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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
08e8 | Problem:
Zanobi e Veronica vanno in piscina assieme e iniziano in contemporanea a nuotare avanti e indietro, a velocità costanti ma diverse, ciascuno nella propria corsia, a partire dallo stesso lato della piscina. Veronica si accorge che, nel momento in cui completa 28 vasche (cioè finisce di percorrere per 28 volte ... | [
"Solution:\n\nLa risposta è $\\mathbf{( C )}$. Sappiamo che Zanobi è più lento di Veronica. In particolare, esiste un fattore $k>1$ tale che $v_{V}=k v_{Z}$, dove $v_{V}$ e $v_{Z}$ sono le velocità di Veronica e Zanobi, rispettivamente.\n\nVeronica completa 28 vasche in tempo $T$. Possiamo calcolare il numero $L_{Z... | Italy | Olimpiadi della Matematica | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | C | |
0dwk | Problem:
Dan je pravokotnik $ABCD$ z oglišči $A(-2,-1)$, $B(1,-1)$, $C(1,3)$, $D(-2,3)$. Izračunaj koordinati središča $S$ in polmer $R$ pravokotniku očrtane krožnice. Nariši sliko. | [
"Solution:\n\nSredišče $S$ pravokotniku očrtane krožnice je hkrati razpolovišče diagonale $AC$, zato je $S\\left(\\frac{x_{A}+x_{C}}{2}, \\frac{y_{A}+y_{C}}{2}\\right)$ oziroma $S\\left(-\\frac{1}{2}, 1\\right)$. S skice je razvidno, da je polmer $R$ enak polovici dolžine diagonale $AC$. Ker je $|AC|=\\sqrt{(x_{C}-... | Slovenia | 4. državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | final answer only | S = (-1/2, 1), R = 5/2 | |
067z | The integers $p$, $q$ and $r$ are primes and their product is equal to $n$. By increasing $p$ and $q$ by 1, then the product $(p+1)(q+1)r$ is equal to $n+138$. Determine all the possible values of $n$. | [
"We have:\n$$\n\\left\\{ \\begin{array}{l} pqr = n \\\\ (p+1)(q+1)r = n+138 \\end{array} \\right\\} \\Leftrightarrow \\left\\{ \\begin{array}{l} pqr = n \\\\ pqr + (p+q)r + r = n+138 \\end{array} \\right\\} \\Leftrightarrow \\left\\{ \\begin{array}{l} pqr = n \\\\ (p+q+1)r = 138 \\end{array} \\right\\}.\n$$\nFrom e... | Greece | 33rd Hellenic Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 138, 258, 854, 2294 | |
073t | Problem:
Let $P(x)$ be a given polynomial with integer coefficients. Prove that there exist two polynomials $Q(x)$ and $R(x)$, again with integer coefficients, such that (i) $P(x) Q(x)$ is a polynomial in $x^{2}$; and (ii) $P(x) R(x)$ is a polynomial in $x^{3}$. | [
"Solution:\n\nLet $P(x)=a_{0}+a_{1} x+a_{2} x^{2}+\\cdots+a_{n} x^{n}$ be a polynomial with integer coefficients.\n\nPart (i) We may write\n$$\nP(x)=a_{0}+a_{2} x^{2}+a_{4} x^{4}+\\cdots+x\\left(a_{1}+a_{3} x^{2}+a_{5} x^{5}+\\cdots\\right)\n$$\nDefine\n$$\nQ(x)=a_{0}+a_{2} x^{2}+a_{4} x^{4}+\\cdots-x\\left(a_{1}+a... | India | INMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0hqn | Problem:
Find $\prod_{n=2}^{\infty}\left(1-\frac{1}{n^{2}}\right)$. | [
"Solution:\n$\\prod_{n=2}^{\\infty}\\left(1-\\frac{1}{n^{2}}\\right) = \\prod_{n=2}^{\\infty} \\frac{n^{2}-1}{n^{2}} = \\prod_{n=2}^{\\infty} \\frac{(n-1)(n+1)}{n \\cdot n}$\n\n$= \\frac{1 \\cdot 3}{2 \\cdot 2} \\cdot \\frac{2 \\cdot 4}{3 \\cdot 3} \\cdot \\frac{3 \\cdot 5}{4 \\cdot 4} \\cdot \\frac{4 \\cdot 6}{5 \... | United States | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | final answer only | 1/2 | |
05k6 | Problem:
Prouver qu'il existe une unique manière de colorier chaque nombre rationnel strictement positif soit en rouge, soit en bleu, de sorte que :
- les nombres $x$ et $x+1$ sont de couleurs différentes ;
- les nombres $x$ et $\frac{1}{x}$ sont de la même couleur;
- le nombre $1$ est colorié en rouge. | [
"Solution:\n\nMontrons tout d'abord qu'un tel coloriage est nécessairement unique. Pour cela, nous allons démontrer par récurrence sur $\\max (a, b)$ que la couleur de la fraction $\\frac{a}{b}$ (supposée écrite sous forme irréductible) est entièrement déterminée. La conclusion est vraie lorsque $\\max (a, b)=1$ pu... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Functional equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
0h8z | Consider a triangle **ABC** and a line that does not coincide with the triangle sides and passes through the point **A**. This line meets altitudes $BH_2$ and $CH_3$ at the points $D_1$ and $E_1$, respectively. By $D_2$ and $E_2$ we denote points that are symmetrical to $D_1$ and $E_1$ with respect to the sides **AB** ... | [
"By $E_3, D_3$ we denote points that are symmetrical to $E_1, D_1$ with respect to **AB** and **AC**, respectively (Fig. 45). It follows from symmetry that $AE_1 = AE_2 = AE_3$, and **CA** is a bisector of an angle that is created by the lines $CE_3$ and $CE_2$. Then we have\n$$\n\\angle(CE_3, E_3A) = \\angle(E_1E_... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid,... | English | proof only | null | |
0foo | Problem:
En el triángulo $ABC$, sea $A'$ el punto simétrico de $A$ respecto del circuncentro $O$ de $ABC$.
a) Probar que la suma de los cuadrados de los segmentos de tangentes trazadas desde $A$ y $A'$ a la circunferencia inscrita en $ABC$ es igual a $4R^2 - 4Rr - 2r^2$.
b) Sea $I$ el incentro del $\triangle ABC$. P... | [
"Solution:\n\na. Si $t$ y $t'$ son las longitudes de los segmentos de tangentes, se tiene\n$$\nt^2 = AI^2 - r^2 \\quad \\text{y} \\quad t'^2 = A'I^2 - r^2\n$$\npor lo tanto,\n\n$$\nt^2 + t'^2 = AI^2 + A'I^2 - 2r^2\n$$\nAplicando el teorema de Apolonio al triángulo $AIA'$ con la mediana $IO$... | Spain | LI Olimpiada matemática Española (Concurso Final) | [
"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 > Distance cha... | null | proof only | null | |
0eze | Problem:
What is the greatest number of sides of a convex polygon that can equal its longest diagonal? | [
"Solution:\nAnswer: 2, except for the equilateral triangle.\nIt is easy to find two. Take the two sides to be $AB$ and $AC$ with angle $BAC = 60^{\\circ}$, and take the other vertices on the minor arc of the circle center $A$ radius $AB$ between $B$ and $C$.\n\nLet the longest diagonal have length $k$. Suppose ther... | Soviet Union | 4th ASU | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 2, except for the equilateral triangle | |
0bam | Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that:
$$
f(xy + x + y) + f(xy - x - y) = 2(f(x) + f(y)), \forall x, y \in \mathbb{R}. \quad (1)
$$
Prove that $f$ fulfills the relation:
$$
f(x + y) = f(x) + f(y), \forall x, y \in \mathbb{R}. \quad (2)
$$ | [
"$x = y = 0$ yields $f(0) = 0$ and $y = 0$ yields $f(-x) = -f(x)$, $x \\in \\mathbb{R}$.\n\n$y = 1 \\Rightarrow f(2x + 1) = 2f(x) + f(1) \\quad (3)$\n\n$y = -1 \\Rightarrow f(2x - 1) = 2f(x) - f(1) \\quad (4)$\n\nDenote $x * y = xy + x + y$ and notice that (ASOC): $x * (y * z) = (x * y) * z$.\nFor $x = 1$, $y * 1 =... | Romania | 2011 CLOCK-TOWER SCHOOL SENIORS COMPETITION | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof only | null | |
0fsh | Problem:
Es sei $ABCD$ ein Rechteck und $P$ sei ein Punkt auf der Geraden $CD$. $M$ und $N$ seien die Mittelpunkte von $AD$ und $BC$. Die Gerade $PM$ schneide $AC$ in $Q$. Zeige, dass $MN$ die Winkelhalbierende des Winkels $\angle QNP$ ist. | [
"Solution:\n\nSei $S$ der Schnittpunkt von $AB$ und $QN$. Wir zeigen im Folgenden, dass $\\frac{AS}{SB} = \\frac{DP}{PC}$ gilt. Daraus folgt offensichtlich, dass $\\triangle MSN$ und $\\triangle MPN$ kongruent sind und somit wären wir fertig.\n\nDie Punkte $S$, $N$ und $Q$ liegen nach Konstruktion auf einer Geraden... | Switzerland | IMO - Selektion | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04v5 | In the domain of non-negative real numbers solve the system of equations
$$
\begin{align*}
\lfloor 3x + 5y + 7z \rfloor &= 7z, \\
\lfloor 3y + 5z + 7x \rfloor &= 7x, \\
\lfloor 3z + 5x + 7y \rfloor &= 7y.
\end{align*}
$$ | [
"The first equation of the given system is fulfilled if and only if the following two conditions are satisfied:\n▷ the number $7z$ is integer,\n▷ $7z \\leq 3x + 5y + 7z < 7z + 1$, i.e. $3x + 5y \\in [0, 1)$.\n\nSimilarly, the second and third equations are fulfilled if and only if the numbers $7x$ and $7y$ are inte... | Czech Republic | 72nd Czech and Slovak Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (0, 0, 0), (1/7, 0, 0), (0, 1/7, 0), (0, 0, 1/7) | |
03z0 | Let set $M = \{1, 2, 3, ..., 50\}$. Find all positive integer $n$, such that there are at least two different elements $a$ and $b$ in any subset with 35 elements of $M$, such that $a + b = n$ or $a - b = n$. (posed by Li Shenghong) | [
"Take $A = \\{1, 2, 3, ..., 35\\}$, then for any $a, b \\in A$,\n$$\na - b \\le 34,\\ a + b \\le 34 + 35 = 69.\n$$\nIn the following, we show that $1 \\le n \\le 69$. Let $A = \\{a_1, a_2, \\dots, a_{35}\\}$, without loss of generality, suppose that $a_1 < a_2 < \\dots < a_{35}$.\n\ni.\nIf $1 \\le n \\le 19$, by\n$... | China | China Southeastern Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 1 <= n <= 69 | |
01ft | Let $ABC$ be a triangle with $AB = AC$. Let $M$ be the midpoint of $BC$. Let circles with diameters $AC$, $BM$ intersect at points $M$, $P$. Let $MP$ intersect $AB$ at $Q$. Let $R$ be a point on $AP$ such that $QR \parallel BP$. Prove that $CP$ bisects $\angle RCB$. | [
"Since $\\{ACMP\\}$ is cyclic, we have $\\angle RPQ = \\angle ACM$. Moreover, $\\angle PQR = 90^\\circ = \\angle CMA$. Hence $\\triangle PQR \\sim \\triangle CMA$. It follows that\n$$\n\\frac{PR}{PQ} = \\frac{CA}{CM}. \\qquad (1)\n$$\nSimilarly, $\\angle PMB = \\angle PAC$ and $\\angle BPM = 90^\\circ = \\angle CPA... | Baltic Way | Baltic Way 2019 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
09tp | Problem:
Gegeven is een positief geheel getal $n$. Bepaal alle positieve reële getallen $x$ met
$$
n x^{2} + \frac{2^{2}}{x+1} + \frac{3^{2}}{x+2} + \ldots + \frac{(n+1)^{2}}{x+n} = n x + \frac{n(n+3)}{2}.
$$ | [
"Solution:\nVoor $1 \\leq i \\leq n$ geldt\n$$\n\\frac{(i+1)^{2}}{x+i} = i+1 + \\frac{(i+1)^{2} - (i+1)(x+i)}{x+i} = i+1 + \\frac{i+1 - (i+1)x}{x+i} = i+1 + \\frac{(i+1)(1-x)}{x+i},\n$$\ndus kunnen we de linkerkant van de gegeven vergelijking herschrijven tot\n$$\nn x^{2} + 2 + 3 + \\ldots + (n+1) + \\frac{2(1-x)}{... | Netherlands | Selectietoets | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | x = 1 | |
095s | Problem:
Funcția $f:[0,1] \rightarrow [0,1]$ este crescătoare pe segmentul $[0,1]$. Demonstrați că există un număr $a \in [0,1]$, astfel încât $f(a)=a$. | [
"Solution:\n\nConsiderăm mulțimea $A=\\{x \\in [0,1] \\mid x \\leq f(x)\\}$. Fie $x_{0}=0$. Cum $f(x)$ este funcție crescătoare pe $[0,1]$, atunci $0 \\leq f(0) \\in [0,1]$. Deci, $0 \\in A \\Rightarrow A \\neq \\varnothing$.\n\nConform definiției $A \\subseteq [0,1]$. Ordonăm crescător elementele mulțimii $A$. Șir... | Moldova | A 62 - A OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Precalculus > Functions"
] | null | proof only | null | |
02yq | Problem:
Um paralelepípedo deve ser construído com a sobreposição de cubinhos de $1~\mathrm{cm}$ de medida de aresta, sendo seu comprimento composto por $n$ cubinhos, sua largura, por $p$ cubinhos e sua altura por $q$ cubinhos.
a) Qual o volume do paralelepípedo?
b) Pintando as faces do paralelepípedo de vermelho, q... | [
"Solution:\n\na) $V = n p q$.\n\nb) Nas faces $n \\times p$, apenas os cubinhos que não estão nas laterais terão apenas uma face pintada de vermelho, ou seja, são $2 \\cdot (n-2) \\cdot (p-2)$ cubinhos. De forma análoga, nas faces $n \\times q$, serão $2 \\cdot (n-2) \\cdot (q-2)$ e nas faces $p \\times q$ são $2 \... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a) V = npq. b) 2[(p−2)(q−2) + (n−2)(p−2) + (n−2)(q−2)]. c) n = 3 and a = 2. | |
0b17 | Problem:
The polynomial $p(x) = x^{2} - 3x + 1$ has zeros $r$ and $s$ and a quadratic polynomial $q(x)$ has leading coefficient $1$ and zeros $r^{3}$ and $s^{3}$. Find $q(1)$. | [
"Solution:\nWe have $r + s = 3$ and $rs = 1$. By Vieta's formulas, we compute\n$$\nr^{3} + s^{3} = (r + s)^{3} - 3rs(r + s) = 27 - 3(1)(3) = 18\n$$\nand $r^{3} s^{3} = 1$. Thus, we obtain $q(x) = (x - r^{3})(x - s^{3}) = x^{2} - (r^{3} + s^{3})x + r^{3} s^{3} = x^{2} - 18x + 1$. Hence, we get $q(1) = -16$."
] | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | -16 | |
0k9k | Problem:
For any real number $\alpha$, define
$$
\operatorname{sign}(\alpha)= \begin{cases}+1 & \text{ if } \alpha>0 \\ 0 & \text{ if } \alpha=0 \\ -1 & \text{ if } \alpha<0\end{cases}
$$
How many triples $(x, y, z) \in \mathbb{R}^3$ satisfy the following system of equations
$$
\begin{aligned}
& x=2018-2019 \cdot \ope... | [
"Solution:\n\nSince $\\operatorname{sign}(x+y)$ can take one of 3 values, $z$ can be one of 3 values: 4037, 2018, or $-1$. The same is true of $x$ and $y$. However, this shows that $x+y$ cannot be $0$, so $z$ can only be 4037 or $-1$. The same is true of $x$ and $y$. Now note that, if any two of $x, y, z$ are $-1$,... | United States | HMMT February 2019 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 3 | |
004u | Dados un número primo $p > 3$ y un número entero $x$, denotamos por $r(x) \in \{0,1,..., p-1\}$ al resto de $x$ módulo $p$. Sean $x_1, x_2,..., x_k$ ($2 < k < p$) números enteros distintos dos a dos módulo $p$ y no divisibles por $p$.
Decimos que un número $a \in \{1,2,..., p-1\}$ es bueno si:
$$
r(a_1) < r(a_2) < \do... | [] | Argentina | XVI Olimpiada Matemática Rioplatense | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | Spanish | proof only | null | |
0i6k | Problem:
For positive integers $L$, let $S_{L} = \sum_{n=1}^{L} \lfloor n / 2 \rfloor$. Determine all $L$ for which $S_{L}$ is a square number. | [
"Solution:\n\nWe distinguish two cases depending on the parity of $L$.\n\nSuppose first that $L = 2k - 1$ is odd, where $k \\geq 1$. Then\n$$\nS_{L} = \\sum_{1 \\leq n \\leq 2k-1} \\left\\lfloor \\frac{n}{2} \\right\\rfloor = 2 \\sum_{0 \\leq m < k} m = 2 \\cdot \\frac{k(k-1)}{2} = k(k-1).\n$$\nIf $k = 1$, this is ... | United States | HMMT 2002 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | L is even or L = 1 | |
08sm | Suppose that the combination $(n; a_1, a_2, \dots, a_n)$ of positive integers satisfy the condition $a_1 + a_2 + \dots + a_n = 2008$. Let for each $k$ with $1 \le k \le n$, $A_k = a_1a_2\cdots a_k$. Determine the maximum possible value that the quantity $A_1 + A_2 + \dots + A_n$ can take. | [
"$\\left[\\frac{47}{2} \\cdot 3^{667} - \\frac{3}{2}\\right]$\n\nFor $(a_1, a_2, \\dots, a_n)$ satisfying the condition of the problem, let us put $A_0 = 1$ and\n$$\nS(a_1, a_2, \\dots, a_n) = A_1 + A_2 + \\dots + A_n.\n$$\n\n**Lemma 1.** For $(a_1, a_2, \\dots, a_n)$ which yields the maximum possible value of $S(a... | Japan | Japan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | (47/2)*3^667 - 3/2 | |
05z3 | Problem:
Soient $ABC$ un triangle, $D, E$ les pieds des hauteurs issues de $A$ et $B$ respectivement. La droite $(DE)$ rencontre le cercle circonscrit à $ABC$ en deux points $P$ et $Q$. Soient $A'$, $B'$ les symétriques de $A$ et $B$ par rapport à $(BC)$ et $(AC)$ respectivement. Montrer que $A'$, $B'$, $P$, $Q$ sont ... | [
"Solution:\n\n\n\nSur la figure, il semble que $H$ l'orthocentre se situe sur le cercle en question. On va donc adopter la stratégie suivante : on va montrer que $A'$ et $B'$ sont sur le cercle circonscrit de $PQH$. De cette manière on aura bien $A'$, $B'$, $P$, $Q$ cocycliques.\n\nSoit $M$... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0383 | Problem:
Find all integers $a, b, c, d$ such that $a c - 3 b d = 5$ and $a d + b c = 6$. | [
"Solution:\nLet $a c - 3 b d = 5$ and $a d + b c = 6$ for some integers $a, b, c, d$. Then\n$$\n(a^2 + 3 b^2)(c^2 + 3 d^2) = (a c - 3 b d)^2 + 3(a d + b c)^2 = 5^2 + 3 \\cdot 6^2 = 133 = 7 \\cdot 19\n$$\nBecause of the symmetry between the pairs $(a, b)$ and $(c, d)$ we consider only the cases when $a^2 + 3 b^2 = 1... | Bulgaria | Spring Mathematical Competition | [
"Number Theory > Algebraic Number Theory > Quadratic forms",
"Number Theory > Algebraic Number Theory > Quadratic fields"
] | null | proof and answer | The solutions are (1,0,5,6), (-1,0,-5,-6), (5,6,1,0), (-5,-6,-1,0), (2,1,4,1), (-2,-1,-4,-1), (4,1,2,1), (-4,-1,-2,-1). | |
03st | In $\triangle PBC$, $\angle PBC = 60^\circ$. The tangent at point $P$ to the circumcircle $w$ of $\triangle PBC$ intersects with line $CB$ at $A$. Points $D$ and $E$ lie on the line segment $PA$ and circle $w$ respectively, such that $\angle DBE = 90^\circ$ and $PD = PE$. $BE$ and $PC$ meet at $F$. It is given that lin... | [
"By the angle bisector theorem, we have\n$$\n\\frac{PF}{FC} \\cdot \\frac{CB}{BA} \\cdot \\frac{AD}{DP} = \\frac{PB}{BC} \\cdot \\frac{BC}{BA} \\cdot \\frac{AB}{PB} = 1.\n$$\nBy the converse of Ceva theorem, the lines $AF$, $BP$ and $CD$ are concurrent.\nSuppose there exists $\\angle D'BF'$ satisfying the condition... | China | China Western Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Tri... | English | proof and answer | (6 + sqrt(3))/11 | |
0ju7 | Problem:
A cao [sic] has 6 legs, 3 on each side. A walking pattern for the cao is defined as an ordered sequence of raising and lowering each of the legs exactly once (altogether 12 actions), starting and ending with all legs on the ground. The pattern is safe if at any point, he has at least 3 legs on the ground and ... | [
"Solution:\n\n# 1 = on ground, 0 = raised, 2 = back on ground\ncache = {}\ndef pangzi(legs):\n if legs == (2,2,2,2,2,2): return 1\n elif legs.count(0) > 3: return 0\n elif legs[0] + legs[1] + legs[2] == 0: return 0\n elif legs[3] + legs[4] + legs[5] == 0: return 0\n elif cache.has_key(legs): return c... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Algorithms"
] | null | final answer only | null | |
08h9 | Problem:
A permutation $\sigma:\{1,2, \ldots, n\} \rightarrow\{1,2, \ldots, n\}$ is called straight if and only if for each integer $k, 1 \leq k \leq n-1$ the following inequality is fulfilled
$$
|\sigma(k)-\sigma(k+1)| \leq 2
$$
Find the smallest positive integer $n$ for which there exist at least 2003 straight permut... | [
"Solution:\nThe main trick is to look where $n$ is positioned. In that idea let us denote by $x_{n}$ the number of all the straight permutations and by $a_{n}$ the number of straight permutations having $n$ on the first or on the last position, i.e. $\\sigma(1)=n$ or $\\sigma(n)=n$. Also let us denote by $b_{n}$ th... | JBMO | Olimpiada Naţională de Matematică | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 16 | |
0491 | Lengths of all sides and diagonals of a rectangle are integers. Prove that its area is an integer divisible by $12$. | [] | Croatia | CroatianCompetitions2011 | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis"
] | null | proof only | null | |
020z | Problem:
A cyclic quadrilateral $A B X C$ has circumcentre $O$. Let $D$ be a point on line $B X$ such that $|A D|=|B D|$. Let $E$ be a point on line $C X$ such that $|A E|=|C E|$. Prove that the circumcentre of triangle $\triangle D E X$ lies on the perpendicular bisector of $O A$. | [
"Solution:\nFirst, note that $\\angle A O X=2 \\angle A B X=2\\left(180^{\\circ}-\\angle A C X\\right)=2 \\angle A C E$ as $A B X C$ is cyclic. Secondly, both $\\triangle D A B$ and $\\triangle E A C$ are isosceles, which implies that $\\angle A E X=\\angle A E C=180^{\\circ}-2 \\angle A C E=180^{\\circ}-\\angle A ... | Benelux Mathematical Olympiad | 13th Benelux Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler ... | null | proof only | null | |
00v3 | Let $A$, $B$, $C$, $D$ be fixed points on this order on a line. Let $\omega$ be a variable circle through $C$ and $D$ and suppose that it meets the perpendicular bisector of $CD$ at the points $X$ and $Y$. Let $Z$ and $T$ be the other points of intersection of $AX$ and $BY$ with $\omega$. Prove that $XY$ passes through... | [
"Let $M$ be the midpoint of $CD$ and let $Q$ and $R$ be the points of intersection of $XT$ and $YZ$ with $CD$ respectively. Since\n$$\n\\angle RZX = \\angle YZX = 90^\\circ = \\angle RMX \\quad \\text{and} \\quad \\angle QTY = \\angle XTY = 90^\\circ = \\angle QMY\n$$\nthe quadrilaterals $XZRM$ and $YTQM$ are cycli... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | English | proof only | null | |
0ind | Problem:
A sequence consists of the digits $122333444455555 \ldots$ such that each positive integer $n$ is repeated $n$ times, in increasing order. Find the sum of the 4501st and 4052nd digits of this sequence. | [
"Solution:\nAnswer: 13. Note that $n$ contributes $n \\cdot d(n)$ digits, where $d(n)$ is the number of digits of $n$. Then because $1+\\cdots+99=4950$, we know that the digits of interest appear amongst copies of two digit numbers. Now for $10 \\leq n \\leq 99$, the number of digits in the subsequence up to the la... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 13 | |
0h34 | Let $f(x) = \frac{x^3}{3x^2 - 3x + 1}$. Compute
$$
f\left(\frac{1}{2012}\right) + f\left(\frac{2}{2012}\right) + \dots + f\left(\frac{2012}{2012}\right).
$$ | [
"Зауважимо, що $f\\left(\\frac{i}{2012}\\right) = \\frac{i^3}{i^3 + (2012 - i)^3}$, $1 \\le i \\le 2012$. Тоді\n$$\nf\\left(\\frac{1006}{2012}\\right) = \\frac{1}{2}, \\quad f\\left(\\frac{2012}{2012}\\right) = 1, \\quad f\\left(\\frac{i}{2012}\\right) + f\\left(\\frac{2012 - i}{2012}\\right) = 1, \\quad 1 \\le i \... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 2013/2 | |
07ve | Segments $AB$ and $CD$ are parallel chords of a circle centre $O$, and $P$ is any point in the plane other than $O$. If $|PA| = |PB|$, prove $|PC| = |PD|$. | [
"A point $X$ is on the perpendicular bisector of $AB$ iff $|XA| = |XB|$. Therefore, the centre $O$ of the circle is on the perpendicular bisectors of $AB$ and $CD$. As $AB \\parallel CD$, this means that the two chords have the same perpendicular bisector. If $|PA| = |PB|$, $P$ must be on this common perpendicular ... | Ireland | IRL_ABooklet_2023 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
05ae | Does there exist an integer $a$ such that
$$
\frac{1}{\sqrt{2024}} < \frac{1}{\sqrt{a+1}} + \frac{1}{\sqrt{a+2}} + \dots + \frac{1}{\sqrt{a+2023}} < \frac{1}{\sqrt{2023}}?
$$ | [
"For each $i = 1, 2, \\dots, 2023$, we have\n$$\n\\frac{1}{\\sqrt{2023^3 + i}} < \\frac{1}{\\sqrt{2023^3}} = \\frac{1}{2023\\sqrt{2023}},\n$$\nthus\n$$\n\\frac{1}{\\sqrt{2023^3+1}} + \\frac{1}{\\sqrt{2023^3+2}} + \\dots + \\frac{1}{\\sqrt{2023^3+2023}} < 2023 \\cdot \\frac{1}{2023\\sqrt{2023}} = \\frac{1}{\\sqrt{20... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | a = 2023^3 | |
0evw | Problem:
Given three arbitrary infinite sequences of natural numbers, prove that we can find unequal natural numbers $m$, $n$ such that for each sequence the $m$th member is not less than the $n$th member. | [
"Solution:\nGiven any infinite sequence of natural numbers, we can find a non-decreasing subsequence (proof below).\n\nSo suppose the three sequences are $a$, $b$, and $c$. Take a non-decreasing subsequence of $a_i$. Suppose it is $a_{i_1}$, $a_{i_2}$, $a_{i_3}$, ... . Now consider the infinite sequence $b_{i_1}$, ... | Soviet Union | 1st ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
07ec | Morteza placed $8$ points on the plane. Then he writes down all the areas of the triangles constructed by these points (a total of $56$ numbers). Prove that he can always assign positive or negative signs to these numbers such that the sum of these numbers becomes zero. | [
"First of all consider $4$ points $A$, $B$, $C$ and $D$.\n\n\n\n\ni) If convex hull of these points is a triangle (without loss of generality let $D$ be the inner point)\n$$\nS_{\\triangle ABC} - S_{\\triangle ABD} - S_{\\triangle ACD} - S_{\\triangle BCD} = 0\n$$\n... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Other"
] | English | proof only | null | |
0gll | Let $ABC$ be an acute triangle with incenter $I$. The line through $I$ parallel to $AC$ intersects $AB$ at $M$; and the line through $I$ parallel to $AB$ intersects $AC$ at $N$. Let the line $MN$ intersect the circumcircle of $ABC$ at $X$ and $Y$. Let $Z$ be the midpoint of arc $BC$ (not containing $A$). Prove that $I$... | [
"It is not hard to see that $AMIN$ is a parallelogram. Furthermore, as $\\angle MAI = \\angle NAI$, $AMIN$ must indeed be a rhombus. It follows that $AI$ is perpendicular to $MN$. Since $AI$ bisects $\\angle BAC$, $A$, $I$, $Z$ are collinear. So we have $ZI$ perpendicular to $XY$.\n\n\n\nLe... | Thailand | The 13th Thailand Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
03yd | Let $n$ be an integer greater than or equal to $3$. For a permutation $p = (x_1, x_2, \dots, x_n)$ of $(1, 2, \dots, n)$, we say that $x_j$ lies in between $x_i$ and $x_k$ if $i < j < k$. (For example, in the permutation $(1, 3, 2, 4)$, $3$ lies in between $1$ and $4$, and $4$ does not lie in between $1$ and $2$.) Set ... | [
"The answer is $2^{n-1}$.\n\nWe first show that $m \\le 2^{n-1}$. We induct on $n$. The base case $n=3$ is trivial. (Indeed, say $3$ does not lie in between $1$ and $2$, then we can have $S = \\{(1, 2, 3), (3, 1, 2), (2, 1, 3), (3, 2, 1)\\}$.) Assume that the statement is true for $n=k$ (where $k \\ge 3$). Now cons... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2^{n-1} | |
0gux | Let $ABC$ be a scalene triangle, $H$ be its orthocenter and $G$ be its centroid. Let $A_b$ and $A_c$ be points on $AB$ and $AC$, respectively, such that $B$, $C$, $A_b$, $A_c$ are cyclic and the points $A_b$, $A_c$, $H$ are collinear. Let $O_a$ be the circumcenter of the triangle $AA_bA_c$. Define $O_b$ and $O_c$ analo... | [
"\nLet us define $B_a$, $B_c$, $C_a$, $C_b$ similarly. Let the circle passing through $B$, $C$, $A_b$, $A_c$ be $\\omega_a$ and let us define $\\omega_b$, $\\omega_c$ analogously. We will start with a lemma that we will use repeatedly.\n\n**Lemma:** Let $ABC$ be a triangle, $O$ be its circu... | Turkey | Team Selection Test for IMO 2024 | [
"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 > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Co... | English | proof only | null | |
00q8 | Let $n \in \mathbb{N}$, $n > 2$, and suppose $a_1, a_2, \dots, a_{2n}$ is a permutation of the numbers $1, 2, \dots, 2n$ such that $a_1 < a_3 < \dots < a_{2n-1}$ and $a_2 > a_4 > \dots > a_{2n}$. Prove that
$$
(a_1 - a_2)^2 + (a_3 - a_4)^2 + \dots + (a_{2n-1} - a_{2n})^2 > n^3.
$$ | [
"Denote $S = (a_1 - a_2)^2 + (a_3 - a_4)^2 + \\dots + (a_{2n-1} - a_{2n})^2$. We have\n$$\n\\begin{aligned}\nS &= \\sum_{i=1}^{2n} i^2 - 2(a_1a_2 + a_3a_4 + \\dots + a_{2n-1}a_{2n}) \\\\\n &= \\frac{n(2n+1)(4n+1)}{3} - 2(a_1a_2 + a_3a_4 + \\dots + a_{2n-1}a_{2n}).\n\\end{aligned}\n$$\nNext, observe that for each $... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
08g2 | Problem:
Alice disegna un cuore sul quaderno di matematica come segue: prima disegna due circonferenze di raggio $1~\mathrm{cm}$ e centri $O_1$, $O_2$, tangenti esternamente. Chiamata $r$ la tangente comune alle due circonferenze passante per il punto di contatto, sceglie poi un punto $P$ su $r$ in modo che si abbia $... | [
"Solution:\n\nLa risposta è (A). Siano $T_1$, $T_2$, $T_3$ i punti in cui le tangenti da $P$ toccano le due circonferenze, come in figura. Osserviamo che, per note proprietà delle tangenti, gli angoli $\\widehat{O_1 T_1 P}$, $\\widehat{O_1 T_2 P}$, $\\widehat{O_2 T_2 P}$, $\\widehat{O_2 T_3 P}$ sono retti. Notiamo ... | Italy | Olimpiadi di Matematica - Febbraio | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | A | |
0dk6 | Let $ABC$ be an acute non-isosceles triangle and inscribed in circle $(O)$ with the median line $A$ intersecting $(O)$ at $D$. Let $G$ be a point symmetrical to $A$ through $BC$. Suppose $GB, GC$ intersect $(O)$ respectively at $E, F$. The inscribed circle of triangle $AEF$ touches $EF$ at $J$.
1. Prove that $GO, GD$ a... | [
"1) Let $T$ be the intersection of the tangent at $A$ of $(O)$ with $BC$. Then, because $ABDC$ is a harmonic quadrilateral, $TD$ is also tangent to $(O)$. It follows that $A, D, G$ both belong to the circle of center $T$, which is also the Apollonius circle at vertex $A$ of triangle $ABC$. On the other hand, $(T)$ ... | Saudi Arabia | Saudi Arabia booklet 2024 | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Advanced Configurations > Po... | English | proof only | null | |
06hn | Let $a_1, a_2, \ldots, a_{24}$ be integers with sum $0$ and satisfying $|a_i| \le i$ for all $i$. Find the greatest possible value of $a_1 + 2a_2 + 3a_3 + \ldots + 24a_{24}$.
設 $a_1, a_2, \ldots, a_{24}$ 為整數, 它們之和為 $0$, 且對所有 $i$ 皆有 $|a_i| \le i$。求 $a_1 + 2a_2 + 3a_3 + \ldots + 24a_{24}$ 的最大可能值。 | [] | Hong Kong | HONG KONG PRELIMINARY SELECTION CONTEST | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English; Chinese | proof and answer | 1432 | |
06te | Consider an infinite sequence $a_{1}, a_{2}, \ldots$ of positive integers with $a_{i} \leqslant 2015$ for all $i \geqslant 1$. Suppose that for any two distinct indices $i$ and $j$ we have $i+a_{i} \neq j+a_{j}$.
Prove that there exist two positive integers $b$ and $N$ such that
$$
\left|\sum_{i=m+1}^{n}\left(a_{i}-b\r... | [
"We visualize the set of positive integers as a sequence of points. For each $n$ we draw an arrow emerging from $n$ that points to $n+a_{n}$; so the length of this arrow is $a_{n}$. Due to the condition that $m+a_{m} \\neq n+a_{n}$ for $m \\neq n$, each positive integer receives at most one arrow. There are some po... | IMO | 56th International Mathematical Olympiad Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
09ph | Problem:
Als we een verzameling punten in de ruimte hebben, mogen we een punt van de verzameling spiegelen in een ander punt van de verzameling en het beeld hiervan toevoegen aan de verzameling.
Als we beginnen met een verzameling bestaande uit zeven van de acht hoekpunten van een kubus, kunnen we dan het achtste hoek... | [
"Solution:\n\nAls we op de getallenlijn een punt $x$ spiegelen in een punt $y$, dan krijgen we als spiegelbeeld het punt $S_{y}(x) = y - (x - y) = 2y - x$. Hetzelfde geldt voor punten in de ruimte: spiegelen we een willekeurig punt $(x_{1}, x_{2}, x_{3})$ in een punt $(y_{1}, y_{2}, y_{3})$, dan krijgen we $S_{(y_{... | Netherlands | UITWERKINGEN TOETS TRAININGSKAMP | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | No | |
094x | Problem:
Define glueing of positive integers as writing their base ten representations one after another and interpreting the result as the base ten representation of a single positive integer.
Find all positive integers $k$ for which there exists an integer $N_{k}$ with the following property: for all $n \geq N_{k}$,... | [
"Solution:\n\nIf $3 \\mid k$, then for any $n \\in \\mathbb{N}$ of the form $n=3k+1$, the sum of the digits of all positive integers up to $n$ gives remainder 1 modulo 3. As a positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3, no matter how we glue $1,2, \\ldots, n$, the resu... | Middle European Mathematical Olympiad (MEMO) | MEMO Szeged | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All positive integers not divisible by 3 | |
0gf5 | 設 $n$ 是一個給定的正整數。某甲和某乙進行一個遊戲：甲決定一個不超過 $n$ 次的整係數多項式 $P(x)$，但是不告訴乙；乙的目標是決定是否存在一個整數 $k$ 使得 $P(x) = k$ 沒有整數解。乙可以進行下述的詢問：乙給甲一個常數 $c$，甲就會告訴乙有幾個整數 $t$ 滿足 $P(t) = c$；每次詢問需要花一塊錢。試問乙至少要付多少錢，才能保證達成他的目標？ | [
"乙至少要付 $n+1$ 元。\n\n我們首先證明：如果對所有的 $c \\in [0, n] \\cap \\mathbb{Z}$, $P(x) = c$ 都有整數解, 則 $P(x) = \\pm x + d$, 其中 $d$ 是某個常數。我們需要以下的引理：\n\n**引理.** 若整係數多項式 $P(x)$ 滿足 $P(a) = k$, $P(b) = k + 1$ 且 $a, b$ 皆為整數, 則 $|a - b| = 1$。\n\n(引理證明：由於 $b - a \\mid P(b) - P(a) = 1$，故 $|a - b| = 1$。)\n\n回到原題。根據假設，分別存在 $a_0, a_1, \\dots... | Taiwan | 2021 年台灣數學奧林匹亞考試試題 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | n+1 | |
079g | Natural numbers $a_1 < a_2 < \dots < a_n$ have this property that for every distinct $i$ and $j$, $a_i$ is divisible by $a_j - a_i$. Prove that for every $i < j$, $i a_j \le j a_i$ | [
"For all distinct $i, j$ that $i < j$, we have $a_j - a_i \\mid a_j$, and\n$$\na_j > a_j - a_1 > a_j - a_2 > \\cdots > a_j - a_i\n$$\nOn the other hand all of the above terms are divisors of $a_j$, so if the divisors of $a_j$ in decreasing order be $a_j > b_1 > b_2 > \\cdots > b_k$, then $a_j - a_i \\le b_i$. Since... | Iran | 27th Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0b4k | Problem:
The product of all real numbers $x$ satisfying $x^{2+\log_{3}(9x)} = \frac{2187}{x^{2}}$ can be written in the form $p/q$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. | [] | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | 730 | |
0fe3 | Problem:
¿Cuáles son los números enteros positivos que se pueden obtener de exactamente 2007 maneras distintas, como la suma de al menos dos números enteros positivos consecutivos? ¿Cuál es el menor de todos ellos?
Ejemplo: el número $9$ se escribe exactamente de dos maneras distintas:
$$
\begin{aligned}
& 9=4+5 \\
& 9... | [
"Solution:\n$$\nN=a+(a+1)+\\ldots+(a+n)=\\frac{(n+1)(2 a+n)}{2} \\Leftrightarrow 2 N=(n+1)(2 a+n)\n$$\nSi $n$ es par $(n+1)$ es impar y $(2 a+n)$ es par.\nSi $n$ es impar $(n+1)$ es par y $(2 a+n)$ es impar.\nSiempre que $2 N$ en su descomposición en factores primos tenga un factor impar distinto de uno, existe una... | Spain | null | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All positive integers N whose odd part has exactly 2008 divisors (equivalently, N has exactly 2008 odd divisors). The smallest such N is 3^250 * 5 * 7 * 11. | |
02ce | Problem:
A distância entre Coco da Selva e Quixajuba é $24~\mathrm{km}$. Dois amigos precisam ir de Quixajuba a Coco da Selva e um terceiro amigo precisa ir de Coco da Selva a Quixajuba. Eles possuem uma bicicleta que inicialmente está em Quixajuba. Cada um deles pode ir caminhando a velocidade de $6~\mathrm{km}/\mathr... | [
"Solution:\nChamaremos de $A$ e $B$ os amigos que estão em Quixajuba e $C$ o que está em Coco da Selva. Nossos personagens podem seguir a seguinte estratégia:\n- Na primeira hora, $A$ vai de bicicleta enquanto $B$ e $C$ irão caminhando. Depois dessa hora, $A$ e $C$ se encontram no quilômetro $18$ (medido desde Quix... | Brazil | Desafios | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof only | null | |
0hj0 | Problem:
Let $ABCDE$ be a convex pentagon with perimeter $1$. Prove that triangle $ACE$ has perimeter less than $1$. | [
"Solution:\n\nBy the triangle inequality, we have $AC < AB + BC$ and $CE < CD + DE$, so $AC + CE + EA < AB + BC + CD + DE + EA$ as desired."
] | United States | Berkeley Math Circle Monthly Contest 1 | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof only | null | |
04rm | Let $l(n)$ denote the greatest odd divisor of any natural number $n$. Find the sum
$$
l(1) + l(2) + l(3) + \cdots + l(2^{2013}). \qquad (\text{Michal Rol\'inek})
$$ | [
"For each natural $k$, the equalities $l(2k) = l(k)$ and $l(2k - 1) = 2k - 1$ are clearly valid. Thus we can add the values $l(n)$ over groups of numbers $n$ lying always between two consecutive powers of $2$. In this way we will prove by induction the formula\n$$\ns(n) = l(2^{n-1} + 1) + l(2^{n-1} + 2) + l(2^{n-1}... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | (4^2013 + 2)/3 | |
0hf2 | Prove that for any integer $n > 1$ there exists such a sequence of positive integers $a_1 \le a_2 \le \dots \le a_m = n$, $m > 1$ such that
$$
5(a_1^2 + a_2^2 + \dots + a_m^2) - 4(a_1a_2 + a_2a_3 + \dots + a_{m-1}a_m) \le 4n^2 + \frac{1}{2}(m+1).
$$ | [
"Rewrite the given inequality in the following way:\n$$\n(a_m - 2a_{m-1})^2 + \\dots + (a_2 - 2a_1)^2 + a_1^2 \\le \\frac{1}{2}(m+1)\n$$\nBuild a sequence $(b_n)$ such that: $b_1 = n$, if $b_i$ is even, then set $b_{i+1}$ to be $\\frac{1}{2}b_i$, otherwise define it as one of the numbers $\\frac{1}{2}(b_i + 1)$, $\... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Other",
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
0b27 | Problem:
A triangle has side lengths $7, 11, 14$. Find the length of its inradius. | [
"Solution:\n\nWe use Heron's formula. The semiperimeter of the triangle is $16$. Thus, the area of the triangle is $$\\sqrt{(16)(16-7)(16-11)(16-14)} = 12 \\sqrt{10}.$$ Since the area of a triangle is the product of the semiperimeter and the inradius, the length of the inradius is $$\\frac{12 \\sqrt{10}}{16} = \\fr... | Philippines | 22nd Philippine 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 > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | 3*sqrt(10)/4 | |
0l5s | Problem:
Let $ABCD$ be a trapezoid such that $AB \parallel CD$, $AD = 13$, $BC = 15$, $AB = 20$, and $CD = 34$. Point $X$ lies inside the trapezoid such that $\angle XAB = 2\angle XBA$ and $\angle XDC = 2\angle XCD$. Compute $XD - XA$. | [
"Solution:\n\nConstruct point $P$ on $AB$ such that $XA = XP$ and point $Q$ on $CD$ such that $XD = XQ$. The angle condition gives $QC = XQ = XD$ and $PB = XP = XA$. Moreover, $ADQP$ is an isosceles trapezoid.\nLet $S$ be the projection of $A$ onto $CD$, and let $T$ be on $CD$ such that $AT... | United States | HMMT February | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | 4 | |
0bdh | Let $f$ and $g$ be two polynomials with integral coefficients such that $\deg f > \deg g$ and $\deg f \ge 2$. If the polynomial $pf+g$ has a rational root for infinitely many primes $p$, prove that $f$ has a rational root. | [
"Since $\\deg f > \\deg g$, $|g(z)/f(z)| < 1$ for all complex numbers $z$ of large enough absolute value. Consequently, as $p$ runs through the infinite set of primes under consideration, the roots of the $pf + g$ all lie in some disc $|z| < R$, where $R$ does not depend on $p$; for if $|z|$ is large enough, then\n... | Romania | 64th NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0jm7 | Problem:
Triangle $A B C$ has sides $A B=14$, $B C=13$, and $C A=15$. It is inscribed in circle $\Gamma$, which has center $O$. Let $M$ be the midpoint of $A B$, let $B^{\prime}$ be the point on $\Gamma$ diametrically opposite $B$, and let $X$ be the intersection of $A O$ and $M B^{\prime}$. Find the length of $A X$. | [
"Solution:\n\nAnswer: $\\frac{65}{12}$\n\nSince $B^{\\prime} B$ is a diameter, $\\angle B^{\\prime} A B = 90^{\\circ}$, so $B^{\\prime} A \\parallel O M$, so $\\frac{O M}{B^{\\prime} A} = \\frac{B M}{B A} = \\frac{1}{2}$. Thus $\\frac{A X}{X O} = \\frac{B^{\\prime} A}{O M} = 2$, so $A X = \\frac{2}{3} R$, where $R ... | United States | HMMT 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"
] | null | proof and answer | 65/12 | |
0647 | Problem:
Es seien ein Dreieck $A B C$ mit Umkreis $\Omega$ sowie Punkte $A_{1}, B_{1}$ und $C_{1}$ auf den Dreiecksseiten $\overline{B C}, \overline{C A}$ und $\overline{A B}$ gegeben, sodass die drei Geraden $A A_{1}, B B_{1}$ und $C C_{1}$ einen Punkt $P$ gemeinsam haben.
Es ist zu zeigen, dass höchstens zwei der dr... | [
"Solution:\n\nWegen $\\measuredangle B P C+\\measuredangle C P A+\\measuredangle A P B=360^{\\circ}=\\left(180^{\\circ}-\\measuredangle B A C\\right)+\\left(180^{\\circ}-\\measuredangle C B A\\right)+\\left(180^{\\circ}-\\measuredangle A C B\\right)$ dürfen wir ohne Beschränkung der Allgemeinheit annehmen, dass $\\... | Germany | Auswahlklausur | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jmh | Problem:
Let $ABC$ be a triangle with sides $AB=6$, $BC=10$, and $CA=8$. Let $M$ and $N$ be the midpoints of $BA$ and $BC$, respectively. Choose the point $Y$ on ray $CM$ so that the circumcircle of triangle $AMY$ is tangent to $AN$. Find the area of triangle $NAY$. | [
"Solution:\n\nLet $G = AN \\cap CM$ be the centroid of $ABC$. Then $GA = \\frac{2}{3} GN = \\frac{10}{3}$ and $GM = \\frac{1}{3} CM = \\frac{1}{3} \\sqrt{8^{2} + 3^{2}} = \\frac{\\sqrt{73}}{3}$. By power of a point, $(GM)(GY) = GA^{2}$, so $GY = \\frac{GA^{2}}{GM} = \\frac{(10/3)^{2}}{\\frac{\\sqrt{73}}{3}} = \\fra... | United States | HMMT 2014 | [
"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"
] | null | proof and answer | 600/73 | |
0eu1 | For three mutually distinct real numbers $a_1$, $a_2$, $a_3$, define three real numbers $b_1$, $b_2$, $b_3$ as follows:
$$
b_j = \left( 1 + \frac{a_j a_i}{a_j - a_i} \right) \left( 1 + \frac{a_j a_k}{a_j - a_k} \right), \quad \{i, j, k\} = \{1, 2, 3\}.
$$
Prove the inequality
$$
1 + |a_1 b_1 + a_2 b_2 + a_3 b_3| \le (1... | [
"Define\n$$\nA = \\frac{a_1 a_2}{a_1 - a_2}, \\quad B = \\frac{a_1 a_3}{a_1 - a_3}, \\quad C = \\frac{a_2 a_3}{a_2 - a_3}.\n$$\nThen\n$$\n\\begin{align*}\n& a_1 b_1 + a_2 b_2 + a_3 b_3 \\\\\n= & a_1 (1+A) (1+B) + a_2 (1-A) (1+C) + a_3 (1-B) (1-C) \\\\\n= & a_1 + a_2 + a_3 + (a_1 - a_2)A + (a_1 - a_3)B \\\\\n& + (a_... | South Korea | Korean Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | Equality holds if and only if all three numbers are nonnegative. | |
0bff | Let $a, b, c \in (0, \frac{\pi}{2})$ so that $2(\tan a + \tan b + \tan c) = 3 \tan a \tan b \tan c$.
a) Prove that $\cos^2 a + \cos^2 b + \cos^2 c \ge 1$.
b) Find the minimum for
$$
\frac{1}{\sin^2 a} + \frac{1}{\sin^2 b} + \frac{1}{\sin^2 c}.
$$ | [] | Romania | Shortlisted Problems for the 64th NMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | 9/2 | |
0d6r | Let $A$ be a point outside the circle $\omega$. Two points $B, C$ lie on $\omega$ such that $A B, A C$ are tangent to $\omega$. Let $D$ be any point on $\omega$ ($D$ is neither $B$ nor $C$) and $M$ the foot of perpendicular from $B$ to $C D$. The line through $D$ and the midpoint of $B M$ meets $\omega$ again at $P$. P... | [
"Let $O$ be the center of $\\omega$ and $Q$ the intersection point of $C O$ and $\\omega$.\nSince $Q D$ and $B M$ are perpendicular to $C D$ then $B M \\parallel Q D$. Because $N$ is the midpoint of $B M$, which implies that $D(M, B, N, Q) = -1$.\nConsider this harmonic quartet with circle $\\omega$, one has $C P B... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | English | proof only | null | |
0ii5 | Problem:
For each positive integer $n$ let $S_{n}$ denote the set $\{1,2,3, \ldots, n\}$. Compute the number of triples of subsets $A, B, C$ of $S_{2006}$ (not necessarily nonempty or proper) such that $A$ is a subset of $B$ and $S_{2006}-A$ is a subset of $C$. | [
"Solution:\n\nLet $A_{o}, B_{o}, C_{o}$ be sets satisfying the said conditions. Note that $1 \\in A_{o}$ implies that $1 \\in B_{o}$ and $1 \\notin S_{2006}-A_{o}$ so that 1 may or may not be in $C_{o}$. Also,\n$1 \\notin A_{o}$ implies that $1 \\in S_{2006}-A_{o} \\subset C_{o}$ while 1 may or may not be in $B_{o}... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | final answer only | 4^{2006} | |
08yw | For every positive integer $n$, let $f(n)$ be the smallest positive integer such that $f(n)$ is prime to $n$ and $f(n) + 1$ is prime to $n + 1$. How many different integers will appear in $\{f(1), f(2), \dots, f(10^{10})\}$? | [
"11\n\nFirst, we prove $f(n) = p-1$ for a positive integer $n$ and the smallest prime $p$ that does not divide $n+1$. For any integer $m$ with $1 \\le m < p-1$, $m+1$ has a prime factor $q$ with $q < p$. Since $q$ must divide $n+1$ from the definition of $p$, $m+1$ cannot be prime to $n+1$, thus $f(n) \\ge p-1$. On... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 11 | |
0c2a | Given the rectangular cuboid $ABCD A'B'C'D'$, let $M$ be the center of the face $ABB'A'$. Denote by $M_1$ the foot of the perpendicular from $M$ to the line $B'C$ and by $M_2$ the foot of the perpendicular from $M$ to $AD'$. Prove that:
a) $[MM_1] = [MM_2]$;
b) the line $d$ of intersection between the planes $(MM_1M_... | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
04y1 | Find all quadruplets $(a, b, c, d)$ of real numbers satisfying the system
$$
\begin{aligned}
(a + b)(a^2 + b^2) &= (c + d)(c^2 + d^2), \\
(a + c)(a^2 + c^2) &= (b + d)(b^2 + d^2), \\
(a + d)(a^2 + d^2) &= (b + c)(b^2 + c^2).
\end{aligned}
$$ | [
"Let us set\n$$\nf(x, y) = (x + y)(x^2 + y^2).\n$$\nWe'll show that for any real $x$ the inequality $y \\ge z$ implies $f(x, y) \\ge f(x, z)$. After subtraction we see that\n$$\nf(x, y) - f(x, z) = \\frac{1}{2}(y - z)((x + y)^2 + (y + z)^2 + (z + x)^2) \\ge 0.\n$$\nMoreover, equality occurs when $y = z$ or $x = y =... | Czech-Polish-Slovak Mathematical Match | Czech-Polish-Slovak Match | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | (t, t, t, t) for any real t | |
0f52 | Problem:
$0 < x, y, z < \pi / 2$. We have $\cos x = x$, $\sin (\cos y) = y$, $\cos (\sin z) = z$. Which of $x, y, z$ is the largest and which the smallest? | [] | Soviet Union | 16th ASU | [
"Precalculus > Functions",
"Precalculus > Trigonometric functions"
] | null | proof and answer | y < x < z | |
0jfd | Let $ABC$ be a scalene triangle with $\angle BCA = 90^\circ$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK = BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL = AC$. The circumcircle ... | [
"Let $T'$ be the point on $AB$ such that $\\angle ACT' = \\angle BCT'$. Let $\\omega$ be the circle centered at $T'$ tangent to segments $AC$ and $BC$, and denote the points of tangency to $AC$ and $BC$ by $M$ and $N$ respectively. Note that $\\omega$ exists because $CT'$ is an angle bisector.\n\n![](attached_image... | United States | IMO Team Selection Test | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry >... | null | proof only | null | |
0g98 | 令 $R$ 與 $Q$ 分別表示實數與有理數所成的集合。試找出所有函數 $f : Q \rightarrow R \setminus \{0\}$ 使得對任意的有理數 $x, y$ 滿足
$$
(f(x))^2 f(2y) + (f(y))^2 f(2x) = 2f(x)f(y)f(x+y).
$$ | [
"令 $y = 0$ 得\n$$\nf^2(x)f(0) + f(0)^2f(2x) = 2f^2(x)f(0).\n$$\n$$\n\\text{故 } f(2x) = \\frac{f^2(x)}{f(0)}, \\quad f(2y) = \\frac{f^2(y)}{f(0)}. \\qquad (1)\n$$\n將 (1) 帶入題設條件，得\n$$\nf(x)f(y)(f(x)f(y) - f(0)f(x+y)) = 0. \\qquad (2)\n$$\n由 (2) 及 $f(x) \\neq 0, f(y) \\neq 0$, 得\n$$\nf(x)f(y) - f(0)f(x+y) = 0 \\qquad (... | Taiwan | 二〇一五數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = b c^x for all rational x, where b ≠ 0 and c > 0 | |
024l | Problem:
Carla escreveu no quadro-negro os números inteiros de $1$ até $21$. Diana deseja apagar alguns deles de tal modo que ao multiplicar os números restantes o resultado seja um quadrado perfeito.
a) Mostre que Diana deve apagar necessariamente os números $11$, $13$, $17$ e $19$ para conseguir seu objetivo.
b) Qu... | [
"Solution:\na) Se Diana decidir não apagar o número $11$ então o produto dos números restantes será da forma $P=11 \\times A$. Como $11$ é o único múltiplo de $11$ dentre os números escritos por Carla, então $A$ é o produto de números não divisíveis por $11$, e logo $A$ não é múltiplo de $11$. Assim, $P$ seria múlt... | Brazil | null | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 5 | |
04yc | For a positive integer $n$, an $n$-configuration is a family of sets $\langle A_{i,j} \rangle_{1\le i,j\le n}$. An $n$-configuration is called *sweet* if for every pair of indices $(i, j)$ with $1 \le i \le n-1$ and $1 \le j \le n$ we have $A_{i,j} \subseteq A_{i+1,j}$ and $A_{j,i} \subseteq A_{j,i+1}$. Let $f(n,k)$ de... | [
"Consider a sweet $n$-configuration $\\langle A_{i,j} \\rangle_{1\\le i,j\\le n}$ with $A_{n,n} \\subset \\{1, 2, \\dots, k\\}$. For any $x \\in \\{1, 2, \\dots, k\\}$ and $i \\in \\{1, 2, \\dots, n\\}$ define\n$$\np_x(i) = |\\{j: x \\in A_{i,j}\\}|.\n$$\nSince $A_{i,j} \\subseteq A_{i,j+1}$ for all suitable $i, j$... | Czech-Polish-Slovak Mathematical Match | CAPS Match 2024 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof and answer | f(2024^2, 2024) is larger | |
03c7 | Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$, that satisfy the inequalities
(i) $f(x + y) \geq f(x) + y$
(ii) $f(f(x)) \leq x$
for all positive $x$ and $y$. | [
"It follows from (i) that $f$ is strictly increasing function. Also, (ii) implies\n$$\nx + y \\geq f(f(x + y)). \\qquad (1)\n$$\n\nFurthermore, (i) gives $f(f(x+y)) \\ge f(f(x)+y)$ and the substitution $x \\to y$ and $y \\to f(x)$ in (i) implies $f(f(x)+y) \\ge f(x)+f(y)$.\nSince $f$ is increasing we have $\\lim_{x... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x for all positive real x | |
031l | Problem:
Let $n$ be a positive integer. Ann writes down $n$ different positive integers. Then Ivo deletes some of them (possibly none, but not all), puts the signs $+$ or $-$ in front of each of the remaining numbers and sums them up. Ivo wins if $2003$ divides the result; otherwise, Ann wins. Who has a winning strate... | [
"Solution:\n\nFor $n \\leq 10$ Ann wins by writing the numbers $1, 2, \\ldots, 2^{n-1}$. Indeed, the result Ivo can get is a non-zero integer between $-1023$ and $1023$, since it has the same sign as the largest remaining number $\\left(2^{j} > 2^{j}-1 = \\sum_{i=0}^{j-1} 2^{i}\\right)$.\n\nFor $n \\geq 11$ the set... | Bulgaria | 52. Bulgarian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Other"
] | null | proof and answer | Ann wins for n ≤ 10; Ivo wins for n ≥ 11. | |
0bbi | Given a positive integer $n = \prod_{i=1}^{s} p_i^{\alpha_i}$, we write $\Omega(n)$ for the total number $\sum_{i=1}^{s} \alpha_i$ of prime factors of $n$, counted with multiplicity. Let $\lambda(n) = (-1)^{\Omega(n)}$ (so, for example, $\lambda(12) = \lambda(2^2 \cdot 3^1) = (-1)^{2+1} = -1$).
Prove the following two... | [
"$\\lambda(n)$ (so $\\lambda$ is a completely multiplicative arithmetic function), hence $\\lambda(p) = -1$ for any prime $p$, and $\\lambda(k^2) = \\lambda(k)^2 = +1$ for all positive integers $k$.\n\nThen from some rank on we only have subsequences $(1, -1, 1, 1, \\dots, 1, -1, 1)$. By \"doubling\" such a subsequ... | Romania | 2011 Fourth ROMANIAN MASTER OF MATHEMATICS | [
"Number Theory > Number-Theoretic Functions",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
002a | Decidir si es posible dividir un cubo en exactamente $100$ cubos más pequeños, no necesariamente iguales. ¿Y en $51$ cubos? | [] | Argentina | XX OLIMPIADA MATEMÁTICA ARGENTINA | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | español | proof and answer | Possible for 100; impossible for 51. | |
09st | Problem:
Zij $k$ een positief geheel getal en geef de som van de cijfers van een positief geheel getal $n$ aan met $s(n)$. Bewijs dat er onder de positieve gehele getallen met $k$ cijfers evenveel getallen $n$ zijn die voldoen aan $s(n)<s(2 n)$ als getallen $n$ die voldoen aan $s(n)>s(2 n)$. | [
"Solution:\n\nWe bewijzen dat er onder de positieve gehele getallen met hoogstens $k$ cijfers evenveel getallen $n$ zijn die voldoen aan $s(n)<s(2 n)$ als getallen die voldoen aan $s(n)>s(2 n)$. Het gevraagde volgt hieruit door dit resultaat voor $k$ en voor $k-1$ te combineren.\n\nWe gaan nu elk getal $n$ met hoog... | Netherlands | IMO-selectietoets III | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Other"
] | null | proof only | null | |
05d0 | Let $n \ge 3$ be any natural number. A real number is written into every vertex of a regular $n$-gon in such a way that numbers in any two neighbouring vertices differ by at most $1$. Find the least non-negative real number $C$ such that, regardless of the choice of the numbers in the vertices, there exist two neighbou... | [
"Denote the vertices of the polygon as $V_1, \\dots, V_n$ and let the real numbers in these vertices be $v_1, v_2, \\dots, v_n$, respectively. Moreover, denote $V_{n+1} = V_1$ and $v_{n+1} = v_1$. Define the value of the side $V_iV_{i+1}$ of the polygon to be $s_i = v_{i+1} - v_i$.\n\nObserve that $v_1, \\dots, v_n... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof and answer | C = 1 if n is even; C = (n−1)/(n+1) if n is odd | |
08zm | We have an equilateral triangle $ABC$ with side length $3$, and points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$, respectively. Given that $BD = 1$ and $\angle ADE = \angle DEF = 60^\circ$, find the length of segment $AF$.
 | [
"$\\frac{7}{9}$\n\nFor the triangles $ABD$, $DCE$, and $EAF$, we have\n$$\n\\angle EDC = \\angle ADC - \\angle ADE = (\\angle DAB + \\angle ABD) - 60^\\circ = \\angle DAB\n$$\nand similarly,\n$$\n\\angle FEA = \\angle DEA - \\angle DEF = (\\angle EDC + \\angle DCE) - 60^\\circ = \\angle EDC\n$$\n\nAlso, since $\\an... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 7/9 | |
0k9e | Problem:
We are given a family $\mathcal{F}$ of functions from the set $\{1, \ldots, n\}$ to itself. A sequence $\left(f_{1}, \ldots, f_{k}\right)$ of functions in $\mathcal{F}$ is said to be $\operatorname{good}$ if $f_{k} \circ f_{k-1} \circ \cdots \circ f_{1}$ is a constant function. Prove that if there exists a go... | [
"Solution:\n\nSuppose there exists a good sequence.\nThen for any two $a, b \\in \\{1, \\ldots, n\\}$ there is some sequence of functions $g_{1}, \\ldots, g_{m}$ such that $g_{m} \\circ g_{m-1} \\circ \\cdots \\circ g_{1}$ maps $a$ and $b$ to the same point. Looking at the images $(a, b), \\left(g_{1}(a), g_{1}(b)\... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
03it | Problem:
The altitudes of a tetrahedron $ABCD$ are extended externally to points $A'$, $B'$, $C'$ and $D'$ respectively, where $AA' = k / h_a$, $BB' = k / h_b$, $CC' = k / h_c$ and $DD' = k / h_d$. Here, $k$ is a constant and $h_a$ denotes the length of the altitude of $ABCD$ from vertex $A$, etc. Prove that the centro... | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
0ibf | Problem:
Kate has four red socks and four blue socks. If she randomly divides these eight socks into four pairs, what is the probability that none of the pairs will be mismatched? That is, what is the probability that each pair will consist either of two red socks or of two blue socks? | [
"Solution:\n\n$3 / 35$\n\nThe number of ways Kate can divide the four red socks into two pairs is $\\binom{4}{2} / 2 = 3$. The number of ways she can divide the four blue socks into two pairs is also $3$. Therefore, the number of ways she can form two pairs of red socks and two pairs of blue socks is $3 \\cdot 3 = ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 3/35 | |
0jyj | Problem:
Let $P$ be a polynomial with integer coefficients such that $P(0)+P(90)=2018$. Find the least possible value for $|P(20)+P(70)|$. | [
"Solution:\n\nFirst, note that $P(x) = x^{2} - 3041$ satisfies the condition and gives $|P(70) + P(20)| = |4900 + 400 - 6082| = 782$.\n\nTo show that $782$ is the minimum, we show $2800 \\mid P(90) - P(70) - P(20) + P(0)$ for every $P$, since $-782$ is the only number in the range $[-782, 782]$ that is congruent to... | United States | HMMT November | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 782 | |
0g4z | Problem:
Sei $n$ eine natürliche Zahl. Roger hat einen quadratischen Garten der Grösse $(2 n+1) \times (2 n+1)$. Er errichtet Zäune, um diesen in rechteckige Beete zu unterteilen. Er möchte genau zwei horizontale $k \times 1$ Beete und genau zwei vertikale $1 \times k$ Beete für jede gerade Zahl $k$ zwischen 1 und $2 ... | [
"Solution:\n\nBetrachte die 4 grössten Rechtecke. Wir wollen zeigen, dass diese zusammen den Rand des Gartens bilden.\nBetrachte zuerst ein einzelnes vertikales $1 \\times 2 n$-Rechteck. Dieses muss mit der kurzen Kante sicher einen Rand berühren (ansonsten wäre oberhalb und unterhalb des Rechtecks eine Fläche mit ... | Switzerland | Zweite Runde 2023 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2^n | |
0cob | Do there exist three distinct nonzero integers $a$, $b$, $c$ such that $a + b + c = 0$, and $a^{13} + b^{13} + c^{13}$ is a square of a positive integer?
Существуют ли три попарно различных ненулевых целых числа, сумма которых равна нулю, а сумма тринадцатых степеней которых является квадратом некоторого натурального ... | [
"For a natural number $t$, the triple $3t$, $-t$, $-2t$ satisfies all conditions except, possibly, the last one. To ensure that $(3t)^{13} + (-t)^{13} + (-2t)^{13} = t^{13}(3^{13} - 1 - 2^{13})$ is a perfect square, it is sufficient to set, for example, $t = 3^{13} - 1 - 2^{13}$.\n\nThus, the triple $3t$, $-t$, $-2... | Russia | Regional round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English; Russian | proof only | null | |
07l9 | Let $a$, $b$, $n$ be positive integers such that
$$
\frac{1}{a} + \frac{1}{b} + \frac{n}{\text{lcm}(a, b)} = \frac{1}{\gcd(a, b)}
$$
(i) Prove that $n$ is odd;
(ii) Find all the possible values of $a$ and $b$ if $n = 2007$. | [
"(i) Let $d = \\gcd(a, b)$. There exist two positive integers $x$ and $y$ which are relatively prime such that $a = dx$, $b = dy$. Using this fact in the initial equality we find $x + y + n = xy$, that is, $(x - 1)(y - 1) = n + 1$. Since $x$ and $y$ are relatively prime numbers, at least one of them is odd. Then $(... | Ireland | Irska | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis,... | English | proof and answer | (i) n is odd. (ii) All solutions are (a, b) = (2d, 2009d), (9d, 252d), (252d, 9d), (2009d, 2d) for any integer d ≥ 1. | |
0gwa | Prove that for any real numbers $x$ and $y$, both taken on the segment $[0; \frac{\pi}{2}]$, the inequality
$$
\cos x + \cos y + |\cos(x + y)| \ge 1
$$
holds. | [
"Для $x, y \\in [0; \\pi/2]$ маємо:\n$$\n\\cos x + \\cos y + |\\cos(x+y)| \\geq \\cos x \\sin y + \\cos y \\sin x + \\cos^2(x+y) = \\\\\n= \\sin(x+y) + \\cos^2(x+y) \\geq \\sin^2(x+y) + \\cos^2(x+y) = 1.\n$$"
] | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
0bm9 | Find all real numbers $x$, $y$, $z$ which verify the equalities
$$
y = \frac{x^3 + 12x}{3x^2 + 4}, \quad z = \frac{y^3 + 12y}{3y^2 + 4}, \quad x = \frac{z^3 + 12z}{3z^2 + 4}.
$$ | [
"We have $y + 2 = \\frac{(x+2)^3}{3x^2+4}$, $y - 2 = \\frac{(x-2)^3}{3x^2+4}$ and the analogous relations.\nIf $y = 2$, then $x = z = 2$. If $y \\ne 2$, then $\\frac{y+2}{y-2} = \\left(\\frac{x+2}{x-2}\\right)^3$, $\\frac{z+2}{z-2} = \\left(\\frac{y+2}{y-2}\\right)^3$ and $\\frac{x+2}{x-2} = \\left(\\frac{z+2}{z-2}... | Romania | 66th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | (-2, -2, -2), (0, 0, 0), (2, 2, 2) | |
0by1 | Triangle $ABC$ has $\angle A = 90^\circ$, $\angle B = 30^\circ$, and $D$ is the foot of the altitude from $A$. Let $E \in (AD)$ be such that $DE = 3AE$ and let $F$ be the foot of perpendicular from $D$ on $BE$.
a) Prove that $AF \perp FC$.
b) Find the measure of the angle $AFB$. | [
"a) The hypothesis gives $AC = \\frac{1}{2}BC$ and $CD = \\frac{1}{2}AC$, hence $CD = \\frac{1}{4}BC$.\nFrom $\\triangle BDE \\sim \\triangle DFE$ follows $\\frac{BD}{DF} = \\frac{DE}{FE}$, that is $\\frac{DE}{BD} = \\frac{EF}{FD}$. Now $\\frac{DC}{DB} = \\frac{1}{3} = \\frac{AE}{DE}$ yields $\\frac{DE}{BD} = \\fra... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 150 degrees | |
0dx2 | Problem:
Dana je funkcija $f(x)=3^{x}$. Določi funkciji $g(x)=f(x+1)$ in $h(x)=4 f(x)-3$ ter izračunaj presečišče njunih grafov. | [
"Solution:\n\nKer je $g(x)=f(x+1)$, dobimo $g(x)$ tako, da v $f(x)$ vstavimo $x+1$ namesto $x$. Tako dobimo $g(x)=3^{x+1}$.\n\nPodobno dobimo $h(x)=4 \\cdot 3^{x}-3$.\n\nPresečišča grafov funkcij $g(x)$ in $h(x)$ dobimo tako, da enačimo enačbi obeh funkcij. Dobimo enačbo\n$$\n3^{x+1}=4 \\cdot 3^{x}-3.\n$$\nEnačbo u... | Slovenia | 6. državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | g(x) = 3^{x+1}, h(x) = 4·3^{x} − 3; intersection point: (1, 9) | |
0jy7 | Problem:
Find the number of pairs of integers $(x, y)$ such that $x^{2}+2 y^{2}<25$. | [
"Solution:\nWe do casework on $y$.\nIf $y=0$, we have $x^{2}<25$, so we get 9 values of $x$.\nIf $y= \\pm 1$, then $x^{2}<23$, so we still have 9 values of $x$.\nIf $y= \\pm 2$, we have $x^{2}<17$, so we have 9 values of $x$.\nIf $y= \\pm 3$, we have $x^{2}<7$, we get 5 values of $x$.\nTherefore, the final answer i... | United States | February 2017 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Other"
] | null | final answer only | 55 | |
0aba | Find $(x+y)^2$, if $\frac{2}{x} - \frac{2}{y} = 1$ and $y-x=1$. | [
"From $\\frac{2}{x} - \\frac{2}{y} = 2\\frac{y-x}{xy} = 2\\frac{1}{xy}$ we have that $xy=2$.\nTherefore,\n$$\n(x+y)^2 = x^2 + 2xy + y^2 = x^2 - 2xy + y^2 + 4xy = (x-y)^2 + 4xy = 1^2 + 4 \\cdot 2 = 9\n$$"
] | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 9 | |
03au | Excircle to the side $BC$ of $\triangle ABC$ is tangent to $BC$, $AB$ and $AC$ at points $D$, $E$ and $F$, respectively. Let $P$ be orthogonal projection of $D$ on $EF$. Prove that point $P$ lies on the circumcircle $k$ of $\triangle ABC$ if and only if $k$ passes through the midpoint of $EF$. | [] | Bulgaria | Selection test for 51. International Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"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 | |
08vi | Determine all the real-valued functions $f$ defined on the real line, for which the following identity is satisfied for any pair of real numbers $x, y$:
$$
f(f(x) - f(y)) = f(f(x)) - 2x^2f(y) + f(y^2).
$$ | [
"Let us label the equation given in the problem by $(\\star)$. We will show that the functions $f$ satisfying the equation $(\\star)$ are given by the following:\n$$\nf(x) = 0, \\quad x^2, \\quad -x^2, \\quad x^2 - 1, \\quad 1 - x^2.\n$$\nIt is easy to check that all of these functions satisfy the equation $(\\star... | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0; f(x) = x^2; f(x) = -x^2; f(x) = x^2 - 1; f(x) = 1 - x^2 | |
05ny | Problem:
Déterminer toutes les fonctions $f$ de $\mathbb{R}$ dans $\mathbb{R}$ telles que, pour tous $x$ et $y$ réels, on ait l'égalité
$$
f(x+y)=f(x-y)+f(f(1-x y))
$$ | [
"Solution:\n\nPour $y=0$ on trouve que $f(x)=f(x)+f(f(1))$ donc $f(f(1))=0$.\n\nPour $x=0$ on trouve $f(y)=f(-y)+f(f(1))$ donc $f(y)=f(-y)$ pour tout $y$.\n\nPour $y=1$ on obtient $f(x+1)=f(x-1)+f(f(1-x))$. En remplaçant $x$ par $-x+2$, il vient $f(-x+3)=f(-x+1)+f(f(x-1))$.\n\nOr, $f(x-1)=f(1-x)$ donc $f(f(1-x))=f(... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x)=0 for all real x | |
0elg | Problem:
Poenostavi izraz v množici realnih števil: $$\left(\frac{1}{x^{4}-27 x}-\frac{1}{x^{4}-3 x^{3}}\right): \frac{x^{n+2}-9 x^{n}}{x^{n+5}+3 x^{n+4}+9 x^{n+3}} ;\ x \neq 0,3,-3.$$ | [
"Solution:\n\nNajprej razstavimo prvi imenovalec $x^{4}-27 x = x(x^{3}-27) = x(x-3)(x^{2}+3 x+9)$ in drugi imenovalec $x^{4}-3 x^{3} = x^{3}(x-3)$ ter razširimo oba ulomka na najmanjši skupni imenovalec in dobimo\n$$\n\\frac{-3 x-9}{x^{3}(x-3)(x^{2}+3 x+9)}.\n$$\nZa tem razstavimo števec tretjega ulomka $x^{n+2}-9 ... | Slovenia | 23. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | -3/(x-3)^2 | |
08z4 | Let $ABC$ be an acute triangle with $AB = 11$, $AC = 10$, and denote the orthocenter by $H$ and the midpoint of $BC$ by $M$. Point $P$ in triangle $ABC$ lies on the circumcircle of triangle $BHC$ and satisfies $\angle ABP = \angle CPM$, $PM = 3$. Find the length of $BC$. | [
"Let the radius of the circumcircles of triangle $ABC$ and $BHC$ be $R$ and $R'$, respectively. From the sine theorem, we have $\\frac{BC}{\\sin \\angle BAC} = 2R$ and $\\frac{BC}{\\sin \\angle BHC} = 2R'$. Also, by $\\angle BHC = \\angle BAC + \\angle ABH + \\angle ACH = \\angle BAC + 2 \\cdot (90^\\circ - \\angle... | Japan | Japan 2022 | [
"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 and answer | 2\sqrt{21} | |
04hi | Let $ABC$ be a triangle such that $|AB| > |AC|$. Let $P$ be the midpoint of the segment $\overline{BC}$ and $S$ the intersection of the bisector of the angle $\angle BAC$ and the segment $\overline{BC}$. The line parallel to the line $AS$, which passes through $P$, intersects the lines $AB$ and $AC$ in points $X$ and $... | [
"Since $|BP| = |CP|$ and\n$$\n\\frac{|BX|}{|BP|} = (\\text{similarity}) = \\frac{|BA|}{|BS|} = (\\text{angle bisector}) = \\frac{|CA|}{|CS|} = (\\text{similarity}) = \\frac{|CY|}{|CP|},\n$$\nwe conclude that $|BX| = |CY|$.\n\nNote that $\\triangle XYA = \\triangle SAC = \\triangle BAS = \\t... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
076n | Let $n$ be a natural number. We define sequences $a_i$ and $b_i$ as follows. We let $a_0 = 1$ and $b_0 = n$. For $i > 0$, we let
$$
(a_i, b_i) = \begin{cases} (2a_{i-1} + 1, b_{i-1} - a_{i-1} - 1) & \text{if } a_{i-1} < b_{i-1}, \\ (a_{i-1} - b_{i-1} - 1, 2b_{i-1} + 1) & \text{if } a_{i-1} > b_{i-1}, \\ (a_{i-1}, b_{i-... | [
"We first note that $a_i + b_i = n + 1$ for all natural numbers $i$. We define three sequences $\\{r_i\\}$, $\\{s_i\\}$, $\\{t_i\\}$ as follows. Let $r_0 = 1$, $s_0 = t_0 = 0$. For $i > 0$, we let\n$$\n(r_i, s_i, t_i) = \\begin{cases} (2r_{i-1}, 2s_{i-1}, 2t_{i-1} + 1) & \\text{if } a_{i-1} < b_{i-1}, \\\\ (2r_{i-1... | India | IND_TSExams | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0ift | Problem:
Triangle $ABC$ has incircle $\omega$ which touches $AB$ at $C_1$, $BC$ at $A_1$, and $CA$ at $B_1$. Let $A_2$ be the reflection of $A_1$ over the midpoint of $BC$, and define $B_2$ and $C_2$ similarly. Let $A_3$ be the intersection of $AA_2$ with $\omega$ that is closer to $A$, and define $B_3$ and $C_3$ simi... | [
"Solution:\n\nNotice that $A_2$ is the point of tangency of the excircle opposite $A$ to $BC$. Therefore, by considering the homothety centered at $A$ taking the excircle to the incircle, we notice that $A_3$ is the intersection of $\\omega$ and the tangent line parallel to $BC$. It follows that $A_1 B_1 C_1$ is co... | United States | Harvard-MIT Mathematics Tournament | [
"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"
] | null | proof and answer | 14/65 |
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