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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0g06 | Problem:
Sei $m$ eine natürliche Zahl. Auf der SMO-Wandtafel steht $2^{m}$ mal die Zahl $1$. In einem Schritt wählen wir zwei Zahlen $a$ und $b$ auf der Tafel und ersetzen sie beide jeweils durch $a+b$. Zeige, dass nach $m 2^{m-1}$ Schritten die Summe der Zahlen mindestens $4^{m}$ beträgt. | [
"Solution:\n\nSei $S$ die Summe der Zahlen nach $m 2^{m-1}$ Zügen. Die Idee ist, zuerst eine untere Schranke für das Produkt der $2^{m}$ Zahlen zu finden. Sei dazu $P_{i}$ das Produkt aller Zahlen nach dem $i$-ten Schritt und $P_{0}=1$. In jedem Schritt werden zwei Zahlen $a, b$ durch $a+b, a+b$ ersetzt. Deren Prod... | Switzerland | SMO - Finalrunde | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0dm2 | Find all non-negative real numbers $c \ge 0$ such that there exists a function $f : (0, +\infty) \to (0, +\infty)$ with the property
$$
f(y^2 f(x) + y + c) = x f(x + y^2)
$$
for all $x, y > 0$. | [
"**Solution.** (Solution of Yousif Bakheet, IMO 2025's team member)\n\nSAUDI ARABIAN IMO Booklet 2025\n---\n## Saudi Booklet 2025 — Page 49\nSolution of IMO Team selection tests\n$$ \\therefore \\text{يوجد و يحقق التساوي } f(x+\\frac{1}{x}) \\ge f(x) \\text{ ($x \\ge 0$)} \\\\ \\text{لما } ... | Saudi Arabia | Saudi Booklet | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | null | |
06eo | The function $f(x, y)$, defined on the set of all nonnegative integers, satisfies
(i) $f(0, y) = y + 1$,
(ii) $f(x + 1, 0) = f(x, 1)$, and
(iii) $f(x + 1, y + 1) = f(x, f(x + 1, y))$.
Find (a) $f(3, 2005)$, and (b) $f(4, 2005)$. | [
"(a) The answer is $f(3, 2005) = 2^{2008} - 3$.\nWe first label the equations as follows:\n$$\nf(0, y) = y + 1, \\qquad (1)\n$$\n$$\nf(x + 1, 0) = f(x, 1), \\qquad (2)\n$$\n$$\nf(x + 1, y + 1) = f(x, f(x + 1, y)). \\qquad (3)\n$$\nBy putting $y = 1$ in (1), we get $f(0, 1) = 2$. By putting $x = 0$ in (2), we get $f... | Hong Kong | IMO HK TST | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | a: 2^{2008} - 3; b: g(2008) - 3, where g(1) = 2 and g(n + 1) = 2^{g(n)} | |
02nv | What is the maximum area of the shadow cast by a unit cube? We understand “area of the shadow” of something as the area of its orthogonal projection in a given plane. | [
"Let $ABCD$ and $EFGH$ be two opposite faces, $AE$, $BF$, $CG$ and $DH$ being edges of the cube, and let $X'$ be the orthogonal projection of point $X$ onto the plane. Notice that $\\{A, G\\}$, $\\{B, H\\}$, $\\{C, E\\}$ and $\\{D, F\\}$ are pairs of opposite vertices. Suppose, without loss of generality, that $A'$... | Brazil | Brazilian Math Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | sqrt(3) | |
0hi2 | A square of size $n \times n$ is given. Some of its $1 \times 1$ cells are marked. It turned out that there is no convex quadrilateral with vertices at these marked points. For each natural number $n \ge 3$, find the largest value of $m$ for which this is possible.
A quadrilateral is called convex if both of its diagon... | [
"We mark two points in the corner cells of the left column, as well as all the points in some non-edge row. Then we have $n + 2$ marked points, none of which form a vertex of a convex quadrilateral (Fig. 6).\n\nWe will show by contradiction that it is not possible to mark more than $n + 2$ points. Suppose at least ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Quadrilaterals",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | n + 2 | |
0eqn | $\frac{3 \times 2016 + 13 \times 2016}{1008} =$
(A) 2 (B) 16 (C) 32 (D) $19 \times 2016$ (E) $6 + 13 \times 2016$ | [
"$\\frac{3 \\times 2016 + 13 \\times 2016}{1008} = \\frac{(3 + 13) \\times 2016}{1008} = (3 + 13) \\times \\frac{2016}{1008} = 16 \\times 2 = 32.$"
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | MCQ | C | |
0d93 | On the table, there are $1024$ marbles and two students $A, B$ alternatively take a positive number of marble(s). The student $A$ goes first, $B$ goes after that and so on. On the first move, $A$ takes $k$ marbles with $1<k<1024$. On the moves after that, $A$ and $B$ are not allowed to take more than $k$ marbles or $0$... | [
"After the first move, there are $1024-k$ marbles remaining and both students $A, B$ cannot take more than $k$ marbles during their turn.\n\nWe shall prove that if $k+1 \\mid 1024-k$ then the student $A$ will have the strategy to win the game.\n\nIndeed, let $1024-k = m(k+1)$ with $m \\in \\mathbb{Z}^{+}$. On each ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 4, 24, 40, 204 | |
0e41 | Problem:
Poišči vsa naravna števila $m$ in $n$, za katera je vsota največjega skupnega delitelja in najmanjšega skupnega večkratnika enaka 101. | [
"Solution:\n\nNaj bo $d$ največji skupni delitelj števil $m$ in $n$. Tedaj je $m = d m_{1}$ in $n = d n_{1}$, števili $m_{1}$ in $n_{1}$ pa sta si tuji. Najmanjši skupni večkratnik $m$ in $n$ je $d m_{1} n_{1}$. Velja\n$$\n101 = d + d m_{1} n_{1} = d\\left(1 + m_{1} n_{1}\\right)\n$$\nKer je $1 + m_{1} n_{1} \\geq ... | Slovenia | 55. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | Ordered pairs: (1, 100), (4, 25), (25, 4), (100, 1). | |
0bim | Let $n \ge 5$ be an integer. Prove that $n$ is prime if and only if for any representation of $n$ as a sum of four positive integers $n = a+b+c+d$, it is true that $ab \ne cd$. | [
"The statement of the problem is equivalent to saying that an integer $n \\ge 5$ is composite if and only if there exists a representation of $n$ as the sum of four positive integers such that $ab = cd$.\n\n($\\Leftrightarrow$) If $ab = cd$ and $n = a+b+c+d$, then\n$$\nan = a^2 + ab + ac + ad = a^2 + cd + ac + ad =... | Romania | 65th NMO Selection Tests for JBMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
095d | Problem:
Două cercuri au o coardă comună $AB$. Prin punctul $B$ se duce o dreaptă care intersectează cercurile în punctele $C$ şi $D$, astfel încât $B$ se află între $C$ şi $D$. Tangentele la cercuri, duse prin punctele $C$ şi $D$, se intersectează într-un punct $E$. Să se compare $AD \cdot AC$ cu $AB \cdot AE$. | [
"Solution:\n\nVom aplica proprietatea măsurii unghiului cu vârful pe cerc, cu o latură tangentă şi alta secantă a cercului şi măsura unghiului înscris în cerc.\nFie $m(\\angle BAD)=\\alpha \\Rightarrow m(\\operatorname{arc} BD)=2\\alpha$, atunci $m(\\angle CDE)=\\alpha$.\nFie $m(\\angle BAC)=\\beta \\Rightarrow m(\... | Moldova | 61-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | AD · AC = AB · AE | |
0g64 | 一無窮數列 $x_1, x_2, \dots$, 有 $x_1 = 1$, 且對任意正整數 $k$, 有
$$
x_{2k} = -x_k, \quad x_{2k-1} = (-1)^{k+1}x_k.
$$
試證: 對任意正整數 $n$, $x_1 + x_2 + \dots + x_n \ge 0$. | [
"定義 $S_n = \\sum_{i=1}^n x_i$。首先觀察到\n$$\nx_{4k-3} = -x_{4k-2} = x_{2k-1}\n$$\n$$\nx_{4k-1} = x_{4k} = -x_{2k} = x_k\n$$\n我們對 $k$ 做數學歸納法證明。對任意 $i \\le 4k$ 都有 $S_i \\ge 0$。當 $k=1$ 時，\n$x_1 = -x_2 = x_3 = x_4 = 1$，命題顯然成立。\n設已知對任意 $i \\le 4k$ 都有 $S_i \\ge 0$，則\n$$\nS_{4k+2} = S_{4k} + x_{4k+1} + x_{4k+2} = S_{4k} \\ge ... | Taiwan | 二〇一一數學奧林匹亞競賽第一階段選訓營,模擬競賽(二) | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
018m | Math competition is held in $8$ different levels of difficulty. The organizing committee has to prepare $5$ problems for each level. The same problem can be used for more than one level, but each two levels can have at most one common problem. What is the least number of problems that is sufficient for the organizers? | [
"$18$ problems are enough. The following table shows how to arrange problems for $8$ levels:\n\n| Level 1 | 1 | 2 | 3 | 4 | 5 |\n|---------|---|---|---|---|---|\n| Level 2 | 1 | 6 | 7 | 8 | 9 |\n| Level 3 | 2 | 6 | 10 | 11 | 12 |\n| Level 4 | 3 | 7 | 10 | 13 | 14 |\n| Level 5 | 4 | 8 | 11 | 13 | 15 |\n| Level 6 | 5... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 18 | |
0l42 | Problem:
Compute the unique real number $x<3$ such that
$$
\sqrt{(3-x)(4-x)}+\sqrt{(4-x)(6-x)}+\sqrt{(6-x)(3-x)}=x
$$ | [
"Solution:\nAnswer: $\\frac{23}{8}=2.875$\n\nLet $a=\\sqrt{3-x}$, $b=\\sqrt{4-x}$, $c=\\sqrt{6-x}$, so $x=ab+bc+ca$.\n\nThen\n$$(a+b)(a+c)=a^2+ab+bc+ca=(3-x)+x=3$$\nLikewise,\n$$(b+a)(b+c)=4$$\n$$(c+a)(c+b)=6$$\nBy multiplying these equations and taking the square root, we get that\n$$(a+b)(b+c)(c+a)=\\sqrt{72}=6\\... | United States | HMMT November | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | 23/8 | |
0jih | Problem:
Let $O_{1}$ and $O_{2}$ be concentric circles with radii $4$ and $6$, respectively. A chord $AB$ is drawn in $O_{1}$ with length $2$. Extend $AB$ to intersect $O_{2}$ in points $C$ and $D$. Find $CD$. | [
"Solution:\n\nLet $O$ be the common center of $O_{1}$ and $O_{2}$, and let $M$ be the midpoint of $AB$. Then $OM \\perp AB$, so by the Pythagorean Theorem, $OM = \\sqrt{4^{2} - 1^{2}} = \\sqrt{15}$. Thus $CD = 2CM = 2\\sqrt{6^{2} - 15} = 2\\sqrt{21}$."
] | United States | HMMT 2014 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2√21 | |
02xe | Problem:
No quadrilátero $ABCD$, temos:
$$
\angle DAB = \angle ABC = \angle BCD = 30^\circ, \quad AB = 4\ \mathrm{cm}, \quad BC = 2\sqrt{3}\ \mathrm{cm}
$$
a) Determine o valor do ângulo $\angle DCA$.
b) Determine o comprimento de $CD$.
c) Encontre a área do quadrilátero $ABCD$.
 | [
"Solution:\n\na) Como $\\frac{BC}{AB} = \\cos 30^\\circ$ e $\\angle ABC = 30^\\circ$, segue que $\\angle ACB = 90^\\circ$. Daí, $\\angle BAC = 60^\\circ$ e $\\frac{AC}{AB} = \\sen 30^\\circ = \\frac{1}{2}$, logo, $AC = \\frac{AB}{2} = 2\\ \\mathrm{cm}$. Consequentemente, $\\angle DAC = 30^\\circ$ e $\\angle DCA = 6... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | ∠DCA = 60°, CD = 1 cm, area = (3√3)/2 cm² | |
06py | Let $f: \mathbb{R} \rightarrow \mathbb{N}$ be a function which satisfies
$$
f\left(x+\frac{1}{f(y)}\right)=f\left(y+\frac{1}{f(x)}\right) \quad \text{ for all } x, y \in \mathbb{R}
$$
Prove that there is a positive integer which is not a value of $f$. | [
"Suppose that the statement is false and $f(\\mathbb{R})=\\mathbb{N}$. We prove several properties of the function $f$ in order to reach a contradiction.\n\nTo start with, observe that one can assume $f(0)=1$. Indeed, let $a \\in \\mathbb{R}$ be such that $f(a)=1$, and consider the function $g(x)=f(x+a)$. By substi... | IMO | 49th International Mathematical Olympiad Spain | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0jxl | Problem:
A box contains three balls, each of a different color. Every minute, Randall randomly draws a ball from the box, notes its color, and then returns it to the box. Consider the following two conditions:
(1) Some ball has been drawn at least three times (not necessarily consecutively).
(2) Every ball has been dra... | [
"Solution:\nAt any time, we describe the current state by the number of times each ball is drawn, sorted in nonincreasing order. For example, if the red ball has been drawn twice and green ball once, then the state would be $(2,1,0)$. Given state $S$, let $P_{S}$ be the probability that the state was achieved at so... | United States | HMMT November | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 13/27 | |
0k48 | Problem:
Triangle $\triangle P Q R$, with $P Q = P R = 5$ and $Q R = 6$, is inscribed in circle $\omega$. Compute the radius of the circle with center on $\overline{Q R}$ which is tangent to both $\omega$ and $\overline{P Q}$. | [
"Solution:\n\nDenote the second circle by $\\gamma$. Let $T$ and $r$ be the center and radius of $\\gamma$, respectively, and let $X$ and $H$ be the tangency points of $\\gamma$ with $\\omega$ and $\\overline{P Q}$, respectively. Let $O$ be the center of $\\omega$, and let $M$ be the midpoint of $\\overline{Q R}$. ... | United States | HMMT November 2018 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 20/9 | |
04ag | Let $P_1, P_2, \dots, P_{2n}$ be a permutation of the vertices of a regular $2n$-gon. Prove that every closed polygonal line that consists of segments
$$
\overline{P_1P_2}, \overline{P_2P_3}, \dots, \overline{P_{2n-1}P_{2n}}, \overline{P_{2n}P_1}
$$
contains at least one pair of parallel segments. | [
"Assign numbers $1, 2, \\ldots, 2n$ to the vertices of the observed $2n$-gon respectively. Let $a_k$ be the number assigned to the vertex $P_k$. Then $a_1, a_2, \\ldots, a_{2n}$ is a permutation of $1, 2, \\ldots, 2n$.\n\n\n\nSegments $\\overline{P_iP_{i+1}}$ and $\\overline{P_jP_{j+1}}$ fo... | Croatia | Hrvatska 2011 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
09kz | Are there ten positive integers such that no number in the set is divisible by another number, and the product of any two numbers is divisible by the remaining numbers in the set? | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | Yes | |
08c2 | Problem:
Un triangolo equilatero è diviso in 9 triangolini come in figura, e su ogni triangolino è inizialmente scritto il numero 0. Marco, per passare il tempo, fa il seguente gioco: ad ogni mossa sceglie 2 triangolini con un lato in comune e somma o sottrae 1 ad entrambi i numeri scritti su questi triangolini (si in... | [
"Solution:\n\nCome in figura, numeriamo i triangolini da $T_{1}$ a $T_{9}$ e coloriamo di rosso i triangoli $T_{1}, T_{2}, T_{4}, T_{5}, T_{7}, T_{9}$, lasciando in bianco gli altri 3. Detto $m_{i}$ il numero scritto nel triangolo $T_{i}$ in un certo momento, dimostriamo che la somma sui triangolini bianchi è ugual... | Italy | Gara di Febbraio | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 0 or 2 | |
0h9t | Find all triple of numbers $(x, y, p)$, where $x, y$ are positive integers and $p$ is a prime number, which satisfy the condition:
$$
y(x^2 + p) - x(y^2 + p) = p.
$$ | [
"Factor the left-hand side of the equation:\n$$\n(yx^2 - xy^2) + (yp - xp) = p \\Rightarrow yx(x - y) - p(x - y) = p \\Rightarrow (yx - p)(x - y) = p.\n$$\nThe last equation is possible in several cases.\n\n**Case 1.** $yx - p = 1$, $x - y = p$. Then, we get a quadratic equation: $x = y + p$\n$$\ny(y + p) - p = 1 \... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | All triples (x, y, p) with p prime given by (x, y) = (p + 1, 1) for any prime p, and the exceptional triple (3, 2, 3). | |
072n | Let $ABCD$ be a parallelogram. A variable line $l$ through the point $A$ intersects the rays $BC$ and $DC$ at $X$ and $Y$ respectively. Let $K$ and $L$ be the centres of the excircles of triangles $ABX$ and $ADY$, touching the sides $BX$ and $DY$ respectively. Prove that the size of $\angle KCL$ does not depend on the ... | [
"Let $\\angle DAX = 2\\gamma$ and $\\angle YAP = 2\\alpha$. Now $K$ being the ex-centre of triangle $ABX$, it lies on the bisectors of $\\angle XAB$ and $\\angle XBP$. Note that $\\angle ABX = 180^\\circ - 2\\alpha - 2\\gamma$ and hence $\\angle XBK = (2\\alpha + 2\\gamma)/2 = \\alpha + \\gamma$. Thus the angles of... | India | Indija TS 2006 | [
"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"
] | null | proof only | null | |
0g6d | 設 $P, Q$ 為三角形 $ABC$ 內兩點，滿足 $\angle BAP = \angle CAQ$, $\angle ACP = \angle BCQ$，且 $\angle CBP = \angle ABQ$。設 $Q_1, Q_2, Q_3$ 分別為 $Q$ 對於 $BC, CA, AB$ 的對稱點。令 $D$ 為 $PQ_1$ 與 $BC$ 之交點，令 $E$ 為 $PQ_2$ 與 $CA$ 之交點，令 $F$ 為 $PQ_3$ 與 $AB$ 交點。試證：$AD, BE, CF$ 三線共點。 | [
"(1) 首先令 $\\angle ACP = \\angle BCQ = \\theta$，則 $\\angle ACQ = \\angle BCP = \\angle C - \\theta$。又因為 $Q_1$ 為 $Q$ 關於邊 $C$ 的對稱點，故 $\\angle QCB = \\angle Q_1CB$ 且 $CQ = CQ_1$。故，$\\angle PCQ_1 = \\angle PCB + \\angle Q_1CB = \\angle ACQ + \\angle BCQ = \\angle C$。\n\n(2) 接著令 $P$ 在 $BC, CA, AB$ 的垂足分別為 $M_1, M_2, M_3$，... | Taiwan | 二〇一二數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
043k | As shown in Fig. 11.1, in a plane rectangular coordinate system $xOy$, the left and right foci of ellipse $\Gamma: \frac{x^2}{2} + y^2 = 1$ are $F_1, F_2$, respectively. Let $P$ be a point on $\Gamma$ in the first quadrant and the extensions of $PF_1, PF_2$ intersect $\Gamma$ at points $Q_1(x_1, y_1), Q_2(x_2, y_2)$, r... | [
"As shown in Fig. 11.1, we find $F_1(-1, 0), Q_2(1, 0)$.\nDenote $P(x_0, y_0)$. By the condition, it follows that $x_0, y_0 > 0$, $y_1 < 0$, $y_2 < 0$.\n\nFig. 11.1\n\nThe equation of line $PF_1$ is $x = \\frac{(x_0 + 1)y}{y_0} - 1$. Substituting it into $\\frac{x^2}{2} + y^2 = 1$ and organ... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 2*sqrt(2)/3 | |
0a9c | Problem:
The integers $1, 2, 3, 4$ and $5$ are written on a blackboard. It is allowed to wipe out two integers $a$ and $b$ and replace them with $a+b$ and $a b$. Is it possible, by repeating this procedure, to reach a situation where three of the five integers on the blackboard are $2009$? | [
"Solution:\n\nThe answer is no. First notice that in each move two integers will be replaced with two greater integers (except in the case where the number $1$ is wiped out). Notice also that from the start there are three odd integers. If one chooses to replace two odd integers on the blackboard, the number of odd... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 23 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | no | |
04mm | In how many ways can we fill a $2018 \times 2018$ board with positive integers so that the sum of numbers in any three consecutive cells in the same row or the same column equals $5$? | [] | Croatia | Croatia_2018 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 21 | |
0523 | The bases of trapezoid $ABCD$ are $AB$ and $CD$, and the intersection point of its diagonals is $P$. Prove that if $\frac{|PA|}{|PD|} = \frac{|PB|}{|PC|}$ then the trapezoid is isosceles. | [
"\nFigure 15\n\nBy assumptions, $\\frac{|PA|}{|PB|} = \\frac{|PD|}{|PC|}$. As the bases $AB$ and $CD$ are parallel, we have also $\\frac{|PA|}{|PB|} = \\frac{|PC|}{|PD|}$ (Fig. 15). Hence $|PC| = |PD|$. Similarity of triangles $APD$ and $BPC$ implies $\\frac{|AD|}{|BC|} = \\frac{|PD|}{|PC|}... | Estonia | Final Round of National Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
00xv | Problem:
Do there exist positive integers $a > b > 1$ such that for each positive integer $k$ there exists a positive integer $n$ for which $a n + b$ is a $k$th power of a positive integer? | [
"Solution:\n\nLet $a = 6$, $b = 3$ and denote $x_{n} = a n + b$. Then we have $x_{l} \\cdot x_{m} = x_{6 l m + 3(l + m) + 1}$ for any natural numbers $l$ and $m$. Thus, any powers of the numbers $x_{n}$ belong to the same sequence."
] | Baltic Way | Baltic Way 1993 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a=6, b=3 | |
09gt | Let $\triangle ABC$ be a triangle. The points $D$ and $E$ lie on the sides $AC$ and $AB$, respectively. Bisector of angle $B$ intersect side $AC$ at a point $P$ and bisector of $\angle ADE$ intersect side $AB$ at a point $Q$. If $BP$ perpendicular to $DQ$, then prove that $PQ \parallel EC$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates"
] | English | proof only | null | |
0842 | Problem:
Sono dati quattro numeri naturali tali che, comunque se ne prendano tre distinti e si sommino, si ottiene un numero maggiore o uguale a $24$. Quante delle seguenti affermazioni sono sicuramente vere?
I - ciascuno dei quattro numeri è maggiore o uguale a $8$;
II - due dei numeri dati hanno somma maggiore o u... | [] | Italy | Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO BIENNIO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | C | |
0619 | Problem:
Die natürlichen Zahlen von $1$ bis $n^{2}$ werden zufällig auf die Felder eines $n \times n$-Quadrats verteilt ($n \geq 2$). Für jedes Paar von Zahlen innerhalb einer Reihe bzw. einer Spalte dividieren wir die größere durch die kleinere Zahl. Der kleinste dieser $n^{2}(n-1)$ Quotienten werde als Charakteristi... | [] | Germany | Auswahlwettbewerb zur IMO 2000 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | (n+1)/n | |
0cuz | Let $ABCD$ be an isosceles trapezoid where $BC \parallel AD$ and $AB \parallel CD$. A circle $\omega$ passes through $B$ and $C$ and meets again the segments $AB$ and $BD$ at $X$ and $Y$, respectively. The tangent to $\omega$ at $C$ meets the ray $AD$ at $Z$. Prove that the points $X$, $Y$, and $Z$ are collinear. | [
"Notice that $CYDZ$ is cyclic.\n\nSince $BC \\parallel AD$, and the line $ZC$ is tangent to the circle $\\omega$, we have $\\angle ADB = \\angle YBC = \\angle YCZ$. Therefore, $\\angle YDZ + \\angle YCZ = 180^\\circ$, that is, the quadrilateral $CYDZ$ is cyclic (see Fig. 11).\n\n\n\nThus, $... | Russia | XLIII Russian mathematical olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
0fng | Let $p$ be an odd positive integer. Find all values of the natural numbers $n \ge 2$ for which holds
$$
\sum_{i=1}^{n} \prod_{j \neq i} (x_i - x_j)^p \geq 0,
$$
where $x_1, x_2, \dots, x_n$ are any real numbers. | [
"Denote by\n$$\nf_n(x_1, x_2, \\dots, x_n) = \\sum_{i=1}^{n} \\prod_{j \\neq i} (x_i - x_j)^p = (x_1 - x_2)^p (x_1 - x_3)^p \\dots (x_1 - x_n)^p\n$$\n$$\n+(x_2-x_1)^p(x_2-x_3)^p\\dots(x_2-x_n)^p+\\dots+(x_n-x_1)^p(x_n-x_2)^p\\dots(x_n-x_{n-1})^p\n$$\nSince $f_2(x_1, x_2) = (x_1 - x_2)^p + (x_2 - x_1)^p = 0$ for all... | Spain | International Mathematical Arhimede Contest | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | Spanish | proof and answer | n = 2, 3, 5 | |
06ol | Let $S$ be a finite set of points in the plane such that no three of them are on a line. For each convex polygon $P$ whose vertices are in $S$, let $a(P)$ be the number of vertices of $P$, and let $b(P)$ be the number of points of $S$ which are outside $P$. Prove that for every real number $x$
$$
\sum_{P} x^{a(P)}(1-x)... | [
"For each convex polygon $P$ whose vertices are in $S$, let $c(P)$ be the number of points of $S$ which are inside $P$, so that $a(P)+b(P)+c(P)=n$, the total number of points in $S$. Denoting $1-x$ by $y$,\n$$\n\\sum_{P} x^{a(P)} y^{b(P)}=\\sum_{P} x^{a(P)} y^{b(P)}(x+y)^{c(P)}=\\sum_{P} \\sum_{i=0}^{c(P)}\\binom{c... | IMO | IMO 2006 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Geometry > Plane Geometry > Combinatorial Geometry > Conve... | English | proof only | null | |
0f1i | Problem:
Given a $7 \times 7$ square subdivided into $49$ unit squares, mark the center of $n$ unit squares, so that no four marks form a rectangle with sides parallel to the square. What is the largest $n$ for which this is possible? What about a $13 \times 13$ square? | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 7x7: 21; 13x13: 52 | |
0kex | Problem:
Alice writes $1001$ letters on a blackboard, each one chosen independently and uniformly at random from the set $S=\{a, b, c\}$. A move consists of erasing two distinct letters from the board and replacing them with the third letter in $S$. What is the probability that Alice can perform a sequence of moves wh... | [
"Solution:\n\nLet $n_{a}$, $n_{b}$, and $n_{c}$ be the number of $a$'s, $b$'s, and $c$'s on the board, respectively. The key observation is that each move always changes the parity of all three of $n_{a}$, $n_{b}$, and $n_{c}$. Since the final configuration must have $n_{a}$, $n_{b}$, and $n_{c}$ equal to $1,0,0$ i... | United States | HMMT February 2020 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 3/4 - 1/(4*3^{999}) | |
0c7b | Let $ABC$ a triangle, $I$ the incenter, $D$ the contact point of the incircle with the side $BC$ and $E$ the foot of the bisector of the angle $A$. If $M$ is the midpoint of the arc $BC$ which contains the point $A$ of the circumcircle of the triangle $ABC$ and $\{F\} = DI \cap AM$, prove that $MI$ passes through the m... | [
"The bisector of the angle $\\hat{A}$ passes through the midpoint of the arc $BC$ which does not contain the point $A$. Denote this point by $S$. $MS$ is the perpendicular bisector of $[BC]$, so $MS \\parallel ID$ (both lines are perpendicular on $BC$).\n\nAlso, $\\angle ASC = \\angle ABC$ and $\\angle SAC = \\angl... | Romania | 70th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circ... | English | proof only | null | |
0crj | Дан выпуклый 7-угольник. Выбираются четыре произвольных его угла и вычисляются их синусы, от остальных трёх углов вычисляются косинусы. Оказалось, что сумма таких семи чисел не зависит от изначального выбора четырёх углов. Докажите, что у этого 7-угольника найдутся четыре равных угла. | [
"Рассмотрим одну из сумм из условия. Затем переставим в ней аргументы одного синуса и одного косинуса (назовём эти аргументы $\\alpha$ и $\\beta$, соответственно); сумма при этом изменится на\n\n$$(\\sin \\beta + \\cos \\alpha) - (\\sin \\alpha + \\cos \\beta) = \\sqrt{2}(\\sin(\\beta - \\pi/4) - \\sin(\\alpha - \\... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
06gm | Let $x$, $y$, $z$, $u$, $v$ and $w$ be integers satisfying $x^2 + y^2 = u^2$, $x^2 + z^2 = v^2$ and $y^2 + z^2 = w^2$. Find an integer $\ell$ so that $518000 < \ell < 518518$ and $\ell$ divides $xyzuvw$. | [
"The answer is $518400$.\n\nWe claim that $3^4 \\cdot 4^4 \\cdot 5^2$ divides $xyzuvw$.\n\nFirstly, observe that $3$ divides $x$ or $3$ divides $y$, since otherwise\n$$\nu^2 = x^2 + y^2 \\equiv 1 + 1 = 2 \\pmod{3},$$\nwhich is impossible. Similarly, $3$ divides $x$ or $z$, and $3$ divides $y$ or $z$. Thus, two of $... | Hong Kong | Year 2012 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | 518400 | |
03ne | Problem:
Let $a$ and $b$ be positive integers such that $a + b^{3}$ is divisible by $a^{2} + 3 a b + 3 b^{2} - 1$. Prove that $a^{2} + 3 a b + 3 b^{2} - 1$ is divisible by the cube of an integer greater than 1. | [
"Solution:\nLet $Z = a^{2} + 3 a b + 3 b^{2} - 1$. By assumption, there is a positive integer $c$ such that $c Z = a + b^{3}$. Noticing the resemblance between the first three terms of $Z$ and those of the expansion of $(a + b)^{3}$, we are led to\n$$\n(a + b)^{3} = a(a^{2} + 3 a b + 3 b^{2}) + b^{3} = a(Z + 1) + b... | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0blo | Prove that the set $\{\sqrt{1}, \sqrt{2}, \sqrt{3}, \dots, \sqrt{2015}\}$ does not contain a non-constant arithmetic sequence of length 45. | [
"If $m, n, p \\in \\mathbb{N}^*$ and $\\sqrt{m}, \\sqrt{n}, \\sqrt{p}$ are in arithmetic progression, then\n$$\np + m + 2\\sqrt{pm} = 4n,\n$$\nhence $\\sqrt{pm}$ is a rational number. It follows that $m = a^2d$, $p = c^2d$, $n = b^2d$, where $a, b, c, d \\in \\mathbb{N}$ and $a + c = 2b$.\n\nTherefore, if one could... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
06zp | Problem:
Let $a_{n}, b_{n}$ be two sequences of integers such that:
(1) $a_{0}=0$, $b_{0}=8$;
(2) $a_{n+2}=2 a_{n+1}-a_{n}+2$, $b_{n+2}=2 b_{n+1}-b_{n}$,
(3) $a_{n}^{2}+b_{n}^{2}$ is a square for $n>0$.
Find at least two possible values for $\left(a_{1992}, b_{1992}\right)$. | [
"Solution:\n$a_{n}$ satisfies a standard linear recurrence relation with general solution $a_{n}=n^{2}+A n+k$. But $a_{0}=0$, so $k=0$. Hence $a_{n}=n^{2}+A n$. If you are not familiar with the general solution, then you can guess this solution and prove it by induction.\n\nSimilarly, $b_{n}=B n+8$. Hence $a_{n}^{2... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (3976032, 7976) and (3960096, -7960) | |
0flc | Problem:
Sea $p$ un número primo y $A$ un subconjunto infinito de los números naturales. Sea $f_{A}(n)$ el número de soluciones distintas de la ecuación $x_{1}+x_{2}+\cdots+x_{p}=n$, con $x_{1}, x_{2}, \ldots, x_{p} \in A$. ¿Existe algún número natural $N$ tal que $f_{A}(n)$ sea constante para todo $n>N$ ? | [
"Solution:\n\nPara demostrar el enunciado procederemos por contradicción. Supongamos que existe un número $N$ para el que se cumpla la propiedad anterior. Como el conjunto $A$ es infinito, tomemos $a \\in A$ mayor que $N$. Vamos a estudiar el valor de $f_{A}(p a)$ y $f_{A}(p a+1)$. Por hipótesis, se cumple que $f_{... | Spain | XLVI Olimpiada Matemática Española Fase nacional | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | No | |
0lgo | Problem:
Let $P(x)$ be a nonconstant polynomial of degree $n$ with rational coefficients which can not be presented as a product of two nonconstant polynomials with rational coefficients. Prove that the number of polynomials $Q(x)$ of degree less than $n$ with rational coefficients such that $P(x)$ divides $P(Q(x))$
... | [
"Solution:\n\nIt is known that an irreducible polynomial $P(x)$ of degree $n$ with rational coefficients has $n$ different complex roots which we denote by $\\alpha_{1}, \\alpha_{2}, \\ldots, \\alpha_{n}$.\n\na) If $P(x)$ divides $P(Q(x))$, then $Q(\\alpha_{k})$ is also a root of $P(x)$ for each $k \\leq n$. It fol... | Zhautykov Olympiad | Zhautykov Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Abstract Algebra > Field Theory"
] | null | proof only | null | |
041b | Given an ellipse equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ($a > b > 0$) in a plane rectangular coordinate system $xOy$, let $A_1, A_2, F_1, F_2$ be its left and right end-points, left and right focuses, respectively, and $P$ be any point on the ellipse different from $A_1, A_2$. Suppose there are points $Q, R$ ... | [
"Let $c = \\sqrt{a^2 - b^2}$. Then $A_1(-a, 0), A_2(a, 0), F_1(-c, 0), F_2(c, 0)$.\n\nDenote $P(x_0, y_0)$, $Q(x_1, y_1)$, $R(x_2, y_2)$, where $\\frac{x_0^2}{a^2} + \\frac{y_0^2}{b^2} = 1$, $y_0 \\neq 0$.\n\nFrom $QA_1 \\perp PA_1, QA_2 \\perp PA_2$, we have\n$$\n\\overrightarrow{A_1Q} \\cdot \\overrightarrow{A_1P... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof and answer | If P has y-coordinate y0, then |QR| = b^2 / |y0| ≥ b, with equality if and only if P = (0, ± b). | |
06eu | Let $a_1, a_2, a_3, \dots$ be a sequence of positive numbers. If there exists a positive number $M$ such that for every $n = 1, 2, 3, \dots$,
$$
a_1^2 + a_2^2 + \dots + a_n^2 < M a_{n+1}^2,
$$
then prove that there exists a positive number $M'$ such that for every $n = 1, 2, 3, \dots$,
$$
a_1 + a_2 + \dots + a_n < M' a... | [
"We say that a sequence $\\{b_n\\}$ of positive real numbers is *good* if there exists a positive constant $N$ such that\n$$\nb_1 + b_2 + \\cdots + b_n < N b_{n+1} \\quad (1)\n$$\nfor any $n \\in \\mathbb{Z}^+$.\nWe first prove the following equivalent conditions for a sequence to be good.\n\n**Claim.** A sequence ... | Hong Kong | CHKMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0j7h | Problem:
Let $p$ be the answer to this question. If a point is chosen uniformly at random from the square bounded by $x=0$, $x=1$, $y=0$, and $y=1$, what is the probability that at least one of its coordinates is greater than $p$? | [
"Solution:\n\nThe probability that a randomly chosen point has both coordinates less than $p$ is $p^{2}$, so the probability that at least one of its coordinates is greater than $p$ is $1 - p^{2}$. Since $p$ is the answer to this question, we have $1 - p^{2} = p$, and the only solution of $p$ in the interval $[0,1]... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | (√5 - 1)/2 | |
0fii | Problem:
Sobre un tablero en forma de triángulo equilátero con un número par de filas $n$ (tal como se indica en la figura), se juega un solitario.
Sobre cada casilla se coloca una ficha. Cada ficha es blanca por un lado y negra por el otro. Inicialmente, sólo una ficha, que está situada en un vértice, tiene la cara n... | [
"Solution:\n\nEn el tablero, hay casillas de tres tipos: vértice, lado, o interiores. Cada una de ellas tiene, respectivamente, dos, cuatro o seis casillas vecinas.\nSi pudiéramos retirar todas las fichas del tablero, habría un momento en que quedaría sobre él una única ficha negra. Esa ficha era inicialmente blanc... | Spain | Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | No | |
08x5 | Let $n$ be an integer greater than or equal to $2$. Find the smallest positive integer $m$ for which there exists a sequence $a_1, a_2, \dots, a_n$ of positive integers satisfying the following two conditions:
* $a_1 < a_2 < \dots < a_n = m$.
* All of the $n-1$ numbers $\frac{a_1^2 + a_2^2}{2}, \dots, \frac{a_{n-1}^2 +... | [
"We will show that smallest possible value for $m$ is $2n^2 - 1$.\nIf we let $a_k = 2k^2 - 1$ for $k = 1, 2, \\dots, n$, then we see that\n$$\n\\frac{a_k^2 + a_{k+1}^2}{2} = \\frac{(2k^2 - 1)^2 + (2(k+1)^2 - 1)^2}{2} = (2k^2 + 2k + 1)^2\n$$\nare all complete squares for each $k = 1, 2, \\dots, n-1$, and we have $m ... | Japan | Japan 2013 Final Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 2n^2 - 1 | |
01o2 | Six teams take part in a football tournament. Each team plays exactly one game with any other team. A team receives 3 points for a win, 1 point for a draw, and 0 point for a loss. After the tournament is over, the teams have 10, 9, 6, 6, 4, and 2 points.
a) Prove that the team taking the second place (i.e. having 9 po... | [
"The following tables contain all possible results of the tournament.\n\n| Wins | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |\n|------|---|---|---|---|---|---|---|---|---|---|\n| Draws | 0 | 1 | 2 | 3 | 4 | 5 | 0 | 1 | 2 | 3 |\n| Losses | 5 | 4 | 3 | 2 | 1 | 0 | 4 | 3 | 2 | 1 |\n| Points | 0 | 1 | 2 | 3 | 4 | 5 | 3 | 4... | Belarus | Belorusija 2012 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | a) The second-place team did not lose to the first-place team. b) No; the result is not uniquely determined. | |
05md | Problem:
Soit $n \geqslant 3$ un nombre entier et $a_{1}, a_{2}, \ldots, a_{n}$ des nombres réels.
a. On suppose que $a_{i}<\max \left(a_{i-1}, a_{i+1}\right)$ pour tout $i \in\{2,3, \ldots, n-1\}$. Montrer que $a_{i}<\max \left(a_{1}, a_{n}\right)$ pour tout $i \in\{2,3, \ldots, n-1\}$.
b. On suppose que $a_{i} \le... | [
"Solution:\n\na. Soit $i \\in\\{1,2, \\ldots, n\\}$ un entier tel que $a_{i}=\\max \\left(a_{1}, \\ldots, a_{n}\\right)$. L'énoncé indique que $i \\notin\\{2,3, \\ldots, n-1\\}$. Cela montre à la fois que $i \\in\\{1, n\\}$, donc que $a_{i}=\\max \\left(a_{1}, a_{n}\\right)$, et que $a_{k}<a_{i}$ si $k \\in\\{2,3, ... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES - ENVOI No. 3 | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Part a: The inequality holds; all interior terms are less than the larger of the two endpoints (the maximum occurs at an endpoint). Part b: No in general; it is true for three terms but false for four or more terms (for example, set both endpoints to one and all interior terms to two). | |
0ibj | Suppose $a_1, \dots, a_n$ are integers whose greatest common divisor is $1$. Let $S$ be a set of integers with the following properties.
a. For $i = 1, \dots, n$, $a_i \in S$.
b. For $i, j = 1, \dots, n$ (not necessarily distinct), $a_i - a_j \in S$.
c. For any integers $x, y \in S$, if $x + y \in S$, then $x - y \i... | [
"We may as well assume that none of the $a_i$ is equal to $0$. We start with the following observations.\n\nd. $0 = a_1 - a_1 \\in S$ by (b).\n\ne. $-s = 0 - s \\in S$ whenever $s \\in S$, by (a) and (d).\n\nf. If $x, y \\in S$ and $x - y \\in S$, then $x + y \\in S$ by (c) and (e).\n\nBy (f) plus strong induction ... | United States | USA IMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
0l16 | A bee is moving in three-dimensional space. A fair six-sided die with faces labeled $A^+$, $A^-$, $B^+$, $B^-$, $C^+$, and $C^-$ is rolled. Suppose the bee occupies the point $(a, b, c)$. If the die shows $A^+$, then the bee moves to the point $(a+1, b, c)$, and if the die shows $A^-$, then the bee moves to the point $... | [
"**Answer (B):** Without loss of generality, assume that the first roll is $A^+$. In order for the bee to traverse four distinct edges of a cube, the second roll cannot be $A^+$ or $A^-$, so there are $4$ rolls ($B^+$, $B^-$, $C^+$, and $C^-$) that are allowed at this stage. Each new roll must represent a perpendic... | United States | AMC 10 A | [
"Geometry > Solid Geometry > 3D Shapes"
] | null | MCQ | B | |
0hfp | The inscribed circle of triangle $ABC$ is tangent to its sides $BC$, $AC$ and $AB$ at points $K$, $L$ and $M$ correspondently. Let $P$ be the point of intersection of the bisector of $\angle BCA$ with the line $MK$. Prove that $AP \parallel LK$. | [
"Denote the inscribed circle by $k$, its center by $S$ (fig. 19), and its angles by $\\alpha, \\beta, \\gamma$. From symmetry, $KL \\perp CP$ and $\\angle LPC = \\angle KPC$. Then with simple calculations we get $\\angle MKB = 90^\\circ - \\frac{1}{2}\\beta$ and $\\angle LKC = 90^\\circ - \\frac{1}{2}\\gamma \\Righ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0474 | There are 8 cards on the table numbered from 1 to 8. Two players, $A$ and $B$, play the following game. In each round:
* Player $A$ selects two cards from the table
* Player $B$, after seeing the two selected cards, chooses one to keep and discards the other
The game consists of four rounds with the restriction that:... | [
"*Proof.* The maximum achievable value of $N$ is $17$.\n\nLet us denote the numbers on the two cards selected by Player $A$ in the $k$-th round as $a_k$ and $b_k$, where $a_k < b_k$. The number selected by Player $B$ is denoted as $c_k$, and the discarded number as $d_k$.\n\nLet $T = d_1 + d_2 + d_3 + d_4$. Then we... | China | 2024 CGMO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 17 | |
07bj | Point $I_b$ is the $B$-excenter of triangle $ABC$. If we denote by $M$ the midpoint of arc $BC$ of the circumcircle of triangle $ABC$ (the one that does not contain vertex $A$), and $MI_b$ intersects the circumcircle of triangle $ABC$ at $T$, prove that $TI_b^2 = TB \times TC$. | [
"Firstly, we prove two lemmas.\n\n**Lemma 1.** In each triangle $ABC$ with altitude $AH$ and circumradius $R$, we have $AC \\cdot AB = 2R \\cdot AH$.\n*Proof.* The proof is easy. □\n\n**Lemma 2.** In triangle $ABC$, let $N$ be the second intersection point of $I_bI_c$ and $\\omega$ (other than $A$), where $I_b$ and... | Iran | Iranian 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 > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles ... | null | proof only | null | |
0ci0 | Let $n$ be a positive integer. We say that a $n \times n$ table is *special* if:
* each cell of the table contains a 2-digit odd positive integer;
* the numbers of the table are pairwise distinct;
* the products of the numbers of each line and the products of the numbers of each column are perfect squares.
Prove that t... | [
"Since $11^2 > 100$, the table cannot contain numbers divisible by squares of prime numbers $p$, with $p \\ge 7$.\nSuppose that there exists a number of the table (situated on line $\\ell$ and column $c$) divisible by a prime $p \\ge 17$. Then there exists one more number on line $\\ell$ (and column $c' \\ne c$) di... | Romania | 74th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 4 | |
03ch | Let $n$ be positive integer and $f(x)$ be a polynomial of degree $n$ with $n$ distinct real positive roots. Are there positive integer $k \ge 2$ and a real polynomial $g(x)$ such that
$$
x(x+1)(x+2)(x+4)f(x) + 1 = (g(x))^k?
$$ | [
"Let $\\alpha_1 < \\alpha_2 < \\dots < \\alpha_n$ be the roots of $f(x)$. Assume that\n$$\nx(x+1)(x+2)(x+4)f(x) + a = g^k(x).\n$$\nNote that $a = b^k = g^k(0)$.\nIf $k \\ge 3$ is odd then the polynomial $g^k(x)-b^k$ has $n+4$ distinct real roots which will be also roots of $g(x)-b$. However, the degree of $g(x)-b$ ... | Bulgaria | BULGARIAN NATIONAL MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | No | |
04x9 | Given the triangle $ABC$, let $k$ be the excircle at the side $BC$. Choose any line $p$ parallel to $BC$ intersecting line segments $AB$ and $AC$ at points $D$ and $E$. Denote by $l$ the incircle of the triangle $ADE$. The tangents from $D$ and $E$ to the circle $k$ not passing through $A$ intersect at $P$. The tangent... | [
"Let $BC$ touch $k$ in $T_k$ and $DE$ touch $l$ in $T_l$. We shall prove the required fixed point is $T_k$.\n\nFirst we show the points $T_k$, $T_l$ and $P$ are collinear. Denote by $U$ and $V$ the points where $EP$ and $DP$ touch $k$, and $M$ and $N$ the points where $EP$ and $DP$ intersect $BC$. Let $T_1$ and $T_... | Czech-Polish-Slovak Mathematical Match | Cesko-Slovacko-Poljsko | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | English | proof only | null | |
02se | Problem:
A figura abaixo é composta por um quadrado e um pentágono regular.
Calcule a soma dos ângulos $a^{\circ}$ e $b^{\circ}$.
Fatos que ajudam. (Você pode usá-los!). A soma dos ângulos internos de um triângulo é sempre igual a $180^{\circ}$. Além disso, a soma dos ângulos internos de um qu... | [
"Solution:\nObserve o triângulo $APM$ na seguinte figura:\n\n\n\nComo $A\\hat{M}P = (180 - a)^{\\circ}$, e a soma dos ângulos internos de um triângulo vale $180^{\\circ}$, então $A\\hat{P}M = (a - 90)^{\\circ}$. Aplicando o mesmo argumento ao triângulo $BQN$, obtemos que $B\\hat{Q}N = (b - ... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 324° | |
0hlj | Problem:
Given three points $A$, $B$, $C$ known to lie on a circle, prove that one can reconstruct the original circle with a straightedge and compass. | [
"Solution:\n\nHere is a suitable procedure: First draw circles centered at $A$ and $B$ with radius $AB$ and join their two points of intersection to create the perpendicular bisector of $AB$. This line contains all points that have the same distance from $A$ and $B$, so the center of a circle through $A$ and $B$ li... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0ilj | Problem:
A cube of edge length $s > 0$ has the property that its surface area is equal to the sum of its volume and five times its edge length. Compute all possible values of $s$. | [] | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Surface Area",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 1, 5 | |
0flv | Problem:
Sean $x$ y $n$ enteros tales que $1 \leq x < n$. Disponemos de $x+1$ cajas distintas y $n-x$ bolas idénticas. Llamamos $f(n, x)$ al número de maneras que hay de distribuir las $n-x$ bolas en las $x+1$ cajas. Sea $p$ número primo, encontrar los enteros $n$ mayores que 1 para los que se verifica que el número p... | [
"Solution:\n\nClaramente $f(n, x)$ es el número de combinaciones con repetición de $x+1$ elementos tomados de $n-x$ en $n-x$. Es decir,\n$$\nf(n, x) = CR(x+1, n-x) = \\binom{(x+1)+(n-x)-1}{n-x} = \\binom{n}{x}\n$$\nVamos a probar que los $n$ buscados son todos los de la forma $p^{a}$ con $a$ entero positivo. Sea $m... | Spain | 48 aME | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | n equals p to the a for some positive integer a | |
01ov | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that
$$
f(x + y) + y \le f(f(f(x)))
$$
for all $x, y \in \mathbb{R}$. | [
"Answer: $f(x) = a - x$ for arbitrary real constant $a$.\n\nFirst set $y = 0$ in the initial inequality\n$$\nf(x + y) + y \\le f(f(f(x))) \\quad (*)\n$$\nthus obtaining\n$$\nf(x) \\leq f(f(f(x))) \\quad \\forall x \\in \\mathbb{R}. \\qquad (1)\n$$\nFurther, set $y = f(f(x)) - x$ in $(*)$. Then\n$$\nf(f(x)) \\leq x ... | Belarus | BelarusMO 2013_s | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All functions are f(x) = a - x for an arbitrary real constant a. | |
00nr | Let $a$, $b$, $c$ be positive real numbers with $a + b + c = 1$.
Prove that
$$
\frac{a}{2a+1} + \frac{b}{3b+1} + \frac{c}{6c+1} \le \frac{1}{2}.
$$
When does equality hold? | [
"We will use the following inequalities:\n$$\n\\frac{a}{2a+1} \\le \\frac{2a+1}{8}, \\quad \\frac{b}{3b+1} \\le \\frac{3b+1}{12} \\quad \\text{and} \\quad \\frac{c}{6c+1} \\le \\frac{6c+1}{24}.\n$$\n\nThey are an immediate consequence of the arithmetic-geometric mean inequality with the pairs of values $1$ and $2a$... | Austria | Austrian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds when a = 1/2, b = 1/3, and c = 1/6. | |
03go | Problem:
A regular pentagon is inscribed in a circle of radius $r$. $P$ is any point inside the pentagon. Perpendiculars are dropped from $P$ to the sides, or the sides produced, of the pentagon.
a) Prove that the sum of the lengths of these perpendiculars is constant.
b) Express this constant in terms of the radius ... | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 5 r cos(π/5) = (5(√5+1)/4) r | |
0jx1 | Problem:
A standard deck of 54 playing cards (with four cards of each of thirteen ranks, as well as two Jokers) is shuffled randomly. Cards are drawn one at a time until the first queen is reached. What is the probability that the next card is also a queen? | [
"Solution:\nSince the four queens are equivalent, we can compute the probability that a specific queen, say the queen of hearts, is right after the first queen. Remove the queen of hearts; then for every ordering of the 53 other cards, there are 54 locations for the queen of hearts, and exactly one of those is afte... | United States | HMMT November | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | proof and answer | 2/27 | |
0gl5 | Let $ABCD$ be a convex quadrilateral with the shortest side $AB$ strictly less than the longest side $CD$. Show that there exists a point $E$ on the segment $CD$ such that for any point $P$ (different from $E$) on the segment $CD$, the length of $O_1O_2$ is constant, where $O_1$ and $O_2$ are circumcenters of the trian... | [
"**Claim:** The point $E$ is the intersection point of the line parallel to $AD$ through $B$ and the line $CD$.\n\nLet $E$ be the intersection point of the line parallel to $AD$ through $B$ and the line $CD$. Firstly, we will show that $E$ is on the segment $CD$. Since $AB \\le AD$ and $BC \\le CD$ (because $AB$ an... | Thailand | Tajland 2014 | [
"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",
"Geometry > Plane Geometry > Miscellaneous > C... | null | proof only | null | |
09lw | Let $A$, $B$, and $C$ be points on line $l$ such that $B$ is between points $A$ and $C$. Points $D$ and $E$ lie on the same side of line $l$ such that $AD = BD$, $BC = EC$, and triangles $\triangle ADB$ and $\triangle ECB$ are similar. The lines $AE$ and $CD$ intersect at point $F$. Prove that the line $CD$ is the angl... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ct3 | Какое из чисел больше: $(100!)!$ или $99!^{100!} \cdot 100!^{99!}$? | [
"Второе число больше.\n\nПусть $a = 99!$. Тогда нам нужно сравнить числа $(100a)!$ и $a^{100a} \\cdot (100a)^a$. Заметим, что\n$$\n\\begin{aligned}\n& 1 \\cdot 2 \\cdot 3 \\cdots a < a^a, \\\\\n& (a+1)(a+2)(a+3) \\cdots 2a < (2a)^a, \\\\\n& (2a+1)(2a+2)(2a+3) \\cdots 3a < (3a)^a, \\\\\n& \\vdots \\\\\n& (99a+1)(99a... | Russia | XL Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0j3q | Problem:
Let $x(t)$ be a solution to the differential equation
$$
\left(x + x'\right)^2 + x \cdot x'' = \cos t
$$
with $x(0) = x'(0) = \sqrt{\frac{2}{5}}$. Compute $x\left(\frac{\pi}{4}\right)$. | [
"Solution:\nAnswer: $\\frac{\\sqrt[4]{450}}{5}$\n\nRewrite the equation as $x^2 + 2x x' + (x x')' = \\cos t$. Let $y = x^2$, so $y' = 2x x'$ and the equation becomes $y + y' + \\frac{1}{2} y'' = \\cos t$.\n\nThe term $\\cos t$ suggests that the particular solution should be in the form $A \\sin t + B \\cos t$. By s... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Equations > ODEs"
] | null | proof and answer | sqrt[4]{450}/5 | |
01dt | Find all pairs $p, q$ of distinct primes, sets $D \subseteq \mathbb{R}$ and functions $f: D \to D$ fulfilling
$$
f^p(x) = x^p \quad \text{and} \quad f^q(x) = x^q
$$
for all $x \in D$. (Here, $f^n$ denotes the $n$'th iterate of $f$.) | [
"**Answer:** For odd $p, q$, the possibilities are $D \\subseteq \\{0, \\pm 1\\}$ and $f(x) = x$. When one of $p, q$ is even, the possibilities are $D \\subseteq \\{0, 1\\}$ and $f(x) = x$.\n\nFrom\n$$\nx^{p^q} = f^{pq}(x) = f^{qp}(x) = x^{q^p}\n$$\nwe have $x^{p^q}(x^{q^p}-x^q) - 1) = 0$ or $x^{q^p}(x^{p^q}-x^p) -... | Baltic Way | Baltic Way 2016 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | If both primes are odd, then D is any subset of {0, −1, 1} and f(x) = x on D. If one of the primes is 2, then D is any subset of {0, 1} and f(x) = x on D. | |
021d | Problem:
Does there exist a function $f\colon \mathbb{R}\to \mathbb{R}$ such that
$$f\big(x^{2} + f(y)\big) = f(x)^{2} - y$$
for all $x,y\in \mathbb{R}$? | [
"Solution:\n\nThere does not exist such a function. Let us suppose by contradiction it does. By substituting $x \\gets 0$, we get\n$$f(f(y)) = f(0)^{2} - y$$\nfor all $y\\in \\mathbb{R}$. Since the right-hand side is bijective, this implies that $f$ is also bijective. Taking $y \\gets 0$ we get\n$$f\\big(x^{2} + f(... | Benelux Mathematical Olympiad | 17th Benelux Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | No | |
0czz | Let $ABC$ be a triangle with circumcenter $O$. Points $P$ and $Q$ are interior to sides $CA$ and $AB$, respectively. Circle $\mathcal{C}$ passes through the midpoints of segments $BP$, $CQ$, $PQ$. Prove that if line $PQ$ is tangent to circle $\mathcal{C}$, then $OP = OQ$. | [
"Let $K$, $L$, $M$, $B'$, $C'$ be the midpoints of $BP$, $CQ$, $PQ$, $CA$, and $AB$, respectively. Since $CA \\parallel LM$, we have $\\widehat{LMP} = \\widehat{QPA}$. Since $\\mathcal{C}$ touches the segment $PQ$ at $M$, we find $\\widehat{LMP} = \\widehat{LKM}$. It follows\n$$\n\\widehat{QPA} = \\widehat{LKM} \\t... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0iaw | Problem:
In trapezoid $ABCD$, $AD$ is parallel to $BC$. $\angle A = \angle D = 45^{\circ}$, while $\angle B = \angle C = 135^{\circ}$. If $AB = 6$ and the area of $ABCD$ is $30$, find $BC$. | [
"Solution:\n\nDraw altitudes from $B$ and $C$ to $AD$ and label the points of intersection $X$ and $Y$, respectively. Then $ABX$ and $CDY$ are $45^{\\circ}$-$45^{\\circ}$-$90^{\\circ}$ triangles with $BX = CY = 3\\sqrt{2}$. So, the area of $ABX$ and the area of $CDY$ are each $9$, meaning that the area of rectangle... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 2√2 | |
0jr0 | Problem:
2015 people sit down at a restaurant. Each person orders a soup with probability $\frac{1}{2}$. Independently, each person orders a salad with probability $\frac{1}{2}$. What is the probability that the number of people who ordered a soup is exactly one more than the number of people who ordered a salad? | [
"Solution:\n\nAnswer:\n$$\n\\frac{\\binom{4030}{2016}}{2^{4030}} \\text{ OR } \\frac{\\binom{4030}{2014}}{2^{4030}} \\text{ OR } \\frac{\\binom{4032}{2016}-2\\binom{4030}{2015}}{2^{4031}}\n$$\n\nSolution 1. Note that total soups $=$ total salads $+1$ is equivalent to total soups $+$ total not-salads $=2016$. So the... | United States | HMMT February 2015 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | binom(4030, 2016) / 2^4030 | |
0ldk | Let $ABC$ be a triangle with $I$ as its incenter and the circle $(I)$ is tangent to $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. Denote $I_b$, $I_c$ as the excenters of triangle $ABC$ with respect to vertices $B$, $C$. Let $P$, $Q$ be the midpoints of segments $I_bE$, $I_cF$. Suppose that $(PAC)$ intersects $AB$ at ... | [
"1)\nSince $EF$ and $I_bI_c$ are both perpendicular to $AI$ then $I_bI_cFE$ is the trapezoid. So $PQ$ is the midline of both trapezoid $I_bI_cFE$ and triangle $AEF$. Thus $P$, $Q$ belong to the radical axis of the degenerate circle ($A$, $0$) and $(I)$. Similarly, $Q$ belongs to the radical axis of ($B$, $0$) and $... | Vietnam | Vietnamese Team Selection Test for IMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"G... | null | proof only | null | |
0imt | Problem:
Prove that for every integer $n$ greater than $1$,
$$
\sigma(n) \phi(n) \leq n^{2}-1
$$
When does equality hold? | [
"Solution:\nNote that\n$$\n\\sigma(m n) \\phi(m n) = \\sigma(m) \\phi(m) \\sigma(n) \\phi(n) \\leq (m^{2}-1)(n^{2}-1) = (m n)^{2} - (m^{2} + n^{2} - 1) < (m n)^{2} - 1\n$$\nfor any pair of relatively prime positive integers $(m, n)$ other than $(1,1)$. Now, for $p$ a prime and $k$ a positive integer, $\\sigma\\left... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | Equality holds if and only if n is prime. | |
023g | Problem:
Três formigas estão paradas em três dos quatro vértices de um retângulo no plano. As formigas se movem no plano uma por vez. A cada vez, a formiga que se move o faz segundo a reta paralela à determinada pelas posições das outras duas formigas. É possível que, após alguns movimentos, as formigas se situem nos ... | [
"Solution:\n\nObserve que, se uma formiga $A$ se movimenta sobre uma reta paralela à reta determinada pelas outras duas formigas $B$ e $C$, então a área do triângulo com vértices sobre as três formigas é invariante, já que a base $BC$ e a medida da altura do triângulo com relação ao lado $BC$ não mudam.\n\nInicialm... | Brazil | Desafios | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
03b8 | Let $k$ be the incircle of a $\triangle ABC$. A line, parallel to $BC$ touches $k$ and intersects the sides $AB$ and $AC$ at points $A_1$ and $A_2$. Define the points $B_1, B_2$ and $C_1, C_2$ in a similar way. Prove that
$$
9(\overline{AA_1} \cdot \overline{AA_2} + \overline{BB_1} \cdot \overline{BB_2} + \overline{CC_... | [
"Let $D, E$ and $F$ be the common points of with the sides $BC, CA$ and $AB$, respectively, and let $AE = AF = x$, $BF = BD = y$ and $CD = CE = z$.\nSince $\\triangle AA_1A_2 \\sim \\triangle ABC$, then $\\frac{AA_1}{AB} = \\frac{AA_2}{AC} = \\frac{P_{AA_1A_2}}{P_{ABC}} = \\frac{x}{x+y+z}$ and hence $AA_1 = \\frac{... | Bulgaria | Bulgaria | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle ... | English | proof only | null | |
0l8f | Let $f: \mathbb{R} \to \mathbb{Z}$ be a function satisfying
$$
f(x - y) - 2f(x) + f(x + y) \geq -1
$$
for all $x, y \in \mathbb{R}$. Find all possible values of the set $\{f(x) \mid x \in \mathbb{R}\}$. | [
"The answer is $\\{a\\}$, $\\{a, a+1\\}$, $\\{a, a+1, a+2, \\dots\\}$, and $\\mathbb{Z}$, for arbitrary $a \\in \\mathbb{Z}$. For constructions, it is not hard to show that if $g: \\mathbb{R} \\to \\mathbb{R}$ is a convex function, then $\\lfloor g \\rfloor$ satisfies the functional equation. Thus $f(x) = a$, $f(x)... | United States | United States of America — TST Selection Test | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | Exactly the sets {a}, {a, a+1}, {a, a+1, a+2, ...}, and Z for arbitrary integer a. | |
0kh9 | Problem:
Compute the number of labelings $f:\{0,1\}^{3} \rightarrow \{0,1, \ldots, 7\}$ of the vertices of the unit cube such that
$$
\left|f\left(v_{i}\right)-f\left(v_{j}\right)\right| \geq d\left(v_{i}, v_{j}\right)^{2}
$$
for all vertices $v_{i}, v_{j}$ of the unit cube, where $d\left(v_{i}, v_{j}\right)$ denotes ... | [
"Solution:\n\nLet $B=\\{0,1\\}^{3}$, let $E=\\{(x, y, z) \\in B : x+y+z$ is even $\\}$, and let $O=\\{(x, y, z) \\in B : x+y+z$ is odd$\\}$. As all pairs of vertices within $E$ (and within $O$) are $\\sqrt{2}$ apart, it is easy to see that $\\{f(E), f(O)\\}=\\{\\{0,2,4,6\\}, \\{1,3,5,7\\}\\}$.\n\n- There are two wa... | United States | HMMT Spring 2021 Guts Round | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | 144 | |
011i | Problem:
Given a triangle $A B C$ with $\angle A=90^{\circ}$ and $|A B| \neq |A C|$. The points $D, E, F$ lie on the sides $B C, C A, A B$, respectively, in such a way that $A F D E$ is a square. Prove that the line $B C$, the line $F E$ and the line tangent at the point $A$ to the circumcircle of the triangle $A B C$... | [
"Solution:\n\nLet $B C$ and $F E$ meet at $P$ (see Figure 3).\n\n\nFigure 3\n\nIt suffices to show that the line $A P$ is tangent to the circumcircle of the triangle $A B C$.\n\nSince $F E$ is the axis of symmetry of the square $A F D E$, we have $\\angle A P E = \\angle B P F$. Moreover, $... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0gp6 | Let $t(n)$ denote the sum of the digits in the binary representation of a positive integer $n$, and let $k \ge 2$ be an integer.
a. Show that there exists a sequence $(a_i)_{i=1}^\infty$ of integers such that $a_m \ge 3$ is an odd integer and $t(a_1 a_2 \cdots a_m) = k$ for all $m \ge 1$.
b. Show that there is an int... | [
"a.\nLet $b_n = \\frac{2^{(k+1)n}k - 1}{2^{(k+1)n} - 1} = 2^{(k+1)n(k-1)} + \\dots + 2^{(k+1)n} + 1$ for $n \\ge 0$ and\n$$\na_n = \\frac{b_n}{b_{n-1}} = \\frac{(2^{(k+1)n}k - 1)(2^{(k+1)n-1} - 1)}{(2^{(k+1)n} - 1)(2^{(k+1)n-1}k - 1)} \\quad \\text{for } n \\ge 1.\n$$\nSince $(2^{(k+1)n} - 1, 2^{(k+1)n-1}k - 1) = 2... | Turkey | Team Selection Test for IMO 2011 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0cyr | Set $A$ consists of 7 consecutive positive integers less than $2011$, while set $B$ consists of 11 consecutive positive integers. If the sum of the numbers in $A$ is equal to the sum of the numbers in $B$, what is the maximum possible element that $A$ could contain? | [
"Let\n$A=\\{x+1, x+2, \\ldots, x+7\\}$ and $B=\\{y+1, y+2, \\ldots, y+11\\}$, where $x<2004$. We have\n$$(x+1)+(x+2)+\\ldots+(x+7)=(y+1)+(y+2)+\\ldots+(y+11)$$\nhence\n$$\n7x+\\frac{7 \\cdot 8}{2}=11y+\\frac{11 \\cdot 12}{2}\n$$\nThis equation is equivalent to\n$$\n7x-11y=38.\n$$\nThe smallest solution in positive ... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 2005 | |
0cth | A quadratic polynomial $f(x) = ax^2 + bx + c$ has no real roots. It is given that $b$ is a rational number, and exactly one of $c$ and $f(c)$ is a rational number. Is it possible for the discriminant of $f(x)$ to be a rational number?
Квадратный трёхчлен $f(x) = ax^2 + bx + c$, не имеющий корней, таков, что коэффициен... | [
"No.\n\nТак как трёхчлен $f(x)$ не имеет корней, то $c = f(0) \\neq 0$ и $f(c) \\neq 0$. Тогда выражение $\\frac{f(c)}{c}$ иррационально как отношение рационального и иррационального чисел. Но $\\frac{f(c)}{c} = \\frac{ac^2 + bc + c}{c} = ac + b + 1$. Так как $b+1$ рационально, то $ac$ — иррационально. Получаем, чт... | Russia | Russian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | English; Russian | proof and answer | No | |
0b8j | Consider the cube $ABCDA'B'C'D'$. The angle bisectors of the angles $\angle A'C'A$ and $\angle A'AC'$ meet $AA'$ and $A'C'$ at points $P$ and $Q$, respectively. The point $M$ is the foot of the perpendicular from $A'$ onto $C'P$ while $N$ is the foot of the perpendicular from $A'$ onto $AS$. The point $O$ is the center... | [
"a) Denote by $T$ and $R$ the intersections of the straight lines $AC'$ and $A'M$, respectively $AC'$ and $A'N$. In the triangle $A'C'T$, $C'M$ is an angle bisector and an altitude, so $A'M = MT$. In the triangle $A'AR$, $AN$ is an angle bisector and an altitude, so $A'N = NR$. The segment $[MO]$ joins the midpoint... | Romania | Romanian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | sqrt(2)/4 | |
02c5 | Problem:
Distância na reta - Cinco pontos estão sobre uma mesma reta. Quando listamos as dez distâncias entre dois desses pontos, da menor para a maior, encontramos $2,4,5,7,8, k, 13,15,17,19$. Qual o valor de $k$ ? | [
"Solution:\n\nSolução 1: - Essa solução é um pouco difícil de escrever porque é feita na base de \"tentativa e erro\". Começamos desenhando uma reta numérica e colocando os pontos $0$ e $19$. Como a primeira distância é $2$, marcamos nossos primeiros três pontos:\n\n\n\nComo temos que ter u... | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 12 | |
09ah | A circle touches $BC$ side of triangle $ABC$ at the point $M$ and intersects sides $AB$ and $AC$ at $D$ and $E$ respectively. If $DE \parallel BC$, prove that $EM = MD$. | [
"Since $BC$ is tangent to the circle, we have $\\angle EMC = \\angle EAM = \\angle EDM$ and $\\angle DMB = \\angle DAM = \\angle DEM$. From $DE \\parallel BC$, we have $\\angle EDM = \\angle DMB$. So $AM$ bisects $\\angle CAB$.\n\n"
] | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ay5 | Problem:
Dominic randomly picks between two words MATHEMATICS and MEMES, each with an equal chance of popping up. From his chosen word, he then randomly draws one letter, with the probability of each letter popping up directly proportional to the number of times it occurs in the word. Given that Dominic drew an M, wha... | [
"Solution:\n\nThe probability of Dominic picking out the letter $M$ from the word MATHEMATICS is $\\frac{2}{11}$. On the other hand, the probability of him picking it from the word MEMES is $\\frac{2}{5}$. This gives him, unconditionally, a probability of $\\frac{1}{2}\\left(\\frac{2}{11}+\\frac{2}{5}\\right)=\\fra... | Philippines | Philippine Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Other",
"Math Word Problems"
] | null | final answer only | 11/16 | |
0948 | Problem:
Let $ABCD$ be a convex quadrilateral such that $AC = BD$ and the sides $AB$ and $CD$ are not parallel. Let $P$ be the intersection point of the diagonals $AC$ and $BD$. Points $E$ and $F$ lie, respectively, on segments $BP$ and $AP$ such that $PC = PE$ and $PD = PF$. Prove that the circumcircle of the triangle... | [
"Solution:\nWithout loss of generality assume that $AP < BP$. Let $X = AB \\cap CD$, $Y = AB \\cap C' D'$, and $Z = CD \\cap C' D'$. Furthermore, denote by $M$ the midpoint of arc $APB$. We will show that the circumcircle of $\\triangle XYZ$ is tangent to the circumcircle of $\\triangle ABP$ at $M$.\n\n | MEMO Team Competition | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
039l | Find all real numbers $a$ and $b$ such that the system
$$
\begin{array}{l@{\quad}l@{\quad}c}
\text{system} & \left\{
\begin{array}{l}
x + a = y + b \\
x^2 - a = 2y
\end{array}
\right.
& \text{has unique solution } (x_0, y_0) \text{ and it satisfies the equality} \\
& x_0^2 + y_0^2 = 1025.
\end{array}
$$ | [
"The given system is equivalent to\n$$\n\\left|\n\\begin{array}{l}\nx^2 - 2x + 2b - 3a = 0 \\\\\nx + a = y + b\n\\end{array}\n\\right.\n$$\nIt has a unique solution $(x_0, y_0)$ if the quadratic equation $x^2 - 2x + 2b - 3a = 0$ has a unique root $x_0$. This means that $D = 1 - 2b + 3a = 0$ and $x_0 = 1$. The condi... | Bulgaria | Spring Mathematical Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | a=5, b=8 or a=-3, b=-4 | |
01pj | Several small circles are arranged inside a unit circle $\Gamma$. The sum of the perimeters of all these small circles is not less than $\pi$ and none of them includes the center of $\Gamma$.
Prove that there exists a concentric with $\Gamma$ circumference intersecting at least two of these small circles. | [
"Let $r_1, r_2, \\dots, r_k$ be radii of small circles. By condition,\n$$\n2\\pi(r_1 + r_2 + \\dots + r_k) \\ge \\pi,\n$$\ni. e.\n$$\nr_1 + r_2 + \\dots + r_k \\ge 1/2. \\quad (*)\n$$\nConsider $360^\\circ$ rotation of $\\Gamma$ (together with all small circles) about its center. Under this rotation each of small c... | Belarus | BelarusMO 2013_s | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
02q4 | Problem:
Observe que
$$
\begin{aligned}
& 1 \times 2 \times 3 \times 4 + 1 = 5^{2} \\
& 2 \times 3 \times 4 \times 5 + 1 = 11^{2} \\
& 3 \times 4 \times 5 \times 6 + 1 = 19^{2}
\end{aligned}
$$
Prove que o produto de quatro inteiros positivos consecutivos, aumentado em uma unidade, é um quadrado perfeito. | [] | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0a6h | Problem:
Some of the 80960 lattice points in a $40 \times 2024$ lattice are coloured red. It is known that no four red lattice points are vertices of a rectangle with sides parallel to the axes of the lattice. What is the maximum possible number of red points in the lattice? | [
"Solution:\n\nLet $a_{1}, a_{2}, a_{3}, \\ldots, a_{2024}$ be the number of red dots in rows $1, 2, 3, \\ldots, 2024$ respectively. So $0 \\leqslant a_{i} \\leqslant 40$ for each $i$.\n\nFor each of the $\\binom{40}{2} = 780$ pairs of columns, there can be at most one row with a red dot in both columns. Therefore w... | New Zealand | NZMO Round One | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2804 | |
0c11 | The sum of 15 consecutive positive integers is a number written with distinct digits, among these being 0, 1, 2 and 4. Find the least possible number among the 15 consecutive numbers. | [] | Romania | 69th Romanian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 676 | |
04kc | Points $M$ and $N$ lie on the sides $\overline{BC}$ and $\overline{CD}$ (respectively) of the square $ABCD$ so that $\angle BMA = \angle NMC = 60^\circ$. Determine $\angle MAN$. (Ukraine 2013) | [] | Croatia | Mathematical competitions in Croatia | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 45° | |
0bne | A *Pythagorean triple* is a solution of the equation $x^2 + y^2 = z^2$ in positive integers such that $x < y$. Given any non-negative integer $n$, show that some positive integer appears in precisely $n$ distinct Pythagorean triples.
AMM Magazine | [
"We show by induction on $n \\ge 0$, that $2^{n+1}$ appears in precisely $n$ distinct Pythagorean triples. Since no Pythagorean triple contains $2$, the assertion holds for $n=0$.\n\nFor the induction step, let $n \\ge 1$, and assume that $2^n$ appears in exactly $n-1$ distinct Pythagorean triples. The latter produ... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0lav | 1. At the point $(1, 1)$ of the coordinate plane $(Oxy)$, there is a cào cào. From that point the cào cào can only jump to other integral points by the rule: from the positive integral point $A$, the cào cào jumps to the positive integral point $B$ if the triangle $OAB$ has area equal to $\frac{1}{2}$.
1/ Find all poi... | [] | Vietnam | IMO2011 Selection | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Linear Algebra > Determinants",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof and answer | All positive integer points (m, n) with gcd(m, n) = 1 are reachable. Moreover, for any such (m, n), there exists a path from (1, 1) to (m, n) consisting of at most |m − n| jumps. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.