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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
07ck | A $(n + 1)$-tuple $(h_1, h_2, \dots, h_{n+1})$ where for every $1 \le i \le n + 1$, $h_i(x_1, x_2, \dots, x_n)$ is an $n$-variable polynomial with real coefficients is called **good** if the following condition holds.
For any $n$ functions $f_1, f_2, \dots, f_n : \mathbb{R} \to \mathbb{R}$, if for all $1 \le i \le n + ... | [
"a) We set an example using induction on $n$. For $n=1$, consider the following pair of polynomials of degree 2\n$$\nh_1(x) = x^2 + x, \\ h_2(x) = -x^2.\n$$\nNow for any function $f: \\mathbb{R} \\to \\mathbb{R}$, if $P_1(x)=h_1(f(x))$ and $P_2(x)=h_2(f(x))$ are polynomials, then $P_1(x) + P_2(x) = f(x)$ is a polyn... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
09a5 | Let $M$ be a point on the circumscribed circle of an acute triangle $ABC$ and distinct from the vertices of $ABC$. $MP$ and $MQ$ are perpendicular lines from $M$ to $AB$ and $AC$ respectively and $H$ is orthocenter of triangle $ABC$. Prove that $PQ$ intersects $HM$ at the midpoint of $HM$. | [
"Let us denote the feet of the perpendicular from $M$ to $AC$ by $R$. Then $P$, $Q$ and $R$ are collinear (Simson's line).\n\n\n\nIf $\\angle CAM = \\varphi_1$, $\\angle ACM = \\varphi_2$ and $\\angle ABC = \\beta$, then $\\varphi_1 + \\varphi_2 = \\beta$.\n\nWe first show that $S_{PHQ} = S... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | Mongolian | proof only | null | |
02j7 | Problem:
O limite de peso que um caminhão pode transportar corresponde a $50$ sacos de areia ou $400$ tijolos. Se este caminhão já contém $32$ sacos de areia, quantos tijolos, no máximo, ele ainda pode carregar?
A) $132$
B) $144$
C) $146$
D) $148$
E) $152$ | [
"Solution:\n\nO enunciado mostra que o peso de $1$ saco de areia é o mesmo que o de $8$ tijolos. Se no caminhão já há $32$ sacos de areia, ele pode carregar ainda $18$ sacos, o que equivale $18 \\times 8 = 144$ tijolos."
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | MCQ | B | |
008u | Let $ABC$ be a triangle such that $A = 105^\circ$ and $B = 45^\circ$. Let $L$ be a point in $BC$ such that $AL$ is the bisector of $BAC$ and $M$ be the midpoint of $AC$. If $AL$ and $BM$ meet at the point $P$, calculate the ratio $\frac{AP}{AL}$. | [] | Argentina | XXIX Olimpíada Matemática Argentina National Round | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | sqrt(2)/2 | |
0gqr | Find all real values of $a$ for which the equation
$$
x(x+1)^3 = (2x+a)(x+a-1)
$$
has four distinct real roots. | [
"Note that $x(x+1)^3 - (2x+a)(x+a-1) = x^4 + 3x^3 + x^2 + x(3-3a) + a - a^2 = (x^2 + 3x + a)(x^2 + 1 - a) = 0$.\n\nIn order to get 4 roots the discriminants of both quadratic polynomials must be positive: $9 - 4a > 0$ and $4a - 4 > 0$.\n\nThe roots $x^2 + 3x + a = 0$ and $x^2 + 1 - a = 0$ should be distinct: let $t... | Turkey | Team Selection Test | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a ∈ (1, 5/4) ∪ (5/4, 2) ∪ (2, 9/4) | |
0dhp | Each pair of vertices of a regular $1001$-gon is joined with a segment, which is either red, or blue, or green. Prove that one can choose $11$ vertices of this $1001$-gon in such a way that they form a convex $11$-gon, in which at least $10$ sides have the same color. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0lc6 | Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying following conditions
i) $\lim_{x \to 0} f(x) = f(0) = 0,$
ii) $f(x)f(x^{1959} + x^{54}) = f(x^{2013} + x)$ for all $x \in \mathbb{R}$. | [] | Vietnam | Vietnamese Mathematical Competitions | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 0 for all real x | |
08sz | How many 3-digit numbers are there that can appear as the top three digits of a 6-digit number, which is a perfect square? | [
"Since $(n+1)^2 - n^2 = 2n + 1$ holds for any positive integer $n$, the difference of any contiguous pair of perfect squares, larger of which is no more than $500^2 = 250000$, does not exceed $2 \\cdot 499 + 1 = 999$. From this fact it follows that for any positive integer $m$ satisfying $100 \\le m < 250$, there e... | Japan | Japan Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Other"
] | English | proof and answer | 650 | |
010q | Problem:
The bisectors of the angles $A$ and $B$ of the triangle $ABC$ meet the sides $BC$ and $CA$ at the points $D$ and $E$, respectively. Assuming that $|AE| + |BD| = |AB|$, determine the size of angle $C$. | [
"Solution:\nLet $F$ be the point of the side $AB$ such that $|AF| = |AE|$ and $|BF| = |BD|$ (see Figure 10).\n\nThe line $AD$ is the angle bisector of $\\angle A$ in the isosceles triangle $AEF$. This implies that $AD$ is the perpendicular bisector of $EF$, whence $|DE| = |DF|$. Similarly we show that $|DE| = |EF|$... | Baltic Way | Baltic Way | [
"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 and answer | 60° | |
03p3 | Assume that $A_1, A_2, \dots, A_8$ are eight points taken arbitrarily on a plane. For a directed line $l$ taken arbitrarily on the plane, assume that projections of $A_1, A_2, \dots, A_8$ on the line are $P_1, P_2, \dots, P_8$ respectively. If the eight projections are pairwise disjoint, they can be arranged as $P_{i_1... | [
"(1) For two parallel and directed lines with the same direction, the order of projections of $A_1, A_2, \\dots, A_8$ must be the same. So, we need only to discuss all directed lines passing through a fixed point $O$.\n\n(2) If a directed line taken is perpendicular to a line joining two given points, then the proj... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | English | proof and answer | 56 | |
0egm | Problem:
Bratje Jure, Klemen, Luka, Miha in Nace so kupili čokolado. Ko so jo odvili, so ugotovili, da je zlomljena na 7 kosov (glej sliko), zato so si teh 7 kosov med sabo razdelili. Jure je pojedel največji kos čokolade. Klemen in Luka sta pojedla enako količino čokolade, toda Klemen je pojedel 3 kose, Luka pa le 1 ... | [
"Solution:\n\nTablica čokolade ima $4 \\cdot 7 = 28$ kvadratnih koščkov. Naj bo en tak košček velik 1 kvadratno enoto. Tedaj je 7 kosov čokolade označenih na sliki po vrsti velikih $\\frac{1 \\cdot 4}{2} = 2$, $\\frac{3 \\cdot 4}{2} = 6$, $\\frac{5 \\cdot 2}{2} = 5$, $\\frac{4 \\cdot 2}{2} = 4$, $\\frac{2 \\cdot 2}... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | MCQ | B | |
0dns | Problem:
Дато је $2n-1$ двоелементних подскупова скупа $\{1,2, \ldots, n\}$. Доказати да се може одабрати $n$ од ових подскупова чија унија садржи не више од $\frac{2}{3}n+1$ елемената.
(Душан Ђукић) | [
"Solution:\n\nПоказаћемо индукцијом по $k\\left(k \\leqslant \\frac{2n-1}{3}\\right)$ да се може избацити $3k$ подскупова тако да је кардиналност уније преосталих не већа од $n-k$.\n\nЗа $k=0$ то је тривијално. Претпоставимо да је $k \\geqslant 1$ и да смо избацили $3(k-1)$ подскупова тако да унија преосталих не са... | Serbia | 10. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
054z | On the sides $BC$, $CA$ and $AB$ of triangle $ABC$, respectively, points $D$, $E$ and $F$ are chosen. Prove that
$$
\frac{1}{2} (BC + CA + AB) < AD + BE + CF < \frac{3}{2} (BC + CA + AB).
$$ | [
"Let us first prove the second inequality. Consider the circle with centre $A$ that passes through the vertex $C$ and consider the extension of the side $CB$ past the vertex $B$ up to this circle (Fig. 5). As a chord lies inside the circle, we have $AD \\le AC$. We also have $BE \\le \\max(BC, BA) = BC$ and $CF \\l... | Estonia | Open Contests | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0h8c | Numbers $1, 2, \ldots, n$ are written in line. Numbers $1$ and $n$ are painted blue, and the rest are painted yellow. Two players - Olesya and Andriy - in turns repaint one of the yellow numbers blue by the following rules: with the first move Olesya (she starts) repaints blue any of the yellow numbers (let us denote i... | [
"We will solve this problem by finding the winning and losing positions. Recall that the position is called a losing position for a player, if after his move he instantly loses, or puts the game in a winning position for another player. Position is a winning one, if the player makes a move and immediately wins, or ... | Ukraine | UkraineMO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Andriy wins if n is a power of two; otherwise Olesya wins. | |
0enr | Prove that for every natural number $n$, there exists a polynomial $p(x)$ with integer coefficients such that $p(1), p(2), \dots, p(n)$ are all distinct powers of $2$. | [
"Suppose that $p_n(x)$ is a polynomial such that $p_n(1), p_n(2), \\dots, p_n(n)$ are distinct powers of $2$. We claim that $\\gcd(p_n(n+1), n!)$ is a power of $2$. Indeed, if there is a prime number $2 < q \\le n$ that divides $p_n(n+1)$, then $q$ divides $p_n(n+1-q)$ (a power of $2$), which is a contradiction.\n\... | South Africa | South-Afrika 2011-2013 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Induction / smoothing"... | null | proof only | null | |
0hd0 | Given a right triangle $ABC$, let $M$ be the midpoint of its hypotenuse $AB$. A perpendicular bisector of $AB$ intersects side $BC$ in point $K$. A perpendicular from $K$ to $CM$ intersects the ray $AC$ in point $P$, which happens not to lie on the segment $AC$. Lines $CM$ and $BP$ intersect in point $T$. Prove that $A... | [
"From $KP \\perp CM$ it follows that $\\angle CPK = \\angle MBC = \\angle ABC$ (fig. 33), hence the quadrilateral $PAKB$ is cyclic. Therefore, $PK$ is a bisector in the triangle $\\triangle CPT$. Clearly, $PK$ is also a height and a median of this triangle.\n\nTherefore, $\\triangle CPT$ is isosceles. In particular... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
0fuz | Problem:
Sei $ABC$ ein spitzwinkliges Dreieck. $M$ und $N$ seien zwei beliebige Punkte auf den Seiten $AB$ respektive $AC$. Die Kreise mit den Durchmessern $BN$ und $CM$ schneiden sich in den Punkten $P$ und $Q$. Zeige, dass die Punkte $P$, $Q$ und der Höhenschnittpunkt des Dreiecks $ABC$ auf einer Geraden liegen. | [
"Solution:\n\nBezeichne die beiden Kreise mit $\\Gamma_{B}$ und $\\Gamma_{C}$. Die Punkte $P$ und $Q$ liegen auf der Potenzlinie von $\\Gamma_{B}$ und $\\Gamma_{C}$, es genügt daher zu zeigen, dass auch der Höhenschnittpunkt $H$ dieselbe Potenz bezüglich beider Kreise besitzt. Bezeichne die Fusspunkte der Höhen von... | Switzerland | SMO Finalrunde | [
"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 > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0aph | Problem:
Graph theory's Four-Color Theorem says that four colors are enough to color the regions in a plane so that no two adjacent regions receive the same color. The theorem was proved in 1976 by Kenneth Appel and Wolfgang Haken, 124 years after the Four-Color Problem was posed.
Fermat's Last Theorem in Number Theo... | [
"Solution:\n\nPoincaré Conjecture"
] | Philippines | Tenth Philippine Mathematical Olympiad | [
"Geometry > Differential Geometry > Manifolds",
"Discrete Mathematics > Graph Theory",
"Number Theory > Other"
] | null | final answer only | Poincaré Conjecture | |
07rt | From a point on the hypotenuse of a right-angled triangle perpendiculars are drawn to the other two sides. If the hypotenuse has length $4$ and the two perpendiculars have length one, find the area of the triangle. | [
"\n\nLet $D$ be the point on the hypotenuse $BC$, and let $DE$ and $DF$ be the perpendiculars on $AC$ and $AB$, respectively. Let $x = |CE|$, $y = |BF|$ and use the standard notation for the side lengths, so that $a^2 = b^2 + c^2$ by Pythagoras.\n\nThe triangles $ABC$, $FBD$ and $EDC$ are s... | Ireland | Irish | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | (1 + sqrt(17)) / 2 | |
01y0 | Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying the equality
$$
f(f(x) + f(y)) = (x + y)f(x + y)
$$
for all real $x$ and $y$. | [
"Answer: $f(x) = 0$ for all $x \\in \\mathbb{R}$.\nConsider several preliminary substitutions to the problem condition\n$$\nf(f(x) + f(y)) = (x + y)f(x + y). \\qquad (1)\n$$\nSubstitution $x = y = 0$ gives the equality $f(2f(0)) = 0$. Substitution $x = 2f(0)$, $y = 0$ gives $f(f(0)) = 0$. Finally, substitution $x =... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | f(x) = 0 for all x in R | |
0fvs | Problem:
In einem Tennistournier haben mindestens drei Spieler teilgenommen. Dabei haben je zwei Spieler genau einmal gegeneinander gespielt, und jeder Spieler hat mindestens ein Match gewonnen. Zeige, dass es drei Spieler $A, B, C$ gibt, sodass $A$ gegen $B$, $B$ gegen $C$ und $C$ gegen $A$ gewonnen hat. | [
"Solution:\n\nWähle einen Spieler $A$, der unter allen Spielern die kleinste Anzahl Matches gewonnen hat. Seien $U$ und $V$ die Mengen der Spieler, gegen die $A$ gewonnen bzw. verloren hat. Nach Annahme sind $U$ und $V$ nicht leer. Hat ein Spieler $B$ aus $U$ gegen einen Spieler $C$ aus $V$ gewonnen, dann erfüllen ... | Switzerland | SMO Finalrunde | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0ghf | 用 $N$ 表示所有正整數形成的集合。找到所有可能的函數 $f: N \to N$, 使得對於所有 $m, n \in N$, 都有 $|f(m+n) - f(m)|$ 整除 $f(n)$。
Denote the set of positive integers by $\mathbb{N}$. Find all functions $f : \mathbb{N} \to \mathbb{N}$ satisfying that for all $m, n \in \mathbb{N}$, $|f(m+n) - f(m)|$ divides $f(n)$. | [
"From the assumptions of the problem, we have the following facts:\n1. $f$ is injective\n2. $f(m+n) \\le f(m) + f(n)$\n3. $|f(m+1) - f(m)| \\le f(1)$\n4. $(f(m+1) - f(1)) \\mid f(m)$\nSince $f$ is injective, $f(m+1) > 3f(1)$ for sufficiently large $m$. Combine the above listed facts, we have\n$$\n2f(1) < f(m+1) - f... | Taiwan | 2023 數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | Chinese (Traditional) | proof and answer | f(n) = c n for some positive integer c | |
0des | For each non-constant integer polynomial $P(x)$, let's define
$$
M_{P(x)} = \max_{x \in [0; 2021]} |P(x)|.
$$
1. Find the minimum value of $M_{P(x)}$ when $\deg P(x) = 1$.
2. Suppose that $P(x) \in \mathbb{Z}[x]$ when $\deg P(x) = n$ and $2 \le n \le 2022$. Prove that $M_{P(x)} \ge 1011$. | [
"1) Since $\\deg P = 1$, put $P(x) = ax + b$ with $a, b \\in \\mathbb{Z}$ and $a \\neq 0$. Note that\n$$\n|P(2021) - P(0)| = |2021a| \\ge 2021 \\Rightarrow \\max\\{|P(2021)|, |P(0)|\\} \\ge \\frac{2021}{2}.\n$$\nThus, $|P(2021) - P(0)| = |2021a| \\ge 2021$ which implies that\n$$\n\\max\\{|P(2021)|, |P(0)|\\} \\ge \... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof and answer | Part 1: 1011. Part 2: For degrees from two to two thousand twenty two, M_P is at least 1011. | |
0be9 | Let $ABC$ be a triangle and let points $M$ and $N$ on the sides $AB$ and $AC$ respectively such that $\angle ABC \equiv \angle ANM$. Let $D$ be the mirror image of point $A$ across $B$, and let $P$ and $Q$ be the midpoints of the line segments $MN$ and $CD$ respectively. Show that points $A$, $P$ and $Q$ are collinear ... | [
"\"$\\Leftrightarrow$\" Notice that triangles $AMN$ and $ACB$ are similar to get $\\frac{AM}{AC} = \\frac{AN}{AB}$ and furthermore $\\frac{AM}{AN} = \\sqrt{2}$. Since $\\frac{AD}{AC} = \\frac{2AB}{AC} = \\sqrt{2}$, one has $\\frac{AM}{AN} = \\frac{AD}{AC} \\Leftrightarrow \\frac{AM}{AD} = \\frac{AN}{AC}$. Consequen... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | AC = AB√2 | |
0jla | Let $n$ be a positive even integer, and let $c_1, c_2, \dots, c_{n-1}$ be real numbers satisfying
$$
\sum_{i=1}^{n-1} |c_i - 1| < 1.
$$
Prove that
$$
2x^n - c_{n-1}x^{n-1} + c_{n-2}x^{n-2} - \dots - c_1x + 2
$$
has no real roots. | [
"Denote $P(x) = 2x^n - c_{n-1}x^{n-1} + c_{n-2}x^{n-2} - \\dots - c_1x + 2$. Since $c_i \\in (0, 2)$ for all $1 \\le i \\le n-1$, it follows $P(x) \\ge 2 > 0$ for all $x \\le 0$. It remains to prove $P(x) > 0$ also for all $x > 0$. Let $c_i = 1 + \\epsilon_i$, so $\\sum_{i=1}^{n-1} |\\epsilon_i| < 1$. Then for $x >... | United States | IMO Team Selection Test | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0h58 | A $9 \times 9$ square is divided into $81$ small $1 \times 1$ squares, $8$ of which are painted black, the rest being white. We cut a fully white rectangle (possibly, a square) out of the big $9 \times 9$ square. What is the maximal area of the rectangle that we can attain regardless of the positions of the black squar... | [
"Cut the square into $9$ smaller $3 \\times 3$ blocks. Since there are only eight black squares, at least one of the blocks doesn't contain any of them. Therefore a white square of area $9$ can always be found.\n\nNext, we show that sometimes it is impossible to find a larger rectangle. Fig. 2 is an example. Here, ... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 9 | |
0jta | Problem:
Let $d$ be a randomly chosen divisor of $2016$. Find the expected value of
$$
\frac{d^{2}}{d^{2}+2016}
$$ | [
"Solution:\nAnswer: $\\frac{1}{2}$\nLet $ab = 2016$. Then\n$$\n\\frac{a^{2}}{a^{2}+2016} + \\frac{b^{2}}{b^{2}+2016} = \\frac{a^{2}}{a^{2}+2016} + \\frac{\\left(\\frac{2016}{a}\\right)^{2}}{\\left(\\frac{2016}{a}\\right)^{2}+2016} = \\frac{a^{2}}{a^{2}+2016} + \\frac{2016}{a^{2}+2016} = 1\n$$\nThus, every divisor $... | United States | HMMT November | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 1/2 | |
03vy | Let $P$ be an interior point of a regular $n$-gon $A_1A_2\cdots A_n$; the lines $A_iP$ meet the regular $n$-gon $A_1A_2\cdots A_n$ at another point $B_i$, where $i = 1, 2, \dots, n$. Prove that
$$
\sum_{i=1}^{n} PA_i \geq \sum_{i=1}^{n} PB_i.
$$ | [
"Denote $t = \\lfloor \\frac{n}{2} \\rfloor + 1$, and let $A_{n+j} = A_j$, $j = 1, 2, \\dots, n$.\nNoting that the distance between any vertex of a regular $n$-gon and a point on its side is not greater than its longest diagonal $d$, we therefore have, for any $1 \\le i \\le n$,\n$$\nA_i P + P B_i = A_i B_i \\le d.... | China | China Western Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
00y3 | Problem:
Let $a \circ b = a + b - a b$. Find all triples $(x, y, z)$ of integers such that $(x \circ y) \circ z + (y \circ z) \circ x + (z \circ x) \circ y = 0$. | [
"Solution:\nNote that\n$$\n(x \\circ y) \\circ z = x + y + z - x y - y z - x z + x y z = (x-1)(y-1)(z-1) + 1.\n$$\nHence\n$$\n(x \\circ y) \\circ z + (y \\circ z) \\circ x + (z \\circ x) \\circ y = 3((x-1)(y-1)(z-1) + 1).\n$$\nNow, if the required equality holds we have $(x-1)(y-1)(z-1) = -1$. There are only four p... | Baltic Way | Baltic Way | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (0, 0, 0), (0, 2, 2), (2, 0, 2), (2, 2, 0) | |
03hb | Problem:
i. Solve the simultaneous inequalities, $x < \frac{1}{4x}$ and $x < 0$; i.e., find a single inequality equivalent to the two given simultaneous inequalities.
ii. What is the greatest integer which satisfies both inequalities $4x + 13 < 0$ and $x^2 + 3x > 16$?
iii. Give a rational number between $\frac{11}{2... | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | i) x < -1/2; ii) -6; iii) 17/37; iv) 100000 = 32*3125; v) 1/2 | |
08rn | The triangle $ABC$ satisfies $BC = 5$, $CA = 7$ and $AB = 8$. Let $O$ be a point inside the triangle $ABC$ such that the triangles $OBC$, $OCA$ and $OAB$ have the same circumradius. Find their common circumradius. | [
"Let $P$, $Q$ and $R$ be the circumcenters of the triangles $OBC$, $OCA$ and $OAB$, respectively, and let $r$ be their common circumradius.\n\nSince $PB = PO = RB = RO = r$, the quadrilateral $BPOR$ is a rhombus and so $\\vec{BR} = \\vec{PO}$. Similarly, $\\vec{CQ} = \\vec{PO}$. So $\\vec{BR} = \\vec{CQ}$, and it f... | Japan | The 16th Japanese Mathematical Olympiad - The First Round | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof and answer | 7/√3 | |
0e3e | Problem:
Miro, Aleš in Lovro so trikrat igrali poker. Prvič je izgubil Miro in je zato moral plačati Alešu in Lovru, vsakemu posebej toliko denarja, kot sta ga imela na začetku igre. Drugič je izgubil Aleš, zato je prav tako moral plačati Miru in Lovru, vsakemu posebej toliko, kot sta ga trenutno imela. Tretjič je izg... | [
"Solution:\n\nNaj bo $x$ začetni znesek Aleša, $y$ začetni znesek Lovra in $z$ začetni znesek Mira. Tabeliramo časovni potek spreminjanja količine denarja pri posameznikih:\n\n| Igra | Miro | Aleš | Lovro |\n|:-----------:|:-------------------:|:------------------... | Slovenia | 10. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | Miro, 15 euros | |
0ijn | Problem:
A sequence of numbers $\{a_n\}$ is given by $a_1 = 1$, $a_{n+1} = 2a_n + \sqrt{3a_n^2 + 1}$ for $n \geq 1$. Prove that each term of the sequence is an integer. | [
"Solution:\n\nFrom the definition of the sequence we see that $\\left(a_{n+1} - 2a_n\\right)^2 = 3a_n^2 + 1$. After simplification we get\n$$\na_{n+1}^2 + a_n^2 - 4a_n a_{n+1} = 1\n$$\nAdding $3a_{n+1}^2$ to both sides of the last equation gives us $\\left(2a_{n+1} - a_n\\right)^2 = 3a_{n+1}^2 + 1$ and after taking... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
01iz | Find the least positive integer $n$ for which it is possible to draw an $n$-gon along the lines of a square grid, whose perimeter is $n$ and whose area is $n$. Here the boundary of the $n$-gon may not visit any point more than once. | [
"Answer: $n = 32$. We start by showing that such a $32$-gon exists.\nIt remains to prove that this is indeed the smallest possible $n$. We will do that by proving that for all $n < 32$ the area $S$ of an $n$-gon with perimeter $P = n$ is smaller than $n$.\n\nAt first we note that $P = n$ if and only if all sides of... | Baltic Way | Baltic Way 2023 Shortlist | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discre... | English | proof and answer | 32 | |
03h7 | Problem:
Find all the real numbers which satisfy the equation $|x+3| - |x-1| = x+1$. (Note: $|a| = a$ if $a \geq 0$; $|a| = -a$ if $a < 0$.) | [
"Solution:\nLet us consider the different cases for $x$ based on the expressions inside the absolute values.\n\n**Case 1:** $x + 3 \\geq 0$ and $x - 1 \\geq 0$\n\nThis means $x \\geq -3$ and $x \\geq 1$, so $x \\geq 1$.\n\nIn this case, $|x+3| = x+3$ and $|x-1| = x-1$.\n\nThe equation becomes:\n$$(x+3) - (x-1) = x+... | Canada | Canadian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | x = -5, -1, 3 | |
05be | Today, on September 23, 2023, twins Mari and Jüri received a total of $5000$ candies for their $10$th birthdays. Starting from this day, their mother allows them both to take candies once per day, such that the amount of candies taken by any child on any day is less than their age in full years (on their birthday, they... | [
"We will show that Mari can avoid taking the last candy. To achieve this, she will take $8$ candies today, and on every following day, she will take candies so that along with Jüri's candies from the previous day, the total is the age of the children on the previous day. Since $2024$ is a leap year, they will be $1... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Mari | |
0gua | Let $n$ be a positive integer and $P, Q$ be polynomials with real coefficients such that $P(x) = x^n Q(1/x)$ and $P(x) \ge Q(x)$ for all real numbers $x$. Prove that $P(x) = Q(x)$ for all real numbers $x$. | [
"We are given that $P(x) = x^n Q(1/x)$ and $P(x) \\ge Q(x)$ for all real $x$.\n\nFirst, note that $P(x) = x^n Q(1/x)$ for all $x \\ne 0$ (since $Q$ is a polynomial, $Q(1/x)$ is defined for $x \\ne 0$).\n\nAlso, $P(x) \\ge Q(x)$ for all real $x$.\n\nLet us substitute $x$ by $1/x$ (for $x \\ne 0$):\n\n$P(1/x) = (1/x)... | Turkey | Team Selection Test for EGMO 2023 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
01bk | $G$ and $G'$ are graphs (without loops and multiple edges) on the same set of 100 vertices. Two vertices are connected by an edge in the graph $G'$ iff the sum of their degrees in the graph $G$ is at least 100. The graph $G'$ has a Hamiltonian cycle (i.e. a cycle of length 100 that passes through all the vertices). Pro... | [
"**Lemma.** Let a graph $H$ has 100 vertices and contains a Hamiltonian path (not cycle) that starts at the vertex $A$ and ends in $B$. If the sum of degrees of vertices $A$ and $B$ is at least 100, then the graph $H$ contains a Hamiltonian cycle.\n**Proof.** Let $N = \\deg A$, then $\\deg B \\ge 100 - N$. Let us n... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Other",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
06uo | Say that an ordered pair $(x, y)$ of integers is an irreducible lattice point if $x$ and $y$ are relatively prime. For any finite set $S$ of irreducible lattice points, show that there is a homogenous polynomial in two variables, $f(x, y)$, with integer coefficients, of degree at least 1, such that $f(x, y)=1$ for each... | [
"First of all, we note that finding a homogenous polynomial $f(x, y)$ such that $f(x, y)= \\pm 1$ is enough, because we then have $f^{2}(x, y)=1$. Label the irreducible lattice points $(x_{1}, y_{1})$ through $(x_{n}, y_{n})$. If any two of these lattice points $(x_{i}, y_{i})$ and $(x_{j}, y_{j})$ lie on the same ... | IMO | International Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathe... | English | proof only | null | |
0hio | Problem:
An integer is called formidable if it can be written as a sum of distinct powers of $4$, and successful if it can be written as a sum of distinct powers of $6$. Can $2005$ be written as a sum of a formidable number and a successful number? Prove your answer. | [
"Solution:\n\nSuppose that $2005 = a + b$, where $a$ is formidable and $b$ is successful. Then $a$ must be a sum of some of the powers of $4$ less than $2005$, namely $1, 4, 16, 64, 256, 1024$. Similarly, $b$ must be a sum of some of the numbers $1, 6, 36, 216, 1296$. So, $a + b$ must be a sum of some distinct entr... | United States | null | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | No | |
01vk | The quadrilateral *ABCD* is inscribed in the parabola $y = x^2$. It is known that $\angle BAD = 90^\circ$, the diagonal *AC* is parallel to the axis *Ox* and *AC* is the bisector of the angle *BAD*.
Find the area of the quadrilateral *ABCD* if the length of the diagonal *BD* is equal to *p*. | [
"Answer: $S = \\frac{p^2}{4} - 1$.\n\nNote that if the points with coordinates $(m, m^2)$ and $(n, n^2)$ belong to the parabola $y = x^2$, then the line passing through them has the equation $y = (m + n)x - mn$. Indeed, since the coordinates of each of the two points satisfy this linear equation, the entire line is... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | p^2/4 - 1 | |
0int | For a polynomial $P(x)$ with integer coefficients, $r(2i - 1)$ (for $i = 1, 2, 3, \dots, 512$) is the remainder obtained when $P(2i - 1)$ is divided by $1024$. The sequence
$$
(r(1), r(3), \dots, r(1023))
$$
is called the *remainder sequence* of $P(x)$. A remainder sequence is called *complete* if it is a permutation o... | [
"Define the polynomials\n$$\n\\begin{align*}\nQ_0(x) &= b_0, \\\\\nQ_1(x) &= b_1 (x + 1), \\\\\nQ_2(x) &= b_2 (x + 1)(x + 3), \\\\\nQ_3(x) &= b_3 (x + 1)(x + 3)(x + 5), \\\\\nQ_4(x) &= b_4 (x + 1)(x + 3)(x + 5)(x + 7), \\\\\nQ_5(x) &= b_5 (x + 1)(x + 3)(x + 5)(x + 7)(x + 9), \\\\\nQ_6(x) &= b_6 (x + 1)(x + 3)(x + 5... | United States | Team Selection Test | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0ehn | Problem:
Določi najmanjši skupni večkratnik in največji skupni delitelj izrazov:
$$
-3 x^{4}+24 x, \quad x^{3}-6 x^{2}+12 x-8, \quad a x^{2}-4 a x+4 a, \quad 3 b x^{2}-12 b
$$
(10 točk) | [
"Solution:\n\nIzraze zapišemo v obliki produkta\n$$\n-3 x^{4}+24 x = -3 x (x^{3}-8) = -3 x (x-2)(x^{2}+2 x+4),\n$$\n$$\nx^{3}-6 x^{2}+12 x-8 = (x-2)^{3},\n$$\n$$\na x^{2}-4 a x+4 a = a(x^{2}-4 x+4) = a(x-2)^{2},\n$$\n$$\n3 b x^{2}-12 b = 3 b (x^{2}-4) = 3 b (x-2)(x+2).\n$$\n\nUgotovimo, da je največji skupni delite... | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Odbirno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | gcd: x - 2; lcm: -3abx (x - 2)^3 (x + 2) (x^2 + 2x + 4) | |
03is | Problem:
Let $\mathbb{R}^n$ be the $n$-dimensional Euclidean space. Determine the smallest number $g(n)$ of points of a set in $\mathbb{R}^n$ such that every point in $\mathbb{R}^n$ is at irrational distance from at least one point in that set. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | g(n) = n + 1 | |
03kv | Problem:
Find all real numbers $x$ such that
$$
x = \left(x - \frac{1}{x}\right)^{1/2} + \left(1 - \frac{1}{x}\right)^{1/2}
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (1 + sqrt(5)) / 2 | |
0837 | Problem:
Trovare tutti i numeri naturali $n$ di tre cifre $(100 \leq n \leq 999)$ che sono uguali al numero formato dalle ultime tre cifre di $n^{2}$. | [
"Solution:\nI numeri richiesti sono $376$ e $625$.\nIl quadrato di $n$ ha le stesse ultime tre cifre di $n$ se e solo se $n^{2}-n=n(n-1)$ è divisibile per $1000=2^{3} \\cdot 5^{3}$. Poiché $n$ ed $n-1$ sono primi fra loro, uno solo dei due è pari ed uno solo dei due può essere divisibile per $5$. Pertanto abbiamo l... | Italy | XIX Gara Nazionale di Matematica | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | 376, 625 | |
0jyn | Problem:
a. Does there exist a finite set of points, not all collinear, such that a line between any two points in the set passes through a third point in the set?
b. Let $ABC$ be a triangle and $P$ be a point. The isogonal conjugate of $P$ is the intersection of the reflection of line $AP$ over the $A$-angle bisecto... | [] | United States | February 2017 | [
"Geometry > Plane Geometry > Combinatorial Geometry > Sylvester's theorem",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Geometric Inequalities > ... | null | final answer only | {'a': 'No', 'b': 'Yes; for example, the excenters are also fixed points', 'c': 'Yes', 'd': 'Yes', 'e': 'Yes', 'f': 'No'} | |
0f2r | Problem:
$ABCD$ is a quadrilateral. $M$ is a point inside it such that $ABMD$ is a parallelogram. $\angle CBM = \angle CDM$. Show that $\angle ACD = \angle BCM$. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06nn | Let $n$ be a positive integer. Show that if $p$ is a prime dividing $5^{4n} - 5^{3n} + 5^{2n} - 5^n + 1$, then $p \equiv 1 \pmod 4$. | [
"Clearly, $p \\neq 2, 5$. Let $m = 5^{4n} - 5^{3n} + 5^{2n} - 5^n + 1$. Then\n$$\n(2 \\cdot 5^{2n} - 5^n + 2)^2 - 5 \\cdot 5^{2n} = 4m \\equiv 0 \\pmod{p}.\n$$\nThis gives $5 \\equiv (5^{-n}(2 \\cdot 5^{2n} - 5^n + 2))^2 \\pmod{p}$. Using the Legendre symbol, we have $\\left(\\frac{5}{p}\\right) = 1$. On the other ... | Hong Kong | IMO HK TST | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity"
] | null | proof only | null | |
0fnd | Sean $A$, $B$ y $C$ los vértices de un triángulo y $P$, $Q$ y $R$ los respectivos pies de las bisectrices trazadas desde esos mismos vértices. Sabiendo que $PQR$ es un triángulo rectángulo en $P$, se te pide probar dos cosas:
a) Que $ABC$ ha de ser obtusángulo.
b) Que en el cuadrilátero $ARPQ$, pese a no ser cíclico, l... | [
"Para resolver el problema utilizaremos dos herramientas fundamentales en los problemas geométricos relativos a triángulos: el teorema de la bisectriz (La bisectriz de un ángulo de un triángulo corta al lado opuesto en dos segmentos de longitudes proporcionales a los otros dos lados del triángulo) y el teorema de S... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Spanish | proof only | null | |
07gk | Let $a$, $b$, $c$, $d$ be pair-wise coprime positive integers not all equal to $1$. Let $f, g: \mathbb{N} \to \{0, 1\}$ be two *multiplicative* functions such that for all positive integers $n$, $f(an + b) = g(cn + d)$. Prove that at least one of these two claims are true:
i. $f(an + b) = g(cn + d) = 0$, for all posit... | [
"Assume that the first claim is false, we then prove the truth of the second claim. Letting $n = p_1^{\\alpha_1} \\dots p_t^{\\alpha_t}$ it follows that $f(n) = f(p_1) \\dots f(p_t)$. Thus, the value of $f$ is only depending upon the values of $f$ at prime points. Hence, it suffices to prove that for all large enou... | Iran | 38th Iranian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof only | null | |
0kmw | Problem:
Eight points are chosen on the circumference of a circle, labelled $P_{1}, P_{2}, \ldots, P_{8}$ in clockwise order. A route is a sequence of at least two points $P_{a_{1}}, P_{a_{2}}, \ldots, P_{a_{n}}$ such that if an ant were to visit these points in their given order, starting at $P_{a_{1}}$ and ending at... | [
"Solution:\n\nHow many routes are there if we are restricted to $n$ available points, and we must use all $n$ of them? The answer is $n 2^{n-2}$: first choose the starting point, then each move after that must visit one of the two neighbors of your expanding region of visited points (doing anything else would preve... | United States | HMMT November 2021 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 8744 | |
0fzq | Problem:
Sei $n$ eine natürliche Zahl. Was ist die maximale Anzahl $1 \times 1$ Quadrate, die man in einem $n \times n$ Quadrat schwarz färben kann, sodass in jedem $2 \times 2$ Quadrat höchstens 2 kleine Quadrate schwarz gefärbt sind? | [
"Solution:\n\n- $n \\equiv 0 \\bmod 2$: Dieser Fall ist relativ simpel: Wir bedecken das Feld mit $2 \\times 2$ Feldern. In jedem dieser Felder darf es gemäss Voraussetzung maximal 2 schwarze Felder haben. Wenn wir immer die unteren zwei Felder schwarz färben, bekommen wir eine gültige Färbung:\n\n![](attached_imag... | Switzerland | IMO-Selektion | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | For even n: n^2/2. For odd n: n(n+1)/2. | |
04zm | Find the last digit of the number $1^1 + 2^2 + 3^3 + \dots + 2011^{2011}$. | [
"Consider the sum modulo $2$ and modulo $5$. As powers of odd numbers are odd and powers of even numbers are even, the number of odd summands equals the number of odd elements in set $\\{1, \\dots, 2011\\}$. As there are an even number of odd elements in this set, the sum given in the problem is even.\n\nConcerning... | Estonia | Selected Problems from the Final Round of National Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | English | proof and answer | 8 | |
0l3c | Problem:
In each cell of a $4 \times 4$ grid, one of the two diagonals is drawn uniformly at random. Compute the probability that the resulting 32 triangular regions can be colored red and blue so that any two regions sharing an edge have different colors. | [
"Solution:\n\nGive each cell coordinates from $(1,1)$ to $(4,4)$.\n\nClaim. The grid has a desired coloring if and only if every vertex not on the boundary meets an even number of edges and diagonals.\n\nProof. If this were not the case, the odd number of regions around the vertex would have to alternate between th... | United States | HMMT February 2024 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 1/512 | |
0g2n | Problem:
Soit $n$ un nombre entier strictement positif. Déterminer s'il existe un nombre réel $\varepsilon>0$ (dépendant de $n$ ) tel que, pour tous nombres réels strictement positifs $x_{1}, x_{2}, \ldots, x_{n}$, on ait
$$
\sqrt[n]{x_{1} x_{2} \cdots x_{n}} \leq (1-\epsilon) \cdot \frac{x_{1}+x_{2}+\cdots+x_{n}}{n}+... | [
"Solution:\n\nIl existe un tel $\\epsilon$ pour tout $n \\in \\mathbb{N}$.\n\nSi un tel $\\epsilon$ existe pour chaque $n$, on s'attend à ce qu'il dépende de $n$. En gardant à l'esprit la chaîne d'inégalités $\\mathrm{AM} \\geq \\mathrm{GM} \\geq \\mathrm{HM}$, on voit que le membre de droite devient plus grand si ... | Switzerland | Selektion | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | Yes; take ε = 1/n. | |
01iu | A robot moves along the lines of an infinitely large square grid. It starts at an intersection and turns $90^\circ$ at each intersection it encounters. At some point it reaches its starting point without having visited any other point twice. What are the possible path lengths of the robot? | [
"*Solution:* See Solution of C2."
] | Baltic Way | Baltic Way 2023 Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Exactly the integers divisible by four other than eight; that is, 4 and all integers at least 12 that are multiples of 4. | |
03er | Points $D$ and $E$ are on sides $BC$ and $AC$ of $\triangle ABC$. Lines $AD$ and $BE$ intersect at point $S$. Point $F$ is on side $AB$ and lines $FE$ and $FD$ intersect line $l$ passing through $C$ and parallel to $AB$ at points $P$ and $Q$. Prove that if $CP = CQ$, then the points $C$, $S$ and $F$ lie on the same lin... | [
"From the similarities $\\triangle PEC \\sim \\triangle FEA$ and $\\triangle CDQ \\sim \\triangle BDF$ we obtain that\n\n$$\n\\frac{CP}{AF} = \\frac{CE}{AE} \\quad \\text{and} \\quad \\frac{CQ}{BF} = \\frac{CD}{BD}.\n$$\nTherefore $\\frac{CP}{CQ} = \\frac{CE}{AE} \\cdot \\frac{AF}{BF} \\cdot \\frac{BD}{DC}$. By con... | Bulgaria | Bulgarian Winter Tournament | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
07mc | A square is inscribed in an acute-angled triangle such that two of its vertices lie on one side of the triangle and one of its vertices on each of the other two sides of the triangle. Prove that the largest such square is that with two of its vertices on the smallest side of the triangle. | [
"Let $DEFG$ be a square inscribed in $\\triangle ABC$ as described in the problem, let $AH$ be the altitude at $A$ and $h_a = |AH|$ its length. By $a$, $b$, $c$ we denote the lengths of the sides of $\\triangle ABC$ in the usual way. Let $x_a$ denote the side length of the square if one of its sides lies on the sid... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0dtz | Find the smallest positive integer $n$, or show no such $n$ exists, such that one can find infinitely many distinct $n$-tuplets of positive rationals $(a_1, a_2, \dots, a_n)$ satisfying:
$$
\sum_{i=1}^{n} a_i \in \mathbb{Z}, \quad \sum_{i=1}^{n} \frac{1}{a_i} \in \mathbb{Z}.
$$ | [
"The smallest such $n$ is $n=3$. For $n=2$, let $a_1 = \\frac{a}{b}$, $a_2 = \\frac{c}{d}$ where $\\gcd(a,b) = \\gcd(c,d) = 1$. Then we must have $bd \\mid ad+bc$, $ac \\mid ad+bc$. Taking mod $d$ in the first condition, we get that $d \\mid bc$ but since $c,d$ are coprime we have that $d|b$. Taking mod $b$, we get... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 3 | |
00en | There is a lamp on each cell of an infinite square grid. Initially, all lamps are *off*. A move consists of choosing either a $3 \times 3$, $4 \times 4$ or $5 \times 5$ square contained on the grid and switch all lamps inside that square from *on* to *off* or vice versa.
a. Prove that for any finite set $S$ of lamps i... | [
"a. It suffices to show an algorithm that leaves exactly one lamp *on* (afterwards, the same algorithm can be repeated using appropriate translations). We will present two such algorithms.\n\n**ALGORITHM 1.** Stacking four $3 \\times 3$ squares we get a $12 \\times 3$ rectangle, and stacking three $4 \\times 4$ squ... | Argentina | Cono Sur Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
01ey | Does there exist a polynomial $f(x)$ with integer coefficients of degree four such that for any integer $k$, the polynomial $f(x) + k$ is either irreducible or has a rational root? | [
"The answer is yes, and our example will be the polynomial $f(x) = x^4 + 2x$. Assume that there exist a constant $k$ such that $f(x) + k$ is reducible but has no rational zeros. This means that $f(x) + k$ factors as the product of two irreducible quadratic polynomials. Since $f(x) + k$ is monic so must the two fact... | Baltic Way | Baltic Way 2019 | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | Yes; for example f(x) = x^4 + 2x. | |
0cby | a) Infinitely many pairs of real numbers $(x, y)$ exist such that $x, y \in [0, \sqrt{3}]$ and the following equality holds: $x \cdot \sqrt{3-y^2} + y \cdot \sqrt{3-x^2} = 3$;
b) **No** pair of rational numbers $(x, y)$ exists such that $x, y \in [0, \sqrt{3}]$ and the following equality holds: $x \cdot \sqrt{3-y^2} +... | [
"a) Any pair $(a, \\sqrt{3-a^2})$, with $a \\in [0, 1]$, is a solution.\n\nb) By squaring the equality $y \\cdot \\sqrt{3-x^2} = 3 - x \\cdot \\sqrt{3-y^2}$, we deduce that $(\\sqrt{3-y^2} - x)^2 = 0$, therefore $x^2 + y^2 = 3$. (1)\nAssume that there are numbers $x, y \\in \\mathbb{Q}_+ \\cap [0, \\sqrt{3}]$, for ... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
02i9 | Problem:
Uma bola de futebol é feita com 32 peças de couro. Dessas peças 12 são pentágonos regulares idênticos e as outras 20 são hexágonos, também regulares e idênticos. Os lados dos pentágonos são iguais aos lados dos hexágonos. Para unir dois lados de duas dessas peças é necessária uma costura. Quantas são as costu... | [
"Solution:\n\nSe somarmos os números de lados de todos os polígonos (20 hexágonos e 12 pentágonos) que compõem a superfície da bola, obteremos um valor que é duas vezes o número de costuras, pois cada costura é lado comum de exatamente dois polígonos. Assim, temos que $2 \\times$ (número de costuras) $= 12 \\times ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | MCQ | C | |
007n | Let $ABCD$ be a trapezoid with $AB \parallel CD$, $AB > CD$, and such that $BC = CD = DA$. Points $E$ and $F$ divide $AB$ into three equal parts; $E$ is between $A$ and $F$. Lines $CF$ and $DE$ intersect at $P$. Prove that $\angle APB = \angle DAB$. | [
"Extend $PA$ and $PB$ to meet $CD$ at $X$ and $Y$ respectively. Since $XY \\parallel AB$, Thales' theorem yields $\\frac{XD}{AE} = \\frac{PD}{PE} = \\frac{CD}{FE}$. Also $AE = FE$, so $XD = CD$, i.e. $D$ is the midpoint of $XC$. In addition $CD = DA$, hence $DA = DX = DC$. Thus triangle $XCA$ is right at $A$, so th... | Argentina | National Olympiad of Argentina | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0hhb | We call a polynomial *mixed* if it has both positive and negative coefficients (we do not consider zero to be either positive or negative). Is it true that the product of two mixed polynomials is always a mixed polynomial? | [
"Here is a counterexample:\n$$\nf(x) = x^5 + x^4 - x^3 - x^2 + x + 1 = (x^4 - x^2 + 1)(x + 1)\n$$\n$$\ng(x) = x^4 - x^3 + 2x^2 - x + 1 = (x^2 - x + 1)(x^2 + 1).\n$$\n\n\\begin{aligned}\nf(x)g(x) &= \\\\\n&= (x^4 - x^2 + 1)(x + 1)(x^2 - x + 1)(x^2 + 1) = \\\\\n&= (x^6 + 1)(x^3 + 1) = x^9 + x^6 + x^3 + 1.\n\\end{alig... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | No | |
0bkl | Let $a$ be a real number in the open interval $(0, 1)$, let $n$ be a positive integer and let $f_n: \mathbb{R} \to \mathbb{R}$, $f_n(x) = x + x^2/n$. Show that
$$
\frac{a(1-a)n^2 + 2a^2n + a^3}{(1-a)^2n^2 + a(2-a)n + a^2} < \underbrace{(f_n \circ \dots \circ f_n)}_{n} (a) < \frac{an + a^2}{(1-a)n + a}.
$$ | [
"Let $a_k = \\underbrace{(f_n \\circ \\cdots \\circ f_n)}_{k}(a)$, $k \\in \\mathbb{N}$, and notice that\n$$\n1/a_{k+1} = 1/a_k - 1/(a_k + n), \\quad k \\in \\mathbb{N},\n$$\nto deduce that $1/a_n = 1/a - \\sum_{k=0}^{n-1} 1/(a_k + n)$, so\n$$\n1/a - n/(a+n) < 1/a_n < 1/a - n/(a_n + n), \\quad (*)\n$$\n$a_n < \\fra... | Romania | 65th NMO Selection Tests for BMO and IMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0b6u | Prove that the real numbers $a$, $b$ fulfill
$$(x + y + z)^3 + a(x + y + z)(xy + xz + yz) + bxyz \ge 0,$$
for all $x, y, z \ge 0$, if and only if there exist $r \ge 0$, $s \ge 0$ such that $a = -4 + r$, $b = 9 - 9r + s$. | [] | Romania | Shortlisted Problems for the Romanian NMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0cad | Problem:
Fie $n \geq 2$ un număr natural. Pe o tablă $n \times n$ se aşază $n$ turnuri astfel încât să nu existe două care să se atace. Toate turnurile se mişcă simultan o dată şi au voie să se mişte doar într-un pătrat adiacent celui în care se află.
Determinaţi toate valorile lui $n$ pentru care există o aşezare a t... | [] | Romania | Primul test de selecţie pentru OBMJ | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | n is even | |
0d15 | Let $ABC$ be an arbitrary triangle. A circle passes through $B$ and $C$ and intersects the lines $AB$ and $AC$ in $D$ and $E$, respectively. The projections of the points $B$ and $E$ on $CD$ are denoted by $B'$ and $E'$, respectively. The projections of the points $D$ and $C$ on $BE$ are denoted by $D'$ and $C'$, respe... | [
"Let $I$ be the intersection point of the lines $BE$ and $CD$. The quadrilaterals $BD'B'D$ and $CE'C'E$ are cyclic, so $\\overline{BDB'} = \\overline{B'D'I}$ and $\\overline{CEC'} = \\overline{IE'C'}$. Since $BDEC$ is also cyclic, $\\overline{BDB'} = \\overline{CEC'}$. It follows that $\\overline{B'D'I} = \\overlin... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0j9o | Problem:
Knot is ready to face Gammadorf in a card game. In this game, there is a deck with twenty cards numbered from $1$ to $20$. Each player starts with a five card hand drawn from this deck. In each round, Gammadorf plays a card in his hand, then Knot plays a card in his hand. Whoever played a card with greater val... | [
"Solution:\nAnswer: $2982$\n\nKnot can only lose if all of his cards are lower than $10$; if not he can win by playing the lowest card that beats Gammadorf's card, or if this is not possible, his lowest card, each turn. There are $\\binom{7}{5} = 21$ losing hands, so he has $\\binom{15}{5} - \\binom{7}{5}$ possible... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 2982 | |
04tu | Let $k$ be a real parameter. Determine the number of real solutions to the system
$$
\begin{align*}
x^2 + kxy + y^2 &= z, \\
y^2 + kyz + z^2 &= x, \\
z^2 + kzx + x^2 &= y,
\end{align*}
$$
in terms of $k$. | [
"We distinguish several cases.\n\nFirst, assume $x = y = z$. Then the whole system reduces to $(k+2)x^2 = x$. Its solution is a triplet $(0, 0, 0)$ for any $k$ and moreover triplet $\\left(\\frac{1}{k+2}, \\frac{1}{k+2}, \\frac{1}{k+2}\\right)$ if $k \\neq -2$.\n\nLet's get back to the original system. Subtracting ... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | Number of real solutions as a function of k:
- For k in (-∞, -2): 8 solutions
- For k = -2: 4 solutions
- For k in (-2, 1 − 2√2): 8 solutions
- For k = 1 − 2√2: 5 solutions
- For k in (1 − 2√2, 1 + 2√2): 2 solutions
- For k = 1 + 2√2: 5 solutions
- For k in (1 + 2√2, ∞): 8 solutions | |
0b8z | $$
x_{n+1} = \left| x_n - \frac{1}{n} \right|, \quad \text{for } n \ge 1.
$$
Prove that it has a finite limit, and calculate it. | [
"Let $N$ be an arbitrary positive integer. There exists a least $k \\in \\mathbb{N}$ such that $x_N < \\frac{1}{N} + \\frac{1}{N+1} + \\dots + \\frac{1}{N+k}$, since the harmonic series is divergent. Then $x_{N+k} = x_N - \\left(\\frac{1}{N} + \\dots + \\frac{1}{N+k-1}\\right) < \\frac{1}{N+k} \\le \\frac{1}{N}$. A... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 0 | |
0b08 | Problem:
Let $(0,0)$, $(10,0)$, $(10,8)$, and $(0,8)$ be the vertices of a rectangle on the Cartesian plane. Two lines with slopes $-3$ and $3$ pass through the rectangle and divide the rectangle into three regions with the same area. If the lines intersect above the rectangle, find the coordinates of their point of in... | [] | Philippines | Philippines Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | (5, 9) | |
07d0 | Find all functions $f: \mathbb{Z}^+ \rightarrow \mathbb{Z}^+$ satisfying the following conditions:
* For all $x, y \in \mathbb{Z}^+$, $f(x) + f(y)$ is divisible by $x + y$.
* For any integer $x \ge 1395$, the inequality $2f(x) \le x^3$ holds. | [
"First note that setting $x = y$ in the first assumption implies $f(x)$ is divisible by $x$ and therefore we can write $f(x) = xg(x)$ where $g(x)$ is always an integer. Now set $g(1) = a$ and $g(2) = b$. We are going to show that for an odd and sufficiently large number $x$, $g(x)$ is equal to $a$. To prove this, w... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | All functions are f(x) = a x for all positive integers x, where a is a positive integer satisfying a ≤ 1395^2 / 2. | |
03k4 | Problem:
A deck of $2n+1$ cards consists of a joker and, for each number between $1$ and $n$ inclusive, two cards marked with that number. The $2n+1$ cards are placed in a row, with the joker in the middle. For each $k$ with $1 \leq k \leq n$, the two cards numbered $k$ have exactly $k-1$ cards between them. Determine ... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Possible for n = 3, 4, 7, 8. Impossible for n = 1, 2, 5, 6, 9, 10. In general, such an arrangement exists iff n ≡ 0 or 3 (mod 4). | |
0a8k | Problem:
In two bowls there are in total $N$ balls, numbered from $1$ to $N$. One ball is moved from one of the bowls into the other. The average of the numbers in the bowls is increased in both of the bowls by the same amount, $x$. Determine the largest possible value of $x$. | [
"Solution:\n\nConsider the situation before the ball is moved from urn one to urn two. Let the number of balls in urn one be $n$, and let the sum of numbers in the balls in that urn be $a$. The number of balls in urn two is $m$ and the sum of numbers $b$. If $q$ is the number written in the ball which was moved, th... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 16 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 1/2 | |
0dxq | Problem:
Označimo z $D$ razpolovišče stranice $AB$ ostrokotnega trikotnika $ABC$. Naj bosta $A'$ in $B'$ taki točki na daljicah $AC$ in $BC$, da sta trikotnika $ADA'$ in $DBB'$ enakokraka s skupnim vrhom v $D$. Pokaži: če je premica $CD$ pravokotna na premico $A'B'$, je trikotnik $ABC$ enakokrak. | [
"Solution:\n\nKer je $D$ razpolovišče stranice $AB$, velja $|AD| = |DB|$. Trikotnika $ADA'$ in $DBB'$ sta enakokraka z vrhom $D$, zato je $|A'D| = |AD| = |BD| = |B'D|$ in je trikotnik $A'DB'$ enakokrak z vrhom $D$.\n\nOznačimo z $E$ presečišče $A'B'$ in $CD$. Če je $DE$ pravokotna na $A'B'$, velja $|A'E| = |EB'|$, ... | Slovenia | 51. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
01nj | Determine all pairs of functions $f, g$, $f : \mathbb{R} \to \mathbb{R}$, $g : \mathbb{R} \to \mathbb{R}$, that satisfy the equality
$$
g(f(x + y)) = f(x) + (2x + y)g(y)
$$
for all real $x, y$. | [] | Belarus | Belorusija 2012 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | Either f(x) = x^2 + c and g(x) = x for an arbitrary real constant c, or f(x) = 0 and g(x) = 0. | |
052l | In a plane there is a triangle $ABC$. Line $AC$ is tangent to circle $c_A$ at point $C$ and circle $c_A$ passes through point $B$. Line $BC$ is tangent to circle $c_B$ at point $C$ and circle $c_B$ passes through point $A$. The second intersection point $S$ of circles $c_A$ and $c_B$ coincides with the incenter of tria... | [
"By the tangent-secant theorem we have $\\angle BCS = \\angle CAS$ and $\\angle ACS = \\angle CBS$ (see fig. 5). The incenter of a triangle is the point of intersection of angle bisectors, therefore $\\angle CAB = 2\\angle CAS = 2\\angle BCS = \\angle BCA$ and $\\angle CBA = 2\\angle CBS = \\angle ACS = \\angle BCA... | Estonia | Open Contests | [
"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 | |
0h78 | Find at least one pair of positive integers $(x, y)$, which satisfies an equality:
$$
\frac{1}{2}(x^2 - y^3) = 2016.
$$ | [
"Rewrite the given condition in the following way:\n$$\nx^2 - y^3 = 2 \\cdot 2016 = 2^6 \\cdot 3^2 \\cdot 7 = 2^6 \\cdot (64 - 1) = 2^{12} - 2^6.\n$$\nNow it is easy to see how to find the solutions."
] | Ukraine | 56th Ukrainian National Mathematical Olympiad, Third Round | [
"Number Theory > Diophantine Equations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | final answer only | (64, 4) | |
0aoa | Problem:
Define $f(x) = \frac{a^{x}}{a^{x} + \sqrt{a}}$ for any $a > 0$. Evaluate
$$
\sum_{i=1}^{2012} f\left(\frac{i}{2013}\right).
$$ | [] | Philippines | AREA STAGE | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1006 | |
0ca1 | Problem:
Fie $n \geq 2$ un număr natural cu proprietatea că mulţimea rădăcinilor de ordin $n$ ale unităţii are mai puţin de $2^{[\sqrt{n}]}-1$ submulţimi cu suma elementelor nulă. Arătaţi că $n$ este prim. (Am notat cu $[x]$ partea întreagă a numărului real $x$.) | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ Etapa a III-a | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
08qb | Problem:
Find all prime numbers $p$ and nonnegative integers $x \neq y$ such that $x^{4}-y^{4}=p\left(x^{3}-y^{3}\right)$. | [
"Solution:\nIf $x=0$ then $y=p$ and if $y=0$ then $x=p$. We will show that there are no other solutions.\n\nSuppose $x, y>0$. Since $x \\neq y$, we have\n$$\np\\left(x^{2}+x y+y^{2}\\right)=(x+y)\\left(x^{2}+y^{2}\\right)\n$$\nIf $p$ divides $x+y$, then $x^{2}+y^{2}$ must divide $x^{2}+x y+y^{2}$ and so it must als... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All solutions are x = 0, y = p or x = p, y = 0 for any prime p. | |
0c9l | Problem:
Determinaţi toate perechile $(m, n)$ de numere naturale impare pentru care
$$
n \mid 3m+1 \text{ şi } m \mid n^{2}+3.
$$ | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | (1,1), (43,13), (49,37) | |
0462 | Given an integer $n > 1$ and an integer $a$ that is coprime with $n$. There is a country consisting of $n$ islands $D_1, D_2, \dots, D_n$. For any two different islands $D_i$ and $D_j$, there is a one-way ferry from $D_i$ to $D_j$ if and only if $ij \equiv ia \pmod{n}$. A tourist hopes to visit as many islands as possi... | [
"Let $x(n) = \\sum_{v_p(n) \\ge 2} 1$ and $y(n) = \\sum_{v_p(n)=1} 1$. Let the maximum number of islands that can be visited be denoted as $k(n)$. Then we have:\n$$\nk(n) = \\begin{cases} 3x(n) + 2y(n) + 1, & v_2(n) \\neq 1, \\\\ 3x(n) + 2y(n), & v_2(n) = 1. \\end{cases}\n$$\nFor $x, y \\in \\mathbb{Z}$ and $m \\in... | China | 2023 Chinese IMO National Team Selection Test | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | Let x(n) = sum over primes p dividing n with v_p(n) ≥ 2 of 1, and y(n) = sum over primes p dividing n with v_p(n) = 1 of 1. Then the maximum number of islands is
k(n) = { 3 x(n) + 2 y(n) + 1, if v_2(n) ≠ 1; 3 x(n) + 2 y(n), if v_2(n) = 1 }. | |
0h7f | 30 children – boys and girls – formed a circle. It occurred that there is no child such that both its neighbors are boys. What is the least possible number of girls there? | [
"Consider a group of consecutive boys. There are less than 3 boys in this group. Also consider a group of consecutive girls. There are at least 2 girls in the group. These groups follow each other hence their amounts are equal. Assume that it may occur that there are 15 boys and 15 girls. Then there are at least 8 ... | Ukraine | 56th Ukrainian National Mathematical Olympiad, Third Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 16 | |
0eqf | If $a \bullet b$ means the value of $ab + a + b$, and $5 \bullet x = 35$, the value of $x$ must be
(A) 5
(B) 7
(C) 9
(D) 12
(E) 15 | [
"By definition of $\\bullet$, we know $5 \\bullet x = 5x + 5 + x = 6x + 5$. If this is $35$, then $6x = 30$, so $x = 5$"
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | A | |
0fwb | Problem:
Ein Paar $(r, s)$ natürlicher Zahlen heisst gut, falls ein Polynom $P$ mit ganzen Koeffizienten und paarweise verschiedene ganze Zahlen $a_{1}, \ldots, a_{r}$ und $b_{1}, \ldots, b_{s}$ existieren, sodass gilt
$$
P\left(a_{1}\right)=P\left(a_{2}\right)=\ldots=P\left(a_{r}\right)=2 \quad \text{ und } \quad P\l... | [
"Solution:\n\nEin Paar $(r, s)$ ist genau dann gut, wenn auch $(s, r)$ gut ist. Dazu ersetze man $P(x)$ einfach durch $7-P(x)$. Ausserdem ist mit $(r, s)$ auch jedes Paar $(u, v)$ mit $u \\leq r, v \\leq s$ gut. Sei nun $(r, s)$ ein gutes Paar und $P$ wie in der Aufgabenstellung. Das Polynom $P(x)-2$ besitzt dann d... | Switzerland | IMO Selektion | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All good pairs are (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2). | |
0ixk | Let $s_1, s_2, s_3, \dots$ be an infinite, non-constant sequence of rational numbers, meaning it is not the case that $s_1 = s_2 = s_3 = \dots$. Suppose that $t_1, t_2, t_3, \dots$ is also an infinite, non-constant sequence of rational numbers with the property that $(s_i - s_j)(t_i - t_j)$ is an integer for all $i$ an... | [
"**Solution 1 (By Gabriel Carroll).** First, we claim there exist $i, j$ such that $(s_i-s_j)(t_i-t_j) \\neq 0$. Indeed, for any fixed $i$, because the sequence $s_1, s_2, \\dots$ is non-constant, there is some $j$ such that $s_j \\neq s_i$. If $t_j \\neq t_i$ the claim follows, so suppose $t_j = t_i$. Because the ... | United States | USAMO 2009 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > A... | null | proof only | null | |
0fu2 | Problem:
Für die positiven reellen Zahlen $a, b, c$ gelte $a b c=1$. Beweise die folgende Ungleichung:
$$
\frac{a b}{a^{5}+a b+b^{5}}+\frac{b c}{b^{5}+b c+c^{5}}+\frac{c a}{c^{5}+c a+a^{5}} \leq 1
$$ | [
"Solution:\n\nEs gilt $a^{5}+b^{5} \\geq a^{2} b^{2}(a+b)$, z.B. nach AM-GM oder Bunching. Daraus folgt mit $a b c=1$\n$$\n\\frac{a b}{a^{5}+a b+b^{5}} \\leq \\frac{a b}{a^{2} b^{2}(a+b)+a b}=\\frac{a b c^{2}}{a^{2} b^{2} c^{2}(a+b)+a b c^{2}}=\\frac{c}{a+b+c}\n$$\nNach dieser und den analogen Abschätzungen ist die... | Switzerland | IMO Selektion | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0c3g | Let $ABCD$ be a cyclic quadrangle, and let its diagonals, $AC$ and $BD$, cross at $X$. Let $I$ be the incenter of the triangle $XBC$, and let $J$ be the center of the circle tangent to the side $BC$ and the extensions of the sides $AB$ and $DC$ beyond $B$ and $C$, respectively. Prove that the line $IJ$ bisects the arc ... | [
"\n\nLet $M$ be the midpoint of the arc $BC$ of the circle $ABCD$, not containing the vertices $A$ and $D$ of the quadrangle.\n\nAn easy angle chase shows the vertices $B$ and $C$, the incenter $K$ of the triangle $ABC$, and its $A$-excenter $K_A$ equally distanced from $M$, so they all lie... | Romania | 69th NMO Selection Tests for BMO and IMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"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 c... | null | proof only | null | |
05zf | Problem:
Quentin et Timothé jouent à un jeu. D'abord, Quentin choisit un nombre premier $p > 2$, puis Timothé choisit un entier strictement positif $n_0$. Quentin choisit alors un entier $n_1 > n_0$ et calcule $s_1 = n_0^{n_1} + n_1^{n_0}$; puis Timothé choisit un entier $n_2 > n_1$ et calcule $s_2 = n_1^{n_2} + n_2^{... | [
"Solution:\n\nNous allons montrer que Timothé a une stratégie gagnante. Notons que Quentin va choisir les $n_{2k+1}$ et Timothé va choisir les $n_{2k}$. On fait d'abord quelques remarques.\n\n- **Remarque 1** : Remarquons d'abord que si l'un des joueurs choisit $n_k \\equiv 0 \\pmod{p}$ et ne gagne pas à cette étap... | France | Préparation Olympique Française de Mathématiques - Envoi 3: Arithmétique | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Timothé has a winning strategy. | |
0egv | Problem:
Koliko je takih trimestnih naravnih števil, pri katerih se poljubni dve števki razlikujeta za vsaj 3?
(A) 80
(B) 88
(C) 92
(D) 100
(E) 648 | [
"Solution:\nObravnavajmo vse možnosti glede na prvo števko števila.\nČe je prva števka enaka $1$, potem moramo drugo in tretjo števko izbrati iz množice šestih zaporednih števil $\\{4,5,6,7,8,9\\}$, tako da se razlikujeta za vsaj $3$. To lahko storimo na $12$ načinov, tj. $47,48,49,58,59,69,74,84,85,94,95,96$. Podo... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | MCQ | D | |
0ev9 | $n$ students with different names are forming a circle and standing toward the center of the circle. Initially $n$ name tags of the students are distributed to them at random. The students repeat the following process:
[process] Every student having one's own name tag gets out of the circle, and the remaining students ... | [
"Let $A$ be a finite set of positive integers and $\\pi$ be a permutation (=bijection) on $A$. A pair $(x, \\pi(x)) \\in A^2$ is called a \"up\" if $x < \\pi(x)$ holds, and $up(\\pi)$ is defined by the number of up's in $\\pi$.\n\nGiven a permutation $\\pi$ on $A = \\{x_1 < x_2 < \\dots < x_m\\}$, define a permutat... | South Korea | Korean Mathematical Olympiad Final Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | n! - 4^n + n 3^n - C(n,2) 2^n + C(n,3) | |
0jz8 | Problem:
Each of the positive integers $a_{1}, a_{2}, \ldots, a_{n}$ is less than $2016$, and the least common multiple of any two is greater than $2016$. Show that
$$
\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}<1+\frac{n}{2016}.
$$ | [
"Solution:\nBy considering multiples of the $a_{i}$ which are less than $2016$ (these don't overlap by condition) we derive\n$$\n\\sum\\left\\lfloor\\frac{2016}{a_{i}}\\right\\rfloor \\leq 2016\n$$\nUpon using the fact that $\\lfloor x\\rfloor>x-1$, we then obtain\n$$\n\\sum\\left(\\frac{2016}{a_{i}}-1\\right)<2016... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
07j5 | Give $n$ black points on the plane such that no three are collinear and the distance between any two of them are pairs distinct. Starting with an arbitrary point and make it red. At each step, we draw the smallest segment such that one of its endpoints is red and the other is black—that has not yet been drawn—and doesn... | [
"The answer is negative. We shall provide the following formations."
] | Iran | 41th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | No | |
08nn | Problem:
Find the largest positive integer $n$ for which the inequality
$$
\frac{a+b+c}{a b c+1}+\sqrt[n]{a b c} \leq \frac{5}{2}
$$
holds for all $a, b, c \in[0,1]$. Here $\sqrt[1]{a b c}=a b c$. | [
"Solution:\nLet $n_{\\max}$ be the sought largest value of $n$, and let $E_{a, b, c}(n)=\\frac{a+b+c}{a b c+1}+\\sqrt[n]{a b c}$. Then $E_{a, b, c}(m)-E_{a, b, c}(n)=\\sqrt[m]{a b c}-\\sqrt[n]{a b c}$ and since $a b c \\leq 1$ we clearly have $E_{a, b, c}(m) \\geq E_{a, b, c}(n)$ for $m \\geq n$. So if $E_{a, b, c}... | JBMO | JBMO Shortlist | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 3 | |
0a70 | Problem:
The positive integer $n$ has the following property: if the three last digits of $n$ are removed, the number $\sqrt[3]{n}$ remains. Find $n$. | [
"Solution:\nIf $x = \\sqrt[3]{n}$, and $y$, $0 \\leq y < 1000$, is the number formed by the three last digits of $n$, we have\n$$\nx^3 = 1000x + y\n$$\nSo $x^3 \\geq 1000x$, $x^2 > 1000$, and $x > 31$. On the other hand, $x^3 < 1000x + 1000$, or $x(x^2 - 1000) < 1000$. The left hand side of this inequality is an in... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 2 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 32768 | |
0190 | Suppose that the quadrilateral $ABCD$ satisfies $\angle ABD = 30^\circ$, $\angle CDB = 20^\circ$ and $\angle BCA = \angle ACD = 40^\circ$. Determine $\angle DAC$. | [
"Let $E$ be the intersection point of the diagonals $AC$ and $BD$. Let $EF$ be the bisector of $\\angle DEC$, with $F$ on $DC$.\n\n\n\nThen the triangles $BEC$ and $CFE$ are congruent ($EFCB$ is a kite). So $EF = EB = EA$, and $ADFE$ is a kite. It follows that $\\angle DAC = 100^\\circ$."
] | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 100° |
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