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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
03mp | Let $p$ be a fixed odd prime. A $p$-tuple $(a_1, a_2, a_3, \dots, a_p)$ of integers is said to be good if
(i) $0 \le a_i \le p-1$ for all $i$, and
(ii) $a_1 + a_2 + a_3 + \dots + a_p$ is not divisible by $p$, and
(iii) $a_1a_2 + a_2a_3 + a_3a_4 + \dots + a_p a_1$ is divisible by $p$.
Determine the number of good $p$-tu... | [
"Let $S$ be the set of all sequences $(b_1, b_2, \\dots, b_p)$ of numbers from the set $\\{0, 1, 2, \\dots, p-1\\}$ such that $b_1 + b_2 + \\dots + b_p$ is not divisible by $p$. We show that $|S| = p^p - p^{p-1}$. For let $b_1, b_2, \\dots, b_{p-1}$ be an arbitrary sequence of numbers chosen from $\\{0, 1, 2, \\dot... | Canada | Kanada 2014 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | p^{p-1} - p^{p-2} | |
03pr | Let $ABCD$ be an inscribed 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 $\angle ABC$ and $\angle ADC$ meet on $AC$. | [
"**Proof** By Simson's Theorem, we know that $P$, $Q$, $R$ are collinear. Moreover, since $\\angle DPC$ and $\\angle DQC$ are right angles, the points $D$, $P$, $Q$, $C$ are concyclic and so $\\angle DCA = \\angle DPQ = \\angle DPR$. Similarly, since $D$, $Q$, $R$, $A$ are concyclic, we have $\\angle DAC = \\angle ... | China | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
091p | Problem:
Let $a, b, c$ be positive real numbers such that
$$
\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2
$$
Prove that
$$
\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2} \geqslant \frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}
$$ | [
"Solution:\nNote that the condition of the problem is equivalent to\n$$\n\\frac{1}{1+a}+\\frac{1}{1+b}+\\frac{1}{1+c}=1\n$$\nWe want to prove that\n$$\n\\begin{gathered}\n\\frac{\\sqrt{a}+\\sqrt{b}+\\sqrt{c}}{2} \\geqslant \\frac{1}{\\sqrt{a}}+\\frac{1}{\\sqrt{b}}+\\frac{1}{\\sqrt{c}} \\\\\n\\Longleftrightarrow \\q... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric ... | null | proof only | null | |
08ef | Problem:
Determinare il numero di terne ordinate $(a, b, c)$ di interi non negativi tali che ciascuno dei numeri $2^{a}$, $2^{b}$, $2^{c}$ sia minore di $10000$ e che il numero $2^{a}+2^{b}+2^{c}$ sia un divisore di $8^{a}+8^{b}+8^{c}$.
(A) 14
(B) 50
(C) 53
(D) 72
(E) 86 | [
"Solution:\n\nLa risposta è $\\mathbf{( E )}$. Iniziamo osservando che vale l'identità algebrica\n$$\nx^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\\left(x^{2}+y^{2}+z^{2}-x y-y z-x z\\right)\n$$\nPonendo $x=2^{a}$, $y=2^{b}$, $z=2^{c}$, la condizione $x+y+z \\mid x^{3}+y^{3}+z^{3}$ è equivalente a $x+y+z \\mid 3 x y z$; in pa... | Italy | Italian Mathematical Olympiad - February Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | E | |
01yx | Let $p \ge 5$ be a prime number. Prove that for each divisor $d > 2$ of $p-1$ it's possible to partition all integers between $1$ and $p-1$ into the sets with $d$ numbers in each in such a way that the sum of squares of all numbers of each set is divisible by $p$.
(Mikhail Karpuk) | [
"Denote the set of all residues modulo $p$ by $\\mathbb{F}_p$. Fermat's little theorem implies that the roots of the polynomial $x^{p-1} - 1$ are all nonzero elements of $\\mathbb{F}_p$. Since $x^{p-1} - 1 = (x^d - 1)(x^{d(m-1)} + x^{d(m-2)} + \\dots + x^d + 1)$ where $p-1 = md$, the polynomial $x^d - 1$ divides $x... | Belarus | SELECTION and TRAINING SESSION | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Abstract Algebra > Group Theory"
] | English | proof only | null | |
01j3 | Let $\triangle ABC$ be a triangle and let $J$ be the center of the excircle opposite to $A$. The reflection of $J$ in $BC$ is $K$. $E$ and $F$ are on $BJ$ and $CJ$, respectively, such that $\angle EAB = \angle CAF = 90^\circ$. Prove that $\angle FKE + \angle FJE = 180^\circ$. | [
"\nLet $JK$ intersect $BC$ at $X$. We'll prove a key claim:\n\n**Claim:** $\\triangle BEK$ is similar to $\\triangle BAX$.\n\n*Proof.* Note that $\\angle EAB = 90^\\circ = \\angle KXB$. Also, since $BJ$ bisects $\\angle CBA$, we get $\\angle ABE = \\angle JBX = \\angle XBK$. Hence $\\triang... | Baltic Way | Baltic Way 2023 Shortlist | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0gdq | Let $x_1, \dots, x_n$ be different real numbers ($n \ge 2$). Prove that
$$
\sum_{1 \le i \le n} \prod_{j \ne i} \frac{1 - x_i x_j}{x_i - x_j} = \begin{cases} 0, & \text{if } n \text{ is even;} \\ 1, & \text{if } n \text{ is odd.} \end{cases}
$$
令 $x_1, \dots, x_n$ 為任意相異的 $n$ 個實數 ($n \ge 2$)。證明下列等式成立：
$$
\sum_{1 \le i ... | [
"Let $G(x_1, \\dots, x_n)$ be the function of the $n$ variables $x_1, \\dots, x_n$ on the LHS of the required identity. Since both sides of the identity are rational functions, it suffices to prove it when all $x_i \\notin \\{-1, 1\\}$. Define\n$$\nf(t) = \\prod_{i=1}^{n} (1 - x_i t),\n$$\nand note that\n$$\nf(x_i)... | Taiwan | 2020 Taiwan IMO 3J | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | null | proof only | null | |
0ivz | Problem:
What is the period of the function $f(x) = \cos(\cos(x))$? | [
"Solution:\n$\\boxed{\\pi}$\n\nSince $f(x)$ never equals $\\cos(1)$ for $x \\in (0, \\pi)$ but $f(0) = \\cos(1)$, the period is at least $\\pi$. However, $\\cos(x+\\pi) = -\\cos(x)$, so $\\cos(\\cos(x+\\pi)) = \\cos(\\cos(x))$."
] | United States | Harvard-MIT November Tournament | [
"Precalculus > Trigonometric functions",
"Precalculus > Functions"
] | null | proof and answer | π | |
01ej | Let $ABC$ be an acute triangle, $H$ its orthocentre, and $M$ the midpoint of $BC$. Furthermore, let $k_1$ and $k_2$ be the circle with diameter $AH$ and the circle with center $M$ that touches the circumcircle of triangle $ABC$ interiorly, respectively. Prove that $k_1$ and $k_2$ are touching circles. | [
"Let $N$ be the midpoint of $AH$ (and of $k_1$), and let $X$ be the image of $H$ with respect to reflection about $M$. Then $X$ lies on the circumcircle of $ABC$, opposite to $A$. As $OM$ and $AH$ are parallel, by the Intercept Theorem, we have $AH = 2OM$. Hence, $AN = OM$, i.e., $ANMO$ is a parallelogram. Let $r_1... | Baltic Way | Baltic Way shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0e9i | For real numbers $a$ and $b$, such that $|a| \neq |b|$ and $a \neq 0$, we have
$$
\frac{a-b}{a^2+ab} + \frac{a+b}{a^2-ab} = \frac{3a-b}{a^2-b^2}.
$$
Determine the value of the expression $\frac{b}{a}$. | [
"Multiplying the equation by $a(a+b)(a-b)$ we get\n$$\n(a-b)^2 + (a+b)^2 = a(3a-b).\n$$\nAfter expanding the terms and moving all the terms to the right-hand side we get\n$$\n0 = a^2 - ab - 2b^2 = (a - 2b)(a + b).\n$$\nSince $a \\neq -b$, we get $a - 2b = 0$ or $a = 2b$. Since $a$ is non-zero, we get $\\frac{b}{a} ... | Slovenia | National Math Olympiad in Slovenia | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1/2 | |
01dr | For which integers $n = 1, \dots, 6$ does the equation
$$
a^n + b^n = c^n + n
$$
have a solution in integers? | [
"**Answer:** Solutions exist for $n = 1, 2, 3$.\n\nFor $n = 6$, we consider the equation $a^6 + b^6 = c^6 + 6$ modulo 13. We always have $x^6 \\equiv 0, 1$ or $-1$ (mod 13) (by Fermat's little theorem or a direct computation). However, then it is evident that $a^6 + b^6 - c^6$ cannot be 6 (mod 13).\n\nFor $n = 5$, ... | Baltic Way | Baltic Way 2016 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | n = 1, 2, 3 | |
0j3o | Problem:
Newton and Leibniz are playing a game with a coin that comes up heads with probability $p$. They take turns flipping the coin until one of them wins with Newton going first. Newton wins if he flips a heads and Leibniz wins if he flips a tails. Given that Newton and Leibniz each win the game half of the time, ... | [
"Solution:\n\nThe probability that Newton will win on the first flip is $p$. The probability that Newton will win on the third flip is $(1-p) p^{2}$, since the first flip must be tails, the second must be heads, and the third flip must be heads. By the same logic, the probability Newton will win on the $(2n+1)^{\\t... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (3 - sqrt(5))/2 | |
0bx3 | Let $a \in (0, 1)$. Solve in $\mathbb{R}$ the equation $a^{[x]} + \log_a\{x\} = x$. | [
"It is clear that $x \\notin \\mathbb{Z}$ and as $a^{[x]} > 0$, $\\log_a\\{x\\} > 0$, we should have $x > 0$. If $x \\in (0, 1)$, then $0 < \\log_a x = x - 1 < 0$, which is absurd. Hence, $x \\in (0, \\infty) \\setminus \\mathbb{N}$.\nLet $f: \\mathbb{R} \\to (0, +\\infty)$, $f(x) = a^x$, which is of course a bijec... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof and answer | All real solutions are x = n + a^n for positive integers n. | |
0dkv | Let $ABC$ be a triangle, its incircle $(I)$ touches its sides $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. The bisector of angle $BIC$ intersects $BC$ at $M$. The line $AM$ intersects $EF$ at $N$. Prove that $DN$ bisects angle $EDF$. | [] | Saudi Arabia | Saudi Booklet | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry >... | null | proof only | null | |
0bxt | In the triangle $ABC$ we denote by $O$ and $I$ the circumcenter and the incenter, respectively. The perpendicular bisectors of the line segments $IA$, $IB$ and $IC$ pairwise intersect, thus defining the triangle $A_1B_1C_1$. Prove that
$$
\vec{OI} = \vec{OA_1} + \vec{OB_1} + \vec{OC_1}.
$$
 | [
"Let $A_1$ be the intersection point of the perpendicular bisectors of the line segments $IB$ and $IC$. Let the angle bisector $AI$ intersect the circumcircle of $ABC$ at $D$. Since $\\angle BID = \\angle DBI$ and $\\angle CID = \\angle DCI$, it follows that $DB = DI = DC$, hence $A_1 = D$. Thus, $A_1$ belongs to t... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0g3v | Problem:
Let $n$ be a positive integer. Prove that there exists a finite sequence $S$ consisting of only zeros and ones, satisfying the following property: For any positive integer $d \geq 2$, when $S$ is interpreted as a number in base $d$, the resulting number is non-zero and divisible by $n$.
Remark: The sequence ... | [
"Solution:\n\nWorking in any base $d \\geq 2$, by pigeonhole, two of the numbers\n$$\n1, 11, 111, \\ldots, \\overbrace{11 \\cdots 11}^{n+1}\n$$\nhave the same residue $\\bmod n$. Therefore their difference\n$$\nD = \\overbrace{11 \\cdots 11}^{t} \\overbrace{00 \\cdots 00}^{k} \\text{ with } k, t \\leq n\n$$\nis div... | Switzerland | IMO Selection | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions... | null | proof only | null | |
0a3z | Johan and Quintijn play the following game.
Before the start of the game, the integers $1$, $2$, $\ldots$, $2024$ are written on a board. The players then each take turns, starting with Johan. On their turn a player must wipe out two integers $a$ and $b$ from the board and write their (possibly negative) difference $a ... | [
"We show that Quintijn has a winning strategy.\nObserve that each move reduces the number of integers by exactly one, so at the start of Johan's turn the number is always even and at the start of Quintijn's turn the number is always odd. Moreover, the number must be at least $2$, otherwise the game would have alrea... | Netherlands | IMO Team Selection Test 2 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Quintijn | |
0gki | Let $\Omega$ be the circumcircle of a triangle $ABC$. Let $D$ be a variable point on the arc $AB$ that does not contain $C$ ($D \neq A, B$) and $E, F$ be the incenters of the triangles $CAD$ and $CBD$, respectively. Find the locus of the second intersection point of the circumcircle of $\triangle DEF$ and $\Omega$ as $... | [
"Consider the following well known Lemma:\n\n*Lemma.* Let $ABC$ be a triangle with incircle $I$. If $M$ is the midpoint of the arc $BC$ of the circumcircle not containing $A$, then $MB = MC = MI$.\n\n*Proof of Lemma.* Let the circumcircle of $DEF$ intersect $\\Omega$ again at $X$. Let $M$ and $N$ be the midpoints o... | Thailand | Thai Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | A single fixed point: let M and N be the midpoints of the arcs opposite the two chosen vertices (not containing the third), and let P be the point where the line through the remaining vertex parallel to MN meets the circumcircle. Then the locus is the second intersection of the circumcircle with the line through P and ... | |
05ow | Problem:
Montrer que $n! = 1 \times 2 \times \cdots \times n$ est divisible par $2^{n-1}$ si et seulement si $n$ est une puissance de $2$. | [
"Solution:\n\nOn définit, pour tout nombre premier $p$, la valuation $p$-adique d'un entier $n$ comme le plus grand entier, noté $v_{p}(n)$, tel que $p^{v_{p}(n)}$ divise $n$. La valuation $p$-adique d'un produit d'entiers est la somme de leurs valuations $p$-adiques. Pour $x$ un nombre réel, on note $\\lfloor x\\r... | France | Envoi 1: Arithmétique | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0g5w | 給定一個平行四邊形, 它的四個角的角度都是 $360°$ 的有理數倍。試問: 是否可能將座標平面上除了原點的每一點著黑白兩色, 使得當某三點和原點構成一個該四邊形的相似形, 則該三點必定不是同色的? | [
"可以。首先若該平行四邊形不是菱形, 可令 $t > 1$ 為長邊和短邊的比值。對平面上的任意點, 若該點與原點距離在 $[t^{2k}, t^{2k+1})$ 內 ($k$ 為任意整數), 則著黑色; 否則著白色。如此平行四邊形最靠近原點的兩個點一定是不同色。\n\n以下考慮平行四邊形是菱形。設其小於或等於 $90°$ 的內角角度為 $2a$, 即 $a \\le 45°$。則其邊與對角線的夾角為 $a$ 或 $90° - a$。現在著色如下: 以圓心及 $x$ 軸方向量仰角, 將區間 $[0°, 90°)$ 分割為 $[0, a)$, $[a, 2a)$, ..., $[ka, 90°)$, 將幅角落在第一個區間的點著黑... | Taiwan | 二〇一一數學奧林匹亞競賽第一階段選訓營,模擬競賽(一) | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04ks | Let $\overline{AB}$ and $\overline{CD}$ be two diameters of the circle $k$ with centre $S$ and let $\angle BAD = 28^\circ$. The circle centred at $A$ passing through the point $S$ meets the circle $k$ at $E$ and $F$ ($D$ and $F$ are on the same side of $AB$). Find $\angle CFS$. (Matija Bašić) | [] | Croatia | Mathematical competitions in Croatia | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 32° | |
0984 | Problem:
Fie matricea $A=\left(\begin{array}{cc}1 & -2 \\ -2 & 1\end{array}\right)$. Determinați $A^{2021}$. | [
"Solution:\n\nObservăm că $A^{2}=\\left(\\begin{array}{cc}5 & -4 \\\\ -4 & 5\\end{array}\\right)$, $A^{3}=\\left(\\begin{array}{cc}13 & -14 \\\\ -14 & 13\\end{array}\\right)$.\nUtilizând metoda inducției matematice, demonstrăm că $A^{n}=\\left(\\begin{array}{ll}a_{n} & b_{n} \\\\ b_{n} & a_{n}\\end{array}\\right)$.... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | [[ (3^{2021} - 1)/2, -(3^{2021} + 1)/2 ], [ -(3^{2021} + 1)/2, (3^{2021} - 1)/2 ]] | |
028b | Problem:
No ponto de ônibus - Um grupo de meninos e meninas aguarda em um ponto pelo ônibus. No primeiro ônibus que passa embarcam somente 15 meninas, e ficam 2 meninos para cada menina no ponto de ônibus. No segundo ônibus que passa, embarcam somente 45 meninos, e ficam 5 meninas para cada menino no ponto de ônibus. ... | [
"Solution:\n\nVamos representar por $M$ o número de meninas e por $H$ o número de meninos que estavam no ponto antes da parada do primeiro ônibus. Depois do embarque das 15 meninas no primeiro ônibus, ficaram no ponto $M-15$ meninas e $H$ meninos. Uma vez que, neste momento, ficam no ponto 2 meninos para cada menin... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 50 boys and 40 girls | |
0l37 | A rectangle has integer length sides and an area of $2024$. What is the least possible perimeter of the rectangle?
(A) $160$ (B) $180$ (C) $222$ (D) $228$ (E) $390$ | [
"Note that $2024 = 44 \\cdot 46$. A $44 \\times 46$ rectangle will have perimeter $2(44 + 46) = 180$. It is straightforward to check the other possible dimensions to show that this gives the rectangle with the least possible perimeter:\n* $23 \\times 88$ gives a perimeter of $2(23 + 88) = 222$.\n* $22 \\times 92$ g... | United States | AMC 10 B | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | MCQ | B | |
0e7n | For every real number $a$, let $[a]$ be the greatest integer that is not greater than $a$. Find all integers $y$ for which there exists a real number $x$ such that $\left[\frac{x+23}{8}\right] = [\sqrt{x}] = y$. | [
"Let $y$ be such a number. Then $\\sqrt{x} \\ge |\\sqrt{x}| = y$. Since $\\sqrt{x} \\ge 0$ we have $y = [\\sqrt{x}] \\ge 0$, so we may square the inequality to get $x \\ge y^2$. Also, $\\frac{x+23}{8} < [\\frac{x+23}{8}]+1 = y+1$, or $x < 8y-15$. This implies $y^2 < 8y-15$, or $(y-3)(y-5) < 0$, which means that $3 ... | Slovenia | National Math Olympiad 2013 - Final Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 4 | |
08vo | There are 4 cards and on each one a positive 1-digit integer is written. By choosing 2 different cards from the 4 and writing down the sum of the 2 numbers written on the chosen cards, we can get 4 distinct integers. Also by choosing 2 distinct cards and writing down the product of the 2 numbers written on the chosen c... | [
"If we assume that the number of distinct integers written on the 4 cards is 1 or 2, then the possible number of the sums of the 2 integers written on 2 chosen cards is at most 3, so this contradicts the assumption that 4 distinct sums can be obtained. Also, if the 4 numbers written on the 4 cards are all distinct,... | Japan | Japan Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (1, 2, 2, 4), (1, 3, 3, 9), (2, 4, 4, 8), (4, 6, 6, 9) | |
0els | If $x_1$, $x_2$ and $x_3$ are the roots of $x^3 - a x^2 + b x - c = 0$, prove that
$$
(x_1 - x_2)^2 + (x_2 - x_3)^2 + (x_3 - x_1)^2 = 2a^2 - 6b.
$$ | [
"Using Vieta's formula, the sum of the roots is $x_1 + x_2 + x_3 = a$, and the sum of the pairwise products is $x_1 x_2 + x_2 x_3 + x_3 x_1 = b$. Starting with the right hand side:\n$$\n\\begin{aligned}\n2a^2 - 6b &= 2(x_1 + x_2 + x_3)^2 - 6(x_1 x_2 + x_2 x_3 + x_3 x_1) \\\\\n&= 2(x_1^2 + x_2^2 + x_3^2 + 2(x_1 x_2 ... | South Africa | South-Afrika 2011-2013 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0gsd | For positive integer $k$, let
$$
R_n = \{-k, -(k-1), \dots, -1, 1, \dots, k-1, k\} \text{ for } n = 2k \text{ and}
$$
$$
R_n = \{-k, -(k-1), \dots, -1, 0, 1, \dots, k-1, k\} \text{ for } n = 2k+1.
$$
A device consists of several balls and red or white ropes connecting some ball pairs. A *labeling* is a coloring of each... | [
"Answer: $m = 2n-1$.\nLet us show that if a device has a good labeling by $R_n$ then it has a sensitive labeling by $R_m$, where $m = 2n-1$. In\n$$\nR_m = R_{2n-1} = \\{-(n-1), -(n-2), \\dots, -1, 0, 1, \\dots, n-2, n-1\\}\n$$\nthere are $n$ non-negative elements. Any good labeling of the device by these $n$ non-ne... | Turkey | Team Selection Test for IMO 2019 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory"
] | English | proof and answer | m(n) = 2n - 1 | |
049n | Let $\overline{AD}$ and $\overline{BE}$ be altitudes of the triangle $ABC$. Given $|AE| = 5$, $|CE| = 3$ and $|CD| = 2$, determine $|BD|$. | [] | Croatia | CroatianCompetitions2011 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof and answer | 10 | |
03y1 | 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. | [
"Let $(a, b) = d$, $a = sd$, $b = td$, $(s, t) = 1$, $t > 1$, then $std^2 = d^k(s^k + t^k)$. So $k \\geq 2$ and $st = d^{k-2}(s^k + t^k)$. Since $(st, s^k + t^k) = 1$, we have $st = d^{k-2}$. Therefore, any prime factor of $st$ can be divided by $d$.\nIf there is a prime factor $p$ of $s$ or $t$ no less than $11$, ... | China | China Southeastern Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 96 | |
0em5 | Find all pairs $(m, n)$ of positive integers for which $6^m + 2^n + 2$ is a perfect square. | [
"If both $m$ and $n$ are greater than one, then $6^m + 2^n + 2 \\equiv_4 0 + 0 + 2 \\equiv_4 2$, which is not a perfect square. So at least one of $m$ and $n$ has to be exactly $1$.\n\n* If $m = 1$, then we want $2^n + 8$ to be a square. This has a solution only for $n = 3$. If $n \\ge 4$ the expression will be div... | South Africa | South-Afrika 2011-2013 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (1, 3) | |
0cb4 | Find the real numbers $x$, $y$, $z > 0$ for which
$$
xyz \le \min \left\{ 4 \left(x - \frac{1}{y}\right), 4 \left(y - \frac{1}{z}\right), 4 \left(z - \frac{1}{x}\right) \right\}.
$$ | [
"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, we get $4x \\ge xyz + \\frac{4}{y} \\ge 2\\sqrt{xyz \\cdot \\frac{4}{y}} = 4\\sqrt{xz}$, so $x \\ge z$.\n\nAnalogously, from $xyz \\le 4 \\left(y - \\frac{1}{z}\\right)$ an... | Romania | THE 73rd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD - SECOND SELECTION TEST | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | x = y = z = sqrt(2) | |
02up | Problem:
Dado um triângulo de lados $a \leq b \leq c$, pela lei dos cossenos temos:
$$
\cos \hat{C}=\frac{a^{2}+b^{2}-c^{2}}{2 a b}
$$
Se o ângulo $\hat{C}$ é obtuso, $\cos \hat{C}<0$. Como $2 a b$ é positivo, isso é o mesmo que $a^{2}+b^{2}-c^{2}<0$. Portanto, para um triângulo ser obtusângulo, o maior lado elevado a... | [
"Solution:\n\nPrimeiro vamos assumir a mesma ordem de lados que o enunciado, ou seja, $a \\leq b \\leq c$. Dividamos o problema em casos a partir do valor de $a$. Para facilitar a análise, lembre-se que os sete primeiros quadrados perfeitos são:\n$$\n1^{2}=1, \\quad 2^{2}=4, \\quad 3^{2}=9, \\quad 4^{2}=16, \\quad ... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof and answer | 8 | |
0bsr | a) Prove that, for every integer $k$, the equation $x^3 - 24x + k = 0$ has at most one integer solution.
b) Prove that the equation $x^3 + 24x - 2016 = 0$ has exactly one integer solution. | [
"a) Suppose that there exist two different integers $m$ and $n$ so that $m^3 - 24m + k = 0$ and $n^3 - 24n + k = 0$.\nSubtracting the above yields $(m-n)(m^2 + mn + n^2 - 24) = 0$. Since $m$ and $n$ are different, $m^2 + mn + n^2 = 24$, whence $(2m+n)^2 + 3n^2 = 96$.\nTherefore $n^2 \\le 32$, hence $n^2 \\in \\{0, ... | Romania | 67th Romanian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 12 | |
0dyv | Problem:
Naj bo $D$ notranja točka stranice $BC$ pravokotnega trikotnika $ABC$ s pravim kotom pri $C$. Trikotniku $ABD$ očrtano krožnico označimo s $\mathcal{K}$. Naj bo $E$ taka točka na $\mathcal{K}$, da je tetiva $DE$ pravokotna na $AB$. Dokaži, da je trikotnik $AEB$ enakokrak z vrhom $B$ natanko tedaj, ko je $CA$ ... | [
"Solution:\n\nOznačimo s $T$ presečišče tetiv $DE$ in $AB$. Vemo, da je trikotnik $DTB$ pravokotni.\n\nDenimo najprej, da je trikotnik $ABE$ enakokrak. Označimo $\\angle AEB = \\angle BAE = \\alpha$. Obodna kota $\\angle EAB$ in $\\angle EDB$ nad tetivo $BE$ sta enaka, zato je tudi $\\angle EDB = \\alpha$. Ker pa j... | Slovenia | 52. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dsu | Consider a polynomial $P(x, y, z)$ in three variables with integer coefficients such that for any real numbers $a, b, c$,
$$
P(a, b, c) = 0 \Leftrightarrow a = b = c.
$$
Find the largest integer $r$ such that for all such polynomials $P(x, y, z)$ and integers $n, m$,
$$
m^r \mid P(n, n+m, n+2m).
$$ | [
"Consider $P(x, y, z) = (y-x)^2 + (z-y)^2$. Then, $P(n, n+m, n+2m) = 2m^2$. Clearly, $r > 2$ does not work. We show that $r = 2$.\n\nIf there exist $(a_1, b_1, c_1)$, $(a_2, b_2, c_2)$ such that $P(a_1, b_1, c_1) > 0$ and $P(a_2, b_2, c_2) < 0$, draw a continuous path from $(a_1, b_1, c_1)$ to $(a_2, b_2, c_2)$, no... | Singapore | Singapore Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 2 | |
0iee | Problem:
Let $P$ be a regular $k$-gon inscribed in a circle of radius $1$. Find the sum of the squares of the lengths of all the sides and diagonals of $P$. | [
"Solution:\nPlace the vertices of $P$ at the $k$th roots of unity, $1, \\omega, \\omega^{2}, \\ldots, \\omega^{k-1}$. We will first calculate the sum of the squares of the lengths of the sides and diagonals that contain the vertex $1$. This is\n$$\n\\begin{aligned}\n\\sum_{i=0}^{k-1} \\left|1-\\omega^{i}\\right|^{2... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | k^2 | |
0jtq | Problem:
In a standard 52 deck of cards, there are 13 cards of each of four suits. Kevin guesses the suit of the top card, and the top card is revealed and discarded. This process continues till there are no cards remaining.
If Kevin always guesses the suit of which there are the most remaining (breaking ties arbitrar... | [
"Solution:\n\nImagine that the cards have been given ranks $1, 2, \\ldots, 13$ and moreover that within each rank the cards have been sorted in ascending order (i.e. Kevin will encounter $1,2, \\ldots, 13$ of hearts in that order).\nThen, observe that Kevin will always guess the last card of rank $r$ correctly, for... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0g9z | 給定正整數 $M$, 可定義數列 $a_0, a_1, a_2, \dots$ 如后: $a_0 = \frac{2M+1}{2}$, 且對所有的 $k = 0, 1, 2, \dots$, 令 $a_{k+1} = a_k \lfloor a_k \rfloor$。
找出所有的正整數 $M$，使得上述定義的數列 $a_0, a_1, a_2, \dots$ 中，至少有一項是整數。
(註：$[x]$ 表示不超過實數 $x$ 的最大整數。)
Determine all positive integers $M$ for which the sequence $a_0, a_1, a_2, \dots$, defined by $a_... | [
"$M$ 可以是任何大於或等於 2 的正整數，即 $M \\ge 2$。\n首先對所有的非負整數 $k$，定義 $b_k = 2a_k$。則有\n$$\nb_{k+1} = 2a_{k+1} = 2a_k \\lfloor a_k \\rfloor = b_k \\left\\lfloor \\frac{b_k}{2} \\right\\rfloor.\n$$\n因為 $b_0$ 是整數 $2M+1$，所以數列 $\\langle b_k \\rangle$ 的每一項都是整數。\n用歸謬法。如果 $\\langle a_k \\rangle$ 的每一項都不是整數，則 $\\langle b_k \\rangle$ 的每一項都... | Taiwan | 二〇一六數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | M ≥ 2 | |
04fg | Let $ABC$ be an isosceles triangle with a base $\overline{AB}$. A point $P$ is chosen on the segment $\overline{AC}$ and a point $Q$ is chosen on the segment $\overline{BC}$ such that $|AP| + |BQ| = |PQ|$. The line parallel with the line $BC$ which passes through the midpoint of the segment $\overline{PQ}$ intersects t... | [
"Let $S$ be the midpoint of the segment $\\overline{PQ}$ and let $M$ be the point on the segment $\\overline{AB}$ such that $MP \\parallel BC$.\n\n\n\nWe have $\\angle PMA = \\angle CBA = \\angle CAB = \\angle PAM$, so the triangle $PAM$ is isosceles and $|PA| = |PM|$. The quadrilateral $PM... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0b2e | Problem:
A $10 \times 1$ rectangular pavement is to be covered by tiles which are either green or yellow, each of width $1$ and of varying integer lengths from $1$ to $10$. Suppose you have an unlimited supply of tiles for each color and for each of the varying lengths. How many distinct tilings of the rectangle are t... | [
"Solution:\n\nNote that the pavement is fixed and cannot be rotated, therefore a tiling is considered distinct from the reverse tiling. Also, note that the restriction that no two consecutive tiles can be of the same color can be addressed simply by treating consecutive tiles of the same color as one tile. Hence, t... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 1022 | |
0gie | 設 $n$ 為正整數。證明：不等式
$$
n \sum_{i=1}^{n} \sum_{j=1}^{n} \sum_{k=1}^{n} \frac{3}{a_i a_j + a_j a_k + a_k a_i} \ge \left( \sum_{j=1}^{n} \sum_{k=1}^{n} \frac{2}{a_j + a_k} \right)^2
$$
對任意正實數 $a_1, a_2, \dots, a_n$ 均成立。
Let $n$ be a positive integer. Prove that the inequality
$$
n \sum_{i=1}^{n} \sum_{j=1}^{n} \sum_{k=1}^{... | [
"熟知（九大於八不等式）\n$$\n9(a_j + a_k)(a_k + a_i)(a_i + a_j) \\geq 8(a_i + a_j + a_k)(a_j a_k + a_k a_i + a_i a_j).\n$$\n所以左式至少為\n$$\n\\begin{aligned}\n& \\frac{8n}{3} \\sum_{i,j,k=1}^{n} \\frac{a_i + a_j + a_k}{(a_j + a_k)(a_k + a_i)(a_i + a_j)} \\\\\n&= \\frac{4n}{3} \\sum_{i,j,k=1}^{n} \\left( \\frac{1}{(a_k + a_i)(a_i ... | Taiwan | IMO 2J, Mock Exam 1 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | Chinese; English | proof only | null | |
0byw | Given a positive integer $m$, determine the smallest integer $n \ge 2$ satisfying the following condition: No matter how the cells of an $n \times n$ array are colored one of $m$ colors, there exist cells $(i, j)$ and $(j, k)$, $i \neq j$ and $j \neq k$, sharing the same color. | [
"Assuming such a coloring exists, let $S_i$ be the set of colors of the off-diagonal cells on row $i$, and notice that the (re)stated condition shows that no $S_i$ contains an $S_j$, $i \\neq j$, so the $S_i$ form an antichain.\n\nConversely, given an antichain $S_1, \\dots, S_n$ of subsets of an $m$-element set, a... | Romania | THE Tenth STARS OF MATHEMATICS COMPETITION | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | C(m, floor(m/2)) + 1 | |
0by8 | Consider the set $A = \{1, 2, 3, \dots, 2017\}$. Determine the number of three element subsets $B \subset A$, which satisfy the following conditions simultaneously:
a) at least two elements in the set $B$ are consecutive positive integers;
b) there is an $a \in B$, such that $3a \in B$. | [
"The sets we search contain elements $a$, $3a$, with $3a \\le 2017$, so $a \\le 672$. For $a = 1$, we have $\\{1, 3\\} \\subset B$, and the possible sets are $\\{1, 2, 3\\}$ and $\\{1, 3, 4\\}$. For $a \\in \\{2, 3, 4, \\dots, 672\\}$ we have the sets $\\{a-1, a, 3a\\}$, $\\{a, a+1, 3a\\}$, $\\{a, 3a-1, 3a\\}$ and ... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics"
] | English | proof and answer | 2686 | |
0iqx | Problem:
Johnny the grad student is typing all the integers from $1$ to $\infty$, in order. The $2$ on his computer is broken however, so he just skips any number with a $2$. What's the $2008$th number he types? | [
"Solution:\n\nAnswer: $3781$\n\nWrite $2008$ in base $9$ as $2671$, and interpret the result as a base $10$ number such that the base $9$ digits $2,3, \\ldots, 8$ correspond to the base $10$ digits $3,4, \\ldots, 9$. This gives an answer of $3781$."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 3781 | |
0are | Problem:
Find all nonempty finite sets $X$ of real numbers with the following property:
$$
x + |x| \in X \quad \text{ for all } x \in X
$$ | [
"Solution:\nLet $X = \\{x_1, x_2, \\ldots, x_n\\}$, $n \\geq 1$, where $x_1 < x_2 < \\cdots < x_n$.\n\nIf $x_n > 0$, then $x_n + |x_n| = 2x_n \\in X$, which is a contradiction because $x_n < 2x_n$ but $x_n$ is the largest element of $X$.\n\nThe contradiction in the previous paragraph implies that $x_n \\leq 0$. If ... | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | proof and answer | All and only the finite subsets of the real numbers less than or equal to zero that contain zero. | |
01j2 | Let $k$ be a non-negative integer and let $A$ be a nonempty finite set. We define a word from the alphabet $A$ of length $k$ as a sequence of elements of $A$ of length $k$. (Notice that by definition the empty sequence is a word of length $0$.)
* Let $B$ be the set of all words from the nine-element alphabet $\{b, @, ... | [
"For $k \\in \\mathbb{N}_0$ we denote by $b_k$ the number of words from the alphabet $\\{b, @, l, t, i, c, w, a, y\\}$ of maximum length $k$, so that $b_{2023} = |B|$. Furthermore, we denote by $m_k$ the number of words of maximum length $k+2$ from the alphabet $\\{m, a, t, h\\}$, that contain the letter $t$ exactl... | Baltic Way | Baltic Way 2023 Shortlist | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Sequences and Series > ... | English | proof only | null | |
06ka | At a mathematical competition $n$ students work on 6 problems each one with three possible answers. After the competition, the Jury found that for every two students the number of the problems for which these students have the same answers is 0 or 2. Find the maximum possible value of $n$. | [
"The maximum possible value of $n$ is $18$.\n\nWe first show that $n \\ge 19$ is impossible. Let $A$, $B$, $C$ be the answers in each problem. For $n \\ge 19$, by the pigeonhole principle, we may assume $\\left\\lfloor \\frac{19}{3} \\right\\rfloor = 6$ students answer $A$ in problem 1. Then by the pigeonhole princ... | Hong Kong | HKG TST | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 18 | |
01g2 | Prove that there are infinitely many different triangles in the coordinate plane whose vertices have integer coordinates and whose side lengths are consecutive integers. | [
"At first we will prove that there are infinitely many triangles whose side lengths are consecutive integers and whose area also is an integer. Let the side lengths of the triangle be $2m-1$, $2m$ and $2m+1$, then by Heron's formula its area is\n$$\nS = \\sqrt{\\frac{6m(2m+2)2m(2m-2)}{16}} = m\\sqrt{3(m^2-1)} \\qua... | Baltic Way | Baltic Way 2019 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Number Theory > Diophantine Equations > Pell's equations"
] | English | proof only | null | |
09wn | Three consecutive vertices $A$, $B$, and $C$ of a regular octagon (8-gon) are the centres of circles that pass through neighbouring vertices of the octagon. The intersection points $P$, $Q$, and $R$ of the three circles form a triangle (see figure).
Prove that triangle $PQR$ is equilateral. | [
"An octagon can be subdivided into six triangles (see figure on the left). Together, the angles of those six triangles add up to the same number of degrees as the eight angles of the octagon. Since the angles of any triangle add up to $180$ degrees, this means that the eight angles of the octagon add up to $6 \\cdo... | Netherlands | Second Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0iah | Problem:
A right circular cone with a height of 12 inches and a base radius of 3 inches is filled with water and held with its vertex pointing downward. Water flows out through a hole at the vertex at a rate in cubic inches per second numerically equal to the height of the water in the cone. (For example, when the heig... | [
"Solution:\nWhen the water in the cone is $h$ inches high, it forms a cone similar to the original, so that its base has radius $h / 4$ and its volume is hence $\\pi h^{3} / 48$. The given condition then states that\n$$\n\\frac{d}{d t}\\left(\\frac{\\pi h^{3}}{48}\\right) = -h \\Rightarrow \\frac{\\pi h^{2}}{16} \\... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Volume"
] | null | proof and answer | 9π/2 seconds | |
0j63 | Problem:
Let $ABCD$ be a rectangle with $AB = 6$ and $BC = 4$. Let $E$ be the point on $BC$ with $BE = 3$, and let $F$ be the point on segment $AE$ such that $F$ lies halfway between the segments $AB$ and $CD$. If $G$ is the point of intersection of $DF$ and $BC$, find $BG$. | [
"Solution:\n\nAnswer: $1$\n\nNote that since $F$ is a point halfway between $AB$ and $AC$, the diagram must be symmetric about the line through $F$ parallel to $AB$. Hence, $G$ must be the reflection of $E$ across the midpoint of $BC$. Therefore, $BG = EC = 1$."
] | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 1 | |
04ce | Determine all values of the real parameter $a$ for which the equation
$$
8^a \sin^2 x = 4 \cdot 2^{\cos^2 x}
$$
has exactly one solution in the interval $\left[-\frac{\pi}{6}, \frac{2\pi}{3}\right]$. | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | a = 2/3 or 17/12 − (1/3) log_2 3 < a < 19/12 | |
01ab | A triangle $ABC$ satisfies $AB < AC$. Let $I$ be the center of the excircle tangent to the side $AC$. Point $P$ lies inside of the angle $BAC$, but outside of the triangle $ABC$ and satisfies $\angle CPB = \angle PBA + \angle ACP$. Prove that $AP \le AI$. | [
"\nLet $D$ be the middle of the arc $BC$, that contains point $A$, of the circumcircle of the triangle $ABC$. Then point $D$ lies on the segment $AI$. Note that points $B$, $A$, $D$, $C$ lie on the circumcircle of the triangle $ABC$ in that order as $AB < AC$. Then note, that the equation $... | Baltic Way | Baltic Way 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions an... | null | proof only | null | |
0hkq | Problem:
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(x)+f(y) \neq 0$ and
$$
\frac{f(x)-f(x-y)}{f(x)+f(x+y)}+\frac{f(x)-f(x+y)}{f(x)+f(x-y)}=0
$$
for all $x, y \in \mathbb{R}$. | [
"Solution:\nAny function of the form $f(x)=c$, where $c$ is a nonzero constant, clearly satisfies the requirements; we will show these are the only solutions. Bringing the original equation to a common denominator yields $2 f(x)^2 - f(x-y)^2 - f(x+y)^2 = 0$ for all $x$, $y$. Define $g(x)=f(x)^2$; clearly $g(x) \\ge... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | All functions of the form f(x) = c for all real x, where c is a nonzero constant. | |
04al | Let $n$ be a positive integer. Prove that there is an odd number of odd numbers among the numbers
$$
\binom{2n+1}{1}, \binom{2n+1}{2}, \dots, \binom{2n+1}{k}, \dots, \binom{2n+1}{n}.
$$ | [
"Let us consider the binomial coefficients $\\binom{2n+1}{k}$ for $1 \\leq k \\leq n$.\n\nRecall that $\\binom{2n+1}{k}$ is odd if and only if, in the binary representation, every bit of $k$ is less than or equal to the corresponding bit of $2n+1$ (Lucas' theorem).\n\nBut we can use a parity argument:\n\nThe sum of... | Croatia | CroatianCompetitions2011 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof only | null | |
072y | Let $A_1, A_2, \dots, A_n$ be arithmetic progressions of integers, each of $k$ terms, such that any two of these arithmetic progressions have at least two common elements. Suppose $b$ of these arithmetic progressions have common difference $d_1$ and the remaining arithmetic progressions have common difference $d_2$, wh... | [
"Let $[d_1, d_2]$ denote the least common multiple of $d_1$ and $d_2$. Let $P_j$ denote the union of all arithmetic progressions with common difference $d_j$, $j = 1, 2$, and let $S = P_1 \\cap P_2$. Then $S$ is an arithmetic progression with common difference $[d_1, d_2]$. Let $y$ be the least element of $P_1$ and... | India | Indija TS 2006 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0fro | Sean $a, b, c, d$ cuatro números reales positivos. Si se cumple
$$
a+b+\frac{1}{ab}=c+d+\frac{1}{cd} \quad \text{y} \quad \frac{1}{a}+\frac{1}{b}+ab=\frac{1}{c}+\frac{1}{d}+cd
$$
demuestra que al menos dos de los valores $a, b, c, d$ son iguales. | [
"**Solución 1:** Sea $u = a+b+\\frac{1}{ab}$ y $v = \\frac{1}{a}+\\frac{1}{b}+ab$. Denotamos $r = \\frac{1}{ab}$. Entonces tenemos que $a+b+r = u$, $ab+br+ra = v$ y $abr = 1$. Por las identidades de Cardano-Vieta, $a, b$ y $r$ son las tres raíces del polinomio $p(x) = x^3-ux^2+vx-1$. Por la misma razón, $c, d$ y $\... | Spain | LVIII Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | Spanish | proof only | null | |
0fhh | Problem:
Un número $N$, múltiplo de $83$, es tal que su cuadrado tiene $63$ divisores. Hallar $N$, sabiendo que es el menor número que cumple las condiciones anteriores. | [
"Solution:\n\nSupongamos $N = 2^{k} \\cdot 3^{p} \\cdot 5^{q} \\cdots 83^{r}$. Entonces $N^{2} = 2^{2k} \\cdot 3^{2p} \\cdot 5^{2q} \\cdots 83^{2r}$, con $r \\neq 0$.\n\nDebe ser $63 = (2k+1)(2p+1)(2q+1) \\cdots (2r+1)$, pero las únicas descomposiciones de $63$ son $63$, $7 \\cdot 9$ y $7 \\cdot 3 \\cdot 3$.\n\nEn ... | Spain | OME 28 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1992 | |
0jfk | Problem:
Let $ABC$ be a triangle with circumcenter $O$ such that $AC = 7$. Suppose that the circumcircle of $AOC$ is tangent to $BC$ at $C$ and intersects the line $AB$ at $A$ and $F$. Let $FO$ intersect $BC$ at $E$. Compute $BE$. | [
"Solution:\n\n$EB = \\frac{7}{2}$\n\n$O$ is the circumcenter of $\\triangle ABC \\Longrightarrow AO = CO \\Longrightarrow \\angle OCA = \\angle OAC$. Because $AC$ is an inscribed arc of circumcircle $\\triangle AOC$, $\\angle OCA = \\angle OFA$. Furthermore, $BC$ is tangent to circumcircle $\\triangle AOC$, so $\\a... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
... | null | proof and answer | 7/2 | |
0iue | Problem:
Lily and Sarah are playing a game. They each choose a real number at random between $-1$ and $1$. They then add the squares of their numbers together. If the result is greater than or equal to $1$, Lily wins, and if the result is less than $1$, Sarah wins. What is the probability that Sarah wins? | [
"Solution:\n\nThe answer is $\\frac{\\pi}{4}$.\n\nIf we let $x$ denote Lily's choice of number and $y$ denote Sarah's, then all possible outcomes are represented by the square with vertices $(-1,-1)$, $(-1,1)$, $(1,-1)$, and $(1,1)$. Sarah wins if $x^{2} + y^{2} < 1$, which is the area inside the unit circle. Since... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Circles"
] | null | final answer only | π/4 | |
0gud | $$
\frac{a^4 + 1}{b^3 + b^2 + b} + \frac{b^4 + 1}{c^3 + c^2 + c} + \frac{c^4 + 1}{a^3 + a^2 + a} \ge 2.
$$ | [
"$$\n\\frac{a^4+1}{b^3+b^2+b} + \\frac{b^4+1}{c^3+c^2+c} + \\frac{c^4+1}{a^3+a^2+a} \\ge 3 \\cdot \\sqrt[3]{\\frac{8}{27}} = 2.\n$$"
] | Turkey | Team Selection Test for JBMO 2023 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
04qf | A point $T$ is chosen inside the triangle $ABC$. Let $A_1, B_1$ and $C_1$ be the reflections of $T$ across the lines $BC, CA$ and $AB$, respectively. The lines $A_1T, B_1T$ and $C_1T$ intersect the circle $k$ circumscribed to the triangle $A_1B_1C_1$ again at $A_2, B_2$ and $C_2$, respectively.
Prove that the lines $AA... | [
"Let $K$ be the intersection of $CC_2$ and $k$.\n\n\n\nSince $CB$ and $CA$ are the bisectors of $\\overline{TA_1}$ and $\\overline{TB_1}$, respectively, the point $C$ is the circumcentre of the triangle $A_1TB_1$. Hence,\n$$\n\\triangle(CA_1, CB) = \\triangle(CB, CT) = \\triangle(B_1A_1, B_... | Croatia | Croatian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0flp | For a given positive integer $n$, find the sum of all positive integers smaller than $10n$ which are not divisible neither by $2$ nor by $5$. | [] | Spain | Spanija 2012 | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 20n^2 | |
00fz | Let $a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ be a sequence of non-negative integers, where $n$ is a positive integer. Let
$$
A_{n} = \frac{a_{1} + a_{2} + \cdots + a_{n}}{n}
$$
Prove that
$$
a_{1}! a_{2}! \ldots a_{n}! \geq \left( \left\lfloor A_{n} \right\rfloor ! \right)^{n}
$$
where $\left\lfloor A_{n} \right\rfloor$ is... | [
"Assume without loss of generality that $a_{1} \\geq a_{2} \\geq \\cdots \\geq a_{n} \\geq 0$, and let $s = \\left\\lfloor A_{n} \\right\\rfloor$. Let $k$ be any (fixed) index for which $a_{k} \\geq s \\geq a_{k+1}$.\nOur inequality is equivalent to proving that\n$$\n\\begin{equation*}\n\\frac{a_{1}!}{s!} \\cdot \\... | Asia Pacific Mathematics Olympiad (APMO) | XIV APMO | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | Equality holds if and only if either all terms are equal, or every term is either zero or one. | |
0fu4 | Problem:
Ein konvexes $n$-Eck zu zwacken bedeutet Folgendes: Man wählt zwei benachbarte Seiten $AB$ und $BC$ aus und ersetzt diese durch den Streckenzug $AM$, $MN$, $NC$, wobei $M \in AB$ und $N \in BC$ beliebige Punkte im Innern dieser Strecken sind. Mit anderen Worten, man schneidet eine Ecke ab und erhält ein $(n+1... | [
"Solution:\n\nBei jedem Zwacken bleibt von den benachbarten Seiten ein Teilsegment positiver Länge übrig, da die gewählten Punkte $M$ und $N$ innere Punkte der entsprechenden Seiten sind. Folglich gibt es in $\\mathcal{P}_n$ stets sechs Seiten, die Teilsegmente der ursprünglichen Seiten von $\\mathcal{P}_6$ sind. W... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0j5l | Problem:
Let $z = \cos \frac{2\pi}{2011} + i \sin \frac{2\pi}{2011}$, and let
$$
P(x) = x^{2008} + 3 x^{2007} + 6 x^{2006} + \ldots + \frac{2008 \cdot 2009}{2} x + \frac{2009 \cdot 2010}{2}
$$
for all complex numbers $x$. Evaluate $P(z) P\left(z^{2}\right) P\left(z^{3}\right) \ldots P\left(z^{2010}\right)$. | [
"Solution:\nAnswer: $2011^{2009} \\cdot \\left(1005^{2011} - 1004^{2011}\\right)$\n\nMultiply $P(x)$ by $x-1$ to get\n$$\nP(x)(x-1) = x^{2009} + 2 x^{2008} + \\ldots + 2009 x - \\frac{2009 \\cdot 2010}{2}\n$$\nor,\n$$\nP(x)(x-1) + 2010 \\cdot 1005 = x^{2009} + 2 x^{2008} + \\ldots + 2009 x + 2010\n$$\nMultiplying b... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 2011^{2009} * (1005^{2011} - 1004^{2011}) | |
05qa | Problem:
Un pays comprend $2018 n+1$ villes, où $n$ est un entier naturel non nul. Certaines paires de villes sont reliées par des lignes directes de chemin de fer, de sorte qu'il y ait au plus une ligne entre deux villes ; chaque ligne va dans les deux sens. La distance entre deux villes $A$ et $B$ est alors le nombr... | [
"Solution:\n\nDans toute la suite, on va bien sûr réinterpréter l'énoncé en termes de graphes, et on va montrer que les entiers $n$ recherchés sont les entiers pairs.\n\nSupposons d'abord que l'on dispose d'un entier $n$ et d'un graphe respectant le critère de l'énoncé. Alors tout sommet est de degré $n$. La somme ... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | All even positive integers n | |
01qa | Given $m, n \in \mathbb{N}$ such that $m > n^{n-1}$ and the numbers $m+1, m+2, \dots, m+n$ are composite.
Prove that there exist distinct primes $p_1, p_2, \dots, p_n$ such that $m+k$ is divisible by $p_k$ for any $k = 1, 2, \dots, n$. | [
"We call the number $m + k$ from the problem condition convenient if it has at least $n$ distinct prime divisors, otherwise we call the number $m + k$ inconvenient. It is easy to see that we may consider inconvenient numbers only. Take one of them, $m + k = q_1^{a_1} q_2^{a_2} \\dots q_l^{a_l}$, where all $q_i$ are... | Belarus | SELECTION and TRAINING SESSION | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0eyb | Problem:
An octagon has equal angles. The lengths of the sides are all integers. Prove that the opposite sides are equal in pairs. | [
"Solution:\n\nExtend the sides to form two rectangles. Let the sides of the octagon have length $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$. Then we can find the rectangle sides. For example, one of the rectangles has opposite sides $a + \\frac{b + h}{\\sqrt{2}}$ and $e + \\frac{d + f}{\\sqrt{2}}$. Hence either $a = e$ ... | Soviet Union | 2nd ASU | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
04xu | Let $ABCD$ be a cyclic quadrilateral with $BC = CD$, let $\omega$ be the circle centered at $C$ tangent to $BD$, and let $I$ be the incenter of $ABD$. Show that the line through $I$ parallel to $AB$ is tangent to $\omega$. | [
"Let $p$ be the line tangent at $D$ to the circumcircle $\\Gamma$ of $ABCD$. Since $C$ is the midpoint of the arc $BD$, we have $\\angle(CD, p) = \\angle CAD = \\angle BAC = \\angle BDC$, and we see that $p$ is tangent to $\\omega$. Similarly, if $E$ is the midpoint of the arc $DA$ of $\\Gamma$, then $p$ is tangent... | Czech-Polish-Slovak Mathematical Match | Cesko-Slovacko-Poljsko 2013 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0a4v | Problem:
Suppose that $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ are real numbers between $0$ and $1$ with sum $s$. Prove that
$$
\sum_{i = 1}^{n} \frac{x_{i}}{s + 1 - x_{i}} + \prod_{i = 1}^{n}(1 - x_{i}) \leq 1.
$$ | [
"Solution:\n\nLet $i$ be arbitrary and consider the set $A = \\{a_{1}, a_{2}, \\ldots, a_{n}\\}$ defined by $a_{i} = s + 1 - x_{i}$ and let $a_{j} = 1 - x_{j}$ for all $j \\neq i$. For example, if $i = 2$ then $A$ would be $\\{1 - x_{1}, s + 1 - x_{2}, 1 - x_{3}, \\ldots, 1 - x_{n}\\}$. The AM-GM inequality on $A$ ... | New Zealand | NZMO Round One | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
07fq | Let $T$ be a triangulation of a convex 100-gon. We construct $P(T)$ by copying the same 100-gon and drawing a diagonal if it was not drawn in $T$, and there is a quadrilateral with this diagonal and two other vertices so that all its sides and the other diagonal are in $T$. Let $f(T)$ be the number of intersections of ... | [
"Call two triangles in $T$ adjacent if they share an edge. First note that for any diagonal $d$ drawn in $P(T)$, there are two adjacent triangles of $T$ such that $d$ is drawn in $P(T)$, because of the quadrilateral formed by these triangles. We call them $d$'s triangles.\nAssume that two diagonals $d_1, d_2$ inter... | Iran | 37th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | minimum 96, maximum 144 | |
0aw0 | Problem:
Find the exact value of $\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{5}\right) + \tan^{-1}\left(\frac{1}{8}\right)$. | [] | Philippines | 19th Philippine Mathematical Olympiad | [
"Precalculus > Trigonometric functions"
] | null | final answer only | π/4 | |
0gb3 | a, b, c, d 是正實數且滿 $a + b + c + d = 4$。試證明：
$$
\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq 4 + (a-d)^2.
$$ | [
"注意到 RHS 如果沒有 $(a-d)^2$ 這項，那麼\n$$\n\\begin{aligned}\n\\sum_{cyc} \\frac{a^2}{b} &\\geq 4 \\\\\n\\Leftrightarrow \\sum_{cyc} \\left(\\frac{a^2}{b} + b\\right) &\\geq 8\n\\end{aligned}\n$$\n上式由四個算幾不等式可以輕易知道成立。回到原題，加上 $(a-d)^2$ 這項後需要證明\n$$\n\\begin{align*}\n\\Leftrightarrow \\sum_{cyc} \\left(\\frac{a^2}{b} + b\\right... | Taiwan | 二〇一七數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0enj | Let $ABCD$ be a square with circumcircle $\Gamma$. Let $M$ be on minor arc $CD$ of $\Gamma$. Let $BD$ and $AM$ intersect at $P$, $CD$ and $AM$ intersect at $R$, $BM$ and $AC$ intersect at $Q$ and $BM$ and $DC$ intersect at $S$. Show that $PS \perp QR$. | [
"Place the square in the coordinate plane with $A(-1, 1)$, $B(1, 1)$, $C(1, -1)$ and $D(-1, -1)$. The equation of $\\Gamma$ is $x^2 + y^2 = 2$, the equation of the line $AC$ is $y = -x$ and the equation of the line $BC$ is $y = x$. Let $M$ have coordinates $(m, n)$, where $m^2 + n^2 = 2$. The lines $AM$ and $BM$ ha... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0bz1 | a) Exhibit a continuous function $f: (0, \infty) \to \mathbb{R}$ such that
$$
\lim_{x \to \infty} \frac{1}{x^2} \int_0^x f(t) \, dt = 1,
$$
but $f(x)/x$ has not a limit as $x \to \infty$.
b) Let $f: (0, \infty) \to \mathbb{R}$ be an increasing function such that
$$
\lim_{x \to \infty} \frac{1}{x^2} \int_0^x f(t) \, dt ... | [
"a) The function $f: (0, \\infty) \\to \\mathbb{R}$, $f(x) = 2x + 2x \\cos x^2$, is clearly continuous,\n$$\n\\lim_{x \\to \\infty} \\frac{1}{x^2} \\int_0^x f(t) \\, dt = \\lim_{x \\to \\infty} \\left( 1 + \\frac{\\sin x^2}{x^2} \\right) = 1,\n$$\nbut $f(x)/x$ has obviously not a limit as $x \\to \\infty$. Another ... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Limits"
] | English | proof and answer | a) One example is f(x) = 2x + 2x cos(x^2). b) Under the given conditions, f(x)/x → 2 as x → ∞. | |
0dgc | Find the number of integer solutions of the equation
$$
\left\lfloor \frac{x}{7} \right\rfloor = \left\lfloor \frac{x}{12} \right\rfloor + \left\lfloor \frac{x}{17} \right\rfloor.
$$ | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | 1428 | |
0jw4 | Problem:
For $n > 1$, consider an $n \times n$ chessboard and place pieces at the centers of different squares.
a. With $2n$ chess pieces on the board, show that there are 4 pieces among them that form the vertices of a parallelogram.
b. Show that there is a way to place $(2n-1)$ chess pieces so that no 4 of them fo... | [
"Solution:\n\na. Since there can be at most $n$ pieces that are leftmost in their rows (some rows may be empty), there are at least $n$ pieces that are not the leftmost in their row. Record the distances (the number of squares) between the leftmost piece and the other pieces on the same row.\n\nThere are then at le... | United States | BAMO-8 and BAMO-12 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
08y8 | Find all the triplets $(x, y, z)$ of positive integers satisfying the following conditions:
$$
x + xy + xyz = 31, \quad x < y < z.
$$ | [
"$(1, 2, 14), (1, 3, 9)$\n\nRewriting the given equation, we get $x(1 + y + yz) = 31$. Since $31$ is a prime, and since $1 + y + yz > 1$, we must have $x = 1$, $1 + y + yz = 31$. Thus, we obtain $y(1 + z) = 30$. From the given inequality, we must have $1 < y < z$ and hence, we conclude that the desired answer is $(... | Japan | 2019 Japan Mathematical Olympiad First Stage | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (1, 2, 14), (1, 3, 9) | |
05ok | Problem:
Soit $a$ et $b$ des entiers strictement positifs. Prouver que si
$$
a+\frac{b}{a}-\frac{1}{b}
$$
est un entier alors c'est un carré. | [
"Solution:\n\nEn multipliant par $a$, on obtient que $a^{2}+b-\\frac{a}{b}$ est un entier, donc $k=\\frac{a}{b}$ est un entier.\n\nDe même, en multipliant par $b$ on voit que $\\frac{b^{2}}{a}$ est un entier. Comme $\\frac{b^{2}}{a}=\\frac{b}{k}$, on a $b=k c$ où $c$ est un entier.\n\nComme $\\frac{b}{a}-\\frac{1}{... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
08lb | Problem:
Integers $1, 2, \ldots, 2n$ are arbitrarily assigned to boxes labeled with numbers $1, 2, \ldots, 2n$. Now, we add the number assigned to the box to the number on the box label. Show that two such sums give the same remainder modulo $2n$. | [
"Solution:\n\nLet us assume that all sums give different remainders modulo $2n$, and let $S$ denote the value of their sum.\n\nFor our assumption,\n$$\nS \\equiv 0 + 1 + \\ldots + 2n - 1 = \\frac{(2n - 1) 2n}{2} = (2n - 1)n \\equiv n \\pmod{2n}\n$$\nBut, if we sum, breaking all sums into its components, we derive\n... | JBMO | 2008 Shortlist JBMO | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
03rl | Nine balls, numbered $1$, $2$, $\ldots$, $9$, are put randomly at $9$ equally spaced points on a circle, each point with a ball. Let $S$ be the sum of the absolute values of the differences of the numbers of all two neighboring balls. Find the probability of $S$ to be the minimum value. (Remark: If one arrangement of t... | [
"Next, we calculate the number of arrangements, which make $S$ the minimum. Along the circle there are two routes from $1$ to $9$, the major arc and the minor arc. For each of them, let $x_1, x_2, \\dots, x_k$ be the numbers of the successive balls on the arc, then\n$$\n\\begin{aligned}\n& |1-x_1| + |x_1-x_2| + \\d... | China | China Mathematical Competition (Jiangxi) | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof and answer | 1/315 | |
09xj | Show that for every positive integer $n$ there exist positive integers $a$ and $b$ with
$$
n \mid 4a^2 + 9b^2 - 1.
$$ | [
"If $n = 1$, all choices of $a$ and $b$ are solutions. Now suppose that $n > 1$ and let $p$ be a prime divisor of $n$. Let $k$ be the number of factors of $p$ in $n$. We give a condition for $a$ and $b$ modulo $p^k$ which guarantees that $p^k \\mid 4a^2 + 9b^2 - 1$. By doing this for every prime divisor of $n$, we ... | Netherlands | IMO Team Selection Test 2 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof only | null | |
02i3 | Problem:
Quantas frações da forma $\frac{n}{n+1}$ são menores do que $\frac{7}{9}$, sabendo que $n$ é um número inteiro positivo?
A) 1
B) 2
C) 3
D) 4
E) 5 | [
"Solution:\n\nAs frações da forma $\\frac{n}{n+1}$, com $n$ inteiro positivo são:\n$$\n\\underbrace{\\frac{1}{2}}_{n=1} ; \\quad \\underbrace{\\frac{2}{3}}_{n=2} ; \\quad \\underbrace{\\frac{3}{4}}_{n=3} ; \\quad \\underbrace{\\frac{4}{5}}_{n=4} ; \\quad \\underbrace{\\frac{5}{6}}_{n=5} \\cdots\n$$\nObserve que est... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | C | |
0fs3 | Problem:
Bestimme alle Funktionen $f: \mathbb{R} \rightarrow \mathbb{R}$, sodass für alle $x, y \in \mathbb{R}$ die folgende Gleichung erfüllt ist:
$$
f\left((x-y)^2\right)=x^2-2 y f(x)+(f(y))^2
$$ | [
"Solution:\n\nWir setzen zuerst verschiedene Dinge für $x$ und $y$ ein:\n$$\n\\begin{aligned}\nx=y=0 & \\Rightarrow f(0)=f(0)^2 \\\\\ny=0 & \\Rightarrow f\\left(x^2\\right)=x^2+f(0)^2 \\\\\nx=0 & \\Rightarrow f\\left(y^2\\right)=-2 y f(0)+f(y)^2 \\\\\nx=y & \\Rightarrow f(0)=x^2-2 x f(x)+f(x)^2=(f(x)-x)^2\n\\end{al... | Switzerland | Vorselektionsprüfung | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = x and f(x) = x + 1 | |
018s | Let $P$ be a point inside a square $ABCD$ such that $PA : PB : PC$ is $1 : 2 : 3$. Determine the angle $\angle BPA$. | [
"*First Solution.* Rotate the triangle $ABP$ by $90^\\circ$ around $B$ such that $A$ goes to $C$ and $P$ is mapped to a new point $Q$. Then $\\angle PBQ = \\angle PBC + \\angle CBQ = \\angle PBC + \\angle ABP = 90^\\circ$. Hence the triangle $PBQ$ is an isosceles right-angled triangle, and $\\angle BQP = 45^\\circ$... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof and answer | 135° | |
01aj | Two circles of the same radius, $K$ and $L$, intersect in two points, one of which is $P$. Denote by $A$ and $B$, respectively, the points diametrically opposite to $P$ on each of $K$ and $L$. Yet another circle of the same radius is brought to pass through $P$, intersecting $K$ and $L$ in the points $X$ and $Y$, respe... | [
"Let $M$ be the third circle, and denote by $Z$ the point on $M$ diametrically opposite to $P$.\nSince $\\angle AX P = \\angle PX Z = 90^\\circ$, the three points $A$, $X$, $Z$ are collinear. Likewise, the three points $B$, $Y$, $Z$ are collinear. Point $P$ is equidistant to the three vertices of triangle $ABZ$, fo... | Baltic Way | Baltic Way 2013 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0a0l | A number is called *nilless* if it is integer and positive and contains no zeros. You can make a positive integer *nilless* by simply omitting the zeros. We denote this with square brackets, for example $[2050] = 25$ and $[13] = 13$. When we multiply, add, and subtract we indicate with square brackets when we omit the ... | [
"1. If $[a \\cdot b] - 1$ is a nilless number, then it follows from $[[a \\cdot b] - 1] = 1$ that $[a \\cdot b] = 2$. This case was covered above in the solution for klas 4 and below. Now assume that $[a \\cdot b] - 1$ is not nilless. The difference is unequal to 1, so $[a \\cdot b] - 1$ is equal to 10, 100, 1000, ... | Netherlands | Dutch Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | [(4, 5), (8, 25), (2, 55), (5, 22), (11, 91), (13, 77), (25, 44)] | |
08a7 | Problem:
Un pilota di aquiloni ha disputato quest'anno un buon campionato, arrivando a podio 16 volte. In ogni gara il primo classificato conquista 10 punti, il secondo 8 e il terzo 5, mentre dal quarto posto in poi non vengono assegnati punti. Con quanti punteggi diversi può aver concluso il campionato?
(A) 153
(B) 8... | [
"Solution:\n\nLa risposta è $(\\mathbf{D})$. Dobbiamo calcolare tutte le possibili somme che si possono ottenere con 16 addendi scelti tra $5$, $8$ e $10$. Scriviamo dunque $n = 5a + 8b + 10c$, dove $a + b + c$ sono interi non negativi la cui somma è $16$. Innanzitutto ricaviamo $a = 16 - b - c$, da cui $n = 80 + 3... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | MCQ | D | |
0d8i | Let $ABCD$ be a quadrilateral inscribed a circle ($O$). Assume that $AB$ and $CD$ intersect at $E$, $AC$ and $BD$ intersect at $K$, and $O$ does not belong to the line $KE$. Let $G$ and $H$ be the midpoints of $AB$ and $CD$ respectively. Let ($I$) be the circumcircle of the triangle $GKH$. Let ($I$) and $(O)$ intersect... | [
"Denote $F$ as the intersection of $AD$, $BC$. Suppose that $FK$ intersects $AB$, $CD$ at $S$, $T$ respectively. Since the harmonic points of complete quadrilateral, we have $(E, T, D, C) = -1$. Then $ET \\cdot EH = ED \\cdot EC$. Similarly, $ES \\cdot EG = EA \\cdot EB$. But $EA \\cdot EB = ED \\cdot EC$ then $ES ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | English | proof only | null | |
00dr | Initially there is a positive integer $N$ written on the blackboard. The following operations are allowed:
* Replace the number by a positive multiple of itself.
* Replace the number by another which has the same digits in a different order (it is allowed for the new number to begin with 0). For example, if $2022$ is w... | [
"First let us observe that rearranging digits does not change its sum, hence it does not change the remainder upon division by $3$. It follows that if $N$ is divisible by $3$ then we will only get numbers divisible by $3$ and hence we will never get $1$.\n\nWe claim that if $N$ is not divisible by $3$ then it is po... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof and answer | All positive integers not divisible by 3 | |
060w | Problem:
Déterminer tous les entiers $x$ tels que $2^{x}+x^{2}+25$ est le cube d'un nombre premier. | [
"Solution:\nSoit $x$ un entier tel que $2^{x}+x^{2}+25$ est le cube d'un nombre premier $p$. Puisque $2^{x}=p^{3}-x^{2}-25$ est un entier, $x$ est positif. De plus $p^{3} \\geqslant 25$ donc $p>2$ et $p$ est impair.\n\nSi $x=0$, alors $2^{x}+x^{2}+25=26$ qui n'est pas un cube. Ainsi, $x \\geqslant 1$. On déduit que... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - Envoi 5 : Pot Pourri | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 6 | |
0go0 | Show that
$$
\sqrt{|AB_1|} + \sqrt{|BC_1|} + \sqrt{|CA_1|} \le \frac{3}{\sqrt{2}}
$$
if the incircle of a triangle $ABC$ touches the sides $BC$, $AC$ and $AB$ at the points $A_1$, $B_1$ and $C_1$, respectively. | [
"Letting $x = |AB_1|$, $y = |BC_1|$, $z = |CA_1|$, we have to show that\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$$\n\\frac{1}{\\sqrt{1+a^2}} + \\frac{1}{\\sqrt{1+b^2}} + \\frac{1}{\\sqrt{1+c^2}} \\le \\frac{3}{\\sqrt{2}}\n$$\nf... | Turkey | Team Selection Test for IMO 2009 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > C... | English | proof only | null | |
0irr | Problem:
The RSA Factoring Challenge, which ended in 2007, challenged computational mathematicians to factor extremely large numbers that were the product of two prime numbers. The largest number successfully factored in this challenge was RSA-640, which has 193 decimal digits and carried a prize of $\$ 20,000$. The n... | [
"Solution:\n\nAnswer: 212\n\nFor more information, see the Wikipedia entry at http://en.wikipedia.org/wiki/RSA_Factoring_Challenge."
] | United States | Harvard-MIT Mathematics Tournament | [
"Math Word Problems"
] | null | final answer only | 212 | |
0ceo | A polygon is tiled with a finite number of triangles whose sides all have an odd length.
a) Prove that, if the polygon is convex, then its perimeter is an integer of the same parity as the number of triangles in the tiling.
b) Does the conclusion still hold if the polygon is not convex? | [
"a) Let $K$ be the polygon under consideration. Since $K$ is convex, the tiling triangles fall into two classes: Those having all edges inside $K$, and those having at least one edge on the boundary of $K$. (If $K$ were not convex, there might also exist triangles having only parts of edges on the boundary of $K$, ... | Romania | Seventeenth Stars of Mathematics Competition | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | For convex polygons: the perimeter is an integer with the same parity as the number of triangles. For nonconvex polygons: No, the conclusion does not always hold (counterexample exists). | |
0gpt | Let $\phi(n)$ denote the number of positive integers less than $n$ that are relatively prime to $n$ where $n$ is a positive integer. Find all pairs $(m, n)$ of positive integers satisfying
$$
2^n + (n - \phi(n) - 1)! = n^m + 1.
$$ | [
"The answer is $(2, 2)$ and $(2, 4)$.\n\nFor $n = 1$, we have $2 + 1 = 2$ which yields a contradiction.\n\nIf $n$ is a prime number, then $\\phi(n) = n - 1$ and hence $2^n = n^m$. Therefore $m = n = 2$.\n\nIf $n = p^2$ where $p$ is a prime number, then $\\phi(n) = p^2 - p$ and we get $2^{p^2} + (p-1)! = p^{2m} + 1$... | Turkey | Team Selection Test for IMO | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / ... | English | proof and answer | (2, 2), (2, 4) | |
08fh | Problem:
Il polinomio $p(x)$ ha la seguente proprietà: per ogni terna di interi $a, b, c$ tali che $a+b+c=2022$ si ha che $p(a)+p(b)+p(c)=p(674)$. Si sa inoltre che $p(0)=-2696$. Quanto vale $p(2022)$ ?
(A) $-2696$
(B) $674$
(C) $5392$
(D) $8088$
(E) Non è possibile determinarlo con i dati forniti. | [
"Solution:\n\nLa risposta è $\\mathbf{(C)}$. Sostituendo $a=b=c=674$ (interi che effettivamente soddisfano $a+b+c=2022$) si ottiene $3p(674)=p(674)$, ovvero $p(674)=0$. Sostituendo allora $a=b=0$ e $c=2022$ otteniamo\n$$\n2p(0)+p(2022)=p(674)=0 \\Rightarrow p(2022)=-2p(0)=5392.\n$$"
] | Italy | Gara di Febbraio | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | MCQ | C | |
070y | Problem:
Some terms are deleted from an infinite arithmetic progression $1, x, y, \ldots$ of real numbers to leave an infinite geometric progression $1, a, b, \ldots$. Find all possible values of $a$. | [
"Solution:\n\nIf $a$ is negative, then the terms in the GP are alternately positive and negative, whereas either all terms in the AP from a certain point on are positive or all terms from a certain point on are negative. So $a$ cannot be negative. If $a$ is zero, then all terms in the GP except the first are zero, ... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | a is any positive integer | |
034m | Problem:
Find the largest possible number $A$ having the following property: if the numbers $1,2, \ldots, 1000$ are ordered in arbitrary way then there exist 50 consecutive numbers with sum not less than $A$. | [
"Solution:\nLet $x_{1}, x_{2}, \\ldots, x_{1000}$ be an arbitrary rearrangement of the numbers $1,2, \\ldots, 1000$. Set\n\n$$\nS_{1} = x_{1} + x_{2} + \\cdots + x_{50}, \\ldots, S_{20} = x_{951} + x_{952} + \\cdots + x_{1000}\n$$\nSince $S_{1} + \\cdots + S_{20} = 500500$, we have $S_{i} \\geq \\frac{500500}{20} =... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 25025 |
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