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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
08lr | Problem:
Determine the greatest number with $n$ digits in the decimal representation which is divisible by $429$ and has the sum of all digits less than or equal to $11$. | [
"Solution:\nLet $A = \\overline{a_n a_{n-1} \\ldots a_1}$ and notice that $429 = 3 \\cdot 11 \\cdot 13$.\n\nSince the sum of the digits $\\sum a_i \\leq 11$ and $\\sum a_i$ is divisible by $3$, we get $\\sum a_i = 3, 6$ or $9$. As $11$ divides $A$, we have\n$$\n11 \\mid a_n - a_{n-1} + a_{n-2} - a_{n-3} + \\ldots\n... | JBMO | 2008 Shortlist JBMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 3003*10^{n-4} | |
07il | Let $G$ be a connected simple graph with vertices $v_1, v_2, \dots, v_{11}$ such that $v_1$ is of degree $2$ and the rest of the vertices are of degree $3$. Assume that for every subset $A$ of vertices with at most $4$ elements the subset of vertices that are not in $A$ but have an edge connecting them to some elements... | [
"We claim that the answer is $12$. We construct the graph as the following; we shall consider two cases:\n\nFirst. If the degree of the zero floor vertex is two. Then, we shall have two vertices in the first floor. Assume now for contradiction, then we would have one vertex in the fifth floor of degree three. There... | Iran | 41th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 12 | |
05iw | Problem:
On dispose de 500 emplacements à billes numérotés de 1 à 500 de gauche à droite. Des billes numérotées de 1 à 500 sont posées dans ces emplacements dans le désordre : leurs numéros ne coïncident pas nécessairement avec les numéros des emplacements. (Dans chaque emplacement il y a exactement une bille.) On dit... | [
"Solution:\n\nMontrons d'abord que si les billes sont disposées dans un ordre plutôt croissant alors on peut les colorier en 5 couleurs de telle sorte que la suite des billes de chaque couleur soit croissante. Pour cela nous allons ordonner les 5 couleurs dans un certain ordre et colorier les billes une par une de ... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0010 | En el triángulo $ABC$ sean $M$ en el lado $AB$ tal que $AM = 2BM$ y $N$ el punto medio del lado $BC$.
Denotamos $O$ al punto de intersección de $AN$ y $CM$. Si el área del triángulo $ABC$ es igual a $30$, calcular el área del cuadrilátero $MBNO$. | [] | Argentina | XIX Olimpíada Matemática Argentina | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles"
] | español | proof and answer | 7 | |
0fry | En el triángulo escaleno $ABC$ con incentro $I$, la recta $AI$ corta de nuevo a la circunferencia circunscrita en el punto $D$, y $J$ es el punto tal que $D$ es el punto medio de $IJ$. Se consideran puntos $E$ y $F$ en la recta $BC$ tales que $IE$ y $JF$ son perpendiculares a $AI$. Se consideran puntos $G$ en $AE$ y $H... | [
"Probaremos que $G$ y $H$ están en la circunferencia circunscrita de $ABC$, y que $GH \\parallel BC$. Esto implicará que los puntos $B, C, H, G$ forman un trapecio isósceles con $BG = CH$, y habremos terminado.\n\n\n\nComenzamos observando que $J$ es el $A$-exincentro de $ABC$, el punto de ... | Spain | LIX Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Circle of Apo... | Spanish | proof only | null | |
0ex1 | Problem:
A circle is inscribed in $ABCD$. $AB$ is parallel to $CD$, and $BC = AD$. The diagonals $AC$, $BD$ meet at $E$. The circles inscribed in $ABE$, $BCE$, $CDE$, $DAE$ have radius $r_1$, $r_2$, $r_3$, $r_4$ respectively. Prove that $1 / r_1 + 1 / r_3 = 1 / r_2 + 1 / r_4$. | [
"Solution:\n\nA necessary and sufficient condition for $ABCD$ to have an inscribed circle is $AB + CD = BC + AD$. So we have $AB + CD = 2AD$, which we use repeatedly. Extend $DC$ to $X$ so that $BX$ is parallel to $EC$. Then $DX = AB + CD = 2AD$ and the triangles $DEC$, $AEB$, $DBX$ are similar. Let $h$ be the perp... | Soviet Union | 4th ASU | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02fw | $ABC$ is a triangle. $D$ is the midpoint of $AB$, $E$ is a point on the side $BC$ such that $BE = 2EC$ and $\angle ADC = \angle BAE$. Find $\angle BAC$. | [
"Let $F$ be the midpoint of $BE$. Thus $DF$ and $AE$ are parallel. Let $M$ be the intersection point of $AE$ and $CD$.\n\n\n\n$E$ is the midpoint of $CF$. Since $ME$ is parallel to $DF$, by Thales theorem $DM = MC$. But triangle $ADM$ is isosceles, so $AM = DM = MC$ and triangle $AMC$ is is... | Brazil | XX OBM | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 90° | |
0fp0 | Determinar todos los números primos positivos $p, q, r, k$ tales que $pq + qr + rp = 12k + 1$. | [] | Spain | XXXI Olimpiada Iberoamericana de Matemáticas | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | Spanish | proof and answer | The only solution is that p, q, r are 2, 3, and 11 in any order, and k = 5. | |
0lb2 | A sequence of integers $(a_n)$ determined by
$$
a_0 = 1,\ a_1 = -1 \quad \text{and} \quad a_n = 6a_{n-1} + 5a_{n-2} \quad \text{for all } n \ge 2.
$$
Show that $a_{2012} - 2010$ is divisible by $2011$. | [
"Consider the sequence of integers $(b_n)$ determined by\n$$\nb_0 = 1,\\ b_1 = -1 \\text{ and } b_n = 6b_{n-1} + 2016b_{n-2} \\text{ for all } n \\ge 2.\n$$\nClearly, for all $n \\ge 0$, we have $a_n \\equiv b_n \\pmod{2011}$. (*)\nThe characteristic equation of sequence $(b_n)$: $x^2 - 6x - 2016 = 0$, or $(x - 48)... | Vietnam | Vijetnam 2011 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof only | null | |
0hox | Problem:
Let $n>3$ be an integer which is not divisible by $3$. Two players $A$ and $B$ play the following game with an $n \times n$ chocolate table. First, player $A$ has to choose and remove one piece of the chocolate, without breaking other pieces. After his move, player $B$ tries to partition the remaining chocola... | [
"Solution:\n\nThe player $A$ has a winning strategy. Imagine that the chocolate is painted in $3$ colors, $R$ (red), $G$ (green), and $B$ (blue) as shown in the table\n\n$$\n\\begin{array}{cccccccc}\nR & G & B & R & G & B & R & \\cdots \\\\\nB & R & G & B & R & G & B & \\cdots \\\\\nG & B & R & G & B & R & G & \\cd... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
05hc | Problem:
Soit $ABC$ un triangle et $\omega$ son cercle inscrit. On note $P, Q, R$ les points de contact de $\omega$ avec $(BC), (CA)$ et $(AB)$. Un cercle passant par $B$ et $C$ est tangent en $X$ à $\omega$, un cercle passant par $C$ et $A$ est tangent en $Y$ à $\omega$ et un cercle passant par $A$ et $B$ est tangent... | [
"Solution:\n\nRappelons d'abord quelques résultats sur la polarité que nous allons utiliser. Soit $\\mathcal{C}$ un cercle de centre $O$ et de rayon $R$. Si $P$ est un point distinct de $O$, la polaire de $P$ par rapport à $\\mathcal{C}$ est la droite formée des points $M$ tels que $\\overrightarrow{OP} \\cdot \\ov... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | null | proof only | null | |
0jtv | Problem:
Denote by $\mathbb{N}$ the positive integers. Let $f: \mathbb{N} \rightarrow \mathbb{N}$ be a function such that, for any $w, x, y, z \in \mathbb{N}$,
$$
f(f(f(z))) f(w x f(y f(z)))=z^{2} f(x f(y)) f(w)
$$
Show that $f(n!) \geq n!$ for every positive integer $n$. | [
"Solution:\n\nIf $f\\left(z_{1}\\right)=f\\left(z_{2}\\right)$, then plugging in $\\left(w, x, y, z_{1}\\right)$ and $\\left(w, x, y, z_{2}\\right)$ yields $z_{1}=z_{2}$. Thus, $f$ is injective. Substitution of $(z, w)$ with $(1, f(f(1)))$ in the main equation yields\n$$\nf(f(f(1)) x f(y f(1)))=f(x f(y))\n$$\nBecau... | United States | HMMT Invitational Competition | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0kle | Problem:
Suppose that $P(x, y, z)$ is a homogeneous degree 4 polynomial in three variables such that $P(a, b, c) = P(b, c, a)$ and $P(a, a, b) = 0$ for all real $a, b$, and $c$. If $P(1,2,3) = 1$, compute $P(2,4,8)$.
Note: $P(x, y, z)$ is a homogeneous degree 4 polynomial if it satisfies $P(k a, k b, k c) = k^{4} P(a,... | [
"Solution:\nSince $P(a, a, b) = 0$, $(x - y)$ is a factor of $P$, which means $(y - z)$ and $(z - x)$ are also factors by the symmetry of the polynomial. So,\n$$\n\\frac{P(x, y, z)}{(x - y)(y - z)(z - x)}\n$$\nis a symmetric homogeneous degree 1 polynomial, so it must be $k(x + y + z)$ for some real $k$. So, the an... | United States | HMMT Spring 2021 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 56 | |
02af | Problem:
Inteiros de 4 algarismos - Sabendo que $a$ é um número natural, e que $4 a^{2}$ e $\frac{4}{3} \times a^{3}$ são números naturais de 4 algarismos, determine $a$. | [
"Solution:\n\nTemos que $1000 \\leq 4 a^{2} < 10000$ e também $1000 \\leq \\frac{4}{3} \\times a^{3} < 10000$.\n\nDe $1000 \\leq 4 a^{2} < 10000$ segue que $250 \\leq a^{2} < 2500$. Sendo $a$ um número inteiro e $15^{2} = 225$, $16^{2} = 256$ e $50^{2} = 2500$, temos que $15 < a < 50$.\n\nDe $1000 \\leq \\frac{4}{3... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 18 | |
0i0h | Problem:
2000 distinct positive integers are written down, and it so happens that the product of any 3 different numbers from this list is a square. Prove that each one of them is a square. | [
"Solution:\n\nLet $a$ be any number in the list; we wish to prove that $a$ is a square. Let $b$, $c$, $d$ be any three other elements of the list, all different from each other. Then, the given implies that there are integers $p$, $q$, $r$ with $a b c = p^{2}$, $a b d = q^{2}$, $a c d = r^{2}$. Multiplying these to... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
05o8 | Problem:
Dans la région reculée de Torturie, le vizir dispose en cercle $n$ condamnés, numérotés dans l'ordre entre 1 et $n$. Implacablement, il envoie au bourreau un condamné sur deux : les condamnés $2,4, \ldots$ et ainsi de suite en tournant autour du cercle, en sautant une personne entre deux supplices, jusqu'au m... | [
"Solution:\n\nOn note $f(n)$ le numéro du survivant. Traitons d'abord le cas où $n$ est une puissance de $2$. Montrons par récurrence que dans ce cas, $f(n)=1$.\n\nSi $n=1$, c'est bien le cas. Si $n=2^{a}$, avec $a \\geq 1$, le vizir élimine d'abord tous les numéros pairs. Il reste alors $\\frac{n}{2}=2^{a-1}$ cond... | France | Envoi 1: Arithmétique | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | If n = 2^a + b with 0 ≤ b < 2^a, then the last remaining position is 2b + 1. | |
0gz3 | Draw the graphic of the equation: $$\frac{x}{|x|} + \frac{|y|}{y} = 2y.$$ | [
"Obviously $x \\neq 0$ and $y \\neq 0$. Now we have to consider three cases:\n\n1st case. $x > 0$, $y > 0$. Hence $2y = 2$, and we have the first ray of the answer.\n\n2nd case. $x < 0$, $y < 0$, analogous to the first case. It gives us the second ray of the answer.\n\n3rd case. $xy < 0$. It follows that the equali... | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Precalculus > Functions"
] | English | final answer only | The graph is the union of two rays: all points with ordinate equal to one and abscissa positive, and all points with ordinate equal to negative one and abscissa negative. | |
0492 | Determine all $a$ such that there exists unique $(x, y) \in \mathbb{R}^2$ satisfying
$$
2^{|x|} + |x| = x^2 + y + a, \quad x^2 + y^2 = 1.
$$ | [
"If the pair $(x, y)$ is the solution of the given system, then the pair $(-x, y)$ is also the solution. We conclude that the unique solution of this system has to be of the form $(0, y)$.\n\nTaking $x = 0$ in the given system we get\n$$\n\\begin{aligned}\n1 &= y + a, \\\\\ny^2 &= 1,\n\\end{aligned}\n$$\nso $y = 1$... | Croatia | Hrvatska 2011 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | a = 0 | |
0jqs | Problem:
$\$ indy has $\$ 100$ in pennies (worth $\$ 0.01$ each), nickels (worth $\$ 0.05$ each), dimes (worth $\$ 0.10$ each), and quarters (worth $\$ 0.25$ each). Prove that she can split her coins into two piles, each with total value exactly $\$ 50$. | [
"Solution:\n\nFirst, observe that if there are pennies in the mix, there must be a multiple of 5 pennies (since $5,10,25 \\equiv 0 \\pmod{5}$), so we can treat each group of 5 pennies as nickels and reduce the problem to the case of no pennies, treated below.\n\nIndeed, suppose there are no pennies and that we are ... | United States | HMMT February 2015 | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0jgc | Problem:
A subset $U \subset \mathbb{R}$ is open if for any $x \in U$, there exist real numbers $a, b$ such that $x \in (a, b) \subset U$. Suppose $S \subset \mathbb{R}$ has the property that any open set intersecting $(0,1)$ also intersects $S$. Let $T$ be a countable collection of open sets containing $S$. Prove tha... | [
"Solution:\n\nIf $S$ is uncountable then we're done, so assume $S$ is countable. We may also assume that the supersets are a chain $V_{1} \\supset V_{2} \\supset V_{3} \\supset \\cdots$ by taking intersections.\n\nWe will use the following fact from point set topology:\nIf $K_{1} \\supset K_{2} \\supset \\cdots$ is... | United States | HMIC | [
"Discrete Mathematics > Logic"
] | null | proof only | null | |
0k6v | Problem:
Consider three hockey pucks lying on a level sheet of ice; the pucks are not collinear. In a move, one may select any of the pucks and hit it so that it passes through the midpoint of the other two pucks. Determine whether it is possible, after $2019$ such moves, for all pucks to be in their original position... | [
"Solution:\n\nIt is not possible. Label the pucks $A$, $B$, $C$. Then the triangle $ABC$ alternates between being oriented clockwise and counterclockwise between each move. Thus, the pucks cannot return to their original positions after an odd number of moves."
] | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Not possible | |
0hce | Sequence of positive integers $a_1, a_2, a_3, \dots$ is defined by $a_{n+1} = a_n^2 + 2018$, where $a_1$ is some positive integer. Prove that in this sequence no more than one number can be a cube of a positive integer number. | [
"Suppose there is more than one number, which is a cube of a positive integer. Let $a_k$ be the smallest of all possible cubes. Then, it gives $0, \\pm 1$ modulo $9$, hence, $a_k^2 \\equiv 0, 1 \\pmod{9}$. From now on, we write all remainders modulo $9$. Let us list all the cases.\n\n$a_k^2 \\equiv 0$ and $a_{k+1} ... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
084i | Problem:
Sia $ABC$ un triangolo rettangolo in $A$, con $AB > AC$; sia $AH$ l'altezza relativa all'ipotenusa. Sulla retta $BC$ si prenda $D$ tale che $H$ sia punto medio di $BD$; sia poi $E$ il piede della perpendicolare condotta da $C$ ad $AD$. Dimostrare che $EH = AH$. | [
"Solution:\n\nIl triangolo $ABD$ è isoscele su $BD$, perché $AH$ è altezza e mediana. $AH$ è pertanto anche bisettrice dell'angolo $B\\widehat{A}D$, e quindi gli angoli $D\\widehat{A}H$, $B\\widehat{A}H$ sono uguali.\n\nGli angoli $A\\widehat{E}C$, $A\\widehat{H}C$ sono retti per costruzione; quindi $E$ ed $H$ appa... | Italy | GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
080g | Problem:
Le dimensioni dello schermo di un televisore sono $60~\mathrm{cm} \times 45~\mathrm{cm}$. Una telecamera inquadra interamente il televisore, e rimanda l'immagine sullo stesso, per cui dentro questo televisore se ne vede un altro e così via. Il televisore più grande che si vede dentro lo schermo ha un'area ugu... | [
"Solution:\n\nLa risposta è 11. L'area dello schermo è $60~\\mathrm{cm} \\times 45~\\mathrm{cm} = 2700~\\mathrm{cm}^2$. Il televisore più grande inquadrato ha un'area uguale a metà dell'area dello schermo. Il secondo ha un'area uguale alla metà del primo, e cioè $1/4 = 1/2^2$ dell'area dello schermo; analogamente, ... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 11 | |
0kh7 | What is the number of terms with rational coefficients among the 1001 terms in the expansion of $$(x^{\sqrt{3}} + y\sqrt{3})^{1000}$$?
(A) 0 (B) 166 (C) 167 (D) 500 (E) 501 | [] | United States | AMC 12 A | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | MCQ | E | |
0a06 | Problem:
Vind alle paren priemgetallen $(p, q)$ waarvoor geldt dat
$$
p\left(p^{2}-p-1\right)=q(2 q+3)
$$ | [
"Solution:\n\nAntwoord: de enige oplossing is $(p, q)=(13,31)$.\n\nAls $p=q$, dan moet ook gelden dat $p^{2}-p-1=2 q+3=2 p+3$. Dit kunnen we ontbinden als $(p-4)(p+1)=0$. Omdat 4 en -1 geen priemgetallen zijn, levert dit dus geen oplossingen op.\n\nIn het andere geval hebben we $p \\mid 2 q+3$ en $q \\mid p^{2}-p-1... | Netherlands | Selectietoets | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | (13,31) | |
0cr5 | On each side of a triangle $ABC$, two points are chosen: points $C_1$ and $C_2$ on side $AB$, points $B_1$ and $B_2$ on $AC$, and points $A_1$ and $A_2$ on $BC$. It happened that the segments $A_1B_2$, $B_1C_2$, and $C_1A_2$ have equal length, meet at one point, and the angle between any two of them is $60^\circ$. Prov... | [
"Заметим, что\n$$\n\\overrightarrow{A_1B_2} + \\overrightarrow{B_2A_1} + \\overrightarrow{B_1C_2} + \\overrightarrow{C_2C_1} + \\overrightarrow{C_1A_2} + \\overrightarrow{A_2A_1} = \\overrightarrow{0}. \\quad (*)\n$$\nПо условию имеем $A_1B_2 = B_1C_2 = C_1A_2$, и угол между любыми двумя из трех прямых $A_1B_2$, $B... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0jv0 | Problem:
Let $ABC$ be a triangle with incenter $I$, incircle $\gamma$ and circumcircle $\Gamma$. Let $M, N, P$ be the midpoints of sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ and let $E, F$ be the tangency points of $\gamma$ with $\overline{CA}$ and $\overline{AB}$, respectively. Let $U$, $V$ be the interse... | [
"Solution:\nLet segments $AI$ and $EF$ meet at $K$. Extending $AK$ to meet the circumcircle again at $Y$, we see that $X$ and $Y$ are diametrically opposite, and it follows that $AX$ and $EF$ are parallel. Therefore the height from $X$ to $\\overline{UV}$ is merely $AK$. Observe that $AE=AF$, so $\\triangle AEF$ is... | United States | HMMT February 2016 | [
"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 | 21*sqrt(3)/8 | |
0hjs | Problem:
In the sequence
$$
77492836181624186886128 \ldots,
$$
all of the digits except the first two are obtained by writing down the products of pairs of consecutive digits. Prove that infinitely many 6 s appear in the sequence. | [
"Solution:\nSince $868$ appears in the sequence and $8 \\cdot 6 = 6 \\cdot 8 = 48$, $4848$ appears later in the sequence. Then $4 \\cdot 8 = 32$, so $3232$ appears somewhere later. And so on:\n$$\n868 \\rightarrow 4848 \\rightarrow 3232 \\rightarrow 666 \\rightarrow 3636 \\rightarrow 1818 \\rightarrow 888 \\rightar... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Other"
] | null | proof only | null | |
0548 | Let $n$ be a natural number such that $n + 1$, $n + 3$, $n + 7$ and $n + 9$ are prime numbers, and $n + 31$, $n + 33$, $n + 37$ and $n + 39$ are also prime numbers. Find the remainder of $n$ divided by 210. | [
"Consider the remainders of $n$ when divided by 2, 3, 5, and 7. In the following tables the left column shows the remainder and the right column shows, which of the given eight numbers cannot be prime, if $n \\ge 7$:\n\n| n mod 2 | divisible by 2 | n mod 3 | divisible by 3 |\n|---------|----------------|---------|-... | Estonia | Estonian Math Competitions | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 190 | |
08ir | Problem:
Prove that there exist two sets $A=\{x, y, z\}$ and $B=\{m, n, p\}$ of positive integers greater than $2003$ such that the sets have no common elements and the equalities $x+y+z=m+n+p$ and $x^{2}+y^{2}+z^{2}=m^{2}+n^{2}+p^{2}$ hold.
Problem:
Demonstraţi că există mulțimi disjuncte $A=\{x, y, z\}$ și $B=\{m, n... | [
"Solution:\nLet $ABC$ be a triangle with $BC=a$, $AC=b$, $AB=c$ and $a<b<c$. Denote by $m_{a}, m_{b}, m_{c}$ the lengths of medians drawn from the vertices $A, B, C$ respectively. Using the formulas\n$$\n4 m_{a}^{2} = 2(b^{2} + c^{2}) - a^{2}, \\quad 4 m_{b}^{2} = 2(a^{2} + c^{2}) - b^{2}, \\quad 4 m_{c}^{2} = 2(a^... | JBMO | 7th JBMO | [
"Geometry > Plane Geometry > Triangles",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
00z4 | Problem:
Prove that
$$
\frac{1995}{2}-\frac{1994}{3}+\frac{1993}{4}-\cdots-\frac{2}{1995}+\frac{1}{1996}=\frac{1}{999}+\frac{3}{1000}+\cdots+\frac{1995}{1996} .
$$ | [
"Solution:\nDenote the left-hand side of the equation by $L$, and the right-hand side by $R$. Then\n$$\n\\begin{aligned}\nL & = \\sum_{k=1}^{1996} (-1)^{k+1} \\left( \\frac{1997}{k+1} - 1 \\right) = 1997 \\cdot \\sum_{k=1}^{1996} (-1)^{k+1} \\cdot \\frac{1}{k+1} = 1997 \\cdot \\sum_{k=1}^{1996} (-1)^k \\cdot \\frac... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0bhs | Let $ABC$ be a triangle and let $M$ be the midpoint of the side $BC$. The circle of radius $MA$ centred at $M$ meets the lines $AB$ and $AC$ again at $B'$ and $C'$, respectively, and the tangents to this circle at $B'$ and $C'$ meet at $D$. Show that the perpendicular bisector of the segment $BC$ bisects the segment $A... | [
"\nLet $A'$ be the antipodal of $A$ in the circle $AB'C'$, and let $A''$ be the point where this circle meets again the line through $A$ parallel to $BC$ (the points $A$ and $A''$ may coincide). Since $M$ is the midpoint of the side $BC$, the lines $AA''$, $AB'$, $AA'$, $AC'$ form a harmoni... | Romania | THE 2014 Twelfth IMAR MATHEMATICAL COMPETITION | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | null | proof only | null | |
0h9h | Let $H$ be an orthocenter of an acute triangle $ABC$, and let $O$ be its circumcenter. The line $AO$ intersects segment $BC$ at $D$. Perpendicular to $BC$ with a foot $D$ intersects altitudes of $ABC$ through $B$ and $C$ at points $X$ and $Y$ respectively. Show that the circumcenter of $\triangle HXY$ is equally distan... | [
"Let $AH_1$, $BH_2$, $CH_3$ be the altitudes of $\\triangle ABC$ (Fig. 29), $M$ is the midpoint of $BC$, and let $O_1$ be the circumcenter of $\\triangle YHX$. We want to show that $\\triangle YHX \\sim \\triangle ABC$.\n\nSince $\\angle BH_3C = \\angle YDC = 90^\\circ$, $\\angle ABC = \\angle HYX$. Similarly, $\\a... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
00ja | Let $a$, $b$, $c$ and $d$ be four integers such that
$7a + 8b = 14c + 28d.$
Prove that $a \cdot b$ is a multiple of $14$. | [
"We consider the equation modulo $2$ and modulo $7$, respectively, and obtain\n$$\n\\begin{aligned}\na &\\equiv 0 \\pmod{2}, \\\\\nb &\\equiv 0 \\pmod{7}.\n\\end{aligned}\n$$\nWe conclude that $a$ is even and $b$ is a multiple of $7$. Therefore, $ab$ is divisible by $2 \\cdot 7$. $\\square$"
] | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0h0z | For the natural number $N = p_1^{a_1} p_2^{a_2} \dots p_n^{a_n}$, written in the canonical form ($p_i$ are distinct primes and $a_i$ are naturals, $1 \le i \le n$), we denote $T(N) = a_1 + a_2 + \dots + a_n$. For some distinct natural $a, b, c, d$ the number $ab + cd$ is divisible by $ac + bd$. Prove that $T(ab + cd) \... | [
"To the contrary, assume that $T(ab + cd) \\le 2$.\n\nProblem 8-3 implies that $T(ac + bd) \\ge 2$. Now, since $(ac + bd) \\mid (ab + cd)$, we have that $T(ab + cd) \\ge T(ac + bd) \\ge 2$, which by the assumption means that $T(ab + cd) = T(ac + bd) = 2$. But then $(ac + bd) = (ab + cd) \\Rightarrow (a - d)(b - c) ... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0l75 | Problem:
Point $P$ is inside triangle $\triangle ABC$ such that $\angle ABP = \angle ACP$. Given that $AB = 6$, $AC = 8$, $BC = 7$, and $\frac{BP}{PC} = \frac{1}{2}$, compute $\frac{|BPC|}{|ABC|}$.
(Here, $[XYZ]$ denotes the area of $\triangle XYZ$.) | [
"Solution:\n\n\nLet the internal and external bisectors of $\\angle BAC$ meet $BC$ at $D$ and $E$. Similarly, let the internal and external bisectors of $\\angle BPC$ meet $BC$ at $D'$ and $E'$. The angle condition implies that $AD\\parallel PD'$ and $AE\\parallel PE'$. Thus, triangles $ADE... | United States | HMMT February 2025 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geom... | null | proof and answer | 7/18 | |
0gyq | Given acute-angled triangle $ABC$. $O$ is the circumcenter, $H$ is the orthocenter and $AH_A$, $BH_B$, $CH_C$ are the altitudes of $\triangle ABC$. Denote by $A_1$, $B_1$, $C_1$ the circumcenters of triangles $BOC$, $COA$ and $AOB$ respectively. Prove that the lines $A_1H_A$, $B_1H_B$, $C_1H_C$ meet at a point, which l... | [
"It is easy to see that point $O$ is the incenter of $\\triangle A_1B_1C_1$, since $OC$ is perpendicular to the tangent of the circumcircle of $\\triangle ABC$ at the point $C$ and is parallel to $H_AH_B$.\n\nTherefore $OC \\perp H_A H_B \\Rightarrow A_1 B_1 \\parallel H_A H_B$. And so $\\Delta A_1 B_1 C_1 \\sim \\... | Ukraine | The Problems of Ukrainian Authors | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
01nm | For each positive integer $k$, let $t(k)$ be the largest odd divisor of $k$.
Determine all positive integers $a$ for which there exists a positive integer $n$ such that all the differences
$t(n+a) - t(n)$, $t(n+a+1) - t(n+1)$, ..., $t(n+2a-1) - t(n+a-1)$ are divisible by $4$. | [] | Belarus | Belorusija 2012 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | English | proof and answer | a = 1 or a = 3 | |
07cu | In triangle $ABC$ let $I_a$ be the $A$-excenter. Let $\omega$ be an arbitrary circle that passes through $A$ and $I_a$ and intersects the extensions of sides $AB$ and $AC$ (extended from $B$ and $C$) at $X$ and $Y$, respectively. Let $S$ and $T$ be points on segments $I_aB$ and $I_aC$, respectively, such that $\widehat... | [
"Clearly, $M$ is a point on $BC$, since $I_a B$ is angle bisector of both angles $\\overline{XBM}$ and $\\overline{XBC}$. Note that\n$$\n\\overline{I_a MC} = 180^\\circ - \\overline{I_a MB} = 180^\\circ - \\overline{I_a XA} = \\overline{CYI_a}.\n$$\nAnd since $\\overline{I_a CM} = \\overline{I_a CY}$, we get $CYI_a... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Radical axi... | null | proof only | null | |
00fs | Let $a_{1}, a_{2}, \cdots$ be a sequence of real numbers satisfying $a_{i+j} \leq a_{i}+a_{j}$ for all $i, j=1,2, \cdots$. Prove that
$$
a_{1}+\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n} \geq a_{n}
$$
for each positive integer $n$. | [
"Letting $b_{i}=a_{i} / i, (i=1,2, \\cdots)$, we prove that\n$$\nb_{1}+\\cdots+b_{n} \\geq a_{n} \\quad(n=1,2, \\cdots)\n$$\nby induction on $n$. For $n=1$, $b_{1}=a_{1} \\geq a_{1}$, and the induction starts. Assume that\n$$\nb_{1}+\\cdots+b_{k} \\geq a_{k}\n$$\nfor all $k=1,2, \\cdots, n-1$. It suffices to prove ... | Asia Pacific Mathematics Olympiad (APMO) | XI APMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0845 | Problem:
Per quali numeri naturali $n$ il prodotto
$$
\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right) \cdots\left(1+\frac{1}{n}\right)
$$
è un numero intero?
(A) Per $n$ dispari
(E) per nessun $n$.
(B) per $n$ pari
(C) per $n$ multiplo di 3
(D) per ogni $n$ | [] | Italy | UNIONE MATEMATICA ITALIANA Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO BIENNIO | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | A | |
0cw9 | Four consecutive integers greater than $100$ are given. Prove that the sum of some three of the given numbers can be represented as the product of three distinct positive integers greater than $1$. | [
"Let $n$, $n+1$, $n+2$, $n+3$ be the given numbers. The sum of the three smallest of them is $3n + 3 = 3(n + 1)$, and the sum of the three largest numbers is $3(n + 2)$. But at least one of the numbers $n+1$ and $n+2$ is even, that is, equal to the product of $2$ and $k$, where $k > 3$. Therefore, this sum can be r... | Russia | Regional round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English; Russian | proof only | null | |
0997 | $I$ нь $ABC$ гурвалжинд багтсан тойргийн төв ба $\Gamma$ нь энэ гурвалжныг багтаасан тойрог. $AI$ шулуун $\Gamma$ тойргийг $A$ ба $D$ цэгт огтолно. $BDC$ нум дээр $E$ цэгийг, $BC$ тал дээр $F$ цэгийг
$$
\angle BAF = \angle CAE < \frac{1}{2} \angle BAC
$$
байхаар сонгов. $IF$ хэрчмийн дундач цэг $G$ байг. $DE$ ба $EI$ ш... | [
"$EI$ ба $\\Gamma$-ийн 2 дахь огтлолцлолын цэг нь $X$, $A$ оройгоос татсан биссектриссийн суурь нь $L$ байг. $G'$ ба $T$ нь харгалзан $DX$-ийн $IF$ ба $AF$-тэй огтлолцох цэгүүд байг. Бид $G = G'$ буюу $IG' = G'F$ гэж баталъя. $AIF$ гурвалжин ба $DX$ шулууны хувьд Менелайн теорем хэрэглэвэл\n$$\n1 = \\frac{G'F}{IG'}... | Mongolia | International Mathematical Olympiad 51 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geom... | Mongolian | proof only | null | |
077w | Let $k \ge 1$ and $N > 1$ be two integers. On a circle are placed $2N + 1$ coins all showing heads. Calvin and Hobbes play the following game. Calvin starts and on his move can turn any coin from heads to tails. Hobbes on his move can turn at most one coin that is next to the coin that Calvin turned just now from tails... | [] | India | INMO_2023 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Calvin wins if and only if 1 ≤ k ≤ ⌈(2N + 1)/3⌉. | |
0euc | Find all functions $f: \mathbb{R}^+ \to \mathbb{R}$ satisfying
(i) $f(1) = 2008$,
(ii) $|f(x)| \le x^2 + 1004^2$, and
$$
(iii) \quad f\left(x + y + \frac{1}{x} + \frac{1}{y}\right) = f\left(x + \frac{1}{y}\right) + f\left(y + \frac{1}{x}\right)
$$
for all $x, y > 0$, where $\mathbb{R}$ is the set of all real numbers an... | [
"Let $u := x + \\frac{1}{y}$, $v := y + \\frac{1}{x}$ for $x > 0, y > 0$. Then we have\n$$\nf(u+v) = f(u) + f(v) \\quad (1)\n$$\nHere $uv = xy + 2 + \\frac{1}{xy} \\ge 4$. First we show that for any $u, v > 0$ with $uv \\ge 4$ can be represented as\n$$\nu = x + \\frac{1}{y}, v = y + \\frac{1}{x} \\quad (2)\n$$\nfor... | South Korea | Korean Mathematical Olympiad Final Round | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | f(x) = 2008x | |
0k29 | Problem:
A function $f:\{1,2,3,4,5\} \rightarrow\{1,2,3,4,5\}$ is said to be nasty if there do not exist distinct $a, b \in\{1,2,3,4,5\}$ satisfying $f(a)=b$ and $f(b)=a$. How many nasty functions are there?
Proposed by: Michael Ren | [
"Solution:\nWe use complementary counting. There are $5^{5}=3125$ total functions. If there is at least one pair of numbers which map to each other, there are $\\binom{5}{2}=10$ ways to choose the pair and $5^{3}=125$ ways to assign the other values of the function for a total of $1250$. But we overcount each time ... | United States | HMMT November 2018 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 1950 | |
08q3 | Problem:
An economist and a statistician play a game on a calculator which does only one operation. The calculator displays only positive integers and it is used in the following way: Denote by $n$ an integer that is shown on the calculator. A person types an integer, $m$, chosen from the set $\{1,2, \ldots, 99\}$ of ... | [
"Solution:\n\nFirst of all, the game finishes because the number on the calculator always decreases. By picking $m \\%$ of a positive integer $n$, players get the number\n$$\n\\frac{m \\cdot n}{100} = \\frac{m \\cdot n}{2^{2} 5^{2}}\n$$\nWe see that at least one of the powers of 2 and 5 that divide $n$ decreases af... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof and answer | 931 | |
074h | Prove the identity:
$$
\sum_{j=0}^{n} \left\{ \binom{3n+2-j}{j} 2^j - \binom{3n+1-j}{j-1} 2^{j-1} \right\} = 2^{3n},
$$
where the second term on the left side is to be interpreted as $0$ for $j = 0$. | [
"Consider $j$-th term in the above sum. It is equal to\n$(\\text{the coefficient of } x^j \\text{ in } (1+2x)^{3n+2-j}) - (\\text{the coefficient of } x^{j-1} \\text{ in } (1+2x)^{3n+1-j})$.\nThis in turn is equal to\n$$\n\\text{the coefficient of } x^{3n+2} \\text{ in } x^{3n+2-j} \\left( (1+2x)^{3n+2-j} - x(1+2x)... | India | Indija TS 2009 | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
00o9 | Let $a$, $b$, $c$, $d$ be real numbers with $0 < a, b, c, d < 1$ and $a + b + c + d = 2$. Show that
$$
\sqrt{(1 - a)(1 - b)(1 - c)(1 - d)} \le \frac{ac + bd}{2}.
$$
Are there infinitely many cases of equality? | [
"Squaring the given inequality and multiplying by $16$, we get\n$$\n(2 - 2a)(2 - 2b)(2 - 2c)(2 - 2d) \\le 4(ac + bd)^2.\n$$\nWe homogenize by replacing the first $2$ in each parenthesis on the left side by $a + b + c + d$ and get the homogeneous inequality\n$$\n(b + d - (a - c))(a + c - (b - d))(b + d + a - c)(a + ... | Austria | AUT_ABooklet_2023 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | Yes; equality occurs when the sum of the squares of the first and third equals the sum of the squares of the second and fourth, for example when the first equals the second and the third equals the fourth equaling one minus the first, with the first between zero and one. | |
0gy5 | a) Consider the square of the numbers $1$, $2$, $\ldots$, $2009$ written one after another in some random order as one number. Is it possible to obtain a perfect square of an integer in this way?
b) Consider the numbers $1$, $2$, $\ldots$, $2009$ written one after another in some random order as one number. Is it poss... | [
"a) Consider remainders of these squares divided by $3$. In the big number which we have generated each square can be represented as $k^2 \\cdot 10^n$, and the big number is the sum of such terms. If $k^2 \\equiv l \\pmod{3}$ then $k^2 \\cdot 10^n \\equiv l \\pmod{3}$. It is well-known that $k^2 \\equiv 0 \\pmod{3}... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0gdg | 班長看老趙當兵太無聊了, 所以派他去排饅頭。有 $n$ 種不同顏色的饅頭, 第 $i$ 種顏色的饅頭有 $a_i$ 顆, 總共有 $S = a_1 + a_2 + \cdots + a_n$ 顆。這 $S$ 顆饅頭兩兩相異。老趙現在可以決定要如何把這 $S$ 顆饅頭排成一排。對於每一個排列, 班長用以下的方式為老趙評分:
(i) 首先, 對於每個 $i \in \{1, 2, \cdots, n\}$, 計算第 $i$ 種顏色的饅頭相鄰的次數, 也就是
$$
c_i = \#\{j: 1 \le j \le S-1, \text{左數第 } j \text{ 顆和第 } j+1 \text{ 顆饅頭都是第 } i \text{ 色}... | [
"首先注意到如果有任何 $a_i = 1$, 則 $c_i$ 恆為 0, 從而得分總和為 0, 必被 $S$ 整除。故以下只討論每種顏色饅頭都至少有 2 顆的狀態, 此時 $S \\ge 2n$。\n\n1. 我們首先用算兩次計算得分總和。為方便解釋, 令 $M_j^i$ 為第 $i$ 色編號第 $j$ 的饅頭, 而一個排列便是將全部 $S$ 個 $M_j^i$ 排成一串。對於每個排列, 其分數等價於從這個排列中, 每種顏色取出相鄰的一對, 總共可以取的方法數。\n\n因此我們可以改為計算每種顏色指定(依序)相鄰一對所面對的排列數:\n- $n$ 對放進 $S$ 格中的方法數為 $\\frac{(S-n)!}{(S-2n)!... | Taiwan | 二〇一九數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0jb9 | Problem:
A frog is at the point $(0,0)$. Every second, he can jump one unit either up or right. He can only move to points $(x, y)$ where $x$ and $y$ are not both odd. How many ways can he get to the point $(8,14)$? | [
"Solution:\nWhen the frog is at a point $(x, y)$ where $x$ and $y$ are both even, then if that frog chooses to move right, his next move will also have to be a step right; similarly, if he moves up, his next move will have to be up.\n\nIf we \"collapse\" each double step into one step, the problem simply becomes ho... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 330 | |
0g65 | 無窮正整數數列 $a_1, a_2, \dots, a_n, \dots$ 滿足對任意正整數 $m, n$ 均有 $m + n$ 整除 $a_m + a_n$, 且對任意正整數 $n$ 有 $a_n \le n^{100}$. 試證: 存在一個多項式 $f(x)$ 使得對任意正整數 $n$, $f(n) = a_n$. | [
"首先對任意 $m < n$, 令 $k = n - m$, 取正整數 $t$ 使得 $tk > n$, 考慮 $tk$ 整除 $a_n + a_{tk-n}$ 及 $(t-1)k$ 整除 $a_m + a_{tk-n}$, 可得 $a_n$ 與 $a_m$ 除以 $k$ 的餘數相同, 亦即 $n - m$ 整除 $a_n - a_m$.\n由拉格朗日插值法, 存在 100 次以內的有理係數多項式 $f(x)$ 使得對於 $k = 1, 2, \\dots, 101$, 均有 $f(k) = a_k$. 以下我們證明 $f(n) = a_n$.\n令 $N$ 為一個使得 $Nf(x)$ 是整係數多項式的一個正整數 $N$. ... | Taiwan | 二〇一一數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof only | null | |
0ezw | Problem:
The real numbers $a$, $b$, $A$, $B$ satisfy $$(B - b)^2 < (A - a)(Ba - Ab).$$ Show that the quadratics $x^2 + ax + b = 0$ and $x^2 + Ax + B = 0$ have real roots and between the roots of each there is a root of the other. | [] | Soviet Union | ASU | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0lf0 | Given an acute, scalene triangle $ABC$ with circumcircle $(O)$. The line passing through $O$ and midpoint $I$ of $BC$ intersects $AB$, $AC$ at $E$, $F$. Let $D$, $G$ be the reflections of $A$ over $O$ and the circumcenter of triangle $AEF$. Let $K$ be the reflection of $O$ over the circumcenter of triangle $OBC$.
a) P... | [
"a) Let $T$ be the projection of $A$ on $GD$, it is well known that $T$ is the second intersection of $(AEF)$ and $(O)$. We observe that $\\triangle TEB \\sim \\triangle TFC$ then\n$$\n\\frac{TB}{TC} = \\frac{BE}{CF}.\n$$\n\n\nOn the other hand,\n$$\n\\frac{BE}{CF} = \\frac{BE}{IB} \\cdot \... | Vietnam | IMO Team Selection Test | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter... | null | proof only | null | |
059j | The bisector of the internal angle at the vertex $B$ of a triangle $ABC$ intersects the circumcircle of the triangle $ABC$ at a point $P$ ($P \neq B$). The line through $P$ perpendicular to the line $AC$ intersects the circumcircle of the triangle $ABC$ at a point $P'$ ($P' \neq P$). Prove that the quadrilateral $APCP'... | [
"As $\\angle ABC = \\angle AP'C$ (see figure), it suffices to show that the quadrilateral $APCP'$ is a square if and only if $\\angle AP'C = 90^\\circ$.\n\nSince $\\angle ABP = \\angle CBP$, we have $AP = CP$. Thus $PP'$ is the perpendicular bisector of the side $AC$. Hence the line $PP'$ passes through the circumc... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0fo2 | Sea $\triangle ABC$ un triángulo y $D$, $E$ y $F$ tres puntos cualesquiera sobre los lados $AB$, $BC$ y $CA$ respectivamente. Llamemos $P$ al punto medio de $AE$, $Q$ al punto medio de $BF$ y $R$ al punto medio de $CD$. Probar que el área del triángulo $\triangle PQR$ es la cuarta parte del área del triángulo $\triangl... | [
"Hagamos primero un dibujo donde queden reflejados los elementos que intervienen en el problema.\n\n\nObservemos que, como $P$, $Q$ y $R$ son los puntos medios de las correspondientes Cevianas $AE$, $BF$ y $CD$, estos puntos se encuentran en los lados del triángulo que determinan los pies d... | Spain | L Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | Spanish | proof only | null | |
0lb3 | Find all positive integer $n$ such that $2^{n+2} \cdot (2^n - 1) - 8 \cdot 3^n + 1$ is a perfect square. | [] | Vietnam | IMO2011 Selection | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | n = 3, 5 | |
0doi | Let $n$ be an integer with $n > 2$ and $a_1, a_2, \dots, a_n \in \mathbb{R}^+$ positive real numbers. Given any positive integers $t, k, p$ with $1 < t < n$, set $m = k+p$. Prove the following inequalities:
$$
1) \quad \frac{a_1^p}{a_2^k + a_3^k + \cdots + a_t^k} + \frac{a_2^p}{a_3^k + a_4^k + \cdots + a_{t+1}^k} + \c... | [
"1) By the rearrangement inequality we have:\n$$\n\\begin{align*} \na_1^m + a_2^m + \\dots + a_n^m &\\ge a_1^p a_2^k + a_2^p a_3^k + \\dots + a_{n-1}^p a_n^k + a_n^p a_1^k \\\\ \na_1^m + a_2^m + \\dots + a_n^m &\\ge a_1^p a_3^k + a_2^p a_4^k + \\dots + a_{n-1}^p a_1^k + a_n^p a_2^k \\\\ \n\\multicolumn{2}{c}{\\text... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof only | null | |
0gw0 | The segments $BC$ and $AD$ are the bases of a trapezoid $ABCD$. Through the point of intersection of its diagonals, a straight line is drawn parallel to the bases. This line intersects the sides $AB$ and $CD$ at points $M$ and $N$ respectively. The segments $DP$ and $AQ$ are the heights of the triangles $DMC$ and $ABN$... | [
"Нехай $K$ — спільна точка прямих $AD$ і $MC$. Оскільки, як легко встановити, $\\frac{KA}{BC} = \\frac{AD}{BC}$, то $KA = AD$, а тому $AP = AK = AD$. Аналогічно доводиться, що $DQ = DA$."
] | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0i7v | Problem:
Bessie the cow is trying to navigate her way through a field. She can travel only from lattice point to adjacent lattice point, can turn only at lattice points, and can travel only to the east or north. (A lattice point is a point whose coordinates are both integers.) $(0,0)$ is the southwest corner of the fi... | [
"Solution: 32\nIn the figure, each point is labeled with the number of ways to reach that point. The numbers are successively computed as follows: The point $(0,0)$ can trivially be reached in 1 way. When Bessie reaches any subsequent point $(x, y)$ (other than a rock), she can arrive either via a northward or an e... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 32 | |
0jz6 | Problem:
Find the number of ordered triples of positive integers $(a, b, c)$ such that
$$
6 a + 10 b + 15 c = 3000
$$ | [
"Solution:\nNote that $6a$ must be a multiple of $5$, so $a$ must be a multiple of $5$. Similarly, $b$ must be a multiple of $3$, and $c$ must be a multiple of $2$.\n\nSet $a = 5A$, $b = 3B$, $c = 2C$. Then the equation reduces to\n$$\n6a + 10b + 15c = 6(5A) + 10(3B) + 15(2C) = 30A + 30B + 30C = 30(A + B + C) = 300... | United States | February 2017 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization"
] | null | final answer only | 4851 | |
0dwu | Problem:
Dan je enakokrak trikotnik $A B C$ z vrhom $A$. Naj bo $D$ razpolovišče stranice $A C$ in $E$ pravokotna projekcija točke $D$ na stranico $B C$. Označimo z $F$ razpolovišče daljice $D E$. Dokaži, da se daljici $B F$ in $A E$ sekata pravokotno natanko tedaj, ko je trikotnik $A B C$ enakostraničen. | [
"Solution:\n\nPredpostavimo najprej, da se daljici $A E$ in $B F$ sekata pravokotno. Naj bo $A'$ nožišče višine na stranico $B C$ ter $G$ presečišče daljic $A E$ in $B F$. Potem je $A G A' B$ tetiven štirikotnik in zato je $\\angle A' A E = \\angle A' A G = \\angle A' B G = \\angle E B F$. Tako sta trikotnika $B E ... | Slovenia | 49. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
02fk | Show that the equation $x^2 + y^2 + z^2 = 3xyz$ has infinitely many solutions in positive integers. | [
"We can regard $x^2 - 3yz \\cdot x + (y^2 + z^2) = 0$ as a quadratic in $x$. So if $1 \\le x \\le y \\le z$ is a solution, then so is $y, z, 3yz - x$.\n\nAlso we have $3y \\ge 3$, so $3yz - x \\ge 3yz - z \\ge 2z > z$. So $1 \\le y \\le z < 3yz - x$, and the new solution has larger largest element.\n\nSo starting w... | Brazil | XVIII OBM | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof only | null | |
05e4 | Problem:
Two different integers $u$ and $v$ are written on a board. We perform a sequence of steps. At each step we do one of the following two operations:
(i) If $a$ and $b$ are different integers on the board, then we can write $a+b$ on the board, if it is not already there.
(ii) If $a, b$ and $c$ are three differ... | [
"Solution:\n\nWe will show that the answer are the integer pairs $(u, v)$ such that $u \\neq 0$, $v \\neq 0$, $\\{u, v\\} \\neq \\{-1,1\\}$ and $u>0$ or $v>0$.\n\nIf $u=0$ or $v=0$, then (i) will never yield a new number and we cannot use (ii) with only two numbers. Hence, if $u=0$ or $v=0$, we cannot reach every p... | European Girls' Mathematical Olympiad (EGMO) | EGMO 2024 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic funct... | null | proof and answer | All distinct integer pairs (u, v) with u and v nonzero, at least one of u or v positive, and not equal to the pair {−1, 1}. | |
03cw | Find the smallest odd prime $p$ for which there exist positive co-prime integers $k$ and $\ell$, such that
$$
4k - 3\ell = 12 \quad \text{and} \quad \ell^2 + \ell k + k^2 \equiv 3 \pmod{p}.
$$ | [
"From $4k = 12 + 3\\ell$, it follows $k = 3k_1$, $k_1 \\in \\mathbb{N}$. Analogously, $\\ell = 4\\ell_1$, $\\ell_1 \\in \\mathbb{N}$. Assuming $p = 3$ gives rise to $3 \\mid k$ and $3 \\mid \\ell^2 + \\ell k + k^2 \\Rightarrow 3 \\mid \\ell \\Rightarrow (k, \\ell) \\ge 3 > 1$, a contradiction with $(k, \\ell) = 1$.... | Bulgaria | Bulgaria 2022 | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequaliti... | null | proof and answer | 11 | |
05k1 | Problem:
Existe-t-il des nombres rationnels positifs ou nuls $x, y$ et $z$ tels que:
$$
x^{5}+2 y^{5}+5 z^{5}=11
$$ | [
"Solution:\nNous allons montrer qu'il n'existe pas de tels rationnels $x, y, z$. On raisonne par l'absurde en supposant qu'il en existe. Soit $d$ le plus petit dénominateur commun de $x, y$ et $z$. On peut alors écrire $x=\\frac{a}{d}$, $y=\\frac{b}{d}$ et $z=\\frac{c}{d}$ pour certains entiers $a, b$ et $c$. L'équ... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null | |
0gm4 | For each positive integer $n$, let
$$
P_n(x) = x^n + x^{n+1} + \dots + x + 1.
$$
Show that for each positive integer $a$, there exists a positive integer $n$ and two polynomials $R(x)$ and $Q(x)$ with integer coefficients such that
$$
P_n(x) = (1 + a x + x^2 R(x)) Q(x).
$$ | [] | Turkey | VIIth NATIONAL MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Number-Theoretic Functions > Möbius inversion"
] | English | proof only | null | |
0ax8 | Problem:
Let $A$ be the set of all two-digit positive integers $n$ for which the number obtained by erasing its last digit is a divisor of $n$. How many elements does $A$ have? | [
"Solution:\nLet $n = 10a + b$. Since $a$ is a divisor of $n$, we infer $a$ divides $b$. Any number $n$ that ends in $0$ is therefore a solution. Assuming $b \\neq 0$, $n$ must be one of the numbers $11, 12, \\ldots, 19, 22, 24, 26, 28, 33, 36, 39, 44, 48, 55, 66, 77, 88, 99$. This gives a total of $32$ positive int... | Philippines | Philippine Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 32 | |
0ecr | Sara is looking for a three-digit positive integer $xyz$ ($z$ are ones, $y$ are tens, and $x$ are hundreds) which satisfies $1 \le x < y < z$ and the property that the sum of the numbers $xyz$, $yzx$, and $zxy$ is a three-digit number which has equal digits. At most
how many such three-digit numbers can Sara find?
(A)... | [
"The sum of the numbers $\\overline{xyz}$, $\\overline{yzx}$, and $\\overline{zxy}$ is equal to $100(x + y + z) + 10(y + z + x) + (z + x + y)$. Since this must be a three-digit number it must hold that $x + y + z < 10$ and therefore it also holds that all the digits of the sum are automatically equal. So we are loo... | Slovenia | National Math Olympiad 2015 – First Round | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | C | |
00v8 | Let $n$ be a fixed natural number and
$$
S_n = \{\overline{c_n c_{n-1} \dots c_1}_{(10)} \mid c_1, \dots, c_{n-1}, c_n \in \{1, 2, 3, 4\}\}.
$$
Are there distinct numbers $x$ and $y$, $x, y \in S_n$, such that $4^n \mid x - y$? | [
"For $n = 1$, the answer is negative. For $n > 1$ we will show that answer is positive. In contrary, since $|S_n| = 4^n$, we see that the set $S_n$ is complete system of remainders modulo $4^n$. Thus,\n$$\n\\sum_{x \\in S_n} x^3 \\equiv \\sum_{i=1}^{4^n} i^3 \\pmod{4^n} \\equiv \\left(\\frac{4^n(4^n+1)}{2}\\right)^... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | No for n = 1; Yes for n > 1. | |
0aso | Problem:
Let $ABC$ be a triangle with $AB = 12$, $BC = 16$, and $AC = 20$. Compute the area of the circle that passes through $C$ and the midpoints of $AB$ and $BC$. | [
"Solution:\n\n$25\\pi$"
] | Philippines | Philippines Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | final answer only | 25π | |
0hqh | Problem:
Given ten points in the plane, show that it is always possible to cover all of them with non-overlapping unit circles. | [
"Solution:\n\nSuppose we place the circles in a hexagonal tiling so they are all tangent to each other. Then each circle has area $\\pi$, and each circle is inscribed in a hexagon. In each hexagon, the distance from the center to each side is $1$. Thus, each of the six equilateral triangles making up the hexagon ha... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof only | null | |
0dcs | Let $P(x)$ be a monic polynomial of degree $100$ with $100$ distinct noninteger real roots. Suppose that each of polynomials $P\left(2x^{2}-4x\right)$ and $P\left(4x-2x^{2}\right)$ has exactly $130$ distinct real roots. Prove that there exist non constant polynomials $A(x), B(x)$ such that $A(x)B(x)=P(x)$ and $A(x)=B(x... | [
"Denote $P(x) = (x - a_1)(x - a_2) \\cdots (x - a_{100})$ with $a_1, a_2, \\ldots, a_{100}$ are real roots of polynomial $P(x)$. So\n$$\nP\\left(2x^{2}-4x\\right) = 0 \\Leftrightarrow 2x^{2}-4x-a_i = 0 \\text{ for } 1 \\leq i \\leq 100.\n$$\nThis equation cannot have $1$ root since $\\Delta = 16 + 8a_i \\neq 0$. So... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | proof only | null | |
0faa | Problem:
$ABCD$ is a square. The points $X$ on the side $AB$ and $Y$ on the side $AD$ are such that $AX \cdot AY = 2 \; BX \cdot DY$. The lines $CX$ and $CY$ meet the diagonal $BD$ in two points. Show that these points lie on the circumcircle of $AXY$. | [] | Soviet Union | 25th ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof only | null | |
005x | Sea $ABC$ un triángulo tal que $\hat{B} = 40°$. Se sabe que hay un punto $P$ de la bisectriz del ángulo $\hat{B}$ que satisface que $BP = BC$ y $\hat{B}AP = 20°$. Determinar las medidas de los ángulos $\hat{A}$ y $\hat{C}$. | [] | Argentina | Argentina 2008 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Spanish | proof and answer | ∠A = 30°, ∠C = 110° | |
017s | Evaluate the sum $\sum_{k=1}^{2010} \gcd(k, 2010) \cos\left(\frac{2\pi k}{2010}\right)$. | [
"Answer: $\\varphi(2010) = 528$ (where $\\varphi(n)$ denotes Euler's totient function).\nIt is well known that\n$$\n\\sum_{k=1}^{n} \\cos\\left(\\frac{2\\pi k}{n}\\right) = 0 \\quad \\text{if } n \\ge 2.\n$$\nWe need to calculate the sum\n$$\nS = \\cos\\left(\\frac{2\\pi}{2010}\\right) + 2\\cos\\left(\\frac{2\\pi \... | Baltic Way | BALTIC WAY | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 528 | |
0cd1 | Let $f_1(x) = a_1x^2 + b_1x + c_1$, $f_2(x) = a_2x^2 + b_2x + c_2$, $f_3(x) = a_3x^2 + b_3x + c_3$ be three quadratic functions with real coefficients and $P_1 : y = f_1(x)$, $P_2 : y = f_2(x)$, $P_3 : y = f_3(x)$ their associated parabolas.
a) Give an example where $P_1 \cap P_2 = \{A\}$, $P_2 \cap P_3 = \{B\}$, $P_3... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | One example: let A = (0, 0), B = (1, √3), C = (2, 0). Take P1: y = x^2 − 2x, P2: y = x^2 + (√3 − 1)x, P3: y = x^2 − (√3 + 3)x + (2√3 + 2). Then P1 ∩ P2 = {A}, P2 ∩ P3 = {B}, P3 ∩ P1 = {C}, and A, B, C form an equilateral triangle of side 2.
For part b: If a1, a2, a3 are nonzero and pairwise distinct and each pair of pa... | |
0ifh | $$
\frac{x^5 - x^2}{x^5 + y^2 + z^2} + \frac{y^5 - y^2}{y^5 + z^2 + x^2} + \frac{z^5 - z^2}{z^5 + x^2 + y^2} \ge 0.
$$ | [
"Note that\n$$\n\\frac{x^5 - x^2}{x^5 + y^2 + z^2} = 1 - \\frac{x^2 + y^2 + z^2}{x^5 + y^2 + z^2}\n$$\nand its cyclic analogous forms. The given inequality is equivalent to\n$$\n\\frac{x^2 + y^2 + z^2}{x^5 + y^2 + z^2} + \\frac{x^2 + y^2 + z^2}{y^5 + z^2 + x^2} + \\frac{x^2 + y^2 + z^2}{z^5 + x^2 + y^2} \\le 3. \\q... | United States | IMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0kq1 | Problem:
Let $ABC$ be a triangle with $AB = 2021$, $AC = 2022$, and $BC = 2023$. Compute the minimum value of $AP + 2BP + 3CP$ over all points $P$ in the plane. | [
"Solution:\nThe minimizing point is when $P = C$. To prove this, consider placing $P$ at any other point $O \\neq C$. Then, by moving $P$ from $O$ to $C$, the expression changes by\n$$\n(AC - AO) + 2(BC - BO) + 3(CC - CO) < OC + 2OC - 3OC = 0\n$$\nby the triangle inequality. Since this is negative, $P = C$ must be ... | United States | HMMT November 2022 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 6068 | |
0bvm | Prove that for any positive integers $n$ and $k$, the number $10^{n+k} - 10^n$ can be written as the sum of $18k$ nonzero perfect squares. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Number Theory > Other"
] | English | proof only | null | |
00wb | Problem:
Positive integers $1, 2, \ldots, 100, 101$ are written in the cells of a $101 \times 101$ square grid so that each number is repeated $101$ times. Prove that there exists either a column or a row containing at least $11$ different numbers. | [
"Solution:\n\nLet $a_{k}$ denote the total number of rows and columns containing the number $k$ at least once. As $i \\cdot (20 - i) < 101$ for any natural number $i$, we have $a_{k} \\geq 21$ for all $k = 1, 2, \\ldots, 101$. Hence $a_{1} + \\cdots + a_{101} \\geq 21 \\cdot 101 = 2121$. On the other hand, assuming... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0i6c | Problem:
Define $\varphi^{k}(n)$ as the number of positive integers that are less than or equal to $n / k$ and relatively prime to $n$. Find $\varphi^{2001}\left(2002^{2}-1\right)$. (Hint: $\varphi(2003)=2002$.) | [
"Solution:\n\n$\\varphi^{2001}\\left(2002^{2}-1\\right)=\\varphi^{2001}(2001 \\cdot 2003)=$ the number of $m$ that are relatively prime to both $2001$ and $2003$, where $m \\leq 2003$. Since $\\varphi(n)=n-1$ implies that $n$ is prime, we must only check for those $m$ relatively prime to $2001$, except for $2002$, ... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 1233 | |
0gwb | A point $M$ lies on the side $BC$ of an equilateral triangle $ABC$ ($M$ is distinct from the vertices). A point $N$ is chosen in such a way that the triangle $BMN$ is also equilateral, and the points $A$ and $N$ belong to the different half-planes with respect to the straight line $BC$. The points $P$, $Q$ and $R$ are ... | [
"Нехай $S$ — така точка, що чотирикутник $PBQS$ є паралелограмом, $L$ — точка перетину прямих $QS$ і $BM$, $F$ — середина $BC$. Тоді нескладно довести, що $PS = QL = FR$, $PF = QS = LR$, $\\angle PSQ = \\angle QLR = \\angle RFP$, а тому трикутники $PSQ$, $QLR$, $RFP$ є рівними. Звідси й випливає твердження задачі."... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0cz1 | Find all primes $q_{1}, q_{2}, q_{3}, q_{4}, q_{5}$ such that $q_{1}^{4} + q_{2}^{4} + q_{3}^{4} + q_{4}^{4} + q_{5}^{4}$ is the product of two consecutive even integers. | [
"Assume that $q_{1}^{4} + q_{2}^{4} + q_{3}^{4} + q_{4}^{4} + q_{5}^{4} = 2k(2k+2)$ for some positive integer $k$. That is\n$$\nq_{1}^{4} + q_{2}^{4} + q_{3}^{4} + q_{4}^{4} + q_{5}^{4} = 4k(k+1)\n$$\nIf $p$ is a prime, $p \\neq 2$, then $p^{4} \\equiv 1 \\pmod{8}$.\nAssume that $q_{1} \\leq q_{2} \\leq q_{3} \\leq... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | q1 = q2 = q3 = q4 = q5 = 2 | |
0c0p | Alina and Bogdan play the following game. They have a heap and $330$ stones in it. They take turns. In one turn it is allowed to take from the heap exactly $1$, exactly $n$ or exactly $m$ stones. The player who takes the last stone wins. Before the beginning Alina says the number $n$, ($1 < n < 10$). After that Bogdan ... | [
"• For the heap initially containing $330$ stones:\n\nBogdan has a winning strategy. One such strategy is the following:\n\n1. If Alina chooses a number $n$ which is not a multiple of $3$, then Bogdan chooses $m = 2$ (or any other number that is not a multiple of $3$ in case Alina has chosen $n = 2$);\n2. If Alina ... | Romania | 69th NMO Selection Tests for JBMO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | For 330 stones: Bogdan. For 2018 stones: Alina. | |
0hwn | Problem:
In the hold of a pirate ship are ten treasure chests lying on pedestals in a circular arrangement. The captain would like to move each chest clockwise by one pedestal. However, the chests are so heavy that the captain and his assistant can only switch two chests at a time. What is the minimum number of switch... | [
"Solution:\n\nThe answer is $9$. It is easy to see that nine moves are sufficient; they can move a single chest counterclockwise one pedestal at a time, until after nine moves all of the other chests have been moved clockwise one pedestal.\n\nSuppose that, at each stage of the game, we draw an arrow from each pedes... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof and answer | 9 | |
04tm | Let $k \in [0, 1]$. Solve the system
$$
\begin{aligned}
k - x^2 &= y, \\
k - y^2 &= z, \\
k - z^2 &= u, \\
k - u^2 &= x
\end{aligned}
$$
in real numbers. | [
"Subtracting the third equation from the first one we get\n$$\nz^2 - x^2 = (z - x)(z + x) = y - u. \\tag{1}\n$$\nSimilarly, the second and the fourth equation imply\n$$\ny^2 - u^2 = (y - u)(y + u) = x - z. \\tag{2}\n$$\nRelations (1) and (2) then imply that $x = z$ holds if and only if $y = u$ holds. We distinguish... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | All solutions are of the form:
1) For all k in [0, 1], the two solutions with all four variables equal: x = y = z = u = (-1 ± √(4k + 1)) / 2.
2) Additionally, for k ≥ 3/4, solutions with x = z, y = u, and x + y = 1 exist. These are:
(x, y, z, u) = ( (1 + √(4k - 3)) / 2, (1 - √(4k - 3)) / 2, (1 + √(4k - 3)) / 2, (1 -... | |
0eoe | Two players alternate placing $2 \times 1$ tiles, with no overlap, on a $5 \times 5$ chessboard until no player can place a tile.
A number of $1 \times 1$ squares remain empty.
a) Prove that there must be an odd number of empty squares remaining.
b) Of the empty squares remaining, are more c... | [
"a. The chessboard contains an odd number of squares and every time a $2 \\times 1$ tile is placed, the number of open squares decreases by $2$, an even number. Hence the number of remaining squares is always an odd number (odd minus even is odd).\n\nb. Before any tile is placed, there are $13$ black squares and $1... | South Africa | South African Mathematics Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) odd number of empty squares; b) more black squares; c) 7 | |
0i2g | Problem:
Square $ABCD$ is drawn. Isosceles triangle $CDE$ is drawn with $E$ a right angle. Square $DEFG$ is drawn. Isosceles triangle $FGH$ is drawn with $H$ a right angle. This process is repeated infinitely so that no two figures overlap each other. If square $ABCD$ has area $1$, compute the area of the entire figur... | [
"Solution:\n\nLet the area of the $n$th square drawn be $S_{n}$ and the area of the $n$th triangle be $T_{n}$. Since the hypotenuse of the $n$th triangle is of length $\\sqrt{S_{n}}$, its legs are of length $l=\\sqrt{\\frac{S_{n}}{2}}$, so $S_{n+1}=l^{2}=\\frac{S_{n}}{2}$ and $T_{n}=\\frac{l^{2}}{2}=\\frac{S_{n}}{4... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 5/2 | |
04ju | Let $n \ge 3$ be a positive integer. Some diagonals are drawn in a convex $n$-gon. We say that a drawn diagonal is *good* if it intersects another drawn diagonal in its interior. Determine the maximal possible number of good diagonals. | [
"Let $M(n)$ be the maximal possible number of good diagonals in a convex $n$-gon.\nWe will show that $M(n) = n - 2$ if $n$ is even and $M(n) = n - 3$ if $n$ is odd. For any $n$, we can draw all $n - 3$ diagonals from one vertex $A$ and one more diagonal joining two vertices adjacent to $A$ if $n > 3$. We show by ma... | Croatia | Croatian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | Maximum equals n minus two when n is even, and n minus three when n is odd. | |
0da1 | Let $ABCD$ be a square inscribed in circle $(O)$. Let $P$ be a point lies on minor arc $CD$ of $(O)$. Line $PB$ intersects $AC$ at $E$. Line $PA$ intersects $DB$ at $F$. Circumcircle of triangle $PEF$ cuts $(O)$ again at $Q$. Prove that $PQ$ is parallel to $CD$. | [
"Let $K$ be the circumcenter of triangle $PEF$. Note that\n$$\n\\angle APB = \\angle ADB = 45^\\circ\n$$\nso triangle $KEF$ is right isosceles at $K$. Thus, quadrilateral $OEKF$ is cyclic.\nHence, $\\angle FOK = \\angle FEK = 45^\\circ$, this means $OK$ is bisector of angle $DOC$, it is also perpendicular bisector ... | Saudi Arabia | Team selection tests for JBMO 2018 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneo... | English | proof only | null | |
0ipo | Problem:
Let $X$ be the point on side $BC$ such that $BX = CD$. Show that the excircle $ABC$ opposite of vertex $A$ touches segment $BC$ at $X$. | [
"Solution:\nLet the excircle touch lines $BC$, $AC$ and $AB$ at $X'$, $Y$ and $Z$, respectively. Using the equal tangent property repeatedly, we have\n$$\nBX' - X'C = BZ - CY = (EY - CY) - (FZ - BZ) = CE - BF = CD - BD.\n$$\nIt follows that $BX' = CD$, and thus $X' = X$. So the excircle touches $BC$ at $X$.\n\n![](... | United States | 11th Annual Harvard-MIT Mathematics Tournament - Team Round: B Division | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
00kc | Let $a_n$ be the sequence defined by some $a_0$ and the recursion
$$
a_{n+1} = a_n + 2 \cdot 3^n
$$
for $n \ge 0$.
Determine all rational values of $a_0$ such that
$$
\frac{a_k^j}{a_j^k}
$$
is an integer for all integers $j$ and $k$ with $0 < j < k$. | [
"We have\n$$\na_n = a_0 + \\sum_{k=0}^{n-1} (a_{k+1} - a_k) = a_0 + \\sum_{k=0}^{n-1} 2 \\cdot 3^k = a_0 + 2 \\frac{3^n - 1}{3-1} = a_0 - 1 + 3^n.\n$$\nIf $a_0 = 1$, then $a_n = 3^n$ and\n$$\n\\frac{a_k^j}{a_j^k} = \\frac{3^{kj}}{3^{jk}} = 1,\n$$\nso $a_0 = 1$ is clearly a solution.\n\nWe now assume that $a_0 \\neq... | Austria | Austria 2014 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | a_0 = 1 | |
0828 | Problem:
Andrea, viaggiando su un tram, incrocia Mafalda che sta camminando sulla stessa strada in direzione opposta. Dopo 10 secondi scende dal tram e la rincorre. Sapendo che la velocità con cui Andrea cammina è doppia di quella di Mafalda e un quinto di quella del tram, quanti secondi impiega a raggiungere l'amica ... | [
"Solution:\n\nLa risposta è (D). Se la velocità di Mafalda è $v$ metri/secondo, sappiamo che quella di Andrea è $2v$ metri/secondo e quella del tram $10v$ metri/secondo. Quando Andrea scende dal tram la sua distanza da Mafalda è dunque $10v$ metri $+ 10 \\times 10v$ metri $= 110v$ metri. Poiché la sua velocità è do... | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
0gvo | A convex quadrilateral $ABCD$ with $BC = CD$, $\angle CBA + \angle DAB > 180^\circ$ is given. Let the points $W$ and $Q$ (different from the quadrilateral's vertices) belong to the sides $BC$ and $DC$ respectively, and let $AD = QD$, $WQ \parallel AD$. It is known that the intersection point $M$ of segments $AQ$ and $B... | [
"Нехай $F$ — точка перетину прямих $BC$ і $AD$ (точки $F$ і $S$ лежать по різні боки від прямої $AB$). Проведемо через точку $D$ пряму $\\ell \\parallel BC$. Позначимо через $S$ точку перетину прямих $WQ$ та $\\ell$. Легко бачити, що $QM$ і $DM$ є бісектрисами зовнішніх кутів трикутника $SDQ$. Звідси випливає, що $... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0cud | The teacher wishes to compose a problem for his students of the following form. He tells them that he thinks of a monic polynomial of degree $2017$ with integer coefficients. Then he tells them $k$ integers $n_1, \dots, n_k$, and finally he tells them the value of the product $P(n_1) \cdot P(n_2) \dots P(n_k)$. By all ... | [
"Answer. $k = 2017$.\n\nIf $k < 2017$, and $P(x)$ is the teacher's polynomial, then the students may find a polynomial $Q(x) = P(x) + (x-n_1)(x-n_2) \\cdots (x-n_k)$ which also satisfies all the conditions. For $k = 2017$, the teacher can, e.g., tell the students that $P(4)P(8)P(12) \\cdots P(4k) = 1$ then the only... | Russia | XLIII Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English; Russian | proof and answer | 2017 | |
0hdi | Consider an isosceles triangle $ABC$, where $AC$ is a base, $P$ is an arbitrary point on $AC$, $T$ is a projection of $P$ onto $BC$. Determine in the ratio in which a symmedian drawn from a vertex $C$ of a triangle $\triangle PBC$ divides $AT$? Symmedian $CS$, $S \in BP$ of a $\triangle PBC$ is a reflection of median $... | [
"Let $BH$ be an altitude of $\\triangle ABC$, then $H$ is a midpoint of $AC$, $BH$ and $PT$ are altitudes of $\\triangle PBC$ (Fig. 39). Recall that symmedian bisects a line segment whose endpoints are feet of the altitudes. Therefore, a symmedian of $\\triangle PBC$ is passing through $M$ which is a midpoint of $H... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem"
] | null | proof and answer | 2:1 | |
079p | A regular dodecahedron is a convex polyhedron such that its faces are regular pentagons. It has 20 vertices and 3 edges connected to each vertex. (As you see in the picture.)
Suppose that we have marked 10 vertices of a regular dodecahedron.
a) Prove that we can rotate the dodecahedron in su... | [
"a) Now consider one of the marked vertices named $A$. We have 10 marked vertices and each of them can lie on its locations in 3 distinct ways, so in $3 \\times 10 - 1 = 29$ rotations (other than the original state) a special vertex lie on the location of $A$ in original state, therefore totally $29 \\times 10 = 29... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Abstract Algebra > Group Theory"
] | English | proof only | null |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.