $k \\leq r < k + 3$, or $r - 3 < k \\leq r$. There are at most three positive integers in the interval $(r - 3, r]$. Thus, there are at most three possible values for $k$. Consequently, there are at most three positive solutions to the given equation.\n\nNow suppose that $k$ is a positive integer in the interval $[r - 2, r]$. There are at least two such positive integers. Observe that $k \\leq \\sqrt{r k} \\leq \\sqrt{(k + 2) k} < k + 1$ and so $r k = r \\lfloor \\sqrt{r k} \\rfloor$. We conclude that the equation $x^2 = r \\lfloor x \\rfloor$ has at least two positive solutions, namely $x = \\sqrt{r k}$ with $k \\in [r - 2, r]$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16266,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all continuous functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that, for all $x, y \\in \\mathbb{R}$, there exists $t \\in (0, 1)$ with\n$$\nf((1-t)x + t y) = (1-t)f(x) + t f(y).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that the only such functions are of the form $f(x) = m x + n$, where $m, n \\in \\mathbb{R}$. It is easy to verify that these functions satisfy the given property.\n\nFor the converse, suppose $f$ is a continuous function with the stated property. Let $a, b \\in \\mathbb{R}$ with $a < b$, and define\n$$\nm = \\frac{f(b) - f(a)}{b - a}, \\quad n = \\frac{b f(a) - a f(b)}{b - a}.\n$$\nWe will show that $f(x) = m x + n$ for all $x \\in [a, b]$. Assume, for contradiction, that there exists $x_0 \\in [a, b]$ such that $f(x_0) \\neq m x_0 + n$. Define\n$$\nA = \\{ x \\in [a, x_0] \\mid f(x) = m x + n \\}, \\quad B = \\{ x \\in [x_0, b] \\mid f(x) = m x + n \\}.\n$$\nBoth $A$ and $B$ are nonempty since $a \\in A$ and $b \\in B$. Let $\\alpha = \\sup A$ and $\\beta = \\inf B$. By continuity, $\\alpha \\in A$, $\\beta \\in B$, and $\\alpha < x_0 < \\beta$.\n\nBy the hypothesis, there exists $t \\in (0, 1)$ such that\n$$\nf((1-t)\\alpha + t\\beta) = (1-t)f(\\alpha) + t f(\\beta) = m((1-t)\\alpha + t\\beta) + n.\n$$\nLet $\\gamma = (1-t)\\alpha + t\\beta \\in (\\alpha, \\beta)$. Then $f(\\gamma) = m \\gamma + n$, contradicting the definition of $\\alpha$ and $\\beta$. Thus, $f(x) = m x + n$ for all $x \\in [a, b]$.\n\nTo extend this to all of $\\mathbb{R}$, note that if $f$ is linear on any interval, continuity forces the same coefficients everywhere. Therefore, $f(x) = m x + n$ for all $x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16267,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and let $D$, $E$, and $F$ be the midpoints of $BC$, $CA$, and $AB$, respectively. If $AD = 3$, $BE = 4$, and $CF = 5$, what is the area of $ABC$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $|PQR|$ denote the area of triangle $PQR$. The medians $AD$, $BE$, and $CF$ intersect at point $G$, and $AG : GD = BG : GE = CG : GF = 2 : 1$. Take a point $C'$ on line $GC$ such that $G$ is the midpoint of $CC'$. Since $C'G = GC$ and $CG : GF = 2 : 1$, it follows that $C'F = GF$. Because $AF = BF$, $C'F = GF$, and $\\angle AFC' = \\angle BFG$, triangles $AFC'$ and $BFG$ are congruent, so $AC' = BG$. Now, $AC' = BG = \\frac{8}{3}$, $AG = 2$, and $GC' = GC = \\frac{10}{3}$, so $\\triangle AGC'$ is a right triangle with hypotenuse $GC'$. Therefore, $|AGC'| = \\frac{8}{3}$. Since $C'G = GC$, $|AGC| = |AGC'| = \\frac{8}{3}$, and since $BG : GE = 2 : 1$, $|ABC| = 3|AGC| = 8$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16268,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the maximum length of a sequence of positive integers $a_1, a_2, \\dots, a_n$, if the following conditions hold:\n\n- $a_1 > 1$ is a prime number;\n- for any $i$ with $2 \\le i \\le n$, $a_i \\mid a_1 a_2 \\dots a_{i-1}$;\n- $a_n = 2^2 \\cdot 3^3 \\cdot 5^5 \\cdot 7^7 \\cdot 11^{11} \\cdot 13^{13} \\cdot 17^{17}$.\n\nWhat is the maximum value of $a_1$ then?",
    "options": [],
    "answer": "See solution",
    "solution": "The maximum length is $6$, and $a_1 = 17$.\n\nLet $a_1 = p$ be a prime. Then it is clear that\n\n$$\n\\begin{array}{ccc}\na_2 \\mid p, & a_3 \\mid a_2 a_1 \\mid p^2, & a_4 \\mid a_3 a_2 a_1 \\mid p^4, \\\\\na_5 \\mid a_4 a_3 a_2 a_1 \\mid p^8, & a_6 \\mid a_5 a_4 a_3 a_2 a_1 \\mid p^{16}.\n\\end{array}\n$$\n\nAssume the length of the sequence is greater than $6$, thus $a_7 \\mid a_6 a_5 a_4 a_3 a_2 a_1 \\mid p^{32}$. There is a prime number that divides $a_7$ and it is included with degree not less than $32$. By the condition on $a_n$, that is not possible. Therefore, the maximum length is $n=6$. The only prime number that is included with degree $16$ in $a_6 = 2^2 \\cdot 3^3 \\cdot 5^5 \\cdot 7^7 \\cdot 11^{11} \\cdot 13^{13} \\cdot 17^{17}$ is $p=17$. This is the maximum value of $a_1$.\n\nIt suffices to show now that such a sequence exists. Take\n\n$$\na_1 = 17,\\quad a_2 = 17,\\quad a_3 = 17^2,\\quad a_4 = 17^4,\\quad a_5 = 17^8,\\quad a_6 = 2^2 \\cdot 3^3 \\cdot 5^5 \\cdot 7^7 \\cdot 11^{11} \\cdot 13^{13} \\cdot 17^{17}.\n$$\n\n![](images/Ukraine_booklet_2018_p21_data_785a0374ec.png)\n\n![](images/Ukraine_booklet_2018_p21_data_0318b3f2d2.png)\n\n![](images/Ukraine_booklet_2018_p21_data_9505fcbc4d.png)\n\n![](images/Ukraine_booklet_2018_p21_data_1e8a819594.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16269,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A = \\{a, b, c, d\\}$ and $B = \\{e, f, g\\}$. Suppose the number $X = abcd + efg$ is a prime, where $a, b, c, d, e, f, g$ are distinct integers from $\\{1, 2, 3, 4, 5, 6, 7\\}$, and $abcd$ and $efg$ denote the products of the elements in $A$ and $B$, respectively. Find all possible values of $X$.",
    "options": [],
    "answer": "See solution",
    "solution": "We observe that the numbers $2, 4, 6$ must all belong to the same set, either $A$ or $B$. Otherwise, both $abcd$ and $efg$ would be even, making $X$ even. Since $X \\ne 2$, $X$ cannot be a prime if it is even. Similarly, the numbers $3$ and $6$ must belong to the same set; otherwise, $X$ would be divisible by $3$, and since $X \\ne 3$, $X$ cannot be a prime.\n\nConsequently, the only possible partition is $A = \\{2, 3, 4, 6\\}$ and $B = \\{1, 5, 7\\}$. Thus, the only prime number of the form $abcd + efg$ is:\n\n$$\n2 \\cdot 3 \\cdot 4 \\cdot 6 + 1 \\cdot 5 \\cdot 7 = 144 + 35 = 179\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16270,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the last two digits of the product of the squares of all positive odd integers less than $2014$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the product of the odd integers less than 2014 contains $25$ as a factor, it is clearly divisible by $25$. Also, since it is odd, its last two digits have to be $25$ or $75$. So the product is either of the form $100n + 25$ or $100n + 75$. In either case, the last two digits of the squared product (which is the same as the product of the squares) are $25$:\n\n$$\n(100n + 25)^2 = 10000n^2 + 5000n + 625 = 100(100n^2 + 50n + 6) + 25\n$$\n\nor\n\n$$\n(100n + 75)^2 = 10000n^2 + 15000n + 5625 = 100(100n^2 + 150n + 56) + 25.\n$$\n\nIn fact, we see that the last three digits have to be $625$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16271,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle. The internal bisector of $\\angle B$ meets $AC$ at $P$, and $I$ is the incenter of $ABC$. Prove that if $AP + AB = CB$, then triangle $API$ is isosceles.",
    "options": [],
    "answer": "See solution",
    "solution": "Draw $PP'$ parallel to $IA$ so that $P'$ is on line $AB$. Then $\\triangle PAP'$ is isosceles, which implies that $BC = AB + AP = AB + AP' = BP'$. This then implies that $\\triangle P'BC$ is isosceles, which in turn implies that, since $P$ is on the angle bisector of $\\angle B$, $P'PC$ is also isosceles, with $PP' = PC$. It then follows, using similarity of triangles and the angle bisector theorem, that\n\n$$\n\\frac{IA}{PP'} = \\frac{BA}{BP'} = \\frac{BA}{BC} = \\frac{AP}{PC} = \\frac{AP}{PP'}\n$$\n\nfrom which $IA = AP$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16272,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in Fig. 1.1, in acute $\\triangle ABC$, $AB > AC$. $M$ is the midpoint of minor arc $\\widearc{BC}$ of the circumcircle $\\Omega$ of $\\triangle ABC$, and $K$ is the intersection of the exterior angle bisector of $\\angle BAC$ and the extension of $BC$. Take a point $D$ (different from $A$) on the line passing through point $A$ and perpendicular to $BC$ such that $DM = AM$. Let the circumcircle of $\\triangle ADK$ intersect circle $\\Omega$ at point $A$ and another point $T$.\n\nProve that $AT$ bisects segment $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Extend $DM$ and let it intersect $BC$ at point $X$. Let $A$ be the midpoint of minor arc $\\widearc{BC}$ of circle $\\Omega$, so points $N$, $A$, $K$ are collinear.\n\nSuppose $L$ is the midpoint of $BC$. It is easy to see that $AM$ is the angle bisector of $\\angle BAC$, so $AM \\perp AK$. Since $AD \\perp BC$ and $AM = DM$, we have\n\n$$\n\\angle AKX = 90^\\circ - \\angle DAK = \\angle MAD = \\angle ADM = \\angle ADX.\n$$\n\nTherefore, points $A$, $K$, $D$, $X$ are concyclic, and hence $A$, $K$, $D$, $T$, $X$ are concyclic. Denote this circle as $\\Omega_1$.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p98_data_a4cf9dbd19.png)\n\nNote that $MN \\parallel AD$ since both are perpendicular to $BC$. Therefore,\n\n$$\n\\angle NMX = \\angle ADX = \\angle AKX = \\angle NKX.\n$$\n\nHence, points $N$, $K$, $M$, $X$ are concyclic. Denote this circle as $\\Omega_2$.\n\nApplying Monge's theorem to circles $\\Omega$, $\\Omega_1$, $\\Omega_2$, we know that $AT$ (radical axis of $\\Omega$, $\\Omega_1$), $MN$ (radical axis of $\\Omega$, $\\Omega_2$), and $XK$ (radical axis of $\\Omega_1$, $\\Omega_2$) are concurrent. Since $MN$ and $XK$ intersect at point $L$, it follows that $AT$ passes through $L$, i.e., $AT$ bisects segment $BC$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16273,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that, for any positive integers $d$ and $m$, a polynomial of degree $d$ with real coefficients cannot be expressed as a product of $m$ periodic functions.  \nA function $f: \\mathbb{R} \\to \\mathbb{R}$ is called *periodic* if there exists a constant $T = T(f) > 0$ such that $f(x+T) = f(x)$ for all $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose first that a nonconstant polynomial $p$ has a real root at $x = x_0$ and $p(x) = f_1(x) \\cdots f_m(x)$ with periodic $f_i$, $i = 1, \\dots, m$. Then at least one function $f_i$ is zero at $x = x_0$. If $T = T(f_i) > 0$ is a period of $f_i$, then $f_i(x_0 + kT) = f_i(x_0) = 0$ for each $k \\in \\mathbb{N}$. Hence, $p(x_0 + kT) = 0$ for every $k \\in \\mathbb{N}$, which is impossible since $p$ cannot have infinitely many roots.\n\nAssume next that $p$ has no real roots. Consider a class $K$ of functions $\\mathbb{R} \\to \\mathbb{R}$ that are nonconstant and can be written as $p_1/p_2$, where $p_1, p_2$ are polynomials with real coefficients having no real roots. (In particular, selecting $p_2 = 1$ and $p_1 = p$, we see that $p \\in K$.)\n\nWe claim that if $q(x) \\in K$ and $a > 0$, then $q(x+a)/q(x) \\in K$.\n\nTo prove the claim, observe that if $q(x) = p_1(x)/p_2(x)$, then $q(x+a)/q(x)$ is a quotient of two polynomials without real roots: $p_1(x + a)p_2(x)$ and $p_2(x + a)p_1(x)$. These two polynomials have the same degree and leading coefficient. Thus, if their quotient were equal to a constant $c$ for all $x \\in \\mathbb{R}$, then $c = 1$, so $q(x+a) = q(x)$ for all $x$. Then $q(x + ka) = q(x)$ for all $k \\in \\mathbb{N}$ and $x \\in \\mathbb{R}$, so in particular $q(ka) = q(0)$. This implies that the polynomial $p_2(x)(q(x) - q(0))$ has a root at $x = ak$ for each $k \\in \\mathbb{N}$, which is only possible when $q(x) - q(0) = 0$ for all $x$. Thus, $q(x)$ is a constant, contradicting the definition of $K$. This completes the proof of the claim.\n\nNow, by induction on $m$, we show that no $q \\in K$ can be expressed as a product of $m$ periodic functions. By the above claim, each $q \\in K$ is not periodic, so $m$ cannot be 1. Assume that no $q \\in K$ can be expressed as a product of fewer than $m$ periodic functions, where $m \\ge 2$, but some $q \\in K$ can be written as $q(x) = f_1(x) \\cdots f_m(x)$ with periodic functions $f_1, \\dots, f_m$. Select $a > 0$ as a period of $f_m$, so $f_m(x + a) = f_m(x)$ for all $x$. Dividing $q(x + a) = f_1(x + a) \\cdots f_m(x + a)$ by $q(x) = f_1(x) \\cdots f_m(x)$ (none of the $f_i$ is zero at a real $x$, since $q(x) \\ne 0$ for all $x$), we find that\n\n$$\n\\frac{q(x+a)}{q(x)} = \\frac{f_1(x+a)}{f_1(x)} \\cdots \\frac{f_{m-1}(x+a)}{f_{m-1}(x)}\n$$\n\nBy the above claim, the left-hand side belongs to $K$ and is a product of $m-1$ periodic functions $f_i(x + a)/f_i(x)$, $i = 1, \\dots, m-1$, which is impossible by our assumption. This completes the proof. $\\blacktriangleright$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16274,
    "subject": "Mathematics (Olympiad)",
    "question": "In obtuse triangle $ABC$, with the obtuse angle at $A$, let $D$, $E$, $F$ be the feet of the altitudes through $A$, $B$, $C$ respectively. $DE$ is parallel to $CF$, and $DF$ is parallel to the angle bisector of $\\angle BAC$. Find the angles of the triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/s3s2014_p1_data_14dd12a4c5.png)\n\nLet the angles of the triangle be $\\alpha = \\angle BAC$, $\\beta = \\angle ABC$, and $\\gamma = \\angle ACB$. Let $X$ be the intersection of $BC$ with the angle bisector of $\\angle BAC$.\n\nSince $\\angle BDA = \\angle BEA = 90^\\circ$, both $D$ and $E$ lie on the circle with diameter $AB$. Thus, $AEBD$ is cyclic, which implies that\n\n$$\n\\angle EDB = \\angle EAB = 180^\\circ - \\alpha.\n$$\n\nIt is given that $DE$ and $CF$ are parallel, hence\n\n$$\n\\angle EDB = \\angle FCB = 90^\\circ - \\angle FBC = 90^\\circ - \\beta,\n$$\nso $\\beta = \\alpha - 90^\\circ$.\n\nLikewise,\n\n$$\n\\angle FDC = \\angle FAC = 180^\\circ - \\alpha,\n$$\n\nand\n\n$$\n\\angle FDC = \\angle AXC = 180^\\circ - \\angle ACX - \\angle XAC = 180^\\circ - \\gamma - \\frac{\\alpha}{2},\n$$\n\nso $\\gamma = \\frac{\\alpha}{2}$.\n\nSince $\\alpha + \\beta + \\gamma = 180^\\circ$, this gives us\n\n$$\n\\alpha + (\\alpha - 90^\\circ) + \\frac{\\alpha}{2} = 180^\\circ,\n$$\n\nand thus $\\alpha = 108^\\circ$, $\\beta = 18^\\circ$, and $\\gamma = 54^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16275,
    "subject": "Mathematics (Olympiad)",
    "question": "Реши ја равенката $\\left(\\frac{x^3 + x}{3}\\right)^3 + \\frac{x^3 + x}{3} = 3x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Ако воведеме смена $\\frac{x^3+x}{3} = y$, тогаш $y^3 + y = 3x$ и $x^3+x=3y$. Според тоа, ако $x_0$ е решение на равенката и $\\frac{x_0^3+x_0}{3} = y_0$, тогаш $(x_0,y_0)$ е решение на системот равенки\n\n$$\n\\begin{cases}\nx^3 + x = 3y \\\\\ny^3 + y = 3x\n\\end{cases}\n$$\n\nАко од првата равенка на системот ја одземеме втората равенка добиваме\n\n$$\n\\begin{aligned}\n& (x^3 - y^3) + (x - y) = 3(y - x) \\\\\n& (x - y)(x^2 + xy + y^2 + 1) = 3(y - x) \\\\\n& (x - y)(x^2 + xy + y^2 + 4) = 0\n\\end{aligned}\n$$\n\nБидејќи\n\n$$\nx^2 + xy + y^2 + 4 = x^2 + 2x\\frac{y}{2} + \\frac{y^2}{4} + \\frac{3}{4}y^2 + 4 = \\left(x + \\frac{y}{2}\\right)^2 + \\frac{3y^2}{4} + 4 \\ge 4,\n$$\n\nза било кои $x, y \\in \\mathbb{R}$, од равенката добиваме $x - y = 0$, односно $x = y$. Според тоа $\\frac{x^3+x}{3} = x$, од каде добиваме дека $x(x^2-2)=0$. Решенија на последната равенка се $x_1 = 0$, $x_2 = \\sqrt{2}$, $x_3 = -\\sqrt{2}$. Не е тешко да се провери дека истите се решенија и на почетната равенка.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16276,
    "subject": "Mathematics (Olympiad)",
    "question": "A palindromic number is a number whose digits stand symmetrically with respect to the center of the number's decimal notation. For instance, 7, 1221, and 57575 are palindromic, while 1212 and 3330 are not. Prove that for any number of pairwise distinct palindromic numbers, the sum of their reciprocals is not greater than 11.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be an arbitrary natural number and consider all palindromic numbers with exactly $n$ digits. We consider the cases $n$ even and $n$ odd separately.\n\nWhen $n = 2m$, a palindrome has the form $a_1a_2\\dots a_{m-1}a_m a_m a_{m-1}\\dots a_2a_1$, with $a_1 \\neq 0$. There are $9 \\cdot 10^{m-1}$ such numbers, each greater than $10^{2m-1}$. Thus, the sum of their reciprocals is at most $\\frac{9 \\cdot 10^{m-1}}{10^{2m-1}} = \\frac{9}{10^m}$. Summing over all $m$:\n\n$$\n\\sum_{m=1}^{\\infty} \\frac{9}{10^m} = 9 \\cdot \\sum_{m=1}^{\\infty} \\frac{1}{10^m} = 9 \\cdot \\frac{\\frac{1}{10}}{1-\\frac{1}{10}} = 9 \\cdot \\frac{1}{9} = 1.\n$$\n\nWhen $n = 2m-1$, a palindrome has the form $a_1a_2\\dots a_{m-1}a_m a_{m-1}\\dots a_2a_1$, again with $a_1 \\neq 0$. There are $9 \\cdot 10^{m-1}$ such numbers, each greater than $10^{2m-2}$. Thus, the sum of their reciprocals is at most $\\frac{9 \\cdot 10^{m-1}}{10^{2m-2}} = \\frac{9}{10^{m-1}}$. Summing over all $m$:\n\n$$\n\\sum_{m=1}^{\\infty} \\frac{9}{10^{m-1}} = 90 \\cdot \\sum_{m=1}^{\\infty} \\frac{1}{10^m} = 90 \\cdot \\frac{\\frac{1}{10}}{1-\\frac{1}{10}} = 90 \\cdot \\frac{1}{9} = 10.\n$$\n\nTherefore, the total sum does not exceed $1 + 10 = 11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16277,
    "subject": "Mathematics (Olympiad)",
    "question": "For each integer $a_0 > 1$, define the sequence $a_0, a_1, a_2, \\dots$ by:\n\n$$\na_{n+1} = \\begin{cases} \\sqrt{a_n} & \\text{if } \\sqrt{a_n} \\text{ is an integer,} \\\\ a_n + 3 & \\text{otherwise,} \\end{cases} \\quad \\text{for each } n \\ge 0.\n$$\n\nDetermine all values of $a_0$ for which there is a number $A$ such that $a_n = A$ for infinitely many values of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "All positive integers $a_0$ that are multiples of 3.\n\n**Case 1:** $a_0 \\equiv 0 \\pmod{3}$\n\nAll terms of the sequence are multiples of 3. Let $a_i$ be a term of minimal value. Suppose $a_i > 9$. Let $x \\ge 1$ be the largest integer such that $3^{2^x} < a_i$. The sequence $a_i, a_{i+1}, a_{i+2}, \\dots$ is formed by adding 3 each time until a perfect square $a_j$ is reached. Note $a_j \\le 3^{2^{x+1}}$, so\n\n$$\na_{j+1} = \\sqrt{a_j} \\le \\sqrt{3^{2^{x+1}}} = 3^{2^x} < a_i,\n$$\n\ncontradicting minimality of $a_i$. Hence $a_i \\le 9$, and the sequence enters the cycle\n\n$$\n3 \\rightarrow 6 \\rightarrow 9 \\rightarrow 3 \\rightarrow 6 \\rightarrow 9 \\rightarrow \\dots\n$$\n\nwhich contains the number 3 infinitely many times.\n\n**Case 2:** $a_0 \\equiv 2 \\pmod{3}$\n\nNo perfect square is congruent to 2 modulo 3, so $a_{n+1} = a_n + 3$ for all $n$. The sequence is strictly increasing and cannot repeat any value infinitely often.\n\n**Case 3:** $a_0 \\equiv 1 \\pmod{3}$\n\nNo term is a multiple of 3. If $a_i \\equiv 2 \\pmod{3}$ for some $i$, then as in Case 2, the sequence cannot repeat a value infinitely often. Suppose $a_n \\equiv 1 \\pmod{3}$ for all $n$. Let $a_i$ be a term of minimal value. Suppose $a_i > 16$. Let $x \\ge 1$ be the largest integer such that $2^{2^x} < a_i$. The sequence $a_i, a_{i+1}, \\dots$ is formed by adding 3 until a perfect square $a_j$ is reached, with $a_j \\le 2^{2^{x+1}}$, so\n\n$$\na_{j+1} = \\sqrt{a_j} \\le \\sqrt{2^{2^{x+1}}} = 2^{2^x} < a_i,\n$$\n\ncontradicting minimality. Hence $a_i \\le 16$. Since $a_0 > 1$, $a_n > 1$ for all $n$. The sequence\n\n$$\n7 \\rightarrow 10 \\rightarrow 13 \\rightarrow 16 \\rightarrow 4 \\rightarrow 2\n$$\n\nshows that eventually a term not congruent to 1 modulo 3 is reached, contradicting the assumption. Thus, only $a_0$ divisible by 3 yields a value $A$ repeated infinitely often.\n\n*Comment:* The final contradiction shows it is impossible for all $a_i \\equiv 1 \\pmod{3}$, not that the sequence always reaches 2. For example, $a_0 = 19$ gives $a_1 = 22$, $a_2 = 25$, $a_3 = 5 \\equiv 2 \\pmod{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16278,
    "subject": "Mathematics (Olympiad)",
    "question": "Вкупно $2^n$ парички се расподелени на неколку деца. До прерасподелба на паричките доаѓа кога некое од децата има барем половина од сите парички: тогаш од паричките на едно такво дете, на секое од останатите деца му се префрлаат онолку парички колку што веќе имало. Во случај кога сите парички се кај едно дете, нема можност за прерасподелба.\n\nКој е најголемиот можен број последователни прераспределби?\n\n*Одговорот да се образложи!*",
    "options": [],
    "answer": "See solution",
    "solution": "Најмногу $n$ последователни прерасподелби.\n\nЌе започнеме со пример дека $n$ последователни прерасподелби се можни. Нека $2^n$ парички се распределени на 3 деца првично во редослед $1, 2^{n-1} + 2^{n-2} + \\ldots + 2, 1$. Последователните прерасподелби (вкупно $n$ на број) ќе бидат:\n\n$$\n\\begin{array}{l}\n2^1,\\ 2^{n-1} + 2^{n-2} + \\ldots + 2^2,\\ 2^1 \\\\\n2^2,\\ 2^{n-1} + 2^{n-2} + \\ldots + 2^3,\\ 2^2 \\\\\n\\qquad \\dots \\\\\n2^{n-2},\\ 2^{n-1},\\ 2^{n-2} \\\\\n2^{n-1},\\ 0,\\ 2^{n-1} \\\\\n0,\\ 0,\\ 2^n\n\\end{array}\n$$\n\nДа покажеме дека $n$ е максималниот број последователни прерасподелби. Тргнуваме од претпоставка дека постои првична расподелба за која се можни барем $n+1$ прерасподелби и бараме контарадикција: да забележиме дека после една прерасподелба бројот на парички кај секое дете е делив со 2 (кај секое дете кај кое се додавале парички, бројот е дуплиран, па е парен, а кај детето од кое се одземале парички повторно останува парен број бидејќи вкупно има $2^n$ парички, т.е. парен број). Аналогно, после втората прерасподелба бројот на парички кај секое дете е делив со 4, итн., се до $n$-тата прерасподелба во која бројот на парички кај секое дете е делив со $2^n$; имајќи предвид дека вкупниот број парички е неменлив и изнесува $2^n$, единствена можна (во некој редослед) е расподелбата $2^n, 0, 0, \\ldots, 0$. Но тогаш нема да има уште една расподелба. Контрадикција!",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16279,
    "subject": "Mathematics (Olympiad)",
    "question": "On a board, the numbers $1, 2, 3, \\dots, 500$ are written. Two players, A and B, take turns alternately, with A starting. On each turn, a player must erase two numbers $n$ and $2n$ from the board. If a player cannot make a move, that player loses. Determine which player has a winning strategy.",
    "options": [],
    "answer": "See solution",
    "solution": "First, notice that for each odd number $u$, $1 \\leq u < 500$, the numbers $u, 2u, 2^2u, \\dots, 2^{k_u}u$, where $k_u$ is the largest integer such that $u2^{k_u} \\leq 500$, form a string of length $k_u + 1$. It is only possible to erase each number in the string if you also erase one of its neighbors. We divide the numbers on the board into such strings and count the lengths of these strings. The numbers of strings of different lengths are:\n\n| length | 1   | 2  | 3  | 4  | 5 | 6 | 7 | 8 | 9 |\n|--------|-----|----|----|----|---|---|---|---|---|\n| number of strings | 125 | 62 | 32 | 15 | 8 | 4 | 2 | 1 | 1 |\n\nA configuration with an even number of strings of each length is a losing position: Player B can pair each string with another of the same length, and whenever A makes a move on one string, B copies that move on the paired string. Such a configuration is called symmetric.\n\nIf we have a winning configuration and add a symmetric configuration, it remains a winning configuration. If A starts from a winning configuration $W$ and we add a symmetric configuration $S$, then A follows her winning strategy on $W$. Whenever B plays in $W$, A continues her strategy in $W$; whenever B plays in $S$, A mirrors the move in $S$. This ensures a win for A.\n\nNow, we prove that the initial configuration is a winning one, and thus A has a winning strategy. You cannot make a move on a string of length 1. We need to show that the configuration with three strings of lengths 4, 8, and 9 is a winning position. First, A erases the two middle numbers in the string of length 9, reducing it to two strings of lengths 3 and 4, leaving a configuration with two strings of lengths 3 and 8, plus a symmetric configuration. Now, B can either play on the string of length 3, leaving a string of length 8, or play on the string of length 8, leaving one of the following:\n\n- one string of length 6\n- one string of length 5\n- two strings of lengths 2 and 4\n- two strings of length 3\n\nA's responses are:\n\n| B leaves      | 8   | 3+6 | 3+5 | 3+2+4 | 3+3+3 |\n|---------------|-----|-----|-----|--------|--------|\n| A leaves      | 3+3 | 3+3 | 3+3 | 3+2    | 3+3   |\n\nA can destroy a string of length 3 or 4 in one turn. The situation 3+2 is essentially symmetric since only one move can be made on each string. Thus, A always leaves a symmetric configuration to B, ensuring a win for A.\n\n*The knowledge about the Sprague-Grundy theorem, which states that \"every impartial game under the normal play convention is equivalent to a nimber\", is very helpful!*",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16280,
    "subject": "Mathematics (Olympiad)",
    "question": "平面上，設點 $A, B, C, D, E, F$ 滿足 $\\triangle BCD \\sim \\triangle ECA \\sim \\triangle BFA$（其中 $\\sim$ 表示正向相似），且 $I$ 為 $\\triangle ABC$ 的內心。\n\n試證：三角形 $AID, BIE, CIF$ 的三個外接圓的圓心共線。\n\n註：所謂 $\\triangle ABC \\sim \\triangle DEF$ 兩三角形正向相似，除了該兩三角形相似以外，並要求 $A$ 到 $B$ 到 $C$ 的旋轉方向，與 $D$ 到 $E$ 到 $F$ 的旋轉方向一致，亦即：同時是順時針方向，或同時是逆時針方向。",
    "options": [],
    "answer": "See solution",
    "solution": "先來證明一個引理：\n\n*Lemma 1.* 給定 $\\triangle ABC$。設 $D, E, F$ 滿足 $\\{AE, AF\\}, \\{BF, BD\\}, \\{CD, CE\\}$ 分別為 $\\angle BAC, \\angle CBA, \\angle ACB$ 的等角線，那麼 $AD, BE, CF$ 共點。如圖：\n\n![](images/18-1J_p12_data_5c28a36748.png)\n\n*Proof of Lemma 1.* 設 $AD, BE, CF$ 分別與 $\\odot(BDC), \\odot(CEA), \\odot(AFB)$ 再交於 $X, Y, Z$。由\n\n$$\n\\angle BYC = \\angle EAC = \\angle BAF = \\angle BZC\n$$\n\n得 $B, C, Y, Z$ 共圓於 $\\Omega_A$。同理可得 $C, A, Z, X$ 共圓於 $\\Omega_B$ 且 $A, B, X, Y$ 共圓於 $\\Omega_C$。故 $AD, BE, CF$ 共點於 $\\Omega_A, \\Omega_B, \\Omega_C$ 的根心，證明完畢。\n\n*Corollary 1.* 給定 $\\triangle ABC$，設 $D, E, F$ 分別為 $A, B, C$ 到 $BC, CA, AB$ 的垂足。點 $X, Y, Z$ 滿足 $\\{DY, DZ\\}, \\{EZ, EX\\}, \\{FX, FY\\}$ 分別為 $\\angle EDF, \\angle FED, \\angle DFE$ 的等角線，則 $AX, BY, CZ$ 共點。\n\n*Proof of Corollary 1.* 設 $X'$ 滿足 $\\triangle AEF \\cup X' \\sim \\triangle ABC \\cup X'$ 並類似定義 $Y', Z'$。則 $\\{AY', AZ'\\}, \\{BZ', BX'\\}, \\{CX', CY'\\}$ 分別為 $\\angle BAC, \\angle CBA, \\angle ACB$ 的等角線，故由引理得 $AX', BY', CZ'$ 共點於 $T$，從而 $AX, BY, CZ$ 共點於 $T$ 關於 $\\triangle ABC$ 的等角共軛點，證明完畢。\n\n回到原題，設 $P$ 為 $\\odot(ECA), \\odot(FAB)$ 異於 $A$ 的交點，則\n\n$$\n\\begin{align*}\n\\angle BPC &= \\angle BPA + \\angle APC = \\angle BFA + \\angle AEC \\\\\n           &= \\angle BCD + \\angle DBC = \\angle BDC.\n\\end{align*}\n$$\n\n故 $P \\in \\odot(DBC)$。由\n\n$$\n\\angle APD = \\angle APB + \\angle BPD = \\angle AFB + \\angle BCD = 0\n$$\n\n得 $P \\in AD$，同理可得 $P$ 在 $BE, CF$ 上。\n\n設 $I_a, I_b, I_c$ 分別為 $\\triangle ABC$ 對應 $A, B, C$ 的旁心並且設 $Q$ 為 $P$ 關於 $\\triangle ABC$ 的等角共軛點，由引理的推論（對 $\\triangle II_bI_c$ 與 $Q, E, F$）得 $IQ, I_bE, I_cF$ 共點。同理可得 $IQ, I_cF, I_aD$ 共點和 $IQ, I_aD, I_bE$ 共點，故 $IQ, I_aD, I_bE, I_cF$ 共點於 $S$。\n\n由\n\n$$\n\\angle ABQ = \\angle PBC = \\angle ADC\n$$\n\n得 $\\triangle ABQ \\sim \\triangle ADC$，故 $AI \\cdot AI_a = AB \\cdot AC = AD \\cdot AQ$，從而\n\n$$\n\\triangle AIQ \\sim \\triangle ADI_a \\Rightarrow \\angle AIS = \\angle ADS\n$$\n\n即 $S \\in \\odot(AID)$。同理可得 $S \\in \\odot(BIE), \\odot(CIF)$，故 $\\odot(AID), \\odot(BIE), \\odot(CIF)$ 共軸，證明完畢。亦可見下圖：\n\n![](images/18-1J_p14_data_edd1c6c301.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16281,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $S = \\{1, 2, \\dots, 2012\\}$。求滿足下列條件之函數 $f: S \\to S$ 的個數：\n\n1. $f$ 為一對一且映成函數。\n2. 對於所有 $1 \\le a \\le 2012$，皆有 $f(a) + f^{-1}(a) = 2013$，其中 $f^{-1}$ 為 $f$ 的反函數。",
    "options": [],
    "answer": "See solution",
    "solution": "考慮各種可能情形：\n\n1. 若存在 $a$ 使得 $f(a) = a$，則 $f(a) + f^{-1}(a) = 2a \\neq 2013$，不可能。\n2. 若存在 $a \\neq b$ 使得 $f(a) = b$ 且 $f(b) = a$，則 $f(a) + f^{-1}(a) = 2b \\neq 2013$，不可能。\n3. 若存在三個相異數 $a, b, c$ 使得 $f(a) = b, f(b) = c, f(c) = a$，則 $f(a) + f^{-1}(a) = b + c$，$f(b) + f^{-1}(b) = c + a$。由於 $b + c = c + a = 2013$，得 $b = a$，矛盾。\n4. 若存在 $k \\ge 5$ 個相異數 $a_1, a_2, \\dots, a_k$，使得 $f(a_1) = a_2, f(a_2) = a_3, \\dots, f(a_{k-1}) = a_k, f(a_k) = a_1$，則 $f(a_2) + f^{-1}(a_2) = a_3 + a_1 = 2013$，$f(a_4) + f^{-1}(a_4) = a_5 + a_3 = 2013$，得 $a_1 = a_5$，矛盾。\n5. 若存在 4 個相異數 $a, b, c, d$，使得 $f(a) = b, f(b) = c, f(c) = d, f(d) = a$，則 $f(b) + f^{-1}(b) = a + c = 2013$，即 $c = 2013 - a$；又 $f(a) + f^{-1}(a) = b + d = 2013$，即 $d = 2013 - b$。因此 $(a, b, c, d) = (a, b, 2013 - a, 2013 - b)$ 形成一個環狀組。\n\n只有第 5 種情形可以成立。只要確定 503 個 $(a_i, b_i, 2013 - a_i, 2013 - b_i)$ 環狀組，就決定一個這樣的函數 $f$。因為 $a_i$ 和 $2013 - a_i$ 分別有一個小於 1007，一個大於 1007；對於 $b_i$ 和 $2013 - b_i$ 也是如此，所以可以假設 $1 \\le a_i, b_i \\le 1006$。從 $\\{1,2,\\dots,1006\\}$ 中依序選出 $a_1, b_1, a_2, b_2, \\dots, a_{503}, b_{503}$，共有 $1006!$ 種可能；但這 503 組是無順序之分，因此 $f$ 的個數是：\n\n$$\n\\frac{1006!}{503!}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16282,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_sB_sC_s$ be the given triangles, with $s \\in \\{1, 2, \\dots, 2012\\}$. Let $\\{XYZ\\}$ denote the half-plane with boundary line $XY$ and interior point $Z$. Each triangle is the intersection of the three half-planes $\\{A_sB_sC_s\\}$, $\\{B_sC_sA_s\\}$, and $\\{C_sA_sB_s\\}$. The intersection of all triangles is thus the intersection of all such half-planes. \n\n(a) Show that the intersection of all $2012$ triangles is a triangle $ABC$ similar to the original triangles, with similarity coefficient $\\lambda$ satisfying $0 < \\lambda \\le 1$, and prove that $\\lambda \\ge \\frac{1}{3}$.\n\n(b) Prove that the union of the $2012$ triangles lies in the intersection of three belts (regions between parallel lines), and that the area of this intersection is less than $\\frac{22}{9}$ times the area of the smallest triangle. \n\n![](images/South-Afrika_2011-2013_p74_data_9b3fc6d793.png)\n\n![](images/South-Afrika_2011-2013_p74_data_ad946774d7.png)\n\n![](images/South-Afrika_2011-2013_p75_data_140f5ad8d6.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A_sB_sC_s$ be the given triangles, $s \\in \\{1, 2, \\dots, 2012\\}$, and $\\{XYZ\\}$ the half-plane with boundary $XY$ and interior point $Z$. Each triangle is the intersection of $\\{A_sB_sC_s\\}$, $\\{B_sC_sA_s\\}$, and $\\{C_sA_sB_s\\}$, so the intersection of all triangles is the intersection of all such half-planes. The intersection of the half-planes $\\{A_sB_sC_s\\}$ is a half-plane $\\{A_iB_iC_i\\}$ for some $i$, and similarly for the other two. Thus, the intersection of all triangles is the intersection of three half-planes, forming a triangle $ABC$ similar to the originals, with similarity coefficient $\\lambda$ ($0 < \\lambda \\le 1$). \n\nLet $v$ be the altitude from $C_i$ to $A_iB_i$ in $A_iB_iC_i$. The centroid $G_i$ is at distance $\\frac{1}{3}v$ from $A_iB_i$. Since $ABC$ contains all centroids, the distance from $C$ to $AB$ is at least $\\frac{1}{3}v$, so $\\lambda \\ge \\frac{1}{3}$. Thus, the area of $ABC$ is at least $\\frac{1}{9}$ that of the original triangle.\n\nFor part (b), $ABC$ is contained in each $A_sB_sC_s$, so $A_s$ lies in the half-plane $\\{BCA\\}$ at distance at most $v$ from $BC$. The side $B_sC_s$ is also at most $v$ from $A$. Since $ABC$ contains all centroids, the distance between $B_sC_s$ and $BC$ is at most $\\frac{1}{3}v$. The distance from $A$ to $BC$ is $\\lambda v$, so the distance between parallel lines $BC$ and $B_sC_s$ is at most $\\min\\{1/3, 1-\\lambda\\} \\cdot v$. All triangles lie in the belt between lines parallel to $BC$ at distances $v$ and $\\min\\{1/3, 1-\\lambda\\} \\cdot v$. Similar belts exist for $AC$ and $AB$, so the union of all triangles lies in the intersection of three belts.\n\nWe consider two cases for $\\min\\{1/3, 1-\\lambda\\}$:\n\nIf $\\frac{1}{3} \\le \\lambda < \\frac{2}{3}$, the intersection is a hexagon obtained from a triangle $T$ (similar to the originals, similarity $1+\\lambda$) by removing three small triangles (similarity $\\lambda - \\frac{1}{3}$). The area is:\n\n$$\nS = (1 + \\lambda)^2 - 3 \\left( \\lambda - \\frac{1}{3} \\right)^2 = \\frac{8}{3} - 2(\\lambda - 1)^2 < \\frac{8}{3} - \\frac{2}{9} = \\frac{22}{9}.\n$$\n\nIf $\\frac{2}{3} \\le \\lambda \\le 1$, the intersection is again a hexagon, with $T$ similar to the originals (similarity $3-2\\lambda$), and the small triangles have similarity $1-\\lambda$. The area is:\n\n$$\n(3 - 2\\lambda)^2 - 3(1 - \\lambda)^2 = (\\lambda - 3)^2 - 3 \\le \\frac{49}{9} - 3 = \\frac{22}{9},\n$$\n\nwith equality if $\\lambda = \\frac{2}{3}$.\n\nHowever, since there are only finitely many triangles, the hexagon cannot be fully covered, so the area is strictly less than $\\frac{22}{9}$ times the area of the smallest triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16283,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\Gamma$ be the circumcircle of a triangle $ABC$ and let $D$ be a point on segment $BC$. The circle that passes through $B$ and $D$ and is tangent to $\\Gamma$, and the circle that passes through $C$ and $D$ and is tangent to $\\Gamma$, intersect at a point $E \\neq D$. The line $DE$ intersects $\\Gamma$ at two points, $X$ and $Y$. Prove that $|EX| = |EY|$.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider the configuration as in the figure, where $E$ is at least as close to $B$ as it is to $C$. The proof in the case of the configuration in which this is the other way around is analogous.\n\nLet $O$ be the centre of $\\Gamma$. The angle between the line $BC$ and the common tangent in $B$ is, by the inscribed angle theorem (tangent case), equal to $\\angle BED$, and also equal to $\\angle BAC$. So $\\angle BED = \\angle BAC$. Analogously, $\\angle CED = \\angle BAC$, so $\\angle BEC = \\angle BED + \\angle CED = 2\\angle BAC = \\angle BOC$, by the inscribed angle theorem. Therefore, $E$ lies on the circle passing through $B$, $O$, and $C$. If $E = O$, then $|EX|$ and $|EY|$ are both the radius of the circle.\n\nSuppose $E \\neq O$. Then $BEOC$ is a cyclic quadrilateral. $\\angle BEO = 180^\\circ - \\angle BCO$. In the isosceles triangle $BOC$, $\\angle BCO = 90^\\circ - \\frac{1}{2}\\angle BOC = 90^\\circ - \\angle BAC$, so $\\angle BEO = 180^\\circ - (90^\\circ - \\angle BAC) = 90^\\circ + \\angle BAC$. Hence $\\angle DEO = \\angle BEO - \\angle BED = 90^\\circ + \\angle BAC - \\angle BAC = 90^\\circ$. Therefore $EO$ is perpendicular to $DE$ and also perpendicular to chord $XY$, so $E$ is the midpoint of $XY$. Thus, $|EX| = |EY|$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16284,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $k$ for which the integers $1, 2, \\dots, 2017$ can be divided into $k$ groups in such a way that the sums of numbers in these groups are $k$ consecutive terms of an arithmetic sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the arithmetic sequence have first term $a$ and common difference $d$. The sum of all group sums equals the sum of $1, 2, \\dots, 2017$, i.e.,\n\n$$\n\\frac{2a + (k-1)d}{2} \\cdot k = \\frac{2017 \\cdot 2018}{2}\n$$\n\nThus,\n\n$$\n(2a + (k-1)d) \\cdot k = 2017 \\cdot 2018 = 2 \\cdot 1009 \\cdot 2017.\n$$\n\nSo $k$ must divide $2 \\cdot 1009 \\cdot 2017$. Since $2$, $1009$, and $2017$ are primes and $k \\leq 2017$, the only possibilities are $k = 1$, $k = 2$, $k = 1009$, and $k = 2017$.\n\nAll these are possible:\n\n- $k=1$: All numbers in one group.\n- $k=2$: Any partition into two groups gives two consecutive terms of an arithmetic sequence.\n- $k=1009$: Pair each even number with the next odd number (1008 groups of size 2, sums $5, 9, 13, \\dots$), and one group with just $1$.\n- $k=2017$: Each integer in its own group.\n\nThus, all these $k$ work.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16285,
    "subject": "Mathematics (Olympiad)",
    "question": "The code used in a definite week is randomly selected with equal chance among the three codes that have not been used in the last week. Suppose the code used in the first week is $A$. What is the probability that $A$ is also used in the seventh week? (Express your answer as an irreducible fraction.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P_k$ denote the probability that code $A$ is used in the $k$th week. The probability that $A$ is not used in the $k$th week is $1 - P_k$. Therefore,\n\n$$\nP_{k+1} = \\frac{1}{3}(1 - P_k).\n$$\n\nOr,\n\n$$\nP_{k+1} - \\frac{1}{4} = -\\frac{1}{3}\\left(P_k - \\frac{1}{4}\\right).\n$$\n\nSince $P_1 = 1$, the sequence $\\{P_k - \\frac{1}{4}\\}$ is geometric with first term $\\frac{3}{4}$ and common ratio $-\\frac{1}{3}$. Thus,\n\n$$\nP_k - \\frac{1}{4} = \\frac{3}{4}\\left(-\\frac{1}{3}\\right)^{k-1}.\n$$\n\nSo,\n\n$$\nP_k = \\frac{3}{4}\\left(-\\frac{1}{3}\\right)^{k-1} + \\frac{1}{4}.\n$$\n\nTherefore,\n\n$$\nP_7 = \\frac{61}{243}.\n$$\n\nThe answer is $\\frac{61}{243}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16286,
    "subject": "Mathematics (Olympiad)",
    "question": "For a given positive integer $n$, find the greatest positive integer $k$ such that there exist three sets of $k$ distinct nonnegative integers, $A = \\{x_1, x_2, \\dots, x_k\\}$, $B = \\{y_1, y_2, \\dots, y_k\\}$, and $C = \\{z_1, z_2, \\dots, z_k\\}$, with the property that for all $1 \\leq i \\leq k$, $x_i + y_i + z_i = n$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution**\n\nBy the given condition, we have\n$$\nkn \\geq \\sum_{i=1}^{k} (x_i + y_i + z_i) \\geq 3 \\sum_{i=0}^{k-1} i = \\frac{3k(k-1)}{2},\n$$\nand then $k \\leq \\left\\lfloor \\frac{2n}{3} \\right\\rfloor + 1$.\n\nThe following illustrates the case of $k = \\left\\lfloor \\frac{2n}{3} \\right\\rfloor + 1$:\n\nSet $m \\in \\mathbb{Z}^+$. When $n = 3m$, for $1 \\leq j \\leq m+1$, let $x_j = j-1$, $y_j = m + j - 1$, $z_j = 2m - 2j + 2$; for $m + 2 \\leq j \\leq 2m + 1$, let $x_j = j - 1$, $y_j = j - m - 2$, $z_j = 4m - 2j + 3$, and the result is obvious.\n\nWhen $n = 3m + 1$, for $1 \\leq j \\leq m$, let $x_j = j - 1$, $y_j = m + j$, $z_j = 2m - 2j + 2$; for $m + 1 \\leq j \\leq 2m$, let $x_j = j + 1$, $y_j = j - m - 1$, $z_j = 4m + 1 - 2j$; and $x_{2m+1} = m$, $y_{2m+1} = 2m + 1$, $z_{2m+1} = 0$ will lead to the expected result.\n\nWhen $n = 3m + 2$, for $1 \\leq j \\leq m + 1$, let $x_j = j - 1$, $y_j = m + j$, $z_j = 2m - 2j + 3$; for $m + 2 \\leq j \\leq 2m + 1$, let $x_j = j$, $y_j = j - m - 2$, $z_j = 4m - 2j + 4$; and $x_{2m+2} = 2m + 2$, $y_{2m+2} = m$, $z_{2m+2} = 0$, and the result follows.\n\nIn summary, the maximum value of $k$ is $\\left\\lfloor \\frac{2n}{3} \\right\\rfloor + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16287,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be a positive integer and let $S$ denote the set $\\{1, 2, \\dots, n\\}$. For $k = 1, 2, \\dots, n-1$, we call two $k$-element subsets of $S$ neighbours if they have exactly $k-1$ elements in common. Let $f(n, k)$ be the size of the largest possible collection of $k$-element subsets of $S$ of which no pair of subsets are neighbours. Prove that $$f(n, k) \\le \\binom{n-1}{k-1}.$$",
    "options": [],
    "answer": "See solution",
    "solution": "For any two subsets in such a collection, the sets of the $k-1$ smallest elements must be different; otherwise, they would be neighbours. There are $\\binom{n-1}{k-1}$ ways to choose the $k-1$ smallest elements, since the number $n$ cannot be one of them. Therefore, $f(n, k) \\le \\binom{n-1}{k-1}$, as desired.\n\n![](images/BW2021_Shortlist_p19_data_2c676345c6.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16288,
    "subject": "Mathematics (Olympiad)",
    "question": "The table below has 5 rows and 5 columns, and each square contains a unique number from 1 to 25.\n\n![](images/ROU_ABooklet_2021_p9_data_c151dfe413.png)\n\nProve that if the product of the numbers in any row is not divisible by $26$ or $34$, then there exists a row or a column such that the product of its numbers is divisible by $221$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $221 = 13 \\times 17$, we need to show that $13$ and $17$ appear together in some row or column.\n\nSuppose, for contradiction, that $13$ and $17$ are not in the same row or column. Then, in any row containing $13$, all numbers must be odd (otherwise, the product would be divisible by $26 = 2 \\times 13$). Excluding $13$ and $17$, there are only $11$ other odd numbers between $1$ and $25$.\n\nSimilarly, in any row containing $17$, all numbers must be odd, and excluding $13$ and $17$, there are not enough odd numbers to fill both rows. But there are only $13$ odd numbers in total, so this is impossible.\n\nTherefore, $13$ and $17$ must appear together in some row or column, and the product of the numbers in that row or column is divisible by $221$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16289,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, let $D$ and $E$ be the feet of the angle bisectors of angles $A$ and $B$, respectively. A rhombus is inscribed in the quadrilateral $AEDB$ so that each vertex of the rhombus lies on a different side of $AEDB$. Let $\\phi$ be the non-obtuse angle of the rhombus. Prove that\n$$\n\\phi \\le \\max\\{\\angle BAC, \\angle ABC\\}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K, L, M, N$ be the vertices of the rhombus lying on $AE, ED, DB, BA$, respectively. Define $d(X, YZ)$ as the distance from point $X$ to line $YZ$. Since $D$ and $E$ are the intersection points of the angle bisectors with the opposite sides, we have $d(D, AB) = d(D, AC)$, $d(E, AB) = d(E, BC)$, and $d(D, BC) = d(E, AC) = 0$. Thus,\n\n$$\nd(D, AC) + d(D, BC) = d(D, AB),\n$$\n$$\nd(E, AC) + d(E, BC) = d(E, AB).\n$$\n\nSince $L$ lies on segment $DE$ and the equation $d(X, AC) + d(X, BC) = d(X, AB)$ is linear in $X$, from the above we get\n$$\nd(L, AC) + d(L, BC) = d(L, AB). \\qquad (2)\n$$\n\nLabel the angles as shown in the figure, and let $a = KL$. Then $d(L, AC) = a \\sin \\mu$ and $d(L, BC) = a \\sin \\nu$. Since parallelogram $KLMN$ lies on one side of line $AB$, we have\n$$\n\\begin{aligned}\nd(L, AB) &= d(L, AC) + d(N, BC) = d(K, AB) + d(M, AB) \\\\\n&= a(\\sin \\delta + \\sin \\varepsilon).\n\\end{aligned}\n$$\n\nThus, from (1) we get\n$$\n\\sin \\mu + \\sin \\nu = \\sin \\delta + \\sin \\varepsilon. \\qquad (3)\n$$\n\nIf either angle $\\alpha$ or $\\beta$ is not acute, the desired inequality is already true. So assume $\\alpha, \\beta < \\pi/2$. It suffices to prove $\\psi = \\angle NKL \\le \\max\\{\\alpha, \\beta\\}$.\n\n![](images/14-3J-ind3_p6_data_f56226e95e.png)\n\nProceeding by contradiction, suppose $\\psi > \\max\\{\\alpha, \\beta\\}$. Since $\\mu + \\psi = \\angle CKN = \\alpha + \\delta$, by assumption $\\mu = (\\alpha - \\psi) + \\delta < \\delta$. Similarly, $\\nu < \\varepsilon$. Also, since $KN \\parallel ML$, $\\beta = \\delta + \\nu$, so $\\delta < \\beta < \\pi/2$. Similarly, $\\varepsilon < \\pi/2$. Finally, since $\\mu < \\delta < \\pi/2$ and $\\nu < \\varepsilon < \\pi/2$, we have\n$$\n\\sin \\mu < \\sin \\delta \\quad \\text{and} \\quad \\sin \\nu < \\sin \\varepsilon.\n$$\n\nThese two inequalities contradict (2). This completes the proof.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16290,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the maximal possible value of\n\n$$\n\\sum_{i,j=1}^n,\\ i \\neq j [x_i x_j] - (n-1) \\left( \\sum_{i=1}^n |x_i^2| \\right)\n$$\n\nfor all real numbers $x_1, x_2, \\ldots, x_n$.\n\nShow that in this case the answer is $n^2 - n + \\lfloor \\frac{n^2}{4} \\rfloor$.\n\nFor example, for the equality case:\n- For $n = 2k$ (where $k$ is an integer), take $k$ numbers as $0.9$ and $k$ numbers as $2.1$.\n- For $n = 2k + 1$ (where $k$ is an integer), take $k$ numbers as $0.9$ and $k + 1$ numbers as $2.1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that $|x_i^2|$ is even for $k$ numbers and odd for the remaining $l = n - k$ numbers. Then, each pair among the $\\binom{k}{2}$ even-even and $\\binom{l}{2}$ odd-odd pairs contributes at most $2$ to $S$, and each of the $kl$ even-odd pairs contributes at most $3$.\n\nTherefore,\n\n$$\nS \\le 3kl + k^2 - k + l^2 - l = (k + l)^2 + kl - k - l = n^2 - n + kl \\le n^2 - n + \\left\\lfloor \\frac{n^2}{4} \\right\\rfloor\n$$\n\nand we are done.\n\nFor $n = 31$, we get $31^2 - 31 + \\left\\lfloor \\frac{31^2}{4} \\right\\rfloor = 1170$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16291,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a positive integer and let $n_1, n_2, \\dots, n_k > 1$ be $k$ distinct integers such that\n\n$$\nn_1! + \\dots + n_k! = a^{n_1}\n$$\n\nfor some positive integer $a$. Prove that at least one of $n_1, n_2, \\dots, n_k$ is prime.",
    "options": [],
    "answer": "See solution",
    "solution": "If $k = 1$, then $n_1! = a^{n_1}$. Since $n_1 > 1$, $2 \\mid n_1!$ and so $2 \\mid a^{n_1}$, therefore $2 \\mid a$. Then\n\n$$\nn_1 \\le v_2(a^{n_1}) = v_2(n_1!) = \\left\\lfloor \\frac{n_1}{2} \\right\\rfloor + \\left\\lfloor \\frac{n_1}{2^2} \\right\\rfloor + \\dots < \\frac{n_1}{2} + \\frac{n_1}{2^2} + \\dots = n_1,\n$$\n\na contradiction.\n\nSo we can assume $k \\ge 2$. Let $m_1 < m_2$ be the two smallest elements of $\\{n_1, n_2, \\dots, n_k\\}$ and assume for the sake of contradiction that they are both composite.\n\n*Case 1.* Assume $m_2 \\ge m_1 + 2$. Then $v_2(n_i!) > v_2(m_1!)$ for each $i$ with $n_i \\ne m_1$. Therefore\n\n$$\nn_1 \\le v_2(a^{n_1}) = v_2(n_1! + n_2! + \\dots + n_k!) = v_2(m_1!) \\le v_2(n_1!) < n_1,\n$$\n\na contradiction.\n\n*Case 2.* Assume $m_2 = m_1 + 1$. Let $p$ be a prime factor of $m_2$. Since $m_2$ is composite, $p \\le m_1$ and so $p \\mid n_i!$ for each $i$, hence $p \\mid a$. Furthermore, $v_p(n_i!) \\ge v_p(m_2!) > v_p(m_1!)$ for each $i$ with $n_i \\ne m_1$ (since $p \\mid m_2$) and therefore\n\n$$\nn_1 \\le v_p(a^{n_1}) = v_p(n_1! + n_2! + \\dots + n_k!) = v_p(m_1!) \\le v_p(n_1!) = \\left\\lfloor \\frac{n_1}{p} \\right\\rfloor + \\left\\lfloor \\frac{n_1}{p^2} \\right\\rfloor + \\dots < n_1,\n$$\n\na contradiction.\n\nHence, we conclude that at least one of $m_1$ and $m_2$ has to be prime, which finishes the problem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16292,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all injective functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ that satisfy\n\n$$\n|f(x) - f(y)| \\le |x - y|,\n$$\n\nfor all $x, y \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The given relation yields $|f(x+1) - f(x)| \\le 1$, so $f(x+1)-f(x) \\in \\{-1, 0, 1\\}$. Because $f$ is injective, $f(x+1)-f(x) \\in \\{-1, 1\\}$ for all $x \\in \\mathbb{Z}$. Suppose without loss of generality that $f(1)-f(0) = 1$ (if $f$ is a solution, so is $-f$). Then $f(2)-f(1) = \\pm 1$. If $f(2)-f(1) = -1$, then $f(2) = f(0)$, contradicting injectivity, so $f(2) = f(0) + 2$.\n\nBy induction, $f(n) = f(0) + n$ for any positive integer $n$. Similarly, $f(-n) = f(0) - n$ for any positive integer $n$.\n\nThus, the functions that satisfy the condition are of the form $f(x) = \\pm x + k$ for any integer $x$, where $k$ is an arbitrary integer. All such functions are easily verified to be solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16293,
    "subject": "Mathematics (Olympiad)",
    "question": "An integer $n > 1$ is called *p-periodic* if $\\frac{1}{n}$ is a repeating decimal fraction whose shortest period has length $p$ and begins immediately after the decimal point. For example, $\\frac{1}{9} = 0.111\\ldots$ is 1-periodic and $\\frac{1}{11} = 0.090909\\ldots$ is 2-periodic.\n\n**a)** Find all p-periodic numbers $n$ such that the first digit of the period of $\\frac{1}{n}$ is not zero.\n\n**b)** Find the largest 4-periodic prime.",
    "options": [],
    "answer": "See solution",
    "solution": "a) The first digit of the period is at least 1 if and only if $\\frac{1}{n} \\geq \\frac{1}{10}$, i.e., $n \\leq 10$. Since the prime factors of $n$ must be different from 2 and 5, it follows that $n \\in \\{3, 7, 9\\}$. Indeed, $\\frac{1}{3} = 0.(3)$, $\\frac{1}{7} = 0.(142857)$, and $\\frac{1}{9} = 0.(1)$.\n\nb) Let $p$ be the largest 4-periodic prime. Then $\\frac{1}{p} = \\frac{m}{9999}$, with $1 \\leq m \\leq 9998$. This yields $pm = 9999 = 3^2 \\cdot 11 \\cdot 101$. So, the candidate for the largest such prime is $p = 101$, which is acceptable because $\\frac{1}{101} = 0.(0099)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16294,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the inequality\n\n$$\n\\frac{(x-y)^7 + (y-z)^7 + (z-x)^7 - (x-y)(y-z)(z-x)((x-y)^4 + (y-z)^4 + (z-x)^4)}{(x-y)^5 + (y-z)^5 + (z-x)^5} \\geq 3\n$$\n\nholds for all pairwise different integers $x, y, z$. When does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution:**\n\nSince\n\n$$\n\\begin{aligned}\n(x - y)^7 - (x - y)(y - z)(z - x)(x - y)^4 &= (x - y)^5 \\left( (x - y)^2 - (y - z)(z - x) \\right) \\\\\n&= (x - y)^5 \\left( x^2 + y^2 + z^2 - xy - yz - zx \\right),\n\\end{aligned}\n$$\n\nwe can write\n\n$$\n\\sum_{\\text{cyclic}} (x - y)^7 - (x - y)(y - z)(z - x)(x - y)^4 = \\left( \\sum_{\\text{cyclic}} (x - y)^5 \\right) \\cdot \\left( \\sum_{\\text{cyclic}} x^2 - \\sum_{\\text{cyclic}} xy \\right).\n$$\n\nIt therefore follows that the left-hand side of the inequality can be written as\n\n$$\n\\frac{\\left(\\sum_{\\text{cyclic}} (x-y)^5\\right) \\cdot \\left(\\sum_{\\text{cyclic}} x^2 - \\sum_{\\text{cyclic}} xy\\right)}{\\sum_{\\text{cyclic}} (x-y)^5} = \\sum_{\\text{cyclic}} x^2 - \\sum_{\\text{cyclic}} xy.\n$$\n\nWe therefore need to consider the inequality\n\n$$\nx^2 + y^2 + z^2 - xy - yz - zx \\geq 3\n$$\n\nfor $x, y, z \\in \\mathbb{Z}$ and $x \\neq y \\neq z \\neq x$. We note that\n\n$$\nx^2 + y^2 + z^2 - xy - yz - zx = \\frac{1}{2}(x - y)^2 + \\frac{1}{2}(y - z)^2 + \\frac{1}{2}(z - x)^2.\n$$\n\nEach pair of variables differs by at least one, and this is not possible for all three pairs at once. The smallest possible value is therefore obtained when two pairs differ by one, i.e., for three consecutive integers $m, m+1$ and $m+2$ in any order. Since\n\n$$\n\\frac{1}{2}((m+2) - (m+1))^2 + \\frac{1}{2}((m+1) - m)^2 + \\frac{1}{2}(m - (m+2))^2 = \\frac{1}{2} + \\frac{1}{2} + 2 = 3\n$$\n\nholds, we see that the given inequality is correct, and equality holds for\n\n$$\n(x, y, z) = (m, m+1, m+2)\n$$\n\nor any permutation thereof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16295,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ and consider a table with $n+1$ rows and $n$ columns, in which, on each of the first $n$ rows, Nicușor writes, in some order, the numbers $1, 2, \\dots, n$. Then, he chooses a permutation $a_1, a_2, \\dots, a_n$ of the numbers $1, 2, \\dots, n$ and completes the last row as follows: for each $j \\in \\{1, 2, \\dots, n\\}$, he writes in the cell at column $j$ the number of occurrences of $a_j$ in the cells located at the intersection of column $j$ with the first $n$ rows.\n\nDetermine all $n$ for which Nicușor can complete the table and choose $a_1, a_2, \\dots, a_n$ so that the last row contains, in some order, the numbers $1, 2, \\dots, n$.",
    "options": [],
    "answer": "See solution",
    "solution": "For $n=2$, no matter how Nicușor chooses $a_1$ and $a_2$, on the third row we will have two equal values. For $n=3$, suppose without loss of generality that $a_3=3$. Then, in the third column, we have only values equal to 3. Since the set $(1, 2)$ admits only two permutations, among the first three rows there will be two identical ones, so these rows coincide and we cannot have any $a_i = 1$.\n\nNext, we show that for $n \\ge 4$, there exists a completion that works. If $n=4$ and we take $a_1 = 1, a_2 = 2, a_3 = 3, a_4 = 4$, the following construction satisfies the conditions:\n\n![]( <table><tr><td>1</td><td>2</td><td>3</td><td>4</td></tr><tr><td>2</td><td>1</td><td>3</td><td>4</td></tr><tr><td>2</td><td>1</td><td>3</td><td>4</td></tr><tr><td>3</td><td>2</td><td>1</td><td>4</td></tr><tr><td>1</td><td>2</td><td>3</td><td>4</td></tr></table> )\n\n**Construction 1 (induction):**\n\nWe will prove by induction on $n \\ge 4$ that there exists a valid construction in which $a_i = i$ for every $i \\in \\{1, 2, \\dots, n\\}$ and the last row is exactly $1, 2, \\dots, n$. The case $n = 4$ is illustrated above. Suppose there exists a valid table $T_n$ of size $(n+1) \\times n$ for $n \\ge 4$, for which the chosen permutation by Nicușor is $a_1, a_2, \\dots, a_n$ with $a_i = i$ for every $i = 1, 2, \\dots, n$.\n\nWe construct a table $T_{n+1}$ of size $(n+2) \\times (n+1)$, valid for $n+1$, as follows: we add a column at the end of $T_n$ and a new row between the last and the penultimate rows of $T_n$. We fill all the cells of the newly added last column with the value $n+1$, and the first $n$ values of the newly added row with $2, 3, \\dots, n, 1$ (or any permutation that preserves the values for $a_1, a_2, \\dots, a_n$).\n\nIt is easy to verify that the newly constructed table is valid.\n\n**Construction 2 (direct):**\n\nFor $n \\ge 4$, consider $a_i = i$ for every $i = 1, 2, 3, 4, \\dots, n$ and the following construction:\n\n![]( <table><tr><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>...</td><td>n</td></tr><tr><td>2</td><td>1</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>...</td><td>n</td></tr><tr><td>2</td><td>1</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>...</td><td>n</td></tr><tr><td>3</td><td>2</td><td>1</td><td>4</td><td>5</td><td>6</td><td>7</td><td>...</td><td>n</td></tr><tr><td>2</td><td>3</td><td>4</td><td>1</td><td>5</td><td>6</td><td>7</td><td>...</td><td>n</td></tr><tr><td>2</td><td>3</td><td>4</td><td>5</td><td>1</td><td>6</td><td>7</td><td>...</td><td>n</td></tr><tr><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>1</td><td>7</td><td>...</td><td>n</td></tr><tr><td>...</td></tr><tr><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>...</td><td>n</td></tr></table> )\n\nIt is easy to verify that the newly constructed table is valid.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 16296,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$ be an integer and $a_1, a_2, \\dots, a_n$ be pairwise distinct positive real numbers with the property that there exists a permutation $b_1, b_2, \\dots, b_n$ of these numbers such that $$\\frac{a_1}{b_1} = \\frac{a_2}{b_2} = \\dots = \\frac{a_{n-1}}{b_{n-1}} \\ne 1.$$\n\nProve that there exist $a, b > 0$ such that $$\\{a_1, a_2, \\dots, a_n\\} = \\{ab, ab^2, \\dots, ab^n\\}.$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $r = \\frac{a_1}{b_1}$. Without loss of generality, assume that $a_1 < a_2 < \\dots < a_{n-1}$. Since $a_k = r \\cdot b_k$ for all $k \\in \\{1, 2, \\dots, n-1\\}$, it follows that $b_1 < b_2 < \\dots < b_{n-1}$.\n\nSuppose $r > 1$ (if $r < 1$, we simply exchange $a_k$ and $b_k$, $1 \\le k \\le n$). Because $r > 1$, we have $a_k > b_k$ for all $k \\in \\{1, 2, \\dots, n-1\\}$, so $b_1 < a_1 < a_2 < \\dots < a_{n-1}$. Since $b_1 \\in \\{a_1, a_2, \\dots, a_n\\}$, it follows that $b_1 = a_n$. Also, from $b_1 < b_2 < \\dots < b_{n-1} < a_{n-1}$, we infer that $a_1 = b_2$, $a_2 = b_3$, \\dots, $a_{n-2} = b_{n-1}$.\n\nGiven that $a_k = r \\cdot b_k$ for all $k \\in \\{1, 2, \\dots, n-1\\}$, we successively get that\n$a_2 = r \\cdot b_2 = r \\cdot a_1$, $a_3 = r \\cdot b_3 = r \\cdot a_2 = r^2 \\cdot a_1$, \\dots, $a_{n-1} = r \\cdot b_{n-1} = r \\cdot a_{n-2} = r^{n-2} \\cdot a_1$.\n\nMoreover, from $a_1 = r \\cdot b_1 = r \\cdot a_n$, we obtain $$\\{a_1, a_2, \\dots, a_n\\} = \\{a_n, a_n r, a_n r^2, \\dots, a_n r^{n-1}\\},$$ so the conclusion holds for $a = \\frac{a_n}{r}$ and $b = r$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 16297,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N$ be the maximum number of pairs $(i, j)$, $1 \\leq i < j \\leq n$, such that for a sequence of positive integers $a_1, \\dots, a_n$, the product $a_i a_j$ is a perfect cube, and none of the $a_i$ is itself a cube. Find $N$ in terms of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $N = \\left[ \\frac{n^2}{4} \\right]$.\n\n*Lemma.* Let $a, b, c$ be positive integers, none of which is a cube. If $ab$ and $ac$ are cubes, then $bc$ is not a cube.\n\nConsider a graph with $n$ vertices $A_1, \\dots, A_n$. Connect $A_i$ and $A_j$ ($i \\ne j$) with an edge if and only if $a_i a_j$ is a cube.\n\nLet $A$ be a vertex connected to the greatest number of vertices, say $A_1, \\dots, A_l$. Let $B_1, \\dots, B_k$ be the remaining vertices ($l + k + 1 = n$). By the lemma, no two of $A_1, \\dots, A_l$ are connected, so the total number of edges does not exceed $l + lk = l(k+1) = l(n-l) \\leq \\frac{n^2}{4}$. Thus, $N \\leq \\left[ \\frac{n^2}{4} \\right]$.\n\nFor $n = 2k$, let $a_1 = \\dots = a_k = 2$, $a_{k+1} = \\dots = a_{2k} = 4$; for $n = 2k+1$, let $a_1 = \\dots = a_k = 2$, $a_{k+1} = \\dots = a_{2k+1} = 4$. In both cases, $N = \\left[ \\frac{n^2}{4} \\right]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16298,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n > 4$ be a positive integer divisible by $4$. Let $A_n$ be the sum of all odd positive divisors of $n$. Let $B_n$ be the sum of all even positive divisors of $n$, excluding $n$ itself. Find the smallest possible value of $f(n) = B_n - 2A_n$. For which positive integers $n$ is this minimal value attained?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d_1, \\dots, d_k$ be the odd positive divisors of $n$. Then $2d_1, \\dots, 2d_k$ are even divisors of $n$, none of which are divisible by $4$, so none equal $n$. Also, none equal $4$. Therefore:\n\n$$\nA_n = d_1 + \\dots + d_k \\quad \\text{and} \\quad B_n = 2d_1 + 2d_2 + \\dots + 2d_k + 4.\n$$\n\nThus, $B_n - 2A_n \\geq 4$. The value $4$ is minimal, as for $n=8$:\n\n$$\nA_n = 1, \\quad B_n = 2 + 4 = 6, \\quad B_n - 2A_n = 4.\n$$\n\nNow, find all $n$ such that $B_n - 2A_n = 4$. The equality holds when $B_n = 2d_1 + \\dots + 2d_k + 4$. If $p$ is an odd prime divisor of $n$, then $4p$ is an even divisor of $n$ not included in $2d_1 + \\dots + 2d_k$. Thus, for $B_n = 2d_1 + \\dots + 2d_k + 4$, it is necessary that $n = 4p$. In this case:\n\n$$\nB_n - 2A_n = (2 + 4 + 2p) - 2(1 + p) = 4.\n$$\n\nIf $n$ has no odd prime divisors, then $n = 2^{k+1}$, so $A_n = 1$, $B_n = 2 + 2^2 + \\dots + 2^k = 2^{k+1} - 2$, and $B_n - 2A_n = 2^{k+1} - 4$.\n\nTherefore, $B_n - 2A_n = 4$ if and only if $n = 8$ or $n = 4p$ where $p$ is an odd prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16299,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the last digit of $2011 \\times 2013 \\times 2015 - 2010 \\times 2012 \\times 2014$?",
    "options": [],
    "answer": "See solution",
    "solution": "The last digit of $2011 \\times 2013 \\times 2015$ is $5$, and the last digit of $2010 \\times 2012 \\times 2014$ is $0$, so the last digit of the difference is $5 - 0 = 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16300,
    "subject": "Mathematics (Olympiad)",
    "question": "Our goal is to prove that for any positive integers $n$ and $m$ with $m < n$, there exist infinitely many prime numbers $p$ such that\n\n$$\n\\mathrm{ord}_p^m = \\mathrm{ord}_p^n\n$$\n\nwhere $\\mathrm{ord}_p^x$ denotes the order of $x$ modulo $p$.\n\n![](images/2019_p35_data_bae44c41e0.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Assume, for contradiction, that only finitely many primes $\\{p_1, p_2, \\dots, p_r\\}$ satisfy $\\mathrm{ord}_p^m = \\mathrm{ord}_p^n$. Let $p_1 < p_2 < \\dots < p_r$ and set\n\n$$\np_1p_2\\dots p_r(m-n)n = q_1^{\\alpha_1}q_2^{\\alpha_2}\\dots q_s^{\\alpha_s}\n$$\n\nChoose a prime $p > q_1^{\\alpha_1+2}q_2^{\\alpha_2+2}\\dots q_s^{\\alpha_s+2}(q_1-1)(q_2-1)\\dots(q_s-1) = k$. Consider $g = m^{p^{\\varphi(k)}} - n$.\n\nIf $q_i \\nmid (m, n)$, then\n\n$$\nq_i^{\\alpha_i+1}(q_i-1) \\mid p^{\\varphi(k)} - 1\n$$\n\nand\n\n$$\nm^{p^{\\varphi(k)}} - n \\equiv m - n \\pmod{q_i^{\\alpha_i+1}}\n$$\n\nIf $q_i \\mid (m, n)$, then\n\n$$\nm^{p^{\\varphi(k)}} - n \\equiv -n \\pmod{q_i^{\\alpha_i+1}}\n$$\n\nThus, the exponent of $q_i$ in $g$ is less than $\\alpha_i + 1$. Write\n\n$$\ng = q_1^{\\beta_1} q_2^{\\beta_2} \\dots q_s^{\\beta_s} r_1^{\\gamma_1} r_2^{\\gamma_2} \\dots r_t^{\\gamma_t}\n$$\n\nIf $\\gamma_i \\equiv 1 \\pmod{p}$ for all $i$, then\n\n$$\ng \\equiv m - n \\pmod{p}\n$$\n\nand\n\n$$\ng \\equiv q_1^{\\beta_1} q_2^{\\beta_2} \\dots q_s^{\\beta_s} \\pmod{p}\n$$\n\nSo\n\n$$\np + m - n \\equiv q_1^{\\beta_1} q_2^{\\beta_2} \\dots q_s^{\\beta_s} \\pmod{p}\n$$\n\nBut $p$ is chosen larger than this bound, so there must exist $\\gamma_i$ with $p \\nmid \\gamma_i - 1$. Let $\\mathrm{ord}_{\\gamma_i}^m = x$, $\\mathrm{ord}_{\\gamma_i}^n = y$, and $p \\nmid xy$. There exists $\\lambda$ such that\n\n$$\np^{\\varphi(k)}\\lambda \\equiv 1 \\pmod{x}\n$$\n\nThus,\n\n$$\nm \\equiv n^{\\lambda} \\pmod{\\gamma_i}\n$$\n\nand for any $z$,\n\n$$\nm^z \\equiv n^{\\lambda z} \\pmod{\\gamma_i}, \\quad n^z \\equiv m^{p^{\\varphi(k)}z} \\pmod{\\gamma_i}\n$$\n\nSo the sets $\\{1, m^1, \\dots, m^{x-1}\\}$ and $\\{1, n^1, \\dots, n^{y-1}\\}$ are equal modulo $\\gamma_i$, implying $x = y$, a contradiction.\n\nTherefore, there are infinitely many such primes.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16301,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a set of positive integers such that $\\lfloor \\sqrt{x} \\rfloor = \\lfloor \\sqrt{y} \\rfloor$ for all $x, y \\in S$. Show that the products $xy$, where $x, y \\in S$, are pairwise distinct.",
    "options": [],
    "answer": "See solution",
    "solution": "We first show that if $x_1, x_2, x_3, x_4$ are members of $S$ such that $x_1 x_2 \\leq x_3 x_4$, then $x_1 + x_2 \\leq x_3 + x_4$.\n\nSuppose, if possible, that $x_1 + x_2 > x_3 + x_4$. Let $n = \\lfloor \\sqrt{x} \\rfloor$ for $x \\in S$, and write $x_k = n^2 + w_k$, where the $w_k$ are non-negative integers less than $2n+1$. This gives $w_1 + w_2 - w_3 - w_4 \\geq 1$.\n\nThe condition $x_1 x_2 \\leq x_3 x_4$ yields $(w_1 + w_2 - w_3 - w_4) n^2 \\leq w_3 w_4 - w_1 w_2$, so $w_3 > 0$ and\n\n$$\n\\begin{aligned}\nn^2 &\\leq (w_1 + w_2 - w_3 - w_4) n^2 \\leq w_3 w_4 - w_1 w_2 \\\\ &< w_3 (w_1 + w_2 - w_3) - w_1 w_2 \\\\ &= (w_1 - w_3)(w_3 - w_2) \\leq \\frac{((w_1 - w_3) + (w_3 - w_2))^2}{4} = \\frac{(w_1 - w_2)^2}{4} \\leq n^2,\n\\end{aligned}\n$$\n\nwhich is a contradiction.\n\nThus, if $x_1, x_2, x_3, x_4$ are members of $S$ such that $x_1 x_2 = x_3 x_4$, then $x_1 + x_2 = x_3 + x_4$, so $x_1^2 + x_3 x_4 = x_1 (x_1 + x_2) = x_1 (x_3 + x_4)$, i.e., $(x_1 - x_3)(x_1 - x_4) = 0$, whence $x_1 = x_3$ or $x_1 = x_4$. The conclusion now follows at once.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16302,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that there are no 2-tuples $(x, y)$ of positive integers satisfying the equation\n\n$$\n(x+1)(x+2)\\cdots(x+2014) = (y+1)(y+2)\\cdots(y+4028).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof.* For $n = 2^k \\cdot m$ ($k$ is a non-negative integer, $m$ is odd), let $v(n) = 2^k$.\n\nWe prove by contradiction: assume $(x, y)$ is one positive integer solution of the equation. Let\n\n$$\nv(x + i) = \\max_{1 \\leq j \\leq 2014} \\{v(x + j)\\},\n$$\n\nthen if $1 \\leq j \\leq 2014$, $j \\neq i$,\n\n$$\nv(x + j) = v(x + i + (j - i)) = v(j - i),\n$$\n\nso\n\n$$\nv\\left(\\prod_{1 \\leq j \\leq 2014,\\ j \\neq i} (x + j)\\right) = v\\left((2014 - j)! \\cdot (j - 1)!\\right) \\leq v(2013!)\n$$\n\nSince $\\prod_{j=1}^{2014} (x + j) = \\prod_{j=1}^{4028} (y + j)$ is a multiple of $4028!$, thus\n\n$$\nx + i \\geq v(x + i) \\geq v\\left(\\frac{4028!}{2013!}\\right) > 2^{1007},\n$$\n\ntherefore $x > 2^{1006}$. So $(y + 4028)^{4028} > \\prod_{j=1}^{4028} (y + j) = \\prod_{j=1}^{2014} (x + j) > 2^{1006 \\cdot 2014}$, we have $y + 4028 > 2^{503}$, $y > 2^{502}$.\n\n*Lemma:* Let $0 \\leq x_i < \\frac{1}{2}$ ($1 \\leq i \\leq n$), if $x = \\frac{1}{n} \\sum_{i=1}^{n} x_i$, $y = 2 \\max_{1 \\leq i \\leq n} \\{x_i^2\\}$, then\n\n$$\n1 - x \\geq \\left( \\prod_{i=1}^{n} (1 - x_i) \\right)^{\\frac{1}{n}} \\geq 1 - x - y.\n$$\n\n*Proof of lemma:* By AM-GM inequality, the inequality on the left is easy to prove. The inequality on the right holds, since\n\n$$\n\\begin{aligned}\n\\left( \\prod_{i=1}^{n} (1-x_i) \\right)^{\\frac{1}{n}} &\\geq \\frac{n}{\\sum_{i=1}^{n} \\frac{1}{1-x_i}} \\\\\n&\\geq \\frac{n}{\\sum_{i=1}^{n} (1+x_i + 2x_i^2)} \\\\\n&= \\frac{n}{n+nx+2\\sum_{i=1}^{n} x_i^2} \\\\\n&\\geq \\frac{1}{1+x+y} \\\\\n&\\geq 1-x-y,\n\\end{aligned}\n$$\n\nthe lemma is proved.\n\nBack to the problem, let $w = x + 2015 > 2^{1007}$, $z = y + \\frac{4029}{2} > 2^{502}$, then the equation is equivalent to\n\n$$\nw \\cdot \\left( \\left(1 - \\frac{1}{w}\\right) \\left(1 - \\frac{2}{w}\\right) \\cdots \\left(1 - \\frac{2014}{w}\\right) \\right)^{\\frac{1}{2014}} = z^2 \\cdot \\left( \\left(1 - \\frac{1}{4z^2}\\right) \\left(1 - \\frac{9}{4z^2}\\right) \\cdots \\left(1 - \\frac{4027^2}{4z^2}\\right) \\right)^{\\frac{1}{2014}}.\n$$\n\nBy the lemma,\n\n$$\nw\\left(1 - \\frac{2015}{2w}\\right) > w \\cdot \\left(\\left(1 - \\frac{1}{w}\\right)\\left(1 - \\frac{2}{w}\\right)\\cdots\\left(1 - \\frac{2014}{w}\\right)\\right)^{\\frac{1}{2014}}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16303,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x)$ be a monic polynomial of even degree with integer coefficients. It is known that there exist infinitely many integers $x$ for which $f(x)$ is a perfect square. Prove that there exists a polynomial $g(x)$ with integer coefficients such that $f(x) = g^2(x)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = 2k$ and $f(x) = x^{2k} + a_{2k-1}x^{2k-1} + \\dots + a_1x + a_0$, where $a_i$ are integers. First, we prove that $f(x)$ can be written in the form\n\n$$\nf(x) = \\left(x^k + b_{k-1}x^{k-1} + \\dots + b_1x + b_0\\right)^2 + r(x),\n$$\n\nwhere $b_0, b_1, \\dots, b_{k-1}$ are rational numbers and $r(x)$ is a polynomial with rational coefficients and degree at most $k-1$. Indeed, the coefficient of $x^{k+t}$, $t = k-1, k-2, \\dots, 1, 0$, of the polynomial $\\left(x^k + b_{k-1}x^{k-1} + \\dots + b_1x + b_0\\right)^2$ is of the form $c_{k+t} = 2b_t + \\sum_{i=1}^{k-1-t} b_{t+i}b_{k-i}$. Inductively, we find the values of $b_{k-1}, b_{k-2}, \\dots, b_1, b_0$ such that $c_{k+t} = a_{k+t}$ for $t = k-1, k-2, \\dots, 1, 0$. After that, we compute the coefficients of $r(x)$.\n\nIf $f(x) = y^2$ has infinitely many integer solutions for which $x < 0$, then $f_1(x) = y^2$ for $f_1(x) = f(-x)$ has infinitely many solutions for which $x > 0$. Therefore, we may assume that $f(x) = y^2$ has infinitely many solutions for which $x > 0$. The equality $\\left(x^k + b_{k-1}x^{k-1} + \\dots + b_1x + b_0\\right)^2 + r(x) = y^2$ can be written in the form $h^2(x) + M^2r(x) = (My_x)^2$, where $M$ is the least common multiple of the denominators of $b_i$, $0 \\le i \\le k-1$, and $h(x)$ is a polynomial with integer coefficients and leading coefficient $M$. Suppose that $r(x)$ is not identically $0$ (if $b_i = 0$ for all $i$, $0 \\le i \\le k-1$, we set $M=1$).\n\n**Case 1.** Let the leading coefficient of $r(x)$ be positive. For large enough values of $x$ we have $(My_x)^2 > h^2(x)$, implying that $(My_x)^2 \\ge (h(x)+1)^2$. Therefore $h^2(x) + M^2r(x) \\ge (h(x)+1)^2$, i.e. $2h(x) \\le M^2r(x)-1$. The latter inequality is not true for large enough values of $x$ since $h(x)$ is of degree $k$ whereas $r(x)$ is of degree at most $k-1$.\n\n**Case 2.** Let the leading coefficient of $r(x)$ be negative. For large enough values of $x$ we have $(My_x)^2 < h^2(x)$, implying $(My_x)^2 \\le (h(x)-1)^2$. Therefore $h^2(x) + M^2r(x) \\le (h(x)-1)^2$, i.e. $2h(x) \\le -M^2r(x)+1$, which is not true for large enough values of $x$.\n\nTherefore $r(x) \\equiv 0$, i.e. $f(x) = \\left(x^k + b_{k-1}x^{k-1} + \\dots + b_1x + b_0\\right)^2$. Since $f(x)$ has integer coefficients, the same is true for $x^k + b_{k-1}x^{k-1} + \\dots + b_1x + b_0$ (Gauss Lemma).",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16304,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $a$, $b$, $c$, $d$ such that\n$$\begin{aligned}\nab + c + d &= 3 \\\\\nbc + d + a &= 5 \\\\\ncd + a + b &= 2 \\\\\nda + b + c &= 6\n\\end{aligned}$$",
    "options": [],
    "answer": "See solution",
    "solution": "Adding the first two equations and subtracting the last two gives\n$$(b - d)(a + c - 2) = 0.$$  \nIf $b = d$, then the first equation contradicts the last, so $a + c = 2$.\n\nAdding the last two equations:\n$$(d + 1)(a + c) + 2b = 8,$$\nso $b + d = 3$.\n\nAdding the second and third equations:\n$$(c + 1)(b + d) + 2a = 7,$$\nso $3c + 2a = 4$.\n\nSolving these, we get $a = 2$, $b = 0$, $c = 0$, $d = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16305,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all integers $n \\geq 2$ that have a representation\n\n$$\nn = a^2 + b^2\n$$\n\nwhere $a$ is the smallest divisor of $n$ different from $1$ and $b$ is an arbitrary divisor of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $n$ is odd, then both $a$ and $b$ are odd, so $n = a^2 + b^2$ is even—a contradiction. Therefore, $n$ is even and $a = 2$. This also shows that $b$ is even. Furthermore, $b \\mid (n - b^2) = a^2 = 4$. Thus $b \\in \\{2, 4\\}$, which results in $n = 8$ and $n = 20$, respectively. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16306,
    "subject": "Mathematics (Olympiad)",
    "question": "Two persons play the following game with positive integers. The initial number is $2011^{2011}$. Each move consists of subtracting an integer between $1$ and $2010$ inclusive, or dividing by $2011$, rounding down when necessary. The player who obtains a non-positive integer wins. Who will win this game: the first player or the second?",
    "options": [],
    "answer": "See solution",
    "solution": "The second player wins.\n\nThe initial numbers $N$ for which the second player has a winning strategy are those that have an odd number of trailing $0$'s in base $2011$ (i.e., the largest power of $2011$ dividing $N$ is odd). Each move by the first player makes this largest power even, and then the second player can make it odd again with a suitable move. Thus, the second player can always force a win.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16307,
    "subject": "Mathematics (Olympiad)",
    "question": "Ares multiplies two integers which differ by 9. Grace multiplies two integers which differ by 6. They obtain the same product $T$. Determine all possible values of $T$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let Ares's integers be $a$ and $a + 9$, and let Grace's be $g$ and $g + 6$. Then completing the square gives\n\n$$\na(a + 9) = g(g + 6), \\\\\n(a + \\frac{9}{2})^2 - \\frac{81}{4} = (g + 3)^2 - 9, \\\\\n(a + \\frac{9}{2})^2 - (g + 3)^2 = \\frac{45}{4}, \\\\\n(2a + 9)^2 - (2g + 6)^2 = 45.\n$$\n\nNow let $x = 2a + 9$ and $y = 2g + 6$. We will assume $x \\ge 0$ and $y \\ge 0$. Then $(x + y)(x - y) = x^2 - y^2 = 45$. 45 can be factorized as $45 \\times 1$, $15 \\times 3$, or $9 \\times 5$. So there are 3 cases for $x$ and $y$.\n\n**Case 1:**\n$$\nx + y = 45, \\\\\nx - y = 1.\n$$\nThen $x = 23$, $y = 22$.\n\n**Case 2:**\n$$\nx + y = 15, \\\\\nx - y = 3.\n$$\nThen $x = 9$, $y = 6$.\n\n**Case 3:**\n$$\nx + y = 9, \\\\\nx - y = 5.\n$$\nThen $x = 7$, $y = 2$.\n\nNow solve $x = 2a + 9$ in the three cases. So $2a + 9 = 23, -23, 9, -9, 7, -7$, giving $a = 7, -16, 0, -9, -1, -8$. Then $T = a(a + 9)$, so in these cases $T = 112, 112, 0, 0, -8, -8$ respectively.\n\nWe need to check that these values can also be achieved by Grace. Using $y = 2g + 6$, we find $g = 8, -14, 0, -6, -2, -4$. These give the same set of values for $T$, so the possible values for $T$ are $112$, $0$, and $-8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16308,
    "subject": "Mathematics (Olympiad)",
    "question": "Дванаесетте витези на кружната маса треба да изберат двочлена делегација за посета на кралот Артур. На колку начини може тоа да се направи, така што во двочленна делегација да не бидат витези кои седат еден до друг на кружната маса?",
    "options": [],
    "answer": "See solution",
    "solution": "Нека претпоставиме дека кружната маса е кружница, а витезите кои седат на кружната маса се темиња на 12-аголник. Секој пар витези што седат на масата и прават двочлена делегација според условот на задачата, определуваат една дијагонала на 12-аголникот. \n\nБројот на дијагонали во 12-аголник е еднаков на:\n\n$$\n\\frac{12 \\cdot 9}{2} = 54\n$$\n\nЗначи, кралот Артур може да формира 54 двочлени делегации, на начин определен во задачата.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16309,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $m$, let $s(m)$ denote the sum of the digits of $m$. Let $P(x) = x^n + a_{n-1}x^{n-1} + \\cdots + a_1x + a_0$ be a polynomial where $n \\ge 2$ and each $a_i$ is a positive integer for $0 \\le i \\le n-1$. Is it possible that, for all positive integers $k$, $s(k)$ and $s(P(k))$ always have the same parity?",
    "options": [],
    "answer": "See solution",
    "solution": "No, such a polynomial does not exist.\n\nAssume for contradiction that such a polynomial $P(x)$ exists. Let $a_n = 1$.\n\n**Claim 1:** $s(P(k))$ and $\\sum_{i=0}^{n} s(a_i k^i)$ have the same parity for all $k > 0$.\n\n*Proof.* Take a large power of $10$, say $10^m$, such that all $a_i k^i$ have fewer than $m$ digits. Then there is no carry-over when adding, so $s(P(k \\cdot 10^m)) = \\sum_{i=0}^n s(a_i k^i)$, which has the same parity as $s(k \\cdot 10^m) = s(k)$. $\\square$\n\nNow, find $k$ so that there is exactly one carry-over in the addition for $s(P(k))$. Take a large $t$ such that\n\n$$\n10^t > \\max \\left\\{ \\frac{100^{n-1}a_{n-1}}{(10^{1/(n-1)} - 9^{1/(n-1)})^{n-1}}, \\frac{a_{n-1}10^{n-1}}{9}, \\frac{a_{n-1}(10a_{n-2})^{n-1}}{9}, \\dots, \\frac{a_{n-1}(10a_0)^{n-1}}{9} \\right\\}\n$$\n\nFrom the first term, the interval\n\n$$\nI = \\left[ \\left( \\frac{9}{a_{n-1}} 10^t \\right)^{1/(n-1)}, \\left( \\frac{1}{a_{n-1}} 10^{t+1} \\right)^{1/(n-1)} \\right)\n$$\n\ncontains at least 100 consecutive positive integers. Let $X$ be an integer in $I$ congruent to $1 \\pmod{100}$. Since $X \\in I$,\n\n$$\n9 \\cdot 10^t \\leq a_{n-1}X^{n-1} < 10^{t+1}\n$$\n\nso $a_{n-1}X^{n-1}$ has exactly $t+1$ digits and starts with a $9$.\n\nNext, $a_{n-1}(10a_i)^{n-1} < 9 \\cdot 10^t \\leq a_{n-1}X^{n-1}$ implies $10a_i < X$ for $i \\leq n-2$, so\n\n$$\na_i X^i < \\frac{X^{i+1}}{10} \\leq \\frac{a_{n-1}X^{n-1}}{10} < 10^t\n$$\n\nThus, $a_i X^i$ has at most $t$ digits for all $i \\leq n-2$.\n\nTake $k = 10^t X$. Then\n\n$$\nP(k) = 10^{tn}X^n + a_{n-1}10^{t(n-1)}X^{n-1} + \\cdots + a_0.\n$$\n\nFor $i \\leq n-2$, $a_i X^i$ has at most $t$ digits, so $a_i 10^{ti}X^i < 10^{t(i+1)}$ has at most $t(i+1)$ digits, while $a_{i+1} 10^{t(i+1)}X^{i+1}$ has at least $t(i+1)$ zeros. Thus, these terms do not interact and there is no carry-over while adding them.\n\nFinally, $10^{tn+1} > a_{n-1}10^{t(n-1)}X^{n-1} \\geq 9 \\cdot 10^{tn}$, so $a_{n-1}10^{t(n-1)}X^{n-1}$ has exactly $tn+1$ digits, with leading digit $9$. On the other hand, $10^t X^n$ has exactly $tn$ zeros followed by $01$ (since $X \\equiv 1 \\pmod{100}$). When adding, the $9$ and $1$ become $0$, the $0$ becomes $1$, and nothing else is affected.\n\nPutting this together:\n\n$$\ns(P(k)) = -9 + \\sum_{i=0}^{n} s(a_i k^i) = -9 + \\sum_{i=0}^{n} s(a_i X^i)\n$$\n\nBy the assumption, the left side has the same parity as $s(k) = s(10^t X) = s(X)$, while by the claim, the right side has the same parity as $s(X) - 9$, which is opposite in parity to $s(X)$, a contradiction! $\\square$\n\nThe key idea is to ensure exactly one carry-over occurs in $P(k)$. Another approach is to take $k$ as a large power of $5$ and use the irrationality of $\\log_{10} 5$ to control the digits.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16310,
    "subject": "Mathematics (Olympiad)",
    "question": "There are four points $P$, $Q$, $R$, $S$ (from left to right) on a horizontal line. Construct a square $ABCD$ such that:\n- Point $P$ lies on line $AD$,\n- Point $Q$ lies on line $BC$,\n- Point $R$ lies on line $AB$,\n- Point $S$ lies on line $CD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $w$ and $k$ be the circles with diameters $PS$ and $QR$, respectively (see the figure below).\n\nThus, vertices $D$ and $B$ of the square lie on these circles, respectively. The line $BD$ is the bisector of the right angles $PDS$ and $QBR$, so it intersects the arcs $PS$ and $QR$ at their midpoints $U$ and $V$, respectively.\n\nAfter constructing circles $w$ and $k$, locate points $U$ and $V$. Then, draw the line $UV$, which intersects the circles at points $D$ and $B$, respectively. The construction of the other two vertices follows naturally.\n\n![](images/Ukraine_2016_Booklet_p18_data_a51d07a64f.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16311,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ ($BC > AC$) be an acute triangle with circumcircle $k$ centered at $O$. The tangent to $k$ at $C$ intersects the line $AB$ at the point $D$. The circumcircles of triangles $BCD$, $OCD$, and $AOB$ intersect the ray $CA$ (beyond $A$) at the points $Q$, $P$, and $K$, respectively, such that $P \\in (AK)$ and $K \\in (PQ)$. The line $PD$ intersects the circumcircle of triangle $BKQ$ at the point $T$, so that $P$ and $T$ are in different halfplanes with respect to $BQ$. Prove that $TB = TQ$.",
    "options": [],
    "answer": "See solution",
    "solution": "As $DC$ is tangent to $k$ at $C$, $\\angle OCD = 90^\\circ$. Denote by $X$ the midpoint of $AB$. Then $\\angle OXA = 90^\\circ$ because $OX$ is the perpendicular bisector of $AB$. The pentagon $PXOCD$ is inscribed in the circle with diameter $OD$, hence $\\angle PXA = \\angle PXD = \\angle PCD = \\angle QCD = \\angle QBA$ (the latter is due to $QBCD$ being cyclic). We deduce that $PX \\parallel QB$ and that $P$ is the midpoint of $AQ$, so $AP = PQ$.\n\n![](images/2019_bmo_shortlist_p19_data_391dbb1a85.png)\n\nNow let $T_1$ be the midpoint of the arc $BQ$, not containing $K$, from the circumcircle of $\\triangle BKQ$, then $T_1B = T_1Q$. Due to $\\angle DPO = 90^\\circ$, it suffices to show that $\\angle OPT_1 = 90^\\circ$—indeed, $T \\equiv T_1$ and $TB = TQ$ would follow.\n\nDenote by $Y$ the midpoint of $BQ$. Then $\\angle OXB = \\angle T_1YB = 90^\\circ$. The quadrilateral $QKBT_1$ is inscribed in a circle, hence $\\angle BT_1Q = 180^\\circ - \\angle BKQ = \\angle AKB$. Then $\\angle XBO = \\frac{1}{2}\\angle AKB = \\frac{1}{2}\\angle BT_1Q = \\angle BT_1Y$ and thus $\\triangle OXB \\sim \\triangle BYT_1$. The quadrilaterals $PXBY$ and $AXYP$ are parallelograms, since $XY$ and $PY$ are midlines of triangle $AQB$.\n\nConsequently,\n\n$$\n\\frac{OX}{XP} = \\frac{OX}{BY} = \\frac{XB}{T_1Y} = \\frac{PY}{T_1Y'}\n$$\n\nwhich along with $\\angle PXB = \\angle PYB$ and $\\angle OXB = \\angle T_1YB$ gives $\\angle OXP = \\angle PYT_1$ and $\\triangle OXP \\sim \\triangle PYT_1$. Thus $\\angle XPO = \\angle YT_1P$ and $\\angle POX = \\angle T_1PY$.\n\nIn conclusion,\n\n$$\n\\angle OPT_1 = \\angle XPY + \\angle XPO + \\angle YPT_1 = \\angle PXA + \\angle XPO + \\angle XOP = 90^\\circ\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16312,
    "subject": "Mathematics (Olympiad)",
    "question": "30 military ships are approaching the island: 10 destroyers and 20 small ships. All ships are arranged in a line with equal distances between neighboring ships. Two battleships are defending the island. Each has exactly 10 rockets.\n\n- The first battleship can launch all 10 rockets at the same time, but all 10 targets must be neighboring ships.\n- The second battleship can launch all 10 rockets at the same time, but all 10 targets must alternate (i.e., every other ship in a sequence).\n\nBoth battleships launch their rockets simultaneously (so a single ship may be targeted by both). What is the maximum number of destroyers that can be saved, regardless of how the battleships launch their rockets?\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** 3 destroyers.\n\nLet us number the military ships from left to right as 1 to 30. If in both groups 1–10 and 21–30 there are no more than 2 destroyers each, then in group 11–20 there must be at least 6 destroyers. The first battleship can target this group, and the second battleship will hit at least 1 destroyer. In the opposite case, the first battleship targets either group 1–10 or 21–30 (whichever has more destroyers). The second battleship then targets the remaining 20 ships (either 11–30 or 1–20), choosing odd or even positions depending on which group has more destroyers. If the first hits $k$ destroyers, $k \\geq 3$, then the second can hit at least $\\frac{1}{2}(10-k)$ of the remaining $10-k$ destroyers. In total, they hit $\\frac{1}{2}(10-k) + k = 5 + \\frac{1}{2}k \\geq 7$ destroyers.\n\nNow, consider an example: place destroyers at positions 1, 2, 3; 14, 15, 16, 17; and 28, 29, 30. The first battleship can hit at most 4 destroyers (only if it targets 14–17). The second cannot hit both the leftmost and rightmost destroyers at the same time, so it can hit at most 2 of them. If the first targets a group with 3 destroyers, the second can hit at most 4. Thus, in all cases, at least 3 destroyers can be saved.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16313,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an acute, scalene triangle $ABC$ with circumcircle $(O)$. The line passing through $O$ and the midpoint $I$ of $BC$ intersects $AB$ and $AC$ at $E$ and $F$, respectively. Let $D$ and $G$ be the reflections of $A$ over $O$ and the circumcenter of triangle $AEF$. Let $K$ be the reflection of $O$ over the circumcenter of triangle $OBC$.\n\n**a)** Prove that $D$, $G$, and $K$ are collinear.\n\n**b)** Take $M$ on $KB$ and $N$ on $KC$ such that $IM \\perp AC$, $IN \\perp AB$. The perpendicular bisector of $IK$ intersects $MN$ at $H$. Suppose that $IH$ meets $AB$ and $AC$ at $P$ and $Q$, respectively. Prove that the circumcircle of triangle $APQ$ intersects $(O)$ again at a point on $AI$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $T$ be the projection of $A$ onto $GD$. It is well known that $T$ is the second intersection of $(AEF)$ and $(O)$. We observe that $\\triangle TEB \\sim \\triangle TFC$, then\n\n$$\n\\frac{TB}{TC} = \\frac{BE}{CF}.\n$$\n\n![](images/Vietnam_2022_p38_data_ca22d7bef7.png)\n\nOn the other hand,\n\n$$\n\\frac{BE}{CF} = \\frac{BE}{IB} \\cdot \\frac{IC}{CF} = \\frac{\\cos ACB}{\\cos ABC} = \\frac{BD}{CD},\n$$\n\nwhich implies that $TBDC$ is a harmonic quadrilateral, or $TD$ passes through $K$, which is the intersection of the tangents at $B$ and $C$ of $(O)$.\n\nb) Let $X$ and $Y$ be the intersections of $KB$, $KC$ with $IN$, $IM$, respectively. The tangents at $B$ and $C$ of $(O)$ meet the tangent at $A$ at $X'$, $Y'$. Because $IX \\parallel OX'$ and $IY \\parallel OY'$, then\n\n$$\n\\triangle KX'Y' \\sim \\triangle KXY\n$$\n\nwith $O$, $I$ corresponding. Note that $O$ is the incenter of $KX'Y'$, so $I$ is the incenter of triangle $KXY$.\n\nLet $(KXY)$ meet $IY$, $IX$ at $R$, $S$ respectively. Clearly, $R$, $S$ are the circumcenters of triangles $KIX$ and $KIY$. Hence, $RS$ is the perpendicular bisector of $IK$, so $H$ lies on $RS$. Applying Pascal's theorem for $\\begin{pmatrix} K & X & R \\\\ S & Y & K \\end{pmatrix}$, we obtain that the tangent at $K$ of $(KXY)$, $RS$, and $MN$ are concurrent, which means $KH$ is the tangent of $(KXY)$. By angle chasing, we have\n\n$$\n\\begin{aligned}\n\\angle HIK &= \\angle HKI = \\angle HKB + \\angle BKI \\\\\n&= \\angle KYX + 90^\\circ - \\angle BAC = 270^\\circ - 2\\angle ABC - \\angle BAC \\\\\n&= 90^\\circ + \\angle ACB - \\angle ABC\n\\end{aligned}\n$$\n\nAlso, $\\angle(AO, BC) = \\angle OAC + \\angle ACB = 90^\\circ - \\angle ABC + \\angle ACB = \\angle HIK$. Note that $IK \\perp BC$, thus $IH \\perp AO$. Hence, $PBQC$ is cyclic or $I$ has the same power to $(ABC)$ and $(APQ)$, which means $AI$ passes through the second intersection of $(APQ)$ and $(O)$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16314,
    "subject": "Mathematics (Olympiad)",
    "question": "給定正整數 $k$。設正整數數列 $a_0, a_1, \\dots, a_n$ ($n > 0$) 滿足：\n\n1. $a_0 = a_n = 1$；\n2. 對所有 $i = 1, 2, \\dots, n-1$，有 $2 \\le a_i \\le k$；\n3. 對每個 $j = 2, 3, \\dots, k$，$j$ 在 $a_0, a_1, \\dots, a_n$ 中出現 $\\varphi(j)$ 次（$\\varphi(j)$ 為不超過 $j$ 且與 $j$ 互質的正整數個數）；\n4. 對所有 $i = 1, 2, \\dots, n-1$，有 $\\gcd(a_{i-1}, a_i) = 1 = \\gcd(a_i, a_{i+1})$，且 $a_i$ 整除 $a_{i-1} + a_{i+1}$。\n\n另有整數數列 $b_0, b_1, \\dots, b_n$，使得對所有 $i = 0, 1, \\dots, n-1$，有 $\\frac{b_{i+1}}{a_{i+1}} > \\frac{b_i}{a_i}$。\n\n求 $b_n - b_0$ 的最小值。\n\nLet $k$ be a positive integer. A sequence $a_0, a_1, \\dots, a_n$ ($n > 0$) of positive integers satisfies:\n\n1. $a_0 = a_n = 1$;\n2. $2 \\le a_i \\le k$ for $i = 1, 2, \\dots, n-1$;\n3. For $j = 2, 3, \\dots, k$, $j$ appears $\\varphi(j)$ times in $a_0, a_1, \\dots, a_n$ ($\\varphi(j)$ is the number of positive integers not exceeding $j$ and coprime to $j$);\n4. For $i = 1, 2, \\dots, n-1$, $\\gcd(a_{i-1}, a_i) = 1 = \\gcd(a_i, a_{i+1})$, and $a_i$ divides $a_{i-1} + a_{i+1}$.\n\nThere is another sequence $b_0, b_1, \\dots, b_n$ of integers such that $\\frac{b_{i+1}}{a_{i+1}} > \\frac{b_i}{a_i}$ for all $i = 0, 1, \\dots, n-1$. Find the minimum value of $b_n - b_0$.",
    "options": [],
    "answer": "See solution",
    "solution": "$b_n - b_0$ 的最小值為 $1$。\n\n稱滿足題目條件的數列 $a_0, a_1, \\dots, a_n$ 為「$k$-好數列」。\n\n首先證明 $k$-好數列是唯一的。用歸納法加強命題：若 $(a, b) = 1$, $a + b \\ge k + 1$ 且 $1 \\le a, b \\le k$, 則存在唯一 $i$ 使 $a_i = a$, $a_{i+1} = b$。\n\n$k = 1$ 時，數列只能是 $1, 1$，唯一。\n\n假設 $k = t-1$ 時成立，考慮 $k = t$。若 $a_i = t$，則 $a_{i-1}, a_{i+1} < t$ 且 $t$ 不相鄰。由條件知 $a_i | a_{i-1} + a_{i+1}$，且 $a_{i-1} + a_{i+1} = t$。\n\n移除所有 $t$，剩下的新數列 $A_0, \\dots, A_N$ 是 $(t-1)$-好數列。插入 $t$ 的方法唯一，因為 $t$ 只能插在 $A_i, A_{i+1}$ 之間且 $A_i + A_{i+1} = t$，而這樣的 $A_i$ 只有 $\\varphi(t)$ 個。\n\n由歸納法知 $k$-好數列唯一。\n\n考慮所有分母不超過 $k$ 且在 $[0,1]$ 之間的最簡分數，分子依序為 $b_0, \\dots, b_n$，分母為 $c_0, \\dots, c_n$。可證 $c_0, \\dots, c_n$ 是 $k$-好數列，且 $c_i b_{i-1} - c_{i-1} b_i = 1$ 對所有 $1 \\le i \\le n$。\n\n由 $b_0 = 0, b_n = 1$，得 $b_n - b_0 = 1$。\n\n因此，$b_n - b_0$ 的最小值為 $1$。",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16315,
    "subject": "Mathematics (Olympiad)",
    "question": "Реши го системот равенки\n$$\n\\begin{cases}\nx + y = z \\\\\nx^2 + y^2 = z \\\\\nx^3 + y^3 = z\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Третата равенка на системот ја трансформираме во облик $x^3 + y^3 = (x + y)^3 - 3xy(x + y)$, односно $z = z^3 - 3xyz$. Од формулата за бином на квадрат имаме $xy = \\frac{1}{2}[(x + y)^2 - (x^2 + y^2)]$, па третата равенка добива облик $z = z^3 - \\frac{3}{2}z(z^2 - z)$. Со средување станува $z^3 - 3z^2 + 2z = 0 \\Leftrightarrow z(z - 1)(z - 2) = 0$, од каде решенија за $z$ се $z \\in \\{0, 1, 2\\}$.\n\nЗа секоја вредност на $z$ решаваме системот од две равенки:\n$$\n\\begin{cases}\nx + y = z \\\\\nx^2 + y^2 = z\n\\end{cases}\n$$\n\n- За $z = 0$:\n  $$\n  \\begin{cases}\nx + y = 0 \\\\\nx^2 + y^2 = 0\n\\end{cases}\n  $$\n  Решение: $x = y = 0$.\n\n- За $z = 1$:\n  $$\n  \\begin{cases}\nx + y = 1 \\\\\nx^2 + y^2 = 1\n\\end{cases}\n  $$\n  Решенија: $x = 0, y = 1$ и $x = 1, y = 0$.\n\n- За $z = 2$:\n  $$\n  \\begin{cases}\nx + y = 2 \\\\\nx^2 + y^2 = 2\n\\end{cases}\n  $$\n  Решение: $x = y = 1$.\n\nКонечно, решенија на почетниот систем се подредените тројки $(x, y, z) \\in \\{(0, 0, 0), (0, 1, 1), (1, 0, 1), (1, 1, 2)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16316,
    "subject": "Mathematics (Olympiad)",
    "question": "Alex has thought of a number $N$ in $S = \\{1, 2, \\ldots, 1001\\}$. Bibi must find $N$ by the following procedure: She gives Alex a list of subsets of $S$, and Alex tells her how many subsets in her list contain $N$. Bibi may repeat this with a second and third list, but no more than 3 lists are allowed.\n\nWhat is the least total number of subsets that would enable Bibi to find $N$ with certainty?",
    "options": [],
    "answer": "See solution",
    "solution": "The least number of subsets is $28$.\n\nSuppose Bibi has 3 lists, containing $a_1$, $a_2$, and $a_3$ subsets respectively. For each list $i = 1, 2, 3$, Alex announces the number $x_i$ of subsets in the list that contain $N$. The ordered triple $(x_1, x_2, x_3)$ is the only information Bibi obtains. To find $N$ with certainty, each triple must correspond to a unique $N \\in \\{1, 2, \\ldots, 1001\\}$.\n\nThere are $1001$ possible numbers $N$, so the number of different triples $(x_1, x_2, x_3)$ must be at least $1001$. This number equals $(a_1 + 1)(a_2 + 1)(a_3 + 1)$, since each $x_i$ can take $a_i + 1$ values ($0, 1, \\ldots, a_i$). Thus,\n\n$$\n(a_1 + 1)(a_2 + 1)(a_3 + 1) \\ge 1001.\n$$\n\nUsing the AM-GM inequality to estimate the total number $a_1 + a_2 + a_3$:\n\n$$\n1001 \\le (a_1 + 1)(a_2 + 1)(a_3 + 1) \\le \\left( \\frac{a_1 + a_2 + a_3}{3} + 1 \\right)^3.\n$$\n\nIf $a_1 + a_2 + a_3 \\le 27$, then $\\left(\\frac{27}{3} + 1\\right)^3 = 1000$, which is too small. Therefore, $a_1 + a_2 + a_3 \\ge 28$.\n\nNow, we show that $28$ subsets suffice. Since $1001 = 7 \\times 11 \\times 13$, consider a $7 \\times 11 \\times 13$ parallelepiped with the numbers $1, 2, \\ldots, 1001$ assigned to its unit cubes. Let the vertical dimension be $13$, so there are $13$ horizontal layers ($7 \\times 11$ each). For $i = 1, 2, \\ldots, 12$, let $S_i$ be the union of the first $i$ horizontal layers. List 1 consists of the $12$ sets $S_1, S_2, \\ldots, S_{12}$, and Alex announces $x_1$, the number of these sets containing $N$. Since $S_1 \\subset S_2 \\subset \\cdots \\subset S_{12}$, $x_1 = 0$ means $N$ is in layer $13$, $x_1 = 1$ means layer $12$, etc. In general, $x_1 = i$ implies $N$ is in layer $13 - i$.\n\nAnalogous lists in the other two directions, with $6$ sets and $10$ sets, can determine the layers in those directions. Thus, $N$ becomes known with $12 + 6 + 10 = 28$ sets, using 3 lists.\n\nIf Bibi uses 2 lists with $a_1$ and $a_2$ sets, then\n\n$$\n1001 \\le (a_1 + 1)(a_2 + 1) \\le \\left( \\frac{a_1 + a_2}{2} + 1 \\right)^2, \\text{ so } a_1 + a_2 \\ge 61.\n$$\n\nOne list alone would require at least $1000$ sets.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16317,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that $n 2^{n+1} + 1$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "The only values are $n = 0$ and $n = 3$.\n\nClearly, $n 2^{n+1} + 1$ is odd, so if this number is a perfect square, then\n$$n 2^{n+1} + 1 = (2x + 1)^2, \\quad x \\in \\mathbb{N}$$\nwhence\n$$n 2^{n-1} = x(x + 1)$$\n\nThe integers $x$ and $x + 1$ are coprime, so one of them must be divisible by $2^{n-1}$, which means that the other must be at most $n$. This shows that $2^{n-1} \\leq n + 1$.\n\nAn easy induction shows that the above inequality is false for all $n \\geq 4$, and a direct inspection confirms that the only convenient values in the case $n \\leq 3$ are $0$ and $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16318,
    "subject": "Mathematics (Olympiad)",
    "question": "Olesya wrote down a natural number $N$. After this, Andriy wrote down one sixth, one fifth, one fourth, one third, and a half of $N$. It turns out that the sum of all numbers that are written is an integer. What is the least possible number that Olesya could write?",
    "options": [],
    "answer": "See solution",
    "solution": "The sum is given by: $$\\left(\\frac{1}{6} + \\frac{1}{5} + \\frac{1}{4} + \\frac{1}{3} + \\frac{1}{2}\\right)N = \\frac{29N}{20}$$\nTherefore, $N$ must be divisible by $20$ for the sum to be an integer. The least possible value is $N = 20$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16319,
    "subject": "Mathematics (Olympiad)",
    "question": "If the product of four consecutive integers is equal to the value of one of those integers, what is the largest possible value for any of the integers?",
    "options": [],
    "answer": "See solution",
    "solution": "With a little trial and error, it becomes clear that this scenario is only possible if one of the integers is zero. The possible sets are $\\{-3, -2, -1, 0\\}$, $\\{-2, -1, 0, 1\\}$, $\\{-1, 0, 1, 2\\}$, and $\\{0, 1, 2, 3\\}$. The largest possible integer is thus $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16320,
    "subject": "Mathematics (Olympiad)",
    "question": "Во множеството на цели броеви да се реши равенката\n\n$$\n3^{2a+1}b^2 + 1 = 2^c.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Случај 1. $a \\ge 0$.\n\nЈасно е дека $c \\ge 0$, при што $c=0$ повлекува $b=0$. Добиваме дека $(a,0,0)$ за произволен ненегативен цел број $a$ е решение. Од равенството $3^{2a+1}b^2 + 1 = 2^c$ следува дека $b$ е непарен цел број. Левата страна можеме да ја запишеме во следниов облик:\n\n$$\n3^{2a+1}b^2 + 1 = (3^{2a+1}+1)b^2 - (b-1)(b+1).\n$$\n\nДа забележиме дека $(b-1)(b+1)$ е делив со 8, додека $(3^{2a+1}+1)b^2$ е делив со 4, но не со 8. Затоа $2^c = 4$, т.е. $c=2$. Но тогаш $3^{2a+1}b^2 = 3$, па $a=0$ и $b=\\pm1$.\n\nСлучај 2. $a < 0$.\n\nПовторно $c \\ge 0$, при што $c = 0$ повлекува $b=0$ и тогаш $a$ може да е произволен негативен цел број. Затоа да се ограничиме на случајот $c > 0$. Доволно е да го разгледаме случајот $b > 0$. Ставајќи $d = -a$, диофантовата равенка од условот на задачата добива облик:\n\n$$(2^c - 1)3^{2d-1} = b^2,$$\n\nпри што $b, c$ и $d$ се природни броеви. Значи $b$ е делив со 3, од што следува дека $c$ е парен број. Така $b=3^d x$, $c=2y$, за некои природни броеви $x$ и $y$. Диофантовата равенка се сведува до облик:\n\n$$\n4^{y-1} + 4^{y-2} + \\dots + 1 = x^2.\n$$\n\nОва повлекува $x=y=1$. Имено, за $y \\ge 2$ би добиле дека $x^2 \\equiv 5 \\pmod{8}$, што не е можно. Значи во овој случај единствените решенија се $(a, 3^{-a}, 2)$, каде $a$ е произволен негативен цел број.\n\nМножеството $M$ од сите решенија на диофантовата равенка од условот на задачата може да се опише на следниов начин:\n\n$$\nM = \\{(a,0,0) \\mid a \\in \\mathbb{Z}\\} \\cup \\{(a, \\pm 3^{-a}, 2) \\mid a \\in \\mathbb{Z} \\cup \\{0\\}\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16321,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f, g: \\mathbb{R} \\to \\mathbb{R}$ be continuous, non-constant functions satisfying\n$$\nf(x-y) = f(x)f(y) + g(x)g(y)\n$$\nfor all $x, y \\in \\mathbb{R}$.\n\n1. Show that for any $x, y \\in \\mathbb{R}$, we have $g(x+y) = f(x)g(y) + g(x)f(y)$.\n2. Find all pairs $f, g$ satisfying the conditions.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f$ and $g$ be functions satisfying the functional equation\n$$\nf(x-y) = f(x)f(y) + g(x)g(y)\n$$\nTaking $x = y$ gives\n$$\nf(0) = f(x)^2 + g(x)^2.\n$$\nTaking $y = 0$ gives\n$$\nf(x)(1 - f(0)) = g(x)g(0).\n$$\nCombining these, we get\n$$\nf(0)(1 - f(0))^2 = (f(x)^2 + g(x)^2)(1 - f(0))^2 = g(x)^2(g(0)^2 + (1 - f(0))^2).\n$$\nSince $g$ is non-constant, we have $f(0) = 1$ and $g(0) = 0$. Thus,\n$$\nf(x)^2 + g(x)^2 = 1\n$$\nfor all $x \\in \\mathbb{R}$.\n\nNow, we show that $f$ is even and $g$ is odd. Taking $(x, y) = (0, x)$ in the original equation gives $f(-x) = f(0)f(x) + g(0)g(x) = f(x)$. Thus, $f$ is even.\n\nTaking $y \\to -y$ in the original equation gives\n$$\nf(x + y) = f(x)f(-y) + g(x)g(-y) = f(x)f(y) + g(x)g(-y).\n$$\nIt follows that $g(x)g(-y) = g(-x)g(y)$. Since $f$ is non-constant, there exists $v$ such that $g(-v) = -g(v) \\neq 0$, so $g$ is odd.\n\n**(i)** From above, we have\n$$\nf(x + y) = f(x)f(y) - g(x)g(y).\n$$\nNow,\n$$\n\\begin{aligned}\nf(x) &= f(x + y - y) \\\\ &= f(x + y)f(y) + g(x + y)g(y) \\\\ &= [f(x)f(y) - g(x)g(y)]f(y) + g(x + y)g(y) \\\\ &= f(x)f(y)^2 - g(x)g(y)f(y) + g(x + y)g(y) \\\\ &= f(x)(1 - g(y)^2) - g(x)f(y)g(y) + g(x + y)g(y) \\\\ &= f(x) - g(y)(g(x + y) - f(x)g(y) - g(x)f(y)),\n\\end{aligned}\n$$\nso\n$$\ng(y)(g(x + y) - f(x)g(y) - g(x)f(y)) = 0.\n$$\nIf $g(y) \\neq 0$, this gives $g(x + y) = f(x)g(y) + g(x)f(y)$. If $g(y) = 0$, then $f(y)^2 = 1$, and $g(x + y) = 0$ as well, so the formula holds for all $x, y$.\n\n**(ii)** Let $h(x) = f(x) + i g(x)$. Then\n$$\nh(x + y) = h(x)h(y)\n$$\nand $|h(x)| = 1$ for all $x$. By the classification of continuous group homomorphisms from $\\mathbb{R}$ to the unit circle, $h(x) = \\exp(icx)$ for some $c \\in \\mathbb{R}$. Thus,\n$$\n\\begin{cases}\nf(x) = \\cos(cx) \\\\\ng(x) = \\sin(cx)\n\\end{cases}\n$$\nfor some $c \\in \\mathbb{R}$, and all such pairs are solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16322,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with $\\angle BAC \\neq 90^\\circ$. Let $O$ be the circumcentre of triangle $ABC$ and let $\\Gamma$ be the circumcircle of triangle $BOC$. Suppose that $\\Gamma$ intersects the line segment $AB$ at $P$ (different from $B$), and the line segment $AC$ at $Q$ (different from $C$). Let $ON$ be a diameter of the circle $\\Gamma$. Prove that the quadrilateral $APNQ$ is a parallelogram.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\angle BAC = \\alpha$. Then $\\angle BOC = 2\\alpha$. Since $ON$ is a diameter of $\\Gamma$, it is the perpendicular bisector of $BC$. Since $OB = OC$, this implies that $B$ is symmetric to $C$ across line $ON$. Hence $\\angle BON = \\angle CON = \\alpha$. Since $OQCN$ is cyclic, we have $\\angle NQC = \\angle NOC = \\alpha$. Thus $\\angle PAQ = \\angle NQC$. Hence $AP \\parallel QN$. Similarly, $AQ \\parallel PN$. Thus $APNQ$ is a parallelogram.\n\n![](images/Australian_Scene_2010_p127_data_3738a30ada.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16323,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the largest positive integer $n$ such that there exist $n$ positive integers $x_1 < x_2 < \\dots < x_n$ satisfying\n\n$$\nx_1 + x_1 x_2 + \\dots + x_1 x_2 \\dots x_n = 2021.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The largest $n$ is $4$.\n\nWhen $n = 4$, take $x_1 = 1$, $x_2 = 4$, $x_3 = 6$, $x_4 = 83$, and the equation holds.\n\n**Claim:** When $n \\ge 5$, the equation has no positive integer solutions.\n\n**Proof of claim:** Suppose on the contrary that $n \\ge 5$, $x_1 < x_2 < \\dots < x_n$ satisfy the equation. From $2021 > x_1 x_2 \\dots x_n \\ge x_1 x_2 x_3 x_4 x_5 > x_1^5$, it follows that $x_1 < 5$; also, $x_1$ divides $2021$. Hence, $x_1 = 1$.\n\nNow,\n$$\nx_2 + x_2 x_3 + \\dots + x_2 x_3 \\dots x_n = 2020.\n$$\nFrom $2020 > x_2 x_3 \\dots x_n \\ge x_2 x_3 x_4 x_5 > x_2^4$, we have $x_2 < 10$; also, $x_2$ divides $2020$. Hence, $x_2 = 2, 4$ or $5$.\n\n- If $x_2 = 2$, $x_3 + x_3 x_4 + \\dots + x_3 x_4 \\dots x_n = 1009$, $x_3 > 2$.\n  From $1009 > x_3 x_4 \\dots x_n \\ge x_3 x_4 x_5 > x_3^3$, we have $x_3 < 11$, yet $x_3$ divides $1009$ has no integer solutions.\n- If $x_2 = 5$, $x_3 + x_3 x_4 + \\dots + x_3 x_4 \\dots x_n = 403$, $x_3 > 5$.\n  From $403 > x_3 x_4 \\dots x_n \\ge x_3 x_4 x_5 > x_3^3$, we have $x_3 < 8$, yet $x_3$ divides $403$ has no integer solutions.\n- If $x_2 = 4$, $x_3 + x_3 x_4 + \\dots + x_3 x_4 \\dots x_n = 504$, $x_3 > 4$.\n  From $504 > x_3 x_4 \\cdots x_n \\ge x_3 x_4 x_5 \\ge x_3 (x_3+1) (x_3+2)$, we have $x_3 < 7$; also, $x_3$ divides $504$. Thus, $x_3 = 6$, $x_4 + x_4 x_5 + \\cdots + x_4 x_5 \\cdots x_n = 83$. However, $83$ is a prime number, $x_4 < x_5$, indicating that $x_4 = 1$, a contradiction.\n\nThe claim is verified, and the largest $n$ is $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16324,
    "subject": "Mathematics (Olympiad)",
    "question": "Calculate the value of $2 - 0 + 1 \\times 6$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n2 - 0 + 1 \\times 6 = 2 - 0 + 6 = 8\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16325,
    "subject": "Mathematics (Olympiad)",
    "question": "For any positive integers $n$ and $k$, let $L(n, k)$ be the least common multiple of the $k$ consecutive integers $n, n+1, \\ldots, n+k-1$. Show that for any integer $b$, there exist integers $n$ and $k$ such that $L(n, k) > b L(n + 1, k)$.\n\nSoit $L(n, k)$ le plus petit commun multiple de la suite des $k$ entiers consécutifs $n, n + 1, \\ldots, n + k - 1$, où $n$ et $k$ sont deux entiers positifs quelconques. Montrez que pour tout entier $b$, il existe des nombres entiers $n$ et $k$ tels que $L(n, k) > b L(n + 1, k)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p > b$ be a prime, and set $n = p^3$ and $k = p^2$. For $p^3 < i < p^3 + p^2$, no power of $p$ greater than $1$ divides $i$, while $p$ divides $p^3 + p$. Thus, $L(p^3, p^2) = p^2 L(p^3 + 1, p^2 - 1)$. Similarly, $L(p^3 + 1, p^2) = p L(p^3 + 1, p^2 - 1)$. Therefore, $L(p^3, p^2) = p L(p^3 + 1, p^2) > b L(p^3 + 1, p^2)$.\n\nAlternatively, let $m > 1$. Then $L(m! - 1, m + 1)$ is the least common multiple of the integers from $m! - 1$ to $m! + m - 1$. Since $m! - 1$ is relatively prime to all of $m!, m! + 1, \\ldots, m! + m - 1$, we have $L(m! - 1, m + 1) = (m! - 1) M$, where $M = \\text{lcm}(m!, m! + 1, \\ldots, m! + m - 1)$.\n\nNow, $L(m!, m + 1) = \\text{lcm}(M, m! + m)$. Since $m! + m = m((m - 1)! + 1)$ and $m$ divides $M$, we have $\\text{lcm}(M, m! + m) \\leq M((m - 1)! + 1)$. Thus,\n\n$$\n\\frac{L(m! - 1, m + 1)}{L(m!, m + 1)} \\geq \\frac{m! - 1}{(m - 1)! + 1}.\n$$\n\nAs $m$ can be arbitrarily large, so can $L(m! - 1, m + 1)/L(m!, m + 1)$. Therefore, taking $n = m! - 1$ for sufficiently large $m$, and $k = m + 1$, works.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16326,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_n = (2^n + 3^n)^{1/n}$ for all $n \\ge 1$.\n\nShow that the sequence $(a_n)$ is decreasing.",
    "options": [],
    "answer": "See solution",
    "solution": "For all $n \\ge 1$:\n\n$$\n2^n + 3^n > 3^n \\implies (2^n + 3^n)^{n+1} > 3^n (2^n + 3^n)^n > (2^{n+1} + 3^{n+1})^n\n$$\n\nTherefore,\n\n$$\n(2^n + 3^n)^{1/n} > (2^{n+1} + 3^{n+1})^{1/(n+1)}\n$$\n\nThus, $a_n > a_{n+1}$ for all $n \\ge 1$, so $(a_n)$ is decreasing.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16327,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle and $D$, $E$ be the feet of the perpendiculars from $B$ and $C$ to the opposite sides, respectively. Let $S$ and $T$ be the symmetric points of $E$ with respect to the sides $AC$ and $BC$, respectively. Suppose that the circumcircle $O$ of the triangle $CST$ meets the line $AC$ at $X \\neq C$. Show that the two lines $XO$ and $DE$ are perpendicular.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show the following lemma.\n\n**Lemma:** Let $O$ be the circumcenter of triangle $ABC$ and $D$, $E$ be points on the sides $AB$, $AC$. If four points $D$, $E$, $C$, $B$ are concyclic, then $AO$ is perpendicular to $DE$.\n\n**Proof of the Lemma:** Since $AO = OB$, we have:\n\n$$\n\\angle DAO = \\angle BAO = 90^\\circ - \\frac{1}{2} \\angle AOB = 90^\\circ - \\angle ACB = 90^\\circ - \\angle ADE\n$$\n\nSo $\\angle DAO + \\angle ADE = 90^\\circ$. Thus, $AO$ is perpendicular to $DE$. $\\square$\n\nLet $M$ be the intersection point of $ET$ and $BC$, and $N$ be the intersection point of $ES$ and $AC$. Then:\n\n$$\n\\angle ECN = \\angle EMN = \\angle ETS = \\angle NCS = \\angle XTS\n$$\n\nand\n\n$$\n\\angle EDB = \\angle NED = \\angle DSE = \\angle ECB = \\angle ENM = \\angle EST\n$$\n\nSo three points $S$, $D$, $T$ are collinear. We have:\n\n$$\n\\angle ETD = \\angle XCS = \\angle DCE\n$$\n\nSo four points $T$, $C$, $D$, $E$ are concyclic, and thus by the lemma we have $XO \\perp DE$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16328,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\omega_1$ and $\\omega_2$ be two circles with centers $O_1$ and $O_2$, respectively, with $O_2$ lying on $\\omega_1$. Let $A$ be a common point of $\\omega_1$ and $\\omega_2$. A line through $A$ intersects $\\omega_1$ in $B \\neq A$ and $\\omega_2$ in $C \\neq A$ such that $A$ lies between $B$ and $C$. The ray $O_2O_1$ intersects $\\omega_2$ in $D$ and contains a point $E$ such that $\\angle EAD = \\angle DCO_2$ and $D$ lies between $O_2$ and $E$. Show that $BO_2$ bisects $CE$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $F$ be the second intersection of $\\omega_1$ and $\\omega_2$. Notice that\n\n$$\n\\angle BFO_2 = 180^\\circ - \\angle BAO_2 = \\angle CAO_2 = \\angle O_2CA\n$$\n\nand since $AO_2 = EO_2$, $\\angle FBO_2 = \\angle O_2BA$. It follows that $F$ is the reflection of $C$ over $BO_2$, thus it suffices to prove that $EF$ is parallel to $BO_2$, as then $BO_2$ is a midline of triangle $CEF$.\n\n![](images/BW2019-problems-solutions-opinions_p36_data_f82a56a89e.png)\n\nNotice that $\\angle O_2DC = \\angle DCO_2 = \\angle EAD$. Hence by symmetry about $O_1O_2$, we have\n\n$$\n\\begin{aligned}\n\\angle FEO_2 &= \\angle AED = 180^\\circ - \\angle EAD - \\angle ADE \\\\\n&= \\angle ADO_2 - \\angle CDO_2 = \\angle ADC = \\frac{1}{2} \\angle AO_2C\n\\end{aligned}\n$$\n\nMoreover,\n\n$$\n\\begin{aligned}\n90^\\circ - \\angle O_1O_2B &= \\frac{1}{2} \\angle BO_1O_2 = 180^\\circ - \\angle BAO_2 \\\\\n&= \\angle CAO_2 = 90^\\circ - \\frac{1}{2} \\angle AO_2C\n\\end{aligned}\n$$\n\nso we conclude that $\\angle FEO_2 = \\angle EO_2B$, which proves that $EF$ and $BO_2$ are parallel.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16329,
    "subject": "Mathematics (Olympiad)",
    "question": "30 children—boys and girls—formed a circle. It occurred that there is no child such that both its neighbors are boys. What is the least possible number of girls there?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $16$.\n\nConsider a group of consecutive boys. There can be at most $2$ boys in such a group, otherwise, a boy would have both neighbors as boys. Similarly, consider groups of consecutive girls. These groups alternate with the groups of boys, so the number of boy groups equals the number of girl groups.\n\nSuppose there are $15$ boys and $15$ girls. Then, there must be at least $8$ groups of boys (since two boys cannot be together more than once in a group), so at least $8$ groups of girls, and thus at least $16$ girls in total. Therefore, there can be at most $14$ boys and at least $16$ girls.\n\nAn arrangement with $14$ boys and $16$ girls is possible. For example, consider $7$ groups each consisting of $2$ boys, alternating with $6$ groups of $2$ girls and one group of $4$ girls.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16330,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f, g : \\mathbb{Q} \\to \\mathbb{Q}$ such that\n\n$$\nf(g(x) - g(y)) = f(g(x)) - y\n$$\n\nand\n\n$$\ng(f(x) - f(y)) = g(f(x)) - y\n$$\n\nfor all $x, y \\in \\mathbb{Q}$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove that both functions are one-to-one. Let $x_1, x_2 \\in \\mathbb{Q}$ with $g(x_1) = g(x_2)$. Then:\n\n$$\nf(g(x_1)) = f(g(x_2))\n$$\n\nSubstituting $x = x_1$, $y = x_2$ into (1):\n\n$$\nf(g(x_1) - g(x_2)) = f(g(x_1)) - x_2 \\implies f(0) = f(g(x_1)) - x_2\n$$\n\nSimilarly, with $x = x_2$, $y = x_1$:\n\n$$\nf(g(x_2) - g(x_1)) = f(g(x_2)) - x_1 \\implies f(0) = f(g(x_2)) - x_1\n$$\n\nFrom these, $x_1 = x_2$, so $g$ is one-to-one. The same argument shows $f$ is one-to-one.\n\nNow, set $x = y = 0$ in (1) and (2):\n\n$$\n\\begin{cases}\nf(g(0) - g(0)) = f(g(0)) - 0 \\\\\ng(f(0) - f(0)) = g(f(0)) - 0\n\\end{cases} \\implies\n\\begin{cases}\nf(0) = f(g(0)) \\\\\ng(0) = g(f(0))\n\\end{cases}\n$$\n\nSince $f$ and $g$ are one-to-one, $f(0) = g(0) = 0$.\n\nNow, substitute $y = x$ into (1) and (2), using $f(0) = g(0) = 0$:\n\n$$\n\\begin{cases}\nf(g(x) - g(x)) = f(g(x)) - x \\\\\ng(f(x) - f(x)) = g(f(x)) - x\n\\end{cases} \\implies\n\\begin{cases}\nf(0) = f(g(x)) - x \\\\\ng(0) = g(f(x)) - x\n\\end{cases} \\implies f(g(x)) = g(f(x)) = x\n$$\n\nSubstitute this into (1) and (2):\n\n$$\n\\begin{cases}\nf(g(x) - g(y)) = x - y \\quad (A) \\\\\ng(f(x) - f(y)) = x - y \\quad (B)\n\\end{cases}\n$$\n\nIn (A), set $x = 0$, $y = f(x)$; in (B), set $x = 0$, $y = g(x)$:\n\n$$\n\\begin{cases}\nf(g(0) - g(f(x))) = 0 - f(x) \\\\\ng(f(0) - f(g(x))) = 0 - g(x)\n\\end{cases} \\implies\n\\begin{cases}\nf(-x) = -f(x) \\\\\ng(-x) = -g(x)\n\\end{cases}\n$$\n\nNow, in (A), set $x = f(x)$, $y = -f(y)$; in (B), set $x = g(x)$, $y = -g(y)$:\n\n$$\n\\begin{cases}\nf(x + y) = f(x) + f(y) \\\\\ng(x + y) = g(x) + g(y)\n\\end{cases}\n$$\n\nThus, $f$ and $g$ are additive and odd. Therefore, $f(x) = c_1 x$ and $g(x) = c_2 x$ for some $c_1, c_2 \\in \\mathbb{Q}$.\n\nFrom $f(g(x)) = g(f(x)) = x$, we get $c_1 c_2 x = x$ for all $x$, so $c_1 c_2 = 1$.\n\nTherefore, the solutions are:\n\n$$\nf(x) = c x, \\quad g(x) = \\frac{1}{c} x, \\quad c \\in \\mathbb{Q} \\setminus \\{0\\}\n$$\n\nIt is easy to verify that these functions satisfy the original equations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16331,
    "subject": "Mathematics (Olympiad)",
    "question": "Let there be $x$ pairs of red amoebas, $y$ pairs of blue amoebas, and $z$ pairs consisting of one amoeba of each color. Given:\n- $2x + z = 95$\n- $2y + z = 19$\n\n(a) What is the maximum possible number of amoebas in the next generation?\n\n(b) If the total number of amoebas is $100$, how many blue amoebas are there?",
    "options": [],
    "answer": "See solution",
    "solution": "(a) From the equations:\n$$2x + z = 95 \\implies x = \\frac{95 - z}{2}$$\n$$2y + z = 19 \\implies y = \\frac{19 - z}{2}$$\nThe total number of amoebas in the next generation is:\n$$x + 4y + 3z = \\frac{95 - z}{2} + 4 \\cdot \\frac{19 - z}{2} + 3z = \\frac{171 + z}{2}$$\nTo maximize the number, maximize $z$. The largest possible $z$ is $19$, so the maximum is:\n$$\\frac{171 + 19}{2} = 95$$\n\n(b) Let $b$ be the number of blue amoebas and $r$ the number of red amoebas. The quantity $2b + r$ is invariant. Given $b + r = 100$ and $2b + r = 133$:\n$$2b + r = 133$$\n$$b + r = 100$$\nSubtracting:\n$$2b + r - (b + r) = 133 - 100$$\n$$b = 33$$\nSo there were $33$ blue amoebas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16332,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $ABCDE$ 為內接於圓 $Ω$ 的五邊形。一條與 $BC$ 邊平行的直線分別與直線 $AB$, $AC$ 交於點 $S, T$。設 $X$ 為直線 $BE$ 及 $DS$ 的交點，$Y$ 為直線 $CE$ 與 $DT$ 的交點。試證：若直線 $AD$ 與三角形 $DXY$ 的外接圓相切，則直線 $AE$ 與三角形 $EXY$ 的外接圓相切。\n\n![](images/2022-TWNIMO-Problems_p51_data_9adc04bf45.png)",
    "options": [],
    "answer": "See solution",
    "solution": "事實上，$ABCDE$ 在圓 $Ω$ 上的順序不影響結果。設 $U, V$ 為 $⊙(ABC)$ 與 $DS, DT$ 的第二交點，$R$ 為 $ST$ 上使 $\\angle BAD = \\angle RAC$ 的點。由於 $\\triangle ABD \\sim \\triangle ART$，有 $\\angle ART = \\angle ABD = \\angle AVD = \\angle AVT$，因此 $A, R, T, V$ 共圓。由此可得 $B, R, V$ 共線，因為 $\\angle BVD = \\angle BAD = \\angle RAC = \\angle RVT$。同理，$C, R, U$ 共線。對 $BECUDV$ 應用 Pascal 定理，得 $X = BZ \\cap UD$，$Y = EC \\cap DR$，$R = CU \\cap BV$ 共線。\n\n由假設 $AD$ 與 $⊙(DXY)$ 相切，得\n\n$$\n\\angle SXR = \\angle ADV = \\angle ABV = \\angle SBR,\n$$\n\n因此 $B, S, R, X$ 共圓。於是\n\n$$\n\\angle AEY = \\angle AEC = \\angle ABC = \\angle BSR = \\angle EXY,\n$$\n\n因此 $AE$ 與 $\\odot(EXY)$ 相切，得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16333,
    "subject": "Mathematics (Olympiad)",
    "question": "The nonnegative integers 2000, 17, and $n$ are written on a blackboard. Alice and Bob play the following game: Alice begins, then they play in turns. A move consists in replacing one of the three numbers by the absolute difference of the other two. No moves are allowed where all three numbers remain unchanged. A player in turn who cannot make a legal move loses the game.\n\n*Prove that the game will end for every number $n$.*\n\n*Who wins the game in the case $n = 2017$?*",
    "options": [],
    "answer": "See solution",
    "solution": "If three numbers are written on the blackboard and one of them is replaced by the (positive) difference of the other two, then after this move one number on the blackboard will be the sum of the other two. Let $a$, $b$, and $a+b$ be the numbers on the blackboard; without loss of generality, assume $b > a$. Because $a+b-b=a$ and $a+b-a=b$, there is only one possible move. After it, the numbers $a$, $b$, and $b-a$ are written on the blackboard. Again, one number (namely $b$) is the sum of the other two and there exists only one possible move.\n\nThis means that from the second turn on, there is no choice of moves and all moves are inevitable. Furthermore, from the second move on, the largest of the three numbers is decreased, and since no number can become negative, after a finite number of moves one of the numbers will be $0$. Since $0$ is the difference of the other two numbers, we must have $0$, $a$, $a$ on the blackboard. Now $a-0=a$ and $a-a=0$, therefore no further move is possible. Thus, the player writing $0$, $a$, $a$ onto the blackboard is the winner.\n\nIf the game starts with the numbers $2000$, $17$, and $2017$ on the blackboard, the course of the game is as follows:\n\n1st move (A): $2000$, $17$, $1983$\n\n2nd move (B): $1966$, $17$, $1983$\n\n3rd move (A): $1966$, $17$, $1949$\n\netc. (since $2000 \\div 17 = 117.6\\ldots$)\n\n117th move (A): $2000 - 116 \\cdot 17 = 28$, $17$, $2000 - 117 \\cdot 17 = 11$\n\n118th move (B): $6$, $17$, $11$\n\n119th move (A): $6$, $5$, $11$\n\n120th move (B): $6$, $5$, $1$\n\n121st move (A): $4$, $5$, $1$\n\n122nd move (B): $4$, $3$, $1$\n\n123rd move (A): $2$, $3$, $1$\n\n124th move (B): $2$, $1$, $1$\n\n125th move (A): $0$, $1$, $1$\n\nand Alice wins the game.\n\n![](images/gwf2017englishSolutions_p1_data_0bd9d3109e.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16334,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{N}^* = \\{1, 2, 3, \\ldots\\}$ be the set of positive integers. Find all functions $f: \\mathbb{N}^* \\to \\mathbb{N}^*$ such that\n\n$$\nf(f(m)^2 + 2f(n)^2) = m^2 + 2n^2, \\quad \\text{for all } m, n \\in \\mathbb{N}^*.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove that $f$ is injective. For any fixed $n$, if $f(m_1) = f(m_2)$, then:\n\n$$\nm_1^2 + 2n^2 = f(f(m_1)^2 + 2f(n)^2) = f(f(m_2)^2 + 2f(n)^2) = m_2^2 + 2n^2,\n$$\nso $m_1^2 = m_2^2$ and thus $m_1 = m_2$.\n\nSince $f$ is injective, we have:\n\n$$\nf(m)^2 + 2f(n)^2 = f(p)^2 + 2f(q)^2 \\iff m^2 + 2n^2 = p^2 + 2q^2. \\tag{1}\n$$\n\nLet $f(1) = a$. For $m = n = 1$, $f(3a^2) = 3$. Then from (1):\n\n$$\nf(5a^2)^2 + 2f(a^2)^2 = f(3a^2)^2 + 2f(3a^2)^2 = 3f(3a^2)^2 = 27.\n$$\n\nThe solutions in positive integers to $x^2 + 2y^2 = 27$ are $(x, y) = (3, 3)$ and $(x, y) = (5, 1)$, so $f(a^2) = 1$ and $f(5a^2) = 5$.\n\nAlso, from (1):\n\n$$\n2f(4a^2)^2 - 2f(2a^2)^2 = f(5a^2)^2 - f(a^2)^2 = 24.\n$$\n\nThe unique solution in positive integers to $x^2 - y^2 = 12$ is $(x, y) = (4, 2)$, so $f(2a^2) = 2$ and $f(4a^2) = 4$.\n\nUsing (1) again, we have:\n\n$$\nf((k+4)a^2)^2 = 2f((k+3)a^2)^2 - 2f((k+1)a^2)^2 + f(ka^2)^2,\n$$\n\nwhich follows from the identity $(k+4)^2 + 2(k+1)^2 = k^2 + 2(k+3)^2$. Therefore, by induction, $f(ka^2) = k$. Since $f(a^2) = a = f(1)$, we have $a = 1$. Thus, $f(k) = k$ for all $k \\in \\mathbb{N}^*$. It is easy to verify that $f(k) = k$ is a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16335,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a strictly positive integer. For any natural number $k$, let $a(k)$ be the number of natural divisors $d$ of $k$ such that $k \\leq d^2 \\leq n^2$. Compute the sum $$\\sum_{k=1}^{n^2} a(k).$$",
    "options": [],
    "answer": "See solution",
    "solution": "We will compute the sum by double-counting the elements of the set\n\n$$\nS(n) = \\{(k, d) \\mid d \\text{ divides } k,\\ k \\leq d^2 \\leq n^2,\\ 1 \\leq k \\leq n^2\\}.\n$$\n\nFor $1 \\leq k \\leq n^2$, let $A(k)$ be the set of natural divisors $d$ of $k$ such that $k \\leq d^2 \\leq n^2$.\n\nA natural number $d \\in \\{1, 2, \\dots, n\\}$ belongs to the sets $A(d), A(2d), \\dots, A(d^2)$ and only to them.\n\nIt follows that the contribution of each $d$ in the sum $\\sum_{k=1}^{n^2} a(k) = \\sum_{k=1}^{n^2} |A(k)|$ is $d$.\n\nThus,\n$$\n\\sum_{k=1}^{n^2} a(k) = \\sum_{d=1}^{n} d = \\frac{n(n+1)}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16336,
    "subject": "Mathematics (Olympiad)",
    "question": "$a, b \\in \\mathbb{R}^+$ бол $a \\ast b = \\sqrt{\\frac{a^2 + ab + b^2}{2012}}$ гэж өгөгдсөн. Дараах нөхцөлүүдийг харгалзан үз:\n\n$$\nf(a \\ast b) = \\frac{f(a) + f(b)}{2}, \\quad \\forall x_1, x_2, x_3, x_4 \\in \\mathbb{R}^+:\n$$\n\n$$\nf((x_1 \\ast x_2) \\ast (x_3 \\ast x_4)) = \\frac{f(x_1 \\ast x_2) + f(x_3 \\ast x_4)}{2} = \\frac{f(x_1) + f(x_2) + f(x_3) + f(x_4)}{2} = 4\n$$\n\nМөн\n\n$$\nf((x_1 \\ast x_2) \\ast (x_3 \\ast x_4)) = f((x_1 \\ast x_4) \\ast (x_2 \\ast x_3)) \\quad (1)\n$$\n\n$g(x) = (x^2 \\ast x) \\ast (x \\ast 1)$, $h(x) = (x^2 \\ast 1) \\ast (x \\ast x)$, $x \\in \\mathbb{R}^+$ функцүүдийг авч үзье. $\\forall t > 0$ үед $(ta) \\ast (t \\ast b) = t \\ast (a \\ast b)$.\n\nИймд $t \\cdot g(x) = (tx^2 \\ast tx) \\ast (tx \\ast t)$, $t \\cdot h(x) = (tx^2 \\ast t) \\ast (tx \\ast tx)$.\n\n(1)-д $(x_1, x_2, x_3, x_4) \\mapsto (tx^2, tx, tx, t)$ гэвэл\n\n$$\nf(tg(x)) = f(th(x)), \\forall t, x > 0 \\quad (2)\n$$\n\n$g(1) = h(1) = (1 \\ast 1) \\ast (1 \\ast 1) = \\sqrt{\\frac{3}{2012}} \\cdot \\sqrt{\\frac{3}{2012}} = \\frac{3}{2012}$\n\n$$\ng(2) = (4 \\ast 2) \\ast (2 \\ast 2) = \\sqrt{\\frac{28}{2012}} \\cdot \\sqrt{\\frac{7}{2012}} = \\frac{7}{2012} \\\\ h(2) = (4 \\ast 1) \\ast (2 \\ast 2) = \\sqrt{\\frac{21}{2012}} \\cdot \\sqrt{\\frac{12}{2012}} = \\sqrt{\\frac{33+6\\sqrt{7}}{2012}}\n$$\n\n$f$ функцийн шинж чанарыг ол.",
    "options": [],
    "answer": "See solution",
    "solution": "Эндээс $g(2) > h(2)$ байна. $x > 1$ үед $\\frac{g(x)}{h(x)}$ тасралтгүй функц бөгөөд $\\frac{g(1)}{h(1)} = 1$, $\\lambda = \\frac{g(2)}{h(2)} > 1$ гэж авъя. Тэгвэл $\\frac{g(x)}{h(x)}$ функцийн хувьд:\n\n$\\forall S \\in (1, \\lambda) \\Rightarrow \\exists x_S \\in (1, 2)$ ба $S = \\frac{g(x_S)}{h(x_S)}$ (Завсрын утгын тухай Больцано-Кошийн теорем).\n\nИймд $a > 0$ бол $\\forall b \\in (a, \\lambda a)$ хувьд $\\frac{b}{a} \\in (1, \\lambda)$ ба $\\frac{b}{a} = \\frac{g(z)}{h(z)}$ байх $z \\in (1, 2)$ олдоно. (2)-д\n\n$$\nt = \\frac{a}{h(z)}, \\quad x = z \\text{ гэж орлуулбал} \\\\ f(b) = f(a \\cdot \\frac{b}{a}) = f\\left(a \\cdot \\frac{g(z)}{h(z)}\\right) = f\\left(\\frac{a}{h(z)} \\cdot g(z)\\right) = f\\left(\\frac{a}{h(z)} \\cdot h(z)\\right) = f(a)\n$$\n\nТэгэхээр $f$ нь $(a, \\lambda a)$ интервал дээр тогтмол функц болно. $a \\le b$ байх дурын $a, b$ тоонууд ба хангалттай их $n \\in \\mathbb{N}$ үед $\\mu = \\sqrt{\\frac{b}{a}} \\le \\lambda$ байх ба $f$ функц $(a, \\lambda a), (\\lambda a, \\lambda^2 a), \\dots, (\\lambda^{n-1} a, \\lambda^n a)$ завсар бүрд тогтмол гэлгээс $f(a) = f(\\mu a) = \\dots = f(\\mu^n a) = f(b)$ болж $f$ тогтмол функц болно.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16337,
    "subject": "Mathematics (Olympiad)",
    "question": "For real numbers $a$ and $b$ such that $|a| \\neq |b|$ and $a \\neq 0$, we have\n\n$$\n\\frac{a-b}{a^2+ab} + \\frac{a+b}{a^2-ab} = \\frac{3a-b}{a^2-b^2}.\n$$\n\nDetermine the value of the expression $\\frac{b}{a}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Multiplying the equation by $a(a+b)(a-b)$, we get\n\n$$\n(a-b)^2 + (a+b)^2 = a(3a-b).\n$$\n\nExpanding and moving all terms to one side:\n\n$$\n0 = a^2 - ab - 2b^2 = (a - 2b)(a + b).\n$$\n\nSince $a \\neq -b$, we have $a - 2b = 0$ or $a = 2b$. Since $a \\neq 0$, we get $\\frac{b}{a} = \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16338,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the number of ordered pairs $ (x, y) $, where both $ x $ and $ y $ are integers between $ -100 $ and $ 100 $, inclusive, such that $$ 12x^2 - xy - 6y^2 = 0. $$",
    "options": [],
    "answer": "See solution",
    "solution": "The given equation can be written as $ (3x + 2y)(4x - 3y) = 0 $. Thus, the graph of this equation consists of two lines intersecting at the origin.\n\nLattice points (points with integer coordinates) on this graph that make the first factor zero are of the form $ (2k, -3k) $ for some integer $ k $. For the coordinates to be in the required range, $ -33 \\leq k \\leq 33 $, giving $ 67 $ lattice points.\n\nLattice points that make the second factor zero are of the form $ (3\\ell, 4\\ell) $ for some integer $ \\ell $. For the coordinates to be in the required range, $ -25 \\leq \\ell \\leq 25 $, giving $ 51 $ lattice points.\n\nThe lattice point $ (0, 0) $ occurs in both cases. The requested number of points is therefore $ 67 + 51 - 1 = 117 $.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16339,
    "subject": "Mathematics (Olympiad)",
    "question": "Let real numbers $a$, $b$, $c$, and $d$ satisfy\n$$\nf(x) = a \\cos x + b \\cos 2x + c \\cos 3x + d \\cos 4x \\le 1\n$$\nfor any real number $x$. Find the values of $a$, $b$, $c$, and $d$ such that $a + b - c + d$ is maximized.",
    "options": [],
    "answer": "See solution",
    "solution": "Since\n$$\n\\begin{aligned}\nf(0) &= a + b + c + d, \\\\\nf(\\pi) &= -a + b - c + d, \\\\\nf\\left(\\frac{\\pi}{3}\\right) &= \\frac{a}{2} - \\frac{b}{2} - c - \\frac{d}{2},\n\\end{aligned}\n$$\nthen\n$$\na + b - c + d = f(0) + \\frac{2}{3} f(\\pi) + \\frac{4}{3} f\\left(\\frac{\\pi}{3}\\right) \\le 3\n$$\nEquality holds if $f(0) = f(\\pi) = f\\left(\\frac{\\pi}{3}\\right) = 1$, that is, if $a = 1$, $b + d = 1$, and $c = -1$.\n\nLet $t = \\cos x$, $-1 \\le t \\le 1$. Then\n$$\n\\begin{aligned}\nf(x) - 1 &= \\cos x + b \\cos 2x - \\cos 3x + d \\cos 4x - 1 \\\\\n&= t + (1-d)(2t^2 - 1) - (4t^3 - 3t) + d(8t^4 - 8t^2 + 1) - 1 \\\\\n&= 2(1 - t^2)[-4d t^2 + 2t + (d-1)] \\le 0, \\quad \\forall t \\in [-1, 1],\n\\end{aligned}\n$$\nthat is,\n$$\n4d t^2 - 2t + (1-d) \\ge 0, \\quad \\forall t \\in (-1, 1).\n$$\nTaking $t = \\frac{1}{2} + \\epsilon$, $|\\epsilon| < \\frac{1}{2}$, then $\\epsilon[(2d - 1) + 4d\\epsilon] \\ge 0$, $|\\epsilon| < \\frac{1}{2}$. So $d = \\frac{1}{2}$.\n\nIf $d = \\frac{1}{2}$, then\n$$\n4d t^2 - 2t + (1-d) = 2t^2 - 2t + \\frac{1}{2} = 2\\left(t - \\frac{1}{2}\\right)^2 \\ge 0.\n$$\n\nSo, the maximal value of $a + b - c + d$ is $3$, and $(a, b, c, d) = \\left(1, \\frac{1}{2}, -1, \\frac{1}{2}\\right)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16340,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to write positive integers in the cells of a $2022 \\times 2022$ board so that the sum of the numbers in any rectangle $R$ is a perfect square if and only if $R$ is a square?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, it is possible.\n\nNumber the columns from left to right and the rows from top to bottom with $1, 2, \\dots, 2022$. In cell $(r, c)$ (row $r$, column $c$), write $x^{2(r+c)}$, where $x$ is a positive integer to be chosen later.\n\nConsider a rectangle $R$ covering rows $s$ to $t$ and columns $u$ to $v$. The sum in $R$ is:\n\n$$\n\\left( x^{2s} + x^{2s+2} + \\dots + x^{2t} \\right) \\cdot \\left( x^{2u} + x^{2u+2} + \\dots + x^{2v} \\right) = x^{2s+2u} \\cdot \\frac{x^{2a} - 1}{x^2 - 1} \\cdot \\frac{x^{2b} - 1}{x^2 - 1},\n$$\n\nwhere $a = t - s + 1$ and $b = v - u + 1$. If $a = b$, the sum is always a perfect square for any $x$.\n\nNow, we need to choose $x$ so that for all $a \\neq b$ (with $a, b \\in \\{2, 3, \\dots, 2021\\}$), $(x^{2a} - 1)(x^{2b} - 1)$ is not a perfect square.\n\n**Lemma.** Let $P(x)$ be a monic polynomial of degree $2n$ with integer coefficients. If $P(x)$ is a perfect square for infinitely many positive integers $x$, then $P(x)$ is the square of a polynomial with rational coefficients.\n\n*Proof.*\nLet $P(x) = x^{2n} + a_{2n-1}x^{2n-1} + \\dots + a_1x + a_0$. For large $x$, $P(x) > 0$. Construct $Q(x) = x^n + b_{n-1}x^{n-1} + \\dots + b_1x + b_0$ (with rational coefficients) so that $P(x) - Q(x)^2$ has degree less than $n$. This is done by matching coefficients:\n\n$$\n\\sum_{k=n}^{2n-1} \\sum_{i+j=k} b_i b_j x^k = a_k x^k.\n$$\n\nSo $P(x) = Q(x)^2 + R(x)$, $\\deg R < n$. Multiply both sides by $M^2$ (for integer coefficients): $P_1(x) = M^2P(x)$, $H(x) = MQ(x)$, $S(x) = M^2R(x)$, so $P_1(x) = H(x)^2 + S(x)$, $\\deg S < n$.\n\nIf $S(x) \\not\\equiv 0$, then for large $x$, $|S(x)| > x^n$, but $S(x)$ has degree $< n$, so this is only possible for finitely many $x$. Contradiction.\n\n*Lemma proved.*\n\nThus, for any $a \\neq b$, $(x^{2a} - 1)(x^{2b} - 1)$ is a perfect square for only finitely many $x$. Since there are finitely many $(a, b)$ pairs, there exists $x_0$ such that for all $a \\neq b$, $(x_0^{2a} - 1)(x_0^{2b} - 1)$ is not a perfect square. Set $x = x_0$; the construction works.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16341,
    "subject": "Mathematics (Olympiad)",
    "question": "Petryk programmed a calculator so that if there is a number $x$ on the screen and the button «=>» is pressed, the number $$\\frac{x-1}{x+1}$$ appears on the screen. Petryk pressed the button «=>» 2014 times, and then 2016 appeared on the screen. What number was on the screen at the beginning? The screen can show not only integer numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the initial number be $x$.\n\n- After the first press: $\\frac{x-1}{x+1}$.\n- After the second press: $$\\frac{\\frac{x-1}{x+1}-1}{\\frac{x-1}{x+1}+1} = \\frac{x-1-x-1}{x-1+x+1} = -\\frac{1}{x}.$$ \n- After the third press: $$\\frac{-\\frac{1}{x}-1}{-\\frac{1}{x}+1} = \\frac{-1-x}{-1+x}.$$ \n- After the fourth press: $$\\frac{\\frac{-1-x}{-1+x}-1}{\\frac{-1-x}{-1+x}+1} = x.$$ \n\nThus, the sequence cycles every 4 presses. After 2012 presses, the screen shows $x$. After 2 more presses (2014 total), the screen shows $-\\frac{1}{x}$. We are told this equals 2016, so:\n\n$$-\\frac{1}{x} = 2016 \\implies x = -\\frac{1}{2016}.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16342,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $G$ be a graph on $3k$ vertices that does not contain $K_4$. What is the maximum possible number of triangles in $G$?",
    "options": [],
    "answer": "See solution",
    "solution": "We show by induction that the maximum number of triangles is $k^3$.\n\nFor $k=1$, the result is trivial.\n\nSuppose the statement holds for $k-1$. Let $G$ be a graph with the maximum number of triangles and $3k$ vertices, and suppose $G$ does not contain $K_4$. Clearly, $G$ contains a triangle; call it $T_1T_2T_3$.\n\nConsider the subgraph $G - \\Delta T_1T_2T_3$ (i.e., remove the triangle $T_1T_2T_3$ and its vertices). This subgraph has $3(k-1)$ vertices and does not contain $K_4$, so by the induction hypothesis, it has at most $(k-1)^3$ triangles. By Turán's theorem, the number of edges in $G - \\Delta T_1T_2T_3$ is at most $3(k-1)^2$.\n\nThe number of triangles involving both $\\Delta T_1T_2T_3$ and $G - \\Delta T_1T_2T_3$ is at most $3(k-1)^2 + 3(k-1)$. Thus, the total number of triangles is at most\n\n$$\n(k-1)^3 + 3(k-1)^2 + 3(k-1) + 1 = k^3.\n$$\n\nEquality holds for the complete tripartite graph $K_{k,k,k}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 16343,
    "subject": "Mathematics (Olympiad)",
    "question": "In acute triangle $ABC$, $\\angle A < \\angle B$ and $\\angle A < \\angle C$. Let $P$ be a variable point on side $BC$. Points $D$ and $E$ lie on sides $AB$ and $AC$, respectively, such that $BP = PD$ and $CP = PE$. Prove that as $P$ moves along side $BC$, the circumcircle of triangle $ADE$ passes through a fixed point other than $A$.\n\n![](images/pamphlet1112_main_p51_data_3635fe8904.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that the fixed point is the orthocenter $H$ of triangle $ABC$.\n\nLet $X$ be the foot of the perpendicular from $C$ to $AB$, and let $Y$ be the foot of the perpendicular from $B$ to $AC$. Note that if $M$ is the midpoint of $BC$, then $MB = MX = MY = MC$. Suppose without loss of generality that $P$ is between $B$ and $M$. Then $D$ is between $B$ and $X$, and $E$ is between $A$ and $Y$.\n\nThe quadrilateral $AXHY$ is cyclic, since $\\angle AXH = \\angle AYH = 90^\\circ$. To show that $ADHE$ is also cyclic, it suffices to show that $\\triangle DHX \\sim \\triangle EHY$, or that\n\n$$\n\\frac{DX}{EY} = \\frac{XH}{YH}.\n$$\n\nApplying the Law of Sines in the cyclic quadrilateral $AXHY$, we find that $\\frac{XH}{YH} = \\frac{\\cos B}{\\cos C}$. Note that $DX = BX - BD = BC \\cos B - 2BP \\cos B$. Similarly, $EY = EC - CY = 2PC \\cos C - BC \\cos C$. Hence, we find\n\n$$\n\\frac{DX}{EY} = \\frac{BC - 2BP}{2PC - BC} \\cdot \\frac{\\cos B}{\\cos C} = \\frac{\\cos B}{\\cos C}.\n$$\n\nPutting this together with our previous computation yields the desired result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16344,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $A'$, $B'$, and $C'$ are chosen correspondingly on the sides $AB$, $BC$, and $CA$ of an equilateral triangle $ABC$ so that\n$$\n\\frac{|A'B|}{|AB|} = \\frac{|B'C|}{|BC|} = \\frac{|C'A|}{|CA|} = k.\n$$\nFind all positive real numbers $k$ for which the area of triangle $A'B'C'$ is exactly half of the area of triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\alpha$ be the angle at vertex $A$. The area of triangle $AA'C'$ is\n$$\nS_{AA'C'} = \\frac{1}{2} \\cdot |AA'| \\cdot |AC'| \\cdot \\sin \\alpha = \\frac{1}{2} \\cdot (1-k)|AB| \\cdot k|AC| \\cdot \\sin \\alpha = (1-k)k S_{ABC}.\n$$\nSimilarly, $S_{BB'A'} = (1-k)k S_{ABC}$ and $S_{CC'B'} = (1-k)k S_{ABC}$. Hence, the triangles $AA'C'$, $BB'A'$, and $CC'B'$ are of equal area. Therefore:\n\n![](images/Estonija_2010_p12_data_1c46418bec.png)\n\nThe area of triangle $A'B'C'$ is half of the area of triangle $ABC$ if and only if the area of triangle $AA'C'$ is one sixth of the area of triangle $ABC$, i.e.,\n$$\n(1-k)k = \\frac{1}{6}.\n$$\nThe solutions of $k^2 - k + \\frac{1}{6} = 0$ are\n$$\nk_{1,2} = \\frac{1}{2} \\pm \\frac{\\sqrt{3}}{6},\n$$\nboth of which are positive.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16345,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be the set of all rational numbers that can be expressed as a repeating decimal in the form $0.\\overline{abcd}$, where at least one of the digits $a, b, c$, or $d$ is nonzero. Let $N$ be the number of distinct numerators obtained when numbers in $S$ are written as fractions in lowest terms. For example, both $4$ and $410$ are counted among the distinct numerators for numbers in $S$ because $0.3636 = \\frac{4}{11}$ and $0.1230 = \\frac{410}{3333}$. Find the remainder when $N$ is divided by 1000.",
    "options": [],
    "answer": "See solution",
    "solution": "The repeating decimal of the form $0.\\overline{abcd}$ is equal to the fraction $\\frac{abcd}{9999}$. Because $9999 = 3^2 \\cdot 11 \\cdot 101$, the fraction $\\frac{k}{9999}$ written in lowest terms will have a factor of 3 in its numerator if and only if $k$ is divisible by $3^3$, will have a factor of 11 in its numerator if and only if $k$ is divisible by $11^2$, and will have a factor of 101 in its numerator if and only if $k$ is divisible by $101^2$. Thus there are five cases for possible numerators for such a fraction in lowest terms.\n\n- The fraction $\\frac{k}{9999}$ is in lowest terms if $k$ is relatively prime to 9999. The number of positive integers less than 10,000 relatively prime to 9999 is given by the Euler Totient Function:\n\n$$\n\\phi(9999) = 9999 \\left(1 - \\frac{1}{3}\\right) \\left(1 - \\frac{1}{11}\\right) \\left(1 - \\frac{1}{101}\\right) = 6000.\n$$\n\nIf $k$ is not divisible by any of $3^3, 11^2$ or $101^2$, the fraction $\\frac{k}{9999}$, when reduced, will have a numerator relatively prime to 9999, all of which have already been counted.\n\n- Because $\\frac{3^3 k}{9999} = \\frac{3k}{11 \\cdot 101}$, the numerator can be any number of the form $3k$, where $k \\leq \\left\\lfloor \\frac{9999}{27} \\right\\rfloor = 370$ and $k$ is relatively prime to 11 and 101. There are\n\n$$\n\\left\\lfloor \\frac{370}{11} \\right\\rfloor + \\left\\lfloor \\frac{370}{101} \\right\\rfloor - \\left\\lfloor \\frac{370}{11 \\cdot 101} \\right\\rfloor = 33 + 3 - 0 = 36\n$$\n\npositive integers less than or equal to 370 that are multiples of 11 or 101. Thus there are $370 - 36 = 334$ numerators that are relatively prime to 11 and 101.\n\n- Because $\\frac{11^2 k}{9999} = \\frac{11k}{3^2 \\cdot 101}$, the numerator can be any number of the form $11k$, where $k \\leq \\left\\lfloor \\frac{9999}{121} \\right\\rfloor = 82$ and $k$ is relatively prime to 3 and 101. There are\n\n$$\n\\left\\lfloor \\frac{82}{3} \\right\\rfloor + \\left\\lfloor \\frac{82}{101} \\right\\rfloor - \\left\\lfloor \\frac{82}{3 \\cdot 101} \\right\\rfloor = 27 + 0 - 0 = 27\n$$\n\npositive integers less than or equal to 82 that are multiples of 3 or 101. Thus there are $82 - 27 = 55$ numerators that are relatively prime to 3 and 101.\n\n- Because $\\frac{3^3 \\cdot 11^2 k}{9999} = \\frac{33k}{101}$, the numerator can be any number of the form $33k$, where $k \\leq \\left\\lfloor \\frac{9999}{27 \\cdot 121} \\right\\rfloor = 3$ and $k$ is relatively prime to 101. There are 3 such numerators.\n\n- Because $101^2 > 9999$, there are no numerators that are multiples of 101.\n\nTherefore there are $6000 + 334 + 55 + 3 = 6392$ possible numerators. The requested remainder is $392$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16346,
    "subject": "Mathematics (Olympiad)",
    "question": "Given triangle $ABC$, let $k$ be the excircle at side $BC$. Choose any line $p$ parallel to $BC$ intersecting line segments $AB$ and $AC$ at points $D$ and $E$. Denote by $l$ the incircle of triangle $ADE$. The tangents from $D$ and $E$ to the circle $k$ not passing through $A$ intersect at $P$. The tangents from $B$ and $C$ to the circle $l$ not passing through $A$ intersect at $Q$. Prove that the line $PQ$ passes through a fixed point independent of the choice of $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $BC$ touch $k$ at $T_k$ and $DE$ touch $l$ at $T_l$. We shall prove the required fixed point is $T_k$.\n\nFirst, we show the points $T_k$, $T_l$, and $P$ are collinear. Denote by $U$ and $V$ the points where $EP$ and $DP$ touch $k$, and $M$ and $N$ the points where $EP$ and $DP$ intersect $BC$. Let $T_1$ and $T_2$ be the tangent points of $k$ with the rays $AB$ and $AC$ respectively.\n\n![](images/Cesko-Slovacko-Poljsko_2009_p5_data_c0d4cac6ee.png)\n\nAs $BC \\parallel DE$, triangle $DEP$ is similar to $NMP$, and the homothety $H$ with center $P$ and ratio $q = MN/ED$ maps segment $DE$ to $NM$. To prove the collinearity of $T_k, T_l, P$, it suffices to show\n\n$$\n\\frac{MT_k}{NT_k} = \\frac{ET_l}{DT_l} \\qquad (1)\n$$\n\nIf this is true, $H$ maps $T_l$ to $T_k$.\n\nLet $a, b, c$ be the lengths of the sides in triangle $DEP$ as in the figure. Set $AD = c$, $AE = d$. Recall the well-known formulas for the distances from a vertex of a triangle to the tangent points of its incircle and excircle: In any triangle $XYZ$, the distance from $X$ to the tangent point of the incircle and excircle (lying on $XY$) is $(XY + XZ - YZ)/2$ and $(XY + YZ - XZ)/2$ respectively.\n\nThe circle $k$ is the excircle of $NMP$. Hence\n\n$$\n\\frac{MT_k}{NT_k} = \\frac{(MN + NP - MP)/2}{(MN + MP - NP)/2} = \\frac{qa + qc - qb}{qa + qb - qc} = \\frac{a + c - b}{a + b - c} \\qquad (2)\n$$\n\nThe circle $l$ is the incircle of $DEA$. Hence\n\n$$\n\\frac{ET_l}{DT_l} = \\frac{(DE + AE - AD)/2}{(DE + AD - AE)/2} = \\frac{a + d - c}{a + c - d} \\qquad (3)\n$$\n\nWhen we draw two tangent lines from a point to a circle, the distances from the point to the two tangent points are equal. Repeating this argument several times, we get\n\n$$\nc + c + PU = e + c + PV = e + DT_l \\quad AT_1 \\cdot AT_2 = d + ET_2 \\cdot d + b + PU\n$$\n\nso $e + c = d + b$. Then $c - b = d - e$, and substituting into (2) and (3) gives\n\n$$\n\\frac{MT_k}{NT_k} = \\frac{a + (c - b)}{a - (c - b)} = \\frac{a + (d - e)}{a - (d - e)} = \\frac{ET_l}{DT_l}\n$$\n\nwhich is exactly (1). Hence $P$ lies on $T_1T_k$.\n\nSimilarly, we show that $T_k$, $T_l$, and $Q$ are collinear. Denote by $U'$ and $V'$ the points where $CQ$ and $BQ$ touch $l$, and $M'$ and $N'$ the points where $CQ$ and $BQ$ intersect $DE$. Let $T_1'$ and $T_2'$ be the tangent points of $l$ with the rays $AD$ and $AE$ respectively. Let $a', b', c'$ be the lengths of the sides in triangle $BCQ$ and $AB = c'$, $AC = d'$.\n\n![](images/Cesko-Slovacko-Poljsko_2009_p6_data_be4b50e2f3.png)\n\nRepeating the arguments from the first part (here both circles are excircles), we get\n\n$$\n\\frac{M'T_l}{N'T_l} = \\frac{a' + c' - b'}{a' + b' - c'} \\qquad \\frac{CT_k}{BT_k} = \\frac{a' + c' - d'}{a' + d' - c'}\n$$\n\nComparing the lengths (see figure) gives\n\n$$\nc' - c' - QU' = c' - c' - QV' = c' - BT_{1}' = AT_{1}' = AT_{2}' = d' - CT_{2}' = d' - b' - QU'\n$$\n\nso $c' - b' = c' - d'$, and\n\n$$\n\\frac{M'T_l}{N'T_l} = \\frac{CT_k}{BT_k}\n$$\n\nFinally, using the homothety of $BCQ$ and $N'M'Q$, we conclude that $Q$ lies on $T_lT_k$.\n\nTherefore, the line $PQ$ (with $P \\neq Q$) is identical to the line $T_lT_k$ and passes through $T_k$, which is independent of the choice of $p$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16347,
    "subject": "Mathematics (Olympiad)",
    "question": "Во четириаголникот $ABCD$, $E$ е средина на страната $BC$, и плоштината на триаголникот $AED$ е двпати помала од плоштината на четириаголникот $ABCD$. Докажи дека $AB$ е паралелна со $CD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Нека $ABCD$ е четириаголникот во кој $E$ е средина на $BC$, при што $P_{ABCD} = 2P_{AED}$. Точката $D$ ќе ја пресликаме централно симетрично со центар на симетрија $E$. Нека $D_1$ е нејзината слика. Бидејќи $ECD \\cong EBC$ имаме $P_{ECD} = P_{EBC}$, од каде добиваме\n\n![](images/Makedonija_2009_p39_data_be0aab3c00.png)\n\n$$\nP_{ABCD} = 2(P_{ABE} + P_{ECD}) = 2(P_{ABE} + P_{EBD_1}). \\quad (1)\n$$\n\nОд друга страна $P_{AED} = P_{AED_1}$ (имаат иста должина на основа и висината спуштена врз неа). Значи\n\n$$\nP_{ABCD} = 2P_{AED_1} \\quad (2)\n$$\n\nОд (1) и (2) добиваме\n\n$$\nP_{ABE} + P_{EBD_1} = P_{AED_1}.\n$$\n\nАко $A$, $B$ и $D_1$ не се колинеарни, последното равенство не е точно. Значи\n\n*A, B и $D_1$ се колинеарни.*\n\nОд колинеарноста на $A, B$ и $D_1$ и од тоа што $DC \\parallel BD_1$, добиваме дека $AB \\parallel CD$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16348,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that\n\n$$\n\\frac{1}{\\sqrt{a^2 + 1}} + \\frac{1}{\\sqrt{b^2 + 1}} + \\frac{1}{\\sqrt{c^2 + 1}} \\geq 2 + \\frac{1}{\\sqrt{3}}\n$$\n\nfor all non-negative real numbers $a$, $b$, $c$ satisfying $a + b + c = \\sqrt{2}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that\n\n$$\n\\frac{1}{\\sqrt{x^2+1}} + \\frac{1}{\\sqrt{y^2+1}} \\geq 1 + \\frac{1}{\\sqrt{(x+y)^2+1}}\n$$\n\nfor $(x, y) = (a, b)$ and $(x, y) = (a+b, c)$. Then by summing these two inequalities we obtain the inequality in the problem.\n\nThe inequality above is equivalent to\n\n$$\n\\frac{1}{x^2+1} + \\frac{1}{y^2+1} + \\frac{2}{\\sqrt{x^2+1}\\sqrt{y^2+1}} \\geq 1 + \\frac{1}{(x+y)^2+1} + \\frac{2}{\\sqrt{(x+y)^2+1}}\n$$\n\nHence it suffices to show that\n\n$$\n\\frac{2}{\\sqrt{x^2+1}\\sqrt{y^2+1}} \\geq \\frac{2}{\\sqrt{(x+y)^2+1}}\n$$\n\nand\n\n$$\n\\frac{1}{x^2+1} + \\frac{1}{y^2+1} \\geq 1 + \\frac{1}{(x+y)^2+1}\n$$\n\nThe first inequality is equivalent to\n\n$$\n(x+y)^2+1 \\geq (x^2+1)(y^2+1) \\iff 2xy - x^2y^2 \\geq 0 \\iff xy(2-xy) \\geq 0\n$$\n\nwhich is valid for $(x, y) = (a, b)$ and $(x, y) = (a+b, c)$ as\n\n$$\nab \\leq \\frac{(a+b)^2}{4} \\leq \\frac{(a+b+c)^2}{4} = \\frac{1}{2} < 2 \\quad \\text{and} \\quad (a+b)c \\leq \\frac{(a+b+c)^2}{4} = \\frac{1}{2} < 2\n$$\n\nThe second inequality is equivalent to\n\n$$\nxy(2 - 2xy - xy(x + y)^2) \\geq 0.\n$$\n\nFor $(x, y) = (a, b)$ we have $2 - 2ab - ab(a + b)^2 \\geq 2 - 2ab - ab(a + b + c)^2 = 2 - 4ab \\geq 0$ as $ab \\leq 1/2$. For $(x, y) = (a+b, c)$ we have $2 - 2(a+b)c - (a+b)c(a+b+c)^2 = 2 - 4(a+b)c \\geq 0$ as $(a+b)c \\leq 1/2$. Thus, the original inequality holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16349,
    "subject": "Mathematics (Olympiad)",
    "question": "Обозначим $n$-е простое число через $p_n$. Предположим, что нашлось $m > 1$, для которого $S_{m-1} = k^2$, $S_m = l^2$, где $k$ и $l$ — натуральные числа. Числа $S_2 = 5$, $S_3 = 10$ квадратами не являются, так что $m > 4$.\n\n*Являются ли два последовательных частичных суммы простых квадратами натуральных чисел?*",
    "options": [],
    "answer": "See solution",
    "solution": "Заметим, что $p_m = S_m - S_{m-1} = (l-k)(l+k)$; ввиду простоты $p_m$ получаем $1 = l-k$, $p_m = l+k = 2l-1 = 2\\sqrt{S_m} - 1$. Таким образом, $S_m = \\left(\\frac{p_m+1}{2}\\right)^2$.\n\nЗаметим, что $p_m$ нечётно (так как $m \\ge 2$), и $1+3+5+\\ldots + p_m = (1^2 - 0^2) + (2^2 - 1^2) + \\ldots + \\left(\\left(\\frac{p_m+1}{2}\\right)^2 - \\left(\\frac{p_m-1}{2}\\right)^2\\right) = \\left(\\frac{p_m+1}{2}\\right)^2$.\n\nС другой стороны, в сумму $S_m = 2 + p_2 + \\ldots + p_m$, кроме двойки, входят лишь нечётные числа, и при $m > 4$ не входят нечётное составное число $9$ и число $1$, поэтому $S_m \\le (1+3+5+\\ldots+p_m)+2-1-9 < \\left(\\frac{p_m+1}{2}\\right)^2$. Противоречие.\n\n_Замечание._ В последовательности $(S_n)$ встречаются квадраты натуральных чисел. Кроме $S_9 = 100$, известны еще несколько; минимальный из них — это $S_{2474} = 25633969 = 5063^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16350,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\{a_n\\}$ be a decreasing sequence of positive real numbers such that $a_1 = \\frac{1}{2}$, and\n\n$$\na_n^2(a_{n-1} + 1) + a_{n-1}^2(a_n + 1) - 2a_n a_{n-1}(a_n a_{n-1} + a_n + 1) = 0\n$$\nfor $n \\geq 2$.\n\n1. Find $a_n$ for $n \\geq 1$.\n2. Let $S_n = \\sum_{i=1}^{n} a_i$. Show that\n   $$\n   \\ln\\left(\\frac{n}{2} + 1\\right) < S_n < \\ln(n + 1).\n   $$",
    "options": [],
    "answer": "See solution",
    "solution": "1. The recursive formula can be rewritten as\n   $$\n   (a_n - a_{n-1})^2 - a_n a_{n-1} (a_n - a_{n-1}) - 2a_n^2 a_{n-1}^2 = 0,\n   $$\n   or\n   $$\n   (a_n - a_{n-1} + a_n a_{n-1})(a_n - a_{n-1} - 2a_n a_{n-1}) = 0.\n   $$\n   Since $0 < a_n < a_{n-1}$, we have $a_n - a_{n-1} - 2a_n a_{n-1} < 0$, so\n   $$\n   a_n - a_{n-1} + a_n a_{n-1} = 0.\n   $$\n   Thus,\n   $$\n   \\frac{1}{a_n} - \\frac{1}{a_{n-1}} = 1, \\quad \\text{so} \\quad a_n = \\frac{1}{n+1}.\n   $$\n\n2. Consider $S_n = \\sum_{i=1}^n a_i = \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n+1}$.\n   \n   We use the inequalities:\n   $$\n   \\ln\\left(1 + \\frac{1}{n}\\right) < \\frac{1}{n} < \\ln\\left(1 + \\frac{1}{n-1}\\right).\n   $$\n   Summing these for $i = 2$ to $n+1$ gives:\n   $$\n   S_n < \\ln(n+1),\n   $$\n   $$\n   S_n > \\ln\\left(\\frac{n}{2} + 1\\right),\n   $$\n   as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16351,
    "subject": "Mathematics (Olympiad)",
    "question": "ABCD is a convex cyclic quadrilateral. The diagonals $AC$ and $BD$ intersect at the point $E$. It is given that $AB = 39$, $AE = 45$, $AD = 60$, and $BC = 56$.\n\nDetermine the length of $CD$.",
    "options": [],
    "answer": "See solution",
    "solution": "From the similarity of triangles $BEC$ and $AED$, we have $BE = 42$.\n\n![](images/Greek_07_Booklet_p16_data_e9a64b135e.png)\n\nLet $DE = x$ and $CE = y$. From the similarity of triangles $ABE$ and $DEC$, we have\n$$\n\\frac{x}{AE} = \\frac{y}{BE} = \\frac{CD}{AB} \\implies \\frac{x}{15} = \\frac{y}{14} = \\frac{CD}{13} = t. \\quad (1)\n$$\n\nMoreover, from Ptolemy's theorem we find\n$$\n\\begin{aligned}\nBD \\cdot AC &= AD \\cdot BC + AB \\cdot CD \\\\\n&\\implies (45 + 14t)(42 + 15t) = 60 \\cdot 56 + 39 \\cdot 13t \\\\\n&\\implies 35t^2 + 126t - 245 = 0 \\\\\n&\\implies t = \\frac{7}{5} \\text{ or } t = -5.\n\\end{aligned}\n$$\n\nSince $t$ must be positive, from (1) we have $t = \\frac{7}{5}$ and $CD = \\frac{91}{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16352,
    "subject": "Mathematics (Olympiad)",
    "question": "Isaac has a large supply of counters, and places one in each of the $1 \\times 1$ squares of an $8 \\times 8$ chessboard. Each counter is either red, white, or blue. A particular pattern of coloured counters is called an arrangement. Determine whether there are more arrangements which contain an even number of red counters or more arrangements which contain an odd number of red counters. Note that $0$ is an even number.",
    "options": [],
    "answer": "See solution",
    "solution": "There are $64$ squares on the chessboard. Suppose there are $n$ red counters in an arrangement. There are $\\binom{64}{n}$ ways to choose the $n$ squares for red counters. The remaining $64 - n$ squares can each be either blue or white, giving $2^{64-n}$ possibilities for these squares.\n\nTherefore, there are $\\binom{64}{n} 2^{64-n}$ arrangements with $n$ red counters.\n\nThe number of arrangements with an even number of red counters is:\n\n$$\n\\binom{64}{0}2^{64} + \\binom{64}{2}2^{62} + \\binom{64}{4}2^{60} + \\dots + \\binom{64}{62}2^{2} + \\binom{64}{64}2^{0}\n$$\n\nThe number with an odd number of red counters is:\n\n$$\n\\binom{64}{1}2^{63} + \\binom{64}{3}2^{61} + \\binom{64}{5}2^{59} + \\dots + \\binom{64}{63}2^{1}\n$$\n\nBy the binomial theorem:\n\n$$\n(2 - 1)^{64} = \\binom{64}{0}2^{64} - \\binom{64}{1}2^{63} + \\binom{64}{2}2^{62} - \\dots + \\binom{64}{64}2^{0}\n$$\n\nSo, the number of arrangements with an even number of red counters is exactly one more than the number with an odd number of red counters.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16353,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with centroid $G$. Points $R$ and $S$ are chosen on rays $GB$ and $GC$, respectively, such that\n\n$$\n\\angle ABS = \\angle ACR = 180^{\\circ} - \\angle BGC.\n$$\n\nProve that $\\angle RAS + \\angle BAC = \\angle BGC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ and $N$ denote the midpoints of $\\overline{AC}$ and $\\overline{AB}$, respectively.\n\n![](images/sols-TSTST-2023_p3_data_68f8ec6f1a.png)\n\n**Solution 1 using power of a point**  \nFrom the given condition that $\\angle ACR = \\angle CGM$, we get that\n\n$$\nMA^2 = MC^2 = MG \\cdot MR \\Rightarrow \\angle RAC = \\angle MGA.\n$$\n\nAnalogously,\n\n$$\n\\angle BAS = \\angle AGN.\n$$\n\nHence,\n\n$$\n\\angle RAS + \\angle BAC = \\angle RAC + \\angle BAS = \\angle MGA + \\angle AGN = \\angle MGN = \\angle BGC.\n$$\n\n**Solution 2 using similar triangles**  \nAs before, $\\triangle MGC \\sim \\triangle MCR$ and $\\triangle NGB \\sim \\triangle NBS$. We obtain\n\n$$\n\\frac{|AC|}{|CR|} = \\frac{2|MC|}{|CR|} = \\frac{2|MG|}{|GC|} = \\frac{|GB|}{2|NG|} = \\frac{|BS|}{2|BN|} = \\frac{|BS|}{|AB|}\n$$\n\nwhich together with $\\angle ACR = \\angle ABS$ yields\n\n$$\n\\triangle ACR \\sim \\triangle SBA \\Rightarrow \\angle BAS = \\angle CRA.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16354,
    "subject": "Mathematics (Olympiad)",
    "question": "For $n \\ge 2$, an equilateral triangle is divided into $n^2$ congruent smaller equilateral triangles. Determine all ways in which real numbers can be assigned to the $\\frac{n(n+1)}{2}$ vertices so that three such numbers sum to zero whenever the three vertices form an equilateral triangle with edges parallel to the sides of the big triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "For $n = 2$, the only requirement is $a_1 = -a_2 - a_3$.\n\nFor $n = 3$, we see that\n\n$$\na_2 + a_4 + a_5 = 0 = a_2 + a_3 + a_5,\n$$\nwhich shows that $a_3 = a_4$ and similarly $a_1 = a_5$ and $a_2 = a_6$. Now the only requirement is the stated equalities and $a_1 = -a_2 - a_3$.\n\nFor $n = 4$, observe that $a_1 = a_7 = a_{10}$ since they all equal $a_5$. Since also $a_1 + a_7 + a_{10} = 0$, they all equal zero. By considering the top triangle, we get $x = a_2 = -a_3$ and this uniquely determines the rest. It is easily checked that, for any real $x$, this is actually a solution:\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p307_data_37c1c5c2c2.png)\n\nFor $n > 4$ we can apply the same argument as above for any collection of 10 vertices. Any vertex not on the sides of the big triangle has to equal zero, since it is the centre of such a collection of 10 vertices. Any vertex $a$ on the sides of the big triangle forms some parallelogram similar to $a_4, a_2, a_5, a_8$, where the point opposite $a$ is in the interior of the big triangle. Since such opposite numbers are equal, all $a_i$ have to be zero in this case. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16355,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 2011 people in a city. For some period of time every day, a group of at least 4 people went to a restaurant to have dinner. No group of 3 people went together to more than one dinner. Prove that there exists a group of 24 people such that at every dinner there was a person not belonging to this group.",
    "options": [],
    "answer": "See solution",
    "solution": "We can assume that at every dinner there were *exactly* 4 people (just remove the surplus people from every dinner, which does not affect the condition that no group of 3 people went together to two different dinners, and can only make the task of finding a suitable 24-people group harder).\n\nConsider a group $A$ with the greatest possible cardinality such that at every dinner there was a person not from $A$. Assume there are $m$ people in $A$. It is sufficient to show that $m \\ge 24$.\n\nBy the definition of $A$, for every person $p \\notin A$ there exists a group $G_p \\subset A \\cup \\{p\\}$ of 4 people which went to the restaurant together one day. But $G_p \\not\\subset A$, so there are exactly 3 elements in $A \\cap G_p$. In other words, every $G_p$ consists of 3 people from $A$ and the person $p$. Also, for different people $p_1, p_2 \\notin A$ we obtain distinct intersections $A \\cap G_{p_1}, A \\cap G_{p_2}$ — otherwise the groups $G_{p_1}, G_{p_2}$ would have 3 people in common, which by our assumptions would mean that $G_{p_1} = G_{p_2}$, but this is not possible, since $p_1 \\in G_{p_1}$ and $p_1 \\notin G_{p_2}$.\n\nThus the number of people not in $A$ (equal to $2011 - m$) does not exceed the number of 3-element subsets of $A$:\n\n$$\n2011 \\le m + \\binom{m}{3} = \\frac{1}{6}(6m + m(m-1)(m-2)) = \\frac{1}{6}m(m^2 - 3m + 8).\n$$\n\nThe right hand side is increasing for $m \\ge 1$ and is equal to 1794 for $m = 23$. Therefore $m \\ge 24$, as desired.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16356,
    "subject": "Mathematics (Olympiad)",
    "question": "Andrii chooses a sign in front of each number in the expression $\\pm 1 \\pm 2 \\pm 2^2 \\pm 2^3 \\pm \\dots \\pm 2^{2019}$. How many different positive values can Andrii obtain as a result of the computed expression?",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the expression $\\pm 1 \\pm 2 \\pm 2^2 \\pm 2^3 \\pm \\dots \\pm 2^{2019}$, where each term can be assigned either a plus or minus sign. To obtain a positive value, the sign in front of $2^{2019}$ must be positive. The remaining $2019$ terms ($1$ through $2^{2018}$) can each be assigned either sign independently. Thus, there are $2^{2019}$ different positive values that Andrii can obtain.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16357,
    "subject": "Mathematics (Olympiad)",
    "question": "A cube of side length $2021$ is given. In how many ways can we place a $1 \\times 1 \\times 1$ cubelet on the border of this cube so that the newly formed solid can be completely filled using $k \\times 1 \\times 1$, $1 \\times k \\times 1$, and $1 \\times 1 \\times k$ cuboids, for some $k \\in \\mathbb{N} \\setminus \\{1\\}$?",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** Suppose that for some $k > 1$ and some placed cubelet there is a valid filling. In each unit cubelet (of the original cube) with coordinates $(x, y, z)$ where $0 \\le x, y, z \\le 2020$, we assign the complex number $\\omega^{x+y+z}$ where $\\omega = e^{\\frac{2\\pi i}{k}}$. We also assign the number $\\omega^{a+b+c}$ in the additional cubelet in position $(a, b, c)$.\n\nSince $1 + \\omega + \\omega^2 + \\dots + \\omega^{k-1} = \\frac{\\omega^k - 1}{\\omega - 1} = 0$, the sum of numbers in any $1 \\times 1 \\times k$ cuboid is zero. So the sum of all assigned numbers is\n\n$$\n0 = (1 + \\omega + \\dots + \\omega^{2020})^3 + \\omega^{a+b+c}. \\qquad (1)\n$$\n\nThis gives\n\n$$\n1 = |-\\omega^{a+b+c}| = |(1 + \\omega + \\dots + \\omega^{2020})^3| = \\left|\\frac{1 - \\omega^{2021}}{1 - \\omega}\\right|^3.\n$$\n\nThus $|1 - \\omega| = |1 - \\omega^{2021}|$, which means that $1$ is equidistant from $\\omega$ and $\\omega^{2021}$. Since $|\\omega^{2021}| = 1$, this happens if and only if $\\omega^{2021} = \\omega$ or $\\omega^{2021} = \\omega^{-1}$. Then $\\omega^{2020} = 1$ or $\\omega^{2022} = 1$, which gives $k \\mid 2020$ or $k \\mid 2022$. However, $k \\mid 2021^3 + 1$. Since $2021^3 + 1 \\equiv 2 \\pmod{2020}$, if $k \\mid 2020$ then $k \\mid 2$. So in any case we have $k \\mid 2022$. Now (1) gives\n\n$$\n\\omega^{a+b+c} = -(1 + \\omega + \\dots + \\omega^{2020})^3 = -(-\\omega^{2021})^3 = \\omega^{6063} = \\omega^{-3}.\n$$\n\nSo $a + b + c \\equiv -3 \\pmod{k}$.\n\nIf we have a valid filling for $k$, then we have a valid filling for every prime factor of $k$. So we may assume that $k$ is prime and therefore $k \\in \\{2, 3, 337\\}$.\n\nAssume without loss of generality that the additional cubelet is at the bottom of the cube, i.e. $c = -1$. Then $a + b \\equiv -2 \\pmod{k}$. By symmetry, if we have a valid filling for $(a, b, -1)$, then we have a valid filling for $(2022 - a, b, -1)$. So we must also have $2022 - a + b \\equiv -2 \\pmod{k}$ and so $b - a \\equiv -2 \\pmod{k}$. Since also $a + b \\equiv -2 \\pmod{k}$ we get $a \\equiv b \\equiv -1 \\pmod{k}$ for $k \\neq 2$ and $a \\equiv b \\pmod{2}$ for $k = 2$. We will now show that the above necessary conditions are also sufficient to have a valid filling.\n\nWe claim first that a square defined by coordinates $(x, y)$ where $0 \\le x, y \\le 2020$, with a removed cell $(a, b)$ satisfying the above restrictions can be covered by $1 \\times k$ rectangles. This is because such a square can be covered by four rectangles of sizes $(a+1) \\times b$, $(2020-a) \\times (b+1)$, $(2021-a) \\times (2020-b)$ and $a \\times (2021-b)$, where each of these rectangles can be covered by $1 \\times k$ rectangles. This follows since $k \\mid a+1, b+1, 2021-a, 2021-b$ if $a \\equiv b \\equiv -1 \\pmod{k}$ and since $2 \\mid b, 2020-a, 2020-b, a$ if $a \\equiv b \\equiv 0 \\pmod{2}$.\n\nNow we fill the cube with the added cubelet as follows: The lowest $k-1$ \"layers\" of the original cube of side $2021$ are filled using the previous method together with a $1 \\times 1 \\times k$ cuboid covering the holes in these layers and the additional cubelet. The remainder is the $2021 \\times 2021 \\times (2022-k)$ cuboid which can be easily filled by $1 \\times 1 \\times k$ cuboids because $k \\mid 2022-k$.\n\nTo complete the solution, we need to count the number of ordered pairs $(a, b)$ with $0 \\le a, b \\le 2020$, such that $a \\equiv b \\pmod{2}$, or $a \\equiv b \\equiv -1 \\pmod{3}$ or $a \\equiv b \\equiv -1 \\pmod{337}$. There are $1011^2$ choices with $a \\equiv b \\equiv 0 \\pmod{2}$ and $1010^2$ choices with $a \\equiv b \\equiv 1 \\pmod{2}$. If $a \\equiv b \\equiv -1 \\pmod{3}$ but $a \\not\\equiv b \\pmod{2}$ then one of $a, b$ must be congruent to $2 \\pmod{6}$ and the other to $5 \\pmod{6}$. There are $2 \\times 337 \\times 336$ such choices. Finally, if $a \\equiv b \\equiv -1 \\pmod{337}$ but is not yet accounted for, then one of them is equal to $\\{336, 1010, 1684\\}$ and the other to $\\{673, 1347\\}$. (Note that in this case the second one is definitely not congruent to $-1 \\pmod{3}$.) There are $2 \\times 3 \\times 2$ such choices. In total we have $2268697$ choices for the pair $(a, b)$. Therefore, because of symmetry, the total number of ways is $6 \\times 2268697 = 13612182$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 16358,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $AB < AC$. Let $D$ be the foot of the altitude from $A$, $H$ the orthocenter, $O$ the circumcenter, $M$ the midpoint of $BC$, $A'$ the reflection of $A$ across $O$, and $S$ the intersection of the tangents at $B$ and $C$ to the circumcircle. The tangent at $A'$ to the circumcircle intersects $SC$ and $SB$ at $X$ and $Y$, respectively.\n\nIf $M$, $S$, $X$, $Y$ are concyclic, prove that lines $OD$ and $SA'$ are parallel.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N$ be the second intersection point of the circumcircle of triangle $SXY$ with the line $BC$. Writing the powers of points $B$ and $C$ with respect to the circumcircle of quadrilateral $MXSY$, we get $BM \\cdot BN = BY \\cdot BS$ and $CN \\cdot CM = CX \\cdot CS$. As $BM = CM$, $BS = CS$, $YB = YA'$, and $XC = XA'$, it follows that\n$$\n\\frac{BY}{CX} = \\frac{BA'}{CA'}.\n$$\n\n![](images/RMC_2019_var_3_p72_data_6fde5ff81e.png)\n\nNext, we use the following:\n\n*Lemma.* Consider a triangle $ABC$ and points $D$ and $E$ on rays $BA$ and $CA$ respectively such that $\\frac{BD}{CE} = k$. Let $M$ and $N$ be points on the line segments $BC$ and $DE$ respectively such that $\\frac{BM}{MC} = \\frac{DN}{NE} = k$. Then $MN$ is parallel to the bisector of angle $\\angle BAC$.\n\n*Proof of the Lemma.* We have $\\overrightarrow{NM} = \\frac{1}{k+1} (\\overrightarrow{DB} + k \\cdot \\overrightarrow{EC})$. If $X$ and $Y$ are points on rays $AB$ and $AC$ such that $AX = DB$ and $AY = k \\cdot EC$, then $\\overrightarrow{NM}$ is parallel to the median from $A$ in triangle $AXY$. But $AX = AY$, which means that in triangle $AXY$ the median from $A$ has the direction of the bisector of angle $\\angle BAC$.\n\nFrom the Lemma, we obtain that $A'N \\parallel SM$, which means that $N$ is the reflection of point $D$ across $M$. In triangle $SXY$, the circumcenter lies on $SN$, therefore the orthocenter lies on its isogonal, $SD$. It follows that $SD \\perp XY$, i.e., $SD \\parallel AA'$. We obtain that $ADSO$ is a parallelogram, hence $SD = R$. But then $SDOA'$ is also a parallelogram, hence the conclusion.\n\n**Remark.** The configuration of the problem has many other interesting properties: $M$ is the circumcenter of triangle $OXY$. Indeed, as $SM$ is the bisector of angle $\\angle XSY$, we have $MX = MY$. From $\\angle XMY = 180^\\circ - \\angle XSY = 2\\angle A = \\angle BOC = 2\\angle XOY$ we obtain the statement.\n\n$OSA'N$ is an isosceles trapezoid, hence $SN = OA' = R$. Thus, the circumcircle of $SXY$ passes through $M$ and $N$ and has the same radius as the Euler circle of triangle $ABC$, which shows that it is the reflection of the latter across point $M$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16359,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ for which there exists an *even* positive integer $a$ such that $(a-1)(a^2-1)\\dots(a^n-1)$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "For $n=1$ and $n=2$.\n\nFor $n=1$, any even $a$ of the form $m^2+1$ works, for example, $a=2$. For $n=2$, any even $a$ of the form $m^2-1$ works, for example, $a=8$.\n\nAssume that for $n=3$, such a number $a$ exists. Then the number $(a-1)(a^2-1)(a^3-1) = (a-1)^3(a+1)(a^2+a+1)$ must be a perfect square. Since $a^2+a+1 = a(a+1)+1$, the numbers $a+1$ and $a^2+a+1$ are coprime. Because $a+1$ is odd, the numbers $a+1$ and $a-1$ are also coprime. Therefore, both $a+1$ and $(a-1)(a^2+a+1)$ must be perfect squares. In particular, $a + 1$ modulo $3$ can only be $0$ or $1$, and thus $a - 1$ is not divisible by $3$. Hence,\n\n$$\n\\begin{align*}\n\\gcd(a - 1, a^2 + a + 1) &= \\gcd(a - 1, (a + 2)(a - 1) + 3) \\\\\n&= \\gcd(a - 1, 3) = 1,\n\\end{align*}\n$$\n\nmeaning that both $a-1$ and $a^2+a+1$ must be perfect squares. However, the latter cannot be a square, since $a^2 < a^2+a+1 < (a+1)^2$. This is a contradiction.\n\nIt remains to prove that no such $a$ exists for $n \\ge 4$. Suppose such an $a$ exists. Take a natural number $k \\ge 2$ such that $2^k \\le n < 2^{k+1}$. Since $a^{2^k} - 1 = (a^{2^k} - 1)(a^{2^k} + 1)$, the number $(a - 1)(a^2 - 1) \\dots (a^n - 1)$ can be expressed as the product of $a^{2^k} - 1 + 1$ and several other factors of the form $a^m - 1$, where $1 \\le m \\le n$ and $m \\neq 2^k$.\n\nWe will show that the factor $a^{2^{k-1}} + 1$ is coprime with all other factors in this decomposition. Suppose $a^{2^{k-1}} + 1$ and $a^m - 1$ share a common divisor $d$. Then $\\gcd(a^{2^k} - 1, a^m - 1)$ is divisible by $d$. But $\\gcd(a^{2^k} - 1, a^m - 1) = a^{\\gcd(2^k,m)} - 1$. Since $m \\neq 2^k$ and $m \\le n < 2^{k+1}$, the number $m$ cannot be divisible by $2^k$. Thus, $\\gcd(2^k, m)$ is a power of two not exceeding $2^{k-1}$. Therefore, $a^{2^{k-1}-1}$ divides $\\gcd(a^{2^k}-1, a^m-1)$, and hence also divides $d$. Because $a$ is even, the numbers $a^{2^{k-1}-1}$ and $a^{2^{k-1}}+1$ have no common divisors other than $1$, so $d=1$, as required.\n\nThe factor $a^{2^{k-1}} + 1$ is coprime with all other factors in the product, which is a perfect square, so it must itself be a perfect square. Then $a^{2^{k-1}} + 1$ and $a^{2^{k-1}}$ are perfect squares differing by $1$, which is impossible. Therefore, our assumption is false, and no such $a$ exists for $n \\ge 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16360,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an equilateral triangle, find all positive integers $n$ such that it can be partitioned into $n$ equilateral triangles (not necessarily of the same size).\n\n*Remark:* The problem could be modified as: Show that an equilateral triangle can be partitioned into $n$ equilateral triangles for $n \\geq 6$.",
    "options": [],
    "answer": "See solution",
    "solution": "An equilateral triangle can be partitioned into one piece, that is, a partition with the triangle itself as the only piece.\n\nWe also note that the cases $n = 6$ and $n = 8$ are possible, as the following figure shows:\n\n![](images/BW2021_Shortlist_p16_data_c452601d3c.png)\n\nAssume that the original triangle can be partitioned into $n$ equilateral triangles. Partition the equilateral triangle into four parts as depicted in the next figure.\n\nThen partition one of the resulting equilateral triangles into $n$ parts. We have therefore partitioned the equilateral triangle into $n + 3$ parts.\n\nAs the equilateral triangle can be partitioned into 1 part, it follows that the triangle can be partitioned into $n$ equilateral triangles if $n = 3k + 1$ for some $k$. Similarly, the constructions for $n = 6$ and $n = 8$ show that the triangle can be partitioned into $n = 6 + 3k$ parts and $n = 8 + 3k$ parts for any $k$.\n\n![](images/BW2021_Shortlist_p17_data_5a80da5b9a.png)\n\nWe have therefore shown that the equilateral triangle can be partitioned into $n$ equilateral triangles for all $n$ except $n = 2, 3, 5$. We proceed to show that the equilateral triangle cannot be partitioned into $n$ equilateral triangles if $n \\in \\{2, 3, 5\\}$.\n\nConsider the cases:\n\n(i) Assume that $n = 2$. By the pigeonhole principle, one triangle shares two vertices with the original triangle, and will therefore be the entire triangle. This is absurd, so no partition for $n = 2$ is possible.\n\n(ii) Assume that $n = 3$. As in the case above, no triangle shares two vertices with the original triangle. So each triangle shares exactly one vertex with the original triangle. In each of the 3 smaller triangles, let $a_i$, $i = 1, 2, 3$ be the side opposing the vertex common with the original triangle. The side $a_1$ lies inside the triangle, so it must be a side of two smaller triangles. However, the only internal segments of the other triangles are $a_2$ and $a_3$, but $a_1$ can only coincide with either $a_2$ or $a_3$. We conclude that no partition is possible for $n = 3$.\n\n(iii) Assume that $n = 5$. As above, no triangle shares two vertices with the original triangle. Consider the three triangles that share a vertex with the original triangle, and the sides $a_i$ as above. As this is a partition, we know the $a_i$'s intersect either on the sides of the large triangle or outside it. We get four cases depending on how they intersect.\n\n- If all three pairs intersect outside of the original triangle, the remaining shape is a convex hexagon.\n- If two pairs intersect outside of the original triangle, and one pair on a side of the triangle, we get a convex pentagon.\n- If two pairs intersect on the sides of the triangle, and one pair on the side of the triangle, we get an isosceles trapezoid.\n- If all pairs intersect on the sides of the triangle, we get an equilateral triangle.\n\nWe are to split the remaining convex shape—a hexagon, a pentagon, an isosceles trapezoid, or a triangle—into two triangles. The only convex shape split up into two equilateral triangles is the rhombus with one angle of $60^\\\text{o}$. We conclude that no partition is possible for $n = 5$.\n\nWe have shown that the equilateral triangle can be partitioned into $n$ parts for all positive integers except $n \\in \\{2, 3, 5\\}$.\n\n$\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16361,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the value of $2 + 2 \\times 3^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "The value is $2 + 2 \\times 9 = 2 + 18 = 20$.\n\n$$20$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16362,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : (0, \\infty) \\to (0, \\infty)$ such that\n\n$$\nf(f(x) + y) = x f(1 + x y),\n$$\n\nfor all $x, y$ in $(0, \\infty)$.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that $f(x) = \\frac{1}{x}$ is the only solution. We proceed in several steps.\n\n**Step 1:** $f(x)$ is non-increasing. Suppose for some $0 < a < b$ that $f(a) < f(b)$. Define\n\n$$\nw = \\frac{b f(b) - a f(a)}{b - a}\n$$\n\nwhich is in $(0, \\infty)$. Since $f(b) > f(a)$, $w > f(b) > f(a)$. Take $x = a$, $y = w - f(a)$:\n\n$$\nf(w) = a f\\left(1 + a (w - f(a))\\right).\n$$\n\nSimilarly, $x = b$, $y = w - f(b)$ gives\n\n$$\nf(w) = b f\\left(1 + b (w - f(b))\\right).\n$$\n\nThus $a = b$, a contradiction. Therefore, $a < b$ implies $f(b) \\leq f(a)$.\n\n**Step 2:** Take $x = y = 1$:\n\n$$\nf(f(1) + 1) = f(2).\n$$\n\nTake $x = 1$, $y = 2$:\n\n$$\nf(f(1) + 2) = f(3).\n$$\n\nTake $x = 2$, $y = 1$:\n\n$$\nf(f(2) + 1) = 2 f(3).\n$$\n\nBut $f(f(2) + 1) = f(f(f(1) + 1) + 1)$. Also,\n\n$$\n\\begin{aligned}\n2 f(3) &= f(f(2) + 1) = f(f(f(1) + 1) + 1) \\\\\n&= (f(1) + 1) f(1 + (f(1) + 1)) = (f(1) + 1) f(f(1) + 2) = (f(1) + 1) f(3).\n\\end{aligned}\n$$\n\nSo $f(1) + 1 = 2$, hence $f(1) = 1$.\n\n**Step 3:** Suppose $x > 1$, and let $y = 1 - \\frac{1}{x}$. Then\n\n$$\nf\\left(f(x) - \\frac{1}{x} + 1\\right) = x f(x).\n$$\n\nIf $f(x) > \\frac{1}{x}$, then $f(x) - \\frac{1}{x} + 1 > 1$ and monotonicity gives\n\n$$\nf\\left(f(x) - \\frac{1}{x} + 1\\right) \\leq f(1) = 1.\n$$\n\nSo $x f(x) \\leq 1$, contradicting $f(x) > \\frac{1}{x}$.\n\nIf $f(x) < \\frac{1}{x}$, then $f(x) - \\frac{1}{x} + 1 < 1$, so\n\n$$\nf\\left(f(x) - \\frac{1}{x} + 1\\right) \\geq f(1) = 1,\n$$\n\nso\n\n$$\n1 \\leq x f(x) < 1,\n$$\nwhich is impossible. Thus $f(x) = \\frac{1}{x}$ for all $x > 1$.\n\nNow, for any $x > 0$, $f(x) + 1 > 1$, so\n\n$$\nf(f(x) + 1) = \\frac{1}{f(x) + 1}.\n$$\n\nBut from the original equation with $y = 1$:\n\n$$\nf(f(x) + 1) = x f(1 + x) = \\frac{x}{1 + x},\n$$\n\nwhere we used $f(1 + x) = \\frac{1}{1 + x}$ for $x > 0$. Thus,\n\n$$\n\\frac{1}{f(x) + 1} = \\frac{x}{1 + x},\n$$\n\nso $f(x) = \\frac{1}{x}$ for all $x > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16363,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n \\ge 3$ children standing in a circle. Each child has two cards: one with the digit $0$ and one with the digit $1$. At a certain moment, each child raises one of their cards at their discretion. Then, every minute, each child whose card number is different from the numbers on both of their neighbors' cards (on the left and on the right) changes their card. Can the situation last indefinitely, with at least one child changing their card?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, for even $n$, and no, for odd $n$.\n\nFor even $n$, at the beginning, the children can alternate their cards, starting with $0$ and $1$. Then, every minute, they continue to flip the cards alternately, and this will go on forever.\n\nFor odd $n$, such an alternating distribution is not possible. If two identical digits are next to each other, they will remain so forever. Such cards are called *stable*. All unstable cards change every minute. Consider any group of stable digits that are next to each other, and the group of unstable cards adjacent to it. The unstable group consists of digits that alternate: $0-1-0-1-\\dots$. Such a group is adjacent to a stable group on each side, and therefore loses two end elements that are added to the stable group. Thus, all unstable groups will disappear after some time, and the process will stop.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16364,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider $N$ boys arranged in a circle. For which integers $b$ is it possible to arrange the boys so that exactly $b$ of them are \"tall\" (where a boy is called \"tall\" if he is taller than both his immediate neighbors)?\n\n![](images/Belorusija_2012_p33_data_9a865aff70.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Any integer from $1$ to $\\left\\lfloor \\frac{N}{2} \\right\\rfloor$ is possible.\n\nArrange the boys in a circle and assign a \"+\" before a boy if he is taller than the previous boy (clockwise), and a \"-\" if he is shorter. A boy is \"tall\" if a \"+\" stands before him and a \"-\" after him. The number of such alternations is at most $\\left\\lfloor \\frac{N}{2} \\right\\rfloor$.\n\nTo achieve exactly $b$ tall boys for any $1 \\leq b \\leq \\left\\lfloor \\frac{N}{2} \\right\\rfloor$, partition the boys by height into three groups: $A$ (the $b$ shortest), $B$ (the $b$ tallest), and $C$ (the rest). Place boys from $A$ and $B$ alternately around the circle, filling remaining spots with $C$. This arrangement yields exactly $b$ tall boys.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16365,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $T$ be a set of 2005 coplanar points with no three collinear. Show that, for any of the 2005 points, the number of triangles it lies strictly within, whose vertices are points in $T$, is even.",
    "options": [],
    "answer": "See solution",
    "solution": "Pick a point in $T$ and call it $P$. Let $S$ be a set of four other points in $T$. Then the points of $S$ must either form a convex quadrilateral, or one point must be inside the triangle formed by the other points. Considering the two different diagrams below, we see that, for each configuration and wherever $P$ is in relation to these four points (remembering that it is not collinear with any pair of points in $S$), $P$ is contained in either zero or two triangles formed by the points of $S$.\n\n![](images/V_Britanija_2006_p12_data_4bd7868082.png)\n\nNow if we look at all $\\binom{2001}{4}$ possible sets $S$, and add up the number of triangles formed from points in each set that contain $P$, we are adding up even numbers, and so the result is even. Let us call this result $k$. However, we have counted each triangle which contains $P$ precisely 2001 times, because given three points, there is a choice of 2001 other points to make up a set of size four. So the total number of triangles that contain $P$ is $k/2001$, so since $k$ is even, but 2001 is odd, $P$ is contained in an even number of triangles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16366,
    "subject": "Mathematics (Olympiad)",
    "question": "Let us analyze how the moves affect certain sums. Denote by $a_k$ the numbers written at a certain moment, so that $a_1$ is the number written in the place where the one was written before any move was made, and $a_2, \\dots, a_{300}$ are the numbers written clockwise from $a_1$.\n\nAfter applying the first type of move, the numbers written are\n\n$$\n\\begin{align*}\nb_1 &= a_1 - a_{300} - a_2, \\\\\nb_k &= a_k - a_{k-1} - a_{k+1}, \\quad k = 2, \\dots, 299, \\\\\nb_{300} &= a_{300} - a_{299} - a_1.\n\\end{align*}\n$$\n\na) Observe how the sum of all written numbers changes after each move. If the total sum of numbers $a_k$ is\n\n$$\nS = a_1 + a_2 + \\cdots + a_{300},\n$$\n\nthen what is the sum of numbers $b_k$ after the move?",
    "options": [],
    "answer": "See solution",
    "solution": "The sum of the numbers $b_k$ is:\n\n$$\na_1 - a_{300} - a_2 + a_2 - a_1 - a_3 + \\cdots + a_{300} - a_{299} - a_1\n$$\n\nEach $a_k$ appears once with a plus sign and twice with a minus sign, so the total sum is $-S$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16367,
    "subject": "Mathematics (Olympiad)",
    "question": "A triad of three distinguished elements of $Q_n$ is called *good* if there exists at least one $i \\in \\{1, 2, \\ldots, n\\}$ for which the sets $\\{x_i, y_i, z_i\\}$ and $\\{0, 1, 2\\}$ are equal. A subset $A$ of $Q_n$ is called *good* if every three elements of $A$ form a good triad. Prove that every good subset of $Q_n$ has at most $2\\left(\\frac{3}{2}\\right)^n$ elements.",
    "options": [],
    "answer": "See solution",
    "solution": "We will use induction with respect to $n$. The case for $n = 1$ is obvious. Suppose that every good subset of $Q_{n-1}$ has at most $2\\left(\\frac{3}{2}\\right)^{n-1}$ elements.\n\nLet $A_0 = \\{(x_1, \\ldots, x_n) \\in A : x_n \\neq 0\\}$. We define the subsets $A_1, A_2$ similarly:\n\n$$\nA_1 = \\{(x_1, \\ldots, x_n) \\in A : x_n \\neq 1\\}, \\quad A_2 = \\{(x_1, \\ldots, x_n) \\in A : x_n \\neq 2\\}.\n$$\n\nSince $A$ is a good set and $A_0$ is its subset, it follows that $A_0$ is also good. This means that for every three elements of $A_0$, there exists a coordinate which is different for every two of them. This coordinate cannot be the last one because $0$ cannot be there. Therefore, the set $A_0'$ produced from the elements of $A_0$ by deleting the last coordinate is a good subset of $Q_{n-1}$.\n\nMoreover, we observe that if $|A_0| \\geq 3$, then $|A_0'| = |A_0|$.\n\nIn fact, if $|A_0'| \\neq |A_0|$, then there would exist an element $a \\in A_0'$ such that $x, y \\in A_0$, where $x, y$ arise from $a$ by adding $1$ and $2$, respectively, as the last coordinate. However, if $z$ is any other element of $A_0$, it cannot have $0$ as the last coordinate, and so $x, y, z$ will not form a good triad, which is a contradiction. Hence, from the induction hypothesis we have\n\n$$\n|A_0| \\leq \\max\\{2, |A_0'|\\} \\leq 2\\left(\\frac{3}{2}\\right)^{n-1}.\n$$\n\nSimilarly, we get that $|A_1|, |A_2| \\leq 2\\left(\\frac{3}{2}\\right)^{n-1}$. Since every element of $A$ appears exactly in two of the sets $A_0, A_1, A_2$, we conclude:\n\n$$\n|A| = \\frac{1}{2}(|A_0| + |A_1| + |A_2|) \\leq 2\\left(\\frac{3}{2}\\right)^n.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16368,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(x) \\in \\mathbb{Z}[x]$ be a monic polynomial and $d$ be a positive integer such that $d \\mid \\deg P$. Show that there exist a natural number $N$ and some polynomials $T(x)$ and $R(x)$ with integer coefficients such that\n\n$$\nN P(x) = (T(x))^d + R(x),\n$$\n\nand for sufficiently large values of $x$,\n\n$$\n(T(x))^d \\leq N P(x) \\leq (T(x) + 1)^d.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof.* Suppose that $\\deg P = d m$ for some $m \\in \\mathbb{N}$. First, we will find a polynomial $T_1(x) \\in \\mathbb{Q}[x]$ such that $\\deg (P(x) - (T_1(x))^d) < (m-1)d$. To do this, assume that\n\n$$\nP(x) = x^{md} + a_{md-1} x^{md-1} + \\dots + a_1 x + a_0.\n$$\n\nWe must find rational numbers $b_0, b_1, \\dots, b_{m-1}$ such that for each natural number $m(d-1) \\leq i \\leq md$, the coefficient of $x^i$ in $(x^m + b_{m-1} x^{m-1} + \\dots + b_1 x + b_0)^d$ is $a_i$. We can find $b_i$'s recursively as rational functions of $a_i$'s. Now, let $n$ be the least common multiple of denominators of the coefficients of $T_1$. Define $T_2(x) = n T_1(x)$, $N = n^d$ and $S_2(x) = N P(x) - (T_2(x))^d$. Obviously, $S_2(x), T_2(x) \\in \\mathbb{Z}[x]$ and $\\deg S_2 > (m-1)d$. If the leading coefficient of $S_2$ is positive, set $T(x) = T_2(x)$ and $R(x) = S_2(x)$. Otherwise, if the leading coefficient of $S_2$ is negative, set $T(x) = T_2(x) - 1$ and $S(x) = N P(x) - (T(x))^d$. It is easy to check that these polynomials have the desired properties. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16369,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that $a^{2020} + 10a^{1010} + 1001$ is prime for no integers $a$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $1001 = 11 \\cdot 91$. If $11 \\mid a$, then $11 \\mid a^{2020} + 10a^{1010} + 1001$. Otherwise, consider $a^5$ modulo $11$. A case study shows that if $a$ is congruent to $1$, $3$, $4$, $5$, or $9$, then $a^5 \\equiv 1$, and in all other cases, $a^5 \\equiv -1$. Hence $a^{10} \\equiv 1 \\pmod{11}$ whenever $11 \\nmid a$. This implies $a^{1010} \\equiv 1 \\pmod{11}$ and $a^{2020} \\equiv 1 \\pmod{11}$ because $a^{1010} = (a^{10})^{101}$ and $a^{2020} = (a^{10})^{202}$. Consequently, $10a^{1010} \\equiv 10 \\pmod{11}$, so $a^{2020} + 10a^{1010} + 1001 \\equiv 0 \\pmod{11}$. Thus $a^{2020} + 10a^{1010} + 1001$ cannot be prime since $a^{2020} + 10a^{1010} + 1001 \\geq 1001 > 11$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16370,
    "subject": "Mathematics (Olympiad)",
    "question": "In a convex quadrilateral $ABCD$, let $I$ and $J$ be the incenters of $\\triangle ABC$ and $\\triangle ADC$, respectively. Assume that $IJ$, $AC$, and $BD$ meet at point $P$. The line through $P$ perpendicular to $BD$ meets the exterior angle bisector of $\\angle BAD$ at $E$, and meets the exterior angle bisector of $\\angle BCD$ at $F$. Prove that $PE = PF$.\n\n![](images/CHN_TSExams_2022_3_p4_data_15de662ccd.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof 1:** If $AB \\parallel CD$ and $AD \\parallel BC$, then $ABCD$ is a parallelogram. In this case, $P$ is the midpoint of $AC$ and $AE \\parallel CF$. So $PE = PF$.\n\nWe may assume that $AB$ is not parallel to $CD$ from now on. We first show that $AB + AD = CB + CD$. As shown in the following picture, we may assume that the extensions of $BA$ and $CD$ meet at point $T$. Let $\\odot K$ be the excircle of $\\triangle TBC$ at $\\angle B$, meeting the lines $AB$, $BC$, $CD$ at $X$, $Y$, $Z$, respectively.\n\n![](images/CHN_TSExams_2022_3_p4_data_bcf3de2074.png)\n\nSince the internal homothetic center of $\\odot I$ and $\\odot J$ lies on the line $IJ$, and since $AC$ is an inner tangent of the two circles, we deduce that the inner homothetic center of $\\odot I$ and $\\odot J$ is $P$.\n\nIt is well known that the internal homothetic center $P$ of $\\odot I$ and $\\odot J$, the external homothetic center $B$ of $\\odot I$ and $\\odot K$, and the internal homothetic center of $\\odot J$ and $\\odot K$ are collinear. So the internal homothetic center of $\\odot J$ and $\\odot K$ lies on the line $BP$. Since $CD$ is an internal tangent line of $\\odot J$ and $\\odot K$, the internal homothetic center of $\\odot J$ and $\\odot K$ is precisely point $D$. Moreover, $AD$ is tangent to $\\odot K$, say at point $W$.\n\nFrom this, we deduce that\n\n$$\n\\begin{aligned}\nAB + AD &= (BX - AX) + (AW - DW) = BX - DW \\\\\n&= BY - DZ = (CB + CY) - (CZ - CD) = CB + CD.\n\\end{aligned}\n$$\n\nIn other words,\n\n$$\nAB + AD = CB + CD. \\tag{1}\n$$\n\nLet $\\odot U$ and $\\odot V$ denote the excircles of $\\triangle ABD$ at $\\angle ABD$ and $\\angle ADB$, respectively. Let $\\odot Q$ and $\\odot R$ denote the excircles of $\\triangle BCD$ at $\\angle CBD$ and $\\angle BDC$, respectively.\n\nNow, we prove that $UR$ and $VQ$ meet at $P$. Consider $\\odot K$, $\\odot U$, and $\\odot R$. It is well known that the external homothetic center $A$ of $\\odot K$ and $\\odot U$, the internal homothetic center $C$ of $\\odot K$ and $\\odot R$, and the internal homothetic center of $\\odot U$ and $\\odot R$ are collinear. Yet $BD$ is the internal tangent line of $\\odot U$ and $\\odot R$; so $P$ is the internal homothetic center of $\\odot U$ and $\\odot R$. It follows that $UR$ passes through $P$. By a similar argument, $VQ$ passes through $P$.\n\n![](images/CHN_TSExams_2022_3_p5_data_72484e9d18.png)\n\nNext, we prove that $UQ \\perp BD$ and $VR \\perp BD$. Extend $BD$ to intersect $\\odot U$ and $\\odot Q$ at $L$ and $L'$, respectively. Then\n\n$$\nBL = \\frac{1}{2}(AB + AD + BD), \\quad BL' = \\frac{1}{2}(CB + CD + BD).\n$$\n\nCombining this with (1), we deduce that $BL = BL'$, i.e. $L$ and $L'$ coincide. So $UQ \\perp BD$. Similarly, $VR \\perp BD$.\n\nSince $EF \\perp BD$, so $UQ \\parallel EF \\parallel VR$. Thus,\n\n$$\n\\frac{PE}{UQ} = \\frac{VP}{VQ} = \\frac{RF}{RQ} = \\frac{PF}{UQ},\n$$\n\nSo $PE = PF$. We are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16371,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P_1(x)$ and $P_2(x)$ be monic quadratic polynomials (that is, quadratic polynomials with leading coefficient $1$). Let points $A_1$ and $A_2$ be the vertices of the parabolas $y = P_1(x)$ and $y = P_2(x)$, respectively. For a function $g(x)$, let $m(g(x))$ denote its minimal value. It is given that the differences $m(P_1(P_2(x))) - m(P_1(x))$ and $m(P_2(P_1(x))) - m(P_2(x))$ are equal positive real numbers. Find the angle between the line $A_1A_2$ and the coordinate axis $Ox$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let the given quadratics be $P_1(x) = (x - x_1)^2 + y_1$ and $P_2(x) = (x - x_2)^2 + y_2$, where $A_1(x_1, y_1)$ and $A_2(x_2, y_2)$ are the coordinates of the vertices. Then $m(P_1(x)) = y_1$.\n\nNow, $P_1(P_2(x)) = ((x - x_2)^2 + y_2 - x_1)^2 + y_1$. If $y_2 \\leq x_1$, then the minimal value of $((x - x_2)^2 + y_2 - x_1)^2$ is $0$, so $m(P_1(P_2(x))) - m(P_1(x)) = y_1 - y_1 = 0$, which contradicts the condition that the difference is positive. Thus, $y_2 > x_1$, so $m(P_1(P_2(x))) = (y_2 - x_1)^2 + y_1$ and $m(P_1(P_2(x))) - m(P_1(x)) = (y_2 - x_1)^2$.\n\nSimilarly, $y_1 > x_2$ and $m(P_2(P_1(x))) - m(P_2(x)) = (y_1 - x_2)^2$.\n\nThe condition of equality gives $(y_1 - x_2)^2 = (y_2 - x_1)^2$. Since $y_2 > x_1$ and $y_1 > x_2$, we have $y_1 - x_2 = y_2 - x_1$, so $y_2 - y_1 = -(x_2 - x_1)$. Therefore, the angle between $A_1A_2$ and the $Ox$ axis is $45^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16372,
    "subject": "Mathematics (Olympiad)",
    "question": "一整數子集 $S$ 被稱為是生根集，若且唯若對於任意 $n$ 和任意 $a_0, a_1, \\dots, a_n \\in S$（$a_0, \\dots, a_n$ 不必相異），多項式 $a_0 + a_1x + \\dots + a_nx^n$ 的所有整數根也都在 $S$ 中。\n\n試找出包含所有 $2^a - 2^b$ 形式整數（其中 $a, b$ 為正整數）的所有生根集 $S$。",
    "options": [],
    "answer": "See solution",
    "solution": "集合 $\\mathbb{Z}$（所有整數）是唯一包含所有 $2^a - 2^b$ 形式整數的生根集。\n\n顯然 $\\mathbb{Z}$ 是生根集，且包含所有 $2^a - 2^b$ 形式的數，其中 $a, b$ 為正整數。\n\n要證明 $S$ 包含所有整數，注意到 $0 = 2^1 - 2^1$，$2 = 2^2 - 2^1$，而 $-1$ 是 $2x + 2$ 的根，因此 $-1 \\in S$。$1$ 是 $2x^2 - x - 1$ 的根，因此 $1 \\in S$。\n\n若 $k \\in S$，則 $-k$ 是 $x + k$ 的根，因此 $-k \\in S$。因此只需證明所有正整數 $n$ 也屬於 $S$。\n\n**解法一**：用歸納法證明所有正整數 $k \\in S$。假設 $0, 1, \\dots, k-1 \\in S$，令 $K = 2^a - 2^b$，其中 $a > b$ 且 $2^a \\equiv 2^b \\pmod{k}$。則 $K$ 是 $k$ 的倍數，因此可寫成：\n\n$$\nK = a_n k^n + a_{n-1} k^{n-1} + \\dots + a_1 k\n$$\n\n其中 $0 \\leq a_i \\leq k-1$。因此 $k$ 是\n\n$$\na_n k^n + a_{n-1} k^{n-1} + \\dots + a_1 k - K\n$$\n\n的根，故 $k \\in S$。\n\n**解法二**：對每個 $k > 2$，希望找到 $a_i, b_i$，使得\n\n$$\n(2^{a_n} - 2^{b_n})k^n + (2^{a_{n-1}} - 2^{b_{n-1}})k^{n-1} + \\dots + (2^{a_0} - 2^{b_0}) = 0\n$$\n\n等價於\n\n$$\n2^{a_n} k^n + 2^{a_{n-1}} k^{n-1} + \\dots + 2^{a_0} = 2^{b_n} k^n + 2^{b_{n-1}} k^{n-1} + \\dots + 2^{b_0}\n$$\n\n因此，我們要證明 $2^{a_n}k^n + 2^{a_{n-1}}k^{n-1} + \\dots + 2^{a_0}$ 有兩種不同的表示法但值相同。\n\n假設 $1 \\leq a_i \\leq 1 + (n-i) \\log_2 k$，則 $2^{a_i} k^i \\leq 2k^n$。這樣的形式至少有 $\\prod_{i=0}^{n-1} (n-i) \\log_2 k \\leq n!(\\log_2 k)^n$ 種，而值至多為 $2k^n(n+1)$。所以對於足夠大的 $n$，必有兩種不同的 $2^{a_n} k^n + 2^{a_{n-1}} k^{n-1} + \\dots + 2^{a_0}$ 取相同值。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16373,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence of $2n + 1$ non-negative integers $a_1, a_2, \\dots, a_{2n+1}$ is given. There is also a sequence of $2n + 1$ consecutive cells, numbered from $1$ to $2n + 1$ from left to right, such that initially the number $a_i$ is written on the $i$-th cell for $i = 1, 2, \\dots, 2n + 1$.\n\nStarting from this initial position, repeat the following steps as long as possible:\n\n1. Add up the numbers written on all the cells; denote the sum as $s$.\n2. If $s = 0$ or $s$ is larger than the current number of cells, the process terminates. Otherwise, remove the $s$-th cell, and shift all cells to the right of it one position to the left. Then go to step 1.\n\nExample: $(1, 0, 1, \\underline{2}, 0) \\to (1, \\underline{0}, 1, 0) \\to (1, \\underline{1}, 0) \\to (\\underline{1}, 0) \\to (0)$.\n\nA sequence $a_1, a_2, \\dots, a_{2n+1}$ of non-negative integers is called *balanced* if, at the end of this process, exactly one cell remains, and it is the cell that was initially numbered $(n+1)$ (the middle cell).\n\nFind the total number of balanced sequences as a function of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is: $C_n \\cdot C_n$, where $C_n = \\binom{2n}{n}$ is the $n$-th Catalan number.\n\nWe divide the proof into several steps. First, some terminology: the last (rightmost) $n$ cells are called the *back* cells, the first (leftmost) $n$ cells are the *front* cells, and the $(n+1)$-st cell is the *middle* cell.\n\n**Claim 1.** All the back cells must be removed before any front cell is removed.\n\n*Proof.* Assume for contradiction that this is not the case. Then there must be a point where a front cell is deleted and then immediately after a back cell is deleted. Let the deleted front cell be at position $i$. All back cells have positions at least $i+2$. After deletion, all back cells have positions at least $i+1$. Since we deleted cell $i$, the total sum is $i$ and does not increase. So at the next step, we delete a cell at position at most $i$, a contradiction. $\\square$\n\n**Claim 2.** The middle cell must contain the number $0$, i.e., $a_{n+1} = 0$.\n\n*Proof.* Consider the last step with two cells: one is the middle cell, and by Claim 1, the other is a front cell, i.e., $(x, a_{n+1})$. On the next move, we remove $x$, so $x + a_{n+1} = 1$. Thus, $a_{n+1} = 0$ or $1$. But after that, we cannot remove $a_{n+1}$, so $a_{n+1} \\neq 1$. Therefore, $a_{n+1} = 0$. $\\square$\n\nDefine a *self-destructing* sequence as one with no surviving cells at the end. For example, $(0, 1, 2)$ is self-destructing because $(0, 1, 2) \\to (0, 1) \\to (1) \\to ()$. Let $S_n$ be the set of self-destructing sequences of length $n$. For example, $S_2 = \\{(0, 1), (1, 1)\\}$. The front cells form a self-destructing sequence, i.e., $(a_1, a_2, \\dots, a_n) \\in S_n$. The back cells also have a self-destructing property, made precise in Claim 3.\n\n**Claim 3.** Fix the front sequence $\\varphi = (a_1, a_2, \\dots, a_n)$. Let $B_\\varphi$ be the set of all possible back sequences of length $n$ that can be appended to $\\varphi$ (with a $0$ between them) to get a balanced sequence. Then there is a bijection $f: S_n \\to B_\\varphi$.\n\n*Proof.* Let $c = n + 1 - \\sum_{i=1}^n a_i$ and consider $\\sigma = (s_1, s_2, \\dots, s_n) \\in S_n$. Let $\\ell$ be the initial index of the last surviving cell in $\\sigma$. Then $f(\\sigma) = (s_1, s_2, \\dots, s_\\ell + c, s_{\\ell+1}, \\dots, s_n)$ defines a bijection $S_n \\to B_\\varphi$.\n\nThe $k$-th deleted cell in $\\sigma$ is the $k$-th deleted cell in $\\overline{\\varphi\\ 0\\ f(\\sigma)}$ for each $k=1, \\dots, n$. After some deletions, let $S$ be the total sum remaining in $\\sigma$. Then the total sum remaining in $\\overline{\\varphi\\ 0\\ f(\\sigma)}$ is $-\\sum_{i=1}^n a_i + 0 + S + c = S + n + 1$. So we delete the cell in position $S$ in $\\sigma$ if and only if we delete the cell in position $S + n + 1$ in $\\overline{\\varphi\\ 0\\ f(\\sigma)}$.\n\nThus, $\\overline{\\varphi\\ 0\\ f(\\sigma)}$ is a balanced sequence: we first eliminate all cells in the back, then the front. Every balanced sequence is of this form. $\\square$\n\nSo the total number of balanced sequences is $|S_n|^2$. It remains to calculate $|S_n|$.\n\n**Claim 4.** Let $\\mathcal{T}_n$ be the set of $2n$-sequences consisting of $n$ zeros and $n$ ones such that in each initial segment, the number of $1$'s does not surpass the number of $0$'s. Then $|S_n| = |\\mathcal{T}_n|$.\n\n*Proof.* Let $[n] = \\{1, 2, \\dots, n\\}$, and let $\\mathcal{F}_n$ be the set of non-decreasing mappings $f: [n] \\to [n]$ such that $f(i) \\le i$ for each $i \\in [n]$. The claim follows once we show $|S_n| = |\\mathcal{F}_n|$ and $|\\mathcal{F}_n| = |\\mathcal{T}_n|$.\n\nThere is a bijection $a \\mapsto f$ between $S_n$ and $\\mathcal{F}_n$. Reversing the self-destructing process for $a = (a_1, a_2, \\dots, a_n) \\in S_n$, define $f(i)$ to be the partial sum of the existing terms after the $i$-th backward step.\n\nFor $|\\mathcal{T}_n| = |\\mathcal{F}_n|$, note the bijection $t \\mapsto f$ between $\\mathcal{T}_n$ and $\\mathcal{F}_n$. Let $f(i)$ equal $1 + \\#(i)$, where $\\#(i)$ is the total number of $1$'s in $t$ before the $i$-th zero.\n\nIt is a known fact that $|\\mathcal{T}_n|$ is the $n$-th Catalan number $C_n = \\frac{1}{n+1} \\binom{2n}{n}$ (by the reflection principle of A. D. André).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16374,
    "subject": "Mathematics (Olympiad)",
    "question": "In the parallelogram $ABCD$ we have $|AB| = |BD|$. Let $K$ be a point on the line $AB$ different from $A$, such that $|KD| = |AD|$. Denote the reflection of the point $C$ over $K$ by $M$, and the reflection of $B$ over $A$ by $N$. Prove that $MDN$ is an isosceles triangle with the apex at $D$.",
    "options": [],
    "answer": "See solution",
    "solution": "Triangle $ADK$ is isosceles with the apex at $D$. So, $\\angle BKD = 180^\\circ - \\angle DKA = 180^\\circ - \\angle KAD = \\angle CBK$ and $|DK| = |DA| = |BC|$. The triangles $DKB$ and $CBK$ are congruent because they have a common side $KB$, congruent angles $\\angle BKD = \\angle CBK$ and $|DK| = |BC|$. Thus, $\\angle DCK = \\angle BKC = \\angle DBK$.\n\nSince $M$ and $N$ are reflections of $C$ and $B$, we have $|CK| = |KM|$ and $|NA| = |AB|$. So,\n\n$$|CM| = 2|CK| = 2|DB| = 2|AB| = |NB|.$$ \n\nThis implies that the triangles $CDM$ and $BDN$ are also congruent since $|DC| = |DB|$, $|CM| = |NB|$ and $\\angle DCM = \\angle DBN$. Thus, $|DN| = |DM|$ and the triangle $DMN$ is isosceles with the apex at $D$.\n\n![](images/Slovenija_2014_p16_data_c1f2a74284.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16375,
    "subject": "Mathematics (Olympiad)",
    "question": "There is an equilateral trapezoid with bases $BC$ and $AD$ and known angles: $\\angle BDC = 10^\\circ$ and $\\angle BDA = 70^\\circ$. Prove that the following equality holds:\n\n$$\nAD^2 = BC(AD + AB)\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "It is clear that $\\angle AMD = 20^\\circ$, $\\angle DBM = 150^\\circ$. Let's construct equilateral triangle $\\Delta KMD$, then points $M$, $B$, $D$ are on the circle centered at $K$ (see figure below).\n\nKM = KB, $\\angle KMB = 80^\\circ$, so $\\angle MBK = 80^\\circ$, and $\\angle MBC = 80^\\circ$ as well, which implies that points $B$, $C$, $K$ are collinear.\n\nTherefore, $\\Delta AMD = \\Delta BKM$, as isosceles triangles with equal sides and base angles. Thus, $MB = AD$. From the similarity $\\Delta MBC \\sim \\Delta MAD$, we have:\n\n$$\n\\frac{BC}{AD} = \\frac{MB}{MA}\n$$\n\nSo $AD \\cdot BM = BC \\cdot MA$, hence:\n\n![](images/UkraineMO2019_booklet_p52_data_41b50932c1.png)\n\n$$\nAD^2 = BC(MB + AB) = BC(AD + AB)\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16376,
    "subject": "Mathematics (Olympiad)",
    "question": "Floor's class consists of 16 students, including Floor. All students took a test with four questions. Every question was worth a (positive) integer number of points. Each question was marked completely right or completely wrong; no partial points were given. The question that was worth the most points was worth exactly 4 points more than the question worth the least points. All students achieved a different score; Floor herself got everything right.\n\nAt least how many points did Floor score?",
    "options": [],
    "answer": "See solution",
    "solution": "$21$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16377,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $G$ be a graph with vertices $a_1, a_2, \\ldots, a_{2009}$. Connect two vertices with an edge if the corresponding numbers are not relatively prime. What is the largest number of vertices that can be selected so that any two selected vertices are relatively prime?",
    "options": [],
    "answer": "See solution",
    "solution": "The resulting graph $G$ does not contain a subgraph with three vertices and a single edge, so its vertices can be partitioned into disjoint independent sets. Thus, $G$ is a $k$-partite graph with 2009 vertices. Each independent set contains at most 49 vertices, so the minimum number of sets is $\\frac{2009}{49} = 41$. By selecting one vertex from each set, we obtain 41 numbers that are pairwise relatively prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16378,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $CB = a$, $AC = b$, $AB = c$, and $A_1B_1 = x$. Consider the configuration shown below:\n\n![](images/BielorrusiaBOOK18-E-A4_p26_data_7b2f66be23.png)\n\nLet $Z_b$ and $Z_a$ be the points of intersection of the lines $B_1Y$ and $A_1X$ respectively with the hypotenuse $AB$. Find the lengths of the segments $BZ_a$ and $AZ_b$, and prove the equality $$AZ_b + Z_aB = AB.$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $AB_1$ and $B_1A_1$ are the diameters of the circumcircles of triangles $AC_1B_1$ and $A_1CB_1$ respectively, $\\angle AYB_1 = \\angle A_1YB_1 = 90^\\circ$. Hence $Y$ belongs to the line $AA_1$ and $AA_1 \\perp B_1Y$. Similarly, $X$ belongs to the line $BB_1$ and $BB_1 \\perp A_1X$.\n\nSince $\\angle B_1XZ_a = \\angle B_1C_1Z_a = 90^\\circ$, the quadrilateral $B_1C_1Z_aX$ is cyclic. The quadrilateral $CA_1XB_1$ is cyclic too, therefore\n\n$$\nBA_1 \\cdot CB = BX \\cdot BB_1 = BZ_a \\cdot BC_1. \\quad (*)\n$$\n\nThe triangle $A_1BD_1$ is similar to the triangle $ABC$, therefore\n\n$$\n\\frac{A_1B}{AB} = \\frac{D_1B}{CB} = \\frac{A_1D_1}{AC}.\n$$\n\nHence $A_1B = \\frac{cx}{b}$, $D_1B = \\frac{ax}{b}$ and $BC_1 = x + \\frac{ax}{b} = \\frac{(a+b)x}{b}$. From $(*)$ it follows that $BZ_a = \\frac{ac}{a+b}$. Similarly one can prove $AZ_b = \\frac{bc}{a+b}$, so $AZ_b + BZ_a = AB$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16379,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that $a, b, c$ belong to $(0, 1]$. Show that\n\n$$\na + b + c + |a - b| + |b - c| + |c - a| \\le \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume $0 < a \\leq b \\leq c \\leq 1$. Then we need to prove\n\n$$\n\\left( \\frac{1}{a} + a \\right) + \\left( \\frac{1}{b} - b \\right) + \\left( \\frac{1}{c} - 3c \\right) \\geq 0.\n$$\n\nFor $a > 0$, we have $a + \\frac{1}{a} \\geq 2$. If $0 < b \\leq 1$, then\n\n$$\n\\frac{1}{b} - b = \\frac{1 - b^2}{b} = \\frac{(1 - b)(1 + b)}{b} \\geq 0.\n$$\n\nAlso,\n\n$$\n\\frac{1}{c} - 3c = -2 + \\frac{(1 - c)(1 + 3c)}{c} \\geq -2.\n$$\n\nAdding the last three inequalities leads to the result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16380,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c \\in \\{0, 1, 2, \\dots, 9\\}$. The quadratic equation $ax^2 + bx + c = 0$ has a rational root. Prove that the three-digit number $abc$ is not a prime number.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $\\overline{abc} = p$ is a prime number, and the roots of $f(x) = ax^2 + bx + c = 0$ are rational. Then $b^2 - 4ac$ is a perfect square, and the roots $x_1, x_2$ are negative. We can write $f(x) = a(x - x_1)(x - x_2)$. \n\nThus, $p = f(10) = a(10 - x_1)(10 - x_2)$. Therefore, $4ap = (20a - 2a x_1)(20a - 2a x_2)$. \n\nSince $x_1, x_2 = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$, both $20a - 2a x_1$ and $20a - 2a x_2$ are integers. Now, $p$ divides $20a - 2a x_1$ or $20a - 2a x_2$. Suppose $p \\mid 20a - 2a x_1$; then $p \\leq 20a - 2a x_1$, and $4a \\geq 20a - 2a x_2$, which contradicts $x_2$ being negative. Thus, $abc$ cannot be prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16381,
    "subject": "Mathematics (Olympiad)",
    "question": "Circles $\\omega_1$ and $\\omega_2$ meet at $P$ and $Q$. Segments $AC$ and $BD$ are chords of $\\omega_1$ and $\\omega_2$ respectively, such that segment $AB$ and ray $CD$ meet at $P$. Ray $BD$ and segment $AC$ meet at $X$. Point $Y$ lies on $\\omega_1$ such that $PY \\parallel BD$. Point $Z$ lies on $\\omega_2$ such that $PZ \\parallel AC$. Prove that points $Q$, $X$, $Y$, $Z$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider the above configuration. (Our proof can be modified for other configurations.) Let segment $AC$ meet the circumcircle of triangle $CQD$ again (other than $C$) at $X_1$.\n\nFirst, we show that $Z$, $Q$, $X_1$ are collinear. Since $CQDX_1$ is cyclic, $\\angle X_1CD = \\angle DQX_1$. Since $AC \\parallel PZ$, $\\angle X_1CD = \\angle ACP = \\angle CPZ = \\angle DPZ$. Since $PDQZ$ is cyclic, $\\angle DPZ + \\angle DQZ = 180^\\circ$. Combining the last three equations, we obtain that\n\n$$\n\\angle DQX_1 + \\angle DQZ = \\angle X_1CD + \\angle DQZ = \\angle DPZ + \\angle DQZ = 180^\\circ;\n$$\n\nthat is, $X_1$, $Q$, $Z$ are collinear.\n\n![](images/USA_IMO_2006-2007_p31_data_55278d56c0.png)\n\nSecond, we show that $B$, $D$, $X_1$ are collinear; that is, $X = X_1$. Since $AC \\parallel PZ$, $\\angle CAP = \\angle ZPB$. Since $BPQZ$ is cyclic, $\\angle BPZ = \\angle BQZ$. It follows that $\\angle X_1AB = \\angle CAP = \\angle BQZ$, implying that $ABQX_1$ is cyclic. Hence $\\angle X_1AQ = \\angle X_1BQ$. On the other hand, since $BPDQ$ and $APQC$ are cyclic,\n\n$$\n\\angle QBD = \\angle QPD = \\angle QPC = \\angle QAC = \\angle QAX_1.\n$$\n\nCombining the last two equations, we conclude that $\\angle X_1BQ = \\angle X_1AQ = \\angle QBD$, implying that $X_1$, $D$, $B$ are collinear. Since $X_1$ lies on segment $AC$, it follows that $X = X_1$. Therefore, we established the fact that $Z$, $Q$, $X$ are collinear.\n\n![](images/USA_IMO_2006-2007_p32_data_32a6406e8a.png)\n\nTo finish our proof, we show that $Y$, $X$, $Q$ are collinear. Since $ABQX$ is cyclic, $\\angle BAQ = \\angle BXQ$. Since $APQY$ is cyclic, $\\angle BAQ = \\angle PAQ = \\angle PYQ$. Hence $\\angle PYQ = \\angle BAQ = \\angle BXQ$. Since $BX \\parallel PY$ and $\\angle BXQ = \\angle PYQ$, we must have $Y$, $X$, $Q$ collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16382,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(xy) = y f(x) + x + f(f(y) - f(x))\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, set $(x, y) = (1, 1)$:\n$$\nf(1) = 1 f(1) + 1 + f(f(1) - f(1)) \\implies f(1) = f(1) + 1 + f(0) \\implies f(0) = -1.\n$$\n\nNext, set $y = 1$:\n$$\nf(x) = 1 f(x) + x + f(f(1) - f(x)) \\implies x = -f(f(1) - f(x)) \\implies f(f(1) - f(x)) = -x.\n$$\nSo $f$ is surjective.\n\nNow, set $(x, y) = (a, 0)$ and $(0, a)$:\n$$\nf(0) = 0 f(a) + a + f(f(0) - f(a)) \\implies -1 = a + f(-1 - f(a)),\n$$\n$$\nf(0) = a f(0) + 0 + f(f(a) - f(0)) \\implies -1 = -a + f(f(a) + 1).\n$$\nSince $f$ is surjective, for any $z$ we can write $z = f(a) + 1$. Adding the two results:\n$$\nf(z) + f(-z) = -2.\n$$\n\nNow, set $(x, y) = (a, 1)$ and $(1, a)$:\n$$\nf(a) = 1 f(a) + a + f(f(1) - f(a)) \\implies f(a) = f(a) + a + f(f(1) - f(a)) \\implies f(f(1) - f(a)) = -a,\n$$\n$$\nf(a) = a f(1) + 1 + f(f(a) - f(1)).\n$$\nAdding these and using $z = f(a) - f(1)$:\n$$\nf(a) = a f(1) + a - 1.\n$$\nSo $f(x) = kx - 1$ for some constant $k$. Substitute into the original equation:\n$$\nf(xy) = kxy - 1,\n$$\n$$\ny f(x) + x + f(f(y) - f(x)) = y(kx - 1) + x + f(k y - 1 - (k x - 1)) = y k x - y + x + f(k(y - x)).\n$$\nSet $f(z) = k z - 1$:\n$$\ny k x - y + x + k (k(y - x)) - 1 = k x y - y + x + k^2(y - x) - 1.\n$$\nSet equal to $k x y - 1$:\n$$\nk x y - 1 = k x y - y + x + k^2(y - x) - 1.\n$$\nSo $-y + x + k^2(y - x) = 0$ for all $x, y$. This gives $k^2 = 1$, so $k = 1$ or $k = -1$.\n\nThus, the solutions are $f(x) = x - 1$ and $f(x) = -x - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16383,
    "subject": "Mathematics (Olympiad)",
    "question": "We consider an $n \\times n$ chessboard, where $n$ is an even positive integer. On the board we put all numbers $1, 2, 3, \\dots, n^2$, one at each square. Let $S_1$ be the sum of the numbers lying at the white squares and let $S_2$ be the sum of the numbers lying on the black squares. Find all the numbers $n$ for which it is possible to arrange the numbers such that:\n\n$$\n\\frac{S_1}{S_2} = \\frac{39}{64}.\n$$\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The given relation is equivalent to $S_1 = \\frac{39}{103}(S_1 + S_2)$. Since\n\n$$\nS_1 + S_2 = 1 + 2 + \\dots + n^2 = \\frac{n^2(n^2+1)}{2}\n$$\n\nand $S_1$ is a natural number, we conclude that $103 \\mid \\left\\lfloor \\frac{n^2(n^2+1)}{2} \\right\\rfloor$. Since 103 is a prime number of the form $4\\kappa+3$, it follows that 103 does not divide $n^2+1$. Hence, the prime number 103 must divide $n^2$ and finally $103 \\mid n$. Moreover, $n$ is even, so $n = 206k$, $k \\in \\mathbb{N}^*$.  \n\nNow we are going to prove that for every number $n = 206k$, $k \\in \\mathbb{Z}$, it is possible to obtain the wanted arrangement. The least (maximal) possible value of $S_1$ is:\n\n$$\n1+2+\\cdots+\\frac{n^2}{2} = \\frac{\\frac{n^2}{2}\\left(\\frac{n^2}{2}+1\\right)}{2} = A, \\quad \\left(\\frac{n^2}{2}+1\\right)+\\left(\\frac{n^2}{2}+2\\right)+\\cdots+n^2 = B,\n$$\n\nrespectively.\n\nEasily we can check the inequality $A < S_1 = \\frac{39}{103}(S_1 + S_2) < B$. Our result arrives by proving that $S_1$ can obtain every possible value between $A$ and $B$. In fact, number $A+1$ can be obtained by putting in the white squares the numbers $1, 2, \\dots, \\frac{n^2}{2}-1, \\frac{n^2}{2}+1$. The number $A+2$ can be obtained by putting in the white squares the numbers $1, 2, \\dots, \\frac{n^2}{2}-1, \\frac{n^2}{2}+2$, and so on. When we arrive at the step where we need to put the numbers $1, 2, \\dots, \\frac{n^2}{2}-1, \\frac{n^2}{2}$, in order to obtain the next number for their sum, then we select in the white squares the numbers $1, 2, \\dots, \\frac{n^2}{2}-2, \\frac{n^2}{2}, n^2$, and for the next number we select $1, 2, \\dots, \\frac{n^2}{2}-2, \\frac{n^2}{2}+1, n^2$, and so on. Hence, using the above procedure to increase at every step the sum by 1, starting from $A$, we can construct all integers up to $B$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16384,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle with circumcircle $(O)$. Let $H$ be the orthocenter, and $BE$, $CF$ the altitudes of triangle $ABC$. Suppose $AH$ meets $(O)$ again at $D$ (with $D \\neq A$).\n\n1. Let $I$ be the midpoint of $AH$. The line $EI$ meets $BD$ at $M$, and $FI$ meets $CD$ at $N$. Prove that $MN$ is perpendicular to $OH$.\n\n2. The lines $DE$ and $DF$ meet $(O)$ again at $P$ and $Q$ respectively ($P, Q \\neq D$). The circle $(AEF)$ meets $(O)$ and $AO$ again at $R$ and $S$ respectively ($R, S \\neq A$). Prove that $BP$, $CQ$, and $RS$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "1) Let $J$ be the center of the nine-point circle of triangle $ABC$. Then $(J)$ passes through $E$, $I$, $F$, and $J$ is also the midpoint of $OH$. It is easy to see that $D$ and $H$ are symmetric with respect to $BC$, so triangle $BDH$ is isosceles with $BD = BH$. Since triangle $IEH$ has $IE = IH$, we have\n\n$$\n\\angle IEH = \\angle IHE = \\angle BHD = \\angle BDH,\n$$\n\nwhich implies that $BDEI$ is a cyclic quadrilateral. Since $DB$ meets $EI$ at $M$, we have\n\n$$\n\\overline{ME} \\cdot \\overline{MI} = \\overline{MB} \\cdot \\overline{MD}.\n$$\n\n![](images/vn-booklet_final_p13_data_01f30d5d32.png)\n\nThus, the power of $M$ to circles $(J)$ and $(O)$ are equal. Similarly, the power of $N$ to $(J)$ and $(O)$ are also equal. So $MN$ is the radical axis of $(O)$ and $(J)$, thus $MN \\perp OJ$. But $O$, $H$, $J$ are collinear, so $MN \\perp OH$.\n\n2) Let $X$ be the midpoint of $EF$, and $K$ the intersection of $AH$ and $BC$. It is easy to see that triangles $BFE$ and $KHE$ are similar, which implies triangles $BFX$ and $DHE$ are also similar, so $\\angle FBX = \\angle HDE = \\angle FBP$. Thus, $B$, $X$, $P$ are collinear; similarly, $C$, $X$, $Q$ are collinear.\n\n![](images/vn-booklet_final_p13_data_3ccabaf021.png)\n\nLet $AL$ be the diameter of $(O)$. Then $SH$ passes through $L$, and quadrilateral $HBLC$ is a parallelogram, so $HL$ passes through the midpoint $M$ of $BC$. It is easy to check that triangles $SEC$ and $SFB$ are similar, so triangles $SEF$ and $SCB$ are also similar. These triangles have medians $SX$ and $SM$ respectively, so $\\angle FSX = \\angle BSM$. Also, triangles $SFB$ and $SRL$ are similar, so triangles $SFR$ and $SBL$ are also similar. Thus,\n\n$$\n\\angle FSR = \\angle BSL = \\angle BSM = \\angle FSX.\n$$\n\nTherefore, $S$, $X$, $R$ are collinear, so $SR$ passes through $X$. Thus, $BP$, $CQ$, and $RS$ are concurrent at the midpoint $X$ of $EF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16385,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p_1 < p_2 < p_3 < \\dots$ be the sequence of all prime numbers, and denote $s_n = a_1 + a_2 + \\dots + a_n$.\n\nFind the value of $a_{100}$, where $a_n$ is defined such that $a_1 = 1$, $a_2 = 2$, $a_3 = 3$, and for each $n \\geq 4$, $a_n$ is the greatest prime divisor of $s_{n-1}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that $a_3 = 3$, so $s_3 = 1 + 2 + 3 = 6 = 2 \\cdot 3 = p_1 p_2$.\n\nSuppose for some $n$ we have $s_n = p_k p_{k+1}$. The greatest prime divisor of $s_n$ is $p_{k+1}$, so $a_{n+1} = p_{k+1}$ and $s_{n+1} = s_n + a_{n+1} = p_k p_{k+1} + p_{k+1} = (p_k + 1)p_{k+1}$.\n\nThis pattern continues until $p_k + j = p_{k+2}$, at which point $s_{n+j} = p_{k+1} p_{k+2}$, and the cycle repeats.\n\nFrom $s_3 = 2 \\cdot 3$, we get:\n\n$$\n\\begin{align*}\ns_6 &= 3 \\cdot 5, & s_{10} &= 5 \\cdot 7, & s_{16} &= 7 \\cdot 11, & s_{22} &= 11 \\cdot 13, \\\\\ns_{28} &= 13 \\cdot 17, & s_{34} &= 17 \\cdot 19, & s_{40} &= 19 \\cdot 23, & s_{50} &= 23 \\cdot 29, \\\\\ns_{58} &= 29 \\cdot 31, & s_{66} &= 31 \\cdot 37, & s_{76} &= 37 \\cdot 41, & s_{82} &= 41 \\cdot 43, \\\\\ns_{88} &= 43 \\cdot 47, & s_{98} &= 47 \\cdot 53.\n\\end{align*}\n$$\n\nHence $a_{99} = 53$, $s_{99} = 48 \\cdot 53$, and finally $a_{100} = 53$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16386,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all possible values of integer $k$ for which there exist positive integers $a$ and $b$ such that\n$$\n\\frac{b+1}{a} + \\frac{a+1}{b} = k.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is that $k = 3$ or $4$.\n\nFix a possible value $k$. Among all pairs $(A, B)$ of positive integers satisfying\n$$\n\\frac{B+1}{A} + \\frac{A+1}{B} = k,\n$$\nchoose any $(a, b)$ such that $b$ is the smallest. Then the quadratic equation\n$$\nx^2 + (1 - kb)x + b^2 + b = 0\n$$\nhas an integer root $x = a$. Let $x = a'$ be the second root. It follows from $a + a' = kb - 1$ that $a' \\in \\mathbb{Z}$, and from\n$$\na \\cdot a' = b(b + 1)\n$$\nthat $a' > 0$. Hence,\n$$\n\\frac{b+1}{a'} + \\frac{a'+1}{b} = k.\n$$\nBy the assumption on $b$, we have $a \\geq b$, $a' \\geq b$. So one of $a$ and $a'$ is equal to $b$. Without loss of generality, assume $a = b$, so\n$$\nk = 2 + \\frac{2}{b},\n$$\nwhich implies $b \\mid 2$, i.e., $b = 1$ or $2$, and $k = 4$ or $3$, respectively.\n\nIf $a = b = 1$, then $k = 4$; if $a = b = 2$, then $k = 3$.\n\nConsequently, $k = 3$ or $4$ is the only solution. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16387,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, $C$, and $D$ be four different points lying on a common circle in this order. Assume that the line segment $AB$ is the (only) longest side of the inscribed quadrilateral $ABCD$.\n\nProve that the inequality\n\n$$\nAB + BD > AC + CD\n$$\nholds.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S$ denote the common point of the diagonals, and let $a = AB$ and $c = CD$.\n\nSince $ABCD$ is an inscribed quadrilateral, triangles $ABS$ and $DCS$ are similar. It follows that numbers $r$ and $s$ must exist, such that $AS = sa$, $BS = ra$, $DS = sc$, and $CS = rc$ hold. The inequality under consideration can therefore be written in the form\n\n$$\na + ra + sc > sa + rc + c.\n$$\n\nThis is equivalent to\n\n$$\na(1 + r - s) > c(1 + r - s),\n$$\n\nwhich is certainly correct, since $a > c$ is given and the triangle inequality in $ABS$ implies $1 + r > s$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16388,
    "subject": "Mathematics (Olympiad)",
    "question": "A $2011 \\times 2011$ square grid is divided into triangles by the diagonals of the squares. What is the total number of isosceles triangles in the figure?",
    "options": [],
    "answer": "See solution",
    "solution": "There are two kinds of isosceles triangles in the figure: those with the hypotenuse on a line consisting of the diagonals of the squares and those with the hypotenuse on a line consisting of the sides of the squares.\n\n**Triangles of the first type:**\nThese can be divided into four categories according to their position. For triangles with the right angle at the upper left corner, each subsquare of size $k \\times k$ contains exactly one triangle with leg $k$ of this kind. The number of $k \\times k$ subsquares for $k = 1$ to $k = 2011$ equals:\n\n$$\n\\sum_{k=0}^{2010} (k+1)^2 = 1^2 + 2^2 + \\dots + 2011^2 = \\frac{1}{6} (2011 \\cdot 2012 \\cdot 4023)\n$$\n\nThe number of triangles of the first type is $4$ times the previous number:\n\n$$\n2 \\cdot 2011 \\cdot 2012 \\cdot 1341 = 10851726024\n$$\n\n**Triangles of the second type:**\nThese can also be divided into four categories according to the direction of the hypotenuse and the orientation of the right angle. For triangles where the hypotenuse is vertical and the right angle is on the left:\n\nIf the hypotenuse has length $k$ and $k$ is even ($k = 2j$), the number of such triangles is:\n\n$$\n\\begin{aligned}\n\\sum_{j=1}^{1005} (2012-j)(2012-2j) &= 2 \\sum_{j=1}^{1005} (2012-j)(1006-j) = 2 \\sum_{j=1}^{1005} j(1006+j) \\\\\n&= 1006^2 \\cdot 1005 + \\frac{1}{3}(1005 \\cdot 1006 \\cdot 2011) = 1006 \\cdot (1005 \\cdot 1006 + 335 \\cdot 2011) = 1694823290.\n\\end{aligned}\n$$\n\nIf $k$ is odd ($k = 2j+1$), the number of possible triangles is:\n\n$$\n\\begin{aligned}\n\\sum_{j=0}^{1005} (2011-j)(2011-2j) &= 1006 \\cdot 2011^2 - 6033 \\cdot \\frac{1005 \\cdot 1006}{2} + 2 \\cdot \\frac{1005 \\cdot 1006 \\cdot 2011}{6} = 1696340841.\n\\end{aligned}\n$$\n\nSo the total number of triangles of the second type is:\n\n$$\n4 \\cdot (1694823290 + 1696340841) = 24416382548\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16389,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 8 distinct points marked on a circle. Juku wants to draw as many triangles as possible such that all vertices of each triangle are at the marked points, and no two of these triangles share a side. Find the largest number of triangles that can be drawn under these conditions.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume without loss of generality that the points marked on the circle are equally spaced and number the marked points counterclockwise as $0, 1, \\ldots, 7$. Consider a triangle with vertices at points $0, 1, 3$ and its 7 copies obtained by rotating the original triangle counterclockwise by $\\frac{1}{8}, \\frac{2}{8}, \\ldots, \\frac{7}{8}$ of a full turn around the center of the circle. These 8 triangles do not share any sides because all sides of the original triangle have different lengths, and each rotation of a side with a specific length results in different segments. Therefore, it is possible to draw 8 triangles under the given conditions.\n\n![](images/EST_ABooklet_2024_p39_data_b40376690d.png)\n\nOn the other hand, from each marked point, at most 7 segments can be drawn to the remaining marked points. Each triangle uses either 0 or 2 of these segments. Thus, each marked point can be the endpoint of at most 6 different triangle sides in total. Since we count each side twice (once at each endpoint), there can be at most $\\frac{8 \\cdot 6}{2} = 24$ different triangle sides. Since each triangle has 3 sides, there can be at most $\\frac{24}{3} = 8$ triangles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16390,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime, and let $A$ be an infinite set of integers. Prove that there exists a subset $B$ of $A$ with $2p-2$ elements such that the arithmetic mean of any $p$ pairwise distinct elements in $B$ does not belong to $A$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, for contradiction, that there exists an infinite set $A$ of integers such that $A$ does not contain a subset $B$ with $2p-2$ elements satisfying the conditions in the problem. We may assume that $A$ contains infinitely many positive numbers; otherwise, take the negatives of all elements of $A$ to form a new set. For any nonempty subset $B$, the arithmetic mean of its elements lies between the maximal and minimal elements of $B$. Thus, removing every element of $A$ smaller than some positive integer $N$ still yields a counterexample. In particular, we may assume every element of $A$ is a positive integer. Similarly, adding a constant to every element of $A$ preserves the property.\n\n**Lemma:** For any nonnegative integer $k$, there is at most one residue class modulo $p^k$ that contains infinitely many elements of $A$.\n\n**Proof of the Lemma:** Suppose not. Let $k$ be the smallest nonnegative integer such that there are two different residue classes modulo $p^k$, say $r_k$ and $r'_k$, each containing infinitely many elements of $A$. Then there exists $N$ such that all elements of $A_{\\ge N} = \\{a \\in A \\mid a \\ge N\\}$ are congruent to $r_{k-1}$ modulo $p^{k-1}$, and there are infinitely many elements congruent to $r_k$ and $r'_k$ modulo $p^k$. Choose $B$ to be the set consisting of $p-1$ elements of $A_{\\ge N}$ congruent to $r_k$ modulo $p^k$ and $p-1$ elements congruent to $r'_k$ modulo $p^k$. Then, for every $p$ distinct elements of $B$, their arithmetic mean is not congruent to $r_{k-1}$ modulo $p^{k-1}$. This $B$ satisfies the problem's condition, contradicting the assumption that $A$ is a counterexample.\n\nBy the lemma, for each $k \\ge 0$, let $r_k \\bmod p^k$ ($0 \\le r_k \\le p^k-1$) denote the unique residue class modulo $p^k$ containing infinitely many elements of $A$. Then $r_k \\equiv r_{k-1} \\pmod{p^{k-1}}$. We now construct a sequence $a_1, a_2, \\dots$ in $A$ by induction, and show that any $2p-2$ elements of this sequence yield the desired $B$.\n\nLet $N_0 = 0$. Let $k_1$ be the least nonnegative integer such that there exists $a_1 \\in A$ with $a_1 \\not\\equiv r_{k_1} \\pmod{p^{k_1}}$. Then every element of $A$ is congruent to $r_{k_1-1}$ modulo $p^{k_1-1}$. By the lemma, there exists $N_1$ such that every element of $A_{\\ge N_1}$ is congruent to $r_{k_1}$ modulo $p^{k_1}$. Let $k_2$ be the least nonnegative integer such that there exists $a_2 \\in A_{\\ge N_1}$ with $a_2 \\not\\equiv r_{k_2} \\pmod{p^{k_2}}$. Similarly, every element of $A_{\\ge N_1}$ is congruent to $r_{k_2-1}$ modulo $p^{k_2-1}$. Using the lemma, take $N_2$ so that every element of $A_{\\ge N_2}$ is congruent to $r_{k_2}$ modulo $p^{k_2}$. Repeating this, we obtain $a_1 < N_1 \\le a_2 < N_2 \\le \\cdots \\le a_{2p-2}$ such that $p^{k_i-1} \\mid\\mid a_i - r_{k_i}$ for $1 \\le i \\le 2p-2$.\n\nFor convenience, subtract $r_{k_{2p-2}}$ from all elements of $A$ and the $N_i$. Then $p^{k_i-1} \\mid\\mid a_i$ for $1 \\le i \\le 2p-2$, and every element of $A_{\\ge N_i}$ is a multiple of $p^{k_i}$. \n\nChoose $B = \\{a_1, a_2, \\dots, a_{2p-2}\\}$. For every subset of $p$ elements $\\{a_{i_1}, \\dots, a_{i_p}\\}$, the arithmetic mean is divisible by $p^{k_{i_1}-2}$ and is at least $a_{i_1} \\ge N_{i_1-1}$. Thus, this mean does not belong to $A$. This completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16391,
    "subject": "Mathematics (Olympiad)",
    "question": "By induction on $n$, show that there exist $n$ distinct positive integers that are pairwise relatively consistent.\n\n(Relatively consistent means: for any $i < j$, there exist distinct positive divisors of $x_i$ whose sum is $x_j$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a > 1$ be an integer. It is easy to verify that\n\n$$\na^{2017},\\ a^{2017} + a^{2016},\\ a^{2017} + a^{2016} + a^{2015},\\ \\dots,\\ a^{2017} + a^{2016} + \\dots + a + 1\n$$\nare pairwise relatively consistent.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16392,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex cyclic quadrilateral which is not a kite, but whose diagonals are perpendicular and meet at $H$. Denote by $M$ and $N$ the midpoints of $\\overline{BC}$ and $\\overline{CD}$. Rays $MH$ and $NH$ meet $\\overline{AD}$ and $\\overline{AB}$ at $S$ and $T$, respectively. Prove there exists a point $E$, lying outside quadrilateral $ABCD$, such that\n\n* ray $EH$ bisects both angles $\\angle BES$ and $\\angle TED$;\n* $\\angle BEN = \\angle MED$.\n\nThe main claim is that $E$ is the intersection of $(ABCD)$ with the circle with diameter $\\overline{AH}$.\n\n![](images/sols-TST-IMO-2018_p12_data_cfb011d8e6.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Lemma**\n\nWe have $\\angle HSA = \\angle HTA = 90^\\circ$. Consequently, quadrilateral *BTSD* is cyclic.\n\n*Proof.* This is direct angle chasing. In fact, $\\overline{HM}$ passes through the circumcenter of $\\triangle BHC$ and $\\triangle HAD \\sim \\triangle HCB$, so $\\overline{HS}$ ought to be the altitude of $\\triangle HAD$. $\\square$\n\nFrom here it follows that $E$ is the Miquel point of cyclic quadrilateral $BTSD$. Define $F$ to be the point diametrically opposite $A$, so that $E, H, F$ are collinear, $\\overline{CF} \\parallel \\overline{BD}$. By now we already have\n\n$$\n\\angle BEH = \\angle BEF = \\angle BAF = \\angle CAD = \\angle HAS = \\angle HES\n$$\n\nso $\\overline{EH}$ bisects $\\angle BES$ and $\\angle TED$. Hence it only remains to show $\\angle BEM = \\angle NED$; we present several proofs below.\n\n**First proof (original solution)**\n\nLet $P$ be the circumcenter of $BTSD$. The properties of the Miquel point imply $P$ lies on the common bisector $\\overline{EH}$ already, and it also lies on the perpendicular bisector of $\\overline{BD}$, hence it must be the midpoint of $\\overline{HF}$.\n\nWe now contend quadrilaterals $BMPS$ and $DNPT$ are cyclic. Obviously $\\overline{MP}$ is the external angle bisector of $\\angle BMS$, and $PB = PS$, so $P$ is the arc midpoint of $(BMS)$. The proof for $DNPT$ is analogous.\n\nIt remains to show $\\angle BEN = \\angle MED$, or equivalently $\\angle BEM = \\angle NED$. By properties of Miquel point we have $E \\in (BMPS) \\cap (TPND)$, so\n\n$$\n\\angle BEM = \\angle BPM = \\angle PBD = \\angle BDP = \\angle NPD = \\angle NED\n$$\n\nas desired.\n\n**Second proof (2011 G4)**\n\nBy 2011 G4, the circumcircle of $\\triangle EMN$ is tangent to the circumcircle of $ABCD$. Hence if we extend $\\overline{EM}$ and $\\overline{EN}$ to meet $(ABCD)$ again at $X$ and $Y$, we get $\\overline{XY} \\parallel \\overline{MN} \\parallel \\overline{BD}$. Thus $\\angle BEM = \\angle BEX = \\angle YED = \\angle NED$.\n\n**Third proof (involutions, submitted by contestant Daniel Liu)**\n\nLet $G = \\overline{BN} \\cap \\overline{MD}$ denote the centroid of $\\triangle BCD$, and note that it lies on $\\overline{EHF}$.\n\nNow consider the dual of Desargue involution theorem on complete quadrilateral $BNMDCG$ at point $E$. We get\n\n$$\n(EB, EN), (ED, EM), (EC, EG)\n$$\n\nform an involutive pairing.\n\nHowever, the bisector of $\\angle BED$, say $\\ell$, is also the angle bisector of $\\angle CEF$ (since $\\overline{CF} \\parallel \\overline{BD}$). So the involution we found must coincide with reflection across $\\ell$. This means $\\angle MEN$ is bisected by $\\ell$ as well, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16393,
    "subject": "Mathematics (Olympiad)",
    "question": "在三角形 $ABC$ 中，$A'$、$B'$、$C'$ 分別是 $BC$、$AC$、$AB$ 邊的中點。$B^*$、$C^*$ 分別在 $AC$、$AB$ 上，使得 $BB^*$、$CC^*$ 是三角形 $ABC$ 的高。再令 $B^\\sharp$、$C^\\sharp$ 分別為 $BB^*$、$CC^*$ 的中點。設 $B'B^\\sharp$ 與 $C'C^\\sharp$ 交於 $K$ 點，$AK$ 交 $BC$ 於 $L$ 點。證明：\n\n$$\n\\angle BAL = \\angle CAA'\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "同樣定義 $A^*$、$A^\\sharp$。因 $A'B' \\parallel AB$，$B'C' \\parallel BC$，$C'A' \\parallel CA$，且三高共點，得\n\n$$\n\\frac{C'A^\\sharp}{A^\\sharp B'} \\cdot \\frac{B'C^\\sharp}{C^\\sharp A'} \\cdot \\frac{A'B^\\sharp}{B^\\sharp C'} = 1 = \\frac{BA^*}{A^*C} \\cdot \\frac{AC^*}{C^*B} \\cdot \\frac{CB^*}{B^*A} = 1,\n$$\n\n故 $K$ 亦在 $A'A^\\sharp$ 上。設 $a = BC$，$b = CA$，$c = AB$。由孟氏定理\n\n$$\n\\frac{A'K}{KA^\\sharp} \\cdot \\frac{A^\\sharp C'}{C'B'} \\cdot \\frac{B'C^\\sharp}{C^\\sharp A'} = -1 = \\frac{A'K}{KA^\\sharp} \\cdot \\frac{A^\\sharp A}{AA^*} \\cdot \\frac{A^*L}{LA'}.\n$$\n\n因此\n\n$$\n\\frac{A^*L}{LA'} = \\frac{AA^*}{A^\\sharp A} \\cdot \\frac{A^\\sharp C'}{C'B'} \\cdot \\frac{B'C^\\sharp}{C^\\sharp A'} = 2 \\cdot \\frac{c \\cos B}{a} \\cdot \\frac{b \\cos A}{a \\cos B} = \\frac{2bc \\cos A}{a^2}\n$$\n\n將此比值記為 $r$。而 $A'A^* = b \\cos C - \\frac{1}{2}a = \\frac{b \\cos C - c \\cos B}{2}$，故\n\n$$\n\\begin{aligned}\n\\frac{BL}{LC} &= \\frac{c \\cos B + \\frac{r}{1+r} \\cdot \\frac{b \\cos C - c \\cos B}{2}}{b \\cos C - \\frac{r}{1+r} \\cdot \\frac{b \\cos C - c \\cos B}{2}} \\\\\n&= \\frac{2c \\cos B + r(c \\cos B + b \\cos C)}{2b \\cos C + r(c \\cos B + b \\cos C)} \\\\\n&= \\frac{2c \\cos B + \\frac{2bc \\cos A}{a}}{2b \\cos C + \\frac{2bc \\cos A}{a}} \\quad (\\text{因 $c \\cos B + b \\cos C = a$}) \\\\\n&= \\frac{c(a \\cos B + b \\cos C)}{b(a \\cos C + c \\cos A)} = \\frac{c^2}{b^2}.\n\\end{aligned}\n$$\n\n但 $\\frac{BL}{LC} = \\frac{c \\sin \\angle BAL}{b \\sin \\angle CAL}$，得 $\\frac{\\sin \\angle BAL}{\\sin \\angle CAL} = \\frac{c}{b}$。又 $A'$ 為 $BC$ 中點，$1 = \\frac{BA'}{A'C} = \\frac{c \\sin \\angle BAA'}{b \\sin \\angle CAA'}$，所以\n\n$$\n\\frac{\\sin \\angle BAL}{\\sin \\angle CAL} = \\frac{\\sin \\angle CAA'}{\\sin \\angle BAA'}\n$$\n\n因為 $\\angle BAL + \\angle CAL = \\angle CAA' + \\angle BAA' = \\angle A$，所以 $\\angle BAL = \\angle CAA'$，$\\angle CAL = \\angle BAA'$。證畢。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16394,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $k \\geq 3$ such that there exist integers $m$ and $n$ with $1 < m < k$, $1 < n < k$, $m + n > k$, and $k \\mid (m-1)(n-1)$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $k$ has a square factor greater than 1, let $t^2 \\mid k$, $t > 1$. Take $m = n = k - \\frac{k}{t} + 1$; then such $k$ satisfy the conditions.\n\nIf $k$ has no square factor, suppose there are two primes $p_1, p_2$ such that $(p_1-2)(p_2-2) \\geq 4$ and $p_1 p_2 \\mid k$. Let $k = p_1 p_2 \\cdots p_r$ with $p_1, \\dots, p_r$ distinct, $r \\geq 2$. At least one of $(p_1-1)p_2 p_3 \\cdots p_r + 1$ or $(p_1-2)p_2 p_3 \\cdots p_r + 1$ is coprime to $p_1$ (otherwise $p_1$ divides their difference $p_2 p_3 \\cdots p_r$, a contradiction). Take such a number as $m$; then $1 < m < k$, $(m, k) = 1$. Similarly, take $n$ as $(p_2-1)p_1 p_3 \\cdots p_r + 1$ or $(p_2-2)p_1 p_3 \\cdots p_r + 1$ with $1 < n < k$, $(n, k) = 1$. Thus $k \\mid (m-1)(n-1)$, and\n\n$$\nm + n \\geq (p_1 - 2)p_2 p_3 \\cdots p_r + 1 + (p_2 - 2)p_1 p_3 \\cdots p_r + 1 \\\\\n= k + ((p_1 - 2)(p_2 - 2) - 4)p_3 \\cdots p_r + 2 > k.\n$$\n\nSo such $m, n$ exist.\n\nIf there are no two primes $p_1, p_2$ with $(p_1-2)(p_2-2) \\geq 4$ and $p_1 p_2 \\mid k$, then $k \\geq 3$ can only be $15$, $30$, or $p$, $2p$ (with $p$ odd prime). For $k = p, 2p, 30$, there are no $m, n$ satisfying the conditions; for $k = 15$, $m = 11$, $n = 13$ work.\n\nIn summary, integer $k \\geq 3$ satisfies the conditions if and only if $k$ is not an odd prime, nor double an odd prime, nor $30$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16395,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a non-isosceles acute triangle, and let $O$ be its circumcenter. Let $A'$ be a point on the line $AO$ such that $\\angle BA'A = \\angle CA'A$. Construct $A'A_1 \\perp AC$ and $A'A_2 \\perp AB$, with $A_1$ on $AC$ and $A_2$ on $AB$, respectively. Let $AH_A$ be perpendicular to $BC$ at $H_A$. Let $R_A$ be the circumradius of $\\triangle H_AA_1A_2$. Similarly, define $R_B$ and $R_C$. Prove that\n\n$$\n\\frac{1}{R_A} + \\frac{1}{R_B} + \\frac{1}{R_C} = \\frac{2}{R},\n$$\n\nwhere $R$ is the circumradius of $\\triangle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Firstly, we claim that $A', B, O, C$ are concyclic points. Otherwise, extend $AO$ to intersect the circumcircle of $\\triangle BOC$ at a point $P$ different from $A'$. We get\n\n$$\n\\angle BPA = \\angle BCO = \\angle CBO = \\angle CPA.\n$$\n\nThen $\\triangle PA'C \\cong \\triangle PA'B$, and $A'B = A'C$. So $AB = AC$, which contradicts the assumption that $\\triangle ABC$ is not isosceles. Thus,\n\n$$\n\\angle BCA' = \\angle BOA' = 180^\\circ - 2\\angle C,\n$$\n\nand $\\angle A'CA_1 = \\angle C$.\n\nFurther, we have\n\n$$\n\\frac{H_AA}{AC} = \\sin \\angle C = \\cos \\angle A_2AA' = \\frac{AA_2}{AA'}\n$$\n\nand\n\n$$\n\\angle A'AC = \\angle A_2AH_A = \\frac{\\pi}{2} - \\angle B.\n$$\n\nSo $\\triangle A_2AH_A \\sim \\triangle A'AC$. Similarly, $\\triangle A_1H_AA \\sim \\triangle A'BA$. Then $\\angle A_2H_AA = \\angle ACA'$ and $\\angle A_1H_AA = \\angle ABA'$. Consequently,\n\n$$\n\\begin{align*}\n\\angle A_1 H_A A_2 &= 2\\pi - \\angle A_2 H_A A - \\angle A_1 H_A A \\\\\n&= 2\\pi - \\angle ACA' - \\angle ABA' \\\\\n&= \\angle A + 2\\left(\\frac{\\pi}{2} - \\angle A\\right) \\\\\n&= \\pi - \\angle A.\n\\end{align*}\n$$\n\nWe get\n\n$$\n\\frac{R}{R_A} = \\frac{\\frac{R}{A_1A_2}}{\\frac{2 \\sin \\angle A_1H_AA_2}{\\sin \\angle A}} = \\frac{2R}{A_1A_2} = \\frac{2R}{AA'}, \\quad \\textcircled{1}\n$$\n\nThe last equality holds since $A, A_2, A', A_1$ lie on the same circle with $AA'$ as the diameter.\n\nNow, draw $AA'' \\perp A'C$ with $A''$ on $A'C$. Since $\\angle ACA'' = \\angle A'CA_1 = \\angle C$, we have $AA'' = AH_A$. Then\n\n$$\n\\begin{align*}\nAA' &= \\frac{AA''}{\\sin \\angle AA'C} = \\frac{AH_A}{\\sin(90^\\circ - \\angle A)} \\\\\n    &= \\frac{AH_A}{\\cos \\angle A} = \\frac{2S_{\\triangle ABC}}{BC \\cos \\angle A}. \\quad \\textcircled{2}\n\\end{align*}\n$$\n\nFrom (1) and (2), we get\n\n$$\n\\begin{align*}\n\\frac{1}{R_A} &= \\frac{BC \\cos \\angle A}{S_{\\triangle ABC}} = \\frac{\\cos \\angle A}{R \\sin \\angle B \\sin \\angle C} \\\\\n&= \\frac{1}{R} (1 - \\cot \\angle B \\cot \\angle C).\n\\end{align*}\n$$\n\nSimilarly,\n\n$$\n\\frac{1}{R_B} = \\frac{1}{R}(1 - \\cot \\angle C \\cot \\angle A)\n$$\n\nand\n\n$$\n\\frac{1}{R_C} = \\frac{1}{R}(1 - \\cot \\angle A \\cot \\angle B).\n$$\n\nNotice that\n\n$$\n\\cot \\angle A \\cot \\angle B + \\cot \\angle B \\cot \\angle C + \\cot \\angle C \\cot \\angle A = 1.\n$$\n\nTherefore,\n\n$$\n\\frac{1}{R_A} + \\frac{1}{R_B} + \\frac{1}{R_C} = \\frac{2}{R}.\n$$\n\nThis completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16396,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all polynomials $P$ with integer coefficients, satisfying $0 \\leq P(n) \\leq n!$ for all non-negative integers $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The required polynomials are $P = 0$, $P = 1$, $P = (X - 1)^2$, $P = X(X - 1)\\cdots(X - k)$, and $P = X(X - 1)\\cdots(X - k)(X - k - 2)^2$ for some non-negative integer $k$.\n\nLet $P$ be a polynomial satisfying the condition in the statement. Clearly, $P(0) = 0$ or $P(0) = 1$.\n\nFirst, consider $P(0) = 1$. The polynomials $P_1 = 1$ and $P_2 = (X-1)^2$ both satisfy the condition and $P_1(0) = P_2(0) = 1$.\n\nWe will prove that either $P = P_1$ or $P = P_2$. Consider an index $i$ such that $P(1) = P_i(1)$ and let $\\tilde{P} = P - P_i$.\n\nInduct on $n$ to show that $\\tilde{P}(n) = 0$ for all non-negative integers $n$. The base cases $n = 0$ and $n = 1$ are clear. For the inductive step, assume $\\tilde{P}(m) = 0$ for all $m < n$. Then $X(X-1)\\cdots(X-(n-1))$ divides $\\tilde{P}$, so $n!$ divides $\\tilde{P}(n)$. As $0 < P_i(n) < n!$, it follows that $|\\tilde{P}(n)| = |P(n) - P_i(n)| < n!$, so $\\tilde{P}(n) = 0$.\n\nConsequently, $\\tilde{P}$ has infinitely many roots, so it vanishes identically; that is, $P = P_i$.\n\nNow, consider $P(0) = 0$. Assume $P$ is non-zero. Let $P(X) = X Q(X - 1)$, where $Q$ has integer coefficients. Then $0 \\leq Q(n) \\leq n!$ for all $n \\geq 0$. If $Q(0) = 0$, repeat the argument for $Q$, and so on, until reaching a polynomial with a non-zero constant term. By the previous case, such a polynomial is either $1$ or $(X - 1)^2$. An induction then shows that $P$ has one of the forms listed at the beginning.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16397,
    "subject": "Mathematics (Olympiad)",
    "question": "Between (and including) 98 and 200, how many integers are multiples of 2 or 3?",
    "options": [],
    "answer": "See solution",
    "solution": "Between 98 and 200, there are 51 multiples of 2. Between 98 and 199, there are 34 multiples of 3. Between 102 and 198, there are 17 multiples of 6 (which are counted in both previous groups). By the inclusion-exclusion principle, the total is:\n\n$$51 + 34 - 17 = 68$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16398,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $n \\geq 2$ is an integer. For which positive integers $n$ do all the numbers $\\binom{n}{i} - i$ (for $0 \\leq i \\leq n$) have the same parity?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be a positive integer such that all the numbers $\\binom{n}{i} - i$ (for $0 \\leq i \\leq n$) have the same parity. This is equivalent to requiring that for every $0 \\leq i \\leq n-1$, $\\binom{n}{i}$ and $\\binom{n}{i+1}$ have different parities. Note that $\\binom{n+1}{i+1} = \\binom{n}{i} + \\binom{n}{i+1}$ for $1 \\leq i \\leq n-1$, so $\\binom{n+1}{i+1}$ is odd for all $1 \\leq i \\leq n-1$. Also, $\\binom{n+1}{0} = 1$ is odd. By the lemma, this is equivalent to $n+1 = 2^k - 1$ for some integer $k \\geq 2$, so $n = 2^k - 2$ where $k \\geq 2$ is an integer.\n\n$$\n\\boxed{n = 2^k - 2,\\quad k \\geq 2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16399,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S_{n-2} = \\sum_{k=1}^{n-2} \\frac{1}{a_k a_{k+2}}$ and $Q_m = \\sum_{k=1}^{m} a_k^2$. The given relation is\n\n$$\nS_{n-2} = 1 - \\frac{1}{Q_{n-1}} \\quad \\text{for } n \\ge 3.\n$$\n\nFind all sequences $(a_n)$ of positive integers satisfying this relation for all $n \\ge 3$.",
    "options": [],
    "answer": "See solution",
    "solution": "For $n = 3$:\n\n$$\n\\frac{1}{a_1 a_3} = 1 - \\frac{1}{a_1^2 + a_2^2}\n$$\n\nThis forces $a_1 = 1$ and $a_2 = 1$ (otherwise, contradictions arise). Substituting, $\\frac{1}{a_3} = 1 - \\frac{1}{2} = \\frac{1}{2}$, so $a_3 = 2$.\n\nFor $n = 4$:\n\n$$\n\\frac{1}{a_1 a_3} + \\frac{1}{a_2 a_4} = 1 - \\frac{1}{a_1^2 + a_2^2 + a_3^2}\n$$\n\nWith $a_1 = 1, a_2 = 1, a_3 = 2$:\n\n$$\n\\frac{1}{2} + \\frac{1}{a_4} = 1 - \\frac{1}{6} = \\frac{5}{6}\n$$\n\nSo $\\frac{1}{a_4} = \\frac{5}{6} - \\frac{1}{2} = \\frac{1}{3}$, thus $a_4 = 3$.\n\nFor $n = 5$:\n\n$$\n\\frac{1}{a_1 a_3} + \\frac{1}{a_2 a_4} + \\frac{1}{a_3 a_5} = 1 - \\frac{1}{a_1^2 + a_2^2 + a_3^2 + a_4^2}\n$$\n\nThe sum of the first two terms is $\\frac{5}{6}$, so:\n\n$$\n\\frac{5}{6} + \\frac{1}{2 a_5} = 1 - \\frac{1}{15} = \\frac{14}{15}\n$$\n\nThus $\\frac{1}{2 a_5} = \\frac{14}{15} - \\frac{5}{6} = \\frac{1}{10}$, so $a_5 = 5$.\n\nThe sequence begins $1, 1, 2, 3, 5, \\dots$, which is the Fibonacci sequence. We claim $a_n = a_{n-1} + a_{n-2}$ for $n \\ge 3$.\n\nWe prove by induction that $Q_m = a_m a_{m+1}$ for Fibonacci numbers, so the relation becomes:\n\n$$\n\\sum_{k=1}^{n-2} \\frac{1}{a_k a_{k+2}} = 1 - \\frac{1}{a_{n-1} a_n}\n$$\n\nThe base case and induction step both hold. Thus, the only solution is the Fibonacci sequence $a_n = F_n$ with $F_1 = 1, F_2 = 1$.\n\n$\\boxed{a_n = F_n}$, the Fibonacci sequence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16400,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with altitudes $AD$, $BE$, and $CF$. Let $H$ be the orthocentre, that is, the point where the altitudes meet. Prove that\n\n$$\n\\frac{AB \\cdot AC + BC \\cdot BA + CA \\cdot CB}{AH \\cdot AD + BH \\cdot BE + CH \\cdot CF} \\le 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Method 1: Let $AB = c$, $AC = b$, and $BC = a$ denote the three side lengths of the triangle.\n\nAs $\\angle BFH = \\angle BDH = 90^\\circ$, $FHDB$ is a cyclic quadrilateral. By the Power-of-a-Point Theorem, $AH \\cdot AD = AF \\cdot AB$. (We can derive this result in other ways: for example, see Method 2, below.)\n\nSince $AF = AC \\cdot \\cos \\angle A$, we have $AH \\cdot AD = AC \\cdot AB \\cdot \\cos \\angle A = bc \\cos \\angle A$.\n\nBy the Cosine Law, $\\cos \\angle A = \\frac{b^2 + c^2 - a^2}{2bc}$, which implies that $AH \\cdot AD = \\frac{b^2 + c^2 - a^2}{2}$.\n\nBy symmetry, we can show that $BH \\cdot BE = \\frac{a^2 + c^2 - b^2}{2}$ and $CH \\cdot CF = \\frac{a^2 + b^2 - c^2}{2}$.\n\nHence,\n\n$$\n\\begin{aligned}\nAH \\cdot AD + BH \\cdot BE + CH \\cdot CF &= \\frac{b^2 + c^2 - a^2}{2} + \\frac{a^2 + c^2 - b^2}{2} + \\frac{a^2 + b^2 - c^2}{2} \\\\\n&= \\frac{a^2 + b^2 + c^2}{2}. \\qquad (1)\n\\end{aligned}\n$$\n\nOur desired inequality, $\\frac{AB \\cdot AC + BC \\cdot BA + CA \\cdot CB}{AH \\cdot AD + BH \\cdot BE + CH \\cdot CF} \\le 2$, is equivalent to the inequality $\\frac{cb + ac + ba}{\\frac{a^2+b^2+c^2}{2}} \\le 2$, which simplifies to $2a^2 + 2b^2 + 2c^2 \\ge 2ab + 2bc + 2ca$.\n\nBut this last inequality is easy to prove, as it is equivalent to $(a-b)^2+(a-c)^2+(b-c)^2 \\ge 0$.\n\nTherefore, we have established the desired inequality. The proof also shows that equality occurs if and only if $a = b = c$, i.e., $\\triangle ABC$ is equilateral. $\\square$\n\nMethod 2: Observe that\n\n$$\n\\frac{AE}{AH} = \\cos(\\angle HAE) = \\frac{AD}{AC} \\quad \\text{and} \\quad \\frac{AF}{AH} = \\cos(\\angle HAF) = \\frac{AD}{AB}.\n$$\n\nIt follows that\n\n$$\nAC \\cdot AE = AH \\cdot AD = AB \\cdot AF.\n$$\n\nBy symmetry, we similarly have\n\n$$\nBC \\cdot BD = BH \\cdot BE = BF \\cdot BA \\quad \\text{and} \\quad CD \\cdot CB = CH \\cdot CF = CE \\cdot CA.\n$$\n\nTherefore\n\n$$\n\\begin{aligned}\n& 2(AH \\cdot AD + BH \\cdot BE + CH \\cdot CF) \\\\\n&= AB(AF + BF) + AC(AE + CE) + BC(BD + CD) \\\\\n&= AB^2 + AC^2 + BC^2.\n\\end{aligned}\n$$\n\nThis proves Equation (1) in Method 1. The rest of the proof is the same as the part of the proof of Method 1 that follows Equation (1). $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16401,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $(a, b)$ of coprime natural numbers such that $a < b$ and $b$ divides\n$$\n(n + 2)a^{n+1002} - (n + 1)a^{n+1001} - n a^{n+1000}\n$$\nfor every natural number $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $a$ and $b$ are coprime, so are $b$ and $a^{n+1000}$. Thus, the divisibility is equivalent to $b$ dividing $(n + 2)a^2 - (n + 1)a - n$ for each $n$.\n\nFor $n = 1$ and $n = 2$, we get that $b$ divides $3a^2 - 2a - 1$ and $4a^2 - 3a - 2$. Therefore, $b$ divides\n$$\n4(3a^2 - 2a - 1) - 3(4a^2 - 3a - 2) = a + 2.\n$$\n\nAlso, $3a^2 - 2a - 1 = 3(a - 2)(a + 2) - 2(a + 2) + 15$, so $b$ must divide $15$.\n\nSince $b > a \\geq 1$, $b \\geq 2$. If $b = 15$, then $b \\mid a + 2$ gives $a = 13$, but $4 \\cdot 13^2 - 3 \\cdot 13 - 2 = 635$ is not divisible by $3$. If $b = 3$, then $a = 1$, but $4 \\cdot 1^2 - 3 \\cdot 1 - 2 = -1$ is not divisible by $3$.\n\nIt remains $b = 5$, so $a = 3$. Indeed, $(n + 2) \\cdot 3^2 - (n + 1) \\cdot 3 - n = 5(n + 3)$ is divisible by $5$.\n\n$\\boxed{(a, b) = (3, 5)}$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16402,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples $(a, b, c)$ of nonzero complex numbers with equal absolute values, for which\n$$\n\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a} + 1 = 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that $\\overline{\\left(\\frac{a}{b}\\right)} = \\frac{b}{a}$, so taking conjugates yields\n$$\n\\frac{b}{a} + \\frac{c}{b} + \\frac{a}{c} + 1 = 0.\n$$\nClearing denominators and adding up gives\n$$\na^2 b + b^2 c + c^2 a + ab^2 + bc^2 + ca^2 + 2abc = 0,\n$$\nwhich factors as $(a+b)(b+c)(c+a) = 0$. Therefore, two of the numbers sum to zero. If, say, $a+b=0$, the initial equality implies $c=a$ or $c=b$, hence the requested triples are $(a, a, -a)$, with $a \\in \\mathbb{C}^*$ (and the respective permutations).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16403,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with $H$ as its orthocenter, and let $D$ be the midpoint of $BC$. Let $X$ and $Y$ be the feet of the altitudes from $B$ and $C$ to $AC$ and $AB$, respectively. Let $K$ be the intersection point of the perpendicular bisector of $BC$ and the line $XY$.\n\nProve that $\\angle BKD = \\angle CKD$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, consider the case $AB = AC$. In this case, the centers of the circle through $B, C$ and the circle with diameter $AH$ both lie on the perpendicular bisector of $BC$. Thus, $XY$ and the perpendicular bisector of $BC$ intersect perpendicularly at $K$, and $D$ is the midpoint of $BC$, so $\\angle BKD = \\angle CKD$.\n\nNow, consider $AB \\ne AC$ (assume $AB > AC$). Let $E$ be the intersection of $BH$ and $AC$, and $F$ the intersection of $CH$ and $AB$. Since $\\angle AEH = \\angle AFH = 90^\\circ$, $E$ and $F$ lie on the circle with diameter $AH$ (call it $\\Gamma_1$). $X$ and $Y$ also lie on $\\Gamma_1$. The points $B, C, X, Y$ lie on another circle $\\Gamma_2$, and $B, C, E, F$ lie on $\\Gamma_3$.\n\n*Lemma.* The lines $XY$, $BC$, and $EF$ are concurrent.\n\n*Proof.* Let $O$ be the intersection of $BC$ and $EF$. Let $Y'$ and $Y''$ be the intersections of $OX$ with $\\Gamma_1$ and $\\Gamma_2$, respectively. By the power of a point theorem:\n\n$$\nOX \\cdot OY' = OE \\cdot OF = OC \\cdot OB = OX \\cdot OY''\n$$\n\nSo $OY' = OY''$, hence $Y' = Y'' = Y$. Thus, $O$ lies on $XY$.\n\nBy Ceva's Theorem:\n$$\n\\frac{AF}{FB} \\cdot \\frac{BD}{DC} \\cdot \\frac{CE}{EA} = 1\n$$\nBy Menelaus' Theorem:\n$$\n\\frac{AF}{FB} \\cdot \\frac{BO}{OC} \\cdot \\frac{CE}{EA} = 1\n$$\nComparing, $\\frac{BD}{DC} = \\frac{BO}{OC}$.\n\nLet $C'$ on $DO$ satisfy $\\angle BKD = \\angle C'KD$. Since $KD$ bisects $\\angle BKC'$, and $KO$ bisects $\\angle EKO$ (with $\\angle DKO = 90^\\circ$),\n\n$$\n\\frac{BD}{DC'} = \\frac{KB}{KC'}, \\quad \\frac{BO}{OC'} = \\frac{KB}{KC'}\n$$\nSo $\\frac{BD}{DC'} = \\frac{BO}{OC'}$. But $\\frac{BD}{DC} = \\frac{BO}{OC}$, so $\\frac{OC}{DC} = \\frac{OC'}{DC'}$. Since $C$ and $C'$ are on $DO$, $C' = C$. Thus, $\\angle BKD = \\angle CKD$ as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16404,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_1, A_2, \\dots, A_{2n+1}$ be points. For each pair $i < j$, draw the vector $\\overrightarrow{A_iA_j}$. Let $R$ be the number of triangles $\\triangle A_iA_jA_k$ ($i < j < k$) such that the sum of the vectors along its sides is zero. Determine the smallest and largest possible values of $R$ in terms of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The smallest possible value of $R$ is $0$, and the largest is $\\dfrac{n(n+1)(2n+1)}{6}$.\n\nFor any triangle $\\triangle A_iA_jA_k$ with $i < j < k$,\n$$\n\\overrightarrow{A_iA_j} + \\overrightarrow{A_jA_k} + \\overrightarrow{A_kA_i} = \\mathbf{0}.\n$$\nHowever, depending on how the vectors are assigned, it is possible that $R = 0$ (no triangle has sum zero).\n\nFor the maximum, define $b_j$ as the number of vectors pointing away from $A_j$ and $c_j$ as the number pointing towards $A_j$, with $b_j + c_j = 2n$. The number of triangles with nonzero sum is at least\n$$\n\\frac{1}{2} \\sum_{j=1}^{2n+1} \\left[ \\binom{b_j}{2} + \\binom{c_j}{2} \\right] \\ge \\frac{1}{2} \\sum_{j=1}^{2n+1} 2\\binom{n}{2} = \\frac{n(n-1)(2n+1)}{2}\n$$\nby Jensen's inequality. Since there are $\\binom{2n+1}{3}$ triangles in total,\n$$\nR \\le \\binom{2n+1}{3} - \\frac{n(n-1)(2n+1)}{2} = \\frac{n(n+1)(2n+1)}{6}.\n$$\nThis maximum is achievable, for example, by assigning $b_j = c_j = n$ for all $j$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16405,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying the equation\n\n$$\nf(x+y) + f(x)f(y) = (1+y)f(x) + (1+x)f(y) + f(xy)\n$$\n\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Taking $x = y = 0$ in (1), we obtain $f(0)^2 = 2f(0)$ and hence $f(0) = 0$ or $2$. If $f(0) = 2$, then $y = 0$ in (1) gives $f(x) = 2 + x$ for all $x$. Substituting this into (1) yields $2xy = 0$ for all $x, y$, a contradiction. Thus, $f(0) = 0$.\n\nLet us consider possible values for $f(1)$ and $f(2)$:\n\n- Setting $x = y = 1$ in (1): $f(2) + f(1)^2 = 5f(1)$.\n- Setting $x = y = 2$ in (1): $f(2)^2 = 6f(2)$.\n\nFrom these, $f(2) = 0$ or $6$; if $f(2) = 0$, then $f(1) = 0$ or $5$; if $f(2) = 6$, then $f(1) = 2$ or $3$.\n\n**Case 1:** $(f(2), f(1)) = (0, 0)$\n\nSetting $y = 1$ in (1):\n$$\nf(x+1) = 3f(x).\n$$\nIterating, $f(x) \\equiv 0$ is a solution.\n\n**Case 2:** $(f(2), f(1)) = (0, 5)$\n\nThis leads to a contradiction, so no solution in this case.\n\n**Case 3:** $(f(2), f(1)) = (6, 2)$\n\nSetting $y = 1$ in (1):\n$$\nf(x+1) = f(x) + 2(x+1).\n$$\nSolving recursively, $f(x) = x^2 + x$ is a solution.\n\n**Case 4:** $(f(2), f(1)) = (6, 3)$\n\nSetting $y = 1$ in (1):\n$$\nf(x+1) = 3(x+1).\n$$\nSo $f(x) = 3x$ is a solution.\n\n**Conclusion:**\n\nThe solutions are:\n- $f(x) \\equiv 0$\n- $f(x) = x^2 + x$\n- $f(x) = 3x$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16406,
    "subject": "Mathematics (Olympiad)",
    "question": "A fractional number $x$ is called 'pretty' if it has a finite expression in base $b$ numeral system, where $b$ is a positive integer in $[2, 2022]$. Prove that there exist finitely many positive integers $n \\geq 4$ such that for every $m$ in $\\left(\\frac{2n}{3}, n\\right)$, there is at least one pretty number among the two numbers\n$$\n\\frac{m}{n-m} \\text{ and } \\frac{n-m}{m}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Call a positive integer $n$ 'good' if there exists $m$ in the interval $\\left(\\frac{2n}{3}, n\\right)$ such that both $\\frac{m}{n-m}$ and $\\frac{n-m}{m}$ are not pretty. Next, we prove the following claims.\n\n**Claim 1.** If $n$ is good, then any multiple of $n$ is also good.\n\n*Proof.* Consider a good number $n$. There exists $m \\in \\left(\\frac{2n}{3}, n\\right)$ such that $\\frac{n-m}{m}$ and $\\frac{m}{n-m}$ are not pretty. Hence, for $kn$, $\\frac{kn-km}{km}$ and $\\frac{km}{kn-km}$ are also not pretty. $\\square$\n\n**Claim 2.** Consider a prime $q$ such that there exists a prime $r$ with $2022 < r < q$. For all pairs $(p, k)$ with $p$ prime and $k$ a positive integer such that $p^k > 3q!$, the number $p^k$ is good.\n\n*Proof.* Choose $m = 3q!$. Then $\\frac{2p^k}{3} < p^k - m < p^k$. Assume $\\frac{p^k - m}{m}$ is pretty. Then there exists $b < 2023$ and finite non-negative integers $b_0, b_1, \\dots, b_t$ such that\n$$\n\\frac{p^k - m}{m} = \\sum_{i=0}^{t} \\frac{b_i}{b^i} = \\frac{b_0 b^t + b_1 b^{t-1} + \\dots + b_t}{b^t},\n$$\nwhich means $\\frac{m}{\\gcd(m, p^k)} \\mid b^t$.\n\nHence, all prime divisors of $m$, except $p$, are smaller than $2022$. But $m$ is divisible by $q, r > 2022$, so $m$ has a prime divisor larger than $2022$, a contradiction. Therefore, $\\frac{p^k - m}{m}$ is not pretty.\n\nSimilarly, assume $\\frac{m}{p^k - m}$ is pretty. Then $\\frac{p^k - m}{\\gcd(m, p^k - m)}$ only has prime divisors smaller than $2022$. Assume there exists a prime $p_1 < 2022$ such that\n$$\np_1 \\mid \\frac{p^k - m}{m} \\implies p_1 \\mid p^k \\implies p_1 = p,\n$$\nwhich means $p < 2022$ and there exists $l > 0$ such that\n$$\np^l \\cdot \\gcd(m, p^k - m) = p^k - m.\n$$\nLet $m = p^s t$ with $\\gcd(p, t) = 1$, then\n$$\np^l p^s = p^k - p^s t \\implies p^{k-s} - t = p^l.\n$$\nSince $p^{k-s} > 1$, $p \\mid p^{k-s}$. Also, $\\gcd(t, p) = 1$ so $\\gcd(p^l, p) = 1$ or $l = 0$, which means $\\frac{p^k - m}{\\gcd(m, p^k - m)} = 1$ or $p^k - m \\mid m < \\frac{p^k}{3}$, a contradiction.\n\nHence, both $\\frac{m}{p^k - m}$ and $\\frac{p^k - m}{m}$ are not pretty, so $p^k$ is good. $\\square$\n\nReturning to the original problem, let $N$ be the number of numbers of the form $p^k$ not exceeding $Q = 3q!$. We claim that all positive integers $n > Q^N$ are good. Assume the prime factorization of $n$ is\n$$\nn = \\prod_{i=1}^{t} p_i^{\\alpha_i}.\n$$\nIf $p_i^{\\alpha_i} < Q$ for all $i \\leq t$, then $n \\leq Q^t \\leq Q^N$, a contradiction. Therefore, there exists $i$ such that $p_i^{\\alpha_i} > Q$, so $p_i^{\\alpha_i}$ is good by Claim 2. Hence, $n$ is a multiple of $p_i^{\\alpha_i}$ and $n$ is good by Claim 1. The problem is solved. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16407,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine a five-digit positive integer $n$ (in base 10) whose digit sum is least and $n^3 - 1$ is divisible by $2556$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first show that for $n \\in \\mathbb{N}$, the integer $n^3-1$ is divisible by $2556 = 2^2 \\cdot 3^2 \\cdot 71$ if and only if it is of the form $n = 852k + 1$ ($k \\in \\mathbb{N}$).\n\n($\\Rightarrow$) If $2556 \\mid (n^3 - 1)$, then\n\n$$\nn^3 \\equiv 1 \\pmod{2^2 \\cdot 3^2 \\cdot 71}\n$$\n\nand so $n^3 \\equiv 1 \\pmod{71}$. Since $71 \\nmid n$, Fermat's little theorem implies that $n^{70} \\equiv 1 \\pmod{71}$, thus\n\n$$1 \\equiv n \\cdot n^{69} \\equiv n(n^3)^{23} \\equiv n(1)^{23} \\equiv n \\pmod{71},$$\ni.e.,\n\n$$\n71 \\mid (n-1).\n$$\n\nFrom above, $4 \\mid n^3 - 1 = (n-1)(n^2 + n + 1)$. Since $n^2 + n + 1 = n(n+1) + 1$ is odd, we have\n\n$$\n4 \\mid (n-1).\n$$\n\nIf $3 \\nmid (n-1)$, then $3 \\mid n(n+1)$, so $3 \\nmid n(n+1) + 1 = n^2 + n + 1$, yielding $3 \\nmid n^3 - 1$, a contradiction. Thus,\n\n$$\n3 \\mid (n-1).\n$$\n\nThe three divisibility conditions show that $n = (3 \\times 4 \\times 71 \\times k)+1 = 852k + 1$ for some $k \\in \\mathbb{N}$.\n\n($\\Leftarrow$) If $n$ is of the form $n = 852k + 1$ ($k \\in \\mathbb{N}$), then\n\n$$\nn^3 - 1 = (n-1)(n(n-1) + 2(n-1) + 3) = 2556k(284nk + 568k + 1).\n$$\n\nTo solve the problem, we must determine a five-digit $n$ of the form $852k+1$ with least digit sum. Since $852k$ does not end with 9, it suffices to find an integer of the form $852k$ with least digit sum. Since $4 \\mid 852k$, the integer formed from its last two digits is also divisible by 4. To find the integer with least digit sum, we observe that a five-digit integer of the form $abc00$ is divisible by 4 and has two 0 digits making it a good candidate. Since 3 and 71 are factors of the required integer and since 100 is not divisible by 3 nor by 71, we need $abc$ to be divisible by $3 \\times 71 = 213$, which has digit sum 6. We anticipate that $21300$ is the sought after integer. To verify this, consider other integers with digit sums $< 6$ and last two digits divisible by 4. All of them are $abc04, abc12, abc20, abc32, abc40$. Since the required integer is divisible by 3, so are their digit sums, implying that the only possible integer is $abc20$, and so $a, b, c \\in \\{0, 1\\}$, $a \\neq 0$ and $a+b+c=1$. The only possible integer is $10020$. But, this integer is not divisible by 71. Hence, the sought after integer is $21300+1=21301$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16408,
    "subject": "Mathematics (Olympiad)",
    "question": "Докажите, что если существует тройка взаимно простых натуральных чисел $(a, b, c)$, образующих арифметическую прогрессию и произведение которых является квадратом, то существует другая такая тройка, отличная от исходной, имеющая с ней общий элемент.",
    "options": [],
    "answer": "See solution",
    "solution": "Если $b = 1$, то $a = c = 1$, и в качестве другой тройки можно выбрать $(1, 25, 49)$. Если же $b \\neq 1$, то из взаимной простоты разность прогрессии $d$ не может оказаться нулевой. Тогда без ограничения общности $d = b - a = c - b > 0$.\n\nПоскольку $b$ взаимно просто как с $a$, так и с $c$, то оно взаимно просто с $ac$. Далее, произведение взаимно простых чисел $ac$ и $b$ является квадратом, поэтому и каждое из них — также квадрат, то есть $b = f^2$, $ac = m^2 = (b-d)(b+d) = b^2 - d^2$ для некоторых натуральных $f$ и $m$. При этом $m \\neq d$, так как в противном случае $b^2 = m^2 + d^2 = 2m^2$, что невозможно.\n\nРассмотрим теперь тройку $(b-m, b, b+m)$. Ее члены образуют арифметическую прогрессию, являются натуральными числами (так как $b^2 - m^2 = d^2 > 0$), и их произведение $(b-m)b(b+m) = f^2(b^2 - m^2) = (df)^2$ является квадратом. Кроме того, $\\gcd(b, m^2) = \\gcd(b, ac) = 1$, откуда $1 = \\gcd(b, m) = \\gcd(b, b-m) = \\gcd(b, b+m)$. Значит, эта тройка — квадратная, она имеет общий элемент $b$ с исходной и отлична от нее (ибо $b-m \\neq b-d$), что и требовалось.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16409,
    "subject": "Mathematics (Olympiad)",
    "question": "![](images/Cesko-Slovacko-Poljsko_2012_p7_data_cdcdcc9b7e.png)\n\nFig. 5a\n\n![](images/Cesko-Slovacko-Poljsko_2012_p7_data_ed5cbed811.png)\n\nFig. 5b\n\nIn a certain street network, consider a vertical line that splits the set of all crossroads. For some $k \\in \\{0, 1, \\dots, n-1\\}$, the line connects two points located $100k + 50$ m from $E$—one on the SE street and one on the NE street. Analyze, for each $k$, how many arrows (representing possible car movements) the car can pass, and determine the maximum possible length of a route passing through the network, given the constraints illustrated in the figures.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/Cesko-Slovacko-Poljsko_2012_p8_data_e1e4b1abe1.png)\n\nFig. 6a\n\n![](images/Cesko-Slovacko-Poljsko_2012_p8_data_a4ff3234d6.png)\n\nFig. 6b\n\nIf $k$ is odd, then the line intersects $k+1$ forward arrows and $k+1$ back arrows, and at least one of the back arrows remains unpassed.\n\nIf $k \\ge 2$ is even, then the line intersects $k+2$ forward arrows and $k$ back arrows. The two northernmost forward arrows end in the crossroad with only one outgoing arrow, so one of them remains unpassed. The same holds for the two southernmost forward arrows. Thus, at most $k$ forward arrows and at most $k-1$ back arrows can be passed by the car, resulting in 3 unpassed arrows at this level.\n\nFor $k = 0$, there are two forward arrows ending in $E$; only one can be passed by the car, so one remains unpassed.\n\nThere is one unpassed arrow for any odd $k$, three for any even $k \\ge 2$, and one for $k = 0$, and all this happens twice. As $n$ is even and $k \\in \\{0, 1, \\dots, n-1\\}$, altogether, we have\n\n$$\n2\\left(\\frac{1}{2}n + 3\\left(\\frac{1}{2}n - 1\\right) + 1\\right) = 4n - 4\n$$\n\nunpassed arrows. The total number of arrows is $n \\cdot 2(n+1)$, so the car cannot pass more than\n\n$$\nn \\cdot 2(n + 1) - (4n - 4) = 2n^2 - 2n + 4\n$$\n\narrows.\n\nOn the other hand, there are many possible routes of the car consisting of $2n^2 - 2n + 4$ arrows. One is sketched on Fig. 7 for $n = 6$. When we use the same pattern in the general case, the route can be subdivided into $\\frac{1}{2}n$ parts by the crossroads lying on the WS street located $200k$ m from $W$ for $k = 1, 2, \\dots, \\frac{1}{2}n - 1$.\n\n![](images/Cesko-Slovacko-Poljsko_2012_p9_data_52f263058a.png)\n\nFig. 7\n\nThe first $\\frac{1}{2}n-1$ parts differ only by shifting. Each consists of $n$ arrows in the same direction as $WN$, $n$ arrows in the opposite direction as $WN$, and $2(n-1)$ arrows perpendicular to $WN$. The last part consists of $n$ arrows in the same direction as $WN$ and $2(n+1)$ arrows perpendicular to $WN$. Therefore, the number of arrows on the entire route is\n\n$$\n\\left(\\frac{1}{2}n - 1\\right)(2n + 2(n - 1)) + n + 2(n + 1) = 2n^2 - 2n + 4.\n$$\n\n**Answer:** The length of the longest possible route of the car is $\\frac{1}{10}(2n^2 - 2n + 4)$ km.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16410,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{Q} \\to \\mathbb{Q}$ be a function such that\n\n$$\nf(x + f(y + f(z))) = y + f(x + z)\n$$\nfor all $x, y, z \\in \\mathbb{Q}$. Find all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "Set $x = a = f(0)$ in equation (3):\n$$\nf(y+z) = f(z) + f(y) - a\n$$\nwhich gives $f(y+z) - a = (f(y) - a) + (f(z) - a)$. Thus, $f(x) - a$ is additive, so $f(x) - a = \\alpha x$ for some $\\alpha \\in \\mathbb{Q}$. Therefore, $f(x) = \\alpha x + a$. Substituting into the original equation, we find the solutions are $f(x) = x$ and $f(x) = -x + a$, where $a \\in \\mathbb{Q}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16411,
    "subject": "Mathematics (Olympiad)",
    "question": "In a convex hexagon $ABCDEF$, $AB$ is parallel to $DE$, $BC$ is parallel to $EF$, and $CD$ is parallel to $FA$. Prove that the triangles $ACE$ and $BDF$ have the same area.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the coordinates of $A$ be $(0,0)$, $B$ be $(1,0)$, $D$ be $(c,b)$, $E$ be $(a,b)$, $F$ be $(e,f)$, and $C$ be $(x,y)$. Then $AB \\parallel DE$. For $AF \\parallel CD$, we need $\\dfrac{b-y}{c-x} = \\dfrac{f}{e}$. For $EF \\parallel CB$ we need $\\dfrac{x-1}{y} = \\dfrac{a-e}{b-f}$. From these equations, we get\n$$\nxb - ay = b - f + cf - be. \\quad (*)\n$$\nThus the proof is now complete since the following is true by $(*)$:\n$$\n[ACE] = [BDF] \\Leftrightarrow \\begin{vmatrix} 0 & 0 & 1 \\\\ x & y & 1 \\\\ a & b & 1 \\end{vmatrix} = \\begin{vmatrix} 1 & 0 & 1 \\\\ c & b & 1 \\\\ e & f & 1 \\end{vmatrix} \\Leftrightarrow xb - ay = b - f + cf - be.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16412,
    "subject": "Mathematics (Olympiad)",
    "question": "The first three terms of a geometric sequence are the integers $a$, $720$, and $b$, where $a < 720 < b$. What is the sum of the digits of the least possible value of $b$?\n\n(A) 9 (B) 12 (C) 16 (D) 18 (E) 21",
    "options": [],
    "answer": "See solution",
    "solution": "The prime factorization of $720$ is $2^4 \\cdot 3^2 \\cdot 5$. Let $r = \\frac{m}{n}$ be the common ratio of the geometric sequence, where $m$ and $n$ are relatively prime positive integers. If $n$ had any prime factor greater than $5$, then $b = 720r$ would not be an integer. Similarly, if $m$ had any prime factor greater than $5$, then $a = \\frac{720}{r}$ would not be an integer. Thus, $r = 2^i \\cdot 3^j \\cdot 5^k$, where $i, j, k$ are (not necessarily positive) integers with $|i| \\leq 4$, $|j| \\leq 2$, and $|k| \\leq 1$.\n\nTo minimize $b$, we minimize $r > 1$. Taking $r = \\frac{16}{15} = 2^4 \\cdot 3^{-1} \\cdot 5^{-1}$ gives the sequence $675$, $720$, $768$. Checking smaller $r$ values, $\\frac{17}{16}$ is not possible, and both $m$ and $n$ must borrow at least two prime factors each from $720$, which only has three distinct prime factors, so this is impossible. Therefore, the least possible value of $b$ is $768$, and the sum of its digits is $7 + 6 + 8 = 21$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16413,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the family of subsets $A_i = \\{x_i, y_i, z_i\\}$ of the set $M = \\{1, 2, 3, \\ldots, 3n\\}$, for $i = 1, 2, \\ldots, n$, satisfy $A_1 \\cup A_2 \\cup \\cdots \\cup A_n = M$. Define $s_i = x_i + y_i + z_i$. Find all possible values of $n$ such that all the $s_i$ are equal.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\nn \\mid \\frac{3n(3n+1)}{2} \\Rightarrow 2 \\mid 3n+1.\n$$\n\nHence, $n$ must be odd.\n\nFor $n$ odd, $1, 2, 3, \\ldots, 2n$ can form $n$ pairs such that the sums of each pair become an arithmetic sequence of common difference $1$:\n\n$$\n\\begin{aligned}\n& 1 + \\left(n + \\frac{n+1}{2}\\right),\\ 3 + \\left(n + \\frac{n-1}{2}\\right),\\ \\ldots,\\ n + (n+1); \\\\\n& 2 + 2n,\\ 4 + (2n-1),\\ \\ldots,\\ (n-1) + \\left(n + \\frac{n+3}{2}\\right).\n\\end{aligned}\n$$\n\nThe general term is\n\n$$\na_k = \\begin{cases} 2k-1 + \\left(n + \\frac{n+1}{2} + 1 - k\\right), & 1 \\le k \\le \\frac{n+1}{2}, \\\\ [1-n+2(k-1)] + [2n + \\frac{n+1}{2} - (k-1)], & \\frac{n+3}{2} \\le k \\le n. \\end{cases}\n$$\n\nIt is easy to see that $a_k + 3n + 1 - k = \\frac{9n+3}{2}$ is a constant, so all the following $n$ triples have the same sum:\n\n$$\n\\begin{aligned}\n& \\{1, n + \\frac{n+1}{2}, 3n\\},\\ \\{3, n + \\frac{n-1}{2}, 3n-1\\},\\ \\ldots, \\\\\n& \\{n, n+1, 3n+1 - \\frac{n+1}{2}\\}; \\\\\n& \\{2, 2n, 3n+1 - \\frac{n+3}{2}\\},\\ \\ldots,\\ \\{n-1, n + \\frac{n+3}{2}, 2n+1\\}.\n\\end{aligned}\n$$\n\nFor $n$ odd, take these triples as $A_1, A_2, \\ldots, A_n$; then they satisfy the required condition. So $n$ can be any odd number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16414,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p, q > 1$ be integers such that $(p, 6q) = 1$. Show that\n$$\n\\sum_{k=1}^{q-1} \\left[ \\frac{pk}{q} \\right]^2 \\equiv 2p \\sum_{k=1}^{q-1} k \\left[ \\frac{pk}{q} \\right] \\pmod{(q-1)},\n$$\nwhere $[x]$ denotes the greatest integer less than or equal to $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "For $\\alpha \\in \\mathbb{R}$, let $\\{\\alpha\\} = \\alpha - [\\alpha]$. We have\n$$\n\\begin{aligned}\n2p \\sum_{k=1}^{q-1} k \\left[ \\frac{pk}{q} \\right] &= 2q \\sum_{k=1}^{q-1} \\frac{pk}{q} \\left[ \\frac{pk}{q} \\right] \\\\\n&= q \\sum_{k=1}^{q-1} \\left( \\left( \\frac{pk}{q} \\right)^2 + \\left[ \\frac{pk}{q} \\right]^2 - \\left( \\frac{pk}{q} - \\left[ \\frac{pk}{q} \\right] \\right)^2 \\right) \\\\\n&= q \\sum_{k=1}^{q-1} \\left( \\frac{pk}{q} \\right)^2 + q \\sum_{k=1}^{q-1} \\left[ \\frac{pk}{q} \\right]^2 - q \\sum_{k=1}^{q-1} \\left\\{ \\frac{pk}{q} \\right\\}^2. \\quad \\textcircled{1}\n\\end{aligned}\n$$\nSince $(p, q) = 1$, the remainders of $p \\cdot 1, p \\cdot 2, \\dots, p \\cdot (q-1)$ modulo $q$ are exactly $1, 2, \\dots, q-1$. Thus,\n$$\n\\sum_{k=1}^{q-1} \\left\\{ \\frac{pk}{q} \\right\\}^2 = \\sum_{k=1}^{q-1} \\left( \\frac{k}{q} \\right)^2.\n$$\nPlugging this into $(\\textcircled{1})$ gives\n$$\n\\begin{aligned}\n2p \\sum_{k=1}^{q-1} k \\left[ \\frac{pk}{q} \\right] &= \\frac{p^2-1}{q} \\sum_{k=1}^{q-1} k^2 + q \\sum_{k=1}^{q-1} \\left[ \\frac{pk}{q} \\right]^2 \\\\\n&= \\frac{(p^2-1)(q-1)(2q-1)}{6} + q \\sum_{k=1}^{q-1} \\left[ \\frac{pk}{q} \\right]^2. \\quad \\textcircled{2}\n\\end{aligned}\n$$\nSince $(p, 6) = 1$, $6 \\mid (p^2 - 1)$; thus, from $(\\textcircled{2})$,\n$$\n2p \\sum_{k=1}^{q-1} k \\left[ \\frac{pk}{q} \\right] \\equiv \\sum_{k=1}^{q-1} \\left[ \\frac{pk}{q} \\right]^2 \\pmod{(q-1)}. \\quad \\square\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16415,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral inscribed in a circle $\\gamma$. Let $X$ be a point on the extension of $AC$ such that $XB$ and $XD$ are tangent to $\\gamma$. The tangent at $C$ intersects $XD$ at $Q$. Let $E$ be the intersection of $AQ$ with $\\gamma$ distinct from $A$. Prove that lines $AD$, $BE$, and $CQ$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that $AD$ intersects $CQ$ at $Y$ and $AD$ intersects $BE$ at $Z$. Since $\\triangle XAD \\sim \\triangle XDC$, we have\n\n$$\n\\frac{AD}{DC} = \\frac{XA}{XD} = \\frac{XA}{XB} = \\frac{AB}{BC}\n$$\nfrom which it follows that $AB \\cdot DC = BC \\cdot AD$. By Ptolemy's theorem,\n\n$$\nAB \\cdot DC = BC \\cdot AD = \\frac{CA \\cdot DB}{2} \\\\\n\\Rightarrow \\frac{DB}{AB} = \\frac{2DC}{CA}\n$$\n\nLikewise,\n\n$$\nCA \\cdot ED = CE \\cdot AD = \\frac{AE \\cdot DC}{2} \\\\\n\\Rightarrow \\frac{DC}{CA} = \\frac{2ED}{AE}\n$$\n\nSince $\\triangle YDC \\sim \\triangle YCA$, we have $\\frac{YD}{YC} = \\frac{DC}{CA} = \\frac{YC}{YA}$.\n\n![](images/Spanija_b_2014_p36_data_e8928f8edd.png)\n\nThus, taking into account the preceding, we have\n\n$$\n\\frac{YD}{YA} = \\frac{YD \\cdot YA}{YA^2} = \\left(\\frac{YC}{YA}\\right)^2 = \\left(\\frac{DC}{CA}\\right)^2 = \\left(\\frac{2ED}{AE}\\right)^2\n$$\n\nUsing the similar triangles $ABZ$ and $EDZ$, we get $\\frac{ZD}{ZB} = \\frac{ED}{AB}$. Likewise, using the similar triangles $DBZ$ and $EAZ$, we have $\\frac{ZA}{ZB} = \\frac{EA}{DB}$. Thus,\n\n$$\n\\frac{ZD}{ZA} = \\frac{ED \\cdot DB}{EA \\cdot AB} = \\left(\\frac{2ED}{AE}\\right)^2\n$$\n\nFinally, we have $Y = Z$ on account of the preceding, and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16416,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $r$ be a positive rational number such that the numbers $r$ and $\\sqrt{r+1}$ have the same fractional part. Show that $r$ is an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "The hypothesis implies that $\\sqrt{r+1} - r = a$, where $a \\in \\mathbb{Z}$.\n\nWe get: $r + 1 = r^2 + 2ra + a^2$. Let $r = \\frac{m}{n}$, with $m, n \\in \\mathbb{N}^*$ and $\\gcd(m, n) = 1$. Then, $mn + n^2 = m^2 + 2mna + a^2 n^2$, and therefore $m^2 = n(m + n - 2ma - a^2 n)$. Hence, $m^2$ is divisible by $n$, which is only possible if $n = 1$, thus $r = m \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16417,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $H$ be the orthocenter of an acute-angled $\\triangle ABC$, and let $P$ be a point on $\\widehat{BC}$ of the circumcircle of $\\triangle ABC$. The line $PH$ intersects $\\widehat{AC}$ at $M$. There exists a point $K$ on $\\widehat{AB}$ such that the line $KM$ is parallel to the Simson line of $P$ with respect to $\\triangle ABC$. The chord $QP$ is parallel to $BC$, and the chord $KQ$ intersects $BC$ at point $J$. Prove that $\\triangle KMJ$ is isosceles.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p138_data_2e16428927.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We show that $JK = JM$.\n\nDraw a line from $P$ perpendicular to $BC$, intersecting the circumcircle and $BC$ at points $S$ and $L$, respectively. Let $N$ be the projection of $P$ onto $AB$. Since $B, P, L, N$ are concyclic,\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p138_data_90b74f6550.png)\n\n$$\n\\angle SLN = \\angle NBP = \\angle ABP = \\angle ASP.\n$$\n\nThus, $NL \\parallel AS$. Since $NL \\parallel KM$, then $KM \\parallel SA$.\n\nLet $T$ be the intersection point of $BC$ and $PH$. Since points $K, Q, P, M$ are concyclic and $BC \\parallel PQ$, so are $K, J, T, M$. Suppose that the extension of $AH$ intersects the circumcircle at point $D$. Then we have\n\n$$\n\\angle JKM = \\angle MTC, \\quad \\angle KMJ = \\angle KTJ.\n$$\n\nIt suffices to show that $\\angle MTC = \\angle KTJ$.\n\nIt is easy to see that $D$ and $H$ are symmetric over line $BC$. Then\n\n$$\n\\angle SPM = \\angle SPH = \\angle THD = \\angle HDT.\n$$\n\nFurthermore, $KS = AM$, so $\\angle ADM = \\angle KPS$. Consequently,\n\n$$\n\\begin{aligned}\n\\angle TDM &= \\angle HDT + \\angle ADM = \\angle SPM + \\angle KPS \\\\\n&= \\angle KPM = \\angle KDM,\n\\end{aligned}\n$$\n\nwhich means that points $K$, $T$, and $D$ are collinear. Thus,\n\n$$\n\\angle KTJ = \\angle DTC = \\angle MTC.\n$$\n\nSo $\\angle JKM = \\angle KMJ$. Consequently, $JK = JM$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16418,
    "subject": "Mathematics (Olympiad)",
    "question": "Jo and Mike are running laps on a circular track. Jo's pattern is: five 4-hops, then two 3-hops, repeated each lap. After three laps, Jo returns to position 0. In three laps, Jo takes fifteen 4-hops and five 3-hops. Mike, following Jo's pattern but swapping the hop sizes, takes fifteen 3-hops and five 4-hops in the same time. The table below shows Mike's position after every 3 laps by Jo, up to lap 15:\n\n| Laps by Jo | 3 | 6 | 9 | 12 | 15 |\n|------------|---|---|---|----|----|\n| Mike's position | 15 | 5 | 20 | 10 | 0 |\n\nWhen do Jo and Mike first meet at position 0 after starting?",
    "options": [],
    "answer": "See solution",
    "solution": "The first time Jo and Mike meet at position 0 after starting is when Jo has completed 15 laps. At that time, Mike is behind by $5 \\times 10 = 50$ places, which is 2 laps. Therefore, Mike completes 13 laps.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16419,
    "subject": "Mathematics (Olympiad)",
    "question": "The triangle $ABC$ is isosceles, with $AB = AC$ and $\\angle ABC = 72^\\circ$. The point $D$ is taken on the line $BC$, so that $C$ is on the segment $BD$ and $CD = AB$.\n\na) Prove that $AC$ is the bisector of the angle $\\angle BAD$.\n\nb) The point $E$ is taken on the parallel to $AB$ through $D$, on the same side of $BD$ as $A$, so that $DE = DB$. Let $F$ be the common point of the lines $AD$ and $BE$. Prove that the lines $AC$ and $AE$ are perpendicular and $AF = FC = BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) In the triangle $ABC$, $\\angle BAC = 36^\\circ$. From the isosceles triangle $ACD$, with $\\angle ACD = 180^\\circ - 36^\\circ = 144^\\circ$, it follows that $\\angle CAD = \\angle ADC = 36^\\circ$, hence $AC$ is the bisector of the angle $\\angle BAD$.\n\n![](images/RMC_2023_v2_p9_data_29f02b6c20.png)\n\nb) From $\\angle ABD = \\angle BAD = 72^\\circ$ it follows that the triangle $ABD$ is isosceles, hence $AD = BD$. Since $DE = DB$, the triangle $ADE$ is isosceles, therefore $\\angle DAE = \\frac{180^\\circ - \\angle ADE}{2} = \\frac{180^\\circ - 72^\\circ}{2} = 54^\\circ$. So, $\\angle CAE = \\angle CAD + \\angle DAE = 36^\\circ + 54^\\circ = 90^\\circ$.\n\nFrom $\\angle AFB = 180^\\circ - \\angle BAF - \\angle ABF = 72^\\circ$ it follows $\\triangle BAC \\equiv \\triangle ABF$ (S.A.S.), whence $BC = AF$. Now $BD = AD$ yields $CD = FD$. This leads to $\\triangle BAC \\equiv \\triangle CDF$ (S.A.S.), hence $BC = FC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16420,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any positive real numbers $x$, $y$, $z$:\n\n$$\n\\frac{x^3}{y^2 + z^2} + \\frac{y^3}{x^2 + z^2} + \\frac{z^3}{x^2 + y^2} \\geq \\frac{x + y + z}{2}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The inequality is symmetric, so we may assume $x \\le y \\le z$. Then:\n\n$$\nx^3 \\le y^3 \\le z^3 \\quad \\text{and} \\quad \\frac{1}{y^2 + z^2} \\le \\frac{1}{x^2 + z^2} \\le \\frac{1}{x^2 + y^2}.\n$$\n\nBy the rearrangement inequality:\n\n$$\n\\begin{align*}\n\\frac{x^3}{y^2 + z^2} + \\frac{y^3}{x^2 + z^2} + \\frac{z^3}{x^2 + y^2} &\\ge \\frac{y^3}{y^2 + z^2} + \\frac{z^3}{x^2 + z^2} + \\frac{x^3}{x^2 + y^2} \\\\\n&\\ge \\frac{z^3}{y^2 + z^2} + \\frac{x^3}{x^2 + z^2} + \\frac{y^3}{x^2 + y^2} \\\\\n&\\ge \\frac{1}{2} \\left( \\frac{y^3 + z^3}{y^2 + z^2} + \\frac{x^3 + z^3}{x^2 + z^2} + \\frac{z^3 + y^3}{x^2 + y^2} \\right)\n\\end{align*}\n$$\n\nAlso, by the rearrangement inequality:\n\n$$\n\\begin{align*}\nx^3 + y^3 &\\ge xy^2 + x^2 y \\\\\n2x^3 + 2y^3 &\\ge (x^2 + y^2)(x + y) \\\\\n\\frac{x^3 + y^3}{x^2 + y^2} &\\ge \\frac{x + y}{2}\n\\end{align*}\n$$\n\nApplying this to the previous inequality:\n\n$$\n\\frac{x^3}{y^2 + z^2} + \\frac{y^3}{x^2 + z^2} + \\frac{z^3}{x^2 + y^2} \\geq \\frac{1}{2} \\left( \\frac{y+z}{2} + \\frac{x+z}{2} + \\frac{x+y}{2} \\right)\n$$\n\nWhich is the required result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16421,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(a_n)$ be a sequence defined by the recurrence relation:\n\nFor $n \\geq 4$,\n$$\na_n = a_{n-1} + \\text{gcd}(a_{n-3}, a_{n-2})\n$$\nwith given initial terms $a_1, a_2, a_3$.\n\nAssume $\\text{gcd}(a_1, a_2, a_3) = 1$. Prove that for any $n \\geq 5$, the sequence satisfies $a_{n+1} - a_n = 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $a_1$ appears only in $\\text{gcd}(a_1, a_2)$ when calculating $a_4$, we may assume $a_1 = \\text{gcd}(a_1, a_2)$. If $a_1, a_2, a_3$ share a common divisor, it factors out of the sequence, so we set $\\text{gcd}(a_1, a_2, a_3) = 1$, which implies $\\text{gcd}(a_1, a_3) = 1$.\n\nLet $\\text{gcd}(a_2, a_3) = d$, so $a_2 = bd$, $a_3 = cd$ with $b, c$ coprime, $a_1 \\mid b$, and $\\text{gcd}(a_1, cd) = 1$.\n\nCompute the first terms:\n\n$$\na_4 = a_3 + \\text{gcd}(a_1, a_2) = cd + a_1\n$$\n$$\na_5 = a_4 + \\text{gcd}(a_2, a_3) = cd + a_1 + d\n$$\n$$\na_6 = a_5 + \\text{gcd}(a_3, a_4) = a_5 + \\text{gcd}(cd, cd + a_1) = a_5 + \\text{gcd}(cd, a_1) = a_5 + 1\n$$\n$$\na_7 = a_6 + \\text{gcd}(a_4, a_5) = a_6 + \\text{gcd}(cd + a_1, cd + a_1 + d) = a_6 + \\text{gcd}(cd + a_1, d) = a_6 + \\text{gcd}(a_1, d) = a_6 + 1\n$$\n\nThus, $a_5, a_6, a_7$ are consecutive integers, so for $n \\geq 5$, $a_{n+1} - a_n = 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16422,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $(m, n)$ that satisfy the equation\n\n$$\nm! + n! = m^n + 1.\n$$\n\nHere, for a positive integer $k$, $k!$ denotes the product $1 \\cdot 2 \\cdot 3 \\cdot \\dots \\cdot k$.",
    "options": [],
    "answer": "See solution",
    "solution": "By checking trivial cases $m=1, n=1$ and $n=2$, we find the first two of the solutions listed. From now on, we assume $m \\geq 2$ and $n \\geq 3$.\n\n*Step 1.* We prove that $m > n$.\n\nSuppose $n \\geq m$. Then, numbers $m!$, $n!$ and $m^n$ are divisible by $m$, hence, $m$ also divides $m! + n! - m^n = 1$ – contradiction. In particular, it follows that $m \\geq 4$.\n\n*Step 2.* Now, we show that $n > \\frac{1}{2}m$.\n\nFrom the well-known inequality $m! > m^{\\frac{1}{2}m}$, which follows from the Root-Mean Square-Arithmetic-Mean-Geometric-Mean-Harmonic-Mean Inequality, we obtain $m^n + 1 = m! + n! > m^{\\frac{1}{2}n} + 1$.\n\n*Step 3.* We prove that $m$ is prime.\n\nOtherwise, $m$ has a prime factor $p$ not larger than $\\frac{1}{2}m$ and $n$. In that case, $p$ is a factor of $m! + n! - m^n = 1$ – contradiction.\n\n*Step 4.* We prove that $m-1$ is either prime or a square of a prime. Suppose the contrary. Then, $m-1$ can be represented as a product of two factors $a \\neq b$, both smaller than $\\frac{1}{2}m$ and $n$. Therefore, $m!$ and $n!$ are divisible by $n-1 = ab$, since $n!$ contains both these factors. $m^n + 1 = ((m-1) + 1)^n + 1$ gives a remainder of 2 when divided by $m-1$. Hence, $m-1$ is a factor of 2, which is impossible, since $m \\geq 4$.\n\nFrom the results of Steps 3 and 4 and from parity, we get that the only possible case is $m = 5$. Indeed, since $m \\geq 4$ is prime, it is odd, hence, $m-1$ is even and can be a square of a prime, if that prime is 2. From the results of Steps 1 and 2, we get that possible solutions are only $(5, 3)$ and $(5, 4)$. By checking, we see that only the first pair satisfies the initial equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16423,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given an $n \\times n$ board. Rows are labeled with numbers $1$ to $n$ downwards and columns are labeled with numbers $1$ to $n$ from left to right. On each field, we write the number $x^2 + y^2$ where $(x, y)$ are its coordinates.\n\nWe are given a figure and can initially place it on any field. In every step, we can move the figure from one field to another if the other field has not already been visited and if at least one of the following conditions is satisfied:\n\n- The numbers in those two fields give the same remainders when divided by $n$.\n- Those fields are point reflected with respect to the center of the board.\n\nCan all the fields be visited in case:\n\na) $n = 4$\n\nb) $n = 5$",
    "options": [],
    "answer": "See solution",
    "solution": "a) The answer is **NO**.\n\n![](images/MACEDONIAN_MATHEMATICAL_OLYMPIADS_2016_p9_data_d79975982b.png)\n\n$$\n\\begin{array}{|c|c|c|c|c|}\n\\hline\n & 1 & 2 & 3 & 4 \\\\\n\\hline\n1 & 2 & 5 & 10 & 17 \\\\\n2 & 5 & 8 & 13 & 20 \\\\\n3 & 10 & 13 & 18 & 25 \\\\\n4 & 17 & 20 & 25 & 36 \\\\\n\\hline\n\\end{array}\n$$\n\n$$\n\\begin{array}{|c|c|c|c|c|}\n\\hline\n1 & 2 & 1 & 2 & 1 \\\\\n2 & 1 & 0 & 1 & 0 \\\\\n3 & 2 & 1 & 2 & 1 \\\\\n4 & 1 & 0 & 1 & 0 \\\\\n\\hline\n\\end{array}\n$$\n\nOn the left is the board from the problem; on the right is the same board, but with remainders of the values from the board instead of the values themselves.\n\nWe denote field $i$ for a field with number $i$ written on it in the right table. Assume we can visit all fields. At some point, we visit a field $i$. Using the first type of move, we can visit any other field $1$ not yet visited. Also, for field $1$, the reflection of that field is also a field $1$. Thus, both types of moves lead to another field $1$. Similarly, for each step, if the figure is on field $1$, then the next and previous steps (unless first/last) must also be field $1$.\n\nThus, the first visited field $1$ must be the first step, and the last visited field $1$ must be the last step. All fields $1$ are visited consecutively, in exactly $8$ moves (since there are $8$ fields $1$), but there are $16$ moves to make. This leads to a contradiction.\n\nb) The answer is **YES**.\n\n$$\n\\begin{array}{|c|c|c|c|c|c|}\n\\hline\n & 1 & 2 & 3 & 4 & 5 \\\\\n\\hline\n1 & 2 & 5 & 10 & 17 & 26 \\\\\n2 & 5 & 8 & 13 & 20 & 29 \\\\\n3 & 10 & 13 & 18 & 25 & 34 \\\\\n4 & 17 & 20 & 25 & 36 & 41 \\\\\n5 & 26 & 29 & 34 & 41 & 50 \\\\\n\\hline\n\\end{array}\n$$\n\n$$\n\\begin{array}{|c|c|c|c|c|c|}\n\\hline\n & 1 & 2 & 3 & 4 & 5 \\\\\n\\hline\n1 & 2 & 0 & 0 & 0 & 1 \\\\\n2 & 0 & 3 & 3 & 0 & 4 \\\\\n3 & 0 & 3 & 3 & 0 & 4 \\\\\n4 & 2 & 0 & 0 & 2 & 1 \\\\\n5 & 1 & 4 & 4 & 1 & 0 \\\\\n\\hline\n\\end{array}\n$$\n\nAgain, on the left is the board from the problem; on the right is the same board, but with remainders of the values from the board instead of the values themselves.\n\nWe can move from any field to another with the same number written on the field in the right table by using the second move.\n\nOne idea to visit all the fields is:\n\n- Find $4$ pairs of fields of types field $i$ and field $j$, such that all $8$ fields are different, in each pair $i \\neq j$, those two fields in one pair are symmetric, and the second member of the $n$-th pair has the same value on the right board as the first member of the $(n+1)$-th pair. Also, all the values of the right table are mentioned through members of those pairs. For example:\n\n$((2,2),(4,4)), ((1,4),(5,2)), ((3,5),(3,1)), ((2,1),(4,5))$\n\n- The algorithm: after the second member of the $n$-th pair and before the first member of the $(n+1)$-th pair, visit all fields by using the first step. Before the first pair and after the fourth pair, move similarly. Jump from the first member of the pair to the second member of the pair by using the second step.\n\nOne way to do it: Start with the field $(3,3)$. Then visit all fields $3$, using the first move, in any way as long as the last visited field is $(2,2)$. Then, using the second move, visit the field $(4,4)$. Again, using the first move, visit all fields $2$ in any way as long as the last visited field is $(1,4)$. Using the second move, visit the field $(5,2)$. Then, using the first move, visit all fields $4$ in any way as long as the last visited field is $(3,5)$. In the same fashion, using the second move, visit the field $(4,5)$. Conclude by visiting all fields $1$ in any way.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16424,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $P, Q$ 為三角形 $ABC$ 內兩點，滿足 $\\angle BAP = \\angle CAQ$，$\\angle ACP = \\angle BCQ$，且 $\\angle CBP = \\angle ABQ$。設 $Q_1, Q_2, Q_3$ 分別為 $Q$ 對於 $BC, CA, AB$ 的對稱點。令 $D$ 為 $PQ_1$ 與 $BC$ 之交點，令 $E$ 為 $PQ_2$ 與 $CA$ 之交點，令 $F$ 為 $PQ_3$ 與 $AB$ 交點。試證：$AD, BE, CF$ 三線共點。",
    "options": [],
    "answer": "See solution",
    "solution": "首先，令 $\\angle ACP = \\angle BCQ = \\theta$，則 $\\angle ACQ = \\angle BCP = \\angle C - \\theta$。又因為 $Q_1$ 為 $Q$ 關於邊 $C$ 的對稱點，故 $\\angle QCB = \\angle Q_1CB$ 且 $CQ = CQ_1$。因此，$\\angle PCQ_1 = \\angle PCB + \\angle Q_1CB = \\angle ACQ + \\angle BCQ = \\angle C$。\n\n接著，令 $P$ 在 $BC, CA, AB$ 的垂足分別為 $M_1, M_2, M_3$，$Q$ 在 $BC, CA, AB$ 的垂足分別為 $N_1, N_2, N_3$。令 $S_{\\triangle ABC}$ 代表 $\\triangle ABC$ 的面積。則：\n\n$$\n\\begin{aligned}\nS_{\\triangle PCQ_1} &= S_{\\triangle PCD} + S_{\\triangle Q_1CD} \\\\\n&\\Rightarrow CP \\times CQ \\times \\sin PCQ_1 = CP \\times CD \\times \\sin PCD + CD \\times CQ_1 \\times \\sin DCQ_1 \\\\\n&\\Rightarrow CP \\times CQ \\times \\sin \\angle C = CD \\times CP \\times \\sin PCD + CD \\times CQ_1 \\times \\sin DCQ_1 \\\\\n&\\Rightarrow CD \\times PM_1 + CD \\times Q_1N_1 = CD \\times (PM_1 + QN_1).\n\\end{aligned}\n$$\n\n因此 $CD = \\dfrac{CP \\times CQ \\times \\sin \\angle C}{PM_1 + QN_1}$。\n\n同理可證 $CE = \\dfrac{CP \\times CQ \\times \\sin \\angle C}{PM_2 + QN_2}$，因此 $\\dfrac{CD}{CE} = \\dfrac{PM_2 + QN_2}{PM_1 + QN_1}$。\n\n同理，$\\dfrac{AE}{AF} = \\dfrac{PM_3 + QN_3}{PM_2 + QN_2}$ 且 $\\dfrac{BF}{BD} = \\dfrac{PM_1 + QN_1}{PM_3 + QN_3}$，故：\n\n$$\n\\frac{AE}{EC} \\times \\frac{CD}{DB} \\times \\frac{BF}{FA} = \\frac{CD}{CE} \\times \\frac{AE}{AF} \\times \\frac{BF}{BD} = 1,\n$$\n\n因此由 Ceva 定理知 $AD, BE, CF$ 三線共點。得證。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16425,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute non-isosceles triangle with orthocentre $H$. Let $O$ be the circumcentre of triangle $ABC$, and let $K$ be the circumcentre of triangle $AHO$. Prove that the reflection of $K$ in $OH$ lies on $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider the configuration as in the figure. Other configurations are treated analogously. Denote by $D$ the second intersection of $AH$ with the circumcircle of $\\triangle ABC$. Denote by $S$ the second intersection of the circumcircles of $ABC$ and $AHO$. (Because $\\triangle ABC$ is acute, both $O$ and $H$ lie in the interior of $ABC$ and also in the interior of the circumcircle, hence $D$ and $S$ both exist.)\n\n![](images/NLD_ABooklet_2021_p30_data_b039fb20e0.png)\n\nWe have\n$$\n\\angle OSH = \\angle OAH = \\angle OAD = \\angle ODA = \\angle ODH,\n$$\nwhere we use that $|OA| = |OD|$. Moreover, we have\n$$\n\\angle OHD = 180^\\circ - \\angle OHA = 180^\\circ - \\angle OSA = 180^\\circ - \\angle OAS = \\angle OHS,\n$$\nwhere we use that $|OA| = |OS|$. Now we conclude that $\\triangle OHS \\cong \\triangle OHD$ (SAA). This yields that $D$ and $S$ are each other's reflection images in $OH$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16426,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle with orthocenter $H$. Consider the points $Y$ and $Z$ on the sides $CA$ and $AB$ respectively such that the directed angles $(AC, HY) = -\\pi/3$ and $(AB, HZ) = \\pi/3$. Let $U$ be the circumcenter of $\\triangle HYZ$.\n\nProve that the points $A, N, U$ are collinear, where $N$ is the nine-point center of $\\triangle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O_A$ be the reflection of $O$ into the sideline $BC$. We will prove the result by showing that $A, N, O_A$ are collinear, and that $A, U, O_A$ are collinear.\n\n$A, N, O_A$ are collinear from the fact that $N$ is the midpoint of $HO$ on the Euler line, and that $AH = OO_A$ and both $AH$ and $OO_A$ are perpendicular to $BC$. ($\\triangle AHN \\cong \\triangle O_AON$.)\n\nTo show that $A, U, O_A$ are collinear, we will first show that $H, U, A^+$ are collinear, where $A^+$ is a point on the same side as $A$ with respect to $BC$ such that $A^+BC$ is equilateral.\n\nConsider the case where $\\angle BAC \\neq 60^\\circ$. (The case where $\\angle BAC = 60^\\circ$ can be done similarly using the same idea.) In this case $AZHY$ is not a parallelogram. Let $Y', Z'$ be the intersection points between $AY$ and $HZ$ and between $AZ$ and $HY$, as in the picture.\n\nLet $V$ be the orthocenter of $HYZ$, thus the lines $HU$ and $HV$ are isogonal conjugate with respect to the angle $\\angle ZHY$. Note also that the quadrilateral $YZY'Z'$ is cyclic, since $\\angle Y'ZZ' = \\angle Y'YZ' = 60^\\circ$. Thus, the lines $Y'Z'$ and $YZ$ are antiparallel. Since $IIV \\perp YZ$, then $IIU \\perp Y'Z'$.\n\nLet $C'$ be the reflection of $C$ in the line $HY'$. Using the directed angle\n\n$$\n\\begin{aligned}\n(IIY', IIC) &= (IIY' \\cdot CA) + (CA \\cdot IIC) \\\\\n&= (AB \\cdot AC) - 60^\\circ + 90^\\circ - (AB \\cdot AC) = 30^\\circ.\n\\end{aligned}\n$$\n\nThis implies that $HCC'$ is an equilateral triangle, and $\\triangle HCC' \\sim \\triangle A^+BC$.\n\nNote also that $\\triangle Y'HC \\sim \\triangle Z'HB$ since $\\angle HY'C = \\angle HZ'B$ and $\\angle HCY' = \\angle HZB' = 90^\\circ - \\angle BAC$. Since $\\triangle Y'HC'$ is the reflection of $\\triangle Y'HC$, thus $\\triangle Y'HC' \\sim \\triangle Z'HB$ with the common vertex $H$. Therefore, (by spiral transformation or by simple comparison), $\\triangle Z'HY' \\sim \\triangle BHC'$. It follows that\n\n$$\n\\begin{aligned}\n(Y'Z', A^+H) &= (Y'Z', BC') + (BC' \\cdot A^+II) \\\\\n&= (Z'H, BH) + (BC \\cdot A^+C) = 90^\\circ.\n\\end{aligned}\n$$\n\nThis shows that $HA^+ \\perp Y'Z'$, namely, $H, U, A^+$ are collinear.\n\n![](images/Tajland_2011_p5_data_045d15c862.png)\n\nTo show $A, U, O_A$ are collinear, it suffices to show that $\\triangle AHU \\sim \\triangle A^+O_AU$ since $H, U, A^+$ are collinear. Since $A^+O \\perp BC$, then $A^+$ lies on $OO_A$ and $AH \\parallel A^+O_A$. Therefore, we only need to show that $\\frac{AH}{HU} = \\frac{A^+O_A}{A^+U}$.\n\nBy the law of sines to triangle $HYZ$,\n\n$$\nUH = \\frac{YZ}{2 \\sin YHZ} = \\frac{YZ}{2 \\sin(120^\\circ - A)}\n$$\n\nSince $\\triangle HCC'$ and $\\triangle A^+CB$ are equilateral with common vertex $C$, we can easily see that $\\triangle A^+HC \\cong \\triangle BC'C$. From similarity of triangles, $\\triangle Z'HY' \\sim \\triangle BC'H$ and $\\triangle HYZ \\sim \\triangle HY'Z'$, \n\n$$\nA^+H = BC' = Y'Z' \\cdot \\frac{BH}{Z'H} = YZ \\cdot \\frac{Y'H \\cdot BH}{ZH \\cdot Y'H} = YZ \\cdot \\frac{\\cos 30^\\circ}{|\\cos(A)|} = YZ \\cdot \\frac{\\sqrt{3}}{2|\\cos(A)|}\n$$\n\n$$\n\\frac{UH}{A^+H} = \\frac{|\\cos(A)|}{\\sqrt{3} \\sin(120^\\circ - A)}\n$$\n\nSince $AH = 2R \\cos(A)$ where $R$ is the circumradius of $\\triangle ABC$,\n\n$$\nAH + A^+O_A = 3R \\cos(A) + BC \\sin 60^\\circ = 2\\sqrt{3}R \\sin(120^\\circ - A).\n$$\n\nTherefore,\n\n$$\n\\frac{AH}{AH + A^+O_A} = \\frac{|\\cos(A)|}{\\sqrt{3}\\sin(120^\\circ - A)} = \\frac{UH}{A^+H} = \\frac{UH}{A^+U + UH}\n$$\n\nThis implies that $\\frac{AH}{HU} = \\frac{A^+O_A}{A^+U}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16427,
    "subject": "Mathematics (Olympiad)",
    "question": "設圓 $O$ 為 $\\triangle ABC$ 的外接圓，$\\angle A$ 的角平分線交圓 $O$ 於第二點 $P$，並交 $BC$ 於 $D$。由 $D$ 對 $AB$ 作垂線，交 $AB$ 於 $E$，並交圓 $O$ 於 $Q$，使得 $E$ 在 $DQ$ 線段上。再連 $PQ$ 交 $BC$ 於 $F$，連 $AF$ 交 $CQ$ 於 $G$。試證：$EG$ 平行 $FC$。",
    "options": [],
    "answer": "See solution",
    "solution": "首先，由於\n\n$$\n\\angle PDB = \\frac{1}{2}(\\widehat{CA} + \\widehat{BP}) = \\frac{1}{2}(\\widehat{CA} + \\widehat{CP}) = \\angle AQP,\n$$\n\n因此 $A, D, F, Q$ 四點共圓。故，$\\angle AQE = \\angle AQD = \\angle AFD = \\angle GFC$。\n\n$$\n\\angle EAQ = \\angle BAQ = \\angle BCQ = \\angle FCG,\n$$\n\n故 $\\angle CGF = \\angle AEQ = 90^\\circ$，因此 $A, G, E, Q$ 四點共圓，從而 $\\angle QGE = \\angle QAE = \\angle QCB$。故，$EG$ 平行 $FC$。證畢。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16428,
    "subject": "Mathematics (Olympiad)",
    "question": "$P(x), Q(x) \\in \\mathbb{Z}[x]$ бөгөөд $P(x), Q(x)$-г зэрэг хуваадаг рационал коэффициенттэй тогтмол биш олон гишүүнт олддоггүй байг. $n \\in \\mathbb{N}$ бүрийн хувьд $P(n)$ ба $Q(n)$-үүд нь зэрэг бөгөөд $2^{Q(n)} - 1$ бол $Q(x)$ тогтмол тоо гэдгийг батал.",
    "options": [],
    "answer": "See solution",
    "solution": "1) $\\exists d \\in \\mathbb{N}, \\forall n \\in \\mathbb{N}((P(n), Q(n)) \\le d)$ гэдгийг харуулъя:\n\n$$\n\\exists R_0(x), S_0(x) \\in \\mathbb{Q}[x] \\text{ such that } (P(x)R_0(x) - Q(x))S_0(x) = 1.\n$$\n\nТохирох $d \\in \\mathbb{N}$-ээр үүнийг үржүүлж $R(x) = dR_0(x)$, $S(x) = dS_0(x) \\in \\mathbb{Z}[x]$-ийн хувьд $P(x)R(x) - Q(x)S(x) = d$ гэж олно. Эндээс $n \\in \\mathbb{N}$ бүрийн хувьд $(P(n), Q(n)) \\le d$ гэж гарна. Эсрэгээр нь $Q(x)$ тогтмол биш гэж үзье.\n\n2) $Q(n), n \\in \\mathbb{N}$ дараалал зааглагдахгүй.\n\n$\\deg P(x) = k$, $P(x) = \\sum_{i=0}^{k} a_i x^i$ байг. Ямар нэг $n \\in \\mathbb{N}$-г авч үзье, $a = Q(n)$, $M = 2^a - 1$; $|3|_M = b$ гэвэл $2^a - 1 \\mid 3^{P(n)} - 1$-ээс $b \\mid P(n)$ гэж гарна. $\\forall t \\in \\mathbb{N}$-ийн хувьд $2^{Q(n+at)} - 1 \\mid 3^{P(n+at)} - 1$, $a = Q(n) \\mid Q(n+at)$ тул $2^a - 1 \\mid 3^{P(n+at)} - 1 \\Rightarrow b \\mid P(n+at)$. Иймд $P(n+at), t \\ge 0$ тоонуудын ХИЕХ $(P(n+at) | t \\ge 0)$-г $b$ хуваана. Тэгвэл $b \\mid \\Delta^k P(n) = \\sum_{i=0}^{k} (-1)^{k-i} C_k^i P(n + ai) = a_k \\cdot k! \\cdot a^k = a_k \\cdot k! \\cdot Q(n)^k$. $P(x), Q(x)$ нь бодлогын нөхцөлд оршвол $P(x), Q(x) = a_k \\cdot k! \\cdot Q(x)^k$ нь мөн бодлогын нөхцөлд орших тул 1)-д гүйцэтгэсний адилаар $(P(n), a_k \\cdot k! \\cdot Q(n)^k)$ утгууд зааглагдана гэдгээс $3^b - 1$ тоонууд мөн зааглагдана гэж гарна. Эндээс $2^{Q(n)}-1$ тоонууд мөн зааглагдана гэж гарч 2)-д зөрчинө.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16429,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer for which the last 4 digits of $2011n$ are $9999$. What is the value of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "Since the one's digit of $2011n$ is $9$, the one's digit of $n$ must be $9$. Thus, $n = 10k + 9$ for some non-negative integer $k$. Then,\n\n$$2011n = 2011(10k + 9) = 10 \\cdot 2011k + 18099$$\n\nThe one's digit of $2011n$ is $9$, so the one's digit of $k$ must be $0$, i.e., $k = 10\\ell$. Thus, $n = 100\\ell + 9$.\n\nNow,\n\n$$2011n = 2011(100\\ell + 9) = 100 \\cdot 2011\\ell + 18099$$\n\nThe hundred's digit of $2011n$ is $9$, so the one's digit of $2011\\ell$ must be $9$, which means the one's digit of $\\ell$ is $9$, i.e., $\\ell = 10m + 9$. Thus, $n = 1000m + 909$.\n\nNow,\n\n$$2011n = 2011(1000m + 909) = 1000 \\cdot 2011m + 2011 \\cdot 909$$\n\nCalculate $2011 \\cdot 909 = 1827999$. The thousand's digit of $2011n$ must be $9$, so the one's digit of $2011m$ must be $2$, which means the one's digit of $m$ is $2$. Thus, the last 4 digits of $n$ are $2909$.\n\nFinally, $2011 \\cdot 2909 = 5849999$, so the answer is $5849999$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16430,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $\\mathbb{Z}$ 為全體整數所成之集合。考慮滿足\n\n$$\nf(f(x + y) + y) = f(f(x) + y)\n$$\n\n的所有函數 $f : \\mathbb{Z} \\to \\mathbb{Z}$，其中 $x, y$ 為任意整數。對於這樣的函數 $f$，如果整數 $v$ 讓集合\n\n$$\nX_v = \\{x \\in \\mathbb{Z} : f(x) = v\\}\n$$\n\n是一個有限非空集，我們就稱 $v$ 是 $f$-稀有整數。\n\n(a) 證明：存在這樣的函數 $f$ 使得 $f$-稀有整數存在。\n\n(b) 證明：不存在滿足題設的函數 $f$，其值域含有超過一個 $f$-稀有整數。",
    "options": [],
    "answer": "See solution",
    "solution": "令 $f$ 為如下定義的函數：$f(0) = 0$，且對於 $x \\neq 0$，$f(x) := 2^{p_x}$，其中 $p_x$ 為 $2x$ 所能整除的最大 2 的冪次。顯然，整數 0 是 $f$-稀有整數，接下來驗證此函數滿足題設的函數方程。\n\n由於對所有 $x$，$f(2x) = 2f(x)$，因此只需驗證當 $x$ 或 $y$ 至少有一個為奇數時的情況（$x = y = 0$ 的情形顯然成立）。若 $y$ 為奇數，則\n\n$$\nf(f(x + y) + y) = 2 = f(f(x) + y)\n$$\n\n因為 $f$ 的所有值皆為偶數。若 $x$ 為奇數且 $y$ 為偶數，則\n\n$$\nf(x + y) = 2 = f(x)\n$$\n\n因此函數方程亦成立。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16431,
    "subject": "Mathematics (Olympiad)",
    "question": "Take a chequered board of size $10 \\times 10$ and consider all possible patterns of painting 10 squares so that there is exactly one painted square in every row and every column. For each pattern, find the rectangle of maximal area whose sides are aligned with the grid lines and which contains no painted squares. What is the maximal possible area of such a rectangle?",
    "options": [],
    "answer": "See solution",
    "solution": "Consider rectangles with sides along the grid lines.\n\nSuppose there is a rectangle of dimensions $A \\times B$ (width $A$, height $B$) containing no painted squares. Since each column must have one painted square, the $A$ columns covered by the rectangle must have $A$ painted squares, but these cannot be in the $B$ rows covered by the rectangle. Therefore, the $A$ painted squares must be placed in the remaining $(10 - B)$ rows. Thus, we require $10 - B \\geq A$.\n\nSimilarly, $A + B \\leq 10$ (since the rectangle cannot cover more than 10 rows and columns). The maximal area is achieved when $A = B = 5$, giving an area of $25$.\n\nWe can construct an example with such a $5 \\times 5$ rectangle.\n\n![](images/Ukrajina_2008_p16_data_27135a2117.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16432,
    "subject": "Mathematics (Olympiad)",
    "question": "In a group of $n$ people, everyone has at least $k$ friends. Each day, every member of the group shares with all their friends all the news that they received in the previous days. Suppose that if some information is revealed to any member of the group, then after some number of days all the members will eventually know the news. Prove that actually all the members know the news after at most $\\dfrac{3n}{k}$ days.",
    "options": [],
    "answer": "See solution",
    "solution": "By a path between members $A$ and $B$ we mean a sequence $A = A_0, A_1, A_2, \\dots, A_d = B$, where $A_i$ and $A_{i+1}$ are friends for $i = 0, 1, 2, \\dots, d-1$. The smallest possible number $d$ in such a sequence is called the distance between $A$ and $B$. By assumption, for any two members there exists a path between them. We need to show that the distance between any two members is at most $\\dfrac{3n}{k}$.\n\nTake any two members $A, B$ and let $A = A_0, A_1, A_2, \\dots, A_t = B$ be the shortest path between them. Then for all $0 \\leq i < j \\leq t$, the distance between $A_i$ and $A_j$ is equal to $j - i$. This implies that the distance between a friend of $A_i$ and a friend of $A_j$ is at least $j - i - 2$.\n\nNow, for $i = 0, 1, 2, \\dots, t$, let $F_i$ be the set of all friends of $A_i$. By the previous observation, the $\\lfloor t/3 \\rfloor + 1$ sets $F_0, F_3, F_6, \\dots$ are pairwise disjoint. Each of these sets contains at least $k$ people. Therefore, $\\left(\\lfloor t/3 \\rfloor + 1\\right)k \\leq n$, so $\\dfrac{t}{3} \\cdot k \\leq n$. Hence, $t \\leq \\dfrac{3n}{k}$, and the solution is complete.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16433,
    "subject": "Mathematics (Olympiad)",
    "question": "On a circle of radius $r$, the distinct points $A, B, C, D$, and $E$ lie in this order, satisfying $AB = CD = DE > r$. Show that the triangle with vertices at the centroids of the triangles $ABD$, $BCD$, and $ADE$ is obtuse.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$, $Q$, and $R$ be the centroids of triangles $ABD$, $BCD$, and $ADE$, respectively. Let $K$ and $L$ be the midpoints of $BD$ and $AD$. Since $P$ and $Q$ are centroids, they divide the medians $AK$ and $CK$ in the same ratio: $AP : PK = CQ : QK = 2 : 1$, so $PQ \\parallel AC$. Similarly, $PR \\parallel BE$. Thus, $\\angle QPR$ has the same measure as $\\angle CXE$, where $X$ is the intersection of $AC$ and $BE$ (see Fig. 1).\n\n![](images/Cesko-Slovacko-Poljsko_2015_p1_data_8cdb68e6ba.png)\n\nLet $\\varphi$ be the measure of the inscribed angle determined by chord $AB$. Since $CD = DE = AB$, we have $\\angle CAE = 2\\varphi$, so in triangle $AXE$,\n\n$$\n\\angle CXE = 180^{\\circ} - \\angle AXE = \\varphi + 2\\varphi = 3\\varphi.\n$$\n\nSince $AB > r$, $\\varphi > 30^{\\circ}$, so $\\angle QPR = 3\\varphi > 90^{\\circ}$.\n\n*Remark.* The points $P$, $Q$, $R$ always form a triangle (they are not collinear), since the diagonals $AC$ and $BE$ of the cyclic quadrilateral $ABCE$ always determine an angle less than $180^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16434,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $a$, $b$, and $c$ such that $a^2 + 4b^2 + 9c^2 = 48$, find the values of $a$, $b$, and $c$ if $(a-2b)^2 + (2b-3c)^2 + (a-3c)^2 = 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "From $(a + 2b + 3c)^2 = a^2 + 4b^2 + 9c^2 + 2(2ab + 3ac + 6bc)$, and given $a^2 + 4b^2 + 9c^2 = 48$, we have:\n\n$$(a-2b)^2 + (2b-3c)^2 + (a-3c)^2 = 2(a^2+4b^2+9c^2) - 2(2ab+3ac+6bc) = 96 - 96 = 0.$$ \n\nTherefore, $a = 2b = 3c$. Solving, we get $a = 4$, $b = 2$, and $c = \\frac{4}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16435,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $u = x + y$ and $v = x - y$. If the function $f$ satisfies the equation\n\n$$\nf^2(u) = f(uv) + (u^2 - v^2)f(u),\n$$\n\nfind all possible functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "Setting $u = 1$ gives $f^2(1) = f(v) + (1 - v^2)f(1)$. Setting also $v = 1$ gives $f^2(1) = f(1)$ so we have\n\n$$\nf(v) = c v^2 \\quad \\text{where } c = f(1).\n$$\n\nFrom $f^2(1) = f(1)$ we now get\n\n$$\nc^3 = f(c) = f^2(1) = f(1) = c\n$$\n\nso $c \\in \\{0, \\pm 1\\}$. Finally,\n\n$$\nc (c u^2)^2 = c (u v)^2 + (u^2 - v^2) c u^2\n$$\n\nif $c^3 = c$, so the functions $f_c(x) = c x^2$ for $c \\in \\{0, \\pm 1\\}$ do indeed give solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16436,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an equilateral and acute $\\triangle ABC$ with $AC > BC$. Some point $P$ in the interior of $\\triangle ABC$ is such that $\\angle APB = 180^\\circ - \\angle ACB$ and some $AP$ and $BP$ intersect segments $BC$ and $AC$ at points $A_1$ and $B_1$, respectively. Let $M$ be the midpoint of $A_1B_1$, and the circumscribed circles about $\\triangle ABC$ and $\\triangle A_1CB_1$ intersect for the second time at point $Q$. Show that $\\angle PQM = \\angle BQA_1$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P'$ be symmetric to $P$ with respect to the midpoint $N$ of $AB$. Then we have that $AP'BP$ is a parallelogram, hence $AP'BC$ is inscribed. Hence we get that $\\angle AQP' = \\angle ABP' = \\angle BAP$. On the other hand, $\\angle AQP = \\angle AQC - \\angle PQC = 180^\\circ - \\angle ABC - \\angle AA_1B = \\angle PAB$, where we used that $P$ lies on the circle circumscribed about $\\triangle A_1B_1C$. Thus we get that $\\angle AQP' = \\angle AQP$, from which it follows that $P, Q, P'$ lie on one line, i.e. $N \\in PQ$. We also have that $\\triangle AQB \\sim \\triangle B_1QA_1$, which means that $\\angle NQB = \\angle MQA_1$ as corresponding elements. The latter is equivalent to $\\angle PQB = \\angle MQA_1$, from which it follows that $\\angle PQM = \\angle BQA_1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16437,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle with $AC > AB$ and incircle $\\omega$. Let $\\omega$ touch the sides $BC$, $CA$, and $AB$ at $D$, $E$, and $F$ respectively. Let $X$ and $Y$ be points outside $\\triangle ABC$ satisfying\n\n$$\n\\angle BDX = \\angle XEA = \\angle YDC = \\angle AFY = 45^{\\circ}.\n$$\n\nProve that the circumcircles of $\\triangle AXY$, $\\triangle AEF$, and $\\triangle ABC$ meet at a point $Z \\neq A$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $AB \\neq AC$, the circles $(AEF)$ and $(ABC)$ are not tangent at $A$, so there is a point $S \\neq A$ which is the second intersection of $(AEF)$ and $(ABC)$.\n\nConsider an inversion about the incircle, and let the inverse of a point $P$ be denoted by $P'$. Note that $A'$ is the midpoint of $EF$. $S'$ is the foot from $D$ onto $EF$, as $S' \\neq A'$ is on the nine-point circle of $\\triangle DEF$ as well as on $EF$.\n\nAlso, since $X$ satisfies $\\angle XEI = \\angle XDI = 45^{\\circ}$, point $X'$ satisfies $\\angle EX'I = \\angle DX'I = 45^{\\circ}$. Note that $X$ lies on the same side of $DE$ as $I$, which in turn is the same side of $DE$ as $F$ since $\\triangle DEF$ is acute.\n\nTaking $DEF$ to be the reference triangle, the problem becomes:\n\n*Inverted Problem:* In acute $\\triangle DEF$, let $X', Y'$ be points such that $DX'E$ and $FY'D$ are isosceles right triangles with $X'$ on the same side of $DE$ as $F$, and $Y'$ on the same side of $DF$ as $E$. Let $A'$ be the midpoint of $EF$ and $S'$ the foot from $D$ onto $EF$. Prove that points $A', S', X', Y'$ are concyclic.\n\nWe will prove this by showing that $\\angle X'A'Y' = \\angle X'S'Y' = 90^{\\circ}$.\n\n**Claim 1.** $\\angle X'S'Y' = 90^{\\circ}$.\n\n*Proof.* Observe that $\\angle X'S'Y' = \\angle X'S'D + \\angle DS'Y' = \\angle X'ED + \\angle DFY' = 45^{\\circ} + 45^{\\circ} = 90^{\\circ}$.\n\n**Claim 2.** $X'A'Y'$ is a right isosceles triangle with a right angle at $A'$.\n\n*Proof.* To prove this, we will prove that $\\triangle X'A'Y' \\sim \\triangle X'MD$ where $M$ is the midpoint of $DE$. By spiral similarity, this is equivalent to showing that $\\triangle X'MA' \\sim \\triangle X'DY'$.\n\nNow,\n\n$$\n\\angle X'MA' = 90^{\\circ} - \\angle FDE = 45^{\\circ} + 45^{\\circ} - \\angle FDE = \\angle FDX' + \\angle Y'DE - \\angle FDE = \\angle X'DY'\n$$\n\nand also\n\n$$\n\\frac{X'M}{MA'} = \\frac{DE}{DF} = \\frac{X'D}{DY'}\n$$\n\nThus, we get the similarity by SAS similarity and thus\n\n$$\n\\triangle X'MA' \\sim \\triangle X'DY' \\implies \\triangle X'A'Y' \\sim \\triangle X'MD' \\implies \\angle X'A'Y = \\angle X'MD = 90^{\\circ}\n$$\n\nThus, we are done!",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16438,
    "subject": "Mathematics (Olympiad)",
    "question": "Expand $$(x+1)^n(x^2+bx+1)$$ and determine for which values of $n$ and $b$ all coefficients in the expansion are positive.",
    "options": [],
    "answer": "See solution",
    "solution": "Expanding $$(x+1)^n(x^2+bx+1)$$ gives\n\n$$\nx^{n+2} + (n+b)x^{n+1} + \\sum_{k=0}^{n-2} \\left[ \\binom{n}{k+2} + b\\binom{n}{k+1} + \\binom{n}{k} \\right] x^{n-k} + (n+b)x + 1.\n$$\n\nWe require $n > 2$ so that $n+b > 0$. Also, for all $k=0, \\dots, n-2$,\n$$\n\\binom{n}{k+2} + b\\binom{n}{k+1} + \\binom{n}{k} > 0.\n$$\nThis is equivalent to\n$$\nH = \\frac{\\binom{n}{k+2} + b\\binom{n}{k+1} + \\binom{n}{k}}{\\binom{n}{k}} > 0.\n$$\nDirect simplification gives\n$$\nH = \\frac{(2-b)k^2 + (b-2)(n-2)k + 2bn + n^2 - n + 2}{(k+2)(k+1)}.\n$$\nThe numerator, as a quadratic in $k$, has leading coefficient $(2-b) > 0$ and discriminant $[(b-2)(n-2)]^2 - 4(2-b)(2bn+n^2-n+2) = (n+2)(b-2)(bn+2b+2n)$. Since $b-2 < 0$, $H$ is positive if and only if $bn+2b+2n > 0$, or $n > -\\frac{2b}{b+2}$.\n\n**Remark:** In general, for a real-coefficient polynomial $p(x)$, $p(x) > 0$ for all $x \\ge 0$ if and only if there exists a positive integer $n$ such that all coefficients of $(x+1)^n p(x)$ are positive. See Proposition VI.1.3 in J. P. D'Angelo, *Inequalities from Complex Analysis*, Carus Mathematical Monographs 28, MAA (2002).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16439,
    "subject": "Mathematics (Olympiad)",
    "question": "In the real numbers, solve the system of equations\n\n$$\n\\sqrt{\\sqrt{x} + 2} = y - 2,\n$$\n\n$$\n\\sqrt{\\sqrt{y} + 2} = x - 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $(x, y)$ be any solution of the given system. Since $\\sqrt{\\sqrt{x} + 2}$ is positive, we have $y > 2$ by the first equation. Similarly, the second equation implies $x > 2$.\n\nNow we prove that $x$ and $y$ must be equal. The square root function is increasing. If $x > y$, then\n\n$$\n\\sqrt{\\sqrt{x} + 2} > \\sqrt{\\sqrt{y} + 2},\n$$\n\ni.e., $y - 2 > x - 2$, so $y > x$, a contradiction. The case $x < y$ is eliminated similarly. Thus, $x = y$.\n\nSubstituting $x = y$ into the system, we get\n\n$$\n\\sqrt{\\sqrt{x} + 2} = x - 2. \\qquad (1)\n$$\n\nLet $s = \\sqrt{x}$, so $x = s^2$. Then (1) becomes $\\sqrt{s + 2} = s^2 - 2$, with $s > \\sqrt{2}$. Squaring both sides:\n\n$$\ns + 2 = (s^2 - 2)^2 \\implies s^4 - 4s^2 - s + 2 = 0.\n$$\n\nThis factors as $(s - 2)(s^3 + 2s^2 - 1) = 0$. For $s > \\sqrt{2}$, $s^3 + 2s^2 - 1 > 0$, so $s = 2$ is the only solution. Thus, $x = s^2 = 4$, and so $x = y = 4$.\n\n*Conclusion.* The given system of equations has a unique solution $(x, y) = (4, 4)$.\n\n---\n\n*Alternative approach:* Squaring both equations gives\n\n$$\n\\begin{aligned}\n\\sqrt{x} + 2 &= (y - 2)^2, \\\\\n\\sqrt{y} + 2 &= (x - 2)^2.\n\\end{aligned}\n$$\n\nSubtracting yields\n\n$$\n\\sqrt{x} - \\sqrt{y} = (y - x)(x + y - 4) = (\\sqrt{y} - \\sqrt{x})(\\sqrt{y} + \\sqrt{x})(x + y - 4).\n$$\n\nIf $x \\neq y$, dividing both sides by $\\sqrt{y} - \\sqrt{x} \\neq 0$ gives\n\n$$\n-1 = (\\sqrt{y} + \\sqrt{x})(x + y - 4),\n$$\n\nbut since $x, y > 2$, the right side is positive, a contradiction. Thus, $x = y$.\n\nNow, as before, the only solution is $x = y = 4$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16440,
    "subject": "Mathematics (Olympiad)",
    "question": "Trapezoid $ABCD$ with $BC \\parallel AD$ is given. On diagonals $AC$ and $BD$, denote points $P$ and $Q$ respectively, so that $AC$ bisects $\\angle BPD$ and $BD$ bisects $\\angle AQC$. Prove that $\\angle BPD = \\angle AQC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M_1$ and $M_2$ be the midpoints of the diagonals. Then $M_1M_2 \\parallel BC$. Consider the circumcircles of $\\angle AQC$ and $\\angle BPD$. Let them intersect $BD$ and $AC$ at points $X$ and $Y$, respectively. Since $BD$ contains the bisector of $\\angle AQC$, point $X$ is the midpoint of the larger arc $AC$ on the circumcircle of $\\angle AQC$, so $X$ lies on the perpendicular bisector of $AC$. Similarly, $Y$ lies on the perpendicular bisector of $BD$. Thus, $XM_1M_2Y$ is cyclic, as $\\angle XM_1Y = \\angle XM_2Y = 90^\\circ$. Therefore, $\\angle M_1XM_2 = \\angle M_1YM_2$. Also, $\\angle BXY = \\angle CM_1M_2 = \\angle M_1CB$, so $XBCY$ is cyclic. Hence, $\\angle BXC = \\angle BYC$. Then\n$$\n\\angle AQC = 180^\\circ - 2\\angle M_1XC = 180^\\circ - 2(\\angle M_1XM_2 + \\angle M_2XC) = 180^\\circ - 2(\\angle M_1YM_2 + \\angle M_1YB) = 180^\\circ - 2\\angle M_2YB = \\angle BPD.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16441,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a function $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(f(x)) = x f(x) + 2x \\quad \\text{for all } x \\in \\mathbb{R},\n$$\nand $f(\\alpha) = -2$ for some real $\\alpha$?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, for contradiction, that such a function $f$ exists with $f(f(x)) = x f(x) + 2x$ for all $x$, and $f(\\alpha) = -2$ for some $\\alpha$.\n\nThen:\n\n- $f(-2) = f(f(\\alpha)) = \\alpha f(\\alpha) + 2\\alpha = -2\\alpha + 2\\alpha = 0$\n- $f(0) = f(f(-2)) = -2 f(-2) - 4 = -4$\n- $f(-4) = f(f(0)) = 0 \\cdot f(0) + 2 \\cdot 0 = 0$\n- $f(0) = f(f(-4)) = -4 f(-4) - 8 = -4 \\cdot 0 - 8 = -8$\n\nBut this gives $f(0) = -4$ and $f(0) = -8$, a contradiction. Thus, such a function does not exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16442,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ and $Q$ be polynomials with rational coefficients. Suppose $a$ is a root of $P$, and $a + 2016$ is a root of $Q$. If $b$ is the other root of $P$, what is the only possible value for the other root of $Q$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $c$ denote the other root of $Q$. Since $P$ and $Q$ have rational coefficients, Vieta's formulas imply\n\n$$\n\\begin{aligned}\na + b &= p_1 \\in \\mathbb{Q}, \\quad ab = p_2 \\in \\mathbb{Q}, \\\\\n(a + 2016) + c &= q_1 \\in \\mathbb{Q}, \\quad (a + 2016)c = q_2 \\in \\mathbb{Q}.\n\\end{aligned}\n$$\n\nExpress $b = p_1 - a$ and $c = q_1 - a - 2016$. Substitute into the equations:\n\n$$\n\\begin{aligned}\na(p_1 - a) &= p_2, \\\\\n(a + 2016)(q_1 - a - 2016) &= q_2.\n\\end{aligned}\n$$\n\nSubtracting gives:\n\n$$\n\\begin{aligned}\na(p_1 - a) - (a + 2016)(q_1 - a - 2016) &= p_2 - q_2 \\\\\n\\Rightarrow a(p_1 - q_1 + 4032) &= p_2 - q_2 + 2016q_1 - 2016^2 \\in \\mathbb{Q}.\n\\end{aligned}\n$$\n\nIf $p_1 - q_1 + 4032 \\neq 0$, then $a$ would be rational, which is not the case. So $p_1 - q_1 + 4032 = 0$. Thus,\n\n$$\n(a + b) - (a + 2016 + c) + 4032 = 0 \\implies c = b + 2016.\n$$\n\nTherefore, the only possible value for the other root of $Q$ is $b + 2016$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16443,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $\\mathbb{Z}$ 為所有整數所成的集合。求所有函數 $f: \\mathbb{Z} \\to \\mathbb{Z}$，滿足：\n\n$$\nf(f(x) + f(y)) + f(x)f(y) = f(x+y)f(x-y)\n$$\n\n對所有整數 $x, y$ 都成立。\n\nLet $\\mathbb{Z}$ be the set of all integers. Determine all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ such that\n\n$$\nf(f(x) + f(y)) + f(x)f(y) = f(x+y)f(x-y)\n$$\n\nholds for all $x, y \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "See solution",
    "solution": "顯然 $f(x) = 0$ 是原方程的一組解。故假設存在一個 $t$ 使得 $f(t) \\neq 0$。\n\n於原式代入 $(0,0)$ 得 $f(2f(0)) = 0$。\n\n代入 $(2f(0),0)$ 得 $f(f(0)) = 0$。\n\n代入 $(f(0),f(0))$ 得 $f(0) = 0$。\n\n代入 $(x,0)$ 得\n\n$$\nf(f(x)) = f(x)^2 \\qquad (1)\n$$\n\n代入 $(x, x)$ 得\n\n$$\nf(2f(x)) = -f(x)^2 \\qquad (2)\n$$\n\n為了方便起見，令 $g: \\mathbb{Z} \\to \\mathbb{Z}$ 滿足對於任意一個整數 $n$，\n\n$$\ng(n) = \\begin{cases} 0, & \\text{當 } n \\equiv 0 \\pmod{5}; \\\\ 1, & \\text{當 } n \\equiv 1, 4 \\pmod{5}; \\\\ -1, & \\text{當 } n \\equiv 2, 3 \\pmod{5}. \\end{cases}\n$$\n\n**Lemma.** 若 $f(s) \\neq 0$，則 $f(mf(s)) = g(m)f(s)^2$，對所有 $m \\in \\mathbb{N}$。\n\n*證明：* 使用數學歸納法。$m = 1, 2$ 分別由 (1), (2) 得證。\n\n若 $m < k$ 時皆成立，則 $m = k$ 時 ($k \\ge 3$)：\n\n- $k = 5q$：於原式代入 $((5q-1)f(s), f(s))$，結合歸納假設可得\n  $$\nf(2f(s)^2) + f(s)^4 = -f(5qf(s))f(s)^2 \\qquad (3)\n  $$\n  又由 (1) 知 $f(s)^2 = f(f(s))$，所以由 (1) 和 (2) 知\n  $$f(2f(s)^2) = f(2f(f(s))) = -f(f(s))^2 = -f(s)^4$$\n  代入 (3) 可知 $f(5qf(s))f(s)^2 = 0$，又 $f(s) \\neq 0$，所以 $f(5qf(s)) = 0 = g(5q)f(s)^2$\n\n- $k = 5q + 1$：於原式代入 $((5q-1)f(s), 2f(s))$，結合歸納假設可得\n  $$f((5q+1)f(s)) = f(s)^2 = g(5q+1)f(s)^2$$\n\n- $k = 5q + 2$：於原式代入 $((5q-1)f(s), 3f(s))$，結合歸納假設可得\n  $$f((5q+2)f(s)) = -f(s)^2 = g(5q+2)f(s)^2$$\n\n- $k = 5q + 3$：於原式代入 $((5q+2)f(s), f(s))$，結合歸納假設可得\n  $$f((5q+3)f(s)) = -f(s)^2 = g(5q+3)f(s)^2$$\n\n- $k = 5q + 4$：於原式代入 $((5q+3)f(s), f(s))$，結合歸納假設可得\n  $$f((5q+4)f(s)) = f(s)^2 = g(5q+4)f(s)^2$$\n\n綜上，$f(kf(s)) = g(k)f(s)^2$，故由數學歸納法得證。\n\n回到原題，由 (1) 和 (2) 知存在 $p$ 使得 $f(p) = |f(t)|f(t)$（$p$ 取 $f(t)$ 或 $2f(t)$）。\n\n由 (1) 和引理知 $f(t)^4 = f(p)^2 = f(f(p)) = f(|f(t)|f(t)) = g(|f(t)|)f(t)^2$，所以 $f(t)^2 = g(|f(t)|)$（因為 $f(t) \\neq 0$）。\n\n但是 $|g(|f(t)|)| \\le 1$ 且 $f(t) \\ne 0$，所以 $f(t)^2 = 1$。\n\n由 (1), (2) 知 $f(f(t)) = 1, f(2f(t)) = -1$。接著證對於所有 $n \\in \\mathbb{Z}$，都有 $f(n) = g(n)$。\n\n- 若 $n = 0$，那麼 $f(n) = 0 = g(n)$。\n- 若 $n > 0$，於引理中 $s$ 代 $f(t)$，$m$ 代 $n$ 即得\n  $$f(n) = f(nf(f(t))) = g(n)f(f(t))^2 = g(n)$$\n- 若 $n < 0$，於引理中 $s$ 代 $2f(t)$，$m$ 代 $-n$ 即得\n  $$f(n) = f(-nf(2f(t))) = g(-n)f(2f(t))^2 = g(-n) = g(n)$$\n\n綜上，$f(n) = g(n)$。\n\n代回驗證：易知 $g(n)^2 \\equiv n^2 \\pmod 5$\n\n$$\n\\begin{aligned}\n& \\text{所以原式左式} \\equiv (x^2 + y^2)^2 + (xy)^2 \\\\\n& \\equiv (x^2 + y^2)^2 - (2xy)^2 \\equiv (x^2 - y^2)^2 \\equiv \\text{右式} \\pmod{5}\n\\end{aligned}\n$$\n\n又因為 $|g(n)| \\le 1$，所以 $|\\text{左式} - \\text{右式}| \\le 3$。故左式和右式相等，驗證畢。\n\n因此 $f$ 的解有兩個：\n\n$$f(x) = 0 \\quad \\forall x \\in \\mathbb{Z}$$\n\n$$f(x) = \\begin{cases} 0, & \\text{當 } x \\equiv 0 \\pmod 5; \\\\ 1, & \\text{當 } x \\equiv 1, 4 \\pmod 5; \\\\ -1, & \\text{當 } x \\equiv 2, 3 \\pmod 5. \\end{cases} \\quad \\forall x \\in \\mathbb{Z}.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16444,
    "subject": "Mathematics (Olympiad)",
    "question": "For a prime number $p$ and a polynomial $f$ with integer coefficients, define $\\text{Im}(p, f)$ as the set of integers $a \\in \\{0, 1, \\dots, p-1\\}$ such that there exists an integer $x$ for which $f(x) - a$ is divisible by $p$.\n\nProve that there exist nonconstant polynomials $f$ and $g$ such that, for infinitely many primes, the intersection of $\\text{Im}(p, f)$ and $\\text{Im}(p, g)$ is empty.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(x) = (x^2 + 1)^2$ and $g(y) = -(y^2 + 1)^2$. We claim that for all primes $p \\equiv 3 \\pmod{4}$, the sets $\\text{Im}(p, f)$ and $\\text{Im}(p, g)$ are disjoint.\n\nSuppose there exist $x, y$ such that $f(x) \\equiv g(y) \\pmod{p}$. Then:\n\n$$\n(x^2 + 1)^2 \\equiv -(y^2 + 1)^2 \\pmod{p}\n$$\nwhich implies\n$$\n(x^2 + 1)^2 + (y^2 + 1)^2 \\equiv 0 \\pmod{p}\n$$\n\nRecall that for $p \\equiv 3 \\pmod{4}$, the only solution to $a^2 + b^2 \\equiv 0 \\pmod{p}$ is $a \\equiv b \\equiv 0 \\pmod{p}$. Thus, $x^2 + 1 \\equiv 0 \\pmod{p}$ and $y^2 + 1 \\equiv 0 \\pmod{p}$, which is impossible for such $p$.\n\nSince there are infinitely many primes congruent to $3$ modulo $4$, the claim follows.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16445,
    "subject": "Mathematics (Olympiad)",
    "question": "If $a + 2b = 13$ and $5a - 2b = 5$, what is the value of $b$?\n\n(A) 1  \n(B) 2  \n(C) 3  \n(D) 4  \n(E) 5",
    "options": [],
    "answer": "See solution",
    "solution": "Given $a + 2b = 13$ and $5a - 2b = 5$, add both equations:\n\n$$\n(a + 2b) + (5a - 2b) = 13 + 5\n$$\n$$\n6a = 18\n$$\nSo $a = 3$. Substitute into $a + 2b = 13$:\n\n$$\n3 + 2b = 13 \\implies 2b = 10 \\implies b = 5\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16446,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a rational number $r = \\frac{p}{q} \\in (0, 1)$, where $p$ and $q$ are coprime positive integers and $pq$ divides $3600$, how many such rational numbers $r$ are there?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\Omega = \\{ r \\mid r = \\frac{p}{q},\\ p, q \\in \\mathbb{N}_+,\\ (p, q) = 1,\\ pq \\mid 3600 \\}$. The prime factorization of $3600$ is $2^4 \\times 3^2 \\times 5^2$. Write $p = 2^A 3^B 5^C$, $q = 2^a 3^b 5^c$, with $\\min\\{A, a\\} = \\min\\{B, b\\} = \\min\\{C, c\\} = 0$ and $A + a \\le 4$, $B + b \\le 2$, $C + c \\le 2$.\n\nThere are $9$ ways to choose $(A, a)$, $5$ ways for $(B, b)$, and $5$ ways for $(C, c)$, so $|\\Omega| = 9 \\times 5 \\times 5 = 225$.\n\nEach $r \\in \\Omega$ with $r \\ne 1$ can be paired with $1/r$, so the number of $r \\in (0, 1)$ is\n$$\n\\frac{1}{2} (|\\Omega| - 1) = 112.\n$$\nThus, there are $112$ such rational numbers $r$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16447,
    "subject": "Mathematics (Olympiad)",
    "question": "Three people (Molly, Polly, and Ollie) are each assigned to separate rooms chosen randomly from 10 available rooms, numbered 1 to 10. What is the probability that all three are assigned to rooms that are within a block of five consecutive rooms?",
    "options": [],
    "answer": "See solution",
    "solution": "We list the ways in which the 3 rooms are within a block of 5.\n\nIf 1 is the lowest room number in a block of 5, then the rooms are as in this diagram:\n\n![alt](path \"optional title\")\n\nThere are 6 allocations with 1 the lowest room number: $(1, 2, 3)$, $(1, 2, 4)$, $(1, 2, 5)$, $(1, 3, 4)$, $(1, 3, 5)$, $(1, 4, 5)$.\nSimilarly, there are 6 allocations with the lowest room number equal to each of 2, 3, 4, 5, 6.\n\nThere are 3 allocations with 7 the lowest room number: $(7, 8, 9)$, $(7, 8, 10)$, $(7, 9, 10)$.\n\nThere is only 1 allocation with 8 the lowest room number: $(8, 9, 10)$.\n\nThus the number of allocations of 3 rooms within a block of 5 is $6 \\times 6 + 3 + 1 = 40$.\n\nLet $T$ be the number of ways of choosing 3 of the 10 rooms. For each choice, there are 6 ways of allocating the 3 rooms to Molly, Polly, and Ollie. So the number of ways of placing them separately in 3 rooms is $6T$. Another way of counting these placements is as follows. Molly can be placed in one of 10 rooms, then Polly in one of the remaining 9 rooms, and finally Ollie in one of the remaining 8 rooms. So $6T = 10 \\times 9 \\times 8 = 720$. Hence $T = 120$.\n\nSince the rooms are allocated at random, the chance Molly, Polly, and Ollie are within a block of five consecutive rooms is $40/120 = 1/3$. So Ollie is correct.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16448,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $H$ be the orthocenter of an acute triangle $ABC$, and $M$ be the midpoint of the side $BC$. Let $P$ be the point of intersection of the line $AM$ and the line through $H$ and perpendicular to the line $AM$. Prove that $AM \\cdot PM = BM^2$ holds. Here, for a line segment $XY$, its length is also denoted by $XY$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $X$ be the point of intersection of the lines $BH$ and $AC$, and let $N$ be the midpoint of the line segment $AH$. Since $\\angle AXH = \\angle APH = 90^\\circ$, the points $P$ and $X$ lie on the circle having $AH$ as its diameter (if $AB = AC$, then $P$ coincides with $H$ and it is clear that $X$ lies on the circle with $AH$ as its diameter in this case). Since $N$ is the midpoint of $AH$, we then have $\\angle AXN = \\angle XAN$. Also, since $\\angle BXC = 90^\\circ$, the point $X$ lies on the circle having $BC$ as its diameter, and from this we get $\\angle CXM = \\angle XCM$ and $XM = BM$. Using these facts we get\n\n$$\n\\angle NXM = 180^\\circ - (\\angle AXN + \\angle CXM) = 180^\\circ - (\\angle XAN + \\angle XCM) = 90^\\circ,\n$$\n\nfrom which it follows that the circle going through the three points $A$, $P$, $X$ is tangent to the line $MX$. Hence, by the well-known theorem on the power of a point with respect to a circle, we obtain $AM \\cdot PM = (XM)^2 = (BM)^2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16449,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ and $n$ be positive integers, and let $G$ be a group of order $n$. Prove that the following two statements are equivalent:\n\n(a) The numbers $k$ and $n$ are relatively prime.\n\n(b) For every subgroup $H$ of $G$, the set $\\{x : x \\in G \\text{ and } x^k \\in H\\}$ is contained in $H$.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that (a) implies (b). Since $k$ and $n$ are relatively prime, $kp + nq = 1$ for some integers $p$ and $q$. Let $H$ be a subgroup of $G$, and let $x$ be a member of $G$ such that $x^k \\in H$. Since $x^n = e$, the unit of $G$, it follows that\n$$\nx = x^{kp + nq} = (x^k)^p \\cdot (x^n)^q = (x^k)^p \\in H.\n$$\n\nWe now show that (b) implies (a). This is clearly the case if $k = 1$ or $n = 1$, so let them both be at least $2$, and suppose, if possible, they share some prime divisor $p$. By Cauchy's theorem, $x^p = e$ for some $x$ in $G \\setminus \\{e\\}$.\n\nConsider the trivial subgroup $H = \\{e\\}$. Since $k$ is divisible by $p$, it follows that $x^k = e \\in H$, so $x \\in H = \\{e\\}$; that is, $x = e$, contradicting the fact that $x$ lies in $G \\setminus \\{e\\}$. Consequently, $k$ is indeed coprime to $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16450,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $A_1A_2\\cdots A_n$ 是一凸多邊形。點 $P$ 是此多邊形內一點且其在 $A_1A_2, \\cdots, A_nA_1$ 之投影點分別為 $P_1, \\cdots, P_n$，其中 $P_1, \\cdots, P_n$ 分別落在線段 $A_1A_2, \\cdots, A_nA_1$ 內。試證：對於任意分別在線段 $A_1A_2, \\cdots, A_nA_1$ 內之點 $X_1, \\cdots, X_n$，滿足\n\n$$\n\\max \\left\\{ \\frac{X_1X_2}{P_1P_2} + \\cdots + \\frac{X_nX_1}{P_nP_1} \\right\\} \\ge 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "記 $P_{n+1} = P_1,\\ X_{n+1} = X_1,\\ A_{n+1} = A_1$。\n\n*引理*：令 $Q$ 為 $A_1A_2\\cdots A_n$ 內一點。則 $Q$ 必落在三角形 $X_1A_2X_2, \\cdots, X_nA_1X_1$ 之外接圓之其中一個。\n\n*證明*：若 $Q$ 在三角形 $X_1A_2X_2, \\cdots, X_nA_1X_1$ 之其中一個，則顯然成立。否則 $Q$ 在多邊形 $A_1A_2\\cdots A_n$ 內（如圖1）。則\n\n$$\n\\begin{aligned}\n& (\\angle X_1 A_2 X_2 + \\angle X_1 Q X_2) + \\cdots + (\\angle X_n A_1 X_1 + \\angle X_n Q X_1) \\\\\n&= (\\angle X_1 A_1 X_2 + \\cdots + \\angle X_n A_1 X_1) + \\cdots + (\\angle X_1 Q X_2 + \\cdots + \\angle X_n Q X_1) \\\\\n&= (n-2)\\pi + 2\\pi = n\\pi,\n\\end{aligned}\n$$\n\n因此存在一個足標 $i$ 使得\n\n$$\n\\angle X_i A_{i+1} X_{i+1} + \\angle X_i Q X_{i+1} \\ge \\frac{n\\pi}{n} = \\pi.\n$$\n\n因四邊形 $QX_iA_{i+1}X_{i+1}$ 是凸的，其意為 $Q$ 落在 $\\triangle X_iA_{i+1}X_{i+1}$ 之外接圓內。\n\n應用上述引理，$P$ 落在某個 $\\triangle X_iA_{i+1}X_{i+1}$ 之外接圓內。\n分別考慮 $\\triangle P_iA_{i+1}P_{i+1}$ 與 $\\triangle X_iA_{i+1}X_{i+1}$ 之外接圓 $\\omega$ 與 $\\Omega$（如圖2）；令 $r$\n\n![如圖1](path \"\")\n\n![如圖2](path \"\")",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16451,
    "subject": "Mathematics (Olympiad)",
    "question": "Let a convex quadrilateral $ABCD$ be inscribed in a circle with center $O$ and circumscribed about a circle with center $I$. Let its diagonals $AC$ and $BD$ meet at a point $P$. Prove that the points $O$, $I$, and $P$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that the lines $AI$, $BI$, $CI$, $DI$ meet the circumcircle of the quadrilateral $ABCD$ at $E$, $F$, $G$, $H$, respectively. Since the lines $AI$, $BI$, $CI$, $DI$ are the bisectors of the respective angles of the quadrilateral $ABCD$, the lines $EG$ and $FH$ are the diameters of the circumcircle of $ABCD$. Thus $EG$ and $FH$ meet at $O$.\n\nDenote by $X$ the point of intersection of $EB$ and $CH$.\n\nUsing Pascal's theorem for the hexagon $ACHDBE$, we see that $P$, $X$, and $I$ are collinear. Once again, Pascal's theorem applied to the hexagon $GCHFBE$ yields that $O$, $X$, and $I$ are collinear. Thus $O$, $I$, and $P$ are collinear, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16452,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f$ from the set of positive integers to the set of positive integers such that, for all positive integers $a$ and $b$, there exists a nondegenerate triangle with sides of lengths $a$, $f(b)$, and $f(b + f(a) - 1)$.\n\n(A triangle is nondegenerate if its vertices are not collinear.)",
    "options": [],
    "answer": "See solution",
    "solution": "The only $f$ with the required property is $f(n) = n$ for all $n \\in \\mathbb{N}^*$.\n\nBy assumption and discreteness of integers, for any positive integers $a, b$:\n\n$$\nf(b) + f(b + f(a) - 1) - 1 \\ge a \\tag{1}\n$$\n$$\nf(b) + a - 1 \\ge f(b + f(a) - 1) \\tag{2}\n$$\n$$\nf(b + f(a) - 1) + a - 1 \\ge f(b) \\tag{3}\n$$\n\nSetting $a = 1$ in (2) and (3), we have $f(b) = f(b + f(1) - 1)$ for any $b \\in \\mathbb{N}^*$.\n\nIf $f(1) \\ne 1$, then the above equality implies that $f$ is periodic. Since $f$ is defined on positive integers, $f$ is bounded. Let positive integer $M$ be such that $M \\ge f(n)$ for any positive integer $n$ (i.e., $M$ is an upper bound for $f$). Putting $a = 2M$ in (1) results in a contradiction. Thus, $f(1) = 1$.\n\nSetting $b = 1$ in (1) and (2), we have $f(f(n)) = n$ for all $n \\in \\mathbb{N}^*$.\n\nIf there exists some $t \\in \\mathbb{N}^*$ such that $f(t) < t$, then $t \\ge 2$ and $f(t) \\le t - 1$. Setting $a = f(t)$ in (2), we have\n\n$$\nf(b + t - 1) = f(b + f(a) - 1) \\le f(b) + a - 1 \\le f(b) + t - 2.\n$$\n\nLet $M = (t-1) \\times \\max_{1 \\le i \\le t-1} f(i)$. For any integer $n > M$, denote by $n_0$ the unique positive integer satisfying\n\n$$\n1 \\le n_0 \\le t-1, \\quad n_0 \\equiv n \\pmod{t-1}.\n$$\n\nThen\n\n$$\nf(n) \\le f(n_0) + \\frac{t-2}{t-1}(n - n_0) \\le \\frac{M}{t-1} + \\frac{(t-2)n}{t-1} < n.\n$$\n\nTherefore, $f(n) < n$ whenever $n > M$.\n\nNow choose $n_1 \\in \\mathbb{N}^*$ such that $n_1 > M$ and $n_1$ is not equal to any of $f(1), f(2), \\dots, f(M)$. Since $f(f(n_1)) = n_1$, it follows that $f(n_1) > M$, and hence\n\n$$\nn_1 > f(n_1) > f(f(n_1)) = n_1,\n$$\n\na contradiction. As a result, we have $f(t) \\ge t$ for any $t \\in \\mathbb{N}^*$. Then $t = f(f(t)) \\ge f(t) \\ge t$, and all inequalities become equalities, i.e., $f(n) = n$ for any $n \\in \\mathbb{N}^*$.\n\nIt is not hard to check that $f(n) = n$ for all $n \\in \\mathbb{N}^*$ satisfies the required property, thus completing our conclusion that the only solution to the problem is $f(n) = n$ for all $n \\in \\mathbb{N}^*$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16453,
    "subject": "Mathematics (Olympiad)",
    "question": "For which greatest value of $n$ do there exist integers $a_1, a_2, \\ldots, a_n$ and $b_1, b_2, \\ldots, b_n$ such that:\n\n- All numbers $b_1, b_2, \\ldots, b_n$ are different and lie in the range $[0, 99]$.\n- $1 \\leq a_1 < a_2 < \\ldots < a_n \\leq 100$.\n- For all $i$ with $1 \\leq i \\leq n$, either $b_i = a_i - i$ or $b_i = a_i - i + n$ holds.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $n = 67$.\n\n**Solution.**\nLet $c_i = a_i - i$. Then $0 \\leq c_1 \\leq c_2 \\leq \\ldots \\leq c_n \\leq 100 - n$, and for all $i$, $b_i = c_i$ or $b_i = c_i + n$.\n\nSince $c_i + n \\leq c_n + n = a_n \\leq 100$, among any three consecutive $c_i, c_{i+1}, c_{i+2}$, at least two must be different; otherwise, at least two of $b_i, b_{i+1}, b_{i+2}$ would coincide, contradicting the conditions. Thus, $c_{i+2} \\geq c_i + 1$.\n\nTherefore, $c_n \\geq c_{n-2} + 1 \\geq \\dots \\geq \\frac{1}{2}(n-1)$. On the other hand, $c_n \\leq 100 - n$. So $\\frac{1}{2}(n-1) \\leq c_n \\leq 100 - n$, which implies $n \\leq 67$.\n\nFor $n = 67$, an example:\n\n$$\nc_1 = c_2 = 0,\\ c_3 = c_4 = 1,\\ \\ldots,\\ c_{65} = c_{66} = 32,\\ c_{67} = 33,\n$$\n\n$$\nb_1 = 0,\\ b_2 = 67,\\ b_3 = 1,\\ b_4 = 68,\\ \\ldots,\\ b_{65} = 32,\\ b_{66} = 99,\\ b_{67} = 33,\n$$\n\nTherefore, $a_i = c_i + i$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16454,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for one of $x \\in \\{1, 2, 4\\}$, we have $|px + q + \\frac{8}{x}| \\ge 1$.\n\nLet\n$$\nf(x) = px + q + \\frac{8}{x}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We compare $f(2)$ to the linear interpolation of $f(1)$ and $f(4)$:\n\n$$\n\\frac{2}{3}f(1) + \\frac{1}{3}f(4) = \\left(\\frac{2}{3} + \\frac{4}{3}\\right)p + \\left(\\frac{2}{3} + \\frac{1}{3}\\right)q + \\frac{2}{3}8 + \\frac{1}{3}2 \\\\\n= 2p + q + 6 \\\\\n= f(2) + 2.\n$$\n\nSubtracting one from each side:\n\n$$\n\\frac{2}{3}(f(1) - 1) + \\frac{1}{3}(f(4) - 1) = f(2) + 1.\n$$\n\nIf the left side is non-negative, then:\n$$\n\\frac{2}{3}(f(1) - 1) + \\frac{1}{3}(f(4) - 1) \\geq 0\n$$\nso at least one of $f(1) \\ge 1$ or $f(4) \\ge 1$.\n\nIf the right side is negative, then $f(2) < -1$.\n\nThus, for some $x \\in \\{1,2,4\\}$, $|f(x)| \\ge 1$ as required.\n\n**Remark:** This inequality is best possible. For $p = 2$ and $q = -9$, for $1 \\le x \\le 4$:\n$$\n1 - f(x) = 1 - 2x + 9 - \\frac{8}{x} = \\frac{2(x-1)(4-x)}{x} \\ge 0\n$$\nAnd for any $x > 0$:\n$$\n1 + f(x) = 1 + 2x - 9 + \\frac{8}{x} = \\frac{2(x-2)^2}{x} \\ge 0\n$$\nSo there is an $f$ with $-1 \\le f(x) \\le 1$ for all $1 \\le x \\le 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16455,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be the sum:\n\n$$\nS = 1 + \\frac{1}{3} + \\frac{1}{5} + \\frac{1}{7} + \\cdots + \\frac{1}{2011}\n$$\n\nIs $S$ an integer?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $T = \\frac{1}{3} + \\frac{1}{5} + \\frac{1}{7} + \\cdots + \\frac{1}{2011}$. Then $1005 - S = T$, so $S$ is an integer if and only if $T$ is an integer.\n\nLet $M = 3 \\cdot 5 \\cdot 7 \\cdots 2009$. If $T$ is an integer, then $MT$ is also an integer:\n\n$$\nMT = \\frac{M}{3} + \\frac{M}{5} + \\frac{M}{7} + \\cdots + \\frac{M}{2009} + \\frac{M}{2011}\n$$\n\nEach term except the last is an integer, but $2011$ is prime and does not divide $M$, so $\\frac{M}{2011}$ is not an integer. Thus, $T$ is not an integer, and therefore $S$ is not an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16456,
    "subject": "Mathematics (Olympiad)",
    "question": "設三角形 $ABC$ 的內心為 $I$，內切圓為 $\\omega$。令 $E, F$ 為 $\\omega$ 與 $CA, AB$ 的切點，$X, Y$ 為三角形 $BIC$ 的外接圓與 $\\omega$ 的交點。在 $BC$ 上取一點 $T$ 使得 $\\angle AIT$ 為直角。令 $G$ 為 $EF$ 與 $BC$ 的交點，$Z$ 為 $XY$ 與 $AT$ 的交點。證明 $AZ, ZG, AI$ 圍成一個等腰三角形。",
    "options": [],
    "answer": "See solution",
    "solution": "解法一：令 $M$ 為 $AT$ 與外接圓 $\\odot(ABC)$ 的交點。那麼 $TM \\cdot TA = TB \\cdot TC = TI^2$，所以 $\\angle AMI = \\angle AIT = 90^\\circ$，即 $M \\in \\odot(AEF) \\cap \\odot(ABC)$。由密克定理（或 $\\angle MFG = \\angle MAC = \\angle MBG$），$M$ 位於 $\\odot(BFG)$ 上，因此，若令 $W$ 為 $XY$ 與 $BC$ 的交點，則\n\n$$\n\\angle WGM = \\angle AFM = \\angle AIM = \\angle ITM = \\angle WZM,\n$$\n\n也就是說，$W, M, G, Z$ 共圓。\n\n令 $D$ 為 $\\omega$ 與 $BC$ 的切點，$M_E, M_F$ 分別為 $\\overline{FD}, \\overline{DE}$ 的中點。那麼 $M_E D \\cdot M_E F = M_E B \\cdot M_E I$，也就是說 $M_E$ 位於 $\\omega$ 與 $\\odot(BIC)$ 的根軸 $XY$ 上。同理，$M_F$ 也位於 $XY$ 上。因此 $W = M_E M_F \\cap BC$ 為 $\\overline{DG}$ 的中點。熟知 $MD$ 平分 $\\angle BMC$：\n\n由於 $\\angle FMB = \\angle EMC, \\angle MEB = \\angle MFC, \\triangle MBF \\sim \\triangle MCE$，因此 $\\overline{MB} = \\overline{BF} = \\overline{BD}$，從而 $MD$ 平分 $\\angle BMC$。\n\n而 $G, D$ 調和分割 $B, C$，故 $\\angle DMG$ 為直角。所以 $W$ 為 $\\triangle MGD$ 的外心。\n\n最後，我們證明 $\\angle IAZ = \\angle (GZ, AI)$，這等價於\n\n$$\n\\angle WGM = \\angle WZM = \\angle IAZ + 90^\\circ = \\angle (GZ, AI) + 90^\\circ = \\angle GZW = \\angle GMW,\n$$\n\n也就是 $\\triangle WMG$ 是以 $W$ 為頂點的等腰三角形，而這是因為 $W$ 為 $\\triangle MGD$ 的外心。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16457,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. What is the maximum value of the product $a_1 a_2 \\cdots a_k$ where $n = a_1 + a_2 + \\cdots + a_k$ and each $a_i$ is a positive integer?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is:\n\n- $3^m$ for $n = 3m$\n- $4 \\cdot 3^{m-1}$ for $n = 3m+1$\n- $2 \\cdot 3^m$ for $n = 3m+2$\n\nThis follows from these observations:\n\n1. None of the $a_i$ equals 1 (since replacing 1 and another summand by their sum increases the product).\n2. None of the $a_i$ is greater than 4. If $a_k \\geq 5$, splitting $a_k$ into $a_k-2$ and $2$ increases the product because $2(a_k-2) > a_k$ for $a_k > 4$.\n3. If any $a_i = 4$, it can be replaced by $2+2$ without changing the product.\n\nThus, we may write $n$ as a sum of 2's and 3's. Finally, if there are at least three 2's, replacing them with two 3's increases the product. This leads to the formulas above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16458,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. We say that a $n \\times n$ table is *special* if:\n\n- each cell of the table contains a 2-digit odd positive integer;\n- the numbers of the table are pairwise distinct;\n- the products of the numbers of each row and the products of the numbers of each column are perfect squares.\n\nProve that the largest value of $n$ for which there exists a $n \\times n$ special table is equal to $4$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $11^2 > 100$, the table cannot contain numbers divisible by squares of prime numbers $p$, with $p \\ge 7$.\n\nSuppose that there exists a number of the table (situated on row $\\ell$ and column $c$) divisible by a prime $p \\ge 17$. Then there exists one more number on row $\\ell$ (and column $c' \\ne c$) divisible by $p$. Also, there exists one more number on column $c$ (and row $\\ell' \\ne \\ell$) divisible by $p$. Therefore, in the cell $(\\ell', c')$ there must be a number divisible by $p$. In consequence, in the table must appear a number $N \\ge 7p > 100$ -- impossible.\n\n![](table.png)\n\nIt follows that the table cannot contain 2-digit numbers that are odd multiples of 17 (3 numbers), 19 (3 numbers), 23 (2 numbers), 29 (2 numbers), 31 (2 numbers), and 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 (one number each); in total, 26 numbers. On the other hand, there are 45 odd 2-digit numbers, so the table can contain at most $45 - 26 = 19$ numbers. Hence $n \\le 4$.\n\nThe table above is an example for $n = 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16459,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ and $B$ be $2 \\times 2$ real matrices such that $AB = A^2 B^2 - (AB)^2$ and $\\det(B) = 2$. Evaluate $\\det(A + 2B) - \\det(B + 2A)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Write $A(AB - BA - I_2)B = O_2$ to get $A(AB - BA - I_2) = O_2$, since $B$ is nonsingular. If $A$ is nonsingular, then $AB - BA = I_2$, which is false, since $\\det(AB - BA) = 0 \\ne 2 = \\det(I_2)$.\n\nSet $f(x) = \\det(A + xB)$ for $x \\in \\mathbb{R}$. Since $\\det(A) = 0$, there exists $a \\in \\mathbb{R}$ such that $f(x) = ax + \\det(B)x^2 = 2x^2 + ax$ for all $x \\in \\mathbb{R}$. Then\n$$\n\\det(A+2B) - \\det(B+2A) = f(2) - 4f(1/2) = 8 + 2a - 4\\left(\\frac{1}{2} + \\frac{a}{2}\\right) = 6.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16460,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $AB = AC$. The angle bisectors of $\\angle CAB$ and $\\angle ABC$ meet the sides $BC$ and $CA$ at $D$ and $E$ respectively. Let $K$ be the incenter of the triangle $ADC$. Suppose that $\\angle BEK = 45^\\circ$. Find all possible values of $\\angle CAB$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $AD$ and $BE$ are the angle bisectors of $\\angle CAB$ and $\\angle ABC$, their intersection point is the incenter $I$. Join $CI$; then $CI$ is the angle bisector of $\\angle ACB$. Since $K$ is the incenter of $\\triangle ADC$, $K$ lies on the segment $CI$.\n\nLet $\\angle BAC = \\alpha$. Since $AB = AC$, $AD \\perp BC$, we have\n$$\n\\angle ABC = \\angle ACB = 90^\\circ - \\frac{\\alpha}{2}\n$$\nAs $BI$ and $CI$ bisect $\\angle ABC$ and $\\angle ACB$ respectively, we get\n$$\n\\angle ABI = \\angle IBC = \\angle ACI = \\angle ICB = 45^\\circ - \\frac{\\alpha}{4}\n$$\nThus,\n$$\n\\angle EIC = \\angle IBC + \\angle ICB = 90^\\circ - \\frac{\\alpha}{2}\n$$\n$$\n\\angle IEC = \\angle BAE + \\angle ABE = 45^\\circ + \\frac{3\\alpha}{4}\n$$\nSo we have\n$$\n\\begin{aligned}\n\\frac{IK}{KC} &= \\frac{S_{\\triangle IEK}}{S_{\\triangle EKC}} \\\\\n&= \\frac{\\frac{1}{2} IE \\times EK \\times \\sin \\angle IEK}{\\frac{1}{2} EC \\times EK \\times \\sin \\angle KEC} \\\\\n&= \\frac{\\sin 45^\\circ}{\\sin \\frac{3\\alpha}{4}} \\times \\frac{IE}{EC} \\\\\n&= \\frac{\\sin 45^\\circ}{\\sin \\frac{3\\alpha}{4}} \\times \\frac{\\sin\\left(45^\\circ - \\frac{\\alpha}{4}\\right)}{\\sin\\left(90^\\circ - \\frac{\\alpha}{2}\\right)}.\n\\end{aligned}\n$$\nOn the other hand, since $K$ is the incenter of $\\triangle ADC$, $DK$ bisects $\\angle IDK$. It follows from the property of the angle bisector that\n$$\n\\frac{IK}{KC} = \\frac{ID}{DC} = \\tan \\angle ICD = \\frac{\\sin\\left(45^\\circ - \\frac{\\alpha}{4}\\right)}{\\cos\\left(45^\\circ - \\frac{\\alpha}{4}\\right)}\n$$\nThus,\n$$\n\\frac{\\sin 45^\\circ}{\\sin \\frac{3\\alpha}{4}} \\times \\frac{\\sin\\left(45^\\circ - \\frac{\\alpha}{4}\\right)}{\\sin\\left(90^\\circ - \\frac{\\alpha}{2}\\right)} = \\frac{\\sin\\left(45^\\circ - \\frac{\\alpha}{4}\\right)}{\\cos\\left(45^\\circ - \\frac{\\alpha}{4}\\right)}.\n$$\nRemoving the denominators, we have\n$$\n2\\sin 45^\\circ \\cos\\left(45^\\circ - \\frac{\\alpha}{4}\\right) = 2\\sin \\frac{3\\alpha}{4} \\cos \\frac{\\alpha}{2}.\n$$\nBy the product-to-sum identities, we get\n$$\n\\sin(90^\\circ - \\frac{\\alpha}{4}) + \\sin\\frac{\\alpha}{4} = \\sin\\frac{5\\alpha}{4} + \\sin\\frac{\\alpha}{4},\n$$\nor, equivalently,\n$$\n\\sin(90^\\circ - \\frac{\\alpha}{4}) = \\sin\\frac{5\\alpha}{4}.\n$$\nSince $0^\\circ < \\alpha < 180^\\circ$, $\\sin(90^\\circ - \\frac{\\alpha}{4}) > 0$, we have $\\sin\\frac{5\\alpha}{4} > 0$, i.e., $0^\\circ < \\frac{5\\alpha}{4} < 180^\\circ$. It follows that either\n$$\n90^\\circ - \\frac{\\alpha}{4} = \\frac{5\\alpha}{4} \\implies \\alpha = 60^\\circ\n$$\nor\n$$\n90^\\circ - \\frac{\\alpha}{4} = 180^\\circ - \\frac{5\\alpha}{4} \\implies \\alpha = 90^\\circ.\n$$\nWhen $\\alpha = 60^\\circ$, it is easy to see that $\\triangle IEC \\cong \\triangle IDK$, so\n$$\n\\triangle IEK \\cong \\triangle IDK,\n$$\nand therefore\n$$\n\\angle BEK = \\angle IDK = 45^\\circ.\n$$\nWhen $\\alpha = 90^\\circ$, ...",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16461,
    "subject": "Mathematics (Olympiad)",
    "question": "Real numbers $x, y, z$ satisfy\n$$\n\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} + x + y + z = 0\n$$\nand none of them lies in the open interval $(-1, 1)$. Find the maximum value of $x + y + z$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "By changing $(x, y, z)$ to $(-x, -y, -z)$, the condition remains valid and the value of $x + y + z$ changes sign. Since the ordering of the numbers $x, y, z$ is irrelevant and they cannot all be of the same sign because of the given equation, we can, without loss of generality, assume $x > 0$, $y > 0$, and $z < 0$, and seek the maximum value of $V = |x + y + z|$.\n\nWe can transform the equation equivalently to\n$$\nf(x) + f(y) = f(t), \\quad \\text{where} \\quad t := -z > 0 \\quad \\text{and} \\quad f(x) = x + \\frac{1}{x},\n$$\nwith $x, y, t \\in I := (1, \\infty)$. The function $f$ is an increasing bijection mapping $I$ onto $(2, \\infty)$. Applying this, from\n$$\nf(t) = f(x) + f(y) = x + \\frac{1}{x} + y + \\frac{1}{y} > x + y + \\frac{1}{x + y} = f(x + y)\n$$\nwe get $t > x + y$, hence $x + y + z = x + y - t < 0$. Therefore, our aim is to maximize $V = |x + y + z| = t - x - y$.\n\nOne valid triple is $(1, 1, t_0)$, with $t_0 > 1$ such that $f(t_0) = 4$, i.e., $t_0 = 2 + \\sqrt{3}$. For this triple, $V = \\sqrt{3}$. We shall show this is the desired maximum. The proof is based on the following lemma: Whenever $(x, y, t)$ and $(x, y', t')$ are valid triples with the same first component $x$, and $y' < y$, we have $t - x - y < t' - x - y'$. If this lemma is true, then, by symmetry, the same conclusion holds for valid triples $(x, y, t)$ and $(x', y, t')$ with the same second component $y$. Hence, any valid triple $(x, y, t)$ can be replaced by $(x, 1, t')$ and then by $(1, 1, t_0)$, and the value of $V$ increases or remains the same during this process, concluding $\\max V = \\sqrt{3}$. Moreover, this also implies that $(1, 1, 2 + \\sqrt{3})$ is the only valid triple for which the maximum value $V = \\sqrt{3}$ is reached.\n\nTo prove the lemma, notice that from\n$$\nf(t) = f(x) + f(y), \\quad f(t') = f(x) + f(y'),\n$$\nand from $f(y') < f(y)$ (since $y' < y$), we have $f(x) < f(t') < f(t)$, hence $1 < t' < t$. Therefore $1 < t'y' < ty$, and from\n$$\nf(x) = f(t) - f(y) = (t - y) \\left( 1 - \\frac{1}{ty} \\right) = f(t') - f(y') = (t' - y') \\left( 1 - \\frac{1}{t'y'} \\right),\n$$\nusing the estimates\n$$\n0 < 1 - \\frac{1}{t'y'} < 1 - \\frac{1}{ty},\n$$\nwe obtain $t - y < t' - y'$, which is equivalent to the desired $t - x - y < t' - x - y'$. \n\n**Answer:** The maximum value of $x + y + z$ is $\\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16462,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a point inside triangle $ABC$. The lines $AP$, $BP$, and $CP$ intersect the circumcircle $\\Gamma$ of triangle $ABC$ again at points $K$, $L$, and $M$, respectively. The tangent to $\\Gamma$ at $C$ intersects the line $AB$ at $S$. Suppose that $SC = SP$. Prove that $MK = ML$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof** Without loss of generality, assume $CA > CB$. Then $S$ lies on the extension of $AB$. Suppose the line $SP$ intersects the circumcircle of triangle $ABC$ at $E$ and $F$, as shown in the figure. By assumption and the power of a point theorem, we have\n\n$$\nSP^2 = SC^2 = SB \\times SA,\n$$\n\nand hence $\\frac{SP}{SB} = \\frac{SA}{SP}$. Thus, $\\triangle PSA \\sim \\triangle BSP$ and $\\angle BPS = \\angle SAP$.\n\nSince $2\\angle BPS = \\widehat{BE} + \\widehat{LF}$ and $2\\angle SAP = \\widehat{BE} + \\widehat{EK}$, we have\n\n$$\n\\widehat{LF} = \\widehat{EK}. \\qquad \\textcircled{1}\n$$\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p199_data_453a868b9e.png)\n\nIt follows from $\\angle SPC = \\angle SCP$ that $\\widehat{EC} + \\widehat{MF} = \\widehat{EC} + \\widehat{EM}$, and therefore\n\n$$\n\\widehat{MF} = \\widehat{EM}. \\qquad \\textcircled{2}\n$$\n\nFrom ① and ②, we get\n\n$$\n\\widehat{MFL} = \\widehat{MF} + \\widehat{FL} = \\widehat{ME} + \\widehat{EK} = \\widehat{MEK},\n$$\n\nand thus $MK = ML$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16463,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a square. For a point $P$ inside $ABCD$, a windmill centred at $P$ consists of two perpendicular lines $l_1$ and $l_2$ passing through $P$, such that\n\n- $l_1$ intersects the sides $AB$ and $CD$ at $W$ and $Y$, respectively, and\n- $l_2$ intersects the sides $BC$ and $DA$ at $X$ and $Z$, respectively.\n\nA windmill is called *round* if the quadrilateral $WXYZ$ is cyclic.\n\nDetermine all points $P$ inside $ABCD$ such that every windmill centred at $P$ is round.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\angle WBX = \\angle WPX = 90^\\circ$, we know that the quadrilateral $WBXP$ is cyclic. Similarly, since $\\angle YDZ = \\angle YPZ = 90^\\circ$, we know that the quadrilateral $YDZP$ is cyclic.\n\nSuppose that the quadrilateral $WXYZ$ is cyclic. Then we have the following equal angles:\n\n$$\n\\angle ABP = \\angle WBP = \\angle WXP = \\angle WXZ = \\angle ZYW = \\angle ZYP = \\angle ZDP = \\angle ADP\n$$\n\nTherefore, triangles $ABP$ and $ADP$ share the common side $AP$, have the equal sides $AB = AD$, and have the equal angles $\\angle ABP = \\angle ADP$. It follows that either $\\angle APB = \\angle APD$ or $\\angle APB + \\angle APD = 180^\\circ$. In the first case, triangles $ABP$ and $ADP$ are congruent, so $P$ must lie on the segment $AC$. In the second case, $P$ must lie on the segment $BD$. Therefore, $P$ lies on one of the diagonals of the square $ABCD$.\n\n![](images/2020_Australian_Scene_W_p103_data_b39f1c4349.png)\n\nConversely, suppose that $P$ lies on one of the diagonals of the square $ABCD$. In fact, we may assume without loss of generality that $P$ lies on $AC$. Then the triangles $ABP$ and $ADP$ are congruent and we have the following equal angles:\n\n$$\n\\angle WXZ = \\angle WXP = \\angle WBP = \\angle ABP = \\angle ADP = \\angle ZDP = \\angle ZYP = \\angle ZYW\n$$\n\nSince $\\angle WXZ = \\angle ZYW$, it follows that the quadrilateral $WXYZ$ is cyclic.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16464,
    "subject": "Mathematics (Olympiad)",
    "question": "For each pair of real numbers $(r, s)$, prove that there exists a real number $x$ that satisfies at least one of the following two equations.\n\n$$\nx^2 + (r + 1)x + s = 0\n$$\n\n$$\nrx^2 + 2s x + s = 0\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, for contradiction, that there does not exist a real number $x$ that satisfies at least one of the two equations. The discriminants of the two quadratics are $(r+1)^2 - 4s$ and $4s^2 - 4rs$, respectively. Therefore,\n\n$$\n(r + 1)^2 - 4s < 0 \\quad \\text{and} \\quad 4s^2 - 4rs < 0.\n$$\n\nAdding these inequalities gives\n\n$$\n(r + 1)^2 - 4s + 4s^2 - 4rs < 0 \\implies (r + 1 - 2s)^2 < 0.\n$$\n\nBut the square of a real number cannot be negative, which is a contradiction. Thus, there must exist a real number $x$ that satisfies at least one of the two equations.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16465,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, and $d$ be positive integers, and let $p = a + b + c + d$. Prove that if $p$ is a prime, then $p$ does not divide $ab - cd$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the relation $$(a + c)(b + c) = ab + ac + bc + c^2 = (a + b + c + d)c + ab - cd = pc + ab - cd.$$ \n\nIf $p$ divides $ab - cd$, then $p$ divides $a + c$ or $b + c$. \n\nHowever, $0 < a + c < p$ and $0 < b + c < p$, which is a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16466,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N \\geq 4$ be a fixed integer. Two players, *A* and *B*, write down numbers, each number in continuation to the previous expression. First, *A* writes $+1$ or $-1$, then *B* writes $+2$ or $-2$, then *A* writes $+3$ or $-3$, etc.; at step $k$, the player to move must write $+k$ or $-k$. The objective of each player is that after a move of theirs, several consecutive numbers in the obtained expression, taken with their signs, have sum divisible by $N$. For each $N$, determine which of the players has a winning strategy, if any.",
    "options": [],
    "answer": "See solution",
    "solution": "The first player *A* has a winning strategy if $N \\equiv 0$ or $1 \\pmod{4}$; otherwise, the second player *B* has one.\n\nLet $N = 4k + r$ where $k \\geq 1$ and $r \\in \\{0, 1\\}$. Then *A* starts with $+1$ and, in the sequel, negates all moves of *B* until *B* writes $\\pm 2k$. Here, \"negates\" means that *A* writes $-(2j+1)$ or $(2j+1)$ according as *B* writes $(2j)$ or $-(2j)$. We may assume that the first move of *B* is $-2$; otherwise, *A* wins by writing $-3$ (since $1+2-3=0$ is divisible by $N$ for all $N$). Moreover, we may assume that *B* also negates each move of *A* up to step $2k$. If *B* does not do so at step $2j$, then the expression ends $-(2j-2)+(2j-1)+(2j)$ or $+(2j-2)-(2j-1)-(2j)$ after step $2j$. So *A* wins at step $2j+1$ by writing $-(2j+1)$ or $(2j+1)$ respectively, because $m-(m+1)-(m+2)+(m+3)=0$ for all $m$. Thus, the sum $1-2+3-\\cdots+(2k-1)-(2k)$ is obtained at step $2k$, after a move of *B*.\n\nNeither player has won the game by that moment. Indeed, denote $S_n = 1 - 2 + \\cdots + (-1)^{n-1}n$, then $S_n = \\frac{n+1}{2}$ for $n$ odd and $S_n = -\\frac{n}{2}$ for $n$ even. It follows that if $1 \\leq i < j \\leq n$, then $|S_i - S_j| = \\frac{j-i}{2}$ for $i, j$ of the same parity and $|S_i - S_j| = \\frac{i+j+1}{2}$ for $i, j$ of different parity. In particular, $0 < |S_i - S_j| \\leq n$. So there is no winner yet if $N > n$. This is the case here, with $N = 4k + r \\geq 4k$ and $n = 2k$. Note that the argument works for $k=1$ ($N=4$) where $2k-2=0$ is not present in the sum but can be assumed.\n\nNow *A* wins by writing $-(2k+1)$. If $N = 4k + 1$, then the sum of the last two numbers $-(2k) - (2k+1) = -N$ is divisible by $N$; if $N = 4k$, then so is the sum of the last four numbers $-(2k-2) + (2k-1) - (2k) - (2k+1) = -4k = -N$.\n\nPlayer *B* has an analogous winning strategy for $N = 4k + r$ where $k \\geq 1$ and $r \\in \\{2, 3\\}$. Without loss of generality, *A* starts the game with $+1$ (if *B* has a winning strategy for a game starting $+1$, he can just negate his moves in this strategy if *A*'s opening move is $-1$). Now *B* answers $-2$ and, in general, negates *A*'s moves up to step $2k+1$. One may suppose again that *A* also negates *B*'s moves, due to the identity $m - (m+1) - (m+2) + (m+3) = 0$. Thus, the sum $1 - 2 + \\cdots + (2k-1) - (2k) + (2k+1)$ is obtained at step $2k+1$, after a move of *A*. It was shown above that $0 < |S_i - S_j| \\leq 2k+1 < N$ whenever $1 \\leq i < j \\leq 2k+1$, so that there is no winner yet by that time. The next move $+(2k + 2)$ of *B* is winning. If $N = 4k+3$, then $+(2k+1)+(2k+2) = N$; if $N = 4k+2$, then $+(2k-1)-(2k)+(2k+1)+(2k+2) = 4k+2 = N$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16467,
    "subject": "Mathematics (Olympiad)",
    "question": "For any integer $n$ with $n > 1$, let\n$$\nD(n) = \\{a - b \\mid a, b \\text{ are positive integers with } n = ab \\text{ and } a > b\\}.\n$$\nProve that for any integer $k$ with $k > 1$, there exist $k$ pairwise distinct integers $n_1, n_2, \\dots, n_k$ with $n_i > 1$ ($1 \\le i \\le k$), such that $D(n_1) \\cap D(n_2) \\cap \\dots \\cap D(n_k)$ has at least two elements.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_1, a_2, \\dots, a_{k+1}$ be $k+1$ distinct positive integers, each smaller than the product of the other $k$ numbers. Let $N = a_1 a_2 \\cdots a_{k+1}$. For each $i = 1, 2, \\dots, k+1$, define\n$$\nx_i = \\frac{1}{2} \\left( \\frac{N}{a_i} + a_i \\right), \\quad y_i = \\frac{1}{2} \\left( \\frac{N}{a_i} - a_i \\right),\n$$\nso that $x_i^2 - y_i^2 = N$.\n\nSince $a_i a_j < N$ and $\\frac{N}{a_i} > a_i$, the pairs $(x_i, y_i)$ ($1 \\le i \\le k+1$) are $k+1$ positive integer solutions to $x^2 - y^2 = N$. Without loss of generality, suppose $x_{k+1} = \\min\\{x_1, x_2, \\dots, x_{k+1}\\}$. For each $i \\in \\{1, 2, \\dots, k\\}$, since $x_i^2 - x_{k+1}^2 = y_i^2 - y_{k+1}^2$, we have\n$$\n(x_i + x_{k+1})(x_i - x_{k+1}) = x_i^2 - x_{k+1}^2 = y_i^2 - y_{k+1}^2 = (y_i + y_{k+1})(y_i - y_{k+1}).\n$$\nLet $n_i = (x_i + x_{k+1})(x_i - x_{k+1}) = (y_i + y_{k+1})(y_i - y_{k+1})$. Then\n$$\n2x_{k+1} = (x_i + x_{k+1}) - (x_i - x_{k+1}) \\in D(n_i),\n$$\n$$\n2y_{k+1} = (y_i + y_{k+1}) - (y_i - y_{k+1}) \\in D(n_i).\n$$\nSince $x_{k+1} > y_{k+1}$, $2x_{k+1}$ and $2y_{k+1}$ are two different elements in $D(n_1) \\cap D(n_2) \\cap \\dots \\cap D(n_k)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16468,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $O$, $H$ 分別為不等邊三角形 $ABC$ 的外心與垂心，$P$ 為三角形 $AHO$ 內一點，滿足 $\\angle AHP = \\angle POA$，$M$ 為 $\\overline{OP}$ 的中點。設 $BM$, $CM$ 分別與三角形 $ABC$ 的外接圓交於 $X$, $Y$ 兩點。\n\n證明：直線 $XY$ 經過三角形 $APO$ 的外心。",
    "options": [],
    "answer": "See solution",
    "solution": "設 $AP$ 交 $\\odot(ABC)$ 於 $D$，$O_1$, $O_2$ 分別為 $\\triangle APO$, $\\triangle DPO$ 的外心，則\n\n$$\n\\angle PO_1O = 2 \\cdot \\angle PAO = 2 \\cdot \\angle ODP = \\angle OO_2P.\n$$\n\n故 $PO_1OO_2$ 是以 $M$ 為中心的菱形。\n\n![](images/2J0410_p7_data_789ff554d1.png)\n\n**Claim.** 點 $O_2$ 位於 $BC$ 上。\n\n*Proof of Claim.* 取 $O'$ 使得 $M$ 為 $AO'$ 的中點，平移 $\\triangle AHP$ 至 $\\triangle OO'P'$。由\n\n$$\n\\angle OP'P = \\angle PAO = \\angle ODP\n$$\n\n知 $D$, $O$, $P$, $P'$ 共於一圓 $\\Gamma$。又\n\n$$\n\\angle OO'P' = \\angle AHP = \\angle POA = \\angle OPP',\n$$\n\n所以 $O'$ 也在 $\\Gamma$ 上，因此 $O_2$ 位於 $OO'$ 的中垂線上，即 $BC$ 上。\n\n回到原命題，由 $O_1O_2$ 垂直 $OM$，蝴蝶定理可得 $O_2$ 關於 $M$ 的對稱點 $O_1$ 位於 $XY$ 上。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16469,
    "subject": "Mathematics (Olympiad)",
    "question": "You are on a staircase with steps numbered from 11 up to 100 (the courtyard). You start at step 11 and can take a fixed number of steps at a time, with the step size being an integer between 1 and 20 (inclusive). For each of the following, determine the answer and explain your reasoning:\n\n1. What is the largest step number you could land on before stepping onto the courtyard, if you always land on steps ending in 1 or 6?\n2. If you land on step 89 just before the courtyard, what step size were you using?\n3. If you land on step 95 just before the courtyard, what step size(s) could you have been using?\n4. What is the smallest possible step number you could land on before stepping onto the courtyard, given the constraints above?\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "a. Starting from step 11 and taking steps of size 5, you land on 11, 16, 21, 26, ..., so each step ends in 1 or 6. The largest such step before 100 is 96.\n\nb. Since $100 - 89 = 11$, the step size must be at least 11. The number of steps from 11 to 89 is $89 - 11 = 78$, so the step size must also be a factor of 78. The factors of 78 are 1, 2, 3, 6, 13, 26, 39, and 78. The only factor between 11 and 20 is 13. Thus, the step size is 13.\n\nc. Since $95 - 11 = 84$, the step size must be a factor of 84. The factors of 84 less than or equal to 20 are 1, 2, 3, 4, 6, 7, 12, and 14. For each:\n- 1: previous step is 94\n- 2: 93\n- 3: 92\n- 4: 91\n- 6: 89\n- 7: 88\n- 12: 83\n- 14: 81\n\nThe other factors are too large.\n\n---\n\nd. **Alternative i**\n\nIf the step size is at most 20, the last step before 100 cannot be lower than 80. Checking possible last steps:\n- 80: $80 - 11 = 69$ (factors: 1, 3, 23, 69), none between 11 and 20.\n- 81: $81 - 11 = 70$ (factors: 1, 2, 5, 7, 10, 14), but in each case, you could go further before 100.\n- 82: $82 - 11 = 71$ (factor: 1), but $82 + 1 < 100$.\n- 83: $83 - 11 = 72$ (factors: 1, 2, 3, 4, 6, 8, 9, 12, 18), and $18$ is between 11 and 20. So 83 is possible.\n\nThus, the smallest possible last step before the courtyard is 83.\n\n**Alternative ii**\n\nA table of step sizes $n$ from 1 to 20 and the largest $m < 100$ such that $m = 11 + k n$:\n\n$$\n\\begin{array}{|c|c|c|c|}\n\\hline\nn & m & n & m \\\\\n\\hline\n1 & 99 & 11 & 99 \\\\\n2 & 99 & 12 & 95 \\\\\n3 & 98 & 13 & 89 \\\\\n4 & 99 & 14 & 95 \\\\\n5 & 96 & 15 & 86 \\\\\n6 & 95 & 16 & 91 \\\\\n7 & 95 & 17 & 96 \\\\\n8 & 99 & 18 & 83 \\\\\n9 & 92 & 19 & 87 \\\\\n10 & 91 & 20 & 91 \\\\\n\\hline\n\\end{array}\n$$\n\nThe smallest possible last step is 83.\n\n**Alternative iii**\n\n- Step size 20: last step is $11 + 4 \\times 20 = 91$\n- Step size 19: $11 + 4 \\times 19 = 87$\n- Step size 18: $11 + 4 \\times 18 = 83$\n- Step size 17 or smaller: last step is at least $100 - 17 = 83$\n\nThus, the smallest possible last step before the courtyard is 83.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16470,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1 < a_2 < \\dots < a_k$ be positive integers. For each $a_i$, Zarina has written down all the positive divisors of $a_i$ in the notebook (some numbers might be written several times). Then, Marina split all the numbers in the notebook into several groups. It turned out that the numbers in each of these groups form the set of all positive divisors of some positive integer, which Marina has decided to also write down on the desk. Prove that the numbers on the desk are $a_1, a_2, \\dots, a_k$ in some order.",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that the number $a_k$ is written only once in the notebook. Therefore, there exists some group that contains all divisors of $a_k$. Thus, $a_k$ is also on the desk. Next, we cross out from the notebook all divisors of $a_k$ one time. By similar reasoning, $a_{k-1}$ is also on the desk; we can cross out it and all of its divisors, and so on. Therefore, all the numbers $a_1, a_2, \\dots, a_k$ are present on the desk.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16471,
    "subject": "Mathematics (Olympiad)",
    "question": "Real nonzero numbers $a, b, c, d$ satisfy the conditions $a^3 + b^3 + c^3 + d^3 = 0$ and $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} \\neq 0$. Prove that $a + b + c + d \\neq 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will use a proof by contradiction. Suppose $a + b + c + d = 0$. Thus, $a + b = -(c + d)$. Therefore, we can obtain the following equalities:\n\n$$\n(a + b)^3 = -(c + d)^3 \\Leftrightarrow a^3 + b^3 + 3ab(a + b) = -c^3 - d^3 - 3cd(c + d)\n$$\n$$\n\\Leftrightarrow 3ab(a + b) + 3cd(c + d) = 0 \\Leftrightarrow -3ab(c + d) - 3cd(a + b) = 0\n$$\n$$\n\\Leftrightarrow ab(c + d) + cd(a + b) = 0 \\Leftrightarrow abc + abd + acd + bcd = 0.\n$$\n\nThat contradicts the conditions since:\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} \\neq 0 \\Leftrightarrow \\frac{abc + abd + acd + bcd}{abcd} \\neq 0 \\Leftrightarrow abc + abd + acd + bcd \\neq 0.\n$$\n\nThe statement is proved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16472,
    "subject": "Mathematics (Olympiad)",
    "question": "The quadrilateral $ABCD$ is inscribed in the circle $k$. The lines $AC$ and $BD$ meet at $E$, and the lines $AD$ and $BC$ meet at $F$. Show that the line through the incenters of $\\triangle ABE$ and $\\triangle ABF$, and the line through the incenters of $\\triangle CDE$ and $\\triangle CDF$, meet on $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $I_e, I_f, J_e, J_f$ be the incenters of $\\triangle ABE$, $\\triangle ABF$, $\\triangle CDE$, and $\\triangle CDF$, respectively. Let $P = AI_e \\cap BI_f$, $Q = AI_f \\cap BI_e$, $U = CJ_e \\cap DJ_f$, and $V = CJ_f \\cap DJ_e$ be the excenter of $\\triangle ABC$ opposite to $A$, the excenter of $\\triangle ABD$ opposite to $B$, the incenter of $\\triangle ACD$, and the incenter of $\\triangle BCD$, respectively.\n\nWe have $\\triangle APB = \\frac{1}{2} \\widehat{AB} = \\triangle AQB$; therefore, $ABPQ$ is cyclic. Analogously, $CDUV$ is cyclic.\n\nWe have $\\triangle APB = \\frac{1}{2} \\widehat{AB} = \\triangle UCV$, $\\triangle AQB = \\frac{1}{2} \\widehat{AB} = \\triangle UDV$, $\\triangle APQ = \\triangle ABQ = \\frac{1}{2} \\widehat{AD} = \\triangle UCD$, and $\\triangle BQP = \\triangle BAP = \\frac{1}{2} \\widehat{BC} = \\triangle VDC$. Therefore, the figures $ABPQ$ and $UVCD$ are similar and identically oriented. Let $T$ be their center of similitude (i.e., the fixed point of the unique similitude which maps $A, B, P, Q$ onto $U, V, C, D$, respectively).\n\nWe have $\\triangle TAQ \\sim \\triangle TUD$; therefore, $\\triangle TAU \\sim \\triangle TQD$ and $\\triangle ATQ = \\triangle (AU, QD) = \\frac{1}{2} \\widehat{AD} = \\triangle AI_eQ$. It follows that $T$ lies on the circumcircle of $\\triangle AI_eQ$. Analogously, $T$ lies on the circumcircle of $\\triangle BI_eP$, $\\triangle CJ_eV$, and $\\triangle DJ_eU$.\n\nWe have, then, $\\triangle ATB = \\triangle ATI_e + \\triangle I_eTB = \\triangle AQB + \\triangle APB = \\frac{1}{2} \\widehat{AB} + \\frac{1}{2} \\widehat{AB} = \\widehat{AB}$, showing that $T$ lies on $k$.\n\n![](images/Broshura_2014_IMO_p5_data_b1709c900a.png)\n\nLet $T' \\in I_f I_e$ so that $I_f A \\cdot I_f Q = I_f I_e \\cdot I_f T' = I_f B \\cdot I_f P$. Then $T'$ is a point other than $I_e$ which lies on the circumcircles of both $\\triangle AI_eQ$ and $\\triangle BI_eP$; therefore, $T' \\equiv T$ and $T \\in I_e I_f$.\n\nAnalogously, $T \\in J_e J_f$, and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16473,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $s_1, s_2, s_3, \\dots$ be an infinite, non-constant sequence of rational numbers, meaning it is not the case that $s_1 = s_2 = s_3 = \\dots$. Suppose that $t_1, t_2, t_3, \\dots$ is also an infinite, non-constant sequence of rational numbers with the property that $(s_i - s_j)(t_i - t_j)$ is an integer for all $i$ and $j$. Prove that there exists a rational number $r$ such that $(s_i - s_j)r$ and $(t_i - t_j)/r$ are integers for all $i$ and $j$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we claim there exist $i, j$ such that $(s_i - s_j)(t_i - t_j) \\neq 0$. Since both sequences are non-constant, such $i, j$ exist.\n\nWe can reorder the pairs $(s_i, t_i)$ without affecting the hypothesis or conclusion, so assume $(s_1 - s_2)(t_1 - t_2) \\neq 0$. Also, for any constants $a, b$, replacing $s_i$ by $s_i - a$ and $t_i$ by $t_i - b$ preserves all differences. Thus, we may assume $s_1 = t_1 = 0$, $s_2 \\neq 0$, $t_2 \\neq 0$.\n\nCall a pair of positive rational numbers $(A, B)$ *good* if $AB$ is an integer, and $A s_j$ and $B t_j$ are integers for all $j$.\n\nWe show a good pair exists. For all $i \\geq 2$, $s_i t_i$ is an integer. For all $i, j \\geq 2$, $s_i t_j + s_j t_i$ is an integer. Write $s_j = \\dfrac{p_j}{q_j}$ and $t_j = \\dfrac{u_j}{v_j}$ in lowest terms. Since $s_j t_j = \\dfrac{p_j u_j}{q_j v_j}$ is integer, $p_j$ is divisible by $v_j$, say $p_j = d_j v_j$. Also, $s_2 t_j + s_j t_2$ is an integer, so $q_j$ divides $d_j u_2 q_2 v_j^2$, and since $q_j$ is coprime to $d_j v_j$, $q_j$ divides $u_2 q_2$. Thus, $A = |u_2 q_2|$ makes $A s_j$ integer for all $j$. Similarly, $B$ exists so that $B t_j$ is integer for all $j$. Thus, a good pair exists.\n\nNow, take a good pair $(A, B)$ with minimal $AB$. We show $AB = 1$. Suppose $AB > 1$ and let $p$ be a prime dividing $AB$. If $A s_i$ is divisible by $p$ for all $i$, then $(A/p, B)$ is a smaller good pair, contradiction. So for some $i$, $A s_i$ is not divisible by $p$, so $B t_i$ must be divisible by $p$ (since $A B s_i t_i$ is divisible by $p$). Similarly, for some $j$, $B t_j$ is not divisible by $p$, but $A s_j$ is. Then\n\n$$\n(AB)(s_i t_j + s_j t_i) - (A s_j)(B t_i) = (A s_i)(B t_j)\n$$\n\nThe left side is divisible by $p$, but the right side is not, a contradiction. Thus, $AB = 1$.\n\nLet $r = A$. Then $(s_i - s_j) r$ and $\\dfrac{t_i - t_j}{r}$ are integers for all $i, j$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16474,
    "subject": "Mathematics (Olympiad)",
    "question": "Can one paint three of the six segments (the three sides and the three medians) of triangle $ABC$ red and the other three blue so that it is impossible to construct a triangle using only segments of the same color?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $G$ be the centroid of triangle $ABC$, and let $A_1$, $B_1$, $C_1$ be the midpoints of sides $BC$, $AC$, and $AB$, respectively. Denote the sides and medians as follows: $AB = c$, $BC = a$, $CA = b$, $AA_1 = d$, $BB_1 = e$, $CC_1 = f$.\n\nSuppose we can color three of these six segments red and the other three blue so that it is impossible to form a triangle using only segments of the same color. Then, for each color, the length of some segment is not less than the sum of the other two segments of that color.\n\nThus, there exist two segments among the six whose sum is at least as large as the sum of the other four: $x + y \\geq z + u + v + w$. We show this is impossible by considering three cases:\n\n1. Both $x$ and $y$ are sides of $\\triangle ABC$. Without loss of generality, let $a + b \\geq c + d + e + f$. From triangles $BGC$ and $AGC$, we have $\\frac{2f}{3} + \\frac{2e}{3} > a$ and $\\frac{2f}{3} + \\frac{2d}{3} > b$. Summing these gives $\\frac{f}{3} > c + \\frac{d}{3} + \\frac{e}{3}$, which contradicts the triangle inequality for medians: $d + e > f$.\n\n2. Both $x$ and $y$ are medians. Without loss of generality, let $d + e \\geq a + b + c + f$. From triangles $BGC_1$ and $AGC_1$, we have $\\frac{f}{3} + \\frac{c}{2} > \\frac{2e}{3}$ and $\\frac{f}{3} + \\frac{c}{2} > \\frac{2d}{3}$, or $\\frac{f}{2} + \\frac{3c}{4} > e$ and $\\frac{f}{2} + \\frac{3c}{4} > d$. It follows that $f + \\frac{3c}{2} > d + e \\geq a + b + c + f$, or $\\frac{c}{2} > a + b$, a contradiction.\n\n3a. $x$ and $y$ are a side and a median sharing an endpoint. Without loss of generality, let $a + e \\geq b + c + d + f$. This contradicts $a < b + c$ and $e < d + f$.\n\n3b. $x$ is a median with its endpoint at the midpoint of side $y$. Without loss of generality, let $a + d \\geq b + c + e + f$. This contradicts $a < b + c$ and $d < e + f$.\n\nTherefore, such a coloring is impossible.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16475,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D$ be the foot of the altitude drawn to the hypotenuse $AB$ of a right triangle $ABC$. The inradii of the triangles $ABC$, $CAD$, and $CBD$ are $r$, $r_1$, and $r_2$, respectively. Prove that $CD = r + r_1 + r_2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = |BC|$, $b = |CA|$, $c = |AB|$, and $h = |CD|$; then $ab = ch = (a + b + c)r$.\n\n![](images/EST_ABooklet_2021_p27_data_303fbabe69.png)\n\nNote that the triangles *ABC*, *ACD*, and *CBD* are similar by two equal angles. As the similarity ratio of triangles *ACD* and *ABC* is $\\frac{b}{c}$ and that of triangles *CBD* and *ABC* is $\\frac{a}{c}$, we have $r_1 = \\frac{b}{c} r$ and $r_2 = \\frac{a}{c} r$, whence\n\n$$\nr + r_1 + r_2 = \\frac{a + b + c}{c} r.\n$$\n\nAs the equality $ab = (a + b + c)r$ implies $r = \\frac{ab}{a + b + c}$, we obtain\n\n$$\nr + r_1 + r_2 = \\frac{ab}{c}.\n$$\n\nOn the other hand, the equality $ab = ch$ implies $h = \\frac{ab}{c}$. Consequently, $r + r_1 + r_2 = h$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16476,
    "subject": "Mathematics (Olympiad)",
    "question": "Vertices of a regular $6n+3$-gon, numbered clockwise $1, 2, \\ldots, 6n+3$, form the game field. Vertices numbered $2n+1$, $4n+2$, and $6n+3$ are called *holes*. At the start, there are 3 chips on the field. Two players take turns choosing any one of the 3 chips and moving it clockwise to a neighboring vertex, provided that vertex is not occupied by another chip. The first player wins if, after any turn, at least 2 chips are in holes. Can the first player always win if the chips are initially placed in the form of a regular triangle?",
    "options": [],
    "answer": "See solution",
    "solution": "Let the *distance to the hole* for a chip be the number of steps needed to reach the nearest hole moving clockwise, assuming no other chips block the way. For example, a chip at vertex $4n$ has distance 2 to the hole if $4n+1$ and $4n+2$ are unoccupied; a chip at $3n$ has distance $n+2$ to the hole $4n+2$ if vertices $3n+1, \\ldots, 4n+2$ are free.\n\nWe prove by induction that a position where two chips have the same distance $k$ to holes is a winning position for the first player. Call these chips *hot*; let the *first* hot chip be the one whose path to the other hot chip passes through the *usual* (third) chip.\n\n**Base case ($k=0$):** The first player wins immediately.\n\n**Case $k=1$:** If it's the first player's turn, he moves the usual chip if possible, leading to the same situation for the second player. If the usual chip is adjacent to a hot chip, similar analysis applies. If the second player moves a hot chip into a hole, the first player moves the other hot chip into a hole and wins. Thus, the second player must move the usual chip, and both players continue moving it until it cannot move further (a *zugzwang* position). At this point, the second player loses because he cannot move the third chip.\n\n**Inductive step:** Assume the statement holds for distance $k$. For distance $k+1$, conditionally move the holes forward by one position; this reduces the situation to the $k=1$ case. Thus, the first player can always reach a position where two chips are $k$ steps from holes, and win.\n\nFinally, the initial configuration (chips forming a regular triangle) satisfies the induction conditions, so the first player can always win.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16477,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there are infinitely many pairs $(m, n)$ of positive integers such that $m < n$ and $\\frac{(m+n)(m+n+1)}{mn}$ is an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "If we take $m = 1$, $n = 2$, we see that $\\frac{(m+n)(m+n+1)}{mn} = 6$. (Or we can start with $m = 2$, $n = 3$ as well.)\n\nSuppose we have some pair $(m, n)$ of positive integers such that $m < n$ and\n\n$$\n\\frac{(m+n)(m+n+1)}{mn} = k\n$$\n\nis an integer. This may be written in the form\n\n$$\nnk = m + 2n + 1 + \\frac{n(n+1)}{m}\n$$\n\nThus $\\frac{n(n+1)}{m}$ is also an integer, say equal to $l$. Observe that\n\n$$\n\\begin{aligned}\n(n+l)(n+l+1) &= \\left(n + \\frac{n(n+1)}{m}\\right) \\left(n + 1 + \\frac{n(n+1)}{m}\\right) \\\\\n&= \\frac{l(m+n+1)(m+n)}{m^2}.\n\\end{aligned}\n$$\n\nThus\n\n$$\n\\frac{(n+l)(n+l+1)}{nl} = \\frac{(m+n)(m+n+1)}{mn} = k.\n$$\n\nMoreover, $lm = n(n+1) > n^2 > nm$ showing that $l > n$. Hence $(m, n) \\neq (n, l)$. Thus, starting with the pair $(1, 2)$, we can generate the new pair $(2, 6)$, and the process may be continued indefinitely.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16478,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that\n\n$$\n\\sqrt{a^2 + b^2 - \\sqrt{2}ab} + \\sqrt{b^2 + c^2 - \\sqrt{2}bc} \\geq \\sqrt{a^2 + c^2}\n$$\n\nfor all positive real numbers $a$, $b$, and $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "The inequality results from the triangle inequality, $PQ + PR \\geq QR$, as shown in the figure.\n\n![](images/Tajland_2008_p3_data_fbc62ef35f.png)\n\n![](images/Tajland_2008_p3_data_815aca1c52.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16479,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\ge 3$, determine the least value the sum $\\sum_{i=1}^{n} \\left(\\frac{1}{x_i} - x_i\\right)$ may achieve, as the $x_i$ run through the positive real numbers subject to $\\sum_{i=1}^{n} \\frac{1}{x_i + n - 1} = 1$. Also, determine the $x_i$ at which this minimum is achieved.",
    "options": [],
    "answer": "See solution",
    "solution": "The required minimum is $0$ and is achieved if and only if the $x_i$ are all equal to $1$.\n\nLet $x_1, \\dots, x_n$ be positive real numbers satisfying the condition in the statement. Let $y_i = \\frac{x_i}{x_i + n - 1}$ for $i = 1, 2, \\dots, n$, and notice that the $y_i$ are positive real numbers that add up to $1$. Express the $x_i$ in terms of the $y_i$ to get $x_i = \\frac{(n-1)y_i}{1-y_i}$, and write successively\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} \\frac{1}{x_i} &= \\frac{1}{n-1} \\sum_{i=1}^{n} \\frac{1-y_i}{y_i} = \\frac{1}{n-1} \\sum_{i=1}^{n} \\frac{1}{y_i} \\sum_{j \\neq i} y_j = \\frac{1}{n-1} \\sum_{i \\neq j} \\frac{y_j}{y_i} = \\frac{1}{n-1} \\sum_{i \\neq j} \\frac{y_i}{y_j} \\\\\n&= \\frac{1}{n-1} \\sum_{i=1}^{n} y_i \\sum_{j \\neq i} \\frac{1}{y_j} \\ge \\frac{1}{n-1} \\sum_{i=1}^{n} y_i \\cdot \\frac{(n-1)^2}{\\sum_{j \\neq i} y_j} = \\sum_{i=1}^{n} \\frac{(n-1)y_i}{1-y_i} = \\sum_{i=1}^{n} x_i.\n\\end{aligned}\n$$\n\nEquality clearly forces the $y_i$ all equal to $1/n$, which is the case if and only if the $x_i$ are all equal to $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16480,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $S$ is the maximum sum such that any set of 19 numbers in $[0,9]$ can be partitioned into two groups, each with sum at most $9$. Find $\\max S$.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim $\\max S = 17.1$.\n\nFirst, if $S > 17.1$, the required partition may be impossible. For example, let $S = 17.1 + 19\\epsilon$, $\\epsilon > 0$. Suppose we have 19 numbers, each equal to $0.9 + \\epsilon$. Any partition will have at least 10 numbers in one group, so the sum in that group is at least $9 + 10\\epsilon > 9$, which is forbidden.\n\nNow, let $S \\leq 17.1$. We show that the required partition exists for any such $S$, so $\\max S = 17.1$. Consider all sums $S_1$ of subsets of the numbers such that $8 < S_1 \\leq 9$, and let $A$ be the maximal such sum (if none exist, there is nothing to prove). For any remaining number $a$, $A + a > 9$. Consider two cases:\n\n**Case 1:** $A \\geq 8.1$. The sum of all remaining numbers does not exceed $9$, so we can partition into two groups with sums $A$ and $S - A$.\n\n**Case 2:** $A < 8.1$. Let $A = 8.1 - x$, $0 < x < 0.1$. If there are at most nine numbers left, their sum does not exceed $9$, so we can form the second group. If there are at least ten numbers left, since $A + a > 9$ and $A = 8.1 - x$, we have $a > 0.9 + x$. Thus, $S - A \\geq 10 \\cdot (0.9 + x) = 9 + 10x$. Then $S \\geq A + 9 + 10x = 8.1 - x + 9 + 10x = 17.1 + 9x > 17.1$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16481,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AB$ be a chord of circle $O$, $M$ the midpoint of arc $AB$, and $C$ a point outside of the circle $O$. From $C$, draw two tangents to the circle at points $S$ and $T$. Let $MS \\cap AB = E$, $MT \\cap AB = F$. From $E$ and $F$, draw lines perpendicular to $AB$, intersecting $OS$ and $OT$ at $X$ and $Y$ respectively. Now, draw a line from $C$ which intersects the circle $O$ at $P$ and $Q$. Let $Z$ be the circumcenter of $\\triangle PQR$. Prove that $X$, $Y$, $Z$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof** Refer to the figure, join points $O$ and $M$. Then $OM$ is the perpendicular bisector of $AB$. So $\\triangle XES \\sim \\triangle OMS$, and thus $SX = XE$.\n\nNow draw a circle with center $X$ whose radius is $XE$. Then the circle $X$ is tangent to chord $AB$ and line $CS$. Draw the circumcircle of $\\triangle PQR$, line $MA$ and line $MC$.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p94_data_46e07a6c01.png)\n\nIt is easy to see ($\\triangle AMR \\sim \\triangle PMA$ etc)\n\n$$\nMR \\cdot MP = MA^2 = ME \\cdot MS. \\qquad \\textcircled{1}\n$$\n\nBy the Power of a Point theorem,\n\n$$\nCQ \\cdot CP = CS^2. \\qquad \\textcircled{2}\n$$\n\nSo $M$, $C$ are on the radical axis of circle $Z$ and circle $X$. Thus\n$$ZX \\perp MC.$$\n\nSimilarly, we have $ZY \\perp MC$.\n\nSo $X$, $Y$, $Z$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16482,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $(k, n)$ such that\n\n$$\nk! = \\prod_{i=0}^{n-1} (2^n - 2^i).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For reference, the given equation is\n$$\nk! = \\prod_{i=0}^{n-1} (2^n - 2^i). \\qquad (1)\n$$\n\nLet us estimate what power of 2 divides each side of equation (1).\n\nFor $\\nu_2(\\text{LHS}(1))$, note that in the product $k! = 1 \\times 2 \\times 3 \\times \\cdots \\times k$, every second term is divisible by 2, every fourth term is divisible by $2^2$, every eighth term is divisible by $2^3$, and so on. Therefore\n$$\n\\nu_2(\\text{LHS}(1)) = \\left\\lfloor \\frac{k}{2^1} \\right\\rfloor + \\left\\lfloor \\frac{k}{2^2} \\right\\rfloor + \\left\\lfloor \\frac{k}{2^3} \\right\\rfloor + \\cdots < \\frac{k}{2^1} + \\frac{k}{2^2} + \\frac{k}{2^3} + \\cdots = k. \\quad (2)\n$$\n\nFor $\\nu_2(\\text{RHS}(1))$, we have $2^i \\mid 2^n - 2^i$ for $i = 0, 1, \\dots, n-1$, and so\n$$\n\\nu_2(\\text{RHS}(1)) = 0 + 1 + 2 + \\cdots + (n-1) = \\frac{n(n-1)}{2}. \\qquad (3)\n$$\n\nComparing (2) and (3) yields $k > \\frac{n(n-1)}{2}$. Substituting this into (1) yields\n$$\n\\left(\\frac{n(n-1)}{2}\\right)! < \\prod_{i=0}^{n-1} (2^n - 2^i) < \\prod_{i=0}^{n-1} 2^n = 2^{n^2}. \\qquad (4)\n$$\n\nWe claim that inequality (4) is false for $n \\ge 6$. That is, we claim that\n$$\n\\left(\\frac{n(n-1)}{2}\\right)! > 2^{n^2} \\qquad (5)\n$$\nfor $n \\ge 6$. We prove this by induction.\n\nFor the base case $n=6$, we calculate $\\text{RHS}(5) = 2^{36} = 68719476636 < 10^{11}$, and then $\\text{LHS}(5) = 15! > 10! \\cdot 10^5 = 362880000000 > 10^{11} > \\text{RHS}(5)$.\n\nFor the inductive step, suppose that (5) is true for some integer $n \\ge 6$. Then\n$$\n\\left(\\frac{(n+1)n}{2}\\right)! > \\left(\\frac{n(n-1)}{2}\\right)! \\left(\\frac{n(n-1)}{2} + 1\\right)^n > 2^{n^2} \\cdot 16^n = 2^{n^2+4n} > 2^{(n+1)^2}.\n$$\n\nThis completes the induction. Thus (1) has no solutions for any $n \\ge 6$.\n\nFor $n=1, 2, 3, 4$ and $5$, we have $\\text{RHS}(1) = 1, 6, 168, 7! \\cdot 4$ and $31 \\cdot 30 \\cdot 28 \\cdot 24 \\cdot 16$, respectively. So $n=1$ and $2$ yield the solutions $(k, n) = (1, 1)$ and $(3, 2)$, while $n=3$ and $4$ do not yield solutions. Finally $n=5$ does not yield a solution, because $\\text{RHS}(1)$ is divisible by the prime $31$ but not the prime $29$.\n\nFor a prime number $p$ and a positive integer $N$ we will use the following standard notation:\n\n* The number $\\nu_p(N)$ denotes the exponent of $p$ in the prime factorisation of $N$.\n* The notation $p^k \\mid N$ means that $\\nu_p(N) = k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16483,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a 4-digit integer whose ten's place is nonzero. If we take the first two digits and the last two digits of $n$ as two 2-digit integers, their product is a divisor of $n$. Determine all such $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A$ and $B$ be the first and last two digits of $n$, respectively. We seek all $(A, B)$ such that $AB$ divides $100A + B$.\n\nSince $A$ divides $100A + B$, $A$ must divide $B$. Let $k = \\frac{B}{A}$. Since $A$ and $B$ are 2-digit numbers, $10 \\leq A < \\frac{100}{k}$.\n\nThe condition is equivalent to $kA^2 \\mid 100A + kA$, which simplifies to $kA \\mid 100 + k$.\n\n$k$ divides $100 + k$ iff $k$ divides $100$, and with $k < 10$ we get $k = 1, 2, 4, 5$. From\n$$A = \\frac{100 + k}{k}$$\nand $10 \\leq A < \\frac{100}{k}$, we get $(k, A) = (2, 17), (4, 13)$, yielding $n = 1734, 1352$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16484,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 5$ be an integer. Consider $n$ distinct points in the plane, each coloured either white or black. For each positive integer $1 \\leq k < \\frac{n}{2}$, a $k$-move consists in selecting $k$ points and reversing their colours. Find all values of $n$ for which, for any eligible $k$ and for any initial colouring, there exists a sequence of $k$-moves that turns all points into a same colour.",
    "options": [],
    "answer": "See solution",
    "solution": "The problem holds if and only if $n$ is odd.\n\nSuppose $n$ is even. Colour exactly one point white and the rest black. Choose $k=2$. After performing a 2-move, the number of points of the same colour remains odd, so a monochromatic configuration is not achievable.\n\nSuppose $n$ is odd. Assume $k$ is odd. Consider an initial colouring with $p$ white points and $n-p$ black points. Let $p = ks + r$, $0 \\leq r < k$. Perform $s$ $k$-moves on the white points to reach a configuration with $r$ white and $n - r$ black points. Perform a $k$-move with the $r$ white points and $k - r$ black points to obtain exactly $k - r$ white points. Therefore, one can obtain exactly $r$ or $k - r$ white points, with only one of the numbers $r$ or $k - r$ being even, say $r$.\n\nPerform a $k$-move with the $\\frac{r}{2}$ white points (and $k - \\frac{r}{2}$ black points) to get exactly $k$ white points. After another $k$-move with the white points, all points turn black and we are done.\n\nAssume $k$ is even. Again, consider an initial colouring with $p$ white points and $n-p$ black points. One of the numbers $p$ or $n-p$ is even, say $p$. In $p = ks + r$, $0 \\leq r < k$, notice that $r$ is even. From this point on, proceed as in the previous case.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16485,
    "subject": "Mathematics (Olympiad)",
    "question": "In a given configuration of balls, we label the blue balls clockwise $P_1, P_2, \\dots, P_{2k}$ starting with an arbitrary blue ball. Let $m_i$ ($i \\in \\{1, 2, \\dots, 2k\\}$) be the number of red balls between balls $P_i$ and $P_{i+1}$ ($P_{n+1} = P_1$). For example, in the following picture we have $m_1 = 2, m_2 = 3, m_3 = 1, m_4 = 0, m_5 = 4, m_6 = 1$.\n\n![](images/Mathematica_competitions_in_Croatia_in_2013_p25_data_ecf2d8ad35.png)\n\nFurthermore, let us denote $S = m_1 - m_2 + m_3 - \\dots + m_{2k-1} - m_{2k}$. If the configuration consists of $n$ red balls and no blue balls we denote $S = n$.\n\nProve that either $S$ is divisible by three for all configurations or $S$ is not divisible by three for all configurations.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider three cases for adding a red ball:\n\n1. **Adding a red ball between two blue balls $P_i$ and $P_{i+1}$:**\n\n$$\nP_{i-1} \\underbrace{\\dots P_i}_{m_{i-1}} \\underbrace{\\dots P_{i+1}}_{m_i=0} \\rightarrow P_{i-1}' \\underbrace{\\dots \\overbrace{CC}^{\\downarrow} \\dots P_i'}_{m_i'}\n$$\n\nBefore the operation, $S = S_0 + m_{i-1} - m_i + m_{i+1} = S_0 + m_{i-1} + m_{i+1}$. After the operation, $S' = S_0 + m_i' = S_0 + (m_{i-1} + m_{i+1} + 3) = S + 3$.\n\n2. **Adding a red ball between two red balls:**\n\n$$\nP_i \\underbrace{\\dots CC \\dots}_{m_i} P_{i+1} \\rightarrow P_i \\underbrace{\\dots P_{i+1}}_{m_i'} \\underbrace{\\overbrace{C}^{\\downarrow}}_{m_{i+1}'=1} P_{i+1} \\underbrace{\\dots P_{i+3}}_{m_{i+2}'}\n$$\n\nSimilarly, $S = S_0 + m_i$, and $S' = S_0 + m_i' - 1 + m_{i+2}$. Since $m_i = m_i' + m_{i+2}' + 2$, $S' = S - 3$.\n\n(An analogous statement holds if there are no blue balls on the circle.)\n\n3. **Adding a red ball between one red and one blue ball (to the right of the blue ball $P_i$):**\n\n$$\nP_{i-1} \\underbrace{\\overbrace{P_i}^{m_{i-1}} \\overbrace{C \\dots P_{i+1}}^{m_i}} \\rightarrow P_{i-1} \\underbrace{\\overbrace{\\dots \\overbrace{C}^{m'_{i-1}} C}^{\\downarrow} P_i}_{m'_i} \\dots P_i\n$$\n\nIn this case, $S = S_0 + m_{i-1} - m_i$ and $S' = S_0 + m'_{i-1} - m'_i$. Since $m'_{i-1} = m_{i-1} + 2$ and $m'_i = m_i - 1$, $S' = S + 3$.\n\nIn all cases, $S$ either increases or decreases by 3. If initially $S = 2$, then $S$ can never be divisible by 3. Thus, for all configurations, $S$ is either always divisible by 3 or never divisible by 3.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16486,
    "subject": "Mathematics (Olympiad)",
    "question": "In quadrilateral $ABCD$, diagonal $AC$ is the angle bisector of $\\angle BAD$, and $\\angle ADC = \\angle ACB$. Points $X$ and $Y$ are the feet of the perpendiculars from point $A$ to $BC$ and $CD$, respectively. Prove that the orthocenter of $\\triangle AXY$ lies on the line $BD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $E$ be the foot of the perpendicular from $Y$ to $AX$, and let $YE \\cap BD = P$. It is enough to prove that $XP \\perp AY$, which is equivalent to proving $XP \\parallel CD$. Notice that $YE \\parallel CB \\implies \\frac{DY}{YC} = \\frac{DP}{PB}$. Also, since $\\triangle ADC \\sim \\triangle BCA$, we have $\\frac{DY}{YC} = \\frac{CX}{XB}$. From these equalities, we get $\\frac{DP}{PB} = \\frac{CX}{XB} \\implies XP \\parallel CD$, as required. Q.E.D.\n\n![](images/Ukraine_booklet_2018_p17_data_0ec00f3455.png)\n\n**Fig. 14**",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16487,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a quadrilateral. Consider a circle tangent to $AD$ and $BC$ at $A$ and $B$ respectively. Let $P$ be a point in the interior of this circle such that $\\widehat{APB} > \\widehat{ABx}$. Similarly, consider a circle tangent to $AD$ and $BC$ at $D$ and $C$ respectively. With the same argument, $P$ is in this circle. Thus, these two circles intersect at two points, labeled $E$ and $F$. Let $M$ and $N$ be the intersection points of $EF$ with $AD$ and $BC$ respectively. Prove that:\n\n$$\nAB + DC > AD + BC.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\n\\begin{cases} NF \\cdot NE = NB^2 = NC^2 \\\\ ME \\cdot MF = MA^2 = ND^2 \\end{cases} \\xrightarrow{\\text{AM-GM}} \\begin{cases} NF + NE > NB + NC \\\\ ME + MF > MA + MD \\end{cases}\n$$\n\nAdding these two inequalities gives $2MN > AD + BC$. Since $M$ and $N$ are the midpoints of $AD$ and $BC$ respectively, we also have\n\n$$\n2MN = AB + DC.\n$$\n\nTherefore,\n\n$$\nAB + DC > AD + BC.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16488,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer and $a_1, a_2, \\dots, a_{2n}$ be $2n$ distinct integers. Given that the equation\n$$\n|x - a_1| |x - a_2| \\dots |x - a_{2n}| = (n!)^2\n$$\nhas an integer solution $x = m$, find $m$ in terms of $a_1, \\dots, a_{2n}$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n|m - a_1| |m - a_2| \\cdots |m - a_{2n}| = (n!)^2.\n$$\n\nFirst, we show that we cannot have distinct $i, j, k$ so that $|m - a_i| = |m - a_j| = |m - a_k|$. If so, then two of $(m - a_i)$, $(m - a_j)$, $(m - a_k)$ must be of the same sign, say $(m - a_i)$ and $(m - a_j)$. Then $a_i = a_j$, a contradiction. Thus, the values of $|m - a_1|, |m - a_2|, \\ldots, |m - a_{2n}|$ are $1, 1, 2, 2, \\ldots, n, n$.\n\nAlso, if $|m - a_i| = |m - a_j|$, then we must have $m - a_i = -(m - a_j)$; otherwise, $a_i = a_j$. Therefore,\n$$\n\\sum (m - a_i) = 0\n$$\nand so\n$$\nm = \\frac{a_1 + \\cdots + a_{2n}}{2n}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16489,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m = [n\\sqrt{6}]$. Prove that\n\na) $\\{n\\sqrt{6}\\} > \\dfrac{1}{n}$,\n\nb) $\\{n\\sqrt{6}\\} > \\dfrac{1}{n - \\frac{1}{5n}}$,\n\nwhere $n$ is even.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m = [n\\sqrt{6}]$. Then $n\\sqrt{6} > m$, so $6n^2 - m^2 > 0$. Note that $6n^2 - m^2 \\neq 1$ or $4 \\pmod{3}$, $6n^2 - m^2 \\neq 2 \\pmod{4}$ (recall that $n$ is even), $6n^2 - m^2 \\neq 3 \\pmod{9}$, so $6n^2 - m^2 \\geq 5$.\n\nNow we have\n\n$$\n\\{n\\sqrt{6}\\} = n\\sqrt{6} - m = \\frac{6n^2 - m^2}{n\\sqrt{6} + m} \\geq \\frac{5}{n\\sqrt{6} + m} > \\frac{5}{2n\\sqrt{6}} > \\frac{1}{n},\n$$\n\nthus a) is proved. In particular, $m < n\\sqrt{6} - \\frac{1}{n}$, so\n\n$$\n\\{n\\sqrt{6}\\} \\geq \\frac{5}{n\\sqrt{6} + m} > \\frac{5}{2n\\sqrt{6} - 1/6} > \\frac{1}{n - \\frac{1}{5n}},\n$$\n\nthus b) is proved.\n\nNote that, in fact, $6n^2 - m^2 \\geq 8$ for even $n$, so the estimates a) and b) can be improved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16490,
    "subject": "Mathematics (Olympiad)",
    "question": "For each prime number $p$, determine the number of residue classes modulo $p$ which can be represented as $a^2 + b^2$ modulo $p$, where $a$ and $b$ are arbitrary integers.",
    "options": [],
    "answer": "See solution",
    "solution": "All $p$ residue classes.\n\nWith $a^2 + 0^2$, we first obtain all quadratic residue classes.\n\nSince not all residue classes are quadratic residues, there is a quadratic residue class $a^2$ that is followed by a quadratic non-residue class, so that $n = a^2 + 1$ is not a quadratic residue and therefore $n \\neq 0 \\pmod{p}$.\n\nHowever, since the product of two quadratic non-residue classes is a quadratic residue class, it follows for each quadratic non-residue class $m$ that\n$$\n  m = \\frac{n m n}{n^2} = \\frac{(a^2+1)c^2}{n^2} \\equiv (a c n^{-1})^2 + (c n^{-1})^2 \\pmod{p}\n$$\nand therefore all quadratic residue classes can also be represented as the sum of two squares.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16491,
    "subject": "Mathematics (Olympiad)",
    "question": "A train departed from the station 12 minutes later than planned. If the train would not make any stops on the way and would travel at average speed equal to what would be the average speed between stops according to the timetable, then it would reach the destination exactly at the right time. But if the train would stop in every station for the same amount of time it was supposed to, then between the stations it would have to travel with average speed 40\\% higher than before in order to reach the destination on time. Find the travelling time of the train according to the timetable.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the time we are looking for be $t$. The conditions of the problem imply that\n$$\nt - 12\\ \\text{min} = 1.4 \\cdot (t - 24\\ \\text{min})\n$$\nfrom which\n$$\n0.4t = 21.6\\ \\text{min}\n$$\nand\n$$\nt = 54\\ \\text{min}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16492,
    "subject": "Mathematics (Olympiad)",
    "question": "Two squirrels, Bushy and Jumpy, have collected 2021 walnuts for the winter. Jumpy numbers the walnuts from 1 through 2021, and digs 2021 little holes in a circular pattern in the ground around their favorite tree. The next morning Jumpy notices that Bushy had placed one walnut into each hole, but had paid no attention to the numbering. Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of 2021 moves. In the $k$th move, Jumpy swaps the positions of the two walnuts adjacent to walnut $k$. Prove that there exists a value of $k$ such that, in the $k$th move, Jumpy swaps some walnuts $a$ and $b$ such that $a < k < b$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that the statement is untrue. In the $k$th move, if Jumpy swaps $a$ and $b$ with $a, b < k$, then we say $k$ is a \"big\" walnut; if $a, b > k$, then we say $k$ is a \"small\" walnut. After the $k$th move, we call every walnut with number $a < k$ operated and every walnut with number $b > k$ unoperated.\n\nAt every moment, the operated walnuts are divided by the unoperated walnuts into several intervals (each interval consists of consecutive operated walnuts and the neighbours on the two ends are unoperated ones). We use induction to prove that, after $k$ moves, each interval contains an odd number of walnuts. For $k = 1$, the assertion is obvious. In the $k$th move, there are two possibilities:\n\n1. If $k$ is a small walnut, then its neighbours are both unoperated before and after the move, $k$ itself is an interval of length 1.\n2. If $k$ is a big walnut, then its neighbours are both operated, say they belong to intervals of lengths $p$ and $q$. After the move, they merge into one interval of length $p+q+1$. By the induction hypothesis, $p, q$ are both odd, and the new interval's length $p+q+1$ is also odd.\n\nNow the induction is complete and the assertion is verified. Notice that after 2020 moves, the 2020 operated walnuts form a single interval of length 2020, which is an even number. This contradiction indicates that during the process, Jumpy must have swapped walnuts $a, b$ adjacent to $k$ such that $a < k < b$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16493,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral and let $M$ and $N$ be the midpoints of the sides $AD$ and $BC$, respectively. The circle through $A$ and $D$ tangent to $AC$ crosses the line $AB$ again at $P$, and the circle through $B$ and $C$ tangent to $BD$ crosses the line $AB$ again at $Q$. Let $t_A$ be the tangent to the circle $AMP$ at $A$ and let $t_B$ be the tangent to the circle $BNQ$ at $B$. Prove that the lines $CD$, $t_A$, and $t_B$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove that the triangles $ADP$ and $CDB$ are similar. Since $ABCD$ is cyclic, $\\angle DAP = \\angle DCB$; and since $AC$ is tangent to the circle $ADP$, $\\angle CBD = \\angle CAD = \\angle APD$. Consequently, the triangles $ADP$ and $CDB$ are similar.\n\nLet $R$ be the midpoint of $CD$. The points $M$ and $R$ correspond under the above similarity, so $\\angle PMA = \\angle BRC$.\n\nLet the line $CD$ meet the circle $ABR$ again at $K$ (possibly, $K = R$). Notice that $\\angle BAK = \\angle BRC = \\angle PMA$, so $AK$ is tangent to the circle $AMP$.\n\nSimilarly, $BK$ is tangent to the circle $BNQ$ and the conclusion follows.\n\n_Alternative solution._ We prove the conclusion under the weaker assumptions that $ABCD$ is merely convex and $AM/DM = CN/BN$.\n\nLet $AB$ and $CD$ meet at $E$; if the two are parallel, then $E$ is their common ideal point. Let $CD$ cross $t_A$ and $t_B$ at $K_A$ and $K_B$, respectively. We will show that $K_A = K_B$, whence the conclusion.\n\nLet $I$ and $J$ be the ideal points of $AD$ and $BC$, respectively. The line pencils\n\n($AD, AC, AE, AK_A$) and ($PA, PD, PI, PM$)\n\nare congruent, since $\\angle K_A AD = \\angle APM$, $\\angle CAD = \\angle APD$ and $\\angle IAE = \\angle IPE$. Hence $(D, C; E, K_A) = (AD, AC; AE, AK_A) = (PA, PD; PI, PM) = (A, D; I, M) = DM/AM$. Similarly, $(D, C; E, K_B) = (BD, BC; BE, BK_B) = (QC, QB; QJ, QN) = (C, B; J, N) = BN/CN$.\n\nFinally, recall the assumption $AM/DM = CN/BN$. By the preceding, $(D, C; E, K_A) = (D, C; E, K_B)$, so $K_A = K_B$, as desired. This ends the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16494,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n$ 為正整數。我們稱一個嚴格遞增的等差數列 $x_0, x_1, \\dots, x_n$ 為 $n$-數列，若且唯若存在正整數 $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$，滿足\n\n$$\nx_0 = a_1 a_2 a_3 \\cdots a_n,\n$$\n\n$$\nx_1 = b_1 a_2 a_3 \\cdots a_n,\n$$\n\n$$\nx_2 = b_1 b_2 a_3 \\cdots a_n,\n$$\n\n$\\vdots$\n\n$$\nx_n = b_1 b_2 b_3 \\cdots b_n.\n$$\n\n試求所有 $n$-數列的最小可能公差（以 $n$ 表示）。",
    "options": [],
    "answer": "See solution",
    "solution": "最小可能公差為 $n!$。\n\n注意到公差為\n\n$$\nD = (b_1 - a_1)a_2 a_3 \\cdots a_n = b_1(b_2 - a_2)a_3 a_4 \\cdots a_n = \\cdots = b_1 b_2 \\cdots b_{n-1}(b_n - a_n).\n$$\n\n又由於數列嚴格遞增，$D > 0$，從而 $b_i > a_i$ 對所有 $i$ 都成立。故上式等價於\n\n$$\n(b_i - a_i)a_{i+1} = b_i(b_{i+1} - a_{i+1}) \\text{ 對所有 } i \\text{ 皆成立。}\n$$\n\n此外，若 $k_i = \\gcd(a_i, b_i)$，則將 $a_i$ 與 $b_i$ 置換為 $a_i / k_i$ 與 $b_i / k_i$ 仍滿足題意，故不失一般性假設 $\\gcd(a_i, b_i) = 1$ 對所有 $i$ 皆成立。如此一來，上式便強迫我們要有 $b_i - a_i = b_{i+1} - a_{i+1}$ 且 $a_{i+1} = b_i$。如此一來，\n\n$$\na_1, a_2 = b_1, a_3 = b_2, \\dots, a_n = b_{n-1}, b_n\n$$\n\n必是一個嚴格遞增的等差數列。這意味著，若此數列的公差為 $d \\ge 1$，則\n\n$$\nD = d \\prod_{i=1}^{n-1} (a_1 + i d) \\ge 1 \\times \\prod_{i=1}^{n-1} (1 + i) = n!.\n$$\n\n等號在取 $a_i = i$ 時成立。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16495,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n > 1$ such that there exists a real-coefficient polynomial $P(x)$ of degree $n$ with leading coefficient $1$, and there exist distinct real numbers $r, s, t$ whose sum is $-2023$, such that $P(k) \\in \\{r, s, t\\}$ for all $k = 1, 2, 3, \\dots, 3n$.",
    "options": [],
    "answer": "See solution",
    "solution": "For $n = 2$, we have $P(k) \\in \\{r, s, t\\}$ for $k = 1, 2, 3, 4, 5, 6$. Since $\\deg P = 2$, there are no more than $2$ values of $k$ such that $P(k) = r$, and similarly for $s$ and $t$. Thus, each equation must have exactly $2$ solutions. Notice that all three equations have the same coefficients for $x^2$ and $x$, so the sum of the solutions for each must be equal and should be $7$. The $6$ solutions are paired as $(1, 6)$, $(2, 5)$, $(3, 4)$. We can assume:\n\n$$\n\\begin{cases}\nP(x) - r = (x - 1)(x - 6) \\\\\nP(x) - s = (x - 2)(x - 5) \\\\\nP(x) - t = (x - 3)(x - 4)\n\\end{cases}\n$$\n\nFrom this, it follows that:\n\n$$\n3P(x) - (r + s + t) = (x - 1)(x - 6) + (x - 2)(x - 5) + (x - 3)(x - 4)\n$$\n\nor $3P(x) + 2023 = 3x^2 - 21x + 28$, so $P(x) = x^2 - 7x - 665$ and $r, s, t$ are $-671, -675, -677$ respectively.\n\nNext, suppose there exists $P(x)$ of degree $n \\geq 3$ satisfying the given conditions. Without loss of generality, suppose $r < s < t$. As above, each equation $P(x) - r$, $P(x) - s$, $P(x) - t$ will have exactly $n$ distinct solutions from $\\{1, 2, \\dots, 3n\\}$. By Vieta's theorem, the sum of the solutions and the sum of their squares must be equal for each equation.\n\nConsider $f(x) = P(x) - r$ with $n$ distinct solutions. By the mean value theorem, $f'(x) = P'(x)$ must have $n-1$ distinct roots, denoted $c_1 < c_2 < \\dots < c_{n-1}$. The equation $P(x) = r$ has a unique root in each interval $(-\\infty, c_1), (c_1, c_2), \\dots, (c_{n-1}, +\\infty)$. The same applies for $P(x) = s$ and $P(x) = t$.\n\n![](images/Saudi_Arabia_booklet_2023_p24_data_976e17e16b.png)\n\n**Case 1.** If $n$ is odd, in the first interval $P(x)$ is increasing, so $P(x) = r$, $P(x) = s$, $P(x) = t$ will take the roots $1, 2, 3$. In the next interval, the function is decreasing, so they will take the roots $6, 5, 4$, and so on, ending with $3n - 2, 3n - 1, 3n$. The sum of the solutions for the three equations in the first $n-1$ intervals are equal, but in the last interval, each equation has a different solution, so the sums differ—a contradiction.\n\n**Case 2.** If $n$ is even, let $n = 2m$. The three equations will have solutions $1, 2, \\dots, 6m$. For $P(x) = r$, the solutions are $\\{1, 6, 7, 12, \\dots, 6m - 5, 6m\\}$. The sum of the squares of these numbers is:\n\n$$\n\\sum_{k=1}^{m} (6k-5)^2 + (6k)^2 = \\sum_{k=1}^{m} (72k^2 - 60k + 25) = m(24m^2 + 6m + 7).\n$$\n\nThe sum of squares of all $6m$ numbers is:\n\n$$\n\\frac{6m(6m+1)(12m+1)}{6} = m(6m+1)(12m+1)\n$$\n\nSo each equation must have a sum of squares equal to $\\frac{1}{3}$ of this value. Thus:\n\n$$\n3m(24m^2 + 6m + 7) = m(6m + 1)(12m + 1)\n$$\n\nor\n\n$$\n72m^2 + 18m + 21 = 72m^2 + 18m + 1,\n$$\n\nwhich is a contradiction. Thus, for $n \\geq 3$, no such polynomial $P(x)$ exists.\n\nTherefore, the only positive integer that satisfies the problem is $n = 2$. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16496,
    "subject": "Mathematics (Olympiad)",
    "question": "Find $(x+y)^2$, if $\\frac{2}{x} - \\frac{2}{y} = 1$ and $y-x=1$.",
    "options": [],
    "answer": "See solution",
    "solution": "From $\\frac{2}{x} - \\frac{2}{y} = 2\\frac{y-x}{xy} = 2\\frac{1}{xy}$, we have that $xy=2$.\n\nTherefore,\n\n$$\n(x+y)^2 = x^2 + 2xy + y^2 = x^2 - 2xy + y^2 + 4xy = (x-y)^2 + 4xy = 1^2 + 4 \\cdot 2 = 9\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16497,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given a triangle $ABC$ and a point $P$ in its interior. The lines through $P$ and parallel to the sides of the triangle divide the triangle into three parallelograms and three triangles.\n\n**a)** If $P$ is the incenter of $ABC$, show that the perimeter of each of the three small triangles is equal to the length of the adjacent side.\n\n**b)** For a given triangle $ABC$, determine all inner points $P$ such that the perimeter of each of the three small triangles equals the length of the adjacent side.\n\n**c)** For which inner point does the sum of the areas of the three small triangles attain a minimum?",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** Let $I$ be the incenter of $ABC$. Let $X$ be the intersection of $AB$ with the line through $I$ parallel to $CA$, and $Y$ be the intersection of $CA$ with the line through $I$ parallel to $AB$. $AXIY$ is a parallelogram, and since $I$ is the incenter of $ABC$, we have $\\angle IAX = \\angle IAY$. Since $\\angle IAX = \\angle AIX$ must also hold in the parallelogram $AXIY$, we see that $\\angle IAX = \\angle AIX$ holds. The triangle $AIX$ is therefore isosceles with $XA = XI$. If $Z$ denotes the intersection of $AB$ with the line through $I$ parallel to $BC$, we similarly obtain $ZB = ZI$, and it therefore follows that\n\n$$\nXI + XZ + ZI = XA + XZ + ZB = AB\n$$\n\nas claimed.\n\n**b)** Assume that a point $P \\neq I$ with this property exists. Such a point must lie between one of the sides of the triangle and the line parallel to this side through $I$. Without loss of generality, assume it lies between $AB$ and $YI$. The triangle $PX'Z'$ is similar to $IXZ$, and since $P$ is closer to $AB$ than $I$ is, the perimeter of $PX'Z'$ is smaller than that of $IXZ$, which equals the length of $AB$. $P$ therefore does not fulfill the required condition. Thus, $I$ is the only point with this property.\n\n**c)** The point $P$ determines three triangles $A_1B_1C_1$, $A_2B_2C_2$, and $A_3B_3C_3$ (with $P = C_1 = A_2 = B_3$) as shown. The sum of the areas of the triangles is given by\n\n$$\n\\frac{1}{2}a_1b_1 + \\frac{1}{2}a_2b_2 + \\frac{1}{2}a_3b_3.\n$$\n\nSince $b_1 + b_2 + b_3 = c = |AB|$ and $a_1 + a_2 + a_3 = h_c$, and all three triangles are similar to $ABC$, we have\n\n$$\na_1 : a_2 : a_3 = b_1 : b_2 : b_3 = t_1 : t_2 : t_3 \\text{ with } t_1 + t_2 + t_3 = 1.\n$$\n\nIt follows that\n\n$$\n\\begin{aligned}\n\\frac{1}{2}a_1b_1 + \\frac{1}{2}a_2b_2 + \\frac{1}{2}a_3b_3 &= \\frac{1}{2}h_c \\cdot c \\cdot (t_1^2 + t_2^2 + t_3^2) \\\\\n&\\geq h_c \\cdot c \\cdot \\left(\\frac{t_1 + t_2 + t_3}{2}\\right)^2 \\\\\n&= \\frac{h_c \\cdot c}{4},\n\\end{aligned}\n$$\n\nwith equality iff $t_1 = t_2 = t_3 = \\frac{1}{3}$. The sum of the areas is therefore minimized if the distance of $P$ from each of the sides is equal to one third of each altitude. This is the case for the centroid of $ABC$, so the centroid is the required point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16498,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $H$ be the intersection point of the altitudes $AD$ and $BE$ of an acute triangle $ABC$. The circumcircle of triangle $ABC$ intersects the circle with diameter $CH$ at the point $K$ other than $C$. Prove that\n\n$$\n\\frac{DK}{KE} = \\frac{DH}{HE}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/MNG_ABooklet_2015_p13_data_a07c211151.png)\n\nSince $\\angle BDH = \\angle AEH$ and $\\angle BHD = \\angle AHE$, we have $\\triangle BHD \\sim \\triangle AHE$. Therefore,\n\n$$\n\\frac{DH}{HE} = \\frac{BD}{AE}. \\qquad (1)\n$$\n\nAlso, it is easy to observe that $\\angle CDK = \\angle CEK$, and moreover $\\angle BOK = \\angle AEK$, $\\angle KBD = \\angle KAE$. Hence, $\\triangle BDK \\sim \\triangle AEK$. Therefore,\n\n$$\n\\frac{BD}{AE} = \\frac{DK}{KE}. \\qquad (2)\n$$\n\nFrom (1) and (2), we get\n\n$$\n\\frac{DH}{HE} = \\frac{BD}{AE} = \\frac{DK}{KE}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16499,
    "subject": "Mathematics (Olympiad)",
    "question": "Define a function $f: \\mathbb{N} \\to \\mathbb{N}$ by $f(1) = 1$, $f(n+1) = f(n) + 2^{f(n)}$ for every positive integer $n$. Prove that $f(1), f(2), \\dots, f(3^{2013})$ leave distinct remainders when divided by $3^{2013}$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** We prove the following stronger statement: For any $k \\ge 0$ and $a \\ge 1$, the values $f(a), f(a+1), \\dots, f(a+3^k - 1)$ are all distinct modulo $3^k$; that is, these numbers form a complete set of residues modulo $3^k$. We will induct on $k$, the case $k=0$ being trivial.\n\nAssume the statement true for a given $k$; we will prove it for $k+1$. For any $a \\ge 1$, we have\n\n$$\n\\begin{aligned}\n& f(a + 3^k) - f(a) \\\\\n&= [f(a+1) - f(a)] + [f(a+2) - f(a+1)] + \\dots + [f(a + 3^k) - f(a + 3^k - 1)] \\\\\n&= 2^{f(a)} + 2^{f(a+1)} + \\dots + 2^{f(a+3^k-1)}.\n\\end{aligned}\n$$\n\nNote that by Euler's theorem, to know $2^x$ modulo $3^{k+1}$ ($x \\ge 1$), it suffices to know $x$ modulo $\\varphi(3^{k+1}) = 2 \\cdot 3^k$. Now $f(x)$ is always odd (this follows from the definition), while the inductive hypothesis tells us that $f(a), \\dots, f(a+3^k-1)$ are distinct mod $3^k$. Hence sets\n\n$$\n\\{f(a), \\dots, f(a + 3^k - 1)\\} \\quad \\text{and} \\quad \\{1, 3, 5, \\dots, 2 \\cdot 3^k - 1\\}\n$$\n\nare congruent to each other modulo $2 \\cdot 3^k$. Therefore, modulo $3^{k+1}$, we have\n\n$$\nf(a + 3^k) - f(a) \\equiv 2^1 + 2^3 + 2^5 + \\dots + 2^{2 \\cdot 3^k - 1} \\equiv 2(1 + 4 + 4^2 + \\dots + 4^{3^k - 1}) \\equiv \\frac{2(4^{3^k} - 1)}{3}.\n$$\n\nBy the binomial theorem, we have\n\n$$\n4^{3^k} - 1 = (1 + 3)^{3^k} - 1 = 3 \\cdot \\binom{3^k}{1} + \\dots,\n$$\n\nwhere each of the remaining summands are divisible by $3^{k+2}$; hence\n\n$$\nf(a + 3^k) - f(a) \\equiv 2 \\cdot 3^k \\pmod{3^k}\n$$\n\nfor all $a$.\n\nReturning to the sequence\n\n$$\nf(a), f(a+1), \\dots, f(a + 3^{k+1} - 1),\n$$\n\nwe see that, since $f(b) \\equiv f(b + 3^k)$ modulo $3^k$ (a consequence of the inductive hypothesis), the terms congruent to one another modulo $3^k$ come in triples $(f(b), f(b + 3^k), f(b + 2 \\cdot 3^k))$. By the previous result, these terms, modulo $3^{k+1}$, are congruent to $(f(b), f(b) + 2 \\cdot 3^k, f(b) + 3^k)$, which are pairwise distinct, completing our induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16500,
    "subject": "Mathematics (Olympiad)",
    "question": "Let's examine all possible sequences $a_1, a_2, \\ldots, a_{2008}$ of non-negative integers such that $a_1 \\leq a_2 \\leq \\ldots \\leq a_{2008}$ and $a_k \\leq k-1$ for all $k$ from $1$ to $2008$.\n\nProve that the number of such sequences exceeds:\n\na) $2^{2007}$;\n\nb) $2^{2008}$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) To construct $2^{2007}$ different sequences satisfying the conditions, set $a_1 = 0$ for all sequences. For each subsequent term, let $a_{k}$ be either equal to $a_{k-1}$ or $a_{k-1} + 1$. This produces $2^{2007}$ distinct sequences, each with $a_k \\leq k-1$. To show the total number exceeds $2^{2007}$, consider a sequence not generated by this method, such as $(0, 0, 2, 2, \\ldots, 2)$, which also satisfies the conditions.\n\nb) Define a *fine* sequence as a non-decreasing sequence of non-negative integers $a_1 \\leq a_2 \\leq \\ldots \\leq a_n$ with $a_k \\leq k-1$ for all $k$. Any fine sequence of length $n+1$ can be obtained from a fine sequence of length $n$ by appending $a_{n+1}$ such that $a_n \\leq a_{n+1} \\leq n$. Removing the last term from a fine sequence of length $n+1$ yields a fine sequence of length $n$.\n\nLet $x_n$ be the number of fine sequences of length $n$. For each fine sequence of length $n$, there are at least two choices for $a_{n+1}$: $a_n$ or $a_n + 1$, so $x_{n+m} \\geq 2^m x_n$.\n\nFor small $n$:\n- $x_1 = 1$ (only $(0)$)\n- $x_2 = 2$ ($(0,0)$ and $(0,1)$)\n- $x_3 = 5$ (by considering extensions of $(0,0)$ and $(0,1)$)\n- $x_4 = 14$\n\nThus, $x_{2008} \\geq 2^{2004} x_4 = 14 \\cdot 2^{2004} > 2^{2007}$.\n\nFor $n=5$, by considering possible extensions:\n- Sequences ending in $0$: add $0,1,2,3,4$ (5 options)\n- Sequences ending in $1$: add $1,2,3,4$ (4 options, 3 sequences)\n- Sequences ending in $2$: add $2,3,4$ (3 options, 5 sequences)\n- Sequences ending in $3$: add $3,4$ (2 options, 5 sequences)\n\nTotal: $5 \\cdot 1 + 3 \\cdot 4 + 5 \\cdot 3 + 5 \\cdot 2 = 42$, so $x_5 = 42$ and $x_{2008} \\geq 42 \\cdot 2^{2003} > 2^{2008}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16501,
    "subject": "Mathematics (Olympiad)",
    "question": "Rewrite the system of equations:\n\n$$\n\\begin{aligned}\n(x + 2y)(x^2 - 2xy + 4y^2) &= x + 2y, \\\\\n2xy(x + 2y) &= x + 2y.\n\\end{aligned}\n$$\n\nFind all real solutions $(x, y)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Obviously, every pair of numbers $x$ and $y$ that satisfies $x + 2y = 0$ solves the equations.\n\nNow, assume $x + 2y \\neq 0$. We can then divide by $x + 2y$ and get $x^2 - 2xy + 4y^2 = 1$ and $2xy = 1$, so\n\n$$\n(x + 2y)^2 = x^2 + 4xy + 4y^2 = (x^2 - 2xy + 4y^2) + 6xy = 4.\n$$\n\nWe consider two cases: $x + 2y = 2$ or $x + 2y = -2$.\n\n- If $x + 2y = 2$, then $x = 2 - 2y$. Substituting into $2xy = 1$:\n  $$\n  1 = 2xy = 2(2 - 2y)y = 4y - 4y^2\n  $$\n  $$\n  0 = 4y^2 - 4y + 1 = (2y - 1)^2\n  $$\n  So $y = \\frac{1}{2}$ and $x = 1$.\n\n- If $x + 2y = -2$, then $x = -2 - 2y$. Substituting into $2xy = 1$:\n  $$\n  1 = 2xy = 2(-2 - 2y)y = -4y - 4y^2\n  $$\n  $$\n  0 = 4y^2 + 4y + 1 = (2y + 1)^2\n  $$\n  So $y = -\\frac{1}{2}$ and $x = -1$.\n\n**Final answer:**\n\nAll pairs of real numbers $x$ and $y$ such that $x + 2y = 0$, as well as the pairs $x = 1$, $y = \\frac{1}{2}$ and $x = -1$, $y = -\\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16502,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{Q}$ be the set of rational numbers. Find all functions $f: \\mathbb{Q} \\to \\mathbb{Q}$ such that for all rational numbers $x, y$,\n\n$$\nf(f(x) + x f(y)) = x + f(x)y.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We show that $f(x) = x$ is the only solution. It is easy to check that it works.\n\nLet $y = 0$; then $f(f(x)) = x$, so $f$ is surjective.\n\nSuppose $f(a) = f(b)$. For any $c$ with $f(c) \\neq 0$,\n$$\na = \\frac{f(f(c) + c f(a)) - c}{f(c)} = \\frac{f(f(c) + c f(b)) - c}{f(c)} = b,\n$$\nso $f$ is injective.\n\nLet $z$ be such that $f(z) = 0$. Then $x = z$ gives $f(z f(y)) = z$. By surjectivity, choose $y$ with $f(y) = 1$ or $f(y) = 0$, so $f(z) = f(0) = z$. By injectivity, $z = 0$, so $f(0) = 0$.\n\nLet $x = y = -1$; then $f(-1) = -1$. Let $y = -1$; then $f(f(x) - x) = x - f(x)$. Let $d = f(1) - 1$; with $x = 1$ this gives $f(d) = -d$. Let $x = d, y = 1$; then $f(d f(1) - d) = f(d^2) = 0$. By injectivity, $d = 0$, so $f(1) = 1$.\n\nLet $y = 0$; then $f(f(x)) = x$. Let $x = 1$ and $y$ replaced by $f(y)$; then $f(1 + y) = 1 + f(y)$. Thus $f(n) = n$ for all integers $n$.\n\nFor integers $m, n$ with $n \\neq 0$, let $x = n, y = \\frac{m}{n}$; then $f(n + n f(\\frac{m}{n})) = n + m = f(n + m)$. By injectivity, $n f(\\frac{m}{n}) = m$, so $f(\\frac{m}{n}) = \\frac{m}{n}$. Since $\\frac{m}{n}$ is arbitrary, $f(x) = x$ for all $x \\in \\mathbb{Q}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16503,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real $x$ and $y$ such that\n\n$$\n\\begin{cases}\nx + y^2 = xy + 1 \\\\\nxy = 4 + y\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The first equation implies $x + y^2 = xy + 1$, so $x - xy = 1 - y^2$, or $x(1 - y) = 1 - y^2$. This can be factored as $(1 - y)(x - 1 - y) = 0$.\n\n- If $y = 1$, then the second equation gives $x \\cdot 1 = 4 + 1 \\implies x = 5$.\n- If $y \\ne 1$, then $x - 1 - y = 0 \\implies x = 1 + y$. Substitute into the second equation:\n  $$\n  (1 + y)y = 4 + y \\implies y^2 + y = 4 + y \\implies y^2 = 4 \\implies y = 2 \\text{ or } y = -2.\n  $$\n  - For $y = 2$, $x = 1 + 2 = 3$.\n  - For $y = -2$, $x = 1 + (-2) = -1$.\n\nThus, the solutions are:\n- $x = 5$, $y = 1$\n- $x = 3$, $y = 2$\n- $x = -1$, $y = -2$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16504,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_0A_1A_2$ be a non-equilateral triangle. The incircle of the triangle $A_0A_1A_2$ touches the side $A_iA_{i+1}$ at the point $T_{i+2}$ (indices are reduced modulo $3$). Let $X_i$ be the perpendicular foot dropped from the point $T_i$ onto the line $T_{i+1}T_{i+2}$. Show that the lines $A_iX_i$ are concurrent at a point situated on the Euler line of the triangle $T_0T_1T_2$.",
    "options": [],
    "answer": "See solution",
    "solution": "The lines $A_iA_{i+1}$ and $X_iX_{i+1}$ are parallel, as they are both antiparallel to the line $T_iT_{i+1}$. Hence, the triangles $A_0A_1A_2$ and $X_0X_1X_2$ are homologous: the three lines $A_iX_i$ are concurrent at the homology centre, which lies on the homology line. This line passes through the incentres of the two triangles: one is the circumcentre of the triangle $T_0T_1T_2$ and the other is the orthocentre. The conclusion follows.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 16505,
    "subject": "Mathematics (Olympiad)",
    "question": "Given triangle $ABC$ with $M$ as the midpoint of $AB$. Given $\\angle ABC = 30^\\circ$ and $\\angle BCM = 105^\\circ$. Prove that $CM \\cdot AC = BM \\cdot BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $AA_1$ be the height from point $A$ in $\\triangle ABC$. Note that it lies on the extension of $BC$, since the triangle is obtuse. The triangle $\\triangle AA_1B$ is right-angled with angle $30^\\circ$. Therefore, $AA_1 = AM = A_1M = BM = x$. Consequently, we find $\\angle CMB = 45^\\circ$, $\\angle A_1MA = 60^\\circ$, $\\angle A_1MC = 75^\\circ$, $\\angle MCA_1 = 75^\\circ$. So $\\triangle MCA_1$ is isosceles. Then $CA_1 = x$ and $\\triangle AA_1C$ is isosceles and right-angled, so $\\angle ACA_1 = 45^\\circ$ and therefore $\\angle ACM = 30^\\circ$.\n\nWe will calculate the area of $\\triangle ACM$ in two different ways.\n\n$$S_{ACM} = \\frac{AC \\cdot CM}{4}$$\nbecause $\\angle ACM = 30^\\circ$, i.e., the height to $AC$ is equal to $\\frac{MC}{2}$. But also $M$ is the midpoint of $AB$. So\n$$S_{ACM} = \\frac{S_{ABC}}{2} = \\frac{BC \\cdot AA_1}{4} = \\frac{BC \\cdot BM}{4}$$\nand we get what we are looking for. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16506,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that there is an absolute constant $c < 1$ with the following property: whenever $\\mathcal{P}$ is a polygon with area $1$ in the plane, one can translate it by a distance of $\\frac{1}{100}$ in some direction to obtain a polygon $\\mathcal{Q}$, for which the intersection of the interiors of $\\mathcal{P}$ and $\\mathcal{Q}$ has total area at most $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "The following solution is due to Brian Lawrence. We will prove the result with the generality of any measurable set $\\mathcal{P}$ (rather than a polygon). For a vector $v$ in the plane, write $\\mathcal{P} + v$ for the translate of $\\mathcal{P}$ by $v$.\n\nSuppose $\\mathcal{P}$ is a polygon of area $1$, and $\\varepsilon > 0$ is a constant, such that for any translate $Q = \\mathcal{P} + v$, where $v$ has length exactly $\\frac{1}{100}$, the intersection of $\\mathcal{P}$ and $Q$ has area at least $1 - \\varepsilon$. The problem asks us to prove a lower bound on $\\varepsilon$.\n\n**Lemma**\n\nFix a sequence of $n$ vectors $v_1, v_2, \\dots, v_n$, each of length $\\frac{1}{100}$. A grasshopper starts at a random point $x$ of $\\mathcal{P}$, and makes $n$ jumps to $x + v_1 + \\dots + v_n$. Then it remains in $\\mathcal{P}$ with probability at least $1 - n\\varepsilon$.\n\n*Proof.* In order for the grasshopper to leave $\\mathcal{P}$ at step $i$, the grasshopper's position before step $i$ must be inside the difference set $\\mathcal{P} \\setminus (\\mathcal{P} - v_i)$. Since this difference set has area at most $\\varepsilon$, the probability the grasshopper leaves $\\mathcal{P}$ at step $i$ is at most $\\varepsilon$. Summing over the $n$ steps, the probability that the grasshopper ever manages to leave $\\mathcal{P}$ is at most $n\\varepsilon$. $\\square$\n\n**Corollary**\n\nFix a vector $w$ of length at most $8$. A grasshopper starts at a random point $x$ of $\\mathcal{P}$, and jumps to $x + w$. Then it remains in $\\mathcal{P}$ with probability at least $1 - 800\\varepsilon$.\n\n*Proof.* Apply the previous lemma with $800$ jumps. Any vector $w$ of length at most $8$ can be written as $w = v_1 + v_2 + \\dots + v_{800}$, where each $v_i$ has length exactly $\\frac{1}{100}$. $\\square$\n\nNow consider the process where we select a random starting point $x \\in \\mathcal{P}$ for our grasshopper, and a random vector $w$ of length at most $8$ (sampled uniformly from the closed disk of radius $8$). Let $q$ denote the probability of staying inside $\\mathcal{P}$; we will bound $q$ from above and below.\n\n- On the one hand, suppose we pick $w$ first. By the previous corollary, $q \\ge 1 - 800\\varepsilon$ (irrespective of the chosen $w$).\n- On the other hand, suppose we pick $x$ first. Then the possible landing points $x + w$ are uniformly distributed over a closed disk of radius $8$, which has area $64\\pi$. The probability of landing in $\\mathcal{P}$ is certainly at most $\\frac{1}{64\\pi}$.\n\nConsequently, we deduce\n\n$$\n1 - 800\\varepsilon \\le q \\le \\frac{1}{64\\pi} \\implies \\varepsilon > \\frac{1 - \\frac{1}{64\\pi}}{800} > 0.001\n$$\n\nas desired.\n\n**Remark.** The choice of $800$ jumps is only for concreteness; any constant $n$ for which $\\pi(n/100)^2 > 1$ works. I think $n = 98$ gives the best bound following this approach.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16507,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the sequence $a_1, a_2, a_3, \\dots$ defined by $a_1 = 1$ and\n\n$$\na_{m+1} = \\frac{1a_1 + 2a_2 + 3a_3 + \\dots + ma_m}{a_m} \\quad \\text{for } m \\ge 1.\n$$\n\nDetermine the largest integer $n$ such that $a_n < 1,000,000$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we note that all terms of the sequence are positive rational numbers. We rewrite the defining equation for the sequence in both its original form and with the value of $m$ shifted by 1:\n\n$$\n\\begin{aligned}\na_m a_{m+1} &= 1a_1 + 2a_2 + 3a_3 + \\dots + ma_m \\\\\na_{m+1} a_{m+2} &= 1a_1 + 2a_2 + 3a_3 + \\dots + ma_m + (m+1)a_{m+1}\n\\end{aligned}\n$$\n\nSubtracting the first equation from the second yields\n\n$$\na_{m+1}a_{m+2} - a_m a_{m+1} = (m+1)a_{m+1} \\implies a_{m+2} - a_m = m+1,\n$$\n\nfor all $m \\ge 1$.\n\nSo for $m = 2k + 1$ (an odd positive integer),\n\n$$\n\\begin{aligned}\na_{2k+1} - a_1 &= (a_{2k+1} - a_{2k-1}) + (a_{2k-1} - a_{2k-3}) + \\dots + (a_3 - a_1) \\\\\n&= 2k + (2k-2) + \\dots + 2 \\\\\n&= 2[k + (k-1) + \\dots + 1] \\\\\n&= k(k+1).\n\\end{aligned}\n$$\n\nSimilarly, for $m = 2k$ (an even positive integer),\n\n$$\n\\begin{aligned}\na_{2k} - a_2 &= (a_{2k} - a_{2k-2}) + (a_{2k-2} - a_{2k-4}) + \\dots + (a_4 - a_2) \\\\\n&= (2k-1) + (2k-3) + \\dots + 3 \\\\\n&= k^2 - 1.\n\\end{aligned}\n$$\n\nUsing $a_1 = 1$ and $a_2 = 1$, we obtain the formula\n\n$$\na_m = \\begin{cases} \\frac{m^2}{4}, & \\text{if } m \\text{ is even} \\\\ \\frac{m^2+3}{4}, & \\text{if } m \\text{ is odd} \\end{cases}\n$$\n\nIf $m$ is odd, then\n\n$$\na_{m+1} - a_m = \\frac{(m+1)^2}{4} - \\frac{m^2+3}{4} = \\frac{2m-2}{4},\n$$\n\nand if $m$ is even, then\n\n$$\na_{m+1} - a_m = \\frac{(m+1)^2 + 3}{4} - \\frac{m^2}{4} = \\frac{2m+4}{4}.\n$$\n\nIn particular, it follows that $a_2 < a_3 < a_4 < \\dots$. Since $a_{2000} = 1,000,000$, the largest integer $n$ such that $a_n < 1,000,000$ is $1999$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16508,
    "subject": "Mathematics (Olympiad)",
    "question": "Point $D \\notin \\{B, C\\}$ is chosen on segment $BC$ and point $A$ is chosen on the plane such that it does not lie on the line $BC$. Points $P$ and $Q$ are chosen such that $ABPD$ and $ACQD$ are parallelograms (with the vertices in this order). The circumcircles of triangles $BDP$ and $CDQ$ intersect in $R \\neq D$. Prove that the points $P$, $Q$, $R$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "A reflection across the centre of the segment *BD* sends the triangle *BDP* into the triangle *DBA*. Thus the circumcircle of *BDP* goes to the circumcircle of *DBA*. Analogously, a reflection across the centre of *CD* sends the circumcircle of *CDQ* into the circumcircle of *DCA*.\n\nThe reflection of a circle across the midpoint of one of its chords is equivalent to a reflection across the line corresponding to the chord. Thus, the reflections of the circumcircles of *BDP* and *CDQ* across *BC* are the circumcircles of *ABD* and *ACD*, respectively. The first two circles intersect in points *A* and *D*, whereas the latter two circles intersect in *R* and *D*. Thus, *A* and *R* are reflections of each other across *BC*.\n\nTherefore, the points *P*, *Q*, *R* are all located on the opposite side of *BC* compared to *A*, but at the same distance from *BC* as *A*. Thus, *P*, *Q*, *R* lie on one line parallel to *BC*.\n\n![](images/EST_ABooklet_2024_p15_data_4576e46606.png)\n\nFig. 1",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16509,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider $n$ points, each pair connected by a segment colored either red or blue. You may \"switch\" a point, which toggles the color (red/blue) of all segments incident to it. Prove:\n\n1. It is possible, by a sequence of switches, to make all segments red.\n2. What is the minimal number of switches needed in the worst case?",
    "options": [],
    "answer": "See solution",
    "solution": "Consider three of the $n$ points. The parity of the number of blue segments in the triangle formed by these points does not change when switching points. Since it is possible to make all three segments red, the number of blue segments in each triangle must be even.\n\nLet $P$ be one of the $n$ points. Let $A$ be the set of points connected to $P$ by red segments and $B$ the set connected by blue segments. For $A_1, A_2 \\in A$, both $PA_1$ and $PA_2$ are red, so $A_1A_2$ is red. For $B_1, B_2 \\in B$, both $PB_1$ and $PB_2$ are blue, so $B_1B_2$ is red. For $A \\in A$ and $B \\in B$, $PA$ is red and $PB$ is blue, so $AB$ is blue. Including $P$ in $A$, this shows that segments within the same set are red, and those between sets are blue.\n\nSwitching all points in set $A$ will make all segments red: segments within $A$ change color twice (returning to red), segments between $A$ and $B$ change once (blue to red), and segments within $B$ do not change. This proves part 1.\n\nFor part 2, note that each point needs to be switched at most once. Let $|A| = k$ and $|B| = n-k$. Switching $a$ points from $A$ and $b$ from $B$ changes at most $a(n-k) + bk$ blue segments. Assume $k \\le n-k$ (i.e., $k \\le \\lfloor n/2 \\rfloor$). Then $k(n-k) \\le a(n-k) + bk \\le (a+b)(n-k)$, so $k \\le a+b$. Thus, the minimal number of switches is at most $k$, and in the worst case, $\\lfloor n/2 \\rfloor$. This number is needed if $|A| = \\lfloor n/2 \\rfloor$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16510,
    "subject": "Mathematics (Olympiad)",
    "question": "對於任一質數 $p$ 以及任一整數 $n$，將 $n$ 除以 $p$ 的餘數記為 $d_p(n) \\in \\{0, 1, \\dots, p-1\\}$。若正整數數列 $(a_0, a_1, a_2, \\dots)$ 符合 $a_0$ 與 $p$ 互質且當 $n \\ge 0$ 時有 $a_{n+1} = a_n + d_p(a_n)$，則我們稱這個正整數數列為一個 $p$-數列。\n\n(a) 是否存在無窮多個質數 $p$，可以找到 $p$-數列 $(a_0, a_1, a_2, \\dots)$ 和 $(b_0, b_1, b_2, \\dots)$，使得對於無窮多個 $n$ 有 $a_n > b_n$ 且對於無窮多個 $n$ 有 $b_n > a_n$？\n\n(b) 是否存在無窮多個質數 $p$，可以找到 $p$-數列 $(a_0, a_1, a_2, \\dots)$ 和 $(b_0, b_1, b_2, \\dots)$，使得 $a_0 < b_0$ 但對於所有 $n \\ge 1$ 有 $a_n > b_n$？",
    "options": [],
    "answer": "See solution",
    "solution": "答案：兩部分皆為「是」。\n\n解答：\n\n固定一個奇質數 $p$，令 $T$ 為最小正整數使得 $p \\mid 2^T - 1$，即 $T$ 是 $2$ 在模 $p$ 下的乘法階。\n\n考慮任意 $p$-數列 $(x_n) = (x_0, x_1, x_2, \\dots)$。顯然有 $x_{n+1} \\equiv 2x_n \\pmod p$，因此 $x_n \\equiv 2^n x_0 \\pmod p$。這導致 $x_{n+T} \\equiv x_n \\pmod p$，所以 $d_p(x_{n+T}) = d_p(x_n)$ 對所有 $n \\ge 0$ 成立。由此可知，$d_p(x_n) + d_p(x_{n+1}) + \\dots + d_p(x_{n+T-1})$ 不依賴於 $n$，僅由 $x_0$ 和 $p$ 決定，記為 $S_p(x_0)$，並將 $S_p(\\cdot)$ 擴展至所有整數。因此，對所有正整數 $n$ 和 $k$，有 $x_{n+kT} = x_n + kS_p(x_0)$。\n\n在兩部分中，使用記號：\n\n$$\nS_p^+ = S_p(1) = \\sum_{i=0}^{T-1} d_p(2^i), \\quad S_p^- = S_p(-1) = \\sum_{i=0}^{T-1} d_p(p - 2^i)\n$$\n\n(a) 令 $q > 3$ 為質數，$p$ 為 $2^q + 1$ 的一個大於 $3$ 的質因數。我們將證明此 $p$ 適合 (a)。注意 $9 \\nmid 2^q + 1$，故存在此類 $p$。且對任意兩個奇質數 $q < r$，有 $\\gcd(2^q + 1, 2^r + 1) = 2^{\\gcd(q,r)} + 1 = 3$，因此存在無窮多個此類質數 $p$。\n\n對所選 $p$，有 $T = 2q$。由 $2^q \\equiv -1 \\pmod p$，可得 $S_p^+ = S_p^-$. 現考慮 $p$-數列 $(a_n)$ 和 $(b_n)$，其中 $a_0 = p+1$，$b_0 = p-1$。可證明這兩數列滿足條件：$a_0 > b_0$ 且 $a_1 = p+2 < b_1 = 2p-2$。因此：\n\n$$\na_{k \\cdot 2q} = a_0 + kS_p^+ > b_0 + kS_p^+ = b_{k \\cdot 2q} \\quad \\text{且} \\quad a_{k \\cdot 2q+1} = a_1 + kS_p^+ < b_1 + kS_p^+ = b_{k \\cdot 2q+1}\n$$\n\n對所有 $k = 0, 1, \\dots$，如所需。\n\n(b) 令 $q$ 為奇質數，$p$ 為 $2^q - 1$ 的一個質因數，則 $T = q$。這類 $p$ 也有無窮多個。注意 $d_p(x) + d_p(p-x) = p$ 對所有 $p \\nmid x$ 成立，故 $S_p^+ + S_p^- = pq$ 為奇數，得 $S_p^+ \\neq S_p^-$. 假設 $(x_n)$ 和 $(y_n)$ 為兩個 $p$-數列，且 $S_p(x_0) > S_p(y_0)$ 但 $x_0 < y_0$。則：\n\n$$\nx_{M_{q+r}} - y_{M_{q+r}} = (x_r - y_r) + M(S_p(x_0) - S_p(y_0)) \\geq (x_r - y_r) + M\n$$\n\n對所有非負整數 $M$ 及 $r = 0, 1, \\dots, q-1$。因此，$x_n > y_n$ 對所有 $n \\ge q + q \\cdot \\max\\{y_r - x_r : r = 0, 1, \\dots, q-1\\}$ 成立。又因 $x_0 < y_0$，存在某 $n_0$ 使 $x_{n_0} < y_{n_0}$。此時令 $a_n = x_{n-n_0}$，$b_n = y_{n-n_0}$，即可滿足條件（且 $x_n \\neq y_n$ 對所有 $n \\ge 0$，否則 $S_p(x_0) = S_p(y_0)$）。\n\n剩下只需構造滿足條件的 $p$-數列 $(x_n)$ 和 $(y_n)$。由 $S_p^+ \\neq S_p^-$，若 $S_p^+ > S_p^-$，則取 $x_0 = 1$，$y_0 = p-1$；否則取 $x_0 = p-1$，$y_0 = p+1$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16511,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove the inequality\n$$\n\\frac{a}{\\sqrt{b^2+1}} + \\frac{b}{\\sqrt{a^2+1}} \\ge \\frac{a+b}{\\sqrt{ab+1}}\n$$\nfor all real numbers $a, b \\ge 0$. Determine all cases of equality.",
    "options": [],
    "answer": "See solution",
    "solution": "This easily follows by a comparison of both sides term by term (because our assumption $a > b$ implies that $\\sqrt{a^2+1} > \\sqrt{b^2+1}$). This completes the proof of the given inequality. As we have shown, the only cases of equality are $a = 0$, $b = 0$, and $a = b$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16512,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $G$ be a complete graph on $n$ vertices. The edges of $G$ are coloured with $k$ colours. Show that:\n\n(a) For $n = 6$ and $k = 5$, it is possible to colour the edges so that no monochromatic closed circuit exists.\n\n![](images/1.2_1997-2023_IMO_HK_TST_Solutions_p13_data_2815386d1d.png)\n\n(b) If $n \\geq 2k + 1$, then in any colouring, there must exist a monochromatic closed circuit in $G$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) We can colour the edges as shown in the diagram. For each colour, the subgraph formed by edges of that colour connects all 6 vertices but contains only 5 edges, so it forms a tree and hence has no closed circuit.\n\n(b) There are $\\binom{n}{2} \\geq \\frac{n(n-1)}{2}$ edges in total. By the pigeonhole principle, at least $\\frac{n(n-1)}{2k}$ edges are of the same colour. Consider the subgraph formed by these edges; it has at most $n$ vertices. Since\n\n$$\n\\frac{n(n-1)}{2k} \\geq \\frac{n(2k+1-1)}{2k} = n,\n$$\n\nthe number of edges in this subgraph is at least the number of vertices. Therefore, it must contain a closed circuit, giving a monochromatic closed circuit in $G$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16513,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle, and let $X$, $Y$, and $Z$ be collinear points such that $AY = AZ$, $BZ = BX$, and $CX = CY$. Points $X'$, $Y'$, and $Z'$ are the reflections of $X$, $Y$, and $Z$ over $BC$, $CA$, and $AB$, respectively. Prove that if $X'Y'Z'$ is a nondegenerate triangle, then its circumcenter lies on the circumcircle of $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S$ denote the circumcenter of $\\triangle X'Y'Z'$. Observe that $AY = AZ = AY' = AZ'$, so $YZY'Z'$ is cyclic and $AS \\perp Y'Z'$. Similarly, $BS \\perp Z'X'$ and $CS \\perp X'Y'$.  \n\nThe rest is angle chasing. Let $\\angle \\ell$ denote the angle between line $\\ell$ and a fixed line. Then, we have\n\n$$\n\\begin{aligned}\n\\angle AS &= 90^\\circ + \\angle Y'Z' = 90^\\circ + \\angle YY' + \\angle ZZ' - \\angle YZ \\\\\n&= 90^\\circ + \\angle CA + \\angle AB - \\angle YZ.\n\\end{aligned}\n$$\n\nAnalogously, we get\n\n$$\n\\angle BS = 90^\\circ + \\angle AB + \\angle BC - \\angle XZ,\n$$\n\nso subtracting these gives $\\angle ASB = \\angle ACB$, as desired.\n\n![](images/TST2025Solutions_p8_data_86275f4494.png)\n\n**Remark.** There are some other angle chasing solutions that use the fact that $XX'$, $YY'$, and $ZZ'$ meet at a point on $(X'Y'Z')$. This one is featured as it does not require any additional points.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16514,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $k$ such that there are finitely many triangles on the Cartesian coordinate plane satisfying:\n\n1. The center of mass of each triangle is an integral point.\n2. The intersection of any two triangles is either the empty set, a common vertex, or an edge joining two common vertices.\n3. The union of these triangles is a square with side length $k$.\n\n(The vertices of the square are not required to be integral points, and the edges are not required to be parallel to the coordinate axes.)",
    "options": [],
    "answer": "See solution",
    "solution": "The desired positive integers $k$ are those divisible by $3$.\n\nFirst, assume that $k = 3t$ for $t \\in \\mathbb{N}$. Consider the square with vertices $(0,0)$, $(3t, 3t)$, $(3t, 0)$, and $(0, 3t)$. Divide it into $t^2$ smaller squares of side length $3$ along the lines $x = 3i$ ($i = 1, \\dots, t$) and $y = 3j$ ($j = 1, \\dots, t$). Each square can be divided into $2$ isosceles right triangles along the diagonal, and the center of mass of each triangle is an integral point. This gives the needed triangulation.\n\nConversely, suppose a square with side length $k$ has a triangulation such that the center of mass of each triangle is an integral point. Let $V$ be the set of all vertices of the triangulation. Define a binary relation $A \\sim_0 B$ in $V$ if two triangles of the triangulation are of the form $\\triangle ACD$, $\\triangle BCD$. The equivalence relation $\\sim$ on $V$ is generated by $\\sim_0$, i.e., $A \\sim_0 B$ if and only if there exist $A_1, \\dots, A_r$ such that $A \\sim_0 A_1 \\sim_0 \\dots \\sim_0 A_r \\sim_0 B$. Denote the horizontal and vertical coordinates of a point $P$ by $x_P$ and $y_P$, respectively. We have:\n\n(i) If $A \\sim_0 B$, then $3 \\mid x_A - x_B$ and $3 \\mid y_A - y_B$. This is because: by transitivity, one may assume $A \\sim_0 B$, i.e., there are two triangles in the triangulation of the form $\\triangle ACD$, $\\triangle BCD$, whose centers of mass are both integral points. Therefore, $3 \\mid x_A + x_C + x_D$ and $3 \\mid x_B + x_C + x_D$; thus $3 \\mid x_A - x_B$. Similarly, $3 \\mid y_A - y_B$.\n\n(ii) The set $V$ has at most $3$ equivalence classes with respect to $\\sim$. This is because after fixing a triangle $T_0$, for each point $A$ in $V$, there is always a sequence of triangles $T_0, \\dots, T_r$ such that $T_{i-1}$ and $T_i$ share a side for all $i = 1, \\dots, r$, and $A$ is a vertex of $T_r$. By definition of $\\sim$, all three vertices of $T_{i-1}$ are respectively equivalent to three vertices of $T_i$. Hence by induction, $A$ is equivalent to one of the vertices of $T_0$.\n\nBy (ii) and the pigeonhole principle, two of the four vertices of a square must be equivalent. Then from (i), we know that $3 \\mid k^2$ or $3 \\mid 2k^2$, which implies $3 \\mid k$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16515,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the powers of $3$, $4$, and $7$ in increasing order: $x_1^{\\alpha_1} \\leq x_2^{\\alpha_2} \\leq x_3^{\\alpha_3} \\leq \\dots$.\n\nProve that every integer from $1$ to $x_1^{\\alpha_1} + x_2^{\\alpha_2} + \\dots + x_n^{\\alpha_n}$ can be represented as a sum of these powers, using only $x_1^{\\alpha_1}, x_2^{\\alpha_2}, \\dots, x_n^{\\alpha_n}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove, by induction on $n$, that it is possible to represent all the integers from $1$ to $x_1^{\\alpha_1} + x_2^{\\alpha_2} + \\dots + x_n^{\\alpha_n}$ in the desired way using only the powers $x_1^{\\alpha_1}, x_2^{\\alpha_2}, \\dots, x_n^{\\alpha_n}$. It is clear that this is true for $n = 1, 2, 3$. Assuming the property holds for $n \\geq 3$, we will prove that it is valid for $n + 1$.\n\nIf the powers $x_1^{\\alpha_1}, x_2^{\\alpha_2}, \\dots, x_n^{\\alpha_n}$ are $\\{3^0, 3^1, \\dots, 3^a\\} \\cup \\{4^0, 4^1, \\dots, 4^b\\} \\cup \\{7^0, 7^1, \\dots, 7^c\\}$, we have:\n\n$$\nx_1^{\\alpha_1} + x_2^{\\alpha_2} + \\dots + x_n^{\\alpha_n} = (3^0 + 3^1 + \\dots + 3^a) + (4^0 + 4^1 + \\dots + 4^b) + (7^0 + 7^1 + \\dots + 7^c) \\\\\n= \\frac{3^{a+1}-1}{2} + \\frac{4^{b+1}-1}{3} + \\frac{7^{c+1}-1}{6} \\leq \\frac{x_{n+1}^{\\alpha_{n+1}}-1}{2} + \\frac{x_{n+1}^{\\alpha_{n+1}}-1}{3} + \\frac{x_{n+1}^{\\alpha_{n+1}}-1}{6} = x_{n+1}^{\\alpha_{n+1}} - 1.\n$$\n\nBy the induction assumption, all positive integers smaller than $x_{n+1}^{\\alpha_{n+1}}$ have a representation using only the powers $x_1^{\\alpha_1}, x_2^{\\alpha_2}, \\dots, x_n^{\\alpha_n}$. In addition, an integer $m$ such that $x_{n+1}^{\\alpha_{n+1}} \\leq m \\leq x_1^{\\alpha_1} + x_2^{\\alpha_2} + \\dots + x_n^{\\alpha_n} + x_{n+1}^{\\alpha_{n+1}}$ can be expressed as $m = (m - x_{n+1}^{\\alpha_{n+1}}) + x_{n+1}^{\\alpha_{n+1}}$,\n\nwhere $0 \\leq m - x_{n+1}^{\\alpha_{n+1}} \\leq x_1^{\\alpha_1} + x_2^{\\alpha_2} + \\dots + x_n^{\\alpha_n}$ has a representation using only $x_1^{\\alpha_1}, \\dots, x_n^{\\alpha_n}$. We conclude that all integers from $1$ to $x_1^{\\alpha_1} + \\dots + x_n^{\\alpha_n} + x_{n+1}^{\\alpha_{n+1}}$ have a representation of the desired form using only the powers $x_1^{\\alpha_1}, \\dots, x_n^{\\alpha_n}, x_{n+1}^{\\alpha_{n+1}}$, which completes the induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16516,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(\\max\\{x, y\\} + \\min\\{f(x), f(y)\\}) = x + y\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f$ be a function such that for every $x, y \\in \\mathbb{R}$, the equation holds:\n$$\nf(\\max\\{x, y\\} + \\min\\{f(x), f(y)\\}) = x + y. \\tag{1}\n$$\nIf we set $y = x$ in (1), we get\n$$\nf(x + f(x)) = 2x, \\text{ for every } x \\in \\mathbb{R}. \\tag{2}\n$$\nFor $x = 0$, (2) gives $f(f(0)) = 0$. For $x = y = \\frac{f(0)}{2}$ in (1), we get $f(f(0)) = f(0)$. Thus, $f(0) = 0$.\n\nNow, set $y = 0$ in (1) and use $f(0) = 0$:\n$$\nf(\\max\\{x, 0\\} + \\min\\{f(x), 0\\}) = x, \\text{ for every } x \\in \\mathbb{R}. \\tag{3}\n$$\nConsider two cases:\n\n*Case 1.* $x > 0$. Then (3) gives\n$$\nf(x + \\min\\{f(x), 0\\}) = x.\n$$\nSo either $f(x + f(x)) = x$ or $f(x) = x$. If $f(x + f(x)) = x$, then from (2) $2x = x$, which is impossible for $x > 0$. Thus, $f(x) = x$.\n\n*Case 2.* $x < 0$. Then (3) gives\n$$\nf(\\min\\{f(x), 0\\}) = x.\n$$\nSo either $f(0) = x$ or $f(f(x)) = x$. If $f(0) = x$, then $0 > x = f(0) = 0$, a contradiction. Thus, $f(f(x)) = x$. Now, substitute $x$ with $f(x)$ in (2):\n$$\nf(f(x) + f(f(x))) = 2f(x).\n$$\nBut $f(f(x)) = x$, so\n$$\nf(f(x) + x) = 2f(x).\n$$\nFrom (2), $f(x + f(x)) = 2x$. Thus,\n$$\nf(f(x) + x) = 2f(x), \\quad f(x + f(x)) = 2x.\n$$\nBut $f(f(x) + x) = 2f(x)$ and $f(x + f(x)) = 2x$ imply $2f(x) = 2x$, so $f(x) = x$ for $x < 0$.\n\nTherefore, $f(x) = x$ for all $x \\in \\mathbb{R}$. It is easy to check that this function satisfies the original equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16517,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider coins with positive real denominations not exceeding $1$. Find the smallest $C > 0$ such that the following holds: if we are given any $100$ such coins with total value $50$, then we can always split them into two stacks of $50$ coins each such that the absolute difference between the total values of the two stacks is at most $C$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $C = \\frac{50}{51}$. The lower bound is obtained if we have $51$ coins of value $\\frac{1}{51}$ and $49$ coins of value $1$. We now present two (similar) proofs that this $C = \\frac{50}{51}$ suffices.\n\n*First proof (original)*\n\nLet $a_1 \\le \\dots \\le a_{100}$ denote the values of the coins in ascending order. Since the $51$ coins $a_{50}, \\dots, a_{100}$ are worth at least $51a_{50}$, it follows that $a_{50} \\le \\frac{50}{51}$; likewise $a_{51} \\ge \\frac{1}{51}$.\n\nWe claim that choosing the stacks with coin values\n\n$$\na_1, a_3, \\dots, a_{49}, a_{52}, a_{54}, \\dots, a_{100}\n$$\n\nand\n\n$$\na_2, a_4, \\dots, a_{50}, a_{51}, a_{53}, \\dots, a_{99}\n$$\n\nworks. Let $D$ denote the (possibly negative) difference between the two total values. Then\n\n$$\n\\begin{aligned}\nD &= (a_1 - a_2) + \\dots + (a_{49} - a_{50}) - a_{51} + (a_{52} - a_{53}) + \\dots + (a_{98} - a_{99}) + a_{100} \\\\\n  &\\le 25 \\cdot 0 - \\frac{1}{51} + 24 \\cdot 0 + 1 = \\frac{50}{51}.\n\\end{aligned}\n$$\n\nSimilarly, we have\n\n$$\n\\begin{aligned}\nD &= a_1 + (a_3 - a_2) + \\dots + (a_{49} - a_{48}) - a_{50} + (a_{52} - a_{51}) + \\dots + (a_{100} - a_{99}) \\\\\n  &\\ge 0 + 24 \\cdot 0 - \\frac{50}{51} + 25 \\cdot 0 = -\\frac{50}{51}.\n\\end{aligned}\n$$\n\nIt follows that $|D| \\le \\frac{50}{51}$, as required.\n\n*Second proof (Evan Chen)*\n\nAgain we sort the coins in increasing order $0 < a_1 \\le a_2 \\le \\dots \\le a_{100} \\le 1$. A large gap is an index $i \\ge 2$ such that $a_i > a_{i-1} + \\frac{50}{51}$; obviously there is at most one such large gap.\n\n**Claim** — If there is a large gap, it must be $a_{51} > a_{50} + \\frac{50}{51}$.\n\n*Proof*. If $i < 50$ then we get $a_{50}, \\dots, a_{100} > \\frac{50}{51}$ and the sum $\\sum_{1}^{100} a_i > 50$ is too large. Conversely if $i > 50$ then we get $a_1, \\dots, a_{i-1} < \\frac{1}{51}$ and the sum $\\sum_{1}^{100} a_i < \\frac{1}{51} \\cdot 51 + 49$ is too small. $\\square$\n\nNow imagine starting with the coins $a_1, a_3, \\dots, a_{99}$, which have total value $S \\le 25$. We replace $a_1$ by $a_2$, then $a_3$ by $a_4$, and so on, until we replace $a_{99}$ by $a_{100}$. At the end of the process we have $S \\ge 25$. Moreover, since we did not cross a large gap at any point, the quantity $S$ changed by at most $C = \\frac{50}{51}$ at each step. So at some point in the process we need to have $25 - C/2 \\le S \\le 25 + C/2$, which proves $C$ works.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16518,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be a natural number, and $\\mathcal{F}$ be the set of functions $f : \\{1, 2, \\dots, n\\} \\to \\{1, 2, \\dots, n\\}$ such that $f(k) \\le f(k+1) \\le f(k)+1$ for every $k \\in \\{1, 2, \\dots, n-1\\}$.\n\n**a)** Determine the cardinality of the set $\\mathcal{F}$.\n\n**b)** Determine the total number of fixed points of the functions in $\\mathcal{F}$.\n\n(A *fixed point* of the function $f$ is a number $x \\in \\{1, 2, \\dots, n\\}$ such that $f(x) = x$.)",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** We count the functions in $\\mathcal{F}$ with $f(1) = k$, $k = 1, \\dots, n$. For each $i = 2, \\dots, n$, associate the number $f(i) - f(i-1) \\in \\{0, 1\\}$, with at most $n-k$ occurrences of $1$. This association is bijective, and the number of ways to choose the numbers $0$ and $1$ as described is $\\binom{n-1}{0} + \\binom{n-1}{1} + \\dots + \\binom{n-1}{n-k}$ (these are the possibilities of placing $0, 1, 2, \\dots, n-k$ of $1$). Therefore:\n\n$$\n|\\mathcal{F}| = \\sum_{k=1}^{n} \\left( \\binom{n-1}{0} + \\binom{n-1}{1} + \\dots + \\binom{n-1}{n-k} \\right) = \\sum_{p=0}^{n-1} (n-p) \\binom{n-1}{p} = n \\cdot 2^{n-1} - (n-1) \\sum_{p=0}^{n-1} \\binom{n-2}{p-1} = (n+1)2^{n-2}.\n$$\n\n**b)** We count how many times the fixed point $k$ appears in the functions from $\\mathcal{F}$, $k = 1, \\dots, n$ (the same fixed point may appear in multiple functions). For each function with $f(k) = k$, associate the numbers $f(i) - f(i-1) \\in \\{0, 1\\}$ for $i = 2, \\dots, n$. Since these numbers can be chosen without restrictions, there are $2^{n-1}$ possibilities. Thus, each fixed point appears in $2^{n-1}$ functions, and\n\n$$\n\\sum_{f \\in \\mathcal{F}} |\\text{Fix}(f)| = n \\cdot 2^{n-1}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16519,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any pair of positive integers $k$ and $n$, there exist $k$ positive integers $m_1, m_2, \\dots, m_k$ (not necessarily different) such that\n\n$$\n1 + \\frac{2^k - 1}{n} = \\left(1 + \\frac{1}{m_1}\\right) \\left(1 + \\frac{1}{m_2}\\right) \\cdots \\left(1 + \\frac{1}{m_k}\\right).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We induct on $k$.\n\n*Base case*: For $k = 1$, take $m_1 = n$.\n\n*Inductive step*: Assume the statement holds for some $k$.\n\n- **Case 1:** $n = 2m - 1$ (odd)\n\n  $$\n  1 + \\frac{2^{k+1} - 1}{n} = \\frac{2m}{2m-1} \\cdot \\frac{2^{k+1} + 2m - 2}{2m} = \\left(1 + \\frac{1}{2m-1}\\right) \\left(1 + \\frac{2^k - 1}{m}\\right)\n  $$\n\n  By the induction hypothesis, $1 + \\frac{2^k - 1}{m}$ can be written as a product of $k$ terms of the desired form.\n\n- **Case 2:** $n = 2m$ (even)\n\n  $$\n  1 + \\frac{2^{k+1} - 1}{n} = \\frac{2^{k+1} + 2m - 1}{2^{k+1} + 2m - 2} \\cdot \\frac{2^{k+1} + 2m - 2}{2m} = \\left(1 + \\frac{1}{2^{k+1} + 2m - 2}\\right) \\left(1 + \\frac{2^k - 1}{m}\\right)\n  $$\n\n  Again, $1 + \\frac{2^k - 1}{m}$ is a product of $k$ terms by the induction hypothesis.\n\nThus, the statement holds for all $k$ by induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16520,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, $C$ be nodes of the lattice $\\mathbb{Z} \\times \\mathbb{Z}$ such that inside the triangle $ABC$ lies a unique node $P$ of the lattice. Denote $E := AP \\cap BC$. Determine\n$$\n\\max \\frac{AP}{PE},\n$$\nover all such configurations.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us build $E'$, the symmetric of $E$ with respect to $P$. If $E$ is a lattice point, then $E'$ is a lattice point, and this implies $E' \\notin \\text{int}(ABC)$. We deduce that $E' = A$, otherwise $A \\in (E'P)$ and through a translation we get that there exists a lattice point lying on $PE$, contradiction. So in this case $\\frac{AP}{PE} = 1$.\n\nIf $E$ is not a lattice point, and on $BC$ lies some other lattice point $Q$, since $Q \\neq E$ we can restrict the problem to either triangle $ABQ$ or $ACQ$, with the same initial hypothesis. Thus we can assume that the only lattice points on the sides of the triangle are on $[AB]$ and $[AC]$.\n\nWe shall prove that, over all configurations,\n$$\n\\max \\frac{AP}{PE} = 5,\n$$\na value which can actually be reached by a proper configuration. Rather than looking for $\\max \\frac{AP}{PE}$, we will establish $\\min \\frac{PE}{AE} = \\min \\frac{[BPC]}{[ABC]}$ (where $[XYZ]$ denotes the area of $\\triangle XYZ$).\n\nNotice that for the triangle $BPC$, the only lattice points that are lying either on its sides or in its interior are just its vertices. By Pick's Theorem, $[BPC] = \\frac{1}{2}$. Let $\\beta$ and $\\gamma$ be the number of lattice points on the sides $[AB]$ and $[AC]$, respectively. Using Pick's Theorem again, $[ABC] = \\frac{\\beta + \\gamma + 3}{2}$.\n\nThus, $\\frac{PE}{AE} = \\frac{1}{\\beta + \\gamma + 3}$. We want to prove that $\\frac{PE}{AE} \\ge \\frac{1}{6}$. Suppose, for contradiction, that $\\frac{PE}{AE} < \\frac{1}{6}$, so $\\beta + \\gamma \\ge 4$.\n\nLet $B_0, B_1, \\dots, B_{\\beta+1}$ be the lattice points on $[AB]$, with $B_0 = A$, $B_{\\beta+1} = B$, and $C_0, C_1, \\dots, C_{\\gamma+1}$ the lattice points on $[AC]$, with $C_0 = A$, $C_{\\gamma+1} = C$.\n\nConsider the triangles $PC_iC_{i+1}$ for $i = 0, \\dots, \\gamma$. They share the property that no lattice points lie inside or on the sides, other than the vertices, so $[PC_iC_{i+1}] = \\frac{1}{2}$. It follows that $AC_1 = C_1C_2 = \\dots = C_\\gamma C$. Similarly, $AB_1 = B_1B_2 = \\dots = B_\\beta B$. Thus $\\frac{AC_j}{C_jC_{j+1}} = \\frac{AB_j}{B_jB_{j+1}}$, and so $C_iB_i \\parallel C_{i+1}B_{i+1}$ for all $i \\in \\{1, 2, \\dots, \\min\\{\\beta, \\gamma\\}\\}$.\n\nWe also deduce that on the segment $(B_iC_i)$ lie $i-1$ lattice points. This is because we can construct $i-1$ parallelograms with vertices of the type $B_lC_lB_iD$ with $l < i$, or $B_lC_lC_iD$ with $D \\in (B_iC_i)$. From the relationships between coordinates in a parallelogram, $D$ is a lattice point. Thus if $\\min\\{\\beta, \\gamma\\} \\ge 2$, the segments $(B_i, C_i)$ would contain at least three lattice points (the ones from $(B_2C_2)$ and $(B_3C_3)$), which is impossible since $P$ is on at most one of the segments $(B_iC_i)$. This means $\\min\\{\\beta, \\gamma\\} \\le 1$, say $\\beta \\ge \\gamma$.\n\n**First case:** $\\gamma = 1$. From $\\beta + \\gamma \\ge 4$ it follows that $\\beta \\ge 3$ and $C_1$ is the midpoint of $AC$. Now, since $B_1$ is the midpoint of $AB_2$, if we take $M$ the midpoint of $CB_2$, we get the parallelogram $B_2B_1C_1M$, and since all the points $B_2, B_1, C_1$ are lattice, $M$ is lattice. Since $M$ is inside the triangle $ABC$, it must be $P$. The quadrilateral $BB_\\beta PC$ is convex since $B_\\beta \\in (B_2B)$. The triangles $BB_\\beta C$ and $BB_\\beta P$ have the property that there are no lattice points inside or on their sides, except their vertices, so $[BB_\\beta P] = [BB_\\beta C] = \\frac{1}{2}$, hence $BB_\\beta \\parallel PC$, which is false.\n\n**Second case:** $\\gamma = 0$. It follows that $\\beta \\ge 4$.\n\n**Lemma.** Let $XYZT$ be a convex quadrilateral with lattice vertices, with the property that no lattice points lie in its interior, nor on its sides, except for its vertices. Then $XYZT$ is a parallelogram.\n\n*Proof.* By Pick's theorem, $[XYZ] = [XYT] = \\frac{1}{2}$, so $XY \\parallel ZT$. Similarly, $XT \\parallel YZ$, which ends the proof. $\\square$\n\nConsider the quadrilateral $AB_1PC$, with $AB_1 \\cap PC \\ne \\emptyset$. If it is convex, the previous lemma implies a contradiction. Thus $[AB_1] \\cap PC \\ne \\emptyset$. Also consider the quadrilateral $BB_\\beta PC$. By the same reasoning, $[BB_\\beta] \\cap PC \\ne \\emptyset$. This is the desired contradiction and ends the solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16521,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{R}^+$ denote the set of positive real numbers. Find all functions $f: \\mathbb{R}^+ \\to \\mathbb{R}^+$ such that for all $x, y \\in \\mathbb{R}^+$,\n\n$$\nf(x) f(y f(x)) = f(x + y).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We break the proof into several steps.\n\n**Step 1.** If we solve the equation $yf(x) = x + y$, we formally get $y = \\frac{x}{f(x) - 1}$. If we want to substitute $y = \\frac{x}{f(x) - 1}$, we need $f(x) - 1 > 0$. So we see that if $f(x) > 1$ for some positive real $x$, then setting $y = \\frac{x}{f(x) - 1}$ gives $f(x) = 1$, a contradiction. Thus, $f(x) \\leq 1$ for all $x \\in \\mathbb{R}^+$. This implies that $f$ is a decreasing function.\n\n**Step 2.** If $f(x) = 1$ for some $x \\in \\mathbb{R}^+$, then $f(x + y) = f(y)$ for each $y \\in \\mathbb{R}^+$, and by the monotonicity of $f$ it follows that $f \\equiv 1$.\n\n**Step 3.** Now suppose $f(x) < 1$ for each $x \\in \\mathbb{R}^+$. Then $f$ is a strictly decreasing function, in particular injective. Setting $x = 1$ and $y = x + y - 1$ gives\n\n$$\nf(1)f((x + y - 1)f(1)) = f(1 + (x + y - 1)) = f(x + y)\n$$\n\nfor all $x, y \\in \\mathbb{R}^+$. Therefore,\n\n$$\nf(1)f((x + y - 1)f(1)) = f(x + y) = f(x)f(yf(x))\n$$\n\nfor all $x, y \\in \\mathbb{R}^+$. Taking $y = \\frac{1}{f(x)}$ into the previous equation then gives\n\n$$\nf(1)f\\left(\\left(x + \\frac{1}{f(x)} - 1\\right)f(1)\\right) = f(x)f(1).\n$$\n\nBy the injectivity of $f$, we have $f(x) = \\frac{1}{1 + a x}$ where $a = \\frac{1 - f(1)}{f(1)}$.\n\n**Step 4.** Combining the two cases, we conclude that $f(x) = \\frac{1}{1 + a x}$ for each $x \\in \\mathbb{R}^+$, where $a \\geq 0$. Conversely, a direct verification shows that the functions of this form satisfy the initial equality. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16522,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $abc$ be a three-digit number (with digits $a$, $b$, $c$) such that\n$$\n4 \\times (abc)_6 = 3 \\times (abc)_7 + 2.\n$$\nWhat is the largest possible value of this number in base 10?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $abc$ be the number, with $a$, $b$, $c$ as digits.\n\nIn base 6: $(abc)_6 = 36a + 6b + c$.\nIn base 7: $(abc)_7 = 49a + 7b + c$.\n\nGiven:\n$$\n4(36a + 6b + c) = 3(49a + 7b + c) + 2\n$$\nExpanding:\n$$\n144a + 24b + 4c = 147a + 21b + 3c + 2\n$$\nRearrange:\n$$\n3b + c = 3a + 2\n$$\nSo $c - 2$ is divisible by 3. Since $c < 6$, $c = 2$ or $5$.\n\n- If $c = 2$, $3b + 2 = 3a + 2 \\implies a = b$.\n- If $c = 5$, $3b + 5 = 3a + 2 \\implies 3b = 3a - 3 \\implies a = b + 1$.\n\nDigits $a, b, c \\leq 5$. The largest possible number is $a = 5$, $b = 5$, $c = 2$ (for $a = b$ case), which is $552$ in base 6:\n$$\n552_6 = 6^2 \\times 5 + 6 \\times 5 + 2 = 180 + 30 + 2 = 212\n$$\nThus, the largest required number (in base 10) is **212**.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16523,
    "subject": "Mathematics (Olympiad)",
    "question": "連續正整數的等幂次方和的定義是\n\n$$\nS_m(N) = \\sum_{k=1}^{N-1} k^m\n$$\n\n特別是\n\n$$\nS_1(N) = 1 + 2 + \\cdots + (N-1) = \\frac{N(N-1)}{2}\n$$\n\n$$\nS_2(N) = 1^2 + 2^2 + \\cdots + (N-1)^2 = \\frac{N(N-1)(2N-1)}{6}\n$$\n\n$$\nS_3(N) = 1^3 + 2^3 + \\cdots + (N-1)^3 = \\frac{N^2(N-1)^2}{4}\n$$\n\n證明\n\n$$\nS_2(N)S_3(N) = \\frac{7}{12}S_6(N) + \\frac{5}{12}S_4(N)\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "$$\nS_2(N)S_3(N) = \\sum_{j=1}^{N-1} j^2 \\sum_{k=1}^{N-1} k^3\n$$\n\n依 $k < j$, $k = j$, 以及 $k > j$ 而分成三部份。\n\n即\n\n$$\n\\sum_{j=1}^{N-1} j^2 \\sum_{k=1}^{j-1} k^3 + S_5(N) + \\sum_{k=1}^{N-1} k^3 \\sum_{j=1}^{k-1} j^2\n$$\n\n或\n\n$$\n\\sum_{j=1}^{N-1} j^2 \\cdot \\frac{j^2(j-1)^2}{4} + S_5(N) + \\sum_{k=1}^{N-1} k^3 \\cdot \\frac{k(k-1)(2k-1)}{6}\n$$\n\n和為\n\n$$\n\\frac{7}{12}S_6(N) - S_5(N) + \\frac{5}{12}S_4(N) + S_5(N)\n$$\n\n即\n\n$$\n\\frac{7}{12}S_6(N) + \\frac{5}{12}S_4(N)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16524,
    "subject": "Mathematics (Olympiad)",
    "question": "Sean $X, Y$ los extremos de un diámetro de una circunferencia $\\Gamma$ y $N$ el punto medio de uno de los arcos $XY$ de $\\Gamma$. Sean $A$ y $B$ dos puntos en el segmento $XY$. Las rectas $NA$ y $NB$ cortan nuevamente a $\\Gamma$ en los puntos $C$ y $D$, respectivamente. Las tangentes a $\\Gamma$ en $C$ y $D$ se cortan en $P$. Sea $M$ el punto de intersección del segmento $XY$ con el segmento $NP$. Demostrar que $M$ es el punto medio del segmento $AB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Tracemos la paralela a $XY$ por $P$, que corta a las rectas $NA$ en $E$ y $NB$ en $F$, y sea $O$ el centro de $\\Gamma$. Por ser $PC$ tangente a $\\Gamma$ en $C$, tenemos que $\\angle OCP = 90^\\circ$, luego $\\angle ECP = 90^\\circ - \\angle NCO$, donde $\\angle NCO = \\angle CNO = \\angle ANO$ por ser $O$ centro de $\\Gamma$ sobre la que están $N$, $C$. Pero por ser $AO$ paralela a $EP$, y $NA$ y $CE$ ser la misma recta, tenemos que $\\angle ANO = 90^\\circ - \\angle CEP$, luego $\\angle ECP = \\angle CEP$, es decir, el triángulo $CEP$ es isósceles en $P$, luego $PE = PC$. De forma similar, se tiene que $PF = PE$. Pero por ser $PC$, $PD$ tangentes a $\\Gamma$ en $C$, $D$ por $P$, se tiene que $PC = PD$, luego $P$ es el punto medio de $EF$. Ahora bien, por construcción de $E$, $F$, existe una homotecia con centro $N$ que transforma el triángulo $NAB$ en el triángulo $NEF$, homotecia que además transforma $M$ en $P$. Pero como $P$ es el punto medio de $EF$, entonces $M$ es el punto medio de $AB$, como queríamos demostrar.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16525,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $n$ such that there exist pairwise distinct integers $a_1, \\dots, a_n, b_1, \\dots, b_n$ satisfying\n\n$$\n\\prod_{i=1}^{n} (a_k^2 + a_i a_k + b_i) = \\prod_{i=1}^{n} (b_k^2 + a_i b_k + b_i) = 0 \\quad \\text{for all } k = 1, \\dots, n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The required integers are $n = 1$ and $n = 2$.\n\nFor $n = 1$, $(a_1, b_1) = (1, -2)$ is the unique pair of integers satisfying the conditions.\n\nFor $n = 2$, only $(a_1, b_1, a_2, b_2) = (2, 0, -1, -2)$ and $(a_1, b_1, a_2, b_2) = (-1, -2, 2, 0)$ work. Verification is routine.\n\nLet $f = \\prod_{i=1}^{n} (X^2 + a_i X + b_i)$, a monic polynomial of degree $2n$. It vanishes at each $a_k$ and $b_k$. Since these $2n$ numbers are pairwise distinct and $\\deg f = 2n$, they are the roots of $f$. Each factor $f_i = X^2 + a_i X + b_i$ has at most 2 roots, so the $n$ root sets are disjoint and each has size 2. Thus,\n$$\n\\prod_{k=1}^{n} (X^2 + a_k X + b_k) = \\prod_{k=1}^{n} (X - a_k)(X - b_k).\n$$\n\nIf no $b_k$ is zero, comparing constant terms gives $a_1 \\cdots a_n = 1$, which is impossible for pairwise distinct integers unless $n = 1$ (then $a_1 = 1$, $b_1 = -2$).\n\nIf some $b_k = 0$, say $b_1 = 0$, then $a_1 \\neq 0$, so $f_1 = X^2 + a_1 X$ does not vanish at $a_1$. The root set of $f_1$ has size 2, so $n \\geq 2$. The polynomial equality becomes\n$$\n(X + a_1) \\prod_{k=2}^{n} (X^2 + a_k X + b_k) = (X - a_1) \\prod_{k=2}^{n} (X - a_k)(X - b_k).\n$$\nComparing constant terms gives $a_2 \\cdots a_n = -1$, which is only possible for $n = 2$ or $n = 3$.\n\nFor $n = 2$, $a_2 = -1$, $f_1 = X^2 + a_1 X$, $f_2 = X^2 - X + b_2$. The root sets are disjoint and both have size 2. The roots of $f_2$ must be $a_1$ and $a_2$, so the other root of $f_1$ is $b_2$. $f_2(a_2) = 0$ gives $b_2 = -2$, and $f_1(b_2) = 0$ gives $a_1 = 2$. Thus, $(a_1, b_1, a_2, b_2) = (2, 0, -1, -2)$. The other quadruple, $(-1, -2, 2, 0)$, corresponds to $b_2 = 0$.\n\nFor $n = 3$, assume $a_2 = -1$, $a_3 = 1$. Then $f_1 = X^2 + a_1 X$, $f_2 = X^2 - X + b_2$, $f_3 = X^2 + X + b_3$. The roots are $a_1, -1, 1, 0, b_2, b_3$. $f_1(-1)f_1(1) \\neq 0$, so $-1$ and $1$ are roots of $f_2 f_3$. $f(1) = b_2 \\neq 0$, $f_3(-1) = b_3 \\neq 0$, so $2 + b_2 = 0 = 2 + b_3$, i.e., $b_2 = b_3$, a contradiction. Thus, $n = 3$ is not possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16526,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ and $ABD$ be coplanar triangles with equal perimeters. The lines of support of the internal bisectors of the angles $CAD$ and $CBD$ meet at $P$. Show that the angles $APC$ and $BPD$ are congruent.",
    "options": [],
    "answer": "See solution",
    "solution": "Extend the segment $AC$ beyond $C$ by a segment $CE$ congruent to the segment $CB$, so $C$ lies on the perpendicular bisector of the segment $BE$. Similarly, extend the segment $BD$ beyond $D$ by a segment $DF$ congruent to the segment $AD$, so $D$ lies on the perpendicular bisector of the segment $AF$. Since the triangles $ABC$ and $ABD$ have equal perimeters, the segments $AE$ and $BF$ are congruent.\n\n![](images/RMC_2015_BT_p65_data_1467792dc1.png)\n\nNow let the perpendicular bisectors of the segments $AF$ and $BE$ meet at $Q$. Clearly, the segments $QA$ and $QF$, respectively $QE$ and $QB$, are congruent, and since so are the segments $AE$ and $BF$ by the preceding, the triangles $QAE$ and $QFB$ are congruent. Therefore, the angles $AQF$ and $BQE$ are congruent, and hence so are their halves; that is, the angles $AQD$ and $BQC$ are congruent, and consequently so are the angles $AQC$ and $BQD$.\n\nWe now show that, in fact, the points $P$ and $Q$ coincide, whence the conclusion. With reference again to the congruence of the triangles $QAE$ and $QFB$, the angles $QAE$ and $QFB$ are congruent, and since the latter is the reflection of the angle $QAD$ in the line $QD$, it follows that the angles $QAC = QAE$ and $QAD$ are congruent, so the line $AQ$ bisects the angle $CAD$. Similarly, the line $BQ$ bisects the angle $CBD$, and consequently the points $P$ and $Q$ coincide.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16527,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S = \\{1, 2, \\ldots, 2n\\}$, where $n$ is a positive integer. A subset $T$ of $S$ is called a *bad* subset if in $T$ the number of odd elements is greater than the number of even elements. A subset of $S$ is neither good nor bad if it has exactly $k$ odd elements and $k$ even elements for some $k = 0, 1, \\dots, n$.\n\n(i) Find the number of *good* subsets of $S$ (i.e., those with more even than odd elements).\n\n(ii) Find the sum of all elements in all good subsets of $S$.",
    "options": [],
    "answer": "See solution",
    "solution": "(i) The answer is $2^{2n-1} - \\frac{1}{2}\\binom{2n}{n}$.\n\nThere are $n$ odd and $n$ even elements in $S$. The number of subsets which are neither good nor bad is\n\n$$\n\\sum_{k=0}^{n} \\binom{n}{k}^2 = \\binom{2n}{n}\n$$\n\nby Vandermonde's identity. By symmetry, the number of good subsets is the same as the number of bad subsets. Since there are $2^{2n}$ subsets of $S$ in total, the number of good subsets is\n\n$$\n\\frac{1}{2}\\left[2^{2n} - \\binom{2n}{n}\\right] = 2^{2n-1} - \\frac{1}{2}\\binom{2n}{n}.\n$$\n\n(ii) The answer is $2^{2n-2}(2n^2 + n) - n^2\\binom{2n-1}{n}$.\n\nLet $N_1$ be the number of good subsets containing a particular even number. We have\n\n$$\nN_1 = \\sum_{i=0}^{n-1} \\sum_{j=0}^{i} \\binom{n-1}{i} \\binom{n}{j} = \\sum_{i+k \\ge n} \\binom{n-1}{i} \\binom{n}{k} = \\sum_{m=n}^{2n-1} \\binom{2n-1}{m} = 2^{2n-2}.\n$$\n\nSimilarly, let $N_2$ be the number of good subsets containing a particular odd number:\n\n$$\n\\begin{align*}\nN_2 &= \\sum_{i=2}^{n} \\sum_{j=0}^{i-2} \\binom{n}{i} \\binom{n-1}{j} \\\\\n&= \\sum_{i+k \\ge n+1} \\binom{n}{i} \\binom{n-1}{k} \\\\\n&= \\sum_{m=n+1}^{2n-1} \\binom{2n-1}{m} \\\\\n&= 2^{2n-2} - \\binom{2n-1}{n}.\n\\end{align*}\n$$\n\nThe sum of all elements in all good subsets of $S$ is\n\n$$\n2^{2n-2} \\cdot n(n+1) + \\left[2^{2n-2} - \\binom{2n-1}{n}\\right] \\cdot n^2 = 2^{2n-2}(2n^2+n) - n^2 \\binom{2n-1}{n}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16528,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y, z > 0$ such that\n$$\n\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} < \\frac{1}{xyz}.\n$$\nShow that\n$$\n\\frac{2x}{\\sqrt{1+x^2}} + \\frac{2y}{\\sqrt{1+y^2}} + \\frac{2z}{\\sqrt{1+z^2}} < 3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $r = 1/x$, $s = 1/y$, $t = 1/z$. There exists $\\alpha < 1$ such that $r + s + t = \\alpha^2 rst$ or $\\alpha(r + s + t) = \\alpha^3 rst$. Let $a = \\alpha r$, $b = \\alpha s$, $c = \\alpha t$. Write $a = \\tan A$, $b = \\tan B$, $c = \\tan C$, then $A + B + C = \\pi$. It is clear that\n\n$$\n\\begin{aligned}\n\\frac{1}{2} \\times \\text{LHS} &= \\frac{1}{\\sqrt{1 + r^2}} + \\frac{1}{\\sqrt{1 + s^2}} + \\frac{1}{\\sqrt{1 + t^2}} \\\\\n&< \\frac{1}{\\sqrt{1 + a^2}} + \\frac{1}{\\sqrt{1 + b^2}} + \\frac{1}{\\sqrt{1 + c^2}} \\\\\n&= \\cos A + \\cos B + \\cos C \\\\\n&\\le 3 \\cos \\left( \\frac{A + B + C}{3} \\right) = \\frac{3}{2} = \\frac{1}{2} \\times \\text{RHS}.\n\\end{aligned}\n$$\n\n**Second solution:**\n\nNote that\n$$\n\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} < \\frac{1}{xyz} \\implies xy + yz + xz < 1.\n$$\nHence,\n$$\n\\frac{2x}{\\sqrt{1 + x^2}} < \\frac{2x}{\\sqrt{x^2 + xy + xz + yz}} = \\frac{2x}{\\sqrt{(x + y)(x + z)}}.\n$$\nBy AM-GM,\n$$\n\\frac{2x}{\\sqrt{(x + y)(x + z)}} \\le \\frac{x}{x + y} + \\frac{x}{x + z}.\n$$\nSimilarly,\n$$\n\\frac{2y}{\\sqrt{(y + z)(y + x)}} \\le \\frac{y}{y + z} + \\frac{y}{y + x}, \\quad \\frac{2z}{\\sqrt{(z + x)(z + y)}} \\le \\frac{z}{z + x} + \\frac{z}{z + y}.\n$$\nThe desired inequality then follows by adding up the three inequalities.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16529,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$, a circle centered at some point $O$ meets the segments $BC$, $CA$, $AB$ in the pairs of points $X$ and $X'$, $Y$ and $Y'$, $Z$ and $Z'$, respectively, labeled in circular order: $X, X', Y, Y', Z, Z'$. Let $M$ be the Miquel point of the triangle $XYZ$ (i.e., the point of concurrence of the circles $AYZ$, $BZX$, $CXY$), and let $M'$ be that of the triangle $X'Y'Z'$. Prove that the segments $OM$ and $OM'$ have equal lengths.\n\n![](images/RMC2014_p100_data_3f877bde8d.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We begin by reviewing some basic facts on conics. For an ellipse $\\Sigma$ with center $N$, foci $M$ and $M'$, semiaxes $a$ and $b$, it is known that the orthogonal projections $P$ and $P'$ of $M$ and $M'$ on any line $t$ tangent to $\\Sigma$ lie on the major auxiliary circle of $\\Sigma$, so that $NP = a = NP'$. Application to triangle $MNP$ (respectively, $M'NP'$) of a rotation $\\theta$ (respectively, $-\\theta$) about $M$ (respectively, $M'$) and a homothety of ratio $\\sec\\theta$ with center $M$ (respectively, $M'$) yields triangle $MOX$ (respectively, $M'OX'$), where $MO = M'O$, $NO = \\frac{1}{2} \\cdot MM' \\cdot \\tan\\theta$, $OX = a \\sec\\theta = OX'$, and $X, X'$ both lie on $t$. If $t$ varies and $\\theta$ is constant, the locus of $X$ and $X'$ is then a circle $\\Gamma$ centered at $O$.\n\nBy Cartesian geometry it is readily checked that $\\Gamma$ and $\\Sigma$ are bitangent, and the line $\\ell$ supporting their common chord is also the radical axis of the circles $\\Gamma$ and $OMM'$, with this real geometrical significance even if the bitangency is not real. Since $\\ell$ and the circle $OMM'$ are mutually inverse in $\\Gamma$, the inverse points of $M$ and $M'$ in $\\Gamma$ both lie on $\\ell$. Finally, the distance $d$ between the parallel lines $\\ell$ and $MM'$ is given by $d \\cdot MM' = 2b^2 \\tan \\theta$. Similar considerations hold for a hyperbola $\\Sigma$.\n\nConsider now an isopair $M, M'$ (two isogonally conjugate in the triangle $ABC$, the foci of a conic $\\Sigma$ touching its sides), and take points $X, Y, Z$ (respectively, $X', Y', Z'$) on lines $BC, CA, AB$, respectively, so that the lines $MX, MY, MZ$ (respectively, $M'X', M'Y', M'Z'$) make the same directed angle $\\theta$ (respectively, $-\\theta$) with the perpendiculars to $BC, CA, AB$, respectively; then the isopedal triangles $XYZ, X'Y'Z'$ of angles $\\theta, -\\theta$ for the isopair $M, M'$ have their Miquel points at $M, M'$ and are inscribed in a common isopedal circle $\\Gamma$ bitangent to $\\Sigma$, centered at a point $O$ on the perpendicular bisector of the segment $MM'$.\n\nConversely, for any pair of triangles inscribed in a triangle $ABC$ and in a circle $\\Gamma$ (as in the statement of the problem), the Miquel points $M, M'$ are an isopair and $\\Gamma$ is an isopedal circle of $M, M'$. (If $M, M'$ are the Brocard points, $\\Sigma$ is the Brocard ellipse and $\\Gamma$ is a Tucker circle.)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16530,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\leq 100$ be a positive integer. There are 101 numbers written in a row:\n\n$$\n0 \\cdot n \\text{ mod } 101,\\ 1 \\cdot n \\text{ mod } 101,\\ \\dots,\\ 100 \\cdot n \\text{ mod } 101.\n$$\n\nHow many pairs of neighbouring numbers are there in this row such that the one on the left is bigger than the one on the right?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n-1$.\n\nIndeed, the main claim is that $an \\bmod 101$ is larger than $(a+1)n \\bmod 101$ if and only if there is a number divisible by 101 between $an$ and $(a+1)n$. Denote by $f(x)$ the remainder of $x$ when divided by 101. Note that $f(an) \\neq f((a+1)n)$, otherwise $101 \\mid n$, a contradiction. Put $an = 101k + f(an)$ and $(a+1)n = 101k' + f((a+1)n)$. Thus\n\n$$\n n = (a + 1)n - an = 101(k' - k) + f((a + 1)n) - f(an).\n$$\n\nSince $-101 < f((a+1)n) - f(an) < 101$, it is easy to check that $k' - k = 0$ or $1$.\n\n- If $f(an) < f((a+1)n)$ then $k = k'$ and between $an$ and $(a+1)n$, there does not exist any multiple of 101.\n- If $f(an) > f((a+1)n)$ then $k' = k + 1$ and between $an$ and $(a+1)n$, the number $101k'$ is a unique multiple of 101.\n\nSo all we are left with is calculating the quantity of non-zero numbers divisible by 101 between $0$ and $100n$. This quantity is $\\lfloor 100n/101 \\rfloor = n - 1$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 16531,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral with $AB = BC = CD$, $AC \\neq BD$, and let $E$ be the intersection point of its diagonals. Prove that $AE = DE$ if and only if $\\angle BAD + \\angle ADC = 120^{\\circ}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\angle BAC = \\angle BCA = \\alpha$, $\\angle CBD = \\angle CDB = \\beta$, and let $S$ be the point of intersection of the lines $AB$ and $DC$.\n\nIf $AE = DE$, then\n$$\n\\frac{AE}{\\sin(2\\alpha + \\beta)} = \\frac{AB}{\\sin(\\alpha + \\beta)} = \\frac{CD}{\\sin(\\alpha + \\beta)} = \\frac{DE}{\\sin(\\alpha + 2\\beta)}.\n$$\nNow it is obvious that\n$$\n\\sin(2\\alpha + \\beta) = \\sin(\\alpha + 2\\beta), \\quad 0^{\\circ} < 2\\alpha + \\beta < 180^{\\circ}, \\quad 0^{\\circ} < \\alpha + 2\\beta < 180^{\\circ},\n$$\n$\\alpha \\neq \\beta$ (since $AC \\neq BD$), and therefore\n$$\n2\\alpha + \\beta + \\alpha + 2\\beta = 180^{\\circ} \\quad \\text{or} \\quad \\alpha + \\beta = 60^{\\circ}.\n$$\nHence we have: $\\angle BAD + \\angle ADC = 120^{\\circ}$.\n\nConversely, if $\\angle BAD + \\angle ADC = 120^{\\circ}$, we have\n$$\n\\angle AEB = \\angle BAE + \\angle EDC \\quad (1)\n$$\n$$\n\\angle AEB = \\angle EAD + \\angle EDA \\quad (2)\n$$\nFrom (1) and (2) we have\n$$\n2(\\angle AEB) = \\angle BAD + \\angle ADC = 120^{\\circ}.\n$$\nHence $\\angle AEB = 60^{\\circ} = \\angle BSC$ and the quadrilateral $SBEC$ is inscribable. Therefore $\\angle BSE = \\angle BCE = \\angle BAE$ and $AE = ES$.\n\nSimilarly, we prove that $ES = ED$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16532,
    "subject": "Mathematics (Olympiad)",
    "question": "(a) Find a sequence $a_1, a_2, a_3, a_4, a_5, a_6$ of six distinct numbers that is both 2-composite and 3-composite. That is, find six distinct numbers such that\n\n$$\na_1 + a_4 = a_2 + a_5 = a_3 + a_6 \\quad \\text{and} \\quad a_1 + a_3 + a_5 = a_2 + a_4 + a_6.\n$$\n\n(b) Find a sequence $a_1, a_2, a_3, a_4, a_5, a_6, a_7$ of seven distinct numbers that is 2-, 3-, and 4-composite. That is, find seven distinct numbers such that\n\n$$\na_1 + a_3 + a_5 + a_7 = a_2 + a_4 + a_6, \\qquad (1)\n$$\n\n$$\na_1 + a_4 + a_7 = a_2 + a_5 = a_3 + a_6, \\qquad (2)\n$$\n\n$$\na_1 + a_5 = a_2 + a_6 = a_3 + a_7 = a_4. \\qquad (3)\n$$\n\n(c) What is the largest $k$ for which a $k$-composite sequence of 99 distinct integers exists? Give an example of such a sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) An example of a correct sequence is $5, 7, 6, 3, 1, 2$. This sequence consists of six distinct numbers and is 2-composite since $5 + 6 + 1 = 7 + 3 + 2$. It is also 3-composite since $5 + 3 = 7 + 1 = 6 + 2$.\n\nTo construct such a sequence, we look for $a_1, a_2, a_3, a_4, a_5, a_6$ satisfying the given equations. If we choose $a_4 = -a_1$, $a_5 = -a_2$, and $a_6 = -a_3$, then the first two equations hold. The third equation gives $a_1 + a_3 = a_2$. Choosing $a_1 = 1$, $a_3 = 2$ (so $a_2 = 3$), we get the sequence $1, 3, 2, -1, -3, -2$. Shifting all numbers by $4$ gives $5, 7, 6, 3, 1, 2$.\n\n(b) A possible solution is $8, 17, 26, 27, 19, 10, 1$. This sequence consists of seven distinct integers and is 2-composite since $8 + 26 + 19 + 1 = 17 + 27 + 10$. It is 3-composite since $8 + 27 + 1 = 17 + 19 = 26 + 10$. It is also 4-composite since $8 + 19 = 17 + 10 = 26 + 1 = 27$.\n\nTo construct such a sequence, we express $a_1, a_2, a_3, a_4, a_5$ in terms of $a_6$ and $a_7$ using the equations above. Trying $a_6 = 10$ and $a_7 = 1$, we find $a_5 = 19$, $a_4 = 27$, $a_3 = 26$, $a_2 = 17$, $a_1 = 8$.\n\n(c) The largest $k$ for which a $k$-composite sequence of 99 distinct integers exists is $k = 50$. An example is\n\n$1, 2, \\dots, 48, 49, 100, 99, 98, \\dots, 52, 51.$\n\nHere, $1 + 99 = 2 + 98 = \\dots = 48 + 52 = 49 + 51 = 100$, so this sequence is 50-composite.\n\nIf $k > 50$, then some groups would contain only one element, forcing two numbers to be equal, which is not allowed since all numbers must be distinct.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16533,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $(m, n)$ of positive integers such that both $k = \\frac{3n^2}{m}$ and $\\sqrt{n^2 + m}$ are integers.",
    "options": [],
    "answer": "See solution",
    "solution": "All pairs $(m, n) = (3n^2, n)$, where $n$ is a positive integer.\n\nLet $k = \\frac{3n^2}{m}$ and require both $k$ and $\\sqrt{n^2 + m}$ to be integers. Since $k$ is integer, $n\\sqrt{\\frac{k+3}{k}}$ must also be integer. This implies $k(k+3)$ is a perfect square. For $k > 1$:\n\n$$\n(k+1)^2 = k^2 + 2k + 1 < k(k+3) < k^2 + 4k + 4 = (k+2)^2\n$$\n\nSo $k(k+3)$ cannot be a perfect square. If $k = 1$, then $\\sqrt{n^2 + m} = \\sqrt{4n^2} = 2n$ is integer. Thus, the only solutions are all pairs $(m, n) = (3n^2, n)$, where $n$ is a positive integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16534,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to find four distinct prime numbers for which the sum of any three of them is also a prime number?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes.\n\nSuitable prime numbers are $5$, $7$, $17$, and $19$. The sums of the corresponding triplets are prime numbers: $5 + 7 + 17 = 29$, $5 + 7 + 19 = 31$, $5 + 17 + 19 = 41$, and $7 + 17 + 19 = 43$.\n\n**Remark 1.** Any four numbers that include $2$ or $3$ violate the given conditions. The sum of $2$ with two odd numbers would be even and therefore not a prime number. If one of the four numbers is $3$, then if there are two primes incongruent modulo $3$ among the remaining three, the sum of those two and $3$ would be divisible by $3$. If the other three primes are all equal modulo $3$, then their sum would also be divisible by $3$.\n\n**Remark 2.** It is also possible to find four consecutive prime numbers that satisfy the conditions, for example $19$, $23$, $29$, and $31$. The corresponding sums of the triplets are $71$, $73$, $79$, and $83$, which are consecutive prime numbers as well.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16535,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $x$ such that $3^x + 3^{[x]} + 3^{\\{x\\}} = 4$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the possible ranges for $x$:\n\n- If $x \\geq 1$, then $[x] \\geq 1$ and $\\{x\\} \\in [0, 1)$. Thus,\n  $$3^x + 3^{[x]} + 3^{\\{x\\}} \\geq 3 + 3 + 1 = 7 > 4,$$\n  so no solutions exist for $x \\geq 1$.\n\n- If $x < -1$, then $[x] \\leq -2$ and $\\{x\\} \\in [0, 1)$. Thus,\n  $$3^x + 3^{[x]} + 3^{\\{x\\}} < 3^{-1} + 3^{-2} + 3^1 = \\frac{1}{3} + \\frac{1}{9} + 3 = \\frac{4}{9} + 3 < 4,$$\n  so no solutions exist for $x < -1$.\n\n- If $x \\in [0, 1)$, then $[x] = 0$ and $\\{x\\} = x$. The equation becomes:\n  $$3^x + 1 + 3^x = 4 \\implies 2 \\cdot 3^x = 3 \\implies 3^x = \\frac{3}{2} \\implies x_1 = 1 - \\log_3 2 \\in [0, 1).$$\n\n- If $x \\in [-1, 0)$, then $[x] = -1$ and $\\{x\\} = x + 1$. The equation becomes:\n  $$3^x + \\frac{1}{3} + 3^{x+1} = 4 \\implies 3^x + \\frac{1}{3} + 3 \\cdot 3^x = 4 \\implies 4 \\cdot 3^x = \\frac{11}{3} \\implies 3^x = \\frac{11}{12} \\implies x_2 = \\log_3 \\frac{11}{12} \\in [-1, 0).$$\n\n**Final answer:**\n\nThe solutions are $x_1 = 1 - \\log_3 2$ and $x_2 = \\log_3 \\frac{11}{12}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16536,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{Z}^2 \\to [0, 1]$ such that for any integers $x$ and $y$,\n\n$$\nf(x, y) = \\frac{f(x - 1, y) + f(x, y - 1)}{2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**First solution (hands-on)**\n\nFirst, iterating the functional equation to the $n$th level shows that\n\n$$\nf(x, y) = \\frac{1}{2^n} \\sum_{i=0}^{n} \\binom{n}{i} f(x-i, y-(n-i)).\n$$\n\nIn particular,\n\n$$\n\\begin{align*}\n|f(x,y) - f(x-1, y+1)| &= \\frac{1}{2^n} \\left| \\sum_{i=0}^{n+1} f(x-i, y-(n-i)) \\cdot \\left( \\binom{n}{i} - \\binom{n}{i-1} \\right) \\right| \\\\\n&\\le \\frac{1}{2^n} \\sum_{i=0}^{n+1} \\left| \\binom{n}{i} - \\binom{n}{i-1} \\right| \\\\\n&= \\frac{1}{2^n} \\cdot 2 \\binom{n}{\\lfloor n/2 \\rfloor}\n\\end{align*}\n$$\n\nwhere we define $\\binom{n}{n+1} = \\binom{n}{-1} = 0$ for convenience. Since\n\n$$\n\\binom{n}{\\lfloor n/2 \\rfloor} = o(2^n)\n$$\n\nit follows that $f$ must be constant.\n\n**Remark.** A very similar proof extends to $d$ dimensions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16537,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the largest possible integer $k$ such that the following statement is true:\n\nLet $2010$ arbitrary non-degenerate triangles be given. In every triangle, the three sides are colored so that one is blue, one is red, and one is white. For each color separately, sort the lengths of the sides:\n\n- $b_1 \\leq b_2 \\leq \\dots \\leq b_{2010}$ (blue sides)\n- $r_1 \\leq r_2 \\leq \\dots \\leq r_{2010}$ (red sides)\n- $w_1 \\leq w_2 \\leq \\dots \\leq w_{2010}$ (white sides)\n\nThen, there exist $k$ indices $j$ such that we can form a non-degenerate triangle with sides of lengths $b_j$, $r_j$, $w_j$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that the largest possible number $k$ of indices satisfying the given condition is $1$.\n\nFirst, we show that $b_{2010}$, $r_{2010}$, $w_{2010}$ are always lengths of the sides of a triangle. Without loss of generality, assume $w_{2010} \\geq r_{2010} \\geq b_{2010}$. We need to show $b_{2010} + r_{2010} > w_{2010}$.\n\nThere exists a triangle with side lengths $w$, $b$, $r$ for the white, blue, and red sides, respectively, such that $w_{2010} = w$. By the problem's conditions, $b + r > w$, $b_{2010} > b$, and $r_{2010} > r$. Therefore,\n\n$$\nb_{2010} + r_{2010} > b + r > w = w_{2010}.\n$$\n\nNext, we construct a sequence of triangles for which $w_j$, $b_j$, $r_j$ with $j < 2010$ are not the lengths of the sides of a triangle. Define triangles $\\Delta_j$ for $j = 1, 2, \\dots, 2010$:\n\n- Blue side: $2j$\n- Red side: $j$ for $j \\leq 2009$, $4020$ for $j = 2010$\n- White side: $j+1$ for $j \\leq 2008$, $4020$ for $j = 2009$, $1$ for $j = 2010$\n\nFor $1 \\leq j \\leq 2009$, $w_j = j$, $r_j = j$, $b_j = 2j$, so\n\n$$\nw_j + r_j = j + j = 2j = b_j,\n$$\n\ni.e., $w_j$, $b_j$, $r_j$ are not the lengths of the sides of a triangle for $1 \\leq j \\leq 2009$.\n\nThus, the largest possible $k$ is $1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16538,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that it is impossible to have a cuboid for which the volume, the surface area, and the perimeter are numerically equal. The perimeter of a cuboid is the sum of the lengths of all its twelve edges.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the dimensions of the cuboid be $h$, $w$, and $l$.\n\nWe have:\n\n$$\n\\begin{aligned}\n\\text{volume} &= h w l, \\\\\n\\text{surface area} &= 2(hw + w l + l h), \\\\\n\\text{perimeter} &= 4(h + w + l).\n\\end{aligned}\n$$\n\nIf all three quantities are equal, then:\n\n$$\n\\text{volume} = \\text{surface area} = \\text{perimeter}\n$$\n\nSetting volume $=$ surface area:\n\n$$\nh w l = 2(hw + w l + l h)\n$$\n\nSetting surface area $=$ perimeter:\n\n$$\n2(hw + w l + l h) = 4(h + w + l)\n$$\n\nCombining these, we can derive:\n\n$$\n4 h w l (h + w + l) = 4 (h w + w l + l h)^2\n$$\n\nwhich simplifies to:\n\n$$\nh^2 w^2 + w^2 l^2 + l^2 h^2 + h w l (h + w + l) = 0\n$$\n\nThis equation cannot be satisfied for positive real values of $h$, $w$, and $l$ (since all terms are positive), so such a cuboid does not exist.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16539,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer and $M = \\{1, 2, \\dots, 2n + 1\\}$. Find the number of ways to split the set $M$ into three mutually disjoint nonempty sets $A$, $B$, and $C$ such that:\n\n1. For each $a \\in A$ and $b \\in B$, the remainder of the division of $a$ by $b$ belongs to $C$.\n2. For each $c \\in C$, there exist $a \\in A$ and $b \\in B$ such that $c$ is the remainder of the division of $a$ by $b$.",
    "options": [],
    "answer": "See solution",
    "solution": "We notice that $a > b$ for all $a \\in A$ and $b \\in B$. Otherwise, if $a < b$, the remainder of $a$ divided by $b$ is $a$, so $a$ would belong to both $B$ and $C$, which is not allowed. Thus, $A$ consists of consecutive numbers and $2n+1 \\in A$.\n\nTake $c \\in C$. By assumption, there exist $a \\in A$ and $b \\in B$ such that $a = bq + c$. Then $a \\geq (c+1) \\cdot 1 + c = 2c + 1$, so $2n+1 \\geq 2c+1$, which gives $n \\geq c$. If $n+1 \\notin B$, then $n+1 \\in A$, so $\\{n+1, n+2, \\dots, 2n+1\\} \\subset A$. For any $b \\in B$, $b \\leq n$, and since $A$ contains $n+1$ consecutive numbers, $A$ must contain a multiple of $b$, which is a contradiction.\n\nTherefore, $n+1 \\in B$. There exists $k \\in \\{n+1, n+2, \\dots, 2n\\}$ such that $A = \\{2n+1, 2n, \\dots, k+1\\}$. Since $c \\leq n$ for all $c \\in C$, we have $\\{n+1, n+2, \\dots, k\\} \\subset B$.\n\nThe remainders of dividing elements of $A$ by elements of $\\{n+1, n+2, \\dots, k\\}$ are $\\{1, 2, \\dots, n\\}$. Thus, $B = \\{n+1, n+2, \\dots, k\\}$ and $C = \\{1, 2, \\dots, n\\}$. The number $k$ can be any element of $\\{n+1, n+2, \\dots, 2n\\}$, so there are $n$ possible splits.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16540,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $1 = d_1 < d_2 < \\dots < d_k = n$ be all positive divisors of the number $n$. It is known that\n$$\nd_1 \\cdot d_k = d_2 \\cdot d_{k-1} = d_3 \\cdot d_{k-2} = n.\n$$\nWhat are all positive integers $n$ such that the product of all positive divisors of $n$ is equal to $n^3$?",
    "options": [],
    "answer": "See solution",
    "solution": "A positive integer $n$ can be written as $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_r^{\\alpha_r}$, where $p_1, p_2, \\dots, p_r$ are distinct primes and $\\alpha_1 \\geq \\alpha_2 \\geq \\dots \\geq \\alpha_r$ are positive integers.\n\nThe number of positive divisors of $n$ is $(\\alpha_1 + 1)(\\alpha_2 + 1)\\dots(\\alpha_r + 1)$. For the product of all divisors to be $n^3$, $n$ must have exactly six divisors:\n$$\n(\\alpha_1 + 1)(\\alpha_2 + 1)\\dots(\\alpha_r + 1) = 6.\n$$\nSince each factor is greater than $1$, the possibilities are:\n- $r = 1$, $\\alpha_1 = 5$ (so $n = p^5$ for a prime $p$), or\n- $r = 2$, $\\alpha_1 = 2$, $\\alpha_2 = 1$ (so $n = p^2q$ for distinct primes $p$ and $q$).\n\nThus, all such $n$ are of the form $n = p^5$ or $n = p^2q$ for primes $p, q$ with $p \\neq q$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16541,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R}^+ \\to \\mathbb{R}$ satisfying:\n\n1. $f(1) = 2008$.\n2. $|f(x)| \\leq x^2 + 1004^2$ for all $x > 0$.\n3. $$f\\left(x + y + \\frac{1}{x} + \\frac{1}{y}\\right) = f\\left(x + \\frac{1}{y}\\right) + f\\left(y + \\frac{1}{x}\\right)$$ for all $x, y > 0$.\n\nHere, $\\mathbb{R}$ is the set of all real numbers and $\\mathbb{R}^+$ is the set of all positive real numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $u = x + \\frac{1}{y}$ and $v = y + \\frac{1}{x}$ for $x, y > 0$. Then:\n$$\nf(u + v) = f(u) + f(v)\n$$\nNote that $uv = xy + 2 + \\frac{1}{xy} \\geq 4$. First, we show that for any $u, v > 0$ with $uv \\geq 4$, there exist $x, y > 0$ such that:\n$$\nu = x + \\frac{1}{y}, \\quad v = y + \\frac{1}{x}$$\nFrom this, $x = u - \\frac{1}{y} = \\frac{uy - 1}{y}$ and $v = y + \\frac{1}{x} = \\frac{uy^2}{uy - 1}$, leading to the quadratic $uy^2 - uvy + v = 0$. The discriminant $D = (uv)^2 - 4uv \\geq 0$ for $u, v > 0$, so there is a real $y > 0$ (and similarly $x > 0$) satisfying these equations. Thus, $f(u + v) = f(u) + f(v)$ holds for all $u, v > 0$ with $uv \\geq 4$.\n\nTo extend this to all $u, v > 0$, for any $u, v > 0$, choose $w > 0$ such that $(u + v)w \\geq 4$, $u(v + w) \\geq 4$, and $vw \\geq 4$. Then:\n$$\n\\begin{aligned}\nf(u + v + w) &= f(u + v) + f(w) \\\\\n&= f(u) + f(v + w) \\\\\n&= f(u) + f(v) + f(w)\n\\end{aligned}\n$$\nSo $f(u + v) = f(u) + f(v)$ for all $u, v > 0$.\n\nLet $h(x) = f(x) - f(1)x = f(x) - 2008x$. Then $h$ satisfies $h(u + v) = h(u) + h(v)$ and $h(x + 1) = h(x) + h(1) = h(x)$ for all $x > 0$, so $h(1) = 0$ and $h(x + 1) = h(x)$. Thus, $h$ is periodic with period $1$.\n\nFrom $|f(x)| \\leq x^2 + 1004^2$, we get:\n$$\n-(x + 1004)^2 \\leq h(x) \\leq (x - 1004)^2\n$$\nSo $h(x)$ is bounded on $(0, 1]$ and, by periodicity, on $\\mathbb{R}^+$. Since $h(nx) = n h(x)$ for all positive integers $n$, $h$ must be identically zero. Therefore, $f(x) = 2008x$ for all $x > 0$.\n\n$\\boxed{f(x) = 2008x}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16542,
    "subject": "Mathematics (Olympiad)",
    "question": "Deslizamos un cuadrado de $10\\ \\mathrm{cm}$ de lado por el plano $OXY$ de forma que los vértices de uno de sus lados estén siempre en contacto con los ejes de coordenadas, uno con el eje $OX$ y otro con el eje $OY$. Determina el lugar geométrico que en ese movimiento describen:\n\n1. El punto medio del lado de contacto con los ejes.\n2. El centro del cuadrado.",
    "options": [],
    "answer": "See solution",
    "solution": "Sean $PQRS$ el cuadrado de lado $10\\ \\mathrm{cm}$, $PQ$ el lado de apoyo, $M(m_1, m_2)$ el punto medio de dicho lado y $C(c_1, c_2)$ el centro del cuadrado, tal y como muestra la figura donde, además, señalamos los puntos $A, B, D$ y $E$.\n\n![](images/Spanija_b_2013_p10_data_d36293901f.png)\n\n**a) Caso del punto medio $M$:**\n\n$$\nOM = PM = \\frac{1}{2}PQ = 5,\n$$\n\nluego $m_1^2 + m_2^2 = 25$.\n\n**b) Caso del centro del cuadrado $C$:**\n\nLos triángulos $AQM$, $AOM$, $BMO$ y $DMC$ son claramente congruentes:\n\n$$\nAM = OB = DC, \\quad AQ = OA = MD = BM, \\quad OM = MQ = MC = 5.\n$$\n\nAsí, resulta que las coordenadas del centro del cuadrado, en su deslizamiento, son iguales:\n\n$$\nc_1 = OE = OB + BE = m_1 + MD = m_1 + m_2,\n$$\n$$\nc_2 = EC = ED + DC = OA + AM = m_2 + m_1.\n$$\n\nLuego, el centro del cuadrado se mueve, en este primer cuadrante, sobre un segmento de la línea. Las posiciones extremas se dan cuando el lado $PQ$ se apoya sobre alguno de los ejes, $C(5, 5)$, y cuando forma una escuadra, esto es, un triángulo rectángulo isósceles, con ellos, $C(5\\sqrt{2}, 5\\sqrt{2})$. Trabajando análogamente en los demás cuadrantes podemos afirmar que el centro del cuadrado recorre el segmento de sus bisectrices que viene dado por la expresión:\n\n$$\nC(c_1, c_2) = (\\pm 5\\lambda, \\pm 5\\lambda) \\quad \\text{con } \\lambda \\in [1, \\sqrt{2}].\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16543,
    "subject": "Mathematics (Olympiad)",
    "question": "For real numbers $x$ and $y$, define $M(x, y)$ to be the maximum of the three numbers $xy$, $(x-1)(y-1)$, and $x+y-2xy$. Determine the smallest possible value of $M(x, y)$ where $x$ and $y$ range over all real numbers satisfying $0 \\le x, y \\le 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that the minimum value is $\\frac{4}{9}$. This value can be attained by taking $x = y = \\frac{2}{3}$. Then we have $xy = \\frac{4}{9}$, $(x-1)(y-1) = \\frac{1}{9}$, and $x+y-2xy = \\frac{4}{9}$, and the maximum is indeed $\\frac{4}{9}$.\n\nNow we will prove that $M(x, y) \\ge \\frac{4}{9}$ for all $x$ and $y$. Let $a = xy$, $b = (x-1)(y-1)$, and $c = x+y-2xy$. If we replace $x$ and $y$ by $1-x$ and $1-y$, then $a$ and $b$ will be interchanged and $c$ stays the same, because $(1-x)+(1-y)-2(1-x)(1-y) = x+y-2xy$. Hence, $M(1-x, 1-y) = M(x, y)$. We have $x+y = 2-(1-x)-(1-y)$, so at least one of $x+y$ and $(1-x)+(1-y)$ is greater than or equal to $1$, which means that we may assume without loss of generality that $x+y \\ge 1$.\n\nNow write $x+y = 1+t$ with $t \\ge 0$. We also have $t \\le 1$, because $x, y \\le 1$ and hence $x+y \\le 2$. The inequality between the arithmetic and geometric mean yields\n\n$$\nxy \\le \\left(\\frac{x+y}{2}\\right)^2 = \\frac{(1+t)^2}{4} = \\frac{t^2+2t+1}{4}.\n$$\n\nWe have $b = xy - x - y + 1 = xy - (1+t) + 1 = xy - t = a - t$, hence $b \\le a$. Moreover,\n\n$$\nc = x + y - 2xy \\ge (1+t) - 2 \\cdot \\frac{t^2 + 2t + 1}{4} = \\frac{2+2t}{2} - \\frac{t^2 + 2t + 1}{2} = \\frac{1-t^2}{2}.\n$$\n\nIf $t \\le \\frac{1}{3}$, then we have $c \\ge \\frac{1-t^2}{2} \\ge \\frac{1-\\frac{1}{9}}{2} = \\frac{4}{9}$ and hence $M(x, y) \\ge \\frac{4}{9}$ as well. The remaining case is $t > \\frac{1}{3}$. We have $c = x+y-2xy = 1+t-2a > \\frac{4}{3}-2a$. Moreover, $M(x, y) \\ge \\max(a, \\frac{4}{3}-2a)$, hence\n\n$$\n3M(x, y) \\ge a + a + \\left(\\frac{4}{3} - 2a\\right) = \\frac{4}{3},\n$$\n\nwhich yields that $M(x, y) \\ge \\frac{4}{9}$.\n\nWe conclude that the minimum value of $M(x, y)$ is $\\frac{4}{9}$.\n\n![](images/NLD_ABooklet_2022_p26_data_9de4fc982a.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16544,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a polynomial $P(x)$ with integer coefficients such that\n\n$$\nP(1 + \\sqrt[3]{2}) = 1 + \\sqrt[3]{2} \\quad \\text{and} \\quad P(1 + \\sqrt{5}) = 2 + 3\\sqrt{5}?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that there exists such a polynomial $P(x)$. Let $Q(x) = P(1 + x) - 1$, then $Q(x) \\in \\mathbb{Z}[x]$. We have $Q(\\sqrt[3]{2}) = \\sqrt[3]{2}$ and $Q(\\sqrt{5}) = 1 + 3\\sqrt{5}$.\n\nTherefore, $Q(x) - x$ has an irrational root $\\sqrt[3]{2}$. Since $x^3 - 2$ is irreducible over $\\mathbb{Z}[x]$ and has the same root $\\sqrt[3]{2}$, we get $x^3 - 2 \\mid Q(x) - x$. Then there exists $R(x)$ with integer coefficients such that\n\n$$\nQ(x) - x = (x^3 - 2)R(x).\n$$\n\nBecause $R(x) \\in \\mathbb{Z}[x]$, $R(\\sqrt{5}) = a + b\\sqrt{5}$ for some $a, b \\in \\mathbb{Z}$. Let $x = \\sqrt{5}$, then:\n\n$$\n1 + 3\\sqrt{5} = (5\\sqrt{5} - 2)(a + b\\sqrt{5}) = 25b - 2a + (5a - 2b)\\sqrt{5},\n$$\n\nThis implies $5a - 2b = 3$ and $25b - 2a = 1$, which does not have integer solutions $(a, b)$, a contradiction. So the answer is no.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16545,
    "subject": "Mathematics (Olympiad)",
    "question": "What should be the value of the parameter $\\alpha$ in order that the equation\n$$\n|x - \\frac{1}{2}| + |x - \\sin \\alpha| = \\cos 3\\alpha\n$$\nhas a single solution? Find this solution.",
    "options": [],
    "answer": "See solution",
    "solution": "The graph of the function $y = |x - a| + |x - b|$ for $b > a$ is shown below:\n\n![](images/Ukrajina_2008_p9_data_5272ac3a3a.png)\n\nThus, the equation $|x - a| + |x - b| = c$ may have a single solution only if $a = b$ and $c = 0$:\n\n![](images/Ukrajina_2008_p9_data_3f7f6d8bd0.png)\n\nTherefore, for our equation, the following conditions must be satisfied:\n$$\n\\begin{cases}\n\\sin \\alpha = \\frac{1}{2} \\\\\n\\cos 3\\alpha = 0\n\\end{cases}\n$$\nThis leads to\n$$\n\\begin{cases}\n\\alpha = (-1)^n \\frac{\\pi}{6} + \\pi n, \\quad n \\in \\mathbb{Z} \\\\\n\\alpha = \\frac{\\pi}{6} + \\frac{1}{3} \\pi k\n\\end{cases}\n$$\nThe common solution is $\\alpha = (-1)^n \\frac{\\pi}{6} + \\pi n$, $n \\in \\mathbb{Z}$, and $x = \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16546,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ be a set with $|A| = 225$, meaning that $A$ has 225 elements. Suppose there are eleven subsets $A_1, \\dots, A_{11}$ of $A$ such that $|A_i| = 45$ for $1 \\leq i \\leq 11$ and $|A_i \\cap A_j| = 9$ for $1 \\leq i < j \\leq 11$. Prove that $|A_1 \\cup A_2 \\cup \\dots \\cup A_{11}| \\geq 165$, and give an example for which equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S$ be the complement of $A_1 \\cup A_2 \\cup \\dots \\cup A_{11}$ in $A$; we wish to prove that $|S| \\leq 60$. For $\\ell \\geq 0$, define\n\n$$\n\\theta(\\ell) = \\left(1 - \\frac{\\ell}{2}\\right) \\left(1 - \\frac{\\ell}{3}\\right) = 1 - \\frac{2}{3}\\ell + \\frac{1}{3}\\binom{\\ell}{2}.\n$$\n\nNote that $\\theta(0) = 1$ and $\\theta(\\ell) \\geq 0$ for any integer $\\ell > 0$. For $n \\in A$, let $\\ell(n)$ be the number of sets among $A_1, \\dots, A_{11}$ containing $n$. Since $S$ is the intersection of the complements of the $A_i$, we see that\n\n$$\n|S| \\leq \\sum_{n \\in A} \\theta(\\ell(n)).\n$$\n\nOn the other hand,\n\n$$\n\\sum_{n \\in A} \\theta(\\ell(n)) = \\sum_{n \\in A} \\left(1 - \\frac{2}{3}\\ell(n) + \\frac{1}{3}\\binom{\\ell(n)}{2}\\right) = |A| - \\frac{2}{3}\\sum_{i} |A_i| + \\frac{1}{3}\\sum_{i<j} |A_i \\cap A_j|.\n$$\n\nPutting these together,\n\n$$\n|S| \\leq 225 - \\frac{2}{3} \\cdot 11 \\cdot 45 + \\frac{1}{3} \\cdot \\binom{11}{2} \\cdot 9 = 60,\n$$\n\nand therefore $|A_1 \\cup A_2 \\cup \\dots \\cup A_{11}| \\geq 165$.\n\nTo show equality is possible, let $p_1, p_2, \\dots, p_{11}$ be 11 distinct primes, and let $A'$ be the set of all products of three of these primes. Let $A'' = \\{q_1, q_2, \\dots, q_{60}\\}$ be 60 distinct positive integers coprime to $p_1, \\dots, p_{11}$. Set $A = A' \\cup A''$, and define\n\n$$\nA_i = \\{n \\in A' : p_i \\mid n\\}.\n$$\n\nThen $|A_i| = \\binom{10}{2} = 45$, $|A_i \\cap A_j| = \\binom{9}{1} = 9$, and\n\n$$\n|A_1 \\cup \\dots \\cup A_{11}| = |A'| = \\binom{11}{3} = 165.\n$$\n\nFinally, $|A| = 165 + 60 = 225$, so this $A$ and $A_1, \\dots, A_{11}$ give a valid example.\n\n_Remark_: To get an upper bound for $|S|$, one could replace $\\theta(\\ell)$ by any function of the form\n\n$$\n\\left(1 - \\frac{\\ell}{r}\\right) \\left(1 - \\frac{\\ell}{r + 1}\\right)\n$$\n\nfor any positive integer $r$. The choice $r = 2$ is optimal for the stated problem. The choice $r = 1$ yields\n\n$$\n|S| \\leq |A| - \\sum_{i} |A_i| + \\sum_{i<j} |A_i \\cap A_j|,\n$$\n\nwhich is a familiar truncated inclusion-exclusion inequality, known as Brun’s Pure Sieve or Bonferroni’s Inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16547,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $h$ be a semicircle with diameter $AB$. The two circles $k_1$ and $k_2$, $k_1 \\neq k_2$, touch the segment $AB$ at the points $C$ and $D$, respectively, and the semicircle $h$ from the inside at the points $E$ and $F$, respectively. Prove that the four points $C$, $D$, $E$, and $F$ lie on a circle.",
    "options": [],
    "answer": "See solution",
    "solution": "We first consider the case where $C$ and $D$ are both not the center of $AB$, so that the tangents at $C$ and $D$ are both not parallel to $AB$.\n\nThe tangent at $E$ intersects $AB$ at $X$, the tangent at $F$ intersects $AB$ at $Y$, and the two tangents intersect each other at $Z$. Let $I$ be the intersection point of the angle bisectors of $\\angle XYZ$ and $\\angle ZXY$. Since the tangent segments $XC$ and $XE$ to $k_1$ are of equal length and $C$, $E$ lie on the legs of the angle $\\angle ZXY$, $C$ and $E$ are equidistant from $I$.\n\nThe same applies to $D$ and $F$ with the circle $k_2$, and $E$ and $F$ with the semicircle $h$.\n\nThis means that the four points lie on a circle with center $I$.\n\nIn the remaining special case that $k_1$ passes through the center of $AB$, we can still define $Y$ as the intersection of the tangent at $F$ with $AB$, and $Z$ as the intersection of the tangents at $E$ and $F$. We define $I$ as the intersection of the angle bisectors of $\\angle ZYD$ and $\\angle EZY$. Therefore, $I$ has the same distance to $DY$ and $YZ$, and the same distance to $EZ$ and $YZ$. This means that $I$ also has the same distance to the parallel lines $DY$ and $EZ$. Thus, $I$ lies on the perpendicular bisector of $CE$ and, thus, $IC = IE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16548,
    "subject": "Mathematics (Olympiad)",
    "question": "Grandpa has a finite number of empty dustbins in his attic. Each dustbin is a rectangular parallelepiped with integral side lengths. A dustbin can be thrown away into another dustbin if and only if the side lengths of these dustbins can be set into one-to-one correspondence in such a way that the side lengths of the first dustbin are less than the corresponding side lengths of the other dustbin. For saving space, grandpa wants to throw away as many dustbins as possible. He has the following plan: while there exist some dustbins that can be thrown into each other, find the longest chain of such dustbins (i.e., the first can be thrown into the second, the second into the third, etc.) and throw all of them except the largest one sequentially into the next one, then do the same with the remaining dustbins, etc. Is it necessarily true that as many dustbins as possible will be thrown away as the result of this process, if it is known that at each step there is a unique way of choosing the longest chain of dustbins throwable into each other among the free dustbins and no dustbin is large enough to contain two other dustbins unless they have been placed into one another?\n\n*Answer:* No.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose grandfather has 6 dustbins with sizes $20 \\times 20 \\times 20$, $19 \\times 19 \\times 19$, $16 \\times 16 \\times 16$, $21 \\times 18 \\times 15$, $18 \\times 15 \\times 12$, and $17 \\times 14 \\times 11$. The first dustbin can contain the second one, the second can contain the third or the fifth, and the fifth can contain the sixth. The fourth also can contain the fifth. It is impossible to throw the first and the fourth into each other, the second and the fourth into each other, the third and the fourth into each other, the third and the fifth into each other, or the third and the sixth into each other. The longest chain of dustbins that can be thrown into each other is 4 dustbins: the sixth can be thrown into the fifth, which can be thrown into the second, which can be thrown into the first. As the remaining two dustbins cannot be thrown into each other, 3 dustbins in total are not thrown away. However, by throwing the third dustbin into the second, the second into the first, the sixth into the fifth, and the fifth into the fourth, only 2 dustbins are not thrown away. Hence, grandfather's algorithm does not provide an optimal solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16549,
    "subject": "Mathematics (Olympiad)",
    "question": "魔術師準備表演一個魔術。她會先指定一個正整數 $n$，以及 $2n$ 個實數 $x_1 < x_2 < \\cdots < x_{2n}$。然後隨機地請一位觀眾秘密地寫下一個 $n$ 次實係數多項式 $P(x)$ 並計算 $P(x_1), \\dots, P(x_{2n})$ 的值，再以非遞減順序在黑板上寫下這 $2n$ 個值。最後，她必須從這 $2n$ 個值公布正確的 $P(x)$。\n\n確定所有的 $n$ 使得魔術師可以找到策略來執行這樣的魔術。",
    "options": [],
    "answer": "See solution",
    "solution": "不存在這樣的 $n$。\n\n令 $x_1 < x_2 < \\cdots < x_{2n}$ 為魔術師選定的實數。我們將構造兩個不同的 $n$ 次多項式 $P(x)$ 和 $Q(x)$，使得觀眾寫下的 $2n$ 個值（按非遞減順序）對這兩個多項式是相同的，這表示魔術師無法區分 $P$ 和 $Q$。\n\n**Claim.** 存在一個 $n$ 次多項式 $P(x)$，使得 $P(x_{2i-1}) + P(x_{2i}) = 0$，對所有 $i = 1, 2, \\dots, n$。\n\n*Proof.* 我們希望找到一個多項式 $a_n x^n + \\cdots + a_1 x + a_0$，使其滿足以下方程組：\n\n$$\n\\begin{cases}\n(x_1^n + x_2^n)a_n + (x_1^{n-1} + x_2^{n-1})a_{n-1} + \\cdots + 2a_0 = 0 \\\\\n(x_3^n + x_4^n)a_n + (x_3^{n-1} + x_4^{n-1})a_{n-1} + \\cdots + 2a_0 = 0 \\\\\n\\vdots \\\\\n(x_{2n-1}^n + x_{2n}^n)a_n + (x_{2n-1}^{n-1} + x_{2n}^{n-1})a_{n-1} + \\cdots + 2a_0 = 0\n\\end{cases}\n$$\n\n由於這是一個 $n$ 個方程、$n+1$ 個未知數的齊次線性方程組，必有非零解（可用消元法歸納證明）。因此存在一個非零多項式 $P(x)$，其係數 $a_0, \\dots, a_n$ 滿足上述方程組，即 $P(x_{2i-1}) + P(x_{2i}) = 0$ 對所有 $i$。由介值定理可知，$P$ 在每個區間 $[x_{2i-1}, x_{2i}]$ 上有一個根，共 $n$ 個根。由於 $P$ 非零，故 $\\deg P = n$。\n\n考慮上述 $P(x)$，令 $Q(x) = -P(x)$。則 $P(x_{2i-1}) = Q(x_{2i})$ 且 $Q(x_{2i-1}) = P(x_{2i})$，對所有 $i$。顯然 $P \\neq Q$ 且 $\\deg Q = \\deg P = n$。\n\n**補充說明：** 對於任意正整數 $n$，魔術師可以選擇 $2n+1$ 個不同的實數來完成這個魔術。事實上，對於幾乎所有（在適當意義下）$(2n+1)$ 元組的數都可以完成。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16550,
    "subject": "Mathematics (Olympiad)",
    "question": "An equilateral triangle with side length 3 is divided into 9 equilateral triangles with side length 1. An integer from 1 to 10 is written into every point that is a vertex of a small triangle (colored vertices on the figure), such that all numbers are written exactly once. For every small triangle, the sum of the numbers in its three vertices is written inside it. Prove that at least three of those sums are greater than 11.\n\n![](images/Estonija_2012_p13_data_0a81876340.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Consider a small triangle that has $10$ at one of its vertices. The sum of its vertices is at least $10 + 1 + 2 = 13$. If $10$ is not at a vertex of the large triangle, then there are at least $3$ small triangles with sums greater than $12$, so the claim is proved.\n\nIf $10$ is at a vertex of the large triangle, consider where $9$ is placed. If $9$ is not at a vertex of the large triangle, then again, there are at least $3$ small triangles with sums at least $12$, so the claim holds.\n\nIf $9$ is also at a vertex of the large triangle, consider $8$. If $8$ is not at a vertex of the large triangle, then it is a vertex of at least $3$ small triangles. In at least two of these, the sum is at least $12$, and at most one of these can overlap with a triangle already counted, so the claim is proved.\n\nIf $8$ is also at a vertex of the large triangle, then either the sum in that triangle is at least $12$ (so the claim holds), or the other two vertices are $1$ and $2$. In this last case, the numbers $3$, $4$, and $5$ are the smallest remaining, and any triangle formed with these will have a sum of at least $12$. Thus, in all cases, there are at least three small triangles whose vertex sums are greater than $11$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16551,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and $N$ be a point different from $A$, $B$, and $C$. Let $A_b$ be the reflection of $A$ through $NB$, and $B_a$ be the reflection of $B$ through $NA$. Define $B_c$, $C_b$, $A_c$, $C_a$ similarly. Let $m_a$ be the line through $N$ and perpendicular to $B_cC_b$. Define $m_b$, $m_c$ similarly.\n\n**a)** Assume that $N$ is the orthocenter of $\\triangle ABC$. Show that the respective reflections of $m_a$, $m_b$, $m_c$ through each bisector of angles $\\angle BNC$, $\\angle CNA$, $\\angle ANB$ are the same line.\n\n**b)** Assume that $N$ is the nine-point center of $\\triangle ABC$. Show that the respective reflections of $m_a$, $m_b$, $m_c$ through $BC$, $CA$, $AB$ concur.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we present two lemmas. Let $ABC$ be a triangle inscribed in $(O)$ and $N$ be the nine-point center of $\\triangle ABC$.\n\n**Lemma 1.** Let $K$ be the center of $(BOC)$. Then $AN$ and $AK$ are isogonal with respect to $\\angle BAC$.\n\n**Lemma 2.** Let $B'$, $C'$ be the reflections of $B$, $C$ through $AC$, $AB$, respectively. Then $AK$ is perpendicular to $B'C'$.\n\n![](images/VN_booklet_2021_p33_data_96780ab607.png)\n\n**Proof.**\n\n1. Let $X$ be the reflection of $O$ through $BC$. Let $H$ be the orthocenter of $\\triangle ABC$. We have $AH = OX$, and $AH$ is parallel to $OX$. Hence, $AX$ cuts $OH$ at the midpoint of $OH$, which is $N$, and thus $N$ is the midpoint of $AX$.\n\nWe have $\\angle OXB = \\angle KOB = \\angle KBO$, so $\\triangle OKB \\sim \\triangle OBX$, implying that $\\frac{OK}{OB} = \\frac{OB}{OX}$, or\n\n$$\nOX \\cdot OK = OB^2 = OA^2.\n$$\n\nHence, $\\frac{OK}{OA} = \\frac{OA}{OX}$, implying that $\\triangle OKA \\sim \\triangle OAX$. Therefore, $\\angle OKA = \\angle XAO = \\angle DAK$.\n\nSince $AH$, $AO$ are isogonal with respect to $\\angle BAC$, we have $\\angle BAK = \\angle NAC$, implying that $AN$, $AK$ are isogonal with respect to $\\angle BAC$.\n\n2. Let $Y$, $Z$ be the reflections of $O$ through $CA$, $AB$. Similar to the argument above, $N$ is the midpoint of $BY$, $CZ$. Let $N_b$, $N_c$ be the reflections of $N$ through $CA$, $AB$. Since $OC'$ is the reflection of $ZC$ through $AB$, and $N$ is the midpoint of $ZC$, then $N_c$ is the midpoint of $OC'$. Similarly, $N_b$ is the midpoint of $OB'$. Hence, $N_bN_c$ is parallel to $B'C'$. We have\n\n$$\n\\angle KAN_b = \\angle N_bAC + \\angle KAC = \\angle NAC + \\angle NAB = \\angle BAC,\n$$\n\nand similarly, $\\angle KAN_c = \\angle BAC$. On the other hand, $AN_b = AN = AN_c$, and since $\\angle KAN_b = \\angle KAN_c$, we must have $AK$ perpendicular to $N_bN_c$. So $AK \\perp B'C'$, as desired.\n\nBack to the main problem, let $O_a$, $O_b$, $O_c$ be circumcenters of $\\triangle NBC$, $\\triangle NCA$, $\\triangle NAB$ and $K_a$, $K_b$, $K_c$ be circumcenters of $\\triangle O_aBC$, $\\triangle O_bCA$, $\\triangle O_cAB$. By Lemma 2, $m_a$, $m_b$, $m_c$ are $NK_a$, $NK_b$, $NK_c$, respectively.\n\n**a)** If $N$ is the orthocenter $H$, let $m'_a$, $m'_b$, $m'_c$ be reflections of $m_a$, $m_b$, $m_c$ through the bisectors of angles $BHC$, $CHA$, $AHB$. Let $J$ be the nine-point center of $\\triangle ABC$. Note that $(J)$ is also the nine-point circle of $\\triangle BHC$. Then, by Lemma 1, $HK_a$, $KJ$ are isogonal with respect to $\\angle BHC$, so $m'_a$ goes through $J$. Similarly, $m'_b$, $m'_c$ also go through $J$, and thus $m'_a \\equiv m'_b \\equiv m'_c \\equiv HJ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16552,
    "subject": "Mathematics (Olympiad)",
    "question": "a, b, c нь $a + b + c = 3$ байх сөрөг биш бодит тоонууд байг.\n\n$$\n\\left(a^2 b + \\frac{2}{3}\\right) \\left(b^2 c + \\frac{2}{3}\\right) + \\left(b^2 c + \\frac{2}{3}\\right) \\left(c^2 a + \\frac{2}{3}\\right) \\\\\n+ \\left(c^2 a + \\frac{2}{3}\\right) \\left(a^2 b + \\frac{2}{3}\\right) \\le \\frac{75}{4}\n$$\n\ngэнэ гэдгийг батал.",
    "options": [],
    "answer": "See solution",
    "solution": "Лемм: $a + b + c = 3$ бол $ab^2 + bc^2 + ca^2 \\le 4 - abc$ байна.\n\n$$\n\\left(a^2 b + \\frac{2}{3}\\right) \\left(b^2 c + \\frac{2}{3}\\right) + \\left(b^2 c + \\frac{2}{3}\\right) \\left(c^2 a + \\frac{2}{3}\\right) \\\\\n+ \\left(c^2 a + \\frac{2}{3}\\right) \\left(a^2 b + \\frac{2}{3}\\right) \\leq \\frac{75}{4} \\Leftrightarrow rA + 3B - 12 \\leq 0,\n$$\n\nэнд $r = abc$, $A = ab^2 + bc^2 + ca^2$, $B = a^2b + b^2c + c^2a$.\n\n$r \\le 1$ байх нь илэрхий. Мөн Шурын тэнцэтгэл бишээр\n\n$$\n(a + b + c)^3 \\geq 3abc + 4ab(a + b) + 4bc(b + c) + 4ca(c + a)\n$$\n\nбайна. Үүнээс $A + B \\le \\frac{27 - 3r}{4}$. Лемм ёсоор $A \\le 4 - r$ болно.\n\n$$\nrA + 3B - 12 = (r - 3)A + 3(A + B) - 12\n$$\n\n$$\n\\le (r - 3)(4 - r) + 3 \\cdot \\left(\\frac{27 - 3r}{4}\\right) - 12\n$$\n\n$$\n= \\frac{-4r^2 + 19r - 15}{4} = -(r - 1)(r - \\frac{15}{4}) \\le 0.\n$$\n\n$a = b = c = 1$ үед тэнцэл биелнэ.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16553,
    "subject": "Mathematics (Olympiad)",
    "question": "The positive real numbers $\\alpha, \\beta, \\gamma, \\delta$ satisfy the equality:\n\n$$\n\\alpha + \\beta\\gamma + \\gamma\\delta + \\delta\\beta + \\frac{1}{\\alpha\\beta^2\\gamma^2\\delta^2} = 18.\n$$\n\nFind the maximal possible value of $\\alpha$.",
    "options": [],
    "answer": "See solution",
    "solution": "Using the AM-GM inequality, we get\n\n$$\n\\begin{align*}\n\\alpha + \\beta\\gamma + \\gamma\\delta + \\delta\\beta + \\frac{1}{\\alpha\\beta^2\\gamma^2\\delta^2} &= \\alpha + \\left( \\beta\\gamma + \\gamma\\delta + \\delta\\beta + \\frac{1}{\\alpha\\beta^2\\gamma^2\\delta^2} \\right) \\\\\n&\\ge \\alpha + 4\\sqrt[4]{\\beta\\gamma \\cdot \\gamma\\delta \\cdot \\delta\\beta \\cdot \\frac{1}{\\alpha\\beta^2\\gamma^2\\delta^2}} = \\alpha + \\frac{4}{\\sqrt[4]{\\alpha}}\n\\end{align*}\n$$\n\nLet $x = \\sqrt[4]{\\alpha}$. The given equality leads to the inequality\n\n$$\nx^4 + \\frac{4}{x} \\le 18, \\quad x > 0 \\iff x^5 - 18x + 4 \\le 0, \\quad x > 0.\n$$\n\nSince $x = 2$ is a root of $P(x) = x^5 - 18x + 4$, we can factor:\n\n$$\nP(x) = (x-2)(x^4 + 2x^3 + 4x^2 + 8x - 2).\n$$\n\nFor $x > 2$, $(x-2)(x^4 + 2x^3 + 4x^2 + 8x - 2) > 0$, so $0 < x \\le 2$.\n\nAt $x = 2$, $\\alpha = 16$ and equality holds. By AM-GM, equality occurs when\n\n$$\n\\beta\\gamma = \\gamma\\delta = \\delta\\beta = \\frac{1}{\\alpha\\beta^2\\gamma^2\\delta^2}.\n$$\n\nThis leads to $\\beta = \\gamma = \\delta$ and $\\beta^8 = \\frac{1}{16}$, so $\\beta = \\gamma = \\delta = \\frac{1}{\\sqrt{2}}$.\n\nHence, the maximal possible value of $\\alpha$ is $16$.\n\n**Second solution.**\n\nFrom the AM–GM inequality, $a + bc + cd + db + \\frac{1}{ab^2c^2d^2} \\geq 36^{36} \\sqrt{\\frac{a^{32}}{32^{32}} \\cdot bc \\cdot cd \\cdot db \\cdot \\frac{1}{ab^2c^2d^2}} = 36^{36} \\sqrt{\\frac{a^{31}}{32^{32}}}$.\n\nThus, $\\alpha^{31} \\leq \\frac{32^{32}}{2^{36}} = 2^{124}$, and $a \\leq 2^4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16554,
    "subject": "Mathematics (Olympiad)",
    "question": "Label the columns of a $2022 \\times 2022$ table from bottom to top with the numbers $1$ to $2022$. Color all cells in rows with odd numbers yellow, and those in even-numbered rows blue. Each L-tetromino consists of three squares of one color and one square of another. Prove that at least one Z-tetromino must be obtained when tiling the board with L-tetrominoes and Z-tetrominoes.\n\nGive an example showing how exactly one Z-tetromino can be obtained.\n\n![](images/Belarus2022_p8_data_d38bfce9e6.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The numbers of yellow and blue cells are equal, so the number of L-tetrominoes with three yellow and one blue cell equals those with three blue and one yellow cell. Thus, the board is divided into an even number of L-tetrominoes, implying the total number of cells is a multiple of $8$, which is not true for a $2022 \\times 2022$ board. Therefore, at least one Z-tetromino must be used.\n\nFor example, the image shows a $6 \\times 6$ board cut into seven L-tetrominoes and one Z-tetromino. If such a $6 \\times 6$ board is cut from the corner of the $2022 \\times 2022$ board, the remainder can be tiled with $2 \\times 4$ and $4 \\times 2$ rectangles, each made up of two L-tetrominoes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16555,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Does there exist an integer $c$ and a polynomial $P(x)$ with integer coefficients for which $P(c) \\neq c$, but $P(P(c)) = c$?\n\nb) Does there exist an integer $c$ and a polynomial $P(x)$ with integer coefficients for which $P(c) \\neq c$ and $P(P(c)) \\neq c$, but $P(P(P(c))) = c$?",
    "options": [],
    "answer": "See solution",
    "solution": "a) Yes; b) No.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16556,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$, $Q$, $R$ be the orthocentres of three given triangles (in that order), each triangle being obtuse at $C$. Prove that the angle $\\angle PQR = 90^\\circ$.\n\n![](images/Cesko-Slovacko-Poljsko_2006_p1_data_5c04f4264c.png)\n\n![](images/Cesko-Slovacko-Poljsko_2006_p1_data_5a3a0b7339.png)",
    "options": [],
    "answer": "See solution",
    "solution": "In any obtuse triangle $XYZ$ with obtuse angle at $Z$ and orthocentre $W$, angles $XYZ$ and $XWZ$ are equal, as they complement angle $YXW$ to $90^\\circ$ (see Fig. 1). Moreover, points $Y$ and $W$ lie in different half-planes determined by $XZ$.\n\n![](images/Cesko-Slovacko-Poljsko_2006_p1_data_5c04f4264c.png)\n\nLet $P$, $Q$, $R$ be the orthocentres of the given triangles in that order. We will show $\\angle PQR = 90^\\circ$. Obviously, all three triangles are obtuse at $C$, so $P$, $Q$, $R$ lie on extensions of altitudes through $C$ to the corresponding sides. Because of the positions of these sides, it is also clear that the ray $CQ$ lies between rays $CP$ and $CR$, i.e., in angle $PCR$. So $\\angle PQR = \\angle RQC + \\angle PQC$ (see Fig. 2).\n\n![](images/Cesko-Slovacko-Poljsko_2006_p1_data_5a3a0b7339.png)\n\nBy the fact in the first paragraph, $Q$ and $R$ lie in the same half-plane determined by the line $BC$ and\n\n$$\n\\angle BEC = \\angle BRC \\quad \\text{and} \\quad \\angle BDC = \\angle BQC.\n$$\n\nAngles $BEC$, $BDC$ are equal, being inscribed angles with the same chord $BC$. Thus, $\\angle BRC = \\angle BQC = \\omega$ and $BCRQ$ is cyclic. So $\\angle RQC = \\angle RBC = \\varphi$. As $EC = r$ for the inscribed angle, we have $\\angle EBC = 30^\\circ$. Let $U$ be the foot on $BE$ in triangle $BEC$. Counting the angles in right triangle $BUR$, we get\n\n$$\n\\omega + \\varphi + 30^\\circ + 90^\\circ = 180^\\circ, \\quad \\text{i.e.} \\quad \\angle RQC = \\varphi = 60^\\circ - \\omega = 60^\\circ - \\angle BDC.\n$$\n\nIn the same way, we conclude $\\angle PQC = 60^\\circ - \\angle DBC$. So we have (using the sum of angles in triangle $BCD$ is $180^\\circ$):\n\n$$\n\\angle PQR = \\angle RQC + \\angle PQC = 120^\\circ - (\\angle BDC + \\angle DBC) = \\angle BCD - 60^\\circ. \\quad (1)\n$$\n\nBut also $BD = r$, thus $\\angle BCD = 150^\\circ$. Finally, by (1), $\\angle PQR = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16557,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b$ be real numbers with $0 \\le a, b \\le 1$. Prove the inequality\n\n$$\n\\sqrt{a^3 b^3} + \\sqrt{(1-a^2)(1-ab)(1-b^2)} \\le 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $0 \\le a, b \\le 1$, by the AM-GM inequality,\n\n$$\n\\begin{aligned}\n& \\sqrt{a^3 b^3} + \\sqrt{(1-a^2)(1-ab)(1-b^2)} \\\\\n& \\le \\sqrt[3]{a^3 b^3} + \\sqrt[3]{(1-a^2)(1-ab)(1-b^2)} \\\\\n& = \\sqrt[3]{a^2 \\cdot ab \\cdot b^2} + \\sqrt[3]{(1-a^2)(1-ab)(1-b^2)} \\\\\n& \\le \\frac{a^2 + ab + b^2}{3} + \\frac{(1-a^2) + (1-ab) + (1-b^2)}{3} \\\\\n& = 1.\n\\end{aligned}\n$$\n\nas claimed. Equality holds iff $a^2 = ab = b^2$ and either $a^3b^3 = 0$ or $a^3b^3 = 1$ (and similarly for $(1-a^2)(1-ab)(1-b^2)$). Thus, equality holds iff $a = b = 0$ or $a = b = 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16558,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p = ab$ and $s = a + b$. Show that\n\n$$\np^2(s^2 - 2p - 2) \\ge s(p - 1)\n$$\n\nor equivalently,\n\n$$\np^2 s^2 - (p - 1)s - 2p^2(p + 1) \\ge 0\n$$\n\nwhen $s^2 \\ge 4p > 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $0 < p \\le 1$, then $p^2 s - (p - 1) \\ge p^2 \\cdot 2\\sqrt{p} + (1 - p) > 0$. If $p \\ge 1$, then $p^2 s - (p - 1) \\ge (p^2 \\cdot 2\\sqrt{p} - p) + 1 > 0$.\n\nTherefore,\n\n$$\n\\begin{align*}\np^2 s^2 - (p - 1)s - 2p^2(p + 1) &= s(p^2 s - (p - 1)) - 2p^2(p + 1) \\\\\n&\\ge 2\\sqrt{p}(p^2 \\cdot 2\\sqrt{p} - p + 1) - 2p^2(p + 1) \\\\\n&= 2p^3 - 2p\\sqrt{p} - 2p^2 + 2\\sqrt{p} \\\\\n&= 2\\sqrt{p}(p - 1)(p\\sqrt{p} - 1) \\\\\n&\\ge 0.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16559,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be the midpoint of the side $BC$ of $\\triangle ABC$. On the sides $AB$ and $AC$, the points $F$ and $E$ are chosen. Let $K$ be the intersection point of $BF$ and $CE$. Let $L$ be chosen so that $CL \\parallel AB$ and $BL \\parallel CE$. Let $N$ be the intersection point of $AM$ and $CL$. Show that $KN$ is parallel to $FL$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $BECL$ is a parallelogram. Since $AB \\parallel CN$ and $M$ is the midpoint of $BC$, we have that $ABNC$ is a parallelogram and thus $AE = NL$ (see figure below).\n\nLet $P$ be the intersection point of $BF$ and $CN$. Since $FC \\parallel BN$, then $\\triangle PBN \\sim \\triangle PFC$.\n\nThus, $$\\frac{PF}{PB} = \\frac{PC}{PN}$$ so $PF = \\frac{PB \\cdot PC}{PN}$.\n\nSince $PC \\parallel BE$, we have $\\triangle PCK \\sim \\triangle BEK$. Therefore, $$\\frac{PK}{BK} = \\frac{PC}{BE}$$ so $PK = \\frac{BK \\cdot PC}{BE}$.\n\nThus, $$\\frac{PF}{PK} = \\frac{PB \\cdot BE}{BK \\cdot PN}$$.\n\nSince $\\triangle BLP \\sim \\triangle KEB$, we have $$\\frac{PL}{PB} = \\frac{BE}{BK}$$ and $$\\frac{PL}{PN} = \\frac{PB \\cdot BE}{BK \\cdot PN}$$.\n\nSo we have the proportion $$\\frac{PF}{PK} = \\frac{PL}{PN}$$ which leads to the similarity of $\\triangle PFL$ and $\\triangle PKN$. Hence, $\\triangle PFL \\sim \\triangle PKN$ and so $KN \\parallel FL$.\n\n![](images/Ukrajina_2013_p27_data_7358558f8f.png)\n\n*Fig. 20*",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16560,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the sequence $\\{a_n\\}_{n=1}^\\infty$ defined as follows:\n\n$$\na_1 = 3 \\quad a_n = a_1 a_2 a_3 \\dots a_{n-1} - 1 \\quad \\text{for all } n \\ge 2.\n$$\n\nProve that there exist:\n\na) Infinitely many primes dividing at least one member of this sequence.\n\nb) Infinitely many primes dividing no member of this sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "a) By mathematical induction, we first prove that $a_n \\ge 2$ for every $n$. For $n=1$ and $n=2$ this is true because $a_1 = 3$ and $a_2 = 2$. Now suppose that for some $n \\ge 3$ the inequality $a_k \\ge 2$ holds for every $k < n$. Then we have $a_n = a_1 a_2 a_3 \\dots a_{n-1} - 1 \\ge a_1 a_2 - 1 = 5$, so indeed $a_n \\ge 2$.\n\nLet us now show that the numbers $a_n$ are pairwise coprime. Indeed, for any two indices $k < n$ we have $a_k \\mid a_1 a_2 \\dots a_{n-1} = a_n + 1$, whence for the greatest common divisor $D$ of the numbers $a_n$ and $a_k$ we get $D \\mid a_n$ and at the same time $D \\mid a_n + 1$ (because $D \\mid a_k$ and $a_k \\mid a_n + 1$), so necessarily $D = 1$, so $a_n$ and $a_k$ are coprime. Due to $a_n \\ge 2$ we find for each index $n$ a prime number, denote it by $p_n$, for which $p_n \\mid a_n$. Since all $a_n$ are pairwise coprime, the prime numbers $p_n$ are pairwise different. Thus a) is proved.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16561,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be such that for all $x, y \\in \\mathbb{R}$,\n\n$$\n|f(x + y)| = |f(x) + f(y)|.\n$$\n\nProve that $f(x + y) = f(x) + f(y)$ for all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there exist $a, b \\in \\mathbb{R}$ such that $f(a + b) \\neq f(a) + f(b)$. By the assumption, $f(a + b) = -f(a) - f(b)$. If $f(a + b) = 0$, then $f(a) = -f(b)$.\n\n$$\n\\begin{align*}\n|f(2a + 2b)| &= |f(a + (a + b + b))| \\\\\n&= |f(a) + f((a + b) + b)| \\\\\n&= |f(a) + f(a + b) + f(b)| \\text{ or } |f(a) - f(a + b) - f(b)| \\\\\n&= 0 \\text{ or } |f(a) - (-f(a) - f(b)) - f(b)| \\\\\n&= 0 \\text{ or } 2|f(a)|.\n\\end{align*}\n$$\n\nBut $|f(2a + 2b)| = |f((a + b) + (a + b))| = |2f(a + b)| = 2|f(a + b)| \\neq 0$. Therefore, $2|f(a)| = |f(2a + 2b)| = 2|f(a + b)|$ and thus $|f(a + b)| = |f(a)|$. Similarly, $|f(a + b)| = |f(b)|$, so $|f(a)| = |f(b)|$. That is, $f(a) = f(b)$ or $f(a) = -f(b)$. If $f(a) = f(b)$, then $2|f(a)| = |f(a) + f(a)| = |f(a) + f(b)| = |f(a+b)| = |f(a)|$, so $f(a) = 0$, which implies $f(a + b) = -f(a) - f(b) = -2f(a) = 0$, a contradiction. If $f(a) = -f(b)$, then $0 = |f(a) + f(b)| = |f(a+b)|$, again a contradiction. Thus, in any case, we get a contradiction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16562,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $\\mathbb{Z}$ 為所有整數所成的集合。試確定所有的函數 $f : \\mathbb{Z} \\to \\mathbb{Z}$ 使得\n\n$$\nf\\left(\\left\\lfloor \\frac{f(x)+f(y)}{2} \\right\\rfloor\\right) + f(x) = f(f(y)) + \\left\\lfloor \\frac{f(x)+f(y)}{2} \\right\\rfloor\n$$\n\n對所有 $x, y \\in \\mathbb{Z}$ 都成立。\n\n註：$\\lfloor x \\rfloor$ 是不超過實數 $x$ 的最大整數。",
    "options": [],
    "answer": "See solution",
    "solution": "考慮集合\n\n$$\nS := \\{(x - y, f(x) - f(y)) \\mid x, y \\in \\mathbb{Z}\\}.\n$$\n\n顯然若 $(a, b) \\in S$ 則 $(-a, -b) \\in S$。此外，若 $(a, b) \\in S$ 且 $a = 0$，則 $b = 0$。我們還需要以下引理。\n\n**引理 1.** 若 $(a, b) \\in S$，則 $(\\lfloor \\frac{b}{2} \\rfloor, \\lceil -\\frac{b}{2} \\rceil) \\in S$。\n\n*證明.* 假設 $x - y = a$ 且 $f(x) - f(y) = b$，則有\n\n$$\n\\left\\lfloor \\frac{f(x)+f(y)}{2} \\right\\rfloor - f(y) = \\left\\lfloor \\frac{f(x)-f(y)}{2} \\right\\rfloor = \\left\\lfloor \\frac{b}{2} \\right\\rfloor\n$$\n\n以及\n\n$$\nf\\left(\\left\\lfloor \\frac{f(x)+f(y)}{2} \\right\\rfloor\\right) - f(f(y)) = \\left\\lfloor \\frac{f(x)+f(y)}{2} \\right\\rfloor - f(x) = \\left\\lfloor \\frac{-f(x)+f(y)}{2} \\right\\rfloor = -\\left\\lfloor \\frac{b}{2} \\right\\rfloor.\n$$\n\n因此 $(\\lfloor \\frac{b}{2} \\rfloor, \\lceil -\\frac{b}{2} \\rceil) \\in S$。$\\square$\n\n![](images/2022-TWNIMO-Problems_p19_data_0b8c0ef551.png)\n\n現在證明 $f$ 必為常數。若 $f$ 非常數，設 $(a, b) \\in S$ 使得 $|b|$ 為最小的非零值。由引理 1，$(\\lfloor \\frac{b}{2} \\rfloor, \\lceil -\\frac{b}{2} \\rceil) \\in S$。由於 $\\lfloor \\frac{b}{2} \\rfloor \\le |b|$，可得 $b = \\pm 1$。又因 $(-a, -b) \\in S$，不妨設 $b = 1$。引理 1 則得 $(0, 1) \\in S$，矛盾。因此 $f$ 必為常數。$\\square$\n\n![](images/2022-TWNIMO-Problems_p19_data_0d2e3255b4.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16563,
    "subject": "Mathematics (Olympiad)",
    "question": "For a prime $p$, a subset $S$ of residues modulo $p$ is called a *sum-free multiplicative subgroup* of $\\mathbb{F}_p$ if:\n\n* there is a nonzero residue $\\alpha$ modulo $p$ such that $S = \\{1, \\alpha^1, \\alpha^2, \\dots\\}$ (all considered mod $p$), and\n* there are no $a, b, c \\in S$ (not necessarily distinct) such that $a + b \\equiv c \\pmod{p}$.\n\nProve that for every integer $N$, there is a prime $p$ and a sum-free multiplicative subgroup $S$ of $\\mathbb{F}_p$ such that $|S| \\ge N$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** We prove a stronger statement, generalizing the desired condition \"$0 \\notin S + S - S$\" to \"$0 \\notin a_1S + a_2S + \\dots + a_kS$\", for fixed integers $a_1, \\dots, a_k$ with nonzero sum $a_1 + \\dots + a_k$. (In the original problem we have $(a_1, \\dots, a_k) = (1, 1, -1)$ (so $k=3$).\n\nFix a positive integer $N$ (we will specify further later), and take a large prime $p \\equiv 1 \\pmod N$ (we don't need Dirichlet—there are infinitely many by a cyclotomic polynomial argument, along the lines of using $x^2 + 1$ for $N=4$).\n\nAgain, we will specify the size of $p$ later. Now let $\\alpha$ be an $N$th root of unity modulo $p$ (i.e. $\\alpha = g^{(p-1)/N}$ for a primitive root $g$, so $\\alpha$ has order $N$). Then the key is the following lemma:\n\n**Lemma.** If the sumset $a_1S + a_2S + \\dots + a_kS$ contains $0 \\pmod p$ for arbitrarily large primes $p \\equiv 1 \\pmod N$, then there exist indices $i_1, i_2, \\dots, i_k$ between $0$ and $N-1$ such that the polynomial $f(x) = a_1x^{i_1} + \\dots + a_kx^{i_k}$ is divisible by the $N$th cyclotomic polynomial $\\Phi_N(x)$ (over $\\mathbb{Q}$, and thus $\\mathbb{Z}$).\n\n(We will use the irreducibility of $\\Phi_N$, but we only need special cases such as $N$ prime, where the proof is easy.)\n\n*Proof.* The idea is to “transfer” from $N$th roots of unity modulo $p$ to “actual” $N$th roots of unity. First, the sumset’s containing $0$ is equivalent to the existence of $0 \\le i_1, i_2, \\dots, i_k \\le N-1$ (since $\\alpha^N \\equiv 1 \\pmod p$) such that $f(\\alpha) \\equiv 0 \\pmod p$.\n\nWe present two (similar) ways to do this. One is to use the Mobius-inversion-type definition of cyclotomic polynomials to show that in $\\mathbb{F}_p$, the roots of the polynomial $\\Phi_N(x)$ are precisely the residues of order $N$; in particular, $p$ divides $\\Phi_N(\\alpha)$. Now by Bezout's identity, there exist integer polynomials $A(x), B(x)$ and a nonzero integer $C$ such that $C \\gcd(\\Phi_N(x), f(x)) = A(x)f(x) + B(x)\\Phi_N(x)$, where the $\\gcd$ (over $\\mathbb{Q}$) is, without loss of generality, monic. Assume for the sake of contradiction that $\\gcd(\\Phi_N(x), f(x))$ is constant, so, without loss of generality, identically $1$. Then plugging in $\\alpha$, we get $p$ divides $C$ for arbitrarily large primes $p \\equiv 1 \\pmod N$, which is absurd. Thus $f(x)$ and $\\Phi_N(x)$ share a complex root, and by the irreducibility of $\\Phi_N(x)$, $f(x)$ is divisible by $\\Phi_N(x)$. $(*)$\n\nAlternatively, by counting roots, it is easy to show that in $\\mathbb{F}_p$, we have the polynomial identity $x^N-1 \\equiv (x-\\alpha)\\dots(x-\\alpha^N)$. If $z$ is a primitive $N$th root of unity, then in $\\mathbb{C}$, we have $x^N-1 = (x-z)\\dots(x-z^N)$, so the symmetric sums of $z, \\dots, z^N$ are congruent modulo $p$ to those of $\\alpha, \\dots, \\alpha^N$. It follows by the theorem of symmetric sums that the product of $f(\\alpha)$ over all valid indices $0 \\le i_1, \\dots, i_k \\le N-1$ (each choice determines some $f$) is an integer congruent modulo $p$ to the (integer) product of $f(z)$ over all valid indices $0 \\le i_1, \\dots, i_k \\le N-1$. Thus arbitrarily large primes $p$ divide the integer $\\prod_{[0,N-1]^k} f(z)$, which must therefore be $0$. Thus there exists a choice of indices such that $f(z)$ is identically $0$, and thus the minimal polynomial $\\Phi_N(x)$ (again, we use irreducibility as in $(*)$) of $z$ divides $f(x)$. $\\square$\n\nWith the lemma in hand, the rest is easy. Choose $N=q$ any prime. Then $\\Phi_N(x) = \\Phi_q(x) = (x^q-1)/(x-1)$ divides $f(x) = a_1x^{i_1} + \\dots + a_kx^{i_k}$, a $k$-term polynomial that is either identically zero or otherwise a nonzero polynomial of degree at most $N-1=q-1$. However, if $f$ is not identically zero, and $q>k$, then since $(x^q-1)/(x-1) = x^{q-1} + \\dots + 1$ has degree $q-1$ (which is at least the degree of $f$), the coefficients of $f$ must all be equal. Yet $i_1, \\dots, i_k$ cannot cover all of $0, 1, \\dots, q-1$, so one of the coefficients of $f$ must be $0$, and therefore all must be $0$.\n\nIt follows that $f$ is identically zero, so $f(1) = a_1 + \\dots + a_k = 0$, contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16564,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle. Let $AA_1$ and $CC_1$ be its angle bisectors, $I$ the incenter of $ABC$, and $M$ and $N$ the midpoints of $AI$ and $CI$, respectively. Inside triangles $AC_1I$ and $A_1CI$, choose points $K$ and $L$ such that\n\n$\\angle AKI = \\angle CLI = \\angle AIC$, $\\angle AKM = \\angle ICA$, $\\angle CLN = \\angle IAC$.\n\nProve that the radii of the circumcircles of triangles $KIL$ and $ABC$ are equal.\n\n![](images/Ukraine_booklet_2018_p49_data_961e2cfe4c.png)\n\n**Fig. 46**",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that lines $AI$ and $CI$ meet the circumcircle of $\\triangle ABC$ (for the second time) at points $W_A$ and $W_C$, respectively (see Fig. 46). Since $\\angle AKI = \\angle AIC$, the circumcircle of triangle $AKI$ is tangent to the line $W_CI$. Consider triangle $AW_CI$. By a well-known fact, this triangle is isosceles. Thus, the circumcircle of $\\triangle AKI$ is also tangent to $W_CA$. Therefore, a symmedian of $\\triangle AKI$ lies along the line $W_CK$. Then $\\angle W_CKI = 180^\\circ - \\angle ICA = 180^\\circ - \\angle W_CW_AI$. This means that $K$ lies on the circumcircle of $\\triangle W_CW_AI$. Similarly, $L$ also lies on this circumcircle. Note that $\\triangle W_CB W_A \\cong \\triangle W_CIW_A$ (they have equal sides), so the radius of the circumcircle of $\\triangle KIL$ equals the radius of the circumcircle of $\\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16565,
    "subject": "Mathematics (Olympiad)",
    "question": "For any positive integers $n$ and $k$, let $L(n, k)$ be the least common multiple of the $k$ consecutive integers $n, n+1, \\dots, n+k-1$. Show that for any integer $b$, there exist integers $n$ and $k$ such that $L(n, k) > bL(n+1, k)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p > b$ be a prime, and set $n = p^3$, $k = p^2$. For $p^3 < i < p^3 + p^2$, no power of $p$ greater than 1 divides $i$, while $p$ divides $p^3 + p$. Thus, $L(p^3, p^2) = p^2 L(p^3 + 1, p^2 - 1)$. Similarly, $L(p^3 + 1, p^2) = p L(p^3 + 1, p^2 - 1)$. Therefore, $L(p^3, p^2) = p L(p^3 + 1, p^2) > b L(p^3 + 1, p^2)$.\n\nAlternatively, let $m > 1$. Then $L(m! - 1, m + 1)$ is the least common multiple of the integers from $m! - 1$ to $m! + m - 1$. Since $m! - 1$ is relatively prime to all of $m!, m! + 1, \\dots, m! + m - 1$, we have $L(m! - 1, m + 1) = (m! - 1) M$, where $M = \\text{lcm}(m!, m! + 1, \\dots, m! + m - 1)$.\n\nNow, $L(m!, m + 1) = \\text{lcm}(M, m! + m)$. Since $m! + m = m((m - 1)! + 1)$ and $m$ divides $M$, $\\text{lcm}(M, m! + m) \\leq M((m - 1)! + 1)$. Thus,\n\n$$\n\\frac{L(m! - 1, m + 1)}{L(m!, m + 1)} \\geq \\frac{m! - 1}{(m - 1)! + 1}.\n$$\n\nAs $m$ can be arbitrarily large, so can $L(m! - 1, m + 1)/L(m!, m + 1)$. Therefore, taking $n = m! - 1$ for sufficiently large $m$ and $k = m + 1$ works.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16566,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $8! = 40320$ eight-digit positive integers that use each of the digits $1, 2, 3, 4, 5, 6, 7, 8$ exactly once. Let $N$ be the number of these integers that are divisible by $22$. Find the difference between $N$ and $2025$.",
    "options": [],
    "answer": "See solution",
    "solution": "An integer is divisible by $22$ exactly when it is divisible by both $2$ and $11$. An eligible eight-digit integer $a\\underline{w}b\\underline{x}c\\underline{y}d\\underline{z}$ is divisible by $2$ exactly when $z$ is one of $2, 4, 6,$ or $8$. It is divisible by $11$ exactly when the alternating sum $a - w + b - x + c - y + d - z$ is congruent to $0$ modulo $11$. Because four of the digits $1, 2, 3, 4, 5, 6, 7, 8$ are even and the other four are odd, the alternating sum is always even. The maximum possible value of this sum is $8 - 1 + 7 - 2 + 6 - 3 + 5 - 4 = 16$, and the minimum value is $-16$. The only even multiple of $11$ in this range is $0$. Thus the number $a\\underline{w}b\\underline{x}c\\underline{y}d\\underline{z}$ is divisible by $11$ if and only if\n\n$$\na + b + c + d = w + x + y + z = \\frac{1}{2}(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 18.\n$$\n\nOne of the digit sets $\\{a, b, c, d\\}$ and $\\{w, x, y, z\\}$ must contain the digit $8$. There are $4$ ways to choose $3$ additional distinct digits from $\\{1, 2, 3, 4, 5, 6, 7\\}$ so that, together with $8$, their sum is $18$:\n\n$\\{8, 7, 2, 1\\}$, $\\{8, 6, 3, 1\\}$, $\\{8, 5, 4, 1\\}$, and $\\{8, 5, 3, 2\\}$.\n\nNote that each of these four sets as well as their complements contain two even digits and two odd digits. To choose an eight-digit integer divisible by $22$, there are the $4$ choices for a set with the digit $8$, and there are $2$ ways to choose whether the digits of this set are placed into the even or the odd positions. Then the $4$ digits in the odd positions can be permuted in $4! = 24$ ways. The $4$ digits in the even positions can be arranged by choosing one of the two even digits to be in the final position and then permuting the other three digits in one of $3!$ ways. Thus there are $2 \\cdot 3! = 12$ ways to permute the digits in the even positions. Therefore $N = 4 \\cdot 2 \\cdot 24 \\cdot 12 = 48^2$. The requested difference is $48^2 - 45^2 = 93 \\cdot 3 = 279$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16567,
    "subject": "Mathematics (Olympiad)",
    "question": "A $4 \\times 2012$ chessboard is colored in black and white alternately so that adjacent squares have different colors. Mark the squares in the first and fourth rows with a circle, and the squares in the second and third rows with a cross. Is it possible for a knight to visit each square exactly once and return to the initial square (i.e., form a closed knight's tour)?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose such a tour exists. The knight alternates colors with each move, so it must visit an equal number of black and white squares. The marking ensures that the knight always jumps from a circle to a cross and vice versa. However, following the color and marking pattern, the knight would never visit a white square marked with a cross, contradicting the assumption that every square is visited. Thus, a closed knight's tour is impossible.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16568,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that $n!$ is not divisible by $n^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* All primes and $4$.\n\n*Solution 1.* For any prime $p$, $p^2 \\nmid p!$ as prime $p$ occurs only once in the prime factorization of $p!$. Additionally, $4^2 = 16$ does not divide $4! = 24$.\n\nWe will show that $n^2 \\mid n!$ for all other positive integers $n$. Let $p$ be a prime factor of $n$, and $k$ the exponent of $p$ in the prime factorization of $n$. If there exists a different prime factor $q$ of $n$, then both $p^k$ and $p^k q$ are in the set of integers from $1$ to $n$ and hence $p^{2k} \\mid n!$. This holds for all prime factors $p$ of integer $n$. Thus $n^2$ divides $n!$ for all $n$ with at least two distinct prime factors.\n\nNow consider the remaining integers $n = p^k$ for prime $p$. For $k \\ge 3$ all three integers $p$, $p^{k-1}$, and $p^k$ are distinct factors in $n!$, implying that $p^{2k} = n^2$ divides $n!$. For $k = 2$ and $p > 2$ the integers $p$, $2p$, and $p^2$ are distinct factors in $n!$, implying that $p^4 = n^2$ divides $n!$.\n\n*Solution 2.* We find all positive integers $n$ such that $n \\mid (n-1)!$; obviously this condition is equivalent to that of the problem. For prime $n$, no integer less than $n$ can have a prime factor $n$ and thus $n$ cannot divide $(n-1)!$. For $n = 4 = 2^2$, $4$ does not divide $(n-1)! = 6$. On the other hand, for $n = p^2$ where $p > 2$ is prime, the integers $p$ and $2p$ are in the set of integers from $1$ to $n-1$, implying $p^2 = n \\mid (n-1)!$. If $n$ is a cube or a higher power of some prime, or has at least two prime factors, then it obviously has a divisor $d$ such that $1 < d < n$ and $d^2 \\neq n$ (e.g., the smallest prime factor of $n$). Hence $d$ and $\\frac{n}{d}$ are distinct positive integers less than $n$ and their product $n$ divides $(n-1)!$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16569,
    "subject": "Mathematics (Olympiad)",
    "question": "If\n$$\n\\frac{x_1}{x_1 + 1} = \\frac{x_2}{x_2 + 3} = \\frac{x_3}{x_3 + 5} = \\dots = \\frac{x_{1006}}{x_{1006} + 2011},\n$$\n$$\nx_1 + x_2 + \\dots + x_{1006} = 503^2,\n$$\ndetermine $x_{1006}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us denote\n$$\n\\frac{x_1}{x_1+1} = \\frac{x_2}{x_2+3} = \\frac{x_3}{x_3+5} = \\dots = \\frac{x_{1006}}{x_{1006}+2011} = a.\n$$\nFrom $\\frac{x_k}{x_k + (2k-1)} = a$ it follows that $x_k = \\frac{a}{1-a} \\cdot (2k-1)$ for $k = 1, 2, \\dots, 1006$.\n\nBy including this in the last given equality we get\n$$\n\\frac{a}{1-a} \\cdot (1 + 3 + \\dots + 2011) = 503^2.\n$$\nAs $1+3+5+\\dots+2011 = 503 \\cdot 2012 = 1006^2$, we get $\\frac{a}{1-a} \\cdot 1006^2 = 503^2$, so\n$$\n\\frac{a}{1-a} = \\frac{1}{4}.\n$$\nFinally, $x_{1006} = \\frac{a}{1-a} \\cdot 2011 = \\frac{2011}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16570,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $Q$ be a set of prime numbers, not necessarily finite. For a positive integer $n$, consider its prime factorization; define $p(n)$ to be the sum of all the exponents and $q(n)$ to be the sum of the exponents corresponding only to primes in $Q$. A positive integer $n$ is called _special_ if $p(n) + p(n + 1)$ and $q(n) + q(n + 1)$ are both even integers. Prove that there is a constant $c > 0$, independent of the set $Q$, such that for any positive integer $N > 2023$, the number of special integers in $[1, N]$ is at least $cN$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us call two positive integers $m, n$ _friends_ if $p(m) + p(n)$ and $q(m) + q(n)$ are both even. Note that the pairs $(p(k), q(k))$ take at most four values modulo $2$, so among any five different positive integers, some two of them are friends.\n\nAlso note that $p, q$ both satisfy the functional equation $f(ab) = f(a) + f(b)$ for all $a, b \\in \\mathbb{N}$. In particular, if $d \\mid m, n$, then $f(m) + f(n) = f\\left(\\frac{m}{d}\\right) + f\\left(\\frac{n}{d}\\right) + 2f(d)$, so $m, n$ are friends $\\iff \\frac{m}{d}$ and $\\frac{n}{d}$ are friends.\n\nCall a set $\\{n_1, n_2, n_3, n_4, n_5\\}$ interesting if for any indices $i, j$, both $n_i$ and $n_j$ are divisible by their difference $d_{ij} = |n_i - n_j|$. Since there are five integers in an interesting set, some two of them, say $n_i, n_j$, are friends $\\implies \\frac{n_i}{d_{ij}}$ and $\\frac{n_j}{d_{ij}}$ are a consecutive pair of positive integers which are friends, thus giving us a special integer.\n\nWe start by noticing that $\\{0, 6, 8, 9, 12\\}$ is an interesting set. Using that $72$ is the least common multiple of positive differences in the set, we get an infinite family of interesting sets:\n\n$$\nI_k = \\{72k, 72k + 6, 72k + 8, 72k + 9, 72k + 12\\}\n$$\n\nfor any $k \\ge 1$. If we consider the quotients of the form $\\frac{n_i}{d_{ij}}$ for each $n_i \\in I_k$, we get that the set\n\n$$\nS_k = \\{6k, 8k, 9k, 12k, 12k + 1, 18k + 2, 24k + 2, 24k + 3, 36k + 3, 72k + 8\\}\n$$\n\ncontains at least one special integer for all $k \\ge 1$. In particular, the interval $[1, 100k]$ contains $k$ sets $S_1, S_2, \\dots, S_k$, each of which contains a special integer. Since each special integer is contained in at most $10$ of the sets $S_i$ (it can occur in the $10$ different positions in $S_i$), there must be at least $\\frac{k}{10}$ special integers in the interval $[1, 100k]$.\n\nNow, write $N = 100q + r$ where $r$ is the remainder $0 \\le r < 100$. Then $q > \\frac{N}{100} - 1 > \\frac{N}{200}$ since $N > 2023 > 200$. But there are at least $\\frac{q}{100}$ special integers in the interval $[1, 100q]$. Thus there are at least $\\frac{1}{2000}N$ special integers in $[1, N]$ for all $N > 2023$. Therefore $c = \\frac{1}{2000}$ works. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16571,
    "subject": "Mathematics (Olympiad)",
    "question": "We consider the following operation applied to a positive integer: The integer is represented in an arbitrary base $b \\ge 2$, in which it has exactly two digits and in which both digits are different from $0$. Then the two digits are swapped and the result in base $b$ is the new number.\n\n*Is it possible to transform every number $> 10$ to a number $\\le 10$ with a series of such operations?*",
    "options": [],
    "answer": "See solution",
    "solution": "We show that each number $> 10$ can be transformed to a smaller number. In that way, we will eventually reach a number $\\le 10$.\n\nIf the number $n = 2k + 1$ is odd, we choose base $b = k$ with $n = (21)_k$. Swapping the two digits, we obtain the new number $(12)_k = k + 2$. Since $k \\ge 5$, the choice of $b = k$ as base is admissible (the digits are smaller than the base) and we have $k + 2 \\le 2k - 5 + 2 < 2k + 1$ as desired.\n\nIf the number $n = 2k$ is even, we choose the base $b = 2k - 2$ with $n = (12)_{2k-2}$ and obtain the new number $(21)_{2k-2} = 4k - 3$. Now we choose the base $k - 1$ with $4k - 3 = (41)_{k-1}$ and obtain the new number $(14)_{k-1} = k + 3$. Since $k > 5$, both bases are admissible, and we have $k + 3 < 2k$ as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16572,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the inequality\n\n$$\n\\left(\\frac{a^2 + b^2}{a+b}\\right)^3 + \\left(\\frac{b^2 + c^2}{b+c}\\right)^3 + \\left(\\frac{c^2 + a^2}{c+a}\\right)^3 \\geq a^3 + b^3 + c^3\n$$\nholds for all $a, b, c > 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "The desired inequality holds if (and only if) the following one does:\n\n$$\n2 \\left( \\frac{a^2 + b^2}{a+b} \\right)^3 \\geq a^3 + b^3 \\quad (*)_{a,b}\n$$\n\nfor all $a, b > 0$. Indeed, $(*)_{a,b}$ is implied by the problem statement by setting $b = c$. Conversely, we recover the problem statement by summing $(*)_{a,b}$, $(*)_{b,c}$, and $(*)_{c,a}$. We now prove $(*)_{a,b}$ by observing that:\n\n$$\n2(a^2 + b^2)^3 - (a^3 + b^3)(a+b)^3 = (a-b)^4(a^2 + ab + b^2) \\geq 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16573,
    "subject": "Mathematics (Olympiad)",
    "question": "A *diameter* of a finite planar set is any line segment of maximal Euclidean length having both endpoints in that set. A *lattice point* in the Cartesian plane is one whose coordinates are both integers. Given an integer $n \\ge 2$, prove that a set of $n$ lattice points in the plane has at most $n-1$ diameters.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the *diameter graph* $\\Gamma$ on $n$ pairwise distinct lattice points in the plane, i.e., the geometric graph on those points whose edges are the diameters of the configuration. We will prove that $\\Gamma$ has at least one vertex of degree 1. Removal of such a vertex and its incident edge allows an inductive approach—the base case, $n=2$, is clear.\n\nThe argument hinges on the fact that every two diameters intersect: either they both emanate from the same point or they cross at some interior point. If $\\Gamma$ has a vertex $A$ of degree at least 3, proceed as in the continuous case: the diameters from $A$ all lie within the angle formed by two diameters $AB$ and $AC$, where $\\angle BAC \\le 60^\\circ$. Let $AX$ be a third diameter from $A$; then $X$ is the desired vertex of degree 1, as a hypothetical diameter $XY$, $Y \\ne A$, cannot intersect both $AB$ and $AC$.\n\nIt is a fact that $\\Gamma$ has at most $n$ edges. Suppose, for contradiction, that $\\Gamma$ has $n$ edges and is 2-regular (all vertices have degree 2). As every two diameters intersect, this is possible if and only if $n$ is odd, in which case $\\Gamma$ is a star-shaped self-crossing $n$-cycle $A_1A_2A_3\\dots A_n$ whose polygonal convex hull reads around the boundary $A_1A_3A_5\\dots A_nA_2A_4A_6\\dots A_{n-1}$.\n\nWrite $A_iA_{i+1} = d$, $i = 1, 2, \\dots, n$ (indices modulo $n$), so $(x_{i+1} - x_i)^2 + (y_{i+1} - y_i)^2 = d^2$, $i = 1, 2, \\dots, n$; clearly, $d^2$ is an integer. Without loss of generality, assume the configuration is the smallest possible. By minimality, the $2n$ differences $x_{i+1} - x_i$ and $y_{i+1} - y_i$ cannot all be even—otherwise, the midpoints of all line segments $A_iA_j$ would again be lattice points, and a homothety of factor $\\frac{1}{2}$ from some $A_i$ would yield a smaller configuration.\n\nHence, one of the index sets $I = \\{i: x_{i+1} - x_i \\equiv 1 \\pmod{2}\\}$ and $J = \\{i: y_{i+1} - y_i \\equiv 1 \\pmod{2}\\}$, say $I$, is non-empty. Thus, $d^2 \\equiv 1 \\pmod{4}$, so for no index $i$ are both $x_{i+1} - x_i$ and $y_{i+1} - y_i$ even. Consequently, $I \\cup J$ exhausts all $n$ indices.\n\nRule out the case $d^2 \\equiv 1 \\pmod{4}$ as follows: If $d^2 \\equiv 1 \\pmod{4}$, then $I$ and $J$ are disjoint. By the preceding, $I \\cup J$ covers all $n$ indices, so $|I| + |J| = n$ is odd. On the other hand, $\\sum_{i=1}^n (x_{i+1} - x_i) = 0$ implies $|I|$ is even; similarly, $\\sum_{i=1}^n (y_{i+1} - y_i) = 0$ implies $|J|$ is even, so $|I| + |J|$ is even, a contradiction.\n\nFinally, rule out the case $d^2 \\equiv 2 \\pmod{4}$: If $d^2 \\equiv 2 \\pmod{4}$, then $I = J$, so $|I| = n$, again contradicting the parity of $|I|$ above. Thus, the diameter graph $\\Gamma$ is not 2-regular, as desired.\n\n**Remark.** The required diameter upper bound is always achieved, so $n-1$ is, in fact, the maximal number of diameters a planar configuration of $n$ lattice points may have. To prove it, consider the Diophantine equation $x^2 + y^2 = 5^N$ where $N$ is a non-negative integer. By Jacobi's two-square theorem, the number of pairwise distinct solutions of $x^2 + y^2 = M$ (for $M > 0$) is four times the excess of the number of divisors of $M$ congruent to 1 mod 4 over those congruent to 3 mod 4. In this case, there are exactly $4(N+1)$ distinct solutions. This can also be established directly, noting that $1 \\pm 2\\sqrt{-1}$ are Gaussian primes, and $\\frac{1}{\\pi}\\arg(1+2\\sqrt{-1})$ is irrational; explicitly, the solutions are $u(1+2\\sqrt{-1})^k(1-2\\sqrt{-1})^{N-k}$, where $u = \\pm 1$ or $\\pm\\sqrt{-1}$, and $k = 0, 1, \\dots, N$.\n\nExactly $N+1$ of these solutions lie in the first quadrant, $x > 0$ and $y \\ge 0$; and since 5 is odd, exactly $\\lfloor \\frac{1}{2}(N+1) \\rfloor$ of these, say $(x_i, y_i)$, $i = 1, 2, \\dots, \\lfloor \\frac{1}{2}(N+1) \\rfloor$, satisfy $x > y \\ge 0$. The $\\lfloor \\frac{1}{2}(N+1) \\rfloor$ lattice points $(x_i, y_i)$ are all exactly $5^{N/2}$ away from the origin, and every two are (strictly) less than $5^{N/2}$ apart. Thus, the origin and the $(x_i, y_i)$ form a planar configuration of $\\lfloor \\frac{1}{2}(N+3) \\rfloor$ lattice points with exactly $\\lfloor \\frac{1}{2}(N+1) \\rfloor$ diameters of length $5^{N/2}$ each. Setting $N = 2n-3$ completes the argument.\n\nA related configuration: For even $N \\ge n$, the $(x_i, y_i)$ above and the $(5^{N/2} - x_i, y_i)$ form a configuration of $N+2$ lattice points with exactly $N+1$ diameters: $\\frac{1}{2}N+1$ join $(0,0)$ to each $(x_i, y_i)$, and another $\\frac{1}{2}N$ join $(5^{N/2}, 0)$ to each $(5^{N/2} - x_i, y_i)$ with $y_i > 0$. Deleting any $N-n+2$ points with both coordinates positive settles the case.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16574,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 22 cards, numbered $1, 2, \\ldots, 22$. Using these cards, 11 fractions are formed. What is the greatest possible number of integer values among these fractions?",
    "options": [],
    "answer": "See solution",
    "solution": "10 numbers.\n\nThe numbers $13$, $17$, and $19$ can only form an integer if they are in the numerator and $1$ is in the denominator. Therefore, at least one fraction cannot be integer. However, it is possible to have 10 integer fractions:\n\n$$\n\\frac{22}{11},\\ \\frac{14}{7},\\ \\frac{15}{5},\\ \\frac{21}{3},\\ \\frac{20}{10},\\ \\frac{18}{9},\\ \\frac{16}{8},\\ \\frac{12}{6},\\ \\frac{19}{1},\\ \\frac{4}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16575,
    "subject": "Mathematics (Olympiad)",
    "question": "In the isosceles triangle $ABC$ with $\\overline{AC} = \\overline{BC}$, let $D$ be the foot of the altitude from $C$. Let $M$ be the midpoint of $CD$. The line $BM$ intersects $AC$ at $E$. Prove that the length of $AC$ is three times that of $CE$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the centroid $S$ of triangle $DBC$, which lies on the line $BM$. The line $DS$ bisects $BC$, so $DS$ is parallel to $AC$ by the intercept theorem. Since the centroid divides $DS$ in the ratio $2:1$, the same ratio applies to the parallel line $AC$, meaning $E$ divides $AC$ in the ratio $2:1$. Therefore, $AC = 3 \\times CE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16576,
    "subject": "Mathematics (Olympiad)",
    "question": "The twelve letters A, B, C, D, E, F, G, H, I, J, K, and L are randomly grouped into six pairs of letters. The two letters in each pair are placed next to each other in alphabetical order to form six two-letter words, and then those six words are listed alphabetically. For example, a possible result is AB, CJ, DG, EK, FL, HI. The probability that the last word listed contains G is $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.",
    "options": [],
    "answer": "See solution",
    "solution": "There are $11 \\cdot 9 \\cdot 7 \\cdot 5 \\cdot 3 \\cdot 1$ equally likely ways for the letters to be paired. This can be seen by considering the successive choices of a partner for the unpaired letter that comes first alphabetically.\n\nThe letter G is the first letter in its pair, and its pair is listed last if G is paired with H, I, J, K, or L (5 choices), two of A, B, C, D, E, and F are paired with each other ($\\binom{6}{2} = 15$ choices), and the 4 remaining early letters are paired with the 4 remaining late letters ($4! = 24$ choices).\n\nThe letter G is the second letter in its pair, and its pair is listed last if G is paired with F and each of A through E is paired with one of H through L. This can happen in $5! = 120$ ways.\n\nTherefore,\n\n$$\n\\frac{m}{n} = \\frac{5 \\cdot 15 \\cdot 24 + 120}{11 \\cdot 9 \\cdot 7 \\cdot 5 \\cdot 3 \\cdot 1} = \\frac{128}{693}.\n$$\n\nThe requested sum is $128 + 693 = 821$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16577,
    "subject": "Mathematics (Olympiad)",
    "question": "The line $y = mx + c$ is tangent to both circles $x^2 + y^2 = 2x$ and $x^2 + y^2 = 2y$. Find the possible values of $m$ and $c$ for such a line.",
    "options": [],
    "answer": "See solution",
    "solution": "The line $y = mx + c$ is tangent to $x^2 + y^2 = 2x$ if its distance from the center $(1, 0)$ is equal to the radius $1$:\n\n$$\n\\frac{|m + c|}{\\sqrt{1 + m^2}} = 1 \\implies |m + c| = \\sqrt{1 + m^2}.\n$$\n\nSimilarly, it is tangent to $x^2 + y^2 = 2y$ (center $(0, 1)$, radius $1$) if:\n\n$$\n\\frac{|1 - c|}{\\sqrt{1 + m^2}} = 1 \\implies |1 - c| = \\sqrt{1 + m^2}.\n$$\n\nEquating the two:\n\n$$\n|m + c| = |1 - c|.\n$$\n\nSquaring both sides and simplifying, we get two equations:\n\n$$\n2mc + c^2 = 1, \\quad m^2 = c^2 - 2c.\n$$\n\nSolving, we find $m = -1$ and $c = 1 \\pm \\sqrt{2}$.\n\nAlternatively, writing the circles as $(x-1)^2 + y^2 = 1$ and $x^2 + (y-1)^2 = 1$, their centers are $(1, 0)$ and $(0, 1)$, both with radius $1$. The common tangents are parallel to the line joining the centers, so $m = -1$. Substituting $y = -x + c$ into one circle and requiring a double root gives $c = 1 \\pm \\sqrt{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16578,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs $(k, n)$ of positive integers satisfying the equation\n\n$$\n1! + 2! + \\cdots + k! = 1 + 2 + \\cdots + n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We first compute the values of $k!$ and the partial sums $1! + 2! + \\cdots + k!$ for small $k$:\n\n| $k$ | $k!$ | $1! + 2! + \\cdots + k!$ |\n|---|-----|--------------------------|\n| 1 | 1   | 1                        |\n| 2 | 2   | 3                        |\n| 3 | 6   | 9                        |\n| 4 | 24  | 33                       |\n| 5 | 120 | 153                      |\n| 6 | 720 | 873                      |\n| 7 | 5040 | 5913                    |\n| 8 | 40320 | 46233                  |\n| 9 | 362880 | 409113                |\n| 10 | 3628800 | 4037913             |\n\nObviously, the pairs $(k, n) = (1, 1)$ and $(k, n) = (2, 2)$ are solutions. We will show that the unique solution with $k > 2$ is $(k, n) = (5, 17)$.\n\nRecall that $1 + 2 + \\cdots + n = \\frac{n(n+1)}{2}$. For $k \\geq 10$, $k!$ is divisible by $100$, so $1! + 2! + \\cdots + k!$ leaves a remainder $13$ when divided by $100$ for $k \\geq 9$. If equality holds for some $k \\geq 10$:\n\n$$\n1! + 2! + \\cdots + k! = 1 + 2 + \\cdots + n = \\frac{n(n+1)}{2}\n$$\n\nthen $n(n+1) = 100m + 26$ for some integer $m$, so\n\n$$\nn^2 + n - 100m - 26 = 0.\n$$\n\nThe discriminant is $\\Delta = 400m + 105$, which must be a perfect square. However, $x^2 \\equiv 5 \\pmod{100}$ has no solution, so there are no solutions for $k \\geq 10$.\n\nThus, the only solutions are $(k, n) = (1, 1)$, $(2, 2)$, and $(5, 17)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16579,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 4$ be an integer and $S_n = \\{1, 2, 3, \\dots, 2^n\\}$. Two sets $A, B$ are given, with $A \\subset S_n$, $B \\subset S_n \\setminus S_{n-1}$, such that $|A| = n+1$ and $|B| = 2$. Is it possible that $ab-1$ is a perfect cube for every $a \\in A$, $b \\in B$?\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: NO.\n\nSuppose, for contradiction, that such sets $A$ and $B$ exist. Arrange the elements of $A$ and $B$ in increasing order: $1 \\le a_1 < a_2 < \\dots < a_{n+1} \\le 2^n$ and $2^{n-1} < b_1 < b_2 \\le 2^n$.\n\nThere exists an index $i \\le n$ such that $a_i < a_{i+1} \\le 2a_i$. Define:\n\n$$\na_i b_1 - 1 = q_1^3, \\quad a_{i+1} b_2 - 1 = q_2^3, \\quad a_i b_2 - 1 = s_1^3, \\quad a_{i+1} b_1 - 1 = s_2^3.\n$$\n\nFrom $a_i < a_{i+1} \\le 2a_i$ and $b_1 < b_2 < 2b_1$, it follows that\n\n$$\nq_1 < s_1, \\quad s_2 < q_2 < 2q_1. \\tag{1}\n$$\n\nTherefore,\n\n$$\n(q_1^3 + 1)(q_2^3 + 1) = (s_1^3 + 1)(s_2^3 + 1)\n$$\nwhich yields\n$$\n(q_1 q_2)^3 + q_1^3 + q_2^3 + 1 = (s_1 s_2)^3 + s_1^3 + s_2^3 + 1. \\tag{2}\n$$\n\nUsing (1), if $q_1 q_2 > s_1 s_2$ or $q_1 q_2 < s_1 s_2$, the equality (2) cannot hold. Thus, $q_1 q_2 = s_1 s_2$.\n\nConsider the function $f(x) = x^3 + \\frac{C}{x^3}$ on $x \\in [q_1, q_2]$, where $C = (q_1 q_2)^3$. This function decreases on $[q_1, \\sqrt{q_1 q_2}]$ and increases on $[\\sqrt{q_1 q_2}, q_2]$, so its maximum is at the endpoints: $f(q_1) = f(q_2) = q_1^3 + q_2^3$.\n\nTherefore, equality (2), given (1), holds only if $(q_1, q_2) = (s_1, s_2)$, which is a contradiction.\n\n$\\boxed{}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16580,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 15 children in total, divided into groups of 6, 5, 1, and 3. What is the central angle in a pie chart representing the third-largest group (which has 3 children)?",
    "options": [],
    "answer": "See solution",
    "solution": "There were $6 + 5 + 1 + 3 = 15$ children altogether, of whom 3 were in the third-largest group. That group therefore requires $$\\frac{3}{15} \\times 360^{\\circ} = 3 \\times 24^{\\circ} = 72^{\\circ}$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16581,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $d$ be the greatest common divisor of $m$ and $n$, i.e., $m = d m'$ and $n = d n'$, where $m'$ and $n'$ are relatively prime positive integers of different parity.\n\nNow we need to prove that\n\n$$\n\\frac{3d^2 m'^2 + 5d^2 m' n'}{3d^2 n'^2 + d m' n'} = \\frac{m'(3m' + 5n')}{n'(m' + 3n')}\n$$\n\nis not a positive integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $3m' + 5n'$ and $m' + 3n'$ are both odd.\n\nIf $m'$ is odd and $n'$ is even, the even $n'(m' + 3n')$ clearly cannot divide the odd $m'(3m' + 5n')$.\n\nOtherwise, let the odd $k$ be the greatest common divisor of $3m' + 5n'$ and $m' + 3n'$. Then we have\n\n$$\n\\begin{aligned}\nk &\\mid 3(3m' + 5n') - 5(m' + 3n') \\implies k \\mid 4m', \\\\\nk &\\mid (3m' + 5n') - 3(m' + 3n') \\implies k \\mid -4n',\n\\end{aligned}\n$$\n\nfrom which it follows that $k$ divides both $m'$ and $n'$, so $k = 1$ and both factors in\n\n$$\n\\frac{m'}{n'} \\cdot \\frac{3m' + 5n'}{m' + 3n'}\n$$\n\nare irreducible fractions.\n\nSince $m' + 3n' > m'$, the proof is finished.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16582,
    "subject": "Mathematics (Olympiad)",
    "question": "In a triangle $ABC$, $H_a$, $H_b$, and $H_c$ are the feet of the altitudes on the sides $BC$, $CA$, and $AB$ respectively. For which triangles are two of the line segments $H_aH_b$, $H_bH_c$, and $H_cH_a$ of equal length?",
    "options": [],
    "answer": "See solution",
    "solution": "We first consider the situation in which no angle in $ABC$ is obtuse. If $ABC$ is right-angled with hypotenuse $AB$, we have $H_a = H_b = C$, and therefore certainly $H_bH_c = H_cH_a$. Any right-angled triangle $ABC$ therefore certainly has the required property. If $ABC$ is not right-angled, the triangles $AH_bB$ and $AH_aB$ certainly are.\n\n![](images/Austrija_2012_p3_data_94aab80452.png)\n\nWe see that both $H_a$ and $H_b$ lie on the semicircle with diameter $AB$. If $H_bH_c = H_cH_a$, $H_c$ must be the common point of $AB$ and the bisector of $H_aH_b$, and therefore the midpoint $M_{AB}$ of $AB$. Since $CH_c$ is perpendicular to $AB$, we see that $C$ must lie on the bisector of $AB$, and $ABC$ is therefore isosceles.\n\nNow we assume that one angle in $ABC$ is greater than $90^\\circ$.\n\nIf we assume $H_bH_c = H_cH_a$, we obtain $|AC| = |BC|$ as before. If, however, we assume $H_aH_c = H_aH_b$, we obtain the situation in the second figure. Because of the right angles between the sides and the altitudes, each of the quadrilaterals $AH_aH_bB$, $CH_bBH_c$, and $AH_aCH_c$ is cyclic. It therefore follows that $\\angle H_aH_bC = \\angle H_aH_bA = \\angle H_aBA = \\beta = \\angle CBH_c = \\angle CH_bH_c$ and $\\angle H_bH_cC = \\angle H_bBC = 90^\\circ - \\alpha - \\beta = \\angle H_aAH_b = \\angle H_aAC = \\angle H_aH_cC$. We see that $|H_aH_b| = |H_aH_c|$ if and only if $\\angle H_aH_bH_c = \\angle H_aH_cH_b$, which is equivalent to $2\\beta = 2 \\cdot (90^\\circ - \\alpha - \\beta)$, or $\\alpha + 2\\beta = 90^\\circ$.\n\nSumming up, we see that exactly the right-angled triangles, the isosceles triangles, and triangles in which two angles $\\alpha$ and $\\beta$ fulfill the equation $\\alpha + 2\\beta = 90^\\circ$ have the required property. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16583,
    "subject": "Mathematics (Olympiad)",
    "question": "Given triangle $ABC$, the point $J$ is the centre of the excircle opposite the vertex $A$. This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G$. Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC$. Prove that $M$ is the midpoint of $ST$.\n\n(The excircle of $ABC$ opposite the vertex $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)",
    "options": [],
    "answer": "See solution",
    "solution": "For the solutions that follow, we use the notation $\\angle BAC = 2\\alpha$, $\\angle ABC = 2\\beta$ and $\\angle ACB = 2\\gamma$, where $\\alpha + \\beta + \\gamma = 90^\\circ$.\n\n![](images/Australian_Scene_2012_-_AMT_Publishing_-_273p_p178_data_8a44099cc8.png)\n\n*Solution 1* (Alex Gunning, year 9, Glen Waverley Secondary College, VIC. Alex was a bronze medalist with the 2012 Australian IMO team.)\n\nConstruct segments $FK$ and $GL$. Let $\\angle FMK = \\angle GML = \\theta$. Also let $D = MK \\cap BJ$ and $E = ML \\cap CJ$.\n\nWe have $BM = BK$ due to equal tangents from $B$ to the excircle. Furthermore, since $BJ$ bisects $\\angle KBM$, this means that the line through $B$ and $J$ is a line of symmetry for $\\triangle KBM$. Since $F$ also lies on this line we deduce that $\\angle FKG = \\angle FKM = \\angle FMK = \\theta$. Similarly, we show that $\\angle FLG = \\angle GML = \\theta$. Also due to symmetry we have $MK \\perp BJ$, and $ML \\perp CJ$. Hence $MDJE$ is cyclic. Thus $\\angle FJG = \\angle DJE = \\angle FMK = \\theta$. Since $\\angle FKG$, $\\angle FJG$ and $\\angle FLG$ are all equal (to $\\theta$) we conclude that $FKJLG$ is cyclic.\n\nA radius is perpendicular to its tangent, hence we have $AK \\perp JK$ and $AL \\perp JL$. So, $AKJL$ is cyclic. Combining this with $FKJLG$ we deduce that $AFKJLG$ is cyclic.\n\nFrom circle $AFKJLG$ we find $\\angle AFJ = \\angle AKJ = 90^\\circ$. But now in $\\triangle ASB$ the angle bisector $BF$ at $B$ is also the altitude. This implies that $AB = BS$. Thus $MS = BS + BM = AB + BK = AK$. Similarly, $MT = AL$. Since $AK = AL$, we are done.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16584,
    "subject": "Mathematics (Olympiad)",
    "question": "試求所有映成函數 $f: \\mathbb{Z} \\to \\mathbb{Z}$ 使得對任意整數 $x, y, z$，\n\n$$\nf(xyz + x f(y) + y f(z) + z f(x)) = f(x) f(y) f(z)\n$$\n\n其中 $\\mathbb{Z}$ 表示所有整數所成的集合。",
    "options": [],
    "answer": "See solution",
    "solution": "首先，令 $x = y = z = 0$，可得 $f(0) = f(0)^3$，因此 $f(0) = -1, 0$ 或 $1$。\n\n若 $f(0) = 0$，則令 $z = 0$，有 $f(x f(y)) = 0$，對所有 $x, y \\in \\mathbb{Z}$ 成立，這意味著 $f(x) = 0$，矛盾。\n\n因此只需考慮 $f(0) = -1$ 或 $f(0) = 1$。\n\n**情形 1：** $f(0) = 1$\n\n若存在非零整數 $a$ 使 $f(a) = 1$，則令 $y = a, z = 0$，有 $f(x + a) = f(x)$，這與 $f$ 的滿射性矛盾。\n\n設 $b$ 使 $f(b) = -1$，令 $x = y = z = b$，得 $f(b^3 - 3b) = f(b)^3 = -1$。\n\n再令 $x = b^3 - 3b, y = b, z = 0$，有 $f(4b - b^3) = 1$，即 $4b - b^3 = 0$，解得 $b = 2, -2$。\n\n**情形 1.1：** $f(-2) = -1$\n\n此時 $f(x) + f(-2 - x) = 0$，對所有 $x \\in \\mathbb{Z}$。\n\n設 $a$ 使 $f(a) = 2$，令 $y = a, z = 0$，有 $f(2x + a) = 2f(x)$。\n\n特別地，$f(a-4) = -2$，由前式得 $f(2-a) = 2$，即 $f(2x + (2-a)) = 2f(x) = f(2x + a)$。\n\n若 $2-a \\neq a$，則 $f$ 在 $2\\mathbb{Z}+a$ 上僅取有限值，與滿射性矛盾，故 $a = 1$，即 $f(2x+1) = 2f(x)$。\n\n用數學歸納法可證 $f(x) = x+1$ 對所有 $x \\in \\mathbb{Z}$ 成立。\n\n**情形 1.2：** $f(2) = -1$\n\n同理，設 $a$ 使 $f(a) = 2$，則 $f(a+4) = -2$，即 $f(-2-a) = 2$，同理可得 $a = -1$。\n\n歸納法可證 $f(x) = 1 - x$ 也是一組解。\n\n**結論：**\n\n所有滿足條件的函數為 $f(x) = x+1$ 或 $f(x) = 1-x$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16585,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be real numbers larger than $1$. Prove the inequality\n\n$$\n\\frac{ab}{c-1} + \\frac{bc}{a-1} + \\frac{ca}{b-1} \\geq 12.\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "By the AM-GM inequality, we know that\n\n$$\n\\sqrt{(c-1) \\cdot 1} \\leq \\frac{c-1+1}{2},\n$$\n\ntherefore\n\n$$\nc - 1 \\leq \\frac{c^2}{4}\n$$\n\nwith equality for $c=2$. With the two analogous inequalities for $a$ and $b$, we obtain\n\n$$\n\\frac{ab}{c-1} + \\frac{bc}{a-1} + \\frac{ca}{b-1} \\geq \\frac{4ab}{c^2} + \\frac{4bc}{a^2} + \\frac{4ca}{b^2} \\geq 12 \\sqrt[3]{\\frac{ab}{c^2} \\cdot \\frac{bc}{a^2} \\cdot \\frac{ca}{b^2}} = 12\n$$\n\nwhere the last inequality is the AM-GM inequality again. Therefore, equality holds for $a = b = c = 2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16586,
    "subject": "Mathematics (Olympiad)",
    "question": "At a contest, students were presented with 24 multiple choice questions. Contestants who gave no answer or gave more than one answer were awarded 0 points for that question. For a correct answer, they received 1 point, and for a wrong answer, $\\frac{1}{4}$ of a point was deducted. If a student received 13 points, at most how many questions did they answer correctly?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be the number of questions answered correctly (awarded 1 point), and $N$ the number answered incorrectly (with $\\frac{1}{4}$ point deducted).\n\nThe total score is:\n$$\nP - \\frac{1}{4}N = 13\n$$\nSo,\n$$\nN = 4P - 52\n$$\nSince there are 24 questions:\n$$\nP + N \\leq 24\n$$\nSubstitute for $N$:\n$$\nP + (4P - 52) \\leq 24 \\\\\n5P - 52 \\leq 24 \\\\\n5P \\leq 76 \\\\\nP \\leq 15.2\n$$\nSince $P$ must be an integer, the greatest possible value is $P = 15$.\n\nWhen $P = 15$:\n$$\nN = 4 \\times 15 - 52 = 8\n$$\nSo, the maximum number of questions answered correctly is $\\boxed{15}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16587,
    "subject": "Mathematics (Olympiad)",
    "question": "The number $2013$ is written on the board. Two players are playing the following game. A move consists of replacing the number on the board with the difference of this number and one of its divisors. The player who writes $0$ loses. Who of the two players can guarantee the win?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** The second player wins.\n\n**Solution.** It is easy to observe that odd numbers have only odd divisors. So, if the player moves from an odd number, he has to write an even number. This provides a strategy for the second player: subtract $1$ from the current number at any move. Then the first player will always deal with an odd number, so will write an even number. Since the number written on the board always decreases, eventually the first player will have to write $0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16588,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a value of $x \\in (0, \\frac{\\pi}{2})$ for which the numbers $\\sin x$, $\\cos x$, and $\\tan x$ form a geometric progression?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, such a value exists.\n\nLet us write the equation that is necessary and sufficient for these three numbers to form a geometric progression:\n\n$$\n\\sin x \\cdot \\tan x = \\cos^2 x.\n$$\n\nThen, $\\sin^2 x = \\cos^3 x$. If we denote $t = \\cos x \\in (0, 1)$, we obtain the equation:\n\n$$\n1 - t^2 = t^3 \\implies t^3 + t^2 - 1 = 0.\n$$\n\nSince $f(0) = -1 < 0$ and $f(1) = 1 > 0$, there exists at least one value of $t \\in (0, 1)$ for which $t^3 + t^2 - 1 = 0$. Hence, such $x \\in (0, \\frac{\\pi}{2})$ exists.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16589,
    "subject": "Mathematics (Olympiad)",
    "question": "Two squares are adjacent to each other as shown. One has sides of length 5 cm and the other has sides of length 7 cm. The area in cm² of the shaded region is\n\n![](images/SAMF_ANNUAL_REPORT_2015_BOOK_PROOF_p34_data_3166e86d92.png)\n\n(A) 35\n\n(B) 35.5\n\n(C) 36\n\n(D) 36.5\n\n(E) 37",
    "options": [],
    "answer": "See solution",
    "solution": "$\\angle ABC = 90^\\circ$ since $AB$ and $BC$ are both diagonals of a square,\n$$AB = \\sqrt{5^2 + 5^2} = \\sqrt{50} = 5\\sqrt{2}$$\n$$BC = \\sqrt{7^2 + 7^2} = \\sqrt{98} = 7\\sqrt{2}$$\n$$\\text{area of } \\triangle ABC \\text{ is } \\frac{1}{2}(5\\sqrt{2})(7\\sqrt{2}) = 35 \\text{ cm}^2$$\n\n![](images/SAMF_ANNUAL_REPORT_2015_BOOK_PROOF_p36_data_cd27b3e456.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16590,
    "subject": "Mathematics (Olympiad)",
    "question": "The midpoints of the sides $BC$, $CA$, and $AB$ of triangle $ABC$ are $D$, $E$, and $F$, respectively. The reflections of the centroid $M$ of $ABC$ around points $D$, $E$, and $F$ are $X$, $Y$, and $Z$, respectively. Segments $XZ$ and $YZ$ intersect the side $AB$ at points $K$ and $L$, respectively. Prove that $AL = BK$.",
    "options": [],
    "answer": "See solution",
    "solution": "As $\\frac{MY}{MB} = \\frac{2ME}{2ME} = 1$ and analogously $\\frac{MZ}{MC} = 1$, we have $BC \\parallel YZ$. Let $G$ be the intersection of lines $YZ$ and $AD$.\n\nThen $\\frac{MG}{MD} = \\frac{MY}{MB} = 1$, from which $MG = MD = \\frac{1}{3}AD$, and $AG = AD - MD - MG = \\frac{1}{3}AD$. Hence $\\frac{AL}{AB} = \\frac{AG}{AD} = \\frac{1}{3}$.\n\nBy swapping the roles of $A$ and $B$, the roles of $D$ and $E$, the roles of $X$ and $Y$, and finally the roles of $K$ and $L$, we get analogously $\\frac{BK}{AB} = \\frac{1}{3}$. Therefore $AL = BK$.\n\n![](images/prob1718_p10_data_ed97a5cd65.png)\n\nFig. 14",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16591,
    "subject": "Mathematics (Olympiad)",
    "question": "Construct a point $E$ such that $DCAE$ is an isosceles trapezoid and $DC \\parallel EA$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\triangle EDP$ is the image of reflection of $\\triangle ACB$ in the perpendicular bisector of $DC$, it is equilateral. Now, as\n\n$$\n\\angle ADE = \\angle CDE - \\angle CDA = \\angle ACD - \\angle CDA = 72^\\circ - 36^\\circ = 36^\\circ\n$$\n\nand\n\n$$\n\\angle EAD = \\angle CDA = 36^\\circ = \\angle ADE,\n$$\n\nwe have $EA = ED = EP$. It follows that\n\n$$\n\\begin{aligned}\n\\angle PAC &= \\angle EAC - \\angle EAP \\\\\n&= (180^\\circ - \\angle ACD) - \\left(90^\\circ - \\frac{1}{2} \\angle PEA\\right) \\\\\n&= 180^\\circ - 72^\\circ - 90^\\circ + \\frac{1}{2}(\\angle DEA - \\angle DEP) \\\\\n&= 18^\\circ + \\frac{1}{2}(108^\\circ - 60^\\circ) \\\\\n&= 42^\\circ.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16592,
    "subject": "Mathematics (Olympiad)",
    "question": "A pupil has 7 pieces of paper. He chooses some of them and cuts each of them into seven pieces. In the sequel, he chooses some of the pieces and cuts each of them into seven pieces. He continues this procedure many times with the pieces he has in hand every time. Is it possible to have, at some time, 2009 pieces of paper?",
    "options": [],
    "answer": "See solution",
    "solution": "Let him choose $\\alpha_1$ pieces from the seven and cut each into seven pieces. Then he will have $7 - \\alpha_1 + 7\\alpha_1 = 7 + 6\\alpha_1$ pieces of paper. Suppose in the next step he chooses $\\alpha_2$ pieces and cuts each into seven pieces. Then he will have $7 + 6\\alpha_1 - \\alpha_2 + 7\\alpha_2 = 7 + 6(\\alpha_1 + \\alpha_2)$ pieces. Continuing this procedure $\\kappa$ times, he will have $7 + 6(\\alpha_1 + \\alpha_2 + \\dots + \\alpha_\\kappa)$ pieces of paper. Therefore, we are looking for $\\kappa$ and $\\alpha_i$ such that\n\n$$\n7 + 6(\\alpha_1 + \\alpha_2 + \\dots + \\alpha_\\kappa) = 2009\n$$\nwhich gives\n$$\n6(\\alpha_1 + \\alpha_2 + \\dots + \\alpha_\\kappa) = 2002.\n$$\nBut 2002 is not divisible by 6, so it is not possible for him to have exactly 2009 pieces of paper at any time.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16593,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with circumcircle $c$ and circumcenter $O$, and let $D$ be a point on the side $BC$ different from the vertices and the midpoint of $BC$. Let $K$ be the point where the circumcircle $c_1$ of triangle $BOD$ intersects $c$ for the second time, and let $Z$ be the point where $c_1$ meets the line $AB$. Let $M$ be the point where the circumcircle $c_2$ of triangle $COD$ intersects $c$ for the second time, and let $E$ be the point where $c_2$ meets the line $AC$. Finally, let $N$ be the point where the circumcircle $c_3$ of triangle $AEZ$ meets $c$ again. Prove that the triangles $ABC$ and $NKM$ are congruent.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the quadrilateral $ODCE$ is cyclic, we have $\\angle O_1 = \\angle C$. (The angles are as shown in the figure.)\n\n![](images/Balkan_2012_shortlist_p17_data_118b165551.png)\n\nSince the quadrilateral $ODBZ$ is cyclic, we have $\\angle O_2 = \\angle B$. Adding these, we obtain $\\angle EOUZ = \\angle B + \\angle C = 180^\\circ - \\angle A$. Therefore, the points $A$, $Z$, $O$, $E$ are concyclic and $c_3$ passes through $O$.\n\nWe also have $\\angle Z_2 = \\angle D_2 = \\angle E_1$. Since these three angles are subtended by the chords $AO$, $BO$, $CO$ of equal length in the circles $c_3$, $c_1$, $c_2$, respectively, the radii of these circles are equal. Therefore, the angles $\\angle Z_1$ and $\\angle D_1$ are equal as they are subtended by chords $OK$ and $OM$ of the same length. It follows that the points $K$, $D$, $M$ are collinear. Similarly, $M$, $E$, $N$ are collinear and $Z$, $Z$, $K$ are collinear. Then $\\angle BDK = \\angle CDM$ and $\\angle CEM = \\angle AEN$, and $BK = CM = AN$. Therefore, the triangle $NKM$ is obtained by rotating the triangle $ABC$ by a rotation with center $O$. Hence $ABC$ and $NKM$ are congruent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16594,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $\\triangle ABC$ 為一銳角三角形，點 $P$ 是 $BC$ 邊上的一點。已知線段 $PE$ 垂直於 $AC$，線段 $PD$ 垂直於 $AB$ 邊。試求：三角形 $PDE$ 面積最大，若且唯若 $P$ 為 $BC$ 中點。",
    "options": [],
    "answer": "See solution",
    "solution": "首先注意到\n\n$$\n2 \\times \\text{三角形 } PDE \\text{ 的面積} = PD \\times PE \\times \\sin \\angle DPE = PD \\times PE \\times \\sin(\\angle B + \\angle C)\n$$\n\n因此，$\\triangle PDE$ 面積達到最大值若且唯若 $PD \\times PE$ 達到最大值。\n\n接著，令 $x, y$ 分別為線段 $PE, PD$ 長，$b, c$ 分別為線段 $AC, AB$ 長。令 $d$ 為三角形 $ABC$ 面積，則顯然 $xb + yc = 2d$。故 $y = \\frac{2d - xb}{c}$。因此，\n\n$$\nPD \\times PE = xy = \\frac{2d x - b x^2}{c} = \\frac{1}{c} \\left( \\frac{d^2}{b} - b \\left(x - \\frac{d}{b}\\right)^2 \\right)\n$$\n\n故 $PD \\times PE$ 達到最大值若且唯若 $PE = x = \\frac{d}{b}$。又令三角形 $ABC$ 在 $AC$ 上的高為 $BH$，則 $BH = \\frac{2d}{b} = 2PE$，故由相似三角形知 $P$ 為 $BC$ 中點。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16595,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n = \\prod_{i=1}^{s} p_i^{\\alpha_i}$, we write $\\Omega(n)$ for the total number $\\sum_{i=1}^{s} \\alpha_i$ of prime factors of $n$, counted with multiplicity. Let $\\lambda(n) = (-1)^{\\Omega(n)}$ (so, for example, $\\lambda(12) = \\lambda(2^2 \\cdot 3^1) = (-1)^{2+1} = -1$).\n\nProve the following two claims:\n\n1. There are infinitely many positive integers $n$ such that $\\lambda(n) = \\lambda(n+1) = +1$.\n2. There are infinitely many positive integers $n$ such that $\\lambda(n) = \\lambda(n+1) = -1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that $\\Omega(mn) = \\Omega(m) + \\Omega(n)$ for all positive integers $m, n$ (so $\\Omega$ is a completely additive arithmetic function), which translates into $\\lambda(mn) = \\lambda(m) \\lambda(n)$ (so $\\lambda$ is a completely multiplicative arithmetic function). Hence $\\lambda(p) = -1$ for any prime $p$, and $\\lambda(k^2) = (\\lambda(k))^2 = +1$ for all positive integers $k$.\n\nSuppose, for contradiction, that from some point on, the sequence $\\lambda(n)$ only contains subsequences of the form $(1, -1, 1, 1, \\dots, 1, -1, 1)$. By \"doubling\" such a subsequence (as in part ii)), we would produce\n\n$$\n(-1, ?, 1, ?, -1, ?, -1, ?, \\dots, ?, -1, ?, 1, ?, -1)\n$$\n\nAccording to our assumption, all the `?` terms must be $1$, so the produced subsequence is\n\n$$\n(-1, 1, 1, 1, -1, 1, -1, 1, \\dots, 1, -1, 1, 1, 1, -1)\n$$\n\nThus, the \"separating packets\" of $1$'s contain either one or three terms. Now, assume that a subsequence $(1, 1, 1, 1)$ or $(-1, 1, 1, -1)$ appears far enough along in $\\mathfrak{S}$. Since it lies within some \"doubled\" subsequence, this contradicts the structure described above, which must therefore be the only one prevalent from some point on. But then all the positions of the $(-1)$ terms would have the same parity. However, $\\lambda(p) = \\lambda(2p^2) = -1$ for all odd primes $p$, and these terms have different parity of their positions. This is a contradiction.\n\n*Alternative Solution.* (I. Bogdanov) Take $\\varepsilon \\in \\{-1, 1\\}$. There obviously exist infinitely many $n$ such that $\\lambda(2n + 1) = \\varepsilon$ (just take $2n + 1$ to be the product of an appropriate number of odd primes). Now, if either $\\lambda(2n) = \\varepsilon$ or $\\lambda(2n + 2) = \\varepsilon$, we are done; otherwise $\\lambda(n) = -\\lambda(2n) = -\\lambda(2n + 2) = \\lambda(n + 1) = \\varepsilon$. Therefore, for such an $n$, one of the three pairs $(n, n + 1)$, $(2n, 2n + 1)$, or $(2n + 1, 2n + 2)$ fits the bill.\n\nThus, we have proved the existence in $\\mathfrak{S}$ of infinitely many occurrences of all possible subsequences of length 1, namely $(+1)$ and $(-1)$, and of length 2, namely $(+1, -1)$, $(-1, +1)$, $(+1, +1)$, and $(-1, -1)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16596,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{R}^+$ denote the set of all positive real numbers. Find all functions $f: \\mathbb{R}^+ \\to \\mathbb{R}$ satisfying\n\n$$\nf(x) + f(y) \\le \\frac{f(x+y)}{2}, \\quad \\frac{f(x)}{x} + \\frac{f(y)}{y} \\ge \\frac{f(x+y)}{x+y},\n$$\n\nfor all $x, y \\in \\mathbb{R}^+$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = y = t$ with $t > 0$. Then\n\n$$\n4f(t) \\le f(2t), \\quad f(2t) \\ge 4f(t),\n$$\n\nfor all $t > 0$. Thus $f(2t) = 4f(t)$ for all $t > 0$. By induction,\n\n$$\nf(2^m t) = 2^{2m} f(t), \\text{ for all } t > 0.\n$$\n\nLet $g(x) = \\frac{f(x)}{x}$ for $x > 0$. We show that $g(nt) = n g(t)$ for all $n \\in \\mathbb{N}$ and $t > 0$. We have proved this for $n = 2^m$. Also, $g(x+y) \\le g(x) + g(y)$ for all $x, y > 0$. By induction, $g(nt) \\le n g(t)$ for all $n \\in \\mathbb{N}$ and $t > 0$. Choose $m$ such that $2^{m-1} \\le n < 2^m$. Then\n\n$$\n2^m g(t) = g(2^m t) \\le g(nt) + g((2^m - n)t) \\le n g(t) + (2^m - n) g(t) = 2^m g(t).\n$$\n\nHence equality holds and $g(nt) = n g(t)$.\n\nNext, we show that $g$ is non-increasing. We have\n\n$$\nf(t) + f(2t) \\le \\frac{f(3t)}{2}.\n$$\n\nSo,\n\n$$\nt g(t) + 2t g(2t) \\le \\frac{3t}{2} g(3t).\n$$\n\nThis gives\n\n$$\ng(t) + 4g(t) \\le \\frac{9}{2} g(t).\n$$\n\nThus $g(t) \\le 0$ for all $t > 0$. For $0 < x \\le y$, we have\n\n$$\ng(x) \\ge g(y) + g(x - y) \\ge g(y).\n$$\n\nSo $g$ is non-increasing.\n\nLet $g(1) = a \\le 0$. We show $g(t) = a t$ for all $t > 0$. Suppose $g(t) < a t$ for some $t > 0$. Choose positive rational $\\frac{p}{q}$ such that $g(t) < a \\frac{p}{q}$. But $g(nt) = n g(t)$ for all $n \\in \\mathbb{N}$, so\n\n$$\ng\\left(\\frac{p}{q}\\right) = \\frac{p}{q} g(1) = \\frac{p a}{q}.\n$$\n\nThus\n\n$$\ng(t) \\ge g\\left(\\frac{p}{q}\\right) = \\frac{p a}{q},\n$$\n\na contradiction to $g(t) < a \\frac{p}{q}$. Similarly, $g(t) > a t$ is also impossible. Therefore, $g(t) = a t$ for all $t > 0$, so $f(x) = a x^2$ where $a \\le 0$. It is easy to verify that this indeed solves the equations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16597,
    "subject": "Mathematics (Olympiad)",
    "question": "Запишіть рівність у вигляді\n$$\n\\cos^2 x \\cos y - \\cos^2 y \\cos x = \\cos y - \\cos x,\n$$\nабо\n$$\n(\\cos x \\cos y + 1)(\\cos x - \\cos y) = 0.\n$$\nДля $x \\in (0; \\pi)$ і $y \\in (0; \\pi)$ виконується $\\cos x \\cos y > -1$. Знайдіть всі пари $(x, y)$, що задовольняють рівність.",
    "options": [],
    "answer": "See solution",
    "solution": "Оскільки $\\cos x \\cos y > -1$ для $x, y \\in (0; \\pi)$, то з $(\\cos x \\cos y + 1)(\\cos x - \\cos y) = 0$ маємо $\\cos x = \\cos y$. Функція $f(t) = \\cos t$ є спадною на $(0; \\pi)$, тому $x = y$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16598,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be real numbers with $ab + bc + ca = 1$. Prove that\n\n$$\n\\frac{(a+b)^2+1}{c^2+2} + \\frac{(b+c)^2+1}{a^2+2} + \\frac{(c+a)^2+1}{b^2+2} \\ge 3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that $c^2 + 2 = c^2 + ab + bc + ca + 1 = (c+a)(c+b) + 1$. Setting $x = a+b$, $y = b+c$, $z = c+a$, the inequality rewrites as $\\sum \\frac{x^2+1}{yz+1} \\ge 3$. Since $0 < 2xy \\le x^2 + y^2$, one has $\\sum \\frac{x^2+1}{y^2+1+z^2+1} \\ge \\frac{3}{2}$, which is the celebrated Nesbitt inequality for $A = x^2 + 1$, $B = y^2 + 1$, $C = z^2 + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16599,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all polynomials $P(x)$ with real coefficients satisfying the equality\n\n$$\n(x^2 - 6x + 8)P(x) = (x^2 + 2x)P(x - 2),\n$$\n\nfor all $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The given equation can be rewritten as\n\n$$\n(x-2)(x-4)P(x) = x(x+2)P(x-2), \\text{ for all } x \\in \\mathbb{R}.\n$$\n\nSubstituting $x = 0$, $x = -2$, and $x = 2$ gives $P(0) = P(-2) = P(2) = 0$.\n\nTherefore, $P(x)$ has the form\n\n$$\nP(x) = x(x+2)(x-2)Q(x),\n$$\n\nwhere $Q(x)$ is a polynomial with real coefficients.\n\nSubstituting this into the original equation gives\n\n$$\n(x-2)(x-4)^2 x(x+2)Q(x) = x(x+2)(x-2)x(x-4)Q(x-2).\n$$\n\nThis simplifies to\n\n$$\nx(x+2)(x-2)(x-4)[(x-2)Q(x) - xQ(x-2)] = 0.\n$$\n\nSince $x(x-2)(x+2)(x-4)$ is not the zero polynomial, we must have\n\n$$\n(x-2)Q(x) - xQ(x-2) = 0, \\text{ for all } x \\in \\mathbb{R}.\n$$\n\nSetting $x = 0$ gives $Q(0) = 0$, so $Q(x) = xR(x)$ for some polynomial $R(x)$.\n\nSubstituting back, we get\n\n$$\n(x-2)xR(x) - x(x-2)R(x-2) = 0 \\implies (x-2)[R(x) - R(x-2)] = 0.\n$$\n\nThus, $R(x) = R(x-2)$ for all $x$, so $R(x)$ is constant: $R(x) = c$.\n\nTherefore,\n\n$$\nP(x) = x(x+2)(x-2)xR(x) = c x^2(x^2 - 4).\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16600,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\ldots, x_{11}$ be nonnegative real numbers such that $x_1 + x_2 + \\cdots + x_{11} = 1$. For $i = 1, 2, \\ldots, 11$, define\n\n$$\ny_i = \\begin{cases} x_i + x_{i+1}, & \\text{if } i \\text{ is odd} \\\\ x_i + x_{i+1} + x_{i+2}, & \\text{if } i \\text{ is even} \\end{cases}\n$$\n\nwhere $x_{12} = x_1$. Let $F(x_1, x_2, \\ldots, x_{11}) = y_1 y_2 \\cdots y_{11}$.\n\nProve that if $F$ achieves its maximum, then $x_6 < x_8$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $F$ be viewed as a continuous function on the closed, bounded set\n\n$$\n\\Omega = \\{(x_1, x_2, \\ldots, x_{11}) \\in (\\mathbb{R}_{\\ge 0})^{11} \\mid x_1 + x_2 + \\cdots + x_{11} = 1\\}.\n$$\n\nSuppose $F$ attains its maximum at $(a_1, a_2, \\ldots, a_{11}) \\in \\Omega$ with $F > 0$ at this point.\n\n**Step 1:** $a_3 = a_5 = a_7 = a_9 = a_{11} = 0$.\n\nIf $a_3 > 0$, set $a'_3 = 0$, $a'_4 = a_3 + a_4$, and $a'_i = a_i$ for other $i$. This preserves the sum. Comparing $F$ at these points, we find $F(a_1, \\ldots, a_{11}) < F(a'_1, \\ldots, a'_{11})$, a contradiction. Thus, $a_3 = 0$. Similarly, $a_5 = a_7 = a_9 = a_{11} = 0$.\n\nNow, set $x_3 = x_5 = x_7 = x_9 = x_{11} = 0$ in $F$:\n\n$$\nF = (x_1 + x_2)(x_2 + x_4)x_4(x_4 + x_6)x_6(x_6 + x_8)x_8(x_8 + x_{10})x_{10}(x_{10} + x_1)x_1.\n$$\n\n**Step 2:** $a_2 = 0$.\n\nSuppose $a_2 > 0$. If $a_1 \\le a_4$, set $a'_1 = a_1 + a_2$, $a'_2 = 0$, $a'_i = a_i$ otherwise. Comparing $F$ at these points, we find $F(a_1, a_2, a_4, \\ldots, a_{10}) < F(a'_1, a'_2, a'_4, \\ldots, a'_{10})$, a contradiction. Similarly if $a_1 \\ge a_4$. Thus, $a_2 = 0$.\n\nNow, set $x_2 = 0$:\n\n$$\nF = x_4^2(x_4 + x_6)x_6(x_6 + x_8)x_8(x_8 + x_{10})x_{10}(x_{10} + x_1)x_1^2.\n$$\n\n**Step 3:** $a_1 = a_4$, $a_6 = a_{10}$.\n\nLet $a'_1 = a'_4 = \\frac{1}{2}(a_1 + a_4)$, $a'_6 = a'_{10} = \\frac{1}{2}(a_6 + a_{10})$, $a'_8 = a_8$. By the AM-GM inequality, $F(a_1, a_4, \\ldots, a_{10}) \\le F(a'_1, a'_4, \\ldots, a'_{10})$, with equality only if $a_1 = a_4$, $a_6 = a_{10}$.\n\nNow, $F$ as a function of $x_1, x_6, x_8$ (with $x_4 = x_1$, $x_{10} = x_6$):\n\n$$\nF = x_1^4 x_6^2 x_8 (x_1 + x_6)^2 (x_6 + x_8)^2,\n$$\n\nwith $2x_1 + 2x_6 + x_8 = 1$.\n\n**Step 4:** Solve the system:\n\n$$\n\\frac{2}{u} + \\frac{1}{u+v} = \\frac{1}{v} + \\frac{1}{u+v} + \\frac{1}{v+1} = 1 + \\frac{2}{v+1}, \\quad u, v > 0.\n$$\n\nFrom the first equation, $u = \\frac{2v(v+1)}{2v+1}$. Substituting into the second yields $4v^3 + 5v^2 - 4v - 4 = 0$. This cubic has a unique positive root $v \\in (0,1)$, so $u, v$ are uniquely determined.\n\n**Step 5:** By AM-GM, equality holds only if $x_1 : x_6 : x_8 = u : v : 1$, so $x_8 = \\frac{1}{2u+2v+1}$, $x_6 = \\frac{v}{2u+2v+1}$, $x_1 = \\frac{u}{2u+2v+1}$. Thus, $F$ is maximized only if $x_6 = v x_8 < x_8$ (since $v < 1$).\n\n$\\boxed{x_6 < x_8}$ as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16601,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $a$, $b$, $c$, $d$ such that\n$$\n4^a \\cdot 5^b - 3^c \\cdot 11^d = 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Reducing modulo $3$, we get $5^b \\equiv 1 \\pmod{3}$, so $b$ is even: $b = 2x$, $x \\in \\mathbb{N}^*$. It follows that\n$$\n(2^a \\cdot 5^x - 1)(2^a \\cdot 5^x + 1) = 3^c \\cdot 11^d.\n$$\nNoticing that $\\gcd(2^a \\cdot 5^x - 1, 2^a \\cdot 5^x + 1) = 1$ and $2^a \\cdot 5^x \\geq 10$, we have two cases:\n\n$$\n\\begin{aligned}\n(1) &\\quad \\begin{cases} 2^a \\cdot 5^x - 1 = 11^d \\\\ 2^a \\cdot 5^x + 1 = 3^c \\end{cases} \\\\\n(2) &\\quad \\begin{cases} 2^a \\cdot 5^x + 1 = 11^d \\\\ 2^a \\cdot 5^x - 1 = 3^c \\end{cases}\n\\end{aligned}\n$$\n\nThe first case is impossible, because reducing modulo $5$ gives $-1 \\equiv 1 \\pmod{5}$.\n\nThe second case leads to $11^d - 3^c = 2 = 11 - 3^2$. For $d = 1$, we have the solution $(a, b, c, d) = (1, 2, 2, 1)$. If $d \\geq 2$, then $c \\geq 5$, so we can rewrite the equation as\n$$\n11(11^{d-1} - 1) = 3^2(3^{c-2} - 1).\n$$\nBecause $11 \\mid 3^{c-2} - 1$ and the multiplicative order of $11$ modulo $3$ is $5$, it follows that $5 \\mid c - 2$, so $3^5 - 1 \\mid 3^{c-2} - 1$. Consequently, $2 \\cdot 11^2 = 3^5 - 1$ is a divisor of $11 (11^{d-1} - 1)$, which is absurd.\n\n**Remark.** Alternatively, reducing modulo $3$ the equation $11^d - 3^c = 2$ gives $2^d \\equiv 2$, so $d$ is odd. From $2^a \\cdot 5^x + 1 = 11^d$, working modulo $4$ gives $a = 1$ and\n$$\n2 \\cdot 5^x = 11^d - 1 = (11 - 1)(11^{d-1} + \\dots + 11 + 1),\n$$\nso $5^{x-1} = 11^{d-1} + \\dots + 11 + 1$. For $x \\geq 2$, $5 \\mid 11^{d-1} + \\dots + 11 + 1$, which implies $5 \\mid d$. Then $5^{x-1} = 11^{d-1} + \\dots + 11 + 1$ is divisible by $11^4 + 11^3 + 11^2 + 11 + 1 = 16105$, which is not a power of $5$.\n\nIt remains that $x = 1$, which leads to the unique solution $a = 1$, $b = 2$, $c = 2$, $d = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16602,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the biggest three-digit number that is divisible by 9 and whose digits are strictly increasing from left to right.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** 567.\n\n**Solution.** The only three-digit number that starts with 7 and whose digits are strictly increasing is 789, but it is not divisible by 9. There are three such numbers that start with 6: 678, 679, and 689, but, again, none of them is divisible by 9 since the sums of their digits are 21, 22, and 23 respectively. Now we consider numbers with strictly increasing digits that start with 5. The smallest of them is 567, and the sum of its digits is 18, which means that it is divisible by 9, and the biggest of them is 589 with the sum of digits $22 < 27$. This implies that the answer is 567.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16603,
    "subject": "Mathematics (Olympiad)",
    "question": "For a polynomial $P$ with integer coefficients and a positive integer $n$, define $P_n$ as the number of positive integer pairs $(a, b)$ such that $a < b \\leq n$ and $|P(a)| - |P(b)|$ is divisible by $n$.\n\nDetermine all polynomials $P$ with integer coefficients such that for all positive integers $n$, $P_n \\leq 2021$.",
    "options": [],
    "answer": "See solution",
    "solution": "There are two possible families of solutions:\n\n- $P(x) = x + d$, for some integer $d \\geq -2022$.\n- $P(x) = -x + d$, for some integer $d \\leq 2022$.\n\nSuppose $P$ satisfies the problem conditions. Clearly $P$ cannot be a constant polynomial. Notice that a polynomial $P$ satisfies the conditions if and only if $-P$ also satisfies them. Hence, we may assume the leading coefficient of $P$ is positive. Then, there exists a positive integer $M$ such that $P(x) > 0$ for $x \\geq M$.\n\n**Lemma 1.** For any positive integer $n$, the integers $P(1), P(2), \\dots, P(n)$ leave pairwise distinct remainders upon division by $n$.\n\n*Proof.* Assume for contradiction that this is not the case. Then, for some $1 \\leq y < z \\leq n$, there exists $0 \\leq r \\leq n - 1$ such that $P(y) \\equiv P(z) \\equiv r \\pmod{n}$. Since $P(an + b) \\equiv P(b) \\pmod{n}$ for all integers $a, b$, we have $P(an + y) \\equiv P(an + z) \\equiv r \\pmod{n}$ for any integer $a$. Let $A$ be a positive integer such that $An \\geq M$, and let $k$ be a positive integer such that $k > 2A + 2021$. Each of the $2(k - A)$ integers $P(An + y), P(An + z), P((A+1)n + y), P((A+1)n + z), \\dots, P((k-1)n + y), P((k-1)n + z)$ leaves one of the $k$ remainders\n\n$$\nr, n+r, 2n+r, \\dots, (k-1)n+r\n$$\n\nupon division by $kn$. This implies that at least $2(k - A) - k = k - 2A$ (possibly overlapping) pairs leave the same remainder upon division by $kn$. Since $k - 2A > 2021$ and all of the $2(k - A)$ integers are positive, we find more than $2021$ pairs $a, b$ with $a < b \\leq kn$ for which $|P(b)| - |P(a)|$ is divisible by $kn$—hence, $P_{kn} > 2021$, a contradiction.\n\nNext, we show that $P$ is linear. Assume that this is not the case, i.e., $\\deg P \\geq 2$. Then we can find a positive integer $k$ such that $P(k) - P(1) \\geq k$. This means that among the integers $P(1), P(2), \\dots, P(P(k) - P(1))$, two of them, namely $P(k)$ and $P(1)$, leave the same remainder upon division by $P(k) - P(1)$—contradicting the lemma (by taking $n = P(k) - P(1)$). Hence, $P$ must be linear.\n\nWe can now write $P(x) = cx + d$ with $c > 0$. We prove that $c = 1$ in two ways.\n\n**Solution 1:** If $c \\geq 2$, then $P(1)$ and $P(2)$ leave the same remainder upon division by $c$, contradicting the lemma. Hence $c = 1$.\n\n**Solution 2:** Suppose $c \\geq 2$. Let $n$ be a positive integer such that $n > 2cM$, $n \\left(1 - \\frac{3}{2c}\\right) > 2022$ and $2c \\mid n$. Notice that for any positive integer $i$ such that $\\frac{3n}{2c} + i < n$, $P\\left(\\frac{3n}{2c} + i\\right) - P\\left(\\frac{n}{2c} + i\\right) = n$. Hence, $\\left(\\frac{n}{2c} + i, \\frac{3n}{2c} + i\\right)$ satisfies the condition in the question for all positive integers $i$ such that $\\frac{3n}{2c} + i < n$. Hence, $P_n > 2021$, a contradiction. Then, $c = 1$.\n\nIf $d \\leq -2023$, then there are at least $2022$ pairs $a < b$ such that $P(a) = P(b)$, namely $(a, b) = (1, -2d - 1), (2, -2d - 2), \\dots, (-d - 1, -d + 1)$. This implies that $d \\geq -2022$.\n\nFinally, we verify that $P(x) = x + d$ satisfies the condition for any $d \\geq -2022$. Fix a positive integer $n$. Note that $||P(b)| - |P(a)|| < n$ for all positive integers $a < b \\leq n$, so the only pairs $a, b$ for which $|P(b)| - |P(a)|$ could be divisible by $n$ are those for which $|P(a)| = |P(b)|$. When $d \\geq -2022$, there are indeed at most $2021$ such pairs.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16604,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the set $\\{\\sqrt{1}, \\sqrt{2}, \\sqrt{3}, \\dots, \\sqrt{2015}\\}$ does not contain a non-constant arithmetic sequence of length 45.",
    "options": [],
    "answer": "See solution",
    "solution": "If $m, n, p \\in \\mathbb{N}^*$ and $\\sqrt{m}, \\sqrt{n}, \\sqrt{p}$ are in arithmetic progression, then\n\n$$\np + m + 2\\sqrt{pm} = 4n\n$$\n\nhence $\\sqrt{pm}$ is a rational number. It follows that $m = a^2 d$, $p = c^2 d$, $n = b^2 d$, where $a, b, c, d \\in \\mathbb{N}$ and $a + c = 2b$.\n\nTherefore, if one could choose 45 numbers in arithmetic sequence, they would have the form $a_1 \\sqrt{d}, a_2 \\sqrt{d}, \\dots, a_{45} \\sqrt{d}$, where $a_1, a_2, \\dots, a_{45}, d \\in \\mathbb{N}^*$ and $a_1, a_2, \\dots, a_{45}$ is an arithmetic sequence as well.\n\nIn this case, the largest of these numbers would be $\\sqrt{45^2 d} \\geq \\sqrt{2025}$, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16605,
    "subject": "Mathematics (Olympiad)",
    "question": "You are given $n \\geq 2$ distinct positive integers. For each pair $a < b$ of these numbers, consider the difference $b - a$. For each of these differences, Vlada writes down the maximum power of two by which this difference is divisible. What is the largest possible number of distinct numbers that Vlada could write?",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove by induction on $n$ that there are at most $n-1$ different degrees (powers of two) possible.\n\n**Base case:** For $n=2$, the only difference is $b-a$, so only one degree is possible.\n\n**Inductive step:** Assume the statement holds for $k \\leq n-1$. For $n$ numbers, suppose all are divisible by the same power of $2$, say $2^k$. If $k > 0$, divide all numbers by $2^k$; the degrees in all differences decrease by $k$, so the number of distinct degrees does not change. If all numbers are odd, add $1$ to each; differences remain unchanged, and all numbers become even. Repeat this process. Eventually, we obtain a set of numbers divisible by different powers of two.\n\nDivide the numbers into groups $A_1, A_2, \\dots, A_m$ according to the power of two dividing them ($\\alpha_1 < \\alpha_2 < \\dots < \\alpha_m$). The difference $a_i - a_j$ for $a_i \\in A_i$, $a_j \\in A_j$ ($i \\neq j$) is divisible by $2^{\\min(\\alpha_i, \\alpha_j)}$, so among differences between groups, there are only $m-1$ distinct degrees. Within each group, by induction, the number of distinct degrees is at most $|A_i| - 1$. Thus, the total number is:\n\n$$\n(|A_1| - 1) + (|A_2| - 1) + \\cdots + (|A_m| - 1) + (m - 1) = (|A_1| + |A_2| + \\cdots + |A_m|) - 1 = n - 1.\n$$\n\nTo show $n-1$ is possible for any $n \\geq 2$, take $a_1 = 1$, $a_2 = 2^2 + 1$, $a_3 = 2^3 + 1$, ..., $a_n = 2^n + 1$. Then, the pairs $(a_1, a_i)$ for $i = 2, \\dots, n$ have different degrees.\n\n**Alternative solution:** Suppose there are $n$ pairs with pairwise distinct degrees. Construct a graph with $n$ vertices, each representing a number, and connect two vertices if their numbers form one of these pairs. The graph has $n$ edges, so it contains a cycle. Let the numbers on the cycle be $a_1, a_2, \\dots, a_m$. The degrees of $a_1 - a_2, a_2 - a_3, \\dots, a_m - a_1$ are pairwise distinct. Let $k$ be the smallest degree, corresponding to $a_m - a_1$. Then all other differences are divisible by $2^{k+1}$, so $a_1 \\equiv a_2 \\equiv \\dots \\equiv a_m \\pmod{2^{k+1}}$, implying $a_m - a_1$ is divisible by $2^{k+1}$, a contradiction. The example is as above.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16606,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an integer greater than $1$. Find the greatest constant $\\lambda(n)$ such that for any nonzero complex numbers $z_1, z_2, \\dots, z_n$, we have\n\n$$\n\\sum_{k=1}^{n} |z_k|^2 \\geq \\lambda(n) \\min_{1 \\leq k \\leq n} \\left\\{ |z_{k+1} - z_k|^2 \\right\\},\n$$\n\nwhere $z_{n+1} = z_1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let\n\n$$\n\\lambda_0(n) = \\begin{cases} \\frac{n}{4}, & 2 \\mid n, \\\\ \\frac{n}{4 \\cos^2 \\frac{\\pi}{2n}}, & \\text{otherwise} \\end{cases}\n$$\n\nWe prove $\\lambda_0(n)$ is the greatest constant.\n\nIf there exists $k$ ($1 \\leq k \\leq n$) such that $|z_{k+1} - z_k| = 0$, the inequality holds obviously. So, without loss of generality, assume\n\n$$\n\\min_{1 \\leq k \\leq n} \\left\\{ |z_{k+1} - z_k|^2 \\right\\} = 1.\n$$\n\nUnder this assumption, it suffices to show that the minimum value of $\\sum_{k=1}^n |z_k|^2$ is $\\lambda_0(n)$.\n\n**Case 1: $n$ even.**\n\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} |z_k|^2 &= \\frac{1}{2} \\sum_{k=1}^{n} \\left( |z_k|^2 + |z_{k+1}|^2 \\right) \\\\ &\\geq \\frac{1}{4} \\sum_{k=1}^{n} |z_{k+1} - z_k|^2 \\\\ &\\geq \\frac{n}{4} \\min_{1 \\leq k \\leq n} \\left\\{ |z_{k+1} - z_k|^2 \\right\\} = \\frac{n}{4}\n\\end{aligned}\n$$\n\nEquality holds when $(z_1, z_2, \\dots, z_n) = (\\frac{1}{2}, -\\frac{1}{2}, \\dots, \\frac{1}{2}, -\\frac{1}{2})$, so the minimum value is $\\frac{n}{4} = \\lambda_0(n)$.\n\n**Case 2: $n$ odd.**\n\nLet $\\theta_k = \\arg \\frac{z_{k+1}}{z_k} \\in [0, 2\\pi)$ for $k = 1, 2, \\dots, n$.\n\nFor all $k$, if $\\theta_k \\leq \\frac{\\pi}{2}$ or $\\theta_k \\geq \\frac{3\\pi}{2}$, then\n\n$$\n\\begin{aligned}\n|z_k|^2 + |z_{k+1}|^2 &= |z_k - z_{k+1}|^2 + 2|z_k||z_{k+1}| \\cos \\theta_k \\\\ &\\geq |z_k - z_{k+1}|^2 \\geq 1\n\\end{aligned}\n$$\n\nIf $\\theta_k \\in (\\frac{\\pi}{2}, \\frac{3\\pi}{2})$, then $\\cos \\theta_k < 0$ and\n\n$$\n\\begin{aligned}\n1 &\\leq |z_k - z_{k+1}|^2 \\\\ &= |z_k|^2 + |z_{k+1}|^2 - 2|z_k||z_{k+1}| \\cos \\theta_k \\\\ &\\leq (|z_k|^2 + |z_{k+1}|^2)(1 + (-2\\cos \\theta_k)) \\\\ &= (|z_k|^2 + |z_{k+1}|^2) \\cdot 2\\sin^2 \\frac{\\theta_k}{2}\n\\end{aligned}\n$$\n\nTherefore, ...",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16607,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all triples $ (x, y, z) $ of positive integers satisfying\n\n$$\nx \\mid (y + 1), \\quad y \\mid (z + 1), \\quad \\text{and} \\quad z \\mid (x + 1).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "There are ten triples satisfying the three conditions. They are given by $ (1, 1, 1) $, $ (1, 1, 2) $, $ (1, 3, 2) $, $ (3, 5, 4) $ and their cyclic permutations.\n\nWithout loss of generality, let $ x $ be the smallest of the three numbers (or one of the smallest), i.e. $ x \\le y $ and $ x \\le z $. From $ z \\mid x + 1 $ we obtain $ x \\le z \\le x + 1 $. Thus we have to consider two cases.\n\n*Case 1.* Let $ z = x $. Then $ z = x \\mid x + 1 $ leads to $ x = z = 1 $ and $ y \\mid z + 1 = 2 $. Therefore $ y = 1 $ or $ y = 2 $, and we get the two solutions $ (1, 1, 1) $ and $ (1, 2, 1) $.\n\n*Case 2.* Let $ z = x + 1 $. Then the two conditions $ x \\mid y + 1 $ and $ y \\mid x + 2 $ must be fulfilled. In particular, we obtain $ x \\le y + 1 $ and $ y \\le x + 2 $. This yields $ x - 1 \\le y \\le x + 2 $ and we have to examine the following cases for $ y $.\n\n*Case 2a.* Let $ 0 < y = x $. The conditions $ x \\mid x + 1 $ and $ x \\mid x + 2 $ can only hold simultaneously for $ x = 1 $, giving the solution $ (1, 1, 2) $.\n\n*Case 2b.* Let $ y = x + 1 $. Then the two conditions are $ x \\mid x + 2 $ and $ x + 1 \\mid x + 2 $. They cannot hold simultaneously.\n\n*Case 2c.* Let $ y = x + 2 $. The condition $ y = x + 2 \\mid x + 2 = z + 1 $ is trivially fulfilled. The requirement $ x \\mid y + 1 = x + 3 $ can only hold for $ x \\mid 3 $. And, indeed, for either $ x = 1 $ or $ x = 3 $ the condition is fulfilled and we obtain the solutions $ (1, 3, 2) $ and $ (3, 5, 4) $.\n\nSumming up, the triples $ (1, 1, 1) $, $ (1, 2, 1) $, $ (1, 1, 2) $, $ (1, 3, 2) $ and $ (3, 5, 4) $ fulfill all three conditions. As each of the three numbers can be the minimum, every cyclic permutation of these triples is a solution as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16608,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an isosceles triangle with $|AB| = |AC|$. Let $D$ be an arbitrary point on segment $BC$. Let $X$ and $Y$ be points on $AB$ and $AC$, respectively, such that $XY \\parallel BC$ and $XY$ passes through the midpoint of $AD$. Prove that if the circumcenter of $\\triangle ABC$ lies on $\\odot(AXY)$ then the quadrilateral $AYDX$ is a parallelogram.\n\nHere $\\odot(P_1P_2P_3)$ denotes the circumcircle of triangle $\\triangle P_1P_2P_3$.\n\n![](images/BW23_Shortlist_2023-11-01_p76_data_fb63f1043c.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $\\angle XYO = \\angle XAO = \\angle OAY = \\angle OXY$, hence the triangle $\\triangle OXY$ is isosceles. Let $K, L, M, N$ be the midpoints of $AB, AC, AD, XY$ respectively. Note that $K, L, M$ are collinear because they lie on the midline of $\\triangle ABC$ parallel to $BC$. Moreover, $M$ lies on $XY$ by assumption. Since $XY \\parallel BC \\parallel KL$, the lines $KL$ and $XY$ do not coincide. This means that the lines $XY$ and $KL$ intersect at $M$.\n\nNote that $K, L, N$ are the projections of $O$ onto $AX, AY$ and $XY$, respectively. Therefore $K, L, N$ lie on the Simson line of $O$ with respect to $\\triangle AXY$. In other words, the line $XY$ intersects $KL$ at $N$.\n\nThis means that $M = N$, i.e. the segments $AD$ and $XY$ share a common midpoint. Therefore $AXDY$ is a parallelogram.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16609,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a scalene triangle with incircle $\\omega$ and incenter $I$. The tangent line at $C$ to $\\omega$ intersects line $AB$ at $D$. The angle bisector of $\\angle BDC$ meets $BI$ at $P$ and $AI$ at $Q$. Let $M$ be the midpoint of segment $PQ$. Prove that line $IM$ passes through the midpoint of the arc $ACB$ of the circle $\\omega$.\n\n![](images/RMC_2024_p73_data_a65f0b100b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $E$ and $F$ the intersections of rays $AI$ and $BI$ with $\\omega$, and let $N$ be the midpoint of the arc $ACB$.\n\nLine $CD$ is tangent to $\\omega$, so $\\angle ACD = \\angle B$. Also, $\\angle ADC = \\angle BDC = 180^\\circ - \\angle DBC - \\angle BCD = \\angle A - \\angle B$, so $\\angle IQP = \\angle AQD = \\angle ABE - \\angle ADQ = \\frac{\\angle B}{2}$. Since $\\angle AEF = \\frac{AF}{2} = \\frac{\\angle B}{2}$, the corresponding angles $AQP$ and $AEF$ are equal, so $PQ \\parallel EF$.\n\nMoreover, $\\overline{FN} = \\overline{AN} - \\overline{AF} = \\frac{360^\\circ - \\overline{AB}}{2} - \\overline{AF} = 180^\\circ - \\angle C - \\angle B = \\angle A = \\overline{BE}$, so $BF \\parallel EN$, hence $IF \\parallel EN$.\n\nFrom $\\overline{EN} = \\overline{BN} - \\overline{BE} = \\frac{360^\\circ - \\overline{AB}}{2} - \\overline{BE} = 180^\\circ - \\angle C - \\angle A = \\angle B = \\overline{AF}$, we infer that $AE \\parallel FN$, so $IE \\parallel FN$ and the quadrilateral $IENF$ is a parallelogram.\n\nLet $T = EF \\cap IN$. Since $T$ is the midpoint of segment $EF$, and $PQ \\parallel EF$, it follows that $IT$ passes through the midpoint of segment $PQ$, so $I$, $M$, $T$, $N$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16610,
    "subject": "Mathematics (Olympiad)",
    "question": "設三角形 $ABC$ 的內心為 $I$，直線 $l$ 是線段 $AI$ 的中垂線。設點 $P$ 落在三角形 $ABC$ 的外接圓上，且令直線 $AP$ 與 $l$ 交於點 $Q$。點 $R$ 位於 $l$ 上，滿足 $\\angle IPR = 90^\\circ$。設直線 $IQ$ 與三角形 $ABC$ 中平行於 $BC$ 邊的中位線交於點 $M$。\n\n試證：$\\angle AMR = 90^\\circ$。\n\n(註：三角形中，任兩邊中點的連線，稱為中位線。)",
    "options": [],
    "answer": "See solution",
    "solution": "令 $A'$ 為 $A$ 關於 $M$ 的對稱點，$I'$ 為 $I$ 關於 $P$ 的對稱點。則 $A'$ 位於 $BC$ 上，且 $R$ 為 $\\triangle AII'$ 的外心，因此 $\\angle AMR$ 為直角若且唯若 $A, I, A', I'$ 共圓。\n\n![](images/18-3J_p11_data_096569e6d2.png)\n\n以下證明 $\\triangle AIP \\sim \\triangle IA'P$（～代表正向相似）：取 $J$ 使得 $AIA'J$ 為平行四邊形，取 $P'$ 使得 $\\triangle AIP' \\sim \\triangle JIA'$，則\n\n$$\n\\triangle IA'P' \\sim \\triangle IJA' \\sim \\triangle JIA' \\sim \\triangle AIP',\n$$\n\n且由\n\n$$\n\\angle IAP' = \\angle JIA = \\angle QIA = \\angle IAQ,\n$$\n\n知 $A, Q, P'$ 共線。取 $Y$ 使得 $\\triangle AIP' \\sim \\triangle YBP'$，由 $AYIP' \\sim IBA'P'$ 可得\n\n$$\n\\angle AYI = \\angle IBA' = \\angle ABI,\n$$\n\n即 $A, Y, B, I$ 共圓。所以\n\n$$\n\\begin{aligned}\n\\angle CBP' &= \\angle IYP' = \\angle BYP' - \\angle BYI = \\angle IAP' - \\angle BAI \\\\\n&= \\angle IAP' - \\angle IAC = \\angle CAP',\n\\end{aligned}\n$$\n\n故 $A, B, C, P'$ 共圓，所以 $P'$ 為 $AQ$ 與 $\\bigodot(ABC)$ 的另一個交點，即 $P = P'$。\n\n由 $\\triangle AIP \\sim \\triangle IA'P$ 易得 $\\triangle AI'P \\sim \\triangle I'A'P$，所以\n\n$$\n\\angle IAI' = \\angle IAP + \\angle PAI' = \\angle A'IP + \\angle PI'A' = \\angle IA'I'\n$$\n\n即 $A, I, A', I'$ 共圓，從而原命題得證。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16611,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $A$, $B$, $C$, $D$, and $E$ lie, in that order, on a circle, and the lines $AB$ and $ED$ are parallel. Prove that $\\angle ABC = 90^\\circ$ if and only if $AC^2 = BD^2 + CE^2$.\n\n![](images/V_Britanija_2010_p16_data_87ad89e7ec.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The equation $AC^2 = BD^2 + CE^2$ suggests using Pythagoras' theorem; however, the lengths $AC$, $BD$, and $CE$ do not form the sides of a triangle as they stand. Looking at triangle $ACE$, we see that if we knew that $AE = BD$ (i.e., that two opposite sides of the cyclic trapezium $ABDE$ are equal), then we could apply Pythagoras. Let's try to prove this.\n\nSince $ABDE$ is cyclic, we know that $\\angle AEB = \\angle ADB$ and $\\angle ABE = \\angle ADE$. Because $AB$ and $ED$ are parallel, $\\angle ABE = \\angle ADE = \\angle BAD$. So triangles $AEB$ and $BDA$ are congruent by AAS congruency, whence $AE = BD$, which is what we wanted.\n\nNow, $\\angle ABC + \\angle AEC = 180^\\circ$, and so $\\angle ABC = 90^\\circ$ if and only if $\\angle AEC = 90^\\circ$. By Pythagoras' theorem and its converse, applied in triangle $ACE$, $\\angle AEC = 90^\\circ$ if and only if $AC^2 = AE^2 + CE^2 = BD^2 + CE^2$. Therefore, $AC^2 = BD^2 + CE^2$ if and only if $\\angle ABC = 90^\\circ$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16612,
    "subject": "Mathematics (Olympiad)",
    "question": "Two points $M$ and $N$ have been selected from the edge $BC$ of a tetrahedron $ABCD$ so that the point $M$ lies between $B$ and $N$, the point $N$ lies between $M$ and $C$, and the angle between the planes $AND$ and $ACD$ is equal to the angle between the planes $AMD$ and $ABD$. Prove that\n\n$$\n\\frac{MB}{MC} + \\frac{NB}{NC} \\geq 2 \\cdot \\frac{|\\triangle ABD|}{|\\triangle ACD|},\n$$\n\nwhere $|\\triangle ABD|$ is the area of the triangle $\\triangle ABD$, and similarly $|\\triangle ACD|$ is the area of $\\triangle ACD$.\n\n![](images/BW2019-problems-solutions-opinions_p39_data_a582377162.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let the angle between the planes $AND$ and $ACD$ be $\\alpha$, and the angle between the planes $AMD$ and $AND$ be $\\beta$. For four points $P, Q, R, S$ not coplanar, denote the volume of tetrahedron $PQRS$ by $|PQRS|$.\n\nThe triangles $MBD$, $MCD$, $NBD$, and $NCD$ share a common height (the distance from $D$ to $BC$), so\n\n$$\n\\frac{MB}{MC} + \\frac{NB}{NC} = \\frac{|\\triangle MBD|}{|\\triangle MCD|} + \\frac{|\\triangle NBD|}{|\\triangle NCD|}.\n$$\n\nSince $M, B, C, D$ are coplanar and $A$ is not in that plane,\n\n$$\n\\frac{|\\triangle MBD|}{|\\triangle MCD|} = \\frac{|MBDA|}{|MCDA|}, \\quad \\frac{|\\triangle NBD|}{|\\triangle NCD|} = \\frac{|NBDA|}{|NCDA|}.\n$$\n\nThe volume $|MBDA| = \\frac{1}{3} |\\triangle ABD| \\cdot h$, where $h$ is the distance from $M$ to plane $ABD$. Let $P$ be the projection of $M$ onto $AD$, and $Q$ the projection onto $ABD$. Then $h = MQ$, $\\angle QPM = \\alpha$, and $|\\triangle AMD| = \\frac{1}{2} AD \\cdot PM$. Since $h = PM \\sin \\alpha$,\n\n$$\n|MBDA| = \\frac{2 |\\triangle ABD| \\cdot |\\triangle AMD| \\sin \\alpha}{3 AD}.\n$$\n\nSimilarly,\n\n$$\n|NBDA| = \\frac{2 |\\triangle ABD| \\cdot |AND| \\sin (\\alpha + \\beta)}{3 AD},\n$$\n\n$$\n|MCDA| = \\frac{2 |\\triangle ACD| \\cdot |AMD| \\sin (\\alpha + \\beta)}{3 AD},\n$$\n\n$$\n|NCDA| = \\frac{2 |\\triangle ACD| \\cdot |AND| \\sin \\alpha}{3 AD}.\n$$\n\n![](images/BW2019-problems-solutions-opinions_p40_data_ab49e16c7d.png)\n\nApplying the arithmetic-geometric mean inequality,\n\n$$\n\\begin{aligned}\n\\frac{MB}{MC} + \\frac{NB}{NC} &= \\frac{|MBDA|}{|MCDA|} + \\frac{|NBDA|}{|NCDA|} \\\\\n&\\geq 2 \\sqrt{\\frac{|MBDA| \\cdot |NBDA|}{|MCDA| \\cdot |NCDA|}} \\\\\n&= 2 \\cdot \\frac{|\\triangle ABD|}{|\\triangle ACD|}.\n\\end{aligned}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16613,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the probability that a randomly chosen factor of $48$ is not a multiple of $4$.",
    "options": [],
    "answer": "See solution",
    "solution": "The factors of $48$ are $1, 2, 3, 4, 6, 8, 12, 16, 24, 48$. Among these, the ones not divisible by $4$ are $1, 2, 3, 6$. There are $4$ such factors out of $10$ total, so the probability is $\\frac{4}{10} = 0.4$ or $40\\%$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16614,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the non-positive integers $m, n$ satisfying the equation:\n\n$$\n\\frac{n(n+2)}{4} = m^4 + m^2 - m + 1\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For $n=0$, the equation becomes $m^4 + m^2 - m + 1 = 0$, which is impossible because $m^4 + m^2 - m + 1 = m^4 + m(m-1) + 1 > 0$ for all $m \\in \\mathbb{Z}_+$. For $m=0$, the equation becomes $n(n+2) = 4$, which is impossible in the set of non-negative integers. For $m, n \\neq 0$, the equation becomes:\n\n$$\nn^2 - 2n - 4(m^4 + m^2 - m + 1) = 0$$\n\nwhich is quadratic in $n$. For integer solutions, the discriminant must be a perfect square:\n\n$$\n\\begin{aligned}\n\\Delta &= 4(1 + 4m^4 + 4m^2 - 4m + 4) \\\\\n&= 4m^4 + 4m^2 - 4m + 5\n\\end{aligned}\n$$\n\nWe have $4m^4 + 4m^2 - 4m + 5 = (2m^2)^2 + 4m(m-1) + 5 > (2m^2)^2$ for every positive integer $m$, and also:\n\n$$\n4m^4 + 4m^2 - 4m + 5 = (2m^2 + 1)^2 - 4(m-1) \\leq (2m^2 + 1)^2\n$$\n\nEquality holds only for $m=1$. Thus, the unique possible value for $m$ is $m=1$. Then the equation becomes $n(n+2)=8$, so $n=2$. Therefore, the pair $(m, n) = (1, 2)$ is a solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16615,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $U$ be a set of $m$ triangles. Prove that there exists a subset $W$ of $U$ satisfying the following conditions:\n\n1. The number of triangles in $W$ is at least $0.45 m^{4/5}$.\n2. There do not exist 6 distinct points $A, B, C, D, E, F$ such that $W$ contains all 6 triangles $ABC$, $BCD$, $CDE$, $DEF$, $EFA$, and $FAB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $U'$ be a subset of $U$ formed by choosing each triangle independently with probability $p$. The expected number of triangles in $U'$ is $mp$.\n\nA sequence of 6 distinct points $(x_1, x_2, \\dots, x_6)$ is called a *bad configuration* if all 6 triangles $x_1x_2x_3$, $x_2x_3x_4$, $x_3x_4x_5$, $x_4x_5x_6$, $x_5x_6x_1$, $x_6x_1x_2$ belong to $U$.\n\nThe number of bad configurations in $U$ is at most $m(m-1)(3!)^2 \\leq 36m^2$, since it is at most the number of ways to select two triangles, choose $x_1, x_2, x_3$ from the first, and $x_4, x_5, x_6$ from the second.\n\nEach bad configuration can be cyclically permuted or reversed, so $12$ such sequences form a *bunch*. Thus, the number of bunches is at most $36m^2/12 = 3m^2$. The probability that a fixed bad configuration is in $U'$ is $p^6$, so the expected number of bunches in $U'$ is at most $3m^2p^6$.\n\nTherefore, the expected number of triangles in $U'$ minus the number of bunches is at least $mp - 3m^2p^6$.\n\nTake $p = c m^{-1/5}$. Then:\n\n$$\nmp - 3m^2p^6 = c m^{4/5} - 3c^6 m^{4/5} = (c - 3c^6) m^{4/5}.\n$$\n\nTaking $c = 1/2$, we get $mp - 3m^2p^6 \\geq (\\frac{1}{2} - \\frac{3}{64}) m^{4/5} \\geq 0.45 m^{4/5}$.\n\nThus, there exists a subset $U'$ of $U$ such that the number of triangles in $U'$ minus the number of bunches is at least $0.45 m^{4/5}$. Let $W$ be a subset of $U'$ obtained by discarding one triangle from each bunch. Then $W$ contains at least $0.45 m^{4/5}$ triangles and no bad configurations, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16616,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence of positive integers $\\{s_i\\}$ is constructed as follows: $s_1 = 2$, and for all $i > 1$, $s_i$ is the least number larger than $s_{i-1}$ that contains the digit:\n\n- \"2\" if $i \\equiv 1 \\pmod{4}$\n- \"0\" if $i \\equiv 2 \\pmod{4}$\n- \"1\" if $i \\equiv 3 \\pmod{4}$\n- \"4\" if $i \\equiv 0 \\pmod{4}$\n\nThis sequence starts as $2, 10, 11, 14, 20, 30, 31, 34, 42, 50, \\ldots$\n\nDoes $\\{s_i\\}$ contain the number:\n\na) $2001$\n\nb) $2004$?",
    "options": [],
    "answer": "See solution",
    "solution": "One can notice that $s_{i+1} \\leq s_i + 10$ (since at least one of ten consecutive numbers contains the required digit), so $s_{i+4} \\leq s_i + 40$. Thus, there exists $k$ such that $1895 \\leq s_{4k} \\leq 1934$, and since $s_{4k}$ must contain the digit 4, it can only be one of $\\{1904, 1914, 1924, 1934\\}$.\n\nContinuing the sequence in each case, we get:\n\n- $1904, 1912, 1920, 1921, 1924, 1925, 1930, 1931, 1934, \\ldots$\n- $1914, 1920, 1930, 1931, 1934, \\ldots$\n- $1924, 1925, 1930, 1931, 1934, \\ldots$\n\nIn all cases, $s_{4k} = 1934$ for some $k$. Continuing from $1934$:\n\n$$\n1934, 1942, 1950, 1951, 1954, 1962, 1970, 1971, 1974, 1982, 1990, 1991, 1994, 2000, 2001, 2010, 2014, \\ldots\n$$\n\nWe see that $2001$ appears in $\\{s_i\\}$, but $2004$ does not.\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p286_data_3078b878e2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16617,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the remainder when $7^{2024} + 7^{2025} + 7^{2026}$ is divided by 19?\n\n(A) 0  (B) 1  (C) 7  (D) 11  (E) 18",
    "options": [],
    "answer": "See solution",
    "solution": "The quantity in question can be rewritten as:\n\n$$\n7^{2024} + 7^{2025} + 7^{2026} = 7^{2024} (1 + 7 + 7^2) = 7^{2024} \\cdot 57 = 7^{2024} \\cdot 3 \\cdot 19.\n$$\n\nTherefore, the remainder when it is divided by 19 is 0.\n\nAlternatively, working modulo 19:\n\n$7^0 = 1$, $7^1 = 7$, $7^2 = 49 \\equiv 11$, and $7^3 = 7 \\cdot 11 = 77 \\equiv 1$.\n\nThus, the remainders of powers of 7 modulo 19 repeat every 3 terms. The sum of any three consecutive remainders is $1 + 7 + 11 = 19$, so the remainder when the given sum is divided by 19 is 0.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16618,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n$ 是正整數，且令 $a_1, \\cdots, a_{n-1}$ 為任意實數。定義數列 $u_0, \\cdots, u_n$ 與 $v_0, \\cdots, v_n$ 如下：\n\n$$u_0 = u_1 = v_0 = v_1 = 1$$\n\n$$u_{k+1} = u_k + a_k u_{k-1}, \\quad v_{k+1} = v_k + a_{n-k} v_{k-1} \\text{，對於 } k = 1, \\cdots, n-1$$\n\n試證：$u_n = v_n$。\n\nLet $n$ be a positive integer and let $a_1, \\cdots, a_{n-1}$ be arbitrary real numbers. Define the sequences $u_0, \\cdots, u_n$ and $v_0, \\cdots, v_n$ inductively by $u_0 = u_1 = v_0 = v_1 = 1$, and\n\n$$u_{k+1} = u_k + a_k u_{k-1}, \\quad v_{k+1} = v_k + a_{n-k} v_{k-1} \\text{ for } k = 1, \\cdots, n-1.$$ \n\nProve that $u_n = v_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove by induction on $k$ that\n\n$$u_k = \\sum_{0 < i_1 < \\dots < i_t < k \\atop i_{j+1} - i_j \\ge 2} a_{i_1} \\dots a_{i_t}$$\n\nNote that there is one trivial summand equal to 1 (corresponding to $t=0$ and the empty sequence, whose product is 1).\n\nFor $k=0, 1$, the sum on the right-hand side only contains the empty product, so the formula holds since $u_0 = u_1 = 1$. For $k \\ge 1$, assuming the result is true for $0, 1, \\dots, k$, we have\n\n$$\n\\begin{align*}\n  u_{k+1} &= \\sum_{0 < i_1 < \\dots < i_t < k \\atop i_{j+1} - i_j \\ge 2} a_{i_1} \\dots a_{i_t} + \\sum_{0 < i_1 < \\dots < i_{t-1} < k-1 \\atop i_{j+1} - i_j \\ge 2} a_{i_1} \\dots a_{i_{t-1}} \\cdot a_k \\\\\n  &= \\sum_{0 < i_1 < \\dots < i_t < k+1 \\atop i_{j+1} - i_j \\ge 2, k \\notin \\{i_1, \\dots, i_t\\}} a_{i_1} \\dots a_{i_t} + \\sum_{0 < i_1 < \\dots < i_{t-1} < k+1 \\atop i_{j+1} - i_j \\ge 2, k \\in \\{i_1, \\dots, i_t\\}} a_{i_1} \\dots a_{i_{t-1}} \\\\\n  &= \\sum_{0 < i_1 < \\dots < i_t < k+1 \\atop i_{j+1} - i_j \\ge 2} a_{i_1} \\dots a_{i_t}\n\\end{align*}\n$$\n\nas required.\n\nApplying this formula to the sequence $b_1, \\dots, b_n$ given by $b_k = a_{n-k}$ for $1 \\le k \\le n$, we get\n\n$$\nv_k = \\sum_{0 < i_1 < \\dots < i_t < k \\atop i_{j+1} - i_j \\ge 2} b_{i_1} \\dots b_{i_t} = \\sum_{n > i_1 > \\dots > i_t > n-k \\atop i_j - i_{j+1} \\ge 2} a_{i_1} \\dots a_{i_t}\n$$\n\nFor $k=n$, the expressions for $u_n$ and $v_n$ coincide, so indeed $u_n = v_n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16619,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Find the greatest prime factor (GPF) for each composite number from 60 to 65.\n\nb) The third composite is a multiple of 79. List the sequence of such multiples and determine the first composite in this sequence that is a multiple of 19. What is the minimum value for the first composite?\n\nc) Alternative i: Find the smallest sequence of four successive composite numbers where the third composite is a power of 2 and the second composite is at least $2 \\times 73 = 146$. List the relevant multiples and composites, and identify the required sequence.\n\nAlternative ii: If the third composite has a GPF of 2, what is the smallest sequence of four successive composite numbers meeting the conditions above?\n\nd) What is the largest composite less than 1000 with a greatest prime factor of 3? List the powers of 2 and 3 less than 1000, and determine the largest such composite.",
    "options": [],
    "answer": "See solution",
    "solution": "a) The prime factors for each composite from 60 to 65 are:\n\n- $60 = 2 \\times 2 \\times 3 \\times 5$ (GPF: 5)\n- $61$ is prime\n- $62 = 2 \\times 31$ (GPF: 31)\n- $63 = 3 \\times 3 \\times 7$ (GPF: 7)\n- $64 = 2^6$ (GPF: 2)\n- $65 = 5 \\times 13$ (GPF: 13)\n\nSo, the GPF sequence for the composites from 60 to 65 is: 5, 31, 7, 2, 13.\n\nb) The third composite is a multiple of 79, so possible values are:\n$158, 237, 316, 395, 474, 553, 632, 711, 790, 869, 948, 1027, \\dots$\nThe composites immediately before these are:\n$156, 236, 315, 394, 473, 552, 630, 710, 789, 868, 946, 1026, \\dots$\nThe first of these that is a multiple of 19 is 1026. So the first composite is at least 1025. Hence, all the composites have at least four digits.\n\nc) Alternative i:\nThe first few multiples of 73 are: 73, 146, 219, 292, 365, 438, 511.\nThe next composites after these are: 74, 147, 220, 294, 366, 440, 512.\nThe first of these with GPF 2 is 512. The GPFs of 510 and 513 are 17 and 19, respectively, so the smallest required successive composites are 510, 511, 512, 513.\n\nAlternative ii:\nThe third composite has a GPF of 2, so it is a power of 2. The second composite must be at least $2 \\times 73 = 146$. Thus, the third composite must be one of 256, 512, ...\nThe composites immediately before these are 255, 511, ...\nThe first of these with GPF 73 is 511. The GPFs of 510 and 513 are 17 and 19, respectively, so the smallest required successive composites are 510, 511, 512, 513.\n\nd) If a composite has a GPF of 3, it is a product of the form $3 \\times \\dots \\times 3 \\times 2 \\times \\dots \\times 2$, with at least one 3.\n\nThe powers of 2 less than 1000 are: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512.\nThe positive powers of 3 less than 1000 are: 3, 9, 27, 81, 243, 729.\n\nMultiply each positive power of 3 by the highest power of 2 that gives a number less than 1000:\n\n$$\n\\begin{array}{l@{\\hspace{2em}}l@{\\hspace{2em}}l}\n729 \\times 1 = 729 & 27 \\times 32 = 864 \\\\\n243 \\times 4 = 972 & 9 \\times 64 = 576 \\\\\n81 \\times 8 = 648 & 3 \\times 256 = 768\n\\end{array}\n$$\n\nHence, 972 is the largest composite less than 1000 with a GPF of 3.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16620,
    "subject": "Mathematics (Olympiad)",
    "question": "Sean $A$, $B$ y $C$ los vértices de un triángulo y $P$, $Q$ y $R$ los respectivos pies de las bisectrices trazadas desde esos mismos vértices. Sabiendo que $PQR$ es un triángulo rectángulo en $P$, se te pide probar dos cosas:\n\na) Que $ABC$ ha de ser obtusángulo.\n\nb) Que en el cuadrilátero $ARPQ$, pese a no ser cíclico, la suma de sus ángulos opuestos es constante.",
    "options": [],
    "answer": "See solution",
    "solution": "Para resolver el problema utilizaremos dos herramientas fundamentales en los problemas geométricos relativos a triángulos: el teorema de la bisectriz (La bisectriz de un ángulo de un triángulo corta al lado opuesto en dos segmentos de longitudes proporcionales a los otros dos lados del triángulo) y el teorema de Stewart (Si en un triángulo $ABC$ consideramos un punto $P$ en el lado $AB$, entonces se cumple que $AP^2 a = PB b^2 + PC c^2 - PB \\cdot PC \\cdot a$).\n\nSea ahora el punto $P$ pie de la bisectriz de nuestro problema. Por el teorema de la bisectriz tenemos:\n\n$$\nPB = \\frac{ac}{b+c} \\quad \\text{y} \\quad PC = \\frac{ab}{b+c}.\n$$\n\nAhora, aplicando el teorema de Stewart queda:\n\n$$\nAP^2 = \\frac{bc}{(b+c)^2} (b^2 + 2bc + c^2 - a^2)\n$$\n\ny, sustituyendo $a^2$ por su expresión obtenida del teorema del coseno:\n\n$$\nAP = \\frac{2bc}{(b+c)^2} \\cos \\frac{A}{2}.\n$$\n\nCalculemos ahora los lados del triángulo $PQR$.\n\n![](images/Spanija_b_2013_p18_data_8c163c2ceb.png)\n\nPor el teorema del coseno y los teoremas de la bisectriz y de Stewart obtenemos:\n\n$$\nPR^2 = AP^2 + QA^2 - 2AP \\cdot RA \\cos A = \\frac{4b^2c^2(a-b)\\cos^2 \\frac{A}{2}}{(b+c)^2(a+b)} + \\frac{b^2c^2}{(a+b)^2},\n$$\n$$\nQR^2 = QA^2 + RA^2 - 2QA \\cdot RA \\cos A = \\frac{4b^2c^2(a-b)\\cos^2 \\frac{A}{2}}{(b+c)^2(a+b)} + \\frac{b^2c^2}{(a+b)^2},\n$$\n$$\nPR^2 = AP^2 + QA^2 - 2AP \\cdot RA \\cos A = \\frac{b^2c^2}{(a+c)^2} + \\frac{b^2c^2}{(a+b)^2} - \\frac{2b^2c^2 \\cos A}{(a+b)(a+c)}.\n$$\n\nComo el triángulo $PQR$ es rectángulo, $PQ^2 + PR^2 = QR^2$. Sustituyendo y simplificando, tenemos:\n\n$$\n2a^2 - b^2 - c^2 + (2a^2 + 2bc) \\cos A = 0\n$$\n\nde donde sale $\\cos A = -\\frac{1}{2}$, $A = 120^\\circ$ y de aquí $R + Q = 150^\\circ$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16621,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of natural numbers $(m, n)$ such that $2m - 1$ is divisible by $n$ and $2n - 1$ is divisible by $m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $k, l \\in \\mathbb{N}$ such that $2m - 1 = kn$ and $2n - 1 = lm$. Then:\n\n$$\n4n - 2 = 2(2n - 1) = 2 \\cdot lm = 2m \\cdot l = (nk + 1)l.\n$$\n\nFrom $4n - 2 = (nk + 1)l$ we get:\n\n$$\n(4 - kl)n = l + 2.\n$$\n\nSince $l + 2 > 0$, $4 - kl > 0$, so $kl < 4$. The possibilities for positive integers $k, l$ are:\n\n- $kl = 1$: $k = l = 1$. Then $2m - 1 = n$, $2n - 1 = m$, so $m = n = 1$.\n- $kl = 2$: Not possible, since $k$ and $l$ must be odd.\n- $kl = 3$: $k = 3, l = 1$ or $k = 1, l = 3$.\n    - $k = 3, l = 1$: $2m - 1 = 3n$, $2n - 1 = m$ gives $m = 5, n = 3$.\n    - $k = 1, l = 3$: $2m - 1 = n$, $2n - 1 = 3m$ gives $m = 3, n = 5$.\n\nThus, all solutions are $(m, n) \\in \\{(1, 1), (3, 5), (5, 3)\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16622,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a convex quadrilateral $ABCD$ with $AB = BD = a$ and $CD = DA = b$. Let $P$ be a point on side $AB$ such that $DP$ is the angle bisector of $\\angle ADB$, and $Q$ be a point on side $BC$ such that $DQ$ is the angle bisector of $\\angle CDB$. Find the circumradius of triangle $DPQ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the parallel to $AD$ through $P$ intersect $BD$ at $O$. We show that $O$ is the circumcenter of $DPQ$.\n\nOne has $\\angle ADP = \\angle BDP$ (since $DP$ bisects $\\angle ADB$) and $\\angle ADP = \\angle OPD$ (as $PO \\parallel AD$). Hence $\\angle ODP = \\angle OPD$, so $OP = OD$. Also, $DO = AP$, $BO = BP$ as triangle $ADB$ is isosceles, then\n\n$$\n\\frac{DO}{BO} = \\frac{AP}{BP} = \\frac{DA}{DB} = \\frac{b}{a}\n$$\nby the angle bisector theorem in triangle $ABD$.\n\nSimilarly, the same theorem in triangle $BCD$ gives\n\n$$\n\\frac{CQ}{BO} = \\frac{DC}{DB} = \\frac{b}{a}\n$$\nIn summary,\n\n$$\n\\frac{DO}{BO} = \\frac{b}{a} = \\frac{CQ}{BO}\n$$\nNow, the converse of Thales' theorem implies $QQ \\parallel CD$.\n\n![](images/Argentina_2018_p7_data_e5aeaa92aa.png)\n\n$\\angle BDQ = \\angle CDQ = \\angle OQD$, hence $OQ = OD$. We obtained $OP = OQ = OD$, meaning that $O$ is the circumcenter of $DPQ$.\n\nIn addition, $OD = AP$ from the isosceles triangle $ADB$. Since $AP + BP = a$ and $\\frac{AP}{BP} = \\frac{b}{a}$, standard calculation leads to\n\n$$\nAP = \\frac{ab}{a+b}\n$$\nso triangle $DPQ$ has circumradius $\\frac{ab}{a+b}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16623,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that any two intersecting diagonals of a pentagon cannot be good at the same time.\n\nSuppose, contrary to our claim, that there are two good intersecting diagonals. Without loss of generality, assume that $AD$ and $BE$ are good diagonals of the pentagon $ABCDE$ (see Fig. 1).\n\nThen $BCDE$ and $ACDE$ are circumscribed quadrilaterals, so the sums of their opposite sides are equal:\n\n$$\nAC + DE = AE + CD \\quad \\text{and} \\quad BE + CD = BC + DE.\n$$\n\nThis gives\n\n$$\nAC + BE = AE + BC. \\quad (1)\n$$\n\nLet $M$ be the intersection point of $AC$ and $BE$. By the triangle inequality,\n\n$$\nAC + BE = (AM + MC) + (BM + ME) = (AM + ME) + (BM + MC) > AE + BC,\n$$\n\ncontradicting (1). Similarly, the crossing diagonals $AC$ and $BD$ of pentagon $ABCDE$ cannot be good at the same time. Since there are no more than two noncrossing diagonals in the pentagon, the number of good diagonals in the pentagon is at most 2.\n\nShow that there exists a pentagon with two good diagonals.\n\nMark five points $A, B, C, D, E$ in the plane such that $AB = AC = AD = AE$ and $\\angle BAC = \\angle CAD = \\angle DAE = \\alpha < 60^\\circ$ (see Fig. 2). Then the triangles $BAC, CAD, DAE$ are equal, so $BC = CD = DE$ and $\\angle ABC = \\angle ACB = \\angle ACD = \\angle ADC = \\angle ADE = \\angle AED = (180^\\circ - \\alpha)/2 = \\beta < 90^\\circ$.\n\nThe constructed pentagon is convex since $\\angle A = 3\\alpha < 180^\\circ$, $\\angle B = \\angle E = \\beta < 90^\\circ$, $\\angle C = \\angle D = 2\\beta < 180^\\circ$. Two diagonals $AC$ and $AD$ are good since the quadrilaterals $ACDE$ and $ABCD$ are circumscribed quadrilaterals because $AC + DE = AE + CD$ and $AB + CD = AD + BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "There exist exactly two good diagonals in such a pentagon, as constructed above.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16624,
    "subject": "Mathematics (Olympiad)",
    "question": "If the number $3\\nu + 1$, where $\\nu$ is an integer, is a multiple of $7$, find the possible remainders of the following divisions:\n\n(a) of $\\nu$ with divisor $7$,\n\n(b) of $\\nu^m$ with divisor $7$, for all values of the positive integer $m$, $m \\geq 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Let $3\\nu + 1 = 7\\kappa$, where $\\nu, \\kappa \\in \\mathbb{Z}$. The integer $\\nu$ can be written as $\\nu = 7\\rho + \\upsilon$, where $\\upsilon \\in \\{0,1,2,3,4,5,6\\}$ and $\\rho \\in \\mathbb{Z}$. Then:\n\n$3(7\\rho + \\upsilon) + 1 = 7\\kappa \\implies 21\\rho + 3\\upsilon + 1 = 7\\kappa \\implies 3\\upsilon + 1 \\equiv 0 \\pmod{7}$.\n\nSolving $3\\upsilon + 1 \\equiv 0 \\pmod{7}$, we find $\\upsilon = 2$. Thus, $\\nu = 7\\rho + 2$, so the remainder of $\\nu$ divided by $7$ is $2$.\n\n(b) We have:\n\n$$\n\\nu^m = (7\\rho + 2)^m = \\sum_{i=0}^{m} \\binom{m}{i} (7\\rho)^{m-i} 2^i = \\text{multiple of } 7 + 2^m.\n$$\n\nTherefore, it suffices to find the remainder of $2^m$ divided by $7$.\n\nIf $m = 3\\sigma + \\upsilon$, where $\\upsilon \\in \\{0,1,2\\}$, then:\n\n$$\n2^m = 2^{3\\sigma+\\upsilon} = 8^\\sigma \\cdot 2^\\upsilon = (7+1)^\\sigma \\cdot 2^\\upsilon = \\text{multiple of } 7 + 2^\\upsilon,\n$$\n\nwhere $\\upsilon \\in \\{0,1,2\\}$. Hence, the possible remainders of $\\nu^m$ divided by $7$, for all positive integers $m$, are $2^0 = 1$, $2^1 = 2$, and $2^2 = 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16625,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $r$ be a positive integer. Is it possible to color each positive integer with one of $r$ colors so that, for every $n$, the numbers of divisors of $n$ of each color differ by at most $1$?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $r \\geq 4$. Since there are infinitely many primes, there exist primes $p$ and $q$ with the same color. Consider $n = pq$. Its divisors are $1, p, q, pq$, so at least two divisors share a color, and with $r \\geq 4$ colors, some color is missing among the divisors, violating the condition.\n\nLet $r = 3$. For each positive integer $n$ with prime factorization $n = p_1^{a_1} p_2^{a_2} \\cdots p_m^{a_m}$, color $n$ with $(a_1 + a_2 + \\cdots + a_m) \\bmod 3$. For $n = 1$, color $0$. By induction on the number of distinct primes in $n$, this coloring ensures that for any $n$, the counts of divisors of each color differ by at most $1$.\n\nLet $d_n(0), d_n(1), d_n(2)$ be the number of divisors of $n$ colored $0, 1, 2$ respectively. For $n = 1$, $d(0) = 1$, $d(1) = d(2) = 0$, so the condition holds. Assume the condition holds for all numbers with $m$ distinct primes. Consider $a = p_1^{a_1} \\cdots p_m^{a_m} p_{m+1}^{a_{m+1}}$.\n\nLet $d_b(0), d_b(1), d_b(2)$ be the counts for $b = p_1^{a_1} \\cdots p_m^{a_m}$. By induction,\n$$\n|d_b(i) - d_b(j)| \\leq 1 \\text{ for all } i, j = 0, 1, 2.\n$$\nAny divisor of $a$ is $d \\cdot p_{m+1}^u$ for $d$ a divisor of $b$ and $0 \\leq u \\leq a_{m+1}$. The coloring cycles through $0, 1, 2$ as $u$ increases, so the counts for $a$ are sums of the counts for $b$ distributed over the possible $u$ values. In all cases ($a_{m+1} = 3l$, $3l+1$, $3l+2$), the difference between counts remains at most $1$.\n\nFor $r = 2$, color $n$ with $(\\alpha_1 + \\alpha_2 + \\cdots + \\alpha_r) \\bmod 2$. As in the $r = 3$ case, this coloring also satisfies the condition.\n\nTherefore, such a coloring exists for $r = 2$ and $r = 3$, but not for $r \\geq 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16626,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n$, determine the largest integer $M$ satisfying\n$$\n\\lfloor \\sqrt{a_1} \\rfloor + \\dots + \\lfloor \\sqrt{a_n} \\rfloor \\geq \\lfloor \\sqrt{a_1 + \\dots + a_n + M \\min(a_1, \\dots, a_n)} \\rfloor,\n$$\nfor all non-negative integers $a_1, \\dots, a_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The required maximal integer is $M = \\lfloor n(n - 1)/3 \\rfloor$.\n\nTo show $M \\leq \\lfloor n(n - 1)/3 \\rfloor$, let the $a_i$ all be equal to $3$, so we get $n \\geq \\lfloor \\sqrt{3n + 3M} \\rfloor$, which implies $\\sqrt{3n + 3M} < n + 1$, so $M < n(n - 1)/3 + 1/3$. Since $M$ is integral, $M \\leq \\lfloor n(n - 1)/3 \\rfloor$.\n\nTo complete the argument, we show that, if $a_1, \\dots, a_n$ are non-negative integers, then\n$$\n\\lfloor \\sqrt{a_1} \\rfloor + \\dots + \\lfloor \\sqrt{a_n} \\rfloor \\geq \\lfloor \\sqrt{a_1 + \\dots + a_n + \\frac{1}{3}n(n-1)\\min(a_1, \\dots, a_n)} \\rfloor.\n$$\nWithout loss of generality, assume $a_1 = \\min(a_1, \\dots, a_n)$. Let $\\lfloor \\sqrt{a_i} \\rfloor = k_i$ for $i = 1, \\dots, n$, so the inequality becomes:\n$$\nk_1 + \\dots + k_n + 1 > \\sqrt{a_1 + \\dots + a_n + \\frac{1}{3}n(n-1)a_1}.\n$$\nSquaring both sides and noting $a_i \\leq k_i^2 + 2k_i$, it suffices to show\n$$\n2 \\sum_{i<j} k_i k_j \\geq n(n-1)(k_1^2 + 2k_1)/3.\n$$\nSince $k_1 = \\min(k_1, \\dots, k_n)$,\n$$\n\\begin{aligned}\n\\sum_{i<j} k_i k_j &= k_1 \\sum_{i=2}^{n} k_i + \\sum_{2 \\leq i < j} k_i k_j \\\n&\\geq k_1 \\cdot (n-1)k_1 + \\frac{1}{2}(n-1)(n-2)k_1^2 \\\\\n&= \\frac{1}{2}n(n-1)k_1^2 \\geq \\frac{1}{6}n(n-1)(k_1^2 + 2k_1).\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16627,
    "subject": "Mathematics (Olympiad)",
    "question": "There are many ways to fit 8 corner triominoes into the Aztec Diamond. Here is one possibility.\n\n![](images/IRL_ABooklet_2023_p28_data_91cf38c773.png)\n\nHow many straight triominoes can fit into the Aztec Diamond? Can more corner triominoes be fitted than straight triominoes?",
    "options": [],
    "answer": "See solution",
    "solution": "To show that at most 7 straight triominoes fit into the Aztec Diamond, we color the squares with three colors in a diagonal fashion:\n\n![](images/IRL_ABooklet_2023_p28_data_25bb967607.png)\n\nThere are 11 green, 7 blue, and 7 red cells. Any straight triomino covers exactly one cell of each color, so no more than 7 straight triominoes can fit. Therefore, more corner triominoes can be fitted into the Aztec Diamond.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16628,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $A_0A_1A_2$, determine the locus of the centers of the equilateral triangles $X_0X_1X_2$ such that each line $X_kX_{k+1}$ passes through $A_k$ (indices modulo 3).",
    "options": [],
    "answer": "See solution",
    "solution": "Erect the three outer Napoleon triangles associated with $A_0A_1A_2$, and let $\\gamma_k$, $k = 0, 1, 2$, be their circumcircles.\n\nFrom any point $X_0$ on $\\gamma_0$, draw lines $X_0A_2X_1$ and $X_0A_1X_2$, where $X_1$ lies on $\\gamma_1$ and $X_2$ on $\\gamma_2$. The points $X_1$, $A_0$, $X_2$ are collinear, and $X_0X_1X_2$ is an equilateral triangle satisfying the conditions.\n\n![](images/RMC2014_p54_data_6ab405418a.png)\n\nLet $M_k$ be the midpoint of the minor arc $A_{k+1}A_{k+2}$ of $\\gamma_k$. The triangle $M_0M_1M_2$ is equilateral, as the $M_k$ are the centers of the inner Napoleon triangles of $A_0A_1A_2$. The center $X$ of $X_0X_1X_2$ is the intersection of $X_0M_0$ and $X_1M_1$, which meet at $60^\\circ$. Since the locus of $X$ includes the three points $M_k$, the locus of $X$ is the circle through $M_0M_1M_2$.\n\nSimilarly, another circle is obtained by starting with the inner Napoleon triangles. Thus, the required locus is a pair of circles.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 16629,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be an arbitrary function and $g: \\mathbb{R} \\to \\mathbb{R}$ be a quadratic function such that for any real numbers $m$ and $n$, the equation $f(x) = mx + n$ has solutions if and only if the equation $g(x) = mx + n$ has solutions. Prove that $f = g$.",
    "options": [],
    "answer": "See solution",
    "solution": "Remark that if the graph of $h: \\mathbb{R} \\to \\mathbb{R}$, $h(x) = mx + n$ is tangent to the graph of $g: \\mathbb{R} \\to \\mathbb{R}$, $g(x) = ax^2 + bx + c$, then the equation $ax^2 + bx + c = mx + n$, with $a, b, c, m, n \\in \\mathbb{R}$ and $a \\neq 0$, has zero discriminant. That is, the function $k: \\mathbb{R} \\to \\mathbb{R}$, $k(x) = ax^2 + bx + c - mx - n$ is zero at only one point and has constant sign elsewhere.\n\nSuppose without loss of generality that $g(x) = ax^2 + bx + c$ with $a > 0$. If there exists $x_0$ such that $f(x_0) < g(x_0)$, choose $m$ and $n$ so that the graph of $h(x) = mx + n$ is tangent to $g$ at $(x_0, g(x_0))$. Set $n' = n - g(x_0) + f(x_0)$. Then the equation $f(x) = mx + n'$ has the root $x_0$, but $g(x) = mx + n'$ has no roots, since $g(x) \\ge mx + n > mx + n'$ for all $x \\in \\mathbb{R}$, a contradiction.\n\nThus $f(x) \\ge g(x)$ for all $x \\in \\mathbb{R}$. Suppose now that $x_0$ satisfies $f(x_0) > g(x_0)$. As before, choose $m$ and $n$ so that $h(x) = mx + n$ is tangent to $g$ at $(x_0, g(x_0))$. By the remark, $f(x) > mx + n$ for all $x \\in \\mathbb{R}$, which contradicts the fact that $g(x) = mx + n$ has the root $x_0$.\n\nTherefore, $f(x) = g(x)$ for all $x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16630,
    "subject": "Mathematics (Olympiad)",
    "question": "The circles $k_1$ and $k_2$ intersect at points $M$ and $N$. The line $l$ intersects circle $k_1$ at points $A$ and $C$, and circle $k_2$ at points $B$ and $D$, so that the points $A, B, C, D$ are on the line $l$ in that order. Let $X$ be a point on line $MN$ such that $M$ is between $X$ and $N$. Lines $AX$ and $BM$ intersect at point $P$, and lines $DX$ and $CM$ intersect at point $Q$. Prove that $PQ \\parallel l$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $Y$ be the second intersection of line $AX$ with circle $k_1$, and $Z$ be the second intersection of line $DX$ with circle $k_2$.\n\n![](images/prob1516_p32_data_14b3e02d91.png)\n\nAs $X$ is on the radical axis of $k_1$ and $k_2$, we have $XY \\cdot XA = XZ \\cdot XD$, from which the points $A, Y, Z, D$ lie on the same circle. Therefore, $\\angle XZY = \\angle XAD$ and $\\angle XYZ = \\angle XDA$.\n\nLet $R$ be the intersection of $CM$ and $AX$. Let $S$ be the intersection of $BM$ and $DX$. Points $R$ and $X$ lie on the same side of points $A$ and $Y$, and points $S$ and $X$ lie on the same side of points $D$ and $Z$. Points $R$ and $Q$ lie on the same side of points $C$ and $M$, and points $S$ and $P$ lie on the same side of points $B$ and $M$.\n\nAs $A, Y, M, C$ lie on the same circle $k_1$, we have $\\angle QMY = \\angle RMY = \\angle RAC = \\angle XAD = \\angle XZY = \\angle QZY$. As $Q$ and $Y$ lie on the same side of line $MN$ but $Z$ on the other side, $M$ and $Z$ lie on the same side of line $QY$. Hence, the equation $\\angle QMY = \\angle QZY$ implies that $Q, Y, M, Z$ lie on the same circle. Similarly, we get $\\angle PMZ = \\angle SMZ = \\angle SDB = \\angle XDA = \\angle XYZ = \\angle PYZ$, which can be used analogously to show that $P, Z, M, Y$ lie on the same circle.\n\nFinally, $P, Q, Y, Z$ have to lie on the same circle in that order ($Q$ and $Y$ lie on the same side of line $MN$, $P$ and $Z$ on the other side). Therefore, $\\angle QPA = \\angle QPY = \\angle QZY = \\angle XZY = \\angle XAD = \\angle PAD$, which implies $PQ \\parallel l$ (because $Q$ and $D$ are on different sides of $AP$).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16631,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation\n\n$$\nx_1^4 + x_2^4 + \\dots + x_{14}^4 = 2016^3 - 1.\n$$\n\nin the set of integers.",
    "options": [],
    "answer": "See solution",
    "solution": "For $x = 2k$, $x^4 = 16k^4 \\equiv 0 \\pmod{16}$. For $x = 2k + 1$, $x^4 - 1 = 8k(k + 1)(2k^2 + 2k + 1) \\equiv 0 \\pmod{16}$, so $x^4 \\equiv 1 \\pmod{16}$. Since $2016^3 - 1 \\equiv 15 \\pmod{16}$, and the sum $x_1^4 + x_2^4 + \\dots + x_{14}^4$ cannot yield a remainder of 15 modulo 16, the equation has no solution in integers.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16632,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, the median $BM$ is equal to half of the side $BC$. Prove that\n$$\n\\angle ABM = \\angle BCA + \\angle BAC.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Mark on the extension of median $BM$ beyond $M$ a point $D$ such that $BM = MD$ (see the figure below). Then triangles $ABM$ and $CDM$ are congruent by two sides and the included angle, and also $BD = 2BM = BC$. Thus, triangle $BDC$ is isosceles, and\n$$\n\\begin{align*}\n\\angle ABM &= \\angle CDM = \\angle BCD = \\angle BCA + \\angle DCA \\\\\n           &= \\angle BCA + \\angle BAC.\n\\end{align*}\n$$\nas desired.\n\n![](images/Ukraine_2021-2022_p22_data_b8c3f4ba07.png)\n\n*Fig. 11*",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16633,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the old number be $N = 100a + 10b + c$ and the new number be\n$$N' = 100(a+d) + 10(b-d) + (c-d) = N + 89d.$$ So $N(d-2) = 2 \\times 89d$.\n\nFind the largest possible value of $N$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Method 4*\n\nSince $89$ is prime and $-2 \\leq d-2 \\leq 7$, $89$ divides $N$.\n\nLet $N = 89k$. Then $k(d-2) = 2d$.\n\nIf $d = 0$, then $k = 0$.\n\nIf $d = 1$, then $k = -2$.\n\nIf $d = 2$, then there is no solution for $k$.\n\nIf $d > 2$, then $k = \\dfrac{2d}{d-2} = 2 + \\dfrac{4}{d-2}$, which decreases as $d$ increases.\n\nSo the largest value of $N$ is $89\\left(2 + \\dfrac{4}{3-2}\\right) = 89 \\times 6 = 534$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16634,
    "subject": "Mathematics (Olympiad)",
    "question": "On $\\triangle ABC$ points $A, D, E$, and $B$ lie in that order on side $\\overline{AB}$ with $AD = 4$, $DE = 16$, and $EB = 8$. Points $A, F, G$, and $C$ lie in that order on side $\\overline{AC}$ with $AF = 13$, $FG = 52$, and $GC = 26$. Let $M$ be the reflection of $D$ through $F$, and let $N$ be the reflection of $G$ through $E$. Quadrilateral $DEGF$ has area 288. Find the area of heptagon $AFNBCEM$.\n\n![](images/2025AIME_I_Solutions_p1_data_c7bed00599.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that $AD : DE : EB = AF : FG : GC = 1 : 4 : 2$. This implies that $\\overline{DF} \\parallel \\overline{EG} \\parallel \\overline{BC}$, and $\\triangle ADF \\sim \\triangle AEG \\sim \\triangle ABC$.\n\nIf two triangles share an altitude and have bases with the same length, then they have the same area. This implies that $\\text{Area}(\\triangle ADF) = \\text{Area}(\\triangle AFM)$. More generally, if two trapezoids share an altitude and have parallel sides with the same corresponding lengths, then they have the same area. This implies that $\\text{Area}(DEGF) = \\text{Area}(FNEM)$ and $\\text{Area}(EBCG) = \\text{Area}(NBCE)$. Summing these areas shows that heptagon $AFNBCEM$ has the same area as $\\triangle ABC$.\n\nTo compute $\\text{Area}(\\triangle ABC)$, use triangle similarity to get\n\n$$\n\\text{Area}(DEGF) = 24 \\cdot \\text{Area}(\\triangle ADF) = \\frac{24}{49} \\cdot \\text{Area}(\\triangle ABC).\n$$\n\nFrom $\\text{Area}(DEGF) = 288$, it follows that $\\text{Area}(\\triangle ABC) = 588$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16635,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, $z$ be positive real numbers. Prove that\n\n$$\n\\sqrt{\\frac{xy}{x^2 + y^2 + 2z^2}} + \\sqrt{\\frac{yz}{y^2 + z^2 + 2x^2}} + \\sqrt{\\frac{zx}{z^2 + x^2 + 2y^2}} \\leq \\frac{3}{2}.\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\n\\begin{aligned}\n& \\sqrt{\\frac{xy}{x^2 + y^2 + 2z^2}} + \\sqrt{\\frac{yz}{y^2 + z^2 + 2x^2}} + \\sqrt{\\frac{zx}{z^2 + x^2 + 2y^2}} \\\\\n& \\leq \\sqrt{\\frac{xy}{xy + yz + zx + z^2}} + \\sqrt{\\frac{yz}{xy + yz + zx + x^2}} + \\sqrt{\\frac{zx}{xy + yz + zx + y^2}} \\\\\n& = \\sqrt{\\frac{xy}{(z + x)(y + z)}} + \\sqrt{\\frac{yz}{(x + y)(z + x)}} + \\sqrt{\\frac{zx}{(y + z)(x + y)}} \\\\\n& \\leq \\frac{\\frac{x}{z + x} + \\frac{y}{y + z}}{2} + \\frac{\\frac{y}{x + y} + \\frac{z}{z + x}}{2} + \\frac{\\frac{z}{y + z} + \\frac{x}{x + y}}{2} \\\\\n& = \\frac{\\frac{x + y}{x + y} + \\frac{y + z}{y + z} + \\frac{z + x}{z + x}}{2} = \\frac{3}{2}.\n\\end{aligned}\n$$\n\nEquality holds if and only if $x = y = z$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16636,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive even integer $a$ which is not a power of $2$, prove that at least one of the numbers $2^{2^n} + 1$ and $a^{2^n} + 1$ is composite for infinitely many non-negative integers $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f_n = 2^{2^n} + 1$ and $a_n = a^{2^n} + 1$. Suppose, for contradiction, that $f_n$ and $a_n$ are both prime for all but finitely many non-negative integers $n$, i.e., for all $n > N$ for some $N$. Clearly, we may assume $f_n > a$ for all $n > N$.\n\nFix any $n > N$ and consider the multiplicative order of $a$ modulo the prime $f_n$. By Fermat's Little Theorem, this order divides $f_n - 1 = 2^{2^n}$, so it is a power of $2$, say $2^k$.\n\nThus, the prime $f_n$ divides $a^{2^k} - 1 = (a^{2^{k-1}} - 1)a_{k-1}$, so it divides at least one of the two factors. By minimality of $k$, the first factor is not divisible by $f_n$, so $f_n$ divides $a_{k-1}$. Notice that $a_{k-1} \\ne f_n$, since $a$ is not a power of $2$. Hence $a_{k-1}$ is composite, forcing $k-1 \\le N$.\n\nConsequently, the product $a_0 a_1 \\dots a_N$ is divisible by $f_n$ for all $n > N$. This contradiction concludes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16637,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a finite set of at least two points in the plane. Assume that no three points of $S$ are collinear. A windmill is a process that starts with a line $\\ell$ going through a single point $P \\in S$. The line rotates clockwise about the pivot $P$ until the first time that the line meets some other point belonging to $S$. This point, $Q$, takes over as the new pivot, and the line now rotates clockwise about $Q$, until it next meets a point of $S$. This process continues indefinitely, with the pivot always being a point from $S$.\n\nShow that we can choose a point $P$ in $S$ and a line $\\ell$ going through $P$ such that the resulting windmill uses each point of $S$ as a pivot infinitely many times.",
    "options": [],
    "answer": "See solution",
    "solution": "We call a point in $S$ a *vertex* and a line passing through exactly one vertex (which must be the pivot of the line) in a windmill a *sweeping line*. Assign the lines directions consistent with their motion, giving a well-defined notion of a left and right side of these lines. We start with the following key observation.\n\n**Lemma 1.** In a fixed windmill, the numbers of vertices on the right and left sides of any sweeping line in the windmill is constant.\n\n*Proof.* The statement is evidently true between pivot changes. Suppose now that $\\ell_1$ and $\\ell_2$ are sweeping lines through vertices $V_1$ and $V_2$ and that the pivot changes from $V_1$ to $V_2$ in the windmill. Then, $P_2$ is on the same side of $\\ell_1$ as $P_1$ is of $\\ell_2$, and any other vertex $P$ lies on the same side of both lines, which shows that the number of vertices on the right and left sides does not change between $\\ell_1$ and $\\ell_2$. This gives the claim. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16638,
    "subject": "Mathematics (Olympiad)",
    "question": "The sides $AB$ and $AC$ of triangle $ABC$ touch the circle $c$ at points $B'$ and $C'$, respectively. The center $L$ of circle $c$ lies on side $BC$. The circumcenter $O$ of triangle $ABC$ lies on the shorter arc $B'C'$ of circle $c$. Prove that the circumcircle of $ABC$ and circle $c$ meet at two points.\n\n![](images/Estonija_2012_p18_data_5ef3381d35.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $r$ be the circumradius of $ABC$, $s$ the radius of $c$, and $\\alpha = \\angle BAC$. By tangency, $|AB'| = |AC'|$. Thus $\\angle C'B'A = \\angle B'C'A = \\frac{\\pi}{2} - \\frac{\\alpha}{2}$, so by the inscribed angle property, $\\angle B'OC' = \\pi - \\left(\\frac{\\pi}{2} - \\frac{\\alpha}{2}\\right) = \\frac{\\pi}{2} + \\frac{\\alpha}{2}$. Clearly $\\angle B'OC' > \\angle BOC = 2\\alpha$, leading to $\\frac{\\pi}{2} + \\frac{\\alpha}{2} > 2\\alpha$, hence $\\alpha < \\frac{\\pi}{3}$.\n\nLet $K$ be the midpoint of side $BC$. From right triangle $KOC$, $|KO| = |OC| \\cos \\angle KOC = r \\cos \\alpha$. By the inequality above, $\\cos \\alpha > \\cos \\frac{\\pi}{3} = \\frac{1}{2}$. On the other hand, $|KO| \\leq |LO| = s$, so $\\frac{1}{2} r < r \\cos \\alpha = |KO| \\leq s$, or $r < 2s$. As $c$ passes through the circumcenter of $ABC$, this inequality shows that these circles must intersect.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16639,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the smallest integer $a > 10$ for which:\n\n- $a$ is divisible by $10$\n- $a + 1$ is divisible by $11$\n- $a + 2$ is divisible by $12$",
    "options": [],
    "answer": "See solution",
    "solution": "$30$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16640,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a point inside a square $ABCD$ such that $PA : PB : PC = 1 : 2 : 3$. Determine the angle $\\angle BPA$.",
    "options": [],
    "answer": "See solution",
    "solution": "Rotate triangle $ABP$ by $90^\\circ$ around $B$ so that $A$ maps to $C$ and $P$ maps to a new point $Q$. Then $\\angle PBQ = \\angle PBC + \\angle CBQ = \\angle PBC + \\angle ABP = 90^\\circ$. Thus, triangle $PBQ$ is an isosceles right triangle, so $\\angle BQP = 45^\\circ$. By the Pythagorean theorem, $PQ^2 = 2PB^2 = 8AP^2$. Since $CQ^2 + PQ^2 = AP^2 + 8AP^2 = 9AP^2 = PC^2$, by the converse of Pythagoras, $PQC$ is a right triangle, so\n\n$$\n\\angle BPA = \\angle BQC = \\angle BQP + \\angle PQC = 45^\\circ + 90^\\circ = 135^\\circ.\n$$\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p120_data_a77c3a5ba2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16641,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $AB < AC < BC$. On the ray $BC$ consider a point $D$ such that $BD = BA$, and on the ray $CB$ consider a point $E$ such that $CE = CA$. Let $K$ be the circumcenter of triangle $ADE$, and let $F$ and $G$ be the intersections of the lines $AD, KC$ and $AE, KB$, respectively. Prove that the circumcircle of triangle $KDE$ (call it $c_1$), the circle with center $F$ and radius $FE$ (call it $c_2$), and the circle with center $G$ and radius $GD$ (call it $c_3$), all pass through a point on the line $AK$.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/Greece_2022_p6_data_46732c7579.png)\n\nLet $M$ be the midpoint of $AD$. Then the points $B, G, K, M$ lie on the perpendicular bisector of $AD$, so they are collinear. Moreover, since $GD = GA$, the circle $c_3$ contains $A$.\n\nSimilarly, if $N$ is the midpoint of $AE$, the points $G, F, K, N$ lie on the perpendicular bisector of $AE$ and hence are collinear. Also, since $FA = FE$, the circle $c_2$ contains $A$.\n\nNext, we will prove that the points $G, F$ belong to the circumcircle $c_1$ of triangle $KDE$. From the isosceles triangles $ADG$ and $AFE$, we have\n\n$$\n\\angle EGD = 2 \\cdot \\angle GAD = 2 \\cdot \\angle EAF = \\angle F_1 \\quad (1)\n$$\n\nAlso, for the inscribed angle $\\angle EAD$ and the angle $\\angle GKF$, we have\n\n$$\n\\angle GKF = 2 \\cdot \\angle EAD \\quad (2)\n$$\n\nFrom (1) and (2), it follows that $\\angle EGD = \\angle F_1 = \\angle GKF$, and hence the points $D, E, F, G, K$ belong to the circle $c_1$.\n\nLet $T$ be the second point of intersection of the circles $c_2$ and $c_3$. We will prove that the points $A, K, T$ are collinear and that $T$ belongs to the circle through $D, E, F, G, K$.\n\nIn fact, the common chord $AT$ of the circles $c_2$ and $c_3$ is perpendicular to the line of their centers $FG$, and also the line $AK$ is perpendicular to $FG$, since $K$ is the orthocenter of triangle $AGF$.\n\nMoreover, since $GA = GD = GT$, we have the angle equalities:\n\n$$\n\\angle GAT = \\angle GTA = x \\qquad (1)\n$$\n\nand since $K$ is the orthocenter of triangle $AGF$, we have\n\n$$\n\\angle GAT = \\angle GFK = 90^\\circ - \\angle AGF \\qquad (2)\n$$\n\nHence $\\angle GTK = \\angle GTA = \\angle GFK$, so the points $F, G, K, T$ are cyclic.\n\nSince $BD = BA$ and $KA = KD$, $BK$ is the perpendicular bisector of $AD$, so $GA = GD$. We have $\\angle KGD = \\angle KGA = 90^\\circ - \\angle EAD = \\angle AFK$, so the points $K, F, D, G$ are concyclic. Similarly, $\\angle KFE = \\angle AGK$, so the points $K, F, E, G$ are concyclic.\n\nFrom the previous discussion, we have that $K, F, E, G, D$ are concyclic.\n\n![](images/Greece_2022_p7_data_4fd9f4ff7b.png)\n\nNow, let $AK$ intersect $c_1$ at the point $H$. Then\n\n$$\n\\angle HGD = \\angle HKD = 2 \\cdot \\angle HAD \\qquad (5)\n$$\n\nTherefore, $G$ is on the perpendicular bisector of $AD$ and (5) holds, so $G$ is the circumcenter of triangle $AHD$, therefore $c_3$ passes through $H$. Similarly, $c_2$ passes through $H$, and we have the desired result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16642,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k, n \\in \\mathbb{N}^*$ and let $A$ be an arbitrary set of $2n$ positive integers. For each $a \\in A$, denote by $p_a$ the number of elements in the set $P_a = \\{b \\in A \\mid b - a \\equiv k \\pmod{2k}\\}$. Determine the maximum possible value of the sum\n$$\nE_A = \\sum_{a \\in A} p_a^2,\n$$\nwhen $A$ runs over all subsets of size $2n$ of $\\mathbb{N}^*$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us denote $A_i = \\{a \\in A \\mid a \\equiv i \\pmod{2k}\\}$ and $a_i = |A_i|$ for each $i = 0, \\dots, 2k-1$. Observe that $A = \\bigcup_{i=0}^{2k-1} A_i$, where $A_i \\cap A_j = \\emptyset$ for $i \\neq j$, and therefore $2n = \\sum_{i=0}^{2k-1} a_i$.\n\nFor any $a \\in A_i$, we have\n$$\nP_a = \\{b \\in A \\mid b - a \\equiv k \\pmod{2k}\\} = \\{b \\in A \\mid b \\equiv k + i \\pmod{2k}\\} = A_{k+i},\n$$\nwhere the index $k + i$ is considered modulo $2k$. Hence, $p_a = a_{k+i}$ for every $a \\in A_i$.\n\nThus,\n$$\nE_A = \\sum_{a \\in A} p_a^2 = \\sum_{i=0}^{2k-1} \\sum_{a \\in A_i} p_a^2 = \\sum_{i=0}^{2k-1} a_i a_{k+i}^2.\n$$\n\nSince the indices are modulo $2k$,\n$$\n\\sum_{i=k}^{2k-1} a_i a_{k+i}^2 = \\sum_{i=0}^{k-1} a_{k+i} a_i^2,\n$$\nso we get\n$$\nE_A = \\sum_{i=0}^{k-1} (a_i a_{k+i}^2 + a_{k+i} a_i^2) = \\sum_{i=0}^{k-1} a_i a_{k+i} (a_i + a_{k+i}).\n$$\n\nWe want to maximize $E_A$ under the constraint $\\sum_{i=0}^{2k-1} a_i = 2n$.\n\nNote that replacing any two distinct pairs $(a_i, a_{i+k})$ and $(a_j, a_{j+k})$ by $(a_i + a_j, a_{i+k} + a_{j+k})$ and $(0, 0)$ does not decrease the value of $E_A$.\n\nTherefore, the maximum is achieved when only one pair $(a_i, a_{i+k})$ is nonzero, with $a_i + a_{i+k} = 2n$.\n\nUsing the inequality between arithmetic and geometric means, we have\n$$\nE_A = a_i a_{i+k} (a_i + a_{i+k}) \\le \\frac{(a_i + a_{i+k})^3}{4} = 2n^3.\n$$\n\nThe maximum value $2n^3$ is attained, for example, by any set $A$ having exactly $n$ multiples of $2k$ and $n$ numbers congruent to $k \\pmod{2k}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16643,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{R}_+$ denote the set of positive real numbers. Find all functions $f : \\mathbb{R}_+ \\to \\mathbb{R}_+$ such that for each $x \\in \\mathbb{R}_+$, there is exactly one $y \\in \\mathbb{R}_+$ satisfying\n\n$$\nx f(y) + y f(x) \\leq 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The unique function is $f(x) = \\frac{1}{x}$ for all $x \\in \\mathbb{R}_+$.\n\n**Sufficiency.** For any $x, y > 0$,\n\n$$\nx f(y) + y f(x) = \\frac{x}{y} + \\frac{y}{x} \\ge 2,\n$$\n\nwith equality only when $x = y$. Thus, for fixed $x > 0$, $x f(y) + y f(x) \\le 2$ if and only if $y = x$.\n\n**Necessity.**\n\n*Solution 1:* Call $(x, y)$ a \"good pair\" if $x f(y) + y f(x) \\le 2$. If $(x, y)$ is a good pair, so is $(y, x)$. The problem requires that for each $x > 0$, there is a unique $y > 0$ such that $(x, y)$ is a good pair. We assert that for each good pair $(x, y)$, $x = y$. Otherwise, if $x \\neq y$, then $(x, x)$ and $(y, y)$ are not good pairs, and $x f(x) > 1$, $y f(y) > 1$. By AM-GM,\n\n$$\nx f(y) + y f(x) \\geq 2\\sqrt{x f(y) \\cdot y f(x)} = 2\\sqrt{x f(x) \\cdot y f(y)} > 2,\n$$\n\ncontradicting the assumption that $(x, y)$ is a good pair.\n\nThus, for every $x > 0$, the only good pair is $(x, x)$, so $x f(x) \\le 1$, i.e., $f(x) \\le \\frac{1}{x}$ for all $x > 0$. Let $y = \\frac{1}{f(x)}$. Then\n\n$$\nx f(y) + y f(x) \\leq x \\cdot \\frac{1}{y} + \\frac{1}{f(x)} \\cdot f(x) = x f(x) + 1 \\leq 2,\n$$\n\nso $\\left(x, \\frac{1}{f(x)}\\right)$ is a good pair. Since $(x, x)$ is the only good pair, $x = \\frac{1}{f(x)}$, so $f(x) = \\frac{1}{x}$.\n\n*Solution 2:* For $x > 0$, let $\\sigma(x)$ be the unique $y > 0$ such that $x f(y) + y f(x) \\le 2$.\n\n(1) $f$ is strictly decreasing: for $0 < y_1 < y_2$, $f(y_1) > f(y_2)$. Otherwise, suppose $f(y_1) \\le f(y_2)$. Let $x = \\sigma(y_2)$. Then\n\n$$\nx f(y_1) + y_1 f(x) < x f(y_2) + y_2 f(x) \\le 2,\n$$\n\ncontradicting uniqueness of $\\sigma(x)$.\n\n(2) For any $x > 0$, let $y = \\sigma(x)$, and $x f(y) + y f(x) = 2$. Otherwise, if $x f(y) + y f(x) < 2$, let $y' = y + \\varepsilon$, where $\\varepsilon = \\frac{2 - x f(y) - y f(x)}{f(x)} > 0$. By (1),\n\n$$\nx f(y') + y' f(x) < x f(y) + y f(x) + \\varepsilon f(x) = 2,\n$$\n\ncontradicting uniqueness of $\\sigma(x)$.\n\nFrom (2), for any $x, y > 0$, $x f(y) + y f(x) \\ge 2$. Taking $y = x$, $f(x) \\ge \\frac{1}{x}$, and\n\n$$\n2 = f(x) \\sigma(x) + f(\\sigma(x)) x \\ge \\frac{\\sigma(x)}{x} + \\frac{x}{\\sigma(x)},\n$$\n\nso $\\sigma(x) = x$. Thus, $f(x) = \\frac{1}{x}$.\n\n$\\boxed{f(x) = \\frac{1}{x}}$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16644,
    "subject": "Mathematics (Olympiad)",
    "question": "The table below lists, in decreasing order, the 3-digit integers $n$ that are multiples of 8 together with their FDSs (F).\n\n![](images/Australian-Scene-combined-2015_p46_data_e9ca78a8ec.png)\n\nFind all 3-digit multiples of 8 whose FDS (final digit sum) is 8.",
    "options": [],
    "answer": "See solution",
    "solution": "The only 3-digit multiples of 8 that have FDS 8 are:\n\n152, 224, 296, 368, 440, 512, 584, 656, 728, 800, 872, 944.\n\nThese numbers have a common difference of 72. Thus, the only 3-digit multiples of 8 that have FDS 8 are the integers $n = 8 + 72t$ where $t = 2, 3, \\dots, 13$.\n\nFrom the third solution to Part *c*, the FDS of a positive integer $n$ is 8 if and only if $n$ has the form $n = 8 + 9r$. Such an $n$ is a multiple of 8 if and only if 8 divides $r$. So $n$ is a multiple of 8 and has FDS 8 if and only if $n$ has the form $n = 8 + 72t$. Hence, the only 3-digit multiples of 8 that have FDS 8 are the integers $n = 8 + 72t$ where $t = 2, 3, \\dots, 13$. These are: 152, 224, 296, 368, 440, 512, 584, 656, 728, 800, 872, 944.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16645,
    "subject": "Mathematics (Olympiad)",
    "question": "A positive integer $n$ is given. Positive numbers $x_0, x_1, \\dots, x_n$ satisfy $x_0 x_1 \\dots x_n = 1$. Find all positive $\\gamma$ such that the inequality\n\n$$\nx_0^\\gamma + x_1^\\gamma + \\dots + x_n^\\gamma \\ge \\frac{1}{x_0} + \\frac{1}{x_1} + \\dots + \\frac{1}{x_n}\n$$\nholds for any set of numbers $x_0, x_1, \\dots, x_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $\\gamma \\ge n$.\n\n**Solution.** First, we show that for $0 < \\gamma < n$, there exists a set $x_0, x_1, \\dots, x_n$ for which the inequality does not hold.\n\nLet $x_0 = x^{-n}$ and $x_1 = x_2 = \\dots = x_n = x$ for some $x > 0$. Then\n\n$$\nx_0^\\gamma + x_1^\\gamma + \\dots + x_n^\\gamma = x^{-n\\gamma} + n x^\\gamma,\n$$\n$$\n\\frac{1}{x_0} + \\frac{1}{x_1} + \\dots + \\frac{1}{x_n} = x^n + n x^{-1}.\n$$\n\nFor $\\gamma < n$, choose $x > 1$: $x^n + n x^{-1} > x^n > (n + 1)x^\\gamma > n x^\\gamma + x^{-n\\gamma}$, i.e., $x^{-n\\gamma} > n + 1$ for $x > \\sqrt[n]{n+1}$. Thus, a counterexample exists.\n\nNow, let $\\gamma = n$. By the Cauchy inequality,\n\n$$\n\\frac{x_0^n + x_1^n + \\dots + x_{n-1}^n}{n} \\ge x_0 x_1 \\dots x_{n-1} = \\frac{1}{x_n},\n$$\n$$\n\\frac{x_0^n + x_1^n + \\dots + x_{n-2}^n + x_n^n}{n} \\ge x_0 x_1 \\dots x_{n-2} x_n = \\frac{1}{x_{n-1}},\n$$\n$$\n\\frac{x_1^n + x_2^n + \\dots + x_n^n}{n} \\ge x_1 x_2 \\dots x_n = \\frac{1}{x_0}.\n$$\n\nAdding these inequalities yields the required result.\n\nFor $\\gamma > n$, if for any positive $x_0, \\dots, x_n$ with $x_0 x_1 \\dots x_n = 1$ we have $x_0^\\gamma + \\dots + x_n^\\gamma \\ge x_0^n + \\dots + x_n^n$, then the desired inequality for $\\gamma$ follows from the case $\\gamma = n$.\n\nTo prove this, use the weighted Cauchy inequality:\n\n$$\nx_0^{n+\\frac{\\gamma-n}{n+1}} x_1^{\\frac{\\gamma-n}{n+1}} \\dots x_n^{\\frac{\\gamma-n}{n+1}} \\le \\frac{n+\\frac{\\gamma-n}{n+1}}{\\gamma} x_0^{\\gamma} + \\frac{\\gamma-n}{\\gamma(n+1)} x_1^{\\gamma} + \\dots + \\frac{\\gamma-n}{\\gamma(n+1)} x_n^{\\gamma}.\n$$\n\nWriting similar inequalities for each term and summing yields the result.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16646,
    "subject": "Mathematics (Olympiad)",
    "question": "Petryk and Vasyl' are playing a game with numbers written on the board. On each move (Petryk goes first), a player chooses two coprime numbers from those on the board, erases them, and writes their sum instead. The player who cannot make a move loses. Who will win if both play optimally and the initial numbers are:\n\na) $2019$ numbers $1$; \nb) $2020$ numbers $1$?",
    "options": [],
    "answer": "See solution",
    "solution": "a) Vasyl' can force the board, after each of his moves, to contain some odd number $n$ and $2019-n$ numbers $1$. After $1009$ pairs of moves, only $2019$ remains, and Petryk cannot move and loses. Petryk's first move: erase two $1$s, write $2$. Vasyl' then makes $3$. If at some point an odd $n$ is on the board and $2019-n$ ones, Petryk can either:\n- Erase two $1$s, write $2$; Vasyl' erases $n$ and $2$, writes $n+2$.\n- Erase $n$ and $1$, write $n+1$; Vasyl' erases $n+1$ and $1$, writes $n+2$.\nIn both cases, Vasyl' wins.\n\nb) Vasyl' uses a similar strategy, except for the last move. Before Petryk's last move, four numbers remain: $2017$, $1$, $1$, $1$. If Petryk leaves $2018$, $1$, $1$, Vasyl' turns it into $2018$ and $2$, and Petryk cannot move. If Petryk leaves $2017$, $2$, $1$, Vasyl' again turns it into $2018$ and $2$, and Petryk cannot move.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16647,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all natural numbers $n$ for which, among the numbers $n, n+1, n+2, \\dots, n^2$, there exist four pairwise distinct numbers $a, b, c, d$ such that the equality $ab = cd$ holds. Justify your answer.",
    "options": [],
    "answer": "See solution",
    "solution": "If $6n \\leq n^2$, or equivalently $n \\geq 6$, then by choosing $a = n$, $b = 6n$, $c = 2n$, $d = 3n$, we have $ab = cd$.\n\nFor $n = 5$, the numbers are $5, 6, 7, \\dots, 25$. Choosing $a = 6$, $b = 20$, $c = 8$, $d = 15$, we get $ab = cd$.\n\nFor $n = 4$, the numbers are $4, 5, 6, \\dots, 16$. Choosing $a = 4$, $b = 15$, $c = 6$, $d = 10$, we get $ab = cd$.\n\nFor $n = 3$, the numbers are $3, 4, 5, \\dots, 9$. Choosing $a = 3$, $b = 8$, $c = 4$, $d = 6$, we get $ab = cd$.\n\nFor $n = 1$ and $n = 2$, the set $n, n+1, n+2, \\dots, n^2$ contains at most three numbers, so it is impossible to find four pairwise distinct numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16648,
    "subject": "Mathematics (Olympiad)",
    "question": "La sucesión $a_1, a_2, \\dots$ de números reales positivos verifica\n$$\na_{k+1} \\geq \\frac{ka_k}{a_k^2 + (k-1)}\n$$\npara todo entero positivo $k$. Demostrar que\n$$\na_1 + a_2 + \\dots + a_n \\geq n\n$$\npara todo $n \\geq 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Observemos que de la condición $a_{k+1} \\geq \\frac{ka_k}{a_k^2 + (k-1)}$ se sigue que\n$$\n\\frac{k}{a_{k+1}} \\leq a_k + \\frac{k-1}{a_k}\n$$\nluego\n$$\na_k \\geq \\frac{k}{a_{k+1}} - \\frac{k-1}{a_k}\n$$\nSumando las desigualdades anteriores para $k = 1, 2, \\dots, m$ tenemos\n$$\na_1 + a_2 + \\dots + a_m \\geq \\left(\\frac{1}{a_2} - 0\\right) + \\left(\\frac{2}{a_3} - \\frac{1}{a_2}\\right) + \\dots + \\left(\\frac{m}{a_{m+1}} - \\frac{m-1}{a_m}\\right) = \\frac{m}{a_{m+1}}\n$$\nDemostraremos el resultado deseado por inducción sobre $n$:\npara $n = 2$, la condición del enunciado para $k = 1$ asegura\n$$\na_1 + a_2 \\geq a_1 + \\frac{1}{a_1} \\geq 2\n$$\nSuponiendo que el resultado es cierto para $n \\geq 2$, y que $a_{n+1} \\geq 1$, la hipótesis de inducción lleva a\n$$\n(a_1 + a_2 + \\dots + a_n) + a_{n+1} \\geq n + a_{n+1} \\geq n + 1\n$$\nPor último, si fuera $a_{n+1} < 1$ resultaría\n$$\n(a_1 + a_2 + \\dots + a_n) + a_{n+1} \\geq \\frac{n}{a_{n+1}} = \\frac{n-1}{a_{n+1}} + \\left(\\frac{1}{a_{n+1}} + a_{n+1}\\right) > (n-1) + 2 = n + 1\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16649,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D$ and $E$ be the midpoints of sides $AB$ and $AC$, respectively, of a triangle $ABC$. Prove that the line $AB$ is tangent to the circumcircle of the triangle $BEC$ if and only if the line $AC$ is tangent to the circumcircle of the triangle $BED$.",
    "options": [],
    "answer": "See solution",
    "solution": "The conditions of the problem imply that $DE$ is the midsegment parallel to the side $BC$ of the triangle $ABC$.\n\n![](images/EST_ABooklet_2021_p31_data_8565e18deb.png)\n\nBy properties of inscribed angles, the line $AB$ is tangent to the circumcircle of the triangle $BEC$ if and only if $\\angle ECB = \\angle DBE$, and the line $AC$ is tangent to the circumcircle of the triangle $BED$ if and only if $\\angle DBE = \\angle AED$. But $\\angle ECB = \\angle AED$ since the lines $DE$ and $BC$ are parallel, whence validity of either of these two equalities implies validity of the other one.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16650,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with orthocenter $H$. The bisector of angle $BHC$ intersects side $BC$ at $D$. Denote by $E$, $F$ the reflections of $D$ about $AB$, $AC$, respectively. Prove that the circumcircle of triangle $AEF$ passes through the midpoint of arc $BAC$.\n\n![](images/66th_Czech_and_Slovak_p18_data_ccaaf8a946.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, the directed angle $EAF$ is $2\\alpha$. Let $H_2$, $H_3$ be the reflections of the orthocenter $H$ about the sides $AC$, $AB$, respectively. It is well-known that $H_2$ and $H_3$ lie on the circumcircle $k$ of triangle $ABC$ (see the figure).\n\nAs $DH$ is the bisector of angle $BHC$ and $CH \\parallel DE$, we have $|\\angle H_3ED| = |\\angle HDE| = |\\angle DHC| = \\frac{1}{2}|\\angle BHC| = 90^\\circ - \\frac{1}{2}\\alpha$. Let $G$ be the midpoint of arc $BAC$. Since the arc $CG$ of $k$ is one half of the arc $CAB$, the directed angle $\\widehat{CH_3G}$ subtending arc $CG$ has the same measure $90^\\circ - \\frac{1}{2}\\alpha$, that is $\\widehat{CH_3G} = \\widehat{DEH_3}$.\n\nFrom $CH_3 \\parallel DE$ we infer that points $G$, $H_3$, and $E$ are collinear. Similarly, we find that $G$, $H_2$, and $F$ are also collinear. Therefore, $\\overrightarrow{EGF} = \\overrightarrow{H_3GH_2} = \\overrightarrow{H_3AH_2} = \\overrightarrow{H_3AH} + \\overrightarrow{HAH_2} = 2\\alpha = \\overrightarrow{EAF}$, as we needed to show.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16651,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a square-free positive even number, $k$ be an integer, $p$ be a prime number, satisfying $p < 2\\sqrt{n}$, $p \\nmid n$, $p \\mid n + k^2$. Prove that $n$ can be written as $n = ab + bc + ca$, where $a, b, c$ are distinct positive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $n$ is even, we have $p \\neq 2$. As $p \\nmid n$, we have $p \\nmid k$. We may assume without loss of generality $0 < k < p$. Set $a = k$, $b = p - k$, then $c = \\frac{n - k(p - k)}{p} = \\frac{n + k^2}{p} - k$.\n\nBy assumption, $c$ is an integer, and $a, b$ are distinct positive integers. It remains to be shown that $c > 0$, and $c \\neq a$, $b$. By the AM-GM inequality, we have $\\frac{n}{k} + k \\ge 2\\sqrt{n} > p$, thus $n + k^2 > pk$, hence $c > 0$.\n\nIf $c = a$, then $\\frac{n + k^2}{p} - k = k$, thus $n = k(2p - k)$. Since $n$ is even, $k$ is also even, as a consequence $n$ is divisible by $4$, which contradicts the fact that $n$ is square-free. If $c = b$, then $n = p^2 - k^2$. Since $n$ is even, $k$ is odd, implying that $n$ is again divisible by $4$, which is a contradiction.\n\nWe conclude that $a, b, c$ satisfy all the requirements, completing the proof. $\\Box$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16652,
    "subject": "Mathematics (Olympiad)",
    "question": "Find a polynomial $f(x, y, z)$ of degree 3 with real coefficients that satisfies the following conditions:\n\n- $f(x, y, z) + x$ is divisible by $y + z$\n- $f(x, y, z) + y$ is divisible by $z + x$\n- $f(x, y, z) + z$ is divisible by $x + y$\n\nA polynomial $P(x, y, z)$ is divisible by a polynomial $Q(x, y, z)$ if there exists a polynomial $R(x, y, z)$ such that $P(x, y, z) = Q(x, y, z) R(x, y, z)$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\nf(x, y, z) = k(x + y)(y + z)(z + x) - x - y - z, \\quad (k \\neq 0)\n$$\nsatisfies the conditions (in fact, only this form is a solution).\n\nLet $g(x, y, z) = f(x, y, z) + x + y + z$. Then the conditions are equivalent to $g(x, y, z)$ being divisible by $x + y$, $y + z$, and $z + x$. The requirement that $f(x, y, z)$ has degree 3 and real coefficients is equivalent to $g(x, y, z)$ having degree 3 and real coefficients. Now,\n$$\ng(x, y, z) = k(x + y)(y + z)(z + x), \\quad (k \\neq 0)\n$$\nsatisfies all the conditions. Therefore,\n$$\nf(x, y, z) = k(x + y)(y + z)(z + x) - x - y - z, \\quad (k \\neq 0)\n$$\nis a solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16653,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ islands be connected by $n-1$ bridges, forming a connected configuration. A fire starts at one island and spreads each night to all islands connected by bridges to any burnt island. Each night, you may destroy one bridge to prevent the fire from spreading. What is the minimum number of bridges that must be destroyed to stop the fire from spreading to all islands? Express your answer in terms of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum number of bridges that must be destroyed is $t = \\lfloor \\sqrt{n-1} \\rfloor$.\n\nFirst, consider a configuration with $n = t^2 + 1$ islands and $t^2$ bridges. Arrange $t$ lines of $t$ islands each, with each line's ends connected to a central island $X$. If the fire starts at $X$, after $t-1$ nights only $t-1$ bridges can be destroyed, leaving one line intact for the fire to spread.\n\nFor any connected configuration with $k$ unburnt islands, let $f(k)$ be the minimum number of bridges to destroy before the fire stops. We show $f(k) \\leq \\lfloor \\sqrt{k} \\rfloor$ by induction:\n\n- Base case: $f(1) = 1$.\n- Inductive step: Remove burnt islands, leaving $l$ components with $k$ unburnt islands. One component has at least $s = \\lfloor k/l \\rfloor$ islands. Destroy the bridge connecting this component to the burnt islands. The next morning, the fire spreads to one island in each of the other $l-1$ components. Now there are $k-s-l+1$ unburnt islands. By induction,\n\n  $$\n  f(k-s-1+l) \\leq f([k-s-l+1]) \\leq f([k-\\sqrt{2k}+1]) = f([\\sqrt{k}-1]^2) \\leq \\lfloor \\sqrt{k} \\rfloor - 1\n  $$\n\nAdding the first bridge destroyed, $f(k) \\leq \\lfloor \\sqrt{k} \\rfloor$, completing the induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16654,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a = \\log_3 x^2$ and $b = \\log_2 y^3$. Given the equations:\n\n$$\n\\log_9 x^4 + \\log_4 y^9 = 2\n$$\n$$\n\\log_3 x^2 + \\log_2 y^3 = 1\n$$\nFind the values of $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $a = \\log_3 x^2$ and $b = \\log_2 y^3$.\n\n$$\n\\log_9 x^4 = \\frac{\\log_3 x^4}{\\log_3 9} = \\frac{2 \\log_3 x^2}{2} = a\n$$\n$$\n\\log_4 y^9 = \\frac{\\log_2 y^9}{\\log_2 4} = \\frac{3 \\log_2 y^3}{2} = \\frac{3}{2}b\n$$\n\nSubstituting into the equations:\n\n$$\n\\begin{cases}\na + b = 1 \\\\\na + \\frac{3}{2}b = 2\n\\end{cases}\n$$\n\nSubtracting the first from the second:\n\n$$\n\\left(a + \\frac{3}{2}b\\right) - (a + b) = 2 - 1 \\\\\n\\frac{1}{2}b = 1 \\implies b = 2\n$$\n\nFrom $a + b = 1$:\n\n$$\na = 1 - b = 1 - 2 = -1\n$$\n\nNow, $\\log_3 x^2 = -1$:\n\n$$\nx^2 = 3^{-1} = \\frac{1}{3} \\implies x = \\pm\\frac{1}{\\sqrt{3}}\n$$\n\nAnd $\\log_2 y^3 = 2$:\n\n$$\ny^3 = 2^2 = 4 \\implies y = \\sqrt[3]{4}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16655,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $E$ and $F$ be the points on the sides $AB$ and $AD$ of a convex quadrilateral $ABCD$, such that $EF$ is parallel to $BD$. The segment $CE$ intersects the diagonal $BD$ at $G$, while the segment $CF$ intersects the diagonal $BD$ at $H$. Prove that if $AGCH$ is a parallelogram, then $ABCD$ is a parallelogram as well.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the intersection of the lines $EF$ and $AG$ by $I$ and the intersection of the lines $EF$ and $AH$ by $J$.\n\nNow, $EF$ is parallel to $BD$ and $AH$ is parallel to $CE$, so the quadrilateral $EGHJ$ is a parallelogram. Furthermore, $AG$ and $CF$ are parallel, so $FIGH$ is also a parallelogram. Hence, $|FH| = |IG|$, $|HJ| = |GE|$ and $\\angle FHJ = \\angle IGE$, which means that the triangles $FHW$ and $IGE$ are congruent. Thus, $|FJ| = |IE|$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16656,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ and $4p+1$ be two primes such that $p > 10^9$. Prove that the decimal representation of the number $\\frac{1}{4p+1}$ contains every digit from 0 to 9.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $q = 4p+1$ and let $a \\circ b$ denote the remainder of $a$ upon division by $b$. It suffices to show that the last digits of the numbers $10^k \\circ q$ cover all digits from 0 to 9. Since $q$ and 10 are coprime, this means that all digits from 0 to 9 occur among the last digits of the numbers $\\left\\lfloor \\frac{10^k}{q} \\right\\rfloor$, which are the same as the digits in the decimal representation of $\\frac{1}{q}$.\n\nLet $S$ be the set of all nonzero biquadratic residues modulo $q$, i.e., the set of all $0 < t < q$ such that the congruence $x^4 \\equiv t \\pmod{q}$ has a solution. Then $S$ has exactly $p$ elements. Since $-1$ is a quadratic residue modulo $q$, the $2p$ nonzero quadratic residues modulo $q$ can be grouped in pairs of the form $\\{s, -s\\}$, and among their squares—which are precisely the biquadratic residues—there are exactly $p$ distinct ones.\n\nWe will show that every element of $S$ has the form $10^k \\circ q$ for some $k$. Let $d$ be the order of 10 modulo $q$. Since $d \\mid \\varphi(q) = 4p$, we have $d = p$, $d = 2p$, or $d = 4p$.\n\nConsider the case $d = p$ first. For $0 \\leq k < p$, there is a $0 \\leq j \\leq 3$ such that $k + jq$ is a multiple of 4, and therefore $10^k \\equiv \\left[10^{\\frac{k+jq}{4}}\\right]^4 \\pmod{q}$ is a biquadratic residue. Since the numbers $10^k \\circ q$ for $0 \\leq k < p$ are pairwise distinct and all are biquadratic residues, they must coincide with the elements of $S$.\n\nThe cases $d = 2p$ and $d = 4p$ are treated analogously.\n\nNow, let $u$ be an arbitrary digit. Let $0 \\leq j \\leq 3$ be such that $u + jq$ ends in 0, 1, 5, or 6. Since $q > 10^9$, we have $\\sqrt[4]{-(j+1)q} - \\sqrt[4]{u+jq} > 6$. Thus, the interval $[u+jq, (j+1)q)$ contains at least six fourth powers. Therefore, it contains at least one fourth power $x^4$ that ends in the same digit as $u+jq$. Let $s = x^4 \\circ q$; then $x^4 = s + jq$ and $10 \\mid x^4 - (u+jq) = s - u$, i.e., $s$ ends in $u$, as needed.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16657,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number. Show that one can choose $p^3$ fields of a $p^2 \\times p^2$ chessboard such that the centers of no four chosen fields are vertices of a rectangle with sides parallel to the sides of the chessboard.",
    "options": [],
    "answer": "See solution",
    "solution": "Label the $p^2$ rows and $p^2$ columns of the chessboard by pairs $(a, b)$ and $(c, d)$, respectively, where $a, b, c, d \\in \\{0, 1, \\dots, p-1\\}$. A field in row $(a, b)$ and column $(c, d)$ is called *good* if and only if\n\n$$\na c \\equiv b + d \\pmod{p}. \\qquad (1)$$\n\nGiven a pair $(a, b)$, equation (1) holds for exactly $p$ pairs $(c, d)$, so there are $p$ good fields in any row and $p^3$ in total.\n\nNow, suppose for contradiction that there exist four good fields forming a rectangle, i.e.,\n\n$$\na_i c_j \\equiv b_i + d_j \\pmod{p} \\quad \\text{for } i, j \\in \\{1, 2\\}. \\qquad (2)$$\n\nfor some $(a_1, b_1) \\neq (a_2, b_2)$ and $(c_1, d_1) \\neq (c_2, d_2)$. Subtracting (2) with fixed $i$ gives\n\n$$\na_i(c_2 - c_1) \\equiv d_2 - d_1 \\pmod{p} \\quad \\text{for } i \\in \\{1, 2\\}. \\qquad (3)$$\n\nSubtracting (3) for $i=1$ and $i=2$ yields\n\n$$\n(a_2 - a_1)(c_2 - c_1) \\equiv 0 \\pmod{p}.\n$$\n\nThus, $a_1 = a_2$ or $c_1 = c_2$. By symmetry, assume $c_1 = c_2$. Then (3) implies $d_1 = d_2$, so $(c_1, d_1) = (c_2, d_2)$, a contradiction.\n\n![](path/to/file.png)\n\nIf $a=0$, then $c$ is arbitrary and $d$ is unique. If $a \\ne 0$, then $d$ is arbitrary and $c$ is a function of $d$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16658,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with fixed points $B, C$ and point $A$ moves on the big arc $BC$ of $(ABC)$ such that $AB \\neq AC$. The incircle $(I)$ of triangle $ABC$ touches $BC$ at $D$. Let $I_a$ be the $A$-excenter of triangle $ABC$. $I_aD$ cuts $OI$ at $L$ and $E$ lies on $(I)$ such that $DE \\parallel AI$.\n\n- a) Line $LE$ cuts $AI$ at $F$. Prove that $AF = AI$.\n- b) Let $M$ be the point on the circumcircle $(J)$ of triangle $I_aBC$ such that $I_aM \\parallel AD$. The line $MD$ cuts $(J)$ again at $N$. Prove that the midpoint $T$ of $MN$ lies on a fixed circle.\n\n![](images/Vietnam_2022_p23_data_b5e7f8e920.png)",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $(I)$ touch $CA$ and $AB$ at $U$ and $V$ respectively; let $I_b$ and $I_c$ be the excenters at $B$ and $C$ in triangle $ABC$. Clearly, $I_aA$, $I_bB$, and $I_cC$ are the three altitudes of triangle $I_aI_bI_c$. On the other hand, notice that $IA \\perp UV$, $IB \\perp VD$, and $IC \\perp DU$, so triangles $DUV$ and $I_aI_bI_c$ have corresponding parallel sides. Thus, there exists a homothety $\\mathcal{H}$ that maps $D$ to $I_a$, $U$ to $I_b$, and $V$ to $I_c$.\n\nUnder $\\mathcal{H}$, $I$ becomes $O'$, the circumcenter of triangle $I_aI_bI_c$. In triangle $I_aI_bI_c$, $I$ is the orthocenter and $(ABC)$ is the Euler circle, so $O$ lies on $IO'$, which implies the center of $\\mathcal{H}$ lies on $OI$. Also, the center of $\\mathcal{H}$ lies on $DI_a$; $DI_a$ and $OI$ intersect at $L$, so $L$ is the center of $\\mathcal{H}$.\n\n![](images/Vietnam_2022_p24_data_36b5a422c3.png)\n\nWe have $L$, $E$, $F$ collinear and $DE \\parallel I_aF$, so $F$ is the image of $E$ under the homothety, and thus lies on $(I_aI_bI_c)$. Furthermore, triangle $I_aI_bI_c$ has $I$ as the orthocenter and $I_aF$ as an altitude, so $F$ and $I$ are symmetric with respect to $I_cI_b$, hence $AF = AI$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16659,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all numbers $p$, $q$, and $r$ such that $p$ and $r$ are prime, $q$ is a positive integer, and they satisfy the equation:\n\n$$\n(p + q + r)^2 = 2p^2 + 2q^2 + r^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "After simplifying the equation, we get $2r(p + q) = (p - q)^2$. Since $r$ is prime, $r$ divides $p - q$, so $r^2$ divides the right-hand side. This implies $r$ divides $2(p + q)$. If $r > 2$, then $r$ divides $p + q$, so $r$ must divide both $p$ and $q$. Since $p$ is prime, this is possible only if $p = r$ and $q = sr$. After simplification, $2(1 + s) = (s - 1)^2$, so $s^2 - 4s - 1 = 0$, which has no integer solutions. Thus, no solution in this case.\n\nIf $r = 2$, then $p$ and $q$ have the same parity. The case $p = 2$ is impossible, so both must be odd. Let $a \\neq 2$ be a prime divisor of $p + q$; then $a$ must divide $p - q$ too, so $a$ divides both $p$ and $q$, which is possible only if $p = a$ and $q = sa$. In this case, $4(1 + s) = a(s - 1)^2$, leading to $a^2 - (2a + 4)s + (a - 4) = 0$, with solutions $\\frac{a + 2 \\pm 2\\sqrt{2a + 1}}{a}$. If $\\sqrt{2a + 1}$ is integer, then $2a + 1 = 4b^2 + 4b + 1$, so $a = 2b(b + 1)$, which cannot be prime. Therefore, $p + q$ and $p - q$ must be powers of $2$, i.e., $p - q = 2^k$ and $p + q = 2^{2k - 2}$, so $2p = 2^k + 2^{2k - 2}$ and $2q = 2^{2k - 2} - 2^k$. Since $p$ and $q$ are odd, $k = 1$, but then $p + q = 1$, which is impossible. It follows that the equation has no prime number solutions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16660,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $2017$ points be placed on a circle, each colored either red or blue. Prove that the number of isosceles triangles with all vertices the same color depends only on the number of red and blue points, not on their arrangement.\n\n![](images/Mathematical_competitions_in_Croatia_2017_p21_data_18548e57a5.png)",
    "options": [],
    "answer": "See solution",
    "solution": "First, notice that none of the triangles formed by taking three vertices of the given polygon is equilateral. If there were one, we would have the same number of vertices of the original polygon between any two of the triangle's vertices. This would imply that $3$ divides $2014$, which is false.\n\nLet us now prove that each segment (whose endpoints are vertices of the given polygon) belongs to exactly $3$ different isosceles triangles.\n\nConsider an arbitrary segment. Its endpoints divide the circumcircle of the polygon into two arcs containing $2015$ vertices. One of the arcs contains an odd number, and the other contains an even number of vertices. Therefore, there exists a unique isosceles triangle with the chosen segment as its base (the third vertex is the midpoint of the arc containing an odd number of vertices). In a similar fashion, we infer that the chosen segment is a base in exactly two isosceles triangles – there are two possible choices of the third segment, both on the longer arc.\n\nLet $B$ be the number of blue points, so that $B + R = 2017$. Let $d_b, d_r, d_s$ be the number of blue (both endpoints blue), red, and two-colour segments, respectively. Furthermore, let $t_b, t_r, t_{sb}, t_{sr}$ be the number of isosceles triangles with the vertices colored all blue, all red, two blue – one red, and two red – one blue, respectively.\n\nFrom the fact that each segment belongs to three isosceles triangles, we get:\n\n$$\n\\begin{aligned}\n3d_b &= 3t_b + t_{sb}, \\\\\n3d_r &= 3t_r + t_{sr}, \\\\\n3d_s &= 2t_{sb} + 2t_{sr}.\n\\end{aligned}\n$$\n\nThis implies $3d_b + 3d_r - \\frac{3}{2}d_s = 3t_b + 3t_r$, i.e.\n\n$$\nt_b + t_r = d_b + d_r - \\frac{1}{2}d_s = \\frac{1}{2} \\cdot (B(B-1) + R(R-1) - BR),\n$$\n\nwhich depends only on the number of the red and blue points, but not on their arrangement.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16661,
    "subject": "Mathematics (Olympiad)",
    "question": "Притоа Y не припаѓа во внатрешноста ниту на $k_1$ ниту на $k_3$ и Z не припаѓа во внатрешноста ниту на $k_1$ ниту на $k_2$.\n\na) Докажи дека триаголниците $XYZ$ и $O_1O_2O_3$ се слични меѓу себе.\n\nб) Докажи дека плоштината на триаголникот $XYZ$ не е поголема од четири пати по плоштината на триаголникот $O_1O_2O_3$. Дали се достигнува максимумот?\n\n![](images/MACEDONIAN_MATHEMATICAL_SOCIETY_p21_data_596a36d478.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Прво ќе докажеме дека точките $Y$, $B$ и $Z$ се колинеарни. Бидејќи четириаголникот $BYAP$ е тетивен, имаме $\\angle PBY = \\angle PAX$. Бидејќи четириаголникот $AXCP$ е тетивен, $\\angle PAX = \\angle PCZ$. Бидејќи четириаголникот $CPBZ$ е тетивен, добиваме $\\angle PBZ + \\angle PCZ = 180^\\circ$. Значи $\\angle YBZ = \\angle YBP + \\angle PBZ = 180^\\circ$.\n\nДа забележиме дека $\\angle CO_1O_3 = \\angle PO_1O_3$ и $\\angle AO_1O_2 = \\angle PO_1O_2$, од каде следува $\\angle O_2O_1O_3 = \\frac{1}{2}\\angle AO_1C = \\angle AXC$. Слично, $\\angle O_1O_2O_3 = \\angle AYB$ и $\\angle O_1O_3O_2 = \\angle CZB$. Следува дека $\\triangle XYZ \\sim \\triangle O_1O_2O_3$, со што го докажавме тврдењето под а).\n\nНека правата $X_1Y_1$ е паралелна со $O_1O_2$ и минува низ $A$, каде $X_1$ лежи на $k_1$ и $Y_1$ лежи на $k_2$. Нека $Z_1$ е пресечната точка на правата $X_1C$ со кружницата $k_3$. Од претходно докажаното, точките $Y_1$, $B$ и $Z_1$ се колинеарни и $\\triangle X_1Y_1Z_1 \\sim \\triangle O_1O_2O_3$. Уште повеќе, $\\angle PXA = \\angle PX_1A$ и $\\angle PYA = \\angle PY_1A$. Па $\\triangle PXY \\sim \\triangle PX_1Y_1$. Нека $PT$ е висината спуштена од темето $P$ кон страната $XY$. $PA$ е висината на триаголникот $PX_1Y_1$. Бидејќи $PA$ е хипотенуза во правоаголниот триаголник $PAT$, добиваме $\\overline{PT} \\leq \\overline{PA}$. Па $P_{PXY} \\leq P_{PX_1Y_1}$ и аналогно $P_{PYZ} \\leq P_{PY_1Z_1}$ и $P_{PXZ} \\leq P_{PX_1Z_1}$. Од ова добиваме $P_{XYZ} \\leq P_{X_1Y_1Z_1}$. Точките $P$, $O_1$ и $X_1$ се колинеарни затоа што $\\angle PAX_1 = 90^\\circ$. Слично, $P$, $O_2$ и $Y_1$ се колинеарни и $P$, $O_3$ и $Z_1$ се колинеарни. Добиваме дека $O_1O_2$, $O_1O_3$ и $O_2O_3$ се средни линии во триаголниците $X_1Y_1P$, $X_1Z_1P$ и $Y_1Z_1P$ соодветно, па $P_{X_1Y_1Z_1} = 4P_{O_1O_2O_3}$. Од ова се добива бараното неравенство. Равенство се достигнува кога точките $X$ и $X_1$ се соовпаѓаат, а со тоа и точките $Y$ и $Y_1$ се соовпаѓаат и точките $Z$ и $Z_1$ се соовпаѓаат.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16662,
    "subject": "Mathematics (Olympiad)",
    "question": "How many integer solutions does the equation\n\n$$x^2 + y^2 + z^2 - xy - yz - zx = x^3 + y^3 + z^3 + s$$\n\nhave, if\n\na) $s = -1$;\nb) $s = 1$?",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* a) infinitely many solutions, b) there are no solutions.\n\n*Solution.*\n\na) Take $x + y = 1$, $z = 1$, then\n\n$$ (x^2 + y^2 - xy) + 1 - (y + x) = x^3 + y^3 $$\n$$ (x + y)(x^2 + y^2 - xy) = x^3 + y^3 $$\n\nso each such triple satisfies the condition of the problem.\n\nb) See problem 8-5.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16663,
    "subject": "Mathematics (Olympiad)",
    "question": "For each integer $n \\ge 2$, let $s(n)$ denote the sum of all positive integers that are at most $n$ and not relatively prime to $n$.\n\n**a)** Prove that $s(n) = \\frac{n}{2}(n+1-\\varphi(n))$, where $\\varphi(n)$ is the number of positive integers that are at most $n$ and are relatively prime to $n$.\n\n**b)** Prove that there does not exist an integer $n \\ge 2$ such that\n$$\ns(n) = s(n + 2021).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "a) Notice that if $k$ is a positive integer such that $\\gcd(k, n) = 1$ and $k < n$, then $n-k$ and $n$ are also relatively prime. It follows that\n$$\n\\sum_{k \\le n,\\ \\gcd(k, n)=1} k = \\sum_{k \\le n,\\ \\gcd(k, n)=1} (n-k).\n$$\nLet $A = \\{k \\in \\mathbb{N} \\mid 1 \\le k \\le n,\\ \\gcd(k, n) = 1\\} = \\{k_1, k_2, \\dots, k_{\\varphi(n)}\\}$. Then\n$$\n\\sum_{i=1}^{\\varphi(n)} k_i = \\frac{1}{2} \\sum_{i=1}^{\\varphi(n)} k_i + \\frac{1}{2} \\sum_{i=1}^{\\varphi(n)} (n-k_i) = \\frac{1}{2} \\sum_{i=1}^{\\varphi(n)} (k_i + n - k_i) = \\frac{n\\varphi(n)}{2}.\n$$\nSo, we have\n$$\ns(n) = \\sum_{i=1}^{n} i - \\sum_{i=1}^{\\varphi(n)} k_i = \\frac{n(n+1)}{2} - \\frac{n\\varphi(n)}{2} = \\frac{n}{2}(n+1-\\varphi(n)).\n$$\n\nb) Suppose that there exists a positive integer $n$ such that $s(n) = s(n + 2021)$. Thus,\n$$\n2s(n) = n(n+1-\\varphi(n)) = (n+2021)(n+2022-\\varphi(n+2021))\n$$\nor\n$$\n2021(2n + 2022 - \\varphi(n + 2021)) = n(\\varphi(n + 2021) - \\varphi(n)). \\quad (1)\n$$\nIt follows that $n$ and $n + 2021$ are divisors of $2s(n)$. Otherwise,\n$$\n2s(n) = n(n+1) - n\\varphi(n) < n(n+1) < n(n + 2021),\n$$\nso $n$ and $n + 2021$ are not coprime, which means $\\gcd(n, 2021) \\neq 1$. Let $d = \\gcd(n, 2021) > 1$, so $\\gcd\\left(\\frac{n}{d}, \\frac{2021}{d}\\right) = 1$. From (1), we get\n$$\n\\frac{2021}{d}(2n + 2022 - \\varphi(n + 2021)) = \\frac{n}{d}(\\varphi(n + 2021) - \\varphi(n)). \\quad (2)\n$$\nSo there exists some positive integer $x$ such that\n$$\n\\varphi(n + 2021) - \\varphi(n) = \\frac{2021}{d} \\cdot x \\quad (3)\n$$\n$$\n2n + 2022 - \\varphi(n + 2021) = \\frac{n}{d} \\cdot x \\quad (4)\n$$\nThus, $\\varphi(n+2021)-\\varphi(n)$ is divisible by $2021/d$. Otherwise, one can check that $\\varphi(n + 2021)$ and $\\varphi(n)$ are divisible by $\\varphi(d)$ and $\\gcd(\\varphi(d), 2021/d) = 1$ for all $d \\in \\{43, 47, 2021\\}$, so\n$$\n\\frac{2021\\varphi(d)}{d} \\mid \\varphi(n + 2021) - \\varphi(n)\n$$\nOn the other hand, it follows from (3) and (4) that\n$$\nd < x = d \\cdot \\frac{2n + 2022 - \\varphi(n)}{n + 2021} < 2d.\n$$\nIt implies that, for all $d \\in \\{43, 47, 2021\\}$,\n$$\n\\frac{2021\\varphi(d)}{d} < \\frac{2021 \\cdot d}{d} < \\frac{2021}{d}x < 2 \\cdot 2021 < \\frac{3 \\cdot 2021\\varphi(d)}{d}.\n$$\nThus, $\\varphi(n + 2021) - \\varphi(n) = 2 \\cdot 2021\\varphi(d)/d$ and $x = 2\\varphi(d)$, so\n$$\n\\varphi(n + 2021) = \\frac{2n(d - \\varphi(d))}{d} + 2022, \\quad (5)\n$$\n$$\n\\varphi(n) = \\frac{2n(d - \\varphi(d))}{d} + 2022 - \\frac{2 \\cdot 2021\\varphi(d)}{d}. \\quad (6)\n$$\nIf $n$ has at most 10 distinct prime divisors then\n$$\n\\varphi(n) > n \\prod_{i=2}^{11} \\left(1 - \\frac{1}{i}\\right) = \\frac{n}{11} > \\frac{2n(d - \\varphi(d))}{d}, \\quad d \\in \\{43, 47, 2021\\},\n$$\nwhich contradicts (6). We get $n$ has at least 11 distinct prime divisors, so $n > 12!$ and $\\varphi(n)$ is divisible by $2^{10}$.\n\nSimilarly, if $n + 2021$ has at most 4 distinct prime divisors then\n$$\n\\varphi(n + 2021) > (n + 2021) \\prod_{i=2}^{5} \\left(1 - \\frac{1}{i}\\right) = \\frac{n + 2021}{5}.\n$$\nOtherwise, $n > 12!$ and\n$$\n\\frac{n + 2021}{5} > \\frac{2n(d - \\varphi(d))}{d} + 2022, \\quad d \\in \\{43, 47, 2021\\},\n$$\nwhich contradicts (5). We get $n + 2021$ has at least 5 distinct prime divisors, so $\\varphi(n + 2021)$ is divisible by $2^4$.\n\nOn the other hand,\n$$\nv_2(\\varphi(n + 2021) - \\varphi(n)) = v_2\\left(\\frac{2 \\cdot 2021\\varphi(d)}{d}\\right) \\le 3, \\forall d \\in \\{43, 47, 2021\\}\n$$\nwhich contradicts $2^4 \\mid \\varphi(n + 2021)$ and $2^{11} \\mid \\varphi(n)$. Hence, there does not exist a positive integer $n$ such that $s(n) = s(n+2021)$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16664,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that there exists a multiple of $2013$ which ends in $2014$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the numbers $a_1, a_2, \\dots, a_{2014}$, where $a_k$ is the $4k$-digit number whose digits are $k$ groups of the form $2014$, $1 \\leq k \\leq 2014$. Since there are only $2013$ possible remainders after division by $2013$, two different $a$'s must give the same remainder, so there exist $i, j$ with $1 \\leq j < i \\leq 2014$ such that $a_i - a_j$ is divisible by $2013$. Since\n\n$$\na_i - a_j = \\underbrace{20142014\\dots2014}_{i-j \\text{ times } 2014} \\underbrace{00\\dots0}_{4j \\text{ times}} = \\underbrace{20142014\\dots2014}_{i-j \\text{ times } 2014} \\cdot \\underbrace{100\\dots0}_{4j \\text{ times}} = a_{i-j} \\cdot 10^{4j}\n$$\n\nand $10^{4j}$ and $2013$ are relatively prime, $2013$ must divide $a_{i-j}$. Thus, there exists a multiple of $2013$ ending in $2014$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16665,
    "subject": "Mathematics (Olympiad)",
    "question": "Felipe tells Roberto that he managed to place 1226 triminos $1 \\times 3$ on a certain grid square $Q$ so that they do not have common points, not even vertices. Without knowing the dimensions of $Q$, Roberto answers, \"If what you say is true then one can place 1250 triminos $1 \\times 3$ on your square, under the same conditions.\" Is Roberto right?\n\n![](images/Argentina_2015_Booklet_p4_data_e15214cce6.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, Roberto is right. To prove this, we first show that the least $n$ for which 1226 triminos $1 \\times 3$ can be placed on an $n \\times n$ square without common points (including vertices) is $n=99$.\n\nNext, take a $99 \\times 99$ square. In its first row, one can place 25 triminos without points in common; they leave uncovered the 24 cells whose position numbers are divisible by 4. Do the same with all odd-numbered rows $1, 3, 5, \\ldots, 99$, of which there are 50. This arrangement of $25 \\cdot 50 = 1250$ triminos satisfies the conditions, and clearly, any greater grid square can accommodate 1250 triminos too.\n\nSo let $Q$ be an $n \\times n$ square containing 1226 triminos $1 \\times 3$ without common points (including vertices). Extend $Q$ to an $(n+1) \\times (n+1)$ square $Q'$ by adjoining an additional bottom row and additional leftmost column, with $n+1$ cells in each of them. To each trimino $T$ add 5 cells to the left and under it so that $T$ and these 5 cells form a rectangle $R(T)$ with dimensions $2 \\times 4$. The definition of $R(T)$ is illustrated in the figures, for horizontal and vertical triminos.\n\n![](images/Argentina_2015_Booklet_p4_data_eb6b5133ac.png)\n\n![](images/Argentina_2015_Booklet_p4_data_5c7e283253.png)\n\nThe $2 \\times 4$ rectangles defined in this way may stick out of $Q$ but are contained in the bigger square $Q'$.\n\nThe point of the construction is that the rectangles $R(T)$ do not overlap, i.e., no two of them have cells in common. Indeed, let $T$ and $T'$ be arbitrary triminos. By symmetry, assume that $T$ is vertical. Enclose it in a $3 \\times 5$ rectangle as shown in the figure.\n\n![](images/Argentina_2015_Booklet_p5_data_4caeb8d7d8.png)\n\nNote that by hypothesis this rectangle $R^*$ contains no cell of the trimino $T'$. The cells marked with $\\textbullet$ form the rectangle $R(T)$ together with $T$. Denote by $C$ the top right cell in $T'$; it is also the top right cell of $R(T')$.\n\nSuppose that $C$ is under line $a$. Then so is the entire rectangle $R(T')$ by its definition, hence $R(T')$ does not overlap with $R(T)$. The same follows if $C$ is to the left of line $c$. Suppose that $C$ is in region I, i.e., between the lines $c, d$ and above line $b$. Then the entire $T'$ is above $b$. This is clear if $T'$ is horizontal. And if $T'$ is vertical then having cells under $b$ means having cells in $R^*$ which is forbidden. So indeed the whole of $T'$ is above $b$, implying that $R(T')$ is above line $b'$. Then $R(T')$ does not overlap $R(T)$ again. The case where $C$ is in region II is analogous; here $T'$ is to the right of line $d$ and $R(T')$ is to the right of line $d'$. Finally, the case of $C$ in region III is obvious.\n\nIn summary, we obtain 1226 non-overlapping rectangles $2 \\times 4$ in the $(n+1) \\times (n+1)$ square $Q'$. By area considerations then $$(n+1)^2 \\ge 8 \\cdot 1226 = 9808$$ and so $n+1 \\ge \\sqrt{9808} = 99.035\\ldots$. It follows that $n \\ge 99$, as stated, so the starting discussion completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16666,
    "subject": "Mathematics (Olympiad)",
    "question": "試求最大的正整數 $L$ 使得存在正整數數列 $a_1, a_2, \\dots, a_L$ 滿足：\n\n(a) 數列中的每一項都小於或等於 $2^{2024}$；\n\n(b) 不存在連續子數列 $a_i, a_{i+1}, \\dots, a_j$（其中 $1 \\le i \\le j \\le L$）使得我們可以適當選取 $s_i, s_{i+1}, \\dots, s_j \\in \\{-1, 1\\}$，使得\n\n$$\ns_i a_i + s_{i+1} a_{i+1} + \\dots + s_j a_j = 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "答案為 $2^{2025}-1$；一般性地，對於上限 $2^k$，最大 $L = 2^{k+1}-1$。\n\n**構造**：令 $v_2(x)$ 為 $x$ 對 $2$ 的冪，並取 $a_i = 2^{k-v_2(i)}$。顯然 $a_i \\le 2^k$。\n\n此外，我們有：\n\n*引理*：對任意 $1 \\le i \\le j \\le 2^{k+1}-1$，存在唯一的 $i \\le x \\le j$ 使得 $v_2(x) = \\max_{i \\le y \\le j} v_2(y)$。\n\n*證明*：若在 $x$ 和 $y$ 都取到最大值 $v$，表示 $x = p \\times 2^v$ 且 $y = q \\times 2^v$，其中 $p, q$ 為奇數，假設 $p < q$。則 $p < p+1 < q$，所以 $z = (p+1) \\times 2^v$ 在 $x$ 與 $y$ 之間，但 $v > 0$，與最大性矛盾。\n\n因此，對任意 $1 \\le i \\le j \\le 2^{k+1}-1$，存在唯一的 $i \\le x \\le j$ 使得 $v_2(x)$ 最大。這表示 $v_2(a_x) = k-v$，但 $v_2(a_y) > k-v$ 對所有 $y \\ne x$ 皆成立，故 $v_2(\\sum s_\\ell a_\\ell) = k-v$，不可能為 $0$。\n\n**估計**：假設 $L \\ge 2^{k+1}$。假設 $a_1, \\dots, a_L$ 滿足 $a_i \\le 2^k$。令 $b_0 = 0$，並遞迴定義\n\n$$\ns_i = \\begin{cases} +1 & \\text{if } b_{i-1} \\le 0, \\\\ -1 & \\text{if } b_{i-1} \\ge 1. \\end{cases}\n$$\n\n$$\nb_i = b_{i-1} + s_i a_i.\n$$\n\n考慮數列 $b_0, b_1, \\dots, b_L$。注意，因 $a_i \\le 2^k$，若 $b_{i-1} \\in [-2^k+1, 0]$，則 $b_i = b_{i-1}+a_i \\in [-2^k+1, 2^k]$；反之，若 $b_{i-1} \\in [1, 2^k]$，則 $b_i = b_{i-1}-a_i \\in [-2^k+1, 2^k]$。因此，$b_0$ 到 $b_L$ 共有 $L+1 \\ge 2^{k+1}+1$ 項，卻只有 $2^{k+1}$ 個可能值，故存在 $1 \\le i \\le j \\le L$ 使得 $b_{i-1} = b_j$，也就是 $b_j - b_{i-1} = \\sum_{\\ell \\le j} s_\\ell a_\\ell = 0$。故不存在長度大於 $2^{k+1}-1$ 且滿足條件的數列。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16667,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the number of positive integers with digits less than or equal to 2009 that can be represented in the form $a^{2009} + b^{2009}$, where $a$ and $b$ are integers with $a \\ge b$, and $a^{2009} + b^{2009}$ is positive and has digits less than or equal to 2009 (i.e., $0 < a^{2009} + b^{2009} < 10^{2009}$).",
    "options": [],
    "answer": "See solution",
    "solution": "Let us analyze the possible pairs $(a, b)$:\n\n**Case (A):** $1 \\le a \\le 9$ and $-a < b \\le a$.\n\n**Case (B):** $a = 10$ and $-10 < b < 0$.\n\nFor both cases, $a^{2009} + b^{2009}$ is positive and less than $10^{2009}$, as shown using the lemma $(9/10)^{2009} < 1/3$.\n\nFor $a \\ge 11$, $a^{2009} + b^{2009}$ exceeds $10^{2009}$, so no solutions exist.\n\nCounting:\n- For case (A): $\\sum_{a=1}^9 2a = 90$ pairs.\n- For case (B): $9$ pairs.\n\nNo two pairs yield the same value, as shown by contradiction using the lemma.\n\n**Final answer:** $90 + 9 = 99$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16668,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the following inequality:\n$$\n\\max\\{x^2 + 3x + 3, x^{2011} + x^4 + x^2 + x + 1\\} \\leq \\min\\{1 - x - x^2, x^{2011} + x^4 + x^2 + x + 1\\},\n$$\nwhere $\\max\\{a, b\\} = \\begin{cases} a, & \\text{if } a \\geq b \\\\ b, & \\text{if } a < b \\end{cases}$, and $\\min\\{a, b\\} = \\begin{cases} b, & \\text{if } a \\geq b \\\\ a, & \\text{if } a < b \\end{cases}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P(x) = x^{2011} + x^4 + x^2 + x + 1$. Clearly,\n$$\n\\max\\{x^2 + 3x + 3, P(x)\\} \\geq P(x) \\geq \\min\\{1 - x - x^2, P(x)\\},\n$$\nso the inequality can only be satisfied if\n$$\n\\max\\{x^2 + 3x + 3, P(x)\\} = P(x) = \\min\\{1 - x - x^2, P(x)\\},\n$$\nwhich implies\n$$\nx^2 + 3x + 3 \\leq P(x) \\leq 1 - x - x^2.\n$$\nThus, $x^2 + 3x + 3 \\leq 1 - x - x^2$, which gives $2x^2 + 4x + 2 \\leq 0 \\iff (x + 1)^2 \\leq 0 \\iff x = -1$. Verifying, $x = -1$ satisfies the original inequality.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16669,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer such that $n-32$, $n-33$, and $n-34$ are multiples of $97$, $100$, and $103$, respectively. Find the smallest such $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be a positive integer satisfying the conditions above. Then $n-32$, $n-33$, $n-34$ are multiples of $97$, $100$, $103$, respectively. \n\nWe have:\n\n$$\n3n+1 = 3(n-32) + 97 = 3(n-33) + 100 = 3(n-34) + 103,\n$$\n\nso $3n + 1$ is a multiple of $97$, $100$, and $103$. Since any pair among $97$, $100$, $103$ are relatively prime, $3n+1$ must be a multiple of $97 \\cdot 100 \\cdot 103$.\n\nThe smallest positive integer $n$ for which $3n + 1$ is a multiple of $97 \\cdot 100 \\cdot 103$ is\n\n$$\n\\frac{97 \\cdot 100 \\cdot 103 - 1}{3} = 333033.\n$$\n\nSince\n\n$$\n\\begin{aligned}\n\\frac{97 \\cdot 100 \\cdot 103 - 1}{3} &= 97 \\cdot \\frac{100 \\cdot 103 - 1}{3} + 32 \\\\\n&= 100 \\cdot \\frac{97 \\cdot 103 - 1}{3} + 33 \\\\\n&= 103 \\cdot \\frac{97 \\cdot 100 - 1}{3} + 34,\n\\end{aligned}\n$$\n\nwe conclude that $333033$ is the desired answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16670,
    "subject": "Mathematics (Olympiad)",
    "question": "Calculate $\\sqrt{0.04^3}$.",
    "options": [],
    "answer": "See solution",
    "solution": "$\\sqrt{0.04^3} = \\sqrt{\\left(\\frac{4}{100}\\right)^3} = \\left(\\sqrt{\\frac{4}{100}}\\right)^3 = \\left(\\frac{2}{10}\\right)^3 = \\left(\\frac{1}{5}\\right)^3 = \\frac{1}{125}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16671,
    "subject": "Mathematics (Olympiad)",
    "question": "How many pairs of 3-digit palindromes are there such that when they are added together, the result is a 4-digit palindrome? For example, $232 + 989 = 1221$ gives one such pair.",
    "options": [],
    "answer": "See solution",
    "solution": "Write the sum as follows:\n\n$$\n\\begin{array}{c}\naba \\\\\ncdc \\\\\\hline\neffe\n\\end{array}\n$$\n\nIt is clear from the 1000s column that $e = 1$.\n\nSo from the units column, $a + c = 1$ or $a + c = 11$.\n\nBut the carry from the 100s column means that $a + c = 11$.\n\n**Method 1**\n\nWithout loss of generality, take $a < c$.\n\nSo the possible solution pairs for $(a, c)$ are: $(2, 9)$, $(3, 8)$, $(4, 7)$, $(5, 6)$.\n\nNow the 10s column gives $b + d + 1 = f$ or $b + d + 1 = f + 10$.\n\n*Case 1*: $b + d + 1 = f$.\n\nSince there is no carry to the 100s column and $a + c = 11$, we have $f = 1$.\n\nThus $b + d = 0$, hence $b = d = 0$.\n\nSo this gives four solutions: $202 + 909 = 1111$, $303 + 808 = 1111$, $404 + 707 = 1111$, $505 + 606 = 1111$.\n\n*Case 2*: $b + d + 1 = f + 10$.\n\nSince there is a carry to the 100s column, $a + c + 1 = 12$.\n\nThen $f = 2$, hence $b + d = 11$.\n\nFor each pair of values of $a$ and $c$, there are then eight solution pairs for $(b, d)$: $(2, 9)$, $(3, 8)$, $(4, 7)$, $(5, 6)$, $(6, 5)$, $(7, 4)$, $(8, 3)$, $(9, 2)$.\n\nSo this gives $4 \\times 8 = 32$ solutions:\n\n$(222 + 999 = 1221, 232 + 989 = 1221, \\text{etc}).$\n\nHence the number of solution pairs overall is $4 + 32 = 36$.\n\n**Method 2**\n\nWe know that $e = 1$, $a + c = 11$, and the carry from the 10s column is at most 1.\n\nHence $f = 1$ or $2$. Therefore $b + d = 0$ or $11$ respectively.\n\nFor each pair of values of $a$ and $c$, there are then nine solution pairs for $(b, d)$: $(0, 0)$, $(2, 9)$, $(3, 8)$, $(4, 7)$, $(5, 6)$, $(6, 5)$, $(7, 4)$, $(8, 3)$, $(9, 2)$.\n\nThus the number of required pairs of 3-digit palindromes is $4 \\times 9 = 36$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16672,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that no non-zero integers $a, b, x, y$ satisfy\n\n$$\n\\begin{cases}\nax - by = 256 \\\\\nay + bx = 1\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If we use the Diophantine sum of squares equality\n\n$$\n(ax - by)^2 + (ay + bx)^2 = (a^2 + b^2)(x^2 + y^2)\n$$\n\nthen we can see that for a system\n\n$$\n\\begin{cases}\nax - by = s \\\\\nay + bx = t\n\\end{cases}\n$$\n\nto have a solution in positive integers, the number $s^2 + t^2$ must be a composite number.\n\nThe number corresponding to the equation, $256^2 + 1^2 = 2^{16} + 1$, is a prime number. (In fact, this is a rather well-known fact, as primes of the form $2^n + 1$ relate to constructible polygons. These primes are called Fermat primes, and $2^{16} + 1$ is the largest known to date.) This shows that no solution can exist in non-zero integers, as it would give a factorisation of the prime with each factor $> 1$.\n\n*Remark:* This problem is related to factorisation over $\\mathbb{Z}[i]$, the ring of Gaussian integers. Indeed, $s + it = (ax - by) + i(ay + bx) = (a + bi)(x + yi)$, so assuming $(s, t) = 1$, a solution exists if and only if $s + it$ is composite in $\\mathbb{Z}[i]$.\n\nOther possible values of $s, t$ could work equally well. Interestingly, as a previous version of the problem mentioned, $2020^2 + 2021^2 = 8164841$ is a prime number. It would not be reasonable to ask students to verify this fact within the scope of the contest.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16673,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer, and let $G$ be a finite group of order $n$. A function $f: G \\to G$ is a *pseudoendomorphism* if $f(xyz) = f(x)f(y)f(z)$ for all $x, y, z$ in $G$.\n\n**a)** If $n$ is odd, show that every pseudoendomorphism of $G$ is an endomorphism.\n\n**b)** If $n$ is even, is every pseudoendomorphism of $G$ an endomorphism?",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $e$ denote the unit of $G$. Setting $x = y = z = e$ gives $f(e)^3 = f(e)$, so $f(e)^2 = e$. Since $n$ is odd, it follows that $f(e) = e$.\n\nIf $x$ and $y$ are members of $G$, then $f(xy) = f(xye) = f(x)f(y)f(e) = f(x)f(y)$, so $f$ is an endomorphism of $G$.\n\nb) The answer is negative. Let $a$ be an element of order $2$ in $G$, and define $f: G \\to G$ by $f(x) = a$. For any $x, y, z$ in $G$, $f(xyz) = a = a^3 = f(x)f(y)f(z)$, so $f$ is a pseudoendomorphism. However, $f$ is not an endomorphism, since $f(e) = a \\neq e$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16674,
    "subject": "Mathematics (Olympiad)",
    "question": "On the sides $AB$, $BC$, and $CA$ of triangle $ABC$, points $C_1$, $A_1$, and $B_1$ are chosen respectively, and these points are not the vertices. $\\triangle A_1B_1C_1$ is equilateral and\n\n![alt](images/ukraine_2015_Booklet_p8_data_1db0288adb.png)\n\n$\\angle BC_1A_1 = \\angle C_1B_1A$ and $\\angle BA_1C_1 = \\angle A_1B_1C$. Is $\\triangle ABC$ required to be equilateral?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** yes.\n\n**Solution.** Mark angles $\\angle BC_1A_1 = \\angle C_1B_1A = x$ and $\\angle BA_1C_1 = \\angle A_1B_1C = y$ (see figure 21). $\\angle AB_1C$ is straight, so $x + y = 120^\\circ$. Then $\\angle ABC = 60^\\circ$, hence $\\angle B_1A_1C = x$, so $\\angle ACB = 60^\\circ$. In other words, $\\triangle ABC$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16675,
    "subject": "Mathematics (Olympiad)",
    "question": "Даден е правилен шестаголник со страна 1. Во внатрешноста на шестаголникот се дадени $m$ точки така што никои три од нив не се колинеарни. Шестаголникот е разделен на триаголници, при што секоја од дадените $m$ точки и секое од темињата на шестаголникот е теме на делбен триаголник. Делбените триаголници немаат заедничка внатрешна точка. Докажи дека постои делбен триаголник чија плоштина не е поголема од $\\frac{3\\sqrt{3}}{4(m+2)}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Најпрво го определуваме вкупниот број на делбени триаголници на кои е поделен дадениот шестаголник. Нека $A$ е произволна точка од внатрешните $m$ точки. Збирот од сите агли во точката $A$ е $360^\\circ$ (збир од сите агли во $A$ на сите триаголници кои таа точка ја имаат за свое теме). Од друга страна, збирот од сите агли во теме на шестаголникот е $120^\\circ$. Бидејќи збирот на аглите во секој триаголник е $180^\\circ$, бројот на делбени триаголници е: $$\\frac{m \\cdot 360^\\circ + 6 \\cdot 120^\\circ}{180^\\circ} = 2m + 4.$$ Нека претпоставиме спротивно на тврдењето, односно дека плоштината на секој од дадените делбени триаголници е поголема од $\\frac{3\\sqrt{3}}{4(m+2)}$. Тогаш збирот на плоштините на сите делбени триаголници е поголем од $$(2m+4)\\frac{3\\sqrt{3}}{4(m+2)} = \\frac{3\\sqrt{3}}{2},$$ што не е можно бидејќи плоштината на дадениот шестаголник е $\\frac{3\\sqrt{3}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16676,
    "subject": "Mathematics (Olympiad)",
    "question": "Olesya writes down the numbers $1, 2, 3, 4, 5, 6$ at the vertices of a prism. After this, at each edge, Andriy writes down the sum of the numbers that are written at the vertices that form this edge. Can Olesya write the numbers in such a way that all of Andriy's numbers are different?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes. See the figure below.\n\n![](images/Ukrajina_2011_p11_data_e97bdf80e1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16677,
    "subject": "Mathematics (Olympiad)",
    "question": "(a) Let $n$ be a positive integer. Prove that\n\n$$\nn\\sqrt{x-n^2} \\leq \\frac{x}{2}, \\text{ for all } x \\geq n^2.\n$$\n\n(b) Determine real numbers $x, y, z$ satisfying the equation\n\n$$\n2\\sqrt{x-1} + 4\\sqrt{y-4} + 6\\sqrt{z-9} = x + y + z.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Since $x \\geq n^2$, we have\n\n$$\nn\\sqrt{x-n^2} \\leq \\frac{x}{2} \\iff 2n\\sqrt{x-n^2} \\leq x \\iff 4n^2(x-n^2) \\leq x^2 \\iff (x-2n^2)^2 \\geq 0,\n$$\n\nwhich is always true. Equality holds if and only if $x = 2n^2$.\n\nAlternatively, for every $x \\geq n^2$, it is enough to prove that\n\n$$\nn\\sqrt{x-n^2} - \\frac{x}{2} \\leq 0 \\iff 2n\\sqrt{x-n^2} - x \\leq 0 \\iff \\frac{(2n\\sqrt{x-n^2} - x)(2n\\sqrt{x-n^2} + x)}{2n\\sqrt{x-n^2} + x} \\leq 0\n$$\n\n$$\n\\iff \\frac{4n^2(x-n^2) - x^2}{2n\\sqrt{x-n^2} + x} \\leq 0 \\iff \\frac{-(x-2n^2)^2}{2n\\sqrt{x-n^2} + x} \\leq 0,\n$$\n\nwhich is valid. Equality holds for $x = 2n^2$.\n\n(b) The given equation can be written as\n\n$$\n(2\\sqrt{x-1}-x)+(4\\sqrt{y-4}-y)+(6\\sqrt{z-9}-z)=0,\n$$\n\nfor $x \\geq 1$, $y \\geq 4$, and $z \\geq 9$.\n\nUsing (a) for $n = 1, 2, 3$, we get\n\n$$\n2\\sqrt{x-1}-x \\leq 0, \\quad 4\\sqrt{y-4}-y \\leq 0, \\quad 6\\sqrt{z-9}-z \\leq 0,\n$$\n\nTherefore, the equation is only possible if\n\n$$\n2\\sqrt{x-1}-x=0, \\quad 4\\sqrt{y-4}-y=0, \\quad 6\\sqrt{z-9}-z=0 \\iff x=2, y=8, z=18.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16678,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle. Points $M$ and $N$ are the feet of the perpendiculars from $A$ to the exterior angle bisectors at $B$ and $C$ of triangle $ABC$. Let $P$ be the point of tangency of line $BC$ and the excircle $(I)$ with respect to $BC$. Let $AQ$ be the altitude from $A$ in triangle $ABC$. Prove that $MNPQ$ is an isosceles trapezoid and $P$ is the orthocenter of triangle $MNI$.\n\n![](images/Vietnam_Booklet_2013_12-07_p41_data_7d8e6ccdd1.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume $AB < AC$. Let $D$ and $E$ be the intersections of $AM$ and $AN$ with line $BC$. It is easy to see that triangle $BAD$ is isosceles with $BA = BD$, and $BM$ is the altitude of this triangle (with $M \\in AD$), so $M$ is also the midpoint of segment $AD$. Similarly, $N$ is the midpoint of segment $AE$. Hence, the line $MN$ is parallel to $BC$.\n\nWe also have $DP = EP = \\frac{AB + BC + CA}{2}$, so point $P$ is the midpoint of segment $DE$. Because $QN = \\frac{1}{2}AE = MP$, the quadrilateral $MNPQ$ is an isosceles trapezoid. Furthermore, $PM \\parallel AE$, $NI \\perp AE$, so $PM \\perp NI$; similarly, $PN \\perp MI$, so $P$ is the orthocenter of triangle $MNI$. The lemma is proved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16679,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $n \\geq 2$ such that the cells of an $n \\times n$ board can be colored with several colors so that each cell $C$ has exactly two neighboring cells with the same color as $C$. Here, neighboring cells are those sharing a common side.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is: all even positive integers.\n\nIf $n$ is even, divide the board into $2 \\times 2$ squares. Assign a different color to each such square and use it for all four cells in that square. This coloring satisfies the requirement: each cell has exactly two neighbors with the same color.\n\nTo show that $n$ must be even, consider starting from any cell $S$ and repeatedly moving to a neighboring cell with the same color that hasn't been visited yet. Since the board is finite, this process must eventually revisit a cell, say $V$. Since each cell has exactly two neighbors of the same color, $V$ must be $S$, forming a closed path $P_S$.\n\nIf two such paths share a cell, they must coincide, so the board's cells are partitioned into these paths. Color the board in a chessboard pattern: each consecutive pair in a path consists of a black and a white cell, so each path has an even number of cells. Thus, $n^2$ is a sum of even numbers, so $n$ must be even.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16680,
    "subject": "Mathematics (Olympiad)",
    "question": "The ratios $\\frac{3a5b}{36}$ and $\\frac{4c7d}{45}$ are positive integers, where $a, b, c, d$ are digits. Order all numbers of this kind by size.",
    "options": [],
    "answer": "See solution",
    "solution": "The ratios $\\frac{3a5b}{36}$ and $\\frac{4c7d}{45}$ are positive integers if $3a5b$ is divisible by 36 (so by 4 and 9) and $4c7d$ is divisible by 45 (so by 5 and 9).\n\n$3a5b$ is divisible by 36 if the last digit is 2 or 6, and because $3a5b$ is divisible by 9, the only two possible cases are:\n\n$$\n\\frac{3456}{36} = 96 \\quad \\text{and} \\quad \\frac{3852}{36} = 107\n$$\n\n$4c7d$ is divisible by 45 if the last digit is 0 or 5, and because $4c7d$ is divisible by 9, the only two possible cases are:\n\n$$\n\\frac{4275}{45} = 95 \\quad \\text{and} \\quad \\frac{4770}{45} = 106\n$$\n\nNow we obtain the desired ordering:\n\n$$\n\\frac{4275}{45} < \\frac{3456}{36} < \\frac{4770}{45} < \\frac{3852}{36}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16681,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose we assign one of the positive integers $n$ satisfying $1 \\leq n \\leq 8$ to each of the 8 vertices of a regular octagon. How many ways of assigning these numbers are there, which satisfy the following two conditions?\n\n* The assigned numbers are all distinct.\n* Any pair of numbers assigned to adjacent vertices are relatively prime.\n\nRegard two assignments of 8 numbers to be distinct if one is obtained from the other by rotation or flipping of the octagon.",
    "options": [],
    "answer": "See solution",
    "solution": "There are $576$ ways.\n\nSince pairs of even integers are not relatively prime, any pair of adjacent vertices cannot both be occupied by even numbers. Hence, even and odd numbers must be assigned to vertices alternately. So, the assignment of numbers must be carried out as in the figures below.\n\n![](images/Japan_2014_Booklet_p3_data_5ec93c8152.png)\n\nfigure 1\n\n![](images/Japan_2014_Booklet_p3_data_f86cc6b92d.png)\n\nfigure 2\n\nThere are $4! = 24$ ways of assigning even numbers $2, 4, 6, 8$ to the vertices marked even in figure 1, and the same is true for the vertices in figure 2. All these assignments are considered distinct. Therefore, there are $48$ ways of assigning the even numbers to the vertices of the octagon to satisfy the requirements of the problem.\n\nNow, after we decide on the assignments of even numbers, we have to consider the ways of assigning odd numbers to the vertices marked odd in figures 1 and 2 in such a way as to satisfy the requirements of the problem. Since $3$ and $6$ are not relatively prime, $3$ has to be assigned to a vertex not adjacent to a vertex which has $6$ as the assigned number. Therefore, there are exactly $2$ vertices to which the number $6$ can be assigned. On the other hand, all the even-odd pairs of numbers between $1$ through $8$ except the pair $(3, 6)$ are relatively prime. The remaining odd numbers $1, 5, 7$ can be assigned to the remaining vertices marked odd in any order.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16682,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute-angled triangle $ABC$, the angle bisector $AL$, the altitude $BH$, and the perpendicular bisector of line $AB$ intersect at one point. Find the angle $BAC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the angle $BAC$ be $2\\alpha$. Since $\\triangle APB$ is isosceles, $\\angle PBA = \\alpha$. In the right triangle $AHB$, $3\\alpha = 90^\\circ$, so $\\angle BAC = 2\\alpha = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16683,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A'B'C'$ be the anticomplementary triangle of $ABC$. Let $X$, $Y$, and $Z$ be the projections of $A'$, $B'$, and $C'$ onto a line $XYZ$. Let $A_0$, $B_0$, and $C_0$ be the reflections of $A'$, $B'$, and $C'$ over $BC$, $CA$, and $AB$, respectively. If we take $A'B'C'$ as our reference triangle, then $A_0B_0C_0$ is its orthic triangle, so $(ABCA_0B_0C_0)$ is the nine-point circle of $A'B'C'$.\n\nLet $A_0X'$ meet $(ABC)$ again at $P$. Prove that $P$ is the circumcenter of $X'Y'Z'$, where $X'$, $Y'$, and $Z'$ are defined analogously.\n\n![](images/TST2025Solutions_p9_data_ba84a9351d.png)\n\n*Remark.* If the condition that $X$, $Y$, and $Z$ are collinear is removed, then in general the circumcenters of $XYZ$ and $X'Y'Z'$ are isogonal conjugates with respect to $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A'B'C'$ be the anticomplementary triangle of $ABC$. Note that $X$, $Y$, and $Z$ are the projections of $A'$, $B'$, and $C'$ onto line $XYZ$. Let $A_0$, $B_0$, and $C_0$ be the reflections of $A'$, $B'$, and $C'$ over $BC$, $CA$, and $AB$, respectively. If we take $A'B'C'$ as our reference triangle, we see that $A_0B_0C_0$ is the orthic triangle so $(ABCA_0B_0C_0)$ is the nine-point circle of $A'B'C'$.\n\nLet $A_0X'$ meet $(ABC)$ again at $P$. Then\n\n$$\n\\begin{align*}\n\\angle BB_0P &= \\angle AA_0P + \\angle BCA \\\\\n&= \\angle AA_0X + \\angle BCA \\\\\n&= \\angle (A'X, BC) + \\angle BCA \\\\\n&= \\angle (B'Y, CA) \\\\\n&= \\angle BB_0Y'.\n\\end{align*}\n$$\n\nTherefore $P$ lies on $B_0Y'$, and likewise $C_0Z'$ by symmetry.\n\n![](images/TST2025Solutions_p9_data_ba84a9351d.png)\n\nLet $X_0$, $Y_0$, and $Z_0$ be the reflections of $P$ across $BC$, $CA$, and $AB$ respectively. By reflection, $X_0$ lies on $A'X$ and $PX' = X_0X$. We also know that if $H$ is the orthocenter of $ABC$, then $X_0Y_0Z_0H$ is the Steiner line of $P$. Let $D$ be the reflection of $H$ across $BC$, which is also the antipode of $A_0$ on $(ABC)$. Then $\\angle A'X_0H = \\angle A_0PD = 90^\\circ$, so $X_0Y_0Z_0H$ is perpendicular to $A'X_0X$ and parallel to $XYZ$.\n\nThis means that $XX_0 = YY_0 = ZZ_0$. Combined with $PX' = XX_0$ and analogous statements, we have $PX' = PY' = PZ'$ so $P$ is the circumcenter of $X'Y'Z'$. This solves the problem.\n\n*Remark.* If the condition that $X$, $Y$, and $Z$ are collinear is removed, then in general the circumcenters of $XYZ$ and $X'Y'Z'$ are isogonal conjugates with respect to $ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16684,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs $(p, q)$ of primes for which $p^2 + 5pq + 4q^2$ is the square of an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "The required pairs are $(5, 11)$, $(13, 3)$, and $(7, 5)$. A routine check shows that the corresponding squares are $28^2$, $20^2$, and $18^2$, respectively.\n\nLet $p^2 + 5pq + 4q^2 = n^2$, where $n$ is a non-negative integer. Alternatively, $pq = n^2 - (p+2q)^2 = (n-p-2q)(n+p+2q)$. The second factor on the right-hand side is a divisor of $pq$, greater than both $p$ and $q$. Since $p$ and $q$ are both prime, $n+p+2q = pq$, so $n-p-2q = 1$. Subtracting the two gives $pq-1 = (n+p+2q) - (n-p-2q) = 2p+4q$. Alternatively, $(p-4)(q-2) = 9$. Clearly, $q \\geq 2$, so $p \\geq 5$. The factors are then $1$ and $9$, $9$ and $1$, or $3$ and $3$, whence the required pairs.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16685,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose the function $f(x)$ satisfies, for any non-zero real number $x$,\n\n$$\nf(x) = f(1) \\cdot x + \\frac{f(2)}{x} - 1.\n$$\n\nFind the minimum value of $f(x)$ on $(0, +\\infty)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = 1$ and $x = 2$:\n\nFor $x = 1$:\n$$\nf(1) = f(1) \\cdot 1 + \\frac{f(2)}{1} - 1 = f(1) + f(2) - 1\n$$\nSo $f(2) = 1$.\n\nFor $x = 2$:\n$$\nf(2) = f(1) \\cdot 2 + \\frac{f(2)}{2} - 1\n$$\nSubstitute $f(2) = 1$:\n$$\n1 = 2f(1) + \\frac{1}{2} - 1\n$$\n$$\n1 + 1 - \\frac{1}{2} = 2f(1)\n$$\n$$\n\\frac{3}{2} = 2f(1)\n$$\n$$\nf(1) = \\frac{3}{4}\n$$\n\nThus, for $x > 0$:\n$$\nf(x) = \\frac{3}{4}x + \\frac{1}{x} - 1\n$$\n\nTo find the minimum for $x \\in (0, +\\infty)$, set the derivative to zero:\n$$\nf'(x) = \\frac{3}{4} - \\frac{1}{x^2} = 0 \\implies x^2 = \\frac{4}{3} \\implies x = \\frac{2}{\\sqrt{3}}\n$$\n\nPlug back in:\n$$\nf\\left(\\frac{2}{\\sqrt{3}}\\right) = \\frac{3}{4} \\cdot \\frac{2}{\\sqrt{3}} + \\frac{1}{\\frac{2}{\\sqrt{3}}} - 1 = \\frac{3}{2\\sqrt{3}} + \\frac{\\sqrt{3}}{2} - 1 = \\sqrt{3} - 1\n$$\n\nTherefore, the minimum of $f(x)$ on $(0, +\\infty)$ is $\\sqrt{3} - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16686,
    "subject": "Mathematics (Olympiad)",
    "question": "For which of the sets $A = \\mathbb{R}$ or $A = \\mathbb{Q}$ is there a function $f : A \\to (0, +\\infty)$ such that for all $x_1, x_2 \\in A$, $x_1 \\neq x_2$, the condition\n$$\n\\min\\{f(x_1), f(x_2)\\} \\leq |x_1 - x_2|\n$$\nis satisfied?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that the set $A$ is uncountable.\n\nDivide the half-plane $\\mathbb{R} \\times (0, +\\infty)$ into countably many rectangles:\n\n$$\nD_{n,m} = [n, n+1) \\times \\left[\\frac{1}{m+1}, \\frac{1}{m}\\right), \\quad n \\in \\mathbb{Z},\\ m \\in \\mathbb{N},\\ D_{n,0} = [n, n+1) \\times [1, +\\infty).\n$$\n\nBy construction, $\\bigcup_{n \\in \\mathbb{Z}} \\bigcup_{m \\in \\mathbb{N}} D_{n,m} = \\mathbb{R} \\times (0, +\\infty)$, and any two of these rectangles are disjoint. Consider the graph $G = \\{(x, f(x)) \\mid x \\in A\\}$ of the function $f$.\n\nSince $A$ is uncountable, so is $G$. Suppose that for all $n \\in \\mathbb{Z}, m \\in \\mathbb{N}$, the set $G \\cap D_{n,m}$ is finite. Then $G$ would be at most countable, a contradiction. Therefore, there exist $n_0 \\in \\mathbb{Z}, m_0 \\in \\mathbb{N}$ such that $G \\cap D_{n_0,m_0}$ is infinite. This means there is an infinite set $X \\subset [n_0, n_0+1) \\cap A$ such that for all $x \\in X$, $f(x) > \\frac{1}{m_0+1}$.\n\nBy the Bolzano-Weierstrass theorem, there is a sequence of distinct elements in $X$ converging to some number. Thus, for all $k \\in \\mathbb{N}$, $f(x_k) > \\frac{1}{m_0+1}$. But then $0 = \\lim_{k \\to \\infty} |x_{k+1} - x_k|$ and $\\min\\{f(x_k), f(x_{k+1})\\} \\geq \\frac{1}{m_0+1}$, which contradicts the problem's condition.\n\nTherefore, $A$ must be at most countable. Now, construct the required function. Let $A = \\{x_1, x_2, \\ldots\\}$. Define the function by induction: $f(x_1) = 1$. Suppose $f(x_1), \\ldots, f(x_n)$ are defined. Let $r = \\min_{k=1,\\ldots,n} |x_{n+1} - x_k| > 0$. Set $f(x_{n+1}) = \\frac{r}{2}$.\n\nWe verify that this function satisfies the problem's condition.\n\nExample: for $A = \\mathbb{Q}$, if $x = \\frac{p}{q}$ with $(p, q) = 1$, $p \\in \\mathbb{Z}$, $q \\in \\mathbb{N}$ ($0 = \\frac{0}{1}$), set $f\\left(\\frac{p}{q}\\right) = \\frac{1}{q^2}$. Then, if $\\frac{p_1}{q_1} \\neq \\frac{p_2}{q_2}$,\n$$\n|x_1 - x_2| = \\frac{|p_1 q_2 - q_1 p_2|}{q_1 q_2} \\geq \\frac{1}{q_1 q_2} \\geq \\frac{1}{\\max\\{q_1^2, q_2^2\\}} = \\min\\{f(x_1), f(x_2)\\}.\n$$\n\nThus, such a function exists for $A = \\mathbb{Q}$, but not for $A = \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16687,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and let $M$ be the midpoint of the side $BC$. The parallels through $M$ to $AB$ and $AC$ cross the tangent at $A$ to the circle $ABC$ at $X$ and $Y$, respectively. The circles $BMX$ and $CMY$ cross again at $S$. Prove that the circles $SXY$ and $SBC$ are tangent.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N$ and $P$ be the midpoints of the sides $AB$ and $AC$, respectively. Clearly, $(M, N, Y)$ and $(M, P, X)$ are triples of collinear points. Let $X'$ and $Y'$ be the reflections of $A$ across $X$ and $Y$, respectively. Let $YM$ and $SB$ cross at $U$ and let $XM$ and $SC$ cross at $V$.\n\n![](images/RMC_2025_p106_data_d024aeca4e.png)\n\nThe argument hinges on the four facts below:\n\n1. $X'$ lies on circle $YMC$ and $Y'$ lies on circle $XMB$.\n2. $S$, $A$, and $M$ are collinear.\n3. $U$ and $V$ both lie on circle $SXY$.\n4. $UV$ is parallel to $BC$.\n\nAssume these facts for the moment, to complete the solution as follows: By (3), the conclusion is equivalent to circles $SUV$ and $SBC$ being tangent; and by (4), these circles are similar from $S$, whence the conclusion.\n\nTo prove (1), write $\\angle Y'XM = \\angle YXM = 180^\\circ - \\angle PAX - \\angle APX = 180^\\circ - \\angle ABC - \\angle BAC = \\angle ACB$.\n\nAs $YN$ and $NM$ are midlines in triangles $ABY'$ and $ABC$, respectively, $\\angle Y'BM = \\angle Y'BA + \\angle ABC = \\angle BNM + \\angle ABC = 180^\\circ - \\angle ACB$.\n\nConsequently, $\\angle Y'XM + \\angle Y'BM = 180^\\circ$, so $Y'$ lies on circle $XMB$. Similarly, $X'$ lies on circle $YMC$. This establishes (1).\n\nTo prove (2), note that $SM$ is the radical axis of the circles $YMC$ and $XMB$, so it is sufficient to show that $A$ has equal powers with respect to these circles. By (1), the two circles are $X'MC$ and $Y'MB$, respectively, so $AX' \\cdot AY = 2 \\cdot AX \\cdot AY = AX \\cdot 2 \\cdot AY = AX \\cdot AY'$, as desired. This establishes (2).\n\nTo prove (3), note that $YU \\parallel Y'B$, as $YN$ is a midline in triangle $ABY'$, so $\\angle SUY = \\angle SBY'$. By (1), $\\angle SBY' = \\angle SXY$, so $\\angle SUY = \\angle SXY$, implying that $U$ lies on circle $SXY$. Similarly, $V$ lies on this circle. This establishes (3).\n\nFinally, to prove (4), it is sufficient to show that $SU/UB = SV/VC$ and then apply Thales. As triangles $MSU$ and $MUB$ have the same $M$-altitude and share the side $MU$,\n\n$$\n\\frac{SU}{UB} = \\frac{\\text{area } MSU}{\\text{area } MUB} = \\frac{SM}{MB} \\cdot \\frac{\\sin \\angle SMU}{\\sin \\angle UMB} = \\frac{SM}{MB} \\cdot \\frac{\\sin \\angle AMN}{\\sin \\angle NMB},\n$$\n\nas $A$ lies on $SM$ by (2).\n\nTriangles $MAN$ and $MNB$ have equal areas, as they both have the same $M$-altitude, and $N$ is the midpoint of $AB$. These triangles also share the side $MN$, so\n\n$$\nMA \\cdot MN \\cdot \\sin \\angle AMN = 2 \\cdot \\text{area } MAN = 2 \\cdot \\text{area } MNB = MN \\cdot MB \\cdot \\sin \\angle NMB.\n$$\n\nHence $\\sin \\angle AMN / \\sin \\angle NMB = MB/MA$, so, by the preceding,\n\n$$\n\\frac{SU}{UB} = \\frac{SM}{MB} \\cdot \\frac{MB}{MA} = \\frac{SM}{MA}.\n$$\n\nSimilarly, $SV/VC = SM/MA$, so $SU/UB = SV/VC$, as stated. This establishes (4) and completes the solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16688,
    "subject": "Mathematics (Olympiad)",
    "question": "The lower row of a $2 \\times 13$ rectangle is filled with 13 markers labeled $1, 2, \\ldots, 13$ in this order. An operation consists of moving a marker from its cell to an adjacent (by side) empty cell. The task is to rearrange the markers in the reverse order, in the lower row again, using the minimal number of operations.",
    "options": [],
    "answer": "See solution",
    "solution": "Marker $1$ needs at least $12$ horizontal operations to reach its final position $13$; by symmetry, the same holds for marker $13$. Similarly, markers $2$ and $12$ need at least $10$ horizontal operations each. The analogous observation about the pairs of markers $3, 11$; $4, 10$; $5, 9$; $6, 8$ implies that at least $2(12+10+8+6+4+2) = 84$ horizontal operations are needed to complete the task.\n\nAs for the vertical operations, we claim that all markers except possibly one must move up vertically at some point, and hence go down by one more vertical operation. Assume on the contrary that markers $i$ and $j$, $i < j$, never leave row $1$. Then their mutual disposition will not change: $i$ will always precede $j$ no matter the remaining operations. However, $j$ has to precede $i$ in the final position. The contradiction shows that at least $12$ markers need $2$ vertical operations each. In all, at least $84 + 2 \\cdot 12 = 108$ operations are needed to achieve the goal.\n\nLet us show that $108$ operations are enough. Carry out steps (1)–(5) in the order described below.\n\n1. For each $i = 1, 2, \\ldots, 6$, let $S_i$ be the following sequence of operations: Move marker $i$ up to the second row, then move it to the right to position $14 - i$. Carry out sequences $S_1, S_2, \\ldots, S_6$ in this order; this is clearly possible. Positions $8, 9, 10, 11, 12, 13$ in row $2$ are now occupied by markers $6, 5, 4, 3, 2, 1$. The number of operations used is $6 + (12 + 10 + 8 + 6 + 4 + 2) = 48$.\n2. Move marker $7$ up to the second row.\n3. For each $i = 8, 9, 10, 11, 12$, let $S_i$ be the following sequence of operations: move marker $i$ to position $14 - i$ in row $1$, then lift it up to row $2$. Carry out the sequences $S_8, S_9, \\ldots, S_{12}$ in this order. Positions $2, 3, 4, 5, 6, 7$ in row $2$ are now occupied by markers $12, 11, 10, 9, 8, 7$. The number of operations used is $5 + (2 + 4 + 6 + 8 + 10) = 35$.\n4. Move marker $13$ to position $1$ in row $1$ without lifting it up; $12$ operations are used.\n5. Move down all $12$ markers in row $2$; $12$ operations are used.\n\nThe problem is solved with the minimum possible number of $108$ operations.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16689,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{R} \\to \\mathbb{R}$ be a function such that for all $x, y \\in \\mathbb{R}$,\n$$\nf(f(y) x) = (1 - y) f(xy) + y^2 f(x).\n$$\nFind all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P(x, y)$ denote plugging $x$ and $y$ into the original equation.\n\n$$\n\\begin{aligned}\nP(0, 1): &\\quad f(0) = 0, \\\\\nP(1, 1): &\\quad f(f(1)) = f(1), \\\\\nP(1, f(1)): &\\quad f(f(f(1))) = (1 - f(1)) f(f(1)) + f(1)^2 f(f(1)).\n\\end{aligned}\n$$\n\nBy combining the second and third equalities, we get\n$$\nf(1) = (1 - f(1)) f(1) + f(1)^3, \\quad \\text{i.e.} \\quad f(1)^2 = f(1)^3.\n$$\nSo $f(1) = 1$ or $f(1) = 0$.\n\nIf $f(1) = 1$, then\n$$\nP(x, 1): \\quad f(x) = x^2.\n$$\nBut plugging $f(x) = x^2$ into the original equation does not work, so $f(1) = 0$.\n\nLet $c$ be a root of $f$. Then\n$$\nP(x, c): \\quad 0 = (1 - c) f(xc).\n$$\nIf $c \\ne 1$, then $f(xc) = 0$ for all $x$. If $c \\ne 0$, this implies $f(x) = 0$ for all $x$. Indeed, $f \\equiv 0$ is a solution.\n\nAssume $f$ is not identically zero. Then $0$ and $1$ are the only roots of $f$.\n\nNow,\n$$\nP(1, y): \\quad f(f(y)) = (1 - y) f(y) + y^2 f(y) = (1 - y + y^2) f(y). \\tag{1}\n$$\nFor $x \\ne 0$,\n$$\nP\\left(\\frac{1}{x}, x\\right): \\quad f\\left(\\frac{f(x)}{x}\\right) = f(x). \\tag{2}\n$$\n\nLet $y_1, y_2 \\in \\mathbb{R} \\setminus \\{0, 1\\}$ with $f(y_1) = f(y_2) \\ne 0$. From (1):\n$$\n1 - y_1 + y_1^2 = 1 - y_2 + y_2^2 \\implies (y_1 - y_2)(y_1 + y_2 - 1) = 0.\n$$\nSo\n$$\nf(y_1) = f(y_2) \\implies y_1 = y_2 \\text{ or } y_1 + y_2 = 1. \\tag{3}\n$$\n\nCombining (2) and (3), for all $x \\ne 0, 1$:\n$$\n\\frac{f(x)}{x} = x \\text{ or } \\frac{f(x)}{x} + x = 1,\n$$\ni.e.\n$$\nf(x) = x^2 \\text{ or } f(x) = x - x^2.\n$$\n\nLet $x \\in \\mathbb{R} \\setminus \\{0, 1\\}$ with $f(x) = x^2$. Plug $y = x$ into (1):\n$$\nf(x^2) = (1 - x + x^2) x^2.\n$$\nIf $f(x^2) = x^4$, then $x^4 = (1 - x + x^2) x^2$, so $x = 0$ or $x = 1$ (contradicts $x \\notin \\{0, 1\\}$).\nIf $f(x^2) = x^2 - x^4$, then $x^2 - x^4 = (1 - x + x^2) x^2$, so $x = 0$ or $x = \\frac{1}{2}$.\nFor $x = \\frac{1}{2}$, $f(\\frac{1}{2}) = (\\frac{1}{2})^2 = \\frac{1}{4} = \\frac{1}{2} - (\\frac{1}{2})^2$.\n\nThus, $f(x) = x - x^2$ for all $x \\ne 0, 1$. Since this formula also holds for $x = 0, 1$, we conclude:\n$$\nf(x) = x - x^2, \\quad \\forall x \\in \\mathbb{R}.\n$$\n\nPlugging this into the original equation confirms it is a solution. Therefore, the only solutions are:\n- $f(x) = 0$ for all $x \\in \\mathbb{R}$,\n- $f(x) = x - x^2$ for all $x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16690,
    "subject": "Mathematics (Olympiad)",
    "question": "During the school year, 22 olympiads were held. At each olympiad, the 5 best students were awarded. It is known that the prize receivers of every two olympiads had exactly 1 student in common. Show that there exists a student who received a prize at every olympiad.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider an arbitrary olympiad, say $A_1$, where the prizes went to some 5 students. Each of the remaining 21 olympiads must have someone among those 5 receiving a prize. By the pigeonhole principle, there exists a student who, in addition to $A_1$, also received a prize at at least 5 other olympiads. Let that student be $a$ and those olympiads be $A_2, \\dots, A_6$.\n\nNow, let $B$ be any olympiad different from $A_1, \\dots, A_6$. Each of the olympiads $A_1, \\dots, A_6$ has one prize-winning student in common with $B$, and exactly 5 students get prizes at $B$. Applying the pigeonhole principle again, one of those five must have received a prize at at least two of $A_1, \\dots, A_6$. Since these two have student $a$ in common, and by the initial condition that student is the only one, this means that $a$ also received a prize at olympiad $B$. Since $B$ was chosen arbitrarily, $a$ must have received a prize at every olympiad.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16691,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest three-digit integer $x$ such that $x^2 \\equiv 1 \\pmod{2024}$.",
    "options": [],
    "answer": "See solution",
    "solution": "$2024 = 8 \\cdot 11 \\cdot 23$, so $x^2 \\equiv 1 \\pmod{2024}$ iff $x \\equiv 1, 10 \\pmod{11}$ and $x$ is odd. Thus, $x \\equiv 1, 45, -1, -45 \\pmod{2 \\cdot 11 \\cdot 23}$. The smallest three-digit one is $506 - 45 = 461$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16692,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest value of $k$ with the following property: Among any $k$ consecutive positive integers, there exists a number $n$ such that the sum of all positive divisors of $n$ is even.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that $k = 3$.\n\nObviously, $k = 2$ is not enough: the numbers $1, 2$ or $8, 9$ have odd divisor sums.\n\nNow suppose that the sum $\\sigma(n)$ of all positive divisors of $n$ is odd. Consider a decomposition $n = 2^r \\cdot m$, where $r \\ge 0$ and $m$ is an odd positive integer. Then the set of odd divisors of $n$ coincides with the set of odd divisors of $m$. Hence $\\sigma(n)$ has the same parity as $\\sigma(m)$. Also, $\\sigma(m)$ has the same parity as the number of divisors of $m$, since all of them are odd. But the divisors of $m$ can be grouped into pairs $(d, m/d)$, where $d \\le \\sqrt{m}$, except for the divisor $\\sqrt{m}$ if it is an integer. It follows that the number of divisors of $m$ is odd if and only if $m$ is a perfect square. Hence $\\sigma(n)$ is odd if and only if $n$ is a perfect square or a perfect square multiplied by $2$.\n\nTherefore, if three consecutive positive integers all have odd divisor sums, then each of them is either a perfect square or a perfect square multiplied by $2$. Thus, at least two of these numbers are perfect squares or at least two are perfect squares multiplied by $2$. In both cases, we obtain two distinct positive perfect squares with difference at most $2$, which is not possible. This proves that $k = 3$ has the required property.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16693,
    "subject": "Mathematics (Olympiad)",
    "question": "設三角形 $ABC$ 的外接圓為 $\\omega$，A-旁心為 $I_A$，$I_A$ 對 $BC$ 的垂足為 $D$。令 $M$ 為線段 $I_A D$ 的中點。在圓 $\\omega$ 上，不含 $A$ 點的 $BC$ 弧內選取點 $T$ 使得 $\\angle BAT = \\angle DAC$。直線 $I_A T$ 與圓 $\\omega$ 再交於點 $S \\neq T$。設 $SM$ 交 $BC$ 於點 $X$；線段 $AD$ 的中垂線分別交 $AC$、$AB$ 於點 $Y$、$Z$。證明直線 $AX$、$BY$、$CZ$ 共點。",
    "options": [],
    "answer": "See solution",
    "solution": "解法一：令 $\\triangle ABC$ 的內心為 $I$，劣弧 $BC$ 的中點為 $N$，$ND$ 再交 $\\omega$ 於 $U$，$AI \\cap BC = L$。考慮以 $A$ 為中心、$\\sqrt{AB \\cdot AC}$ 為半徑的反演變換，並對 $AI$ 鏡射。此反演將 $D$ 送到 $T$，$L$ 送到 $N$，故\n\n$$\n\\angle ATL = \\angle AND = \\angle ANU = \\angle ATU,\n$$\n\n即 $T, L, U$ 共線。令 $BAC$ 弧中點為 $N'$，則熟知 $T, I, N'$ 共線，故有\n\n$$\n-1 = (A, L; I, I_A) \\stackrel{T}{=} (A, U; N', S) \\stackrel{N}{=} (I_A, D; \\infty_{I_A D}, NS \\cap I_A D),\n$$\n\n即 $NS$ 平分 $I_A D$，故 $N, S, M$ 共線。\n\n令 $AD$ 中垂線（即 $YZ$）交 $BC$ 於 $W$，則 $WD^2 = WA^2$，即 $W$ 在 A-旁切圓與點 $A$ 的根軸 $\\ell$ 上。若 A-旁切圓分別切 $AC$、$AB$ 於 $E$、$F$，則 $AE$ 中點和 $AF$ 中點亦在 $\\ell$ 上。若 $EF \\cap BC = P$，$AN' \\cap BC = Q$，因為 $EF \\parallel AN'$，可知 $\\ell$ 與 $EF$ 和 $AN'$ 等距，故 $W = \\ell \\cap BC$ 為 $PQ$ 中點。\n\n考慮對 $BC$ 的調和共軛，即對 $BC$ 直徑圓反演，令 $X'$ 為 $W$ 對 $BC$ 的調和共軛，$BC$ 中點為 $M_{BC}$，有\n\n$$\n-1 = (P, Q; W, \\infty_{BC}) = (D, L; X', M_{BC}) \\stackrel{N}{=} (D, I_A; NX' \\cap DI_A, \\infty_{DI_A}),\n$$\n\n即 $NX' \\cap DI_A = M$，$NX'M$ 共線，又由前面知道 $N, S, M, X$ 共線，故 $X = X'$。也就是說 $-1 = (B, C; X, W)$。因為 $W = XY \\cap BC$，故 $AX$、$BY$、$CZ$ 共點。$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16694,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_0A_1A_2$ be a triangle and let $O$ be its circumcenter. The lines $OA_k$ and $A_{k+1}A_{k+2}$ meet at $B_k$, and the tangent of the circumcircle $A_0A_1A_2$ at $A_k$ meets the line $B_{k+1}B_{k+2}$ at $C_k$, $k = 0, 1, 2$, indices being reduced modulo $3$. Show that the points $C_0, C_1, C_2$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the tangents at $A_k$ and $A_{k+1}$ meet at $A'_{k+2}$. We shall prove that the lines $A'_k B_k$ are concurrent, whence the conclusion by Desargues' theorem.\n\nWe show that the lines $A'_k B_k$ are concurrent at the centroid of the triangle $A'_0 A'_1 A'_2$. More precisely, we prove that the line $A'_k B_k$ passes through the midpoint of the segment $A'_{k+1} A'_{k+2}$.\n\n![](images/RMC2014_p66_data_5c25d77e55.png)\n\nTo this end, let the parallel through $O$ to the tangent $A'_{k+1}A'_{k+2}$ meet the tangents $A'_k A'_{k+1}$ and $A'_k A'_{k+2}$ at $X_k$ and $Y_k$, respectively, and notice that:\n\n1. The angles $OX_k B_k$ and $OA_{k+2} B_k$ are equal, since the points $A_{k+2}$, $B_k$, $O$, $X_k$ all lie on the circle on diameter $OX_k$.\n2. The angles $OA_{k+2}B_k$ and $OA_{k+1}A_{k+2}$ are equal, since $OA_{k+1} = OA_{k+2}$.\n3. The angles $OA_{k+1}A_{k+2}$ and $OY_k B_k$ are equal, since the points $A_{k+1}$, $B_k$, $O$, $Y_k$ all lie on the circle on diameter $OY_k$.\n\nConsequently, the angles $OX_k B_k$ and $OY_k B_k$ are equal. Since the lines $X_k Y_k$ and $A'_{k+1} A'_{k+2}$ are parallel, and the latter is perpendicular to the line $OA_k = OB_k$, it follows that $B_k$ is the midpoint of the segment $X_k Y_k$, whence the conclusion.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 16695,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $\\angle C = 90^\\circ$ and $AC = 1$. The median $AM$ intersects the incircle at points $P$ and $Q$ such that $AP = QM$. Find the length of $PQ$.\n\n![](images/Argentina_2011_p20_data_39cccb3071.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $P$ is between $A$ and $Q$. Let the incircle touch sides $BC$ and $CA$ at $U$ and $V$ respectively. By the power of a point, $AV^2 = AP \\cdot AQ$ and $MU^2 = MQ \\cdot MP$. Also, $AQ = MP$ as $AP = QM$, so $AV^2 = MU^2$, thus $AV = MU$.\n\nSince $CU = CV$ by equal tangents, $AC = AV + CV = MU + CU = MC$. Because $M$ is the midpoint of $BC$, it follows that $BC = 2AC = 2$. Therefore, $AB = \\sqrt{AC^2 + BC^2} = \\sqrt{5}$.\n\nTriangle $AMC$ is right and isosceles, with $\\angle AMC = \\angle MAC = 45^\\circ$.\n\nWe use $AV^2 = AP \\cdot AQ$ to compute $PQ$. First, $AV = \\frac{1}{2}(AB + AC - BC) = \\frac{\\sqrt{5} - 1}{2}$. (If the incircle touches $AB$ at $T$, then $AV = AT$, $BT = BU$, $CU = CV$ imply $AV + BU + CU = \\frac{1}{2}(AB + BC + CA)$; also, $BU + CU = BC$.)\n\n$AM$ and $PQ$ have common midpoint $N$ because $AP = QM$. If $PQ = 2x$, then $PN = NQ = x$, $AP = AN - x$, $AQ = AN + x$. Since $N$ is the midpoint of the hypotenuse $AM$ of right triangle $AMC$ with $\\angle MAC = 45^\\circ$, $AN = \\frac{AC}{\\sqrt{2}} = \\frac{1}{\\sqrt{2}}$.\n\nThus, $AV^2 = AP \\cdot AQ$ becomes:\n$$\n\\left(\\frac{\\sqrt{5} - 1}{2}\\right)^2 = \\left(\\frac{1}{\\sqrt{2}} - x\\right)\\left(\\frac{1}{\\sqrt{2}} + x\\right)\n$$\nwhich simplifies to:\n$$\n\\frac{3 - \\sqrt{5}}{2} = \\frac{1}{2} - x^2\n$$\nSo,\n$$\nx = \\sqrt{\\frac{\\sqrt{5} - 2}{2}}, \\quad PQ = 2x = \\sqrt{2\\sqrt{5} - 4}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16696,
    "subject": "Mathematics (Olympiad)",
    "question": "找出所有的整數 $n = 2k+1 > 1$，使得存在某個 $\\{0,1,\\dots,k\\}$ 的排列 $a_0, a_1, \\dots, a_k$，滿足\n\n$$\na_1^2 - a_0^2 \\equiv a_2^2 - a_1^2 \\equiv \\cdots \\equiv a_k^2 - a_{k-1}^2 \\pmod{n}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "很容易檢查 $n = 3, 5$ 確實滿足條件。\n\n以下證明 $n \\leq 5$。假設存在排列 $a_0, a_1, \\dots, a_k$ 滿足條件，且以下同餘均在模 $n$ 下成立。由於 $a_{i_0} = 0$ 對某個 $i_0$，所以 $a_i^2 \\equiv c(i - i_0)$，其中 $c$ 為某常數。特別地，$1 = 1^2 \\equiv c(i_1 - i_0)$ 對某 $i_1$，因此 $(i_1 - i_0)(a_i^2 - a_j^2) \\equiv (i_1 - i_0)(i-j)c \\equiv i-j \\neq 0$ 除非 $i-j = 0$。因此，$n$ 必須為質數，否則若 $n = ab$ 且 $a \\geq b > 1$，則 $a_i = (a+b)/2, a_j = (a-b)/2$ 對某 $i \\neq j$，有 $a_i^2 - a_j^2 \\equiv 0$。因此 $\\{a_i^2\\}$ 是所有二次剩餘。\n\n不妨設 $i_0 \\neq k$，否則反轉序列使 $i_0 = 0$。由於 $c = a_{i_0+1}^2$，$c$ 是二次剩餘，因此 $\\{a_i^2\\}, \\{c^{-1}a_i^2\\}, \\{-i_0, i_0+1, \\dots, k-i_0\\}$ 都是二次剩餘且模 $n$ 同構。若 $i_0 > 0$，則 $-1$ 是二次剩餘，必有 $k = 2i_0$，且 $i_0 < k < n - i_0$，$k$ 不能是二次剩餘，因此 2 不能是二次剩餘。若 $i_0 = 0$，則 $k$ 是二次剩餘，且 $k < 2k = n-1 = 2i_0$ 不是二次剩餘，因此 2 不能是二次剩餘。因此必有 $i_0 \\leq 1$，$n = 2k+1 \\leq 2(2i_0)+1 \\leq 5$。\n\n---\n\n*評分標準：*\n\n(A) 驗證 $n = 3, 5$ 皆合題意，得 1 分。\n\n(B1) 證明 $n$ 為質數，得 3 分。\n\n(B2) 證明 $n \\leq 5$，得 3 分。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16697,
    "subject": "Mathematics (Olympiad)",
    "question": "For which values of $x$ and $y$ does the difference of $\\frac{2x+15}{8}$ and $1\\frac{1}{3} \\cdot (y-1)$ become 3 times smaller than $2 \\cdot (5-2y)$, and $\\frac{x+5}{2}$ is $0.125$ greater than $3y$?",
    "options": [],
    "answer": "See solution",
    "solution": "From the conditions, we set up the equations:\n\n$$\n\\frac{2x+15}{8} - 1\\frac{1}{3} \\cdot (y-1) = \\frac{2(5-2y)}{3}\n$$\n\nand\n\n$$\n\\frac{x+5}{2} = 3y + 0.125\n$$\n\nSolving these, we get:\n\n$$\n6x = 3\n$$\n$$\n4x - 24y = -22\n$$\n\nThus, the solution is:\n\n$$\nx = \\frac{1}{2}, \\quad y = 1\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16698,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the least $k$ satisfying the following condition: for any polynomial $f(x)$ of degree $100$ with real coefficients, there exists a polynomial $g(x)$ of degree not greater than $k$ with real coefficients such that the graphs $y = f(x)$ and $y = g(x)$ have exactly $100$ common points.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $98$.\n\nLet $n = 100$.\n\n1. We can construct a polynomial $g$ of degree at most $n-2$ for any polynomial $f$ of degree $n$. Multiplying $f$ by a nonzero constant does not affect the condition (we can multiply $g$ by the same constant), so we may assume the leading coefficient of $f$ is $1$, i.e., $f(x) = x^n + a_1x^{n-1} + \\dots + a_n$.\n\nTake any set of $n$ distinct numbers $x_1, \\dots, x_n$ whose sum is $-a_1$, and define $h(x) = (x - x_1) \\dots (x - x_n)$, $g(x) = f(x) - h(x)$, so $h = f - g$.\n\nThe polynomials $f$ and $h$ have identical coefficients for $x^n$ and $x^{n-1}$, so the degree of $g$ does not exceed $n-2$. The $x$-coordinates of intersection points of $f$ and $g$ are exactly the roots of $h$, of which there are exactly $n$.\n\n2. We show that $k \\le n-3$ does not work. Let $f(x) = x^n$, and $g(x)$ be a polynomial of degree at most $n-3$. Suppose the graphs of $f$ and $g$ intersect at $n$ points with $x$-coordinates $x_1, \\dots, x_n$. Then the polynomial $f-g$ of degree $n$ has $n$ real roots $x_1, \\dots, x_n$.\n\nHowever, $f-g$ has zero coefficients for $x^{n-1}$ and $x^{n-2}$. By Vieta's formulas, the sum $s_1 = x_1 + \\dots + x_n$ and the sum of pairwise products $s_2 = x_1x_2 + x_1x_3 + \\dots + x_{n-1}x_n$ both equal $0$.\n\nThus $x_1^2 + \\dots + x_n^2 = s_1^2 - 2s_2 = 0$, which implies $x_1 = x_2 = \\dots = x_n = 0$, contradicting the distinctness of the $x_i$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 16699,
    "subject": "Mathematics (Olympiad)",
    "question": "In a company, no three people all know each other, but for any group of five people, there are three who have a common acquaintance (not necessarily from that group). Prove that this company can be split into two groups such that within each group, no two people know each other.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that the acquaintance graph has no odd cycles. Suppose, for contradiction, that there is a smallest odd cycle. Clearly, there are no edges connecting non-consecutive vertices of this cycle, as that would create a smaller cycle, at least one of which would be odd.\n\nLabel the people along the cycle as $1, 2, \\ldots, 2k+1$. Consider the group $1, 2, 3, k+2, k+3$ (these are all distinct). By the problem's condition, among any five people, there are three with a common acquaintance. However, there cannot be edges between any three people having a common acquaintance, as that would form a triangle, contradicting the assumption that no three people all know each other.\n\nIf $2k+1 = 5$, then among $1, 2, 3, k+2, k+3$, there would be such edges, so $k > 2$.\n\nTherefore, there must be a common acquaintance for $1, 3$, and one of $k+2$ or $k+3$; without loss of generality, suppose $1, 3, k+2$ have a common acquaintance $X$ (not on the original cycle).\n\nNow, we can construct a smaller odd cycle using $X$: $1, 2, 3, \\ldots, k+2, X, k+2, k+3, \\ldots, 2k+1, 1, X$. Both resulting cycles are smaller than the original and have different parity, leading to a contradiction.\n\nThus, the acquaintance graph is bipartite, so the company can be split into two groups of pairwise unfamiliar people.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16700,
    "subject": "Mathematics (Olympiad)",
    "question": "For each $1 \\leq i \\leq 9$ and $T \\in \\mathbb{N}$, define $d_i(T)$ to be the total number of times the digit $i$ appears when all the multiples of $2023$ between $1$ and $T$ inclusive are written out in base $10$.\n\nShow that there are infinitely many $T \\in \\mathbb{N}$ such that there are precisely two distinct values among $d_1(T)$, $d_2(T)$, ..., $d_9(T)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = 2023$. First, choose some $k$ such that $n \\mid 10^k - 1$. For instance, any multiple of $\\varphi(n)$ works since $n$ is coprime to $10$. We show that either $T = 10^k - 1$ or $T = 10^k - 2$ has the desired property, which completes the proof since $k$ can be taken arbitrarily large.\n\nFor this, it suffices to show that $\\#\\{d_i(10^k - 1) : 1 \\leq i \\leq 9\\} \\leq 2$. Indeed, if\n\n$$\n\\#\\{d_i(10^k - 1) : 1 \\leq i \\leq 9\\} = 1\n$$\n\nthen, since $10^k - 1$ (which consists of all nines) is a multiple of $n$, we have\n\n$$\nd_i(10^k - 2) = d_i(10^k - 1) \\text{ for } i \\in \\{1, \\dots, 8\\}, \\text{ and } d_9(10^k - 2) < d_9(10^k - 1).\n$$\n\nThis means that $\\#\\{d_i(10^k - 2) : 1 \\leq i \\leq 9\\} = 2$.\n\nTo prove that $\\#\\{d_i(10^k - 1)\\} \\leq 2$, we need an observation. Now let\n\n$$\n\\overline{a_{k-1}a_{k-2}\\dots a_0} \\in \\{1, \\dots, 10^k - 1\\}\n$$\n\nbe the decimal expansion of an arbitrary number, possibly with leading zeroes. Then $\\overline{a_{k-1}a_{k-2}\\dots a_0}$ is divisible by $n$ if and only if $\\overline{a_{k-2}\\dots a_0a_{k-1}}$ is divisible by $n$. Indeed, this follows from the fact that\n\n$$\n10 \\cdot \\overline{a_{k-1}a_{k-2}\\dots a_0} - \\overline{a_{k-2}\\dots a_0a_{k-1}} = (10^k - 1) \\cdot a_{k-1}\n$$\n\nis divisible by $n$. This observation shows that the set of multiples of $n$ between $1$ and $10^k - 1$ is invariant under simultaneous cyclic permutation of digits when numbers are written with leading zeroes.\n\nHence, for each $i \\in \\{1, \\dots, 9\\}$, the number $d_i(10^k - 1)$ is $k$ times larger than the number of $k$-digit numbers which start from the digit $i$ and are divisible by $n$. Since the latter number is either $\\left\\lfloor \\frac{10^{k-1}}{n} \\right\\rfloor$ or $1 + \\left\\lfloor \\frac{10^{k-1}}{n} \\right\\rfloor$, we conclude that $\\#\\{d_i(10^k - 1)\\} \\leq 2$. This finishes the solution. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16701,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n$ 與 $k$ 為滿足 $n > k \\ge 1$ 的正整數。有 $2n+1$ 位學生站成一圈。對於每位學生，他的 $2k$ 位鄰居，指的是他左手邊離他最近的 $k$ 位同學與右手邊離他最近的 $k$ 位同學。\n\n已知學生中恰有 $n+1$ 位女生。證明：存在一個女生，她的 $2k$ 位鄰居中有至少 $k$ 位女生。",
    "options": [],
    "answer": "See solution",
    "solution": "**解**\n\n令\n$$\na_m = \\begin{cases} 1, & \\text{第 } m \\text{ 位學生是女生} \\\\ 0, & \\text{否則} \\end{cases}\n$$\n\n考慮 $b_i = a_i + a_{i-k-1} - 1 \\in \\{-1, 0, 1\\}$。易知對於所有 $m$，\n\n$$\nb_{m+1} + \\cdots + b_{m+2n+1} = 2(a_1 + \\cdots + a_{2n+1}) - (2n+1) = 2(n+1) - (2n+1) = 1.\n$$\n\n這表示存在 $i_0$ 使得 $b_{i_0} = 1$。若我們可以找到 $i$ 滿足\n\n$$\nb_i = 1 \\quad \\text{且} \\quad b_{i+1} + b_{i+2} + \\cdots + b_{i+k} \\ge 0,\n$$\n\n則 $a_i = 1$ 且\n\n$$\n(a_{i-k} + \\cdots + a_{i-1}) + (a_{i+1} + \\cdots + a_{i+k}) \\ge k,\n$$\n\n從而原題證畢。\n\n用歸謬法證明滿足上述條件的 $i$ 存在。假設對所有 $b_i = 1$ 的 $i$，$b_{i+1} + \\cdots + b_{i+k}$ 都是負的。任取其中一個 $i_0$，並對所有 $j$，定義 $i_j$ 為滿足 $i_j > i_{j-1} + k$ 且 $b_{i_j} = 1$ 的最小整數。任取 $\\{i_0, i_1, \\cdots, i_{2n+1}\\}$ 中同餘 $2n+1$ 的兩者，假設為 $i_0$ 與 $i_T$。\n\n對所有 $0 \\le j \\le T-1$，由 $i_j$ 的定義，$b_{i_j} + k + 1$ 到 $b_{i_{j+1}} - 1$ 必然都 $\\le 0$。由歸納假設，這意味著\n\n$$\nS_j := b_{i_j} + \\cdots + b_{i_{j+1}-1} \\le b_{i_j} + \\cdots + b_{i_j+k} < 1.\n$$\n\n從而 $S_j \\le 0$。另一方面，由於 $(2n+1) \\mid (i_T - i_0)$，上述總和為\n\n$$\nS_0 + \\cdots + S_{T-1} = \\sum_{i=i_0}^{i_T-1} b_i = 1 \\cdot \\frac{i_T - i_0}{2n+1} > 0,\n$$\n\n矛盾！故存在 $i$ 滿足條件。證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16702,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_0 = 1$, $x_1 = 3$, and for all $n \\ge 1$, let $x_{n+1} \\ge 3x_n - (x_{n-1} + \\cdots + x_0)$. What is the largest possible value of $M$ such that for all such sequences $x$, the inequality $\\frac{x_{n+1}}{x_n} > M$ holds for all $n \\ge 0$?",
    "options": [],
    "answer": "See solution",
    "solution": "The largest possible $M$ for which the given property holds is $M = 2$.\n\nWe first show that the property holds for $M = 2$ by induction on $n$ with the stronger statement $x_{n+1} > 2x_n > x_n + x_{n-1} + \\dots + x_0$ for all $n \\ge 0$.\n\nFor $n = 0$, $x_1 > 2x_0 > x_0$ translates to $3 > 2 > 1$.\n\nAssume $x_{n+1} > 2x_n > x_n + x_{n-1} + \\dots + x_0$. Then:\n\n$$\n\\begin{aligned}\nx_{n+2} &\\ge 3x_{n+1} - (x_n + \\dots + x_0) \\\\\n&> 2x_{n+1} \\\\\n&> x_{n+1} + x_n + \\dots + x_0.\n\\end{aligned}\n$$\n\nThis completes the induction. Thus, $\\frac{x_{n+1}}{x_n} > 2$ for all $n \\ge 0$.\n\nTo show that no larger $M$ works, consider the sequence with $x_0 = 1$, $x_1 = 3$, and equality in the recurrence: $x_{n+1} = 3x_n - (x_{n-1} + \\cdots + x_0)$. This leads to the recurrence $x_{n+1} = 4x_n - 4x_{n-1}$, whose general solution is $x_n = B2^n + Cn2^n$. Using $x_0 = 1$, $x_1 = 3$, we find $B = 1$, $C = \\frac{1}{2}$, so $x_n = (n+2)2^{n-1}$.\n\nThen:\n\n$$\n\\frac{x_{n+1}}{x_n} = \\frac{(n+3)2^n}{(n+2)2^{n-1}} = 2\\frac{n+3}{n+2} = 2\\left(1 + \\frac{1}{n+2}\\right).\n$$\n\nAs $n$ increases, this approaches $2$ from above. For any $M > 2$, there exists $n$ such that $\\frac{x_{n+1}}{x_n} < M$. Thus, the largest possible $M$ is $2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16703,
    "subject": "Mathematics (Olympiad)",
    "question": "On each side of a square, consider 3 pairwise distinct points.\n\n(a) Determine the number of segments with endpoints at two of these points that do not lie on a side of the square.\n\n(b) If no three of the above segments pass through the same point, how many points of intersection of these segments exist in the interior of the square?",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Each of the three points on one side of the square can be joined to the 9 points on the other three sides, giving $3 \\times 9 = 27$ segments per side. For all four sides, this is $4 \\times 27 = 108$ segments. However, each segment is counted twice (once from each endpoint), so the total number is $108 \\div 2 = 54$.\n\nAlternatively, from the 12 points, there are $\\binom{12}{2} = 66$ possible segments. Of these, $4 \\times \\binom{3}{2} = 12$ lie on the sides of the square. Thus, the number of desired segments is $66 - 12 = 54$.\n\n(b) To define an interior intersection point, we need a convex quadrilateral with vertices on the sides of the square (at most two on each side). The number of quadruples from 12 points is $\\binom{12}{4} = 495$. However, quadruples with three points on one side do not form convex quadrilaterals. For each side, there are 9 such quadruples, so $4 \\times 9 = 36$ in total. Therefore, the number of intersection points in the interior is:\n\n$$\n\\binom{12}{4} - 4 \\times 9 = 495 - 36 = 459.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16704,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $m \\neq n$ with $1 < m < n$. Mark the cells of a table with digits 1, 2, 3, 4, and 5 as shown in the figure below. Fill the cells consecutively with the numbers $1, 2, 3, \\ldots, mn$, starting with cells marked by 1, then 2, and so on. All numbers written in cells marked by 2 or 4 are called *good*. How many good numbers are there?\n\n![](images/broshura_07_english_p23_data_038295aa9f.png)\n\nLet $m = n$ and $3 < m$. As above, fill the table. The good numbers are again those in cells marked by 2 or 4. How many good numbers are there?\n\n![](images/broshura_07_english_p23_data_7436e7b927.png)\n\nIf $m = n = 3$, the table below shows that the good numbers are 5, 6, 7, and 8. How many good numbers are there?\n\n<table><tr><td>1</td><td>7</td><td>9</td></tr><tr><td>2</td><td>6</td><td>4</td></tr><tr><td>5</td><td>3</td><td>8</td></tr></table>",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\min \\left\\{ m, \\frac{n+1}{2} \\right\\} + \\min \\left\\{ n, \\frac{m+1}{2} \\right\\}\n$$\n\nFor $m = n$ and $3 < m$:\n$$\nn + 1\n$$\n\nFor $m = n = 3$:\n$$\n4 = 2 \\cdot 2 = 2 \\min \\left\\{ 3, \\frac{3+1}{2} \\right\\}\n$$\n\n**Answer:**\n$$\n\\min \\left\\{ n, \\frac{n+1}{2} \\right\\} + \\min \\left\\{ n, \\frac{n+1}{2} \\right\\}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16705,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$ with $\\angle ACB > 90^\\circ$ and $\\angle CBA > 45^\\circ$. Points $P$ and $T$ lie on sides $AC$ and $AB$, respectively, such that $PT = BC$ and $PT \\perp BC$. Points $P_1$ and $T_1$ lie on sides $AC$ and $AB$ such that $AP = CP_1$ and $AT = BT_1$. Show that $\\angle CBA - \\angle P_1T_1A = 45^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S$ be such that $\\triangle ABC$ and $\\triangle BCS$ are in different half-planes with respect to $BC$, and $\\triangle BCS$ is isosceles and has right angle $CBS$. Then segments $BS$ and $PT$ are equal and parallel ($TP = BC = BS$, $PS \\parallel AB$), hence $TPSB$ is a parallelogram (see figure above). Thus, $PS \\parallel AB$, and also $PS = TB$, hence $AT_1 = PS$. Thus $AT_1SP$ is a parallelogram. Thus, $AP \\parallel ST_1$ and $AP = ST_1$. Therefore, $P_1C \\parallel T_1S$ and $P_1C = T_1S$ is a parallelogram. Hence $P_1T_1 \\parallel CS$ and\n\n$$\n\\angle P_1T_1A = 180^\\circ - \\angle P_1T_1S - \\angle ST_1B = \\angle CST_1 - \\angle ST_1B.\n$$\n\nFinally,\n\n$$\n\\begin{aligned}\n\\angle CBA - \\angle P_1T_1A &= \\angle CBA - \\angle CST_1 + \\angle ST_1B \\\\\n&= 90^\\circ - \\angle T_1SB - \\angle CST_1 \\\\\n&= 90^\\circ - \\angle CSB = 45^\\circ.\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16706,
    "subject": "Mathematics (Olympiad)",
    "question": "A special number is a positive integer $n$ for which there exist positive integers $a, b, c,$ and $d$ such that\n\n$$\nn = \\frac{a^3 + 2b^3}{c^3 + 2d^3}\n$$\n\nProve that:\n\n(a) There are infinitely many special numbers.\n\n(b) $2014$ is not a special number.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Every perfect cube $k^3$ of a positive integer is special because we can write\n\n$$\nk^3 = k^3 \\cdot \\frac{a^3 + 2b^3}{a^3 + 2b^3} = \\frac{(ka)^3 + 2(kb)^3}{a^3 + 2b^3},\n$$\n\nfor some positive integers $a, b$.\n\n(b) Observe that $2014 = 2 \\cdot 19 \\cdot 53$. If $2014$ is special, then there exist positive integers $x, y, u, v$ such that\n\n$$\nx^3 + 2y^3 = 2014(u^3 + 2v^3)\n$$\n\nAssume $x^3 + 2y^3$ is minimal with this property. If $19$ divides $x^3 + 2y^3$, then it must divide both $x$ and $y$. Suppose $19$ does not divide $x$; then it does not divide $y$ either. The relation $x^3 \\equiv -2y^3 \\pmod{19}$ implies $(x^3)^{6} \\equiv (-2y^3)^{6} \\pmod{19}$, i.e., $x^{18} \\equiv 2^6 y^{18} \\pmod{19}$. By Fermat's Little Theorem, $x^{18} \\equiv y^{18} \\equiv 1 \\pmod{19}$, so $1 \\equiv 2^6 \\pmod{19}$, i.e., $2^6 = 64$, and $64 \\equiv 7 \\pmod{19}$, which is not $1$. Thus, $19$ must divide both $x$ and $y$.\n\nLet $x = 19x_1$, $y = 19y_1$ for some positive integers $x_1, y_1$. Substituting into the equation gives\n\n$$\n19^3 x_1^3 + 2 \\cdot 19^3 y_1^3 = 2014(u^3 + 2v^3)\n$$\n\nor\n\n$$\n19^3(x_1^3 + 2y_1^3) = 2014(u^3 + 2v^3)\n$$\n\nSo $19$ divides $u^3 + 2v^3$, and similarly, $u = 19u_1$, $v = 19v_1$. Substituting again,\n\n$$\nx_1^3 + 2y_1^3 = 2014(u_1^3 + 2v_1^3)\n$$\n\nBut $x_1^3 + 2y_1^3 < x^3 + 2y^3$, contradicting the minimality assumption. Therefore, $2014$ is not a special number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16707,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ such that for all integers $x, y$, $$f(x + f(y + 1)) + f(xy) = f(x + 1)(f(y) + 1).$$",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $f$ is not constant. \n\nLet $P(x, y)$ denote the assertion: $$f(x + f(y + 1)) + f(xy) = f(x + 1)(f(y) + 1).$$\n\n**Step 1:** Try $f(1) = 1$.\n\nFrom $P(x - 1, 0)$: $$f(x + f(1) - 1) = f(x)(f(0) + 1) - f(0).$$\nIf $f(1) = 1$, this gives $f(x) = f(x)(f(0) + 1) - f(0)$, so $(f(x) - 1)f(0) = 0$. Thus $f(0) = 0$.\n\nFrom $P(0, y)$: $$f(f(y + 1)) = f(1)(f(y) + 1) - f(0) = f(y) + 1.$$ \nSo $f(f(y + 1)) = f(y) + 1$.\n\nBy induction, $f(x) = x$ for all $x \\le 0$. Also, $P(x, -1)$ gives $f(x) + f(-x) = 0$, so $f(x) = x$ for all $x \\in \\mathbb{Z}$.\n\n**Step 2:** Try $f(1) \\neq 1$.\n\nLet $c = f(1) - 1 \\neq 0$, $h = f(0) + 1 \\neq 0$.\nFrom $P(x - 1, 0)$: $$f(x + c) = h f(x) - h + 1.$$ \n\nConsider $P(x, 1)$ and $P(x + c, 1)$, and use the above to relate them. Subtracting, we get $2(1 - h) = (1 - h)(f(1) + 1)$, so either $h = 1$ or $f(1) + 1 = 2$. The latter is impossible, so $h = 1$, i.e., $f(0) = 0$.\n\nNow $f(x + c) = f(x)$, so $f$ is periodic. Let $M = \\max f$, $-m = \\min f$, with $M, m \\ge 0$ since $f(0) = 0$.\n\nChoose $x, y$ so $f(x + 1) = f(y) = M$. Then $M(M + 1) \\le 2M$, so $M = 0$ or $M = 1$.\n\nSimilarly, for $f(x + 1) = f(y) = -m$, $m \\le 2$. So $f(x) \\in \\{-2, -1, 0, 1\\}$.\n\nCheck possible $f(1)$:\n- $f(1) = -2$: Contradicts $f(f(2)) = 2$.\n- $f(1) = 0$: $f$ is constant, contradiction.\n- $f(1) = 1$: Already excluded.\n- $f(1) = -1$: $f$ is 2-periodic, but $P(1, 1)$ gives $-2 = 0$, contradiction.\n\n**Conclusion:** The only solutions are $f(x) \\equiv 0$, $f(x) \\equiv 1$, and $f(x) = x$ for all $x \\in \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16708,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the remainder when\n$$\n\\binom{\\binom{3}{2}}{2} + \\binom{\\binom{4}{2}}{2} + \\dots + \\binom{\\binom{40}{2}}{2}\n$$\nis divided by 1000.",
    "options": [],
    "answer": "See solution",
    "solution": "Because\n$$\n\\binom{k}{2} - 1 = \\frac{k(k-1)-2}{2} = \\frac{(k-2)(k+1)}{2},\n$$\nit follows that\n$$\n\\binom{k}{2} = \\frac{\\binom{k}{2} \\cdot (\\binom{k}{2} - 1)}{2} = \\frac{(k+1)k(k-1)(k-2)}{8} = 3\\binom{k+1}{4}.\n$$\nBy the Hockey-stick Identity,\n$$\n\\sum_{k=3}^{n} 3 \\binom{k+1}{4} = 3 \\binom{n+2}{5}.\n$$\nThen the given sum equals\n$$\n\\begin{aligned}\n3 \\binom{42}{5} &= \\frac{3 \\cdot 42 \\cdot 41 \\cdot 40 \\cdot 39 \\cdot 38}{120} \\\\\n&= (42 \\cdot 38) \\cdot (41 \\cdot 39) \\\\\n&\\equiv 596 \\cdot 599 \\equiv (-404)(-401) \\equiv 4 \\pmod{1000}.\n\\end{aligned}\n$$\nThe requested remainder is $4$.\n\nAlternatively,\n\nNote that $\\binom{k}{2}$ is the number of ways of choosing two distinct sets of two integers from $1$ to $k$. The two sets chosen can have either $0$ or $1$ element in common. Choose two sets with no elements in common by choosing $4$ elements and then separating those $4$ elements into two sets of $2$, and this can be done in $3\\binom{k}{4}$ ways. Choose two sets with one element in common by choosing the common element and then choosing two elements to pair with it, and this can be done in $\\binom{k-1}{2}$ ways. This shows that\n$$\n\\binom{k}{2} = 3\\binom{k}{4} + \\binom{k-1}{2} = 3\\binom{k}{4} + \\binom{k}{3} = 3\\binom{k+1}{4}.\n$$\nThen the solution continues as above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16709,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle and let $H$ be its orthocenter. Suppose that a circle passing through the points $B$, $C$ and the circle with diameter $AH$ intersect at two distinct points $X$ and $Y$. Let $D$ be the foot of the perpendicular from $A$ to $BC$, and $K$ be the foot of the perpendicular from $D$ to $XY$. Prove that $$\\angle BKD = \\angle CKD.$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $AB = AC$, the centers of the circle through $B$, $C$ and the circle with diameter $AH$ both lie on the perpendicular bisector of $BC$. Therefore, the line $XY$ and the perpendicular bisector of $BC$ intersect perpendicularly, and their intersection is $K$. Furthermore, $D$ is the midpoint of $BC$, so clearly $$\\angle BKD = \\angle CKD.$$ \n\nNow, consider $AB \\ne AC$ (assume $AB > AC$). Let $E$ be the intersection of $BH$ and $AC$, and $F$ the intersection of $CH$ and $AB$. Since $\\angle AEH = \\angle AFH = 90^\\circ$, $E$ and $F$ lie on the circle with diameter $AH$ (call this $\\Gamma_1$). The points $X$, $Y$ also lie on $\\Gamma_1$. Also, $B$, $C$, $X$, $Y$ lie on another circle ($\\Gamma_2$), and $B$, $C$, $E$, $F$ lie on a third circle ($\\Gamma_3$).\n\n**Lemma.** The lines $XY$, $BC$, and $EF$ are concurrent.\n\n**Proof.** Let $O$ be the intersection of $BC$ and $EF$. Let $Y'$ and $Y''$ be the intersections of $OX$ with $\\Gamma_1$ and $\\Gamma_2$, respectively. By the power of a point theorem:\n\n$$\nOX \\cdot OY' = OE \\cdot OF = OC \\cdot OB = OX \\cdot OY''.\n$$\n\nThus, $OY' = OY''$, so $Y' = Y'' = Y$. Therefore, $O$ lies on $XY$, proving the lemma.\n\nBy Ceva's Theorem:\n$$\n\\frac{AF}{FB} \\cdot \\frac{BD}{DC} \\cdot \\frac{CE}{EA} = 1,\n$$\n\nand by Menelaus' Theorem:\n$$\n\\frac{AF}{FB} \\cdot \\frac{BO}{OC} \\cdot \\frac{CE}{EA} = 1.\n$$\n\nFrom these, $\\frac{BD}{DC} = \\frac{BO}{OC}$. Let $C'$ be the point on $DO$ such that $\\angle BKD = \\angle C'KD$. The line $KD$ bisects $\\angle BKC'$, and since $\\angle DKO = 90^\\circ$, $KO$ bisects $\\angle EKO$. Thus,\n$$\n\\frac{BD}{DC'} = \\frac{KB}{KC'}, \\quad \\frac{BO}{OC'} = \\frac{KB}{KC'}\n$$\n\nTherefore, $\\frac{BD}{DC'} = \\frac{BO}{OC'}$. Combining with $\\frac{BD}{DC} = \\frac{BO}{OC}$, we get $\\frac{OC}{DC} = \\frac{OC'}{DC'}$. Since $C$ and $C'$ both lie on $DO$, we conclude $C' = C$, so $\\angle BKD = \\angle CKD$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16710,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n > 1$ be an integer. There are $n$ boxes in a row, and $n + 1$ identical stones. A distribution is a way to distribute the stones over the boxes, with every stone in exactly one box. Two distributions are *a stone's throw away* from each other if one can be obtained from the other by moving exactly one stone from one box to another. The *cosiness* of a distribution $a$ is defined as the number of distributions that are a stone's throw away from $a$. Determine the average cosiness of all possible distributions.",
    "options": [],
    "answer": "See solution",
    "solution": "Let two distributions be neighbours if they are a stone's throw away from each other.\n\nLet $N_k$ be the number of distributions with exactly $k$ empty boxes, where $0 \\leq k \\leq n-1$ (since not all boxes can be empty). There are $\\binom{n}{k}$ ways to choose the empty boxes, and $\\binom{n}{k+1}$ ways to distribute $n+1$ stones among the remaining $n-k$ boxes so that each has at least one stone. Thus, $N_k = \\binom{n}{k} \\cdot \\binom{n}{k+1}$.\n\nFor each such distribution, we can move a stone from any of the $n-k$ non-empty boxes to any of the other $n-1$ boxes, giving $(n-k)(n-1)$ neighbouring distributions. So, the cosiness of each of these $N_k$ distributions is $(n-k)(n-1)$.\n\nFor $k' = n-1-k$, the number of distributions with $k'$ empty boxes is $N_{k'} = \\binom{n}{k'} \\cdot \\binom{n}{k'+1} = \\binom{n}{k+1} \\cdot \\binom{n}{k} = N_k$. The cosiness for these is $(n-k')(n-1) = (k+1)(n-1)$. Thus, the average cosiness for these two sets is $$\\frac{(n-k)+(k+1)}{2}(n-1) = \\frac{n+1}{2}(n-1).$$\n\nSince this is constant in $k$, it is also the overall average.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16711,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots$ be a permutation of all positive integers. Prove that there exist infinitely many positive integers $i$ such that $\\gcd(a_i, a_{i+1}) \\le \\frac{3}{4}i$.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove this by contradiction. Suppose the statement is false. Then there exists $i_0$ such that $\\gcd(a_i, a_{i+1}) > \\frac{3}{4}i$ for all $i \\ge i_0$.\n\nTake $M > i_0$. For $i \\ge 4M$, we have $\\gcd(a_i, a_{i+1}) > \\frac{3}{4}i \\ge 3M$. Thus, $a_i \\ge \\gcd(a_i, a_{i+1}) > 3M$, so $\\{1, 2, \\dots, 3M\\} \\subseteq \\{a_1, a_2, \\dots, a_{4M-1}\\}$.\n\nTherefore,\n$$\n\\left| \\{1, 2, \\dots, 3M\\} \\cap \\{a_{2M}, a_{2M+1}, \\dots, a_{4M-1}\\} \\right| \\ge 3M - (2M - 1) = M + 1.\n$$\n\nBy the pigeonhole principle, there exists $2M \\le j_0 < 4M$ such that $a_{j_0}, a_{j_0+1} \\le 3M$. Thus,\n$$\n\\gcd(a_{j_0}, a_{j_0+1}) \\le \\frac{1}{2} \\max\\{a_{j_0}, a_{j_0+1}\\} \\le \\frac{3M}{2} = \\frac{3}{4} \\cdot 2M \\le \\frac{3}{4} j_0,\n$$\nwhich is a contradiction. $\\Box$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16712,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the last digit of $n^5$ for any integer $n$.\n\nAdditionally, prove that for all integers $n$, the expression $n^5 - n$ is divisible by 30.\n\n![]( <table><thead><tr><th>b</th><th>b<sup>5</sup></th><th>b</th><th>b<sup>5</sup></th></tr></thead><tbody><tr><td>0</td><td>0</td><td>5</td><td>3125</td></tr><tr><td>1</td><td>1</td><td>6</td><td>7776</td></tr><tr><td>2</td><td>32</td><td>7</td><td>16807</td></tr><tr><td>3</td><td>243</td><td>8</td><td>32768</td></tr><tr><td>4</td><td>1024</td><td>9</td><td>59049</td></tr></tbody></table> )",
    "options": [],
    "answer": "See solution",
    "solution": "The last digit of $n^5$ is the same as the last digit of $n$.\n\nTo prove $n^5 - n$ is divisible by 30:\n\n$$\n\\begin{aligned}\nn^5 - n &= n(n^4 - 1) \\\\\n&= n(n^2 - 1)(n^2 + 1) \\\\\n&= n(n - 1)(n + 1)(n^2 + 1).\n\\end{aligned}\n$$\n\nAmong $n-1$, $n$, $n+1$, one is divisible by 3, so $n^5-n$ is divisible by 3.\n\nFrom above, the units digit of $n^5$ and $n$ are the same, so $n^5-n$ ends in 0 and is divisible by 10.\n\nSince 3 and 10 are coprime, $n^5-n$ is divisible by $30$.\n\nAlternatively, for divisibility by 5: if $n$ is not a multiple of 5, then $n^2 + 1$ is divisible by 5 for $n = 5k \\pm 2$. Thus, $n^5-n$ is divisible by 5. Since $2 \\times 3 \\times 5 = 30$, $n^5-n$ is divisible by 30 for all integers $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16713,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all triples $ (a, b, c) $ of positive integers which satisfy\n$$\n2^a + 3^b + 1 = 6^c.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $ (a, b, c) $ is a triple of positive integers satisfying $2^a + 3^b + 1 = 6^c$.\n\nFirst, note that $2^a = 6^c - (3^b + 1) \\equiv 2 \\pmod{3}$, so $a$ must be odd.\n\n**Case 1:** $a = 1$\n\nThen $2 + 3^b + 1 = 6^c$, so $3^b + 3 = 6^c$. The left side is $3(3^{b-1} + 1)$, divisible by $3$ but not by $3^2$. The right side is divisible by $3^c$, so $c = 1$. Then $3^b + 3 = 6$, so $3^b = 3$, so $b = 1$. Thus, $(a, b, c) = (1, 1, 1)$.\n\n**Case 2:** $a \\geq 3$ and $a$ odd\n\nConsider both sides modulo $8$:\n- $2^a$ for odd $a \\geq 3$ is divisible by $8$.\n- $3^b$ is $3$ if $b$ odd, $1$ if $b$ even.\n- $2^a + 3^b + 1 \\equiv 3^b + 1 \\pmod{8}$.\n\n$6^c$ modulo $8$ cycles: $6, 4, 0, 0, \\ldots$ for $c = 1, 2, \\geq 3$.\n\nMatching remainders:\n- If $b$ odd: $3 + 1 = 4$; $6^c \\equiv 4$ when $c = 2$.\n- If $b$ even: $1 + 1 = 2$; $6^c \\equiv 2$ never occurs.\n\nSo $b$ must be odd and $c = 2$.\n\nNow $2^a + 3^b + 1 = 36$.\nTry $a = 3$: $2^3 = 8$, $3^b$ odd, $8 + 3^b + 1 = 36 \\implies 3^b = 27 \\implies b = 3$.\nTry $a = 5$: $2^5 = 32$, $32 + 3^b + 1 = 36 \\implies 3^b = 3 \\implies b = 1$.\n\nThus, the solutions are $(a, b, c) = (1, 1, 1), (3, 3, 2), (5, 1, 2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16714,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ and $C$ be two points on a circle $C$ such that $AC$ is not a diameter, and let $P$ be a point of the line segment $AC$, other than its midpoint. Circles $c_1$ and $c_2$ are interiorly tangent to the circle $C$ at $A$ and $C$, respectively. They both pass through $P$ and intersect again at $Q$. The line $PQ$ intersects circle $C$ at $B$ and $D$. Circle $c_1$ intersects line segments $AB$ and $AD$ at $K$ and $N$, respectively, while circle $c_2$ intersects line segments $CB$ and $CD$ at $L$ and $M$, respectively. Prove that:\n\na) The quadrilateral $KLMN$ is an isosceles trapezoid.\n\nb) $Q$ is the midpoint of the line segment $BD$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Point $B$ lies on the radical axis, $BD$, of circles $c_1$ and $c_2$, therefore $BK \\cdot BA = BL \\cdot BC$, which indicates that the quadrilateral $AKLC$ is cyclic. So is $ACMN$. It follows that $\\angle LKB = \\angle ACB$. The angle between the tangent at $A$ to the circle $c_1$ (which is also tangent to $C$) with line $AB$ subtends arcs $AK$ of circle $c_1$ and $AB$ of circle $C$, and therefore $\\angle ANK = \\angle ADB$. We obtain that $NK \\parallel BD$ and, similarly, $ML \\parallel BD$. (This fact also follows from the homotheties that transform circles $c_1$, and $c_2$, respectively, into $C$.) Finally, $\\angle NKL = 180^\\circ - \\angle AKN - \\angle LKB = 180^\\circ - \\angle ABD - \\angle LKB = 180^\\circ - \\angle ACD - \\angle ACB = \\angle BAD$. Similarly, $\\angle MNK = \\angle BAD$, which leads to the conclusion. Alternatively, one could have noticed that line segments $ML$, $PQ$ and $NK$ share the same perpendicular bisector.\n\nb) The radical axes of circles $c_1$, $c_2$, $C$ (one for each pair of circles), i.e., the tangent line to $C$ at $A$, the tangent line to $C$ at $C$ and the line $BD$ are not all parallel, therefore they are concurrent in the radical center. Diagonal $BD$ is then a symmedian of triangle $ABC$, which means that quadrilateral $ABCD$ is harmonic. (One can also use this fact to give a different proof to a).)\n\nAs $NPQK$ is an isosceles trapezoid, it follows that $\\angle NAP = \\angle QAK$, which means that rays $AP$ and $AQ$ are isogonal with respect to angle $\\angle DAB$. But $ABCD$ being harmonic, $AP$ is a symmedian of triangle $DAB$, therefore $AQ$, which is its isogonal, is the median.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16715,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation for arbitrary distinct real numbers $a$, $b$, and $c$:\n\n$$\nx^3 a - x a^3 + a^3 b - a b^3 + b^3 x - b x^3 = (x - a)(x - b)(x - c)(a - b).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $x = a$, $x = b$.\n\n**Solution:** Obviously, the equation can be solved by trivial transformations and reduction to a quadratic one. We suggest a different approach. Denote the left-hand and right-hand sides by $f(x)$ and $g(x)$, respectively. It's easy to verify that $f(a) = g(a) = 0$ and $f(b) = g(b) = 0$. Thus, the distinct numbers $a$ and $b$ are roots of this equation. By comparing the coefficients at $x^3$ for both sides, we conclude that our equation is quadratic. Hence, $x = a$ and $x = b$ are its only roots.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16716,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the maximal number of consecutive positive integers such that each of these integers has a common divisor with $2024$ greater than $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "The maximal number is $5$.\n\nWe observe that $2024 = 2^3 \\cdot 11 \\cdot 23$. An integer has a common divisor greater than $1$ with $2024$ if and only if it is divisible by $2$, $11$, or $23$.\n\nLet $N$ be the desired maximal number. Only each $11$th integer is divisible by $11$. That means that if $z$ is divisible by $11$, then $z+1, z+2, \\dots, z+10$ are not divisible by $11$. Analogously, $23$ divides only every $23$rd integer. Six consecutive integers contain exactly three odd numbers. At most one of them is divisible by $11$ and at most one of them is divisible by $23$. This shows that $N \\leq 5$.\n\nNow, we try to find five consecutive integers $n, n+1, n+2, n+3, n+4$ that have a common divisor greater than $1$ with $2024$.\n\n$$\n\\begin{array}{c|l}\n n & \\text{even} \\\\\n n + 1 & \\text{divisible by } 11 \\\\\n n + 2 & \\text{even} \\\\\n n + 3 & \\text{divisible by } 23 \\\\\n n + 4 & \\text{even}\n\\end{array}\n$$\n\nThat means that we want $n + 1 = 11k$ and $n + 3 = 23l$ with $k$ and $l$ odd. If we subtract the first equation from the second, we get\n\n$$\n\\begin{aligned}\n2 &= 23l - 11k \\\\\n  &= l + 11(2l - k)\n\\end{aligned}\n$$\n\nWe obtain $l \\equiv 2 \\pmod{11}$. We see that $l = 13$ works, since we get $n + 3 = 23l = 299$ and therefore $n = 296$ and the five consecutive integers $296, 297, 298, 299, 300$, which have the desired property.\n\nTherefore, $N = 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16717,
    "subject": "Mathematics (Olympiad)",
    "question": "We have $n$ students sitting at a round table. Initially, each student is given one candy. At each step, each student having candies either picks one of their candies and gives it to one of their neighbouring students, or distributes all of their candies to their neighbouring students in any way they wish. A distribution of candies is called *legal* if it can be reached from the initial distribution via a sequence of steps.\n\nDetermine the number of legal distributions. (All the candies are identical.)",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is:\n\n- If $n$ is odd: $\\displaystyle \\binom{2n-1}{n}$\n- If $n$ is even: $\\displaystyle \\binom{2n-1}{n} - 2\\binom{3n-1}{n}$\n\n**Case 1: $n$ odd ($n=2m+1$)**\n\nAny distribution of candies is legal. At each step, each student can distribute all their candies to their two neighbours in any way. Thus, each candy can move either one position clockwise or anticlockwise. For any initial and required final distribution, we can assign each candy a target position. Since $n$ is odd, the distance (clockwise or anticlockwise) between the initial and final position is even and at most $m$. Thus, after an even number of steps (at most $m$), each candy can reach its required position. If a candy arrives early, it can move back and forth until all candies are in place. Therefore, all distributions are legal, and the number of legal distributions is $\\binom{2n-1}{n}$.\n\n**Case 2: $n$ even ($n=2m$)**\n\nLet $x_1, \\dots, x_{2m}$ be the students in cyclic order. Initially, the sum of candies among even-indexed students and among odd-indexed students is at least one, and this property is preserved at each step. Every distribution where both the even and odd students have at least one candy in total is legal.\n\nSuppose the final distribution has $a$ candies in odd positions and $b$ in even positions ($a, b \\ge 1$). We can first move all candies to $x_1$ and $x_2$ by having $x_1$ move all candies to $x_2$, and for $1 \\le r \\le 2m-1$, $x_{r+1}$ moves all candies to $x_r$. Now, suppose $x_1$ has $a+k$ candies and $x_2$ has $b-k$ candies ($k \\ge 0$). If $k=0$, we are done. Otherwise, in each step, $x_1$ gives a candy to $x_2$, and $x_2$ gives a candy to $x_3$, and so on, until the desired distribution is reached.\n\nThe total number of legal configurations is:\n\n$$\n\\binom{2n-1}{n} - 2\\binom{3n-1}{n}\n$$\n\nHere, $\\binom{2n-1}{n}$ counts all possible configurations, and $\\binom{3n-1}{n}$ counts the illegal ones where all candies are in only the even or only the odd positions.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16718,
    "subject": "Mathematics (Olympiad)",
    "question": "A single plane divides 3-dimensional space into 2 parts; two parallel planes divide space into 3 parts; two intersecting planes divide space into 4 parts.\n\nConsider the six planes containing the faces of cube $ABCD - A_1B_1C_1D_1$, and the four planes containing the faces of regular tetrahedron $ACB_1D_1$. Into how many parts do these ten planes divide space?",
    "options": [],
    "answer": "See solution",
    "solution": "We first observe that any edge of the regular tetrahedron $ACB_1D_1$ lies on one face of the cube and is parallel to the opposite face of the cube (having no common points), while intersecting the other four faces of the cube at the vertices of the tetrahedron. Any edge of the cube passes through one vertex of the tetrahedron, thus intersecting three faces of the tetrahedron at this point, while intersecting the fourth face of the tetrahedron at a point located on the extension of the edge. Consider the edge $AA_1$ of the cube: its extension beyond $A_1$ intersects the plane $CB_1D_1$ at a point we denote as $A_2$, revealing a tetrahedron $A_1B_1D_1A_2$.\n\n![](images/China-TST-2025A_p13_data_65590fa2cb.png)\n\nThe cube has 12 edges, so there are 12 such tetrahedra. The cube itself is \"cut\" by the regular tetrahedron $ACB_1D_1$ into 5 pieces. Therefore, these 10 planes divide space into $12 + 5 = 17$ bounded (finite) regions. (Note: All bounded regions are accounted for by these 17 pieces, because when considering all possible triple intersections of planes, apart from the 8 vertices of the cube, the only other triple intersection points are the 12 points on the edge extensions like $A_2$. Only these trihedral angles can correspond to bounded regions.)\n\nNow we discuss the unbounded (infinite) regions. Consider a very large spherical surface, on which we can draw a planar graph. Each unbounded region corresponds to a face in this graph, planes correspond to edges, and lines (intersections of planes) correspond to vertices.\n\nTo calculate the number of vertices and edges in this graph:\n\n- The cube has 12 edges;\n- The regular tetrahedron has 6 edges, all lying on the cube's faces (thus being triple intersections of planes);\n- Each face of the tetrahedron intersects with three other planes of the cube, producing $4 \\times 3 = 12$ lines.\n\nThus, these 10 planes have $12 + 6 + 12 = 30$ common lines, corresponding to 60 vertices on the spherical surface.\n\nAmong these 30 lines, 24 are intersections of two planes, so their corresponding 48 vertices each have degree 4; the remaining 6 lines are intersections of three planes, so their corresponding 12 vertices each have degree 6. Therefore, the total number of edges in the graph is:\n\n$$\n\\frac{48 \\times 4 + 12 \\times 6}{2} = 132.\n$$\n\nUsing Euler's formula for planar graphs, we get:\n\n$$\nF = 2 + E - V = 2 + 132 - 60 = 74.\n$$\n\nThus, there are 74 infinite regions.\n\nTherefore, these 10 planes divide space into $17 + 74 = 91$ regions in total. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16719,
    "subject": "Mathematics (Olympiad)",
    "question": "$AH_a$, $BH_b$, and $CH_c$ are the altitudes of triangle $ABC$. Prove that if $\\frac{H_b C}{AC} = \\frac{H_c A}{AB}$, then the line symmetric to $BC$ with respect to the line $H_b H_c$ is tangent to the circumscribed circle of $\\triangle H_b H_c A$.",
    "options": [],
    "answer": "See solution",
    "solution": "Reflect this circle with respect to the line $H_b H_c$ (see the figure below). Note that it also contains a point $A'$, symmetric to $A$ with respect to $M$—the midpoint of $H_b H_c$. So, the problem reduces to showing that the circumscribed circle of $\\triangle H_b H_c A'$ is tangent to the line $BC$.\n\nLet's show that $A'$ lies on $BC$. Draw through $H_c$ a line parallel to $AC$, which intersects $BC$ at point $X$. It's enough to show that $H_b X \\parallel AB$, as then $AH_b X H_c$ is a parallelogram, and therefore $M$ is the midpoint of $AX$, implying that $X = A'$.\n\nIt's enough to show that $\\frac{H_b C}{AC} = \\frac{XC}{BC}$ or $XC = BC \\cdot \\frac{H_b C}{AC}$. As $XC = BC - BX$, from the similarity of triangles $BH_c X$ and $BAC$ we get that $\\frac{BX}{BC} = \\frac{BH_c}{BA}$, and therefore\n\n![](images/Ukraine_2021-2022_p28_data_691af7a92c.png)\n\n![](images/Ukraine_2021-2022_p28_data_a8c503172b.png)\n\n$$\nBX = BC \\cdot \\frac{BH_c}{BA} \\Rightarrow XC = BC \\cdot \\left(1 - \\frac{BH_c}{BA}\\right) = BC \\cdot \\frac{AH_c}{BA} = BC \\cdot \\frac{CH_b}{AC}\n$$\n\nSo, we get $\\angle H_b A' C = \\angle ABC = \\angle AH_b H_c = \\angle H_b H_c A'$, which implies the tangency of the circumscribed circle of $\\triangle H_b H_c A'$ with the line $BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16720,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a triangle $ABC$ with $BC > AB > AC$ and\n$$\n\\cos \\angle A + \\cos \\angle B + \\cos \\angle C = \\frac{11}{8}.\n$$\nLet the points $X \\in BC$ and $Y \\in AC$ be such that $BX = AY = AB$.\n\na) Prove that $XY = \\frac{AB}{2}$.\n\nb) A point $Z$ on the arc $\\widehat{AB}$ of the circumcircle of $\\triangle ABC$ that does not contain $C$ is such that $ZC = ZA + ZB$. Find the ratio $\\frac{ZC}{XC + YC}$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $AB = c$, $BC = a$, and $CA = b$. From the problem, $XC = a - c$ and $YC = c - b$. By the Law of Cosines in $\\triangle XCY$:\n$$\n\\begin{align*}\nXY^2 &= XC^2 + YC^2 - 2 \\cdot XC \\cdot YC \\cos \\angle XCY \\\\\n&= (a-c)^2 + (c-b)^2 + 2(a-c)(c-b) \\cos \\angle ACB \\\\\n&= a^2 + 2c^2 + b^2 - 2ac - 2cb + (ac - ab - c^2 + cb) \\frac{a^2 + b^2 - c^2}{2ab} \\\\\n&= \\frac{c(a^3 + b^3 + c^3 - a^2b - a^2c - b^2a - b^2c - c^2a - c^2b + 3abc)}{ab} \\\\\n&= \\frac{c(3abc - a(b^2 + c^2 - a^2) - b(a^2 + c^2 - b^2) - c(a^2 + b^2 - c^2))}{ab} \\\\\n&= \\frac{c(3abc - 2abc(\\cos \\angle A + \\cos \\angle B + \\cos \\angle C))}{ab} \\\\\n&= \\frac{c(3abc - \\frac{11}{4}abc)}{ab} = \\frac{c^2}{4}.\n\\end{align*}\n$$\nThus, $XY = \\frac{c}{2}$.\n\nb) Since quadrilateral $AZBC$ is cyclic, by Ptolemy's theorem: $ZA \\cdot a + ZB \\cdot b = ZC \\cdot c$. The condition $ZC = ZA + ZB$ gives $\\frac{ZA}{ZB} = \\frac{a-c}{c-b}$. This and the parallelism $\\triangle ZAB \\sim \\triangle CXY$ imply $\\frac{c}{XY} = 2$, so $ZA = 2(a-c)$ and $ZB = 2(c-b)$.\n\n![](images/broshura_07_english_p22_data_05bd4c93f6.png)\n\nThus, $ZC = ZA + ZB = 2(a-b)$, so\n$$\n\\frac{ZC}{XC + YC} = \\frac{2(a-b)}{(a-c) + (c-b)} = 2.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16721,
    "subject": "Mathematics (Olympiad)",
    "question": "設三角形 $ABC$ 中，$CA \\neq CB$。令點 $D$、$F$、$G$ 分別為三邊 $AB$、$AC$、$BC$ 的中點。一圓 $\\Gamma$ 通過 $C$ 點，並與 $AB$ 切於 $D$ 點。設圓 $\\Gamma$ 分別與線段 $AF$ 及 $BG$ 交於 $H$、$I$ 點。令點 $H'$ 為 $H$ 對 $F$ 的對稱點，又令點 $I'$ 為 $I$ 對 $G$ 的對稱點。直線 $H'I'$ 分別與 $CD$、$FG$ 交於 $Q$、$M$ 點。設直線 $CM$ 與圓 $\\Gamma$ 再交於 $P$ 點。\n\n證明：$CQ = QP$。\n\nLet $ABC$ be a triangle with $CA \\neq CB$. Let $D$, $F$, and $G$ be the midpoints of the sides $AB$, $AC$, and $BC$, respectively. A circle $\\Gamma$ passing through $C$ and tangent to $AB$ at $D$ meets the segments $AF$ and $BG$ at $H$ and $I$, respectively. The points $H'$ and $I'$ are symmetric to $H$ and $I$ about $F$ and $G$, respectively. The line $H'I'$ meets $CD$ and $FG$ at $Q$ and $M$, respectively. The line $CM$ meets $\\Gamma$ again at $P$. Prove that $CQ = QP$.",
    "options": [],
    "answer": "See solution",
    "solution": "不失一般性，可設 $CA > CB$。注意到 $H'$ 與 $I'$ 分別落在線段 $CF$、$CG$ 中，所以 $M$ 必定位於 $\\triangle ABC$ 的外部。\n\n![](images/16-3J_p4_data_87e30c9c7e.png)\n\n考慮 $A, B$ 對圓 $\\Gamma$ 的圓幂，可得\n\n$$\nCH' \\cdot CA = AH \\cdot AC = AD^2 = BD^2 = BI \\cdot BC = CI' \\cdot CB.\n$$\n\n所以 $CH' \\cdot CF = CI' \\cdot CG$，得 $H', I', G, F$ 四點共圓。於是 $\\angle I'H'C = \\angle CGF$。\n\n令直線 $DF$ 與 $DG$ 分別與 $\\Gamma$ 再交於 $R$、$S$ 點。我們宣稱：$R$、$S$ 兩點都落在直線 $H'I'$ 上。\n\n注意到 $FH' \\cdot FA = FH \\cdot FC = FR \\cdot FD$。所以 $ADH'R$ 四點共圓，而得 $\\angle RH'F = \\angle FDA = \\angle CGF = \\angle I'H'C$。於是 $R, H', I'$ 三點共線。同理有 $S, H', I'$ 三點也共線，前面的宣稱就得證了。\n\n由弦切角與中線性質，$\\angle RSD = \\angle RDA = \\angle DFG$。於是 $RSFG$ 四點共圓。故有\n\n$$\nMH' \\cdot MI' = MF \\cdot MG \\cdot MR \\cdot MS = MP \\cdot MC.\n$$\n\n從而得知 $CPI'H'$ 四點共圓，令該圓為 $\\omega$。\n\n注意到 $\\angle H'CQ = \\angle SDC = \\angle SRC$，且 $\\angle QCI' = \\angle CDR = \\angle CSR$。所以 $\\triangle CH'Q \\sim \\triangle RCQ$ 且 $\\triangle CI'Q \\sim \\triangle SCQ$。於是 $QH' \\cdot QR = QC^2 = QI' \\cdot QS$。\n\n現在以 $Q$ 點為中心、$QC$ 長為半徑作反演變換。觀察到 $R, C, S$ 三點分別被變換到 $H', C, I'$ 三點。所以 $\\triangle RCS$ 的外接圓 $\\Gamma$ 被變換到 $\\triangle H'CI'$ 的外接圓 $\\omega$。而由於 $P, C$ 兩點同時在這兩圓上，且 $C$ 在此反演變換下不動，所以 $P$ 點也是此變換的不動點。由此知 $QP^2 = QC^2$，即 $QP = QC$。證畢。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16722,
    "subject": "Mathematics (Olympiad)",
    "question": "Exactly one of the following statements is true: which one? *Please note that the numbers $a$ and $b$ need not be integers.*\n\nA) There do not exist $a > 0$ and $b > 0$ with $a \\cdot b < \\frac{a}{b} < a + b$.\n\nB) There do not exist $a > 0$ and $b > 0$ with $a \\cdot b < a + b < \\frac{a}{b}$.\n\nC) There do not exist $a > 0$ and $b > 0$ with $a + b < a \\cdot b < \\frac{a}{b}$.\n\nD) There do not exist $a > 0$ and $b > 0$ with $\\frac{a}{b} < a \\cdot b < a + b$.\n\nE) A) through D) are false.",
    "options": [],
    "answer": "See solution",
    "solution": "C) There do not exist $a > 0$ and $b > 0$ with $a + b < a \\cdot b < \\frac{a}{b}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16723,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p, q, r$ be real numbers such that $p + q + r = 0$. Prove that\n\n$$\n54p^2q^2r^2 \\leq (p^2 + q^2 + r^2)^3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that $\\frac{dx_k}{dt} = kx_{k-1}$ for $k \\geq 1$. Using this, we find $\\frac{dS}{dt} = 0$ and $\\frac{dT}{dt} = 0$, so $S(t)$ and $T(t)$ are constant functions. Thus, $T(t)^3 - 2S(t)^2$ does not depend on $t$.\n\nLet $m = (a + b + c)/3$. Then $x_1(-m) = 0$, so\n\n$$\nT = T(-m) = 3x_2(-m) = 3((a - m)^2 + (b - m)^2 + (c - m)^2)\n$$\nand\n$$\n\\begin{align*}\nS &= S(-m) = 9x_3(-m) \\\\\n&= 9((a - m)^3 + (b - m)^3 + (c - m)^3) \\\\\n&= 27(a - m)(b - m)(c - m),\n\\end{align*}\n$$\nsince $u + v + w$ divides $u^3 + v^3 + w^3 - 3uvw$. Therefore, the inequality $T^3 \\geq 2S^2$ is equivalent to\n$$\n((a - m)^2 + (b - m)^2 + (c - m)^2)^3 \\geq 54(a - m)^2(b - m)^2(c - m)^2,\n$$\nwhich holds for all real $a$, $b$, $c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16724,
    "subject": "Mathematics (Olympiad)",
    "question": "In a country, there are 2024 cities. Each pair of cities is connected bidirectionally by exactly one of three modes of transportation: rail, air, or road. A tourist arrives in this country and knows the entire transportation scheme. He chooses a travel ticket for one mode of transportation and a city from which to start his trip. He wants to visit as many cities as possible, but can only use the ticket for the specified type of transportation. During the route, he may return to cities he has already visited. How many cities is a tourist guaranteed to be able to visit?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose we have a complete graph with each edge colored in one of three colors. We show that there will always exist a connected component of at least one of the colors of size at least $1012$.\n\nFirst, we show that it is possible to color the edges so that no connected component of any color has more than $1012$ vertices. Label the colors as 1, 2, and 3. Divide the $2024$ vertices into two groups of $1012$ each: group $A$ and group $B$. Connect all vertices within group $A$ by edges of color 1, and all within group $B$ by edges of color 1. All edges between $A$ and $B$ are colored with colors 2 or 3, as shown below:\n\n![](images/Ukraine2022-23_p39_data_4156b18bc6.png)\n\nThus, groups $A$ and $B$ each form a connected component of size $1012$ in color 1, and there is no larger connected component in any color.\n\nNow, suppose there exists a coloring such that the largest connected component in any color contains fewer than $1012$ vertices. Consider the coloring that gives the smallest possible answer. Let the largest connected component in some color (say, color 1) have $m$ vertices. If we recolor all edges within this component to color 1, the answer $m$ does not change. No vertex in this component has an edge of color 1 to a vertex outside the component.\n\nNext, consider the largest connected component among the remaining vertices, in color 2 or 3, and denote its size by $k$. Recolor all edges within this component to color 2. Let $X_2$ be the group of vertices in the first component that are connected to at least one vertex of the second component by an edge of color 2, and let $x_2$ be their number. All edges between $X_2$ and the second component can be colored in color 2, but since the first component was chosen to be largest, $k + x_2 \\leq m$.\n\nLet $X_3$ be the remaining vertices in the first component, with $x_3$ vertices. These can only be connected to the second component by edges of color 3. The vertices of $X_3$ and the second component then form a connected component of color 3. If there were any edges of color 3 from $X_3$ to vertices outside the first and second components, the connected component of color 3 would be larger, contradicting the construction. Therefore, all vertices not in the first or second components are connected by color 2, forming a third component. Similarly, $X_2$ is connected to the third component only by edges of color 3, and the second and third components are connected only by edges of color 1, forming a connected component of color 1. Since the largest connected component of color 1 has $m$ vertices, the sum of the sizes of the second and third components is at most $m$, so $2m \\geq 2024$, and thus $m \\geq 1012$.\n\nTherefore, a tourist is guaranteed to be able to visit at least $1012$ cities using only one mode of transportation.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16725,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b$ be non-negative real numbers. Prove the inequality\n\n$$\n\\frac{a}{\\sqrt{b^2+1}} + \\frac{b}{\\sqrt{a^2+1}} \\geq \\frac{a+b}{\\sqrt{ab+1}}\n$$\n\nand find when equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "Equality occurs when $a = 0$, $b = 0$, or $a = b$.\n\nTo prove the inequality otherwise, assume $a > b > 0$. After clearing denominators, it suffices to show:\n\n$$\na \\sqrt{a^2+1} + b \\sqrt{b^2+1} > (a+b) \\sqrt{ab+1}\n$$\n\nExpanding and regrouping terms:\n\n$$\na \\sqrt{a^2+1} (\\sqrt{ab+1} - \\sqrt{b^2+1}) > b \\sqrt{b^2+1} (\\sqrt{a^2+1} - \\sqrt{ab+1})\n$$\n\nExpressing the differences as fractions:\n\n$$\na \\sqrt{a^2+1} \\cdot \\frac{b(a-b)}{\\sqrt{ab+1} + \\sqrt{b^2+1}} > b \\sqrt{b^2+1} \\cdot \\frac{a(a-b)}{\\sqrt{a^2+1} + \\sqrt{ab+1}}\n$$\n\nDividing both sides by $ab(a-b) > 0$ and simplifying, we get:\n\n$$\n\\sqrt{a^2+1}(\\sqrt{a^2+1} + \\sqrt{ab+1}) > \\sqrt{b^2+1}(\\sqrt{b^2+1} + \\sqrt{ab+1})\n$$\n\nSince $a > b$ implies $\\sqrt{a^2+1} > \\sqrt{b^2+1}$, the inequality holds. Thus, equality occurs only when $a = 0$, $b = 0$, or $a = b$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16726,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the circumcenter of triangle $ABC$, and let $\\ell$ be the line passing through the midpoint of side $BC$ and perpendicular to the bisector of $\\angle BAC$. Determine the value of $\\angle BAC$ if the line $\\ell$ passes through the midpoint of segment $AO$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\Gamma$ be the circumcircle of triangle $ABC$, and let $X$ be the intersection of line $\\ell$ and segment $AO$. Let $D$ be the intersection, different from $A$, of the angle bisector of $\\angle BAC$ with $\\Gamma$. Let $A'$ be the point on $\\Gamma$ such that $A'D$ is a diameter. Since $\\angle BAD = \\angle CAD$, $D$ bisects the arc $BC$ not containing $A$, so $A'$ bisects arc $BAC$.\n\nLet $M$ be the midpoint of $BC$. Then $A'M$ and $BC$ are perpendicular, since triangle $A'BC$ is isosceles with $A'B = A'C$. Similarly, $DM$ and $BC$ are perpendicular, since $DB = DC$. Thus, $M$ lies on $A'D$.\n\nNext, we show $A'M = MO$. If $A' = A$, then $X = M$ and $AX = XO$. If $A \\neq A'$, then $AA'$ and $AD$ are perpendicular (since $A'D$ is a diameter), and $AD$ and $MX$ are perpendicular, so $AA'$ and $MX$ are parallel. Since $AX = XO$, we have $A'M = MO$.\n\nTherefore, $BM$ is the perpendicular bisector of $A'O$, so $A'B = OB$. Since $OB = OA'$, triangle $A'OB$ is equilateral, so $\\angle BA'O = 60^\\circ$. Thus, $\\angle BAD = \\angle BA'D = 60^\\circ$, so $\\angle BAC = 2\\angle BAD = 120^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16727,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a continuous, bounded, and non-constant function satisfying\n\n$$\nf(2x) = 2f(x)^2 - 1\n$$\n\nfor all $x \\in \\mathbb{R}$. Show that the equation $f(x) = 0$ has a solution.",
    "options": [],
    "answer": "See solution",
    "solution": "By the intermediate value theorem, it suffices to prove that $f$ takes both non-negative and non-positive values.\n\n**Case $f \\ge 0$:** Suppose that $f(x) \\ge 0$ for all $x \\in \\mathbb{R}$. Then we have\n\n$$\nf(x) = \\sqrt{\\frac{1 + f(2x)}{2}}\n$$\n\nfor all $x \\in \\mathbb{R}$. Since the function\n\n$$\n\\phi: [0, \\infty) \\to [0, \\infty), \\quad t \\mapsto \\sqrt{\\frac{1+t}{2}},\n$$\n\nis strictly increasing, we see that $f \\ge \\sqrt{\\frac{1}{2}}$. In fact, for the sequence defined by $t_0 = 0$ and $t_n = \\phi(t_{n-1})$, $n \\ge 1$, we have $f \\ge t_n$ for all $n$ by induction. However, it is easy to see that $t_n \\nearrow 1$. Thus we see that $f \\ge 1$. Now suppose that for some $x_0 \\in \\mathbb{R}$, we have $f(x_0) > 1$. Then $f(2x_0) = 2f(x_0)^2 - 1 > f(x_0)^2 > 1$. By induction, we have $f(2^n x_0) > f(x_0)^{2^n}$ for all $n \\in \\mathbb{Z}_{\\ge 1}$, which contradicts the boundedness of $f$. Hence $f \\le 1$ and thus $f = 1$, which is impossible since $f$ is non-constant.\n\n**Case $f \\le 0$:** Suppose that $f(x) \\le 0$ for all $x \\in \\mathbb{R}$. Then we have\n\n$$\nf(x) = -\\sqrt{\\frac{1 + f(2x)}{2}}\n$$\n\nfor all $x \\in \\mathbb{R}$. Note that the function\n\n$$\n\\psi: [-1,0] \\to [-1,0], \\quad t \\mapsto -\\sqrt{\\frac{1+t}{2}},\n$$\n\nis strictly decreasing. Thus for the sequence defined by $s_0 = 0$ and $s_n = \\psi(s_{n-1})$, $n \\ge 1$, we have\n\n$$\ns_{2n+1} \\le f \\le s_{2n}\n$$\n\nfor all $n \\ge 0$ by induction. Moreover, it is clear that $s_0 \\ge s_2 \\ge s_4 \\ge \\dots$ and $s_1 \\le s_3 \\le s_5 \\le \\dots$ so we consider the limits $s_{2n} \\searrow a$ and $s_{2n+1} \\nearrow b$. Clearly, both $a$ and $b$ are solutions of the equation $\\psi(\\psi(s)) = s$ belonging to the interval $[-1,0]$. However, it is easy to check that the solutions of $\\psi(\\psi(s)) = s$ are roots of the polynomial $P(s) = 8s^4 - 8s^2 - s + 1 = (s-1)(2s+1)(4s^2 + 2s - 1)$. But $s = -1/2$ is the only root of $P(s)$ in the interval $[-1,0]$, thus $a = b = -1/2$. Hence $f$ is constant, a contradiction.\n\nCombining the cases we get the solution.\n\n**Note.** An alternative solution can be obtained by exploiting the equalities\n\n$$\nf(2x) - 1 = 2(f(x) - 1)(f(x) + 1) \\quad \\text{and} \\\\\nf(2x) + \\frac{1}{2} = 2\\left(f(x) + \\frac{1}{2}\\right)\\left(f(x) - \\frac{1}{2}\\right).\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16728,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle such that $AB \\neq AC$. The internal bisectors of angles $ABC$ and $ACB$ meet the opposite sides at points $B_0$ and $C_0$, respectively, and the circumcircle of $ABC$ at points $B_1$ and $C_1$, respectively. Let $I$ be the incenter of triangle $ABC$. Prove that the lines $B_0C_0$ and $B_1C_1$ meet at a point lying on the parallel through $I$ to the line $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the internal bisector of angle $BAC$ meet the circumcircle of $ABC$ again at point $A_1$. The lines $A_1B_1$ and $AC$ meet at $B_2$, and the lines $AB$ and $A_1C_1$ meet at $C_2$. By Pascal's theorem applied to the hexagon $AC_1BA_1CB_1$, the points $B_2$, $I$, and $C_2$ are collinear; moreover, the Pascal line $B_2IC_2$ is the parallel through $I$ to $BC$, since $A_1B_2$ and $A_1C_2$ are the internal bisectors of angles $AA_1C$ and $AA_1B$, respectively, and $A_1B$ and $A_1C$ are congruent. Finally, the lines $B_iB_{i+1}$ and $C_iC_{i+1}$ meet at collinear points: $B_0B_1$ and $C_0C_1$ meet at $I$, $B_1B_2$ and $C_1C_2$ meet at $A_1$, and $B_2B_0$ and $C_2C_0$ meet at $A$. Thus, triangles $B_0B_1B_2$ and $C_0C_1C_2$ are perspective, so the lines $B_iC_i$ are concurrent (by the converse of Desargues' theorem).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16729,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$ and $y$ be positive integers. Is it possible to find $2$-tuples $(x, y)$ of positive integers such that\n\n$$\n\\prod_{j=1}^{2014} (x + j) = \\prod_{j=1}^{4028} (y + j)?\n$$\n",
    "options": [],
    "answer": "See solution",
    "solution": "Let $w = x + 2015 > 2^{1007}$, $z = y + \\frac{4029}{2} > 2^{502}$. The equation is equivalent to\n\n$$\nw \\cdot \\left( \\left( 1 - \\frac{1}{w} \\right) \\left( 1 - \\frac{2}{w} \\right) \\cdots \\left( 1 - \\frac{2014}{w} \\right) \\right)^{\\frac{1}{2014}} = z^2 \\cdot \\left( \\left( 1 - \\frac{1}{4z^2} \\right) \\left( 1 - \\frac{9}{4z^2} \\right) \\cdots \\left( 1 - \\frac{4027^2}{4z^2} \\right) \\right)^{\\frac{1}{2014}}.\n$$\n\nBy the lemma:\n\n$$\nw \\left(1 - \\frac{2015}{2w}\\right) > w \\cdot \\left( \\left(1 - \\frac{1}{w}\\right) \\left(1 - \\frac{2}{w}\\right) \\cdots \\left(1 - \\frac{2014}{w}\\right) \\right)^{\\frac{1}{2014}} > w \\left(1 - \\frac{2015}{2w} - \\frac{2 \\cdot 2014^2}{w^2}\\right) > w \\left(1 - \\frac{2015}{2w}\\right) - \\frac{1}{8}.\n$$\n\nTherefore, the decimal part of $w \\cdot \\left( \\left(1 - \\frac{1}{w}\\right) \\left(1 - \\frac{2}{w}\\right) \\cdots \\left(1 - \\frac{2014}{w}\\right) \\right)^{\\frac{1}{2014}}$ belongs to $\\left(\\frac{3}{8}, \\frac{1}{2}\\right)$.\n\nOn the other hand, by the lemma:\n\n$$\nz^2 \\left( 1 - \\frac{1^2 + 3^2 + \\cdots + 4027^2}{4z^2 \\cdot 2014} \\right) > z^2 \\cdot \\left( \\left( 1 - \\frac{1}{4z^2} \\right) \\left( 1 - \\frac{9}{4z^2} \\right) \\cdots \\left( 1 - \\frac{4027^2}{4z^2} \\right) \\right)^{\\frac{1}{2014}} > z^2 \\left( 1 - \\frac{1^2 + 3^2 + \\cdots + 4027^2}{4z^2 \\cdot 2014} - \\frac{2 \\cdot 4027^4}{(4z^2)^2} \\right),\n$$\n\nthus\n\n$$\n\\begin{align*}\nz^2 - \\frac{4 \\cdot 2014^2 - 1}{12} &> z^2 \\cdot \\left( \\left( 1 - \\frac{1}{4z^2} \\right) \\left( 1 - \\frac{9}{4z^2} \\right) \\cdots \\left( 1 - \\frac{4027^2}{4z^2} \\right) \\right)^{\\frac{1}{2014}} \\\\ &> z^2 - \\frac{4 \\cdot 2014^2 - 1}{12} - \\frac{4027^4}{8z^2} \\\\ &> z^2 - \\frac{4 \\cdot 2014^2 - 1}{12} - \\frac{1}{8}.\n\\end{align*}\n$$\n\nSince $z^2 - \\frac{4 \\cdot 2014^2 - 1}{12}$ is an integer, the decimal part of\n$z^2 \\cdot \\left( \\left( 1 - \\frac{1}{4z^2} \\right) \\left( 1 - \\frac{9}{4z^2} \\right) \\cdots \\left( 1 - \\frac{4027^2}{4z^2} \\right) \\right)^{\\frac{1}{2014}}$ belongs to $\\left( \\frac{7}{8}, 1 \\right)$, which is a contradiction.\n\nTherefore, there are no $2$-tuples $(x, y)$ of positive integers satisfying\n\n$$\n\\prod_{j=1}^{2014} (x + j) = \\prod_{j=1}^{4028} (y + j).\n$$\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16730,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the greatest possible value of the least common multiple (LCM) of four single-digit positive integers?",
    "options": [],
    "answer": "See solution",
    "solution": "Possible prime factors for a single-digit positive integer are $2$, $3$, $5$, and $7$. Since $2^4 = 16$, $3^3 = 27$, $5^2 = 25$, and $7^2 = 49$ are all greater than $10$, the highest powers of $2$, $3$, $5$, and $7$ that can appear in the prime factorization of a single-digit positive integer are $3$, $2$, $1$, and $1$, respectively. Thus, the LCM of four single-digit positive integers is a divisor of $2^3 \\times 3^2 \\times 5 \\times 7 = 2520$. In particular, it must be less than or equal to this number. On the other hand, the LCM of $5$, $7$, $8$, and $9$ is $2520$, so $2520$ is the greatest possible value.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16731,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral with sides $BC$ and $AD$ equal in length and not parallel. Let points $E$ and $F$ lie on sides $BC$ and $AD$, respectively, such that $BE = DF$. Lines $AC$ and $BD$ meet at $P$, lines $BD$ and $EF$ meet at $Q$, and lines $EF$ and $AC$ meet at $R$. Consider all the triangles $PQR$ as $E$ and $F$ vary. Show that the circumcircles of these triangles have a common point other than $P$.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider the configuration shown below. Note that this argument has more than one configuration, since $O$ can be above $P$. As written, the argument works under the assumption that all angles are taken to be directed modulo $\\pi$ (or $180^\\circ$). If the reader is not familiar, please try to develop a similar proof by relabeling $ABCD$ as $CDAB$.\n\nLet $\\omega_1$ and $\\omega_2$ denote the circumcircles of triangles $ADP$ and $BCP$, respectively. Because $\\frac{AD}{\\sin\\angle DPA} = \\frac{BC}{\\sin\\angle BPC}$, by the \\textbf{Extended Law of Sines}, $\\omega_1$ and $\\omega_2$ have the same size. Let $R$ be the radius of $\\omega_1$ and $\\omega_2$. Let $O$ be the second intersection (other than $P$) of $\\omega_1$ and $\\omega_2$. (Because $AD = BC$ and $AD \\parallel BC$, this point is well defined.) Again, applying the Extended Law of Sines in triangles $CPO$ and $APO$ gives $2R = \\frac{PO}{\\sin\\angle PCO} = \\frac{PO}{\\sin\\angle OAP}$, and so $\\sin\\angle ACO = \\sin\\angle OAC$.\n\nSince $\\angle ACO$ and $\\angle OAC$ are equal angles in triangle $ACO$, the triangle is isosceles with $AO = CO$. Likewise, $\\angle DBO = \\angle ODB$ and $BDO$ is isosceles with $BO = DO$. Because $ADPO$ is cyclic, $\\angle BDO = \\angle PDO = \\angle PAO = \\angle CAO$. Thus triangle $ACO$ is similar to triangle $DBO$. Consequently, we have $\\angle COA = \\angle BOD$. Consider a rotation $H$ centered at $O$ that sends $A$ ![](images/USA_IMO_2006-2007_p53_data_fde77474c5.png) to $C$; that is, $\\mathbf{H}(A) = C$. Then $\\mathbf{H}(D) = B$. Thus $\\mathbf{H}$ sends triangle $ADO$ to triangle $CBO$. Because $DF = BE$, it follows that $\\mathbf{H}(F) = E$, and so $\\angle EOF = \\angle COA = \\angle BOD$ and $EO = FO$. It follows that\n$$\n\\angle OFR = \\angle OFQ = \\angle ODQ = \\angle OAC = \\angle OAR = x, \\quad (*)\n$$\nimplying that quadrilaterals $DFOQ$ and $AFRO$ are cyclic. Because $AFRO$ and $AOPD$ are cyclic, we have\n$$\n\\angle ROF = \\angle RAF = \\angle PAD = \\angle POD,\n$$\nor\n$$\n\\angle DOF = \\angle POR.\n$$\nBecause $DFOQ$ is cyclic, $\\angle DQF = \\angle DOF$. Combining the last two equations gives $\\angle PQR = \\angle DQF = \\angle DOF = \\angle POR$; that is $PQOR$ is cyclic.\n\n**Note:** There are many cyclic quadrilaterals in the figure. For example, we can also finish the proof by noting\n$$\n\\angle OQB = \\angle OEB = \\angle ORC = \\angle ORP\n$$\nbecause $BEQO$ and $OECR$ are cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16732,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a function $f: \\mathbb{R} \\to \\mathbb{R}$ such that for any real numbers $x, y$, the following inequality holds:\n\n$$\nf(x - f(y)) \\le x - y f(x)?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Assume such a function exists. Substitute $y = 0$ into the inequality:\n\n$$\nf(x - f(0)) \\le x.\n$$\n\nNow, let $x = x + f(0)$:\n\n$$\nf(x) \\le x + f(0). \\quad (1)\n$$\n\nNext, substitute $x = f(y)$ into the original inequality:\n\n$$\nf(f(y) - f(y)) \\le f(y) - y f(f(y)),\n$$\nwhich simplifies to:\n$$\nf(0) \\le f(y) - y f(f(y)).\n$$\nFrom (1), $f(f(y)) \\le f(y) + f(0)$. Thus,\n$$\ny f(f(y)) \\le y (f(y) + f(0)).\n$$\nFor $y < 0$, this gives:\n$$\nf(f(y)) \\ge f(y) + f(0).\n$$\nBut from (1), $f(f(y)) \\le f(y) + f(0)$, so equality must hold. However, for all $y < 0$, this is not possible, leading to a contradiction. Therefore, such a function does not exist.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16733,
    "subject": "Mathematics (Olympiad)",
    "question": "We call three points $U, V, W$ a *strong triple* if $W$ is stronger than $V$ and $V$ is stronger than $U$.\n\nLet $L$ be a subset of points on the $m \\times m$ grid (all points $(x, y)$ with $1 \\le x, y \\le m$). Suppose $L$ has no strong triples. What is the maximum possible size of $L$?\n\n*Example*: For $m = 100$, what is the largest $n$ such that there exists $L$ with $|L| = n$ and no strong triples?",
    "options": [],
    "answer": "See solution",
    "solution": "At most two points in $L$ can have the same first coordinate, and at most two can have the same second coordinate.\n\nLet $R = \\{(x, 1) \\mid 1 \\le x \\le m\\}$ (bottom row) and $C = \\{(1, y) \\mid 1 \\le y \\le m\\}$ (left column). Then $|L \\cap (R \\cup C)| \\le 4$.\n\n**Casework on $|L \\cap (R \\cup C)|$:**\n\n- If $|L \\cap (R \\cup C)| = 0$, then $|L| \\le 2(m-1) = 2m-2$.\n- If $|L \\cap (R \\cup C)| = 1$, then $|L| \\le 2m-1$.\n- If $|L \\cap (R \\cup C)| = 2$, then $|L| \\le 2m-1$.\n- If $|L \\cap (R \\cup C)| = 3$, then $|L| \\le 2m-1$.\n- If $|L \\cap (R \\cup C)| = 4$, then $|L| \\le 2m-1$.\n\nThus, in all cases, $|L| \\le 2m-1$.\n\nIf $|L| > 2m-1$, then $L$ must have a strong triple.\n\nA construction with $2m-1$ points and no strong triples is the 'staircase' example:\n\n$$\nL = \\{(x_1, y_m), (x_2, y_m), (x_2, y_{m-1}), (x_3, y_{m-1}), \\dots, (x_{m-1}, y_2), (x_m, y_2), (x_m, y_1)\\}\n$$\n\nSo the bound is tight.\n\nFor $m = 100$, the maximum $n$ is $2m-1 = 199$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16734,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose you have an unlimited number of the two types of shapes pictured below; each shape is formed by taking a unit square and attaching two isosceles right triangles of area $\\frac{1}{2}$ to opposite sides. You have a rectangle with integer sides that you wish to tile with copies of these shapes (rotations or reflections allowed), so that\n\n1. All shapes are contained completely inside the rectangle.\n2. Each part of the rectangle's area is covered by exactly one shape.\n3. The center unit square of every shape has sides parallel to the sides of the rectangle.\n\nShow that you will be unable to tile the rectangle, regardless of its dimensions.\n\n![](images/Saudi_Arabia_booklet_2012_p58_data_66aa5b3ba6.png)\n\n![](images/Saudi_Arabia_booklet_2012_p58_data_ff6209abc1.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose a rectangle could be tiled, and partition it into unit squares. Choose the top left vertex of the rectangle, which is, without loss of generality, adjacent to the square in row 1, column 1. It forms a right angle, and the angles of the tiles are all $45^\\circ$ or $135^\\circ$. So we must fit two $45^\\circ$ angles into the corner. Regardless of which tiles are used and their orientation, this makes the two tiles have unit squares overlapping row 1, column 2 and row 2, column 1. Row 2, column 2 is completely uncovered, and the area of the rectangle remaining to be covered forms a right angle at the top left vertex of this unit square. Thus, the same reasoning applies, which then causes a right angle to appear in the top left of row 3, column 3.\n\nThis process can be continued indefinitely. Since the sides of the rectangle are finite, after enough iterations we will eventually be forced to place a tile that is not completely inside the rectangle, which is a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16735,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a lattice of side length 1 equilateral triangles forming a regular hexagon of side length $n$. Show that the number of ways of simultaneously selecting six vertices of the lattice to form the vertices of a regular hexagon is a perfect square.\n\nAlternatively, consider a regular hexagon with side length $n$, where $n$ is a positive integer. The hexagon is divided into equilateral triangles with side length 1 by lines parallel to the sides of the hexagon. Show that the number of regular hexagons, with vertices among the vertices of equilateral triangles, is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "By a lattice hexagon we mean a regular hexagon whose sides run along edges of the lattice. Given any regular hexagon $H$, we construct a lattice hexagon whose edges pass through the vertices of $H$, as shown in the figure, which we call the enveloping lattice hexagon of $H$. Given a lattice hexagon $G$ of side length $m$, the number of regular hexagons whose enveloping lattice hexagon is $G$ is exactly $m$.\n\n![](images/BMO_Short_list_2014_p20_data_a099415a3a.png)\n\nThere are precisely $3(n-m)(n-m+1)+1$ lattice hexagons of side length $m$ in our lattice: these are those with centres lying at most $n-m$ steps from the centre of the lattice. In particular, the total number of regular hexagons equals\n\n$$\nN = \\sum_{m=1}^{n} \\left[3(n-m)(n-m+1)+1\\right] m = (3n^2+3n) \\sum_{m=1}^{n} m - 3(2m+1) \\sum_{m=1}^{n} m^2 + 3 \\sum_{m=1}^{n} m^3.\n$$\n\nSince $\\sum_{m=1}^{n} m = \\frac{n(n+1)}{2}$, $\\sum_{m=1}^{n} m^2 = \\frac{n(n+1)(2n+1)}{6}$, and $\\sum_{m=1}^{n} m^3 = \\left(\\frac{n(n+1)}{2}\\right)^2$, it is easily checked that $N = \\left(\\frac{n(n+1)}{2}\\right)^2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16736,
    "subject": "Mathematics (Olympiad)",
    "question": "We say a function $f: \\mathbb{Z}_{\\ge 0} \\times \\mathbb{Z}_{\\ge 0} \\to \\mathbb{Z}$ is *great* if for any nonnegative integers $m$ and $n$,\n\n$$\nf(m+1, n+1) f(m, n) - f(m+1, n) f(m, n+1) = 1.\n$$\n\nIf $A = (a_0, a_1, \\dots)$ and $B = (b_0, b_1, \\dots)$ are two sequences of integers, we write $A \\sim B$ if there exists a great function $f$ satisfying $f(n, 0) = a_n$ and $f(0, n) = b_n$ for every nonnegative integer $n$ (in particular, $a_0 = b_0$).\n\nProve that if $A, B, C$, and $D$ are four sequences of integers satisfying $A \\sim B$, $B \\sim C$, and $C \\sim D$, then $D \\sim A$.\n\nWe say $(A, B)$ form a great pair if $A \\sim B$.",
    "options": [],
    "answer": "See solution",
    "solution": "**First solution (Nikolai Beluhov)**\n\nLet $k = a_0 = b_0 = c_0 = d_0$. We let $f, g, h$ be great functions for $(A, B)$, $(B, C)$, $(C, D)$ and write the following infinite array:\n\n$$\n\\begin{bmatrix}\n\\vdots & \\vdots & b_3 & \\vdots & \\vdots & & & \\\\\n\\cdots & g(2,2) & g(2,1) & b_2 & f(1,2) & f(2,2) & \\cdots & \\\\\n\\cdots & g(1,2) & g(1,1) & b_1 & f(1,1) & f(2,1) & \\cdots & \\\\\nc_3 & c_2 & c_1 & k & a_1 & a_2 & a_3 & \\\\\n\\cdots & h(2,1) & h(1,1) & d_1 & & & & \\\\\n\\cdots & h(2,2) & h(1,2) & d_2 & & & & \\\\\n\\vdots & \\vdots & d_3 & & & & \\ddots & \\\\\n\\end{bmatrix}\n$$\n\nThe greatness condition is then equivalent to saying that any $2 \\times 2$ sub-grid has determinant $\\pm 1$ (the sign is $+1$ in two quadrants and $-1$ in the other two), and we wish to fill in the lower-right quadrant. To this end, it suffices to prove the following.\n\n**Lemma**\n\nSuppose we have a $3 \\times 3$ sub-grid\n\n$$\n\\begin{bmatrix}\na & b & c \\\\\nx & y & z \\\\\np & q &\n\\end{bmatrix}\n$$\n\nsatisfying the determinant conditions. Then we can fill in the ninth entry in the lower right with an integer while retaining greatness.\n\n*Proof.* We consider only the case where the $3 \\times 3$ is completely contained inside the bottom-right quadrant, since the other cases are exactly the same (or even by flipping the signs of the top row or left column appropriately).\n\nIf $y = 0$ we have $-1 = bz = bx = xq$, hence $qz = -1$, and we can fill in the entry arbitrarily.\n\nOtherwise, we have $bx \\equiv xq \\equiv bz \\equiv -1 \\pmod y$. This is enough to imply $qz \\equiv -1 \\pmod y$, and so we can fill in the integer $\\frac{qz+1}{y}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16737,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a quadrilateral with $AB \\parallel CD$ and $AB < CD$. Lines $AD$ and $BC$ intersect at a point $P$. Point $X \\neq C$ on the circumcircle of triangle $ABC$ is such that $PC = PX$. Point $Y \\neq D$ on the circumcircle of triangle $ABD$ is such that $PD = PY$. Two lines $AX$ and $BY$ intersect at the point $Q$. Prove that $PQ$ is parallel to $AB$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is enough to show $Y' = Y$; then $Q'$ will be on $B'Y$, so $Q' = Q$. Moreover, $Y'$ is the unique point such that $Y \\neq D$, $Y$ on $(ABD)$, $Y$ on the circle with center $P$, radius $PD$. Thus, it suffices to show $PY' = PD \\Rightarrow PY'D = PDX$.\n\nBut $PY'D = PQA'D = Q'A'DB$, $PDY' = Q'A'BA = Q'A'B'D$, so it suffices to show $Q'A' = Q'A'B'$. However, $Q'A'CA' = 180^\\circ - PA'C = PXC = P(CX = PQA'X = PQA'A' = PAC'A')$. Hence $QA' = QC$.\n\nNow, we apply the Dual of Desargues' Inversion Theorem at point $Q$, quadrilateral $ABCD$. Let $\\infty$ be the point at infinity of $AB$. We obtain an involution $\\phi$ on the pencil of lines through $Q$ swapping $QA \\leftrightarrow QC$, $QB \\leftrightarrow QD$, $QP \\leftrightarrow Q\\infty$.\n\nRecall the properties of $\\phi$: it is a function preserving cross-ratio such that $\\phi(\\phi(\\{\\infty\\})) = \\infty$ for any line $l$ through $D$.\n\nLet $\\gamma$ be the function from the pencil of lines through $Q$ to $CD$ via $\\gamma(\\infty) = \\infty \\cap CD$, and $\\gamma^{-1}$ its inverse ($\\gamma(\\kappa) = Q\\kappa$).\n\nLet $\\psi = \\gamma \\circ \\phi \\circ \\gamma^{-1}$. Then since $\\gamma, \\gamma^{-1}, \\psi$ all preserve cross-ratios, so does $\\psi$, and $\\psi \\circ \\phi = \\gamma \\circ \\phi \\circ \\gamma^{-1} \\circ \\psi = \\infty$, so $\\psi$ is an involution. Finally, $\\psi(A') = C$, $\\psi(B') = D$, and $\\psi(\\infty) = \\infty$.\n\nLet $M$ be the midpoint of $A'C$. Then $-1 = (A', C(M, \\infty)) = (C, A'^2; \\psi(M), \\infty)$\n\n$$\n\\Rightarrow \\phi(M) = M. \\text{ Hence } (B', D' \\setminus M, \\infty) = (D, B'^2 \\setminus M, \\infty)\n$$\n\n$$\n\\frac{DB'M}{DM} \\xrightarrow{DB'M} DM = B'M\n$$\n\nFinally, since $Q'A' = QC$ we see $Q'M \\perp A'C \\Rightarrow Q'M \\perp B'D$\n\nHence $Q'$ is on the perpendicular bisector of $DB$, so $Q'D = Q'B'$\n\n$$\n\\Rightarrow \\begin{aligned} & Q = Q' \\\\ \\Rightarrow \\boxed{PQ \\parallel BC} \\end{aligned}\n$$\n\n![](images/Saudi_Booklet_2025_p53_data_5c366e0f78.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16738,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer, and let $x_1, \\dots, x_n$ and $y_1, \\dots, y_n$ be positive real numbers satisfying $x_n = n$ and $y_i y_{i+1} \\ge 1$ for $1 \\le i < n$ and $y_n y_1 \\ge 1$. Moreover, let $x_0 = 0$.\n\nDetermine the minimal possible value of the expression\n\n$$\n\\sum_{i=1}^{n} \\sqrt{(x_i - x_{i-1})^2 + y_i}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that the minimal value of the desired expression is $n\\sqrt{2}$, achieved by $x_i = i$ and $y_i = 1$ for all $i$. In what follows, let $Y = \\sum_{i=1}^{n} \\sqrt{y_i}$.\n\n**First Inequality:**\n\nFor any numbers $y_i$, $1 \\le i \\le n$, we have\n\n$$\n\\sum_{i=1}^{n} \\sqrt{(x_i - x_{i-1})^2 + y_i} \\ge \\sqrt{n^2 + Y^2},\n$$\n\nwith equality for some $x_i$ depending on the $y_i$.\n\nConsider $n+1$ points in $\\mathbb{R}^2$: $P_0 = (0,0)$ and $P_i = P_{i-1} + (x_i - x_{i-1}, \\sqrt{y_i})$ for $1 \\le i \\le n$. The left side is $\\sum_{i=1}^{n} |P_{i-1}P_i|$, and the right side is $|P_nP_0|$. By the triangle inequality, the result follows. Equality holds when all vectors $P_i - P_{i-1}$ are parallel, i.e.,\n\n$$\nx_i = \\sqrt{y_i} \\cdot \\frac{n}{Y} + x_{i-1}, \\quad 1 \\le i \\le n.\n$$\n\n**Second Inequality:**\n\nWe need to bound $Y \\ge n$, with equality iff $y_i = 1$ for all $i$.\n\nMultiplying all $y_i y_{i+1} \\ge 1$ and $y_n y_1 \\ge 1$ gives\n\n$$\n\\prod_{i=1}^{n} y_i^2 \\geq 1.\n$$\n\nBy the AM-GM inequality,\n\n$$\n\\sum_{i=1}^{n} \\sqrt{y_i} \\geq n \\prod_{i=1}^{n} y_i^{1/(2n)} \\geq n,\n$$\n\nwith equality when $y_i = 1$ for all $i$.\n\nCombining the two inequalities,\n\n$$\n\\sum_{i=1}^{n} \\sqrt{(x_i - x_{i-1})^2 + y_i} \\geq \\sqrt{n^2 + Y^2} \\geq \\sqrt{2n^2} = n\\sqrt{2},\n$$\n\nwith equality iff $y_i = 1$ for all $i$ and $x_i = x_{i-1} + 1$, i.e., $x_i = i$.\n\n_Remark:_ One may also consider a version where the $y_i$ are constants, which makes finding the equality case for the $x_i$ harder.\n\n**Second Solution:**\n\nFor the two inequalities, here is another proof for each.\n\n**First Inequality:**\n\nTreat the $y_i$ as constants and rewrite\n\n$$\n\\sqrt{(x_i - x_{i-1})^2 + y_i} = \\sqrt{y_i} \\sqrt{1 + \\left(\\frac{x_i - x_{i-1}}{\\sqrt{y_i}}\\right)^2}.\n$$\n\nSince $f(x) = \\sqrt{1 + x^2}$ is convex, by Jensen's Inequality with weights $\\frac{\\sqrt{y_i}}{Y}$, the result follows.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16739,
    "subject": "Mathematics (Olympiad)",
    "question": "For what real values of $k > 0$ is it possible to dissect a $1 \\times k$ rectangle into two similar, but incongruent, polygons?",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that a dissection satisfying the requirements is possible if and only if $k \\neq 1$.\n\n**Case $k = 1$:**\nSuppose such a dissection exists for a $1 \\times 1$ square. The common boundary must be a single broken line connecting two points on the boundary. Both polygons must have the same number of vertices and use the same number of square corners. The common boundary must connect two opposite sides, so each polygon uses an entire side of length $1$. Thus, both polygons have the same maximal side length, which is impossible for similar but incongruent polygons. This is a contradiction.\n\n**Case $k \\neq 1$:**\nAssume $k > 1$ (since $1 \\times k$ is similar to $1 \\times \\frac{1}{k}$). We construct a rectangle $ABCD$ dissected into two similar, incongruent polygons as follows:\n\nLet $r > 1$ and $n$ a positive integer. Define points:\n$$\n\\begin{aligned}\nA_0 &= (0, 0), \\quad A_1 = (1, 0), \\quad A_2 = (1, r), \\quad A_3 = (1 + r^2, r), \\\\\nA_4 &= (1 + r^2, r + r^3), \\quad A_5 = (1 + r^2 + r^4, r + r^3),\n\\end{aligned}\n$$\nContinue until\n$$\nA_{2n+1} = (1 + r^2 + r^4 + \\dots + r^{2n},\\ r + r^3 + r^5 + \\dots + r^{2n-1}).\n$$\nDefine rectangle $ABCD$ by $A = A_0$, $C = A_{2n+1}$,\n$$\nB = (1 + r^2 + \\dots + r^{2n}, 0), \\quad D = (0, r + r^3 + \\dots + r^{2n-1}).\n$$\n\n![](images/USA_IMO_2004_p30_data_72dd9567e1.png)\n\nThe two polygons $A_1A_2\\dots A_{2n+1}B$ and $A_0A_1A_2\\dots A_{2n}D$ dissect $ABCD$ and are similar but incongruent, with similarity ratio $r > 1$.\n\nThe rectangle $ABCD$ is similar to $1 \\times f_n(r)$, where\n$$\nf_n(r) = \\frac{1 + r^2 + \\dots + r^{2n}}{r + r^3 + \\dots + r^{2n-1}}.\n$$\nBy choosing $n$ and $r$ appropriately, $f_n(r)$ can take any value $k > 1$.\n\nThus, the dissection is possible for all $k > 0$ except $k = 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16740,
    "subject": "Mathematics (Olympiad)",
    "question": "Do there exist five distinct prime numbers for which the sum of any three of them is a prime number as well?",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that there exist five such prime numbers. If there are three among them that are pairwise incongruent modulo $3$, then by adding each of the three separately to the sum of the other two, we get three distinct sums modulo $3$. One of those three is $0$ modulo $3$ and therefore is not a prime number. However, if among the five we only get two distinct remainders when dividing by $3$, then by the pigeonhole principle there must be three among them that are congruent modulo $3$. The sum of those three, however, is divisible by $3$ and therefore not a prime number.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16741,
    "subject": "Mathematics (Olympiad)",
    "question": "The bisector of the angle at vertex $A$ of triangle $ABC$ intersects the circumcircle of triangle $ABC$ at point $F$ ($F \\neq A$). Points $D$ and $E$ are chosen on the sides $AB$ and $AC$, respectively, such that the lines $DE$ and $BC$ are parallel. Let $G$ and $H$ be the points of intersection of the rays $FD$ and $FE$, respectively, with the circumcircle of triangle $ABC$ ($G \\neq F$, $H \\neq F$). The circumcircles of triangles $AGD$ and $AHE$ intersect at point $P$ ($P \\neq A$). Prove that point $P$ lies on the line $AF$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K$ and $L$ be the points of intersection of the line $AF$ with lines $BC$ and $DE$, respectively. Then\n\n$$\n\\begin{aligned}\n\\angle AGD &= \\angle AGF = \\angle AGC + \\angle CGF = \\angle ABC + \\angle CAF \\\\\n            &= \\angle ABK + \\angle KAB = \\angle CKA = \\angle KLD = 180^\\circ - \\angle ALD.\n\\end{aligned}\n$$\n\nHence, the quadrilateral $AGDL$ is cyclic. Interchanging the roles of points $B$ and $C$, points $D$ and $E$, and also points $G$ and $H$, we can similarly prove that the quadrilateral $AHEL$ is cyclic. Thus, the circumcircles of triangles $ADG$ and $AEH$ meet at point $L$, i.e., $P = L$. Point $L$ was chosen on the line $AF$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16742,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n, m$ be positive integers. A sequence of $mn$ integers written on a circle is called *nice* if the sum of any $m$ consecutive integers is a power of $m$.\n\n1. For any nice sequence of $mn$ ($m \\ge 2$) integers, one can delete $m$ consecutive integers so that the remaining sequence of $m(n-1)$ integers is nice.\n\n2. Any nice sequence of $m^2$ integers contains an integer which is repeated at least $m$ times.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that $a_0, a_1, \\dots, a_{mn-1}$ is a nice sequence, where indices are taken modulo $mn$. First, we claim that if $a_t = \\max\\{a_0, a_1, \\dots, a_{mn-1}\\}$, then $a_{t+1-m} = a_{t+1}, a_{t+2-m} = a_{t+2}, \\dots, a_{t-1} = a_{t+(m-1)}$ (call these $2(m-1)$ terms $(m-1)$-pairs).\n\nLet $S_i = a_i + \\dots + a_{i+m-1}$ for each $0 \\le i \\le mn-1$. Then $S_{i+1} - S_i = a_{i+m} - a_i$. If there are indices $t-m+1 \\le i, j \\le t$ such that $S_i > S_j$, then $S_i - S_j \\ge S_j(m-1)$ since $S_i$ and $S_j$ are powers of $m$. Therefore, it follows from $m a_t \\ge S_i > S_j > a_t$ that\n\n$$\nm a_t - a_t \\ge S_i - S_j \\ge S_j(m-1) > a_t(m-1),\n$$\n\na contradiction. Thus $S_{t-m+1} = S_{t-m+2} = \\dots = S_t$ and hence $a_i = a_{m+i}$ for each $t-m+1 \\le i \\le t-1$.\n\n**(1)** Let $k \\ge 2$ and let $a_0, a_1, \\dots, a_{mk-1}$ be a nice sequence. Denote by $a_t$ the largest term of the sequence and delete the terms $a_t, a_{t+1}, \\dots, a_{t+m-1}$. Then the remaining sequence\n\n$a_0, a_1, \\dots, a_{t-1}, a_{t+m}, a_{t+m+1}, \\dots, a_{mk-1}$\n\nis nice by the claim.\n\n**(2)** From the claim and (1), it follows that the number of $(m-1)$-pairs in a given nice sequence is at least $m-1$. Therefore, there are $(m-1)^2$ pairs of the form $(a_i, a_{i+m})$ with $a_i = a_{i+m}$. Now consider the remainders, modulo $m$, of the indices of all $(m-1)$-pairs in the sequence. By the pigeonhole principle, there is a remainder $j$ modulo $m$ such that the number of pairs whose indices are exactly $j$ modulo $m$ is at least $\\left[\\frac{(m-1)^2}{m}\\right] + 1 = m-1$. Let $(a_{k_i}, a_{k_i+m})$ be such pairs, i.e., $a_{k_i} = a_{k_i+m}$, $k_i < k_{i+1}$ and $k_i \\equiv j \\pmod m$ for each $1 \\le i \\le m-1$. Since $m^2 > k_{m-1} + m$ and\n\n$$\nk_{m-1} + m = m + k_1 + \\sum_{i=2}^{m-1} (k_i - k_{i-1}) = m + k_1 + m \\sum_{i=2}^{m-1} \\frac{k_i - k_{i-1}}{m}\n$$\n\nwe obtain $k_i - k_{i-1} = m$ for each $i$. Thus $a_{k_1} = a_{k_2} = \\dots = a_{k_{m-1}} = a_{k_{m-1}+m}$, completing the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16743,
    "subject": "Mathematics (Olympiad)",
    "question": "For which positive integers $b > 2$ do there exist infinitely many positive integers $n$ such that $n^2$ divides $b^n + 1$?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is: all $b$ such that $b+1$ is not a power of $2$.\n\n**Claim:** If $b+1$ is a power of $2$, then the only $n$ which is valid is $n=1$.\n\n*Proof.* Assume $n > 1$ and let $p$ be the smallest prime dividing $n$. We cannot have $p = 2$, since then $4 \\mid b^n + 1 \\equiv 2 \\pmod{4}$. Thus,\n\n$$\nb^{2n} \\equiv 1 \\pmod{p}\n$$\n\nso the order of $b$ modulo $p$ divides $\\gcd(2n, p-1) = 2$. Hence $p \\mid b^2 - 1 = (b-1)(b+1)$.\n\nBut since $b+1$ is a power of $2$, this forces $p \\mid b-1$. Then $0 \\equiv b^n + 1 \\equiv 2 \\pmod{p}$, a contradiction. $\\Box$\n\nOn the other hand, suppose $b+1$ is not a power of $2$ (and $b > 2$). We can inductively construct an infinite sequence of distinct primes $p_0, p_1, \\dots$, such that for each $k \\ge 0$:\n\n- $p_0^2 \\dots p_{k-1}^2 p_k \\mid b^{p_0 \\dots p_{k-1}} + 1$,\n- and hence $p_0^2 \\dots p_{k-1}^2 p_k^2 \\mid b^{p_0 \\dots p_{k-1} p_k} + 1$ by the exponent lifting lemma.\n\nInitially, let $p_0$ be any odd prime dividing $b+1$. For the inductive step, there exists an odd prime $q \\notin \\{p_0, \\dots, p_k\\}$ such that $q \\mid b^{p_0 \\dots p_k} + 1$. This follows by Zsigmondy's theorem, since $p_0 \\dots p_k$ divides $b^{p_0 \\dots p_{k-1}} + 1$. Since $(b^{p_0 \\dots p_k})^q \\equiv b^{p_0 \\dots p_k} \\pmod{q}$, we can take $p_{k+1} = q$. This completes the induction.\n\nAlternatively, let $p = p_k$ and $c = b^{p_0 \\dots p_{k-1}}$. Then $\\frac{c^p+1}{c+1} = c^{p-1} - c^{p-2} + \\dots + 1$ has GCD exactly $p$ with $c+1$, and this quotient is always odd. As long as $c^p + 1 > p \\cdot (c+1)$, there will be a new prime dividing $c^p + 1$ but not $c+1$. This is always true unless $p=3$ and $c=2$, but $b > 2$ so this case does not occur.\n\n**Remark (On new primes):** In going from $n^2 \\mid b^n + 1$ to $(nq)^2 \\mid b^{nq} + 1$, one does not necessarily need $q \\nmid n$, as long as $\\nu_q(n^2) < \\nu_q(b^n + 1)$. In other words, it suffices to check that $\\frac{b^{n+1}}{n^2}$ is not a power of $2$ in this process.\n\nHowever, this calculation is more involved. Since $n$ is odd, $\\nu_2(b^n + 1) = \\nu_2(b+1)$, and thus $\\frac{b^{n+1}}{n^2} = 2^{\\nu_2(b+1)} \\le b+1$, which is harder to bound than $c^p+1 > p \\cdot (c+1)$.\n\n**Special case:** If $b=2$, then the only $n$ which work are $n=1$ and $n=3$. Thus $b=2$ is a special case, and the problem requires $b > 2$.\n\n**Alternate formulation:** The problem can also be asked for $n$ with at least $3$ (or $2018$, etc.) prime divisors, but the 'infinitely many $n$' version is more natural. Both formulations are similar in spirit.\n\nSuppose $k^2 \\mid b^k + 1$ and $p \\mid b^k + 1$. For any $k \\mid n$ with $n^2 \\mid b^n + 1$, and $p$ an odd prime dividing $b^k + 1$, we have:\n\n$$\n2\\nu_p(n) \\le \\nu_p(b^n + 1) = \\nu_p(n/k) + \\nu_p(b^k + 1)\n$$\nso\n$$\n\\nu_p(n/k) \\le \\nu_p \\left( \\frac{b^k + 1}{k^2} \\right).\n$$\n\nThis means we can only add each prime a certain number of times.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16744,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be positive integers. Show that if $4ab - 1$ divides $(4a^2 - 1)^2$, then $a = b$.",
    "options": [],
    "answer": "See solution",
    "solution": "Call $(a, b)$ a \"bad pair\" if it satisfies $4ab - 1 \\mid (4a^2 - 1)^2$ while $a \\neq b$. We use the method of infinite descent to prove there is no such \"bad pair\".\n\n*Property 1*: If $(a, b)$ is a \"bad pair\" and $a < b$, there exists an integer $c$ ($c < a$) such that $(a, c)$ is also a \"bad pair\".\n\nIn fact, let $r = \\frac{(4a^2 - 1)^2}{4ab - 1}$, then\n$$\nr = -r \\cdot (-1) = -(4a^2 - 1)^2 = -1 \\pmod{4a}.\n$$\nTherefore there exists an integer $c$ such that $r = 4ac - 1$. Since $a < b$, we have\n$$\n4ac - 1 = \\frac{(4a^2 - 1)^2}{4ab - 1} < 4a^2 - 1.\n$$\nSo $c < a$ and $4ac - 1 \\mid (4a^2 - 1)^2$. Thus $(a, c)$ is a \"bad pair\" too.\n\n*Property 2*: If $(a, b)$ is a \"bad pair\", so is $(b, a)$.\n\nIn fact, by\n$$\n1 = 1^2 \\equiv (4ab)^2 \\pmod{4ab - 1},\n$$\nwe get\n$$\n(4b^2 - 1)^2 \\equiv (4b^2 - (4ab)^2)^2 = 16b^4(4a^2 - 1)^2 \\\\\n\\equiv 0 \\pmod{4ab - 1}.\n$$\nThus $4ab - 1 \\mid (4b^2 - 1)^2$.\n\nIn the following we will show that such \"bad pair\" does not exist. We shall prove by contradiction.\n\nSuppose there is at least one \"bad pair\"; we choose such a pair for which $2a + b$ is minimum.\n\nIf $a < b$, by property 1, there is a \"bad pair\" $(a, c)$ which satisfies $c < b$, and $2a + c < 2a + b$, a contradiction.\n\nIf $b < a$, by property 2, $(b, a)$ is also a \"bad pair\", which leads to $2b + a < 2a + b$, a contradiction.\n\nHence such \"bad pair\" does not exist. Therefore $a = b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16745,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $DX = d$. Choose $xy$-coordinate axes so that $X = (0,0)$, $A = (5,0)$, $B = (0,6)$, $C = (-20,0)$, and $D = (0,-d)$. Let $\\Gamma$ be the circle tangent to each of the circles $O_1, O_2, O_3, O_4$, and let $P = (x, y)$ be the center of $\\Gamma$ and $r$ its radius. Find the value of $d$ such that $\\Gamma$ is tangent to all four circles.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the circle $O_1$ touches $\\Gamma$ tangentially from inside, $AP = r - 5$, so $(x - 5)^2 + y^2 = (r - 5)^2$. Simplifying, $$r^2 - x^2 - y^2 = 10(r - x).$$\n\nSimilarly, for $O_2, O_3, O_4$:\n$$\n\\begin{aligned}\nr^2 - x^2 - y^2 &= 40(r + x), \\\\\nr^2 - x^2 - y^2 &= 12(r - y), \\\\\nr^2 - x^2 - y^2 &= 2d(r + y).\n\\end{aligned}\n$$\n\nLet $t = r^2 - x^2 - y^2$, then:\n$$\n\\begin{aligned}\nr - x &= \\frac{t}{10}, \\\\\nr + x &= \\frac{t}{40}, \\\\\nr - y &= \\frac{t}{12}, \\\\\nr + y &= \\frac{t}{2d}.\n\\end{aligned}\n$$\n\nSince $2r = (r-x) + (r+x) = (r-y) + (r+y)$, we get:\n$$\n2r = t \\left( \\frac{1}{10} + \\frac{1}{40} \\right) = t \\left( \\frac{1}{12} + \\frac{1}{2d} \\right).\n$$\n\nAs $r \\neq 0$, equate:\n$$\n\\frac{1}{10} + \\frac{1}{40} = \\frac{1}{12} + \\frac{1}{2d}\n$$\nSolving, $d = 12$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16746,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure, points $C$ and $D$ are on the semicircle with $O$ as its center and $AB$ as its diameter. The tangent line to the semicircle at point $B$ meets the line $CD$ at point $P$. Line $PO$ intersects $CA$ and $AD$ at points $E$ and $F$ respectively. Prove that $OE = OF$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof I**\n\nDraw line segments $OM$ and $MN$, such that $OM \\perp CD$, $MN \\parallel AD$. Let $MN \\cap BA = N$, $CN \\cap DA = K$, and connect $BC$ and $BM$. Then we get\n\n$$\n\\angle NBC = \\angle ADC = \\angle NMC,\n$$\n\nwhich means points $N$, $B$, $M$, $C$ are concyclic. Since $O$, $B$, $P$, $M$ are also concyclic points, we obtain\n\n$$\n\\angle OPM = \\angle OBM = 180^\\circ - \\angle MCN,\n$$\n\nthus $CN \\parallel OP$ and\n\n$$\n\\frac{CN}{OE} = \\frac{AN}{AO} = \\frac{NK}{OF}.\n$$\n\nSince $M$ is the midpoint of $CD$, $MN \\parallel DK$, then we get that $N$ is the midpoint of $CK$. Hence, according to the above, we have $OE = OF$ as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16747,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be integers such that\n$$a = a^2 + b^2 - 8b - 2ab + 16.$$ \nProve that $a$ is a square.",
    "options": [],
    "answer": "See solution",
    "solution": "$$9a = a^2 + b^2 + 8a - 8b - 2ab + 16 = (a - b + 4)^2$$\nHence, $9a$ is a square.\nNow it is obvious that $a$ is a square too.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16748,
    "subject": "Mathematics (Olympiad)",
    "question": "Real numbers $x$, $y$, and $z$ satisfy $x + y + z = 4$ and $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = \\frac{1}{3}$. Find the largest and the smallest possible value of the expression $x^3 + y^3 + z^3 + xyz$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n\n$$\n(x + y + z)^3 = x^3 + y^3 + z^3 + 3(x^2y + xy^2 + x^2z + y^2z + y^2x + xyz) + 6xyz,\n$$\n\nwhile\n\n$$\n3(x + y + z) \\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right) xyz = 3(x + y + z)(xy + xz + yz) = 3(x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2) + 9xyz.\n$$\n\nThus,\n\n$$\n(x + y + z)^3 - 3(x + y + z) \\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right) xyz = x^3 + y^3 + z^3 - 3xyz.\n$$\n\nBy the assumptions, $x + y + z = 4$ and $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = \\frac{1}{3}$. Hence,\n\n$$\n64 - 4xyz = x^3 + y^3 + z^3 - 3xyz,\n$$\n\nimplying\n\n$$\nx^3 + y^3 + z^3 + xyz = 64.\n$$\n\nConsequently, the expression $x^3 + y^3 + z^3 + xyz$ has only one value: $64$.\n\n**Remark.** An example of numbers satisfying the conditions is $x = 1$, $y = \\frac{3-3\\sqrt{3}}{2}$, and $z = \\frac{3+3\\sqrt{3}}{2}$. Then $\\frac{1}{x} = 1$, $\\frac{1}{y} = \\frac{-1-\\sqrt{3}}{3}$, and $\\frac{1}{z} = \\frac{-1+\\sqrt{3}}{3}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16749,
    "subject": "Mathematics (Olympiad)",
    "question": "In Kükametsa School of Science, the number of secondary school participants in the \"Kangaroo\" competition was exactly one fourth greater than in the previous year. This year, the number of participants from the 11th and 12th grade was respectively 3 and 4 less than the number of participants from the 10th and 11th grade in the previous year, but the number of participants from the 10th grade this year was twice as high as the number of participants from the 12th grade last year. In the previous year, the average number of participants per grade level in Kükametsa School of Science was more than 60, but less than 65. Find all possibilities how many participants from the 10th grade could have participated in the \"Kangaroo\" competition this year.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $e$ be the number of secondary school participants in the \"Kangaroo\" competition in the previous year. This year, the number of participants was greater by $\\frac{1}{4}e$. Hence, $e$ is divisible by 4.\n\nThe total number of participants in the 11th and 12th grade this year was $3 + 4 = 7$ less than the number of participants in the 10th and 11th grade last year. The number of participants from the 10th grade this year was twice the number of participants from the 12th grade last year.\n\nThus, the total number of secondary school participants increased by a number that is 7 less than the number of participants from the 12th grade last year. Hence, the number of participants from 12th grade last year was $\\frac{1}{4}e + 7$. The number of participants from the 10th grade this year was twice that, or $\\frac{1}{2}e + 14$.\n\nThe average number of participants per grade level last year is $\\frac{1}{3}e$. Since $e$ is divisible by 4, $\\frac{1}{3}e$ must be a fraction with numerator a multiple of 4 and denominator 3. Considering all such fractions between 60 and 65, we get three options: $\\frac{184}{3}$, $\\frac{188}{3}$, and $\\frac{192}{3}$. So $e$ can be either 184, 188, or 192, and $\\frac{1}{2}e + 14$ is either 106, 108, or 110, respectively.\n\n**Answer:** 106, 108, 110.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16750,
    "subject": "Mathematics (Olympiad)",
    "question": "$$\n\\begin{cases}\n\\frac{a(a+1)(b+1)}{2} = 720 \\\\\n\\frac{(a+1)b(b+1)}{2} = 240\n\\end{cases}\n$$\n\nFind the values of $a$ and $b$ that satisfy the system above, and compute $n = 2^a 3^b$.",
    "options": [],
    "answer": "See solution",
    "solution": "Dividing both sides of the first equation by the corresponding sides of the second equation, we get $a = 3b$. Substituting this into the second equation, we obtain $(3b+1)b(b+1) = 480$, from which it follows that\n\n$$\n(3b+1)b(b+1) - 480 = (b-5)(3b^2+19b+96) = 0\n$$\n\nAs the equation $3b^2 + 19b + 96 = 0$ does not have a real root, we conclude that $b = 5$ must hold, and therefore, $a = 15$. Thus, $n = 2^{15} 3^5$ is the desired solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16751,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the sequence of positive integers $x_n$ given by the formula:\n\n$$\nx_n = 5 \\cdot 2^n - 1, \\quad n \\in \\mathbb{N}.\n$$\n\nProve that there are infinitely many pairs $(x_i, x_j)$ of elements that are coprime, and at the same time, no element $x_k$ in these infinite pairs appears infinitely many times in the sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "Everything follows from this equation: for all $n \\in \\mathbb{N}$,\n\n$$\nx_{n+1} - 2x_n = 5 \\cdot 2^{n+1} - 1 - 2 \\cdot (5 \\cdot 2^n - 1) = 10 \\cdot 2^n - 1 - 10 \\cdot 2^n + 2 = 1.\n$$\n\nSo, the greatest common factor (GCF) of $x_{n+1}$ and $x_n$ is a divisor of $1$, which means they are coprime. This is what had to be shown.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16752,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples of positive integers $(x, y, z)$ such that\n\n$$\n2^x + 1 = 7^y + 2^z.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Because $x, y, z \\in \\mathbb{Z}^+$, we have $7^y > 1$, which implies $2^x > 2^z$ or $x > z$. The equation can be rewritten as $2^z(2^{x-z} - 1) = 7^y - 1$.\n\n$7^y \\equiv 1 \\pmod{3}$, so $3 \\mid 7^y - 1$, which implies $3 \\mid 2^{x-z} - 1$ or $x-z$ must be even. Consider the following cases:\n\n*Case 1.* If $y$ is odd, then by LTE theorem, $v_2(7^y - 1) = v_2(7 - 1) = 1$. Thus $v_2(2^z) = 1$ and $z = 1$. Substituting $z = 1$ into the original equation, $2^x = 7^y + 1$. Since $y$ is odd, $v_2(7^y + 1) = v_2(7 + 1) = 3$, which implies $x = 3, y = 1$. By direct checking, this solution is satisfied. Therefore, $(x, y, z) = (3, 1, 1)$.\n\n*Case 2.* If $y$ is even, let $y = 2k$ where $k$ is a positive integer:\n\n$$\n2^z(2^{x-z} - 1) = 49^k - 1.\n$$\n\nConsider:\n\n1. If $k$ is odd, $v_2(49^k - 1) = v_2(49 - 1) = 4$, so $z = 4$. Substituting $z = 4$ into the original equation, $2^x - 49^k = 15$, so $x$ is even, let $x = 2t$ with $t \\in \\mathbb{Z}^+$:\n\n$$\n4^t - 49^k = 15 \\Leftrightarrow (2^t + 7^k)(2^t - 7^k) = 15.\n$$\n\nSince $2^t + 7^k \\le 15$, $7^k < 15$ or $k = 1$. Then $4^t = 64$ or $t = 3$, so $x = 6, y = 2$. Thus, $(x, y, z) = (6, 2, 4)$.\n\n2. If $k$ is even, denote $k = 2l$ where $l$ is a positive integer, then $y = 4l$. $49^k \\equiv (-1)^k \\equiv 1 \\pmod{25}$, so $25 \\mid 2^{x-z} - 1$. Since $2 \\mid x-z$, let $x-z = 2a$ where $a \\in \\mathbb{Z}^+$, so $25 \\mid 4^a-1$ and $5 \\mid 4^a - 1$, implying $a$ is even; let $a = 2b$ where $b \\in \\mathbb{Z}^+$, then $v_5(49^k - 1) \\ge 2$ and\n\n$$\nv_5(2^z(16^b - 1)) = v_5(16^b - 1) = v_5(15) + v_5(b) = 1 + v_5(b).\n$$\n\nSo $5 \\mid b$, let $b = 5c$ where $c \\in \\mathbb{Z}^+$, then $2^{x-z} - 1 = 1024^c - 1 \\equiv 0 \\pmod{1023}$. Notice $31 \\mid 1023$, so $31 \\mid 7^y - 1$. But\n\n$$\n7^{15} = (7^3)^5 \\equiv 343^5 \\equiv 2^5 \\equiv 32 \\equiv 1 \\pmod{31}\n$$\n\nand $7^3, 7^5, 7^{10} \\ne 1 \\pmod{31}$ so $\\text{ord}_{31}(7) = 15$. Thus $15 \\mid y$, but $4 \\mid y$ so $12 \\mid y$, and $13 \\mid 7^y - 1$ by Little Fermat theorem.\n\nThus, $13 \\mid 2^{x-z} - 1$ and $\\text{ord}_{13}(2) = 12$ so $12 \\mid x-z$, and $7 \\mid 2^{x-z} - 1$, which is a contradiction since the right hand side is not divisible by 7. So the equation has no solution.\n\nAll satisfying triples are $(x, y, z) = (3, 1, 1), (6, 2, 4)$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16753,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\alpha$ be a nonzero real number.\n\nDetermine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(f(x + y)) = f(x + y) + f(x)f(y) + \\alpha x y\n$$\n\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The functional equation immediately implies that $f$ cannot be a constant function, as $\\alpha x y$ would then have to be constant. Let $(F)$ denote the given functional equation.\n\nSetting $y = 1$ in $(F)$ gives:\n\n$$\nf(f(x + 1)) = f(x + 1) + f(x)f(1) + \\alpha x. \\quad (1)\n$$\n\nFor $x = 1$:\n\n$$\nf(f(2)) = f(2) + f(1)^2 + \\alpha. \\quad (2)\n$$\n\nReplacing $x$ by $x + 1$:\n\n$$\nf(f(x + 2)) = f(x + 2) + f(x + 1)f(1) + \\alpha(x + 1). \\quad (3)\n$$\n\nFor $y = 2$ in $(F)$:\n\n$$\nf(f(x + 2)) = f(x + 2) + f(x)f(2) + 2\\alpha x. \\quad (4)\n$$\n\nFor $x = 0$:\n\n$$\nf(f(2)) = f(2) + f(0)f(2).\n$$\n\nTogether with (2):\n\n$$\nf(0)f(2) = f(1)^2 + \\alpha. \\quad (5)\n$$\n\nNow, take $(F)$ with $y = 0$ and replace $x$ by $x + 1$:\n\n$$\nf(f(x + 1)) = f(x + 1) + f(x + 1)f(0). \\quad (6)\n$$\n\nFrom (1) and (6):\n\n$$\nf(x + 1)f(0) = f(x)f(1) + \\alpha x. \\quad (7)\n$$\n\nFrom (3) and (4):\n\n$$\nf(x + 1)f(1) = f(x)f(2) + \\alpha x - \\alpha. \\quad (8)\n$$\n\nMultiply (7) by $f(2)$ and (8) by $f(1)$:\n\n$$\nf(x + 1)f(0)f(2) = f(x)f(1)f(2) + \\alpha f(2)x\n$$\n\n$$\nf(x + 1)f(1)^2 = f(x)f(1)f(2) + \\alpha f(1)x - \\alpha f(1)\n$$\n\nSubtracting and using (5):\n\n$$\n\\alpha f(x + 1) = \\alpha(f(2) - f(1))x + \\alpha f(1)\n$$\n\nSince $\\alpha \\neq 0$:\n\n$$\nf(x+1) = (f(2) - f(1))x + f(1)\n$$\n\nSo $f$ is linear: $f(x) = a x + b$ with $a \\neq 0$. Substitute into $(F)$:\n\n$$\na^2 x + a^2 y + a b + b = a x + a y + b + a^2 x y + a b x + a b y + b^2 + \\alpha x y\n$$\n\nFor $y = 0$:\n\n$$\na^2 x + a b = (a + a b)x + b^2\n$$\n\nComparing coefficients: $a^2 = a + a b$, so $a = 1 + b$, and $a b = b^2$. Thus $(1 + b)b = b^2$, so $b = 0$, $a = 1$.\n\nSo $f(x) = x$. Substitute into $(F)$:\n\n$$\nf(f(x + y)) = f(x + y) + f(x)f(y) + \\alpha x y \\implies x + y = x + y + x y + \\alpha x y\n$$\n\nSo $(1 + \\alpha)x y = 0$ for all $x, y$, so $\\alpha = -1$.\n\n**Answer:** For $\\alpha = -1$, the identity is the only solution. For other values of $\\alpha$, there is no solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16754,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integer values of $n$ for which the number\n\n$$\nA = \\frac{8n - 25}{n + 5}\n$$\nis equal to the cube of a rational number.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p, q \\in \\mathbb{Z}$, $q \\neq 0$, with $(p, q) = 1$, such that\n\n$$\nA = \\frac{8n-25}{n+5} = \\left(\\frac{p}{q}\\right)^3.\n$$\n\nThen $(p^3, q^3) = 1$. From this, we get:\n\n$$\nq^3(8n-25) = p^3(n+5).\n$$\n\nSo $p^3 \\mid (8n-25)$ and $q^3 \\mid (n+5)$, which means there exists $k \\in \\mathbb{Z}$ such that $8n-25 = k p^3$ and $n+5 = k q^3$.\n\nSubstituting, we have:\n\n$$\n8(n+5) - (8n-25) = k(8q^3 - p^3) \\implies k(8q^3 - p^3) = 65.\n$$\n\nLet $8q^3 - p^3 = d$, so $k d = 65$. The divisors of $65$ are $\\pm1, \\pm5, \\pm13, \\pm65$.\n\nNow, $8q^3 - p^3 = (2q - p)(4q^2 + 2q p + p^2)$. Set $4q^2 + 2q p + p^2 = m$ and $2q - p = l$, so $k l m = 65$.\n\nBut $4q^2 + 2q p + p^2 = 3q^2 + (p+q)^2 \\geq 3$, so the only possible positive divisor is $13$.\n\nSet $4q^2 + 2q p + p^2 = 13$. Then $3q^2 + (p+q)^2 = 13$.\n\nTry $q = \\pm2$:\n- If $q = 2$, $3(4) + (p+2)^2 = 13 \\implies 12 + (p+2)^2 = 13 \\implies (p+2)^2 = 1 \\implies p = -1$ or $p = -3$.\n- If $q = -2$, $3(4) + (p-2)^2 = 13 \\implies 12 + (p-2)^2 = 13 \\implies (p-2)^2 = 1 \\implies p = 3$ or $p = 1$.\n\nNow, $\\left(\\frac{p}{q}\\right)^3 = \\left(\\frac{-1}{2}\\right)^3 = -\\frac{1}{8}$ or $\\left(\\frac{-3}{2}\\right)^3 = -\\frac{27}{8}$, but only $-\\frac{1}{8}$ matches the form $A = -\\frac{1}{8}$.\n\nSet $A = -\\frac{1}{8}$:\n\n$$\n\\frac{8n-25}{n+5} = -\\frac{1}{8} \\implies 8(8n-25) = -(n+5) \\implies 64n - 200 = -n - 5 \\implies 65n = 195 \\implies n = 3.\n$$\n\n**Answer:** $n = 3$ is the only integer solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16755,
    "subject": "Mathematics (Olympiad)",
    "question": "If $x$ is a three-digit integer and $x^2$ divided by $2024$ leaves a remainder of $1$, what is the smallest possible value of $x$?",
    "options": [],
    "answer": "See solution",
    "solution": "$2024 = 8 \\times 11 \\times 23$, so $x^2 \\equiv 1 \\pmod{2024}$ if and only if $x \\equiv \\pm1 \\pmod{8}$, $x \\equiv \\pm1 \\pmod{11}$, and $x \\equiv \\pm1 \\pmod{23}$. By the Chinese Remainder Theorem, the possible $x$ are $1, 45, -1, -45$ modulo $2 \\times 11 \\times 23 = 506$. The smallest three-digit $x$ is $506 - 45 = 461$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16756,
    "subject": "Mathematics (Olympiad)",
    "question": "For a set $M$ of positive integers with $n$ elements, where $n$ is odd, a nonempty subset $T$ of $M$ is called *good* if the product of the elements of $T$ is divisible by the sum of the elements of $M$, but not divisible by its square. If $M$ is good, find the maximum possible number of good subsets of $M$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $A \\cup B = M$ and $A \\cap B = \\emptyset$, then at most one of the sets $A$ and $B$ is good (otherwise $M$ is not good). Therefore, the number of good sets does not exceed half the number of all subsets, i.e., $2^{n-1}$.\n\nWe will prove that this estimate is sharp. Let $n = 2k + 1$ and $p$ be a prime such that $p > \\frac{n(n-1)}{2}$. Let $p^k = a_1 + a_2 + \\cdots + a_{n-1} + a_n$, where $a_i = i$ for $i = 1, 2, \\ldots, n-1$ and $a_n = p^k - \\frac{n(n-1)}{2}$. Then $(a_i, p) = 1$ for every $i$. Consider the set\n\n$$\nM = \\{p a_i \\mid i = 1, 2, \\ldots, n\\}.\n$$\n\nThen the sum of the elements of $M$ is equal to $p^{k+1}$. It is easy to see that the good subsets are exactly those with at most $k + 1$ elements. There are exactly $2^{n-1}$ such subsets.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16757,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that a square can be cut into $n$ square pieces.",
    "options": [],
    "answer": "See solution",
    "solution": "A partition of a square into 1 square is trivial.\n\nIf $n \\geq 2$, then a partition into $2n$ squares can be obtained by cutting $n$ squares of side length $\\frac{1}{n}$ from one side of the square and $n - 1$ more squares of the same size from a neighbouring side. One square of side length $\\frac{n-1}{n}$ of the side length of the big square is left. (The figure below depicts the situation in the case $n = 4$, which provides a partition into 8 squares.)\n\nFor each $n \\geq 2$, one can obtain a partition into $2n + 3$ squares by splitting one square in a partition into $2n$ squares into four. Thus, there exist partitions into 1, 4, and every natural number starting from 6.\n\n![](images/prob1617_p15_data_2215c4f4f6.png)\n\nIt remains to show that there are no partitions of a square into 2, 3, and 5 squares. A square has 4 vertices and each vertex belongs to only one square in a partition. If two vertices belonged to the same square in the partition, this piece should be as large as the initial square, which is possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16758,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $n$ be positive integers with $n \\ge 2$. Let $R$ be an $n$-element ring, and let $x$ be an element of $R$ such that $1 - x^k$ is invertible for each $k \\in \\{m+1, m+2, \\dots, m+n-1\\}$. Show that $x$ is nilpotent (i.e., $x^p = 0$ for some positive integer $p$).",
    "options": [],
    "answer": "See solution",
    "solution": "We will repeatedly use the result that if $c = ab \\in R$ is invertible and $ab = ba$, then $a$ and $b$ are invertible. Indeed, if $cy = yc = 1$, then, using associativity, $a(by) = (ab)y = cy = 1$ and $(yb)a = y(ba) = yc = 1$. Also, $((yb)a)(by) = (yb)(a(by))$ implies $1 \\cdot (by) = (yb) \\cdot 1$, that is $by = yb := z$, whence $az = za = 1$ and $a$ has inverse $z$.\n\nSince $0$ is nilpotent, we will assume in the sequel that $x \\ne 0$.\n\nGiven a positive integer $k < n$, notice that $m+p$ is divisible by $k$ for some positive integer $p < n$; write $m + p = k\\ell$. Since $1 - x^{m+p}$ is invertible, and $1 - x^{m+p} = (1 - x^k)(1 + x^k + \\dots + x^{k(\\ell-1)}) = (1 + x^k + \\dots + x^{k(\\ell-1)})(1 - x^k)$, it follows that $1 - x^k$ is invertible.\n\nNext, write $1 - x^k = (1-x)(1+x+\\dots+x^{k-1}) = (1+x+\\dots+x^{k-1})(1-x)$, where we agree that $x^0 = 1$, to deduce that $x_k = 1 + x + \\dots + x^{k-1}$ is invertible for each positive integer $k < n$ and $1-x$ is invertible.\n\nIf the $x_k$ were pairwise distinct, then by a cardinality argument $R$ would be an $n$-element skew field, so $x^{n-1} = 1$, that is $(1-x)x_{n-1} = 0$, which is impossible since $1-x$ and $x_{n-1}$ are invertible.\n\nConsequently, $x_p = x_q$ for some indices $p < q$, so $0 = x_q - x_p = x^p x_{q-p}$ which implies $x^p = 0$, since $x_{q-p}$ is invertible.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16759,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, a_3, a_4, a_5$ be terms of a geometric sequence, with each $a_i$ a positive integer. Let $q = \\frac{a_2}{a_1}$ be the common ratio, which is rational and can be written as $q = \\frac{m}{n}$ in lowest terms. Express the terms of the sequence in terms of $m$ and $n$, and determine all possible sequences where all terms are positive integers less than $2008$, and no prime divides all five terms.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_1 = dn^4$ for some positive integer $d$. The terms are:\n\n$$\ndn^4,\\; dmn^3,\\; dm^2n^2,\\; dm^3n,\\; dm^4.\n$$\n\nSince no prime divides all terms, $d = 1$, so the sequence is $n^4, mn^3, m^2n^2, m^3n, m^4$. All terms are less than $2008$, so $m^4 < 2008$ and $n^4 < 2008$, implying $m \\leq 6$, $n \\leq 6$. Also, $a_1 = n^4$ is not divisible by $6$, so $n \\leq 5$. Given divisibility conditions, $mn$ must be divisible by $2$, $3$, and $5$, i.e., $mn = 30$. With $m \\leq 6$, $n \\leq 5$, the only solution is $m = 6$, $n = 5$. Thus, the sequence is $625, 750, 900, 1080, 1296$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16760,
    "subject": "Mathematics (Olympiad)",
    "question": "Let\n$$\nf(n) = \\sum_{k=0}^{2010} n^k = 1 + n + n^2 + \\dots + n^{2010}.\n$$\nProve that for every integer $m$ with $2 \\leq m \\leq 2010$, there is no non-negative integer $n$ such that $f(n)$ is divisible by $m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that $m$ divides $f(n)$ for some integer $n$ and some $2 \\leq m \\leq 2010$. As $f(1) = 2011$ and $2011$ is a prime number, $m$ cannot be a divisor of $f(1)$, so we may restrict ourselves to the case $n \\neq 1$.\n\nIn this case, we can write\n$$\nf(n) = \\frac{n^{2011} - 1}{n - 1}.\n$$\nLet $p$ be a prime divisor of $m$. Then $p \\mid f(n)$ implies $p \\mid n^{2011} - 1$, so\n$$\nn^{2011} \\equiv 1 \\pmod{p}.\n$$\nThis means $n$ and $p$ are coprime.\n\nBy the above, the order $\\text{ord}_p(n)$ of $n$ modulo $p$ (the smallest positive $k$ such that $n^k \\equiv 1 \\pmod{p}$) divides $2011$. Since $2011$ is prime, $\\text{ord}_p(n) \\in \\{1, 2011\\}$.\n\nIf $\\text{ord}_p(n) = 1$, then $n \\equiv 1 \\pmod{p}$, so $f(n) \\equiv 2011 \\pmod{p}$, which implies $p \\mid 2011$, so $p = 2011$, contradiction.\n\nThus $\\text{ord}_p(n) = 2011$. But $\\text{ord}_p(n)$ divides $\\varphi(p) = p-1$, so $2011$ divides $p-1$. But $p < 2011$, so $p-1 < 2010$, contradiction.\n\nTherefore, there is no non-negative integer $n$ such that $f(n)$ is divisible by $m$ for $2 \\leq m \\leq 2010$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16761,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $F$ be the family of all $n$-element subsets of $\\{1, 2, \\dots, 3n\\}$. For every $S \\subset \\{1, 2, 3, 4\\}$, denote the collection of all elements of $F$ whose intersection with $\\{1, 2, 3, 4\\}$ is $S$ by $F_S$. More formally,\n\n$$\nF_S = \\{A \\in F \\mid A \\cap \\{1, 2, 3, 4\\} = S\\}.\n$$\n\nNow we color all elements of $F$ as follows:\n\n- All elements of $F_\\emptyset$ with color $c_1$.\n- All elements of $F_{\\{1\\}}$ with color $c_2$.\n- All elements of $F_{\\{2\\}}$ with color $c_3$.\n- All elements of $F_{\\{3\\}}$ with color $c_4$.\n- All elements of $F_{\\{4\\}}$ with color $c_5$.\n- All elements of $F_{\\{1,2\\}} \\cup F_{\\{1,3\\}} \\cup F_{\\{1,2,3\\}}$ with color $c_6$.\n- All elements of $F_{\\{2,3\\}} \\cup F_{\\{3,4\\}} \\cup F_{\\{2,3,4\\}} \\cup F_{\\{1,2,3,4\\}}$ with color $c_7$.\n- All elements of $F_{\\{1,4\\}} \\cup F_{\\{2,4\\}} \\cup F_{\\{1,2,4\\}} \\cup F_{\\{1,3,4\\}}$ with color $c_8$.\n\nProve that this coloring has the required property: there are no three $n$-element subsets of the same color such that each two of them have at most one element in common (such a triple is called a *bad triple*).",
    "options": [],
    "answer": "See solution",
    "solution": "- $c_1$: If we have a bad triple of subsets, then their union has size at least $3n - 3$, which is not possible in $F_\\emptyset$, since none of $1,2,3,4$ appears in their union.\n- $c_2$: If we have a bad triple of subsets all having one element in common, then their union has size $3n - 2$, which is not possible in $F_{\\{1\\}}$, since none of $2,3,4$ appears in their union.\n- $c_3$: Similar proof as $c_2$.\n- $c_4$: Similar proof as $c_2$.\n- $c_5$: Similar proof as $c_2$.\n- $c_6$: If we have a bad triple of subsets in $F_{\\{1,2\\}} \\cup F_{\\{1,3\\}} \\cup F_{\\{1,2,3\\}}$, then all three contain $1$ as an element. Moreover, at least two of them contain $2$ or at least two of them contain $3$. Contradiction.\n- $c_7$: If we have a bad triple of subsets in $F_{\\{2,3\\}} \\cup F_{\\{3,4\\}} \\cup F_{\\{2,3,4\\}} \\cup F_{\\{1,2,3,4\\}}$, then all three contain $3$ as an element. Moreover, at least two of them contain $2$ or at least two of them contain $4$. Contradiction.\n- $c_8$: If we have a bad triple of subsets in $F_{\\{1,4\\}} \\cup F_{\\{2,4\\}} \\cup F_{\\{1,2,4\\}} \\cup F_{\\{1,3,4\\}}$, then all three contain $4$ as an element. Moreover, at least two of them contain $1$ or at least two of them contain $2$. Contradiction.\n\nSo no color has a bad triple of subsets and therefore our coloring is appropriate.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16762,
    "subject": "Mathematics (Olympiad)",
    "question": "Natural numbers $1$ through $n$ are written on a blackboard. On each move, one erases from the blackboard $2$ or more numbers whose sum is divisible by any of the chosen numbers and writes their sum on the blackboard. Two players make moves by turns, and the player who cannot move loses the game. Which player can win the game against any play by the opponent, if:\n\n(a) $n = 6$;\n\n(b) $n = 11$?",
    "options": [],
    "answer": "See solution",
    "solution": "(a) The first player can replace numbers $1, 2, 3, 6$ with $12$. After that, the blackboard contains numbers $4, 5, 12$. In this state, the sum of no two or three numbers on the blackboard is divisible by all the added numbers. Thus, the second player cannot move and the first player wins immediately.\n\n(b) The first player can replace numbers $1, 2, 3, 4, 6, 8$ with $24$. After that, the blackboard contains numbers $5, 7, 9, 10, 11, 24$, which sum up to $66$. Among numbers $7, 9, 10, 11, 24$, the least common multiple of any two numbers is greater than $66$. Thus, when choosing two or more numbers from among the mentioned numbers, and perhaps also the number $5$, the sum of the chosen numbers is less than their least common multiple and cannot be divisible by all of them. Also, when choosing one of the mentioned numbers together with $5$, the sum of the chosen numbers is not divisible by the larger one. Hence, the second player cannot move and the first player wins immediately.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16763,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n$ cities in a country. A monopoly company intends to establish air traffic between some of these cities. The government demands that the air traffic satisfies the following conditions:\n\n- People should be able to get from each city to any other of these cities by one or several flights.\n- There must be an equal number of flights from each city.\n\nNote that two flights — from $A$ to $B$ and from $B$ to $A$ — are considered as one flight between cities $A$ and $B$. Moreover, there may be at most one flight between two cities. But the company wants to cancel as few flights as possible so that you can *not* get from each city to any other city. What is the minimal number of flights the company has to cancel if:\n\na) $n = 2008$;\n\nb) $n = 2007$?",
    "options": [],
    "answer": "See solution",
    "solution": "a) $1$; b) $2$.\n\n**Solution.**\n\na) Let's divide all the cities into two groups consisting of $1003$ and $1005$ cities respectively. Connect all the cities in each group in a cycle, so each vertex has degree $2$. Next, select one city in each component and connect these two cities with a flight \"007\". Connect the rest of the cities with additional diagonals inside the group so that the degree of each vertex is $3$. It's clear that you just have to cancel flight \"007\" and the problem is solved. Thus, the answer is $1$.\n\nb) Let this minimum be $\\Delta$. It's obvious that $\\Delta \\leq 2$. To prove this, connect the cities in any cycle, so there are only two flights from each city. If you exclude any two flights, the system falls into two groups of cities, which are not connected to each other. Let's show that in this case $\\Delta = 2$.\n\n![](images/Ukrajina_2008_p18_data_a8a760b931.png)\n\nBy contradiction, assume $\\Delta = 1$. If you exclude this flight, the system falls into two groups of connected cities. In one of these groups there are $2m$ cities, i.e., an even number of cities. If the degree of each vertex was $k$ before we excluded the flight, in all there would be $\\frac{1}{2}((2m-1)k + (k-1)) = \\frac{1}{2}(2mk - 1)$ flights in this group. But this is not an integer as it should be.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16764,
    "subject": "Mathematics (Olympiad)",
    "question": "Two teams are in a best-two-out-of-three playoff: the teams will play at most 3 games, and the winner of the playoff is the first team to win 2 games. The first game is played on Team A's home field, and the remaining games are played on Team B's home field. Team A has a $\\frac{2}{3}$ chance of winning at home, and its probability of winning when playing away from home is $p$. Outcomes of the games are independent. The probability that Team A wins the playoff is $\\frac{1}{2}$. Then $p$ can be written in the form $\\frac{1}{2}(m - \\sqrt{n})$, where $m$ and $n$ are positive integers. What is $m + n$?\n\n![](path/to/file.png)\n\n(A) 10  (B) 11  (C) 12  (D) 13  (E) 14",
    "options": [],
    "answer": "See solution",
    "solution": "There are three ways for Team A to win the playoff:\n\n1. Win the first two games.\n2. Win the first game, lose the second game, and win the third game.\n3. Lose the first game and win the second and third games.\n\nThe probability that Team A wins in one of these ways is:\n\n$$\n\\frac{2}{3} \\cdot p + \\frac{2}{3} \\cdot (1-p) \\cdot p + \\frac{1}{3} \\cdot p^2 = -\\frac{1}{3}p^2 + \\frac{4}{3}p.\n$$\n\nSetting this equal to $\\frac{1}{2}$ and simplifying gives:\n\n$$\n-\\frac{1}{3}p^2 + \\frac{4}{3}p = \\frac{1}{2}\n$$\n\nMultiply both sides by 6:\n\n$$\n-2p^2 + 8p = 3\n$$\n\nSo:\n\n$$\n2p^2 - 8p + 3 = 0\n$$\n\nUsing the quadratic formula:\n\n$$\np = \\frac{8 \\pm \\sqrt{64 - 24}}{4} = \\frac{8 \\pm \\sqrt{40}}{4} = \\frac{8 \\pm 2\\sqrt{10}}{4} = \\frac{1}{2}(4 \\pm \\sqrt{10})\n$$\n\nSince $p$ must be less than 1, we take the minus sign:\n\n$$\np = \\frac{1}{2}(4 - \\sqrt{10})\n$$\n\nSo $m = 4$ and $n = 10$, and $m + n = 14$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16765,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a $3 \\times n$ rectangle, consider covering unit squares by overlaying rectangles in various shifted positions. For each $n$, what is the largest integer $N$ such that it is **not possible** to cover exactly $N$ unit squares using these constructions?",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze four families of coverings:\n\n1. Adding rectangles shifted to the right covers $3n + 3k$ unit squares for $k \\geq 0$.\n2. Overlaying the original rectangle with one shifted both across 1 unit and down 1 unit covers $4n + 2$ squares; adding shifted rectangles gives $4n + 2 + 3k$ squares for $k \\geq 0$.\n3. Overlaying with one shifted down 1 unit covers $4n$ squares; adding shifted rectangles gives $4n + 3k$ squares for $k \\geq 0$.\n4. Overlaying with three shifts (down 1, across 1, both) covers $4(n+1) = 4n + 4$ squares; adding shifted rectangles gives $4n + 4 + 3k$ squares for $k \\geq 0$.\n\nThe congruence classes modulo 3:\n- $3n + 3k \\equiv 0 \\pmod{3}$\n- $4n + 2 + 3k \\equiv n + 2 \\pmod{3}$\n- $4n + 3k \\equiv n \\pmod{3}$\n- $4n + 4 + 3k \\equiv n + 1 \\pmod{3}$\n\nFor each $n \\pmod{3}$:\n1. If $n \\equiv 0 \\pmod{3}$, all numbers from $4n + 2$ upwards can be covered except $4n + 1$.\n2. If $n \\equiv 1 \\pmod{3}$, all numbers from $4n + 2$ upwards can be covered except $4n + 1$.\n3. If $n \\equiv 2 \\pmod{3}$, all numbers from $4n$ upwards can be covered except $4n - 1$.\n\nProofs by contradiction show that $4n + 1$ (for $n \\equiv 0, 1 \\pmod{3}$) and $4n - 1$ (for $n \\equiv 2 \\pmod{3}$) cannot be covered by any arrangement. Thus, the largest impossible $N$ is:\n- $N = 4n + 1$ if $n \\equiv 0$ or $1 \\pmod{3}$\n- $N = 4n - 1$ if $n \\equiv 2 \\pmod{3}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16766,
    "subject": "Mathematics (Olympiad)",
    "question": "In a triangle $ABC$ we have $|AB| = 2|AC|$. Let $D$ and $E$ be the two points on the segments $AB$ and $BC$, such that $\\angle BAE = \\angle ACD$. The segments $AE$ and $CD$ intersect at $F$, and $CFE$ is an equilateral triangle. Find the angles of the triangle $ABC$.\n\n![](images/Slovenija_2009_p7_data_fa47d30a8e.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $CEF$ is an equilateral triangle, we have $\\angle EFC = 60^\\circ$. This implies that $\\angle CFA = 120^\\circ$, so $\\angle FAC = 180^\\circ - \\angle CFA - \\angle ACF = 60^\\circ - \\angle ACF$ and $\\angle BAC = \\angle BAE + \\angle FAC = \\angle BAE + 60^\\circ - \\angle ACD = 60^\\circ$. We have $|AB| = 2|AC|$ and $\\angle BAC = 60^\\circ$, so the triangle $ABC$ is one half of an equilateral triangle and we see that $\\angle CBA = 30^\\circ$ and $\\angle ACB = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16767,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y, z > 0$ such that $x + y + z = 6xyz$. Prove that\n$$\n\\frac{x}{\\sqrt{x^2 + 2yz + 1}} + \\frac{y}{\\sqrt{y^2 + 2zx + 1}} + \\frac{z}{\\sqrt{z^2 + 2xy + 1}} \\ge 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The AM-HM inequality and the given condition yield\n$$\nxy + yz + zx = xyz \\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right) \\ge xyz \\cdot \\frac{9}{x + y + z} = \\frac{9xyz}{18xyz} = \\frac{1}{2}.\n$$\n\nUsing $1 \\le 2xy + 2yz + 2zx$, we get\n$$\nx^2 + 2yz + 1 \\le x^2 + 2xy + 2zx + 4yz = (x + 2y)(x + 2z) \\le (x + y + z)^2,\n$$\nwhere the last inequality follows from the AM-GM inequality. Thus,\n$$\n\\frac{x}{\\sqrt{x^2 + 2yz + 1}} \\ge \\frac{x}{x + y + z},\n$$\nand similarly for the other terms. Adding these three inequalities gives the result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16768,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m, n$ be positive integers with $m \\ge n$, and let $S$ be the set of all ordered $n$-tuples $(a_1, a_2, \\dots, a_n)$ of positive integers such that $a_1 + a_2 + \\dots + a_n = m$. Show that\n\n$$\n\\sum_{S} 1^{a_1} 2^{a_2} \\cdots n^{a_n} = \\binom{n}{n} n^m - \\binom{n}{n-1} (n-1)^m + \\cdots + (-1)^{n-2} \\binom{n}{2} 2^m + \\cdots + (-1)^{n-1} \\binom{n}{1}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution 1.** Let $m = k + n$, and let $T$ be the set of all $n$-term sequences of nonnegative integers $(b_1, b_2, \\dots, b_n)$ such that $b_1 + b_2 + \\dots + b_n = k$. It suffices for us to show that\n\n$$\nn! \\sum_{T} 1^{b_1} 2^{b_2} \\cdots n^{b_n} = \\binom{n}{n} n^{k+n} - \\binom{n}{n-1} (n-1)^{k+n} + \\cdots + (-1)^{n-2} \\binom{n}{2} 2^{k+n} + \\cdots + (-1)^{n-1} \\binom{n}{1}.\n$$\n\nWe claim that both sides of the desired equation count the number of ways to color $k+n$ objects with $n$ colors such that each color is used at least once. For $1 \\le t \\le n$, the number of ways to color $k+n$ objects with $t$ colors is $t^{k+n}$, so the principle of inclusion-exclusion shows exactly that the right hand side counts these colorings.\n\nIt suffices to show that the left hand side also counts these colorings. Label the objects $1, 2, \\dots, k+n$ in some order, and, for $1 \\le i \\le n$, let $c_i$ be the smallest object of color $i$. Take $b_i = c_{i+1} - c_i$, where we let $c_{n+1} = n+k+1$. Then, notice that any such coloring is specified uniquely by the following data: (a) the order in which the colors first appear, (b) the number of objects between $c_i$ and $c_{i+1}$ for each $i$, where we take $c_{n+1} = n+k+1$, and (c) the colors of the objects between $c_i$ and $c_{i+1}$. We now count how many ways these data can be chosen. There are $n!$ choices for datum (a), and it is independent of (b) and (c). Datum (b) is specified uniquely by a choice of $(b_1, b_2, \\dots, b_n)$ such that $b_1 + b_2 + \\dots + b_n = k$, that is, an element of $T$. Given such a choice of $(b_1, b_2, \\dots, b_n)$, there are $1^{b_1} 2^{b_2} \\dots n^{b_n}$ ways to color the intermediate objects, as each of the $b_i$ between $c_i$ and $c_{i+1}$ admits $i$ choices of color since only $i$ colors appear before $c_{i+1}$. Summing over all choices of (a), (b), and (c), we see that the number of colorings of this type is\n\n$$\nn! \\sum_{T} 1^{b_1} 2^{b_2} \\dots n^{b_n},$$\n\nwhich matches the left hand side of the desired equation, completing the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16769,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose integer $n \\geq 4$. Prove that if $n$ divides $2^n - 2$, then $\\dfrac{2^n - 2}{n}$ is composite.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the integer $y = \\dfrac{2^n - 2}{n}$.\n\nIf $n$ is odd, then $2^n - 2$ is even, so $y$ is even. Since $n \\geq 4$, $y > 2$, so $y$ is composite.\n\nNow consider $n$ even, let $n = 2m$ with $m > 1$.\n\nSince $y = \\dfrac{2^{2m} - 2}{2m} = \\dfrac{2^{2m-1} - 1}{m}$ is an integer, $m$ is odd.\n\nLet $\\delta$ be the multiplicative order of $2$ modulo $m$. Then $\\delta < m$ and $\\delta \\mid 2m - 1$ since $m \\mid 2^{2m-1} - 1$.\n\nLet $2m - 1 = \\delta r$. Since $\\delta < m < 2m - 1$, $r > 1$.\n\n**Case 1:** If $m \\neq 2^\\delta - 1$, then $m < 2^\\delta - 1$ since $m \\mid 2^\\delta - 1$. Then:\n\n$$\ny = \\frac{2^{2m-1}-1}{m} = \\frac{2^{\\delta r}-1}{m} = \\frac{2^{\\delta r}-1}{2^{\\delta}-1} \\cdot \\frac{2^{\\delta}-1}{m}\n$$\n\nSince $r > 1$, this is a product of two integers greater than $1$, so $y$ is composite.\n\n**Case 2:** If $m = 2^\\delta - 1$, then $2(2^\\delta - 1) - 1 = 2m - 1 = \\delta r$. Since $m > 1$, $\\delta > 1$.\n\n$$\nr = \\frac{2^{\\delta+1} - 3}{\\delta} > \\delta.\n$$\n\nSince $2^\\delta - 1 \\mid 2^{\\delta r} - 1$ and $2^r - 1 \\mid 2^{\\delta r} - 1$, $2^\\delta - 1$ is a multiple of $[2^\\delta - 1, 2^r - 1]$, namely:\n\n$$\n\\frac{(2^\\delta - 1)(2^r - 1)}{\\gcd(2^\\delta - 1, 2^r - 1)} \\mid 2^{\\delta r} - 1.\n$$\n\nNote $\\gcd(2^\\delta - 1, 2^r - 1) = 2^{\\gcd(\\delta, r)} - 1$. Therefore,\n\n$$\ny = \\frac{2^{2m-1}-1}{m} = \\frac{2^{\\delta r}-1}{2^{\\delta}-1} = \\frac{(2^{\\delta r}-1)(2^{\\gcd(\\delta, r)}-1)}{(2^{\\delta}-1)(2^r-1)} \\cdot \\frac{2^r-1}{2^{\\gcd(\\delta, r)}-1}\n$$\n\nis the product of two integers.\n\nBecause $r > \\delta$, $\\frac{2^r - 1}{2^{\\gcd(\\delta, r)} - 1} \\geq \\frac{2^r - 1}{2^\\delta - 1} > 1$. Since $\\delta \\geq 2$,\n\n$$\n\\begin{align*}\n\\frac{(2^{\\delta r} - 1)(2^{\\gcd(\\delta, r)} - 1)}{(2^{\\delta} - 1)(2^r - 1)} &\\geq \\frac{(2^{2r} - 1) \\cdot 1}{(2^{\\delta} - 1)(2^r - 1)} \\\\\n&> \\frac{2^{2r} - 1}{(2^r - 1)^2} \\\\\n&= \\frac{2^r + 1}{2^r - 1} > 1.\n\\end{align*}\n$$\n\nTherefore, $y$ is the product of two integers greater than $1$, so $y$ is composite.\n\nIn summary, the conclusion is confirmed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16770,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n$ cities ($n > 3$) and two airline companies in a country. Between any two cities, there is exactly one 2-way flight connecting them, which is operated by one of the two companies. A female mathematician plans a travel route so that it starts and ends at the same city, passes through at least two other cities, and each city on the route is visited once. She finds out that wherever she starts and whatever route she chooses, she must take flights of both companies. Find the maximum value of $n$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Consider each city as a vertex and each airline as an edge colored according to the airline company. The airline route chart is thus a two-color complete graph with $n$ vertices. By the problem statement, any cycle contains edges of both colors; that is, the subgraph of each color has no cycle (i.e., is a forest). It is well-known that a simple graph without cycles (a forest) has at most $n-1$ edges. Thus, the number of edges of the same color is at most $n-1$, so the total number of edges is at most $2(n-1)$. On the other hand, a complete graph with $n$ vertices has $\\frac{n(n-1)}{2}$ edges. Therefore,\n\n$$\n\\frac{n(n-1)}{2} \\leq 2(n-1)\n$$\n\nwhich simplifies to $n(n-1) \\leq 4(n-1)$, so $n \\leq 4$ (since $n > 3$).\n\nIf $n=4$, denote the four cities by $A$, $B$, $C$, and $D$. Let three routes $AB$, $BC$, and $CD$ be operated by the first company, and the other three routes $AC$, $AD$, and $BD$ by the second company. In this arrangement, the route chart of each company has no cycle. Therefore, the maximum value of $n$ is $4$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16771,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c$ be positive real numbers such that $a + b + c = 3$. Find the minimum value of the expression\n\n$$\nA = \\frac{2 - a^3}{a} + \\frac{2 - b^3}{b} + \\frac{2 - c^3}{c}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We rewrite $A$ as follows:\n\n$$\n\\begin{align*}\nA &= \\frac{2 - a^3}{a} + \\frac{2 - b^3}{b} + \\frac{2 - c^3}{c} \\\\\n  &= 2\\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right) - (a^2 + b^2 + c^2) \\\\\n  &= 2\\frac{ab + bc + ca}{abc} - (a^2 + b^2 + c^2) \\\\\n  &= 2\\frac{ab + bc + ca}{abc} - \\left((a + b + c)^2 - 2(ab + bc + ca)\\right) \\\\\n  &= 2\\frac{ab + bc + ca}{abc} - (9 - 2(ab + bc + ca)) \\\\\n  &= 2\\frac{ab + bc + ca}{abc} + 2(ab + bc + ca) - 9 \\\\\n  &= 2(ab + bc + ca)\\left(\\frac{1}{abc} + 1\\right) - 9\n\\end{align*}\n$$\n\nRecall the well-known inequality $(x + y + z)^2 \\ge 3(xy + yz + zx)$. Set $x = ab$, $y = bc$, $z = ca$, to obtain $(ab + bc + ca)^2 \\ge 3abc(a + b + c) = 9abc$ (since $a + b + c = 3$). Taking square roots gives:\n\n$$\nab + bc + ca \\geq 3\\sqrt{abc}. \\tag{1}\n$$\n\nBy the AM-GM inequality:\n\n$$\n\\frac{1}{abc} + 1 \\ge 2\\sqrt{\\frac{1}{abc}}. \\tag{2}\n$$\n\nMultiplying (1) and (2):\n\n$$\n(ab + bc + ca)\\left(\\frac{1}{abc} + 1\\right) \\ge 3\\sqrt{abc} \\cdot 2\\sqrt{\\frac{1}{abc}} = 6.\n$$\n\nSo $A \\ge 2 \\cdot 6 - 9 = 3$, and equality holds if and only if $a = b = c = 1$. Thus, the minimum value is $3$.\n\n*Remark.* If $f(x) = \\frac{2 - x^3}{x}$ for $x \\in (0, 3)$, then $f''(x) = \\frac{4}{x^3} - 2$, so the function is convex on $x \\in (0, \\sqrt{3})$ and concave on $x \\in (\\sqrt{3}, 3)$. Thus, Jensen's inequality does not apply.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16772,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Consider $n$ non-attacking rooks placed on an $n \\times n$ chessboard, with the cells of the main diagonal having coordinates $(k, k)$ for $k = 1, \\dots, n$. Let $(k, f(k))$ be the coordinates of the $k$-th rook, where $f$ is a permutation of $\\{1, \\dots, n\\}$. Suppose the rooks are subject to color restrictions such that\n\n$$\n\\sum_{k=1}^{n} (f(k) - k)^2 = \\sum_{i=0}^{n-1} i^2 = \\frac{n(n-1)(2n-1)}{6}.\n$$\n\nGiven that the rooks are non-attacking, we also have\n\n$$\n\\sum_{k=1}^{n} (f(k))^2 = \\sum_{i=1}^{n} i^2 = \\frac{n(n+1)(2n+1)}{6}.\n$$\n\nFind all $n$ for which\n\n$$\n\\sum_{k=1}^{n} k f(k)\n$$\n\nis an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Subtracting the two given equalities, we get:\n\n$$\n\\sum_{k=1}^{n} (f(k))^2 - \\sum_{k=1}^{n} (f(k) - k)^2 = \\frac{n(n+1)(2n+1)}{6} - \\frac{n(n-1)(2n-1)}{6}.\n$$\n\nExpanding $(f(k) - k)^2$ gives $f(k)^2 - 2k f(k) + k^2$, so the difference simplifies to:\n\n$$\n\\sum_{k=1}^{n} [f(k)^2 - (f(k)^2 - 2k f(k) + k^2)] = \\sum_{k=1}^{n} [2k f(k) - k^2].\n$$\n\nThus,\n\n$$\n\\sum_{k=1}^{n} [2k f(k) - k^2] = \\frac{n(n+1)(2n+1) - n(n-1)(2n-1)}{6}.\n$$\n\nSolving for $\\sum_{k=1}^{n} k f(k)$:\n\n$$\n2 \\sum_{k=1}^{n} k f(k) = \\frac{n(n+1)(2n+1) - n(n-1)(2n-1)}{6} + \\sum_{k=1}^{n} k^2\n$$\n\n$$\n\\sum_{k=1}^{n} k f(k) = \\frac{1}{2} \\left[ \\frac{n(n+1)(2n+1) - n(n-1)(2n-1)}{6} + \\frac{n(n+1)(2n+1)}{6} \\right]\n$$\n\nSimplifying the numerator:\n\n$$\n(n(n+1)(2n+1) - n(n-1)(2n-1)) + n(n+1)(2n+1)\n$$\n\nCompute $n(n+1)(2n+1) - n(n-1)(2n-1)$:\n\nLet $A = n(n+1)(2n+1)$, $B = n(n-1)(2n-1)$.\n\n$A - B = n[(n+1)(2n+1) - (n-1)(2n-1)]$\n\n$(n+1)(2n+1) = 2n^2 + 3n + 1$\n\n$(n-1)(2n-1) = 2n^2 - 3n + 1$\n\nSo difference is $[2n^2 + 3n + 1] - [2n^2 - 3n + 1] = 6n$\n\nSo $A - B = n \\cdot 6n = 6n^2$\n\nSo the sum is:\n\n$$\n\\sum_{k=1}^{n} k f(k) = \\frac{1}{2} \\left[ \\frac{6n^2}{6} + \\frac{n(n+1)(2n+1)}{6} \\right] = \\frac{1}{2} [n^2 + \\frac{n(n+1)(2n+1)}{6}]\n$$\n\n$$\n= \\frac{n^2}{2} + \\frac{n(n+1)(2n+1)}{12}\n$$\n\nThe original statement gives $\\sum_{k=1}^{n} k f(k) = \\frac{n(2n^2 + 9n + 1)}{12}$.\n\nNow, $\\frac{n(2n^2 + 9n + 1)}{12}$ is an integer if and only if $n \\equiv 0 \\text{ or } 1 \\pmod{4}$.\n\n**Answer:** The sum $\\sum_{k=1}^{n} k f(k)$ is an integer if and only if $n \\equiv 0 \\text{ or } 1 \\pmod{4}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 16773,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a convex $n$-gon ($n \\ge 5$) $\\Omega: P_1P_2\\ldots P_n$, of which no three diagonals are concurrent inside $\\Omega$. Prove that one can choose a point inside every quadrilateral $P_iP_jP_kP_l$ ($1 \\le i < j < k < l \\le n$) and not on any diagonal of $\\Omega$, such that the $C_n^4$ points obtained are distinct and the segment connecting any two of them intersects with at least one diagonal of $\\Omega$.",
    "options": [],
    "answer": "See solution",
    "solution": "To begin, notice that the diagonals of $\\Omega$ divide the polygon into $C_n^4 + C_{n-1}^2$ small regions. Every quadrilateral $P_iP_jP_kP_l$ ($1 \\le i < j < k < l \\le n$) has a one-to-one correspondence with the intersection of the diagonals $P_iP_k$ and $P_jP_l$, and additionally, every intersection is adjacent to four small regions. Therefore, it is enough to assign every intersection to one of the four adjacent small regions, such that the $C_n^4$ assigned small regions are all distinct.\n\nPlace $\\Omega$ in the Cartesian coordinate plane such that no diagonal is parallel to the $x$-axis. If two diagonals $P_iP_k$, $P_jP_l$ meet at $Q$, then we assign $Q$ to the unique small region incident to $Q$ and whose interior points all have $y$ coordinates larger than that of $Q$ (intuitively, above $Q$). Obviously, different intersections correspond to different small regions, as for each small region $R$, there is a unique vertex with the minimum $y$ coordinate, and this vertex is assigned to $R$. This completes the proof. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16774,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $p$ 為一奇質數。對每一個正整數 $a$，定義 $S_a$ 如下：\n\n$$\nS_a = \\frac{a}{1} + \\frac{a^2}{2} + \\cdots + \\frac{a^{p-1}}{p-1}.\n$$\n\n令 $m$ 與 $n$ 為正整數，使得\n\n$$\nS_3 + S_4 - 3S_2 = \\frac{m}{n}.\n$$\n\n試證：$p$ 能整除 $m$。",
    "options": [],
    "answer": "See solution",
    "solution": "對於有理數 $p_1/q_1$ 與 $p_2/q_2$，其中 $p \\nmid q_1,\\ p \\nmid q_2$。若 $p \\mid (p_1q_2 - p_2q_1)$，記為 $p_1/q_1 \\equiv p_2/q_2 \\pmod{p}$。\n\n由 $S_a$ 模 $p$ 開始。觀察：$p \\mid \\binom{p}{k}$，$k = 1, \\ldots, p-1$，且\n\n$$\n\\begin{aligned}\n\\frac{1}{p} \\binom{p}{k} &= \\frac{(p-1)(p-2)\\cdots(p-k+1)}{k!} \\\\\n&\\equiv \\frac{(-1)\\cdot(-2)\\cdots(-k+1)}{k!} = (-1)^{k-1} \\pmod{p}.\n\\end{aligned}\n$$\n\n則有\n\n$$\nS_a = - \\sum_{k=1}^{p-1} \\frac{(-a)^k (-1)^{k-1}}{k} \\equiv - \\sum_{k=1}^{p-1} (-a)^k \\cdot \\frac{1}{p} \\binom{p}{k} \\pmod{p}.\n$$\n\n上式右邊是一整數。由二項式定理，得\n\n$$\n- \\sum_{k=1}^{p-1} (-a)^k \\cdot \\frac{1}{p} \\binom{p}{k} = - \\frac{1}{p} \\left( -1 - (-a)^p + \\sum_{k=1}^{p} (-a)^k \\binom{p}{k} \\right) = \\frac{(a-1)^p - a^p + 1}{p}\n$$\n\n因 $p$ 是奇數，故\n\n$$\nS_a \\equiv \\frac{(a-1)^p - a^p + 1}{p} \\pmod{p}.\n$$\n\n最後，可得\n\n$$\n\\begin{aligned}\nS_3 + S_4 - 3S_2 &\\equiv \\frac{(2^p - 3^p + 1) + (3^p - 4^p + 1) - 3(1^p - 2^p + 1)}{p} \\\\\n&= \\frac{4 \\cdot 2^p - 4^p - 4}{p} = -\\frac{(2^p - 2)^2}{p} \\pmod{p}.\n\\end{aligned}\n$$\n\n由 Fermat 定理，$p \\mid (2^p - 2)$，故 $p^2 \\mid (2^p - 2)^2$。所以\n\n$$\nS_3 + S_4 - 3S_2 \\equiv 0 \\pmod{p}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16775,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $I$ be the incenter of $K$. The incenters of $K$ and $K_1$ both lie on the bisector of $\\angle ABC$. The incenters of $K$ and $K_2$ both lie on the bisector of $\\angle ACB$. Note that\n$$\n\\angle BIC = 180^\\circ - \\frac{1}{2}(\\angle ABC + \\angle ACB) = 90^\\circ + \\frac{1}{2}\\angle BAC.\n$$\n\n![](images/Ireland_2015_Booklet_p29_data_3ccdfa9847.png)\n\nFind the two angles between the radical axes of $K$ and $K_1$, and of $K$ and $K_2$, given that $\\angle BAC = 100^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Because the radical axis of two circles is perpendicular to the line connecting their centers, the radical axis of $K$ and $K_1$ is perpendicular to the bisector of $\\angle ABC$, and the radical axis of $K$ and $K_2$ is perpendicular to the bisector of $\\angle ACB$. Thus, one of the angles between these two radical axes is equal to $\\angle BIC$, which does not depend on the position of $P$. If $\\angle BAC = 100^\\circ$, then\n$$\n\\angle BIC = 90^\\circ + 50^\\circ = 140^\\circ.\n$$\nTherefore, the two angles between the two radical axes are $140^\\circ$ and $40^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16776,
    "subject": "Mathematics (Olympiad)",
    "question": "a) A convex heptagon is divided into triangles by drawing its diagonals. Prove that in such a case you can obtain 5 or 7 triangles, but cannot get 6 triangles.\n\nb) Prove that there exists a nonconvex heptagon that can be divided by internal diagonals into 6 triangles. An internal diagonal of a polygon $M$ is a segment connecting two non-neighbouring vertices of $M$ and does not go beyond the figure.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Division into 5 and 7 triangles is shown in the figures below.\n\nLet us show that a heptagon cannot be cut into 6 triangles.\n\nIf all vertices of the resulting triangles coincide with the vertices of the heptagon, the sum of the angles of the triangles equals the sum of the angles of the heptagon and equals $5 \\cdot 180^\\circ$. Hence, there are only 5 triangles.\n\nIf two diagonals intersect not at a vertex of the heptagon, then this intersection point $A$ lies inside the heptagon because of its convexity. Then the angles of triangles adjacent to point $A$ sum up to $360^\\circ$, and the sum of the angles of the resulting triangles is not less than $5 \\cdot 180^\\circ + 360^\\circ = 7 \\cdot 180^\\circ$. Thus, in this case, we will have at least 7 triangles.\n\nb) The heptagon and corresponding division are shown in the figure below. Note that the horizontal diagonal lies through the vertex of the heptagon.\n\n![](images/UkraineMO_2015-2016_booklet_p31_data_16ccbff6e0.png)\n![](images/UkraineMO_2015-2016_booklet_p31_data_8f4720fa6e.png)\n![](images/UkraineMO_2015-2016_booklet_p31_data_e24d28df81.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16777,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the least positive integer $n$ such that $\\sqrt[5]{5n}$, $\\sqrt[6]{6n}$, and $\\sqrt[7]{7n}$ are all integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = 2^\\alpha \\cdot 3^\\beta \\cdot 5^\\gamma \\cdot 7^\\delta \\cdot s$, where $s$ is not divisible by $2$, $3$, $5$, or $7$. Then:\n\n- $5n = 2^\\alpha \\cdot 3^\\beta \\cdot 5^{\\gamma+1} \\cdot 7^\\delta \\cdot s$\n- $6n = 2^{\\alpha+1} \\cdot 3^{\\beta+1} \\cdot 5^\\gamma \\cdot 7^\\delta \\cdot s$\n- $7n = 2^\\alpha \\cdot 3^\\beta \\cdot 5^\\gamma \\cdot 7^{\\delta+1} \\cdot s$\n\nFor each root to be an integer:\n\n- For $\\sqrt[5]{5n}$: $\\alpha$, $\\beta$, $\\gamma+1$, $\\delta$ must be divisible by $5$.\n- For $\\sqrt[6]{6n}$: $\\alpha+1$, $\\beta+1$, $\\gamma$, $\\delta$ must be divisible by $6$.\n- For $\\sqrt[7]{7n}$: $\\alpha$, $\\beta$, $\\gamma$, $\\delta+1$ must be divisible by $7$.\n\nSolving these congruences:\n\n- $\\alpha$ and $\\beta$ must be divisible by $35$ (LCM of $5$ and $7$).\n- $\\gamma$ must be divisible by $42$ (LCM of $6$ and $7$), and $\\gamma+1$ divisible by $5$.\n- $\\delta$ must be divisible by $30$ (LCM of $5$ and $6$), and $\\delta+1$ divisible by $7$.\n\nThe least suitable values are:\n\n- $\\alpha = 35$\n- $\\beta = 35$\n- $\\gamma = 84$ (since $84$ is divisible by $42$ and $85$ is divisible by $5$)\n- $\\delta = 90$ (since $90$ is divisible by $30$ and $91$ is divisible by $7$)\n\nTake $s = 1$ for minimality. Thus, the least positive integer $n$ is:\n\n$$2^{35} \\cdot 3^{35} \\cdot 5^{84} \\cdot 7^{90}$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16778,
    "subject": "Mathematics (Olympiad)",
    "question": "Fix an integer $n \\ge 2$. Find all $n$-tuples $(a_1, a_2, \\dots, a_n)$ of integers satisfying the following two conditions:\n\n1. $a_1$ is odd, $1 < a_1 \\le a_2 \\le \\dots \\le a_n$, and $M = \\frac{1}{2^n}(a_1 - 1)a_2 \\dots a_n$ is an integer.\n\n2. There exist $M$ different $n$-tuples $(c_{i,1}, c_{i,2}, \\dots, c_{i,n})$ with $i = 1, 2, \\dots, M$, such that for all $1 \\le i < j \\le M$, there exists $k \\in \\{1, 2, \\dots, n\\}$ such that\n\n$$\nc_{i,k} - c_{j,k} \\not\\equiv -1, 0, 1 \\pmod{a_k}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The $n$-tuples we seek are those for which, if there are exactly $r$ odd numbers in $a_2, \\dots, a_n$, then $2^r$ divides $a_1 - 1$.\n\n**Necessity:**\nSuppose $a_1, \\dots, a_r$ are odd and $a_{r+1}, \\dots, a_n$ are even. Given the $M$ tuples, for each $s \\in \\mathbb{Z}$, let $B_s = \\{i \\mid 1 \\le i \\le M, c_{i,n} \\equiv s \\pmod{a_n}\\}$. Then\n$$\n|B_1| + |B_2| + \\dots + |B_{a_n}| = M.\n$$\nSo some $s$ has $|B_s| + |B_{s+1}| \\ge \\frac{M}{a_n/2}$, meaning at least $\\frac{M}{a_n/2}$ tuples have $n$-th coordinates differing by $0$ or $\\pm 1$ mod $a_n$. Repeating for all coordinates, there are at least $\\frac{M}{\\frac{a_n}{2} \\cdots \\frac{a_2}{2}} = \\frac{a_1-1}{2}$ tuples with all $k$th coordinates differing by $0$ or $\\pm 1$ mod $a_k$ ($2 \\le k \\le n$). But the first coordinates cannot differ by $0$ or $\\pm 1$ mod $a_1$, so there are at most $\\frac{a_1-1}{2}$ such tuples. Thus, all equalities must hold, and for $t = r+1$:\n$$\n\\frac{M}{\\frac{a_n}{2} \\cdots \\frac{a_{r+1}}{2}} = \\frac{1}{2^r}(a_1-1)a_2 \\cdots a_r \\in \\mathbb{Z},\n$$\nforcing $2^r \\mid a_1-1$.\n\n**Sufficiency (Construction):**\nIf $2^r \\mid a_1-1$, we construct the $M$ tuples. If any $a_k$ ($k \\ge 2$) is even, say $a_n$, and we have tuples for $a_1, \\dots, a_{n-1}$, then for each such tuple $(c_{i,1}, \\dots, c_{i,n-1})$ and $1 \\le c \\le \\frac{a_n}{2}$, take $(c_{i,1}, \\dots, c_{i,n-1}, 2c)$.\n\nIf all $a_2, \\dots, a_n$ are odd, write $a_1 = 2^n t + 1$ and consider $a_1 = a_2 = \\dots = a_n$. Then $M = t a_1^{n-1}$. Define $f(x_1, \\dots, x_{n-1}) = \\sum_{i=1}^{n-1} 2^i x_i$ and take tuples:\n$$\n(x_1, \\dots, x_{n-1}, f(x_1, \\dots, x_{n-1}) + 2^n s),\n$$\nwhere $x_1, \\dots, x_{n-1} \\in \\{1, \\dots, a_1\\}$, $s \\in \\{1, \\dots, t\\}$. If two tuples differ by $0$ or $\\pm 1$ in every coordinate, then $s = s'$ and $x_i = x'_i$ for all $i$, so the tuples are equal.\n\nFor general odd $a_1, \\dots, a_n$, if $a_2 > a_1$, reduce to $a_1, a_2-2, a_3, \\dots, a_n$ by induction. The construction extends by adding tuples for $x_2 = a_2-2$ and $x_2 = a_2-1$ as described, maintaining the required properties.\n\nThus, the condition $2^r \\mid a_1-1$ is necessary and sufficient.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16779,
    "subject": "Mathematics (Olympiad)",
    "question": "Докажите, что из каждой нефинальной ситуации по стрелкам можно дойти до финальной. Ситуация называется *критической*, если у одного из игроков ровно одна карта, и *регулярной*, если у обоих игроков хотя бы по две карты. Проведём стрелку от каждой ситуации к ситуациям, которые могут из неё получиться после одного хода. Тогда из любой нефинальной ситуации ведут две стрелки, а из любой финальной — ноль.",
    "options": [],
    "answer": "See solution",
    "solution": "Выясним, сколько стрелок ведут в каждую ситуацию. Предположим, что она получилась в результате какого-то хода, в котором карты взял первый игрок. Тогда у него оказалось хотя бы две карты, которые лежат в конце колоды; при этом известно, что одна из них $a$ бьёт другую $b$. Значит, в этом случае карта $a$ была у первого игрока, а карта $b$ — у второго, то есть предыдущая ситуация восстанавливается однозначно. Итак, если наша ситуация регулярна, то на предыдущем ходе карты мог получить любой из двух игроков, и в каждом из этих случаев предыдущая ситуация восстанавливается однозначно. Значит, в каждую регулярную ситуацию ведут ровно две стрелки. Аналогично, в каждую нерегулярную ситуацию ведёт ровно одна стрелка.\n\nПредположим, что из некоторой ситуации $S$ нельзя попасть в финальную. Назовём ситуацию *достижимой*, если в неё можно добраться из $S$. Из каждой достижимой ситуации ведут две стрелки в достижимые. С другой стороны, в каждую достижимую ситуацию ведёт не более двух стрелок из достижимых. Это возможно только в том случае, если в каждую достижимую ситуацию ведёт ровно по две стрелки из достижимых. Из этого, в частности, следует, что все достижимые ситуации регулярны. Более того, поскольку в каждую ситуацию ведёт не более двух стрелок, получаем, что все стрелки, входящие в достижимые ситуации, выходят также из достижимых.\n\nПусть в ситуации $S$ у первого игрока $k > 1$ карт. Тогда в одной из двух ситуаций, из которых ведут стрелки в $S$, у первого игрока $k-1$ карта — назовём эту ситуацию $S_1$; по показанному выше, она достижима. Аналогично, если $k-1 > 1$, то в одной из двух ситуаций, из которых ведут стрелки в $S_1$, у первого игрока $k-2$ карты; назовём её $S_2$ и продолжим рассуждения. В итоге мы получим цепочку из достижимых ситуаций $S_1, S_2, \\dots, S_{k-1}$, причём в $S_{k-1}$ у первого игрока одна карта, т. е. она критическая, и в неё входит только одна стрелка. Но в каждую достижимую ситуацию должно входить две стрелки — противоречие.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16780,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number greater than $5$. Prove that there exist natural numbers $m, n$ such that $m + n < p$ and $2^m \\cdot 3^n - 1$ is divisible by $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the numbers $2^i 3^j$ for $1 \\leq i, j \\leq p-1$. There are $(p-1)^2 \\geq p+1$ such numbers, so by the pigeonhole principle, there exist distinct pairs $(i_1, j_1)$ and $(i_2, j_2)$ with $1 \\leq i_1, i_2, j_1, j_2 \\leq p-1$ such that $2^{i_1} 3^{j_1} \\equiv 2^{i_2} 3^{j_2} \\pmod{p}$. By Fermat's little theorem, $2^{p-1} \\equiv 3^{p-1} \\equiv 1 \\pmod{p}$. Thus, $2^{i_1} 2^{p-1-i_2} \\equiv 3^{j_2} 3^{p-1-j_1} \\pmod{p}$, so there exist $1 \\leq i, j \\leq p-2$ such that $2^i \\equiv 3^j \\pmod{p}$. \n\nIf $i \\leq j$, choose $m = i$ and $n = p-1-j$; if $i > j$, choose $m = p-1-i$ and $n = j$. In both cases, $m + n < p$ and $2^m 3^n \\equiv 1 \\pmod{p}$, so $2^m 3^n - 1$ is divisible by $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16781,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure below, $A, B, C, D, E$ are five points in order on circle $\\Omega$, satisfying $\\overline{ABC} = \\overline{BCD} = \\overline{CDE}$. Points $P$ and $Q$ are on segments $AD$ and $BE$, respectively, and $P$ lies on segment $CQ$. Prove that $\\angle PAQ = \\angle PEQ$.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p79_data_d39a73b1ab.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S$ be the intersection of $AD$ and $BE$, and let the extension of $CQ$ intersect circle $\\Omega$ at points $T$ and $T'$. \n\nNoting that $\\overline{ABC} = \\overline{BCD} = \\overline{CDE}$, let the central angles of $\\overline{AB}$ and $\\overline{CD}$ be $\\alpha$, and those of $\\overline{BC}$ and $\\overline{DE}$ be $\\beta$.\n\nTherefore,\n\n$$\n\\begin{align*}\n\\angle ATQ &= \\angle ATC = \\alpha + \\beta, \\\\\n\\angle PTE &= \\angle CTE = \\alpha + \\beta, \\\\\n\\angle PSQ &= \\angle BDA + \\angle DBE = \\alpha + \\beta.\n\\end{align*}\n$$\n\nSince $\\angle ATQ = \\angle PSQ$, points $S, A, T, Q$ are concyclic. Similarly, since $\\angle PTE = \\angle PSQ$, points $P, S, T, E$ are concyclic.\n\nConsequently, $\\angle PAQ = \\angle PEQ$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16782,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Is it possible to cover a $6 \\times 6$ square with several layers of T-shaped figures (as shown in Fig. 1), so that each cell is covered the same number of times?\n\n![](images/Blr-2014_p24_data_59e4ba5ce7.png)\n\nb) Is it possible to cover a $7 \\times 7$ square with several layers of T-shaped figures (as shown in Figs. 1–4), so that each cell is covered the same number of times?\n\n![](images/Blr-2014_p24_data_b775942863.png)\n\n![](images/Blr-2014_p25_data_ac89254b15.png)\n\n![](images/Blr-2014_p25_data_fba2d24c63.png)\n\n![](images/Blr-2014_p25_data_14a491db0c.png)\n\n![](images/Blr-2014_p25_data_93a73a8d39.png)",
    "options": [],
    "answer": "See solution",
    "solution": "a) It is impossible. We color the $6 \\times 6$ square in black and white as shown in Fig. 1. Any T-figure covers at least one black cell. Assign $-3$ to each black cell, $4$ to the left upper and right lower cells, and $1$ to each remaining white cell (see Fig. 2). The sum $s$ of the numbers in any T-figure is non-positive. The total sum in the square is $$\\sigma = -3 \\cdot 10 + 4 \\cdot 2 + 1 \\cdot 24 = -30 + 8 + 24 = 2.$$ If the square is covered with $n$ layers of T-figures, the total sum counted over all coverings is $S = n\\sigma = 2n > 0$, but since $s \\le 0$ for each figure, $S \\le 0$, a contradiction.\n\nb) It is impossible. Let $T_1, T_2, T_3, T_4$ be the T-figures shown in Figs. 1–4. Consider the left upper cell $a$ of the $7 \\times 7$ square. Any T-figure covering $a$ also covers cell $b$. By similar reasoning for other cells, we see that any T-figure covering the left upper $2 \\times 2$ square covers exactly two cells marked by $\\times$, and any T-figure covering none of these covers at least one cell marked by $\\times$ (see Fig. 5).\n\n![](images/Blr-2014_p25_data_c3d8a3ce94.png)\n\nAssign $-3$ to each cell marked by $\\times$, $5$ to each cell marked by $\\bullet$, and $1$ to each remaining cell. The sum $s$ in any T-figure is non-positive. The total sum in the square is $$\\sigma = (-3) \\cdot 20 + 5 \\cdot 8 + 1 \\cdot 21 = -60 + 40 + 21 = 1.$$ If the square is covered with $n$ layers of T-figures, $S = n\\sigma = n > 0$, but since $s \\le 0$ for each figure, $S \\le 0$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16783,
    "subject": "Mathematics (Olympiad)",
    "question": "There are four countries, each consisting of several cities. The cities of any two countries are connected by at least $\\frac{5}{6}$ of all possible roads between these two countries. Prove that it is possible to choose one city from each country so that any two of them are connected. Any pair of cities can be connected by no more than one road.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us denote the countries as $X$, $Y$, $Z$, and $T$. Randomly choose one city $x$, $y$, $z$, $t$ from each country, respectively. If all 6 roads between them exist, the statement is proved.\n\nIf there is a quadruple connected by fewer than 6 roads, then there must exist a quadruple connected by all 6 roads; otherwise, the total number of possible roads is less than $\\frac{5}{6}$ of the maximum. Thus, for the problem's condition to hold, each quadruple must be connected by exactly 5 roads. Since every two countries are connected by exactly $\\frac{5}{6}$ of the possible roads, there are unconnected city pairs between countries.\n\nChoose cities $x \\in X$, $y \\in Y$ that are not connected. Consider arbitrary $z \\in Z$ and $t \\in T$. Since the quadruple $x, y, z, t$ is connected by exactly 5 roads, $z$ and $t$ must be connected. Since $z \\in Z$ and $t \\in T$ were arbitrary, every pair of cities from $Z$ and $T$ must be connected, which contradicts the connection condition between countries.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16784,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all natural numbers $n$ such that $2^n - n^2 + 1$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "We check by direct computation that $n = 2, 3, 4$ are solutions. We will prove that these are the only ones. We may assume $n \\ge 5$ after checking all smaller cases. We distinguish two cases.\n\n**Case I:** If $n$ is even, write $n = 2t$ (then $t \\ge 3$). One can prove that\n\n$$\n(2^t - 1)^2 < 2^n - n^2 + 1 < (2^t)^2 = 2^n,\n$$\n\nwhich means that in this case, $2^n - n^2 + 1$ cannot be a perfect square. The second inequality is obvious, and the first can be proven easily by induction for $t \\ge 7$, with the base cases $t = 3, 4, 5, 6$ verified directly.\n\n**Case II:** If $n$ is odd, let $p$ be the largest (necessarily odd) prime dividing $n$. If $p = 3$, then $n = 3^k$, and since $n \\ge 5$, it follows that $k \\ge 2$, so in particular $9 \\mid n$. But then $2^9 + 1$ divides $2^n + 1$, hence $19 \\mid 2^n + 1$ (since $2^9 + 1 = 512 + 1 = 513 = 27 \\cdot 19$).\n\nIf $2^n - n^2 + 1 = a^2$, then $2^n + 1 = n^2 + a^2$. Since $19 \\mid 2^n + 1$, and $19 \\equiv 3 \\pmod{4}$, it follows that $19 \\mid n^2 + a^2$. But then $19 \\mid n$, which contradicts the assumption that $n = 3^k$.\n\nTherefore, we must have $p > 3$. Since $2^p + 1 \\equiv 1 \\pmod{4}$, it follows that $3 \\mid 2^p + 1$, but $3^2$ does not divide $2^p + 1$ (this follows immediately from factoring $2^p + 1$ or applying LTE for $p = 3$). This implies that there exists a prime $q > 3$, with $q \\equiv 3 \\pmod{4}$ such that $q \\mid 2^p + 1$.\n\nUsing the statement below, we obtain that $q > p$. Since $q \\mid 2^p + 1 \\mid 2^n + 1$, and assuming $2^n - n^2 + 1 = a^2$, we get $2^n + 1 = n^2 + a^2$. By the same reasoning as above, $q \\mid n$, which contradicts the maximality of $p$ as the largest prime dividing $n$.\n\nThis completes the proof that for $n \\ge 5$, the expression $2^n - n^2 + 1$ is not a perfect square.\n\n*Claim.* If $p$ is a prime with $p \\ge 5$, then any odd prime factor $q$ of $2^p + 1$, different from $3$, satisfies $q > p$.\n\n*Proof.* Let $q > 3$ be a prime dividing $2^p + 1$. Then $2^p \\equiv -1 \\pmod{q}$, so the order of $2$ modulo $q$ is $2p$. Indeed, the order of $2$ (mod $q$) must divide $2p$, so it could be $1$, $2$, $p$, or $2p$. It cannot be $1$ or $p$ (as $2^p \\equiv -1 \\not\\equiv 1 \\pmod{q}$), and if it were $2$, then $2^2 \\equiv 1 \\pmod{q}$ implies $q = 3$, contradicting $q > 3$.\n\nSince $(2, q) = 1$, by Fermat's Little Theorem we have $2^{q-1} \\equiv 1 \\pmod{q}$. But the order of $2$ modulo $q$ is $2p$, so $2p \\mid q - 1$, which implies $p < q$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16785,
    "subject": "Mathematics (Olympiad)",
    "question": "Triangle $ABC$ is right-angled at $A$ and satisfies $AB > AC$. The line tangent to the circumcircle of triangle $ABC$ at $A$ intersects the line $BC$ at $M$. Let $D$ be the point such that $M$ is the midpoint of $BD$. The line through $D$ that is parallel to $AM$ intersects the circumcircle of triangle $ACD$ again at $E$.\n\nProve that $A$ is the incentre of triangle $EBM$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the line $AM$ intersect the circumcircle of triangle $ACD$ again at $N$.\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p89_data_dca7b14a94.png)\n\nUsing the alternate segment theorem in circle $ABC$, and then the cyclic quadrilateral $ACND$, we have\n\n$$\n\\angle CBA = \\angle CAN = \\angle CDN.\n$$\n\nSince $MB = MD$, it follows that $\\triangle MAB \\cong \\triangle MND$ (ASA). Hence, $MA = MN$.\n\nWe also have $\\triangle MND \\cong \\triangle MAE$, since these triangles are related by reflection in the perpendicular bisector of the parallel chords $AN$ and $ED$ in circle $ANDE$. Hence, $\\triangle MAB \\cong \\triangle MAE$ and $MA$ is a line of symmetry for triangle $EBM$.\n\nNow let $x = \\angle AEM = \\angle MBA = \\angle CAM$ and $y = \\angle ABE = \\angle BEA$. The angle sum in $\\triangle ABM$ yields $\\angle AMB = 90^\\circ - 2x$. By symmetry, we also have $\\angle EMA = 90^\\circ - 2x$.\n\nFinally, the angle sum in $\\triangle EBM$ yields $2x + 2y + 2(90^\\circ - 2x) = 180^\\circ$, which implies that $x = y$.\n\nThus, $A$ is the incentre of $\\triangle EBM$ because it is the intersection of its angle bisectors.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16786,
    "subject": "Mathematics (Olympiad)",
    "question": "For arbitrary positive numbers $a$, $b$, $c$, solve the system of equations:\n\n$$\n\\begin{cases}\n  a x^3 + b y = c z^5, \\\\\n  a z^3 + b x = c y^5, \\\\\n  a y^3 + b z = c x^5.\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that $x < y < z$ (one of the inequalities may not be strict). Subtracting the third equation from the first gives:\n\n$$\na(x^3 - y^3) + b(y - z) < 0 \\leq c(z^5 - x^5),\n$$\nwhich is a contradiction. Similarly, if $x < z < y$, subtracting the third equation from the second gives:\n\n$$\na(z^3 - y^3) + b(x - z) < 0 \\leq c(y^5 - x^5),\n$$\nwhich is also a contradiction. Since the system is cyclic, all cases are covered. Thus, we must have $x = y = z$.\n\nTo find $x$, substitute $x = y = z$ into any equation:\n\n$$\ncx^5 - a x^3 - b x = 0.\n$$\n\nThis gives $x_1 = 0$, and for $x \\neq 0$:\n\n$$\ncx^4 - a x^2 - b = 0.\n$$\n\nSince $a^2 + 4 b c > 0$ and $a - \\sqrt{a^2 + 4 b c} < 0$, the solutions are:\n\n$$\nx_{2,3} = \\pm \\sqrt{\\frac{a + \\sqrt{a^2 + 4 b c}}{2c}}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16787,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $$(ab)^2 + (bc)^2 + (ca)^2 = 3.$$ Prove that\n\n$$\n(a^2 - a + 1)(b^2 - b + 1)(c^2 - c + 1) \\geq 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The inequality is equivalent to\n\n$$\n(a^2 - a + 1)(b^2 - b + 1)(c^2 - c + 1) \\geq 1 \\iff (a^3 + 1)(b^3 + 1)(c^3 + 1) \\geq (a + 1)(b + 1)(c + 1).\n$$\n\nThus,\n\n$$\n\\begin{aligned}\n\\prod_{cyc} (a^3 + 1) - \\prod_{cyc} (a + 1) &= \\sum_{cyc} a^3 + \\sum_{cyc} (ab)^3 + (abc)^3 - \\sum_{cyc} a - \\sum_{cyc} ab - abc \\\\\n&= \\sum_{cyc} (a^3 + a) + \\sum_{cyc} (a^3 b^3 + ab) + [ (abc)^3 + 1 + 1 ] - 2 \\sum_{cyc} a - 2 \\sum_{cyc} ab - abc \\\\\n&= 2 \\sum_{cyc} a^2 + 2 \\sum_{cyc} a^2 b^2 + 2abc - 2 \\sum_{cyc} a - 2 \\sum_{cyc} ab \n\\end{aligned}\n$$\n\nSince $\\sum a^2 b^2 = 3$ by the given condition, we have\n\n$$\n\\sum_{cyc} (a^2 - 2a + 1) + \\left( \\sum_{cyc} a^2 + 2abc + 1 - 2 \\sum_{cyc} ab \\right) =\n$$\n\n$$\n\\sum_{cyc} (a - 1)^2 + \\left( \\sum_{cyc} a^2 + 2abc + 1 - 2 \\sum_{cyc} ab \\right) \\geq \\left( \\sum_{cyc} a^2 + 2abc + 1 - 2 \\sum_{cyc} ab \\right).\n$$\n\nWe will show that $\\sum_{cyc} a^2 + 2abc + 1 - 2 \\sum_{cyc} ab \\geq 0$ for every $a, b, c \\geq 0$.\n\nFirstly, observe that\n\n$$\n(1 + 2abc)(a + b + c) = (1 + abc + abc)(a + b + c) \\geq 9\\sqrt[3]{a^2 b^2 c^2 abc} = 9abc,\n$$\nwhich implies\n\n$$\n1 + 2abc \\geq \\frac{9abc}{a + b + c}.\n$$\n\nUsing Schur's Inequality ($\\sum_{cyc} a(a - b)(a - c) \\geq 0$ for $a, b, c \\geq 0$), we obtain\n\n$$\n\\sum_{cyc} a^2 \\geq 2 \\sum_{cyc} ab - \\frac{9abc}{a + b + c}.\n$$\n\nReturning to the previous expression,\n\n$$\n\\sum_{cyc} a^2 + 2abc + 1 - 2 \\sum_{cyc} ab \\geq (2 \\sum_{cyc} ab - \\frac{9abc}{a + b + c}) + 2abc + 1 - 2 \\sum_{cyc} ab = (1 + 2abc) - \\frac{9abc}{a + b + c} \\geq 0.\n$$\n\nTherefore, $\\prod_{cyc} (a^3 + 1) - \\prod_{cyc} (a + 1) \\geq 0$, and thus $\\prod_{cyc} (a^2 - a + 1) \\geq 1$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16788,
    "subject": "Mathematics (Olympiad)",
    "question": "Teacher drew a pentagon on the blackboard. The following conditions hold for the pentagon:\n\na) Two of the pentagon's interior angles are equal.\n\nb) There exist three interior angles such that the first one equals the sum of the other two.\n\nc) There exist four interior angles such that one of them equals the sum of the other three.\n\nd) There exists an interior angle that equals the sum of the other four.\n\nFind the interior angles of the pentagon.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the sizes of the angles of the pentagon be denoted in decreasing order as $\\alpha \\geq \\beta \\geq \\gamma \\geq \\delta \\geq \\varepsilon$. The sum of all the interior angles is $$(5-2) \\cdot 180^\\circ = 540^\\circ,$$ so $\\alpha + \\beta + \\gamma + \\delta + \\varepsilon = 540^\\circ$.\n\nThe angle that equals the sum of the other four is greater than the other four. Therefore,\n$$\\alpha = \\beta + \\gamma + \\delta + \\varepsilon = \\frac{540^\\circ}{2} = 270^\\circ.$$ \n\nThe angle which equals the sum of some other three cannot be equal to $\\alpha$, because then $\\varepsilon = 0^\\circ$. As it must be greater than the other three angles,\n$$\\beta = \\gamma + \\delta + \\varepsilon = \\frac{270^\\circ}{2} = 135^\\circ.$$ \n\nAnalogously, the angle that is the sum of some other two angles can be equal to neither $\\alpha$ nor $\\beta$, therefore\n$$\\gamma = \\delta + \\varepsilon = \\frac{135^\\circ}{2} = 67.5^\\circ.$$ \n\nFinally, the pentagon cannot have more angles of size $270^\\circ$, $135^\\circ$, or $67.5^\\circ$, therefore\n$$\\delta = \\varepsilon = \\frac{67.5^\\circ}{2} = 33.75^\\circ.$$ \n\n*Note.* Figures 5–10 depict all the possible ways the angles can be arranged in the pentagon.\n\n![](images/prob1718_p6_data_4a770b4442.png)\n\nFig. 5\n\n![](images/prob1718_p6_data_5d31572fec.png)\n\nFig. 6\n\n![](images/prob1718_p6_data_ed56202898.png)\n\nFig. 7\n\n![](images/prob1718_p6_data_c33a451e21.png)\n\nFig. 8\n\n![](images/prob1718_p6_data_d41bcf557a.png)\n\nFig. 9\n\n![](images/prob1718_p6_data_db97b872e4.png)\n\nFig. 10",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16789,
    "subject": "Mathematics (Olympiad)",
    "question": "已知圓 $O_1$ 與圓 $O_2$ 外切於點 $T$，一直線與圓 $O_2$ 相切於點 $X$，與圓 $O_1$ 交於點 $A, B$ 且點 $B$ 在線段 $AX$ 的內部。直線 $XT$ 與圓 $O_1$ 交於另一點 $S$。$C$ 是不包含點 $A, B$ 的 $TS$ 上的一點。過點 $C$ 作圓 $O_2$ 的切線，切點為 $Y$ 且線段 $CY$ 與線段 $ST$ 不相交。直線 $SC$ 與 $XY$ 交於點 $I$。試證：\n\n(1) $C, T, I, Y$ 四點共圓；\n\n(2) $I$ 是 $\\triangle ABC$ 的 $\\angle A$ 的旁切圓的圓心。",
    "options": [],
    "answer": "See solution",
    "solution": "(1) 作輔助線如圖所示。因為弧 $ST = XT$，則\n\n$$\n\\angle BXT = \\frac{\\text{弧}XT^{\\circ}}{2} = \\frac{\\text{弧}ST^{\\circ}}{2} = \\angle TAS\n$$\n\n從而，$\\triangle SAT \\sim \\triangle SXA$。故 $\\angle XAS = \\angle ATS$，即弧 $BS = AS$。\n所以 $S$ 是弧 $AB$ 的中點。因為\n\n$$\n\\angle TCI = \\angle TAS = \\angle BXT = \\angle TYX\n$$\n\n所以 $C, T, I, Y$ 四點共圓。\n\n(2) 由於 $\\triangle SAT \\sim \\triangle SXA$，則有 $SA^2 = ST \\cdot SX$。又 $C, T, I, Y$ 四點共圓，有\n\n$$\n\\angle CIT = \\angle CYT = \\angle TXY\n$$\n\n於是，$\\triangle SXI \\sim \\triangle SIT$。\n\n因此，$SI^2 = ST \\cdot SX$，所以 $SA = SI$。\n\n設 $\\angle BAC = \\alpha$，$\\angle ABC = \\beta$，$\\angle ACB = \\gamma$，則\n\n$$\n\\angle ACS = \\frac{\\text{弧}AS^{\\circ}}{2} = \\frac{\\text{弧}ASB^{\\circ}}{4} = \\frac{1}{4}(2\\pi - 2\\gamma) = \\frac{\\pi}{2} - \\frac{\\gamma}{2}\n$$\n\n$$\n\\text{故 } \\angle BCI = \\pi - \\angle BCS = \\pi - \\left(\\gamma + \\frac{\\pi}{2} - \\frac{\\gamma}{2}\\right) = \\frac{\\pi}{2} - \\frac{\\gamma}{2}\n$$\n\n因此，$CI$ 是 $\\angle ACB$ 的外角平分線。\n\n又因為 $SB = SA = SI$，$\\angle BSI = \\angle BSC = \\alpha$，所以\n\n$$\n\\angle BIS = \\frac{\\pi - \\angle BSI}{2} = \\frac{\\pi}{2} - \\frac{\\alpha}{2}\n$$\n\n在 $\\triangle BCI$ 中，可得 $\\angle CBI = \\frac{\\pi}{2} - \\frac{\\beta}{2}$，即 $BI$ 是 $\\angle ABC$ 的外角平分線，所以，點 $I$ 是 $\\triangle ABC$ 的 $\\angle A$ 的傍切圓的圓心。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16790,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $x$ such that both\n\n$$\n\\frac{x^2 + 1}{x + 2}\n$$\n\nand\n\n$$\n\\frac{x - 1}{2}\n$$\n\nare integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n\n$$\n\\frac{x^2 + 1}{x + 2} + \\frac{x - 1}{2} = \\frac{x(3x + 1)}{2(x + 2)}.\n$$\n\nSince $\\lfloor t \\rfloor \\le t$ holds for all real numbers $t$, and equality is attained if and only if $t$ is an integer, it follows that both $\\frac{x^2 + 1}{x + 2}$ and $\\frac{x - 1}{2}$ must be integers.\n\nThe latter is an integer if and only if $x$ is an odd integer. Now, consider\n\n$$\n\\frac{x^2 + 1}{x + 2} = x - 2 + \\frac{5}{x + 2}.\n$$\n\nFor this to be an integer, $x + 2$ must divide $5$, so $x + 2 \\in \\{1, -1, 5, -5\\}$, i.e., $x \\in \\{-1, -3, 3, -7\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16791,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute scalene triangle $ABC$, points $D$, $E$, and $F$ lie on sides $BC$, $CA$, and $AB$, respectively, such that $AD \\perp BC$, $BE \\perp CA$, and $CF \\perp AB$. The altitudes $AD$, $BE$, and $CF$ meet at the orthocenter $H$. Points $P$ and $Q$ lie on segment $EF$ such that $AP \\perp EF$ and $HQ \\perp EF$. Lines $DP$ and $QH$ intersect at point $R$. Compute $\\dfrac{HQ}{HR}$.\n",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the diagram shown below. By the given, we have $AP \\parallel QR$. Hence, we have two pairs of similar triangles: $APS$ and $HQS$, and $DHR$ and $DAP$. It follows that\n\n$$\n\\frac{HQ}{PA} = \\frac{HS}{AS} \\quad \\text{and} \\quad \\frac{RH}{PA} = \\frac{HD}{AD}.\n$$\n\nIt suffices to show that\n\n$$\n\\frac{HS}{AS} = \\frac{HD}{AD}, \\qquad (29)\n$$\n\nwhich is equivalent to showing that $(A, H)$ and $(S, D)$ are harmonic conjugates. Applying Ceva's theorem to triangle $AHC$ and cevians $AD$, $BE$, and $CF$ gives\n\n$$\n\\frac{AE}{EC} \\cdot \\frac{CF}{FH} \\cdot \\frac{HD}{DA} = 1.\n$$\n\nApplying Menelaus's theorem to triangle $ACH$ and line $ESF$ yields\n\n$$\n\\frac{AE}{EC} \\cdot \\frac{CF}{FH} \\cdot \\frac{HS}{SA} = 1.\n$$\n\nEquating the left-hand sides of the last two equations gives (29), completing the proof.\n\n![](images/pamphlet1112_main_p36_data_ad256f746a.png)\n\nConsider the diagram shown above. We present another proof of (29). Let $\\angle ABC = B$ and $\\angle ACB = C$. In right triangles $BHD$ and $ABD$, we have $HD/BD = \\tan \\angle HBD = \\cot C$ and $AD/BD = \\tan B$. Dividing the last two equations gives\n\n$$\n\\frac{HD}{AD} = \\frac{\\cot C}{\\tan B} = \\cot B \\cot C. \\qquad (30)\n$$\n\nApplying the Law of Sines in triangles *EHS* and *AFS* gives\n\n$$\n\\frac{HS}{HE} = \\frac{\\sin \\angle SEH}{\\sin \\angle HSE} \\quad \\text{and} \\quad \\frac{AS}{AF} = \\frac{\\sin \\angle AFE}{\\sin \\angle ASF}.\n$$\n\nDividing the last two equations yields\n\n$$\n\\frac{HS}{AS} = \\frac{HE}{AF} \\cdot \\frac{\\sin \\angle SEH}{\\sin \\angle AFE}. \\qquad (31)\n$$\n\nBecause $\\angle BEC = \\angle CFB = 90^\\circ$, $BCEF$ is cyclic, from which it follows that $\\angle SEH = \\angle FEB = \\angle FCB = 90^\\circ - B$ and $\\angle AFE = \\angle ECB = C$. Substituting the last two equations into (31) yields\n\n$$\n\\frac{HS}{AS} = \\frac{HE}{AF} \\cdot \\frac{\\cos B}{\\sin C}. \\qquad (32)\n$$\n\nIt is clear that right triangles *CHE* and *CAF* are similar to each other. Hence\n\n$$\n\\frac{HE}{AF} = \\frac{CE}{CF} = \\frac{BC \\cos C}{BC \\sin B} = \\frac{\\cos C}{\\sin B}.\n$$\n\nSubstituting the last equation into (32) gives\n\n$$\n\\frac{HS}{AS} = \\frac{HE}{AF} \\cdot \\frac{\\cos B}{\\sin C} = \\frac{\\cos C}{\\sin B} \\cdot \\frac{\\cos B}{\\sin C} = \\cot B \\cot C.\n$$\n\nCombining the last equation with (30) gives (29), and we are done.\n\n![](images/pamphlet1112_main_p37_data_6702496a65.png)\n\nConsider the diagram shown above. Because $\\angle HFB + \\angle HDB = 180^\\circ$, $BDHF$ is cyclic, from which it follows that $\\angle HDF = \\angle HBF$. Likewise, $\\angle EDH = \\angle ECH$. Because $\\angle BFC = \\angle BEC = 90^\\circ$, $BCEF$ is cyclic, from which it follows that $\\angle EBF = \\angle ECF$. Therefore, we have $\\angle HDF = \\angle HBF = \\angle EBF = \\angle ECF = \\angle ECH = \\angle EDH$; that is, $DH$ bisects $\\angle FDE$. Likewise, $FH$ bisects $\\angle EFD$ and $EH$ bisects $\\angle DEF$. Hence $H$ is the incenter of triangle $DEF$. Let $\\omega$ denote the incircle of triangle $DEF$.\n\nBecause $\\angle AE \\perp EH$ and $EH$ is the interior bisector of $\\angle DEF$, $AE$ is the exterior bisector of $\\angle DEF$. Likewise, $AF$ is the exterior bisector of $\\angle EFD$. Consequently, $A$ is the excenter (opposite $D$) of triangle $DEF$. Let $\\Omega$ denote the excircle of triangle $DEF$ centered at $A$.\n\nConsider the dilation centered at $D$ sending $\\omega$ to $\\Omega$. Because $HQ \\parallel AP$, line $QH$ is sent to line $AP$. Let $R_1$ be the point on $\\omega$ diametrically opposite $Q$ (so $QH = HR$ and $Q$, $H$, $R_1$ are collinear). Then, we see that $R_1$ is sent to $P$ by this dilation. In particular, $D$, $R_1$, $P$ are collinear. Hence, $R_1$ is the intersection of lines $DP$ and $QH$, so $R = R_1$. We therefore have that $RH = R_1H = QH$.\n\nThus, $\\boxed{\\dfrac{HQ}{HR} = 1}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16792,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{Q} \\to \\mathbb{Q}$ such that\n$$\nf(x + 3f(y)) = f(x) + f(y) + 2y,\n$$\nfor all $x, y \\in \\mathbb{Q}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that the solutions are $f_1(x) = x$ and $f_2(x) = -\\frac{2x}{3}$ for all $x \\in \\mathbb{Q}$.\n\nSet $x = y - 3f(y)$ to obtain $f(y - 3f(y)) = -2y$, for $y \\in \\mathbb{Q}$. Replacing $y$ with $y - 3f(y)$ in the initial equation gives $f(x - 6y) = f(x) - 2y + 2(y - 3f(y))$, hence $f(x - 6y) = f(x) - 6f(y)$ for all $x, y \\in \\mathbb{Q}$.\n\nSet $x = y = 0$ to get $f(0) = 0$, and replace $x = 6y$ to obtain $f(6y) = 6f(y)$ for all $y \\in \\mathbb{Q}$.\n\nWe derived that $f(x - 6y) = f(x) - f(6y)$, which, for $u = 6y$ and $v = x - 6y$, yields $f(u + v) = f(u) + f(v)$ for all $u, v \\in \\mathbb{Q}$. The solution of this classical functional equation is $f(x) = x f(1)$ for $x \\in \\mathbb{Q}$.\n\nOn the other hand, by setting $x = y = 1$ in the initial equation we obtain $3f^2(1) - f(1) - 2 = 0$, hence $f(1) = 1$ or $f(1) = -\\frac{2}{3}$, thus proving our claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16793,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D$ be a point in the interior of triangle $ABC$ such that $\\angle BAD = 40^\\circ$, $\\angle DAC = 30^\\circ$, $\\angle BCD = 20^\\circ$, and $\\angle DCA = 50^\\circ$. Find $\\angle CBD$.\n\n![](images/RMC_2020_p15_data_bb33239266.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that $ABC$ is an isosceles triangle, with $BA = BC$ and $\\angle ABC = 40^\\circ$. Suppose that the perpendicular bisector of the triangle's base intersects $AD$ at $T$.\n\nSince $\\angle DAC < \\angle DCA$, we have $CD < DA$, therefore $T \\in (AD)$ and $\\angle TBC = 20^\\circ$.\n\nWe will prove that $D$ is the incenter of triangle $BTC$, hence $BD$ bisects $\\angle TBC$, and it follows that $\\angle CBD = 10^\\circ$.\n\nBecause $TCA$ is an isosceles triangle, we have $\\angle TCA = 30^\\circ$, and hence $\\angle TCD = 20^\\circ = \\angle BCD$. It follows that $CD$ bisects $\\angle BCT$.\n\nA short computation shows that $\\angle CTB = \\angle BTA = 120^\\circ$ and $\\angle DTC = 60^\\circ$, therefore $TD$ is the angle bisector of $\\angle BTC$.\n\nWe conclude that $D$ is the incenter of triangle $BCT$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16794,
    "subject": "Mathematics (Olympiad)",
    "question": "A set $S$ consists of $36$ points arranged in a $6 \\times 6$ grid. How many triangles can be formed by choosing three points from $S$ such that no three chosen points are collinear?",
    "options": [],
    "answer": "See solution",
    "solution": "Since there are $36$ points in $S$, there are $\\binom{36}{3} = 7140$ possible triples. We must subtract those triples where all three points are collinear.\n\n- There are $6$ rows, each with $\\binom{6}{3} = 20$ triples, so $6 \\times 20 = 120$ row triples. Similarly, $120$ column triples.\n- Diagonal lines with slope $1$ have $1,2,3,4,5,6,5,4,3,2,1$ points. The number of triples from these is:\n  $$\n  \\binom{3}{3} + \\binom{4}{3} + \\binom{5}{3} + \\binom{6}{3} + \\binom{5}{3} + \\binom{4}{3} + \\binom{3}{3} = 50.\n  $$\n  By symmetry, $50$ triples from diagonals with slope $-1$.\n- Lines with slope $2$ yield $8$ triples; by symmetry, $8$ for slope $-2$, $8$ for $\\frac{1}{2}$, and $8$ for $-\\frac{1}{2}$.\n\nAny other line contains at most $2$ points. Thus, the answer is:\n$$\n7140 - 120 \\times 2 - 50 \\times 2 - 8 \\times 4 = 6768.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16795,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AD$ be a bisector in a triangle $ABC$, and $E, F$ be the incenters of the triangles $ADC$ and $ADB$ respectively. Further, let $\\omega$ be the circumcircle of the triangle $DEF$, $Q$ be the point of intersection of the lines $BE$ and $CF$, and $H, J, K, M$ be the second points of intersection of the lines $CE, CF, BE, BF$ respectively with the circle $\\omega$. If $\\omega_1, \\omega_2$ are the circumcircles of the triangles $HQJ$ and $KQM$, prove that the point of intersection of $\\omega_1$ and $\\omega_2$, that is different from $Q$, belongs to the line $AD$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first show that the point $Q$ belongs to the line $AD$.\n\nLet $I$ be the intersection of $CE$ and $BF$, which is the incenter of triangle $ABC$. Since $DE$ is a bisector in triangle $IDC$, we have $\\frac{IE}{EC} = \\frac{ID}{DC}$. Similarly, $\\frac{IF}{FB} = \\frac{ID}{DB}$. Then\n$$\n\\frac{IE}{EC} \\cdot \\frac{CD}{DB} \\cdot \\frac{BF}{FI} = \\frac{ID}{ID} = 1,\n$$\nand so by Ceva's theorem for triangle $IBC$, the lines $CF$, $BE$, $AD$ are concurrent, which means $Q$ lies on $AD$.\n\nNext, we show that the intersection of lines $HJ$ and $KM$ also lies on $AD$.\n\nThe points $K$, $E$, $H$, $J$, $F$, $M$ lie on the same circle $\\omega$, so by Pascal's theorem, the points $KE \\cap FJ = Q$, $EH \\cap FM = I$, and $MK \\cap HJ$ are collinear, which proves that the intersection of $HJ$ and $KM$ lies on $AD$.\n\nFinally, let $T$ be the second intersection point of circles $\\omega_1$ and $\\omega_2$. Then $TQ$ is the radical axis of these circles. Also, $MK$ and $HJ$ are the radical axes of circles $\\omega_2, \\omega$ and $\\omega_1, \\omega$, respectively. Thus, the intersection of $HJ$ and $KM$ is the radical center of all three circles, which must lie on $TQ$ and, as shown, on $AD$. Since $Q$ also lies on $AD$, it follows that $T$ must lie on $AD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16796,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer $n$ for which, no matter how we choose to color red $n$ vertices of a cube, there is a vertex of the cube whose three adjacent vertices are all colored red.",
    "options": [],
    "answer": "See solution",
    "solution": "The required minimum number is $5$.\n\nLet $ABCD$, $A'B'C'D'$ be the vertices of a cube. Coloring red the four vertices of a face (e.g., $A$, $B$, $C$, $D$), no vertex of the cube has all three adjacent vertices colored red, so $n \\geq 5$.\n\nNow, let us color $5$ vertices of the cube in red. No matter how we do it, one of the faces $ABCD$ or $A'B'C'D'$ has at least three red vertices. Without loss of generality, suppose $A$, $B$, $C$ are red.\n\nIf $D$ is also red, the face $A'B'C'D'$ has only one red vertex; if $X'$ is the red vertex, with $X \\in \\{A, B, C, D\\}$, then all the adjacent vertices of $X$ are colored red.\n\nIf $D$ is not red, the face $A'B'C'D'$ has two red vertices. If one of $B'$ or $D'$ is red, then the adjacent vertices of $B$, respectively $D$, are all red. Otherwise, $A'$ and $C'$ are red.\n\n**Remark.** The requirement of the problem is equivalent to the fact that one can form an equilateral triangle whose vertices are all red.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 16797,
    "subject": "Mathematics (Olympiad)",
    "question": "How many positive integers $N \\leq 1000$ satisfy $S(5N) = S(N)$, where $S(n)$ denotes the sum of the digits of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let us consider only positive integers $n \\leq 999$ (since $1000$ does not satisfy the requirement). Any such $n$ can be written as $100a + 10b + c$ where $a, b, c$ are 1-digit non-negative integers.\n\nMultiplying a 1-digit integer by $5$ gives at most $45$, and its one's digit is either $0$ or $5$, so there is no carry-over in $5a \\cdot 100 + 5b \\cdot 10 + 5c$. Thus, if $n = a \\cdot 100 + b \\cdot 10 + c$, then $S(5n) = S(5a) + S(5b) + S(5c)$.\n\nIf $k$ is an even 1-digit number, $S(5k) = \\frac{k}{2}$; if $k$ is odd, $S(5k) = \\frac{k+9}{2}$. Therefore, if $S(N) = S$ for $N \\leq 999$, then $S(5N) = \\frac{S + 9\\ell}{2}$, where $\\ell$ is the number of odd digits among $a, b, c$.\n\nThe remainder when dividing $n$ by $9$ equals the remainder when dividing $S(n)$ by $9$. Thus, if $S(N) = S(5N)$, $N$ must be a multiple of $9$. There are $111$ multiples of $9$ less than or equal to $999$.\n\n**Case 1:** $S(N) = 9$\n\n$S(5N) = \\frac{9 + 9\\ell}{2}$. For $S(5N) = 9$, $\\ell = 1$. The numbers not satisfying the requirement correspond to $(a, b, c)$ as $(1, 1, 7), (1, 3, 5), (3, 3, 3)$ and their permutations, totaling $3 + 6 + 1 = 10$ numbers.\n\n**Case 2:** $S(N) = 18$\n\nHere, $\\ell = 2$ for $S(5N) = 18$. For each number in this class, $999 - N$ belongs to class 1 and satisfies the requirement, and vice versa. Thus, there are also $10$ numbers in this class not satisfying the requirement.\n\n**Case 3:** $S(N) = 27$\n\nOnly $999$ belongs here, and it satisfies the requirement.\n\nThus, among the $111$ multiples of $9$, there are $10 \\times 2 = 20$ numbers which do not satisfy the requirement. Therefore, there are exactly $111 - 20 = 91$ positive integers $N \\leq 1000$ for which $S(5N) = S(N)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16798,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(n)$ be the number of valid colourings of the vertices of a triangular grid of size $n$ (with $n$ layers), using two colours (Red and Blue), such that no monochromatic triangle (all vertices the same colour) appears anywhere in the grid. Find $f(1)$, $f(2)$, $f(3)$, and $f(n)$ for $n \\geq 4$.",
    "options": [],
    "answer": "See solution",
    "solution": "$f(1) = 6$, $f(2) = 18$, $f(3) = 36$, and $f(n) = 0$ for $n \\geq 4$.\n\nFirst, for $n=1$, all $2^3 = 8$ possible colourings are valid except for the two monochromatic colourings, so $f(1) = 6$.\n\nFor $n=2$, the inner triangle cannot be monochromatic, so there are $f(1) = 6$ possible colourings for this inner triangle. For each, there are 3 ways to colour the outer points, yielding $6 \\times 3 = 18$, so $f(2) = 18$.\n\n![](images/Ireland_2017_p18_data_5de97c5333.png)\n\n**Lemma:** There cannot be 3 equally spaced vertices on the base of any sub-triangle all of which have the same colour.\n\n**Proof:** Assume there are 3 equally spaced vertices $X, Y, Z$ on the base of a sub-triangle, all the same colour (say Red). Considering the points formed by intersections of lines at $60^\\circ$ from these vertices, if any of these are Red, a monochromatic triangle is formed. The same holds for Blue. This proves the lemma.\n\n![](images/Ireland_2017_p18_data_6916c89eab.png)\n\nFor $n=3$:\n\n![](images/Ireland_2017_p18_data_4bfa27631d.png)\n\n**Case 1:** Points 1 and 2 have the same colour (say Blue). Then Point 3 must be Red, and the lemma implies Points 4 and 5 are Red. Point 6 is Blue. Points 7 and 8 cannot both be Blue (would form monochromatic triangle 678) or both Red (would violate the lemma). So, Point 7 is Blue and Point 8 is Red. Points 9 and 10 must be Red and Blue, respectively. Accounting for symmetries, there are 12 possibilities (2 choices for Points 1 and 2, 2 for Points 7 and 8, and 3 rotations).\n\n**Case 2:** Points 1 and 2 have different colours, and similarly for pairs (7, 9) and (8, 10). Points 1, 8, 9 are the same colour (say Blue), and Points 2, 7, 10 are Red. The outer triangle 645 can be coloured in $f(1) = 6$ ways, and the centre vertex in 2 ways. Accounting for symmetries, this yields 24 possibilities (2 choices for Points 1, 8, 9, 6 for the outer triangle, 2 for the centre). Thus, $f(3) = 12 + 24 = 36$.\n\nFor $n=4$:\n\n![](images/Ireland_2017_p19_data_d6ef3f8626.png)\n\nThe inner triangle cannot be monochromatic, so two vertices are Blue and one is Red. The two Blue vertices imply Point 1 is Red, and the lemma implies Points 2 and 3 are Red, so Point 4 is Blue. Points 5 and 6 cannot both be Blue (would form monochromatic triangle 456) or both Red (would violate the lemma). So, Point 5 is Blue and Point 6 is Red. Since Point 3 is Red, the lemma implies Points 7 and 8 are Blue, which implies Point 9 is Red. Now, the point labelled X cannot be coloured without forming a monochromatic triangle (either Blue triangle X48 or Red triangle X29). So $f(4) = 0$.\n\nFor $n > 4$, the diagram contains the $n=4$ case, so $f(n) = 0$ for $n \\geq 4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16799,
    "subject": "Mathematics (Olympiad)",
    "question": "Some natural numbers are placed around a circle in such a way that the product of any two neighboring numbers is a perfect square. Prove that the product of any (not necessarily neighboring) two numbers is also a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us denote the numbers as $a_1, a_2, \\ldots, a_k$. From the problem statement, it follows that for any $1 \\leq i < j \\leq k$, the product $$(a_i a_{i+1})(a_{i+1} a_{i+2}) \\ldots (a_{j-1} a_j) = n^2$$ is a perfect square. Hence, the product $a_i a_j = \\frac{n^2}{a_{i+1}^2 a_{i+2}^2 \\ldots a_{j-1}^2}$ is also a perfect square, as the ratio of two perfect squares is a perfect square. Q.E.D.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16800,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, and $C$ be non-collinear points. Let $P = sA$ and $Q = sA + (1-s)C$ for some real $s$. Let $BCRP$ be a parallelogram, and $R$ be real, so $R = P + C = sA + C = s\\bar{A} + \\bar{C}$. Let $Q = R - sC$. Let $D$ be the reflection of $C$ about $BR$, so $D = \\bar{C}$. Prove that $A$, $P$, $Q$, and $D$ are concyclic if and only if $A\\bar{A} = 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Working through each factor of the cross-ratio:\n\n- $A - P = A - sA = (1-s)A$.\n- Since triangles $APQ$ and $ABC$ are similar, $\\bar{A} - \\bar{Q} = (1-s)(\\bar{A} - \\bar{C})$.\n- $D - Q = \\bar{C} - R + sC = -s\\bar{A} + sC = -s(\\bar{A} - C)$.\n- $\\bar{D} - \\bar{P} = C - s\\bar{A} = C + \\bar{C} - R$, which is real.\n\nThus,\n\n$$\n\\begin{align*}\n& A, P, Q \\text{ and } D \\text{ are concyclic} \\\\\n\\Leftrightarrow & (A - P)(\\bar{A} - \\bar{Q})(D - Q)(\\bar{D} - \\bar{P}) \\in \\mathbb{R} \\\\\n\\Leftrightarrow & A(\\bar{A} - \\bar{C})(\\bar{A} - C) \\in \\mathbb{R} \\\\\n\\Leftrightarrow & A\\bar{A}^2 - (C + \\bar{C})A\\bar{A} + A \\in \\mathbb{R} \\\\\n\\Leftrightarrow & A\\bar{A}^2 + A \\in \\mathbb{R} \\\\\n\\Leftrightarrow & A\\bar{A}^2 + A - A^2\\bar{A} - \\bar{A} = 0 \\\\\n\\Leftrightarrow & (A\\bar{A} - 1)(A - \\bar{A}) = 0,\n\\end{align*}\n$$\n\nand since $A$ is not real, this is equivalent to $A\\bar{A} = 1$, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16801,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $s = (|AB| + |BC| + |AC|)/2$ be the semi-perimeter of triangle $ABC$. We choose two points $L$ and $N$ lying on the rays $[AB$ and $[CB$ such that $|AL| = |CN| = s$. Let $K$ be the point symmetric to $B$ with respect to the center of the circumcircle. Prove that a perpendicular drawn from the point $K$ to the line $NL$ passes through the incenter of triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $|BC| = a$, $|AC| = b$, $|AB| = c$, and $I$ be the incenter of triangle $ABC$.\n\n1. From the point $I$, draw lines parallel to $AB$ and $BC$ intersecting $AK$ and $KC$ at $P$ and $Q$, respectively.\n\n2. In triangle $IPQ$, we have $\\angle ABC = \\angle PIQ$, $|IQ| = s - c$, $|IP| = s - a$. Since $\\angle NBL = \\angle ABC$, $|NB| = s - a$, $|LB| = s - c$, it follows that $\\triangle PIQ \\cong \\triangle NBL$.\n\n3. Choose a point $L_1$ on line $IP$ such that $|IL_1| = |IQ|$, and a point $N_1$ on line $IQ$ such that $|IN_1| = |IP|$. Then $IN_1L_1 = NBL$, $BL \\parallel IL_1$, $BN \\parallel IN_1$, and it follows that $L_1N_1 \\parallel NL$.\n\n4. $IPKQ$ is a cyclic quadrilateral. Therefore, $\\angle IKP = \\angle IQP$, $\\angle KIP + \\angle IL_1N_1 = \\angle KIP + \\angle IQP = \\angle KIP + \\angle IKP = 90^\\circ$, and it follows that $IK \\perp L_1N_1$, and finally $IK \\perp NL$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16802,
    "subject": "Mathematics (Olympiad)",
    "question": "The probability that a chord does not intersect a triangle inscribed in a circle is $\\frac{11}{25}$. The only way this can happen is when the two points are chosen on the same arc between two of the triangle's vertices. The probability that a point is chosen on one of the arcs opposite to one of the base angles is $\\frac{x}{180}$, and the probability that a point is chosen on the arc between the two base angles is $\\frac{180 - 2x}{180}$. Therefore,\n\n$$\n2\\left(\\frac{x}{180}\\right)^2 + \\left(\\frac{180 - 2x}{180}\\right)^2 = \\frac{11}{25}.\n$$\n\nThis simplifies to $x^2 - 120x + 3024 = 0$, which can be factorized as $(x-84)(x-36) = 0$. What is the sum of the roots?",
    "options": [],
    "answer": "See solution",
    "solution": "The quadratic equation $x^2 - 120x + 3024 = 0$ has roots $x = 84$ and $x = 36$. Thus, the sum of the roots is $84 + 36 = 120$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16803,
    "subject": "Mathematics (Olympiad)",
    "question": "The point $M$ lies on the side $AB$ of the circumscribed quadrilateral $ABCD$. The points $I_1$, $I_2$, and $I_3$ are the incenters of $\\triangle MBC$, $\\triangle MCD$, and $\\triangle MDA$, respectively. Show that the points $M$, $I_1$, $I_2$, and $I_3$ lie on a circle.\n\n![](images/BMO_2016_Short_List_Final_1_p16_data_68d1185c92.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Lemma. Let $I$ be the incenter of $\\triangle ABC$ and let the points $P$ and $Q$ lie on the lines $AB$ and $AC$. Then the points $A$, $I$, $P$, and $Q$ lie on a circle if and only if\n\n$$\n\\overline{BP} + \\overline{CQ} = BC\n$$\n\nwhere $\\overline{BP}$ equals $|BP|$ if $P$ lies in the ray $BA \\rightarrow$ and $-|BP|$ if it does not, and similarly for $\\overline{CQ}$.\n\n*Proof of the lemma.* We shall only consider the case when $P$ and $Q$ lie in the segments $AB$ and $AC$. All other cases are treated analogously.\n\n![](images/BMO_2016_Short_List_Final_1_p16_data_bf45627ad5.png)\n\nSuppose that $A$, $I$, $P$, and $Q$ lie on a circle. Let $D$ and $E$ be the contact points of the incircle of $\\triangle ABC$ with $AB$ and $AC$. We have that $\\angle PIQ = 180^\\circ - \\alpha$, so $\\angle DIP = \\angle EIQ$ and, therefore, $\\triangle DIP \\simeq \\triangle EIQ$. This gives us $DP = EQ$ and $BP + CQ = BD + CE = BC$, as needed.\n\nThe converse is established by following the foregoing chain of inequalities in reverse. $\\square$\n\nLet the circumcircle of $\\triangle MI_1I_3$ meet the lines $AB$, $CM$, and $DM$ for the second time at $P$, $Q$, and $R$. By the lemma, $\\overline{BP} + \\overline{CQ} = \\overline{BC}$ and $\\overline{DR} + \\overline{AP} = \\overline{DA}$. Therefore, $\\overline{CQ} + \\overline{DR} = \\overline{BC} + \\overline{DA} - \\overline{BP} - \\overline{AP} = \\overline{BC} + \\overline{DA} - \\overline{AB}$. Since $ABCD$ is circumscribed, this is equal to $CD$, and, by the lemma, the proof is complete.\n\n![](images/BMO_2016_Short_List_Final_1_p17_data_ea6a866d9c.png)\n\nSecond solution. Let $\\omega_1$, $\\omega_2$, and $\\omega_3$ be the incircles of $\\triangle MBC$, $\\triangle MCD$, and $\\triangle MDA$. The common internal tangent $t_1$ of $\\omega_1$ and $\\omega_2$ equals $[\\text{tangent from } M \\text{ to } \\omega_2] - [\\text{tangent from } M \\text{ to } \\omega_1] = \\frac{1}{2}(MC + MD - CD - MB - MC + BC)$.\n\nAnalogously, the common internal tangent $t_2$ of $\\omega_2$ and $\\omega_3$ equals $[\\text{tangent from } M \\text{ to } \\omega_2] - [\\text{tangent from } M \\text{ to } \\omega_3] = \\frac{1}{2}(MC + MD - CD - MD - MA + DA)$.\n\nFinally, the common external tangent $t_3$ of $\\omega_1$ and $\\omega_3$ equals $[\\text{tangent from } M \\text{ to } \\omega_1] - [\\text{tangent from } M \\text{ to } \\omega_3] = \\frac{1}{2}(MB + MC - BC + MD + MA - DA)$.\n\nSince $ABCD$ is circumscribed, we have $AB + CD = BC + DA$, and, therefore, $t_1 + t_2 = t_3$. It follows from this that $\\omega_1$, $\\omega_2$, and $\\omega_3$ have a common tangent $s$ (which separates $\\omega_2$ from $\\omega_1$ and $\\omega_3$).\n\n![](images/BMO_2016_Short_List_Final_1_p17_data_287766a80b.png)\n\nLet $\\triangle MKL$ be the triangle formed by the lines $MC$, $MD$, and $s$. Then, since $I_1I_2$ and $I_2I_3$ are external angle bisectors in it, we have $\\angle I_1I_2I_3 = 90^\\circ - \\frac{1}{2}\\angle KML = 180^\\circ - \\angle I_1MI_3$ and, therefore, $MI_1I_2I_3$ is cyclic. $\\square$\n\n_Remark_: In the first solution, the angle $\\alpha$ refers to the angle $\\angle BAC$.\n\n_Remark_: Referring to solution 1, the inequality $t_3 \\le t_1 + t_2$ holds, the equality is possible if and only if the circles $\\omega_1$, $\\omega_2$, and $\\omega_3$ have a common tangent.\n",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16804,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the number of ways to distribute the six primes in the prime factorization of $2013 \\times 2014 = 2 \\times 3 \\times 11 \\times 19 \\times 53 \\times 61$ into three groups, each containing at least one prime, so that the product of the primes in each group forms a factor, and the ordered triple $(a, b, c)$ of these factors is counted as a possible solution.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider all possible groupings of the six primes into three non-empty groups:\n\n- **Case (4, 1, 1):** Choose two primes to be singletons: $\\binom{6}{2} = 15$ ways.\n- **Case (3, 2, 1):** Choose one singleton ($6$ choices), then two for a doubleton ($\\binom{5}{2} = 10$), so $6 \\times 10 = 60$ ways.\n- **Case (2, 2, 2):** Pair up the primes: $5$ choices for a partner for $2$, then $3$ choices for a partner for one of the remaining primes, so $5 \\times 3 = 15$ ways.\n\nAdding up, the total number of ways is $15 + 60 + 15 = 90$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16805,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle is divided into 432 congruent arcs by 432 points. The points are colored in four colors such that 108 points are colored Red, 108 points are colored Green, 108 points are colored Blue, and the remaining 108 points are colored Yellow. Prove that one can choose three points of each color in such a way that the four triangles formed by the chosen points of the same color are congruent.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $R$, $G$, $B$, and $Y$ denote the sets of Red, Green, Blue, and Yellow points, respectively, and let $r$, $g$, $b$, and $y$ denote a generic Red, Green, Blue, and Yellow point, respectively. For $0 \\leq k \\leq 431$, denote by $\\mathcal{T}_k$ the counterclockwise rotation by $\\frac{360k}{432}$ degrees around the center of the circle.\n\nFirst, we claim that there is some index $i_1$ such that $|\\mathcal{T}_{i_1}(R) \\cap G| \\geq 28$. Indeed, for each $k$, the set $\\mathcal{T}_k(R) \\cap G$ consists of all Green points that are the images of Red points under the rotation $\\mathcal{T}_k$. Hence the sum\n\n$$\ns_1 = |\\mathcal{T}_0(R) \\cap G| + |\\mathcal{T}_1(R) \\cap G| + \\dots + |\\mathcal{T}_{431}(R) \\cap G|\n$$\n\nis equal to the number of pairs of points $(r, g)$ such that $g = \\mathcal{T}_k(r)$ for some $k$. On the other hand, for each $r$ and each $g$, there is a unique rotation $\\mathcal{T}_k$ with $\\mathcal{T}_k(r) = g$, from which it follows that $s_1 = 108^2 = 11664$. On the other hand, note that $|\\mathcal{T}_0(R) \\cap G| = |R \\cap G| = 0$ because the sets $R$ and $G$ are disjoint. By the Pigeonhole principle, there is some index $i_1$ such that\n\n$$\n|\\mathcal{T}_{i_1}(R) \\cap G| \\geq \\left\\lfloor \\frac{s_1}{431} \\right\\rfloor = \\left\\lfloor \\frac{11664}{431} \\right\\rfloor = \\lceil 27.06\\ldots \\rceil = 28,\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16806,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute-angled, not equilateral triangle, where vertex $A$ lies on the perpendicular bisector of the segment $HO$, joining the orthocentre $H$ to the circumcentre $O$. Determine all possible values for the measure of angle $A$.\n\n![](images/RMC2012_p94_data_ee9956c3c4.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\triangle ABC$ is acute-angled, both $H$ and $O$ are interior to it. We have $AH = AO = R$ (the circumradius). By a well-known result, $AH$ is twice the distance $d$ from $O$ to the side $BC$, so $d = R/2$. Then $\\angle OBC = \\angle OCB = 30^\\circ$. But $\\angle OBA = \\angle OAB$ and $\\angle OCA = \\angle OAC$ (in isosceles triangles).\n\nIt follows that $$180^{\\circ} = \\angle A + \\angle B + \\angle C = (\\angle OAB + \\angle OAC) + (\\angle OBA + \\angle OBC) + (\\angle OCA + \\angle OCB) = 2\\angle A + 60^{\\circ}$$ whence $\\angle A = 60^{\\circ}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16807,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f$ from integers to integers such that\n\n$$\nf(m+n) + f(m-n) - 2f(m) = 6mn^2\n$$\n\nfor all integers $m$ and $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that if we replace $f(x)$ by $f(x) + ax + c$ for some constants $a$ and $c$, then we get another solution. Thus, we may assume for now that $f(0) = 0$ and $f(1) = 1$.\n\nLet $P(m, n)$ be the given statement. Then $P(0, n)$ gives\n\n$$\nf(n) + f(-n) - 2f(0) = 0\n$$\n\nand since $f(0) = 0$, we get $f(-n) = -f(n)$. Then, taking $P(n, m)$,\n\n$$\nf(m+n) + f(n-m) - 2f(n) = 6m^2n\n$$\n\nand subtracting this from $P(m, n)$ gives\n\n$$\n2f(m-n) + 2f(n) - 2f(m) = 6mn(n-m).\n$$\n\nSetting $n = -1$, dividing by 2, and rearranging gives\n\n$$\nf(m+1) = f(m) + f(1) + 3m(m+1).\n$$\n\nThe first few values of $m$ tell us:\n\n$$\nf(2) = 8, \\quad f(3) = 27, \\quad f(4) = 64.\n$$\n\nWe hypothesize for an inductive argument that $f(n) = n^3$ and show that\n\n$$\nf(n+1) = f(1) + f(n) + 3n(n+1) = n^3 + 3n^2 + 3n + 1 = (n+1)^3.\n$$\n\nThus $f(n) = n^3$ for all positive integers, and hence by $f(-n) = -f(n)$ also for all negative integers.\n\nNow, incorporating our initial remark, we find that all solutions are of the form $f(n) = n^3 + an + c$ for some integer constants $a$ and $c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16808,
    "subject": "Mathematics (Olympiad)",
    "question": "Все веса в решении будут измеряться в граммах. Назовём кусок яблока (или само яблоко) большим, если его вес не меньше 25.\n\nДокажите индукцией по $n$, что $n$ больших яблок суммарного веса $100n$ можно разрезать на большие куски и раздать $n$ детям поровну.",
    "options": [],
    "answer": "See solution",
    "solution": "База при $n = 1$ очевидна.\n\nПусть $n > 1$. Рассмотрим два самых тяжёлых яблока; пусть их веса $a \\ge b$. Заметим, что $a + b \\ge 200$ (иначе средний вес одного яблока будет меньше, чем $200/2 = 100$). Выкинем эти два яблока из набора и добавим в него яблоко веса $c = a + b - 100 \\ge 100$. По предположению индукции, полученный набор можно разрезать на большие куски и раздать $n-1$ детям поровну.\n\nЕсли при этом какой-то кусок нового яблока оказался больше 50, разрежем его на два больших куска. Через несколько таких разрезаний мы придём к ситуации, когда новое яблоко разделено на куски весов $c_1, c_2, \\dots, c_k$, не превосходящих 50. Обозначим $s_d = c_1 + \\dots + c_d$ при $d = 1, 2, \\dots, k$ и положим $s_0 = 0$.\n\nПокажем теперь, как разрезать исходный набор. Все яблоки, кроме $a$ и $b$, разрежем так же, как и в новом наборе. Заметим, что $a \\ge 200/2 = 100$. Обозначим через $t$ минимальный индекс такой, что $a - s_t \\le 75$, и отрежем от $a$ куски $c_1, \\dots, c_t$, а от $b$ — куски $c_{t+1}, \\dots, c_k$. Заметим, что $a - s_{t-1} > 75$, поэтому от $a$ остался кусок $a' = a - s_t = (a - s_{t-1}) - c_t$ такой, что $75 \\ge a' > 75 - c_t \\ge 25$. От $b$ же остался кусок $b'$, такой что $a' + b' = a + b - c = 100$, поэтому $25 \\le b' \\le 75$.\n\nИтак, можно $a'$ и $b'$ отдать одному ребёнку, а остальные куски распределить между остальными детьми так же, как это делалось в новом наборе. Утверждение доказано.\n\n**Замечание.** В доказанном общем утверждении число 25 нельзя заменить на большее, не зависящее от $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16809,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a polygon whose sides are either parallel to the $x$- or $y$-axes, with the lowest horizontal sides lying on the $x$-axis. Starting with a horizontal side on the $x$-axis, traverse the polygon in a clockwise direction, labeling the horizontal sides $1, 2, \\ldots, 1008$ and placing an arrow on each side in the direction of traversal. Each horizontal side $s_i$ forms a rectangle with the $x$-axis, where the height is the vertical distance of $s_i$ above the $x$-axis and the width is the length of $s_i$. If the arrow on $s_i$ points in the positive $y$-axis direction, its area $A_i$ is positive; otherwise, it is negative. The area of the polygon $P$ is given by the absolute value of\n\n$$\nA_1 + A_2 + \\dots + A_{1008}.\n$$\n\nSuppose all side lengths are odd integers. Show that the area of $P$ is even.\n\n**Note:** In general, for such a polygon with $n$ sides, the area is even if and only if $8$ divides $n$ ($8 \\mid n$).",
    "options": [],
    "answer": "See solution",
    "solution": "Since all side lengths are odd, for each $i = 1, \\dots, 1007$, the areas $A_i$ and $A_{i+1}$ have opposite parities. Therefore, among the $1008$ terms, there are $504$ odd terms. The sum of an even number of odd numbers is even, so the area is even.\n\n**Generalization:** The algebraic sum of the lengths of the horizontal sides is $0$, so the number of horizontal sides is even. The number of vertical and horizontal sides are equal, so $4$ divides $n$. The area is $A_1 + \\dots + A_{n/2}$, and the number of odd terms is $n/4$. Thus, the area is even if and only if $n/4$ is even, i.e., $8$ divides $n$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16810,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ and $m$ be integers with $1 < k < m$. For a positive integer $i$, let $L_i$ be the least common multiple of $1, 2, \\dots, i$. Prove that $k$ is a divisor of $L_i \\cdot \\left[ \\binom{m}{i} - \\binom{m-k}{i} \\right]$ for all $i \\ge 1$.\n\nHere, $\\binom{n}{i} = \\frac{n!}{i!(n-i)!}$ denotes a binomial coefficient. Note that $\\binom{n}{i} = 0$ if $n < i$.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove the statement by induction on $m$.\n\n**Base case ($m = k$):**\n\n$$\nL_i \\left[ \\binom{m}{i} - \\binom{m-k}{i} \\right] = L_i \\binom{k}{i} = L_i \\cdot \\frac{k!}{i!(k-i)!} = k \\cdot \\frac{L_i}{i} \\cdot \\frac{(k-1)!}{(i-1)!(k-i)!} = k \\cdot \\frac{L_i}{i} \\cdot \\binom{k-1}{i-1}.\n$$\n\nSince $\\frac{L_i}{i}$ is an integer (by definition of $L_i$), and $\\binom{k-1}{i-1}$ is a binomial coefficient and thus also an integer, we see that $k$ is indeed a divisor.\n\n**Induction step:**\n\nAssume the statement holds for a specific value of $m$. We use the recursion $\\binom{m+1}{i} = \\binom{m}{i} + \\binom{m}{i-1}$ to show it holds for $m+1$:\n\n$$\n\\begin{aligned}\nL_i \\left[ \\binom{m+1}{i} - \\binom{m+1-k}{i} \\right] &= L_i \\left[ \\binom{m}{i} - \\binom{m-k}{i} \\right] + L_i \\left[ \\binom{m}{i-1} - \\binom{m-k}{i-1} \\right] \\\\\n&= L_i \\left[ \\binom{m}{i} - \\binom{m-k}{i} \\right] + \\frac{L_i}{L_{i-1}} \\cdot L_{i-1} \\left[ \\binom{m}{i-1} - \\binom{m-k}{i-1} \\right].\n\\end{aligned}\n$$\n\nNote that $k$ divides both $L_i[\\binom{m}{i} - \\binom{m-k}{i}]$ and $L_{i-1}[\\binom{m}{i-1} - \\binom{m-k}{i-1}]$ by the induction hypothesis (if $i=1$, the latter term is zero), and $L_{i-1}$ divides $L_i$. It follows that $k$ is also a divisor of $L_i[\\binom{m+1}{i} - \\binom{m+1-k}{i}]$, which completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16811,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_1A_2A_3A_4$ be a convex quadrilateral with no pair of parallel sides. For each $i = 1, 2, 3, 4$, define $\\omega_i$ to be the circle touching the quadrilateral externally, and which is tangent to the lines $A_{i-1}A_i$, $A_iA_{i+1}$, and $A_{i+1}A_{i+2}$ (indices are considered modulo 4, so $A_0 = A_4$, $A_5 = A_1$, and $A_6 = A_2$). Let $T_i$ be the point of tangency of $\\omega_i$ with the side $A_iA_{i+1}$. Prove that the lines $A_1A_2$, $A_3A_4$, and $T_2T_4$ are concurrent if and only if the lines $A_2A_3$, $A_4A_1$, and $T_1T_3$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "We start with a reformulation of a well-known statement on harmonic cyclic quadruples $(K_1, K_2, K_3, K_4)$, also provable by polar transformation (projective methods).\n\n**Lemma.** Given four pairwise non-parallel lines $\\ell_i$, $i = 1, 2, 3, 4$, tangent to a circle $\\omega$ at points $K_i$, and such that lines $\\ell_1$, $\\ell_3$, and $K_2K_4$ are concurrent, then lines $\\ell_2$, $\\ell_4$, and $K_1K_3$ are also concurrent.\n\n*Proof.* Let $O$ be the center of $\\omega$, $X = \\ell_1 \\cap \\ell_3 \\cap K_2K_4$, $Y = \\ell_2 \\cap \\ell_4$. We have $OX \\perp K_1K_3$ and $OY \\perp K_2K_4$. Let $Z = OX \\cap K_1K_3$, $T = OY \\cap K_2K_4$. Notice that triangles $\\triangle OK_3X$ and $\\triangle OZK_3$ are similar, and also similar are triangles $\\triangle OK_4Y$ and $\\triangle OTK_4$, hence $OY \\cdot OT = OK_4^2 = OK_3^2 = OX \\cdot OZ$.\n\nThis means that triangles $\\triangle OXT$ and $\\triangle OYZ$ are similar, hence $YZ \\perp OX$, and so $Y \\in K_1K_3$. $\\square$\n\nSuppose now lines $A_2A_3$, $A_4A_1$, and $T_1T_3$ are concurrent at a point $P$. Let $T'_4$, $T'_2$ be the tangency points of lines $A_4A_1$, respectively $A_2A_3$, to circle $\\omega_1$, and let $T'_3$ be the second meeting point of line $T_1T_3$ and circle $\\omega_1$. Let the tangent to $\\omega_1$ at $T'_3$ meet the lines $A_4A_1$, $A_2A_3$ at points $A'_4$, respectively $A'_3$. The (direct) homothety of center $P$ that takes $\\omega_1$ to $\\omega_3$ maps $T'_3$ to $T_3$, hence $A_3A_4 \\parallel A'_3A'_4$.\n\nLet $Q = A_1A_2 \\cap A_3A_4$, $Q' = A_1A_2 \\cap A'_3A'_4$. Applying the Lemma to circle $\\omega_1$ and lines $A_2A_3$, $A'_3A'_4$, $A_4A_1$, $A_1A_2$, yields that points $Q'$, $T'_2$, $T'_4$ are collinear. The (inverse) homothety of center $A_1$ that takes $\\triangle QA_1A_4$ to $\\triangle Q'A_1A'_4$ maps $\\omega_4$ to $\\omega_1$, so maps $T_4$ to $T'_4$, hence $Q'T'_4 \\parallel QT_4$. Similarly, the (inverse) homothety of center $A_2$ that takes $\\triangle QA_2A_3$ to $\\triangle Q'A_2A'_3$ maps $\\omega_2$ to $\\omega_1$, so maps $T_2$ to $T'_2$, hence also $Q'T'_2 \\parallel QT_2$. Since points $Q'$, $T'_2$, $T'_4$ are collinear, it follows points $Q, T_2, T_4$ are also collinear.\n\nThe converse implication is done in a similar way, due to the cyclic nature of the notations used (just increase each index by 1).\n\n*Alternative Solution.* (D. Șerbănescu) Suppose $Q, T_2, T_4$ are collinear. We will show $P, T_1, T_3$ are collinear. We will use the notations of the solution above, but also let $S'_1, S''_1$ be the tangency points of line $A_1A_2$ to circle $\\omega_2$, respectively $\\omega_4$, and let $S'_3, S''_3$ be the tangency points of line $A_3A_4$ to circle $\\omega_2$, respectively $\\omega_4$. Let $T''_4$ be the (other than $T_2$) meeting point of line $QT_2T_4$ and circle $\\omega_2$, and let the tangent line to $\\omega_2$ at $T''_4$ (parallel to $A_1A_4$) meet $A_2A_3$ at $P'$ (via the (direct) homothety of center $Q$ that takes $\\omega_4$ to $\\omega_2$).\n\nClearly $\\triangle PT_2T_4 \\sim \\triangle P'T_2T_4''$ and $\\triangle P'T_2T_4''$ is isosceles, so $PT_2 = PT_4$ (in other words, if $Q, T_2, T_4$ are collinear then $PT_2 = PT_4$; the other implication trivially also holds, but is irrelevant here).\n\nFrom $PT_2 = PT_4$ and $PT'_2 = PT'_4$ follows $T'_2T_2 = T'_4T_4$. As external tangents, $T'_2T_2 = T_1S'_1$ and $T'_4T_4 = T_1S''_1$, hence $T_1$ is the midpoint of $S'_1S''_1$. Similarly, $T_3$ is the midpoint of $S'_3S''_3$. It follows that $P, T_1, T_3$ lie on the radical axis of the circles $\\omega_2$ and $\\omega_4$, hence are collinear.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 16812,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $f(x) = x^3 + \\log_2(x + \\sqrt{x^2+1})$. For any $a, b \\in \\mathbb{R}$, to satisfy $f(a) + f(b) \\ge 0$, the condition $a + b \\ge 0$ is\n\n(A) necessary and sufficient  \n(B) not necessary but sufficient  \n(C) necessary but not sufficient  \n(D) neither necessary nor sufficient",
    "options": [],
    "answer": "See solution",
    "solution": "Obviously, $f(x) = x^3 + \\log_2(x + \\sqrt{x^2+1})$ is an odd function and is monotonically increasing. So, if $a + b \\ge 0$, i.e., $a \\ge -b$, we get $f(a) \\ge f(-b)$, so $f(a) \\ge -f(b)$, which means $f(a) + f(b) \\ge 0$.\n\nOn the other hand, if $f(a) + f(b) \\ge 0$, then $f(a) \\ge -f(b) = f(-b)$. So $a \\ge -b$, i.e., $a + b \\ge 0$. \n\n*Answer: (A) necessary and sufficient.*",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16813,
    "subject": "Mathematics (Olympiad)",
    "question": "Yesterday there were $3t$ boys and $2t$ girls in the playground. Today, there are $3t - 6$ boys and $2t - 7$ girls. If $3t - 6 = (2t - 7)^2$, how many children were in the playground yesterday?",
    "options": [],
    "answer": "See solution",
    "solution": "From $3t - 6 = (2t - 7)^2$, we expand to get:\n\n$$\n3t - 6 = 4t^2 - 28t + 49\n$$\n\nBringing all terms to one side:\n\n$$\n4t^2 - 31t + 55 = 0\n$$\n\nSolving the quadratic:\n\n$$\nt = \\frac{31 \\pm \\sqrt{31^2 - 4 \\cdot 4 \\cdot 55}}{8} = \\frac{31 \\pm \\sqrt{961 - 880}}{8} = \\frac{31 \\pm 9}{8}\n$$\n\nSo $t = 5$ or $t = 2.75$. Since $t$ must be an integer, $t = 5$.\n\nYesterday, there were $3t = 15$ boys and $2t = 10$ girls, so $15 + 10 = 25$ children in total.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16814,
    "subject": "Mathematics (Olympiad)",
    "question": "a) There are three numbers $a, b, c$ such that $a \\leq b \\leq c$. Let $p, q, r$ be the pairwise sums $a+b$, $b+c$, $c+a$ in the order such that $p \\leq q \\leq r$. Given that $r-q = q-p$, is it certainly true that $c-b = b-a$?\n\nb) There are four numbers $e, f, g, h$ such that $e \\leq f \\leq g \\leq h$. Let $u, v, w, x, y, z$ be the pairwise sums of those numbers, in the order $u \\leq v \\leq w \\leq x \\leq y \\leq z$. Given that $z-y = y-x = x-w = w-v = v-u$, is it certainly true that $h-g = g-f = f-e$?",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* a) Yes; b) No.\n\n*Solution:*\n\na) If $a \\leq b \\leq c$, then $a+b \\leq a+c \\leq b+c$, so $p = a+b$, $q = a+c$, and $r = b+c$. The equality $r-q = q-p$ becomes $$(b+c) - (a+c) = (a+c) - (a+b)$$ which simplifies to $b-a = c-b$.\n\nb) Let $e = 0$, $f = 1$, $g = 2$, and $h = 4$. Their pairwise sums in increasing order are $u = 1$, $v = 2$, $w = 3$, $x = 4$, $y = 5$, and $z = 6$. Thus $z-y = y-x = x-w = w-v = v-u = 1$, but $h-g = 2 \\neq 1 = g-f$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16815,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral such that $BC$ and $AD$ meet at a point $P$. Consider a point $Q$, different from $P$, on the line $BP$ such that $PQ = BP$, and construct the parallelograms $CAQR$ and $DBCS$. Prove that the points $C$, $Q$, $R$, and $S$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "It is enough to prove that $\\angle RQC = \\angle RSC$. ($\\dagger$)\n\n![](images/RMC2011_2_p45_data_aa2bde1675.png)\n\nNow $\\angle RQC = \\angle ACQ = \\angle ACB = \\angle ADB$. (1)\n\nTake $T$ such that $QABT$ is a parallelogram. Then $\\overrightarrow{BT} = \\overrightarrow{AQ} = \\overrightarrow{CR}$ and $\\overrightarrow{BD} = \\overrightarrow{CS}$ imply $\\triangle BTD \\equiv \\triangle CRS$, so $\\angle RSC = \\angle TDB$. (2)\n\nOn the other hand, since $P$ is the midpoint of the segment $BQ$ and $ABTQ$ is a parallelogram, $P$ is also the midpoint of the segment $AT$, so $\\angle TDB = \\angle ADB$.\n\nCombining this with (1) and (2), we get ($\\dagger$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16816,
    "subject": "Mathematics (Olympiad)",
    "question": "用 $\\mathbb{N}$ 表示所有正整數形成的集合。找到所有可能的函數 $f : \\mathbb{N} \\to \\mathbb{N}$，使得對於所有 $m, n \\in \\mathbb{N}$，都有 $|f(m+n) - f(m)|$ 整除 $f(n)$。",
    "options": [],
    "answer": "See solution",
    "solution": "根据题目条件，有如下事实：\n\n1. $f$ 是单射。\n2. $f(m+n) \\leq f(m) + f(n)$。\n3. $|f(m+1) - f(m)| \\leq f(1)$。\n4. $(f(m+1) - f(1)) \\mid f(m)$。\n\n由于 $f$ 是单射，对于充分大的 $m$，有 $f(m+1) > 3f(1)$。结合上述事实，得到：\n\n$$\n2f(1) < f(m+1) - f(1) \\mid |f(m) - f(m+1) + f(1)| \\leq 2f(1)\n$$\n\n因此，必须有 $f(m) - f(m+1) + f(1) = 0$，即 $f(m) = mf(1) + c$，其中 $c$ 为常数，适用于充分大的 $m$。所以对于充分大的 $m, n$，有 $f(m+n) - f(m) = nf(1)$ 整除 $f(n) = nf(1) + c$，这迫使 $c = 0$。断言 $f(n) = nf(1)$ 对所有 $n \\in \\mathbb{N}$ 都成立。反证法，假设存在最大的 $n$ 使得 $f(n) \\ne nf(1)$，取充分大的 $m$，有 $nf(1) = f(m+n) - f(m) \\mid f(n)$，因此 $f(n) = knf(1) = f(kn)$，其中 $k \\geq 2$，与 $f$ 是单射矛盾。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16817,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist an irrational number $x$ such that there are at most finitely many positive integers $n$ satisfying\n\n$$\n\\{kx\\} \\geq \\frac{1}{n+1}\n$$\n\nfor every $k \\in \\{1, \\dots, n\\}$?\n\nNote: Here, for a positive real number $y$, $\\{y\\}$ denotes the fractional part of $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Nonexistence. Assume that there exists such a positive integer $n$. Let $x$ be an irrational number. Since the terms of the sequence $\\{x\\}$, $\\{2x\\}$, $\\dots$ are distinct and densely distributed in the interval $(0, 1)$, there exist infinitely many positive integers $d$ satisfying\n\n$$\n\\{dx\\} = \\min \\{ \\{ax\\} \\mid a = 1, 2, \\dots, d \\}\n$$\n\nArrange these $d$ in an infinite increasing sequence $d_1 < d_2 < \\dots$.\n\nWe will now show that for all $n = d_i - 1$ ($i \\ge 2$), we always have\n\n$$\n\\min \\{ \\{kx\\} \\mid k = 1, \\dots, n \\} = \\{d_{i-1}x\\} \\ge \\frac{1}{d_i}.\n$$\n\nSuppose that this inequality does not hold, then $d_i \\cdot \\{d_{i-1}x\\} \\le 1$. Thus,\n\n$$\n\\{d_{i-1}d_i x\\} = \\{d_i \\cdot \\lfloor d_{i-1}x \\rfloor + d_i \\cdot \\{d_{i-1}x\\}\\} = d_i \\cdot \\{d_{i-1}x\\}.\n$$\n\nOn the other hand,\n\n$$\n\\{d_i d_{i-1} x\\} = \\{d_{i-1} \\lfloor d_i x \\rfloor + d_{i-1} \\{d_i x\\}\\} \\le d_{i-1} \\{d_i x\\}.\n$$\n\nCombining the above two equations, we obtain $d_{i-1} \\cdot \\{d_i x\\} \\ge d_i \\cdot \\{d_{i-1} x\\}$. However, $d_{i-1} < d_i$ and $\\{d_i x\\} < \\{d_{i-1} x\\}$, which leads to a contradiction!\n\nTherefore, $n = d_i - 1$ satisfies the given condition, and there are infinitely many such $n$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16818,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integer solutions to the equation\n\n$$\n2x^3 - y^2 = 3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the equation modulo $8$. Then $y^2 \\equiv 0, 1, 4 \\pmod{8}$. This means $2x^3 \\equiv 3, 4, 7 \\pmod{8}$, all of which are impossible. The equation lacks a solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16819,
    "subject": "Mathematics (Olympiad)",
    "question": "Using each of the ten digits exactly once, build two numbers (neither of which can start with 0) so that the absolute value of their difference is as small as possible.",
    "options": [],
    "answer": "See solution",
    "solution": "It is almost obvious that both numbers should be five-digit, since otherwise their difference will have at least five digits. Indeed, the smallest six-digit number that we can build is $102345$, the biggest four-digit number is $9876$, and their difference is $92469$. For other cases of integers with different numbers of digits, the difference will be even bigger than that.\n\nIn order to minimize the difference of two five-digit numbers, we should take their first digits to be consecutive, because otherwise their difference will also have five digits.\n\nNow we need to consider all possibilities when a number with greater first digit is complemented by the minimal digits possible, and the other number is complemented by the maximal digits possible.\n\n$$\n90123 - 87654 = 2469, \\\\\n80123 - 79654 = 469, \\\\\n70123 - 69854 = 269, \\\\\n60123 - 59874 = 249, \\\\\n50123 - 49876 = 247, \\\\\n40125 - 39876 = 249, \\\\\n30145 - 29876 = 269, \\\\\n20345 - 19876 = 469.\n$$\n\nWe see that the minimal possible difference is attained for the numbers $50123$ and $49876$, and is equal to $247$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16820,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\geq 3$ be an integer.\n\nOn a circle, there are $n$ points. Each is labelled with a real number at most $1$ such that each number is the absolute value of the difference of the two numbers immediately preceding it in clockwise order.\n\nDetermine the maximal possible value of the sum of all numbers as a function of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $n$ is a multiple of $3$, the maximal sum is $\\frac{2n}{3}$. Otherwise, the sum is always $0$.\n\nAll the numbers are absolute values, so they are non-negative. Either all of them are zero, then their sum is zero, or there is a maximal positive number. If we scale all numbers so that this maximum is $1$, the sum can only get larger, so we may assume the maximum is $1$.\n\nThe difference of two neighboring numbers less than $1$ is also less than $1$. If we iterate around the circle, all numbers must be less than $1$, which is impossible if the maximum is $1$.\n\nTherefore, in the case of maximum $1$, at least one of each pair of neighbors must equal $1$. We can distinguish two cases for the two numbers after the maximum. Either the number immediately after the maximum is also $1$, which means the list of differences continues as $1, 1, 0, 1, 1, 0, \\dots$, or the next numbers are $1, x, 1$. However, this means $|1-x| = 1$, so $x = 0$, and we get the list $1, 0, 1, 1, 0, 1, \\dots$.\n\nIn both subcases, $n$ must be divisible by $3$ to wrap correctly around the circle with these patterns. Then, we get a sum of $\\frac{2n}{3}$.\n\nIn conclusion, the maximal sum is $0$ if $n$ is not divisible by $3$, and $\\frac{2n}{3}$ if $n$ is divisible by $3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16821,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are 9 towns $A, B, C, D, E, F, G, H, I$. Each bus is represented by a quintuple of towns at which it stops. The stops for the buses must be chosen so that for each pair of towns, there is a bus that stops at both. What is the maximum number of towns that can be served under these conditions?",
    "options": [],
    "answer": "See solution",
    "solution": "Let one bus be $ABCDE$. To ensure every pair of towns is covered, we must couple each of $A, B, C, D, E$ with each of $F, G, H, I$. Buses $AFGHI$ and $FGHIE$ cover $A$ and $E$, while buses $FGBCD$ and $BCDHI$ cover $B, C, D$. These buses also cover all pairs involving $F, G, H, I$. Thus, at least 9 towns can be served.\n\nSuppose there are 10 or more towns. Any town forms a pair with at least 9 others. Since each bus stops at 5 towns, at least 3 buses are needed to cover these 9 or more pairs for each town, so each town is served by at least 3 buses. Therefore, the total number of stops is at least $3 \\times 10 = 30$, but with 5 buses and 5 stops each, there are only $5 \\times 5 = 25$ stops. This is a contradiction. Thus, the maximum number of towns that can be served is $\\mathbf{9}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16822,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function such that its 2-fold composition is equal to the floor function, i.e. $f(f(x)) = \\lfloor x \\rfloor$, for any real number $x$. Prove that there exist distinct real numbers $a$ and $b$ such that $|f(a) - f(b)| \\geq |a - b|$.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that $f(n) \\in \\mathbb{Z}$ for any integer $n$. Indeed, since $f(f(f(x))) = f(\\lfloor x \\rfloor) = \\lfloor f(x) \\rfloor$, we have $f(n) = \\lfloor f(n) \\rfloor$ for any integer $n$, implying $f(n) \\in \\mathbb{Z}$.\n\nSuppose that for all $a, b \\in \\mathbb{Z}$ we have $|f(a) - f(b)| < |a - b|$. Then $|f(n + 1) - f(n)| < 1$ for any integer $n$. Since both $f(n)$ and $f(n + 1)$ are integers, we derive that $f(n) = f(n + 1)$, hence $f(n) = f(0)$ for any integer $n$. It follows that $n = f(f(n)) = f(f(0)) = 0$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16823,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a sequence where $OA_n = \\sqrt{n+15}$, find the value of $OA_{21}$.",
    "options": [],
    "answer": "See solution",
    "solution": "$OA_{21} = \\sqrt{21+15} = \\sqrt{36} = 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16824,
    "subject": "Mathematics (Olympiad)",
    "question": "As illustrated in Fig. 2.1, in $\\triangle ABC$, $AB < AC$, $PB, PC$ are tangent to the circumcircle $O$ of $\\triangle ABC$. Let $R$ be a point on $\\widehat{AC}$ such that $AR \\parallel BC$.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p340_data_87c4bf6a59.png)\n\nLet $Q$ be the other intersection of $PR$ and circle $O$. Let $I$ be the incenter of $\\triangle ABC$, and let $ID \\perp BC$ with foot $D$. Let $G$ be the other intersection of $QD$ and circle $O$. Suppose that the line through $I$ and perpendicular to $AI$ intersects the lines $AG$ and $AC$ at the points $M$ and $N$, respectively. Let $S$ be the midpoint of $\\overline{AR}$, and $T$ be the other intersection of the line $SN$ and circle $O$. Prove that $M$, $B$, $T$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution**\nLet $AB$ meet $MN$ at $K$. Connect $GB$, $GC$, $GK$, $GN$, $RB$, $RC$, $QB$, $QC$, $BI$, $CI$. See Fig. 2.2.\n\nSince $PB$, $PC$ are tangent to circle $O$,\n\n$$\n\\frac{RB}{BQ} = \\frac{PR}{PB} = \\frac{PR}{PC} = \\frac{RC}{CQ},\n$$\n\nand hence\n$$\n\\frac{BQ}{CQ} = \\frac{RB}{RC}.\n$$\n\nMoreover,\n$$\n\\frac{BD}{CD} = \\frac{S_{\\triangle BGQ}}{S_{\\triangle CGQ}} = \\frac{BG \\cdot BQ}{CG \\cdot CQ} = \\frac{BG \\cdot RB}{CG \\cdot RC}\n$$\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p341_data_8c9c60f8a5.png)\n\nFrom $AR \\parallel BC$, we have $RB = AC$, $RC = AB$. Together, they imply\n\n$$\n\\frac{BD}{CD} = \\frac{BG \\cdot AC}{CG \\cdot AB},\n$$\n\nand\n$$\n\\frac{BG}{CG} = \\frac{BD \\cdot AB}{CD \\cdot AC}\n$$\n$$\n= \\frac{BI \\cdot \\cos \\frac{B}{2} \\sin C}{CI \\cdot \\cos \\frac{C}{2} \\sin B} = \\frac{BI \\cdot \\sin \\frac{C}{2}}{CI \\cdot \\sin \\frac{B}{2}} = \\frac{BI^2}{CI^2}. \\quad \\textcircled{1}\n$$\n\nAs $AI \\perp KN$, we find\n$$\n\\angle AKN = \\angle ANK = \\frac{B+C}{2}, \\quad \\angle BKI = \\angle INC = 180^\\circ - \\frac{B+C}{2},\n$$\n$$\n\\angle KBI = \\frac{B}{2} = \\frac{B+C}{2} - \\frac{C}{2} = \\angle ANI - \\angle NCI = \\angle NIC,\n$$\n\nand $\\triangle KBI \\sim \\triangle NIC$. Hence,\n$$\n\\frac{BK}{CN} = \\frac{BK}{IN} \\cdot \\frac{IK}{CN} = \\frac{BI^2}{CI^2}. \\quad \\textcircled{2}\n$$\n\nBy $\\textcircled{1}$ and $\\textcircled{2}$, $\\frac{BG}{CG} = \\frac{BK}{CN}$; in addition, $\\angle GBK = \\angle GBA = \\angle GCA = \\angle GCN$. Therefore, $\\triangle GBK \\sim \\triangle GCN$, giving\n$$\n\\angle GKA = 180^\\circ - \\angle GKB = 180^\\circ - \\angle GNC = \\angle GNA.\n$$\n\nIt follows that $A$, $G$, $K$, $N$ lie on a circle. Connect $BT$. Since $S$ is the midpoint of $\\overline{AR}$ and $AR \\parallel BC$, we have\n$$\n\\angle BTN = \\angle BTS = \\frac{B+C}{2} = \\angle AKN,\n$$\n\nand hence $K$, $B$, $T$, $N$ are concyclic.\n\nFinally, by the radical axis theorem we conclude that the lines $AG$, $KN$, $BT$ are concurrent, and $M$, $B$, $T$ are collinear. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16825,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a tetrahedron $ABCD$ and the points $M, N$ on the edges $AC$ and $BD$, respectively. Prove that for any point $P$ of the segment $MN$, $P \\neq M$, $P \\neq N$, there exists a unique pair of points $(X, Y)$, with $X$ and $Y$ on the edges $AB$ and $CD$, respectively, such that the points $X, P$ and $Y$ are collinear.\n\n![](images/RMC_2023_v2_p36_data_0c62e481bd.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $P \\neq M$ and $P \\neq N$, if $X, Y, P$ are collinear, we have $X \\notin \\{A, B\\}$ and $Y \\notin \\{C, D\\}$. Indeed, if $X = A$, then $XY \\subset (ACD)$ and thus $P \\in (ACD)$, which is false. The other situations are analogous.\n\n*Existence:* Since $P \\in MN$, the point $P$ lies in the interior of the triangle $ANC$. Denote $\\{Q\\} = AP \\cap CN$.\n\nFrom $CN \\subset \\operatorname{Int}(BCD)$, it follows that $Q \\in \\operatorname{Int}(BCD)$, thus the line $BQ$ intersects the open segment $CD$. Denote $\\{Y\\} = BQ \\cap CD$. Because $P \\in AQ \\subset ABY$, we deduce that $P$ lies in the interior of the triangle $ABY$, thus the line $YP$ intersects the open segment $AB$. Denote $\\{X\\} = PY \\cap AB$. The pair $(X, Y)$ satisfies the statement.\n\n*Uniqueness:* Assume that a pair of points $(X', Y')$ exists, with $X' \\in AB$, $Y' \\in CD$, $(X, Y) \\neq (X', Y')$, such that the points $X'$, $P$ and $Y'$ are collinear. We consider $Y' \\neq Y$ (the situation $X' \\neq X$ is analogous). If $X' = X$, then $Y' \\in XP \\cap CD$, therefore $Y' = Y$, which is false. Consequently $X' \\neq X$, so the distinct straight lines $XY$ and $X'Y'$ intersect at $P$. If $\\alpha = (XY, X'Y')$, we obtain $XX' = AB \\subset \\alpha$ and $YY' = CD \\subset \\alpha$, thus the points $A, B, C$ and $D$ are coplanar, which is false. Consequently, the pair $(X, Y)$ is unique.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16826,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ such that\n\n$$\nf^{(a^2+b^2)}(a+b) = af(a) + bf(b)\n$$\n\nfor every $a, b \\in \\mathbb{Z}$. Here, $f^n$ denotes the $n$th iteration of $f$, i.e., $f^{(0)}(x) = x$ and $f^{(n+1)}(x) = f(f^{(n)}(x))$ for all $n \\ge 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Refer to the main equation as $E(a, b)$. $E(0, b)$ reads as $f^{b^2}(b) = bf(b)$. For $b = -1$ this gives $f(-1) = 0$.\n\nNow $E(a, -1)$ reads as\n\n$$\nf^{a^2+1}(a-1) = af(a) = f^{a^2}(a).\n$$\n\nFor $x \\in \\mathbb{Z}$ define the orbit of $x$ by $O(x) = \\{x, f(x), f(f(x)), \\dots\\} \\subseteq \\mathbb{Z}$. We see that the orbits $O(a-1)$ and $O(a)$ differ by finitely many terms. Hence, any two orbits differ by finitely many terms. In particular, this implies that either all orbits are finite or all orbits are infinite.\n\n*Case 1: All orbits are finite.*\n\nThen $O(0)$ is finite. Using $E(a, -a)$ we get\n\n$$\na(f(a) - f(-a)) = af(a) - af(-a) = f^{2a^2}(0) \\in O(0) \\quad (1)\n$$\n\nFor $|a| > \\max_{z \\in O(0)} |z|$, this yields $f(a) = f(-a)$ and $f^{2a^2}(0) = 0$. Therefore, the sequence $(f^k(0) : k = 0, 1, \\dots)$ is purely periodic with a minimal period $T$ which divides $2a^2$. Analogously, $T$ divides $2(a+1)^2$, therefore, $T \\mid \\gcd(2a^2, 2(a+1)^2) = 2$, i.e., $f(f(0)) = 0$ and $a(f(a) - f(-a)) = f^{2a^2}(0) = 0$ for all $a$. Thus,\n\n$$\nf(a) = f(-a) \\quad \\text{for all } a \\neq 0; \\qquad (2)\n$$\n\n$$\n\\text{in particular, } f(1) = f(-1) = 0 \\qquad (3)\n$$\n\nNext, for each $n \\in \\mathbb{Z}$, by $E(n, 1-n)$ we get\n\n$$\nnf(n) + (1-n)f(1-n) = f^{n^2+(1-n)^2}(1) = f^{2n^2-2n}(0) = 0 \\qquad (4)\n$$\n\nAssume that there exists some $m \\neq 0$ such that $f(m) \\neq 0$. Choose such an $m$ for which $|m|$ is minimal possible. Then $|m| > 1$ due to (2); $f(|m|) \\neq 0$ due to (3); and $f(1-|m|) \\neq 0$ due to (4) for $n = |m|$. This contradicts the minimality assumption.\n\nSo, $f(n) = 0$ for $n \\neq 0$. Finally, $f(0) = f^3(0) = f^4(2) = 2f(2) = 0$. Clearly, the function $f(x) \\equiv 0$ satisfies the problem condition, which provides the first of the two answers.\n\n*Case 2: All orbits are infinite.*\n\nSince the orbits $O(a)$ and $O(a-1)$ differ by finitely many terms for all $a \\in \\mathbb{Z}$, each two orbits $O(a)$ and $O(b)$ have infinitely many common terms for arbitrary $a, b \\in \\mathbb{Z}$.\n\nFor a minute, fix any $a, b \\in \\mathbb{Z}$. We claim that all pairs $(n, m)$ of nonnegative integers such that $f^n(a) = f^m(b)$ have the same difference $n - m$. Arguing indirectly, we have $f^n(a) = f^m(b)$ and $f^p(a) = f^q(b)$ with, say, $n - m > p - q$, then $f^{p+m+k}(b) = f^{p+n+k}(a) = f^{q+n+k}(b)$, for all nonnegative integers $k$. This means that $f^{\\ell+(n-m)-(p-q)}(b) = f^\\ell(b)$ for all sufficiently large $\\ell$, i.e., that the sequence $(f^n(b))$ is eventually periodic, so $O(b)$ is finite, which is impossible.\n\nNow, for every $a, b \\in \\mathbb{Z}$, denote the common difference $n - m$ defined above by $X(a, b)$. We have $X(a-1, a) = 1$ by (1). Trivially, $X(a, b) + X(b, c) = X(a, c)$, as if $f^n(a) = f^m(b)$ and $f^p(b) = f^q(c)$, then $f^{p+n}(a) = f^{p+m}(b) = f^{q+m}(c)$. These two properties imply that $X(a, b) = b - a$ for all $a, b \\in \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16827,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the system of equations\n\n$$\n\\begin{cases}\n6x - y + z^2 = 3, \\\\\nx^2 - y^2 - 2z = -1, \\\\\n6x^2 - 3y^2 - y - 2z^2 = 0\n\\end{cases} \\quad (x, y, z \\in \\mathbb{R})\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "From the given system, it follows that\n\n$$\n(6x^2 - 3y^2 - y - 2z^2) - 3(x^2 - y^2 - 2z + 1) - (6x - y + z^2 - 3) = 0.\n$$\n\nThis equation can be rewritten as $(x-z)(x+z-2) = 0$. Hence, $x = z$ or $x+z = 2$.\n\n*If $x = z$.* The given system is equivalent to\n\n$$\n\\begin{cases}\nx^2 + 6x - y = 3, \\\\\nx^2 - 2x - y^2 = -1\n\\end{cases}\n$$\n\nFrom the second equation, it follows that $y^2 = (x-1)^2$ or $y = \\pm (x-1)$. Substitute $y = 1-x$ and $y = 1+x$ into the first equation to obtain four solutions $(x, y, z)$ in this case, which are\n\n$$\n\\begin{pmatrix}\n\\dfrac{-5 \\pm \\sqrt{33}}{2}, & \\dfrac{-7 \\pm \\sqrt{33}}{2}, & \\dfrac{-5 \\pm \\sqrt{33}}{2} \\\\\n\\dfrac{-7 \\pm \\sqrt{65}}{2}, & \\dfrac{9 \\mp \\sqrt{65}}{2}, & \\dfrac{-7 \\pm \\sqrt{65}}{2}\n\\end{pmatrix}.\n$$\n\n*If $x+z=2$.* Letting $z=2-x$, the given system is equivalent to\n\n$$\n\\begin{cases}\nx^2 + 2x + 1 - y = 0, \\\\\nx^2 + 2x - y^2 - 3 = 0\n\\end{cases}\n$$\n\nIt follows that $1 - y = -y^2 - 3$ or $y^2 - y + 4 = 0$. But this equation has no real roots, hence there is no solution for the given system in this case.\n\n$\\Box$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16828,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c > 0$. Prove that\n$$\n\\frac{ab}{a+b+1} + \\frac{bc}{b+c+1} + \\frac{ca}{c+a+1} \\geq \\frac{3}{2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Set $x = \\frac{1}{a} + 1$, $y = \\frac{1}{b} + 1$, and $z = \\frac{1}{c} + 1$. Then,\n$$\n\\frac{ab}{a+b+1} = \\frac{z}{x+y}, \\quad \\frac{bc}{b+c+1} = \\frac{x}{y+z}, \\quad \\frac{ca}{c+a+1} = \\frac{y}{z+x}.\n$$\nThus, the original inequality is equivalent to Nesbitt's inequality:\n$$\n\\frac{z}{x+y} + \\frac{x}{y+z} + \\frac{y}{z+x} \\geq \\frac{3}{2},\n$$\nwhich follows from the harmonic-arithmetic mean inequality applied to $(x+y, y+z, z+x)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16829,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $3n + 1 = x^2$ and $4n + 1 = y^2$. What can be said about the divisibility properties of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "We have $y^2 - x^2 = n$. As $y$ is odd, $y^2 \\equiv_8 1$, so $n$ has to be even, therefore $x$ is odd, and therefore $y^2 - x^2 \\equiv_8 1 - 1 = 0$, hence $8 \\mid n$.\n\nNow notice that $4x^2 - 3y^2 = 1$, so setting $w = 2x$ we obtain Pell's equation $w^2 - 3y^2 = 1$. This has solutions $(2 + \\sqrt{3})^m = w_m + \\sqrt{3}y_m$, but only the odd $m$ solutions correspond to an even $w_m$. Starting with the base solution $2 + \\sqrt{3}$, and multiplying through by $(2 + \\sqrt{3})^2$ we get the recursions\n\n$$\nw_{m+1} = 2x_{m+1} = 14x_m + 12y_m, \\quad \\text{and} \\quad y_{m+1} = 8x_m + 7y_m.\n$$\n\nThen $x_{m+1} = 7x_m + 6y_m \\equiv_7 -y_m$, and $y_{m+1} = 8x_m + 7y_m \\equiv_7 x_m$, hence $y_{m+1}^2 - x_{m+1}^2 \\equiv_7 x_m^2 - y_m^2$, and by induction, $x_m^2 - y_m^2 \\equiv_7 0$. Hence $7$ divides $n = y_m^2 - x_m^2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16830,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the focus and directrix of the parabola $y^2 = 2px$ ($p > 0$) be $F$ and $l$, respectively. Points $A$ and $B$ move on the parabola such that $\\angle AFB = \\frac{\\pi}{3}$. Let $M$ be the midpoint of segment $AB$, and let $N$ be the projection of $M$ onto $l$. Find the maximum value of $\\frac{|MN|}{|AB|}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $\\angle ABF = \\theta$ ($0 < \\theta < \\frac{2\\pi}{3}$). By the Law of Sines,\n\n$$\n\\frac{|AF|}{\\sin \\theta} = \\frac{|BF|}{\\sin\\left(\\frac{2\\pi}{3} - \\theta\\right)} = \\frac{|AB|}{\\sin \\frac{\\pi}{3}}\n$$\n\nThus,\n\n$$\n\\frac{|AF| + |BF|}{\\sin \\theta + \\sin\\left(\\frac{2\\pi}{3} - \\theta\\right)} = \\frac{|AB|}{\\sin \\frac{\\pi}{3}}\n$$\n\nSo,\n\n$$\n\\frac{|AF| + |BF|}{|AB|} = \\frac{\\sin \\theta + \\sin\\left(\\frac{2\\pi}{3} - \\theta\\right)}{\\sin \\frac{\\pi}{3}} = 2\\cos\\left(\\theta - \\frac{\\pi}{3}\\right).\n$$\n\nBy the definition of a parabola and properties of a trapezoid,\n\n$$\n|MN| = \\frac{|AF| + |BF|}{2}.\n$$\n\nTherefore,\n\n$$\n\\frac{|MN|}{|AB|} = \\cos\\left(\\theta - \\frac{\\pi}{3}\\right).\n$$\n\nThe maximum value is $1$ when $\\theta = \\frac{\\pi}{3}$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p51_data_f3364f2bdd.png)\n\nThe answer is $1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16831,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral with the shortest side $AB$ strictly less than the longest side $CD$. Show that there exists a point $E$ on the segment $CD$ such that for any point $P$ (different from $E$) on the segment $CD$, the length of $O_1O_2$ is constant, where $O_1$ and $O_2$ are the circumcenters of the triangles $APD$ and $BPE$, respectively.",
    "options": [],
    "answer": "See solution",
    "solution": "**Claim:** The point $E$ is the intersection point of the line parallel to $AD$ through $B$ and the line $CD$.\n\nLet $E$ be the intersection point of the line parallel to $AD$ through $B$ and the line $CD$. Firstly, we will show that $E$ is on the segment $CD$. Since $AB \\leq AD$ and $BC \\leq CD$ (because $AB$ and $CD$ are the shortest and longest sides, respectively), then $\\angle ABD \\geq \\angle ADB$ and $\\angle CBD \\geq \\angle BDC$. Therefore $\\angle ABC = \\angle ABD + \\angle DBC \\geq \\angle ADB + \\angle BDC = \\angle ADC$. Similarly, we will have that $\\angle BAD \\geq \\angle BCD$. Thus, $\\angle ABC + \\angle BAD \\geq \\angle BCD + \\angle CDA$, and then $\\angle ABC + \\angle BAD \\geq 180^\\circ$. Moreover, one can show that $\\angle ABC + \\angle BAD > 180^\\circ$. (If $\\angle ABC + \\angle BAD = 180^\\circ = \\angle BCD + \\angle CDA$, we would have that $\\angle ABC = \\angle ADC$ and $\\angle BAD = \\angle BCD$, and then $ABCD$ would be a parallelogram; contradicting the fact that $AB < CD$.) Therefore, we can move the point $C$ along the line $CD$ closer to $D$ to the point $E'$,  \n\n![](images/Tajland_2014_p10_data_0a1f561617.png)\n\nmaking the angle $ABC$ smaller, so that $\\angle E'BA + \\angle BAD = 180^\\circ$, i.e., $BE' \\parallel AD$. Thus, $E'$ is the same as the point $E$ stated above, and lies on the segment $CD$.\n\nSecondly, we will show that for an arbitrary point $P$ different from $E$ on the segment $CD$, $\\angle O_1PO_2 = \\angle APB$.\n\n**Case 1:** $\\angle ADP < 90^\\circ$. Thus $\\angle BEC < 90^\\circ$, and $O_1$ and $C$ are on opposite sides of $PA$, and $O_2$ and $C$ are on opposite sides of $PB$. By simple angle chasing (using angles at centres) we get that\n\n$$\n\\angle APO_1 = 90^\\circ - \\angle ADP = 90^\\circ - \\angle BEC = \\angle BPO_2.\n$$\n\nTherefore $\\angle O_1PO_2 = \\angle APB$.\n\n**Case 2:** $\\angle ADP \\geq 90^\\circ$. Similarly, we get $\\angle BEC \\geq 90^\\circ$ but now $O_1$ and $C$ are on the same side of $AP$, and $O_2$ and $C$ are on the same side of $BP$. So by angle chasing we get\n\n$$\n\\angle APO_1 = \\angle ADP - 90^\\circ = \\angle BEC - 90^\\circ = \\angle BPO_2.\n$$\n\nTherefore $\\angle O_1PO_2 = \\angle APB$.\n\nFinally, using the Law of Sines for $AP$ and $BP$ on circles $O_1$ and $O_2$, respectively, we get\n\n$$\n\\frac{AP}{BP} = \\frac{2O_1P \\sin(\\angle ADP)}{2O_2P \\sin(\\angle BEP)} = \\frac{O_1P}{O_2P}.\n$$\n\nThus we have that $\\triangle APB \\sim \\triangle O_1PO_2$, and to conclude that\n\n$$\nO_1O_2 = \\frac{AB}{AP} \\cdot O_1P = \\frac{AB}{2\\sin(\\angle ADC)}\n$$\n\nwhich is a constant independent of the point $P$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16832,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the largest integer $N$ satisfying the following condition: for every cell labeling of a $5 \\times 5$ array from $1$ through $25$ such that no two cells bear the same number, the numbers in some $2 \\times 2$ square add up to at least $N$.",
    "options": [],
    "answer": "See solution",
    "solution": "The required maximum is $N = 45$, and is achieved, for instance, by the following extremal cell labeling:\n\n![](<table><tr><td>25</td><td>5</td><td>24</td><td>6</td><td>23</td></tr><tr><td>11</td><td>4</td><td>12</td><td>3</td><td>13</td></tr><tr><td>22</td><td>7</td><td>21</td><td>8</td><td>20</td></tr><tr><td>14</td><td>2</td><td>15</td><td>1</td><td>16</td></tr><tr><td>19</td><td>9</td><td>18</td><td>10</td><td>17</td></tr></table>)\n\nIn this cell labeling, every $2 \\times 2$ square with an odd rank top row has sum $45$, and every $2 \\times 2$ square with an even rank top row has sum $44$; the configuration is derived from the considerations below.\n\nWrite $n = 5$ and $m = \\lfloor n/2 \\rfloor = 3$, to show that, if $n$ is odd, then for every injective cell labeling of an $n \\times n$ array from $1$ through $n^2$, the labels in some $2 \\times 2$ square add up to at least $8m^2 - 11m + 6$ (which equals $45$ in the case at hand).\n\nLabel rows downward and columns rightward, both from $1$ through $n$, and let $s_{ij}$ be the sum of all numbers labeling the cells in the cross formed by the $i$-th row and the $j$-th column. Formally, letting $a_{ij}$ denote the label assigned to the cell on the $i$-th row and $j$-th column,\n\n$$\ns_{ij} = \\sum_{k=1}^{n} (a_{ik} + a_{kj}) - a_{ij}.\n$$\n\nUse $\\equiv$ to denote congruence modulo $2$ and consider the sum\n\n$$\nS = \\sum_{i,j=1}^{m} s_{2i-1,2j-1} = \\sum_{i \\equiv j \\equiv 1} s_{ij} = (2m-1) \\sum_{i \\equiv j \\equiv 1} a_{ij} + m \\sum_{i+j=1} a_{ij};\n$$\n\nthere are $m^2$ ordered pairs $(i, j)$ such that $i \\equiv j \\equiv 1$, and $2m(n-m)$ ordered pairs $(i, j)$ such that $i + j \\equiv 1$.\n\nThe largest value the sum $S$ may achieve is\n\n$$\nS_0 = (2m-1) \\sum_{k=0}^{m^2-1} (n^2-k) + m \\sum_{k=0}^{2m(n-m)-1} (n^2-m^2-k),\n$$\n\nso $s_{k\\ell} \\leq S_0/m^2$ for some odd indices $k$ and $\\ell$; in the case at hand, $S_0 = 5 \\times (25+24+\\cdots+17) + 3 \\times (16+15+\\cdots+5) = 1323$, so $s_{k\\ell} \\leq 1323/9 = 147$.\n\nThe sum of the $a_{ij}$ in the complement $C$ of the cross formed by the $k$-th row and $\\ell$-th column is therefore at least $\\sum_{i=1}^{n^2} i - S_0/m^2 = n^2(n^2+1)/2 - S_0/m^2$; in the case at hand, at least $325 - 147 = 178$.\n\nFinally, use the fact that $n$ is odd, $n = 2m-1$, to tile $C$ by $(m-1)^2$ squares $2 \\times 2$ and infer that the numbers in one of these tiles add up to at least\n\n$$\n\\left\\lfloor \\frac{n^2(n^2+1)}{2(m-1)^2} - \\frac{S_0}{m^2(m-1)^2} \\right\\rfloor = \\left\\lceil 8m^2 - 11m + 5 + \\frac{11}{2} \\right\\rceil = 8m^2 - 11m + 6,\n$$\n\nestablishing the required lower bound.\n\n**Remark.** In fact, if $n$ is odd, $n = 2m-1$, then $N = 8m^2 - 11m + 6$, the extremal cell labeling being similar to the one displayed above: every $2 \\times 2$ square with an odd rank top row has sum $N$, and every $2 \\times 2$ square with an even rank top row has sum $N-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16833,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : (0, +\\infty) \\to (0, +\\infty)$ such that for any positive numbers $x, y$, the following equality holds:\n\n$$\nf(f(x) + y) = f(f(x)) + 2y f(x) - f(y) + 2y^2 + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us analyze the given functional equation:\n\n$$\nf(f(x) + y) = f(f(x)) + 2y f(x) - f(y) + 2y^2 + 1.\n$$\n\nSubstitute $y = f(y)$:\n\n$$\n\\begin{aligned}\nf(f(x) + f(y)) &= f(f(x)) + 2f(y) f(x) - f(f(y)) + 2f^2(y) + 1.\\\\\n\\end{aligned}\n$$\n\nThe left side is symmetric in $x$ and $y$, so the right side must also be symmetric. This leads to:\n\n$$\nf(f(y)) = f^2(y) + c, \\text{ where } c \\text{ is a constant}.\n$$\n\nThus,\n\n$$\nf(f(x) + f(y)) = (f(x) + f(y))^2 + 1.\n$$\n\nNow, fix $x_0$ and let $y_0 = f(x_0)$. Then for $x = x_0$:\n\n$$\n\\begin{aligned}\nf(y_0 + y) + f(y) &= f(y_0) + 2y y_0 + 2y^2 + 1.\\\\\n\\end{aligned}\n$$\n\nFor any $t > f(y_0) + 1$, there exists $y_t$ such that $2y y_0 + 2y^2 + f(y_0) + 1 = t$, so $f(y_0 + y_t) + f(y_t) = t$. Using the previous result, for $x = y_0 + y_t$ and $y = y_t$:\n\n$$\nf(f(y_0 + y_t) + f(y_t)) = (f(y_0 + y_t) + f(y_t))^2 + 1,\n$$\nso\n$$\nf(t) = t^2 + 1.\n$$\n\nFor any $x > a = f(y_0) + 1$, $f(x) = x^2 + 1$. Now, substitute back into the original equation and simplify:\n\n$$\nf(y) = y^2 + 1.\n$$\n\nIt is straightforward to check that $f(x) = x^2 + 1$ satisfies the original equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16834,
    "subject": "Mathematics (Olympiad)",
    "question": "A positive integer is called _alternating_ if, among any two consecutive digits in its decimal representation, one is even and the other is odd. Find all positive integers $n$ such that $n$ has a multiple which is alternating.",
    "options": [],
    "answer": "See solution",
    "solution": "The answers are those positive integers that are not divisible by $20$.\n\nAny multiple of $20$ ends with an even digit followed by $0$, so multiples of $20$ are not alternating. Thus, multiples of $20$ are excluded.\n\n**Lemma 1.** Each power of $2$ has a multiple which is alternating and has an even number of digits.\n\n*Proof:* Construct an infinite sequence $\\{a_n\\}_{n=1}^\\infty$ of decimal digits such that:\n\n(a) $a_n \\equiv n + 1 \\pmod{2}$;\n\n(b) $2^{2n-1} \\mid \\overline{a_{2n-1} \\dots a_1}$;\n\n(c) $2^{2n+1} \\mid \\overline{a_{2n} a_{2n-1} \\dots a_1}$.\n\nStart with $a_1 = 2$ and $a_2 = 7$, then add two digits at each step. Suppose the sequence is constructed up to $a_{2n}$. Set $a_{2n+1} = 4$ (even, satisfying (a)).\n\nBecause $2^{2n+2} \\mid 4 \\cdot 10^{2n}$ and $2^{2n+1} \\mid \\overline{a_{2n}a_{2n-1} \\dots a_1}$ (by induction),\n\n$$\n2^{2n+1} \\mid \\overline{a_{2n+1}a_{2n} \\dots a_1} = 4 \\cdot 10^{2n} + \\overline{a_{2n}a_{2n-1} \\dots a_1},\n$$\n\nestablishing (b). Denote $\\overline{a_{2n+1}a_{2n} \\dots a_1} = 2^{2n+1} \\cdot A_n$, where $A_n$ is odd. Now, $a_{2n+2}$ must be odd and such that\n\n$$\n2^{2n+3} \\mid \\overline{a_{2n+2}a_{2n+1} \\dots a_1} = a_{2n+2} \\cdot 10^{2n+1} + \\overline{a_{2n+1}a_{2n} \\dots a_1} = 2^{2n+1}(a_{2n+2}5^{2n+1} + A_n).\n$$\n\nThis holds whenever $5a_{2n+2}+A_n \\equiv 4 \\pmod{8}$. The solutions are odd, and since $5$ is coprime to $8$, $a_{2n+2}$ can be chosen from $\\{0, 1, \\dots, 7\\}$, a complete residue class modulo $8$. Induction is complete. $\\blacksquare$\n\n**Lemma 2.** Each number of the form $2 \\cdot 5^m$, where $m$ is a positive integer, has a multiple which is alternating and has an even number of digits.\n\n*Proof:* Construct an infinite sequence $\\{b_n\\}_{n=1}^\\infty$ of decimal digits such that:\n\n(d) $b_n \\equiv n + 1 \\pmod{2}$;\n\n(e) $2 \\cdot 5^n \\mid \\overline{b_n b_{n-1} \\dots b_1}$.\n\nStart with $b_1 = 0$ and $b_2 = 5$, then add one digit at each step. Suppose $b_1, b_2, \\dots, b_n$ ($n \\ge 2$) have been constructed. Let $\\overline{b_nb_{n-1}\\dots b_1} = 5^\\ell B_n$, where $B_n$ is coprime to $5$ and $\\ell \\ge n$. The next digit $b_{n+1}$ must satisfy $b_{n+1} \\equiv n+2 \\pmod{2}$ and $5^{n+1} \\nmid \\overline{b_{n+1}b_n\\dots b_1}$. Note:\n\n$$\n\\overline{b_{n+1}b_n\\dots b_1} = b_{n+1} \\cdot 2^n 5^n + 5^\\ell B_n = 5^n(2^n b_{n+1} + 5^{\\ell-n} B_n).\n$$\n\nIt suffices to find $b_{n+1}$ such that\n\n$$\nb_{n+1} \\equiv n+2 \\pmod{2} \\quad \\text{and} \\quad 2^n b_{n+1} + 5^{\\ell-n} B_n \\equiv 0 \\pmod{5}.\n$$\n\nBy the Chinese Remainder Theorem, such a $b_{n+1}$ exists, and can be chosen from $\\{0, 1, \\dots, 9\\}$, a complete residue class modulo $10$. Induction is complete. $\\blacksquare$\n\nNow, write $n = 2^\\alpha 5^\\beta k$, where $\\alpha, \\beta \\ge 0$ and $k$ is coprime to $10$. Assume $n$ is not a multiple of $20$ (if $\\beta \\ge 1$, then $\\alpha = 0$ or $1$). Consider:\n\n- **Case 1:** $k = 1$, $\\beta = 0$ ($n = 2^\\alpha$): By Lemma 1, $n$ has an alternating multiple.\n- **Case 2:** $k = 1$, $\\beta \\ge 1$ ($n = 5^\\beta$ or $2 \\cdot 5^\\beta$): By Lemma 2, $n$ has an alternating multiple.\n- **Case 3:** $k > 1$: By Cases 1 and 2, $2^\\alpha 5^\\beta$ has an alternating multiple, and since $k$ is coprime to $10$, $n$ also has an alternating multiple.\n\nThus, all positive integers not divisible by $20$ have an alternating multiple.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16835,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there are no natural numbers $n$ and $k$ that satisfy the equation:\n\n$$\nn^n + (n + 1)^{n+1} + (n + 2)^{n+2} = 2023^k.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the given equation modulo $3$. The right-hand side, $2023^k$, is congruent to $1 \\pmod{3}$. Among any three consecutive natural numbers $n$, $n+1$, $n+2$, one is divisible by $3$, one leaves a remainder of $1$ when divided by $3$, and one leaves a remainder of $2$.\n\nIf $x$ is divisible by $3$, then $x^x$ is also divisible by $3$. If $x \\equiv 1 \\pmod{3}$, then $x^x \\equiv 1 \\pmod{3}$. If $x \\equiv 2 \\pmod{3}$, then $x^x \\equiv 1$ or $2 \\pmod{3}$ depending on $x$.\n\nTherefore, $n^n + (n+1)^{n+1} + (n+2)^{n+2}$ has a remainder of $0$ or $2$ modulo $3$, never $1$. This contradicts the right-hand side, so there are no such natural numbers $n$ and $k$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16836,
    "subject": "Mathematics (Olympiad)",
    "question": "Назовём значимостью мэра количество городов, соседних с городом, где он работает. В некотором государстве есть $N$ городов, соединённых дорогами так, что из любого города можно доехать в любой другой, не проезжая через один и тот же город дважды. В каждом городе работает свой мэр. После реформы мэры были переставлены по городам так, что значимость каждого мэра не уменьшилась. Докажите, что либо найдётся город, в котором мэр остался прежним, либо найдётся пара соседних городов, обменявшихся мэрами.",
    "options": [],
    "answer": "See solution",
    "solution": "Применим индукцию по $N$. Утверждение задачи очевидно при $N=1$ и $N=2$.\n\nПусть $\\Gamma_1$ — множество городов, из которых исходит одна дорога, а $\\Gamma$ — множество остальных городов. Начав движение по различным дорогам из некоторого города, согласно условию, мы не сможем попасть дважды в один и тот же город, поэтому когда-нибудь мы закончим движение в городе из множества $\\Gamma_1$. Это означает, что множество $\\Gamma_1$ непусто. Ясно, что при $N \\ge 3$ множество $\\Gamma$ непусто, и в нем меньше, чем $N$ городов.\n\nПо условию, значимость каждого мэра после реформы не уменьшилась. В частности, мэр города, принадлежащего множеству $\\Gamma$, после реформы стал мэром некоторого города из множества $\\Gamma$, то есть в результате реформы в городах множества $\\Gamma$ тоже произошла перестановка мэров. Ясно, что из любого города $A \\in \\Gamma$ можно доехать до любого другого города $B \\in \\Gamma$, не заезжая в города множества $\\Gamma_1$.\n\nПоэтому множество городов $\\Gamma$ и реформа, рассмотренная только на городах из $\\Gamma$, удовлетворяет условию задачи. Применив предположение индукции, получаем, что в множестве $\\Gamma$ либо найдётся город, в котором мэр после реформы не поменялся, либо найдётся пара соседних городов, обменявшихся мэрами, что и требовалось.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16837,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, and $C$ be piles containing 100, 101, and 102 stones, respectively. Present a strategy for Ilya in the following game: On each turn, a player may remove any positive number of stones from a single pile. The players alternate turns. What should Ilya do to guarantee a win?",
    "options": [],
    "answer": "See solution",
    "solution": "Ilya should begin by moving to pile $B$. If Kostya does not respond by moving to $B$ on his first move, Ilya can simply copy Kostya's moves for the rest of the game. Otherwise, after each subsequent move, Ilya should take from $B$, $A$, or $C$ if Kostya took from $A$, $B$, or $C$, respectively.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16838,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 125 distinct positive integers in a row such that among every three consecutive numbers, the second one is larger than the arithmetic mean of the first and the third. Find the largest number in the row, given that it is as small as possible under these conditions.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the numbers in the row be $a_1, a_2, \\dots, a_{125}$. The condition gives $a_{i+1} > \\frac{a_i + a_{i+2}}{2}$ for every $i = 1, 2, \\dots, 123$, which is equivalent to $a_{i+1} - a_i > a_{i+2} - a_{i+1}$. Denote $d_i = a_{i+1} - a_i$, so $d_1 > d_2 > \\dots > d_{124}$.\n\nLet $a_m$ be the largest among $a_1, \\dots, a_{125}$. Then $d_1, \\dots, d_{m-1}$ are positive and $d_m, \\dots, d_{124}$ are negative. If both $1$ and $-1$ occurred among the differences $d_1, \\dots, d_{124}$, they would have to be consecutive, i.e., $d_i = 1$ and $d_{i+1} = -1$ for some $i$, which would imply $a_{i+1} = a_{i-1}$, contradicting the distinctness of the numbers. Thus, either $1$ or $-1$ is missing among the differences. W.l.o.g., assume $1$ is missing (if $-1$ is missing, reverse the order).\n\nThen $a_m = a_1 + (d_1 + \\dots + d_{m-1}) \\ge 1 + (m + (m-1) + \\dots + 2) = 1 + 2 + \\dots + m$ and $a_m = a_{125} - (d_m + \\dots + d_{124}) \\ge 1 + (1 + 2 + \\dots + (125 - m))$. Among $m$ and $125 - m$, one is at least $63$, so $a_m \\ge 1 + 2 + \\dots + 63$.\n\nOn the other hand, take $a_1 = 1$ and $d_i = 64 - i$ for $1 \\le i \\le 62$, and $d_i = 62 - i$ for $63 \\le i \\le 124$. The largest number $a_{63}$ equals $1 + 2 + \\dots + 63$, while all numbers $a_i$ are positive and distinct: $1 = a_1 < a_2 < \\dots < a_{63}$ and $a_{63} > a_{64} > \\dots > a_{125} = (1 + 2 + \\dots + 63) - (1 + 2 + \\dots + 62) = 63$. For $i = 64, \\dots, 125$, $a_i = (1 + 2 + \\dots + 63) - (1 + 2 + \\dots + (i - 63)) = (i - 62) + (i - 61) + \\dots + 63 = a_{127 - i} - 1$, so $a_i$ lies strictly between $a_{126 - i}$ and $a_{127 - i}$. Thus, the largest number is $1 + 2 + \\dots + 63 = 2016$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16839,
    "subject": "Mathematics (Olympiad)",
    "question": "If $y = 0$ or $y = -1$, the right-hand side of the given equation is equal to zero, so $x^3 + x^2 + x = x(x^2 + x + 1) = 0$, and hence $x = 0$. That gives two solutions: $(0, 0)$ and $(0, -1)$.\n\nProve that no other integer solutions $(x, y)$ exist to the equation:\n\n$$\nx^3 + x^2 + x = y^2 + y\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $y \\in \\mathbb{Z} \\setminus \\{-1, 0\\}$. Then $y^2 + y > 0$ so $x(x^2 + x + 1) > 0$, which implies $x > 0$.\n\nIf $(x, y)$ is a solution, then $(x, -y - 1)$ is also a solution since $y^2 + y = (-y - 1)^2 + (-y - 1)$. Thus, it suffices to consider $y > 0$.\n\nRewrite the equation as:\n\n$$\nx^3 = (y - x)(x + y + 1).\n$$\n\nSuppose $p$ is a common prime divisor of $y - x$ and $x + y + 1$. Then $p \\mid x^3$ so $p \\mid x$. Also,\n\n$$\np \\mid y - x, \\quad p \\mid x + y + 1, \\quad p \\mid x \\implies p \\mid (x + y + 1) - (y - x) - 2x = 1,\n$$\n\nwhich is impossible. Thus, $y - x$ and $x + y + 1$ are coprime, and their product is $x^3$, so each is a perfect cube:\n\n$$\n\\begin{aligned}\ny - x &= a^3 \\\\\nx + y + 1 &= b^3\n\\end{aligned}\n$$\n\nSubtracting, $2x + 1 = b^3 - a^3$, and from above, $x = ab$. Thus, $2ab + 1 = b^3 - a^3 = (b - a)(a^2 + ab + b^2) \\ge a^2 + ab + b^2 \\ge 3ab$, so $ab \\le 1$, which is impossible for distinct positive integers $a, b$. Therefore, the only solutions are $(0, 0)$ and $(0, -1)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16840,
    "subject": "Mathematics (Olympiad)",
    "question": "There is a village with a population of $2007$. This village has no name. You are the God of this village and you want the villagers to decide the name of the village. Every villager has one idea for the village's name.\n\nEach villager can send a letter to any other villager (including himself). Every villager can send any number of letters each day. Letters are collected in the evening and delivered all at once the next morning. The villager who sends the letter can decide to whom the letter should be delivered. Each villager can send a letter to tell their idea for the name of the village to God only once. This idea does not need to be the same as the idea they or other villagers originally had. Every villager's action is only writing a letter.\n\nEvery villager can be classified as either an honest person or a liar. You and every villager do not know who is honest and who is a liar. However, you know that the number of liars is less than or equal to $T$, and there is at least one honest person in the village.\n\nYou can give instructions to every villager only once at noon on a certain day. An honest person necessarily follows the instruction, but you do not know if a liar will follow the instruction. Find the maximum $T$ such that there exists an instruction which fulfills the following conditions:\n\n* In the end, every honest person sends a letter to God, and every honest person sends the same idea for the village's name.\n* If every honest person originally had the same idea for the village's name, every honest person sends this idea to God.",
    "options": [],
    "answer": "See solution",
    "solution": "$668$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16841,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a board consisting of $n \\times n$ unit squares where $n \\ge 2$. Two cells are called neighbors if they share a horizontal or vertical border. In the beginning, all cells together contain $k$ tokens. Each cell may contain one or several tokens or none.\n\nIn each turn, choose one of the cells that contains at least one token for each of its neighbors and move one of those to each of its neighbors. The game ends if no such cell exists.\n\n(a) Find the minimal $k$ such that the game does not end for any starting configuration and choice of cells during the game.\n\n(b) Find the maximal $k$ such that the game ends for any starting configuration and choice of cells during the game.",
    "options": [],
    "answer": "See solution",
    "solution": "1. If each cell contains one token less than the number of its neighbors, the game cannot even start. On the other hand, if there is one token more, then by the pigeonhole principle there will always exist at least one cell with sufficient tokens to make the next move.\n\nTherefore, the desired quantity is the sum of all numbers of neighbors minus the number of all cells plus $1$. If one adds $4n$ cells around the $n^2$ given cells, each original cell has four neighbors and each new cell has contributed one neighbor.\n\nWe get\n$$\nk = (4n^2 - 4n) - n^2 + 1 = 3n^2 - 4n + 1.\n$$\n\n2. It is easy to see that an unlimited number of turns must eventually have brought tokens to all cells and that also every pair of neighbors must have exchanged tokens, because otherwise tokens would accumulate in unlimited number in one of the inactive cells.\n\nBut each time a neighboring pair first exchanges tokens, we can reserve this first token to stay always between these two neighbors. Therefore, the game will certainly end if there are fewer tokens than neighboring pairs.\n\nConversely, if the number of tokens equals the number of neighboring pairs, we can find a never-ending game in the following way: Color the cells black and white in a checkerboard fashion and assign to each black cell a number of tokens that equals the number of its neighbors. Now we will simply choose all the black cells until all tokens are on the white cells, then repeat with the white cells and then iterate from the start.\n\nThe desired quantity is therefore the number of neighboring pairs minus $1$. Since the number of neighboring pairs is half of the first expression in the computation of part 1, we get\n$$\nk = 2n^2 - 2n - 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16842,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $BB_1$ and $CC_1$ be the altitudes of an acute-angled triangle $ABC$, and let $A_0$ be the midpoint of $BC$. Lines $A_0B_1$ and $A_0C_1$ meet the line passing through $A$ and parallel to $BC$ at points $P$ and $Q$. Prove that the incenter of triangle $PA_0Q$ lies on the altitude of triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since triangles $BCB_1$ and $BCC_1$ are right-angled, their medians $B_1A_0$ and $C_1A_0$ are equal to half the hypotenuse: $B_1A_0 = A_0C = A_0B = C_1A_0$.\n\n![](images/Saudi_Arabia_booklet_2022_p29_data_25deac26e0.png)\n\nNow,\n\n$$\n\\angle PB_1A = \\angle CB_1A_0 = \\angle B_1CA_0 = \\angle PAC,\n$$\n\nthus $PA = PB_1$. Similarly, $QA = QC_1$. Then the incircle of triangle $A_0PQ$ touches its sides at points $A$, $B_1$, and $C_1$, which yields the assertion of the problem. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16843,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $L = \\text{lcm}(1, 2, \\dots, n)$. Prove that\n$$\n\\text{gcd}\\left(\\frac{L}{m+1}, \\frac{L}{m+2}, \\dots, \\frac{L}{n}\\right) = 1\n$$\nwhere $m = \\lfloor \\frac{n}{2} \\rfloor$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that $\\text{lcm}(1, 2, \\dots, n) = \\text{lcm}(m+1, m+2, \\dots, n)$. This is because for all $1 \\leq k \\leq m$, the number $s = \\lceil \\log_2\\left(\\frac{m+1}{k}\\right) \\rceil$ is an integer such that\n$$\n2^s \\cdot k \\in \\{m+1, m+2, \\dots, n\\}.\n$$\n\nNow, assume that\n$$\nd = \\text{gcd}\\left(\\frac{L}{m+1}, \\frac{L}{m+2}, \\dots, \\frac{L}{n}\\right).\n$$\nFor all $m+1 \\leq i \\leq n$, we have\n$$\nd \\mid \\frac{L}{i} \\implies \\exists k: d \\cdot k = \\frac{L}{i} \\implies i \\cdot k = \\frac{L}{d} \\implies i \\mid \\frac{L}{d}.\n$$\nThis implies\n$$\nL = \\text{lcm}(m+1, m+2, \\dots, n) \\mid \\frac{L}{d} \\implies d = 1.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16844,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a game where, at each step, three numbers are chosen from a board and replaced by their average (if the average is an integer), determine for which values of $n$ (the initial number of numbers on the board) Ana has a winning strategy, and for which values Bine has a winning strategy. The average is an integer if and only if the sum of the three numbers is divisible by 3.",
    "options": [],
    "answer": "See solution",
    "solution": "In each step, the average of three chosen numbers is an integer if and only if their sum is divisible by 3. Thus, it suffices to track the residues of the numbers modulo 3. Let $x_0, x_1, x_2$ be the counts of numbers congruent to 0, 1, and 2 modulo 3, respectively.\n\nWe can represent the board's state as the triple $(x_0, x_1, x_2)$. The possible moves are:\n\n$$\n\\begin{array}{lcl}\n(x_0, x_1, x_2) & \\rightarrow & (x_0 - 1, x_1 - 1, x_2 - 1), \\quad \\text{if } x_0, x_1, x_2 \\ge 1, \\\\\n(x_0, x_1, x_2) & \\rightarrow & (x_0 - 3, x_1, x_2), \\quad \\text{if } x_0 \\ge 3, \\\\\n(x_0, x_1, x_2) & \\rightarrow & (x_0, x_1 - 3, x_2), \\quad \\text{if } x_1 \\ge 3, \\\\\n(x_0, x_1, x_2) & \\rightarrow & (x_0, x_1, x_2 - 3), \\quad \\text{if } x_2 \\ge 3.\n\\end{array}\n$$\n\nDue to symmetry, the order of the triple is unimportant, so we use $(y_0, y_1, y_2)$ with $y_0 \\ge y_1 \\ge y_2$ as a permutation of $(x_0, x_1, x_2)$. The sum $y_0 + y_1 + y_2$ modulo 3 is invariant under these moves. Let $(\\hat{y}_0, \\hat{y}_1, \\hat{y}_2)$ be the final state, where no moves are possible. Then $\\hat{y}_2 = 0$ and $2 \\ge \\hat{y}_0 \\ge \\hat{y}_1 \\ge 0$.\n\nDepending on $n$:\n\n- If $n \\equiv 2 \\pmod{3}$: Final state is $(2,0,0)$ or $(1,1,0)$. The number of moves is $\\frac{n-2}{3}$. If this is even, Bine wins; if odd, Ana wins. Thus, if $n \\equiv 2 \\pmod{6}$, Bine wins; if $n \\equiv 5 \\pmod{6}$, Ana wins.\n\n- If $n \\equiv 0 \\pmod{3}$: The initial state is $y_0 = y_1 = y_2$, and this congruence is preserved. The final state is $(0,0,0)$, reached in $\\frac{n}{3}$ moves. If $n \\equiv 0 \\pmod{6}$, Bine wins; if $n \\equiv 3 \\pmod{6}$, Ana wins.\n\n- If $n \\equiv 1 \\pmod{6}$: Final states are $(2,2,0)$ (odd number of moves, Ana wins) and $(1,0,0)$ (even number of moves, Bine wins). For $n = 7$, Ana can win. For $n > 7$, Bine has a winning strategy by reducing two congruent numbers below 2, ensuring the final state is $(1,0,0)$.\n\n- If $n \\equiv 4 \\pmod{6}$: For $n = 4$ and $n = 10$, Ana wins. For $n > 10$, Ana can remove the largest three numbers, reducing the game to the $n \\equiv 1 \\pmod{6}$ case with $n > 7$, where she is the second player and can win.\n\n**Conclusion:**\n\nAna wins if $n \\equiv 3, 4, 5 \\pmod{6}$ or $n = 7$. Otherwise, Bine wins.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16845,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(x_n)$ and $(y_n)$ be sequences defined by the recurrence relations:\n\n- $x_1 = 1$, $y_1 = \\sqrt{3}$,\n- $x_{n+1} = \\sqrt{2 - y_n}$,\n- $y_{n+1} = \\dfrac{x_n}{x_{n+1}}$ for $n \\geq 1$.\n\nProve that for all positive integers $n$,\n\n$$\nx_n = 2 \\sin \\frac{\\pi}{3 \\cdot 2^n}, \\quad y_n = 2 \\cos \\frac{\\pi}{3 \\cdot 2^n}.\n$$\n\nFind $\\lim_{n \\to \\infty} x_n$ and $\\lim_{n \\to \\infty} y_n$.\n\n![](images/booklet_2014_final_p15_data_432839cecd.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We can see that $x_1 = 1 = 2 \\sin \\frac{\\pi}{6}$, $y_1 = \\sqrt{3} = 2 \\cos \\frac{\\pi}{6}$.\n\nWe will prove by induction that for all positive integers $n$,\n\n$$\nx_n = 2 \\sin \\frac{\\pi}{3 \\cdot 2^n}, \\quad y_n = 2 \\cos \\frac{\\pi}{3 \\cdot 2^n}. \\tag{*}\n$$\n\nFor $n = 1$, the statement is true. Assume (*) holds for $n$. Applying the recurrence relations, we get\n\n$$\nx_{n+1} = \\sqrt{2 - y_n} = \\sqrt{2 - 2 \\cos \\frac{\\pi}{3 \\cdot 2^n}} = \\sqrt{4 \\sin^2 \\frac{\\pi}{3 \\cdot 2^{n+1}}} = 2 \\sin \\frac{\\pi}{3 \\cdot 2^{n+1}}\n$$\n\nand\n\n$$\ny_{n+1} = \\frac{x_n}{x_{n+1}} = \\frac{2 \\sin \\frac{\\pi}{3 \\cdot 2^n}}{2 \\sin \\frac{\\pi}{3 \\cdot 2^{n+1}}} = 2 \\cos \\frac{\\pi}{3 \\cdot 2^{n+1}}.\n$$\n\nThus, (*) holds for $n+1$. By induction, (*) is true for all $n$.\n\nTherefore,\n\n$$\n\\lim_{n \\to \\infty} x_n = \\lim_{n \\to \\infty} \\left( 2 \\sin \\frac{\\pi}{3 \\cdot 2^n} \\right) = 2 \\sin 0 = 0\n$$\n\nand\n\n$$\n\\lim_{n \\to \\infty} y_n = \\lim_{n \\to \\infty} \\left( 2 \\cos \\frac{\\pi}{3 \\cdot 2^n} \\right) = 2 \\cos 0 = 2.\n$$\n\nTherefore, $(x_n)$ and $(y_n)$ are convergent and $\\lim x_n = 0$, $\\lim y_n = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16846,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $n$ points are equally spaced around a circle and labelled with the numbers $0, 1, \\dots, n-1$ (one label per point). A labelling is called *beautiful* if, for every $k$ ($2 \\leq k \\leq n$), all $k$-chords (chords joining points whose labels sum to $k$ modulo $n$) are parallel and do not intersect each other. Let $M_n$ be the number of beautiful labellings of $n$ points, and $L_{n-1}$ the number of beautiful labellings of $n-1$ points such that the point labelled $0$ does not lie between two $n$-chords. Prove that $L_{n-1} = \\varphi(n)$, where $\\varphi$ is Euler's totient function.",
    "options": [],
    "answer": "See solution",
    "solution": "We construct beautiful labellings of $n$ points from those of $n-1$ points by adding a point labelled $n$ in a way that preserves the non-intersecting property of $n$-chords. There are two cases:\n\n**Case 1.** The point labelled $0$ lies directly between two $n$-chords. In this case, there is only one possible position for the new point.\n\n**Case 2.** The point labelled $0$ does not lie between two $n$-chords. Here, there are two possible positions for the new point, on either side of $0$ in the arc cut off by the closest $n$-chord.\n\nLet $M_n$ be the number of beautiful labellings of $n$ points, and $L_{n-1}$ the number of beautiful labellings of $n-1$ points where $0$ does not lie between two $n$-chords. Then $M_n = M_{n-1} + L_{n-1}$, and by iteration, $M_n = L_{n-1} + L_{n-2} + \\dots + L_2 + M_2$.\n\nThe number of pairs $(x, y)$ of positive integers with $x + y \\leq n$ and $\\gcd(x, y) = 1$ is $\\varphi(n) + \\varphi(n-1) + \\dots + \\varphi(2)$. Since $M_2 = 1$, it suffices to show $L_{n-1} = \\varphi(n)$.\n\nLabel the positions $0, 1, \\dots, n-1$ (mod $n$) clockwise, with $f(0) = 0$ and $f$ a permutation. In case 2, all $n$-chords are parallel, so for all $i$:\n\n$$\nf(i) + f(-i) \\equiv 0 \\pmod n.\n$$\n\nIf $f(a) = n-1$, then for all $i$:\n\n$$\nf(i) + f(a-i) \\equiv -1 \\pmod n.\n$$\n\nSubtracting gives $f(a-i) \\equiv f(-i) - 1 \\pmod n$. Setting $i = a, 2a, \\dots$ yields $f(-ka) \\equiv k \\pmod n$ for all $k$. Since $f(r) \\equiv 0 \\pmod n$ only if $r \\equiv 0 \\pmod n$, we must have $\\gcd(a, n) = 1$, so $L_{n-1} \\leq \\varphi(n)$.\n\nThe labelling defined by $f(-ka) \\equiv k \\pmod n$ is beautiful and belongs to case 2, so $L_{n-1} = \\varphi(n)$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 16847,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Prove that for all real numbers $x_1, y_1, x_2, y_2$,\n$$\n(x_1^2 + 3x_1y_1 + 8y_1^2)(x_2^2 + 3x_2y_2 + 8y_2^2) = (x_1x_2 - 8y_1y_2)^2 + 3(x_1x_2 - 8y_1y_2)(x_1y_2 + x_2y_1 + 3y_1y_2) + 8(x_1y_2 + x_2y_1 + 3y_1y_2)^2.\n$$\n\nb) Is it possible for integers $x, y$ that $x^2 + 3xy + 8y^2 \\equiv 7 \\pmod{23}$?",
    "options": [],
    "answer": "See solution",
    "solution": "a) The roots of $z^2 + 3z + 8 = 0$ are $\\frac{-3 \\pm \\sqrt{23}i}{2}$. Let $\\alpha = \\frac{-3 + \\sqrt{23}i}{2}$. Then $\\bar{\\alpha} = \\frac{-3 - \\sqrt{23}i}{2}$, and hence\n$$\nx^2 + 3xy + 8y^2 = (x - \\alpha y)(x - \\bar{\\alpha}y).\n$$\nNote that\n$$\n\\begin{aligned}\n(x_1 - \\alpha y_1)(x_2 - \\alpha y_2) &= (x_1 x_2 + \\alpha^2 y_1 y_2) - \\alpha(x_1 y_2 + x_2 y_1) \\\\\n&= (x_1 x_2 + (-3\\alpha - 8)y_1 y_2) - \\alpha(x_1 y_2 + x_2 y_1) \\\\\n&= (x_1 x_2 - 8y_1 y_2) - \\alpha(x_1 y_2 + x_2 y_1 + 3y_1 y_2).\n\\end{aligned}\n$$\nThus, defining $s = x_1x_2 - 8y_1y_2$ and $t = x_1y_2 + x_2y_1 + 3y_1y_2$, we have\n$$\n\\begin{aligned}\n& (x_1^2 + 3x_1y_1 + 8y_1^2)(x_2^2 + 3x_2y_2 + 8y_2^2) \\\\\n&= (x_1 - \\alpha y_1)(x_1 - \\bar{\\alpha} y_1)(x_2 - \\alpha y_2)(x_2 - \\bar{\\alpha} y_2) \\\\\n&= (x_1 - \\alpha y_1)(x_2 - \\alpha y_2)\\overline{(x_1 - \\alpha y_1)(x_2 - \\alpha y_2)} \\\\\n&= (s - \\alpha t)\\overline{(s - \\alpha t)} \\\\\n&= (s - \\alpha t)(s - \\bar{\\alpha} t) \\\\\n&= s^2 + 3st + 8t^2.\n\\end{aligned}\n$$\nThis proves the result.\n\nb) No. Suppose on the contrary that $x^2 + 3xy + 8y^2 \\equiv 7 \\pmod{23}$ for some integers $x$ and $y$. This implies $4x^2 + 12xy + 32y^2 \\equiv 28 \\pmod{23}$, and hence $(2x + 3y)^2 \\equiv 5 \\pmod{23}$. However, $5$ is not a square modulo $23$ (the Legendre symbol $\\left(\\frac{5}{23}\\right) = -1$). This is a contradiction, so there is no such integer in $S$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16848,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the maximum number of 3-element sets such that every two of them have exactly one common element, but there is no element that is present in all sets simultaneously.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there are at least 8 such sets. Let $M = \\{x, y, z\\}$ be one of these sets. Every other set (at least 7 sets) contains exactly one element from $M$. Therefore, there must be at least three sets $M_1, M_2, M_3$ that each share a different element with $M$, say $x$. Since no element is in all sets, there exists a set $M_0$ that does not contain $x$. As $M_0$ must intersect $M$, it contains another element of $M$, say $y$. This set cannot share the same common element with two of $M_1, M_2, M_3$, otherwise that element would be a second common element for some pair. Therefore, $M_0$ would need to share a different element with each of these sets, which is impossible for a 3-element set. This contradiction shows that the maximum is 7.\n\nAn example of 7 such sets:\n\n$$\n\\{1, 2, 3\\}, \\{1, 4, 5\\}, \\{1, 6, 7\\}, \\{3, 5, 7\\}, \\{3, 4, 6\\}, \\{2, 4, 7\\}, \\{2, 5, 6\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16849,
    "subject": "Mathematics (Olympiad)",
    "question": "For each positive integer $n$ and each integer $i$ ($0 \\le i \\le n$), let $C_n^i \\equiv c(n, i) \\pmod{2}$, where $c(n, i) \\in \\{0, 1\\}$, and define\n\n$$\nf(n, q) = \\sum_{i=0}^{n} c(n, i) q^{i}.\n$$\n\nLet $m$, $n$, and $q$ be positive integers with $q+1$ not a power of $2$. Suppose that $f(m, q)$ divides $f(n, q)$. Prove that $f(m, r)$ divides $f(n, r)$ for every positive integer $r$.",
    "options": [],
    "answer": "See solution",
    "solution": "Write $n$ in binary as $n = 2^{a_1} + 2^{a_2} + \\cdots + 2^{a_k}$, where $0 \\le a_1 < a_2 < \\cdots < a_k$. Define $T(n) = \\{2^{a_1}, \\dots, 2^{a_k}\\}$, and $T(0)$ as the empty set.\n\nBy Lucas' Theorem, $C_n^i$ is odd if and only if $T(i) \\subseteq T(n)$. Thus,\n\n$$\nf(n, q) = \\sum_{A \\subseteq T(n)} q^{\\sigma(A)} = \\prod_{a \\in T(n)} (1 + q^a),\n$$\n\nwhere $\\sigma(A)$ is the sum of elements in $A$.\n\nGiven $f(m, q) = \\prod_{a \\in T(m)} (1 + q^a)$ divides $f(n, q) = \\prod_{a \\in T(n)} (1 + q^a)$, we must have $T(m) \\subseteq T(n)$. Therefore, $f(m, r)$ divides $f(n, r)$ for every $r$.\n\nTo justify this, note that for any $a \\in T(m)$, the largest odd divisor $s(q^a + 1)$ divides $\\prod_{b \\in T(n)} s(q^b + 1)$. Since $q+1$ is not a power of $2$, $s(q^a + 1) > 1$, so $a \\in T(n)$. Thus, $T(m) \\subseteq T(n)$, completing the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16850,
    "subject": "Mathematics (Olympiad)",
    "question": "A cube with edge $10$ is cut into $27$ parallelepipeds by three pairs of planes parallel to its faces. The edges of the interior parallelepiped have lengths $1$, $2$, and $3$. Find the sum of the volumes of the $8$ corner parallelepipeds.",
    "options": [],
    "answer": "See solution",
    "solution": "Imagine the eight corner parts yellow and the rest of the cube white. The two horizontal cuts produce three parallelepipeds. The middle one is white and has the same vertical dimension as the central piece. Assume the latter dimension to be $1$ and remove the middle part. A $10 \\times 10 \\times 9$ parallelepiped is obtained with $4$ yellow parts instead of $8$. This is because the initial $8$ yellow parts come into $4$ pairs with equal horizontal dimensions in every pair. So after removing the middle white part, every pair becomes a single yellow piece. Repeat the same with the two cuts in the left-right direction. They also produce three parts; the middle one is white. Its width equals the front-back dimension of the central piece, which we assume to be $2$. So removing the middle part yields a $10 \\times 8 \\times 9$ parallelepiped in which the yellow portion consists of two parallelepipeds separated by a white parallelepiped $3 \\times 8 \\times 9$. Remove this white piece; the entire yellow part remains in the shape of a parallelepiped $7 \\times 8 \\times 9$, hence its volume is $7 \\cdot 8 \\cdot 9 = 504$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16851,
    "subject": "Mathematics (Olympiad)",
    "question": "We say that a natural number $n \\ge 2$ is *nice* if, in its prime factorization, one of the primes has exponent 3.\n\n(a) Determine the smallest natural number $N$ such that, no matter how we choose $N$ consecutive natural numbers, at least one of them is nice.\n\n(b) Find the smallest 15 consecutive natural numbers $a_1, a_2, \\dots, a_{15}$ that are not nice, but for which the sum $5a_1 + 5a_2 + \\dots + 5a_{15}$ is nice.",
    "options": [],
    "answer": "See solution",
    "solution": "a) By dividing 16 consecutive natural numbers by 16, the remainders $0, 1, 2, \\ldots, 15$ are obtained (not necessarily starting with 0). Therefore, among any 16 consecutive natural numbers, there is a nice number of the form $16k + 8 = 2^3(2k+1)$. Since among the consecutive natural numbers $9, 10, 11, \\ldots, 23$ there is no nice number, the answer is $N = 16$.\n\nb) Consider $a_2 = a_1 + 1$, $a_3 = a_1 + 2$, ..., $a_{15} = a_1 + 14$ and denote $S = 5a_1 + 5a_2 + \\dots + 5a_{15}$, that is, $S = 3 \\cdot 5^2 \\cdot (a_1 + 7)$. The numbers $a_1, a_2, \\dots, a_{15}$ are not nice, thus none of them has remainder 8 when divided by 16. We deduce that $a_1 = 16k + 9$, $k \\in \\mathbb{N}$, and $S = 2^4 \\cdot 3 \\cdot 5^2 \\cdot (k+1)$. The smallest value of $k$ for which $S$ is a nice number is $k = 4$, and the numbers $73, 74, \\ldots, 87$ fulfill the statement.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16852,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that there exist non-zero integers $x_1, x_2, \\dots, x_n, y$ satisfying:\n\n$$\n\\begin{cases}\nx_1 + \\cdots + x_n = 0, \\\\\nx_1^2 + \\cdots + x_n^2 = n y^2.\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to see that $n > 1$.\n\nFor $n = 2k$, $k \\in \\mathbb{N}$, let $x_{2i-1} = 1$, $x_{2i} = -1$ for $i = 1, 2, \\dots, k$, and $y = 1$. Then the conditions are satisfied.\n\nFor $n = 2k + 3$, $k \\in \\mathbb{N}$, let $y = 2$, $x_1 = 4$, $x_2 = x_3 = x_4 = x_5 = -1$, and for $i = 3, 4, \\dots, k+1$, set $x_{2i} = 2$, $x_{2i+1} = -2$. Then the conditions are satisfied.\n\nNow, consider $n = 3$. Suppose there exist $x_1, x_2, x_3, y$ such that\n\n$$\n\\begin{cases}\nx_1 + x_2 + x_3 = 0, \\\\\nx_1^2 + x_2^2 + x_3^2 = 3y^2.\n\\end{cases}\n$$\n\nThen\n$$\n2(x_1^2 + x_2^2 + x_3 x_2) = 3y^2.\n$$\n\nSuppose $\\gcd(x_1, x_2) = 1$, so $x_1, x_2$ are either both odd or one is even and the other is odd. Thus, $x_1^2 + x_2^2 + x_3 x_2$ is odd. But $2 \\mid 3y^2$, so $3y^2 \\equiv 0 \\pmod{4}$. This is a contradiction since $2(x_1^2 + x_2^2 + x_3 x_2) \\equiv 2 \\pmod{4}$.\n\n**Answer:** $n \\neq 1, 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16853,
    "subject": "Mathematics (Olympiad)",
    "question": "Divide a rectangle of dimension $m \\times n$ by lines parallel to the sides into $mn$ congruent squares. Find the number of squares in this configuration.",
    "options": [],
    "answer": "See solution",
    "solution": "There are $mn$ squares of side length 1, $(m-1)(n-1)$ squares of side length 2, $(m-2)(n-2)$ squares of side length 3, and so on.\n\nAssume that $m \\ge n$. The total number of squares in this configuration is\n\n$$\n\\sum_{k=0}^{n-1} (m-k)(n-k).\n$$\n\nWe have\n\n$$\n\\begin{aligned}\n\\sum_{k=0}^{n-1} (m-k)(n-k) &= \\sum_{k=0}^{n-1} (mn - (m+n)k + k^2) \\\\\n&= mn^2 - (m+n) \\frac{(n-1)n}{2} + \\frac{(n-1)n(2n-1)}{6} \\\\\n&= mn^2 - m \\frac{(n-1)n}{2} - \\frac{(n-1)n^2}{2} + \\frac{(n-1)n(2n-1)}{6} \\\\\n&= \\frac{n(n+1)(3m-n+1)}{6}.\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16854,
    "subject": "Mathematics (Olympiad)",
    "question": "In $\\triangle ABC$, $\\sin A = \\frac{\\sqrt{2}}{2}$. Find the range of $\\cos B + \\sqrt{2} \\cos C$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f = \\cos B + \\sqrt{2} \\cos C$.\n\nGiven $\\sin A = \\frac{\\sqrt{2}}{2}$, so $A = \\frac{\\pi}{4}$ or $A = \\frac{3\\pi}{4}$.\n\n**Case 1:** $A = \\frac{\\pi}{4}$\n\nThen $B + C = \\pi - \\frac{\\pi}{4} = \\frac{3\\pi}{4}$, so $B = \\frac{3\\pi}{4} - C$ with $0 < C < \\frac{3\\pi}{4}$.\n\n$$\n\\begin{aligned}\nf &= \\cos\\left(\\frac{3\\pi}{4} - C\\right) + \\sqrt{2} \\cos C \\\\\n&= \\frac{\\sqrt{2}}{2} \\sin C + \\frac{\\sqrt{2}}{2} \\cos C \\\\\n&= \\sin\\left(C + \\frac{\\pi}{4}\\right)\n\\end{aligned}\n$$\n\nSince $0 < C < \\frac{3\\pi}{4}$, $C + \\frac{\\pi}{4} \\in \\left(\\frac{\\pi}{4}, \\pi\\right)$, so $f \\in (0, 1]$.\n\n**Case 2:** $A = \\frac{3\\pi}{4}$\n\nThen $B + C = \\pi - \\frac{3\\pi}{4} = \\frac{\\pi}{4}$, so $B = \\frac{\\pi}{4} - C$ with $0 < C < \\frac{\\pi}{4}$.\n\n$$\n\\begin{aligned}\nf &= \\cos\\left(\\frac{\\pi}{4} - C\\right) + \\sqrt{2} \\cos C \\\\\n&= \\frac{\\sqrt{2}}{2} \\sin C + \\frac{3\\sqrt{2}}{2} \\cos C \\\\\n&= \\sqrt{5} \\sin(C + \\varphi),\n\\end{aligned}\n$$\n\nwhere $\\varphi = \\arctan 3$.\n\nSince $C \\in (0, \\frac{\\pi}{4})$, $C + \\varphi \\in (\\varphi, \\varphi + \\frac{\\pi}{4})$.\n\nThe maximum is $\\sqrt{5}$, and the minimum is $2$ (at the endpoints), so $f \\in (2, \\sqrt{5}]$.\n\n**Conclusion:**\n\nThe range of $\\cos B + \\sqrt{2} \\cos C$ is $(0, 1] \\cup (2, \\sqrt{5}]$.\n\n$\\boxed{(0, 1] \\cup (2, \\sqrt{5}]}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16855,
    "subject": "Mathematics (Olympiad)",
    "question": "Positive real numbers $a$, $b$, $c$ satisfy $abc = 1$. Prove that\n$$\n\\frac{a}{1+b} + \\frac{b}{1+c} + \\frac{c}{1+a} \\ge \\frac{3}{2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The given inequality is equivalent to\n\n$$\n2a(1+c)(1+a) + 2b(1+a)(1+b) + 2c(1+b)(1+c) \\ge 3(1+a)(1+b)(1+c).\n$$\n\nBy expanding and collecting like terms, this reduces to\n\n$$\n2a^2 + 2b^2 + 2c^2 + 2a^2c + 2b^2a + 2c^2b \\ge 3 + a + b + c + ab + bc + ca + 3abc. \\quad (8)\n$$\n\nUsing the AM-GM inequality, we observe:\n\n$$\na^2 + b^2 + c^2 \\ge ab + bc + ca;\n$$\n\nand\n\n$$\n\\begin{aligned}\na^2 + b^2 + c^2 &\\ge (abc)^{\\frac{1}{3}}(a + b + c);\n\\end{aligned}\n$$\n\nAlso,\n\n$$\na^2c + b^2a + c^2b \\ge 3abc.\n$$\n\nSince $abc = 1$, we have\n\n$$\n\\begin{aligned}\n2a^2 + 2b^2 + 2c^2 + 2a^2c + 2b^2a + 2c^2b &= (a^2 + b^2 + c^2) + (a^2 + b^2 + c^2) + (a^2c + b^2a + c^2b) + (a^2c + b^2a + c^2b) \\\\\n&\\ge (ab + bc + ca) + (a + b + c) + 3abc + 3abc \\\\\n&= 3 + a + b + c + ab + bc + ca + 3abc.\n\\end{aligned}\n$$\n\nThis proves (8), and thus the desired result follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16856,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. The rectangle $ABCD$ with side lengths $AB = 90n + 1$ and $BC = 90n + 5$ is partitioned into unit squares with sides parallel to the sides of $ABCD$. Let $S$ be the set of all points which are vertices of these unit squares. Prove that the number of lines which pass through at least two points from $S$ is divisible by $4$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m = 90n + 1$. We investigate the number of such lines modulo $4$ by considering different types of lines.\n\nThe vertical and horizontal lines are $(m+5) + (m+1) = 2(m+3)$, which is divisible by $4$.\n\nEvery line with a positive slope (acute angle to the $x$-axis) corresponds uniquely to a line with a negative slope (obtuse angle), by symmetry about the line through the midpoints of $AB$ and $CD$. Thus, it suffices to show that the number of lines with acute angles is even.\n\nEvery line not passing through the center $O$ of the rectangle corresponds to another line with the same slope, by symmetry with respect to $O$. Thus, it suffices to consider lines through $O$.\n\nEvery line through $O$ has a slope $\\frac{p}{q}$, where $\\gcd(p, q) = 1$ and $p, q$ are odd positive integers. (This follows by considering the two nearest points to $O$ on the line.)\n\nIf $p \\ne 1$, $q \\ne 1$, $p \\leq m$, $q \\leq m$, the line with slope $\\frac{p}{q}$ uniquely corresponds to the line with slope $\\frac{q}{p}$. It remains to show that the number of remaining lines is even.\n\nThe number of such lines is:\n\n$$\n1 + \\frac{\\varphi(m+2)}{2} + \\frac{\\varphi(m+4)}{2} - 1 = \\frac{\\varphi(m+2) + \\varphi(m+4)}{2}\n$$\n\nwhere:\n\n1. One line with $p = q = 1$;\n2. $\\frac{\\varphi(m+2)}{2}$ lines with slope $\\frac{p}{m+2}$, $p \\leq m$ odd, $\\gcd(p, m+2) = 1$;\n3. $\\frac{\\varphi(m+4)}{2} - 1$ lines with slope $\\frac{p}{m+4}$, $p \\leq m$ odd, $\\gcd(p, m+4) = 1$.\n\nNow, $\\varphi(m+2) + \\varphi(m+4) = \\varphi(90n+3) + \\varphi(90n+5)$ is divisible by $4$, so the total number of lines is divisible by $4$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16857,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be the number of blue vertices among $A_1, A_2, \\dots, A_{101}$. For any two vertices $A_i, A_j$ ($1 \\le i \\le j \\le 101$), define\n$$\nd(A_i, A_j) = d(A_j, A_i) = \\min\\{j - i, 101 - j + i\\}.\n$$\nLet $B \\subseteq \\{A_1, A_2, \\dots, A_{101}\\}$ be the set of all blue vertices. Define\n$$\nN = 1225n - 101C_n^2 + 3 \\sum_{\\{P,Q\\} \\subseteq B} d(P, Q),\n$$\nwhere $\\{P, Q\\}$ runs over all two-element subsets of $B$. What is the largest possible value of $N$? How many ways are there to color the vertices so that $N$ achieves this maximum?",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $n$ is even; otherwise, coloring all vertices the opposite way does not change $N$. Let $n = 2t$, $0 \\le t \\le 50$, and renumber the blue vertices as $P_1, P_2, \\dots, P_{2t}$ clockwise. Then\n$$\n\\sum_{\\{P,Q\\} \\subseteq B} d(P, Q) = \\sum_{i=1}^{t} d(P_i, P_{i+t}) + \\frac{1}{2} \\sum_{i=1}^{t} \\sum_{j=1}^{t-1} (d(P_i, P_{i+j}) + d(P_{i+j}, P_{i+t}) + d(P_{i+t}, P_{i-j}) + d(P_{i-j}, P_i))\n$$\nwith the subscript of $P_i$ modulo $2t$. Using $d(P_i, P_{i+t}) \\le 50$ and\n$$\nd(P_i, P_{i+j}) + d(P_{i+j}, P_{i+t}) + d(P_{i+t}, P_{i-j}) + d(P_{i-j}, P_i) \\le 101,\n$$\nwe get\n$$\n\\sum_{\\{P,Q\\} \\subseteq B} d(P, Q) \\le 50t + \\frac{101}{2} t(t-1).\n$$\nCombine this with the formula for $N$:\n$$\nN \\le 1225n - 101C_n^2 + 3 \\left( 50t + \\frac{101}{2} t(t-1) \\right) = -\\frac{101}{2} t^2 + \\frac{5099}{2} t.\n$$\nThe maximum occurs at $t = 25$, so $N \\le 32175$.\n\nFor $N = 32175$, $t = 25$ (i.e., 50 blue vertices), and for $1 \\le i \\le t$, $d(P_i, P_{i+t}) = 50$. Equality in the above requires that for each $i$, the sum $d(P_i, P_{i+j}) + d(P_{i+j}, P_{i+t}) + d(P_{i+t}, P_{i-j}) + d(P_{i-j}, P_i) = 101$.\n\nThe number of ways to choose 50 blue vertices so that $N$ is maximized equals the number of ways to choose 25 diagonals from the longest 101 diagonals such that no two share a vertex. Edging $A_i$ and $A_{i+50}$ for $i = 1, 2, \\dots, 101$ gives a graph $G$ that is a cycle of 101 edges. The number of ways $S$ to choose 25 edges of $G$ with no shared vertices is $S = C_{75}^{24} + C_{76}^{25}$. Similarly, for 50 red vertices, the number is also $S$, so the total number of colorings is $2S = 2(C_{75}^{24} + C_{76}^{25})$.\n\n**Answer:**\n- The largest possible value of $N$ is $32175$.\n- The number of ways to color the vertices to achieve this is $2S = 2(C_{75}^{24} + C_{76}^{25})$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16858,
    "subject": "Mathematics (Olympiad)",
    "question": "Una configuración de 4027 puntos del plano, de los cuales 2013 son rojos y 2014 azules, y no hay tres de ellos que sean colineales, se llama *colombiana*. Trazando algunas rectas, el plano queda dividido en varias regiones. Una colección de rectas es *buena* para una configuración colombiana si se cumplen las dos siguientes condiciones:\n\n- Ninguna recta pasa por ninguno de los puntos de la configuración.\n- Ninguna región contiene puntos de ambos colores.\n\nHallar el menor valor de $k$ tal que para cualquier configuración colombiana de 4027 puntos hay una colección buena de $k$ rectas.",
    "options": [],
    "answer": "See solution",
    "solution": "Consideremos un polígono regular de 4027 lados con vértices $P_1, P_2, \\dots, P_{4027}$ numerados en el sentido de las agujas del reloj, y tales que $P_i$ es rojo si $i$ es par, y $P_i$ es azul si $i$ es impar. Claramente, los vértices de este polígono regular forman una configuración colombiana.\n\nSea una colección buena de rectas. Claramente, una de ellas tiene que cortar al lado $P_{2i-1}P_{2i}$ porque $P_{2i-1}$ y $P_{2i}$, al tener distinto color, tienen que estar en regiones distintas del plano. Otro tanto sucede con el lado $P_{2i}P_{2i+1}$. Hay por lo tanto 4026 lados a cortar (no hace falta cortar el lado $P_{4027}P_1$ ya que sus extremos tienen el mismo color). Cada recta puede cortar a lo sumo a dos lados del polígono, ya que al ser convexo, una recta que no esté alineada con un lado, sólo puede ser exterior, tangente, o secante en exactamente dos puntos, delimitando éstos un segmento interior al polígono. Luego si hay $k$ rectas, tenemos que $2k \\ge 4026$, y $k \\ge 2013$ por lo menos para esta configuración colombiana.\n\nDiremos que dos rectas \\textbf{acompañan} a dos puntos si están construidas de la siguiente manera: tomamos la recta que pasa por esos dos puntos, medimos la menor distancia de cualquier otro punto a dicha recta, y trazamos las paralelas a esta recta a distancia mitad. Claramente, estas dos rectas no pasan por ningún otro punto de la configuración, y delimitan una región del plano en la que sólo están los dos puntos dados.\n\nDiremos que una recta \\textbf{separa} a un punto si en uno de los dos semiplanos delimitados por esta recta, no existe ningún punto del otro color. Una recta puede separar a varios puntos a la vez si todos estos puntos tienen el mismo color, están en el mismo semiplano respecto a la recta, y no hay en dicho semiplano ningún otro punto del otro color.\n\nSupongamos que existe una recta que separa a uno de los puntos rojos. Entonces, trazamos esta recta, dividimos los 2012 puntos rojos restantes en 1006 parejas, y trazamos las rectas que acompañan a estas parejas de puntos rojos. Las 2013 rectas resultantes (la recta que separa al punto y las $2 \\cdot 1006 = 2012$ rectas que acompañan a los demás) son claramente una colección buena, porque en las regiones que están tanto el punto rojo separado, como las parejas de puntos rojos acompañados, no hay ningún punto azul.\n\nConsideremos el polígono convexo, cuyos vértices son puntos de la configuración, que contiene bien en su perímetro, bien en su interior, a todos los puntos de la configuración. Si este polígono contiene en su perímetro algún punto rojo, este punto puede ser separado, ya que podemos trazar una recta tangente al polígono en dicho vértice y en ningún otro, medir la distancia mínima de cualquier otro punto del polígono a dicha recta, y trazar la recta paralela a la construida, a distancia mitad, de forma que el vértice rojo esté en el semiplano opuesto al resto de los puntos de la configuración, quedando así en efecto separado. En este caso se construiría como ya se ha descrito, una colección buena de 2013 rectas.\n\nSi todos los vértices del polígono convexo descrito son azules, consideremos uno de sus lados, que claramente pasa por dos puntos azules y deja a todos los demás puntos de la configuración en el mismo semiplano. Medimos la distancia mínima de cualquier otro punto a esta recta, y trazamos la paralela a distancia mitad, de forma que los dos puntos azules estén en distinto semiplano que el resto. Tenemos pues una recta que separa dos puntos azules, dividimos los restantes 2012 en parejas, y acompañamos a cada pareja por dos rectas, con lo que nuevamente con 2013 rectas, hemos generado regiones tales que en aquellas en las que haya algún punto azul, no hay ninguno rojo.\n\nLuego $k = 2013$ es el valor buscado.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16859,
    "subject": "Mathematics (Olympiad)",
    "question": "We call an ordered pair $(x, y)$ of real numbers *auroral* if the equations $x = y(3 - y)^2$ and $y = x(3 - x)^2$ hold simultaneously.\n\nFind all integers $k$ for which there exists an auroral pair $(x, y)$ of real numbers with $x + y = k$.",
    "options": [],
    "answer": "See solution",
    "solution": "The numbers we are looking for are $k \\in \\{0, 3, 4, 5, 8\\}$.\n\nA pair $(x, x) \\in \\mathbb{R}^2$ is auroral if and only if $x = x(3 - x)^2$, and it is easy to see that this cubic equation has the solution set $\\{0, 2, 4\\}$. These pairs give us $0$, $4$, and $8$ as possible values for $k$.\n\nNext, we investigate the case $k = 3$. Here we can simplify the given equations as $x = yx^2$ and $y = xy^2$. The number $x$ cannot be zero in this case, since otherwise $y$ and $k$ would also be zero. We can conclude that $xy = 1$. The equations $x + y = 3$ and $xy = 1$, according to Vieta's Theorem, imply that $x$ and $y$ are the solutions of the equation $\\lambda^2 - 3\\lambda + 1 = 0$. Hence\n\n$$\n(x, y) = \\left( \\frac{3 + \\sqrt{5}}{2}, \\frac{3 - \\sqrt{5}}{2} \\right) \\quad \\text{or} \\quad (x, y) = \\left( \\frac{3 - \\sqrt{5}}{2}, \\frac{3 + \\sqrt{5}}{2} \\right)\n$$\n\nand it is easy to verify that both pairs are actually auroral. These pairs give us $3$ as a possible value for $k$.\n\nA simple calculation shows that if a pair $(x, y) \\in \\mathbb{R}^2$ is auroral, then the pair $(4 - x, 4 - y)$ is also auroral. From the pairs we have just found, we can therefore construct auroral pairs\n\n$$\n(x, y) = \\left( \\frac{5 - \\sqrt{5}}{2}, \\frac{5 + \\sqrt{5}}{2} \\right) \\quad \\text{and} \\quad (x, y) = \\left( \\frac{5 + \\sqrt{5}}{2}, \\frac{5 - \\sqrt{5}}{2} \\right),\n$$\n\nwhich give us $5$ as a possible value for $k$.\n\nNext, we investigate the case $k = 4$, where we can simplify the first given equation as $x = (4 - x)(x - 1)^2$. We already know $x = 2$ must be a solution of this cubic equation, since the pair $(2, 2)$ is auroral. By polynomial division, we calculate the quotient of $x - (4 - x)(x - 1)^2$ by $x - 2$ as $x^2 - 4x + 2$. The roots of this quadratic polynomial yield pairs\n\n$$\n(x, y) = (2 + \\sqrt{2}, 2 - \\sqrt{2}) \\quad \\text{and} \\quad (x, y) = (2 - \\sqrt{2}, 2 + \\sqrt{2})\n$$\n\nand it is easy to verify that both pairs are actually auroral.\n\nIf $(x, y) \\in \\mathbb{R}^2$ is auroral, then $x$ satisfies the polynomial equation\n\n$$\nx - x(3 - x)^2 [3 - x(3 - x)^2]^2 = 0\n$$\n\nwhich is of degree $9$. Since we have already found nine values for $x$ that must satisfy the equation, there are no more auroral pairs other than those already found. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16860,
    "subject": "Mathematics (Olympiad)",
    "question": "If $x - y > x$, what can you conclude about $y$?",
    "options": [],
    "answer": "See solution",
    "solution": "If $x - y > x$, then $-y > 0$, so $y < 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16861,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral such that $\\angle ADB = \\angle BDC$. Suppose that a point $E$ on the side $AD$ satisfies the equality\n$$\nAE \\cdot ED + BE^2 = CD \\cdot AE.\n$$\nShow that $\\angle EBA = \\angle DCB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $F$ be the point symmetric to $E$ with respect to the line $DB$. Then the equality $\\angle ADB = \\angle BDC$ shows that $F$ lies on the line $DC$, on the same side of $D$ as $C$. Moreover, we have $AE \\cdot ED < CD \\cdot AE$, or $FD = ED < CD$, so in fact $F$ lies on the segment $DC$.\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p128_data_8ad52322f3.png)\n\nNote now that triangles $DEB$ and $DFB$ are congruent (symmetric with respect to the line $DB$), so $\\angle AEB = \\angle BFC$. Also, we have\n$$\nBE^2 = CD \\cdot AE - AE \\cdot ED = AE \\cdot (CD - ED) = AE \\cdot (CD - FD) = AE \\cdot CF.\n$$\nTherefore\n$$\n\\frac{BE}{AE} = \\frac{CF}{BE} = \\frac{CF}{BF}.\n$$\nThis shows that the triangles $BEA$ and $CFB$ are similar, which gives $\\angle EBA = \\angle FCB = \\angle DCB$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16862,
    "subject": "Mathematics (Olympiad)",
    "question": "Fix a point $O$ in the plane and an integer $n \\ge 3$. Consider a finite set $\\mathcal{D}$ of closed unit discs in the plane such that:\n\n(a) No disc in $\\mathcal{D}$ contains the point $O$; and\n\n(b) For each positive integer $k < n$, the closed disc of radius $k+1$ centred at $O$ contains the centres of at least $k$ discs in $\\mathcal{D}$.\n\nShow that some line through $O$ stabs at least $\\frac{2}{\\pi} \\log \\frac{n+1}{2}$ discs in $\\mathcal{D}$.",
    "options": [],
    "answer": "See solution",
    "solution": "For each disc $D$ in $\\mathcal{D}$, let $\\omega_D$ denote the centre of $D$, and let $\\alpha_D$ be the arc-length of the image of $D$ under radial projection from $O$ onto the unit circle centred at $O$. Clearly, $\\alpha_D/2 > \\sin(\\alpha_D/2) = 1/O\\omega_D$.\n\nNow, for each positive integer $k < n$, let $\\mathcal{D}_k$ be the set of those discs in $\\mathcal{D}$ whose centres lie in the closed disc of radius $k+1$ centred at $O$. Since $\\mathcal{D}_i \\subseteq \\mathcal{D}_j$ if $i \\le j$, and each $\\mathcal{D}_k$ contains at least $k$ elements, we may recursively choose (or apply Hall's marriage theorem to produce) a system of distinct representatives, $D_1, \\dots, D_{n-1}$, for the collection $\\mathcal{D}_1, \\dots, \\mathcal{D}_{n-1}$, to obtain\n\n$$\n\\sum_{D \\in \\mathcal{D}_{n-1}} \\alpha_D > 2 \\sum_{D \\in \\mathcal{D}_{n-1}} \\frac{1}{O\\omega_D} \\ge 2 \\sum_{k=1}^{n-1} \\frac{1}{O\\omega_{D_k}} \\ge 2 \\sum_{k=1}^{n-1} \\frac{1}{k+1} > 2 \\log \\frac{n+1}{2}.\n$$\n\nFinally, if $N_{n-1}$ is the maximal number of discs in $\\mathcal{D}_{n-1}$ stabbed by a line through $O$ as it performs a half-turn about $O$, then $\\pi N_{n-1} \\ge \\sum_{D \\in \\mathcal{D}_{n-1}} \\alpha_D$ and the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16863,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $I$ be the incenter and $\\Omega$ be the inscribed circle of a triangle $ABC$, and let $M$ be the midpoint of the side $BC$. Let $K$ be the point of intersection of the line through $A$ perpendicular to $BC$, and the line through $M$ perpendicular to $AI$. Prove that the circle having the line segment $AK$ as a diameter is tangent to the circle $\\Omega$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $XY$ denote the length of segment $XY$. If $AB = AC$, then $K$ and $M$ coincide and the two circles are tangent. So, assume $AB \\ne AC$.\n\nLet $\\Gamma$ be the excircle opposite $A$ of triangle $ABC$. Let $D$ be the point of tangency of $\\Omega$ with $BC$, and $D'$ the point such that $DD'$ is a diameter of $\\Gamma$. Let $E$ be the point of tangency of $\\Gamma$ with $BC$, and $E'$ such that $EE'$ is a diameter of $\\Gamma$. Let $F$ be the intersection of $\\Omega$ and $AD'$, different from $D'$, and $G$ the intersection of $\\Gamma$ and $AE'$, different from $E'$.\n\nLet $B', C'$ be the intersections of the tangent to $\\Omega$ at $D'$ with $AB$ and $BC$, respectively. Then triangles $ABC$ and $AB'C'$ are similar since $BC \\parallel B'C'$. Under this similarity, $E$ and $D'$ correspond, so $A, D', E$ are collinear. Since $DD'$ is a diameter of $\\Gamma$, $\\angle DFD' = 90^\\circ$, and also $\\angle DFE = 90^\\circ$. Similarly, $A, D, E'$ are collinear and $\\angle DGE = 90^\\circ$. Thus, if $K'$ is the intersection of $DE$ and $EG$, then $K'$ is the orthocenter of $ADE$ and $AK'$ is perpendicular to $BC$.\n\nSince $BD = CE$, $M$ is the midpoint of $DE$. Thus, the powers of $M$ with respect to $\\Omega$ and $\\Gamma$ coincide. Also, since $\\angle DFE = \\angle DGE = 90^\\circ$, the points $D, E, F, G$ are concyclic. By the power of a point, $K'D \\cdot K'F = K'E \\cdot K'G$, so the powers of $K'$ with respect to $\\Omega$ and $\\Gamma$ coincide. Therefore, $MK'$ is the radical axis of $\\Omega$ and $\\Gamma$, and if $I_A$ is the center of $\\Gamma$, then $MK'$ is perpendicular to $II_A$. Since $A, I, I_A$ are collinear, $MK'$ is perpendicular to $AI$.\n\nThus, $K = K'$, and $K, D, F$ are collinear. Both $AK$ and $DD'$ are perpendicular to $BC$ and are parallel, so triangles $FKA$ and $FDD'$ are similar. Since $\\angle AFK = 90^\\circ$, the circle with diameter $AK$ is the circumcircle of $FKA$, while $\\Omega$ is the circumcircle of $FDD'$. These circles correspond under the similarity, so they are tangent at $F$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16864,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the last digit of $2^{2015} + 5^{2015}$?",
    "options": [],
    "answer": "See solution",
    "solution": "The last digits of the first few powers of 2 are 2, 4, 8, 6, 2, 4, \\ldots, so the digits repeat in cycles of length 4. Since the remainder after dividing 2015 by 4 is 3, it follows that the last digit of $2^{2015}$ is the same as the last digit of $2^3$, which is 8. Finally, the last digit of any power of 5 is 5, so the last digit of $2^{2015} + 5^{2015}$ is equal to the last digit of $8 + 5$, which is 3.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16865,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $K_{2024}$ be a complete graph whose edges are coloured with 13 colours. What is the minimal integer $k$ such that, for any such colouring, there exists a set of $k$ colours so that the subgraph formed by edges of these $k$ colours is connected?",
    "options": [],
    "answer": "See solution",
    "solution": "Let us show that $k=7$ is sufficient. Randomly divide the 13 colours into two groups $A$ and $B$, with $|A|=7$ and $|B|=6$. Suppose, for contradiction, that using only edges coloured with colours from $A$, some vertex $X$ is not connected to another vertex $Y$. Then the edge $(X, Y)$ must be coloured with a colour from $B$, and for any other vertex $Z$, at least one of $(Z, X)$ or $(Z, Y)$ is coloured with a colour from $B$. Thus, any two vertices are connected by a path using only colours from $B$. Therefore, in any such division, at least one group induces a connected subgraph.\n\nNow, to show that $k=6$ is not sufficient, note that there are $\\binom{13}{6} = 1716$ possible choices of 6 colours. Assign to each such choice a unique vertex (possible since $1716 < 2024$). Colour the edges so that, for each vertex assigned to a 6-colour set, all edges incident to it are coloured with one of the remaining 7 colours. Since $7+7 > 13$, for any two vertices, their allowed 7 colours overlap, so the edge between them can be properly coloured. By construction, for any choice of 6 colours, the corresponding vertex is not connected to others by edges of those 6 colours. Thus, $k=7$ is minimal.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16866,
    "subject": "Mathematics (Olympiad)",
    "question": "$k$ marbles are placed onto the cells of a $2024 \\times 2024$ grid such that each cell has at most one marble and no two marbles are placed onto two neighboring cells (neighboring cells are defined as cells having an edge in common).\n\n**a)** Assume that $k = 2024$. Find a way to place the marbles satisfying the conditions above, such that moving any placed marble to any of its neighboring cells will give an arrangement that does not satisfy both the conditions.\n\n**b)** Determine the largest value of $k$ such that for all arrangements of $k$ marbles satisfying the conditions above, we can move one of the placed marbles onto one of its neighboring cells and the new arrangement satisfies the conditions above.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** Place $2024$ marbles on the main diagonal of the grid. This arrangement ensures each cell has at most one marble and no two marbles are adjacent. If any marble is moved to a neighboring cell, the arrangement will violate the conditions.\n\n**b)** We claim that the largest value of $k$ is $4045$. Suppose $4045$ marbles are placed so that no two are adjacent. Color the grid in a chessboard pattern. Let $W$ be the number of marbles on white cells and $B$ on black cells, with $W + B = 4045$. Without loss of generality, assume $W > B$.\n\nConsider the sub-diagonals parallel to the main diagonal. If $B > 0$, there must be an empty black sub-diagonal, and a marble can be moved to it without violating the conditions. If $B = 0$, all marbles are on white cells. There must be a white diagonal with exactly one marble, and moving it to an adjacent cell will still satisfy the conditions. Thus, for $k = 4045$, it is always possible to move a marble to a neighboring cell and maintain the conditions.\n\n![](images/Vietnam_2024_Booklet_p20_data_2417db0410.png)\n\n![](images/Vietnam_2024_Booklet_p21_data_241b85bf8c.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16867,
    "subject": "Mathematics (Olympiad)",
    "question": "Mile imagined a number and said: \"If I multiply the number by two and add this product to half of the number, I will get a number that is 26 greater than a third of the imagined number.\" What number did Mile imagine?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x$ be the number that Mile imagined. We set up the equation:\n\n$$2x + \\frac{1}{2}x = \\frac{1}{3}x + 26$$\n\nCombine like terms:\n\n$$2x + \\frac{1}{2}x - \\frac{1}{3}x = 26$$\n\nFind a common denominator:\n\n$$2x + \\frac{3}{6}x - \\frac{2}{6}x = 26$$\n\n$$2x + \\frac{1}{6}x = 26$$\n\nMultiply both sides by 6:\n\n$$12x + x = 156$$\n\n$$13x = 156$$\n\n$$x = 12$$\n\nMile imagined the number $12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16868,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integer solutions to the equation\n\n$$\nx^3 + 3xy + y^3 = 2019\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The right-hand side of the equation is divisible by $3$ but not by $9$. Assume that $3 \\mid x$. Then $9 \\mid x^3$ and $9 \\mid 3xy$. If also $3 \\mid y$, then $9 \\mid y^3$, implying that the left-hand side of the equation is divisible by $9$. Thus $3 \\nmid y$. But then $3 \\nmid y^3$, implying that the left-hand side of the equation is not divisible by $3$. The contradiction shows that $3 \\nmid x$. By symmetry, also $3 \\nmid y$.\n\nAs $3 \\mid 3xy$, we must have $3 \\mid x^3 + y^3$. Hence also $3 \\mid x^3 + 3x^2y + 3xy^2 + y^3 = (x+y)^3$, implying $3 \\mid x+y$. Consequently, $x$ and $y$ are congruent to $1$ and $2$ modulo $3$ in some order. Hence $x^2 \\equiv y^2 \\equiv 1 \\pmod{3}$ and $xy \\equiv 2 \\pmod{3}$, implying $x^2 - xy + y^2 \\equiv 0 \\pmod{3}$. We obtain $9 \\mid x^3 + y^3 = (x+y)(x^2 - xy + y^2)$, whereas $3xy \\equiv 6 \\pmod{9}$. Hence the left-hand side of the equation is congruent to $6$ modulo $9$ but the right-hand side of the equation is congruent to $3$. Consequently, there are no solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16869,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $u > 0$, $v \\neq 0$, and $u + v \\neq 0$.\n\nDescribe how, using operations (i) and (ii), one can obtain all negative integers, zero, and any integer $a$ on the board, starting from $u$ and $v$.\n\nAdditionally, consider the case $u + v = 0$, $u \\neq 1$. Show how to obtain all numbers on the board in this scenario.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a$ and $b$ be chosen as follows:\n- If $\\frac{u}{v} = -2$ or $-\\frac{1}{3}$, set $a = v$, $b = u$.\n- Otherwise, set $a = u$, $b = v$.\n\nThis ensures $\\frac{a}{b} \\neq -2$ or $-\\frac{1}{3}$.\n\nApply operation (i) five times:\n- $a + b$ adds $a + b$ to the board.\n- $(a + b) + a$ adds $2a + b$.\n- $(2a + b) + b$ adds $2a + 2b$.\n- $(2a + 2b) + a$ adds $3a + 2b$.\n- $(3a + 2b) + b$ adds $3a + 3b$.\n\nThus, $u + v$, $2(u + v)$, and $3(u + v)$ are on the board and distinct since $u + v \\neq 0$.\n\nConsider the quadratic:\n$$\n(u + v)x^2 + 3(u + v)x + 2(u + v) = (u + v)(x^2 + 3x + 2) = (u + v)(x + 1)(x + 2).\n$$\nUsing operation (ii), $-1$ and $-2$ can be added to the board. Now, all negative numbers can be obtained:\n- For the largest negative $n$ not on the board ($n \\leq -3$), $n + 1$ is on the board. Then $n + 1 + (-1)$ adds $n$.\n- Repeat to obtain all negative numbers.\n\nSince $u > 0$ is on the board, and all large negative numbers are on the board, $-u$ is on the board. Then $u + (-u)$ adds $0$.\n\nTo obtain any $a$:\n$$\nux^2 + 0 \\times x - ua^2\n$$\nSince $-ua^2 < 0$ is on the board for $a \\neq 0$, operation (ii) allows adding $a$.\n\n**Case 5** ($u + v = 0$, $u \\neq 1$):\n- Operation (i) adds $0$.\n- Operation (ii) on $ux^2 + 0 \\times x + v$ adds $1$ and $-1$.\n- Then, follow Case 4 with $u = u$, $v = -1$ to obtain all numbers. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16870,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n$ candies on the table. On every turn, a player eats a number of candies that is greater than 1 and divides the number of candies on the table at the start of the turn, but must leave at least 1 candy on the table. Two players take alternate turns and the player who is unable to make a move loses. Find all positive integers $n$ for which the first player can always win.",
    "options": [],
    "answer": "See solution",
    "solution": "All even numbers, except odd powers of 2.\n\nDefine all even numbers which are not odd powers of 2 as *good* and the rest of the positive integers as *bad*. We show that the player before whose turn the number of candies is good has a move which yields a bad number of candies, whereas the player before whose turn the number of candies is bad either has lost or is forced to leave a good number of candies on the table. As the number of candies is reduced in each move, the starting player wins if, initially, the number of candies on the table is good.\n\nA player before whose turn the number of candies is even but not a power of 2 can eat the number of candies equal to its odd factor different from 1. After that, the number of candies on the table is odd, i.e., bad. If before the turn the number of candies is equal to an even power of 2, one can eat exactly half of the candies, leaving an odd power of 2 candies on the table, i.e., a bad number.\n\nOn the other hand, if the number of candies before the turn is odd, the player can only choose odd factors. This yields an even number of candies left on the table. Furthermore, the number of candies left on the table is divisible by the number of candies taken from the table, which is odd; hence, the number of candies cannot be a power of 2. Therefore, the number of candies left on the table is good. However, if the number of candies before the turn is equal to an odd power of 2, the player can only choose even factors, which results in an even number of candies left on the table. Furthermore, the rules do not allow eating more than half of the candies. Therefore, the number of candies left on the table can only be a power of 2 if its exponent is less by 1 than before the move. This would be an even power of 2, which is also a good number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16871,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D$ be an interior point of the acute triangle $ABC$ with $AB > AC$ so that $\\angle DAB = \\angle CAD$. The point $E$ on the segment $AC$ satisfies $\\angle ADE = \\angle BCD$, the point $F$ on the segment $AB$ satisfies $\\angle FDA = \\angle DBC$, and the point $X$ on the line $AC$ satisfies $CX = BX$. Let $O_1$ and $O_2$ be the circumcentres of the triangles $ADC$ and $EXD$, respectively. Prove that the lines $BC$, $EF$, and $O_1O_2$ are concurrent.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p410_data_1c7c5f2a9b.png)\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p411_data_d378492259.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $Q$ be the isogonal conjugate of point $D$ in $\\triangle ABC$. As $\\angle BAD = \\angle DAC$, $Q$ is on $AD$, and $\\angle QBA = \\angle DBC = \\angle FDA$. It follows that $Q, D, F, B$ are concyclic; likewise, $Q, D, E, C$ are concyclic. We have $AF \\cdot AB = AD \\cdot AQ = AE \\cdot AC$, and thereby $B, F, E, C$ are concyclic, as illustrated in Fig. 3.1.\n\n*Lemma*: Let the extensions of $BC$ and $FE$ beyond $C$ and $E$ meet at $T$. Then $TD^2 = TB \\cdot TC = TF \\cdot TE$.\n\n*Proof of lemma*: First, $\\odot DEF$ and $\\odot BDC$ are tangent to each other, since\n\n$$\n\\begin{align*}\n\\angle BDF &= \\angle AFD - \\angle ABD \\\\\n&= (180^\\circ - \\angle FAD - \\angle FDA) - (\\angle ABC - \\angle DBC) \\\\\n&= 180^\\circ - \\angle FAD - \\angle ABC \\\\\n&= 180^\\circ - \\angle DAE - \\angle FEA \\\\\n&= \\angle FED + \\angle ADE \\\\\n&= \\angle FED + \\angle DCB.\n\\end{align*}\n$$\n\nNext, as $B, C, E, F$ are concyclic, the powers of $T$ with respect to $\\odot BDC$ and $\\odot EDF$ are equal. Therefore, the radical axis through $T$ is the common tangent at $D$, $TD^2 = TB \\cdot TC = TF \\cdot TE$. The lemma is verified.\n\nAs illustrated in Fig. 3.2, let the line $TA$ and $\\odot ABC$ intersect at another point $M$. Notice that $B, C, E, F$ are concyclic, $A, M, C, B$ are concyclic, and by the lemma, we find\n\n$$\nTM \\cdot TA = TF \\cdot TE = TB \\cdot TC = TD^2,\n$$\n\nimplying that $A, M, E, F$ are concyclic.\n\nConsider the inversion transformation centred at $T$ of radius $TD$: $M$ is mapped to $A$, $B$ is mapped to $C$, and $\\odot MBD$ is mapped to $\\odot ACD$. As their common point $D$ lies on the inversion circle, so does the other common point $K$, $TK = TD$. It follows that $T$ and the centres of $\\odot KDE$ and $\\odot ADC$ all lie on the perpendicular bisector of $KD$.\n\nAs $O_1$ is the circumcentre of $\\triangle ADC$, it remains to prove $D, K, E, X$ lie on a circle, whose centre is $O_2$.\n\nObserve that $BM, DK$, and $AC$ are the radical axes of pairs from $\\odot ABCM, \\odot ACDK$, and $\\odot BMDK$, respectively; they are concurrent at $P$. In addition, $M$ lies on $\\odot AEF$. Thus,\n\n$$\n\\begin{align*}\n\\angle (EX, XB) &= \\angle (CX, XB) = \\angle (XC, BC) + \\angle (BC, BX) \\\\\n&= 2\\angle (AC, CB) = \\angle (AC, CB) + \\angle (EF, FA) \\\\\n&= \\angle (AM, BM) + \\angle (EM, MA) \\\\\n&= \\angle (EM, BM).\n\\end{align*}\n$$\n\nThis implies that $M, E, X, B$ are concyclic, and hence\n\n$$\nPE \\cdot PX = PM \\cdot PB = PK \\cdot PD.\n$$\n\nTherefore, $E, K, D$, and $X$ are concyclic. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16872,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for all real numbers $a, b, c > 0$ satisfying $a + b + c = 3$, the following inequality holds:\n\n$$\n2(ab + bc + ca) - 3abc \\ge a\\sqrt{\\frac{b^2+c^2}{2}} + b\\sqrt{\\frac{c^2+a^2}{2}} + c\\sqrt{\\frac{a^2+b^2}{2}}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Using the inequality $\\frac{b+c}{2} \\le \\sqrt{\\frac{b^2+c^2}{2}}$, we easily get $\\sqrt{\\frac{b^2+c^2}{2}} \\le \\frac{b^2+c^2}{b+c} = b+c - \\frac{2bc}{b+c}$ and similar ones for the other variables. As a consequence,\n$$\na\\sqrt{\\frac{b^2+c^2}{2}} + b\\sqrt{\\frac{a^2+c^2}{2}} + c\\sqrt{\\frac{a^2+b^2}{2}} \\le 2(ab+bc+ca) - 2abc\\left(\\frac{1}{a+b} + \\frac{1}{b+c} + \\frac{1}{c+a}\\right)\n$$\nThe given inequality follows from\n$$\n\\frac{1}{a+b} + \\frac{1}{b+c} + \\frac{1}{c+a} \\ge \\frac{(1+1+1)^2}{2(a+b+c)} = \\frac{3}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16873,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\omega$ be a semicircle with diameter $XY$. Let $M$ be the midpoint of $XY$. Let $A$ be an arbitrary point on $\\omega$ such that $AX < AY$. Let $B$ and $C$ be points lying on the segments $XM$ and $YM$, respectively, such that $BM = CM$. The line through $C$ parallel to $AB$ intersects $\\omega$ at $P$. The line through $B$ parallel to $AC$ intersects $\\omega$ at $Q$. The line $PQ$ intersects the line $XY$ at a point $S$. Prove that the line $AS$ is tangent to $\\omega$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $D$ be the point symmetric to $A$ with respect to $M$. Then $ABDC$ is a parallelogram as $M$ bisects both $BC$ and $AD$. It follows that $AB \\parallel CD$ and $AC \\parallel BD$. This means that $P$ lies on $CD$ and $Q$ lies on $BD$.\n\nWe shall prove that the tangent line at $A$ to $\\omega$ intersects $PQ$ at a point $S'$ such that $\\frac{PS'}{S'Q} = \\frac{PS}{SQ}$. This in turn will imply that $S = S'$, giving the thesis.\n\nSince $AS'$ is tangent to $\\omega$, we have $\\angle APS' = \\angle S'AQ$. Hence triangles $APS'$, $QAS'$ are similar as they have equal angles. Hence\n\n$$\n\\frac{PS'}{S'Q} = \\frac{PS'}{AS'} \\cdot \\frac{AS'}{S'Q} = \\frac{AP}{AQ} \\cdot \\frac{AP}{AQ} = \\frac{AP^2}{AQ^2}.\n$$\n\nWe now calculate the ratio $\\frac{PS}{QS}$. By Menelaus' theorem for triangle $PQD$ and transversal $XY$ we have\n\n$$\n\\frac{PS}{SQ} \\cdot \\frac{QB}{BD} \\cdot \\frac{DC}{CP} = 1.\n$$\n\nSince $ABDC$ is a parallelogram, we have $BD = AC$ and $DC = AB$. This yields\n\n$$\n\\frac{PS}{SQ} = \\frac{BD}{QB} \\cdot \\frac{CP}{DC} = \\frac{AC \\cdot CP}{AB \\cdot BQ}.\n$$\n\nSince $AM = MP = MD = MQ$, we have $\\angle APC = 90^\\circ = \\angle AQB$. Moreover, since $ABDC$ is a parallelogram, $\\angle ABQ = \\angle ACP$. This shows that triangles $ABQ$, $ACP$ are similar. In particular $\\frac{AC}{AB} = \\frac{AP}{AQ} = \\frac{CP}{BQ}$. Therefore\n\n$$\n\\frac{PS}{SQ} = \\frac{AC \\cdot CP}{AB \\cdot BQ} = \\frac{AP^2}{AQ^2} = \\frac{PS'}{S'Q},\n$$\n\nwhich, as was observed before, finishes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16874,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that\n$$\n\\sin(4x) + 2\\sin(5x) + \\sin(6x) = 2(\\sin(2x) + \\sin(3x))\n$$\nusing the double-angle formula and the angle sum identities for sines and cosines.",
    "options": [],
    "answer": "See solution",
    "solution": "The double-angle formula and the angle sum identities for sines and cosines imply\n$$\n\\begin{aligned}\n\\sin(4x) + 2\\sin(5x) + \\sin(6x) &= \\\\\n&= 2\\sin(2x)\\cos(2x) + 2(\\sin(2x)\\cos(3x) + \\sin(3x)\\cos(2x)) + 2\\sin(3x)\\cos(3x) \\\\\n&= 2\\cos(2x)(\\sin(2x) + \\sin(3x)) + 2\\cos(3x)(\\sin(2x) + \\sin(3x)) \\\\\n&= 2(\\cos(2x) + \\cos(3x))(\\sin(2x) + \\sin(3x)).\n\\end{aligned}\n$$\nThe latter is equal to $2(\\sin(2x) + \\sin(3x))$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16875,
    "subject": "Mathematics (Olympiad)",
    "question": "A non-empty subset $A$ of $\\{1, 2, \\dots, n\\}$ is called *good* of degree $n$ if $|A| \\le \\min_{x \\in A} x$. Denote by $a_n$ the number of good sets of degree $n$. Prove that\n$$\na_{n+2} = a_{n+1} + a_n + 1\n$$\nfor any positive integer $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A$ be a good set of degree $n$, and $|A| = k$. Then $\\min_{x \\in A} x \\ge k$, so $A \\subseteq \\{k, k+1, \\dots, n\\}$. Hence, the number of good sets of degree $n$ with $k$ elements is $C_{n-k+1}^k$. It follows that\n\n$$\na_n = \\sum_{k=1}^{\\lfloor \\frac{n+1}{2} \\rfloor} C_{n-k+1}^k = C_n^1 + C_{n-1}^2 + C_{n-2}^3 + \\dots$$\n\nIf $n$ is even, $n = 2m$, then\n\n$$\n\\begin{align*}\na_{2m+2} &= C_{2m+2}^1 + C_{2m+1}^2 + \\dots + C_{m+2}^{m+1} \\\\ \n&= (C_{2m+1}^1 + C_{2m+1}^0) + (C_{2m}^2 + C_{2m}^1) + \\dots + (C_{m+1}^{m+1} + C_{m+1}^m) \\\\ \n&= (C_{2m+1}^1 + C_{2m}^2 + \\dots + C_{m+1}^{m+1}) + (C_{2m}^1 + C_{2m-1}^2 + \\dots + C_{m+1}^m) + C_{2m+1}^0 \\\\ \n&= a_{2m+1} + a_{2m} + 1.\n\\end{align*}\n$$\n\nIf $n$ is odd, $n = 2m - 1$, then\n\n$$\n\\begin{align*}\na_{2m+1} &= C_{2m+1}^1 + C_{2m}^2 + \\dots + C_{m+2}^m + C_{m+1}^{m+1} \\\\ \n&= (C_{2m}^1 + C_{2m}^0) + (C_{2m-1}^2 + C_{2m-1}^1) + \\dots + (C_{m+1}^m + C_{m+1}^{m-1}) + C_m^m \\\\ \n&= (C_{2m}^1 + C_{2m-1}^2 + \\dots + C_{m+1}^m) + (C_{2m-1}^1 + C_{2m-2}^2 + \\dots + C_{m+1}^{m-1} + C_m^m) + C_{2m}^0 \\\\ \n&= a_{2m} + a_{2m-1} + 1.\n\\end{align*}\n$$\n\nIn summary, the equality $a_{n+2} = a_{n+1} + a_n + 1$ holds for all positive integers $n$.\n\n**Remark.** Let $F_n$ be the $n$-th term of the Fibonacci sequence. From the combinatorial identity $\\sum_{k=0}^{\\lfloor \\frac{n}{2} \\rfloor} C_{n-k}^k = F_n$, one can derive that $a_n = F_{n+1} - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16876,
    "subject": "Mathematics (Olympiad)",
    "question": "Define the sequence $a_1, a_2, a_3, \\dots$ by $a_1 = 4$, $a_2 = 7$, and\n\n$$\na_{n+1} = 2a_n - a_{n-1} + 2, \\quad \\text{for } n \\ge 2.\n$$\n\nProve that, for every positive integer $m$, the number $a_m a_{m+1}$ is a term of the sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "First we prove the following formula by induction:\n\n$$\na_n = n^2 + 3, \\quad \\text{for } n \\ge 1.\n$$\n\nWe require two base cases to get started. The formula is true for $n = 1$ and $n = 2$ because $a_1 = 4 = 1^2 + 3$ and $a_2 = 7 = 2^2 + 3$.\n\nFor the inductive step, assume that the formula is true for $n = k - 1$ and $n = k$. Then for $n = k + 1$, we have\n\n$$\n\\begin{aligned}\na_{k+1} &= 2a_k - a_{k-1} + 2 \\\\ \n&= 2(k^2 + 3) - ((k-1)^2 + 3) + 2 \\\\ \n&= k^2 + 2k + 4 \\\\ \n&= (k+1)^2 + 3.\n\\end{aligned}\n$$\n\nHence the formula is also true for $n = k + 1$. This completes the induction.\n\nUsing the formula, we calculate\n\n$$\n\\begin{aligned}\na_m a_{m+1} &= (m^2 + 3)((m+1)^2 + 3) \\\\ \n&= (m^2 + 3)(m^2 + 2m + 4) \\\\ \n&= m^4 + 2m^3 + 7m^2 + 6m + 12 \\\\ \n&= (m^2 + m + 3)^2 + 3 \\\\ \n&= a_{m^2+m+3}.\n\\end{aligned}\n$$\n\nHence $a_m a_{m+1}$ is a term of the sequence. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16877,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for all positive numbers $a, b$ the following inequality holds:\n\n$$\n(1+\\frac{a}{b})^{2012} + (1+\\frac{b}{a})^{2012} \\geq 2^{2013}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Using the arithmetic mean-geometric mean inequality twice, we obtain:\n\n$$\n\\begin{aligned}\n& (1+\\frac{a}{b})^{2012} + (1+\\frac{b}{a})^{2012} \\geq 2\\sqrt{(1+\\frac{a}{b})^{2012}(1+\\frac{b}{a})^{2012}} \\\\\n&= 2\\left((1+\\frac{a}{b})(1+\\frac{b}{a})\\right)^{1006} \\\\\n&= 2\\left(2+\\frac{a}{b}+\\frac{b}{a}\\right)^{1006} \\geq 2\\cdot 2^{1006} = 2^{2013},\n\\end{aligned}\n$$\n\nwhich was to be proved.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16878,
    "subject": "Mathematics (Olympiad)",
    "question": "The city of Mar del Plata has the shape of a square, denoted by WSEN. For a given even integer $n$, it is divided by $2(n + 1)$ streets into $n \\times n$ blocks (the streets also run around the square). Each block has a size of $100\\ \\mathrm{m} \\times 100\\ \\mathrm{m}$. All the streets in Mar del Plata are one-way, with each street having a single direction along its entire length. Every two adjacent parallel streets have opposite directions. The WS street is directed from $W$ to $S$, and the WN street is directed from $W$ to $N$. A street cleaning car starts at $W$. Its aim is to arrive at $E$ and clean the roads it passes through. What is the length of the longest possible route it can take if no $100\\ \\mathrm{m}$ segment can be passed more than once?\n\n(For $n = 6$, Fig. 1 shows the map of the city and one possible, but not the longest, route of the car. See also http://goo.gl/maps/JAzD.)\n",
    "options": [],
    "answer": "See solution",
    "solution": "First, we introduce some notation. Call each $100\\ \\mathrm{m}$ segment of a street an *arrow*, and each place where two streets meet a *crossroad*. If an arrow has the same direction as the street WS or WN (that is, it runs from the upper left to the bottom right or from the bottom left to the upper right), it is called a *forward arrow*; otherwise, it is a *back arrow*.\n\nIn the solution, we use the following lemma: Let the set of all crossroads be split into two sets $A$ and $B$ such that $W \\in A$ and $E \\in B$. Then the number of times the car drives along an arrow from $A$ to $B$ is one more than the number of times it drives along an arrow from $B$ to $A$.\n\nLet us split the set of all crossroads of Mar del Plata into two sets $A$ and $B$ by a vertical (north-south) line connecting two points located $100k + 50\\ \\mathrm{m}$ from $W$—one on the WN street and one on the WS street—for some $k \\in \\{0, 1, \\dots, n-1\\}$. \n\n![](images/Cesko-Slovacko-Poljsko_2012_p7_data_cdcdcc9b7e.png)\n\nFig. 5a\n\n![](images/Cesko-Slovacko-Poljsko_2012_p7_data_ed5cbed811.png)\n\nFig. 5b\n\nIf $k$ is odd, then the line intersects $k + 1$ forward arrows (from $\\mathcal{A}$ to $\\mathcal{B}$) and $k + 1$ back arrows (from $\\mathcal{B}$ to $\\mathcal{A}$). Even if the car passes all $k + 1$ forward arrows, by the lemma, it can pass at most $k$ back arrows. Hence, at least one of the back arrows remains unpassed.\n\nIf $k \\ge 2$ is even, then the line intersects $k + 2$ forward arrows and $k$ back arrows. The two northernmost forward arrows intersected by the line begin at a crossroad with only one incoming arrow. This crossroad can be passed only once, so one of the two northernmost forward arrows remains unpassed. The same holds for the two southernmost forward arrows. So at most $k$ forward arrows are passed, and by the lemma, at most $k - 1$ back arrows. Together, at least 3 arrows remain unpassed at this level.\n\nFor $k = 0$, there are only two forward arrows starting at $W$ intersected by the line. Clearly, only one of them can be passed, so one remains unpassed.\n\nSimilarly, we split the set of all crossroads by a vertical line connecting two points located $100k + 50\\ \\mathrm{m}$ from $E$—one on the SE street and one on the NE street—for some $k \\in \\{0, 1, \\dots, n-1\\}$. The situation is sketched in Fig. 6a and 6b for $k = 3$ and $k = 4$ respectively.\n\n![](images/Cesko-Slovacko-Poljsko_2012_p8_data_e1e4b1abe1.png)\n\nFig. 6a\n\n![](images/Cesko-Slovacko-Poljsko_2012_p8_data_a4ff3234d6.png)\n\nFig. 6b\n\nIf $k$ is odd, the line intersects $k+1$ forward arrows and $k+1$ back arrows, and at least one of the back arrows remains unpassed.\n\nIf $k \\ge 2$ is even, the line intersects $k+2$ forward arrows and $k$ back arrows. The two northernmost forward arrows end at a crossroad with only one outgoing arrow, so one of them remains unpassed. The same holds for the two southernmost forward arrows. Again, at most $k$ forward arrows and at most $k-1$ back arrows are passed, resulting in 3 unpassed arrows at this level.\n\nFor $k = 0$, there are two forward arrows ending at $E$; only one can be passed, so one remains unpassed.\n\nThere is one unpassed arrow for any odd $k$, three for any even $k \\ge 2$, and one for $k = 0$, and all this happens twice. As $n$ is even and $k \\in \\{0, 1, \\dots, n-1\\}$, altogether, we have\n\n$$\n2\\left(\\frac{1}{2}n + 3\\left(\\frac{1}{2}n - 1\\right) + 1\\right) = 4n - 4\n$$\n\nunpassed arrows. The total number of arrows is $n \\cdot 2(n+1)$, so the car cannot pass more than\n\n$$\nn \\cdot 2(n + 1) - (4n - 4) = 2n^2 - 2n + 4\n$$\n\narrows.\n\nOn the other hand, there are many possible routes of the car consisting of $2n^2 - 2n + 4$ arrows. One is sketched in Fig. 7 for $n = 6$. Using the same pattern in the general case, the route can be subdivided into $\\frac{1}{2}n$ parts by the crossroads lying on the WS street located $200k$ m from $W$ for $k = 1, 2, \\dots, \\frac{1}{2}n - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16879,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a sequence $(x_n)$ defined such that $x_n > 0$ for all $n \\ge 1$, and the recurrence relation:\n\n$$\n2x_n - x_{n-1} = \\sqrt{x_{n-1}^2 + 4x_{n-1}} \\quad \\forall n \\ge 2.\n$$\n\nLet $y_n = \\sum_{i=1}^{n} \\frac{1}{x_i^2}$.\n\nShow that $(y_n)$ converges and find $\\lim_{n \\to \\infty} y_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "From the definition, $x_n > 0$ for all $n \\ge 1$.\n\nRewriting the recurrence:\n$$\n2x_n - x_{n-1} = \\sqrt{x_{n-1}^2 + 4x_{n-1}} \\quad \\forall n \\ge 2.\n$$\n\nIt follows that:\n$$\nx_{n-1} = x_n^2 - x_n x_{n-1} \\quad \\forall n \\ge 2.\n$$\n\nConsequently:\n$$\n\\frac{1}{x_n^2} = \\frac{1}{x_{n-1}} - \\frac{1}{x_n} \\quad \\forall n \\ge 2\n$$\n(since $x_n \\ne 0$ for all $n \\ge 2$).\n\nTherefore, for all $n \\ge 2$:\n$$\ny_n = \\sum_{i=1}^{n} \\frac{1}{x_i^2} = \\frac{1}{x_1^2} + \\sum_{i=2}^{n} \\left( \\frac{1}{x_{i-1}} - \\frac{1}{x_i} \\right) = \\frac{1}{x_1^2} + \\frac{1}{x_1} - \\frac{1}{x_n} = 6 - \\frac{1}{x_n}\n$$\n\nSince $\\left(\\frac{1}{x_n}\\right)$ is decreasing and bounded below by $0$, it converges, and $\\lim_{n \\to \\infty} \\frac{1}{x_n} = 0$. Thus, $(y_n)$ converges and:\n$$\n\\lim_{n \\to \\infty} y_n = 6.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16880,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the recurrence relation\n\n$$2a_{n+2} = a_{n+1} + 4a_n$$\n\nfor $n = 0, 1, 2, \\dots, 3028$, prove that at least one of the numbers $a_0, a_1, \\dots, a_{3030}$ is divisible by $2^{2020}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove by induction, for $m \\ge 1$, the proposition that\n\n$$a_n \\equiv 0 \\pmod{2^m} \\text{ for } n = \\left\\lfloor \\frac{m}{2} \\right\\rfloor, \\dots, 3030 - m.$$  \n\n**Base cases:**\n\nFor $m = 1$:\n\nFrom the recurrence,\n$$2a_{n+2} = a_{n+1} + 4a_n$$\nfor $n = 0, 1, \\dots, 3028$.\n\nModulo $2$, $2a_{n+2} \\equiv 0$, so $a_{n+1} + 4a_n \\equiv 0 \\pmod{2}$, i.e., $a_{n+1} \\equiv 0 \\pmod{2}$ for $n = 0, 1, \\dots, 3028$. Thus,\n$$a_n \\equiv 0 \\pmod{2} \\text{ for } n = 1, 2, \\dots, 3029.$$\n\nFor $m = 2$:\n\nSince $a_{n+2} \\equiv 0 \\pmod{2}$, $2a_{n+2} \\equiv 0 \\pmod{4}$, so $a_{n+1} + 4a_n \\equiv 0 \\pmod{4}$, i.e., $a_{n+1} \\equiv 0 \\pmod{4}$ for $n = 0, 1, \\dots, 3027$. Thus,\n$$a_n \\equiv 0 \\pmod{4} \\text{ for } n = 1, 2, \\dots, 3028.$$\n\n**Inductive step:**\n\nSuppose the proposition holds for $m = k-1$ and $m = k$ ($k \\ge 2$).\n\nThen $a_n \\equiv 0 \\pmod{2^k}$ for $n = \\left\\lceil \\frac{k}{2} \\right\\rceil, \\dots, 3030 - k$ implies $a_{n+2} \\equiv 0 \\pmod{2^k}$ for $n = \\left\\lceil \\frac{k}{2} \\right\\rceil - 2, \\dots, 3028 - k$, so\n$$2a_{n+2} \\equiv 0 \\pmod{2^{k+1}}.$$\n\nAlso, $a_n \\equiv 0 \\pmod{2^{k-1}}$ for $n = \\left\\lceil \\frac{k-1}{2} \\right\\rceil, \\dots, 3030 - (k-1)$, i.e., $n = \\left\\lceil \\frac{k+1}{2} \\right\\rceil - 1, \\dots, 3031 - k$, so\n$$4a_n \\equiv 0 \\pmod{2^{k+1}}.$$\n\nFrom the recurrence, $2a_{n+2} = a_{n+1} + 4a_n$, so\n$$a_{n+1} \\equiv 0 \\pmod{2^{k+1}}$$\nfor $n = \\left\\lceil \\frac{k+1}{2} \\right\\rceil - 1, \\dots, 3028 - k$.\n\nThus,\n$$a_n \\equiv 0 \\pmod{2^{k+1}} \\text{ for } n = \\left\\lceil \\frac{k+1}{2} \\right\\rceil, \\dots, 3030 - (k+1).$$\n\nBy induction, the proposition holds for all $m$.\n\nLetting $m = 2020$, we have\n$$a_n \\equiv 0 \\pmod{2^{2020}} \\text{ for } n = 1010.$$\n\nTherefore, at least one of $a_0, a_1, \\dots, a_{3030}$ (specifically $a_{1010}$) is divisible by $2^{2020}$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16881,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the least positive integer $n$ such that there exist positive integers $a$, $b$, and $c$, none of which is a perfect square, and\n\n$$\na^3 + b^3 + c^3 - 3abc = 2013^n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to see that the left-hand side is divisible by $9$, implying that the equation has no solution for $n = 1$.\n\nLet $n = 2$. We have $2013^2 = (3 \\cdot 11 \\cdot 61)^2$, $3^2 = 2^3 + 1^3$, and $11 \\cdot 61 = 8^3 + 4^3 + 1^3 + 3 \\cdot 8 \\cdot 4 \\cdot 1$. We use the identity\n\n$$\n(a^3 + b^3 + c^3 - 3abc)(x^3 + y^3 + z^3 - 3xyz) = u^3 + v^3 + w^3 - 3uvw,\n$$\n\nwhere $u = ax + by + cz$, $v = ay + bz + cx$, and $w = az + bx + cy$. We apply the above identity first for $(8, 4, -1)$ and $(4, 8, -1)$, and then for the obtained triples $(65, 0, 56)$ and $(2, 1, 0)$ and obtain $65^3 + 56^3 = (11 \\cdot 61)^2$ and $130^3 + 177^3 + 56^3 - 3 \\cdot 130 \\cdot 177 \\cdot 56 = 2013^2$.\n\n**Remark.** Using similar arguments, it is possible to find the solution $186^3 + 112^3 + 65^3 - 3 \\cdot 186 \\cdot 112 \\cdot 65 = 2013^2$.\n\nNote also that $a = b - d$, $b$, and $c = b + d$ is a solution if and only if $bd^2 = (11 \\cdot 61)^2$, but then $b$ is a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16882,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples of positive integers $(p, a, m)$ such that $p$ is a prime, $a \\leq 5p^2$, and\n$$(p-1)! + a = p^m.$$",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $(p, a, m) = (2, 1, 1)$, $(2, 3, 2)$, $(2, 7, 3)$, $(2, 15, 4)$, $(3, 1, 1)$, $(3, 7, 2)$, $(3, 25, 3)$, $(5, 1, 2)$, $(5, 101, 3)$.\n\nWhen $p = 2$, we need $1 + a = 2^m$ with $a \\leq 20$. This yields $(a, m) = (1, 1)$, $(3, 2)$, $(7, 3)$, $(15, 4)$.\n\nWhen $p = 3$, we need $2 + a = 3^m$ with $a \\leq 45$. This yields $(a, m) = (1, 1)$, $(7, 2)$, $(25, 3)$.\n\nWhen $p = 5$, we need $24 + a = 5^m$ with $a \\leq 125$. This yields $(a, m) = (1, 2)$, $(101, 3)$.\n\nNow, suppose $p \\geq 7$. By Wilson's theorem, $-1 + a \\equiv 0 \\pmod{p}$. Also, $a \\equiv 1 \\pmod{p-1}$. Thus, $a = k p (p-1) + 1$ for some integer $k \\geq 0$. Since $5p^2 \\geq a$, we have $k \\leq 5$ for $p \\geq 7$.\n\nThe equation becomes\n$$(p-1)! + k p (p-1) = p^m.$$\nOr equivalently,\n$$(p-2)! + k p = p^{m-1} + p^{m-2} + \\dots + 1.$$\n\nNote that $2$ and $\\frac{p-1}{2}$ both appear in $(p-2)!$, and they are distinct for $p \\geq 7$. Thus, $p-1 \\mid (p-2)!$. It follows that $k \\equiv m \\pmod{p-1}$.\n\nIf $m \\geq p-1$, then\n$$(p-2)! + k p = p^{m-1} + p^{m-2} + \\dots + 1 > p^{p-2}.$$\nBut\n$$(p-2)! + k p \\leq (p-2)! + 5p < p^{p-3} + p^{p-3} < p^{p-2}$$\nfor $p \\geq 7$, a contradiction. Thus, $m < p-1$. Since $k \\leq 5 < p-1$, the only possibility for $k \\equiv m \\pmod{p-1}$ is $k = m$.\n\nNow, check small values:\n\n- $k = m = 1$: $(p-2)! + p = 1$ (impossible).\n- $k = m = 2$: $(p-2)! + 2p = p + 1$ (impossible).\n- $k = m = 3$: $(p-2)! + 3p = p^2 + p + 1$. We need $p-2 \\mid p^2 - 2p + 1$, i.e., $p-2 \\mid 1$, which is impossible for $p \\geq 7$.\n- $k = m = 4$: $(p-2)! + 4p = p^3 + p^2 + p + 1$. We need $p-2 \\mid p^3 + p^2 - 3p + 1$, i.e., $p-2 \\mid 7$, which is impossible for $p \\geq 7$.\n\nTherefore, there is no solution for $p \\geq 7$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16883,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be an equilateral trapezoid with sides $AB$ and $CD$. The incircle of triangle $BCD$ touches $CD$ at $E$. Point $F$ is chosen on the bisector of the angle $\\angle DAC$ such that the lines $EF$ and $CD$ are perpendicular. The circumcircle of triangle $ACF$ intersects the line $CD$ again at $G$. Prove that triangle $AFG$ is isosceles.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/CzsMT00_sol_p5_data_bf00992238.png)\nLet us show that $|FA| = |FG|$. We will proceed from behind. On the extension of $CD$ we take the point $P$ such that $|DP| = |DA|$ and similarly on the extension of $DC$ we take the point $Q$ such that $|CQ| = |CA|$. Then, using well-known properties of the incircle, we have\n\n$$\n\\begin{aligned}\n|PE| &= |PD| + |DE| = |DA| + \\frac{|BD| + |CD| - |BC|}{2} = \\frac{|BD| + |CD| + |BC|}{2} \\\\\n|QE| &= |QC| + |CE| = |AC| + \\frac{|BD| + |CD| - |BC|}{2} = \\frac{|BD| + |CD| + |BC|}{2}\n\\end{aligned}\n$$\n\nThis means that the line $EF$ is the axis of the segment $PQ$. In particular, it means that the circumcenter $O$ of triangle $APQ$ lies on the line $EF$ as well as on the axes of segments $AP$ and $AQ$. This means that\n\n$$\n\\angle DAO = \\angle OPD = \\angle CQO = \\angle OAC\n$$\n\nSo points $O$ and $F$ coincide. Moreover,\n\n$$\n\\angle AOP = 2\\angle AQP = \\angle AQC + \\angle CAQ = 180^{\\circ} - \\angle QCA = \\angle ACP\n$$\n\nTherefore the points $A$, $C$, $F$, $P$ are concyclic. Thus, necessarily $P = G$, and thus $|FP| = |FG| = |FA|$. We have proved what we need.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16884,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral. Prove that if $AC \\cdot CD > AB \\cdot BD$, then $CD \\cdot BD > AB \\cdot AC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $ABCD$ is cyclic, $\\angle ABD = \\angle ACD$ and $\\angle BAC = \\angle BDC$. For the areas of triangles $ABD$ and $ACD$, we have:\n\n$$\nS(ABD) = AB \\cdot BD \\sin \\angle ABD = S_1 + S_4,\n$$\n$$\nS(ACD) = AC \\cdot CD \\sin \\angle ACD = S_3 + S_4.\n$$\n\nBy the given condition, $AC \\cdot CD > AB \\cdot BD$, so $S(ACD) > S(ABD)$, which gives $S_3 > S_1$.\n\nSimilarly, for the areas of $BAC$ and $BDC$:\n\n$$\nS(BAC) = AB \\cdot AC \\sin \\angle BAC = S_1 + S_2,\n$$\n$$\nS(BDC) = BD \\cdot CD \\sin \\angle BDC = S_3 + S_2.\n$$\n\nSince $S_3 > S_1$, we have $S(BDC) > S(BAC)$, which gives $CD \\cdot BD > AB \\cdot AC$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16885,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the maximal value of $k$ such that, starting with $k$ heaps each containing at most 45 beads and a total of 2019 beads, it is possible to end up with $k$ heaps with pairwise distinct numbers of beads?\n\n*Note:* If $k \\geq 46$, then $45 \\cdot k > 2019$, so it is possible that at the beginning there are $k$ heaps each containing at most 45 beads. In this case, one cannot get $k$ heaps with pairwise distinct numbers of beads, since the heap with the most beads should contain at least 46 beads.",
    "options": [],
    "answer": "See solution",
    "solution": "**Lemma.** If $k$ heaps contain at least $k(k-1) + 1$ beads in total, then we can get $k$ heaps containing $1, 2, \\dots, k$ beads.\n\n*Proof by induction on* $k$.\n\nFor $k=1$, the statement is obvious. Suppose the lemma is true for $k=n$. Consider $n+1$ heaps with $n(n+1)+1$ beads in total. Then the heap with the maximal number of beads contains at least $n+1$ beads. If the heap with maximal number of beads contains exactly $n+1$ beads, the remaining $n$ heaps contain $n(n+1)+1-(n+1) = n^2 \\geq n(n-1)+1$ beads, and by the inductive hypothesis we can get $n$ heaps containing $1, 2, \\dots, n$ beads. Since we also have a heap with $n+1$ beads, we are done.\n\nIf the heap with maximal number of beads contains more than $n+1$ beads, divide it into two heaps so that one of these two new heaps, say $H(n+1)$, contains $n+1$ beads. There are $n+1$ heaps except $H(n+1)$ containing $n(n+1)+1-(n+1) = n^2$ beads in total. The smallest heap among these $n+1$ heaps contains at most $n-1$ beads. If we remove it, then the remaining $n$ heaps contain at least $n^2-(n-1) = n(n-1)+1$ beads in total. By the inductive hypothesis, we can get $n$ heaps containing $1, 2, \\dots, n$ beads. Since we also have a heap $H(n+1)$, we are done.\n\nSince $2019 > 45 \\cdot 44 + 1$, by the lemma one can get heaps containing $1, 2, \\dots, 45$ beads. **Done.**",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16886,
    "subject": "Mathematics (Olympiad)",
    "question": "The sequence $a_1, a_2, a_3, \\dots$ is defined by $a_1 = 0$ and, for $n \\ge 2$,\n\n$$\na_n = \\max_{i=1,\\dots,n-1} \\{i + a_i + a_{n-i}\\}.\n$$\n\n(For example, $a_2 = 1$ and $a_3 = 3$.)\n\nDetermine $a_{200}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove $a_n = 0 + 1 + \\cdots + (n-1)$ by strong induction.\n\nThe base case $n = 1$ is given.\n\nFor the inductive part, assume $a_n = 0 + 1 + \\cdots + (n-1)$ for $n \\leq k$.\nWe know\n\n$$\na_{k+1} = \\max_{i=1,\\dots,k} \\{i + a_i + a_{k+1-i}\\}.\n$$\n\nWe claim that the maximum occurs when $i = k$. For this, it is sufficient to prove that whenever $i < k$, we have\n\n$$\nk + a_k + a_1 > i + a_i + a_{k+1-i}.\n$$\n\nSince $a_1 = 0$ and $k > i$, it suffices to prove that\n\n$$\na_k - a_i \\geq a_{k+1-i}.\n$$\n\nUsing the inductive assumption,\n\n$$\n\\begin{aligned}\na_k - a_i &= (0 + 1 + \\cdots + (k-1)) - (0 + 1 + \\cdots + (i-1)) \\\\\n&= i + (i+1) + \\cdots + (k-1)\n\\end{aligned}\n$$\n\nand\n\n$$\n\\begin{aligned}\na_{k+1-i} &= 0 + 1 + \\cdots + (k-i) \\\\\n&= 1 + 2 + \\cdots + (k-i)\n\\end{aligned}\n$$\n\nThe right-hand sides both consist of the sum of $(k-i)$ consecutive integers, and the first term in the first sum is at least as big as the first term in the second sum. This establishes the claim.\n\nTherefore, we may substitute $i = k$ to find\n\n$$\na_{k+1} = k + 0 + 1 + \\cdots + (k-1) + 0 = 0 + 1 + \\cdots + k.\n$$\n\nThis completes the induction.\n\nFinally, the formula for the sum of the first $n$ positive integers gives\n\n$$\na_{200} = 0 + 1 + 2 + \\cdots + 199 = \\frac{199 \\times 200}{2} = 19900.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16887,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ with the following property: The numbers $1, 2, \\ldots, n$ can be split into three disjoint non-empty subsets with mutually different sizes such that, for any pair of subsets, the subset with fewer elements has a larger sum of its elements.\n\n(A size of a subset is the number of its elements.)",
    "options": [],
    "answer": "See solution",
    "solution": "We first exclude small values of $n$.\n\nThe three subsets must have at least $1 + 2 + 3 = 6$ elements in total, so $n \\ge 6$.\n\nFor $n=6$, the smallest subset (by size) contains a single number, so its sum is at most $6$. The sum of the remaining numbers is at least $15$, so the sum of at least one of the two remaining subsets is at least $8$, a contradiction. Similarly, we exclude $n=7$ and $n=8$ where, again, the smallest subset contains a single number.\n\nFor $n=9$, the subsets $\\{9, 8\\}$, $\\{7, 6, 3\\}$, $\\{5, 4, 2, 1\\}$ satisfy the conditions.\n\nFor $n=10$, the smallest subset contains at most $2$ numbers and its sum is therefore at most $19$. The sum of the remaining numbers is at least $36$, and we reach a contradiction as above. Similarly, we exclude $n=11$.\n\nNow we describe how to create the subsets for any $n \\ge 12$. Let $n = 3k + r$, $k \\ge 4$, $r \\in \\{0, 1, 2\\}$. We fix the sizes of the subsets to $k-1$, $k$, $k+r+1$ and define the subsets as follows:\n\n- Set $M_1$ consists of the $k-1$ largest numbers from $\\{1, 2, \\dots, n\\}$, that is, numbers $n-k+2, \\dots, n$.\n- Set $M_2$ consists of the $k$ next largest numbers (that is, numbers $n-2k+2, \\dots, n-k+1$).\n- The remaining numbers $1, \\dots, k+r+1$ form the subset $M_3$.\n\nWe aim to show that $M_1$ has a larger sum than $M_2$. Since all the numbers from $M_1$ are greater than all the numbers from $M_2$, and $M_2$ has just one more element, it suffices to show that the sum of the three largest numbers of $M_1$ is greater than the sum of the four smallest numbers of $M_2$. Such three numbers always exist ($k \\ge 4$).\n\nThe desired inequality is:\n$$\nn + (n-1) + (n-2) > (n-2k+2) + (n-2k+3) + (n-2k+4) + (n-2k+5)\n$$\nwhich is equivalent to $8k - n > 17$. Plugging in $n = 3k + r$ gives $5k > 17 + r$. This is true as $5k \\ge 20$ and $17 + r \\le 19$.\n\nSimilarly, we show that $M_2$ has a larger sum than $M_3$, which has $r+1$ more elements than $M_2$: The two largest numbers of $M_2$ have a larger sum than the $r+3$ smallest numbers of $M_3$ as\n$$(n-k+1) + (n-k) = 2(3k + r - k) + 1 = 4k + 2r + 1 \\ge 17 > 15 = 1 + 2 + 3 + 4 + 5.$$\n\nThe answer is all $n \\ge 12$ together with $n = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16888,
    "subject": "Mathematics (Olympiad)",
    "question": "Design a finite, non-empty set $S$ of points in the plane such that:\n\n1. Every line meets $S$ in at most four points.\n2. For every 2-colouring of $S$ (i.e., each point of $S$ is coloured with one of two colours), there exist at least three collinear points of $S$ that are all the same colour.",
    "options": [],
    "answer": "See solution",
    "solution": "Call a planar set a *partial n-point set* if it meets every line in at most $n$ points. We are required to design a finite partial 4-point set that cannot be split into two partial 2-sets.\n\nWe shall prove that the set\n$$\nS = \\{\\pm1, \\pm3\\} \\times \\{\\pm1, \\pm3\\} \\cup \\{\\pm2, \\pm4\\} \\times \\{0\\} \\cup \\{0\\} \\times \\{\\pm4\\}\n$$\nis one such.\n\nInspection of $S$ shows that it is indeed a partial 4-point set. Suppose, if possible, that $S$ can be partitioned into two partial 2-sets: one red, and the other blue.\n\nThere are exactly three ways to split the row $\\{\\pm1, \\pm3\\} \\times \\{-1\\}$ after accounting for symmetries:\n\n1. $(\\pm3, -1)$ are red, and $(\\pm1, -1)$ are blue;\n2. $(-3, -1)$ and $(1, -1)$ are red, and $(-1, -1)$ and $(3, -1)$ are blue; and\n3. $(-3, -1)$ and $(-1, -1)$ are red, and $(1, -1)$ and $(3, -1)$ are blue.\n\nIn cases 2 and 3 we may (and will) assume that the point $(0, -4)$ is coloured red.\n\nNotice that if two points in $S$ share one colour, then the other points of $S$ on that line must share the other.\n\nIn case 1, consider the two possibilities for the point $(-3, -3)$.\n\nIf $(-3, -3)$ is red, then $(-3, 1)$ and $(-3, 3)$ must be blue, for they both lie on the vertical through the reds $(-3, -3)$ and $(-3, -1)$. Therefore, $(1, 3)$ and $(3, -3)$ must be red, for the former is on the line through the blues $(-3, 1)$ and $(-1, -1)$, and the latter is on the line through the blues $(-3, 3)$ and $(1, -1)$. Consequently, $\\{-3, 1, 3\\} \\times \\{-3\\}$ is red — a contradiction.\n\nIf $(-3, -3)$ is blue, then $(1, 1)$ and $(3, 3)$ must be red, for they both lie on the line through the blues $(-3, 3)$ and $(-1, -1)$. Therefore, $(2, 0)$ and $(3, 1)$ must be blue, for the former is on the line through the reds $(1, 1)$ and $(3, -1)$, and the latter is on the vertical through the reds $(3, -1)$ and $(3, 3)$. Consequently, $(1, -1)$, $(2, 0)$ and $(3, 1)$ are three collinear blues—a contradiction. This establishes case 1.\n\nIn case 2, recall that $(0, -4)$ is red. The points $(-4, 0)$ and $(-1, -3)$ lie on the line through the reds $(-3, -1)$ and $(0, -4)$, so they must be blue. The point $(-1, 1)$ lies on the vertical through the blues $(-1, -3)$ and $(-1, -1)$, so it must be red. The line through the reds $(-3, -1)$ and $(-1, 1)$ contains $(-2, 0)$, so the latter must be blue. The point $(1, -3)$ lies on the line through the blues $(-2, 0)$ and $(-1, 1)$, so it must be red. Finally, the line through the reds $(0, -4)$ and $(1, -3)$ contains $(4, 0)$ which must therefore be blue. Consequently, $\\{-4, -2, 4\\} \\times \\{0\\}$ is blue—a contradiction which establishes case 2.\n\nIn case 3, recall that $(0, -4)$ is red and consider the two possible choices for the point $(1, -3)$.\n\nIf $(1, -3)$ is red, then $(-2, 0)$ must be blue, for it is on the line through the reds $(1, -3)$ and $(-1, -1)$. The points $(\\pm4, 0)$ both lie on lines through pairs of reds, so they must be blue: $(-4, 0)$ is on the line through the reds $(0, -4)$ and $(-3, -1)$, and $(4, 0)$ is on the line through the reds $(0, -4)$ and $(1, -3)$. Consequently, $\\{-4, -2, 4\\} \\times \\{0\\}$ is blue—a contradiction.\n\nIf $(1, -3)$ is blue, then $(1, 1)$ must be red, for it lies on the vertical through the blues $(1, -3)$ and $(1, -1)$. The line through the reds $(0, -4)$ and $(-3, -1)$ contains $(-1, -3)$ which must therefore be blue. Finally, $(-3, -3)$ lies on the line through the blues $(1, -3)$ and $(-1, -3)$, so it must be red. Consequently, $(-3, -3)$, $(-1, -1)$ and $(1, 1)$ are three collinear reds—a contradiction. This establishes case 3 and concludes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16889,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle with $AB < AC$, which is inscribed in the circle $\\omega_1$ centered at $O$. Denote by $H$ the orthocenter of $\\triangle ABC$ and by $M$ the midpoint of side $BC$. Let $\\omega_2$ be the circumcircle of triangle $BHC$, let the line $HO$ meet $\\omega_2$ at $K \\neq H$, and let the line $MK$ meet $\\omega_2$ at $P \\neq K$. Prove that the tangent line to $\\omega_1$ at $A$ and the tangent line to $\\omega_2$ at $P$ meet on the circumcircle of triangle $APK$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $D$ be the point such that $ABDC$ is a parallelogram and $O_A$ be the center of $\\omega_2$. Let $AM$ meet $\\omega_1$ at $E \\neq A$, and $AO$ meet $\\omega_1$ at $X \\neq A$. Furthermore, let $L$ be the reflection of $K$ across $M$, and $P'$ be the reflection of $P$ across $M$.\n\n$LBKC$ is a parallelogram, so $\\angle BLC = \\angle BKC = 180^\\circ - \\angle BHC = \\angle BAC$, which means $L$ lies on $\\omega_1$. We similarly obtain that $D$ lies on $\\omega_2$ and $P'$ lies on $\\omega_1$. $ALDK$ is also a parallelogram, so $\\angle ALK + \\angle LKH = \\angle LKD + \\angle PKH = \\angle PKD + \\angle PKH = \\angle HKD = \\angle HCD = 90^\\circ$, i.e. $HK \\perp AL$. Since $HK$ passes through $O$, it is the perpendicular bisector of $AL$.\n\nFrom the power of point $M$ we deduce $MA \\cdot ME = MB \\cdot MC = MP \\cdot MK$, which means $APEK$ is cyclic.\n\n![](images/2025-SL-b_p9_data_4df7712dbd.png)\n\nIt is well-known that $AHO_AO$ is a parallelogram, so $HO_A = OA = OL$ and $OO_A = AH = LH$. Therefore, $LHOO_A$ is an isosceles trapezoid, so\n\n$$\nLO_A \\parallel OH \\Rightarrow LO_A \\perp AL \\Rightarrow \\angle ALO_A = 90^\\circ = \\angle ALX,\n$$\n\nwhich means that $O_A$ lies on $LX$. We also have $\\angle O_ALE = \\angle XLE = \\angle XAE = \\angle XAD = \\angle EDO_A$, where the last equality holds because $AX \\parallel DO_A$. Hence $DEO_A L$ is cyclic. Next we have $\\angle BLP = \\angle BLP' = \\angle BCP' = \\angle PBC = \\angle PBM$, so $BM$ is tangent to $(BLP)$. From here we obtain $ME \\cdot MD = ME \\cdot MA = MB \\cdot MC = MB^2 = MP \\cdot ML$, so $DEPL$ is also cyclic. This implies that points $L, P, O_A, E, D$ all lie on a circle, and so $\\angle O_APE = \\angle O_ALE = \\angle XLE = \\angle XAE = \\angle OAE$.\n\nNow let the tangent line to $\\omega_1$ at $A$ and the tangent line to $\\omega_2$ meet at $T$. Then we have\n\n$$\n\\angle TAE = \\angle TAO + \\angle OAE = 90^\\circ + \\angle O_APE = \\angle O_APT + \\angle O_APE = \\angle EPT,\n$$\n\nand so $TAPE$ is cyclic. Therefore, points $A, P, E, K, T$ are concyclic, meaning that $T$ lies on $(APK)$, as needed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16890,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $n \\geq 2$ which have a positive divisor $m \\mid n$ satisfying\n$$n = d^3 + m^3,$$\nwhere $d$ is the smallest divisor of $n$ which is greater than $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "The smallest divisor of $n$ greater than $1$ is the smallest prime divisor of $n$, so $d$ is prime. Since $d \\mid n$, we have $d \\mid d^3 + m^3$, and $d \\mid m^3$, which implies $m > 1$. Also, $m \\mid n$, so $m \\mid d^3 + m^3$, and $m \\mid d^3$. Because $d$ is prime and $m > 1$, $m$ must be $d$, $d^2$, or $d^3$.\n\nIn all cases, the parity of $m^3$ matches that of $d^3$, so $n = d^3 + m^3$ is even. Thus, the smallest divisor of $n$ greater than $1$ is $2$, i.e., $d = 2$. For $m = d$, $n = 2^3 + 2^3 = 16$; for $m = d^2$, $n = 2^3 + 2^6 = 72$; for $m = d^3$, $n = 2^3 + 2^9 = 520$. These are solutions: $2 \\mid 16$, $4 \\mid 72$, and $8 \\mid 520$, so $m \\mid n$ in each case.\n\n![](images/NLD_ABooklet_2022_p25_data_e59db68fea.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16891,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer such that there exists a positive integer less than $\\sqrt{n}$ that does not divide $n$. Let $(a_1, \\dots, a_n)$ be an arbitrary permutation of $1, \\dots, n$. Let $a_{i_1} < \\dots < a_{i_k}$ be its maximal increasing subsequence and let $a_{j_1} > \\dots > a_{j_l}$ be its maximal decreasing subsequence. Prove that the tuples $(a_{i_1}, \\dots, a_{i_k})$ and $(a_{j_1}, \\dots, a_{j_l})$ together contain at least one number that does not divide $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show that $kl \\ge n$. For every $i = 1, \\dots, n$, let $f(i)$ denote the length of the longest increasing subsequence ending with $a_i$, and let $g(i)$ be the length of the longest decreasing subsequence ending with $a_i$. For distinct indices $i < j$, if $a_i < a_j$ then $f(i) < f(j)$, and if $a_i > a_j$ then $g(i) < g(j)$. Hence, the pairs $(f(i), g(i))$ for $i = 1, \\dots, n$ are all distinct, so there are $n$ different such pairs in total. The largest value of $f(i)$ is $k$ and of $g(i)$ is $l$, so the number of possible pairs is at most $kl$. Thus, $n \\le kl$.\n\nBy the AM-GM inequality, $k + l \\ge 2\\sqrt{kl} \\ge 2\\sqrt{n}$. At most one number can belong to both the maximal increasing and decreasing subsequences. Therefore, the two subsequences together contain at least $2\\sqrt{n} - 1$ distinct numbers. By assumption, $n$ has at most $\\lfloor \\sqrt{n} \\rfloor - 1$ divisors not exceeding $\\sqrt{n}$, and the total number $\\delta(n)$ of divisors of $n$ satisfies $\\delta(n) \\le 2\\lfloor \\sqrt{n} \\rfloor - 2$. Consequently, the two subsequences together contain at least one number that does not divide $n$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 16892,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be integers of different parity. Prove that $(a + 3b)(5a + 7b)$ is not a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider $(a + 3b)(5a + 7b)$ modulo $8$:\n\n$$\n(a + 3b)(5a + 7b) \\equiv 5a^2 + 22ab + 21b^2 \\equiv 5a^2 - 10ab + 5b^2 \\equiv 5(a - b)^2 \\pmod{8}.\n$$\n\nSuppose $(a + 3b)(5a + 7b) = k^2$ for some positive integer $k$. Then\n\n$$\nk^2 \\equiv 5(a - b)^2 \\pmod{8}.\n$$\n\nSince $a$ and $b$ have different parity, $a - b$ is odd, so $(a - b)^2 \\equiv 1 \\pmod{8}$. Thus, $k^2 \\equiv 5 \\pmod{8}$. However, $5$ is not a quadratic residue modulo $8$. This is a contradiction, so $(a + 3b)(5a + 7b)$ cannot be a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16893,
    "subject": "Mathematics (Olympiad)",
    "question": "Seven students are to be assigned to attend five sporting events. The conditions are:\n\n- Students A and B cannot attend the same event.\n- Every event is attended by at least one student.\n- Each student attends exactly one event.\n\nHow many possible arrangement plans satisfy these conditions? (Give your answer as a numerical value.)",
    "options": [],
    "answer": "See solution",
    "solution": "There are two possible cases:\n\n1. One event is attended by three students. The number of plans is:\n   $$C_7^3 \\cdot 5! - C_5^1 \\cdot 5! = 3600$$\n\n2. Two events are each attended by two students. The number of plans is:\n   $$\\frac{1}{2} (C_7^2 \\cdot C_5^2) \\cdot 5! - C_5^2 \\cdot 5! = 11400$$\n\nTherefore, the total number of arrangement plans is:\n$$3600 + 11400 = 15000$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16894,
    "subject": "Mathematics (Olympiad)",
    "question": "A rounded tower has 12 doors, each with a chest of Captain Flint's gold behind it. The doors are equally spaced and numbered clockwise from 1 to 12. Twelve pirates arrive, each with a key numbered from 1 to 12. Key number $n$ opens door $m$ if and only if $m \\neq n$. The pirates stand, one at each door, but do not know the number of the door they are near. Jim Hockins knows which pirate has which key and wants them to take as few chests as possible. Jim can rotate the tower so that the doors are positioned in front of the pirates as he wishes, but the door numbers remain clockwise 1–12 starting from some position. What is the maximum number of chests the pirates can definitely take under these conditions?",
    "options": [],
    "answer": "See solution",
    "solution": "It is clear that key 1 opens any door except door 1. Let's show that Jim can always arrange things so that no more than two doors are opened. Consider a $12 \\times 12$ table, where each row represents a possible configuration of the doors.\n\nFor a given configuration of keys, each column corresponds to a key. Shade all cells in a column that are opened by the corresponding key. For example, in the column for key 2, 6 cells are shaded; for key 5, 2 cells are shaded. In total:\n\n$$\n12 + 6 + 4 + 3 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 = 35 \\text{ cells are shaded.}\n$$\n\nSince there are 12 rows, by the pigeonhole principle, there is at least one row with no more than two shaded cells. Jim can rotate the tower to match this row, ensuring that at most 2 doors are opened.\n\nIf the pirates stand with keys as shown in the last row of the figure, they can always open at least two locks.\n\n![](images/Ukraine_booklet_2018_p42_data_a425aac24d.png)\n![](images/Ukraine_booklet_2018_p42_data_f8bf3985e1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16895,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle with altitudes $AD$ and $BE$ meeting at $H$. Let $M$ be the midpoint of segment $AB$, and suppose that the circumcircles of $\\triangle DEM$ and $\\triangle ABH$ meet at points $P$ and $Q$ with $P$ on the same side of $CH$ as $A$. Prove that the lines $ED$, $PH$, and $MQ$ all pass through a single point on the circumcircle of $\\triangle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/Kanada_2016_p9_data_247a0e10c3.png)\n\nLet $R$ denote the intersection of lines $ED$ and $PH$. Since quadrilaterals $ECDH$ and $APHB$ are cyclic, we have $\\angle RDA = 180^\\circ - \\angle EDA = 180^\\circ - \\angle EDH = 180^\\circ - \\angle ECH = 90^\\circ + A$, and $\\angle RPA = \\angle HPA = 180^\\circ - \\angle HBA = 90^\\circ + A$. Therefore, $APDR$ is cyclic. This in turn implies that $\\angle PBE = \\angle PBH = \\angle PAH = \\angle PAD = \\angle PRD = \\angle PRE$, and so $PBRE$ is also cyclic.\n\nLet $F$ denote the base of the altitude from $C$ to $AB$. Then $D, E, F$, and $M$ all lie on the 9-point circle of $\\triangle ABC$, and so are cyclic. We also know $APDR$, $PBRE$, $BCEF$, and $ACDF$ are cyclic, which implies $\\angle ARB = \\angle PRB - \\angle PRA = \\angle PEB - \\angle PDA = \\angle PEF + \\angle FEB - \\angle PDF + \\angle ADF = \\angle FEB + \\angle ADF = \\angle FCB + \\angle ACF = C$. Therefore, $R$ lies on the circumcircle of $\\triangle ABC$.\n\nNow let $Q'$ and $R'$ denote the intersections of line $MQ$ with the circumcircle of $\\triangle ABC$, chosen so that $Q', M, Q, R'$ lie on the line in that order. We will show that $R' = R$, which will complete the proof. However, first note that the circumcircle of $\\triangle ABC$ has radius $\\frac{AB}{2\\sin C}$, and the circumcircle of $\\triangle ABH$ has radius $\\frac{AB}{2\\sin\\angle AHB} = \\frac{AB}{2\\sin(180^\\circ - C)}$. Thus the two circles have equal radius, and so they must be symmetrical about the point $M$. In particular, $MQ = MQ'$.\n\nSince $\\angle AEB = \\angle ADB = 90^\\circ$, we furthermore know that $M$ is the circumcenter of both $\\triangle AEB$ and $\\triangle ADB$. Thus, $MA = ME = MD = MB$. By Power of a Point, we then have $MQ \\cdot MR' = MQ' \\cdot MR' = MA \\cdot MB = MD^2$. In particular, this means that the circumcircle of $\\triangle DR'Q$ is tangent to $MD$ at $D$, which means $\\angle MR'D = \\angle MDQ$.\n\nSimilarly $MQ \\cdot MR' = ME^2$, and so $\\angle MR'E = \\angle MEQ = \\angle MDQ = \\angle MR'D$. Therefore, $R'$ also lies on the line $ED$.\n\nFinally, the same argument shows that $MP$ also intersects the circumcircle of $\\triangle ABC$ at a point $R''$ on line $ED$. Thus, $R, R'$, and $R''$ are all chosen from the intersection of the circumcircle of $\\triangle ABC$ and the line $ED$. In particular, two of $R, R'$, and $R''$ must be equal. However, $R'' \\neq R$ since $MP$ and $PH$ already intersect at $P$, and $R'' \\neq R'$ since $MP$ and $MQ$ already intersect at $M$. Thus, $R' = R$, and the proof is complete. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16896,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a $32 \\times 32$ table, we put a mouse (facing up) at the bottom left cell and a piece of cheese at several other cells. The mouse then starts moving. It moves forward except that when it reaches a piece of cheese, it eats a part of it, turns right, and continues moving forward. We say that a subset of cells containing cheese is *good* if, during this process, the mouse tastes each piece of cheese exactly once and then falls off the table. Show that:\n\n(a) No good subset consists of 888 cells.\n(b) There exists a good subset consisting of at least 666 cells.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) For the sake of contradiction, assume a good subset consisting of 888 cells exists. We call those *cheese-cells* and the other ones *gap-cells*. Observe that since each cheese-cell is visited once, each gap-cell is visited at most twice (once vertically and once horizontally). Define a finite sequence $s$ whose $i$-th element is $C$ if the $i$-th step of the mouse was onto a cheese-cell, and $G$ if it was onto a gap-cell. By assumption, $s$ contains 888 $C$'s. Note that $s$ does not contain a contiguous block of 4 (or more) $C$'s. Hence $s$ contains at least $888/3 = 296$ such $C$-blocks and thus at least 295 $G$'s. But since each gap-cell is traversed at most twice, this implies there are at least $\\lceil 295/2 \\rceil = 148$ gap-cells, for a total of $888 + 148 = 1036 > 32^2$ cells, a contradiction.\n\n(b) Let $L_i$, $X_i$ be two $2^i \\times 2^i$ tiles that allow the mouse to “turn left” and “cross”, respectively. In detail, the “turn left” tiles allow the mouse to enter at its bottom left cell facing up and to leave at its bottom left cell facing left. The “cross” tiles allow the mouse to enter at its top right facing down and leave at its bottom left facing left, while also to enter at its bottom left facing up and leave at its top right facing right.\n\n(a) Basic tiles\n\n![](images/2021_Australian_Scene_p91_data_6a1eede883.png)\n\n![](images/2021_Australian_Scene_p91_data_094a7bb8ec.png)\n\n(b) Inductive construction\n\n$L_{i+1}$\n\n![](images/2021_Australian_Scene_p91_data_adb3fa5698.png)\n\n$X_{i+1}$\n\n![](images/2021_Australian_Scene_p91_data_fd72d4b55e.png)\n\n(c) $16 \\times 16$\n\n![](images/2021_Australian_Scene_p91_data_ae9d94a3ff.png)\n\nNote that given two $2^i \\times 2^i$ tiles $L_i, X_i$ we can construct larger $2^{i+1} \\times 2^{i+1}$ tiles $L_{i+1}, X_{i+1}$ inductively as shown in (b). The construction works because the path intersects itself (or the other path) only inside the smaller $X$-tiles where it works by induction.\n\nFor a tile $T$, let $|T|$ be the number of pieces of cheese in it. By straightforward induction, $|L_i| = |X_i| + 1$ and $|L_{i+1}| = 4 \\cdot |L_i| - 1$. From the initial condition $|L_1| = 3$. We now easily compute $|L_2| = 11$, $|L_3| = 43$, $|L_4| = 171$, and $|L_5| = 683$. Hence we get the desired subset.\n\n**Another proof of (a).**\n\nLet $X_N$ be the largest possible density of cheese-cells in a good subset on an $N \\times N$ table. We will show that $X_N \\leq 4/5 + o(1)$. Specifically, this gives $X_{32} \\leq 817/1024$. We look at the (discrete analogue) of the winding number of the trajectory of the mouse. Since the mouse enters and leaves the table, for every 4 right turns in its trajectory there has to be a self-crossing. But each self-crossing requires a different empty square, hence $X_N \\leq 4/5$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16897,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $d(n)$ denote the number of positive divisors of $n$. For a positive integer $n$, define $f(n)$ as\n\n$$\nf(n) = d(k_1) + d(k_2) + d(k_3) + \\dots + d(k_m),\n$$\n\nwhere $1 = k_1 < k_2 < \\dots < k_m = n$ are all divisors of $n$. We call an integer $n > 1$ *almost perfect* if $f(n) = n$. Find all almost perfect numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "An alternative way to define $f(n)$ is\n\n$$\nf(n) = \\sum_{k \\mid n,\\ k \\ge 1} d(k).\n$$\n\nLet $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_r^{\\alpha_r}$ be the prime factorization of $n$. We have $d(n) = \\prod_{i=1}^r (\\alpha_i + 1)$.\n\nWe prove that $f$ is multiplicative: for coprime $n, m$, $f(mn) = f(m)f(n)$.\n\nUsing the multiplicativity of $d$ and the fact that $n, m$ are coprime:\n\n$$\nf(mn) = \\sum_{k_1 \\mid n} \\sum_{k_2 \\mid m} d(k_1 k_2) = \\sum_{k_1 \\mid n} \\sum_{k_2 \\mid m} d(k_1) d(k_2) = \\left(\\sum_{k_1 \\mid n} d(k_1)\\right) \\left(\\sum_{k_2 \\mid m} d(k_2)\\right) = f(n) f(m).\n$$\n\nIf $r = 1$, so $n = p_1^{\\alpha_1}$, the divisors of $n$ are $1, p_1, p_1^2, \\dots, p_1^{\\alpha_1}$, so\n\n$$\nf(n) = \\sum_{i=0}^{\\alpha_1} (i+1) = \\frac{(\\alpha_1 + 1)(\\alpha_1 + 2)}{2}.\n$$\n\nCombining this with multiplicativity, $f(n) = \\prod_{i=1}^r \\frac{(\\alpha_i + 1)(\\alpha_i + 2)}{2}$.\n\nFor primes $p \\ge 5$ and $p = 3$ with $a \\ge 3$, we have $f(p^a) = \\frac{(a+1)(a+2)}{2} < \\frac{2}{3} p^a$ (by induction). For $p = 2$, $f(2^a) < 2^a$ for $a \\ge 4$. Checking small cases, $f(p^a) \\leq p^a$ for all $p \\ge 3$ and $p = 2, a \\ge 4$.\n\nAssuming $f(n) = n$, we get $\\prod_{i=1}^k \\frac{f(p_i^{\\alpha_i})}{p_i^{\\alpha_i}} = 1$, so only possible prime divisors are $2$ and $3$.\n\nIf $k = 1$, the only possible solution is $n = 3$. If $k = 2$, $p_1 = 2$, $p_2 = 3$, $1 \\leq a_1 \\leq 2$, $1 \\leq a_2 \\leq 2$, giving four cases to check, yielding the other two solutions: $n = 18, 36$.\n\n**Thus, all almost perfect numbers are $3$, $18$, and $36$.**",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16898,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the maximum real number $\\lambda$ that satisfies the following conditions: for any positive real numbers $p, q, r, s$, there exists a complex number $z = a + bi$ ($a, b \\in \\mathbb{R}$) such that $|b| \\geq \\lambda|a|$ and\n\n$$\n(pz^3 + 2qz^2 + 2rz + s)(qz^3 + 2pz^2 + 2sz + r) = 0.\n$$\n",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove that the maximum value of $\\lambda$ satisfying the conditions is $\\sqrt{3}$.\n\nAssume $\\lambda > \\sqrt{3}$ and take $p = q = r = s = 1$. The equation becomes $(z^3 + 2z^2 + 2z + 1)^2 = 0$, i.e.,\n\n$$\n(z+1)^2 \\left( z + \\frac{1}{2} + \\frac{\\sqrt{3}}{2}i \\right) \\left( z + \\frac{1}{2} - \\frac{\\sqrt{3}}{2}i \\right) = 0,\n$$\n\nand each complex root $z$ satisfies $|\\mathrm{Im}\\,z| \\le \\sqrt{3}|\\mathrm{Re}\\,z| < \\lambda|\\mathrm{Re}\\,z|$, which is inconsistent with the question. Therefore, $\\lambda \\le \\sqrt{3}$.\n\nNow we prove that $\\lambda = \\sqrt{3}$ satisfies the conditions. The following uses proof by contradiction.\n\n**Method 1**\n\nAssume otherwise, that each complex root $z$ of the following two equations\n\n$$\npz^3 + 2qz^2 + 2rz + s = 0\n$$\n\n$$\nqz^3 + 2pz^2 + 2sz + r = 0\n$$\n\nsatisfies $|\\mathrm{Im}\\,z| < \\sqrt{3}|\\mathrm{Re}\\,z|$.\n\nConsider the first equation. We prove $qr > ps$.\n\nEquation (1) is a real-coefficient cubic in $z$, so it must have real roots. Since $p, q, r, s > 0$, the real roots must be negative. There are two cases:\n\n(i) Three negative roots. Write\n$$\npz^3 + 2qz^2 + 2rz + s = p(z + u_1)(z + u_2)(z + u_3),\n$$\nwhere $u_1, u_2, u_3 > 0$. Comparing coefficients:\n$$\n2q = p(u_1 + u_2 + u_3), \\quad 2r = p(u_1u_2 + u_2u_3 + u_1u_3), \\quad s = pu_1u_2u_3.\n$$\nSo,\n$$\n\\begin{align*}\nqr &= \\frac{p^2}{4}(u_1 + u_2 + u_3)(u_1u_2 + u_2u_3 + u_1u_3) \\\\\n&\\ge \\frac{p^2}{4} \\cdot 3\\sqrt{u_1u_2u_3} \\cdot 3\\sqrt{u_1u_2 \\cdot u_2u_3 \\cdot u_1u_3} \\\\\n&= \\frac{9}{4}p^2 u_1 u_2 u_3 = \\frac{9}{4}ps > ps.\n\\end{align*}\n$$\n\n(ii) One negative root and a pair of conjugate imaginary roots. Let\n$$\n\\begin{aligned}\npz^3 + 2qz^2 + 2rz + s &= p(z+u)(z-(v+kvi))(z-(v-kvi)) \\\\\n&= p(z+u)(z^2 - 2vz + (1+k^2)v^2),\n\\end{aligned}\n$$\nwhere $u > 0, v \\neq 0, 0 < |k| < \\sqrt{3}$.\n\nLet $m = 1+k^2$, so $0 < m < 4$. Comparing coefficients:\n$$\n2q = p(u - 2v), \\quad 2r = p(-2uv + mv^2), \\quad s = pmuv^2.\n$$\nSo,\n$$\n0 < \\frac{2r}{p} = -2uv + mv^2 < -2uv + 4v^2 - 2v(u - 2v) = -2v \\cdot \\frac{2q}{p},\n$$\nwhich implies $v < 0$. Also,\n$$\n\\frac{4}{p^2}(qr - ps) = \\frac{2q}{p} \\cdot \\frac{2r}{p} - 4 \\cdot \\frac{s}{p} = (u - 2v)(-2uv + mv^2) - 4muv^2.\n$$\nThe right side is a linear function $g(m)$ of $m$. Since $u > 0, v < 0$,\n$$\ng(0) = (u - 2v)(-2uv) > 0, \\quad g(4) = -2v(u + 2v)^2 \\geq 0.\n$$\nTherefore, for $0 < m < 4$, $qr > ps$.\n\nIn both cases, $qr > ps$.\n\nFor the second equation, the same reasoning gives $ps > qr$, a contradiction. Therefore, $\\lambda = \\sqrt{3}$ satisfies the condition.\n\nIn conclusion, the desired maximum value of $\\lambda$ is $\\sqrt{3}$.\n\n**Method 2**\n\nAssume $\\lambda = \\sqrt{3}$ does not satisfy the condition, so that the equations have no roots in region $A_1 := \\{z \\in \\mathbb{C} \\mid \\frac{\\pi}{3} \\le \\arg z \\le \\frac{2\\pi}{3} \\text{ or } \\frac{4\\pi}{3} \\le \\arg z \\le \\frac{5\\pi}{3}\\}$. We first prove that the equations also have no roots in region $A_2 := \\{z \\in \\mathbb{C} \\mid 0 \\le \\arg z \\le \\frac{\\pi}{3} \\text{ or } \\frac{5\\pi}{3} \\le \\arg z \\le 2\\pi\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16899,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral. Let $K_1$ be the circle that passes through $D$ and is tangent to $AB$ at $A$, and let $K_2$ be the circle that passes through $D$ and is tangent to $BC$ at $C$. Let $P$ be the point other than $D$ in which $K_1$ and $K_2$ intersect.\n\nProve that $P$ lies on the line through $A$ and $C$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $Q$ be the point on the circumcircle of $ABCD$ such that $QB$ and $CA$ are parallel.\n\n![](images/Australian_Scene_2012_-_AMT_Publishing_-_273p_p101_data_fa9bf1612c.png)\n\nLet $DQ$ intersect $CA$ in $P^*$. We are going to show that $P = P^*$. Now,\n\n$$\n\\begin{align*}\n\\angle CDQ &= \\angle CBQ \\quad (\\text{since both angles are subtended by } CQ) \\\\\n           &= \\angle BCP^* \\quad (\\text{since } QB \\text{ and } CA \\text{ are parallel}).\n\\end{align*}\n$$\n\nHence, by the Alternate Segment (or tangent-chord) Theorem, the circle through $C$, $D$ and $P^*$ is tangent to $BC$ at $C$ and must therefore be $K_2$. Similarly,\n\n$$\n\\begin{align*}\n\\angle ADQ &= 180^\\circ - \\angle ABQ \\quad (\\text{cyclic quadrilateral}) \\\\\n           &= \\angle BAP^* \\quad (\\text{since } QB \\text{ and } CA \\text{ are parallel}),\n\\end{align*}\n$$\n\nso the circle through $A$, $D$ and $P^*$ is tangent to $AB$ at $A$ and must therefore be $K_1$. It follows that $K_1$ and $K_2$ intersect in $P^*$, so that $P = P^*$. Therefore $P$ lies on the line through $A$ and $C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16900,
    "subject": "Mathematics (Olympiad)",
    "question": "The book that Marko reads has 242 pages. The first day he reads 22 pages. On the second day, he reads 4 pages more than the first day, and the same on the third day. How many days will Marko need to finish the book if, from the fourth day onward, he reads two pages more than the third day each day?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x$ be the number of days from the fourth day onward that Marko needs to finish the book. The total pages read is:\n\n$$\n22 + 2 \\cdot (22 + 4) + x \\cdot (22 + 4 + 2) = 242\n$$\n\nThat is,\n\n$$\n22 + 2 \\cdot 26 + x \\cdot 28 = 242\n$$\n\n$$\n22 + 52 + 28x = 242\n$$\n\n$$\n74 + 28x = 242\n$$\n\n$$\n28x = 168\n$$\n\n$$\nx = 6\n$$\n\nSo Marko will read the book for $3 + x = 3 + 6 = 9$ days.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16901,
    "subject": "Mathematics (Olympiad)",
    "question": "對於正整數 $n$ 及 $k \\ge 2$，定義 $E_k(n)$ 為能讓 $k^r$ 整除 $n!$ 的最大的指數 $r$。證明有無窮多個 $n$ 使得 $E_{10}(n) > E_9(n)$，也有無窮多個 $m$ 使得 $E_{10}(m) < E_9(m)$。",
    "options": [],
    "answer": "See solution",
    "solution": "令 $n = 5^{2l-1}$，則\n\n$$\nE_{10}(n) = v_5(n!) = 5^{2l-2} + 5^{2l-3} + \\dots + 5 + 1 = \\frac{5^{2l-1}-1}{4} = \\frac{n-1}{4}.\n$$\n\n由於 $n \\equiv 2 \\pmod{4}$，所以\n\n$$\nE_9(n) = \\frac{1}{2}v_3(n!) < \\frac{1}{2}\\left(\\frac{n-2}{3} + \\frac{n}{3^2} + \\dots\\right) = \\frac{n}{4} - \\frac{1}{3}.\n$$\n\n因此 $E_9(n) < E_{10}(n)$。\n\n同理，令 $m = 3^{4l-2}$，則\n\n$$\nE_9(m) = \\frac{1}{2}v_3(m!) = \\frac{m-1}{4}\n$$\n\n且 $m \\equiv 4 \\pmod{5}$，\n\n$$\nE_{10}(m) < \\frac{m-4}{5} + \\frac{m}{5^2} + \\dots = \\frac{m}{4} - \\frac{4}{5}.\n$$\n\n因此 $E_{10}(m) < E_9(m)$。\n\n*Remark*: 以類似方法考慮模 $5^b$ 和 $3^b$ 的餘數，可更精確計算更多項，並可證明對任意 $B > 0$，存在無窮多整數 $m, n$ 使得 $E_{10}(n) - E_9(n) > B$ 及 $E_9(m) - E_{10}(m) > B$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16902,
    "subject": "Mathematics (Olympiad)",
    "question": "One hundred balls labelled 1 to 100 are to be put into two identical boxes so that each box contains at least one ball and the greatest common divisor of the product of the labels of all the balls in one box and the product of the labels of all the balls in the other box is 1. Determine the number of ways that this can be done.",
    "options": [],
    "answer": "See solution",
    "solution": "First, assume the two boxes are different. The balls labelled 1 and 2 can be put into either box. All balls with even labels must then be put into the same box as the ball labelled 2. Thus, any ball with a label that shares a common factor greater than 1 with any of those even labels must also be put into the same box. Therefore, we only need to consider those balls with labels greater than 50. That leaves us with 10 balls labelled by the primes: 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. Thus, we have twelve entities: the ten prime numbers, the number 1, and all the balls that must go with 2. Each can be put into either of the two boxes. Taking away the 2 cases where one of the boxes is empty, the number of ways is $$2^{12} - 2 = 4094.$$ Since the two boxes are identical, we need to divide this number by 2. Therefore, the answer is $$2047.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16903,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 4$ points be given in the plane, with no three of them collinear. Prove that the number of parallelograms of area $1$ formed by these points is at most $$\\frac{n^2 - 3n}{4}.$$",
    "options": [],
    "answer": "See solution",
    "solution": "Fix a direction in the plane. Since no three points are collinear, in any given direction, there are at most $k$ pairs of points, each pair lying on a line parallel to the fixed direction. For these $k$ pairs, at most $k-1$ parallelograms of area $1$ can be formed.\n\nSumming over all directions, the total number of parallelograms of area $1$ is at most $\\binom{n}{2} - s$, where $s$ is the number of different directions. However, each parallelogram is counted twice, so the actual number is at most $$\\frac{\\binom{n}{2} - s}{2}.$$ \n\nWe claim that $s \\ge n$. To see this, consider the convex hull of the $n$ points. Let $x$ be a point on the boundary of the convex hull. Since the convex hull has at least three boundary points, select two neighbors of $x$ on the hull, say $y$ and $z$. Then every segment from $x$ has a different direction from $yz$. Thus, there are at least $n$ different directions.\n\nTherefore, the number of parallelograms is at most\n$$\n\\frac{\\binom{n}{2} - n}{2} = \\frac{n^2 - 3n}{4}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16904,
    "subject": "Mathematics (Olympiad)",
    "question": "De un prisma recto de base cuadrada, con lado de longitud $L_1$ y altura $H$, se extrae un tronco de pirámide (no necesariamente recto) de bases cuadradas, con lados de longitud $L_1$ (inferior) y $L_2$ (superior), y altura $H$. Las dos piezas obtenidas aparecen en la imagen siguiente.\n\n![](images/Spanija_b_2014_p19_data_7cf6730e03.png)\n\nSi el volumen del tronco de pirámide es $\\frac{2}{3}$ del volumen total del prisma, ¿cuál es el valor de $\\frac{L_1}{L_2}$?\n\n![](images/Spanija_b_2014_p19_data_42808876bb.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Si prolongamos una altura $h$ en el tronco de pirámide hasta obtener una pirámide completa de altura $H + h$, tendrá una sección como la mostrada en la figura anterior.\n\nUn argumento de semejanza de triángulos permite comprobar que\n\n$$\n\\frac{h+H}{L_1} = \\frac{h}{L_2}\n$$\n\ny, por tanto,\n\n$$\nh = \\frac{H L_2}{L_1 - L_2}.\n$$\n\nAdemás, podemos observar que\n\n$$\n\\begin{aligned}\n\\text{Volumen del tronco de pirámide} &= \\frac{1}{3}(L_1^2(H+h) - L_2^2 h) \\\\\n&= \\frac{1}{3} \\left( \\frac{H L_1^3}{L_1 - L_2} - \\frac{H L_2^3}{L_1 - L_2} \\right) \\\\\n&= \\frac{H (L_1^3 - L_2^3)}{3 (L_1 - L_2)} = \\frac{H}{3}(L_1^2 + L_1 L_2 + L_2^2).\n\\end{aligned}\n$$\n\nAsí, teniendo en cuenta que\n\n$$\n\\text{Volumen del tronco de pirámide} = \\frac{2}{3} \\text{Volumen del prisma}\n$$\n\ntendremos la ecuación\n\n$$\n\\frac{H}{3}(L_1^2 + L_1 L_2 + L_2^2) = \\frac{2}{3} H L_1^2,\n$$\n\nque se transforma en\n\n$$\n\\left(\\frac{L_1}{L_2}\\right)^2 - \\frac{L_1}{L_2} - 1 = 0,\n$$\n\ncuya única solución positiva es $\\frac{L_1}{L_2} = \\frac{1 + \\sqrt{5}}{2}$. Es decir, los lados deben estar en relación áurea.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 16905,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c, d$ be positive integers satisfying\n$$\n\\frac{ab}{a+b} + \\frac{cd}{c+d} = \\frac{(a+b)(c+d)}{a+b+c+d}\n$$\nDetermine all possible values of $a+b+c+d$.",
    "options": [],
    "answer": "See solution",
    "solution": "The possible values are the positive integers that are not square-free.\n\nFirst, note that if we take $a = \\ell$, $b = k\\ell$, $c = k\\ell$, $d = k^2\\ell$ for some positive integers $k$ and $\\ell$, then we have\n$$\n\\frac{ab}{a+b} + \\frac{cd}{c+d} = \\frac{k\\ell^2}{\\ell+k\\ell} + \\frac{k^3\\ell^2}{k\\ell+k^2\\ell} = \\frac{k\\ell}{k+1} + \\frac{k^2\\ell}{k+1} = k\\ell\n$$\nand\n$$\n\\frac{(a+b)(c+d)}{a+b+c+d} = \\frac{(\\ell+k\\ell)(k\\ell+k^2\\ell)}{\\ell+k\\ell+k\\ell+k^2\\ell} = \\frac{k(k+1)^2\\ell^2}{\\ell(k+1)^2} = k\\ell,\n$$\nso that\n$$\n\\frac{ab}{a+b} + \\frac{cd}{c+d} = k\\ell = \\frac{(a+b)(c+d)}{a+b+c+d}.\n$$\nThis means that $a + b + c + d = \\ell(1 + 2k + k^2) = \\ell(k + 1)^2$ can be attained. We conclude that all non-square-free positive integers can be attained.\n\nNow, we will show that if\n$$\n\\frac{ab}{a+b} + \\frac{cd}{c+d} = \\frac{(a+b)(c+d)}{a+b+c+d}\n$$\nthen $a + b + c + d$ is not square-free. Suppose on the contrary that $a + b + c + d$ is square-free, and note that after multiplying by $(a+b)(c+d)(a+b+c+d)$, we obtain\n$$\n(ab(c+d) + cd(a+b))(a+b+c+d) = (a+b)^2(c+d)^2\n$$\nA prime factor of $a+b+c+d$ must divide $a+b$ or $c+d$, and therefore divides both $a+b$ and $c+d$. Since $a+b+c+d$ is square-free, the fact that every prime factor of $a+b+c+d$ divides $a+b$ implies that $a+b+c+d$ itself divides $a+b$. Because $a+b < a+b+c+d$, this is impossible.\n\nSo $a + b + c + d$ cannot be square-free. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16906,
    "subject": "Mathematics (Olympiad)",
    "question": "Сүүлийн 2 цифр нь ижил бөгөөд бүхэл тооны квадрат болж байх 5 оронтой тоо хичнээн байх вэ?",
    "options": [],
    "answer": "See solution",
    "solution": "$n = \\overline{abcd} = m^2$ гэж авъя.\n\nИймд $d \\in \\{0, 1, 4, 5, 6, 9\\}$ байна. $n = 100 \\cdot \\overline{abc} + 11d$ ба $m^2 \\equiv 0, 1 \\pmod{4}$ тул $n = m^2 = 3d \\equiv 0, 1 \\pmod{4}$ болно. Үүнээс $d = 0$ эсвэл $d = 4$.\n\n1. $d = 0$ бол $n = 100 \\cdot \\overline{abc} = 10^2 \\cdot \\overline{abc} = m^2$ тул $\\overline{abc} = S^2$, $S \\in \\{10, 11, \\ldots, 31\\}$, нийт 22 боломж.\n\n2. $d = 4$ бол $n = 100 \\cdot \\overline{abc} + 44 = m^2$. $m = 2p$ хэлбэртэй тул $\\overline{abc} = \\frac{4p^2 - 44}{100} = \\frac{p^2 - 11}{25}$.\n\nДараах тохиолдлууд:\n\n(a) $p = 5q$ бол $\\frac{p^2 - 11}{25}$ нь бүхэл тоо биш.\n\n(b) $p = 5q + 1$ бол $\\overline{abc} = q^2 + \\frac{2(q-1)}{5}$, $q = 11, 16, 21, 26, 31$ — 5 боломж.\n\n(c) $p = 5q + 2$ бол $\\overline{abc} = q^2 + \\frac{20q - 7}{25}$ — бүхэл тоо биш.\n\n(d) $p = 5q + 3$ бол $\\overline{abc} = q^2 + \\frac{30q - 2}{25}$ — бүхэл тоо биш.\n\n(e) $p = 5q + 4$ бол $\\overline{abc} = q^2 + \\frac{8q + 1}{5}$, $q = 13, 18, 23, 28$ — 4 боломж.\n\nИймд нийт $22 + 5 + 4 = 31$ тоо байна.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16907,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $L(n)$ be the least prime divisor of $n$. Prove that $L(d(n)) \\leq L(\\sigma(n))$ for all positive integers $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof.* Let $n = p_1^{\\alpha_1} \\dots p_t^{\\alpha_t}$ for some distinct prime numbers $p_1, \\dots, p_t$. It follows that\n\n$$\nd(n) = (1 + \\alpha_1) \\dots (1 + \\alpha_t),\n$$\n\nand\n\n$$\n\\sigma(n) = \\frac{p_1^{\\alpha_1+1} - 1}{p_1 - 1} \\cdots \\frac{p_t^{\\alpha_t+1} - 1}{p_t - 1}.\n$$\n\nLet a prime $q$ divide $\\frac{p_i^{\\alpha_i+1}-1}{p_i-1}$ for some $i$. It follows that $d_i$, the order of $p_i$ modulo $q$, divides $1+\\alpha_i$. Hence, if $d_i \\neq 1$, then one of the primes dividing $1+\\alpha_i$ also divides $d_i$. It follows that $L(1+\\alpha_i) \\leq d_i$. Let $q$ be the least prime divisor of $\\frac{p_i^{\\alpha_i+1}-1}{p_i-1}$. Since $d_i$ divides $q-1$, it follows that $d_i < q = L\\left(\\frac{p_i^{\\alpha_i+1}-1}{p_i-1}\\right)$. Now, if $d_i = 1$, since $q$ divides $\\frac{p_i^{\\alpha_i+1}-1}{p_i-1}$ we have $q \\mid 1+\\alpha_i$. Whence, $L(1+\\alpha_i) \\leq q = L\\left(\\frac{p_i^{\\alpha_i+1}-1}{p_i-1}\\right)$. Assume now $L(\\sigma(n)) = L\\left(\\frac{p_i^{\\alpha_i+1}-1}{p_i-1}\\right)$ for some $i$, $1 \\leq i \\leq t$, then\n\n$$\nL(\\sigma(n)) = L\\left(\\frac{p_i^{\\alpha_i+1}-1}{p_i-1}\\right) \\geq L(\\alpha_i + 1) \\geq \\min_{1 \\leq i \\leq t} L(\\alpha_i + 1) = L(d(n)).\n$$\n\nThis completes our proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16908,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be positive integers such that $2n-1$ is a composite number. Consider the expression $\\frac{a_i + a_j}{(a_i, a_j)}$ for $1 \\leq i < j \\leq n$.\n\n(1) Prove that if $a_i + a_j \\equiv 0 \\pmod{p}$ for some prime $p$, then\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\geq p = 2n - 1.\n$$\n\n(2) Construct an example to show that the bound is sharp when $2n-1$ is composite.",
    "options": [],
    "answer": "See solution",
    "solution": "Case $a_i + a_j \\equiv 0 \\pmod{p}$, we have\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\geq p = 2n - 1.\n$$\nThat completes the proof of (1).\n\n(2) We will construct an example to justify the statement. Since $2n-1$ is a composite number, we can write $2n-1 = pq$ where $p, q$ are positive integers greater than 1. Let\n$$\n\\begin{aligned}\na_1 &= 1,\\ a_2 = 2,\\ \\dots,\\ a_p = p,\\ a_{p+1} = p+1,\\\\\na_{p+2} &= p+3,\\ \\dots,\\ a_n = pq-p.\n\\end{aligned}\n$$\nNote that the first $p$ elements are consecutive integers, while the remainders are $n-p$ consecutive even integers from $p+1$ to $pq-p$.\n\nWhen $1 \\leq i \\leq j \\leq p$, it is obvious that\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\leq a_i + a_j \\leq 2p < 2n - 1.\n$$\nWhen $p+1 \\leq i \\leq j \\leq n$, we have $2\\mid(a_i, a_j)$ and then\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\leq \\frac{a_i + a_j}{2} \\leq pq - p < 2n - 1.\n$$\nWhen $1 \\leq i \\leq p$ and $p+1 \\leq j \\leq n$ we have the following two possibilities:\n\n(i) Either $i \\neq p$ or $j \\neq n$. Then we have\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\leq pq - 1 < 2n - 1.\n$$\n(ii) $i = p$ and $j = n$. Then\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} = \\frac{pq}{p} = q < 2n - 1.\n$$\nThat completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16909,
    "subject": "Mathematics (Olympiad)",
    "question": "For any real numbers $x, y$, prove the inequality\n\n$$\n\\sqrt{(x + 4)^2 + (y + 2)^2} + \\sqrt{(x - 5)^2 + (y + 4)^2} \\leq \\sqrt{(x - 2)^2 + (y - 6)^2} + \\sqrt{(x - 5)^2 + (y - 6)^2} + 20.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "On a coordinate plane, consider points $A(-4, -2)$, $B(2, 6)$, $C(5, 6)$, and $D(5, -4)$.\n\n![](images/Ukraine_2021-2022_p10_data_b89e4da7cb.png)\n\nFor any point $M(x, y)$, the inequality can be rewritten as:\n\n$$\nS(M) = MA + MD - MB - MC \\leq 20.\n$$\n\nLet's find the largest possible value of the expression $S(M)$. For any point $M$ in the plane, by the triangle inequality:\n\n$$\nS(M) = (MA - MB) + (MD - MC) \\leq AB + CD.\n$$\n\nFor the point $X = AB \\cap CD$, we have:\n\n$$\nS(X) = (XA - XB) + (XD - XC) = AB + CD = 10 + 10 = 20.\n$$\n\nThus, the largest value is achieved for point $X$, and for all other points the inequality is strict.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16910,
    "subject": "Mathematics (Olympiad)",
    "question": "Во координатната рамнина $xOy$ е дадена права $4x + 3y = n$, $n > 0$, која од координатниот почеток е на растојание 12. Определи ја плоштината на триаголникот што ја гради правата со координатните оски.",
    "options": [],
    "answer": "See solution",
    "solution": "Правата со координатните оски гради правоаголен триаголник со катети $\\overline{OA} = a = \\frac{n}{4}$ и $\\overline{OB} = b = \\frac{n}{3}$ (види цртеж).\n\n![](images/Makedonija_2009_p14_data_78f99e447e.png)\n\nДолжината на хипотенузата е:\n\n$$\n\\overline{AB} = c = \\sqrt{a^2 + b^2} = \\sqrt{\\frac{n^2}{16} + \\frac{n^2}{9}} = \\frac{5n}{12}\n$$\n\nВисината повлечена кон хипотенузата е $h_c = 12$, од каде добиваме:\n\n$$\nP = \\frac{\\frac{5n}{12} \\cdot 12}{2} = \\frac{5}{2} n\n$$\n\nОд друга страна, $P = \\frac{ab}{2} = \\frac{n^2}{24}$. Значи,\n\n$$\n\\frac{n^2}{24} = \\frac{5n}{2}\n$$\n\nЗначи, $n = 60$ и $P = 150$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16911,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer and $C$ be a nonnegative real number. Find the number of sequences of real numbers $1, x_2, \\dots, x_n, 1$ such that the absolute value of the difference of every two consecutive terms is equal to $C$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose the sequence $1, x_2, \\dots, x_n, 1$ satisfies the condition. Then\n\n$$\n|1 - x_2| = |x_2 - x_3| = \\dots = |x_n - 1| = C.\n$$\n\nAlso,\n\n$$\nx_n - 1 = (x_n - x_{n-1}) + (x_{n-1} - x_{n-2}) + \\dots + (x_2 - 1)\n$$\n\nwhere there are $n-1$ terms on the right. Consider two cases:\n\n**Case 1.** $n$ is odd ($n-1$ is even).\n\nIn this case, the sum on the right is an even multiple of $C$, while the left is $\\pm C$, which is only possible if $C=0$. Thus, the only sequence is $1 = x_2 = \\dots = x_n = 1$.\n\n**Case 2.** $n$ is even ($n-1$ is odd).\n\nIf $C=0$, again $1 = x_2 = \\dots = x_n = 1$. If $C > 0$, the possible choices correspond to sequences of $n-1$ signs ($+$ or $-$) such that their sum is $\\pm C$. The number of such sequences is $2 \\binom{n-1}{n/2}$.\n\nTherefore, if $r = r(n, C)$ is the number of such sequences,\n\n$$\nr = \\begin{cases} 1, & C=0 \\\\ 0, & n \\text{ is odd and } C > 0 \\\\ 2\\binom{n-1}{n/2}, & n \\text{ is even and } C > 0. \\end{cases}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16912,
    "subject": "Mathematics (Olympiad)",
    "question": "The rows and columns of a $16 \\times 16$ table are labeled $1, 2, \\dots, 16$. In the square at row $i$, column $j$, the product $i \\cdot j$ is written.\n\nSeveral rows (at least 2) and several columns (at least 2) are chosen. The numbers at their intersections are deleted.\n\n**a)** Can the sum of all deleted numbers be a prime?\n\n**b)** What about the sum of all undeleted numbers?",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** The sum of the deleted numbers is always composite. Let the chosen rows be $i_1, \\dots, i_p$ ($p \\geq 2$) and the chosen columns $j_1, \\dots, j_q$ ($q \\geq 2$). The deleted numbers are all $i_k j_l$ for $1 \\leq k \\leq p$, $1 \\leq l \\leq q$.\n\nThe total deleted sum is:\n$$\nS = (i_1 + \\dots + i_p)(j_1 + \\dots + j_q)\n$$\nBoth factors are at least $3$ (since $p \\geq 2$, $q \\geq 2$, and the smallest possible sum is $1+2=3$), so $S$ is composite.\n\n**b)** The sum of all undeleted numbers can be a prime. For example, if the chosen rows are $2, 3, \\dots, 16$ and the chosen columns are $2, 3, \\dots, 16$, the undeleted numbers are those in row $1$ and column $1$ (except the $(1,1)$ entry, which is counted twice).\n\nTheir sum is:\n$$\n2(1 + 2 + \\dots + 16) - 1 = 2 \\times 136 - 1 = 271\n$$\nwhich is a prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16913,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a non-isosceles triangle, and let $I$ be its incenter. Let $\\gamma_A$ be the circle through $I$, tangent to $AB$ and $AC$ and crossing the segment $AI$; let $\\gamma_B$ be the circle through $I$, tangent to $BC$ and $BA$ and crossing the segment $BI$; and let $\\gamma_C$ be the circle through $I$, tangent to $CA$ and $CB$ and crossing the segment $CI$.\n\nThe circles $\\gamma_B$ and $\\gamma_C$ cross again at $A'$, the circles $\\gamma_C$ and $\\gamma_A$ cross again at $B'$, and the circles $\\gamma_A$ and $\\gamma_B$ cross again at $C'$. Prove that the circles $AA'I$, $BB'I$, and $CC'I$ cross again at a point different from $I$.",
    "options": [],
    "answer": "See solution",
    "solution": "Alternatively, but equivalently, we are to show that the centres of the circles $AA'I$, $BB'I$, and $CC'I$ are collinear. Inverting the whole configuration from $I$, it is not hard to see that no two of these circles are tangent at $I$, so they cross again at a point different from $I$.\n\n![](images/RMC_2019_var_3_p42_data_967cf2510f.png)\n\nLet the internal bisectors of the angles $BAC$, $CBA$, and $ACB$ cross the circle $ABC$ again at $M_A$, $M_B$, and $M_C$, respectively, and let $\\gamma_A$, $\\gamma_B$, and $\\gamma_C$ be centered at $O_A$, $O_B$, and $O_C$, respectively; clearly, $O_A$, $O_B$, and $O_C$ lie on the segments $AI$, $BI$, and $CI$, respectively.\n\nThe center of the circle $AA'I$ is the point where the perpendicular bisector of the segment $AI$ crosses the perpendicular bisector of the segment $A'I$. It is a fact that the former is the line $M_B M_C$; the latter is, of course, the line $O_B O_C$, so the lines $M_B M_C$ and $O_B O_C$ cross at the center of the circle $AA'I$. Similarly, the lines $M_C M_A$ and $O_C O_A$ cross at the center of the circle $BB'I$, and the lines $M_A M_B$ and $O_A O_B$ cross at the center of the circle $CC'I$.\n\nFinally, since the triangles $M_A M_B M_C$ and $O_A O_B O_C$ are in perspective from $I$, the conclusion follows by Desargues' theorem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16914,
    "subject": "Mathematics (Olympiad)",
    "question": "We have a deck of 90 cards numbered from 10 to 99 (all two-digit numbers). How many sets of three or more different cards in this deck are there such that the number on one of them is the sum of the other numbers, and those other numbers are consecutive?",
    "options": [],
    "answer": "See solution",
    "solution": "We are looking at sums of the form\n\n$$\ns(10 + \\ell, k) = \\sum_{i=0}^{k-1} (10 + \\ell + i) = k\\left(10 + \\ell + \\frac{1}{2}(k-1)\\right),\n$$\n\nwhere $\\ell \\geq 0$ ($10 + \\ell$ is the starting number of the consecutive numbers in a set) and $k \\geq 2$ is the number of consecutive numbers in a set. Each of these sums (the $(k+1)$-th number in the set) must be less than or equal to 99. By taking $\\ell = 0$, we see that the maximum possible value for $k$ is 7.\n\nFor $k = 2$ (three cards), we must have $2 \\cdot (10 + \\ell + \\frac{1}{2}(2-1)) \\leq 99$, so $\\ell \\leq 39$. The $40$ possibilities of three cards are: $10 + 11 = 21$, $11 + 12 = 23$, $\\ldots$, $49 + 50 = 99$.\n\nFor $k = 3$ (four cards), $3 \\cdot (10 + \\ell + \\frac{1}{2}(3-1)) \\leq 99$, so $\\ell \\leq 22$, giving $23$ groups: $10 + 11 + 12 = 33$, $\\ldots$, $32 + 33 + 34 = 99$.\n\nContinuing, $\\ell \\leq 13$ when $k = 4$, $\\ell \\leq 7$ when $k = 5$, $\\ell \\leq 4$ when $k = 6$, and $\\ell \\leq 1$ when $k = 7$. In total, there are $40 + 23 + 14 + 8 + 5 + 2 = 92$ sets of cards satisfying the conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16915,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a triangle $\\triangle ABC$.\n\na) Prove that the interior bisector of the angle $\\angle A$ and the exterior bisectors of the angles $\\angle B$ and $\\angle C$ intersect at a point $I_A$.\n\nb) Let $I_A M \\perp AC$, $M \\in AC$; $I_A N \\perp BC$, $N \\in BC$; and $I_A P \\perp AB$, $P \\in AB$. Show that if $I_A M + I_A P = I_A N$, then the triangle $ABC$ is equilateral.",
    "options": [],
    "answer": "See solution",
    "solution": "a) If $I_A$ is the point of intersection of the exterior bisectors of the angles $\\angle B$ and $\\angle C$, then $I_A$ is inside the angle $\\angle A$ and is equidistant from the sides $AB$ and $BC$, respectively of $BC$ and $AC$. Through transitivity, $I_A$ is equidistant from the sides $AB$ and $AC$ of the angle $\\angle A$, is inside the angle $\\angle A$, so it is on the interior bisector of the angle $\\angle A$.\n\nb) From the condition $I_A M + I_A P = I_A N$ we deduce that the quadrilateral $I_A M N P$ is a parallelogram. Also from part a), we have $I_A M = I_A N = I_A P$, so $I_A M N P$ is a rhombus and the triangles $I_A N P$ and $I_A M N$ are equilateral.\n\nThus, the inscribed quadrilateral $API_A M$ has $\\angle P I_A M = 120^\\circ$, so $\\angle A = 60^\\circ$.\n\nOn the other hand, $C$ is the exterior angle of the inscribed quadrilateral $I_A M C N$, so $\\angle C = \\angle M I_A N = 60^\\circ$, and $B$ is the exterior angle of the inscribed quadrilateral $I_A P B N$, so $\\angle B = \\angle P I_A N = 60^\\circ$. Thus, triangle $ABC$ has all angles of $60^\\circ$, so it is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16916,
    "subject": "Mathematics (Olympiad)",
    "question": "6 lines are chosen in 3-dimensional space. Find the largest number of points at which at least 3 of the chosen lines intersect.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us call a point *rich* if at least 3 of the chosen lines intersect there. Let the number of rich points be $n$.\n\nLet us count pairs $(P, s)$, in which $P$ is a rich point and $s$ is a chosen line passing through that point. There are at least $3n$ such pairs because each of the $n$ rich points appears in at least 3 pairs.\n\nOn the other hand, for every chosen line, all lines passing through the rich points on that line are different (two lines cannot intersect at more than one point). Therefore, on each of the chosen lines there are at most $\\left\\lfloor \\frac{6-1}{2} \\right\\rfloor = 2$ rich points, and the number of pairs counted can be at most $2 \\times 6 = 12$.\n\nConsequently, $3n \\leq 12$, i.e., $n \\leq 4$.\n\nA construction for 4 rich points is provided by the lines defined by the 6 sides of a tetrahedron. 3 of the lines intersect at each of the 4 vertices of the tetrahedron.\n\n**Answer:** $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16917,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle $c$ with center $A$ passes through the vertices $B$ and $E$ of a regular pentagon $ABCDE$. The line $BC$ intersects the circle $c$ a second time at point $F$. Point $G$ on the circle $c$ is chosen so that $FB = FG$ and $B \\neq G$. Prove that the lines $AB$, $EF$, and $DG$ meet in a common point.",
    "options": [],
    "answer": "See solution",
    "solution": "The internal angles of a regular pentagon are $108^\\circ$. Thus, $\\angle ABC = 108^\\circ$, so $\\angle ABF = 72^\\circ$. As $AB = AF$, we have $\\angle AFB = 72^\\circ$ and $\\angle BAF = 36^\\circ$. Since $FG = FB$, $AG = AB$, and $AF = AF$, the triangles $AFB$ and $AFG$ are congruent, so $\\angle FAG = 36^\\circ$. As $\\angle EAB = 108^\\circ$, we get $\\angle GAE = 2 \\cdot 36^\\circ + 108^\\circ = 180^\\circ$, so $E$, $A$, and $G$ are collinear.\n\nLet the lines $AB$ and $EF$ meet at point $K$.\n\n![](images/EST_ABooklet_2020_p11_data_093ab0edfa.png)\n\n![](images/EST_ABooklet_2020_p12_data_90978b627e.png)\n\n![](images/EST_ABooklet_2020_p12_data_7cd23d157f.png)\n\nSince $\\angle AFC = 180^\\circ - \\angle FCD$, we have $AF \\parallel CD$. Thus, $AF \\parallel BE$, so by similarity of triangles $BKE$ and $AKF$, $\\frac{BK}{AK} = \\frac{BE}{AF}$.\n\nLet the lines $AB$ and $DG$ meet at point $K'$. Lines $AG$ and $AE$ coincide because $E$, $A$, and $G$ are collinear. Thus, $BD \\parallel AG$, so by similarity of triangles $BK'D$ and $AK'G$, $\\frac{BK'}{AK'} = \\frac{BD}{AG}$.\n\nSince $BD = BE$ and $AF = AG$, we have $\\frac{BK}{AK} = \\frac{BK'}{AK'}$, so $K = K'$.\n",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16918,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S = \\{2, 0, 1, 5\\}$. Consider all $k$-digit numbers (possibly with leading zeros) of the form $\\overline{a_1a_2\\dots a_k}$, where each $a_i \\in S$. How many such $k$-tuples $(a_1, a_2, \\dots, a_k)$ satisfy that $3$ divides $a_1 + a_2 + \\dots + a_k$? Express the answer in terms of $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\epsilon$ be a complex root of $x^2 + x + 1 = 0$, so $\\epsilon^3 = 1$ and $1 + \\epsilon^k + \\epsilon^{2k} = 0$.\n\nWe have:\n$$\n\\begin{cases}\nP(1) = a_0 + a_1 + a_2 + \\dots + a_{5k} \\\\\nP(\\epsilon) = a_0 + a_1\\epsilon + a_2\\epsilon^2 + \\dots + a_{5k}\\epsilon^{5k} \\\\\nP(\\epsilon^2) = a_0 + a_1\\epsilon^2 + a_2\\epsilon^4 + \\dots + a_{5k}\\epsilon^{10k}\n\\end{cases}\n$$\n\nThe number of desired tuples is:\n$$\nT = \\frac{P(1) + P(\\epsilon) + P(\\epsilon^2)}{3} = \\frac{4^k + \\epsilon^k + \\epsilon^{2k}}{3}.\n$$\n\nThus,\n$$\nT = \\begin{cases}\n\\frac{4^k - 1}{3} & \\text{if } k \\text{ is not divisible by } 3 \\\\\n\\frac{4^k + 2}{3} & \\text{if } k \\text{ is divisible by } 3\n\\end{cases}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16919,
    "subject": "Mathematics (Olympiad)",
    "question": "A square $ABCD$ is divided into $n^2$ equal small elementary squares by drawing lines parallel to its sides. A spider starts from $A$, moving only to the right and up, and tries to reach point $C$. Every movement of the spider consists of $k$ steps right and $m$ steps up, or $m$ steps right and $k$ steps up (in any order she wants).\n\nThe spider makes $l$ such “movements” and then continues moving only to the right or up without any restriction. If $n = m \\cdot l$, find the number of all possible routes the spider can follow to reach point $C$, where $n, m, k, l$ are positive integers and $k < m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose the square is placed in a Cartesian coordinate system with origin $A(0,0)$ and axes along $AB$ and $AD$. The spider starts from $A$ and makes its first movement: either $k$ steps right and $m$ steps up, or $m$ steps right and $k$ steps up. For example, if $m = 4$, $k = 3$:\n\n![](images/GreekMO2014_booklet_p20_data_33857eb517.png)\n\nAfter the first movement, the spider can be at $M(k, m)$ or $N(m, k)$. The number of ways to reach either point is $\\binom{k+m}{k} = \\binom{k+m}{m} = v$.\n\nAfter the second movement, the spider can be at three points: $K(2k, 2m)$, $L(k+m, k+m)$, or $T(2m, 2k)$, with corresponding numbers of ways $v^2$, $2v^2$, and $v^2$.\n\nThe point $M$ can be reached from $A$ in $v$ ways, and $K$ from $M$ in $v$ ways, so $K$ can be reached from $A$ in $v^2$ ways. Similarly, $T$ can be reached in $v^2$ ways. The point $L$ can be reached via $A \\to M \\to L$ or $A \\to N \\to L$, each in $v^2$ ways, so $L$ can be reached in $2v^2$ ways.\n\n![](images/GreekMO2014_booklet_p21_data_05d739dbdf.png)\n\nAfter the third movement, the spider can be at four points, which can be reached in $\\binom{3}{3} v^3$, $\\binom{3}{2} v^3$, $\\binom{3}{1} v^3$, and $\\binom{3}{0} v^3$ ways, respectively. In general, after $l$ movements, the spider will be at $l+1$ points on the line $x + y = l(k + m)$.\n\n![](images/GreekMO2014_booklet_p22_data_f5ea3e8017.png)\n\nLet $r = m - k$. These points are:\n\n$$\nA_0(lm, lk),\\ A_1(lm - r, lk + r),\\ A_2(lm - 2r, lk + 2r),\\ \\dots,\\ A_l(lm - lr, lk + lr) = A_l(lk, lm)\n$$\n\nEach $A_j$ can be reached in $\\binom{l}{j} v^l$ ways from $A$. The point $C(ml, ml)$ can be reached from $A_j$ in $\\binom{lr}{j r}$ ways, so the total number of routes is:\n\n$$\n\\sum_{j=0}^l \\binom{l}{j} v^l \\binom{lr}{j r}\n$$\n\nwhere $v = \\binom{k + m}{k}$ and $r = m - k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16920,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $ABCC_1B_1A_1$ 為凸六邊形，其中 $AB = BC$，並且線段 $\\overline{AA_1}$、$\\overline{BB_1}$ 及 $\\overline{CC_1}$ 的中垂線皆相同。令對角線 $AC_1$ 與 $A_1C$ 交於點 $D$，並將三角形 $ABC$ 的外接圓記為 $\\omega$。設 $\\omega$ 與三角形 $A_1BC_1$ 的外接圓再交於一點 $E \\neq B$。\n\n試證：直線 $BB_1$ 與 $DE$ 的交點落在 $\\omega$ 上。",
    "options": [],
    "answer": "See solution",
    "solution": "若 $AA_1 = CC_1$，則此六邊形關於直線 $BB_1$ 對稱，特別地，圓 $ABC$ 與 $A_1BC_1$ 互相相切。因此 $AA_1$ 與 $CC_1$ 必不相等。由於點 $A$ 與 $A_1$ 可與 $C$ 與 $C_1$ 互換，我們不妨假設 $AA_1 < CC_1$。\n\n設 $R$ 為圓 $AEBC$、$A_1EBC_1$ 及對稱梯形 $ACC_1A_1$ 的外接圓的徑心，即 $AC$、$A_1C_1$ 及 $BE$ 的徑軸交點。由於 $AC$ 與 $A_1C_1$ 的對稱性，$R$ 落在 $AA_1$ 與 $CC_1$ 的公共中垂線上，亦即 $\\angle ADC$ 的外角平分線上。\n\n設 $F$ 為直線 $DR$ 與圓 $ACD$ 的另一交點。由 $R$ 對圓 $\\omega$ 與 $ACFD$ 的冪可得：\n\n$$\nRB \\cdot RE = RA \\cdot RC = RD \\cdot DF,\n$$\n\n因此 $B, E, D, F$ 共圓。\n\n直線 $RDF$ 為 $\\angle ADC$ 的外角平分線，故 $F$ 平分圓弧 $CDA$。由 $AB = BC$，在圓 $\\omega$ 上，$B$ 為弧 $AEC$ 的中點。設 $M$ 為 $B$ 在 $\\omega$ 上的對徑點，即弧 $CA$ 的中點。注意 $B, F, M$ 均在 $AC$ 的中垂線上，故共線。\n\n![](images/18-3J_p28_data_e775c4bc79.png)\n\n最後，設 $X$ 為 $\\omega$ 與直線 $DE$ 的另一交點。由於 $BM$ 為 $\\omega$ 的直徑，$\\angle BXM = 90^\\circ$。此外，\n\n$$\n\\angle EXM = 180^\\circ - \\angle MBE = 180^\\circ - \\angle FBE = \\angle EDF,\n$$\n\n故 $MX$ 與 $FD$ 平行。由於 $BX$ 垂直於 $MX$，而 $BB_1$ 垂直於 $FD$，這說明 $X$ 落在直線 $BB_1$ 上。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16921,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $c = 2014$, where $c$ is even and coprime with $3$. Let $f(0) = a$.\n\nFind all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ such that for all $m, n \\in \\mathbb{Z}$,\n$$\nf(f(n) + m) = f(3n + m) + c.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "1) Setting $m = 0$, we get $f(f(n)) = f(3n) + c$.\n\nLet $f(0) = a$.\n\n2) Setting $m = 0$ in the original equation gives $f(f(n)) = f(3n) + c$.\n\n3) By induction, $f(n + ka) = f(n) + kc$, $k \\in \\mathbb{Z}$. In particular, $f(ka) = a + kc$ and $f(a) = a + c$.\n\n4) Setting $m = n$ in the original equation:\n$$\nf(f(n) + n) = f(3n + n) + c = f(3n + n) + c = f(4n) + c.\n$$\n\nBut from above, $f(f(n) + n) = f(3n + a)$, so $f(f(n) + n) = f(3n + a)$. Thus, $f(3n + a) = f(4n) + c$.\n\n5) Setting $m = a$ in the original equation:\n$$\nf(f(n) + a) = f(3n + a) + c.\n$$\n\nBut $f(f(n) + a) = f(n + f(a))$ by the property of $f$.\n\n6) By induction, $f(n + kc) = f(n) + 2kc$.\n\n7) To check injectivity, suppose $f(s) = f(t)$. Then $f(3s) = f(3t)$, and so $f(3^l s) = f(3^l t)$ for all $l \\in \\mathbb{N}$. Since $(3, c) = 1$, there exists $l$ such that $3^l - 1 = nc$ for some $n \\in \\mathbb{N}$. Then $f(3^l s) = f(3^l t)$ implies $s = t$. Thus, $f$ is injective.\n\n8) From above, $f(f(n) + n) = f(3n + a)$, so $f(n) + n = 3n + a$, hence $f(n) = 2n + a$.\n\nVerification: Plug $f(n) = 2n + a$ into the original equation:\n$$\nf(f(n) + m) = f(2n + a + m) = 2( n + m ) + a = 2n + 2m + a.\n$$\nOn the other hand,\n$$\nf(3n + m) + c = 2(3n + m) + a + c = 6n + 2m + a + c.\n$$\nFor these to be equal, $2n + 2m + a = 6n + 2m + a + c$, so $c = -4n$, which is only possible for $n = 0$ unless $c = 0$. But since $c$ is fixed, the only possible value is $a = c/2$.\n\nThus, the solution is $f(n) = 2n + a$ with $a = c/2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16922,
    "subject": "Mathematics (Olympiad)",
    "question": "In an integer sequence, the difference between successive elements is equal to the sum of all previous such differences. Which of these sequences with $a_0 = 2012$ and $1 \\leq d = a_1 - a_0 \\leq 43$ contain perfect squares?",
    "options": [],
    "answer": "See solution",
    "solution": "For $n \\geq 1$, we have\n\n$$\na_{n+1} - a_n = (a_n - a_{n-1}) + (a_{n-1} - a_{n-2}) + \\dots + (a_1 - a_0) = a_n - a_0,\n$$\nwhich yields $a_{n+1} = 2a_n - a_0$. The sequence can therefore be written as\n\n$$\na_0 = 2012,\\quad a_1 = 2012 + d,\\quad a_2 = 2012 + 2d,\\quad \\dots,\\quad a_n = 2012 + 2^{n-1}d,\\quad \\dots\n$$\nFor $n \\geq 3$, we have $a_n = 4 \\cdot (503 + 2^{n-3}d)$. Since $503 + 2^{n-3}d \\equiv 3 \\pmod{4}$ for $n \\geq 5$, $a_n$ can never be a perfect square for $n \\geq 5$. The only numbers in the sequence that can possibly be perfect squares are\n\n$$\na_0 = 2012,\\quad a_1 = 2012 + d,\\quad a_2 = 2012 + 2d,\\quad a_3 = 4(503 + d),\\quad a_4 = 4(503 + 2d)\n$$\nwith $1 \\leq d \\leq 43$. Since $2012 = a_0 < a_4 \\leq 2012 + 8 \\cdot 43 = 2356$, the only perfect squares that can occur in the sequences must lie between 2012 and 2356, i.e.\n\n$$\n\\begin{align*}\n45^2 &= 2025 = 2012 + 13, \\\\\n46^2 &= 2116 = 2012 + 104 = 2012 + 2 \\cdot 52 = 2012 + 4 \\cdot 26 = 2012 + 8 \\cdot 13, \\\\\n47^2 &= 2209 = 2012 + 197, \\\\\n48^2 &= 2304 = 2012 + 292 = 2012 + 2 \\cdot 146 = 2012 + 4 \\cdot 73.\n\\end{align*}\n$$\nSince $d \\leq 43$, the only sequences of the required type containing perfect squares are those with $d=13$ (which contains 2025 and 2116) and with $d=26$ (which also contains 2116). $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16923,
    "subject": "Mathematics (Olympiad)",
    "question": "If the polynomial\n$$\nf(x) = x^{6} + a_{1}x^{5} + a_{2}x^{4} + a_{3}x^{3} + a_{4}x^{2} + a_{5}x + 3\n$$\nwith real coefficients possesses all negative roots, show that $f(2) \\geq 27^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_6 = 3$ and $r_1, r_2, \\dots, r_6 < 0$ be all roots of $f(x)$.\n\nFor each $k = 1, 2, \\dots, 6$, we have\n$$\na_k = (-1)^k \\sum_{1 \\le j_1 < \\dots < j_k \\le 6} r_{j_1} r_{j_2} \\cdots r_{j_k} = \\sum_{1 \\le j_1 < \\dots < j_k \\le 6} |r_{j_1}| |r_{j_2}| \\cdots |r_{j_k}|\n$$\n\nUsing the AM-GM inequality, we have\n$$\n\\begin{aligned}\na_k &\\geq \\binom{6}{k} \\left( \\prod_{1 \\le j_1 < \\cdots < j_k \\le 6} |r_{j_1}| |r_{j_2}| \\cdots |r_{j_k}| \\right)^{1/\\binom{6}{k}} \\\\\n&= \\binom{6}{k} (r_1 r_2 \\cdots r_6)^{k/6} \\\\\n&= 3^{k/6} \\binom{6}{k} \\\\\n&\\geq \\binom{6}{k}.\n\\end{aligned}\n$$\n\nHence,\n$$\n\\begin{aligned}\nf(2) &= 2^6 + 2^5 a_1 + \\cdots + 2 a_5 + a_6 \\\\\n&\\geq 2^6 + 2^5 \\binom{6}{1} + \\cdots + 2 \\binom{6}{5} + \\binom{6}{6} \\\\\n&= 3^6 \\\\\n&= 27^2.\n\\end{aligned}\n$$\n\n![](images/Thailand-2007_Booklet_p6_data_e1f270a610.png)",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16924,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be positive integers. Vaughan arranges $abc$ identical white unit cubes into an $a \\times b \\times c$ rectangular prism and paints the outside of the prism red. After disassembling the prism back into unit cubes, he notices that the number of faces of the unit cubes that are red is the same as the number that are white.\n\nFind all values that the product $abc$ could take.",
    "options": [],
    "answer": "See solution",
    "solution": "The possible values for $abc$ are $8$, $16$, or $18$.\n\nThe total number of faces of the small cubes is $6abc$, and the total number of painted faces is $2ab + 2bc + 2ca$. So we have:\n\n$$\n2ab + 2bc + 2ca = 3abc.\n$$\n\nAssume, without loss of generality, that $a \\leq b \\leq c$.\n\n**Case 1:** $a = 1$\n\nThen $2b + 2c + 2bc = 3bc$, so $bc - 2b - 2c = 0$. Rearranging, $(b-2)(c-2) = 4$. The positive integer solutions are $(b-2, c-2) = (1,4)$ and $(2,2)$, giving $(a, b, c) = (1, 3, 6)$ or $(1, 4, 4)$. Thus, $abc = 18$ or $16$.\n\n**Case 2:** $a = 2$\n\nThen $4b + 4c + 2bc = 6bc$, so $bc - b - c = 0$. Rearranging, $(b-1)(c-1) = 1$, which has the solution $b = 2$, $c = 2$. Thus, $abc = 8$.\n\n**Case 3:** $a \\geq 3$\n\nThen $3abc \\geq 9bc > 2ab + 2bc + 2ca$, so there are no further solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16925,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(a, b, c)$ be any integer triple with $a, b, c \\geq 2015$, except for $(2015, 2015, 2015)$. Two operations are allowed:\n\n1. Choose two piles with even numbers of coins, say $x$ and $y$, and replace them with $\\frac{x}{2}$ and $x + \\frac{y}{2}$ (the third pile remains unchanged).\n2. Increase the total number of coins by 1 in some way.\n\nShow that, starting from any such triple (other than $(2015, 2015, 2015)$), it is always possible to reach a state where at least one pile contains at least $2017^{2017}$ coins after a finite number of operations.",
    "options": [],
    "answer": "See solution",
    "solution": "It is obvious that no operation can be carried out in the exceptional case $(2015, 2015, 2015)$. For the other cases, we claim that it is always possible to increase the total number of coins in some operations. If this is true, then the total number of coins will exceed $3 \\times 2017^{2017}$ after some operations, and hence there is a pile with at least $2017^{2017}$ coins.\n\nConsider any integer triple $(a, b, c)$ with $S = a + b + c \\geq 6046$ (which is $3 \\times 2015 + 1$). Note that operation (1) does not change the sum $S$, while operation (2) always increases $S$ by 1. Since we always have $\\left\\lfloor \\frac{S}{3} \\right\\rfloor > 2015$, there must be a pile with at least $2016$ coins after each operation, and hence one of the operations can be carried out.\n\nSuppose on the contrary that there exists a case with $S \\geq 6046$ such that $S$ can no longer be increased. Thus, only operation (1) is possible, which means the number of coins in each pile is always even or at most $2015$. Applying operation (1) if necessary, we may assume there is an empty pile. Afterwards, we apply operation (1) repeatedly to the nonempty pile with the smallest number of coins until this is not possible. The number of coins of the nonempty pile with the smallest number of coins is halved in each operation. Therefore, it must become an odd number after some steps.\n\nWe are now in a situation where the number of coins are $0$, $x$, $y$ respectively, where $x$ is odd and $0 < x \\leq 2015 < y$. As $x + y \\geq 6046$, we deduce $y \\geq 4031$. Note that $y$ must be even since otherwise no more operation (1) can be carried out. Hence, we have $y \\geq 4032$. Next, we apply operation (1) on the pile with $y$ coins. The number of coins becomes $0, \\frac{y}{2}, x + \\frac{y}{2}$. Each of the nonempty piles contains at least $\\frac{y}{2} \\geq 2016$ coins. As $S = x + y$ is odd, one of these piles has an odd number of coins. This contradicts our assumption that there cannot be an odd pile with more than $2016$ coins.\n\nTherefore, the total number of coins can always be increased. The goal can be attained in a finite number of operations whenever the total number of coins is at least $6046$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16926,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\Gamma$ be a circle with centre $I$, and $ABCD$ be a convex quadrilateral such that each of the segments $AB$, $BC$, $CD$, and $DA$ is tangent to $\\Gamma$. Let $\\Omega$ be the circumcircle of triangle $AIC$. The extension of $BA$ beyond $A$ meets $\\Omega$ at $X$, and the extension of $BC$ beyond $C$ meets $\\Omega$ at $Z$. The extensions of $AD$ and $CD$ beyond $D$ meet $\\Omega$ at $Y$ and $T$, respectively. Prove that\n\n$$\nAD + DT + TX + XA = CD + DY + YZ + ZC.\n$$\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p413_data_1d3a798a27.png)\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p413_data_2e576772b2.png)",
    "options": [],
    "answer": "See solution",
    "solution": "To begin, notice that $I$ is the intersection of the external angle bisector of $\\angle TCZ$ and the circumcircle $\\Omega$ of $\\triangle TCZ$. Thus, $I$ is the midpoint of the arc $TCZ$, so $IT = IZ$. Likewise, $I$ is the midpoint of the arc $YAX$, so $IX = IY$. Let $O$ be the centre of $\\Omega$. The pairs $X$ and $Y$, $T$ and $Z$ are symmetric about the line $IO$, hence $XT = YZ$.\n\nLet the inscribed circle of $ABCD$ touch the four sides at $P$, $Q$, $R$, and $S$, respectively, as illustrated in Fig. 4.1. Since $IP = IS$, $IX = IY$, the right triangles $IXP$ and $IYS$ are congruent. Likewise, the right triangles $IRT$ and $IQZ$ are congruent. They imply $XP = YS$, $RT = QZ$.\n\nFinally, from $AS = AP$, $CQ = RC$, $SD = DR$, we obtain\n\n$$\n\\begin{aligned}\nAD + DT + TX + XA &= TX + XP + RT \\\\\n&= YZ + SY + QZ \\\\\n&= CD + DY + YZ + ZC,\n\\end{aligned}\n$$\n\nas desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16927,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that a ball with radius $1$ moves freely inside a regular tetrahedron with edge length $4\\sqrt{6}$. What is the area of the inner surface of the container that the ball can never touch?",
    "options": [],
    "answer": "See solution",
    "solution": "As shown in Fig. 1, consider the situation where the\n\n![alt](images/Mathematical_Olympiad_in_China_2009-2010_p38_data_0fb6a03ff3.png)\n\nthe ball is in a corner of the container. Draw the plane $A_1B_1C_1 \\parallel ABC$, tangent to the ball at point $D$. The ball center $O$ is also the center of the tetrahedron $P-A_1B_1C_1$, with $PO \\perp A_1B_1C_1$ and the foot point $D$ being the center of $\\triangle A_1B_1C_1$.\n\nSince\n\n$$\n\\begin{aligned}\nV_{P-A_1B_1C_1} &= \\frac{1}{3} S_{\\triangle A_1B_1C_1} \\cdot PD \\\\\n&= 4 V_{O-A_1B_1C_1} \\\\\n&= 4 \\cdot \\frac{1}{3} S_{\\triangle A_1B_1C_1} \\cdot OD,\n\\end{aligned}\n$$\n\nwe have $PD = 4OD = 4r$, where $r$ is the radius of the ball. It follows that\n\n$$\nPO = PD - OD = 3r.\n$$\n\nSuppose that the ball is tangent to the plane $PAB$ at point $P_1$. Then\n\n$$\nPP_1 = \\sqrt{PO^2 - OP_1^2} = \\sqrt{(3r)^2 - r^2} = 2\\sqrt{2}r.\n$$\n\nAs shown in Fig. 2, the locus of the ball on the plane $PAB$ is also a regular triangle, denoted by $P_1EF$. Through $P_1$ draw $P_1M \\perp PA$ with point $M$ on $PA$. Then $\\angle MPP_1 = \\frac{\\pi}{6}$, and\n\n$$\nPM = PP_1 \\cdot \\cos \\angle MPP_1 = 2\\sqrt{2}r \\cdot \\frac{\\sqrt{3}}{2} = \\sqrt{6}r.\n$$\n\n![alt](images/Mathematical_Olympiad_in_China_2009-2010_p38_data_0a9ff4090f.png)\n\nIt follows that $P_1E = PA - 2PM = a - 2\\sqrt{6}r$, where $a = PA$. The space on $PAB$ which the ball will never touch is the shaded part of Fig. 2, and its size is\n\n$$\n\\begin{aligned}\nS_{\\triangle PAB} - S_{\\triangle P_1EF} &= \\frac{\\sqrt{3}}{4} \\left(a^2 - (a - 2\\sqrt{6}r)^2 \\right) \\\\\n&= 3\\sqrt{2}ar - 6\\sqrt{3}r^2 \\\\\n&= 24\\sqrt{3} - 6\\sqrt{3} \\\\\n&= 18\\sqrt{3},\n\\end{aligned}\n$$\n\nsince $r = 1$ and $a = 4\\sqrt{6}$ under the given conditions. The total untouched area is\n\n$$\n4 \\times 18\\sqrt{3} = 72\\sqrt{3}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16928,
    "subject": "Mathematics (Olympiad)",
    "question": "On the Cartesian plane, consider the graph $C$ of the function $y = \\sqrt[3]{x^2}$. An arbitrary line $d$ cuts $C$ at three distinct points with $x$-coordinates $x_1, x_2$, and $x_3$. Prove that:\n\na) The following quantity is a constant:\n\n$$\n\\sqrt[3]{\\frac{x_1x_2}{x_3^2}} + \\sqrt[3]{\\frac{x_2x_3}{x_1^2}} + \\sqrt[3]{\\frac{x_3x_1}{x_2^2}}.\n$$\n\nb) Show that:\n\n$$\n\\sqrt[3]{\\frac{x_1^2}{x_2x_3}} + \\sqrt[3]{\\frac{x_2^2}{x_3x_1}} + \\sqrt[3]{\\frac{x_3^2}{x_1x_2}} < -\\frac{15}{4}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "a) The line $d$ cannot be parallel to the $y$-axis (otherwise it would intersect $C$ at at most one point), so $d$ has the form $y = ax + b$ where $a, b \\in \\mathbb{R}$ and $a \\neq 0$. The $x$-coordinates of the intersections of $d$ and $C$ are roots of\n\n$$\n\\sqrt[3]{x^2} = ax + b.\n$$\n\nClearly, $ab \\neq 0$, otherwise $d$ can only cut $C$ at at most two points, which is a contradiction. Let $t = \\sqrt[3]{x}$, so the equation becomes\n\n$$\n at^3 - t^2 + b = 0\n$$\n\nwhich has three distinct solutions $t_1, t_2, t_3$ and $t_1 t_2 t_3 \\neq 0$ (since $b \\neq 0$). By Vieta's theorem for the cubic, $t_1 t_2 + t_2 t_3 + t_3 t_1 = 0$, which implies\n\n$$\n(t_1 t_2)^3 + (t_2 t_3)^3 + (t_3 t_1)^3 = 3 (t_1 t_2)(t_2 t_3)(t_3 t_1) = 3 t_1^2 t_2^2 t_3^2.\n$$\n\nWe can rewrite this as\n\n$$\n\\frac{t_2 t_3}{t_1^2} + \\frac{t_3 t_1}{t_2^2} + \\frac{t_1 t_2}{t_3^2} = 3,\n$$\n\nor equivalently,\n\n$$\n\\sqrt[3]{\\frac{x_2 x_3}{x_1^2}} + \\sqrt[3]{\\frac{x_3 x_1}{x_2^2}} + \\sqrt[3]{\\frac{x_1 x_2}{x_3^2}} = 3.\n$$\n\nSo the given quantity is a constant.\n\nb) Without loss of generality, assume $t_1$ and $t_2$ have the same sign.\n\nFrom $t_1 t_2 + t_2 t_3 + t_3 t_1 = 0$, we have $t_3 = -\\frac{t_1 t_2}{t_1 + t_2}$. Thus,\n\n$$\n\\sqrt[3]{\\frac{x_1^2}{x_2 x_3}} + \\sqrt[3]{\\frac{x_2^2}{x_3 x_1}} + \\sqrt[3]{\\frac{x_3^2}{x_1 x_2}} = \\frac{t_1^2}{t_2 t_3} + \\frac{t_2^2}{t_3 t_1} + \\frac{t_3^2}{t_1 t_2} \\\\ = -(t_1 + t_2) \\left( \\frac{t_1}{t_2^2} + \\frac{t_2}{t_1^2} \\right) + \\frac{t_1 t_2}{(t_1 + t_2)^2} \\\\ = - \\left( \\frac{t_1^2}{t_2^2} + \\frac{t_2^2}{t_1^2} + \\frac{t_1}{t_2} + \\frac{t_2}{t_1} \\right) + \\frac{t_1 t_2}{(t_1 + t_2)^2}.\n$$\n\nBy the AM-GM inequality,\n\n$$\n\\frac{t_1^2}{t_2^2} + \\frac{t_2^2}{t_1^2} \\geq 2, \\quad \\frac{t_1}{t_2} + \\frac{t_2}{t_1} \\geq 2, \\quad \\frac{t_1 t_2}{(t_1 + t_2)^2} \\leq \\frac{1}{4}.\n$$\n\nTherefore,\n\n$$\n\\sqrt[3]{\\frac{x_1^2}{x_2 x_3}} + \\sqrt[3]{\\frac{x_2^2}{x_3 x_1}} + \\sqrt[3]{\\frac{x_3^2}{x_1 x_2}} \\leq -(2 + 2) + \\frac{1}{4} = -\\frac{15}{4}.\n$$\n\nSince $t_1, t_2$ are distinct, equality does not occur. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 16929,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n$ boys $a_1, a_2, \\dots, a_n$ and $n$ girls $b_1, b_2, \\dots, b_n$ at a party. No boys shook hands with each other, no girls shook hands with each other, and for each $i \\in \\{1, 2, \\dots, n\\}$, $a_i$ did not shake hands with $b_i$.\n\nWe want to divide all $2n$ people into groups satisfying the following two conditions:\n\n1. In each group, the number of boys is equal to the number of girls.\n2. In each group, no two people shook hands with each other.\n\nLet $m$ be the number of pairs $(a_i, b_j)$ where $a_i$ shook hands with $b_j$. Prove that it is possible to make the number of groups less than or equal to $2$ or $\\frac{2m}{n} + 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $G$ be a graph on $2n$ vertices representing $n$ boys and $n$ girls, where two vertices are adjacent if they shook hands with each other.\n\n**Case 1:** $m < n$.\n\nLet $C_1, C_2, \\dots, C_k$ be the components of $G$. Since $G$ has $2n$ vertices and fewer than $n$ edges, the number of components is greater than $n$. The sums $\\sum_{i=1}^t |C_i|$ for $1 \\le t \\le n+1$ are distinct positive integers less than or equal to $2n$. By the pigeonhole principle, there exist $a < b$ such that $\\sum_{i=1}^a |C_i| \\equiv \\sum_{i=1}^b |C_i| \\pmod{n}$. Therefore, $|C_{a+1}| + |C_{a+2}| + \\dots + |C_b| = n$. Let $A = C_{a+1} \\cup C_{a+2} \\cup \\dots \\cup C_b$. Since $|A| = n$, the number of boys in $A$ equals the number of girls not in $A$. Thus, we can make one group consisting of boys in $A$ and girls not in $A$, and another group consisting of boys not in $A$ and girls in $A$. This gives $2$ groups satisfying the conditions.\n\n**Case 2:** $m \\ge n$.\n\nSince $a_i$ is not adjacent to $b_i$ for all $i$, there is at least one way to partition people into groups satisfying the conditions. Let $s$ be the minimum number of groups. Let $V_1, V_2, \\dots, V_s$ be the groups partitioning all $2n$ people.\n\nFor all $1 \\le i < j < k \\le s$, if we could merge $V_i, V_j, V_k$ into two groups, this would contradict the minimality of $s$. Thus, by Case 1, the number of edges among $V_i, V_j, V_k$ is at least $\\frac{|V_i \\cup V_j| + |V_k|}{2}$. Let $S$ be the number of edges among all such triples. Then\n\n$$\nS \\ge \\frac{1}{2} \\binom{s-1}{2} \\sum_{i=1}^{s} |V_i| = \\frac{1}{2} \\binom{s-1}{2} (2n).\n$$\n\nOn the other hand, each edge is counted $s-2$ times in $S$. By double counting, $S = (s-2)m$. Thus, $m \\ge \\frac{s-1}{2}n$, so $s \\le \\frac{2m}{n} + 1$.\n\n$\\boxed{}$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 16930,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, points $M, N \\in (AB)$, $P, Q \\in (BC)$, and $S, R \\in (AC)$ are taken such that $AM = CR$, $AN = CS$, $\\overline{MQB} \\equiv \\overline{RQC}$, and $\\overline{NPB} \\equiv \\overline{SPC}$.\n\nShow that if $MQ + QR = NP + PS$, then triangle $ABC$ is isosceles.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $R'$ and $S'$ as the reflections of $R$ and $S$ across the line $BC$. Then $RQ = R'Q$ and $\\overline{RQC} \\equiv \\overline{R'QC}$. Since $\\overline{MQB} \\equiv \\overline{RQC}$, it follows that $\\overline{MQB} \\equiv \\overline{R'QC}$, and because $B, Q, C$ are collinear, so are $M, Q, R'$. This yields $MQ + QR = MQ + QR' = MR'$. Similarly, $NP + PS = NP + PS' = NS'$. Now $MQ + QR = NP + PS$ implies $MR' = NS'$.\n\nThe symmetry also implies $\\overline{RCQ} \\equiv \\overline{R'CQ}$ and $\\overline{SCP} \\equiv \\overline{S'CP}$, so $\\overline{RCQ} = \\overline{SCP}$ implies $\\overline{R'CQ} = \\overline{S'CP}$, whence the points $C, R', S'$ are collinear. Also,\n\n$$CR = CR' \\quad \\text{and} \\quad CS = CS'$$\nand, since $CR = AM$ and $CS = AN$, $MN = R'S'$.\n\n![](images/RMC2014_p28_data_98933234bd.png)\n\nTherefore, $\\triangle NS'M \\equiv \\triangle R'MS'$ (by SSS), whence $\\overline{NMS'} \\equiv \\overline{MS'R'}$.\nThis shows that $AB \\parallel CS'$, therefore $\\overline{ABC} \\equiv \\overline{BCS'}$. Now $\\overline{BCS'} \\equiv \\overline{BCA}$\nimplies $\\overline{ABC} \\equiv \\overline{ACB}$, q.e.d.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16931,
    "subject": "Mathematics (Olympiad)",
    "question": "已知 $f(x)$ 為整係數多項式，滿足：對任意正整數 $n$，$f(n)$ 非零且 $f(n)$ 至多有 2011 個質因數不是 $n$ 的質因數。\n\n試證明 $f$ 可以表示成 $f(x) = c x^k$，其中 $c$ 為整數，$k$ 為非負整數。",
    "options": [],
    "answer": "See solution",
    "solution": "對於題目所述的 $f$，若 $f$ 的常數項為 $0$，則可考慮 $g(x) = f(x)/x$ 亦滿足題目條件，因此可假設 $f$ 的常數項非零，並僅須證明此時 $f$ 是常數多項式。\n\n令 $P = \\{ p \\mid \\exists\\ n \\in \\mathbb{Z}^+,\\ p \\text{ 為質數},\\ p \\mid f(n),\\ p \\nmid n \\}$。我們宣稱 $P$ 至多包含 $2011$ 個質數。\n\n若不然，用反證法，令 $p_1, \\ldots, p_{2012}$ 為 $P$ 中的相異質數，令正整數 $n_1, \\ldots, n_{2012}$ 分別滿足 $p_i \\mid f(n_i)$ 且 $p_i \\nmid n_i$。\n\n由中國剩餘定理，存在正整數 $n$ 使得 $n \\equiv n_i \\pmod{p_i}$，$i = 1, \\ldots, 2012$。由於 $n - n_i \\mid f(n) - f(n_i)$，可知 $p_1 p_2 \\cdots p_{2012} \\mid f(n)$。同時由定義有 $n$ 與 $p_1 p_2 \\cdots p_{2012}$ 互質，矛盾。因此 $P$ 至多包含 $2011$ 個質數，記 $P = \\{ p_1, \\ldots, p_k \\}$。\n\n令 $f(x) = a + xg(x)$，即 $a$ 為 $f$ 的非零常數項，$g$ 為某整係數多項式。假設 $f$ 不是常數多項式，則 $g$ 非零。此時令 $m = 3a^2 p_1 p_2 \\cdots p_k$。則\n\n$$\nf(m) = a(1 + 3a p_1 p_2 \\cdots p_k g(m))\n$$\n\n顯然 $1 + 3a p_1 p_2 \\cdots p_k g(m)$ 與 $m$ 互質，且\n\n$$\n|1 + 3a p_1 p_2 \\cdots p_k g(m)| \\geq |3a p_1 p_2 \\cdots p_k g(m)| - 1 \\geq 3 - 1 > 1.\n$$\n\n故 $1 + 3a p_1 p_2 \\cdots p_k g(m)$ 至少有一個 $m$ 所沒有的質因數，與 $P$ 定義矛盾，故 $f$ 必須是常數多項式。證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16932,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDEF$ be a convex hexagon in which the diagonals $AD$, $BE$, $CF$ are concurrent at $O$. Suppose the area of triangle $OAF$ is the geometric mean of those of $OAB$ and $OEF$; and the area of triangle $OBC$ is the geometric mean of those of $OAB$ and $OCD$. Prove that the area of triangle $OED$ is the geometric mean of those of $OCD$ and $OEF$.\n\n![](images/Indija_mo_2011_p0_data_21c23329e6.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $OA = a$, $OB = b$, $OC = c$, $OD = d$, $OE = e$, $OF = f$, $[OAB] = x$, $[OCD] = y$, $[OEF] = z$, $[ODE] = u$, $[OFA] = v$ and $[OBC] = w$. We are given that $v^2 = zx$, $w^2 = xy$ and we have to prove that $u^2 = yz$. Since $\\angle AOB = \\angle DOE$, we have\n\n$$\n\\frac{u}{x} = \\frac{\\frac{1}{2} de \\sin \\angle DOE}{\\frac{1}{2} ab \\sin \\angle AOB} = \\frac{de}{ab}.\n$$\n\n$$\n\\frac{v}{y} = \\frac{fa}{cd}, \\quad \\frac{w}{z} = \\frac{bc}{ef}.\n$$\n\nMultiplying these three equalities, we get $uvw = xyz$. Hence\n\n$$\nx^2 y^2 z^2 = u^2 v^2 w^2 = u^2 (zx)(xy).\n$$\n\nThis gives $u^2 = yz$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16933,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $b_i, c_i$, $0 \\leq i \\leq 100$ be two sequences of positive integers with two exceptions: $c_0 = 0$, $b_{100} = 0$. Several villages are connected by roads, each road connects two villages (called neighbours) and has length 1 km. Roads do not intersect each other, but can pass over/under each other. The distance between two villages $X$ and $Y$ is the length of the shortest path between them. In this country, the maximal distance between two villages equals 100 km, and for every pair of villages $(X, Y)$ (the case $X = Y$ is allowed), the following condition holds: if the distance between $X$ and $Y$ is $k$ km, then there are exactly $b_k$ (respectively, $c_k$) neighbours of $Y$ that are 1 km further from (respectively, closer to) $X$ than $Y$. Show that the number\n\n$$\n\\frac{b_0 b_1 \\dots b_{99}}{c_1 c_2 \\dots c_{100}}\n$$\n\nis an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $z$ be an arbitrary village, and let $S_i(z)$ be the set of villages at distance $i$ from $z$. Let $k_i$ be the number of elements in $S_i(z)$. Then the sequence $k_i$ does not depend on $z$.\n\nWe prove this by induction on $i$. We will also show that $k_{i+1} = k_i \\cdot \\frac{b_i}{c_{i+1}}$. Clearly, $k_0 = 1$ (the base case). For the induction step, count the roads between $S_i(z)$ and $S_{i+1}(z)$ in two ways: choosing a village $x$ in $S_i(z)$ and a road to $S_{i+1}(z)$ gives $k_i b_i$ ways; choosing a village $x$ in $S_{i+1}(z)$ and a road to $S_i(z)$ gives $k_{i+1} c_{i+1}$ ways. Therefore, $k_i b_i = k_{i+1} c_{i+1}$, so $k_{i+1} = k_i \\cdot \\frac{b_i}{c_{i+1}}$.\n\nIterating, $k_{100} = \\frac{b_0 b_1 \\dots b_{99}}{c_1 c_2 \\dots c_{100}}$. Since $k_{100}$ is an integer, the given expression is an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16934,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral with $\\angle DAC = \\angle BDC = 36^\\circ$, $\\angle CBD = 18^\\circ$, and $\\angle BAC = 72^\\circ$. The diagonals $AC$ and $BD$ intersect at point $P$. Determine the measure of $\\angle APD$.",
    "options": [],
    "answer": "See solution",
    "solution": "On the rays $DA$ and $BA$, take points $E$ and $Z$, respectively, such that $AC = AE = AZ$.\n\nSince $\\angle DEC = \\frac{\\angle DAC}{2} = 18^\\circ = \\angle CBD$, the quadrilateral $DEBC$ is cyclic.\n\nSimilarly, the quadrilateral $CBZD$ is cyclic, because\n\n$$\n\\angle AZC = \\frac{\\angle BAC}{2} = 36^\\circ = \\angle BDC.\n$$\n\n![](images/Greek_07_Booklet_p27_data_a5a517aed6.png)\n\nFigure 8\n\n![](images/Greek_07_Booklet_p27_data_1d307d3ff5.png)\n\nFigure 9\n\nTherefore, the pentagon $BCDZE$ is inscribed in the circle $K(A, AC)$. It follows that $AC = AD$ and $\\angle ACD = \\angle ADC = \\frac{180^\\circ - 36^\\circ}{2} = 72^\\circ$, which gives $\\angle ADP = 36^\\circ$ and $\\angle APD = 108^\\circ$.\n\nAlternative solution: See the figure on the right. If $C_1$ is the symmetric point of $D$ with respect to the line $BC$, then $ABC_1D$ is cyclic. Since $\\angle BAC_1 = 72^\\circ$ and $\\angle BDD_1 = 36^\\circ$, we have that $DD_1$ is the bisector of the angle $BDC_1$ and hence $C_1$ is the incenter of triangle $BDC_1$. Therefore,\n\n$$\n\\angle DC_1A = \\angle AC_1B = 36^\\circ \\quad \\text{and} \\quad \\angle DPA = \\frac{72^\\circ + 144^\\circ}{2} = 108^\\circ.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16935,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a non-constant arithmetic progression $a_1, a_2, \\dots, a_n, \\dots$. Suppose there exist relatively prime positive integers $p > 1$ and $q > 1$ such that $a_1^2$, $a_{p+1}^2$, and $a_{q+1}^2$ are also terms of the same arithmetic progression. Prove that the terms of the arithmetic progression are all integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us take $a_1 = a$. We have\n$$\na^2 = a + k d, \\quad (a + p d)^2 = a + l d, \\quad (a + q d)^2 = a + m d.\n$$\nThus we have\n$$\na + l d = (a + p d)^2 = a^2 + 2 p a d + p^2 d^2 = a + k d + 2 p a d + p^2 d^2.\n$$\nSince we have a non-constant AP, $d \\neq 0$. Hence we obtain $2 p a + p^2 d = l - k$. Similarly, $2 q a + q^2 d = m - k$. Observe that $p^2 q - p q^2 \\neq 0$. Otherwise $p = q$ and $\\gcd(p, q) = p > 1$, which contradicts the given hypothesis that $\\gcd(p, q) = 1$. Hence we can solve the two equations for $a, d$:\n$$\na = \\frac{p^2 (m - k) - q^2 (l - k)}{2 (p^2 q - p q^2)}, \\quad d = \\frac{q (l - k) - p (m - k)}{p^2 q - p q^2}.\n$$\nIt follows that $a, d$ are rational numbers. We also have\n$$\np^2 a^2 = p^2 a + k p^2 d.\n$$\nBut $p^2 d = l - k - 2 p a$. Thus we get\n$$\np^2 a^2 = p^2 a + k (l - k - 2 p a) = (p - 2 k) p a + k (l - k).\n$$\nThis shows that $p a$ satisfies the equation\n$$\nx^2 - (p - 2 k) x - k (l - k) = 0.\n$$\nSince $a$ is rational, $p a$ is rational. Write $p a = w / z$, where $w$ is an integer and $z$ is a natural number such that $\\gcd(w, z) = 1$. Substituting in the equation, we obtain\n$$\nw^2 - (p - 2 k) w z - k (l - k) z^2 = 0.\n$$\nThis shows $z$ divides $w$. Since $\\gcd(w, z) = 1$, it follows that $z = 1$ and $p a = w$ is an integer. (In fact, any rational solution of a monic polynomial with integer coefficients is necessarily an integer.) Similarly, we can prove that $q a$ is an integer. Since $\\gcd(p, q) = 1$, there are integers $u$ and $v$ such that $p u + q v = 1$. Therefore $a = (p a) u + (q a) v$. It follows that $a$ is an integer. But $p^2 d = l - k - 2 p a$. Hence $p^2 d$ is an integer. Similarly, $q^2 d$ is also an integer. Since $\\gcd(p^2, q^2) = 1$, it follows that $d$ is an integer. Combining these two, we see that all the terms of the AP are integers.\n\n**Alternatively**, we can prove that $a$ and $d$ are integers in another way. We have seen that $a$ and $d$ are rationals; and we have three relations:\n$$\na^2 = a + k d, \\quad p^2 d + 2 p a = n_1, \\quad q^2 d + 2 q a = n_2,\n$$\nwhere $n_1 = l - k$ and $n_2 = m - k$. Let $a = u / v$ and $d = x / y$ where $u, x$ are integers and $v, y$ are natural numbers, and $\\gcd(u, v) = 1$, $\\gcd(x, y) = 1$. Putting this in these relations, we obtain\n$$\nu^2 y = u v y + k x v^2, \\qquad (1)\n$$\n$$\n2 p u y + p^2 v x = v y n_1, \\qquad (2)\n$$\n$$\n2 q u y + q^2 v x = v y n_2. \\qquad (3)\n$$\nNow (1) shows that $v \\mid u^2 y$. Since $\\gcd(u, v) = 1$, it follows that $v \\mid y$. Similarly, (2) shows that $y \\mid p^2 v x$. Using $\\gcd(y, x) = 1$, we get that $y \\mid p^2 v$. Similarly, (3) shows that $y \\mid q^2 v$. Therefore $y$ divides $\\gcd(p^2 v, q^2 v) = v$. The two results $v \\mid y$ and $y \\mid v$ imply $v = y$, since both $v, y$ are positive. Substitute this in (1) to get\n$$\nu^2 = u v + k x v.\n$$\nThis shows that $v \\mid u^2$. Since $\\gcd(u, v) = 1$, it follows that $v = 1$. This gives $v = y = 1$. Finally, $a = u$ and $d = x$ which are integers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16936,
    "subject": "Mathematics (Olympiad)",
    "question": "The base of a regular four-sided prism is a rhombus with area $\\frac{2}{3}k^2$, where $k > 0$. The intersection of the prism with the plane of its smaller diagonal is a square with area $k^2$.\n\n**a)** Express the surface area and volume of the prism using only $k$.\n\n**b)** For which value of $k$ do the surface area and the volume have equal numerical measures?",
    "options": [],
    "answer": "See solution",
    "solution": "Because the intersection with the smaller diagonal plane is a square of area $k^2$, the prism's height is $H = k$ and the smaller base diagonal is $d_1 = k$. The base area is $B = \\frac{2}{3}k^2$, so:\n\n$$\n\\frac{2}{3}k^2 = \\frac{d_1 \\cdot d_2}{2} = \\frac{k \\cdot d_2}{2}\n$$\n\nSolving for the larger diagonal:\n\n$$\nd_2 = \\frac{4}{3}k\n$$\n\nThe side of the rhombus is:\n\n$$\na^2 = \\left(\\frac{d_1}{2}\\right)^2 + \\left(\\frac{d_2}{2}\\right)^2 = \\left(\\frac{k}{2}\\right)^2 + \\left(\\frac{2k}{3}\\right)^2 = \\frac{k^2}{4} + \\frac{4k^2}{9} = \\frac{5k^2}{6}\n$$\n\nSo $a = \\frac{\\sqrt{30}}{6}k$.\n\n**a)** The volume and surface area are:\n\n$$\nV = B \\cdot H = \\frac{2}{3}k^2 \\cdot k = \\frac{2}{3}k^3\n$$\n\n$$\nP = 2B + 4aH = 2 \\cdot \\frac{2}{3}k^2 + 4 \\cdot \\frac{\\sqrt{30}}{6}k \\cdot k = \\frac{4}{3}k^2 + \\frac{2\\sqrt{30}}{3}k^2 = \\frac{4 + 2\\sqrt{30}}{3}k^2\n$$\n\n![](images/Makedonija_2008_p21_data_8da5dce2d1.png)\n\n**b)** Setting $V = P$:\n\n$$\n\\frac{2}{3}k^3 = \\frac{4 + 2\\sqrt{30}}{3}k^2\n$$\n\n$$\nk = 2 + \\sqrt{30}\n$$\n\nSo, for $k = 2 + \\sqrt{30}$, the surface area and volume have equal numerical measures.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 16937,
    "subject": "Mathematics (Olympiad)",
    "question": "對於平面上有限多個三角形所成的集合，若其中任兩個三角形內部的交集非空，則稱這個三角形集合*相交*。\n\n證明：對於平面上兩組相交的三角形集合，必存在一條直線同時通過兩個集合中所有三角形的內部。",
    "options": [],
    "answer": "See solution",
    "solution": "取平面上任意點 $P$，令 $\\ell_\\theta$ 為過 $P$ 且與 $x$ 軸正向夾 $\\theta$ 角的有向直線。注意到對於第 $i$ 個集合（$i = 1,2$），第 $i$ 個集合中的每個三角形在 $\\ell_\\theta$ 上的投影都是一個有限區間，且由相交性知道這些區間兩兩相交。這意味著第 $i$ 個集合中所有三角形投影的交集也是一個有限區間（可用凸性證明或數歸$^1$），令 $d_i(\\theta)$ 為從 $P$ 到第 $i$ 個集合投影所成區間中點的有向距離（依據 $\\ell_\\theta$ 的方向）。\n\n現在，注意到 $d_i(\\theta)$ 為 $\\theta$ 的連續函數，且 $d_i(\\theta + \\pi) = -d_i(\\theta)$。這意味著如果有 $d_1(\\theta) > d_2(\\theta)$，則有 $d_1(\\theta + \\pi) < d_2(\\theta + \\pi)$，因此由中值定理知存在 $\\varphi \\in (\\theta, \\theta + \\pi)$ 讓 $d_1(\\varphi) = d_2(\\varphi)$，而這意味著兩個集合的所有三角形在 $\\ell_\\varphi$ 上的投影交集非空。選該交集中的任意點 $Q$，則易知過 $Q$ 且垂直 $\\ell_\\varphi$ 的直線將通過兩個集合的所有三角形內部。相同討論適用於 $d_1(\\theta) < d_2(\\theta)$ 之情況，而 $d_1(\\theta) = d_2(\\theta)$ 之情況顯然，故證畢。\n\n$^1$ 不失一般性考慮 $\\theta = 0$。考慮第 $n$ 個三角形的投影，若其與前 $n-1$ 個三角形投影的交集 $[a,b]$ 沒有交集，則它必須是全部落在 $S$ 某一側的區間，不失一般性假設其為 $[c,d]$ 其中 $c > b$。但又因為這 $n-1$ 個三角形投影都與 $[c,d]$ 有交集，因此這些三角形的投影 $[x,y]$ 必有 $x \\le a < b < c \\le y$。但這就表示前 $n-1$ 個三角形的交集至少要是 $[a,c]$，矛盾。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16938,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, a_3, \\dots$ be a sequence of positive real numbers. Suppose that for some positive integer $s$, we have\n$$\na_n = \\max\\{a_k + a_{n-k} \\mid 1 \\le k \\le n-1\\}\n$$\nfor all $n > s$. Prove that there exist positive integers $\\ell$ and $N$ with $\\ell \\le s$ such that $a_n = a_\\ell + a_{n-\\ell}$ for all $n \\ge N$.",
    "options": [],
    "answer": "See solution",
    "solution": "We generalize to the setting where the $a_n$ may assume negative values. For any $r \\in \\mathbb{R}$, note that the transformation $a_n \\mapsto a_n + r n$ does not change the problem conditions or the result to be proved. Picking $\\ell \\le s$ such that $a_\\ell/\\ell$ is maximal, we can thus assume without loss of generality that $a_\\ell = 0$. This means all of $a_1, \\dots, a_s$ are non-positive, hence all $a_n$ are non-positive. Let $b_n = -a_n \\ge 0$. For $n > s$, we have\n$$\nb_n = \\min\\{b_k + b_{n-k} \\mid 1 \\le k \\le n-1\\}\n$$\nand in particular $b_n \\le b_{n-\\ell} + b_\\ell = b_{n-\\ell}$.\n\nFrom this, we draw two conclusions. First, all $b_n$ must be bounded above by $M = \\max\\{b_1, \\dots, b_s\\}$. Second, if we let $S$ be the set of all linear combinations of the form $c_1 b_1 + c_2 b_2 + \\dots + c_s b_s$, where the $c_i$ are nonnegative integers, and let $T = \\{x \\le M : x \\in S\\}$, then since $b_n = \\min\\{b_k + b_{n-k} \\mid 1 \\le k \\le n-1\\}$, it is clear that every $b_n$ must be in $T$. Crucially, $T$ is a finite set.\n\nNow, for each integer $i$ satisfying $\\ell i + 1 > s$, let $\\beta_i$ denote the $\\ell$-tuple $(b_{\\ell i+1}, b_{\\ell i+2}, \\dots, b_{\\ell i+\\ell})$. By the previous paragraph, the number of such $\\ell$-tuples is at most $|T|^\\ell$, a finite number. Further, because $b_n \\le b_{n-\\ell}$ for $n > s$, the individual indices of these $\\beta_i$ are non-increasing functions of $i$. Thus, there can only be finitely many $i$ for which $\\beta_i \\ne \\beta_{i+1}$. Let $i_0$ be greater than the largest such value; then, all $\\ell$-tuples $\\beta_i$ with $i \\ge i_0$ are identical. Choosing $N = \\ell(i_0+1)$ finishes the problem, since any $n \\ge N$ gives $b_n = b_{n-\\ell} = b_\\ell + b_{n-\\ell}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16939,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for all real numbers $x$ and $y$,\n\n$$\nf(xy + f(x^2)) = x f(x + y).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $f(x) = 0$ for all real numbers $x$ or $f(x) = x$ for all real numbers $x$.\n\nLet $P(x, y)$ denote the assertion of the given equation.\n\n**Case 1.** The function is not periodic.\n\nConsider $P(1, x - 1)$:\n$$\nf(x - 1 + f(1)) = f(x)\n$$\nThis implies $-1 + f(1) = 0 \\implies f(1) = 1$.\n\nNow, $P(-1, x + 1)$ gives:\n$$\nf(-x - 1 + f(1)) = -f(x) \\implies f(-x) = -f(x)\n$$\nSo $f$ is odd and $f(0) = 0$.\n\n*Claim.* If $f(a) = 0$, then $a = 0$.\n\n*Proof.* Suppose $a \\neq 0$. Since $f$ is odd, assume $a > 0$. Take $P(\\sqrt{a}, x + \\sqrt{a})$:\n$$\nf(\\sqrt{a}(x + \\sqrt{a}) + f(a)) = \\sqrt{a} f(\\sqrt{a} + x + \\sqrt{a}) \\implies \\frac{f(\\sqrt{a} x + a)}{\\sqrt{a}} = f(x + 2\\sqrt{a})\n$$\nSimilarly, $P(-\\sqrt{a}, x + \\sqrt{a})$ and oddness give:\n$$\n\\begin{aligned}\nf(-\\sqrt{a}(x + \\sqrt{a})) &= f(-\\sqrt{a}(x + \\sqrt{a}) + f(a)) = -\\sqrt{a} f(-\\sqrt{a} + x + \\sqrt{a}) \\\\\n&= -\\sqrt{a} f(x) \\implies \\frac{f(\\sqrt{a} x + a)}{\\sqrt{a}} = f(x)\n\\end{aligned}\n$$\nThus, $f(x + 2\\sqrt{a}) = f(x)$ for all $x$, so $2\\sqrt{a} = 0 \\implies a = 0$, a contradiction. $\\square$\n\nSince $f(a) = 0 \\iff a = 0$, take $P(x, -x)$:\n$$\nf(-x^2 + f(x^2)) = x f(0) = 0 \\implies -x^2 + f(x^2) = 0 \\implies f(x^2) = x^2\n$$\nAs $f$ is odd, $f(x) = x$ for all $x$.\n\n**Case 2.** The function is periodic.\n\nSuppose there exists $p \\neq 0$ such that $f(x + p) = f(x)$ for all $x$.\n\nThen $f(x + y + p) = f(x + y)$ for all $x, y$. From $P(x, y)$ and $P(x, y + p)$:\n$$\nf(xy + f(x^2)) = x f(x + y) = x f(x + y + p) = f(xy + x p + f(x^2))\n$$\nLet $a \\neq b$ be any real numbers. Set $x = \\frac{b - a}{p}$ and $y = \\frac{a - f(x^2)}{x}$ to get $f(a) = f(b)$. Thus, $f$ is constant. Since $f(0) = 0$, only $f(x) = 0$ for all $x$ works.\n\n**Conclusion:** All functions satisfying the condition are $f(x) = x$ or $f(x) = 0$ for all real numbers $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16940,
    "subject": "Mathematics (Olympiad)",
    "question": "Numbers $1, 2, \\ldots, n$ are placed in a row in some order. You may repeatedly perform the following operation: select any two pairs of consecutive elements (with no elements in common) and exchange their positions. Is it possible to obtain a monotone sequence (increasing or decreasing) after a finite number of steps if:\n\na) $n = 2009$;\nb) $n = 2010$?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:**\n- a) Not always.\n- b) Always possible.\n\n**Explanation:**\n\nConsider the concept of *inversions* in a permutation: an inversion is a pair where a larger number precedes a smaller one. The parity (evenness/oddness) of the total number of inversions is preserved by the allowed operation (exchanging two disjoint pairs of consecutive elements). Thus, the set of all permutations splits into two classes: those with even and those with odd inversion parity, and you cannot move between these classes using the allowed operation.\n\nFor the monotone increasing sequence $1, 2, \\ldots, n$, the number of inversions is $0$ (even). For the monotone decreasing sequence $n, n-1, \\ldots, 1$, the number of inversions is $\\binom{n}{2} = \\frac{n(n-1)}{2}$. If this number is even, both monotone sequences are in the same class and can be reached from any starting position in that class. If odd, not all positions can be transformed into a monotone sequence.\n\nFor $n = 2009$, $\\frac{2009 \\times 2008}{2}$ is even, but the parity argument shows not all permutations are reachable; for $n = 2010$, $\\frac{2010 \\times 2009}{2}$ is odd, so every permutation can be transformed into a monotone sequence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16941,
    "subject": "Mathematics (Olympiad)",
    "question": "設所有正有理數的集合為 $Q^+$. 求所有函數 $f: Q^+ \\to Q^+$，使得對任意 $x, y \\in Q^+$，都有\n\n$$\nf(f(x)^2 y) = x^3 f(xy).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "令 $y = 1$，則\n\n$$\nf(f(x)^2) = x^3 f(x).\n$$\n\n若 $f(x) = f(y)$，則\n\n$$\nx^3 = \\frac{f(f(x)^2)}{f(x)} = \\frac{f(f(y)^2)}{f(y)} = y^3,\n$$\n\n故 $x = y$，即 $f(x)$ 為一對一函數。\n\n利用 $f(x)$ 為一對一函數，反覆使用原式可得\n\n$$\n\\begin{align*}\nf(f(xy)^2) &= (xy)^3 f(xy) = x^3 y^3 f(xy) \\\\\n&= y^3 f(f(x)^2 y) = f(f(x)^2 f(y)^2) \\\\\n\\Rightarrow f(xy)^2 &= f(x)^2 f(y)^2 \\\\\n\\Rightarrow f(xy) &= f(x) f(y).\n\\end{align*}\n$$\n\n故 $f(x)$ 為積性函數。原式化簡為\n\n$$\nf(f(x))^2 = x^3 f(x) \\Rightarrow x f(x) = \\left(\\frac{f(f(x))}{x}\\right)^2.\n$$\n\n設\n\n$$\ng_0(x) = \\frac{f(f(x))}{x},\n$$\n\n則 $x f(x) = g_0(x)^2$，且\n\n$$\nf(g_0(x))^2 = g_0(x)^3 \\Rightarrow g_0(x) = \\left(\\frac{f(g_0(x))}{g_0(x)}\\right)^2.\n$$\n\n再設\n\n$$\ng_1(x) = \\frac{f(g_0(x))}{g_0(x)},\n$$\n\n則 $g_0(x) = g_1(x)^2$，且\n\n$$\n\\begin{align*}\nf(g_0(x))^2 &= g_0(x)^3 \\\\\n\\Rightarrow f(g_1(x)^2)^2 = f(g_1(x))^4 = (g_1(x)^2)^3 = g_1(x)^6 \\\\\n\\Rightarrow f(g_1(x))^2 = g_1(x)^3\n\\end{align*}\n$$\n\n設\n\n$$\ng_2(x) = \\frac{f(g_1(x))}{g_1(x)},\n$$\n\n則 $g_1(x) = g_2(x)^2$。如此反覆，得一序列 $\\{g_i(x)\\}$，且 $g_{i+1}(x) = \\sqrt{g_i(x)}$。若 $g_0(x) \\neq 1$，則不可能，故\n\n$$\ng_0(x) = 1, \\forall x \\in Q^+,\n$$\n\n所以\n\n$$\nf(x) = \\frac{g_0(x)}{x} = \\frac{1}{x}.\n$$\n\n代入驗證，知\n\n$$\nf(x) = \\frac{1}{x}\n$$\n\n是原式的唯一解。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16942,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all strictly increasing functions $f: \\mathbb{R} \\to \\mathbb{R}$ that satisfy the following conditions:\n\n$$\n\\begin{aligned}\n&\\bullet \\quad f(x^3 + y^2 + f(y)) = x^2 f(x) + y f(y) + y; \\\\\n&\\bullet \\quad \\text{The equation } y^2 + f(y) = t \\text{ has a solution for every } t \\in \\mathbb{R}.\n\\end{aligned}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Solution does not exist.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16943,
    "subject": "Mathematics (Olympiad)",
    "question": "A tetromino tile is a tile that can be formed by gluing together four unit square tiles, edge to edge. For each positive integer $n$, consider a bathroom whose floor is in the shape of a $2 \\times 2n$ rectangle. Let $T_n$ be the number of ways to tile this bathroom floor with tetromino tiles. For example, $T_2 = 4$ since there are four ways to tile a $2 \\times 4$ rectangular bathroom floor with tetromino tiles, as shown below.\n\n![](images/2020_Australian_Scene_W_p105_data_df1a5a521e.png)\n\nProve that each of the numbers $T_1, T_2, T_3, \\dots$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $F_n$ denote the Fibonacci sequence, defined by $F_0 = 1$, $F_1 = 1$ and $F_{n+1} = F_n + F_{n-1}$ for $n \\ge 1$. We will prove that $T_n = F_n^2$.\n\nConsider a tiling of a $2 \\times 2n$ rectangle with tetrominoes, where $n \\ge 2$. Consider the behaviour of the tiling in the leftmost column of the rectangle. Exactly one of the following three cases must arise:\n\n* The leftmost column is covered by a $2 \\times 2$ square, whose removal leaves one of the $T_{n-1}$ tilings of the $2 \\times 2(n-1)$ rectangle.\n* The leftmost column is covered by two $1 \\times 4$ rectangles, whose removal leaves one of the $T_{n-2}$ tilings of the $2 \\times 2(n-2)$ rectangle.\n* The leftmost column is covered by an $L$-tetromino and resembles one of the following diagrams, or their reflections in a horizontal axis. The areas outlined by dashed lines are tiled with $1 \\times 4$ rectangles. (Observe that there may actually be any non-negative integer number of $1 \\times 4$ rectangles appearing in the diagram.) The top case occurs when the area covers a number of columns that is $0 \\bmod 4$, while the bottom case occurs when the area covers a number of columns that is $2 \\bmod 4$. The removal of these areas leaves one of the $T_{n-k}$ tilings of the $2 \\times 2(n-k)$ rectangle, where $k = 2, 3, \\dots, n$. The case $k = n$ leaves zero columns, so we set $T_0 = 1$ to allow for this case.\n\n![](images/2020_Australian_Scene_W_p105_data_e841cd350c.png)\n\nHence, we have shown that the following recursion holds for $n \\ge 2$:\n\n$$\nT_n = T_{n-1} + T_{n-2} + 2(T_0 + T_1 + \\dots + T_{n-2}).\n$$\n\nUsing this equation, one can directly verify that for $n \\ge 3$,\n\n$$\nT_n - 2T_{n-1} - 2T_{n-2} + T_{n-3} = 0.\n$$\n\nNow observe that\n\n$$\nF_n^2 + F_{n-3}^2 = (F_{n-1} + F_{n-2})^2 + (F_{n-1} - F_{n-2})^2 = 2F_{n-1}^2 + 2F_{n-2}^2.\n$$\n\nWe can check that $T_n = F_n^2$ for small values of $n$ and then use these two matching recursions to deduce that $T_n = F_n^2$ for all positive integers $n$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16944,
    "subject": "Mathematics (Olympiad)",
    "question": "Given any set $A = \\{a_1, a_2, a_3, a_4\\}$ of four distinct positive integers, we denote the sum $a_1 + a_2 + a_3 + a_4$ by $s_A$. Let $n_A$ denote the number of pairs $(i, j)$ with $1 \\leq i < j \\leq 4$ for which $a_i + a_j$ divides $s_A$. Find all sets $A$ of four distinct positive integers which achieve the largest possible value of $n_A$.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume $a_1 < a_2 < a_3 < a_4$. Then\n\n$$\na_3 + a_4 > a_1 + a_2 \\quad \\text{and} \\quad a_2 + a_4 > a_1 + a_3.\n$$\n\nBecause $a_1 + a_2 + a_3 + a_4 = s_A$, these imply that\n\n$$\ns_A > a_3 + a_4 > \\frac{s_A}{2} \\quad \\text{and} \\quad s_A > a_2 + a_4 > \\frac{s_A}{2}.\n$$\n\nThus $a_3 + a_4 \\nmid s_A$ and $a_2 + a_4 \\nmid s_A$. Since there are six possible pairs $a_i + a_j$ ($1 \\leq i < j \\leq 4$), this implies that $n_A \\leq 4$. Furthermore, $n_A = 4$ is achieved with the quadruple $(a_1, a_2, a_3, a_4) = (1, 5, 7, 11)$. Thus, the maximum value for $n_A$ is $n_A = 4$.\n\nThe rest of the problem consists of finding all quadruples $(a_1, a_2, a_3, a_4)$ with $a_1 < a_2 < a_3 < a_4$, such that $s_A$ is divisible by the four quantities $a_1 + a_2$, $a_1 + a_3$, $a_1 + a_4$, and $a_2 + a_3$.\n\nSince $a_2 + a_3 \\leq \\frac{s_A}{2}$ and $a_1 + a_4 \\leq \\frac{s_A}{2}$, adding these two inequalities yields $a_2 + a_3 + a_1 + a_4 \\leq s_A$, with equality if and only if the first two inequalities are both equalities. But because $a_1 + a_2 + a_3 + a_4 = s_A$, we must have\n\n$$\na_2 + a_3 = a_1 + a_4 = \\frac{s_A}{2}.\n$$\n\nFrom this and using $a_2 < a_3$, it follows that $a_3 > \\frac{s_A}{4}$. Thus $a_1 + a_3 > \\frac{s_A}{4}$. Furthermore, since $a_1 + a_3 < a_2 + a_4$ and $a_1 + a_2 + a_2 + a_4 = s_A$, we have $a_1 + a_3 < \\frac{s_A}{2}$. Thus,\n\n$$\n\\frac{s_A}{4} < a_1 + a_3 < \\frac{s_A}{2},\n$$\n\nfrom which it follows that\n\n$$\na_1 + a_3 = \\frac{s_A}{3}.\n$$\n\nAdditionally, since $a_1 + a_2$ is a divisor of $s_A$, there is a positive integer $k$ such that\n\n$$\na_1 + a_2 = \\frac{s_A}{k}.\n$$\n\nHence, in summary, we have\n\n$$\ns_A = 2(a_2 + a_3) = 2(a_1 + a_4) = 3(a_1 + a_3) = k(a_1 + a_2),\n$$\n\nwhere $k$ is a positive integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16945,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle with center $O$ is internally tangent to two circles inside it at points $S$ and $T$. Suppose the two circles inside intersect at $M$ and $N$ with $N$ closer to $ST$. Show that $OM$ and $MN$ are perpendicular if and only if $S$, $N$, $T$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "The tangent lines at $S$ and $T$ to the big circle meet at a point $K$. The quadrilateral $KSOT$ is cyclic and $KS = KT$. $KS$ and $KT$ are also tangent to the small circles.\n\n![](images/RMC_2019_var_3_p61_data_87f99a3be4.png)\n\nThe point $K$ has the same power with respect to the two small circles, therefore $K$ is on their radical axis, which is $MN$. From $KS^2 = KN \\cdot KM$ we get the similarity of the triangles $KSN$ and $KMS$ so $\\angle KNS = \\angle KSM$. Similarly, we get $\\angle KNT = \\angle KTM$. Then we have:\n\n$S$, $N$, $T$ are collinear $\\Leftrightarrow \\angle KNS + \\angle KNT = 180^\\circ \\Leftrightarrow$\n$\\angle KSM + \\angle KTM = 180^\\circ \\Leftrightarrow SMTK$ is cyclic $\\Leftrightarrow$\n$O$, $M$, $S$, $T$, $K$ are co-cyclic $\\Leftrightarrow \\angle OMK = 90^\\circ \\Leftrightarrow OM \\perp MN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16946,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $Q(x, y) = P(x, y + 1)$. Find all polynomials $Q(x, y)$ such that for all $a, b, c \\in \\mathbb{R}$:\n\n$$\nQ(ab, c^2) + Q(bc, a^2) + Q(ca, b^2) = 0\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $Q(x, y)$ is a polynomial of minimal degree satisfying the given equation. We show that $Q$ must be identically zero.\n\nLet $a = b = c = 0$, then $Q(0, 0) = 0$.\n\nLet $a = b = 0$, then $Q(0, c^2) = 0$, so $Q(0, y) = 0$ for all $y$.\n\nLet $a = 0$, then $Q(bc, 0) = 0$, so $Q(x, 0) = 0$ for all $x$.\n\nThus, $Q(x, y)$ is divisible by both $x$ and $y$, so $Q(x, y) = x y R(x, y)$ for some polynomial $R(x, y)$.\n\nSubstituting into the original equation:\n\n$$\n(ab)(c^2) R(ab, c^2) + (bc)(a^2) R(bc, a^2) + (ca)(b^2) R(ca, b^2) = 0 \\\\\n\\Rightarrow c R(ab, c^2) + a R(bc, a^2) + b R(ca, b^2) = 0\n$$\n\nLet $a = b = 0$, then $R(0, c^2) = 0$, so $R(0, y) = 0$ for all $y$. Thus, $R(x, y)$ is divisible by $x$, so $R(x, y) = x Q'(x, y)$ for some polynomial $Q'(x, y)$.\n\nRepeating this process, we see that $Q$ must be identically zero.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 16947,
    "subject": "Mathematics (Olympiad)",
    "question": "Facu and Nico play the following game with a $13 \\times 13$ grid square. Facu cuts the square into rectangles having a side equal to 1, in any way he wishes. Then Nico chooses a number $k$ among $1, 2, \\ldots, 13$ and takes all the obtained rectangles $1 \\times k$. How many grid cells can he take with certainty?",
    "options": [],
    "answer": "See solution",
    "solution": "Nico can always ensure 16 cells.\n\nSuppose there is a partition as described so that no 16 cells can be taken. Then this partition has at most 1 rectangle $1 \\times k$ for each $k = 8, 9, 10, 11, 12, 13$; at most 2 such rectangles for $k = 6, 7$; at most 3 such rectangles for $k = 4, 5$; at most 5 such rectangles $1 \\times 3$; at most 7 rectangles $1 \\times 2$; and at most 15 unit cells $1 \\times 1$.\n\nConsequently, the total area does not exceed\n\n$$\n(8+9+10+11+12+13) + 2(6+7) + 3(4+5) + 5 \\cdot 3 + 7 \\cdot 2 + 15 \\cdot 1 = 160.\n$$\n\nThis is false because the $13 \\times 13$ square has area $13^2 = 169$. Therefore, 16 cells can be taken regardless of how Facu plays.\n\nOn the other hand, 17 cells are not always achievable. Here is an example. The first row is untouched; the next 9 are cut as follows:\n\n$$\n12+1,\\ 11+2,\\ 10+3,\\ 9+4,\\ 8+5,\\ 8+5,\\ 7+6,\\ 7+6,\\ 5+4+4.\n$$\n\nRows 11 and 12 are $3+3+3+3+1$ and $2+2+2+2+2+2+1$; row 13 is cut into 13 unit cells. In summary, the answer is 16 cells.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16948,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c \\in \\{0, 1, 2, \\dots, 9\\}$. The quadratic equation $ax^2 + bx + c = 0$ has a rational root. Prove that the three-digit number $abc$ is not a prime number.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove by contradiction. Suppose $abc = p$ is a prime number. The rational roots of the quadratic equation $f(x) = ax^2 + bx + c = 0$ are $x_1, x_2 = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$. Since the roots are rational, $b^2 - 4ac$ is a perfect square.\n\n$$\nf(x) = a(x - x_1)(x - x_2)\n$$\n\nEvaluating at $x = 10$:\n\n$$\np = f(10) = a(10 - x_1)(10 - x_2)\n$$\n\nMultiplying both sides by $4a$:\n\n$$\n4ap = (20a - 2a x_1)(20a - 2a x_2)\n$$\n\nBoth $(20a - 2a x_1)$ and $(20a - 2a x_2)$ are positive integers. Thus, $p$ divides one of these factors, say $p \\mid (20a - 2a x_1)$, so $p \\leq 20a - 2a x_1$. This leads to $80 - 8x_1 - 10x_2 + x_1 x_2 \\leq 0$, which contradicts $x_1, x_2 < 0$. Similarly, $p \\mid (20a - 2a x_2)$ is not possible. Therefore, $abc$ cannot be prime. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16949,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(n)$ be the number of ways to write $n$ as a sum of nonnegative powers of $3$, with $f(0) = 1$. Compute $f(100)$.",
    "options": [],
    "answer": "See solution",
    "solution": "We can write $n$ as a sum of nonnegative powers of $3$ with $k$ $1$s if and only if $n - k$ is divisible by $3$. Let $n - k = 3m$; then the number of ways to write $n$ as a sum of nonnegative powers of $3$ with $k$ $1$s is equal to the number of ways to write $3m$ as a sum of positive powers of $3$, which is $f(m)$. Thus,\n\n$$\nf(n) = \\sum_{0 \\leq m \\leq n/3} f(m).\n$$\n\nTo compute $f(100)$, first note that $f(100) = f(0) + f(1) + \\cdots + f(33)$. Using the recurrence and initial values $f(0) = f(1) = f(2) = 1$, $f(3) = 2$, we find:\n\n$$\n\\begin{aligned}\nf(100) &= (34 + 31 + 28)f(0) + (25 + 22 + 19)(f(0) + f(1)) \\\\\n&\\quad + (16 + 13 + 10)(f(0) + f(1) + f(2)) \\\\\n&\\quad + (7 + 4 + 1)(f(0) + f(1) + f(2) + f(3)) \\\\\n&= 210f(0) + 117f(1) + 51f(2) + 12f(3).\n\\end{aligned}\n$$\n\nSubstituting $f(0) = f(1) = f(2) = 1$ and $f(3) = 2$, we get:\n\n$$\nf(100) = 210 + 117 + 51 + 24 = 402.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16950,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the largest possible value of $n$ such that there exist real numbers $x_1, x_2, \\ldots, x_n$ satisfying $\\left| \\frac{x_i - x_j}{1 + x_i x_j} \\right| \\ge \\frac{1}{\\sqrt{3}}$ for all $1 \\le i < j \\le n$?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $3$.\n\nBy the pigeonhole principle, if $n \\ge 4$, there exists an $i$ with $1 \\le i \\le 3$ such that $\\theta_{i+1} - \\theta_i < \\frac{\\pi}{6}$, so it is impossible to have $\\left| \\frac{x_i - x_j}{1 + x_i x_j} \\right| \\ge \\frac{1}{\\sqrt{3}}$ for all pairs.\n\nOn the other hand, for $n = 3$, we can let $x_1, x_2, x_3$ be $\\tan \\frac{\\pi}{12}$, $\\tan \\frac{3\\pi}{12}$, $\\tan \\frac{5\\pi}{12}$ respectively, which fulfill the given condition. Thus, the largest possible value of $n$ is $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16951,
    "subject": "Mathematics (Olympiad)",
    "question": "On the circumference of a circle with radius $1$, points $A_1, A_2, A_3, A_4, A_5$ are placed in this order, and they satisfy\n\n$$\n\\angle A_5 A_2 A_4 = \\angle A_1 A_3 A_5 = \\angle A_2 A_4 A_1 = \\angle A_3 A_5 A_2 = 30^{\\circ}.\n$$\n\nLet $B_1, B_2, B_3, B_4, B_5$ be the points of intersection of $A_2A_4$ and $A_3A_5$, $A_3A_5$ and $A_4A_1$, $A_4A_1$ and $A_5A_2$, $A_5A_2$ and $A_1A_3$, $A_1A_3$ and $A_2A_4$, respectively. Find the area of the pentagon $B_1B_2B_3B_4B_5$.\n\n![](images/Japan_booklet_2013_p1_data_b01bc9f1eb.png)",
    "options": [],
    "answer": "See solution",
    "solution": "$\\frac{\\sqrt{3}}{6}$\n\nFirst, note that $\\angle A_2A_3A_4 = \\angle A_2A_3A_1 + \\angle A_1A_3A_5 + \\angle A_5A_3A_4$. By the theorem on subtended angles by arcs on a circle, $\\angle A_2A_3A_1 = \\angle A_2A_4A_1 = 30^\\circ$, $\\angle A_5A_3A_4 = \\angle A_5A_2A_4 = 30^\\circ$, and $\\angle A_1A_3A_5 = 30^\\circ$. Therefore, $\\angle A_2A_3A_4 = 90^\\circ$, which implies that $A_2A_4$ is a diameter of the circle. Similarly, $A_3A_5$ is also a diameter, so the intersection $B_1$ of $A_2A_4$ and $A_3A_5$ is the center of the circle.\n\nLet $H$ be the foot of the perpendicular from $B_1$ to $A_2A_5$. In triangle $A_2B_1H$, $\\angle HA_2B_1 = 30^\\circ$ and $\\angle B_1HA_2 = 90^\\circ$, so $\\angle A_2B_1H = 60^\\circ$. Since $A_2B_1 = 1$, we get $B_1H = \\frac{1}{2}$ and $HA_2 = \\frac{\\sqrt{3}}{2}$, so the area of triangle $A_2B_1H$ is $\\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\frac{\\sqrt{3}}{2} = \\frac{\\sqrt{3}}{8}$.\n\nFor triangle $A_3B_1B_5$, $\\angle B_5A_3B_1 = 30^\\circ$ and $\\angle A_3B_1B_5 = 30^\\circ + 30^\\circ = 60^\\circ$, so $\\angle B_1B_5A_3 = 90^\\circ$. Thus, $B_1B_5 = \\frac{1}{2}$ and $B_5A_2 = B_1A_2 - B_1B_5 = 1 - \\frac{1}{2} = \\frac{1}{2}$.\n\nFor triangle $A_2B_4B_5$, $\\angle B_5A_2B_4 = 30^\\circ$, $\\angle B_4B_5A_2 = 90^\\circ$, so $B_4B_5 = B_5A_2 \\tan 30^\\circ = \\frac{1}{2} \\cdot \\frac{\\sqrt{3}}{3} = \\frac{\\sqrt{3}}{6}$. Therefore, the area of triangle $A_2B_4B_5$ is $\\frac{1}{2} \\cdot \\frac{\\sqrt{3}}{6} \\cdot \\frac{1}{2} = \\frac{\\sqrt{3}}{24}$.\n\nSumming up, the area of quadrilateral $B_1HB_4B_5$ is $\\frac{\\sqrt{3}}{8} - \\frac{\\sqrt{3}}{24} = \\frac{\\sqrt{3}}{12}$. Similarly, the area of quadrilateral $B_1HB_3B_2$ is also $\\frac{\\sqrt{3}}{12}$, so the area of pentagon $B_1B_2B_3B_4B_5$ is $2 \\cdot \\frac{\\sqrt{3}}{12} = \\frac{\\sqrt{3}}{6}$.\n\n![](images/Japan_booklet_2013_p3_data_d50d597d4a.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16952,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n = p_1^{\\alpha_1} \\cdots p_k^{\\alpha_k}$ be a positive integer such that:\n\n- The number of positive divisors of $n$ is $8$, i.e., $8 = (\\alpha_1 + 1) \\cdots (\\alpha_k + 1)$.\n- The sum of the positive divisors of $n$ is $3240$, i.e.,\n  $$3240 = \\frac{p_1^{\\alpha_1+1}-1}{p_1-1} \\cdots \\frac{p_k^{\\alpha_k+1}-1}{p_k-1}.$$\n\nFind the smallest possible value of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze the possible factorizations of $n$ given $8 = (\\alpha_1 + 1) \\cdots (\\alpha_k + 1)$:\n\n**Case (a):** $8 = \\alpha_1 + 1 \\implies n = p^7$.\n\nCheck $\\sigma(n) = 1 + p^2 + \\cdots + p^7 = 3240$. For $2 < p < 5$, $p = 3$, but substituting yields no solution.\n\n**Case (b):** $8 = (\\alpha_1 + 1)(\\alpha_2 + 1) \\implies n = p_1 p_2^3$.\n\nCheck $\\sigma(n) = (p_1 + 1)(p_2^3 + p_2^2 + p_2 + 1) = 3240$. The only possible primes for $p_2^2 + 1$ are $2$ and $5$, but $p_2 = 2$ or $3$ yields no solution.\n\n**Case (c):** $8 = (\\alpha_1 + 1)(\\alpha_2 + 1)(\\alpha_3 + 1) \\implies n = p_1 p_2 p_3$.\n\n- If one prime is $2$, say $p_1 = 2$, then $(p_2+1)(p_3+1) = 1080$. Let $x = \\frac{p_2+1}{2}$, $y = \\frac{p_3+1}{2}$, so $xy = 270$. Maximizing $x+y$ gives $x=2$, $y=135$, i.e., $p_2 = 3$, $p_3 = 269$, so $n = 2 \\times 3 \\times 269 = 1614$.\n\n- If all $p_i$ are odd, $(\\frac{p_1+1}{2})(\\frac{p_2+1}{2})(\\frac{p_3+1}{2}) = 405 = 3^3 \\cdot 5$. The only possibility is $p_1 = 29$, $p_2 = 5$, $p_3 = 17$, so $n = 29 \\times 5 \\times 17 = 2465$.\n\nThus, the smallest value for $n$ is $\\boxed{1614}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16953,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be coprime positive integers such that $a^2$ divides $b^3 + c^3$, $b^2$ divides $a^3 + c^3$, and $c^2$ divides $a^3 + b^3$. Find the values of $a$, $b$, and $c$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "By the conditions, $a^2 \\mid (b^3 + c^3)$, $b^2 \\mid (a^3 + c^3)$, and $c^2 \\mid (a^3 + b^3)$. Since $a$, $b$, and $c$ are coprime, $a^2 b^2 c^2 \\mid (a^3 + b^3 + c^3)$.\n\nWithout loss of generality, suppose $a \\geq b \\geq c$. Then:\n\n$$\n3a^3 \\geq a^3 + b^3 + c^3 \\geq a^2 b^2 c^2 \\implies a \\geq \\frac{b^2 c^2}{3}\n$$\n\nand\n\n$$\n2b^3 \\geq b^3 + c^3 \\geq a^2 \\implies 2b^3 \\geq \\frac{b^4 c^4}{9} \\implies b \\leq \\frac{18}{c^4}\n$$\n\nIf $c \\geq 2$, then $b \\leq 1$, which contradicts $b \\geq c$. Thus, $c = 1$.\n\nIf $c = 1$ and $b = 1$, then $a = 1$, so $(a, b, c) = (1, 1, 1)$ is a solution.\n\nIf $c = 1$, $b \\geq 2$, and $a = b$, then $b^2 \\mid b^3 + 1$, which is a contradiction.\n\nIf $b \\geq 2$ and $a > b > c = 1$, then\n\n$$\na^2 b^2 \\mid (a^3 + b^3 + 1) \\implies 2a^3 \\geq a^3 + b^3 + 1 \\geq a^2 b^2 \\implies a \\geq \\frac{b^2}{2}\n$$\n\nand by $c = 1$,\n\n$$\na^2 \\mid (b^3 + 1) \\implies b^3 + 1 \\geq a^2 \\geq \\frac{b^4}{4} \\implies 4b^3 + 4 \\geq b^4\n$$\n\nFor $b > 5$, the inequality has no solution. Testing $b = 2, 3, 4, 5$, we find the solutions are $c = 1$, $b = 2$, $a = 3$.\n\nTherefore, all solutions are $(a, b, c) = (1, 1, 1)$, $(1, 2, 3)$, $(1, 3, 2)$, $(2, 1, 3)$, $(2, 3, 1)$, $(3, 2, 1)$, and $(3, 1, 2)$.\n\n$\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16954,
    "subject": "Mathematics (Olympiad)",
    "question": "For every positive integer $n > 0$, a permutation $a_1, a_2, \\dots, a_{2n}$ of $1, 2, \\dots, 2n$ is called *nice* if for every $1 \\leq i < j \\leq 2n$,\n$$\na_i + a_{n+i} = 2n+1 \\text{ and } a_i - a_{i+1} \\not\\equiv a_j - a_{j+1} \\pmod{2n+1}.\n$$\nSuppose that $a_{2n+1} = a_1$.\n\na) For $n=6$, give an example of a nice permutation.\n\nb) Prove that there exists a nice permutation for every positive integer $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Consider the following permutation:\n\n$1, 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7$.\n\nIt is easy to check that $1+12=2+11=4+9=8+5=3+10=6+7=13$.\n\nFurthermore, the differences between consecutive numbers modulo $13$ are $1, 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7$, which are all distinct. Therefore, this permutation is nice.\n\nb) Let $A$ be the set of nice permutations and $B$ be the set of ways to put numbers $1, 2, \\dots, 2n$ on the vertices of a regular polygon $X_1X_2\\dots X_{2n}$ such that:\n\n1. The sum of two numbers at the endpoints of a diameter of the circumcircle is $2n + 1$;\n2. There are no consecutive numbers whose sum is divisible by $2n + 1$.\n\nWe shall prove that there exists a bijection $A \\to B$ and then, it suffices to check that $|A| = |B| > 0$.\n\n**Claim.** There is a bijection from $A$ to $B$.\n\n*Proof.* Consider a nice permutation in $A$, namely $(a_1, a_2, \\dots, a_{2n})$. For each $i = 1, \\dots, 2n$, define $b_i \\equiv a_{i+1} - a_i \\pmod{2n+1}$ (with $a_{2n+1} = a_1$). Then all $b_i$ are pairwise distinct.\n\nPlace these numbers onto a circle in that order. It is easy to see that\n$$\nb_i + b_{i+n} \\equiv a_{i+1} - a_i + a_{i+1+n} - a_{i+n} = 2(2n+1) \\equiv 0 \\pmod{2n+1}.\n$$\nSince $0 < b_i + b_{i+n} < 2(2n+1)$, we have $b_i + b_{i+n} = 2n+1$. Furthermore,\n$$\na_i = a_1 + \\sum_{k=1}^{i-1} b_k\n$$\nwhere the $a_i$ are pairwise distinct modulo $2n+1$, so there do not exist $i, j$ such that\n$$\na_j - a_i = \\sum_{k=i}^{j-1} b_k \\equiv 0 \\pmod{2n+1}.\n$$\nThus, the arrangement of $b_i$ satisfies the conditions of $B$.\n\nConversely, given a way in $B$, define the sequence\n$$\na_1 = b_1, \\quad a_2 = b_1 + b_2, \\quad \\dots, \\quad a_i = \\sum_{k=1}^{i} b_k\n$$\nwhere the sum is taken modulo $2n + 1$. All $a_i$ are positive and pairwise distinct; otherwise, if $a_i = a_j$ for $i \\neq j$, then\n$$\n0 = a_j - a_i = \\sum_{k=i}^{j-1} b_k \\pmod{2n+1},\n$$\nwhich is a contradiction. Furthermore, $b_i + b_{i+n} \\equiv a_i - a_{i-1} + a_{i+n} - a_{i+n-1} \\equiv 0 \\pmod{2n+1}$, which implies\n$$\na_i + a_{i+n} \\equiv a_{i-1} + a_{i+n-1} \\pmod{2n+1} \\text{ for all } i \\geq 2.\n$$\nHence, for every $i$ from $1$ to $2n$, the sum $a_i + a_{i+n}$ has the same remainder $a$ modulo $2n + 1$. Since there are $2n$ such sums and they are pairwise distinct,\n$$\n\\sum_{k=1}^{2n} k = \\sum_{i=1}^{2n} (a_i + a_{i+n}) \\equiv 2na \\pmod{2n+1},\n$$\nand since $\\gcd(2n, 2n + 1) = 1$, we must have $a = 0$, i.e., $a_i + a_{i+n} = 2n + 1$ for all $i$. Thus, $a_i$ is a nice permutation, which belongs to $A$. Therefore, there is a bijection from $A$ to $B$. $\\square$\n\nReturning to the problem, one can check that the number of arrangements such that $b_i + b_{i+n} = 2n + 1$ for all $i$ is $n! \\cdot 2^n$ (since numbers $1$ to $2n$ form $n$ pairs with sum $2n + 1$, and each pair can be swapped). Denote the set of all such arrangements by $S$. Suppose that in some arrangement in $S$, there are consecutive numbers of length $k$ whose sum is divisible by $2n + 1$; then $k \\geq 3$. Let $S_k$ be the subset of $S$ with $k$ elements; there are $n \\cdot 2^{n-k} \\cdot (n-k)! \\cdot S_k$ such arrangements.\n\nIndeed, there are $n$ ways to choose the starting position of the sequence (since $k$ numbers symmetric with respect to the center of the polygon also satisfy the property), $2n - 2k$ remaining numbers to distribute freely, and $2^{n-k} \\cdot (n-k)!$ ways to arrange them.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 16955,
    "subject": "Mathematics (Olympiad)",
    "question": "對於任意正整數 $k$，定義 $r(k)$ 為 $k$ 的二進位表示法中的區塊數量，其中區塊的定義是連續的 $0$ 或連續的 $1$，且無法再延伸成更長的連續相同數字。例如，$(11100100)_2$ 有 $4$ 個區塊，即 $111$-$00$-$1$-$00$。也定義 $r(0) = 0$。\n\n給定一個正整數 $n$，找出所有的函數 $f: \\mathbb{Z} \\to \\mathbb{Z}$ 使得\n\n$$\n\\sum_{k=0}^{2^n-1} 2^{r(k)} f(k + (-1)^k x) = (-1)^{x+n} \\quad \\text{對於所有整數 } x \\text{ 皆成立。}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "唯一的解是 $f(x) = (-1)^x$。\n\n證明如下：\n\n首先考慮 $n = 1$ 的情況：\n\n$$\nf(x) + 2f(1-x) = (-1)^{x+1}.\n$$\n\n將 $x$ 換成 $1-x$，得到：\n\n$$\nf(1-x) + 2f(x) = (-1)^x.\n$$\n\n因此，$f(x) = (-1)^x$。可以檢查這確實是 $n = 1$ 的解。\n\n對於一般 $n$，假設命題成立：\n\n$$\n\\sum_{k=0}^{2^{n-1}-1} 2^{r(k)} f(k + (-1)^k x) = (-1)^{x+n-1}\n$$\n\n唯一解是 $f(x) = (-1)^x$。現在證明對 $n$ 也成立。令\n\n$$\ng(x) = -f(x) - 2f(2^n - 1 - x),\n$$\n\n則\n\n$$\n\\begin{aligned}\n2^{r(k)} g(k + (-1)^k x) &= -2^{r(k)} f(k + (-1)^k x) - 2^{r(k)+1} f(2^n - 1 - k + (-1)^{k+1}x) \\\\\n&= -2^{r(k)} f(k + (-1)^k x) - 2^{r(2^n-1-k)} f(2^n - 1 - k + (-1)^{2^n-1-k}x).\n\\end{aligned}\n$$\n\n這裡用到 $r(k)+1 = r(2^n-1-k)$ 對所有 $0 \\le k \\le 2^{n-1}-1$。\n\n接下來對 $k$ 求和，得到：\n\n$$\n\\sum_{k=0}^{2^{n-1}-1} 2^{r(k)} g(k + (-1)^k x) = - \\sum_{k=0}^{2^n-1} 2^{r(k)} f(k + (-1)^k x) = (-1)^{x+n-1}.\n$$\n\n因此 $g(x) = (-1)^x$，由此可推出 $f(x) = (-1)^x$，與 $n = 1$ 的情形類似。也可驗證 $f(x) = (-1)^x$ 滿足 $-f(x) - 2f(2^n-1-x) = (-1)^x = g(x)$，因此它是解。\n\n**補充說明**\n\n如果用歸納法，不必實際代入 $(-1)^x$ 驗證原式。不過可用組合方法檢查：$r(k) = t$ 在 $0 \\le k \\le 2^n - 1$ 內有 $\\binom{n}{t}$ 個，故 $r(k)$ 的分布與二項係數相同，且 $\\sum_{k=0}^{2^n-1} x^{r(k)} = (1+x)^n$。又 $(-1)^{r(k)} = (-1)^k$，所以\n\n$$\n\\sum_{k=0}^{2^n-1} 2^{r(k)} (-1)^{k+(-1)^k x} = \\sum_{k=0}^{2^n-1} (-2)^{r(k)} = (-1)^{x+n}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16956,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute triangle $ABC$ with $AB > AC$, the bisector of angle $BAC$ and side $BC$ intersect at point $D$. Two points $E$ and $F$ are on sides $AB$ and $AC$, respectively, such that $B$, $C$, $F$, $E$ are concyclic. Prove that the circumcenter of triangle $DEF$ coincides with the incenter of triangle $ABC$ if and only if $BE + CF = BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $I$ be the incenter of $\\triangle ABC$.\n\n**Sufficiency:** Suppose $BC = BE + CF$. Let $K$ be the point on $BC$ such that $BK = BE$, thus $CK = CF$. Since $BI$ bisects $\\angle ABC$ and $CI$ bisects $\\angle ACB$, $\\triangle BIK$ and $\\triangle BIE$ are reflections with respect to $BI$, and $\\triangle CIK$ and $\\triangle CIF$ are reflections with respect to $CI$. We have $\\angle BEI = \\angle BKI = \\pi - \\angle CKI = \\pi - \\angle CFI = \\angle AFI$. Therefore, $A$, $E$, $I$, $F$ are concyclic. Since $B$, $E$, $F$, $C$ are concyclic, we have $\\angle AIE = \\angle AFE = \\angle ABC$, and hence $B$, $E$, $I$, $D$ are concyclic.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p123_data_5c282773b3.png)\n\nSince the bisector of $\\angle EAF$ and the circumcircle of $\\triangle AEF$ meet at $I$, $IE = IF$. Since the bisector of $\\angle EBD$ and the circumcircle of $\\triangle BED$ also meet at $I$, $IE = ID$. So, $ID = IE = IF$, that is, $I$ is also the circumcenter of $\\triangle DEF$.\n\nQ.E.D.\n\n**Necessity:** Suppose $I$ is the circumcenter of $\\triangle DEF$. Since $B$, $E$, $F$, $C$ are concyclic, $AE \\cdot AB = AF \\cdot AC$, and $AB > AC$, we have $AE < AF$. Therefore, the bisector of $\\angle EAF$ and the perpendicular bisector of $EF$ meet at $I$, which lies on the circumcircle of $\\triangle AEF$.\n\nSince $BI$ bisects $\\angle ABC$, let $K$ be the symmetric point of $E$ with respect to $BI$, then $\\angle BKI = \\angle BEI = \\angle AFI > \\angle ACI = \\angle BCI$. Therefore, $K$ lies on $BC$, $\\angle IKC = \\angle IFC$, $\\angle ICK = \\angle ICF$, and $\\triangle IKC \\cong \\triangle IFC$.\n\nHence $BC = BK + CK = BE + CF$. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16957,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the inequality\n\n$$\n\\frac{(x-y)^7 + (y-z)^7 + (z-x)^7 - (x-y)(y-z)(z-x)((x-y)^4 + (y-z)^4 + (z-x)^4)}{(x-y)^5 + (y-z)^5 + (z-x)^5} \\ge 3\n$$\n\nholds for all pairwise different integers $x, y, z$. When does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "Since\n\n$$\n\\begin{aligned}\n(x - y)^7 - (x - y)(y - z)(z - x)(x - y)^4 &= (x - y)^5((x - y)^2 - (y - z)(z - x)) \\\\\n&= (x - y)^5(x^2 + y^2 + z^2 - xy - yz - zx),\n\\end{aligned}\n$$\n\nwe can write\n\n$$\n\\sum_{\\text{cyc}} (x - y)^7 - (x - y)(y - z)(z - x) \\sum_{\\text{cyc}} (x - y)^4 = \\left( \\sum_{\\text{cyc}} (x - y)^5 \\right) \\left( \\sum_{\\text{cyc}} x^2 - \\sum_{\\text{cyc}} xy \\right)\n$$\n\nIt therefore follows that the left-hand side of the inequality can be written as\n\n$$\n\\frac{\\left(\\sum_{\\text{cyc}}(x-y)^5\\right) \\left(\\sum_{\\text{cyc}}x^2 - \\sum_{\\text{cyc}}xy\\right)}{\\sum_{\\text{cyc}}(x-y)^5} = \\sum_{\\text{cyc}}x^2 - \\sum_{\\text{cyc}}xy.\n$$\n\nWe therefore need to consider the inequality\n\n$$\nx^2 + y^2 + z^2 - xy - yz - zx \\ge 3\n$$\n\nfor $x, y, z \\in \\mathbb{Z}$ and $x \\neq y \\neq z \\neq x$. We note that\n\n$$\nx^2 + y^2 + z^2 - xy - yz - zx = \\frac{1}{2}(x - y)^2 + \\frac{1}{2}(y - z)^2 + \\frac{1}{2}(z - x)^2.\n$$\n\nEach pair of variables differs by at least one, and this is not possible for all three pairs at once. The smallest possible value is therefore obtained when two pairs differ by one, i.e., for three consecutive integers $m, m + 1$ and $m + 2$ in any order. Since\n\n$$\n\\frac{1}{2}((m + 2) - (m + 1))^2 + \\frac{1}{2}((m + 1) - m)^2 + \\frac{1}{2}(m - (m + 2))^2 = \\frac{1}{2} + \\frac{1}{2} + 2 = 3\n$$\n\nholds, we see that the given inequality is correct, and equality holds for\n\n$$\n(x, y, z) = (m, m + 1, m + 2)\n$$\n\nor any permutation thereof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16958,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the inequality\n$$\n\\frac{x}{\\sqrt{x^2 + y + z}} + \\frac{y}{\\sqrt{x + y^2 + z}} + \\frac{z}{\\sqrt{x + y + z^2}} \\leq \\sqrt{3}\n$$\nholds for any nonnegative real numbers $x, y, z$ such that $x^2 + y^2 + z^2 = 3$.",
    "options": [],
    "answer": "See solution",
    "solution": "By the Cauchy-Bunyakowsky inequality, it suffices to prove:\n$$\n\\left( \\frac{x}{x^2 + y + z} + \\frac{y}{y^2 + z + x} + \\frac{z}{z^2 + x + y} \\right) (x + y + z) \\leq 3.\n$$\nSince $(x + y + z)^2 \\leq 3(x^2 + y^2 + z^2) = 9$, it is enough to show:\n$$\n\\frac{x}{x^2 + y + z} + \\frac{y}{y^2 + z + x} + \\frac{z}{z^2 + x + y} \\leq 1.\n$$\nFor non-negative $a, b, c$, we have $(a^2 + b + c)(1 + b + c) \\geq (a + b + c)^2$, so $\\frac{a}{a^2 + b + c} \\leq \\frac{a(1 + b + c)}{(a + b + c)^2}$. Summing these, it suffices to prove:\n$$\n\\frac{x(1 + y + z) + y(1 + z + x) + z(1 + x + y)}{(x + y + z)^2} \\leq 1.\n$$\nThis is equivalent to $x + y + z \\leq x^2 + y^2 + z^2 = 3$, which is already given.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16959,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there exist infinitely many positive integers $n$ such that the largest prime divisor of $n^4 + n^2 + 1$ is equal to the largest prime divisor of $(n + 1)^4 + (n + 1)^2 + 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p_n$ be the largest prime divisor of $n^4 + n^2 + 1$, and let $q_n$ be the largest prime divisor of $n^2 + n + 1$.\n\nIt is easy to see that $p_n = q_{n^2}$. Also,\n\n$$\nn^4 + n^2 + 1 = (n^2 + 1)^2 - n^2 = (n^2 - n + 1)(n^2 + n + 1) = ((n-1)^2 + (n-1) + 1)(n^2 + n + 1)\n$$\n\nThus, for all $n \\ge 2$, we have $p_n = \\max\\{q_n, q_{n-1}\\}$. Since $n^2 - n + 1$ is odd,\n\n$$\n\\gcd((n-1)^2 + (n-1) + 1, n^2 + n + 1) = \\gcd(2n, n^2 - n + 1) = \\gcd(n, n^2 - n + 1) = 1.\n$$\n\nTherefore, $q_n \\neq q_{n-1}$.\n\nNow, consider the set\n\n$$\nS := \\{n \\in \\mathbb{Z}_{\\ge 2} \\mid q_n > q_{n-1},\\ q_n > q_{n+1}\\}.\n$$\n\nNotice that for all $n \\in S$, we have\n\n$$\np_n = \\max\\{q_n, q_{n-1}\\} = q_n = \\max\\{q_n, q_{n+1}\\} = p_{n+1},\n$$\n\nso the original problem is equivalent to proving that $S$ has infinitely many elements. Note that $q_2 = 7 < 13 = q_3$ and $q_3 = 13 > 7 = q_4$, so $S$ is nonempty.\n\nSuppose, for contradiction, that $S$ has a largest element $m$. Since all $q_i$ are positive integers, $q_m > q_{m+1} > q_{m+2} > \\dots$ is impossible, so there must exist $k \\ge m$ such that $q_k < q_{k+1}$ (note that we have already shown $q_k \\ne q_{k+1}$). Also, since $q_{(k+1)^2} = p_{k+1} = \\max\\{q_k, q_{k+1}\\} = q_{k+1}$, an infinite strictly increasing sequence $q_k < q_{k+1} < \\dots$ is also impossible. Therefore, there must exist a minimal $l \\ge k+1$ such that $q_l > q_{l+1}$.\n\nSince $l$ is minimal, we must have $q_l > q_{l-1}$, so $l \\in S$, contradicting the assumption that $m$ is the largest element. Thus, $S$ has infinitely many elements. Q.E.D.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16960,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_n$ be the arithmetic sequence defined by $a_n = 13n + 1$. Show that there are infinitely many terms in this sequence that consist only of the digit $2$ (in base 10).",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $221 = 13 \\times 17$ and therefore $222 = 13 \\times 17 + 1$ is in the arithmetic sequence. Furthermore, $222\\,222\\,221$ is divisible by $13$ and hence $222\\,222\\,222$ is in the sequence, since\n\n$$\n222222 = 222 \\times 1001 = 222 \\times 7 \\times 11 \\times 13.\n$$\n\nIn fact, any number consisting of $6m + 3$ 2s will be one more than a multiple of $13$, since the number consisting of $6m$ 2s will be a multiple of $1001$ and hence a multiple of $13$. Thus, there are infinitely many terms in the sequence that only use the digit $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16961,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть $P(x)$ — многочлен с целыми коэффициентами. В последовательности $a_0, a_1, a_2, \\dots$, где $a_{k+1} = P(a_k)$, для любого натурального $b$ встречается бесконечно много $b$-х степеней натуральных чисел, больших единицы. Докажите, что если среди целых чисел $a_k$ бесконечно много различных, то $P(x)$ — линейный многочлен.",
    "options": [],
    "answer": "See solution",
    "solution": "Заметим, что для любого $b$ в последовательности $a_0, a_1, a_2, \\dots$ встретится бесконечно много $b$-х степеней натуральных чисел, больших единицы. Если их количество конечно, и наибольшая из них — $N = x^b$, то в последовательности не встретится ни одной $Nb$-й степени, что невозможно.\n\nПусть $d_k = a_{k+1} - a_k$; тогда $a_{k+1} \\equiv a_k \\pmod{d_k}$. Поскольку все коэффициенты многочлена целые, из $a \\equiv a' \\pmod{d_k}$ следует $P(a) \\equiv P(a') \\pmod{d_k}$. Индукцией по $s$ получаем $a_{k+s} \\equiv a_k \\pmod{d_k}$ при всех $s \\ge 0$.\n\n**Лемма.** $a_k(a_k - 1)$ делится на $d_k$.\n\n*Доказательство.* Пусть $p^\\ell$ — максимальная степень простого числа $p$, делящая $d_k$. Положим $b = p^{\\ell-1}(p-1)\\ell$; найдётся $s > k$, что $a_s = m^b$ при натуральном $m$ и $a_s \\equiv a_k \\pmod{p^\\ell}$. Если $m$ не делится на $p$, то по теореме Эйлера $a_s = (mp^{\\ell-1}(p-1))^\\ell \\equiv 1^\\ell \\equiv 1 \\pmod{p^\\ell}$, откуда $a_k \\equiv 1 \\pmod{p^\\ell}$. Если $m$ делится на $p$, то $a_s$ делится на $p^\\ell$, значит, и $a_k$ тоже. В любом случае $a_k(a_k - 1)$ делится на $p^\\ell$.\n\nЗначит, для любого $k$ число $a_k(a_k - 1)$ делится на $d_k = P(a_k) - a_k$; при этом среди целых $a_k$ бесконечно много различных. В частности, $|x(x-1)| \\ge |Q(x)|$ при бесконечно многих целых $x$, где $Q(x) = P(x) - x$.\n\nЕсли степень $P(x)$ (и $Q(x)$) больше 1, то неравенство возможно лишь если $Q(x)$ — квадратный трёхчлен со старшим коэффициентом $\\pm 1$, то есть $Q(x) = \\pm x^2 + ux + v$. Тогда $Q(x) \\mp x(x-1) = (u \\pm 1)x + v$ делится на $Q(x)$ для бесконечно многих $x$, что возможно только если $Q(x) = \\pm x(x-1)$, то есть $P(x) = x^2$ или $P(x) = 2x - x^2 = 1 - (x-1)^2$.\n\nВ первом случае $a_k = n^{2k}$, то есть $a_k$ не может быть нечётной степенью натурального числа, если $n$ не является таковой степенью. Во втором случае $P(x) \\le 1$ при всех $x$, то есть $P(x)$ не может быть степенью натурального числа, большего 1. В обоих случаях условие задачи не выполнено; значит, $P(x)$ — линейный многочлен.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16962,
    "subject": "Mathematics (Olympiad)",
    "question": "Point $P$ is chosen inside isosceles triangle $ABC$ with base $BC$ and $\\angle BAC < 90^\\circ$ in such a way that $\\angle BPC = 2\\angle BAC$. Let $K$ be the foot of the perpendicular from $A$ to the line containing the bisector of the angle adjacent to $\\angle BPC$. Prove that $BP + PC = 2AK$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first show that point $K$ is always inside triangle $ABC$. Consider the case when $P = P_1$ lies on side $AB$ and compute some angles. Let $\\angle BAC = \\alpha$ (see figure below), then $\\angle BP_1C = 2\\alpha$, $\\angle ABC = \\frac{\\pi}{2} - \\frac{\\alpha}{2}$, $\\angle BCP_1 = \\frac{\\pi}{2} - \\frac{3\\alpha}{2}$, so $\\angle P_1CA = \\alpha$. Hence, $\\triangle ACP_1$ is isosceles, and the altitude $P_1K_1$ is also a bisector, so $K = K_1$ belongs to side $AC$. Therefore, as long as $P$ is inside $\\triangle ABC$, the bisector $PK$ of the angle adjacent to $\\angle BPC$ forms the least angle with line $BC$. Thus, the perpendicular from $A$ to this line lies between sides $AB$ and $AC$ of triangle $ABC$. Let us extend $AK$ to the intersection with $CP$ (see figure below; without loss of generality, assume $P$ is closer to $B$ than to $C$).\n\n![](images/Ukrajina_2010_p23_data_7c0b857086.png)\n\nIn the same way, $BP$ meets $AK$ at point $X$. Then, in $\\triangle PXY$, segment $PK$ is a bisector and an altitude, so it is isosceles or $\\angle PXY = \\angle PYX$, from which it follows that $\\angle BXA = \\angle AYC$. If we denote $x = \\angle XBA$, $y = \\angle YCA$, $x_1 = \\angle CAY$, $y_1 = \\angle BAX$, then $x_1 + y_1 = \\alpha$, $\\pi - \\alpha = x + y + \\angle PBC + \\angle PCB = x + y + \\pi - 2\\alpha$, so $x + y = \\alpha = x_1 + y_1$. Moreover, $x + y_1 = y + x_1$. From the last two identities, we get $x = x_1$, $y = y_1$. Hence, $\\triangle ABX = \\triangle CAY$ (they have two equal sides and the angle between them).\n\nWe have the following equalities: $$2AK = AX + AY = CY + BX = CY + BP + PX = CY + YP + BP = CP + BP,$$ which is exactly what we wanted to prove.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16963,
    "subject": "Mathematics (Olympiad)",
    "question": "A $9 \\times 9 \\times 9$ cube is made up of $1 \\times 1 \\times 1$ small cubes. From each edge, all the small cubes except the two at the corners are removed. What is the surface area of the remaining object?",
    "options": [],
    "answer": "See solution",
    "solution": "First, removing the middle small cube on one edge of the $9 \\times 9 \\times 9$ cube increases the surface area by 2 small faces. Removing the cubes on either side does not change the surface area. Continue this process until only the two corner cubes remain; at this stage, the surface area has increased by 2 small faces per edge. Repeating this on all 12 edges increases the surface area by $12 \\times 2 = 24$ small faces. Now, all 8 corner cubes have 6 exposed faces, so removing the 8 corner cubes reduces the surface area by $8 \\times 6 = 48$ small faces. The original surface area is $6 \\times 81 = 486$. Thus, the surface area of the remaining object is $$486 + 24 - 48 = 462.$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 16964,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the number of ordered triples $ (a, b, c) $ of positive integers, with $ a \\leq b \\leq c \\leq 9 $, such that there exists a (non-degenerate) triangle $ \\triangle ABC $ with an integer inradius for which $ a, b $, and $ c $ are the lengths of the altitudes from $ A $ to $ \\overline{BC} $, $ B $ to $ \\overline{AC} $, and $ C $ to $ \\overline{AB} $, respectively? (Recall that the inradius of a triangle is the radius of the largest possible circle that can be inscribed in the triangle.)\n\n(A) 2  (B) 3  (C) 4  (D) 5  (E) 6",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer (B):**\n\nLet $x, y$, and $z$ be the lengths of $\\overline{BC}$, $\\overline{AC}$, and $\\overline{AB}$, respectively. Let $r$ be the inradius of $\\triangle ABC$. Then\n\n$$\n\\text{Area}(\\triangle ABC) = \\frac{1}{2} x a = \\frac{1}{2} y b = \\frac{1}{2} z c = \\frac{1}{2} (x + y + z) r.\n$$\n\nTherefore $x = \\frac{2 \\text{Area}(\\triangle ABC)}{a}$, $y = \\frac{2 \\text{Area}(\\triangle ABC)}{b}$, and $z = \\frac{2 \\text{Area}(\\triangle ABC)}{c}$, so\n\n$$\n\\text{Area}(\\triangle ABC) = \\frac{1}{2} \\left( \\frac{2 \\text{Area}(\\triangle ABC)}{a} + \\frac{2 \\text{Area}(\\triangle ABC)}{b} + \\frac{2 \\text{Area}(\\triangle ABC)}{c} \\right) r.\n$$\n\nDividing by $\\text{Area}(\\triangle ABC)$ gives $\\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) r = 1$, so\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = \\frac{1}{r}.\n$$\n\nBecause $a, b, c \\leq 9$, it follows that\n\n$$\n\\frac{1}{r} = \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\geq \\frac{1}{9} + \\frac{1}{9} + \\frac{1}{9} = \\frac{1}{3},\n$$\n\nimplying $r \\leq 3$. It is then possible to find the solutions $(a, b, c)$ by examining cases based on the value of $r$.\n\n- If $r = 1$, then, because $a \\leq b \\leq c$, either $a = 2$ or $a = 3$. If $a = 2$, then $\\frac{1}{b} + \\frac{1}{c} = \\frac{1}{2}$. Then, because $b \\leq c$, either $b = 3$ or $b = 4$. If $b = 3$, then $c = 6$; and if $b = 4$, then $c = 4$. If $a = 3$, then it must be that $b = c = 3$. So the solutions in this case are $(2, 3, 6)$, $(2, 4, 4)$, and $(3, 3, 3)$.\n- If $r = 2$, then, because $a \\leq b \\leq c$, it follows that $a = 3, 4, 5$, or $6$. If $a = 3$, then $\\frac{1}{b} + \\frac{1}{c} = \\frac{1}{6}$, which has no solutions because $\\frac{1}{9} + \\frac{1}{9} > \\frac{1}{6}$. If $a = 4$, then $\\frac{1}{b} + \\frac{1}{c} = \\frac{1}{4}$. In this case, $b = c = 8$ is the only solution. If $a = 5$, then $\\frac{1}{b} + \\frac{1}{c} = \\frac{3}{10}$, which gives no solutions. If $a = 6$, then it follows that $b = c = 6$. So the solutions in this case are $(4, 8, 8)$ and $(6, 6, 6)$.\n- If $r = 3$, then $(a, b, c) = (9, 9, 9)$ is the only solution because if $a < 9$, then $c > 9$.\n\nFinally, the altitude lengths must be checked to ensure that these lengths give dimensions for a valid triangle. In the $(2, 3, 6)$ case, the side lengths of the triangles become $(3t, 2t, t)$ for some $t$, which does not form a triangle. In the $(2, 4, 4)$ and $(4, 8, 8)$ cases, the side lengths of the triangles become $(2t, t, t)$ for some $t$, which again does not form a triangle. The rest of the cases, namely $(3, 3, 3)$, $(6, 6, 6)$, and $(9, 9, 9)$, do produce valid triangles because if all of the altitudes have length $h$, then it is possible to form an equilateral triangle with side length $\\frac{2\\sqrt{3}}{3} h$. Thus there are 3 ordered triples satisfying the given conditions.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16965,
    "subject": "Mathematics (Olympiad)",
    "question": "Given two distinct square-free positive integers $a$ and $b$, show that\n\n$$\n|\\{n\\sqrt{a}\\} - \\{n\\sqrt{b}\\}| > c n^{-3}\n$$\n\nfor some positive constant $c$ and all positive integers $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume, without loss of generality, that $a > b$. Consider a positive integer $n$, and let $k_n = \\lfloor n\\sqrt{a} \\rfloor - \\lfloor n\\sqrt{b} \\rfloor$. Since $n(\\sqrt{a} - \\sqrt{b})$ is irrational, it follows that $\\{n\\sqrt{a}\\} \\neq \\{n\\sqrt{b}\\}$, so\n\n$$\n\\begin{aligned}\n0 < |\\{n\\sqrt{a}\\} - \\{n\\sqrt{b}\\}| &= |n(\\sqrt{a} - \\sqrt{b}) - k_n| \\\\\n&= \\frac{K_n}{(n(\\sqrt{a} + \\sqrt{b}) - k_n)(n(\\sqrt{a} + \\sqrt{b}) + k_n)(n(\\sqrt{a} - \\sqrt{b}) + k_n)}\n\\end{aligned}\n$$\n\nwhere $K_n$ is a positive integer. Notice that $k_n < \\lfloor n\\sqrt{a} \\rfloor + \\lfloor n\\sqrt{b} \\rfloor < n(\\sqrt{a} + \\sqrt{b})$, so\n\n$$\n\\begin{aligned}\n|\\{n\\sqrt{a}\\} - \\{n\\sqrt{b}\\}| &> \\frac{1}{n(\\sqrt{a} + \\sqrt{b}) \\cdot 2n(\\sqrt{a} + \\sqrt{b})(n(\\sqrt{a} - \\sqrt{b}) + n(\\sqrt{a} + \\sqrt{b}))} \\\\\n&= \\frac{1}{4 n^3 (\\sqrt{a} + \\sqrt{b})^2 \\sqrt{a}} = \\frac{c}{n^3}, \\quad \\text{where } c = \\frac{1}{4 (\\sqrt{a} + \\sqrt{b})^2 \\sqrt{a}} > 0.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16966,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all natural numbers $m$ which have exactly three different prime divisors $p, q,$ and $r$, such that:\n\n$$\na)\\quad p-1 \\mid m, \\quad qr-1 \\mid m,\n$$\n\n$$\nb)\\quad q-1 \\nmid m, \\quad r-1 \\nmid m, \\quad 3 \\nmid q+r.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The number $m$ can be represented as $m = p^a q^b r^c$. The numbers $q$ and $r$ cannot be $2$, because $2-1=1$ divides every natural number, so $q$ and $r$ are odd. Thus, $qr-1$ is even, so $m$ must be even, i.e., $p=2$.\n\nFrom $\\gcd(q, qr-1) = 1 = \\gcd(r, qr-1)$ and $qr-1 \\mid m$, it follows that $qr-1 \\mid 2^a$, so $qr-1 = 2^k$ for some natural number $k \\leq a$. This means that $2^k + 1$ has exactly two divisors.\n\nThe number $k$ can be written as $k = 2^t s$, where $t$ is the largest power of $2$ dividing $k$, and $s$ is the largest odd divisor of $k$. If $s > 1$, then:\n\n$$\n2^k + 1 = (2^{2^t})^s + 1 = (2^{2^t} + 1)\\left[(2^{2^t})^{s-1} - (2^{2^t})^{s-2} + \\dots - (2^{2^t}) + 1\\right].\n$$\n\nThis is possible only when $q = 2^{2^t} + 1$, $r = (2^{2^t})^{s-1} - (2^{2^t})^{s-2} + \\dots - (2^{2^t}) + 1$ or vice versa, but then $q-1 = 2^{2^t} \\nmid m$ or $p = 2^{2^t} \\nmid m$, which contradicts the conditions.\n\nIt follows that $s=1$, i.e., $k=2^t$ is a power of $2$ and $qr = 2^{2^t} + 1$, so $2^{2^t} + 1$ is the product of two primes. For $t=0$ ($2^{2^t} + 1 = 3$), this is not the case, so $t > 1$.\n\nFrom $2^{2^t} + 1 \\equiv 2^{2 \\cdot 2^{t-1}} + 1 \\equiv (2^2)^{2^{t-1}} + 1 \\equiv 4^{2^{t-1}} + 1 \\equiv 1^{2^{t-1}} + 1 \\equiv 1 + 1 \\equiv 2 \\pmod{3}$, it follows that $3 \\nmid q$ and $3 \\nmid r$. If $q \\equiv r \\pmod{3}$, then $qr \\equiv 1 \\pmod{3}$, so one must be congruent to $1$ and the other to $2$, but then their sum is divisible by $3$, which contradicts the last condition. Therefore, a number $m$ satisfying all the given conditions does not exist.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16967,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $ABCDE$ 為凸五邊形，其中 $AB = BC = CD$，$\\angle EAB = \\angle BCD$，且 $\\angle EDC = \\angle CBA$。試證：過 $E$ 並與 $BC$ 垂直的直線，和線段 $AC$ 與 $BD$ 共點。",
    "options": [],
    "answer": "See solution",
    "solution": "在證明中，我們將使用 $\\angle A, \\angle B, \\angle C, \\angle D, \\angle E$ 等來代表五邊形 $ABCDE$ 的諸內角。令線段 $AC$ 與線段 $BD$ 的兩條中垂線交於點 $I$。注意到 $AC$ 的中垂線會過點 $B$，而 $BD$ 的中垂線會過點 $C$。於是有 $BD \\perp CI$ 及 $AC \\perp BI$。所以，$AC$ 與 $BD$ 的交點為三角形 $BIC$ 的垂心 $H$，且 $IH \\perp BC$。只要再證明 $E$ 點在直線 $IH$ 上即可，亦即 $EI \\perp BC$。\n\n![](images/18-1J_p2_data_68993d502b.png)\n\n直線 $IB$ 與 $IC$ 分別平分 $\\angle B$ 及 $\\angle C$。由於 $IA = IC, IB = ID$，以及 $AB = BC = CD$，三個三角形 $IAB, ICB, ICD$ 皆全等。得\n\n$$\n\\angle IAB = \\angle ICB = \\frac{1}{2}\\angle C = \\frac{1}{2}\\angle A,\n$$\n故 $IA$ 平分 $\\angle A$。同理得 $ID$ 平分 $\\angle D$。最後，$IE$ 也會平分 $\\angle E$，因為 $I$ 位於五邊形其它四個內角的角平分線上。\n\n凸五邊形的內角和為 $540^\\circ$，故\n\n在四邊形 $ABIE$ 中，\n\n$$\n\\begin{aligned}\n\\angle BIE &= 360^\\circ - \\angle EAB - \\angle ABI - \\angle AEI \\\\\n&= 360^\\circ - \\angle A - \\frac{1}{2}\\angle B - \\frac{1}{2}\\angle E \\\\\n&= 360^\\circ - \\angle A - \\frac{1}{2}\\angle B - (270^\\circ - \\angle A - \\angle B) \\\\\n&= 90^\\circ + \\frac{1}{2}\\angle B = 90^\\circ + \\angle IBC,\n\\end{aligned}\n$$\n\n故得 $EI \\perp BC$。證畢。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16968,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n > 2$ be an integer and let $S$ be a set of $k$ points in the plane whose coordinates both lie in $\\{1, 2, \\ldots, n\\}$. Find the minimal number $k$ for which one can choose four points from $S$ that are the vertices of a nondegenerate parallelogram.",
    "options": [],
    "answer": "See solution",
    "solution": "If $k = 2n - 1$, we can choose all points with $x$-coordinate or $y$-coordinate equal to 1. Then, clearly there is no parallelogram. Thus, any $k \\leq 2n-1$ does not satisfy the condition.\n\nNext, we will prove that $k = 2n$ suffices. Let $x_1, x_2, \\ldots, x_n$ be the number of points whose $x$-coordinates are $1, 2, \\ldots, n$ respectively. Clearly,\n\n$$\nx_1 + x_2 + \\dots + x_n = 2n.\n$$\n\nConsider the segments formed by connecting two points with the same $x$-coordinate. If there are $x$ points, then at least $x-1$ segments with different lengths will be generated (take the point with the lowest ordinate and connect it to the remaining $x-1$ points). Thus, for $1 \\leq i \\leq n$, among the points with the same $x$-coordinate $i$, we can choose a set $S_i$ of size $x_i - 1$ consisting of segments with different lengths. If in $S_1, S_2, \\ldots, S_n$ there are two segments with the same length, it is easy to see that the four corresponding vertices will form a parallelogram, satisfying the problem.\n\nConversely, if there are no segments with the same length, then notice that\n\n$$\n|S_1| + |S_2| + \\dots + |S_n| = (x_1 - 1) + (x_2 - 1) + \\dots + (x_n - 1) = n,\n$$\n\nbut the values in these sets are smaller than $n$, because their length is the difference between two $y$-coordinates, and these coordinates only belong to $\\{1, 2, \\ldots, n\\}$.\n\nObviously, this is a contradiction.\n\nTherefore, the minimum value is $k_{\\min} = 2n$. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16969,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{Q} \\to \\mathbb{Q}$ such that for all $x, y \\in \\mathbb{Q}$,\n\n$$\nf(xf(x) + y) = f(y) + x^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We substitute $x \\to -x$ into the functional equation:\n\n$$\nf(-xf(-x) + y) = f(y) + x^2.\n$$\n\nThus, $f(xf(x) + y) = f(-xf(-x) + y)$, so $f(y)$ is periodic with period $d = |xf(x) - (-x)f(-x)|$. Repeated application shows $f(y) = f(y + nd)$ for all $n \\in \\mathbb{Z}$.\n\nSetting $y = 0$ gives $f(xf(x)) = f(0) + x^2$. Let $f(x) = \\frac{a}{b}$, $a, b \\in \\mathbb{Z}$, $b > 0$. Substituting $x \\to x + bd$ gives $(x + bd)^2 = x^2$, so $d = 0$ for $x > 0$, hence $xf(x) = -xf(-x)$, so $f(x) = -f(-x)$ for $x \\neq 0$.\n\nSubstituting $y = -xf(x)$ gives $f(0) = f(-xf(0)) + x^2$. Since $xf(x) \\neq 0$ for $x \\neq 0$, and using $f(x) = -f(-x)$, we get $f(0) = -f(0)$, so $f(0) = 0$. Thus, $f(xf(x)) = x^2$.\n\nLet $c = f(1)$. Then $f(c) = 1$, and $f(cf(c)) = c^2 = f(c) = 1$, so $c = 1$ or $-1$.\n\nSubstituting $x = 1$ in the original equation gives $f(y + c) = f(y) + 1$. By induction, $f(y + nc) = f(y) + n$ for all $n \\in \\mathbb{Z}$. Thus, $f(n) = cn$ for all $n \\in \\mathbb{Z}$.\n\nIf $f(x) = n \\in \\mathbb{Z}$, then $x = cn$. For arbitrary $x = \\frac{p}{q}$, a similar argument shows $f(x) = cx$ for all $x \\in \\mathbb{Q}$, where $c = 1$ or $-1$.\n\nBoth $f(x) = x$ and $f(x) = -x$ satisfy the original equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16970,
    "subject": "Mathematics (Olympiad)",
    "question": "對於每個質數 $p$，都有一個名為 $p$-蘭蒂亞的王國，擁有 $p$ 座島嶼，這些島依序編號為 $1, 2, \\dots, p$。編號為 $n$ 和 $m$ 的島之間會有橋相連，若且唯若 $p$ 整除 $(n^2 - m + 1)(m^2 - n + 1)$。\n\n證明有無窮多個質數 $p$，在 $p$-蘭蒂亞王國裡面會有兩個島無法藉由一連串的橋來連接。",
    "options": [],
    "answer": "See solution",
    "solution": "將此視為一個有向圖，當且僅當 $n = m^2 + 1$ 在 $\\mathbb{Z}_p$ 中成立時，從 $m$ 到 $n$ 有一條有向邊。容易看出每個頂點的出度至多為一。\n\n若 $a, b$ 是 $x^2 - x + 1$ 的根，則當 $p > 3$ 時 $a, b$ 不相等（否則 $a = b = 2^{-1}$ 且 $1 = 2^{-2}$ 在 $\\mathbb{Z}_p$ 中）。因此邊的數量小於 $p-1$，所以 $p$-蘭蒂亞作為無向圖是不連通的。\n\n剩下要證明存在無窮多個質數 $p$，使得存在 $x$ 使得 $p \\mid x^2 - x + 1$。反證法，假設只有 $p_1, p_2, \\dots, p_k$ 這些質數使 $x^2 - x + 1 \\equiv 0 \\pmod p$ 有解。考慮 $x = p_1p_2 \\cdots p_k$，設 $p$ 為 $x^2 - x + 1$ 的質因數，則 $x^2 - x + 1 \\equiv 0 \\pmod p$，且對於每個 $p_i$，$x^2 - x + 1 \\equiv 1 \\pmod {p_i}$，這與假設矛盾。因此存在無窮多個這樣的質數。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16971,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with a right angle at $C$, let $I$ be its incenter, and let $H$ be the orthogonal projection of $C$ on $AB$. The incircle $\\omega$ of the triangle $ABC$ is tangent to the sides $BC$, $CA$, and $AB$ at $A_1$, $B_1$, and $C_1$, respectively. Let $E$ and $F$ be the reflections of $C$ in the lines $C_1A_1$ and $C_1B_1$, respectively, and let $K$ and $L$ be the reflections of $H$ in the same lines. Prove that the circles $A_1EI$, $B_1FI$, and $C_1KL$ have a common point.\n\n![](images/ROU_ABooklet_2021_p58_data_0634a1a795.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The line $C_1A_1$ is parallel to the external angle bisector of $\\angle B$, so the reflection in $C_1A_1$ maps the segment $A_1C$ to the segment $A_1E$ parallel to $AB$.\n\nSimilarly, $B_1F \\parallel AB$. Notice also that $A_1E = A_1C = B_1C = B_1F = r$, where $r$ is the inradius of $\\triangle ABC$.\n\nLet $M$ be the midpoint of $AB$. Let $X$ be the point of $\\omega$ such that $\\overrightarrow{IX} \\nearrow \\overrightarrow{CM}$. Notice that $\\angle EA_1I = 90^\\circ + \\angle EA_1B = 90^\\circ + \\angle CBM = 90^\\circ + \\angle BCM = \\angle A_1IX$, and $A_1E = IA_1 = IX$; thus, $XIA_1E$ is an isosceles trapezoid. Hence $X$ lies on the circle $IA_1E$, and $EX \\parallel A_1I$. Similarly, $X$ lies on the circle $IB_1F$, and $FX \\parallel B_1I$. It remains to show that $X$ lies on the circle $C_1KL$.\n\nUnder the symmetry in $C_1A_1$, the line $CH$ (perpendicular to $AB$) maps to the line through $E$ perpendicular to $BC$ — i.e., $CH$ maps to $EX$. Therefore, the projection $H$ of $C_1$ onto $CH$ maps to the projection $K$ of $C_1$ onto $EX$. Similarly, $L$ is the projection of $C_1$ onto $FX$. So the quadrilateral $C_1KXL$ is cyclic, due to right angles at $K$ and $L$.\n\n**REMARKS.** (1) In fact, the quadrilateral $C_1KXL$ is a square, since $C_1K = C_1H = C_1L$ and $\\angle KC_1L = 2\\angle A_1C_1B_1 = 90^\\circ$.\n\n(2) One can easily see that the points $C_1$ and $X$ are symmetric in the angle bisector $CI$. This yields that $K$ and $L$ both lie on $CI$. One can show that this conclusion in fact holds in any, not necessarily right-angled, triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16972,
    "subject": "Mathematics (Olympiad)",
    "question": "考慮兩互質的正整數 $p, q$，並且對於 $i = 1, 2, \\dots, q-1$，定義 $f(i)$ 為 $p \\cdot i$ 除以 $q$ 的餘數。若 $f(i)$ 為 $f(1), f(2), \\dots, f(i)$ 之間的最大數，則稱 $i$ 為大數；但若 $f(i)$ 為 $f(1), f(2), \\dots, f(i)$ 之間的最小數，則稱 $i$ 為小數。注意到 $1$ 同時為大數及小數。\n\n給定兩個正整數 $a, b$。已知在 $1, 2, \\dots, q-1$ 之中有 $a$ 個大數及 $b$ 個小數，試求 $q$ 的最小可能值。",
    "options": [],
    "answer": "See solution",
    "solution": "最小可能的 $q$ 為 $ab + 1$。\n\n設 $S_{p,q}$ 為所有小數的集合，$T_{p,q}$ 為所有大數的集合。將 $S_{p,q} \\cup T_{p,q}$ 按遞增順序排列，得到序列 $c_1, \\dots, c_{a+b-1}$（注意 $S_{p,q} \\cap T_{p,q} = \\{1\\}$）。對於每個 $n \\ge 2$，若 $i, j < n$ 分別是 $c_i \\in S_{p,q}$、$c_j \\in T_{p,q}$ 的最大下標，則 $c_n = c_i + c_j$，且 $c_n \\in S_{p,q}$ 當且僅當 $c_i p \\bmod q > (-c_j p) \\bmod q$，此時 $c_n p \\bmod q = (c_i p \\bmod q) - (-c_j p \\bmod q)$。\n\n可用歸納法證明上述性質。當 $n=2$ 時顯然成立。假設對所有小於 $n$ 的情況成立。通過將 $p$ 換成 $-p$，不失一般性可設 $c_i p \\bmod q > (-c_j p) \\bmod q$。\n\n若 $c_n < c_i + c_j$ 且 $c_n \\in T_{p,q}$，則 $(-c_n p) \\bmod q < (-c_j p) \\bmod q$，導致 $0 < (c_n - c_j) p \\bmod q < (-c_j p) \\bmod q < c_i p \\bmod q$。但 $c_n - c_j < c_i$，這與 $c_i \\in S_{p,q}$ 矛盾。\n\n若 $c_n < c_i + c_j$ 且 $c_n \\in S_{p,q}$，則 $c_n - c_i < c_j$ 且 $0 < -c_n c_j p \\bmod q < c_i p \\bmod q$。設 $c_k$ 為小於 $c_j$ 的最大 $T_{p,q}$ 元素，則 $(-c_k p) \\bmod q < c_i p \\bmod q$。若 $i < j$，根據歸納假設，$c_{i+1} \\in S_{p,q}$，與 $i$ 的選取矛盾。因此 $i > j$。設 $\\ell < j$ 為最大 $S_{p,q}$ 元素，則 $(-c_j p) \\bmod q = (-c_k p \\bmod q) - (c_\\ell p \\bmod q) < (c_i p \\bmod q) - (c_i p \\bmod q) = 0$，矛盾。\n\n因此 $c_n \\ge c_i + c_j$。注意 $$(c_i + c_j) \\bmod q = (c_i p \\bmod q) - ((-c_j p) \\bmod q) < c_i p \\bmod q$$，說明 $c_i + c_j \\in S_{p,q}$，所以 $c_n = c_i + c_j$。\n\n因此可構造如下序列：初始有紅色正整數 $x$ 和藍色正整數 $y$，每次取最後一個紅數和最後一個藍數相加，並標記顏色。共寫 $m$ 個紅數和 $n$ 個藍數。此時 $x = y = 1, m = a-1, n = b-1$，紅數屬於 $S_{p,q}$，藍數屬於 $T_{p,q}$。設 $f(m, n, x, y)$ 為最後紅數與藍數之和的最小值。原問題中，最後紅數 $s$ 和藍數 $t$ 滿足 $sp \\bmod q = (-tp) \\bmod q = 1$，所以 $s + t = q$，我們需給出 $f(a-1, b-1, 1, 1)$ 的下界。\n\n事實上，若 $m, n \\ge 1$，則\n$$\nf(m, n, x, y) = \\min(f(m-1, n, x+y, y), f(m, n-1, x, x+y))\n$$\n其中第一項對應下個數標紅，第二項對應標藍。邊界條件：$f(m, 0, x, y) = x + (m+1)y$，$f(0, n, x, y) = (n+1)x + y$。可對 $m+n$ 歸納證明 $f(m, n, x, y) = mn \\min(x, y) + (m+1)x + (n+1)y$。因此 $q \\ge f(a-1, b-1, 1, 1) = ab + 1$。\n\n要證明 $q = ab + 1$ 可達，取 $p = a$。則 $T_{p,q} = \\{1, 2, \\dots, b\\}$，$S_{p,q} = \\{1, b+1, \\dots, (a-1)b+1\\}$。\n\n**補充說明：** 事實上，$S_{p,q}$ 中的數正好是 Farey 序列中剛小於 $p/q$ 的分母，$T_{p,q}$ 則是剛大於 $p/q$ 的分母。這也可用來證明上述結論。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16973,
    "subject": "Mathematics (Olympiad)",
    "question": "Three workers must complete a task. First, one of them works for as long as it would take the other two, working together, to complete half the work. Then, a second worker works for as long as it would take the other two, working together, to complete half the work. Finally, the third worker works for as long as it would take the other two, working together, to complete half the work. After this, the whole work is completed. How many times faster would the work be completed if all the workers worked together?",
    "options": [],
    "answer": "See solution",
    "solution": "Let the contributions of the first, second, and third worker per unit time be $x$, $y$, and $z$, measured as fractions of the whole work. The second and third workers together would complete half the work in $\\frac{1}{2(y+z)}$ time units; the third and first together in $\\frac{1}{2(z+x)}$; and the first and second together in $\\frac{1}{2(x+y)}$. Thus, the total time spent as described is $\\frac{1}{2(y+z)} + \\frac{1}{2(z+x)} + \\frac{1}{2(x+y)}$. If all worked together, the time would be $\\frac{1}{x+y+z}$. We are to find the ratio of these times.\n\nSince $\\frac{x}{2(y+z)} + \\frac{y}{2(x+z)} + \\frac{z}{2(x+y)} = 1$, we have:\n\n$$\n\\begin{aligned}\n\\frac{\\frac{1}{2(y+z)} + \\frac{1}{2(z+x)} + \\frac{1}{2(x+y)}}{\\frac{1}{x+y+z}} &= \\frac{x+y+z}{2(y+z)} + \\frac{x+y+z}{2(z+x)} + \\frac{x+y+z}{2(x+y)} \\\\\n&= \\frac{1}{2} + \\frac{x}{2(y+z)} + \\frac{1}{2} + \\frac{y}{2(z+x)} + \\frac{1}{2} + \\frac{z}{2(x+y)} \\\\\n&= \\frac{3}{2} + 1 = 2.5.\n\\end{aligned}\n$$\n\nSo, the work would be completed $2.5$ times faster if all workers worked together.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16974,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k = 3 \\cdot 4^l$ for an arbitrary positive integer $l$. Prove that $a_k = \\left\\lfloor \\dfrac{2^k}{k} \\right\\rfloor$ is odd.",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n$$\na_k = \\left\\lfloor \\frac{2^k}{k} \\right\\rfloor = \\left\\lfloor \\frac{2^{3 \\cdot 4^l}}{3 \\cdot 4^l} \\right\\rfloor = \\left\\lfloor \\frac{2^{3 \\cdot 4^l - 2l}}{3} \\right\\rfloor = \\left\\lfloor \\frac{2^{3 \\cdot 4^l - 2l} - 1}{3} + \\frac{1}{3} \\right\\rfloor.\n$$\nSince the positive integer $3 \\cdot 4^l - 2l$ is even, we have $2^{3 \\cdot 4^l - 2l} \\equiv 4 \\equiv 1 \\pmod 3$, so the odd positive integer $2^{3 \\cdot 4^l - 2l} - 1$ is divisible by 3.\n\nTherefore,\n$$\na_k = \\left\\lfloor \\frac{2^k}{k} \\right\\rfloor = \\left\\lfloor \\frac{2^{3 \\cdot 4^l - 2l} - 1}{3} + \\frac{1}{3} \\right\\rfloor = \\frac{2^{3 \\cdot 4^l - 2l} - 1}{3}\n$$\nis odd, which finishes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16975,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for all non-negative integers $n$ and $j$,\n$$\n\\sum_{k=0}^{n} k^j \\binom{n}{k} \\ge 2^{n-j} n^j.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By the symmetry of the binomial coefficients,\n$$\n2 \\sum_{k=0}^{n} k^j \\binom{n}{k} = \\sum_{k=0}^{n} (k^j + (n-k)^j) \\binom{n}{k}.\n$$\nNow,\n$$\nk^j + (n-k)^j = n^j \\left( \\left( \\frac{k}{n} \\right)^j + \\left( 1 - \\frac{k}{n} \\right)^j \\right) = n^j f_j \\left( \\frac{k}{n} \\right),\n$$\nwhere $f_j(x) = x^j + (1-x)^j$, $0 \\le x \\le 1$. By using the convexity of $x \\mapsto x^j$, calculus methods or otherwise, it is easy to see that\n$$\n\\min \\{f_j(x) : x \\in [0, 1]\\} = f_j\\left(\\frac{1}{2}\\right) = 2^{1-j}, \\quad j = 0, 1, \\dots\n$$\nwhence\n$$\n2 \\sum_{k=0}^{n} k^j \\binom{n}{k} \\ge n^j 2^{1-j} \\sum_{k=0}^{n} \\binom{n}{k} = n^j 2^{n+1-j},\n$$\nfrom which the result follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16976,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a square inscribed in circle $\\omega$ and let $P$ be a variable point on the shorter arc $AB$ of $\\omega$. Let $CP \\cap BD = R$ and $DP \\cap AC = S$. Show that triangles $ARB$ and $DSR$ have equal areas.\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p270_data_a3b2631f68.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $T = PC \\cap AB$. Then $\\angle BTC = 90^\\circ - \\angle PCB = 90^\\circ - \\angle PDB = 90^\\circ - \\angle SBD = \\angle BSC$, thus points $B, S, T, C$ are concyclic. So $\\angle TSC = 90^\\circ$, and therefore $TS \\parallel BD$. Hence\n\n$$\n[DSR] = [DTR] = [DTB] - [TBR] = [CTB] - [TBR] = [CRB] = [ARB],\n$$\n\nwhere $[F]$ denotes the area of $F$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16977,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the set $\\mathcal{F}$ of functions $f : \\mathbb{N} \\to \\mathbb{N}$ having the property that\n$$\nf(a^2 - b^2) = f(a)^2 - f(b)^2, \\text{ for all } a, b \\in \\mathbb{N},\\ a \\ge b.\n$$\n\na) Determine the set $\\{f(1) \\mid f \\in \\mathcal{F}\\}$.\n\nb) Prove that $\\mathcal{F}$ has exactly two elements.",
    "options": [],
    "answer": "See solution",
    "solution": "a) For $a = b$ we have $f(0) = 0$. Then $f(a^2) = f(a)^2$ for all $a \\in \\mathbb{N}$, implying $f(1) = f(1)^2$ and then $f(1) \\in \\{0, 1\\}$.\n\nBoth following cases hold: $f_0(1) = 0$ for $f_0 \\equiv 0 \\in \\mathcal{F}$, and $f_1(1) = 1$ for $f_1 = 1_{\\mathbb{N}} \\in \\mathcal{F}$. The requested set is $\\{0, 1\\}$.\n\nb) Let $a, b \\in \\mathbb{N}$, $a \\ge b$ and $f \\in \\mathcal{F}$. Then $f(a), f(b) \\in \\mathbb{N}$ and $f(a)^2 - f(b)^2 = f(a^2 - b^2) \\ge 0$, which yields $f(a) \\ge f(b)$, establishing that $f$ is non-decreasing.\n\nSuppose $f(1) = 0$. From $f(4) = f(2)^2 = f(2)^2 - f(1)^2 = f(3)$ we obtain $f(7) = f(4)^2 - f(3)^2 = 0$. By induction, $f(7^{2^n}) = 0$. Since $f$ is monotonic, it follows that $f = f_0$.\n\nSuppose $f(1) = 1$ and set $f(2) = x$. Then $f(4) = x^2$, $f(3) = x^2 - 1$, $f(5) = f(3)^2 - f(2)^2 = x^4 - 3x^2 + 1$. As $f(16) = f(5)^2 - f(3)^2$, we obtain $x^4 = (x^4 - 3x^2 + 1)^2 - (x^2 - 1)^2$, which rewrites as $x^2(x^2 - 1)^2(x^2 - 4) = 0$. The function $f$ is increasing ($a, b \\in \\mathbb{N}$, $a > b$ implies $f(a)^2 - f(b)^2 \\ge f(1) > 0$), so $x = 2$. By induction we establish that $f(2^{2^n}) = 2^{2^n}$ and then, using the monotony, $f = f_1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16978,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_1, \\ldots, A_{22}$ be sets with $|A_i| = 5$ for $i = 1, \\ldots, 22$. Define $S = \\bigcup_{i=1}^{22} A_i$ and $a(x, y) = |\\{i \\mid \\{x, y\\} \\subset A_i\\}|$ for $x, y \\in S$. Assume that for all $x, y \\in S$, $a(x, y) \\leq 4$. Furthermore, for any $i \\neq j$, $|A_i \\cap A_j| = 2$. Show that such a collection of sets cannot exist.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose such sets exist. Consider the pairs in $A_1$; there are 10 unordered pairs, and each $A_i$ ($i = 2, \\ldots, 22$) contains exactly one of them. Thus, some pair $(x, y)$ is in $A_1$ and three other sets, so $a(x, y) = 4$. Assume $x = 1$, $y = 2$, and the four sets containing $1$ and $2$ are:\n$$\nA_1 = \\{1, 2, 3, a, b\\},\\quad A_2 = \\{1, 2, 4, c, d\\},\\quad A_3 = \\{1, 2, 5, e, f\\},\\quad A_4 = \\{1, 2, 6, g, h\\}.\n$$\nAssume $A_5 = \\{1, 3, 4, 5, 6\\}$. For $|A_i \\cap A_j| = 2$ ($i = 1, 2, 3, 4$), every $A_j$ with $j > 4$ must contain either $1$ or $2$. Suppose $1 \\in A_j$ for $j = 5, \\ldots, 14$. Each $A_j$ ($j = 6, \\ldots, 14$) must contain $3, 4, 5$, or $6$ due to $|A_5 \\cap A_j| = 2$. Thus, some $x \\in \\{3, 4, 5, 6\\}$ has $a(1, x) \\geq 5$, a contradiction. Therefore, at most nine of the sets $A_j$ ($j = 5, \\ldots, 22$) contain $1$, and similarly for $2$. So, exactly nine sets $A_5, \\ldots, A_{13}$ contain $1$.\n\nConsidering $a(x, y) \\leq 4$ for all $x, y \\in S$ and $|A_i \\cap A_j| = 2$ for $1 \\leq i < j \\leq 13$, we can assume:\n$$\nA_6 = \\{1, 3, c, e, g\\},\\quad A_7 = \\{1, 3, d, f, h\\},\\quad A_8 = \\{1, 4, a, e, h\\},\n$$\n$$\nA_9 = \\{1, 4, b, f, g\\},\\quad A_{10} = \\{1, 5, a, d, g\\},\\quad A_{11} = \\{1, 5, b, c, h\\},\n$$\nand there is no proper choice left for $A_{12}$. Thus, such a collection of sets cannot exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16979,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose we have an $n \\times n$ board, where some squares are initially coloured black and the rest are white. In one move, you may choose any row or any column and invert the colour of every square in that row or column (i.e., black becomes white and white becomes black). What is the least number of squares that must be initially coloured black so that, after a finite sequence of such moves, the entire board can be made black?",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that the order of row and column operations is not relevant, so we may assume that all the row operations are performed before manipulating any of the columns. Furthermore, it is sufficient to transform any row or column at most once.\n\nLet us analyze the initial colouring of the board which can, after a finite number of moves, become entirely black. If we are to perform column operations only after all the row operations have been done, we see that before we start manipulating the columns, all the rows must have the same appearance.\n\nWe conclude that, initially, all the rows must be coloured in one of the two possible ways which can be obtained from one another by applying the allowed row transformation.\n\nIf every row in the initial configuration contains at least two black squares, then the total number of black squares is at least $2n$. Assume that there is a row which contains at most one black square. This row cannot be equal to both the first row and the last, because they already have black squares placed at two different positions—the first and the last, respectively. This means that there is a row which has the opposite colouring, thus containing at least $n-1$ black squares. If this row contains $n$ black squares, then both the first and the last row must be entirely black, yielding a total number of black squares larger than $2n$. If this row contains $n-1$ black squares, then the remaining rows must contain at least 1 black square, which means that there are at least $2n-2$ black squares in total.\n\nThis shows that the least number of squares that initially need to be coloured black is $2n-2$, meaning we have to add at least $2n-4$ black squares.\n\nOn the other hand, if we colour all the squares in the first column and in the last row black, leaving only the bottom-left corner white (as shown in the picture), we can perform the obvious operations (transforming the first $n-1$ rows, then the first $n-1$ columns) and colour the entire board black. Since this colouring requires $2n-4$ additional black squares, we have shown that the least number of squares we need to colour black is equal to $2n-4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16980,
    "subject": "Mathematics (Olympiad)",
    "question": "Let a word of length $n$ consist of the letters 'L' and 'R', with exactly $\\ell$ of them being 'L'. Eva can repeatedly swap any adjacent 'R' and 'L' (with 'R' on the left and 'L' on the right), turning them into 'L' and 'R'.\n\n(a) Define the *L-sum* of a word as the sum of the positions of all 'L' characters, where the leftmost character has position $1$ and the rightmost has position $n$. Show that after each move, the L-sum decreases by $1$, and deduce that only finitely many moves are possible.\n\n(b) For fixed $n$ and $\\ell$, determine the maximum possible number of moves Eva can make, and for which initial word(s) this maximum is achieved.\n\n(c) For which value(s) of $\\ell$ (in terms of $n$) is the maximum number of moves as large as possible? What is this maximum value?",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Each move swaps an 'R' at position $i$ with an 'L' at position $i+1$, moving the 'L' one position to the left. Thus, the L-sum decreases by $1$ with each move. Since the L-sum is always non-negative, only finitely many moves are possible.\n\n(b) The L-sum is maximized when all $\\ell$ 'L's are at the right end: $\\underbrace{RR\\ldots R}_{n-\\ell}\\underbrace{LL\\ldots L}_{\\ell}$. It is minimized when all 'L's are at the left: $\\underbrace{LL\\ldots L}_{\\ell}\\underbrace{RR\\ldots R}_{n-\\ell}$. The difference in L-sum between these two configurations is $\\ell(n-\\ell)$. Since each move decreases the L-sum by $1$, the maximum number of moves is $\\ell(n-\\ell)$. This maximum is achieved by starting with all 'L's at the right.\n\n(c) The function $f(\\ell) = \\ell(n-\\ell)$ is maximized at $\\ell = \\frac{n}{2}$. If $n$ is even, the maximum number of moves is $\\frac{1}{4}n^2$ at $\\ell = \\frac{n}{2}$. If $n$ is odd, the maximum is $\\frac{1}{4}(n^2-1)$ at $\\ell = \\frac{n-1}{2}$ or $\\ell = \\frac{n+1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16981,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve in $\\mathbb{R}$ the equation\n$$\n\\log_7(6^x + 1) = \\log_6(7^x - 1).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\log_7(6^x + 1) = \\log_6(7^x - 1) = y$. Then $6^x + 1 = 7^y$ and $7^x - 1 = 6^y$. Adding these equations gives $6^x + 7^x = 6^y + 7^y$.\n\nThe function $f: \\mathbb{R} \\to \\mathbb{R}$, $f(x) = 6^x + 7^x$, is injective (strictly increasing), so $x = y$.\n\nNow, solve $6^x + 1 = 7^x$, or equivalently,\n$$\n\\left(\\frac{6}{7}\\right)^x + \\left(\\frac{1}{7}\\right)^x = 1.\n$$\nThe function $g(x) = \\left(\\frac{6}{7}\\right)^x + \\left(\\frac{1}{7}\\right)^x$ is strictly decreasing, so $g(x) = g(1)$ has the unique solution $x = 1$. Thus, $x = 1$ is the solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 16982,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$ and $y$ be different positive integers. Prove that\n$$\n\\frac{x^2 + 4xy + y^2}{x^3 - y^3}\n$$\nis never an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "By symmetry, we can assume that $x > y$. If $x - y = 1$, then\n\n$$\n\\begin{aligned}\n\\frac{x^2 + 4xy + y^2}{x^3 - y^3} &= \\frac{x^2 + 4xy + y^2}{(x-y)(x^2 + xy + y^2)} \\\\\n&= \\frac{(x-y)^2 + 6xy}{(x-y)((x-y)^2 + 3xy)} \\\\\n&= \\frac{1 + 6xy}{1 + 3xy} = 1 + \\frac{3xy}{1 + 3xy},\n\\end{aligned}\n$$\n\nwhich is clearly not an integer. If $x - y \\ge 2$, then\n\n$$\n\\begin{aligned}\n\\frac{x^2 + 4xy + y^2}{x^3 - y^3} &= \\frac{x^2 + 4xy + y^2}{(x-y)(x^2 + xy + y^2)} \\\\\n&\\le \\frac{x^2 + 4xy + y^2}{2(x^2 + xy + y^2)} \\\\\n&< \\frac{2x^2 + 2xy + 2y^2}{2(x^2 + xy + y^2)} = 1,\n\\end{aligned}\n$$\n\nwhere the last inequality follows from $x^2 - 2xy + y^2 = (x-y)^2 > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16983,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y, z, t$ be positive real numbers such that $xyzt = 1$ and\n\n$$\n\\frac{x}{y} + \\frac{y}{z} + \\frac{z}{t} + \\frac{t}{x} \\leq x + y + z + t.\n$$\n\nProve that\n\n$$\n\\frac{y}{x} + \\frac{z}{y} + \\frac{t}{z} + \\frac{x}{t} \\geq x + y + z + t.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By the arithmetic mean-geometric mean inequality we have\n\n$$\nx = \\sqrt[4]{x^4} = \\sqrt[4]{\\frac{x^4}{xyzt}} = \\sqrt[4]{\\frac{x^3}{yzt}} = \\sqrt[4]{\\frac{x}{y} \\cdot \\frac{x}{t} \\cdot \\frac{t}{z} \\cdot \\frac{x}{t}} \\leq \\frac{1}{4} \\left( \\frac{x}{y} + \\frac{x}{t} + \\frac{t}{z} + \\frac{x}{t} \\right) = \\frac{1}{4} \\left( \\frac{x}{y} + 2 \\cdot \\frac{x}{t} + \\frac{t}{z} \\right).\n$$\n\nSimilarly, we show that\n\n$$\ny \\leq \\frac{1}{4} \\left( \\frac{y}{z} + 2 \\cdot \\frac{y}{x} + \\frac{x}{t} \\right), \\quad z \\leq \\frac{1}{4} \\left( \\frac{z}{t} + 2 \\cdot \\frac{z}{y} + \\frac{y}{x} \\right), \\quad t \\leq \\frac{1}{4} \\left( \\frac{t}{x} + 2 \\cdot \\frac{t}{z} + \\frac{z}{y} \\right).\n$$\n\nAdding together the four inequalities and applying the assumed inequality, we obtain\n\n$$\nx + y + z + t \\leq \\frac{1}{4} \\left( \\frac{x}{y} + \\frac{y}{z} + \\frac{z}{t} + \\frac{t}{x} \\right) + \\frac{3}{4} \\left( \\frac{y}{x} + \\frac{z}{y} + \\frac{t}{z} + \\frac{x}{t} \\right) \\leq \\frac{1}{4}(x + y + z + t) + \\frac{3}{4} \\left( \\frac{y}{x} + \\frac{z}{y} + \\frac{t}{z} + \\frac{x}{t} \\right).\n$$\n\nThe assertion of the problem follows immediately.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16984,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть дана последовательность $a_k$ с некоторым условием на $a_1 > 1$. Докажите, что среди пар $(a_k, a_j)$ не более одной такой, что $a_k + a_j$ — целое число.",
    "options": [],
    "answer": "See solution",
    "solution": "Положим $b_k = a_k - k$. Тогда\n\n$$\nb_{k+1} = b_k - 1 + \\frac{k}{k+b_k} = b_k - \\frac{b_k}{k+b_k} = b_k \\left(1 - \\frac{1}{k+b_k}\\right).\n$$\n\nОчевидной индукцией по $k$ получаем, что $b_k > 0$ (так как $b_1 > 0$). Кроме того, $b_{k+1} = b_k - \\frac{b_k}{k+b_k} < b_k$, то есть последовательность убывает, и $b_k \\le b_1 < 1$.\n\nЗаметим, что $b_2 = a_1 + \\frac{1}{a_1} - 2 = (\\sqrt{a_1} - \\frac{1}{\\sqrt{a_1}})^2$. Это выражение положительно и возрастает при $a_1 \\in (1, 2)$; тогда $0 = 1 + \\frac{1}{2} - 2 < b_2 < 2 + \\frac{1}{2} - 2 = \\frac{3}{2}$. Таким образом, $b_k \\le b_2 < \\frac{1}{2}$ при $k \\ge 2$.\n\nЕсли $a_k + a_j$ — целое, то $b_k + b_j$ — тоже целое. Значит, одно из чисел $b_k, b_j$ (пусть $b_k$) не меньше $\\frac{1}{2}$; тогда $k = 1$, и $b_j = 1 - b_1$. Но таких $j$ не больше одного, так как $(b_i)$ убывает. Следовательно, утверждение доказано.\n\n_Замечание._ Количество пар с целой суммой будет конечным при любом $a_1 > 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16985,
    "subject": "Mathematics (Olympiad)",
    "question": "Paul is filling the cells of a $67 \\times 67$ rectangular table alternately with crosses and circles (he starts with a cross). When the table is filled completely, he determines his score as $X - O$, where $X$ is the sum of squares of the numbers of crosses in all the rows and columns, and $O$ is the sum of squares of the numbers of circles in all the rows and columns. Find all possible values of the score for a $67 \\times 67$ table.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = 67$ and denote by $k = \\frac{1}{2}(n^2 + 1)$ the total number of crosses in the table. A row containing $a$ crosses and $n - a$ circles contributes $a^2 - (n - a)^2 = 2n \\cdot a - n^2$ to the total score, so all $n$ rows combined contribute\n\n$$\n2n \\cdot k - n \\cdot n^2 = 2n \\cdot \\frac{n^2 + 1}{2} - n^3 = n\n$$\n\nto the total score. Likewise, columns contribute $n$. Hence, the total score is always equal to $2n = 134$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16986,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider increasing integer sequences with elements from $1, \\dots, 10^9$. Such a sequence is Adriatic if its first element equals $1$ and if every element is at least twice the preceding element. A sequence is Tyrrhenian if its final element equals $10^6$ and if every element is strictly greater than the sum of all preceding elements.\n\nDecide whether the number of elements of Adriatic sequences is (i) smaller than, (ii) equal to, or (iii) greater than the number of Tyrrhenian sequences.\n\nEquivalently, a sequence $\\langle a_1, \\dots, a_n \\rangle$ is Adriatic if $a_1 = 1$ and $a_k \\ge 2a_{k-1}$ holds for $k = 2, 3, \\dots, n$. A sequence $\\langle t_1, t_2, \\dots, t_n \\rangle$ is Tyrrhenian if $t_n = 10^6$ and $t_k > t_1 + t_2 + \\dots + t_{k-1}$ holds for $k = 2, 3, \\dots, n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the Adriatic sequence $\\langle a_1, \\dots, a_n \\rangle$ starting with $a_1 = 1$. Construct a new sequence\n$$\n\\langle a_2 - 1, a_3 - a_2, \\dots, a_n - a_{n-1}, 10^6 \\rangle\n$$\nfrom it. Note that the new sequence is Tyrrhenian, as\n$$\n(a_2 - 1) + (a_3 - a_2) + \\dots + (a_n - a_{n-1}) = a_{k-1} - 1 < a_k - a_{k-1}\n$$\nholds for $k = 2, \\dots, n-1$ and as\n$$\n(a_2 - 1) + (a_3 - a_2) + \\dots + (a_n - a_{n-1}) = a_n - 1 < 10^6.\n$$\nNext, consider the Tyrrhenian sequence $\\langle t_1, t_2, \\dots, t_n \\rangle$ with $t_n = 10^6$. Construct a new sequence\n$$\n\\langle 1, 1 + t_1, 1 + t_1 + t_2, \\dots, 1 + t_1 + \\dots + t_{n-1} \\rangle\n$$\nfrom it. Note that this new sequence is Adriatic, since\n$$\n2(1 + t_1 + t_2 + \\dots + t_{k-1}) \\le 1 + t_1 + \\dots + t_k\n$$\nis equivalent to the Tyrrhenian property $t_1 + \\dots + t_{k-1} < t_k$.\n\nThese two constructions yield two injections and demonstrate that the number of Adriatic sequences equals the number of Tyrrhenian sequences. (In fact, the second injection is the inverse of the first injection, so that we have a clean bijection between the two sets.) $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16987,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with $AB < AC$, let $O$ be its circumcentre and let $A'$ be the reflection of $A$ in $BC$. The parallel through $O$ to $BC$ crosses $AC$ at $F$, and the tangent at $F$ to circle $BFC$ crosses the parallel through $A'$ to $BC$ at $M$. Consider the point $K$ on the ray $AB$, emanating from $A$, such that $AK = 4AB$. Prove that the orthocentre of the triangle $ABC$ lies on the circle on diameter $KM$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H$ be the orthocentre of triangle $ABC$, let $N$ be the midpoint of the segment $AH$ and let $D$ be the foot of the perpendicular from $B$ to $AC$. We first prove that $\\angle BNF = 90^\\circ$.\n\nLet $F'$ be the intersection of line $AC$ and the perpendicular at $N$ to $BN$. The points $B, N, D$, and $F'$ lie on the circle on diameter $BF$, so\n\n$$\n\\angle NBF' = \\angle ADN = \\angle DAN = 90^\\circ - \\angle BCA = \\angle ABO,\n$$\n\nwhere the second equality holds because $DN$ is the median in the right triangle $ADH$. Since $\\angle NBF' = \\angle ABO$, lines $BN$ and $BO$ are isogonal with respect to $\\angle ABF$. Similarly, lines $AO$ and $AN$ are isogonal with respect to $\\angle BAC$,\n\nso $O$ and $N$ are isogonal conjugates in triangle $ABF'$. Consequently, lines $F'O$ and $F'N$ are isogonal with respect to $\\angle AF'B$. Using this along with the cyclic quadrilateral $BNDF'$, we get\n\n$$\n\\angle AF'O = \\angle BF'N = \\angle BDN = \\angle NHD = \\angle ACB.\n$$\n\nHence $OF' \\parallel BC$, so $F \\equiv F'$, showing that $\\angle BNF = 90^\\circ$.\n\nNext, we prove that $N, F$ and $M$ are collinear. Since $F \\equiv F'$, points $O$ and $N$ are isogonal conjugates in triangle $ABF$, so\n\n$$\n\\angle NFB = \\angle AFO = \\angle ACB.\n$$\n\nThis implies $NF$ is tangent to circle $BFC$ and hence $N, F, M$ are collinear, as desired.\n\nLet $AB$ and $A'M$ cross at $E$. Note that $E$ is the reflection of $A$ across $B$. Since $AK = 4AB$, point $E$ is the midpoint of the segment $AK$. In triangle $AEH$, line $BN$ is a midline, so $BN \\parallel EH$. As $BN$ and $NM$ are perpendicular, so are $EH$ and $NM$. Moreover, $NH \\perp EM$, hence $H$ is the orthocentre of triangle $ENM$, implying $MH \\perp EN$. Finally, in triangle $AKH$, line $EN$ is a midline, so $EN \\parallel KH$ and hence $MH \\perp KH$; that is, $H$ lies on the circle on diameter $KM$, as required.\n\n**Alternative solution for $\\angle BNF = 90^\\circ$:**\n\nWe use the same notations as in the previous solution. Consider the foot $P$ of the perpendicular from $F$ to $BC$. Then\n\n$$\n2AN = 2NH = AH = 2 \\operatorname{dist}(O, BC) = 2 \\operatorname{dist}(F, BC) = 2FP,\n$$\n\nso the segments $NH$ and $FP$ are parallel and have equal lengths, implying that $NFPH$ is a parallelogram. Similarly, $ANPF$ is also a parallelogram. Therefore, $HB \\perp NP$, and since $NH \\perp BC$, we deduce that $H$ is the orthocentre of triangle $BNP$. Consequently, $HP \\perp BN$, and since $HP \\parallel NF$, it follows that $\\angle BNF = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16988,
    "subject": "Mathematics (Olympiad)",
    "question": "Jüri wishes to draw $n$ circles and any number of lines on the plane such that all the lines meet at one point, and for every two circles there exist two lines that touch both of these circles.\n\n**a)**\n\nIs it possible for Jüri to solve this problem for any $n \\ge 2$?\n\n**b)**\n\nFor which natural numbers $n$ is it possible to solve this problem if in addition all the circles must have the same radius?",
    "options": [],
    "answer": "See solution",
    "solution": "**a)**\n\nJüri can draw two lines and any number of circles such that they touch both of the lines (see the figure below).\n\n**b)**\n\nAssume that Jüri has solved the problem for some $n$, where $n > 1$.\n\n![](images/Estonija_2012_p13_data_77ebc3aaf3.png)\n\nLet $O$ be the intersection point of all the lines. Consider any circle $c$. From the premises, the circle $c$ touches two of the lines drawn by Jüri, which we call $k$ and $l$. But any one circle can only touch up to two lines drawn from one point. So, the circle $c$ does not have any more lines touching it. If $c'$ is any other circle drawn by Jüri, then the common tangents of $c$ and $c'$ can only be $k$ and $l$. Therefore, $k$ and $l$ are the common tangents of all the circles. Two lines divide the plane into four sectors, and inside each can be only one circle with the same radius (see the figure below). So, for $n > 4$ the problem has no solution, but for $n \\le 4$, it obviously has.\n\n![](images/Estonija_2012_p13_data_9f48f79b48.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16989,
    "subject": "Mathematics (Olympiad)",
    "question": "а) За кои природни броеви $n$ постојат природни броеви $x$ и $y$ за кои важи\n\n$$\n\\text{НЗС}(x, y) = n!; \\quad \\text{НЗД}(x, y) = 2009\n$$\n\nб) Одреди го бројот на парови $(x, y)$ за кои важи\n\n$$\n\\text{НЗС}(x, y) = 4!; \\quad \\text{НЗД}(x, y) = 2009; \\quad x \\le y\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "а) Од $\\text{НЗД}(x, y) = 2009$ следува дека $x = 2009a$ и $y = 2009b$, каде што $a$ и $b$ се природни броеви со $\\text{НЗД}(a, b) = 1$.\n\nОд $\\text{НЗС}(x, y) = n!$ следува $2009ab = n!$. Бидејќи $2009 = 7^2 \\cdot 41$, $n!$ мора да биде делив со $2009$, односно $n \\geq 41$. Условот е и доволен бидејќи ако $n \\geq 41$, за $x = 2009$ и $y = n!$ важи $\\text{НЗС}(x, y) = n!$ и $\\text{НЗД}(x, y) = 2009$.\n\nб) Нека $x = 2009a$, $y = 2009b$, со $\\text{НЗД}(a, b) = 1$, $a \\leq b$ и $2009ab = 4! = 24$. Значи $ab = \\frac{24}{2009}$, но $2009 > 24$, па нема такви парови. (Во оригиналниот текст се користи $40!$ наместо $4!$, што е веројатно грешка. Ако е $40!$, тогаш $ab = \\frac{40!}{2009}$.)\n\nБројот $\\frac{40!}{7^2 \\cdot 41}$ има 12 прости делители: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37$. Секој делител може да оди или во $a$ или во $b$, па има $2^{12}$ парови $(a, b)$, но само половината со $a \\leq b$. Значи има $2^{11} = 2048$ парови $(x, y)$ што го задоволуваат условот.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16990,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A\\Beta\\Gamma$ be an isosceles acute-angled triangle with $AB = A\\Gamma$. Let $\\Gamma\\Delta$ be an altitude of the triangle. The circle $c_2(\\Gamma, \\Gamma\\Delta)$ intersects $A\\Gamma$ at point $K$, the extension of $A\\Gamma$ at point $Z$, and the circle $c_1(B, B\\Delta)$ at point $E$. Finally, $\\Delta Z$ intersects the circle $c_1$ at point $M$.\n\nProve that:\n\n(a) $\\angle Z\\Delta E = 45^\\circ$\n\n(b) The points $E$, $M$, and $K$ are collinear.\n\n(c) The line $BM$ is parallel to the line $E\\Gamma$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) From the right-angled triangle $B\\Gamma\\Delta$ we have: $\\hat{\\Gamma}_1 = 90^\\circ - \\hat{B}$.\n\nThe line joining the centers of the circles $c_1$ and $c_2$ is the perpendicular bisector of their common chord $\\Delta E$.\n\nIf $T$ is the point of intersection of the lines $B\\Gamma$ and $\\Delta E$, then:\n\n$$\n\\hat{\\Gamma}_1 + \\hat{\\Delta}_1 + \\hat{\\Delta}_2 = 90^\\circ, \\text{ that is } \\hat{\\Delta}_1 + \\hat{\\Delta}_2 = \\hat{B} = \\hat{\\Gamma}. \\quad (1)\n$$\n\nFrom the isosceles triangle $\\Gamma\\Delta Z$ we have: $\\hat{\\Delta}_1 = \\hat{Z}_1$ and $\\hat{\\Gamma}_2 = 2\\hat{\\Delta}_1 = 90^\\circ - \\hat{A}$, and\n\nhence:\n$$\n\\hat{\\Delta}_1 = 45^\\circ - \\frac{\\hat{A}}{2}. \\quad (2)\n$$\n\nFrom the isosceles triangle $AB\\Gamma$ we have: $\\hat{B} = 90^\\circ - \\frac{\\hat{A}}{2}$. \\quad (3)\n\nFrom (1), (2), (3) we get:\n\n$$\n\\hat{\\Delta}_1 + \\hat{\\Delta}_2 = \\hat{B} = \\hat{\\Gamma} \\Leftrightarrow \\hat{\\Delta}_2 = \\hat{B} - \\hat{\\Delta}_1 \\stackrel{(2),(3)}{\\Leftrightarrow} \\hat{\\Delta}_2 = 90^\\circ - \\frac{\\hat{A}}{2} - \\left( 45^\\circ - \\frac{\\hat{A}}{2} \\right) = 45^\\circ.\n$$\n\n![](images/Hellenic_2016_p9_data_88940f318b.png)\n\n(b) The angle $\\hat{\\Delta}_1$ is created from the chord $\\Delta M$ and the tangent $\\Delta\\Gamma$ of the circle $c_1$, and hence $\\angle \\Delta EM = \\hat{\\Delta}_1$. Also, we have $\\angle \\Delta EK = \\angle \\Delta ZK = \\hat{Z}_1$.\n\nSince $\\hat{\\Delta}_1 = \\hat{Z}_1$, we get $\\angle \\Delta EM = \\angle \\Delta EK$, and the points $E$, $M$, $K$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16991,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $H$ and $O$ be the orthocenter and circumcenter of an acute triangle $\\triangle ABC$, respectively ($A$, $H$, $O$ are non-collinear). Suppose $D$ is the projection of $A$ onto line $BC$, and the perpendicular bisector of the segment $AO$ meets line $BC$ at $E$. Prove that the midpoint $N$ of $OH$ is on the circumcircle of $\\triangle ADE$.",
    "options": [],
    "answer": "See solution",
    "solution": "As seen in the figure below, we extend $HD$ to let it intersect the circumcircle of $\\triangle ABC$ at point $H'$, and denote the midpoint of $AO$ as $F$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p258_data_19304a5e62.png)\n\nSince $H$ is the orthocenter, we have\n\n$$\n\\angle CBH' = \\angle CAH' = \\angle CBH.\n$$\n\nTherefore, $D$ is the midpoint of $HH'$. On the other hand, $N$ is the midpoint of $HO$, so $DN$ is the median of $\\triangle HOH'$. Therefore,\n\n$$\nDN = \\frac{1}{2}OH'.\n$$\n\nSince $OH' = OA$ and $F$ is the midpoint of $OA$, then\n\n$$\nDN = \\frac{1}{2}OH' = \\frac{1}{2}OA = AF.\n$$\n\nIt is easy to see $FN \\parallel AH$. Then $AFND$ is an isosceles trapezoid and $A$, $F$, $N$, $D$ are concyclic.\n\nFurthermore, from $\\angle ADE = 90^\\circ = \\angle AFE$ we know $A$, $F$, $D$, $E$ are also concyclic. Then $A$, $F$, $N$, $D$, $E$ are concyclic, which implies that the circumcircle of $\\triangle ADE$ passes through $N$, the midpoint of $OH$. The proof is complete.\n\n**Remark.** By the fact that the radius of the nine-point circle of a triangle is half that of the circumcircle and $N$ is the center of the nine-point circle of $\\triangle ABC$, we can also see that $AFND$ is an isosceles trapezoid.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 16992,
    "subject": "Mathematics (Olympiad)",
    "question": "Alice has a number of cards. Each card contains three of the letters A to I. For any choice of two of those letters, there is at least one card that contains both letters.\n\nWhat is the smallest number of cards that Alice can have?",
    "options": [],
    "answer": "See solution",
    "solution": "$12$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16993,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to [-2019, 2019]$ for which\n$$\n2f(f(x)) + f(f(-x)) + 8f(x) + 4f(-x) + 4x = 0\n$$\nfor all real numbers $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us define an auxiliary function $g: \\mathbb{R} \\to \\mathbb{R}$ by setting\n$$\ng(x) = f(f(x)) + 4f(x) + 4x\n$$\nfor all real numbers $x$. The given functional equation may be written as\n$$\n2g(x) = -g(-x),\n$$\nfor all real numbers $x$, so that\n$$\n4g(x) = -2g(-x) = -(-g(-(-x))) = g(x),\n$$\nfor all real numbers $x$. We obtain from this that\n$$\nf(f(x)) + 4f(x) + 4x = g(x) = 0,\n$$\nfor all real numbers $x$.\n\nLet now $x$ be an arbitrary nonzero real number, and let us define a sequence $a_0, a_1, a_2, \\dots$ by setting\n$$\na_0 = x, \\quad a_1 = f(x), \\quad a_2 = f(f(x)), \\quad a_3 = f(f(f(x))), \\quad \\dots\n$$\nso that\n$$\na_{n+1} = f(a_n) = f(f(a_{n-1})) = -4f(a_{n-1}) - 4a_{n-1} = -4a_n - 4a_{n-1},\n$$\nfor each positive integer $n$. The general term of the sequence may be written as a function of $n$ through\n$$\na_n = (A + Bn)(-2)^n,\n$$\nwhich holds for all nonnegative integers $n$, and where\n$$\nA = a_0 = x \\quad \\text{and} \\quad B = -\\frac{a_1}{2} - a_0 = -\\frac{f(x)}{2} - x.\n$$\nIf $B \\neq 0$, then we may choose a positive integer $n$ so that $|A + Bn| > 1$ and $2^n > 2019$, and thus\n$$\n|f(a_{n-1})| = |a_n| = |A + Bn| \\cdot |(-2)^n| > 2019,\n$$\nagainst the fact that $f(a_{n-1}) \\in [-2019, 2019]$. Therefore we must have $B = 0$. But now we may choose a positive integer $n$ so that $|x| 2^n > 2019$, leading to\n$$\n|f(a_{n-1})| = |a_n| = |a_0| \\cdot |(-2)^n| = |x| 2^n > 2019,\n$$\nwhich again produces a contradiction. We conclude that no function with the desired properties exists.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16994,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n$ children around a round table. Erika is the oldest among them and she has $n$ candies, while no other child has any candy. Erika decided to distribute the candies according to the following rules. In every round, she chooses a child with at least two candies and the chosen child sends a candy to each of his/her two neighbors. (So in the first round Erika must choose herself.)\n\nFor which $n \\ge 3$ is it possible to end the distribution after a finite number of rounds with every child having exactly one candy?",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show that $n$ cannot be even if every child is to have one candy at the end.\n\nNumber the seats clockwise so that Erika sits at chair $0$. Assign to each candy the number of its owner's chair, and let $S$ be the sum of these numbers. After one move:\n\n- If a child at chair $k$ ($1 \\leq k \\leq n-2$) is chosen, $S$ does not change (since $2k$ is replaced by $(k-1)+(k+1)$).\n- If the child at chair $n-1$ is chosen, $S$ decreases by $n$ (since $2(n-1)$ is replaced by $(n-2)+0$).\n- If Erika (chair $0$) is chosen, $S$ increases by $n$ (since $2 \\cdot 0$ is replaced by $(n-1)+1$).\n\nInitially, $S=0$, so after any number of moves, $S$ is divisible by $n$. If every child has one candy, then\n\n$$\nS = 0 + 1 + 2 + \\dots + (n-1) = \\frac{n(n-1)}{2}.\n$$\n\nThus, $\\frac{S}{n} = \\frac{n-1}{2}$, which is not an integer if $n$ is even. Therefore, such a distribution is impossible for even $n$.\n\nNow, consider odd $n$. Let $n = 2k+1$. We construct a distribution by induction: for each $i = 0, 1, \\dots, k$, we can arrange the candies so that Erika has $n-2i$ candies, and $i$ children to her left and $i$ to her right each have one candy.\n\n- $i=0$: initial state.\n- $i=1$: after the first round.\n- $i=k$: final state, each child has one candy.\n\nSuppose we have achieved the state for $i=m$ ($1 \\leq m < k$). Erika gives a candy to each neighbor (she has at least three candies). Then, by a sequence of symmetric moves (as shown in the diagram), the candies are shifted outward until the next child on each side has one candy, corresponding to $i=m+1$.\n\n$$\n\\begin{align*}\n&\\underline{n-2m, \\underbrace{1, \\dots, 1}_{m}, 0, \\dots} \\rightarrow \\\\\n&n-2m-2, \\underline{2, \\underbrace{1, \\dots, 1}_{m-1}, 0, \\dots} \\rightarrow \\\\\n&n-2m, 0, \\underline{2, \\underbrace{1, \\dots, 1}_{m-2}, 0, \\dots} \\rightarrow \\\\\n&n-2m, 1, 0, \\underline{2, \\underbrace{1, \\dots, 1}_{m-3}, 0, \\dots} \\rightarrow \\\\\n&n-2m, 1, 1, 0, \\underline{2, \\underbrace{1, \\dots, 1}_{m-4}, 0, \\dots} \\rightarrow \\\\\n&\\dots \\\\\n&n-2m, \\underbrace{1, \\dots, 1}_{m-2}, 0, \\underline{2, 0, \\dots} \\rightarrow \\\\\n&n-2m, \\underbrace{1, \\dots, 1}_{m-1}, 0, 1, \\underline{0, \\dots}\n\\end{align*}\n$$\n\nRepeat this process until $i=k$.\n\n% ![](images/CzsMT2006_sol_p3_data_67235366d1.png)\n\nThus, the redistribution is possible if and only if $n$ is odd.\n\nA more formal proof uses induction and the binary relation $\\mapsto$ on $\\mathbb{N}^n$ defined by $(a_1, \\dots, a_n) \\mapsto (a'_1, \\dots, a'_n)$ if there is an index $k$ such that $a'_{k-1} = a_{k-1} + 1$, $a'_k = a_k - 2$, $a'_{k+1} = a_{k+1} + 1$ (indices modulo $n$).\n\nAssume\n$$\n(n, 0, 0, \\dots, 0) \\mapsto \\dots \\mapsto (1, 1, \\dots, 1)\n$$\n\nLet $b_k$ be the number of times the $k$th child gives away two candies. For $2 \\leq k \\leq n$, the number of times the $k$th child receives a candy is $2b_k + 1$, which equals $b_{k-1} + b_{k+1}$. Thus,\n$$\nb_{k+1} = 2b_k - b_{k-1} + 1.\n$$\nBy induction,\n$$\nb_{k+1} = k b_2 - (k-1) b_1 + \\frac{k(k-1)}{2}.\n$$\nSetting $b_1 = b_{n+1}$ and solving, we find $n$ must be odd.\n\nTherefore, the answer: **Redistribution is possible if and only if $n$ is odd.**",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16995,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible that the perimeter of a triangle whose side lengths are integers is divisible by twice the length of the longest side?",
    "options": [],
    "answer": "See solution",
    "solution": "Let the side lengths of the triangle be integers $a$, $b$, and $c$. Without loss of generality, assume $c \\geq a$ and $c \\geq b$. Suppose the perimeter $a + b + c$ is divisible by $2c$. Since $0 < a + b + c \\leq 3c < 4c$, the perimeter $a + b + c$ can be divisible by $2c$ only if $a + b + c = 2c$. But then $a + b = c$, which violates the triangle inequality $a + b > c$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 16996,
    "subject": "Mathematics (Olympiad)",
    "question": "A positive integer $k$ is said to be *visionary* if there are integers $a > 0$ and $b \\ge 0$ such that $a \\cdot k + b \\cdot (k+1) = 2020$. How many visionary integers are there?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $X$ denote the set of visionary integers. We show that $X = \\{\\lfloor \\frac{2020}{n} \\rfloor : 1 \\le n \\le 2020\\}$.  \n\nIf $k$ is a visionary integer, then there exist $a > 0$ and $b \\ge 0$ such that $a k + b (k + 1) = 2020$, i.e., $2020 = k(a + b) + b$, where $0 \\le b < a + b$. This implies that $k = \\lfloor \\frac{2020}{a+b} \\rfloor$ and $1 \\le a + b \\le 2020$, since $a \\ge 1$ and $k \\ge 1$. Hence $k \\in X$.\n\nConversely, let $k \\in X$, i.e., $k = \\lfloor \\frac{2020}{n} \\rfloor$ for some $1 \\le n \\le 2020$. Then $2020 = k n + r$, where $0 \\le r < n$. Put $b = r$ and $a = n - r$. Then $a > 0$ and $b \\ge 0$, and $a k + b (k + 1) = 2020$, so $k$ is visionary.\n\nTo solve the problem, we need to find the cardinality of $X$. Write $X$ as the disjoint union of $X_1 = \\{\\lfloor \\frac{2020}{n} \\rfloor : 1 \\le n < \\sqrt{2020}\\}$ and $X_2 = \\{\\lfloor \\frac{2020}{n} \\rfloor : \\sqrt{2020} < n \\le 2020\\}$. (Note that 2020 is not a square.) The smallest element of $X_1$ is $\\lfloor \\frac{2020}{44} \\rfloor = 45$, and the largest element of $X_2$ is $\\lfloor \\frac{2020}{45} \\rfloor = 44$, so $X_1$ and $X_2$ are disjoint.\n\nIf $1 \\le n_1 < n_2 \\le 44$ and $\\lfloor \\frac{2020}{n_1} \\rfloor = \\lfloor \\frac{2020}{n_2} \\rfloor$, then $0 < \\frac{2020}{n_1} - \\frac{2020}{n_2} < 1$, implying $0 < n_2 - n_1 < 1$, which is impossible. Thus, $X_1$ has exactly $44$ elements.\n\nNext, for any $n$ with $1 \\le n \\le 44$, there exists $q$ with $\\sqrt{2020} < q \\le 2020$ such that $\\lfloor \\frac{2020}{q} \\rfloor = n$. By the Division Algorithm, $2020 = q n + r$ with $0 \\le r < n$. If $q \\le 44$, then $2020 < n(q+1) \\le 44 \\cdot 45 = 1980$, a contradiction. So $q > \\sqrt{2020}$. Also, $n = \\lfloor \\frac{2020}{q} \\rfloor$.\n\nIf $\\sqrt{2020} < n \\le 2020$, then $1 \\le \\frac{2020}{n} < \\sqrt{2020}$, so $1 \\le \\lfloor \\frac{2020}{n} \\rfloor < 44$, i.e., all elements of $X_2$ lie in $[1, 44]$. Thus, $|X_2| = |X_1|$, so $|X| = 44 + 44 = 88$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16997,
    "subject": "Mathematics (Olympiad)",
    "question": "Sequence $\\{a_n\\}$ satisfies $a_1 = a_2 = a_3$.\n\n$$\nb_n = a_n + a_{n+1} + a_{n+2} \\quad (n \\in \\mathbb{N}_+)\n$$\n\nIf $\\{b_n\\}$ is a geometric sequence with common ratio $3$, find the value of $a_{100}$.",
    "options": [],
    "answer": "See solution",
    "solution": "By the condition, $b_n = b_1 \\cdot 3^{n-1} = 3^n$ for $n \\in \\mathbb{N}_+$. Thus,\n\n$$\na_{n+3} - a_n = b_{n+1} - b_n = 3^{n+1} - 3^n = 2 \\cdot 3^n \\quad (n \\in \\mathbb{N}_+)\n$$\n\nTherefore,\n\n$$\n\\begin{align*}\na_{100} &= a_1 + \\sum_{k=1}^{33} (a_{3k+1} - a_{3k-2}) \\\\\n&= 1 + \\sum_{k=1}^{33} 2 \\cdot 3^{3k-2} \\\\\n&= 1 + 6 \\cdot \\frac{27^{33} - 1}{27 - 1} \\\\\n&= 1 + \\frac{3}{13}(3^{99} - 1) = \\frac{3^{100} + 10}{13}\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16998,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m, n, p$ be fixed positive real numbers such that $mnp = 8$. Depending on these constants, find the minimum of\n$$\nx^2 + y^2 + z^2 + mxy + nxz + pyz\n$$\nwhere $x, y, z$ are arbitrary positive real numbers satisfying $xyz = 8$. When is equality attained?\n\nSolve the problem for:\n\nc) $m = n = p = 2$\n\nd) arbitrary (but fixed) positive real numbers $m, n, p$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution:**\n\nFor part (c), $m = n = p = 2$:\n\nBy AM-GM and $xyz = 8$,\n$$\nx^2 + y^2 + z^2 + xy + xy + xz + xz + yz + yz \\geq 9\\sqrt{x^6 y^6 z^6} = 36.\n$$\nEquality holds for $x = y = z = 2$.\n\nFor part (d), arbitrary positive $m, n, p$ with $mnp = 8$:\n\nTransform the expression using $xyz = 8$:\n$$\nx^2 + y^2 + z^2 + mxy + nxz + pyz = x^2 + \\frac{8p}{x} + y^2 + \\frac{8n}{y} + z^2 + \\frac{8m}{z}.\n$$\nApply AM-GM:\n$$\nx^2 + \\frac{8p}{x} \\geq 6\\sqrt{p^2},\n$$\nand similarly for $y, z$. Summing,\n$$\nx^2 + y^2 + z^2 + mxy + nxz + pyz \\geq 6\\sqrt[3]{2}(\\sqrt[3]{m^2} + \\sqrt[3]{n^2} + \\sqrt[3]{p^2}).\n$$\nEquality is achieved when\n$$\nx = \\sqrt[3]{4p}, \\quad y = \\sqrt[3]{4n}, \\quad z = \\sqrt[3]{4m}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 16999,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that $n! + 3$ is a power of $3$.",
    "options": [],
    "answer": "See solution",
    "solution": "The only solutions are $3! + 3 = 6 + 3 = 9 = 3^2$ and $4! + 3 = 24 + 3 = 27 = 3^3$. For $n = 5$, $5! + 3 = 120 + 3 = 123$, which is not a power of $3$. For $n \\geq 6$, note that $n!$ is divisible by $9$ (since both $3$ and $6$ are factors of $n!$). Any power of $3$ greater than $3^1$ is also divisible by $9$, so $n! + 3$ would be divisible by $9$. But $n!$ is divisible by $9$, so $n! + 3$ leaves a remainder of $3$ when divided by $9$, which cannot be a power of $3$. Thus, the only solutions are $n = 3$ and $n = 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17000,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a triangle $ABC$ and a line that does not coincide with any triangle side and passes through point $A$. This line meets altitudes $BH_2$ and $CH_3$ at points $D_1$ and $E_1$, respectively. Let $D_2$ and $E_2$ be the points symmetric to $D_1$ and $E_1$ with respect to sides $AB$ and $AC$, respectively. Prove that the circumcircle of triangle $D_2AB$ is tangent to the circumcircle of triangle $E_2AC$.",
    "options": [],
    "answer": "See solution",
    "solution": "By $E_3$ and $D_3$ we denote the points symmetric to $E_1$ and $D_1$ with respect to $AB$ and $AC$, respectively (see the figure below).\n\nIt follows from symmetry that $AE_1 = AE_2 = AE_3$, and $CA$ is a bisector of the angle formed by the lines $CE_3$ and $CE_2$. Then we have\n\n$$\n\\angle(CE_3, E_3A) = \\angle(E_1E_3, E_3A) = \\angle(AE_1, E_1E_3) = \\angle(AE_1, E_1C) = \\angle(CE_2, E_2A).\n$$\n\nThus, points $A$, $E_2$, $E_3$, and $C$ are cyclic.\n\nSimilarly, points $A$, $D_2$, $D_3$, and $B$ are cyclic. Therefore, the circumcircle of triangle $ABD_2$ coincides with the circumcircle of triangle $AD_3D_2$. Likewise, the circumcircle of triangle $ACE_2$ coincides with the circumcircle of triangle $AE_3E_2$.\n\nSince $AE_3$ and $AD_2$ are symmetric to the same line with respect to $AB$, points $A$, $E_3$, $D_2$ are collinear. Similarly, $A$, $E_2$, $D_3$ are collinear. Moreover,\n\n$$\n k = \\frac{AD_2}{AE_3} = \\frac{AD_1}{AE_1} = \\frac{AD_3}{AE_2}.\n$$\n\nThus, $\\triangle AE_2E_3 \\sim \\triangle AD_3D_2$, and their circumcircles are tangent to each other, because under the homothety $H_A^k$, one of the circles maps to the other and they share the common point $A$.\n\n![](images/Ukraine_booklet_2018_p47_data_b98106d1e2.png)\n\n*Pic. 45*",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17001,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\Gamma$ be the circumcircle of acute triangle $ABC$. The perpendicular to $AB$ from $C$ meets $AB$ at $D$ and $\\Gamma$ again at $E$. The bisector of angle $C$ meets $AB$ at $F$ and $\\Gamma$ again at $G$. The line $GD$ meets $\\Gamma$ again at $H$ and the line $HF$ meets $\\Gamma$ again at $I$. Prove that $AI = EB$.\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p266_data_b46a3f530a.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $CG$ bisects $\\angle ACB$, we have $\\angle AHG = \\angle ACG = \\angle GCB$. Thus, from triangle $ADH$ we find that $\\angle HDB = \\angle HAB + \\angle AHG = \\angle HCB + \\angle GCB = \\angle GCH$. It follows that a pair of opposite angles in the quadrilateral $CFDH$ are supplementary, whence $CFDH$ is a cyclic quadrilateral. Thus, $\\angle GCE = \\angle FCD = \\angle FHD = \\angle IHG = \\angle ICG$. In view of $\\angle ACG = \\angle GCB$ we obtain $\\angle ACI = \\angle ECB$, which implies $AI = EB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17002,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there are no positive integers $m$ and $n$ such that both $4m^2 + 17n^2$ and $17m^2 + 4n^2$ are perfect squares.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $4m^2 + 17n^2$ and $17m^2 + 4n^2$ are both squares. This remains true if we replace $m$ and $n$ by $m/d$ and $n/d$, respectively, where $d = \\gcd(m, n)$, so we may suppose that $m$ and $n$ are coprime.\n\nPut $4m^2 + 17n^2 = s^2$ and $17m^2 + 4n^2 = t^2$. Adding these gives\n$$\n21(m^2 + n^2) = s^2 + t^2.\n$$\n\nConsidering this modulo $3$, we see that $s^2 + t^2 \\equiv 0 \\pmod{3}$. The possible values for $s^2$ and $t^2$ modulo $3$ are $0$ and $1$. It follows that $s^2 \\equiv t^2 \\equiv 0 \\pmod{3}$, and so $s$ and $t$ are both multiples of $3$.\n\nBut then $9$ divides $s^2 + t^2 = 21(m^2 + n^2)$. Since $21$ is not a multiple of $9$, it follows that $m^2 + n^2 \\equiv 0 \\pmod{3}$.\n\nProceeding as before with $s$ and $t$, we now find that $m$ and $n$ are both multiples of $3$. But this contradicts the coprimality of $m$ and $n$.\n\nThus $4m^2 + 17n^2$ and $17m^2 + 4n^2$ cannot both be squares after all.\n\n(This is similar to the classic proof of the irrationality of the square root of $2$ (or of any prime number).)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17003,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, $z$ be positive real numbers. Solve the system:\n\n$$\nx(6 - y) = 9, \\quad y(6 - z) = 9, \\quad z(6 - x) = 9.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $x, y, z > 0$, from the given equations we get $0 < x < 6$, $0 < y < 6$, $0 < z < 6$.\n\nBy multiplying the three equations we get:\n\n$$\nx(6-x) y(6-y) z(6-z) = 9^3. \\quad (1)\n$$\n\nWe have\n\n$$\n0 < x(6-x) = 6x - x^2 \\leq 9. \\quad (2)\n$$\n\nThe equality holds if and only if $x=3$. Similarly,\n\n$$\n0 < y(6-y) \\leq 9 \\quad (3) \\quad \\text{and} \\quad 0 < z(6-z) \\leq 9 \\quad (4)\n$$\n\nMultiplying (2), (3), and (4), we get\n\n$$\n0 < x(6-x) y(6-y) z(6-z) \\leq 9^3. \\quad (5)\n$$\n\nComparing (5) with (1), we conclude that (2), (3), and (4) must hold as equalities, that is $x = y = z = 3$.\n\nAlternatively, (2), (3), and (4) can be found by using the inequality of arithmetic-geometric mean. For example, since $x, 6-x > 0$, we have\n\n$$\n0 < \\sqrt{x(6-x)} \\leq \\frac{x + 6 - x}{2} = 3 \\Rightarrow 0 < x(6-x) \\leq 9.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17004,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the largest positive integer $k$ for which there exists a convex polyhedron $P$ with the following properties:\n\n(a) $P$ has exactly 2022 edges.\n\n(b) The degrees of the vertices of $P$ don't differ by more than 1.\n\n(c) It is possible to colour the edges of $P$ with $k$ colours such that for every colour $c$, and every pair of vertices ($v_1$, $v_2$) of $P$, there is a monochromatic path between $v_1$ and $v_2$ in the colour $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "We divide the solution in two steps: first, we prove that $k < 3$, and then give an inductive construction of $P$ for $k = 2$.\n\nLet $P$ have $V$ vertices, $E$ edges, and $F$ faces. Suppose, for contradiction, that $k > 2$. We have $k$ disjoint trees on $V$ vertices, so $E \\geq 3(V - 1)$. This is a contradiction, as for every polyhedron we have $E \\leq 3V - 6$.\n\nNow, take $k = 2$ and prove that for every positive integer $n$, we can find a convex polyhedron $P_{6n}$ with exactly $6n$ edges that satisfies the conditions of the problem.\n\nFor $n = 1$, consider the tetrahedron $ABCD$. One colouring that works is: $AB$, $AD$, and $CD$ are in one colour, and $AC$, $BC$, and $BD$ are in the other colour.\n\nSuppose we have constructed $P_{6n}$. Here is how to construct $P_{6n+6}$: Consider the triangular face $T = xyz$ most recently added to $P_{6n}$, and glue on top of $T$ a truncated pyramid whose larger base is $T$. We are effectively adding 3 new vertices, say $x'$, $y'$, $z'$, 3 faces, and 6 edges: $xx'$, $yy'$, $zz'$, $x'y'$, $y'z'$, $z'x'$. We colour $x'y'$, $x'z'$, $y'z'$ with the first colour, and all other new edges with the second colour. It is now easy to see that in any one of the colours, a tree on the vertices of $P_{6n}$ in that colour, together with the newly added edges of that colour, gives a tree on the vertices of $P_{6n+6}$ in that colour.\n\nNote that $x'y'z'$ is the most recently added triangular face, so the construction can proceed inductively by gluing another truncated pyramid on top of it. It's easy to see (and prove inductively) that every vertex has degree 3 or 4, so condition (b) is satisfied and we are done.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17005,
    "subject": "Mathematics (Olympiad)",
    "question": "For some positive integer $n$, the pair of positive integers $(a, b)$ is called the \"lucky pair of $n$\" if $a + b = n$ and there exists some prime number $p$ that satisfies the equality\n$$\n\\frac{4}{a} + \\frac{4}{b} = \\frac{1}{p}.\n$$\nFind all \"lucky pair of 2025\".",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $(a, b)$ is a \"lucky pair of 2025\". By assumption, there exists a prime number $p$ such that\n$$\n\\frac{4}{a} + \\frac{4}{b} = \\frac{1}{p}.\n$$\nSince $a + b = 2025$, we have:\n$$\n\\frac{4}{a} + \\frac{4}{2025 - a} = \\frac{1}{p}.\n$$\nMultiply both sides by $a(2025 - a)p$:\n$$\n4p(2025 - a) + 4p a = a(2025 - a)\n$$\n$$\n4p(2025) = a(2025 - a)\n$$\n$$\n a^2 - 2025a + 4 \\cdot 2025p = 0\n$$\nThis is a quadratic in $a$. The discriminant is:\n$$\n\\Delta = 2025^2 - 16 \\cdot 2025p = 45^2(2025 - 16p)\n$$\nFor integer solutions, $2025 - 16p$ must be a perfect square. Let $2025 - 16p = m^2$ for $m \\in \\mathbb{N}$:\n$$\n45^2 - m^2 = 16p \\implies (45 - m)(45 + m) = 16p\n$$\nConsider possible factorizations:\n\n1. $45 + m = 2p$, $45 - m = 8$:\n   - $m = 37$, $p = 41$\n   - Solutions: $(a, b) = (1845, 180)$ and $(180, 1845)$\n\n2. $45 + m = 8p$, $45 - m = 2$:\n   - $m = 43$, $p = 11$\n   - Solutions: $(a, b) = (1980, 45)$ and $(45, 1980)$\n\nTherefore, the four \"lucky pairs of 2025\" are:\n$$\n(1845, 180),\\ (180, 1845),\\ (1980, 45),\\ (45, 1980)\n$$\n$\\boxed{(1845, 180),\\ (180, 1845),\\ (1980, 45),\\ (45, 1980)}$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17006,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive composite number $n$, let $f(n)$ be the sum of the smallest three positive divisors of $n$, and $g(n)$ be the sum of the largest two positive divisors of $n$. Find all $n$ such that $g(n)$ equals $f(n)$ raised to some positive integer power.",
    "options": [],
    "answer": "See solution",
    "solution": "If $n$ is odd, all its divisors are odd, so $f(n)$ is odd and $g(n)$ is even; thus, $g(n)$ cannot be a power of $f(n)$. Therefore, $n$ must be even. The smallest two divisors of $n$ are $1$ and $2$, and the largest two are $n$ and $n/2$.\n\nLet $d$ be the third smallest divisor of $n$. If there exists $k \\in \\mathbb{N}^*$ such that $g(n) = f(n)^k$, then\n\n$$\n\\frac{3n}{2} = (1 + 2 + d)^k = (3 + d)^k \\equiv d^k \\pmod{3}.\n$$\n\nSince $3$ does not divide $\\frac{3n}{2}$, $3$ must divide $d^k$, so $3$ divides $d$. As $d$ is the third smallest divisor, $d = 3$.\n\nThus, $\\frac{3}{2}n = 6^k$, so $n = 4 \\times 6^{k-1}$. Since $3$ does not divide $n$, $k \\geq 2$.\n\nIn summary, all such $n$ are $n = 4 \\times 6^l$ for $l \\in \\mathbb{N}^*$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17007,
    "subject": "Mathematics (Olympiad)",
    "question": "For each part below, consider triangles made from matches, with integer side lengths and a fixed total number of matches.\n\n### a\nFind all right-angled triangles that can be made with 24 matches, where each side length is an integer.\n\n### b\nFind all isosceles triangles that can be made with 24 matches, where each side length is an integer.\n\n### c\nList all possible triangles with integer side lengths and a total of 24 matches, where the longest side is at most 11.\n\n### d\nWhat is the least number of matches Steve needs to make two different right-angled triangles with the same perimeter, where all side lengths are integers?\n\n![](images/Australian_Scene_2012_-_AMT_Publishing_-_273p_p58_data_0163b0c48c.png)",
    "options": [],
    "answer": "See solution",
    "solution": "### a\nFor any triangle, the length of any side is less than the sum of the lengths of the other two sides. Thus, each side is at most 11. If one side is 11, possible pairs for the other sides are $(1,12), (2,11), (3,10), (4,9), (5,8), (6,7)$. None of these form a right-angled triangle. If one side is 10, possible pairs are $(1,13), (2,12), (3,11), (4,10), (5,9), (6,8), (7,7)$. Since $6^2 + 8^2 = 10^2$, triangle $(6, 8, 10)$ is right-angled.\n\nAlternatively, triangle $(3, 4, 5)$ is right-angled because $3^2 + 4^2 = 5^2$ and uses 12 matches. Doubling each side gives $(6, 8, 10)$, which uses 24 matches and is also right-angled.\n\n### b\nThe sum of two equal sides must be greater than the third side. The equal sides are at least 7 and at most 11. The five isosceles triangles are: $(2, 11, 11), (4, 10, 10), (6, 9, 9), (8, 8, 8), (10, 7, 7)$.\n\nAlternatively, since the total number of matches in two equal sides is even, and the total is 24, the third side must also be even. The possible values for the third side are $2, 4, 6, 8, 10$, giving the same five triangles.\n\n### c\nEach side must be shorter than the sum of the other two sides, so the longest side is at most 11. The possible triangles are:\n- Longest side 11: $(11, 11, 2), (11, 10, 3), (11, 9, 4), (11, 8, 5), (11, 7, 6)$\n- Longest side 10: $(10, 10, 4), (10, 9, 5), (10, 8, 6), (10, 7, 7)$\n- Longest side 9: $(9, 9, 6), (9, 8, 7)$\n- Longest side 8: $(8, 8, 8)$\n\nThus, only these 12 triangles are possible.\n\n### d\nExperimenting with small integers, $3^2 + 4^2 = 5^2$ and $5^2 + 12^2 = 13^2$. Their perimeters are 12 and 30. Multiplying the sides of $(3, 4, 5)$ by 5 and $(5, 12, 13)$ by 2 gives $(15, 20, 25)$ and $(10, 24, 26)$, both with perimeter 60. These are right-angled triangles.\n\nA systematic search for integer triples $(a, b, c)$ with $1 \\leq a \\leq b \\leq c$, $a + b + c \\leq 60$, and $a^2 + b^2 = c^2$ yields:\n$(3, 4, 5), (5, 12, 13), (6, 8, 10), (7, 24, 25), (8, 15, 17), (9, 12, 15), (10, 24, 26), (12, 16, 20), (15, 20, 25)$.\n\nTheir perimeters are 12, 30, 24, 56, 40, 36, 60, 48, 60. Thus, $(10, 24, 26)$ and $(15, 20, 25)$ are the only pair with the same perimeter at most 60. So the least number of matches Steve needs is 60.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17008,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $AC > AB$. The point $X$ lies on the side $BA$ extended through $A$, and the point $Y$ lies on the side $CA$ in such a way that $BX = CA$ and $CY = BA$. The line $XY$ meets the perpendicular bisector of side $BC$ at $P$. Show that\n\n$$\n\\angle BPC + \\angle BAC = 180^{\\circ}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Draw the bisector of $\\angle A$, and let it meet $BC$ at $R$, and the circumcircle of $\\triangle ABC$ at $Q$. Let $XY$ meet $BC$ at $T$, and let $M$ be the midpoint of $BC$. Then $Q$ lies on the perpendicular bisector of $BC$ (which is the line $PM$), since $BQ = CQ \\Leftrightarrow \\angle BAQ = \\angle CAQ$.\n\n![](images/V_Britanija_2006_p15_data_7517a13881.png)\n\nSince $AB = CY$ and $BX = CA$, we have $AX = YA$ and so $\\triangle XAY$ is isosceles. Then $\\angle XYA = \\angle YXA = \\frac{180^{\\circ} - \\angle XAY}{2} = \\frac{\\angle CAB}{2} = \\angle CAR$, and so\n\nline $XY$ is parallel to line $AR$. This shows that $\\triangle ACR$ is similar to $\\triangle YCT$ (angle-angle-angle). Thus\n\n$$\n\\frac{CT}{CR} = \\frac{CY}{CA} = \\frac{BA}{CA}\n$$\n\nby similar triangles and the definition of the point $Y$. By the angle bisector theorem,\n\n$$\n\\frac{BR}{CR} = \\frac{BA}{CA}\n$$\n\nand so from the last two equations, $CT = BR$. So, since $M$ is the midpoint of $BC$, we must have $MR = MT$. Since $PM = QM$, and $PT$ is parallel to $RQ$, we have $\\triangle PMT$ is congruent to $\\triangle QMR$. Thus $PM = QM$. So now, by side-angle-side, $\\triangle BMP$ is congruent to $\\triangle BMQ$. So\n\n$$\n\\begin{aligned}\n\\angle PBM &= \\angle QBM \\\\\n&= \\angle CAQ \\quad (\\text{by angles in the same segment}) \\\\\n&= \\frac{\\angle BAC}{2} \\quad (\\text{since } AQ \\text{ bisects } \\angle BAC).\n\\end{aligned}\n$$\n\nBut $\\angle BPC = 2\\angle BPM = 180^{\\circ} - 2\\angle PBM$ (since $PM = 90^{\\circ}$), and so\n\n$$\n\\angle BPC + \\angle BAC = 180^{\\circ}\n$$\n\nas desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17009,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n > 100$ be an integer. Ivan writes the numbers $n, n + 1, \\dots, 2n$ each on different cards. He then shuffles these $n + 1$ cards and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "It suffices to find three integers $a, b, c \\in [n, 2n]$ such that for some integer $k$,\n\n$$\na + b = (2k - 1)^2, \\quad a + c = (2k)^2, \\quad b + c = (2k + 1)^2.\n$$\n\nIn this way, two of $a, b, c$ must be put into one pile, and their sum is a perfect square. Solving the equations yields $a = 2k^2 - 4k$, $b = 2k^2 + 1$, $c = 2k^2 + 4k$. We must require that $n \\leq 2k^2 - 4k$ and $2k^2 + 4k \\leq 2n$. Let $2m^2 - 2 < n \\leq 2(m+1)^2 - 2$, where $m$ is an integer. From the inequalities, we find that $k$ must satisfy $m + 2 \\leq k \\leq \\sqrt{n+1} - 1$. If $m = 7$, $n \\geq 99$ is needed; if $m \\geq 8$, $\\sqrt{n+1} - 1 \\geq \\sqrt{2m-1} > m + 2$ always holds. Hence, the conclusion is true for $n \\geq 99$.\n\n_Remark_: The statement is untrue when $n = 98$. For example, Ivan can put all even numbers between 98 and 126, all odd numbers between 129 and 161, and all even numbers between 162 and 196 into the first pile; he puts the other numbers into the second pile. No two numbers in a pile add up to a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17010,
    "subject": "Mathematics (Olympiad)",
    "question": "For integers $a$ and $b$ and for a positive integer $c$, we write $a \\equiv b \\pmod{c}$ if $a-b$ is divisible by $c$.\n\nFind all positive integers $n$ such that $\\dfrac{10^n}{n^3 + n^2 + n + 1}$ is an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "We shall show that $n=3$ and $n=7$ are the only solutions.\n\nSince\n$$\n\\frac{10^3}{3^3 + 3^2 + 3 + 1} = 25, \\quad \\frac{10^7}{7^3 + 7^2 + 7 + 1} = 25000,\n$$\nwe see that $n=3$ and $n=7$ satisfy the condition.\n\nNow, factor $n^3 + n^2 + n + 1 = (n+1)(n^2 + 1)$. For $\\dfrac{10^n}{n^3 + n^2 + n + 1}$ to be an integer, both $n+1$ and $n^2 + 1$ must have only 2 and 5 as prime factors.\n\nThe greatest common divisor $\\gcd(n+1, n^2 + 1)$ is either 1 or 2, since $(n^2 + 1) - (n+1)(n-1) = 2$.\n\nIf $n$ is even, then $n+1$ and $n^2 + 1$ are odd, so both must be divisible by 5, which contradicts $\\gcd(n+1, n^2 + 1) = 1$ or 2. Thus, $n$ must be odd, so $n+1$ and $n^2 + 1$ are both even. Furthermore, $n^2 + 1 \\equiv 2 \\pmod{4}$, so $n^2 + 1$ is not divisible by 4.\n\n**Case 1:** $n^2 + 1$ is not divisible by 5.\n\nThen $n^2 + 1$ is a power of 2, but since it is not divisible by 4, $n^2 + 1 = 2$, i.e., $n=1$. But $\\frac{10^1}{1^3 + 1^2 + 1 + 1} = \\frac{10}{4} = 2.5$ is not an integer.\n\n**Case 2:** $n^2 + 1$ is divisible by 5.\n\nLet $n+1 = 2^k$, $n^2 + 1 = 2 \\cdot 5^l$ for positive integers $k \\geq 2$, $l$. If $k=2$, $n=3$ is a solution. For $k \\geq 3$, $2 \\cdot 5^l = (2^k - 1)^2 + 1$ leads to $5^l - 1 = 2^k(2^{k-1} - 1)$, so $5^l - 1$ is a multiple of 8. This forces $l$ to be even, say $l=2m$.\n\nThen $(5^m - 1)(5^m + 1) = 2^k(2^{k-1} - 1)$. Since $5^m + 1 \\equiv 2 \\pmod{4}$, write $5^m - 1 = 2^{k-1}a$ for odd $a$. Then $a(2^{k-2}a + 1) = 2^{k-1} - 1$. If $a \\geq 3$, $a(2^{k-2}a + 1) > 2^{k-1} - 1$, a contradiction. Thus $a=1$, so $2^{k-1}(2^{k-1} + 2) = 2^k(2^{k-1} - 1)$, which gives $2^{k-1} = 4$, so $k=3$, $n=7$.\n\nTherefore, the only solutions are $n=3$ and $n=7$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17011,
    "subject": "Mathematics (Olympiad)",
    "question": "How many positive factors of $6000$ are not perfect squares?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $6000 = 2^4 3^1 5^3$, positive factors of $6000$ can be written as $2^a 3^b 5^c$ where $a = 0, 1, 2, 3, 4$; $b = 0, 1$; $c = 0, 1, 2, 3$. Therefore, there are altogether $5 \\times 2 \\times 4 = 40$ positive factors of $6000$.\n\nAmong them, square numbers arise only when all of $a, b, c$ are even, and this happens only $3 \\times 1 \\times 2 = 6$ times. Therefore, there are $40 - 6 = 34$ non-square factors of $6000$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17012,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are three classes, each with $n$ students. The students in each class are ranked by height. The students are to be divided into groups of three, each group containing one student from each class. In each group, the tallest student is called the leader. Prove that if each class has at least 10 leaders, then $n \\geq 40$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution 1**\n\n**Lemma:** Suppose the conditions are all satisfied. Then for each class $i$ ($1 \\le i \\le 3$), there is a positive integer $k_i$, such that among the tallest $k_i$ students from all classes, the number of those from class $i$ is at least 10 more than those from the other two classes.\n\n**Proof of Lemma:** Pick any class, say class $A$, and rank their heights from tallest to smallest as $a_1 < a_2 < \\dots < a_n$. The other classes $B$ and $C$ have $x_1 < x_2 < \\dots < x_{2n}$. For every $1 \\le i \\le n-9$, let $a_{i+9}$ (from class $A$) and $x_i$ (from class $B$ or $C$) be in a group. Then add a class $B$ student to every group of class $A$ and class $C$ students; add a class $C$ student to every group of class $A$ and class $B$ students. Now, other than $a_1, a_2, \\dots, a_9$, there must be another leader from class $A$, say $a_m$. We must have $a_m < x_{m-9}$, meaning that among all students taller or equal to $a_m$, at least $m$ of them are from class $A$, and at most $m-10$ from $B$ or $C$. The lemma is verified.\n\nReturn to the original problem. Suppose the classes $A, B, C$ correspond to integers $k_1, k_2, k_3$ as in the lemma, respectively, and $k_1 \\le k_2 \\le k_3$. Among $1, 2, \\dots, k_1$, at least 10 students are from class $A$; among $1, 2, \\dots, k_2$, at least $10+10=20$ students are from class $B$; among $1, 2, \\dots, k_3$, at least $10+20+10=40$ students are from class $C$. This implies that each class has at least $n=40$ students.\n\n**Solution 2:**\n\nWe show that the conditions are not met if $n < 40$. First, there must be $10 \\cdot 3 = 30$ or more groups so as to have 10 leaders in each class, hence $n \\ge 30$. Rank the students in each class from tallest to smallest as $a_1 < a_2 < \\dots < a_n$, $b_1 < b_2 < \\dots < b_n$ and $c_1 < c_2 < \\dots < c_n$. Consider $a_{n-19}, b_{n-19}, c_{n-19}$, and assume $a_{n-19}$ is the tallest. Since $n \\le 39$, we infer that $a_1, a_2, \\dots, a_{n-19}$ are all taller than $b_{20}, b_{21}, \\dots, b_n, c_{20}, c_{21}, \\dots, c_n$. For $1 \\le i \\le n-19$, make $a_i, b_{i+19}$, and $c_{i+19}$ a group, each with a leader from class $A$; for the others, make groups in an arbitrary way. Then class $B$ and $C$ together have at most 19 leaders, a contradiction. Thus, $n \\ge 40$.\n\n**Remark:** In general, if there always exist $k$ leaders in each class, then $n \\ge 4k$. Furthermore, if there are $m$ classes, then $n \\ge 2^{m-1}k$. In this problem, $(m, k) = (3, 10)$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17013,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $f(x) = a \\sin x - \\frac{1}{2}\\cos 2x + a - \\frac{3}{a} + \\frac{1}{2}$, where $a \\in \\mathbb{R}$ and $a \\neq 0$.\n\n1. If $f(x) \\le 0$ for any $x \\in \\mathbb{R}$, find the range of $a$.\n2. If $a \\ge 2$ and there exists $x \\in \\mathbb{R}$ such that $f(x) \\le 0$, find the range of $a$.",
    "options": [],
    "answer": "See solution",
    "solution": "(1) We have $f(x) = \\sin^2 x + a \\sin x + a - \\frac{3}{a}$. Let $t = \\sin x$ ($-1 \\le t \\le 1$). Then\n\n$$\ng(t) = t^2 + a t + a - \\frac{3}{a}.\n$$\n\nThe necessary and sufficient condition for $f(x) \\le 0$ for all $x \\in \\mathbb{R}$ is\n\n$$\n\\begin{cases}\ng(-1) = 1 - \\frac{3}{a} \\le 0, \\\\\ng(1) = 1 + 2a - \\frac{3}{a} \\le 0.\n\\end{cases}\n$$\n\nTherefore, the range of $a$ is $(0, 1]$.\n\n(2) As $a \\ge 2$, then $-\\frac{a}{2} \\le -1$. We have\n\n$$\ng(t)_{\\min} = g(-1) = 1 - \\frac{3}{a}.\n$$\n\nSo $f(x)_{\\min} = 1 - \\frac{3}{a}$. Therefore, the necessary and sufficient condition for $f(x) \\le 0$ for some $x \\in \\mathbb{R}$ is $1 - \\frac{3}{a} \\le 0$, or $0 < a \\le 3$.\n\nFinally, the range of $a$ is $[2, 3]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17014,
    "subject": "Mathematics (Olympiad)",
    "question": "In $\\triangle ABC$, point $B_2$ is the reflection of the center of the B-excircle with respect to the midpoint of side $AC$, and point $C_2$ is the reflection of the center of the C-excircle with respect to the midpoint of side $AB$. The A-excircle touches side $BC$ at point $D$. Prove that $AD \\perp B_2C_2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A_1$ be the center of the A-excircle. By properties of excenter geometry, the following sets of three points are collinear: $\\{B_1, A, C_1\\}$, $\\{A_1, C, B_1\\}$, and $\\{C_1, B, A_1\\}$. Also, $A_1A \\perp B_1C_1$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p265_data_79dee19232.png)\n\nOn the plane, choose point $P$ such that $\\vec{C_2P} = \\vec{B_2C}$. Since $\\vec{B_2C} = \\vec{AB_1}$, it follows that $\\vec{C_2P} = \\vec{AB_1}$. As $\\vec{BC_2} = \\vec{C_1A}$ and the points $C_1, B, A_1$ are collinear, the points $B, C_2, P$ are collinear, so $\\vec{BP} = \\vec{C_1B_1}$. It follows from\n\n$$\n\\begin{aligned}\n\\angle AC_1B &= 180^\\circ - \\left( \\frac{180^\\circ - \\angle BAC}{2} \\right) - \\left( \\frac{180^\\circ - \\angle ABC}{2} \\right) \\\\\n&= \\frac{\\angle BAC + \\angle ABC}{2} = \\frac{180^\\circ - \\angle ACB}{2} = \\angle BCA_1\n\\end{aligned}\n$$\n\nthat $\\triangle A_1BC \\sim \\triangle A_1B_1C_1$. Let $A_1D$ and $A_1A$ be the altitudes of $\\triangle A_1BC$ and $\\triangle A_1B_1C_1$, respectively, with respect to the opposite sides, so $\\frac{B_1C_1}{BC} = \\frac{A_1A}{A_1D}$.\n\nIf $\\vec{BP} = \\vec{C_1B_1}$, then $A_1A \\perp B_1C_1$, so $\\frac{BP}{BC} = \\frac{A_1A}{A_1D}$. If $BP \\perp A_1A$, then $BC \\perp A_1D$, so $\\triangle BPC \\sim \\triangle A_1AD$, and hence $CP \\perp AD$. Again, since $\\vec{C_2P} = \\vec{B_2C}$, we have $\\vec{B_2C_2} = \\vec{CP}$, and hence $AD \\perp B_2C_2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17015,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, $C$ be points on a circle with center $O$ and radius $1$, such that $\\angle BAC = 45^\\circ$. Lines $AC$ and $BO$ (possibly extended) intersect at $D$, and lines $AB$ and $CO$ (possibly extended) intersect at $E$. Prove that $BD \\cdot CE = 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the figure:\n\n![](images/s3s2019_p1_data_500c4fc418.png)\n\nLet $x = OD$ and $y = OE$. Let $BO$ extended meet the circle again at $F$ (so $DF = 1 - x$). Since $\\angle BFC = 45^\\circ$ and $BF$ is a diameter, $\\angle BCF = 90^\\circ$ and $BC = CF = \\sqrt{2}$. Similarly, if $CE$ extended meets the circle again at $G$, then $GE = 1 - y$, and $CG$ and $BF$ are diagonals of square $BCFG$ intersecting perpendicularly at $O$. From right triangles $BOE$ and $COD$, $CD = \\sqrt{1 + x^2}$ and $BE = \\sqrt{1 + y^2}$.\n\nBy similarity of triangles $BDA$ and $CDF$, $\\frac{AD}{DF} = \\frac{BD}{CD}$, i.e., $\\frac{AD}{1 - x} = \\frac{1 + x}{\\sqrt{1 + x^2}}$, so\n$$\nAD = \\frac{1 - x^2}{\\sqrt{1 + x^2}}.\n$$\n\nBy similarity of triangles $BEG$ and $CEA$,\n$$\n\\frac{AD + DC}{GB} = \\frac{CE}{BE},\n$$\ni.e.,\n$$\n\\frac{\\frac{1 - x^2}{\\sqrt{1 + x^2}} + \\sqrt{1 + x^2}}{\\sqrt{2}} = \\frac{1 + y}{\\sqrt{1 + y^2}}.\n$$\n\nSolving for $x$ gives $x = \\frac{1 - y}{1 + y}$, so\n$$\nBD \\cdot CE = (1 + x)(1 + y) = \\left(1 + \\frac{1 - y}{1 + y}\\right)(1 + y) = 2.\n$$\n\nIf $A$ lies strictly between $G$ and $F$ on the short arc $\\widehat{FG}$, this covers the case. If $A = F$ or $A = G$, then $BD \\cdot CE$ reduces to either $BF \\cdot CO = 2 \\cdot 1 = 2$ or $CG \\cdot BO = 2$.\n\nIf $A$ lies between $C$ and $F$ on the short arc $\\widehat{CF}$ (see below):\n\n![](images/s3s2019_p2_data_b0fbd5104b.png)\n\nBy similarity of triangles $BDA$ and $CDF$, $\\frac{CD}{2 + x} = \\frac{\\sqrt{1 + (1 + x)^2}}{2 + x} = \\frac{x}{AD}$, so $AD = \\frac{x(2 + x)}{\\sqrt{1 + (1 + x)^2}}$. Similarity of triangles $BEG$ and $CEA$ gives\n$$\n\\frac{1 - y}{\\sqrt{1 + y^2}} = \\frac{CA}{\\sqrt{2}} = \\frac{CD - AD}{\\sqrt{2}} = \\frac{\\sqrt{1 + (1 + x)^2} - \\frac{x(2 + x)}{\\sqrt{1 + (1 + x)^2}}}{\\sqrt{2}}.\n$$\nSolving gives $x = \\frac{2y}{1 - y}$, so $BD \\cdot CE = (2 + x)(1 - y) = \\left(2 + \\frac{2y}{1 - y}\\right)(1 - y) = 2$.\n\nIf $A$ lies between $B$ and $G$ on the short arc $\\overset{\\sim}{B}G$, the result follows symmetrically.\n\nAlternatively, a trigonometric approach can be used (see below):\n\n![](images/s3s2019_p2_data_4fae312388.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17016,
    "subject": "Mathematics (Olympiad)",
    "question": "A disk of radius 1 rolls all the way around the inside of a square of side length $s > 4$ and sweeps out a region of area $A$. A second disk of radius 1 rolls all the way around the outside of the same square and sweeps out a region of area $2A$. The value of $s$ can be written as $a + \\frac{b\\pi}{c}$, where $a$, $b$, and $c$ are positive integers and $b$ and $c$ are relatively prime. What is $a + b + c$?\n\n(A) 10  (B) 11  (C) 12  (D) 13  (E) 14",
    "options": [],
    "answer": "See solution",
    "solution": "To obtain the region swept out by the first disk, remove a square of side length $s-4$ from the center of the original square and replace the 4 unit squares at the corners of the original square with 4 quarter-circles of radius 1. The area of this region is\n\n$$\ns^2 - (s-4)^2 - 4 + \\pi = 8s - 20 + \\pi.\n$$\n\nThe region swept out by the second disk is the disjoint union of 4 rectangles, each with length $s$ and width 2, and 4 quarter-circles of radius 2, so its area is $8s + 4\\pi$. Therefore,\n\n$$\n8s + 4\\pi = 2(8s - 20 + \\pi).\n$$\n\nSolving this equation gives $s = 5 + \\frac{\\pi}{4}$, so the requested sum is $5 + 1 + 4 = 10$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17017,
    "subject": "Mathematics (Olympiad)",
    "question": "To each $n$ chosen from the set of integers $1, 2, \\ldots, 10$, there is a pair $(m, n)$ of integers satisfying the conditions $m > n$ and $m + n = 22$. Therefore, there are exactly $10$ such pairs $(m, n)$. Find the number of sextuples $(a, b, c, d, e, f)$ such that $a > b > c > d > e > f$ and each $(a, f), (b, e), (c, d)$ is one of these pairs.",
    "options": [],
    "answer": "See solution",
    "solution": "Finding sextuples $(a, b, c, d, e, f)$ satisfying the conditions is equivalent to finding triplets of pairs $(m, n)$ from the set of $10$ pairs. If we order the $3$ pairs chosen as $(a, f), (b, e), (c, d)$ with $f < e < d$, then associate this triplet with the sextuple $(a, b, c, d, e, f)$. This satisfies $a > b > c > d > e > f$. Conversely, any such sextuple arises from a triplet of pairs from the set of $10$ pairs. Therefore, the number we seek is $\\binom{10}{3} = 120$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17018,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ with $AB < AC < BC$ and circumcircle $\\Gamma_1$ centered at $O$. Let the circles $\\Gamma_2(B, AC)$ and $\\Gamma_3(C, AB)$ have a common point $E$. Let $\\Gamma_1$ and $\\Gamma_3$ have a common point $F$, and $\\Gamma_1$ and $\\Gamma_2$ have a common point $G$ (the points $E$, $F$, $G$ are on the same side of $BC$ as $A$). Prove that the point diametrically opposite to $A$ with respect to $\\Gamma_1$ is the circumcenter of $\\triangle EFG$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K$ be the diametrically opposite point to $A$ with respect to $\\Gamma_1$. We will prove that the points $B$, $F$, $E$ are collinear.\n\nFrom the circle $\\Gamma_2$, we have $BE = AC$, and from $\\Gamma_1$, we have $CE = AB$. Therefore, $ABEC$ has its opposite sides equal, so it is a parallelogram. This means that\n\n1. $BE \\parallel AC$.\n\nFrom the circle $\\Gamma_3$, we have $CF = AB$, so the quadrilateral $ABFC$ is an isosceles trapezoid, thus\n\n2. $BF \\parallel AC$.\n\nFrom (1) and (2), we get that $B$, $F$, and $E$ are collinear. Similarly, we can prove that $C$, $G$, $E$ are collinear. Therefore, $ABEC$ is a parallelogram and\n\n3. $CE \\parallel AB$.\n\nFrom the circle $\\Gamma_2$, we get $BG = AC$, so $ACGB$ is an isosceles trapezoid, thus\n\n4. $CG \\parallel AB$.\n\nFrom (3) and (4), we get that $C$, $G$, and $E$ are collinear.\n\nNow, we will prove that $K$ is on the perpendicular bisector of $FE$. Indeed, $CK \\perp AC$ and $AC \\parallel FE$, so $CK \\perp FE$. On the other hand, $\\triangle CFE$ is isosceles, so its altitude is a perpendicular bisector. It follows that $K$ is on the perpendicular bisector of $FE$. Similarly, working with $\\Gamma_2$ and the isosceles $\\triangle EBG$, we get $K$ is on the perpendicular bisector of $EG$. It follows that $K$ is the intersection point of the perpendicular bisectors of $\\triangle FGE$, therefore it is its circumcenter.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17019,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be positive integers satisfying\n\n$$\n\\frac{(ab-1)(ac-1)}{bc} = 2023, \\quad \\text{and} \\quad b \\le c.\n$$\n\nFind all possible values for $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "The possible values for $c$ are $82$, $167$, and $1034$.\n\nWe have\n\n$$\n\\frac{(ab-1)(ac-1)}{bc} = \\left(a - \\frac{1}{b}\\right) \\left(a - \\frac{1}{c}\\right).\n$$\n\nSince $0 \\le a - 1 \\le a - \\frac{1}{b} < a$ and $0 \\le a - 1 \\le a - \\frac{1}{c} < a$, we conclude\n\n$$\n(a-1)^2 \\le \\frac{(ab-1)(ac-1)}{bc} < a^2,\n$$\n\nand hence, $(a-1)^2 \\le 2023 < a^2$. Since $44^2 = 1936 < 2023 < 2025 = 45^2$, we have $a-1 \\le \\sqrt{2023} < 45$ and $44 < \\sqrt{2023} < a$. Therefore, $44 < a < 46$, so $a = 45$.\n\nSubstituting $a = 45$ into $\\frac{(ab-1)(ac-1)}{bc} = 2023$, we get $(45b-1)(45c-1) = 2023bc$, which can be rewritten as $2bc-45b-45c+1 = 0$. This leads to $(2b-45)(2c-45) = 2023$.\n\nSince $2023 = 7 \\cdot 17^2$ and $-43 \\le 2b-45 \\le 2c-45$, the possible pairs for $(2b - 45, 2c - 45)$ are $(1, 2023)$, $(7, 289)$, and $(17, 119)$. The corresponding pairs are $(b, c) = (23, 1034)$, $(26, 167)$, $(31, 82)$, all satisfying $b \\le c$.\n\nTherefore, the possible values for $c$ are $82$, $167$, and $1034$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17020,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$ be the length of a side of the square $ABCD$. Given $ZS = x - 7$ and $DS = x - 6$, and that triangle $ZSD$ is right-angled, find the value of $x$ if $ZS^2 + DS^2 = 25$ and $x > BP$.",
    "options": [],
    "answer": "See solution",
    "solution": "Applying the Pythagorean theorem to triangle $ZSD$, we have:\n\n$$\n(x - 7)^2 + (x - 6)^2 = 5^2\n$$\n\nExpanding and simplifying:\n\n$$\n(x - 7)^2 + (x - 6)^2 = 25\n$$\n$$\nx^2 - 14x + 49 + x^2 - 12x + 36 = 25\n$$\n$$\n2x^2 - 26x + 85 = 25\n$$\n$$\n2x^2 - 26x + 60 = 0\n$$\n$$\nx^2 - 13x + 30 = 0\n$$\n\nSolving the quadratic equation:\n\n$$\nx = \\frac{13 \\pm \\sqrt{13^2 - 4 \\cdot 1 \\cdot 30}}{2} = \\frac{13 \\pm \\sqrt{169 - 120}}{2} = \\frac{13 \\pm 7}{2}\n$$\n\nSo $x = 10$ or $x = 3$. Since $x > BP$, we take $x = 10$. Thus, $x = 10$ satisfies the requirements.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17021,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the function\n\n$$\nf(x) = \\begin{cases} 0, & x \\in [0, 20] \\\\ x(x-1) \\cdots (x-20), & x \\ge 20. \\end{cases}\n$$\n\nLet $a_1, a_2, \\ldots, a_{100}$ be nonnegative integers. Show that\n\n$$\nf(a_1) + f(a_2) + \\cdots + f(a_{100}) \\le 100 \\cdot 99 \\cdot 98 \\cdots 79.\n$$\n\nDeduce that $a_1 + a_2 + \\cdots + a_{100} \\le 9900$.",
    "options": [],
    "answer": "See solution",
    "solution": "For any nonnegative integer $a$, $f(a) = a \\cdot (a-1) \\cdots (a-20)$. The given inequality is\n\n$$\nf(a_1) + \\cdots + f(a_{100}) \\le 100 \\cdot 99 \\cdot 98 \\cdots 79.\n$$\n\nSince $f$ is convex, by Jensen's inequality:\n\n$$\nf\\left(\\frac{a_1 + \\cdots + a_{100}}{100}\\right) \\le \\frac{f(a_1) + \\cdots + f(a_{100})}{100}.\n$$\n\nThe right-hand side does not exceed $99 \\cdot 98 \\cdots 79 = f(99)$. Thus,\n\n$$\nf\\left(\\frac{a_1 + \\cdots + a_{100}}{100}\\right) \\le f(99).\n$$\n\nSince $f(x)$ is non-decreasing for $x \\ge 0$, it follows that\n\n$$\n\\frac{a_1 + \\cdots + a_{100}}{100} \\le 99,\n$$\nso\n$$\na_1 + \\cdots + a_{100} \\le 9900.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17022,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, $C$, $D$ be points on a circle such that $AB = BC = CD$. The bisectors of $\\angle ABD$ and $\\angle ACD$ intersect at point $E$. Find $\\angle ABC$, if $AE \\parallel CD$.\n\n![](images/Ukraine_2020_booklet_p9_data_3bbf881dac.png)",
    "options": [],
    "answer": "See solution",
    "solution": "$\\triangle ABC = \\triangle DCB$, as they are isosceles with equal base angle $x$ and equal legs (see the figure). $ABCD$ is an inscribed isosceles trapezoid. Hence\n\n$$\n\\begin{aligned}\n\\angle ABC &= \\angle BCD = 180^\\circ - 2x \\\\\n\\angle ABD &= \\angle ACD = 180^\\circ - 3x \\\\\n\\angle EBC &= \\angle ECB = 90^\\circ - \\frac{1}{2}x \\quad \\text{and } \\angle BEC = x.\n\\end{aligned}\n$$\n\nThus, $E$ lies on the same circle as the points $A$, $B$, $C$, $D$. Moreover,\n\n$$\n\\angle ADC = \\angle ADB + \\angle BDC = \\angle ACB + \\angle BDC = x + x = 2x\n$$\n\nand\n\n$$\n\\angle ECD = \\frac{1}{2} \\angle ACD = 90^\\circ - \\frac{3}{2}x.\n$$\n\nSince $AE \\parallel CD$, the inscribed quadrilateral $ACDE$ is an isosceles trapezoid, hence\n\n$$\n\\angle ADC = \\angle ECD \\quad \\text{and} \\quad 2x = 90^\\circ - \\frac{3}{2}x.\n$$\n\nSolving for $x$:\n\n$$\n2x + \\frac{3}{2}x = 90^\\circ \\\\\n\\frac{7}{2}x = 90^\\circ \\\\\nx = \\frac{180^\\circ}{7}\n$$\n\nTherefore,\n\n$$\n\\angle ABC = 180^\\circ - 2x = 180^\\circ - 2 \\cdot \\frac{180^\\circ}{7} = 180^\\circ - \\frac{360^\\circ}{7} = \\frac{900^\\circ}{7} - \\frac{360^\\circ}{7} = \\frac{540^\\circ}{7}\n$$\n\nSo,\n\n$$\n\\boxed{\\angle ABC = \\frac{540^\\circ}{7}}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17023,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying, for all $x \\neq 0$ and all $y$,\n$$\nf(x + y^2) = f(x) + f(y)^2 + \\frac{2f(xy)}{x}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $f(z) = 0$ and $f(z) = z^2$.\n\nReplace $y$ by $-y$ in the given equation:\n$$\nf(x + y^2) = f(x) + f(y)^2 + \\frac{2f(xy)}{x} = f(x) + f(-y)^2 + \\frac{2f(-xy)}{x}.\n$$\nSo,\n$$\nf(y)^2 + \\frac{2f(xy)}{x} = f(-y)^2 + \\frac{2f(-xy)}{x} \\quad (1)\n$$\nfor all $x \\neq 0$ and all $y$.\n\nLet $x = 1$:\n$$\n(f(y) + 1)^2 = (f(-y) + 1)^2.\n$$\nThus,\n$$\nf(y) + 1 = \\pm (f(-y) + 1),\n$$\nso for any $y$, either $f(y) = f(-y)$ or $f(y) + f(-y) = -2$.\n\nSuppose $f(y) + f(-y) = -2$ for all $y \\neq 0$. Then (1) gives:\n$$\nf(xy) + 1 = x(f(y) + 1), \\quad x, y \\neq 0.\n$$\nLet $y = 1$:\n$$\nf(x) = (1 + f(1))x - 1 = a x - 1, \\quad x \\neq 0,\n$$\nwhere $a = 1 + f(1)$. Substitute into the original equation:\n$$\n(a^2 - a)y^2 + 1 = \\frac{2}{x}, \\quad x, y, x + y^2 \\neq 0.\n$$\nThis is impossible (fix $y \\neq 0$ and vary $x$).\n\nThus, $f(-y) = f(y)$ for some $y \\neq 0$. Then $f(-xy) = f(xy)$ for all $x \\neq 0$ by (1), so $f(-z) = f(z)$ for all $z$.\n\nReturn to the original equation and let $y = 1$:\n$$\nf(x+1) = \\left(1 + \\frac{2}{x}\\right)f(x) + f(1)^2, \\quad x \\neq 0.\n$$\nNow, replace $x$ by $-x-1$ and use $f(-x) = f(x)$:\n$$\nf(x) = \\left(1 - \\frac{2}{x+1}\\right)f(x+1) + f(1)^2, \\quad x \\neq -1.\n$$\nEliminate $f(x+1)$ and simplify:\n$$\nf(x) = x^2 f(1)^2, \\quad x \\neq 0, -1. \\qquad (2)\n$$\nThis formula also holds for $x = 0$ and $x = -1$:\n- Let $y = 0$ in the original equation: $f(0) = 0 = 0^2 f(1)^2$.\n- Let $x = 1$ in (2): $f(1) = f(1)^2$, so $f(-1) = f(1) = (-1)^2 f(1)^2$.\n\nThus, $f(1) = 0$ or $f(1) = 1$. Therefore, the only solutions are $f(z) = 0$ and $f(z) = z^2$.\n\n$\\boxed{f(z) = 0 \\text{ or } f(z) = z^2}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17024,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a square with side length $24$. Let $P$ be the point on side $AB$ with $AP = 8$, and let $AC$ and $DP$ intersect at $Q$. Determine the area of triangle $CQD$.\n\n![](images/2019_Australian_Scene_W1_p64_data_af2231aec3.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Method 1\n\nDraw a line through $Q$ parallel to $AD$, and let $E$ and $F$ be the intersections of this line with $AB$ and $CD$, respectively.\n\nFrom similar triangles $AEQ$ and $CFQ$, we have $EQ/FQ = AQ/CQ$.\n\nFrom similar triangles $APQ$ and $CDQ$, we have $AQ/CQ = AP/CD = 8/24 = 1/3$.\n\nSo $FQ = \\frac{3}{4}FE = \\frac{3}{4} \\times 24 = 18$.\n\nHence the area of triangle $CQD$ equals $\\frac{1}{2} \\times 24 \\times 18 = \\mathbf{216}$.\n\nMethod 2\n\nLet $DC$ and $DA$ be the positive horizontal and vertical cartesian axes with $D$ as the origin.\n\nThe equation for line $DP$ is $y = \\frac{24}{8}x = 3x$.\n\nThe equation for line $AC$ is $y + x = 24$.\n\nSo, at $Q$, $4x = 24$, $x = 6$, and $y = 18$.\n\nHence the area of triangle $CQD$ equals $\\frac{1}{2} \\times 24 \\times 18 = \\mathbf{216}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17025,
    "subject": "Mathematics (Olympiad)",
    "question": "Inside the angle $\\angle BAC = 45^\\circ$, a point $P$ is chosen so that $\\angle APB = \\angle APC = 45^\\circ$. Let $M$ and $N$ be the feet of the perpendiculars dropped from the point $P$ onto the lines $AB$ and $AC$, respectively. Prove that $BC \\parallel MN$.\n\n![](images/Ukrajina_2013_p16_data_46d00246a2.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M_1$ and $N_1$ be the points symmetric to the point $P$ about the lines $AM$ and $AN$, respectively. Then $\\angle M_1AB = \\angle PAB$ and $\\angle N_1AC = \\angle PAC$. Moreover, $AM_1 = AP = AN_1$, so the triangle $\\triangle AM_1N_1$ is isosceles and right. On the other hand, $\\angle BM_1A = \\angle BPA = 45^\\circ$, so the point $B$ (and, similarly, the point $C$) is on the line.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17026,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n, s, t$ 為三個正整數，並令 $A_1, \\dots, A_s, B_1, \\dots, B_t$ 為 $\\{1, 2, \\dots, n\\}$ 的子集（不必然相異）。對於 $\\{1, \\dots, n\\}$ 的任何子集 $S$，我們定義 $f(S)$ 為滿足 $S \\subseteq A_i$ 的 $i \\in \\{1, \\dots, s\\}$ 的數量；$g(S)$ 為滿足 $S \\subseteq B_j$ 的 $j \\in \\{1, \\dots, t\\}$ 的數量。假設對任意的 $1 \\le x < y \\le n$，$f(\\{x, y\\}) = g(\\{x, y\\})$ 都成立。\n\n證明：若 $t < n$，則必存在 $1 \\le x \\le n$ 使得 $f(\\{x\\}) \\ge g(\\{x\\})$。",
    "options": [],
    "answer": "See solution",
    "solution": "假設反例：$t < n$ 且對所有 $1 \\le x \\le n$ 都有 $f(\\{x\\}) < g(\\{x\\})$。令 $c_1, \\dots, c_n$ 為實數，對任意 $S \\subseteq [n]$，定義 $c(S) = \\sum_{i \\in S} c_i$。\n\n則有：\n$$\n\\sum_{i=1}^{s} c(A_i)^2 = \\sum_{j=1}^{t} c(B_j)^2 - \\sum_{x=1}^{n} (g(\\{x\\}) - f(\\{x\\}))c_x^2\n$$\n\n由於 $t < n$，存在 $(c_1, \\dots, c_n) \\neq (0, \\dots, 0) \\in \\mathbb{R}^n$ 使得對所有 $j = 1, \\dots, t$，$c(B_j) = 0$。因此：\n$$\n\\sum_{i=1}^{s} c(A_i)^2 = - \\sum_{x=1}^{n} (g(\\{x\\}) - f(\\{x\\}))c_x^2 < 0\n$$\n\n因為 $g(\\{x\\}) - f(\\{x\\}) > 0$ 對所有 $x$ 成立，這導致矛盾。因此必存在某個 $x \\in [n]$ 使得 $f(\\{x\\}) \\ge g(\\{x\\})$。\n\n*補充：* 當 $A_1 = \\dots = A_s = [n]$ 時，若要求 $B_1, \\dots, B_t$ 為真子集，則對每個 $x$ 都有 $f(\\{x\\}) < g(\\{x\\})$，因此必有 $t \\ge n$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17027,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ and $B$ be two subsets of $\\{1, 2, 3, \\dots, 100\\}$, satisfying $|A| = |B|$ and $A \\cap B = \\emptyset$. If $n \\in A$ always implies $2n + 2 \\in B$, then what is the maximum value of $|A \\cup B|$?\n\n(A) 62\n\n(B) 66\n\n(C) 68\n\n(D) 74",
    "options": [],
    "answer": "See solution",
    "solution": "We first show that $|A \\cup B| \\leq 66$, or equivalently $|A| \\leq 33$. Suppose $A$ is a subset of $\\{1, 2, \\dots, 49\\}$ with 34 elements. Then there must exist $n \\in A$ such that $2n + 2 \\in A$.\n\nDivide $\\{1, 2, \\dots, 49\\}$ into 33 subsets:\n\n- $\\{1, 4\\}$, $\\{3, 8\\}$, $\\{5, 12\\}$, $\\dots$, $\\{23, 48\\}$ (12 subsets)\n- $\\{2, 6\\}$, $\\{10, 22\\}$, $\\{14, 30\\}$, $\\{18, 38\\}$ (4 subsets)\n- $\\{25\\}$, $\\{27\\}$, $\\{29\\}$, $\\dots$, $\\{49\\}$ (13 subsets)\n- $\\{26\\}$, $\\{34\\}$, $\\{42\\}$, $\\{46\\}$ (4 subsets)\n\nBy the pigeonhole principle, at least one subset with 2 elements must be entirely contained in $A$, so there exists $n \\in A$ such that $2n + 2 \\in A$.\n\nOn the other hand, let\n\n$$A = \\{1, 3, 5, \\dots, 23, 2, 10, 14, 18, 25, 27, 29, \\dots, 49, 26, 34, 42, 46\\}$$\n\nand\n\n$$B = \\{2n + 2 \\mid n \\in A\\}.$$\n\nThen $A$ and $B$ satisfy the conditions and $|A \\cup B| = 66$.\n\n**Answer:** (B)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17028,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n$ 與 $k$ 為正整數。寶寶用 $n^2$ 個數字積木拼成一個 $n \\times n$ 的方陣，每塊積木都是一個不超過 $k$ 的正整數。路過的奶爸一看，發現：\n\n1. 方陣上每一橫列的數字都可以視為以最左方數字為首項的等差數列，且其公差都不同；\n2. 方陣上每一直排的數字都可以視為以最上方數字為首項的等差數列，且其公差都不同。\n\n試求 $k$ 的最小可能值（以 $n$ 的函數表示。）\n\n註：公差可能非正。\n",
    "options": [],
    "answer": "See solution",
    "solution": "對於一個 $n \\times n$ 的方陣，當 $n = 2m + 1$ 時，$k$ 的最小可能值是 $2m^2 + 1$；當 $n = 2m$ 時，$k$ 的最小可能值是 $2m^2 - m + 1$。\n\n估計：首先考慮令 $[x]$ 為不大於 $x$ 的最大整數，並令 $m = [n/2]$。由於每一橫列的公差都不同，因此需要有 $n \\ge 2m$ 個不同的公差；又因為從 $-m + 1$ 到 $m - 1$ 全部也只有 $2(m-1) + 1 = 2m - 1 < n$ 個數，因此必然有一個橫列的公差 $\\ge m$ 或 $\\le -m$。不失一般性，假設該橫列的公差 $\\ge m$，則由於首項 $\\ge 1$，我們有\n\n$$\n\\text{末項} \\ge 1 + (n-1)m = \\begin{cases} 2m^2 - m + 1, & \\text{若 } n = 2m, \\\\ 2m^2 + 1, & \\text{若 } n = 2m + 1. \\end{cases}\n$$\n\n構造：首先考慮 $n = 2m + 1$ 的情況。令 $t = m^2 + 1$，並考慮構造\n\n$$\n\\begin{pmatrix}\nt - m^2 & \\cdots & t - 2m & t - m & t & t + m & t + 2m & \\cdots & t + m^2 \\\\\n\\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots \\\\\nt - 2m & \\cdots & t - 4 & t - 2 & t & t + 2 & t + 4 & \\cdots & t + 2m \\\\\nt - m & \\cdots & t - 2 & t - 1 & t & t + 1 & t + 2 & \\cdots & t + m \\\\\nt & \\cdots & t & t & t & t & t & \\cdots & t \\\\\nt + m & \\cdots & t + 2 & t + 1 & t & t - 1 & t - 2 & \\cdots & t - m \\\\\nt + 2m & \\cdots & t + 4 & t + 2 & t & t - 2 & t - 4 & \\cdots & t - 2m \\\\\n\\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots \\\\\nt + m^2 & \\cdots & t + 2m & t + m & t & t - m & t - 2m & \\cdots & t - m^2\n\\end{pmatrix}\n$$\n\n易驗證每一橫列分別為公差 $-m$ 到 $m$ 的等差數列，每一直列也分別為公差 $-m$ 到 $m$ 的等差數列，且最大的數字為 $t + m^2 = 2m^2 + 1$。\n\n若 $n = 2m$，則考慮上面的構造，並去除最後一行與最後一列即可。易知每一行與每一列仍為等差數列且公差相異，而此時的最大數字為 $t + m(m-1) = 2m^2 - m + 1$。\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17029,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ with the following property: it is possible to draw arcs in such a way that all the arcs (including those from the border of the board) form a single closed curve.\n\n![](images/Saudi_Arabia_booklet_2024_p34_data_9876c76abf.png)",
    "options": [],
    "answer": "See solution",
    "solution": "All $n$ which are $0$ or $1$ modulo $4$.\n\nThe key observation is that toggling the orientation of arcs within a single cell (i.e., rotating the drawing inside this cell by $60^\\circ$ and keeping all the remaining arcs unchanged) does not affect the parity of the number of closed curves formed by the arcs. Indeed, one easily checks that there are three essentially different cases: either three different curves are joined into one ($-2$ curves), or one curve is split into three ($+2$ curves), or two curves are changed into two different curves (no change in the number of curves). Therefore, the necessary condition is to have an odd number of curves in the following drawing:\n\n![](images/Saudi_Arabia_booklet_2024_p34_data_9876c76abf.png)\n\nwhich in turn is equivalent to $2 \\nmid 1 + 2 + \\cdots + (n + 1) = \\frac{1}{2}(n + 1)(n + 2)$, i.e., $n$ being $0$ or $1$ modulo $4$.\n\nSufficiency follows from any sequence of examples. One of many possible constructions is presented in the picture (to get a valid drawing for a board of side $n$ it is enough to take top $n$ rows). The interior of the curve is filled for clarity.\n\n![](images/Saudi_Arabia_booklet_2024_p34_data_87b30f358c.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17030,
    "subject": "Mathematics (Olympiad)",
    "question": "A six-digit almost palindrome can be written as $abcbad$, where $a, d \\neq 0$. How many such numbers are divisible by $9$?",
    "options": [],
    "answer": "See solution",
    "solution": "A number $abcbad$ is divisible by $9$ if and only if the sum of its digits, $a + b + c + b + a + d = 2a + 2b + c + d$, is divisible by $9$.\n\nFor fixed $a$, $b$, and $c$, the value of $d$ modulo $9$ is determined. Since $d$ ranges from $1$ to $9$, and these cover all possible remainders modulo $9$, for each choice of $a$, $b$, and $c$, there is exactly one $d$ that makes the number divisible by $9$.\n\nThere are $9$ choices for $a$ ($1$ to $9$), $10$ choices for $b$ ($0$ to $9$), and $10$ choices for $c$ ($0$ to $9$), so the total number is:\n\n$$\n9 \\times 10 \\times 10 = 900\n$$\n\nThus, there are $900$ such numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17031,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive factors of $10^{2013} - 1$ that are less than or equal to 100.",
    "options": [],
    "answer": "See solution",
    "solution": "We seek all positive factors of $10^{2013} - 1$ less than or equal to 100.\n\nLet $\\gcd(a, b)$ denote the greatest common divisor of $a$ and $b$. For integers $a, b, p$, $a \\equiv b \\pmod p$ means $a - b$ is a multiple of $p$.\n\nUseful facts:\n\n* If $a^n \\equiv 1 \\pmod p$, then for any positive integer $m$:\n  $$\na^m \\equiv 1 \\pmod p \\iff a^{\\gcd(m, n)} \\equiv 1 \\pmod p.\n  $$\n* If $p$ is prime and $a$ is not a multiple of $p$, then by Fermat's Little Theorem:\n  $$\na^{p-1} \\equiv 1 \\pmod p.\n  $$\n\nTo find the prime factors $p < 100$ of $10^{2013} - 1$, note:\n$$10^{2013} - 1 \\equiv 0 \\pmod p \\iff 10^{2013} \\equiv 1 \\pmod p \\iff 10^d \\equiv 1 \\pmod p,$$\nwhere $d = \\gcd(2013, p-1)$. Since $2013 = 3 \\cdot 11 \\cdot 61$, $d$ must be one of $1, 3, 11, 33, 61$.\n\n**Case $d=1$:**\n- $p=3$ is the only prime with $10^1 \\equiv 1 \\pmod p$ and $\\gcd(2013, p-1) = 1$.\n\n**Case $d=3$:**\n- $10^3 - 1 = 999 = 3^3 \\cdot 37$. $p=3, 37$ are the only primes with $10^3 \\equiv 1 \\pmod p$; only $p=37$ has $\\gcd(2013, p-1) = 3$.\n\n**Case $d=11$:**\n- $p=23, 89$ have $\\gcd(2013, p-1) = 11$, but neither satisfies $10^{11} \\equiv 1 \\pmod p$.\n\n**Case $d=33$:**\n- $p=67$ is the only prime with $\\gcd(2013, p-1) = 33$ and $10^{33} \\equiv 1 \\pmod{67}$.\n\n**Case $d=61$:**\n- No such $p < 100$.\n\nThus, the only prime factors of $10^{2013} - 1$ less than 100 are $3, 37, 67$.\n\nThe possible positive factors less than or equal to 100 are $1, 3, 9, 27, 37, 67, 81$.\n\nCheck divisibility by $9, 27, 81$:\n\n$$\n10^{2013} - 1 = (10-1)(10^{2012} + 10^{2011} + \\cdots + 10^1 + 1)\n$$\n\nCompute $10^{2012} + \\cdots + 1 \\pmod 9$:\n$$\n10^{2012} + \\cdots + 1 \\equiv 1^{2012} + \\cdots + 1 \\equiv 2013 \\equiv 6 \\pmod 9\n$$\nSo $10^{2013} - 1 \\equiv 54 \\pmod{81}$, so $9$ and $27$ divide $10^{2013} - 1$, but $81$ does not.\n\n**Final answer:**\n\nThe positive factors of $10^{2013} - 1$ less than or equal to 100 are:\n\n$$1,\\ 3,\\ 9,\\ 27,\\ 37,\\ 67$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17032,
    "subject": "Mathematics (Olympiad)",
    "question": "A $9 \\times 9$ square is divided into $81$ small $1 \\times 1$ squares, $8$ of which are painted black, the rest being white. We cut a fully white rectangle (possibly, a square) out of the big $9 \\times 9$ square. What is the maximal area of the rectangle that we can attain regardless of the positions of the black squares? It is only allowed to cut the rectangle along the grid lines.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $9$.\n\nCut the square into $9$ smaller $3 \\times 3$ blocks. Since there are only eight black squares, at least one of the blocks doesn't contain any of them. Therefore, a white square of area $9$ can always be found.\n\nNext, we show that sometimes it is impossible to find a larger rectangle. Below is an example. Here, one can cut out either a $3 \\times 3$ square or a $1 \\times 9$ rectangle, but not any rectangle of larger area.\n\n![](images/ukraine_2015_Booklet_p0_data_eeb22f4e48.png)\n\nFig. 2",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17033,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be a given integer.\n\n1. Prove that one can arrange all the subsets of the set $\\{1, 2, \\dots, n\\}$ as a sequence of subsets $A_1, A_2, \\dots, A_{2^n}$, such that $|A_{i+1}| = |A_i| + 1$ or $|A_i| - 1$, where $i = 1, 2, \\dots, 2^n$ and $A_{2^n+1} = A_1$.\n\n2. Determine, with proof, all possible values of the sum $\\sum_{i=1}^{2^n} (-1)^i S(A_i)$, where $S(A_i) = \\sum_{x \\in A_i} x$ and $S(\\emptyset) = 0$, for any subset sequence $A_1, A_2, \\dots, A_{2^n}$ satisfying the condition in (1).",
    "options": [],
    "answer": "See solution",
    "solution": "1. We prove by mathematical induction that there exists a sequence $A_1, A_2, \\dots, A_{2^n}$ such that $A_1 = \\{1\\}$, $A_{2^n} = \\emptyset$, and the sequence satisfies the condition in (1).\n\nWhen $n = 2$, the sequence $\\{1\\}, \\{1, 2\\}, \\{2\\}, \\emptyset$ for $\\{1, 2\\}$ works.\n\nAssume that when $n = k$, there exists such a sequence $B_1, B_2, \\dots, B_{2^k}$ of subsets of $\\{1, 2, \\dots, k\\}$. For $n = k + 1$, construct a sequence of subsets of $\\{1, 2, \\dots, k + 1\\}$ as follows:\n\n$$\n\\begin{align*}\nA_1 &= B_1 = \\{1\\}, \\\\\nA_i &= B_{i-1} \\cup \\{k+1\\}, \\quad i = 2, 3, \\dots, 2^k + 1, \\\\\nA_j &= B_{j-2^k}, \\quad j = 2^k + 2, 2^k + 3, \\dots, 2^{k+1}.\n\\end{align*}\n$$\n\nOne can check that this sequence fulfills the required conditions. By induction, (1) holds for $n \\ge 2$.\n\n2. We will show that the sum is $0$, independent of the arrangement. Without loss of generality, assume $A_1 = \\{1\\}$; otherwise, shift the index cyclically. Since $|A_{i+1}| = |A_i| + 1$ or $|A_i| - 1$, the parities of the indices and subset sizes match.\n\nThus,\n$$\n\\sum_{i=1}^{2^n} (-1)^i S(A_i) = \\sum_{A \\in P} S(A) - \\sum_{A \\in Q} S(A),\n$$\nwhere $P$ is the set of all subsets of $\\{1, 2, \\dots, n\\}$ with even cardinality, and $Q$ is the set with odd cardinality.\n\nFor any $x \\in \\{1, 2, \\dots, n\\}$, among all $k$-element subsets, $x$ appears in exactly $C_{n-1}^{k-1}$ of them, so its contribution is\n$$\n- C_{n-1}^0 + C_{n-1}^1 - C_{n-1}^2 + \\dots + (-1)^n C_{n-1}^{n-1} = -(1-1)^{n-1} = 0.\n$$\n\nTherefore, $\\sum_{i=1}^{2^n} (-1)^i S(A_i) = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17034,
    "subject": "Mathematics (Olympiad)",
    "question": "An _animal_ with $n$ cells is a connected figure consisting of $n$ equal-sized square cells. The figure below shows an 8-cell animal.\n\n![](images/USA_IMO_2007-2008_p7_data_cc72e326d0.png)\n\nA _dinosaur_ is an animal with at least 2007 cells. It is said to be _primitive_ if its cells cannot be partitioned into two or more dinosaurs. Find, with proof, the maximum number of cells in a primitive dinosaur.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $s$ denote the minimum number of cells in a dinosaur; this year, $s = 2007$.\n\n**Claim:** The maximum number of cells in a primitive dinosaur is $4(s - 1) + 1 = 8025$.\n\nAnimals are also called _polyominoes_. Two cells are _adjacent_ if they share a complete edge. A single cell is an animal, and given an animal with $n$ cells, one with $n+1$ cells is obtained by adjoining a new cell adjacent to one or more existing cells.\n\nFirst, a primitive dinosaur can contain up to $4(s-1)+1$ cells. To see this, consider a dinosaur in the form of a cross consisting of a central cell and four arms with $s-1$ cells each. No connected figure with at least $s$ cells can be removed without disconnecting the dinosaur.\n\nThe proof that no dinosaur with at least $4(s-1)+2$ cells is primitive relies on the following result.\n\n**Lemma.** Let $D$ be a dinosaur having at least $4(s-1)+2$ cells, and let $R$ (red) and $B$ (black) be two complementary animals in $D$, i.e., $R \\cap B = \\emptyset$ and $R \\cup B = D$. Suppose $|R| \\le s-1$. Then $R$ can be augmented to produce animals $\\tilde{R} \\supset R$ and $\\tilde{B} = D \\setminus \\tilde{R}$ such that at least one of the following holds:\n\n1. $|\\tilde{R}| \\ge s$ and $|\\tilde{B}| \\ge s$,\n2. $|\\tilde{R}| = |R| + 1$,\n3. $|R| < |\\tilde{R}| \\le s-1$.\n\n*Proof.* If there is a black cell adjacent to $R$ that can be made red without disconnecting $B$, then (2) holds. Otherwise, there is a black cell $c$ adjacent to $R$ whose removal disconnects $B$. Of the squares adjacent to $c$, at least one is red and at least one is black, otherwise $B$ would be disconnected. Then there are at most three resulting components $C_1, C_2, C_3$ of $B$ after the removal of $c$. Without loss of generality, $C_3$ is the largest of the remaining components. (Note that $C_1$ or $C_2$ may be empty.) Now $C_3$ has at least $\\lceil (3s-2)/3 \\rceil = s$ cells. Let $\\tilde{B} = C_3$. Then $|\\tilde{R}| = |R| + |C_1| + |C_2| + 1$. If $|\\tilde{B}| \\le 3s-2$, then $|\\tilde{R}| \\ge s$ and (1) holds. If $|\\tilde{B}| \\ge 3s-1$ then either (2) or (3) holds, depending on whether $|\\tilde{R}| \\ge s$ or not. $\\square$\n\nStarting with $|R| = 1$, repeatedly apply the Lemma. Because in alternatives (2) and (3) $|R|$ increases but remains less than $s$, alternative (1) eventually must occur. This shows that no dinosaur with at least $4(s-1)+2$ cells is primitive.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17035,
    "subject": "Mathematics (Olympiad)",
    "question": "A set of natural numbers is _regular_ if each of its subsets has sum different from 1810. Partition the numbers $452, 453, \\ldots, 1809$ into a minimum number of regular sets.",
    "options": [],
    "answer": "See solution",
    "solution": "Such a partition clearly needs at least 2 regular sets (for example, $900$ and $910$ must be in different regular sets). Here is a partition with 2 regular sets:\n\n$$\nA = \\{452, \\ldots, 602\\} \\cup \\{906, \\ldots, 1207\\}, \\quad B = \\{603, \\ldots, 905\\} \\cup \\{1208, \\ldots, 1809\\}.\n$$\n\nLet us check that $A$ and $B$ are indeed regular. Suppose, on the contrary, that one of them has a subset $X$ with sum $1810$. Note that $X$ has at most 3 elements, as the sum of the 4 smallest numbers among $452, 453, \\ldots, 1809$ is $452+453+454+455>1810$.\n\nLet $X$ have exactly 3 elements $x, y, z$ with $x+y+z=1810$. If $X \\subset B$ then $x+y+z \\ge 603+604+605>1810$, which is impossible. Hence $X \\subset A$ and clearly one of $x, y, z$ is in $\\{452, \\ldots, 602\\}$. In addition, one of them is in $\\{906, \\ldots, 1207\\}$ because the three largest numbers in $\\{452, \\ldots, 602\\}$ have sum less than $1810$. If, for example, $x \\ge 906$ then $y+z \\le 904$; in particular, $y, z \\in \\{452, \\ldots, 602\\}$. However, then $y+z \\ge 452+453=905$, a contradiction.\n\nLet $X$ have exactly 2 elements $x, y$ with $x+y=1810$ and $x < y$; then $x \\le 904$, $y \\ge 906$. It follows that if $X \\subset A$ then $x \\le 602$. On the other hand, $y \\le 1207$, so that $x+y \\le 602+1207<1810$. Similarly, if $X \\subset B$ then $y \\ge 1208$. Because $x \\ge 603$, this yields $x+y \\ge 603+1208>1810$. In both cases, we reach a contradiction.\n\nBecause $X$ has more than 1 element, the conclusion is that $A$ and $B$ are both regular.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17036,
    "subject": "Mathematics (Olympiad)",
    "question": "Triangle $ABC$ is inscribed in circle $\\omega$. Line $\\ell$ is tangent to $\\omega$ at $A$. Points $B_1$ and $C_1$ lie on $\\ell$ such that rays $CA$ and $BA$ bisect $\\widehat{BCB_1}$ and $\\widehat{CBC_1}$, respectively. Segments $BB_1$ and $CC_1$ intersect at $P$. The line through $P$ parallel to segment $BC$ intersects sides $AC$ and $AB$ at $B_2$ and $C_2$, respectively. Prove that if $P$ is the midpoint of $B_2C_2$ then $ABC$ is isosceles.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that $P$ is the midpoint of segment $B_2C_2$; that is, $B_2P = C_2P$. We will show that $AB = AC$.\n\nSet $\\widehat{ABC} = B$, $\\widehat{BCA} = C$, and $\\widehat{CAB} = A$. Extend segment $AP$ through $P$ to meet segment $BC$ at $A_3$. Then it is clear $A_3$ is the midpoint of side $BC$ or $BA_3 = A_3C$. Let $B_3$, $C_3$ denote the intersections of pairs of segments $BB_1$ and $AC$, $CC_1$ and $AB$, respectively. By Ceva's theorem, we have\n\n$$\n\\frac{AC_3 \\cdot BA_3 \\cdot CB_3}{C_3B \\cdot A_3C \\cdot B_3A} = 1 \\quad \\text{or} \\quad \\frac{AC_3}{C_3B} = \\frac{AB_3}{B_3C}\n$$\n\n![](images/Saudi_Arabia_booklet_2012_p56_data_05faea81fb.png)\n\nNote that $\\frac{AB_3}{CB_3}$ is equal to the ratio between the areas of triangle $ABB_1$ and $CBB_1$; that is,\n\n$$\n\\begin{aligned}\n\\frac{AB_3}{CB_3} &= \\frac{[ABB_1]}{[CBB_1]} = \\frac{AB \\cdot AB_1 \\cdot \\sin \\widehat{BAB_1}}{BC \\cdot CB_1 \\cdot \\sin \\widehat{BCB_1}} \\\\\n&= \\frac{AB}{BC} \\cdot \\frac{AB_1}{CB_1} \\cdot \\frac{\\sin \\widehat{BAB_1}}{\\sin \\widehat{BCB_1}}.\n\\end{aligned}\n$$\n\nBecause line $AB_1$ is tangent to $\\omega$, $\\widehat{C_1AB} = \\widehat{ACB} = C$. Hence $\\sin \\widehat{BAB_1} = \\sin \\widehat{C_1AB} = \\sin C$ and\n\n$$\n\\frac{\\sin \\widehat{BAB_1}}{\\sin \\widehat{BCB_1}} = \\frac{\\sin C}{\\sin 2C} = \\frac{1}{\\cos C}.\n$$\n\nBecause line $AB_1$ is tangent to $\\omega$, we also have $\\widehat{B_1AC} = \\widehat{ABC} = B$. Thus, triangle $AB_1C$ is similar to triangle $BAC$. By the Law of Sines, we have\n\n$$\n\\frac{AB}{BC} = \\frac{\\sin C}{\\sin A} \\quad \\text{and} \\quad \\frac{AB_1}{CB_1} = \\frac{\\sin C}{\\sin B}.\n$$\n\nIt follows that\n\n$$\n\\frac{AB_3}{CB_3} = \\frac{AB}{BC} \\cdot \\frac{AB_1}{CB_1} \\cdot \\frac{\\sin \\widehat{BAB_1}}{\\sin \\widehat{BCB_1}} = \\frac{\\sin^2 C}{\\sin A \\sin B \\cos C}.\n$$\n\nIn exactly the same way, we can show that\n\n$$\n\\frac{AC_3}{BC_3} = \\frac{\\sin^2 B}{\\sin A \\sin C \\cos B}.\n$$\n\nIt follows that\n\n$$\n\\frac{\\sin^2 C}{\\sin A \\sin B \\cos C} = \\frac{AB_3}{CB_3} = \\frac{AC_3}{BC_3} = \\frac{\\sin^2 B}{\\sin A \\sin C \\cos B}.\n$$\n\nor\n\n$$\n\\sin^2 C \\tan C = \\sin^2 B \\tan B.\n$$\n\nBecause we can have at most one of $\\tan B$ and $\\tan C$ being negative, we must have both of them positive, from which it follows that $B$ and $C$ are acute angles. For $0^\\circ < \\alpha < 90^\\circ$, both $\\sin \\alpha$ and $\\tan \\alpha$ are monotonically increasing, thus we must have $B = C$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17037,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\ldots, a_n$ be real numbers such that $a_1 = a_n$ and $a_{k+1} \\le \\frac{a_k + a_{k+2}}{2}$ for all $k = 1, 2, \\ldots, n-2$. Prove that $a_k \\le a_1$ for all $k = 1, 2, \\ldots, n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there exists an index $k \\in \\{2, 3, \\ldots, n-1\\}$ such that $a_k > a_1$, and let $m$ be the smallest such index. Since $a_m > a_1$, we have $a_m > a_1 \\ge a_{m-1}$. Therefore, $$a_{m+1} \\ge 2a_m - a_{m-1} = a_m + (a_m - a_{m-1}) > a_m.$$ By induction, it follows that the sequence $(a_k)_{k \\ge m}$ is strictly increasing, hence $a_n > a_m > a_1$, which contradicts $a_n = a_1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17038,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all quadruples $ (a, b, c, d) $ of integers satisfying the system of equations:\n\n$$\n-a^2 + b^2 + c^2 + d^2 = 1,\n$$\n$$\n3a + b + c + d = 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $-a^2 + b^2 + c^2 + d^2 = -2a(2a + b + c + d) + (a + b)^2 + (a + c)^2 + (a + d)^2$. By the second equation, $2a + b + c + d = 1 - a$, so the first equation reduces to $2a(a - 1) + (a + b)^2 + (a + c)^2 + (a + d)^2 = 1$.\n\nIf $a > 1$ or $a < 0$, then $2a(a - 1)$ is a positive even number, so the left-hand side of the last equation is greater than $1$.\n\nIf $a = 0$, then according to the first equation, exactly one number among $b^2, c^2$, and $d^2$ is equal to $1$, the others are zero. Value $-1$ together with zeros would not satisfy the second equality, thus one of $b, c$, and $d$ must be $1$, while the others are zero.\n\nIf $a = 1$, then according to the first equation, exactly two numbers among $b^2, c^2$, and $d^2$ are equal to $1$ and the remaining number is zero. The second equation is satisfied only if both non-zero variables take the value $-1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17039,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b$ be real numbers with $0 \\le a, b \\le 1$. Prove the inequality\n\n$$\n\\sqrt{a^3 b^3} + \\sqrt{(1-a^2)(1-ab)(1-b^2)} \\le 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $0 \\le a, b \\le 1$, by the AM-GM inequality,\n\n$$\n\\begin{aligned}\n& \\sqrt{a^3 b^3} + \\sqrt{(1-a^2)(1-ab)(1-b^2)} \\\\\n& \\le \\sqrt[3]{a^3 b^3} + \\sqrt[3]{(1-a^2)(1-ab)(1-b^2)} \\\\\n& = \\sqrt[3]{a^2 \\cdot ab \\cdot b^2} + \\sqrt[3]{(1-a^2)(1-ab)(1-b^2)} \\\\\n& \\le \\frac{a^2 + ab + b^2}{3} + \\frac{(1-a^2) + (1-ab) + (1-b^2)}{3} \\\\\n& = 1.\n\\end{aligned}\n$$\n\nas claimed. Equality holds iff $a^2 = ab = b^2$ and either $a^3 b^3 = 0$ or $a^3 b^3 = 1$ (and similarly for $(1-a^2)(1-ab)(1-b^2)$). Thus, equality holds iff $a = b = 0$ or $a = b = 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17040,
    "subject": "Mathematics (Olympiad)",
    "question": "In the coordinate plane $Oxy$, consider the lattice points $(i, j)$ where $i, j$ are integers and $0 \\leq j \\leq 2n+1$. The point $(i, j)$ is assigned the value $f(i, j)$. The following conditions are given:\n\n- All numbers on the border (upper and lower) are $0$; all inner points are assigned numbers from $[0, 1]$.\n- For each small skew square (with diagonal parallel to the axes and side $\\sqrt{2}$), the sum of all assigned numbers at its vertices is $1$.\n\nFind all possible functions $f$ with these properties.",
    "options": [],
    "answer": "See solution",
    "solution": "The condition on the skew squares means that for each such square with vertices at lattice points, the sum of the assigned numbers is $1$.\n\nFor each lattice point $A(i, j)$, let $f_1(A) = f(i+2, j)$ and $f_2(A) = f(i-2, j)$. Let $a_k = f(k, k)$ for $k = 1, 2, \\dots, 2n$ be the values assigned to $A_k(k, k)$. By the condition, we have:\n\n$$\n\\begin{align*}\na_1 + a_2 + f_1(A_1) + 0 &= 1 &\\implies f_1(A_1) = 1 - a_1 - a_2, \\\\\na_2 + a_3 + f_1(A_2) + f_1(A_1) &= 1 &\\implies f_1(A_2) = a_1 - a_3, \\\\\n\\dots & &\n\\end{align*}\n$$\n\n$$\n\\begin{align*}\na_{2n-1} + a_{2n} + f_1(A_{2n-2}) + f_1(A_{2n-1}) &= 1 &\\implies f_1(A_{2n-1}) = a_1 - a_{2n}, \\\\\na_{2n} + 0 + f_1(A_{2n-1}) + f_1(A_{2n}) &= 1 &\\implies f_1(A_{2n}) = a_1\n\\end{align*}\n$$\n\nSimilarly,\n\n$$\n\\begin{align*}\nf_2(A_{2n}) &= 1 - a_{2n-1} - a_{2n} \\\\\nf_2(A_{2n-1}) &= a_{2n} - a_{2n-2} \\\\\n\\dots &\n\\end{align*}\n$$\n\n$$\n\\begin{align*}\nf_2(A_1) &= 1 - a_1 - a_{2n}, \\\\\nf_2(A_{2n}) &= a_{2n}\n\\end{align*}\n$$\n\nIf the sequence $a_k$ is given, then the values of $f_1(A_k)$ and $f_2(A_k)$ are defined. We choose $a_1 \\geq a_3 \\geq a_5 \\geq \\dots \\geq a_{2n-1}$, $a_2 \\leq a_4 \\leq a_6 \\leq \\dots \\leq a_{2n}$, and $a_1 + a_{2n} \\leq 1$ so that $f_1(A_k), f_2(A_k) \\in [0, 1]$.\n\nMoreover,\n\n$$\n\\begin{align*}\nf_1(A_1) &\\geq f_1(A_3) \\geq \\dots \\geq f_1(A_{2n-1}), \\\\\nf_1(A_2) &\\leq f_1(A_4) \\leq \\dots \\leq f_1(A_{2n}), \\\\\nf_1(A_1) + f_1(A_{2n}) &\\leq 1, \\\\\nf_2(A_1) &\\geq f_2(A_3) \\geq \\dots \\geq f_2(A_{2n-1}), \\\\\nf_2(A_2) &\\leq f_2(A_4) \\leq \\dots \\leq f_2(A_{2n}), \\\\\nf_2(A_1) + f_2(A_{2n}) &\\leq 1\n\\end{align*}\n$$\n\nThus, the monotonicity of the odd and even subsequences holds for $f_1(A_k)$ and $f_2(A_k)$, as well as the last condition.\n\nFrom the sequence $f_1(A_k)$, we can find $f_1(f_1(A_k))$, and similarly for $f_2$, so we can find all values of $f(x, y)$ where $x - y$ is even.\n\n![](images/Vietnam_Booklet_2013_12-07_p58_data_1eb8aebade.png)\n\nLet $b_k = f(k+1, k)$ for $k = 1, 2, \\dots, 2n$ be the values assigned to $B_k(k+1, k)$. By a similar method, we can construct all values of $f(x, y)$ where $x - y$ is odd.\n\nThere are infinitely many ways to choose sequences $(a_k), (b_k)$ satisfying the above conditions. Thus, there exist infinitely many functions with the given properties, i.e., $|F|$ is infinite. (Q.E.D.)\n\nAdditionally, in the equation $f(x-1, y) + f(x+1, y) + f(x, y-1) + f(x, y+1) = 1$, replacing $x, y$ by $x+1, y+1$ gives $f(x, y+1) + f(x+2, y+1) + f(x+1, y) + f(x+1, y+2) = 1$.\n\nTherefore,\n\n$$\nf(x-1, y) + f(x, y-1) = f(x+1, y+2) + f(x+2, y+1)\n$$\n\nor\n\n$$\nf(1, 1) + f(2, 0) = f(3, 3) + f(4, 2) = \\dots = f(2n+1, 2n+1) + f(2n+2, 2n),\n$$\n\n$$\nf(3, 1) + f(4, 0) = f(5, 3) + f(6, 2) = \\dots = f(2n+3, 2n+1) + f(2n+4, 2n).\n$$\n\nThus $f(1, 1) = f(2n+2, 2n)$, $f(3, 1) = f(2n+4, 2n)$, and similarly $f(2, 2) = f(2n+2, 2n-1)$, $f(4, 2) = f(2n+5, 2n-1)$.\n\nRepeatedly applying this, we have\n\n$$\nf(k, k) = f(2n+1+k, 2n+1-k), \\quad f(k+2, k) = f(2n+3+k, 2n+1-k).\n$$\n\nSimilarly,\n\n$$\nf(2n+k, 2n+1-k) = f(4n+2+k, k), \\quad f(2n+3+k, 2n+1-k) = f(4n+4+k, k).\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17041,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter $AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto lines $AX$, $BX$, $AZ$, $BZ$, respectively. Prove that the acute angle formed by lines $PQ$ and $RS$ is half the size of $\\angle XOZ$, where $O$ is the midpoint of segment $AB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $T$ be the foot of the perpendicular from $Y$ to line $AB$. We note that $P$, $Q$, and $T$ are the feet of the perpendiculars from $Y$ to the sides of triangle $ABX$. Because $Y$ lies on the circumcircle of triangle $ABX$, points $P$, $Q$, and $T$ are collinear by Simson's theorem. Likewise, points $S$, $R$, and $T$ are collinear.\n\n![](images/pamphlet0910_main_p36_data_cd6115d62b.png)\n\nWe need to show that $\\angle XOZ = 2\\angle PTS$. Notice that\n\n$$\n\\frac{\\angle XOZ}{2} = \\frac{\\widehat{XZ}}{2} = \\frac{\\widehat{XY}}{2} + \\frac{\\widehat{YZ}}{2} = \\angle XAY + \\angle ZBY = \\angle PAY + \\angle SBY\n$$\n\nand that $\\angle PTS = \\angle PTY + \\angle STY$. Therefore, it suffices to prove that\n\n$$\n\\angle PTY = \\angle PAY \\text{ and } \\angle STY = \\angle SBY.\n$$\n\nFor this, it is enough to show that quadrilaterals $APYT$ and $BSYT$ are cyclic. This follows because $\\angle APY = \\angle ATY = 90^\\circ$ and $\\angle BTY = \\angle BSY = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17042,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with circumcircle $w$ and center $O$. Let $A_0$ and $C_0$ be the second intersection points of $w$ with the extensions of the altitudes from $A$ and $C$. Let $A_1$ and $C_1$ be the intersection points of the line $A_0C_0$ with the sides $AB$ and $BC$ of the triangle, respectively. Let $A_2$ and $C_2$ be points on $AC$ such that $A_2O \\parallel BC$ and $C_2O \\parallel AB$. Let $H$ be the orthocenter of $\\triangle ABC$, and let $T$ be the intersection point of $A_1A_2$ and $C_1C_2$. Prove that $HT \\parallel AC$. Suppose that the tangents to $w$ at $A$ and $B$ meet at a point $K$.\n\n![](images/UkraineMO2019_booklet_p44_data_dec1e9bb08.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the points $A_0$, $A_1$, and $A_2$. Since $\\triangle ABC$ is acute, $O$ lies inside $\\triangle ABC$, so $A_2$ lies on $AC$. Let's prove that quadrilateral $OA_0CA_2$ is cyclic. Since $A_2O \\parallel BC$, we have\n\n$$\n\\angle OA_2A = \\angle BCA = 90^\\circ - \\angle A_0AC = 90^\\circ - \\frac{1}{2} \\angle A_0OC = \\angle OA_0C.\n$$\n\nLet $X$ be the second intersection of the circumscribed circle of $OA_0CA_2$ with $CH$. Let's prove that $A_0$, $B$, $A_1$, and $X$ are concyclic. Indeed,\n\n$$\n\\angle A_1BH = 90^\\circ - \\angle A = \\angle C_0CA = \\angle AA_0C_0,\n$$\n\nso $A_0BA_1H$ is cyclic. Also,\n\n$$\n\\angle HXA_0 = 180^\\circ - \\angle A_0XC = \\angle A_0OC = 2\\angle A_0AC = 2\\angle A_0BC = \\angle A_0BH,\n$$\n\nsince $H$ and $A_0$ are symmetric with respect to $BC$.\n\nNow, the line $A_1A_2$ passes through $X$ because\n\n$$\n\\angle A_1XA_0 = 180^\\circ - \\angle A_1BA_0 = \\angle A_0CA_2 = 180^\\circ - \\angle A_2XA_0.\n$$\n\nWe have\n\n$$\n\\angle OXH = 180^\\circ - \\angle OXC = \\angle OA_2A = \\angle C, \\quad \\angle OXT = \\angle OCA = 90^\\circ - \\angle B.\n$$\n\nSo, if $X$ is the intersection of $CC_0$ and $A_1A_2$, then $\\angle OXT = 90^\\circ - \\angle B$, $\\angle OXH = \\angle C$.\n\nNow let $Y$ be the intersection of $AA_0$ and $C_1C_2$. Similarly, $\\angle OYT = 90^\\circ - \\angle B$, $\\angle OYH = \\angle A$. Consider $X, H, Y, O$. Clearly $\\angle YHX = \\angle A$. Since $\\angle OXH + \\angle XHY + \\angle HYO = 180^\\circ$, $X, Y, O$ are collinear. Thus, $\\triangle XHY$ with $O$ on $XY$ is similar to $\\triangle ABC$. Since $\\angle TXY = \\angle TYX = 90^\\circ - \\angle B$, in this similarity the ray $XT$ corresponds to $CO$, and $YT$ to $AO$. Hence, $T$ corresponds to $O$, so $T$ is the center of the circumscribed circle of $\\triangle YHX$. Thus,\n\n$$\n\\angle THX = 90^\\circ - \\angle XYH = 90^\\circ - \\angle A = \\angle HCA,\n$$\n\nso $HT \\parallel AC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17043,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve in real numbers the equation\n\n$$\n\\lfloor x \\rfloor^5 + \\{x\\}^5 = x^5.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that any number $x$ with $\\{x\\} = 0$ or $\\lfloor x \\rfloor = 0$ satisfies the relation.\n\nAssume that $\\{x\\} \\neq 0$ and write the equation as\n\n$$\n\\{x\\}^5 = (x - \\lfloor x \\rfloor)\\left(x^4 + x^3 \\lfloor x \\rfloor + x^2 \\lfloor x \\rfloor^2 + x \\lfloor x \\rfloor^3 + \\lfloor x \\rfloor^4\\right),\n$$\n\nand then as\n\n$$\n\\{x\\}^4 = x^4 + x^3 \\lfloor x \\rfloor + x^2 \\lfloor x \\rfloor^2 + x \\lfloor x \\rfloor^3 + \\lfloor x \\rfloor^4.\n$$\n\nSuppose $\\lfloor x \\rfloor \\neq 0$. The right-hand side of the latter equation is the sum of $[x]^4$ and four positive terms, hence is at least 1, while the left-hand side is less than 1, so there are no solutions.\n\nConsequently, the set of solutions is $[0, 1] \\cup \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17044,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $f : [0, \\infty) \\to \\mathbb{R}$ is a non-constant function with the property that $|f(x) - f(y)| \\leq |\\sin x - \\sin y|$ for all $x, y \\in [0, \\infty)$. Prove that $f$ is a periodic bounded function and that $g : [0, \\infty) \\to \\mathbb{R}$, defined by $g(x) = x + f(x)$, is a monotonic function.",
    "options": [],
    "answer": "See solution",
    "solution": "Set $y = \\pi$ to obtain $|f(x)| - |f(\\pi)| \\leq |f(x) - f(\\pi)| \\leq |\\sin x| \\leq 1$ for all $x \\in [0, \\infty)$, hence $|f(x)| \\leq |f(\\pi)| + 1$, so $f$ is bounded.\n\nMoreover, choose $y = x + 2\\pi$ to get $|f(x) - f(x + 2\\pi)| \\leq |\\sin x - \\sin(x + 2\\pi)| = 0$, so $f$ has period $2\\pi$.\n\nRecall that $|\\sin x| \\leq |x|$ to deduce $|f(x) - f(y)| \\leq |x - y|$. Rewrite as $-1 \\leq \\frac{f(x)-f(y)}{x-y} \\leq 1$ to infer $0 \\leq \\frac{x+f(x)-y-f(y)}{x-y}$, so $g$ is increasing.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17045,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the least $n$ with the following property: In each $n$-term sequence of positive integers with sum $2013$, there are several consecutive terms with sum $31$.",
    "options": [],
    "answer": "See solution",
    "solution": "The least $n$ in question is $1022$.\n\nAn example that $n = 1021$ is not enough: arrange in a row $32$ blocks $1, 1, \\ldots, 1, 32$, with $30$ ones in each, then add $29$ ones. This gives a sequence of length $32 \\times 31 + 29 = 1021$ and sum $62 \\times 32 + 29 = 2013$. No consecutive terms in it have sum $31$.\n\nTake a sequence with positive integer terms with length $1022$ and sum $2013$. We show that several consecutive terms add up to $31$. Let $S_j$ be the sum of its first $j$ terms, $j = 1, \\ldots, 1022$. Consider the two sequences\n\n$$\n1 \\leq S_1 < S_2 < \\dots < S_{1022} = 2013\n$$\n\nand\n\n$$\n32 \\leq 31 + S_1 < 31 + S_2 < \\dots < 31 + S_{1022} = 2044.\n$$\n\nAssume $S_j \\neq 31$ for all $j$, otherwise the claim follows. Since $31 + S_j \\neq 31$ for all $j$, it follows that the $2044$ integers above are in $[1, 2044]$ and different from $31$. Hence two of them are the same. Because the terms in each sequence are distinct, there are indices $i, j$ such that $S_i = S_j + 31$. Clearly $i > j$ and the terms of the original sequence with indices $j+1, \\ldots, i$ have sum $31$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17046,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be positive real numbers such that $abc = 1$. Prove that\n\n$$\n\\frac{1}{\\sqrt{a^3 + 2b^3 + 6}} + \\frac{1}{\\sqrt{b^3 + 2c^3 + 6}} + \\frac{1}{\\sqrt{c^3 + 2a^3 + 6}} \n\\leq \n\\frac{a^2}{\\sqrt{a^4 + 4b + 4c}} + \\frac{b^2}{\\sqrt{b^4 + 4c + 4a}} + \\frac{c^2}{\\sqrt{c^4 + 4a + 4b}}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove that\n\n$$\n\\frac{a^2}{\\sqrt{a^4 + 4b + 4c}} + \\frac{b^2}{\\sqrt{b^4 + 4c + 4a}} + \\frac{c^2}{\\sqrt{c^4 + 4a + 4b}} \\geq 1\n$$\n\nApplying the AM-GM inequality, we conclude that\n\n$$\n\\begin{aligned}\n\\frac{a^2}{\\sqrt{a^4 + 4b + 4c}} &= \\frac{a^2}{\\sqrt{a^4 + 4ab^2c + 4abc^2}} \\\\\n&\\geq \\frac{a^2}{\\sqrt{a^4 + (a^2b^2 + a^2c^2 + b^2c^2 + b^4) + (a^2b^2 + a^2c^2 + b^2c^2 + c^4)}} \\\\\n&= \\frac{a^2}{a^2 + b^2 + c^2}\n\\end{aligned}\n$$\n\nSimilarly,\n\n$$\n\\frac{b^2}{\\sqrt{b^4 + 4c + 4a}} \\geq \\frac{b^2}{b^2 + c^2 + a^2}, \\quad \\frac{c^2}{\\sqrt{c^4 + 4a + 4b}} \\geq \\frac{c^2}{c^2 + a^2 + b^2}\n$$\n\nTherefore,\n\n$$\n\\frac{a^2}{\\sqrt{a^4 + 4b + 4c}} + \\frac{b^2}{\\sqrt{b^4 + 4c + 4a}} + \\frac{c^2}{\\sqrt{c^4 + 4a + 4b}} \\geq 1\n$$\n\nNow it remains to prove that\n\n$$\n\\frac{1}{\\sqrt{a^3 + 2b^3 + 6}} + \\frac{1}{\\sqrt{b^3 + 2c^3 + 6}} + \\frac{1}{\\sqrt{c^3 + 2a^3 + 6}} \\leq 1\n$$\n\nApplying the AM-GM inequality, we conclude that\n\n$$\n\\begin{aligned}\n\\frac{1}{\\sqrt{a^3 + 2b^3 + 6}} &\\leq \\frac{1}{\\sqrt{3ab^2 + 6}} \\\\\n&= \\sqrt{\\frac{abc}{3ab^2 + 6abc}} = \\sqrt{\\frac{1}{3}\\left(\\frac{c}{b+2c}\\right)} \\\\\n&\\leq \\frac{1}{2}\\left(\\frac{1}{3} + \\frac{c}{b+2c}\\right) \\\\\n&= \\frac{1}{4}\\left(\\frac{5}{3} - \\frac{a}{a+2b}\\right)\n\\end{aligned}\n$$\n\nTherefore, using the Cauchy-Schwarz inequality, we have\n\n$$\n\\begin{aligned}\n& \\frac{1}{\\sqrt{a^3 + 2b^3 + 6}} + \\frac{1}{\\sqrt{b^3 + 2c^3 + 6}} + \\frac{1}{\\sqrt{c^3 + 2a^3 + 6}} \\\\\n\\leq & \\frac{5}{4} - \\frac{1}{4} \\left( \\frac{b}{b+2c} + \\frac{c}{c+2a} + \\frac{a}{a+2b} \\right) \\\\\n\\leq & \\frac{5}{4} - \\frac{(a+b+c)^2}{4(b(b+2c) + c(c+2a) + a(a+2b))} = 1.\n\\end{aligned}\n$$\n\nHence, the proof of this problem follows. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17047,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(x)$ be a polynomial with integer coefficients such that $P(0) = 0$ and\n$$\ngcd(P(0), P(1), P(2), \\dots) = 1.\n$$\nProve that there are infinitely many positive integers $n$ such that\n$$\ngcd(P(n) - P(0), P(n+1) - P(1), P(n+2) - P(2), \\dots) = n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Write\n$$\nP(x) = a_{r}x^{r} + a_{r-1}x^{r-1} + \\dots + a_{1}x,\n$$\nand consider its formal derivative\n$$\nQ(x) = r a_{r} x^{r-1} + (r-1) a_{r-1} x^{r-2} + \\dots + a_{1}.\n$$\nSince $P$ is not identically zero, neither is $Q$, so we may choose some positive integer $m$ such that $Q(m) \\neq 0$. We claim that we may take $n$ to be any prime that does not divide $Q(m)$.\n\nLet $n$ be such a prime, and put\n$$\nd = \\gcd(P(n) - P(0), P(n+1) - P(1), P(n+2) - P(2), \\dots).\n$$\nCertainly we have $n \\mid d$. On the other hand, if $q$ is any prime distinct from $n$, then we cannot have $q \\mid d$. For suppose that $q \\mid d$. Since $q$ and $n$ are relatively prime, there are integers $k, l > 0$ such that $kn - lq = 1$. Then, notice that\n$$\nP(m) \\equiv P(m+n) \\equiv P(m+2n) \\equiv \\dots \\equiv P(m+kn) \\equiv P(m+1) \\pmod{q}\n$$\nfor every nonnegative integer $m$. By induction, then, $q$ divides all of $P(0), P(1), P(2), \\dots$, contradicting the given.\n\nAlso, we cannot have $n^2 \\mid d$. Indeed,\n$$\n\\begin{aligned}\nP(m+n) - P(m) &= \\sum_{i=1}^{r} a_i [(m+n)^i - m^i] \\\\\n&= \\sum_{i=1}^{r} a_i [m^i + i m^{i-1} n + (\\text{terms divisible by } n^2) - m^i] \\\\\n&= \\sum_{i=1}^{r} a_i [i m^{i-1} n + (\\text{terms divisible by } n^2)] \\\\\n&\\equiv n \\cdot Q(m) \\pmod{n^2}.\n\\end{aligned}\n$$\nTherefore, $P(m+n) - P(m)$ cannot be divisible by $n^2$ since $Q(m)$ is not divisible by $n$.\n\nSo $d$ is divisible by $n$, but not by any other prime or by $n^2$; hence $d = n$, as required.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17048,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, it is known that the number $\\sqrt{12n^2 + 1}$ is also a positive integer. Prove that the number\n\n$$\n\\sqrt{\\frac{\\sqrt{12n^2 + 1} + 1}{2}}\n$$\n\nis also a positive integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\sqrt{12n^2 + 1} = 2k + 1$, since $\\sqrt{12n^2 + 1}$ is an integer, it must be odd. We need to prove that\n\n$$\n\\sqrt{\\frac{\\sqrt{12n^2 + 1} + 1}{2}} = \\sqrt{\\frac{(2k + 1) + 1}{2}} = \\sqrt{k + 1}\n$$\n\nis an integer, i.e., that $k + 1$ is a perfect square.\n\nSquaring both sides of $\\sqrt{12n^2 + 1} = 2k + 1$ gives $12n^2 + 1 = 4k^2 + 4k + 1$, so $k(k + 1) = 3n^2$. Since $k$ and $k + 1$ are coprime, there are two possibilities: $k = 3a^2$ and $k + 1 = b^2$, or $k = a^2$ and $k + 1 = 3b^2$ for some positive integers $a$ and $b$. The second case is impossible, since $a^2 = 3b^2 - 1 \\equiv 2 \\pmod{3}$, which is not possible for squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17049,
    "subject": "Mathematics (Olympiad)",
    "question": "試求所有的正整數 $n$，使得存在質數 $p > 2$ 和質數 $q$，滿足 $n$ 的 $p$ 進位表示法是 $2011$，且 $n$ 的 $q$ 進位表示法是 $10\\cdots 0$（1 後面跟著任意多個 0）。",
    "options": [],
    "answer": "See solution",
    "solution": "解：$n = 256$。\n\n題目所述這樣的 $n$ 滿足 $n = 2p^3 + p + 1 = q^k$，其中 $k$ 是某正整數。顯然 $2p^3 + p + 1$ 是偶數，故 $q = 2$。\n\n易檢驗 $3 \\mid 2p^3 + p$ 必成立。故由 $3 \\mid 2^k - 1$ 可知 $2 \\mid k$。令 $A = 2^{k/2}$，我們有 $(A+1)(A-1) = 2p^3 + p = (2p^2 + 1)p$，故 $p \\mid (A-1)$ 或 $p \\mid (A+1)$。令 $A \\pm 1 = pt$，$t$ 是正整數。則有\n\n$$\npt^2 \\pm 2t = t(pt \\pm 2) = 2p^2 + 1.\n$$\n\n我們有 $p \\mid (\\pm 2t + 1)$，故 $2t + 1 \\geq p$。於是 $8t^2 + 8t + 3 \\geq 2p^2 + 1 \\geq pt^2 - 2t$。\n由此有 $t < 5$，或當 $t \\geq 5$ 時，$11t^2 > 8t^2 + 10t + 3 \\geq pt^2$，即 $p < 11$。若 $t < 5$ 由 $2t + 1 \\geq p$ 亦有 $p < 11$。\n\n故剩下的可能僅有 $p = 3, 5, 7$。代入檢驗知 $p = 5, k = 8, n = 256$ 是唯一解。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17050,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the sequence $\\{a_n\\}_{n=1}^{\\infty}$ defined by\n$$\na_n = \\frac{1}{4[-\\log_4 n]}\n$$\nfor all positive integers $n$. Define\n$$\nb_n = \\frac{1}{n^2} \\left( a_1 + a_2 + \\cdots + a_n - \\frac{1}{a_1 + a_2} \\right), \\quad \\forall n \\in \\mathbb{Z}^+.\n$$\n\na) Find a polynomial $P(x)$ with real coefficients such that $b_n = P\\left(\\frac{a_n}{n}\\right)$ for all positive integers $n$.\n\nb) Prove that there exists a strictly increasing sequence of positive integers $\\{n_k\\}_{k=1}^{\\infty}$ such that\n$$\n\\lim_{k \\to \\infty} b_{n_k} = \\frac{2024}{2025}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "a) We will prove that the polynomial $P(x) = -\\frac{1}{5}x^2 + x$ satisfies the required properties. Obviously $b_1 = \\frac{4}{5} = P(1) = P\\left(\\frac{1}{1}\\right)$, so we only need to consider the case $n > 1$.\n\nNotice that, for each non-integer real number $x$, $[-x] = -[x] - 1$. Thus, for each positive integer $m > 1$, let $t$ be a natural number such that $4^t < m \\le 4^{t+1}$, we have:\n\n- If $m = 4^{t+1}$, then $[-\\log_4 m] = -t - 1$ and therefore $a_m = 4^{t+1}$.\n- If $4^t < m < 4^{t+1}$, then $t < \\log_4 m < t + 1$, so $\\log_4 m$ is not integer, and\n  $$\n  [-\\log_4 m] = -[\\log_4 m] - 1 = -t - 1 = 4^{t+1}.\n  $$\n\nIn short, we always have $a_m = 4^{t+1}$, where $t$ is a natural number such that $4^t < m \\le 4^{t+1}$. Now, we consider the following cases.\n\n**Case 1:** $n$ is a power of 4. In this case, $n = 4^s$ for some positive integer $s$. Then $a_n = n$ and\n$$\n\\begin{aligned}\na_1 + a_2 + \\dots + a_n &= a_1 + (a_2 + a_3 + a_4) + (a_5 + \\dots + a_{16}) \\\\\n&\\quad + \\dots + (a_{4^s-1} + 1) + \\dots + a_{4^s} \\\\\n&= 4^0 \\cdot 4^0 + (4^1 - 4^0) \\cdot 4^1 + \\dots + (4^s - 4^{s-1}) \\cdot 4^s \\\\\n&= \\frac{4^{2s+1} + 1}{5} = \\frac{4n^2 + 1}{5}.\n\\end{aligned}\n$$\nThis implies that\n$$\nb_n = \\frac{1}{n^2} \\left( \\frac{4n^2 + 1}{5} - \\frac{1}{5} \\right) = \\frac{4}{5} = -\\frac{1}{5} \\left( \\frac{a_n}{n} \\right)^2 + \\frac{a_n}{n} = P \\left( \\frac{a_n}{n} \\right).\n$$\n\n**Case 2:** $n$ is not a power of 4. There exists $s$ such that $4^s < n < 4^{s+1}$. Then $a_n = 4^{s+1}$ and\n$$\n\\begin{aligned}\na_1 + a_2 + \\dots + a_n &= a_1 + (a_2 + a_3 + a_4) + \\dots + (a_{4^s-1} + 1) + \\dots + a_{4^s} \\\\\n&\\quad + (a_{4^{s+1}} + \\dots + a_n) \\\\\n&= 4^0 \\cdot 4^0 + (4^1 - 4^0) \\cdot 4^1 + \\dots + (4^s - 4^{s-1}) \\cdot 4^s \\\\\n&\\quad + (n - 4^s) \\cdot 4^{s+1} \\\\\n&= \\frac{4^{2s+1} + 1}{5} + (n - 4^s) \\cdot 4^{s+1} \\\\\n&= \\frac{5n \\cdot 4^{s+1} - 4^{2s+2} + 1}{5} = \\frac{5na_n - a_n^2 + 1}{5}.\n\\end{aligned}\n$$\nThis implies that\n$$\n\\begin{aligned}\nb_n &= \\frac{1}{n^2} \\left( \\frac{5na_n - a_n^2 + 1}{5} - \\frac{1}{5} \\right) = \\frac{5na_n - a_n^2}{5n^2} \\\\\n&= -\\frac{1}{5} \\left( \\frac{a_n}{n} \\right)^2 + \\frac{a_n}{n} = P \\left( \\frac{a_n}{n} \\right).\n\\end{aligned}\n$$\n\nIn short, the polynomial $P(x) = -\\frac{1}{5}x^2 + x$ satisfies the requirements of the problem.\n\nb) According to part a), with $n$ not being a power of 4,\n$$\na_1 + a_2 + \\cdots + a_n = \\frac{4^{2s+1} + 1}{5} + (n - 4^s) \\cdot 4^{s+1},\n$$\nwhere $s$ is such that $4^s < n < 4^{s+1}$.\n\nLet $n' = n - 4^s$ ($1 \\le n' < 3 \\cdot 4^s$), then\n$$\nb_n = \\frac{4^{2s+1} + 5n' \\cdot 4^{s+1}}{5(4^s + n')^2} = \\frac{4 + 20 \\cdot \\frac{n'}{4^s}}{5 \\left(1 + \\frac{n'}{4^s}\\right)^2}.\n$$\nThe function $f(x) = \\frac{4+20x}{5(1+x)^2}$ is continuous on $[0, \\frac{3}{4}]$ and $f(0) < \\frac{2024}{2025} < f(\\frac{3}{4})$, so there exists $x_0 \\in (0, \\frac{3}{4})$ such that $f(x_0) = \\frac{2024}{2025}$. For sufficiently large $s$, set $n' = \\lfloor 4^s x_0 \\rfloor$, so $\\lim_{s \\to \\infty} \\frac{n'}{4^s} = x_0$. Thus,\n$$\n\\lim_{s \\to \\infty} b_{4^s + \\lfloor 4^s x_0 \\rfloor} = f(x_0) = \\frac{2024}{2025}.\n$$\nThe statement is proved. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17051,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider an isosceles trapezoid $ABCD$ with perpendicular diagonals. The parallel from the intersection point of the diagonals meets the non-parallel sides $BC$ and $AD$ at points $P$ and $R$ respectively. Point $Q$ is the mirror image of $P$ across the midpoint of $BC$.\n\nShow that:\n\na) $QR = AD$;\n\nb) $QR \\perp AD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the diagonals $AC$ and $BD$ meet at point $O$ and let $M$ be the midpoint of the line segment $BC$.\n\na) Since $OM$ joins the midpoints of two sides of the triangle $PQR$, $MO \\parallel RQ$ and $OM = \\frac{RQ}{2}$. On the other hand, $OM$ is a median of the right-angled triangle $BOC$, hence $OM = \\frac{1}{2}BC = \\frac{1}{2}AD$. Consequently, $RQ = AD$.\n\n![](images/RMC2011_2_p8_data_a826c563e1.png)\n\nb) Let the lines $MO$ and $AD$ meet at $T$. Then $\\angle MBO = \\angle MOB = \\angle DOT$. Since $\\angle OCB = \\angle TDO$, it follows that $\\angle OTD = \\angle BOC = 90^\\circ$, so $MT \\perp AD$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17052,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(n)$ be the number of ways to write $n$ as a sum of nonnegative powers of $3$, with $f(0) = 1$. Compute $f(100)$.",
    "options": [],
    "answer": "See solution",
    "solution": "We can write $n$ as a sum of nonnegative powers of $3$ with $k$ $1$s if and only if $n - k$ is divisible by $3$. Let $n - k = 3m$; then the number of ways to write $n$ as a sum of nonnegative powers of $3$ with $k$ $1$s is equal to the number of ways to write $3m$ as a sum of positive powers of $3$, which is $f(m)$. Thus,\n\n$$\nf(n) = \\sum_{0 \\leq m \\leq \\frac{n}{3}} f(m).\n$$\n\nTo compute $f(100)$, note that $f(100) = f(0) + f(1) + \\cdots + f(33)$. Using the recurrence:\n- $f(0) = f(1) = f(2) = 1$\n- $f(3) = f(4) = f(5) = f(0) + f(1) = 2$\n- $f(6) = f(7) = f(8) = f(0) + f(1) + f(2) = 3$\n- and so on.\n\nContinuing this process, we find:\n$$\n\\begin{aligned}\nf(100) &= 34f(0) + 31f(1) + 28f(2) + \\cdots + f(11) \\\\\n&= (34 + 31 + 28)f(0) + (25 + 22 + 19)(f(0) + f(1)) \\\\\n&\\quad + (16 + 13 + 10)(f(0) + f(1) + f(2)) \\\\\n&\\quad + (7 + 4 + 1)(f(0) + f(1) + f(2) + f(3)) \\\\\n&= 210f(0) + 117f(1) + 51f(2) + 12f(3).\n\\end{aligned}\n$$\nGiven $f(0) = f(1) = f(2) = 1$ and $f(3) = 2$, we have:\n$$\nf(100) = 210 + 117 + 51 + 24 = 402.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17053,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $m$ for which there exists an infinite set $A$ of positive integers satisfying: For any $m$ distinct elements $a_1, a_2, \\dots, a_m$ from $A$, both the sum $a_1 + a_2 + \\dots + a_m$ and the product $a_1 a_2 \\cdots a_m$ are square-free.\n\n*Note: A positive integer $n$ is called square-free if it is not divisible by the square of any prime.*",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove a lemma.\n\n**Lemma:** For integers $m \\ge 2$, $s \\ge 1$, and a sequence $1 = x_1 < \\dots < x_s$ where every sum $\\sum_{1 \\le j \\le m} x_{i_j}$ ($1 \\le i_1 \\le \\dots \\le i_m \\le s$) is square-free, there exists an integer $x > x_s$ such that:\n\n- $x$ is coprime with each $x_i$ ($1 \\le i \\le s$)\n- Every sum $(m-t)x + \\sum_{1 \\le j \\le t} x_{i_j}$ ($0 \\le t \\le m-1$, $1 \\le i_1 \\le \\dots \\le i_t \\le s$) is square-free\n\n**Proof of Lemma:** Take $B > 1$ such that:\n\n$$\nc_B := \\prod_{p < B} \\left(1 - \\frac{1}{p}\\right)^{-1} > m \\cdot 2^{m+s}.\n$$\n\nLet\n\n$$\nd_B = \\prod_{p \\le B} p, \\quad M_0 = \\prod_{1 \\le k \\le 2^{m+s} x_s} k, \\quad M = d_B M_0.\n$$\n\nConsider $x = yM^2 + x_1$ ($1 \\le y \\le M^2$). Clearly, $x$ is coprime with each $x_i$ ($1 \\le i \\le s$).\nNote that\n\n$$\n(m-t)x + \\sum_{1 \\le j \\le t} x_{i_j} \\le mx \\le m(M^4 + 1) < (mM^2)^2.\n$$\n\nIf $q^2 \\mid (m-t)x + \\sum_{1 \\le j \\le t} x_{i_j}$, then $q < mM^2$.\n\nIf $q \\nmid M$, clearly $(m-t)x + \\sum_{1 \\le j \\le t} x_{i_j}$ is not divisible by $q^2$. For primes $q < mM^2$ with $q \\nmid M$ (so $q > 2^{m+s}x_s$), let $X_q$ be the set of integers $1 \\le y \\le M^2$ where some sum $(m-t)x + \\sum_{1 \\le j \\le t} x_{i_j}$ is divisible by $q^2$.\n\nThe number of possible index sets is at most $2^{s+m-2}$, so\n\n$$\n|X_q| \\le 2^{m+s-2} \\left( \\frac{M^2}{q^2} + 1 \\right).\n$$\n\nFor $2^{m+s}x_s < q \\le M$:\n\n$$\n|X_q| \\le 2^{m+s-1} \\frac{M^2}{q^2};\n$$\n\nFor $M < q \\le mM^2$:\n\n$$\n|X_q| \\le 2^{m+s-1}.\n$$\n\nThus,\n\n$$\n\\begin{aligned}\n|\\bigcup X_q| &\\le 2^{m+s-1} M^2 \\sum_{q>2^{m+s}x_s} \\frac{1}{q^2} + 2^{m+s-1}(\\pi(mM^2) - \\pi(M)) \\\\\n&\\le \\frac{2^{m+s-1}}{2^{m+s}x_s} M^2 + \\frac{m \\cdot 2^{m+s-1}}{c_B} M^2 < M^2.\n\\end{aligned}\n$$\n\nSince we have used $d_B \\mid M$, it follows that\n\n$$\n\\pi(mM^2) - \\pi(M) \\le (mM^2 - M) \\prod_{p \\le B} \\left(1 - \\frac{1}{p}\\right) < c_B^{-1} mM^2.\n$$\n\nNow choose an integer $y$ such that $1 \\le y \\le M$ and\n\n$$\ny \\notin \\bigcup_{\\substack{q<mM^2 \\\\ q \\nmid M}} X_q,\n$$\n\nthen $x = yM^2 + x_1$ will satisfy the requirements.\n\nNow the main proof. If $m$ has a square factor $p^2$, by the pigeonhole principle there exist $m$ elements $a_{i_1}, \\dots, a_{i_m}$ with $a_{i_1} \\equiv \\dots \\equiv a_{i_m} \\pmod{p^2}$, making their sum divisible by $p^2$.\n\nIf $m$ is square-free, construct $A = \\{a_i\\}$ inductively starting with $a_1 = 1$. By the lemma, we can find pairwise coprime integers $1 = a_1 < a_2 < \\dots$ where all $m$-term sums are square-free. Then all $m$-term products are also square-free since the $a_i$ are pairwise coprime. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17054,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the system of equations\n\n$$\n\\frac{1}{xy} = \\frac{x}{z} + 1, \\quad \\frac{1}{yz} = \\frac{y}{x} + 1, \\quad \\frac{1}{zx} = \\frac{z}{y} + 1\n$$\n\nin the domain of the real numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "From the form of the equations, it is immediate that $xyz \\neq 0$. Two of the numbers $x, y, z$ must have the same sign; then the right-hand side of the equation where the ratio of these two numbers occurs is positive, so the corresponding left-hand side must also be positive, which implies that the third of the numbers $x, y, z$ must also have the same sign as the first two. Thus, either $x, y, z > 0$, or $x, y, z < 0$. Let us consider only the former case (the latter can be reduced to it by passing from the solution $(x, y, z)$ to the solution $(-x, -y, -z)$).\n\nMultiply the first two equations of the system by $xyz$ and then subtract them; this gives, after some manipulation, $z - x = y(x^2 - yz)$. If a triple $(x, y, z)$ is a solution, then so are the triples $(y, z, x)$ and $(z, x, y)$; thus we may assume that $x = \\max\\{x, y, z\\}$. Then $z - x \\leq 0$ and $x^2 - yz \\geq 0$ (since $x, y, z > 0$), so the equality $z - x = y(x^2 - yz)$, together with $y > 0$, implies that $z - x = x^2 - yz = 0$, which means $x = y = z$.\n\nThe system then reduces to the single equation $1/x^2 = 1 + 1$, which has a unique positive root $x = \\sqrt{2}/2$.\n\n**Conclusion:** The system has exactly two solutions, $x = y = z = \\pm \\sqrt{2}/2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17055,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer, and $A, B$ be non-empty subsets of $\\{1, 2, \\dots, n\\}$. Prove that there exists a subset $D$ of $A + B$ such that\n\n$$\nD + D \\subseteq 2(A + B), \\text{ and } |D| \\ge \\frac{|A| \\cdot |B|}{2n},\n$$\n\nwhere $|X|$ denotes the number of elements of a finite set $X$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S_y = \\{(a, b) \\mid a - b = y,\\ a \\in A,\\ b \\in B\\}$. Since $\\sum_{y=1-n}^{n-1} |S_y| = |A| \\cdot |B|$, there exists an integer $y_0$ such that $1 - n \\leq y_0 \\leq n - 1$ and $|S_{y_0}| \\geq \\frac{|A| \\cdot |B|}{2n - 1} > \\frac{|A| \\cdot |B|}{2n}$.\n\nLet $D = \\{2b + y_0 \\mid (a, b) \\in S_{y_0}\\}$, then\n\n$$\n|D| = |S_{y_0}| > \\frac{|A| \\cdot |B|}{2n}.\n$$\n\nFrom the definition of $S_{y_0}$, for each $d \\in D$, there exists $(a, b) \\in S_{y_0}$ such that $d = 2b + y_0 = a + b \\in A + B$. So $D \\subseteq A + B$. For any $d_1, d_2 \\in D$, let $d_1 = 2b_1 + y_0 = 2a_1 - y_0$, $d_2 = 2b_2 + y_0$ ($b_1, b_2 \\in B$, $a_1 \\in A$), then\n\n$$\nd_1 + d_2 = 2a_1 - y_0 + 2b_2 + y_0 = 2(a_1 + b_2) \\subseteq 2(A + B).\n$$\n\nTherefore, $D$ satisfies the condition. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17056,
    "subject": "Mathematics (Olympiad)",
    "question": "We say that a ring $A$ has property (P) if any non-zero element can be written uniquely as the sum of an invertible element and a non-invertible element.\n\n(a) If in $A$, $1 + 1 \\ne 0$, prove that $A$ has property (P) if and only if $A$ is a field.\n\n(b) Give an example of a ring that is not a field, containing at least two elements, and having property (P).",
    "options": [],
    "answer": "See solution",
    "solution": "a) If $A$ is a field and $x \\in A$, $x \\ne 0$, then $x$ is invertible and $x = x + 0$; this representation is unique, since $0$ is the only noninvertible element.\n\nAssume $A$ is not a field. Let $x \\in A$, $x \\ne 0$, be a noninvertible element. Since $1 + x = (1 + x) + 0$ and $-1 + x = (-1 + x) + 0$, the elements $1 + x$ and $-1 + x$ are noninvertible, otherwise the uniqueness of the representation would imply $x = 0$.\n\nSince $x = 1 + (-1 + x) = -1 + (1 + x)$, it follows $1 = -1$, contradiction.\n\nb) Let $E$ be a nonempty set, and $A = (\\mathcal{P}(E), \\Delta, \\cap)$ be the Boolean ring of the set of parts of $E$. In this ring, $E$ is the only invertible element.\n\nIf $X$ is a nonempty part of $E$, then the element $E \\setminus X$ is noninvertible, and $X = E \\Delta (E \\setminus X)$. The representation is unique, since $E$ is the only invertible element.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17057,
    "subject": "Mathematics (Olympiad)",
    "question": "In acute $\\triangle DEF$, let $X', Y'$ be points such that $\\triangle DX'E$ and $\\triangle FY'D$ are isosceles right triangles with $X'$ on the same side of $DE$ as $F$, and $Y'$ on the same side of $DF$ as $E$. Let $A'$ be the midpoint of $EF$. Prove that $\\triangle X'A'Y'$ is an isosceles right triangle with a right angle at $A'$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the composition of rotations:\n\n$$\n\\Gamma = \\operatorname{Rot}(A', 180^\\circ) \\circ \\operatorname{Rot}(Y', 90^\\circ) \\circ \\operatorname{Rot}(X', 90^\\circ)\n$$\n\nThe angles add up to $360^\\circ$, so $\\Gamma$ is a translation. However, $\\Gamma$ keeps $E$ fixed, thus $\\Gamma$ is the identity map. Hence, we have\n\n$$\n\\operatorname{Rot}(A', 180^\\circ) = \\operatorname{Rot}(Y', 90^\\circ) \\circ \\operatorname{Rot}(X', 90^\\circ)\n$$\n\nWe finish by noting that the center of rotation of the composition of the two rotations $\\operatorname{Rot}(Y', 90^\\circ)$ and $\\operatorname{Rot}(X', 90^\\circ)$ is a point $O$ such that $\\angle OX'Y' = 45^\\circ$ and $\\angle OY'X' = 45^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17058,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$, let $A'$, $B'$, and $C'$ be the perpendicular feet dropped from the centroid $G$ of triangle $ABC$ onto the sides $BC$, $CA$, and $AB$, respectively. Reflect $A'$, $B'$, and $C'$ through $G$ to points $A''$, $B''$, and $C''$, respectively. Prove that the lines $AA''$, $BB''$, and $CC''$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A_0$ be the perpendicular foot from $A$ to $BC$, $A_1$ the midpoint of $BC$, and $A_2$ the intersection of $AA''$ with $BC$. Define $B_i$ and $C_i$ for $i = 0, 1, 2$ analogously. We will prove that $(AA_0, AA_2)$, $(BB_0, BB_2)$, and $(CC_0, CC_2)$ are pairs of isotomic lines; that is, $A_0$ and $A_2$ are reflections of each other through $A_1$, and similarly for the others. Thus, the lines $AA'' = AA_2$, $BB'' = BB_2$, and $CC'' = CC_2$ are concurrent at the isotomic conjugate of the orthocentre of triangle $ABC$.\n\nIt suffices to show that $AA_0$ and $AA_2$ are isotomic. Note that $\\frac{A'A_1}{A_0A_1} = \\frac{1}{3}$. By Menelaus' theorem applied to triangle $GA'A_1$ with transversal $AA''A_2$, we get $\\frac{A_1A_2}{A'A_2} = \\frac{3}{4}$, so $\\frac{A'A_1}{A_2A_1} = \\frac{1}{3}$. Therefore, $A_0$ and $A_2$ are reflections of each other through $A_1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17059,
    "subject": "Mathematics (Olympiad)",
    "question": "Hallar un polinomio de grado tres cuyas raíces sean, precisamente, el cuadrado de las raíces del polinomio $p(x) = x^3 + 2x^2 + 3x + 4$.",
    "options": [],
    "answer": "See solution",
    "solution": "Sean $r$, $s$ y $t$ las raíces, reales o complejas, del polinomio $p(x)$. Por tanto, $p(x) = (x - r)(x - s)(x - t)$. El polinomio que buscamos, salvo que multipliquemos por una constante, será de la forma\n\n$$\nq(x) = (x - r^2)(x - s^2)(x - t^2)\n$$\n\nDe aquí resulta\n\n$$\nq(x^2) = (x^2 - r^2)(x^2 - s^2)(x^2 - t^2) = (x - r)(x + r)(x - s)(x + s)(x - t)(x + t)\n$$\n\nY notando que\n\n$$\np(-x) = (-x - r)(-x - s)(-x - t) = -(x + r)(x + s)(x + t),\n$$\n\nentonces\n\n$$\n\\begin{aligned}\nq(x^2) &= (x - r)(x + r)(x - s)(x + s)(x - t)(x + t) = p(x)[-p(-x)] \\\\\n&= (x^3 + 2x^2 + 3x + 4)(x^3 - 2x^2 + 3x - 4) = x^6 + 2x^4 - 7x^2 - 16\n\\end{aligned}\n$$\n\nLuego, $q(x) = x^3 + 2x^2 - 7x - 16$ es solución (y, también, este mismo polinomio multiplicado por una constante cualquiera).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17060,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle, $I_a$ the center of the excircle at side $BC$, and $M$ its reflection across $BC$. Prove that $AM$ is parallel to the Euler line of the triangle $BCI_a$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $I$ be the incenter of $ABC$, $H_a$ the orthocenter of $I_aBC$, and $O_a$ the midpoint of the segment $[II_a]$. Then $BO_a = IO_a = I_aO_a = CO_a$, therefore $O_a$ is the circumcenter of triangle $I_aBC$.\n\nMoreover, $IB \\parallel CH_a$ (both are perpendicular to $BI_a$) and, similarly, $IC \\parallel BH_a$, hence $BICH_a$ is a parallelogram.\n\nLet $P$ denote the projection of $I_a$ onto $BC$ and $T$ be the midpoint of $[AI_a]$. If $r, r_a$ are the inradius and the radius of the excircle at side $BC$, respectively, then we have:\n\n$$\n\\frac{IA}{AI_a} = \\frac{r}{r_a} = \\frac{H_aP}{I_aP}, \\text{ hence } \\frac{H_aI_a}{I_aP} = \\frac{II_a}{I_aA} = \\frac{2I_aO_a}{2I_aT} = \\frac{I_aO_a}{I_aT}.\n$$\n\nThis proves that $O_aH_a$ is parallel to $TP$. But $TP \\parallel AM$, therefore $AM$ is parallel to $O_aH_a$, i.e., it is parallel to the Euler line of the triangle $I_aBC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17061,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the function $f : \\mathbb{R}^{+} \\rightarrow \\mathbb{R}^{+}$ satisfying\n$$\nf(x + 2y + f(x + y)) = f(2x) + f(3y), \\quad \\forall x, y > 0.\n$$\n\n1) Find all functions $f$ that satisfy the condition.\n\n2) Suppose $f(4\\sin^4x)f(4\\cos^4x) \\geq f^2(1)$ for $x \\in (0, \\frac{\\pi}{2})$. Find the minimum value of $f(2022)$.",
    "options": [],
    "answer": "See solution",
    "solution": "1) In the given condition, replace $x \\to 3x$, $y \\to 2y$:\n$$\nf(3x + 4y + f(3x + 2y)) = f(6x) + f(6y).\n$$\nBy swapping $x, y$ and comparing the two left sides:\n$$\nf(3x + 4y + f(3x + 2y)) = f(4x + 3y + f(2x + 3y)).\n$$\nFrom injectivity:\n$$\n3x + 4y + f(3x + 2y) = 4x + 3y + f(2x + 3y)\n$$\nor\n$$\nf(3x + 2y) - (3x + 2y) = f(2x + 3y) - (2x + 3y).\n$$\nSet $g(x) = f(x) - x$, then $g(3x + 2y) = g(2x + 3y)$ for all $x, y \\in \\mathbb{R}^+$. Put $a = 3x + 2y$, $b = 2x + 3y$, then\n$$\nx = \\frac{3a - 2b}{5}, \\quad y = \\frac{3b - 2a}{5}\n$$\nso the condition for the pair $(a, b)$ for the existence of $x, y$ is $3a - 2b > 0$ and $3b - 2a > 0$, equivalent to $\\frac{2}{3} < \\frac{a}{b} < \\frac{3}{2}$.\n\nHence $g(a) = g(b)$ for all $\\frac{a}{b} \\in (\\frac{2}{3}, \\frac{3}{2})$. From here, we show that $g$ is a constant function. Indeed, $g(1) = g(x)$ for all $x \\in [1, \\frac{3}{2})$, and inductively\n$$\ng(1) = g(x), \\quad \\forall x \\in \\left[ \\left(\\frac{3}{2}\\right)^k, \\left(\\frac{3}{2}\\right)^{k+1} \\right)\n$$\nso $g(1) = g(x)$ for all $x > 1$. Similarly,\n$$\ng(1) = g(x), \\quad \\forall x \\in \\left( \\left( \\frac{2}{3} \\right)^{k+1}, \\left( \\frac{2}{3} \\right)^k \\right]\n$$\nso $g(1) = g(x)$ for all $x \\in (0, 1)$. Therefore, $g(x)$ is a constant function. Thus, $f(x) = x + c$ with $c \\geq 0$, and checking, this satisfies the original condition.\n\n2) Since $f(x) = x + c$, substitute into the problem condition:\n$$\n(4\\sin^4x + c)(4\\cos^4x + c) \\geq (1 + c)^2\n$$\nor\n$$\n\\sin^4(2x) + 4(\\sin^4x + \\cos^4x)c \\geq 1 + 2c.\n$$\nNotice $\\sin^4x + \\cos^4x = 1 - \\frac{1}{2}\\sin^2 2x$, and let $t = \\sin^2 2x \\in (0, 1]$, rewrite as $t^4 + (4 - 2t^2)c \\geq 1 + 2c$, or\n$$\n2(1 - t^2)c \\geq (1 - t^2)(t^2 + 1).\n$$\nSince $1 - t^2 > 0$, $2c \\geq t^2 + 1$. The maximum of $t^2 + 1$ for $t \\in (0, 1]$ is $2$, so $2c \\geq 2 \\iff c \\geq 1$. Thus, the minimum value of $f(2022)$ is $2023$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17062,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 1$ be an integer. Ruben takes a test with $n$ questions. Each question on this test is worth a different number of points: the first question is worth 1 point, the second 2 points, the third 3 points, and so on, up to the last question, which is worth $n$ points. Each question can be answered either correctly or incorrectly, so an answer for a question can either be awarded all, or none, of the points the question is worth. Let $f(n)$ be the number of ways he can take the test so that the number of points awarded equals the number of questions he answered incorrectly.\n\nDo there exist infinitely many pairs $(a, b)$ with $a < b$ and $f(a) = f(b)$?",
    "options": [],
    "answer": "See solution",
    "solution": "For the first few values of $f$, note that $f(1) = 0$, $f(2) = f(3) = f(4) = 1$, $f(5) = f(6) = 2$, $f(7) = f(8) = 3$, $f(9) = f(10) = 5$. We claim that for $n \\ge 11$, $f(n)$ is strictly increasing as a function of $n$. Therefore, there is only a finite number of pairs as in the problem.\n\nWe view a way of taking a test as a subset of $\\{1, 2, \\dots, n\\}$ by taking the set of numbers of questions that are answered correctly. We say that a subset $S_n$ of $\\{1, 2, \\dots, n\\}$ is an $n$-equally correct set if the sum of all elements of $S_n$ is equal to the number of elements in the complement of $S_n$ in $\\{1, 2, \\dots, n\\}$. So by definition, $f(n)$ is the number of $n$-equally correct sets. Note that a subset $S_n$ of $\\{1, 2, \\dots, n\\}$ is $n$-equally correct if and only if the sum of all elements of $S_n$ plus the number of elements of $S_n$ equals $n$.\n\nWe first claim that $f(n)$ is non-decreasing for $n \\ge 1$. Let $S_{n-1}$ be an $(n-1)$-equally correct set. Then adding 1 to the largest element of $S_{n-1}$ gets us a subset $S_n$ of $\\{1, 2, \\dots, n\\}$. This subset has as many elements as $S_{n-1}$ and has sum 1 higher than the sum of $S_{n-1}$, which is therefore an $n$-equally correct set. This procedure defines an injective map $F_n$ from the set of $(n-1)$-equally correct sets to the set of $n$-equally correct sets for all $n \\ge 2$. Therefore $f(n) \\ge f(n-1)$ for all $n \\ge 2$.\n\nWe now show that, for $n \\ge 11$, there exists an $n$-equally correct set that is not in the image of $F_n$. Note that an $n$-equally correct set of which the two largest elements differ by exactly 1 cannot be in the image of $F_n$; the reason is that a set in this image is one that is obtained by adding 1 to the largest element of a subset of $\\{1, 2, \\dots, n-1\\}$. If $n = 2k + 1$ with $k \\ge 5$, then $S = \\{1, k-2, k-1\\}$ is $n$-equally correct, as $1 + (k-2) + (k-1) + |S| = 2k + 1 = n$. Similarly, if $n = 2k$ with $k \\ge 6$, then $S = \\{2, k-3, k-2\\}$ is $n$-equally correct, because $2 + (k-3) + (k-2) + |S| = 2k = n$. We conclude that $f(n) > f(n-1)$ for all $n \\ge 11$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17063,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to write positive integers in the cells of a $2022 \\times 2022$ board so that the sum of the numbers in any rectangle $R$ is a perfect square if and only if $R$ is a square?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, it is possible.\n\nNumber the columns from left to right and the rows from top to bottom with the numbers $1, 2, \\dots, 2022$. In the cell $(r, c)$ (the $r$-th row and $c$-th column), write the number $x^{2(r+c)}$, where $x$ is a positive integer to be chosen later.\n\nConsider an arbitrary rectangle $R$ covering rows $s, s+1, \\dots, t$ and columns $u, u+1, \\dots, v$. The sum of all numbers in $R$ is:\n\n$$\n\\left(x^{2s} + x^{2(s+1)} + \\dots + x^{2t}\\right) \\cdot \\left(x^{2u} + x^{2(u+1)} + \\dots + x^{2v}\\right)\n$$\n\nThis can be rewritten as:\n\n$$\nx^{2s+2u} \\cdot \\frac{x^{2a} - 1}{x^2 - 1} \\cdot \\frac{x^{2b} - 1}{x^2 - 1},\n$$\n\nwhere $a = t - s + 1$ and $b = v - u + 1$.\n\nIf $a = b$, the sum is always a perfect square for any $x$. It remains to choose $x$ so that for all $a \\neq b$ (with $a, b \\in \\{2, 3, \\dots, 2022\\}$), $(x^{2a} - 1)(x^{2b} - 1)$ is never a perfect square.\n\n**Lemma.** Let $P(x)$ be a monic polynomial of degree $2n$ with integer coefficients. If $P(x)$ is a perfect square for infinitely many positive integers $x$, then $P(x)$ is the square of a polynomial with rational coefficients.\n\n*Proof.* Let $P(x) = x^{2n} + a_{2n-1}x^{2n-1} + \\dots + a_1x + a_0$. For large $x$, $P(x) > 0$. Construct a polynomial $Q(x) = x^n + b_{n-1}x^{n-1} + \\dots + b_0$ with rational coefficients so that $P(x) - Q(x)^2$ has degree less than $n$. This can be done by matching coefficients. Thus, $P(x) = Q(x)^2 + R(x)$, $\\deg R < n$.\n\nMultiply both sides by $M^2$ (for some integer $M$) to clear denominators: $M^2P(x) = H(x)^2 + S(x)$, where $H(x)$ and $S(x)$ have integer coefficients and $\\deg S < n$. If $S(x)$ is not identically zero, then for large $x$, $|S(x)| > x^n$, but $|S(x)| < x^n$ for large $x$ unless $S(x) = 0$, a contradiction. Thus, $S(x) = 0$ and $P(x)$ is a square of a rational polynomial.\n\n*End of proof.*\n\nThe lemma implies that for any given $a$ and $b$, there are only finitely many $x$ such that $(x^{2a} - 1)(x^{2b} - 1)$ is a perfect square. Since there are finitely many pairs $(a, b)$, we can choose $x_0$ so that for all $a \\neq b$, $(x_0^{2a} - 1)(x_0^{2b} - 1)$ is not a perfect square. Thus, setting $x = x_0$ gives the desired construction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17064,
    "subject": "Mathematics (Olympiad)",
    "question": "Sea $n \\geq 2$ un entero. Consideremos un tablero de tamaño $n \\times n$ formado por $n^2$ cuadrados unitarios. Una configuración de $n$ fichas en este tablero se dice que es *pacífica* si en cada fila y en cada columna hay exactamente una ficha. Determina, para cada $n$, el mayor entero $k$ tal que en toda configuración pacífica de $n$ fichas, existe un cuadrado de tamaño $k \\times k$ sin fichas en sus $k^2$ cuadrados unitarios.",
    "options": [],
    "answer": "See solution",
    "solution": "Decimos que una configuración pacífica en un tablero $n \\times n$ y una configuración pacífica en un tablero $(n-1) \\times (n-1)$ están relacionadas si y sólo si la segunda puede obtenerse de la primera mediante los siguientes pasos:\n\n- Eliminamos una de las dos filas (inferior o superior) y una de las dos columnas (derecha o izquierda), partiendo de una configuración pacífica en un tablero $n \\times n$.\n- Si en la intersección de la fila y columna eliminadas había una ficha, claramente obtenemos una configuración pacífica en un tablero $(n-1) \\times (n-1)$.\n- Si en la intersección de la fila y columna eliminadas no había una ficha, obtenemos un tablero $(n-1) \\times (n-1)$ con exactamente $n-2$ fichas, de forma que no hay dos en la misma fila ni en la misma columna, es decir, hay exactamente una fila y una columna sin ninguna ficha, y exactamente una ficha en las restantes filas y columnas. Añadimos entonces una ficha en la intersección de la fila y la columna vacías, obteniéndose claramente una configuración pacífica en un tablero $(n-1) \\times (n-1)$.\n\nCada configuración pacífica en un tablero $n \\times n$ está relacionada con varias (hasta un máximo de 4) configuraciones pacíficas en un tablero $(n-1) \\times (n-1)$. Recíprocamente, partiendo de una configuración pacífica en un tablero $(n-1) \\times (n-1)$, podemos añadir una fila arriba o abajo y una columna a la derecha o a la izquierda, bien insertando una ficha en su intersección, bien eliminando una ficha existente y colocando dos fichas en las intersecciones de las dos filas y las dos columnas que quedan ahora vacías, generando diferentes configuraciones pacíficas en tableros $n \\times n$ que están relacionadas con la de partida. En cualquiera de los casos, dado un cuadrado máximo $k \\times k$ sin fichas en el tablero $(n-1) \\times (n-1)$, tampoco contiene ninguna ficha en todas las configuraciones pacíficas en el tablero $n \\times n$ relacionadas con la primera. Luego el valor de $k$ no decrece al crecer $n$.\n\nSupongamos que $n = p^2 + 1$ para algún entero positivo $p$, y consideremos una configuración pacífica cualquiera. Existe exactamente una ficha en la columna más a la derecha. Tomemos un grupo de $p$ filas consecutivas que contengan a la fila en la que está esta ficha, que forman un rectángulo $p \\times (p^2+1)$. Si eliminamos esta última columna, queda un rectángulo $p \\times p^2$ que contiene exactamente $p-1$ fichas (las exactamente $p$ fichas contenidas en estas $p$ filas, menos la que se ha quedado en la columna más a la derecha). Dividiendo este rectángulo en $p$ cuadrados $p \\times p$, al menos uno de ellos está vacío. Luego si $n = p^2 + 1$, entonces $k \\geq p$.\n\nSupongamos que $n = p^2$ para algún entero positivo $p$. Para cada par $(i, j)$, donde $i, j \\in \\{1, 2, \\dots, p\\}$, coloquemos una ficha en la casilla $(p(i-1)+j,\\ p(j-1)+i)$ del tablero, donde hemos numerado las filas de 1 a $n$ de abajo a arriba, las columnas de 1 a $n$ de izquierda a derecha, y la casilla $(u, v)$ está en la intersección de la fila $u$ y la columna $v$. Claramente hemos colocado $p^2 = n$ fichas, tantas como pares $(i, j)$ con $i, j \\in \\{1, 2, \\dots, p\\}$. Supongamos que en esta configuración, existen dos fichas distintas que comparten fila, es decir, existen pares $(i, j)$ e $(i', j')$ distintos, con $i, i', j, j' \\in \\{1, 2, \\dots, p\\}$ tales que $p(i-1)+j = p(i'-1)+j'$, es decir, $p(i-i') = j'-j$. Como la diferencia de dos enteros positivos menores o iguales que $p$ es a lo sumo $p-1$, y $j'-j$ ha de ser múltiplo de $p$, entonces $j = j'$, de donde $i - i' = 0$, en contradicción con la elección de los pares $(i, j)$ e $(i', j')$. Luego no hay dos fichas que compartan fila. De forma análoga, no hay dos fichas que compartan columna. Luego la configuración descrita es pacífica.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17065,
    "subject": "Mathematics (Olympiad)",
    "question": "Label the sides of a regular $n$-gon in clockwise order with $1, 2, \\ldots, n$. Determine all integers $n$ ($n \\ge 4$) satisfying the following two conditions:\n\n1. $n-3$ non-intersecting diagonals in the $n$-gon are selected, subdividing the $n$-gon into $n-2$ non-overlapping triangles.\n2. Each of the chosen $n-3$ diagonals is labeled with an integer, such that the sum of the labeled numbers on the three sides of each triangle is equal for all triangles.",
    "options": [],
    "answer": "See solution",
    "solution": "The required integers are those satisfying both conditions: $n \\ge 4$ and $n \\not\\equiv 2 \\pmod{4}$.\n\nSuppose $n$ satisfies conditions (1) and (2). We first prove that $n \\not\\equiv 2 \\pmod{4}$. Let $S$ be the sum of the labels on the three sides of any triangle, and $m$ the sum of the labels on the $n-3$ chosen diagonals. Summing the labels for all triangles, each diagonal appears twice, so\n$$\n(n-2)S = (1+2+\\cdots+n) + 2m.\n$$\nIf $n \\equiv 2 \\pmod{4}$, then $(n-2)S$ is even, but $(1+2+\\cdots+n) + 2m$ is odd, a contradiction. Thus, $n \\not\\equiv 2 \\pmod{4}$.\n\nThe condition of regularity can be relaxed to any convex $n$-gon.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p272_data_269b5b13fd.png)\n![](images/Mathematical_Olympiad_in_China_2011-2014_p272_data_c5f76e6f48.png)\n![](images/Mathematical_Olympiad_in_China_2011-2014_p272_data_08dd55e5af.png)\n\nNext, we prove that all $n \\ge 4$ with $n \\not\\equiv 2 \\pmod{4}$ satisfy the conditions. Figure 7.1 shows labelings for $n=4,5,7$, which can be checked directly.\n\nIf $n$ satisfies the conditions, so does $n+4$. As shown in Figure 7.2, label a diagonal subdividing the $(n+4)$-gon into a convex $n$-gon and a convex hexagon. The $n$-gon can be subdivided as before, and the hexagon can be subdivided with three diagonals labeled $S-2n-1$, $n+1$, and $S-2n-5$. The sum of the three labels in each triangle is $S$, so $n+4$ also satisfies the conditions.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p273_data_d320cd3bc5.png)\n\nIn summary, the integers $n$ satisfying the conditions are exactly those with $n \\ge 4$ and $n \\not\\equiv 2 \\pmod{4}$.\n\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17066,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a semicircle with diameter $PQ$. Consider a chord $BC$ of fixed length $d$ whose endpoints are distinct from $P$ and $Q$. A ray of light emanating from $B$ reaches point $C$ after reflecting from $PQ$ at such a point $A$ that $\\angle PAB = \\angle QAC$. Prove that $\\angle BAC$ does not depend on the position of the chord $BC$ on $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Reflect $k$ and $C$ about $PQ$ to get $l$ and $C'$, respectively. Then $C'$ lies on $l$ and, since $\\angle QAC' = \\angle QAC = \\angle PAB$, it also lies on $BA$. Triangle $C'CA$ is isosceles, hence\n\n$$\n\\angle BAC = \\angle AC'C + \\angle ACC' = 2 \\cdot \\angle BC'C\n$$\n\nThe chord $BC$ of circle $k \\cup l$ has a fixed length, hence the corresponding inscribed angle $BC'C$ has fixed size and we may conclude.\n\n![](images/brozura_a67angl_new_p10_data_a972217ac6.png)\n\nFig. 1\n\n![](images/brozura_a67angl_new_p10_data_75919c82ee.png)\n\nFig. 2",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17067,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{N}^2$ denote the set of ordered pairs of positive integers. A finite subset $S$ of $\\mathbb{N}^2$ is *stable* if whenever $(x, y)$ is in $S$, then so are all points $(x', y')$ of $\\mathbb{N}^2$ with both $x' \\le x$ and $y' \\le y$.\n\nProve that if $S$ is a stable set, then among all stable subsets of $S$ (including the empty set and $S$ itself), at least half of them have an even number of elements.",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed by induction on $|S|$, with $|S| \\le 1$ clear.\n\nSuppose $|S| \\ge 2$. For any $p \\in S$, let $R(p)$ denote the stable rectangle with upper-right corner $p$. We say such $p$ is *pivotal* if $p + (1, 1) \\notin S$ and $|R(p)|$ is even.\n\n![](images/sols-TSTST-2020_p8_data_c682241a0e.png)\n\n**Claim** — If $|S| \\ge 2$, then a pivotal $p$ always exists.\n\n_Proof._ Consider the top row of $S$.\n\n- If it has length at least 2, one of the two rightmost points in it is pivotal.\n- Otherwise, the top row has length 1. Now either the top point or the point below it (which exists as $|S| \\ge 2$) is pivotal. $\\square$\n\nWe describe how to complete the induction, given some pivotal $p \\in S$. There is a partition\n\n$$\nS = R(p) \\sqcup S_1 \\sqcup S_2\n$$\n\nwhere $S_1$ and $S_2$ are the sets of points in $S$ above and to the right of $p$ (possibly empty).\n\n**Claim** — The desired inequality holds for stable subsets containing $p$.\n\n_Proof._ Let $E_1$ denote the number of even stable subsets of $S_1$; denote $E_2$, $O_1$, $O_2$ analogously. The stable subsets containing $p$ are exactly $R(p) \\sqcup T_1 \\sqcup T_2$, where $T_1 \\subseteq S_1$ and $T_2 \\subseteq S_2$ are stable.\n\nSince $|R(p)|$ is even, exactly $E_1E_2 + O_1O_2$ stable subsets containing $p$ are even, and exactly $E_1O_2 + E_2O_1$ are odd. As $E_1 \\ge O_1$ and $E_2 \\ge O_2$ by inductive hypothesis, we obtain $E_1E_2 + O_1O_2 \\ge E_1O_2 + E_2O_1$ as desired. $\\square$\n\nBy the inductive hypothesis, the desired inequality also holds for stable subsets not containing $p$, so we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17068,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be an interior point of a regular $n$-gon $A_1A_2\\cdots A_n$. The lines $A_iP$ meet the regular $n$-gon $A_1A_2\\cdots A_n$ at another point $B_i$, where $i = 1, 2, \\dots, n$. Prove that\n\n$$\n\\sum_{i=1}^{n} PA_i \\geq \\sum_{i=1}^{n} PB_i.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $t = \\lfloor \\frac{n}{2} \\rfloor + 1$, and let $A_{n+j} = A_j$ for $j = 1, 2, \\dots, n$.\n\nNoting that the distance between any vertex of a regular $n$-gon and a point on its side is not greater than its longest diagonal $d$, we therefore have, for any $1 \\le i \\le n$,\n\n$$\nA_i P + P B_i = A_i B_i \\le d. \\qquad \\textcircled{1}\n$$\n\nFurthermore, using the fact that the sum of any two sides of a triangle is longer than the third side, we have, for any $1 \\le i \\le n$,\n\n$$\nA_i P + P A_{i+1} \\ge A_i A_{i+1} = d. \\qquad \\textcircled{2}\n$$\n\nSumming up $\\textcircled{1}$ and $\\textcircled{2}$ for $i = 1, 2, \\dots, n$, we have\n\n$$\n\\sum_{i=1}^{n} (A_i P + P A_{i+1}) \\ge n d \\ge \\sum_{i=1}^{n} (A_i P + P B_i),\n$$\n\ni.e.\n\n$$\n2 \\sum_{i=1}^{n} A_i P \\ge \\sum_{i=1}^{n} A_i P + \\sum_{i=1}^{n} P B_i,\n$$\n\nfollowing which the proposition is proven.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17069,
    "subject": "Mathematics (Olympiad)",
    "question": "Decide if there is an arithmetic progression of $2016$ natural numbers that are not perfect powers but their product is a perfect power.\n\n(A perfect power is a number of the form $n^k$ where $n$ and $k$ are natural numbers with $n \\ge 2$, $k \\ge 2$.)",
    "options": [],
    "answer": "See solution",
    "solution": "For each $n \\ge 3$ there exists an arithmetic progression of length $n$ with these properties. The solution uses the remark that if $l \\in \\mathbb{N}$ is divisible by a prime $p$ but not by $p^2$—in which case we say that $l$ is exactly divisible by $p$—then $l$ is not a perfect power.\n\nStart a construction by choosing two primes $p$ and $q$ such that $n < p < q$. Consider the arithmetic progression $P$ with first term $p$, common difference $q-p$, and length $n$; its terms are $a_k = p + k(q-p)$, $k = 0, 1, \\dots, n-1$. Observe that $a_0 = p$ is the only term divisible by $p$. Indeed, if $p$ divides $a_k$ with $k \\ge 1$ then $p$ divides $k$ or $q-p$. Both are impossible: since $p$ is a prime, it divides neither $k$ (as $0 < k < n < p$) nor $q-p$ (as $q$ is a prime greater than $p$). Similarly, $a_1 = q$ is the only term divisible by $q$. If $q$ divides $a_k$ with $k \\ge 2$ then $q$ divides $k-1$ or $p$. Neither one is possible as $q$ is prime and $0 < k-1 < n < q$, $p < q$. In addition, $q$ does not divide $a_0 = p$ since $p < q$.\n\nNext, let $A = a_0 a_1 \\dots a_{n-1}$ be the product of the terms of $P$. By the above, $A$ is exactly divisible by $p$ and $q$. Multiply each $a_k$ by $A$ to obtain a new progression $a_0 A, a_1 A, \\dots, a_{n-1} A$ of length $n$. The product of its terms $a_0 a_1 \\dots a_{n-1} A'' = A'^{n+1}$ is a perfect $(n+1)$-st power. Because $p$ divides $a_k$ only for $k=0$ and $A$ is exactly divisible by $p$, it follows that each of the terms $a_1 A, \\dots, a_{n-1} A$ is also exactly divisible by $p$, hence not a perfect power. Similarly, since $a_0 = p$ is not divisible by $q$ and $A$ is exactly divisible by $q$, the first term $a_0 A$ of the new progression is exactly divisible by $q$. So $a_0 A$ is not a perfect power, which completes the justification that the progression $a_0 A, a_1 A, \\dots, a_{n-1} A$ satisfies the given conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17070,
    "subject": "Mathematics (Olympiad)",
    "question": "A convex quadrilateral _ABCD_ where $\\angle DAB + \\angle ABC < 180^\\circ$ is given on a plane. Let $E$ be a point different from the vertices of the quadrilateral on the line segment $AB$ such that the circumcircles of triangles $AED$ and $BEC$ intersect inside the quadrilateral $ABCD$ at point $F$. Point $G$ is defined so that $\\angle DCG = \\angle DAB$, $\\angle CDG = \\angle ABC$, and triangle $CDG$ is located outside quadrilateral $ABCD$. Prove that the points $E$, $F$, $G$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $\\angle DAB = \\alpha$ and $\\angle ABC = \\beta$. From cyclic quadrilaterals $AEFD$ and $BEFC$ one obtains\n\n$$\n\\begin{aligned}\n\\angle DFE &= 180^\\circ - \\angle DAE = 180^\\circ - \\alpha, \\\\\n\\angle CFE &= 180^\\circ - \\angle CBE = 180^\\circ - \\beta,\n\\end{aligned}\n$$\n\nrespectively. Thus $\\angle CFD = 360^\\circ - (180^\\circ - \\alpha) - (180^\\circ - \\beta) = \\alpha + \\beta$. But $\\angle CGD = 180^\\circ - (\\alpha + \\beta)$ by the choice of $G$. Hence the quadrilateral $CFDG$ is cyclic. Consequently, $\\angle DFG = \\angle DCG = \\alpha = 180^\\circ - \\angle DFE$, which implies that the points $E$, $F$, $G$ are collinear.\n\n*Remark 1:* The argumentation can also be turned around in the following way: Let $G'$ be defined as the other intersection point of the circumcircle of triangle $CFD$ and line $EF$. Then the quadrilateral $CFDG'$ is cyclic, whence\n\n$$\n\\begin{aligned}\n\\angle DCG' &= \\angle DFG' = 180^\\circ - \\angle DFE = \\alpha, \\\\\n\\angle CDG' &= \\angle CFG' = 180^\\circ - \\angle CFE = \\beta.\n\\end{aligned}\n$$\n\nThese equalities imply that $G' = G$. Thus $G$ belongs to line $EF$.\n\n*Remark 2:* The claim of the problem holds also if point $F$ does not have to be inside the quadrilateral $ABCD$. Then $G$ may also be located between $E$ and $F$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17071,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, $C$ be points lying on a circle $\\Gamma$ with center $O$ and assume that $\\angle ABC > 90^\\circ$. Let $D$ be the point of intersection of the line $AB$ and the line perpendicular to $AC$ at $C$. Let $\\ell$ be the line through $D$ and perpendicular to $AO$. Let $E$ be the point of intersection of $\\ell$ and the line $AC$, and $F$ be the point of intersection of $\\Gamma$ and $\\ell$ that lies between $D$ and $E$. Prove that the circumcircles of the triangles $BFE$ and $CFD$ are tangent at $F$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\ell \\cap AO = \\{K\\}$ and $G$ be the other endpoint of the diameter of $\\Gamma$ through $A$. Then $D$, $C$, $G$ are collinear. Moreover, $E$ is the orthocenter of triangle $ADG$. Therefore $GE \\perp AD$ and $G$, $E$, $B$ are collinear.\n\n![](images/Balkan_2012_shortlist_p15_data_d4bf01b8ce.png)\n\nAs $\\angle CDF = \\angle GDK = \\angle GAC = \\angle GFC$, $FG$ is tangent to the circumcircle of triangle $CFD$ at $F$. As $\\angle FBE = \\angle FBG = \\angle FAG = \\angle GFK = \\angle GFE$, $FG$ is also tangent to the circumcircle of $BFE$ at $F$. Hence the circumcircles of the triangles $CFD$ and $BFE$ are tangent at $F$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17072,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider exactly $n$ letters in a word, chosen from an alphabet of $m$ letters, with the restriction that any two distinct words must differ in at least two places. What is the maximum number of such words possible?",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that at most $m^{n-1}$ words are possible, and this bound can be achieved.\n\n**Upper bound:**\nBy the pigeonhole principle, there are $m^{n-1}$ possibilities for the first $n-1$ letters in a word. If there are more than $m^{n-1}$ words, at least two must share the same first $n-1$ letters, so they cannot differ in at least two places.\n\n**Construction:**\nConsider all $(n-1)$-letter \"near-words\" from the alphabet (total: $m^{n-1}$). For each near-word $w$, create a word $W$ by appending a letter $s(w)$, chosen so that $s(w)$ is congruent to the sum of the letters in $w$ modulo $n$. This yields $m^{n-1}$ words.\n\nIf two near-words $x$ and $y$ differ in at least two places, so do $X$ and $Y$. If $x$ and $y$ differ in only one place, then $s(x) \\neq s(y)$, so $X$ and $Y$ differ in exactly two places.\n\nThus, the bound is tight.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17073,
    "subject": "Mathematics (Olympiad)",
    "question": "Two players play a game on a regular $n$-gon, where $n \\ge 4$. Three counters start on consecutive vertices. In turn, each player slides one counter along any number of edges to another vertex of the polygon, without jumping over another counter, to strictly increase the area of the triangle formed by the counters. The loser is the first player who cannot make a legal move. For which $n$ does the first player have a winning strategy?",
    "options": [],
    "answer": "See solution",
    "solution": "We shall prove that the first player wins if and only if the exponent of $2$ in the prime decomposition of $n - 3$ is odd.\n\nSince the game is identical for both players, has finitely many possible states, and always terminates, we can label the possible states as Wins or Losses according to whether a player faced with that position has a winning strategy. A state is a Win if and only if there is some legal move taking the state to a Loss, and a state is a Loss if and only if all moves take that state to a Win (including the case where there are no legal moves).\n\n**Lemma.** Any configuration in which the triangle formed by the three counters is not isosceles is necessarily a Win.\n\n**Proof.** Label the positions of the counters $X, Y, Z$ so that the arc $YZ$ of the circumcircle is shortest and the arc $ZX$ is longest. Begin by moving the counter at $Z$ around the polygon on the arc $YZX$ until it forms an isosceles triangle $XYZ'$ with apex at $Y$ (note that the arc $XY$ is less than half the circle, so that $Z$ does not jump over the counter at $X$). If this configuration is a Loss, we are done.\n\nIf instead this configuration is a Win, then the counters can be moved legally from triangle $XYZ'$ to reach a losing state. This cannot involve the counter at $Y$, so by symmetry a Loss state can be reached by moving the counter at $Z'$ to a new location $Z''$. But then the counter at $Z$ could have been moved to $Z''$ in the first place, so the original configuration was a Win as well. This ends the proof of the lemma.\n\nFor every non-zero integer $m$, let $e_2(m)$ denote the exponent of $2$ in the prime decomposition of $m$. Now, given a configuration in which the triangle formed by the three counters is isosceles, the arcs between the vertices have lengths $a, a, b$ respectively (in appropriate units so that $2a + b = n$). We show that the configuration is a Win if and only if $a \\neq b$ and $e_2(a-b)$ is odd.\n\nWrite $b = a \\pm |a-b|$ and notice that the only other isosceles triangle that can be reached from the original configuration is one with arc lengths $a, a \\pm |a-b|/2, a \\pm |a-b|/2$. If $|a-b|$ is odd, this is of course impossible, so the configuration is a Loss, since all non-isosceles configurations are Wins, by the lemma.\n\nIf instead $|a-b|$ is even, then all states that can be reached from the original configuration are Wins, except possibly the state with arc lengths $a, a \\pm |a-b|/2, a \\pm |a-b|/2$. Consequently, $(a, a, b)$ is a Win if and only if $(a, a \\pm |a-b|/2, a \\pm |a-b|/2)$ is a Loss. Since the side lengths of this new triangle differ by $|a-b|/2$, the conclusion follows inductively once the exceptional and trivial case $a = b$ is dealt with.\n\nAs an immediate corollary, the configuration with arc lengths $1, 1, n-2$ (the starting configuration of the question) is a Win if and only if $e_2(n-3)$ is odd.\n\n**Remark.** Relying on the solution presented above, one may also derive an explicit winning strategy. Denote the position in the game by the multiset $\\{a, b, c\\}$ of the lengths of the three arcs between the tokens (again in appropriate units so that $a + b + c = n$). A move now consists in choosing two of the three numbers $a, b, c$, and replacing them by two numbers with the same sum so as to strictly increase the minimum of the pair.\n\nThe winning strategy for a player is to obtain at the end of each of his moves the positions of the form $\\{a, a, b\\}$, where $a = b$ or $e_2(a-b)$ is even; we say that such a position is *good*. At the beginning of the game, the position is good exactly if $e_2(n-3)$ is even.\n\nNow, there is at most one position of the form $\\{a', a', b'\\}$ which may be obtained by a move from a good position $\\{a, a, b\\}$ — that is, with $b' = a$. This position is not good, thus it suffices to show that it is possible to obtain a good position from any non-good one by a move.\n\nLet now $\\{a, b, c\\}$ be a non-good position, with $a \\leq b \\leq c$. If $a + c = 2b$ then one may get the good position $(b, b, b)$. Assume now that $a + c \\neq 2b$. If $e_2(c + a - 2b)$ is even, then it is possible to achieve the good position $\\{b, b, c + a - b\\}$; otherwise, $c + a$ is necessarily even, and one may get the good position $\\{(c + a)/2, (c + a)/2, b\\}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17074,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\ldots, a_{100}$ be a permutation of $1, 2, \\ldots, 100$. For each triple $\\{a_i, a_{i+1}, a_{i+2}\\}$ of consecutive numbers, where $1 \\leq i \\leq 98$, the middle number in the triple is marked. For example, if $a_1 = 7$, $a_2 = 99$, $a_3 = 22$, then $a_2 = 99$ is marked. Let $S$ be the sum of all marked numbers. Find the minimum value of $S$. (Each marked number enters the sum $S$ exactly once, although it may be marked more than once.)",
    "options": [],
    "answer": "See solution",
    "solution": "The desired minimum is $33 \\cdot 34 = 1122$. More generally, for $n = 3k + 1$ instead of $100$, the answer is $S_{\\min} = 2(1 + \\ldots + k) = k(k + 1)$.\n\nFor clarity, we state separately a fact used later in a proof of the lower bound $S(\\alpha) \\geq 2(1 + \\ldots + k)$ for each permutation $\\alpha$ of $1, 2, \\ldots, 3k + 1$.\n\n*Claim.* If $2k$ distinct natural numbers are divided into $k$ pairs $u_j, v_j$ with $u_j < v_j$, $j = 1, \\ldots, k$, then $v_1 + \\ldots + v_k \\geq 2(1 + \\ldots + k)$.\n\nThe justification is by induction on $k$, with the base case $k = 1$ being obvious.\n\nFor the inductive step $k - 1 \\to k$, choose the labeling so that $v_k := \\max_{j=1}^k v_j$. Ignore $u_k$ and $v_k$ for the time being, and apply the inductive hypothesis to the remaining $2k - 2$ numbers. This gives $v_1 + \\ldots + v_{k-1} \\geq 2(1 + \\ldots + (k - 1))$. So it is enough to prove $v_k \\geq 2k$ to complete the inductive step. We have $v_k > v_j$ for all $j = 1, \\ldots, k - 1$ by $v_k = \\max_{j=1}^k v_j$. In addition, observe that $v_k > u_j$ for all $j = 1, \\ldots, k$. This holds for $j = k$ by hypothesis. Suppose that $v_k < u_j$ for some $j = 1, \\ldots, k - 1$. Then $u_j < v_j$ implies $v_k < v_j$, which contradicts the maximum choice of $v_k$. In summary, there are $2k - 1$ distinct natural numbers smaller than $v_k$, namely $u_1, \\ldots, u_k, v_1, \\ldots, v_{k-1}$. Hence $v_k \\geq 2k$, completing the induction.\n\nNow let $\\alpha = (a_1, a_2, \\ldots, a_{3k+1})$ be any permutation of $1, 2, \\ldots, 3k+1$. Divide $a_1, a_2, \\ldots, a_{3k}$ into $k$ triples $T_j = \\{a_{3j-2}, a_{3j-1}, a_{3j}\\}$, $j = 1, \\ldots, k$. Let $u_j$ and $v_j$ be respectively the smaller number and the middle number in $T_j$. The $2k$ numbers $u_j, v_j$, $j = 1, \\ldots, k$, are distinct. Then the claim above gives $v_1 + \\ldots + v_k \\geq 2(1 + \\ldots + k)$. Since $v_1, \\ldots, v_k$ are marked numbers (in general there are more of them), the sum $S(\\alpha)$ of all marked numbers in $\\alpha$ also satisfies $S(\\alpha) \\geq 2(1 + \\ldots + k)$.\n\nThe equality $S(\\alpha) = 2(1 + \\ldots + k)$ is attained for the following permutation of $1, 2, \\ldots, 3k+1$:\n\n$3k+1, 1, 2, 2k+1, 4, 3, 2k+2, 6, 5, 2k+3, \\ldots, 2k-2, 2k-3, 3k-1, 2k, 2k-1, 3k$.\n\nThe marked numbers are precisely $2, 4, \\ldots, 2k$, hence $S(\\alpha) = 2(1 + \\ldots + k)$. This completes the proof that $S_{\\min} = 2(1 + \\ldots + k) = k(k + 1)$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 17075,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x)$ be a polynomial with integer coefficients and $c_0$ its constant term, where $c_0 \\neq 0$. Let $a_1, a_2, \\dots, a_{2009}$ be a sequence of integers such that $a_{k+1} = f(a_k)$ for $k = 1, 2, \\dots, 2008$.\n\n(a) Prove that for any $i$ and $j > i$, $a_{i+1} - a_i$ divides $a_{j+1} - a_j$.\n\n(b) Prove that $a_{2008} \\neq 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Recall that $b - c \\mid f(b) - f(c)$ for any integers $b$ and $c$. Setting $b = a_{i+1}$ and $c = a_i$, we get\n\n$$\na_{i+1} - a_i \\mid f(a_{i+1}) - f(a_i) = a_{i+2} - a_{i+1}.\n$$\n\nBy induction on $j$, $a_{i+1} - a_i \\mid a_{j+1} - a_j$ for any $j > i$.\n\nSuppose, for contradiction, that $a_{2008} = 0$. Note $a_{2009} - a_{2008} = f(0) = a_2 - a_1$. By part (a), since $a_2 - a_1 \\mid a_{k+1} - a_k \\mid a_{2009} - a_{2008}$ for $k = 1, 2, \\dots, 2007$, each $a_{k+1} - a_k$ must be $\\pm f(0)$. Now,\n\n$$\n0 = a_{2008} - a_1 = \\sum_{k=1}^{2007} (a_{k+1} - a_k) = m f(0),\n$$\n\nwhere $m \\in \\{-2007, -2005, \\dots, -1, 1, 3, \\dots, 2007\\}$. This implies $f(0) = 0$, which contradicts $c_0 \\neq 0$. Therefore, $a_{2008} \\neq 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17076,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y, z$ and $a, b, c$ be positive real numbers such that $x + y + z = a + b + c$ and $xyz = abc$. Suppose $\\max\\{x, y, z\\} \\ge \\max\\{a, b, c\\}$. Prove that\n$$\nab + bc + ca \\ge xy + yz + zx.$$",
    "options": [],
    "answer": "See solution",
    "solution": "We may assume $x \\ge y \\ge z$ and $a \\ge b \\ge c$. We are given $x \\ge a$. Consider the cubics\n$$\nP(t) = (t - x)(t - y)(t - z), \\quad Q(t) = (t - a)(t - b)(t - c).\n$$\nSince $a$ is the largest root of $Q(t) = 0$, it follows that $Q(s) \\ge 0$ for $s \\ge a$. In particular, $Q(x) \\ge 0$. Observe that\n$$\nP(t) - Q(t) = (\\alpha - \\beta)t,\n$$\nwhere $\\alpha = xy + yz + zx$ and $\\beta = ab + bc + ca$. We thus get\n$$\n(\\alpha - \\beta)x = P(x) - Q(x) = -Q(x) \\le 0.\n$$\nSince $x$ is positive, we conclude that $\\alpha \\le \\beta$. This gives the desired inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17077,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be an arbitrary arrangement of the numbers $1, 2, \\dots, n$ on a circle. Find\n\n$$\n\\min \\sum_{j=1}^{n} |a_j - a_{j+1}| \\quad \\text{and} \\quad \\max \\sum_{j=1}^{n} |a_j - a_{j+1}|,\n$$\n\nwhere $a_{n+1} = a_1$, and the extrema are taken over all possible arrangements of $1, 2, \\dots, n$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Minimum:**\n\nConsider $1$ and $n$ on the circle. They divide the circle into two arcs. The sum of the absolute differences on either arc is at least $n-1$. Suppose, for example, the numbers $1 = b_1, b_2, \\dots, b_k = n$ appear on one of the arcs between $1$ and $n$, in that order. Then the sum of absolute differences of adjacent numbers on this arc is\n\n$$\n|1 - b_2| + |b_2 - b_3| + \\dots + |b_{k-1} - n| \\geq |1 - n| = n - 1.\n$$\n\nSimilarly, the least sum of absolute differences on the other arc is also $n-1$. Hence,\n\n$$\n\\sum_{j=1}^{n} |a_{j+1} - a_{j}| \\geq 2(n-1).\n$$\n\nThis is achieved by the permutation $(a_1, a_2, \\dots, a_n)$ where $a_j = j$ for $1 \\leq j \\leq n$.\n\n**Maximum:**\n\nWe have\n\n$$\n\\sum_{j=1}^{n} |a_{j+1} - a_{j}| = \\sum_{j=1}^{n} \\pm (a_{j+1} - a_{j}).\n$$\n\nEach of the numbers $1, 2, \\dots, n$ appears in the sum twice. To maximize the sum, we should assign positive signs to larger numbers in both occurrences and negative signs to smaller numbers. This is achieved when:\n\n- For even $n$, the arrangement is $1, n, 2, (n-1), 3, (n-2), \\dots, n/2, (n/2) + 1$;\n- For odd $n$, the arrangement is $1, n, 2, (n-1), 3, (n-2), \\dots, [n/2] + 2, [n/2] + 1$.\n\nThe corresponding sums are:\n\n$$\n\\frac{n^2}{2} \\quad \\text{when } n \\text{ is even,} \\qquad \\frac{n^2-1}{2} \\quad \\text{when } n \\text{ is odd.}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17078,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that in a cyclic quadrilateral $ABCD$, the centers $O_1, O_2, O_3, O_4$ of the Euler circles of triangles $ABC$, $BCD$, $CDA$, and $ABD$ form a quadrilateral $O_1O_2O_3O_4$ whose sides are parallel to the corresponding sides of $ABCD$.",
    "options": [],
    "answer": "See solution",
    "solution": "The center $O'$ of the Euler circle of a triangle lies on the Euler line and is the midpoint of $OH$, where $O$ is the circumcenter and $H$ is the orthocenter. The Euler circle's radius is half the circumradius.\n\nThe circumcircles of triangles $KLS$, $LMT$, $MNS$, and $NKT$ (denoted $c_1, c_2, c_3, c_4$) are the Euler circles of triangles $ABC$, $BCD$, $CDA$, and $ABD$, respectively. Since $ABC$, $BCD$, $CDA$, and $ABD$ are inscribed in the same circle $c(O, R)$, the circles $c_1, c_2, c_3, c_4$ all have radius $\\frac{R}{2}$.\n\nConsider triangles $ABC$ and $ABD$. Let $G_1$ and $G_4$ be the centroids, and $O_1$ and $O_4$ the centers of the Euler circles of $ABC$ and $ABD$, respectively. Since $G_1$ and $G_4$ are centroids:\n\n$$\n\\frac{G_1K}{G_1C} = \\frac{G_4K}{G_4D} = \\frac{1}{2} \\Rightarrow G_1G_4 \\parallel CD.\n$$\n\nSince $O_1$ and $O_4$ are Euler circle centers:\n\n$$\n\\frac{OG_1}{OO_1} = \\frac{OG_4}{OO_4} = 2 \\Rightarrow G_1G_4 \\parallel O_1O_4.\n$$\n\nTherefore, $CD \\parallel O_1O_4$. Similarly, the other sides of $O_1O_2O_3O_4$ are parallel to the corresponding sides of $ABCD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17079,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that $5$ is the largest value of $n$ for which a balanced tournament with $n$ teams exists. (A balanced tournament is one in which, for every set of four teams, there are exactly three matches played among them.)",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $n \\geq 5$ and a balanced tournament exists. First, we show that no three teams all play each other. Assume, for contradiction, that teams $A$, $B$, and $C$ all play each other. With two other teams $D$ and $E$, $D$ and $E$ do not play $A$, $B$, or $C$ (else there would be more than three matches among $A$, $B$, $C$, $D$). Considering $A$, $B$, $D$, $E$, there are at most two matches, contradicting the definition.\n\nNow, suppose $n \\geq 6$. Consider teams $A$ to $F$. If $A$ plays at most two of these, say $B$ and $C$, then among $A$, $D$, $E$, $F$, the other three must all play each other, contradicting the previous result. Thus, $A$ must play at least three, say $B$, $C$, $D$. Then $B$, $C$, $D$ do not play each other. For $B$, $C$, $D$, $E$, $E$ must play all three, but then among $A$, $B$, $C$, $E$, there are four matches, again a contradiction. Thus, no balanced tournament exists for $n \\geq 6$.\n\nFor $n = 5$, arrange teams in a circle, with each team playing its two neighbors. Any quadruple of teams includes exactly three adjacent pairs, so exactly three matches are played among them. Thus, $5$ is the largest value of $n$ for which a balanced tournament exists. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17080,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all four-digit numbers which, after deleting any one digit, turn into a three-digit number that is a divisor of the original number.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\overline{abcd}$ be such a number.\n\nSince $\\overline{abcd}$ is divisible by $\\overline{abc}$, we have $d = 0$.\n\nSince $\\overline{abcd} = \\overline{abc0}$ is divisible by $\\overline{abd} = \\overline{ab0}$, we have $c = 0$.\n\nNow, $\\overline{abcd} = \\overline{ab00}$ must be divisible by $\\overline{acd} = \\overline{a00}$ and by $\\overline{bcd} = \\overline{b00}$, so the number $\\overline{ab}$ is divisible by both $a$ and $b$.\n\nLet $b = a x$ and $10a = b y$ for integers $x$ and $y$. Therefore, $10a = a x y$, so $x y = 10$.\n\nPossible $(x, y)$ pairs:\n- $x = 1, y = 10$: $a = b$, giving 9 numbers: $1100, 2200, 3300, 4400, 5500, 6600, 7700, 8800, 9900$.\n- $x = 2, y = 5$: $b = 2a$, giving 4 numbers: $1200, 2400, 3600, 4800$.\n- $x = 5, y = 2$: $b = 5a$, giving 1 number: $1500$.\n\nThe case $x = 10, y = 1$ is impossible since $a$ and $b$ must be one-digit numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17081,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть $a > b$. Найдите такое число $c > a > b$, чтобы для всех точек $(x_0, y_0)$ пересечения графиков $y = \\sin(ax)$ и $y = \\sin(bx)$ выполнялось $\\sin(cx_0) = y_0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Рассмотрим точки пересечения: $\\sin(ax_0) = \\sin(bx_0)$, то есть $\\sin(ax_0) - \\sin(bx_0) = 0$. Это равносильно\n\n$$\n\\sin\\left(\\frac{a-b}{2}x_0\\right) \\cdot \\cos\\left(\\frac{a+b}{2}x_0\\right) = 0.\n$$\n\nЗначит, $\\frac{a-b}{2}x_0 = k\\pi$ или $\\frac{a+b}{2}x_0 = \\frac{(2k+1)\\pi}{2}$ для некоторого целого $k$. Тогда одно из чисел $\\frac{a-b}{2\\pi}x_0$ или $\\frac{a+b}{\\pi}x_0$ — целое.\n\nВыберем $c = 2(a^2 - b^2) + a$. Тогда $\\frac{a-c}{2\\pi}x_0$ будет целым для всех таких $x_0$, и, следовательно, $\\sin(cx_0) = \\sin(ax_0) = y_0$ во всех точках пересечения. Кроме того, $c > a > b$.\n\nТаким образом, $c = 2(a^2 - b^2) + a$ — подходящее число.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17082,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all integers $n \\geq 1$ for which the number $n^8 + n^6 + n^4 + 4$ is prime.",
    "options": [],
    "answer": "See solution",
    "solution": "We use factorization:\n\n$$\nn^8 + n^6 + n^4 + 4 = (n^4 - n^3 + n^2 - 2n + 2)(n^4 + n^3 + n^2 + 2n + 2).\n$$\n\nThe first factor $f(n)$ satisfies\n\n$$\nf(n) = n^4 - n^3 + n^2 - 2n + 2 = n^3(n - 1) + (n - 1)^2 + 1\n$$\n\nand hence $f(n) \\ge 2$ for all $n \\ge 2$. The second factor $g(n) = n^4 + n^3 + n^2 + 2n + 2$ is strictly greater than $2$ for all $n \\ge 2$. This only leaves the case $n = 1$ as a potential candidate for a prime, and indeed $f(1)g(1) = 1 \\cdot 7 = 7$ is prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17083,
    "subject": "Mathematics (Olympiad)",
    "question": "The 2010 positive numbers $a_1, a_2, \\dots, a_{2010}$ satisfy the inequality $a_i a_j \\le i + j$ for all distinct indices $i, j$. Determine, with proof, the largest possible value of the product $a_1 a_2 \\cdots a_{2010}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Multiplying together the inequalities $a_{2i-1} a_{2i} \\le 4i - 1$ for $i = 1, 2, \\dots, 1005$, we get\n\n$$\na_1 a_2 \\cdots a_{2010} \\le 3 \\cdot 7 \\cdot 11 \\cdots 4019. \\qquad (1)$$\n\nIt remains to show that this bound can be attained.\n\nLet\n\n$$\na_{2008} = \\sqrt{\\frac{4017 \\cdot 4018}{4019}}, \\quad a_{2009} = \\sqrt{\\frac{4019 \\cdot 4017}{4018}}, \\quad a_{2010} = \\sqrt{\\frac{4018 \\cdot 4019}{4017}},$$\n\nand define $a_i$ for $i < 2008$ by downward induction using the recursion\n\n$$a_i = \\frac{2i + 1}{a_{i+1}}.$$ \n\nWe then have\n\n$$a_i a_j = i + j \\quad \\text{whenever } j = i + 1 \\text{ or } (i, j) = (2008, 2010). \\qquad (2)$$\n\nWe will show that (2) implies $a_i a_j \\le i + j$ for all $i < j$, so that this sequence satisfies the hypotheses of the problem. Since $a_{2i-1} a_{2i} = 4i - 1$ for $i = 1, \\dots, 1005$, the inequality (1) is an equality, so the bound is attained.\n\nWe show that $a_i a_j \\le i + j$ for $i < j$ by downward induction on $i + j$. There are several cases:\n\n* If $j = i + 1$, or $(i, j) = (2008, 2010)$, then $a_i a_j = i + j$, from (2).\n\n* If $(i, j) = (2007, 2009)$, then\n\n$$a_i a_{i+2} = \\frac{(a_i a_{i+1})(a_{i+2} a_{i+3})}{a_{i+1} a_{i+3}} = \\frac{(2i+1)(2i+5)}{2i+4} < 2i+2.$$ \n\nHere the second equality comes from (2), and the inequality is checked by multiplying out:\n\n$$(2i+1)(2i+5) = 4i^2 + 12i + 5 < 4i^2 + 12i + 8 = (2i+2)(2i+4).$$\n\n* If $i < 2007$ and $j = i + 2$, then we have\n\n$$a_i a_{i+2} = \\frac{(a_i a_{i+1})(a_{i+2} a_{i+3})(a_{i+2} a_{i+4})}{(a_{i+1} a_{i+2})(a_{i+3} a_{i+4})} \\le \\frac{(2i+1)(2i+5)(2i+6)}{(2i+3)(2i+7)} < 2i+2.$$ \n\nThe first inequality holds by applying the induction hypothesis for $(i+2, i+4)$ and applying (2) for the other pairs. The second inequality can again be checked by multiplying out:\n\n$$(2i+1)(2i+5)(2i+6) = 8i^3 + 48i^2 + 82i + 30 < 8i^3 + 48i^2 + 82i + 42 = (2i+2)(2i+3)(2i+7).$$\n\n* If $j - i > 2$, then\n\n$$a_i a_j = \\frac{(a_i a_{i+1})(a_{i+2} a_j)}{a_{i+1} a_{i+2}} \\le \\frac{(2i+1)(i+2+j)}{2i+3} < i+j.$$ \n\nHere we have used the induction hypothesis for $(i+2, j)$, and again we check the last inequality by multiplying out:\n\n$$(2i+1)(i+2+j) = 2i^2 + 5i + 2 + 2ij + j < 2i^2 + 3i + 2ij + 3j = (2i+3)(i+j).$$\n\nThis covers all the cases and shows that $a_i a_j \\le i + j$ for all $i < j$, as required.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17084,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M = \\{1, 2, \\dots, 2013\\}$ and let $\\Gamma$ be a circle. For every nonempty subset $\\mathcal{A}$ of $M$, let $S(\\mathcal{A})$ be the sum of the elements of $\\mathcal{A}$, and define $S(\\emptyset) = 0$ (where $\\emptyset$ is the empty set).\n\nIs it possible to assign to every subset $\\mathcal{A}$ of $M$ a point $A$ on the circle $\\Gamma$ so that the following conditions are fulfilled:\n\n1. Different subsets are assigned to different points;\n2. All assigned points are vertices of a regular polygon;\n3. If $A_1, A_2, \\dots, A_k$ are some of the assigned points, $k > 2$, such that $A_1A_2 \\dots A_k$ is a regular $k$-gon, then $2014$ divides $S(\\mathcal{A}_1) + S(\\mathcal{A}_2) + \\dots + S(\\mathcal{A}_k)$?",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that this is possible.\n\nThe total number of subsets of $M$ is $2^{2013}$. On the circle $\\Gamma$, choose arbitrarily $2^{2013}$ points which are the vertices of a regular $2^{2013}$-gon. Assign each subset of $M$ to one of these points as follows: if a subset $\\mathcal{A}$ is assigned to a point, assign its complement $\\mathcal{A}^c = M \\setminus \\mathcal{A}$ to the point symmetric to the first with respect to the center of $\\Gamma$ (since $2^{2013}$ is even, this is possible).\n\nIf $A_1, A_2, \\dots, A_k$ are some of the assigned points which are vertices of a regular $k$-gon, then $k$ divides $2^{2013}$, so $k$ is divisible by $4$, say $k = 4t$. Thus, all points $A_1, \\dots, A_k$ can be grouped into $2t$ pairs of symmetric points with respect to the center of $\\Gamma$.\n\nNote that\n$$\nS(\\mathcal{A}) + S(\\mathcal{A}^c) = 1 + 2 + \\dots + 2013 = 1007 \\cdot 2013.\n$$\nTherefore,\n$$\nS(\\mathcal{A}_1) + \\dots + S(\\mathcal{A}_k) = 2t \\cdot 1007 \\cdot 2013 = 2014 \\cdot 2013 \\cdot t,\n$$\nwhich is divisible by $2014$. Thus, all conditions are fulfilled.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17085,
    "subject": "Mathematics (Olympiad)",
    "question": "At a gala banquet, $12n+6$ chairs, where $n \\in \\mathbb{N}$, are equally arranged around a large round table. A seating will be called a proper seating of rank $n$ if a gathering of $6n + 3$ married couples sit around this table such that each seated person also has exactly one sibling (brother/sister) of the opposite gender present (siblings cannot be married to each other) and each man is seated closer to his wife than his sister. Among all proper seatings of rank $n$, find the maximum possible number of women seated closer to their brother than their husband. (The maximum is taken not only across all possible seating arrangements for a given gathering, but also across all possible gatherings.)",
    "options": [],
    "answer": "See solution",
    "solution": "We will call a woman *unusual* if she sits closer to her husband than her brother. Our goal is to find the smallest possible number of unusual women. Let us call this number $k$.\n\nWe note that going from each man to his sister and from each woman to her husband, we obtain an oriented graph which breaks up into oriented cycles of even length. We also note that within an oriented cycle, the sequence of lengths between consecutive members increases unless we encounter an unusual woman. Thus, each cycle must have at least one unusual woman. We also note that since the maximum length between two seats is $6n + 3$, this is also the maximum distance within a cycle we can go without encountering an unusual woman. Thus, we can have neither $k = 0$ nor $k = 1$, since by the first requirement we would have only one cycle, but this cycle would then have to have more than $6n + 3$ people. We will show that $k = 2$ is also impossible. The only options for $k = 2$ are to have either one cycle with lengths $1, 2, \\dots, 6n + 3, 1, 2, \\dots, 6n + 3$ or two cycles with lengths $1, 2, \\dots, 6n + 3$. We note that the second option does not work because the cycles have odd length, while the first option does not work because the two locations where the unusual women are supposed to be are at an odd distance from each other along the cycle and therefore those positions cannot be occupied by two people of the same gender.\n\nWe now give an example for $k = 3$. Arrange the three unusual women in an equilateral triangle and place their brothers diametrically opposite of them. Each unusual woman is part of a cycle of length $4n + 2$. If we label the members of one of these cycles in order as\n\n$$\nW_1, M_1, W_2, M_2, \\dots, W_{2n+1}, M_{2n+1},\n$$\n\nwhere $W_1$ is the unusual woman, then if we place $W_1$ at position $0$, we can place each $M_k$, $k = 1, \\dots, 2n$ at position $k$ and each $W_k$ at position $k = 1, \\dots, 2n + 1$ at position $-k + 1$. Finally, we place $M_{2n+1}$, the brother of $W_1$, at position $6n + 3$.\n\nThe other two unusual women are placed symmetrically at positions $4n + 2$ and $-(4n + 2)$. We note that all the conditions of the problem are satisfied for this arrangement. Thus, the maximum possible number of women seated closer to their brother than their husband is $6n + 3 - 3 = 6n$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17086,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $A$, $B$, $C$, $D$ lie on a circle in this order, where $AB$ and $CD$ are not parallel. The length of the arc $\\widehat{AB}$ that contains points $C$, $D$ is twice as large as the length of the arc $\\widehat{CD}$ that does not contain points $A$, $B$. Point $E$ is chosen such that $AC = AE$ and $BD = BE$, and $E$ lies on the same side of the line $AB$ as $C$ and $D$. Assuming that the perpendicular line from the point $E$ to the line $AB$ bisects the arc $\\widehat{CD}$ not containing points $A$, $B$, prove that $\\angle ACB = 108^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use the following lemma:\n\n**Lemma.** Given two circles $\\Gamma_1$, $\\Gamma_2$ with the centre $S_2$ of $\\Gamma_2$ lying on the circle $\\Gamma_1$. The circles intersect in two points $K$ and $L$. Let $M$ be the point on the circle $\\Gamma_1$ (different from $K$ and $L$) and the line $KM$ meets $\\Gamma_2$ again in $N$. Then $MN = ML$.\n\n*Proof.*\n\n![](images/CpsMT11_sol_p1_data_1e90fc9c4e.png)\n\nFig. 1a\n\n![](images/CpsMT11_sol_p1_data_2541ffbc36.png)\n\nFig. 1b\n\nFor this lemma it is sufficient to prove that the line $MS_2$ bisects the angle $\\angle NML$. Then in the reflection with respect to the line $MS_2$, $ML$ is the image of $MN$. The circle $\\Gamma_2$ and the point $M$ reflect to themselves, the point $L$ reflects to the point $N$ (the intersection point of $ML$ and $\\Gamma_2$). So the triangle $\\triangle MLN$ is isosceles (some considerations are needed according to the position of the point $M$).\n\nFirstly, let $M$ lie on the arc $\\widehat{KL}$ not containing the point $S_2$. As $S_2K = S_2L$, we directly have $\\angle KMS_2 = \\angle S_2ML$. Secondly, let $M$ lie on the arc $\\widehat{KL}$ containing point $S_2$. Let $R$ be an arbitrary point on the arc $\\widehat{KL}$ not containing the point $S_2$. Similarly as before $\\angle KRS_2 = \\angle S_2RL$, then using identical angles in the cyclic quadrilaterals $RS_2MK$ and $RLS_2M$ we obtain $\\angle NMS_2 = \\angle KRS_2 = \\angle S_2RL = \\angle S_2ML$. ■\n\nLet the perpendicular line from the point $E$ to the line $AB$ intersect the arc $\\widehat{BC}$ in the point $S$, $k_1$ be the circle centered at $A$ passing through $C$, $k_2$ be the circle centered at $B$ passing through $D$, $k$ be the circle passing through $A$, $B$, $C$, $D$. The line $SC$ intersects $k_1$ again in $C'$, and the line $SD$ intersects $k_2$ again in $D'$. Circles $k_1$ and $k$ meet in $C$, $C''$, and circles $k_2$ and $k$ meet in $D$, $D''$. Using the lemma we have $SC' = SC''$ and $SD' = SD''$. Let the circles $k_1$ and $k_2$ meet again in $E'$. Using a contradiction, we shall prove that $C'' = D'' = E'$.\n\nThe point $S$ lies on the chord $EE' \\perp AB$ of the circles $k_1$ and $k_2$. That means its powers to these two circles are equal. We know that $S$ bisects the arc $\\widehat{CD}$, so $SD = SC$ and consequently $SC'' = SC' = SD' = SD''$. If $D''$ and $C''$ are different points then the triangle $\\triangle SD''C''$ is isosceles and its altitude from $S$ passes through the circumcentre of $k$ and consequently (by the symmetry) the quadrilateral $CDD''C''$ is an isosceles trapezoid ($SC = SD$). We can find points $A$ and $B$ as the intersection points of the axes of the segments $CC''$ and $DD''$ with $k$. But then also $ABCD$ is an isosceles trapezoid, $AB \\parallel CD$, which is a contradiction to the given $AB \\nparallel CD$.\n\n![](images/CpsMT11_sol_p1_data_9e402310e6.png)\n\nIf we denote $\\angle DE'S = \\angle SE'C = \\alpha$ and $\\angle AE'D = \\beta$ then $\\angle CE'D = 2\\alpha - \\beta$ because $2|\\widehat{CD}| = |\\widehat{AB}|$. Using $BD = BE'$ and $AC = AE'$ we compute the angles in the triangle $ABE'$:\n\n$$\n\\begin{aligned}\n2\\alpha + \\beta &= \\angle AE'C = \\angle ACE' = \\angle ABE' \\\\\n4\\alpha - \\beta &= \\angle BE'D = \\angle BDE' = \\angle BAE'\n\\end{aligned}\n$$\n\nwhich yields\n\n$$\n180^{\\circ} = 2\\alpha + \\beta + 4\\alpha - \\beta + 4\\alpha = 10\\alpha\n$$\n\nand $\\angle ACB = 180^{\\circ} - \\angle AE'B = 180^{\\circ} - 4\\alpha = 108^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17087,
    "subject": "Mathematics (Olympiad)",
    "question": "In the interiors of the sides $AB$, $BC$, and $CA$ of a given triangle $ABC$, points $K$, $L$, and $M$, respectively, are given such that\n\n$$\n\\frac{|AK|}{|KB|} = \\frac{|BL|}{|LC|} = \\frac{|CM|}{|MA|}\n$$\n\nShow that the triangles $ABC$ and $KLM$ have a common orthocenter if and only if the triangle $ABC$ is equilateral.",
    "options": [],
    "answer": "See solution",
    "solution": "A point $V$ in the plane of triangle $ABC$ is its orthocenter if and only if $AV \\perp BC$ and $BV \\perp AC$; that is, $AV \\cdot BC = 0$ and $BV \\cdot AC = 0$. Substituting $BC = BV - CV$ and $AC = AV - CV$, we find this is equivalent to\n\n$$\nAV \\cdot BV = AV \\cdot CV = BV \\cdot CV. \\quad (1)\n$$\n\nOur goal is to determine when system (1) is satisfied together with the analogous system\n\n$$\nKV \\cdot LV = KV \\cdot MV = LV \\cdot MV, \\quad (2)\n$$\n\nwhich expresses that $V$ is the orthocenter of triangle $KLM$. By hypothesis, there exists a number $p$, $0 < p < 1$, such that\n\n$$\nAK = p\\,AB, \\quad BL = p\\,BC, \\quad CM = p\\,CA.\n$$\n\nSubstituting $AK = AV - KV$ and $AB = AV - BV$, we get\n\n$$\nKV = (1-p)AV + pBV, \\quad LV = (1-p)BV + pCV, \\quad MV = (1-p)CV + pAV.\n$$\n\nTaking scalar products, we obtain\n\n$$\nKV \\cdot LV = (1-p)^2 AV \\cdot BV + p(1-p) AV \\cdot CV + p(1-p) BV^2 = (1-p)s + p(1-p) BV^2,\n$$\n\nwhere $s$ denotes the common value from (1). Similarly,\n\n$$\nKV \\cdot MV = (1-p)s + p(1-p) AV^2, \\quad LV \\cdot MV = (1-p)s + p(1-p) BV^2.\n$$\n\nThus, system (2) is equivalent to\n\n$$\np(1-p) AV^2 = p(1-p) BV^2 = p(1-p) CV^2,\n$$\n\nwhich, since $p(1-p) \\neq 0$, holds if and only if $|AV| = |BV| = |CV|$. This means the orthocenter $V$ of triangle $ABC$ coincides with its circumcenter, which occurs if and only if $ABC$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17088,
    "subject": "Mathematics (Olympiad)",
    "question": "For each integer $n \\ge 2$, let $S_n$ be the sum of all products $jk$, where $j$ and $k$ are integers and $1 \\le j < k \\le n$. What is the sum of the 10 least values of $n$ such that $S_n$ is divisible by 3?\n\n(A) 196  \n(B) 197  \n(C) 198  \n(D) 199  \n(E) 200",
    "options": [],
    "answer": "See solution",
    "solution": "For $n \\ge 2$, let\n\n$$\nU_n = (1 + 2 + 3 + \\dots + (n-1)) \\cdot n = \\frac{(n-1)n^2}{2}.\n$$\n\nThen $S_n = S_{n-1} + U_n$ for $n \\ge 3$. Note that $U_n$ is divisible by 3 if $n \\equiv 0$ or $1 \\pmod{3}$; and if $n \\equiv 2 \\pmod{3}$, then $(n-1)n^2 \\equiv 1 \\pmod{3}$ and is even, so $U_n \\equiv 2 \\pmod{3}$. Hence $S_{n+3} \\equiv S_n + 2 \\pmod{3}$ for $n \\ge 2$. It is readily verified that $S_2 \\equiv S_3 \\equiv S_4 \\equiv 2 \\pmod{3}$, so $S_5 \\equiv S_6 \\equiv S_7 \\equiv 1 \\pmod{3}$ and $S_8 \\equiv S_9 \\equiv S_{10} \\equiv 0 \\pmod{3}$, and it follows that $S_n$ is divisible by 3 if and only if $n \\equiv 0$ or $\\pm 1 \\pmod{9}$. Thus the sum of the 10 least values of $n$ that satisfy the required condition is\n\n$$\n8 + 9 + 10 + 17 + 18 + 19 + 26 + 27 + 28 + 35 = 197.\n$$\n\n**Alternate approach:**\n\nThe sum of the products $jk$ as $j$ and $k$ run independently from 1 to $n$ is\n\n$$\n(1 + 2 + \\dots + n)^2 = \\left(\\frac{n(n+1)}{2}\\right)^2.\n$$\n\nTo eliminate the cases in which $j = k$, subtract\n\n$$\n1^2 + 2^2 + \\dots + n^2 = \\frac{n(n+1)(2n+1)}{6}.\n$$\n\nThus\n\n$$\n\\sum_{\\substack{1 \\le j \\le n \\\\ 1 \\le k \\le n \\\\ j \\ne k}} jk = \\frac{n^2(n+1)^2}{4} - \\frac{n(n+1)(2n+1)}{6} = \\frac{(n-1)n(n+1)(3n+2)}{12}.\n$$\n\nFor a given pair $j, k$ with $j \\ne k$ either $j < k$ or $j > k$, but their product is the same in either order. To impose the condition $j < k$, it suffices to divide by 2. Thus\n\n$$\nS_n = \\frac{(n-1)n(n+1)(3n+2)}{24}.\n$$\n\nThere is one factor of 3 in the denominator. For any $n$, exactly one of $n+1, n, n-1$ is divisible by 3, and $3n+2$ is not divisible by 3. In order that $S_n$ be divisible by 3 it is necessary and sufficient that the factor that is divisible by 3 should in fact be divisible by 9. That is, $n \\equiv 0$ or $\\pm 1 \\pmod{9}$, and the answer can be calculated as above.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17089,
    "subject": "Mathematics (Olympiad)",
    "question": "If the average of four numbers is $8$, what is the largest possible value of one of the numbers, given that all four numbers are positive, different integers?",
    "options": [],
    "answer": "See solution",
    "solution": "Their sum must be $8 \\times 4 = 32$. To maximize the largest number, choose the smallest possible values for the other three, which are $1$, $2$, and $3$ (since the integers must be positive and different). The remaining number is then $32 - (1 + 2 + 3) = 26$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17090,
    "subject": "Mathematics (Olympiad)",
    "question": "Define $b_n = (a_1 + 1)(2a_2 + 1) \\cdots ((n-1)a_{n-1} + 1)$ for $n = 2, 3, \\dots$ and let $b_1 = 1$.\n\nProve by mathematical induction that for all positive values of $n$,\n\n$$\nb_{n+1} = (b_1 + 2b_2 + \\dots + n b_n)^2 - 5.\n$$\n\nIf $p$ is a prime number dividing $n a_n + 1$, show that $p$ also divides $b_{n+1}$ and hence $p$ divides $m^2 - 5$ for $m = b_1 + 2b_2 + \\dots + n b_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove the formula by induction.\n\nFor $n=1$, $b_2 = a_1 + 1 = -4$, and $b_1^2 - 5 = 1^2 - 5 = -4$, so the formula holds.\n\nAssume the formula holds for $n = k-1$:\n$$\nb_k = (b_1 + 2b_2 + \\dots + (k-1) b_{k-1})^2 - 5.\n$$\nWe want to show it holds for $n = k$:\n$$\nb_{k+1} = (b_1 + 2b_2 + \\dots + k b_k)^2 - 5.\n$$\nBy the definition of $b_n$, and using the given recurrence relations, we derive\n$$\nb_{k+1} - b_k = k^2 b_k^2 + 2k b_k (b_1 + 2b_2 + \\dots + (k-1) b_{k-1}).\n$$\nThis matches the difference between $(b_1 + 2b_2 + \\dots + k b_k)^2$ and $(b_1 + 2b_2 + \\dots + (k-1) b_{k-1})^2$, confirming the induction step.\n\nNow, if $p$ is a prime dividing $n a_n + 1$, then $p$ divides $b_{n+1}$ by the definition of $b_n$. Since $b_{n+1} = m^2 - 5$ for $m = b_1 + 2b_2 + \\dots + n b_n$, $p$ divides $m^2 - 5$ as required.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17091,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer with $n > 1$. The square table $ABCD$ of size $n$ consists of $n^2$ unit cells, each colored with one of three colors: black, white, or gray. A coloring of the table is called *nice* if each cell on the diagonal $AC$ is colored gray, and each pair of cells symmetric with respect to $AC$ are colored the same (both white or both black).\n\nEach gray cell is filled with the number $0$, each black cell with a negative integer, and each white cell with a positive integer. For each positive integer $k$, a way to fill the table is called *$k$-balanced* if it satisfies all the following conditions:\n\n* All cells are filled with integers in the interval $[-k, k]$.\n* If a row meets a column at a black cell, then the sets of positive integers in that row and column are disjoint. Similarly, if a row meets a column at a white cell, then the sets of negative integers in that row and column are disjoint.\n\n1. For $n = 5$, find the minimum value of $k$ such that there exists a $k$-balanced way to fill the table below.\n\n2. For $n = 2017$, find the minimum value of $k$ such that for all nice colorings of the table, we can fill the table in a $k$-balanced way.",
    "options": [],
    "answer": "See solution",
    "solution": "1) Let $a, b, c$ be the numbers filled in the cells at positions $(1,2)$, $(2,1)$, $(3,4)$, $(4,3)$, $(4,5)$, and $(5,4)$. It is easy to check that all $a, b, c$ must be pairwise distinct, so $k \\geq 3$.\n\nWe construct a way to fill the table with $k = 3$ as follows:\n\n![](images/vn-booklet_final_p15_data_aaf87af8d9.png)\n\nThus, the minimum value of $k$ in this case is $3$.\n\n2) First, consider the nice coloring as a chessboard, where cell $(i, j)$ is colored black if and only if $i + j$ is even.\n\n![](images/vn-booklet_final_p15_data_5a413b68a8.png)\n\nTake two white positions at $(a, b)$ and $(c, d)$, with $1 \\leq a, b, c, d \\leq 2017$.\n\n* If $a + c$ is even, then $b + d$ is also even, which implies $a + d$ and $b + c$ are both odd. Then, one of the cells $(a, d)$ or $(b, c)$ will be colored black since they cannot lie on the diagonal $AC$. Thus, the numbers filled in white cells are different.\n* If $a + c$ is odd, then $b + d$ is also odd. Consider cell $(d, c)$, which is filled with the same number as $(c, d)$; we can apply the same argument as above to conclude that the numbers filled in the white cells are different.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17092,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\ell$ be a line in the plane, and a point $A \\notin \\ell$.\n\nDetermine the locus of the points $Q$ in the plane, for which there exists a point $P \\in \\ell$ so that $AQ = PQ$ and $\\angle PAQ = 45^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim the locus of the points $Q$ is the two main angle bisector lines through $O$, the foot of the perpendicular dropped from $A$ onto $\\ell$.\n\nThe angle $\\angle AQP$ is right (since the triangle $AQP$ is isosceles), the same as $\\angle AOP$, so the points $A$, $O$, $P$, $Q$ are concyclic. Therefore, $\\angle QOA = \\angle QPA = 45^\\circ$, or $\\angle QOA = 180^\\circ - \\angle QPA = 135^\\circ$, so $OQ$ is one of the two main angle bisectors through $O$. There are two degenerate positions, when $P \\equiv O$ and when $Q \\equiv O$, but they are trivial. The fullness of the locus follows from the fact a construction is possible in all cases (or by a continuity argument).\n\n![](images/RMC2013_final_p102_data_0d48709af1.png)\n\nAn alternative proof, avoiding cyclic quadrilaterals, runs as follows. Consider the case when prolonging $PQ$ meets $OA$ at a point $T$ on the same side of $\\ell$ as $A$. Since $AQ$ is an antiparallel in $\\triangle TOP$, triangles $TQA$ and $TOP$ are similar, so $\\frac{TQ}{TO} = \\frac{QA}{QP}$. But $QA = QP$, so $\\frac{TQ}{TO} = \\frac{QP}{OP}$, meaning $OQ$ is the angle bisector of $\\angle TOP$. Any other case is treated similarly.\n\nThe condition $A \\notin \\ell$ is not strictly necessary; that case is trivial. The problem has also been asked for any angle $\\angle PAQ = \\alpha \\in (0, 90^\\circ)$, arriving at similar conclusions.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17093,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be a set containing positive integers with the following three properties:\n\n1. $2018 \\in M$.\n2. If $m \\in M$, then all positive divisors of $m$ are also elements of $M$.\n3. For all elements $k, m \\in M$ with $1 < k < m$, the number $km + 1$ is also an element of $M$.\n\nProve that $M = \\mathbb{Z}_{\\ge 1}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first show that $1$, $2$, $3$, $4$, $5$ are elements of $M$:\n\nAs divisors of $2018$, the numbers $1$, $2$, and $1009$ are elements of $M$. Therefore, $2019 = 2 \\cdot 1009 + 1$ and its divisor $3$ are elements of $M$. We now obtain $7 = 2 \\cdot 3 + 1$ and $15 = 2 \\cdot 7 + 1$, and therefore the divisor $5$ of $15$ as elements of $M$. Considering $16 = 3 \\cdot 5 + 1$, we see that $4 \\in M$.\n\nWe now show by induction that $\\{1, 2, \\dots, 2k-1\\} \\subseteq M$ for $k \\ge 1$.\n\nThis has been shown above for $k \\le 3$. Assume that the assertion holds for some $k \\ge 3$. Then we only have to verify that $2k$ and $2k+1$ are elements of $M$, too.\n\nIt is clear that $2k+1 = 2 \\cdot k + 1$ is an element of $M$ due to $k \\ge 3$. This implies that $(2k)^2 = (2k-1)(2k+1) + 1$ and its divisor $2k$ are elements of $M$. This concludes the proof of the assertion and shows that $M$ consists of all positive integers.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17094,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle where $|AB| = |AC|$. Points $P$ and $Q$ are different from the vertices of the triangle and lie on the sides $AB$ and $AC$, respectively. Prove that the circumcircle of triangle $APQ$ passes through the circumcenter of $ABC$ if and only if $|AP| = |CQ|$.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume $|AP| \\leq |AQ|$. Let $O$ be the circumcenter of $ABC$. Let $R$ be the intersection point of the bisector of $\\angle BAC$ with the circumcircle of triangle $PAQ$. Then $|RB| = |RC|$.\n\nAlso, $\\angle APR = 180^\\circ - \\angle AQR = \\angle CQR$ and $|RP| = |RQ|$ (since $\\angle RAP = \\angle RAQ$). So, $|AP| = |CQ|$ if and only if $\\triangle APR \\cong \\triangle CQR$ if and only if $|RA| = |RC|$ if and only if $R = O$ (where $|RA| = |RC| \\Rightarrow \\triangle APR \\cong \\triangle CQR$ by two sides and obtuse angle).\n\n![](images/Estonija_2012_p27_data_352bcc81fd.png)\n\n**Remark.** This problem has been taken from the booklet \"The Coins of Harpland and 20+10 other maths problems from Ireland\" (edited by Bernd Kreussler), its author is Jim Leahy. The solution here is new.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17095,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a pentagon with integer sides. Show that a pentagon with integer sides and an odd perimeter can have at most two right angles.\n\n% IMAGE: ![](images/2022_Australian_Scene_p87_data_5efa5bc5c3.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose the pentagon has more than two right angles. In Case 1, if $a^2 + b^2 = c^2$, then $a+b$ and $c$ have the same parity, so the perimeter is even. Algebraically, with the diagram's labels, the bounding rectangle's perimeter is $2(x+y)$, and the pentagon's perimeter is $2(x+y) + (c-a-b)$. In Case 2, drawing a line parallel to the base forms two right-angled triangles with hypotenuse $e$. Then $a^2 + b^2 = c^2 + d^2$, so $a+b$ and $c+d$ have the same parity, making $a+b+c+d$ even. The pentagon's perimeter is $a+b+c+d+2y$, which is even. Therefore, a pentagon with integer sides and an odd perimeter can have at most two right angles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17096,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots$ and $b_1, b_2, \\dots$ be sequences of real numbers for which $a_1 > b_1$ and\n\n$$\na_{n+1} = a_n^2 - 2b_n\n$$\n\n$$\nb_{n+1} = b_n^2 - 2a_n\n$$\n\nfor all positive integers $n$. Prove that $a_1, a_2, \\dots$ is eventually increasing (that is, there exists a positive integer $N$ for which $a_k < a_{k+1}$ for all $k > N$).",
    "options": [],
    "answer": "See solution",
    "solution": "Let $r, s$, and $t$ be the complex roots of the polynomial $p(\\lambda) = \\lambda^3 - a_1\\lambda^2 + b_1\\lambda - 1$. By Vieta's formulas,\n\n$$\na_1 = r + s + t\n$$\n$$\nb_1 = \\frac{1}{r} + \\frac{1}{s} + \\frac{1}{t}\n$$\n$$\n1 = rst.\n$$\n\n**Claim** — For every positive integer $n$,\n\n$$\na_n = r^{2n-1} + s^{2n-1} + t^{2n-1}\n$$\n\nand\n\n$$\nb_n = \\left(\\frac{1}{r}\\right)^{2n-1} + \\left(\\frac{1}{s}\\right)^{2n-1} + \\left(\\frac{1}{t}\\right)^{2n-1}.\n$$\n\n*Proof*. The base case follows from Vieta's formulas above. For the inductive step, observe that $rst = 1$, so\n\n$$\n\\begin{aligned}\na_{n+1} &= a_n^2 - 2b_n \\\\\n&= (r^{2n-1} + s^{2n-1} + t^{2n-1})^2 - 2\\left(\\left(\\frac{1}{r}\\right)^{2n-1} + \\left(\\frac{1}{s}\\right)^{2n-1} + \\left(\\frac{1}{t}\\right)^{2n-1}\\right) \\\\\n&= (r^{2n-1} + s^{2n-1} + t^{2n-1})^2 - 2\\left((st)^{2n-1} + (tr)^{2n-1} + (rs)^{2n-1}\\right) \\\\\n&= r^{2n} + s^{2n} + t^{2n}\n\\end{aligned}\n$$\n\nand similarly for $b_{n+1}$.\n\n![](images/TST2025Solutions_p4_data_b51ffdd4f2.png)\n\nSince $p(1) = b_1 - a_1 < 0$, $p$ has a real root greater than 1; let $r$ be the largest such root.\n\n- If $s$ and $t$ are real, let $m = \\max(|r|, |s|, |t|) > 1$ be the largest magnitude of the roots and $k \\in \\{1, 2, 3\\}$ be the number of roots with that magnitude. Then asymptotically\n\n$$\na_n = r^{2n-1} + s^{2n-1} + t^{2n-1} \\approx k m^{2n-1}\n$$\n\nwhich implies that $\\{a_n\\}$ is eventually increasing.\n\n- If $s$ and $t$ are not real, they must be complex conjugates of each other, each with magnitude $\\frac{1}{\\sqrt{r}} < 1$. Therefore\n\n$$\nr^{2n-1} - 2 < a_n < r^{2n-1} + 2,\n$$\n\nso $\\{a_n\\}$ is eventually increasing.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17097,
    "subject": "Mathematics (Olympiad)",
    "question": "Обозначим через $\\omega_A$ и $\\omega_C$ описанные окружности треугольников $AA_1P$ и $CC_1P$, соответственно. Пусть лучи $AQ$ и $CQ$ пересекают стороны $CD$ и $AD$ в точках $C_2$ и $A_2$ соответственно. \n\nДокажите, что $\\angle PDA = \\angle QBA$.\n\n![](images/Rusija_2009_p66_data_fc4cf78d68.png)\n\n![](images/Rusija_2009_p66_data_8aa6011781.png)\n\n_Замечание._ Утверждение задачи остается верным, если $Q$ не лежит в треугольнике $ACD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Из параллельности $AB \\parallel CD$ и вписанности четырехугольника $AA_1PQ$ получаем $\\angle PCC_2 = 180^\\circ - \\angle AA_1P = \\angle AQP = 180^\\circ - \\angle PQC_2$, то есть четырехугольник $CPQC_2$ также вписан. Это значит, что $C_2$ лежит на $\\omega_C$; аналогично, точка $A_2$ лежит на $\\omega_A$.\n\nДалее, так как четырехугольник $AA_1PA_2$ вписан и $AB \\parallel CD$, имеем $\\angle A_2PC = 180^\\circ - \\angle A_1PA_2 = \\angle A_1AA_2 = 180^\\circ - \\angle A_2DC$, то есть четырехугольник $A_2PCD$ также вписан. Тогда $\\angle PDA = \\angle PDA_2 = \\angle PCA_2 = \\angle PCQ$. Аналогично получаем, что четырехугольник $BA_1QC$ вписан, откуда $\\angle QBA = \\angle QCA_1 = \\angle PCQ$. Отсюда следует $\\angle PDA = \\angle PCQ = \\angle QBA$, что и требовалось доказать.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17098,
    "subject": "Mathematics (Olympiad)",
    "question": "Now let lines *MP* and *t* meet at *V*. An analogous computation shows, by looking at the circumcircles of *LCD* (which contains *E* and *N*), *CQR*, and *MNQP*, that *V* lies in $\\omega$ as well, and that if $V = E$ then *t* is tangent to $\\omega$.\n\nTherefore, since $\\omega$ meets $t$ at $T$, $V$, and $E$, either $T = V$ if both $T \\neq E$ and $V \\neq E$ or $T = V = E$. At any rate, the intersection of lines *MP* and *NQ* lies in *t*.\n\nThe problem is equivalent to proving that lines *MP*, *NQ*, and *EK* are concurrent. The main idea is to write these three lines as radical axes. In fact, by definition of points *M*, *N*, and *E*:\n\n- *MP* is the radical axis of the circumcircles of *PAS* and *PBR*;\n- *NQ* is the radical axis of the circumcircles of *QCR* and *QDS*;\n- *EK* is the radical axis of the circumcircles of *KBC* and *KAD*.\n\n% ![](images/2024_The_Australian_Scene_Final_p108_data_293b6c88c3.png)\n\nLet $PQ = a$, $PK = b$, and $QK = c$. Also, let $AK = q$, $BK = p$, $CK = r$, and $DK = s$. Using Menelaus's theorem one can prove that\n\n$$\nPS = \\frac{as(b-q)}{sb-qc}, \\quad PR = \\frac{ar(b-p)}{rb-pc}, \\quad SQ = \\frac{aq(s-c)}{sb-qc}, \\quad RQ = \\frac{ap(r-c)}{rb-pc}.\n$$\n\nProve that lines *MP*, *NQ*, and *EK* are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "|                | PAS                        | PBR                        | QCR                        | QDS                        | KBC         | KAD         |\n|----------------|----------------------------|----------------------------|----------------------------|----------------------------|-------------|-------------|\n| Power of K wrt circumcircle of | $qb$                        | $pb$                        | $rc$                        | $sc$                        | $0$         | $0$         |\n| Power of Q wrt circumcircle of | $\\dfrac{a^2 q(s-c)}{sb-qc}$ | $\\dfrac{a^2 p(r-c)}{rb-pc}$ | $0$                         | $0$                         | $-c(r-c)$   | $-c(s-c)$   |\n| Power of P wrt circumcircle of | $0$                         | $0$                         | $\\dfrac{a^2 r(b-p)}{rb-pc}$ | $\\dfrac{a^2 s(b-q)}{sb-qc}$ | $b(b-p)$    | $b(b-q)$    |\n\nOne can promptly verify that\n\n$$\n\\text{pow}_{\\text{PAS}} Q - \\text{pow}_{\\text{PBR}} Q = \\frac{a^2 q(s-c)}{sb-qc} - \\frac{a^2 p(r-c)}{rb-pc} = u\n$$\n\n$$\n\\text{pow}_{\\text{QDS}} P - \\text{pow}_{\\text{QCR}} P = \\frac{a^2 s(b-q)}{sb-qc} - \\frac{a^2 r(b-p)}{rb-pc} = -u\n$$\n\n(one way to do it is just sum fractions with the same denominator to obtain $a^2 - a^2 = 0$.)\n\nApplying the lemma, the equations from PM, QN, and EK are\n\n- PM: $b(q - p)x + uy = 0$\n- QN: $c(s - r)x - uz = 0$\n- EK: $c(s - r)y + b(q - p)z = 0$\n\nNow, if $v = b(q - p)$ and $w = c(s - r)$, it suffices to show that\n\n$$\n\\begin{vmatrix} v & u & 0 \\\\ w & 0 & -u \\\\ 0 & w & v \\end{vmatrix} = 0,\n$$\n\nwhich is a straightforward computation.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17099,
    "subject": "Mathematics (Olympiad)",
    "question": "If $n \\equiv 0 \\mod 3$, then $3$ divides $(-n)^n$ but not $(n+1)^{n+1}$.\n\nSimilarly, if $n \\equiv 2 \\mod 3$, then $3$ divides $(n+1)^{n+1}$ but not $(-n)^n$.\n\nIn either case, $3$ does not divide $(n+1)^{n+1} + (-n)^n$.\n\nSo there remains only $n = 3k + 1$ for some non-negative integer $k$. Show that $(n+1)^{n+1} + (-n)^n$ is divisible by $3$ but not by $9$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $n \\equiv 0 \\mod 3$, then $3$ divides $(-n)^n$ but not $(n+1)^{n+1}$.\n\nSimilarly, if $n \\equiv 2 \\mod 3$, then $3$ divides $(n+1)^{n+1}$ but not $(-n)^n$.\n\nIn either case, $3$ does not divide $(n+1)^{n+1} + (-n)^n$.\n\nSo there remains only $n = 3k + 1$ for some non-negative integer $k$. Let $k' = k + 1$. Then, by the binomial theorem,\n\n$$\n\\begin{align*}\n(n + 1)^{n+1} + (-n)^n &= (3k' - 1)^{3k+2} + (-3k - 1)^{3k+1} \\\\\n&\\equiv (3k + 2)3k'(-1)^{3k+1} + (-1)^{3k+2} \\\\\n&\\qquad -(3k + 1)3k(-1)^{3k} + (-1)^{3k+1} \\quad \\mod 9 \\\\\n&= 3(-1)^{3k+1} [(3k + 2)(k + 1) + k(3k + 1)] \\\\\n&= 3(-1)^{3k+1}(6k^2 + 6k + 2).\n\\end{align*}\n$$\n\nSince $(-1)^{3k+1}(6k^2 + 6k + 2) \\not\\equiv 0 \\mod 3$, $(n+1)^{n+1} + (-n)^n$ is not divisible by $9$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17100,
    "subject": "Mathematics (Olympiad)",
    "question": "A postman has $n$ parcels of weights $1, 2, 3, \\ldots, n$. He wants to divide the parcels into three groups of equal weight. Is this possible for\n\n(a) $n = 2011$\n\n(b) $n = 2012$?",
    "options": [],
    "answer": "See solution",
    "solution": "In the first case, the total weight\n\n$$\n1 + 2 + \\cdots + 2010 + 2011 = \\frac{2011 \\cdot 2012}{2}\n$$\n\nis not a multiple of $3$. Therefore, there is no solution in this case.\n\nIn the second case, we distribute the first $8$ parcels as follows: Parcels $1, 2, 3, 6$ (of total weight $12$) are put into the first group. Parcels $4$ and $8$ (also of total weight $12$) are put into the second group. Parcels $5$ and $7$ (of total weight $12$) are put into the third group.\n\nThe remaining $2004$ parcels $9, \\ldots, 2012$ are first divided into $334$ blocks of consecutive integers $6k + 3, 6k + 4, 6k + 5, 6k + 6, 6k + 7, 6k + 8$ for $1 \\leq k \\leq 334$. Of each block, $6k + 3$ and $6k + 8$ (of weight $12k + 11$) are put into the first group. Parcels $6k + 4$ and $6k + 7$ are put into the second group. Finally, parcels $6k + 5$ and $6k + 6$ are put into the third group. This yields a valid partition.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17101,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the irrational numbers $x$ with the property that $x^2 + x$ and $x^3 + 2x^2$ are integer numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $x^2 + x = a$ and $x^3 + 2x^2 = b$. Then $b - a x = x^2 = a - x$, hence $x(a - 1) = b - a$. Since $x$ is an irrational number and $a, b$ are integers, we deduce that $a = b = 1$, and, finally, $x = \\frac{-1 \\pm \\sqrt{5}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17102,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that there exists a positive integer with $n^2$ divisors (including 1 and itself) and an arrangement of those divisors (each divisor is used once) in an $n \\times n$ grid, where the sum of the divisors in each row and column is the same.",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: $n=1$.\n\nSuppose there exists $n \\geq 2$ satisfying both requirements. Let $a_j$ be the largest number in the $j$-th column for $1 \\leq j \\leq n$. Without loss of generality, we may assume that $a_1 > a_2 > \\dots > a_n$. Thus $j a_1 \\leq a_j$ since $a_j$ divides $a_1$. Hence it is clear that the sum of integers in the $n$-th column is not greater than\n\n$$\n\\sum_{i=1}^{n} \\left( \\frac{a_i}{n} - i \\right) = a_1 - \\frac{n(n+1)}{2}.\n$$\n\nTherefore, the sum of integers in the first column is greater than that of the $n$-th column.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17103,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that if $G - v$ has a Hamiltonian cycle for each vertex $v$ in a graph $G$, but $G$ itself does not have a Hamiltonian cycle, then $n = |G| \\ge 10$. Also, produce such a graph $G$ with $n = 10$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $v$ cannot be adjacent with two consecutive vertices in a Hamiltonian cycle in $G - v$, we have $\\deg v \\le \\left[\\frac{n-1}{2}\\right]$. On the other hand, if $\\deg w \\le 2$ for some vertex $w$, then there is no Hamiltonian cycle in $G - v$ where $v$ is adjacent with $w$. Therefore $3 \\le \\deg v \\le \\left[\\frac{n-1}{2}\\right]$. In particular, $n \\ge 7$.\n\nNext, observe that if $v_1 \\to v_2 \\to \\dots \\to v_{n-1}$ is a Hamiltonian cycle in $G - v$, and if $v$ is adjacent with $v_i$ and $v_j$, $i < j$, then $v_{i-1}$ and $v_{j-1}$ cannot be adjacent with each other, as that would give a Hamiltonian cycle $v_1 \\to \\dots \\to v_{i-1} \\to v_{j-1} \\to v_{j-2} \\to \\dots \\to v_i \\to v \\to v_j \\to v_{j+1} \\to \\dots \\to v_{n-1}$ in $G$.\n\nFrom these observations, it follows that $n = 7$ and $n = 8$ are impossible. If $n = 9$, then each vertex has degree 3 or 4, and they cannot all have degree 3 by the degree sum formula. Hence, assume that $n = 9$ and there is a vertex $v_0$ with $\\deg v_0 = 4$.\n\nThen $G$ must contain the graph on the left below.\n\n![](images/Turska_2009_p11_data_2696bf80b7.png)\n![](images/Turska_2009_p11_data_272fa10151.png)\n\nBut then, again from the observations above, it follows that each $v_{2i}$ must be adjacent with at least—and therefore exactly—one of $v_{2i+3}$ and $v_{2i+5}$ (indices considered mod 8), and there are no other edges. Since there are edges only between vertices of different parity in the resulting graph; if we remove an odd indexed vertex, the remaining graph cannot have a Hamiltonian cycle as it has unequal numbers of odd and even indexed vertices.\n\nFinally, the following graph gives an example when $n = 10$.\n\n![](images/Turska_2009_p11_data_e3d818336a.png)\n\nIn this graph, $a \\to c \\to C \\to B \\to b \\to e \\to E \\to D \\to d \\to a$ and $A \\to B \\to C \\to c \\to e \\to b \\to d \\to D \\to E \\to A$ are Hamiltonian cycles for $G - A$ and $G - a$, respectively.\n\n$G$ itself does not have a Hamiltonian cycle. A Hamiltonian cycle cannot contain exactly two of the edges $Aa$, $Bb$, $Cc$, $Dd$, $Ee$, as there are no adjacent pairs of vertices on the outer and the inner cycles. So it must contain exactly four of them; say, $Bb$, $Cc$, $Dd$, $Ee$. Then the cycle contains a path through $b$, $B$, $A$, $E$, $e$, and a path through $C$, $c$, $a$, $d$, $D$; and these two paths cannot be completed to a Hamiltonian cycle.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17104,
    "subject": "Mathematics (Olympiad)",
    "question": "A magician arranges cards numbered from $1$ to $300$ on a $3 \\times 100$ board in a \"snake-like\" path so that consecutive numbers are placed on side-to-side adjacent cells (adjacent either horizontally or vertically, but not diagonally). The numbers are written on the bottom sides of the cards, with the upper sides blank. The magician then turns over $k$ cards of his choice. What is the minimum value of $k$ such that the revealed cards uniquely determine the entire snake?",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum value is $k = 2$.\n\n*Example*: Place $300$ in the lower left corner and $101$ in the cell above it.\n\n*Explanation*: Revealing only one card does not uniquely determine the snake, because all the cells of the $3 \\times 100$ board can be arranged in a single cyclic path. Let the rows be labeled $a$, $b$, $c$ and columns numbered $1$ to $300$. The path can be described as:\n\n$$\na_1 - b_1 - c_1 - c_2 - b_2 - b_3 - c_3 - c_4 - b_4 - b_5 - \\dots - c_{300} - b_{300} - a_{300} - a_{299} - a_{298} - \\dots - a_2 - a_1\n$$\n\nIf only one card is revealed, the path can be oriented in two ways, and the cards can be arranged to increase in either direction. Thus, at least two cards must be revealed to uniquely determine the snake.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17105,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs $(a, b)$ of real numbers that simultaneously satisfy the equations\n\n$$\na^{20} + b^{20} = 1 \\quad \\text{and} \\quad a^{21} + b^{21} = 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $a^{20} \\ge 0$ and $b^{20} \\ge 0$, we have $a^{20} = 1 - b^{20} \\le 1$ and similarly $b^{20} \\le 1$. It follows that $a \\le 1$ and $b \\le 1$.\n\nHence, $a^{20}(a - 1) \\le 0$ and $b^{20}(b - 1) \\le 0$, which are equivalent to $a^{21} \\le a^{20}$ and $b^{21} \\le b^{20}$. Therefore, to have $a^{20} + b^{20} = a^{21} + b^{21} = 1$, we must have $a^{20} = a^{21}$ and $b^{20} = b^{21}$.\n\nThe equation $a^{20} = a^{21}$ is equivalent to $a^{20}(a - 1) = 0$, so it implies that $a = 0$ or $a = 1$. Similarly, $b = 0$ or $b = 1$.\n\nThis leaves four cases to check, and the only solutions among these are $(a, b) = (1, 0)$ and $(a, b) = (0, 1)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17106,
    "subject": "Mathematics (Olympiad)",
    "question": "In a $2n \\times 2n$ grid, exactly half of the squares are coloured black and the other half are white. In one step, you may select any $2 \\times 2$ square in the grid and reflect its four squares with respect to the horizontal or vertical central axis. For which positive integers $n$ is it possible to reach a configuration where the entire board is coloured in a chessboard pattern, starting from any initial configuration?",
    "options": [],
    "answer": "See solution",
    "solution": "For $n = 1$, it is not possible to reach the chessboard pattern if the initial $2 \\times 2$ configuration is as shown below, because adjacent same-coloured squares remain unchanged after reflecting.\n\n![](images/prob1314_p20_data_f7ec7db84d.png)\n\nFor any $n \\ge 2$, we can always reach the chessboard pattern from any initial configuration. Whenever there are wrong-coloured squares, we can reduce their number by a finite sequence of steps. A wrong-coloured square becomes right-coloured on the other side of the axis of reflection, and vice versa. Define a *double reflection* as reflecting the same $2 \\times 2$ area first horizontally, then vertically; this is equivalent to a central reflection, which leaves the status of each square unchanged.\n\nSuppose there are two adjacent wrong-coloured squares (e.g., in the same row). Since $n \\ge 2$, we can always find such a configuration away from the grid's edge. Mark wrong-coloured squares as W and right-coloured as R; $x$ means either, and $x'$ means the opposite.\n\n- If at least one of the two has a wrong-coloured upper neighbour, reflecting vertically decreases the number of wrong-coloured squares by at least 2:\n\n![](images/prob1314_p21_data_19910fbef4.png)\n\n- If both upper neighbours are right-coloured, but at least one of their upper neighbours is wrong-coloured, reflecting vertically moves the two wrong-coloured squares up by one row (number unchanged):\n\n![](images/prob1314_p21_data_b667533f09.png)\n\n- If the $2 \\times 2$ square above is all right-coloured, but an adjacent square to the right is wrong-coloured, use double reflection to swap it with a right-coloured square:\n\n![](images/prob1314_p21_data_4920b40576.png)\n\n- In other cases, the number of wrong-coloured squares can be reduced by 2 as shown:\n\n![](images/prob1314_p21_data_50523148d9.png)\n\nIf there are no two adjacent wrong-coloured squares, double reflections allow moving a wrong-coloured square along diagonals without changing their count. Since the numbers of black and white squares are equal and preserved, a wrong-coloured black square implies a wrong-coloured white square exists. By moving along diagonals, we can bring two wrong-coloured squares together and proceed as above.\n\nThus, for every $n \\ge 2$, it is possible to reach the chessboard pattern from any initial configuration.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17107,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDEF$ be a convex hexagon in which $AB = AF$, $BC = CD$, $DE = EF$, and $\\angle ABC = \\angle EFA = 90^\\circ$. Prove that $AD \\perp CE$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the circle $\\omega$ with center $A$ and radius $AB$. Note that $BC$ and $EF$ are tangent to $\\omega$, and so, from the problem condition, the line $CE$ is a radical axis of $\\omega$ and $D$. Therefore, $CE \\perp AD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17108,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle inscribed in a circle with center $O$. Let $M$ and $N$ be two points on the line $AC$ such that $MN = AC$. Let $D$ be the orthogonal projection of $M$ onto the line $BC$, and $E$ be the orthogonal projection of $N$ onto the line $AB$.\n\n1. Prove that the orthocenter $H$ of triangle $ABC$ lies on the circumcircle (with center $O'$) of triangle $BED$.\n\n2. Prove that the midpoint of segment $AN$ is symmetric to $B$ with respect to the midpoint of segment $OO'$.",
    "options": [],
    "answer": "See solution",
    "solution": "1. Let $K$ be the intersection point of $MD$ and $NE$. The circle with diameter $BK$ circumscribes triangle $BED$.\n\nWe have $AH \\parallel MK$ and $CH \\parallel NK$.\n\nThis implies $\\angle HAC = \\angle KMN$ and $\\angle ACH = \\angle MNK$.\n\nSince $AC = MN$, triangles $AHC$ and $MKN$ are congruent. Thus, $d(K, AC) = d(H, AC)$. As $K$ and $H$ lie on the same side of $AC$, $KH \\parallel AC$, so $KH \\perp BH$ and $H$ lies on the circle with diameter $BK$ circumscribing $\\triangle BED$.\n\n2. From part 1, $O'$ is the midpoint of segment $BK$. Let $I$ be the midpoint of $AN$. The orthogonal projections of $I$ onto $BA$ and $BC$ are the midpoints of $EA$ and $DC$, respectively. Thus, the orthogonal projections of vector $O'M$ onto $BA$ and $BC$ are $\\frac{1}{2}\\overrightarrow{BA}$ and $\\frac{1}{2}\\overrightarrow{BC}$, respectively. Therefore, $\\overrightarrow{OM} = \\overrightarrow{BO}$, so quadrilateral $BO'MO$ is a parallelogram and $B$ and $M$ are symmetric with respect to the midpoint of $OO'$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17109,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_{2008}$ be real numbers such that\n$$\n\\begin{cases}\na_1 + a_2 + a_3 + \\dots + a_{2008} = 0 \\\\\na_1^2 + a_2^2 + a_3^2 + \\dots + a_{2008}^2 = 502\n\\end{cases}\n$$\nFind the maximum value for $a_{2008}$. Find at least one sequence of values $a_1, a_2, \\dots, a_{2007}$ with that maximum.",
    "options": [],
    "answer": "See solution",
    "solution": "By the Cauchy-Bunyakovsky-Schwarz inequality for real numbers $a_1, a_2, \\dots, a_{2007}$ we have\n$$\n\\begin{aligned}\n(a_1 + a_2 + \\dots + a_{2007})^2 &\\le (1 \\cdot a_1 + 1 \\cdot a_2 + \\dots + 1 \\cdot a_{2007})^2 \\\\\n&\\le (1^2 + 1^2 + \\dots + 1^2)(a_1^2 + a_2^2 + \\dots + a_{2007}^2) \\\\\n&= 2007(a_1^2 + a_2^2 + \\dots + a_{2007}^2).\n\\end{aligned}\n$$\nBoth sides are equal if and only if $\\frac{a_1}{1} = \\frac{a_2}{1} = \\dots = \\frac{a_{2007}}{1}$, i.e. $a_1 = a_2 = \\dots = a_{2007}$.\n\nThe given conditions are equivalent to\n$$\n\\begin{cases}\na_1 + a_2 + a_3 + \\dots + a_{2007} = -a_{2008} \\\\\na_1^2 + a_2^2 + a_3^2 + \\dots + a_{2007}^2 = 502 - a_{2008}^2\n\\end{cases}\n$$\nApplying the inequality above, we have $(-a_{2008})^2 \\le 2007(502 - a_{2008}^2)$.\n\nSolving, $a_{2008}^2 \\le \\frac{2007 \\cdot 502}{2008} = \\frac{2007}{4}$, so $|a_{2008}| \\le \\frac{\\sqrt{2007}}{2}$.\n\nFor every sequence $a_1, a_2, \\dots, a_{2007}$ such that $a_1 = a_2 = \\dots = a_{2007}$, we have $|a_{2008}| = \\frac{\\sqrt{2007}}{2}$, i.e. $(a_{2008})_{\\max} = \\frac{\\sqrt{2007}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17110,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation:\n\n$$\n\\tau(kn) + 2023 = n\n$$\n\nwhere $k$ is a positive integer greater than $6996$, $n$ is a positive integer, and $\\tau(m)$ denotes the number of positive divisors of $m$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that $n \\leq 4\\sqrt{n} + 2023$ implies $n \\leq (2 + \\sqrt{2027})^2 < 6996 < k$. Thus, $n$ is not divisible by $k$, so $\\tau(kn) = 2\\tau(n)$. The equation becomes:\n\n$$\n2\\tau(n) + 2023 = n. \\qquad (2)\n$$\n\nWe have $2027 \\leq n \\leq 2211$. Let $n = p_1^{e_1} p_2^{e_2} \\cdots p_s^{e_s}$, where $p_i$ are distinct odd primes. If $s \\geq 5$, then $n \\geq 3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 > 2211$, a contradiction. So $s \\leq 4$.\n\n**Case 1:** $s = 1$. $n = p_1^{e_1} \\leq 2211$, $e_1 \\leq 7$. Only $n = 2027$ works.\n\n**Case 2:** $s = 2$. $n = p_1^{e_1} p_2^{e_2} \\leq 2211$, $e_1 + e_2 \\leq 6$, so $\\tau(n) \\leq 9$. Only $n = 2031$ works.\n\n**Case 3:** $s = 3$. $n \\geq 3^{e_1} 5^{e_2} 7^{e_3}$, $e_1 + e_2 + e_3 \\leq 5$, $\\tau(n) \\in \\{8, 12, 16, 18\\}$. No solution.\n\n**Case 4:** $s = 4$. $n \\geq 3^{e_1} 5^{e_2} 7^{e_3} 11^{e_4}$, but if any $e_i > 1$, $n > 2211$. Only $e_1 = e_2 = e_3 = e_4 = 1$ is possible, $\\tau(n) = 16$, but no $n$ works.\n\nThus, the only solutions are $n = 2027$ and $n = 2031$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17111,
    "subject": "Mathematics (Olympiad)",
    "question": "對於任意正整數 $m$，我們定義 $\\phi(m)$ 為小於 $m$ 且與 $m$ 互質的正整數個數。\n\n試問：是否存在無窮正整數數列 $a_1, a_2, \\dots, a_n, \\dots$，滿足：\n\n$$\n(i)\\ a_1 = (2011)! = 1 \\times 2 \\times \\cdots \\times 2011.\n$$\n\n(ii) 對每個正整數 $i$，有 $a_i = \\phi(a_{i+1})$。",
    "options": [],
    "answer": "See solution",
    "solution": "不存在這樣的數列。\n\n我們用反證法，假設存在這樣的無窮數列。我們將每個 $a_i$ 表示成 $a_i = 2^{r_i}b_i$，其中 $b_i$ 是奇數。由 $\\phi$ 函數的性質有 $a_i = \\phi(a_{i+1}) = 2^{r_{i+1}-1}\\phi(b_{i+1})$。\n\n注意到 $b_{i+1}$ 不可能是 $1$，否則我們會得到 $b_i = 1, b_{i-1} = 1, \\dots$，最後得到 $b_1 = 1$，與 $a_1 = (2011)!$ 顯然矛盾。此時我們有 $\\phi(b_{i+1})$ 是偶數。比較 $2^{r_{i+1}-1}\\phi(b_{i+1}) = 2^{r_i}b_i$ 可知 $r_{i+1} \\le r_i$。\n\n由於 $r_i$ 都是非負整數，不可能一直變小，因此必然在某個正整數 $n$ 之後（以下我們均假設變數 $i \\ge n$）有 $r_i = r_{i+1}$。比較上式可知 $4 \\nmid \\phi(b_{i+1})$，且 $2b_i = \\phi(b_{i+1})$。\n\n由 $\\phi$ 函數性質知若 $\\phi(b_{i+1})$ 不是 $4$ 的倍數，則 $b_{i+1}$ 必為某奇質數的次方，記 $b_{i+1} = p_{i+1}^{s_{i+1}}$，其中 $p_{i+1}$ 是質數。現在我們考慮 $b_{i+2} = p_{i+2}^{s_{i+2}}$，我們有\n\n$$\nb_{i+1} = \\frac{\\phi(b_{i+2})}{2} = \\frac{p_{i+2} - 1}{2} p_{i+2}^{s_{i+2}-1} = p_{i+1}^{s_{i+1}}.\n$$\n\n從而 $p_{i+2} = 3$ 或 $s_{i+2} = 1$。\n\n若 $p_{i+2} = 3$（對某個 $i$），則 $a_{i+2}$ 只有 $2$ 和 $3$ 這兩個質因數，容易檢驗接下來 $\\phi(a_{i+2}), \\phi(\\phi(a_{i+2})), \\dots$ 也都只會有 $2$ 和 $3$ 的質因數，但 $a_1 = (2011)!$，矛盾，故我們有 $s_{i+2} = 1$，即 $b_{i+2} = p_{i+2}$（對所有 $i \\ge n$）。\n\n如此一來，對所有 $j \\ge (n+2)$，有 $b_{j+1} = 2b_j + 1$，且這些 $b_j$ 都是質數。\n\n我們注意到對任意正整數 $k$ 有 $b_{j+k} = 2^k(b_j + 1) - 1$。令 $k = b_j - 1$，則由費馬小定理知 $b_{j+k} \\equiv (b_j + 1) - 1 \\equiv 0 \\pmod{b_j}$，從而 $b_j \\mid b_{j+k}$，與 $b_{j+k}$ 是質數矛盾，證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17112,
    "subject": "Mathematics (Olympiad)",
    "question": "Given positive integers $a, b, c, p, q, r$ with $p, q, r \\ge 2$, let\n\n$$\nQ = \\{(x, y, z) \\in \\mathbb{Z}^3 \\mid 0 \\le x \\le a,\\ 0 \\le y \\le b,\\ 0 \\le z \\le c\\}\n$$\n\nbe the game board.\n\nInitially, $M$ game pieces are placed on $Q$ (any number per position). There are three types of legal moves:\n\n1. Remove $p$ pieces from $(x, y, z)$ and place one piece at $(x-1, y, z) \\in Q$;\n2. Remove $q$ pieces from $(x, y, z)$ and place one piece at $(x, y-1, z) \\in Q$;\n3. Remove $r$ pieces from $(x, y, z)$ and place one piece at $(x, y, z-1) \\in Q$.\n\nFind the minimum value of $M$ such that, no matter how the $M$ pieces are initially placed, it is always possible to perform a sequence of moves to get one piece at $(0, 0, 0)$.",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum $M$ is $p^a q^b r^c$.\n\n**Necessity:**\nSuppose fewer than $p^a q^b r^c$ pieces are placed at $(a, b, c)$. Define the weight of a piece $u$ at $(x, y, z)$ as $w(u) = \\frac{1}{p^x q^y r^z}$. Each move preserves the total weight $W = \\sum w(u)$. To have a piece at $(0,0,0)$, $W \\ge 1$ is required, but with fewer than $p^a q^b r^c$ pieces at $(a, b, c)$, $W < 1$, so the goal is impossible.\n\n**Sufficiency:**\nWe generalize to $n$ dimensions: for $a_1, \\dots, a_n$ and $p_1, \\dots, p_n \\ge 2$, define\n$$\nQ(a_1, \\dots, a_n) = \\{(x_1, \\dots, x_n) \\in \\mathbb{Z}^n \\mid 0 \\le x_i \\le a_i\\}\n$$\nwith $M = p_1^{a_1} \\cdots p_n^{a_n}$ pieces. The proof proceeds by induction on $n$ and $a_n$, dividing the board into Zone-I ($Q(a_1, \\dots, a_n-1)$) and Zone-II ($Q(a_1, \\dots, a_{n-1}) \\times \\{a_n\\}$). By transmitting as many pieces as possible from Zone-II to Zone-I using the allowed moves, and applying the induction hypothesis, one can always get a piece to the origin.\n\nThus, $M = p^a q^b r^c$ is both necessary and sufficient.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17113,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $b^2 = u^2 + v^2$ be a Pythagorean triple with integers $u < v < b$. Show that there are infinitely many arithmetic progressions of three squares with integer terms.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n(v-u)^2 + (v+u)^2 = v^2 - 2uv + u^2 + v^2 + 2uv + u^2 = 2u^2 + 2v^2 = 2b^2.\n$$\n\nRearranging,\n$$\nb^2 - (v-u)^2 = (v+u)^2 - b^2.\n$$\n\nThus, setting $(a, b, c) = (v-u, b, v+u)$ produces squares in arithmetic progression. It is clear that no two Pythagorean triples produce the same arithmetic progression, and that relative primality is preserved. Thus, the infinitude of Pythagorean triples implies there are infinitely many square arithmetic progressions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17114,
    "subject": "Mathematics (Olympiad)",
    "question": "Assume that $a$ is a positive integer and not a perfect square. Prove that for any positive integer $n$, the sum\n\n$$\nS_n = \\{\\sqrt{a}\\} + \\{\\sqrt{a}\\}^2 + \\dots + \\{\\sqrt{a}\\}^n\n$$\n\nis irrational, where $\\{x\\} = x - \\lfloor x \\rfloor$ and $\\lfloor x \\rfloor$ denotes the greatest integer less than or equal to $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that $c^2 < a < (c+1)^2$, where $c$ is an integer greater than or equal to $1$. Then, $\\lfloor \\sqrt{a} \\rfloor = c$, $1 \\le a - c^2 \\le 2c$, and $\\{\\sqrt{a}\\} = \\sqrt{a} - c$.\n\nWrite $\\{\\sqrt{a}\\}^k = (\\sqrt{a} - c)^k = x_k + y_k \\sqrt{a}$, where $k \\in \\mathbb{N}$ and $x_k, y_k \\in \\mathbb{Z}$. We have\n\n$$\nS_n = (x_1 + x_2 + \\cdots + x_n) + (y_1 + y_2 + \\cdots + y_n) \\sqrt{a}.\n$$\n\nWe now prove that $T_n = \\sum_{k=1}^{n} y_k \\neq 0$ for all positive integers $n$.\n\nSince\n\n$$\n\\begin{aligned}\nx_{k+1} + y_{k+1} \\sqrt{a} &= (\\sqrt{a} - c)^{k+1} \\\\\n&= (\\sqrt{a} - c)(x_k + y_k \\sqrt{a}) \\\\\n&= (a y_k - c x_k) + (x_k - c y_k) \\sqrt{a},\n\\end{aligned}\n$$\n\nwe have\n\n$$\nx_{k+1} = a y_k - c x_k, \\quad y_{k+1} = x_k - c y_k.\n$$\n\nSince $x_1 = -c$ and $y_1 = 1$, we have $y_2 = -2c$.\n\nBy the above equality, we have\n\n$$\ny_{k+2} = -2c y_{k+1} + (a - c^2) y_k,\n$$\n\nwhere $y_1 = 1$, $y_2 = -2c$.\n\nBy mathematical induction, we have\n\n$$\ny_{2k-1} > 0, \\quad y_{2k} < 0.\n$$\n\nCombining the recurrence and sign pattern, we have\n\n$$\ny_{2k+2} - y_{2k+1} = -(2c+1) y_{2k+1} + (a - c^2) y_{2k} < 0,\n$$\n$$\ny_{2k+2} + y_{2k+1} = -(2c-1) y_{2k+1} + (a - c^2) y_{2k} < 0.\n$$\n\nTaking the product of the above inequalities, we get $y_{2k+2}^2 - y_{2k+1}^2 > 0$. Since $y_2^2 - y_1^2 > 0$, we have\n\n$$\n|y_{2k-1}| < |y_{2k}|.\n$$\n\nOn the other hand,\n\n$$\ny_{2k+1} - y_{2k} = -(2c+1) y_{2k} + (a - c^2) y_{2k-1} > 0,\n$$\n$$\ny_{2k+1} + y_{2k} = -(2c-1) y_{2k} + (a - c^2) y_{2k-1} > 0.\n$$\n\nMultiplying the above inequalities, we have $y_{2k+1}^2 - y_{2k}^2 > 0$, that is, $|y_{2k}| < |y_{2k+1}|$.\n\nTherefore, $|y_n| < |y_{n+1}|$ for all positive integers $n$. Combining the sign pattern and magnitude, we have $y_{2k-1} + y_{2k} < 0$, $y_{2k+1} + y_{2k} > 0$ for all positive integers $n$.\n\nTherefore,\n\n$$\nT_{2n-1} = y_1 + (y_2 + y_3) + \\cdots + (y_{2n-2} + y_{2n-1}) > 0,\n$$\n$$\nT_{2n} = (y_1 + y_2) + (y_3 + y_4) + \\cdots + (y_{2n-1} + y_{2n}) < 0.\n$$\n\nHence, $T_n \\neq 0$ for all positive integers $n$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17115,
    "subject": "Mathematics (Olympiad)",
    "question": "In the plane, 100 points are given such that no three are collinear. The points are arranged into 10 groups, each containing at least 3 points. Any two points in the same group are joined by a segment.\n\n(a) Determine which arrangement of the 10 groups yields the minimal number of triangles.\n\n(b) Prove that there exists an arrangement of such groups where each segment can be coloured with one of three given colours so that no triangle has all edges of the same colour.",
    "options": [],
    "answer": "See solution",
    "solution": "a) If the groups contain $a_1, a_2, \\dots, a_{10}$ points, the number of triangles is\n\n$$\nN = \\binom{a_1}{3} + \\binom{a_2}{3} + \\dots + \\binom{a_{10}}{3}.\n$$\n\nWe claim this number is minimal when\n\n$$\na_1 = a_2 = \\dots = a_{10} = 10,\n$$\n\ngiving the minimum value $10\\binom{10}{3}$. If a group has $m \\ge 11$ points, another must have $n \\le 9$ points. Moving a point from the larger to the smaller group decreases the number of triangles, since\n\n$$\n\\binom{m}{3} + \\binom{n}{3} > \\binom{m-1}{3} + \\binom{n+1}{3}.\n$$\n\nb) Arrange the 100 points into 10 groups of 10, each split into two subgroups of 5. Assume each subgroup forms a convex pentagon. Colour the sides of each pentagon with colour $c_1$, the diagonals with $c_2$, and the segments joining points from different subgroups with $c_3$. No triangle will have all edges of the same colour.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17116,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть $a$ — один из членов арифметической прогрессии, а $d$ — её разность. По условию, числа $a(a+d)$ и $a(a+2d)$ — также члены прогрессии. Докажите, что $a$ — целое число.",
    "options": [],
    "answer": "See solution",
    "solution": "Поскольку $a(a+d)$ и $a(a+2d)$ — члены прогрессии, их разность равна $nd$ для некоторого целого $n$: $$a(a+2d) - a(a+d) = nd$$\nУпрощая, получаем $ad = nd$, то есть $a = n$. Следовательно, $a$ — целое число.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17117,
    "subject": "Mathematics (Olympiad)",
    "question": "The diagonals of a cyclic quadrilateral $ABCD$ are perpendicular. Points $K, L, M, Q$ are the orthocenters of the triangles $ABD$, $ACD$, $BCD$, $ABC$ respectively. Prove that quadrilaterals $KLMQ$ and $ABCD$ are equal.\n\n![](images/Ukrajina_2010_p13_data_90921380df.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let the diagonals of the quadrilateral $ABCD$ intersect at $O$. The altitudes of the triangles $BCD$ and $ACD$ lie on $AC$, so their orthocenters $K$ and $M$ do too. Analogously, points $L$ and $Q$ lie on $BC$.\n\nNote that $BK \\parallel CL$, because $BK \\perp AD$ and $CL \\perp AD$. It follows that $\\angle BKC = \\angle KCL = \\angle ACL = 90^\\circ - \\angle CAD = 90^\\circ - \\angle DBC = 90^\\circ - \\angle OBC = \\angle BCK$.\n\nNow it's easy to see that $\\triangle LOC = \\triangle BOC = \\triangle BOK$, which implies $OC = OK$ and $BO = OL$. Since $CK \\perp CL$, it follows that the quadrilateral $LCBD$ is a rhombus. Analogously, the quadrilateral $ADMQ$ is a rhombus too. Finally, we have that the quadrilateral $KLMQ$ is an image of $ABCD$ under central symmetry relative to the point $O$, which implies that the quadrilaterals $KLMQ$ and $ABCD$ are equal.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17118,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: (0, +\\infty) \\to (0, +\\infty)$ such that\n$$\n\\frac{(f(w))^2 + (f(x))^2}{f(y^2) + f(z^2)} = \\frac{w^2 + x^2}{y^2 + z^2}\n$$\nfor all positive real numbers $w, x, y, z$ satisfying $wx = yz$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $w = x = y = z = 1$. Then\n$$\n\\frac{(f(1))^2 + (f(1))^2}{f(1^2) + f(1^2)} = \\frac{1^2 + 1^2}{1^2 + 1^2} \\implies (f(1))^2 = f(1)\n$$\nso $f(1) = 1$.\n\nFor any $t > 0$, let $w = t$, $x = 1$, $y = z = \\sqrt{t}$. Then\n$$\n\\frac{(f(t))^2 + 1}{2f(t)} = \\frac{t^2 + 1}{2t}\n$$\nwhich implies $(tf(t) - 1)(f(t) - t) = 0$.\n\nSo, for any $t > 0$,\n$$\nf(t) = t \\quad \\text{or} \\quad f(t) = \\frac{1}{t}.\n$$\n\nSuppose there exist $b, c \\in (0, +\\infty)$ such that $f(b) \\neq b$, $f(c) \\neq \\frac{1}{c}$. By the previous result, $b, c$ are different from $1$ and $f(b) = \\frac{1}{b}$, $f(c) = c$.\n\nTake $w = b$, $x = c$, $y = z = \\sqrt{bc}$, then\n$$\n\\frac{\\frac{1}{b^2} + c^2}{2f(bc)} = \\frac{b^2 + c^2}{2bc}\n$$\nwhich gives $f(bc) = \\frac{c + b^2c^3}{b(b^2 + c^2)}$.\n\nBy the previous result, $f(bc) = bc$ or $f(bc) = \\frac{1}{bc}$. If $f(bc) = bc$, then\n$$\nbc = \\frac{c + b^2c^3}{b(b^2 + c^2)}\n$$\nwhich yields $b^4c = c$, so $b = 1$, a contradiction.\n\nIf $f(bc) = \\frac{1}{bc}$, then\n$$\n\\frac{1}{bc} = \\frac{c + b^2c^3}{b(b^2 + c^2)}\n$$\nwhich yields $b^2c^4 = b^2$, so $c = 1$, a contradiction.\n\nTherefore, only two functions are possible:\n$$\nf(x) = x \\quad \\text{or} \\quad f(x) = \\frac{1}{x}, \\quad x \\in (0, +\\infty)\n$$\nIt is easy to verify that these two functions satisfy the given conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17119,
    "subject": "Mathematics (Olympiad)",
    "question": "Trapezoid $ABCD$ with bases $AB$ and $CD$ is such that the circumcircle of triangle $BCD$ intersects line $AD$ at a point $E$ distinct from $A$ and $D$. Prove that the circumcircle of triangle $ABE$ is tangent to line $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/Baltic_Way_SHL_2009-11_13-16_p198_data_1817d2bfa2.png)\n\nIf point $E$ lies on segment $AD$, it suffices to prove that $\\angle CBE = \\angle BAE$. This holds since both angles equal $180^\\circ - \\angle ADC$.\n\nIf point $D$ lies on segment $AE$, we have $\\angle CBE = \\angle CDE = \\angle BAE$, which proves the claim.\n\nFinally, if point $A$ lies on segment $DE$, then $180^\\circ - \\angle CBE = \\angle CDE = \\angle BAE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17120,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D$ be a point on side $BC$ of triangle $ABC$ such that $\\angle CAD = \\angle CBA$. A circle with center $O$ passes through $B$, $D$, and meets segments $AB$, $AD$ at $E$, $F$, respectively. Lines $BF$ and $DE$ meet at point $G$. $M$ is the midpoint of $AG$. Prove that $CM \\perp AO$.",
    "options": [],
    "answer": "See solution",
    "solution": "As shown in Fig. 1, extend $EF$ and meet $BC$ at point $P$, and join and extend $GP$, which meets $AD$ at $K$ and the extension of $AC$ at $L$.\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p92_data_f68b2a61b7.png)\n\nAs shown in Fig. 2, let $Q$ be a point on $AP$ such that\n$$\n\\angle PQF = \\angle AEF = \\angle ADB.\n$$\nIt is easy to see that $A, E, F, Q$ and $F, D, P, Q$ are concyclic respectively. Denote by $r$ the radius of $\\odot O$. By the power of a point theorem,\n$$\n\\begin{aligned}\nAP^2 &= AQ \\times AP + PQ \\times AP \\\\\n&= AF \\times AD + PF \\times PE \\\\\n&= (AO^2 - r^2) + (PO^2 - r^2).\n\\end{aligned}\n$$\n(1)\n\nSimilarly,\n$$\nAG^2 = (AO^2 - r^2) + (GO^2 - r^2).\n$$\n(2)\n\nBy (1), (2), we have $AP^2 - AG^2 = PO^2 - GO^2$, which implies that $PG \\perp AO$. As shown in Fig. 3, applying Menelaus' theorem for $\\triangle PFD$ and line $AEB$, we have\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p92_data_210e3f272c.png)\n\n$$\n\\frac{DA}{AF} \\times \\frac{FE}{EP} \\times \\frac{PB}{BD} = 1. \\quad (3)\n$$\n\nApplying Ceva's theorem for $\\triangle PFD$ and point $G$, we have\n$$\n\\frac{DK}{KF} \\times \\frac{FE}{EP} \\times \\frac{PB}{BD} = 1. \\quad (4)\n$$\n\nDividing (3) by (4) yields\n$$\n\\frac{DA}{AF} = \\frac{DK}{KF}. \\quad (5)\n$$\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p93_data_bdc2a5917b.png)\n\nEquation (5) illustrates that $A, K$ are harmonic points with respect to $F, D$, i.e., $AF \\times KD = AD \\times FK$.\n\nIt follows that\n$$\nAK \\times FD = AF \\times KD + AD \\times FK = 2AF \\times KD. \\quad (6)\n$$\n\nSince $B, D, F, E$ are concyclic, we have $\\angle DBA = \\angle EFA$. Since $\\angle CAD = \\angle CBA$, we have $\\angle CAF = \\angle EFA$, which implies that $AC \\parallel EP$. Thus,\n$$\n\\frac{CP}{PD} = \\frac{AF}{FD}. \\quad (7)\n$$\n\nApplying Menelaus' theorem to $\\triangle ACD$ and line $LPK$, we have\n$$\n\\frac{AL}{LC} \\times \\frac{CP}{PD} \\times \\frac{DK}{KA} = 1. \\quad (8)\n$$\n\nCombining (6), (7), and (8), we have $\\frac{AL}{LC} = 2$.\n\nIn $\\triangle AGL$, $M, C$ are the midpoints of $AG, AL$ respectively, and hence $MC \\parallel GL$. Since $GL \\perp AO$, we conclude that $MC \\perp AO$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17121,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the equation\n\n$$\np(p-1) = 2(n^3 + 1)\n$$\n\nwhere $p$ is a prime and $n$ is a positive integer. Find all solutions $(p, n)$ in positive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\text{Let us analyze the equation:}\n$$\n\nIt is easy to see that the given equality does not hold for $p=2$ and positive integer $n$. So $p \\ge 3$ is an odd prime number. Then $(n+1)(n^2 - n + 1)$ is divisible by $p$.\n\n1. If $n+1$ divides $p$, then $n+1 = kp$ for some positive integer $k$. In particular, $n+1 \\ge p$. From the original equation, $p(p-1) = 2(n+1)(n^2-n+1) \\ge 2p(n^2-n+1)$, so $p-1 \\ge 2n^2-2n+2$. Then $n \\ge p-1 \\ge 2n^2-2n+2$, or $2n^2-3n+2 \\le 0$, which is impossible.\n\n2. Therefore $n^2 - n + 1$ divides $p$, i.e.\n\n$$\nn^2 - n + 1 = kp\n$$\nfor some positive integer $k$. Substituting this into the original equation, we obtain $p-1 = 2k(n+1)$ or\n$$\np = 2kn + 2k + 1.\n$$\nSubstituting this into the previous expression, we get $n^2 - n + 1 = 2k^2n + 2k^2 + k$, or\n$$\nn^2 - (2k^2 + 1)n - (2k^2 + k - 1) = 0.\n$$\nThe discriminant $D$ of this quadratic equation with respect to $n$ is $D = (2k^2 + 1)^2 + 4(2k^2 + k - 1)$. For integer solutions, $D$ must be a perfect square. After analysis, we find $k = 3$ is the only solution.\n\nThen the equation becomes $n^2 - 19n - 20 = 0$, so $n = 20$. From the earlier formula, $p = 2 \\cdot 3 \\cdot 20 + 2 \\cdot 3 + 1 = 127$, and $p$ is indeed a prime number.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17122,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $3n^2$ be the number of vertices of a simple graph $G$ (where $n \\geq 2$ is an integer). If the degree of each vertex is not greater than $4n$, there exists at least one vertex with degree $1$, and there exists a route of length not greater than $3$ between any two vertices, prove that the minimum number of edges of $G$ is $\\frac{7}{2}n^2 - \\frac{3}{2}n$.\n\nA route between two distinct vertices $u$ and $v$ with length $k$ is a sequence of vertices $u = v_0, v_1, \\dots, v_k = v$, where $v_i$ and $v_{i+1}$, $i = 0, 1, \\dots, k-1$, are adjacent.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p137_data_2336842d75.png)",
    "options": [],
    "answer": "See solution",
    "solution": "For any two distinct vertices $u$ and $v$, define the distance between $u$ and $v$ as the shortest length of a route between them. Consider a graph $G^*$ with vertex set $\\{x_1, x_2, \\dots, x_{3n^2-n}, y_1, y_2, \\dots, y_n\\}$, where $y_i$ and $x_i$ are adjacent ($1 \\leq i \\leq n$), $x_i$ and $x_j$ are not adjacent ($1 \\leq i < j \\leq 3n^2 - n$), and $x_i$ and $y_j$ are adjacent if and only if $i \\equiv j \\pmod{n}$. Thus, the degree of each $x_i$ is $1$, and the degree of each $y_i$ does not exceed\n\n$$\nn - 1 + \\frac{3n^2 - n}{n} = 4n - 2.\n$$\n\nIt is easy to see that the distance between $x_i$ and $x_j$ is not greater than $3$. So $G^*$ satisfies the conditions of the problem. $G^*$ has $N = 3n^2 - n + \\binom{n}{2} = \\frac{7}{2}n^2 - \\frac{3}{2}n$ edges.\n\nNow, we show that any graph $G = G(V, E)$ satisfying the conditions must have at least $N$ edges. Let $X \\subseteq V$ be the set of vertices with degree $1$, $Y \\subseteq (V \\setminus X)$ be the set of vertices adjacent to $X$, and $Z \\subseteq V \\setminus (X \\cup Y)$ be the set of vertices adjacent to $Y$. Let $W = V \\setminus (X \\cup Y \\cup Z)$. The following properties hold:\n\n*Property 1.* Any two vertices in $Y$ are adjacent. If $y_1, y_2 \\in Y$, there exist $x_1, x_2 \\in X$ adjacent to $y_1$ and $y_2$ respectively; since the distance between $x_1$ and $x_2$ is at most $3$, $y_1$ and $y_2$ must be adjacent.\n\n*Property 2.* The distance between any vertex in $W$ and any vertex in $Y$ is $2$. If $w_0 \\in W$ and $y_0 \\in Y$ and the distance is greater than $2$, let $x_0 \\in X$ be adjacent to $y_0$; then the distance between $w_0$ and $x_0$ would be greater than $3$, a contradiction. Thus, each vertex in $W$ is adjacent to some vertex in $Z$.\n\nLet $x, y, z, w$ be the sizes of $X, Y, Z, W$ respectively. Counting edges: there are $\\binom{y}{2}$ edges between $Y$, $x$ edges from $X$ to $Y$, at least $z$ edges from $Z$ to $Y$, and at least $w$ edges from $W$ to $Z$. If $y \\geq n$, then\n\n$$\n|E| \\geq \\binom{y}{2} + x + z + w = 3n^2 + \\binom{y}{2} - y \\geq 3n^2 + \\binom{n}{2} - n = N.\n$$\n\nIf $y \\leq n-1$, since each vertex degree is at most $4n$,\n\n$$\n\\begin{align*}\nx + z &\\leq y(4n - (y - 1)) = y(4n + 1 - y) \\\\\n&\\leq (n - 1)(3n + 2) = 3n^2 - n - 2, \\\\\nw &\\geq 3n^2 - y - y(4n + 1 - y) \\geq 3.\n\\end{align*}\n$$\n\nSelect $P \\in W$ adjacent to as few vertices in $Z$ as possible, say $a > 0$ (by Property 2). Let $N_P \\subseteq Z$ be these $a$ vertices. Counting edges: $\\binom{y}{2}$ between $Y$, $x$ from $X$ to $Y$, at least $y$ from $N_P$ to $Y$, at least $z - a$ from $Z \\setminus N_P$ to $Y$, and at least $aw$ from $W$ to $Z$:\n\n$$\n|E| \\geq \\binom{y}{2} + x + y + z - a + aw = 3n^2 - 1 + \\binom{y}{2} + (a - 1)(w - 1).\n$$\n\nIf $a > 1$,\n\n$$\n|E| \\geq 3n^2 - 1 + \\binom{y}{2} + (w - 1) \\geq 3n^2 - 2 + \\binom{y}{2} + 3n^2 - y - y(4n + 1 - y) > N.\n$$\n\nIf $a = 1$, since each vertex in $W$ has degree at least $2$, we add at least $w/2$ edges:\n\n$$\n|E| \\geq 3n^2 - 1 + \\binom{y}{2} + \\frac{1}{2}w \\geq 3n^2 - 1 + \\binom{y}{2} + \\frac{1}{2}(3n^2 - y - y(4n + 1 - y)) > N.\n$$\n\nTherefore, the minimum number of edges is $N = \\frac{7}{2}n^2 - \\frac{3}{2}n$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17123,
    "subject": "Mathematics (Olympiad)",
    "question": "For a polynomial $P \\in \\mathbb{R}[x]$, let $f(P) = n$ if $n$ is the smallest positive integer such that\n\n$$\n(\\forall x \\in \\mathbb{R}) \\underbrace{(P(P(\\dots P(x))\\dots))}_{n} > 0,\n$$\n\nand $f(P) = 0$ if such an integer $n$ does not exist. Does there exist a polynomial $P \\in \\mathbb{R}[x]$ of degree $2014^{2015}$ such that $f(P) = 2015$?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, such a polynomial exists. In fact, we can prove a more general result:\n\nLet $s$ be an even integer and $t > 1$ be any integer. Then, for some constant $c > 0$, the polynomial $P(x) = (x+1)^s + c - 1$ (which has degree $s$) satisfies $f(P) = t$.\n\nThe polynomial $P$ is strictly increasing on $[-1, \\infty)$. Let $x_k(c)$ be the minimal value of the $k$-fold composition $\\underbrace{P(P(\\dots P(x))\\dots)}_{k}$ for fixed $c > 0$.\n\nNote that $x_1(c) = c - 1 > -1$ and $x_{k+1}(c) = P(x_k(c))$, so $x_k(c)$ is strictly increasing in $k$. Consider the equation $x_{t-1}(c) = 0$ (with $c$ as the unknown). The leading coefficient of $x_{t-1}(c)$ is $1$ and the constant term is $-1$ (provable by induction on $t$), so this polynomial has a positive root. Let $c_0$ be such a root. Then, for $P(x) = (x+1)^s + c_0 - 1$, we have $x_{t-1}(c_0) = 0$, so $x_t(c_0) = P(0) = c_0 > 0$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17124,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $C \\subset S$ and consider a mapping $F: \\mathfrak{S} \\to \\mathfrak{S}$ (where $\\mathfrak{S}$ is the set of all subsets of $S$) such that for all $A, B \\in \\mathfrak{S}$,\n\n$$\nF(F(A) \\cup B) = A \\cap F(B)\n$$\n\nand $F(\\emptyset) = C$.\n\nHow many such mappings $F$ exist when $S$ is a set with 6 elements?",
    "options": [],
    "answer": "See solution",
    "solution": "(2k-1)(2k-3)\\ldots1. Therefore, the answer we seek for the problem is\n\n$$\n\\binom{6}{0} 2^6 + \\binom{6}{2} 2^4 + \\binom{6}{4} 3 \\cdot 2^2 + \\binom{6}{6} 5 \\cdot 3 = 64 + 240 + 180 + 15 = 499\n$$\n\n*Remark*: More generally, if $S = \\{1, 2, \\dots, n\\}$, the arguments above for $n=6$ apply. If $a_k$ is the number of such functions $F$ for general $n$, then\n\n$$\na_{k+2} = 2a_{k+1} + (k+1)a_k.\n$$\n\nThis recurrence holds because, when calculating $a_{k+2}$, the number of ways to pair $k+2$ is $k+1$ (with $a_k$ ways for the rest), and if $k+2$ is left unpaired, there are $a_{k+1}$ ways for the rest, regardless of whether $k+2$ is in $C$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17125,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all non-constant functions $f : \\mathbb{R} \\to \\mathbb{R}$ that satisfy\n\n$$\nf(2xy + x) = f(xy + x) + f(x)f(y)\n$$\n\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = y = 0$:\n\n$$\nf(0) = f(0) + f(0)^2 \\implies f(0) = 0.\n$$\n\nLet $x = 1$, $y = -1$:\n\n$$\nf(-1) = f(0) + f(1)f(-1) \\implies f(-1) = f(1)f(-1).\n$$\n\nIf $f(-1) \\neq 0$, then $f(1) = 1$. If $f(-1) = 0$, then for $y = -1$:\n\n$$\nf(-x) = f(0) + f(x)f(-1) = 0,\n$$\nwhich is not non-constant. So $f(1) = 1$.\n\nLet $x = y = -1$:\n\n$$\nf(1) = f(0) + f(-1)^2 \\implies 1 = 0 + f(-1)^2 \\implies f(-1) = \\pm 1.\n$$\n\nIf $f(-1) = 1$, try $y = -\\frac{1}{2}$:\n\n$$\nf\\left(\\frac{x}{2}\\right) = f(0) + f(x)f\\left(-\\frac{1}{2}\\right) = f(x)f\\left(-\\frac{1}{2}\\right).\n$$\n\nLet $x = -1$:\n\n$$\nf\\left(-\\frac{1}{2}\\right) = f(-1)f\\left(-\\frac{1}{2}\\right) \\implies f\\left(-\\frac{1}{2}\\right) = 0.\n$$\n\nThen $f\\left(\\frac{x}{2}\\right) = 0$ for all $x$, which is constant, a contradiction. So $f(-1) = -1$.\n\nNow, substitute $y = -1$:\n\n$$\nf(-x) = f(0) + f(x)f(-1) = -f(x),\n$$\nso $f$ is odd.\n\nNow, consider $y \\to -y$ and $y \\to y-1$ and use oddness:\n\n$$\nf(x)f(y) = f(2xy - x) - f(xy - x) = f(xy) + f(x)f(y-1) - f(xy - x). \\quad (*)\n$$\n\nFor positive integers $y$:\n\n$$\n\\begin{aligned}\nf(xy) - f(x)f(y) &= f(x(y-1)) - f(x)f(y-1) \\\\\n&= f(x(y-2)) - f(x)f(y-2) \\\\\n&= \\cdots = f(0) - f(x)f(0) = 0.\n\\end{aligned}\n$$\n\nSo $f(xy) = f(x)f(y)$ for all $x \\in \\mathbb{R}$, $y \\in \\mathbb{Z}^+$.\n\nIn $(*)$, let $x = 1$, $y = 2$:\n\n$$\nf(3) - f(1) = f(2)^2.\n$$\n\nLet $y = 1$ in the original equation:\n\n$$\nf(2x + 1) = f(x + 1) + f(x).\n$$\n\nLet $x = 1$:\n\n$$\nf(3) = f(2) + f(1) \\implies f(2) = f(3) - f(1) = f(2)^2.\n$$\n\nSo $f(2) = 2$ or $f(2) = 0$. If $f(2) = 0$, then $f(2x) = f(2)f(x) = 0$, which is constant, a contradiction. So $f(2) = 2$ and $f(2x) = 2f(x)$ for all $x$.\n\nBy induction, $f(n) = n$ for all positive integers $n$.\n\nNow, replace $y \\to \\frac{y}{x}$ in the original equation (for $x \\neq 0$):\n\n$$\nf(2y + x) = f(x + y) + f(x)f\\left(\\frac{y}{x}\\right).\n$$\n\nSwitch $x$ and $y$:\n\n$$\nf(2x + y) = f(x + y) + f(y)f\\left(\\frac{x}{y}\\right).\n$$\n\nAdd these two:\n\n$$\nf(2x + y) + f(2y + x) = 2f(x + y) + f(x)f\\left(\\frac{y}{x}\\right) + f(y)f\\left(\\frac{x}{y}\\right).\n$$\n\nBut from earlier, $f(x + y) = f(x)f\\left(\\frac{y}{x}\\right) + f(y)f\\left(\\frac{x}{y}\\right)$ for $x, y \\neq 0$.\n\nSo:\n\n$$\nf(2x + y) + f(2y + x) = 3f(x + y).\n$$\n\nLet $a, b \\in \\mathbb{R}$, and choose $c$ such that $c > 4|a| + 4|b|$ (so $c \\neq 2a, 2b$):\n\n$$\nf(c) + f(a + b) = f(a + b + c) = f(c + a) + f(b) = f(c) + f(a) + f(b).\n$$\n\nThus, $f(a + b) = f(a) + f(b)$ for all $a, b \\in \\mathbb{R}$, so $f$ is additive.\n\nSince $f(xy) = f(x)f(y)$ and $f$ is additive and non-constant, the only solution is $f(x) = x$ for all $x \\in \\mathbb{R}$.\n\nIt is easy to check that $f(x) = x$ satisfies the original equation.\n\n**Answer:** $f(x) = x$ for all $x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17126,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $N$ 為正整數。怪怪國有 $N$ 座城堡，其中每對城堡之間至多只有一條道路。每條道路上有至多 4 名守衛。為了節省人事開銷，怪怪國王頒布以下命令：\n\n1. 如果三座城堡之間都互有道路，則其中任何一條道路都不能有 4 名守衛；\n2. 如果四座城堡之間都互有道路，則從其中任何一座城堡出發，通往另外三座城堡的三條道路不能全部都站有 3 名守衛。\n\n證明：在此命令下，怪怪國在道路上的守衛總數不超過 $N^2$。\n\n註：只證出守衛總數不超過 $cN^2$，其中 $c > 1$ 與 $N$ 無關，將依 $c$ 值來給分。",
    "options": [],
    "answer": "See solution",
    "solution": "以城堡為點集 $V$，道路為邊集 $E$，守衛數量為邊權重，構成一個加權圖。我們將對這個圖做一系列操作，使得國王命令持續被滿足，且守衛數量不減。\n\n定義 $W(ab)$ 為 $ab$ 邊上的權重，而 $W(a) = \\sum_{ab \\in E} W(ab)$ 為點 $a$ 所有連邊的權重總和。此外，對任何不相連的兩點 $a$ 和 $b$，定義它們兩點等價，若且唯若對於任何其他 $c \\in V$，$ac$ 連邊且 $W(ac) = t$ 若且唯若 $bc$ 連邊且 $W(bc) = t$。\n\n現在考慮以下操作：若有 $a$、$b$ 兩點不相鄰也不等價，假設 $W(a) \\geq W(b)$，則對於所有 $c \\neq a, b$，將 $W(bc)$ 改成 $W(ac)$。注意到：\n\n1. 由於 $W(a) \\geq W(b)$，所以在此操作下，$\\sum_{v \\in V} W(v)$ 不減；\n2. 由於 $ab$ 不相連，$a$ 與 $b$ 不可能同時屬於命令中所提及的三座城堡或四座城堡，故操作後仍符合命令；\n3. 每次操作都讓等價類的總類數嚴格遞減，故操作至多 $N-1$ 次後停止。\n\n注意到操作結束後的圖中，任兩個不相鄰的點一定等價，故其必為一個完全 $k$ 部加權圖。又注意到，若有任何一條邊的權重為 4，則依據命令 (1)，$k$ 必然為 2，而完全二部圖的總邊數至多為 $(N/2)^2$，從而總權重 $\\leq 4 \\times (N/2)^2 = N^2$。\n\n故假設沒有任何一條邊的權重為 4，且 $k$ 部的點數各為 $n_1, n_2, \\dots, n_k$。注意到命令 (2) 強迫每部至多只能對另外兩部連出權重為 3 的邊。故，若我們考慮每一部僅保留一個點的子圖，且只保留權重為 3 的邊，則此子圖的每個點的 degree 都至多為 2，從而其必為若干不相連的鍊和／或環的聯集。而對於每個環 $c_1c_2\\cdots c_s$，其對應在原圖上的總權重為\n\n$$\n3(n_{c_1}n_{c_2} + \\cdots + n_{c_s}n_{c_1}) \\leq 3(n_{c_1}^2 + \\cdots + n_{c_s}^2)\n$$\n\n其中的不等式為排序不等式。故在所有其他邊權重都是 2 的情況下，原圖總權重\n\n$$\n\\begin{aligned}\n& 2 \\times \\#\\{ab : W(a, b) = 2\\} + 3 \\times \\#\\{ab : W(a, b) = 3\\} \\\\\n& \\leq 2 \\times \\#\\{ab\\} + \\#\\{ab : W(a, b) = 3\\} \\\\\n& \\leq 2 \\sum_{i \\neq j} n_i n_j + \\sum_{i=1}^k n_i^2 = \\left(\\sum_{i=1}^k n_i\\right)^2 = N^2.\n\\end{aligned}\n$$\n\n對應不同 $cN^2$ 的建議評分標準：\n\n- $c = 1$: 7pt\n- $c = 1 + o(1)$: $7^{-pt}$\n- $c < 1.5$: 2pt\n- $c = 1.5$: 1pt\n- $c > 1.5$: 0pt",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17127,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\in \\mathbb{N}^*$. Does there exist a permutation $\\sigma$ of $\\{1,2,\\dots,n\\}$ such that\n$$\n\\sqrt{\\sigma(1) + \\sqrt{\\sigma(2) + \\sqrt{\\cdots + \\sqrt{\\sigma(n)}}}} \\in \\mathbb{Q}?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose for some $n \\in \\mathbb{N}^*$ there exists a permutation $\\sigma$ such that\n$$\n\\sqrt{\\sigma(1) + \\sqrt{\\sigma(2) + \\sqrt{\\cdots + \\sqrt{\\sigma(n)}}}} = r_1 \\in \\mathbb{Q}.\n$$\nSquaring both sides, $\\sqrt{\\sigma(2) + \\sqrt{\\sigma(3) + \\cdots + \\sqrt{\\sigma(n)}}}$ is also rational. Repeating, for each $k \\in \\{1,2,\\dots,n\\}$,\n$$\nr_k := \\sqrt{\\sigma(k) + \\sqrt{\\sigma(k+1) + \\cdots + \\sqrt{\\sigma(n)}}} \\in \\mathbb{Q}.\n$$\nLet $a_k := \\sqrt{n + \\sqrt{n + \\cdots + \\sqrt{n}}}$ ($k$ times). By induction, $a_k < \\sqrt{n+1}$ for all $k$, so\n$$\nr_1 < a_n < \\sqrt{n+1}.\n$$\nLet $\\ell$ be a positive integer with $\\ell^2 \\leq n < (\\ell+1)^2$. Then for some $i$, $\\sigma(i) = \\ell^2$.\n\n**Case 1: $i \\neq n$**\n$$\n\\ell < \\sqrt{\\ell^2 + \\sqrt{\\sigma(i+1) + \\cdots + \\sqrt{\\sigma(n)}}} < \\sqrt{n+1} < \\ell+2\n$$\nSo,\n$$\n\\sqrt{\\ell^2 + \\sqrt{\\sigma(i+1) + \\cdots + \\sqrt{\\sigma(n)}}} = \\ell+1\n$$\nSquaring,\n$$\n2\\ell+1 = \\sqrt{\\sigma(i+1) + \\cdots + \\sqrt{\\sigma(n)}} < \\sqrt{n+1} < \\ell+2\n$$\nwhich implies $\\ell < 1$, a contradiction.\n\n**Case 2: $i = n$**\n- If $\\ell > 1$, then $\\ell^2 - 1$ is among $\\{\\sigma(1),\\dots,\\sigma(n-1)\\}$. For some $j < n$, $\\sigma(j) = \\ell^2 - 1$.\n  Similarly,\n  $$\n  \\ell < \\sqrt{\\ell^2 - 1 + \\sqrt{\\sigma(j+1) + \\cdots + \\sqrt{\\ell^2}}} < \\sqrt{n+1} < \\ell+2\n  $$\n  So,\n  $$\n  \\sqrt{\\ell^2 - 1 + \\sqrt{\\sigma(j+1) + \\cdots + \\sqrt{\\ell^2}}} = \\ell+1\n  $$\n  Squaring,\n  $$\n  2\\ell+2 = \\sqrt{\\sigma(j+1) + \\cdots + \\sqrt{\\ell^2}} < \\sqrt{n+1} < \\ell+2\n  $$\n  which is impossible.\n- If $\\ell = 1$, then $n \\in \\{1,2,3\\}$. Checking all cases, only $n=1$ and $n=3$ allow such a permutation:\n  - For $n=1$: $\\sqrt{1} = 1$.\n  - For $n=3$: $\\sqrt{2 + \\sqrt{3} + \\sqrt{1}} = 2$.\n  - For $n=2$: no such permutation exists.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 17128,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Prove that $12n + 13$ and $13n + 14$ are coprime for every natural $n$.\n\nb) Find the number of pairs $(a, b)$ of natural numbers for which there exists a natural number $n$ such that $\\frac{a}{b} = \\frac{12n + 13}{13n + 14}$ and $17a + 19b < 2024$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) If $d$ is a common divisor of $12n + 13$ and $13n + 14$, then $d$ divides $13(12n + 13)$ and $12(13n + 14)$. Thus, $d$ divides $13(12n + 13) - 12(13n + 14) = 1$, so $d = 1$. Therefore, $12n + 13$ and $13n + 14$ are coprime.\n\nb) The relation $\\frac{a}{b} = \\frac{12n + 13}{13n + 14}$ leads to $a(13n + 14) = b(12n + 13)$, so $13n + 14 \\mid b(12n + 13)$. Since $12n + 13$ and $13n + 14$ are coprime, $13n + 14$ divides $b$. Let $b = k(13n + 14)$ for $k \\in \\mathbb{N}^*$, then $a = k(12n + 13)$.\n\nSubstitute $a$ and $b$ into $17a + 19b < 2024$:\n$$k(451n + 487) < 2024$$\nSo $451n + 487 < 2024$, which gives $n \\leq 3$.\n\n- For $n = 0$: $k \\leq 4$, so $(a, b) \\in \\{(13, 14), (26, 28), (39, 42), (52, 56)\\}$.\n- For $n = 1$: $k \\leq 2$, so $(a, b) \\in \\{(25, 27), (50, 54)\\}$.\n- For $n = 2, 3$: $k = 1$, so $(a, b) \\in \\{(37, 40), (49, 53)\\}$.\n\nIn total, there are $8$ pairs.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17129,
    "subject": "Mathematics (Olympiad)",
    "question": "We consider permutations $f$ on the set $N$ of non-negative integers, i.e., bijective mappings $f$ from $N$ to $N$, with the following properties:\n\nFor all $x \\in N$, we have $f(f(x)) = x$ and $|f(x) - x| \\leq 3$.\n\nFurthermore, for all integers $n > 42$,\n\n$$\nM(n) = \\frac{1}{n+1} \\sum_{j=0}^{n} |f(j) - j| < 2.011\n$$\n\nProve that there exist infinitely many integers $K$ such that $f$ maps the set $\\{n \\mid 0 \\leq n \\leq K\\}$ onto itself.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there do not exist infinitely many such $K$. Then there must exist some $K_0$ such that for all $K > K_0$, there exists an $n$ with $n \\leq K < f(n)$. Since $|f(x) - x| \\leq 3$, such an $n$ can only be $K$, $K-1$, or $K-2$.\n\nThe same must hold for $K = f(n)$, and so on. This means that, from some $K_0$ on, the function must map “up” into each interval $[n, f(n)]$, and also “up” out of each such interval. In order for this to be possible, we must have $f(n) - n \\geq 3$, and it therefore follows that $|f(n) - n| = 3$ must hold for all $n > K > K_0$.\n\nIf this is the case, then\n\n$$\nM(n) = \\frac{1}{n+1} \\sum_{j=0}^{n} |f(j) - j| = \\frac{1}{n+1} \\left( \\sum_{j=0}^{K} |f(j) - j| + 3(n-K) \\right) = 3 - \\frac{C}{n+1}\n$$\n\nwhere\n\n$$\nC = 3K + 3 - \\sum_{j=0}^{K} |f(j) - j| = \\sum_{j=0}^{K} (3 - |f(j) - j|) \\geq 0.\n$$\n\nTherefore, there exists a $K_1$ such that $M(n) = 3 - \\frac{C}{n+1} > 2.011$ for $n > K_1$, which contradicts the assumption $M(n) < 2.011$. Thus, there must exist infinitely many $K$ with the required properties.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 17130,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n = m + r$ where $r > 0$. For $k \\ge 2$, show that\n\n$$\n\\frac{m!}{m^k (m-k)!} < \\frac{n!}{n^k (n-k)!}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\n\\frac{m!}{m^k (m-k)!} < \\frac{(m+r)!}{(m+r)^k (m+r-k)!}\n$$\n\nwhich is equivalent to\n\n$$\n\\left(1 + \\frac{r}{m}\\right)^k < \\frac{(m+r)(m+r-1) \\cdots (m+r-k+1)}{m(m-1) \\cdots (m-k+1)}\n$$\n\nThe right side equals\n\n$$\n\\left(1 + \\frac{r}{m}\\right) \\left(1 + \\frac{r}{m-1}\\right) \\cdots \\left(1 + \\frac{r}{m-k+1}\\right)\n$$\n\nSince $1 + \\frac{r}{m}$ is less than each of $1 + \\frac{r}{m-1}, \\ldots, 1 + \\frac{r}{m-k+1}$, the last inequality is valid. Hence, the first inequality is valid.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17131,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 4 identical fair dice. Denote $x_i$ ($1 \\leq x_i \\leq 6$) as the number of dots on the face appearing on the $i$-th die, for $1 \\leq i \\leq 4$.\n\n1. Find the number of possible tuples $(x_1, x_2, x_3, x_4)$.\n2. Find the probability that there exists a number $x_j$ such that $x_j$ is equal to the sum of the remaining numbers.\n3. Find the probability that we can divide $x_1, x_2, x_3, x_4$ into 2 groups that have the same sum.",
    "options": [],
    "answer": "See solution",
    "solution": "1. By the multiplication principle, the answer is $6^4 = 1296$.\n\n2. There are 4 ways to choose $i$ such that $x_i$ equals the sum of the remaining numbers. For each $i$, we choose $x_i$ first, and by the stars-and-bars method, there are $\\binom{x_i-1}{2}$ ways to choose the other 3 numbers. For $x_i = 3, 4, 5, 6$:\n\n$$\n\\binom{2}{2} + \\binom{3}{2} + \\binom{4}{2} + \\binom{5}{2} = 1 + 3 + 6 + 10 = 20\n$$\n\nSo there are $4 \\times 20 = 80$ tuples satisfying the condition, and the probability is $\\frac{80}{1296} = \\frac{5}{81}$.\n\n3. Let $S_1, S_2, S_3$ be the sets of tuples satisfying\n$$\nx_1 + x_2 = x_3 + x_4, \\quad x_1 + x_3 = x_2 + x_4, \\quad x_1 + x_4 = x_2 + x_3.\n$$\n\nConsider $x_1 + x_2 = x_3 + x_4$, which is equivalent to\n$$\n(x_1 - 1) + (x_2 - 1) + (6 - x_3) + (6 - x_4) = 10.\n$$\nLet $y_1 = x_1 - 1$, $y_2 = x_2 - 1$, $y_3 = 6 - x_3$, $y_4 = 6 - x_4$, then\n$$\n\\begin{cases} 0 \\le y_1, y_2, y_3, y_4 \\le 5, \\\\ y_1 + y_2 + y_3 + y_4 = 10. \\end{cases}\n$$\n\nLet $A$ be the set of tuples $(y_1, y_2, y_3, y_4)$ with $y_i \\ge 0$ and $y_1 + y_2 + y_3 + y_4 = 10$. By stars-and-bars,\n$$\n|A| = \\binom{13}{10} = 286.\n$$\n\nLet $A_i$ be the set where $y_i \\ge 6$. For $y_1 \\ge 6$:\n$$\n(y_1 - 6) + y_2 + y_3 + y_4 = 4 + 4 - 1 \\choose 4 = 35.\n$$\nSo $|A_1| = |A_2| = |A_3| = |A_4| = 35$, and these are disjoint. The number of tuples satisfying the original bounds is\n$$\n|A| - (|A_1| + |A_2| + |A_3| + |A_4|) = 146.\n$$\nThus, $|S_1| = 146$.\n\nThe intersection $S_1 \\cap S_2$ is the set where $x_1 = x_4$, $x_2 = x_3$, so $6^2 = 36$ tuples. Similarly,\n$$\n|S_2 \\cap S_3| = |S_3 \\cap S_1| = 36.\n$$\nFor $S_1 \\cap S_2 \\cap S_3$, $x_1 = x_2 = x_3 = x_4$, so 6 tuples.\n\nBy inclusion-exclusion:\n$$\n\\begin{aligned}\n|S| &= |S_1 \\cup S_2 \\cup S_3| \\\\\n&= (146 \\times 3) - (36 \\times 3) + 6 = 336.\n\\end{aligned}\n$$\n\nIncluding part 2, the number of tuples where the numbers can be divided into two groups with equal sum is $336 + 80 = 416$, so the probability is $\\frac{416}{1296} = \\frac{26}{81}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17132,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{R}^+ \\to \\mathbb{R}^+$ be a function satisfying\n$$\n(f(xy))^2 = f(x^2)f(y^2)\n$$\nfor all $x, y \\in \\mathbb{R}^+$ with $x^2y^3 > 2008$.\n\nProve that $(f(xy))^2 = f(x^2)f(y^2)$ for all $x, y \\in \\mathbb{R}^+$.",
    "options": [],
    "answer": "See solution",
    "solution": "Define $\\lambda(x, y) = \\frac{2008}{x^2y^3}$ for all $x, y > 0$.\n\nWe can see that $\\left(\\frac{x}{\\lambda(x, y)}\\right)^2 (y \\cdot \\lambda(x, y))^3 = 2008$. Thus,\n$$\n\\left(\\frac{x}{z}\\right)^2 (y z)^3 > 2008 \\text{ for every } z > \\lambda(x, y).\n$$\nIt follows that $(f(xy))^2 = f(x^2/z^2)f(y^2z^2)$ for all $z > \\lambda(x, y)$.\n\nNow for any positive real numbers $x$ and $y$, we choose\n$$\nz > \\max\\{\\lambda(x, x), \\lambda(y, y), \\lambda(x, y), \\lambda(y, x)\\}.\n$$\nIt follows that\n$$\n(f(x^2))^2 = f(x^2/z^2) f(x^2z^2) \\quad (\\text{since } z > \\lambda(x, x))\n$$\n$$\n(f(y^2))^2 = f(y^2/z^2) f(y^2z^2) \\quad (\\text{since } z > \\lambda(y, y))\n$$\n$$\n(f(xy))^2 = f(x^2/z^2) f(y^2z^2) \\quad (\\text{since } z > \\lambda(x, y))\n$$\n$$\n(f(xy))^2 = f(y^2/z^2) f(x^2z^2) \\quad (\\text{since } z > \\lambda(y, x))\n$$\nWe can see that\n$$\n\\begin{aligned}\nf(xy)^2 \\cdot f(xy)^2 &= (f(x^2/z^2) f(y^2z^2))(f(y^2/z^2) f(x^2z^2)) \\\\\n&= (f(x^2/z^2) f(x^2z^2))(f(y^2/z^2) f(y^2z^2)) \\\\\n&= (f(x^2))^2 (f(y^2))^2\n\\end{aligned}\n$$\nThus $(f(xy))^2 = f(x^2)f(y^2)$ for all $x, y > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17133,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f$ be a function such that for all positive real numbers $x, y, z$, the following holds:\n\n$$\nf(x + f(y) + f^2(z)) = z + f(y) + f^2(x)\n$$\n\nwhere $f^2(x) = f(f(x))$, $f^3(x) = f(f(f(x)))$, etc. Find all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show that $f$ must be injective. Suppose $f(z_1) = f(z_2)$. Replacing $z$ by $z_1$ or $z_2$ in (1) leads to the same left-hand side, so the right-hand sides must be equal: $z_1 + f(y) + f^2(x) = z_2 + f(y) + f^2(x)$, which implies $z_1 = z_2$.\n\nNext, we make the right-hand side of (1) equal to $z + f^2(y) + f^2(x)$ in two ways to use injectivity. Replacing $y$ by $f(y)$ gives $f(x + f^2(y) + f^2(z)) = z + f^2(y) + f^2(x)$. Replacing $x$ by $y$ and $y$ by $f(x)$ gives $f(y + f^2(x) + f^2(z)) = z + f^2(x) + f^2(y)$. Since the right-hand sides are equal and $f$ is injective, it follows that\n\n$$\nx + f^2(y) + f^2(z) = y + f^2(x) + f^2(z),\n$$\n\ni.e., $f^2(x) - x = f^2(z) - z$ for all positive $x, z$.\n\nThus, $f^2(x) - x = c$ is a constant independent of $x$. Setting $t = 1 + f(1) + f^2(1)$, equation (1) with $x = y = z = 1$ gives $f(t) = t$, so $f^2(t) = t$ and $f^2(t) - t = 0$. Therefore, $c = 0$ and we have shown\n\n$$\nf^2(x) = x \\quad \\text{for all } x > 0. \\qquad (23)\n$$\n\nPick any $d > 0$ and put $x = d$, $y = f(d)$, $z = d$ in (1), then use (23) to obtain $f(3d) = 3d$. This means that\n\n$$\nf(x) = x \\quad \\text{for all } x > 0. \\qquad (24)\n$$\n\nThis function clearly satisfies (1).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17134,
    "subject": "Mathematics (Olympiad)",
    "question": "A segment *AB* and its midpoint *K* are given. An arbitrary point *C*, different from *K*, is chosen on the perpendicular to $AB$ through $K$. Let $N$ be the intersection of $AC$ and the line passing through $B$ and the midpoint of the segment $CK$. Let $U$ be the intersection of $AB$ with the line that passes through $C$ and the midpoint $L$ of the segment $BN$. Prove that the ratio of the areas of the triangles $CNL$ and $BUL$ doesn't depend on the choice of point $C$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of the segment $CK$. From Menelaus' theorem for the triangle $AKC$ and the line $BN$ we have\n\n$$\n\\frac{\\overline{CN}}{\\overline{NA}} \\cdot \\frac{\\overline{AB}}{\\overline{BK}} \\cdot \\frac{\\overline{KM}}{\\overline{MC}} = 1.\n$$\n\nFrom this we get $\\overline{NA} = 2\\overline{NC}$, from which it follows that $\\overline{AC} = 3\\overline{NC}$. Hence $P_{BNC} = \\frac{1}{3}P_{ABC}$. From Menelaus' theorem for the triangle $ABN$ and the line $CU$ we have\n\n$$\n\\frac{\\overline{AU}}{\\overline{UB}} \\cdot \\frac{\\overline{BL}}{\\overline{LN}} \\cdot \\frac{\\overline{NC}}{\\overline{CA}} = 1.\n$$\n\n![](images/MACEDONIAN_MATHEMATICAL_OLYMPIADS_2016_p23_data_6f55d099d0.png)\n\nTherefore we get $\\overline{AU} = 3\\overline{UB}$. Therefore $U$ is the midpoint of the segment $BK$. It follows that $P_{BUC} = \\frac{1}{4}P_{ABC}$. Let $x = P_{CNL}$ and $y = P_{BLU}$. Since $L$ is the midpoint of $BN$, we have $P_{BLC} = x$. Now\n\n$$\nx + y = P_{BLC} + P_{BLU} = P_{BUC} = \\frac{1}{4}P_{ABC},\n$$\n\non the other hand we have\n\n$$\n2x = P_{CNL} + P_{BLC} = P_{BNC} = \\frac{1}{3}P_{ABC}.\n$$\n\nIf we divide these two equalities we get\n\n$$\n\\frac{1}{2} + \\frac{y}{2x} = \\frac{3}{4}, \\text{ hence } \\frac{y}{x} = \\frac{1}{2},\n$$\n\nfrom where we get the required statement.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17135,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ and $m$ be integers with $n \\geq m$. Prove that\n$$\n\\sum_{k=m}^{n} \\frac{1}{k^2} + \\sum_{k=m}^{n} \\frac{1}{k^3} \\geq m \\left( \\sum_{k=m}^{n} \\frac{1}{k^2} \\right)^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Fix $n$; we'll induct downwards on $m$.\n\n**Base case:** For $n = m$, the left side is $\\frac{1}{m^2} + \\frac{1}{m^3} = \\frac{m+1}{m^3}$, and the right side is $m \\left( \\frac{1}{m^2} \\right)^2 = \\frac{1}{m^3}$. Since $\\frac{m+1}{m^3} \\geq \\frac{1}{m^3}$, the base case holds.\n\n**Inductive step:** Suppose the result holds for $m+1$. Let\n$$\nA = \\sum_{k=m+1}^{n} \\frac{1}{k^2}, \\quad B = \\sum_{k=m+1}^{n} \\frac{1}{k^3}.\n$$\nWe know $A + B \\geq (m+1)A^2$ by the inductive hypothesis. We want to show\n$$\n\\left(A + \\frac{1}{m^2}\\right) + \\left(B + \\frac{1}{m^3}\\right) \\geq m \\left(A + \\frac{1}{m^2}\\right)^2.\n$$\nIndeed,\n$$\n\\begin{align*}\n&\\left(A + \\frac{1}{m^2}\\right) + \\left(B + \\frac{1}{m^3}\\right) - m \\left(A + \\frac{1}{m^2}\\right)^2 \\\\\n&= A + B + \\frac{m+1}{m^3} - mA^2 - \\frac{2A}{m} - \\frac{1}{m^3} \\\\\n&= \\left(A + B - (m+1)A^2\\right) + \\left(A - \\frac{1}{m}\\right)^2 \\geq 0,\n\\end{align*}\n$$\nso the result follows by induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17136,
    "subject": "Mathematics (Olympiad)",
    "question": "Dos circunferencias $C$ y $C'$ son secantes en dos puntos $P$ y $Q$. La recta que une los centros corta a $C$ en $R$ y a $C'$ en $R'$, la que une $P$ y $R'$ corta a $C$ en $X \\neq P$ y la que une $P$ y $R$ corta a $C'$ en $X' \\neq P$. Si los tres puntos $X$, $Q$, $X'$ están alineados, se pide:\n\n1. Hallar el ángulo $\\angle XPX'$.\n\n2. Demostrar que $(d + r - r')(d - r + r') = rr'$, donde $d$ es la distancia entre los centros de las circunferencias y $r$ y $r'$ sus radios.",
    "options": [],
    "answer": "See solution",
    "solution": "(i) Sean $F$ y $F'$ los puntos diametralmente opuestos a $R$ y $R'$ en $C$ y $C'$, respectivamente. Por el Teorema del ángulo inscrito se tiene que $\\angle PFQ = \\angle PXQ = \\alpha$ que, por simetría, es el doble de $\\angle PFR$, luego $\\angle PFR = \\alpha/2$. Como el triángulo $PFR$ es rectángulo en $P$ (al ser $FR$ diámetro de $C$), deducimos que $\\angle PRF = \\pi/2 - \\alpha/2$. Similarmente, $\\angle PR'F' = \\pi/2 - \\beta/2$, donde $\\beta = \\angle PX'Q$. Por otro lado, considerando el triángulo $XPX'$, $\\angle XPX' = \\pi - \\alpha - \\beta$, luego sumando los ángulos del triángulo $PRR'$,\n\n$$\n\\left(\\frac{\\pi}{2} - \\frac{\\alpha}{2}\\right) + \\left(\\frac{\\pi}{2} - \\frac{\\beta}{2}\\right) + (\\pi - \\alpha - \\beta) = \\pi,\n$$\n\nes decir, $\\alpha + \\beta = 2\\pi/3$ y $\\angle XPX' = \\pi/3$.\n\n![](images/Spanija_b_2016_p17_data_258aadef50.png)\n\n(ii) Consideramos el triángulo $OPO'$, donde $O$ y $O'$ son los centros de $C$ y $C'$, respectivamente. Nuevamente, por el Teorema del ángulo inscrito, el ángulo central $\\angle POR$ es $2\\angle PFR = \\alpha$ y similarmente $\\angle PO'R' = 2\\angle PF'R' = \\beta$, luego $\\angle OPO' = \\pi/3$. Los lados del triángulo $OPO'$ son los radios $r$ y $r'$ y la distancia $d$ entre los centros, por tanto el resultado se sigue directamente del Teorema del coseno: $d^2 = r^2 + r'^2 - rr'$, que es equivalente a la relación dada en el enunciado.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17137,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all two-digit positive integers $\\overline{ab}$, with $a < b$, which are equal to the sum of the integers from $a$ to $b$ ($a$ and $b$ included).",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $27$ and $15$.\n\nWe start by noticing that $\\overline{ab} = a + (a+1) + \\dots + b \\leq 1 + 2 + \\dots + 9 = 45$, which yields $a \\leq 4$.\n\nIf $a = 4$, then $\\overline{ab} \\geq 45 = 1 + 2 + \\dots + 9$, leading to $a = 1$ – impossible.\n\nIf $a = 3$, then $3 + 4 + \\dots + (b-1) + b = \\overline{3b} = 30 + b$, leading to $1 + 2 + \\dots + (b-1) = 33$, whence $(b-1) \\cdot b = 66$. This equation has no solution $b \\in \\{4, 5, \\dots, 9\\}$.\n\nIf $a = 2$, then $2 + 3 + \\dots + (b-1) + b = \\overline{2b} = 20 + b$, leading to $1 + 2 + \\dots + (b-1) = 21$, whence $(b-1) \\cdot b = 42$. This equality is true for $b = 7$, so $\\overline{ab} = 27$ is a solution.\n\nIf $a = 1$, then $1 + 2 + \\dots + (b-1) + b = \\overline{1b} = 10 + b$, whence $1 + 2 + \\dots + (b-1) = 10$, i.e. $(b-1) \\cdot b = 20$. This equality is true for $b = 5$, so $\\overline{ab} = 15$ is the other solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17138,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all non-negative integers $(a, b)$ such that\n\n$$\na + 2b - b^2 = \\sqrt{2a + a^2 + |2a + 1 - 2b|}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $2a + 1$ and $2b$ have different parity, $|2a + 1 - 2b| \\neq 0$, hence $|2a + 1 - 2b| \\ge 1$. This leads to\n\n$$\n\\sqrt{2a + a^2 + |2a + 1 - 2b|} \\ge \\sqrt{2a + a^2 + 1} = a + 1.\n$$\n\nOn the other hand, $a + 2b - b^2 = a + 1 - (b - 1)^2 \\le a + 1$. This shows that the given equality is possible if and only if\n\n$$\na + 2b - b^2 = a + 1 = \\sqrt{2a + a^2 + |2a + 1 - 2b|}.\n$$\n\nThe first equality yields $b = 1$ and the second leads to $|2a + 1 - 2b| = 1$. It follows $1 = |2a - 1|$, so $a = 0$ or $a = 1$. The required pairs $(a, b)$ are $(0, 1)$ and $(1, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17139,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle AB\\Gamma$ be a triangle inscribed in the circle $C(O, R)$, with $AB < A\\Gamma < B\\Gamma$. Let $E, Z, K$ be the midpoints of the sides $B\\Gamma, A\\Gamma, AB$, respectively, and $K$ the foot of the altitude from vertex $A$. On the diameters $AB$ and $A\\Gamma$, outside the triangle, draw semicircles $(c_1)$ and $(c_2)$, respectively. The lines $ZK$ and $KZ$ intersect the semicircle $(c_1)$ at points $\\Pi$ and $\\Sigma$ (respectively), and the lines $EK$ and $KE$ intersect the semicircle $(c_2)$ at points $P$ and $T$, respectively. The lines $\\Pi\\Sigma$ and $PT$ meet at $M$. Prove that:\n\n(a) $\\Pi P$ and $\\Sigma T$ intersect at point $A$.\n\n(b) $\\Pi P$ and $M\\Delta$ intersect on the circle $C(O, R)$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) The quadrilateral $AKB\\Sigma$ is a rectangle, because:\n\n1. $ZA = ZB = Z\\Sigma$ (radii of the semicircle $(c_1)$),\n2. $ZA = ZB = ZK$ ($ZK$ is the median corresponding to the hypotenuse of the right-angled triangle $AKB$),\n3. $\\angle AKB = 90^\\circ$.\n\nHence $A\\Sigma \\parallel KB$ and $AT \\parallel K\\Gamma$, so the points $A, \\Sigma, T$ are collinear.\n\n![](images/Hellenic_2016_p17_data_5c04e676f0.png)\n\n![](images/Hellenic_2016_p17_data_6173473422.png)\n\nThe triangle $EAP$ is isosceles ($ZA = Z\\Pi$ as radii of $(c_1)$).\n\nThus:\n\n$$\n\\angle ZA\\Pi = \\angle Z\\Pi A = \\hat{x}, \\quad \\angle AZ\\Delta = 2\\hat{x}.\n$$\n\nSimilarly, the triangle $ZAP$ is isosceles, so\n\n$$\n\\angle EAP = \\angle EPA = \\hat{y}, \\quad \\angle AE\\Delta = 2\\hat{y}.\n$$\n\nFrom the parallelogram $AE\\Delta Z$, we have:\n\n$$\n2\\hat{x} = 2\\hat{y} \\Leftrightarrow \\hat{x} = \\hat{y}, \\quad 2\\hat{x} + \\hat{A} = 180^{\\circ} \\quad (1)\n$$\n\nThat is, $\\hat{x} + \\hat{A} + \\hat{y} = 180^{\\circ}$, so the points $A, \\Pi, P$ are collinear.\n\n(b) First, we prove that the triangle $M\\Sigma T$ is isosceles. It suffices to show:\n\n$$\n\\angle \\Sigma_I = \\angle T_I = \\hat{x} = \\hat{y}.\n$$\n\nFrom the inscribed quadrilateral $AB\\Pi\\Sigma$:\n\n$$\n\\angle \\Sigma_I = \\angle B_I = \\hat{x}, \\quad \\angle Z_I = \\angle Z_2 = 2\\hat{x}.\n$$\n\nFrom the inscribed quadrilateral $A\\Gamma TP$:\n\n$$\n\\angle T_I = \\angle \\Gamma_I = \\hat{y}, \\quad \\angle E_I = \\angle E_2 = 2\\hat{y}.\n$$\n\nTherefore, from the parallelogram $AE\\Delta Z$:\n\n$$\n2\\hat{x} = 2\\hat{y} \\Leftrightarrow \\hat{x} = \\hat{y}.\n$$\n\nFrom part (a), the quadrilateral $B\\Gamma T\\Sigma$ is a rectangle, and since $M\\Sigma T$ is isosceles, $M\\Delta$ is the perpendicular bisector of $\\Sigma T$ and $B\\Gamma$, so it passes through $O$.\n\nIf $M\\Delta$ intersects the circle $C(O, R)$ at $N$, then $N$ is the midpoint of the major arc $B\\Gamma$, so $\\angle NA\\Gamma = 90^\\circ - \\frac{\\hat{A}}{2}$.\n\nSince $A, \\Pi, P$ are collinear, $\\angle PA\\Gamma = 90^\\circ - \\frac{\\hat{A}}{2}$ (from (1)).\n\n![](images/Hellenic_2016_p18_data_00e9a2c39b.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17140,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all infinite sequences $a_1, a_2, a_3, \\dots$ of positive integers such that:\n\n1. $a_{nm} = a_n a_m$ for all positive integers $n, m$.\n2. There are infinitely many positive integers $n$ such that $\\{1, 2, \\dots, n\\} = \\{a_1, a_2, \\dots, a_n\\}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(n)$ denote the sequence, i.e., $f(n) = a_n$.\n\n**Fact 1:** $f(1) = 1$.\n\n*Proof.* From (1), $f(1) = f(1)f(1) = f(1)^2$. Since $f(1)$ is a positive integer, $f(1) = 1$.\n\n**Fact 2:** $f$ is bijective.\n\n*Proof.*\n- *Injectivity:* Suppose $f(a) = f(b)$ for $a \\neq b$. For some $n > \\max(a, b)$ with $\\{1, 2, \\dots, n\\} = \\{f(1), \\dots, f(n)\\}$, the set $\\{f(1), \\dots, f(n)\\}$ would have fewer than $n$ elements, a contradiction.\n- *Surjectivity:* For any $c \\in \\mathbb{N}$, for some $n \\geq c$ with $\\{1, 2, \\dots, n\\} = \\{f(1), \\dots, f(n)\\}$, $c$ must appear among $f(1), \\dots, f(n)$.\n\n**Fact 3:** $n$ is prime if and only if $f(n)$ is prime.\n\n*Proof.*\n- If $n$ is prime but $f(n)$ is not, then $f(n) = f(a)f(b) = f(ab)$ for $a, b > 1$, so $n = ab$, contradiction.\n- If $f(n)$ is prime but $n$ is not, then $n = ab$ for $a, b > 1$, so $f(n) = f(a)f(b)$, not prime.\n\n**Fact 4:** If $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_k^{\\alpha_k}$, then\n$$\nf(n) = f(p_1)^{\\alpha_1} f(p_2)^{\\alpha_2} \\dots f(p_k)^{\\alpha_k}\n$$\nis the unique factorization of $f(n)$.\n\n**Fact 5:** For all $y < x$ in $\\mathbb{N}$, there exists $n_0$ such that for all $n > n_0$,\n$$\ny^{n+1} < x^n.\n$$\n\n*Proof.* For large $n$, $x^n$ grows faster than $y^{n+1}$.\n\n**Fact 6:** For all primes $p$, $f(p) \\leq p$.\n\n*Proof.* Let $p_1, \\dots, p_n$ be the first $n$ primes. For large $N$, $\\{1, \\dots, N\\} = \\{f(1), \\dots, f(N)\\}$ and $N \\geq p_n^{n_0}$. Consider $\\alpha$ maximal with $p_n^{\\alpha} \\leq N$. The only $\\alpha$-th powers of primes in $\\{1, \\dots, N\\}$ are $p_1^{\\alpha}, \\dots, p_n^{\\alpha}$. Thus $f(p_n^{\\alpha}) = p_i^{\\alpha}$ for some $i \\leq n$, so $f(p_n) = p_i \\leq p_n$.\n\n**Fact 7:** For all $n$, $f(n) = n$.\n\n*Proof.* By induction and injectivity, $f(p_n) = p_n$ for all primes $p_n$. By Fact 4, $f(n) = n$ for all $n$.\n\n**Conclusion:** The only such sequence is $a_n = n$ for all $n \\geq 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17141,
    "subject": "Mathematics (Olympiad)",
    "question": "There is a $5 \\times 5$ grid. Write the integers $1, 2, \\ldots, 16$ (each used only once) in the upper left $4 \\times 4$ subgrid (■). For each of the 4 rows, write the sum of the four numbers in that row at the right end of the row. Similarly, for each of the 4 columns, write the sum of the four numbers in that column at the bottom of the column. The lower right cell remains empty. \n\nFind the maximum integer $m$ such that, after filling the grid as above, there exists a way to choose two numbers $a, b$ from the rightmost column and also from the bottom row so that $|a - b| \\geq m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the numbers in the $4 \\times 4$ grid be $a_{i,j}$, and let $A_1, \\ldots, A_4$ be the row sums, $B_1, \\ldots, B_4$ the column sums. Assume $A_1$ is the minimum and $A_4$ is the maximum among the $A_i$, and similarly for $B_1$ and $B_4$ among the $B_j$. Then $m \\leq A_4 - A_1$ and $m \\leq B_4 - B_1$, so\n\n$$\n\\begin{aligned}\nm &\\leq \\frac{(A_4 - A_1) + (B_4 - B_1)}{2} \\\\\n  &= \\frac{1}{2}(a_{4,4} - a_{1,1}) + \\frac{1}{2}(a_{4,4} + a_{4,2} + a_{4,3} + a_{2,4} + a_{3,4} - a_{1,1} - a_{1,2} - a_{1,3} - a_{2,1} - a_{3,1}) \\\\\n  &\\leq \\frac{1}{2}(16 - 1) + \\frac{1}{2}(16 + 15 + 14 + 13 + 12 - 1 - 2 - 3 - 4 - 5) \\\\\n  &= 35\n\\end{aligned}\n$$\n\nThere exists a way of writing if $m = 35$ (see table below). So the maximum $m$ is $35$.\n\n![](path/to/file.png)\n\n![](path/to/file.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17142,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that\n\n$$\n\\sum_{n=0}^{1006} \\frac{2012!}{(n!(1006-n)!)^2}\n$$\n\nis a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n\n$$\n\\sum_{n=0}^{1006} \\frac{2012!}{(n!(1006-n)!)^2} = \\frac{2012!}{(1006!)^2} \\sum_{n=0}^{1006} \\left( \\frac{1006!}{n!(1006-n)!} \\right)^2\n$$\n\nSince $\\binom{1006}{n} = \\binom{1006}{1006-n}$, we have\n\n$$\n\\sum_{n=0}^{1006} \\left(\\binom{1006}{n}\\right)^2 = \\sum_{n=0}^{1006} \\binom{1006}{n} \\binom{1006}{1006-n} = \\binom{2012}{1006},\n$$\n\nby the Vandermonde identity. Therefore,\n\n$$\n\\sum_{n=0}^{1006} \\frac{2012!}{(n!(1006-n)!)^2} = \\left(\\binom{2012}{1006}\\right)^2\n$$\n\nwhich is a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17143,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider $N$ boys arranged in a circle. For which integers $b$ is it possible to arrange the boys so that exactly $b$ of them are \"tall\" (where a boy is called \"tall\" if he is taller than both his immediate neighbors)?\n\n![](images/Blr2012_p16_data_13ff59cc33.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Any integer $b$ with $1 \\leq b \\leq \\left\\lfloor \\frac{N}{2} \\right\\rfloor$ is possible.\n\n**Explanation:**\n\nAssign a \"+\" before a boy if he is taller than the previous boy (clockwise), and a \"-\" if he is shorter. A boy is \"tall\" if a \"+\" stands before him and a \"-\" after him. The number of such alternations is at most $\\left\\lfloor \\frac{N}{2} \\right\\rfloor$.\n\nTo construct an arrangement with exactly $b$ tall boys ($1 \\leq b \\leq \\left\\lfloor \\frac{N}{2} \\right\\rfloor$):\n\n- Number the boys by height: $1$ (shortest) to $N$ (tallest).\n- Partition into three groups:\n  - $A$: shortest $b$ boys ($1,2,\\ldots,b$)\n  - $B$: tallest $b$ boys ($N-b+1,\\ldots,N$)\n  - $C$: the remaining boys\n- Place boys from $A$ at positions $2k-1$ ($k=1,\\ldots,b$), $B$ at $2k$ ($k=1,\\ldots,b$), and $C$ in the remaining spots.\n\nThis yields exactly $b$ tall boys.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17144,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all sequences $a_0, a_1, a_2, \\dots$ of positive real numbers such that\n\n$$\na_{n^2 + m^2} = a_n^n a_m^m\n$$\n\nfor all $n, m$.",
    "options": [],
    "answer": "See solution",
    "solution": "These sequences are given by $a_n = c^n$ for any positive real $c$.\n\n**Proof:**\n\nEvidently, $a_n = c^n$ satisfies the equation. To prove that there are no other solutions, we show that $a_1$ determines the entire sequence. This proves the claim because any $a_1$ may be obtained by a choice of $c$.\n\nTo this end, notice\n\n$$\na_p^p a_q^q = a_n^n a_m^m\n$$\n\nfor any $p, q, n, m$ such that\n\n$$\np^2 + q^2 = n^2 + m^2,\n$$\n\nwhich determines $a_p$ in terms of $a_q, a_n, a_m$ when the latter relation holds. This is the case, in particular, when $p$ is odd and\n\n$$\nq = \\frac{p-5}{2}, \\quad n = p-2, \\quad m = \\frac{p+3}{2}\n$$\n\nor $p$ is even and\n\n$$\nq = \\frac{p-10}{2}, \\quad n = p-4, \\quad m = \\frac{p+6}{2}.\n$$\n\nSince $q, n, m$ are non-negative and less than $p$ when $p \\ge 5$ and $p \\ge 10$, respectively, the entire sequence is thus given recursively when $a_n$ is known for $n = 0, 1, 2, 3, 4, 6, 8$.\n\nNow let $a_1$ be given. We have\n\n$$\na_0 = a_0^0 a_0^0 = 1.\n$$\n\nThe given equation determines any of $a_k, a_n, a_m$ in terms of the two others when\n\n$$\nk = n^2 + m^2.\n$$\n\nWith\n\n$$\n\\begin{align*}\n(n, m) &= (1, 1), (2, 0), (2, 1), (5, 0), \\\\\n(k, n) &= (25, 4), \\\\\n(n, m) &= (5, 5), \\\\\n(k, n) &= (50, 1), \\\\\n(n, m) &= (3, 0), (3, 1), (10, 0), (2, 2), \\\\\n(k, n) &= (100, 8)\n\\end{align*}\n$$\n\nwe thus get in turn $a_n$ for $n = 2, 4, 5, 25, 3, 50, 7, 9, 10, 100, 8, 6$. This proves the claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17145,
    "subject": "Mathematics (Olympiad)",
    "question": "The number of integral points (i.e., the points whose $x$- and $y$-coordinates are both integers) within the area (not including the boundary) enclosed by the right branch of the hyperbola $x^2 - y^2 = 1$ and the line $x = 100$ is \\underline{\\hspace{2cm}}.",
    "options": [],
    "answer": "See solution",
    "solution": "By symmetry, we only need to consider the part of the area above the $x$-axis. Suppose the line $y = k$ intercepts the right branch of the hyperbola and the line $x = 100$ at points $A_k$ and $B_k$ ($k = 1, 2, \\dots, 99$), respectively. Then the number of integral points within the segment $A_k B_k$ is $99 - k$. Therefore, the number of integral points within the area above the $x$-axis is\n$$\n\\sum_{k=1}^{99} (99 - k) = 99 \\times 49 = 4851.\n$$\nFinally, we obtain the total number of integral points within the whole area as $2 \\times 4851 + 98 = 9800$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17146,
    "subject": "Mathematics (Olympiad)",
    "question": "Two squares overlap so that their overlap has area $1\\ \\text{cm}^2$. The perimeter of the final (combined) shape is $32\\ \\text{cm}$. What are the possible side lengths of the two squares?\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p39_data_6ccf325b45.png)\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p39_data_cff1d4b7e8.png)\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p39_data_fad6e17c32.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let the side lengths of the two squares be $a$ and $b$ with $a \\leq b$.\n\nSince the overlap has area $1\\ \\text{cm}^2$, it must be a $1 \\times 1$ square in the corner of each overlapping square.\n\nThe perimeter of the final shape is $32\\ \\text{cm}$, and the perimeter of the overlapping square is $4\\ \\text{cm}$, so the sum of the perimeters of the original two squares is $36\\ \\text{cm}$.\n\nSince the perimeter of a square is $4$ times its side length, $4a + 4b = 36$, so $a + b = 9$.\n\nThe possible integer pairs $(a, b)$ with $a \\leq b$ and both at least $2$ (since the overlap is $1 \\times 1$) are:\n- $a = 2$, $b = 7$\n- $a = 3$, $b = 6$\n- $a = 4$, $b = 5$\n\nThus, the possible side lengths for the two squares are $(2, 7)$, $(3, 6)$, and $(4, 5)$ centimeters.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17147,
    "subject": "Mathematics (Olympiad)",
    "question": "н-ийн сондгой нэмэгдүүнүүд нь л тэнцүү байж болох хуваалтуудын тоо нь н-ийн нэмэгдүүн бүр $3$-аас илүүгүй давтагдаж болох хуваалтын тоотой тэнцүү болохыг батал.",
    "options": [],
    "answer": "See solution",
    "solution": "n-ийн сондгой нэмэгдэхүүнүүд нь л тэнцүү байж болох хуваалтын тоог $a_n$ гэе. Түүний үүсгэгч функцийг бичвэл:\n\n$$\nf(x) = 1 + \\sum_{k=0}^{\\infty} a_k x^k = (1 + x^2)(1 + x^4)(1 + x^6)\\dots (1 + x + x^2 + x^3 + \\dots) \\\\\n(1 + x^3 + x^6 + x^9 + \\dots)(1 + x^5 + x^{10} + \\dots)\n$$\n\n$$\n= \\frac{1 - x^4}{1 - x^3} \\cdot \\frac{1 - x^8}{1 - x^4} \\cdot \\frac{1 - x^{12}}{1 - x^6} \\dots \\cdot \\frac{1}{1 - x^3} \\cdot \\frac{1}{1 - x^5}\n$$\n\n$$\n= \\frac{1}{(1 - x^2)(1 - x^6)(1 - x^{10})\\dots} \\cdot \\frac{1}{(1 - x)(1 - x^3)(1 - x^5)\\dots}\n$$\n\nнөгөө талаас, n-ийн нэмэгдэхүүн бүр $3$-аас олонгүй давтагдаж болох хуваалтын тоог $b_n$ гэе. Түүний үүсгэгч функцийг бичвэл:\n\n$$\nf(x) = 1 + \\sum_{k=0}^{\\infty} b_k x^k = (1 + x + x^2 + x^3)(1 + x^2 + x^4 + x^6)(1 + x^3 + x^6 + x^9)\\dots\n$$\n\n$$\n= \\frac{1 - x^4}{1 - x^3} \\cdot \\frac{1 - x^8}{1 - x^2} \\cdot \\frac{1 - x^{12}}{1 - x^3} \\cdot \\frac{1 - x^{16}}{1 - x^4} \\dots\n$$\n\n$$\n= \\frac{(1 - x^2)(1 - x^3)(1 - x^5)\\dots}{1} \\cdot \\frac{(1 - x^2)(1 - x^6)(1 - x^{10})\\dots}{1}\n$$\n\nИймд $f(x) = F(x)$ буюу $a_n = b_n$ гэж батлагдав.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17148,
    "subject": "Mathematics (Olympiad)",
    "question": "On the unit square $ABCD$, a point $E$ is given on $CD$ such that $|\\angle BAE| = 60^\\circ$. Let $X$ be an arbitrary interior point of the segment $AE$. Let $Y$ be the intersection of the line through $X$ perpendicular to $BX$ with the line $BC$. What is the least possible length of $BY$?\n\n![](Fig1.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the Thales circle over $BY$, circumscribed about triangle $BYX$. This circle contains $X$ and is tangent to $AB$ at $B$. Among all such circles, the one tangent to $AE$ at $X$ has the least diameter (call this circle $k$). This circle is inscribed in the equilateral triangle $AA'F$, where $A'$ is the reflection of $A$ across $B$, and $F$ lies on the ray $BC$. The center of $k$ is the centroid of triangle $AA'F$, which is equilateral with side length $2$. Thus, the diameter of $k$ is $BY = \\frac{2}{3}\\sqrt{3}$, and the corresponding $X$ is the center of $AF$, which lies on $AE$ since $|AX| = 1 < |AE|$.\n\n**Answer:** The least possible length of $BY$ is $\\frac{2}{3}\\sqrt{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17149,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a positive integer. Show that for any $k \\geq 4022$, the number $10^k + 1$ raised to the $2011$th power, that is, $(10^k + 1)^{2011}$, is a number whose decimal representation begins with the digit $1$ followed by at least $k - 1 - 2010$ zeros.",
    "options": [],
    "answer": "See solution",
    "solution": "A number starts with the digit $1$ followed by $r$ zeros if and only if it can be written as $10^{r+s} + A$, with $0 \\leq A < 10^s$. \n\nFor $k > r \\geq 0$, we have:\n$$\n10^{r+s} + 10^k A + A < 10^{r+s} + 10^{k+s} + 10^s < 10^{k+s+1}.\n$$\nThus,\n$$(10^k + 1)(10^{r+s} + A) = 10^{k+r+s} + 10^{r+s} + 10^k A + A$$\nis a number with at least $r-1$ zeros after its leading digit $1$.\n\nBy induction, for $k > r \\geq 0$ and any $1 \\leq m < r$, $(10^k + 1)^m (10^{r+s} + A)$ has at least $r - m$ zeros after the leading digit $1$.\n\nLet $A = 1$, $s = 1$, $r = k-1$. Then $(10^k + 1)^{2011} = (10^k + 1)^{2010}(10^k + 1)$ has at least $k - 1 - 2010$ zeros after the leading digit $1$, provided $k > 2011$.\n\nTherefore, any $k \\geq 4022$ gives a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17150,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A = \\{a_1, a_2, \\dots, a_n\\}$ be a set of positive integers with $n \\ge 1$ elements. The set $A$ is called *good* if, for any two distinct subsets $X$ and $Y$ of $A$ ($X \\ne Y$), the difference $S(X) - S(Y)$ is not divisible by $2^n$. Here, $S(X) = \\sum_{a \\in X} a$ denotes the sum of the elements in the subset $X \\subseteq A$, and $S(\\emptyset) = 0$ by definition.\n\nDetermine the number of good sets $A$ such that all elements in $A$ are less than $2^n$.\n\n![](images/MNG2024_p28_data_18a590a5f8.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $2^{n(n-1)/2}$.\n\nLet $N = 2^n$ and define $v_2(a) = s$ if and only if $2^s \\mid a$ and $2^{s+1} \\nmid a$, for a positive integer $a$. Let $v_2(A) = \\{v_2(a) \\mid a \\in A\\}$ for the set $A$.\n\n**Claim:** A set $A$ is good if and only if $v_2(A) = \\{0, 1, \\dots, n-1\\}$.\n\nFirst, we show that sets of the form $A_n = \\{2^i b_i \\mid b_i \\text{ odd},\\ 0 \\le i \\le n-1\\}$ are good. For $n=1$, this is trivial. For $n \\ge 1$, assume $A_n$ is good:\n\n$$\n\\{S(X) \\bmod N \\mid X \\subseteq A_n\\} = \\{0, 1, \\dots, N-1\\} \\bmod N,\n$$\n\nand\n\n$$\n\\{S(X) \\bmod 2N \\mid X \\subseteq A_{n+1}\\} = \\{S(X), S(X) + N \\bmod 2N \\mid X \\subseteq A_n\\}.\n$$\n\nThis implies\n\n$$\n\\{S(X) \\bmod 2N \\mid X \\subseteq A_{n+1}\\} = \\{0, 1, \\dots, 2N-1\\} \\bmod 2N.\n$$\n\nHence, $A_{n+1}$ is good.\n\nAssume that $A = \\{a_1, a_2, \\dots, a_n\\}$ is good, and $M = 2^N - 1$. Since $2^N \\equiv 1 \\pmod M$ and\n\n$$\n\\{S(X) \\bmod N \\mid X \\subseteq A\\} = \\{0, 1, \\dots, N-1\\} \\bmod N,\n$$\n\nwe have\n\n$$\n\\prod_{j=1}^{n} (2^{a_j} + 1) = \\sum_{X \\subseteq A} 2^{S(X)} \\equiv \\sum_{k=0}^{N-1} 2^k \\equiv 0 \\pmod M.\n$$\n\nNote $M = \\prod_{i=0}^{n-1} (2^{2^i} + 1)$. Therefore, for any $0 \\le i \\le n-1$, there exists $1 \\le j \\le n$ such that $d = \\gcd(2^{2^i} + 1, 2^{a_j} + 1) \\ne 1$. Thus $2^{2^i} \\equiv 2^{a_j} \\equiv -1 \\pmod d$, and $2^{2^{i+1}} \\equiv 2^{2^{a_j}} \\equiv 1 \\pmod d$, implying $2^{i+1} \\mid 2^{a_j}$. Since $2^s \\equiv 1 \\pmod d$ where $s \\le i+1$, $s > i$. Therefore, $s = i+1$, and $a_j = 2^i b_i$, where $b_i$ is odd. Thus, $\\{0, 1, \\dots, n-1\\} \\subseteq v_2(A)$, and $|v_2(A)| = n$.\n\nThus, the number of good sets such that all elements are less than $2^n$ is:\n\n$$\n2^{n-1} \\times 2^{n-2} \\times \\dots \\times 2^0 = 2^{\\frac{n(n-1)}{2}}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17151,
    "subject": "Mathematics (Olympiad)",
    "question": "I accidentally decreased a number by $60\\%$ instead of increasing it by $60\\%$. This incorrect value now needs to be increased by $k\\%$ to get to the correct value. What is the value of $k$?",
    "options": [],
    "answer": "See solution",
    "solution": "I ended up with $40\\%$ of the number instead of $160\\%$. So I need to multiply this new result by $4$, or add it three times to itself, which is an increase of $300\\%$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17152,
    "subject": "Mathematics (Olympiad)",
    "question": "Call a tuple $$(a_1, \\dots, a_n)$$ of real numbers *stable* if the sums $a_1 + a_2 + \\dots + a_k$, as well as the sums $a_k + a_{k+1} + \\dots + a_n$, where in both cases $0 < k \\le n$, are either all negative or all non-negative.\n\nFor instance, the tuple $(3, -1, 2)$ is stable, since:\n\n$$\n\\begin{array}{lclcl}\n3 & \\ge & 0, & 2 & \\ge 0, \\\\\n3 + (-1) & \\ge & 0, & (-1) + 2 & \\ge 0, \\\\\n3 + (-1) + 2 & \\ge & 0; & 3 + (-1) + 2 & \\ge 0.\n\\end{array}\n$$\n\nProve that in any stable tuple with at least 3 terms where all terms are alternately negative and non-negative (it is unknown whether the first term is negative or non-negative), there exist 3 consecutive terms that together (without reordering) form a stable tuple on their own.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider terms whose absolute value is minimal in the tuple. If there exists a negative such element, denote it $a_i$, then the sum of $a_i$ and any neighbor is non-negative. Thus $a_i$ is neither the first nor the last in the tuple because of the stability of the tuple. But then both $a_{i-1} + a_i$ and $a_i + a_{i+1}$ are non-negative, as well as $a_{i-1} + a_i + a_{i+1}$, hence $a_{i-1}$, $a_i$, $a_{i+1}$ together form a stable subtuple. If all elements with minimal absolute value are non-negative, then let $a_i$ be any of them. Analogously to the previous case, both $a_{i-1} + a_i$ and $a_i + a_{i+1}$ are negative, as well as $a_{i-1} + a_i + a_{i+1}$, hence $a_{i-1}$, $a_i$, $a_{i+1}$ together form a stable tuple.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17153,
    "subject": "Mathematics (Olympiad)",
    "question": "How many quadruples $(a, b, c, d)$ of positive integers lying between 1 and 9 (inclusive) are there which satisfy $0 < b - a < c - b < d - c$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = b-a$, $y = c-b$, $z = d-c$. Then $x, y, z$ are integers satisfying $0 < x < y < z$. Furthermore, since $x + y + z = d - a \\leq 9 - 1 = 8$, the triple $(x, y, z)$ must be one of the following:\n\n$$\n(1, 2, 3),\\quad (1, 2, 4),\\quad (1, 2, 5),\\quad (1, 3, 4)\n$$\n\nFor $(x, y, z) = (1, 2, 3)$, $(a, b, c, d) = (a, a+1, a+3, a+6)$, so for $a = 1, 2, 3$, we get 3 quadruples. For $(x, y, z) = (1, 2, 4)$, $a = 1, 2$ gives 2 quadruples. For $(x, y, z) = (1, 2, 5)$ and $(1, 3, 4)$, only $a = 1$ works, giving 1 quadruple each. Therefore, the total number is $3 + 2 + 1 + 1 = 7$ quadruples $(a, b, c, d)$ satisfying the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17154,
    "subject": "Mathematics (Olympiad)",
    "question": "The bisector of the angle $\\angle ABC$ of triangle $ABC$ intersects the circumcircle of triangle $ABC$ at point $K$. Point $N$ belongs to segment $AB$ and $NK \\perp AB$. Let $P$ be the midpoint of segment $NB$. Consider the line through $P$ that is parallel to $BC$ and intersects line $BK$ at point $T$. Prove that the line $NT$ passes through the midpoint of segment $AC$.\n\n![](images/UkraineMO_2015-2016_booklet_p24_data_b37dbe9921.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M = NT \\cap AC$. Note that since $BK$ is the bisector of $\\angle ABC$, $\\angle KBC = \\angle KBA = \\alpha$. Since $PT$ is parallel to $BC$, $\\angle KBC = \\angle PTB$. Thus, $PT = PB = PN$. It follows that $\\triangle BNT$ is right-angled with hypotenuse $BN$. Since $\\triangle BNK$ is also right-angled, $\\angle KNT = \\alpha$. Moreover, $\\angle KAC = \\angle KBC = \\alpha$. Thus, quadrilateral $KANM$ is cyclic. Since $\\angle ANK = 90^\\circ$, $AK$ is a diameter of the circumcircle of $KANM$. Hence, $\\angle AMK = 90^\\circ$. Since $\\triangle AKC$ is isosceles, the altitude $KM$ is also a median. Hence, $AM = MC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17155,
    "subject": "Mathematics (Olympiad)",
    "question": "There are exactly $K$ positive integers $b$ with $5 \\leq b \\leq 2024$ such that the base-$b$ integer $2024_b$ is divisible by $16$ (where $16$ is in base ten). What is the sum of the digits of $K$?\n\n(A) 16  (B) 17  (C) 18  (D) 20  (E) 21",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that $2024_b = 2b^3 + 2b + 4 = 2(b + 1)(b^2 - b + 2)$. Consider the residue classes of this number modulo $8$.\n\nIf $b \\equiv 7 \\pmod{8}$, then $b + 1 \\equiv 0 \\pmod{8}$. If $b \\equiv 3 \\pmod{8}$, then $b^2 - b + 2 \\equiv 0 \\pmod{8}$. In each case, $2024_b$ is divisible by $16$.\n\nIn all other cases, $2024_b$ is not divisible by $16$. For $b \\equiv 0, 2, \\text{ or } 4 \\pmod{8}$, $b + 1$ is odd, and $b^2 - b + 2 \\equiv b + 2 \\pmod{8}$, so $2024_b$ is divisible by no power of $2$ greater than $2^3$. If $b \\equiv 1 \\pmod{5}$, then $b + 1$ and $b^2 - b + 2$ are both odd multiples of $2$, so $2024_b$ is divisible by $8$, but not by $16$.\n\nBecause $2024 = 253 \\times 8$, there are $253 \\times 3 = 759$ positive integers $b \\leq 2024$ that are congruent to $3$, $6$, or $7$ modulo $8$. The number $3$ must be excluded from this total, because the problem statement requires $b$ to be at least $5$. Thus $K = 759 - 1 = 758$, and the sum of the digits of $K$ is $7 + 5 + 8 = 20$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17156,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all polynomials $f(x)$ with nonnegative integer coefficients such that $f(2) = 2017$ and $f(1) \\leq 7$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose one of the coefficients of $f_0(x) = f(x)$, say the coefficient of $x^k$, is at least 2. Define $f_1(x) = f_0(x) - 2x^k + x^{k+1}$. Then $f_1(2) = f_0(2) = 2017$ and $f_1(1) < f_0(1) = 7$. Also, the coefficients of $f_1$ are nonnegative integers. Repeating the same process, since the sequence $f_0(1), f_1(1), \\dots$ of nonnegative integers is strictly decreasing, the process must end at some step. Then we obtain a polynomial $g(x)$ with coefficients 0 or 1 such that $g(2) = 2017$ and $g(1) \\leq 7$.\n\nSince the binary representation of $2017$ is $1111100001_{(2)}$, we know that $g$ is uniquely determined, and is given by\n\n$$\ng(x) = x^{10} + x^9 + x^8 + x^7 + x^6 + x^5 + 1.\n$$\n\nAs $g(1) = 7$, the equality $g(1) \\leq 7$ holds. This means we do not need to carry out any process, and $f$ must be the same as $g$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17157,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $a, b, c > 1$ and $(a^2b)^{\\log_a c} = a \\cdot (ac)^{\\log_a b}$ is satisfied. Then the value of $\\log_c(ab)$ is ______.",
    "options": [],
    "answer": "See solution",
    "solution": "Taking the logarithm of the original equation with respect to base $a$ on both sides, we get\n\n$$\n\\log_a c \\cdot (2 + \\log_a b) = 1 + \\log_a b \\cdot (1 + \\log_a c)\n$$\n\nSimplifying the above equation gives $2\\log_a c = 1 + \\log_a b$. Therefore, $c^2 = ab$, and then $\\log_c(ab) = \\log_c c^2 = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17158,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\Gamma_2$ be a circle contained in the interior of another circle $\\Gamma_1$ on the plane. Prove that there exists a point $P$ on the plane satisfying the following conditions: if $\\ell$ is a line that does not contain $P$, that intersects $\\Gamma_1$ at two different points $A, B$, and that intersects $\\Gamma_2$ at two different points $C, D$ (so that $A, C, D, B$ lie in order on $\\ell$), then $\\angle APC = \\angle DPB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the centers of the two circles by $O_1, O_2$, and the radii by $r_1, r_2$, respectively, where $r_1 > r_2$. We first show that there are two points $P, Q$ on the ray $O_1O_2$ such that $O_1P \\cdot O_1Q = r_1^2$ and $O_2P \\cdot O_2Q = r_2^2$.\n\nOne can choose a point $K$ on the ray $O_1O_2$ such that\n$$\nO_1K = \\frac{O_1O_2^2 + r_1^2 - r_2^2}{2O_1O_2}.\n$$\nSince $O_1O_2 \\le r_1 - r_2$, we have $O_1K \\ge r_1$. We choose two points $P, Q$ on the line $O_1O_2$ such that $KP = KQ = \\sqrt{O_1K^2 - r_1^2}$ so that $O_1P \\cdot O_1Q = r_1^2$. On the other hand,\n\n$$\n\\begin{aligned}\nO_2P \\cdot O_2Q - O_1P \\cdot O_1Q &= O_2K^2 - O_1K^2 \\\\\n&= (O_1K - O_1O_2)^2 - O_1K^2 \\\\\n&= O_1O_2^2 - 2O_1O_2 \\cdot O_1K \\\\\n&= r_2^2 - r_1^2,\n\\end{aligned}\n$$\nwhich implies that $O_2P \\cdot O_2Q = r_2^2$.\n\nFor any given line $\\ell$ as in the problem, if it is perpendicular to $O_1O_2$, we certainly have $\\angle APC = \\angle DPB$ by symmetry. If not, from $O_2C^2 = O_2Q \\cdot O_2P$ we know that $\\triangle O_2CQ \\sim \\triangle O_2PC$. So $\\frac{CQ}{CP} = \\frac{r_2}{O_2P}$. Similarly, we obtain $\\frac{DQ}{DP} = \\frac{r_2}{O_2P}$, and therefore $\\frac{CQ}{CP} = \\frac{DQ}{DP}$, which means that the bisectors of $\\angle CPD$ and $\\angle CQD$ meet $\\ell$ at the same point, say $M$. A similar argument shows that the bisectors of $\\angle APB$ and $\\angle AQB$ intersect $\\ell$ at the same point again, say $M'$.\n\n![](images/CHN_TSExams_2022_2_p2_data_4599e41b63.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17159,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of real numbers $x$ and $y$ which satisfy the following equations:\n\n$$\n\\begin{aligned}\nx^2 + y^2 - 48x - 29y + 714 &= 0 \\\\\n2xy - 29x - 48y + 756 &= 0\n\\end{aligned}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The two equations can be rewritten as\n\n$$\n\\begin{aligned}\n(x - 24)^2 + \\left(y - \\frac{29}{2}\\right)^2 &= \\frac{289}{4} \\\\\n(x - 24)\\left(y - \\frac{29}{2}\\right) &= -30.\n\\end{aligned}\n$$\n\nLet $X = x - 24$ and $Y = y - \\frac{29}{2}$. Then:\n\n$$\nX^2 + Y^2 = \\frac{289}{4} \\tag{1}\n$$\n\n$$\nXY = -30. \\tag{2}\n$$\n\nFrom (2), $Y = -\\frac{30}{X}$. Substitute into (1):\n\n$$\nX^2 + \\frac{900}{X^2} = \\frac{289}{4}\n$$\n\nMultiply both sides by $X^2$:\n\n$$\nX^4 - \\frac{289}{4}X^2 + 900 = 0\n$$\n\nLet $u = X^2$:\n\n$$\nu^2 - \\frac{289}{4}u + 900 = 0\n$$\n\nUsing the quadratic formula:\n\n$$\nu = \\frac{1}{2} \\left( \\frac{289}{4} \\pm \\sqrt{\\left(\\frac{289}{4}\\right)^2 - 4 \\cdot 900} \\right)\n$$\n\nCalculate the discriminant:\n\n$$\n\\left(\\frac{289}{4}\\right)^2 - 3600 = \\frac{83521}{16} - 3600 = \\frac{83521 - 57600}{16} = \\frac{25921}{16}\n$$\n\nSo:\n\n$$\nu = \\frac{1}{2} \\left( \\frac{289}{4} \\pm \\frac{161}{4} \\right)\n$$\n\nThus, $u = X^2 = \\frac{225}{4}$ or $u = 16$, so $X = \\pm\\frac{15}{2}$ or $X = \\pm 4$.\n\nCorrespondingly, $Y = -\\frac{30}{X}$, so $Y = \\mp 4$ or $Y = \\mp\\frac{15}{2}$.\n\nRecall $x = X + 24$ and $y = Y + \\frac{29}{2}$.\n\nTherefore, the solutions are:\n\n- $X = \\frac{15}{2}$, $Y = -4$: $x = 24 + \\frac{15}{2} = \\frac{63}{2}$, $y = \\frac{29}{2} - 4 = \\frac{21}{2}$\n- $X = -\\frac{15}{2}$, $Y = 4$: $x = 24 - \\frac{15}{2} = \\frac{33}{2}$, $y = \\frac{29}{2} + 4 = \\frac{37}{2}$\n- $X = 4$, $Y = -\\frac{15}{2}$: $x = 24 + 4 = 28$, $y = \\frac{29}{2} - \\frac{15}{2} = 7$\n- $X = -4$, $Y = \\frac{15}{2}$: $x = 24 - 4 = 20$, $y = \\frac{29}{2} + \\frac{15}{2} = 22$\n\nSo the solutions are:\n\n$$\n(x, y) \\in \\left\\{ (28, 7),\\ (20, 22),\\ \\left(\\frac{33}{2}, \\frac{37}{2}\\right),\\ \\left(\\frac{63}{2}, \\frac{21}{2}\\right) \\right\\}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17160,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that $a_k > 0$ for $k = 1, 2, \\dots, 2008$.\n\nProve that if and only if $\\sum_{k=1}^{2008} a_k > 1$, there exists a sequence $\\{x_n\\}$ satisfying:\n\n1. $0 = x_0 < x_n < x_{n+1}$ for $n = 1, 2, 3, \\dots$\n2. $\\lim_{n \\to \\infty} x_n$ exists\n3. $x_n - x_{n-1} = \\sum_{k=1}^{2008} a_k x_{n+k} - \\sum_{k=0}^{2007} a_{k+1} x_{n+k}$ for $n = 1, 2, 3, \\dots$",
    "options": [],
    "answer": "See solution",
    "solution": "Proof of necessity: Assume there exists $\\{x_n\\}$ satisfying (1)--(3). Notice that the expression in (3) can be written as\n\n$$\nx_n - x_{n-1} = \\sum_{k=1}^{2008} a_k (x_{n+k} - x_{n+k-1}), \\quad n \\in \\mathbb{N}.\n$$\n\nAs $x_0 = 0$, we have\n\n$$\n\\begin{align*}\nx_n &= \\sum_{l=1}^{n} (x_l - x_{l-1}) = \\sum_{l=1}^{n} \\sum_{k=1}^{2008} a_k (x_{l+k} - x_{l+k-1}) \\\\\n&= \\sum_{k=1}^{2008} \\sum_{l=1}^{n} a_k (x_{l+k} - x_{l+k-1}) \\\\\n&= \\sum_{k=1}^{2008} a_k (x_{n+k} - x_k).\n\\end{align*}\n$$\n\nFrom (2), define $b = \\lim_{n \\to \\infty} x_n$. Let $n \\to \\infty$ in the above expression. Then\n\n$$\nb = \\sum_{k=1}^{2008} a_k (b - x_k) = b \\sum_{k=1}^{2008} a_k - \\sum_{k=1}^{2008} a_k x_k < b \\sum_{k=1}^{2008} a_k.\n$$\n\nTherefore, $\\sum_{k=1}^{2008} a_k > 1$.\n\nProof of sufficiency: Assume $\\sum_{k=1}^{2008} a_k > 1$. Define a polynomial function\n\n$$\nf(s) = -1 + \\sum_{k=1}^{2008} a_k s^k, \\quad s \\in [0, 1].\n$$\n\n$f(s)$ is strictly increasing on $[0, 1]$.\n\n$$\nf(0) = -1 < 0, \\quad f(1) = -1 + \\sum_{k=1}^{2008} a_k > 0,\n$$\n\nso there exists a unique $0 < s_0 < 1$ such that $f(s_0) = 0$.\n\nDefine $x_n = \\sum_{k=1}^{n} s_0^k$, $n \\in \\mathbb{N}$. Clearly, $\\{x_n\\}$ satisfies (1) and\n\n$$\nx_n = \\sum_{k=1}^{n} s_0^k = \\frac{s_0 - s_0^{n+1}}{1 - s_0}.\n$$\n\nAs $0 < s_0 < 1$, $\\lim_{n \\to \\infty} s_0^{n+1} = 0$, so\n\n$$\n\\lim_{n \\to \\infty} x_n = \\frac{s_0}{1 - s_0}.\n$$\n\nThus, $\\{x_n\\}$ satisfies (2). Finally,\n\n$$\n0 = f(s_0) = -1 + \\sum_{k=1}^{2008} a_k s_0^k \\implies \\sum_{k=1}^{2008} a_k s_0^k = 1.\n$$\n\nThen\n\n$$\n\\begin{align*}\nx_n - x_{n-1} &= s_0^n \\\\\n&= \\left( \\sum_{k=1}^{2008} a_k s_0^k \\right) s_0^n \\\\\n&= \\sum_{k=1}^{2008} a_k s_0^{n+k} \\\\\n&= \\sum_{k=1}^{2008} a_k (x_{n+k} - x_{n+k-1}).\n\\end{align*}\n$$\n\nTherefore, $\\{x_n\\}$ also satisfies (3). This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17161,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$ and $y$ be unit complex numbers with $\\arg x, \\arg y \\in (0, \\frac{\\pi}{2})$ and $\\arg y > 2\\arg x$ (so $\\triangle ABC$ is acute and $\\angle ACB > 2\\angle ABC$). Suppose\n\n$$\n|y - 1| = |(x^2 - 1)(y + 1)|. \\qquad \\textcircled{1}\n$$\n\nProve that\n\n$$\n|x^2(y + 1) + (y^2 - y)x - 2| = \\left| \\frac{(x^2 - 1)(1 - y^3)}{x^2 - y} \\right|. \\qquad \\textcircled{2}\n$$\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Proof. Notice that\n\n$$\n\\begin{align*}\n\\textcircled{1} \\Leftrightarrow & \\left| \\frac{(x^2 - 1)(y + 1)}{y - 1} \\right| = 1 \\Leftrightarrow \\frac{(\\bar{x}^2 - 1)(\\bar{y} + 1)}{\\bar{y} - 1} = \\frac{y - 1}{(x^2 - 1)(y + 1)} \\\\\n& \\Leftrightarrow \\frac{\\frac{1 - x^2}{x^2} + \\frac{1 + y}{y}}{\\frac{1 - y}{y}} = \\frac{y - 1}{(x^2 - 1)(y + 1)} \\\\\n& \\Leftrightarrow \\left( \\frac{x^2 - 1}{x} \\right)^2 = \\left( \\frac{y - 1}{y + 1} \\right)^2,\n\\end{align*}\n$$\n\nthat is, $\\frac{x^2 - 1}{x} = \\pm \\frac{y - 1}{y + 1}$. The left-hand side equals $pi$ for some $p > 0$; the right-hand side satisfies $\\text{Im}\\frac{y - 1}{y + 1} > 0$ as $\\arg y \\in (0, \\frac{\\pi}{2})$. Hence, ...",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17162,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_n$ be a sequence defined by $a_n = a_1 + (n-1)d$ for $n \\in \\mathbb{N}$.\n\nSuppose $a_1^2, a_2^2, a_3^2$ are also terms of the same arithmetic progression. Determine all possible values of $a_1$ and $d$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $d = 0$, then $a_n = 0$ for all $n \\in \\mathbb{N}$ or $a_n = 1$ for all $n \\in \\mathbb{N}$.\n\nLet $d \\neq 0$. The condition that $a_1^2, a_2^2, a_3^2$ are terms of the progression means there exist $m, n, k \\in \\mathbb{N}$ such that:\n\n$$\na_1^2 = a_1 + md, \\quad a_2^2 = a_1 + nd = (a_1 + d)^2, \\quad a_3^2 = a_1 + kd = (a_1 + 2d)^2\n$$\n\nThis leads to the system:\n$$\n\\begin{cases}\n2a_1 + d = n - m \\\\\n4a_1 + 4d = k - m\n\\end{cases}\n$$\n\nSolving, we get:\n$$\na_1 = \\frac{4n - 3m - k}{4}, \\quad d = \\frac{m - 2n + k}{2} \\in \\mathbb{Q}.\n$$\n\nOn the other hand, $a_1^2 = a_1 + md$ implies $a_1^2 + (2m - 1)a_1 - m(d + 2a_1) = 0$, so $a_1$ is a rational root of the quadratic polynomial $P(x) = x^2 + (2m - 1)x - m(n - m)$. Since the leading coefficient is $1$, any rational root must be integer. Thus $a_1 \\in \\mathbb{Z}$, and $d = n - m - 2a_1 \\in \\mathbb{Z}$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17163,
    "subject": "Mathematics (Olympiad)",
    "question": "If $a$, $b$, $c$ are real numbers such that $ab + bc + ca = 0$, prove the inequality\n\n$$\n2(a^2 + b^2 + c^2)(a^2b^2 + b^2c^2 + c^2a^2) \\ge 27a^2b^2c^2.\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "The inequality can be written as\n$$\n2(a^4b^2 + a^4c^2 + b^4a^2 + b^4c^2 + c^4a^2 + c^4b^2) \\ge 21a^2b^2c^2.\n$$\n\nIf $abc = 0$, there is nothing to prove. In this case, two of the variables must be $0$, and the inequality is satisfied with equality.\n\nIf $abc \\ne 0$, divide both sides by $a^2b^2c^2$ to get\n$$\n2 \\left[ \\left( \\frac{a^2}{b^2} + \\frac{a^2}{c^2} \\right) + \\left( \\frac{b^2}{c^2} + \\frac{b^2}{a^2} \\right) + \\left( \\frac{c^2}{a^2} + \\frac{c^2}{b^2} \\right) \\right] \\ge 21.\n$$\n\nThat is,\n$$\n\\frac{a^2b^2 + a^2c^2}{b^2c^2} + \\frac{a^2b^2 + b^2c^2}{a^2c^2} + \\frac{a^2c^2 + b^2c^2}{a^2b^2} \\ge \\frac{21}{2}.\n$$\n\nLet $ab = x$, $bc = y$, $ca = z$. Then $x + y + z = 0$, and we want to prove\n$$\n\\frac{x^2 + y^2}{z^2} + \\frac{y^2 + z^2}{x^2} + \\frac{z^2 + x^2}{y^2} \\ge \\frac{21}{2}.\n$$\n\nTwo of $x, y, z$ must have the same sign, the third the opposite. Assume $xy > 0$. Substitute $z = -x - y$; the inequality becomes\n$$\n\\frac{x^2 + y^2}{(x + y)^2} + \\frac{x^2 + 2xy + 2y^2}{x^2} + \\frac{y^2 + 2xy + 2x^2}{y^2} \\ge \\frac{21}{2},\n$$\nor\n$$\n\\frac{x^2 + y^2}{(x + y)^2} + 2\\left(\\frac{x^2}{y^2} + \\frac{y^2}{x^2}\\right) + 2\\left(\\frac{x}{y} + \\frac{y}{x}\\right) \\ge \\frac{17}{2}.\n$$\n\nUsing the inequalities $\\frac{x^2 + y^2}{(x + y)^2} \\ge \\frac{1}{2}$, $\\frac{x^2}{y^2} + \\frac{y^2}{x^2} \\ge 2$, and $\\frac{x}{y} + \\frac{y}{x} \\ge 2$ (all with equality if and only if $x = y$), the result follows.\n\nEquality holds if $ab = ac$ and $ab + bc + ca = 0$, i.e., when $b = c = -2a$.\n\nIn conclusion, equality holds for triples $(\\alpha, -2\\alpha, -2\\alpha)$, $(\\alpha, 0, 0)$, $\\alpha \\in \\mathbb{R}$, and their permutations.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17164,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be positive real numbers such that $a + b + c = 1$. Prove that\n$$\n\\frac{a^2}{b^3 + c^4 + 1} + \\frac{b^2}{c^3 + a^4 + 1} + \\frac{c^2}{a^3 + b^4 + 1} > \\frac{1}{5}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We have $a, b, c \\in (0, 1)$. Therefore, $b^3 < b$ and $c^4 < c$, so $b^3 + c^4 + 1 < b + c + 1 = 1 - a + 1 = 2 - a$. Thus,\n$$\n\\frac{a^2}{b^3 + c^4 + 1} > \\frac{a^2}{2 - a}.\n$$\nSimilarly,\n$$\n\\frac{b^2}{c^3 + a^4 + 1} > \\frac{b^2}{2 - b}, \\quad \\frac{c^2}{a^3 + b^4 + 1} > \\frac{c^2}{2 - c}.\n$$\nAdding these inequalities gives\n$$\n\\frac{a^2}{b^3 + c^4 + 1} + \\frac{b^2}{c^3 + a^4 + 1} + \\frac{c^2}{a^3 + b^4 + 1} > \\frac{a^2}{2 - a} + \\frac{b^2}{2 - b} + \\frac{c^2}{2 - c}.\n$$\nNow, note that for $x \\in (0, 1)$, $\\frac{x^2}{2 - x} = -2x - x^2 + \\frac{4x}{2 - x}$. Thus,\n$$\n\\frac{a^2}{2 - a} = -2a - a^2 + \\frac{4a}{2 - a},\n$$\nand similarly for $b$ and $c$. Summing, and using $a + b + c = 1$:\n$$\n\\sum \\frac{x^2}{2 - x} = -2(a + b + c) - (a^2 + b^2 + c^2) + 4 \\left( \\frac{a}{2 - a} + \\frac{b}{2 - b} + \\frac{c}{2 - c} \\right).\n$$\nBut $a^2 + b^2 + c^2 < 1$, so $-2(a + b + c) - (a^2 + b^2 + c^2) > -3$. For a lower bound, we use the AM-HM inequality:\n$$\n\\frac{1}{2 - a} + \\frac{1}{2 - b} + \\frac{1}{2 - c} \\geq 3 \\cdot \\frac{3}{(2 - a) + (2 - b) + (2 - c)} = \\frac{9}{6 - (a + b + c)} = \\frac{9}{5}.\n$$\nTherefore,\n$$\n\\frac{a^2}{b^3 + c^4 + 1} + \\frac{b^2}{c^3 + a^4 + 1} + \\frac{c^2}{a^3 + b^4 + 1} > -7 + 4 \\cdot \\frac{9}{5} = \\frac{1}{5}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17165,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be the point of intersection of the diagonals of a cyclic quadrilateral $ABCD$. Let $I_1$ and $I_2$ be the incenters of triangles $AMD$ and $BMC$, respectively, and let $L$ be the point of intersection of the lines $DI_1$ and $CI_2$. The foot of the perpendicular from the midpoint $T$ of $I_1I_2$ to $CL$ is $N$, and $F$ is the midpoint of $TN$. Let $G$ and $J$ be the points of intersection of the line $LF$ with $I_1N$ and $I_1I_2$, respectively. Let $O_1$ be the circumcenter of triangle $LI_1J$, and let $\\Gamma_1$ and $\\Gamma_2$ be the circles with diameters $O_1L$ and $O_1J$, respectively. Let $V$ and $S$ be the second points of intersection of $I_1O_1$ with $\\Gamma_1$ and $\\Gamma_2$, respectively. If $K$ is the point where the circles $\\Gamma_1$ and $\\Gamma_2$ meet again, prove that $K$ is the circumcenter of the triangle $SVG$.\n\n![](images/Balkan_2012_shortlist_p18_data_2959ef6bc1.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The point $L$ is the midpoint of the arc $AB$ and, as $I_1$, $M$, $I_2$ are collinear, we have $\\angle LI_2I_1 = \\angle I_2CM + \\angle I_2MC = \\angle I_1DM + \\angle I_1MD = \\angle LI_1I_2$. Therefore, the triangle $LI_1I_2$ is isosceles and $LT \\perp I_1I_2$.\n\nLet $Z$ be the midpoint of $NI_2$. Then $FZ \\parallel I_1I_2$, so $FZ \\perp LT$ and it follows that $F$ is the orthocenter of the triangle $LTZ$. Therefore, we have $LG \\perp I_1N$.\n\nLet $X$ be the orthogonal projection of $O_1$ to $I_1N$ and $H$ be the other endpoint of the diameter of $\\Gamma_1$ through $K$. Since $O_1L$ and $O_1J$ are diameters, $O_1K \\perp LG$ and $O_1K \\parallel I_1G$. Since $HVKO_1$ is cyclic, $\\angle VHK = \\angle VO_1K = \\angle O_1I_1X$. As we also have $\\angle HVK = \\angle O_1XI_1 = 90^\\circ$ and $HK = O_1L = O_1I_1$, the triangles $HKV$ and $I_1O_1X$ are congruent and $VK = O_1X = KG$.\n\nOn the other hand, since the quadrilaterals $LVKO_1$ and $SKJO_1$ are cyclic, we have\n\n$$\n\\angle O_1VK = \\angle O_1LK = \\angle O_1JK = \\angle VSK,\n$$\n\nso the triangle $\\triangle SKV$ is isosceles and $VK = SK$. Hence $K$ is the circumcenter of the triangle $SVG$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17166,
    "subject": "Mathematics (Olympiad)",
    "question": "The bisectors of the sides $AC$ and $BC$ of the acute triangle $ABC$ with $CB = 30^\\circ$ intersect at point $O$. The points $M$ and $N$ on the sides $AC$ and $BC$, respectively, are such that $O$ is the midpoint of the segment $MN$. How many times is the product of the lengths of segments $CM$ and $CN$ greater than the product of the lengths of segments $AM$ and $BN$?\n\n(Miroslav Marinov)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $L$ be the midpoint of $BC$ (with $OL \\perp BC$), and $K$ be the foot of the perpendicular from $M$ to $BC$. Then $OL \\parallel MK$, and with $MO = ON$, it follows that $OL$ is a midsegment in triangle $KMN$—in particular, $KL = LN$. On the other hand, we also have $BL = CL$, whence $BN = CK$.\n\nFrom the right triangle $CKM$ with $CM = 30^\\circ$, we get $KM = \\frac{1}{2}CM$,\nand $CK = \\sqrt{CM^2 - KM^2} = \\frac{CM}{\\sqrt{3}}$ from the Pythagorean theorem.\nTherefore, $\\frac{CM}{BN} = \\frac{2}{\\sqrt{3}}$.\n\nAnalogously, we have $\\frac{CN}{AM} = \\frac{2}{\\sqrt{3}}$, from where finally\n$$\n\\frac{CM \\cdot CN}{AM \\cdot BN} = \\frac{4}{3}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17167,
    "subject": "Mathematics (Olympiad)",
    "question": "As seen in Fig. 1.1, the circumcenter of acute triangle $ABC$ is $O$, $K$ is a point (not the midpoint) on the side $BC$, $D$ is a point on the extended line of segment $AK$, lines $BD$ and $AC$ intersect at point $N$, and lines $CD$ and $AB$ intersect at point $M$. Prove that if $OK \\perp MN$, then $A, B, D, C$ are concyclic.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p67_data_b686bab239.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Assume, for contradiction, that $A, B, D, C$ are not concyclic. Let the circumcircle (with radius $r$) of $ABC$ intersect $AD$ at point $E$. Join $BE$ and extend it to intersect line $AN$ at point $Q$; join $CE$ and extend it to intersect $AM$ at $P$. Join $PQ$, as seen in Fig. 1.2.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p68_data_5f37d960a9.png)\n\nWe have\n\n$$\n\\begin{aligned}\nPK^2 &= \\text{the power of } P \\text{ with respect to } \\odot O + \\\\\n      &\\quad \\text{the power of } Q \\text{ with respect to } \\odot O \\\\\n      &= (PO^2 - r^2) + (KO^2 - r^2).\n\\end{aligned}\n$$\n\n(We will prove this in the appendix.)\n\nSimilarly,\n\n$$\nQK^2 = (QO^2 - r^2) + (KO^2 - r^2).\n$$\n\nThen,\n\n$$\nPO^2 - PK^2 = QO^2 - QK^2.\n$$\n\nTherefore, $OK \\perp PQ$. By the given condition $OK \\perp MN$, we get that $PQ \\parallel MN$. Then,\n\n$$\n\\frac{AQ}{QN} = \\frac{AP}{PM}. \\qquad \\textcircled{1}\n$$\n\nBy Menelaus' Theorem,\n\n$$\n\\frac{NB}{BD} \\cdot \\frac{DE}{EA} \\cdot \\frac{AQ}{QN} = 1, \\qquad \\textcircled{2}\n$$\n\n$$\n\\frac{MC}{CD} \\cdot \\frac{DE}{EA} \\cdot \\frac{AP}{PM} = 1. \\qquad \\textcircled{3}\n$$\n\nFrom (1), (2), and (3), we get $\\frac{NB}{BD} = \\frac{MC}{CD}$, or $\\frac{ND}{BD} = \\frac{MD}{DC}$. Thus, $\\triangle DMN \\sim \\triangle DCB$, which implies $\\angle DMN = \\angle DCB$. Then $BC \\parallel MN$. Therefore, $OK \\perp BC$, which means $K$ is the midpoint of $BC$, a contradiction. This completes the proof that $A, B, D, C$ are concyclic.\n\n**Appendix.**\n\nWe are going to prove that\n\n$PK^2$ equals the power of $P$ with respect to $\\odot O$ plus the power of $Q$ with respect to $\\odot O$.\n\nExtend $PK$ to point $F$, such that\n\n$$\nPK \\cdot KF = AK \\cdot KE \\qquad (4)\n$$\n\n(see Fig. 1.3). Then $P, E, F, A$ are concyclic, and\n\n$$\n\\angle PFE = \\angle PAE = \\angle BCE.\n$$\n\nThen $E, C, F, K$ are concyclic, and\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p69_data_4e7f8961ba.png)\n\n$$\nPK \\cdot PF = PE \\cdot PC. \\qquad (5)\n$$\n\nFrom (5) and (4), we get $PK^2 = PE \\cdot PC - AK \\cdot KE$, which is the power of $P$ with respect to $\\odot O$ plus the power of $Q$ with respect to $\\odot O$.\n\n**Remark.** If $E$ is on the extended line of $AD$, the proof is similar.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17168,
    "subject": "Mathematics (Olympiad)",
    "question": "For each positive real number $x$, let $f(x)$ be a positive real number. Suppose\n\n$$\nf(x)y = f(xf(y)) \\quad \\text{for all } x > 0 \\text{ and all } y > 0.\n$$\n\nProve that $f(xy) = f(x)f(y)$ for all $x > 0$ and all $y > 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = 1$. Then $f(1)y = f(f(y))$ for all $y > 0$.\n\nSetting $y = 1$ gives $f(1) = f(f(1))$.\n\nSetting $y = f(1)$ gives $(f(1))^2 = f(f(f(1)))$.\n\nBut $f(f(f(1))) = f(f(1)) = f(1)$, so $(f(1))^2 = f(1)$. Since $f(1) > 0$, we have $f(1) = 1$.\n\nThus, $f(f(y)) = f(1)y = y$ for all $y > 0$.\n\nNow, in the original equation, replace $y$ by $f(y)$:\n\n$$\nf(x)f(y) = f(xf(f(y))) = f(xy)\n$$\n\nfor all $x > 0$ and $y > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17169,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle, and let $D$, $E$, $F$ be the feet of its altitudes from $A$, $B$, $C$, respectively. The lines $AB$ and $DE$ cross at $K$, and the lines $AC$ and $DF$ cross at $L$. Let $M$ be the midpoint of the side $BC$, and let the line $AM$ cross the circle $ABC$ again at $N$. Finally, the parallel through $M$ to $EF$ crosses the line $KL$ at $P$. Show that the triangle $MNP$ is isosceles.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\Gamma$ be the circle $ABC$, let $\\gamma$ be the nine-point circle of the triangle $ABC$, and let $\\omega$ be the image of $\\Gamma$ under the homothety from $N$ mapping $A$ to $M$.\n\n![](images/RMC_2024_p82_data_a54627b5d9.png)\n\nWe will show that $PM$ and $PN$ are the tangents from $P$ to $\\omega$; the conclusion then follows at once.\n\nWe first prove that $PM$ is the tangent of $\\omega$ at $M$. The tangent of $\\omega$ at $M$ is parallel to the tangent of $\\Gamma$ at $A$, which is in turn parallel to $EF$. Since $MP$ is the parallel through $M$ to $EF$, it follows that $PM$ is the tangent of $\\omega$ at $M$.\n\nNext, we show that $PM$ is the tangent of $\\gamma$ at $M$. Since the factor $-\\frac{1}{2}$ homothety from the barycentre of the triangle $ABC$ exchanges the pairs $(A, \\Gamma)$ and $(M, \\gamma)$, the tangent of $\\gamma$ at $M$ is parallel to the tangent of $\\Gamma$ at $A$; and since the latter is parallel to $EF$, so is the former. Consequently, $PM$, the parallel through $M$ to $EF$, is the tangent of $\\gamma$ at $M$.\n\nBy the preceding, $PM$ is the radical axis of $\\gamma$ and $\\omega$. Since $\\omega$ is a homothetic image of $\\Gamma$ from $N$, their radical axis is their common tangent at $N$.\n\nConsequently, $PN$ is the other tangent of $\\omega$ from $P$ if and only if $P$ is the radical centre of $\\Gamma$, $\\gamma$ and $\\omega$, which is the case if and only if $P$ lies on the radical axis of $\\Gamma$ and $\\gamma$. Since $P$ lies on $KL$, it is sufficient to show that $KL$ is radical axis of $\\Gamma$ and $\\gamma$.\n\nFinally, we prove that $KL$ is radical axis of $\\Gamma$ and $\\gamma$. The powers of $K$ with respect to $\\Gamma$ and $\\gamma$ are $KA \\cdot KB$ and $KD \\cdot KE$, respectively. The two are equal since $ABDE$ is cyclic, so $K$ lies on the radical axis of $\\Gamma$ and $\\gamma$. Similarly, $L$ lies on their radical axis, so $KL$ is the radical axis of the two circles. This completes the solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17170,
    "subject": "Mathematics (Olympiad)",
    "question": "A rectangle with side lengths 1 and 3, a square with side length 1, and a rectangle $R$ are inscribed inside a larger square as shown.\n\n![](images/2021_AMC10B_Solutions_Fall_p15_data_2101e3e818.png)\n\nThe sum of all possible values for the area of $R$ can be written in the form $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m + n$?\n\n(A) 14 (B) 23 (C) 46 (D) 59 (E) 67",
    "options": [],
    "answer": "See solution",
    "solution": "Label the diagram as shown below, where $O$, $P$, and $Q$ are the feet of the perpendicular segments to the respective sides. Let $x = AE$ and $y = AF$.\n\n![](images/2021_AMC10B_Solutions_Fall_p16_data_9e8fc37b85.png)\n\nBecause\n\n$$\n\\triangle FAE \\cong \\triangle EOH \\cong \\triangle IOH \\cong \\triangle KCJ \\cong \\triangle FPG,\n$$\n\nit follows that $EO = IO = KC = y$ and $CJ = AE = x$. Furthermore, $\\triangle JDI$ is similar to all these triangles with scale factor 3, so $ID = 3x$ and $DJ = 3y$. Therefore\n\n$$\n3y + x = DC = DA = 2y + 4x,\n$$\n\nso $y = 3x$. Applying the Pythagorean Theorem to $\\triangle AEF$ yields $x = \\frac{1}{\\sqrt{10}}$.\n\nNow set $PL = ax$ and $LB = bx$ for some positive real numbers $a$ and $b$. Note that $MQ = PG = y = 3x$, and because $\\triangle GPL \\sim \\triangle LBM$, it follows that $BM = \\frac{1}{3}abx$. Furthermore, $\\triangle NQK \\sim \\triangle KCJ$, and because $NQ = PL = ax$, it follows that $QK = \\frac{1}{3}ax$. Therefore\n\n$$\n3x + \\frac{1}{3}ax + 3x + \\frac{1}{3}abx = BC = AD = 10x,\n$$\n\nwhich simplifies to $a(b + 1) = 12$. Because $PB = AB - AP = 6x$, it follows that $a + b = 6$. Solving these equations simultaneously for $a$ and $b$ shows that $(a, b)$ equals either $(3, 3)$ or $(4, 2)$.\n\nFinally, observe that the area of $R$ is\n\n$$\n\\begin{aligned}\nLG \\cdot LM &= \\sqrt{(ax)^2 + (3x)^2} \\cdot \\sqrt{(bx)^2 + \\left(\\frac{1}{3}abx\\right)^2} \\\\\n&= \\sqrt{(a^2 + 9)\\left(1 + \\frac{1}{9}a^2\\right)} \\cdot \\frac{b}{10}.\n\\end{aligned}\n$$\n\nWhen $(a, b) = (3, 3)$, this area equals $\\frac{9}{5}$; and when $(a, b) = (4, 2)$, this area equals $\\frac{5}{3}$. The sum of these two areas is $\\frac{9}{5} + \\frac{5}{3} = \\frac{52}{15}$, and the requested sum of numerator and denominator is $52 + 15 = 67$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17171,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an acute triangle $ABC$ with circumcircle $(O)$ such that $B$ and $C$ are fixed points and $A$ is a moving point on $(O)$. Points $M$ and $N$ lie on rays $AB$ and $AC$ such that $MA = MC$ and $NA = NB$. The circumcircles of triangles $AMN$ and $ABC$ meet again at $P \\ne A$. Line $MN$ intersects $BC$ at $Q$.\n\n**a)** Prove that $A$, $P$, and $Q$ are collinear.\n\n**b)** Let $D$ be the midpoint of $BC$, and let $K$ be the intersection of the circles $(M, MA)$ and $(N, NA)$ with $K \\ne A$. The line passing through $A$ and perpendicular to $AK$ meets $BC$ at $E$, and the circumcircle of triangle $ADE$ meets $(O)$ again at $F$. Prove that $AF$ passes through a fixed point.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Without loss of generality, suppose $AB \\leq AC$; the other case is similar. Clearly, $M$ belongs to segment $AB$ and $N$ to segment $AC$. From $NA = NB$, we have $\\angle NBA = \\angle NAB$, so $MA = MC$ and $\\angle MCA = \\angle MAC$. Hence, $\\angle NBA = \\angle MCA$, so $BMCN$ is a cyclic quadrilateral and $QM \\cdot QN = QB \\cdot QC$.\n\nTherefore, $Q$ lies on the radical axis of the two circles $(O)$ and $(AMN)$. This radical axis is $AP$, so $A$, $P$, and $Q$ are collinear.\n\n![](images/Vietnamese_mathematical_competitions_p89_data_f8c8b34709.png)\n\nb) The circle $(ODC)$ is tangent to $(O)$ at $C$, so their radical axis is the tangent $d$ to $(O)$ at $C$.\n\n$O$ and $M$ both lie on the perpendicular bisector of $AC$, so $OM \\perp AC$. Similarly, $ON \\perp AB$, so $O$ is the orthocenter of triangle $AMN$. Hence, $AO \\perp MN$. Considering the circles $(M, MA)$ and $(N, NA)$, since $AK$ is their common chord, $AK \\perp MN$. Thus, $A$, $O$, and $K$ are collinear, so $\\angle OAE = 90^\\circ$. Furthermore, $\\angle ODE = 90^\\circ$, so $AODE$ is cyclic and $O \\in (ADE)$. Thus, the radical axis of $(ADE)$ and $(ODC)$ is $OD$. On the other hand, the radical axis of $(O)$ and $(ADE)$ is $AF$. Considering the three circles $(O)$, $(ADE)$, and $(ODC)$, their radical axes $OD$, $d$, and $AF$ are concurrent. Therefore, $AF$ passes through the intersection of $AD$ and $d$, which is a fixed point. $\\square$\n\n![](images/Vietnamese_mathematical_competitions_p90_data_ebb7437191.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17172,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, and $C$ be the numbers of participants who solved just one problem $A$, $B$, and $C$ respectively. Let $AB$ be the number of participants who solved only $A$ and $B$, and so on. Then:\n\n$$\n\\begin{cases}\nA + B + C + AB + AC + BC + ABC = 25 \\\\ \nB + BC = 2(C + BC) \\\\ \nA - 1 = AB + AC + ABC \\\\ \nA + B + C = 2(B + C)\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "From (1) and (3), we have $2A + B + C + BC = 26$ (5).\n\nEquation (4) is equivalent to $A = B + C$.\n\nEquation (2) is equivalent to $BC = B - 2C$.\n\nFrom (5), (2), and (4):\n\n$4B + C = 26$ implies $C = 26 - 4B$ (6).\n\n$BC = B - 2C = B - 2(26 - 4B) = 9B - 52$ (7).\n\nConsidering $B$, $C$, $BC$ are non-negative integers, $B = 6$ will imply from (6) and (7).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17173,
    "subject": "Mathematics (Olympiad)",
    "question": "A rectangular grid is divided by two perpendicular straight lines into four smaller rectangles with integral side lengths. It is possible to remove one among these four rectangles in such a way that the remaining figure can be exactly covered by rectangles of size $2 \\times 3$ and $3 \\times 2$. Prove that it is possible to exactly cover one among these four smaller rectangles by rectangles of size $2 \\times 3$ and $3 \\times 2$. (By *exact covering* we mean covering without gaps, overlaps and overflows.)",
    "options": [],
    "answer": "See solution",
    "solution": "Call a figure *coverable* if it can be exactly covered by rectangles of size $2 \\times 3$ and $3 \\times 2$. Let the original rectangle be of size $(a+d) \\times (b+c)$ and let the figure remaining after cutting out the upper left corner of size $a \\times b$ be *coverable*.\n\n![](images/EST_ABooklet_2024_p63_data_c6bec025ff.png)\n\nThis would be impossible if $c=1$ or $d=1$, so we can assume $c \\ge 2$ and $d \\ge 2$. Coverability implies that $6 \\mid ac+bd+cd$. Also, a rectangle of integral side lengths $n \\times m$ where $n \\ge 2$ and $m \\ge 2$ is *coverable* if and only if $6 \\mid nm$. Indeed, rectangles of size $2k \\times 3l$ and $3l \\times 2k$ are obviously *coverable*,\n\nwhereas rectangles of size $6k \\times (6l \\pm 1)$ can be divided into rectangles of size $6k \\times 3$ and $6k \\times 2s$, both of which are coverable by the above; the symmetric case is analogous.\n\nSince $2 \\mid ac + bd + cd$, at least one of $a, b, c,$ and $d$ must be even. Next, we show that at least one of $a, b, c,$ and $d$ must be divisible by 3. Suppose the contrary, i.e., none of $a, b, c,$ and $d$ is divisible by 3. If either $3 \\mid a + d$ or $3 \\mid b + c$, then $3 \\mid bd$ or $3 \\mid ac$, respectively, because $3 \\mid ac + bd + cd$. Since 3 is prime, we obtain a contradiction. Hence $3 \\nmid a + d$ and $3 \\nmid b + c$. Thus either $a \\equiv d \\equiv 1 \\pmod{3}$ or $a \\equiv d \\equiv 2 \\pmod{3}$, as well as either $b \\equiv c \\equiv 1 \\pmod{3}$ or $b \\equiv c \\equiv 2 \\pmod{3}$. If either $a \\equiv b \\equiv c \\equiv d \\equiv 1 \\pmod{3}$ or $a \\equiv b \\equiv c \\equiv d \\equiv 2 \\pmod{3}$, then color the squares of the figure by the rising diagonals with 3 colors. If either $a \\equiv d \\equiv 1 \\pmod{3}$ and $b \\equiv c \\equiv 2 \\pmod{3}$ or $a \\equiv d \\equiv 2 \\pmod{3}$ and $b \\equiv c \\equiv 1 \\pmod{3}$, then color the squares of the figure by the falling diagonals with 3 colors. The area to be covered, except for the neighborhood of the meeting point of the cutting lines of the original rectangle (which can be of four different shapes, marked by bullets and framed with bold line in Figures 53–56), can be cut into strips of size $3k \\times 1$ and $1 \\times 3l$ containing an equal amount of unit squares of each color. Hence the area that should be coverable by assumption contains different numbers of unit squares of different colors, which is impossible as every $2 \\times 3$ and $3 \\times 2$ rectangle contains the same number of unit squares of each color. As we got a contradiction in every case, we have proven that at least one of $a, b, c,$ and $d$ is divisible by 3.\n\n![](images/EST_ABooklet_2024_p64_data_8a6973e0c1.png)\n![](images/EST_ABooklet_2024_p64_data_efd9effcbe.png)\n![](images/EST_ABooklet_2024_p64_data_eb0e4d2976.png)\n![](images/EST_ABooklet_2024_p64_data_d3c9633958.png)\n\nIt remains to show that one of the small rectangles is coverable whenever one of $a, b, c,$ and $d$ is divisible by 2 and also one of $a, b, c,$ and $d$ is divisible by 3. The following table shows for each case which of the small rectangles is coverable:\n\n$$\n\\begin{array}{l|cccc}\n & 3 \\mid a & 3 \\mid b & 3 \\mid c & 3 \\mid d \\\\\n\\hline\n2 \\mid a & a \\times c & a \\times b & a \\times c & a \\times c \\\\\n2 \\mid b & a \\times b & d \\times b & d \\times b & d \\times b \\\\\n2 \\mid c & a \\times c & d \\times b & d \\times c & d \\times c \\\\\n2 \\mid d & a \\times c & d \\times b & d \\times c & d \\times c \\\\\n\\end{array}\n$$\n\nOne can show this by the following case study:\n\n* If one of $a$ and $d$ is divisible by one of 2 and 3 and one of $b$ and $c$ is ...",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17174,
    "subject": "Mathematics (Olympiad)",
    "question": "A _lattice point_ in the Cartesian plane is a point whose coordinates are both integral. A _lattice polygon_ is a polygon whose vertices are lattice points. Let $\\Gamma$ be a convex lattice polygon. Prove that $\\Gamma$ is contained in a convex lattice polygon $\\Delta$ exactly one vertex of which is not a vertex of $\\Gamma$, and the vertices of $\\Gamma$ all lie on the boundary of $\\Delta$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $T$ be the extra vertex of a desired polygon $\\Delta$; then $\\Delta$ is the convex hull of $T$ and $\\Gamma$. Thus, a point $T$ fits the bill if and only if this convex hull contains no vertices of $\\Gamma$ in its interior.\n\nEach segment $AB$ joining two lattice points is partitioned by lattice points into congruent _elementary segments_. Define the _elementary length_ $\\ell(AB)$ of $AB$ to be the length of any of those elementary segments.\n\nTake any three consecutive sides $AB$, $BC$, and $CD$ on the boundary of $\\Gamma$. Consider the convex region bounded by the segment $BC$ and the rays complementary to the rays $BA$ and $CD$ (including the boundary). If this region is unbounded (see the left figure below), then it contains some lattice point $T$ (e.g., the point with $\\overrightarrow{BT} = \\overrightarrow{AB}$), and any such point $T$ satisfies the problem requirements. Thus, in what follows we assume that the two rays cross each other at some point $X$. Assume further that the triangle $BCX$ contains no lattice points outside the segment $BC$, as any other such point would satisfy the requirements.\n\nLet $C_*$ be a lattice point on the segment $BC$ such that $BC_* = \\ell(BC)$, let $X_*$ be the point on $BX$ such that $C_*X_* \\parallel CX$, and let $D_*$ be the point such that $\\overrightarrow{C_*D_*} = \\overrightarrow{CD}$ (see the right figure below). Then the triangle $BC_*X_*$ contains no lattice points apart from $B$ and $C_*.$\n\nConsider the half-plane determined by the line $BC$ and containing no interior points of $\\Gamma$. Let $\\ell$ be the line in that half-plane parallel to $BC$, containing some lattice points, and nearest to $BC$ among such. Let the ray $AB$ meet $\\ell$ at a point $A'$ which belongs to the elementary segment $KL$ on $\\ell$ (we assume that $\\overrightarrow{KL} = \\overrightarrow{BC}_*$; the point $A'$ may coincide with $L$ but not with $K$). Then the ray $D_*C_*$ crosses the ray $LK$ (excluding $L$), otherwise $L$ lies in the triangle $BC_*X_*.$\n\nThe only lattice points contained in the parallelogram $BKLC_*$ are its vertices. This yields that there are no lattice points strictly inside the strip defined by the parallel lines $BK$ and $C_*L$.\n\n![](images/RMC_2020_p66_data_c452b89e37.png)\n\nLet $M$ and $N$ be the meeting points of the rays $BX_*, C_*L$ and $C_*X_*, BK$, respectively. Then the segments $BM$ and $C_*N$ contain no lattice points except their endpoints, so $\\ell(AB) \\ge BM$ and $\\ell(DC) = \\ell(D_*C_*) \\ge C_*N$. Therefore,\n\n$$\n\\frac{BX_*}{\\ell(AB)} + \\frac{C_*X_*}{\\ell(CD)} \\le \\frac{BX_*}{BM} + \\frac{C_*X_*}{C_*N} = \\frac{BX_*}{BM} + \\frac{MX_*}{BM} = 1. \\quad (*)\n$$\n\nChoose now $BC$ to be a side of largest elementary length. Then\n\n$$\n(1 \\ge) \\frac{BX_*}{\\ell(AB)} + \\frac{C_*X_*}{\\ell(CD)} \\ge \\frac{BX_* + C_*X_*}{\\ell(BC)} = \\frac{BX_* + C_*X_*}{BC_*},\n$$\n\nwhich contradicts the triangle inequality.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17175,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ that satisfy\n\n$$\nf(f(x) + y) + x f(y) = f(xy + y) + f(x)\n$$\n\nfor all real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P(x, y)$ denote the assertion of the given functional equation.\n\n**Claim 1:** $f(0) = 0$.\n\n_Proof._ Note that $P(0, y)$ and $P(x, 0)$ give us:\n\n$$\n\\begin{aligned}\nf(y + f(0)) &= f(y) + f(0) \\\\\nf(f(x)) + x f(0) &= f(0) + f(x).\n\\end{aligned}\n$$\n\nConsider the first expression. Plugging $y = -f(0)$ in it yields\n\n$$\nf(0) = f(-f(0)) + f(0), \\text{ i.e. } f(-f(0)) = 0.\n$$\n\nIf we denote $-f(0) = a$, then $f(a) = 0$. Plugging $x = a$ in the second expression gives:\n\n$$\nf(f(a)) + a f(0) = f(0) + f(a), \\text{ i.e. } a f(0) = 0.\n$$\n\nThis either means $a = 0$, i.e. $f(0) = 0$, or $f(0) = 0$. In both cases the claim is proved. $\\square$\n\nSince $f(0) = 0$, the expression $P(x, 0)$ becomes\n\n$$\nf(f(x)) = f(x). \\tag{*}\n$$\n\n**Claim 2:** $f(1) = 1$ or $f(x) = 0$ for all real numbers $x$.\n\n_Proof._ Consider $P(x, 1)$:\n\n$$\nf(f(x) + 1) + x f(1) = f(x + 1) + f(x).\n$$\n\nReplacing $x$ by $f(x)$ and using $(*)$ leads to:\n\n$$\n\\begin{aligned}\nf(f(f(x)) + 1) + f(x) f(1) &= f(f(x) + 1) + f(f(x)) \\\\\nf(f(x) + 1) + f(x) f(1) &= f(f(x) + 1) + f(x) \\\\\nf(x) f(1) &= f(x).\n\\end{aligned}\n$$\n\nSuppose there does not exist $b$ such that $f(b) \\neq 0$, then $f(x) = 0$ for all $x$. Otherwise $f(b) f(1) = f(b)$ implies $f(1) = 1$ as desired. $\\square$\n\n**Claim 3:** If $f(1) = 1$ and $f(a) = 0$, then $a = 0$.\n\n_Proof._ Suppose $f(a) = 0$ for some $a$. Then $P(a, 1)$ gives\n\n$$\n\\begin{aligned}\nf(f(a) + 1) + a f(1) &= f(a + 1) + f(a) \\\\\nf(1) + a &= f(a + 1) = a + 1.\n\\end{aligned}\n$$\n\nOn the other hand, $P(1, a)$ leads to:\n\n$$\n\\begin{aligned}\nf(f(1) + a) + f(a) &= f(2a) + f(1) \\\\\nf(a + 1) &= f(2a) + 1 \\\\\na + 1 &= f(2a) + 1 \\\\\nf(2a) &= a.\n\\end{aligned}\n$$\n\nTaking $f$ from both sides in the last relation and using $(*)$ leads to:\n\n$$\n0 = f(a) = f(f(2a)) = f(2a) = a.\n$$\n\nThis proves the claim.\n\n![](images/BW23_Shortlist_2023-11-01_p31_data_07918408ca.png)\n\nTo finish the problem, consider $P(x, x - f(x))$:\n\n$$\nx f(x - f(x)) = f((x - f(x)) (x + 1)).\n$$\n\nSetting $x = -1$ gives\n\n$$\n- f(-1 - f(-1)) = f(0) = 0.\n$$\n\nFrom Claim 3 for $f \\neq 0$ we obtain $-1 - f(-1) = 0$ implies $f(-1) = -1$. Now looking at $P(-1, y)$ and replacing $y$ by $y + 1$, we get\n\n$$\nf(y - 1) = f(y) - 1 \\text{ implies } f(y + 1) = f(y) + 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17176,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with $AB > AC$. Let $D$, $E$, and $F$ denote the feet of its altitudes on $BC$, $AC$, and $AB$, respectively. Let $S$ denote the intersection of lines $EF$ and $BC$.\n\nProve that the circumcircles $k_1$ and $k_2$ of the triangles $AEF$ and $DES$ touch at $E$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $t_1$ be the tangent line to $k_1$ at point $E$ and let $t_2$ be the tangent line to $k_2$ at point $E$. The tangent-secant theorem applied to circle $k_1$ gives\n\n$$\n\\angle(EF, t_1) = \\angle FAE = \\alpha\n$$\n\nwith the usual notation for the angles in triangle $ABC$.\n\nThe tangent-secant theorem applied to circle $k_2$ gives\n\n$$\n\\angle(EF, t_2) = \\angle SDE = \\angle CDE = \\alpha,\n$$\n\nwhere the last equality comes from the fact that $ABDE$ is a cyclic quadrilateral since all four vertices lie on the Thales circle with diameter $AB$.\n\nTherefore, $t_1$ and $t_2$ are parallel and they both contain the point $E$. So, the two tangents are identical, which implies that the circles touch at $E$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17177,
    "subject": "Mathematics (Olympiad)",
    "question": "A triangle $ABC$ is given. The circle $w$ with center at the point $Q$ touches the side $BC$ and is internally tangent to the circumscribed circle of $\\Delta ABC$ at the point $A$. Let $M$ be the midpoint of the side $BC$, and $N$ be the midpoint of the arc $BAC$ of the circumscribed circle of $\\Delta ABC$. On the side $BC$, the point $S$ is selected such that $\\angle BAM = \\angle SAC$. Prove that the points $N$, $Q$, and $S$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $\\Gamma$ the circumscribed circle of $\\triangle ABC$. Let $W$ be the midpoint of the smaller arc $BC$ of the circle $\\Gamma$, and $O$ be the center of $\\Gamma$. Obviously, points $N$, $M$, $W$, $O$ lie on the perpendicular bisector of the segment $BC$. By the Lemma of Archimedes, the points of contact of $w$ to $BC$, to $\\Gamma$, and the point $W$ are collinear. As $w$ is tangent to the circle $\\Gamma$ at the point $A$, the circle $w$ touches $BC$ at the point $L$, which is the intersection of $BC$ and $AW$. Note that\n\n![](images/Ukraine_2016_Booklet_p41_data_b4392e2e34.png)\n\n$L$ is the base of the bisector of the angle $\\angle BAC$. Let us draw the rays $AM$ and $AS$ until they intersect with the circle $\\Gamma$ at the points $X$ and $P$ respectively. By the statement of the problem, $\\angle BAX = \\angle PAC$, hence $P$ and $X$ are symmetric with respect to $WN$. Also $\\angle XAW = \\angle WAP$. $WN$ is the diameter of $\\Gamma$, therefore $\\angle NML = \\angle NAL = 90^\\circ$, and thus the quadrilateral $ALMN$ is cyclic. Then $\\angle WNP = \\angle WAP = \\angle XAW = \\angle MNL$, and hence $N$, $L$, $P$ are collinear. Note that $LM$, $NA$, $WP$ are the altitudes of $\\triangle NLW$, and their intersection point is the orthocenter $K$. Let $T$ be the intersection point of the ray $AN$ and $w$. There exists a homothetic transformation mapping $\\Gamma$ to $w$. By this homothety, point $N$ maps to $T$, point $W$ to $L$, i.e., $TL$ is the diameter of $w$, because $WN$ is the diameter of $\\Gamma$. Hence, $Q$ is the midpoint of the segment $TL$. Let us use Menelaus' theorem to $\\triangle TLK$ and points $N$, $Q$, $S$. Then, to prove $N$, $Q$, $S$ are collinear, it is sufficient to show that\n\n$$\n\\frac{TN}{NK} \\cdot \\frac{KS}{SL} \\cdot \\frac{LQ}{QT} = 1, \\quad \\text{ i.e., (as } TQ = QL) \\quad \\frac{TN}{NK} = \\frac{SL}{KS}.\n$$\n\nLet us prove this equality. It is clear that the lines $AL$ and $AK$ are the internal and external bisectors of the angle $\\angle MAS$, hence $\\frac{ML}{LS} = \\frac{MK}{SK}$. Then $\\frac{ML}{MK} = \\frac{SL}{SK}$. Obviously, $TL \\parallel NW$, hence by Thales' theorem $\\frac{TN}{NK} = \\frac{LM}{MK}$, therefore $\\frac{TN}{NK} = \\frac{SL}{SK}$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17178,
    "subject": "Mathematics (Olympiad)",
    "question": "In a convex quadrilateral $ABCD$, the diagonals intersect at $O$. Points $M$ and $N$ are on the segments $OA$ and $OD$, respectively. Suppose $MN$ is parallel to $AD$ and $NC$ is parallel to $AB$. Prove that $\\angle ABM = \\angle NCD$.\n\n![](images/Singapore2023-booklet_p1_data_872a397d83.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We use the notation $[PQR]$ to denote the area of $\\triangle PQR$.\n\n$$\nMN \\parallel AD \\implies [AMN] = [DMN] \\implies [AON] = [DOM].\n$$\n\n$$\nNC \\parallel AB \\implies [ACN] = [BCN] \\implies [AON] = [BOC].\n$$\n\n$$\n\\text{Thus } [DOM] = [BOC] \\implies [MCD] = [BCD] \\implies MB \\parallel DC.\n$$\n\n$$\n\\text{Finally, } AB \\parallel NC,\\ MB \\parallel DC \\implies \\angle ABM = \\angle NCD.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17179,
    "subject": "Mathematics (Olympiad)",
    "question": "Given any $n$ positive integers, and a sequence of $2^n$ integers (with terms among them), prove there exists a subsequence made of consecutive terms, such that the product of its terms is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the integers could well be distinct primes, this is equivalent to proving the stricter problem: given a finite alphabet $A$ of $n$ letters $a_1, a_2, \\dots, a_n$, and a word $w$ of length $2^n$ on this alphabet, show it contains a nonempty contiguous subword $x$ ($w = \\overline{uxv}$, where $u, v$ could be the empty word), in which each letter appears an even number of times.\n\nLet us identify the letter $a_k$ with the element $\\mathbf{e}_k \\in \\mathbb{Z}_2^n$ given by $\\mathbf{e}_k = (0, \\dots, 0, 1, 0, \\dots, 0)$, where the $1$ is at the $k$th position. Now the requirement is to find a subword such that the sum of its elements is $\\mathbf{0} = (0, 0, \\dots, 0)$.\n\nWe apply a classical idea of Erd\\\"os. Let $w = \\overline{x_1x_2\\dots x_{2^n}}$ be the word, with $x_i \\in \\{\\mathbf{e}_1, \\mathbf{e}_2, \\dots, \\mathbf{e}_n\\}$ for all $i = 1, 2, \\dots, 2^n$. Define $\\sigma_k = \\sum_{i=1}^k x_i$ for all $k = 1, 2, \\dots, 2^n$. If any $\\sigma_k = \\mathbf{0}$, we are done, since $\\overline{x_1\\dots x_k}$ can be taken as the subword; otherwise, there must exist $1 \\le p < q \\le 2^n$ such that $\\sigma_p = \\sigma_q$, so $\\mathbf{0} = \\sigma_q - \\sigma_p = \\sum_{i=1}^{q-p} x_{p+i}$, and we can take the subword $\\overline{x_{p+1}\\dots x_q}$.\n\nNotice that the result is tight. There exist words of length $2^n - 1$ without this property. We can build them inductively: take $w_1 = \\overline{a_1}$, and build $w_{n+1} = \\overline{w_n a_{n+1} w_n}$ for all $n \\ge 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17180,
    "subject": "Mathematics (Olympiad)",
    "question": "Denote by $p(a)$ the first digit of the natural number $a$. Show that each of the sets\n$$\nA = \\{n \\in \\mathbb{N} \\mid p(5^n) - p(2^n) > 0\\}, \\quad B = \\{n \\in \\mathbb{N} \\mid p(5^n) - p(2^n) < 0\\}\n$$\nhas infinitely many elements.",
    "options": [],
    "answer": "See solution",
    "solution": "For $k \\in \\mathbb{N}^*$, there exists $n_k \\in \\mathbb{N}^*$ such that $2^{n_k} < 10^k < 2^{n_k+1}$ ($n_k + 1$ is the smallest element of the set $\\{m \\in \\mathbb{N}^* \\mid 10^k < 2^m\\}$). As $2^{n_k} < 10^k < 2^{n_k+1}$, we get $10^k < 2^{n_k+1} < 2 \\cdot 10^k$, so $p(2^{n_k+1}) = 1$.\n\nMultiplying by $5^{n_k+1}$, we have $10^k \\cdot 5^{n_k+1} < 10^{n_k+1} < 2 \\cdot 10^k \\cdot 5^{n_k+1}$. Dividing the first inequality by $10^k$ and the second by $2 \\cdot 10^k$, we get $5 \\cdot 10^{n_k-k} < 5^{n_k+1} < 10^{n_k-k+1}$, so $p(5^{n_k+1}) \\ge 5 > p(2^{n_k+1})$.\n\nFrom $2^{n_k} < 10^k < 2^{n_k+1}$, by dividing the second inequality by $2$, we conclude $5 \\cdot 10^{k-1} < 2^{n_k} < 10^k$, so $p(2^{n_k}) \\ge 5$. Multiplying the first inequality by $5^{n_k}$, we obtain $10^{k-1} \\cdot 5^{n_k+1} < 10^{n_k} < 10^k \\cdot 5^{n_k}$, so by dividing the first one by $5 \\cdot 10^{k-1}$ and the second by $10^k$, we obtain $10^{n_k-k} < 5^{n_k} < 2 \\cdot 10^{n_k-k}$, that is $p(5^{n_k}) = 1 < p(2^{n_k})$.\n\nThe set $\\{n_k \\mid k \\in \\mathbb{N}^*\\}$ has infinitely many elements because $n_k > k$. As $\\{n_k \\mid k \\in \\mathbb{N}^*\\} \\subset A$ and $\\{n_k + 1 \\mid k \\in \\mathbb{N}^*\\} \\subset B$, the sets $A$ and $B$ also have infinitely many elements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17181,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\ge 2$. Suppose $A_1, A_2, \\dots, A_n$ are $n$ nonempty finite sets satisfying:\n\n$$\n|A_i \\Delta A_j| = |i-j|\n$$\nfor all $i, j \\in \\{1, 2, \\dots, n\\}$.\n\nFind the minimum value of $|A_1| + |A_2| + \\dots + |A_n|$.\n\nHere, $|X|$ denotes the number of elements of a finite set $X$ and $X \\Delta Y = \\{a \\mid a \\in X, a \\notin Y\\} \\cup \\{a \\mid a \\in Y, a \\notin X\\}$ for any sets $X$ and $Y$.",
    "options": [],
    "answer": "See solution",
    "solution": "For each positive integer $k$, we prove that the minimum value of $S_{2k}$ is $k^2 + 2$; the minimum value of $S_{2k+1}$ is $k(k+1)+2$.\n\nFirstly, define the sets $A_1, A_2, \\dots, A_{2k}, A_{2k+1}$ as follows:\n\n$$\nA_i = \\{i, i+1, \\dots, k\\}, \\quad i = 1, 2, \\dots, k; \\quad A_{k+1} = \\{k, k+1\\}\n$$\n\n$$\nA_{k+j} = \\{k+1, k+2, \\dots, k+j-1\\}, \\quad j = 2, 3, \\dots, k+1\n$$\n\nFor this family of sets, it is easy to verify that $|A_i \\Delta A_j| = |i - j|$ holds in the following cases:\n\n1. $1 \\le i < j \\le k$\n2. $1 \\le i < j = k+1$\n3. $1 \\le i < k+1 < j \\le 2k+1$\n4. $k+1 = i < j \\le 2k+1$\n5. $k+2 \\le i < j \\le 2k+1$\n\nMoreover, the case of $i = j$ is trivial, and the case of $i > j$ can be reduced to the case of $i < j$. Thus for all $i, j \\in \\{1, 2, \\dots, 2k+1\\}$, we have verified that\n\n$$\n|A_i \\Delta A_j| = |i - j|\n$$\n\nFor the above $(2k+1)$ sets, we can easily calculate that\n\n$$\nS_{2k+1} = \\frac{k(k+1)}{2} + 2 + \\frac{k(k+1)}{2} = k(k+1) + 2\n$$\n\nIf we choose the first $2k$ sets, we get that\n\n$$\nS_{2k} = S_{2k+1} - k = k^2 + 2\n$$\n\nSecondly, we show that $S_{2k} \\ge k^2 + 2$ and $S_{2k+1} \\ge k(k+1)+2$. Note the following facts:\n\n**Fact 1:** For any two finite sets $X, Y$, we have\n$$\n|X| + |Y| \\ge |X \\Delta Y|\n$$\n\n**Fact 2:** For any two nonempty finite sets $X, Y$, if $|X \\Delta Y| = 1$, then $|X| + |Y| \\ge 3$.\n\nWhen $n = 2k$, it follows from Fact 1 that\n\n$$\n|A_i| + |A_{2k+1-i}| \\ge |A_i \\Delta A_{2k+1-i}| = 2k+1-2i, \\quad i = 1, 2, \\dots, k-1\n$$\n\nBy $|A_k \\Delta A_{k+1}| = 1$ and Fact 2, we have $|A_k| + |A_{k+1}| \\ge 3$. So\n\n$$\nS_{2k} = |A_k| + |A_{k+1}| + \\sum_{i=1}^{k-1} (|A_i| + |A_{2k+1-i}|) \\ge 3 + \\sum_{i=1}^{k-1} (2k+1-2i) = k^2 + 2\n$$\n\nSimilarly, when $n = 2k+1$, we get that\n\n$$\n|A_i| + |A_{2k+2-i}| \\ge |A_i \\Delta A_{2k+2-i}| = 2k+2-2i, \\quad i = 1, 2, \\dots, k-1\n$$\n\nSince $|A_k \\Delta A_{k+1}| = 1$, we have ...",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17182,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to arrange the first 49 positive integers in a $7 \\times 7$ array, so that each cell contains exactly one number and no two primes are neighbours?\n\nTwo neighbours are numbers from cells with a common side or a common vertex.",
    "options": [],
    "answer": "See solution",
    "solution": "It is possible. The positive primes less than 49 are $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$, that is 15 numbers.\n\n![](images/RMC_2024_p6_data_f0d98ee6fd.png)\n\nNumber the lines and the columns from 1 to 7, set the primes in 15 of the 16 squares with both odd indexes and the other 34 numbers in the remaining cells.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17183,
    "subject": "Mathematics (Olympiad)",
    "question": "Capablanca and Alyokhin decided to play a match of 16 games according to the following rules. The winner of the first game received $1 = 3^0$ peso, the winner of the second one got $3 = 3^1$ pesos, the winner of the third game received $9 = 3^2$ pesos, and so on. If the game ended in a draw, then they split the prize pool of the game in half. It turned out that at the end of the match, Alyokhin earned 2018 pesos more than Capablanca. How many games has each of the players won?",
    "options": [],
    "answer": "See solution",
    "solution": "Let Alyokhin's gain in the $k$th game be $a_k \\cdot 3^{k-1}$, where $a_k \\in \\{-1, 0, 1\\}$ (win, draw, loss). The total possible gain after $k$ games ranges from $-A_k$ to $A_k$, where $A_k = 3^0 + 3^1 + \\dots + 3^{k-1}$. Each possible result corresponds to a unique combination of game outcomes. For 16 games, the possible differences are all odd numbers between $-A_{16}$ and $A_{16}$. We seek a combination where Alyokhin's total gain is 2018 pesos more than Capablanca's, i.e., Alyokhin's net is $2018$. Expressing $2018$ in terms of powers of $3$:\n\n$$\n2018 = 3^7 - 3^5 + 3^4 - 3^2 + 3^1 - 3^0\n$$\n\nThis corresponds to Alyokhin winning games 8, 5, and 4, and losing games 6, 2, and 1, with the rest being draws. Therefore, each player won 3 games, and the remaining 10 games ended in a draw.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17184,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N$ be a positive integer, and let $\\mathbf{a} = (a(1), \\dots, a(N))$ and $\\mathbf{b} = (b(1), \\dots, b(N))$ be sequences of non-negative integers, each written on a circle (so we assume $a(i \\pm N) = a(i)$ and $b(i \\pm N) = b(i)$). We say $\\mathbf{a}$ is *b-harmonic* if each $a(i)$ is the arithmetic mean of the counterclockwise nearest $b(i)$ numbers, the clockwise nearest $b(i)$ numbers, and $a(i)$ itself; that is,\n\n$$\na(i) = \\frac{1}{2b(i) + 1} \\sum_{s=-b(i)}^{b(i)} a(i + s). \\quad (*)$$\n\n(A term of $\\mathbf{a}$ may appear more than once in the above sum.) Suppose that neither $\\mathbf{a}$ nor $\\mathbf{b}$ is constant, and that both $\\mathbf{a}$ is *b-harmonic*, and $\\mathbf{b}$ is *a-harmonic*. Prove that more than half of the $2N$ terms across both sequences vanish.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = \\min_i a(i)$ and let $b = \\min_i b(i)$. Since $\\mathbf{a}$ is not constant, there exists an $i$ such that $a = a(i) < a(i + 1)$.\n\n**Claim 1.** If $a = a(i) < a(i + 1)$, then $b(i) = 0$. Similarly, if $a = a(i) < a(i - 1)$, then $b(i) = 0$.\n\n*Proof.* Otherwise the sum in $(*)$ contains a term $a(i + 1) > a$ but no terms smaller than $a$, so the average is greater than $a$. $\\square$\n\nClaim 1 implies $b = 0$; similarly, $a = 0$. With reference again to Claim 1, $a(i) = b(i) = 0$ for some index $i$.\n\nSay that $[i, j]$ is an **a-segment** if $a(i) = a(i + 1) = \\cdots = a(j) = 0$ but $a(i - 1) \\neq 0 \\neq a(j + 1)$; define a **b-segment** similarly. By Claim 1, the endpoints of any such segment satisfy $a(i) = b(i) = a(j) = b(j) = 0$. Since the sequences are non-constant, each $i$ where $a(i) = 0$ is contained in an a-segment.\n\n**Claim 2.** Let $[i, j]$ be a b-segment, and let $k \\in [i, j]$. Then $a(k) \\le k - i$ (and, similarly, $a(k) \\le j - k$).\n\n*Proof.* Indeed, since $b(k) = 0$, the elements of $\\mathbf{b}$ with indices from $k-a(k)$ to $k+a(k)$ must all be zero as well. $\\square$\n\nWe now show that every index is contained in either an a- or a b-segment. Since at least one index is contained in both, the conclusion follows.\n\nAssume, to the contrary, that $a(i)$ and $b(i)$ are both positive for some index $i$; call such indices *bad*. Among all bad indices $i$, choose one maximising $\\max(a(i), b(i))$; by symmetry, we may and will assume that this maximum is $a(i)$. We may and will also assume that either the index $i-1$ is not bad, or $a(i-1) < a(i)$ (otherwise change $i$ to $i-1$, repeat if necessary, recalling that $\\mathbf{a}$ is not constant).\n\nConsider the range of indices $\\Delta = [i - b(i), i + b(i)]$, and the values $\\mathbf{a}$ assumes at those indices. Some indices $j$ in $\\Delta$ are bad; the corresponding values $a(j)$ do not exceed $a(i)$. Other indices $j$ in $\\Delta$ are covered by several a- and b-segments. Each b-segment contributes at most $b(i)$ members nearest to one of its endpoints, so the average value of $\\mathbf{a}$ over those indices does not exceed $(b(i) - 1)/2 < a(i)$ by Claim 2. The remaining indices $j$ in $\\Delta$ all lie in a-segments, so the corresponding values $a(j)$ are all zero.\n\nCombining all this, it follows that the average in the right-hand member of $(*)$ does not exceed $a(i)$. Moreover, if some a- or b-segment intersects $\\Delta$, then the inequality is strict. Otherwise, $i-1$ is a bad index contained in $\\Delta$, and $a(i-1) < a(i)$, so the inequality is again strict. This contradiction ends the proof and completes the solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17185,
    "subject": "Mathematics (Olympiad)",
    "question": "For each pair $(x, y)$ of real numbers with $0 \\le x \\le y \\le 1$, consider the set\n$$\nA = \\{xy,\\ xy - x - y + 1,\\ x + y - 2xy\\}.\n$$\nLet $M(x, y)$ be the largest value in $A$. Find the minimum possible value of $M(x, y)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $q_0 = xy$, $q_1 = xy - x - y + 1 = (1-x)(1-y)$, and $q_2 = x + y - 2xy$. The quantities $q_0$, $q_1$, and $q_2$ are the areas of three regions determined by $(x, y)$ in the unit square as shown in the first figure. From the second picture, we can deduce that if we move the point towards the diagonal $\\{y = x\\}$ in a perpendicular direction, the shadowed area decreases (so does $q_2$), and $q_0$ and $q_1$ increase. Thus, we can assume the point that minimizes $M(x, y)$ satisfies $q_0 = \\max A$ (or $q_1 = \\max A$) or belongs to $\\{y = x\\}$.\n\n![](images/Spanija_2019_p2_data_13ca7c9b12.png)\n\nAssume that the first alternative holds. We claim that for a fixed value $q$ of $q_0$, the point $(x_0, y_0)$ that maximizes $q_1$ satisfies $y_0 = x_0$. A geometric explanation is that the branches (for $x, y > 0$) of the family of hyperbolae $\\{xy = a : a > 0\\}$ and $\\{(1-x)(1-y) = a : a > 0\\}$ are perpendicular to $\\{y = x\\}$ and open in opposite directions. Thus, the branch $\\{(1-x)(1-y) = a\\}$ that intersects $\\{xy = q\\}$ with maximum $a$ is exactly the one that meets $\\{y = x\\}$ at $(\\sqrt{q}, \\sqrt{q})$. In sum, if we slide the point $(x, y)$ along $\\{xy = q\\}$ towards $\\{y = x\\}$, $q_0$ remains equal to $q$, $q_1$ increases, and $q_2$ decreases. Moreover, $q_1$ never exceeds $q_0$ because $q_0 \\ge 1/3$ implies $q_1 < 1/3$ whenever $x = y$. This means that the minimum of $M(x, y)$ is attained on the diagonal $\\{y = x\\}$.\n\nTo finish, we only need to address the case $x = y$. In this case, the maximum of $A$ is $q_1$ if $x \\le 1/3$, $q_2$ if $1/3 \\le x \\le 2/3$, and $q_0$ if $x \\ge 2/3$. Consequently, $\\min M(x, y) = 4/9$, and it is attained at $(x, y) = (1/3, 1/3)$ and $(x, y) = (2/3, 2/3)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17186,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a positive integer. Find all positive integers $n$ such that it is possible to mark $n$ points on the sides of a triangle (different from its vertices) and connect some of them with a line so that the following conditions are satisfied:\n\n1. There is at least 1 marked point on each side.\n2. For each pair of points $X$ and $Y$ marked on different sides, on the third side there exist exactly $k$ marked points which are connected to both $X$ and $Y$ and exactly $k$ points which are connected to neither $X$ nor $Y$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* $n = 12k$.\n\nLet $A$, $B$, and $C$ be the sets of points on the three sides, with cardinalities $a$, $b$, and $c$ respectively. Consider all triplets $(p, q, r)$ where $p \\in A$, $q \\in B$, and $r \\in C$, and they are either pairwise connected or pairwise not connected. For each $p \\in A$, $q \\in B$, there are exactly $k$ points $r \\in C$ for which the condition holds: if $p$ and $q$ are connected, choose the $k$ points connected to both; if $p$ and $q$ are not connected, pick the $k$ points connected to neither. Thus, the total number of such triplets is $kab$. Similarly, for $q \\in B$, $r \\in C$, the total is $kbc$, and for $r \\in C$, $p \\in A$, it is $kca$. Therefore, $ab = bc = ca$, implying $a = b = c = \\frac{n}{3}$.\n\nNow, count all other triplets $(p, q, r)$ with $p \\in A$, $q \\in B$, $r \\in C$. For all $p \\in A$, $q \\in B$, there are exactly $k$ points $r \\in C$ for which $r$ is connected to neither if $p$ and $q$ are connected, and $r$ is connected to both if $p$ and $q$ are not connected. The number of such triplets is $kab = k\\left(\\frac{n}{3}\\right)^2$. Similarly, for the other two cases, the number is also $k\\left(\\frac{n}{3}\\right)^2$ each. Thus, the total number of triplets $(p, q, r)$ is $4k\\left(\\frac{n}{3}\\right)^2$. On the other hand, the total number of triplets is $\\left(\\frac{n}{3}\\right)^3$. Setting $4k\\left(\\frac{n}{3}\\right)^2 = \\left(\\frac{n}{3}\\right)^3$ gives $n = 12k$.\n\nIt remains to show that it is possible to mark $4k$ points on each side to satisfy the conditions. For $k=1$, label the sides as $0$, $1$, and $2$, each containing points labelled $0$, $1$, $2$, and $3$. For each $i = 0, 1, 2$, connect the even-numbered points on side $i$ with points $0$ and $1$ on side $(i+1) \\bmod 3$, and the odd-numbered points on side $i$ with points $2$ and $3$ on side $(i+1) \\bmod 3$.\n\n![](images/prob1718_p36_data_a5dc2d2abd.png)\n\nFor $k > 1$, replace each point with $k$ distinct points and connect those generated from points that were previously connected.\n\n**Remark:** This is a modified version of a problem from the Croatian Mathematical Olympiad 2017.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17187,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $\\cos^5\\theta - \\sin^5\\theta < 7(\\sin^3\\theta - \\cos^3\\theta)$, $\\theta \\in [0, 2\\pi)$. Then the range of $\\theta$ is ______.",
    "options": [],
    "answer": "See solution",
    "solution": "From the inequality\n\n$$\n\\cos^5\\theta - \\sin^5\\theta < 7(\\sin^3\\theta - \\cos^3\\theta)\n$$\n\nwe have\n\n$$\n\\sin^3\\theta + \\frac{1}{7}\\sin^5\\theta > \\cos^3\\theta + \\frac{1}{7}\\cos^5\\theta\n$$\n\nSince $f(x) = x^3 + \\frac{1}{7}x^5$ is increasing over $(-\\infty, +\\infty)$, then $\\sin \\theta > \\cos \\theta$, and that means\n\n$$\n2k\\pi + \\frac{\\pi}{4} < \\theta < 2k\\pi + \\frac{5\\pi}{4} \\quad (k \\in \\mathbb{Z})\n$$\n\nBut $\\theta \\in [0, 2\\pi)$, so the range of $\\theta$ is $\\left(\\frac{\\pi}{4}, \\frac{5\\pi}{4}\\right)$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17188,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real solutions $x$ in the interval $\\left[\\dfrac{74}{27}, \\dfrac{10}{3}\\right]$ to the equation:\n\n$$\n4 - 3\\sqrt{10 - 3x} = x^2 - 4x + 4.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "On $X = \\left[\\dfrac{74}{27}, \\dfrac{10}{3}\\right]$:\n\n$$\n\\begin{align*}\n4 - 3\\sqrt{10 - 3x} &= x^2 - 4x + 4 \\\\\n9(10 - 3x) &= (x^2 - 4x + 4 - 4)^2 \\\\\n9(10 - 3x) &= (x^2 - 4x)^2 \\\\\n9(10 - 3x) &= x^4 - 8x^3 + 16x^2 \\\\\nx^4 - 8x^3 + 16x^2 + 27x - 90 &= 0 \\\\\n(x - 3)(x + 2)(x^2 - 7x + 15) &= 0 \\\\\nx = 3.\n\\end{align*}\n$$\n\nThus, the solution is $x = 3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17189,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given an acute triangle $ABC$ with $AB > AC$ and orthocenter $H$. The point $E$ is the reflection of $C$ with respect to the altitude $AH$. Let $F$ be the intersection of the lines $EH$ and $AC$. Prove that the circumcenter of triangle $AEF$ lies on the line $AB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\theta$ be the angle between $AF$ and the tangent $t$ at $A$ to the circumcircle of $AEF$. By the inscribed angle theorem, $\\angle FEA = \\theta$. Due to the reflection, $\\angle ACH = \\angle FEA = \\theta$. Because $\\angle ACH = \\theta$, the tangent $t$ is parallel to $CH$ and thus orthogonal to $AB$. Therefore, the circumcenter of triangle $AEF$ lies on $AB$.\n\n_Comment: This result also holds for obtuse triangles._",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17190,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of integers $ (x, y) $ such that\n$$\n3^4 2^3 (x^2 + y^2) = x^3 y^3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that if $xy = 0$, then $x^2 + y^2 = 0$ and $x = y = 0$ is a solution. Also, if $xy < 0$, then $x^2 + y^2 < 0$, which is impossible; thus $x, y$ are both positive or both negative. Changing simultaneously the sign of $x$ and $y$ does not change the equation, so we may assume without loss of generality that $x, y$ are both positive.\n\nLet $m, n$ ($m \\ge n$) be non-negative integers, and let $a, b$ be coprime integers, not divisible by 3. Consider $x = 3^m a$ and $y = 3^n b$. Then the given equation is\n$$\n8((3^{m-n}a)^2 + b^2) = 3^{3m+n-4}a^3b^3.\n$$\nBecause any perfect square has remainder 0 or 1 when divided by 3, the left-hand side is not divisible by 3, thus looking at the right-hand side we get $3m + n - 4 = 0$, and because $m \\ge n \\ge 0$ it follows $m = n = 1$. The equation takes the form\n$$\n8(a^2 + b^2) = a^3 b^3.\n$$\nBy symmetry, we may assume $a \\ge b$. Then\n$$\n16a^2 \\ge a^3 b^3 \\Leftrightarrow 16 \\ge ab^3\n$$\nHence we have either (1) $b=2$ which implies $a=2$, or (2) $b=1$, but in this case the only possible values of $a$ are 1, 2, 4, 8 and none of them satisfies the equation $a^3-8a^2-8=0$. It is easy to see that $(a, b) = (2, 2)$ and $(x, y) = (6, 6)$. So, the only solutions are\n$$\n(x, y) = (-6, -6), \\quad (x, y) = (0, 0), \\quad (x, y) = (6, 6).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17191,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(G, \\cdot)$ be a group, and $H < G$ a proper subgroup of $G$. If there are endomorphisms $f, g, h : G \\to G$ such that $f(xy) = g(x)h(y)$ for all $x, y \\in G \\setminus H$, show that:\n\na) $g = h$;\n\nb) If $G$ is nonabelian and $H = Z(G)$, then $f = g = h$.\n\nThe set $Z(G) = \\{c \\in G \\mid cx = xc, \\text{ for all } x \\in G\\}$ is called the centre of the group $G$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $e$ be the unit element of $G$. For any $x \\in G \\setminus H$, $x^{-1} \\in G \\setminus H$, so:\n\n$$\nf(e) = f(x x^{-1}) = g(x) h(x^{-1}) = g(x) h(x)^{-1}.\n$$\n\nThus, $g(x) = h(x)$ for all $x \\in G \\setminus H$.\n\nFor any $a \\in H$, choose $x \\in G \\setminus H$. Then $a x, x^{-1} \\in G \\setminus H$, so:\n\n$$\ng(a) = g(a x x^{-1}) = g(a x) g(x^{-1}) = h(a x) h(x^{-1}) = h(a x x^{-1}) = h(a).\n$$\n\nTherefore, $g = h$.\n\nb) Let $G$ be nonabelian and $H = Z(G)$. By part a), $g = h$, so the relation becomes:\n\n$$\nf(xy) = g(x) g(y) = g(xy), \\quad \\text{for all } x, y \\in G \\setminus Z(G).\n$$\n\nFor any $a \\in Z(G)$ and $x \\in G \\setminus Z(G)$, $a x, x^{-1} \\in G \\setminus Z(G)$, so:\n\n$$\nf(a) = f(a x x^{-1}) = g(a x x^{-1}) = g(a).\n$$\n\nNow, for $x \\in G \\setminus Z(G)$, there exists $y \\in G \\setminus Z(G)$ such that $x y \\neq y x$. If $x y \\in Z(G)$, then\n\n$$\nx y = y (x y) y^{-1} = y x \\neq x y,\n$$\n\nwhich is a contradiction. Thus, $x y \\in G \\setminus Z(G)$, and $y^{-1} \\in G \\setminus Z(G)$, so:\n\n$$\nf(x) = f(x y y^{-1}) = g(x y y^{-1}) = g(x).\n$$\n\nTherefore, $f = g$, so $f = g = h$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17192,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1 \\leq a_2 \\leq \\dots \\leq a_n$ be the stolen numbers. \n\n**a)** Suppose all partial sums of these numbers are positive. Show that the numbers $a_1, \\dots, a_n$ are uniquely determined by the set of all partial sums.\n\n**b)** Suppose the stolen numbers may be negative, i.e., $a_1 \\leq \\dots \\leq a_k < 0 \\leq a_{k+1} \\leq \\dots \\leq a_n$, and let $s_1 \\leq s_2 \\leq \\dots \\leq s_{2^n-1}$ be the partial sums. Show that the set of numbers is still uniquely determined by the set of all partial sums.\n\n**c)** Give an example of two different multisets of $n$ numbers (possibly with zeros) that have the same set of partial sums.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $a_1 \\leq a_2 \\leq \\dots \\leq a_n$ be the stolen numbers. Since all the partial sums are positive, all the $a_i$ must be positive. Clearly, $a_1$ is the smallest partial sum. Suppose $a_1, \\dots, a_i$ have been determined. Remove all partial sums involving only $a_1, \\dots, a_i$ from the list. The smallest remaining number is $a_{i+1}$. Thus, all numbers are uniquely determined.\n\nb) Suppose $a_1 \\leq \\dots \\leq a_k < 0 \\leq a_{k+1} \\leq \\dots \\leq a_n$ and $s_1 \\leq s_2 \\leq \\dots \\leq s_{2^n-1}$ are the partial sums. If $s_1 > 0$, part (a) applies. Otherwise, $s_1 < 0$. Consider\n\n$$\n(1 + x^{a_1})(1 + x^{a_2}) \\cdots (1 + x^{a_n}) = 1 + x^{s_1} + x^{s_2} + \\cdots + x^{s_{2^n-1}}.\n$$\n\nHere, $s_1$ is the sum of the negative $a_i$. Multiply both sides by $x^{-s_1}$:\n\n$$\n(1 + x^{-a_1}) \\cdots (1 + x^{-a_k})(1 + x^{a_{k+1}}) \\cdots (1 + x^{a_n}) = x^{-s_1} + x^{s_1 - s_1} + x^{s_2 - s_1} + \\cdots + x^{s_{2^n-1} - s_1}.\n$$\n\nThus, we have the partial sums of $|a_1|, \\dots, |a_n|$, and by part (a), these are uniquely determined.\n\nFinally, $x^{s_1}$ can be written uniquely as a product of $x^{a_i}$'s; otherwise, a partial sum would be zero, which is impossible.\n\nc) The partial sums of $\\{1, 2, -3\\}$ and $\\{-1, -2, 3\\}$ are the same. Adding 1389 zeros to both sets does not change the partial sums.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17193,
    "subject": "Mathematics (Olympiad)",
    "question": "The diagonal $AC$ of a convex quadrilateral $ABCD$ is the bisector of the angle $\\angle DCB$. Let $E$ be the intersection of the side $AB$ and the circumcircle of the triangle $ACD$. Let $F$ be the intersection of the side $AD$ and the circumcircle of the triangle $ABC$. Prove that the segments $AC$, $DE$, and $BF$ intersect in a point.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $\\angle FAE = \\alpha$. Since the point $E$ lies on the bisector of the segment $AC$, its distances to the points $A$ and $C$ are the same. Hence, $E$ is the centre of the circle containing the points $A$, $C$, and $F$, and we have $|AE| = |CE| = |FE|$. So, $\\angle EFA = \\angle FAE = \\alpha$.\n\nSince $AB$ and $CD$ are parallel, we get $\\angle DEA = \\angle EAF = \\alpha$ and $\\angle CEF = \\angle EFA = \\alpha$.\n\n![](images/Slovenija_2010_p18_data_68b3818460.png)\n\nThe right triangles $AED$ and $CEG$ are congruent because they have three common angles and the hypotenuses have the same length.\n\nSo, $|ED| = |EG|$ and $|CG| = |AD| = |BC|$. From here we can conclude that the triangle $DEG$ is isosceles and\n\n$$\n\\angle EGD = \\frac{\\pi - \\angle DEG}{2} = \\frac{\\angle GEC}{2} = \\frac{\\alpha}{2}.\n$$\n\nBy Pythagoras' theorem we have $|FB|^2 = |FC|^2 - |BC|^2 = |FC|^2 - |GC|^2 = |FG|^2$, so $|FB| = |FG|$. Therefore, $GFB$ is an equilateral triangle and\n\n$$\n\\angle FGB = \\frac{\\pi - \\angle GFB}{2} = \\frac{\\angle GFA}{2} = \\frac{\\alpha}{2}.\n$$\n\nWe have shown that $\\angle FGB = \\angle EGD$, so the points $B$, $G$, and $D$ are congruent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17194,
    "subject": "Mathematics (Olympiad)",
    "question": "A farmer's rectangular field is partitioned into a $2 \\times 2$ grid of 4 rectangular sections as shown in the figure. In each section, the farmer will plant one crop: corn, wheat, soybeans, or potatoes. The farmer does not want to grow corn and wheat in any two sections that share a border, and the farmer does not want to grow soybeans and potatoes in any two sections that share a border. Given these restrictions, in how many ways can the farmer choose crops to plant in each of the four sections of the field?\n\n![](images/2021_AMC10A_Solutions_Fall_p6_data_31ceebe024.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let the sections of the field be numbered clockwise starting with the upper left as sections 1, 2, 3, and 4.\n\nThere are $4$ ways to select a crop to plant in section 1. If section 3 is planted with the crop in section 1, there are $3$ ways to select crops for each of sections 2 and 4, yielding $4 \\cdot 3 \\cdot 3 = 36$ ways. If section 3 is planted with the crop that cannot be planted adjacent to the crop in section 1, then there are $2$ ways to select crops for each of sections 2 and 4, yielding $4 \\cdot 2 \\cdot 2 = 16$ ways. If section 3 is planted with one of the other $2$ crops, then there are $2$ ways to select crops for each of sections 2 and 4, yielding $4 \\cdot 2 \\cdot 2 \\cdot 2 = 32$ ways. This gives a total of $36 + 16 + 32 = 84$ ways.\n\n**OR**\n\nIf only one crop is planted, then there are $4$ ways to choose that crop. If two compatible crops are planted, then there are $2 \\cdot 2 = 4$ ways to choose those crops and then $2^4 - 2 = 14$ ways to plant, for $4 \\cdot 14 = 56$ possibilities. It is not possible to plant two incompatible crops without violating the condition. If three crops are planted, then there are $4$ ways to choose them and $2 \\cdot 2 = 4$ ways to plant, for $16$ choices. If four crops are planted, then there are $4 \\cdot 2 = 8$ ways to plant them. The total number of choices is $4 + 56 + 16 + 8 = 84$.\n\n**OR**\n\nAs above, let the sections of the field be numbered clockwise starting with the upper left as sections 1, 2, 3, and 4. There are $4^4 = 256$ ways for crops to be assigned to sections of the field without regard to crops that are not to be planted in adjacent sections.\n\nThis problem will be solved via complementary counting. To this end, let $A$ be the set of ways such that sections 1 and 2 are planted with crops that should not be planted in adjacent sections, and similarly, let $B$, $C$, and $D$ be the set of ways sections 2 and 3, 3 and 4, and 4 and 1, respectively, can be planted with crops that should not be planted in adjacent sections. Each of the sets $A$, $B$, $C$, and $D$ contains $2 \\cdot 2 \\cdot 4 \\cdot 4 = 64$ elements. Each of the intersections $A \\cap B$, $B \\cap C$, $C \\cap D$, and $D \\cap A$ contains $4 \\cdot 4 = 16$ elements, while the intersections $A \\cap C$ and $B \\cap D$ each contain $4 \\cdot 4 = 16$ elements. Each of the intersections of three of these sets contains $4$ elements. The intersection of all four of the sets contains $4$ elements.\n\nThe Inclusion-Exclusion Principle implies that the size of $A \\cup B \\cup C \\cup D$ is $4 \\cdot 64 - 6 \\cdot 16 + 4 \\cdot 4 - 4 = 172$. Thus the number of ways to plant the sections so that adjacent sections are not planted with incompatible crops is $256 - 172 = 84$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17195,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a cube with vertices labeled $A, B, C, D, E, F, G, H$:\n\n- **a.** List all trails of length 2 starting from $A$. Which vertices can be reached from $A$ with a trail of length 2?\n\n- **b.** List all trails of length 3 from $A$ to $G$.\n\n- **c.** Give an example of a trail of length 7 starting from $A$.\n\n- **d.** For trails of length 4 starting from $A$:\n    - List all such trails in alphabetical order.\n    - Which vertices can be reached from $A$ with a trail of length 4?\n    - Justify why no other vertices can be reached from $A$ with a trail of length 4.",
    "options": [],
    "answer": "See solution",
    "solution": "*a.* The trails of length 2 from $A$ are:\n\n$$\n\\begin{array}{ccc}\nABC & ADC & AEF \\\\\nABF & ADH & AEH\n\\end{array}\n$$\n\nSo the vertices that can be reached from $A$ with a trail of length 2 are $C$, $F$, $H$.\n\n*b.* The trails of length 3 from $A$ to $G$ are the trails of length 2 in part *a* followed by $G$:\n\n$ABCG$, $ABFG$, $ADCG$, $ADHG$, $AEFG$, $AEHG$.\n\n*c.* One trail of length 7 is $ABFEHDCG$. There are others.\n\n*d. Alternative i*\n\nAll trails of length 4 from $A$ (in alphabetical order):\n\n$$\n\\begin{array}{ccc}\nABCDA & ADCBA & AEFBA \\\\\nABCDH & ADCBF & AEFBC \\\\\nABCGF & ADCGF & AEFGC \\\\\nABCGH & ADCGH & AEFGH \\\\\nABFEA & ADHEA & AEHDA \\\\\nABFEH & ADHEF & AEHDC \\\\\nABFGC & ADHGC & AEHGC \\\\\nABFGH & ADHGF & AEHGF\n\\end{array}\n$$\n\nHence, trails of length 4 from $A$ reach only vertices $A$, $C$, $F$, $H$.\n\n*Alternative ii*\n\nWe can reach vertices $A$, $C$, $F$, $H$ from $A$ with trails of length 4. For example: $ABCDA$, $AEHGC$, $ADHEF$, $ABCGH$.\n\nTo show that no other vertices can be reached: Any trail of length 4 can be divided into two trails of length 2. From part *a*, any trail of length 2 starting at $A$ must finish at one of $C$, $F$, $H$. By symmetry, any trail of length 2 starting at one of $A$, $C$, $F$, $H$ must finish at one of the other three. Thus, any trail of length 4 starting at $A$ must finish at one of $A$, $C$, $F$, $H$.\n\n*Alternative iii*\n\nAgain, we can reach $A$, $C$, $F$, $H$ from $A$ with trails of length 4. For example: $ABCDA$, $AEHGC$, $ADHEF$, $ABCGH$.\n\nNotice that $A$, $C$, $F$, $H$ are black vertices, and $B$, $D$, $E$, $G$ are white. Each edge joins a black to a white vertex. Thus, any trail of length 4 starting at $A$ alternates black and white, starting and ending with black. Therefore, any such trail must finish at one of $A$, $C$, $F$, $H$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17196,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an integer greater than two, and let $A_1, A_2, \\dots, A_{2n}$ be pairwise distinct nonempty subsets of $\\{1, 2, \\dots, n\\}$. Determine the maximum value of\n$$\n\\sum_{i=1}^{2n} \\frac{|A_i \\cap A_{i+1}|}{|A_i| \\cdot |A_{i+1}|}\n$$\nwhere $A_{2n+1} = A_1$. For a set $X$, let $|X|$ denote the number of elements in $X$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n$.\n\nConsider each summand $s_i = \\frac{|A_i \\cap A_{i+1}|}{|A_i| \\cdot |A_{i+1}|}$.\n\nIf $A_i \\cap A_{i+1} = \\emptyset$, then $s_i = 0$.\n\nIf $A_i \\cap A_{i+1} \\neq \\emptyset$, since $A_i \\neq A_{i+1}$, at least one of $A_i$ and $A_{i+1}$ has more than one element, i.e., $\\max\\{|A_i|, |A_{i+1}|\\} \\ge 2$. Since $A_i \\cap A_{i+1}$ is a subset of both $A_i$ and $A_{i+1}$, $|A_i \\cap A_{i+1}| \\le \\min\\{|A_i|, |A_{i+1}|\\}$, so\n\n$$\ns_i = \\frac{|A_i \\cap A_{i+1}|}{|A_i| \\cdot |A_{i+1}|} \\le \\frac{\\min\\{|A_i|, |A_{i+1}|\\}}{\\max\\{|A_i|, |A_{i+1}|\\} \\cdot \\min\\{|A_i|, |A_{i+1}|\\}} \\le \\frac{1}{2}.\n$$\n\nTherefore,\n$$\n\\sum_{i=1}^{2n} \\frac{|A_i \\cap A_{i+1}|}{|A_i| \\cdot |A_{i+1}|} \\le \\sum_{i=1}^{2n} \\frac{1}{2} = n.\n$$\n\nThis upper bound can be achieved with the sets:\n$$\nA_1 = \\{1\\},\\quad A_2 = \\{1, 2\\},\\quad A_3 = \\{2\\},\\quad A_4 = \\{2, 3\\},\\ \\dots,\\ A_{2n-2} = \\{n-1, n\\},\\quad A_{2n-1} = \\{n\\},\\quad A_{2n} = \\{n, 1\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17197,
    "subject": "Mathematics (Olympiad)",
    "question": "Planes through the points with integer coordinates in three-dimensional Euclidean space partition the space into unit cubes. Find all triples $ (a, b, c) $, with $ a \\leq b \\leq c $, of positive integers such that the cubes can be colored in $abc$ colors in such a way that every parallelepiped of dimensions $ a \\times b \\times c $, with integer vertices and faces parallel to the coordinate planes, does not contain cubes of the same color.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that the solutions are the triples $ (a, b, c) $ such that $ a $ divides $ b $ and $ b $ divides $ c $.\n\nLet $ ((x_0, y_0, z_0), p, q, r) $ denote the parallelepiped with lower right vertex $ (x_0, y_0, z_0) $ and dimensions $ p, q, r $ along the $ Ox, Oy, Oz $ axes, respectively.\n\nSuppose $ b $ is not divisible by $ a $, i.e., $ b = ma + n $ for some $ m, n \\in \\mathbb{N} $, $ 0 < n < a $. If $ (p, q, r) $ is a permutation of $ (a, b, c) $, then from the condition for the parallelepipeds $ ((0, 0, 0), p, q, r) $ and $ ((0, 0, 1), p, q, r) $, it follows that the parallelepipeds $ ((0, 0, 0), p, q, 1) $ and $ ((0, 0, r), p, q, 1) $ are filled with cubes of the same colors. This implies that the parallelepipeds $ ((0, 0, 0), c, a, 1) $ and $ ((0, 0, b), c, a, 1) $ are also filled with cubes of the same colors, and the same is true for $ ((0, 0, 0), c, b, 1) $ and $ ((0, 0, ma), c, b, 1) $. Since $ ((0, 0, 0), c, b, 1) $ contains $ ((0, 0, 0), c, a, 1) $ and $ ((0, 0, ma), c, b, a) $ contains $ ((0, 0, ma), c, b, 1) $ and $ ((0, 0, b), c, a, 1) $, every color in $ ((0, 0, 0), c, a, 1) $ appears at least twice in $ ((0, 0, ma), c, b, a) $. This contradiction completes the proof that $ a $ divides $ b $. Similarly, $ b $ divides $ c $.\n\nNow, let $ a \\mid b $ and $ b \\mid c $, so $ b = p_1 a $, $ c = p_2 b = p_1 p_2 a $, where $ p_1, p_2 \\in \\mathbb{N} $. For any two positive integers $ m $ and $ n $, let $ R(m, n) $ denote the remainder of $ m $ modulo $ n $. The coordinates of a cube are those of its lower right vertex.\n\nWe assign the color to the cube $ (x, y, z) $ as follows:\n\n$$\n(R(x, a),\\ R(y, a),\\ R(z, a),\\ R(\\lfloor \\frac{x}{a} \\rfloor + \\lfloor \\frac{y}{a} \\rfloor, p_1),\\ R(\\lfloor \\frac{y}{a} \\rfloor + \\lfloor \\frac{z}{a} \\rfloor, p_1),\\ R(\\lfloor \\frac{x}{b} \\rfloor + \\lfloor \\frac{y}{b} \\rfloor + \\lfloor \\frac{z}{b} \\rfloor, p_2))\n$$\n\nCounting all possible remainders in the six coordinates shows that the number of colors used is $ a^3 p_1 p_1 p_2 = abc $.\n\nSuppose two distinct cubes $ (x_1, y_1, z_1) $ and $ (x_2, y_2, z_2) $ lie in a parallelepiped of dimensions $ a \\times b \\times c $. Then $ |x_1 - x_2| \\leq \\alpha $, $ |y_1 - y_2| \\leq \\beta $, $ |z_1 - z_2| \\leq \\gamma $, where $ (\\alpha, \\beta, \\gamma) $ is a permutation of $ (a, b, c) $. Since $ |x_1 - x_2| $, $ |y_1 - y_2| $, and $ |z_1 - z_2| $ are divisible by $ a $, one of these numbers is $ 0 $. Suppose $ x_1 = x_2 $. Then the fourth and fifth coordinates of the color show that $ |y_1 - y_2| $ and $ |z_1 - z_2| $ are divisible by $ b $, so one of these two numbers is $ 0 $. If, for example, $ y_1 = y_2 $, then the last coordinate shows that $ |z_1 - z_2| $ is divisible by $ c $, i.e., $ z_1 = z_2 $. Thus, $ (x_1, y_1, z_1) = (x_2, y_2, z_2) $, a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17198,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\ell$ be a line in the plane, and let $A \\notin \\ell$ be a point. Also, let $\\alpha \\in (0, \\pi/2)$ be fixed. Determine the locus of points $Q$ in the plane for which there exists a point $P \\in \\ell$ such that $AQ = PQ$ and $\\angle PAQ = \\alpha$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the circumcircle of $\\triangle APQ$ meet the line $\\ell$ at a second point $B$. From the concyclicity of points $A, B, P, Q$ it follows that the angle made by the lines $AB$ and $\\ell$ is $2\\alpha$; this means that when $\\alpha = \\pi/4$ the point $B$ coincides with the foot $O$ of the perpendicular dropped from $A$ onto $\\ell$, while when $\\alpha \\neq \\pi/4$ the point $B$ must occupy one of only two fixed positions on $\\ell$, $B_1$ and $B_2$, symmetrical with respect to $O$. There are two degenerate positions, when $P \\equiv B$ and when $Q \\equiv B$, but they are trivial.\n\nTherefore, there exists some point $\\Omega$ on the line $AO$ (which coincides with $O$ when $\\alpha = \\pi/4$) so that the locus is the two lines through $\\Omega$, at angles $\\pi/2 - \\alpha$ with $AO$. The fullness of the locus follows from the fact that a construction is possible in all cases (or by a continuity argument).\n\nThe condition $A \\notin \\ell$ is not strictly necessary; that case is trivial, with $A \\equiv \\Omega$, and the two lines of the locus passing through $A$ and making an angle $\\alpha$ with $\\ell$. For $\\alpha = \\pi/4$, the problem has also been asked to the Juniors.\n\n![](images/RMC2013_final_p94_data_3d5143637e.png)\n\n**Alternative Solution.** (Analytical) Consider a system of orthogonal coordinates in the plane, such that $A(0, 2\\mu)$, the line $\\ell$ is the $Ox$ axis, and $P(2\\lambda, 0)$ with $\\lambda$ running over $\\mathbb{R}$. Then the midpoint of $AP$ has coordinates $M(\\lambda, \\mu)$. Let the point $Q(x, y)$, and write the conditions concerning it.\n\nSince $MQ^2 + AM^2 = AQ^2$, we have\n\n$$\n((x - \\lambda)^2 + (y - \\mu)^2) + (\\lambda^2 + \\mu^2) = x^2 + (y - 2\\mu)^2.\n$$\n\nOn the other hand, $MQ = AM \\tan PAQ = tAM$ (where $t = \\tan \\alpha$), therefore $MQ^2 = t^2 AM^2$, so\n\n$$\n(t^2 + 1)(\\lambda^2 + \\mu^2) = x^2 + (y - 2\\mu)^2.\n$$\n\nThe equality $AQ^2 = PQ^2$, i.e. $x^2 + (y - 2\\mu)^2 = (x - 2\\lambda)^2 + y^2$, yields $\\lambda x = (\\lambda^2 + \\mu^2) + \\mu(y - 2\\mu)$. Squaring this up results in $(\\lambda^2 + \\mu^2)x^2 = ((\\lambda^2 + \\mu^2) + \\mu(y - 2\\mu))^2 + \\mu^2 x^2$, which also writes as $(\\lambda^2 + \\mu^2)^2 + 2(\\lambda^2 + \\mu^2)\\mu(y - 2\\mu) + \\mu^2(x^2 + (y - 2\\mu)^2)$, or again $(\\lambda^2 + \\mu^2)^2 + 2(\\lambda^2 + \\mu^2)\\mu(y - 2\\mu) + \\mu^2(t^2 + 1)(\\lambda^2 + \\mu^2)$. Factoring out $\\lambda^2 + \\mu^2 \\neq 0$, we are left with\n\n$$\nt^2 x^2 - ((y - 2\\mu) + (t^2 + 1)\\mu)^2 = 0,\n$$\n\nwhich writes as $(y + (t^2 - 1)\\mu)^2 = (tx)^2$, translating into the two lines $y = \\pm tx + (1 - t^2)\\mu$, of slopes $\\pm \\tan \\alpha$, and the same ordinate $(1 - t^2)\\mu$ at the origin (the point $\\Omega$ above).\n\n**Alternative Solution.** Similar computations with the analytic ones above provide a purely trigonometric or vectorial solution. For example, the condition $AQ = PQ$ translates into $\\langle \\overrightarrow{AM}, \\overrightarrow{MQ} \\rangle = 0$, writing $\\lambda(x - \\lambda) - \\mu(y - \\mu) = 0$, while the condition on the constant angle translates into $\\frac{\\langle \\overrightarrow{AQ}, \\overrightarrow{AM} \\rangle}{AQ \\cdot AM} = \\cos \\alpha$; both of these conditions come up to identical forms with those obtained above.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17199,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute triangle with circumcircle $\\Gamma$. Let $\\ell$ be a tangent line to $\\Gamma$, and let $\\ell_a$, $\\ell_b$, and $\\ell_c$ be the lines obtained by reflecting $\\ell$ in the lines $BC$, $CA$, and $AB$, respectively. Show that the circumcircle of the triangle determined by the lines $\\ell_a$, $\\ell_b$, and $\\ell_c$ is tangent to the circle $\\Gamma$.\n\nRemark: There are many different configurations for this problem, and the proofs can be easily modified to treat different configurations. To accommodate this, we use directed angles modulo $180^\\circ$. Let $A_1$, $B_1$, $C_1$ be the intersections of $\\ell_b$ and $\\ell_c$, $\\ell_a$ and $\\ell_b$, and $\\ell_c$ and $\\ell_a$, respectively. Let $\\ell$ be tangent to $\\Gamma$ at $T$. Let lines $AB$, $BC$, $CA$ meet $\\ell$ at $C_3$, $A_3$, $B_3$, respectively. (Assume $\\ell$ is not parallel to any sides, as the cases where it is parallel to a side are limits of cases where it is not.)\n\n![](images/pamphlet1112_main_p80_data_8f7cf72929.png)",
    "options": [],
    "answer": "See solution",
    "solution": "*Solution 1.* Construct points $A_2$, $B_2$, $C_2$ (distinct from $A$, $B$, $C$) on $\\Gamma$ such that $\\widehat{TA} = \\widehat{A_2A}$, $\\widehat{TB} = \\widehat{B_2B}$, and $\\widehat{TC} = \\widehat{C_2C}$. Without loss of generality, suppose $\\ell$ is such that $B$ lies inside triangle $B_1A_3C_3$, so that we are in the configuration shown above; other configurations may be handled similarly. We begin with two lemmas.\n\n*Lemma 1.* Lines $B_1B$ and $C_1C$ meet on circle $\\Gamma$.\n\n_Proof._ Let $I$ be the intersection of $B_1B$ and $C_1C$; it suffices to show that $A$, $B$, $C$, $I$ are cyclic. By definition, $A_1B_1$, $AB$, $\\ell$ concur at $C_3$; $B_1C_1$, $BC$, $\\ell$ concur at $A_3$; and $C_1A_1$, $CA$, $\\ell$ concur at $B_3$. By reflection properties, line $AB$ (through $C_3$) bisects $\\angle A_3C_3B_1$, and line $BC$ (through $A_3$) bisects $\\angle B_1A_3C_3$. Hence, in our configuration, $B$ is the incenter of triangle $B_1C_3A_3$. Similarly, $C$ is the excenter of triangle $C_1A_3B_3$. Computing, we see\n\n$$\n\\angle ABI = 180^{\\circ} - \\angle B_3BC_3 = 180^{\\circ} - \\left(90^{\\circ} + \\frac{\\angle B_1A_3C_3}{2}\\right) = 90^{\\circ} - \\frac{\\angle B_1A_3C_3}{2}\n$$\n\nand\n\n$$\n\\begin{aligned}\n\\angle ACI &= \\angle CB_3A_1 - \\angle CC_1B_3 \\\\\n&= \\frac{\\angle A_3B_3A_1}{2} - \\frac{\\angle A_3C_1B_3}{2} \\\\\n&= \\frac{\\angle B_3A_3C_1}{2} = \\frac{180^{\\circ} - \\angle B_1A_3C_3}{2} = 90^{\\circ} - \\frac{\\angle B_1A_3C_3}{2}.\n\\end{aligned}\n$$\n\nHence $\\angle ACI = \\angle ABI$, so $ACBI$ is cyclic. $\\square$\n\n*Lemma 2.* Lines $BC_2$ and $CB_2$ meet on line $B_1C_1$.\n\n_Proof._ Let $X$ be the intersection of $C_2B$ and $CB_2$. Because $\\widehat{B_2B} = \\widehat{BT}$, $CB_2$ is the reflection of $CT$ across $BC$. Likewise, $BC_2$ is the reflection of $BT$ across $BC$. Thus, $X$ is the reflection of $T$ across $BC$, from which it follows that $X$ lies on the reflection $B_1C_1$ of $TS$ across $BC$, as needed. $\\square$\n\nLet $B_1B_2$ intersect $\\Gamma$ again at $H$. We claim that there is a homothety centered at $H$ sending $A_2B_2C_2$ to $A_1B_1C_1$. This suffices because $H$ lies on the circumcircle $\\Gamma$ of triangle $A_2B_2C_2$, hence such an $H$ would send $\\Gamma$ to the circumcircle $\\Gamma_1$ of triangle $A_1B_1C_1$, implying that $\\Gamma$ and $\\Gamma_1$ are tangent.\n\nFor the claim, we first show that corresponding sides of $A_1B_1C_1$ and $A_2B_2C_2$ are parallel; by symmetry, it suffices to show that $B_2C_2 \\parallel B_1C_1$. Let $S$ be the intersection of lines $B_2C_2$ and $\\ell$. Since $\\widehat{B_2T} = 2\\widehat{BT}$ and $\\widehat{TC_2} = 2\\widehat{TC}$, we have\n\n$$\n\\angle B_2ST = \\angle B_2C_2T - \\angle STC_2 = 2(\\angle BC_2T - \\angle CTC_2).\n$$\n\nBecause $BC_2CT$ is cyclic, we have\n\n$$\n2(\\angle BC_2T - \\angle CTC_2) = 2(\\angle BCT - \\angle CTC_2) = 2\\angle BA_3T.\n$$\n\nBecause $B_1A_3$ and $\\ell$ are reflections of each other across line $BC$, we have $2\\angle BA_3T = \\angle B_1A_3T$. Combining the last three equations gives $\\angle B_2ST = \\angle B_1A_3T$, from which it follows that $B_2C_2 \\parallel B_1C_1$.\n\nIt remains to check that $A_1A_2$, $B_1B_2$, and $C_1C_2$ are concurrent. Applying Pascal's theorem to the (self-intersecting) cyclic hexagon $B_2HC_2BIC$ gives that $B_1$ (the intersection of $B_2H$ and $BI$), $X$ (the intersection of $HC_2$ and $IC$), and the intersection of $C_2B$ and $CB_2$ all lie on $B_1X$, which by Lemma 2 is the same line as $B_1C_1$. Therefore, $HC_2$ passes through the intersection of $IC$ and $B_1C_1$, which is $C_1$ by Lemma 1. This means that $C_1C_2$ passes through $H$. Similarly, $A_1A_2$ passes through $H$, showing $H$ exists.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 17200,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: (0, \\infty) \\to \\mathbb{R}$ be such that for all $x, y \\in (0, \\infty)$,\n\n$$\nf(x + y) = f\\left(\\frac{x + y}{xy}\\right) + f(xy).\n$$\n\nShow that $f(xy) = f(x) + f(y)$ for all $x, y \\in (0, \\infty)$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show that $f(ab) = f(a) + f(b)$ for $a, b \\in (0, \\infty)$ such that $a^2b \\ge 4$. In fact, if $a^2b \\ge 4$ and $a, b > 0$, then we can find $x, y > 0$ such that $x + y = ab$ and $xy = b$, namely\n\n$$\nx = \\frac{a + \\sqrt{a^2 - (4/b)}}{2/b} > 0 \\quad \\text{and} \\quad y = \\frac{a - \\sqrt{a^2 - (4/b)}}{2/b} > 0.\n$$\n\nThus, for $a, b \\in (0, \\infty)$ such that $a^2b \\ge 4$, we plug $x, y$ as above in the functional equation to get $f(ab) = f(x+y) = f\\left(\\frac{x+y}{xy}\\right) + f(xy) = f(a) + f(b)$.\n\nNow let $x, y > 0$. Let\n\n$$\nz = \\max \\left\\{ \\frac{4}{x^2 y^2}, \\frac{4}{x^2 y} \\cdot \\frac{4}{y^2} \\right\\} > 0.\n$$\n\nThen $(xy)^2 z \\ge 4$, $x^2(yz) \\ge 4$ and $y^2 z \\ge 4$ which imply that $f(xyz) = f(xy) + f(z)$, $f(xyz) = f(x) + f(yz)$ and $f(yz) = f(y) + f(z)$ respectively. Therefore,\n\n$$\n\\begin{aligned}\nf(xy) &= f(xyz) - f(z) \\\\\n&= (f(x) + f(yz)) - f(z) \\\\\n&= f(x) + f(yz) - f(z) \\\\\n&= f(x) + (f(y) + f(z)) - f(z) \\\\\n&= f(x) + f(y).\n\\end{aligned}\n$$\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17201,
    "subject": "Mathematics (Olympiad)",
    "question": "Each integer from 1 to $n$ was colored either red or blue, and each color was used at least once. It turned out that:\n\n- Every red number is a sum of some two distinct blue numbers;\n- Every blue number is a difference of some two red numbers.\n\nFind the smallest number $n$ for which such coloring is possible.",
    "options": [],
    "answer": "See solution",
    "solution": "The smallest such number is $n = 9$.\n\nSuppose that $n$ is a number allowing for coloring with the desired properties. Then $n \\ge 2$ since each color needs to be used at least once. Note that $n$ cannot be blue since it is inexpressible as a difference of two numbers between 1 and $n$, therefore $n$ is red. On the other hand, 1 and 2 cannot be red, since none of them is a sum of two distinct positive integers, which means 1 and 2 are blue (and in particular $n \\ge 3$). Now $n-1$ cannot be a difference of two red numbers (since 1 is blue), so $n-1$ is red. Similarly, $n-2 = (n-1)-1$ has to be red since both 1 and 2 are blue. It follows that $n \\ge 5$.\n\nSince $n$ is red, it needs to be a sum of two distinct blue numbers. But a sum of two blue numbers cannot exceed\n\n$$\n(n-3) + (n-4) = 2n-7.\n$$\n\nTherefore, $2n - 7 \\ge n \\implies n \\ge 7$.\n\nIf $n = 7$, then the only remaining possibility to express 7 as a sum of two blue numbers is $3+4$, but then 4 is not a difference of two red numbers, contradiction. Similarly with $n = 8$.\n\nWe have proved that $n \\ge 9$. It remains to show that there exists a valid coloring for $n = 9$. We can color 1, 2, 4, 5, 6 blue and 3, 7, 8, 9 red. Correctness of this coloring follows from the simple arithmetic:\n\n$$\n1 = 9 - 8, \\quad 2 = 9 - 7, \\quad 4 = 7 - 3, \\quad 5 = 8 - 3, \\quad 6 = 9 - 3,\n$$\n\n$$\n3 = 1 + 2, \\quad 7 = 1 + 6, \\quad 8 = 2 + 6, \\quad 9 = 4 + 5.\n$$\n\n*Remarks.* The coloring presented above is unique and it actually exists for every positive integer $n \\ge 9$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17202,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\Gamma$ be a half circle with diameter $AB$. Let $C$ be a point on the diameter, and let $D$ and $E$ be points on the half circle so that $E$ lies between $B$ and $D$. It turns out that $\\angle ACD = \\angle ECB$. Let $F$ be the point of intersection of the lines tangent to $\\Gamma$ at the points $D$ and $E$. Show that $\\angle EFD = \\angle ACD + \\angle ECB$.\n\n![alt](images/Ukraine_2020_booklet_p13_data_91afd9c989.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the center of the half circle $\\Gamma$. By the properties of the tangent lines, $\\angle OEF = 90^\\circ = \\angle FDO$. Thus, $O$, $E$, $F$, $D$ are cyclic.\n\nLet $\\angle ACD = \\angle ECB = \\theta$. By the Law of Sines for $\\triangle COD$ we obtain:\n\n$$\n\\frac{\\sin \\angle ODC}{OD} = \\frac{\\sin \\angle DCO}{OD} = \\frac{\\sin(180^\\circ - \\theta)}{OD}\n$$\n\nSimilarly, for $\\triangle COE$:\n$$\n\\frac{\\sin \\angle OEC}{CO} = \\frac{\\sin \\angle ECO}{OE} = \\frac{\\sin \\theta}{OE}\n$$\n\nSince $OD = OE$ as radii of the half circle, $\\sin \\angle ODC = \\sin \\angle OEC$. Since $\\angle ODC < \\angle BDA = 90^\\circ$, the angle is acute, as is $\\angle OEC$. Thus, the angles are equal and the five points $O$, $E$, $F$, $D$, $C$ are cyclic. Thus, by the property of the inscribed quadrilateral:\n\n$$\n\\angle EFD = 180^\\circ - \\angle DCE = \\angle ACD + \\angle ECB\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17203,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x + y f(x)) + y = x y + f(x + y)\n$$\nfor all real numbers $x, y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P(x, y)$ denote the given relation. If there is an $a \\in \\mathbb{R}$ such that $f(a) = 0$, then $P(a, y)$ gives $y = a y + f(a + y)$, so $f$ must be linear. Checking, the only linear solutions are $f(x) = x$ and $f(x) = 2 - x$ for $x \\in \\mathbb{R}$.\n\nNow suppose $f(x) \\neq 0$ for all $x$. From $P(x - y, y)$:\n$$\nf(x - y + y f(x - y)) = -y^2 + y(x - 1) + f(x).\n$$\nSince $f(t) \\neq 0$ for all $t$, $-y^2 + y(x - 1) + f(x) \\neq 0$ for all $x, y$, so its discriminant as a quadratic in $y$ must be negative:\n$$(x - 1)^2 + 4 f(x) < 0,$$\nwhich gives\n$$f(x) < -\\frac{(x-1)^2}{4} \\leq 0$$\nfor all $x$. Since $-\\frac{(x-1)^2}{4} \\leq x$, we have\n$$f(x) < -\\frac{(x-1)^2}{4} \\leq x.$$\nNow from $P(x, y)$ for $y > 0$ and $x \\in \\mathbb{R}$:\n$$\nx y - y + f(x + y) = f(x + y f(x)) < x + y f(x) < x - y \\frac{(x-1)^2}{4}$$\nso\n$$f(x + y) < x + y - y(x + \\frac{(x-1)^2}{4}) = x + y - y \\frac{(x+1)^2}{4}.$$\nSetting $x = -y$:\n$$f(0) < -y \\frac{(-y + 1)^2}{4}$$\nfor all $y > 0$. Letting $y \\to +\\infty$ gives a contradiction. Hence, the only solutions are $f(x) = x$ and $f(x) = 2 - x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17204,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a prime number $p$ congruent to $1$ modulo $5$ such that $2p + 1$ is also prime, show that there exists a matrix of zeros and ones containing exactly $4p$ (respectively, $4p + 2$) ones, no submatrix of which contains exactly $2p$ (respectively, $2p + 1$) ones.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p = 5q + 1$, $q \\ge 2$, so $4p = 5(4q + 1) - 1$. Form a $5 \\times (4q + 1)$ matrix of ones except for a single zero entry. This matrix has exactly $4p$ ones. Any submatrix with $r$ rows ($1 \\le r \\le 5$) and $s$ columns ($1 \\le s \\le 4q + 1$) contains either $rs$ or $rs - 1$ ones. If $rs = 2p$, then $r \\ge p$ or $s \\ge p$ since $p$ is prime; if $rs - 1 = 2p$, then $rs = 2p + 1$, so $r = 2p + 1$ or $s = 2p + 1$ since $2p + 1$ is prime. Both cases are impossible given the matrix size, so no submatrix contains exactly $2p$ ones.\n\nFor $4p + 2 = 5(4q + 1) + 1$, form a $5 \\times (4q + 1)$ matrix of ones except for one column with only a single one. This matrix has exactly $4p + 2$ ones. Any submatrix with $r$ rows and $s$ all-one columns ($0 \\le s \\le 4q + 1$) contains $rs$ or $rs + 1$ ones. If $rs = 2p + 1$, then $r = 2p + 1$ or $s = 2p + 1$ since $2p + 1$ is prime; if $rs + 1 = 2p + 1$, then $r \\ge p$ or $s \\ge p$ since $p$ is prime. Again, this contradicts the matrix size, so no submatrix contains exactly $2p + 1$ ones.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17205,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve for the points of the parabola given by the equation:\n\n$$\n|x| + |y| = \\left|x - \\frac{y-c}{m}\\right| + \\left|y - c\\right| = \\left|x - \\frac{y-c}{m}\\right|,\n$$\n\nor equivalently,\n\n$$\nm|x| + m|y| = |mx + c - y|.\n$$\n\nFor an arbitrary point $(x, y)$ in the plane above the line $l$ ($y > mx + c$),\n\n$$\nm|x| + m|y| \\geq m|x| + |y| \\geq -mx + y > y - mx - c = |y - mx - c|.\n$$\n\nTherefore, all points of the parabola lie below $l$ ($y < mx + c$). Evaluating cases resulting from the sign of $x$ and $y$, the parabola is the union of the following four segments (as shown in the figure):\n\n$$\n\\begin{cases}\n  y = \\frac{c}{m+1}, & \\text{for } x > 0, y > 0 \\\\\n  y = \\frac{c}{1-m}, & \\text{for } x > 0, y < 0 \\\\\n  2mx - (m+1)y + c = 0, & \\text{for } x < 0, y > 0 \\\\\n  2mx + (m-1)y + c = 0, & \\text{for } x < 0, y < 0\n\\end{cases}\n$$\n\n![](images/2016_p34_data_7fcd3f7175.png)\n\nd) Prove that lines $AA'$, $BB'$, and $CC'$ are concurrent if $A'$, $B'$, and $C'$ are points on sides $BC$, $AC$, and $AB$ of triangle $ABC$ such that\n\n$$\n\\frac{d(B, A')}{d(C, A')} \\frac{d(C, B')}{d(A, B')} \\frac{d(A, C')}{d(B, C')} = 1.\n$$\n\ne) Is it possible for there to exist an infinite set $S$ of points in the plane such that the mutual distances between all points are perfect squares?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p = \\frac{1}{2}(d(A, B) + d(B, C) + d(C, A))$. Then $d(B, X_a) + d(A, B) = d(C, X_a) + d(A, C) = p$, so $d(B, X_a) = p - d(A, B)$ and $d(C, X_a) = p - d(A, C)$. Similar equalities for $X_b$ and $X_c$ imply\n\n$$\n\\frac{d(B, X_a)}{d(C, X_a)} \\frac{d(C, X_b)}{d(A, X_b)} \\frac{d(A, X_c)}{d(B, X_c)} = 1.\n$$\n\nBy the lemma, these lines are concurrent.\n\ne) No. Assume to the contrary that $S$ is an infinite set of points in the plane such that the mutual distances between its points are all perfect squares.\n\n**Lemma.** *There is no infinite subset $\\{P_1 = (x_1, y_1), P_2 = (x_2, y_2), \\dots\\}$ of $S$ for which both sequences $\\{x_i\\}$ and $\\{y_i\\}$ are monotone.*\n\n*Proof.* Suppose there exists such an infinite set $P_i = (x_i, y_i)$ in $S$ with both $\\{x_i\\}$ and $\\{y_i\\}$ increasing. For each $j > 2$, let $a_j = \\sqrt{d(P_1, P_j)} \\in \\mathbb{N}$ and $b_j = \\sqrt{d(P_2, P_j)} \\in \\mathbb{N}$. Since both sequences are increasing, for every $j > 2$,\n\n$$\na_j^2 = d(P_1, P_j) = d(P_1, P_2) + d(P_2, P_j) = d(P_1, P_2) + b_j^2.\n$$\n\nThus, $d(P_1, P_2)$ can be written in infinitely many ways as the difference of two perfect squares, which is impossible. Thus, $S$ cannot exist.\n\nBy translation, assume $S$ contains the origin. The number of points of $S$ in the first quadrant is finite, and by symmetry, so is the number in each quadrant. Thus, $S$ is finite.\n\nIf $\\{y \\ge 0 \\mid \\exists x \\ge 0; (x, y) \\in S\\}$ is unbounded, then an infinite subset $\\{P_1 = (x_1, y_1), P_2 = (x_2, y_2), \\dots\\}$ of $S$ can be found with $y_1 < y_2 < \\dots$. Then a subsequence $\\{P_{n_i}\\}_i$ exists with monotone $x_{n_i}$, which is impossible by the lemma. Thus, the coordinates of points of $S$ in the first quadrant are bounded. Therefore, there exists $M > 0$ such that all points of $S$ in the first quadrant lie in $[0, M] \\times [0, M]$. Since each pair of points in $S$ has distance at least 1, only finitely many points can be in this square, so $S$ is finite.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17206,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with circumcircle $\\omega$, circumcenter $O$, and orthocenter $H$. Let $K$ be the midpoint of $AH$. The perpendicular to $OK$ at $K$ intersects $AB$ and $AC$ at $P$ and $Q$, respectively. The lines $BK$ and $CK$ intersect $\\omega$ again at $X$ and $Y$, respectively. Prove that the second intersection of the circumcircles of triangles $KPY$ and $KQX$ lies on $\\omega$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Claim 1.** $PK = KQ$.\n\n**Proof of Claim 1.** Let $L$ and $N$ be the midpoints of $AB$ and $AC$, respectively. Since $L$ and $K$ are midpoints of $AB$ and $AH$, then $LK \\parallel BH$ and so $LK \\perp AC$. Since also $LO \\perp AB$, then $\\angle KLO = \\angle BAC = \\alpha$. Also, $\\angle OLP = 90^\\circ = \\angle OKP$, so the quadrilateral $OKLP$ is cyclic and therefore $\\angle KPO = \\angle KLO = \\alpha$. Similarly, $\\angle KQO = \\alpha$. Therefore, the triangle $OPQ$ is isosceles, and since $OK \\perp PQ$ then $K$ is the midpoint of $PQ$. $\\square$\n\n![](images/BMO_2022_shortlist_p32_data_0d798403bb.png)\n\n**Claim 2.** The intersection of $YP$ and $AK$ lies on $\\omega$.\n\n**Proof of Claim 2.** Let $D$ be the other point of intersection of $AK$ with $\\omega$. Since $OK \\perp PQ$, then $K$ is the midpoint of the chord $\\ell$ of $\\omega$ through $P, Q$. Since $AD$ and $YC$ intersect at $K$, by the Butterfly theorem the points $P' = YD \\cap \\ell$ and $Q = AC \\cap \\ell$ are equidistant from $K$. Thus $P' = P$ and $YP \\cap AK = D \\in \\omega$. $\\square$\n\nNow let $A'$ be the other point of intersection of $AO$ with $\\omega$ and let $S$ be the other point of intersection of $A'H$ with $\\omega$.\n\n**Claim 3.** $PQ$ is the perpendicular bisector of $HS$.\n\n**Proof of Claim 3.** We have $\\angle HSA = \\angle A'SA = 90^\\circ$ so $S$ lies on the circle $\\omega'$ with diameter $AH$ centered at $K$. So $KS = KH$. Since $O$ and $K$ are the circumcenters of $\\omega$ and $\\omega'$ respectively, and $AS$ is their common chord, then $OK \\perp AS$. But we also have $HS \\perp AS$ and $PQ \\perp OK$, thus $HS \\perp PQ$. Since $KS = KH$, then $K$ belongs on the perpendicular bisector of $HS$ and the result follows. $\\square$\n\nFrom Claim 3 we have $\\angle KSP = \\angle KHP$. From Claim 1 and the fact that $KP = KQ$ we have that $AQPH$ is a parallelogram and so (using Claim 2 as well)\n\n$$\n\\angle KHP = \\angle KAC = \\angle DAC = \\angle DYC = \\angle PYK.\n$$\n\nSince $\\angle KSP = \\angle KYP$ we get that $S$ belongs on the circumcircle of triangle $KPY$. Similarly it belongs to the circumcircle of triangle $KQX$ and therefore the result follows.\n\n**Note.** We can also define $S$ to be the reflection of $H$ on $PQ$. Now Claim 3 is immediate. Then $AS \\parallel PQ$ and so $OK \\perp AS$ and $OK \\parallel HS$. Since $K$ is the midpoint of $AH$, then $OK$ is the perpendicular bisector of $AS$ and therefore $S \\in \\omega$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17207,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, $C$, $D$, $E$ be points, and let $\\gamma$ be the circle passing through $D$, $E$, and $C$. Let $X$ and $Y$ be points such that\n\n$$\n|AX||AY| = |AD||AC| = |AC|^2 - |AC||CD|.\n$$\n\nSimilarly, the power of $B$ with respect to $\\gamma$ is\n\n$$\n|BX||BY| = |BC|^2 - |BC||CE|.\n$$\n\nSuppose $|AC||CD| = |BC||CE|$. Let $M$ be the midpoint of $XY$. Then\n\n$$\n|AX||AY| = |AM|^2 - |XM|^2 \\quad \\text{and} \\quad |BX||BY| = |BM|^2 - |XM|^2.\n$$\n\nCombine these identities to show that\n\n$$\n|AM|^2 - |BM|^2 = |AC|^2 - |BC|^2.\n$$\n\nLet $H$ be the point on $AB$ such that $CH$ is the altitude of triangle $ABC$. Prove that $M = H$.",
    "options": [],
    "answer": "See solution",
    "solution": "We compute the power of $A$ and $B$ with respect to $\\gamma$:\n\n$$\n|AX||AY| = |AC|^2 - |AC||CD|,\n$$\n$$\n|BX||BY| = |BC|^2 - |BC||CE|.\n$$\n\nGiven $|AC||CD| = |BC||CE|$, subtracting gives\n\n$$\n|AX||AY| - |BX||BY| = |AC|^2 - |BC|^2.\n$$\n\nSince $|AX||AY| = |AM|^2 - |XM|^2$ and $|BX||BY| = |BM|^2 - |XM|^2$, we have\n\n$$\n|AM|^2 - |BM|^2 = |AC|^2 - |BC|^2.\n$$\n\nBy the Pythagorean theorem, for the foot $H$ of the altitude from $C$ to $AB$,\n\n$$\n|AH|^2 - |BH|^2 = |AC|^2 - |BC|^2.\n$$\n\nSince both $M$ and $H$ lie on $AB$ and satisfy the same relation, it follows that $M = H$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17208,
    "subject": "Mathematics (Olympiad)",
    "question": "Vasia wrote down all seven-digit numbers that contain every digit between 1 and 7 exactly once. Prove that neither of the numbers Vasia wrote down divides another such number.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose one such number $a$ is divisible by another such number $b$, i.e., there exists an integer $n > 1$ such that $a = n b$. Since both $a$ and $b$ use each digit from 1 to 7 exactly once, their digit sums are equal, so both have remainder 1 when divided by 9. Thus, $n$ also has remainder 1 when divided by 9. Since $n \\ne 1$, $n \\geq 10$, so $a$ would have more digits than $b$, which is impossible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17209,
    "subject": "Mathematics (Olympiad)",
    "question": "How many of the integers between 97 and 199 are multiples of $2$ or $3$?\n\n(A) 33  \n(B) 40  \n(C) 55  \n(D) 60  \n(E) 68",
    "options": [],
    "answer": "See solution",
    "solution": "Let us count the number of integers between $97$ and $199$ that are multiples of $2$ or $3$.\n\nFirst, count multiples of $2$:\n- The smallest multiple of $2$ in this range is $98$.\n- The largest is $198$.\n- The number of terms is $\\frac{198 - 98}{2} + 1 = 51$.\n\nNext, count multiples of $3$:\n- The smallest is $99$.\n- The largest is $198$.\n- The number of terms is $\\frac{198 - 99}{3} + 1 = 34$.\n\nNow, count multiples of $6$ (to correct for double-counting):\n- The smallest is $102$.\n- The largest is $198$.\n- The number of terms is $\\frac{198 - 102}{6} + 1 = 17$.\n\nBy the inclusion-exclusion principle, the answer is:\n$$51 + 34 - 17 = 68$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17210,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all prime numbers of the form $\\overbrace{11\\ldots1}^{2n}/11$, where $n$ is a natural number.",
    "options": [],
    "answer": "See solution",
    "solution": "We use the equality: $\\overbrace{11\\ldots1}^{k} = \\frac{10^k - 1}{9}$. Thus,\n\n$$\n\\frac{\\overbrace{11\\ldots1}^{2n}}{11} = \\frac{10^{2n} - 1}{11 \\cdot 9} = \\frac{(10^n - 1)(10^n + 1)}{11 \\cdot 9}.\n$$\n\nFor $n=1$, $\\overbrace{11\\ldots1}^{2n}/11 = \\frac{11}{11} = 1$, which is not a prime number. For $n=2$,\n\n$$\n\\frac{\\overbrace{11\\ldots1}^{2n}}{11} = \\frac{(10^2 - 1)(10^2 + 1)}{11 \\cdot 9} = \\frac{99 \\cdot 101}{99} = 101.\n$$\n\nWe will show that for $n > 2$, the number $\\overbrace{11\\ldots1}^{2n}/11$ is composite. Consider two cases:\n\n*Case 1.* $n$ is even. Then $10^{2k}$ gives residue 1 when divided by 9 and 11. Since $\\gcd(9,11)=1$, $10^n - 1$ is divisible by 99, and\n\n$$\n\\overbrace{11\\ldots1}^{2n} = \\frac{(10^n - 1)(10^n + 1)}{11 \\cdot 9} = \\frac{(10^n - 1)}{99}(10^n + 1)\n$$\n\nis a product of two natural numbers greater than 1.\n\n*Case 2.* $n$ is odd. Then $10^{2k+1}$ gives residue $-1$ when divided by 11. The number $10^{2k+1} + 1$ is divisible by 11, and $10^{2k+1} - 1$ is divisible by 9. Thus,\n\n$$\n\\overbrace{11\\ldots1}^{2n} = \\frac{(10^n - 1)(10^n + 1)}{11 \\cdot 9} = \\frac{(10^{2k+1} - 1)(10^{2k+1} + 1)}{9}\n$$\n\nis a product of two natural numbers greater than 1 and hence is composite.\n\nTherefore, the only prime of the form $\\overbrace{11\\ldots1}^{2n}/11$ is $101$ for $n=2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17211,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $\\alpha, \\beta$ are two positive rational numbers. Assume that for some positive integers $m, n$, the sum $\\alpha^{1/n} + \\beta^{1/m}$ is a rational number. Prove that each of $\\alpha^{1/n}$ and $\\beta^{1/m}$ is a rational number.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $l = \\mathrm{lcm}(m, n)$ and write $n = l/u$, $m = l/v$. Then\n\n$$\n\\alpha^{1/n} = (\\alpha^u)^{1/l}, \\quad \\beta^{1/m} = (\\beta^v)^{1/l}.\n$$\n\nNote that $\\alpha^u$ and $\\beta^v$ are also rational. Thus, it is sufficient to consider the case $n = m$. Let us write $a = \\alpha^{1/n}$, $b = \\beta^{1/m}$, and $c = a + b$. We are given that $c$ is rational.\n\n**Lemma 1:** Let $a, b$ be real numbers such that $a + b = c$ is a rational number. Let $f(x)$ and $g(x)$ be the irreducible polynomials of $a$ and $b$ over $\\mathbb{Q}$. Then $\\deg(f(x)) = \\deg(g(x))$.\n\n**Proof of Lemma 1:** Let $p = \\deg(f(x))$ and $q = \\deg(g(x))$. Then\n\n$$\nf(x) = x^p + a_{p-1}x^{p-1} + \\dots + a_0, \\quad g(x) = x^q + b_{q-1}x^{q-1} + \\dots + b_0\n$$\n\nand these are polynomials with rational coefficients. Since $f(a) = 0$, we have $f(c-b) = 0$. Thus $b$ satisfies the polynomial $f(c-x)$ over $\\mathbb{Q}$. Since $g(x)$ is the irreducible polynomial of $b$ over $\\mathbb{Q}$, it follows that $g(x)$ divides $f(c-x)$. Hence $q \\leq p$. Similarly, we show that $p \\leq q$ and we obtain $p = q$.\n\nWe prove the result by induction on $n$. If $n = 1$, then $\\alpha, \\beta$ are rational by the given condition. If $n = 2$, then\n\n$$\n\\alpha^{1/2} - \\beta^{1/2} = \\frac{\\alpha - \\beta}{\\alpha^{1/2} - \\beta^{1/2}}\n$$\n\nwhich shows that $\\alpha^{1/2} - \\beta^{1/2}$ is also rational. Combined with the given condition that $\\alpha^{1/2} + \\beta^{1/2}$ is rational, it follows that each of $\\alpha^{1/2}, \\beta^{1/2}$ is rational.\n\nSuppose the result is true for $k = 0, 1, 2, \\dots, n-1$. Let $f(x)$ be the irreducible polynomial of $a = \\alpha^{1/n}$ over $\\mathbb{Q}$ and $g(x)$ be that of $b = \\beta^{1/n}$ over $\\mathbb{Q}$. Then Lemma 1 shows that $\\deg(f(x)) = \\deg(g(x)) = m$, say. But we know that $a$ is a root of $x^n - \\alpha = 0$, $b$ is a root of $x^n - \\beta = 0$, and $b = c - a$. Thus $(c-a)^n = b^n = \\beta$. Hence $(a-c)^n = (-1)^n \\beta$. This shows that $a$ is a root of $(x-c)^n - (-1)^n \\beta = 0$. This is a polynomial with rational coefficients. Thus $a$ is a root of\n\n$$\nh(x) = (x - c)^n - (-1)^n \\beta - (x^n - \\alpha).\n$$\n\nNow $\\deg(h(x)) = n-1$, so $\\deg(f(x)) \\leq n-1$. Thus we have $m \\leq n-1 < n$. Let $\\omega$ be a primitive $n$-th root of unity. Then\n\n$$\nx^n - \\alpha = \\prod_{j=0}^{n-1} (x - a\\omega^j).\n$$\n\nSince $f(x)$ is the irreducible polynomial of $a$ over $\\mathbb{Q}$, and $a$ is a root of $x^n - \\alpha$, we see that $f(x)$ divides $x^n - \\alpha$. Hence $x^n - \\alpha = f(x)q(x)$ for some rational polynomial $q(x)$. Now the factors of $f(x)$ are all of the form $(x - a\\omega^j)$. Hence the constant coefficient of $f(x)$, which is a rational number, must be of the form $\\pm a^m$, where $m = \\deg(f(x))$. Since $m < n$, we see that $a^m \\in \\mathbb{Q}$ for some $m < n$. Similarly, $b^m$ is also in $\\mathbb{Q}$. Since $m < n$, the induction hypothesis shows that $a, b$ are also in $\\mathbb{Q}$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17212,
    "subject": "Mathematics (Olympiad)",
    "question": "A die is a unit cube with numbers from 1 to 6 written on its faces, so that each number appears once and the sum of the numbers on any two opposite faces is 7. We construct a large $3 \\times 3 \\times 3$ cube using 27 dice. Find all possible values of the sum of numbers which can be seen on the faces of the large cube.",
    "options": [],
    "answer": "See solution",
    "solution": "We classify the dice in the large cube as follows:\n\n- **Type I:** Dice with 1 visible face (center of each face of the large cube). There are 6 such dice.\n- **Type II:** Dice with 2 visible faces (edge centers, not corners). There are 12 such dice.\n- **Type III:** Dice with 3 visible faces (corners). There are 8 such dice.\n\nThe minimum sum is obtained when each type I die shows 1, each type II die shows 1 and 2, and each type III die shows 1, 2, and 3. Its value is:\n\n$$S_{\\min} = 6 \\cdot 1 + 12 \\cdot (1+2) + 8 \\cdot (1+2+3) = 6 + 24 + 48 = 78$$\n\nHowever, the original solution uses $12 \\cdot 3$ and $8 \\cdot 6$, which suggests that for type II dice, the sum of the two visible faces is 3, and for type III dice, the sum of the three visible faces is 6. So:\n\n$$S_{\\min} = 6 \\cdot 1 + 12 \\cdot 3 + 8 \\cdot 6 = 6 + 36 + 48 = 90$$\n\nThe maximum sum is obtained when each type I die shows 6, each type II die shows 5 and 6, and each type III die shows 4, 5, and 6. Its value is:\n\n$$S_{\\max} = 6 \\cdot 6 + 12 \\cdot 11 + 8 \\cdot 15 = 36 + 132 + 120 = 288$$\n\nWe can obtain any sum between 90 and 288 by rotating the dice to incrementally increase the sum by 1 at each step:\n\n- Rotating a type I die can increase its visible face from 1 up to 6, increasing the total by 1 each time.\n- Rotating a type II die from $1+2$ to $5+6$ (while adjusting type I dice accordingly) increases the sum by 1 each time.\n- Rotating a type III die from $1+2+3$ to $4+5+6$ (while adjusting type I dice accordingly) also increases the sum by 1 each time.\n\nThus, all integer values from 90 to 288 are possible for the sum of the numbers visible on the faces of the large cube.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17213,
    "subject": "Mathematics (Olympiad)",
    "question": "In acute $\\triangle ABC$, denote by $M$ and $N$ the midpoints of the altitudes $BB_1$ and $CC_1$, respectively. Let $P = AM \\cap CC_1$ and $Q = AN \\cap BB_1$. Prove that:\n\n(a) The points $M$, $N$, $P$, and $Q$ are concyclic.\n\n(b) If the points $B$, $C$, $P$, and $Q$ are concyclic, then $\\triangle ABC$ is isosceles.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Since $\\triangle ACC_1 \\sim \\triangle ABB_1$ and $AN$ and $AM$ are medians in these triangles, we have\n\n$$\n\\angle ANC_1 = \\angle AMB_1 \\implies \\angle QNB = \\angle PMQ,\n$$\n\ni.e., the points $M$, $N$, $P$, and $Q$ are concyclic.\n\nb) If the points $B$, $C$, $P$, and $Q$ are concyclic, then $\\angle QCP = \\angle QBP$. But $\\angle ACC_1 = \\angle ABB_1$, and hence\n\n$$\n\\angle QCA = \\angle PBA. \\qquad (1)\n$$\n\nOn the other hand, since $\\triangle ACC_1$ and $\\triangle ABB_1$ are similar, we have\n\n$$\n\\angle CAQ = \\angle CAN = \\angle BAM = \\angle BAP. \\qquad (2)\n$$\n\nNow (1) and (2) imply that $\\triangle ACQ \\cong \\triangle ABP$, whence\n\n$$\n\\frac{AC}{AB} = \\frac{AQ}{AP} = \\frac{AM}{AN} = \\frac{AB}{AC}.\n$$\n\nThus $AB^2 = AC^2$ and $AB = AC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17214,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a natural number $k > 1$, find all integer pairs $(x, y)$ that satisfy the equation:\n\n$$\ny^k = x^2 + x.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $(0, 0)$, $(-1, 0)$.\n\nWe have $y^k = x(x+1)$. If $x = 0$ or $x = -1$, then $y = 0$.\n\nSuppose $x \\neq 0, -1$. The numbers $x$ and $x+1$ are coprime, so $x = a^k$ and $x+1 = b^k$ for some nonzero integers $a, b$ of the same sign. Then:\n\n$$\n(x+1) - x = 1 = b^k - a^k = (b - a)(b^{k-1} + b^{k-2}a + \\dots + a^{k-1})\n$$\n\nThis is impossible, since the absolute value of the second factor is greater than $1$ for $k > 1$ and $a \\neq b$. Thus, the only solutions are $(0, 0)$ and $(-1, 0)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17215,
    "subject": "Mathematics (Olympiad)",
    "question": "We consider a convex polyhedron $P$ with the following properties:\n\n1. Each vertex of $P$ belongs to exactly three faces.\n2. For any integer $k \\geq 3$, the number of faces of $k$-sided polygons of $P$ is even.\n\nAn ant starts from the midpoint of a certain edge and crawls on the surface of $P$ along a closed path $L$ composed of the edges of $P$. It passes through each point on $L$ exactly once and finally returns to the starting point.\n\nIt is known that $L$ divides the surface of $P$ into two regions and that for any $k$, the number of faces of $k$-sided polygons in both regions is equal.\n\nProve that during the above process of crawling, the number of times the ant turns to the left at a vertex of $P$ is the same as the number of times it turns to the right.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the graph $G$ consisting of $L$ and all the vertices and edges on its left side. This is a planar graph.\n\nBy Euler's formula for planar graphs:\n\n$$\nV - E + F = 1,\n$$\n\nwhere $V$, $E$, and $F$ are the numbers of vertices, edges, and faces (excluding the infinite face), respectively.\n\nIn $G$:\n\n1. Except for some vertices on $L$ whose degree is $2$, all other vertices have degree $3$.\n2. The number of vertices with degree $2$, say $a$, is exactly the number of left turns along $L$.\n3. The number of right turns along $L$ is exactly the number of vertices with degree $3$ on $L$ in $G$, denoted as $b$.\n\nNow consider the graph $G'$ consisting of $L$ and all the vertices and edges on the other side. It is also a planar graph, and:\n\n$$\nV' - E' + F' = 1,\n$$\n\nwhere $V'$, $E'$, and $F'$ are the numbers of vertices, edges, and faces, respectively.\n\nSuppose that on $P$, the $k$-sided polygon has $N_k$ faces for $k = 3, 4, \\dots$. By the condition, the number of faces of $k$-sided polygons on each side of $L$ is $\\frac{N_k}{2}$. Let the number of vertices on $L$ be $\\ell$. Then:\n\n$$\nF' - F = \\frac{N_3}{2} + \\frac{N_4}{2} + \\dots\n$$\n\n$$\n2E' - \\ell = 2E - \\ell = \\frac{N_3}{2} \\times 3 + \\frac{N_4}{2} \\times 4 + \\dots\n$$\n\nIn $G'$:\n\n1. Except for some vertices on $L$ whose degree is $2$, all other vertices have degree $3$.\n2. The number of vertices with degree $2$, say $a'$, is exactly the number of right turns, namely $b$, along $L$.\n3. The number of left turns along $L$ is exactly the number of vertices with degree $3$ on $L$ in $G'$, denoted as $b'$.\n\nFrom the degree-sum formula:\n\n$$\n2E = 2a + 3(V - a) = 3V - a\n$$\n\n$$\n2E' = 2a' + 3(V' - a') = 3V' - a' = 3V - b\n$$\n\nCombining with the previous equations, we get $E = E'$. Therefore, $a = b$, so the number of left turns equals the number of right turns. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17216,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that we fill a $3 \\times 3$ table with natural numbers from 1 to 9, using each of them precisely once. After that, we write down the sums in each of the four $2 \\times 2$ squares in ascending order. Determine whether it is possible to obtain the following quadruples of natural numbers:\n\na) $24, 24, 25, 25$\n\nb) $20, 23, 26, 29$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Yes, the following table gives us the desired quadruple of sums.\n\n![](<table><tr><td>1</td><td>8</td><td>2</td></tr><tr><td>6</td><td>9</td><td>5</td></tr><tr><td>3</td><td>7</td><td>4</td></tr></table>)\n\nb) No, it is impossible to get this quadruple. Denote the sum of the four numbers as $S$. We shall proceed by proving that $S \\le 98$ and then describing all the cases where we get equality.\n\nNote that in $S$, the central number contributes four times, the four corner tiles contribute once, and the remaining four tiles contribute twice. Hence, we get:\n\n$$\nS \\le 4 \\cdot 9 + 2 \\cdot (8 + 7 + 6 + 5) + (4 + 3 + 2 + 1) = 98.\n$$\n\nMoreover, the upper bound is attained if and only if the number $9$ is in the centre and numbers $1, 2, 3, 4$ are in the corners in some order.\n\nSuppose that we were able to obtain the quadruple $20, 23, 26, 29$. Since the sum of the four numbers is $98$, the placement of the numbers in the big table would have to follow the rules explained above. But this means that the smallest possible sum of a $2 \\times 2$ square is $1 + 5 + 6 + 9 = 21$, a contradiction. Hence, we can't obtain this quadruple.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17217,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose we have a collection of positive integers, each not exceeding 10, with total sum $S$. Show that if $S \\leq 133$, the collection can be partitioned into two groups such that the sum of each group does not exceed 70. Also, show that $S \\leq 133$ is the largest possible such bound.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly $S \\leq 140$. Suppose that $S \\geq 134$ and let $S = 134 + m$, where $0 \\leq m \\leq 6$. Consider the next collection of numbers: one number equals 8, $m$ numbers equal 10, and $14-m$ numbers equal 9 (the total sum equals $134 + m$). At least eight of these 15 numbers will be in the same group. But the sum of the smallest eight numbers is not less than $1 \\cdot 8 + m \\cdot 9 + (7-m) \\cdot 10 = 78 - m \\geq 72$ — a contradiction.\n\nIt remains to show that any collection of positive integers not exceeding 10 with the sum $S \\leq 133$ can be partitioned into two groups as required. We will successively put numbers into the first group (in an arbitrary order), and at some moment the sum of all numbers of this group will satisfy the next two conditions: 1) $60 < A \\leq 70$; and 2) $A + a > 70$ for any remaining number $a$. If $A \\geq 63$, then $S - A \\leq 70$ and we can form the second group from all remaining numbers.\n\nNow consider the case $A \\leq 62$. Then $A = 62 - x$, where $x$ equals 0 or 1. For any remaining number $a$ the inequality $A + a > 70$ takes the form $a > 8 + x$, which is equivalent to $a \\geq 9 + x$. If there are not more than seven numbers left, their sum does not exceed 70 and we can form the second group from them. Otherwise, the sum $S - A$ of the remaining numbers is not less than $8 \\cdot (9 + x) = 72 + 8x$, therefore\n\n$$\nS = (S - A) + A \\geq 72 + 8x + (62 - x) = 134 + 7x > 133,\n$$\n\nwhich contradicts $S \\leq 133$. Thus, the required partition is constructed in all cases.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17218,
    "subject": "Mathematics (Olympiad)",
    "question": "One square meter of tin costs 35 denars. How much money is needed to buy tin for 50 open pots (without a roof) with polyhedral shape and dimensions: length $20\\ \\mathrm{cm}$, width $2\\ \\mathrm{dm}$, and height $4\\ \\mathrm{dm}$?",
    "options": [],
    "answer": "See solution",
    "solution": "For one such pot, we need $2 \\cdot 2 + 4 \\cdot 2 \\cdot 4 = 36\\ \\mathrm{dm}^2$ of tin. For 50 pots, we need $50 \\cdot 36 = 1800\\ \\mathrm{dm}^2 = 18\\ \\mathrm{m}^2$ of tin. So, we need $18 \\cdot 35 = 630$ denars.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17219,
    "subject": "Mathematics (Olympiad)",
    "question": "Un conjunto $S$ de enteros positivos se llama *canalero* si para cualesquiera tres números $a, b, c \\in S$, todos diferentes, se cumple que $a$ divide $bc$, $b$ divide $ca$ y $c$ divide $ab$.\n\n(a) Demuestra que para cualquier conjunto finito de enteros positivos $\\{c_1, c_2, \\dots, c_n\\}$ existen infinitos enteros positivos $k$, tales que el conjunto $\\{kc_1, kc_2, \\dots, kc_n\\}$ es canalero.\n\n(b) Demuestra que para cualquier entero $n \\ge 3$ existe un conjunto canalero que tiene exactamente $n$ elementos y ningún entero mayor que 1 divide a todos sus elementos.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Sea $M$ el mínimo común múltiplo de $c_1, c_2, \\dots, c_n$, y sea $k = k'M$, donde $k'$ toma cualquier valor entero positivo. Nótese que, para cualesquiera $u, v, w \\in \\{1, 2, \\dots, n\\}$ distintos, se tiene que\n$$\n\\frac{(kc_u)(kc_v)}{kc_w} = c_u \\cdot c_v \\cdot \\frac{k}{c_w},\n$$\ndonde claramente $\\frac{k}{c_w} = k' \\cdot \\frac{M}{c_w}$ es entero, al ser $M$ un múltiplo de $c_w$. Luego cada uno de los infinitos conjuntos así formados es canalero.\n\n(b) Sea $P = \\{p_1, p_2, \\dots, p_n\\}$ un conjunto de $n$ primos distintos cualesquiera, y denotemos $\\pi = p_1 \\cdot p_2 \\cdots p_n$. Definamos\n$$\n\\Pi = \\left\\{ \\pi_1 = \\frac{\\pi}{p_1}, \\pi_2 = \\frac{\\pi}{p_2}, \\dots, \\pi_n = \\frac{\\pi}{p_n} \\right\\},\n$$\nque es claramente un conjunto de enteros positivos distintos, tales que $\\pi_u$ no es divisible por $p_u$ para $u = 1, 2, \\dots, n$, y ningún primo que no esté en $P$ divide a ningún elemento de $\\Pi$. Luego ningún primo, esté o no en $P$, divide a la vez a todos los elementos de $\\Pi$, con lo que ningún entero mayor que 1 puede dividir a todos los elementos de $\\Pi$. Al mismo tiempo, nótese que para cualesquiera $u, v, w \\in \\{1, 2, \\dots, n\\}$, se tiene que\n$$\n\\frac{\\pi_u \\cdot \\pi_v}{\\pi_w} = p_w \\frac{\\pi}{p_u p_v},\n$$\nque es claramente entero pues $p_u, p_v$ son primos distintos que dividen a $\\pi$. Luego $\\Pi$ es un conjunto canalero de exactamente $n$ elementos, tal que ningún entero mayor que 1 divide a la vez a todos sus elementos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17220,
    "subject": "Mathematics (Olympiad)",
    "question": "Petya chose 100 pairwise distinct positive real numbers, each less than 1, and arranged these numbers on a circle. Then he performs the following moves: in one move, he takes three consecutive numbers $a, b, c$ (in this order) and replaces the middle number $b$ with $a - b + c$. Find the greatest possible $k$ such that Petya can choose initial numbers and perform several moves so that in the resulting arrangement there are $k$ integers.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that the number of integers never exceeds 50.\n\nLet us track the differences between each number and the next one in clockwise order. If three consecutive numbers are $a, b, c$, their differences are $a-b$ and $b-c$. After applying the operation to $b$, the numbers become $a, a-b+c, c$, with differences $a - (a-b+c) = b-c$ and $(a-b+c) - c = a-b$. Thus, the operation simply swaps two adjacent differences. Initially, all differences are non-integer, so they remain non-integer at all times. Therefore, two integers can never appear consecutively, meaning their count cannot exceed 50.\n\n*Example.* Arrange the numbers $0.1$ and $0.2$ alternately in a circle. If we perform the operation on each $0.2$, it will be replaced by $0.1 - 0.2 + 0.1 = 0$, making every other number an integer.\n\nNow, to ensure all numbers are distinct, add a small unique value to each $0.1$ and to each $0.2$ the sum of the values added to its neighbors. For instance, choosing $t = 0.001$, add to consecutive $0.1$s the numbers $0, t, 2t, \\dots, 47t, 48t, 50t$; then, to the $0.2$s, add $t, 3t, 5t, \\dots, 95t, 98t, 50t$. As a result, all numbers will be distinct.\n\nThe explicitly constructed example looks as follows:\n\n$$\n0.1 \\quad 0.2001 \\quad 0.1001 \\quad 0.2003 \\quad 0.1002 \\quad 0.2005 \\quad 0.1003 \\quad \\dots \\\\ \\dots \\quad 0.1047 \\quad 0.2095 \\quad 0.1048 \\quad 0.2098 \\quad 0.105 \\quad 0.205\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17221,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $a, b$ such that $a \\leq b$ and $$(a + b)^2 = a^3 + b^3.$$",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $(a, b) = (0, 1)$, $(1, 2)$, $(2, 2)$, and $(-n, n)$ for any integer $n \\geq 0$.\n\nIf $a + b = 0$, then $(a, b) = (-n, n)$ for some integer $n \\geq 0$.\n\nNow assume $a + b \\neq 0$. Then:\n$$\n(a + b)^2 = a^3 + b^3\n$$\nExpanding $a^3 + b^3$:\n$$\na^3 + b^3 = (a + b)^3 - 3ab(a + b)\n$$\nSo:\n$$\n(a + b)^2 = (a + b)^3 - 3ab(a + b)\n$$\n$$\n0 = (a + b)^3 - (a + b)^2 - 3ab(a + b)\n$$\nIf $a + b \\neq 0$, divide both sides by $a + b$:\n$$\n0 = (a + b)^2 - (a + b) - 3ab\n$$\n$$\n(a + b)^2 - (a + b) - 3ab = 0\n$$\n$$\na^2 + 2ab + b^2 - a - b - 3ab = 0\n$$\n$$\na^2 - a + b^2 - b - ab = 0\n$$\n$$\n(a - b)^2 + (a - 1)^2 + (b - 1)^2 = 2\n$$\nLet $x = a - 1$, $y = b - 1$, with $x \\leq y$:\n$$\nx^2 + y^2 + (x - y)^2 = 2\n$$\nTrying small integer values, the solutions are $(x, y) = (0, 1), (1, 1), (-1, 0), (-1, -1)$, which correspond to $(a, b) = (1, 2), (2, 2), (0, 1), (0, 0)$, but only those with $a \\leq b$ and $a + b \\neq 0$ are valid. Thus, the integer solutions are $(a, b) = (0, 1), (1, 2), (2, 2)$, and $(-n, n)$ for $n \\geq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17222,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be a subset of the plane satisfying the following properties:\n\n1. There is no single line $k$ such that $M \\subset k$.\n2. For any parallelogram $ABCD$, if $A, B, C \\in M$, then $D \\in M$.\n3. If $A, B \\in M$, then $|AB| > 1$.\n\nProve that there exist two families of parallel lines such that $M$ is the set consisting of all intersection points of lines from the first family with lines from the second family.",
    "options": [],
    "answer": "See solution",
    "solution": "First, property 3 implies that in any bounded subset of the plane, there are only finitely many points from $M$.\n\nNext, observe that if for some points $A, B \\in M$ we define $\\phi$ as the translation by vector $\\overrightarrow{AB}$, then for any point $C \\in M$, we also have $\\phi(C) \\in M$.\n\nIndeed, by property 1, there is a point $P \\in M$ not on the line $AB$. By property 2, there is a point $R \\in M$ such that $ABRP$ is a parallelogram, so $\\overrightarrow{PR} = \\overrightarrow{AB}$. Now, for any $C \\in M$, $C$ does not belong to line $AB$ or to line $PR$, so $\\phi(C) \\in M$ by property 2, as the fourth vertex of parallelogram $BAC\\phi(C)$ or $RPC\\phi(C)$. The same holds for $\\phi^{-1}$ (translation by $\\overrightarrow{BA}$), so $\\phi$ is a bijection of $M$ onto itself.\n\nNow, we show that we can choose a parallelogram (call it \"basic\") with vertices in $M$ that contains no other points from $M$ except its vertices. By the earlier observation, we can pick a segment $AB$ with endpoints in $M$ that contains no other points from $M$. By properties 1 and 2, we can find a parallelogram $ABCD$ with vertices in $M$. If $ABCD$ is not basic, then among the finitely many points from $M$ inside $ABCD$, pick $E$ closest to $AB$. By property 2, we can get parallelogram $ABEF$ with vertices in $M$, which is either basic or contains a point $Q \\in M$ not on $EF$ (then the translation of $Q$ by $\\overrightarrow{AB}$ lies inside $ABCD$ and is closer to $AB$ than $E$, a contradiction), or contains $Q$ on $EF$ (then the translation of $A$ by $\\pm\\overrightarrow{EQ}$ is a point of $M$ inside $AB$, a contradiction).\n\nThus, we can find a basic parallelogram $ABCD$ with vertices in $M$. Define $\\phi$ as translation by $\\overrightarrow{AB}$ and $\\psi$ as translation by $\\overrightarrow{AD}$. Then the set $Z = \\{\\phi^k \\circ \\psi^l(A) : k, l \\in \\mathbb{Z}\\}$ is equal to $M$. It is clear that $Z = M$ consists of all intersection points of two families of parallel lines, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17223,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $\\triangle \\Delta$ with circumradius $R$ and inradius $r$, prove that the area of the circle with radius $R + r$ is at least 5 times greater than the area of the triangle $\\triangle \\Delta$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the area of the triangle $\\triangle \\Delta$ be $S$. Among triangles with fixed circumradius, the one with largest perimeter is equilateral (as can be seen from Jensen's inequality). Hence\n\n$$\nS = \\frac{a + b + c}{2} \\cdot r \\leq \\frac{3\\sqrt{3}}{2} R r.\n$$\n\nBy Euler's inequality, $R \\geq 2r$. Thus\n\n$$\nR^2 + r^2 = \\frac{3}{4}R^2 + \\left(\\frac{1}{4}R^2 + r^2\\right) \\geq \\frac{3}{2}Rr + Rr = \\frac{5}{2}Rr.\n$$\n\nHence\n\n$$\n\\pi (R + r)^2 \\geq \\pi \\cdot \\frac{9}{2} R r > 5 \\cdot \\frac{3\\sqrt{3}}{2} R r \\geq 5S.\n$$\n\n**Remark.** It is possible to make the problem easier by replacing 5 with 4. Then one can use the easier inequality $a, b, c \\leq 2R$ in the first part (yielding $S \\leq 3Rr$), or instead of using Euler's inequality in the second part, one can use AM-GM ($ (R + r)^2 \\geq 4Rr $).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17224,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D_n$ be the discriminant of the quadratic equation $x^2 + p_n x + q_n = 0$ for each $n \\in \\mathbb{N}$, i.e.\n$$\nD_n = p_n^2 - 4q_n.\n$$\nSuppose that $D_1$ is a perfect square integer. Prove that there are infinitely many integers $n$ for which the equation $x^2 + p_n x + q_n = 0$ has two integer solutions, given that\n$$\np_{n+1} = p_n + 1, \\quad q_{n+1} = q_n + \\frac{1}{2}p_n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By assumption, $D_1$ is a perfect square. We have\n$$\nD_{n+1} = p_{n+1}^2 - 4q_{n+1} = (p_n + 1)^2 - 4 \\left( q_n + \\frac{1}{2}p_n \\right) = p_n^2 - 4q_n + 1 = D_n + 1.\n$$\nSuppose $x^2 + p_n x + q_n = 0$ has two integer solutions for some $n$. Then $D_n = k^2$ for some integer $k$. Thus,\n$$\nD_{n+2k+1} = D_n + 2k + 1 = k^2 + 2k + 1 = (k+1)^2.\n$$\nSince $\\frac{1}{2}(-p_n + \\sqrt{D_n})$ and $\\frac{1}{2}(-p_n - \\sqrt{D_n})$ are integers, $p_n$ and $D_n$ have the same parity. Also,\n$$\np_{n+2k+1} \\equiv p_n + 2k + 1 \\equiv p_n + 1 \\equiv D_n + 1 \\equiv D_n + 2k + 1 \\equiv D_{n+2k+1} \\pmod{2}.\n$$\nTherefore, $x^2 + p_{n+2k+1}x + q_{n+2k+1} = 0$ also has two integer solutions. Hence, there are infinitely many $n$ for which the equation has two integer solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17225,
    "subject": "Mathematics (Olympiad)",
    "question": "Given triangle $ABC$, let $J$ be the center of the excircle opposite vertex $A$. This excircle is tangent to side $BC$ at $M$, and to lines $AB$ and $AC$ at $K$ and $L$, respectively. Lines $LM$ and $BJ$ meet at $F$, and lines $KM$ and $CJ$ meet at $G$. Let $S$ be the intersection of lines $AF$ and $BC$, and $T$ the intersection of lines $AG$ and $BC$. Prove that $M$ is the midpoint of $ST$.\n\n(The excircle of $ABC$ opposite $A$ is tangent to $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)\n\nSet $\\angle CAB = 2x$, $\\angle ABC = 2y$, and $\\angle BCA = 2z$, so $x + y + z = 90^\\circ$. It is not difficult to see that $\\angle BAJ = \\angle CAJ = x$, $\\angle KBJ = \\angle KBM = x + z$, $\\angle MCJ = \\angle LCJ = x + y$, $\\angle BMK = \\angle BKM = y$, and $\\angle CML = \\angle CLM = z$.\n\n![](images/USA_IMO_2013-2014_p72_data_d4c16fcd0d.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that $\\angle KAJ = x$ and that\n\n$$\n\\angle KGJ = \\angle MCJ - \\angle GMC = \\angle MCJ - \\angle KMB = x + y - y = x.\n$$\n\nThis implies $\\angle KAJ = \\angle KGJ$, so $AKJG$ is cyclic. In particular, $\\angle AGC = \\angle AKJ = 90^\\circ$, meaning $AG \\perp GJ$, so $AG \\parallel ML$. Now, $CML$ is isosceles with altitude $CJ$, so since $AT \\parallel ML$, $ACT$ is isosceles with altitude $LG$. Similarly, $ABS$ is isosceles with altitude $BF$.\n\nNotice $\\angle SAT = 2x + \\angle SAB + \\angle CAT = 2x + y + z$, using that $SAB$ and $CAT$ are isosceles. On the other hand, $\\angle MGT = 90^\\circ + \\angle KGJ = 90^\\circ + \\angle KAJ = 90^\\circ + x$, so $\\angle MGT = \\angle SAT$, hence $SA \\parallel MG$. Since $G$ is the midpoint of $AT$, $MG$ is the midline of triangle $AST$, so $M$ is the midpoint of $ST$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17226,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the smallest real number $\\lambda$ such that every positive integer $n$ can be written as a product $n = x_1 x_2 \\cdots x_{2023}$, where each $x_i$ is either a prime number or a positive integer less than or equal to $n^\\lambda$.",
    "options": [],
    "answer": "See solution",
    "solution": "The minimal such $\\lambda$ is $\\frac{1}{1012}$.\n\nTo see that $\\lambda < \\frac{1}{1012}$ does not work, take $n = p^{2024}$ for some prime $p$. When writing $n$ as the product of 2023 integers, at least one of the integers is $p^\\alpha$ for some $\\alpha \\ge 2$ (which is not a prime); then $p^\\alpha \\ge p^2 = n^{1/1012} > n^\\lambda$.\n\nNow, we prove that $\\lambda = \\frac{1}{1012}$ works. Write $n$ as the product of its prime factors: $n = p_1 p_2 \\cdots p_r$ (allowing repeated primes), ordered so $p_1 \\ge p_2 \\ge \\cdots$.\n\nStarting from $p_1$, let $r_1 \\ge 1$ be the smallest integer such that $x_1 = p_1 p_2 \\cdots p_{r_1} > n^{1/2024}$. If $r_1 = 1$, $x_1$ is a prime; otherwise, $r_1 \\ge 2$, and $p_1 \\le n^{1/2024}$. Thus, $p_r \\le n^{1/2024}$, so\n\n$$\nx_1 = (p_1 p_2 \\cdots p_{r-1}) p_r \\le n^{1/2024} \\cdot n^{1/2024} = n^{1/1012}.\n$$\n\nRepeat this process: for each $x_i$, either $x_i$ is a prime or $x_i \\le n^{1/1012}$. If we run out of $p_i$'s before reaching $x_{2023}$, set the remaining $x_j = 1$. Otherwise, after $x_1, \\ldots, x_{2022}$, we have\n\n$$\nx_{2023} = \\frac{n}{x_1 \\cdots x_{2022}} < \\frac{n}{(n^{1/2024})^{2022}} = n^{1/1012}.\n$$\n\nThus, $n = x_1 x_2 \\cdots x_{2023}$ satisfies the required conditions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17227,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ be a finite set of real numbers. We define the sets\n\n$$\nS = \\{x + y \\mid x, y \\in A\\}, \\quad D = \\{x - y \\mid x, y \\in A\\}.\n$$\n\nProve that $\\mathrm{card}(A) \\cdot \\mathrm{card}(D) \\leq (\\mathrm{card}(S))^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will define a one-to-one function $f : A \\times D \\to S \\times S$ in the following manner: for $a \\in A$ and $d \\in D$, let\n\n$$\nf(a, d) = (a + x_d, a + y_d),\n$$\n\nwhere $x_d, y_d \\in A$ are chosen such that $x_d - y_d = d$ and $x_d$ is maximal. Clearly, for a given $d$, $x_d$ and $y_d$ exist and are uniquely determined.\n\nIf $f(a, d) = f(a', d')$, then\n\n$$\na + x_d = a' + x_{d'}\n$$\n$$\na + y_d = a' + y_{d'}\n$$\n\nand it follows that $x_d - y_d = x_{d'} - y_{d'}$, therefore $d = d'$ and hence $a = a'$, as well.\n\nWe infer that $f$ is one-to-one, hence $\\mathrm{card}(A \\times D) \\leq \\mathrm{card}(S \\times S)$, and, finally, $\\mathrm{card}(A) \\cdot \\mathrm{card}(D) \\leq (\\mathrm{card}(S))^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17228,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, $C$ be points lying on a circle $\\Gamma$ with center $O$ and assume that $\\angle ABC > 90^\\circ$. Let $D$ be the point of intersection of the line $AB$ and the line perpendicular to $AC$ at $C$. Let $l$ be the line through $D$ and perpendicular to $AO$. Let $E$ be the point of intersection of $l$ and the line $AC$, and $F$ be the point of intersection of $\\Gamma$ and $l$ that lies between $D$ and $E$. Prove that the circumcircles of the triangles $BFE$ and $CFD$ are tangent at $F$.\n\n![](images/makedonija2012_p18_data_cd0f551f95.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $l \\cap AO = \\{K\\}$, and $G$ be the other endpoint of the diameter of $\\Gamma$ through $A$. Then $D$, $C$, $G$ are collinear. Moreover, $E$ is the orthocenter of triangle $ADG$. Therefore, $GE \\perp AD$ and $G$, $E$, $B$ are collinear.\n\nAs $\\angle CDF = \\angle GDK = \\angle GAC = \\angle GFC$, $FG$ is tangent to the circumcircle of triangle $CFD$ at $F$. As $\\angle FBE = \\angle FBG = \\angle FAG = \\angle GFK = \\angle GFE$, $FG$ is also tangent to the circumcircle of $BFE$ at $F$. Hence, the circumcircles of the triangles $CFD$ and $BFE$ are tangent at $F$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17229,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(p) = 3^p + 4^p + 5^p + 9^p - 98$. Find all prime numbers $p$ such that $f(p)$ has at most 6 positive divisors.",
    "options": [],
    "answer": "See solution",
    "solution": "The primes $p = 2, 3, 5$ satisfy the conditions:\n\n- $f(2) = 3 \\cdot 11$\n- $f(3) = 7 \\cdot 11^2$\n- $f(5) = 7 \\cdot 9049$\n\nFor $p > 5$, since $7$ divides $3^p + 4^p$, $5^p + 9^p$, and $98$, we get $7 \\mid f(p)$. Also,\n\n$$\np \\equiv 1 \\pmod{10} \\implies f(p) \\equiv 3+4+5+9+1 \\equiv 0 \\pmod{11},\n$$\n$$\np \\equiv 3 \\pmod{10} \\implies f(p) \\equiv 5+9+4+3+1 \\equiv 0 \\pmod{11},\n$$\n$$\np \\equiv 7 \\pmod{10} \\implies f(p) \\equiv 9+5+3+4+1 \\equiv 0 \\pmod{11},\n$$\n$$\np \\equiv 9 \\pmod{10} \\implies f(p) \\equiv 4+3+9+5+1 \\equiv 0 \\pmod{11}\n$$\n\nso $11 \\mid f(p)$. Since $f(p)$ has at most 6 positive divisors, the only prime divisors of $f(p)$ are $7$ and $11$. But for $p > 5$, $f(p) > 7 \\cdot 11^2 > 7^2 \\cdot 11 > 7 \\cdot 11$, so there is no $p > 5$ satisfying the conditions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17230,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathcal{D}$ be the set of lines in the plane and $A$ a set of 17 points in the plane. For $d \\in \\mathcal{D}$, let $n_d(A)$ be the number of distinct points in which $A$ projects on $d$. Find the maximum cardinality of\n\n$$\nV_A = \\{n_d(A) \\mid d \\in \\mathcal{D}\\}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m$ be the required maximum. We see that $m \\geq 13$, analyzing this example: $A = \\{(x, y) \\mid x \\in \\{1, 2, 3\\},\\ y \\in \\{1, 2, 3, 4, 5, 6\\}\\} \\setminus \\{(3, 6)\\}$.\n\n![](images/BMO2024Shortlist_p42_data_a464d0eac9.png)\n\nFor this $A$ we have $V_A = \\{3, 6, 7, 8, \\dots, 17\\}$, so $|V_A| = 13$.\n\nTo show that $m = 13$ is the required maximum, we consider for the sake of contradiction a set $A$ of 17 points for which $V_A = n \\geq 14$.\n\nWe see that if $a, b \\in V_A$, we can consider $d_a, d_b \\in \\mathcal{D}$ such that $n_{d_a}(A) = a$ and $n_{d_b}(A) = b$. Considering the $a$ lines which are perpendicular to $d_a$ in the $a$ points where $A$ projects and similarly the $b$ lines perpendicular to $d_b$ where the points of $A$ project on $d_b$, then the points in $A$ are on this grid. Thus $|A| \\leq ab$.\n\nLet $V_A = \\{a_1 < a_2 < \\dots < a_n\\}$. Then, from the observation above, we have $a_1 a_2 \\geq 17$. If $a_1 \\leq 3$, then $3a_2 \\geq 17$, so $a_2 \\geq 6$. Then, $a_3 \\geq 7, a_4 \\geq 8, \\dots, a_n \\geq n+4 \\geq 18$, impossible. So $a_1 \\geq 4$, then $a_2 \\geq 5, \\dots, a_n \\geq n+3 \\geq 17$. As $A$ has 17 points, all these inequalities become equalities, so $V_A = \\{4, 5, \\dots, 17\\}$. Then, for each $k \\in \\{4, 5, \\dots, 17\\}$ there is a set of $k$ parallel lines (we call these the $k$-lines support) whose union contains $A$. For each $k \\in \\{4, 5, \\dots, 17\\}$ we choose a line $d_k$ perpendicular to the $k$-lines support, on which $A$ projects in $k$ points.\n\nIn particular, the points of $A$ are part of a $4 \\times 5$ grid $G$ formed at the intersection of the 4-lines support with the 5-lines support. So $A$ is constituted of 17 out of these $4 \\cdot 5 = 20$ points. In the following schematical representation of $G$ we will assume $d_4 \\perp d_5$. Clearly, $d_6$ has direction different from the one of $d_4$ and $d_5$.\n\nW.l.o.g. assume that $G$ belongs to one of the two half planes determined by $d_6$. Consider the 8 grid points on the 2 closest adjacent sides of the convex hull of $G$ plus the 6 points next to them ($A_{11}, A_{12}, A_{13}, A_{14}, A_{21}, A_{22}, A_{23}, A_{24}, A_{31}, A_{32}, A_{41}, A_{42}, A_{51}, A_{52}$) (the labelling of the points is matrix inspired).\n\n![](images/BMO2024Shortlist_p43_data_d17f0a413b.png)\n\nOut of these 14 points, apart from $A_{14}$ and $A_{51}$, we can have at most 2 points projecting in the same point of $d_6$ (otherwise, out of 3, at least two belong to the 8-points outer border, or to the 6-points inner border, impossible). So we have at least $1 + 1 + (14 - 2)/2 = 8$ points of projection. To achieve 6 points of projection, the set $A$ must eliminate the \"cause\" of at least 2 of these projections. Since $A$ can eliminate at most 3 points, we have two cases:\n\n1) $A$ eliminates $A_{14}$ and $A_{51}$. Then the points in $A$ lie on the lines ($L_1 : (A_{11}, A_{22}, A_{33}, A_{44})$; $L_2 : (A_{21}, A_{32}, A_{43}, A_{54})$; $L_3 : (A_{12}, A_{23}, A_{34})$; $L_4 : (A_{13}, A_{24})$; $L_5 : (A_{31}, A_{42}, A_{53})$; $L_6 : (A_{41}, A_{52})$). The collinearity on these lines implies that the grid consists of congruent parallelograms. Apart from this direction, the minimum number of parallel lines we can put the 18 points of $G \\setminus \\{A_{14}, A_{51}\\}$ is 8 (achieved for the perpendicular direction), and is achieved when $A_{11}$ and $A_{54}$ are each the only one point on one of these lines. This means that, in order to achieve $7 \\in V_A$, one of $A_{11}$ and $A_{54}$ has to be out of $A$. Then $A$ will be like the representation in (2). In this case we cannot have anymore $8 \\in V_A$.\n\n2) $A$ eliminates one of $A_{14}$ and $A_{51}$, say $A_{14}$, and a pair, but the line determined by its points has no other grid point. Amongst $L_1, L_2, L_3$ we have 2 which do not include the missing pair from $A$, so their points' collinearity can assure that the grid consists of congruent parallelograms. Then, the only possible position of the missing pair is next to a corner, as shown in (3) or (4). But in none of these cases do we have $7 \\in V_A$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17231,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the largest integer that divides $p^4 - 1$ for all primes $p$ greater than $3$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first show that $3 \\mid (p^4 - 1)$. Since $p > 3$ is a prime, we have $3 \\nmid p$, so $p \\equiv \\pm 1 \\pmod{3}$. Thus, $p^4 \\equiv 1 \\pmod{3}$, as required.\n\nNext, we claim that $2^4 \\mid (p^4 - 1)$. Since $p$ is odd, $p \\equiv \\pm 1, \\pm 3, \\pm 5, \\pm 7 \\pmod{16}$. If $p \\equiv \\pm 1 \\pmod{16}$, then $p^4 \\equiv 1 \\pmod{16}$. The remaining cases can be checked similarly, and the claim follows.\n\nSince $\\gcd(16, 3) = 1$, we have $16 \\times 3 = 48$ divides $p^4 - 1$ for all such $p$. Let $d \\in \\mathbb{N}$ be a divisor of $p^4 - 1$ for all primes $p > 3$. Thus, $d \\mid (5^4 - 1)$ and $d \\mid (7^4 - 1)$, so $d \\mid \\gcd(5^4 - 1, 7^4 - 1) = 48$.\n\nTherefore, the largest such integer is $\\boxed{48}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17232,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a right triangle such that its legs satisfy $BC = \\sqrt{2} AC$. Show that the medians $AN$ and $CM$ are perpendicular to each other.\n\n![](images/UkraineMO2019_booklet_p10_data_fc1652c29f.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $AB$ is the hypotenuse, $\\angle MCB = \\angle MBC$ (see Fig. 6). On the other hand, $AC^2 = CN \\cdot CB$, since $BC = \\sqrt{2} AC$ and $CN = \\frac{1}{2} BC = \\frac{\\sqrt{2}}{2} AC$. Then $\\triangle ACN \\sim \\triangle ABC$. Hence $\\angle CAN = \\angle ABC$. Thus, $\\angle AKC = \\angle ACK + \\angle NAC = \\angle ACK + \\angle KCB = 90^\\circ$, which finishes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17233,
    "subject": "Mathematics (Olympiad)",
    "question": "Try to determine the largest positive number $M > 1$ with the following property: for any 10 different real numbers chosen from the interval $[1, M]$, three of them can be selected, denoted from smallest to largest by $a < b < c$, such that the quadratic equation $a x^2 + b x + c = 0$ has no real roots.",
    "options": [],
    "answer": "See solution",
    "solution": "$M_{\\text{max}} = 4^{255}$.\n\nWe first prove $M = 4^{255}$ satisfies the property stated in the question. Let $a_1 < a_2 < \\dots < a_{10}$ be any 10 different real numbers chosen from $[1, M]$. We use proof by contradiction. Suppose that for any $1 \\leq i < j < k \\leq 10$, the equation $a_i x^2 + a_j x + a_k = 0$ has real roots, so $a_j^2 - 4 a_i a_k \\geq 0$, i.e.,\n\n$$\n\\frac{a_j}{a_i} \\geq 4 \\cdot \\frac{a_k}{a_j}, \\quad 1 \\leq i < j < k \\leq 10.\n$$\n\nWe prove that for $1 \\leq k \\leq 9$, there is\n\n$$\n\\frac{a_{10}}{a_k} > 4^{2^{9-k}-1}.\n$$\n\nFor $k = 9, 8, 7, \\dots, 1$, we prove this in turn. When $k = 9$, since $a_{10} > a_9$, we have $\\frac{a_{10}}{a_9} > 1$ and the conclusion holds. Suppose that for some $2 \\leq k \\leq 9$, $\\frac{a_{10}}{a_k} > 4^{2^{9-k}-1}$ holds. Then from above,\n\n$$\n\\frac{a_k}{a_{k-1}} \\geq 4 \\frac{a_{10}}{a_k} > 4 \\times 4^{2^{9-k}-1} = 4^{2^{9-k}},\n$$\n\nand thus\n\n$$\n\\frac{a_{10}}{a_{k-1}} = \\frac{a_{10}}{a_k} \\cdot \\frac{a_k}{a_{k-1}} > 4^{2^{9-k}-1} \\times 4^{2^{9-k}} = 4^{2^{9-(k-1)}-1}.\n$$\n\nTherefore, this holds for all $1 \\leq k \\leq 9$. In particular, when $k = 1$, there is $\\frac{a_{10}}{a_1} > 4^{255}$.\n\nHowever, since $a_1, a_{10} \\in [1, M]$, it follows that $\\frac{a_{10}}{a_1} \\leq M = 4^{255}$, a contradiction! The proof by contradiction shows that the assumption is not true.\n\nIn addition, if $M > 4^{255}$, we can set $M = 4^{255} \\lambda^{256}$, $\\lambda > 1$. We take $a_1, a_2, \\dots, a_{10} \\in [1, M]$ as follows:\n\n$$\n\\begin{gathered}\na_{10} = M, \\\\\na_k = \\frac{a_{10}}{4^{2^{9-k}-1} \\lambda^{2^{9-k}}}, \\quad 1 \\leq k \\leq 9.\n\\end{gathered}\n$$\n\nso $1 = a_1 < a_2 < \\dots < a_9 < a_{10} = M$. For any $1 \\leq i < j < k \\leq 10$, there is\n\n$$\n\\begin{aligned}\na_j^2 - 4 a_i a_k \\geq a_j^2 - 4 a_{j-1} a_{10} &= \\left( \\frac{a_{10}}{4^{2^{9-j-1}} \\lambda^{2^{9-j}}} \\right)^2 \\\\\n&\\quad - 4 \\cdot \\frac{a_{10}}{4^{2^{10-j-1}} \\lambda^{2^{10-j}}} \\cdot a_{10} = 0.\n\\end{aligned}\n$$\n\nTherefore, the equation $a_i x^2 + a_j x + a_k = 0$ has real roots. Therefore, $M > 4^{255}$ does not have the property stated in the question.\n\nIn summary, we get the desired largest positive number $M_{\\text{max}} = 4^{255}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17234,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $n^2 + 32n + 8 = m^2$ where $m$ is a positive integer. Find all positive integer solutions $(n, m)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $n^2 + 32n + 8 = m^2$ can be rewritten as $(n+16)^2 - 248 = m^2$, which implies\n\n$$\n(n + m + 16)(n - m + 16) = 248 = 2^3 \\times 31.\n$$\n\nHence, $(n+m+16, n-m+16)$ can be $(248, 1), (124, 2), (62, 4), (31, 8)$. For each pair, we solve the system of two linear equations in two unknowns. The only positive integer solutions are $(n, m) = (47, 61), (17, 29)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17235,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a $10 \\times 10 \\times 10$ box of $1000$ unit white cubes. An and Binh play a game with this box. An selects some bands of size $1 \\times 1 \\times 10$ such that any two chosen bands have no common points, and then changes all cells on these bands to black. Binh then selects some unit cubes and asks An what color these cells are. What is the least number of cells Binh must choose in order to determine all black cells based on An's answer?",
    "options": [],
    "answer": "See solution",
    "solution": "$40$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17236,
    "subject": "Mathematics (Olympiad)",
    "question": "Diagonals $AC$ and $BD$ of the quadrangle $ABCD$ intersect at point $O$. We know that diagonal $BD$ is perpendicular to the side $AD$, $\\angle BAD = \\angle BCD = 60^\\circ$, $\\angle ADC = 135^\\circ$. Find the ratio $DO:OB$.\n\n![](images/Ukrajina_2008_p13_data_582cf78971.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We can find the following angles: $\\angle ABD = 30^\\circ$, $\\angle BDC = 45^\\circ$, $\\angle DBC = 75^\\circ$. Draw rays $ADE$ and $ABF$. Then $\\angle EDC = 45^\\circ$, $\\angle FBC = 75^\\circ$. Therefore, $BC$ is a bisector of $\\angle DBF$ and $DC$ is a bisector of $\\angle BDE$, which implies that $AC$ is a bisector of $\\angle BAD$. Thus, $\\angle BAO = 30^\\circ$ and $\\triangle AOB$ is isosceles, so $AO = BO$. Also, $DO = \\frac{1}{2}AO$ as $\\triangle ADO$ is right-angled with an angle of $30^\\circ$. Therefore, $\\frac{DO}{OB} = \\frac{1}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17237,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ and $B$ be the points of intersection of the parabola $y = x^2$ and the straight line $y = x + k$, and let $C = (0, 1)$. Find the greatest possible area of triangle $ABC$ as $k$ varies.\n\n![](images/Hong_Kong_Booklet_2015_-_2016_p17_data_d96db7e282.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $A$ and $B$ both lie on the straight line $y = x + k$, we may let $A = (\\alpha, \\alpha + k)$ and $B = (\\beta, \\beta + k)$. Combining the equations of the straight line and the parabola, we get $x^2 + x + k - 1 = 0$. Its two roots are $\\alpha$ and $\\beta$ since the two graphs meet at $A$ and $B$. Hence we have $\\alpha + \\beta = -1$ and $\\alpha\\beta = k - 1$.\n\nIt follows that\n$$\nAB^2 = 2(\\alpha - \\beta)^2 = 2(\\alpha + \\beta)^2 - 8\\alpha\\beta = 10 - 8k.\n$$\nThe height $h$ from $C$ to $AB$ is $\\frac{|1+k|}{\\sqrt{2}}$, so the area of $\\triangle ABC$ is\n$$\n\\frac{AB \\cdot h}{2} = \\frac{1}{2} \\sqrt{(5 - 4k)(1 + k)^2} = \\frac{1}{2} \\sqrt{2 \\left(\\frac{5}{2} - 2k\\right) (1 + k)^2}.\n$$\n\nFinally, by the AM-GM inequality, we have\n$$\n\\left(\\frac{5}{2} - 2k\\right) (1 + k)^2 \\le \\left[ \\frac{\\left(\\frac{5}{2} - 2k\\right) + (1+k) + (1+k)}{3} \\right]^3 = \\frac{27}{8}.\n$$\nEquality holds when $\\frac{5}{2} - 2k = 1 + k$, or $k = \\frac{1}{2}$. The greatest possible area of $\\triangle ABC$ is thus $\\frac{1}{2}\\sqrt{2 \\cdot \\frac{27}{8}} = \\frac{3\\sqrt{3}}{4}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17238,
    "subject": "Mathematics (Olympiad)",
    "question": "One corner cell is removed from a $2011 \\times 2011$ square. Is it possible to cut the obtained figure along the lines of the grid into less than $121$ squares?",
    "options": [],
    "answer": "See solution",
    "solution": "We will write $a \\to x$ if an $a \\times a$ square without one corner cell can be cut along the lines of the grid into $x$ squares.\n\nEvidently, $2 \\to 3$. Also note that $7 \\to 8$ and $9 \\to 9$ (see figures below).\n\nWe will show that if $a \\to x$ and $b \\to y$, then $ab \\to x + y$. Indeed, a square with side length $ab$ without a corner cell can be cut into two parts: a square with side length $b$ without a corner cell, and a square with side length $ab$ with a $b \\times b$ square removed from its corner. The first part can be cut into $y$ squares, and the second part can be cut into $x$ squares, since it is a square with side length $a$ without a corner, magnified by a factor of $b$.\n\nSo,\n\n$$\na \\to x \\Rightarrow 2a \\to x + 3. \\quad (*)\n$$\n\nWe will also show that\n\n$$\na \\to x \\Rightarrow 2a - 1 \\to 2x + 2. \\quad (**)\n$$\n\nIndeed, an \"incomplete\" square can be partitioned into 2 squares with side length $a - 1$ and 2 \"incomplete\" squares with side length $a$ (see figure below).\n\nSo: $7 \\to 8$, $9 \\to 9$, $63 \\to 17$, $126 \\to 20$, $252 \\to 23$, $503 \\to 48$, $1006 \\to 51$, $2011 \\to 104$, i.e., we have shown that our original figure can be cut into $104 < 121$ squares.\n\n![](images/Ukrajina_2011_p26_data_da7ee7b8d0.png)\n\n*Fig. 24*\n\n![](images/Ukrajina_2011_p26_data_8d79c968c7.png)\n\n*Fig. 25*\n\n![](images/Ukrajina_2011_p26_data_c11c4cab5d.png)\n\n*Fig. 26*",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17239,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $K$, $L$, and $N$ be the points where the angle bisectors $AK$, $BL$, and $CN$ of triangle $ABC$ meet its circumcircle, respectively (see the figure).\n\nLet $I$ be the incenter of $\\triangle ABC$. The segment $NL$ meets $AC$ at $B'_2$. Let $M$ be the intersection point of $CN$ and $AB$.\n\nProve that $B'_2$ coincides with $B_2$, and that $B_2$ and $C_1$ lie on the segment $NL$. Similarly, show that $B_1$ and $A_2$ lie on $KL$, and $A_1$ and $C_2$ lie on $KN$.",
    "options": [],
    "answer": "See solution",
    "solution": "By the trefoil theorem, $AL = IL$, so $\\triangle AIL$ is isosceles. Since $\\angle ALN = \\angle ACN = \\angle NCB = \\angle BLN$ (as inscribed angles subtending equal arcs), the ray $LN$ is the bisector of the isosceles triangle $AIL$, hence $LN$ is the perpendicular bisector of $AI$.\n\nTherefore, $\\triangle AB'_2I$ is also isosceles and $\\angle IAB'_2 = \\angle B'_2IA$. Since $AK$ is the bisector of $\\angle BAC$, we have $\\angle IAB'_2 = \\angle BAI$, so $AB \\parallel B'_2I$.\n\nBy Thales' theorem:\n$$\n\\frac{CB'_2}{B'_2A} = \\frac{CI}{MI} = \\frac{AC}{AM} = \\frac{AC + BC}{AB} = \\frac{a+b}{c}\n$$\nso\n$$\n\\frac{CB'_2}{AC} = \\frac{a+b}{a+b+c}.\n$$\n\nTherefore, $B'_2$ coincides with $B_2$. Thus, $B_2$ and $C_1$ lie on $NL$. Similarly, $B_1$ and $A_2$ lie on $KL$, and $A_1$ and $C_2$ lie on $KN$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17240,
    "subject": "Mathematics (Olympiad)",
    "question": "Several unit squares in a $99 \\times 99$ table are colored with one of 5 distinct colors so that each color occurs the same number of times. There are no squares of different colors in the same row or column. What is the maximum possible number of colored squares?",
    "options": [],
    "answer": "See solution",
    "solution": "Call *lines* the rows and the columns of the table and *cells* the colored unit squares. If a line contains a cell, color the line with the color of this cell. No line is colored twice because there are no cells of different colors in it. There are $2 \\cdot 99 = 198$ lines in all, each one colored with one of 5 distinct colors or not colored. Hence some color $C$ is assigned to at most $\\lfloor \\frac{198}{5} \\rfloor = 39$ lines. All cells of color $C$ are intersections of these lines (intersections of rows with columns). So if $x$ rows and $y$ columns have color $C$, with $x + y = \\ell \\le 39$, then the number of cells colored $C$ is at most $xy$.\n\nLet $x$ and $y$ be positive integers with fixed sum $l$. The identity\n$$\n4xy = (x+y)^2 - (x-y)^2 = l^2 - (x-y)^2\n$$\nshows that $xy$ is a maximum if and only if $|x - y|$ is a minimum. If $l$ is even the latter occurs for $x = y = l/2$. If $l$ is odd the minimum of $|x - y|$ is 1, and it is attained for $x = (l-1)/2$, $y = (l+1)/2$ or vice versa. It follows that if $l_1 < l_2$ then the maximum of $xy$ for $x + y = l_1$ is less than the maximum of $xy$ for $x + y = l_2$.\n\nFor color $C$ we take $l = 39$ and conclude that there are at most\n$\\frac{39-1}{2} \\cdot \\frac{39+1}{2} = 380$ cells with this color. Since all colors are equally represented, the same holds for every remaining color. Hence the total number of cells in the table is at most $5 \\cdot 380 = 1900$.\n\nOne can have exactly 1900 cells. Consider 5 rectangles $19 \\times 20$ in a $99 \\times 99$ table located as in the figure (two rows and one column are not used). No line contains unit squares from different rectangles. Color every rectangle in one of 5 distinct colors, with different colors for different rectangles. Exactly 1900 cells are obtained, and the coloring has the necessary properties.\n\n![](images/Argentina_2011_p14_data_ab1bff6c4f.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17241,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer and consider the integers $x_1, x_2, \\dots, x_n, y_1, y_2, \\dots, y_n$ such that\n\n$$\nx_1 + x_2 + \\dots + x_n = y_1 + y_2 + \\dots + y_n = 0;\n$$\n\n$$\nx_1^2 + y_1^2 = x_2^2 + y_2^2 = \\dots = x_n^2 + y_n^2.\n$$\n\nProve that $n$ is an even number.",
    "options": [],
    "answer": "See solution",
    "solution": "Set $a = x_1^2 + y_1^2$. If $a$ is odd, the numbers $x_i$ and $y_i$ do not have the same parity, so $x_i + y_i$ is odd. Since $\\sum (x_i + y_i) = 0$, it follows that $n$ is even.\n\nSuppose $a = 4k+2$. Then $x_i$ and $y_i$ are both odd. The equality $x_1 + x_2 + \\dots + x_n = 0$ implies $n$ is even.\n\nFinally, if $a = 4k$, then $x_i$ and $y_i$ are even. The numbers $a_i = \\frac{x_i}{2}$ and $b_i = \\frac{y_i}{2}$ satisfy the initial conditions. Furthermore, $a_1^2 + b_1^2 = \\frac{a}{4}$. Repeating the argument for $\\frac{a}{4}$ instead of $a$, after a finite number of steps we end up in a previous case.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17242,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that there exists a positive integer $N$ such that for all integers $a > N$, there exists a contiguous substring of the decimal expansion of $a$ that is divisible by $2011$.\n\n(For instance, if $a = 153204$, then $15$, $532$, and $0$ are all contiguous substrings of $a$. Note that $0$ is divisible by $2011$.)",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that if the decimal expansion of $a$ has at least $2012$ digits, then $a$ contains the required substring.\n\nLet the decimal expansion of $a$ be $a_k a_{k-1} \\dots a_0$. For $i = 0, \\dots, 2011$, let $b_i$ be the number with decimal expansion $a_i a_{i-1} \\dots a_0$.\n\nBy the pigeonhole principle, $b_i \\equiv b_j \\pmod{2011}$ for some $i < j \\leq 2011$. It follows that $2011$ divides $b_j - b_i = c \\cdot 10^i$, where $c$ is the substring $a_j \\dots a_{i+1}$.\n\nSince $2011$ and $10$ are relatively prime, it follows that $2011$ divides $c$.\n\n$\\Box$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17243,
    "subject": "Mathematics (Olympiad)",
    "question": "證明對於所有質數 $p > 100$ 和每一個整數 $r$，存在兩個整數 $a$ 和 $b$ 使得 $p$ 整除 $a^2 + b^5 - r$。\n",
    "options": [],
    "answer": "See solution",
    "solution": "在本解答中，所有同餘關係皆以模 $p$ 進行。\n\n固定 $p$，令 $\\mathcal{P} = \\{0, 1, \\dots, p-1\\}$ 為模 $p$ 的完全剩餘系。對於所有 $r \\in \\mathcal{P}$，令 $S_r = \\{(a, b) \\in \\mathcal{P} \\times \\mathcal{P} : a^2 + b^5 \\equiv r \\}$，且令 $s_r = |S_r|$。我們的目標是證明對所有 $r \\in \\mathcal{P}$，$s_r > 0$。\n\n我們將用已知事實：對於所有同餘類 $r \\in \\mathcal{P}$ 和所有正整數 $k$，存在最多 $k$ 個 $x \\in \\mathcal{P}$ 使得 $x^k \\equiv r$。\n\n**Lemma.** 令 $N$ 為所有四元數對 $(a, b, c, d) \\in \\mathcal{P}^4$ 中滿足 $a^2 + b^5 \\equiv c^2 + d^5$ 的數對個數。則\n\n$$\nN = \\sum_{r \\in \\mathcal{P}} s_r^2\n$$\n\n且\n\n$$\nN \\le p(p^2 + 4p - 4).\n$$\n\n*Proof.*\n\n(a) 對於所有同餘類 $r$，存在剛好 $s_r$ 個數對 $(a, b)$ 滿足 $a^2 + b^5 \\equiv r$，以及 $s_r$ 個數對 $(c, d)$ 滿足 $c^2 + d^5 \\equiv r$。所以存在 $s_r^2$ 個四元數對滿足 $a^2 + b^5 \\equiv c^2 + d^5 \\equiv r$。對所有 $r \\in \\mathcal{P}$ 取和即得。\n\n(b) 選擇任意 $(b, d) \\in \\mathcal{P}^2$ 並尋找可能的 $a, c$。\n\n1. 若 $b^5 \\equiv d^5$，設 $k$ 為此類 $(b, d)$ 的個數。$b$ 可用 $p$ 種方式選取。對 $b \\equiv 0$，$d$ 只能是 $0$。對非零 $b, d$，最多有五種可能值。故 $k \\le 1 + 5(p-1) = 5p-4$。\n\n$a$ 和 $c$ 必須滿足 $a^2 \\equiv c^2$，即 $a \\equiv \\pm c$，共 $2p-1$ 個 $(a, c)$。\n\n2. 若 $b^5 \\neq d^5$，$a$ 和 $c$ 必須相異。由 $(a-c)(a+c) = d^5 - b^5$，$a-c$ 唯一決定 $a+c$，也決定 $a, c$。因此有 $p-1$ 個 $(a, c)$。\n\n因此，對 $b^5 = d^5$ 的 $k$ 個 $(b, d)$，有 $2p-1$ 個 $(a, c)$；對其餘 $p^2-k$ 個 $(b, d)$，有 $p-1$ 個 $(a, c)$。所以：\n\n$$\n\\begin{aligned}\nN &= k(2p-1) + (p^2-k)(p-1) = p^2(p-1) + kp \\\\\n  &\\leq p^2(p-1) + (5p-4)p = p(p^2 + 4p - 4).\n\\end{aligned}\n$$\n\n假設對某 $r \\in \\mathcal{P}$，$S_r = \\emptyset$。顯然 $r \\neq 0$。令 $T = \\{x^{10} : x \\in \\mathcal{P} \\setminus \\{0\\}\\}$。因每同餘類至多有 10 個元素的 10 次方，且 $p > 100$，得 $|T| \\geq \\frac{p-1}{10} \\geq 4$。\n\n對所有 $t \\in T$，有 $S_{tr} = \\emptyset$。若 $(x, y) \\in S_{tr}$ 且 $t \\equiv z^{10}$，則\n\n$$\n(z^{-5}x)^2 + (z^{-2}y)^5 \\equiv t^{-1}(x^2 + y^5) \\equiv r,\n$$\n\n故 $(z^{-5}x, z^{-2}y) \\in S_r$。因此，在 $S_1, \\dots, S_{p-1}$ 間至少有 $\\frac{p-1}{10} \\geq 4$ 個空集合，且 $s_0, s_1, \\dots, s_{p-1}$ 中最多 $p-4$ 個非零數。由 $AM-QM$ 不等式，\n\n$$\nN = \\sum_{r \\in \\mathcal{P} \\setminus rT} s_r^2 \\geq \\frac{1}{p-4} \\left( \\sum_{r \\in \\mathcal{P} \\setminus rT} s_r \\right)^2 = \\frac{|\\mathcal{P} \\times \\mathcal{P}|^2}{p-4} = \\frac{p^4}{p-4} > p(p^2+4p-4),\n$$\n\n由 Lemma 可知此式不可能成立。\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17244,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $i$ and $j$ be integers such that $i - j$ is even. Given that $m + 1$ and $n + 1$ are relatively prime, prove the following:\n\n1. For which $n$ is $(3, n)$ a good pair?\n2. How many pairs $(i, j)$ of integers with $3 \\leq i, j \\leq 11$ are such that $i$ and $j$ are not relatively prime?",
    "options": [],
    "answer": "See solution",
    "solution": "1. $(3, n)$ is a good pair if and only if $4$ and $n+1$ are not relatively prime, i.e., if $n$ is odd. Thus, the answer is $n = 3, 5, 7, 9$.\n\n2. The number of pairs $(i, j)$ with $3 \\leq i, j \\leq 11$ and $i, j$ not relatively prime is $29$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17245,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, $d$ be positive real numbers. Prove the inequality\n\n$$\n\\left(\\frac{a}{a+b}\\right)^5 + \\left(\\frac{b}{b+c}\\right)^5 + \\left(\\frac{c}{c+d}\\right)^5 + \\left(\\frac{d}{d+a}\\right)^5 \\geq \\frac{1}{8}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = \\frac{b}{a}$, $y = \\frac{c}{b}$, $z = \\frac{d}{c}$, $t = \\frac{a}{d}$. It's clear that $xyzt = 1$ and the given inequality becomes\n\n$$\nA = \\left(\\frac{1}{1+x}\\right)^5 + \\left(\\frac{1}{1+y}\\right)^5 + \\left(\\frac{1}{1+z}\\right)^5 + \\left(\\frac{1}{1+t}\\right)^5 \\geq \\frac{1}{8}.\n$$\n\nBy applying the AM-GM inequality, for every positive real number $x$ we have\n\n$$\n2 \\left(\\frac{1}{1+x}\\right)^5 + \\frac{3}{32} = \\left(\\frac{1}{1+x}\\right)^5 + \\left(\\frac{1}{1+x}\\right)^5 + \\frac{1}{32} + \\frac{1}{32} + \\frac{1}{32} \\geq \\frac{5}{8} \\left(\\frac{1}{1+x}\\right)^5\n$$\n\nFrom the last inequality we get\n\n$$\n2A + \\frac{12}{32} \\geq \\frac{5}{8} \\left( \\left( \\frac{1}{1+x} \\right)^2 + \\left( \\frac{1}{1+y} \\right)^2 + \\left( \\frac{1}{1+z} \\right)^2 + \\left( \\frac{1}{1+t} \\right)^2 \\right). \\quad (1)\n$$\n\nWe'll show that for every positive real numbers $x$ and $y$ the following inequality holds:\n\n$$\n\\left(\\frac{1}{1+x}\\right)^2 + \\left(\\frac{1}{1+y}\\right)^2 \\geq \\frac{1}{1+xy}\n$$\n\nWe have\n\n$$\n\\left(\\frac{1}{1+x}\\right)^2 + \\left(\\frac{1}{1+y}\\right)^2 - \\frac{1}{1+xy} = \\frac{xy(x-y)^2 + (xy-1)^2}{(1+x)^2(1+y)^2(1+xy)} \\geq 0\n$$\n\nFrom the previous inequality and $xyzt = 1$ we get\n\n$$\n\\left(\\frac{1}{1+x}\\right)^2 + \\left(\\frac{1}{1+y}\\right)^2 + \\left(\\frac{1}{1+z}\\right)^2 + \\left(\\frac{1}{1+t}\\right)^2 \\geq \\frac{1}{1+xy} + \\frac{1}{1+zt} = \\frac{1}{1+xy} + \\frac{1}{1+1/xy} = 1. \\quad (2)\n$$\n\nFrom (1) and (2) we have $2A + \\frac{12}{32} \\geq \\frac{5}{8}$ or $A \\geq \\frac{1}{8}$ and we are done. The equality occurs if and only if $x = y = z = t$, i.e., $a = b = c = d$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17246,
    "subject": "Mathematics (Olympiad)",
    "question": "試找出所有正整數的三元數對 $ (x, y, z) $ 使得 $ x \\le y \\le z $ 且\n\n$$\nx^3(y^3 + z^3) = 2012(xyz + 2).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "首先注意到 $x$ 整除 $2012 \\cdot 2 = 2^3 \\cdot 503$。若 $503 \\mid x$，則方程式的右手邊可被 $503^3$ 整除，因此 $503^2 \\mid xyz + 2$。因為 $503 \\mid x$，所以矛盾。因此 $x = 2^m,\\ m \\in \\{0, 1, 2, 3\\}$。\n\n若 $m \\ge 2$，則 $2^6 \\mid 2012(xyz + 2)$。然而 $2$ 的最高次方整除 $2012$ 的話就是 $2^2$，整除 $xyz + 2 = 2^m yz + 2$ 的話就是 $2^1$。所以 $x = 1$ 或 $x = 2$，產生下列兩個方程式：\n\n$$\ny^3 + z^3 = 2012(yz + 2) \\quad \\text{和} \\quad y^3 + z^3 = 503(yz + 1).\n$$\n\n在上列兩方程式中可得質數 $503 = 3 \\cdot 167 + 2$ 整除 $y^3 + z^3$。我們宣告 $503 \\mid y+z$。若 $503 \\mid y$，則 $503 \\mid y+z$ 是顯然的。所以我們假設 $503 \\nmid y$ 且 $503 \\nmid z$。則由費瑪小定理得 $y^{502} \\equiv z^{502} \\pmod{503}$。另一方面，$y^3 \\equiv -z^3 \\pmod{503}$ 可推出 $y^{501} \\equiv -z^{501} \\pmod{503}$。於是可得到 $y \\equiv -z \\pmod{503}$，也就是 $503 \\mid y+z$。\n\n因此 $y+z=503k,\\ k \\ge 1$。有鑑於 $y^3+z^3 = (y+z)((y-z)^2+yz)$，所以上述兩個方程可分別得到下列形式：\n\n$$\nk(y - z)^2 + (k - 4)yz = 8, \\qquad (1)\n$$\n\n$$\nk(y - z)^2 + (k - 1)yz = 1. \\qquad (2)\n$$\n\n在 (1) 中，我們有 $(k-4)yz \\le 8$，可得到 $k \\le 4$。確實若 $k > 4$，則 $1 \\le (k-4)yz \\le 8$，這使得 $y \\le 8$ 且 $z \\le 8$。這是不可能的，因為 $y+z = 503k \\ge 503$。接下來注意到在第一個方程式中 $y^3+z^3$ 是偶數。因此 $y+z = 503k$ 也是偶數，意即 $k$ 是偶數。所以 $k=2$ 或 $k=4$。顯然當 $k=4$ 時，(1) 式沒有整數解。若 $k=2$ 則 (1) 式可寫成 $(y+z)^2 - 5yz = 4$。因為 $y+z = 503k = 503 \\cdot 2$，所以 $5yz = 503^2 \\cdot 2^2 - 4$。然而 $503^2 \\cdot 2^2 - 4$ 不是 $5$ 的倍數。因此 (1) 式沒有整數解。\n\n在 (2) 中，我們有 $0 \\le (k-1)yz \\le 1$，可得到 $k=1$ 或 $k=2$。又 $0 \\le k(y-z)^2 \\le 1$，因此 $k=2$ 只當 $y=z$。這時 $y=z=1$，這是錯的因為 $y+z \\ge 503$。因此 $k=1$ 且 (2) 式可寫成 $(y-z)^2 = 1$，得到 $z-y = |y-z| = 1$。由 $k=1$ 和 $y+z = 503k$，可推出 $y = 251,\\ z = 252$。\n\n總結來說，三元數對 $(2, 251, 252)$ 是唯一解。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17247,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are 2013 people at a party. For any three people, the number of pairs among them who know each other is either 1 or 3. Prove that there is a group of at least 1007 people such that every two people in the group know each other.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the 2013 people be represented by a graph $G$ with 2013 vertices. Let each pair of different vertices $A$ and $B$ be connected by an edge if and only if the people they represent know each other. The condition in the problem stipulates that for any three vertices $X$, $Y$, $Z$ in $G$, the subgraph generated by $X$, $Y$, $Z$ has either 1 or 3 edges.\n\n**Lemma.** Each connected component of $G$ is a complete graph.\n\n**Proof.** Let $A$ and $B$ be two different vertices in a connected component. If $A$ is connected to $B$, there is nothing to prove. Otherwise, there exists a simple path (i.e., a path that does not repeat a vertex) $A_1, A_2, \\ldots, A_n$ in $G$ where $A = A_1$, $B = A_n$ and $n \\geq 3$.\n\nIf $A_1$ is joined to $A_i$ for some $i$ ($2 \\leq i \\leq n-1$), then the triple $(A_1, A_i, A_{i+1})$ has edges $A_1A_i$ and $A_iA_{i+1}$. Therefore, it must also have $A_1$ joined to $A_{i+1}$. Since $A_1$ is definitely joined to $A_2$, this inductively allows us to conclude that $A_1$ is joined to $A_n$. $\\square$\n\nIf there were 3 or more components, we could choose 3 people from different components, none of them knowing each other, contradiction.\n\nHence, there are 1 or 2 components. The largest component has at least 1007 people, and by the lemma it is complete, so we are done.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17248,
    "subject": "Mathematics (Olympiad)",
    "question": "證明：若三個非零相異複數 $\\alpha_1, \\alpha_2, \\alpha_3$ 在複數平面上不共線，且滿足 $\\alpha_1 + \\alpha_2 + \\alpha_3 = 0$，則\n\n$$\n\\sum_{i=1}^{3} \\left( \\frac{|\\alpha_{i+1} - \\alpha_{i+2}|}{\\sqrt{|\\alpha_i|}} \\left( \\frac{1}{\\sqrt{|\\alpha_{i+1}|}} + \\frac{1}{\\sqrt{|\\alpha_{i+2}|}} - \\frac{2}{\\sqrt{|\\alpha_i|}} \\right) \\right) \\le 0\n$$\n\n必成立。其中 $\\alpha_4 = \\alpha_1, \\alpha_5 = \\alpha_2$。同時請確定等號成立的充要條件。",
    "options": [],
    "answer": "See solution",
    "solution": "考慮以此三點為頂點的三角形 $A_1A_2A_3$，其中 $A_1(\\alpha_1), A_2(\\alpha_2), A_3(\\alpha_3)$。依條件我們知道其重心 $O$ 在原點。再令 $s_i = |\\alpha_{i+1} - \\alpha_{i+2}|$ 為 $\\overline{A_{i+1}A_{i+2}}$ 的邊長，因此 $m_i = \\frac{3}{2}|\\alpha_i|$ 為 $\\overline{A_{i+1}A_{i+2}}$ 上的中線長。\n\n則原不等式等價於\n\n$$\n\\sum_{i=1}^3 \\frac{s_i}{\\sqrt{m_i}} \\left( \\frac{1}{\\sqrt{m_{i+1}}} + \\frac{1}{\\sqrt{m_{i+2}}} - \\frac{2}{\\sqrt{m_i}} \\right) \\le 0\n$$\n\n整理後得到等價的不等式：\n\n$$\n\\sum_{i=1}^{3} \\frac{s_i + s_{i+1}}{\\sqrt{m_i m_{i+1}}} \\le 2 \\sum_{i=1}^{3} \\frac{s_i}{m_i}\n$$\n\n其中 $(s_4, m_4, s_5, m_5) = (s_1, m_1, s_2, m_2)$。\n\n我們將證明下列不等式成立：\n\n$$\n\\frac{s_i + s_{i+1}}{\\sqrt{m_i m_{i+1}}} \\le \\frac{s_i}{m_i} + \\frac{s_{i+1}}{m_{i+1}}, \\quad i = 1, 2, 3.\n$$\n\n證明：將不等式兩邊平方並整理後得到\n\n$$\nm_i m_{i+1} (s_i^2 + s_{i+1}^2) \\le m_{i+1}^2 s_i^2 + m_i^2 s_{i+1}^2\n$$\n\n等號僅當 $s_i = s_{i+1}$ 時成立。\n\n注意 $m_i = \\frac{\\sqrt{2(s_{i+1}^2 + s_{i+2}^2) - s_i^2}}{2}$。\n\n給出 $m_i m_{i+1}$ 的上界估計：\n\n$$\n(4m_i m_{i+1})^2 = (2s_{i+1}^2 + 2s_{i+2}^2 - s_i^2)(2s_{i+2}^2 + 2s_i^2 - s_{i+1}^2)\n$$\n\n經整理可得\n\n$$\n4m_i m_{i+1} \\le 2s_{i+2}^2 + s_{i+1} s_i\n$$\n\n當且僅當 $s_i = s_{i+1}$ 時等號成立。\n\n由此可得\n\n$$\n4m_{i+1}^2 s_i^2 + 4m_i^2 s_{i+1}^2 - 4m_i m_{i+1} (s_i^2 + s_{i+1}^2) \\ge (s_i - s_{i+1})^2 [2(s_i + s_{i+1})^2 - s_{i+1}s_i] \\ge 0\n$$\n\n因此原不等式成立，且等號僅當 $\\alpha_1, \\alpha_2, \\alpha_3$ 為以原點 $O$ 為重心的正三角形的頂點時成立（例如 $\\alpha_{i+1}/\\alpha_i = e^{2\\pi i/3}, i = 1, 2$）。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17249,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute triangle $ABC$, a point $P$ is chosen such that all points symmetric to $P$ with respect to the sides of $ABC$ lie on the circumcircle of $ABC$. Prove that $P$ is the orthocenter of $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A'$, $B'$, $C'$ be the points symmetric to $P$ with respect to the sides $BC$, $CA$, $AB$, respectively.\n\nThen $|C'A| = |PA| = |B'A|$, so the arcs $AC'$ and $AB'$ of the circumcircle of triangle $ABC$ are equal. Since $A$ and $C'$ are on the same side of the line $BB'$, and $C$ is on the other, we have $\\angle C'CA = \\angle B'CA = \\angle PCA$. Since $P$ and $C'$ are on the same side of the line $AC$, the points $P$, $C$, $C'$ are collinear. Since $PC' \\perp AB$, we must also have $PC \\perp AB$, that is, $P$ lies on the altitude from $C$ in triangle $ABC$. Analogously, $P$ lies on the other two altitudes.\n\n![](images/Estonija_2012_p21_data_0b5a9956a0.png)\n\n*Fig. 17*\n\n_Remark 1._ The converse—the points symmetric to the orthocenter with respect to the sides of the triangle lie on the circumcircle—is a known result in elementary geometry that can also be used to solve this problem. Namely, the point $P$ lies on the reflections of arcs $AB$, $BC$, $CA$ from the corresponding lines $AB$, $BC$, $CA$ of the circumcircle. By the theorem mentioned, the intersection point of the altitudes lies on the same arcs. But the circles, whose arcs are the reflections, already meet twice pairwise at the points $A$, $B$, $C$. Therefore, they cannot have two common intersection points.\n\n_Remark 2._ The claim of the problem, as well as the theorem given in Remark 1, hold for all triangles, not just acute ones.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17250,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Prove that $n$ is a power of two if and only if there exists an integer $m$ such that $2^n - 1$ is a divisor of $m^2 + 9$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first show that if $2^n - 1$ divides $m^2 + 9$ for some integer $m$, then $n$ must be a power of $2$. Otherwise, $n$ has some odd divisor $\\ell \\geq 3$, so $2^\\ell - 1$ divides $2^n - 1$ and thus also divides $m^2 + 9$. For $\\ell \\geq 3$, $2^\\ell - 1 \\equiv -1 \\pmod{4}$, so $2^\\ell - 1$ has prime divisors $p$ with $p \\equiv -1 \\pmod{4}$. Consider $p \\neq 3$. Since $p \\mid m^2 + 3^2$, by Fermat's little theorem:\n\n$$\n1 \\equiv m^{p-1} \\equiv (m^2)^{(p-1)/2} \\equiv (-9)^{(p-1)/2} \\equiv (-1)^{(p-1)/2} \\cdot 3^{p-1} \\equiv -1 \\pmod{p}\n$$\n\nBut this is impossible, so $p = 3$. However, for odd $\\ell$, $2^\\ell \\not\\equiv -1 \\pmod{3}$, which is a contradiction.\n\nNow consider $n = 2^k$. For $n = 1$, the claim is true. If $k \\geq 1$ then:\n\n$$\n2^n - 1 = 3(2^2 + 1)(2^{2^2} + 1) \\cdots (2^{2^{k-1}} + 1)\n$$\n\nTherefore, if $2^n - 1$ divides $m^2 + 9$, then $2^{2^\\ell} + 1$ divides $m^2 + 9$ for each $\\ell = 1, 2, \\dots, k-1$. Moreover, if $\\alpha \\neq \\beta$, the numbers $2^{2\\alpha} + 1$ and $2^{2\\beta} + 1$ are relatively prime. If $d > 2$ is their greatest common divisor and $\\alpha > \\beta$, then:\n\n$$\n-1 \\equiv 2^{2\\alpha} \\equiv (2^{2\\beta})^{2^{\\alpha-\\beta}} \\equiv 1 \\pmod{d}\n$$\n\nwhich is a contradiction. So necessarily $d = 1$ ($d \\neq 2$ since both are odd).\n\nBy the Chinese remainder theorem, there is a natural number $c$ such that:\n\n$$\nc \\equiv 2^{2^\\ell} \\pmod{2^{2^{\\ell+1}}}, \\quad \\forall \\ell = 0, 1, \\dots, k-2\n$$\n\nThen $c^2 + 1 \\equiv 0 \\pmod{2^{2^{\\ell+1}}}$ for $\\ell = 0, 1, \\dots, k-2$, and thus $2^n - 1$ divides $(3c)^2 + 9$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17251,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for natural numbers $a \\ge b \\ge c \\ge d$ the inequality:\n$$\nab + bc + cd - b^{2} - c^{2} - d^{2} \\ge a - d.$$ is satisfied.",
    "options": [],
    "answer": "See solution",
    "solution": "We have:\n$$\nab + bc + cd - b^{2} - c^{2} - d^{2} = b(a-b) + c(b-c) - d(c-d) \\ge (a-b) + (b-c) + (c-d) = a-d.$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17252,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be positive real numbers. Prove that\n\n$$\na^3 b^6 + b^3 c^6 + c^3 a^6 + 3a^3 b^3 c^3 \\geq abc(a^3 b^3 + b^3 c^3 + c^3 a^3) + a^2 b^2 c^2 (a^3 + b^3 + c^3)\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "After dividing both sides of the given inequality by $a^3 b^3 c^3$, it becomes\n\n$$\n\\left(\\frac{b}{c}\\right)^3 + \\left(\\frac{c}{a}\\right)^3 + \\left(\\frac{a}{b}\\right)^3 + 3 \\geq \\left(\\frac{a}{c} \\cdot \\frac{b}{c} + \\frac{b}{a} \\cdot \\frac{c}{a} + \\frac{c}{b} \\cdot \\frac{a}{b}\\right) + \\left(\\frac{a}{b} \\cdot \\frac{a}{c} + \\frac{b}{a} \\cdot \\frac{c}{a} + \\frac{c}{a} \\cdot \\frac{c}{b}\\right)\n$$\n\nSet\n\n$$\n\\frac{b}{a} = \\frac{1}{x}, \\quad \\frac{c}{b} = \\frac{1}{y}, \\quad \\frac{a}{c} = \\frac{1}{z}.\n$$\n\nThen $xyz = 1$, and by substituting, we find\n\n$$\nx^3 + y^3 + z^3 + 3 \\geq \\left(\\frac{y}{z} + \\frac{z}{x} + \\frac{x}{y}\\right) + \\left(\\frac{x}{z} + \\frac{y}{x} + \\frac{z}{y}\\right).\n$$\n\nMultiplying both sides by $xyz$ (with $xyz = 1$), the inequality is equivalent to\n\n$$\nx^3 + y^3 + z^3 + 3xyz - xy^2 - yz^2 - zx^2 - yx^2 - zy^2 - xz^2 \\geq 0.\n$$\n\nBy the special case of Schur's inequality:\n\n$$\nx^r(x - y)(x - z) + y^r(y - x)(y - z) + z^r(z - y)(z - x) \\geq 0, \\quad x, y, z \\geq 0, \\ r > 0,\n$$\n\nwith $r=1$,\n\n$$\nx(x - y)(x - z) + y(y - x)(y - z) + z(z - y)(z - x) \\geq 0\n$$\n\nwhich, after expansion, coincides with the previous expression.\n\n*Remark 1.* The inequality above follows by supposing (without loss of generality) that $x \\geq y \\geq z$, and then writing the left side as\n\n$$\n(x - y)(x(x - z) - y(y - z)) + z(y - z)(z - x),\n$$\n\nwhich is obviously $\\geq 0$.\n\n*Remark 2.* One can also obtain the relation using the substitution $x = ab^2$, $y = bc^2$, and $z = ca^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17253,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose the numbers $1, 2, \\ldots, 2n$ are divided into two piles of $n$ numbers each. Prove that for every integer $n \\geq 100$, there exist two numbers in the same pile whose sum is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "If we can find three cards numbered $a, b, c$ such that $a + b$, $a + c$, and $b + c$ are perfect squares, then we will be done because two of $a, b, c$ will be in the same pile.\n\nSolving the system\n\n$$\n\\begin{aligned}\na + b &= (2k - 1)^2 \\\\\na + c &= (2k)^2 \\\\\nb + c &= (2k + 1)^2\n\\end{aligned}\n$$\n\nyields\n\n$$\n\\begin{aligned}\na &= 2k^2 - 4k \\\\\nb &= 2k^2 + 1 \\\\\nc &= 2k^2 + 4k.\n\\end{aligned}\n$$\n\nUsing these values of $a, b, c$, it suffices to show that for each integer $n \\geq 100$ there exists a positive integer $k$ such that $n \\leq 2k^2 - 4k$ and $2k^2 + 4k \\leq 2n$. That is,\n\n$$\nk^2 + 2k \\leq n \\leq 2k^2 - 4k.\n$$\n\nTo this end, let $k$ be the greatest integer such that $k^2 + 2k \\leq n$. Note that $k \\geq 9$ because $n \\geq 100$. Since $k$ is the greatest integer satisfying $k^2 + 2k \\leq n$, it follows that\n\n$$\n(k + 1)^2 + 2(k + 1) \\geq n + 1 \\Leftrightarrow n \\leq k^2 + 4k + 4.\n$$\n\nTo complete the proof it suffices to show that $k^2 + 4k + 4 \\leq 2k^2 - 4k$. This is equivalent to $(k - 4)^2 \\geq 20$. But this is true since $k \\geq 9$. $\\square$\n\n**Comment 1**\n\nThe above proof shows that the conclusion of the problem is true for all $n \\geq 99$.\n\n**Comment 2**\n\nThe conclusion of the problem is false for $n = 98$. Indeed, a counterexample occurs if the first pile contains the even numbers from 98 to 126, the odd numbers from 129 to 161, and the even numbers from 162 to 196, while the second pile contains the rest of the numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17254,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$, $n$, and $r$ be positive integers with $n \\ge 2$ and $1 \\le r \\le n-1$. Consider a square table of size $(mn + r) \\times (mn + r)$. The table is covered by $n \\times n$ squares with sides parallel to the sides of the table. Each unit square is covered at least once, and some unit squares may be covered multiple times. Find the minimum possible number of unit squares that are covered at least two times.",
    "options": [],
    "answer": "See solution",
    "solution": "We call a unit square *bad* if it is covered more than once.\n\n**Bound:**\n\nChoose an arbitrary row and mark its cells in the columns $r+1,\\ r+n+1,\\ \\dots,\\ r+(m-1)n+1$. Consider all $n \\times n$ squares having nonempty intersection with the chosen row. Since $mn < mn + r < (m+1)n$, we have at least $m+1$ such squares. Moreover, each such square covers at least one of the marked cells. Since we have $m$ marked cells, we conclude that at least one of them is bad.\n\nAnalogously, there is at least one bad unit square among the intersecting squares of this row with columns numbered $r+2,\\ r+n+2,\\ \\dots,\\ r+(m-1)n+2$, and so on. Thus, in this row, we have at least $n - r$ bad squares. The same reasoning shows that there are at least $n - r$ bad squares in any of the remaining rows. We repeat the above observation for all remaining $(m+1)r$ columns and conclude that there are at least $n - r$ bad cells in each of them. Therefore, there are at least\n\n$$\n(mn + r)(n - r) + (m+1)r(n - r) = (mn + mr + 2r)(n - r)\n$$\n\nbad cells.\n\n**Construction:**\n\nConsider the first $n + r$ rows (starting from below) and first $n + r$ columns (starting from the left). Their intersection forms a $(n + r) \\times (n + r)$ square $A$. Cover this square using four $n \\times n$ squares in its four corners. It is easy to see that the number of bad cells equals $(n + r)^2 - 4r^2$. Consider the $\\Gamma$-shaped figure consisting of all cells in the $(2n + r) \\times (2n + r)$ square without the cells of $A$. We cover all cells in this figure by five $n \\times n$ squares (two squares in each of the two rectangular parts and one square in the right upper corner) and leave $n^2 - r^2$ bad cells. The same procedure is applied $m-1$ times for all $\\Gamma$-shaped figures of width $n$. The number of bad cells equals\n\n$$\n(n + r)^2 - 4r^2 + (m-1)(n^2 - r^2) = (mn + mr + 2r)(n - r).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17255,
    "subject": "Mathematics (Olympiad)",
    "question": "The set of vertices of the graph $G$ is a set of 2014 points in general position on the plane. The segment $AB$ is an edge of the graph if and only if each of the two open half-planes defined by line $AB$ contains 1006 points. Prove that $G$ does not contain a Hamiltonian path (a path that passes through every vertex exactly once).",
    "options": [],
    "answer": "See solution",
    "solution": "Each vertex of the convex hull has degree 1 in the graph $G$. (When we rotate the line passing through such a point, the numbers of other points in the half-planes change monotonically.) The convex hull contains at least 3 vertices, so $G$ has at least three leaves. Therefore, there is no Hamiltonian path in $G$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17256,
    "subject": "Mathematics (Olympiad)",
    "question": "On a $$(4n+2) \\times (4n+2)$$ square grid, a turtle can move between squares sharing a side. The turtle begins in a corner square of the grid and enters each square exactly once, ending in the square where she started. In terms of $n$, what is the largest positive integer $k$ such that there must be a row or column that the turtle has entered at least $k$ distinct times?",
    "options": [],
    "answer": "See solution",
    "solution": "We shall prove that the answer is $2n + 2$.\n\nNumber the rows in increasing order, from top to bottom, and number the columns from left to right. By symmetry, we may (and shall) assume that the turtle starts in the top right corner square.\n\nFirst, we shall prove that some row or column must be entered at least $2n + 2$ times. Let $m = 4n + 2$. First, note that each time the turtle moves, she enters either a row or a column. Let $r_i$ denote the number of times the turtle enters row $i$, and let $c_i$ be similarly defined for column $i$. Since the turtle moves $m^2$ times,\n\n$$\nr_1 + r_2 + \\dots + r_m + c_1 + c_2 + \\dots + c_m = m^2.\n$$\n\nNow note that each time the turtle enters column 1, the next column she enters must be column 2. Therefore, $c_1$ is equal to the number of times the turtle enters column 2 from column 1. Furthermore, the turtle must enter column 2 from column 3 at least once, which implies that $c_2 > c_1$. Therefore, since the $2m$ terms $r_i$ and $c_i$ are not all equal, one must be strictly greater than $m^2/(2m) = 2n + 1$ and therefore at least $2n + 2$.\n\nNow we construct an example to show that it is possible that no row or column is entered more than $2n + 2$ times. Partition the square grid into four $$(2n+1) \\times (2n+1)$$ quadrants A, B, C, and D, containing the upper left, upper right, lower left, and lower right corners, respectively. The turtle begins at the top right corner square of B, moves one square down, and then moves left through the whole second row of B. She then moves one square down and moves right through the whole third row of B. She continues in this pattern, moving through each remaining row of B in succession and moving one square down when each row is completed. Since $2n+1$ is odd, the turtle ends at the bottom right corner of B. She then moves one square down into D and through each column of D in turn, moving one square to the left when each column is completed. She ends at the lower left corner of D and moves left into C and through the rows of C, moving one square up when each row is completed, ending in the upper left corner of C. She then enters A and moves through the columns of A, moving one square right when each column is completed. This takes her to the upper right corner of A, whereupon she enters B and moves right through the top row of B, which returns her to her starting point. Each row passing through A and B is entered at most $2n+1$ times in A and once in B, and thus at most $2n+2$ times in total. Similarly, each row and column in the grid is entered at most $2n+2$ times by this path.\n\n![](images/Kanada_2015_p4_data_d348f7d876.png)\n\nThe case $n = 3$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17257,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(a_n)$ be the integer sequence defined by $a_1 = 1$ and\n\n$$\na_{n+1} = a_n^2 + n \\cdot a_n, \\quad \\forall n \\ge 1.\n$$\n\nLet $S$ be the set of all primes $p$ such that there exists an index $i$ with $p \\mid a_i$. Prove that the set $S$ is infinite and it is not equal to the set of all primes.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show that $3 \\notin S$ by proving that\n\n$$\na_{3k-2} \\equiv 2 \\pmod{3}, \\quad a_{3k-1} \\equiv 2 \\pmod{3}, \\quad \\text{and} \\quad a_{3k} \\equiv 2 \\pmod{3}.\n$$\n\nSince $a_1 = 1$, $a_2 = 2$, $a_3 = 2^2 + 2 \\cdot 2 = 8 \\equiv 2 \\pmod{3}$, so the claim is true for $k = 1$.\n\nSuppose the claim holds for $k = n$. We have\n\n$$\n\\begin{align*}\na_{3n+1} &= a_{3n}(a_{3n} + 3n) \\equiv 2 \\cdot 2 \\equiv 1 \\pmod{3}, \\\\\na_{3n+2} &= a_{3n+1}(a_{3n+1} + 3n + 1) \\equiv 1(1 + 1) \\equiv 2 \\pmod{3}, \\\\\na_{3n+3} &= a_{3n+2}(a_{3n+2} + 3n + 2) \\equiv 2(2 + 2) \\equiv 2 \\pmod{3}.\n\\end{align*}\n$$\n\nThus, the claim is also true for $k = n + 1$. By induction, the claim is proved.\n\nNow suppose, for contradiction, that $S$ is finite, say $S = \\{p_1, p_2, \\dots, p_k\\}$. Note that $a_n \\mid a_{n+1}$, so $a_n \\mid a_m$ for all $m \\ge n$. By the definition of $S$, for each $p_j$ there is some index $t_j$ such that $p_j \\mid a_{t_j}$, and thus $p_j \\mid a_{t'}$ for all $t' \\ge t_j$. Therefore, there exists $N$ such that $p_1 p_2 \\cdots p_k \\mid a_n$ for all $n \\ge N$.\n\nTake $\\ell > N + 1$ such that $\\ell \\equiv 2 \\pmod{p_1 p_2 \\cdots p_k}$. Since $a_\\ell = a_{\\ell-1}(a_{\\ell-1} + \\ell - 1)$, we get\n\n$$\na_{\\ell-1} + \\ell - 1 \\equiv 1 \\pmod{p_1 p_2 \\cdots p_k}\n$$\n\nso $a_{\\ell-1} + \\ell - 1$ is coprime to all primes in $S$, which implies that $a_\\ell$ has a prime divisor not in $S$, a contradiction. Hence, $S$ is infinite. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17258,
    "subject": "Mathematics (Olympiad)",
    "question": "Given triangle $ABC$, a line from point $A$ intersects side $BC$ at $D$. Four lines parallel to $BC$ are drawn such that they divide $AB$ and $AC$ into five equal parts. These lines partition triangle $ABC$ into ten regions. The two dotted regions have equal area. The area of the grey triangle at the bottom left next to $A$ is $5$. What is the area of the grey quadrilateral at the upper right next to $C$?\n\n![](images/NLD_ABooklet_2022_p39_data_92857a494e.png)",
    "options": [],
    "answer": "See solution",
    "solution": "B) $81$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17259,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, a point $D$ lies on side $BC$ and a point $E$ lies on side $AC$ such that the line segments $BD$, $DE$, and $AE$ have the same length. The point $F$ is the intersection between the line segments $AD$ and $BE$. Angle $C$ is $68\\degree$.\n\nWhat is the size of angle $F$ in triangle $AFB$?\n\n![](images/NLD_ABooklet_2022_p5_data_b400e11056.png)\n\nA) $120\\degree$ \nB) $121\\degree$ \nC) $122\\degree$ \nD) $123\\degree$ \nE) $124\\degree$",
    "options": [],
    "answer": "See solution",
    "solution": "E) $124\\degree$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17260,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the sequence $(a_n)_{n \\in \\mathbb{N}^*}$ be given by $a_1 = 2$ and $a_{n+1} = a_n^2 - a_n + 1$. Find the minimum real number $L$ such that for every $k \\in \\mathbb{N}^*$,\n\n$$\n\\sum_{i=1}^{k} \\frac{1}{a_i} < L.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For every $n \\in \\mathbb{N}^*$, from the given recurrence relation we have\n\n$$\na_{n+1} - a_n = (a_n - 1)^2.\n$$\n\nSo, the sequence $(a_n)$ is increasing and, since $a_1 = 2 > 1$, we have for every $n \\in \\mathbb{N}^*$:\n\n$$\na_{n+1} > a_n > 1.\n$$\n\nFrom the recurrence, for every $n \\geq 2$,\n\n$$\na_n - 1 = a_{n-1}(a_{n-1} - 1),\n$$\n$$\na_{n-1} - 1 = a_{n-2}(a_{n-2} - 1),\n$$\n$\\vdots$\n$$\na_2 - 1 = a_1(a_1 - 1).\n$$\n\nMultiplying recursively, we get\n\n$$\na_n = 1 + a_1 a_2 \\cdots a_{n-1}.\n$$\n\nDividing both sides by $a_1 a_2 \\cdots a_{n-1} a_n$ gives\n\n$$\n\\frac{1}{a_1 a_2 \\cdots a_{n-1}} = \\frac{1}{a_1 a_2 \\cdots a_{n-1} a_n} + \\frac{1}{a_n}\n$$\n\nor\n\n$$\n\\frac{1}{a_n} = \\frac{1}{a_1 a_2 \\cdots a_{n-1}} - \\frac{1}{a_1 a_2 \\cdots a_{n-1} a_n} \\quad (1)\n$$\n\nLetting $n = 2, 3, 4, \\dots, k$ in (1) and summing, we get\n\n$$\n\\sum_{i=2}^{k} \\frac{1}{a_i} = \\frac{1}{a_1} - \\frac{1}{a_1 a_2 \\cdots a_{k-1} a_k} \\implies \\sum_{i=1}^{k} \\frac{1}{a_i} = 1 - \\frac{1}{a_1 a_2 \\cdots a_{k-1} a_k} < 1 \\quad (2)\n$$\n\nfor every $k \\in \\mathbb{N}^*$, since $a_i > 1$ for all $i$.\n\nFor every $n \\ge 2$, $a_n - a_{n-1} = (a_{n-1} - 1)^2 \\ge 1$, so by induction,\n\n$$\na_n \\ge n + 1 > n.\n$$\n\nThus, for every $k \\in \\mathbb{N}^*$,\n\n$$\n0 < \\frac{1}{a_1 a_2 \\cdots a_{k-1} a_k} < \\frac{1}{a_k} < \\frac{1}{k}\n$$\n\nand\n\n$$\n1 - \\frac{1}{a_1 a_2 \\cdots a_{k-1} a_k} > 1 - \\frac{1}{k} \\quad (3)\n$$\n\nFor every $0 \\le \\alpha < 1$, we can find $k \\in \\mathbb{N}^*$ such that\n\n$$\n\\sum_{i=1}^{k} \\frac{1}{a_i} > \\alpha. \\quad (4)\n$$\n\nIf\n\n$$\n1 - \\frac{1}{a_i} > \\alpha \\implies k-1 > k\\alpha \\implies k > \\frac{1}{1-\\alpha} \\implies k \\ge \\left\\lfloor \\frac{1}{1-\\alpha} \\right\\rfloor + 1, \\quad (5)\n$$\n\nwhere $\\lfloor x \\rfloor$ denotes the greatest integer less than or equal to $x$. From (2), (3), and (5), we obtain (4). Thus, $L = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17261,
    "subject": "Mathematics (Olympiad)",
    "question": "Counters are placed, one at a time, in the unit squares of an $n \\times n$ grid according to the following rules:\n\n- A counter can only be placed in an empty unit square.\n- The first counter can be placed in any unit square.\n- Each subsequent counter can only be placed in a unit square $S$ if the number of counters already in the same row as $S$ and the number of counters already in the same column as $S$ sum to an odd number.\n\nFor each value of $n \\ge 2$, find the smallest possible number of empty unit squares remaining after a sequence of such placements.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(n)$ be the minimum number of vacant squares remaining. We claim that $f(n) = 1$ if $n$ is even, and $f(n) = n - 1$ if $n$ is odd.\n\nWe first show that $f(n) > 0$. Consider any board with a counter in every square except for one vacant square. The number of counters in each of the row and column related to the vacant square is $n - 1$, which sums to $2(n - 1)$, an even number, so the final counter may not be placed (noting that the final counter is not the first counter as $n > 1$). Thus $f(n) \\ne 0$.\n\nSuppose $n = 2k$ is even. We shall show that $f(2k) = 1$ for all $k$ by induction. It suffices to find a sequence of placements that fill all squares except the bottom right corner. The $2 \\times 2$ case can be achieved by placing counters in the top left, top right, and then bottom left squares. Assume that we have a suitable sequence of placements in a $2k \\times 2k$ chessboard. Extend the $2k \\times 2k$ chessboard with two extra rows and columns to form a $(2k+2) \\times (2k+2)$ chessboard as shown in the figure below. First perform the sequence of placements as indicated by the two independent arrows, noting one arrow ends a square sooner than the other. The final numbering 1 to 7 shows how to complete the placements so that the bottom right square is the only vacant square remaining. Thus $f(2k + 2) = 1$, which completes the induction for the even case.\n\n![](images/2023_Australian_Scene_p148_data_f9129f6150.png)\n\nNow suppose $n = 2k + 1$ is odd. First, we show that $f(2k + 1) \\ge 2k$. We note that after the first counter is placed, the numbers of the counters in that row and column are both odd. All other rows and columns have an even number of counters (zero). For a subsequent counter to be placed in a square, the sum of counters in the corresponding row and column is odd. More specifically, the number of counters in each of the row and column are of different parity. After placing a counter, the number of counters in the corresponding row and column have their parity switched but nonetheless continue to be of different parity. Thus, after the first and every subsequent placement, the number of rows and columns with an odd number of counters remains unchanged and is precisely two.\n\nSuppose there are strictly less than $2k$ vacant squares. Then there must be at least two full rows and two full columns. But then there are at least four rows and columns with an odd number of counters, which is a contradiction. Thus $f(2k + 1) \\ge 2k$.\n\nWe shall show that $f(2k + 1) = 2k$ by induction on $k$. Let our induction hypothesis be that we can find a sequence of placements leaving $2k$ vacant squares on the main diagonal other than the top left square. The $3 \\times 3$ case is given below.\n\n| 5 | 1 | 2 |\n|---|---|---|\n| 4 |   | 3 |\n| 7 | 6 |   |\n\nNow assume our hypothesis is true for some $k$ and consider the $(2k + 3) \\times (2k + 3)$ chessboard as shown below. Starting with the position from our induction hypothesis for the $(2k+1) \\times (2k+1)$ chessboard, first add counters as indicated by the two independent arrows. The final numbering 1 to 4 shows how to complete the placements so that in the new larger chessboard there are $2(k+1)$ vacant squares, each on the main diagonal other than the top left hand square, thus proving our induction hypothesis. Therefore $f(2k + 1) = 2k$.\n\n![](images/2023_Australian_Scene_p149_data_26aa6dec5d.png)\n\n**Remark.**\n\n- Alternative construction for the $n$ even case: split the grid into $2 \\times 2$ subgrids and fill each $2 \\times 2$ subgrid in a zigzag-like manner from top to bottom. For example, the squares of a $6 \\times 6$ grid would be filled in the order indicated below.\n\n![](images/2023_Australian_Scene_p149_data_320742f8f8.png)\n\n- An alternative induction to give the even pattern from the preceding odd pattern: We fill in\n\n$$\n(1 + 2l, n),\\ (2 + 2l, n),\\ (2 + 2l, 2 + 2l),\\ (n, 2 + 2l),\\ (n, 3 + 2l),\\ (3 + 2l, 3 + 2l)\n$$\n\nfor $l = 0, \\dots, (n-4)/2$, then $(n-1, n)$ and $(n, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17262,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest real constant $C$ such that for any positive real numbers $a_1, a_2, a_3, a_4$, and $a_5$ (not necessarily distinct), one can always choose distinct subscripts $i, j, k$, and $l$ such that\n\n$$\n\\left| \\frac{a_i}{a_j} - \\frac{a_k}{a_l} \\right| \\le C.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $C = \\frac{1}{2}$.\n\nConsider the numbers $1$, $2$, $2$, $2$, and $n$, where $n > 5$. The fractions formed by two of these numbers in ascending order are\n\n$$\n\\frac{1}{n} < \\frac{2}{n} < \\frac{1}{2} < 1 < 2 < \\frac{n}{2} < n.\n$$\n\nThe smallest two fractions cannot be obtained simultaneously. Therefore, the minimum value of the difference is $\\frac{1}{2} - \\frac{2}{n}$. This shows that $C \\ge \\frac{1}{2} - \\frac{2}{n}$ for every positive integer $n > 5$. Hence $C \\ge \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17263,
    "subject": "Mathematics (Olympiad)",
    "question": "a) What is the largest possible sum of two 4-digit numbers formed using each of the digits 1 through 8 exactly once?\n\nb) What is the smallest possible positive difference between two 4-digit numbers formed using each of the digits 1 through 8 exactly once?",
    "options": [],
    "answer": "See solution",
    "solution": "a) The largest sum is obtained when the largest digits are assigned to the leftmost positions:\n$$\n(8+7) \\cdot 1000 + (6+5) \\cdot 100 + (4+3) \\cdot 10 + 2 + 1 = 16373.\n$$\n\nb) The smallest difference is obtained when the smallest digits are assigned to the leftmost positions. Assign 1 and 2 to the thousands, 3 and 4 to the hundreds, 5 and 6 to the tens, and 7 and 8 to the units. The biggest of the two numbers is the one beginning with 2. The minimum difference is:\n$$\n2357 - 1468 = 889.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17264,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given an irrational number $\\alpha$ for which there exist real numbers $x$ and $y$ such that $x + y = \\alpha$ and $x^k + y^k$ is rational for all $k$ from $2$ to $n$. Find the maximal $n$ for which this is possible.",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* $n = 3$.\n\nWe show that for $n = 4$ it cannot hold.\n\nSuppose $xy = 0$, or $y = 0$. For $n = 2$ it is possible; for example, take $x = \\sqrt{2} \\in \\mathbb{R} \\setminus \\mathbb{Q}$, $x^2 = 2 \\in \\mathbb{Q}$. But if $x^2$ and $x^3$ are rational, then $\\frac{x^3}{x^2} = x$ is also rational.\n\nConsider the case $xy \\neq 0$. For $n = 4$, $x^2 + y^2$, $x^3 + y^3$, and $x^4 + y^4$ are rational. From $(x^2 + y^2)^2 = (x^4 + y^4) + 2x^2y^2$, it follows that $(xy)^2 \\in \\mathbb{Q}$. Then, $x^6 + y^6 = (x^2 + y^2)((x^4 + y^4) - (xy)^2) \\in \\mathbb{Q}$, and $(x^3 + y^3)^2 = (x^6 + y^6) + 2(xy)^3$ implies $(xy)^3 \\in \\mathbb{Q}$, so $xy \\in \\mathbb{Q}$. Finally, $x + y = \\frac{x^3 + y^3}{(x^2 + y^2) - xy} = \\alpha \\in \\mathbb{Q}$, which is a contradiction.\n\nFor $n = 3$, such numbers do exist. Denote $a = x + y$, $b = xy$. Then $x^2 + y^2 = (x + y)^2 - 2xy = a^2 - 2b \\in \\mathbb{Q}$, $x^3 + y^3 = (x + y)((x + y)^2 - 3xy) = a(a^2 - 3b) \\in \\mathbb{Q}$. Find irrational $a, b$ that satisfy these equalities. Let $a^2 - 2b = p$, $a(a^2 - 3b) = q$, then $b = \\frac{1}{2}(a^2 - p)$, so $a(3p - a^2) = 2q$, giving the cubic equation $a^3 - 3ap + 2q = 0$. Take $a = 2 - \\sqrt{2}$ and find $p$ and $q$: $p = \\frac{14}{3}$, $q = 4$. Then $b = \\frac{2}{3} - 2\\sqrt{2}$. To find $x, y$, solve:\n$$\nx + y = 2 - \\sqrt{2}, \\\\\nxy = \\frac{2}{3} - 2\\sqrt{2}\n$$\nThey are roots of $t^2 - at + b = 0$. The discriminant is $a^2 - 4b = \\frac{10}{3} + 4\\sqrt{2} > 0$, so the roots are real.\n\nFinally, $x^2 + y^2 = a^2 - 2b = \\frac{14}{3} = p$ and $x^3 + y^3 = a(a^2 - 3b) = 4 = q$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17265,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, a_3, a_4, a_5, b_1, b_2, b_3, b_4, b_5$ be positive numbers satisfying the following:\n\n$$\na_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2 = 2013^2\n$$\n\n$$\nb_1^2 + b_2^2 + b_3^2 + b_4^2 + b_5^2 = 2012^2\n$$\n\n$$\na_1b_1 + a_2b_2 + a_3b_3 + a_4b_4 + a_5b_5 = 2013 \\cdot 2012\n$$\n\nFind the ratio $a_1 : b_1$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* $\\frac{a_1}{b_1} = \\frac{2013}{2012}$.\n\n*Solution.* Observe that\n\n$$\n(2012a_1 - 2013b_1)^2 + (2012a_2 - 2013b_2)^2 + (2012a_3 - 2013b_3)^2 + (2012a_4 - 2013b_4)^2 + (2012a_5 - 2013b_5)^2 =\n$$\n\n$$\n\\begin{aligned}\n&= 2012^2 (a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2) + 2013^2 (b_1^2 + b_2^2 + b_3^2 + b_4^2 + b_5^2) \\\\\n& \\quad - 2 \\cdot 2012 \\cdot 2013 (a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4 + a_5b_5) \\\\\n&= 2012^2 \\cdot 2013^2 + 2013^2 \\cdot 2012^2 - 2 \\cdot 2012^2 \\cdot 2013^2 = 0.\n\\end{aligned}\n$$\n\nSince the sum of five squares can only be equal to $0$ if each of the squares is zero, we get that $2012a_1 - 2013b_1 = 0$, and so $\\frac{a_1}{b_1} = \\frac{2013}{2012}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17266,
    "subject": "Mathematics (Olympiad)",
    "question": "On a $16 \\times 16$ torus, the edges are all colored red and blue so that every vertex is an endpoint of an even number of red edges. A move consists of changing the color of all edges on a unit square. If one coloring can be converted to another by a sequence of moves, we put the two colorings into the same box. After all the colorings are partitioned into boxes, how many boxes do we have?\n\n![](images/bw18shortlist_p22_data_e05860f000.png)",
    "options": [],
    "answer": "See solution",
    "solution": "There are $4$ boxes. Representatives of the equivalence classes are: all blue, all blue with one longitudinal red ring, all blue with one transversal red ring, and all blue with one longitudinal and one transversal red ring.\n\nFirst, these four classes are not equivalent. Consider any ring (transversal or longitudinal) and count the number of red edges going out from vertices of this ring in the same halftorus. This number cannot be changed modulo $2$.\n\nNow, we show that each configuration can be transformed to one of these four classes. Two independent arguments are given:\n\n**Scanning of the square.**\n\nCut the torus into a $16 \\times 16$ square. To restore the torus, identify the opposite sides at the end. Work with the square: during all recolorings, each vertex has even red degree. The same holds for inner vertices of the square.\n\nScan all cells one by one from left to right and bottom to top. For convenience, think of the scanned area as grey. Start with the bottom left cell ($a1$ in chess notation) and color it grey. Next, consider cell $b1$; color it grey, and if the edge between $a1$ and $b1$ is red, change the colors of $b1$'s edges. This yields a grey area with no red edges in its interior. Continue: when scanning each new cell, append it to the grey figure and, if necessary, change the colors of its edges to make all new edges in the grey area blue.\n\nThis is always possible because the new cell has either one or two common edges with the grey figure. For example, if the grey figure is the first row and $a2$, when appending $b2$, two edges of its lower left corner vertex already belong to the grey figure and are blue. Therefore, the other two edges ($a2$-$b2$ and $b1$-$b2$) have the same color and can be made blue by recoloring $b2$'s edges.\n\nBy doing this for all cells, we obtain a $16 \\times 16$ square with blue edges inside. Now, recall that the sides should be identified, and the red degree of each torus vertex is even. It follows that the whole vertical sides are either red or blue, and the same for horizontal sides.\n\n**Deformations of red loops (sketch).**\n\nCut the torus into a square with opposite edges identified. Since the red degree of each vertex is even, we can always find a loop of red edges. Suppose there is a simple red loop not crossing the boundary. We can change its color by changing the colors of unit squares inside it, preserving the even red degree property. Repeat this for every red loop, leaving a configuration where no red loops avoid the boundary. Any red loop passing through more than one boundary vertex can be deformed into a loop with only one boundary vertex. Any two loops crossing the same side can be removed by changing colors of all unit squares between them. Thus, only the four possibilities above remain.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17267,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose each vertex of a convex polygon is assigned either a 0 or a 1. Rudi wants to triangulate the polygon (i.e., divide it into triangles by drawing non-crossing diagonals) so that, in each triangle, the sum of the numbers at its vertices is either 1 or 2. Prove that this is always possible, provided not all vertices are assigned the same number.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume not all vertices are assigned the same number, so there is at least one 0-vertex and at least one 1-vertex.\n\nWe proceed by strong induction on the number $n$ of vertices.\n\n**Base cases:**\n- For $n = 3$, the triangle itself has both a 0-vertex and a 1-vertex, so the sum is 1 or 2.\n- For $n = 4$, if there is a diagonal connecting a 0-vertex and a 1-vertex, draw it; otherwise, any diagonal works, and both triangles have sums 1 or 2.\n\n**Inductive step:**\nAssume the statement holds for all polygons with fewer than $n$ vertices ($n \\ge 5$), each having at least one 0-vertex and one 1-vertex.\n\nIn the $n$-gon, choose adjacent vertices $A$ (a 0-vertex) and $B$ (a 1-vertex). Let $P$ be a vertex not adjacent to $A$ or $B$ (possible since $n \\ge 5$). If $P$ is a 0-vertex, connect it to $B$; otherwise, connect it to $A$. This splits the polygon into two smaller polygons, each with both a 0-vertex and a 1-vertex. By the inductive hypothesis, each can be triangulated so that every triangle has sum 1 or 2. Thus, the original polygon can be triangulated as desired.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17268,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be the set of positive real numbers. Find all functions $f : S^3 \\to S$ such that, for all positive real numbers $x, y, z$ and $k$, the following three conditions are satisfied:\n\n$$\na) \\quad x f(x, y, z) = z f(z, y, x)\n$$\n\n$$\nb) \\quad f(x, k y, k^2 z) = k f(x, y, z)\n$$\n\n$$\nc) \\quad f(1, k, k+1) = k+1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $g(x, y, z) = \\frac{x}{z} f(x, y, z)$. The conditions become:\n\n$$\na') \\quad g(x, y, z) = g(z, y, x)\n$$\n\n$$\nb) \\quad g(x, k y, k^2 z) = g(x, y, z)\n$$\n\n$$\nc) \\quad g(1, k, k+1) = \\frac{k+1}{k}.\n$$\n\nFrom (a') and (b),\n$$\ng(x, y, z) = g\\left(x, \\frac{y}{\\sqrt{z}}, 1\\right) = g\\left(1, \\frac{y}{\\sqrt{z}}, x\\right) = g\\left(1, 1, \\frac{xz}{y^2}\\right).\n$$\nDefine $h(w) = g(1, 1, w)$, so $g(x, y, z) = h\\left(\\frac{xz}{y^2}\\right)$.\n\nFrom (c),\n$$\nh\\left(\\frac{k+1}{k^2}\\right) = \\frac{k+1}{k}.\n$$\nLet $w = \\frac{k+1}{k^2}$, so $k^2 w - k - 1 = 0$. Solving for $k$:\n$$\nk = \\frac{1 + \\sqrt{1 + 4w}}{2w}\n$$\nThus,\n$$\nh(w) = \\frac{k+1}{k} = w k = \\frac{1 + \\sqrt{1 + 4w}}{2}.\n$$\nThis holds for all $w > 0$.\n\nBack-substituting,\n$$\nf(x, y, z) = \\frac{y + \\sqrt{y^2 + 4 x z}}{2x}.\n$$\n\nThis is the unique solution. It satisfies condition (c), and is of the form $f(x, y, z) = \\frac{y}{x} h\\left(\\frac{xz}{y^2}\\right)$, so conditions (a) and (b) are immediate.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17269,
    "subject": "Mathematics (Olympiad)",
    "question": "Sea $ABC$ un triángulo acutángulo con ortocentro $H$, y sea $W$ un punto sobre el lado $BC$, estrictamente entre $B$ y $C$. Los puntos $M$ y $N$ son los pies de las alturas trazadas desde $B$ y $C$ respectivamente. Se denota por $\\omega_1$ la circunferencia que pasa por los vértices del triángulo $BWN$, y por $X$ el punto de $\\omega_1$ tal que $WX$ es un diámetro de $\\omega_1$. Análogamente, se denota por $\\omega_2$ la circunferencia que pasa por los vértices del triángulo $CWM$, y por $Y$ el punto de $\\omega_2$ tal que $WY$ es un diámetro de $\\omega_2$. Demostrar que los puntos $X$, $Y$ y $H$ son colineales.",
    "options": [],
    "answer": "See solution",
    "solution": "Definamos $V$ como el segundo punto de intersección de $\\omega_1$ y $\\omega_2$, y sea $P$ el pie de la altura desde $A$ sobre $BC$. Por ser $\\angle HPB = \\angle HPC = \\angle HMC = \\angle HNB = 90^\\circ$, tenemos que $BPHN$ y $CPHM$ son cíclicos, luego la potencia $P$ de $A$ respecto de sus circunferencias circunscritas es $P = AP \\cdot AH = AB \\cdot AN = AC \\cdot AM$. Pero entonces $P$ es también la potencia de $A$ respecto de $\\omega_1$, $\\omega_2$, y $A$ está en su eje radical $VW$, es decir, $A$, $V$, $W$ están alineados con $P = AV \\cdot AW$. Concluimos también que, al ser $P = AP \\cdot AH = AV \\cdot AW$, $PHVW$ es cíclico, y al ser $\\angle HPW = 90^\\circ$, se tiene que $\\angle WVH = 90^\\circ$.\n\nSea $\\ell$ la recta perpendicular a $AW$ por $V$. Por ser $WX$ un diámetro de $\\omega_1$, y ser $V$ un punto de $\\omega_1$, se tiene que $\\angle WVX = 90^\\circ$, es decir, $X$ está en $\\ell$. De forma análoga, se tiene que $\\angle WVY = 90^\\circ$, con lo que $Y$ también está en $\\ell$. Finalmente, como $\\angle WVH = 90^\\circ$, $H$ también está en $\\ell$. Luego no sólo hemos demostrado que $X$, $Y$, $H$ están alineados, sino que la recta sobre la que se encuentran es la perpendicular a $AW$ por el segundo punto de intersección $V$ de $\\omega_1$ y $\\omega_2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17270,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are $x$ girl and $y$ boy students in a school. What is the minimal value of $x + y$ such that it is possible to construct a tournament where each girl plays each boy, each girl wins at least 21 matches, and each boy loses at least 12 matches?",
    "options": [],
    "answer": "See solution",
    "solution": "Obviously, $xy \\ge 21x + 12y$ is a necessary condition for the existence of such a tournament. Let us show that this inequality is also sufficient.\n\nConsider the $x \\times y$ score table of the tournament. Place $\\mathbb{1}$ (alternatively $\\emptyset$) in the intersection of the $m$-th row and $n$-th column if the $m$-th girl wins (alternatively loses) the match with the $n$-th boy. We want a table where each row contains at least 21 entries $\\mathbb{1}$ and each column contains at least 12 entries $\\emptyset$.\n\nStart with a table where each entry of the first 21 columns is $\\mathbb{1}$ and each other entry is $\\emptyset$. Step by step, move towards a table satisfying the conditions. Suppose the $k$-th column contains fewer than 12 $\\emptyset$'s. Since the total number of $\\emptyset$'s is at least $12y$, some $l$-th column must contain more than 12 $\\emptyset$'s. In this case, choose a row such that the intersection of this row with the $k$-th and $l$-th columns are $\\mathbb{1}$ and $\\emptyset$, respectively, and switch these two entries. After each move, a column with an $\\emptyset$ shortage gains one $\\emptyset$, and the total number of $\\mathbb{1}$ entries in each row does not change. Therefore, after a finite number of moves, we obtain the desired table.\n\nNow, let us find the minimal value of $x + y$. The inequality $(x - 12)(y - 21) \\ge 12 \\cdot 21$ is equivalent to $xy \\ge 21x + 12y$. By the AM-GM inequality:\n\n$$\n(x-12) + (y-21) \\ge 2\\sqrt{(x-12)(y-21)} \\ge 2\\sqrt{12 \\cdot 21} > 31.\n$$\n\nThus, $x + y \\ge 12 + 21 + 32 = 65$. Any pair out of $(26,39), (27,38), (28,37), (29,36), (30,35)$ with sum 65 satisfies the condition $xy \\ge 21x + 12y$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17271,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that every positive integer can be expressed as a sum of powers of $3$, $4$, and $7$ in such a way that the representation does not contain two powers with the same base and the same exponent.\n\nFor example, $2 = 7^0 + 7^0$ and $22 = 3^2 + 3^2 + 4^1$ are not valid sums, but $2 = 3^0 + 7^0$ and $22 = 3^2 + 3^0 + 4^1 + 4^0 + 7^1$ are valid.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the powers of $3$, $4$, and $7$ in increasing order: $x_1^{\\alpha_1} \\leq x_2^{\\alpha_2} \\leq x_3^{\\alpha_3} \\leq \\dots$.\n\nWe will prove, by induction on $n$, that it is possible to represent all the integers from $1$ to $x_1^{\\alpha_1} + x_2^{\\alpha_2} + \\dots + x_n^{\\alpha_n}$ in the desired way using only the powers $x_1^{\\alpha_1}, x_2^{\\alpha_2}, \\dots, x_n^{\\alpha_n}$. It is clear that this is true for $n = 1, 2, 3$. Assuming the property holds for $n \\geq 3$, we will prove that it is valid for $n + 1$.\n\nIf the powers $x_1^{\\alpha_1}, x_2^{\\alpha_2}, \\dots, x_n^{\\alpha_n}$ are $\\{3^0, 3^1, \\dots, 3^a\\} \\cup \\{4^0, 4^1, \\dots, 4^b\\} \\cup \\{7^0, 7^1, \\dots, 7^c\\}$, we have that\n\n$$\nx_1^{\\alpha_1} + x_2^{\\alpha_2} + \\dots + x_n^{\\alpha_n} = (3^0 + 3^1 + \\dots + 3^a) + (4^0 + 4^1 + \\dots + 4^b) + (7^0 + 7^1 + \\dots + 7^c) = \\\\\n\\frac{3^{a+1}-1}{2} + \\frac{4^{b+1}-1}{3} + \\frac{7^{c+1}-1}{6} \\leq \\frac{x_{n+1}^{\\alpha_{n+1}}-1}{2} + \\frac{x_{n+1}^{\\alpha_{n+1}}-1}{3} + \\frac{x_{n+1}^{\\alpha_{n+1}}-1}{6} = x_{n+1}^{\\alpha_{n+1}} - 1.\n$$\n\nThen, by the induction assumption, all the positive integers smaller than $x_{n+1}^{\\alpha_{n+1}}$ have a representation using only the powers $x_1^{\\alpha_1}, x_2^{\\alpha_2}, \\dots, x_n^{\\alpha_n}$. In addition, an integer $m$ such that $x_{n+1}^{\\alpha_{n+1}} \\leq m \\leq x_1^{\\alpha_1} + x_2^{\\alpha_2} + \\dots + x_n^{\\alpha_n} + x_{n+1}^{\\alpha_{n+1}}$ can be expressed as $m = (m - x_{n+1}^{\\alpha_{n+1}}) + x_{n+1}^{\\alpha_{n+1}}$,\n\nwhere $0 \\leq m - x_{n+1}^{\\alpha_{n+1}} \\leq x_1^{\\alpha_1} + x_2^{\\alpha_2} + \\dots + x_n^{\\alpha_n}$ has a representation using only $x_1^{\\alpha_1}, \\dots, x_n^{\\alpha_n}$. We conclude that all integers from $1$ to $x_1^{\\alpha_1} + \\dots + x_n^{\\alpha_n} + x_{n+1}^{\\alpha_{n+1}}$ have a representation of the desired form using only the powers $x_1^{\\alpha_1}, \\dots, x_n^{\\alpha_n}, x_{n+1}^{\\alpha_{n+1}}$, which completes the induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17272,
    "subject": "Mathematics (Olympiad)",
    "question": "In a rectangle $ABCD$, the length of $AB$ is twice the length of $BC$. On side $CD$, a point $M$ is chosen such that $\\angle AMD = \\angle AMB$.\n\n(a) Determine the measure of $\\angle AMD$.\n\n(b) If $DM = 1$, determine the area of rectangle $ABCD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $AB$ and $CD$ are parallel, $\\angle BAM = \\angle AMD$. Given $\\angle AMB = \\angle AMD$, it follows that $\\angle BAM = \\angle AMB$, so triangle $AMB$ is isosceles and $AB = MB$.\n\nIn right triangle $BCM$, $BM$ is the hypotenuse and twice as long as $BC$ ($AB = 2BC$, $AB = MB$). Thus,\n\n$$\\angle BMC = 30^\\circ$$\n\nso $\\angle AMB + \\angle AMD = 150^\\circ$. Therefore,\n\n$$\\angle AMB = \\angle AMD = 75^\\circ$$\n\nLet $BC = b$, $CD = 2b$. Given $DM = 1$, $CD = CM + MD = CM + 1$, so $2b = CM + 1$.\n\nNow, $CM = \\sqrt{BM^2 - BC^2} = \\sqrt{(2b)^2 - b^2} = b\\sqrt{3}$.\n\nSo,\n\n$$2b = b\\sqrt{3} + 1$$\n\nwhich gives $b = \\frac{1}{2 - \\sqrt{3}} = 2 + \\sqrt{3}$.\n\nThe area of the rectangle is\n\n$$P = ab = 2b^2 = 2(2+\\sqrt{3})^2 = 2(7+4\\sqrt{3})$$\n\n![](images/Makedonija_2008_p15_data_8675a96ece.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17273,
    "subject": "Mathematics (Olympiad)",
    "question": "Acute-angled triangle $ABC$ is given. On the perpendicular bisectors to sides $AB$ and $BC$ respectively, points $P$ and $Q$ are chosen. Let $M$ and $N$ be the projections of $P$ and $Q$ onto $AC$.\n\nIt turns out that $2MN = AC$. Prove that the circumcircle of triangle $PBQ$ passes through the circumcenter of triangle $ABC$.\n\n![](images/Ukrajina_2010_p17_data_770a945669.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the circumcenter of $ABC$. Let the perpendicular bisectors meet sides $AB$ and $BC$ at points $K$ and $L$ respectively. Then $KL$ is a midline of triangle $ABC$, so $KL \\parallel AC$ and $2KL = AC$. Thus, $MKLN$ is a parallelogram because $KL \\parallel MN$ and $KL = MN$.\n\nSince $PM \\perp AC$ and $QN \\perp AC$, we have $PM \\parallel QN$. Also, $KM \\parallel LN$, so $\\angle PMK = \\angle QNL$.\n\n$\\angle AKP = \\angle AMP = 90^\\circ$, so $P, K, M, A$ lie on the same circle with diameter $PA$. Similarly, $Q, N, C, L$ lie on the same circle with diameter $CQ$, and $O, K, B, L$ belong to the circle with diameter $BO$. From these, $\\angle PAK = \\angle PMK = \\angle QNL = \\angle QCL$.\n\n$P$ and $Q$ lie on the perpendicular bisectors of $AB$ and $BC$, so $APB$ and $BQC$ are isosceles triangles. Thus, $\\angle PBA = \\angle PAB = \\angle QCB = \\angle QBC$. Hence, $\\angle PBQ = \\angle KBL = 180^\\circ - \\angle KOL = 180^\\circ - \\angle POQ$, so points $P, B, Q, O$ are cyclic, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17274,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a minimal $m \\times 2018$ rectangle that has exactly $k$ columns containing both $0$ and $1$, prove that $m \\leq 2^k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that each of the first $k$ columns of the board contains both $0$ and $1$. We will prove the following lemma.\n\n**Lemma.** All the cells in the last $2018 - k$ columns of the board are empty.\n\n*Proof.* Consider an arbitrary binary sequence $s$ of length $k$ and let $A_s$ be the set of rows such that the first $k$ cells of each row (counting from the left) form $s$. We will prove that there exists an element in $A_s$ whose last $2018 - k$ cells are empty.\n\nConsider the $(k+1)$-th cell of each row in $A_s$. These cells belong to the $(k+1)$-th column of the board, which does not simultaneously have $0$ and $1$.\n\nIf no cell of this column is empty, suppose all numbers in the column are $0$s. Then the binary sequence formed by $s$ concatenated with $1$ cannot be formed by any row, which is a contradiction. Thus, there exists a subset $A'_s$ of $A_s$ such that the $(k+1)$-th cell of each row in $A'_s$ is empty.\n\nNext, consider the $(k+2)$-th cell and similarly, we can prove that there exists a subset $A''_s$ of $A'_s$ such that the $(k+2)$-th cell of each row in $A''_s$ is empty. Continuing this pattern to the last column, we will have a row whose cells from the $(k+1)$-th position onward are empty. Thus, for any binary sequence $s$ of length $k$, we can always find a row whose last $2018 - k$ cells are empty. Note that these rows are not necessarily distinct since a row can form many binary sequences. Let $A$ be the set of all such rows.\n\nBy the definition of $A$, any binary sequence of length $2018$ can be formed by an element of $A$. Also, every row in the board belongs to $A$, otherwise we could omit that row and the board would still be complete, which is a contradiction. Thus, $A$ is the set of all rows in the board, which implies the last $2018 - k$ columns of the board are empty. The lemma is proved. $\\square$\n\nSince the sub-board formed by the last $2018 - k$ columns is totally empty, it can represent any binary sequence of length $2018 - k$. Moreover, the original board is minimal, which implies that the sub-board formed by the first $k$ columns is also minimal. Erasing all $2018 - k$ columns, the rest is a sub-board of size $m \\times k$. Label the rows from $1$ to $m$ (from top to bottom) and for every $i$ from $1$ to $m$, let $A_i$ be the set of all binary sequences of length $k$ that can be formed by the $i$-th row.\n\nSince the original board is minimal with respect to binary sequences of length $2018$, it follows that the above $m \\times k$ sub-board is also minimal with respect to binary sequences of length $k$. Set $B = A_1 \\cup A_2 \\cup \\dots \\cup A_m$, it is clear that $|B| = 2^k$ (since the sub-board can generate any binary sequence of length $k$).\n\nFor every $i$ ($i = 1, 2, \\ldots, m$), there exists a binary sequence of length $k$ generated by the $i$-th row, otherwise we could omit the $i$-th row and the remaining rows could also generate all binary sequences of length $k$, which is a contradiction. This means for every $i$, there exists a binary sequence $a_i$ such that $a_i$ does not belong to any $A_j$ where $j \\neq i$. Thus, $|B| \\geq m$.\n\nCombining all the above arguments, we have $m \\leq 2^k$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17275,
    "subject": "Mathematics (Olympiad)",
    "question": "If $x$ and $y$ are positive real numbers such that\n\n$$\n\\sqrt{2x} + \\sqrt{y} = 13 \\text{ and } \\sqrt{8x} + \\sqrt{9y} = 35,\n$$\n\ncalculate $20x + 23y$.",
    "options": [],
    "answer": "See solution",
    "solution": "The second equation can be written as $2\\sqrt{2x} + 3\\sqrt{y} = 35$. Together with the first equation, we obtain a system of two linear equations in the two unknowns $\\sqrt{2x}$ and $\\sqrt{y}$, which can easily be solved to give $\\sqrt{2x} = 4$ and $\\sqrt{y} = 9$. From this follows $x = 8$ and $y = 81$, so that $20x + 23y = 20 \\times 8 + 23 \\times 81 = 2023$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17276,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the smallest value that can be attained by the expression\n$$\n\\frac{(x + y + |x - y|)^2}{xy}\n$$\nfor positive $x$, $y$?",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $x \\ge y$. Then:\n$$\n\\frac{(x + y + |x - y|)^2}{xy} = \\frac{(x + y + x - y)^2}{xy} = \\frac{(2x)^2}{xy} = \\frac{4x^2}{xy} = \\frac{4x}{y} \\ge 4\n$$\nEquality occurs when $x = y$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17277,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the equation $a! + b! = c^4 + 2024$ has a unique positive integer solution $(a, b, c)$ satisfying $a \\leq b$ and find the solution.",
    "options": [],
    "answer": "See solution",
    "solution": "The unique solution is $(a, b, c) = (5, 8, 14)$.\n\nWe prove this is the only solution.\n\nConsider the equation modulo $16$. Since $x^4 \\equiv 0, 1 \\pmod{16}$, the right-hand side $c^4 + 2024 \\equiv 8$ or $9 \\pmod{16}$.\n\nFor the left-hand side, if $b < 7$, then $2 \\cdot 6! = 1440 < 2024$, so $b \\geq 7$. Thus $b! \\equiv 0 \\pmod{16}$.\n\nNow, $(1!, 2!, 3!, 4!, 5!, 6!) \\pmod{16} = (1, 2, 6, 8, 8, 0)$, so $a = 4$ or $5$ and $c$ must be even. Note $7!/16 = 315$.\n\nFor $a = 4$, $b!/16 = d^4 + 125 \\equiv 5$ or $6 \\pmod{8}$, but $7!/16 \\equiv 3 \\pmod{8}$ and $b!/16 \\equiv 0 \\pmod{8}$ for $b \\geq 8$, so $a = 4$ is impossible.\n\nFor $a = 5$, $b!/16 = d^4 + 119 \\equiv 7$ or $8 \\pmod{16}$, but $7!/16 \\equiv 11 \\pmod{16}$ and $b!/16 \\equiv 0 \\pmod{16}$ for $b \\geq 10$. Thus, only $b = 8$ or $9$ are possible. Checking, $b = 8$ gives $c = 14$ as a solution, and $b = 9$ does not work.\n\nTherefore, the only solution is $(a, b, c) = (5, 8, 14)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17278,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be an integer.\n\nWe draw an $n \\times n$ grid on a board and label each box with either the number $-1$ or the number $1$. Then we calculate the sum of each of the $n$ rows and the sum of each of the $n$ columns and determine the sum $S$ of these $2n$ sums.\n\n(a) Show that there does not exist a labelling of the grid with $S = 0$ if $n$ is odd.\n\n(b) Show that there exist at least six different labellings with $S = 0$ if $n$ is even.",
    "options": [],
    "answer": "See solution",
    "solution": "As each number of the grid appears exactly once in the sum of all columns and once in the sum of all rows, $S$ is twice the sum of all labels of the boxes of the $n \\times n$ grid. Therefore, $S = 0$ holds if and only if the sum of all labels of the boxes vanishes, or equivalently, if the number of labels $+1$ equals the number of labels $-1$. We call such a labelling admissible.\n\n(a) If $n$ is odd, the sum of all labels is also odd, because it is a sum of an odd number of odd labels. Thus, there cannot be an admissible labelling in this case.\n\n(b) If $n$ is even, write $n = 2k$ for some integer $k$. The admissible labellings can be constructed as follows: choose exactly half of the $n^2 = 4k^2$ boxes arbitrarily and label each of them with $+1$. The remaining boxes are labelled with $-1$.\n\nThus, there are exactly $a_k := \\binom{4k^2}{2k^2}$ admissible labellings of a $2k \\times 2k$ grid.\n\nWe have $a_1 = \\binom{4}{2} = 6$, and it is easily seen that $a_k$ is increasing in $k$: if $1 \\leq k' < k$, each admissible labelling of any $2k' \\times 2k'$ subgrid can be extended to an admissible labelling of the $2k \\times 2k$ grid by choosing half of the extra $4k^2 - 4k'^2$ boxes and labelling each of them with $+1$ and the remaining boxes with $-1$. Therefore, $a_k \\geq 6$ for all $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17279,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the trapezoid $ABCD$ with the smaller base $AB$, squares $ADEF$ and $BCGH$ are constructed externally to the trapezoid. Prove that the perpendicular bisector of $AB$ passes through the midpoint of $FH$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $I$ be the midpoint of $AB$ and $M$, $Q$ be points on the base $CD$, such that $DM = QC = AI$.\n\nObviously, $AIMN$ and $BIQC$ are parallelograms, so $IM = AD$, $IM \\parallel AD$ and $IQ = BC$, $IQ \\parallel BC$.\n\nConstruct squares $IMNP$ and $IQRS$, externally to the triangle $IMQ$.\n\nLet $U$ and $V$ be two points such that $IPUS$ is a parallelogram and $V$ is the projection of $I$ onto $CD$.\n\n![](images/RMC_2023_v2_p65_data_945a992d4f.png)\n\nSince $\\angle PIS + \\angle PIM + \\angle MIQ + \\angle QIS = 360^\\circ$ and $IPUS$ is a parallelogram, we infer that $\\angle PIS + \\angle MIQ = 180^\\circ = \\angle IPU + \\angle PIS$, so $\\angle IPU = \\angle MIQ$. As $IP = IM$ and $PU = IS = IQ$, triangles $IPU$ and $MIQ$ are congruent (S.A.S.), so $\\angle PIU = \\angle IMQ$.\n\nTriangle $IMV$ is right-angled at $V$, so $\\angle MIV = 90^\\circ - \\angle IMV = 90^\\circ - \\angle PIU$. It follows that $\\angle MIV + \\angle PIU + \\angle MIP = 180^\\circ$, therefore $U$, $I$, $V$ are collinear.\n\nHence, $UI$ is the perpendicular bisector of segment $AB$. Since $IPUS$ is a parallelogram, the line $UI$ passes through the midpoint $J$ of the diagonal $PS$.\n\nAs $\\angle DAF = \\angle MIP = 90^\\circ$ and $AD \\parallel IM$, it follows that $AF \\parallel IP$. But $AF = IP$, so $AIPF$ is a parallelogram, which means that $FP \\parallel AI$ and $FP = AI$. Analogously, we prove that $SH \\parallel IB$ and $SH = IB$. Since $I$ is the midpoint of $AB$, we infer that $FP \\parallel SH$ and $FP = SH$, so $FPHS$ is a parallelogram. Therefore, the diagonals $FH$ and $PS$ have the same midpoint $J$, so $UI$ (the perpendicular bisector of $AB$) passes through the midpoint of $FH$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17280,
    "subject": "Mathematics (Olympiad)",
    "question": "Charlie has a thick book of $n$ pages in which, after opening, the first page on the right shows the page number $1$. Then all the pages are numbered continuously so that if you open the book just anywhere, you see an even page number on the left and an odd page number on the right. The very last page before the back cover has an even number. Unfortunately, someone tore a whole sheet of paper out of the book. Charlie counts the total number of digits of the page numbers on the $n - 2$ remaining pages and arrives at a total of $2024$ digits.  \n(Note that this is not the sum of the page numbers.)\n\nWhat is $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "$712$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17281,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, let $d(n)$ be the number of positive divisors of $n$, and let $\\varphi(n)$ be the number of positive integers not exceeding $n$ which are coprime with $n$. Prove that there exists an integer $n$ for which\n\n$$\n\\frac{\\varphi(d(n))}{d(\\varphi(n))} > C.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Fix $N > 1$, let $p_1, \\dots, p_k$ be all primes between $1$ and $N$ and $p_{k+1}, \\dots, p_{k+s}$ be all primes between $N+1$ and $2N$. Since for $j \\leq k+s$ all prime divisors of $p_j - 1$ do not exceed $N$, we have\n\n$$\n\\prod_{j=1}^{k+s} (p_j - 1) = \\prod_{i=1}^{k} p_i^{c_i}\n$$\n\nwith some fixed exponents $c_1, \\dots, c_k$. Choose a huge prime number $q$ and consider a number\n\n$$\nn = (p_1 \\cdots p_k)^{q-1} \\cdot (p_{k+1} \\cdots p_{k+s})\n$$\n\nThen\n\n$$\n\\varphi(d(n)) = \\varphi(q^k \\cdot 2^s) = q^{k-1}(q-1)2^{s-1}\n$$\n\nand\n\n$$\nd(\\varphi(n)) = d \\left( (p_1 \\cdots p_k)^{q-2} \\prod_{i=1}^{k+s} (p_i - 1) \\right) = d \\left( \\prod_{i=1}^{k} p_i^{q-2+c_i} \\right) = \\prod_{i=1}^{k} (q-1+c_i)\n$$\n\nSo\n\n$$\n\\frac{\\varphi(d(n))}{d(\\varphi(n))} = \\frac{q^{k-1}(q-1)2^{s-1}}{\\prod_{i=1}^{k} (q-1+c_i)} = 2^{s-1} \\cdot \\frac{q-1}{q} \\cdot \\prod_{i=1}^{k} \\frac{q}{q-1+c_i}\n$$\n\nwhich can be made arbitrarily close to $2^{s-1}$ by choosing $q$ large enough. It remains to show that $s$ can be arbitrarily large, i.e., that there can be arbitrarily many primes between $N$ and $2N$.\n\nThis follows, for instance, from the well-known fact that $\\sum \\frac{1}{p} = \\infty$, where the sum is taken over the set $\\mathbb{P}$ of prime numbers. Indeed, if, for some constant $C$, there were always at most $C$ primes between $2^{\\ell}$ and $2^{\\ell+1}$, we would have\n\n$$\n\\sum_{p \\in \\mathbb{P}} \\frac{1}{p} = \\sum_{\\ell=0}^{\\infty} \\sum_{p \\in \\mathbb{P}_{[2^{\\ell}, 2^{\\ell+1}]}} \\frac{1}{p} \\leq \\sum_{\\ell=0}^{\\infty} \\frac{C}{2^{\\ell}} < \\infty\n$$\n\nwhich is a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17282,
    "subject": "Mathematics (Olympiad)",
    "question": "有 $n$ 種不同顏色的饅頭，第 $i$ 種顏色的饅頭有 $a_i$ 顆，總共有 $S = a_1 + a_2 + \\cdots + a_n$ 顆。這 $S$ 顆饅頭兩兩相異。老趙可以決定如何把這 $S$ 顆饅頭排成一排。對於每一個排列，班長用以下方式為老趙評分：\n\n1. 對每個 $i \\in \\{1, 2, \\cdots, n\\}$，計算第 $i$ 種顏色的饅頭相鄰的次數：\n\n$$\nc_i = \\#\\{j: 1 \\le j \\le S-1, \\text{左數第 } j \\text{ 顆和第 } j+1 \\text{ 顆饅頭都是第 } i \\text{ 色}\\}\n$$\n\n2. 老趙的分數定義為 $c_1 \\times c_2 \\times \\cdots \\times c_n$。\n\n若 $S$ 是合數，證明所有排列的得分總和是 $S$ 的倍數。",
    "options": [],
    "answer": "See solution",
    "solution": "首先，若有任何 $a_i = 1$，則 $c_i$ 恆為 0，得分總和為 0，必被 $S$ 整除。以下只討論每種顏色饅頭都至少有 2 顆的情況，此時 $S \\ge 2n$。\n\n1. 用雙重計數法計算得分總和。令 $M_j^i$ 為第 $i$ 色編號第 $j$ 的饅頭，一個排列即是將全部 $S$ 個 $M_j^i$ 排成一串。每個排列的分數等價於：\n\n每種顏色取出相鄰的一對，總共可以取的方法數。\n\n每種顏色指定（依序）相鄰一對，放進 $S$ 格中的方法數為 $\\frac{(S-n)!}{(S-2n)!}$。\n\n剩下 $S - 2n$ 個位置共有 $(S - 2n)!$ 種排法。\n\n因此，每種顏色指定（依序）相鄰一對的排列數為 $(S - 2n)!$，而每種顏色指定（依序）相鄰一對共有 $\\prod_i a_i (a_i - 1)$ 種，得分總和為：\n\n$$\nP = (S-n)!\\,(a_1)(a_1-1)(a_2)(a_2-1)\\cdots(a_n)(a_n-1)\n$$\n\n2. 證明 $S \\mid P$。\n\n*Lemma*: 如果 $S$ 是合數，且 $S \\nmid \\lfloor \\frac{S}{2} \\rfloor!$，則 $S = 4$ 或 $9$。\n\n證明：若 $S$ 不是質數的平方，設 $S = pq, p \\neq q$，則 $p, q$ 都在 $\\lceil \\frac{S}{2} \\rceil$ 內，所以 $S \\mid \\lceil \\frac{S}{2} \\rceil!$。\n\n若 $S = p^2$ 且 $p \\ge 5$，則 $p, 2p$ 都在 $\\lceil \\frac{S}{2} \\rceil$ 內，所以 $S \\mid \\lceil \\frac{S}{2} \\rceil!$。\n\n所以反例只剩 $S = 4$ 或 $9$。\n\n因為 $(S-n) \\ge \\lceil \\frac{S}{2} \\rceil$，只需討論 $S = 4$ 或 $9$。\n\n當 $S = 4$，若 $n = 1$ 得分是 72，若 $n = 2$ 且 $a_1 = a_2 = 2$，得分是 8。\n\n當 $S = 9$，注意到 $9 \\mid 6!$，所以 $n > 3$ 即 $n = 4$。\n\n將 9 拆成四個不小於 2 的正整數之和，必有一個是 3，所以 $(9-4)!(a_1)(a_1-1)(a_2)(a_2-1)\\dots(a_4)(a_4-1)$ 是 9 的倍數，證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17283,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the sequence $(u_n)$ defined by:\n\n$$u_1 = 52$$\n\nand\n\n$$u_{n+1} = u_n^2 + 9u_n, \\quad \\forall n \\geq 1.$$ \n\nProve that every term of this sequence is not a perfect square, but there exist infinitely many terms of the sequence which, when divided by $2^{2023}$, leave a remainder that is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to see that this sequence satisfies $u_n \\geq 52$ for all $n \\geq 1$. Suppose there exists $n$ such that $u_{n+1}$ is a perfect square, then there exists $k \\in \\mathbb{Z}^+$ such that $u_n^2 + 9u_n = k^2$, or equivalently $4u_n^2 + 36u_n = 4k^2$. We have\n\n$$\n(2u_n + 7)^2 < 4u_n^2 + 36u_n < (2u_n + 9)^2.\n$$\n\nThis implies that $2k = 2u_n + 8$ or $k = u_n + 4$. But then $u_n^2 + 9u_n = (u_n + 4)^2$ leads to $u_n = 16$, which is a contradiction since $u_n \\geq 52$. Thus, there is no perfect square number in this sequence.\n\nNow consider the sequence $(v_n)$ with $v_1 = 4$, $v_{n+1} = v_n^2 + 9v_n$. Then $v_2 = 52$, so it's easy to see that $u_n = v_{n+1}$ for all $n \\geq 1$. We will prove that the remainder of this sequence when divided by $m = 2^{2023}$ is periodic from the first term.\n\nLet $r_n$ be the remainder of $v_n$ when divided by $m$. Since $v_n$ is even and $m$ is even, we also have $r_n$ even for all $n \\geq 1$. Since these remainders only belong to $\\{0, 1, \\dots, m-1\\}$ and the sequence has infinite terms, there exist distinct indices $n, k \\in \\mathbb{Z}^+$ with $n < k$ such that $r_n = r_k \\pmod m$. From there, it is easy to see that $v_{n+1} \\equiv v_{k+1} \\pmod m$, which implies $r_{n+1} = r_{k+1}$ and so on; the remainder of the sequence when divided by $m$ is periodic. On the other hand,\n\n$$\nv_{n-1}^2 + 9v_{n-1} \\equiv v_{k-1}^2 + 9v_{k-1} \\pmod m\n$$\n\nso $m \\mid (v_{k-1} - v_{n-1})(v_{k-1} + v_{n-1} + 9)$. Note that $v_{k-1}, v_{n-1}$ are both even, so $v_{k-1} + v_{n-1} + 9$ is odd, and $\\gcd(2^{2023}, v_{k-1} + v_{n-1} + 9) = 1$. This shows $m \\mid v_{k-1} - v_{n-1}$ or $r_{n-1} = r_{k-1}$. Therefore, going backwards, there will be $r_1 = r_{k-n}$, which is a cyclic sequence from the first term. Therefore, in the sequence $(v_n)$, there will be infinitely many terms divided by $m$ with remainder 4. It's easy to see that this is also true for the sequence $(u_n)$ because the two sequences differ only by one index. Since 4 is a perfect square number, it can be concluded that there exist infinitely many terms of $(u_n)$ which, when divided by $2^{2023}$, leave a perfect square remainder. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17284,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integer solutions to the equation\n\n$$\nx^2(x^2 + 1) = 21^y - 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Rewrite the equation as $x^2(x^2 + 1) + 1 = 21^y$, or $x^4 + x^2 + 1 = 21^y$. This factors as $(x^2 - x + 1)(x^2 + x + 1) = 21^y$.\n\nSince $x^2 - x + 1$ and $x^2 + x + 1$ are coprime, and $0 < x^2 - x + 1 < x^2 + x + 1$, the only possibilities are:\n\n- $x^2 - x + 1 = 1$, $x^2 + x + 1 = 21^y$\n- $x^2 - x + 1 = 3^y$, $x^2 + x + 1 = 7^y$\n\n**Case 1:** $x^2 - x + 1 = 1$ gives $x = 0$ or $x = 1$. Plugging into the original equation, only $x = y = 0$ works.\n\n**Case 2:** $x^2 - x + 1 = 3^y$, $x^2 + x + 1 = 7^y$. Subtracting, $2x = 7^y - 3^y$, so $x$ is integer only for small $y$. Testing $y = 1$ gives $x^2 - x + 1 = 3$, so $x^2 - x - 2 = 0$, so $x = 2$ or $x = -1$. But $x^2 + x + 1 = 7$, so $x^2 + x - 6 = 0$, so $x = 2$ or $x = -3$. The only common integer solution is $x = 2$.\n\nSimilarly, for $x = -2$, $x^2 - x + 1 = 7$, $x^2 + x + 1 = 3$, which does not match the powers. But checking $x = -2$ in the original equation: $(-2)^2((-2)^2 + 1) = 4 \\times 5 = 20 = 21^1 - 1$, so $x = -2$, $y = 1$ is also a solution.\n\nThus, the integer solutions are:\n\n- $x = 0$, $y = 0$\n- $x = 2$, $y = 1$\n- $x = -2$, $y = 1$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17285,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all infinite sequences $a_1, a_2, \\dots$ of positive integers such that for any positive integers $n$ and $k$\n\n$$\n\\frac{a_n + a_{n+1} + \\dots + a_{n+k-1}}{k} \\quad \\text{and} \\quad \\sqrt[k]{a_n a_{n+1} \\dots a_{n+k-1}} \\quad \\text{are integers.}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/Saudi_Booklet_2025_p41_data_60a32e958a.png)\n\n![](images/Saudi_Booklet_2025_p42_data_0a82399ce8.png)\n\n![](images/Saudi_Booklet_2025_p42_data_9e2bcb14bc.png)\n\n![](images/Saudi_Booklet_2025_p42_data_a038043e28.png)\n\n![](images/Saudi_Booklet_2025_p43_data_66072370fe.png)\n\n![](images/Saudi_Booklet_2025_p43_data_0407f7ca84.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17286,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $m, n \\ge 2$ 為整數，$X = \\{1, 2, \\dots, n\\}$，且 $X_1, X_2, \\dots, X_m$ 為 $m$ 個 $X$ 的相異非空子集（其兩兩交集不必然非空）。一個函數 $f : X \\to \\{1, 2, \\dots, n+1\\}$ 是好棒棒的，若且唯若存在指標 $k$ 使得\n\n$$\n\\sum_{x \\in X_k} f(x) > \\sum_{x \\in X_i} f(x) \\text{ 對於所有 } i \\neq k \\text{ 皆成立。}\n$$\n\n證明好棒棒函數的總數不少於 $n^n$。",
    "options": [],
    "answer": "See solution",
    "solution": "對於 $Y \\subset X$ 與任何定義域在 $X$ 上的函數 $f$，我們定義 $f(Y) = \\sum_{x \\in Y} f(x)$。則 $f$ 是好棒棒的等價於 $f(X_i)$ 在唯一的指標 $i \\in \\{1, 2, \\dots, m\\}$ 時取到最大值。\n\n令 $\\mathcal{G}$ 為所有函數 $g : X \\to \\{1, 2, \\dots, n\\}$ 所成集合，注意到 $|\\mathcal{G}| = n^n$。現在，對於每一個 $g \\in \\mathcal{G}$，取任一極大化 $g(X_\\ell)$ 的指標 $\\ell$，並定義\n\n$$\nf_g(x) := \\begin{cases} g(x) + 1, & x \\in X_\\ell, \\\\ g(x), & x \\notin X_\\ell. \\end{cases}\n$$\n\n若能證明 (1) $f_g$ 好棒棒且 (2) $f_g$ 全相異，便證明好棒棒函數至少有 $n^n$ 個。\n\n(1) **$f_g$ 好棒棒**\n\n注意到 $f_g(X_i) = g(X_i) + |X_i \\cap X_\\ell|$ 對於所有 $i$ 皆成立。讓我們證明 $f_g(X_i)$ 在唯一的指標 $\\ell$ 取得最大值：任取 $i \\neq \\ell$。注意到我們必有 $X_\\ell \\not\\subset X_i$（否則 $g(X_\\ell) < g(X_i)$，與 $g(X_\\ell)$ 的最大性矛盾），而這意味著 $|X_\\ell| > |X_i \\cap X_\\ell|$，因此\n\n$$\nf_g(X_\\ell) = g(X_\\ell) + |X_\\ell| \\ge g(X_i) + |X_\\ell| > g(X_i) + |X_i \\cap X_\\ell| = f_g(X_i).\n$$\n\n故 $f_g(X_i)$ 在唯一的指標 $\\ell$ 取得最大值，亦即 $f_g$ 好棒棒。\n\n(2) **$f_g$ 全相異**\n\n注意到前項說明 $f_g(X_\\ell)$ 是唯一最大值，從而我們有\n\n$$\ng(x) := \\begin{cases} f_g(x) - 1, & x \\in X_\\ell, \\\\ f_g(x), & x \\notin X_\\ell. \\end{cases}\n$$\n\n這表示 $f_g$ 與 $g$ 為一一對應，因此 $f_g$ 全相異。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17287,
    "subject": "Mathematics (Olympiad)",
    "question": "Draw a semicircle with radius 1. Let $A$ and $B$ be the endpoints of the arc. Take two points $C$ and $D$ on arc $AB$ so that $A$, $C$, $D$, and $B$ are on the arc in this order.\n\nLet $2a = AC$, $2b = AD$, $\\angle CAB = \\alpha$, $\\angle DAB = \\beta$. Find $a$ in terms of $\\cos \\alpha$ and $\\beta$, and find the area of triangle $ACD$.\n\nNext, take points $A_1, A_2, \\dots, A_n$ on arc $AB$ with $AA_i = 2a_i$, and consider the sum of the areas of triangles $AA_iA_{i+1}$. Show that the left side of the given inequality does not exceed $\\pi/4$, which is half the area of the semicircle.\n\nFinally, prove that $c$ cannot be smaller than $\\pi/4$ by dividing the arc $AB$ into $2^n$ equal parts and considering the area not covered by the polygon $A_0A_1 \\cdots A_{2^n}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = \\cos \\alpha$ and $b = \\cos \\beta$. The area of triangle $ACD$ is:\n\n$$\n\\frac{1}{2}(2a)(2b) \\sin(\\alpha - \\beta) = 2ab(\\sin \\alpha \\cos \\beta - \\cos \\alpha \\sin \\beta) = 2ab(b\\sqrt{1-a^2} - a\\sqrt{1-b^2}) = 2f(a, b).\n$$\n\nBy taking points $A_1, A_2, \\dots, A_n$ on arc $AB$ with $AA_i = 2a_i$ and summing the areas of triangles $AA_iA_{i+1}$, the total does not exceed $\\pi/4$, half the area of the semicircle.\n\nTo show $c$ cannot be smaller than $\\pi/4$, divide arc $AB$ into $2^n$ equal parts: $A = A_0, A_1, \\dots, A_{2^n} = B$, with $AA_i = 2a_i$. Let $S_n$ be the part of the semicircle not covered by polygon $A_0A_1 \\cdots A_{2^n}$, consisting of $2^n$ congruent figures $P_n$. Since $2|P_{n+1}| < \\frac{1}{2}|P_n|$, $|S_{n+1}| < \\frac{1}{2}|S_n|$, so $|S_n| < \\frac{1}{2^{n-1}}|S_1|$. If $c < \\pi/4$, for large $n$ the uncovered area is less than $\\pi/4 - c$, which is impossible. Thus, the minimum value of $c$ is $\\pi/4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17288,
    "subject": "Mathematics (Olympiad)",
    "question": "We say that the integer $x$, $1 \\leq x \\leq 100$, is a member of family $n$ if $x$ has exactly $n$ positive divisors. Which family has the greatest number of members?",
    "options": [],
    "answer": "See solution",
    "solution": "If $x = p_1^{k_1} \\cdots p_m^{k_m}$ is the prime factorization of $x$, then $x$ has $(k_1 + 1) \\cdots (k_m + 1)$ positive divisors. Let family $n$ have $M(n)$ members. We find $M(n)$ for $n = 1, \\dots, 6$.\n\nClearly, $M(1) = 1$.\n\nFamily 2 consists of the primes not exceeding 100, of which there are 25; so $M(2) = 25$.\n\nFamily 3 is the set of numbers $p^2$ where $p$ is a prime. Its members are $2^2, 3^2, 5^2, 7^2$, hence $M(3) = 4$.\n\nThe members of family 4 are the numbers $pq$ where $p, q$ are distinct primes, and the cubes $p^3$ with $p$ a prime. If $pq$ is a member of the first kind with $p < q$, then $p \\in \\{2, 3, 5, 7\\}$. For $p = 2$ there are 14 members: $2 \\cdot 3, 2 \\cdot 5, \\dots, 2 \\cdot 47$. Similarly, $p = 3$, $p = 5$, and $p = 7$ give rise to 9, 5, and 2 members respectively. So far we have $14 + 9 + 5 + 2 = 30$ members of family 4. There are two more, the cubes $2^3$ and $3^3$. Hence $M(4) = 32$.\n\nFamily 5 contains only numbers of the form $p^4$ with $p$ a prime. There are two of them in $[1, 100]$, $2^4$ and $3^4$, so $M(5) = 2$.\n\nThe members of family 6 are the numbers $p^2q$ where $p, q$ are distinct primes, and the numbers of the form $p^5$ with $p$ a prime. If $p^2q$ is a member of the first kind, then again $p \\in \\{2, 3, 5, 7\\}$. For $p = 2$ we obtain 8 members: $2^2 \\cdot 3, 2^2 \\cdot 5, \\dots, 2^2 \\cdot 23$. Likewise, $p = 3$, $p = 5$, and $p = 7$ contribute 4, 2, and 1 members respectively. So far we have $8 + 4 + 2 + 1 = 15$ members of family 6. There is one more, the fifth power $2^5$. Hence $M(6) = 16$.\n\nSo families $1, 2, \\dots, 6$ contain $1 + 25 + 4 + 32 + 2 + 16 = 80$ members in all, and $M(4) = 32$. Since there are only 20 numbers left in $[1, 100]$, family 4 has the greatest number of members.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17289,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle such that $AB = AC$. Point $K$ lies on the altitude drawn from vertex $A$, and point $L$ is chosen on the line $BK$ such that $AL \\parallel BC$. Prove that if $KC \\perp CL$, then point $L$ lies on the bisector of the external angle at vertex $C$ of triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $AL \\parallel BC$ and $AK \\perp BC$, we have $\\angle KAL = 90^\\circ$. As $KAL$ and $KCL$ are both right angles, points $A$ and $C$ lie on a circle with diameter $KL$. By the inscribed angle theorem, $\\angle KCA = \\angle KLA$. Also, $\\angle KLA = \\angle KBC$, and by the symmetry of the isosceles triangle, $\\angle KBC = \\angle KCB$. Therefore, $\\angle KCA = \\angle KCB$, implying that $KC$ bisects the internal angle at vertex $C$ of triangle $ABC$. Since $KCL$ is a right angle, $CL$ bisects the corresponding external angle.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17290,
    "subject": "Mathematics (Olympiad)",
    "question": "The altitudes $AD$ and $BE$ of acute triangle $ABC$ intersect at $H$. Let $F$ be the intersection of $AB$ and a line that is parallel to the side $BC$ and goes through the circumcenter of $ABC$. Let $M$ be the midpoint of $AH$. Prove that $\\angle CMF = 90^\\circ$.\n\n![](images/MNG_ABooklet_2017_p20_data_380b4e3605.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the circumcenter of $\\triangle ABC$, and $N$ be the midpoint of $BC$. Let $P$ be a point on $BC$ so that $KF \\perp BC$, and $K$ be the intersection of $AB$ and $HP$. Then $FP = \\frac{AH}{2}$ since $ON = \\frac{AH}{2}$ and $FP = ON$. Thus it follows from $FP \\parallel AH$ that $FP$ is the mid-segment of $\\triangle AKH$. Moreover, $MP \\parallel AK$ and $CH \\perp MP$. Thus $PH \\perp MC$ since $H$ is the orthocenter of $\\triangle CMP$. Hence $PH \\perp MC$ which implies $FM \\perp MC$ since $PH \\parallel FM$. Hence $\\angle CMF = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17291,
    "subject": "Mathematics (Olympiad)",
    "question": "The medians $AD$, $BE$, and $CF$ of triangle $ABC$ intersect at $G$. Let $P$ be a point lying in the interior of the triangle, not belonging to any of its medians. The line through $P$ parallel to $AD$ intersects the side $BC$ at $A_1$. Similarly, define the points $B_1$ and $C_1$. Prove that\n\n$$\n\\overrightarrow{A_1D} + \\overrightarrow{B_1E} + \\overrightarrow{C_1F} = \\frac{3}{2} \\overrightarrow{PG}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us draw through point $P$ parallels to the triangle's sides. If the parallels to $AB$ and $AC$ intersect the side $BC$ at points $A_B$ and $A_C$, then the triangles $ABC$ and $PA_BA_C$ are similar, and since $PA_1$ is parallel to the median $AD$, it follows that $A_1$ is the midpoint of the line segment $A_BA_C$. With the notations in the figure, we have\n\n$$\n2\\overrightarrow{A_1D} = \\overrightarrow{A_1B} + \\overrightarrow{A_1C} = \\overrightarrow{PA_B} + \\overrightarrow{PA_C}.\n$$\n\nSimilarly for the other medians. We deduce that the sum $2(\\overrightarrow{A_1D} + \\overrightarrow{B_1E} + \\overrightarrow{C_1F})$ equals\n\n$$\n\\vec{v} = \\overrightarrow{PA_B} + \\overrightarrow{PA_C} + \\overrightarrow{PB_C} + \\overrightarrow{PB_A} + \\overrightarrow{PC_A} + \\overrightarrow{PC_B}.\n$$\n\nBut $\\overrightarrow{PA_B} + \\overrightarrow{PB_A} = \\overrightarrow{PA}$, and adding up all similar equalities yields\n\n$$\n\\vec{v} = \\overrightarrow{PA} + \\overrightarrow{PB} + \\overrightarrow{PC} = 3\\overrightarrow{PG},\n$$\n\nhence the conclusion.\n\n![](images/RMC2014_p15_data_9bc5d5e494.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17292,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of single-digit numbers $(m, n)$ such that the last digit of $m^3$ is $n$, and the last digit of $n^3$ is $m$. For example, if $m$ ends with 2 and $m^3$ ends with $n$, and $n$ ends with 8 and $n^3$ ends with $m$, find such pairs. Then, generalize this process to construct pairs of $l$-digit numbers $(m, n)$ with the same property: the last $l$ digits of $m^3$ are $n$, and the last $l$ digits of $n^3$ are $m$. Prove that such pairs exist for every $l$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us show by examples the transition from a single-digit number to a two-digit number:\n\n$$\n\\begin{align*}\nm_1 = 2,\\quad m_1^3 = 8 &\\rightarrow 10 \\cdot 0 + 8 \\Rightarrow A = 0,\\\\\nn_1 = 8,\\quad n_1^3 = 512 &\\rightarrow 10 \\cdot 1 + 2 \\Rightarrow B = 1.\\\\\n3x \\rightarrow -(2A+B) = -1 &\\rightarrow 9 \\Rightarrow x = 3,\\\\\n3x \\rightarrow -(2B+A) = -2 &\\rightarrow 8 \\Rightarrow y = 6.\n\\end{align*}\n$$\n\nWe obtain the numbers:\n\n$$\nm_2 = \\overline{xm_1} = 32,\\quad m_2^3 = 32^3 = 32768,\\\\\nn_2 = \\overline{yn_1} = 68,\\quad n_2^3 = 68^3 = 314432,\n$$\n\nwhich satisfy the statement.\n\nThe transition from a two-digit number to a three-digit number:\n\n$$\n\\begin{align*}\nm_2 = 32,\\quad m_2^3 = 32768 &\\rightarrow 100 \\cdot 7 + 68 \\Rightarrow A = 7,\\\\\nn_2 = 68,\\quad n_2^3 = 314432 &\\rightarrow 100 \\cdot 4 + 32 \\Rightarrow B = 4.\n\\end{align*}\n$$\n\n$$\n3x \\rightarrow -(2A+B) = -18 \\rightarrow 2 \\Rightarrow x = 4,\\\\\n3x \\rightarrow -(2B+A) = -15 \\rightarrow 5 \\Rightarrow y = 5.\n$$\n\nWe obtain the numbers:\n\n$$\nm_3 = \\overline{xm_2} = 432,\\quad m_3^3 = 432^3 = 80621568,\\\\\nn_3 = \\overline{yn_2} = 568,\\quad n_3^3 = 568^3 = 183250432,\n$$\nwhich, too, satisfy the statement.\n\n**Alternative solution.**\n\nLet us prove that such a pair exists for every value of $l$. We show that there is a positive integer $x$ such that $2^l \\mid x^2 - 1$ and $5^l \\mid x^2 + 1$.\n\nFor the first divisibility, set $x \\equiv 1 \\pmod{2^l}$. Using mathematical induction for $l$, we show that for any $l$ there exists $x$ such that $5^l \\mid x^2 + 1$. If $l = 1$, set $x = 2$.\n\nSuppose for some $l$, the statement holds. Consider $l+1$: if $x$ works for $l$, i.e. $x^2 + 1 = t \\cdot 5^l$, then set $x_1 = x + k \\cdot 5^l$; so\n\n$$\nx_1^2 + 1 = x^2 + k^2 \\cdot 5^{2l} + 2xk \\cdot 5^l + 1 = t \\cdot 5^l + k^2 \\cdot 5^{2l} + 2xk \\cdot 5^l = 5^l(t + k^2 \\cdot 5^l + 2xk).\n$$\n\nThere exists $k$ such that the expression in parentheses is divisible by 5; we choose such $x_1$ for this $k$, and the statement is proved.\n\nBy the Chinese remainder theorem, we select $x$ which satisfies both conditions. Then, the last $l$ digits of the pair $x$ and $x^3$ satisfy the statement. By construction, both numbers have no more than $l$ digits. They have different last digits: otherwise, $x^3 - x$ would be divisible by 5, but $x^3 - x = x(x^2 - 1) \\equiv -2x \\pmod{5}$, so $x$ would be divisible by 5, which is impossible if $x^2 + 1 = t \\cdot 5^l$.\n\nIt remains to show that $x - x^3 \\equiv 0 \\pmod{10^l}$, which is obvious by construction, and also\n\n$$\n(x^3)^3 - x = x^9 - x = x(x^4 + 1)(x^2 + 1)(x^2 - 1) \\equiv 0 \\pmod{10^l},\n$$\n\nwhich is also clear.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17293,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $F$ be a point, and a line through $F$ intersects the line $AD$ again at $H$ and the line $CF$ again at $K$. Prove that\n$$\n\\frac{FD \\times HK}{FH \\times DK} = 3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $AF = x$, $BF = y$, $CD = z$. By Stewart's theorem,\n\n$$\n\\begin{aligned}\nAD^2 &= \\frac{BD}{BC} \\times AC^2 + \\frac{CD}{BC} \\times AB^2 - BD \\times DC \\\\\n&= \\frac{y(x+z)^2 + z(x+y)^2}{y+z} - yz \\\\\n&= x^2 + \\frac{4xyz}{y+z}\n\\end{aligned}\n$$\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p175_data_02f92bab83.png)\n\nBy the power of a point theorem,\n\n$$\n\\begin{aligned}\nAH &= \\frac{AF^2}{AD} = \\frac{x^2}{AD}, \\\\\nHD &= AD - AH = \\frac{AD^2 - x^2}{AD} = \\frac{4xyz}{AD(y+z)}.\n\\end{aligned}\n$$\n\nSimilarly, $KF = \\frac{4xyz}{CF(x+y)}$. From $\\triangle CDK \\sim \\triangle CFD$, $DK = \\frac{DF \\times CD}{CF} = \\frac{DF}{CF} \\times z$.\n\nFrom $\\triangle AFH \\sim \\triangle ADF$, $FH = \\frac{DF \\times AF}{AD} = \\frac{DF}{AD} \\times x$. Using the cosine theorem,\n\n$$\n\\begin{aligned}\nDF^2 &= BD^2 + BF^2 - 2BD \\cdot BF \\cos B \\\\\n&= 2y^2 \\left(1 - \\frac{(y+z)^2 + (x+y)^2 - (x+z)^2}{2(x+y)(y+z)}\\right) \\\\\n&= \\frac{4xy^2z}{(x+y)(y+z)}.\n\\end{aligned}\n$$\n\nSo,\n$$\n\\frac{KF \\times HD}{FH \\times DK} = \\frac{\\frac{4xyz}{CF(x+y)} \\times \\frac{4xyz}{AD(y+z)}}{\\frac{DF}{AD} \\times \\frac{DF}{CF} z} = \\frac{16xy^2z}{DF^2(x+y)(y+z)} = 4.\n$$\n\n$D, K, H, F$ are concyclic, so by Ptolemy's theorem:\n\n$$\nKF \\times HD = DF \\times HK + FH \\times DK,\n$$\n\nand thus\n$$\n\\frac{FD \\times HK}{FH \\times DK} = 3.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17294,
    "subject": "Mathematics (Olympiad)",
    "question": "The sequence of positive integers $\\{a_n\\}_{n \\ge 1}$ satisfies $a_n \\le 1000^{1000^{\\sqrt{\\log_2 n}}}$ for all $n \\ge 1$. It is known that any number appears in the sequence at most 5000 times. Prove that there are infinitely many primes, each of which divides at least one term of the sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/Saudi_Booklet_2025_p47_data_57b0623783.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17295,
    "subject": "Mathematics (Olympiad)",
    "question": "Let a finite decimal fraction be given. Juku starts appending digits to this fraction in such a way that each new digit equals the remainder of the sum of all digits existing so far divided by 10. For example, if the initial fraction is $27.35$, then the digits added to the end are $7$, $4$, $8$, etc.\n\nProve that the infinite decimal fraction obtained this way represents a rational number.",
    "options": [],
    "answer": "See solution",
    "solution": "It suffices to show that the infinite decimal fraction is periodic. For that, note that each new digit except for the first digit is congruent to twice the previous digit modulo $10$. Indeed, let $a_1, \\dots, a_{k-1}, a_k$ be the existing digits at some moment, where $a_k$ is already added by Juku. Then the next digit $a_{k+1}$ satisfies\n\n$$\na_{k+1} \\equiv a_1 + \\dots + a_k = (a_1 + \\dots + a_{k-1}) + a_k \\equiv a_k + a_k = 2a_k \\pmod{10}.\n$$\n\nHence, each new digit is uniquely determined by the last existing digit. As there are only a finite number of different digits, some digit must be added repeatedly. According to the fact just proven, all following digits are repeated as well.\n\n**Remark:** This solution can be reformulated without mentioning the fact that each new digit except for the first is congruent to twice the previous one. After noting that Juku must add some digit repeatedly, denote their position numbers in the decimal fraction by $m$ and $n$. Hence, the digits before the $m$th digit and the digits before the $n$th digit sum up to congruent numbers modulo $10$. Adding the $m$th and the $n$th digit, respectively, to the sums maintains the congruence. This means that the next digits are also equal. Thus, the digits start repeating periodically.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17296,
    "subject": "Mathematics (Olympiad)",
    "question": "A grasshopper is sitting at an integer point in the Euclidean plane. Each second, it jumps to another integer point in such a way that the jump vector is constant. A hunter, who knows neither the starting point of the grasshopper nor the jump vector (but knows that the jump vector is constant each second), wants to catch the grasshopper. Each second, the hunter can choose one integer point in the plane and, if the grasshopper is there, he catches it. Can the hunter always catch the grasshopper in a finite amount of time?",
    "options": [],
    "answer": "See solution",
    "solution": "The hunter can catch the grasshopper. Here is a strategy:\n\nLet $f$ be any bijection between the set of positive integers and the set $\\{((x, y), (u, v)) : x, y, u, v \\in \\mathbb{Z}\\}$. Denote\n\n$$\nf(t) = ((x_t, y_t), (u_t, v_t))\n$$\n\nOn the $t$-th second, the hunter should search at the point $(x_t + t u_t, y_t + t v_t)$.\n\nAssume the grasshopper starts at $(x', y')$ and the jump vector is $(u', v')$. Then, at second $t$, the grasshopper is at $(x' + t u', y' + t v')$. Let\n\n$$\nt' = f^{-1}((x', y'), (u', v'))\n$$\n\nBy the hunter's strategy, at second $t'$, he searches at $(x_{t'} + t' u_{t'}, y_{t'} + t' v_{t'})$, which is $(x' + t' u', y' + t' v')$—the grasshopper's position at that time. Thus, the hunter will catch the grasshopper in finite time.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17297,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $x$ such that $\\lfloor \\frac{7}{x} \\rfloor + \\lfloor \\frac{x}{7} \\rfloor = 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $x$ is negative, both $\\lfloor \\frac{7}{x} \\rfloor$ and $\\lfloor \\frac{x}{7} \\rfloor$ are negative, so there are no solutions.\n\nIf $x$ is positive, both terms are non-negative integers, and the only way for them to sum to $2$ is if both are equal to $1$, or one is equal to $0$ and the other equal to $2$. We consider these cases:\n\n- **Case 1:** Both terms are equal to $1$:\n  $$\n  \\lfloor \\frac{7}{x} \\rfloor = 1 \\implies 1 \\leq \\frac{7}{x} < 2\n  $$\n  $$\n  \\lfloor \\frac{x}{7} \\rfloor = 1 \\implies 1 \\leq \\frac{x}{7} < 2\n  $$\n  The only way both inequalities are satisfied is if $x = 7$.\n\n- **Case 2:** $\\lfloor \\frac{7}{x} \\rfloor = 2$ and $\\lfloor \\frac{x}{7} \\rfloor = 0$:\n  $$\n  2 \\leq \\frac{7}{x} < 3 \\implies \\frac{7}{3} < x \\leq \\frac{7}{2}\n  $$\n  $$\n  0 \\leq \\frac{x}{7} < 1 \\implies 0 < x < 7\n  $$\n  The intersection is $\\frac{7}{3} < x \\leq \\frac{7}{2}$.\n\n- **Case 3:** $\\lfloor \\frac{7}{x} \\rfloor = 0$ and $\\lfloor \\frac{x}{7} \\rfloor = 2$:\n  $$\n  0 \\leq \\frac{7}{x} < 1 \\implies x > 7\n  $$\n  $$\n  2 \\leq \\frac{x}{7} < 3 \\implies 14 \\leq x < 21\n  $$\n  The intersection is $14 \\leq x < 21$.\n\n**Final answer:**\n\nThe possible values for $x$ are:\n- $\\frac{7}{3} < x \\leq \\frac{7}{2}$,\n- $x = 7$,\n- $14 \\leq x < 21$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17298,
    "subject": "Mathematics (Olympiad)",
    "question": "A and B are two convex polygons without common points. None of them is fully contained inside the other one. Prove that there exists such a line $l$ that does not intersect each of the polygons and A and B lie on the different sides of $l$.\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p207_data_3bdd83f8a0.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Take a point $P$ on $A$ and a point $Q$ on $B$ such that $|PQ|$ is minimal possible. We will prove that the perpendicular bisector of $PQ$ fits the definition of $l$.\n\nDraw a circle $\\omega$ with center $P$ and radius $|PQ|$, and denote the perpendicular bisector of $PQ$ as $t$. Suppose that some segment $s$ of $B$ that passes through $Q$ intersects $t$. Then the line on which $s$ lies intersects $\\omega$ in some point $R$ other than $Q$. Then there is some point on the segment $RQ$ that belongs to $s$ and is closer to $P$ than $Q$ is, contradiction. Hence none of the $B$ segments that pass through $Q$ intersect $t$. Since $B$ is convex, none of the other $B$ segments also intersect $t$, thus $B$ does not intersect $t$. Similarly we prove that $A$ also does not intersect $t$, and the proof is complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17299,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that $A \\subset \\{0, 1, \\dots, 9\\}$ is a set such that every positive integer can be written as a sum of two nonnegative integers consisting only of digits from $A$. At least how many elements can set $A$ have?",
    "options": [],
    "answer": "See solution",
    "solution": "Five digits are enough. For instance, the set $A = \\{0, 1, 2, 3, 6\\}$ satisfies the requirement, because every digit can be written as a sum of two digits from $A$.\n\nSuppose that there is a working set $A$ consisting of only four digits. Then their ten pairwise sums would have to give all possible remainders modulo $10$. However, if $A$ contains $k$ even and $4-k$ odd digits, then these digits will form only $k(4-k) \\leq 4$ odd pairwise sums, which is a contradiction.\n\nTherefore, the answer is $5$. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17300,
    "subject": "Mathematics (Olympiad)",
    "question": "In a mathematical competition, some students are friends, and friendship is always mutual. Prove that there exists a subset $M$ of the students such that every member of $M$ has at most three friends in $M$, and every non-member of $M$ has at least four friends in $M$.",
    "options": [],
    "answer": "See solution",
    "solution": "We start with the set $M_0 = \\emptyset$ and modify it step by step until it satisfies the two desired conditions.\n\nThere are two types of improving steps:\n\n1. If there is a student $x \\in M$ who has $\\geq 4$ friends in $M$, remove $x$ from $M$.\n2. If there is a student $y \\notin M$ who has $\\leq 3$ friends in $M$, add $y$ to $M$.\n\nThese steps may repeat for the same student, but the process must eventually terminate. For a set $M$, let $p(M)$ denote the number of ordered pairs $(x, y) \\in M \\times M$ for which $x$ and $y$ are friends. Define\n\n$$\nf(M) := 7|M| - p(M).\n$$\n\nThis function is integer-valued, bounded above by $7$ times the total number of students, and starts at $f(M_0) = 0$. Each improving step strictly increases $f(M)$:\n\n- If $x \\in M$ has $\\geq 4$ friends in $M$, removing $x$ decreases $|M|$ by $1$ and $p(M)$ by at least $8$, so $f(M)$ increases.\n- If $y \\notin M$ has $\\leq 3$ friends in $M$, adding $y$ increases $|M|$ by $1$ and $p(M)$ by at most $6$, so $f(M)$ increases.\n\nSince $f(M)$ is bounded above and strictly increases with each step, the process terminates with a set $M$ satisfying the required conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17301,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all non-negative integers $n$ smaller than $128^{97}$ which have exactly 2019 positive divisors.",
    "options": [],
    "answer": "See solution",
    "solution": "Numbers with exactly 2019 positive divisors are either of the form $p^{2018}$ or $p^{672} \\cdot q^2$ for distinct prime numbers $p$ and $q$.\n\nThe number $128^{97}$ can be written as\n\n$$\n128^{97} = (2^7)^{97} = 2^{679}.\n$$\n\nAs $p$ is at least 2, the number $p^{2018}$ is greater than $2^{679}$ and therefore, the case $n = p^{2018}$ is impossible. Thus we have $n = p^{672} \\cdot q^2$ with $p^{672} \\cdot q^2 < 2^{679}$. Hence $p = 2$ and as $q^2 < 2^7 = 128$, $q$ is one of the primes 3, 5, 7, or 11.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17302,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $\\alpha$ and $\\beta$ are real numbers. Consider the set $\\left\\{ \\frac{1}{\\alpha}, \\frac{1}{\\beta}, 1 \\right\\}$ in $\\mathbb{Q}$. Show that if this set is linearly dependent over $\\mathbb{Q}$, then there exist integers $a, b, c$ with $\\gcd(a, b, c) = 1$ such that\n\n$$\n\\frac{a}{\\alpha} + \\frac{b}{\\beta} = c.\n$$\n\nFurthermore, show that $a, b > 0$ and $c = 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = ak + u$, where $a, u$ are positive integers, and choose $k$ so that\n\n$$\nt = \\frac{\\frac{n}{\\beta} - v}{a} = \\frac{ak + u - v\\beta}{a\\beta} = \\frac{k}{\\beta} - \\frac{v\\beta - u}{a\\beta}\n$$\n\nhas decimal part $\\{t\\} < \\min\\left\\{\\frac{1}{ab}, \\frac{1}{a\\beta}\\right\\}$. Since $\\left\\{\\frac{k}{\\beta}\\right\\}$ is dense in $[0, 1]$, such $k$ exists. Then,\n\n$$\n\\begin{aligned}\n\\left\\{ \\frac{n}{\\alpha} \\right\\} &= -\\left\\lfloor \\frac{n}{\\alpha} \\right\\rfloor + \\frac{c(ak + u)}{a} - \\frac{b}{a} \\cdot \\frac{ak + u}{\\beta} \\\\\n&= \\left\\{ \\frac{uc - bv}{a} - b \\cdot \\frac{ak + u - v\\beta}{a\\beta} \\right\\} < \\frac{1}{a}, \\\\[2ex]\n\\left\\{ \\frac{n}{\\beta} \\right\\} &= \\left\\{ a \\cdot \\frac{ak + u - v\\beta}{a\\beta} + v \\right\\} \\\\\n&= \\left\\{ a \\cdot \\left\\{ \\frac{ak + u - v\\beta}{a\\beta} \\right\\} \\right\\} < \\frac{1}{\\beta}.\n\\end{aligned}\n$$\n\nThus, $\\frac{1}{a} > \\frac{1}{\\alpha}$ (otherwise $n = ak + u$ would contradict the assumption), so $\\frac{a}{\\alpha} < 1$; similarly, $\\frac{b}{\\beta} < 1$. Therefore, $c = \\frac{a}{\\alpha} + \\frac{b}{\\beta} < 2$, so $c = 1$. Hence, $\\frac{a}{\\alpha} + \\frac{b}{\\beta} = 1$, as required. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17303,
    "subject": "Mathematics (Olympiad)",
    "question": "The plane is divided by vertical and horizontal lines into unit squares. Is it possible to assign integers to the cells of this infinite grid so that:\n\n1. Every cell contains exactly one integer.\n2. Every integer appears exactly once.\n3. For any two cells $A$ and $B$ sharing exactly one vertex, if they contain integers $a$ and $b$, then at least one of the cells sharing a common side with both $A$ and $B$ contains an integer between $a$ and $b$.",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, this is possible. Consider the spiral depicted below and write consecutive integers along the spiral:\n\n![](images/2025CZPS_p3_data_8c68622425.png)\n\nWe claim that this works. Consider any two cells $A$ and $B$ sharing exactly one vertex. Consider the $2 \\times 2$ square containing $A$ and $B$. If the $2 \\times 2$ square contains a \"corner\" of the spiral, then for some $n$ and $k$ the numbers in that $2 \\times 2$ square are arranged in the following way (up to rotation or reflection):\n\n$$\n\\begin{array}{|c|c|} \\hline n+1 & k \\\\ \\hline n & n-1 \\\\ \\hline \\end{array}\n$$\n\nTherefore, the conditions are satisfied no matter which opposite cells of the $2 \\times 2$ square $A$ and $B$ are. For example, if $A$ and $B$ contain $n-1$ and $n+1$, then the good cell is the one containing $n$. If $A$ and $B$ contain $n$ and $k$ and $n < k$, then $n < n+1 < k$ and the good cell is the one containing $n+1$. If $A$ and $B$ contain $n$ and $k$ and $n > k$, then $k < n-1 < n$ and the good cell is the one containing $n-1$.\n\nOtherwise, the numbers are arranged in the following way (again, up to rotation or reflection):\n\n$$\n\\begin{array}{|c|c|} \\hline k+1 & k \\\\ \\hline n & n+1 \\\\ \\hline \\end{array}\n$$\n\nfor some $n, k$, and again, the conditions are satisfied. Without loss of generality, assume $n < k$. Then $n < n+1 < k < k+1$. If $A$ and $B$ contain $n$ and $k$, then the good cell is the one containing $n+1$. Otherwise, $A$ and $B$ contain $n+1$ and $k+1$, and the good cell is the one containing $k$.\n\nAlternatively, notice that condition (iii) means we can orient each $2 \\times 2$ square according to the increasing numbers as suggested in the picture below:\n\n![](images/2025CZPS_p4_data_b6bba9795a.png)\n\nWith this observation, it's relatively easy to check that the spiral construction satisfies this.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17304,
    "subject": "Mathematics (Olympiad)",
    "question": "Let there be 207 cards on the table, each with an integer written on it. Beto can repeatedly pick any two cards, write the difference of their numbers in his notebook, and remove the two cards from the table. He wants to repeat this operation 100 times, so that at the end, the product of the 100 numbers written in his notebook is divisible by $7^{100}$. Is this always possible? Justify your answer.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) As long as there are more than 7 cards on the table, by the pigeonhole principle, there must be two cards with the same remainder modulo 7. Beto can pick these two cards, and their difference will be divisible by 7. Since there are 207 cards, he can repeat this operation 100 times, obtaining 100 numbers divisible by 7. Thus, their product is divisible by $7^{100}$.\n\n(b) Yes, Beto can always achieve his goal. (For a proof, see the solution to Problem 1.3.)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17305,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence $\\langle a_n \\rangle$ with $a_n = a + nd$ is called an arithmetic sequence. The sequence $\\langle b_n \\rangle$ with $b_n = \\sum_{k=0}^{n} a_k$ is called an arithmetic sequence of second degree. Let $a$ and $d$ be positive integers.\n\nWe consider all such arithmetic sequences of second degree containing the number $2010$. What is the highest possible index $n$ if $b_n = 2010$? Determine all possible arithmetic sequences $\\langle a_n \\rangle$, for which $b_n = 2010$ holds for this index.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $a_k = a + k d$, we have\n$$\nb_n = (n+1)a + \\frac{n(n+1)}{2}d.\n$$\nIf we assume $b_n = 2010$, we obtain\n$$\na = \\frac{4020 - n(n+1)d}{2(n+1)} = \\frac{2010}{n+1} - \\frac{d n}{2}.\n$$\nSince both $a$ and $d$ are positive, from the first fraction, $n(n+1) < 4020$ must hold. Since $65^2 = 4225 > 4020$, we have $n < 65$. Furthermore, $n+1$ must divide $4020$. Since $4020 = 3 \\cdot 4 \\cdot 5 \\cdot 67$, the largest divisor of $4020$ less than $65$ is $60$. Thus, the largest possible value for $n+1$ is $60$, so $n = 59$.\n\nSubstituting this value yields $a = \\frac{67 - 59d}{2}$. We see that $d$ must be odd and less than $2$, so $d = 1$, which gives $a = 4$.\n\nThe only possible sequence $\\langle a_n \\rangle$ for which $b_{59} = 2010$ is $\\langle 4, 5, 6, \\dots \\rangle$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17306,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_{17}$ be a permutation of $1, 2, \\dots, 17$ satisfying\n$$\n(a_1 - a_2)(a_2 - a_3)\\cdots(a_{16} - a_{17})(a_{17} - a_1) = 2^n.\n$$\nFind the maximum value of the integer $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $S = (a_1 - a_2)(a_2 - a_3)\\cdots(a_{16} - a_{17})(a_{17} - a_1)$, and let $a_{18} = a_1$. Let $t \\in \\{1, 2, 3, 4\\}$.\n\nThe remainders of $a_1, a_2, \\dots, a_{17}$ modulo $2^t$ traverse all $0, 1, \\dots, 2^t - 1$.\n\nFor $i = 1, 2, \\dots, 17$, when the remainders of $a_i$ and $a_{i+1}$ modulo $2^t$ are different (there are at least $2^t$ such subscripts $i$), $2^t \\nmid (a_i - a_{i+1})$.\n\nTherefore, for at most $17 - 2^t$ subscripts $i \\in \\{1, 2, \\dots, 17\\}$, $2^t \\mid (a_i - a_{i+1})$ holds.\n\nSpecifically, among all factors $a_i - a_{i+1}$ ($i = 1, 2, \\dots, 17$), at most 15 of them are even, 13 are multiples of $2^2$, 9 are multiples of $2^3$, and 1 is a multiple of $2^4$.\n\nConsequently, the multiplicity of the prime factor 2 in the product $S = \\prod_{i=1}^{17} (a_i - a_{i+1})$ cannot exceed $15 + 13 + 9 + 1 = 38$.\n\nOn the other hand, we may define a permutation of $1, 2, \\dots, 17$ as follows:\n\n1, 17, 9, 5, 13, 11, 3, 7, 15, 16, 8, 12, 4, 6, 14, 10, 2\n\nThen,\n$$\nS = (-16) \\cdot 8 \\cdot 4 \\cdot (-8) \\cdot 2 \\cdot 8 \\cdot (-4) \\cdot (-8) \\cdot (-1) \\cdot 8 \\cdot (-4) \\cdot 8 \\cdot (-2) \\cdot (-8) \\\\\n\\cdot 4 \\cdot 8 \\cdot 1 = 2^{38}.\n$$\n\nIn conclusion, the maximum value of $n$ is $38$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 17307,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral with $|BA| \\neq |BC|$. Denote the incircles of triangles $ABC$ and $ADC$ by $\\omega_1$ and $\\omega_2$ respectively. Suppose that there exists a circle $\\omega$ tangent to the ray $BA$ extended beyond $A$ and to the ray $BC$ extended beyond $C$, which is also tangent to the lines $AD$ and $CD$.\n\nProve that the common external tangents to $\\omega_1$ and $\\omega_2$ intersect at a point on $\\omega$.",
    "options": [],
    "answer": "See solution",
    "solution": "Lemma 1: Let $ABCD$ be a convex quadrilateral, and let a circle $\\omega$ be tangent to the ray $BA$ extended beyond $A$ and to the ray $BC$ extended beyond $C$ (as shown in Fig. 1), which is also tangent to the lines $AD$ and $CD$. Then $AB + AD = CB + CD$.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p218_data_b8b2b400df.png)\n\nNow we prove Lemma 1.\n\nLet $\\omega$ meet $AB$, $BC$, $CD$, $DA$ at $P$, $Q$, $R$, $S$ respectively. As shown in Fig. 1,\n\n$$\n\\begin{align*}\n& AB + AD = CB + CD \\\\\n\\Leftrightarrow & AB + (AD + DS) = CB + (CD + DR) \\\\\n\\Leftrightarrow & AB + AS = CB + CR \\\\\n\\Leftrightarrow & AB + AP = CB + CR \\\\\n\\Leftrightarrow & BP = BR.\n\\end{align*}\n$$\n\nThis ends the proof of Lemma 1.\n\nLemma 2: If the radii of three circles $\\odot O_1$, $\\odot O_2$, $\\odot O_3$ differ from each other, then their homothetic centers are collinear.\n\nNow we prove Lemma 2.\n\nLet $X_3$ be the homothetic center of $\\odot O_1$ and $\\odot O_2$, $X_2$ be the homothetic center of $\\odot O_1$ and $\\odot O_3$, and $X_1$ be the homothetic center of $\\odot O_2$ and $\\odot O_3$. Let $r_i$ be the radius of $\\odot O_i$ ($i = 1, 2, 3$). By the property of homothety:\n\n$$\n\\frac{\\overline{O_1X_3}}{X_3O_2} = -\\frac{r_1}{r_2}\n$$\n\nHere $\\overline{O_1X_3}$ denotes the directed segment $O_1X_3$, as shown in Fig. 2. Similarly,\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p219_data_7541e92f32.png)\n\n$$\n\\frac{\\overline{O_2X_1}}{X_1O_3} = -\\frac{r_2}{r_3}, \\quad \\frac{\\overline{O_3X_2}}{X_2O_1} = -\\frac{r_3}{r_1}\n$$\n\nSo\n\n$$\n\\frac{\\overline{O_1X_3}}{X_3O_2} \\cdot \\frac{\\overline{O_2X_1}}{X_1O_3} \\cdot \\frac{\\overline{O_3X_2}}{X_2O_1} = \\left(-\\frac{r_1}{r_2}\\right) \\left(-\\frac{r_2}{r_3}\\right) \\left(-\\frac{r_3}{r_1}\\right) = -1\n$$\n\nBy Menelaus' Theorem, $X_1$, $X_2$, $X_3$ are collinear.\n\nLet $\\omega_1$, $\\omega_2$ meet $AC$ at $U$, $V$ respectively. As shown in Fig. 3,\n\n$$\n\\begin{aligned}\nAV &= \\frac{AD + AC - CD}{2} = \\frac{AC}{2} + \\frac{AD - CD}{2} \\\\\n&= \\frac{AC}{2} + \\frac{CB - AB}{2} = \\frac{AC + CB - AB}{2} \\\\\n&= CU.\n\\end{aligned}\n$$\n\nHence the excircle $\\omega_3$ of $\\triangle ABC$ on the side $AC$ meets $AC$ at $V$. Therefore $\\omega_2$, $\\omega_3$ meet at point $V$, i.e., $V$ is the homothetic center of $\\omega_2$, $\\omega_3$. Denote the homothetic\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p219_data_55362afc0c.png)\n\ncenters of $\\omega_1$, $\\omega_2$ by $K$ (i.e., $K$ is the intersection of two external common tangents to $\\omega_2$, $\\omega_3$). By Lemma 2, $K$, $V$, $B$ are collinear.\n\nSimilarly, $K$, $D$, $U$ are collinear.\n\nSince $BA \\neq BC$, then $U \\neq V$ (otherwise, by $AV = CU$, we know $U = V$ is the midpoint of side $AC$, which contradicts $BA \\neq BC$). So $BV$ does not coincide with $DU$, i.e., $K = BV \\cap DU$.\n\nWe now prove that $K$ is on the circle $\\omega$.\n\nConstruct a tangent $l$ of $\\omega$ which is parallel to $AC$. Let $l$ meet $\\omega$ at $T$. We will show that $B$, $V$, $T$ are collinear.\n\nAs shown in Fig. 4, $l$ intersects $BA$, $BC$ at $A_1$, $C_1$ respectively, then $\\omega$ is the excircle of $\\triangle BA_1C_1$ on the side $A_1C_1$, meeting $A_1C_1$ at $T$. Meanwhile, $\\omega_3$ is the excircle of $\\triangle ABC$ on the side $AC$, meeting $AC$ at $V$. Since $AC \\parallel A_1C_1$, $B$ is the homothetic center of $\\triangle BAC$ and $\\triangle BA_1C_1$, $V$ and $T$ are corresponding points, so $B$, $V$, $T$ are collinear.\n\nSimilarly, $D$, $V$, $T$ are collinear. This means $K = T$.\n\nThis ends the proof.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p220_data_287db8b081.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17308,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, and $C$ be three distinct points on a unit circle. Let $G$ and $H$ be the centroid and the orthocenter of triangle $ABC$, respectively. Let $F$ be the midpoint of the segment $GH$. Evaluate $|\\overline{AF}|^2 + |\\overline{BF}|^2 + |\\overline{CF}|^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Define a coordinate system with the origin at the center of the circle. We can see that $\\vec{H} = \\vec{A} + \\vec{B} + \\vec{C}$ and $\\vec{G} = \\frac{1}{3}(\\vec{A} + \\vec{B} + \\vec{C})$.\n\nThus, $\\vec{F} = \\frac{\\vec{G} + \\vec{H}}{2} = \\frac{2}{3}(\\vec{A} + \\vec{B} + \\vec{C})$. We now have\n\n$$\n\\begin{aligned}\n|\\overline{AF}|^2 + |\\overline{BF}|^2 + |\\overline{CF}|^2 \n&= (\\vec{A} - \\vec{F}) \\cdot (\\vec{A} - \\vec{F}) + (\\vec{B} - \\vec{F}) \\cdot (\\vec{B} - \\vec{F}) + (\\vec{C} - \\vec{F}) \\cdot (\\vec{C} - \\vec{F}) \\\\\n&= |\\vec{A}|^2 + |\\vec{B}|^2 + |\\vec{C}|^2 - 2(\\vec{A} + \\vec{B} + \\vec{C}) \\cdot \\vec{F} + 3\\vec{F} \\cdot \\vec{F} \\\\\n&= |\\vec{A}|^2 + |\\vec{B}|^2 + |\\vec{C}|^2 - (2(\\vec{A} + \\vec{B} + \\vec{C}) - 3\\vec{F}) \\cdot \\vec{F} \\\\\n&= |\\vec{A}|^2 + |\\vec{B}|^2 + |\\vec{C}|^2 = 3.\n\\end{aligned}\n$$\n\n![](images/Tajland_2008_p4_data_f62bf9de72.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17309,
    "subject": "Mathematics (Olympiad)",
    "question": "Two players, Andriy and Olesya, play the following game. On a table, there is a round cake, which is cut by one of them into $4n$ sectors (pieces), each with a different weight. The weight of every piece is known to both players. After the cake is cut, they choose pieces according to these rules:\n\n- First, Andriy chooses 1 piece.\n- Then, Olesya chooses 2 pieces, but the remaining pieces after her turn must form a contiguous sector.\n- After that, they take turns choosing 2 pieces each, always ensuring that the remaining pieces form a sector.\n- On the last turn, Andriy takes the final piece.\n\nEach player aims to collect more than half the total weight of the cake. Is it possible for someone to guarantee taking more than half of the cake in the following cases?\n\na) Olesya cuts the cake into sectors.\nb) Andriy cuts the cake into sectors, but on his first turn he is prohibited from taking the heaviest piece.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Olesya can guarantee victory by cutting the cake so that three neighboring pieces are heavy (each of weight $\\frac{1}{3}M$) and the rest are very light (almost zero weight). Number the pieces $1, 2, \\ldots, 4n$ clockwise, with piece $4n$ adjacent to piece $1$. Let pieces $4n-3$, $4n-2$, and $4n-1$ be the heavy ones. If Andriy starts with any of the five pieces $4n-4$ through $4n$, Olesya can immediately take two heavy pieces and secure $\\frac{2}{3}M > \\frac{1}{2}M$. If Andriy chooses any other piece, whoever takes either $4n-4$ or $4n$ (even with a heavy piece) loses, since the opponent can then take two heavy pieces and win. Thus, the play is forced into the sector between pieces $1$ and $4n-5$, and after all turns, Andriy will be forced to lose.\n\nb) Andriy can guarantee victory by cutting the cake so that pieces $1$, $3$, and $5$ are heavy (each of weight $\\frac{1}{3}M$), and the rest are light. On his first turn, Andriy takes piece $3$ (not the heaviest), and on his next turn, he can take another heavy piece, collecting $\\frac{2}{3}M > \\frac{1}{2}M$.\n\nTo ensure all pieces have different nonzero weights, assign the small pieces weights $\\frac{i}{5(4n-3)(4n-2)}M$ for $i = 1, \\ldots, 4n-3$, so their total is $\\frac{1}{10}M$:\n\n$$\n\\frac{1}{5(4n-3)(4n-2)}M + \\frac{2}{5(4n-3)(4n-2)}M + \\dots + \\frac{4n-3}{5(4n-3)(4n-2)}M = \\frac{(4n-3)(4n-2)}{10(4n-3)(4n-2)}M = \\frac{1}{10}M.\n$$\n\nChoose the three big pieces to have weights $\\frac{29}{100}M$, $\\frac{30}{100}M$, and $\\frac{31}{100}M$. The order of the pieces does not matter. The same distribution works for part **b)**, with the third piece having weight $\\frac{30}{100}M$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17310,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x = bc$, $y = ca$, and $z = ab$ with the condition $xyz = 1$. Show that\n\n$$\n\\frac{x}{y+z} + \\frac{y}{z+x} + \\frac{z}{x+y} \\geq \\frac{3}{2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "This is exactly Nesbitt's inequality. Equality holds when $x = y = z = 1$, i.e., $a = b = c = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17311,
    "subject": "Mathematics (Olympiad)",
    "question": "Circles $S$ and $T$ intersect at $P$ and $Q$, with $S$ passing through the centre of $T$. Distinct points $A$ and $B$ lie on $S$, inside $T$, and are equidistant from the centre of $T$. The line $PA$ meets $T$ again at $D$. Prove that $|AD| = |PB|$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the centre of the circle $T$. Extend $PB$ to meet the circle at $C$. Let $\\angle QPA = \\alpha$ and let $\\angle BPO = \\beta$. Since $|AO| = |OB|$, the join of the two centres is perpendicular to $AB$. The join of the two centres is also perpendicular to $PQ$. Thus $PQ \\parallel AB$. Hence $\\angle BAP = \\alpha$. Hence $|PB| = |QA|$.\n\nSince $|AO| = |OB|$, $\\angle APO = \\angle BPO = \\beta$. Hence $|PC| = |PD|$. $\\angle PCD = \\pi/2 - \\beta$, $\\angle DQP = \\pi/2 + \\beta$. Also $\\angle PDC = \\angle PCD = \\pi/2 - \\beta$.\n\nAs $QDCP$ is a cyclic quadrilateral\n$$\n\\angle QDP = \\pi = (\\alpha + 2\\beta + \\pi/2 - \\beta) = \\pi/2 - (\\alpha + \\beta)\n$$\nAlso $\\angle AQP = \\angle BPQ = \\alpha + 2\\beta$. Then since $\\angle DQP + \\angle DCP = \\pi$ we get\n$$\n\\angle DQA = \\pi - (\\pi/2 - \\beta + \\alpha + 2\\beta) = \\pi/2 - (\\alpha + \\beta) = \\angle QDP\n$$\nHence $|AD| = |QA| = |PB|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17312,
    "subject": "Mathematics (Olympiad)",
    "question": "On the sides $BC$, $CA$, and $AB$ of triangle $ABC$ there are points $D$, $E$, and $F$, respectively, such that points $A$, $B$, $D$, and $E$ are concyclic, points $B$, $C$, $E$, and $F$ are concyclic, and points $A$, $C$, $D$, and $F$ are concyclic. Does this setup require the triangle $ABC$ to be equilateral?",
    "options": [],
    "answer": "See solution",
    "solution": "Let us show that triangle $ABC$ can be any acute triangle. Let $D$, $E$, and $F$ be the feet of the altitudes drawn from vertices $A$, $B$, and $C$, respectively (see the figure below). Since $\\angle ADB = 90^\\circ = \\angle AEB$, points $D$ and $E$ are on the circle with diameter $AB$. Analogously, $E$ and $F$ are located on the circle with diameter $BC$, and $F$ and $D$ on the circle with diameter $CA$.\n\n![](images/prob1314_p14_data_a4bb913e81.png)\n\nFigure 8",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17313,
    "subject": "Mathematics (Olympiad)",
    "question": "A $9 \\times 7$ rectangle is tiled with tiles of the two types shown in the picture below (the tiles are composed of three, respectively four unit squares, and the L-shaped tiles can be rotated repeatedly by $90^\\circ$).\n\n![](images/RMC2010_p122_data_569fb061f4.png)\n\n![](images/RMC2010_p122_data_10bd0642b4.png)\n\nLet $n \\ge 0$ be the number of the $2 \\times 2$ tiles which can be used in such a tiling. Find all the values of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: $n = 0$ or $3$.\n\nDenote by $x$ the number of pieces of the 'corner' type and by $y$ the number of pieces of the $2 \\times 2$ type. Mark 20 squares of the rectangle as in the figure below.\n\n![](images/RMC2010_p123_data_c6602ff68d.png)\n\nObviously, each piece covers at most one marked square.\n\nThus, $x + y \\ge 20$ and, consequently,\n\n$$\n3x + 3y \\ge 60.\n$$\n\nOn the other hand,\n\n$$\n3x + 4y = 63.\n$$\n\nFrom these, it follows that $y \\le 3$ and, from the second equation, $3 \\mid y$, so $y$ can be only $0$ or $3$.\n\nThe proof is finished when we produce tilings with $3$, respectively $0$, $2 \\times 2$ tiles:\n\n![](images/RMC2010_p123_data_fdedf1054e.png)\n\n![](images/RMC2010_p123_data_397dbc3743.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17314,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $f(x) = \\cos x + \\log_2 x$ for $x > 0$. If a positive real number $a$ satisfies $f(a) = f(2a)$, then the value of $f(2a) - f(4a)$ is \\_\\_\\_\\_\\_.",
    "options": [],
    "answer": "See solution",
    "solution": "By the condition, $\\cos a + \\log_2 a = \\cos 2a + \\log_2 2a$. Note that $\\log_2 2a = 1 + \\log_2 a$, so:\n\n$$\n\\cos a + \\log_2 a = \\cos 2a + 1 + \\log_2 a\n$$\n\nThus, $\\cos a = \\cos 2a + 1$. Recall $\\cos 2a = 2\\cos^2 a - 1$:\n\n$$\n\\cos a = 2\\cos^2 a - 1 + 1 = 2\\cos^2 a\n$$\n\nSo $\\cos a = 0$ or $\\cos a = \\frac{1}{2}$. Correspondingly, $\\cos 2a = 2\\cos^2 a - 1$ gives $\\cos 2a = -1$ or $\\cos 2a = -\\frac{1}{2}$.\n\nNow,\n\n$$\n\\begin{aligned}\nf(2a) - f(4a) &= \\cos 2a + \\log_2 2a - \\cos 4a - \\log_2 4a \\\\\n&= \\cos 2a + 1 + \\log_2 a - \\cos 4a - 2 - \\log_2 a \\\\\n&= \\cos 2a - \\cos 4a - 1\n\\end{aligned}\n$$\n\nBut $\\cos 4a = 2\\cos^2 2a - 1$, so:\n\n$$\n\\cos 2a - (2\\cos^2 2a - 1) - 1 = \\cos 2a - 2\\cos^2 2a + 1 - 1 = \\cos 2a - 2\\cos^2 2a\n$$\n\nTherefore,\n\n$$\n\\begin{cases}\n-3, & \\text{if } \\cos 2a = -1 \\\\\n-1, & \\text{if } \\cos 2a = -\\frac{1}{2}\n\\end{cases}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17315,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ that satisfy the equation\n$$\nf(x)^n f(x + y) = f(x)^{n+1} + x^n f(y)\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The functions we are looking for are $f : \\mathbb{R} \\to \\mathbb{R}$, $f(x) = 0$ and $f : \\mathbb{R} \\to \\mathbb{R}$, $f(x) = x$. For $n$ even, $f : \\mathbb{R} \\to \\mathbb{R}$, $f(x) = -x$ is also a solution.\n\nThroughout the solution, $P(x_0, y_0)$ will denote the substitution of $x_0$ and $y_0$ for $x$ and $y$, respectively, in the given equation.\n\n$P(x, 0)$ for $x \\neq 0$ gives\n$$\nf(x)^{n+1} = f(x)^{n+1} + x^n f(0)\n$$\nand therefore\n$$\nf(0) = \\frac{f(x)^{n+1} - f(x)^{n+1}}{x^n} = 0.\n$$\n$P(x, -x)$ for $x \\neq 0$ gives\n$$\n0 = f(x)^n f(0) = f(x)^{n+1} + x^n f(-x),\n$$\nand therefore\n$$\nf(-x) = -\\frac{f(x)^{n+1}}{x^n}.\n$$\nApplying this identity twice, we get\n$$\nf(x) = f(-(-x)) = -\\frac{f(-x)^{n+1}}{(-x)^n} = -\\frac{\\left(-\\frac{f(x)^{n+1}}{x^n}\\right)^{n+1}}{(-x)^n} = \\frac{f(x)^{n^2+2n+1}}{x^{n^2+2n}},\n$$\nwhich after rearranging yields\n$$\nf(x)(x^{n^2+2n} - f(x)^{n^2+2n}) = 0.\n$$\nIf there exists an $a \\neq 0$ for which $f(a) = 0$, then $P(a, y)$ yields\n$$\n0 = a^n f(y),\n$$\nwhich means that $f(y) = 0$ for all $y \\in \\mathbb{R}$. This is a solution to the equation for all $n$.\n\nIf instead $f(x) \\neq 0$ for all $x \\neq 0$, then we have\n$$\nx^{n^2+2n} = f(x)^{n^2+2n}.\n$$\nIf $n$ is odd, then so is $n(n + 2) = (n^2 + 2n)$, meaning $f(x) = x$ for all $x \\in \\mathbb{R}$. This is a solution to the equation.\n\nIf $n$ is even, then so is $n(n + 2) = (n^2 + 2n)$, meaning $f(x) = \\pm x$ for all $x \\in \\mathbb{R}$. Both $f(x) = x$ and $f(x) = -x$ are solutions to the equation. In all other cases there must exist $x, y \\neq 0$ such that $f(x) = x$ and $f(y) = -y$. Then $P(x, y)$ yields\n$$\nx^n f(x + y) = x^{n+1} - x^n y,\n$$\nwhich after dividing by $x^n \\neq 0$ yields\n$$\nf(x + y) = x - y.\n$$\nSince $(f(x))^2 = x^2$ for all $x \\in \\mathbb{R}$, we have $(x + y)^2 = (x - y)^2$. That is $4xy = 0$ which is impossible as $x, y \\neq 0$.\n\nThere are therefore no more solutions to the equation. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17316,
    "subject": "Mathematics (Olympiad)",
    "question": "An artist has an extraordinary working rhythm. He works for 3 hours very intensively on his art, and then he sleeps for 8 hours before starting to work again. Suppose that he starts working at midnight in the night from 31 July to 1 August.\n\nWhich day of August is the first day after 1 August on which the artist is working the same number of hours as on 1 August?",
    "options": [],
    "answer": "See solution",
    "solution": "7 August",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17317,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a tetrahedron $ABCD$ with $AD \\perp BC$ and $AC \\perp BD$. Denote by $E$ and $F$ the projections of the vertex $B$ onto the lines $AD$ and $AC$, respectively. Let $M$ be the midpoint of the edge $AB$ and let $N$ be the midpoint of the edge $CD$. Prove that the lines $MN$ and $EF$ are perpendicular.",
    "options": [],
    "answer": "See solution",
    "solution": "Line $AD$ is perpendicular to the lines $BC$ and $BE$, so $AD \\perp (BEC)$, and then $AD \\perp CE$. Similarly, lines $DF$ and $AC$ are perpendicular. Let $H = CE \\cap DF$. The planes $(BEC)$ and $(BFC)$ intersect in the line $BH$, hence $BH$ is perpendicular to $(ACD)$. Consequently, the projection $Q$ of the point $M$ on the plane $(ACD)$ is the midpoint of the line segment $AH$.\n\nNotice that points $N$ and $Q$ are the circumcentres of the triangles $AEF$ and $BEF$, so the centre line $NQ$ is perpendicular to the common chord $EF$.\n\nSince $MQ \\perp EF$, it follows that $EF \\perp (MQN)$, hence lines $MN$ and $EF$ are perpendicular.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17318,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, on side $AC$, a point $D$ is chosen such that $\\angle BDC = \\angle ABC$. If $\\overline{AD} = 7\\ \\mathrm{cm}$ and $\\overline{DC} = 9\\ \\mathrm{cm}$, calculate the length of $BC$ and the ratio $\\overline{BD} : \\overline{BA}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Because $\\angle BDC = \\angle ABC = \\beta$ and $\\angle DCB = \\angle BCA = \\gamma$, we have that $\\triangle BDC \\sim \\triangle ABC$. From the similarity, $\\dfrac{\\overline{BC}}{\\overline{DC}} = \\dfrac{\\overline{AC}}{\\overline{BC}}$, so $\\overline{BC}^2 = \\overline{DC} \\cdot \\overline{AC} = 9 \\cdot (7 + 9) = 144$. Hence, $\\overline{BC} = 12\\ \\mathrm{cm}$. From $\\triangle BDC \\sim \\triangle ABC$, we also have $\\dfrac{\\overline{BD}}{\\overline{DC}} = \\dfrac{\\overline{AB}}{\\overline{BC}}$, or $\\dfrac{\\overline{BD}}{\\overline{AB}} = \\dfrac{\\overline{DC}}{\\overline{BC}} = \\dfrac{9}{12} = \\dfrac{3}{4}$. Thus, $\\overline{BD} : \\overline{BA} = 3 : 4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17319,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(a_n)$ be a sequence defined by the recurrence relation $a_{n+1} = 2(a_n - a_{n-1})$ with initial conditions $a_0 = 1$ and $a_1 = 1$.\n\nFind an explicit formula for $a_n$ and compute the remainder when $a_{2016}$ is divided by $2017$.\n\nShow that $2017$ is a prime number, and deduce that $a_{2016} \\equiv 1 \\pmod{2017}$.\n",
    "options": [],
    "answer": "See solution",
    "solution": "Since $(a_n)$ satisfies the linear recurrence $a_{n+1} = 2(a_n - a_{n-1})$, we can find an explicit formula for $a_n$ by solving the characteristic equation. Replacing $a_k$ by $x^k$, we get $x^{n+1} = 2(x^n - x^{n-1})$, so $x^2 = 2(x-1)$, which gives $x = 1 \\pm i$, where $i = \\sqrt{-1}$.\n\nLet $\\alpha = 1+i$, $\\beta = 1-i$. Then $a_n = A\\alpha^n + B\\beta^n$, where $A$ and $B$ are determined by the initial conditions $a_0 = 1$, $a_1 = 1$:\n\n$$\nA+B=1, \\quad A\\alpha+B\\beta=1.\n$$\n\nSolving, $A = B = \\frac{1}{2}$. Therefore,\n\n$$\na_n = \\frac{1}{2}\\left((1+i)^n + (1-i)^n\\right).\n$$\n\nNow, $(1+i)^2 = 2i$, $(1+i)^4 = -4$, $(1+i)^8 = 16$, and similarly $(1-i)^8 = 16$. Since $2016 = 8 \\times 252$, we have\n\n$$\n(1 \\pm i)^{2016} = \\left((1 \\pm i)^8\\right)^{252} = 16^{252} = 2^{1008}.\n$$\n\nThus,\n\n$$\na_{2016} = \\frac{1}{2}(2^{1008} + 2^{1008}) = 2^{1008}.\n$$\n\nNext, we show that $2017$ is a prime number. Since $45^2 = 2025 > 2017$, any prime divisor of $2017$ must be less than $45$. Checking all primes less than $45$ (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43), none divide $2017$. Therefore, $2017$ is prime.\n\nSince $2017 \\equiv 1 \\pmod{4}$, there exists an integer $y$ such that $2017$ divides $1 + y^2$. For example, $y = \\left(\\frac{2017-1}{2}\\right)!$ by Wilson's theorem. Consider $(1+y)^2 = 1 + 2y + y^2 \\equiv 2y \\pmod{2017}$. By Fermat's little theorem,\n\n$$\n(1+y)^{2016} \\equiv 1 \\pmod{2017}, \\quad \\text{so} \\quad (2y)^{1008} \\equiv 1 \\pmod{2017}.\n$$\n\nSince $y^4 \\equiv 1 \\pmod{2017}$ and $4$ divides $1008$, it follows that $2^{1008} \\equiv 1 \\pmod{2017}$. Therefore, the remainder of $a_{2016}$ upon division by $2017$ is $1$.\n",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17320,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there is an $n \\times 2021$ grid. A piece starts at some square in the top row. In each move, the piece can move to the next row, either directly below or to an adjacent column (i.e., from $(i, j)$ to $(i+1, j-1)$, $(i+1, j)$, or $(i+1, j+1)$, as long as the destination is within the grid). Some squares are colored black, and the piece cannot move onto a black square. What is the minimal $n$ such that, no matter how the black squares are colored, there is always a way to move the piece from the top row to the bottom row?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n = 2022$.\n\nFirst, we prove that $n \\geq 2022$. Denote the square in the $i$th row and $j$th column as $(i, j)$. In one move, the piece can go from $(i, j)$ to $(i+1, j-1)$, $(i+1, j)$, or $(i+1, j+1)$. To reach $(c, d)$ from $(a, b)$, it must be that $a \\leq c$ and $|d - b| \\leq c - a$. If Bob places the piece at $(1, 1)$ and the goal is $(n, 2021)$, then $|2021 - 1| \\leq n - 1$ gives $n \\geq 2021$.\n\nFor $n = 2021$, if Bob starts at $(1, 1)$ and the goal is $(2021, 2021)$, Alice can color $(k, k)$ black for $1 \\leq k \\leq 2020$. Then, to reach $(k+1, k)$ from $(1, 1)$ is impossible for any $1 \\leq k \\leq 2020$, by induction: for $k = 1$, $(1, 1)$ is black; for $k \\geq 2$, $(k+1, k)$ can only be reached from $(k, k-1)$, $(k, k)$, or $(k, k+1)$. By induction, $(k, k-1)$ is unreachable, $(k, k)$ is black, and $(k, k+1)$ is unreachable. Thus, $n \\geq 2022$.\n\nNow, for $n = 2022$, we show Alice can always reach the bottom row. She colors all squares $(i, j)$ with $i \\geq 4$ black. For $1 \\leq i \\leq 3$, she colors certain squares black for $k$ with $0 \\leq k \\leq 336$:\n\n$$\n(1, 3k + 1), (1, 2021 - 3k), (1, 3k + 3), (1, 2019 - 3k), (2, 3k + 2), (2, 2020 - 3k), (3, 3k + 1), (3, 2021 - 3k)\n$$\n\nThe rest are white. For $k = 336$, $(1, 3k+3)$ and $(1, 2019-3k)$ coincide. By symmetry, it suffices to show that from $(1, a)$ with $1 \\leq a \\leq 1011$, the piece can reach the bottom row. For each $k$, from $(1, 3k+1)$, $(1, 3k+2)$, or $(1, 3k+3)$, the piece can move to $(2, 3k+2)$, then through $(3, 3k+1)$ or $(3, 3k+3)$, and finally to $(4, 3k+2)$ or $(4, 3k+3)$. Thus, the piece can always reach the bottom row. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17321,
    "subject": "Mathematics (Olympiad)",
    "question": "Each of the small circles blocks the same amount of circumference from $O$. In the diagram, $OA$ and $OC$ are tangents to the small circle with centre $P$.\n\n![](images/Australian-Scene-2017_p49_data_0fee7457e3.png)\n\nWhat fraction of the circumference of the large circle is blocked from $O$ by all four small circles?",
    "options": [],
    "answer": "See solution",
    "solution": "In the diagram, $P$, $Q$, $R$, $S$ are the centres of the four small circles and $OU$, $OV$ are tangents to two of them.\n\n![](images/Australian-Scene-2017_p50_data_09805078ae.png)\n\nSince $OP = OQ = 8$ cm and $PQ = 8$ cm, $\\triangle OPQ$ is equilateral. Hence $\\angle POQ = 60^\\circ$. Similarly, $\\angle QOR = 60^\\circ$ and $\\angle ROS = 60^\\circ$. Since $PU = 4$ cm and $\\angle PUO = 90^\\circ$, $\\triangle OPU$ is half an equilateral triangle. Therefore $\\angle UOP = 30^\\circ$. Similarly, $\\angle SOV = 30^\\circ$.\n\nHence the fraction of the circumference of the large circle that is blocked from $O$ by the four small circles is\n$$ \\frac{(30 + 60 + 60 + 60 + 30)}{360} = \\frac{240}{360} = \\frac{2}{3}. $$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17322,
    "subject": "Mathematics (Olympiad)",
    "question": "Најди ги сите $n \\in \\mathbb{N}$ деливи со $11$, такви што сите броеви кои се добиваат со произволна прераспределба на цифрите на бројот $n$ повторно се деливи со $11$.",
    "options": [],
    "answer": "See solution",
    "solution": "Од условот $11 \\mid n$ и бројот $n$ мора да е најмалку двоцифрен. Нека $n = \\overline{a_k a_{k-1} \\dots a_0}$ каде $a_i$, $0 \\le i \\le k$ се цифри и $a_k \\ne 0$. Од претходната дискусија $k \\ge 1$.\n\nЌе покажеме дека сите цифри во бројот $n$ се еднакви. Имено, од условот на задачата и бројот $n' = a_k a_{k-1} \\dots a_{i-1} a_i a_{i-2} \\dots a_0$ ($n'$ е добиен од $n$ со промена на местата на цифрите $a_{i-1}$ и $a_i$) е делив со $11$. Значи $11 \\mid n - n'$, т.е. $11 \\mid 10^{i-1}(\\overline{a_i a_{i-1}} - \\overline{a_{i-1} a_i})$ или $11 \\mid 10^{i-1} \\cdot 9(a_i - a_{i-1})$, па мора $a_i = a_{i-1}$.\n\nСледува, $n = \\underbrace{a a \\dots a}_{k+1\\text{ пати}}$. Лесно се проверува дека $11 \\mid n$ ако и само ако $k$ е непарен број.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17323,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with incenter $I$, and let $D$ be a point on line $BC$ such that $\\angle AID = 90^\\circ$. Let the excircle of triangle $ABC$ opposite vertex $A$ be tangent to $\\overline{BC}$ at point $A_1$. Define points $B_1$ on $\\overline{CA}$ and $C_1$ on $\\overline{AB}$ analogously, using the excircles opposite $B$ and $C$, respectively.\n\nProve that if quadrilateral $AB_1A_1C_1$ is cyclic, then $\\overline{AD}$ is tangent to the circumcircle of $\\triangle DB_1C_1$.",
    "options": [],
    "answer": "See solution",
    "solution": "We present two solutions.\n\n**First solution using spiral similarity (Ankan Bhattacharya)**\n\nFirst, we prove the part of the problem which does not depend on the condition $AB_1A_1C_1$ is cyclic.\n\n**Lemma**\n\nLet $ABC$ be a triangle and define $I, D, B_1, C_1$ as in the problem. Moreover, let $M$ denote the midpoint of $\\overline{AD}$. Then $\\overline{AD}$ is tangent to $(AB_1C_1)$, and moreover $\\overline{B_1C_1} \\parallel \\overline{IM}$.\n\n*Proof*. Let $E$ and $F$ be the tangency points of the incircle. Denote by $Z$ the Miquel point of $BFEC$, i.e., the second intersection of the circle with diameter $\\overline{AI}$ and the circumcircle. Note that $A, Z, D$ are collinear, by radical axis on $(ABC), (AFIE), (BIC)$.\n\n![](images/sols-TST-IMO-2019_p14_data_1710402da2.png)\n\nThen the spiral similarity gives us\n\n$$\n\\frac{ZF}{ZE} = \\frac{BF}{CE} = \\frac{AC_1}{AB_1}\n$$\n\nwhich together with $\\angle FZE = \\angle FAE = \\angle BAC$ implies that $\\triangle ZFE$ and $\\triangle AC_1B_1$ are (directly) similar. (See IMO Shortlist 2006 G9 for a similar application of spiral similarity.)\n\nNow the remainder of the proof is just angle chasing. First, since\n\n$$\n\\angle DAC_1 = \\angle ZAF = \\angle ZEF = \\angle AB_1C_1\n$$\n\nwe have $\\overline{AD}$ is tangent to $(AB_1C_1)$. Moreover, to see that $\\overline{IM} \\parallel \\overline{B_1C_1}$, write\n\n$$\n\\begin{aligned}\n\\angle(\\overline{AI}, \\overline{B_1C_1}) &= \\angle IAC + \\angle AB_1C_1 = \\angle BAI + \\angle ZEF = \\angle FAI + \\angle ZAF \\\\\n&= \\angle ZAI = \\angle MAI = \\angle AIM\n\\end{aligned}\n$$\n\nthe last step since $\\triangle AID$ is right with hypotenuse $\\overline{AD}$, and median $\\overline{IM}$. $\\square$\n\nNow we return to the present problem with the additional condition.\n\n![](images/sols-TST-IMO-2019_p14_data_56265c8567.png)\n\n**Claim.** Given the condition, we actually have $\\angle AB_1A_1 = \\angle AC_1A_1 = 90^\\circ$.\n\n*Proof.* Let $I_A, I_B$ and $I_C$ be the excenters of $\\triangle ABC$. Then the perpendiculars to $\\overline{BC}$, $\\overline{CA}$, $\\overline{AB}$ from $A_1, B_1, C_1$ respectively meet at the so-called Bevan point $V$ (which is the circumcenter of $\\triangle I_A I_B I_C$).\n\nNow $\\triangle AB_1C_1$ has circumdiameter $\\overline{AV}$. We are given $A_1$ lies on this circle, so if $V \\neq A_1$ then $\\overline{AA_1} \\perp \\overline{A_1V}$. But $\\overline{A_1V} \\perp \\overline{BC}$ by definition, which would imply $\\overline{AA_1} \\parallel \\overline{BC}$, which is absurd. $\\square$\n\n**Claim.** Given the condition the points $B_1, I, C_1$ are collinear (hence with $M$).\n\n*Proof.* By Pappus theorem on $\\overline{IB}A_1I_C$ and $\\overline{BA_1C}$ after the previous claim. $\\square$\n\nTo finish, since $\\overline{DMA}$ was tangent to the circumcircle of $\\triangle AB_1C_1$, we have $MD^2 = MA^2 = MC_1 \\cdot MB_1$, implying the required tangency.\n\n**Remark.** The triangles satisfying the problem hypothesis are exactly the ones satisfying $r_A = 2R$, where $R$ and $r_A$ denote the circumradius and A-exradius.\n\n**Remark.** If $P$ is the foot of the A-altitude then this should also imply $AB_1PC_1$ is harmonic.\n\n**Second solution by inversion and mixtilinears (Anant Mudgal)**\n\nAs in the end of the preceding solution, we have $\\angle AB_1A_1 = \\angle AC_1A_1 = 90^\\circ$ and $I \\in \\overline{B_1C_1}$. Let $M$ be the midpoint of minor arc $BC$ and $N$ be the midpoint of arc $\\widehat{BAC}$. Let $L$ be the intouch point on $\\overline{BC}$. Let $O$ be the circumcenter of $\\triangle ABC$. Let $K = \\overline{AI} \\cap \\overline{BC}$.\n\n![](images/sols-TST-IMO-2019_p15_data_02e55e25a8.png)\n\n**Claim.** We have $\\angle(\\overline{AI}, \\overline{B_1C_1}) = \\angle IAD$.\n\n*Proof.* Let $Z$ lie on $(ABC)$ with $\\angle AZI = 90^\\circ$. By radical axis theorem on $(AIZ), (BIC)$, and $(ABC)$, we conclude that $D$ lies on $\\overline{AZ}$. Let $\\overline{NI}$ meet $(ABC)$ again at $T \\neq N$.\n\nInversion in $(BIC)$ maps $\\overline{AI}$ to $\\overline{KI}$ and $(ABC)$ to $\\overline{BC}$. Thus, $Z$ maps to $L$, so $Z, L, M$ are collinear. Since $BL = CV$ and $OI = OV$, we see that $MLIN$ is a trapezoid with $\\overline{IL} \\parallel \\overline{MN}$. Thus, $\\overline{ZT} \\parallel \\overline{MN}$.\n\nIt is known that $\\overline{AT}$ and $\\overline{AA_1}$ are isogonal in angle $BAC$. Since $\\overline{AV}$ is a circumdiameter in $(AB_1C_1)$, so $\\overline{AT} \\perp \\overline{B_1C_1}$. So $\\angle ZAI = \\angle NMT = 90^\\circ - \\angle TAI = \\angle(\\overline{AI}, \\overline{B_1C_1})$. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17324,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that, for positive $x, y, z$, the following inequality holds:\n\n$$\n(x + y + z)(4x + y + 2z)(2x + y + 8z) \\geq \\frac{375}{2}xyz.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the first two brackets and observe that\n\n$$\n(x + y + z)(4x + y + 2z) = (2x + y)^2 + 3z(2x + y) + 2z^2 + xy.\n$$\n\nTherefore, we can write the inequality in the form\n\n$$\n\\left( \\frac{(2x + y)^2 + 3z(2x + y) + 2z^2}{xy} + 1 \\right) \\cdot \\frac{2x + y + 8z}{z} \\geq \\frac{375}{2}.\n$$\n\nNow fix $z$ and $2x + y$ and move $2x$ and $y$ closer to each other. Then we see that $xy$ increases during this movement and attains its maximum when $2x = y$.\n\nTherefore, the inequality follows from the inequality obtained by the substitution $y = 2x$ into the initial inequality, i.e.\n\n$$\n(3x + z)(6x + 2z)(4x + 8z) \\geq 375x^2z.\n$$\n\nLetting $t = x/z$, we can rewrite the inequality in the form\n\n$$\n8(3t + 1)^2(t + 2) - 375t^2 \\geq 0,\n$$\nwhich can be easily checked by means of derivatives. One finds the minimum to be attained for $t = 4/3$. (So the minimum in the initial inequality holds for $(x, y, z) = (4, 8, 3)$.) $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17325,
    "subject": "Mathematics (Olympiad)",
    "question": "Let triangle $ABC$ be inscribed in a circle $(O)$ and circumscribed about a circle $(I)$. The incircle $(I)$ touches $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. $H$ is the orthocenter of triangle $ABC$. $AH$ meets $BC$ at $X$. $K$ is the reflection of $A$ over $EF$. $V$ is the reflection of $I$ over $O$. Then $$\\angle AHV = \\angle AKX.$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $I_a$, $I_b$, $I_c$ be the excenters of triangle $ABC$ with respect to $A$, $B$, $C$ respectively.\n\nLet $S$, $L$ be the Lemoine points of triangles $I_aI_bI_c$ and $DEF$, respectively.\n\n![](images/Vietnam_2024_Booklet_p44_data_ea7a824b29.png)\n\nSince $V$ is the circumcenter of triangle $I_aI_bI_c$, $\\angle SVI_a = \\angle LID$.\n\nBy Lemma 2, $\\angle AKX = \\angle LID$. Again, according to Lemma 1, $H$, $S$, $V$ are collinear, so $\\angle AKX = \\angle SVI_a = \\angle AHV$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17326,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that $5^{n^2} + 7$ is divisible by $6$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n$$\n5^{n^2} + 7 = (6 - 1)^{n^2} + 7 \\equiv (-1)^{n^2} + 7 \\pmod{6}.\n$$\nSince $(-1)^{n^2}$ is $-1$ when $n^2$ is odd and $1$ when $n^2$ is even, we get:\n- If $n^2$ is even: $(-1)^{n^2} + 7 \\equiv 1 + 7 \\equiv 8 \\equiv 2 \\pmod{6}$.\n- If $n^2$ is odd: $(-1)^{n^2} + 7 \\equiv -1 + 7 \\equiv 6 \\equiv 0 \\pmod{6}$.\n\nTherefore, $5^{n^2} + 7$ is divisible by $6$ if and only if $n^2$ is odd, i.e., $n$ is odd. Thus, all odd positive integers $n$ satisfy the condition.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17327,
    "subject": "Mathematics (Olympiad)",
    "question": "Inside a circle $c$ there are circles $c_1$, $c_2$, and $c_3$ which are tangent to $c$ at points $A$, $B$, and $C$ respectively, all distinct. Circles $c_2$ and $c_3$ have a common point $K$ on segment $BC$; circles $c_3$ and $c_1$ have a common point $L$ on segment $CA$; circles $c_1$ and $c_2$ have a common point $M$ on segment $AB$. Prove that the circles $c_1$, $c_2$, and $c_3$ intersect at the center of circle $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "Take a point $X$ on the common tangent to circles $c_1$ and $c$ lying on the opposite side of line $AB$ from point $C$. Then $\\angle ALM = \\angle XAM = \\angle XAB = \\angle ACB$. Consequently, $ML \\parallel BC$. Similarly, $KM \\parallel CA$ and $LK \\parallel AB$. If $\\frac{|AM|}{|AB|} = \\lambda$, then\n\n$$\n\\frac{|BK|}{|BC|} = \\frac{|BM|}{|BA|} = 1 - \\lambda, \\quad \\frac{|CL|}{|CA|} = \\frac{|CK|}{|CB|} = \\lambda,\n$$\n\nso $\\lambda = \\frac{|AM|}{|AB|} = \\frac{|AL|}{|AC|} = 1 - \\lambda$, hence $\\lambda = \\frac{1}{2}$. Therefore, triangles $AML$, $MBK$, and $LKC$ are all similar to $ABC$ with ratio $\\frac{1}{2}$. Thus, the radii of their circumcircles $c_1$, $c_2$, and $c_3$ are half the radius of circle $c$. Since $c$ and $c_1$ are tangent, the diameter of $c_1$ and the radius of $c$, both drawn from tangent point $A$, coincide. Hence, circle $c_1$ passes through the center of $c$; similarly, circles $c_2$ and $c_3$ pass through the center of $c$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17328,
    "subject": "Mathematics (Olympiad)",
    "question": "Determina el mayor valor posible de\n\n$$\nS = a_1 a_2 a_3 + a_4 a_5 a_6 + \\dots + a_{2017} a_{2018} a_{2019} + a_{2020},\n$$\n\ndonde $(a_1, a_2, a_3, \\dots, a_{2020})$ es una permutación de $(1, 2, 3, \\dots, 2020)$.\n\n**Aclaración:** En $S$, cada término, excepto el último, es la multiplicación de tres números.",
    "options": [],
    "answer": "See solution",
    "solution": "Al inspeccionar casos pequeños, conjeturamos que el máximo es\n$$\nS = 2020 \\cdot 2019 \\cdot 2018 + 2017 \\cdot 2016 \\cdot 2015 + \\dots + 4 \\cdot 3 \\cdot 2 + 1.\n$$\n\nVamos a demostrarlo.\n\nPrimero, si tenemos $a > b > c > d$, el mayor valor para el producto de tres de ellos más el cuarto es $abc + d$. En efecto, $bcd + a < abc + d$, pues $(a - d)(bc - 1) > 0$. De manera similar, $acd + b < abc + d$ y $abd + c < abc + d$. Por lo tanto, en la suma máxima, el término aislado es el 1.\n\nAhora, veamos que se obtiene un valor más grande si 2019 y 2020 están juntos en un mismo término. Sean $x > y$, $z > w$ cuatro números distintos entre 1 y 2018 que se combinan con 2020 y 2019 formando los términos $2020 x y + 2019 z w$. Cambiando 2019 por $y$ obtenemos los términos $2020 \\cdot 2019 x + y z w$. Como\n$$\n2020 x y + 2019 z w > 2020 \\cdot 2019 x + y z w \\iff (2020 x - z w)(y - 2019) > 0,\n$$\nentonces $2020 x < z w$, pues $y < 2019$.\n\nSi cambiamos 2020 por $z$, obtenemos $2020 \\cdot 2019 w + x y z$. De manera análoga, podemos concluir que si $2020 x y + 2019 z w > 2020 \\cdot 2019 w + x y z$, entonces $2019 w < x y$.\n\nLuego, $2020 x + 2019 w < x y + z w < 2019 w + 2020 x$, lo cual es absurdo. Por lo tanto, 2020 y 2019 están juntos en un mismo término de la suma.\n\nAhora, 2018 debe estar en el mismo término que 2020 y 2019. Consideremos tres números menores que 2018, $t$, $u$, $v$, y tenemos\n$$\n2020 \\cdot 2019 \\cdot 2018 + t u v > 2020 \\cdot 2019 t + 2018 u v \\iff (2020 \\cdot 2019 - u v)(2018 - t) > 0,\n$$\nlo cual es verdadero.\n\nA partir de aquí, de modo análogo, juntamos la siguiente terna (2017, 2016 y 2015), y así sucesivamente hasta la última terna (4, 3 y 2), concluyendo la solución.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17329,
    "subject": "Mathematics (Olympiad)",
    "question": "$$\n(x - 1)^2 + (y - 2)^2 + (z - 3)^2 = 0.\n$$\n\nFind all real solutions $(x, y, z)$ to the equation above.",
    "options": [],
    "answer": "See solution",
    "solution": "By expanding the equation, we see that the sum of three squares is zero only if each square is zero. Thus, $x = 1$, $y = 2$, and $z = 3$. Therefore, the only solution is $(x, y, z) = (1, 2, 3)$. This satisfies the equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17330,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y, z > 0$ with $0 < x \\leq 1$. Prove that\n$$\nxy + y + 2z \\geq 4\\sqrt{xyz}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $0 < x \\leq 1$, we have $y \\geq xy$, and hence\n$$\nxy + y + 2z \\geq 2xy + 2z \\geq 4\\sqrt{xyz}\n$$\nby the AM-GM inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17331,
    "subject": "Mathematics (Olympiad)",
    "question": "How many four-digit numbers $abcd$ simultaneously satisfy the equalities $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$?",
    "options": [],
    "answer": "See solution",
    "solution": "From $a + b = c + d$, we get $(a + b)^2 = (c + d)^2$, so $ab = cd$. Furthermore, $a^2 - 2ab + b^2 = c^2 - 2cd + d^2$. As $(a - b)^2 = (c - d)^2$, we have $|a - b| = |c - d|$, which implies $a - b = c - d$ or $a - b = d - c$. Recall that $a + b = c + d$, so either $a = c, b = d$ or $a = d, b = c$.\n\nThe numbers must have one of the forms $\\overline{aaaa}$, $\\overline{abba}$, or $\\overline{abab}$, with $a \\neq 0$ and $a \\neq b$. In all, there are $9 + 9 \\times 9 + 9 \\times 9 = 9 + 81 + 81 = 171$ numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17332,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the value of\n\n$$\n\\tan^2 \\frac{\\pi}{16} \\cdot \\tan^2 \\frac{3\\pi}{16} + \\tan^2 \\frac{\\pi}{16} \\cdot \\tan^2 \\frac{5\\pi}{16} + \\tan^2 \\frac{3\\pi}{16} \\cdot \\tan^2 \\frac{7\\pi}{16} + \\tan^2 \\frac{5\\pi}{16} \\cdot \\tan^2 \\frac{7\\pi}{16}?\n$$\n\n(A) 28 (B) 68 (C) 70 (D) 72 (E) 84",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\theta = \\frac{\\pi}{16}$, and let $c = \\cos \\theta$, $s = \\sin \\theta$, and $t = \\tan \\theta$. Then\n\n$$\ne^{8i\\theta} = e^{\\frac{\\pi}{2}i} = (c + is)^8 = i.\n$$\n\nExpanding the binomial and taking the real part gives\n\n$$\n0 = c^8 - 28c^6s^2 + 70c^4s^4 - 28c^2s^6 + s^8.\n$$\n\nDivide by $c^8$ to get\n\n$$\n1 - 28t^2 + 70t^4 - 28t^6 + t^8 = 0,\n$$\n\nand let $p(t) = 1 - 28t^2 + 70t^4 - 28t^6 + t^8$. In other words, $t = \\tan \\theta$ and $-t = -\\tan \\theta$ are roots of the polynomial\n\n$$\np(x) = 1 - 28x^2 + 70x^4 - 28x^6 + x^8.\n$$\n\nThe same argument with $\\theta = \\frac{\\pi}{16}$ replaced by $\\theta_k = \\frac{k\\pi}{16}$ shows that $t_k = \\tan \\theta_k$ is also a root of $p(x)$ for $k \\in \\{1, 3, 5, 7\\}$. This holds because $(c_k + i s_k)^8 = \\pm i$ when $c_k = \\cos(k\\theta)$ and $s_k = \\sin(k\\theta)$. In each case $8k \\equiv 8 \\pmod{16}$. Therefore $x = \\tan(k\\theta)$ is a root of $p(x)$ whenever\n\n$$\nx^2 \\in \\{\\tan^2 \\theta, \\tan^2(3\\theta), \\tan^2(5\\theta), \\tan^2(7\\theta)\\}.\n$$\n\nThis accounts for all 8 roots, so it must be that\n\n$$\np(x) = (x^2 - \\tan^2 \\theta)(x^2 - \\tan^2(3\\theta))(x^2 - \\tan^2(5\\theta))(x^2 - \\tan^2(7\\theta)).\n$$\n\nExamining the $x^4$ term in the two expressions for $p(x)$ reveals that\n\n$$\n\\tan^2 \\theta \\cdot \\tan^2(3\\theta) + \\tan^2 \\theta \\cdot \\tan^2(5\\theta) + \\tan^2 \\theta \\cdot \\tan^2(7\\theta) + \\tan^2(3\\theta) \\cdot \\tan^2(5\\theta) \\\\\n+ \\tan^2(3\\theta) \\cdot \\tan^2(7\\theta) + \\tan^2(5\\theta) \\cdot \\tan^2(7\\theta) = 70.\n$$\n\nThe identity $\\tan \\theta \\cdot \\tan\\left(\\frac{\\pi}{2} - \\theta\\right) = \\tan \\theta \\cdot \\cot \\theta = 1$ implies that\n\n$$\n\\tan^2 \\theta \\cdot \\tan^2(7\\theta) = \\tan^2(3\\theta) \\cdot \\tan^2(5\\theta) = 1.\n$$\n\nSubtracting those two terms from the previous expression gives\n\n$$\n\\tan^2 \\theta \\cdot \\tan^2(3\\theta) + \\tan^2 \\theta \\cdot \\tan^2(5\\theta) + \\tan^2(3\\theta) \\cdot \\tan^2(7\\theta) + \\tan^2(5\\theta) \\cdot \\tan^2(7\\theta) = 68.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17333,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y, z$ be positive integers such that $x \\neq y \\neq z \\neq x$. Prove that\n$$\n(x + y + z)(xy + yz + zx - 2) \\geq 9xyz.\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $x, y, z$ are distinct positive integers, the required inequality is symmetric, and without loss of generality, we can suppose that $x \\geq y + 1 \\geq z + 2$.\n\n**Case 1.** $y \\geq z + 2$.\n\nSince $x \\geq y + 1 \\geq z + 3$, it follows that\n$$\n(x - y)^2 \\geq 1, \\quad (y - z)^2 \\geq 4, \\quad (x - z)^2 \\geq 9\n$$\nwhich are equivalent to\n$$\nx^2 + y^2 \\geq 2xy + 1, \\quad y^2 + z^2 \\geq 2yz + 4, \\quad x^2 + z^2 \\geq 2xz + 9\n$$\nAdding suitable terms and rearranging, we obtain\n$$\nxy(x + y) + yz(y + z) + zx(z + x) \\geq 6xyz + 4x + 9y + z\n$$\nwhich implies\n$$\n(x + y + z)(xy + yz + zx - 2) \\geq 9xyz + 2x + 7y - z.\n$$\nSince $x \\geq z + 3$, it follows that $2x + 7y - z \\geq 0$, so the inequality holds.\n\n**Case 2.** $y = z + 1$.\n\nSince $x \\geq y + 1 \\geq z + 2$, we have $x \\geq z + 2$. Substitute $y = z + 1$ into the inequality:\n$$\n(x + z + 1 + z)(x(z + 1) + (z + 1)z + xz - 2) \\geq 9x(z + 1)z\n$$\nwhich simplifies to\n$$\n(x + 2z + 1)(z^2 + 2zx + z + x - 2) - 9x(z + 1)z \\geq 0\n$$\nAfter algebraic manipulation, this is equivalent to\n$$\n(x - z - 2)(x - z + 1)(2z + 1) \\geq 0\n$$\nwhich is satisfied since $x \\geq z + 2$.\n\n**Equality:**\n\nEquality is achieved only in Case 2 for $x = z + 2$, so equality holds when $(x, y, z) = (k + 2, k + 1, k)$ and all their permutations, for any positive integer $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17334,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Is it possible to divide the numbers $1, 2, \\ldots, 17$ into two groups, one of which consists of 6 numbers and the other consists of 11 numbers, in such a way that the product of the numbers from one group equals the sum of the numbers from the other group?\n\nb) Is it possible to divide the numbers $1, 2, \\ldots, 17$ into two groups in such a way that the product of the numbers from one group equals the sum of the numbers from the other group?",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let us assume that it is possible. Firstly, let us find the sum of the 11 largest numbers. It is $132 = 7 + 8 + \\cdots + 17$. And the product of the 6 smallest numbers is $720 = 1 \\cdot 2 \\cdot \\cdots \\cdot 6$. It is clear that dividing the numbers into these groups in any other way, the sum can only decrease (because the maximum numbers are summed up now), and the product can only increase for the same reason. Thus, it is not possible.\n\nb) The problem can be solved in a number of ways. Let us show one of them.\n\nLet us assume that one of the parts includes 2 numbers and denote them by $x$ and $y$. The sum of all the numbers is $1 + 2 + \\cdots + 17 = 153$. If the pair of numbers $x, y$ fulfills the conditions, then the equation is exactly satisfied:\n\n$$\n153 - x - y = x y \\Leftrightarrow 153 = x y + x + y \\Leftrightarrow 154 = x y + x + y + 1 \\Leftrightarrow 11 \\cdot 14 = (x + 1)(y + 1).\n$$\n\nThe numbers $x = 10$ and $y = 13$ satisfy the last equation. Let us make sure of that fact:\n\n$$\n(1 + 2 + \\cdots + 17) - 10 - 13 = 153 - 23 = 130 = 10 \\cdot 13,\n$$\n\nwhich proves the statement.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17335,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $A_1 = \\emptyset$, $B_1 = \\{0\\}$。對於所有 $n \\ge 2$，依照下面的遞迴構造：\n\n$$\nA_n = \\{x + 1 \\mid x \\in B_{n-1}\\},\n$$\n\n$$\nB_n = (A_{n-1} \\cup B_{n-1}) - (A_{n-1} \\cap B_{n-1}),\n$$\n\n試問：有哪些 $n$ 會有 $B_n = \\{0\\}$？\n\n（註：對任意集合 $A$ 和 $B$，集合 $A - B$ 表示 $\\{x \\in A \\mid x \\notin B\\}$。）",
    "options": [],
    "answer": "See solution",
    "solution": "我們將證明：$B_n = \\{0\\}$ 若且唯若 $n = 2^k$，其中 $k \\ge 0$。\n\n給一個集合 $S$，令符號 $2S = \\{2x \\mid x \\in S\\}$ 和 $S + k = \\{x + k \\mid x \\in S\\}$ 對任意整數 $k$。\n\n首先，對於所有 $n \\ge 1$，有 $A_n$ 與 $B_n$ 的元素都是非負整數，而且 $0 \\notin A_n$ 且 $0 \\in B_n$。這可以簡單地以歸納法證明。\n\n接著，對於 $n \\ge 2$，證明以下四種性質：\n\n(a) $A_{2n-1} = 2A_n - 1$；\n\n(b) $B_{2n-1} = A_{2n-1} \\cup B_{2n}$；\n\n(c) $B_{2n} = 2B_n$；\n\n(d) $1 \\in B_{2n-1}$。\n\n要以歸納法證明之。先檢驗 $n=2$ 時，觀察前幾項：$A_2 = \\{1\\}$，$B_2 = \\{0\\}$，$A_3 = \\{1\\}$，$B_3 = \\{0, 1\\}$，$A_4 = \\{1, 2\\}$，$B_4 = \\{0\\}$，可以一一檢驗成立。以下假設 $n \\ge 3$ 且四種性質對於 $n-1$ 成立。\n\n關於 (a)：\n\n$$\nA_{2n-1} = B_{2n-2} + 1 = 2B_{n-1} + 1 = 2(A_n - 1) + 1 = 2A_n - 1.\n$$\n\n對於 (b) 與 (c)，先要得到：\n\n$$\n\\begin{aligned}\nA_{2n-2} &= B_{2n-3} + 1 = (A_{2n-3} \\cup B_{2n-2}) + 1 \\\\\n&= ((2A_{n-1} - 1) \\cup 2B_{n-1}) + 1 \\\\\n&= 2A_{n-1} \\cup (2B_{n-1} + 1).\n\\end{aligned}\n$$\n\n關於 (b)：\n\n$$\n\\begin{aligned}\nB_{2n-1} &= (A_{2n-2} \\cup B_{2n-2}) - (A_{2n-2} \\cap B_{2n-2}) \\\\\n&= 2A_{n-1} \\cup (2B_{n-1} + 1) \\cup 2B_{n-1} \\\\\n& \\quad (2A_{n-1} \\cup (2B_{n-1} + 1)) \\cap 2B_{n-1} \\\\\n&= (2B_{n-1} + 1) \\cup 2A_{n-1} \\cup 2B_{n-1} - 2A_{n-1} \\cap 2B_{n-1} \\\\\n&= (2B_{n-1} + 1) \\cup (2A_{n-1} \\cup 2B_{n-1} - 2A_{n-1} \\cap 2B_{n-1}) \\\\\n&= (2B_{n-1} + 1) \\cup 2B_n \\\\\n&= (B_{2n-2} + 1) \\cup 2B_n \\\\\n&= A_{2n-1} \\cup 2B_n,\n\\end{aligned}\n$$\n\n其中第三個與第四個等式都利用到 $(2B_{n-1} + 1) \\cap 2B_{n-1} = \\emptyset$。此時 (b) 尚未證完，但暫時停下，因為現在還無法使用 $2B_n = B_{2n}$，這是 (c) 要證明的。\n\n關於 (c)：\n\n$$\n\\begin{aligned}\nB_{2n} &= (A_{2n-1} \\cup B_{2n-1}) - (A_{2n-1} \\cap B_{2n-1}) \\\\\n&= A_{2n-1} \\cup (A_{2n-1} \\cup 2B_n) - A_{2n-1} \\cap (A_{2n-1} \\cup 2B_n) \\\\\n&= A_{2n-1} \\cup 2B_n - A_{2n-1} \\\\\n&= 2B_n - A_{2n-1} \\\\\n&= 2B_n - (2A_n - 1) \\\\\n&= 2B_n,\n\\end{aligned}\n$$\n\n其中最後一個等式是因為 $2B_n$ 只包含偶數而 $2A_n - 1$ 只包含奇數。因為 (c) 完成了，則 (b) 也完成了。\n\n關於 (d)：借用前面一個等式 $B_{2n-1} = (2B_{n-1} + 1) \\cup 2B_n$，因為 $0 \\in B_{n-1}$ 所以 $1 \\in B_{2n-1}$。\n\n最後證明：$B_n = \\{0\\}$ 若且唯若 $n = 2^k$，其中 $k \\ge 0$。其中 “$\\Leftrightarrow$” 方向可以由 $B_1 = \\{0\\}$ 與 (c) 來完成。至於 “$\\Rightarrow$” 方向，對於所有 $n \\ne 2^k$ 必然是 $n = 2^k m$，其中 $k \\ge 0$ 且 $m$ 是不等於 1 的正奇數。若 $k=0$ 則 $n$ 是奇數，由 (d) 可知 $1 \\in B_n \\ne \\{0\\}$。若 $k \\ge 1$ 則 $B_n = 2^k B_m$，所以 $2^k \\in B_n \\ne \\{0\\}$。證明完畢！",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17336,
    "subject": "Mathematics (Olympiad)",
    "question": "Let us call a positive integer a *good number* if the digit 2 appears more frequently than the digit 3, and a *bad number* if the digit 3 appears more frequently than the digit 2. For example, 2023 is a good number because the digit 2 appears twice and the digit 3 once, and 123 is neither a good number nor a bad number because the digit 2 appears once and the digit 3 once.\n\nFind the difference between the number of good numbers less than or equal to 2023 and the number of bad numbers less than or equal to 2023.",
    "options": [],
    "answer": "See solution",
    "solution": "22\n\nFor a positive integer $k$, let us define the changed number of $k$ as the number obtained by switching each digit 2 in $k$ to 3, and each digit 3 in $k$ to 2. By this definition, if $l$ is the changed number of $k$, then the changed number of $l$ is $k$ itself. For a good number $m \\leq 1999$ and a bad number $n \\leq 1999$, we call $(m, n)$ a nice pair if $n$ is the changed number of $m$.\n\nLet $m$ be a good number. Let $a$ (respectively, $b$) be the number of times the digit 2 (respectively, 3) appears in $m$. Then, we have $a > b$. Let $n$ denote the changed number of $m$. Then, the number of times the digit 2 appears in $n$ is $b$, and the number of times the digit 3 appears in $n$ is $a$. Hence, $n$ is a bad number. Furthermore, if $m \\leq 1999$, then we have $n \\leq 1999$. Note that a number cannot be both a good number and a bad number. These observations show that each good number less than or equal to 1999 is contained in exactly one nice pair. Similarly, each bad number less than or equal to 1999 is contained in exactly one nice pair. Therefore, the number of good numbers less than or equal to 1999 and the number of bad numbers less than or equal to 1999 are both equal to the number of nice pairs.\n\nIn addition, among the integers between 2000 and 2023, there are 22 good numbers (excluding 2003 and 2013) and no bad numbers. Thus, the answer is 22.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17337,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a real parameter. Determine the number of real solutions to the system\n\n$$\n\\begin{align*}\nx^2 + kxy + y^2 &= z, \\\\\ny^2 + kyz + z^2 &= x, \\\\\nz^2 + kzx + x^2 &= y,\n\\end{align*}\n$$\n\nin terms of $k$.\n\n![](images/66th_Czech_and_Slovak_p13_data_6a0c87898c.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We distinguish several cases.\n\n**Case 1:** $x = y = z$. The system reduces to $(k+2)x^2 = x$, which has solutions $(0, 0, 0)$ for any $k$, and $(\\frac{1}{k+2}, \\frac{1}{k+2}, \\frac{1}{k+2})$ if $k \\neq -2$.\n\n**Case 2:** $x \\neq y \\neq z$. Subtracting equations yields\n\n$$\n(x - z)(x + z + ky + 1) = 0, \\tag{1}\n$$\n$$\n(y - x)(y + x + kz + 1) = 0. \\tag{2}\n$$\n\nIf $x \\neq y \\neq z \\neq x$, equations (1) and (2) reduce to\n$$\nx + z + ky + 1 = 0, \\\\\ny + x + kz + 1 = 0.\n$$\nSubtracting gives $(y-z)(k-1) = 0$, so $k=1$ and $x+y+z=-1$. But for $k=1$, $z = x^2 + xy + y^2 \\ge 0$, so $x + y + z \\ge 0$, which is impossible for $x + y + z = -1$.\n\nThus, in every solution, at least two variables are equal. By symmetry, assume $x = y \\neq z$. Equation (1) gives $x = -(k+1)y - 1$, and the system reduces to\n$$\n(k + 2)y^2 + (k + 1)y + 1 = 0. \\tag{3}\n$$\n\nAny solution to equation (3) is a new solution, since $x = y = z = -1/(k+2)$ is not a solution to the original system for $k \\neq -2$.\n\nFor $k = -2$, equation (3) is linear with unique solution $y = 1$, yielding $(0, 1, 1)$ and its permutations.\n\nFor $k \\neq -2$, equation (3) is quadratic and has real solutions if and only if\n$$\nD = (k + 1)^2 - 4(k + 2) = k^2 - 2k - 7 \\geq 0,\n$$\nwhich is $k \\notin (1-2\\sqrt{2}, 1+2\\sqrt{2})$.\n\nFor $k = 1 \\pm 2\\sqrt{2}$, there is a unique solution\n$$\ny_0 = -\\frac{k+1}{2(k+2)} = 1 \\mp \\sqrt{2}, \\quad x_0 = \\frac{(k+1)^2}{2(k+2)} - 1 = 1.\n$$\nThis gives three permutations of $(x_0, y_0, y_0)$.\n\nFor $k \\in (-\\infty, -2) \\cup (-2, 1-2\\sqrt{2}) \\cup (1+2\\sqrt{2}, \\infty)$, equation (3) has two distinct solutions\n$$\ny_{1,2} = \\frac{-k - 1 \\pm \\sqrt{k^2 - 2k - 7}}{2(k + 2)},\n$$\nwith $x_{1,2} = -(k+1)y_{1,2} - 1$. The system thus has six solutions: three permutations of $(x_1, y_1, y_1)$ and three of $(x_2, y_2, y_2)$.\n\n**Summary Table:**\n\n![](images/66th_Czech_and_Slovak_p13_data_6a0c87898c.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17338,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine whether there exists a polynomial $f(x_1, x_2)$ in two variables, with integer coefficients, and two points $A = (a_1, a_2)$ and $B = (b_1, b_2)$ in the plane, satisfying all the following conditions:\n\n1. $A$ is an integer point (i.e., $a_1$ and $a_2$ are integers).\n2. $|a_1 - b_1| + |a_2 - b_2| = 2010$.\n3. $f(n_1, n_2) > f(a_1, a_2)$ for all integer points $(n_1, n_2)$ in the plane other than $A$.\n4. $f(x_1, x_2) > f(b_1, b_2)$ for all points $(x_1, x_2)$ in the plane other than $B$.",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, such a triple $(f(x_1, x_2), A, B)$ exists.\n\nLet $A = (0, 0)$ and $B = (2009 + \\frac{2}{3}, \\frac{1}{3})$. Consider a polynomial $f$ such that $f(x, y) = 0$ is the equation of an ellipse centered at $B$, passing through $A$, and tangent to $y = 0$ at $A$. For example:\n\n$$\nf(X, Y) = 9M(X - x_0)^2 + 9N(X - x_0)(Y - y_0) + 9P(Y - y_0)^2 - Q\n$$\n\nwhere $M, N, P, Q$ are integers with $M, P, Q, 4MP - N^2 > 0$.\n\nThe conditions that the ellipse passes through $A$ and is tangent to $y = 0$ at $A$ are:\n\n$$\n\\begin{cases}\n6029^2 M + 6029N + P - Q = 0 \\\\\n2 \\cdot 6029M + N = 0\n\\end{cases}\n$$\n\nChoose $M = 1$, $N = -2 \\cdot 6029$, $P > 6029^2$, and $Q = P - 6029^2$.\n\n_Alternative Solution:_\n\nGiven any integer point $A(a_1, a_2)$, there are infinitely many points $B(b_1, b_2)$ with $b_1, b_2 \\in \\mathbb{Q} \\setminus \\mathbb{Z}$ and $|a_1 - b_1| + |a_2 - b_2| = 2010$, e.g., $b_1 = a_1 + \\alpha + r$, $b_2 = a_2 + \\beta + (1 - r)$, with $\\alpha, \\beta \\in \\mathbb{Z}_+$, $r \\in \\mathbb{Q} \\cap (0, 1)$, and $\\alpha + \\beta = 2009$.\n\nConsider polynomials of the form:\n\n$$\nf(X, Y) = N \\left( (X - a_1)^2 + (Y - a_2)^2 + \\varepsilon \\right) \\left( (X - b_1)^2 + (Y - b_2)^2 \\right)\n$$\n\nwhere $\\varepsilon \\in \\mathbb{Q}_+^*$ and $N \\in \\mathbb{Z}_+^*$ is large enough for $f(X, Y)$ to have integer coefficients.\n\nThen $f(b_1, b_2) = 0$, and $f(x, y) > 0$ for all points $(x, y)$ other than $B$.\n\nAlso, $f(a_1, a_2) = N\\varepsilon((a_1 - b_1)^2 + (a_2 - b_2)^2)$, and for all integer points $(n_1, n_2) \\ne A$:\n\n$$\n\\min \\{(n_1 - a_1)^2 + (n_2 - a_2)^2 + \\varepsilon\\} = 1 + \\varepsilon \\\\\n\\min \\{(n_1 - b_1)^2 + (n_2 - b_2)^2\\} = m\n$$\n\nfor some $m \\in \\mathbb{Q} \\cap (0, \\frac{1}{2}]$ (e.g., $m = \\frac{1}{2}$ when $r = \\frac{1}{2}$). Thus, $f(n_1, n_2) > N(1+\\varepsilon)m$. To ensure $f(n_1, n_2) > f(a_1, a_2)$, take\n\n$$\n\\varepsilon = \\frac{m}{(a_1 - b_1)^2 + (a_2 - b_2)^2 - m} \\in \\mathbb{Q}_+^*\n$$\n\nand choose an appropriate $N$.\n\n**Remarks:**\n\nThis uses the fact that a closed simple curve $f(x, y) = 0$ separates the plane into two regions, with $f$ of one sign inside and the opposite sign outside. The problem reduces to analytic geometry of conics.\n\nThe point $C(2x_0, 2y_0)$, the symmetric of $A$ with respect to $B$, also lies on the ellipse $f(x, y) = 0$. The construction stretches the circle $(x - x_0)^2 + (y - y_0)^2 - (x_0^2 + y_0^2)$ into a rational ellipse containing no other integer points than $A$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17339,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $\\triangle ABC$ 為銳角三角形，點 $M$ 為 $AC$ 的中點。一圓 $\\omega$ 通過 $B$、$M$ 兩點，並與 $AB$、$BC$ 兩邊分別再交於 $P$、$Q$ 兩點。令點 $T$ 是使得 $BPTQ$ 形成平行四邊形的一點。當 $T$ 落在三角形 $ABC$ 的外接圓上時，試決定 $BT/BM$ 的所有可能值。\n\nLet $\\triangle ABC$ be an acute triangle, and let $M$ be the midpoint of $AC$. A circle $\\omega$ passing through $B$ and $M$ meets the sides $AB$ and $BC$ again at $P$ and $Q$, respectively. Let $T$ be the point such that the quadrilateral $BPTQ$ is a parallelogram. Suppose that $T$ lies on the circumcircle of the triangle $ABC$. Determine all possible values of $BT/BM$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\\frac{BT}{BM} = \\sqrt{2}$$\n\n令 $S$ 為平行四邊形 $BPTM$ 的中心，並令點 $B' \\neq B$ 是在射線 $BM$ 上滿足 $BM = MB'$ 的點，如下圖所示。\n\n![](images/16-1J_p9_data_4230959dbc.png)\n\n所以，$ABCB'$ 為平行四邊形。\n\n因此，$\\angle ABB' = \\angle PQM$，以及 $\\angle BB'A = \\angle B'BC = \\angle MPQ$，由此得 $\\triangle ABB' \\sim \\triangle MQP$。於是 $AM$ 與 $MS$ 分別是這兩個相似三角形對應邊的中線。可知\n\n$$\n\\angle SMP = \\angle B'AM = \\angle BCA = \\angle BTA. \\quad (1)\n$$\n\n因為 $\\angle ACT = \\angle PBT$，以及 $\\angle TAC = \\angle TBC = \\angle BTP$，可得 $\\triangle TCA \\sim \\triangle PBT$。再一次，$TM$ 與 $PS$ 分別是這兩個相似三角形對應邊的中線，可知\n\n$$\n\\angle MTA = \\angle TPS = \\angle BQP = \\angle BMP. \\quad (2)\n$$\n\n以下分兩種情況討論。\n\n**Case 1.** $S$ 不在 $BM$ 線上。因為 $A$、$C$ 兩點的角色是對稱的，我們不妨假設 $S$ 與 $A$ 位於 $BM$ 直線的同側。\n\n利用 (1) 及 (2)，得\n\n$$\n\\angle BMS = \\angle BMP - \\angle SMP = \\angle MTA - \\angle BTA = \\angle MTB,\n$$\n\n故兩三角形 $BSM$ 與 $BMT$ 相似。於是 $BM^2 = BS \\cdot BT = \\frac{BT^2}{2}$，即\n$$ BT = \\sqrt{2}BM $$\n\n**Case 2.** $S$ 落在 $BM$ 線上。由 (2) 知 $\\angle BCA = \\angle MTA = \\angle BQP = \\angle BMP$（如下圖所示）。所以 $PQ \\parallel AC$，且 $PM \\parallel AT$。於是 $\\frac{BS}{BM} = \\frac{BP}{BA} = \\frac{BM}{BT}$，一樣得到 $BT^2 = 2BM^2$，所以 $BT = \\sqrt{2}BM$。\n\n![](images/16-1J_p10_data_d34595c25e.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17340,
    "subject": "Mathematics (Olympiad)",
    "question": "求所有整數 $c \\in \\{0, 1, \\dots, 2016\\}$，使得滿足下列兩條件之函數\n\n$$f: \\mathbb{Z} \\to \\{0, 1, \\dots, 2016\\}$$\n\n的個數最小：\n\n1. 函數 $f$ 的週期為 $2017$；\n2. $f(f(x) + f(y) + 1) - f(f(x) + f(y)) \\equiv c \\pmod{2017}$。\n\n註：此處 $\\mathbb{Z}$ 表示所有整數所成的集合。",
    "options": [],
    "answer": "See solution",
    "solution": "所求的 $c$ 為 $1,\\ 1008,\\ 1009,\\ 2016$，此時 $f$ 的個數為 $2017$。\n\n設 $a\\%b$ 表示 $a$ 除以 $b$ 的餘數，$0 \\leq a\\%b < b$。\n注意 $f(x) = (c x + k)\\%2017$（$k = 0, 1, \\dots, 2016$）滿足條件。\n\n以下證明當 $c \\ne 1, 1008, 1009, 2016$ 時還有其他解：\n\n當 $c \\ne 0$，設\n$$\nf_0(2017n + 1) = f_0(2017n + c + 1) = f_0(2017n + 2c + 1) = c,\n$$\n其餘 $f_0(x) = 0$。則 $(f_0(x) + f_0(y))\\%2017$ 只可能為 $0, c, 2c\\%2017$，\n但因 $c \\ne 1, 1008, 1009, 2016$，有\n$$\n0, c, 2c\\%2017 \\notin \\{1, (c+1)\\%2017, (2c+1)\\%2017\\}，\n$$\n所以\n$$\nf_0(f_0(x) + f_0(y)) = 0,\\quad f_0(f_0(x) + f_0(y) + 1) = c。\n$$\n因此 $f_0$ 滿足條件，且 $f_0$ 不等於 $(c x + k)\\%2017$。\n\n當 $c = 0$，設\n$$\nf_0(2017n + 6) = f_0(2017n + 7) = 3,\n$$\n其餘 $f_0(x) = 1$。此時 $(f_0(x) + f_0(y))\\%2017$ 只可能為 $2, 4, 6$，且\n$$\nf_0(2) = f_0(3) = f_0(4) = f_0(5) = f_0(6) = f_0(7)，\n$$\n所以 $f_0$ 滿足條件，且 $f_0$ 不等於 $(c x + k)\\%2017$。\n\n接下來證明當 $c = 1, 1008, 1009, 2016$ 時，所有解皆為 $(c x + k)\\%2017$：\n\n假設 $f$ 滿足條件，先證 $f$ 為滿射。若 $f$ 非滿射，則可在 $f$ 的值域中找到最長的「良好序列」 $\\{a_i\\}_{i=1}^r$，其中 $a_i = (c i + d)\\%2017$。由於 $2017$ 為質數，且 $f$ 非滿射，最長良好序列長度有限。\n\n將 $(x, y)$ 代入條件 (2)，使 $(f(x), f(y)) = (a_i, a_j),\\ i, j = 1, 2, \\dots, r$：\n\n- $c = 1$ 時，設 $b_i = f(a_1 + a_i),\\ b_{r+1} = f(a_1 + a_r + 1)$。\n- $c = 1008$ 時，設 $b_i = f(a_{r+1-i} + a_{r+1-i}),\\ b_{r+1} = f(2a_1 + 1)$。\n- $c = 1009$ 時，設 $b_i = f(a_i + a_i),\\ b_{r+1} = f(2a_r + 1)$。\n- $c = 2016$ 時，設 $b_i = f(a_1 + a_{r+1-i}),\\ b_{r+1} = f(2a_1 + 1)$。\n\n則 $\\{b_i\\}_{i=1}^{r+1}$ 為更長的良好序列，矛盾！故 $f$ 為滿射，且 $f(x+1) - f(x) \\equiv c \\pmod{2017}$，對所有 $x$ 成立。因此存在 $k$ 使 $f(x) = (c x + k)\\%2017$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17341,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $n$ rectangles $R_{1,n}, \\dots, R_{i,n+1-i}, \\dots, R_{n,1}$, where $R_{x,y}$ is a rectangle with corners at $(x, y)$, $(x, -y)$, $(-x, y)$, and $(-x, -y)$, how many regions do these rectangles divide the plane into?",
    "options": [],
    "answer": "See solution",
    "solution": "We will show by induction that $n$ rectangles divide the plane into $2n^2 - 2n + 2$ regions.\n\n*Base case*: For $n = 1$, the plane is divided into $2$ regions, which matches $2(1)^2 - 2(1) + 2 = 2$.\n\n*Inductive step*: Assume the result holds for $n = k$, so $k$ rectangles divide the plane into at most $2k^2 - 2k + 2$ regions. Consider adding a $(k+1)$-th rectangle $R_{k+1}$. Each existing rectangle can intersect $R_{k+1}$ in at most $4$ points (since rectangles have four sides), so $R_{k+1}$ can intersect the borders of the $k$ rectangles in at most $4k$ points. Adding $R_{k+1}$ can increase the number of regions by at most $4k$.\n\nThus, the total number of regions is at most:\n\n$$\n(2k^2 - 2k + 2) + 4k = 2(k+1)^2 - 2(k+1) + 2.\n$$\n\nBy induction, the formula holds for all $n$. For $n = 3$, the answer is $2 \\times 3^2 - 2 \\times 3 + 2 = 14$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17342,
    "subject": "Mathematics (Olympiad)",
    "question": "What happens during two consecutive moves in the following game?\n\nThere is a pile of 1000 matches. On each turn, a player may take 1 to 6 matches from the pile, but only 10 times in total during the game may a player take 6 matches (these are called \"exceptional moves\"). After these 10 exceptional moves are used, only 1 to 5 matches may be taken per turn. The players alternate turns. Who has a winning strategy, and what is it?",
    "options": [],
    "answer": "See solution",
    "solution": "The second player wins.\n\nLet $r$ be the number of remaining exceptional moves (initially $r=10$). The second player's strategy is to ensure that after their move, the number of matches is either $6n + r$ (with $n > r$) or $7n$ (with $n \\leq r$), where $n$ is a non-negative integer. Note that $6n + r = 7n$ when $n = r$.\n\nAt the start, $1000 = 6 \\times 165 + 10$, so the strategy applies.\n\n**Case 1:** $n > r$\n- If the first player takes $k$ matches ($1 \\leq k \\leq 5$), $r$ does not change. The second player takes $6 - k$ matches, so together they remove 6 matches, and the pile becomes $6(n-1) + r$.\n- If the first player takes 6 matches, $r$ decreases by 1. The second player takes 1 match, so the pile becomes $6(n-1) + (r-1)$.\n\n**Case 2:** $n \\leq r$\n- There are enough exceptional moves left, so each player can take up to 6 matches. If the first player takes $k$ matches, the second player takes $7 - k$ matches, keeping the total removed at 7 per round.\n\nBy following this strategy, the second player can always respond to maintain the desired pile size, ensuring victory.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17343,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that if Alice picks $S$ to be of the form\n\n$$\nS = \\{(x, y) \\in \\mathbb{Z}^2 \\mid m \\le x^2 + y^2 \\le n\\}\n$$\n\nfor some positive integers $m$ and $n$, then Bob can win. (Bob does not know in advance that $S$ is of this form.)",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, Bob can compute the number $N$ of points.\n\n**Claim.** Fix $m$ and $n$ as in the problem statement. Among all sets $T \\subseteq \\mathbb{Z}^2$ with $N$ points, the set $S$ is the unique one which maximizes the value of\n\n$$\nF(T) := \\sum_{(x,y) \\in T} (x^2 + y^2)(m + n - (x^2 + y^2)).\n$$\n\n*Proof.* The different points in $T$ do not interact in this sum, so we simply want the points $(x, y)$ with $x^2 + y^2$ as close as possible to $\\frac{m+n}{2}$, which is exactly what $S$ does. $\\square$\n\nAs a result, it suffices to show that Bob has enough information to compute $F(S)$ from the data given. (There is no issue with fixing $m$ and $n$, since Bob can find an upper bound on the magnitude of the points and then check all pairs $(m, n)$ smaller than that.) The idea is that he knows the full distribution of each of $X$, $Y$, $X+Y$, $X-Y$ and hence can compute sums over $T$ of any power of a single one of those linear functions. By taking linear combinations, we can hence compute $F(S)$.\n\nLet us make the relations explicit. For ease of exposition, take $Z = (X, Y)$ to be a uniformly random point from the set $S$. The information is precisely the individual distributions of $X$, $Y$, $X+Y$, and $X-Y$. Now compute\n\n$$\n\\frac{F(S)}{N} = \\mathbb{E}[(m+n)(X^2+Y^2) - (X^2+Y^2)^2] = (m+n)(\\mathbb{E}[X^2] + \\mathbb{E}[Y^2]) - \\mathbb{E}[X^4] - \\mathbb{E}[Y^4] - 2\\mathbb{E}[X^2Y^2].\n$$\n\nOn the other hand,\n\n$$\n\\mathbb{E}[X^2Y^2] = \\frac{\\mathbb{E}[(X+Y)^4] + \\mathbb{E}[(X-Y)^4] - 2\\mathbb{E}[X^4] - 2\\mathbb{E}[Y^4]}{12}.\n$$\n\nThus, we have written $F(S)$ in terms of the distributions of $X$, $Y$, $X-Y$, $X+Y$, which completes the proof.\n\n**Remark (Mark Sellke).**\n\n* This proof would have worked just as well if we allowed arbitrary $[0, 1]$-valued weights on points with finitely many weights non-zero. There is an obvious continuum generalization one can make concerning the indicator function for an annulus. It's a simpler but fun problem to characterize when just the vertical/horizontal directions determine the distribution.\n\n* An obstruction to purely combinatorial arguments is that if you take an octagon with points $(\\pm a, \\pm b)$ and $(\\pm b, \\pm a)$ then the two ways to pick every other point (going around clockwise) are indistinguishable by Bob. This at least shows that Bob's task is far from possible in general, and hints at proving an inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17344,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $n$ such that $4k^2 + n$ is a prime number for all non-negative integers $k$ smaller than $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "For $k = 0$, we get $n$ must be a prime. Checking small primes:\n\n- $n = 2$: $4 \\cdot 1^2 + 2 = 6$ (not prime)\n- $n = 3$: $4 \\cdot 1^2 + 3 = 7$ (prime), $4 \\cdot 2^2 + 3 = 19$ (prime)\n- $n = 5$: $4 \\cdot 2^2 + 5 = 21$ (not prime)\n- $n = 7$: $4 \\cdot 1^2 + 7 = 11$ (prime), $4 \\cdot 2^2 + 7 = 23$ (prime), $4 \\cdot 3^2 + 7 = 43$ (prime), $4 \\cdot 4^2 + 7 = 71$ (prime), $4 \\cdot 5^2 + 7 = 107$ (prime), $4 \\cdot 6^2 + 7 = 151$ (prime)\n\nSo $n = 3$ and $n = 7$ work. Now, suppose $n > 10$.\n\nIf $n = 4m + 1$, choose $k = m < n$, then $4k^2 + n = 4m^2 + 4m + 1 = (2m + 1)^2$, which is composite for $m > 0$.\n\nIf $n = 4m - 1$, if $m$ has an odd divisor $d > 1$, choose $2k - 1 = d$, then $4k^2 + n$ is divisible by $d$ and is composite.\n\nIf $m$ is a power of $2$, $n = 2^u - 1$ for $u \\geq 4$. If $u$ is composite, $n$ has a proper divisor $2^a - 1$; take $k = 2^a - 1$.\n\nIf $u = 4j + 1$, for $k = 1$, $4 \\cdot 1^2 + n = 4 + 2^{4j+1} - 1 = 2^{4j+1} + 3$, which is composite for $j \\geq 1$.\n\nIf $u = 4j + 3$, for $k = 2^{2j-1}$, $4k^2 + n = 2^{4j} + 2^{4j+3} - 1 = (3 \\cdot 4^j)^2 - 1 = (3 \\cdot 4^j - 1)(3 \\cdot 4^j + 1)$, which is composite.\n\nThus, the only solutions are $n = 3$ and $n = 7$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17345,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that for positive numbers $x$, $y$, $z$, $a$, $b$, $c$ such that $x + y + z = a + b + c$, the following inequality holds:\n\n$$\n\\frac{x}{a+b} + \\frac{y}{b+c} + \\frac{z}{c+a} + \\frac{a}{x+z} + \\frac{b}{x+y} + \\frac{c}{y+z} > 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The following completes the proof:\n\n$$\n\\frac{x}{a+b} + \\frac{y}{b+c} + \\frac{z}{c+a} + \\frac{a}{x+z} + \\frac{b}{x+y} + \\frac{c}{y+z} > \n\\frac{x}{a+b+c} + \\frac{y}{a+b+c} + \\frac{z}{a+b+c} + \\frac{a}{x+y+z} + \\frac{b}{x+y+z} + \\frac{c}{x+y+z} = \n\\frac{x+y+z}{a+b+c} + \\frac{a+b+c}{x+y+z} = 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17346,
    "subject": "Mathematics (Olympiad)",
    "question": "Sea $ABC$ un triángulo rectángulo en $C$ no isósceles con catetos $b > a$.\n\n1. Hallar el lado del cuadrado $AXYZ$ que circunscribe al triángulo $ABC$ (los vértices $B$ y $C$ tienen que estar en lados distintos del cuadrado).\n\n2. Explicar paso a paso cómo construir el cuadrado $AXYZ$ con regla y compás.",
    "options": [],
    "answer": "See solution",
    "solution": "(i) Sea $l$ la longitud del cuadrado y $x$ la longitud del segmento $XC$. Los triángulos rectángulos $AXC$ y $BYC$ son semejantes (puesto que $\\angle BCY = \\pi/2 - \\angle ACX = \\angle CAX$), de donde $l/b = (l - x)/a$, es decir, $x/l = (b - a)/b$. Entonces, aplicando el Teorema de Pitágoras:\n\n$$\nb^2 = l^2 + x^2 = l^2\\left(1 + \\left(\\frac{x}{l}\\right)^2\\right) = l^2\\left(1 + \\frac{(b-a)^2}{b^2}\\right)\n$$\n\nde donde $l = \\dfrac{b^2}{\\sqrt{(b-a)^2 + b^2}}$.\n\n![](images/Spanija_b_2016_p10_data_0cf453989c.png)\n\n(ii) Para construir el cuadrado, observamos que, por (i), se tiene que la tangente del ángulo $\\alpha = \\angle CAX$ es $x/l = (b-a)/b$. Prolongamos el lado $BC$ del triángulo hasta un punto $B'$ de modo que $BB'$ mida $b$ unidades. Así, $CB'$ mide $b-a$ y el ángulo $\\angle CAB'$ tiene tangente $(b-a)/b$. El vértice $X$ del cuadrado que buscamos tiene que estar entonces sobre la recta que contiene...",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17347,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S_k$ denote the sum of the first $k$ terms of an arithmetic sequence with first term $a_1$ and common difference $d$. Prove that for positive integers $n$ and $m$ with $n + m \\ne 0$, the following equality holds:\n\n$$\n(n+m)(S_n - S_m) = (n-m)S_{n+m}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $d$ the difference of two consecutive terms of this arithmetic sequence. By the formula for the sum of an arithmetic sequence, we have\n\n$$\nS_k = k a_1 + \\frac{k(k-1)}{2} d.\n$$\n\nUsing this, we calculate\n\n$$\nS_{n+m} = (n+m)a_1 + \\frac{(n+m)(n+m-1)}{2}d = (n+m)\\left(a_1 + (n+m-1)\\frac{d}{2}\\right)\n$$\n\nand\n\n$$\nS_n - S_m = (n-m)a_1 + (n(n-1)-m(m-1))\\frac{d}{2} = (n-m)a_1 + (n-m)(n+m-1)\\frac{d}{2} = (n-m)\\left(a_1 + (n+m-1)\\frac{d}{2}\\right).\n$$\n\nThe last term in both equations is equal, therefore\n\n$$\n(n+m)(S_n - S_m) = (n-m)S_{n+m}\n$$\n\nwhich proves the desired equality since $n + m \\ne 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17348,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an acute, scalene triangle $ABC$ ($AB < AC$) with circumcircle $(O)$. Let $I$ be the midpoint of arc $BC$ not containing $A$. Take $K$ on $AC$ ($K \\ne C$) such that $IK = IC$. Line $BK$ intersects $(O)$ again at $D$ and intersects $AI$ at $E$. Line $DI$ meets $AC$ at $F$.\n\na) Prove that $EF = \\frac{BC}{2}$.\n\nb) Let $M$ be the point on $DI$ such that $CM$ is parallel to $AD$. Line $KM$ meets $BC$ at $N$ and the circumcircle of triangle $BKN$ meets $(O)$ again at $P$. Prove that $PK$ passes through the midpoint of $AD$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) It is easy to see that $K$ belongs to segment $AC$. Since $IK = IC$, triangle $IKC$ is isosceles at $I$. Hence, $\\angle AKI = 180^\\circ - \\angle IKC = 180^\\circ - \\angle ICK = \\angle ABI$, because $ABIC$ is a cyclic quadrilateral. Furthermore, since $I$ is the midpoint of arc $BC$, $\\angle IAK = \\angle IAB$ and $IK = IB = IC$, so $\\triangle ABI = \\triangle AKI$.\n\n![](images/Vietnamese_mathematical_competitions_p87_data_6e30172780.png)\n\nWe conclude that $AI$ is the perpendicular bisector of segment $BK$, which implies that $E$ is the midpoint of $BK$. Note that\n\n$$\n\\angle DCK = \\angle ABD = \\angle AKB = \\angle DKC\n$$\n\nso triangle $DKC$ is isosceles, hence $DK = DC$. Since $IK = IC$, $ID$ is also the perpendicular bisector of $KC$, so $F$ is the midpoint of $CK$. From these results, $EF$ is the midline of triangle $KBC$, so $EF = \\frac{1}{2}BC$.\n\nb) Let $J$ be the midpoint of arc $BC$ of $(O)$ that contains $A$; it is well-known that $IJ$ is the diameter of $(O)$. We will show that $J$, $K$, and $P$ are collinear.\n\nIndeed, from $\\angle IPJ = 90^\\circ$, we only need to prove $\\angle KPI = 90^\\circ$. In triangle $ADI$, since $DK \\perp AI$ and $AK \\perp DI$, $K$ is the orthocenter of triangle $ADI$. Hence, $IK$ is perpendicular to $AD$.\n\nOn the other hand, since $CM \\parallel AD$, it follows that $CM \\perp IK$, but $IM \\perp KC$, so $M$ is the orthocenter of triangle $IKC$. Therefore,\n\n$$\n\\angle MKC = 90^\\circ - \\angle KCI = 90^\\circ - (\\angle ACB + \\frac{1}{2}\\angle BAC), \\text{ so}\n$$\n\n$$\n\\angle KNB = \\angle NKC + \\angle NCK = 90^\\circ - \\frac{1}{2}\\angle BAC.\n$$\n\nWe have $\\angle KPI = \\angle KPB + \\angle BPI = \\angle KNB + \\angle IAB = 90^\\circ - \\frac{1}{2}\\angle BAC + \\frac{1}{2}\\angle BAC = 90^\\circ$. Hence, $J$, $P$, and $K$ are collinear.\n\nWe also have $AJ \\perp AI$, $KD \\perp AI$ so $AJ \\parallel KD$, and $DJ \\perp ID$, $AK \\perp ID$ so $DJ \\parallel AK$. Therefore, $AJDK$ is a parallelogram and the line $PK$ bisects segment $AD$. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17349,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a < c < b$ be real numbers and let $f : [a, b] \\to \\mathbb{R}$ be a function continuous at $c$. Show that if $f$ is the derivative of a function $F_a$ on $[a, c)$, and also the derivative of a function $F_b$ on $(c, b]$, then $f$ is the derivative of some function $F$ on the entire interval $[a, b]$.",
    "options": [],
    "answer": "See solution",
    "solution": "The function $f$ has primitives on the interval $[a, b]$ if and only if the function $g = f + 1 - f(c)$ has primitives on this interval. So, we can assume that $f(c) = 1$. Since $f$ is continuous at $c$, there exists $[\\alpha, \\beta] \\subset [a, b]$ such that $c \\in (\\alpha, \\beta)$ and $0 < f(x) < 2$, for every $x \\in [\\alpha, \\beta]$.\n\nLet $F_a$, respectively $F_b$, be a primitive of $f$ on $[a, c)$, respectively $(c, b]$. Because $F_a$ and $F_b$ are increasing on the intervals $[\\alpha, c)$, respectively $(c, \\beta]$, it follows that the limits $\\lim_{x \\to c} F_a(x) = p$ and $\\lim_{x \\to c} F_b(x) = q$ exist. Also they are finite, because $F_a$ and $F_b$ are bounded on these intervals (for instance, if $x \\in (\\alpha, c)$, then $F_a(x) - F_a(\\alpha) = f(\\xi)(x - \\alpha)$ for some $\\xi \\in (\\alpha, x)$ and $f$ is bounded on $[\\alpha, x] \\subset [\\alpha, \\beta]$).\n\nThe function\n\n$$\nF : [a, b] \\to \\mathbb{R}, \\quad F(x) = \\begin{cases} F_a(x) - p, & \\text{if } x \\in [a, c) \\\\ 0, & \\text{if } x = c \\\\ F_b(x) - q, & \\text{if } x \\in (c, b] \\end{cases}\n$$\n\nis a primitive of $f$ on $[a, b]$.\n\n_Alternative Solution._ It is enough to show the limits $\\lim_{x \\to c} F_a(x) = p$ and $\\lim_{x \\to c} F_b(x) = q$ exist and are finite. But $|f|$ is bounded by some value $M$ in a neighborhood $(c - \\varepsilon, c + \\varepsilon)$ of $c$, since $f$ (therefore $|f|$ also) is continuous at $c$.\n\nFor $x, x' \\in (c - \\varepsilon, c)$ we will thus have $\\frac{F_a(x) - F_a(x')}{x - x'} = f(\\xi)$ at some point $\\xi \\in (x, x')$, hence $|F_a(x) - F_a(x')| < M\\varepsilon$, which implies the existence of the finite limit $\\lim_{x \\to c} F_a(x) = p$. An identical reasoning holds for $F_b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17350,
    "subject": "Mathematics (Olympiad)",
    "question": "Each term of an infinite sequence $a_1, a_2, a_3, \\dots$ is equal to $0$ or $1$. For each positive integer $n$:\n\n1. $a_n + a_{n+1} \\neq a_{n+2} + a_{n+3}$,\n2. $a_n + a_{n+1} + a_{n+2} \\neq a_{n+3} + a_{n+4} + a_{n+5}$.\n\nProve that if $a_1 = 0$, then $a_{2020} = 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, there can never be three consecutive terms $0,1,0$. For if there were, then the next term must be $0$ (else $0+1 = 0+1$), then $0$ (else $1+0 = 0+1$), then $1$ (else $0+0 = 0+0$). But then we have six consecutive terms $0,1,0,0,0,1$, a contradiction since $0+1+0 = 0+0+1$.\n\nThe same argument with $0$s and $1$s reversed shows there can never be any three consecutive terms $1,0,1$.\n\nConsider the sequence as consisting of a block of $0$s, then a block of $1$s, then a block of $0$s, and so on. If a block (other than the first) has size $1$, then we have $3$ consecutive terms $1,0,1$ or $0,1,0$, a contradiction. If a block (other than the first) has size $2$, then we have $4$ consecutive terms $0,1,1,0$ or $1,0,0,1$, a contradiction since $0+1 = 1+0$ and $1+0 = 0+1$. And if a block has size $4$ or more, then we have $4$ consecutive terms $0,0,0,0$ or $1,1,1,1$, a contradiction since $0+0 = 0+0$ and $1+1 = 1+1$.\n\nThus the first block has length at most $3$, and every subsequent block has length exactly $3$. Accordingly, for all $i$, $a_{i+3}$ is different from $a_i$. Since $2020-1=2019$ is an odd multiple of $3$, $a_1$ is different from $a_{2020}$. So $a_1 = 0$ implies $a_{2020} = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17351,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n > 1$ be a given integer and $A$ be an infinite set of positive integers satisfying: for any prime $p$ with $p \\nmid n$, there exist infinitely many elements of $A$ not divisible by $p$.\n\nProve that for any integer $m > 1$ with $(m, n) = 1$, there exists a finite subset of $A$ whose sum of elements, say $S$, satisfies $S \\equiv 1 \\pmod{m}$ and $S \\equiv 0 \\pmod{n}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose a prime $p$ satisfies $p^a \\mid m$. By the given conditions, there exists an infinite subset $A_1$ of $A$ such that $p$ is coprime to every element in $A_1$.\n\nBy the pigeonhole principle, there is an infinite subset $A_2 \\subset A_1$ such that $x \\equiv a \\pmod{mn}$ for each $x \\in A_2$, where $a$ is a positive integer and $p \\nmid a$.\n\nSince $(m, n) = 1$, we have $(p^a, \\frac{mn}{p^a}) = 1$. By the Chinese Remainder Theorem, the system\n\n$$\n\\begin{cases}\nx \\equiv a^{-1} \\pmod{p^a} \\\\\nx \\equiv 0 \\pmod{\\frac{mn}{p^a}}\n\\end{cases}\n$$\n\nhas infinitely many solutions. Take one such $x$. Define $B_p$ as the set of the first $x$ elements in $A_2$, and $S_p$ as their sum. Then $S_p \\equiv a x \\pmod{mn}$, so\n\n$$\nS_p \\equiv 1 \\pmod{p^a}, \\quad S_p \\equiv 0 \\pmod{\\frac{mn}{p^a}}.\n$$\n\nSuppose $m = p_1^{a_1} \\cdots p_k^{a_k}$. For each $p_i$ ($1 \\leq i \\leq k$), select a finite subset $B_i \\subset A \\setminus (B_1 \\cup \\cdots \\cup B_{i-1})$ such that the sum $S_{p_i}$ of its elements satisfies\n\n$$\nS_{p_i} \\equiv 1 \\pmod{p_i^{a_i}}, \\quad S_{p_i} \\equiv 0 \\pmod{\\frac{mn}{p_i^{a_i}}}.\n$$\n\nLet $B = \\bigcup_{i=1}^k B_i$ and $S = \\sum_{i=1}^k S_{p_i}$. Then $S \\equiv 1 \\pmod{p_i^{a_i}}$ for all $i$, and $S \\equiv 0 \\pmod{n}$. Thus, $B$ is the required subset.\n\nAlternatively, divide every element in $A$ by $mn$ and let the remainders which occur infinitely often be $\\alpha_1, \\alpha_2, \\dots, \\alpha_k$. We claim that\n\n$$\n(\\alpha_1, \\alpha_2, \\dots, \\alpha_k, m) = 1.\n$$\n\nOtherwise, if $p \\mid (\\alpha_1, \\alpha_2, \\dots, \\alpha_k, m)$, then $p \\nmid n$ since $(m, n) = 1$. By the given, there are infinitely many elements of $A$ not divisible by $p$, but by the definition of $\\alpha_i$, only finitely many such elements exist—a contradiction. Thus, the claim holds.\n\nConsequently, there exist integers $x_1, \\dots, x_k, y$ such that $\\alpha_1 x_1 + \\cdots + \\alpha_k x_k - y m = 1$. Choose $r$ such that $r n \\equiv 1 \\pmod{m}$. Then\n\n$$\n\\alpha_1(r n x_1) + \\cdots + \\alpha_k(r n x_k) = r n + r m n y.\n$$\n\nSelect $r x_i$ elements from $A$ with remainder $\\alpha_i$ modulo $mn$. The set of all these elements is the required subset.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17352,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $(a, b)$ of integers satisfying the equation\n\n$$\n3(a^2 + b^2) - 7(a + b) = -4.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** The given equation is equivalent to\n$$\n3(a^2 + b^2) - 7(a + b) = -4.\n$$\nRewriting,\n$$\n3(a^2 + b^2) - 7(a + b) + 4 = 0.\n$$\nOr,\n$$\n3a^2 - 7a + 3b^2 - 7b + 4 = 0.\n$$\nGroup terms:\n$$\n3a^2 - 7a + 3b^2 - 7b = -4.\n$$\nNow, complete the square for $a$ and $b$:\n\nFor $a$:\n$$\n3a^2 - 7a = 3\\left(a^2 - \\frac{7}{3}a\\right) = 3\\left[\\left(a - \\frac{7}{6}\\right)^2 - \\left(\\frac{7}{6}\\right)^2\\right] = 3\\left(a - \\frac{7}{6}\\right)^2 - \\frac{49}{12}\n$$\nSimilarly for $b$:\n$$\n3b^2 - 7b = 3\\left(b - \\frac{7}{6}\\right)^2 - \\frac{49}{12}\n$$\nSo the equation becomes:\n$$\n3\\left(a - \\frac{7}{6}\\right)^2 + 3\\left(b - \\frac{7}{6}\\right)^2 - \\frac{49}{6} + 4 = 0\n$$\n$$\n3\\left(a - \\frac{7}{6}\\right)^2 + 3\\left(b - \\frac{7}{6}\\right)^2 = \\frac{49}{6} - 4\n$$\n$$\n3\\left(a - \\frac{7}{6}\\right)^2 + 3\\left(b - \\frac{7}{6}\\right)^2 = \\frac{25}{6}\n$$\n$$\n\\left(a - \\frac{7}{6}\\right)^2 + \\left(b - \\frac{7}{6}\\right)^2 = \\frac{25}{18}\n$$\nAlternatively, as in the original solution, multiply both sides of the original equation by $3$:\n$$\n3(a^2 + b^2) - 7(a + b) = -4\n$$\n$$\n3a^2 - 7a + 3b^2 - 7b = -4\n$$\nAdd $49/12$ to both $a$ and $b$ terms as above, or, as in the original, set $x = 6a - 7$, $y = 6b - 7$.\n\nThen $a = \\frac{x + 7}{6}$, $b = \\frac{y + 7}{6}$, and the equation becomes:\n$$\n3\\left(\\left(\\frac{x + 7}{6}\\right)^2 + \\left(\\frac{y + 7}{6}\\right)^2\\right) - 7\\left(\\frac{x + 7}{6} + \\frac{y + 7}{6}\\right) = -4\n$$\nMultiply both sides by $6$:\n$$\n\\frac{3}{6}\\left((x + 7)^2 + (y + 7)^2\\right) - 7(x + 7 + y + 7) = -24\n$$\nBut as in the original, the equation reduces to\n$$\n(x)^2 + (y)^2 = 50\n$$\nSo $x$ and $y$ are integers such that $x^2 + y^2 = 50$.\n\nThe integer solutions for $x$ and $y$ are $(\\pm 7, \\pm 1)$, $(\\pm 5, \\pm 5)$, and permutations.\n\nNow, $a = \\frac{x + 7}{6}$ and $b = \\frac{y + 7}{6}$ must be integers, so $x$ and $y$ must be congruent to $-7 \\pmod{6}$.\n\nChecking possible values:\n- $x = 5$: $a = (5 + 7)/6 = 2$\n- $x = -1$: $a = (-1 + 7)/6 = 1$\n- $x = -7$: $a = (-7 + 7)/6 = 0$\n\nSo possible $(a, b)$ pairs are $(2, 2)$, $(1, 0)$, and $(0, 1)$.\n\n**Answer:** The integer solutions are $(a, b) = (2, 2), (1, 0), (0, 1)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17353,
    "subject": "Mathematics (Olympiad)",
    "question": "在 $999 \\times 999$ 的棋盤中，有些格子塗成白色，剩下的格子塗成紅色。考慮三個格子所形成的序列 $$(C_1, C_2, C_3)$$，其中 $C_1, C_2$ 在同一列、$C_2, C_3$ 在同一行，且 $C_1, C_3$ 為白色，$C_2$ 為紅色。令滿足這些條件的序列總數為 $T$。求 $T$ 的最大可能值。",
    "options": [],
    "answer": "See solution",
    "solution": "我們證明：在 $n \\times n$ 的棋盤中，最多有 $\\frac{4n^4}{27}$ 個滿足題設的序列（下稱可行序列）。\n\n設第 $i$ 列與第 $j$ 行分別有 $a_i, b_j$ 個白色方格，並設 $R$ 是紅色方格所成的集合。對於坐標為 $(i, j)$ 的紅色方格，會有 $a_i b_j$ 個可行序列 $(C_1, C_2, C_3)$，其中 $C_2 = (i, j)$。因此\n\n$$\nT = \\sum_{(i,j) \\in R} a_i b_j.\n$$\n\n利用不等式 $2ab \\le a^2 + b^2$ 可得\n\n$$\nT \\le \\frac{1}{2} \\sum_{(i,j) \\in R} (a_i^2 + b_j^2) = \\frac{1}{2} \\sum_{i=1}^{n} (n-a_i)a_i^2 + \\frac{1}{2} \\sum_{j=1}^{n} (n-b_j)b_j^2, \\quad (1)\n$$\n\n這是因為在第 $i$ 列有 $n-a_i$ 個紅色方格，在第 $j$ 行有 $n-b_j$ 個紅色方格。現在我們求 (1) 式的最大值。\n\n根據算幾不等式可得\n\n$$\n(n-x)x^2 = \\frac{1}{2}(2n-2x) \\cdot x \\cdot x \\le \\frac{1}{2}\\left(\\frac{2n}{3}\\right)^3 = \\frac{4n^3}{27},\n$$\n\n其中等號成立的充要條件是 $x = \\frac{2n}{3}$。將所有式子併在一起，得\n\n$$\nT \\le \\frac{n}{2} \\cdot \\frac{4n^3}{27} + \\frac{n}{2} \\cdot \\frac{4n^3}{27} = \\frac{4n^4}{27}.\n$$\n\n若 $n = 999$，則滿足每一列、每一行各有 $x = 666$ 個白色方格的任意著色法皆能得到上述 $T$ 的最大值，因為上面的不等式都成為等式。例如：\n\n若 $i - j \\equiv 1, 2, \\dots, 666 \\pmod{999}$，就將方格 $(i, j)$ 塗成白色，否則塗成紅色，就是一種可行的塗法。\n\n綜上所述，$T$ 可以獲得的最大值為 $\\frac{4 \\cdot 999^4}{27}$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17354,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence $a_1, a_2, a_3, \\ldots$ of positive integers is such that $a_{n+1}$ is the last digit of $a_n^n + a_{n-1}$ for all $n > 2$. Is it always true that for some $n_0$ the sequence $a_{n_0}, a_{n_0+1}, a_{n_0+2}, \\ldots$ is periodic?",
    "options": [],
    "answer": "See solution",
    "solution": "Since for $n > 2$, we actually consider the sequence mod $10$, and $\\varphi(10) = 4$, we have that the recursive formula itself has a period of $4$. Furthermore, the subsequent terms of the sequence are uniquely determined by two consecutive terms. Therefore, if there exist integers $n_0 > 2$ and $k > 0$ such that $a_{n_0} = a_{n_0+4k}$ and $a_{n_0+1} = a_{n_0+4k+1}$, then the sequence is periodic from $a_{n_0}$ on with period $4k$.\n\nConsider the pairs $(a_{2+4j}, a_{3+4j})$ for $0 \\leq j \\leq 100$. Since there are at most $100$ possible different among these, there have to exist $0 \\leq j_1 < j_2 \\leq 100$ such that $a_{2+4j_1} = a_{2+4j_2}$ and $a_{3+4j_1} = a_{3+4j_2}$. Choosing $n_0 := 2 + 4j_1$, we are done.\n\n**Remark:** Note that if $a_1, a_2$ both are between $0$ and $9$, then the recursion is invertible and the sequence can be extended to the left. By invertibility, it then follows that the original sequence is actually periodic with the choice $n_0 = 1$. Similar arguments show that, in general, $n_0$ can be chosen to be $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17355,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n$. Show that for any positive real number $x$, we have the inequality:\n\n$$\n\\frac{x^n (x^{n+1} + 1)}{x^n + 1} \\le \\left( \\frac{x+1}{2} \\right)^{2n+1}\n$$\n\nWhen does equality occur?",
    "options": [],
    "answer": "See solution",
    "solution": "The inequality can be proved by induction on $n$. Equality occurs if and only if $x = 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17356,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine whether $712! + 1$ is a prime number.",
    "options": [],
    "answer": "See solution",
    "solution": "It is composite.\n\nWe will show that $719$ is a prime factor of the given number. All congruences are considered modulo $719$.\n\nBy Wilson's theorem, $718! \\equiv -1 \\pmod{719}$. Furthermore,\n$$\n713 \\cdot 714 \\cdot 715 \\cdot 716 \\cdot 717 \\cdot 718 \\equiv (-6)(-5)(-4)(-3)(-2)(-1) \\equiv 720 \\equiv 1 \\pmod{719}.\n$$\nHence, $712! \\equiv -1 \\pmod{719}$, so $712! + 1$ is divisible by $719$. Note that $719$ is the smallest prime greater than $712$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17357,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N$ be a positive integer. Assume the integers $w_i$ for $1 \\leq i \\leq N^3$, and $x_j, y_j, z_j$ for $1 \\leq j \\leq 2N^2$, satisfy $1 \\leq w_i \\leq 2$ and $1 \\leq x_j, y_j, z_j \\leq N$. Prove that there exist integers $g(v) \\leq h(v)$ for $v \\in \\{w, x, y, z\\}$ such that\n\n$$\n\\sum_{j=g(w)}^{h(w)} w_j = \\sum_{j=g(x)}^{h(x)} x_j = \\sum_{j=g(y)}^{h(y)} y_j = \\sum_{j=g(z)}^{h(z)} z_j.\n$$\n",
    "options": [],
    "answer": "See solution",
    "solution": "We restore symmetry to the problem by supposing the existence of positive integers $a, b, c, d$ such that\n\n$$\n\\begin{aligned}\n1 \\leq x_1, \\dots, x_{bcd} \\leq a, &\\quad 1 \\leq y_1, \\dots, y_{acd} \\leq b, \\\\\n1 \\leq z_1, \\dots, z_{abd} \\leq c, &\\quad 1 \\leq w_1, \\dots, w_{abc} \\leq d.\n\\end{aligned}\n$$\n\nThe original problem follows if we take $a = b = c = N$ and $d = 2$.\n\nThere will now be no loss of generality in assuming $\\sum w_j$ to be the smallest of the four sums. For every integer $1 \\leq m \\leq abc$, there will thus exist a number $p(m)$ such that\n\n$$\n0 \\leq \\sum_{j=1}^{p(m)} x_j - \\sum_{i=1}^{m} w_i \\leq a - 1.\n$$\n\nConsider the numbers\n\n$$\n\\sum_{j=1}^{p(m)} x_j - \\sum_{i=1}^{m} w_i, \\quad 0 \\leq m \\leq abc.\n$$\n\nBecause these $abc+1$ numbers take one of $a$ possible values, the Pigeonhole Principle asserts that at least $bc+1$ of them are equal. Denote the set of these $m$ by $S_1$.\n\nNow proceed similarly for the numbers $y_j$. For every integer $m \\in S_1$, there will exist a number $q(m)$ such that\n\n$$\n0 \\leq \\sum_{j=1}^{q(m)} y_j - \\sum_{i=1}^{m} w_i \\leq b - 1.\n$$\n\nThese numbers, $bc+1$ or more, thus take on one of $b$ possible values, so at least $c+1$ of them are equal. Denote the set of these $m$ by $S_2$.\n\nFinally, with the aid of the $z$-variables, construct a set $S_3$, consisting of at least two elements $l$ and $m$, $l < m$, for which the sums\n\n$$\n0 \\leq \\sum_{j=1}^{r(m)} z_j - \\sum_{i=1}^{m} w_i \\leq c - 1.\n$$\n\nare equal.\n\nNow, $l$ and $m$ belong to $S_1$, so that\n\n$$\n\\sum_{j=1}^{p(l)} x_j - \\sum_{i=1}^{l} w_i = \\sum_{j=1}^{p(m)} x_j - \\sum_{i=1}^{m} w_i,\n$$\n\nwhich is equivalent to\n\n$$\n\\sum_{i=l+1}^{m} w_i = \\sum_{j=p(l)+1}^{p(m)} x_j.\n$$\n\nSimilarly, we show\n\n$$\n\\sum_{i=l+1}^{m} w_i = \\sum_{j=p(l)+1}^{p(m)} x_j = \\sum_{j=q(l)+1}^{q(m)} y_j = \\sum_{j=r(l)+1}^{r(m)} z_j.\n$$\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17358,
    "subject": "Mathematics (Olympiad)",
    "question": "A $9 \\times 12$ rectangle is partitioned into unit squares. The centres of all the unit squares, except for the four corner squares and the eight squares sharing a common side with one of them, are coloured red. Is it possible to label these red centres $C_1, C_2, \\dots, C_{96}$ in such a way that the following two conditions are both fulfilled:\n\n1. The distances $C_1C_2, C_2C_3, \\dots, C_{95}C_{96}, C_{96}C_1$ are all equal to $\\sqrt{13}$.\n2. The closed broken line $C_1C_2 \\dots C_{96}C_1$ has a centre of symmetry.",
    "options": [],
    "answer": "See solution",
    "solution": "Such a broken line does not exist.\n\nTo show this, color the red point squares in a chess pattern (black and white), so that every two red points at distance $1$ lie in squares of different color. It is easy to see that any two red points at distance $\\sqrt{13}$ lie on squares of different colors, so black and white alternate along the broken line. Also, the center of symmetry of the line must coincide with that of the set of points, and thus with that of the rectangle.\n\nConsider now the points $A(2,2)$ and $B(8,11)$ (as usual, the point $(i,j)$ is the center of the unit square in the $i$-th row and the $j$-th column). The line can be divided in two parts – one leading from $A$ to $B$, and the other from $B$ to $A$. If they are symmetric to each other, each of them must consist of $96 \\div 2 = 48$ edges. So an even number of edges connect $A$ to $B$, hence $A$ and $B$ must lie in squares of the same color, which is not true.\n\nSo, each part is symmetric to itself (since the symmetrical of the part leading from $A$ to $B$ can only be the other part, case dismissed above, or itself; and same for the part leading from $B$ to $A$), and each part contains an odd number of edges. Since the edges can be divided in symmetric pairs, each part must contain some edge symmetric to itself. Only two such edges are possible: one joining $(4,5)$ and $(6,8)$, and the other joining $(6,5)$ and $(4,8)$.\n\nConsider now the point $(2,2)$. It can only be joined to $(5,4)$ and $(4,5)$, so the line must include these two edges. A similar consideration for the points $(8,2)$, $(8,11)$, and $(2,11)$ shows that the line must include the edges $(4,5)-(2,2)-(5,4)-(8,2)-(6,5)-(4,8)-(2,11)-(5,9)-(8,11)-(6,8)-(4,5)$. But this is a closed broken line that does not contain all the points, a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17359,
    "subject": "Mathematics (Olympiad)",
    "question": "Нека $a$, $b$, $c$ се страните, а $\\alpha$, $\\beta$, $\\gamma$ соодветните агли во триаголникот $ABC$ со плоштина $P$. Докажи дека важи равенството\n\n$$\na^2 (\\sin 2\\beta + \\sin 2\\gamma) + b^2 (\\sin 2\\gamma + \\sin 2\\alpha) + c^2 (\\sin 2\\alpha + \\sin 2\\beta) = 12P.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Ќе ги прегрупираме собироците на левата страна од равенството во облик\n\n$$\n(a^2 \\sin 2\\beta + b^2 \\sin 2\\alpha) + (b^2 \\sin 2\\gamma + c^2 \\sin 2\\beta) + (c^2 \\sin 2\\alpha + a^2 \\sin 2\\gamma).\n$$\n\nСо примена на синусната теорема и формула за синус од двоен агол, ќе трансформираме збировите во заградите. Од $\\frac{a}{\\sin \\alpha} = \\frac{b}{\\sin \\beta}$, односно од\n\n$$\na \\sin \\beta = b \\sin \\alpha, \\text{ со замена, првиот збир добива облик}\n$$\n\n$$\n\\begin{aligned}\na^2 \\sin 2\\beta + b^2 \\sin 2\\alpha &= 2a^2 \\sin \\beta \\cos \\beta + 2b^2 \\sin \\alpha \\cos \\alpha \\\\\n&= 2ab \\sin \\alpha \\cos \\beta + 2ab \\sin \\beta \\cos \\alpha \\\\\n&= 2ab (\\sin \\alpha \\cos \\beta + \\sin \\beta \\cos \\alpha) \\\\\n&= 2ab \\sin(\\alpha + \\beta) \\\\\n&= 2ab \\sin(\\pi - \\gamma) \\\\\n&= 2ab \\sin \\gamma \\\\\n&= 4P\n\\end{aligned}\n$$\n\nАналогно, $b^2 \\sin 2\\gamma + c^2 \\sin 2\\beta = 4P$ и $c^2 \\sin 2\\alpha + a^2 \\sin 2\\gamma = 4P$. Собирајќи ги изразите се добива\n\n$$\n(a^2 \\sin 2\\beta + b^2 \\sin 2\\alpha) + (b^2 \\sin 2\\gamma + c^2 \\sin 2\\beta) + (c^2 \\sin 2\\alpha + a^2 \\sin 2\\gamma) = 12P, \\text{ што требаше да се докаже.}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17360,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose the equilateral triangle $PQR$ with side length $a$ covers points $A$, $B$, and $C$ (inside or on the boundary). What is the minimum value of $a$ such that this is possible, given that $\\angle ABC, \\angle ACB > 60^\\circ$?",
    "options": [],
    "answer": "See solution",
    "solution": "By translation, we can place one of $A$, $B$, $C$ (say $A$) on the boundary of $\\triangle PQR$. By rotation about $A$, we can place $B$ or $C$ on the boundary. If neither is a vertex, assume all lie on $PQ$ or $PR$. Take $P$ as the homothetic center and rescale $\\triangle PQR$ so all points fall on the boundary or become a vertex. Since $\\angle ABC, \\angle ACB > 60^\\circ$, $B$ and $C$ cannot be vertices; there is symmetry between $B$ and $C$. Thus, consider four cases:\n\n1. $A$ is a vertex, say $A = P$. Let $PH \\perp QR$ with foot $H$. The angle between $PH$ and one of $PB$, $PC$ is $\\leq 10^\\circ$. Thus, $PH \\geq 2 \\sin^2 80^\\circ$, so\n\n$$\na = PQ \\geq \\frac{2 \\sin^2 80^\\circ}{\\sin 60^\\circ}.\n$$\n\n2. $A$ and $B$ lie on the same side, say $PQ$, with $PA \\leq PB$. Let $PH \\perp QR$ at $H$. The angle between $AC$ and $PH$ is $10^\\circ$. Similarly,\n\n$$\na \\geq \\frac{2 \\sin^2 80^\\circ}{\\sin 60^\\circ}.\n$$\n\n3. $B$ and $C$ lie on the same side, say $QR$. Let $PH \\perp QR$ at $H$. The lines $AC$ and $PH$ cross at $10^\\circ$. Similarly,\n\n$$\na \\geq \\frac{2 \\sin^2 80^\\circ}{\\sin 60^\\circ}.\n$$\n\n4. $A$, $B$, $C$ are all on different sides. Let $A \\in PQ$, $B \\in QR$, $C \\in RP$, and $\\vartheta = \\angle RBC \\le \\angle RCB$, $\\vartheta \\le 60^\\circ$.\n\n(i) If $\\vartheta \\le 20^\\circ$, the angle between $AB$ and the altitude on $QR$ does not exceed $10^\\circ$. Thus,\n\n$$\na \\geq \\frac{2 \\sin^2 80^\\circ}{\\sin 60^\\circ}.\n$$\n\n(ii) If $40^\\circ \\le \\vartheta \\le 60^\\circ$, the angle between $AC$ and the altitude on $PQ$ does not exceed $10^\\circ$. Thus,\n\n$$\na \\geq \\frac{2 \\sin^2 80^\\circ}{\\sin 60^\\circ}.\n$$\n\n(iii) If $20^{\\circ} \\le \\vartheta \\le 40^{\\circ}$,\n\n$$\n\\begin{aligned}\na &= QR = QB + BR \\\\\n  &= \\frac{2 \\sin 80^{\\circ} \\sin(20^{\\circ} + \\vartheta)}{\\sin 60^{\\circ}} + \\frac{2 \\sin 20^{\\circ} \\sin(60^{\\circ} + \\vartheta)}{\\sin 60^{\\circ}} \\\\\n  &\\ge \\frac{2 \\sin 80^{\\circ} \\sin 40^{\\circ} + 2 \\sin 20^{\\circ} \\sin 80^{\\circ}}{\\sin 60^{\\circ}} \\\\\n  &= \\frac{2 \\sin^2 80^{\\circ}}{\\sin 60^{\\circ}}.\n\\end{aligned}\n$$\n\nIn summary, the minimum is $a = \\frac{2 \\sin^2 80^{\\circ}}{\\sin 60^{\\circ}}$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17361,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $a, b > 0$. The equation\n$$\n\\sqrt{|x|} + \\sqrt{|x+a|} = b\n$$\nfor $x$ has exactly three different real solutions, namely $x_1, x_2, x_3$, and $x_1 < x_2 < x_3 = b$. Then the value of $a+b$ is \\_\\_\\_\\_\\_.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $t = x + \\frac{a}{2}$. Then the equation\n$$\n\\sqrt{|t - \\frac{a}{2}|} + \\sqrt{|t + \\frac{a}{2}|} = b\n$$\nfor $t$ has exactly three different real solutions $t_i = x_i + \\frac{a}{2}$ ($i = 1, 2, 3$).\n\nSince $f(t) = \\sqrt{|t - \\frac{a}{2}|} + \\sqrt{|t + \\frac{a}{2}|}$ is an even function, the three real solutions of $f(t) = b$ are symmetrically distributed about the origin, so $b = f(0) = \\sqrt{2a}$. Next, we find the real solutions of $f(t) = \\sqrt{2a}$.\n\nWhen $|t| \\leq \\frac{a}{2}$,\n$$\nf(t) = \\sqrt{\\frac{a}{2} - t} + \\sqrt{\\frac{a}{2} + t} = \\sqrt{a + \\sqrt{a^2 - 4t^2}} \\leq \\sqrt{2a}\n$$\nand equality holds if and only if $t = 0$.\n\nWhen $|t| > \\frac{a}{2}$, $f(t)$ is monotonic: for $t > \\frac{a}{2}$, $f(t)$ increases, and for $t < -\\frac{a}{2}$, $f(t)$ decreases. When $t = \\frac{5a}{8}$ and $t = -\\frac{5a}{8}$, $f(t) = \\sqrt{2a}$.\n\nThus, $f(t) = \\sqrt{2a}$ has exactly three real solutions: $t_1 = -\\frac{5}{8}a$, $t_2 = 0$, $t_3 = \\frac{5}{8}a$.\n\nGiven $x_3 = b$ and $x_3 = t_3 - \\frac{a}{2} = \\frac{5a}{8} - \\frac{a}{2} = \\frac{a}{8}$, so $b = \\frac{a}{8}$. Also, $b = \\sqrt{2a}$, so $\\frac{a}{8} = \\sqrt{2a}$, which gives $a = 128$.\n\nTherefore,\n$$\na + b = 128 + 16 = 144.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17362,
    "subject": "Mathematics (Olympiad)",
    "question": "A crazy physicist discovered a new kind of particle which he called an imon, after some of them mysteriously appeared in his lab. Some pairs of imons in the lab can be entangled, and each imon can participate in many entanglement relations. The physicist has found a way to perform the following two kinds of operations with these particles, one operation at a time.\n\n1. If some imon is entangled with an odd number of other imons in the lab, then the physicist can destroy it.\n\n2. At any moment, he may double the whole family of imons in his lab by creating a copy $I'$ of each imon $I$. During this procedure, the two copies $I'$ and $J'$ become entangled if and only if the original imons $I$ and $J$ are entangled, and each copy $I'$ becomes entangled with its original imon $I$; no other entanglements occur or disappear at this moment.\n\nProve that the physicist may apply a sequence of such operations resulting in a family of imons, no two of which are entangled.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us model the imons as vertices of a graph, with an edge between two vertices if the corresponding imons are entangled. We will also color the vertices. We say a graph $G$ is a *good graph* if any two adjacent vertices have different colors.\n\n**Lemma.** Given a good graph colored with $n$ colors, it is possible to perform a sequence of operations to obtain a good graph with $n-1$ colors (for $n > 1$).\n\n**Proof.** First, repeatedly perform operation (1) until all vertices have even degree. The resulting graph is still a good graph.\n\nNext, perform operation (2). If the original imon $I$ is colored with color $k$, color its copy $I'$ with color $k+1 \\pmod n$. It is easy to see that the new graph is still a good graph.\n\nNow, in this new graph, all vertices have odd degree, so we can perform operation (1) on all imons colored with color $n$. Since these imons are not adjacent to each other (because the graph is good), repeatedly performing operation (1) will remove all imons of color $n$, resulting in a good graph with $n-1$ colors. QED.\n\nNow, suppose we start with $n$ imons; they can be colored with $n$ colors to form a good graph. By the lemma above, we can reduce the number of colors step by step until we have a single-color good graph, i.e., a graph with no edges (no entangled pairs). QED.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17363,
    "subject": "Mathematics (Olympiad)",
    "question": "Positive integers $a$ and $b$ satisfy $$(a+b)(a-b) = 2023$$ and $$\\frac{a}{b} + \\frac{b}{a} = \\frac{25}{12}$$. Find the value of $a$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $2023 = 7 \\times 17^2$, we have from the first equation $$(a+b)(a-b) = 1 \\times 2023 = 7 \\times 289 = 17 \\times 119.$$ Since $a+b > a-b$, we have the following three pairs of simultaneous equations:\n\n- $a+b=2023$ and $a-b=1$. Adding gives $a=1012$, $b=1011$, then $$\\frac{a}{b} + \\frac{b}{a} < 1 + \\frac{1}{1011} + 1 < 2.001 < \\frac{25}{12}.$$\n- $a+b=289$ and $a-b=7$. Adding gives $a=148$, $b=141$, then $$\\frac{a}{b} + \\frac{b}{a} < 1 + \\frac{7}{141} + 1 < 2.05 < \\frac{25}{12}.$$\n- $a+b=119$ and $a-b=17$. Adding gives $a=68$, $b=51$.\n\nOnly the last of these $(a, b)$ pairs satisfy the second given equation, so $a = \\mathbf{68}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17364,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{R}^+ \\to \\mathbb{R}^+$ be a function satisfying the following properties:\n\n1. $f$ is strictly increasing.\n2. $f(y) > 0$ for all $y \\in \\mathbb{R}^+$.\n3. For all $x, y \\in \\mathbb{R}^+$, $f(x + y) = f(x) + f(y)$.\n4. $f(f(x)) = x$ for all $x \\in \\mathbb{R}^+$.\n\nDetermine all such functions $f$.\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p159_data_bd94196202.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We use the given properties to determine $f$.\n\nFrom property 3, $f(x + y) = f(x) + f(y)$ for all $x, y \\in \\mathbb{R}^+$. This is Cauchy's functional equation on $\\mathbb{R}^+$, and since $f$ is strictly increasing (property 1), the only solution is $f(x) = cx$ for some $c > 0$.\n\nFrom property 4, $f(f(x)) = x$ for all $x > 0$. Substituting $f(x) = cx$ gives $f(f(x)) = f(cx) = c(cx) = c^2 x$. Setting this equal to $x$ yields $c^2 x = x$, so $c^2 = 1$, and since $c > 0$, $c = 1$.\n\nTherefore, $f(x) = x$ for all $x \\in \\mathbb{R}^+$.\n\nThis function is strictly increasing and positive for all $x > 0$, and satisfies all the given properties.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17365,
    "subject": "Mathematics (Olympiad)",
    "question": "Can the number $2010$ be written as the sum of two squares? If not, list all possible representations of $2010$ as the sum of three squares.",
    "options": [],
    "answer": "See solution",
    "solution": "The number $2010$ is a multiple of $3$ but not of $9$. The sum of two squares can be a multiple of $3$ only if both squares are multiples of $3$, which means they must also be multiples of $9$. Therefore, $2010$ cannot be written as the sum of two squares.\n\nAll possible representations of $2010$ as the sum of three squares are:\n\n$$\n\\begin{aligned}\n2010 &= 1^2 + 28^2 + 35^2 \\\\\n     &= 4^2 + 25^2 + 37^2 \\\\\n     &= 5^2 + 7^2 + 44^2 \\\\\n     &= 5^2 + 31^2 + 32^2 \\\\\n     &= 7^2 + 19^2 + 40^2 \\\\\n     &= 11^2 + 17^2 + 40^2 \\\\\n     &= 16^2 + 23^2 + 35^2 \\\\\n     &= 19^2 + 25^2 + 32^2\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17366,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ be a 20-element set of residue classes modulo 397. Prove that for any non-negative integer $n$, there exist $x_1, x_2, x_3, x_4 \\in A$ such that $x_1 \\neq x_2$ and\n\n$$\n(x_1 - x_2)n \\equiv x_3 - x_4 \\pmod{397}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that 397 is a prime. If $n$ is divisible by 397, the statement is trivial.\n\nFix an arbitrary positive integer $n$ not divisible by 397. If for $x_1, x_2, x_3, x_4 \\in A$, where $(x_1, x_4)$ and $(x_2, x_3)$ are different ordered pairs, we have\n\n$$\nx_1 n + x_4 \\equiv x_2 n + x_3 \\pmod{397},\n$$\n\nthen we are done (if $x_1 = x_2$ then $x_3 = x_4$ due to this equivalence). Otherwise, this equivalence is impossible and therefore the map $(a, b) \\mapsto an + b$ is injective from $A \\times A$ to $\\mathbb{Z}_{397}$. But this is impossible since $|A \\times A| = 400 > 397$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 17367,
    "subject": "Mathematics (Olympiad)",
    "question": "Each of the 50 students in a class sent Christmas cards to exactly 25 other students. Prove that at least two of the 50 students each received a card from the other.",
    "options": [],
    "answer": "See solution",
    "solution": "There are $\\frac{50 \\cdot 49}{2} = 1225$ distinct pairs of students. The total number of cards sent is $50 \\cdot 25 = 1250$, which is greater than the number of distinct pairs. Therefore, at least one pair of students must have sent cards to each other.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17368,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 100 participants and 25 juries in an oral examination. Each participant likes at least 10 juries.\n\n**a)** Prove that it is possible to choose 7 juries such that every participant likes at least one of these juries.\n\n**b)** Prove that there exists an arrangement such that each participant is asked by exactly one jury he likes, and each jury asks no more than 10 participants.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)**\n\nLet $A_i$ be the jury member who has the largest number of participants wanting him to ask, after removing all the students that are interested in $A_1, \\dots, A_{i-1}$. Let the number of remaining students wanting $A_i$ to ask be $a_i$. By assumption,\n\n$$\na_i \\geq \\frac{10(100 - a_1 - \\dots - a_{i-1})}{25 + 1 - i}.\n$$\n\nThis leads to\n\n$$\na_1 + a_2 + \\dots + a_n\n$$\n\n$$\n\\begin{aligned}\n&\\geq \\left[ a_1 + a_2 + \\dots + a_{n-1} + \\frac{10 \\cdot [100 - (a_1 + a_2 + \\dots + a_{n-1})]}{26-n} \\right] \\\\\n&= \\left[ \\frac{1000 + (25-n)(a_1 + a_2 + \\dots + a_{n-1})}{26-n} \\right].\n\\end{aligned}\n$$\n\nSet $a_1 + a_2 + \\dots + a_n = b_n$, hence\n\n$$\nb_n \\geq \\left\\lfloor \\frac{1000 + (25 - n)b_{n-1}}{26 - n} \\right\\rfloor.\n$$\n\nWe can compute that $b_1 \\geq 40$, $b_2 \\geq 65$, $b_3 \\geq 81$, $b_4 \\geq 90$, $b_5 \\geq 95$, $b_6 \\geq 98$, and $b_7 \\geq 100$, which means it is possible to choose 7 juries such that every participant likes at least one of them.\n\n**b)**\n\nArrange a schedule so that each jury is required to ask at most 10 participants. Suppose that after the arrangement, there exists a student $A$ who has not been asked by any examiner. Since $A$ likes at least 10 examiners, consider 10 of these juries. Each of these juries must have already asked 10 students, so there are at least $10 \\times 10 = 100$ students who have been asked. This is a contradiction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17369,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples $(x, y, z)$ of real numbers which are solutions of the system:\n\n$$\n\\begin{cases}\nx^2 + y^2 + 25z^2 = 6xz + 8yz \\\\\n3x^2 + 2y^2 + z^2 = 240\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The first equation can be rewritten as:\n\n$$\n\\begin{aligned}\nx^2 - 6xz + 9z^2 + y^2 - 8yz + 16z^2 &= 0 \\\\\n(x - 3z)^2 + (y - 4z)^2 &= 0 \\\\\nx - 3z = 0 \\quad \\text{and} \\quad y - 4z = 0 \\\\\nx = 3z \\quad \\text{and} \\quad y = 4z.\n\\end{aligned}\n$$\n\nTherefore, all triples satisfying the first equation are:\n\n$$\n(x, y, z) = (3t, 4t, t), \\quad t \\in \\mathbb{R}.\n$$\n\nSubstituting into the second equation:\n\n$$\n3(3t)^2 + 2(4t)^2 + t^2 = 240 \\\\\n3 \\cdot 9t^2 + 2 \\cdot 16t^2 + t^2 = 240 \\\\\n27t^2 + 32t^2 + t^2 = 240 \\\\\n60t^2 = 240 \\\\\nt^2 = 4 \\\\\nt = \\pm 2.\n$$\n\nFor $t = 2$, $(x, y, z) = (6, 8, 2)$. For $t = -2$, $(x, y, z) = (-6, -8, -2)$.\n\nThus, the solutions are $(6, 8, 2)$ and $(-6, -8, -2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17370,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be $n$ non-negative real numbers.\n\nProve that\n\n$$\n\\frac{1}{1+a_1} + \\frac{a_1}{(1+a_1)(1+a_2)} + \\dots + \\frac{a_1 a_2 \\dots a_{n-1}}{(1+a_1)(1+a_2)\\dots(1+a_n)} \\le 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_0 = 1$. We prove the following identity:\n\n$$\n\\sum_{k=1}^{n} \\prod_{j=1}^{k} \\frac{a_{j-1}}{1+a_j} = 1 - \\prod_{j=1}^{n} \\frac{a_j}{1+a_j}\n$$\n\nby induction on $n$.\n\nIt is evident that the identity is true for $n = 1$. Suppose it is true for $n-1$, $n \\ge 2$, then for $n$,\n\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} \\prod_{j=1}^{k} \\frac{a_{j-1}}{1+a_j} &= \\sum_{k=1}^{n-1} \\prod_{j=1}^{k} \\frac{a_{j-1}}{1+a_j} + \\prod_{j=1}^{n} \\frac{a_{j-1}}{1+a_j} \\\\\n&= 1 - \\prod_{j=1}^{n-1} \\frac{a_j}{1+a_j} + \\prod_{j=1}^{n} \\frac{a_{j-1}}{1+a_j} \\\\\n&= 1 - \\prod_{j=1}^{n} \\frac{a_j}{1+a_j}.\n\\end{aligned}\n$$\n\nSince $a_j \\ge 0$, we have $0 \\le \\frac{a_j}{1+a_j} < 1$, so $\\prod_{j=1}^{n} \\frac{a_j}{1+a_j} \\ge 0$. Thus,\n\n$$\n\\sum_{k=1}^{n} \\prod_{j=1}^{k} \\frac{a_{j-1}}{1+a_j} \\le 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17371,
    "subject": "Mathematics (Olympiad)",
    "question": "For what real values $a$ and $b$ does the maximum among $3a^2 + 2b$ and $3b^2 + 2a$ attain its minimum value?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M(a, b) = \\max\\{3a^2 + 2b,\\ 3b^2 + 2a\\}$. Then $M(a, b) \\geq 3a^2 + 2b$ and $M(a, b) \\geq 3b^2 + 2a$. Adding these, $2M(a, b) \\geq 3a^2 + 2b + 3b^2 + 2a$.\n\nWe can rewrite:\n\n$$\n\\frac{2}{3}M(a, b) + \\frac{2}{9} \\geq \\left(a + \\frac{1}{3}\\right)^2 + \\left(b + \\frac{1}{3}\\right)^2 \\geq 0\n$$\n\nThus, $M(a, b) \\geq -\\frac{1}{3}$, and equality holds when $a = b = -\\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17372,
    "subject": "Mathematics (Olympiad)",
    "question": "A rectangular sheet of grid paper with dimensions $59 \\times 133$ must be divided into a maximum number of pieces by making two cuts. Before cutting, it is allowed to fold the sheet along grid lines as many times as desired. The folded sheet is placed on the table and then cut twice along grid lines that are visible after the folding. What is the maximum number of pieces that can be obtained in this way?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $l$ and $m$ be grid lines in the initial rectangle that are cut by the first cut. There was a folding that made them coincide, hence $l \\parallel m$ (a folding cannot make perpendicular lines coincide). Also, $l$ and $m$ are separated by at least one line of folding which is intact. So the grid lines cut by one cut are all parallel, interior, and no two of them are adjacent. Let their direction be perpendicular to a side of odd length $d$. Of the $d-1$ interior grid lines in this direction, at most $\\frac{d-1}{2}$ can be cut. So the cut yields at most $\\frac{d-1}{2} + 1 = \\frac{d+1}{2}$ pieces. For $d = 59$ we obtain $29$ lines and $30$ pieces. The same reasoning applies to the second cut. It must be made in the other direction to reach a maximum number of parts. At most $\\frac{133-1}{2} = 66$ grid lines can be cut, so each piece from the first cut decomposes into at most $67$ parts. Thus, the total number of parts does not exceed $30 \\times 67 = 2010$.\n\nFor an example with $2010$ parts, label $1, 2, \\ldots, 132$ the interior grid lines parallel to side $59$. Fold along every line with odd label in a bandoneón-like fashion to obtain a rectangle $2 \\times 59$. Do the same in the other direction. Flattening down gives a $2 \\times 2$ square whose middle lines contain all grid lines that are not lines of folding. Two cuts along the middle lines yield $(66+1)(29+1) = 2010$ parts.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17373,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose we have three natural numbers $a$, $b$, and $c$ such that one of the values\n\n$$\n\\gcd(a, b) \\cdot \\mathrm{lcm}(b, c), \\quad \\gcd(b, c) \\cdot \\mathrm{lcm}(c, a), \\quad \\gcd(c, a) \\cdot \\mathrm{lcm}(a, b)\n$$\n\nis equal to the product of the other two. Prove that one of the numbers $a$, $b$, or $c$ is a multiple of a different one.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume that\n\n$$\nP = \\gcd(a, b) \\cdot \\mathrm{lcm}(b, c) = \\gcd(b, c) \\cdot \\mathrm{lcm}(c, a) \\cdot \\gcd(c, a) \\cdot \\mathrm{lcm}(a, b).\n$$\n\nUsing the relation $\\gcd(x, y) \\cdot \\mathrm{lcm}(x, y) = xy$, we get:\n\n$$\n(abc)^2 = (ab) \\cdot (bc) \\cdot (ca) = \\gcd(a, b) \\cdot \\mathrm{lcm}(a, b) \\cdot \\gcd(b, c) \\cdot \\mathrm{lcm}(b, c) \\cdot \\gcd(c, a) \\cdot \\mathrm{lcm}(c, a) = P^2,\n$$\n\nwhich gives $abc = P$ after taking square roots. Using the same relation again:\n\n$$\n\\gcd(a, b) \\cdot \\mathrm{lcm}(b, c) = P = a \\cdot (bc) = a \\cdot \\gcd(b, c) \\cdot \\mathrm{lcm}(b, c),\n$$\n\nwhich simplifies to $\\gcd(a, b) = a \\cdot \\gcd(b, c)$. Since the greatest common divisor is a divisor, $a \\mid \\gcd(a, b) \\mid b$, so $b$ is a multiple of $a$, as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17374,
    "subject": "Mathematics (Olympiad)",
    "question": "When you line up 4020 stones, 2010 of them white and 2010 black, along a horizontal straight line, how many ways are there to arrange them so that the number of pairs of stones with a white stone lying to the right of a black stone is odd?",
    "options": [],
    "answer": "See solution",
    "solution": "Let us consider two types of configurations:\n\n1. For each $i$, the stones at positions $(2i - 1)$ and $2i$ are of the same color.\n2. There exists an $i$ such that the stones at positions $(2i - 1)$ and $2i$ have different colors.\n\nThe total number of ways to arrange 4020 stones (2010 white, 2010 black) is $${4020 \textbf{C}_{2010}}$$. The number of ways for type (1) is $${2010 \textbf{C}_{1005}}$$, so type (2) has $${4020 \textbf{C}_{2010} - 2010 \textbf{C}_{1005}}$$ ways.\n\nNone of the type (1) arrangements satisfy the requirement, because the number of pairs with a white stone to the right of a black stone is always even. For type (2), exactly half of the arrangements have an odd number of such pairs. This is because for each configuration A of type (2), swapping the first pair of consecutive stones of different colors yields another configuration B, and exactly one of A or B has an odd number of such pairs.\n\nTherefore, the answer is:\n\n$$\n\\frac{4020 \\textbf{C}_{2010} - 2010 \\textbf{C}_{1005}}{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17375,
    "subject": "Mathematics (Olympiad)",
    "question": "A hexagonal table containing 2012 columns is given, as in the drawing. Each of the odd columns contains 2012 hexagons, and each of the even ones contains 2013 hexagons. In each hexagon in the $i$-column the number $i$ is written. The allowed changes of the numbers in the table are the following: we pick three neighboring hexagons, we rotate the numbers, and: if the rotation is clockwise then we decrease the three numbers by $1$, and if the rotation is counterclockwise then we increase the numbers by $1$ (see drawing). At most how many zeroes can be obtained in the table using the above-defined steps?\n\n![](images/makedonija2012_p16_data_be6cb0c27d.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The sum of all numbers in the table is congruent to $2$ modulo $3$. With each step, this sum changes by $+3$ or $-3$, so it always leaves a remainder of $2$ when divided by $3$.\n\nIf we have a figure as in the right drawing, we can obtain a zero after finitely many steps in the place of the number $d$ as follows. If $d$, $c + 1$, or $b - 1$ is divisible by $3$, we do it by rotating the numbers $b$, $c$, and $d$. If $b$ is divisible by $3$:\n\n![](images/makedonija2012_p16_data_f35cd34c38.png)\n![](images/makedonija2012_p16_data_a694aa256b.png)\n\nIn this way, in the place of the number $c$ we get a number with residue $-1$ modulo $3$ and implement the first case. If $b + 1$ is divisible by $3$:\n\nIn this way, in the place of $c$ we get a number with residue $-1$ modulo $3$ and implement the first case. With this procedure, we can make all numbers equal $0$ except three of them, e.g., the second and third of the second column and the second from the third column. The procedure for this is the following (see drawing), where the numbers give the order in which we obtain the zeroes in the hexagons.\n\n![](images/makedonija2012_p16_data_0a7a420f07.png)\n![](images/makedonija2012_p16_data_b529c9271f.png)\n\nWith the following sequence of steps, we can decrease an arbitrary number from these three numbers by $3$, or analogously, increase it by $3$ with the reverse sequence of steps.\n\n![](images/makedonija2012_p16_data_4766cbd084.png)\n\n![](images/makedonija2012_p17_data_338735fd74.png)\n\nSymmetrically, we can increase $c$ by $2$ and $e$ by $1$. (We need to implement the above\n\n![](images/makedonija2012_p17_data_6b687651b9.png)\n\nprocedure for the figure:\n\nAfter three clockwise rotations we get:\n\n![](images/makedonija2012_p17_data_93bb1029ec.png)\n\nThis means that the remaining three numbers can be reduced to $0$, $1$, or $2$. Since their sum equals $2$ modulo $3$, it follows that two of them must be equal (they are either two $0$'s and one $2$, or two $1$'s and one $0$, or two $2$'s and one $1$). In each of these cases, the two that are equal can be reduced to $0$ and the remaining one to $2$. Therefore, all the elements of the table except one can be made to equal $0$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17376,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be real numbers different from $0$, and let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function defined by\n$$\nf(x) = \\begin{cases}\n  ax, & x \\in \\mathbb{Q} \\\\\n  bx, & x \\in \\mathbb{R} \\setminus \\mathbb{Q}\n\\end{cases}\n$$\nProve that $f$ is injective if and only if $f$ is surjective.",
    "options": [],
    "answer": "See solution",
    "solution": "The function $f$ is injective if and only if $\\frac{a}{b} \\in \\mathbb{Q}$.\n\nSuppose $f$ is injective and $\\frac{a}{b} \\in \\mathbb{R} \\setminus \\mathbb{Q}$. Then $f\\left(\\frac{a}{b}\\right) = a = f(1)$, a contradiction.\n\nSuppose now that $\\frac{a}{b} \\in \\mathbb{Q}$ and let $x_1, x_2$ such that $f(x_1) = f(x_2)$. If $x_1, x_2 \\in \\mathbb{R} \\setminus \\mathbb{Q}$ or $x_1, x_2 \\in \\mathbb{Q}$, then $x_1 = x_2$. If $x_1 \\in \\mathbb{R} \\setminus \\mathbb{Q}$ and $x_2 \\in \\mathbb{Q}$, then $ax_2 = bx_1$ and consequently $\\frac{a}{b} = \\frac{x_1}{x_2} \\in \\mathbb{R} \\setminus \\mathbb{Q}$, a contradiction. Hence $f$ is injective.\n\nThe function $f$ is surjective if and only if $\\frac{a}{b} \\in \\mathbb{Q}$.\n\nSuppose $f$ is surjective and let $x$ with $f(x) = b$. If $x \\in \\mathbb{R} \\setminus \\mathbb{Q}$ then $x = 1$, a contradiction. Therefore $x$ is rational and so $ax = b$, implying $\\frac{a}{b} = \\frac{1}{x} \\in \\mathbb{Q}$. Conversely, suppose $\\frac{a}{b} \\in \\mathbb{Q}$ and let $y \\in \\mathbb{R}$. Since $\\frac{y}{a} \\in \\mathbb{R} \\setminus \\mathbb{Q}$ and $\\frac{y}{b} \\in \\mathbb{Q}$ cannot occur simultaneously, we have either $f\\left(\\frac{y}{a}\\right) = y$ or $f\\left(\\frac{y}{b}\\right) = y$. Hence $f$ is surjective.\n\nThe claim now follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17377,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the collection of lines of the form $y = (k+n)x + (k-n)$ on a plane, where $k$ and $n$ are any integers. Is there a point with integer coordinates that doesn't belong to any of these lines?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = 1$. Then, for any line in the collection, the $y$-coordinate is $y = (k+n) \times 1 + (k-n) = 2k$. Thus, all such points have even $y$-coordinates. Therefore, the point $(1, 1)$ does not belong to any of these lines.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17378,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle APB$ have $AP = 3$, $BP = 7$, and $\\angle APB = \\theta$. Let $S$ be the area of $\\triangle APB$. Find the minimum value of $AP \\times BP$ given that $S = \\frac{21}{2}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The area is given by $S = \\frac{1}{2} \\times AP \\times BP \\times \\sin \\theta = \\frac{21}{2}$. Since $\\sin \\theta$ is positive, $AP \\times BP$ is minimized when $\\sin \\theta$ is maximized, i.e., $\\sin \\theta = 1$. Thus, $AP \\times BP = 21$ is the minimum value.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17379,
    "subject": "Mathematics (Olympiad)",
    "question": "$\\angle AKI = \\angle CLI = \\angle AIC$, $\\angle AKM = \\angle ICA$, $\\angle CLN = \\angle IAC$. Prove that the radii of the circumcircles of triangles $KIL$ and $ABC$ are equal.\n\n![](images/Ukraine_booklet_2018_p49_data_961e2cfe4c.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that lines $AI$ and $CI$ meet the circumcircle of $\\triangle ABC$ (for the second time) at points $W_A$ and $W_C$, respectively. Since $\\angle AKI = \\angle AIC$, the circumcircle of triangle $AKI$ is tangent to the line $W_C I$. Consider triangle $AW_C I$. By a well-known fact, this triangle is isosceles. Therefore, the circumcircle of $\\triangle AKI$ is also tangent to $W_C A$. Thus, a symmedian of $\\triangle AKI$ lies on the line $W_C K$. Then $\\angle W_C KI = 180^\\circ - \\angle ICA = 180^\\circ - \\angle W_C W_A I$. This means that $K$ lies on the circumcircle of $\\triangle W_C W_A I$. Similarly, $L$ also lies on this circumcircle. Note that $\\triangle W_C B W_A \\cong \\triangle W_C I W_A$ (they have equal sides), so the radius of the circumcircle of $\\triangle KIL$ equals the radius of the circumcircle of $\\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17380,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Consider sequences $a_0, a_1, \\dots, a_n$ such that $a_i \\in \\{1, 2, \\dots, n\\}$ for each $i$ and $a_n = a_0$.\n\n(a) Call such a sequence *good* if for all $i = 1, 2, \\dots, n$, $a_i - a_{i-1} \\not\\equiv i \\pmod n$. Suppose that $n$ is odd. Find the number of good sequences.\n\n(b) Call such a sequence *great* if for all $i = 1, 2, \\dots, n$, $a_i - a_{i-1} \\not\\equiv i, 2i \\pmod n$. Suppose that $n$ is an odd prime. Find the number of great sequences.",
    "options": [],
    "answer": "See solution",
    "solution": "**First Solution:** The answer is $(n-1)^n - (n-1)$ for part (a) and $(n-1)((n-2)^{n-1} - 1)$ for part (b).\n\n(a) Observe that the number of good sequences is the same for any choice of $a_0$. For fixed $a_0$, call the condition $a_i - a_{i-1} \\not\\equiv i \\pmod n$ *condition (i)*. Now let $S_i$ be the set of sequences $A = \\{a_i\\}_{i=1}^n, a_i \\in \\{1, 2, \\dots, n\\}$ such that $A$ satisfies conditions (1), (2), \\dots, $(n-i)$ and fails to satisfy conditions $(n-i+2)$, $(n-i+3)$, \\dots, $(n)$. Note that there is no constraint on whether or not $A$ satisfies condition $(n-i+1)$. Finally, let $F$ be the set of sequences that fails all conditions (1), (2), \\dots, $(n)$. Then we claim that the number of good sequences starting with $a_0$ is\n\n$$\nm = |S_1| - |S_2| + |S_3| - \\dots + |S_n| - |F|.\n$$\n\nConsider the following table, representing the conditions on sequences in $S_i$.\n\n$$\n\\begin{array}{c|ccccccc}\nS_1 & (1) & (2) & (3) & \\dots & (n-2) & (n-1) & \\\\\n\\hline\nS_2 & (1) & (2) & (3) & \\dots & (n-2) & \\overline{(n)} & \\\\\n\\hline\nS_3 & (1) & (2) & (3) & \\dots & \\overline{(n-1)} & \\overline{(n)} & \\\\\n\\hline\n\\vdots & \\vdots & \\vdots & \\vdots & \\dots & \\vdots & \\vdots & \\vdots \\\\\n\\hline\nS_n & \\overline{(2)} & \\overline{(3)} & \\dots & \\overline{(n-2)} & \\overline{(n-1)} & \\overline{(n)} & \\\\\n\\hline\nF & (1) & (2) & (3) & \\dots & \\overline{(n-2)} & \\overline{(n-1)} & \\overline{(n)}\n\\end{array}\n$$\n\nConditions that appear as $(j)$ in row $i$ are satisfied by sequences in $S_i$, and those that appear as $\\overline{(j)}$ are failed.\n\nNow take any sequence $A$. If $A$ is good, then $A$ is counted exactly once in $|S_1|$ and is not counted anywhere else. Otherwise, if $A$ is not good, let (i) be the first condition that $A$ fails. If $i > 1$, then $A$ belongs to exactly the sets $S_{n-1}, S_{n-2}$. As one of $|S_{n-1}|, |S_{n-2}|$ is added and one is subtracted, $A$ has no net contribution to $m$. Finally, if $i = 1$, then $A$ belongs to $S_n$ and $F$, so again $A$ does not contribute to $m$. Therefore, $m$ is indeed the number of good sequences.\n\nWe are left with computing the sizes of the $S_i$'s and $F$. This is easy, as the failure of the final $i-1$ conditions determines the terms $a_{n-1}, a_{n-2}, \\dots, a_{n-i+1}$, whereas there are just $n-1$ ways to satisfy each of the conditions (1), (2), \\dots, $(n-i)$, producing $(n-1)^{n-i}$ subsequences $a_0, a_1, \\dots, a_{n-i}$. As there is no constraint on $a_{n-i+1} - a_{n-i}$, we have found all sequences in $S_i$. Hence, $|S_i| = (n-1)^{n-i}$. As for $F$, there is exactly one way to fail every single condition, because $1+2+\\dots+(n-1) = \\frac{n(n-1)}{2}$ is divisible by $n$. Therefore, $|S_n| - F = 0$ and\n\n$$\n\\begin{align*}\nm &= |S_1| - |S_2| + \\dots - |S_{n-1}| \\\\\n&= (n-1)^{n-1} - (n-1)^{n-2} + \\dots - (n-1) \\\\\n&= \\frac{(n-1)^n - (n-1)}{n}.\n\\end{align*}\n$$\n\nAs there were originally $n$ choices for $a_0$, multiplying by $n$ gives the desired answer.\n\n(b) We can use the same argument as in part (a), but we must recompute the values of $|S_i|$ and $|F|$. Note that for $1 \\le i \\le n-1$, there are exactly two ways to fail condition (i) (because $n$ is odd) while condition (n) is degenerate and reduces to the single restriction $a_n - a_{n-1} \\not\\equiv 0 \\pmod{n}$. Hence, we have\n\n$$\n|S_1| = (n-2)^{n-1} \\quad \\text{and} \\quad |S_i| = 2^{i-2}(n-2)^{n-i},\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17381,
    "subject": "Mathematics (Olympiad)",
    "question": "Define $f(A) = \\max(A)$ when $A$ is finite. We say two subsets $A, B$ of $\\mathbb{N}$ are equivalent if $A \\Delta B$ is finite. This is an equivalence relation. By the Axiom of Choice, select an element $S_A$ from each equivalence class. For any proper subset $A$, let $S_A$ be the selected element from the class of $A$, so $A \\Delta S_A$ is finite. Define $f(A) = \\max(A \\Delta S_A)$ when $A \\neq S_A$, and define $f(S_A)$ arbitrarily. Prove that this function is an $A$-predicator for all subsets $A \\subseteq \\mathbb{N}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x$ be a natural number such that $x \\notin A$. Since $A$ and $A \\cup \\{x\\}$ are equivalent, $S_A = S_{A \\cup \\{x\\}}$. Thus,\n\n$$\nf(A \\cup \\{x\\}) = f((A \\cup \\{x\\}) \\Delta S_A) = \\max((A \\Delta S_A) \\Delta \\{x\\}).\n$$\n\nFor $x > \\max(A \\Delta S_A)$, we have $(A \\Delta S_A) \\Delta \\{x\\} = (A \\Delta S_A) \\cup \\{x\\}$, so $f(A \\cup \\{x\\}) = x$. Therefore, the function $f$ is an $A$-predicator for all subsets $A \\subseteq \\mathbb{N}$. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17382,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle AB\\Gamma$ be an acute-angled triangle with circumcircle $c(O, R)$. From the midpoint $\\Delta$ of the side $B\\Gamma$, we draw a line perpendicular to $AB$ which meets $AB$ at $E$. If the line $AO$ intersects the line $\\varepsilon$ at $Z$, prove that the points $A$, $Z$, $\\Delta$, $\\Gamma$ are cyclic.\n\n![](images/Greece-IMO2019finalbook_p12_data_afb72e9246.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The external angle $\\mathrm{E}\\hat{Z}A$ of the quadrilateral $AZ\\Delta\\Gamma$ belongs to the right triangle $AEZ$, with the acute angle $\\mathrm{E}\\hat{A}Z = \\omega$ equal to the angle $A\\hat{B}O$, since $OA = OB$. Hence, $\\mathrm{E}\\hat{Z}A = 90^\\circ - \\omega$.\n\nLet the extension of the radius $BO$ intersect the circle $c(O, R)$ at $H$. Then $A\\hat{\\Gamma}H = A\\hat{B}H = \\omega$ and\n$$90^\\circ = B\\hat{\\Gamma}H = B\\hat{\\Gamma}A + A\\hat{\\Gamma}H \\Rightarrow B\\hat{\\Gamma}A = 90^\\circ - A\\hat{\\Gamma}H = 90^\\circ - \\omega.$$ \n\nHence $\\mathrm{E}\\hat{Z}A = B\\hat{\\Gamma}A$, and the quadrilateral $AZ\\Delta\\Gamma$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17383,
    "subject": "Mathematics (Olympiad)",
    "question": "We know that a certain number $A$ has exactly 2018 positive integer factors (including 1 and $A$ itself), and it is divisible by 2018. Prove that $A$ is not divisible by $2018^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A = p_1^{k_1} p_2^{k_2} \\dots p_n^{k_n}$ be the canonical prime factorization of $A$, where $p_i$ are distinct primes and $k_i$ are positive integers for $i = 1, \\dots, n$. The total number of positive integer factors of $A$ is\n\n$$\n(k_1 + 1)(k_2 + 1)\\dots(k_n + 1) = 2018.\n$$\n\nSince each $(k_i + 1) \\geq 2$, there are only two cases:\n\n1. $k_1 + 1 = 2018$.\n2. $(k_1 + 1)(k_2 + 1) = 2 \\cdot 1009$.\n\nThere are no other possibilities, since 2018 has only two prime factors.\n\nIn the first case, $A = p_1^{2017}$, but then $A$ cannot be divisible by both 2 and 1009, so this is impossible.\n\nThus, $(k_1 + 1)(k_2 + 1) = 2 \\cdot 1009$, so (without loss of generality) $k_1 = 1$, $k_2 = 1008$. Therefore, $A = p_1 p_2^{1008}$. Since $A$ is divisible by both 2 and 1009, $A = 2 \\cdot 1009^{1008}$ or $A = 1009 \\cdot 2^{1008}$.\n\n![](images/Ukraine_booklet_2018_p13_data_254230e1a4.png)\n\nHowever, neither of these numbers is divisible by $2018^2$. Q.E.D.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17384,
    "subject": "Mathematics (Olympiad)",
    "question": "For any positive real numbers $a$, $b$, $c$, and $d$, prove that\n\n$$\n\\frac{(a+b)^2 + (c+d)^2}{(a^2 + c^2)(b + d) + (a + c)(b^2 + d^2)} \\leq \\frac{1}{a + c} + \\frac{1}{b + d}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "It suffices to prove\n\n$$\n\\left( \\frac{a^2 + c^2}{a + c} + \\frac{b^2 + d^2}{b + d} \\right) (a + b + c + d) \\geq (a + b)^2 + (c + d)^2.\n$$\n\nBy the Cauchy-Schwarz and Minkowski inequalities, we have\n\n$$\n\\left( \\frac{a^2 + c^2}{a + c} + \\frac{b^2 + d^2}{b + d} \\right) (a + b + c + d) \\geq \\left( \\sqrt{a^2 + c^2} + \\sqrt{b^2 + d^2} \\right)^2 \\geq (a + b)^2 + (c + d)^2.\n$$\n\nIt is easy to check that equality holds if and only if $ad = bc$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17385,
    "subject": "Mathematics (Olympiad)",
    "question": "Applying $R$ followed by $L$ to $\\frac{p}{q}$, we get $\\frac{p+q}{p+2q}$. We now do this 5 times starting with $1 = 1/1$.\n\n$$\n\\begin{align*}\nL(R(1/1)) &= 2/3 \\\\\nL(R(2/3)) &= 5/8 \\\\\nL(R(5/8)) &= 13/21 \\\\\nL(R(13/21)) &= 34/55 \\\\\nL(R(34/55)) &= 89/144\n\\end{align*}\n$$\n\nSince 89 is prime, what is the required number?",
    "options": [],
    "answer": "See solution",
    "solution": "Starting with $1 = 1/1$, we get:\n\n$$\n\\begin{array}{ll}\nR(1/1) = 2/1 & L(2/1) = 2/3 \\\\\nR(2/3) = 5/3 & L(5/3) = 5/8 \\\\\nR(5/8) = 13/8 & L(13/8) = 13/21 \\\\\nR(13/21) = 34/21 & L(34/21) = 34/55 \\\\\nR(34/55) = 89/55 & L(89/55) = 89/144\n\\end{array}\n$$\n\nSince 89 is prime, the required number is $89 + 144 = 233$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17386,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, point $P$ divides side $AB$ in the ratio $\\frac{AP}{PB} = \\frac{1}{4}$. The perpendicular bisector of segment $PB$ intersects side $BC$ at point $Q$. If $\\text{area}(PQC) = \\frac{4}{25}\\,\\text{area}(ABC)$ and $AC = 7$, find $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $\\text{area}(ABC) = S$, then $\\text{area}(APC) = \\frac{AP}{AB} S = \\frac{1}{5} S$. As $\\text{area}(PQC) = \\frac{4}{25} S$, we have $\\text{area}(PQB) = S - \\frac{1}{5} S - \\frac{4}{25} S = \\frac{16}{25} S$.\n\nOn the other hand,\n\n$$\n\\text{area}(PBQ) = \\frac{BQ}{BC} \\cdot \\text{area}(PBC) = \\frac{BQ}{BC} \\cdot \\frac{BP}{BA} \\cdot S = \\frac{BQ}{BC} \\cdot \\frac{4}{5} S.\n$$\n\nHence, $\\frac{16}{25} S = \\frac{BQ}{BC} \\cdot \\frac{4}{5} S$, which implies $\\frac{BQ}{BC} = \\frac{4}{5}$. Because $\\frac{BP}{BA} = \\frac{4}{5}$, it follows that $PQ \\parallel AC$, so triangles $ABC$ and $PBQ$ are similar. But $PQ = BQ$ as $Q$ belongs to the perpendicular bisector of $PB$. Therefore, $BC = AC = 7$.\n\n![](images/Argentina_2011_p11_data_6fe1653adf.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17387,
    "subject": "Mathematics (Olympiad)",
    "question": "We consider permutations $f$ on the set $N$ of non-negative integers, i.e., bijective mappings $f$ from $N$ to $N$, with the following properties:\n\nFor all $x \\in N$, we have $f(f(x)) = x$ and $|f(x) - x| \\leq 3$.\n\nFurthermore, for all integers $n > 42$, we have\n\n$$\nM(n) = \\frac{1}{n+1} \\sum_{j=0}^{n} |f(j) - j| < 2.011\n$$\n\nProve that there exist infinitely many integers $K$ such that $f$ maps the set $\\{n \\mid 0 \\leq n \\leq K\\}$ onto itself.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there do not exist infinitely many such $K$. Then there must exist some $K_0$ such that for all $K > K_0$, there exists an $n$ with $n \\leq K < f(n)$. Since $|f(x) - x| \\leq 3$ must always hold, such an $n$ can only be $K$, $K-1$, or $K-2$.\n\nThe same must hold for $K = f(n)$, and so on. This means that, from some $K_0$ on, the function must map “up” into each interval $[n, f(n)]$, and also “up” out of each such interval. In order for this to be possible, we must have $f(n) - n \\geq 3$, and it therefore follows that $|f(n) - n| = 3$ must hold for all $n > K > K_0$. If this is the case, we have\n\n$$\nM(n) = \\frac{1}{n+1} \\sum_{j=0}^{n} |f(j) - j| = \\frac{1}{n+1} \\left( \\sum_{j=0}^{K} |f(j) - j| + 3(n-K) \\right) = 3 - \\frac{C}{n+1}\n$$\n\nwhere\n\n$$\nC = 3K + 3 - \\sum_{j=0}^{K} |f(j) - j| = \\sum_{j=0}^{K} (3 - |f(j) - j|) \\geq 0.\n$$\n\nIt therefore follows that there exists a $K_1$ such that $M(n) = 3 - \\frac{C}{n+1} > 2.011$ for $n > K_1$, which contradicts the assumption $M(n) < 2.011$. This is not possible, and we see that an infinite number of $K$ with the required properties exist, as claimed. QED.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 17388,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\alpha = \\frac{1+\\sqrt{5}}{2}$. Find all continuous functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that, for all $x, y, z \\in \\mathbb{R}$,\n\n$$\n\\begin{aligned}\n& f(\\alpha x + y) + f(\\alpha y + z) + f(\\alpha z + x) \\\\\n&= \\alpha f(x + y + z) + 2f(x) + 2f(y) + 2f(z).\n\\end{aligned}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $(*)$ be the given functional equation.\n\nSetting $x = y = z = 0$ in $(*)$, we have $f(0) = 0$.\n\nSetting $y = z = 0$ in $(*)$ and simplifying, we have\n\n$$\nf(\\alpha x) = \\alpha^2 f(x) \\text{ for all } x \\in \\mathbb{R}.\n$$\n\nLetting $z = 0$ in the functional equation, we get\n\n$$\nf(\\alpha x + y) + f(\\alpha y) + f(x) = \\alpha f(x + y) + 2f(x) + 2f(y).\n$$\n\nUsing $f(\\alpha y) = \\alpha^2 f(y) = (\\alpha + 1)f(y)$, it follows that\n\n$$\nf(\\alpha x + y) = \\alpha f(x + y) + f(x) + (1 - \\alpha)f(y).\n$$\n\nThen $(*)$ simplifies to\n\n$$\nf(x + y) + f(y + z) + f(z + x) = f(x + y + z) + f(x) + f(y) + f(z).\n$$\n\nPutting $z = -y$ in the above equation, we have\n\n$$\nf(x + y) + f(x - y) = 2f(x) + f(y) + f(-y).\n$$\n\nLet $f_c(x) = \\frac{f(x) + f(-x)}{2}$ and $f_o(x) = \\frac{f(x) - f(-x)}{2}$. Then\n\n$$\nf_c(x + y) + f_c(x - y) = 2f_c(x) + 2f_c(y)\n$$\n\n$$\nf_o(x + y) + f_o(x - y) = 2f_o(x)\n$$\n\nwhich respectively are the quadratic and the Jensen functional equations with the continuous solutions $f_c(x) = ax^2$ and\n\n$$\nf_o(x) = bx \\text{ (note that } f_o(0) = 0\\text{). Thus, } f(x) = ax^2 + bx.\n$$\n\nSubstituting $f(x) = ax^2 + bx$ into $(*)$, we can see that $b = 0$.\n\nThus $f(x) = ax^2$ is the only continuous solution.\n\n![](images/Thailand-2007_Booklet_p9_data_d7390d2e7c.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17389,
    "subject": "Mathematics (Olympiad)",
    "question": "Positive integers $m$ and $n$ are such that the numbers\n$$\n\\frac{m^2 + 2n}{n^2 - 2m} \\quad \\text{and} \\quad \\frac{n^2 + 2m}{m^2 - 2n}\n$$\nare integers.\n\n(a) Show that $|m-n| \\le 2$.\n\n(b) Find all pairs $(m, n)$ fulfilling the hypothesis.",
    "options": [],
    "answer": "See solution",
    "solution": "a) $m^2 - 2n$ divides $n^2 + 2m > 0$, hence $m^2 - 2n \\le n^2 + 2m$, that is $(m-1)^2 \\le (n+1)^2$, so $m \\le n + 2$. In the same way, $n \\le m + 2$; consequently $|m-n| \\le 2$.\n\nb) It is enough to consider only the case $n \\ge m$, so $n \\in \\{m, m+1, m+2\\}$.\n\n*Case 1*: $n = m$. Then\n$$\n\\frac{n+2}{n-2} = 1 + \\frac{4}{n-2}\n$$\nis an integer, whence $n \\in \\{1, 3, 4, 6\\}$. We get the pairs $(n, m) \\in \\{(1, 1), (3, 3), (4, 4), (6, 6)\\}$.\n\n*Case 2*: $n = m + 1$. Then\n$$\n\\frac{m^2+2n}{n^2-2m} = 1 + \\frac{2m+1}{m^2+1}\n$$\nand\n\n- if $m \\ge 3$, then $0 < \\frac{2m+1}{m^2+1} < 1$, so $\\frac{2m+1}{m^2+1}$ is not an integer;\n- if $m = 1$, then $\\frac{2m+1}{m^2+1} = \\frac{3}{2}$ is not an integer;\n- if $m = 2$ ($n = 3$) then $\\frac{n^2+2m}{m^2-2n} = \\frac{13}{-2}$ is not an integer.\n\nTherefore, there are no solutions in this case.\n\n*Case 3*: $n = m + 2$. Then $\\frac{m^2+2n}{n^2-2m} = 1$ and\n$$\n\\frac{n^2+2m}{m^2-2n} = 1 + \\frac{8m+8}{m^2-2m-4}\n$$\nmust be an integer, so $m^2 - 2m - 4 \\le 8m + 8$, that is $m(m - 10) \\le 12$, whence $m \\le 11$.\n\nChecking the remaining possibilities for this case we get the pairs $(2, 4), (3, 5), (4, 6)$.\n\nSo, the final answer is\n$$(m, n) \\in \\{(1, 1), (3, 3), (4, 4), (6, 6), (2, 4), (4, 2), (3, 5), (5, 3), (4, 6), (6, 4)\\}.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17390,
    "subject": "Mathematics (Olympiad)",
    "question": "愛麗絲被告知正整數 $n$ 與 $k$，其中 $n \\ge 2k+1$。有 $1$ 到 $n$ 號的牌各一張，隨機洗混後背面朝下組成牌庫。每回合，愛麗絲依序進行以下動作：\n\n1. 她翻開牌庫頂的一張牌，面朝上放在桌面上。\n2. 接著，如果她還沒在任何牌上簽名，此時她可以在最新翻開的那張牌上簽名（或不簽）。\n\n遊戲在 $2k+1$ 回合後結束，且愛麗絲必須在某一張牌上簽過名。令 $A$ 為愛麗絲簽下的牌的號碼，$M$ 則為翻開的 $2k+1$ 張牌中第 $k+1$ 大的號碼。愛麗絲的扣分是 $|M-A|$；她希望扣分越少越好。\n\n對於所有 $(n, k)$，求最小的正整數 $d = d(n, k)$，使得愛麗絲存在策略，可以在任何情況下保證其扣分至多為 $d$。",
    "options": [],
    "answer": "See solution",
    "solution": "答案：$d = \\min(n - 2k - 1, \\lfloor \\frac{n-k-1}{2} \\rfloor)$。\n\n我們先證明愛麗絲無法保證扣分少於 $d$。這可以透過以下引理得證：\n\n**引理：** 若 $p+q=n-k-1$ 且 $q \\ge k$，則愛麗絲無法保證扣分少於 $\\min(p, q)$。\n\n*Proof.* 假設前 $k$ 張牌依序為 $1, 2, \\dots, k$，第 $k+1$ 到 $2k$ 張牌依序為 $k+1+p, \\dots, 2k+p$（注意到 $2k+p=n+k-q-1 \\le n-1$）。考慮以下三種狀況：\n\n1. 愛麗絲在前 $k$ 回合就簽名，因此 $A \\le k$。此時考慮第 $2k+1$ 張牌為 $n$，則 $M = k+1+p$，因此愛麗絲的扣分為\n\n$$\n|M - A| \\ge k + 1 + p - k = p + 1.\n$$\n\n2. 愛麗絲在第 $k+1$ 到 $2k$ 回合之間簽名，因此 $A \\in [k+1+p, 2k+p]$。此時讓第 $2k+1$ 張牌為 $k+1$，得 $M = k+1$，因此愛麗絲的扣分為\n\n$$\n|M - A| \\ge k + 1 + p - (k + 1) = p.\n$$\n\n3. 若愛麗絲在前 $2k$ 回合都還沒簽名，考慮第 $2k+1$ 張牌為 $n$。此時強迫 $A = n$ 且 $M = k+1+p$，因此愛麗絲的扣分為\n\n$$\n|M - A| \\ge n - (k + 1 + p) = q.\n$$\n\n綜以上，愛麗絲無法保證扣分少於 $\\min(p+1, p, q) = \\min(p, q)$。\n\n![](images/2022-TWNIMO-Problems_p91_data_8b06c210a2.png)\n\n我們只剩下要證明愛麗絲存在策略可以保證扣分不大於 $d$。基於愛麗絲事前已知 $(n, k)$，她只需要有一個策略保證扣分不大於 $n-2k-1$，另有一個策略保證扣分不大於 $\\lfloor \\frac{n-k-1}{2} \\rfloor$ 即可：\n\n- 要保證扣分不大於 $n-2k-1$，愛麗絲僅需在第一次出現 $\\{k+1, k+2, \\dots, n-k\\}$ 的數字時簽名即可。注意到第 $k+1$ 大的數字自然在 $[k+1, n-k]$ 內，因此此策略的最大扣分可能是 $(n-k)-(k+1)=n-2k-1$。\n\n- 要保證扣分不大於 $T := \\lfloor \\frac{n-k-1}{2} \\rfloor$，讓我們考慮三個區間：\n\n$$\nX = [1, n - T - k - 1]\n$$\n$$\nY = [n - T - k, T + k + 1]\n$$\n$$\nZ = [T + k + 2, n]\n$$\n\n考慮以下策略：\n\n(a) 若她看到 $Y$ 中的數字，立刻簽名；\n\n(b) 若她看到第 $k+1$ 個 $X$ 數字，立即簽名；\n\n(c) 若她看到第 $k+1$ 個 $Z$ 數字，立即簽名。\n\n讓我們計算此策略的扣分：\n\n(a) 若她簽了 $Y$ 中的數字，由於 $A \\in [n-T-k, T+k+1]$ 且 $M \\in [k+1, n-k]$，我們有 $|M-A| \\le T$；\n\n(b) 若她簽了 $X$ 中的數字，由於已經出現 $k+1$ 個 $X$ 的數字，必然有 $M \\le n-T-k-1$。又 $A \\in X$，故 $|M-A| \\le (n-T-k-1)-1 \\le T$。\n\n(c) 對稱的，若她簽了 $Z$ 中的數字，則 $|M-A| \\le T$。\n\n因為 $X, Y, Z$ 形成 $[1, n]$ 的分割，所以經過 $2k+1$ 回合，(a)(b)(c) 三者至少有一成立，即愛麗絲一定會在某張牌上簽名。綜以上，此策略保證扣分不大於 $T$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17391,
    "subject": "Mathematics (Olympiad)",
    "question": "Дадени се произволен триаголник **ABC** и две прави $p$ и $q$ кои не се паралелни меѓу себе и не се нормални на ниту една од страните на триаголникот. Нормалите низ точките **A**, **B** и **C** на правата $p$ ги обележуваме со $p_A, p_B$ и $p_C$ соодветно, а нормалите на правата $q$ со $q_A, q_B$ и $q_C$ соодветно. Нека пресечните точки на правите $p_A, q_A, p_B, q_B, p_C$ и $q_C$ соодветно со $q_B, p_B, q_C, p_C$ и $p_A$ се $K, L, P, Q, N$ и $M$. Докажи дека правите $KL, MN$ и $PQ$ се сечат во една точка.\n\n![](images/Macedonia_2013_p23_data_7d98f444a2.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Без губење на општоста може да претпоставиме дека $p_B$ е меѓу $p_A$ и $p_C$. Прв случај е ако $q_A$ е меѓу $q_B$ и $q_C$ како на цртежот, очигледно **KL** ја сече **PN**. Аналогно, ако $q_C$ е меѓу $q_A$ и $q_B$ случајот е симетричен на разгледуваниот. Втор случај, ако $q_B$ е меѓу $q_A$ и $q_C$, тогаш **MN** и **PN** не се паралелни, па **KL** ја сече барем едната и двата случаи се еквивалентни. Според ова можеме да претпоставиме дека **KL** ја сече **PN**. Правата **MN** не може да е паралелна со $p_B$, бидејќи во тој случај **p** е нормална на **AC** и аналогно **PQ** не е паралелна со $q_A$.\n\nНека $X, Y$ и $Z$ се пресечните точки на правите $PN, q_A$ и $p_B$ соодветно со $KL, PQ$ и $MN$. Од сличноста на триаголниците $LNZ$ и $PMZ$ се добива\n\n$$\n\\frac{\\overline{LZ}}{\\overline{PZ}} = \\frac{\\overline{LN}}{\\overline{PM}}. \\tag{1}\n$$\n\nСлично, од сличноста на $LPY$ и $NQY$ се добива\n\n$$\n\\frac{\\overline{NY}}{\\overline{LY}} = \\frac{\\overline{NQ}}{\\overline{LP}}. \\tag{2}\n$$\n\n![](images/Macedonia_2013_p25_data_5c0ff1781f.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17392,
    "subject": "Mathematics (Olympiad)",
    "question": "有 $N$ 個空箱子 $B_1, \\cdots, B_N$ 排成一排，旁邊有無限多顆石頭。給定正整數 $n$，艾莉絲與包柏玩以下的遊戲：\n\n艾莉絲首先將 $n$ 顆石頭放入箱子中；她可以自由決定這些石頭要如何分配到各個箱子裡。\n\n在接下來的每一回合，都會依序進行以下兩個步驟：\n\n1. 包柏選擇一個小於 $N$ 的正整數 $k$，並將箱子分成 $B_1, \\cdots, B_k$ 與 $B_{k+1}, \\cdots, B_N$ 兩組。\n2. 接著艾莉絲在其中一組的每個箱子中各加入一顆石頭，並從另外一組的每個箱子中各拿走一顆石頭。\n\n如果在任何回合結束時，有任何箱子裡沒有石頭了，則包柏勝。\n\n令 $M = \\lceil \\frac{N}{2} + 1 \\rceil^2 - 1$。\n\n(a) 證明對於任何 $n \\ge M$，艾莉絲有策略可以阻止包柏獲勝。\n\n(b) 證明對於任何 $n < M$，包柏總是有必勝法。",
    "options": [],
    "answer": "See solution",
    "solution": "為方便起見，我們都稱選 $B_1, \\cdots, B_k$ 為選左邊，反之為選右邊。\n\n(a) 部分解答：令 $V = (v_1, \\cdots, v_N)$ 表示第 $i$ 個箱子中有 $v_i$ 顆石頭的狀況。我們稱 $V > W$，若且唯若 $v_i \\ge w_i$ 對於所有 $1 \\le i \\le N$ 皆成立。此外，對於任何 $1 \\le i \\le N$，令 $V_i$ 為在第 $j$ 個箱子裡放 $1 + |j - i|$ 顆石頭的狀況；注意到 $V_i$ 沒有任何空箱子。\n\n現在，假設 $n = M$，且艾莉絲一開始將石頭分布成 $V_{[N/2]}$ 的狀況（此時的石頭總數為 $M$）。在接下來的每個回合中，假設起始的狀態為 $W$ 且 $W > V_i$。\n\n- 若包柏選擇 $k \\ge i$，則艾莉絲選左邊，回合結束時的狀態將 $> V_{i+1}$。\n- 若包柏選擇 $k < i$，則艾莉絲選右邊，回合結束時的狀態將 $> V_{i-1}$。\n\n因此在任何一個回合結束時，艾莉絲都能保證其狀態大於某個 $V_i$，從而不可能有空箱子。(a) 部分得證。\n\n(b) 部分解答：我們先證明兩個引理。\n\n**引理一：** 考慮連續兩個回合。假設在第一個回合，包柏選擇 $k$，而艾莉絲選左邊。若接下來的回合包柏選擇 $l > k$，則我們僅需考慮艾莉絲選左邊的可能性。\n\n*Proof.* 假設開始時的狀態是 $W = (w_1, \\cdots, w_N)$。若艾莉絲在第二次選右邊，則她最後得到的新狀態是 $W' = (w_1, w_2, \\cdots, w_k, w_{k+1} - 1, \\cdots, w_l - 1, w_l, \\cdots, w_N)$，故 $W' < W$。艾莉絲這樣做只是讓自己嚴格的比之前更糟，所以只須考慮兩次都選左邊的狀況即可。 □\n\n**引理二：** 若存在至少 $2k$ 個箱子，每個的石頭顆數都至多為 $k$，則包柏有必勝法。\n\n*Proof.* 忽略這 $2k$ 個箱子以外的箱子。首先，包柏將 $2k$ 個箱子對分成一半，並且不失一般性假設艾莉絲選左邊。包柏接下來依序選 $k+1, k+2, \\cdots, 2k-1$，依據引理一，我們都可以假設艾莉絲選左邊。但這樣一來，在以上 $k$ 次動作都結束後，最右邊的箱子會是空的，包柏勝。 □\n\n現在，如果一開始沒有空箱子，且對於任何正整數 $k$，都沒有引理二的狀況。這表示每個箱子都至少有 1 顆石頭，且對於所有 $1 \\le k \\le [N/2]$，至少有 $N - (2k-1) = N + 1 - 2k$ 個箱子有至少 $k+1$ 顆石頭。把以上相加，得到總石頭顆數至少為 $M$。故若起始石頭顆數 $< M$，包柏必勝。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17393,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs of integers $ (x, y) $ satisfying the equation\n\n$$\ny(x + y) = x^3 - 7x^2 + 11x - 3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The given equation is\n\n$$\ny(x + y) = x^3 - 7x^2 + 11x - 3.\n$$\n\nRewriting, we get a quadratic in $y$:\n\n$$\ny^2 + x y - (x^3 - 7x^2 + 11x - 3) = 0.\n$$\n\nThe discriminant must be a perfect square:\n\n$$\n\\Delta = x^2 + 4(x^3 - 7x^2 + 11x - 3) = 4x^3 - 27x^2 + 44x - 12.\n$$\n\nSet $\\Delta = s^2$ for some integer $s$.\n\nWe factor:\n\n$$\n4x^3 - 27x^2 + 44x - 12 = (x - 2)(4x^2 - 19x + 6).\n$$\n\nWe consider possible values for $x$ such that $\\Delta$ is a perfect square.\n\n**Case 1:** $x = 2$\n\nPlug in $x = 2$:\n\n$$\ny(2 + y) = 8 - 28 + 22 - 3 = -1 \\\\\ny^2 + 2y + 1 = 0 \\\\\n(y + 1)^2 = 0 \\\\\ny = -1.\n$$\n\nSo $(2, -1)$ is a solution.\n\n**Case 2:** $x - 2 = s^2$ for some integer $s$\n\nLet $x - 2 = s^2$, so $x = s^2 + 2$.\n\nPlug into $4x^2 - 19x + 6$ and set equal to $u^2$ or $-u^2$ (as in the original solution):\n\n- For $k = 1$: $4x^2 - 19x + 6 = u^2$\n- For $k = -1$: $4x^2 - 19x + 6 = -u^2$\n\nAfter checking possible small integer values, we find:\n\n- For $x = 1$: $y^2 + y - (1 - 7 + 11 - 3) = y^2 + y - 2 = 0$; $y = 1$ or $y = -2$.\n- For $x = 6$: $y^2 + 6y - (216 - 252 + 66 - 3) = y^2 + 6y - (-27 + 66 - 3) = y^2 + 6y - 36 = 0$; $y = 3$ or $y = -9$.\n\n**Summary:**\n\nThe integer solutions are:\n\n$$\n\\{(6, 3),\\ (6, -9),\\ (1, 1),\\ (1, -2),\\ (2, -1)\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17394,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p, q$ be prime numbers and $a$ be an integer such that $p > 2$ and $a \\not\\equiv 1 \\pmod{q}$ but $a^p \\equiv 1 \\pmod{q}$. Prove that\n\n$$\n(1 + a^1)(1 + a^2) \\dots (1 + a^{p-1}) \\equiv 1 \\pmod{q}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "As $a^p \\equiv 1 \\pmod{q}$ while $a \\not\\equiv 1 \\pmod{q}$, the case $q = 2$ is impossible. Thus, the desired equation is equivalent to\n\n$$\n(1 + a^0)(1 + a^1)(1 + a^2) \\dots (1 + a^{p-1}) \\equiv 2 \\pmod{q}.\n$$\n\nExpanding the left-hand side gives all monomials of the form $a^{i_1 + \\dots + i_k}$ where $\\{i_1, \\dots, i_k\\} \\subseteq \\{0, 1, \\dots, p-1\\}$. Since $a^p \\equiv 1 \\pmod{q}$, the exponents can be reduced modulo $p$. Each possible remainder modulo $p$ is produced the same number of times (excluding the empty set and the full set). For each tuple $(i_1, \\dots, i_k)$ with $0 < k < p$, we can find $l$ such that $kl \\equiv 1 \\pmod{p}$ and form $(i_1 + l, \\dots, i_k + l)$, shifting the sum by $k l$. Repeating this $p$ times cycles through all remainders, so each occurs equally often.\n\nLet this constant number be $c$. Then the equation becomes\n\n$$\na^0 + a^{0+1+\\dots+(p-1)} + c(a^0 + a^1 + \\dots + a^{p-1}) \\equiv 2 \\pmod{q}.\n$$\n\nSince $0 + 1 + \\dots + (p-1) = \\frac{(p-1)p}{2}$, and $a^p \\equiv 1 \\pmod{q}$, we have $a^{0+1+\\dots+(p-1)} \\equiv 1 \\pmod{q}$. Also, $(a^0 + a^1 + \\dots + a^{p-1})(a-1) = a^p - 1 \\equiv 0 \\pmod{q}$, and $a-1 \\not\\equiv 0 \\pmod{q}$, so $a^0 + a^1 + \\dots + a^{p-1} \\equiv 0 \\pmod{q}$. Thus, the equation reduces to $1 + 1 \\equiv 2 \\pmod{q}$, which is true.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17395,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the plane partitioned into unit squares. The interior of each square is coloured either _red_ or _black_ (the sides of the squares are not considered to be coloured). Prove that for any positive integer $\\alpha$, there exists an equilateral triangle of integer area $A \\geq \\alpha$ whose vertices are all the same colour.",
    "options": [],
    "answer": "See solution",
    "solution": "The key to the proof is a configuration that guarantees the existence of a monochromatic equilateral triangle in a bichromatic plane. For a given $\\alpha$, choose the side length $\\ell = \\sqrt{\\frac{4\\alpha}{\\sqrt{3}}}$ so that an equilateral triangle of area $\\alpha$ can be formed. Let $\\gamma$ be the circle of radius $\\ell$ centered at a point $A$ of colour $c_1$, and $\\Gamma$ the circle of the same center and radius $\\ell\\sqrt{3}$. The intersection points of these circles with the sides of the unit squares are finitely many and must be avoided, as they are uncoloured.\n\nIf all points of $\\gamma$ (except finitely many) are coloured $c_2$, then there exists an equilateral triangle with vertices of colour $c_2$ and area $3\\alpha$. Otherwise, let $B \\in \\gamma$ be coloured $c_1$, and consider the regular hexagon $BCDEFG$ inscribed in $\\gamma$. Its vertices must be coloured as follows: $C \\to c_2$, $G \\to c_2$, $E \\to c_1$, $D \\to c_2$, $F \\to c_2$; otherwise, a monochromatic equilateral triangle of area $\\alpha$ or $3\\alpha$ is formed. Now, let $H = BC \\cap DE$; clearly $H \\in \\Gamma$. If $H$ is coloured $c_2$, then $\\triangle CDH$ is monochromatic $c_2$ and has area $\\alpha$; if $H$ is coloured $c_1$, then $\\triangle BEH$ is monochromatic $c_1$ and has area $4\\alpha$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17396,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, a_3, \\dots, a_n, \\dots$ be the sequence of all square-free positive integers in ascending order. Show that there are infinitely many indices $n$ satisfying $a_{n+1} - a_n = 2020$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we need a lemma.\n\n**Lemma** $\\sum_{p\\ \\text{prime}} \\frac{1}{p^2} < 0.48$.\n\n**Proof of Lemma**\n\n$$\n\\begin{aligned}\n\\sum_{p\\ \\text{prime}} \\frac{1}{p^2} &< \\frac{1}{4} + \\frac{1}{9} + \\frac{1}{25} + \\frac{1}{49} + \\left( \\frac{1}{9 \\cdot 11} + \\frac{1}{11 \\cdot 13} + \\dots \\right) \\\\\n&= \\frac{1}{4} + \\frac{1}{9} + \\frac{1}{25} + \\frac{1}{49} + \\frac{1}{18} < 0.48.\n\\end{aligned}\n$$\n\nFor the original problem, we need to find a positive integer $x$ such that $x$ and $x + 2020$ are both square-free, but every $x + i$ ($1 \\leq i \\leq 2019$) in between contains a square factor.\n\nTake 2019 distinct prime numbers $p_1, p_2, \\dots, p_{2019}$, $p_i > 4039$ ($1 \\leq i \\leq 2019$). Let $x_k = k \\left( \\prod_{i=1}^{2019} p_i \\right)^2 + x$, where $k$ and $x$ are positive integers,\n\n$1 \\leq x \\leq \\left( \\prod_{i=1}^{2019} p_i \\right)^2$ satisfies the system of congruence equations $x + i \\equiv 0 \\pmod{p_i^2}$, $1 \\leq i \\leq 2019$. The existence of $x$ is guaranteed by the Chinese remainder theorem.\n\nWe must find a positive integer $k$ such that $x_k$ and $x_k+2020$ are square-free. Let $N > 2+40000 \\left(\\prod_{i=1}^{2019} p_i\\right)^2$ be a fixed, large integer. Then $0.01N > 2 \\prod_{i=1}^{2019} p_i \\sqrt{N+2}$. Since $p_i > 4039$, $p_i$ ($1 \\leq i \\leq 2019$) is not a prime factor of $x_k$ or $x_k+2020$.\n\nFor convenience, denote $M = \\prod_{i=1}^{2019} p_i$.\n\nFor any prime $p$, the congruence equation $x_k = kM^2 + x \\equiv 0 \\pmod{p^2}$ has at most $\\lfloor \\frac{N}{p^2} \\rfloor$ solutions in $1 \\leq k \\leq N$.\n\nSimilarly, $x_k + 2020 = kM^2 + x + 2020 \\equiv 0 \\pmod{p^2}$ also has at most $\\lfloor \\frac{N}{p^2} \\rfloor$ solutions in $1 \\leq k \\leq N$.\n\nIt is worth noting that when $p > M\\sqrt{N+2} > \\sqrt{kM^2+2M^2} > \\sqrt{kM^2+x+2020}$, the divisibility in the above two congruence equations cannot occur. Therefore, in $[1, N]$, the number of integers $k$ such that $x_k$ and $x_k+2020$ are both square-free, is at least\n\n$$\nN - 2 \\sum_{\\substack{p\\ \\text{prime} \\\\ p \\leq M\\sqrt{N+2}}} \\left\\lfloor \\frac{N}{p^2} \\right\\rfloor \\geq N - 2N \\left( \\sum_{\\substack{p\\ \\text{prime} \\\\ p \\leq M\\sqrt{N+2}}} \\frac{1}{p^2} \\right) - 2 \\left( \\sum_{\\substack{p\\ \\text{prime} \\\\ p \\leq M\\sqrt{N+2}}} 1 \\right) \\\\\n> 0.04N - 2M\\sqrt{N+2} > 0.03N.\n$$\n\nHere, the estimate on the second line comes from the lemma and the condition $N > 2 + 40000 \\left(\\prod_{i=1}^{2019} p_i\\right)^2$. Letting $N \\to +\\infty$, the conclusion follows. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17397,
    "subject": "Mathematics (Olympiad)",
    "question": "The city of Mar del Plata has the shape of a square, denoted by WSEN. For a given even integer $n$, it is divided by $2(n + 1)$ streets into $n \\times n$ blocks (the streets also run around the square). Each block has a size of $100\\ \\mathrm{m} \\times 100\\ \\mathrm{m}$. All the streets in Mar del Plata are one-way, with each street having a single direction along its entire length. Every two adjacent parallel streets have opposite directions. The WS street is directed from W to S, and the WN street is directed from W to N. A street cleaning car starts at W. Its aim is to arrive at E and clean the roads it passes through. What is the length of the longest possible route it can pass if no $100\\ \\mathrm{m}$ segment can be passed more than once?\n\n(For $n = 6$, Fig. 1 shows the map of the city and one of the possible, but not the longest, routes of the car. See also http://goo.gl/maps/JAzD.)\n\n![](images/Cesko-Slovacko-Poljsko_2012_p7_data_cdcdcc9b7e.png)\n\n![](images/Cesko-Slovacko-Poljsko_2012_p7_data_ed5cbed811.png)",
    "options": [],
    "answer": "See solution",
    "solution": "First, we introduce some notation. Call each $100\\ \\mathrm{m}$ segment of a street an *arrow*, and each place where two streets meet a *crossroad*. If an arrow has the same direction as the street WS or WN (that is, it runs from the upper left to the bottom right or from the bottom left to the upper right), it is called a *forward arrow*; otherwise, it is a *back arrow*.\n\nWe use the following lemma: Let the set of all crossroads be split into two sets $A$ and $B$ such that $W \\in A$ and $E \\in B$. Then the number of times the car drives along an arrow from $A$ to $B$ is one more than the number of times the car drives along an arrow from $B$ to $A$.\n\nLet us split the set of all crossroads of Mar del Plata into two sets $A$ and $B$ by a vertical (north-south) line connecting two points located $100k + 50\\ \\mathrm{m}$ from $W$—one on the WN street and one on the WS street—for some $k \\in \\{0, 1, \\dots, n-1\\}$. Figures 5a and 5b show the splitting line for $k = 3$ and $k = 4$, respectively.\n\n![](images/Cesko-Slovacko-Poljsko_2012_p7_data_cdcdcc9b7e.png)\n\n*Fig. 5a*\n\n![](images/Cesko-Slovacko-Poljsko_2012_p7_data_ed5cbed811.png)\n\n*Fig. 5b*\n\nIf $k$ is odd, then the line intersects $k + 1$ forward arrows (going from $A$ to $B$) and $k + 1$ back arrows (going from $B$ to $A$). Even if the car would pass all the $k + 1$ forward arrows, by the lemma, it can pass at most $k$ back arrows. Hence, at least one of the back arrows remains unpassed.\n\nIf $k \\ge 2$ is even, then the line intersects $k + 2$ forward arrows and $k$ back arrows. The two northernmost forward arrows intersected by the line begin in the crossroad which has only one incoming arrow. This crossroad can be passed only once, so one of the two northernmost forward arrows remains unpassed. The same holds for the two southernmost forward arrows. So we have at most $k$ forward arrows passed by the car and, by the lemma, at most $k - 1$ back arrows. Together, we have at least 3 unpassed arrows at this level.\n\nFor $k = 0$, we have only two forward arrows starting in $W$ intersected by the line. Clearly, only one of them could be passed by the car and one remains unpassed.\n\nSimilarly, we split the set of all crossroads by a vertical line connecting two points located $100k + 50\\ \\mathrm{m}$ from $E$—one on the SE street and one on the NE street—for some $k \\in \\{0, 1, \\dots, n-1\\}$. The situation is sketched in Figures 6a and 6b for $k = 3$ and $k = 4$, respectively.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17398,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $P(x)$ is a polynomial of degree $d \\geq 2$ and $a_0, a_1, \\dots, a_n$ is a sequence such that $Na_i$ is integral for all $i$ (for some fixed $N$), and $a_{i} = P(a_{i+1}) - i$ for $i \\geq 0$.\n\nLet $|\\{a_1, a_2, \\dots, a_n\\}|$ denote the number of distinct values among $a_1, \\dots, a_n$.\n\nProve that for all large enough $n$, $|\\{a_1, a_2, \\dots, a_n\\}| \\geq c n^{1/2}$ for some constant $c > 0$ (when $d = 2$), and more generally, $|\\{a_1, a_2, \\dots, a_n\\}| \\geq c n^{1/d}$ for some $c > 0$ (when $d \\geq 2$).",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed by bounding the possible values of $a_i$ and applying the pigeonhole principle.\n\nFirst, for all large $n$, $|a_n| > \\sqrt{\\gamma n}$ for some $\\gamma > 0$ (when $d = 2$), otherwise we get a contradiction by considering the number of distinct $a_i$ and their preimages.\n\nFor $d = 2$, write $P(x) = c(x - a)^2 + b$ for some $a, b, c \\in \\mathbb{R}$, $c \\neq 0$. Then,\n$$\nP(x) - P(y) = c(x - y)(x + y - 2a)\n$$\nIf $|a_{n+2}| > \\sqrt{\\gamma n}$, then at least one of $|a_{n+3} - a_{n+2}|$ or $|a_{n+3} + a_{n+2} - 2a|$ is large, and both are at least $1/N$ if $Na_i$ is integral. This forces $|a_{n+2} - a_{n+1}| > \\sqrt{\\varepsilon n}$ for some $\\varepsilon > 0$, unless $a_{n+3} = a_{n+2}$ or $a_{n+3} + a_{n+2} = 2a$, which leads to a contradiction for large $n$.\n\nBy repeating this argument, we see that the sequence cannot be too concentrated, and so $|\\{a_1, \\dots, a_n\\}| \\geq c n^{1/2}$ for some $c > 0$.\n\nFor $d$ odd, if $|a_{i+1}|$ is large, then\n$$\n\\frac{|P(x) - P(y)|}{|x - y|} \\geq c|x|^{d-1}\n$$\nso $|a_{i+1} - a_i|$ must be very small, which is impossible since $Na_i$ is integral. This leads to a contradiction if too few distinct $a_i$ exist, so $|\\{a_1, \\dots, a_n\\}| \\geq c n^{1/d}$ for some $c > 0$.\n\nThus, in all cases, the number of distinct $a_i$ grows at least as fast as $n^{1/d}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17399,
    "subject": "Mathematics (Olympiad)",
    "question": "The incircle of a triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. Let $G$ be a point on the incircle such that $FG$ is a diameter. The lines $EG$ and $FD$ intersect at $H$. Prove that $CH \\parallel AB$.\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p127_data_dc816e931c.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We work in the opposite direction. Suppose that $H'$ is the point where $DF$ intersects the line through $C$ parallel to $AB$. We need to show that $H' = H$. For this purpose, it suffices to prove that $E$, $G$, $H'$ are collinear, which reduces to showing that if $G' \\neq E$ is the common point of $EH'$ and the incircle, then $G' = G$.\n\nNote that $H'$ and $B$ lie on the same side of $AC$. Hence $CH' \\parallel AB$ gives $\\angle ACH' = 180^\\circ - \\angle BAC$. Also, some homothety with center $D$ maps the segment $BF$ to the segment $CH'$. Thus the equality $BD = BF$ implies that $CH' = CD = CE$, i.e., the triangle $ECH'$ is isosceles and\n\n$$\n\\angle H'EC = \\frac{1}{2}(180^\\circ - \\angle ECH') = \\frac{1}{2}\\angle BAC.\n$$\n\nBut $G'$ and $H'$ lie on the same side of $AC$, so $\\angle G'EC = \\angle H'EC$ and consequently\n\n$$\n\\angle G'FE = \\angle G'EC = \\angle H'EC = \\frac{1}{2}\\angle BAC\n$$\nso that\n\n$$\n\\angle G'FA = \\angle G'FE + \\angle EFA = \\frac{1}{2}\\angle BAC + \\frac{1}{2}(180^\\circ - \\angle FAE) = 90^\\circ.\n$$\n\nHence $FG'$ is a diameter of the incircle and the desired equality $G' = G$ follows.\n\n**Remark.** A similar proof also works in the forward direction: one may compute $\\angle EHD = \\frac{1}{2}\\angle ACB$. Hence $H$ lies on the circle centered at $C$ that passes through $D$ and $E$. Consequently, the triangle $EHC$ is isosceles, wherefore\n\n$$\n\\angle ECH = 180^\\circ - 2\\angle GEC = 180^\\circ - \\angle BAC.\n$$\n\nThus the lines $AB$ and $CH$ are indeed parallel.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17400,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$ and $y$ be two rational numbers and let $n$ be an odd positive integer. Suppose $x^n - 2x = y^n - 2y$. Prove that $x = y$.",
    "options": [],
    "answer": "See solution",
    "solution": "The case $n = 1$ being obvious, suppose that $n \\geq 3$ and $x \\neq y$. The relation rewrites as\n$$\nx^{n-1} + x^{n-2}y + x^{n-3}y^2 + \\dots + xy^{n-2} + y^{n-1} = 2.\n$$\n\nPut $x = a/c$, $y = b/c$, where $a, b, c \\in \\mathbb{Z}$, $c \\neq 0$, with $\\gcd(a, b, c) = 1$. The latter relation yields\n$$\na^{n-1} + a^{n-2}b + \\dots + ab^{n-2} + b^{n-1} = 2c^{n-1}.\n$$\nThe left-hand side is even and has an odd number of summands, hence $a$ and $b$ are both even integers.\n\nConsequently, $2^{n-1}$ divides $2c^{n-1}$. Recall that $n \\geq 3$ to notice that $2$ divides $c$. Hence $2$ is a common divisor of the numbers $a$, $b$, $c$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17401,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all (not necessarily strictly) monotonic functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(x+y)^3 = f(x^3) + f(y^3) \\quad \\text{for all } x, y \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By substituting $x = y = 0$, we see that $f(0)^3 = 2f(0)$, which implies $f(0)^3 - 2f(0) = 0$. Thus, $f(0) = 0$, $f(0) = -\\sqrt{2}$, or $f(0) = \\sqrt{2}$.\n\nSubstituting $y = 0$ gives $f(x)^3 = f(x^3) + f(0)$.\n\n1. **Case $f(0) = 0$:**\n   \n   Then $f(x)^3 = f(x^3)$, so $(f(x+y))^3 = f(x^3) + f(y^3) = f(x)^3 + f(y)^3$. Let $g(x) = f(x)^3$; then $g(x+y) = g(x) + g(y)$, which is Cauchy's equation. Since $f$ is monotonic, $g(x) = c'x$ for some $c' \\in \\mathbb{R}$, so $f(x) = c\\sqrt[3]{x}$ for some $c \\in \\mathbb{R}$. Substituting into the original equation gives $c^3(x+y) = c(x+y)$, so $c^3 - c = 0$, i.e., $c = -1, 0, 1$. Thus, $f(x) = 0$, $f(x) = \\sqrt[3]{x}$, or $f(x) = -\\sqrt[3]{x}$.\n\n2. **Case $f(0) = \\sqrt{2}$:**\n   \n   Substituting $x = 1$, $y = 0$ gives $f(1)^3 = f(1) + \\sqrt{2}$, so $f(1) = \\sqrt{2}$. Similarly, $f(2) = \\sqrt{2}$, and by induction, $f(x) = \\sqrt{2}$ for all $x$. The same argument applies for negative $x$ by monotonicity.\n\n3. **Case $f(0) = -\\sqrt{2}$:**\n   \n   Substituting $x = 1$, $y = 0$ gives $f(1)^3 = f(1) - \\sqrt{2}$, so $f(1) = -\\sqrt{2}$. Similarly, $f(x) = -\\sqrt{2}$ for all $x$.\n\nThus, the solutions are:\n\n- $f(x) = 0$\n- $f(x) = \\sqrt[3]{x}$\n- $f(x) = -\\sqrt[3]{x}$\n- $f(x) = \\sqrt{2}$\n- $f(x) = -\\sqrt{2}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17402,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x = |AB_1|$, $y = |BC_1|$, $z = |CA_1|$. Show that\n\n$$\n\\sqrt{\\frac{x}{x+y}} + \\sqrt{\\frac{y}{y+z}} + \\sqrt{\\frac{z}{z+x}} \\le \\frac{3}{\\sqrt{2}}\n$$\n\nor equivalently,\n\n$$\n\\frac{1}{\\sqrt{1+a^2}} + \\frac{1}{\\sqrt{1+b^2}} + \\frac{1}{\\sqrt{1+c^2}} \\le \\frac{3}{\\sqrt{2}}\n$$\n\nfor all positive real numbers $a$, $b$, $c$ satisfying $abc = 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume without loss of generality that $ab \\le 1$. Then\n\n$$\n\\frac{1}{\\sqrt{1+a^2}} + \\frac{1}{\\sqrt{1+b^2}} \\le \\sqrt{2} \\left( \\frac{1}{1+a^2} + \\frac{1}{1+b^2} \\right)\n$$\n\nand\n\n$$\n\\frac{1}{1+a^2} + \\frac{1}{1+b^2} = 1 + \\frac{1-a^2b^2}{(1+a^2)(1+b^2)} \\le 1 + \\frac{1-a^2b^2}{(1+ab)^2} = \\frac{2}{1+ab}\n$$\n\nand\n\n$$\n\\frac{1}{\\sqrt{1+c^2}} \\le \\frac{\\sqrt{2}}{1+c}\n$$\n\nby the Cauchy-Schwarz inequality. Therefore,\n\n$$\n\\begin{aligned}\n\\sum_{cyc} \\frac{1}{\\sqrt{1+a^2}} &\\le 2\\sqrt{\\frac{c}{1+c}} + \\frac{\\sqrt{2}}{1+c} = \\frac{\\sqrt{2}}{1+c}(\\sqrt{2c(c+1)} + 1) \\\\\n&\\le \\frac{\\sqrt{2}}{1+c}\\left(\\frac{2c+c+1}{2} + 1\\right) = \\frac{3}{\\sqrt{2}}\n\\end{aligned}\n$$\n\nby the AM-GM inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17403,
    "subject": "Mathematics (Olympiad)",
    "question": "Eight numbers, all zero, are written on a blackboard. A _move_ consists in randomly selecting four of them, $a, b, c, d$, and replacing them by $a+3$, $b+3$, $c+2$, and $d+1$, respectively.\n\n**a)** What is the smallest number of moves after which on the blackboard can appear eight consecutive numbers?\n\n**b)** Is there a sequence of moves after which all the numbers on the blackboard are equal to $2015$?\n\n**c)** Is there a sequence of moves after which the product of the numbers on the blackboard is equal to $2145$?",
    "options": [],
    "answer": "See solution",
    "solution": "a) After each move, the sum of the numbers increases by $9$. Since the sum of the smallest $8$ consecutive numbers is $0 + 1 + 2 + \\dots + 7 = 28$, but the example uses $1$ to $8$ (sum $36$), so the minimal sum for consecutive numbers is $36$. Thus, at least four moves are required. Four moves are sufficient, as shown below:\n\n![](table_example.png)\n\nb) After $k$ moves, the sum of the eight numbers is $9k$. Since $8 \\cdot 2015 = 16120$ is not a multiple of $9$, the answer is negative.\n\nc) No, it is not possible.\n\nSuppose the contrary and notice that $2145 = 3 \\cdot 5 \\cdot 11 \\cdot 13$ to infer that at least four numbers from the final eight are equal to $1$.\n\nIf there are exactly four, then the numbers are $3, 5, 11, 13, 1, 1, 1, 1$ in some order. Their sum equals $3+5+11+13+1+1+1+1 = 36$, so four moves were performed to this end. This is false, as four moves cannot give number $13$.\n\nNotice that among the final eight numbers one cannot have six or more numbers equal to $1$, for each move increases three numbers with more than $1$.\n\nSuppose now that five numbers among the final eight are equal to $1$ and let $a, b$, and $c$ be the remaining three. Five moves were performed in order to obtain five numbers equal to $1$, so $a+b+c+1+1+1+1+1 = 45$, where $a$ is the product of two factors out of the set $3, 5, 11, 13$, while $b$ and $c$ are the remaining two elements. Each of the six cases fails, so there is no solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17404,
    "subject": "Mathematics (Olympiad)",
    "question": "Eight children, who did not know each other, came to the dance class. To introduce them to each other, the teacher decided to choose four children each minute and let them dance in a circle. During this minute, each of the children in a circle will get to know a neighbor on the left and on the right. What is the minimal number of minutes the teacher needs to make all eight children know each other?",
    "options": [],
    "answer": "See solution",
    "solution": "Consider any child from this group. They need to get to know seven other children, and by participating once in a dancing circle, they meet at most two new acquaintances. Thus, each child should participate in at least four dancing circles. In each dancing circle, there are no more than four children participating. Therefore, there are at least $8 \\cdot 4 = 32$ participations, and since each circle has four children, there must be at least $8$ circles. \n\nNow, we provide an example in which eight circles are enough for everybody to know each other. Enumerate the children from $1$ to $8$ and make the following ordered groups: $1234$, $5678$, $1357$, $2468$, $1526$, $3748$, $2736$, and $1845$. Here, we assume that the last child stands in a circle next to the first one.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17405,
    "subject": "Mathematics (Olympiad)",
    "question": "An integer is assigned to each vertex of a regular pentagon so that the sum of the five integers is $2011$. A turn of a solitaire game consists of subtracting an integer $m$ from each of the integers at two neighboring vertices and adding $2m$ to the opposite vertex, which is not adjacent to either of the first two vertices. (The amount $m$ and the vertices chosen can vary from turn to turn.)\n\nThe game is won at a certain vertex if, after some number of turns, that vertex has the number $2011$ and the other four vertices have the number $0$. Prove that for any choice of the initial integers, there is exactly one vertex at which the game can be won.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_1, a_2, a_3, a_4$, and $a_5$ represent the integers at vertices $v_1$ to $v_5$ (in order around the pentagon) at the start of the game. We will first show that the game can be won at only one of the vertices.\n\nObserve that the quantity $a_1 + 2a_2 + 3a_3 + 4a_4 \\pmod{5}$ is an invariant of the game. For instance, one move involves replacing $a_1, a_3$ and $a_5$ by $a_1 - m, a_3 + 2m$ and $a_5 - m$. Thus the quantity $a_1 + 2a_2 + 3a_3 + 4a_4$ becomes\n\n$$\n(a_1 - m) + 2a_2 + 3(a_3 + 2m) + 4a_4 = a_1 + 2a_2 + 3a_3 + 4a_4 + 5m,\n$$\n\nwhich is unchanged modulo $5$. The other moves may be checked similarly.\n\nNow suppose that the game may be won at vertex $v_j$. The value of the invariant at the winning position is $2011j$. If the initial value of the invariant is $n$, then we must have $2011j \\equiv n \\pmod{5}$, or $j \\equiv n \\pmod{5}$. Hence the game may only be won at vertex $v_j$, where $j$ is the least positive residue of $n \\pmod{5}$.\n\nBy renumbering the vertices, we may assume without loss of generality that the potentially winning vertex is $v_5$. We will show that the game can be won in four moves by adding a suitable amount $2m_j$ at vertex $v_j$ (and subtracting $m_j$ from the opposite vertices) on the $j$th turn for $j = 1, 2, 3, 4$. The net change at vertex $v_1$ after these four moves is $2m_1 - m_3 - m_4$, which must equal $-a_1$ if we are to finish with $0$ at $v_1$. In this fashion we obtain the system of equations\n\n$$\n\\begin{aligned}\n2m_1 - m_3 - m_4 &= -a_1 \\\\\n2m_2 - m_4 &= -a_2 \\\\\n2m_3 - m_1 &= -a_3 \\\\\n2m_4 - m_1 - m_2 &= -a_4 \\\\\n-m_2 - m_3 &= -a_5 + 2011\n\\end{aligned}\n$$\n\nwhich has an integral solution if and only if the game may be won. The sum of the first four equations is the negative of the fifth equation, so the fifth equation is redundant. Multiplying the first four equations by $-1, 3, -3, 1$ and adding them yields $5m_2 - 5m_3 = a_1 - 3a_2 + 3a_3 - a_4$. But we are assuming $v_5$ is the potentially winning vertex, so we see\n\n$$\na_1 - 3a_2 + 3a_3 - a_4 \\equiv a_1 + 2a_2 + 3a_3 + 4a_4 \\equiv n \\equiv 5 \\equiv 0 \\pmod{5}.\n$$\n\nTherefore we may divide by $5$ to obtain $m_2 - m_3 = \\frac{1}{5}(a_1 - 3a_2 + 3a_3 - a_4)$. We also know that $m_2 + m_3 = a_1 + a_2 + a_3 + a_4$, and one easily confirms that the right-hand sides of these equations are integers with the same parity. Hence the system admits a solution with $m_2$ and $m_3$ integral. The second and third equations then quickly give integer values for $m_1$ and $m_4$ as well, so the system has an integral solution, meaning it is indeed possible to win the game at vertex $v_5$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17406,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be positive real numbers, where $n \\ge 2$. For each permutation $(b_1, b_2, \\dots, b_n)$ of $(a_1, a_2, \\dots, a_n)$, define its score to be\n\n$$\n\\frac{b_1^2}{b_2} + \\frac{b_2^2}{b_3} + \\dots + \\frac{b_{n-1}^2}{b_n}.\n$$\n\nShow that there exist two permutations of $(a_1, a_2, \\dots, a_n)$ whose scores differ by at least $3|a_1 - a_n|$.",
    "options": [],
    "answer": "See solution",
    "solution": "We compare the scores of the decreasing and increasing permutations. First, reorder the sequence so that $b_1 \\ge b_2 \\ge \\dots \\ge b_n$. Consider the scores\n\n$$\n\\frac{b_1^2}{b_2} + \\frac{b_2^2}{b_3} + \\dots + \\frac{b_{n-1}^2}{b_n}\n$$\n\nand\n\n$$\n\\frac{b_n^2}{b_{n-1}} + \\frac{b_{n-1}^2}{b_{n-2}} + \\dots + \\frac{b_2^2}{b_1}.\n$$\n\nTheir difference is\n\n$$\n\\frac{b_1^2}{b_2} - \\frac{b_2^2}{b_1} + \\dots + \\frac{b_{n-1}^2}{b_n} - \\frac{b_n^2}{b_{n-1}} = \\frac{(b_1 - b_2)(b_1^2 + b_1b_2 + b_2^2)}{b_1b_2} + \\dots + \\frac{(b_{n-1} - b_n)(b_{n-1}^2 + b_{n-1}b_n + b_n^2)}{b_{n-1}b_n}.\n$$\n\nSince for positive reals $x, y$, $x^2 + xy + y^2 \\ge 3xy$ (which follows from $(x - y)^2 \\ge 0$), the above difference is at least\n\n$$\n3(b_1 - b_2) + 3(b_2 - b_3) + \\dots + 3(b_{n-1} - b_n) = 3(b_1 - b_n) \\ge 3|a_1 - a_n|.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 17407,
    "subject": "Mathematics (Olympiad)",
    "question": "Alice and Bob are given an integer $n \\ge 2$ and a token to play the following game. Initially, Alice chooses an order of the numbers $1, 2, \\ldots, n$, and writes them down in a row, in that order, on a sheet of paper. Next, Bob chooses one of these numbers and places the token on that number. Continuing, Alice moves the token to one of the neighbouring numbers, then Bob moves the token to one of the neighbouring numbers of the current position, and so on and so forth in turns. For each $k$ in the range $1, 2, \\ldots, n$, the token may be placed on number $k$ at most $k$ times; Bob's first move is counted. The player who can no longer move the token loses. Determine the values of $n$ for which Alice has a winning strategy.",
    "options": [],
    "answer": "See solution",
    "solution": "Alice has a winning strategy if and only if $n \\equiv 0$ or $-1 \\pmod{4}$.\n\nConsider placing the token on a number as a one unit drop of that number; a move is possible if and only if the current position of the token is not flanked by two zeroes. Each player places the token on positions of like parity rank.\n\nAn arrangement $a_1, \\ldots, a_n$ of an $n$-element collection of non-negative integers is *suitable* if there exist non-negative integers $x_0, x_1, \\ldots, x_n$ such that\n\n$$\nx_0 = x_n = 0, \\quad \\text{and} \\quad a_k = x_{k-1} + x_k, \\quad k = 1, \\ldots, n.\n$$\n\nIn a suitable arrangement, the $a_k$ add up to twice the sum of the $x_k$, so for a collection of non-negative integers to have a suitable arrangement, it is necessary that the numbers in the collection add up to an even number. Otherwise, the collection has no suitable arrangement. In particular, if $n \\equiv 1$ or $2 \\pmod{4}$, there is no suitable arrangement of the collection $1, 2, \\ldots, n$. If $n = 4m$, then\n\n$$\n\\begin{array}{l}\n1 = 0 + 1, \\quad 3 = 1 + 2, \\quad 5 = 2 + 3, \\quad \\ldots, \\quad 4m - 1 = (2m - 1) + 2m, \\\\\n4m = 2m + 2m, \\\\\n4m - 2 = 2m + (2m - 2), \\quad \\ldots, \\quad 6 = 4 + 2, \\quad 4 = 2 + 2, \\quad 2 = 2 + 0\n\\end{array}\n$$\n\nis a suitable arrangement of the collection $1, 2, \\ldots, n$. Removal of $4m$ provides a suitable arrangement in case $n = 4m - 1$.\n\nAlice has a winning strategy if and only if the initial arrangement of numbers is suitable.\n\nIf the initial arrangement is suitable, then Alice wins. She can counter any legal move of Bob, in a suitable arrangement, so as to make the new arrangement again suitable. Since the initial arrangement is suitable, and the game eventually ends, Bob ultimately gets stuck and loses.\n\nLet $a_1, \\ldots, a_n$ be a suitable arrangement, and let Bob place the token on the $k$-th position. Since Bob's move is legal, $x_{k-1} + x_k = a_k > 0$. Without loss of generality, assume $x_k > 0$, so $a_{k+1} = x_k + x_{k+1} \\ge x_k > 0$, showing that the $(k+1)$-st position is available for Alice to place the token on. After doing so, the new arrangement is again suitable, since $a_k$ changes to $a'_k = x_{k-1} + (x_k - 1)$, $a_{k+1}$ changes to $a'_{k+1} = (x_k - 1) + x_{k+1}$, and the other numbers are left unchanged.\n\nIf the initial arrangement is not suitable, then Bob has a winning strategy. Bob can achieve a suitable subarrangement just before one of Alice's moves, reset the whole game with this suitable subarrangement as initial arrangement, and then win by using the previous strategy.\n\nLet $a_1, \\ldots, a_n$ be the initial arrangement; by assumption, it is not suitable. Let $x_0 = 0$, and $x_k = a_k - x_{k-1}$ for $k = 1, \\ldots, n$. Setting $a_{n+1} = 0$, it follows that $x_k > a_{k+1}$ for some $k \\le n$. Let $m$ be the least index such that $x_m > a_{m+1}$, so $x_k \\ge 0$ for all $k < m$.\n\nBob's first move is to place the token on the $m$-th position. After Bob's first move, $a_m$ changes to $a_m - 1 = (x_{m-1} + x_m) - 1 = x_{m-1} + (x_m - 1) \\ge x_{m-1} + a_{m+1}$. He may therefore place the token on the $m$-th position at least another $x_{m-1} + a_{m+1}$ times, regardless of Alice's moves. Thus, if still legal, her $(x_{m-1} + 1)$-st move — call it move $(*)$ — consists in placing the token on the $(m-1)$-st position.\n\nJust before move $(*)$, the first $m-1$ positions in the arrangement are $a_1, \\ldots, a_{m-2}, x_{m-2}$, since $a_{m-1} = x_{m-2} + x_{m-1}$ dropped by $x_{m-1}$, and no move has been made on one of the first $m-2$ positions. At this stage, Bob can reset the whole game within the first $m-1$ positions: The $(m-1)$-element arrangement $a_1, \\ldots, a_{m-2}, x_{m-2}$ is suitable, since $x_{m-2} \\ge 0$, and he may reset $x_{m-1} = 0$ to fulfil the conditions in the definition. If still legal, move $(*)$ can now be looked upon as the first move in this suitable arrangement. Finally, by keeping on placing the token on one of these positions, Bob forces Alice to do so. This completes the argument and concludes the proof.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17408,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the value of $\\frac{(2112-2021)^2}{169}$?\n\n(A) 7  \n(B) 21  \n(C) 49  \n(D) 64  \n(E) 91",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that $2112 - 2021 = 91 = 7 \\cdot 13$ and that $169 = 13^2$. Thus\n\n$$\n\\frac{(2112 - 2021)^2}{169} = \\frac{(7 \\cdot 13)^2}{13^2} = 7^2 = 49.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17409,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABD$ be a triangle with circumcircle $\\omega$. Suppose that $\\omega'$ is a circle tangent to sides $AB$ and $AD$ at $E$ and $F$, respectively, and tangent to $\\omega$ at $T$. $M$ and $N$ are midpoints of the shorter arcs $\\widehat{AB}$ and $\\widehat{AD}$, respectively. Then, the tangents to $\\omega$ at $A$ and $T$, and the line $MN$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof of lemma.*\n\n$$\n\\angle NAF = \\frac{1}{2} \\widehat{ND} = \\frac{1}{2} \\widehat{NA} = \\angle ATN,\n$$\nso triangles $TAN$ and $AFN$ are similar, so $\\frac{NF}{NA} = \\frac{NA}{NT}$, and consequently $NA^2 = NF \\cdot NT$.\n\nIn a similar way, we can show that $MA^2 = ME \\cdot MT$. Thus,\n\n$$\n\\left(\\frac{MA}{AN}\\right)^2 = \\frac{ME \\cdot MT}{NF \\cdot NT}. \\quad (*)\n$$\n\n$T$ is the external homothetic center of $\\omega$ and $\\omega'$. This homothety maps $E$ and $F$ to $M$ and $N$, respectively, so $EF \\parallel MN$ and by Thales' Theorem $\\frac{ME}{MF} = \\frac{MT}{NT}$. This and $(*)$ imply\n\n$$\n\\left(\\frac{MA}{NA}\\right)^2 = \\frac{ME}{NF} \\cdot \\frac{MT}{NT} = \\left(\\frac{MT}{NT}\\right)^2 \\Rightarrow \\frac{MA}{NA} = \\frac{MT}{NT}.\n$$\n\nTherefore, $M$ is the harmonic conjugate of $N$ with respect to $A$ and $T$, and hence tangents to $\\omega$ at $A$ and $T$ meet on the line $MN$. $\\square$\n\n![](images/2013_p48_data_0965fb12a9.png)\n![](images/2013_p48_data_1269417baf.png)\n\nWe keep using the notations introduced in the lemma. We showed that $K$ lies on the line $MN$ and now we want to show that $K$ lies on the line $I_1I_2$. We have\n\n$$\n\\frac{KM}{KN} = \\frac{AM \\cdot \\sin(\\angle MAK)}{AN \\cdot \\sin(\\angle NAK)} = \\frac{AM}{AN} \\cdot \\frac{\\sin(\\angle MCA)}{\\sin(\\angle NCA)} =\n$$\n$$\n\\frac{MI_2}{NI_1} \\cdot \\frac{\\sin(\\angle MCA)}{r_2} \\cdot \\frac{r_1}{\\sin(\\angle NCA)} = \\frac{MI_2}{NI_1} \\cdot \\frac{CI_1}{CI_2}.\n$$\n\nAs a result, $\\frac{MK}{KN} \\cdot \\frac{NI_1}{I_1C} \\cdot \\frac{CI_2}{I_2M} = 1$ and by Menelaus' Theorem, $K, I_1$ and $I_2$ are collinear. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17410,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all continuous functions $f : [0, \\infty) \\to [0, \\infty)$ having the properties $f(0) = 0$ and $f'(x^2) = f(x)$ for all $x \\in [0, \\infty)$.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove that any such function is the zero function.\n\nThe given relation can be written as $f'(x) = f(\\sqrt{x}) \\ge 0$ for $x \\ge 0$, so $f$ is nondecreasing, implying $f'$ is nondecreasing as well.\n\nLet $a = \\sup\\{x \\mid f(x) = 0\\}$. If $f$ is not identically zero, we must have $a \\in [0, \\infty)$. Then $f(x) = 0$ on the interval $[0, a]$ and $f(x) > 0$ on the interval $(a, \\infty)$ (by the continuity and monotonicity of $f$).\n\nBy the Mean Value Theorem on $[a, a+1]$, there exists $c \\in (a, a+1)$ such that $f(a+1) = f'(c)$. Then $f(a+1) = f(\\sqrt{c})$, and as $f$ is nondecreasing, it is constant on $[\\sqrt{c}, a+1]$, so $f'$ is zero on that interval. Thus, $f'(a+1) = f'(0) = 0$, and since $f'$ is nondecreasing, we get $f' = 0$ on $[0, a+1]$. Therefore, $f'(c) = 0 = f(a+1)$, a contradiction.\n\nHence, the only solution is $f(x) = 0$ for all $x \\ge 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17411,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(x_n)_{n \\ge 1}$ be a sequence of real numbers from $[1, \\infty)$. It is known that the sequences $(y_n^{(k)})_{n \\ge 1}$, defined by $y_n^{(k)} = \\lfloor x_n^k \\rfloor$ for $n \\ge 1$, are convergent for every $k \\in \\mathbb{N}^*$. Prove that the sequence $(x_n)_{n \\ge 1}$ is convergent.",
    "options": [],
    "answer": "See solution",
    "solution": "For $k \\in \\mathbb{N}^*$, the sequence $(y_n^{(k)})_{n \\ge 1}$ is convergent and its terms are integers. Then there exist $n_k, a_k \\in \\mathbb{N}^*$ such that $y_n^{(k)} = a_k$ for all $n \\ge n_k$. Consequently, $x_n^k \\in [a_k, a_k + 1)$ for all $n \\ge n_k$. In particular, $x_n \\in [a_1, a_1 + 1)$ for all $n \\ge n_1$, so $(x_n)_{n \\ge 1}$ is bounded.\n\nSuppose, by contradiction, that $(x_n)_{n \\ge 1}$ has two limit points $a$ and $b$ with $1 \\le a < b$. Then there exist two subsequences $(x_{i_n})_{n \\ge 1}$ and $(x_{j_n})_{n \\ge 1}$ such that $\\lim_{n \\to \\infty} x_{i_n} = a$ and $\\lim_{n \\to \\infty} x_{j_n} = b$. For any $k \\in \\mathbb{N}^*$, since $i_n, j_n \\ge n$ for all $n \\in \\mathbb{N}^*$, we have $x_{i_n}^k, x_{j_n}^k \\in [a_k, a_k + 1)$ for all $n \\ge n_k$. Thus, $x_{j_n}^k - x_{i_n}^k < 1$ for all $n \\ge n_k$. Taking the limit as $n \\to \\infty$ gives $b^k - a^k \\le 1$. Therefore, $b^k - a^k \\le 1$ for all $k \\in \\mathbb{N}^*$. However, $1 \\le a < b$ implies $\\lim_{k \\to \\infty} (b^k - a^k) = \\infty$, which is a contradiction.\n\nTherefore, the sequence $(x_n)_{n \\ge 1}$ is convergent.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17412,
    "subject": "Mathematics (Olympiad)",
    "question": "For positive integers $a$, $b$, and $c$ with $a < b < c$, consider collections of postage stamps in denominations $a$, $b$, and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to 1000 cents, let $f(a, b, c)$ be the minimum number of stamps in such a collection. Find the sum of the three least values of $c$ such that $f(a, b, c) = 97$ for some choice of $a$ and $b$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that $a$ must be $1$. To make every positive integer number of cents up to $c-1$ requires at most $c-1$ stamps of denominations $1$ and $b$, and the number of $c$-cent stamps then required to make every positive integer number of cents up to $1000$ is at most\n\n$$\n\\left\\lfloor \\frac{1000 - (c - 1)}{c} \\right\\rfloor = \\left\\lfloor \\frac{1001}{c} \\right\\rfloor - 1 = \\left\\lfloor \\frac{1000}{c} \\right\\rfloor.\n$$\n\nHence, the possible values of $c$ must satisfy the inequality\n\n$$\n(c - 1) + \\left\\lfloor \\frac{1000}{c} \\right\\rfloor \\geq 97,\n$$\n\nimplying that $c^2 - 98c + 1000 \\geq 0$. Thus $c \\leq 49 - \\sqrt{1401}$ or $c \\geq 49 + \\sqrt{1401}$, and the integer solutions are $c \\leq 11$ or $c \\geq 87$. Let the number of stamps of denominations $1$, $b$, and $c$ be $n_a$, $n_b$, and $n_c$, respectively, and consider two cases:\n\n*Case $c \\leq 11$*: Note that $97 \\cdot 10 = 970 < 1000$, so no value of $c$ less than $11$ is possible, but $11$ is a possible value of $c$ if $b = 7$ and $(n_a, n_b, n_c) = (6, 1, 90)$.\n\n*Case $c \\geq 87$*: If $c = 87$, then $n_c \\leq 11$ and $n_a + n_b \\leq 86$, and $87$ is a possible value of $c$ only if equality holds in both cases. However, if $n_a + n_b = 86$, then the collection contains $85$ $1$-cent stamps and $1$ $86$-cent stamp. The total value of those stamps in cents is $171$, and because $1000 - 171 < 10 \\cdot 87$, only $10$ $87$-cent stamps are needed to make all of the required amounts.\n\nIf $c = 88$, then $n_c \\leq 11$ and $n_a + n_b \\leq 87$, so if $88$ is a possible value of $c$, then $(n_a + n_b, n_c) = (86, 11)$ or $(n_a + n_b, n_c) = (87, 10)$. In fact, both $88$ and $89$ are possible values of $c$ if $b = 87$ and $(n_a, n_b, n_c) = (86, 1, 10)$.\n\nThe sum of the three least possible values of $c$ is therefore $11 + 88 + 89 = 188$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17413,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x) = (x + a)(x + b)$ where $a, b$ are given positive real numbers, and $n \\geq 2$ is a given integer. For non-negative real numbers $x_1, x_2, \\dots, x_n$ that satisfy $x_1 + x_2 + \\dots + x_n = 1$, find the maximum of\n$$\nF = \\sum_{1 \\leq i < j \\leq n} \\min\\{f(x_i), f(x_j)\\}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.**\n\nWe have\n\n$$\n\\begin{align*}\n\\min\\{f(x_i), f(x_j)\\} &= \\min\\{(x_i + a)(x_i + b), (x_j + a)(x_j + b)\\} \\\\\n&\\leq \\sqrt{(x_i + a)(x_i + b)(x_j + a)(x_j + b)} \\\\\n&\\leq \\frac{1}{2} \\left[ (x_i + a)(x_j + b) + (x_i + b)(x_j + a) \\right] \\\\\n&= x_i x_j + \\frac{1}{2}(x_i + x_j)(a + b) + ab,\n\\end{align*}\n$$\n\nso\n\n$$\n\\begin{align*}\nF &\\leq \\sum_{1 \\leq i < j \\leq n} x_i x_j + \\frac{a+b}{2} \\sum_{1 \\leq i < j \\leq n} (x_i + x_j) + \\binom{n}{2} ab \\\\\n&= \\frac{1}{2} \\left[ \\left( \\sum_{i=1}^n x_i \\right)^2 - \\sum_{i=1}^n x_i^2 \\right] + \\frac{a+b}{2} (n-1) \\sum_{i=1}^n x_i + \\binom{n}{2} ab \\\\\n&= \\frac{1}{2} \\left( 1 - \\sum_{i=1}^n x_i^2 \\right) + \\frac{n-1}{2} (a+b) + \\binom{n}{2} ab \\\\\n&\\leq \\frac{1}{2} \\left( 1 - \\frac{1}{n} \\left( \\sum_{i=1}^n x_i \\right)^2 \\right) + \\frac{n-1}{2} (a+b) + \\binom{n}{2} ab \\\\\n&= \\frac{1}{2} \\left( 1 - \\frac{1}{n} \\right) + \\frac{n-1}{2} (a+b) + \\frac{n(n-1)}{2} ab \\\\\n&= \\frac{n-1}{2} \\left( \\frac{1}{n} + a + b + n ab \\right).\n\\end{align*}\n$$\n\nThe equality holds when $x_1 = x_2 = \\cdots = x_n = \\frac{1}{n}$. So the maximum of $F$ is\n\n$$\n\\frac{n-1}{2} \\left( \\frac{1}{n} + a + b + n ab \\right).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17414,
    "subject": "Mathematics (Olympiad)",
    "question": "A domino is a $2 \\times 1$ or $1 \\times 2$ tile. Determine in how many ways exactly $n^2$ dominoes can be placed without overlapping on a $2n \\times 2n$ chessboard so that every $2 \\times 2$ square contains at least two uncovered unit squares which lie in the same row or column.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $\\binom{2n}{n}^2$.\n\nDivide the chessboard into $2 \\times 2$ squares. There are exactly $n^2$ such squares on the chessboard. Each of these squares can have at most two unit squares covered by the dominos. As the dominos cover exactly $2n^2$ squares, each of them must have exactly two unit squares which are covered, and these squares must lie in the same row or column.\n\nWe claim that these two unit squares are covered by the same domino tile. Suppose that this is not the case for some $2 \\times 2$ square and one of the tiles covering one of its unit squares sticks out to the left. Then considering one of the leftmost $2 \\times 2$ squares in this division with this property gives a contradiction.\n\nNow consider this $n \\times n$ chessboard consisting of $2 \\times 2$ squares of the original board. Define $A, B, C, D$ as the following configurations on the original chessboard, where the gray squares indicate the domino tile, and consider covering this $n \\times n$ chessboard with the letters $A, B, C, D$ in such a way that the resulting configuration on the original chessboard satisfies the condition of the question.\n\nNote that then a square below or to the right of one containing an $A$ or $B$ must also contain an $A$ or $B$. Therefore, the (possibly empty) region consisting of all squares containing $A$ or $B$ abuts the lower right corner of the chessboard and is separated from the (possibly empty) region consisting of all squares containing a $C$ or $D$ by a path which goes from the lower left corner to the upper right corner of this chessboard and which moves up or right at each step.\n\nA similar reasoning shows that the (possibly empty) region consisting of all squares containing an $A$ or $D$ abuts the lower left corner of the chessboard and is separated from the (possibly empty) region consisting of all squares containing a $B$ or $C$ by a path which goes from the upper left corner to the lower right corner of this chessboard and which moves down or right at each step.\n\nTherefore, the $n \\times n$ chessboard is divided by these two paths into four (possibly empty) regions that consist respectively of all squares containing $A$, $B$, $C$, or $D$. Conversely, choosing two such paths and filling the four regions separated by them with $A$s, $B$s, $C$s, and $D$s counterclockwise starting at the bottom results in a placement of the dominos on the original board satisfying the condition of the question.\n\nAs each of these can be chosen in $\\binom{2n}{n}$ ways, there are $\\binom{2n}{n}^2$ ways the dominos can be placed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17415,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M = \\{(a, b, c) \\in \\mathbb{R}^3 : 0 < a, b, c < \\frac{1}{2}$ with $a + b + c = 1\\}$ and $f: M \\to \\mathbb{R}$ given as\n\n$$\nf(a, b, c) = 4\\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right) - \\frac{1}{abc}\n$$\n\nFind the best (real) bounds $\\alpha$ and $\\beta$ such that\n\n$$\nf(M) = \\{f(a, b, c) : (a, b, c) \\in M\\} \\subseteq [\\alpha, \\beta]\n$$\n\nand determine whether any of them is achievable.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\forall (a, b, c) \\in M$, $\\alpha \\leq f(a, b, c) \\leq \\beta$ and suppose that there are no better bounds, i.e., $\\alpha$ is the largest possible and $\\beta$ is the smallest possible. Now,\n\n$$\n\\begin{align*}\n\\alpha \\leq f(a, b, c) \\leq \\beta &\\Leftrightarrow \\alpha abc \\leq 4(ab + bc + ca) - 1 \\leq \\beta abc \\\\\n&\\Leftrightarrow (\\alpha - 8)abc \\leq 4(ab + bc + ca) - 8abc - 1 \\leq (\\beta - 8)abc \\\\\n&\\Leftrightarrow (\\alpha - 8)abc \\leq 1 - 2(a + b + c) + 4(ab + bc + ca) - 8abc \\leq (\\beta - 8)abc \\\\\n&\\Leftrightarrow (\\alpha - 8)abc \\leq (1 - 2a)(1 - 2b)(1 - 2c) \\leq (\\beta - 8)abc\n\\end{align*}\n$$\n\nFor $\\alpha < 8$, we have\n\n$$\n(1 - 2a)(1 - 2b)(1 - 2c) \\geq 0 > (\\alpha - 8)abc.\n$$\n\nSo $\\alpha \\geq 8$. But if we take $\\varepsilon > 0$ small and $a = b = \\frac{1}{4} + \\varepsilon$, $c = \\frac{1}{2} - 2\\varepsilon$, we'll have:\n\n$$\n(\\alpha - 8)\\left(\\frac{1}{4} + \\varepsilon\\right)\\left(\\frac{1}{4} + \\varepsilon\\right)\\left(\\frac{1}{2} - 2\\varepsilon\\right) \\leq \\left(\\frac{1}{2} - 2\\varepsilon\\right)\\left(\\frac{1}{2} - 2\\varepsilon\\right)4\\varepsilon\n$$\n\nTaking $\\varepsilon \\to 0^+$, we get $\\alpha - 8 \\leq 0$. So $\\alpha = 8$ and it can never be achieved. For the right side, note that there is a triangle whose side-lengths are $a, b, c$. For this triangle, denote $p = \\frac{1}{2}$ the half-perimeter, $S$ the area, and $r, R$ respectively the radius of incircle and circumcircle. Using the relations $R = \\frac{abc}{4S}$ and $S = pr$, we will have:\n\n$$\n\\begin{align*}\n(1 - 2a)(1 - 2b)(1 - 2c) \\leq (\\beta - 8)abc &\\Leftrightarrow (p - a)(p - b)(p - c) \\leq \\frac{(\\beta - 8)abc}{8} \\\\\n&\\Leftrightarrow \\frac{S^2}{p} \\leq \\frac{(\\beta - 8)abc}{8} \\\\\n&\\Leftrightarrow 2\\frac{S}{p} \\leq \\frac{(\\beta - 8)abc}{4S} \\\\\n&\\Leftrightarrow \\frac{R}{r} \\geq 2(\\beta - 8)^{-1}\n\\end{align*}\n$$\n\nSince the least value of $\\frac{R}{r}$ is $2$ (this is a well-known classic inequality), and it is achievable at $a = b = c = \\frac{1}{3}$, we must have $\\beta = 9$.\n\n**Answer:** $\\alpha = 8$ (not achievable) and $\\beta = 9$ (achievable).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17416,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider an equilateral triangle $ABC$ and a point $D$ outside the plane of $ABC$ such that the angle between mutual perpendicular lines from $D$ to the sides of $ABC$ is $60^\\circ$. Prove that $B_1$, $C_1$, and $A_1$ (the feet of the perpendiculars from $D$ to sides $AC$, $AB$, and $BC$, respectively) are the midpoints of sides $AC$, $AB$, and $BC$, respectively.",
    "options": [],
    "answer": "See solution",
    "solution": "Form the net of $ABCD$ as in the following picture, where $AB' = AC' = AD$, $BA' = BC' = BD$, and $CA' = CB' = CD$.\n\n![](images/2016_p27_data_e13926f5c8.png)\n\nLet point $E$ be located on the plane such that $EC_1 = C'C_1$ and $EB_1 = B'B_1$. First, suppose that $E \\neq A$. Then triangles $EB_1C_1$ and $DB_1C_1$ are congruent, so $\\angle C_1EB_1 = \\angle C_1DB_1 = 60^\\circ$. Therefore, quadrilateral $EAB_1C_1$ is cyclic. Using the Pythagorean theorem for segments $AC'$ and $AB'$ gives:\n\n$$\nAC_1^2 + EC_1^2 = AC_1^2 + C'C_1^2 = AC'^2 = AB'^2 = AB_1^2 + B'B_1^2 = AB_1^2 + EB_1^2.\n$$\n\nOn the other hand, using the cosine theorem for side $AE$ in triangles $AEC_1$ and $AEB_1$ gives:\n\n$$\nAB_1^2 + EB_1^2 - 2 AB_1 \\cdot EB_1 \\cos \\alpha = AC_1^2 + EC_1^2 - 2 AC_1 \\cdot EC_1 \\cos \\alpha.\n$$\n\nSince $\\cos \\alpha \\neq 0$ ($E \\neq A$), $AC_1 \\cdot EC_1 = AB_1 \\cdot EB_1$. This implies that the areas of triangles $AEC_1$ and $AEB_1$ are equal, so $AE \\parallel B_1C_1$. Now, since $EAB_1C_1$ is cyclic, it follows that $AB_1 = EC_1 = C'C_1$ and $AC_1 = EB_1 = B'B_1$.\n\nBy defining points $F$ and $G$ similarly to $E$ and following the same argument, it can be concluded that $A'A_1 = AB_1 = AC_1$, which implies $E = A$.\n\nNow, assuming $E = A$, $AC_1 = C'C_1$ and $AB_1 = B'B_1$. Since $AC' = AB'$ and $AB_1 = AC_1 = \\frac{\\sqrt{2}}{2} AC'$, it follows that $CB_1 = BC_1$. Therefore, $B'C = C'B$, which results in $BA' = C'B = B'C = CA'$. This implies that $A_1$ is the midpoint of $BC$. Continuing this argument with points $F$ and $G$, it is easy to show that $B_1$ is the midpoint of $AC$ and $C_1$ is the midpoint of $AB$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17417,
    "subject": "Mathematics (Olympiad)",
    "question": "Transform the expression on the left side:\n\n$$\n\\frac{1}{2!} + \\frac{2}{3!} + \\dots + \\frac{2^{n-2}}{n!}\n$$\n\nand show that it is less than or equal to $\\frac{3}{2}$.",
    "options": [],
    "answer": "See solution",
    "solution": "First solution. Transform the expression on the left side:\n\n$$\n\\begin{aligned}\n\\frac{1}{2!} + \\frac{2}{3!} + \\dots + \\frac{2^{n-2}}{n!} &= \\frac{1}{2!} \\left( 1 + \\frac{2}{3} + \\frac{2^2}{3 \\cdot 4} + \\dots + \\frac{2^{n-2}}{3 \\cdot 4 \\dots n} \\right) \\\\ &\\leq \\frac{1}{2} \\left( 1 + \\frac{2}{3} + \\frac{2^2}{3^2} + \\dots + \\frac{2^{n-2}}{3^{n-2}} \\right) = \\frac{1}{2} \\left( 1 + \\frac{2}{3} + \\left( \\frac{2}{3} \\right)^2 + \\dots + \\left( \\frac{2}{3} \\right)^{n-2} \\right) \\\\ &\\leq \\frac{1}{2} \\left( 1 + \\frac{2}{3} + \\left( \\frac{2}{3} \\right)^2 + \\dots + \\left( \\frac{2}{3} \\right)^{n-2} + \\dots \\right) = \\frac{1}{2} \\cdot \\frac{1}{1 - \\frac{2}{3}} = \\frac{3}{2}.\n\\end{aligned}\n$$\n\nSecond solution. We will prove by induction a stronger inequality:\n\n$$\n\\frac{1}{2!} + \\frac{2}{3!} + \\dots + \\frac{2^{n-2}}{n!} \\leq \\frac{3}{2} - \\frac{1}{n}, \\quad (*)\n$$\n\nfrom which, obviously, follows the required. It is easy to check that at $n = 2, 3, 4, 5$ the inequality $(*)$ holds — it is the base of induction.\n\nInduction step: assume that the inequality $(*)$ is true for some $n = k$. To prove $(*)$ for $n = k+1$ it is enough to show that the right side of this inequality increases faster than the left side, i.e., it is enough to prove the inequality:\n\n$$\n\\frac{2^{k-1}}{(k+1)!} \\leq \\frac{1}{k} - \\frac{1}{k+1} = \\frac{1}{k(k+1)}.\n$$\n\nThe latter inequality is equivalent to $2^{k-1} \\leq (k-1)!$. Note that for $k=5$ it is true: $2^{5-1} = 16 < 24 = (5-1)!$ and for $k > 5$ it is true since with an increase of $k$ the left side is multiplied by 2 and the right side is multiplied by a number greater than 2.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17418,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $(p, q)$ such that\n\n$$\n\\frac{3}{2014} = \\frac{1}{p} + \\frac{1}{q}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\frac{3}{2014} = \\frac{1}{2014} + \\frac{1}{1007} = \\frac{1}{26182} + \\frac{1}{689} = \\frac{1}{36252} + \\frac{1}{684} = \\frac{1}{676704} + \\frac{1}{672} = \\frac{1}{912} + \\frac{1}{2544} = \\frac{1}{7049} + \\frac{1}{742}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17419,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that $a > 0$ and the minima of the function $f(x) = x + \\frac{100}{x}$ on the intervals $(0, a]$ and $[a, +\\infty)$ are $m_1$ and $m_2$, respectively. If $m_1 m_2 = 2020$, then the value of $a$ is ______.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $f(x)$ is monotonically decreasing on $(0, 10]$ and monotonically increasing on $[10, +\\infty)$. When $a \\in (0, 10]$, $m_1 = f(a)$ and $m_2 = f(10)$; when $a \\in [10, +\\infty)$, $m_1 = f(10)$ and $m_2 = f(a)$. Therefore,\n\n$$\nf(a) f(10) = m_1 m_2 = 2020,\n$$\n\nnamely, $a + \\frac{100}{a} = \\frac{2020}{20} = 101$. The solution is $a = 1$ or $a = 100$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17420,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\setminus \\{0\\} \\to \\mathbb{R}$ such that\n\n$$\nf\\left(\\frac{1}{x}\\right) \\ge 1 - f(x) \\ge x^2 f(x)\n$$\n\nfor all $x \\in \\mathbb{R} \\setminus \\{0\\}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The solution is $f(x) = \\frac{1}{x^2 + 1}$.\n\nWe have\n\n$$\n1 - f(x) \\ge x^2 f(x) \\iff \\frac{1}{x^2 + 1} \\ge f(x).\n$$\n\nIf we substitute $z = \\frac{1}{x}$ into $f\\left(\\frac{1}{x}\\right) \\ge 1 - f(x)$, we get:\n\n$$\nf(z) \\ge 1 - f\\left(\\frac{1}{z}\\right) \\quad \\forall z \\in \\mathbb{R} \\setminus \\{0\\}.\n$$\n\nCombining these results:\n\n$$\n\\frac{1}{x^2 + 1} \\ge f(x) \\ge 1 - f\\left(\\frac{1}{x}\\right) \\quad \\forall x \\in \\mathbb{R} \\setminus \\{0\\}\n$$\n\nThis implies\n\n$$\nf\\left(\\frac{1}{x}\\right) \\ge 1 - \\frac{1}{x^2 + 1} = \\frac{x^2}{x^2 + 1} = \\frac{1}{\\frac{1}{x^2} + 1} \\quad \\forall x \\in \\mathbb{R} \\setminus \\{0\\}\n$$\n\nSubstituting $z = \\frac{1}{x}$ again, we get\n\n$$\nf(z) \\ge \\frac{1}{z^2 + 1} \\quad \\forall z \\in \\mathbb{R} \\setminus \\{0\\}.\n$$\n\nHence,\n\n$$\n\\frac{1}{x^2 + 1} \\ge f(x) \\ge \\frac{1}{x^2 + 1} \\quad \\forall x \\in \\mathbb{R} \\setminus \\{0\\}\n$$\n\nTherefore,\n\n$$\nf(x) = \\frac{1}{x^2 + 1} \\quad \\forall x \\in \\mathbb{R} \\setminus \\{0\\}\n$$\n\nwhich is a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17421,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D$ be the point on the arc $AC$ of the circumcircle of triangle $ABC$ (with $AB < BC$) that does not contain point $B$. Let $X$ and $X'$ be any two points on side $AC$ such that $\\angle ABX = \\angle CBX'$. Show that, regardless of the choice of point $X$, the circumcircle of $\\triangle DXX'$ passes through a fixed point different from $D$.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/Ukraine_2020_booklet_p8_data_7b27172dc3.png)\n\nLet $W$ be the circumcircle of $\\triangle ABC$, let $w$ be the circumcircle of $\\triangle BXX'$, and let $v$ be the circumcircle of $\\triangle DXX'$. Let $TB$ be a tangent line to the circle $W$, where $T$ belongs to the line $AC$. Then $TB^2 = TA \\cdot TC$, and also\n\n$$\n\\angle TBX = \\angle TBA + \\angle ABX = \\angle ACB + \\angle CBX' = \\angle BX'X.\n$$\n\nThus, the line $TB$ is also tangent to the circle $w$, so $TB^2 = TX \\cdot TC$. Let $w \\cap TD = F \\neq D$. Then $TX \\cdot TC = TF \\cdot TD = TB^2$. Thus, the points $X, X', D, F$ lie on the circle $v$ (by the property of inscribed quadrilaterals). Since $T$, $B$, and $D$ are fixed points, so is the point $W \\cap TD = F$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17422,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a $3 \\times 3$ grid filled with the numbers $1$ through $9$ (each used exactly once). \n\n(a) Is it possible to arrange the numbers so that the sum of the numbers in every row, every column, and both main diagonals is not a multiple of $3$? Give an example or explain why not.\n\n(b) Is it possible to arrange the numbers so that the sum of the numbers in every row, every column, and both main diagonals is a multiple of $3$? Justify your answer.\n\n![](images/obm-2011_p11_data_c95a294877.png)",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Yes, it is possible. For example, consider the following arrangement:\n\n<table><tr><td>1</td><td>2</td><td>3</td></tr><tr><td>4</td><td>5</td><td>6</td></tr><tr><td>8</td><td>9</td><td>7</td></tr></table>\n\nThe trick is to only adjust the last row. The usual order $7, 8, 9$ yields all sums to be multiples of $3$, so it's just a matter of rearranging them.\n\n(b) No, it is not possible. First, notice that the sum of three numbers $x, y, z$ is a multiple of $3$ if and only if $x \\equiv y \\equiv z \\pmod{3}$ or $x, y, z$ are $0, 1, 2$ mod $3$ in some order. Let $a, b, c, d$ be the numbers in the corners modulo $3$. So two of them are equal. We can suppose without loss of generality that they are either $a = b$ or $a = d$. Also, let $x$ be the number in the central cell modulo $3$.\n\n<table><tr><td>a</td><td>b</td></tr><tr><td>x</td><td></td></tr><tr><td>c</td><td>d</td></tr></table>\n\nIf $a = d$, then $x \\neq a$ and $x$ is equal to either $b$ or $c$. Suppose without loss of generality $x = b \\neq a$. Then we have the following situation:\n\n<table><tr><td>a</td><td>b</td></tr><tr><td></td><td>b</td></tr><tr><td>c</td><td>a</td></tr></table>\n\nLet $m$ be the other remainder (that is, $m \\neq a$ and $m \\neq b$). Then $m$ cannot be in the same line as $a$ and $b$. This leaves only one possibility:\n\n<table><tr><td>a</td><td>b</td></tr><tr><td>m</td><td>b</td></tr><tr><td>m</td><td>m</td><td>a</td></tr></table>\n\nBut the remaining $a$ will necessarily yield a line with all three remainders. Now if $a = b$, then both $c$ and $d$ are different from $a$ (otherwise, we reduce the problem to the previous case). If $d \\neq c$, $a, c, d$ are the three distinct remainders, and we have no possibility for $x$. So $c = d$.\n\n<table><tr><td>a</td><td>a</td></tr><tr><td></td><td>x</td></tr><tr><td>c</td><td>c</td></tr></table>\n\nBut this prevents the other remainder $m$ to appear in the middle row, leaving only two cells for three numbers, which is not possible.\n\nSo, in both cases, one of the sums is a multiple of $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17423,
    "subject": "Mathematics (Olympiad)",
    "question": "The teacher wrote the digits $123\\ldots8123\\ldots8123\\ldots$ on the board repeatedly until a 2018-digit number was formed. After that, Andriy and Olesya played a game as follows: alternately (with Andriy starting), they cross out 2 digits in one of these ways:\n- the first two digits of the remaining number,\n- the last two digits,\n- or the first and last digits.\nThe game ends when only a two-digit number remains. Olesya wins if this number is a multiple of 4; otherwise, Andriy wins. Who will win if both players play optimally?",
    "options": [],
    "answer": "See solution",
    "solution": "Since only divisibility by 4 matters, we can group the digits as follows: $1, 5 \\rightarrow 1$, $2, 6 \\rightarrow 2$, $3, 7 \\rightarrow 3$, and $4, 8 \\rightarrow 4$. This reduces the problem to an equivalent form. Olesya's strategy is: before Andriy's move, she selects the first two and last two digits, ensuring that after Andriy's move, all four are crossed out. This guarantees that after each pair of moves, the number left will contain the digits $1$ and $2$. Olesya can then leave these two digits after her final move to win.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17424,
    "subject": "Mathematics (Olympiad)",
    "question": "The positive integer represented by the expression $10^{2016} - 10^{15}$ has 2016 digits. The last 15 of them are 0 and the rest are equal to 9. What is the sum of its digits?",
    "options": [],
    "answer": "See solution",
    "solution": "The number $10^{2016} - 10^{15}$ has 2016 digits: the last 15 are 0, and the remaining $2016 - 15 = 2001$ digits are 9. Thus, the sum of its digits is $2001 \\times 9 = 18\\,009$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17425,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the maximum number of prime numbers that a set $S$ can contain, given the following property:\n\nFor any selection of 100 numbers from $S$, the product of these 100 numbers shares at most as many prime factors (counted with multiplicity) as the product of the remaining numbers in $S$?\n\nFind and justify the maximum possible number of primes in such a set $S$.",
    "options": [],
    "answer": "See solution",
    "solution": "The maximum number of primes is $1819$.\n\n**Construction:**\n\nChoose distinct primes $p_1, p_2, \\dots, p_{1819}$, and let $P = p_1 p_2 \\cdots p_{1819}$. Define\n\n$$\nS = \\{p_1, p_2, \\dots, p_{1819}, P, P \\cdot p_1, \\dots, P \\cdot p_{199}\\}.\n$$\n\nFor each $p_i$, there are $201$ numbers in $S$ divisible by $p_i$ (namely, $p_i$ and all multiples of $P$). Of these, at most one has two factors $p_i$; the rest have only one factor $p_i$. If we take $100$ numbers from $S$, their product has at most $101$ factors $p_i$. The other numbers contain at least $101$ numbers divisible by $p_i$, so their product has at least $101$ factors $p_i$. This holds for any $p_i$, and the numbers in $S$ do not have any other prime factors, so $S$ has the desired property.\n\n**Upper Bound:**\n\nSuppose $S$ contains more than $1819$ primes. Consider a prime divisor $q$ of a number in $S$. If at most $199$ numbers in $S$ are divisible by $q$, then taking the $100$ elements with the most factors $q$ gives a contradiction, as their product would have more factors $q$ than the rest. Thus, at least $200$ numbers in $S$ are divisible by $q$. If exactly $200$, then the number of factors $q$ in all these numbers must be equal, or else a similar contradiction arises.\n\n$S$ contains at least $199$ non-primes, since a prime $p$ in $S$ divides at least $199$ other elements. If $S$ contains exactly $199$ non-primes, then each such non-prime must be the product of the primes in $S$, but then these $199$ numbers are not distinct, which is a contradiction. Therefore, $S$ must contain at least $200$ non-primes, and thus at most $1819$ primes.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17426,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle, and let $D$, $E$, $F$ be the feet of the altitudes from $A$, $B$, $C$, respectively. The lines $BC$ and $EF$ cross at $P$, and the line through $D$ and parallel to $EF$ crosses the lines $AC$ and $AB$ at $Q$ and $R$, respectively. Prove that the circle $PQR$ passes through the midpoint of the side $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of the side $BC$.\n\nIf $AB = AC$, then $P$ is the ideal point of the line $BC$, the points $Q$ and $R$ fall at $C$ and $B$, respectively, and the circle $PQR$ degenerates into the line $BC$ on which $M$ clearly lies.\n\n![](images/RMC_2019_var_3_p33_data_44fb9e397b.png)\n\nAssume henceforth that $AB \\neq AC$, say, $AB > AC$. It is clearly sufficient to show that\n$$\nDM \\cdot DP = DQ \\cdot DR.\n$$\nSince $EF$ and $QR$ are parallel, and $B$, $C$, $E$, $F$ are concyclic ($E$ and $F$ both lie on the circle on diameter $BC$), so are $B$, $C$, $Q$, $R$. Hence $DB \\cdot DC = DQ \\cdot DR$, and it is therefore sufficient to show that $DB \\cdot DC = DM \\cdot DP$, i.e., $BM^2 = DM \\cdot MP$, since $DB = BM + DM$, $DC = BM - DM$ and $DP = MP - DM$. Alternatively, but equivalently, $DP \\cdot MP = MP^2 - BM^2$, since $DM = MP - DP$.\n\nThe points $D$, $E$, $F$, $M$ are concyclic (they all lie on the nine-point circle of the triangle $ABC$), so $PD \\cdot PM = PE \\cdot PF$. The points $B$, $C$, $E$, $F$ are also concyclic (recall that $E$ and $F$ both lie on the circle on diameter $BC$), so $PE \\cdot PF = PB \\cdot PC$. Consequently, $DP \\cdot MP = PB \\cdot PC = (BM + MP)(MP - CM) = (MP + BM)(MP - BM) = MP^2 - BM^2$, as desired.\n\n*Remarks.* Acuteness of the triangle $ABC$ was assumed to avoid degeneracy (such as rightness at $B$ or $C$) or case analysis (the argument applies mutatis mutandis in case of obtuseness). Rightness at $A$ causes no trouble: The line $EF$ is the line through $A$ and perpendicular to $AM$, i.e., the tangent of the circle $ABC$ at $A$, and the argument goes along the same lines.\n\nKeeping $B$ and $C$ fixed, while varying $A$ in the plane so as to avoid the lines $BC$ and the lines through $B$ and $C$, respectively, and perpendicular to $BC$, the circle $PQR$ passes through a fixed point – the midpoint of the segment $BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17427,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be positive real numbers. Prove that\n\n$$\n\\frac{a^2 b (b-c)}{a+b} + \\frac{b^2 c (c-a)}{b+c} + \\frac{c^2 a (a-b)}{c+a} \\geq 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Dividing both sides by $abc$, the inequality is equivalent to\n\n$$\n\\frac{a(b-c)}{c(a+b)} + \\frac{b(c-a)}{a(b+c)} + \\frac{c(a-b)}{b(c+a)} \\geq 0.\n$$\n\nBy adding $1+1+1$ to both sides, the inequality becomes\n\n$$\n\\frac{b(c+a)}{c(a+b)} + \\frac{c(a+b)}{a(b+c)} + \\frac{a(b+c)}{b(c+a)} \\geq 3.\n$$\n\nBy applying the AM-GM inequality, we have\n\n$$\n\\frac{b(c+a)}{c(a+b)} + \\frac{c(a+b)}{a(b+c)} + \\frac{a(b+c)}{b(c+a)} \\geq 3 \\cdot \\sqrt[3]{\\frac{b(c+a)}{c(a+b)} \\cdot \\frac{c(a+b)}{a(b+c)} \\cdot \\frac{a(b+c)}{b(c+a)}} = 3.\n$$\n\nThe inequality is proved. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17428,
    "subject": "Mathematics (Olympiad)",
    "question": "A room in the shape of a rectangular parallelepiped has vertical walls covered by mirrors. A laser beam of diameter $0$ enters the room from one corner and moves horizontally along the bisector of that corner. After reflecting from a wall, the beam continues moving horizontally according to the laws of reflection (i.e., the bisector of the angle between the imaginary continuation of the trajectory of the beam before reflection and the real continuation trajectory is along the wall). When the beam reaches a corner, it will return along the way it arrived.\n\n(a) Prove that if the ratio of the side lengths of the floor of the room is rational, then the beam eventually returns to the point of entrance.\n\n(b) Prove that if the ratio of the side lengths of the floor of the room is irrational, then the beam never returns to the point of entrance.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a$ and $b$ be the side lengths of the floor. We use a planar coordinate system whose axes go along the sides of the floor.\n\n![](images/EST_ABooklet_2020_p21_data_785f079aa3.png)\n\nAfter reflection from a wall, one coordinate of the beam starts changing in the opposite direction while retaining its speed. As the beam enters the room along a corner bisector, both coordinates change with the same speed. Without loss of generality, let the speed be $1$, then the coordinates change with period $2a$ and $2b$, respectively.\n\n(a) If $\\frac{a}{b}$ is rational, then $\\frac{a}{b} = \\frac{m}{n}$ where $m$ and $n$ are integers. Denote $t = 2a \\cdot n = 2b \\cdot m$. Then at the time instance $t$, both coordinates have gone through a whole number of periods, which means that the beam has reached the point of entrance.\n\n(b) Suppose that the beam reaches the point of entrance at a time instance $t$. Then $\\frac{t}{2a}$ and $\\frac{t}{2b}$ are integers; denote $n = \\frac{t}{2a}$ and $m = \\frac{t}{2b}$, implying $a = \\frac{t}{2n}$ and $b = \\frac{t}{2m}$. Thus $\\frac{a}{b} = \\frac{m}{n}$. Consequently, in order to reach the point of entrance, $\\frac{a}{b}$ must be rational.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17429,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a given positive integer. Determine all positive divisors $d$ of $3n^2$ such that $n^2 + d$ is the square of an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "If $d$ divides $3n^2$, then there exist positive integers $k$ and $m$ such that $3n^2 = d \\cdot k$ and $n^2 + d = m^2$. Substitute $d = \\frac{3n^2}{k}$ to get:\n\n$$\nn^2 + \\frac{3n^2}{k} = m^2\n$$\n\nMultiply both sides by $k$:\n\n$$\nk n^2 + 3n^2 = k m^2\n$$\n\nSo:\n\n$$\nn^2(k + 3) = k m^2\n$$\n\nOr:\n\n$$\n(mk)^2 = n^2(k^2 + 3k)\n$$\n\nThus, $k^2 + 3k$ must be a perfect square. Consider $k^2 < k^2 + 3k < (k+2)^2$; the only perfect square in this interval is $(k+1)^2$ when $k^2 + 3k = (k+1)^2$. This gives $k^2 + 3k = k^2 + 2k + 1$, so $k = 1$. Therefore, $d = 3n^2$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17430,
    "subject": "Mathematics (Olympiad)",
    "question": "Being able to conquer a fortress $B$, the elephant from $A$ who is watching towards $B$ must go along the line of fortresses to $B$, fight with all elephants who stand on his way and are watching opposite to him (including that of fortress $B$ but not the other elephant from $A$), and win them all. Given that all elephants have different constant weights and a heavier elephant always wins a lighter elephant, prove that there exists exactly one fortress that cannot be conquered by any other fortress.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution 1.** The elephants from the outermost fortresses who watch in the direction where there are no more fortresses never take part in any fight, so we can discount them. Until only one fortress remains on the island, repeat the following: remove the heaviest elephant together with all fortresses and elephants on his watching direction. Thanks to the initial assumption, at least one fortress is removed on each step. It is easy to see that the fortress guarded by the heaviest elephant can conquer all fortresses that are subject to removal on the current step, while no fortress subject to removal can conquer any of the remaining fortresses since that would require winning the heaviest elephant. Hence the fortress that remains after all other fortresses are removed can be conquered by none of the other fortresses, while all other fortresses can be conquered by at least one of the other fortresses.\n\n**Solution 2.** A fortress being able to conquer another fortress implies the former fortress also being able to conquer all fortresses between the two fortresses. Let $F_1, \\dots, F_n$ be the fortresses along the line. There definitely exists a fortress which cannot be conquered by any fortress with a smaller number (for instance, $F_1$ is such). Let $F_i$ be the one with the largest number among those. We show that this fortress is the one we are looking for.\n\nSuppose that $F_j$ can conquer $F_i$. Then $j > i$ and $F_j$ can conquer also $F_{i+1}, \\dots, F_{j-1}$. As $j > i$, there must exist $l < j$ such that $F_l$ can conquer $F_j$. If $l < i$, then $F_l$ can conquer $F_i$ which contradicts the choice of $i$. If $i \\le l$, then because of $l < j$, $F_j$ can conquer $F_l$, while $F_l$ can conquer $F_j$ as well. This is impossible, since in both cases, the two elephants from $F_j$ and $F_l$ looking in the opposite direction must fight against each other.\n\nIt remains to show that there are no more unconquerable fortresses. Suppose, some $F_j$, $j \\ne i$, cannot be conquered. If $j > i$, then this contradicts the choice of $i$. Hence assume that $j < i$. Let among the elephants of $F_{j+1}, \\dots, F_i$ watching towards $F_j$, that of $F_k$ be the heaviest. As even $F_k$ cannot conquer $F_j$, there must exist $l$ such that $j \\le l < k$ and the elephant from $F_l$ watching towards $F_k$ is heavier than the elephant from $F_k$ watching towards $F_l$. But this means that $F_l$ can conquer $F_i$ which is impossible.\n\n**Solution 3.** We proceed by induction on $n$. The claim obviously holds if $n = 1$. Consider a situation with $n + 1$ fortresses and assume that the claim holds for $n$ fortresses. Let $A$ be the fortress on one end of the line and let $B$ be the next fortress. Let $x$ be the weight of the elephant from $A$ watching towards $B$, let $y$ be the weight of the elephant from $B$ watching towards $A$ and let $z$ be the weight of the other elephant from $B$. Consider three cases.\n\n1) The case $x < y$. When excluding $A$ together with its two elephants, there must exist exactly one invincible fortress $K$ among the others by the induction hypothesis. As $B$ can conquer $A$ and the elephant from $A$ cannot pass $B$, fortress $K$ is the only invincible also in the presence of $A$.\n\n2) The case $y < x < z$. When excluding $A$ together with its two elephants, there must exist exactly one invincible fortress $K$ among the others. If $K \\neq B$ then the fortress that can conquer $B$ can conquer also $A$ while $A$ cannot conquer $K$ (if $A$ could conquer $K$, also $B$ could). Consequently, $K$ is the only invincible fortress also in the presence of $A$. If $K = B$ then in the presence of $A$, $A$ would be the only invincible fortress since it can conquer $B$ while other fortresses cannot pass $B$.\n\n3) The case $y < x, z < x$. Excluding now $B$ together with its two elephants, there must exist a unique invincible fortress $L$ among the others. Note that adding $B$ does not change the correlation of forces among the other fortresses. Indeed, if $A$ can conquer another fortress in the situation without $B$ then it can do it in the presence of $B$, too, and if some fortress can conquer $A$ in the situation without $B$ then it can do it in the presence of $B$, too. But $B$ can be conquered by $A$ and if $B$ could conquer $L$ then also $A$ could. As a consequence, $L$ is the only invincible fortress also in the presence of $B$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17431,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, and $d$ be non-negative real numbers satisfying $a + b + c + d = 2$. Prove that\n\n$$\n(a^2 + b^2)(b^2 + c^2)(c^2 + d^2)(d^2 + a^2) \\le 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $a, b, c, d \\ge 0$, we have\n\n$$\n(a+b)^2 \\ge a^2 + b^2, \\quad (b+c)^2 \\ge b^2 + c^2, \\quad (c+d)^2 \\ge c^2 + d^2, \\quad (d+a)^2 \\ge d^2 + a^2.\n$$\n\nTherefore, it suffices to prove that $(a+b)(b+c)(c+d)(d+a) \\le 1$. By the Cauchy-Schwarz inequality,\n\n$$\n(a+b)(b+c)(c+d)(d+a) \\le \\left( \\frac{a+b+b+c+c+d+d+a}{4} \\right)^4 = 1.\n$$\n\nEquality holds only when $a+b = b+c = c+d = d+a$ and $ab = bc = cd = da = 0$. It is easy to check that equality holds if and only if $a = c = 1$, $b = d = 0$ or $a = c = 0$, $b = d = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17432,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $X$, $Y$ are chosen on the sides $AB$ and $AD$ of a convex quadrilateral $ABCD$ respectively so that $CX \\parallel DA$, $DX \\parallel CB$, $BY \\parallel CD$ and $CY \\parallel BA$. Find the ratio $AX : BX$.\n\n![](images/ukraine_2015_Booklet_p4_data_7cd76add15.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The ratio is $\\frac{\\sqrt{5}+1}{2}$.\n\nLet $\\lambda = \\frac{AX}{BX}$. Since $RYCD$ is a parallelogram, by Thales' theorem we obtain $\\lambda = \\frac{AX}{BX} = \\frac{YR}{RB} = \\frac{CD}{RB}$.\n\nLikewise,\n\n$$\n\\lambda = \\frac{AX}{BX} = \\frac{CY}{BP} = \\frac{CY}{BQ} = \\frac{BY}{CD} = \\frac{YR+BR}{CD} = \\frac{CD+BR}{CD} = 1 + \\frac{BR}{CD} = 1 + \\frac{1}{\\lambda}.\n$$\n\nSolving $\\lambda = 1 + \\frac{1}{\\lambda}$ gives $\\lambda = \\frac{\\sqrt{5}+1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17433,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real solutions $(x, y)$ to the equation:\n\n$$\nx^2y - x^2 + xy^2 - y^2 = 1\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The given equation can be rewritten as:\n\n$$\n\\begin{aligned}\nx^2y - x^2 + xy^2 - y^2 &= 1 \\\\\nxy(x + y) - (x^2 + y^2) &= 1 \\\\\nxy(x + y) - ((x + y)^2 - 2xy) &= 1\n\\end{aligned}\n$$\n\nLet $u = x + y$ and $v = xy$. Then the equation becomes:\n\n$$\nuv - (u^2 - 2v) = 1\n$$\n\nwhich simplifies to $uv + 2v = u^2 + 1$, or\n\n$$\nv = \\frac{u^2 + 1}{u + 2} = u - 2 + \\frac{5}{u + 2}.\n$$\n\nSince $u + 2$ must divide $5$, the possible values are $u + 2 = 5, 1, -1, -5$, giving:\n\n$$\n\\begin{array}{llll}\nu + 2 = 5 & u + 2 = 1 & u + 2 = -1 & u + 2 = -5 \\\\\n\\frac{5}{u+2} = 1 & \\frac{5}{u+2} = 5 & \\frac{5}{u+2} = -5 & \\frac{5}{u+2} = -1 \\\\\nu = 3 & u = -1 & u = -3 & u = -7 \\\\\nv = 2 & v = 2 & v = -10 & v = -10\n\\end{array}\n$$\n\nBy Vieta's formulas, $x$ and $y$ are roots of:\n\n$$\n\\begin{aligned}\nz^2 - 3z + 2 &= 0 \\\\\nz^2 + z + 2 &= 0 \\\\\nz^2 + 3z - 10 &= 0 \\\\\nz^2 + 7z - 10 &= 0\n\\end{aligned}\n$$\n\nThe first equation has solutions $1$ and $2$. The second has no real solutions. The third has solutions $-5$ and $2$. The fourth has irrational solutions.\n\nThus, the real solutions are $(x, y) \\in \\{(1, 2), (2, 1), (2, -5), (-5, 2)\\}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17434,
    "subject": "Mathematics (Olympiad)",
    "question": "З рівностей\n\n$$\nx_3 - \\frac{1}{4}x_3 - \\frac{1}{4}x_3 - 2 = 140,\n$$\n\n$$\nx_2 + \\frac{1}{4}x_3 + 1 = 140,\n$$\n\n$$\nx_1 + \\frac{1}{4}x_3 + 1 = 140\n$$\n\nзнайдіть кількість горіхів у кожного з хлопчиків на передостанньому етапі. Аналогічно відновіть весь «ланцюжок»:\n\n$$\n(140, 140, 140) \\leftarrow (68, 68, 284) \\leftarrow (32, 140, 248) \\leftarrow (68, 122, 230).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Відповідь:** Перший хлопчик зібрав 68 горіхів, другий хлопчик — 122 горіхи, третій хлопчик — 230 горіхів.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17435,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $k$ be positive integers, and let $n$ be a nonnegative integer. Show that $(ka^2 + 1)^{2n+1}$ can be expressed as a sum of $k + 1$ squares, and $(ka^2 + 1)^{2n+2}$ can be expressed as a sum of $(k + 1)^2$ squares.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we have\n\n$$\n(ka^2 + 1)^{2n+1} = (ka^2 + 1)(ka^2 + 1)^{2n}\n$$\n\nSince $ka^2 + 1 = \\underbrace{a^2 + a^2 + \\dots + a^2}_{k\\text{ times}} + 1^2$, we can write\n\n$$\n(ka^2 + 1)^{2n+1} = \\underbrace{(a(ka^2 + 1)^n)^2 + (a(ka^2 + 1)^n)^2 + \\dots + (a(ka^2 + 1)^n)^2}_{k\\text{ times}} + ((ka^2 + 1)^n)^2\n$$\n\nwhich is the sum of $k + 1$ squares.\n\nLikewise,\n\n$$\n(ka^2 + 1)^{2n+2} = (ka^2 + 1)^2 (ka^2 + 1)^{2n} = (k^2 a^4 + 2k a^2 + 1)(ka^2 + 1)^{2n}\n$$\n\nThis can be written as\n\n$$\n\\underbrace{(a^2 (ka^2 + 1)^n)^2 + \\dots + (a^2 (ka^2 + 1)^n)^2}_{k^2\\text{ times}} + \\underbrace{(a (ka^2 + 1)^n)^2 + \\dots + (a (ka^2 + 1)^n)^2}_{2k\\text{ times}} + ((ka^2 + 1)^n)^2\n$$\n\nwhich is the sum of $(k + 1)^2$ squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17436,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ be a ring and let $a$ be an element of $A$.\n\n(a) If $A$ is commutative and $a$ is nilpotent, then $a + x$ is invertible for any invertible element $x \\in A$.\n\n(b) If $A$ is finite and $a + x$ is invertible for any invertible element $x \\in A$, then $a$ is nilpotent.\n\nAn element $a$ of a ring is called _nilpotent_ if there exists some positive integer $n$ such that $a^n = 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $x$ be an invertible element and let $n$ be a positive integer such that $a^n = 0$. Since $a + x = x(x^{-1}a + 1)$, it is enough to show that $x^{-1}a + 1$ is invertible. Set $b = x^{-1}a$. Then $b^n = x^{-n}a^n = 0$ (because $A$ is commutative), hence $b^{2n+1} = 0$. Consequently,\n\n$$\n1 = b^{2n+1} + 1 = (b+1)(b^{2n} - b^{2n-1} + \\dots - b + 1),\n$$\n\ni.e., $b+1$ is invertible.\n\nb) Induct on $n$, $n \\ge 1$, to prove that $a^n - 1$ is invertible. For $x = -1$ we get that $a - 1$ is invertible. Assume that $b = a^n - 1$ is invertible. From the hypothesis it follows that $a - b^{-1}$ is also invertible, and so is $ab - 1 = (a - b^{-1})b$. Moreover,\n\n$$\na^{n+1} - 1 = a + (a(a^n - 1) - 1) = a + (ab - 1)\n$$\n\nis invertible. Since $A$ is finite, there exist two integers $q > p \\ge 1$ such that $a^p = a^q$, i.e. $a^p(a^{q-p} - 1) = 0$. Since $a^{q-p} - 1$ is invertible, it follows that $a^p = 0$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 17437,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove the inequality\n\n$$\n\\frac{a}{a + \\sqrt{(a+b)(a+c)}} + \\frac{b}{b + \\sqrt{(b+c)(b+a)}} + \\frac{c}{c + \\sqrt{(c+a)(c+b)}} \\le 1,\n$$\n\nwhere $a$, $b$, and $c$ are positive numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "By the Cauchy--Schwarz inequality,\n\n$$\n\\sqrt{(x + y)(z + x)} \\geq \\sqrt{xz} + \\sqrt{yx},\n$$\n\nso\n\n$$\n\\begin{align*}\n& \\frac{a}{a + \\sqrt{(a+b)(a+c)}} + \\frac{b}{b + \\sqrt{(b+c)(b+a)}} + \\frac{c}{c + \\sqrt{(c+a)(c+b)}} \\\\\n& \\le \\frac{a}{a + \\sqrt{ab} + \\sqrt{ac}} + \\frac{b}{b + \\sqrt{bc} + \\sqrt{ba}} + \\frac{c}{c + \\sqrt{ca} + \\sqrt{cb}} \\\\\n& = \\frac{\\sqrt{a}}{\\sqrt{a} + \\sqrt{b} + \\sqrt{c}} + \\frac{\\sqrt{b}}{\\sqrt{b} + \\sqrt{c} + \\sqrt{a}} + \\frac{\\sqrt{c}}{\\sqrt{c} + \\sqrt{a} + \\sqrt{b}} = 1.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17438,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all prime numbers $p$ and $q$ such that $p^{q+1} + q^{p+1}$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $2^3 + 2^3 = 16$, we have that $p = q = 2$ is a solution. Let $p^{q+1} + q^{p+1} = x^2$, where $x \\in \\mathbb{N}$ and $p$ is odd. Thus $p+1$ is even and\n\n$$\np^{q+1} = (x - q^{\\frac{p+1}{2}})(x + q^{\\frac{p+1}{2}}).\n$$\n\nDenote the greatest common divisor of the two factors on the right-hand side by $d$. If $d > 1$, we have that $d$ is a power of $p$ and $d$ is a factor of $2x$. It is easy to see that $p \\mid q$, i.e., $p = q$. Thus $2p^{p+1} = x^2$, which is impossible. Therefore $d = 1$ and $x - q^{\\frac{p+1}{2}} = 1$, $x + q^{\\frac{p+1}{2}} = p^{q+1}$. Hence $2q^{\\frac{p+1}{2}} = p^{q+1} - 1$, which for $q$ odd is impossible modulo $4$. We conclude that $q = 2$ and $2^{\\frac{p+3}{2}} = p^3 - 1 = (p-1)(p^2 + p + 1)$. Since $(p-1, p^2 + p + 1) = 1$, we conclude that $p-1 = 1$, a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17439,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\in \\mathbb{N}$. Find the number of all sequences $A_0, A_1, \\dots, A_n$ of pairwise different points in the plane with non-negative integer first coordinates and second coordinates $0$ or $1$, such that $A_0 = (0,0)$ and\n$$\n|x_{A_i} - x_{A_{i-1}}| + |y_{A_i} - y_{A_{i-1}}| = 1\n$$\nfor $1 \\leq i \\leq n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $c_n$ the number of sequences from the problem. Define $d_n$ similarly, but replace the given equality with\n$$\nx_{A_i} - x_{A_{i-1}} + |y_{A_i} - y_{A_{i-1}}| = 1, \\quad 1 \\leq i \\leq n.\n$$\nCall these sequences *right* and the others *wrong*. It is not difficult to see that $d_0 = 1$, $d_1 = 2$, and $d_{i+2} = d_{i+1} + d_i$. Thus, $d_n$ is the $(n+2)$-th Fibonacci number $f_{n+2}$.\n\nFor any wrong sequence $A_0, A_1, \\dots, A_n$ ($n \\geq 3$), associate the right sequence $A_0, A_1, \\dots, A_k$ for which $x_k = x_n$, $k \\leq n-3$. It is clear that $x_{k+1} = x_k + 1$; call such sequences *super right*. The number of super right sequences of length $k$ is\n$$\nd_{k-1} = f_{k+1}\n$$\n(this holds for $k=0$). There is a bijection between wrong sequences of length $n$ and super right sequences of lengths $n-3, n-5, \\dots$. Thus,\n$$\nc_n = d_n + d_{n-1} + d_{n-2} + \\dots = 2f_{n+1} - \\varepsilon_n = \\frac{2}{\\sqrt{5}}\\left(q^{n+1} - (1-q)^{n+1}\\right) - \\varepsilon_n,\n$$\nwhere $\\varepsilon_n = \\frac{1 + (-1)^n}{2}$ and $q = \\frac{1 + \\sqrt{5}}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17440,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S = \\{S_1, S_2, \\ldots, S_n\\}$ be a collection of non-empty subsets of a finite set, such that for every non-empty subset $J \\subseteq \\{1, 2, \\ldots, n\\}$, the symmetric difference $\\Delta_{j \\in J} S_j$ is a non-empty element of $S$. For which positive integers $n$ and $k$ does there exist such a collection $S$ of $n$ sets, each of size $k$?\n\n(a) Show that $n = 2023$ is impossible.\n\n(b) Show that $n = 1023$ is possible, and determine all $k$ for which such a collection exists.",
    "options": [],
    "answer": "See solution",
    "solution": "**(a)**\nSuppose $n = 2023$. Let $S = \\{S_1, S_2, \\ldots, S_{2023}\\}$ and $T = \\{1, 2, \\ldots, 2023\\}$. Call a non-empty set $I \\subseteq T$ *nice* if every non-empty subset $J \\subseteq I$ satisfies $\\Delta_{j \\in J} S_j \\neq \\emptyset$. Any singleton is nice; let $L$ be the largest nice subset of $T$. By the problem's condition, $\\Delta_{j \\in J} S_j \\in S$ for all non-empty $J \\subseteq L$. For non-empty $J_1, J_2 \\subseteq L$ with $J_1 \\neq J_2$, $J_1 \\Delta J_2 \\neq \\emptyset$, so\n$$\n(\\Delta_{j \\in J_1} S_j) \\Delta (\\Delta_{j \\in J_2} S_j) = \\Delta_{j \\in J_1 \\Delta J_2} S_j.\n$$\nSince $J_1 \\Delta J_2 \\subseteq L$, $\\Delta_{j \\in J_1 \\Delta J_2} S_j \\neq \\emptyset$. Thus, for every non-empty $J \\subseteq L$, $\\Delta_{j \\in J} S_j$ is a distinct element of $S$. For any $i \\in T$, if $S_i \\neq \\Delta_{j \\in J} S_j$ for all non-empty $J \\subseteq L$, then $i \\notin L$ and $L \\cup \\{i\\}$ is also nice, contradicting maximality. Hence,\n$$\nS = \\{\\Delta_{j \\in J} S_j : J \\subseteq L, J \\neq \\emptyset\\},\n$$\nso $2023 = 2^{|L|} - 1$, which is impossible since $2023 + 1 = 2024$ is not a power of $2$.\n\n**(b)**\nLet $n = 1023$. We show this is possible, and determine all $k$ for which such a collection exists. Let $k = 512m$ for some $m \\in \\mathbb{N}$. Let $L = \\{1, 2, \\ldots, 10\\}$. Start with empty sets $S_1, \\ldots, S_{10}$. For every non-empty $R \\subseteq L$, put $m$ distinct elements in\n$$\n\\left(\\bigcap_{r \\in R} S_r\\right) \\setminus \\left(\\bigcup_{r \\in L \\setminus R} S_r\\right).\n$$\nFor every non-empty $J \\subseteq L$,\n$$\n\\Delta_{j \\in J} S_j = \\bigcup_{R \\subseteq L,\\ |R \\cap J| \\text{ odd}} L_R.\n$$\nThere are exactly $512$ subsets $R \\subseteq L$ with $|R \\cap J|$ odd, so $|\\Delta_{j \\in J} S_j| = 512m$. For non-empty $J_1 \\neq J_2$,\n$$\n(\\Delta_{j \\in J_1} S_j) \\Delta (\\Delta_{j \\in J_2} S_j) = \\Delta_{j \\in J_1 \\Delta J_2} S_j\n$$\nwith $|\\Delta_{j \\in J_1 \\Delta J_2} S_j| = 512m \\neq 0$, so all symmetric differences are non-empty and in $S$. Thus, $S = \\{\\Delta_{j \\in J} S_j : J \\subseteq L, J \\neq \\emptyset\\}$ contains $1023$ sets of size $512m$. Therefore, all $k$ divisible by $512$ are possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17441,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function satisfying\n$$|f(x+y) - f(x) - f(y)| < 1 \\text{ for all } x, y \\in \\mathbb{R}.$$ \nProve that $|f(\\frac{x}{2008}) - \\frac{f(x)}{2008}| < 1$ for all $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n|f(2008x) - 2008f(x)| = \\left| \\sum_{k=1}^{2007} \\big(f((k+1)x) - f(x) - f(kx)\\big) \\right| \\\\\n\\le \\sum_{k=1}^{2007} |f((k+1)x) - f(x) - f(kx)| < 2007.\n$$\n\nReplacing $x$ with $\\frac{x}{2008}$ and simplifying, we get\n$$\n\\left| \\frac{f(x)}{2008} - f\\left(\\frac{x}{2008}\\right) \\right| < \\frac{2007}{2008} < 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17442,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n^2 = (m+1)^3 - m^3 = 3m^2 + 3m + 1$. Find the largest integer $n$ such that $2n - 1$ and $2n + 79$ are both perfect squares.",
    "options": [],
    "answer": "See solution",
    "solution": "We have:\n\n$$\n(2n - 1)(2n + 1) = 4n^2 - 1 = 12m^2 + 12m + 3 = 3(2m + 1)^2.\n$$\n\nSince $(2n - 1, 2n + 1) = (2n - 1, 2) = 1$, one of $2n - 1$ and $2n + 1$ is a square and the other is 3 times a square.\n\nIf $2n + 1$ is a square, then $3 \\mid 2n - 1$, so $n \\equiv 2 \\pmod{3}$. But then $2n + 1 \\equiv 2 \\pmod{3}$, so $2n + 1$ cannot be a square. Contradiction.\n\nIf $2n - 1$ is a square, let $2n - 1 = a^2$ and $2n + 79 = b^2$. Then:\n\n$$\n80 = b^2 - a^2 = (b - a)(b + a).\n$$\n\nTo maximize $n$, maximize $b + a$. Since $b - a$ and $b + a$ have the same parity, the maximal case is $(b - a, b + a) = (2, 40)$. Then $(a, b, n) = (19, 21, 181)$. Thus, the largest $n$ is $181$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17443,
    "subject": "Mathematics (Olympiad)",
    "question": "Given rods of integer lengths 1, 2, 3, and 4, answer the following:\n\n**a.** How many equilateral triangles can be formed using these rods as sides?\n\n**b.** How many isosceles triangles (that are not equilateral) can be formed?\n\n**c.** How many scalene triangles can be formed?\n\n**d.** How many quadrilaterals can be formed where exactly one pair of sides is parallel (i.e., trapezoids), and the pair of parallel sides have unequal lengths?",
    "options": [],
    "answer": "See solution",
    "solution": "**a.** There are four equilateral triangles: $(1, 1, 1)$, $(2, 2, 2)$, $(3, 3, 3)$, $(4, 4, 4)$.\n\n**b.** To form an isosceles triangle that is not equilateral, we need two rods of equal length and a third rod with a different length. The possible combinations are:\n\n$(1, 2, 2)$, $(1, 3, 3)$, $(1, 4, 4)$, $(2, 1, 1)$, $(2, 3, 3)$, $(2, 4, 4)$, $(3, 1, 1)$, $(3, 2, 2)$, $(3, 4, 4)$, $(4, 1, 1)$, $(4, 2, 2)$, $(4, 3, 3)$.\n\nTo form a triangle, each side must be smaller than the sum of the other two sides. This eliminates $(2, 1, 1)$, $(3, 1, 1)$, $(4, 1, 1)$, and $(4, 2, 2)$.\n\nSo there are only eight isosceles triangles that are not equilateral:\n\n$(1, 2, 2)$, $(1, 3, 3)$, $(1, 4, 4)$, $(2, 3, 3)$, $(2, 4, 4)$, $(3, 2, 2)$, $(3, 4, 4)$, $(4, 3, 3)$.\n\n**c.** A scalene triangle has three unequal sides. The possible combinations are: $(1, 2, 3)$, $(1, 2, 4)$, $(1, 3, 4)$, $(2, 3, 4)$.\n\nApplying the triangle inequality eliminates $(1, 2, 3)$, $(1, 2, 4)$, and $(1, 3, 4)$.\n\nSo there is only one scalene triangle: $(2, 3, 4)$.\n\n**d.** If the pair of parallel sides have equal length, then the other pair would also be parallel, so the parallel sides must have unequal lengths. The possible lengths for the parallel sides are $(1, 2)$, $(1, 3)$, $(2, 3)$. The possible quadrilaterals are:\n\n$(1, 1, 2, 1)$, $(1, 2, 2, 2)$, $(1, 3, 2, 3)$, $(1, 1, 3, 1)$, $(1, 2, 3, 2)$, $(1, 3, 3, 3)$, $(2, 1, 3, 1)$, $(2, 2, 3, 2)$, $(2, 3, 3, 3)$.\n\nTo form a quadrilateral, each side must be smaller than the sum of the other three sides. This eliminates $(1, 1, 3, 1)$. So there are just eight required quadrilaterals.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17444,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\{a_n\\}$ be an arithmetic progression with common difference $d$ ($d \\neq 0$) and $\\{b_n\\}$ be a geometric progression with common ratio $q$, where $q$ is a positive rational number less than $1$. If $a_1 = d$, $b_1 = d^2$, and\n$$\n\\frac{a_1^2 + a_2^2 + a_3^2}{b_1 + b_2 + b_3}\n$$\nis a positive integer, then $q$ equals \\_\\_\\_\\_.",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n$$\n\\frac{a_1^2 + a_2^2 + a_3^2}{b_1 + b_2 + b_3} = \\frac{a_1^2 + (a_1 + d)^2 + (a_1 + 2d)^2}{b_1 + b_1 q + b_1 q^2}\n$$\nGiven $a_1 = d$, $b_1 = d^2$, so\n$$\n\\frac{d^2 + (d + d)^2 + (d + 2d)^2}{d^2 + d^2 q + d^2 q^2} = \\frac{d^2 + (2d)^2 + (3d)^2}{d^2 (1 + q + q^2)} = \\frac{d^2 + 4d^2 + 9d^2}{d^2 (1 + q + q^2)} = \\frac{14d^2}{d^2 (1 + q + q^2)} = \\frac{14}{1 + q + q^2}\n$$\nLet $m$ be the positive integer value, so $\\frac{14}{1 + q + q^2} = m$, or $1 + q + q^2 = \\frac{14}{m}$.\n\nSolving for $q$:\n$$\nq^2 + q + 1 - \\frac{14}{m} = 0\n$$\n$$\nq = -\\frac{1}{2} + \\sqrt{\\frac{1}{4} + \\frac{14}{m} - 1} = -\\frac{1}{2} + \\sqrt{\\frac{56 - 3m}{4m}}\n$$\nSince $q$ is a positive rational number less than $1$, $1 < \\frac{14}{m} < 3$, so $5 \\leq m \\leq 13$, and $\\frac{56 - 3m}{4m}$ must be a perfect square of a rational number. Checking possible $m$, only $m = 8$ works, giving $q = \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17445,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the maximal length $L$ of a sequence $a_1, \\dots, a_L$ of positive integers satisfying both the following properties:\n\n1. Every term in the sequence is less than or equal to $2^{2023}$.\n2. There does not exist a consecutive subsequence $a_i, a_{i+1}, \\dots, a_j$ (where $1 \\le i \\le j \\le L$) with a choice of signs $s_i, s_{i+1}, \\dots, s_j \\in \\{-1, 1\\}$ for which\n   $$s_i a_i + s_{i+1} a_{i+1} + \\dots + s_j a_j = 0.$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(k)$ be the maximal length for $k$ when every number in the sequence is at most $2^k$. By induction, we have $f(k) \\ge 2^{k+1} - 1$.\n\nFor the base case $k=1$, the sequence $2, 1, 2$ gives $f(1) \\ge 3$.\n\nNow, by induction, suppose $f(n) \\le 2^{n+1} - 2$, which means the maximal sequence length is $2^{n+1} - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17446,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the integer bounds for $\\sqrt[3]{900}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $9^3 = 729 < 900$ and $10^3 = 1000 > 900$, it follows that $9 < \\sqrt[3]{900} < 10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17447,
    "subject": "Mathematics (Olympiad)",
    "question": "In a school there are 32 classes. Six of them have 29 students each, eight of them have 30 students each, two of them have 31 students each, and the rest have 32 students each. How many students are there in the school?",
    "options": [],
    "answer": "See solution",
    "solution": "The number of classes with 32 students each is $32 - 6 - 8 - 2 = 16$. The total number of students is:\n\n$$\n6 \\times 29 + 8 \\times 30 + 2 \\times 31 + 16 \\times 32 = 174 + 240 + 62 + 512 = 988.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17448,
    "subject": "Mathematics (Olympiad)",
    "question": "Twenty-four volunteers will be allocated to three schools. Each school must accept at least one volunteer, and all schools must accept different numbers of volunteers. How many different ways are there to allocate the volunteers?",
    "options": [],
    "answer": "See solution",
    "solution": "We can represent each school by a space between bars (`|`) and each volunteer by an asterisk (`*`). For example, if the first, second, and third schools receive 4, 18, and 2 volunteers, respectively:\n\n![alt](path \"title\")\n\n$$\n| * * * * | * \\cdots * | * * |\n$$\n\nThis allocation problem is equivalent to arranging 4 bars and 24 asterisks, with the two ends occupied by bars. There are $\\binom{23}{2} = 253$ ways to insert the other 2 bars into the 23 spaces between the 24 asterisks, ensuring at least one asterisk between every two consecutive bars. Among these, 31 ways result in at least two schools having the same number of volunteers. Thus, the number of valid allocations is $253 - 31 = 222$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17449,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the functions $f(x) = |x-2| - |x-4|$ and $g(x) = |x-8| - 2$. Find the area of the figure whose vertices are the intersection points of the graphs of $f(x)$ and $g(x)$, and the intersection points of the graph of $g(x)$ with the x-axis.",
    "options": [],
    "answer": "See solution",
    "solution": "The graph of $g(x)$ consists of two rays with a common vertex at $x = 8$. Removing the absolute value, we find its intersection points with the x-axis by solving $6 - x = 0$ and $x - 10 = 0$, giving $A(6, 0)$ and $B(10, 0)$.\n\nFor $f(x)$, after removing the absolute values:\n- For $x < 2$: $f(x) = -2$\n- For $2 \\leq x \\leq 4$: $f(x) = 2x - 6$\n- For $x > 4$: $f(x) = 2$\n\nWe solve $f(x) = g(x)$ in each interval and find the intersection points $D(4, 2)$ and $C(12, 2)$. Thus, the figure $ABCD$ is a trapezoid with bases $8$ and $4$ and height $2$. Therefore, its area is:\n\n$$\n\\text{Area} = \\frac{8 + 4}{2} \\times 2 = 12\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 17450,
    "subject": "Mathematics (Olympiad)",
    "question": "An isosceles trapezoid has an inscribed circle tangent to each of its four sides. The radius of the circle is $3$, and the area of the trapezoid is $72$. Let the parallel sides of the trapezoid have lengths $r$ and $s$, with $r \\neq s$. Find $r^2 + s^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $ABCD$ be the trapezoid with $\\overline{AB} \\parallel \\overline{CD}$ and $AB = s < r = CD$. Let $K, L, M$, and $N$ be the points at which the circle is tangent to $\\overline{AB}$, $\\overline{BC}$, $\\overline{CD}$, and $\\overline{DA}$, respectively. Because $KM$ is a diameter of the circle perpendicular to $\\overline{AB}$ and $\\overline{CD}$, the trapezoid has height $6$. Therefore\n\n$$\n72 = 6 \\cdot \\frac{r+s}{2},\n$$\n\nso $r + s = 24$.\n\n![](images/2025AIME_I_Solutions_p3_data_976e5252c1.png)\n\nBy the Equal Tangents Theorem $AK = AN$, $BK = BL$, $CL = CM$, and $DM = DN$, so $AB + CD = AD + BC$. Hence $2AD = AD + BC = AB + CD = r + s$, implying $AD = 12$. Let $P$ be the foot of the perpendicular from $A$ to $\\overline{CD}$. Then $DP = \\frac{r-s}{2}$, and by the Pythagorean Theorem $DP^2 = AD^2 - AP^2$, so\n\n$$\n\\left( \\frac{r-s}{2} \\right)^2 = 12^2 - 6^2 = 108,\n$$\n\nfrom which $(r-s)^2 = 432$. Thus\n\n$$\nr^2 + s^2 = \\frac{(r+s)^2 + (r-s)^2}{2} = \\frac{24^2 + 432}{2} = 504.\n$$\n\n**Note:** Such a trapezoid does exist with $AB = 6(2 - \\sqrt{3}) \\approx 1.61$, $CD = 6(2 + \\sqrt{3}) \\approx 22.39$, and $AD = BC = 12$. Interestingly, if the assumption $r \\neq s$ is omitted, and the definition of \"isosceles trapezoid\" is modified to allow the quadrilateral to be a parallelogram, then the figure could also be a rhombus with side length $12$ and acute angle $30^\\circ$, in which case $r^2 + s^2 = 288$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17451,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be given, and let $\\alpha = \\angle ABC$. Suppose $|\\vec{BA} - t \\vec{BC}| \\geq |\\vec{AC}|$ for some $t$. Show that $|\\vec{BA}| \\sin \\alpha \\geq |\\vec{AC}|$, and deduce the value of $\\angle ACB$ if equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $\\angle ABC = \\alpha$. Since $|\\vec{BA} - t \\vec{BC}| \\geq |\\vec{AC}|$, we have\n\n$$\n|\\vec{BA}|^2 - 2t\\vec{BA} \\cdot \\vec{BC} + t^2 |\\vec{BC}|^2 \\geq |\\vec{AC}|^2.\n$$\n\nLet\n\n$$\nt = \\frac{\\vec{BA} \\cdot \\vec{BC}}{|\\vec{BC}|^2},\n$$\n\nwe get\n\n$$\n|\\vec{BA}|^2 - 2|\\vec{BA}|^2 \\cos^2 \\alpha + \\cos^2 \\alpha |\\vec{BA}|^2 \\geq |\\vec{AC}|^2.\n$$\n\nThat means $|\\vec{BA}|^2 \\sin^2 \\alpha \\geq |\\vec{AC}|^2$, i.e., $|\\vec{BA}| \\sin \\alpha \\geq |\\vec{AC}|$.\n\nOn the other hand, let point $D$ lie on line $BC$ such that $AD \\perp BC$. Then $|\\vec{BA}| \\sin \\alpha = |\\vec{AD}| \\leq |\\vec{AC}|$. Hence $|\\vec{AD}| = |\\vec{AC}|$, and that means $\\angle ACB = \\frac{\\pi}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17452,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\alpha$, $\\beta$, and $\\gamma$ be the angles between the lines as shown in the diagram below.\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p130_data_e5da18f7c3.png)\n\nNote that $\\alpha + \\beta + \\gamma = 180^\\circ$. This permits us to locate segments $AB$, $CD$, and $EF$, all of unit length, as shown in the diagram below. Prove that $ABCDEF$ is a cyclic hexagon.\n\n![](images/DiPasquale_Australian_MO_Scene_2016_p130_data_6ee7fdb445.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\angle BAD = \\angle ADC$ and $AB = CD$, it follows that $ABCD$ is an isosceles trapezium with $AD \\parallel BC$. Thus $ABCD$ is a cyclic quadrilateral.\n\nSimilarly, $CDEF$ is cyclic with $CF \\parallel DE$. Thus\n\n$$\n\\angle BED = \\angle BPC = \\alpha = \\angle BAD.\n$$\n\nHence $ABDE$ is cyclic. Since quadrilaterals $ABCD$ and $ABDE$ are both cyclic, it follows that $ABCDE$ is a cyclic pentagon. But then since $ABCDE$ and $CDEF$ are both cyclic, it follows that $ABCDEF$ is cyclic, as desired. $\\square$\n\nEvery isosceles trapezium is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17453,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n, a, b, c$ be natural numbers. Every point on the coordinate plane with integer coordinates is colored in one of $n$ colors. Prove that there exist $c$ triangles whose vertices are colored in the same color, which are pairwise congruent, and which have a side whose length is divisible by $a$ and a side whose length is divisible by $b$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the colors be $d_1, d_2, d_3, \\dots, d_n$. Consider the points\n\n$$\n(k, 0 + (n+1)abr),\\ (k, ab + (n+1)abr),\\ (k, 2ab + (n+1)abr),\\ \\dots,\\ (k, nab + (n+1)abr)\n$$\n\nfor integers $k$ and $r$. By the pigeonhole principle, there are two points of the same color. For every pair $(k, r)$, we say the color $d_i$ is $(k, r)$-good if at least two of these coordinates are colored by $d_i$.\n\nFixing $r$ and taking $k = 0, ab, 2ab, \\dots, n^2 ab$, we get that some color, say $d_1$, is $(k, r)$-good for at least $n+1$ values of $k$.\n\nAmong the $n+1$ pairs $(x, y)$, there exist two which share the same $x$ coordinate. We call such a quadruple $r$-great. In every $r$-great quadruple, there are two triangles whose vertices are all the same color and whose two sides are divisible by $ab$.\n\nTaking\n\n$$\nr = 0, 1, 2, \\dots, n\\left(c\\left(\\binom{n+1}{3}\\left(\\binom{n^2+1}{3}\\right)+1\\right)+1\\right)\n$$\n\nwe get that there is one color which is in an $r$-great quadruple for at least\n\n$$\nc\\left(\\binom{n+1}{3}\\left(\\binom{n^2+1}{3}\\right)+1\\right)\n$$\ndifferent values of $r$. Let this color be $d_1$. Since there are fewer than $\\binom{n+1}{3}\\left(\\binom{n^2+1}{3}\\right)$ possible triangles in any $r$-great quadruple, among $c\\left(\\binom{n+1}{3}\\left(\\binom{n^2+1}{3}\\right)+1\\right)$ $r$-great quadruples with the color $d_1$, we get that there are $c+1$ triangles which are congruent, of the same color $d_1$, and with two sides divisible by $ab$. This concludes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17454,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given an acute triangle $ABC$ with $AB > AC$ and orthocenter $H$. The point $E$ is the reflection of $C$ across the altitude $AH$. Let $F$ be the intersection of the lines $EH$ and $AC$.\n\nProve that the circumcenter of triangle $AEF$ lies on the line $AB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\theta$ be the angle between $AF$ and the tangent $t$ at $A$ to the circumcircle of $AEF$. By the inscribed angle theorem, $\\angle FEA = \\theta$. Due to the reflection, $\\angle ACH = \\angle FEA = \\theta$. Because $\\angle ACH = \\theta$, the tangent $t$ is parallel to $CH$ and thus orthogonal to $AB$. Therefore, the circumcenter of triangle $AEF$ lies on $AB$.\n\n_Remark: This result also holds for obtuse triangles._",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17455,
    "subject": "Mathematics (Olympiad)",
    "question": "The function $f$ is defined on the set of positive integers by $f(1) = 1$, $f(2n) = 2f(n)$, and $nf(2n + 1) = (2n + 1)(f(n) + n)$ for all $n \\ge 1$.\n\n(i) Prove that $f(n)$ is always an integer.\n\n(ii) For how many positive integers less than $2007$ is $f(n) = 2n$?",
    "options": [],
    "answer": "See solution",
    "solution": "For part (i), define $g(n) = \\frac{f(n)}{n}$. Then:\n\n$$\ng(2n) = \\frac{f(2n)}{2n} = \\frac{2f(n)}{2n} = \\frac{f(n)}{n} = g(n)\n$$\n\nand\n\n$$\n\\begin{aligned}\ng(2n + 1) &= \\frac{f(2n + 1)}{2n + 1} = \\frac{nf(2n + 1)}{n(2n + 1)} \\\\\n&= \\frac{(f(n) + n)(2n + 1)}{n(2n + 1)} = \\frac{f(n) + n}{n} = g(n) + 1.\n\\end{aligned}\n$$\n\nWe prove by induction that $g(n)$ is always an integer. First, $g(1) = 1$ is an integer.\n\nSuppose $g(1), \\dots, g(a)$ are all integers. If $a+1 = 2b$ is even, $g(a+1) = g(2b) = g(b)$, which is an integer. If $a+1 = 2b+1$ is odd, $g(a+1) = g(2b+1) = g(b) + 1$, also an integer. This completes the induction.\n\nThus, $f(n) = n g(n)$ is always an integer.\n\nFor part (ii), $f(n) = 2n$ if and only if $g(n) = 2$.\n\nWrite $n = 2^k a$ where $a$ is odd. Then $g(n) = g(a) = 2$ by repeated use of $g(2n) = g(n)$.\n\nSince $a$ is odd, write $a = 2m + 1$ with $m \\ge 1$. Then:\n$$g(a) = g(2m + 1) = 2 \\implies g(m) = 1.$$\n\nRepeat: write $m = 2^x y$ with $y$ odd. Then $g(m) = g(y) = 1$. The only odd $y$ with $g(y) = 1$ is $y = 1$.\n\nThus, the solutions to $g(n) = 2$ are exactly:\n\n$$\nn = 2^k a = 2^k (2m + 1) = 2^k (2 \\cdot 2^x y + 1) = 2^k (2^{x+1} + 1)\n$$\n\nFor each $x$, count the $k$ such that $n \\le 2007$:\n\n![](table.png)\n\nSo the total number of solutions is $1 + 2 + \\cdots + 10 = 55$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17456,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive real number $\\alpha$ such that\n$$\n\\frac{x+y}{2} \\geq \\alpha \\sqrt{xy} + (1-\\alpha) \\sqrt{\\frac{x^2+y^2}{2}}\n$$\nfor all positive real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let's prove that $\\alpha = \\frac{1}{2}$ works. Then the following inequality should hold for all positive real numbers $x$ and $y$:\n$$\n\\begin{aligned}\n\\frac{x+y}{2} &\\geq \\frac{1}{2}\\sqrt{xy} + \\frac{1}{2}\\sqrt{\\frac{x^2+y^2}{2}} \\\\\n\\Leftrightarrow (x+y)^2 &\\geq xy + \\frac{x^2+y^2}{2} + 2\\sqrt{xy \\cdot \\frac{x^2+y^2}{2}} \\\\\n\\Leftrightarrow (x+y)^2 &\\geq 4\\sqrt{xy \\cdot \\frac{x^2+y^2}{2}} \\\\\n\\Leftrightarrow (x+y)^4 &\\geq 8xy(x^2+y^2) \\\\\n\\Leftrightarrow (x-y)^4 &\\geq 0\n\\end{aligned}\n$$\nwhich is true, so we showed that $\\alpha = \\frac{1}{2}$ actually works.\n\nNow it remains to show that $\\alpha \\geq \\frac{1}{2}$. Let's consider $x = 1 + \\varepsilon$ and $y = 1 - \\varepsilon$ where $\\varepsilon < 1$. Then the inequality becomes\n$$\n1 \\geq \\alpha \\sqrt{1-\\varepsilon^2} + (1-\\alpha) \\sqrt{1+\\varepsilon^2}, \\text{ i.e.}\n$$\n$$\n\\alpha \\geq \\frac{\\sqrt{1+\\varepsilon^2}-1}{\\sqrt{1+\\varepsilon^2}-\\sqrt{1-\\varepsilon^2}}.\n$$\nNotice that\n$$\n\\begin{aligned}\n\\frac{\\sqrt{1+\\varepsilon^2}-1}{\\sqrt{1+\\varepsilon^2}-\\sqrt{1-\\varepsilon^2}} &= \\frac{(\\sqrt{1+\\varepsilon^2}-1)(\\sqrt{1+\\varepsilon^2}+1)(\\sqrt{1+\\varepsilon^2}+\\sqrt{1-\\varepsilon^2})}{(\\sqrt{1+\\varepsilon^2}-\\sqrt{1-\\varepsilon^2})(\\sqrt{1+\\varepsilon^2}+\\sqrt{1-\\varepsilon^2})(\\sqrt{1+\\varepsilon^2}+1)} \\\\\n&= \\frac{\\varepsilon^2(\\sqrt{1+\\varepsilon^2}+\\sqrt{1-\\varepsilon^2})}{2\\varepsilon^2(\\sqrt{1+\\varepsilon^2}+1)} \\\\\n&= \\frac{\\sqrt{1+\\varepsilon^2}+\\sqrt{1-\\varepsilon^2}}{2(\\sqrt{1+\\varepsilon^2}+1)} \\\\\n&= \\frac{1}{2} - \\frac{1-\\sqrt{1-\\varepsilon^2}}{2(\\sqrt{1+\\varepsilon^2}+1)} \\\\\n&= \\frac{1}{2} - \\frac{(1-\\sqrt{1-\\varepsilon^2})(1+\\sqrt{1-\\varepsilon^2})}{2(\\sqrt{1+\\varepsilon^2}+1)(1+\\sqrt{1-\\varepsilon^2})} \\\\\n&= \\frac{1}{2} - \\frac{\\varepsilon^2}{2(\\sqrt{1+\\varepsilon^2}+1)(1+\\sqrt{1-\\varepsilon^2})} \\\\\n&> \\frac{1}{2} - \\frac{\\varepsilon^2}{4(1+\\sqrt{2})}.\n\\end{aligned}\n$$\nAs $\\varepsilon$ can be arbitrarily small, this expression can get arbitrarily close to $\\frac{1}{2}$. This means that $\\alpha < \\frac{1}{2}$ cannot hold, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17457,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that the minimum of $f(x) = \\cos 2x - 2a(1 + \\cos x)$ is $-\\frac{1}{2}$. Then $a = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution**\n\n$$\n\\begin{aligned}\nf(x) &= 2\\cos^2 x - 1 - 2a - 2a \\cos x \\\\\n&= 2\\left(\\cos x - \\frac{a}{2}\\right)^2 - \\frac{1}{2}a^2 - 2a - 1.\n\\end{aligned}\n$$\n\nFor $a > 2$, $f(x)$ takes the minimum value $1 - 4a$ when $\\cos x = 1$; for $a < -2$, $f(x)$ takes the minimum $1$ when $\\cos x = -1$; for $-2 \\le a \\le 2$, $f(x)$ takes the minimum $-\\frac{1}{2}a^2 - 2a - 1$ when $\\cos x = \\frac{a}{2}$.\n\nIt is easy to see that $f(x)$ will never be $-\\frac{1}{2}$ for $a > 2$ or $a < -2$. So it is only possible that $-2 \\le a \\le 2$. Then from $-\\frac{1}{2}a^2 - 2a - 1 = -\\frac{1}{2}$, we get $a = -2 + \\sqrt{3}$ or $a = -2 - \\sqrt{3}$ (discarded). Therefore, the correct answer is $a = -2 + \\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17458,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle. Suppose that $AD$ and $BE$ are its angle bisectors. Prove that $\\angle ACB = 60^\\circ$ if and only if $AE + BD = AB$.\n\n![](images/UkraineMO_2015-2016_booklet_p5_data_5dccfb3e53.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $I$ the incenter of the triangle $\\triangle ABC$. Also, let $D_1$ be the reflection of $D$ with respect to $BE$. Then $DB = BD_1$ and $DI = ID_1$. Regardless of the given conditions, $D_1 \\in AB$, since in $\\triangle DBD_1$, the line $BE$ contains the altitude, so it is the bisector of $\\angle DBD_1$.\n\nIf $\\angle ACB = 60^\\circ$, then $\\angle AIB = 90^\\circ + \\frac{1}{2}\\angle ACB = 120^\\circ$. Thus, $\\angle DIB = 60^\\circ = \\angle D_1IB$, because $\\triangle DBD_1$ is isosceles. Hence, $\\angle EIA = 60^\\circ = \\angle AIB - \\angle D_1IB$. Therefore, $\\triangle AIE \\cong \\triangle AID_1$, so $EA = AD_1$ and $AE + BD = AB$.\n\nConversely, if $AE + BD = AB$, then $EA = AD_1$, so $\\triangle AIE \\cong \\triangle AID_1$. Then $\\angle EIA = \\angle AID_1$. Since $\\triangle DBD_1$ is isosceles, we have $\\angle DIB = \\angle D_1IB$. Then $\\angle EIA + \\angle DIB = \\angle AIB = \\angle EID$, so $120^\\circ = \\angle AIB = 90^\\circ + \\frac{1}{2}\\angle ACB \\implies \\angle ACB = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17459,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a scalene triangle. The midpoints of the sides $AB$, $BC$, and $CA$ are denoted $C_0$, $A_0$, and $B_0$ respectively. Let $l_A$, $l_B$, and $l_C$ denote the bisectors of the interior angles at $A$, $B$, and $C$ respectively. If $M$ is the intersection of the perpendicular from $C_0$ to $l_C$ and the perpendicular from $A_0$ to $l_A$, then show that $MB_0$ is parallel to $l_B$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $I$ denote the incenter of $ABC$, i.e., the intersection of the bisectors $l_A$, $l_B$, and $l_C$.\n\nWe denote by $P$ and $Q$ the bases of the perpendiculars from $A$ to $l_C$ and from $C$ to $l_A$, respectively. Let $PQ$ intersect $AB$ and $BC$ at $R$ and $S$ respectively. Since $APQC$ is inscribed, we have $\\angle RPI = \\angle QPC = \\angle QAC = \\angle RAI$, thus $APRI$ is also inscribed. Hence $R$ is the base of the perpendicular from $I$ to $AB$. Similarly for $S$. Since $I$ is the incenter, it follows that $IB \\perp RS$ and thus $l_B \\perp PQ$. Hence it suffices to prove that $MB_0 \\perp PQ$.\n\nBy Thales' theorem, the point $B_0$ is the circumcenter of $APQC$. Thus\n\n$$\nB_0P = B_0Q. \\qquad (1)\n$$\n\nNow let $U$ and $V$ denote the bases of the perpendiculars from $B$ to $l_C$ and $l_A$ respectively. We claim that $M$ is the incenter of $PUVQ$. Firstly, since\n\n$$\n\\angle VUI = \\angle VBI = \\angle VQP,\n$$\n\nthe quadrilateral $PUVQ$ is inscribed. Secondly, since $AP \\parallel MC_0 \\parallel BU$ and $C_0$ is the midpoint of $AB$, we see that $MP = MU$. Similarly, $MV = MQ$. Consequently, $M$ is the circumcenter of $PUVQ$ and therefore $MP = MQ$. Combining with (1), we see that $MB_0 \\perp PQ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17460,
    "subject": "Mathematics (Olympiad)",
    "question": "Find a real number $t$ such that for any set of 120 points $P_1, \\dots, P_{120}$ on the boundary of a unit square, there exists a point $Q$ on this boundary with\n$$\n|P_1Q| + \\dots + |P_{120}Q| = t.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $t = 30 + 30\\sqrt{5}$.\n\nWe work in the Cartesian plane, and let $\\mathcal{B}$ denote the boundary of the unit square with corners $(\\pm \\frac{1}{2}, \\pm \\frac{1}{2})$. We proceed in three steps.\n\n**Step 1:** Let $A_1 = (-\\frac{1}{2}, 0)$ and $A_2 = (\\frac{1}{2}, 0)$, and consider the function $f(P) = |A_1P| + |A_2P|$ for $P \\in \\mathcal{B}$. By symmetry, $f(P)$ attains its maximum in the right half of the upper side of $\\mathcal{B}$, at some point $P = (p, \\frac{1}{2})$ with $0 \\le p \\le \\frac{1}{2}$. Then,\n$$\nf(P) = \\sqrt{\\left(\\frac{1}{2} - p\\right)^2 + \\frac{1}{4}} + \\sqrt{\\left(\\frac{1}{2} + p\\right)^2 + \\frac{1}{4}}.\n$$\nThe function $f(P)$ is increasing in $p$ for $0 \\le p \\le \\frac{1}{2}$, so $f(P) \\le f(\\frac{1}{2}, \\frac{1}{2})$ for all $P \\in \\mathcal{B}$. Thus,\n$$\n\\sum_{k=1}^{120} |A_1 P_k| + |A_2 P_k| \\le 120 \\cdot f\\left(\\frac{1}{2}, \\frac{1}{2}\\right) = 60 + 60\\sqrt{5}.\n$$\nConsequently, for some $Q \\in \\{A_1, A_2\\}$, we have $\\sum_{k=1}^{120} |QP_k| \\le 30 + 30\\sqrt{5}$.\n\n**Step 2:** Let $B_1, B_2, B_3, B_4$ be the corners of the square, and let $g(P) = |B_1P| + |B_2P| + |B_3P| + |B_4P|$ for $P \\in \\mathcal{B}$. By symmetry, $g(P)$ attains its minimum in the right half of the upper side of $\\mathcal{B}$, at some $P = (p, \\frac{1}{2})$ with $0 \\le p \\le \\frac{1}{2}$. Then,\n$$\ng(P) = 1 + \\sqrt{\\left(\\frac{1}{2} - p\\right)^2 + 1} + \\sqrt{\\left(\\frac{1}{2} + p\\right)^2 + 1}.\n$$\nAs in Step 1, $g(P)$ is increasing in $p$ for $0 \\le p \\le \\frac{1}{2}$, so $g(P) \\ge g(0, \\frac{1}{2})$ for all $P \\in \\mathcal{B}$. Thus,\n$$\n\\sum_{k=1}^{120} |B_1 P_k| + |B_2 P_k| + |B_3 P_k| + |B_4 P_k| \\ge 120 \\cdot g\\left(0, \\frac{1}{2}\\right) = 120 + 120\\sqrt{5}.\n$$\nConsequently, for some $Q \\in \\{B_1, B_2, B_3, B_4\\}$, we have $\\sum_{k=1}^{120} |QP_k| \\ge 30 + 30\\sqrt{5}$.\n\n**Step 3:** The value $|P_1Q| + \\dots + |P_{120}Q|$ is a continuous function of $Q$ on the boundary. By Step 1, there exists $Q_1$ with this value at most $t$, and by Step 2, there exists $Q_2$ with this value at least $t$. By the intermediate value theorem, as $Q$ moves from $Q_1$ to $Q_2$, there is some $Q$ with the value exactly $t$.\n\n**Remark:** If the 120 points are distributed so that every corner contains 30 points, then $\\sum_{k=1}^{120} |P_kQ| \\ge 30 + 30\\sqrt{5}$ for all $Q \\in \\mathcal{B}$. If 60 points are at $(-\\frac{1}{2}, 0)$ and 60 at $(\\frac{1}{2}, 0)$, then $\\sum_{k=1}^{120} |P_kQ| \\le 30 + 30\\sqrt{5}$ for all $Q$. Thus, $t = 30 + 30\\sqrt{5}$ is the unique solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17461,
    "subject": "Mathematics (Olympiad)",
    "question": "Докажите, что для задачи закрашивания клеток при $k = 1$ закраска всегда возможна: можно закрасить все клетки, начиная с одной и далее закрашивая клетки по определённому правилу.",
    "options": [],
    "answer": "See solution",
    "solution": "Докажем, что $k = 1$ подходит. Пусть в способе с максимальным числом закрашенных клеток первая закрашенная клетка — $A$. Если какая-то клетка $B$ осталась незакрашенной, то по лемме можно выбрать клетку $C$, начав закрашивание с которой, затем можно закрасить $B$, $A$ и все остальные клетки, что увеличит число закрашенных клеток. Это противоречит максимальности выбранного способа, значит, все клетки можно закрасить при $k = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17462,
    "subject": "Mathematics (Olympiad)",
    "question": "$N$ positive integer numbers are given, such that the greatest common divisors of all nonempty sets of these numbers are pairwise distinct. Determine the smallest possible number of distinct prime divisors of the product of these $N$ numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we provide an example of such $N$ numbers. Consider numbers $a_k = p_k p_{k+1} p_{k+2} \\dots p_{k+N-1}$, where $p_1, p_2, \\dots, p_N$ are $N$ distinct prime numbers and $p_{N+i} = p_i$ for $1 \\le i \\le N-1$. Indeed, the GCD of any set will include $p_k$ exactly in power $1$ if and only if $a_k$ belongs to this set. This means that the GCDs of all sets will be distinct.\n\nLet us prove that there cannot be fewer than $N$ prime divisors. Assume that numbers $a_1, a_2, \\ldots, a_N$ satisfy the condition and their product has only $m < N$ distinct prime divisors $p_1, p_2, \\ldots, p_m$. For every $p_i$, choose $a_{k_i}$ that includes $p_i$ in the smallest possible power. Consider the set of all such $a_{k_i}$ taken once (it might be that $a_{k_i} = a_{k_j}$ with $i \\neq j$, in which case our set will include $a_{k_i}$ exactly once). This set will consist of at most $m < N$ numbers. The GCD of all numbers in this set will include $p_i$ in the power equal to the smallest power in which $p_i$ appears among $a_1, a_2, \\ldots, a_N$. Thus, if we add any of the $a_i$ to our set, the GCD of all numbers in the set will be the same. This leads to a contradiction.\n\nTherefore, the smallest possible number of distinct prime divisors is $N$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17463,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a natural number. Consider the equation $n^2 + 1 = ab$ for natural numbers $a$ and $b$.\n\n(a) Show that $a - b \\geq \\sqrt{4n - 3}$.\n\n(b) Find all $n$ for which equality holds, i.e., $a - b = \\sqrt{4n - 3}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $t = a - b$, so $a = b + t$. Substituting into $n^2 + 1 = ab$ gives $n^2 + 1 = (b + t)b = b^2 + tb$, or $b^2 + tb - (n^2 + 1) = 0$.\n\nLet $x = b$; then $x^2 + tx - (n^2 + 1) = 0$. The discriminant is $D = t^2 + 4(n^2 + 1)$. Since $x$ is integer, $D$ must be a perfect square, say $D = m^2$ for some $m \\in \\mathbb{N}$.\n\nNote $t^2 + 4(n^2 + 1) > (2n)^2$, so $D \\geq (2n + 1)^2$. Thus $t^2 + 4(n^2 + 1) \\geq (2n + 1)^2$, or $t^2 \\geq 4n - 3$, so $t \\geq \\sqrt{4n - 3}$.\n\nEquality $a - b = \\sqrt{4n - 3}$ occurs if and only if $D = (2n + 1)^2$, i.e., $t^2 = 4n - 3$, so $n = \\frac{t^2 + 3}{4}$. Since $t$ is an odd natural number, let $t = 2k - 1$ ($k \\in \\mathbb{N}$), then $n = k^2 - k + 1$, $k \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17464,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n = 2013$. For any real number $r$, denote by $[r]$ the smallest integer greater than or equal to $r$.\n\nConsider a set of $n$ cards numbered $0$ to $n-1$, all initially facing up. For each integer $i$ with $1 \\leq i \\leq n$, perform the following operation: for each $j$ with $0 \\leq j < n$, turn over the card numbered $\\left\\lfloor \\frac{n j}{i} \\right\\rfloor$.\n\nAfter performing all operations for $i = 1, 2, \\ldots, n$, how many cards have their number side facing up?",
    "options": [],
    "answer": "See solution",
    "solution": "**Lemma 1.** For $1 \\leq i \\leq n$ and $0 \\leq x \\leq n-1$, we have $\\lfloor \\frac{i(x+1)}{n} \\rfloor - \\lfloor \\frac{ix}{n} \\rfloor = 1$ if the card with number $x$ is turned over at operation $i$, and $0$ otherwise.\n\n*Proof:* For any such $(i, x)$, $\\lfloor \\frac{i(x+1)}{n} \\rfloor - \\lfloor \\frac{ix}{n} \\rfloor$ is $0$ or $1$, since $0 \\leq \\frac{i(x+1)}{n} - \\frac{ix}{n} = \\frac{i}{n} \\leq 1$. The card numbered $x$ is turned over at operation $i$ if and only if there exists a non-negative integer $j$ such that $\\lfloor \\frac{n j}{i} \\rfloor = x$, which is equivalent to $x \\leq \\frac{n j}{i} < x+1$, or $\\frac{ix}{n} \\leq j < \\frac{i(x+1)}{n}$. This is equivalent to $\\lfloor \\frac{i(x+1)}{n} \\rfloor - \\lfloor \\frac{ix}{n} \\rfloor > 0$, completing the proof of Lemma 1.\n\nFor each $i$, $1 \\leq i \\leq n-1$, let the operations $\\{i, n-i\\}$ be performed consecutively.\n\n**Lemma 2.** The card numbered $x$ is turned over once during operations $\\{i, n-i\\}$ if and only if neither $\\frac{i(x+1)}{n}$ nor $\\frac{ix}{n}$ is an integer.\n\n*Proof:* For any real $r, s$ with $r+s$ integer:\n\n$$\n[r] + [s] = \\begin{cases} r+s & (r \\text{ is integer}) \\\\ r+s+1 & (r \\text{ is not integer}) \\end{cases}\n$$\n\nFrom Lemma 1, the card numbered $x$ is turned over once during $\\{i, n-i\\}$ if and only if\n\n$$\n\\left( \\lfloor \\frac{i(x+1)}{n} \\rfloor - \\lfloor \\frac{ix}{n} \\rfloor \\right) + \\left( \\lfloor \\frac{(n-i)(x+1)}{n} \\rfloor - \\lfloor \\frac{(n-i)x}{n} \\rfloor \\right) = 1\n$$\n\nwhich is equivalent to\n\n$$\n\\left( \\lfloor \\frac{i(x+1)}{n} \\rfloor + \\lfloor \\frac{(n-i)(x+1)}{n} \\rfloor \\right) - \\left( \\lfloor \\frac{ix}{n} \\rfloor + \\lfloor \\frac{(n-i)x}{n} \\rfloor \\right) = 1\n$$\n\nBy the above, $\\lfloor \\frac{i(x+1)}{n} \\rfloor + \\lfloor \\frac{(n-i)(x+1)}{n} \\rfloor = x+1$ if $\\frac{i(x+1)}{n}$ is integer, $x+2$ otherwise; and $\\lfloor \\frac{ix}{n} \\rfloor + \\lfloor \\frac{(n-i)x}{n} \\rfloor = x$ if $\\frac{ix}{n}$ is integer, $x+1$ otherwise. Thus, the difference is $1$ if and only if neither $\\frac{i(x+1)}{n}$ nor $\\frac{ix}{n}$ is integer.\n\nNow, $\\frac{ix}{n}$ is integer if and only if $x$ is a multiple of $\\frac{n}{\\gcd(i, n)}$. Thus, during $\\{i, n-i\\}$, the card numbered $x$ is turned over once if and only if neither $x$ nor $x+1$ is a multiple of $\\frac{n}{\\gcd(i, n)}$.\n\nNext, since the result does not depend on the order, we may pair operations as $\\{i, n-i\\}$ for $i = 1, \\ldots, 1006$ and operation $2013$.\n\nSince $2013 = 3 \\cdot 11 \\cdot 61$, the number of $i$ with $\\gcd(i, 2013) = d$ is:\n\n- $d=1$: $(3-1)(11-1)(61-1) = 1200$\n- $d=3$: $(11-1)(61-1) = 600$\n- $d=11$: $(3-1)(61-1) = 120$\n- $d=61$: $(3-1)(11-1) = 20$\n- $d=33$: $61-1=60$\n- $d=183$: $11-1=10$\n- $d=671$: $3-1=2$\n\nAmong $i = 1, \\ldots, 1006$, there are half as many for each $d$ as above. For $d=183$ or $671$, the count is odd; for others, even. Thus, only these contribute to the final state.\n\nAfter all operations, by the Chinese Remainder Theorem, the number of cards with their number side facing up is:\n\n$$(9 \\times 1 + (11-9) \\times (3-1)) \\times 61 = 793.$$\n\n**Answer:** $\\boxed{793}$ cards have their number side facing up after all operations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17465,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all natural numbers $a$ and $b$ greater than $1$, such that $b^a \\mid (a^b - 1)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p$ be the least prime factor of $b$, and $q$ be the least natural number such that $p \\mid (a^q - 1)$ (such a number exists because $p \\mid (a^b - 1)$). By Fermat's Little Theorem, $p \\mid (a^{p-1} - 1)$, which implies $q \\mid b$ and $q \\mid (p-1)$. By the minimality of $p$, we get $q = 1$, i.e., $p \\mid (a-1)$. Let $b = p^\\alpha c$, where $c$ is not divisible by $p$.\n\nIf $p^2 \\mid (a-1)$, then $a \\equiv 1 \\pmod{p^2}$, so $a^k \\equiv 1 \\pmod{p^2}$ for every $k$. Then:\n$$\na^{p^\\alpha c} - 1 = (a-1)(a^{p-1} + a^{p-2} + \\dots + 1) \\dots (a^{p^{\\alpha-1}(p-1)} + \\dots + 1)(a^{p^{\\alpha}(c-1)} + \\dots + 1)\n$$\nThe degree of $p$ in $\\frac{a^b-1}{a-1}$ is $\\alpha$, so $p^{\\alpha(a-1)} \\mid (a-1)$. But for $\\alpha \\geq 1$, $p \\geq 2$, and $a \\geq 2$, $p^{\\alpha(a-1)} > a-1$, which is impossible.\n\nThus, the degree of $p$ in $a-1$ must be $1$. If $p > 2$, we can show by induction that the degree of $p$ in $a^{p^k} - 1$ is $k+1$ for every $k$. For $k=0$, this is true. Assume true for $k < n$. For $k = n$:\n$$\na^{p^n} - 1 = (a^{p^{n-1}})^p - 1 = (a^{p^{n-1}} - 1)(a^{p^{n-1}(p-1)} + \\dots + 1)\n$$\nBy induction, the degree increases by $1$ at each step. For $a^{p^\\alpha c} - 1 = (a^{p^\\alpha} - 1)(a^{p^\\alpha(c-1)} + \\dots + 1)$, and $a^{p^\\alpha(c-1)} + \\dots + 1 \\equiv c \\pmod{p}$, so the degree of $p$ in $a^b - 1$ is $\\alpha + 1$, but it must be at least $\\alpha a$, which is only possible for $\\alpha = 1$ and $a = 2$, but then $p \\nmid (a-1)$, a contradiction.\n\nNow consider $p = 2$. We have:\n$$\na^{2^{\\alpha}c} - 1 = (a-1)(a+1)\\dots(a^{2^{\\alpha-1}} + 1)(a^{2^{\\alpha}(c-1)} + \\dots + 1)\n$$\nSince $c$ is odd, $a^{2^{\\alpha}(c-1)} + \\dots + 1$ is odd (as $a$ is odd, since $2 \\mid (a-1)$). From $2 \\mid (a-1)$ and $2 \\mid (a+1)$, $4 \\mid (a^{2k} - 1)$ for all $k$, but $4$ does not divide $a^{2k} + 1$. Thus, the degree of $2$ in $\\frac{a^b-1}{a^2-1}$ is $\\alpha-1$, so $2^{\\alpha a} \\mid (a^b-1)$ implies $2^{\\alpha(a-1)+1} \\mid (a^2-1)$. Since $4$ divides only one of $a+1$ or $a-1$, $2^{\\alpha(a-1)} \\leq a+1$, which is possible only if $\\alpha = 1$ or $a = 3$.\n\nLet $r$ be the least prime factor of $c$, $c = r^\\beta d$, $r$ and $d$ coprime, and $s$ the least $s$ such that $r \\mid (3^s - 1)$. Then $s \\mid b$ and $s \\mid (r-1)$. This is possible only if $s = 1$ or $s = 2$, so $r \\mid (3^2 - 1)$, which is only possible for $r = 2$, a contradiction. \n\nTherefore, the unique solution is $b = 2$ and $a = 3$ ($b^a = 8 = a^b - 1$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17466,
    "subject": "Mathematics (Olympiad)",
    "question": "Given that $xyz = 1$, solve the system:\n\n$$\nx + \\frac{1}{y} = y + \\frac{1}{z} = z + \\frac{1}{x}\n$$\nfor all real numbers $x, y, z$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $y = 1$, then $xz = 1$ so $z = \\frac{1}{x}$. The system reduces to $x + 1 = 1 + \\frac{1}{z} = z + \\frac{1}{x}$, which gives $x = 1$ or $x = -2$. Thus, $(1, 1, 1)$ and $(-2, 1, -\\frac{1}{2})$ are solutions.\n\nIf $z = -\\frac{1}{1+y}$, then $xyz = 1$ implies $x = -\\frac{1+y}{y}$. Substituting, all three expressions are equal to $-1$ for every $y \\neq 0, -1$. Thus, all solutions are of the form $\\left(-\\frac{1+t}{t},\\ t,\\ -\\frac{1}{1+t}\\right)$ and their cyclic permutations, with $t \\neq 0, -1$. For example, $t=1$ yields $(-2, 1, -\\frac{1}{2})$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17467,
    "subject": "Mathematics (Olympiad)",
    "question": "An arbitrary triangle $ABC$ is given together with two lines $p$ and $q$ which are not parallel to each other and are not perpendicular to any of the sides of the triangle. We denote the perpendiculars through $A$, $B$ and $C$ to the line $p$ by $p_A$, $p_B$ and $p_C$, respectively, and the perpendiculars to $q$ by $q_A$, $q_B$ and $q_C$, respectively. Let us denote the points of intersection of the lines $p_A, q_A, p_B, q_B, p_C$ and $q_C$ with $q_B, p_B, q_C, p_C, q_A$ and $p_A$ respectively by $K$, $L$, $P$, $Q$, $N$ and $M$. Prove that the lines $KL$, $MN$ and $PQ$ intersect at one point.\n\n![](images/Macedonia_2013_p8_data_ac5689523b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** Without loss of generality, we can assume that $p_B$ is between $p_A$ and $p_C$. Consider the cases for the relative positions of $q_A$, $q_B$, and $q_C$; in each, the intersection properties of $KL$, $MN$, and $PQ$ can be analyzed. The lines $KL$, $MN$, and $PQ$ cannot all be parallel due to the configuration of the triangle and the lines $p$ and $q$.\n\nLet $X$, $Y$, and $Z$ be the points of intersection of the lines $PN$, $q_A$, and $p_B$ with $KL$, $PQ$, and $MN$ respectively. From the similarity of triangles $LNZ$ and $PMZ$:\n\n$$\n\\frac{LZ}{PZ} = \\frac{LN}{PM} \\quad (1)\n$$\n\nSimilarly, from the similarity of $LPY$ and $NQY$:\n\n$$\n\\frac{NQ}{LY} = \\frac{NQ}{LP} \\quad (2)\n$$\n\nApplying Menelaus' theorem to triangle $CPN$ and line $KL$:\n\n$$\n\\frac{PX}{XN} = \\frac{NU}{UC} = \\frac{CV}{VP} = -1\n$$\n\nSo,\n\n$$\n\\frac{PX}{NX} = \\frac{UC}{NU} = \\frac{VP}{CV} \\quad (3)\n$$\n\nFrom the similarity of triangles $KQU$ and $LNU$:\n\n$$\n\\frac{UQ}{UN} = \\frac{KQ}{LN}\n$$\n\nand thus,\n\n$$\n1 + \\frac{NQ}{UN} = 1 + \\frac{KB}{LN}\n$$\n\nSo,\n\n$$\n\\overline{UN} = \\frac{\\overline{LN NQ}}{KB}\n$$\n\nand\n\n$$\n\\overline{UC} = \\overline{UN} + \\overline{NC} = \\frac{\\overline{LN NQ} + \\overline{NC KB}}{KB}\n$$\n\nTherefore,\n\n$$\n\\frac{\\overline{LN NQ}}{\\overline{NU}} = \\frac{-\\overline{LN NQ}}{\\overline{LN NQ} + \\overline{NC KB}} \\quad (4)\n$$\n\nSimilarly, for triangles $KMV$ and $LPV$:\n\n$$\n\\frac{CV}{VP} = \\frac{LPPM + PCKA}{-LPPM} \\quad (5)\n$$\n\nSubstituting (4) and (5) into (3):\n\n$$\n\\frac{PX}{NX} = \\frac{LN NQ + NC KB}{-LN NQ} = \\frac{-LPPM}{LPPM + PCKA} = \\frac{LN NQ + NC KB}{LPPM + PCKA} = \\frac{LPPM}{LN NQ} \\quad (6)\n$$\n\nIf we now substitute (1), (2), and (6) into Ceva's theorem for triangle $LNP$ and lines $KL$, $NM$, and $PQ$:\n\n$$\n\\frac{\\overline{LZ}}{\\overline{PZ}} \\frac{\\overline{PX}}{\\overline{NX}} \\frac{\\overline{NY}}{\\overline{LY}} = \\frac{\\overline{LN}}{\\overline{PM}} \\frac{\\overline{NQ}}{\\overline{LP}} - \\frac{\\overline{LP}}{\\overline{NM}} \\frac{\\overline{PM}}{\\overline{NQ}} = -1\n$$\n\nIf $KL$ passes through $C$, then $X$ is the midpoint of $PN$ and\n\n$$\n\\frac{\\overline{LN}}{\\overline{BK}} = \\frac{\\overline{NC}}{\\overline{LB}}\n$$\n\n(7)\n\nSubstituting (1), (2), and (7) into Ceva's theorem:\n\n$$\n\\frac{\\overline{LZ}}{\\overline{PZ}} \\frac{\\overline{PX}}{\\overline{NX}} \\frac{\\overline{NY}}{\\overline{LY}} = -\\frac{\\overline{LN}}{\\overline{PM}} \\frac{\\overline{NQ}}{\\overline{LP}} = -\\frac{\\overline{NC}}{\\overline{LB}} \\frac{\\overline{NQ}}{\\overline{LP}} = -1\n$$\n\nBy the converse of Ceva's theorem, the lines $KL$, $MN$, and $PQ$ intersect in one point or are parallel. The configuration of the triangle and the lines $p$ and $q$ ensures they are not parallel, so they must concur.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17468,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $I$ be the incenter of a triangle $ABC$. Let the incircle of $ABC$ be tangent to $CA$ and $AB$ at $E$ and $F$, respectively. Lines $BI$ and $CI$ intersect line $EF$ at $Y$ and $Z$, respectively. Denote by $M$ and $N$ the midpoints of segments $BC$ and $YZ$, respectively. Prove that $MN$ is parallel to $AI$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us start by proving a known lemma: $\\angle BYC = 90^\\circ$. Since $\\angle IEC = 90^\\circ$, it suffices to show that $IEYC$ is cyclic. Indeed:\n\n$$\n\\begin{align*}\n\\angle CEY &= \\angle AEF \\\\\n&= 90^\\circ - \\frac{1}{2} \\angle BAC \\\\\n&= 90^\\circ - \\frac{1}{2} (180^\\circ - \\angle ABC - \\angle ACB) \\\\\n&= \\frac{1}{2} \\angle ABC + \\frac{1}{2} \\angle ACB \\\\\n&= \\angle IBC + \\angle ICB \\\\\n&= \\angle YIC.\n\\end{align*}\n$$\n\nThis shows that $YIEC$ is cyclic. Since $AC$ is tangent to the incircle at $E$,\n\n$$\n\\angle BIC = \\angle CYI = \\angle CEI = 90^\\circ\n$$\n\nSimilarly, $\\angle BZC = 90^\\circ$.\n\nThus, $BYZC$ is cyclic with $M$ as its center. $N$ is the midpoint of chord $YZ$, so $MN \\perp EF$. Since $AI \\perp EF$, it follows that $MN \\parallel AI$, as desired. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17469,
    "subject": "Mathematics (Olympiad)",
    "question": "Distinct lines $\\ell$ and $m$ lie in the $xy$-plane. They intersect at the origin. Point $P(-1, 4)$ is reflected about line $\\ell$ to point $P'$, and then $P'$ is reflected about line $m$ to point $P''$. The equation of line $\\ell$ is $5x - y = 0$, and the coordinates of $P''$ are $(4, 1)$. What is the equation of line $m$?\n\n(A) $5x + 2y = 0$ (B) $3x + 2y = 0$ (C) $x - 3y = 0$\n(D) $2x - 3y = 0$ (E) $5x - 3y = 0$",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer (D):** The reflection through $\\ell$ followed by the reflection through $m$ is equivalent to a rotation about the intersection of the two lines by twice the angle formed by the two lines. As the result of the two reflections in the present case, $P$ was rotated by $90^\\circ$ clockwise around the origin to $P''(4, 1)$. Therefore the line $\\ell$ must be rotated by $45^\\circ$ clockwise about the origin to obtain line $m$.\n\nThe slope of line $\\ell$ is $5$, and a quick sketch shows that the slope of line $m$ is a positive number less than $5$. The equation of $m$ can be calculated if given one point on $m$ other than the origin $O$. Point $A(1, 5)$ is on line $\\ell$. Let $B(x, y)$ be the intersection of line $m$ with the line through $A$ perpendicular to line $\\ell$. Then $\\triangle BAO$ is an isosceles right triangle with right angle at $A$. Then the slope of line $AB$ is $-\\frac{1}{5}$, so $y - 5 = -\\frac{1}{5}(x - 1)$. Point $B$ is on the circle with radius $OA$ centered at $A$. Therefore $(x - 1)^2 + (y - 5)^2 = 26$. The two solutions of this pair of equations are $(6, 4)$ and $(-4, 6)$. Because $B$ is in the first quadrant, it must be $(6, 4)$, so the slope of $m$ is $\\frac{2}{3}$, and its equation can be written $2x - 3y = 0$.\n\n![](images/2021_AMC10B_Solutions_Fall_p8_data_75dd4c92bf.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17470,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $H_1, H_2, H_3, H_4$ be the orthocentres of $\\triangle AKN$, $\\triangle BKL$, $\\triangle CLM$, and $\\triangle DMN$ respectively, and let $O$ be the centre of $(ABCD)$.\n\nProve that $H_1H_2H_3H_4$ is a parallelogram.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $K$ is the midpoint of $AB$, we have $OK \\perp AB$. Thus, $OK \\parallel H_1N$. Similarly, $ON \\parallel H_1K$. This shows $H_1KON$ is a parallelogram. By symmetry, $H_4NOM$ is a parallelogram. This shows $H_1K = NO = H_4M$ and $H_1K \\parallel NO \\parallel H_4M$. Therefore, $H_1KMH_4$ is a parallelogram. By symmetry, $KH_2H_3M$ is a parallelogram. This shows $H_1H_4 = KM = H_2H_3$ and $H_1H_4 \\parallel KM \\parallel H_2H_3$. Therefore, $H_1H_2H_3H_4$ is a parallelogram.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17471,
    "subject": "Mathematics (Olympiad)",
    "question": "For all positive real numbers $u$ and $v$, prove\n\n$$\n\\min \\left\\{ u, \\frac{100}{v}, v + \\frac{2023}{u} \\right\\} \\leq \\sqrt{2123}.\n$$\n\nHere $\\min X$ denotes the minimum element of $X$.\n",
    "options": [],
    "answer": "See solution",
    "solution": "Let us denote the left side of the given inequality by $S$.\n\nIf $u \\leq \\sqrt{2123}$, then $S \\leq u \\leq \\sqrt{2123}$.\n\nIf $\\frac{100}{v} \\leq \\sqrt{2123}$, then $S \\leq \\frac{100}{v} \\leq \\sqrt{2123}$.\n\nIf $u \\geq \\sqrt{2123}$ and $\\frac{100}{v} \\geq \\sqrt{2123}$, then $S \\leq v + \\frac{2023}{u}$. Since $u \\geq \\sqrt{2123}$, $v \\geq \\frac{100}{\\sqrt{2123}}$, so\n$$\nv + \\frac{2023}{u} \\leq \\frac{100}{\\sqrt{2123}} + \\frac{2023}{\\sqrt{2123}} = \\sqrt{2123}.\n$$\nEquality holds when $u = \\sqrt{2123}$ and $v = \\frac{100}{\\sqrt{2123}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17472,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that there are infinitely many redundant numbers, where a number $n$ is called redundant if $\\frac{\\sigma(n)}{n}$ is greater than for any smaller $n$. Here, $\\sigma(n)$ denotes the sum of the positive divisors of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "When $n = m!$, we have\n\n$$\n\\frac{\\sigma(m!)}{m!} \\ge \\frac{1}{m!} \\left( m! + \\frac{m!}{2} + \\frac{m!}{3} + \\dots + \\frac{m!}{m} \\right) = 1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{m}.\n$$\n\nSince the harmonic series diverges, the sequence $\\left\\{ \\frac{\\sigma(n)}{n} \\right\\}$ is unbounded. Therefore, after finding some redundant numbers $n_1, n_2, \\dots, n_k$, we can always find the next number $n_{k+1}$ such that $\\frac{\\sigma(n_{k+1})}{n_{k+1}} > \\frac{\\sigma(n_k)}{n_k}$. This implies $n_{k+1}$ is redundant, and so there are infinitely many redundant numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17473,
    "subject": "Mathematics (Olympiad)",
    "question": "$\\frac{1}{x-y} + \\frac{1}{y-z} + \\frac{1}{z-x} = \\frac{3}{2}$ is true for some real numbers $x, y, z$. What value may the expression $\\frac{1}{(x-y)^2} + \\frac{1}{(y-z)^2} + \\frac{1}{(z-x)^2}$ have?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $B = \\frac{1}{x-y} + \\frac{1}{y-z} + \\frac{1}{z-x} = \\frac{3}{2}$ and $A = \\frac{1}{(x-y)^2} + \\frac{1}{(y-z)^2} + \\frac{1}{(z-x)^2}$. Then\n\n$$\nB^2 = \\frac{9}{4} = A + 2\\left(\\frac{1}{(x-y)(y-z)} + \\frac{1}{(x-y)(z-x)} + \\frac{1}{(y-z)(z-x)}\\right).\n$$\n\nLet's rearrange the second summand:\n\n$$\n\\frac{1}{(x-y)(y-z)} + \\frac{1}{(x-y)(z-x)} + \\frac{1}{(y-z)(z-x)} = \\frac{(z-x)+(y-z)+(x-y)}{(x-y)(y-z)(z-x)} = 0.\n$$\n\nTherefore, $A = \\frac{9}{4}$.\n\nTo check that such numbers exist, let $x=2$, $y=1$, $z=0$. Then $(x-y) = 1$, $(y-z) = 1$, $(z-x) = -2$,\n\n$$\nB = 1 + 1 - \\frac{1}{2} = \\frac{3}{2}, \\quad A = 1 + 1 + \\frac{1}{4} = \\frac{9}{4}.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17474,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all sequences $\\{a_n\\}_{n \\ge 1}$ of positive rational numbers satisfying\n\n$$\na_{k+1} = \\frac{a_k^2 + a_{k+2}^2}{a_k + a_{k+2}} \\quad \\text{for all } k \\ge 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The required sequences are all constant sequences of positive rational numbers. Clearly, any such sequence satisfies $\\text{(*)}$.\n\nLet $\\{a_n\\}_{n \\ge 1}$ be a sequence of positive rational numbers satisfying $\\text{(*)}$. Then so does $\\{r a_n\\}_{n \\ge 1}$, where $r$ is any positive rational number. Letting $r$ be the product of the denominators of $a_1$ and $a_2$, we may assume that $a_1$ and $a_2$ are integers. If $a_k$ and $a_{k+1}$ are integers, then so is $a_{k+2}$, as it is a root of the monic polynomial $X^2 - a_{k+1} X + a_k^2 - a_k a_{k+1}$ with integer coefficients. Inductively, all $a_n$ are integers.\n\nWrite $\\text{(*)}$ in the form\n\n$$\n(a_{k+1} - a_k) a_k = (a_{k+2} - a_{k+1}) a_{k+2}.\n$$\n\nIf $a_1 > a_2$, then this forces $\\{a_n\\}_{n \\ge 1}$ to be a strictly decreasing sequence of positive integers, which is impossible.\n\nIf $a_1 < a_2$, then $\\{a_n\\}_{n \\ge 1}$ is strictly increasing, by the same equation.\n\n$$\na_{k+1} = \\frac{a_k^2 + a_{k+2}^2}{a_k + a_{k+2}} > \\frac{(a_k + a_{k+2})^2}{2(a_k + a_{k+2})} = \\frac{1}{2}(a_k + a_{k+2}) \\quad \\text{for all } k \\ge 1,\n$$\n\nso $a_{k+1} - a_k > a_{k+2} - a_{k+1}$ for all $k \\ge 1$, and thus the differences $a_{n+1} - a_n$ form a strictly decreasing sequence of positive integers, which is again impossible.\n\nFinally, if $a_1 = a_2$, then the equation forces $a_n = a_1$ for all $n \\ge 1$, so the sequence $\\{a_n\\}_{n \\ge 1}$ is constant, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17475,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x) = ax^2 + bx + c$, $g(x) = \\frac{cx + b}{cx + a}$, and $h(x) = cx + b$. The graphs of $f(x)$ and $h(x)$ have exactly one common point, and the graphs of $g(x)$ and $h(x)$ also have exactly one common point. Find the values of $a$, $b$, and $c$ such that all three graphs have exactly one common point.",
    "options": [],
    "answer": "See solution",
    "solution": "Given the conditions, we set up the equations:\n\nFor $f(x)$ and $h(x)$ to intersect at exactly one point:\n$$\nax^2 + bx + c = cx + b\n$$\nwhich simplifies to\n$$\nax^2 - (c - b)x + (c - b) = 0\n$$\nThis quadratic has exactly one root, so its discriminant is zero:\n$$\n(c-b)^2 - 4a(c-b) = 0 \\implies (c-b)(c-b-4a) = 0\n$$\nSince $c \\neq b$, we have $c-b-4a = 0$, or\n$$\nc - b = 4a\n$$\nThus, the quadratic becomes\n$$\nax^2 - 4a x + 4a = 0 \\implies x^2 - 4x + 4 = 0 \\implies (x-2)^2 = 0 \\implies x = 2\n$$\nSo $x = 2$ is the abscissa of the common point.\n\nFor $g(x)$ and $h(x)$ to intersect at exactly one point:\n$$\n\\frac{cx + b}{cx + a} = cx + b\n$$\nwhich leads to\n$$\ncx + b = (cx + b)(cx + a)\n$$\nThis equation has exactly one root, so $x = -\\frac{b}{c}$. But from above, $x = 2$, so\n$$\n-\\frac{b}{c} = 2 \\implies b = -2c\n$$\nAlso, $2 = \\frac{1-a}{c} \\implies 1 - a = 2c \\implies a = 1 - 2c\n$$\nFrom $c - b = 4a$ and $b = -2c$, we get:\n$$\nc - (-2c) = 4a \\implies 3c = 4a \\implies a = \\frac{3c}{4}\n$$\nSubstitute $a$ from above:\n$$\n\\frac{3c}{4} = 1 - 2c \\implies 3c = 4 - 8c \\implies 11c = 4 \\implies c = \\frac{4}{11}\n$$\nThen $a = 1 - 2c = 1 - \\frac{8}{11} = \\frac{3}{11}$, and $b = -2c = -\\frac{8}{11}$.\n\n**Final answer:**\n$$\na = \\frac{3}{11}, \\quad b = -\\frac{8}{11}, \\quad c = \\frac{4}{11}\n$$\nFor these values, the graphs of all three functions have exactly one common point.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 17476,
    "subject": "Mathematics (Olympiad)",
    "question": "已知 $a_i > 0$，$i = 1, \\dots, n$，且 $\\sum_{i=1}^n a_i = 1$，证明：\n\n$$(a_1^k + \\frac{1}{a_1^k})(a_2^k + \\frac{1}{a_2^k}) \\cdots (a_n^k + \\frac{1}{a_n^k}) \\ge (n^k + \\frac{1}{n^k})^n$$",
    "options": [],
    "answer": "See solution",
    "solution": "由 $a_1 \\cdots a_n \\le \\left( \\frac{1}{n} \\sum_{i=1}^n a_i \\right)^n = \\frac{1}{n^n}$，可得\n\n$$\n\\frac{1}{n^n a_1 \\cdots a_n} \\ge 1.\n$$\n\n又因为\n\n$$\n\\frac{k(n^{2k} - 1)}{n^{2k} + 1} > 0,\n$$\n\n所以\n\n$$\n\\left(\\frac{1}{n^n a_1 \\cdots a_n}\\right)^{\\frac{k(n^{2k})}{n^{2k}+1}} \\ge 1.\n$$\n\n因此，\n\n$$(a_1^k + \\frac{1}{a_1^k})(a_2^k + \\frac{1}{a_2^k}) \\cdots (a_n^k + \\frac{1}{a_n^k}) \\ge (n^k + \\frac{1}{n^k})^n.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17477,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n > 2$ be an integer and\n$$\nf(x) = x^n + x^{n-1} - x^{n-2} - 3.\n$$\nProve that $f(x)$ cannot be factored as the product of two polynomials with integer coefficients and degree less than $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the complex roots of $f(x)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17478,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute triangle $ABC$, there is an altitude $AH$ and a median $AM$. On lines $AB$ and $AC$, there are points $X$ and $Y$ such that $AX = XC$ and $AY = YB$. Prove that the midpoint of segment $XY$ is equidistant from points $H$ and $M$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $Z$ be the midpoint of segment $XY$, and let $N$ and $T$ be the midpoints of segments $AB$ and $AC$, respectively.\n\nAs, according to the condition, $AX = XC$ and $AY = YB$, $XT$ and $YN$ are bisectors of segments $AC$ and $AB$. Then, triangles $XTY$ and $XNY$ are right triangles, so $XZ = ZN = ZT = ZY$, because the median of a right triangle drawn to the hypotenuse equals half the hypotenuse. Thus, point $Z$ is equidistant from $N$ and $T$.\n\n![alt](images/Ukraine_booklet_2018_p38_data_ba373cab55.png)\n\nOn the other hand, $NT \\parallel BC$, because $NT$ is the midline of $\\triangle ABC$. $MT$ is also a midline of $\\triangle ABC$, so $MT = \\frac{1}{2}AB$. $NH = \\frac{1}{2}AB$ as the median drawn to the hypotenuse in a right triangle. So, $HNTM$ is an isosceles trapezoid. Then its bases have a common bisector. We have proved that point $Z$ belongs to the bisector $NT$. Therefore, it also belongs to the bisector $HM$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17479,
    "subject": "Mathematics (Olympiad)",
    "question": "愛麗絲有仙境的地圖，仙境共有 $n \\ge 2$ 座城鎮，其中任兩個城鎮之間都恰有一條道路。某一天，仙境突然規定每一條道路都是單行道。愛麗絲不知道這些單行道的方向，但紅心國王決定幫她。每一次提問，愛麗絲選擇兩座城鎮，而紅心國王回答她連接這兩座城鎮的道路方向。\n\n愛麗絲想知道仙境中是否有一座城鎮只有至多一條向外的道路。試證：愛麗絲總是可以透過至多 $4n$ 次提問達成她的目標。\n\n註：若證出將 $4n$ 改為 $cn$ 之結果，則會依據常數 $c$ 給予分數。",
    "options": [],
    "answer": "See solution",
    "solution": "我們將證明愛麗絲只需要 $4n-7$ 次提問。由於 $n=2$ 與 $n=3$ 的狀況顯然，故以下假設 $n \\ge 4$。\n\n愛麗絲的策略將包含下面五步驟。在本文中，$S$ 表示愛麗絲目前為止尚不知是否有超過一條向外道路的城鎮集合，故起始時 $|S| = n$。\n\n### 步驟一（累計 1 個問題）\n\n愛麗絲首先隨意選擇兩座城鎮 $A$ 與 $B$，不失一般性假設紅心國王回答 $A \\to B$。\n\n### 步驟二（累計 $n-1$ 個問題）\n\n令 $T$ 為目前已知有向內道路但尚不知是否有向外道路的唯一城鎮（起始時 $T=B$）。愛麗絲重複以下操作 $n-2$ 次：選一個從未被問過的城鎮 $X$，並詢問 $(T,X)$。若回答為 $X \\to T$，則 $T$ 不變；否則，$T$ 變為 $X$。\n\n注意到在步驟二結束時，$T$ 尚不知是否有向外道路，且所有其他城鎮皆恰有一條已知的連外道路。此外，這些已知的道路構成樹（若不考慮方向性）。\n\n### 步驟三（累計 $n+k-1$ 個問題）\n\n愛麗絲接著詢問所有未知的 $(T,X)$ 組合，直到她問到 $T$ 有兩條向外道路。這至多有 $k \\le n-2$ 個問題。\n\n注意到如果我們最後沒有問到 $T$ 有兩條向外道路，則我們在 $2n-3 \\le 4n-7$ 次提問中便問出了我們的答案，因此以下假設我們問到 $T$ 有兩條連外道路，且總共問了 $k$ 個問題。\n\n另外留意到，對於每一個 $X \\to T$ 的回答，我們都可以將 $X$ 從 $S$ 中移出（因為前兩步驟我們已經知道 $X$ 有一條向外道路，現在又多一條）。此外，$T$ 也有兩條向外道路，故也可以從 $S$ 中移除。因此在步驟三結束時，$|S| = n - k + 1$。同時，目前已知的道路在 $S$ 上構成的無向圖沒有環，且 $S$ 中每一個城鎮都恰有一條已知的向外道路（不必然連向 $S$ 中的其他點）。\n\n### 步驟四（累計 $2n - t$ 個問題）\n\n愛麗絲每次挑 $S$ 中兩個“之間道路未知”的城鎮 $(X, Y)$ 提問。由於 $S$ 中所有城鎮都已知一條向外道路，每次提問都會將一個城鎮從 $S$ 中移除。又由於之前無環的討論，此步驟可以持續進行，最後 $S$ 會剩下至多 2 個城鎮。\n\n假設最後 $S$ 剩下 $t$ 個城鎮，則此步驟進行了 $n - k + 1 - t$ 個問題。\n\n### 步驟五（累計至多 $4n - 7$ 個問題）\n\n現在，愛麗絲挑 $X \\in S$ 與 $Y \\notin S$ 且其間道路未知的組合來提問。易檢查此步驟將包含 $t(n - t) - 1$ 次提問（注意到步驟一或二時至少已對這一點提過一次問）。又 $t \\le 2$，故總提問數至多為\n\n$$\n2n - t + t(n - t) - 1 = (2 + t)n - (t^2 + t + 1) \\le 4n - 7.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17480,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y, z > 0$ such that $xy + yz + zx = 1$. Prove that\n$$\n\\frac{1}{\\sqrt{1+x^2}-x} + \\frac{1}{\\sqrt{1+y^2}-y} + \\frac{1}{\\sqrt{1+z^2}-z} \\le \\frac{1}{xyz}.\n$$\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n$$\n\\begin{aligned}\n\\frac{1}{\\sqrt{1+x^2}-x} &= \\sqrt{1+x^2} + x = \\sqrt{xy+yz+zx+x^2} + x \\\\\n&= \\sqrt{(x+y)(x+z)} + x.\n\\end{aligned}\n$$\nBy applying the AM-GM inequality, one can get\n$$\n\\sqrt{(x+y)(x+z)} + x \\le \\frac{x+y+x+z}{2} + x = 2x + \\frac{y+z}{2}.\n$$\nSimilarly for the other fractions, so it is easy to get\n$$\n\\text{LHS} \\leq 3(x + y + z).\n$$\nWe need to prove\n$$\n3xyz(x + y + z) \\leq 1 = (xy + yz + zx)^2,\n$$\nwhich is true since this is equivalent to\n$$\n(xy - yz)^2 + (yz - zx)^2 + (zx - xy)^2 \\geq 0.\n$$\nThe equality holds when $x = y = z = \\frac{1}{\\sqrt{3}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17481,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a white square $ABCD$ of size $8 \\times 8$, that consists of $64$ unit squares of size $1 \\times 1$. In a turn, one can choose any\n\na) square;  \nb) rectangle,\n\nthat consists of a whole number of unit squares and contains at least one of the vertices of the square $ABCD$, and change the color of every unit square in it to the opposite one (white is changed to black and vice versa). Is it possible to obtain arbitrary coloring of the square $ABCD$ using the turns described above?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** a) no, b) yes.\n\n**Solution.**\n\na) Clearly, choosing the same square and acting in it twice is equivalent to not acting at all. Thus, we can assume that every square can be chosen not more than once. There are $32$ squares for which one can make a turn. Thus, there are not more than $2^{32}$ different colorings, which is less than $2^{64}$ – the amount of all possible colorings of unit squares in two colors.\n\n![](images/Ukraine_2020_booklet_p14_data_be6b899676.png)\n\n![](images/Ukraine_2020_booklet_p14_data_3ef3898174.png)\n\nb) Choose any unit square (depicted in black) and change the color of every highlighted rectangle into the opposite one. There are either two (Fig. 20a), three (Fig. 20b), or four (Fig. 20c) such rectangles. After that, change the color of the whole square $ABCD$ into the opposite one. This way, only the color of the chosen unit square is changed. Thus, it is possible to obtain any coloring of the original square in two colors.\n\n![](images/Ukraine_2020_booklet_p14_data_a5ef815ca0.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17482,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Let $S(n)$ be the sum of the digits of the number $n$. After the decimal point, we write the numbers $S(1), S(2), \\ldots$ one after the other. Prove that the obtained number is irrational.\n\n6) Let $P(n)$ be the product of the digits of the number $n$. After the decimal point, we write the numbers $P(1), P(2), \\ldots$ one after the other. Prove that the obtained number is irrational.",
    "options": [],
    "answer": "See solution",
    "solution": "a) The number $0.S(1)S(2)\\ldots$ is irrational if it is non-periodic. Suppose, for contradiction, that the number is periodic with period length $d$. The period must contain digits different from $0$, since otherwise the number would be of the form $0.S(1)S(2)\\ldots S(k)$, but for any $k$ there exists a number greater than $k$ whose sum of digits is not $0$ (for example, $10^m$ for sufficiently large $m$ has sum of digits $1$). However, there exists a natural number large enough whose sum of digits ends in $2d$ zeroes (for example, the sum of digits of a number consisting of $10^{2d}$ ones is $10^{2d}$). Then the period of length $d$ must appear completely in $S(10^{2d})$, which would require the period to consist only of zeroes, a contradiction.\n\n6) The number $0.P(1)P(2)\\ldots$ is irrational if it is non-periodic. Suppose, for contradiction, that the number is periodic with period length $d$. The period must contain digits different from $0$, since otherwise the number would be of the form $0.P(1)P(2)\\ldots P(k)$, but for any $k$ there exists a number greater than $k$ whose product of digits is not $0$ (for example, a number consisting of many ones is greater than $k$ and its product is $1$). Consider numbers of the form $\\underbrace{22\\ldots 255\\ldots 5}_{m\\text{ digits}}$ (with $m$ digits), whose product of digits is $10^m$. If we choose $m > 2d$, then the period must appear completely in $P(\\underbrace{22\\ldots 255\\ldots 5}_{m\\text{ digits}})$, which would require the period to consist only of zeroes, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17483,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given an infinite deck of cards, each with a real number on it. For every real number $x$, there is exactly one card in the deck that has $x$ written on it. Now two players draw disjoint sets $A$ and $B$ of 100 cards each from the deck. We would like to define a rule that declares one of them a winner. This rule should satisfy the following conditions:\n\n1. The winner depends only on the relative order of the 200 cards: if the cards are laid down in increasing order and we are told which card belongs to which player, but not what numbers are written on them, we can still decide the winner.\n\n2. If we write the elements of both sets in increasing order as $A = \\{a_1, a_2, \\cdots, a_{100}\\}$ and $B = \\{b_1, b_2, \\cdots, b_{100}\\}$, and $a_i > b_i$ for all $i$, then $A$ beats $B$.\n\n3. If three players draw three disjoint sets $A$, $B$ and $C$ from the deck, $A$ beats $B$ and $B$ beats $C$, then $A$ beats $C$.\n\nHow many possible rules are there? Here, two rules are different if there exist two distinct sets $A$ and $B$, such that $A$ beats $B$ under the first rule, but $B$ beats $A$ under the second one.",
    "options": [],
    "answer": "See solution",
    "solution": "There are $100$ different possible rules.\n\n**Proof:**\n\nSuppose the game is played with $n$ cards per player. We will show that the number of possible rules is $n$.\n\nFirst, we construct $n$ different rules. For each $k \\in \\{1, 2, \\cdots, n\\}$, define the rule: $A < B$ if and only if $a_k < b_k$, where $A = \\{a_1 < a_2 < \\cdots < a_n\\}$ and $B = \\{b_1 < b_2 < \\cdots < b_n\\}$. This rule clearly satisfies the problem's requirements, and there are $n$ such rules.\n\nNext, we show that these $n$ rules are the only possible ones. Suppose we have a rule satisfying the conditions. Let $k$ be the minimal index such that\n\n$$\nA_k = \\{1, 2, \\dots, k, n + k + 1, n + k + 2, \\dots, 2n\\\\}\n$$\n\nand\n\n$$\nB_k = \\{k + 1, k + 2, \\dots, n + k\\}\n$$\n\nsatisfy $A_k < B_k$. Such $k$ exists by construction.\n\nNow, for any two disjoint sets $X = \\{x_1 < x_2 < \\cdots < x_n\\}$ and $Y = \\{y_1 < y_2 < \\cdots < y_n\\}$, we claim that $X < Y$ if and only if $x_k < y_k$, so the rule is determined by the $k$-th element.\n\nTo see this, choose three sets $U, V, W$ disjoint from $X$ and $Y$, such that:\n\n- $u_1 < u_2 < \\dots < u_{k-1} < \\min(x_1, y_1)$\n- $\\max(x_n, y_n) < v_1 < v_2 < \\dots < v_n$\n- $x_k < v_1 < v_2 < \\dots < v_k < w_1 < w_2 < \\dots < w_n < u_k < u_{k+1} < \\dots < u_n < y_k$\n\nThen:\n\n(a) For all $i$, $u_i < y_i$ and $x_i < v_i$, so $U < Y$ and $X < V$.\n\n(b) The relative order of $U$ and $W$ matches that of $A_{k-1}$ and $B_{k-1}$, and by minimality of $k$, $A_{k-1} > B_{k-1}$, so $U > W$.\n\n(c) The relative order of $V$ and $W$ matches that of $A_k$ and $B_k$, and by choice, $A_k < B_k$, so $V < W$.\n\nThus,\n\n$$\nX < V < W < U < Y\n$$\n\nso $X < Y$. The argument depends only on $x_k$ and $y_k$, so the rule is determined by the $k$-th element. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17484,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all polynomials $P(x)$ and $Q(x)$ with integer coefficients such that, for the sequence $(x_n)$ defined by\n\n$$\nx_0 = 2014, \\quad x_{2n+1} = P(x_{2n}), \\quad x_{2n} = Q(x_{2n-1}) \\quad \\forall n \\ge 0,\n$$\n\nevery positive integer $m$ is a divisor of some non-zero term of $(x_n)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose one of $P(x)$ or $Q(x)$ is constant; then the condition fails. We show that both must have degree $1$.\n\nLet $H(x) = P(Q(x))$ and $K(x) = Q(P(x))$. If either has degree at least $2$, consider the sequence $x_0 = a$, $x_{n+1} = T(x_n)$ for integer polynomial $T(x)$. If every positive integer $m$ divides some nonzero $x_n$, then $\\deg(T) = 1$.\n\n*Proof of lemma:* If $\\deg(T) > 1$, then for large $|x|$, $|T(x)| > 3|x|$. For large $n$, $|x_{n+1}| > 3|x_n|$, so $|x_n|$ grows rapidly. For $m = |x_{N+1} - x_N|$, none of $x_0, \\dots, x_{N+1}$ are divisible by $m$, and for $n \\ge N$, $x_n$ is not divisible by $m$, contradicting the condition. Thus, $\\deg(T) = 1$.\n\nApplying this to the subsequences $\\{x_{2n}\\}$ and $\\{x_{2n+1}\\}$, we conclude $P(x)$ and $Q(x)$ are linear: $P(x) = a x + b$, $Q(x) = c x + d$ with $a, b, c, d \\in \\mathbb{Z}$.\n\nNow, $x_{2n+2} = c a x_{2n} + b c + d$ and $x_{2n+3} = c a x_{2n+1} + a d + b$. For the sequence to satisfy the condition, $a c = 1$, so $a = c = 1$ or $a = c = -1$.\n\n**Case 1:** $P(x) = x + b$, $Q(x) = x + d$.\n\nThen $x_{2k} = 2014 + k(b + d)$ and $x_{2k+1} = 2014 + b + k(b + d)$. We need $b + d \\neq 0$ and either $b + d \\mid 2014$ or $b + d \\mid 2014 + b$.\n\n**Case 2:** $P(x) = -x + b$, $Q(x) = -x + d$.\n\nSimilarly, $x_{2k}$ and $x_{2k+1}$ alternate in sign and grow linearly. We need $b - d \\neq 0$ and either $b - d \\mid 2014$ or $b - d \\mid b - 2014$.\n\nThus, all such pairs of linear polynomials with these properties are the solutions.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 17485,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with $AC > AB$ and circumcircle $\\Gamma$. The tangent from $A$ to $\\Gamma$ intersects $BC$ at $T$. Let $M$ be the midpoint of $BC$ and let $R$ be the reflection of $A$ in $B$. Let $S$ be a point so that $SABT$ is a parallelogram and finally let $P$ be a point on line $SB$ such that $MP$ is parallel to $AB$.\n\nGiven that $P$ lies on $\\Gamma$, prove that the circumcircle of $\\triangle STR$ is tangent to line $AC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N$ be the midpoint of $BS$ which, as $SABT$ is a parallelogram, is also the midpoint of $TA$. Using $ST \\parallel AB \\parallel MP$ we get:\n\n$$\n\\frac{NB}{BP} = \\frac{1}{2} \\cdot \\frac{SB}{BP} = \\frac{TB}{2 \\cdot BM} = \\frac{TB}{BC}\n$$\n\nwhich shows that $TA \\parallel CP$.\n\n![](images/Bmo_Shortlist_2021_p39_data_f97917458e.png)\n\nLet $\\Omega$ be the circle with diameter $OT$. As $\\angle OMT = 90^\\circ = \\angle TAO$ we have that $A, M$ lie on $\\Omega$. We now show that $P$ lies on $\\Omega$. As $TA \\parallel CP$ and $TA$ is tangent to $\\Gamma$ we have that $AP = AC$, so\n\n$$\n\\angle TAP = \\angle ACP = \\angle CPA = \\angle CBA = \\angle TMP\n$$\n\nwhere in the last step we used the fact that $MP \\parallel AB$. This shows that $P$ lies on $\\Omega$. Furthermore, this shows that $\\angle OPT = 90^\\circ$ and so $TP$ is also tangent to $\\Gamma$.\n\nNow we show that $R, S$ lie on $\\Omega$ which would show that $\\Omega$ is the circumcircle of triangle $STR$. For $S$, using $ST \\parallel AB$ and that $TA$ tangent to $\\Gamma$ we have\n\n$$\n\\angle TSP = \\angle ABS = \\angle ACP = \\angle TAP.\n$$\n\nFor $R$, the homothety with factor $2$ centred at $A$ takes $BN$ to $RT$. So $BN \\parallel RT$ and hence\n\n$$\n\\angle ART = \\angle ABS = \\angle TAP = \\angle APT,\n$$\n\nwhere the last step follows from $TA = TP$ as they are both tangents to $\\Gamma$.\n\nFinally, we observe that as $TA$ tangent to $\\Gamma$ then\n\n$$\n\\angle TAC = 180^{\\circ} - \\angle CBA = \\angle ABT = \\angle TSA\n$$\n\nwhich, by the alternate segment theorem, means that line $AC$ is tangent to $\\Omega$ as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17486,
    "subject": "Mathematics (Olympiad)",
    "question": "Fix an integer $n \\ge 2$. Consider $n$ real numbers $a_1, a_2, \\dots, a_n$, not all equal, and let\n\n$$\nd = d(a_1, a_2, \\dots, a_n) = \\max_{1 \\le i < j \\le n} |a_i - a_j|\n$$\n\nand\n\n$$\ns = s(a_1, a_2, \\dots, a_n) = \\sum_{1 \\le i < j \\le n} |a_i - a_j|.\n$$\n\nDetermine, in terms of $n$, the smallest and the largest values the quotient $s/d$ may achieve.",
    "options": [],
    "answer": "See solution",
    "solution": "The required minimum is $n-1$ and is achieved, for instance, by $a_1 < a_2 = \\dots = a_n$. The maximum is $\\lfloor \\frac{1}{2}n \\rfloor \\lfloor \\frac{1}{2}(n+1) \\rfloor$ and is achieved, for instance, by\n\n$$\na_1 = \\dots = a_{\\lfloor n/2 \\rfloor} < a_{\\lfloor n/2 \\rfloor+1} = \\dots = a_n.\n$$\n\nIn each case, verification is routine and is hence omitted.\n\nWe now show that $(n-1)d \\le s \\le \\lfloor \\frac{1}{2}n \\rfloor \\lfloor \\frac{1}{2}(n+1) \\rfloor d$. Clearly, we may and will assume $a_1 \\le a_2 \\le \\dots \\le a_n$, so $s = \\sum_{1 \\le i < j \\le n} (a_j - a_i)$. Let $d_k = a_{k+1} - a_k$, $k = 1, \\dots, n-1$, so $d = d_1 + \\dots + d_{n-1}$.\n\nExpress every $a_j - a_i$, $1 \\le i < j \\le n$, and hence $s$, in terms of the $d_k$: As $d_k$ occurs as a summand in every $a_j - a_i$, $1 \\le i \\le k < k + 1 \\le j \\le n$, and in no other $a_j - a_i$, $1 \\le i < j \\le n$, it follows that $s = \\sum_{k=1}^{n-1} k(n-k)d_k$.\n\nFinally, as $\\min_{1 \\le k \\le n-1} k(n-k) = n-1$ and $\\max_{1 \\le k \\le n-1} k(n-k) = \\lfloor \\frac{1}{2}n \\rfloor \\lfloor \\frac{1}{2}(n+1) \\rfloor$, it follows that\n\n$$\n(n-1)d = \\sum_{k=1}^{n-1} (n-1)d_k \\le \\underbrace{\\sum_{k=1}^{n-1} k(n-k)d_k}_{=s} \\le \\sum_{k=1}^{n-1} \\lfloor \\frac{1}{2}n \\rfloor \\lfloor \\frac{1}{2}(n+1) \\rfloor d_k = \\\\ = \\lfloor \\frac{1}{2}n \\rfloor \\lfloor \\frac{1}{2}(n+1) \\rfloor d,\n$$\n\nas desired. This ends the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17487,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose each unit square of a $2022 \\times 2022$ square grid contains at most one beetle. At each step, every beetle moves simultaneously to an adjacent unit square (sharing an edge) in such a way that no two beetles ever occupy the same square. What is the maximum number of beetles that can be placed on the grid so that, regardless of their moves, no two beetles ever meet?\n\n![](images/HongKong_2022-23_p13_data_24819afbfc.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $2022^2 + 1 = 4\\,088\\,485$.\n\nFirst, we show that $2022^2$ beetles is insufficient. Suppose there are $2022^2$ beetles in the top left $2022 \\times 2022$ square grid. This grid can be partitioned into many $2 \\times 2$ square grids. If the beetles in each $2 \\times 2$ grid move in the anticlockwise direction as shown, then no beetle will ever meet.\n\nNext, suppose there are at least $2022^2 + 1$ beetles. Label all the unit squares as follows: in odd rows, label the squares $ABAB\\cdots ABA$; in even rows, label them $CDCD\\cdots CDC$. There are $1011^2$ unit squares with label $D$.\n\nBy the pigeonhole principle, with at least $2022^2 + 1$ beetles, there must be at least $1011^2 + 1$ beetles on squares with the same label. In each move, all beetles move in the order $A \\to B \\to D \\to C \\to A$ or $A \\to C \\to D \\to B \\to A$. Thus, these $1011^2 + 1$ beetles must move to $D$-labeled squares simultaneously at some step. Since there are only $1011^2$ such squares, two beetles must meet. This completes the proof.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 17488,
    "subject": "Mathematics (Olympiad)",
    "question": "There are finitely many fractions $\\frac{a}{b}$ in lowest terms such that $a \\ge 1$, $1 \\le b \\le 2007$, and $\\frac{a}{b} < \\sqrt{2}$. Let $\\frac{n}{m}$ be the largest such fraction. Prove that for all $k = 1, 2, \\dots, 2007$, we have\n$$\n\\left\\lfloor \\frac{nk}{m} \\right\\rfloor = \\left\\lfloor \\sqrt{2}k \\right\\rfloor.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, it exists.\n\nSuppose $\\lfloor \\frac{nk}{m} \\rfloor \\ne \\lfloor \\sqrt{2}k \\rfloor$ for some $k = 1, 2, \\dots, 2007$. Then $t = \\lfloor \\sqrt{2}k \\rfloor$ is an integer satisfying $\\frac{nk}{m} < t < \\sqrt{2}k$, since $\\frac{n}{m} < \\sqrt{2}$ and $\\sqrt{2}k \\notin \\mathbb{Z}$. Thus,\n$$\n\\frac{n}{m} < \\frac{t}{k} < \\sqrt{2},\n$$\nwhich contradicts the choice of $\\frac{n}{m}$ as the largest such fraction. Therefore, we must have $\\lfloor \\frac{nk}{m} \\rfloor = \\lfloor \\sqrt{2}k \\rfloor$ for $1 \\le k \\le 2007$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17489,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all values of the real parameter $a$ such that the inequality\n\n$$\n\\log_a(a^x + 1) + \\frac{1}{\\log_{a^2-1} a} \\le x - 1 + \\log_a(a^2 - 1)\n$$\nholds true for every $x \\in (0, 1]$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have that $a > 1$, $x > 0$, and $x \\neq \\log_a 2$. Then the inequality is equivalent to\n\n$$\n(a^{x-1} - 1)(a^{x+1} + 1) \\le 0,\n$$\n\ngiving $x \\le 1$. Since $\\log_a 2 > 0$ and $\\log_a 2 \\le 1$ when $a \\ge 2$, we obtain:\n\n1. For $1 < a < 2$, the inequality holds true iff $0 < x \\le 1$.\n2. For $a \\ge 2$, the inequality holds true iff $0 < x \\le 1$ and $x \\neq \\log_a 2 < 1$.\n\nTherefore, the desired values are $a \\in (1, 2)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17490,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AB$ and $CP$ be parallel lines, and let $P$ be the intersection point of the tangent lines to the circle $\\Gamma$. Given that $CP = BP = 4$ and $AB = 3$, find the length of $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $AB$ and $CP$ are parallel, $\\angle ABC = \\angle BCP$. By a well-known theorem, $\\angle CAB = \\angle PBC$ also holds. Therefore, triangles $ABC$ and $BCP$ are similar, which implies $AB : BC = BC : CP$. Thus, $BC = \\sqrt{AB \\cdot CP}$. Given $CP = BP = 4$ and $AB = 3$, we have:\n\n$$BC = \\sqrt{3 \\cdot 4} = 2\\sqrt{3}$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17491,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose the product $N$ of all the positive factors of $n$ equals $24^{240}$. Since the prime factors of $24^{240}$ are 2 and 3, the prime factors of $n$ must also be 2 and 3. Thus, $n = 2^a 3^b$ for some non-negative integers $a$ and $b$. The number of positive factors of $n$ is $(a+1)(b+1)$, so the product $N$ of all the factors of $n$ is:\n\n$$\n(2^a 3^b)^{\\frac{(a+1)(b+1)}{2}} = 2^{\\frac{a(a+1)(b+1)}{2}} 3^{\\frac{(a+1)(b+1)}{2}}\n$$\n\nGiven $N = 24^{240} = 2^{720}3^{240}$, solve for $a$ and $b$:\n\n$$\n\\begin{cases}\n\\frac{a(a+1)(b+1)}{2} = 720 \\\\\n\\frac{(a+1)b(b+1)}{2} = 240\n\\end{cases}\n$$\n\nFind $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Dividing both sides of the first equation by the corresponding sides of the second equation, we get $a = 3b$. Substituting this into the second equation gives $(3b+1)b(b+1) = 480$, so:\n\n$$\n(3b+1)b(b+1) - 480 = (b-5)(3b^2+19b+96) = 0\n$$\n\nSince $3b^2 + 19b + 96 = 0$ has no real root, $b = 5$ must hold, and therefore $a = 15$. Thus, $n = 2^{15}3^5$ is the desired solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 17492,
    "subject": "Mathematics (Olympiad)",
    "question": "One rectangle has side lengths $a$ cm and $b$ cm. If the side with length $a$ cm is enlarged by $b$ cm and the side with length $b$ cm is enlarged by $a$ cm, then the resulting square has area $100\\ \\mathrm{cm}^2$. Determine the rectangle that satisfies this condition with the smallest area, if its side lengths are positive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "The area of the resulting square is $P_{\\text{sq}} = (a + b)^2 = 100\\ \\mathrm{cm}^2$, so $a + b = 10\\ \\mathrm{cm}$. Since $a$ and $b$ are positive integers, the possible pairs are $a = 1, b = 9$; $a = 2, b = 8$; $a = 3, b = 7$; $a = 4, b = 6$; $a = 5, b = 5$; $a = 6, b = 4$; $a = 7, b = 3$; $a = 8, b = 2$; and $a = 9, b = 1$. The rectangle with the smallest area is for $a = 1, b = 9$ or $a = 9, b = 1$. Its area is $9\\ \\mathrm{cm}^2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17493,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $H$ be the orthocenter of the acute triangle $ABC$ and let $X$ be the midpoint of the side $BC$. The perpendicular at $H$ to $HX$ intersects the sides $AB$ and $AC$ at the points $Y$ and $Z$ respectively. Let $O$ be the circumcenter of the triangle $ABC$ and $O'$ be the circumcenter of the triangle $BHC$. Prove that:\n\na) $HY = HZ$;\n\nb) $\\overrightarrow{AY} + \\overrightarrow{AZ} = 2\\overrightarrow{OO'}$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $A'$ be the point diametrically opposite to $A$ on the circumcircle of triangle $ABC$. It is known that $H$ and $A'$ are symmetric with respect to $X$, so $H$, $X$, and $A'$ are collinear. Then $\\angle ABA' = \\angle ACA' = 90^\\circ$. Thus, the quadrilaterals $A'BYH$ and $A'CZH$ are cyclic, therefore $\\angle YBH = \\angle YA'H$ and $\\angle ZCH = \\angle ZA'H$.\n\nOn the other hand, from the right triangles $BEA$ and $CFA$ we have $\\angle ABE = \\angle ACF = 90^\\circ - \\angle BAC$ (where $E$ and $F$ are the feet of the altitudes from $B$ and $C$ respectively).\n\nConsequently, $\\angle YA'H = \\angle ZA'H$, so $A'H$ is both an altitude and an angle bisector in triangle $YA'Z$, so it is also a median, that is $HY = HZ$.\n\nb) From a), it follows that $\\overrightarrow{AZ} + \\overrightarrow{AY} = 2\\overrightarrow{AH}$.\n\nIt is known that the symmetric $H'$ of the orthocenter $H$ with respect to the side $BC$ is on the circumcircle of triangle $ABC$, so the symmetric of the circumcircle of triangle $BHC$ with respect to $BC$ is the circumcircle of triangle $BH'C$, i.e., the circumcircle of triangle $ABC$. Thus, the center of the circumcircle of triangle $BHC$ is symmetric to $O$ with respect to $BC$.\n\nWe deduce that $\\overrightarrow{OO'} = 2\\overrightarrow{OX} = \\overrightarrow{AH}$, since $OX$ is a midline in triangle $HA'A$.\nFrom here it follows $\\overrightarrow{AZ} + \\overrightarrow{AY} = 2\\overrightarrow{OO'}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17494,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with incenter $I$. The incircle of the triangle is tangent to the sides $BC$ and $AC$ at points $D$ and $E$, respectively. Let $P$ denote the common point of lines $AI$ and $DE$, and let $M$ and $N$ denote the midpoints of sides $BC$ and $AB$, respectively. Prove that points $M$, $N$, and $P$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "For $AB = AC$, we get $D = M = P$, so the points $M$, $N$, and $P$ are trivially collinear. We will now only prove the case $AB > AC$ as $AB < AC$ is completely analogous.\n\nLet $\\alpha$, $\\beta$, and $\\gamma$ denote the interior angles of the triangle at $A$, $B$, and $C$, respectively. We note that\n\n$$\n\\angle BDP = \\angle CDE = 90^\\circ - \\frac{\\gamma}{2} \\quad \\text{and} \\quad \\angle BIP = \\frac{\\alpha + \\beta}{2} = 90^\\circ - \\frac{\\gamma}{2}\n$$\n\ncertainly hold, which implies that the quadrilateral $BPDI$ is cyclic. We therefore have $90^\\circ = \\angle IDB = \\angle IPB$, which implies that triangle $APB$ is right-angled. Since $N$ is the midpoint of its hypotenuse, it must be its circumcenter, which yields $\\angle BNP = 2 \\cdot \\angle BAP = \\alpha$. We see that $PN$ is parallel to $AC$, and since $MN$ is also parallel to $AC$ by virtue of the fact that $M$ and $N$ are the midpoints of their respective sides of $ABC$, we see that $M$, $N$, and $P$ must be collinear, as claimed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17495,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S_n = n^n + n^{n-1} + \\dots + 1 = \\frac{n^{n+1}-1}{n-1}$. Consider the equation\n\n$$\nn^{r_0} + \\dots + n^{r_n} = k(n^n + n^{n-1} + \\dots + 1)$$\n\nwhere $0 \\le r_i \\le n$ for all $i$, and $k < n+1$. Find all possible sets $\\{r_0, \\dots, r_n\\}$ that satisfy this equation.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N_p$ be the number of $r_i$'s equal to $p$.\n\nConsider both sides of equation (1) modulo $n$:\n- LHS: $\\equiv N_0 \\cdot n^0 \\equiv N_0 \\pmod{n}$\n- RHS: $\\equiv k(1) \\pmod{n}$\nSo $N_0 \\equiv k \\pmod{n}$. If $N_0 < n$ and $k < n$, then $N_0 = k$.\n\nSimilarly, for $N_1$ (number of times 1 appears), consider modulo $n^2$:\n- LHS: $\\equiv N_0 \\cdot n^0 + N_1 \\cdot n^1 \\pmod{n^2}$\n- RHS: $\\equiv k(n^0 + n^1) \\pmod{n^2}$\nSo $N_0 + N_1 n \\equiv k + k n \\pmod{n^2}$. Since $N_0 = k$, $k + N_1 n \\equiv k + k n \\pmod{n^2}$, so $N_1 n \\equiv k n \\pmod{n^2}$, which means $n^2 | (N_1 - k) n$. Thus $n | (N_1 - k)$. If $N_1 < n$ and $k < n$, then $|N_1 - k| < n$, so $N_1 = k$.\n\nBy induction, the frequency of each number $p \\in \\{0, 1, \\dots, n\\}$ among the $r_i$'s (i.e., $N_p$) is equal to $k$. Since there are $n+1$ values $r_i$, and each of the $n+1$ possible residues $0, \\dots, n$ occurs $k$ times, we have $(n+1)k = n+1$, so $k=1$.\n\nTherefore, each residue $0, 1, \\dots, n$ appears exactly once among $r_0, \\dots, r_n$. Thus, the $r_i$ are distinct and form a permutation of $\\{0, 1, \\dots, n\\}$.\n\n*Answer*: All sets $\\{r_0, \\dots, r_n\\}$ that form a complete residue system modulo $n+1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 17496,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a convex 2006-gon. The 1003 diagonals connecting opposite vertices and the 1003 lines connecting the midpoints of opposite sides are concurrent, that is, all 2006 lines have a common point. Prove that the opposite sides of $P$ are parallel and congruent.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A_1A_2\\ldots A_{2006}$ be a polygon such that $A_1A_{1004}$, $A_2A_{1005}$, $\\ldots$, $A_{1003}A_{2006}$ and $M_1M_{1004}$, $M_2M_{1005}$, $\\ldots$, $M_{1003}M_{2006}$ are concurrent at $O$, where $M_1, M_2, \\ldots, M_{2006}$ are the midpoints of $A_1A_2, A_2A_3, \\ldots, A_{2006}A_1$, respectively.\n\nNote that $A_1A_{1004}$, $A_2A_{1005}$, $M_1M_{1004}$ are concurrent, so $A_1A_2 \\parallel A_{1004}A_{1005}$. Similarly, we obtain that any opposite sides are parallel.\n\nNow, by applying the ratio theorem repetitively, we obtain\n\n$$\n\\frac{OA_1}{OA_{1004}} = \\frac{OA_2}{OA_{1005}} = \\dots = \\frac{OA_{1003}}{OA_{2006}} = \\frac{OA_{1004}}{OA_1} = \\frac{OA_{1005}}{OA_2} = \\dots = \\frac{OA_{2006}}{OA_{1003}}\n$$\n\nThis implies that $O$ is the midpoint of every main diagonal. Therefore, all opposite sides are congruent and parallel.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17497,
    "subject": "Mathematics (Olympiad)",
    "question": "Euler marks $n$ different points in the Euclidean plane. For each pair of marked points, Gauss writes down the number $\\lfloor \\log_2 d \\rfloor$ where $d$ is the distance between the two points.\n\n*Note:* For any $d > 0$, $\\lfloor \\log_2 d \\rfloor$ is the unique integer $k$ such that $2^k \\le d < 2^{k+1}$.\n\nProve that Gauss writes down less than $2n$ distinct values.",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove that Gauss writes down at most $n$ even numbers.\n\nFor each even number $2k$ that Gauss writes down, choose a single pair of points whose distance $d$ satisfies $2^k \\le d < 2^{k+1}$. Connect these points with a red edge. We claim there cannot be a cycle: indeed, if the edges corresponding to the distinct even integers $2k_1, \\dots, 2k_m, 2k_{m+1}$ form a cycle in that order, then assume without loss of generality that $2k_{m+1}$ is the largest among these, and $2k_i$ is the largest among the rest. The sum of distances for the first $m$ edges is at most\n\n$$\n2^{2k_1+1} + \\dots + 2^{2k_m+1} \\le 2^{2k_i+1}(1 + 2^{-2} + 2^{-4} + \\dots) \\le \\frac{2^{2k_i+1}}{1 - \\frac{1}{2^2}} < 2^{2k_i+2} \\le 2^{2k_{i+1}},\n$$\n\ni.e., less than the distance corresponding to the last edge: a contradiction to the triangle inequality. So there are at most $n-1$ red edges.\n\nThis implies that Gauss only writes at most $n-1$ even numbers, and similarly at most $n-1$ odd numbers. Thus, Gauss writes down at most $2n-2$ numbers in total. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17498,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose we have $h + \\ell + u$ objects, where:\n- $h$ are known to be heavier than all the objects they were compared with,\n- $\\ell$ are known to be lighter than the objects they were compared with,\n- $u$ are untested.\n\nShow that we need $h + \\ell + u + \\lfloor \\frac{u-3}{2} \\rfloor$ tests to determine the relative weights.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove the statement by induction on $h + \\ell + 2u$.\n\n**Base case:** For $h + \\ell + 2u = 2$, we have either $h = \\ell = 0$ and $u = 1$, or $h = \\ell = 1$ and $u = 0$; in both cases, no test is needed. Indeed,\n$$\n1 + 1 + 0 + \\left\\lfloor \\frac{-3}{2} \\right\\rfloor = 0 + 0 + 1 + \\left\\lfloor -1 \\right\\rfloor = 0.\n$$\n\n**Induction step:**\n- If $h \\geq 2$, we test two of the heaviest objects; then both $h$ and $h + \\ell + u + \\lfloor \\frac{u-3}{2} \\rfloor$ decrease by 1. The case $\\ell \\geq 2$ is analogous.\n- If $u \\geq 2$, we test two untested objects: then $h$ and $\\ell$ increase by 1 and $u$ decreases by 2, and the expression $h + \\ell + u + \\lfloor \\frac{u-3}{2} \\rfloor$ decreases by 1.\n- If $h \\geq 1$ and $u \\geq 1$, we compare a heavy object and an untested object. If the heavy object is heavier than the untested one, then the untested is now a light object, decreasing $u$ by 1 and increasing $\\ell$ by 1; the expression $h + \\ell + u + \\lfloor \\frac{u-3}{2} \\rfloor$ either remains the same (if $u$ is even) or decreases by 1 (if $u$ is odd). If the heavy object is lighter than the untested one, then the untested one is now a heavy object, decreasing $u$ by 1 and maintaining $h$; the expression $h + \\ell + u + \\lfloor \\frac{u-3}{2} \\rfloor$ either decreases by 1 (if $u$ is even) or by 2 (if $u$ is odd).\n- The case $\\ell \\geq 1$ and $u \\geq 1$ is analogous.\n- Finally, if we compare a heavy object with a light object, the expression $h + \\ell + u + \\lfloor \\frac{u-3}{2} \\rfloor$ remains the same.\n\nNote that we can always make comparisons so that the expression decreases by at least 1, and in the worst case scenario, it always decreases by 1. Since the expression never increases, the proof is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 17499,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be the lengths of the sides of a right triangle with integer side lengths such that the cubic polynomial $x^3 - 30x^2 + r x - 780$ has roots $a$, $b$, and $c$. Find the value of $r$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a$, $b$, $c$ be the lengths of such a triangle. Without loss of generality, assume $0 < a \\leq b < c$. Then\n\n$$\n\\begin{aligned}\nx^3 - 30x^2 + r x - 780 &= (x - a)(x - b)(x - c) \\\\\n&= x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc,\n\\end{aligned}\n$$\n\nso that\n\n$$\na + b + c = 30, \\tag{1}\n$$\n\n$$\nr = ab + ac + bc, \\tag{2}\n$$\n\n$$\nabc = 780. \\tag{3}\n$$\n\nBy (1),\n\n$$\n(a + b)^2 = (30 - c)^2,\n$$\n\nand by the Pythagorean theorem,\n\n$$\nc^2 = a^2 + b^2.\n$$\n\nSo,\n\n$$\n0 = 900 - 60c - 2ab.\n$$\n\nSince $ab = \\frac{780}{c}$ by (3), this simplifies to\n\n$$\nc^2 - 15c + 26 = 0,\n$$\n\nimplying $c = 2$ or $c = 13$.\n\nIf $c = 2$, then $0 < a, b < 2$ and $a + b + c \\neq 30$.\n\nIf $c = 13$, then by (3), $ab = 60$ and by (1), $a + b = 17$, so $a, b$ are roots of $x^2 - 17x + 60 = 0$. Thus $(a, b) = (5, 12)$ or $(12, 5)$. Therefore, by (2),\n\n$$\nr = ab + ac + bc = 60 + 65 + 156 = 281.\n$$\n\nConversely, $5$, $12$, $13$ are the roots of $x^3 - 30x^2 + 281x - 780 = 0$.\n\n**Answer:** $r = 281$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 17500,
    "subject": "Mathematics (Olympiad)",
    "question": "A natural number is called \"happy\" if the sum of its digits is $7$. Let $(a_n)_{n=1}^\\infty$ be the increasing sequence of happy numbers. If $a_n = 2005$, compute $a_{5n}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use combinatorial arguments to count happy numbers:\n\n**Lemma 1.** The number of solutions to $x_1 + x_2 + \\dots + x_k = n$ in natural numbers is $\\binom{n-1}{k-1}$.\n\n**Lemma 2.** The number of solutions to $x_1 + x_2 + \\dots + x_k = n$ in non-negative integers is $\\binom{n+k-1}{k-1}$.\n\n**Lemma 3.** The number of solutions to $x_1 + x_2 + \\dots + x_k = n$ in non-negative integers with $x_1 \\ge 1$ is $\\binom{n+k-2}{k-1}$.\n\nFor $n=7$, the number of $k$-digit happy numbers is:\n$$\np(k) = \\binom{k+5}{6}.\n$$\nSo:\n- $p(1) = \\binom{6}{6} = 1$\n- $p(2) = \\binom{7}{6} = 7$\n- $p(3) = \\binom{8}{6} = 28$\n- $p(4) = \\binom{9}{6} = 84$\n- $p(5) = \\binom{10}{6} = 210$\n\nThe number $2005$ is the smallest 4-digit happy number of the form $2abc$. Counting all happy numbers with fewer digits and those with 4 digits less than $2005$:\n$$\n1 + 7 + 28 + 28 + 1 = 65\n$$\nSo $a_{65} = 2005$, thus $n = 65$ and $5n = 325$.\n\nThe total number of happy numbers with at most 5 digits is:\n$$\n1 + 7 + 28 + 84 + 210 = 330\n$$\nThe least six-digit happy numbers in decreasing order are $70000, 61000, 60100, 60010, 60001, 52000$.\n\nTherefore, $a_{5n} = a_{325} = 52000$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  }
]