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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0aqd | Problem:
For what value(s) of $k$ will the lines $2x + 7y = 14$ and $kx - y = k + 1$ intersect in the first quadrant? | [] | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | k < -3 or k > 1/6 | |
0l4a | Problem:
Let $P$ be a point inside isosceles trapezoid $A B C D$ with $A B \parallel C D$ such that
$$
\angle P A D = \angle P D A = 90^{\circ} - \angle B P C
$$
If $P A = 14$, $A B = 18$, and $C D = 28$, compute the area of $A B C D$. | [
"Solution:\n\n\n\nLet $Q$ be the circumcenter of $\\triangle B P C$. Thus, $\\angle Q B C = \\angle Q C B = 90^{\\circ} - \\angle B P C$, and so $\\triangle P A D$ and $\\triangle Q B C$ are congruent. This means that $P Q$, $A B$, and $C D$ share the common perpendicular bisector.\n\nWe no... | United States | HMMT November | [
"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"
] | null | proof and answer | 345√3 | |
0ixa | Problem:
Tanks has a pile of 5 blue cards and 5 red cards. Every morning, he takes a card and throws it down a well. What is the probability that the first card he throws down and the last card he throws down are the same color? | [
"Solution:\n\nOnce he has thrown the first card down the well, there are 9 remaining cards, and only 4 have the same color as the card that was thrown down. Therefore, the probability that the last card he throws down has the same color is $\\frac{4}{9}$."
] | United States | Harvard-MIT November Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 4/9 | |
0bvr | The real numbers $a$ and $b$ fulfill the conditions $a, b \ge 0$ and $2a\sqrt{b+1} + 2b\sqrt{a+1} = (\sqrt{a} + \sqrt{b})(a+b+1)$. Prove that $a+b=1$. | [
"Let $a, b \\ge 0$ and $2a\\sqrt{b+1} + 2b\\sqrt{a+1} = (\\sqrt{a} + \\sqrt{b})(a+b+1)$.\n\nLet us denote $x = \\sqrt{a}$ and $y = \\sqrt{b}$, so $a = x^2$, $b = y^2$, $a+1 = x^2+1$, $b+1 = y^2+1$.\n\nSubstitute into the equation:\n\n$2x^2\\sqrt{y^2+1} + 2y^2\\sqrt{x^2+1} = (x + y)(x^2 + y^2 + 1)$\n\nDivide both si... | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
076r | Consider a non-constant arithmetic progression $a_1, a_2, \dots, a_n, \dots$. Suppose there exist relatively prime positive integers $p > 1$ and $q > 1$ such that $a_1^2, a_{p+1}^2$ and $a_{q+1}^2$ are also the terms of the same arithmetic progression. Prove that the terms of the arithmetic progression are all integers... | [
"Let us take $a_1 = a$. We have\n$$\na^2 = a + kd, \\quad (a + pd)^2 = a + ld, \\quad (a + qd)^2 = a + md.\n$$\nThus we have\n$$\na + ld = (a + pd)^2 = a^2 + 2pad + p^2d^2 = a + kd + 2pad + p^2d^2.\n$$\nSince we have non-constant AP, we see that $d \\neq 0$. Hence we obtain $2pa + p^2d = l - k$. Similarly, we get $... | India | IND_National | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
06t9 | Let $n$ be a fixed positive integer. Find the maximum possible value of
$$
\sum_{1 \leqslant r<s \leqslant 2 n}(s-r-n) x_{r} x_{s}
$$
where $-1 \leqslant x_{i} \leqslant 1$ for all $i=1,2, \ldots, 2 n$. | [
"Let $Z$ be the expression to be maximized. Since this expression is linear in every variable $x_{i}$ and $-1 \\leqslant x_{i} \\leqslant 1$, the maximum of $Z$ will be achieved when $x_{i}=-1$ or $1$. Therefore, it suffices to consider only the case when $x_{i} \\in\\{-1,1\\}$ for all $i=1,2, \\ldots, 2 n$.\nFor $... | IMO | 56th International Mathematical Olympiad Shortlisted Problems | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | n(n-1) | |
0cqr | A student got $17$ marks during a week, each one is $2$, $3$, $4$, or $5$. The arithmetic mean of these marks is an integer. Prove that he could not get each mark at least thrice. | [
"Suppose the contrary. Then the student received each of the marks $2$, $3$, $4$, $5$ at least three times. Take three marks of each kind; the sum of these $12$ marks is $42$. Since each of the remaining five marks is at least $2$ and at most $5$, the sum of all $17$ marks is at least $42 + 5 \\cdot 2 = 52$ and at ... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
08dy | Problem:
Su un enorme foglio a quadretti, Marco considera un rettangolo lungo $2020$ quadretti e alto $2$. A questo punto vuole unire con $2020^{2}$ segmenti ciascuno dei $2020$ centri dei quadretti nella fila inferiore del rettangolo a ciascuno dei centri dei quadretti della fila superiore. Inoltre, vuole che se due ... | [
"Solution:\n\nIndichiamo i centri usando le coordinate dei quadretti, così che i centri della riga inferiore sono i punti $(0,0), \\ldots, (2019,0)$ e quelli della riga superiore sono $(0,1), \\ldots, (2019,1)$.\n\na.\nConsideriamo l'insieme di segmenti $S$ che collegano le seguenti coppie di punti:\n(i) $(0,0)-(k,... | Italy | Olimpiadi della Matematica | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 4039 | |
0ftx | Problem:
Beweise für alle $a_{1}, \ldots, a_{n}>0$ die folgende Ungleichung und bestimme alle Fälle, in denen das Gleichheitszeichen steht:
$$
\sum_{k=1}^{n} k a_{k} \leq \binom{n}{2} + \sum_{k=1}^{n} a_{k}^{k}
$$ | [
"Solution:\n\nFür $1 \\leq k \\leq n$ gilt nach AM-GM\n$$\na_{k}^{k} + (k-1) = a_{k}^{k} + \\underbrace{1 + \\ldots + 1}_{k-1} \\geq k \\cdot \\sqrt[k]{a_{k}^{k} \\cdot 1 \\cdots 1} = k a_{k}\n$$\nAddiert man diese $n$ Ungleichungen, dann folgt\n$$\n\\sum_{k=1}^{n} a_{k}^{k} + \\sum_{k=1}^{n} (k-1) \\geq \\sum_{k=1... | Switzerland | SMO Finalrunde | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Equality holds exactly when all entries from the second through the last are equal to one, while the first entry is any positive real number. | |
00qg | Let $ABCD$ be a trapezium inscribed in a circle $k$ with diameter $AB$. A circle with center $B$ and radius $BE$, where $E$ is the intersection point of the diagonals $AC$ and $BD$ meets $k$ at points $K$ and $L$. If the line, perpendicular to $BD$ at $E$, intersects $CD$ at $M$, prove that $KM \perp DL$. | [
"Since $AB \\parallel CD$, we have that $ABCD$ is isosceles trapezium. Let $O$ be the center of $k$ and $EM$ meets $AB$ at point $Q$. Then, from the right angled triangle $BEQ$, we have $BE^2 = BO \\cdot BQ$. Since $BE = BK$, we get $BK^2 = BO \\cdot BQ$ (1). Suppose that $KL$ meets $AB$ at $P$. Then, from the righ... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"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 | |
0g0u | Problem:
Sei $A B C$ ein rechtwinkliges Dreieck mit $\angle A C B=90^{\circ}$ und $M$ der Mittelpunkt von $A B$. Sei $G$ ein beliebiger Punkt auf der Strecke $M C$ und $P$ ein Punkt auf der Geraden $A G$, sodass $\angle C P A=\angle B A C$ gilt. Weiter sei $Q$ ein Punkt auf der Geraden $B G$, sodass $\angle B Q C=\ang... | [
"Solution:\n\nSei $H$ der Höhenfusspunkt von $C$. Wir wollen zeigen, dass der Umkreis vom Dreieck $A G Q$ die Strecke $A B$ in $H$ schneidet.\n\nSimple Winkeljagd liefert, dass $\\angle B A C=\\angle H B C$. Mit der Umkehrung des Tangentenwinkelsatzes folgt, dass die Gerade $B C$ tangential an den Umkreis $k$ vom D... | Switzerland | SMO - Finalrunde | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0ch7 | For any non-zero natural number $n$ consider the set
$$
A = \{n^2, n^2 + 1, n^2 + 2, \dots, (n+1)^2\}.
$$
Find the numbers $a, b, c \in A$, $a < b < c$, knowing that $b$ is the geometric mean of the numbers $a$ and $c$. | [
"From $b^2 = ac$ it follows that $\\frac{b}{a} = \\frac{c}{b} = \\frac{x}{y}$, where $x, y \\in \\mathbb{N}^*, (x, y) = 1$ and $x > y$. Therefore, $b = \\frac{a x}{y}$, $c = a \\left(\\frac{x}{y}\\right)^2$.\nSince $c = a \\left(\\frac{x}{y}\\right)^2 \\in \\mathbb{N}^*$ and $(x, y) = 1$, it follows that $\\frac{a}... | Romania | 74th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a = n^2, b = n(n + 1), c = (n + 1)^2 | |
06xi | Let $k \geqslant 2$ be an integer. Determine all sequences of positive integers $a_{1}, a_{2}, \ldots$ for which there exists a monic polynomial $P$ of degree $k$ with non-negative integer coefficients such that
$$
P\left(a_{n}\right)=a_{n+1} a_{n+2} \cdots a_{n+k}
$$
for every integer $n \geqslant 1$. | [
"Answer: The sequence $\\left(a_{n}\\right)$ must be an arithmetic progression consisting of positive integers with common difference $d \\geqslant 0$, and $P(x)=(x+d) \\cdots(x+k d)$.\n\nCommon remarks. The following arguments and observations are implicit in the solutions given below.\nSuppose the sequence $\\lef... | IMO | International Mathematical Olympiad Shortlist | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Exactly the arithmetic progressions of positive integers with nonnegative common difference. If the common difference is zero, the sequence is constant and the polynomial is the pure power of degree k. If the common difference is a positive integer d, then the polynomial equals the product of k consecutive linear terms... | |
0kz1 | Problem:
A four-digit integer in base 10 is friendly if its digits are four consecutive digits in any order. A four-digit integer is shy if there exist two adjacent digits in its representation that differ by 1. Compute the number of four-digit integers that are both friendly and shy. | [
"Solution:\nThere are 24 friendly numbers with digits $d_{1}=a, d_{2}=a+1, d_{3}=a+2, d_{4}=a+3$, for any $1 \\leq a \\leq 6$, and 18 with $a=0$. Out of these, only the numbers\n$$\n\\overline{d_{2} d_{4} d_{1} d_{3}}, \\quad \\text{ and } \\overline{d_{3} d_{1} d_{4} d_{2}}\n$$\nare not shy, none of which has a le... | United States | HMMT November 2024 | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | final answer only | 148 | |
0ed0 | Jure wrote positive integers 1 through 2015 on a whiteboard. Urška then inspected the written numbers from smallest to largest respectively and erased each number that was not divisible by 3. From the numbers still left on the whiteboard she then erased from smallest to largest each number that was not divisible by $3^... | [
"After Urška inspected all the written numbers for the first time only numbers divisible by 3 were left on the whiteboard. After her second inspection only numbers divisible by $3^2$ were left. After the third inspection only numbers divisible by $3^3$ were left, and so on. The largest power of the number 3 smaller... | Slovenia | National Math Olympiad 2015 – First Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1458 | |
0e1f | Let $x$ be a real number such that $\cos(2x) + \cos(3x) = 1$. Show that $2\sin(2x) + 2\sin(3x) = \sin(4x) + 2\sin(5x) + \sin(6x)$. | [
"The double-angle formula and the angle sum identity for sines and cosines imply\n$$\n\\begin{aligned}\n\\sin(4x) + 2\\sin(5x) + \\sin(6x) &= \\\\\n&= 2\\sin(2x)\\cos(2x) + 2(\\sin(2x)\\cos(3x) + \\sin(3x)\\cos(2x)) + 2\\sin(3x)\\cos(3x) \\\\\n&= 2\\cos(2x)(\\sin(2x) + \\sin(3x)) + 2\\cos(3x)(\\sin(2x) + \\sin(3x))... | Slovenia | National Math Olympiad | [
"Precalculus > Trigonometric functions"
] | null | proof only | null | |
0kbc | Problem:
For how many positive integers $n \leq 1000$ does the equation in real numbers
$$
x^{\lfloor x\rfloor}=n
$$
have a positive solution for $x$? (For a real number $x$, $\lfloor x\rfloor$ denotes the largest integer that is not greater than $x$.) | [
"Solution:\nIf $\\lfloor x\\rfloor=0$, then $x^{\\lfloor x\\rfloor}=1=1^{\\lfloor 1\\rfloor}$, so we can safely ignore this case, as it does not introduce new solutions.\nIf $\\lfloor x\\rfloor=k$ for some $k>0$, $x \\mapsto x^{\\lfloor x\\rfloor}=x^{k}$ is a continuous and increasing function on the interval $[k, ... | United States | HMMO | [
"Precalculus > Functions"
] | null | proof and answer | 412 | |
03vt | For a given positive integer $n$, find the greatest positive integer $k$, such that there exist three sets of $k$ distinct nonnegative integers, $A = \{x_1, x_2, \dots, x_k\}$, $B = \{y_1, y_2, \dots, y_k\}$ and $C = \{z_1, z_2, \dots, z_k\}$ with $x_j + y_j + z_j = n$ for any $1 \leq j \leq k$. | [
"By the given condition, we have\n$$\nkn \\geq \\sum_{i=1}^{k} (x_i + y_i + z_i) \\geq 3 \\sum_{i=0}^{k-1} i = \\frac{3k(k-1)}{2},\n$$\nand then $k \\leq \\lfloor \\frac{2n}{3} \\rfloor + 1$.\n\nThe following illustrates the case of $k = \\lfloor \\frac{2n}{3} \\rfloor + 1$:\n\nSet $m \\in \\mathbb{Z}^+$. \n\nWhen ... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | floor(2n/3) + 1 | |
0bgq | Problem:
Pe o hârtie este scrisă la început o listă de numere naturale distincte. O continuare a listei înseamnă alegerea a două numere dintre cele existente şi scrierea pe listă a celui mai mic multiplu comun al acestora, cu condiţia ca el să nu fie deja scris. Spunem că lista $s$-a închis dacă nu mai există nicio co... | [] | Romania | Olimpiada Naţională de Matematică Etapa Judeţeană | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1023 | |
067g | Let $c(O, R)$ be a circle and let $A, B$ be two points on the circle such that $R < AB < 2R$. The circle $c_1(A, r)$, $0 < r < R$, meets the circle $c(O, R)$ at points $C$ and $D$ ($C$ belongs to the small arch $\widehat{AB}$). From point $B$ we draw the tangents $BE$ and $BF$ to the circle $c_1(A, r)$, such that $E$ l... | [
"$AEBF$ is cyclic ($BE$ and $BF$ are tangents and: $A\\hat{E}B = A\\hat{F}B = 90^\\circ$). Let $c_2$ be its circumcircle. The segment $CD$ is the common chord of the circles ($c$) and ($c_1$) and the segment $AB$ is the common chord of the circles ($c$) and ($c_2$). Finally, $EF$ is the common chord of the circles ... | Greece | 31st Hellenic Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Polar tria... | English | proof only | null | |
0f6b | Problem:
$ABCDE$ is a convex pentagon. $A'$ is chosen so that $B$ is the midpoint of $AA'$, $B'$ is chosen so that $C$ is the midpoint of $BB'$ and so on. Given $A'$, $B'$, $C'$, $D'$, $E'$, how do we construct $ABCDE$ using ruler and compasses? | [] | Soviet Union | 19th ASU | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0d6l | Given three numbers $x$, $y$, $z$, and set $x_{1} = |x - y|$, $y_{1} = |y - z|$, $z_{1} = |z - x|$. From $x_{1}$, $y_{1}$, $z_{1}$, form in the same fashion the numbers $x_{2}$, $y_{2}$, $z_{2}$, and so on. It is known that $x_{n} = x$, $y_{n} = y$, $z_{n} = z$ for some $n$. Find all possible values of $(x, y, z)$. | [
"First, consider 3 sequences $\\left(x_{n}\\right)$, $\\left(y_{n}\\right)$, $\\left(z_{n}\\right)$, $n \\geq 1$ with\n$$\nx_{1} = |x - y|, \\quad y_{1} = |y - z|, \\quad z_{1} = |z - x| \\text{ and } \n\\left\\{\n\\begin{array}{l}\nx_{n+1} = \\left|x_{n} - y_{n}\\right| \\\\\ny_{n+1} = \\left|y_{n} - z_{n}\\right|... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | All triples that are permutations of (a, a, 0) with a ≥ 0 (including (0, 0, 0)). | |
00be | All numbers from $1$ to $16$ are written in a $4 \times 4$ board, one in each cell. We calculate all the differences between two numbers that occupy adjacent cells and call *value* of the board the greatest of these differences.
Which is the smallest value that a board can have?
Show a board... | [
"The smallest value that a board can have is $4$, and an example of a board having this value is the following:\n\n| 1 | 2 | 3 | 4 |\n|---|---|---|---|\n| 5 | 6 | 7 | 8 |\n| 9 | 10 | 11 | 12 |\n| 13 | 14 | 15 | 16 |\n\nLet us now show that the value of any board is at least $4$. Assume there is a board where the va... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 4 | |
01md | Given an $n \times n$ square table. Exactly one beetle sits in each cell of the table. At 12.00 all beetles creep to some neighboring cell (two cells are neighboring if they have the common side).
Find the greatest number of cells which can become empty (i.e. without beetles) if
a) $n = 8$;
b) $n = 9$.
(Problem Commi... | [
"a) Answer: $44$.\n(Solution of A. Zhuk, O. Volod'ko.) Let's mark $20$ cells of the table as it is shown in Fig. 1. It is obvious that all beetles from the marked cells after their moves must occupy different cells. That is, at least $20$ cells will remain occupied, so at most $64 - 20 = 44$ cells can become empty.... | Belarus | Selection and Training Session | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) 44; b) 56 | |
02tj | Problem:
Dizemos que um número inteiro positivo $n$ é espertinho se existirem números inteiros positivos $a, b, c$ e $d$, não necessariamente distintos, tais que:
$$
n=\frac{a^{2}-b^{2}}{c^{2}+d^{2}}
$$
Por exemplo, 12 é espertinho, pois:
$$
12=\frac{16^{2}-4^{2}}{4^{2}+2^{2}}
$$
Mostre que todos os números inteiros p... | [
"Solution:\n\nPodemos reescrever a equação da seguinte forma:\n$$\n\\begin{aligned}\nn & =\\frac{a^{2}-b^{2}}{c^{2}+d^{2}} \\\\\nn\\left(c^{2}+d^{2}\\right) & =(a+b)(a-b)\n\\end{aligned}\n$$\nSe conseguirmos inteiros positivos tais que $a+b=n$ e $a-b=c^{2}+d^{2}$, teremos uma solução para a equação. Se $n$ é par, b... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Other"
] | null | proof only | null | |
057d | A circle $c$ with center $A$ passes through the vertices $B$ and $E$ of a regular pentagon $ABCDE$. The line $BC$ intersects the circle $c$ the second time at point $F$. Point $G$ on the circle $c$ is chosen in such a way that $FB = FG$ and $B \neq G$. Prove that the lines $AB$, $EF$ and $DG$ meet in a common point. | [
"The internal angles of a regular pentagon have size $108^\\circ$. Thus $\\angle ABC = 108^\\circ$, implying $\\angle ABF = 72^\\circ$ (Fig. 7). As $AB = AF$, we have $\\angle AFB = 72^\\circ$ and $\\angle BAF = 36^\\circ$. Since $FG = FB$, $AG = AB$ and $AF = AF$, the triangles $AFB$ and $AFG$ are equal, implying ... | Estonia | Open Contests | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem"
] | null | proof only | null | |
03yr | Let $A = \{a_1, a_2, a_3, a_4\}$. Suppose the set of sums of all the elements in every ternary subset of $A$ is $B = \{-1, 3, 5, 8\}$. Then $A = \underline{\hspace{2cm}}$. | [
"Obviously, every element of $A$ appears three times in all the ternary subsets. Then we have\n$$3(a_1 + a_2 + a_3 + a_4) = (-1) + 3 + 5 + 8 = 15,$$\nor $a_1 + a_2 + a_3 + a_4 = 5$. Therefore, the four elements of $A$ are $5 - (-1) = 6$, $5 - 3 = 2$, $5 - 5 = 0$, $5 - 8 = -3$, respectively.\nThe answer is $A = \\{-... | China | China Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | {-3, 0, 2, 6} | |
01zg | Let $ABC$ be a triangle such that $\angle CAB = 30^\circ$ and $\angle ACB = 60^\circ$. An arbitrary point $D$ is chosen on the extension of the ray $AB$ beyond point $B$. The point $E$ lies on the extension of the ray $CB$ beyond the point $B$ such that $\angle BDE = 60^\circ$. The lines $AC$ and $DE$ intersect at $F$.... | [
"Let $G$ be the point on the ray $BC$ such that $BG = 3BC$. We will show that $G$ lies on the circumcircle of the triangle $AEF$. Angles $\\angle ECF$ and $\\angle ACG$ are equal $120^\\circ$ as they are adjacent to $\\angle ACB = 60^\\circ$. Since $\\angle BED = 30^\\circ$ and $\\angle FCE = 120^\\circ$, $\\angle ... | Belarus | Belarus2022 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
09nq | A positive integer $A$ is called *good* if the exponent of each prime divisor of $A$ is at least two. Find the largest positive integer which cannot be expressed as a sum of different good integers (the number of summands can be one). | [] | Mongolia | MMO2025 Round 3 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | English | proof and answer | 23 | |
0d0a | A square of side $1$ is decomposed into $9$ equal squares of sides $\frac{1}{3}$ and the one in the center is painted black. The remaining eight squares are analogously divided into nine squares each, and the squares in the centers are painted in black. Prove that after $1000$ steps the total area of the black region e... | [
"The first step gives rise to one black square of area $\\left(\\frac{1}{3}\\right)^2 = \\frac{1}{9}$. After the second step we obtain eight more squares of side $\\frac{1}{9}$, the black region increasing thus by $\\frac{8}{9^2}$. In the same manner, the third step increases the black area by $8^2 = 64$ black squa... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
03nv | Problem:
Let $ABCDE$ be a convex pentagon such that the five vertices lie on a circle and the five sides are tangent to another circle inside the pentagon. There are $\binom{5}{3} = 10$ triangles which can be formed by choosing 3 of the 5 vertices. For each of these 10 triangles, mark its incenter. Prove that these 10... | [
"Solution:\n\nLet $I$ be the incenter of pentagon $ABCDE$. Let $I_A$ denote the incenter of triangle $EAB$ and $I_a$ the incenter of $DAC$. Define $I_B, I_b, I_C, I_c, I_D, I_d, I_E, I_e$ similarly.\n\nWe will first show that $I_A I_B I_C I_D I_E$ are concyclic. Let $\\omega_A$ be the circle with center at the midp... | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Quadrilaterals > Cycli... | null | proof only | null | |
0bph | Problem:
Határozd meg azokat a $z$ komplex számokat, amelyekre:
$$
|z|+|z-5 i|=|z-2 i|+|z-3 i|
$$
Problem:
Să se determine numerele complexe $z$ pentru care are loc relaţia
$$
|z|+|z-5 i|=|z-2 i|+|z-3 i|
$$ | [
"Solution:\nAvem\n$$\n|z-2 i|=\\left|\\frac{2}{5}(z-5 i)+\\frac{3}{5} z\\right| \\leq \\frac{2}{5}|z-5 i|+\\frac{3}{5}|z|\n$$\nAnalog\n$$\n|z-3 i|=\\left|\\frac{3}{5}(z-5 i)+\\frac{2}{5} z\\right| \\leq \\frac{3}{5}|z-5 i|+\\frac{2}{5}|z|\n$$\nde unde\n$$\n|z|+|z-5 i| \\geq|z-2 i|+|z-3 i|\n$$\nEgalitatea are loc at... | Romania | Olimpiada Naţională de Matematică, Etapa Judeţeană şi a Municipiului Bucureşti | [
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | z = a i with a ≤ 0 or a ≥ 5 | |
0ady | Шаховската табла е поделена на 64 единечни квадрати. Најди го бројот на сите квадрати на шаховската табла, кои се формирани од единечните квадрати. | [
"Бројот на квадратите $1 \\times 1$ на шаховската табла е $8^2 = 64$. Квадрати $2 \\times 2$ ги има $7^2 = 49$, а $3 \\times 3$ има $6^2 = 36$, $4 \\times 4$ има $5^2 = 25$, $5 \\times 5$ има $4^2 = 16$, $6 \\times 6$ има $3^2 = 9$, $7 \\times 7$ има $2^2 = 4$ и $8 \\times 8$ има $1^2 = 1$. Според тоа на таблата им... | North Macedonia | Републички натпревар по математика за основно образование | [
"Discrete Mathematics > Combinatorics"
] | Macedonian, English | final answer only | 204 | |
0bbq | The triangle $ABC$ and the points $M \in (BC)$, $N \in (AC)$, $P \in (AB)$ fulfill the conditions $\angle BMP \equiv \angle CNM \equiv \angle APN$ and $BM = CN = AP$.
Prove that the triangle $ABC$ is equilateral. | [
"We start noticing that $m(\\overline{PMN}) = 180^\\circ - \\overline{BMP} - \\overline{NMC} = 180^\\circ - \\overline{CNM} - \\overline{NMC} = \\overline{NCM}$ and, in the same way, $\\overline{MNP} = \\overline{NAP}$, so\n$$\n\\Delta ABC \\sim \\Delta NPM. \\qquad (1)\n$$\n\n\n\nSuppose n... | Romania | 2011 CLOCK-TOWER SCHOOL JUNIORS COMPETITION | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof only | null | |
0e3z | Problem:
Določi vsa cela števila $x$, za katera je število $9x^{2}-40x+39$ potenca praštevila. (Naravno število $m$ je potenca praštevila, če je $m=p^{a}$ za neko praštevilo $p$ in nenegativno celo število $a$.) | [
"Solution:\n\nNaj bo $9x^{2}-40x+39=p^{n}$ za neko praštevilo $p$ in nenegativno celo število $n$. Iz\n$$\np^{n}=9x^{2}-40x+39=(9x-13)(x-3)\n$$\nsledi $9x-13=p^{k}$ in $x-3=p^{l}$ ali $9x-13=-p^{k}$ in $x-3=-p^{l}$ za neki števili $k$ in $l$, kjer je $0 \\leq l<k$ in $n=k+l$.\n\nRešimo najprej sistem enačb $9x-13=p... | Slovenia | 55. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | -4, 1, 4, 5 | |
0690 | Determine the smallest integer $n$, for which there exists integers $x_1, x_2, \dots, x_n$ and positive integers $a_1, a_2, \dots, a_n$ so that:
$$
x_1 + x_2 + \dots + x_n = 0, \quad a_1x_1 + a_2x_2 + \dots + a_nx_n > 0, \quad a_1^2x_1 + a_2^2x_2 + \dots + a_n^2x_n < 0.
$$ | [] | Greece | 20th Mediterranean Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 4 | |
0ds3 | The incircle of $\triangle ABC$ touches the sides $BC$, $CA$, $AB$, at $D$, $E$, $F$ respectively. A circle through $A$ and $B$ encloses $\triangle ABC$ and intersects the line $DE$ at points $P$ and $Q$. Prove that the midpoint of $AB$ lies on the circuncircle of $\triangle PQF$. | [
"Let $M$ be the midpoint of $AB$. If $DE \\parallel AB$, then $\\triangle ABC$ is isosceles with $CA = CB$, and $F$ coincides with $M$.\n\n\n\nConsider the case where $DE \\parallel AB$. Let the lines $DE$ and $AB$ intersect at $X$. By Menelaus theorem, $\\frac{AX}{XB} = \\frac{BD}{CD} = \\... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous... | null | proof only | null | |
0dev | Let a finite set of integers be given, such that each of its elements can be written as a sum of some two elements (not necessarily distinct) from the same set. For such a set, we say that it is of 'safety of order $n$' if it does not contain a subset of $n$ or less elements whose sum is $0$. Prove that there exist set... | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | English | proof only | null | |
0eup | Around a table $2n$ people are seated and $m$ cookies are distributed to them. These people can pass cookies under the following rules:
- One can only pass cookies to his/her neighbors.
- One can pass a cookie only if he or she eats one cookie.
Let $A$ be one of these people. Find the minimum $m$ such that no matter ... | [
"We will prove that the minimum number $m$ of cookies is $2^n$. Let us write $A_{-n+1}, \\dots, A_0(= A), A_1, A_2, \\dots, A_{n-1}, A_n$ to denote the $2n$ people in the counterclockwise order.\n\nFirst let us show that if $m < 2^n$, then there is a way to distribute $m$ cookies so that $A$ cannot get a cookie. Le... | South Korea | 23rd Korean Mathematical Olympiad Final Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 2^n | |
08yl | A $4 \times 4$ grid made up of $16$ $1 \times 1$ squares is given. Suppose we want to color each of the squares by using one of the four colors, red, blue, yellow or green. How many ways of painting all of the boxes are there if the following requirement must be met:
For any row and any column one of the following 3 co... | [
"When we say color in this solution we mean red, blue, yellow and green.\nAn $n$-tuple $(c_1, c_2, \\dots, c_n)$ of colors (in which the same color can appear any number of times) is called consistent if every color appearing appears an even number of times, or every color appearing appears an odd number of times. ... | Japan | 2019 Japan Mathematical Olympiad First Stage | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 262144 | |
05lm | Problem:
Soit $P$ un polynôme à coefficients entiers, de degré $n$, avec $n \leq 10$. On suppose que $|P(10)-P(0)|<1000$ et que, pour tout $k \in\{1, \ldots, 10\}$, il existe un entier $m$ tel que $P(m)=k$. Montrer que, pour tout entier $k$ il existe un entier $m$ tel que $P(m)=k$. | [
"Solution:\n\nPour $i \\in\\{1, \\ldots, 10\\}$, on désigne par $c_{i}$ un entier tel que $P\\left(c_{i}\\right)=i$. Pour $i \\in\\{1, \\ldots, 9\\}$, puisque $P$ est à coefficients entiers, l'entier $c_{i+1}-c_{i}$ divise $P\\left(c_{i+1}\\right)-P\\left(c_{i}\\right)=i+1-i=1$, d'où $c_{i+1}-c_{i}= \\pm 1$. De plu... | France | Pays-Bas | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
0k2d | Problem:
Let $S$ be a subset with four elements chosen from $\{1,2, \ldots, 10\}$. Michael notes that there is a way to label the vertices of a square with elements from $S$ such that no two vertices have the same label, and the labels adjacent to any side of the square differ by at least $4$. How many possibilities ar... | [
"Solution:\nLet the four numbers be $a, b, c, d$ around the square. Assume without loss of generality that $a$ is the largest number, so that $a > b$ and $a > d$. Note that $c$ cannot be simultaneously smaller than one of $b, d$ and larger than the other because, e.g. if $b > c > d$, then $a > b > c > d$ and $a \\g... | United States | HMMT November 2018 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 36 | |
0exs | Problem:
Two players alternately choose the sign for one of the numbers $1, 2, \ldots, 20$. Once a sign has been chosen it cannot be changed. The first player tries to minimize the final absolute value of the total and the second player to maximize it. What is the outcome (assuming both players play perfectly)?
Examp... | [
"Solution:\n\nAnswer: $30$.\n\nThe second player can play the following strategy:\n\n(1) If the first player plays $2n - 1$ for $1 \\leq n \\leq 9$, then he replies $2n$ with the opposite sign.\n\n(2) If the first player plays $2n$ for $1 \\leq n \\leq 9$ then he replies $2n - 1$ with the opposite sign.\n\n(3) If t... | Soviet Union | 6th ASU | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 30 | |
0ce5 | Let $ABCDEF$ be a convex hexagon. The diagonals $AC$ and $BD$ cross at $P$, the diagonals $AE$ and $DF$ cross at $Q$, and the line $PQ$ crosses the sides $BC$ and $EF$ at $X$ and $Y$, respectively. Prove that the length of the segment $XY$ does not exceed the sum of the lengths of one of the diagonals through $P$ and o... | [
"The problem at hand is a special case of the following situation: Let $U$ be a point on the diagonal $AD$ and let $V = UP \\cap BC$ and $W = UQ \\cap EF$. Then $UV \\le \\max(AC, BD)$ and, similarly, $UW \\le \\max(AE, DF)$, so $VW \\le UV + UW \\le \\max(AC, BD) + \\max(AE, DF)$. In particular, if $U = AD \\cap P... | Romania | THE 73rd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS - SECOND SELECTION TEST | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometr... | null | proof only | null | |
0712 | Problem:
$ABC$ is a triangle. The incircle has center $I$ and touches the sides $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. The rays $BI$ and $CI$ meet the line $EF$ at $P$ and $Q$ respectively. Show that if $DPQ$ is isosceles, then $ABC$ is isosceles. | [
"Solution:\n\n\n\n$AF = AE$, so $\\angle AFE = 90^{\\circ} - A / 2$. Hence $\\angle BFP = 90^{\\circ} + A / 2$. But $\\angle FBP = B / 2$, so $\\angle FPB = C / 2$. But $BFP$ and $BDP$ are congruent ($BF = BD$, $BP$ common, $\\angle FBP = \\angle FDP$), so $\\angle DPB = C / 2$ and $\\angle... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0eo4 | Given that
$$
\frac{a-b}{c-d} = 2 \quad \text{and} \quad \frac{a-c}{b-d} = 3
$$
for certain real numbers $a$, $b$, $c$, $d$, determine the value of
$$
\frac{a-d}{b-c}
$$ | [
"Set $x = c - d$ and $y = b - d$. We have\n$$\na - d = (c - d) + (a - c) = x + 3y\n$$\nand\n$$\na - d = (b - d) + (a - b) = y + 2x,\n$$\nhence $x + 3y = y + 2x$, which implies $x = 2y$. Now we get $a - d = 5y$ and\n$$\nb - c = (b - d) - (c - d) = y - x = -y,\n$$\nso\n$$\n\\frac{a-d}{b-c} = -5.",
"We are given tha... | South Africa | The South African Mathematical Olympiad Third Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | -5 | |
03qy | Determine the maximum value of constant $\lambda$ such that
$$
u + v + w \ge \lambda,$$
where $u, v$ and $w$ are positive real numbers with $u\sqrt{vw} + v\sqrt{wu} + w\sqrt{uv} \ge 1$. | [
"**Solution I**\n\nBy the AM-GM inequality and the given condition, we have\n$$\nu \\cdot \\frac{v+w}{2} + v \\cdot \\frac{w+u}{2} + w \\cdot \\frac{u+v}{2} \\ge u\\sqrt{vw} + v\\sqrt{wu} + w\\sqrt{uv} \\ge 1,$$\nor\n$$uv + vw + wu \\ge 1.$$ \nSince\n$$(u-v)^2 + (v-w)^2 + (w-u)^2 \\ge 0, \\quad u^2 + v^2 + w^2 \\ge... | China | China Girls' Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | sqrt(3) | |
0hie | You are given a set of $m$ positive integers such that they all give distinct remainders modulo some positive integer $n$. Prove that for any positive integer $k \leq m$ this set can be partitioned into $k$ nonempty subsets such that the sums of the numbers in these subsets are also distinct modulo $n$. | [
"Let these numbers be $a_1, a_2, ..., a_m$. It is enough to show that you can choose some two of these numbers $a_i, a_j$ (with $i < j$) so that all the numbers $a_1, a_2, ..., a_{i-1}, a_{i+1}, ..., a_{j-1}, a_{j+1}, ..., a_m, a_i + a_j$ give distinct remainders when divided by $n$, then we can combine the numbers... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
040l | Suppose two regular triangular pyramids $P$-$ABC$ and $Q$-$ABC$ sharing the same base are inscribed in the same sphere. If the angle between the side-face and the base of $P$-$ABC$ is $45^\circ$, then the tangent value of the angle between the side-face and the base of $\triangle ABC$ is ______. | [
"As seen in Fig. 5.1, connecting $PQ$, then $PQ$ is perpendicular to plane $ABC$ with the foot point $H$ being the center of $\\triangle ABC$. The center of the sphere $O$ is also on $PQ$. Connect and extend $CH$ to let it intersect with $AB$ at point $M$. $M$ is then the midpoint of $AB$, and $CM \\perp AB$. It is... | China | China Mathematical Competition | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing... | English | proof and answer | 4 | |
0hxk | Problem:
Compute $1 \cdot 2 + 2 \cdot 3 + \cdots + (n-1)n$. | [
"Solution:\nLet $S = 1 \\cdot 2 + 2 \\cdot 3 + \\cdots + (n-1)n$. We know $\\sum_{i=1}^{n} i = \\frac{n(n+1)}{2}$ and $\\sum_{i=1}^{n} i^{2} = \\frac{n(n+1)(2n+1)}{6}$. So\n$$\nS = 1(1+1) + 2(2+1) + \\cdots + (n-1)n = \\left(1^{2} + 2^{2} + \\cdots + (n-1)^{2}\\right) + (1 + 2 + \\cdots + (n-1))\n$$\n$$\n= \\frac{(... | United States | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | (n-1)n(n+1)/3 | |
01mr | Points $A_1$ and $B_1$ are marked on the sides $AC$ and $BC$ of a triangle $ABC$ so that $AA_1 + BB_1 = AB$. Let $I$ be incenter of the triangle $ABC$, and $O$ be circumcenter of the triangle $A_1B_1C$.
Prove that the lines $OI$ and $A_1B_1$ are perpendicular.
(I. Voronovich) | [] | Belarus | 61st Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
04xl | Polynomial $P(x)$ with integer coefficients satisfies the following condition: for every polynomials $F(x)$, $G(x)$, $Q(x)$ with integer coefficients, if
$$
P(Q(x)) = F(x) \cdot G(x)
$$
then either $F(x)$ or $G(x)$ is a constant polynomial. Prove that $P(x)$ has to be a constant polynomial. | [
"For the sake of contradiction suppose that $P(x)$ is not constant and consider the case when $P(x)$ is a linear polynomial. It means that $P(x) = a x + b$ for some $a, b \\in \\mathbb{Z}$, where $a \\neq 0$. Let $Q(x) = a x^2 + (b+1)x$. Then\n$$\nP(Q(x)) = a(a x^2 + (b+1)x) + b = a^2 x^2 + a(b+1)x + b = (a x + b)(... | Czech-Polish-Slovak Mathematical Match | 11-th Czech-Slovak-Polish Match, 2011 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | English | proof only | null | |
0a86 | Problem:
Let $n$ be a positive integer. Count the number of numbers $k \in \{0,1,2, \ldots, n\}$ such that $\binom{n}{k}$ is odd. Show that this number is a power of two, i.e. of the form $2^{p}$ for some nonnegative integer $p$. | [
"Solution:\nThe number of odd binomial coefficients $\\binom{n}{k}$ equals the number of ones on the $n$th line of the Pascal Triangle $\\bmod\\ 2$:\n\n\n\n(We count the lines so that the uppermost line is line $0$.) We notice that line $1$ has two copies of line $0$, lines $2$ and $3$ cont... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 12 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2^{number of ones in the binary representation of n} | |
091o | Problem:
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the equality
$$
y^{2} f(x)+x^{2} f(y)+x y=x y f(x+y)+x^{2}+y^{2}
$$
holds for all $x, y \in \mathbb{R}$, where $\mathbb{R}$ is the set of real numbers. | [
"Solution:\nSubstituting $y=0$ we find that $x^{2} f(0)=x^{2}$ holds for all real numbers $x$ which implies $f(0)=1$.\nLet us introduce a new function $g: \\mathbb{R} \\rightarrow \\mathbb{R}$ given by $g(x)=f(x)-1$. Equation from the problem becomes\n$$\ny^{2} g(x)+x^{2} g(y)=x y g(x+y)\n$$\nwhile $g(0)=0$.\nDenot... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = c x + 1 for all real c | |
0c9u | Problem:
Să se determine inelele nenule finite, cu unitate, în care suma tuturor elementelor este un element inversabil. | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ | [
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Abstract Algebra > Group Theory"
] | null | proof and answer | The only such ring is the field with two elements. | |
0kpp | Problem:
How many ways are there to arrange the numbers $1,2,3,4,5,6$ on the vertices of a regular hexagon such that exactly 3 of the numbers are larger than both of their neighbors? Rotations and reflections are considered the same. | [
"Solution:\n\nLabel the vertices of the hexagon $a b c d e f$.\nThe numbers that are larger than both of their neighbors can't be adjacent, so assume (by rotation) that these numbers take up slots $a c e$. We also have that $6$ and $5$ cannot be smaller than both of their neighbors, so assume (by rotation and refle... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 8 | |
0gkz | Let $ABCD$ be a convex quadrilateral. Let $M$ and $N$ be the midpoints of $AB$ and $CD$, respectively. Let $P$ be the point on the line $CD$ such that $MP$ is perpendicular to $CD$. Let $Q$ be the point on the line $AB$ such that $NQ$ is perpendicular to $AB$. Prove that $AD$ is parallel to $BC$ if and only if $\frac{A... | [
"Denote the area of $A$ by $[A]$. Observe that\n\n$$\n[ABCD] = [ADM] + [DMC] + [MCB] = \\frac{1}{2}[ABD] + \\frac{1}{2}CD \\cdot MP + \\frac{1}{2}[ABC]\n$$\n$$\n[ABCD] = [ADN] + [ANB] + [NBC] = \\frac{1}{2}[ADC] + \\frac{1}{2}AB \\cdot NQ + \\frac{1}{2}[BCD].\n$$\n\nThus,\n$$\n\\begin{aligned}\nCD \\cdot MP - AB \\... | Thailand | The 10th Thailand Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof only | null | |
027x | Problem:
Na cidade de Oropis existe um lago em forma de triângulo com cada um dos três lados sendo parte do perímetro de um terreno em forma de quadrado com áreas $370~\mathrm{m}^2$, $116~\mathrm{m}^2$ e $74~\mathrm{m}^2$, como na primeira figura a seguir. O prefeito de Oropis, Arnaldo, deseja calcular a área do lago,... | [
"Solution:\n\n(a) Veja que os lados do lago ao quadrado resultam em $74$, $116$ e $370$. As condições que devem ser atendidas são as equações oriundas de três aplicações do Teorema de Pitágoras:\n$$\n\\begin{aligned}\na^2 + c^2 & = 74 \\\\\nb^2 + d^2 & = 116 \\\\\n(a+b)^2 + (c+d)^2 & = 370\n\\end{aligned}\n$$\nPode... | Brazil | null | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | a=5, b=4, c=7, d=10; area of the lake = 11 | |
0dym | Let $D$ be the midpoint of the segment $AB$ and denote the centre of gravity of the triangle $ABC$ by $T$. Find the lengths of the sides given that $|AD| = 3$, $|DT| = 5$ and $|TA| = 4$. | [
"Since $D$ is the midpoint of $AB$ and $|AD| = 3$, we have $|AB| = 6$. Let $E$ be the midpoint of $BC$ and let $F$ be the midpoint of $AC$. The sides of the triangle $ADT$ satisfy Pythagoras's theorem, so $ADT$ is a right triangle. The ratio in which the centre of gravity divides the median is $2:1$ and since $|AT|... | Slovenia | Slovenija 2008 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | AB = 6, BC = 12√2, AC = 6√5 | |
08h5 | Problem:
Let $ABC$ be an acute scalene triangle with circumcenter $O$. Let $D$ be the foot of the altitude from $A$ to the side $BC$. The lines $BC$ and $AO$ intersect at $E$. Let $s$ be the line through $E$ perpendicular to $AO$. The line $s$ intersects $AB$ and $AC$ at $K$ and $L$, respectively. Denote by $\omega$ t... | [
"Solution:\n\n\nLet us denote angles of triangle $ABC$ with $\\alpha, \\beta, \\gamma$ in a standard way. By basic angle chasing we have\n$$\n\\angle BAD = 90^\\circ - \\beta = \\angle OAC \\text{ and } \\angle CAD = \\angle BAO = 90^\\circ - \\gamma\n$$\nUsing the fact that lines $AE$ and ... | JBMO | null | [
"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 > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Co... | null | proof only | null | |
05o0 | Problem:
Soit $I$ le centre du cercle inscrit à un triangle $ABC$. Soit $D$ le point diamétralement opposé à $A$ sur le cercle circonscrit. On suppose que le point $E$ de la demi-droite $[BA)$ et le point $F$ de la demi-droite $[CA)$ satisfont la condition
$$
BE = CF = \frac{AB + BC + CA}{2}.
$$
Montrer que $(EF) \per... | [
"Solution:\n\n\n\nNotons $a, b, c$ les longueurs des côtés, $r$ le rayon du cercle inscrit et $p = (a + b + c)/2$. Comme $[AD]$ est un diamètre, les angles $\\widehat{ABD}$ et $\\widehat{ACD}$ sont droits donc\n$$\n\\begin{aligned}\nDE^2 - DF^2 &= (DB^2 + BE^2) - (DC^2 + CF^2) = DB^2 - DC^2... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasin... | null | proof only | null | |
0iou | Problem:
Call a year ultra-even if all of its digits are even. Thus $2000$, $2002$, $2004$, $2006$, and $2008$ are all ultra-even years. They are all $2$ years apart, which is the shortest possible gap. $2009$ is not an ultra-even year because of the $9$, and $2010$ is not an ultra-even year because of the $1$.
a. In... | [
"Solution:\n\na. The longest possible gap has length $1112$. There are four valid examples of two ultra-even years that far apart with no ultra-even years between them: $888$ to $2000$, $2888$ to $4000$, $4888$ to $6000$, or $6888$ to $8000$. ($8888$ to $10000$ is not acceptable as $10000$ is not ultra-even).\n\nTo... | United States | Bay Area Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) 1112; for example, 888 to 2000. b) 12; for example, 8 to 20. | |
0cb9 | Solve in real numbers the equation $2^{x-1} + 2^{\frac{1}{\sqrt{x}}} = 3$. | [
"We notice that $x > 0$. The given equation can be written as $2^x + 2 \\cdot 2^{\\frac{1}{\\sqrt{x}}} = 6$.\n\nFrom the AM-GM inequality we have:\n$$\n2^x + 2^{\\frac{1}{\\sqrt{x}}} + 2^{\\frac{1}{\\sqrt{x}}} \\ge 3 \\cdot \\sqrt[3]{2^x \\cdot 2^{\\frac{1}{\\sqrt{x}}} \\cdot 2^{\\frac{1}{\\sqrt{x}}}} = 3 \\cdot 2^... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 1 | |
0dpw | Integers $x$, $y$, $z$, $t$ satisfy $x^2 + y^2 = z^2 + t^2$ and $xy = 2zt$. Prove that $xyzt = 0$. | [
"Assume that $xyzt \\neq 0$.\nIf three of the numbers $x$, $y$, $z$, $t$ have a common divisor $d$, the remaining number is also a multiple of $d$. The division of all the four numbers by $d$ preserves the validity of our equations. Since $x$, $y$, $z$, $t$ are not zero, we can divide them by their greatest common ... | Silk Road Mathematics Competition | SILK ROAD MATHEMATICS COMPETITION XX | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common ... | English | proof only | null | |
07ln | Suppose $a$, $b$, $c$ are positive numbers such that
$$
(a + b + c) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) < 5 + 3\sqrt{2} .
$$
Prove that $a$, $b$, $c$ are the side lengths of an acute-angled triangle. | [
"Suppose this is false. Then, without loss of generality, we can assume that $a^2 \\ge b^2 + c^2$. Since\n$$\n(a+b+c) \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) = 3 + a \\left( \\frac{1}{b} + \\frac{1}{c} \\right) + \\frac{1}{a} (b+c) + \\frac{b}{c} + \\frac{c}{b},\n$$\nand\n$$\n\\frac{b}{c} + \\fr... | Ireland | Irska | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0ggo | 一正整數是回文數若且唯若它由左而右讀或由右而左讀是一樣的, 例如 $1221$ 或 $74847$ 都是回文數。一 $n$ 位數正整數 $k$ 以十進位寫成 $\overline{a_{n-1}a_{n-2}\cdots a_0}$。證明若 $k$ 是回文數, 則 $k^2$ 是回文數若且唯若 $a_0^2 + a_1^2 + \cdots + a_{n-1}^2 < 10$。 | [
"($\\Rightarrow$) **Step 1.** 用窮舉法討論 $k^2$ 的首位與末位。首先由於 $10^{n-1} \\le k < 10^n$ 可知 $10^{2n-2} \\le k^2 < 10^{2n}$，因此知 $k^2$ 為 $2n-1$ 位數或 $2n$ 位數。\n\n若 $k^2$ 為 $2n$ 位數, 則窮舉得以下表格:\n\n| $a_{n-1} = a_0$ 的值 | $k^2$ 的可能首位數 | $k^2$ 的末位數 |\n|---|---|---|\n| 3 | 1 | 9 |\n| 4 | 1, 2 | 6 |\n| 5 | 2, 3 | 5 |\n| 6 | 3, 4 | 6 |\... | Taiwan | 2022 數學奧林匹亞競賽第二階段選訓營, 獨立研究(一) | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Modular Arithmetic",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | Chinese; English | proof only | null | |
00gr | Consider the function $f: \mathbb{N}_0 \rightarrow \mathbb{N}_0$, where $\mathbb{N}_0$ is the set of all non-negative integers, defined by the following conditions:
(i) $f(0)=0$,
(ii) $f(2 n)=2 f(n)$ and
(iii) $f(2 n+1)=n+2 f(n)$ for all $n \geq 0$.
a. Determine the three sets $L:=\{n \mid f(n)<f(n+1)\}$, $E:=\{n \mid... | [
"(a) Let\n$$\nL_1:=\\{2 k: k>0\\}, \\quad E_1:=\\{0\\} \\cup\\{4 k+1: k \\geq 0\\}, \\quad \\text{ and } \\quad G_1:=\\{4 k+3: k \\geq 0\\} .\n$$\nWe will show that $L_1=L$, $E_1=E$, and $G_1=G$. It suffices to verify that $L_1 \\subseteq L$, $E_1 \\subseteq E$, and $G_1 \\subseteq G$ because $L_1$, $E_1$, and $G_1... | Asia Pacific Mathematics Olympiad (APMO) | XX Asian Pacific Mathematics Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | L = {2k : k > 0}, E = {0} ∪ {4k + 1 : k ≥ 0}, G = {4k + 3 : k ≥ 0}; and a_k = k·2^(k−1) − 2^k + 1 for all k ≥ 0. | |
0iqi | Problem:
You have a die with faces labelled $1$ through $6$. On each face, you draw an arrow to an adjacent face, such that if you start on a face and follow the arrows, after $6$ steps you will have passed through every face once and will be back on your starting face. How many ways are there to draw the arrows so th... | [
"Solution:\n\nAnswer: $32$\n\nThere are $4$ choices for where to go from face $1$. Consider the $4$ faces adjacent to $1$. We can visit either $1$, $2$, or $3$ of them before visiting the face opposite $1$. If we only visit one of these adjacent faces, we have $4$ choices for which one, then we visit face $6$, oppo... | United States | 1st Annual Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Geometry > Solid Geometry > Other 3D problems"
] | null | final answer only | 32 | |
0jkm | Problem:
$ABC$ is a triangle such that $BC = 10$, $CA = 12$. Let $M$ be the midpoint of side $AC$. Given that $BM$ is parallel to the external bisector of $\angle A$, find area of triangle $ABC$. (Lines $AB$ and $AC$ form two angles, one of which is $\angle BAC$. The external bisector of $\angle A$ is the line that bi... | [
"Solution:\n\nAnswer: $8 \\sqrt{14}$\n\nSince $BM$ is parallel to the external bisector of $\\angle A = \\angle BAM$, it is perpendicular to the angle bisector of $\\angle BAM$. Thus $BA = BM = \\frac{1}{2} BC = 6$.\n\nBy Heron's formula, the area of $\\triangle ABC$ is therefore\n$$\n\\sqrt{(14)(8)(4)(2)} = 8 \\sq... | United States | HMMT 2014 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 8 sqrt(14) | |
0330 | Problem:
Solve in integers the equation
$$
2^{a} + 8 b^{2} - 3^{c} = 283
$$ | [
"Solution:\nIt is easy to see that $a, c \\geq 0$. Since $3^{c}$ is congruent to $1$ or $3$ modulo $8$, then $0 \\leq a \\leq 2$. If $a=0$ or $a=1$, then $2$ divides $3^{c}$ or $8$ divides $3^{c}+1$, a contradiction.\n\nLet $a=2$, i.e., $8 b^{2} - 3^{c} = 279$. The cases $c=0,1$ are impossible and hence $c \\geq 2$... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (a,b,c) = (2,6,2) and (2,-6,2) | |
046t | A *state* refers to a way to place the numbers $1, 2, \ldots, 99$ on the vertices of a given regular $99$-gon, with exactly one number on each vertex and every number appearing exactly once. Two states are considered *equivalent* if one of them can be obtained from the other by rotating the regular $99$-gon (on the pla... | [
"The smallest $N = 2401$.\n\nLet $\\alpha^*$ be the state where the numbers $1, 2, \\ldots, 99$ are arranged counterclockwise at the vertices, and $\\beta^*$ be the state where $1, 2, \\ldots, 99$ are arranged clockwise on the circle. We prove that it takes at least $2401$ operations to transform $\\alpha^*$ into a... | China | Chinese Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, ex... | English | proof and answer | 2401 | |
0e9s | Let $\mathcal{K}$ be a circle with centre $O$ and radius $r$. What is the upper bound for the length of $|OT|$ given that $T$ is the point in the plane such that there exists an equilateral triangle $ABT$, where $A$ and $B$ lie on the circle $\mathcal{K}$? | [
"Let $M$ be the intersection of the segments $OT$ and $AB$. Then $|OT| = |OM| + |MT|$. Write $|AB| = |AT| = |BT| = a$. We know that $AB \\perp OT$. By Pythagoras' theorem we have $|OM| = \\sqrt{r^2 - |BM|^2} = \\sqrt{r^2 - \\frac{a^2}{4}}$ and $|MT| = \\sqrt{a^2 - \\frac{a^2}{4}} = \\frac{\\sqrt{3}a}{2}$. We can no... | Slovenia | National Math Olympiad in Slovenia | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2r | |
0h3h | Is it possible to paint each cell of an $8 \times 8$ table with one of 16 colors so that for each two colors there are two cells painted with these colors and having a common side? | [
"In the table, there are 112 unit segments that separate its neighboring cells. But from 16 given colors, one can form 120 pairs of colors, and for each of these pairs, there must be two neighboring cells, which is impossible.\n\n*Answer:* It is not possible."
] | Ukraine | Ukrainian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | It is not possible. | |
014q | Problem:
Determine the maximal size of a set of positive integers with the following properties:
(1) The integers consist of digits from the set $\{1,2,3,4,5,6\}$.
(2) No digit occurs more than once in the same integer.
(3) The digits in each integer are in increasing order.
(4) Any two integers have at least one digi... | [
"Solution:\n\nAssociate with any $a_{i}$ the set $M_{i}$ of its digits. By (1), (2), and (3), the numbers are uniquely determined by their associated subsets of $\\{1,2, \\ldots, 6\\}$. By (4), the sets are intersecting. Partition the 64 subsets of $\\{1,2, \\ldots, 6\\}$ into 32 pairs of complementary sets $(X, \\... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 32 | |
05oj | Problem:
Soit $S$ l'ensemble des nombres à deux chiffres qui ne contiennent pas le chiffre $0$. Deux nombres de $S$ sont dits amis si leurs plus grands chiffres sont égaux, et si la différence entre leurs plus petits chiffres est égale à $1$. Par exemple, $68$ et $85$ sont amis, $78$ et $88$ sont amis, mais $58$ et $7... | [
"Solution:\n\nRéponse : $45$. On peut prendre pour $T$ l'ensemble des nombres dont le plus petit chiffre est impair.\n\nRéciproquement, si $x=\\overline{a b}$ avec $1 \\leqslant b < a \\leqslant 9$ alors $x$ et $x+1$ sont amis. Si $x=\\overline{a b}$ avec $2 \\leqslant a < b \\leqslant 9$ et $a$ pair, alors $x$ et ... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 45 | |
0cfx | Let $(A, +, \cdot)$ be a finite ring and $a, b \in A$ with $ab + ba = 0$. If there is an $n \in \mathbb{N}^*$, $n \neq 2 \pmod 4$, such that $a^n - b^n$ is invertible, prove that $a - b$ and $a + b$ are invertible.
Mihai Opincariu | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Abstract Algebra > Ring Theory"
] | English | proof only | null | |
0dwo | Problem:
Naj bo $[x]$ največje celo število, ki ni večje od $x$, in naj bo $\{x\}=x-[x]$. Poišči vsa pozitivna realna števila $x$, za katera velja
$$
20\{x\}+0.5[x]=2005
$$ | [
"Solution:\n\nPomnožimo enačbo z $2$. Tedaj jo lahko zapišemo kot $40\\{x\\}=4010-[x]=a$, kjer je $a$ neko celo število. Zato je $\\{x\\}=\\frac{a}{40}$ za $0 \\leq a \\leq 39$. Velja $[x]=4010-a$, zato imamo $40$ rešitev oblike $x=4010-a+\\frac{a}{40}$, kjer je $a=0,1, \\ldots, 39$."
] | Slovenia | 49. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | All solutions are x = 4010 − a + a/40 for integers a = 0, 1, …, 39. | |
0l1i | Problem:
A deck of 100 cards is labeled $1, 2, \ldots, 100$ from top to bottom. The top two cards are drawn; one of them is discarded at random, and the other is inserted back at the bottom of the deck. This process is repeated until only one card remains in the deck. Compute the expected value of the label of the rema... | [
"Solution:\nNote that we can just take averages: every time you draw one of two cards, the expected value of the resulting card is the average of the expected values of the two cards. This average must be of the form\n$$\n2^{\\bullet} \\cdot 1 + 2^{\\bullet} \\cdot 2 + 2^{\\bullet} \\cdot 3 + \\cdots + 2^{\\bullet}... | United States | HMMT February 2024 Guts Round | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 467/8 | |
0hek | The managers of two companies are 1000 km apart. The segment that connects them contains $n$ points, in each of which sits a mathematician. Every second each mathematician moves to a point which is a middle of a segment that connects him/her and the closest mathematician or manager. If there is more than one such point... | [
"Let $S_0 = \\{0, x_1, x_2, \\dots, x_n, 1000\\}$ be the initial locations of mathematicians and managers, where $0$ and $1000$ represent the locations of managers, and $x_1 < x_2 < \\dots < x_n$ represent the locations of mathematicians. Similarly, let $S_i$ denote the positions of everyone in $i$ seconds. Note th... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0b0y | Problem:
In acute triangle $ABC$, $M$ and $N$ are the midpoints of sides $AB$ and $BC$, respectively. The tangents to the circumcircle of triangle $BMN$ at $M$ and $N$ meet at $P$. Suppose that $AP$ is parallel to $BC$, $AP = 9$ and $PN = 15$. Find $AC$. | [
"Solution:\n\nExtend rays $PM$ and $CB$ to meet at $Q$. Since $AP \\parallel QC$ and $M$ is the midpoint of $AB$, triangles $AMP$ and $BMQ$ are congruent. This gives $QM = MP = PN = 15$ and $QB = AP = 9$. Observe that the circumcircle of triangle $BMN$ is tangent to $QM$, so by power of a point, we compute $QM^{2} ... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 20 sqrt(2) | |
0bxn | Given a positive integer $n$, does there exist a planar polygon and a point in its plane such that every line through that point meets the boundary of the polygon at exactly $2n$ points? | [
"The answer is in the affirmative. To describe the configuration, fix a coordinate frame and let $a_0, a_1, \\dots, a_{4n-1}$ be real numbers such that $a_0 > 0 > a_2 > a_{4n-2} > a_4 > a_{4n-4} > \\dots > a_{2n-2} > a_{2n+2} > a_{2n}$, and $a_{2n+1} < a_{2n+3} < \\dots < a_{4n-1} < 0 < a_1 < a_3 < \\dots < a_{2n-1... | Romania | THE Fourteenth IMAR MATHEMATICAL COMPETITION | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0ega | Problem:
Naj bo $P$ poljubna točka na daljici $AB$ z dolžino $\sqrt{27}\ \mathrm{cm}$. Nad daljico $AP$ z dolžino $x$ je konstruiran kvadrat $APDE$, nad daljico $BD$ pa enakostranični trikotnik $BCD$. Nariši skico.
a) Izrazi ploščino petkotnika $ABCDE$ z $x$.
b) Določi $x$, da bo ploščina petkotnika $ABCDE$ najmanjš... | [
"Solution:\n\nPloščina $S$ petkotnika $ABCDE$ je vsota ploščin kvadrata $APDE$ ($S_1$), pravokotnega trikotnika $PBD$ ($S_2$) in enakostraničnega trikotnika $BCD$ ($S_3$).\n\nPloščine vseh treh likov izrazimo z $x$:\n\nPloščina kvadrata je $S_1 = x^2$,\n\nploščina pravokotnega trikotnika je $S_2 = \\frac{x \\cdot (... | Slovenia | 17. tekmovanje dijakov srednjih tehniških in strokovnih šol v znanju matematike | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | S(x) = ((1 + √3)/2) x^2 + ((3√3 − 9)/2) x + (27√3)/4; minimizing at x = (−9 + 6√3)/2 cm | |
03l0 | Problem:
Let $ABC$ be an equilateral triangle of altitude $1$. A circle with radius $1$ and center on the same side of $AB$ as $C$ rolls along the segment $AB$. Prove that the arc of the circle that is inside the triangle always has the same length. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
00rg | Let $a, b, c$ and $d$ be real numbers such that $a + b + c + d = 2$ and $ab + bc + cd + da + ac + bd = 0$.
Find the minimum value and the maximum value of the product $abcd$. | [
"Let's find the minimum first.\n$$\na^2 + b^2 + c^2 + d^2 = (a + b + c + d)^2 - 2(ab + bc + cd + da + ac + bd) = 4\n$$\nBy AM-GM, $4 = a^2 + b^2 + c^2 + d^2 \\ge 4\\sqrt{|abcd|} \\Rightarrow 1 \\ge |abcd| \\Rightarrow abcd \\ge -1$.\nNote that if $a = b = c = 1$ and $d = -1$, then $abcd = -1$.\n\nWe'll find the max... | Balkan Mathematical Olympiad | BMO 2016 Short List Final | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | minimum -1, maximum 1/4 | |
04kb | An $8 \times 8 \times 8$ cube consists of $512$ small grey cubes. Three sides of the big cube are painted in white, and the other three sides are painted in red. If each of the eight small cubes in the vertices of the big cube has at least one white and at least one red side, how many small cubes have at least one red ... | [] | Croatia | Mathematical competitions in Croatia | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 56 | |
0egu | Problem:
Če polinom $p(x)$ delimo s polinomom $x-18$, je ostanek pri deljenju enak 20. Če polinom $p(x)$ delimo s polinomom $x-20$, je ostanek pri deljenju enak 18. Koliko je ostanek pri deljenju, če polinom $p(x)$ delimo s polinomom $(x-20)(x-18)$ ?
(A) 2018
(B) $-x-2$
(C) $x+2$
(D) $-x+38$
(E) $x+38$ | [
"Solution:\n\nIz podatkov sklepamo, da je $p(x) = (x-18) r(x) + 20$ za nek polinom $r(x)$. Od tod sledi $p(18) = 20$. Podobno sklepamo, da je $p(20) = 18$. Naj bo $s(x)$ ostanek pri deljenju polinoma $p(x)$ z $(x-20)(x-18)$. Tedaj je $p(x) = (x-20)(x-18) t(x) + s(x)$ za nek polinom $t(x)$ in $s(x) = a x + b$ za nek... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | D | |
0fn7 | Estudia si existe una sucesión estrictamente creciente de enteros $0 = a_0 < a_1 < a_2 < \dots$, que cumple las dos condiciones siguientes:
* Todo número natural puede ser escrito como suma de dos términos, no necesariamente distintos, de la sucesión.
* Para cada entero positivo $n$, se cumple que $a_n > \frac{n^2}{16}... | [] | Spain | Olimpiada Matemática Española | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | Spanish | proof and answer | No existe tal sucesión. | |
0fbs | Problem:
En el triángulo $ABC$, el área $S$ y el ángulo $C$ son conocidos. Hallar el valor de los lados $a$ y $b$ para que el lado $c$ sea lo más corto posible. | [
"Solution:\n\nPor una parte\n$$\nc^{2} = a^{2} + b^{2} - 2ab \\cos C = (a-b)^{2} + 2ab(1-\\cos C)\n$$\ny por otra\n$$\nS = \\frac{1}{2} ab \\sen C \\Rightarrow ab = \\frac{2S}{\\sen C}.\n$$\nEntonces,\n$$\nc^{2} = (a-b)^{2} + \\frac{4S(1-\\cos C)}{\\sen C}\n$$\nserá mínimo cuando $a = b = \\sqrt{\\frac{2S}{\\sen C}... | Spain | null | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | a = b = sqrt(2S/sen C) | |
0j79 | Problem:
The integer $843301$ is prime. The primorial of a prime number $p$, denoted $p \#$, is defined to be the product of all prime numbers less than or equal to $p$. Determine the number of digits in $843301\#$. Your score will be
$$
\left.\max \left\{\left\lvert\, 60\left(\frac{1}{3}-\left|\ln \left(\frac{A}{d}\r... | [
"Solution:\n\nAnswer: $365851$\n\nRemark: $843301\\#-1$ is the largest known prime number of the form $p\\#-1$, where $p$ is prime."
] | United States | Harvard-MIT November Tournament | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 365851 | |
0kyt | Problem:
Across all polynomials $P$ such that $P(n)$ is an integer for all integers $n$, determine, with proof, all possible values of $P(i)$, where $i^{2} = -1$. | [
"Solution:\n\nWe claim the answer is every complex number $a+b i$ where $a$ and $b$ are rationals whose simplified denominators are not multiples of any prime congruent to $1$ modulo $4$. The proof consists of two main steps: proving that powers of $p \\equiv 1 \\bmod 4$ can't appear in the denominator, and showing... | United States | HMMT February 2024 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Intermediate Algebra > Complex numbers",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theo... | null | proof and answer | All values of the form a + b i where a and b are rational numbers whose reduced denominators are not divisible by any prime congruent to 1 modulo 4. | |
0c78 | On a line, there are 51 positive integers whose sum is 100. Prove that, for all positive integers $k$, $1 \le k \le 99$, one can find either a succession of numbers on the line whose sum is $k$, or a succession of numbers whose sum is $100-k$. | [
"Let $a_1, a_2, \\dots, a_{51}$ be the 51 numbers with $a_1 + a_2 + \\dots + a_{51} = 100$. On a circle of total length 100 we place 100 points such that the length of the arc between any two neighboring points (of these 100) is equal to 1. We fix one of these points and denote it by $A_1$. Then, we mark on the cir... | Romania | The DANUBE Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0ijj | Problem:
A fat coin is one which, when tossed, has a $\frac{2}{5}$ probability of being heads, $\frac{2}{5}$ of being tails, and $\frac{1}{5}$ of landing on its edge. Mr. Fat starts at $0$ on the real line. Every minute, he tosses a fat coin. If it's heads, he moves left, decreasing his coordinate by $1$; if it's tail... | [
"Solution:\n\nFor $n \\in \\mathbb{Z}$, let $a_n$ be the fraction of the time Mr. Fat spends at $n$. By symmetry, $a_n = a_{-n}$ for all $n$.\n\nFor $n > 0$, we have\n$$\na_n = \\frac{2}{5} a_{n-1} + \\frac{2}{5} a_{n+1},\n$$\nor\n$$\na_{n+1} = \\frac{5}{2} a_n - a_{n-1}.\n$$\nThis Fibonacci-like recurrence can be ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 1/3 | |
0hwg | Problem:
Let $a$ and $b$ be real numbers. Prove that the polynomial
$$
P(x) = x^{3} + (2a + 1)x^{2} + (2a^{2} + 2a - 3)x + b
$$
does not have three distinct rational roots. | [
"Solution:\n\nAssume the contrary, and suppose that the polynomial has rational roots $u, v, w$. Then by Viète's formulas,\n$$\n\\begin{aligned}\nu + v + w & = -(2a + 1) \\\\\nu v + v w + w u & = 2a^{2} - 2a - 3\n\\end{aligned}\n$$\nSquaring the first equation and subtracting twice the second yields\n$$\n\\begin{al... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
065y | Solve in the set of integers the equation
$$
x^3 y^2 (2y - x) = x^2 y^4 - 36.
$$ | [
"After manipulation equation becomes\n$$\n\\begin{align*} x^2 y^2 (x-y)^2 - 6^2 &= 0, \\quad x, y \\in \\mathbb{Z} \\\\\n&\\Leftrightarrow [xy(x-y)-6][xy(x-y)+6] = 0 \\\\\n&\\Leftrightarrow xy(x-y) = 6, \\quad x, y \\in \\mathbb{Z} \\text{ or } xy(x-y) = -6, \\quad x, y \\in \\mathbb{Z} \\\\\n&\\Leftrightarrow xy(x... | Greece | 28th Hellenic Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | [[3, 2], [2, 3], [-2, -3], [-3, -2], [3, 1], [1, 3], [-1, -3], [-3, -1], [-2, 1], [1, -2], [-1, 2], [2, -1]] | |
0cvv | Let $a \neq 1$ be a positive real number. Prove that the sequence $x_1, x_2, \dots$, where $x_n = 2^n (2^n\sqrt{a} - 1)$, is strictly decreasing. | [
"Пусть $t = \\sqrt{a}$. Заметим, что $t \\neq 1$. Тогда $x_{n+1} = 2^{n+1}(t-1)$ и $x_n = 2^n(t^2 - 1)$. Следовательно,\n$$\n\\begin{align*}\nx_n - x_{n+1} &= 2^n(t^2 - 1) - 2^{n+1}(t-1) = \\\\\n&= 2^n(t^2 - 2t + 1) = 2^n(t-1)^2 > 0,\n\\end{align*}\n$$\nчто и требовалось доказать.\n\nЗамечание. При других подходах ... | Russia | Regional round | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English; Russian | proof only | null | |
0c37 | Triangle $ABC$ has the property that there exists a unique point $X$ on the line segment $BC$ such that $AX^2 = BX \cdot CX$. Prove that $AB + AC = BC\sqrt{2}$. | [
"Let $T$ be the reflection of $A$ with respect to $X$. By the converse of the Power of a Point Theorem, it follows that the quadrilateral $ABTC$ is cyclic. If the line parallel to $BC$ through $T$ intersects the circumcircle of $ABC$ again at $U$, and lines $AU$ and $BC$ meet at $Y$, then $Y \\in [BC]$ and $BY \\cd... | Romania | 69th NMO Selection Tests for JBMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Reflection",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > D... | null | proof only | null | |
0jir | Problem:
Compute the prime factorization of $159999$. | [
"Solution:\n\n$159999 = 160000 - 1 = 20^{4} - 1 = (20 - 1)(20 + 1)(20^{2} + 1) = 19 \\cdot 21 \\cdot 401$\n\nNow, $21 = 3 \\cdot 7$, so:\n\n$159999 = 3 \\cdot 7 \\cdot 19 \\cdot 401$"
] | United States | HMMT 2014 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 3 * 7 * 19 * 401 | |
0a19 | A positive integer $a$ consists of four digits, three of which are equal to each other. The square of $a$ consists of seven digits, all of which are different. The number $b$ arises by reading number $a$ from right to left. It turns out that $b$ is greater than $a$. In addition, it turns out that $b^2$ is exactly equal... | [
"1113"
] | Netherlands | Dutch Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | 1113 | |
0c95 | We consider the cube $ABCD EFGH$ with side $a$ cm, $a > 0$ and the points $K$, $L$ on the segments $AC$, respectively $EG$ such that $CK = EL = \frac{AC}{4}$. Let $M$ be the midpoint of $AE$ and the points $N$, $P$ on the segment $HF$ such that
$$
HN = FP = \frac{(\sqrt{2} - 1)a}{2} \text{ cm.}
$$
Knowing that lines $M... | [
"Triangles $\\Delta MEP$ and $\\Delta MEN$ are congruent, therefore $MN = MP$.\nDenote by $O_1$ the midpoint of $HF$. It follows that $MO_1 = \\frac{AC}{2} = \\frac{a\\sqrt{3}}{2}$ and, since $NP = a$, the triangle $MNP$ must be equilateral.\nLet $O$ be the midpoint of $AC$, $M_1$ the midpoint of $AO$ and $\\{L_1\\... | Romania | Romanian Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
05ur | Problem:
Soit $ABCD$ un rectangle. Soit $\omega$ le demi-cercle de diamètre $[BC]$, de sorte que le point $A$ et le demi-cercle $\omega$ sont situés du même côté par rapport au segment $[BC]$. Le cercle de centre $B$ et de rayon $AB$ recoupe le demi-cercle $\omega$ au point $E$. La droite $(AE)$ recoupe le demi-cercle... | [
"Solution:\n\nTout d'abord, l'énoncé nous indique que $BA = BE$, de sorte que le triangle $ABE$ est isocèle en $B$. En outre, puisque $(AB)$ est tangente à $\\omega$, on sait que\n$$\n\\widehat{EBA} = \\widehat{EFB} = \\widehat{AFB} .\n$$\nComme les triangles $ABE$ et $AFB$ ont l'angle $\\widehat{BAE}$ en commun, i... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null |
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