id stringlengths 4 4 | problem_markdown stringlengths 36 3.59k | solutions_markdown listlengths 0 10 | images images listlengths 0 15 | country stringclasses 58
values | competition stringlengths 3 108 ⌀ | topics_flat listlengths 0 12 | language stringclasses 18
values | problem_type stringclasses 4
values | final_answer stringlengths 1 1.22k ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0fbz | Problem:
Alrededor de una mesa circular están sentadas seis personas. Cada una lleva un sombrero. Entre cada dos personas hay una mampara de modo que cada una puede ver los sombreros de las tres que están enfrente, pero no puede ver el de la persona de su izquierda ni el de la de su derecha ni el suyo propio. Todas sa... | [
"Solution:\n\nNumeramos las personas en el orden en que van respondiendo, con lo que la persona 1 ve los sombreros de las personas $3,4,5$, la persona 2 los de las personas $4,5,6$, y la persona 3 los de las personas $5,6,1$.\n\nSupongamos que ni la persona 1 ni la persona 2 han podido responder \"Sí\". Los sombrer... | Spain | null | [
"Discrete Mathematics > Logic"
] | null | proof only | null | |
0l9x | Let $\triangle ABC$ be a triangle with two fixed vertices $B, C$ and the vertex $A$ is variable. Let $H$ and $G$ be the orthocenter and the centroid of the triangle $ABC$ respectively. Find the locus of $A$ such that the midpoint $K$ of the segment $HG$ moves on the line $BC$. | [
"Choose orthogonal Cartesian coordinate system $Oxy$ where $O$ is the midpoint of the segment $BC$ and $Oy$ is the line $BC$. Denote by $2a > 0$ the length of the segment $BC$. The coordinates of the vertices $B$ and $C$ are $B(-a, 0)$ and $C(a, 0)$. Suppose that $A$ has the coordinates $A(x_0, y_0)$ ($y_0 \\neq 0$... | Vietnam | Vijetnam 2007 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | The locus is the hyperbola x^2/a^2 - y^2/(3a^2) = 1, excluding the points B and C. | |
0gcz | 設 $n$ 為正整數, $A$ 和 $B$ 為互質正整數且
$$
\left(\frac{n(n+1)}{2}\right)! \cdot \prod_{k=1}^{n} \frac{k!}{(2k)!} = \frac{B}{A}.
$$
證明 $A$ 是2的幂次。 | [
"只須證明對所有質數 $p \\ge 3$, 均有\n$$\n\\sum_{i=1}^{\\infty} \\left( \\left[ \\frac{n(n+1)}{2p^i} \\right] + \\sum_{k=1}^{n} \\left( \\left[ \\frac{k}{p^i} \\right] - \\left[ \\frac{2k}{p^i} \\right] \\right) \\right) \\ge 0\n$$\n\n注意到 $[2x] = [x] + [x + \\frac{1}{2}]$, 那麼只要對每個正整數 $i$ 證明\n$$\n\\left[ \\frac{n(n+1)}{2P} \\r... | Taiwan | 二〇一九數學奧林匹亞競賽第一階段選訓營 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
052r | During the schoolyear $22$ olympiads were held. At each one $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 got a prize at every olympiad. | [
"Look at an arbitrary olympiad, let that be $A_1$, where the prizes went to some $5$ students. Each of the remaining $21$ olympiads had to have someone among those $5$ receiving a prize. By pigeonhole principle there exists a student who in addition to $A_1$ also got a prize at at least $5$ olympiads. Let that stud... | Estonia | Open Contests | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0dy0 | The terms $a_1, a_2, a_3, a_4, a_5$ of a geometric sequence are positive integers. All are less than $2008$, $a_2$ is divisible by $5$, $a_3$ is divisible by $4$, $a_4$ is divisible by $3$, $a_1$ is not divisible by $6$ and no prime number divides all five of them. Determine the terms of this sequence. | [
"Since $a_1, a_2, a_3, a_4, a_5$ are terms of a geometric sequence, we can write $a_i = a_1 \\cdot q^{i-1}$ for $i = 2, 3, 4, 5$ and for some real number $q$. But $q = \\frac{a_2}{a_1}$ is a quotient of two positive integers, so $q$ is rational. Write $q = \\frac{m}{n}$ as a reduced fraction. The terms of the seque... | Slovenia | Slovenija 2008 | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof and answer | 625, 750, 900, 1080, 1296 | |
0ivl | Problem:
Penta chooses 5 of the vertices of a unit cube. What is the maximum possible volume of the figure whose vertices are the 5 chosen points? | [
"Solution:\n\nThe answer is $\\frac{1}{2}$.\n\nLabel the vertices of the cube $A, B, C, D, E, F, G, H$, such that $ABCD$ is the top face of the cube, $E$ is directly below $A$, $F$ is directly below $B$, $G$ is directly below $C$, and $H$ is directly below $D$.\n\nWe can obtain a volume of $\\frac{1}{2}$ by taking ... | United States | Harvard-MIT November Tournament | [
"Geometry > Solid Geometry > Volume"
] | null | proof and answer | 1/2 | |
0am8 | Problem:
A square with an area of $40~\mathrm{m}^2$ is inscribed in a semicircle. The area of the square that could be inscribed in the circle with the same radius is
(a) $100~\mathrm{m}^2$
(b) $120~\mathrm{m}^2$
(c) $80~\mathrm{m}^2$
(d) $140~\mathrm{m}^2$ | [] | Philippines | Qualifying Round | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | MCQ | a | |
0ldw | Given an acute, scalene triangle $ABC$ with circumcircle $(O)$, orthocenter $H$ and centroid $G$. Let $H_a$, $H_b$ and $H_c$ be the feet of altitudes from $A$, $B$ and $C$ in triangle $ABC$ and $D$, $E$ and $F$ be the midpoints of $BC$, $CA$ and $AB$ in that order. Let the rays $GH_a$, $GH_b$ and $GH_c$ intersect $(O)$... | [
"a) Let the line passing through $A$ and parallel to $BC$ meet $(O)$ at $A_0$. We have $AA_0CB$ is an isosceles trapezoid so $A_0C = AB = 2FH_a$. We also have $\\angle FH_aB = \\angle FBH_a = \\angle A_0CB$ then $FH_a \\parallel A_0C$. But $GC = 2GC$, it follows that $G$, $A_0$ and $H_a$ are collinear.\n\n![](attac... | Vietnam | VMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometr... | English | proof only | null | |
0e8w | In the triangle $ABC$, the bisectors of the angles $\angle BAC$ and $\angle CBA$ intersect at the point $T$. Let $\gamma$ denote the size of the angle $\angle ACB$. What is the size of the angle $\angle ATB$?
(A) $2\gamma$
(B) $180^\circ - \gamma$
(C) $360^\circ - 4\gamma$
(D) $60^\circ + \ga... | [
"We have $\\angle ATB = 180^\\circ - \\angle BAT - \\angle TBA = 180^\\circ - \\frac{\\alpha}{2} - \\frac{\\beta}{2} = 180^\\circ - \\frac{1}{2}(\\alpha + \\beta)$. Since $\\alpha + \\beta + \\gamma = 180^\\circ$, we get $\\alpha + \\beta = 180^\\circ - \\gamma$. So, $\\angle ATB = 180^\\circ - \\frac{1}{2}(180^\\c... | Slovenia | National Math Olympiad 2013 - First Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | E | |
09j6 | A class has 30 students and they sit in three groups of 10 students each. At the beginning of each month, the teacher swaps the students' seats. What is the minimum number of months needed in order for every pair of students to sit in the same group for at least a month?
(Proposed by Otgonbayar Uuye) | [
"Let us say a pair of students is friends if they have sat in one group for at least a month.\nNow we prove that four months is not enough. First note that for integers $x \\ge 0$, $y \\ge 0$, $z \\ge 0$ such that $x + y + z = 10$, we have\n$$\nf(x, y, z) = \\binom{x}{2} + \\binom{y}{2} + \\binom{z}{2} \\ge 12.\n$$... | Mongolia | Mongolian Mathematical Olympiad Round 3 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 5 | |
06sa | Let $n$ be a positive integer, and consider a sequence $a_{1}, a_{2}, \ldots, a_{n}$ of positive integers. Extend it periodically to an infinite sequence $a_{1}, a_{2}, \ldots$ by defining $a_{n+i}=a_{i}$ for all $i \geqslant 1$. If
$$
a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n} \leqslant a_{1}+n
$$
and
$$
a... | [
"First, we claim that\n$$\n\\begin{equation*}\na_{i} \\leqslant n+i-1 \\quad \\text{ for } i=1,2, \\ldots, n . \\tag{3}\n\\end{equation*}\n$$\nAssume contrariwise that $i$ is the smallest counterexample. From $a_{n} \\geqslant a_{n-1} \\geqslant \\cdots \\geqslant a_{i} \\geqslant n+i$ and $a_{a_{i}} \\leqslant n+i... | IMO | International Mathematical Olympiad Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0jd5 | Problem:
a. A number is written at each edge of a cube. The cube is called magic if:
(i) For every face, the four edges around it have the same sum.
(ii) For every vertex, the three edges meeting at it have the same sum. (The face and vertex sums may be different.)
Determine if there exists a magic cube using
(a) the ... | [
"Solution:\n\n(a) The answer is no. If the cube is magic, then every triple of edges abutting a vertex has of course the same average. Since every edge belongs to the same number of vertices (two), these vertex averages are the same as the average of all the numbers on the cube, which is $13/2$. But it is impossibl... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Graph Theory",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) no; b) yes | |
08d9 | Problem:
Sull'isola dei Cavalieri e dei Furfanti, i Cavalieri dicono la verità tranne quando si sbagliano ed i Furfanti mentono sempre. Durante una riunione, 40 isolani si siedono attorno a un grande tavolo rotondo e ciascuno dice: "Io sono vicino a un Cavaliere e a un Furfante". Sapendo che 3 Cavalieri presenti si sb... | [
"Solution:\n\nLa risposta è $27$.\n\nCominciamo con qualche osservazione di base. Non possono esserci solo Cavalieri al tavolo, perché dovrebbero sbagliarsi tutti e non solo 3 di loro. Se guardiamo un Furfante o, equivalentemente, un Cavaliere che sbaglia, accanto a lui siedono o due Furfanti o due Cavalieri.\n\nPo... | Italy | Gara di Febbraio | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 27 | |
0587 | Juku claims that if the sum of the squares of all digits of a natural number is divisible by $3$ then the number itself is divisible by $3$. Is Juku's claim always true? | [
"The sum of the squares of the digits of the number $112$ is $6$ which is divisible by $3$, while the number $112$ is not divisible by $3$."
] | Estonia | Estonian Math Competitions | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | proof only | null | |
0h4r | Using each of ten digits exactly once build two numbers (none of them can start with $0$) so that the absolute value of their difference is smallest possible.
**Answer:** $50123$ and $49876$. | [
"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 num... | Ukraine | Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 50123 and 49876 | |
07i6 | There are $n \ge 3$ points in a plane. No three points are collinear. Prove one can choose an ordering $P_1, P_2, \dots, P_n$ for these points so that for all $1 < i < n$ the angle $\angle P_{i-1}P_iP_{i+1}$ is acute. | [
"We prove a stronger statement; there is an ordering $P_1, P_2, \\dots, P_n$ such that for every $1 \\le i < n$, the angle $\\angle P_{i-1}P_iP_{i+1}$ is acute.\n\nAssume that $P_nP_{n-1}$ is the longest segment between any two of these points. Now we start with $P_{n-1}$ and at each step we choose a point $P$ such... | Iran | 40th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0ffl | Problem:
Sea $ABC$ un triángulo equilátero, y $\mathcal{E}$ el conjunto de todos los puntos contenidos en los tres segmentos $AB$, $BC$ y $CA$ (con $A$, $B$ y $C$ incluidos). Determinar si es cierto que para cada partición de $\mathcal{E}$ en dos conjuntos disjuntos, por lo menos uno de los dos conjuntos contiene los ... | [] | Spain | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Yes | |
0477 | For a positive integer $n$, a subset $S$ of $\{1,2,\dots,n\}$ is called an $n$-good set if for any elements $x,y$ in $S$ (they can be the same), if $x+y \le n$, then $x+y \in S$.
For a positive integer $n$, define $r_n$ as the smallest real number such that for any positive integer $m \le n$, there exists an $n$-good s... | [
"*Proof.* We will show that $\\alpha = 2 - \\sqrt{2}$ satisfies the requirement. We will prove this in two steps.\n\nFirst, we prove that $r_n \\ge \\alpha n - 4$. For this, we only need to consider the case where $n > 4$. Let $m = \\lfloor \\frac{n}{\\sqrt{2}} \\rfloor < n$, and let $d = n - m$. Consider an $n$-go... | China | The 65th IMO China National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
0f2q | Problem:
Let $x_n = (1 + \sqrt{2} + \sqrt{3})^{n}$. We may write $x_n = a_n + b_n \sqrt{2} + c_n \sqrt{3} + d_n \sqrt{6}$, where $a_n, b_n, c_n, d_n$ are integers. Find the limit as $n$ tends to infinity of $b_n / a_n$, $c_n / a_n$, $d_n / a_n$. | [
"Solution:\n\nLet $x_n = (1 + \\sqrt{2} + \\sqrt{3})^n$. Consider the conjugates of $1 + \\sqrt{2} + \\sqrt{3}$ under the field automorphisms fixing $\\mathbb{Q}$:\n\nLet $\\alpha_1 = 1 + \\sqrt{2} + \\sqrt{3}$\nLet $\\alpha_2 = 1 + \\sqrt{2} - \\sqrt{3}$\nLet $\\alpha_3 = 1 - \\sqrt{2} + \\sqrt{3}$\nLet $\\alpha_4... | Soviet Union | ASU | [
"Number Theory > Algebraic Number Theory > Algebraic numbers"
] | null | proof and answer | lim b_n/a_n = 1/sqrt(2), lim c_n/a_n = 1/sqrt(3), lim d_n/a_n = 1/sqrt(6) | |
0686 | The positive integer $n$ is such that $n^2-9$ has exactly 6 positive divisors. Prove that $\text{gcd}(n-3, n+3)=1$. | [
"In order the positive integer $n$ to have exactly 6 positive divisors, it must be of the form $n^2-9 = q^5$ or $n^2-9 = p^2q$, where $p$, $q$ are primes mutually different.\n\nIn the first case we have $n^2-9 = q^5 \\Rightarrow (n-3)(n+3) = q^5$ (1)\nIt gives that\n$$\nn-3 = q^s \\text{ and } n+3 = q^t, \\text{ wi... | Greece | Selection Examination | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0546 | a) Find the largest number that is the greatest common divisor of some four different two-digit numbers.
b) Find the largest number that is the least common multiple of some four different two-digit numbers. | [
"a) Let $d$ be the greatest common divisor of some four different two-digit numbers. Since all these numbers are divisible by $d$, the least possible candidates of these four numbers are $d$, $2d$, $3d$, $4d$. Hence $4d < 100$, thus $d \\le 24$. On the other hand, the greatest common divisor of $24$, $48$, $72$ and... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | a) 24; b) 99*98*97*95 | |
0g74 | 令 $n$ 為正整數並給定無限長的週期字串 $W = \dots x_{-1}x_0x_1x_2\dots$, 其中 $x_i$ 可以是字母 $a$ 或 $b$ 對所有整數 $i$, 而 $W$ 的最小週期 $p$ 是大於 $2^n$ (即 $x_i = x_{i+p}$ 對於所有 $i$, 而這樣的 $p$ 不可能再小了)。我們稱一個長度 $\le p$ 的非空的子字串 $U$ 是左右逢源的 (左右逢源以 FU 當作縮寫), 如果 $aU, bU, Ua$ 與 $Ub$ 這四種字串都出現在 $W$ 中。試證明: $W$ 至少包含 $n$ 個 FU 子字串。 | [
"對於字串 $U$, 以 $|U|$ 來表示該字串的長度。這裡所提到的子串都是非空而且長度 $\\le p$, 而給定的 $W$ 可以視為一個長度為 $p$ 的圓周字串 $x_0x_1x_2\\dots x_{p-1}$; 因為這樣的 $W$ 是有限的, 所以我們可以定義任意子字串 $U = a_0a_1\\dots a_{k-1}$ 出現在 $W$ 的次數 (multiplicity)\n$$\nm(U) = |\\{i|0 \\le i \\le p-1 \\text{ 且 } a_0a_1\\cdots a_{k-1} = x_i x_{i+1} \\cdots x_{i+k-1}\\}|\n$$\n其中所有的足標都需... | Taiwan | 二〇一二數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0hgb | $2n$ distinct positive integers are given. What's 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? | [
"Let $p_1, p_2, \\dots, p_{2n-1}$ be distinct prime integers larger than $2$. Then for a set $(1, p_1-1, p_2-1, \\dots, p_{2n-1}-1)$ it's not possible to form $n$ such pairs, as no number can be paired with number $1$.\n\nFrom the other side, we can always form at least $n-1$ pairs from the numbers of the same pari... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | n-1 | |
0f5p | Problem:
$ABC$ is a triangle and $P$ is any point. The lines $PA$, $PB$, $PC$ cut the circumcircle of $ABC$ again at $A'$, $B'$, $C'$ respectively. Show that there are at most eight points $P$ such that $A'B'C'$ is congruent to $ABC$. | [] | Soviet Union | 18th ASU | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangen... | null | proof only | null | |
02rq | Problem:
Decide, with proof, if each of the following statement is true.
a. If $(a_n)_{n \ge 1}$ is a decreasing sequence with positive terms such that $\sum a_n = +\infty$ then there exists a decreasing sequence $(b_n)_{n \ge 1}$ with positive terms with $b_n \le a_n$ for all $n$, $\sum b_n = +\infty$ and $\lim(n b_n... | [] | Brazil | Brazilian Math Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a is false; b is false | |
0g77 | 設 $P$ 為銳角三角形 $ABC$ 內部一點，且 $P$ 到三頂點的距離分別為 $d_A, d_B, d_C$，到三邊的垂直距離分別為 $d_1, d_2, d_3$。試證：
$$
d_A + d_B + d_C \geq 2(d_1 + d_2 + d_3).
$$ | [
"如圖所示，$PD \\perp BC$，$PE \\perp CA$，$PF \\perp AB$。\n\n\n因為 $\\angle AEP + \\angle AFP = 90^\\circ + 90^\\circ = 180^\\circ$，所以 $A, E, P, F$ 共圓，且 $AP = d_A$ 為該圓直徑，於是可得：$\\angle EPF = 180^\\circ - \\angle A = \\angle B + \\angle C$。\n\n利用正弦定理 $EF = d_A \\sin A$，及餘弦定理可推得\n$$\n\\begin{aligned}... | Taiwan | 二〇一三數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle... | null | proof only | null | |
03yc | There are $2n$ real numbers $a_1, a_2, \dots, a_n, r_1, r_2, \dots, r_n$ satisfying $a_1 \le a_2 \le \dots \le a_n$ and $0 \le r_1 \le r_2 \le \dots \le r_n$. Prove that $\sum_{i=1}^n \sum_{j=1}^n a_i a_j \min(r_i, r_j) \ge 0$. | [
"Since\n$$\n\\sum_{i=1}^{n} \\sum_{j=1}^{n} a_i a_j \\min(r_i, r_j) = \\sum_{j=1}^{n} a_1 a_j \\min(r_1, r_j) + \\sum_{j=1}^{n} a_2 a_j \\min(r_2, r_j) \\\\\n+ \\dots + \\sum_{j=1}^{n} a_k a_j \\min(r_k, r_j) + \\dots \\\\\n+ \\sum_{j=1}^{n} a_n a_j \\min(r_n, r_j),\n$$\nits $k$-th term is\n$$\n\\sum_{j=1}^{n} a_k ... | China | China Southeastern Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof only | null | |
06sz | A card deck consists of 1024 cards. On each card, a set of distinct decimal digits is written in such a way that no two of these sets coincide (thus, one of the cards is empty). Two players alternately take cards from the deck, one card per turn. After the deck is empty, each player checks if he can throw out one of hi... | [
"Let us identify each card with the set of digits written on it. For any collection of cards $C_{1}, C_{2}, \\ldots, C_{k}$ denote by their sum the set $C_{1} \\triangle C_{2} \\triangle \\cdots \\triangle C_{k}$ consisting of all elements belonging to an odd number of the $C_{i}$'s. Denote the first and the second... | IMO | 55th International Mathematical Olympiad Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | All nonempty first moves; if the first player starts with the empty card, the second player can force a win. | |
0h5l | The equal segments $AB$ and $CD$ intersect in the point $O$ and are divided into the ratio $AO:OB=CO:OD=1:2$. The lines $AD$ and $BC$ intersect in the point $M$. Prove that $DM = MB$. | [
"Let the length of the segments be $AB=CD=3a$, then $AO=CO=a$ and $OB=OD=2a$. Since $\\angle AOD=\\angle COB$ as vertical (fig/ 17), then $\\triangle AOD=\\triangle COB$. Hence $\\angle ADO=\\angle CBO$. As $\\triangle BOD$ is isosceles, then $\\angle BDO=\\angle DBO$, consequently $\\angle MDB=\\angle MBD$ as the ... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round (Second Tour) | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0bm1 | Prove that, for every integer $n \ge 2$,
$$
\sum_{k=2}^{n} \frac{1}{\sqrt[k]{(2k)!}} \ge \frac{n-1}{2n+2}.
$$ | [
"We use induction on $n$. In case $n = 2$ the relation is an equality. We notice now that when going from $n-1$ to $n$ the right member increases with\n$$\n\\frac{n-1}{2n+2} - \\frac{n-2}{2n} = \\frac{1}{n(n+1)},\n$$\nso it is enough to prove that\n$$\n\\frac{1}{\\sqrt[n]{(2n)!}} \\ge \\frac{1}{n(n+1)}.\n$$\nThis i... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0ert | Determine the value of $\frac{13 \times 13 \times 12}{1 + 13 + 13 + 12}$ | [
"The fraction is $\\frac{13 \\times 13 \\times 12}{39} = \\frac{13 \\times 13 \\times 12}{3 \\times 13} = \\frac{13 \\times 12}{3} = 13 \\times 4 = 52$"
] | South Africa | South African Mathematics Olympiad Second Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 52 | |
0bh9 | $$
a(a + 2b)^x + b(b + 2c)^x + c(c + 2a)^x \le (a + b + c)^{x+1}.
$$ | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
0eoq | The last digit of $2^{2015} + 5^{2015}$ is
(A) 1 (B) 3 (C) 5 (D) 7 (E) 9 | [
"The last digits of the first few powers of $2$ are $2$, $4$, $8$, $6$, $2$, $4$, ..., so the digits repeat in cycles of length $4$. Since the remainder after dividing $2015$ by $4$ is $3$, it follows that the last digit of $2^{2015}$ is the same as the last digit of $2^3$, which is $8$. Finally, the last digit of ... | South Africa | South African Mathematics Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | B | |
0deu | Let $I$ be the incenter of the triangle $ABC$. Let $X$ lie on segment $AB$, such that $\angle AIX = 90^\circ$. The circumcircle of triangle $BIX$ intersects the circumcircle of triangle $ABC$ at point $Y \neq B$ lying on the same side of $AB$ as point $C$. Prove that $YX$ is the bisector of angle $AYB$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0hem | Find the largest positive integer $n$ such that it has exactly 4 positive integer divisors (including $1$ and $n$), the sum $S$ of which satisfies the condition $40 \le S \le 42$. | [
"In the case $n = pq$, $S = 1 + p + q + pq = (1+p)(1+q)$. Suppose $p < q$.\nLet us consider possible cases.\nIf $S = 40$, then $(p+1)(q+1) = 40$. Since $p+1 \\ge 3$, the following cases are possible:\n$$\np+1=4, q+1=10 \\Rightarrow p=3, q=9 \\Rightarrow q \\text{ is not prime.}\n$$\n$$\np+1=5, q+1=8 \\Rightarrow p=... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)"
] | null | proof and answer | 27 | |
07fd | We are given a natural number $d$. Find all open intervals of maximum length $I \subseteq \mathbb{R}$ such that for all real numbers $a_0, a_1, \dots, a_{2d-1}$ inside the interval $I$, the polynomial $P(x) = x^{2d} + a_{2d-1}x^{2d-1} + \dots + a_1x + a_0$ has no real roots. | [
"The answer is $I = (1, 1 + \\frac{1}{d})$.\n\nAssume that the desired interval is of the form $(b, c)$. For some $q, r$ in $(b, c)$, put $a_i = q$ for odd and $a_i = r$ for even $i$, where $0 \\le i \\le 2d - 1$. Then\n$$\nP(-1) = 1 - dq + dr.\n$$\nSince $P(x)$ has no real root, we must have $P(-1) > 0$. Thus, $q ... | Iran | 37th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | I = (1, 1 + 1/d) | |
03l2 | Problem:
At 12:00 noon, Anne, Beth and Carmen begin running laps around a circular track of length three hundred meters, all starting from the same point on the track. Each jogger maintains a constant speed in one of the two possible directions for an indefinite period of time. Show that if Anne's speed is different f... | [
"Solution:\n\nBy rotating the frame of reference we may assume that Anne has speed zero, that Beth runs at least as fast as Carmen, and that Carmen's speed is positive. If Beth is no more than twice as fast as Carmen, then both are at least $100$ meters from Anne when Carmen has run $100$ meters. If Beth runs more ... | Canada | Canadian Mathematics Olympiad | [
"Math Word Problems"
] | null | proof only | null | |
069j | The sequence $\alpha_0, \alpha_1, \alpha_2, \ldots, \alpha_\nu, \ldots$, $\nu \in \mathbb{N}$, with integer terms, not necessarily different, has the following properties:
(α) $0 \le \alpha_i \le i$, for every integer $i \ge 0$,
(β) $\binom{\kappa}{\alpha_0} + \binom{\kappa}{\alpha_1} + \dots + \binom{\kappa}{\alpha_... | [
"We will prove using induction with respect to $\\kappa$ that every initial part $\\alpha_0, \\alpha_1, \\alpha_2, \\ldots, \\alpha_\\kappa$ of the sequence consists of the following integers (not necessarily with the turn of the terms of the sequence)\n$0, 1, \\ldots, \\ell-1, 0, 1, \\ldots, \\kappa-\\ell$, for so... | Greece | SELECTION EXAMINATION 2019 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
08iv | Problem:
The positive real numbers $x$, $y$ and $z$ satisfy the relation $x + y + z \geq 1$. Prove the inequality
$$
\frac{x \sqrt{x}}{y + z} + \frac{y \sqrt{y}}{x + z} + \frac{z \sqrt{z}}{x + y} \geq \frac{\sqrt{3}}{2}
$$ | [] | JBMO | The first selection test for IMO 2003 and BMO 2003 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0k8c | Problem:
Yannick is playing a game with 100 rounds, starting with 1 coin. During each round, there is a $n \%$ chance that he gains an extra coin, where $n$ is the number of coins he has at the beginning of the round. What is the expected number of coins he will have at the end of the game? | [
"Solution:\n\nLet $X_{i}$ be the random variable which is the number of coins at the end of round $i$. Say that $X_{0}=1$ for convenience. Fix $i>0$ and some positive integer $x$. Conditioning on the event $X_{i-1}=x$, there are only two cases with positive probability. In particular,\n$$\n\\Pr\\left[X_{i}=x+1 \\mi... | United States | HMMT February 2019 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 1.01^100 | |
03xe | The range of $f(x) = \sqrt{x-5} - \sqrt{24-3x}$ is ______. | [
"It is easy to see that $f(x)$ is increasing on its domain $[5, 8]$. Therefore, its range is $[-3, \\sqrt{3}]$."
] | China | China Mathematical Competition | [
"Precalculus > Functions"
] | English | final answer only | [-3, √3] | |
0ezz | Problem:
A polygon $P$ has an inscribed circle center $O$. If a line divides $P$ into two polygons with equal areas and equal perimeters, show that it must pass through $O$. | [
"Solution:\nLet the line divide $P$ into two polygons $P_1$ and $P_2$ with equal areas and equal perimeters. Suppose, for contradiction, that the dividing line $\\ell$ does not pass through $O$.\n\nLet $r$ be the radius of the inscribed circle. The inscribed circle is tangent to every side of $P$.\n\nLet $d$ be the... | Soviet Union | ASU | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0cam | Problem:
Fie $ABC$ un triunghi dreptunghic în $A$, astfel încât $A'$ este mijlocul ipotenuzei, $M$ mijlocul înălțimii $AD$, $D \in BC$ și $\{P\} = BM \cap AA'$.
Dacă notăm $\alpha = m(\widehat{PCB})$, să se demonstreze că
$$
\operatorname{tg} \alpha = \sin C \cdot \cos C
$$ | [
"Solution:\nAplicăm teorema lui Menelaus în triunghiul $AA'D$, cu punctele $B, M, P$ coliniare:\n$$\n\\frac{BA'}{BD} \\cdot \\frac{MD}{MA} \\cdot \\frac{PA}{PA'} = 1 \\Leftrightarrow \\frac{PA}{PA'} = \\frac{BD}{BA'} = \\frac{2BD}{BC} = \\frac{2BD \\cdot BC}{BC^2} = \\frac{2AB^2}{BC^2} = 2 \\sin^2 C \\quad (1)\n$$\... | Romania | Olimpiada Națională de Matematică | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0icv | Problem:
You would like to provide airline service to the 10 cities in the nation of Schizophrenia, by instituting a certain number of two-way routes between cities. Unfortunately, the government is about to divide Schizophrenia into two warring countries of five cities each, and you don't know which cities will be in ... | [
"Solution:\nEach city $C$ must be directly connected to at least 6 other cities, since otherwise the government could put $C$ in one country and all its connecting cities in the other country, and there would be no way out of $C$. This means that we have 6 routes for each of 10 cities, counted twice (since each rou... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 30 | |
0173 | Two players, Maker and Breaker are playing the following game: Maker starts, and the players take turns choosing distinct numbers from the set $\{0, 1, \dots, 10\}$. Maker wins, if some of his chosen numbers form a strictly increasing arithmetic progression of length four, Breaker wins, if she manages to prevent this. ... | [
"We observe first that there are only two arithmetic progression of the desired kind that avoid the pair $\\{4, 7\\}$, namely $(0, 1, 2, 3)$ and $(0, 3, 6, 9)$. Therefore, if Breaker manages to choose these as her first two moves, she wins: The two progressions have only two common points, $0$ and $3$. Thus, after ... | Baltic Way | BALTIC WAY | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | Breaker has a winning strategy. | |
067e | We color the numbers $1, 2, 3, \ldots, 20$ with two colors, white and blank, in such a way that both colors are used. Find the number of ways we can perform this coloring if the product of white numbers and the product of blank numbers have maximal common divisor equal to $1$. (P. Bregiannis) | [
"Number $1$ can be colored in two ways, white or blank. Number $2$ also can be colored white or blank. Then all even numbers $2, 4, 6, 8, 10, 12, 14, 16, 18, 20$ have to be colored with the color of $2$.\nAlso all numbers having common divisor greater than $1$ with the above numbers must be colored with the color o... | Greece | 31st Hellenic Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 62 | |
0fgz | Problem:
El programa de una asignatura consta de $n$ preguntas; el examen consiste en desarrollar una de esas preguntas, elegida al azar. Un alumno sólo se sabe una pregunta, pero puede repetir el examen $n$ veces. Expresar, en función de $n$, la probabilidad $p_{n}$ de que el alumno apruebe el examen. ¿Crece o decrec... | [
"Solution:\n\nSi $n=1$, es $p_{1}=1$, si $n=2$, es $p_{2}=\\frac{1}{2}+\\frac{1}{2} \\cdot \\frac{1}{2}$; si $n=3$, tenemos $p_{3}=\\frac{1}{3}+\\frac{2}{3} \\cdot \\frac{1}{3}+\\frac{2}{3} \\cdot \\frac{2}{3} \\cdot \\frac{1}{3}$; En general\n$$\n\\begin{aligned}\n& p_{n}=\\frac{1}{n}+\\frac{n-1}{n} \\cdot \\frac{... | Spain | OME 25 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | p_n = 1 - ((n - 1)/n)^n; it decreases with n; lim p_n = 1 - e^{-1}; the greatest lower bound is 1 - e^{-1}. | |
0g7d | 試找出所有正整數的三元數對 $(x, y, z)$ 使得 $x \le y \le z$ 且
$$
x^3(y^3 + z^3) = 2012(xyz + 2).
$$ | [
"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 $x = 2^m, m \\in \\{0, 1, 2, 3\\}$。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$$\... | Taiwan | 二〇一三數學奧林匹亞競賽第一階段選訓營 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (2, 251, 252) | |
0amv | Problem:
The quartic (4th-degree) polynomial $P(x)$ satisfies $P(1)=0$ and attains its maximum value of $3$ at both $x=2$ and $x=3$. Compute $P(5)$. | [
"Solution:\n\nConsider the polynomial $Q(x) = P(x) - 3$. Then $Q(x)$ has zeros and maximum value $0$ at $x=2, 3$. These conditions imply that $Q(x)$ has the form\n$$\nQ(x) = A(x-2)^2(x-3)^2\n$$\nThat is, its graph looks like\n\n\n\nbecause the values of $Q(x)$ should grow larger and larger ... | Philippines | AREA STAGE | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | -24 | |
0jnt | Problem:
For how many triples $(x, y, z)$ of integers between $-10$ and $10$ inclusive do there exist reals $a, b, c$ that satisfy
$$
\begin{gathered}
a b = x \\
a c = y \\
b c = z ?
\end{gathered}
$$ | [
"Solution:\nAnswer: $4061$\nIf none of $x, y, z$ are zero, then there are $4 \\cdot 10^{3} = 4000$ ways, since $x y z$ must be positive. Indeed, $(a b c)^{2} = x y z$. So an even number of them are negative, and the ways to choose an even number of $3$ variables to be negative is $4$ ways. If one of $x, y, z$ is $0... | United States | HMMT November 2015 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Discrete Mathematics > Other"
] | null | final answer only | 4061 | |
0dnd | Problem:
На страницама $BC$ и $AC$ троугла $ABC$ дате су тачке $D$ и $E$, редом. Нека је $F$ ($F \neq C$) тачка пресека кружнице описане око троугла $CED$ и праве која садржи тачку $C$ и паралелна је са правом $AB$. Нека је $G$ тачка пресека праве $FD$ и странице $AB$, а $H$ тачка на правој $AB$ таква да је $\varangle... | [
"Solution:\n\nКако је $\\varangle AGD = 180^\\circ - \\varangle CFD = \\varangle CED = 180^\\circ - \\varangle AED$, тачке $A$, $E$, $D$, $G$ су концикличне. Одавде је $\\varangle DAG = \\varangle DEG$.\n\nМеђутим, тада је $\\varangle DHB = \\varangle DAG - \\varangle HDA = \\varangle DEG - \\varangle BEG = \\varan... | Serbia | 8. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0g11 | Problem:
Soit $n \geq 2$ un nombre naturel. Pour un sous-ensemble $F$ à $n$ éléments de $\{1, \ldots, 2 n\}$, on définit $m(F)$ comme le minimum de tous les $\operatorname{kgV}(x, y)$, où $x$ et $y$ sont deux éléments distincts de $F$. Trouver la valeur maximale que peut atteindre $m(F)$. | [
"Solution:\n\nNous montrons d'abord que $m(F)$ atteint son maximum pour $F=\\{n+1, \\ldots, 2 n\\}$. En effet soit $G$ un autre sous-ensemble et $x \\leq n$ un élément de $G$.\n\na. Si $2 x$ est aussi un élément de $G$, alors $m(G) \\leq \\operatorname{kgV}(x, 2 x)=2 x$. D'autre part comme tout élément de $F$ est s... | Switzerland | SMO - Finalrunde | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Maximum m(F) equals 3(n+1) if n is odd and 3(n+2) if n is even, with the exception n = 4 where it is 24. | |
02pi | Problem:
Distribuímos nos vértices de um bloco retangular oito números dentre $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$ de tal forma que a soma dos números de uma face qualquer seja igual a $18$.
a. Quais os números descartados na distribuição?
b. Exiba uma possível distribuição. | [
"Solution:\n\n(a) Como o bloco possui seis faces, a soma dos números em todas as faces é $18 \\times 6 = 108$, mas o número atribuído a cada vértice é contado três vezes nesta soma. Portanto, a soma dos números distribuídos é $108/3 = 36$. Como a soma de todos os números de $1$ a $10$ é igual a $55$, a soma dos doi... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | a) The discarded numbers are 9 and 10. b) One valid placement using the numbers 1 through 8 is: bottom face corners in cycle 1, 8, 5, 4 and top face directly above them 7, 2, 3, 6 (so the vertical pairs are 1–7, 8–2, 5–3, 4–6). Each face then sums to 18. | |
0hc3 | Show that for the positive $x$, $y$, $z$ the following inequality holds:
$$
\frac{x^8+1}{x^4} + \frac{y^8+1}{y^4} + \frac{z^8+1}{z^4} \geq 2 \cdot \left( \frac{x}{z} + \frac{z}{y} + \frac{y}{x} \right).
$$ | [
"Due to inequality of arithmetic and geometric means applied several times:\n$$\n\\left( x^4 + \\frac{1}{y^4} \\right) + \\left( y^4 + \\frac{1}{z^4} \\right) + \\left( z^4 + \\frac{1}{x^4} \\right) \\geq 2 \\cdot \\left( \\frac{x^2}{y^2} + \\frac{y^2}{z^2} + \\frac{z^2}{x^2} \\right) = \\\\\n= \\left( \\frac{x^2}{... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0de9 | Determine if there exists pairwise distinct positive integers $a_1, a_2, ..., a_{101}, b_1, b_2, ..., b_{101}$ satisfying the following property: for each non-empty subset $S$ of $\{1, 2, ..., 101\}$ the sum $\sum_{i \in S} a_i$ divides $100! + \sum_{i \in S} b_i$. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | No, such integers do not exist. | |
0g5i | 令 $x_1, \cdots, x_{100}$ 為非負實數, 滿足對 $i = 1, \cdots, 100$, 都有
$$
x_i + x_{i+1} + x_{i+2} \le 1,
$$
(其中我們記 $x_{101} = x_1, x_{102} = x_2$)。試求下式 $S$ 的最大可能值:
$$
S = \sum_{i=1}^{100} x_i x_{i+2}.
$$ | [
"答 $\\frac{25}{2}$。\n令 $x_{2i} = 0, x_{2i-1} = \\frac{1}{2}$ for all $i = 1, \\cdots, 50$。則我們有\n$$\nS = 50\\left(\\frac{1}{2}\\right)^2 = \\frac{25}{2}.\n$$\n故只須證 $S \\le \\frac{25}{2}$ 對所有滿足題設之 $x_i$。\n考慮 $1 \\le i \\le 50$。由題設得\n$$\nx_{2i-1} \\le 1 - x_{2i} - x_{2i+1}, \\quad x_{2i+2} \\le 1 - x_{2i} - x_{2i+1}.\... | Taiwan | 二〇一一數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 25/2 | |
05a0 | Let $\omega$ be a circle with center $O$ and diameter $AB$. A circle with center $B$ intersects $\omega$ at $C$ and $AB$ at $D$. The line $CD$ intersects $\omega$ at the point $E$ ($E \neq C$). The intersection of lines $OE$ and $BC$ is $F$.
a. Prove that the triangle $OBF$ is isosceles.
b. Find the ratio $\frac{FB}{... | [
"(a) Let $\\angle BCD = \\alpha$, then the equal radii $BC = BD$ give us $\\angle BDC = \\alpha$ (Fig. 2) and $\\angle CBD = 180^\\circ - 2\\alpha$. Then $\\angle OBF = \\angle CBD = 180^\\circ - 2\\alpha$. But on the other hand $\\angle BOE = 2\\angle BCE = 2\\angle BCD = 2\\alpha$ and therefore $\\angle BOF = 180... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 4 | |
0dyx | Problem:
Poišči vsa realna števila $x$, za katera je vrednost izraza
$$
\sqrt{1-x^{2}}+\sqrt{5x-x^{2}}
$$
celo število. | [
"Solution:\nOcenimo vrednost izraza. Očitno je $1-x^{2} \\leq 1$, vrednost $5x-x^{2}$ pa je omejena z $5x-x^{2}=\\frac{25}{4}-\\left(x-\\frac{5}{2}\\right)^{2} \\leq \\frac{25}{4}$, zato je\n$$\n\\sqrt{1-x^{2}}+\\sqrt{5x-x^{2}} \\leq \\sqrt{1}+\\sqrt{\\frac{25}{4}}=1+\\frac{5}{2}=3+\\frac{1}{2}\n$$\nPo drugi strani... | Slovenia | 52. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x = 0, x = 9/41, x = 1 | |
07ni | Prove that there is a positive integer, not divisible by $10$, whose $2011$-th power has in its decimal expansion (at least) $2011$ consecutive zeros immediately after its non-zero leading digit. | [
"A number starts with the digit $1$ followed by $r$ zeros if and only if it can be written in the form $10^{r+s} + A$, with $0 \\le A < 10^s$. If $k > r \\ge 0$ we have $10^{r+s} + 10^k A + A < 10^{r+s} + 10^{k+s} + 10^s < 10^{k+s+1}$. Hence $(10^k + 1)(10^{r+s} + A) = 10^{k+r+s} + 10^{r+s} + 10^k A + A$ is a numbe... | Ireland | Ireland | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
016k | Problem:
Let $n$ be a given positive integer. Show that we can choose numbers $c_{k} \in \{-1,1\}$ $(1 \leq k \leq n)$ such that
$$
0 \leq \sum_{k=1}^{n} c_{k} \cdot k^{2} \leq 4
$$ | [] | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0g25 | Problem:
Zeige, dass keine Funktion $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ existiert, sodass für alle $x, y \in \mathbb{R}_{>0}$ gilt:
$$
f(x f(x)+y f(y))=x y
$$ | [
"Solution:\nNehme an, $f$ sei eine Lösung der Gleichung. Wir sehen zuerst, dass die rechte Seite alle möglichen Werte in $\\mathbb{R}^{+}$ annehmen kann, also ist $f$ surjektiv. Dann existiert ein $c \\in \\mathbb{R}^{+}$ mit $f(c)=1 / 2$. Mit $x=y=c$ erhalten wir:\n$$\nf\\left(c \\cdot \\frac{1}{2}+c \\cdot \\frac... | Switzerland | SMO - Finalrunde | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
0lev | For every pair of positive integers $n, m$ with $n < m$, denote $s(n, m)$ as the number of positive integers in the range $[n, m]$ that are coprime with $m$. Find all positive integers $m \ge 2$ such that $m$ satisfies these conditions
i) $\frac{s(n, m)}{m-n} \ge \frac{s(1, m)}{m}$ for all $n = 1, 2, \dots, m-1$.
ii) ... | [
"Firstly, we prove that if $m$ satisfies the first condition, then $m$ has only one prime divisor. Assume that $m$ has at least 2 prime divisors, let $p$ be the smallest prime divisor of $m$ and $p_1, p_2, \\dots, p_k$ be the remaining prime divisors of $m$. We have\n$$\n\\frac{\\varphi(m)}{m} = \\left(1 - \\frac{1... | Vietnam | VMO | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 7, 17, 289 | |
0eh0 | Problem:
Naj za realni števili $x$ in $y$ velja $\frac{x}{x+y}=101$. Kolikšna je vrednost izraza $\frac{y-x}{y}$ ?
(A) 1,02
(B) 100
(C) 201
(D) 2,01
(E) 1,01 | [
"Solution:\n\nEnakost pomnožimo z imenovalcem in izrazimo $x=-1,01 y$, to vstavimo v izraz $\\frac{y-x}{y}=$ $\\frac{y+1,01 y}{y}=\\frac{2,01 y}{y}=2,01$."
] | Slovenia | Državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | D | |
0k9t | Problem:
Let $S$ be the set of all nondegenerate triangles formed from the vertices of a regular octagon with side length $1$. Find the ratio of the largest area of any triangle in $S$ to the smallest area of any triangle in $S$. | [
"Solution:\nBy a smoothing argument, the largest triangle is that where the sides span $3$, $3$, and $2$ sides of the octagon respectively (i.e. it has angles $45^{\\circ}$, $67.5^{\\circ}$, and $67.5^{\\circ}$), and the smallest triangle is that formed by three adjacent vertices of the octagon. Scaling so that the... | United States | HMMT November 2019 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Jensen/smoothing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 3+2√2 | |
0cno | Rational numbers $a$ and $b$ satisfy the equality
$$
a^3 b + ab^3 + 2a^2 b^2 + 2a + 2b + 1 = 0.
$$
Prove that $1 - ab$ is a square of a rational number. (R. Zhenodarov) | [
"First solution. Transform the original expression:\n$$\n\\begin{aligned}\n0 &= a^3 b + ab^3 + 2a^2 b^2 + 2a + 2b + 1 = ab(a + b)^2 + 2(a + b) + 1 \\\\\n&= (ab - 1)(a+b)^2 + (a+b)^2 + 2(a+b) + 1 = (ab - 1)(a+b)^2 + (a+b+1)^2.\n\\end{aligned}\n$$\nNotice that $a + b \\neq 0$, because otherwise $1 = 0$.\nTherefore, $... | Russia | Russian mathematical olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English; Russian | proof only | null | |
0eo5 | 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. | [
"\n\nWe denote the angles of the triangle by $\\alpha = \\angle BAC$, $\\beta = \\angle ABC$ and $\\gamma = \\angle ACB$. $X$ is the intersection of $BC$ with the angle bisector of $\\angle BAC$. Since $\\angle BDA = \\angle BEA = 90^\\circ$, both $D$ and $E$ lie on the circle with diameter... | South Africa | The South African Mathematical Olympiad Third Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | ∠A = 108°, ∠B = 18°, ∠C = 54° | |
0jgq | Problem:
Let $ABC$ be an isosceles triangle with $AB = AC$. Let $D$ and $E$ be the midpoints of segments $AB$ and $AC$, respectively. Suppose that there exists a point $F$ on ray $\overrightarrow{DE}$ outside of $ABC$ such that triangle $BFA$ is similar to triangle $ABC$. Compute $\frac{AB}{BC}$. | [
"Solution:\n\n$\\boxed{\\sqrt{2}}$\n\nLet $\\alpha = \\angle ABC = \\angle ACB$, $AB = 2x$, and $BC = 2y$, so $AD = DB = AE = EC = x$ and $DE = y$. Since $\\triangle BFA \\sim \\triangle ABC$ and $BA = AC$, we in fact have $\\triangle BFA \\cong \\triangle ABC$, so $BF = BA = 2x$, $FA = 2y$, and $\\angle DAF = \\al... | United States | HMMT November 2013 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | sqrt(2) | |
01y9 | The set $\{2, 3, 4, \ldots, 2020\}$ is partitioned into triples. In each triple $(a, b, c)$ the numbers were arranged in the ascending order, i.e. $a < b < c$, and the difference $|b - \frac{a+c}{2}|$ is called the error of this triple. Find the maximal possible sum of errors of all $673$ triples. | [
"Note that if we shift all numbers in the set by the same value, the errors will remain the same. Hence we can consider the partitions of the set $\\{1, 2, \\ldots, 2019\\}$, moreover, we will solve the generalized problem by changing $2019$ with an arbitrary number $3n$, $n \\in \\mathbb{N}$.\n\nConsider the parti... | Belarus | BY 2020-2021 tst for Navid | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 395808 | |
036z | Problem:
Let $\mathbb{Q}^{+}$ be the set of positive rational numbers. Find all functions $f: \mathbb{Q}^{+} \rightarrow \mathbb{R}$ such that $f(1)=1$, $f(1 / x)=f(x)$ for any $x \in \mathbb{Q}^{+}$ and $x f(x)=(x+1) f(x-1)$ for any $x \in \mathbb{Q}^{+}, x>1$. | [
"Solution:\nLet $x=\\frac{p}{q}$, where $p$ and $q$ are coprime positive integers. We shall prove by induction on $n=p+q \\geq 2$ that $f(x)$ is uniquely determined. This is true for $n=2$ (since $f(1)=1$).\n\nSuppose that it is true for all integers less than a given $n \\geq 3$ and consider $x=\\frac{p}{q}$, wher... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | For x = p/q in lowest terms, f(x) = (p + q) / 2. | |
0hd7 | Out of three expressions $\frac{x}{y}$, $\frac{x^2+x}{y^2+y}$ and $\frac{x^2+2}{y^2+2}$, for some integer $x, y$ all three are defined, two take the same integer value, and the remaining one takes a different integer value. For which pairs of integers $x, y$ is this possible?
(Bohdan Rublyov) | [
"Obviously, $|x| > |y|$, for $x \\neq 0$, since otherwise, the first fraction will not be an integer. If the first two expressions are equal, then we have the equality:\n$$\n\\frac{x}{y} = \\frac{x^2+x}{y^2+y} \\Rightarrow xy^2 + xy = yx^2 + xy \\Rightarrow xy(y-x) = 0.\n$$\nFor $x \\neq y$ and possible range of va... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (2,1) and (-4,1) | |
0gjy | Let $\alpha = \frac{1 + \sqrt{5}}{2}$. Find all continuous functions $f: \mathbb{R} \to \mathbb{R}$ such that, for all $x, y, z \in \mathbb{R}$,
$$
\begin{aligned}
& f(\alpha x + y) + f(\alpha y + z) + f(\alpha z + x) \\
&= \alpha f(x + y + z) + 2f(x) + 2f(y) + 2f(z).
\end{aligned}
$$ | [
"Let (*) be the given functional equation.\nSetting $x = y = z = 0$ in (*), we have $f(0) = 0$.\nSetting $y = z = 0$ in (*) and simplifying, we have\n$$\nf(\\alpha x) = \\alpha^2 f(x) \\text{ for all } x \\in \\mathbb{R}.\n$$\nLetting $z = 0$ in the functional equation, we get\n$$\nf(\\alpha x + y) + f(\\alpha y) +... | Thailand | Thai Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All continuous solutions are f(x) = a x^2 for any real constant a. | |
0l2x | Problem:
Determine, with proof, whether there exist positive integers $x$ and $y$ such that $x+y$, $x^{2}+y^{2}$, and $x^{3}+y^{3}$ are all perfect squares. | [
"Solution:\n\nTake $(x, y) = (184, 345)$. Then $x+y = 23^{2}$, $x^{2}+y^{2} = 391^{2}$, and $x^{3}+y^{3} = 6877^{2}$."
] | United States | HMMT February 2024 | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Other"
] | null | proof and answer | (184, 345) | |
0kue | Problem:
Suppose $a$, $b$, $c$, and $d$ are pairwise distinct positive perfect squares such that $a^{b} = c^{d}$. Compute the smallest possible value of $a + b + c + d$. | [
"Solution:\nNote that if $a$ and $c$ are divisible by more than one distinct prime, then we can just take the prime powers of a specific prime. Thus, assume $a$ and $c$ are powers of a prime $p$. Assume $a = 4^{x}$ and $c = 4^{y}$. Then $x b = y d$.\n\nBecause $b$ and $d$ are squares, the ratio of $x$ to $y$ is a s... | United States | HMMT February 2023 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 305 | |
05zg | Problem:
Soient $x$ et $y$ deux réels positifs. Montrer que
$$
\left(x^{2}+x+1\right)\left(y^{2}+y+1\right) \geqslant 9 x y .
$$
Quels sont les cas d'égalité? | [
"Solution:\n\nOn applique l'inégalité arithmético-géométrique pour avoir\n$$\n\\begin{gathered}\nx^{2}+x+1 \\geqslant 3 \\sqrt[3]{x^{3}} = 3x \\\\\ny^{2}+y+1 \\geqslant 3 \\sqrt[3]{y^{3}} = 3y\n\\end{gathered}\n$$\nEn multipliant ces deux inégalités, qui sont à termes positifs, on obtient l'inégalité de l'énoncé. C... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 2 : AlgèBre | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | x=y=1 | |
04l5 | A number written only with digits $2$ and $3$ is called *happy*. Therefore, the happy numbers are $2$, $3$, $22$, $23$, $32$, $33$, $222$, $223$, $232$, $233$, $322$, $323$, $332$, $333$, ... Determine the $2050$th happy number. | [] | Croatia | Mathematical competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 22222222233 | |
0d46 | There are $2015$ coins on a table. For $i=1,2, \ldots, 2015$ in succession, one must turn over exactly $i$ coins. Prove that it is always possible either to make all of the coins face up or to make all of the coins face down, but not both. | [
"We start by proving that it is always possible either to make all of the coins face up or to make all of the coins face down.\nNotice that if we proceed by choosing the numbers $i=1,2, \\ldots, 2015$ in this order or in any other order, the final result will depend only on the $i$ coins chosen at each step. For th... | Saudi Arabia | SAMC | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English, Arabic | proof only | null | |
03yy | Let $3n^2$ be the vertex number of a simple graph $G$ (integer $n \ge 2$). 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 with length not greater than $3$ between any two vertices. Prove that the minimum number of edges of $G$ is $\frac{... | [
"For any two distinct vertices $u$ and $v$, we say that the distance between $u$ and $v$ is the shortest length of the route between $u$ and $v$. 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 ($1 \\le i < j \\le n$), $x_i$ and ... | China | China National Team Selection Test | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 7/2 n^2 − 3/2 n | |
0dkb | Let $x$, $y$, $z$ be real numbers such that $x \ge y \ge z \ge 0$ and $2x + y + 2z = 5$. Prove that
$$
5 \le x^2 + z^2 + xy + yz + zx \le \frac{25}{4}.
$$
When does the equality case hold? | [
"Let $P = x^2 + z^2 + xy + yz + zx$ then we have\n$$\n25 = (2x + y + 2z)^2 = 4x^2 + y^2 + 4z^2 + 4xy + 4yz + 4zx = 4P + y^2 + 4zx.\n$$\nNote that $y^2 + 4zx \\ge 0$ so $4P \\le 25$ which implies that $P \\le \\frac{25}{4}$. The equality case is $y = z = 0$ and $x = \\frac{5}{2}$.\n\nContinue, note that $(y - x)(y -... | Saudi Arabia | Saudi Arabia booklet 2024 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Lower bound equality holds when x = y = z = 1. Upper bound equality holds when y = z = 0 and x = 5/2. | |
07qk | The sequence $(a_n)$ is defined as follows: $a_0 = 1$, $a_1 = 1$, and
$$
a_{n+1} = 2(a_n - a_{n-1}),
$$
for all positive integers $n$. Determine, with proof, the remainder of $a_{2016}$ on division by $2017$. | [
"Since $(a_n)$ satisfies the linear recurrence $a_{n+1} = 2(a_n - a_{n-1})$, an explicit formula for $a_n$ can be determined in the usual manner. Formally replacing $a_k$ in the recurrence by $x^k$, we obtain $x^{n+1} = 2(x^n - x^{n-1})$ and solving for nonzero $x$, we get $x^2 = 2(x-1)$ and $x = 1 \\pm i$, where $... | Ireland | Irish Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Complex numbers",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / F... | null | proof and answer | 1 | |
0ipm | Problem:
Find the number of distinct primes dividing $1 \cdot 2 \cdot 3 \cdots 9 \cdot 10$. | [
"Solution:\nA prime divides this product if and only if it divides one of the multiplicands, so prime divisors of this product must be less than $10$. There are $4$ primes less than $10$, namely, $2, 3, 5$, and $7$."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 4 | |
08oc | Problem:
Let $n$ be a positive integer, and let $x_{1}, \ldots, x_{n}, y_{1}, \ldots, y_{n}$ be positive real numbers such that $x_{1}+\ldots+x_{n}=y_{1}+\ldots+y_{n}=1$. Show that
$$
\left|x_{1}-y_{1}\right|+\ldots+\left|x_{n}-y_{n}\right| \leq 2-\min_{1 \leq i \leq n} \frac{x_{i}}{y_{i}}-\min_{1 \leq i \leq n} \frac{... | [
"Solution:\nUp to reordering the real numbers $x_{i}$ and $y_{i}$, we may assume that $\\frac{x_{1}}{y_{1}} \\leq \\ldots \\leq \\frac{x_{n}}{y_{n}}$. Let $A=\\frac{x_{1}}{y_{1}}$ and $B=\\frac{x_{n}}{y_{n}}$, and $S=\\left|x_{1}-y_{1}\\right|+\\ldots+\\left|x_{n}-y_{n}\\right|$. Our aim is to prove that $S \\leq 2... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0hhe | In a country there are $2024$ cities, each two of which are connected bidirectionally by exactly one of three modes of transportation: rail, air, or road. A tourist arrives in this country and has the entire transportation scheme. He chooses a travel ticket for one of the modes of transportation and the city from which... | [
"Suppose we have a complete graph with each edge colored in one of three colors. We show that there will always exist a connected component of at least one of the colors of size at least $1012$. First, we show that it is possible to color the edges in such a way that connected components of size more than $1012$ ve... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1012 | |
05vk | Problem:
Soit $x$, $y$ et $z$ trois nombres réels tels que $x^{2}+y^{2}+z^{2}=1$. Trouver les valeurs minimale et maximale possibles du nombre réel $x y+y z-z x$. | [
"Solution:\n\nEn vertu de l'identité\n$$\n(x-y+z)^{2}=x^{2}+y^{2}+z^{2}-2(x y+y z-z x)=1-2(x y+y z-z x),\n$$\nil s'agit ici de trouver les valeurs extrêmes que peut prendre le nombre $(x-y+z)^{2}$.\nTout d'abord, puisque $(x-y+z)^{2} \\geqslant 0$, avec égalité lorsque $x=y=1 / \\sqrt{2}$ et $z=0$, on a bien\n$$\nx... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | minimum = -1, maximum = 1/2 | |
08hm | Problem:
Prove that
$$
\lim_{n \rightarrow +\infty} \frac{\ln \left(1 + 2e + 4e^{4} + 6e^{9} + \ldots + 2n e^{n^{2}}\right)}{n^{2}} = 1.
$$ | [
"Solution:\nLet $S_n = 1 + 2e + 4e^{4} + 6e^{9} + \\ldots + 2n e^{n^{2}} = \\sum_{k=1}^{n} 2k e^{k^{2}} + 1$.\n\nFor large $n$, the last term $2n e^{n^{2}}$ dominates the sum. We estimate $S_n$:\n\nFor $k = n$, $2n e^{n^{2}}$ is much larger than all previous terms, since $e^{k^{2}}$ grows very rapidly.\n\nLet us co... | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0aa3 | Problem:
Find all $a \in \mathbb{R}$ for which there exists a function $f: \mathbb{R} \rightarrow \mathbb{R}$, such that
(i) $f(f(x))=f(x)+x$, for all $x \in \mathbb{R}$,
(ii) $f(f(x)-x)=f(x)+a x$, for all $x \in \mathbb{R}$. | [
"Solution:\n$a=\\frac{1 \\pm \\sqrt{5}}{2}$.\nFrom (i) we get $f(f(f(x))-f(x))=f(x)$. On the other hand (ii) gives\n$$\nf(f(f(x))-f(x))=f(f(x))+a f(x)\n$$\nThus we have $(1-a) f(x)=f(f(x))$. Now it follows by (i) that $(1-a) f(x)=f(x)+x$, and hence $f(x)=-\\frac{1}{a} x$, since $a=0$ obviously does not give a solut... | Nordic Mathematical Olympiad | Nordic Mathematical Contest | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | (1 + sqrt(5))/2, (1 - sqrt(5))/2 | |
01vu | The vertices of the regular $n$-gon are marked. Two players play the following game: they, in turn, select a vertex and connect it by a segment to either the adjacent vertex or the center of the $n$-gon. The winner is a player if after his move it is possible to get any vertex from any other vertex moving along segment... | [
"Answer: for odd $n$ first player wins, and for even $n$ the second player wins.\n\n**Let $n$ be even.** Let's describe the winning strategy of the second player. First, similar to the solution of E1. of the problem of Category B, note that we can assume that the first player with his first move connected two adjac... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Odd number of vertices: first player wins. Even number of vertices: second player wins. | |
0ipa | Problem:
Determine the number of 8-tuples of nonnegative integers $\left(a_{1}, a_{2}, a_{3}, a_{4}, b_{1}, b_{2}, b_{3}, b_{4}\right)$ satisfying $0 \leq a_{k} \leq k$, for each $k=1,2,3,4$, and
$$
a_{1} + a_{2} + a_{3} + a_{4} + 2 b_{1} + 3 b_{2} + 4 b_{3} + 5 b_{4} = 19.
$$ | [
"Answer: 1540 Same as Combinatorics Test problem 10."
] | United States | 11th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof and answer | 1540 | |
0jgp | Problem:
Let $ABC$ be a triangle with $AB=13$, $BC=14$, $CA=15$. Company XYZ wants to locate their base at the point $P$ in the plane minimizing the total distance to their workers, who are located at vertices $A$, $B$, and $C$. There are $1$, $5$, and $4$ workers at $A$, $B$, and $C$, respectively. Find the minimum p... | [
"Solution:\n\nWe want to minimize $1 \\cdot PA + 5 \\cdot PB + 4 \\cdot PC$. By the triangle inequality, $(PA + PB) + 4(PB + PC) \\geq AB + 4 BC = 13 + 56 = 69$, with equality precisely when $P = [AB] \\cap [BC] = B$."
] | United States | HMMT November 2013 | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 69 | |
09ew | Initially a needle with length $l$ was in vertical position. Can one transfer the needle to horizontal position by passing no more than $2014$ lattice points? | [
"By Chinese remainder theorem the system\n$$\n\\left\\{\n\\begin{array}{l}\nn \\equiv p_0 \\pmod{p_0^2} \\\\\nn + 1 \\equiv p_1 \\pmod{p_1^2} \\\\\n\\text{................} \\\\\nn + k \\equiv p_k \\pmod{p_k^2}\n\\end{array}\n\\right.\n$$\nhas $n$ positive integer solutions for $\\forall k \\in \\mathbb{N}$, where ... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | Yes | |
0eeh | Problem:
Poišči vse pare tujih celih števil $x$ in $y$, ki rešijo enačbo
$$
4 x^{3}+y^{3}=3 x y^{2}
$$ | [
"Solution:\n\nŠtevilo $x$ deli $4 x^{3}$ in $3 x y^{2}$, zato mora deliti tudi $y^{3}$. Ker pa sta števili $x$ in $y$ tuji, je to mogoče le, če je $x=1$ ali $x=-1$.\n\nČe je $x=1$, dobimo enakost $4+y^{3}=3 y^{2}$, ki jo preoblikujemo v $4=y^{2}(3-y)$. Torej $y^{2}$ deli $4$, zato je bodisi $y^{2}=1$ ali $y^{2}=4$.... | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (1, -1), (1, 2), (-1, 1), (-1, -2) | |
0k37 | Problem:
David and Evan each repeatedly flip a fair coin. David will stop when he flips a tail, and Evan will stop once he flips 2 consecutive tails. Find the probability that David flips more total heads than Evan. | [
"Solution:\n\nWe can find the values of the functions $D(h)$ and $E(h)$, the probabilities that David and Evan, respectively, flip exactly $h$ heads. It is easy to see that $D(h) = 2^{-h-1}$. In order to find $E(h)$, we note that each sequence must end with the flips HTT (unless Evan flips only 2 heads). We disrega... | United States | HMMT November 2018 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 1/5 | |
097m | Problem:
Să se demonstreze, că pentru oricare numere pozitive $a, b, c$ este adevărată inegalitatea
$$
\frac{a+b-2c}{b+c} + \frac{b+c-2a}{c+a} + \frac{c+a-2b}{a+b} \geq 0
$$
Când are loc egalitatea? | [
"Solution:\n\nDupă înmulţirea cu $(a+b)(b+c)(c+a)$ se obţine inegalitatea\n$$\n(a+b)(c+a)(a+b-2c) + (a+b)(b+c)(b+c-2a) + (b+c)(c+a)(c+a-2b) \\geq 0\n$$\nechivalentă celei din enunţ. Pentru fiecare din cei trei termeni se deschid parantezele:\n$$\n\\begin{aligned}\n& (a+b)(c+a)(a+b-2c) = a^3 + 2a^2b - 2ac^2 - 2bc^2 ... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a = b = c | |
07cd | Let $ABC$ be a triangle with altitudes $AD$, $BF$ and $CE$. Let $P$ and $Q$ be two points on line $EF$ such that $EP = DF$ ($E$ is between $P$ and $F$) and $QF = DE$ ($F$ is between $E$ and $Q$). Assuming that the perpendicular bisector of $DQ$ intersects $AB$ at $X$ and the perpendicular bisector of $DP$ intersects $A... | [
"It is claimed that $XY$ is the perpendicular bisector of $EF$. To show it, first a lemma is proved.\n\n**Lemma.** Let $D$ be a point on the extension of side $AB$ of triangle $ABC$ such that $A$ is between $B$ and $D$, and $AD = BC$. Let $O$ be the intersection point of external angle bisector of vertex $B$ in tri... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
08h6 | Problem:
Let $M$ be a subset of the set of 2021 integers $\{1,2,3, \ldots, 2021\}$ such that for any three elements (not necessarily distinct) $a, b, c$ of $M$ we have $|a+b-c|>10$. Determine the largest possible number of elements of $M$. | [
"Solution:\nThe set $M=\\{1016,1017, \\ldots, 2021\\}$ has 1006 elements and satisfies the required property, since $a, b, c \\in M$ implies that $a+b-c \\geqslant 1016+1016-2021=11$.\n\nWe will show that this is optimal.\n\nSuppose $M$ satisfies the condition in the problem. Let $k$ be the minimal element of $M$. ... | JBMO | null | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 1006 | |
08wr | Let $O$ be the circum-center of a triangle $ABC$. When points $D$ and $E$ were chosen on the line segments $AB$ and $AC$, respectively, the mid-point of the line segment $DE$ coincided with the point $O$. If $AD = 8$, $BD = 3$ and $AO = 7$, determine the value of $CE$. Here for a line segment $XY$ we denote also by $XY... | [
"\\boxed{\\frac{4\\sqrt{21}}{7}}\n\nLet $P(Q)$ be the point of intersection of the line $DE$ and the circum-circle of the triangle $ABC$, which lies on the opposite side (on the same side, respectively) as the point $O$ with respect to the line $AB$. Let $x = OD = OE$, $y = AE$ and $z = EC$. Then, we have $OP = OQ ... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 4*sqrt(21)/7 | |
0fp6 | Let $n \ge 1$ be a positive integer. Consider a pile of $3^n$ coins, one of which is fake. Suppose that all coins are either white or black and that if the fake coin is white, it is lighter than the others, and if the fake is black, it is heavier than the others. Furthermore, assume that the number of white coins and t... | [
"For each $n \\ge 1$, let $P(n)$ be the statement to be proven. We will argue by induction. Indeed,\n\n* **Base step:** For $n=1$, consider $3^1=3$ coins, and without loss of generality, suppose that two are black and one is white. Put a black on each pan and set the white aside. If the scale balances, then white c... | Spain | BARCELONA TECH MATHCONTEST | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Algorithms"
] | Spanish | proof only | null | |
0kgg | Problem:
A domino has a left end and a right end, each of a certain color. Alice has four dominos, colored red-red, red-blue, blue-red, and blue-blue. Find the number of ways to arrange the dominos in a row end-to-end such that adjacent ends have the same color. The dominos cannot be rotated. | [
"Solution:\n\nWithout loss of generality assume that the left end of the first domino is red. Then, we have two cases:\n\nIf the first domino is red-red, this forces the second domino to be red-blue. The third domino cannot be blue-red, since the fourth domino would then be forced to be blue-blue, which is impossib... | United States | HMMT November 2021 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 4 | |
0aes | Бројот 1 000 000 пртстави го како производ на два броја, во чиј запис не се појавува ниту една нула. | [
"Бројот 1 000 000 можеме да го претставиме како $1\\,000\\,000 = 10 \\cdot 10 \\cdot 10 \\cdot 10 \\cdot 10 \\cdot 10 = (2 \\cdot 5) \\cdot (2 \\cdot 5) \\cdot (2 \\cdot 5) \\cdot (2 \\cdot 5) \\cdot (2 \\cdot 5) \\cdot (2 \\cdot 5)$. Ако $x$ и $y$ се природни броеви такви што $xy = 1\\,000\\,000$ и во записот на б... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | Macedonian, English | final answer only | 15625 * 64 | |
0e54 | Find all non-zero integers $a$, different from $4$, such that $\frac{a}{a-4} + \frac{2}{a}$ is an integer as well. | [
"If $\\frac{a}{a-4} + \\frac{2}{a} = \\frac{a^2+2a-8}{a(a-4)}$ is an integer, then $a(a-4)$ divides $a^2+2a-8$. So, $a$ divides $a^2+2a-8$, and $a$ divides $8$. All integer divisors of $8$ different from $4$ are $1$, $2$, $8$, $-1$, $-2$, $-4$ and $-8$. We check all seven cases and see that the value of the express... | Slovenia | National Math Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 2 and -4 | |
071w | Problem:
Determine the least positive value taken by the expression $a^{3}+b^{3}+c^{3}-3 a b c$ as $a, b, c$ vary over all positive integers. Find also all triples $(a, b, c)$ for which this least value is attained. | [
"Solution:\n\nWe observe that\n$$\nQ = a^{3} + b^{3} + c^{3} - 3 a b c = \\frac{1}{2}(a + b + c)\\left((a - b)^{2} + (b - c)^{2} + (c - a)^{2}\\right)\n$$\nSince we are looking for the least positive value taken by $Q$, it follows that $a, b, c$ are not all equal. Thus $a + b + c \\geq 1 + 1 + 2 = 4$ and $(a - b)^{... | India | INMO | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | Least value is 4; attained exactly by the triples that are permutations of (1, 1, 2). |
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