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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
07q9 | Assume $ABCD$ is a convex quadrilateral such that the triangles $ABD$, $BCD$, $CDA$, and $ABC$ have the same area. Prove that $ABCD$ is a parallelogram. | [
"Let $B'$ and $D'$ on $AC$ be the feet of the altitudes of the triangles $\\triangle ABC$ and $\\triangle ADC$. Because these two triangles have the same area, we get $|BB'| = |DD'|$. This implies that the two right triangles $BMB'$ and $DMD'$, which have equal angles at $M$, are congruent. In particular, $|BM| = |... | Ireland | Ireland | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
06tl | Determine all triples $(a, b, c)$ of positive integers for which $a b - c$, $b c - a$, and $c a - b$ are powers of $2$.
Explanation: A power of $2$ is an integer of the form $2^{n}$, where $n$ denotes some nonnegative integer. | [
"It can easily be verified that these sixteen triples are as required. Now let $(a, b, c)$ be any triple with the desired property. If we would have $a = 1$, then both $b - c$ and $c - b$ were powers of $2$, which is impossible since their sum is zero; because of symmetry, this argument shows $a, b, c \\geqslant 2$... | IMO | 56th International Mathematical Olympiad Shortlisted Problems | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | All permutations of (2,2,2), (2,2,3), (3,5,7), and (2,6,11). | |
02sr | Problem:
Sergio pediu para Ivan pensar em um número inteiro positivo. Depois, pediu para Ivan calcular a soma de seus algarismos e, finalmente, elevar ao quadrado o resultado. Sem falar o número em que pensou inicialmente, Ivan contou que obteve como resultado final $x$. Mostre a Sergio como chegar às seguintes conclu... | [
"Solution:\n\na) Se Ivan tivesse pensado em um número com 3 ou menos algarismos, a soma de seus algarismos seria no máximo $9+9+9=27$. Então o número final de Ivan $x$ seria no máximo $27^{2}=729$.\n\nb) Se Ivan tivesse pensado em um número com 4 algarismos, digamos $\\overline{a b c d}$, então $x=(a+b+c+d)^{2}$. D... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 1, 81, 169, 256 | |
0ekp | Problem:
Poišči vse pare naravnih števil $k$ in $n$, za katere ima ulomek $\frac{4^{k-1}}{n}$ v decimalnem zapisu obliko $0, \bar{n}$, kjer $\bar{n}$ označuje periodo. Na primer, če je $n=720$, tedaj je $0, \bar{n}=0,720720720 \ldots$ | [
"Solution:\n\nIz navodil naloge sledi $\\frac{4^{k-1}}{n}=0, \\bar{n}$. Označimo z $m$ število števk števila $n$. Če enačbo pomnožimo z $10^{m}$, dobimo $\\frac{4^{n}-1 \\cdot 10^{m}}{n}=n, \\bar{n}$. Prvo enačbo odštejemo od druge, da dobimo $\\frac{4^{k-1} \\cdot 10^{m}}{n}-\\frac{4^{k-1}}{n}=n$, kar lahko poenos... | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | k=1, n=3; k=2, n=6 | |
0b8m | Let $x$ be a real number. Prove that $x$ is an integer if and only if
$$
[x] + [2x] + [3x] + \dots + [nx] = \frac{n([x] + [nx])}{2}
$$
holds for all positive integers $n$ (here, $[a]$ denotes the integer part (floor function) of the real number $a$). | [
"If $x \\in \\mathbb{Z}$, then $[kx] = k[x]$ for every $k \\in \\mathbb{N}^*$.\n\nFor the converse, notice that the hypothesis implies, for all $n$,\n$$\nn([x] + [nx]) + 2[(n+1)x] = (n+1)([x] + [(n+1)x]).\n$$\nThis comes to $n[nx] = [x] + (n-1)[(n+1)x]$, $\\forall n \\ge 1$. Replacing $n$ with $n+1$ and subtracting... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
02vs | Problem:
João escreveu todas as potências de $2$, $3$ e $5$ maiores que $1$ e menores que $2017$ em uma folha de papel. Em seguida, ele realizou todos os produtos possíveis de dois números distintos dessa folha e os escreveu em outra folha de papel. Qual a quantidade de inteiros que João registrou na segunda folha? | [
"Solution:\n\nInicialmente, devemos encontrar as potências de $2$, $3$ e $5$ registradas na primeira folha. Como $2^{10}<2017<2^{11}$, $3^{6}<2017<3^{7}$ e $5^{4}<2017<5^{5}$, as potências escritas na primeira folha podem ser divididas em três conjuntos:\n$$\nP_{2} = \\{2^{1}, 2^{2}, \\ldots, 2^{10}\\}, \\quad P_{3... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Other"
] | null | final answer only | 155 | |
04ox | Find all complex numbers $z$ such that $z^2 = \frac{1}{z} + \frac{1}{\bar{z}}$. | [] | Croatia | Croatian Mathematical Society Competitions | [
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | 2^{1/3} | |
0gpp | Let $P$ be the set of all $2012$ tuples $(x_1, x_2, \dots, x_{2012})$, where $x_i \in \{1, 2, \dots, 20\}$ for each $1 \le i \le 2012$. The set $A \subset P$ is said to be decreasing if for each $(x_1, x_2, \dots, x_{2012}) \in A$ any $(y_1, y_2, \dots, y_{2012})$ satisfying $y_i \le x_i$ ($1 \le i \le 2012$) also belo... | [
"The answer is $\\frac{1}{20^{2012}}$.\nLet us treat more general case when $P$ is the set of all $n$ tuples. If $A = B = P$ then\n$f(A, B) = \\frac{1}{20^n}$. We prove that $f(A, B) \\le \\frac{1}{20^n}$ by induction over $n$.\n\n$n = 1$. Suppose that $A = \\{1, 2, \\dots, a+c\\}$, $B = \\{20-b-c+1, \\dots, 20\\}$... | Turkey | 20th Turkish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof and answer | 1/20^2012 | |
0kco | Problem:
There exist several solutions to the equation
$$
1 + \frac{\sin x}{\sin 4x} = \frac{\sin 3x}{\sin 2x}
$$
where $x$ is expressed in degrees and $0^{\circ} < x < 180^{\circ}$. Find the sum of all such solutions. | [
"Solution:\nWe first apply sum-to-product and product-to-sum:\n$$\n\\frac{\\sin 4x + \\sin x}{\\sin 4x} = \\frac{\\sin 3x}{\\sin 2x}\n$$\n$$\n2 \\sin(2.5x) \\cos(1.5x) \\sin(2x) = \\sin(4x) \\sin(3x)\n$$\nFactoring out $\\sin(2x) = 0$,\n$$\n\\sin(2.5x) \\cos(1.5x) = \\cos(2x) \\sin(3x)\n$$\nFactoring out $\\cos(1.5... | United States | HMMT February 2020 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof and answer | 320° | |
09gv | Let $A$ be the number defined as connecting the decimal representation of $8^{n+1}$ to the end of the decimal representation of $8^n$. Prove that there exist infinitely many natural numbers $n$ such that $A$ is divisible by $2016$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Other"
] | English | proof only | null | |
089q | Problem:
Anacleto ha appena finito di mangiare una tavoletta di cioccolato, e inizia a giocare con la carta in cui era avvolto, un rettangolo di lati $360~\mathrm{mm}$ e $300~\mathrm{mm}$. Decide di far una sola piega rettilinea in modo che, una volta piegata la carta, un vertice del rettangolo si trovi esattamente a ... | [
"Solution:\n\nLa risposta è 325. Sia $ABCD$ il rettangolo che costituisce la carta del cioccolato; supponiamo che $AB$ sia lungo $360~\\mathrm{mm}$ e che Anacleto faccia combaciare il punto $A$ con il punto medio del lato $BC$, che chiameremo $M$. Per fare ciò deve piegare lungo l'asse del segmento $AM$, che inters... | Italy | Progetto Olimpiadi della Matematica - Gara di Febbraio | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 325 | |
0533 | In an isosceles right triangle $ABC$ the right angle is at vertex $C$. On the side $AC$ points $K$, $L$ and on the side $BC$ points $M$, $N$ are chosen so that they divide the corresponding side into three equal segments. Prove that there is exactly one point $P$ inside the triangle $ABC$ such that $\angle KPL = \angle... | [
"\nFigure 3\n\nWithout loss of generality let the points on the side $AC$ be in the order $A$, $K$, $L$, $C$ and on the side $BC$ in the order $C$, $M$, $N$, $B$ (see Fig. 3). Choose the point $P$ so that the quadrilateral $LCMP$ is a square. Then $|KL| = |LC| = |LP|$ and $|MN| = |CM| = |MP... | Estonia | Open Contests | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0i4y | Problem:
A math professor stands up in front of a room containing 100 very smart math students and says, "Each of you has to write down an integer between 0 and 100, inclusive, to guess 'two-thirds of the average of all the responses.' Each student who guesses the highest integer that is not higher than two-thirds of t... | [
"Solution:\nSince the average cannot be greater than $100$, no student will write down a number greater than $\\frac{2}{3} \\cdot 100$. But then the average cannot be greater than $\\frac{2}{3} \\cdot 100$, and, realizing this, each student will write down a number no greater than $\\left(\\frac{2}{3}\\right)^{2} \... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Logic"
] | null | final answer only | 0 | |
02vg | Problem:
Considere um triângulo acutângulo $A B C$ e quadrados $A B Z X$ e $A C W Y$, construídos externamente sobre os seus lados. Os pontos $M$ e $P$ são os pontos médios dos segmentos $B C$ e $X Y$, respectivamente; os pontos $Q$ e $N$ são os centros dos quadrados $A B Z X$ e $A C W Y$, respectivamente; e os pontos... | [
"Solution:\n\na) Uma base média de um triângulo é um segmento ligando dois pontos médios de seus lados. Esse segmento é paralelo ao lado oposto e possui metade do seu comprimento. Sabendo isso, temos $R M=\\frac{A C}{2}, S M=\\frac{A B}{2}$ e $\\angle B R M=\\angle B A C=\\angle M S C$. Além disso, como $Q$ e $N$ s... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
00o4 | Determine all triples $(a, b, c)$ of positive integers such that
$$
a! + b! = 2^{c!}.
$$ | [
"*Answer.* The only solutions are $(1, 1, 1)$ and $(2, 2, 2)$.\n\nWe can assume without loss of generality that $a \\le b$.\n\n* For $a = b = 1$, we get $c = 1$, which gives the solution $(1, 1, 1)$.\n\n* For $a = 1$ and $b > 1$, the left-hand side is bigger than $1$ and odd, therefore, it cannot be a power of $2$ ... | Austria | AUT_ABooklet_2023 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (1, 1, 1) and (2, 2, 2) | |
0cfc | Let $O$ and $I$ be the circumcenter, respectively the incenter of the triangle $ABC$. The incircle is tangent to the sides $BC$, $AC$ and $AB$ at the points $A'$, $B'$, respectively $C'$, and $G'$ is the centroid of the triangle $A'B'C'$. If $G'O = 7G'I$, prove that triangle $ABC$ is equilateral.
Marian Ursărescu | [] | Romania | 74th NMO Shortlisted Problems | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geomet... | English | proof only | null | |
02xp | Problem:
Um número é dito TOP se possui 5 algarismos e quando o produto entre o $1^{\circ}$ e o $5^{\circ}$ é igual a soma do $2^{\circ}$, $3^{\circ}$ e $4^{\circ}$. Por exemplo, $12.338$ é TOP, pois possui 5 algarismos e $1 \cdot 8 = 2 + 3 + 3$.
a) Qual o valor de $a$ para que $23.4a8$ seja TOP?
b) Quantos números ... | [
"Solution:\n\na) Temos que $2 \\cdot 8 = 3 + 4 + a$, segue que $a = 9$.\n\nb) Seja $1b.cd2$ um número TOP. Temos que $b + c + d = 2$, sendo que todas as possibilidades $(b, c, d)$ são $(0,0,2)$, $(0,1,1)$, $(0,2,0)$, $(1,0,1)$, $(1,1,0)$, $(2,0,0)$, ou seja, são 6 números TOP.\n\nc) Seja $9e.fgh$ um número TOP. Vam... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a = 9; b = 6; c = 112 | |
0g95 | 令 $Q^+$ 表示正有理數所成的集合。試找出所有函數 $f : Q^+ \to Q^+$ 滿足
$f(1) = 1$ 與
$f(x+n) = f(x) + n f(\frac{1}{x})$ 對所有的正整數 $n$ 與所有的正有理數 $x$ 皆成立。
Let $Q^+$ be the set of all positive rational numbers. Find all functions $f : Q^+ \to Q^+$ satisfying $f(1) = 1$ and
$f(x+n) = f(x) + n f(\frac{1}{x})$ for all positive integers $n$ and $\for... | [
"首先，對一個正整數對 $(a, b)$ (其中 $a \\neq b$)，定義對他的一次操作為將較大的數換成他除以較小的數所得餘數。(例如 $(2, 5)$ 變成 $(2, 1)$, $(5, 2)$ 變成 $(1, 2)$)\n令 $g: Q^+ \\to N_0$ 使得對所有正有理數 $\\frac{p}{q}$ (其中 $p, q$ 為互質正整數), 皆有:\n對數對 $(p, q)$ 操作 $g(\\frac{p}{q})$ 遍後其中一者會變為 1 並且另一者不為 0.\n首先, 由於 $p, q$ 互質, 故對所有正有理數來說, $g$ 皆存在。又一旦其中一者變為 1 了, 那麼下次的操作會使另一者變為 0, 且... | Taiwan | 二〇一五數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | For x = p/q in lowest terms with p, q positive integers, f(x) = p. | |
0l6k | Problem:
Let $\triangle ABC$ be an equilateral triangle with side length $4$. Across all points $P$ inside triangle $\triangle ABC$ satisfying $[PAB] + [PAC] = [PBC]$, compute the minimum possible length of $PA$.
(Here, $[XYZ]$ denotes the area of triangle $\triangle XYZ$.) | [
"Solution:\n\nThe area condition implies $[ABC] = 2[PBC]$. Hence, $P$ lies on the $A$-midline of $\\triangle ABC$. Therefore, the minimum possible value of $PA$ is the distance from $A$ to this midline. This is achieved by taking $P$ to be the foot of the perpendicular from $A$ to the $A$-m... | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | sqrt(3) | |
03ht | Problem:
Each of the 36 line segments joining 9 distinct points on a circle is coloured either red or blue. Suppose that each triangle determined by 3 of the 9 points contains at least one red side. Prove that there are four points such that the 6 segments connecting them are all red. | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
0ge7 | 設 $S$ 為一個有 $n \ge 3$ 個正整數的集合，且集合中任兩相異數字之總和都不在 $S$ 中。證明必可將 $S$ 中的元素排成 $a_1, a_2, \dots, a_n$，使得對於所有 $i = 2, 3, \dots, n-1$，$a_{i-1} + a_{i+1}$ 和 $a_{i-1} - a_{i+1}$ 都不會被 $a_i$ 整除。
Let $S$ be a set consisting of $n \ge 3$ positive integers, none of which is a sum of two other distinct members of $S$. Prove that the elem... | [
"We prove by induction. Let $a = \\max S$ and use the inductive hypothesis to find an ordering $b_1, b_2, \\dots, b_{n-1}$ of $S \\setminus \\{a\\}$ so that $b_i$ does not divide $b_{i-1} \\pm b_{i+1}$ for all $2 \\le i \\le n-1$. Observe that\n$$\n|b_j - b_{j+1}| < a \\neq b_j + b_{j+1} < 2a,\n$$\nso that $a$ divi... | Taiwan | 2021 數學奧林匹亞競賽第二階段選訓營, 國際競賽實作(一) | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
08ss | Let $O$ be the circumcenter of an acute triangle $ABC$. A circle passing through points $A, O$ intersects lines $AB$ and $AC$ at points $P, Q$ distinct from $A$, respectively. If the lengths of the line segments $PQ$ and $BC$ are equal, what is the magnitude of the angle formed by the lines $PQ$ and $BC$ and less than ... | [
"Let us denote by $(XYZ)$ the directed angle determined by the 3 points $X, Y, Z$. More precisely, we set $(XYZ)$ to be equal to $\\alpha$ if the line $XY$ comes on top of the line $YZ$ when $XY$ is rotated around $Y$ counterclockwise by $\\alpha$ degrees. When an integral multiple of $180^\\circ$ is added to such ... | Japan | Japan 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 > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coor... | English | proof and answer | 60° | |
0e4k | Let $M$, $N$ and $P$ be the points on the sides $AB$, $BC$ and $CA$ of the triangle $ABC$, such that the quadrilateral $CPMN$ is a parallelogram. The lines $AN$ and $MP$ meet at $R$, the lines $BP$ and $MN$ meet at $S$ and the lines $AN$ and $BP$ meet at $Q$. Let $T$ be the image of $Q$ under the orthogonal projection ... | [] | Slovenia | Selection Examinations for the IMO | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0das | Some $n>2$ lamps are cyclically connected: lamp $1$ with lamp $2$, ..., lamp $k$ with $k+1$, ..., lamp $n$ with lamp $1$. At the beginning, all lamps are off. When one pushes the switch of a lamp, that lamp and the two ones connected to it change status (from off to on, or vice-versa). Determine the number of configura... | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Vectors",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | English | proof and answer | 2^n if 3 does not divide n; 2^{n-2} if 3 divides n | |
0b6h | Let $p$ be a positive number and $f : [0, 1] \to \mathbb{R}$ be a continuous concave function, such that $f(0) = 1$. Show that
$$
(p+1) \int_{0}^{1} x^{2p} f(x) \, dx + \frac{2p-1}{8p+4} \le \left( \int_{0}^{1} f(x) \, dx \right)^{2}
$$
and find when the equality is possible. | [] | Romania | Shortlisted Problems for the Romanian NMO | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | Equality holds if and only if f(x) = 1 − x on [0, 1]. | |
0cuq | Let $P(x)$ be a polynomial of degree $n \ge 2$ with nonnegative coefficients. Let $a$, $b$, and $c$ be the side lengths of some triangle. Prove that the numbers $\sqrt[n]{P(a)}$, $\sqrt[n]{P(b)}$, and $\sqrt[n]{P(c)}$ are also the side lengths of some triangle.
Пусть $P(x)$ — многочлен степени $n \ge 2$ с неотрицатель... | [
"Assume that $a \\ge b \\ge c$. Set $G(x) = P(x)/x^n$ and notice that $G(a) \\le G(b) \\le G(c)$.\n\nLet, without loss of generality, $a \\ge b \\ge c$; these three positive numbers are the side lengths of a triangle if and only if $a < b + c$. Since the coefficients of $P(x)$ are nonnegative, we have $P(a) \\ge P(... | Russia | XLIII Russian mathematical olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English; Russian | proof only | null | |
02tf | Problem:
Um decorador distribuirá flores em oito pontos ao redor de um arranjo quadrado de flores, como indicado na figura abaixo. Ele quer fazer isso de modo tal que, em cada lado do arranjo, as pessoas vejam sempre a mesma quantidade de flores. No exemplo abaixo, temos o total de 11 flores e em cada um dos 4 lados d... | [
"Solution:\n\na) A soma das flores vistas nos lados é $4 \\cdot 9=36$. Como as flores nos cantos são vistas por dois lados e as flores no meio dos lados são vistas apenas uma vez, podemos escrever:\n$$\n2C+M=36\n$$\nonde $C$ e $M$ indicam as quantidades de flores nos cantos e no meio. Consequentemente, $C+M=36-C \\... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) 36; b) 24 | |
0klz | Problem:
In a group of 50 children, each of the children in the group have all of their siblings in the group. Each child with no older siblings announces how many siblings they have; however, each child with an older sibling is too embarrassed, and says they have 0 siblings.
If the average of the numbers everyone say... | [
"Solution:\n\nFor $i \\geq 1$, let $a_{i}$ be the number of families that have $i$ members in the group. Then, among each family with $i$ children in the group, the oldest child will say $i-1$, and the rest will say $0$. Thus, the sum of all the numbers said will be $a_{2} + 2a_{3} + 3a_{4} + 4a_{5} + \\cdots = 50 ... | United States | HMMT Spring 2021 Guts Round | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 26 | |
052s | Circles $c_1, c_2$ with centers $O_1, O_2$, respectively, intersect at points $P$ and $Q$ and touch circle $c$ internally at points $A_1$ and $A_2$, respectively. Line $PQ$ intersects circle $c$ at points $B$ and $D$. Lines $A_1B$ and $A_1D$ intersect circle $c_1$ the second time at points $E_1$ and $F_1$, respectively... | [
"\nFigure 6\n\nSolution:\n\nLet the radii of $c_1, c_2$ and $c$ be $r_1, r_2$ and $r$, respectively. Homothety of ratio $\\frac{r}{r_1}$ with center $A_1$ takes circle $c_1$ to circle $c$ and points $E_1, F_1$ to points $B, D$, respectively. Thus it takes line $E_1F_1$ to line $BD$. Analogo... | Estonia | Open Contests | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
04pb | Cities are connected by one-way airline routes. Each two cities $A$ and $B$ are connected by exactly one route: either from $A$ to $B$, or from $B$ to $A$. Prove that there exists a city from which it is possible to reach any other city with at most one layover. | [] | Croatia | Croatian Mathematical Society Competitions | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
05fq | Problem:
Déterminer tous les couples d'entiers $(n, p)$ strictement positifs où $p$ est un nombre premier et tels que $n+p$ soit une puissance de $n$.
Une puissance de $n$ est de la forme $n^k$ pour $k$ entier naturel. | [
"Solution:\n\nSoit $(p, n)$ un éventuel couple solution. On dispose d'un entier naturel $k$ tel que\n$$\nn + p = n^k\n$$\nTout d'abord, si $k=0$, alors $n+p=1$. Or $p$ étant premier, $p>1$ et $n \\geqslant 0$ donc $n+p=1<p \\leqslant p+n$ ce qui est absurde. On a donc $k \\geqslant 1$.\n\nOn peut alors réécrire l'é... | France | ENVOi 3 : ARITHMÉTIQUE | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (2, 2) | |
0kln | Problem:
A chord is drawn on a circle by choosing two points uniformly at random along its circumference. This is done two more times to obtain three total random chords. The circle is cut along these three lines, splitting it into pieces. The probability that one of the pieces is a triangle is $\frac{m}{n}$, where $m,... | [
"Solution:\nInstead of choosing three random chords, we instead first choose 6 random points on the circle and then choosing a random pairing of the points into 3 pairs with which to form chords. If the chords form a triangle, take a chord $C$. Any other chord $C^{\\prime}$ must have its endpoints on different side... | United States | HMMT November 2021 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Geometry > Plane Geometry > Circles"
] | null | proof and answer | 115 | |
0bcv | The right prism $ABCA'B'C'$, with $AB = AC = BC = a$, has the property that there exists a unique point $M \in (BB')$ so that $AM \perp MC'$. Find the measure of the angle of the straight line $AM$ and the plane $(ACC')$.
Mircea Fianu, Cristian Mangra
 | [
"We first prove that $M$ is the midpoint of the edge $[BB']$. Indeed, if this is not the case, denote $M'$ the reflection of $M$ across the midpoint of $[BB']$. Then $\\Delta MAB \\equiv \\Delta M'C'B'$ and $\\Delta M'AB \\equiv \\Delta MC'B'$ imply $[MA] \\equiv [M'C']$ and $[M'A] \\equiv [MC']$. Therefore $\\Delt... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 45° | |
04np | A Gaussian integer is a complex number whose real and imaginary parts are both integers. Determine the largest positive integer $n$ for which there is a set of $n$ Gaussian integers such that the squares of their absolute values are consecutive positive integers. (Matko Ljulj) | [
"If a complex number $z = x + yi$ is Gaussian integer, then $|z|^2 = x^2 + y^2$ is the sum of two squares of integers.\nThe square of an even integer is divisible by 4, while the square of an odd integer gives remainder 1 when divided by 4. Therefore, the sum of squares of two integers can give remainder 0, 1 or 2 ... | Croatia | Croatia_2018 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | 3 | |
0fns | Let $ABCD$ be a cyclic quadrilateral inscribed in a circle $\gamma$. Let $X$ be a point on the extension of $AC$ such that $XB$ and $XD$ are tangent to $\gamma$. The tangent at $C$ intersects $XD$ at $Q$. Let $E$ be the intersection of $AQ$ with $\gamma$ distinct from $A$. Prove that lines $AD$, $BE$, and $CQ$ are conc... | [
"Suppose that $AD$ intersects $CQ$ at $Y$ and $AD$ intersects $BE$ at $Z$. Since $\\triangle XAD \\sim \\triangle XDC$, then we have $\\frac{AD}{DC} = \\frac{XA}{XD} = \\frac{XA}{XB} = \\frac{AB}{BC}$\n\n---\n\nfrom which follows $AB \\cdot DC = BC \\cdot AD$. By Ptolemy's theorem,\n$$\nAB \\cdot DC = BC \\cdot AD ... | Spain | International Mathematical Arhimede Contest | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Spanish | proof only | null | |
01as | Find all pairs $(x, y)$ of integers such that $y^3 - 1 = x^4 + x^2$. | [
"If $x = 0$, we get the solutions $(x, y) = (0, \\pm 1)$. These solutions will turn out to be the only ones. From now on, assume $x \\neq 0$. We add $1$ to both sides and factor: $y^3 = x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)$. We show that the factors $x^2 + x + 1$ and $x^2 - x + 1$ are co-prime. Assume that a ... | Baltic Way | Baltic Way 2013 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Algebraic Number Theory > Unique factorization"
] | null | proof and answer | (0, 1) | |
0076 | Hallar todas las potencias perfectas que terminan con los dígitos 2, 0, 0, 8, en ese orden.
ACLARACIÓN: Se llama potencia perfecta a un número de la forma $a^k$ donde $a$ y $k$ son enteros positivos y $k \geq 2$. Por ejemplo, $6^2$; $2^7$; $100^3$. | [] | Argentina | Argentina 2009 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | Spanish | proof and answer | All such perfect powers are cubes with base congruent to 1002 modulo 2500; equivalently, they are exactly the numbers of the form (1002 + 2500 t)^3 for integers t with 1002 + 2500 t > 0. | |
01tc | Let $m$ and $n$ be positive integers such that $m > n$. Define $x_k = \frac{m + k}{n + k}$ for $k = 1, 2, \ldots, n + 1$.
Prove that if all the numbers $x_1, x_2, \ldots, x_{n+1}$ are integers,
then $x_1 x_2 \ldots x_{n+1} - 1$ is divisible by an odd prime.
(IMO-2015 Shortlist, Problem N3) | [] | Belarus | 66th Belarusian 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 only | null | |
02pk | Problem:
A sequência de números $t_{1}, t_{2}, t_{3}, \ldots$ está definida por
$$
\left\{\begin{array}{l}
t_{1}=2 \\
t_{n+1}=\frac{t_{n}-1}{t_{n}+1}
\end{array}\right.
$$
para cada inteiro positivo $n$. Encontrar $t_{2011}$. | [
"Solution:\n\nCalculemos os primeiros termos da sequência:\n$$\n\\begin{aligned}\n& t_{2}=\\frac{2-1}{2+1}=\\frac{1}{3} \\\\\n& t_{3}=\\frac{1/3-1}{1/3+1}=-\\frac{1}{2} \\\\\n& t_{4}=\\frac{-1/2-1}{-1/2+1}=-3 \\\\\n& t_{5}=\\frac{-3-1}{-3+1}=2\n\\end{aligned}\n$$\nAssim, os primeiros cinco termos da sequência são $... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | -1/2 | |
064w | If $v$ is a positive integer and the number $v(v+3)$ is a perfect square, prove that $v$ is not a multiple of 3. | [
"Let $v(v+3) = w^2$, $w \\in \\mathbb{Z}$. We assume that: $v = 3\\kappa$, $\\kappa \\in \\mathbb{N}^*$. Then we have:\n$$\nv(v+3) = 3\\kappa(3\\kappa+3) = 9\\kappa(\\kappa+1) = w^2 \\Leftrightarrow 3 \\cdot 3\\kappa(\\kappa+1) = w^2 \\Rightarrow 3|w^2\n$$\nfrom which, since 3 is a prime, it follows that $3|w$.\n$$... | Greece | Selection Examination for Juniors | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
062y | Problem:
Es bezeichne $\mathbb{Q}^{+}$ die Menge der positiven rationalen Zahlen. Eine Funktion $f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}$ heiße elastisch, wenn für alle $x, y \in \mathbb{Q}^{+}$ die Ungleichung
$$
f(x)+f(y) \geq 4 f(x+y)
$$
gilt.
a. Man zeige: Ist $f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}$ e... | [
"Solution:\n\na.\nEs gilt\n$$\n4 f(x)+f(y)+f(z) \\stackrel{(*)}{\\geq} 4(f(x)+f(y+z)) \\stackrel{(*)}{\\geq} 16 f(x+y+z) .\n$$\nAddiert man dazu die zyklisch vertauschten Versionen $f(x)+4 f(y)+f(z) \\geq \\ldots$ und $f(x)+f(y)+4 f(z) \\geq \\ldots$, erhält man $6(f(x)+f(y)+f(z)) \\geq 48 f(x+y+z)$ und nach Divisi... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | yes | |
0537 | Find all pairs of integers $(a, b)$ satisfying $(a+1)(b-1) = a^2 b^2$. | [
"Answer: $(0,1)$ and $(-1,0)$.\n\nSince $a$ and $a+1$ are coprime, $a^2$ and $a+1$ are also coprime. Similarly $b^2$ and $b-1$ are coprime. Hence the equality can hold only in the case $a+1 = \\pm b^2$ and $b-1 = \\pm a^2$, where the signs in both equations are the same.\n\nLet both signs be pluses. Then from the f... | Estonia | Open Contests | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (0,1) and (-1,0) | |
0etm | If $x$ and $y$ are positive real numbers such that
$$
\sqrt{2x} + \sqrt{y} = 13 \text{ and } \sqrt{8x} + \sqrt{9y} = 35,
$$
calculate $20x + 23y$. | [
"The second equation can be written as $2\\sqrt{2x} + 3\\sqrt{y} = 35$. Together with the first equation, we obtain a system of two linear equations in the two unknowns $\\sqrt{2x}$ and $\\sqrt{y}$, which can easily be solved to give $\\sqrt{2x} = 4$ and $\\sqrt{y} = 9$. From this follows $x = 8$ and $y = 81$, so t... | South Africa | The South African Mathematical Olympiad Third Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 2023 | |
054s | In the mathematics circle, Juku raised a hypothesis that, for every integer $n > 4$, at least one out of the two largest integers that are less than $\frac{n}{2}$ is relatively prime to $n$. Is Juku's hypothesis valid? | [
"If $n$ is odd then the largest integer that is less than $\\frac{n}{2}$ is $\\frac{n-1}{2}$. Let $d$ be a common divisor of numbers $\\frac{n-1}{2}$ and $n$. Then $d$ is a common divisor of numbers $n-1$ and $n$, implying that $d = 1$. Hence $\\frac{n-1}{2}$ and $n$ are relatively prime, meaning that the hypothesi... | Estonia | National Olympiad Final Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
090y | Problem:
Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(x f(y)+2 y)=f(x y)+x f(y)+f(f(y))
$$
holds for all real numbers $x$ and $y$. | [
"Solution:\n\nThe functional equation has two solutions, $f(x) \\equiv 0$ and $f(x) \\equiv 2 x$.\n\nSetting $x=0$ and $y=0$ in the functional equation yields $f(f(0))=0$. So there is at least one zero point of $f$. Let $a$ be any of them. Setting $y=a$ gives us $f(2 a)=f(a x)+f(0)$. If $a \\neq 0$, then $f$ is a c... | Middle European Mathematical Olympiad (MEMO) | MEMO Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0 for all x, and f(x) = 2x for all x | |
0h16 | A circle touches the side $BC$ of a triangle $ABC$ at the vertex $B$ and intersects the side $AC$ at the vertex $A$ and a point $E$. Another circle touches the side $BC$ at the vertex $C$ and intersects the side $AB$ at the vertex $A$ and a point $D$. $F$ is the point of intersection of the line segments $BE$ and $CD$.... | [
"Denote by $G$ the second point of intersection of the circles and join $G$ with all the vertices of the triangle $ABC$ (fig. 42). Then we have: $\\angle GBC = \\angle BAG = \\alpha$ (the angle subtended by $BG$ of the first circle). On the other hand, $\\angle GCD = \\angle BAG = \\alpha$ (the angle subtended by $... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
0gdb | 試求所有正整數 $n$，使得我們可以在 $n \times n$ 棋盤的 $n^2$ 個方格中，各放入 $\uparrow, \downarrow, \leftarrow, \rightarrow$ 四個箭號中的其中一個，讓以下三個條件皆被滿足：
(1) 從任何一格作為起點出發，按照箭號的方向走，我們都會回到起點，且途中不會走出棋盤。
(2) 除了最上與最下兩橫列之外，任一列中 $\uparrow$ 和 $\downarrow$ 的箭號一樣多（但不同列之間的數量可能不同。）
(3) 除了最左與最右兩直排之外，任一排中 $\leftarrow$ 和 $\rightarrow$ 的箭號一樣多（但不同排之間的數量可能不同。） | [
"唯一可能是 $n=2$ (此時箭號形成順時鐘或逆時鐘的環。)\n\n顯然最上橫列不能有 $\\uparrow$, 假設其有 $k$ 個 $\\downarrow$. 由條件 (1), 這表示第二橫列必須有 $k$ 個 $\\uparrow$, 但又由條件 (2), 我們知第二橫列同時須有 $k$ 個 $\\downarrow$. 依相同論證, 我們知道第 3 到第 $n-1$ 橫列都必須有 $k$ 個 $\\uparrow$ 與 $k$ 個 $\\downarrow$, 而最下橫列必須有 $k$ 個 $\\uparrow$ 與 0 個 $\\downarrow$. 換言之, 全棋盤中 $\\uparrow$ 和 $\\do... | Taiwan | 二〇一九數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 2 | |
07s3 | There are $2^{2018}$ buckets placed in a circle. Their positions are numbered from $1$ to $2^{2018}$ according to their order in the circle. Initially only buckets at positions $2018$ and $2019$ are full of water, the others are empty.
Before the game starts, John is allowed to fill as many additional buckets as he wi... | [
"If we let $k = 2^n$, the positions John visits during his journey are the positive integer multiples of $k$. Because $k \\mid 2^{2018}$, he will arrive at position $2^{2018}$ and then visit the same positions in the circle for each subsequent time he moves around the circle.\n\nWe will say \"John carries empty-ful... | Ireland | Irish | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Yes. Pre-fill so that all positions whose number minus one has an odd count of ones in binary are full and the others are empty; equivalently, fill all type minus one positions and leave type plus one positions empty. | |
0eop | If $a + 2b = 13$ and $5a - 2b = 5$, the value of $b$ is
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5 | [
"Given $a + 2b = 13$ and $5a - 2b = 5$, we can add both left-hand sides and both right-hand sides to find $6a = 18$. Thus $a = 3$, and then since $a + 2b = 13$, we must have $2b = 10$, i.e. $b = 5$."
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | E | |
0iu7 | Problem:
Points $A$ and $B$ lie on circle $\omega$. Point $P$ lies on the extension of segment $A B$ past $B$. Line $\ell$ passes through $P$ and is tangent to $\omega$. The tangents to $\omega$ at points $A$ and $B$ intersect $\ell$ at points $D$ and $C$ respectively. Given that $A B=7$, $B C=2$, and $A D=3$, compute... | [
"Solution:\n\nSay that $\\ell$ is tangent to $\\omega$ at point $T$. Observing equal tangents, write\n\n$$\nC D = C T + D T = B C + A D = 5.\n$$\n\nLet the tangents to $\\omega$ at $A$ and $B$ intersect each other at $Q$. Working from Menelaus applied to triangle $C D Q$ and line $A B$ gives\n$$\n\\begin{aligned}\n... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 9 | |
08wa | Determine all the real valued functions $f$ defined on the real line for which the following identity is satisfied for every pair of real numbers $x, y$:
$$
f(f(x+y)f(x-y)) = x^2 - y f(y).
$$ | [
"If we substitute $(0,0)$ for $(x, y)$ into the given identity, we get $f((f(0))^2) = 0$. And applying this fact after substituting $(0, (f(0))^2)$ for $(x, y)$ in the given identity, we obtain the fact that $f(0) = 0$ must hold.\n\nLet $t$ be an arbitrary non-zero real number, and substitute $(t, t)$ for $(x, y)$ ... | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = x for all real x | |
09b5 | Let $n$ and $d$ be positive integers satisfying $d < \frac{n}{2}$. There are $n$ boys and $n$ girls in a school. Each boy has at most $d$ girlfriends and each girl has at most $d$ boyfriends. Prove that one can introduce some of them to make each boy have exactly $2d$ girlfriends and each girl have exactly $2d$ boyfrie... | [
"Let us define a bipartite graph $G$ as usual way. The boys and girls are vertices of the graph and pairs of friends are edges. It suffices to prove that the bipartite complement of this graph has a $(n - 2d)$ regular subgraph with $2n$ vertices.\n\n**Lemma.** Let $G$ be a bipartite graph. $V(G) = A \\cup B$ and al... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Menger's theorem / max-flow, min-cut",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof only | null | |
0hmt | Problem:
A group of mathematicians is lost in a forest. The forest has a shape of an infinite strip that is 1 mile wide. Prove that they can choose a path that will guarantee them a way out and that is at most $2 \sqrt{2}$ miles long.
Remark. The mathematicians have no device for orientation and no maps. All they kno... | [
"Solution:\n\nStarting from their initial point $A$, the mathematicians should first move $\\sqrt{2}$ miles in any direction. If they didn't get to the exit, they have arrived at point $B$. Then they should turn by $90^{\\circ}$ and walk for another $\\sqrt{2}$ miles to the point $C$. We claim that they did reach t... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 2 sqrt(2) miles | |
030n | Problem:
Na figura, $\angle ABC = 100^\circ$, $\angle FAC = 3 \angle ECB$ e $\angle GCA = 3 \angle DAB$. Determine a medida do ângulo agudo na interseção das bissetrizes internas dos triângulos $\triangle ADB$ e $\triangle CEB$ relativo aos ângulos $D$ e $E$.
 | [
"Solution:\n\nVamos marcar na figura os pontos de intersecção e os ângulos que nos interessam, sendo $x$ o ângulo da intersecção das bissetrizes que procuramos.\n\n\n\nPela soma dos ângulos internos do triângulo $\\triangle ABC$, temos:\n$$\n\\begin{aligned}\n100^\\circ + (180^\\circ - \\al... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 35° | |
03nn | Problem:
Let $n \geq 2$ be some fixed positive integer and suppose that $a_{1}, a_{2}, \ldots, a_{n}$ are positive real numbers satisfying $a_{1}+a_{2}+\cdots+a_{n}=2^{n}-1$.
Find the minimum possible value of
$$
\frac{a_{1}}{1}+\frac{a_{2}}{1+a_{1}}+\frac{a_{3}}{1+a_{1}+a_{2}}+\cdots+\frac{a_{n}}{1+a_{1}+a_{2}+\cdots+... | [
"Solution:\nWe claim the the minimum possible value of this expression is $n$. Observe that by AM-GM, we have that\n$$\n\\begin{aligned}\n\\frac{a_{1}}{1} &+ \\frac{a_{2}}{1+a_{1}}+\\cdots+\\frac{a_{n}}{1+a_{1}+a_{2}+\\cdots+a_{n-1}} \\\\\n&=\\frac{1+a_{1}}{1}+\\frac{1+a_{1}+a_{2}}{1+a_{1}}+\\cdots+\\frac{1+a_{1}+a... | Canada | Canadian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | n | |
01u8 | Let $p = \overline{abc}$ be the decimal representation of a three-digit prime number $p$.
Prove that the quadratic equation $ax^2 + bx + c = 0$ has no real roots. | [
"Since $abc$ is the decimal representation of a three-digit prime number $p$, we have $a \\neq 0$, $c \\neq 0$. Suppose, contrary to our claim, that there exists a rational root of the equation $ax^2+bx+c=0$. Then the discriminant of this equation is a perfect square, i.e. $b^2-4ac=n^2$, where $n$ is a positive int... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
05mb | Problem:
Soient $\Gamma$ un cercle de centre $O$, et $(d)$ une droite qui n'intersecte pas $\Gamma$. On appelle $P$ le projeté orthogonal de $O$ sur $(d)$. Soit $Q$ un point variable sur la droite $(d)$, et $\left(t_{1}\right)$ et $\left(t_{2}\right)$ les tangentes à $\Gamma$ passant par $Q$. On note $A$ et $B$ les pr... | [
"Solution:\n\nOn note $X$ l'intersection de $(AB)$ et $(OP)$, $S$ et $T$ les points où $\\left(t_{1}\\right)$ et $\\left(t_{2}\\right)$ touchent $\\Gamma$, et $Y$ le point d'intersection de $(ST)$ et $(OP)$. Une première remarque qu'on peut faire est que $O, P, Q, S$ et $T$ sont cocycliques sur le cercle de diamètr... | France | French Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Misce... | null | proof only | null | |
0hg8 | Consider the circumscribed circle of an obtuse triangle $ABC$ with an obtuse angle $B$. Tangents to this circle at points $A$ and $B$ meet at point $P$, and the perpendicular to the line $BC$ at point $B$ intersects $AC$ at point $K$. Prove, that $PA = PK$.
(Danylo Khilko) | [] | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0f00 | Problem:
$ABCD$ is a rectangle. $M$ is the midpoint of $AD$ and $N$ is the midpoint of $BC$. $P$ is a point on the ray $CD$ on the opposite side of $D$ to $C$. The ray $PM$ intersects $AC$ at $Q$. Show that $MN$ bisects the angle $PNQ$. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0d5y | Let $p$ be an odd prime number. Prove that there exists a unique integer $k$ such that $0 \leq k \leq p^{2}$ and $p^{2}$ divides $k(k+1)(k+2) \cdots(k+p-3)-1$. | [
"Assume that there exists an integer $k$ such that $0 \\leq k \\leq p^{2}$ and $p^{2}$ divides $k(k+1)(k+2) \\cdots(k+p-3)-1$. Because $p$ does not divide any of the $p-2$ consecutive integers $k,(k+1),(k+2), \\ldots,(k+p-3)$, either $k \\equiv 1$ or $2 \\bmod p$. If $k \\equiv 2 \\bmod p$ then, by Wilson's theorem... | Saudi Arabia | SAMC 2015 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English, Arabic | proof only | null | |
06r6 | Let $P_{1}, \ldots, P_{s}$ be arithmetic progressions of integers, the following conditions being satisfied:
(i) each integer belongs to at least one of them;
(ii) each progression contains a number which does not belong to other progressions.
Denote by $n$ the least common multiple of steps of these progressions; let ... | [
"First, we prove the key lemma, and then we show how to apply it to finish the solution.\nLet $n_{1}, \\ldots, n_{k}$ be positive integers. By an $n_{1} \\times n_{2} \\times \\cdots \\times n_{k}$ grid we mean the set $N= \\left\\{\\left(a_{1}, \\ldots, a_{k}\\right): a_{i} \\in \\mathbb{Z}, 0 \\leq a_{i} \\leq n_... | IMO | 51st IMO Shortlisted Problems | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
02fo | There are $n$ boys $B_1, B_2, \dots, B_n$ and $n$ girls $G_1, G_2, \dots, G_n$. Each boy ranks the girls in order of preference, and each girl ranks the boys in order of preference. Show that we can arrange the boys and girls into $n$ pairs so that we cannot find a boy and a girl who prefer each other to their partners... | [
"In round $k$ each unpaired boy $B$ in turn proposes to the girl $G$ he ranks highest amongst those he has not yet proposed to. If $G$ is not yet paired, or if she prefers $B$ to the boy she is currently paired with, then we pair $B$ and $G$ (and remove any existing pair for $G$). We claim that this algorithm termi... | Brazil | XVIII OBM | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0hc6 | Determine the positive integer $n$, for which the following holds:
$$
n^2 = 2 \cdot (20^4 + 19^4 + 39^4).
$$ | [
"Consider the following expression:\n$$\nx^4 + y^4 + (x + y)^4 = 2x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + 2y^4,\n$$\nthus\n$$\n2(x^4 + y^4 + (x+y)^4) = 4(x^4 + 2x^3y + 3x^2y^2 + 2xy^3 + y^4) = (2(x^2 + xy + y^2))^2.\n$$\nHence for $x = 20$, $y = 19$ the following holds:\n\n$$\nn^2 = (2(x^2 + xy + y^2))^2 \\text{ or } n = 2... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | 2282 | |
04do | Only one divisor of the number $3^{12} - 1$ is larger than $70$ and smaller than $80$. Find it. (Hong Kong) | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 73 | |
06n8 | Someone extracted $n$ consecutive digits in the infinite decimal representation of $\frac{1}{13}$ and found that the sum of the $n$ digits is equal to $2023$. Find the value of $n$. | [
"Answer: $449$\n\nThe number is the recurring decimal $0.\\dot{0}7692\\dot{3}$. Note that $0 + 7 + 6 + 9 + 2 + 3 = 27$ and $2023 = 27 \\times 75 - 2$. Hence there should be $75$ periods of '307692' with the last '2' removed.\n\nThe answer is thus $75 \\times 6 - 1 = 449$."
] | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 449 | |
0daw | Find the smallest positive integer $n$ with the following property: After painting black exactly $n$ cells of a $7 \times 7$ board there always exists a $2 \times 2$ square with at least three black cells. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 29 | |
0h9j | Is it possible to draw a square with area of $50$ units on a graphing paper using a ruler that can connect two edges with a line? | [
"Consider a $10 \\times 10$ square and connect the midpoints of the sides as shown in Fig. 17. The square obtained clearly has area that is a half of the original square's area, that is $\\frac{1}{2} \\cdot 10 \\cdot 10 = 50$.\n\n"
] | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | Yes | |
0ivh | Problem:
A circle having radius $r_{1}$ centered at point $N$ is tangent to a circle of radius $r_{2}$ centered at $M$. Let $l$ and $j$ be the two common external tangent lines to the two circles. A circle centered at $P$ with radius $r_{2}$ is externally tangent to circle $N$ at the point at which $l$ coincides with c... | [
"Solution:\nSuppose the lines are parallel. Draw the other tangent line to $N$ and $P$—since $M$ and $P$ have the same radius, it is tangent to all three circles. Let $j$ and $k$ meet circle $N$ at $A$ and $B$, respectively. Then by symmetry we see that $\\angle ANM = \\angle MNP = \\angle PNB = 60^{\\circ}$ since ... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 3 | |
0kzk | Problem:
Compute the sum of the three smallest positive integers $n$ for which
$$
\frac{1+2+3+\cdots+(2024 n-1)+2024 n}{1+2+3+\cdots+(4 n-1)+4 n}
$$
is an integer. | [
"Solution:\nWe simplify the expression as follows:\n$$\n\\begin{aligned}\n\\frac{(2024 n)(2024 n+1) / 2}{(4 n)(4 n+1) / 2} & = \\frac{506 \\cdot (2024 n+1)}{4 n+1} \\\\\n& = \\frac{506 \\cdot (506 \\cdot (4 n+1) - 505)}{4 n+1} \\\\\n& = 506^{2} - \\frac{506 \\cdot 505}{4 n+1} \\\\\n& = 506^{2} - \\frac{2 \\cdot 5 \... | United States | HMMT November 2024 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 89 | |
0a0p | In the left square, rectangle $ABFE$ is drawn. In the right square, rectangle $PSRQ$ is drawn, where the sides $QR$ and $PS$ are parallel to the diagonal $AC$ of the square. The sides of both squares have length $6$. It turns out that the rectangles $ABFE$ and $PSRQ$ have exactly the same dimensions.
![](attached_imag... | [
"$6\\sqrt{2} - 6$"
] | Netherlands | Dutch Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | final answer only | 6√2 - 6 | |
0938 | Problem:
Let $\mathbb{Q}^{+}$ denote the set of all positive rational numbers and let $\alpha \in \mathbb{Q}^{+}$. Determine all functions $f: \mathbb{Q}^{+} \rightarrow (\alpha,+\infty)$ satisfying
$$
f\left(\frac{x+y}{\alpha}\right)=\frac{f(x)+f(y)}{\alpha}, \quad \text{ for all } x, y \in \mathbb{Q}^{+} .
$$ | [
"Solution:\nAnswer. For $\\alpha=2$ the solutions of our functional equation are given by $f(x)=A x+B$ for all $x \\in \\mathbb{Q}^{+}$, where either $A>0$ and $B \\geq 2$ or $A=0$ and $B>2$. For $\\alpha \\neq 2$ there are no solutions.\n\nBy putting $x=y$ in the given functional equation we get $f\\left(\\frac{2 ... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | For α = 2: f(x) = A x + B with either A > 0 and B ≥ 2, or A = 0 and B > 2. For α ≠ 2: no solutions exist. | |
0in6 | Problem:
Let $N$ be the number of ordered pairs $(x, y)$ of integers such that
$$
x^2 + x y + y^2 \leq 2007
$$
Remember, integers may be positive, negative, or zero!
a. Prove that $N$ is odd.
b. Prove that $N$ is not divisible by $3$. | [
"Solution:\n\na.\nIf $(x, y)$ is a pair of integers that satisfies the inequality, then $(-x, -y)$ is also such a pair, since\n$$\n(-x)^2 + (-x)(-y) + (-y)^2 = x^2 + x y + y^2.\n$$\nSo we can match up pairs of solutions to the inequality, $(x, y) \\leftrightarrow (-x, -y)$. Every solution will be paired with a diff... | United States | 9th Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | null | proof only | null | |
0cvz | Two monic quadratic trinomials $f(x)$ and $g(x)$ are given, each of them has two distinct real roots. Assume that $f(1) = g(2)$ and $g(1) = f(2)$. Find all possible values of the sum of all four roots of the trinomials. | [
"**Первое решение.** Пусть $f(x) = x^2 + a x + b$, $g(x) = x^2 + c x + d$. Тогда условия задачи запишутся в виде\n$$\n1 + a + b = 4 + 2c + d \\text{ и } 4 + 2a + b = 1 + c + d.\n$$\nВычитая из первого равенства второе, получаем $-3 - a = 3 + c$, то есть $a + c = -6$. Но по теореме Виета $-a$ — это сумма корней перв... | Russia | Regional round | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English; Russian | proof and answer | 6 | |
0j7d | Problem:
Order any subset of the following twentieth century mathematical achievements chronologically, from earliest to most recent. If you correctly place at least six of the events in order, your score will be $2(n-5)$, where $n$ is the number of events in your sequence; otherwise, your score will be zero. Note: if... | [
"Solution:\n\nAnswer: IAJCBMHODEFLNGK\n\nThe dates are as follows:\nA). Axioms for Set Theory published by Zermelo 1908\nB). Category Theory introduced by Mac Lane and Eilenberg 1942-1945\nC). Collatz Conjecture proposed 1937\nD). Erdos number defined by Goffman 1969\nE). First United States delegation sent to Inte... | United States | Harvard-MIT November Tournament | [
"Math Word Problems"
] | null | final answer only | IAJCBMHODEFLNGK | |
0727 | Problem:
Suppose $p$ is a prime greater than $3$. Find all pairs of integers $(a, b)$ satisfying the equation
$$
a^{2} + 3 a b + 2 p(a + b) + p^{2} = 0
$$ | [
"Solution:\nWe write the equation in the form\n$$\na^{2} + 2 a p + p^{2} + b(3 a + 2 p) = 0\n$$\nHence\n$$\nb = \\frac{-(a + p)^{2}}{3 a + 2 p}\n$$\nis an integer. This shows that $3 a + 2 p$ divides $(a + p)^{2}$ and hence also divides $(3 a + 3 p)^{2}$. But, we have\n$$\n(3 a + 3 p)^{2} = (3 a + 2 p + p)^{2} = (3... | India | INMO 2004 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | Let p > 3 be prime.
- If p = 3k + 1, with k = (p − 1)/3, the solutions are (a, b) = (−2k − 1, k^2), (−p, 0), (−(k + 1)p, k^2).
- If p = 3k − 1, with k = (p + 1)/3, the solutions are (a, b) = (1 − 2k, −k^2), (−p, 0), ((k − 1)p, −k^2). | |
0gd6 | 有一條小河, 其一側的露營基地裡插著 4038 根營柱。其中有些營柱之間有繩子連接, 每條繩子連接兩根不同的營柱, 兩根營柱之間至多只有一條繩子。假設當我們把其中任意 2019 根營柱移到河的對岸, 都會剛好有 $k$ 條繩子通過河面。試求 $k$ 的所有可能值。 | [
"答案是 $0, 2019, 2018 \\times 2019, 2019^2$。\n\n考慮圖 $G(V, E)$，點集 $V$ 和邊集 $E$ 分別對應到營柱和繩子。對集合 $S, T \\subset V$，令\n$$\nE(S, T) = \\left| \\left\\{ (s,t) \\in E : s \\in S, t \\in T \\right\\} \\right|,\n$$\n也就是連接兩群之間的邊數。為簡化符號, 當 $S$ 只有 $a$ 一個點時, 我們簡寫此集合為 $a$, 依此類推。\n\n1. **Lemma:** 對所有相異 $a, b, c, d \\in V$，都有\n$$\nE(a... | Taiwan | 二〇一九數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 0, 2019, 2018*2019, 2019^2 | |
08t1 | 8 coins are placed on a line from left to right.
In the sequel, we keep on repeating the following action:
Choose at random a coin showing a head and satisfying the following condition:
Condition: Either there is no coin on its right showing a tail, or there is no coin on its left showing a tail.
Suppose we keep on rep... | [
"It is clear that the conclusion would be the same if we follow the following procedure instead of the procedure specified in the statement of the problem:\nFirst we assign one of the numbers $1, 2, \\ldots, 8$ to the coins already lined up in a random way (so that each of $8!$ ways of assigning the numbers has the... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | proof and answer | 621/140 | |
0j1s | Problem:
Place eight rooks on a standard $8 \times 8$ chessboard so that no two are in the same row or column. With the standard rules of chess, this means that no two rooks are attacking each other. Now paint 27 of the remaining squares (not currently occupied by rooks) red.
Prove that no matter how the rooks are ar... | [
"Solution:\n\nLook at the $\\binom{8}{2} = 28$ pairs of rooks. (Ignore the coloring for now.) Each pair of rooks determines a pair of empty squares in the usual way: take the other two vertices of the rectangle (with sides parallel to the edge of the chessboard, of course) having our given pair of rooks as two vert... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0kvl | Problem:
Suppose $ABCD$ is a convex quadrilateral with $\angle ABD = 105^{\circ}$, $\angle ADB = 15^{\circ}$, $AC = 7$, and $BC = CD = 5$. Compute the sum of all possible values of $BD$. | [
"Solution:\n\nLet $O$ be the circumcenter of triangle $ABD$. By the inscribed angle theorem, $\\angle AOC = 90^{\\circ}$ and $\\angle BOC = 60^{\\circ}$. Let $AO = BO = CO = x$ and $CO = y$. By the Pythagorean theorem on triangle $AOC$,\n$$\nx^{2} + y^{2} = 49\n$$\nand by the Law of Cosines on triangle $BOC$,\n$$\n... | United States | HMMT February 2023 | [
"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 | sqrt(291) | |
0cmf | The Mumbo-Jumbo tribe lives at a river. Once two members of a tribe — a young warrior Mumbo and a wise shaman Jumbo — departed to the neighboring tribe with an urgent news. Mumbo ran at $11$ km/h to the nearest location of the rafts, took a raft and drove it to the neighboring tribe. Jumbo went at $6$ km/h to another r... | [
"**Answer.** $26$ km/h.\n\nLet the location of the Mumbo-Jumbo tribe be $O$, the raft location to which Mumbo ran be $M$, and the raft location to which Jumbo went be $U$. Obviously, $M$ is upstream from $O$, and $U$ is downstream.\n\nLet the distances from $O$ to $M$ and $U$ be $x$ and $y$ km respectively ($x < y$... | Russia | Euler olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English; Russian | proof and answer | 26 km/h | |
02oi | Problem:
Uma grade de pontos com 10 linhas e 10 colunas é dada. Cada ponto é colorido de vermelho ou de azul. Sempre que dois pontos da mesma cor são vizinhos em uma mesma linha ou coluna, eles são ligados por um segmento da mesma cor dos pontos. Se dois pontos são vizinhos mas de cores diferentes, são ligados por um ... | [
"Solution:\n\nInicialmente, observe que existem 9 segmentos em cada linha e em cada coluna, de modo que existem $9 \\times 10 + 9 \\times 10 = 180$ segmentos no total.\nSeja $A$ o número de segmentos azuis e $V$ o número de segmentos vermelhos. Então $A + V + 98 = 180$, de modo que $A + V = 82$, já que existem 98 s... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 37 | |
0jkg | Problem:
Fix positive integers $m$ and $n$. Suppose that $a_{1}, a_{2}, \ldots, a_{m}$ are reals, and that pairwise distinct vectors $v_{1}, \ldots, v_{m} \in \mathbb{R}^{n}$ satisfy
$$
\sum_{j \neq i} a_{j} \frac{v_{j}-v_{i}}{\left\|v_{j}-v_{i}\right\|^{3}}=0
$$
for $i=1,2, \ldots, m$.
Prove that
$$
\sum_{1 \leq i<j ... | [
"Solution:\n\nAnswer: N/A Since $v_{i} \\cdot\\left(v_{j}-v_{i}\\right)+v_{j} \\cdot\\left(v_{i}-v_{j}\\right)=-\\left\\|v_{j}-v_{i}\\right\\|^{2}$ for any $1 \\leq i<j \\leq m$, we have\n$$\n0=\\sum_{i=1}^{m} a_{i} v_{i} \\cdot 0=\\sum_{i=1}^{m} a_{i} v_{i} \\sum_{j \\neq i} a_{j} \\frac{v_{j}-v_{i}}{\\left\\|v_{j... | United States | HMMT 2014 HMIC | [
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
0l9n | In plane, let be given an octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ such that no three of its diagonals are concurrent. Each point of intersection of every two diagonals of the octagon is called a cross.
Consider the convex quadrilaterals whose vertices are vertices of the given octagon. Each such quadrilateral is called a su... | [
"Suppose that there are $n$ crosses which can be coloured so that the numbers $s(i, k)$ are all equal. Denote by $s$ the common value of these numbers.\nSince for each couple of vertices $A_i, A_k$ of the octagon, there exist exactly $s$ subquadrilaterals admitting $A_i, A_k$ as vertices and admitting a coloured cr... | Vietnam | Vietnam Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | English | proof and answer | 14 | |
0cwz | Numbers $1, 2, 3, \ldots, 60$ are written in a row (in this order). Igor and Ruslan take turns in making moves; Igor starts; by one move the player puts one of the signs of operation $+$, $-$, $\times$ between some pair of adjacent numbers. When a sign is placed between every two adjacent numbers we calculate the value... | [
"Let us replace all numbers in the sequence with their remainders modulo $3$; this will not change the game's outcome.\n\nWe obtain the sequence $1, 2, 0, \\ldots, 1, 2, 0$. The gaps between the numbers are numbered from left to right from $1$ to $59$.\nOn his first move, Igor places a \"$-$\" sign in the $30$th ga... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Other"
] | Russian | proof and answer | Igor | |
0hum | Problem:
For positive integers $n$, define
$$
a_{n} = 3^{n} + 6^{n} - 2^{n}.
$$
Find, with proof, all primes that do not divide any of the numbers $a_{1}, a_{2}, a_{3}, \ldots$. | [
"Solution:\n\nAnswer: 2 and 3. It is clear that, for $n \\geq 1$, the terms $6^{n}$ and $2^{n}$ are even while $3^{n}$ is odd, so $2 \\nmid a_{n}$. Similarly, $3^{n}$ and $6^{n}$ are divisible by $3$ but $2^{n}$ is not, so $3 \\nmid a_{n}$.\n\nLet $p \\geq 5$ be a prime. We claim that $a_{p-2} \\equiv 0 \\pmod{p}$.... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 2 and 3 | |
0eyd | Problem:
Given a regular tetrahedron $ABCD$, prove that it is contained in the three spheres on diameters $AB$, $BC$ and $AD$. Is this true for any tetrahedron? | [
"Solution:\n\nLet the tetrahedron have side $1$. Then the center $O$ is a distance $1 / \\sqrt{8}$ from the center of each of the spheres, so it is contained in each of the spheres. We now use convexity.\n\nTwo circles with diameters two of the sides of a triangle cover the triangle (consider the foot of the altitu... | Soviet Union | 2nd ASU | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | False in general. | |
06dj | $\triangle ABC$ is isosceles with $AB = AC$. $D$ is a point on the minor arc $AC$ on the circumcircle of $\triangle ABC$. Let $E$ be the foot of the perpendicular from $A$ to $BD$. Prove that $BE = ED + DC$. | [
"(Archimedes' broken-chord theorem) Let $F$ be the foot of perpendicular from $A$ to $CD$. Note that\n$$\n\\angle ADE = \\angle ADB = \\angle ACB = \\angle ABC = \\angle ADF.\n$$\nAlso, we have $\\angle AED = \\angle AFD$ and $AD = AF$. This implies $\\triangle ADE \\cong \\triangle ADF$, and hence $AE = AF$ and $D... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0j5o | Problem:
Consider a cube $A B C D E F G H$, where $A B C D$ and $E F G H$ are faces, and segments $A E$, $B F$, $C G$, $D H$ are edges of the cube. Let $P$ be the center of face $E F G H$, and let $O$ be the center of the cube. Given that $A G=1$, determine the area of triangle $A O P$. | [
"Solution:\n\nAnswer: $\\frac{\\sqrt{2}}{24}$\n\nFrom $A G=1$, we get that $A E=\\frac{1}{\\sqrt{3}}$ and $A C=\\frac{\\sqrt{2}}{\\sqrt{3}}$. We note that triangle $A O P$ is located in the plane of rectangle $A C G E$. Since $O P \\parallel C G$ and $O$ is halfway between $A C$ and $E G$, we get that $[A O P]=\\fr... | United States | Harvard-MIT November Tournament | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(2)/24 | |
06yp | Problem:
Find all integer solutions to: $a+b+c=24$, $a^{2}+b^{2}+c^{2}=210$, $a b c=440$. | [
"Solution:\n\n$ab+bc+ca = \\left((a+b+c)^2 - (a^2+b^2+c^2)\\right)/2 = 183$, so $a$, $b$, $c$ are roots of the cubic $x^3 - 24x^2 + 183x - 440 = 0$. But it easily factorises as $(x-5)(x-8)(x-11)=0$, so the only solutions are permutations of $(5,8,11)$."
] | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | All permutations of (5, 8, 11). | |
071b | Problem:
The sequence $a_{n}$ is defined as follows: $a_{1}=56$, $a_{n+1}=a_{n}-1/a_{n}$. Show that $a_{n}<0$ for some $n$ such that $0<n<2002$. | [
"Solution:\n\nNote that whilst $a_{n}$ remains positive we have $a_{1}>a_{2}>a_{3}>\\ldots>a_{n}$. Hence if $a_{m}$ and $a_{m+n}$ are in this part of the sequence, then $a_{m+1}=a_{m}-1/a_{m}$, $a_{m+2}=a_{m+1}-1/a_{m+1}<a_{m+1}-1/a_{m}=a_{m}-2/a_{m}$. By a trivial induction $a_{m+n}<a_{m}-n/a_{m}$.\n\nIf we use on... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0067 | Determine si los enteros positivos se pueden partir en 12 subconjuntos disjuntos tales que, para cada $k = 1, 2, \ldots,$ los números $k, 2k, \ldots, 12k$ pertenecen a distintos subconjuntos. | [] | Argentina | XVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | Spanish | proof and answer | Yes. Partition by assigning each number to the class equal to its residue modulo thirteen after dividing out all factors of thirteen; then for every number its first twelve multiples occupy all twelve distinct classes. | |
01rk | Do there exist numbers $a, b \in \mathbb{R}$ and a surjective function $f: \mathbb{R} \to \mathbb{R}$ such that
$$
f(f(x)) = b x f(x) + a
$$
for all real $x$? | [
"(Solution by A. Asanau, Y. Dubovik, Y. Laurenau, V. Vityaz.) Suppose that such $a$, $b$, $f$ exist. Obviously $b \\neq 0$. Further, since $f$ is surjective, there exists a $\\lambda$ such that $f(\\lambda) = 0$. Set $x = \\lambda$ in\n$$\nf(f(x)) = b x f(x) + a, \\tag{1}\n$$\nthen $f(0) = f(f(\\lambda)) = b \\lamb... | Belarus | SELECTION and TRAINING SESSION | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | No; there are no real a, b and surjective f satisfying the equation. | |
031k | Problem:
Let $\mathbb{R}$ be the set of real numbers. Find all $a>0$ such that there exists a function $f: \mathbb{R} \rightarrow \mathbb{R}$ with the following two properties:
a) $f(x)=a x+1-a$ for any $x \in [2,3)$;
b) $f(f(x))=3-2 x$ for any $x \in \mathbb{R}$. | [
"Solution:\nSetting $h(x)=f(x+1)-1$, it is easy to see that the conditions for $f(x)$ are equivalent to $h(x)=a x$ for any $x \\in [1,2)$ and $h(h(x))=-2 x$ for any $x \\in \\mathbb{R}$. Then $h(-2 x)=h(h(h(x)))=-2 h(x)$; in particular, $h(0)=0$.\n\nIt follows by induction that $h\\left(4^{n} x\\right)=4^{n} h(x)$ ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | a = 2 * 4^k for some integer k | |
0bz0 | Determine all ways to color in green and red the numbers $1, 2, \ldots, 10$ such that the following conditions are fulfilled:
a) the number $5$ is colored red;
b) if the numbers $x$ and $y$ have different colors and $x + y \le 10$, then the number $x + y$ must be colored green;
c) if the numbers $x$ and $y$ have dif... | [
"If $1$ is colored red, then for any green $a > 1$, we have $1 \\cdot a = a$ must be colored red, a contradiction. Hence all numbers must be red in this case.\n\nConsider now that $1$ is colored green.\n\nIf $2$ is red, then $1 + 2 = 3$ must be green, so $2 + 3 = 5$ must also be green, contradiction. So $2$ must be... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Two colorings: (1) all numbers red; (2) red numbers: 5 and 10, and all other numbers green. | |
0gum | We say that an integer sequence $\{a_1, a_2, \dots\}$ is good if for some function $f: \mathbb{Z}^+ \to \mathbb{Z}^+$ and all positive integers $i, j, n$
$$
a_i \equiv a_j \pmod{n} \Leftrightarrow i \equiv j \pmod{f(n)}
$$
Find all good sequences. | [
"Given a good sequence, consider the following for all $n$ for fixed $i$, $j$, $k$:\n$$\nn \\mid a_{i+k} - a_i \\Leftrightarrow f(n) \\mid k \\Leftrightarrow n \\mid a_{j+k} - a_j.\n$$\nThus one has\n$$\n|a_{i+k} - a_i| = |a_{j+k} - a_j|\n$$\nfor all $i$, $j$, $k$. Now consider two cases separately:\n\na. If $a_1 =... | Turkey | Team Selection Test for IMO 2023 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | Exactly the following sequences are good:
1) Two-periodic sequences: a_i equals c_0 for even indices and c_1 for odd indices (c_0 and c_1 may be equal). A valid choice is f(n) = 1 if n divides c_0 − c_1, and f(n) = 2 otherwise.
2) Arithmetic progressions: a_i = k i + c for fixed integers k and c. A valid choice is f(n)... | |
030a | Problem:
Seja $ABCD$ um quadrilátero tal que $AC = BC + CD$. Se $\angle BCD = 120^{\circ}$, $\overline{CA}$ é bissetriz e $AB = x$, qual o valor de $BD$, em função de $x$? | [
"Solution:\n\nVamos analisar a figura:\n\n\nAplicando a Lei dos Cossenos no triângulo $ABC$, temos:\n$$\n\\begin{aligned}\n& x^{2} = b^{2} + (b + c)^{2} - 2 \\cdot b \\cdot (b + c) \\cdot \\cos 60^{\\circ} \\\\\n& x^{2} = b^{2} + b^{2} + 2 b c + c^{2} - 2 b^{2} \\frac{1}{2} - 2 b c \\frac{1... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | x | |
0g6o | 設 $\triangle DEF$ 為 $\triangle ABC$ 的內接正三角形, $D_1, D_2, D_3$ 分別為 $BD, DC$ 和 $BC$ 的中點。$E_1, E_2, E_3$ 分別為 $CE, EA$ 和 $CA$ 的中點, $F_1, F_2, F_3$ 分別為 $AF, FB$ 和 $AB$ 的中點。在 $\triangle ABC$ 外部取三點 $P, Q, R$, 使得 $\triangle PD_1D_2, \triangle QE_1E_2, \triangle RF_1F_2$ 皆為正三角形。設 $\triangle PDD_3, \triangle QEE_3, \triangle RFF_... | [
"同理 $M_2 = \\frac{1}{3}\\left(\\frac{C+A}{2} + E + Q\\right)$, $M_3 = \\frac{1}{3}\\left(\\frac{A+B}{2} + F + R\\right)$。\n\n現在，考慮 $\\omega = \\cos \\frac{2\\pi}{3} + i \\sin \\frac{2\\pi}{3}$，注意到 $\\omega^3 = 1$ 且 $1+\\omega+\\omega^2=0$。由題設，$\\triangle DEF$ 為正三角形，故 $D+\\omega E+\\omega^2 F = 0$。又 $\\triangle PD_1... | Taiwan | 二〇一二數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0888 | Problem:
Giulio scrive un polinomio $P_{1}(x)$ e un altro polinomio $P_{2}(x)$, prodotto di fattori di primo grado, avente grado strettamente maggiore del precedente. Eseguendo la divisione di $P_{2}(x)$ per $P_{1}(x)$, si ottiene resto $0$. Indicando con $Q(x)$ il quoziente di tale divisione, quale delle seguenti aff... | [
"Solution:\n\nLa risposta è $(\\mathbf{C})$. $P_{2}$ si può scrivere come prodotto di fattori di primo grado, quindi esso sarà del tipo\n$$\n\\left(x-\\alpha_{1}\\right)^{m_{1}} \\ldots\\left(x-\\alpha_{k}\\right)^{m_{k}}\n$$\ndove $\\alpha_{1}, \\ldots, \\alpha_{k}$ sono tutte e sole le soluzioni dell'equazione $P... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | C | |
02xh | Problem:
Cinco amigas são titulares de um time de vôlei. Suas camisas são numeradas nas costas com os 5 primeiros ímpares positivos. Ana é a número 1; Bia é a número 3; Cátia é a número 5; Dani é a número 7; e Esmeralda é a número 9. Durante os treinos as cinco amigas fazem filas para formar números com suas camisas, ... | [
"Solution:\n\na) Como são 5 algarismos para permutar, temos $5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 120$ números diferentes.\n\nb) Se Cátia não pode ser a primeira, temos $4 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 96$ números diferentes.\n\nc) Esmeralda e Bia lado a lado serão consideradas como uma só, podendo troca... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | a) 120; b) 96; c) 48; d) 600 | |
07qw | Solve the equations
$$
a + b + c = 0, \quad a^2 + b^2 + c^2 = 1, \quad a^3 + b^3 + c^3 = 4abc
$$
for $a$, $b$, and $c$. | [
"So, suppose $a$, $b$, $c$ satisfy the given equations, and eliminate $c$, say. Then, from the first, we deduce that\n$$\n\\begin{aligned}\na^3 + b^3 + c^3 &= a^3 + b^3 - (a + b)^3 \\\\\n&= a^3 + b^3 - (a^3 + 3a^2b + 3ab^2 + b^3) \\\\\n&= -3ab(a + b) \\\\\n&= 3abc.\n\\end{aligned}\n$$\nThis and the third equation f... | Ireland | Ireland_2017 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | All permutations of (1/sqrt(2), -1/sqrt(2), 0). |
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