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
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values | final_answer stringlengths 1 1.22k ⌀ |
|---|---|---|---|---|---|---|---|---|---|
052q | Let $x$ and $y$ be different positive integers. Prove that $\frac{x^2+4xy+y^2}{x^3-y^3}$ is never an integer. | [
"By symmetry we can assume that $x > y$. If $x - y = 1$, then\n$$\n\\begin{aligned} \\frac{x^2 + 4xy + y^2}{x^3 - y^3} &= \\frac{x^2 + 4xy + y^2}{(x-y)(x^2 + xy + y^2)} = \\frac{(x-y)^2 + 6xy}{(x-y)((x-y)^2 + 3xy)} = \\\\ &= \\frac{1 + 6xy}{1 + 3xy} = 1 + \\frac{3xy}{1 + 3xy}, \\end{aligned}\n$$\nwhich is clearly n... | Estonia | Open Contests | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0i00 | Problem:
You want to sort the numbers $54321$ using block moves. In other words, you can take any set of numbers that appear consecutively and put them back in at any spot as a block. For example, $653421 \rightarrow 426531$ is a valid block move for 6 numbers. What is the minimum number of block moves necessary to ge... | [
"Solution:\n\nHere is a sequence of 3 moves that works: $54321 \\rightarrow 32541 \\rightarrow 34125 \\rightarrow 12345$.\n\nBut how do we know we can't do it in 2 moves? From any position there are 20 possible permutations via block moves, 16 from moving a block of size 1 and 4 from moving a block of size 2. One c... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 3 | |
04ly | Prove that for all positive real numbers $a$, $b$ and $c$ the following inequality holds
$$
\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} + \sqrt{\frac{ab+bc+ca}{a^2+b^2+c^2}} \ge \frac{5}{2}.
$$ | [
"Because of simplicity let $A = a^2 + b^2 + c^2$ and $B = ab + bc + ca$. The CSB inequality gives us\n$$\n\\left( \\frac{a}{b+c} + \\frac{b}{c+a} + \\frac{c}{a+b} \\right) (a(b+c) + b(c+a) + c(a+b)) \\ge (a+b+c)^2,\n$$\ni.e.\n$$\n\\frac{a}{b+c} + \\frac{b}{c+a} + \\frac{c}{a+b} \\ge \\frac{A+2B}{2B} = \\frac{A}{2B}... | Croatia | Croatian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
04hq | Ten chairs are arranged around a round table and marked with numbers $1$ to $10$ successively (in such a way that chairs $1$ and $10$ are also adjacent), and a knight is sitting in each chair. In the beginning, every knight has an even number of coins. Simultaneously, each knight gives half of his coins to his left nei... | [
"Let us denote by $2x_1, 2x_2, \\dots, 2x_{10}$ the number of coins that the knights sitting in chairs $1, 2, \\dots, 10$ had in the beginning, respectively. We have to determine $2x_8$.\n\nWe have a system of equations: $x_{10} + x_2 = 22$, $x_1 + x_3 = 24$, $x_2 + x_4 = 26$, $x_3 + x_5 = 28$, $\\dots$, $x_8 + x_{... | Croatia | Croatia Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 46 | |
0ktx | Problem:
There are $800$ marbles in a bag. Each marble is colored with one of $100$ colors, and there are eight marbles of each color. Anna draws one marble at a time from the bag, without replacement, until she gets eight marbles of the same color, and then she immediately stops.
Suppose Anna has not stopped after d... | [
"Solution:\n\nIn order to not stop after $699$ marbles, the last $101$ marbles must consist of $2$ marbles of one color, and one marble from each other color. Since each of these marbles is equally likely to be the next to be drawn, and we stop after drawing the next marble as long as it's not one of the two of the... | United States | HMMT February 2023 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 99/101 | |
024a | Problem:
Cururu é um sapo estranho, que se desloca apenas com dois tipos de saltos:
Tipo I: $10~\mathrm{cm}$ para o Leste e $30~\mathrm{cm}$ para o Norte;
Tipo II: $20~\mathrm{cm}$ para Oeste e $40~\mathrm{cm}$ para o Sul.
a. Como Cururu faz para chegar a um ponto situado a $190~\mathrm{cm... | [
"Solution:\n\nA cada $x$ saltos do tipo I, o sapo se desloca $10x~\\mathrm{cm}$ para o Leste e $30x~\\mathrm{cm}$ para o Norte e, a cada $y$ saltos do tipo II, o sapo se desloca $20y~\\mathrm{cm}$ para o Oeste e $40y~\\mathrm{cm}$ para o Sul. Assim, ao final de $x$ saltos do tipo I e $y$ do tipo II, o sapo se deslo... | Brazil | Nível 2 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a) 57 jumps of type I and 19 jumps of type II (in any order). b) No, it is impossible. | |
0j02 | There are $n$ students standing in a circle, one behind the other. The students have heights $h_1 < h_2 < \dots < h_n$. If a student with height $h_k$ is standing directly behind a student with height $h_{k-2}$ or less, the two students are permitted to switch places. Prove that it is not possible to make more than $\b... | [
"(By Kiran Kedlaya). Let $h_i$ also denote the student with height $h_i$. We prove that for $1 \\le i < j \\le n$, $h_j$ can switch with $h_i$ at most $j - i - 1$ times. We proceed by induction on $j - i$, the base case $j - i = 1$ being evident because $h_i$ is not allowed to switch with $h_{i-1}$.\n\nFor the indu... | United States | USAMO 2010 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
08r5 | A dodecahedron and a vertex $X$ of the dodecahedron are given. An ant started from $X$, walked along the edges of the dodecahedron, passing each of the vertices of the dodecahedron except $X$ just once, and returned to $X$. How many such routes exist? We consider a route and its reversal to be different. | [
"Take an arbitrary route. For each face, count the number of the edges on the face which are on the route. Since they can be neither less than $3$ nor more than $4$, and they add up to $2 \\times 20 = 40$, there exists exactly $4$ faces with $4$ of their edge on the route.\n\nNow, assume that the ant passed consecu... | Japan | The 16th Japanese Mathematical Olympiad - The First Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Graph Theory",
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof and answer | 60 | |
0h6m | Let $O$ be a circumcenter of a right triangle. The circle with smaller radius and center at point $O$ is tangent to the greater cathetus and the height of the triangle from the right angle.
Find the acute angles of the right triangle and the relation between the radii of the circumcircle and the other circle.
(Bogdan R... | [
"The acute angles are $30^\\circ$ and $60^\\circ$; the ratio of the radii is $2:1$.\n\nLet the right triangle be $CMD$ (see fig. 16).\nLet $ON$ be the radius of the smaller circle that is tangent to the cathetus.\nThus $\\triangle ODN = \\triangle OMN$, and $\\triangle AMO = \\triangle OMN$, therefore, $2AM = MN + ... | Ukraine | UkraineMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | null | proof and answer | Acute angles: 30° and 60°; radii ratio: 2:1 | |
05ag | Pieces of cardboard of dimensions $1 \times 4$ are placed on a $10 \times 10$ grid in such a way that each piece covers exactly 4 adjacent unit squares (either horizontally or vertically) and no two pieces touch each other side-to-side, edge-to-edge, or corner-to-corner. Find the largest possible number of cardboard pi... | [
"Let the pieces of cardboard be placed on the grid as required. Since each piece covers exactly 4 unit squares and no two pieces touch, there is at least a 1-unit wide space between every two pieces. Therefore, if we draw a half-unit wide \"no-go zone\" around each piece, the areas covered by the pieces and their n... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | 12 | |
0c0x | Two continuous functions $f, g: [0, 1] \to [0, 1]$ are such that $f \circ g = g \circ f$, and one of them is nondecreasing. Prove that there exists $a \in [0, 1]$ so that $f(a) = a = g(a)$. | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
0kwa | Problem:
Isabella writes the expression $\sqrt{d}$ for each positive integer $d$ not exceeding $8!$ on the board. Seeing that these expressions might not be worth points on HMMT, Vidur simplifies each expression to the form $a \sqrt{b}$, where $a$ and $b$ are integers such that $b$ is not divisible by the square of a ... | [
"Solution:\n\nLet $\\sqrt{n}$ simplify to $a_{n} \\sqrt{b_{n}}$, and replace $8!$ by $x$. First, notice that $\\sum_{n \\leq x} a_{n}$ is small $\\left(O\\left(x^{3 / 2}\\right)\\right.$ in particular) because each term cannot exceed $\\sqrt{x}$. On the other hand, $\\sum_{n \\leq x} b_{n}$ will be large; we have $... | United States | HMMT November 2023 | [
"Number Theory > Number-Theoretic Functions > Möbius inversion",
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Other"
] | null | final answer only | 534834652 | |
036c | Problem:
Consider the inequality $\sqrt{x} + \sqrt{2 - x} \geq \sqrt{a}$, where $a$ is a real number.
a) Solve the inequality for $a = 3$.
б) Find all $a$, for which the set of solutions of the inequality is a segment (possibly, a point) of length less than or equal to $\sqrt{3}$. | [] | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) x ∈ [1 − sqrt(3)/2, 1 + sqrt(3)/2].
b) a ∈ [3, 4]. | |
0bfu | Given a prime $p \ge 5$, show that there exist at least two distinct primes $q$ and $r$ in the range $2, 3, \dots, p-2$ such that $q^{p-1} \not\equiv 1 \pmod{p^2}$ and $r^{p-1} \not\equiv 1 \pmod{p^2}$. | [
"(1) An improper integer greater than 1 has at least one improper prime divisor; and\n(2) If $k$ is an integer coprime to $p$ and $n$ is a proper integer, then $kp - n$ is improper.\nThe first claim follows from the fact that the product of two proper integers is again proper. For the second, notice that $p$ does n... | Romania | The Tenth IMAR Mathematical Competition | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
04bz | Solve $\sin x \cdot \cos 2x \cdot \cos 4x = 1$. | [
"We have $|\\sin x| \\leq 1$, $|\\cos 2x| \\leq 1$, $|\\cos 4x| \\leq 1$. The product $\\sin x \\cdot \\cos 2x \\cdot \\cos 4x$ can only be $1$ if each factor is $1$ or $-1$, and their product is $1$.\n\nBut $|\\sin x| = 1$ only when $\\sin x = 1$ or $\\sin x = -1$.\n\nCase 1: $\\sin x = 1$\nThen $x = \\dfrac{\\pi}... | Croatia | Mathematica competitions in Croatia | [
"Precalculus > Trigonometric functions"
] | English | proof and answer | x = 3π/2 + 2kπ, k ∈ ℤ | |
05hz | Problem:
On considère une rangée de cases numérotées $0,1, \ldots, k$ de gauche à droite où, pour chaque $i \geqslant 1$, la case numéro $i$ contient $x_{i}$ jetons. Il n'y a initialement aucun jeton sur la case numéro $0$. À tour de rôle, Alice et Bob jouent alors selon les règles suivantes :
- Bob choisit un ensembl... | [
"Solution:\n\nDans tout ce qui suit, on peut clairement supposer que Bob ne choisit jamais $S=\\emptyset$ ou $S$ l'ensemble de tous les jetons non encore éliminés, sans quoi Alice gagne immédiatement. Du coup, le nombre de jetons diminue strictement après chaque fois qu'Alice joue, ce qui assure que le jeu se termi... | France | Olympiades Françaises de Mathématiques, Envoi No. 6 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Alice has a winning strategy when the halving-weighted sum is less than one. Yes: when the halving-weighted sum is at least one, Bob has a winning strategy. | |
07b9 | Prove that for each natural number $k > 1$, only a finite number of $k$-tuples of consecutive natural numbers exist such that their product is a perfect square. | [
"If $k$ is even, using the lemma proved in the solution of the 7th problem of the Third Round, we can say that there exist an integer $a$ and some polynomial $Q(x) \\in \\mathbb{Z}[x]$ such that\n$$\n(Q(x))^2 \\le a^2(x+1)(x+2)\\cdots(x+k) < (Q(x)+1)^2, \\quad \\text{for large values of } x.\n$$\nSince $Q(x)$ is a ... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0958 | Problem:
Se consideră funcţia strict monotonă $f: \mathbb{R} \rightarrow \mathbb{R}$. Să se determine toate funcţiile $g: \mathbb{R} \rightarrow \mathbb{R}$, astfel încât
$$
f(g(x+2017)) \geq f(x) \geq f(g(x)+2017),\; \forall x \in \mathbb{R}
$$ | [
"Solution:\nConsiderăm funcţia $f$ strict crescătoare. Din condiţiile problemei rezultă\n$$\n\\left\\{\n\\begin{array}{l}\ng(x+2017) \\geq x, \\\\\nx \\geq g(x)+2017,\n\\end{array}\n\\right.\n$$\nadică\n$$\n\\left\\{\n\\begin{array}{l}\nx \\leq g(x+2017) \\\\\nx \\geq g(x)+2017\n\\end{array}\n\\right.\n$$\nÎn prima... | Moldova | Olimpiada de Matematică a Republicii Moldova | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | g(x) = x - 2017 for all real x | |
047z | In convex quadrilateral $ABCD$, $AB \perp AD$ and $AD = DC$. Let point $E$ lie inside segment $BC$, and point $F$ lie on the extension of $DE$ such that $\angle ABF = \angle DEC > 90^\circ$. Let $O$ be the circumcenter of triangle $CDE$. Let $P$ be a point on the extension of $FO$ such that $FP = FB$. Let segment $BP$ ... | [
"As shown in the figure, draw a line through $D$ parallel to $AB$, intersecting $BF$ at $K$. Then:\n$$\n\\angle DKB = 180^\\circ - \\angle ABF = 180^\\circ - \\angle DEC = \\angle DEB,\n$$\nso points $B, K, E, D$ are concyclic.\nSince $AB \\perp AD$ and $DK \\parallel AB$, we have $AD = BK \\cdot \\sin \\angle ABK$... | China | 2025 International Mathematical Olympiad China National Team Selection Test | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point... | English | proof only | null | |
08c3 | 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"
] | null | proof and answer | n = 0 or 2 | |
00hq | Let $ABC$ be a right triangle with $\angle B = 90^{\circ}$. Point $D$ lies on the line $CB$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $AD$ and let $F$ be the second intersection point of the circumcircle of $\triangle ACD$ and the circumcircle of $\triangle BDE$. Prove that as $D$ varies, the lin... | [
"Let the line $EF$ intersect the line $BC$ at $P$ and the circumcircle of $\\triangle ACD$ at $G$ distinct from $F$. We will prove that $P$ is the fixed point.\nFirst, notice that $\\triangle BED$ is isosceles with $EB = ED$. This implies $\\angle EBC = \\angle EDP$.\nThen, $\\angle DAG = \\angle DFG = \\angle EBC ... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0fsm | Problem:
Einem Kreis ist ein konvexes Viereck $ABCD$ einbeschrieben. Zeige, dass die Sehne, welche die Mittelpunkte der beiden Bogen $\widehat{AB}$ und $\widehat{CD}$ verbindet, senkrecht steht auf der Sehne, welche die beiden Bogenmittelpunkte von $\widehat{BC}$ und $\widehat{DA}$ miteinander verbindet. | [
"Solution:\n\nLemma. Sei $ABCD$ ein konvexes Sehnenviereck und $P$ der Diagonalenschnittpunkt. Es gilt\n$$\n\\angle APB = \\angle ACB + \\angle CBD\n$$\noder in Worten ausgedrückt: der Winkel $\\angle APB$ ist so gross wie die Summe der beiden Peripheriewinkel über $AB$ und $CD$.\n\nBeweis des Lemmas. Aussenwinkels... | Switzerland | IMO - Selektion | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
097n | Problem:
Să se arate, că pentru oricare număr natural nenul $n$ există un şir de $2 n+1$ numere naturale consecutive astfel, încât suma pătratelor primelor $n+1$ dintre ele să fie egală cu suma pătratelor următoarelor $n$ dintre ele. Să se verifice, dacă există un asemenea şir cu numărul din mijloc egal cu 2020. | [
"Solution:\nFie $k$ numărul din mijlocul acestui şir, $0<k<n$. Şirul are forma\n$$\nk-n, k-(n-1), k-(n-2), \\ldots, k-1, k, k+1, \\ldots, k+(n-2), k+(n-1), k+n\n$$\nConform condiţiei,\n$$\n(k-n)^2+[k-(n-1)]^2+[k-(n-2)]^2+\\ldots+(k-1)^2+k^2=(k+1)^2+\\ldots+[k+(n-1)]^2+(k+n)^2\n$$\nDupă ridicarea la pătrat şi reduce... | Moldova | Olimpiada Republicană la Matematică, Ziua a doua | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | k = 2 n (n+1); no sequence with middle term 2020 exists | |
0kuz | Problem:
Ara and Bea play a game where they take turns putting numbers from $1$ to $5$ into the cells of the $X$-shaped diagram on the right. Each number must be played exactly once, and a cell cannot have more than one number placed in it. Ara's goal is for the two diagonals of the $X$ diagram to have the same sum wh... | [
"Solution:\n\na.\nSuppose Ara goes first. He can begin by placing a $5$ in the center of the board. This is the only space shared by both diagonals, so when the game is over, the two diagonal sums are guaranteed to add up to $1+2+3+4+5+5=20$.\n\nFor her first move, Bea must play some number $n$ in one of the four c... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
05c5 | Every night, Juku listens to exactly $14$ songs from a playlist containing exactly $100$ songs. Every time a song ends, the next song is chosen from among all $100$ songs with equal probability (the same song may also repeat). Prove that, on more than half of all nights, Juku listens some song more than once. | [
"We show that the probability of listening to the same song multiple times is greater than $0.5$, which is equivalent to the probability of listening to $14$ distinct songs being less than $0.5$. The latter probability can be expressed as\n$$\n1 \\cdot 0.99 \\cdot 0.98 \\cdot 0.97 \\cdot 0.96 \\cdot 0.95 \\cdot 0.9... | Estonia | Estonian Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Permutations"
] | English | proof only | null | |
0krh | Problem:
a) Find the expected number of flips to get two heads in a row.
b) Find the expected number of flips to get heads followed by tails. | [
"Solution:\na.\nLet $x$ be the expected number of flips. There is a $\\frac{1}{4}$ chance that she gets two heads right away. There is a $\\frac{1}{2}$ chance the first flip is tails, in which case she is basically starting over after the first flip, so it will take an expected $x+1$ flips total. Finally, there is ... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | a) 6; b) 4 | |
0j55 | Problem:
Let $ABCD$ be a cyclic quadrilateral, and suppose that $BC = CD = 2$. Let $I$ be the incenter of triangle $ABD$. If $AI = 2$ as well, find the minimum value of the length of diagonal $BD$. | [
"Solution:\n\n\n\nLet $T$ be the point where the incircle intersects $AD$, and let $r$ be the inradius and $R$ be the circumradius of $\\triangle ABD$. Since $BC = CD = 2$, $C$ is on the midpoint of arc $BD$ on the opposite side of $BD$ as $A$, and hence on the angle bisector of $A$. Thus $... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometri... | null | proof and answer | 2√3 | |
02ra | When two red amoebas join, the result in one blue amoeba; when a red amoeba and a blue amoeba join, they turn into three red amoeba; and when two blue amoeba join, they become four red amoeba. Fernando observes a test tube with initially $b$ blue amoebas and $r$ red amoebas.
Determine, in terms of $b$ and $r$, all poss... | [
"If the number of blue amoebas is $b$ and the number of red amoebas is $r$ then $2b + r$ is invariant: indeed, whenever one blue amoeba appears/disappears, two red amoebas disappear/appear.\n\nIf there is only one amoeba, then it will never change. Otherwise, change all blue amoebas into red amoebas: this is possib... | Brazil | Brazilian Math Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | All reachable configurations are exactly those of the form: k blue amoebas and (2b + r − 2k) red amoebas, where k is any integer with 0 ≤ k ≤ b + floor(r/2). Equivalently, the possible total counts are 2b + r − k for such k. Here 2b + r is invariant under the allowed operations. | |
02ar | Problem:
Triângulos com lados inteiros - Quantos triângulos existem cujos lados são números inteiros e o perímetro é $12$?
(a) $1$
(b) $3$
(c) $5$
(d) $7$
(e) $9$ | [
"Solution:\n\nPara que três números $a$, $b$, $c$ sejam os comprimentos dos lados do triângulo, cada um deles deve ser maior que a diferença e menor que a soma dos outros dois.\n\nSejam $a \\leq b \\leq c$ os comprimentos dos lados do triângulo. Assim, $c < a + b$.\n\nAgora, somando $c$ a ambos os membros temos que... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | MCQ | b | |
0iz0 | Problem:
How many ways are there to place pawns on an $8 \times 8$ chessboard, so that there is at most 1 pawn in each horizontal row? Express your answer in the form $p_{1}^{e_{1}} \cdot p_{2}^{e_{2}} \cdots$, where the $p_{i}$ are distinct primes and the $e_{i}$ are positive integers. | [
"Solution:\n\n$3^{16}$\n\nIf there is at most 1 pawn in each row, then each row of the chessboard may have either 0 or 1 pawn somewhere in the row. There is 1 case if there are no pawns in the row. There are 8 possible cases if there is 1 pawn in the row, one case for each square in the row. Hence for each row, the... | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 3^16 | |
00qk | Let $p$ be a prime number and $x_1, x_2, \dots, x_p$ be integers. Show that if
$$
x_1^n + x_2^n + \dots + x_p^n \equiv 0 \pmod{p}
$$
for all positive integers $n$ then $x_1 \equiv x_2 \equiv \dots \equiv x_p \pmod{p}$. | [
"Letting $n = p - 1$, we have $x_i^{p-1} \\equiv 0$ or $1 \\pmod{p}$. Therefore, the congruence $x_1^n + x_2^n + \\dots + x_p^n \\equiv 0 \\pmod{p}$ is true when either all $x_i$ are divisible by $p$ or no $x_i$ is divisible by $p$.\n\nOn the other hand, if no $x_i$ is divisible by $p$ we have\n$$\n\\sum_{i=1}^{p} ... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
05b3 | In an acute triangle $ABC$ with $AB < AC$, the altitudes $BE$ and $CF$ intersect at $H$. The tangent to the circumcircle of $ABC$ at $A$ intersects the circumcircle of $AEF$ at $T \neq A$. The circumcircles of $TBE$ and $TCF$ intersect at $K \neq T$. Prove that $\angle KHB = \angle ABC$. | [
"From $\\angle HEA = 90^\\circ = 180^\\circ - 90^\\circ = 180^\\circ - \\angle HFA$ (Fig. 5) we deduce that $AEHF$ is cyclic. By definition, $T$ lies on this circle as well, so\n$$\n\\angle THE = 180^\\circ - \\angle TAE = 180^\\circ - \\angle TAC = \\angle ABC,\n$$\nwhere the final equality is due to the tangent-c... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadril... | English | proof only | null | |
0ctr | A positive integer $N$ is presented in the form $N = a_1 - a_2 = b_1 - b_2 = c_1 - c_2 = d_1 - d_2$, where $a_1$ and $a_2$ are perfect squares, $b_1$ and $b_2$ are perfect cubes, $c_1$ and $c_2$ are perfect 5th powers, and $d_1$ and $d_2$ are perfect 7th powers. Under the conditions above, determine whether two of the ... | [
"No.\nSet $N = (3^2 - 2^2)^{105}(3^3 - 2^3)^{70}(3^5 - 2^5)^{126}(3^7 - 2^7)^{120}$. Then\n$$N = M_2^2(3^2 - 2^2) = M_3^3(3^3 - 2^3) = M_5^5(3^5 - 2^5) = M_7^7(3^7 - 2^7),$$\nwhere none of the $M_i$ is divisible by $2$ or $3$.\n\n$N = (3M_2)^2 - (2M_2)^2 = (3M_3)^3 - (2M_3)^3 = (3M_5)^5 - (2M_5)^5 = (3M_7)^7 - (2M_... | Russia | Russian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English; Russian | proof and answer | No | |
0c0v | Let $ABC$ be a triangle, let $I$ be its incenter, let $\Omega$ be its circumcircle, and let $\omega$ be the circle tangent to the sides $AB$ and $AC$, and internally tangent to $\Omega$. Let $D, E$ and $T$ be the points of contact of $\omega$ and $AB, AC$ and $\Omega$, respectively, let the line $IT$ cross $\omega$ aga... | [
"\n\nThe points $D, E, M, N$ are concyclic if and only if $\\angle DEP = \\angle BMD$. Read from $\\omega$ to infer that $\\angle DEP = \\angle DTP = \\angle ADP$.\nWe now show that $\\angle DTP = \\angle ATE$. To this end, consider the homothety from $T$ mapping $\\omega$ onto $\\Omega$. T... | Romania | 69th NMO Selection Tests for BMO and IMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Brocard ... | null | proof and answer | the A-excenter of triangle ABC | |
01rv | After division of a positive integer $n$ by $29$, $41$, and $59$ one has three nonzero remainders such that their sum is equal to $n$.
Find all possible values of $n$. | [
"Let, by condition,\n$$\nn = 29q_1 + r_1 = 41q_2 + r_2 = 59q_3 + r_3 = r_1 + r_2 + r_3,\n$$\n$r_1 < 29$, $r_2 < 41$, $r_3 < 59$. From these inequalities it follows that $59q_3 = r_1 + r_2 \\le 28 + 40 = 68$, so $q_3 = 1$. Then\n$$\nr_1 + r_2 = 59. \\quad (1)\n$$\n\nFurther, $41q_2 = r_1 + r_3 \\le 28 + 58 = 86$, so... | Belarus | FINAL ROUND | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 79 and 114 | |
0h48 | Positive numbers $a$ and $b$ satisfy the equality $a + b + a^{-1} + b^{-1} = 5$. Prove that $3\sqrt{a + b} \ge a + b + 2$. | [
"Let $u = a + b$ and $v = \\frac{1}{a} + \\frac{1}{b}$. Then $u + v = 5$. As is known, $uv \\ge 4$. Therefore, $5 = u + v \\ge u + \\frac{4}{u}$. Hence, $u^2 - 5u + 4 \\le 0$. Thus, $1 \\le \\sqrt{u} \\le 2$, that is, $(\\sqrt{u} - 1)(\\sqrt{u} - 2) \\le 0$, and $3\\sqrt{u} \\ge u + 2$."
] | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0g2d | Problem:
Seien $a$, $b$, $c$, $d$ reelle Zahlen. Beweise:
$$
\left(a^{2}-a+1\right)\left(b^{2}-b+1\right)\left(c^{2}-c+1\right)\left(d^{2}-d+1\right) \geq \frac{9}{16}(a-b)(b-c)(c-d)(d-a)
$$ | [
"Solution:\n\nBeachte, dass $a^{2}-a+1=\\left(\\frac{a}{2}-1\\right)^{2}+\\left(\\frac{\\sqrt{3}}{2} a\\right)^{2}>0$ für alle $a \\in \\mathbb{R}$ gilt. Mit Cauchy-Schwarz und dann der Dreiecksungleichung erhalten wir\n$$\n\\begin{aligned}\n\\sqrt{\\left(a^{2}-a+1\\right)\\left(b^{2}-b+1\\right)} & =\\sqrt{\\left(... | Switzerland | SMO-Selektion | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
09lm | Consider a tournament with $n$ participants, where $n = 2m - 1$ and $m \ge 1$. This tournament has no draws, and each pair of students plays against each other exactly once. The students are listed based on their number of wins, with those having an equal number of wins being listed alphabetically. What is the maximum ... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | floor((3(m-1))/2) | |
08pt | Problem:
Let $X Y$ be a chord of a circle $\Omega$, with center $O$, which is not a diameter. Let $P, Q$ be two distinct points inside the segment $X Y$, where $Q$ lies between $P$ and $X$. Let $\ell$ the perpendicular line dropped from $P$ to the diameter which passes through $Q$. Let $M$ be the intersection point of ... | [
"Solution by PSC. At first, we will allow $P$ and $Q$ to coincide, and we will prove the inequality in this case. Let the perpendicular from $Q$ to $O Q$ meet $\\Omega$ at $B$ and $C$. Then, we have that $Q B=Q C$. We will show that\n$$\nB Q \\cdot X Y \\geq 2 Q X \\cdot Q Y\n$$\nBy the power of a point Theorem we ... | JBMO | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0k8f | Problem:
Convex hexagon $A B C D E F$ is drawn in the plane such that $A C D F$ and $A B D E$ are parallelograms with area $168$. $A C$ and $B D$ intersect at $G$. Given that the area of $A G B$ is $10$ more than the area of $C G B$, find the smallest possible area of hexagon $A B C D E F$. | [
"Solution:\n\nSince $A C D F$ and $A B D E$ have area $168$, triangles $A B D$ and $A C D$ (which are each half a parallelogram) both have area $84$. Thus, $B$ and $C$ are the same height away from $A D$, and since $A B C D E F$ is convex, $B$ and $C$ are on the same side of $A D$. Thus, $B C$ is parallel to $A D$,... | United States | HMMT February 2019 February 16, 2019 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 196 | |
0g86 | 令 $n$ 是一正整數且令 $a_1, \cdots, a_{n-1}$ 為任意實數。定義數列 $u_0, \cdots, u_n$ 與 $v_0, \cdots, v_n$ 如下：
$$
u_0 = u_1 = v_0 = v_1 = 1,\ u_{k+1} = u_k + a_k u_{k-1},\ v_{k+1} = v_k + a_{n-k} v_{k-1}\ \text{ 對於 } k = 1, \cdots, n-1.$$
試證：$u_n = v_n$.
Let $n$ be a positive integer and let $a_1, \cdots, a_{n-1}$ be arbitrary real numb... | [
"We prove by induction on $k$ that\n$$\nu_k = \\sum_{0 < i_1 < \\dots < i_t < k \\atop i_{j+1} - i_j \\ge 2} a_{i_1} \\dots a_{i_t} \\quad (1)$$\nNote that we have one trivial summand equal to 1 (which corresponds to $t=0$ and the empty sequence, whose product is 1).\n\nFor $k=0, 1$ the sum on the right-hand side o... | Taiwan | 2014 International Mathematical Olympiad Second Stage Training Camp | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
06xz | Determine all real numbers $\alpha$ such that the number
$$
\lfloor\alpha\rfloor+\lfloor 2 \alpha\rfloor+\cdots+\lfloor n \alpha\rfloor
$$
is a multiple of $n$ for every positive integer $n$. (Here $\lfloor z\rfloor$ denotes the greatest integer less than or equal to $z$.) | [
"Answer: All even integers satisfy the condition of the problem and no other real number $\\alpha$ does so.\n\nSolution 1. First we will show that even integers satisfy the condition. If $\\alpha=2 m$ where $m$ is an integer then\n$$\n\\lfloor\\alpha\\rfloor+\\lfloor 2 \\alpha\\rfloor+\\cdots+\\lfloor n \\alpha\\rf... | IMO | IMO2024 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | All even integers | |
0410 | Find all prime numbers $p$, for which there are infinitely many positive integers $n$, such that: $p \mid n^{n+1} + (n+1)^n$. | [] | China | China Western Invitational Mathematical Competition | [
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof and answer | All odd primes | |
0fbx | Problem:
Se consideran 17 enteros positivos tales que ninguno de ellos tiene un factor primo mayor que 7. Demuestra que, al menos, el producto de dos de estos números es un cuadrado perfecto. | [
"Solution:\n\nTodos estos 17 números se pueden descomponer en factores primos de la forma $2^{a} 3^{b} 5^{c} 7^{d}$, donde sus exponentes $a, b, c, d$ son enteros no negativos.\n\nSi dos números de este tipo, es decir $2^{a} 3^{b} 5^{c} 7^{d}$ y $2^{a'} 3^{b'} 5^{c'} 7^{d'}$ se multiplican, su producto es $2^{a+a'}... | Spain | null | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
00wo | Problem:
An equilateral triangle is divided into $25$ congruent triangles enumerated with numbers from $1$ to $25$. Prove that one can find two triangles having a common side and with the difference of the numbers assigned to them greater than $3$. | [
"Solution:\n\nDefine the distance between two small triangles to be the minimal number of steps one needs to move from one of the triangles to the other (a step here means transition from one triangle to another having a common side with it). The maximum distance between two small triangles is $8$ and this maximum ... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0a6g | Problem:
Let $\omega$ be the incircle of scalene triangle $ABC$. Let $\omega$ be tangent to $AB$ and $AC$ at points $X$ and $Y$. Construct points $X'$ and $Y'$ on line segments $AB$ and $AC$ respectively such that $AX' = XB$ and $AY' = YC$. Let line $CX'$ intersect $\omega$ at points $P, Q$ such that $P$ is closer to $... | [
"Solution:\nLet $a, b, c$ be the sidelengths $BC, AC, AB$ respectively, and let $s$ be the semiperimeter of triangle $ABC$ (i.e. let $s = \\frac{a + b + c}{2}$). Since $X$ and $Y$ are the points of contact of the incircle we get $AX = AY$. Similarly $BX = BZ$ and $CY = CZ$ where $Z$ is the point of tangency between... | New Zealand | NZMO Round One | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nin... | null | proof only | null | |
0i9b | Problem:
How many 2-digit positive integers have an even number of positive divisors? | [
"Solution:\n\nAn integer has an odd number of divisors precisely if it is a square. So we take the 90 2-digit numbers $10, 11, \\ldots, 99$ and remove the 6 squares $4^{2}, 5^{2}, \\ldots, 9^{2}$, for a total of 84."
] | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | final answer only | 84 | |
02d9 | Show that for any positive integer $n > 2$ we can find $n$ distinct positive integers such that the sum of their reciprocals is $1$. | [
"We have $\\frac{1}{2} + \\frac{1}{3} + \\frac{1}{6} = 1$.\n\nNow we have $\\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\cdots + \\frac{1}{2^n} + \\frac{1}{2^n} = 1$, replace the second $\\frac{1}{2^n}$ by $\\left(\\frac{1}{2} + \\frac{1}{3} + \\frac{1}{6}\\right) \\frac{1}{2^n}$ to get $n + 3$ terms ($n \\ge 1$):... | Brazil | II OBM | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof only | null | |
0a84 | Problem:
Let $C_{1}$ and $C_{2}$ be two circles intersecting at $A$ and $B$. Let $S$ and $T$ be the centres of $C_{1}$ and $C_{2}$, respectively. Let $P$ be a point on the segment $AB$ such that $|AP| \neq |BP|$ and $P \neq A, P \neq B$. We draw a line perpendicular to $SP$ through $P$ and denote by $C$ and $D$ the po... | [
"Solution:\n\n(See Figure 10.) The power of the point $P$ with respect to the circles $C_{1}$ and $C_{2}$ is $PA \\cdot PB = PC \\cdot PD = PE \\cdot PF$. Since $SP$ is perpendicular to the chord $CD$, $P$\n\n\n\nFigure 10.\n\nhas to be the midpoint of $CD$. So $PC = PD$. In a similar manne... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 12 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
0fnc | Una **configuración** es un conjunto finito $S$ de puntos del plano entre los cuales no hay tres colineales y a cada punto se le asigna algún color, de modo que si un triángulo cuyos vértices están en $S$ tiene un ángulo mayor o igual a $120^\circ$, entonces exactamente dos de sus vértices son de un mismo color. Hallar... | [
"El número máximo de puntos que puede tener una configuración es $25$. Primero observamos que dados $6$ puntos del plano, entre los cuales no hay tres colineales, se determina al menos un triángulo con un ángulo de medida mayor o igual a $120^\\circ$. En efecto, si la envolvente convexa de los puntos es un hexágono... | Spain | Olimpiada Iberoamericana de Matemáticas | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combi... | Spanish | proof and answer | 25 | |
04vr | Let $ABCDE$ be a convex pentagon such that $AB = BC$, $AE = DE$, $AC \perp AD$ and $CD \parallel BE$. Prove that triangles $ABC$ and $ADE$ have equal areas. (Patrik Bak) | [] | Czech Republic | School Round | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
09go | Let $n$ be a natural number. Prove that $2^n + 3^n + 5^n$ is not a perfect square. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
09i2 | Find the smallest positive integer $n$ with the following property:
For any sequence $a_1, a_2, \dots, a_{2021}$ of real numbers satisfying
$$
0 < a_1, a_2, \dots, a_{2021} < 2 \quad \text{and} \quad a_1 + a_2 + \dots + a_{2021} = 2021,
$$
there is a sequence $b_1, b_2, \dots, b_n$ of real numbers satisfying
$$
0 < b_1... | [
"Answer: $n = 2021$.\nFirst we show that $n = 2021$ is the minimum.\nLet $N = 2021$ and $m = n + N$ and suppose that the sequence $c_1, c_2, \\dots, c_m$ satisfies\n$$\n\\begin{cases} c_1 + c_2 + \\dots + c_l \\le l, & \\text{for all } 1 \\le l \\le m \\text{ odd} \\\\ c_1 + c_2 + \\dots + c_l \\ge l, & \\text{for ... | Mongolia | Round 3 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2021 | |
02iz | Problem:
Quantos números entre $1$ e $601$ são múltiplos de $3$ ou múltiplos de $4$?
A) $100$
B) $150$
C) $250$
D) $300$
E) $430$ | [
"Solution:\n\nPara achar o número de múltiplos de $3$ compreendidos de $1$ a $601$, basta usar o algoritmo da divisão e escrever $601 = 200 \\times 3 + 1$. Isso mostra que $3 \\times 1, 3 \\times 2, \\ldots, 3 \\times 200$ são os múltiplos de $3$ de $1$ a $601$, ou seja, temos $200$ destes múltiplos. Do mesmo modo ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | MCQ | D | |
0jkf | Problem:
Bob writes a random string of 5 letters, where each letter is either $A$, $B$, $C$, or $D$. The letter in each position is independently chosen, and each of the letters $A$, $B$, $C$, $D$ is chosen with equal probability. Given that there are at least two $A$'s in the string, find the probability that there a... | [
"Solution:\n\nThere are $\\binom{5}{2} 3^{3} = 270$ strings with $2$ $A$'s. There are $\\binom{5}{3} 3^{2} = 90$ strings with $3$ $A$'s.\nThere are $\\binom{5}{4} 3^{1} = 15$ strings with $4$ $A$'s. There is $\\binom{5}{5} 3^{0} = 1$ string with $5$ $A$'s.\n\nThe desired probability is $\\frac{90+15+1}{270+90+15+1}... | United States | HMMT 2014 | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 53/188 | |
0b59 | Problem:
Pentru orice număr natural nenul $n$, suma $1+\frac{1}{2}+\ldots+\frac{1}{n}$ se scrie sub forma $\frac{p_{n}}{q_{n}}$ unde $p_{n}$ şi $q_{n}$ sunt numere naturale prime între ele.
a. Arătați că $p_{67}$ nu se divide cu 3.
b. Determinați toate numerele naturale nenule $n$, pentru care $p_{n}$ este divizibil... | [] | Romania | 53. Bulgarian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a) p_67 is not divisible by 3.
b) The numerator p_n is divisible by 3 exactly for n = 2 and for all n congruent to 7 modulo 9. | |
068i | Let $ABC$ be a triangle with $AB < AC$, inscribed in a circle $c$ of center $O$. Let $G$ be its barycenter and $D, E, F$ the foot of the altitudes from $A, B, C$, respectively. If the rays $AG, GD$ intersect $c$ at $M, N$ respectively, prove that the points $F, E, M, N$ are cocyclic. | [
"Let $K$ be the midpoint of $BC$ and $P$ the second intersection of $GD$ with $c$. Then $c$ and the Euler's circle are homothetic with center $G$ and ratio $-2$, so\n$$\n\\frac{GP}{GD} = \\frac{GA}{GK} = 2 \\Rightarrow GP = 2GD.\n$$\nFrom the power of a point theorem we have\n$$\nGM \\cdot GA = GN \\cdot GP \\Right... | Greece | Selection Examination | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | proof only | null | |
0gwx | Prove that inequality
$$ \frac{x}{\sqrt{x^2 + y + z}} + \frac{y}{\sqrt{x + y^2 + z}} + \frac{z}{\sqrt{x + y + z^2}} \le \sqrt{3} $$
is true for any random nonnegative real numbers $x, y, z$ which obey equality $x^2 + y^2 + z^2 = 3$. | [
"Cauchy-Bunyakowsky inequality implies that you just have to prove the following inequality:\n$$\n\\left( \\frac{x}{x^2 + y + z} + \\frac{y}{y^2 + z + x} + \\frac{z}{z^2 + x + y} \\right) (x + y + z) \\le 3. \\text{ And since} \\\\ (x + y + z)^2 \\le 3(x^2 + y^2 + z^2) = 9, \\text{ it's enough to prove that}\n$$\n$... | Ukraine | Ukrajina 2008 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0kk2 | Problem:
Let $X$ be the number of sequences of integers $a_{1}, a_{2}, \ldots, a_{2047}$ that satisfy all of the following properties:
- Each $a_{i}$ is either $0$ or a power of $2$.
- $a_{i} = a_{2i} + a_{2i+1}$ for $1 \leq i \leq 1023$.
- $a_{1} = 1024$.
Find the remainder when $X$ is divided by $100$. | [
"Solution:\nThis problem can be visualized as a complete binary tree with $2048$ leaves, such that each node contains the sum of its two children. Let $f(p)$ be the number of ways to fill in a binary tree with $2^{p}$ leaves and the root having value $2^{p}$. We want $f(10)$.\n\nSince all values must be a power of ... | United States | HMMT November 2021 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 15 | |
04pt | A semicircle is correctly placed in a larger semicircle if their diameters are parallel, the endpoints of the diameter of the smaller semicircle lie on the arc of the larger semicircle, and the arc of the smaller semicircle touches the diameter of the larger semicircle.
A sequence of semicir... | [] | Croatia | Croatian Mathematical Society Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | π/3 | |
0akz | Let $ABC$ be a triangle such that $AB < AC$. Let $D$ be a point on the segment $BC$ such that $BD < CD$. The angle bisectors of $\angle ADB$ and $\angle ADC$ meet the segments $AB$ and $AC$ at $E$ and $F$ respectively. Let $\omega$ be the circumcircle of $\triangle AEF$ and $M$ be the midpoint of $EF$. The ray $AD$ mee... | [] | North Macedonia | Team Selection Test for BMO | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous"
] | English | proof only | null | |
0dwg | Problem:
Ana je izbrala števke $1,2,3,4,5,6,7$ in $9$. Odločila se je, da bo oblikovala skupine s po 4 dvomestnimi praštevili in da bo za vsako skupino praštevil uporabila vse izbrane števke. Kolikšna je vsota praštevil posamezne skupine? | [
"Solution:\n\nAna lahko z izbranimi števkami oblikuje skupino praštevil $\\{23,41,59,67\\}$. Da ne bi iskali vseh možnih skupin praštevil, razmišljajmo drugače. Dvomestno praštevilo se ne more končati z nobeno izmed števk $2,4,5$ oziroma $6$, zato te števke nastopajo na mestu desetic. Kakor koli Ana oblikuje skupin... | Slovenia | 48. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 190 | |
0bvc | The continuous functions $f, g: [0, \infty) \to [0, \infty)$ have the properties:
(i) $f(0) = g(0) = 0$;
(ii) $g(x) \neq 0$, for every $x > 0$;
(iii) $f(x + g(f(x))) = f(x)$, for every $x \ge 0$.
Show that $f(x) = 0$, for every $x \ge 0$. | [
"Let $A = \\{x \\ge 0 : f(x) = 0\\}$. By (i), $0 \\in A$.\n\nLet $x \\in A$. Then $f(x) = 0$.\nBy (iii), $f(x + g(f(x))) = f(x)$. Since $f(x) = 0$, $g(f(x)) = g(0) = 0$ (by (i)).\nSo $f(x + 0) = f(x)$, i.e., $f(x) = f(x)$, which is trivial.\n\nBut this does not help us extend $A$ directly. Instead, let us try to sh... | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof only | null | |
0fpq | En un triángulo $ABC$ la bisectriz por $A$, la mediana por $B$ y la altura por $C$ son concurrentes y además la bisectriz por $A$ y la mediana por $B$ son perpendiculares. Si el lado $AB$ mide una unidad, hallar cuánto miden los otros dos lados. | [
"Sean $P$, $M$ y $Q$ los pies de la bisectriz por $A$, la mediana por $B$ y la altura por $C$, respectivamente, que se cortan en el punto $X$. En el triángulo $ABM$ la bisectriz por $A$, $AX$, es perpendicular a $BM$ (puesto que por hipótesis la mediana y la bisectriz de $ABC$ son perpendiculares), por tanto $\\ang... | Spain | LII Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triang... | Spanish | proof and answer | 2 and sqrt(33)/3 | |
0kzg | A list of 9 real numbers consists of $1$, $2.2$, $3.2$, $5.2$, $6.2$, and $7$, as well as $x$, $y$, and $z$ with $x \le y \le z$. The range of the list is $7$, and the mean and the median are both positive integers. How many ordered triples $(x, y, z)$ are possible?
(A) 1 (B) 2 (C) 3 (D) 4 (E) infinitely many | [
"Because the range is $7$, the values of $x$, $y$, and $z$ are in the interval $[0, 8]$. Because the median is an integer, it is one of $x$, $y$, or $z$ and is either $3$, $4$, $5$, or $6$. The sum of the list is $s = 24.8 + x + y + z$, which is between $24.8 + 0 + 0 + 3 = 27.8$ and $24.8 + 6 + 8 + 8 = 46.8$. Becau... | United States | AMC 10 B | [
"Statistics > Mathematical Statistics",
"Math Word Problems"
] | null | MCQ | C | |
089b | Problem:
Sia $ABCD$ un quadrato. Si descriva il luogo di punti $P$ del piano diversi da $A, B, C, D$ per i quali
$$
\widehat{APB} + \widehat{CPD} = 180^{\circ} \text{.}
$$ | [
"Solution:\n\nRicordiamo dapprima che, dati due punti $A, B$ del piano e fissato un angolo $\\alpha$, il luogo dei punti $Q$ tali che $\\widehat{AQB} = \\alpha$ è costituito dall'unione di due archi di circonferenza, uno per ogni semipiano individuato dalla retta $AB$.\n\nIdentifichiamo innanzitutto il nostro luogo... | Italy | Cesenatico | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof and answer | The locus is exactly the union of the two diagonals AC and BD and the two arcs AB and CD of the circumcircle of the square, excluding the vertices. | |
0gba | 令 $N$ 表示所有正整數所成的集合。試求所有一對一函數 $f: N \to N$ 使得
$$
f^{f(a)}(b) f^{f(b)}(a) = (f(a+b))^2
$$
成立, 對所有的正整數 $a, b$. 此處 $f^k(n)$ 表示 $\underbrace{f(f(\dots f(n)\dots))}_{k}$ | [
"答: $f(n) = n+1$, 對所有的正整數 $n$.\n\nLet $f$ be such a solution.\n\n**First step.** The pre-image of 1 is empty.\n\nObviously, if there is an integer $x$ so that $f(x) = 1$. Set $a = b = x$ into (1), then\n$$\nf(2x)^2 = f^{f(x)}(x) f^{f(x)}(x) = f(x)^2 = 1\n$$\nThis implies $f(2x) = 1$. But, it's impossible since $f$ ... | Taiwan | 二〇一七數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | f(n) = n+1 for all positive integers n | |
0dfp | Let $f(x) = x^2 - 6x + 5$. Draw on the plane the set of pairs $(x, y)$ that satisfy the following system of inequalities
$$
\begin{cases}
f(x) + f(y) \leq 0 \\
f(x) - f(y) \geq 0
\end{cases} .
$$ | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0jct | Problem:
A parking lot consists of 2012 parking spots equally spaced in a line, numbered 1 through 2012. One by one, 2012 cars park in these spots under the following procedure: the first car picks from the 2012 spots uniformly randomly, and each following car picks uniformly randomly among all possible choices which m... | [
"Solution:\n$\\frac{1}{2062300}$\n\nWe see that for 1 to be the last spot, 2 must be picked first (with probability $\\frac{1}{n}$), after which spot $n$ is picked. Then, cars from 3 to $n-1$ will be picked until there are only gaps of 1 or 2 remaining. At this point, each of the remaining spots (including spot 1) ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 1/2062300 | |
071g | Problem:
A $A \subset \{1,2,3, \ldots, 49\}$ does not contain six consecutive integers. Find the largest possible value of $|A|$. How many such subsets are there (of the maximum size)? | [
"Solution:\n\nWe must exclude at least one element of each of the 8 sets $\\{1,2, \\ldots, 6\\}, \\{7, \\ldots, 12\\}, \\{13, \\ldots, 18\\}, \\ldots, \\{43, \\ldots, 48\\}$. So $|A| \\leq 41$. But a value of 41 is certainly possible, for example, exclude $2,8,14, \\ldots, 44$.\n\nThe largest excluded element must ... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | Maximum size = 41; Number of maximum subsets = 495 | |
0emv | Let $a_1, a_2, \dots, a_n$ be real numbers such that
$$
a_m + a_{m+1} + \dots + a_n \ge m + (m+1) + \dots + n
$$
for every $m = 1, 2, \dots, n$. Prove that
$$
a_1^2 + a_2^2 + \dots + a_n^2 \ge \frac{n(n+1)(2n+1)}{6}.
| [
"Make the substitution $a_k = b_k + k$ for $k = 1, 2, \\dots, n$. The conditions of the problem then become $b_m + b_{m+1} + \\dots + b_n \\ge 0$ for every $m = 1, 2, \\dots, n$, while the required inequality becomes\n$$\n\\begin{aligned}\n\\frac{n(n+1)(2n+1)}{6} &\\le (b_1+1)^2 + (b_2+2)^2 + \\dots + (b_n+n)^2 \\\... | South Africa | South-Afrika 2011-2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
07cp | There are 6 points on the plane such that no three of them are collinear. We know that among every 4 points of them, there exists a point that its power with respect to the circle passing through the other three points is a constant value $k$ (power of a point in the interior of a circle has a negative value). Prove th... | [
"In any quadruple of the points, consider the point which has the power $k$ with respect to the circle passing through the other three, name these points **good** points. We claim that there are two quadruples with the same **good** point, and two other common points.\nThere are $\\binom{6}{4} = 15$ quadruples, eac... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | k = 0 and all six points lie on a common circle | |
0557 | Juku thought of a 3-digit number that, when reversing the order of the digits, stays the same 3-digit number. Juku noticed that when adding 2016 to that number, the 4-digit number that arises is again the same 4-digit number when reading the digits from right to left. What number did Juku think of? | [
"Let the number be $aba$ and let the number we get by adding $2016$ be $cddc$. Clearly $c$ can only be $2$ or $3$.\n\nIf $c = 2$ then by the ones digit the only possibility is $a = 6$, and we have a carry from the ones to the tens digit. By the tens digit then $b + 1 + 1 = d$ or $b + 1 + 1 = d + 10$. The second opt... | Estonia | Open Contests | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 646 | |
0bhq | Find all polynomials $P, Q$ with rational coefficients, with the property that the polynomial $P + aQ$ is irreducible for every rational $a$. | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | null | proof and answer | Exactly the pairs (P, Q) over the rationals where either Q is the zero polynomial and P is irreducible, or P has degree one and Q is a constant polynomial. | |
080f | Problem:
In un cubo di lato $12$, $P$ e $Q$ sono i centri di due facce che hanno in comune lo spigolo $AB$. Qual è il volume del tetraedro che ha per vertici i punti $A$, $B$, $P$, $Q$? | [
"Solution:\n\nLa risposta è $72$. Infatti il tetraedro $ABPQ$ può essere pensato come una piramide di base $ABP$ e altezza $QM$, dove $M$ è il punto medio di $AB$. Se il lato del cubo è $12$, l'area di base sarà $\\frac{144}{4} = 36$ e l'altezza sarà $\\frac{12}{2} = 6$. Il volume del tetraedro sarà dunque $\\frac{... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Geometry > Solid Geometry > Volume"
] | null | proof and answer | 72 | |
0act | In a circle with diameter $d$ the chords $AB$ and $CD$ are perpendicular and intersect at a point $E$ distinct from the centre of the circle. Prove that $\overline{AE}^2 + \overline{BE}^2 + \overline{CE}^2 + \overline{DE}^2 = d^2$. | [
"$AB$ intersects the circumference in a point $M$. The line, which passes through $M$ and is parallel to $CD$, intersects the circumference in a point $N$. Because $\\angle MDC = 90^\\circ$, the segment $MC$ passes through the centre of the circle, which means that $\\angle MNC = 90^\\circ$, we conclude that the qu... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
0g08 | Problem:
Trouver tous les ensembles finis et non-vides $A$ de fonctions $f: \mathbb{R} \rightarrow \mathbb{R}$ tels que :
Pour tous $f_{1}, f_{2} \in A$, il existe $g \in A$ telle que pour tout $x, y \in \mathbb{R}$
$$
f_{1}\left(f_{2}(y)-x\right)+2 x=g(x+y)
$$ | [
"Solution:\n\nPour commencer, on montre que $A$ est ensemble de fonctions surjectives. En effet, pour $f \\in A$ et $f_{1}=f_{2}=f$ on a, avec $x=-y$,\n$$\nf(f(y)+y)=g(0)+2 y\n$$\net donc $f$ est surjective.\n\nDe plus, avec $x=0$, on obtient que si $f_{1}, f_{2} \\in A$ alors $f_{1} \\circ f_{2} \\in A$.\nComme $A... | Switzerland | IMO-Selektion | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | the singleton set containing only the identity function | |
01d8 | There are $n$ students at a school. It is known that, for any two students $A$ and $B$, either $A$ loves $B$ or $B$ loves $A$ (but not both). (A student may thus be in love with several other students.) A *love triangle* is a configuration of three students $A$, $B$, $C$ such that $A$ loves $B$, $B$ loves $C$ and $C$ l... | [
"**Answer:** $\\frac{n(n^2-1)}{24}$ if $n$ is odd and $\\frac{n(n^2-4)}{24}$ if $n$ is even.\n\nLet $a_i$ be the number of students that student $i$ loves ($1 \\le i \\le n$). Then the number of love triangles is\n$$\nT = \\binom{n}{3} - \\binom{a_1}{2} - \\dots - \\binom{a_n}{2}.\n$$\nThis is explained as follows.... | Baltic Way | Baltic Way 2016 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | n(n^2−1)/24 if n is odd, and n(n^2−4)/24 if n is even | |
0bp7 | Problem:
Determinaţi cel mai mic număr natural $n$ pentru care, oricum am colora în roşu $n$ dintre vârfurile unui cub, există un vârf al cubului care are cele trei vârfuri alăturate roşii. | [
"Solution:\n\nVom demonstra că numărul căutat este $5$.\n\nFie $A B C D A' B' C' D'$ un cub. Colorând în roşu cele $4$ vârfuri ale unei feţe a cubului, fie ele de exemplu $A, B, C, D$, nu există niciun vârf care să aibă cele trei vârfuri alăturate roşii, deci $n \\geq 5$.\n\nAcum, pentru $n=5$, oricum am colora $n$... | Romania | Primul test de selecţie pentru OBMJ | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 5 | |
0fn6 | Obtén los dos valores enteros de $x$ más próximos a $2013^\circ$, tanto por defecto como por exceso, que cumplen esta ecuación trigonométrica:
$$
2^{\sin^2 x} + 2^{\cos^2 x} = 2\sqrt{2}
$$ | [
"Aplicando la desigualdad entre las medias aritmética y geométrica resulta\n$$\n2^{\\sin^2 x} + 2^{\\cos^2 x} \\geq 2\\sqrt{2^{\\sin^2 x} \\cdot 2^{\\cos^2 x}} = 2\\sqrt{2^{\\sin^2 x + \\cos^2 x}} = 2\\sqrt{2}\n$$\nLa igualdad se alcanza cuando $2^{\\sin^2 x} = 2^{\\cos^2 x}$. Es decir, cuando $\\sin^2 x = \\cos^2 ... | Spain | Olimpiada Matemática Española | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | Spanish | proof and answer | 1935°, 2025° | |
0iir | Problem:
Find all real solutions $(x, y)$ of the system $x^{2}+y=12=y^{2}+x$. | [
"Solution:\nWe have $x^{2}+y = y^{2}+x$ which can be written as $(x-y)(x+y-1)=0$.\n\nThe case $x = y$ yields $x^{2}+x-12=0$, hence $(x, y) = (3, 3)$ or $(-4, -4)$.\n\nThe case $y = 1-x$ yields $x^{2}+1-x-12 = x^{2}-x-11=0$ which has solutions $x = \\frac{1 \\pm \\sqrt{1+44}}{2} = \\frac{1 \\pm 3\\sqrt{5}}{2}$. The ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (3, 3), (-4, -4), ((1 + 3*sqrt(5))/2, (1 - 3*sqrt(5))/2), ((1 - 3*sqrt(5))/2, (1 + 3*sqrt(5))/2) | |
0l57 | Let $\triangle ABC$ be a right triangle with $\angle A = 90^\circ$ and $BC = 38$. There exist points $K$ and $L$ inside the triangle such that
$$
AK = AL = BK = CL = KL = 14.
$$
The area of the quadrilateral $BKLC$ can be expressed as $n\sqrt{3}$ for some positive integer $n$. Find $n$. | [
"Because $AL = CL = KL$, the circumcenter of $\\triangle ACK$ is $L$. Similarly, the circumcenter of $\\triangle ABL$ is $K$.\nBecause $\\triangle AKL$ is equilateral, $\\angle ALK = \\angle AKL = 60^\\circ$, and it follows that $\\angle ACK = 30^\\circ = \\angle ABL$.\nBecause\n$$\n\\angle BAK = \\angle BAC - \\an... | United States | AIME II | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angl... | null | final answer only | 104 | |
07ik | Given a positive integer $n$, we say two sets $A$ and $B$ of integers are congruent modulo $n$ if and only if the residues of elements of $A$ modulo $n$ are the same as the residues of elements of $B$ modulo $n$, and we write $A \equiv B \pmod{n}$. For example,
$$
\{3, 5, 8\} \equiv \{-4, 0, 9, 15\} \pmod{3}
$$
We say ... | [
"We shall prove that this can only happen whenever $n$ is prime. We shall prove the following lemma;\n\n**Lemma 1.** Let $p$ be a prime number and $x_1, \\cdots, x_r$ be pair-wise distinct (mod $p$) and $y_1, \\cdots, y_r$ be distinct positive integers. Then, there is a polynomial $P(x)$ with integer coefficients s... | Iran | 41th Iranian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All prime numbers greater than four | |
0g3h | Problem:
For each prime $p$, somewhere in the multiverse there exists a kingdom consisting of $p$ islands numbered from $1$ to $p$ with a bridge between any pair of them. When Jana visits a kingdom, coronavirus restrictions mean she must obey the following rule: Directly after visiting island $m$, she can only cross o... | [
"Solution:\n\nNote that the divisibility condition is symmetrical in $m$ and $n$. In other words: If Jana can cross over from island $m$ to island $n$, she can also go the other way. Let us define a graph $G$ as follows: Each of the islands $1,2, \\ldots, p$ represents a vertex and we draw an edge between two verti... | Switzerland | IMO Selection | [
"Discrete Mathematics > Graph Theory",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n"
] | null | proof only | null | |
08o7 | Problem:
Let $a$, $b$ and $c$ be positive real numbers such that $a b c = \frac{1}{8}$. Prove the inequality
$$
a^{2} + b^{2} + c^{2} + a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} \geq \frac{15}{16}
$$
When does equality hold? | [
"Solution:\nBy using the Arithmetic-Geometric Mean Inequality for 15 positive numbers, we find that\n$$\n\\begin{aligned}\n& a^{2} + b^{2} + c^{2} + a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = \\\\\n& \\quad = \\frac{a^{2}}{4} + \\frac{a^{2}}{4} + \\frac{a^{2}}{4} + \\frac{a^{2}}{4} + \\frac{b^{2}}{4} + \\frac{b^{2}}... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Equality holds if and only if a = b = c = 1/2. | |
0df4 | Find all triples $(a, b, c)$ of real numbers satisfying
$$
a + b + c = 1 \quad \text{and} \quad 3(a + bc) = 4(b + ca) = 5(c + ab).
$$ | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | (a, b, c) = (1/2, 1/3, 1/6) and (a, b, c) being any permutation of (1, 1, -1). | |
087a | Problem:
Quanti interi positivi $n$ hanno la proprietà che la loro rappresentazione in base 2 coincide con la rappresentazione in base 3 di $2 n$ ?
(A) Nessuno
(B) 1
(C) 2
(D) più di 2, ma in numero finito
(E) infiniti. | [
"Solution:\n\nLa risposta è (C). Sia $n$ un intero positivo. Siano $a_{k}, a_{k-1}, \\ldots, a_{0}$ le $k+1$ cifre, da sinistra a destra, della rappresentazione di $n$ in base 2. Ognuno degli $a_{i}$ assumerà un valore tra 0 e 1, e possiamo supporre senza perdita di generalità che $a_{k}=1$. Perciò $n=2^{k} a_{k}+2... | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | MCQ | (C) | |
0589 | (a) Find the largest number expressible as the difference of two two-digit numbers obtained from each other by changing the order of digits.
(b) The same question with three-digit instead of two-digit numbers. | [
"(a) Let the given two-digit number be $\\overline{ab}$. The only number that can be obtained by changing the order of digits is $\\overline{ba}$. The difference of these numbers is $(10a + b) - (10b + a) = 9(a - b)$. To obtain the largest difference, $a$ must be as large as possible and $b$ as small as possible. S... | Estonia | Estonian Math Competitions | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (a) 72; (b) 801 | |
02do | Show that there are only finitely many solutions to
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{1983}
$$
in positive integers. | [
"Suppose without loss of generality that $a \\leq b \\leq c$. We must have $a \\leq 3 \\cdot 1983$, otherwise $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} < \\frac{1}{1983}$. So there are only finitely many possible values for $a$.\n\nNow consider the number of solutions for fixed $a$. We have $\\frac{1}{b} + \\frac... | Brazil | V OBM | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
08an | Problem:
Una scacchiera quadrata $(2n+1) \times (2n+1)$, con $n > 0$, è colorata in modo tale che ogni casella sia bianca o nera. Una casella è detta speciale se ci sono almeno altre $n$ caselle dello stesso colore nella sua riga, e almeno altre $n$ caselle dello stesso colore nella sua colonna.
a. Dimostrare che esi... | [] | Italy | Olimpiade Italiana di Matematica | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2n+1 | |
06im | Let $\ell(n)$ denote the greatest odd factor of any natural number $n$. Find the sum
$$ \ell(1) + \ell(2) + \ell(3) + \cdots + \ell(2^{2015}). $$ | [
"Note that $\\ell(n) = d$ if and only if $n = 2^k d$ for some nonnegative integer $k$ and odd integer $d$. Let $a_d$ be the number of positive integers $n$ with $\\ell(n) = d$ and $1 \\le n \\le 2^{2015}$. Therefore, for any odd $d \\le 2^{2015}$, $a_d$ is 1 plus the largest integer $k$ such that $2^k d \\le 2^{201... | Hong Kong | 1997-2023 IMO HK TST | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | null | proof and answer | (2^{4030} + 2)/3 | |
0c99 | Let $D$ be a point in the interior of triangle $ABC$ such that $\angle BAD = 40^\circ$, $\angle DAC = 30^\circ$, $\angle BCD = 20^\circ$ and $\angle DCA = 50^\circ$. Find $\angle CBD$.
 | [
"Observe that $ABC$ is an isosceles triangle, with $BA = BC$ and $\\angle ABC = 40^\\circ$. Suppose that the perpendicular bisector of the triangle's base intersects $AD$ at $T$.\nSince $\\angle DAC < \\angle DCA$, we have $CD < DA$ therefore $T \\in (AD)$ and $\\angle TBC = 20^\\circ$.\n\nWe will prove that $D$ is... | Romania | RMC 2020 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof and answer | 10° | |
00s5 | The acute-angled triangle $ABC$ with circumcenter $O$ is given. The midpoints of the sides $BC$, $CA$ and $AB$ are $D$, $E$ and $F$ respectively. An arbitrary point $M$ on the side $BC$, different from $D$, is chosen. The straight lines $AM$ and $EF$ intersect at the point $N$ and the straight line $ON$ cuts again the ... | [
"The straight lines $DO$, $EO$ and $FO$ are the perpendicular bisectors of the sides $BC$, $CA$ and $AB$ respectively. It follows that $[OM]$ is the diameter of the circumscribed circle of the triangle $ODM$ and $MP \\perp ON$. The point $O$ is the orthocenter of the triangle $DEF$ (see the picture).\nLet $O_1$ be ... | Balkan Mathematical Olympiad | BMO 2017 | [
"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 | |
01co | Let $P$ be a real polynomial of degree $2015$ and $Q$ a real quadratic polynomial. Could it be that the polynomial $P(Q(x))$ has precisely the roots
$$
-2014, -2013, \dots, -2, -1, 1, 2, \dots, 2014, 2015, 2016?
$$ | [
"The values of $Q$ at the $4030$ points indicated in the problem need be a subset of the zeroes of $P$. But these are at most $2015$ in number, and $Q$, being quadratic, assumes any given value at most twice. Therefore, the $4030$ numbers can be split into $2015$ pairs $(p_i, q_i)$, for which $Q(p_i) = Q(q_i)$ runs... | Baltic Way | Baltic Way 2015 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | No | |
03m9 | Problem:
Show that there exists a positive integer $N$ such that for all integers $a > N$, there exists a contiguous substring of the decimal expansion of $a$ that is divisible by $2011$. (For instance, if $a = 153204$, then $15$, $532$, and $0$ are all contiguous substrings of $a$. Note that $0$ is divisible by $2011$... | [
"Solution:\nWe claim that if the decimal expansion of $a$ has at least $2012$ digits, then $a$ contains the required substring. Let the decimal expansion of $a$ be $a_{k} a_{k-1} \\ldots a_{0}$. For $i = 0, \\ldots, 2011$, let $b_{i}$ be the number with decimal expansion $a_{i} a_{i-1} \\ldots a_{0}$. Then by the p... | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof only | null | |
07in | Let $n$ be a positive integer. $2n$ persons are around a round table. Let $a \ge n$ be an integer. We call two persons *friends* if and only if there are exactly $a$ persons between one of the arcs passing between them. For each positive integer $n$ find all possible values of $a$ for which, we can choose $n$ persons, ... | [
"We claim that independent of the way we shall choose $n$ persons, the parity of the number of friendship relations among them is equal to the parity of $n - a$. We shall prove the following lemma;\n\n**Lemma 1.** *If we replace one person with her antipodal the parity of number of friendship relations among the pe... | Iran | 41th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | All integers a ≥ n with n ≡ a + 1 (mod 2). | |
03lf | Problem:
Find all ordered triples $(x, y, z)$ of real numbers which satisfy the following system of equations:
$$
\left\{\begin{array}{l}
x y = z - x - y \\
x z = y - x - z \\
y z = x - y - z
\end{array}\right.
$$ | [
"Solution:\nSubtracting the second equation from the first gives $x y - x z = 2 z - 2 y$. Factoring $y - z$ from each side and rearranging gives\n$$\n(x + 2)(y - z) = 0\n$$\nso either $x = -2$ or $z = y$.\n\nIf $x = -2$, the first equation becomes $-2 y = z + 2 - y$, or $y + z = -2$. Substituting $x = -2$, $y + z =... | Canada | Canadian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (-2, 0, -2), (-2, -2, 0), (0, 0, 0), (0, -2, -2), (-1, -1, -1) | |
00w7 | Problem:
Do there exist $1990$ relatively prime numbers such that all possible sums of two or more of these numbers are composite numbers? | [
"Solution:\n\nSuch numbers do exist. Let $M = 1990!$ and consider the sequence of numbers $1 + M, 1 + 2M, 1 + 3M, \\ldots$ For any natural number $2 \\leq k \\leq 1990$, any sum of exactly $k$ of these numbers (not necessarily different) is divisible by $k$, and hence is a composite number. It remains to show that ... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | Yes | |
0e8e | Each hexagon of an infinite honeycomb contains one natural number. This number equals the average number of five of the six numbers written in the six neighbouring hexagons. Prove that all the numbers in the honeycomb are the same.
 | [
"Since all the numbers in the honeycomb are positive integers their minimum exists. Let us denote it by $n$. It suffices to show the following: if one of the hexagons contains the number $n$, then all of the neighbouring hexagons also contain $n$. Since there exists at least one hexagon containing $n$, this will im... | Slovenia | National Math Olympiad 2013 - Final Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0dd7 | Find all positive integers $a, b, c$ and prime number $p$ such that
$$
73p^2 + 6 = 9a^2 + 17b^2 + 17c^2.
$$ | [
"Suppose that $b \\le c$, we distinguish two cases regarding the value of $p$\n\n1. if $p \\ne 2$ implies $p$ is odd, so $p^2 \\equiv 1 \\pmod 8$. It follows that\n$$\na^2 + b^2 + c^2 \\equiv 9a^2 + 17b^2 + 17c^2 \\equiv 73p^2 + 6 \\equiv 7 \\pmod 8. \\quad (2)\n$$\nOtherwise, $a^2 + b^2 + c^2 \\equiv 0, 1, 2, 3, 4... | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | (a, b, c, p) = (1, 1, 4, 2) and (1, 4, 1, 2) |
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