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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0eqj | A palindromic number is a number that reads the same backwards as it does forwards.
For example: $15751$ and $909$.
a) What is the smallest palindromic number greater than $2016$?
b) What is the largest palindromic number less than $2016$? | [
"a) $2112$, since the next smaller palindrome is $2002$, which is less than $2016$.\n\nb) $2002$, since $2112$ is the next palindrome."
] | South Africa | South African Mathematics Olympiad Third Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | final answer only | a) 2112; b) 2002 | |
0cmg | Find all positive integers $n$ such that there exist nonzero real numbers $a, b, c, d$ for which the polynomial
$$
(ax + b)^{1000} - (cx + d)^{1000}
$$
after expanding all brackets (and collecting terms) has exactly $n$ nonzero coefficients. | [
"The answers are $1001$, $1000$, and $500$.\n\nIt is clear that there exist such polynomials with $1001$ and $1000$ nonzero coefficients (for example, $(2x+2)^{1000} - (x+1)^{1000}$ and $(2x+1)^{1000} - (x+1)^{1000}$).\n\nSuppose that in our polynomial there are two coefficients equal to zero — for $x^i$ and $x^j$ ... | Russia | Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English; Russian | proof and answer | 1001, 1000, 500 | |
0f3t | Problem:
There are several settlements around Big Lake. Some pairs of settlements are directly connected by a regular shipping service. For all $A \neq B$, settlement $A$ is directly connected to $X$ iff $B$ is not directly connected to $Y$, where $B$ is the next settlement to $A$ counterclockwise and $Y$ is the next ... | [
"Solution:\n\nSuppose there are $n$ settlements $A_1, A_2, \\ldots, A_n$ in counterclockwise order around the lake. We will use cyclic indices, so that $A_{n + 1}$ means $A_1$, and so on. WLOG $A_1$ has a direct service to $A_2$. Then it follows that $A_2$ does not have a direct service to $A_3$, so $A_3$ does have... | Soviet Union | ASU | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
08wh | Let $ABC$ be an acute triangle for which $\angle BAC = 30^\circ$. Take a point $X$ inside of the triangle $ABC$ in such a way that $\angle XBC = \angle XCB = 30^\circ$. Also, take points $P$ and $Q$ on the straight lines $BX$ and $CX$, respectively, in such a way that $AP = BP$ and $AQ = CQ$ are satisfied. Let $M$ be t... | [
"Let $B'$ be the point symmetric to $B$ with respect to $P$, and $C'$ be the point symmetric to $C$ with respect to $Q$. Since $B$ is the mid-point of the line segment $PP'$ and $M$ is the mid-point of the line segment $BC$ we see that $PM // BC'$, and similarly, we have $QM // C'B$. Thus, it suffices to show that ... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nin... | English | proof only | null | |
0ho1 | Problem:
In $\triangle A B C$, let $O$ be the circumcenter, $D$ the foot of the altitude from $A$ to $B C$, and $E$ the foot of the altitude from $B$ to $A C$. Show that $D E \perp C O$. | [
"Solution:\n\nBy AA, $\\triangle A C D \\sim \\triangle B C E$, as $\\angle C$ is shared and both are right triangles. Thus $A C / C D = B C / C E$. It follows by SAS that\n$$\n\\triangle A B C \\sim \\triangle D E C \\Longrightarrow \\angle C D E = \\angle B A C.\n$$\nSince $O$ is the circumcenter of $\\triangle A... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"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 | |
08nb | Problem:
Let $ABC$ be an isosceles triangle with $AB = AC$. Let also $c(K, KC)$ be a circle tangent to the line $AC$ at point $C$ which intersects the segment $BC$ again at an interior point $H$. Prove that $HK \perp AB$. | [
"Solution:\nLet lines $KH$, $AB$ intersect at $M$ (Figure 5a).\nFrom the quadrilateral $KMAC$ we have\n$$\n\\angle KMA = 360^{\\circ} - \\angle A - \\angle ACK - \\angle CKM = 360^{\\circ} - \\angle A - 90^{\\circ} - (180^{\\circ} - 2\\angle KCH) = 90 - \\angle A + 2\\angle KCH = 90 - \\angle A + 2(90^{\\circ} - \\... | JBMO | JBMO Shortlist | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07z6 | Problem:
Sia $ABCD$ un tetraedro generico di cui si conosce la lunghezza $a$ dello spigolo $AB$ e l'area $S$ della proiezione del tetraedro su un piano perpendicolare alla retta per $A$ e $B$.
Determinare il volume del tetraedro. | [
"Solution:\n\nSia $\\pi$ il piano passante per $A$ perpendicolare allo spigolo $AB$ e siano $C'$, $D'$ le proiezioni di $C$ e $D$ su $\\pi$. La proiezione del tetraedro $ABCD$ su $\\pi$ è il triangolo $AC'D'$.\n\n\n\nIl volume del tetraedro $ABCD$ è uguale a quello del tetraedro $ABC'D$ in ... | Italy | XIII GARA NAZIONALE di MATEMATICA | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | a*S/3 | |
0ks3 | Problem:
Alice is once again very bored in class. On a whim, she chooses three primes $p, q, r$ independently and uniformly at random from the set of primes of at most $30$. She then calculates the roots of $p x^{2} + q x + r$. What is the probability that at least one of her roots is an integer? | [
"Solution:\n\nSince all of the coefficients are positive, any root $x$ must be negative. Moreover, by the rational root theorem, in order for $x$ to be an integer we must have either $x = -1$ or $x = -r$. So we must have either $p r^{2} - q r + r = 0 \\Longleftrightarrow p r = q - 1$ or $p - q + r = 0$. Neither of ... | United States | HMMT November 2022 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 3/200 | |
0k1h | Problem:
Suppose $\triangle ABC$ has lengths $AB = 5$, $BC = 8$, and $CA = 7$, and let $\omega$ be the circumcircle of $\triangle ABC$. Let $X$ be the second intersection of the external angle bisector of $\angle B$ with $\omega$, and let $Y$ be the foot of the perpendicular from $X$ to $BC$. Find the length of $YC$. | [
"Solution:\n\nExtend ray $\\overrightarrow{AB}$ to a point $D$. Since $BX$ is an angle bisector, we have $\\angle XBC = \\angle XBD = 180^{\\circ} - \\angle XBA = \\angle XCA$, so $XC = XA$ by the inscribed angle theorem. Now, construct a point $E$ on $BC$ so that $CE = AB$. Since $\\angle BAX \\cong \\angle BCX$, ... | United States | HMMT February 2018 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 13/2 | |
0ek9 | Problem:
Kateri izraz je ekvivalenten danemu izrazu
$$
\frac{\sqrt[3]{4 a^{2} \cdot \sqrt[3]{a b} \cdot (a+b)^{0}}}{\left(8 b^{-2} \sqrt{a}\right)^{-\frac{2}{3}}}
$$
kjer $a, b, a+b \neq 0$?
(A) $2^{\frac{8}{3}} a^{\frac{10}{9}} b^{-\frac{11}{9}}$
(B) $2^{\frac{2}{3}} a^{\frac{5}{6}} b^{\frac{4}{9}}$
(C) $2^{-3} a^... | [
"Solution:\n\nIzraz zapišemo s potencami in dobimo\n$$\n\\frac{4^{\\frac{1}{3}} \\frac{2}{3} a^{\\frac{1}{9}} b^{\\frac{1}{9}} \\cdot 1}{8^{-\\frac{2}{3}} b^{\\frac{4}{3}} a^{-\\frac{1}{3}}}\n$$\nŠtevili $4$ in $8$ zapišemo kot potenco števila $2$ in izraz zapišemo brez ulomka ter dobimo\n$$\n2^{\\frac{2}{3}} \\cdo... | Slovenia | 22. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | MCQ | A | |
0e6o | What is the sum of all the real numbers that solve the equation $|x - 2011| + |x - 2012| = 3$?
(A) 2011 (B) 2012 (C) 2013 (D) 4021 (E) 4023 | [
"There are only two such real numbers: $2010$ and $2013$. Their sum is equal to $4023$."
] | Slovenia | National Math Olympiad 2012 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | E | |
0jf4 | Problem:
Find all triples of real numbers $(a, b, c)$ such that $a^{2}+2 b^{2}-2 b c=16$ and $2 a b-c^{2}=16$. | [
"Solution:\nAnswer: $(4,4,4), (-4,-4,-4)$ (need both, but order doesn't matter)\n\n$a^{2}+2 b^{2}-2 b c$ and $2 a b-c^{2}$ are both homogeneous degree 2 polynomials in $a, b, c$, so we focus on the homogeneous equation $a^{2}+2 b^{2}-2 b c=2 a b-c^{2}$, or $(a-b)^{2}+(b-c)^{2}=0$. So $a=b=c$, and $a^{2}=2 a b-c^{2}... | United States | HMMT November 2013 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (4, 4, 4) and (-4, -4, -4) | |
01z5 | A point $P$ is marked inside the square $ABCD$, and the points $K$, $L$, $M$ and $N$ are marked on its sides $AB$, $BC$, $CD$ and $DA$ respectively. The lines $KP$, $LP$, $MP$ and $NP$ intersect the sides $CD$, $DA$, $AB$ and $BC$ at the points $K_1$, $L_1$, $M_1$ and $N_1$ respectively. It turned out that
$$
\frac{KP}... | [
"Since the sides $AB$ and $CD$ of a square are parallel, the cross-lying angles are equal: $\\angle PKM_1 = \\angle PK_1M$ and $\\angle PM_1K = \\angle PMK_1$. Therefore the triangles $PKM_1$ and $PK_1M$ are similar and have equal ratios $KP : K_1P = M_1P : MP$. Denote $KP : K_1P = k$, then $MP : PM_1 = 1/k$. Simil... | Belarus | Belarus2022 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0gzc | a) Let's look at a table of 3 rows and 2008 columns. In the first row, a random integer is written in each cell of the table, in a non-decreasing order. The numbers of the second row are obtained in the following way: under every number $A$ from the first row is written a number $B$, which equals the number of numbers ... | [
"a.\n1) If $A_{i+1} = A_i$, then the quantity of numbers from the first row smaller than $A_i$ and situated to the left of $A_i$ is the same as the quantity of numbers from the first row smaller than $A_{i+1}$ and situated to the left of $A_{i+1}$. As number $A_i$ can't increase this quantity, that's why $B_{i+1} =... | Ukraine | The Problems of Ukrainian Authors | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 2^{2007} | |
0101 | Problem:
Prove that the arithmetic mean $a$ of $x_{1}, \ldots, x_{n}$ satisfies
$$
\left(x_{1}-a\right)^{2}+\cdots+\left(x_{n}-a\right)^{2} \leqslant \frac{1}{2}\left(\left|x_{1}-a\right|+\cdots+\left|x_{n}-a\right|\right)^{2} .
$$ | [
"Solution:\n\nDenote $y_{i}=x_{i}-a$. Then $y_{1}+y_{2}+\\cdots+y_{n}=0$. We can assume $y_{1} \\leqslant y_{2} \\leqslant \\cdots \\leqslant y_{k} \\leqslant 0 \\leqslant y_{k+1} \\leqslant \\cdots \\leqslant y_{n}$. Let $y_{1}+y_{2}+\\cdots+y_{k}=-z$, then $y_{k+1}+\\cdots+y_{n}=z$ and\n$$\n\\begin{aligned}\ny_{1... | Baltic Way | Baltic Way 1997 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0aq1 | Problem:
Find the largest three-digit number such that the number minus the sum of its digits is a perfect square. | [
"Solution:\n919\nLet $abc$ be a three-digit number such that the difference between the number and the sum of its digits is a perfect square; that is,\n$$\n(100a + 10b + c) - (a + b + c) = 99a + 9b = 9(11a + b)\n$$\nis a perfect square. To maximize the number $100a + 10b + c$, we set $a = 9$, $b = 1$, and $c = 9$."... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 919 | |
0cfd | The acute triangle $ABC$ has the altitude $AD$ and $X$ is a point inside the segment $AD$. The circles $C_1$ and $C_2$ have their centres on the line $BC$ and contain the points $B$ and $X$, respectively $C$ and $X$. The circle $C_1$ meets again the altitude from $B$ of the triangle $ABC$ at $N$ and the side $AB$ at $M... | [] | Romania | 74th NMO Shortlisted Problems | [
"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",
"Geometry > Plane Geometry > Advanced Configurations > Isogon... | English | proof only | null | |
03v5 | Find the smallest constant $a > 1$, such that for any point $P$ inside a square $ABCD$ there exist two triangles among $\triangle PAB$, $\triangle PBC$, $\triangle PCD$, $\triangle PDA$, with the ratio between their areas belonging to the interval $[a^{-1}, a]$. (Posed by Li Weigu) | [
"$a_{\\min} = \\frac{1+\\sqrt{5}}{2}$.\n\nWrite $\\varphi = \\frac{1+\\sqrt{5}}{2}$. We may assume that each edge has length $\\sqrt{2}$. For any point $P$ inside the square $ABCD$, let $S_1, S_2, S_3, S_4$ denote the area of $\\triangle PAB$, $\\triangle PBC$, $\\triangle PCD$, $\\triangle PDA$ respectively; we ca... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | (1+sqrt{5})/2 | |
08p2 | Problem:
Let $a, b, c$ be positive real numbers such that $a b c=8$. Prove that
$$
\frac{a b+4}{a+2}+\frac{b c+4}{b+2}+\frac{c a+4}{c+2} \geq 6
$$ | [
"Solution:\nWe have $a b+4=\\frac{8}{c}+4=\\frac{4(c+2)}{c}$ and similarly $b c+4=\\frac{4(a+2)}{a}$ and $c a+4=\\frac{4(b+2)}{b}$. It follows that\n$$\n(a b+4)(b c+4)(c a+4)=\\frac{64}{a b c}(a+2)(b+2)(c+2)=8(a+2)(b+2)(c+2)\n$$\nso that\n$$\n\\frac{(a b+4)(b c+4)(c a+4)}{(a+2)(b+2)(c+2)}=8\n$$\nApplying AM-GM, we ... | JBMO | Junior Balkan Mathematics Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0j8c | Let $ABC$ be an acute scalene triangle inscribed in circle $\Omega$. Circle $\omega$, centered at $O$, passes through $B$ and $C$ and intersects sides $AB$ and $AC$ at $E$ and $D$, respectively. Point $P$ lies on major arc $\widehat{BAC}$ of $\Omega$. Prove that lines $BD$, $CE$, $OP$ are concurrent if and only if tria... | [
"We first prove the “if” direction by assuming that $P$ lies on $\\Omega$ and triangles $PBD$ and $PCE$ have the same incenter $I$. We consider the diagram shown above. (We can adjust our proof easily for other possible configurations.) Because $PI$ bisects both $\\angle BPD$ and $\\angle EPC$, $\\angle EPB = \\ang... | United States | Team Selection Test | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangl... | null | proof only | null | |
046z | For any two points $A(x_1, y_1)$ and $B(x_2, y_2)$ in the coordinate plane, define
$$
d(A, B) = |x_1 - x_2| + |y_1 - y_2|.
$$
Let $P_1, P_2, \dots, P_{2023}$ be 2023 pairwise different points in the coordinate plane. Denote
$$
\lambda = \frac{\max_{1 \le i < j \le 2023} d(P_i, P_j)}{\min_{1 \le i < j \le 2023} d(P_i, P... | [
"**Proof Method 1:**\n(1) For $k = 1, 2, \\dots, 2023$, let the coordinates of $P_k$ be $(x_k, y_k)$, and denote $u_k = x_k + y_k$, $v_k = x_k - y_k$. Let $D = \\max_{1 \\le i < j \\le 2023} d(P_i, P_j)$. Then, for any $1 \\le i, j \\le 2023$,\n$$\n|u_i - u_j| = |(x_i - x_j) + (y_i - y_j)| \\le |x_i - x_j| + |y_i -... | China | 22nd Chinese Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 44 | |
0ams | Problem:
The numbers from $1$ to $36$ can be written in a counterclockwise spiral as follows:
| 31 | 30 | 29 | 28 | 27 |
| :---: | :---: | :---: | :---: | :---: |
| 32 | 13 | 12 | 11 | 10 |
| 33 | 14 | 3 | 2 | 9 |
| 34 | 15 | 4 | 1 | 8 |
| 35 | 16 | 5 | 6 | 7 |
| 36 | 17 | 18 | 19 | 20 |
In the figure above, all the... | [
"Solution:\n\nThe closest perfect square to $2015$ is $2025 = 45^2$ which means that only the rightmost side will be incomplete while the required diagonal would still have a total of $45$ entries.\n\nLooking at the values on the diagonal, we see that the numbers on the diagonal above $1$ have a common second diffe... | Philippines | 18th PMO Area Stage | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 30405 | |
0d0e | Define the sequence $x_1, x_2, \dots$, by $x_1 = \frac{1}{6}$ and
$$
x_{n+1} = \frac{n+1}{n+3} \left( x_n + \frac{1}{2} \right),
$$
for every $n \ge 1$. Find $x_{2011}$. | [
"We have\n$$\nx_2 = \\frac{2}{4} \\left( \\frac{1}{6} + \\frac{1}{2} \\right) = \\frac{2}{6}, \\quad x_3 = \\frac{3}{5} \\left( \\frac{2}{6} + \\frac{1}{2} \\right) = \\frac{3}{6}, \\text{ etc.}\n$$\nWe will prove by induction that for every $n \\ge 1$, we have $x_n = \\frac{n}{6}$.\nIndeed, assuming $x_n = \\frac{... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 2011/6 | |
0304 | Problem:
Uma escola tem 100 alunos e 100 armários numerados de 1 a 100. Inicialmente, todos os armários estão fechados. O primeiro aluno passa e abre todos os armários; o segundo passa e fecha todos os de números pares; o terceiro passa e muda a posição de todos os múltiplos de 3, ou seja, os que estão abertos ele fec... | [
"Solution:\n\nOs armários que ficam abertos são aqueles com números que possuem uma quantidade ímpar de divisores e isso só acontece com os números que são quadrados perfeitos. De 1 a 100, temos 1, 4, 9, 16, 25, 36, 49, 64, 81 e 100, que são quadrados perfeitos, ou seja, são 10 quadrados perfeitos. Portanto, o núme... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 90 | |
01v4 | We call a coloring of an $m \times n$ table ($m, n \ge 5$) in three colors a *good coloring* if the following two conditions are satisfied:
1) Each cell has the same number of neighboring cells of two other colors;
2) Each corner cell has no neighboring cells of its color.
Find all pairs $(m, n)$ ($m, n \ge 5$) for w... | [
"Answer: all $(m, n)$ such that $6$ divides both of them."
] | Belarus | Selection and Training Session | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | All pairs where both dimensions are multiples of six. | |
0fh5 | Problem:
Sean $x$ e $y$ dos números reales positivos. Probar que la expresión
$$
A=\sqrt{x}+\sqrt{y}+\sqrt{x y}
$$
se puede escribir en la forma
$$
B=\sqrt{x}+\sqrt{y+x y+2 y \sqrt{x}}
$$
y comparar los números
$$
L=\sqrt{3}+\sqrt{10+2 \sqrt{3}} \quad \text{y} \quad M=\sqrt{5+\sqrt{22}}+\sqrt{8-\sqrt{22}+2 \sqrt{15-3 \... | [
"Solution:\nTenemos\n$$\n\\begin{aligned}\nB & =\\sqrt{x}+\\sqrt{y} \\sqrt{1+x+2 \\sqrt{x}}= \\\\\n& =\\sqrt{x}+\\sqrt{y} \\sqrt{(1+\\sqrt{x})^{2}}= \\\\\n& =\\sqrt{x}+\\sqrt{y}(1+\\sqrt{x})= \\\\\n& =\\sqrt{x}+\\sqrt{y}+\\sqrt{x y}=A\n\\end{aligned}\n$$\n\nEn lo que se refiere a la segunda parte, transformando el ... | Spain | OME 26 | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | L = M | |
07bl | Consider a perfectly straight land having a canyon in the form of an infinite strip of width $\omega$. A polyhedron of diameter (maximum distance between any two points on or inside the polyhedron) $d$ is placed on one side of the canyon and a cavity of radius $d$ on the other. We want to roll this polyhedron and drop ... | [
"We claim that it is possible to roll the polyhedron in each direction, as far as we want.\nFirstly, we draw a line from a point inside the face of polyhedron touching the ground to the destination point. Since the polyhedron is convex, this line intersects one of edges of that face. We can roll the polyhedron arou... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0bes | In the acute-angled triangle $ABC$, with $AB \neq AC$, $D$ is the foot of the angle bisector of angle $A$, and $E$, $F$ are the feet of the altitudes from $B$ and $C$, respectively. The circumcircles of triangles $DBF$ and $DCE$ intersect for the second time at $M$. Prove that $ME = MF$. | [
"\n\nSolution:\n\nTriangles $AEF$ and $ABC$ are similar, therefore $AF \\cdot AB = AE \\cdot AC$. It follows the point $A$ is on the radical axis of the two circumcircles, hence $M \\in AD$. We have that $m(\\angle EMF) = 360^\\circ - (180^\\circ - m(\\angle FBD)) - (180^\\circ - m(\\angle ... | Romania | 64th NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ij5 | For integral $m$, let $p(m)$ be the greatest prime divisor of $m$. By convention, we set $p(\pm 1) = 1$ and $p(0) = \infty$. Find all polynomials $f$ with integer coefficients such that the sequence $\{p(f(n^2)) - 2n\}_{n \ge 0}$ is bounded above. (In particular, this requires $f(n^2) \ne 0$ for $n \ge 0$.) | [
"The polynomial $f$ has the required properties if and only if\n$$\nf(x) = c(4x - a_1^2)(4x - a_2^2)\\cdots(4x - a_k^2), \\quad (*)\n$$\nwhere $a_1, a_2, \\dots, a_k$ are odd positive integers and $c$ is a nonzero integer. It is straightforward to verify that polynomials given by $(*)$ have the required property. I... | United States | USAMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | All such polynomials are exactly f(x) = c ∏_{i=1}^k (4x − a_i^2), where c is a nonzero integer and a_i are odd positive integers. | |
08qd | Problem:
Find all positive integers $x, y, z$ such that
$$
45^{x}-6^{y}=2019^{z}
$$ | [
"Solution:\nWe define $v_{3}(n)$ to be the non-negative integer $k$ such that $3^{k} \\mid n$ but $3^{k+1} \\nmid n$. The equation is equivalent to\n$$\n3^{2x} \\cdot 5^{x} - 3^{y} \\cdot 2^{y} = 3^{z} \\cdot 673^{z}\n$$\nWe will consider the cases $y \\neq 2x$ and $y = 2x$ separately.\n\nCase 1. Suppose $y \\neq 2... | 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 > Factorization techniques"
] | null | proof and answer | (2, 1, 1) | |
0cky | Let $ABC$ be a triangle with $\angle BAC = 40^\circ$ and $\angle ABC = 80^\circ$. Denote $I$ its incircle. Prove that $AI = BC$.
 | [
"Let $\\{D\\} = BI \\cap AC$. Then $\\angle BAD = \\angle ABD = 40^\\circ$, so $\\triangle ABD$ is isosceles, yielding $AD = BD$ (1).\n\n*First construction.* Draw the bisector $BM$ of $\\angle DBA$, with $M \\in AC$. Then $\\angle ABM = \\angle IAB = 20^\\circ$ and $AB$ is a common side, hence $\\triangle IAB \\eq... | Romania | 75th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0dci | A sequence $(a_{1}, a_{2}, \ldots, a_{k})$ consisting of pairwise different cells of an $n \times n$ board is called a cycle if $k \geq 4$ and cell $a_{i}$ shares a side with cell $a_{i+1}$ for every $i=1,2, \ldots, k$, where $a_{k+1}=a_{1}$. We will say that a subset $X$ of the set of cells of a board is malicious if ... | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Other"
] | English | proof and answer | All real numbers C ≥ 1/4 | |
07mu | In a tournament with $N$ players, $N < 10$, each player plays once against each other player scoring 1 point for a win and 0 points for a loss. Draws do not occur. In a particular tournament only one player ended with an odd number of points and was ranked fourth. Determine whether or not this is possible. If so, how m... | [
"If only one player has an odd score, the total number of points won must be odd. This leaves two possibilities, $N = 6$ and $N = 7$.\n\nConsider first the case $N = 6$. The player with an odd score must have won 1, 3 or 5 games. If 5, the player would have come first. If 1, the player would have come 5th or 6th. T... | Ireland | Ireland | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | Yes; with seven players, the fourth-place player has 3 wins. | |
0elj | Problem:
Dana je krožnica $\mathcal{K}$ s premerom $AB$. Točki $C$ in $D$ ležita na krožnici $\mathcal{K}$ na istem bregu premice $AB$, tako da je premica $BD$ simetrala kota $\angle CBA$. Tetivi $AC$ in $BD$ se sekata v točki $E$. Koliko je dolžina daljice $DE$, če je $|AE|=169$ in $|CE|=119$? | [
"Solution:\n\nKer je premica $BE$ simetrala kot $\\angle CBA$, je po izreku o simetrali kota trikotnika $\\frac{|AB|}{|BC|}=\\frac{|AE|}{|CE|}=\\frac{169}{119}$ oziroma $|AB|=\\frac{169}{119}|BC|$. Po Talesovem izreku o kotu v polkrogu je $\\angle ACB=\\angle ADB=90^{\\circ}$, zato po Pitagorovem izreku velja $|AB|... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 65 | |
0234 | Problem:
Num grupo de 20 pessoas, algumas pessoas trocam apertos de mão.
a) Contamos quantos apertos de mão cada pessoa deu e somamos todos esses números. Mostre que o resultado é par.
b) É possível que num grupo de 99 pessoas cada pessoa tenha dado exatamente 3 apertos de mão? | [
"Solution:\n\na) Um aperto de mão é dado entre duas pessoas. Logo, quando somamos os apertos de mão de todas as pessoas, cada aperto de mão é contado duas vezes! Logo, a soma de quantas vezes cada pessoa apertou a mão de alguém é par, pois é o dobro de algum número.\n\nb) Como são 99 pessoas, se cada uma apertasse ... | Brazil | null | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | a) even; b) no | |
067f | A square $ABCD$ is divided into $n^2$ equal small elementary squares, by drawing parallel lines to its sides. A spider starts from $A$ moving only to the right and up and tries to approach point $C$. Every movement of the spider consists of $k$ steps right and $m$ steps up or of $m$ steps right and $k$ steps up (which ... | [
"We suppose that the square is placed into a Cartesian system of coordinates with origin $A(0,0)$ and axes on the sides $AB$ and $AD$, as in the figure 7. Let the spider start from $A$ and make its first movement ($m$ steps right and $k$ steps up or $m$ steps up and $k$ right). In the figure you can see the case $m... | Greece | SELECTION EXAMINATION | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | Let v = binom(k+m, m) and r = m - k. The total number of routes is v^l * sum_{i=0}^{l} binom(l, i) * binom(lr, i r). In particular, if r = 1 (i.e., m - k = 1), this simplifies to binom(k+m, m)^l * binom(2l, l). | |
0jt0 | Problem:
Victor has four red socks, two blue socks, and two green socks in a drawer. He randomly picks two of the socks from the drawer, and is happy to see that they are a matching pair. What is the probability the pair was red? | [
"Solution:\n\nThere are $\\binom{4}{2} = 6$ possible red pairs, and then $\\binom{2}{2} = 1$ blue pair, and finally $\\binom{2}{2} = 1$ green pair.\n\nSo there were a total of $6 + 1 + 1 = 8$ possible pairs, of which there were $6$ red ones. So the answer is $\\frac{3}{4}$."
] | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 3/4 | |
0hre | Problem:
Find the minimal natural number $n$ with the following property: It is possible to tile the plane with squares whose side lengths belong to the set $\{1,2, \ldots, n\}$ so that no two squares with the same side length touch along a segment of an edge.
Remark. Squares with the same side length can touch at a v... | [
"Solution:\nThe answer is $n=5$. The desired tiling is shown in Figure 1. It is formed by translation from the L-shaped region in bold borders. Since none of the five squares in this region border on squares of like side length, neither does any square in the infinite tiling.\n\nTo show that $n \\leq 4$ does not wo... | United States | Berkeley Math Circle Monthly Contest | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 5 | |
04ki | Prove that among any three positive integers we can choose two, say $a$ and $b$, so that the number $a^3b - ab^3$ is a multiple of $10$. | [] | Croatia | Mathematical competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0k6a | Problem:
Let $a$, $b$, $c$, $d$ be real numbers such that
$$
\min (20x+19, 19x+20) = (a x + b) - |c x + d|
$$
for all real numbers $x$. Find $a b + c d$. | [
"Solution:\nIn general, $\\min(p, q) = \\frac{p+q}{2} - \\left|\\frac{p-q}{2}\\right|$. Letting $p = 20x + 19$ and $q = 19x + 20$ gives $a = b = 19.5$ and $c = d = \\pm 0.5$. Then the answer is $19.5^2 - 0.5^2 = 19 \\cdot 20 = 380$."
] | United States | HMMT November 2019 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | final answer only | 380 | |
06be | Show that $\cos \frac{\pi}{7}$ is not of the form $p + \sqrt{q} + \sqrt[3]{r}$, where $p, q$ and $r$ are rational numbers. | [
"Let $\\alpha = \\cos \\frac{\\pi}{7}$. As $\\cos \\frac{4\\pi}{7} + \\cos \\frac{3\\pi}{7} = 0$, we have\n$$\n2(2\\alpha^2 - 1)^2 - 1 + (4\\alpha^3 - 3\\alpha) = 0.\n$$\nThis is the same as $(\\alpha + 1)(8\\alpha^3 - 4\\alpha^2 - 4\\alpha + 1) = 0$. Clearly, $\\alpha \\ne -1$. Therefore, $\\alpha$ is a root of\n$... | Hong Kong | 1997-2023 IMO HK TST | [
"Number Theory > Algebraic Number Theory > Algebraic numbers",
"Algebra > Abstract Algebra > Field Theory",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | null | proof only | null | |
07me | Suppose $a, b, c$ are the side lengths of a triangle $ABC$. Show that
$$
x = \sqrt{a(b+c-a)}, \quad y = \sqrt{b(c+a-b)}, \quad z = \sqrt{c(a+b-c)}
$$
are the side lengths of an acute-angled triangle $XYZ$, with the same area as $ABC$, but with a smaller perimeter, unless $ABC$ is equilateral. | [
"Let $2s = a+b+c$, and denote by $\\Delta$ the area of $\\triangle ABC$. Then $x = \\sqrt{2a(s-a)}$, etc., and $\\Delta^2 = s(s-a)(s-b)(s-c)$. To show that $XYZ$ is acute, because of the cosine theorem, we must show $x^2, y^2, z^2$ satisfy the triangle inequality. But, e.g.,\n$$\nx^2 < y^2 + z^2 \\iff a(s-a) < b(s-... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0971 | Problem:
Demonstrați că orice mulțime alcătuită din 112 numere naturale mai mici decât 1000, conține două numere diferența cărora este un număr de trei cifre de forma $\overline{a a a}$, $a \geq 1$. | [
"Solution:\n\nFie $X=\\{x_{1}, x_{2}, \\ldots, x_{112}\\}$ unde $x_{i} \\in \\mathbb{N}$ și $x_{i}<1000$, $\\forall i=\\overline{1,112}$.\n\nÎmpărțind cu rest $x_{i}$ la $111$, obținem $x_{i}=111 \\cdot q_{i}+r_{i}$, unde $q_{i}$ - cifră și $r_{i} \\in\\{0,1, \\ldots, 110\\}$.\n\nConform principiului Dirichlet exis... | Moldova | Olimpiada Republicană la Matematică, A doua zi | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0biq | Let $m$ be a positive integer and let $A$, respectively $B$, be an alphabet with $m$, respectively $2m$ letters. Let $n$ be an even integer greater than or equal to $2m$. Let $a_n$ be the number of words of length $n$ made of letters from $A$ such that every letter in $A$ occurs a positive even number of times. Let $b_... | [
"Deletion of all bars in a word in $B^n$ produces a word in $A^n$. Now let $\\alpha$ be a word in $A^n$ and let $a_i$ occur $k_i$ times in $\\alpha$; the $k_i$ are positive even integers which add up to $n$. Since there are exactly $2^{k_i-1}$ distinct ways to bar $a_i$ an odd number of times in $\\alpha$, the prei... | Romania | 65th NMO Selection Tests for BMO and IMO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof and answer | 2^(n - m) | |
0lfp | In the space, there is a convex polyhedron $D$ such that for every vertex of $D$, there are an even number of edges passing through that vertex. We choose a face $F$ of $D$. Then we assign each edge of $D$ a positive integer such that for all faces of $D$ different from $F$, the sum of the numbers assigned on the edges... | [
"We have the following two observations.\n\n**Lemma 1.** The surface of any convex polyhedron $D$ can be projected to a plane so that the faces of $D$ are in one-to-one correspondence with the faces of a planar graph $G$ and edges of the $D$ are in one-to-one correspondence with the edges of $G$.\n\n*Proof.* First,... | Vietnam | Vietnamese MO | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Solid Geometry > Other 3D problems",
"Number Theory > Other"
] | English | proof only | null | |
0bu2 | Problem:
Fie matricele $A, B, C, D \in \mathcal{M}_{n}(\mathbb{C}), n \geq 2$ și $k \in \mathbb{R}$ astfel încât $A C + k B D = I_{n}$ și $A D = B C$. Demonstrați că $C A + k D B = I_{n}$ și $D A = C B$. | [
"Solution:\n\nPentru început considerăm $k \\neq 0$. Fie $w \\in \\mathbb{C}$ astfel încât $w^{2} = -k$. Considerăm matricele $X = A + w B$, $Y = C - w D$, $Z = A - w B$ și $U = C + w D$. Ipoteza conduce la $X Y = I_{n}$ și $Z U = I_{n}$.\n\nAtunci $Y X = I_{n}$ și $U Z = I_{n}$.\n\nDeducem $(C A + k D B) - w (D A ... | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
07zj | Problem:
Siano $a_{1}, a_{2}, a_{3}, a_{4}$ quattro numeri interi distinti e sia $P(x)$ un polinomio a coefficienti interi tale che
$$
P\left(a_{1}\right)=P\left(a_{2}\right)=P\left(a_{3}\right)=P\left(a_{4}\right)=1 .
$$
a. Dimostrare che non esiste nessun numero intero $n$ tale che $P(n)=12$.
b. Esistono un polino... | [
"Solution:\n\nConsideriamo il polinomio $Q(x)=P(x)-1$. La condizione $(*)$ assicura che $Q\\left(a_{1}\\right)=Q\\left(a_{2}\\right)=Q\\left(a_{3}\\right)=Q\\left(a_{4}\\right)=0$ e, per il teorema di Ruffini, $Q(x)$ è divisibile per ciascuno dei fattori $(x-a_{i})$ ($1 \\leq i \\leq 4$). Poiché gli $a_{i}$ sono di... | Italy | GARA NAZIONALE di MATEMATICA | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | a: No such integer exists. b: No, such a polynomial and integer do not exist. | |
0jqn | Problem:
Consider all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfying
$$
f(f(x)+2x+20)=15
$$
Call an integer $n$ good if $f(n)$ can take any integer value. In other words, if we fix $n$, for any integer $m$, there exists a function $f$ such that $f(n)=m$. Find the sum of all good integers $x$. | [
"Solution:\n\nAnswer: $-35$\n\nFor almost all integers $x$, $f(x) \\neq -x-20$. If $f(x) = -x-20$, then\n$$\nf(-x-20+2x+20) = 15 \\Longrightarrow -x-20 = 15 \\Longrightarrow x = -35\n$$\nNow it suffices to prove that the $f(-35)$ can take any value.\n$f(-35) = 15$ in the function $f(x) \\equiv 15$. Otherwise, set $... | United States | HMMT November 2015 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | -35 | |
0ave | Problem:
Let $f$ be a real-valued function such that
$$
f(x-f(y))=f(x)-x f(y)
$$
for any real numbers $x$ and $y$. If $f(0)=3$, determine $f(2016)-f(2013)$. | [] | Philippines | 19th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | final answer only | 6048 | |
0dig | Let $ABC$ be an acute triangle, $D$ be the foot of altitude from $A$ to $BC$. We constructed the two squares $ABKL$, $ACMN$ outside the triangle. Prove that the lines $AD$, $BM$, $KC$ are concurrent at one point. | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0hah | On sides $AB$, $BC$, $AC$ of triangle $ABC$ with $\angle BAC = 120°$ there are points $M$, $K$, $N$ respectively so, that $\triangle MKN$ is equilateral, and $AM = 2,017$, $AN = 2,018$. Baron Munchausen assures, that $\triangle MKN$ has the smallest perimeter of all equilateral triangles, that have exactly one vertex o... | [
"According to the condition, all the angles of $\\triangle MKN$ equal to $60°$ (Fig. 39). As $\\angle BAC + \\angle MKN = 180°$, quadrilateral $AMKN$ is cyclic, so\n$$\n\\angle KAC = \\angle KMN = 60° = \\angle MNK = \\angle BAK.\n$$\nLet's consider points $M_1$ and $N_1$ – projections of point $K$ on sides $AB$ an... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | No | |
0cm7 | Problem:
Prove that for every positive integer $n$ there exists a (not necessarily convex) polygon with no three collinear vertices, which admits exactly $n$ different triangulations.
(A triangulation is a dissection of the polygon into triangles by interior diagonals which have no common interior points with each othe... | [] | Romanian Master of Mathematics (RMM) | Romanian Master of Mathematics Competition | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
058i | Xavier and Olivier are playing tic-tac-toe in the rectangular grid of size $3 \times 3$ with modified rules. On every move, a player chooses an empty square and writes his token into it. Players take turns alternately, with Xavier starting. The player who is the first to occupy any three squares that either are all in ... | [
"Note that every two unit squares uniquely determine a unit square that constitutes a winning triple together with the given two squares. Label all unit squares except the middle square clockwise with numbers 1 through 8 (Fig. 1); let Xavier initially play into the middle square and if Olivier then moves into the s... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Xavier | |
0fqb | Problem:
Se han coloreado 46 cuadrados unitarios de una cuadrícula $9 \times 9$. ¿Hay, en la cuadrícula, alguna figura del tipo
(no necesariamente con la orientación que muestra el dibujo) con las tres casillas coloreadas? | [
"Solution:\n\nLa respuesta es afirmativa. En efecto, probaremos que si coloreamos como indica el enunciado, se puede encontrar siempre una figura del tipo dado con las tres casillas coloreadas.\nA tal fin, dividamos la cuadrícula $9 \\times 9$ en dieciséis cuadrados $2 \\times 2$, tres figuras como la siguiente:\n!... | Spain | null | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | Yes | |
01io | Find the smallest positive real number $\alpha$, such that
$$
\frac{x+y}{2} \geq \alpha\sqrt{xy} + (1-\alpha)\sqrt{\frac{x^2+y^2}{2}}
$$
for all positive real numbers $x$ and $y$. | [
"Let's prove that $\\alpha = \\frac{1}{2}$ works. Then the following inequality should hold for all positive real numbers $x$ and $y$:\n$$\n\\begin{aligned}\n\\frac{x+y}{2} &\\ge \\frac{1}{2}\\sqrt{xy} + \\frac{1}{2}\\sqrt{\\frac{x^2+y^2}{2}} \\\\\n\\Leftrightarrow (x+y)^2 &\\ge xy + \\frac{x^2+y^2}{2} + 2\\sqrt{xy... | Baltic Way | Baltic Way 2023 Shortlist | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 1/2 | |
0ece | In a company led by several directors they have a safe which is locked by six locks. Each director has three keys which unlock three different locks. With each key he can unlock exactly one lock. No two directors can unlock the same three locks and no two directors together can unlock the safe. At most how many directo... | [
"The company is led by at most **10** directors.\nDenote the locks and the corresponding keys with numbers $1$ through $6$. Each director has a set of $3$ from $3$ different locks and no two directors have the same sets. Therefore we first count how many different sets of $3$ keys there exist. For a specific set of... | Slovenia | National Math Olympiad 2015 – Final Round | [
"Discrete Mathematics > Combinatorics",
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | 10 | |
01ll | In an acute-angled triangle $ABC$ the orthocenter is $H$. $I_H$ is the incenter of $\triangle BHC$. The bisector of $\angle BAC$ intersects the perpendicular from $I_H$ to the side $BC$ at point $K$. Let $F$ be the foot of the perpendicular from $K$ to $AB$.
Prove that $2KF + BC = BH + HC$.
(A. Voidelevich) | [
"First, easy counting of angles shows that $AK \\parallel HI_H$. By condition, $KI_H \\parallel AH$, so we conclude that $AKIH_H$ is a parallelogram. Hence $AK = HI_H$. Let $T$ be the tangency point of the incircle of $\\triangle BHC$ with the side $BH$, then $I_HT \\perp HT$; thus from the above $\\triangle EKA = ... | Belarus | Selection and Training Session | [
"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",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | English | proof only | null | |
0i3d | Let $ABCD$ be a convex quadrilateral such that $\angle ABC = \angle ADC = 135^\circ$ and
$$
AC^2 \cdot BD^2 = 2AB \cdot BC \cdot CD \cdot DA.
$$
Prove that the diagonals of quadrilateral $ABCD$ are perpendicular. | [
"**First Solution.**\n\nLet $\\alpha$ be the angle formed by diagonals $AC$ and $BD$, where $0^\\circ < \\alpha \\le 90^\\circ$. We calculate the area of quadrilateral $ABCD$ in two ways. We obtain\n$$\n[ABCD] = \\frac{1}{2}(AC \\cdot BD \\sin \\alpha)\n$$\nand\n$$\n\\begin{align*} \n[ABCD]... | United States | USA IMO | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Equations... | English | proof only | null | |
0dx7 | Problem:
Poenostavi izraz $\left((1-x) \sqrt{\frac{1+x}{1-x}}+1-x\right) \cdot\left(\sqrt{\frac{1+x}{1-x}}-1\right)$, če je $-1 \leq x<1$. | [
"Solution:\n\nČe uporabimo v prvem oklepaju pravilo $a \\cdot \\sqrt{b}=\\sqrt{a^{2} b}$, ki velja za $a \\geq 0$, dobimo\n$$\n((1-x) \\sqrt{\\frac{1+x}{1-x}}+1-x) = \\left(\\sqrt{\\frac{(1-x)^{2}(1+x)}{1-x}}+1-x\\right)\n$$\nkrajšamo in dobimo\n$$\n(\\sqrt{(1-x)(1+x)}+1-x) \\left(\\sqrt{\\frac{1+x}{1-x}}-1\\right)... | Slovenia | 6. državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 2x | |
02hc | Professor Piraldo takes part in soccer matches with a lot of goals and judges a match in his own peculiar way. A match with score of $m$ goals to $n$ goals, $m \ge n$, is *tough* when $m \le f(n)$, where $f(n)$ is defined by $f(0) = 0$ and, for $n \ge 1$, $f(n) = 2n - f(r) + r$, where $r$ is the largest integer such th... | [
"First note that if $n$ is written in the Fibonacci basis (as a sum of distinct Fibonacci numbers containing no neighbors), then the representation of $f(n)$ is just that of $n$ with a 0 in the end. The proof goes by induction: it is true for $n=0$. Suppose $n > 0$. Then $r$ will be such that $f(r) = n$, if the las... | Brazil | Brazilian Math Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
09gx | Гүдгэр $n$-өнцөгтийн талууд ба диагоналиудыг $n-1$ өнгөөр буджээ. Хэрэв оройгоос гарсан талууд ба диагоналиуд нь бүгд өөр өнгөтэй байвал уг оройг “сайн” орой гэж нэрлэе. Сайн орой хамгийн олондоо хэдэн ширхэг байх вэ? | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n if n is even; n−1 if n is odd | |
04rw | In the real numbers, solve the following system of equations:
$$
\begin{aligned}
\sin^2 x + \cos^2 y &= \tan^2 z, \\
\sin^2 y + \cos^2 z &= \tan^2 x, \\
\sin^2 z + \cos^2 x &= \tan^2 y.
\end{aligned}
$$ | [
"Substituting $\\cos^2 x = a$, $\\cos^2 y = b$, $\\cos^2 z = c$ leads to the system\n$$\n\\begin{cases}\n 1 - a + b = \\frac{1}{c} - 1, \\\\\n 1 - b + c = \\frac{1}{a} - 1, \\\\\n 1 - c + a = \\frac{1}{b} - 1,\n\\end{cases} \\quad (1)\n$$\nwhere $a, b, c \\in (0, 1)$.\nAdding these equations together yields\n$$\n\\... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | (x, y, z) = (pi/4 + k*pi/2, pi/4 + l*pi/2, pi/4 + m*pi/2) for any integers k, l, m | |
0g8i | 甲、乙兩人玩以下的數字遊戲：從甲開始，兩個人輪流自 $1$ 到 $9$ 的數字中不重複地選一個數字出來，並且把選出的數字由左至右依序排成一個七位數（即 $\overline{A_1B_2A_3B_4A_5B_6A_7}$）。如果排出來的七位數是某個完全七次方數的末七位數字，則甲獲勝；否則的話，乙獲勝。
請問誰有必勝策略？
Alice and Bob play a game. Starting from Alice, two guys take turns to choose a digit from $1$ to $9$ without repetition, and put them from the leftmost to t... | [
"甲有必勝策略。由以下引理知，只要甲讓 $A_7 \\in \\{1,3,7,9\\}$，甲必便獲勝；而基於乙最多只能擋其中三個，故甲必勝。\n\n引理\n若 $0 < a < 10^7$ 且 $\\text{gcd}(a, 10) = 1$, 則 $x^7 \\equiv a \\pmod{10^7}$ 有解。\n\n證明\n我們先證明：\n\nClaim. 若 $\\text{gcd}(m, 10) = \\text{gcd}(n, 10) = 1$ 且 $10^7|m^7 - n^7$, 則 $10^7|m - n$.\n\nPf of Claim. 注意到\n$$\nm^7 - n^7 = (m-n)(m^6 + m... | Taiwan | 二〇一四年國際數學奧林匹亞競賽第三階段選訓營 獨立研究(一) | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | Alice | |
0eim | Problem:
Poišči vse polinome stopnje $n \geq 1$, ki imajo vse ničle racionalne, vsak izmed njihovih $n+1$ koeficientov pa je enak 1 ali -1. | [
"Solution:\n\nNaj bo $p(x) = a_n x^n + \\cdots + a_0$ iskani polinom. Če je $\\frac{k}{m}$ neka njegova racionalna ničla, tedaj $k$ deli $a_0$ in $m$ deli $a_n$. Ker pa sta $a_0$ in $a_n$ oba enaka 1 ali -1, sta tudi $k$ in $m$ enaka 1 ali -1. Torej so vse ničle polinoma $p(x)$ enake 1 ali -1. Hkrati je tudi vodiln... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | ±(x−1), ±(x+1), ±(x^3 + x^2 − x − 1), ±(x^3 − x^2 − x + 1) | |
0es2 | The value of $2 - 0 + 1 \times 6$ is
(A) $-4$
(B) $4$
(C) $8$
(D) $9$
(E) $18$ | [
"$2 - 0 + 1 \\times 6 = 2 - 0 + 6 = 8$"
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | C | |
02mg | Problem:
Um caminho triangular - Janete passeia por um caminho de forma triangular $\triangle ABC$, com o lado $AB$ medindo $1992~\mathrm{m}$. Ela gasta 24 minutos para percorrer esse lado $AB$ e, depois, com a mesma velocidade, ela percorre o outro lado $BC$ seguido da hipotenusa $CA$ em 2 horas e 46 minutos. Qual é ... | [] | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles"
] | null | final answer only | 6745 m | |
04le | Let $\triangle ABC$ be an acute-angled triangle and let $BCD$, $ACE$ and $ABF$ be equilateral triangles drawn on the outside of triangle $ABC$. Let $M$ be the midpoint of the segment $\overline{BD}$, and let $O$ be the centre of the triangle $ACE$. Prove that $|AM| : |OF| = \frac{\sqrt{3}}{2}$. | [] | Croatia | Mathematical competitions in Croatia | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthoce... | null | proof only | sqrt(3)/2 | |
0hbx | In triangle $ABC$, by $T_A, T_B, T_C$ are denoted tangency points of excircles of $\triangle ABC$ tangent to sides $BC, AC$ and $AB$, respectively. Let $O$ be the center of circumscribed circle of $\triangle ABC$, and $I$ be the center of its inscribed circle. It is known that $OI \parallel AC$. Prove that $\angle T_A ... | [
"Let $I_A, I_B, I_C$ be the centers of excircles, $I_1$ be the point, symmetrical to $I$ with respect to point $O$ (fig. 29). Let us look at $\\Delta I_A I_B I_C$. $I$ is its orthocenter, since bisectors of outer and inner angles are perpendicular. It is also clear that $O$ is the center of Euler's circle of this t... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0kq2 | Problem:
Let $ABCD$ be an isosceles trapezoid such that $AB = 17$, $BC = DA = 25$, and $CD = 31$. Points $P$ and $Q$ are selected on sides $AD$ and $BC$, respectively, such that $AP = CQ$ and $PQ = 25$. Suppose that the circle with diameter $PQ$ intersects the sides $AB$ and $CD$ at four points which are vertices of a ... | [
"Solution:\nLet the midpoint of $PQ$ be $M$; note that $M$ lies on the midline of $ABCD$. Let $B'$ and $C'$ be a translate of $BC$ (parallel to $AB$ and $CD$) so that $M$ is the midpoint of $B'$ and $C'$. Since $MB' = MC' = 25/2 = MP = MQ$, $B'$ and $C'$ are one of the four intersections of the circle with diameter... | United States | HMMT February | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 168 | |
0e8n | Let $x$, $y$, $z$ and $w$ be nonnegative real numbers such that $x + y + z + w = 1$. Find the greatest and least value of the expression
$$
(x + 3y)(y + 2z) + (y + 3z)(z + 2w) + (z + 3w)(w + 2x) + (w + 3x)(x + 2y).
$$ | [
"After eliminating parentheses and combining like terms in the expression we get\n$$\n3(x^2 + y^2 + z^2 + w^2) + 7(xy + yz + zw + wx) + 4(xz + yw),\n$$\nwhich can be further transformed to\n$$\n3(x + y + z + w)^2 + (xy + yz + zw + wx) - 2(xz + yw).\n$$\nBecause $x + y + z + w = 1$, the expression given in the probl... | Slovenia | Selection Examinations for the IMO 2013 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | least value 5/2, greatest value 13/4 | |
07ud | There are 28 towns on the island of Mathematia. Each pair of towns is either connected by a single road, or is not connected. It turns out that for any two towns $A$ and $B$ that have the same number of roads connected to them, there is no road that connects $A$ to $B$. Determine, with proof, the maximum number of road... | [
"The problem can be restated as follows: Given a graph on 28 vertices, where no two vertices of the same degree $i$ are connected by an edge ($1 \\le i \\le 27$), determine (with proof) the maximum number of edges in the graph.\nFor $0 \\le i \\le 27$, let $a_i$ denote the number of vertices of degree $i$. Thus the... | Ireland | IRL_ABooklet | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 322 | |
0eic | Problem:
Krožnici $\mathcal{K}_1$ in $\mathcal{K}_2$ s središčema $O_1$ in $O_2$ se od zunaj dotikata. Naj bo $p$ skupna tangenta krožnic $\mathcal{K}_1$ in $\mathcal{K}_2$, ki ne poteka skozi njuno dotikališče. Dokaži, da je $p$ tangenta tudi krožnice s premerom $O_1 O_2$. | [
"Solution:\n\nNaj bo $S$ razpolovišče daljice $O_1 O_2$ ter $A$ dotikališče krožnic $\\mathcal{K}_1$ in $\\mathcal{K}_2$. Dotikališči premice $p$ s krožnicama $\\mathcal{K}_1$ in $\\mathcal{K}_2$ označimo zaporedoma z $B_1$ in $B_2$, presečišče premice $p$ s tangento na krožnici $\\mathcal{K}_1$ in $\\mathcal{K}_2$... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02zv | Problem:
a) As letras $A$, $B$ e $C$ podem ser dispostas em linha de 6 formas distintas:
$$
A B C,\ A C B,\ B A C,\ B C A,\ C A B,\ C B A
$$
Note que em 3 delas a letra $A$ aparece à esquerda da letra $B$:
$$
A B C,\ A C B,\ C A B
$$
Dispondo as letras $A$, $B$, $C$ e $D$ em linha de todas as 24 formas distintas possív... | [
"Solution:\na) Para cada palavra de 4 letras em que $A$ está à esquerda de $B$, trocando a posição dessas duas letras, obtemos uma palavra de 4 letras em que $A$ está à direita da letra $B$:\n$$\nC \\mathbf{A B} D \\leftrightarrow C \\mathbf{B A} D\n$$\nAgrupando as 24 palavras em 12 pares desse tipo, podemos concl... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | a) 12; b) 16!/2^8 | |
0dqq | Let $a_1, a_2, \dots$ be a non-constant sequence of positive integers such that $m-n \mid a_m - a_n$ for all distinct pairs of positive integers $m, n$. Prove that there is an infinite set of primes such that each divides $a_n$ for some $n$. | [
"Let $P$ be the union of the sets of the prime divisors of $a_n$, $n \\ge 1$. Denote by $v_p(n)$ the highest power of $p$ that divides $n$. Suppose on the contrary, that $|P|$ is finite. Let $A = \\{\\prod_{p \\in P} p^k : k \\in \\mathbb{Z}, k > v_p(a_1) \\ \\forall p \\in P\\}$. Then $|A| = \\infty$. Let $t \\in ... | Singapore | Singapore International Mathematical Olympiad Committee National Team Selection Test | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
03o2 | Problem:
Jane writes down 2024 natural numbers around the perimeter of a circle. She wants the 2024 products of adjacent pairs of numbers to be exactly the set $\{1!, 2!, \ldots, 2024!\}$. Can she accomplish this? | [
"Solution:\n\nGiven any prime $p$ and positive integer $x$, let $v_{p}(x)$ denote the highest power of $p$ dividing $x$. We claim that Jane cannot write 2024 such numbers as that would imply that $1! \\cdot 2! \\cdots 2024!$ is the square of the product of the 2024 numbers. Let $p$ be a prime and $k$ be a natural n... | Canada | CMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | No | |
0anp | Problem:
Find all positive values of $a$ for which the equation $x^{2} - a x + 1 = 0$ has roots that differ by $1$. | [
"Solution:\nLet the roots be $r_1$ and $r_2$ with $r_1 - r_2 = 1$ (without loss of generality, $r_1 > r_2$).\n\nFrom Vieta's formulas:\n$$\n\\begin{align*}\nr_1 + r_2 &= a \\\\\nr_1 r_2 &= 1\n\\end{align*}\n$$\n\nSince $r_1 - r_2 = 1$, we can write:\n$$\nr_1 = r_2 + 1\n$$\n\nSubstitute into the sum:\n$$\n(r_2 + 1) ... | Philippines | Area Stage | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | sqrt(5) | |
07d5 | By angle, we mean its vertex together with its two rays. Find the maximum value of $n$ such that we can put $n$ $60^\circ$ angles on the plane, in a way that each pair of which has 4 intersection points. | [
"Assume that two $60^\\circ$ angles with vertices $A$ and $B$ have four intersection points $M$, $N$, $P$ and $Q$ as shown in the figure.\n\nWe have\n$$\n\\widehat{AKB} = \\widehat{KAP} + \\widehat{APB} + \\widehat{PBK} = 30^\\circ + \\widehat{APB} + 30^\\circ > 60^\\circ\n$$\nAnd\n$$\n\\wi... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 2 | |
03dw | At an IMO prep camp there are 14 participants. Each participant has at least $k$ favorite numbers. The organizers want to give to each student a T-shirt with one of their favorite numbers on it. Find the smallest $k$ for which this is always plausible under the restriction:
a) the participants form a circle and the T-... | [
"a) Obviously $k=1$ fails (e.g., whenever a couple of neighbors share their favorite number). We will show that $k=2$ works. If all of them share the same favorite numbers, say 1 and 2, then we can easily solve the problem via providing 7 T-shirts of each kind, since 14 is even. If this is not the case, WLOG we can... | Bulgaria | Bulgaria 2022 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | a) 2; b) 4 | |
0e5d | There are $n$ chameleons standing in a circle. Some of them are colored red and the others are colored green. Every minute, some of the chameleons change their color from red to green or vice versa in accordance with the following rule: a chameleon changes its color if and only if both its neighbours have the same colo... | [
"Because at the end the chameleons have the same color as at the beginning, they must have changed their color an even number of times, hence $0$ times, $2$ times, ... or $2n$ times.\n\nFirst suppose there exists a chameleon that has never changed its color. Its neighbours then have had the same color every minute,... | Slovenia | Selection Examinations for the IMO 2012 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
016t | Let $n$ be an integer with $n \ge 3$. Consider all dissections of a convex $n$-gon into triangles by $n-3$ diagonals that have no common points inside the polygon, and all colourings of the triangles with black and white so that triangles with a common side are always of a different colour. Find the least possible numb... | [
"$\\lfloor \\frac{n-1}{3} \\rfloor$.\n\nLet $f(n)$ denote the minimum number of black triangles in an $n$-gon. It is clear that $f(3) = 0$ and that $f(n)$ is at least 1 for $n = 4, 5, 6$. It is easy to see that for $n = 4, 5, 6$ there is a coloring with only one black triangle, so $f(n) = 1$ for $n = 4, 5, 6$.\n\nF... | Baltic Way | BALTIC WAY | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | floor((n-1)/3) | |
05mc | Problem:
Soit $ABC$ un triangle acutangle tel que $AC > AB$, $D$ sur $BC$ tel que $AD = AB$ et $D$ différent de $B$. Soit $\Gamma$ le cercle circonscrit à $ABC$, $\Delta$ la tangente à $\Gamma$ en $C$, $E$ l'intersection de $(AD)$ et $\Delta$.
Montrer que $CD^{2} = AD \cdot DE - BD \cdot DC$. | [
"Solution:\n\n\n\nOn a $\\widehat{EDC} = \\widehat{ADB} = \\widehat{CBA}$ et $\\widehat{DCE} = \\widehat{BAC}$ donc $ABC$ et $CDE$ sont semblables. On en déduit que $\\frac{AB}{CD} = \\frac{BC}{DE}$, donc\n$$\nAD \\cdot DE - BD \\cdot DC = AB \\cdot DE - BD \\cdot DC = BC \\cdot CD - BD \\c... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
019j | Let $a_1 = 1$, $a_2 = 10$ and $a_{n+1} = 2a_n + 3a_{n-1}$, $n > 1$. Then the infinite sum
$$
P(x) = \sum_{i=1}^{\infty} a_i x^i,
$$
is defined and finite for $x \in ] -\frac{1}{3}, \frac{1}{3}[$. Find all $y \in \mathbb{Z}$ such that
$$
P\left(\frac{1}{y}\right) \in \mathbb{Z}.
$$ | [
"Notice that $a_{n+1} + a_n = 3(a_n + a_{n-1})$, and let $b_n = a_n + a_{n-1}$, $b > 1$. Then $b_2 = 11$ and $b_n = 11 \\cdot 3^{n-2}$, $n > 1$. Using this gives\n$$\n\\begin{align*}\n(1+x)P(x) &= a_1 + \\sum_{i=2}^{\\infty} a_i x^i + \\sum_{i=2}^{\\infty} a_{i-1} x^i = \\\\\n&= a_1 x + \\sum_{i=2}^{\\infty} b_i x^... | Baltic Way | Baltic Way 2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Generating functions",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | -8 | |
0a33 | Floor's class consists of $16$ students, including Floor. All students took a test with four questions. Every question was worth a (positive) integer number of points. Each question was marked completely right or completely wrong; no partial points were given. The question that was worth the most points was worth exact... | [] | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 21 | |
0evr | Find the smallest positive real number $p (\le 1)$ such that the inequality
$$
\sum_{i=1}^{2024} x_i (y_{2025-i} - y_{2024-i}) \ge 1 - p
$$
holds for all real numbers $0 \le x_1 \le x_2 \le \dots \le x_{2024} \le 1$ and $0 = y_0 \le y_1 \le y_2 \le \dots \le y_{2024} \le 1$ satisfying
$$
\sum_{i=1}^{2024} x_i = \sum_{i... | [
"If $x_1 = x_2 = \\dots = x_{2024} = y_1 = y_2 = \\dots = y_{2024} = p$, then $p^2 \\ge 1-p$, that is, $p \\ge \\frac{1+\\sqrt{5}}{2}$. So, $p$ must be at least $\\frac{1+\\sqrt{5}}{2}$. We prove that the condition holds for $p = \\frac{1+\\sqrt{5}}{2}$, which implies that the answer is $\\frac{1+\\sqrt{5}}{2}$.\nL... | South Korea | The 37th Korean Mathematical Olympiad Final Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | (sqrt(5) - 1)/2 | |
0enz | Let the sequence $(u_n)$ be defined recursively by
$$
u_0 = 0,\ u_1 = 1,\ u_n = 2011u_{n-1} - u_{n-2} \quad \text{for } n \ge 2.$$
Find all the values of $n$ for which $u_n$ is prime. | [
"For generality, let $x = 2011$. Expanding the first few terms of the sequence with the recurrence $u_n = x u_{n-1} - u_{n-2}$ we find\n$$\n\\begin{aligned}\nu_0 &= 0, \\quad u_1 = 1, \\quad u_2 = x \\\\\nu_3 &= x^2 - 1 = u_2^2 - u_1^2 \\\\\nu_4 &= x^3 - 2x = 2u_3u_2 - x u_2^2 \\\\\nu_5 &= x^4 - 3x^2 + 1 = u_3^2 - ... | South Africa | South-Afrika 2011-2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 2 | |
01kq | Prove that
$$
\frac{a^2}{a+b} + \frac{b^2}{b+c} \ge \frac{3a+2b-c}{4}
$$
for all positive real $a, b, c$.
(D. Pirshtuk) | [
"$$\n(2x - y)^2 \\ge 0 \\iff \\frac{x^2}{y} \\ge \\frac{4x - y}{4}\n$$\nfor any positive $x, y$.\n\nTherefore,\n$$\n\\frac{a^2}{a+b} + \\frac{b^2}{b+c} \\ge \\frac{4a - (a+b)}{4} + \\frac{4b - (b+c)}{4} = \\frac{3a + 2b - c}{4},\n$$\nas required."
] | Belarus | 60th Belarusian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0etd | Let $ABC$ be a triangle with $\angle ABC \ne 90^\circ$ and $AB$ its shortest side. Denote by $H$ the intersection of the altitudes of triangle $ABC$. Let $K$ be the circle through $A$ with centre $B$. Let $D$ be the other intersection of $K$ and $AC$. Let $K$ intersect the circumcircle of $BCD$ again at $E$. If $F$ is ... | [
"Consider Figure 2:\n\n\nFigure 2\n\nWe note that $H$ must also be on $V$, the circumcircle of triangle $BDC$. This is because triangles $BLH$ and $BMD$ are similar (note that triangle $ABD$ is iscoceles, with $BA = BD$, and $BM$ is a perpendicular bisector of $AD$, that also bisects $\\ang... | South Africa | The South African Mathematical Olympiad, Third Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point ... | English, Afrikaans | proof only | null | |
0fy1 | Problem:
Betrachte drei Steine auf der reellen Geraden mit ganzzahligen Koordinaten. In einem Schritt kann man zwei dieser Steine auswählen, den einen um $1$ nach links und den anderen um $1$ nach rechts verschieben. Für welche Anfangspositionen lassen sich alle drei Steine mit einer geeigneten Folge von Schritten auf... | [
"Solution:\n\nAntwort: Dies ist genau dann der Fall, wenn die Summe $S$ der Koordinaten der drei Steine durch $3$ teilbar ist.\n\nOffenbar ändert sich $S$ bei einem Schritt nicht. Sind am Schluss alle Steine auf demselben Punkt $n$, dann muss $S = 3 n$ also durch $3$ teilbar sein.\n\nSei umgekehrt $S = 3 n$. Wähle ... | Switzerland | SMO Finalrunde | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Exactly those initial positions where the sum of the three coordinates is divisible by three. | |
0g0y | Problem:
Finde alle natürlichen Zahlen $n$, sodass gilt:
$$
\sum_{\substack{d \mid n \\ 1 \leq d < n}} d^{2} = 5(n+1)
$$ | [
"Solution:\nWir betrachten $n$ gerade und $n$ ungerade einzeln.\n\nFall 1: $n$ ist gerade:\n$n$ kann nicht $2$ sein. Falls $n > 2$ ist, hat $n$ die Teiler $1$, $2$ und $n / 2$. Es folgt, dass $1 + 4 + (n / 2)^{2} \\leq 5(n+1)$. Umformen liefert $n \\leq 20$. Durchtesten der Fälle ergibt die Lösung $n = 16$.\n\nFall... | Switzerland | IMO-Selektion | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 16 | |
05ob | Problem:
Soit $n \geqslant 2$ un nombre entier. On dispose les $n^{2}$ nombres entiers $1,2, \ldots, n^{2}$ dans les $n^{2}$ cases d'un échiquier $n \times n$ comme suit : la première ligne contient les nombres $1,2, \ldots, n$ (de gauche à droite), la deuxième ligne contient les nombres $n+1, n+2, \ldots, 2 n$ et ain... | [
"Solution:\n\nTout d'abord, on vérifie que la parité de la somme des nombres écrits ne change pas lorsqu'on effectue une étape. La somme des $n^{2}$ premiers entiers vaut $n^{2}(n^{2}+1)/2$, qui est paire si et seulement si $n$ est pair. Comme on veut obtenir la configuration où il n'y a que des 0 (dont la somme de... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Algorithms",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | n must be even; the minimal number of steps is 3n^2/4. | |
035c | Problem:
The point $K$ on the edge $AB$ of the cube $ABCD A_{1}B_{1}C_{1}D_{1}$ is such that the angle between the line $A_{1}B$ and the plane $(B_{1}CK)$ is equal to $60^{\circ}$. Find $\tan \alpha$, where $\alpha$ is the angle between the planes $(B_{1}CK)$ and $(ABC)$. | [
"Solution:\n\nWe may assume that the edges of the cube have length $1$. Denote $M = A_{1}B \\cap KB_{1}$ and set $KB = x$. Since $\\triangle KBM \\sim \\triangle A_{1}MB_{1}$ it follows that $\\frac{MB}{A_{1}M} = x$. Now using the identity $A_{1}M + MB = \\sqrt{2}$, we get $MB = \\frac{x \\sqrt{2}}{x+1}$.\n\nDenote... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(5) | |
0edd | Given an acute triangle $ABC$ let $D$ denote the foot of the altitude from $A$, and let $H$ be the orthocentre. Drop the tangent from the point $B$ to the circle centered at $H$ with radius $HD$ and let $P$, distinct from $D$, be the point where the tangent touches the circle. Drop the tangent from the point $C$ to the... | [
"There are two different configurations of this problem depending on whether the point $P$ lies inside the triangle $ABC$ or outside, so we use directed angles in our solution.\nLet $K_1$ denote the circle centered at $H$ with radius $HD$. Let $K_2$ denote the circle centered at $A$ with radius $AD$. Let $E$ and $F... | Slovenia | Slovenija 2016 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0kzu | Problem:
Compute the number of ways to shade in some subset of the 16 cells in a $4 \times 4$ grid such that each of the 25 vertices of the grid is a corner of at least one shaded cell. | [
"\nObserve that every corner cell must be shaded, as they are the only cells incident to the four corners of the grid. Furthermore, for each side of the grid, the midpoint of that side is incident to exactly two cells; at least one must be shaded. Finally, at least one of the four central c... | United States | HMMT November 2024 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 1215 | |
01i7 | Let $ABC$ be a triangle with circumcircle $\Gamma$ and circumcenter $O$. Denote by $M$ the midpoint of $BC$. Point $D$ is the reflection of $A$ over $BC$, and $E$ is the intersection of $\Gamma$ and ray $MD$. Let $S$ be the circumcenter of triangle $ADE$. Prove that $A, E, M, O$, and $S$ are concyclic. | [
"**Solution.** Take $S$ to be the point such that $CPSQ$ is a parallelogram, as seen in figure 20. For points $X, Y, Z$ let $\\text{rot } XYZ$ denote the morphism on translations induced by the rotation that takes line $XY$ to line $XZ$, modulo half turn. As $ABCD$ is a cyclic quadrilateral it follows that $\\text{... | Baltic Way | Baltic Way 2021 Shortlist | [
"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 > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometr... | null | proof only | null | |
0asl | Problem:
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function that satisfies the functional equation
$$
f(x-y)=2009 f(x) f(y)
$$
for all $x, y \in \mathbb{R}$. If $f(x)$ is never zero, what is $f(\sqrt{2009})$? | [
"Solution:\n$\\frac{1}{2009}$"
] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 1/2009 | |
0e09 | Find all even positive integers $n$ such that $-53 < \frac{2009}{53-n} < 53 - n$. | [
"If $n < 53$, then $\\frac{2009}{53-n} > 0 > -53$. The inequality $\\frac{2009}{53-n} < 53-n$ is equivalent to $2009 < (53-n)^2$ or $\\sqrt{2009} < 53-n$. Since $\\sqrt{2009} > 44$, we have $53-n > 44$ or $9 > n$. The possible values of $n$ are $2$, $4$, $6$ and $8$, and for any of those we have $53-n \\ge 45 > \\s... | Slovenia | National Math Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 2, 4, 6, 8, 92, 94, 96 | |
02b8 | Problem:
Se dois lados de um triângulo medem $5$ e $7$ cm, então o terceiro lado não pode medir quantos centímetros?
(a) $11$
(b) $10$
(c) $6$
(d) $3$
(e) $1$ | [
"Solution:\n\nA opção correta é (e).\n\nLembre que, num triângulo, a soma de dois lados quaisquer deve ser maior que o terceiro lado. Como $1+5$ não é maior do que $7$, o terceiro lado não pode medir $1~\\mathrm{cm}$."
] | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | MCQ | (e) | |
0649 | Problem:
Gegeben ist eine positive ganze Zahl $n$ und ein Spielbrett, das aus $n+1$ nebeneinander angeordneten quadratischen Feldern besteht, die von links nach rechts von $0$ bis $n$ nummeriert sind. Zu Beginn des Spiels befinden sich $n$ Spielsteine auf dem Feld Nr. $0$ und die anderen Felder sind leer.
Ein geduldi... | [
"Solution:\n\nDie Spielsteine sind ununterscheidbar und haben alle dasselbe Start- und Zielfeld. Daher kann der geduldige Spieler eine Vorschrift erfinden, die ihm angibt, welchen der Steine auf dem jeweils ausgewählten Feld er ziehen soll.\n\nEine mögliche Vorschrift erhält er, indem er die Steine fest von $1$ bis... | Germany | 1. Auswahlklausur | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
01s1 | The sides $AB$ and $CD$ of the trapezoid $ABCD$ ($BC \parallel AD$) are the diameters of the circles $\Gamma_{AB}$ and $\Gamma_{CD}$. $\Gamma_{AB}$ meets the segments $AC$ and $BD$ at points $M$ and $N$, respectively. $\Gamma_{CD}$ meets the segments $AC$ and $BD$ at points $K$ and $L$, respectively.
Prove that $NK \pa... | [
"**Construct the circumference $\\Gamma_{BC}$ with the diameter $BC$.** $\\Gamma_{BC}$ passes through $M$ and $L$. Indeed, $\\angle AMB = 90^\\circ$ ($AB$ is the diameter of $\\Gamma_{AB}$) then $\\angle BMC = 180^\\circ - \\angle AMB = 90^\\circ$, so $M$ belongs to $\\Gamma_{BC}$. Similarly\n$$\n\\angle DLC = 90^\... | Belarus | FINAL ROUND | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0amt | Problem:
Find the last two digits of $0! + 5! + 10! + 15! + \cdots + 100!$.
(a) 00
(b) 11
(c) 21
(d) 01 | [] | Philippines | QUALIFYING STAGE | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | MCQ | 21 |
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