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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0062 | ¿Cuál es el mayor número de casillas que puede colorearse en un tablero de $7 \times 7$ de manera que todo subtablero de $2 \times 2$ posea a lo más $2$ casillas coloreadas? | [] | Argentina | XIX Olimpiada de Matemática de Países del Cono Sur | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | 28 | |
04c0 | Let $D$ be a point on the side $\overline{BC}$ of triangle $ABC$. Denote $\alpha_1 = \angle DAB$ and $\alpha_2 = \angle CAD$. Prove the equality
$$
\frac{\sin(\alpha_1 + \alpha_2)}{|AD|} = \frac{\sin \alpha_1}{|AC|} + \frac{\sin \alpha_2}{|AB|}.
$$ | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
0701 | Problem:
$ABC$ is an acute-angled triangle. $P$ is a point inside its circumcircle. The rays $AP$, $BP$, $CP$ intersect the circle again at $D$, $E$, $F$. Find $P$ so that $DEF$ is equilateral. | [
"\n\n$PAB$ and $PED$ are similar, so $DE / AB = PD / PB$. Similarly, $DF / AC = PD / PC$, so $DE / DF = (AB / AC)(PC / PB)$. Thus we need $PB / PC = AB / AC$. So $P$ must lie on the circle of Apollonius, which is the circle we constructed with center $X$. Similarly, it must lie on the circl... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | P is the common intersection of the three Apollonius circles for triangle ABC, equivalently the point satisfying PB/PC = AB/AC (and the analogous conditions for the other pairs); take the intersection lying inside the triangle. | |
050e | Two circles lie completely outside each other. Let $A$ be the point of intersection of internal common tangents of the circles and let $K$ be the projection of this point onto their external common tangent. The tangents, different from the common tangent, to the circles through point $K$ meet the circles at $M_1$ and $... | [
"Let $L_1$ and $L_2$ be the points of tangency of the external common tangent of the circles, $N_1$ and $N_2$ be the points of tangency of an internal common tangent, and $O_1$ and $O_2$ be the centers of the two circles (see Fig. 18).\n\nAs all the lines $O_1L_1$, $AK$, and $O_2L_2$ are perpendicular to the line $... | Estonia | IMO Team Selection Contest | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
04ny | There are 300 contestants at the competition. Each pair of the contestants is either acquainted (knows each other) or unfamiliar with each other, and there are no three contestants who all know each other. Determine the maximum possible $n$ such that the following conditions hold:
* Every contestant is acquainted with ... | [
"The maximum possible $n$ is $200$.\n\nLet us assume that there is a contestant, say $X$, who knows $201$ other contestants and let those $201$ contestants make up a set $S$. There must exist contestants who know exactly $1$, $2$, $\\dots$, $200$ other contestants.\n\nWe will say that the contestant *has degree* $m... | Croatia | Croatia_2018 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 200 | |
09ro | Problem:
Zij $ABCD$ een koordenvierhoek met $|AD| = |BD|$. Zij $M$ het snijpunt van $AC$ en $BD$. Zij $I$ het middelpunt van de ingeschreven cirkel van $\triangle BCM$. Zij $N$ het tweede snijpunt van $AC$ met de omgeschreven cirkel van $\triangle BMI$. Bewijs dat $|AN| \cdot |NC| = |CD| \cdot |BN|$. | [
"Solution:\n\nOplossing I. Zij $\\alpha = \\angle DAB$. Omdat $|AD| = |BD|$, is dan ook $\\angle ABD = \\alpha$. Vanwege de omtrekshoekstelling vinden we ook $\\angle ACD = \\alpha$, terwijl de koordenvierhoekstelling geeft dat $\\angle BCD = 180^{\\circ} - \\alpha$. Dus $\\angle BCA = 180^{\\circ} - 2\\alpha$. De ... | Netherlands | Selectietoets | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line,... | null | proof only | null | |
09ou | Find all natural numbers $n$ such that $(n + 1)!(n + 2)! = (2n)!$. Here $m! = 1 \cdot 2 \cdot \dots \cdot m$. | [] | Mongolia | MMO2025 Round 2 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | n = 5 | |
01jz | The side $AB$ is the least side in a triangle $ABC$. Points $M$ and $N$ are marked on the rays $CA$ and $CB$ respectively so that $CM = MB$, $CN = NA$. Let $O$ be a circumcenter of the triangle $ABC$.
Prove that $A$, $B$, $N$, $M$, and $O$ are concyclic. | [
"Let $P$ and $Q$ be the midpoints of $BC$ and $AC$ respectively. Since $CM = MB$ and $CN = NA$, we see that $M$ and $N$ lie on the perpendicular bisectors of the sides $BC$ and $AC$ respectively. Since $AB$ is the smallest side, the distance between $A$ and $B$ is less than the distance between $A$ and $C$, so $A$ ... | Belarus | 60th Belarusian 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 > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
0igp | Problem:
A cube with side length $2$ is inscribed in a sphere. A second cube, with faces parallel to the first, is inscribed between the sphere and one face of the first cube. What is the length of a side of the smaller cube? | [
"Solution:\nFirst note that the long diagonal of the cube has length $2\\sqrt{3}$, so the radius of the sphere is $\\sqrt{3}$. Let $x$ be the side length of the smaller cube. Then the distance from the center of the sphere to the far face of the smaller cube is $1 + x$, while the distance from the center of the far... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 2/3 | |
091w | Problem:
Let $a, b, c$ be positive real numbers such that
$$
a+b+c=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}
$$
Prove that
$$
2(a+b+c) \geq \sqrt[3]{7 a^{2} b+1}+\sqrt[3]{7 b^{2} c+1}+\sqrt[3]{7 c^{2} a+1}
$$
Find all triples $(a, b, c)$ for which equality holds. | [
"Solution:\nFrom the AM-GM inequality, we obtain that\n$$\n\\sqrt[3]{7 a^{2} b+1}=2 \\cdot \\sqrt[3]{a \\cdot a \\cdot\\left(\\frac{7 b}{8}+\\frac{1}{8 a^{2}}\\right)} \\leq \\frac{2}{3}\\left(a+a+\\frac{7 b}{8}+\\frac{1}{8 a^{2}}\\right)\n$$\nWe have analogous upper bounds for $\\sqrt[3]{7 b^{2} c+1}$ and $\\sqrt[... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | (1, 1, 1) | |
0jke | Problem:
Let $\omega$ be a fixed circle with radius $1$, and let $BC$ be a fixed chord of $\omega$ such that $BC = 1$. The locus of the incenter of $ABC$ as $A$ varies along the circumference of $\omega$ bounds a region $\mathcal{R}$ in the plane. Find the area of $\mathcal{R}$. | [
"Solution:\nAnswer: $\\pi\\left(\\frac{3-\\sqrt{3}}{3}\\right)-1$\n\nWe will make use of the following lemmas.\n\nLemma 1: If $ABC$ is a triangle with incenter $I$, then $\\angle BIC = 90 + \\frac{A}{2}$.\n\nProof: Consider triangle $BIC$. Since $I$ is the intersection of the angle bisectors, $\\angle IBC = \\frac{... | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / C... | null | proof and answer | pi*(3 - sqrt(3))/3 - 1 | |
03a6 | A regular heptagon $ABCDEFG$ is given. The sides $AB$, $BC$, $CD$, $DE$, $EF$, $FG$ and $GA$ are called opposite to the vertices $E$, $F$, $G$, $A$, $B$, $C$ and $D$, respectively. If $M$ is an interior point of $ABCDEFG$, we say that a line through $M$ and a vertex of $ABCDEFG$ intersects the boundary of $ABCDEFG$ at ... | [
"If $AM$ intersects the segment $DE$ in an interior point (i.e. we get a good point) then $M$ is interior for the triangle $ADE$. The number of the good points which can be assigned to a fixed point $M$ is therefore equal to the number of the triangles amongst $ADE$, $BEF$, $CFG$, $DGA$, $EAB$, $FBC$ and $GCD$ whic... | Bulgaria | Fall Mathematical Competition | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0ja0 | Problem:
If $x^{x} = 2012^{2012^{2013}}$, find $x$. | [
"Solution:\nAnswer: $2012^{2012}$\nWe have\n$$\n2012^{2012^{2013}} = 2012^{2012 \\cdot 2012^{2012}} = \\left(2012^{2012}\\right)^{2012^{2012}}.\n$$\nThus, $x = 2012^{2012}$."
] | United States | HMMT November 2012 | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | 2012^{2012} | |
0lea | A sequence $(a_n)$ is defined by $a_1 = 5$, $a_2 = 13$ and
$$
a_{n+2} = 5a_{n+1} - 6a_n, \forall n \ge 2.
$$
a) Prove that $\gcd(a_n, a_{n+1}) = 1$ for all positive integers $n$.
b) Prove that if $p$ is the prime divisor of $a_{2k}$ then $p-1$ is divisible by $2^{k+1}$ for all non-negative integers $k$. | [
"a) It is easy to find the general formula of $(a_n)$, which is\n$$\na_n = 2^n + 3^n, \\forall n \\in \\mathbb{Z}^+.\n$$\nSuppose that there exists $n \\ge 1$ that $a_n, a_{n+1}$ have common prime divisor $p$. Clearly, $\\gcd(p, 6) = 1$. We have\n$$\n\\begin{cases} p|2^n + 3^n, \\\\ p|2^{n+1} + 3^{n+1}, \\end{cases... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof only | null | |
0ac9 | One school has less than $400$ students in $6$th grade. They are divided in several classes. Six of them have equal number of students and together they have more than $150$ students. In the remaining classes there are $15\%$ more students than in these six classes together. How many students of $6$th grade are there i... | [
"Let $n$ be the total number of students in the six classes that have equal number of students. So $6 \\mid n$. In the remaining classes there are $15\\%$ more students than in these six classes together, so the number of students in the remaining classes is $0.15n$ more than $n$, i.e., $n + 0.15n = 1.15n$.\n\nThe ... | North Macedonia | Macedonian Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 387 | |
0jsp | Problem:
Alice and Bob play a game on a circle with $8$ marked points. Alice places an apple beneath one of the points, then picks five of the other seven points and reveals that none of them are hiding the apple. Bob then drops a bomb on any of the points, and destroys the apple if he drops the bomb either on the poi... | [
"Solution:\n\nLet the points be $0, \\ldots, 7 \\pmod{8}$, and view Alice's reveal as revealing the three possible locations of the apple. If Alice always picks $0,2,4$ and puts the apple randomly at $0$ or $4$, by symmetry Bob cannot achieve more than $\\frac{1}{2}$. Here's a proof that $\\frac{1}{2}$ is always po... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 1/2 | |
01wl | A point $P$ is chosen in the interior of the side $BC$ of the triangle $ABC$. The points $D$ and $E$ are symmetric to $P$ with respect to the vertices $B$ and $C$ respectively. The circumcircles of the triangles $ABE$ and $ACD$ intersect at the points $A$ and $X$. The ray $AB$ intersects the segment $XD$ at the point $... | [
"First we will prove that point $X$ lies on the line $AP$. Let the line $AP$ intersect the circumcircle of the triangle $ACD$ at points $A$ and $Y$.\n\n\n\nSince $PC \\cdot PD = 2PB \\cdot PC = PB \\cdot PE$, the equality $AP \\cdot PY = PC \\cdot DP$ implies $AP \\cdot PY = PB \\cdot PE$, ... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
04bh | An isosceles triangle has the basis of length $a$, legs of length $b$, and the circumradius $R$. Prove that the equality $a^2R^2 + b^4 = 4b^2R^2$ holds, regardless of whether the triangle is acute, right or obtuse. | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
0ids | Problem:
You have a $10 \times 10$ grid of squares. You write a number in each square as follows: you write $1,2,3, \ldots, 10$ from left to right across the top row, then $11,12, \ldots, 20$ across the second row, and so on, ending with $100$ in the bottom right square. You then write a second number in each square, ... | [
"Solution:\n\nThe number in the $i$th row, $j$th column will receive the numbers $10(i-1)+j$ and $10(j-1)+i$, so the question is how many pairs $(i, j)$ ($1 \\leq i, j \\leq 10$) will have\n$$\n101 = [10(i-1)+j] + [10(j-1)+i] \\quad \\Leftrightarrow \\quad 121 = 11i + 11j = 11(i+j).\n$$\nNow it is clear that this i... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 10 | |
09j7 | Are there positive integers $A$, $B$ and $C$ such that $A$, $B$, $C$ have exactly $550$ common divisors and $A$, $B$ have exactly $2000$ common divisors and $A$, $C$ have exactly $1440$ common divisors? | [] | Mongolia | Mongolian Mathematical Olympiad Round 1 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | No | |
09fa | Let $M$ be the centroid and $O$ be the circumcenter of a triangle $ABC$. Take a point $K$ on the line $OM$ in such a way that $\angle KAB = \angle ABC$ and the points $K$, $C$ are on the same side of the line $AB$. Prove that $\angle KCB = \angle ABC$. | [
"Let $P$, $Q$ be the midpoints of the segments $AB$, $BC$ respectively. Let $F := CK \\cap AB$ and let $E := AK \\cap BC$. Since $\\angle BAK = \\angle ABC$, the triangle $AEB$ is an isosceles triangle. Hence $EP$ is an altitude of the triangle $\\triangle AEB$. Therefore $O \\in EP$.\n\nBy assumption, we have $AQ ... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Pappus 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 | |
0l5d | Problem:
Let $f:\{1,2,3,\ldots ,9\} \to \{1,2,3,\ldots ,9\}$ be a permutation chosen uniformly at random from the $9!$ possible permutations. Compute the expected value of $\underbrace{f(f(\cdots f(f(1))\cdots))}_{2025\ f\ s}$. | [
"Solution:\nWe first compute the probability that $f(1) = 1$. Note that $f(1) = 1$ if and only if $1$ is part of a cycle whose length divides $2025$.\nWe claim that for any given $k$, the probability that $1$ is in a cycle of length $k$ is $\\frac{1}{9}$. Indeed, the probability that $f(1) \\neq 1$ is $\\frac{8}{9}... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 7/2 | |
0jdr | Problem:
David has a unit triangular array of 10 points, 4 on each side. A looping path is a sequence $A_{1}, A_{2}, \ldots, A_{10}$ containing each of the 10 points exactly once, such that $A_{i}$ and $A_{i+1}$ are adjacent (exactly 1 unit apart) for $i=1,2, \ldots, 10$. (Here $A_{11}=A_{1}$.) Find the number of loop... | [
"Solution:\n\nAnswer: $60$\n\nThere are $10 \\cdot 2$ times as many loop sequences as loops. To count the number of loops, first focus on the three corners of the array: their edges are uniquely determined. It's now easy to see there are $3$ loops (they form \"$V$-shapes\"), so the answer is $10 \\cdot 2 \\cdot 3 =... | United States | HMMT November 2013 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 60 | |
0gap | 給定一圓及圓上的四個點 $B$, $C$, $X$, $Y$，設 $A$ 為線段 $BC$ 中點，$Z$ 為線段 $XY$ 中點。過 $B$, $C$ 分別作垂直 $BC$ 的直線 $L_1$, $L_2$。設過 $X$ 且垂直 $AX$ 的直線分別交 $L_1$, $L_2$ 於 $X_1$, $X_2$ 兩點，過 $Y$ 且垂直 $AY$ 的直線分別交 $L_1$, $L_2$ 於 $Y_1$, $Y_2$ 兩點。令 $X_1Y_2$ 與 $X_2Y_1$ 相交於 $P$ 點。證明: $\angle AZP = 90^\circ$. | [
"作 $A$ 對 $X_2Y_1$ 垂足 $D$ 點。因為 $\\angle AYY_1 = \\angle ABY_1 = \\angle AXX_2 = \\angle ACX_2 = 90^\\circ$，所以 $DX_2XAC$ 五點共圓，$DY_1YAB$ 五點亦共圓。考慮此兩圓與一開始的給定圓等三圓的根心，得 $AD$, $BY$, $CX$ 三線共點。設此共點為 $S$，且有\n$$\nSA \\cdot SD = SB \\cdot SY = SC \\cdot SX.\n$$\n考慮以 $S$ 為中心，$SA \\cdot SD$ 為反演幂的變換。此變換將 $AD$, $BY$, $CX$ 互換，且 $BA... | Taiwan | 二〇一七數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0f5z | Problem:
The squares of a $1983 \times 1984$ chess board are colored alternately black and white in the usual way. Each white square is given the number $1$ or the number $-1$. For each black square the product of the numbers in the neighbouring white squares is $1$. Show that all the numbers must be $1$. | [] | Soviet Union | 18th ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
06mt | Let $ABCD$ be a quadrilateral inscribed in a circle $\Gamma$ such that $AB = BC = CD$. Let $M$ and $N$ be the midpoints of $AD$ and $AB$ respectively. The line $CM$ meets $\Gamma$ again at $E$. Prove that the tangent at $E$ to $\Gamma$, the line $AD$ and the line $CN$ are concurrent. | [
"Let $P$ be the intersection of the tangent at $B$ to $\\Gamma$ and the line $AD$. Then we have\n$$\n\\angle ABP = \\angle ACB = \\angle BAC.\n$$\nThis shows $BP // CA$. Note that $CB // AP$ since $ABCD$ is an isosceles trapezoid. Therefore, $PBCA$ is a parallelogram. As $N$ is the midpoint of $AB$, the diagonal $C... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0l4y | Problem:
Determine, with proof, all possible values of $\gcd (a^{2} + b^{2} + c^{2}, abc)$ across all triples of positive integers $(a, b, c)$. | [
"Solution:\nFirst, we show that no other $n$ work. If there does exist prime $p \\equiv 3$ (mod $4$) such that $\\nu_{p}(n) = 1$, then $p \\mid abc$; without loss of generality, assume $p$ divides $a$. Then, $p^{2} \\mid a^{2}$ and $p \\mid a^{2} + b^{2} + c^{2}$, so $p \\mid b^{2} + c^{2}$. Since $p$ is $3$ modulo... | United States | HMMT February 2025 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | All positive integers n such that for every prime p congruent to 3 modulo 4, the exponent of p in n is not equal to 1 (i.e., each such prime either does not divide n or divides it with exponent at least 2); primes equal to 2 or congruent to 1 modulo 4 may appear with any nonnegative exponent. | |
001g | Sea $ABC$ un triángulo tal que $A\hat{C}C = 2B\hat{A}A$; además, si $D$ denota al punto del lado $BC$ tal que $AD$ es bisectriz del ángulo $C\hat{A}B$, se tiene que $CD=AB$. Calcular las medidas de los ángulos del triángulo $ABC$. | [] | Argentina | XIX Olimpíada Matemática Argentina | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | español | proof and answer | A = 72°, B = 72°, C = 36° | |
0673 | Find all polynomials $P(x)$ with real coefficients satisfying the equality
$$
(x^2 - 6x + 8)P(x) = (x^2 + 2x)P(x - 2),
$$
for all $x \in \mathbb{R}$. | [
"The given equation is written as\n$$\n(x-2)(x-4)P(x) = x(x+2)P(x-2), \\text{ for all } x \\in \\mathbb{R}, \\quad (1)\n$$\nand so for $x = 0, -2$ and $4$ we obtain: $P(0) = P(-2) = P(2) = 0$.\nTherefore the polynomial $P(x)$ takes the form:\n$$\nP(x) = x(x+2)(x-2)Q(x), \\quad (2)\n$$\nwhere $Q(x)$ is a polynomial ... | Greece | 31st Hellenic Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | P(x) = c x^2 (x^2 - 4) for any real constant c | |
0bva | Let $P$ be a point in the interior of the triangle $ABC$, and let the lines $AP$, $BP$, $CP$ meet the sides $BC$, $CA$, $AB$ respectively at the points $D, E, F$. Let the circles on diameters $BC$ and $AD$ meet at points $a$ and $a'$; the circles on diameters $CA$ and $BE$ meet at points $b$ and $b'$; and the circles o... | [
"\n\nLet $A$, $B$, $C$ and $P$ have position vectors $\\mathbf{a}$, $\\mathbf{b}$, $\\mathbf{c}$ and\n$$\n\\mathbf{p} = \\frac{\\alpha}{\\alpha + \\beta + \\gamma} \\mathbf{a} + \\frac{\\beta}{\\alpha + \\beta + \\gamma} \\mathbf{b} + \\frac{\\gamma}{\\alpha + \\beta + \\gamma} \\mathbf{c},... | Romania | Fifteenth IMAR Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates"
] | null | proof only | null | |
03qg | Let $a_0, a_1, a_2, \dots, a_n, \dots$ be a sequence of numbers satisfying $(3 - a_{n+1}) \cdot (6 + a_n) = 18$, and $a_0 = 3$. Then $\sum_{i=0}^{n} \frac{1}{a_i}$ equals ________. | [
"Set $b_n = \\frac{1}{a_n}$, $n = 0, 1, 2, \\dots$, then $(3 - \\frac{1}{b_{n+1}})(6 + \\frac{1}{b_n}) = 18$, namely,\n$$\n3b_{n+1} - 6b_n - 1 = 0.\n$$\nHence $b_{n+1} = 2b_n + \\frac{1}{3}$, or $b_{n+1} + \\frac{1}{3} = 2(b_n + \\frac{1}{3})$. So $\\{b_n + \\frac{1}{3}\\}$ is a geometric progression with common ra... | China | China Mathematical Competition (Hainan) | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | (1/3)(2^{n+2} - n - 3) | |
0ja9 | Problem:
Let $N$ be a three-digit integer such that the difference between any two positive integer factors of $N$ is divisible by $3$. Let $d(N)$ denote the number of positive integers which divide $N$. Find the maximum possible value of $N \cdot d(N)$. | [
"Solution:\n\nWe first note that all the prime factors of $n$ must be $1$ modulo $3$ (and thus $1$ modulo $6$). The smallest primes with this property are $7, 13, 19, \\ldots$. Since $7^{4} = 2401 > 1000$, the number can have at most $3$ prime factors (including repeats). Since $7 \\cdot 13 \\cdot 19 = 1729 > 1000$... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 5586 | |
0i7l | Problem:
Simplify $\sqrt[2003]{2 \sqrt{11}-3 \sqrt{5}} \cdot \sqrt[4006]{89+12 \sqrt{55}}$. | [
"Solution:\nNote that $(2 \\sqrt{11}+3 \\sqrt{5})^{2}=89+12 \\sqrt{55}$. So, we have\n$$\n\\begin{aligned}\n\\sqrt[2003]{2 \\sqrt{11}-3 \\sqrt{5}} \\cdot \\sqrt[4006]{89+12 \\sqrt{55}} & =\\sqrt[2003]{2 \\sqrt{11}-3 \\sqrt{5}} \\cdot \\sqrt[2003]{2 \\sqrt{11}+3 \\sqrt{5}} \\\\\n& =\\sqrt[2003]{(2 \\sqrt{11})^{2}-(3... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Other"
] | null | final answer only | -1 | |
0i65 | Problem:
Professor Moriarty has designed a "prime-testing trail.” The trail has 2002 stations, labeled $1, \ldots, 2002$. Each station is colored either red or green, and contains a table which indicates, for each of the digits $0, \ldots, 9$, another station number. A student is given a positive integer $n$, and then... | [
"Solution:\n\nNo, this is impossible. Suppose on the contrary that such a trail has been designed. Since there are infinitely many primes, we can choose a prime $p$ with more than $2002$ decimal digits. When we test $p$ on the trail, we will visit some station more than once. Write $p$ as $10^{N} A + 10^{m} B + C$ ... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0g9w | 設 $n$ 為正整數。一條東西向的路上從東到西有 $n$ 個城鎮。每個城鎮都派出兩頭犀牛，一隻從城鎮往東出發，另一隻從城鎮往西出發（犀牛不會轉向），而這 $2n$ 頭犀牛的大小都不一樣。當兩頭犀牛面對面相撞時，大隻的會把小隻的撞出道路；但如果一隻犀牛從背後撞向另一頭犀牛，不論犀牛的大小，從背後被撞的犀牛都會被撞出道路。
假設有兩個城鎮 $A$ 和 $B$，其中 $B$ 在 $A$ 的東邊。如果 $A$ 的東進犀牛可以一路抵達 $B$ 而把其間的犀牛全部撞飛，則稱 $A$ 城輾過 $B$ 城。反之，如果 $B$ 的西進犀牛可以一路抵達 $A$ 並把其間的犀牛全部撞飛，則稱 $B$ 輾過 $A$。
證明：恰有一個城鎮不會被任何其他城鎮輾... | [
"對 $n$ 歸納。當 $n=1$ 時顯然成立。\n\n假設原命題在 $n \\le N$ 時都成立。當 $n = N + 1$ 時，除卻最西邊的西進犀牛和最東邊的東進犀牛（牠們沒有功能），考慮剩餘的 $2N$ 隻犀牛中最大者；不失一般性，假設最大隻的犀牛為從西邊數來第 $k$ 座城鎮的東進犀牛，其中 $k < N + 1$（因為我們不考慮最東邊城鎮的東進犀牛）。\n\n顯然，在第 $k$ 座城鎮以東的城鎮都會被第 $k$ 座城鎮輾過；且第 $k$ 座城鎮以東的城鎮都被這頭最大犀牛擋住，而不能輾過第 $k$ 座城鎮及其西邊的任何城鎮。所以可以丟棄第 $k$ 座城鎮以東的諸城鎮。而基於 $k \\le N$，由歸納假設，剩下的... | Taiwan | 二〇一六數學奧林匹亞競賽第二階段選訓營 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
03jh | Problem:
On a large, flat field $n$ people are positioned so that for each person the distances to all the other people are different. Each person holds a water pistol and at a given signal fires and hits the person who is closest. When $n$ is odd show that there is at least one person left dry. Is this always true whe... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | When the number of people is odd, at least one person is left dry. When the number is even, this is not always true; there are configurations where every person is hit (for example, arranging the people in well-separated mutual nearest neighbor pairs). | |
0dpe | Let $a$, $b$ and $c$ be real numbers satisfying
$$
|(a - b)(b - c)(c - a)| = 1.
$$
Find minimum value of $|a| + |b| + |c|$. | [] | Silk Road Mathematics Competition | XV Silk Road Mathematics Competition | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 2^{2/3} | |
0hgi | In one magic country there are only banknotes of nominal 3, 25 and 80 hryvnyas. Businessman Victor ate in a restaurant of this country for 2024 days in a row, and each day he paid (without change) exactly 1 hryvnya more than the previous one. Is it possible that he paid exactly a million banknotes? | [
"Suppose that he paid the sum of $S$ UAH with $k$ banknotes, among which there are $a$ of 3 UAH, $b$ of 25 UAH, and $c$ of 80 UAH. Then\n$$\nS = 3a + 25b + 80c \\equiv 3a + 3b + 3c = 3k \\pmod{11}.\n$$\nIf $n+1$ is the sum that he paid in the first day, then in the $i$-th day he paid a sum of $n+i$ with precisely $... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | No | |
064p | A triangle $AB\Gamma$ is given with $AB < A\Gamma$. Let $I$ be the point of intersection of its bisectors. Bisector $A\Delta$ meets the circumcircle $C$ of the triangle $B\Gamma$ at the point $N$ with $N \neq I$.
(i) Determine the angles of the triangle $B\Gamma N$ with respect to the angles of the triangle $AB\Gamma$,... | [
"(i)\n$$\n\\widehat{\\Gamma BN} = \\widehat{\\Gamma IN} = \\frac{\\hat{A}}{2} + \\frac{\\hat{\\Gamma}}{2} = \\frac{180^\\circ - \\hat{B}}{2} = 90^\\circ - \\frac{\\hat{B}}{2}\n$$\n$$\n\\widehat{B\\Gamma N} = \\widehat{B\\Gamma IN} = \\frac{\\hat{A}}{2} + \\frac{\\hat{B}}{2} = \\frac{180^\\circ - \\hat{\\Gamma}}{2} ... | Greece | 24th Hellenic 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 and answer | (i) ∠ΓBN = 90° − (∠B)/2, ∠BΓN = 90° − (∠Γ)/2, ∠BNΓ = 90° − (∠A)/2.
(ii) The center of C is the point M where the angle bisector from A meets the circumcircle of triangle ABΓ; equivalently, M lies on line IN (the diameter of C) and on the perpendicular bisector of BI, hence M is the center. | |
0dyo | In a quadrilateral $ABCD$ let $K$ be a point inside the triangle $ABD$ such that triangles $ABD$ and $KCD$ are similar. Prove that triangles $BCD$ and $AKD$ are similar as well. | [
"Since triangles $ABD$ and $KCD$ are similar, we have $\\angle ADB = \\angle KDC$ and $\\frac{|DA|}{|DB|} = \\frac{|DK|}{|DC|}$. We see that\n$$\n\\angle ADK = \\angle ADB - \\angle BDK = \\\\\n= \\angle KDC - \\angle BDK = \\angle BDC\n$$\nand since $\\frac{|DA|}{|DK|} = \\frac{|DB|}{|DC|}$, we conclude that trian... | Slovenia | Slovenija 2008 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08tv | 8 cubes each of edge length 1 are assembled to form a cube of edge length 2. How many straight lines are there in the space which go through at least 2 points, which are vertices of some of the 8 small cubes?
 | [
"There are $3^3 = 27$ points in the space which are the vertices of the 8 small cubes assembled to form the cube of the edge length 2. There are $\\frac{27!}{2!25!} = 351$ ways of choosing 2 different points from these 27 and each chosen pair will determine a straight line in the space going through this pair. Howe... | Japan | Japan Junior Mathematical Olympiad First Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 253 | |
06xq | Let $ABCD$ be a cyclic quadrilateral with $\angle BAD < \angle ADC$. Let $M$ be the midpoint of the arc $CD$ not containing $A$. Suppose there is a point $P$ inside $ABCD$ such that $\angle ADB = \angle CPD$ and $\angle ADP = \angle PCB$.
Prove that lines $AD$, $PM$, $BC$ are concurrent. | [
"Let $X$ and $Y$ be the intersection points of $AM$ and $BM$ with $PD$ and $PC$ respectively. Since $ABCM D$ is cyclic and $CM = MD$, we have\n$$\n\\angle XAD = \\angle MAD = \\angle CBM = \\angle CBY.\n$$\nCombining this with $\\angle ADX = \\angle YCB$, we get $\\angle DXA = \\angle BYC$, and so $\\angle PXM = \\... | IMO | International Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
030t | Problem:
În triunghiul scalen ascuțitunghic $A B C$ se notează cu $D$ piciorul bisectoarei din $A$ și cu $E$ piciorul înălțimii din $A$. Mediatoarea segmentului $A D$ intersectează semicercurile de diametre $A B$ și $A C$, construite în exteriorul triunghiului $A B C$, în $X$, respectiv $Y$. Demonstrați că punctele $X... | [] | Brazil | Al patrulea baraj de selecție pentru OBMJ | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof only | null | |
02qf | Problem:
André, Bianca, Carlos e Dalva querem sortear um livro entre si. Para isto, colocam 3 bolas brancas e 1 preta em uma caixa e combinam que, em ordem alfabética de seus nomes, cada um tirará uma bola, sem devolvê-la à caixa. Aquele que tirar a bola preta ganhará o livro.
a) Qual é a probabilidade de que André ga... | [
"Solution:\n\na) Para André ganhar o livro ele deve retirar a bola preta. Como a caixa contém quatro bolas das quais apenas uma é preta, a probabilidade de ele retirar a bola preta é $\\frac{1}{4}$.\n\nUma outra solução aparece na $2^{\\mathrm{a}}$ solução do item $b$ ).\n\nb)\n1a solução: Para Dalva ganhar o livro... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | a) 1/4, b) 1/4, c) 5/14, d) 1/7 | |
0drm | Let $n > 3$ be a given integer. Find the largest integer $d$ (in terms of $n$) such that for any set $S$ of $n$ integers, there are four distinct (but not necessarily disjoint) nonempty subsets, the sum of the elements of each of which is divisible by $d$. | [
"$d \\ge n$ is not possible. To see this, take a set $S$ of $n$ integers so that each element of $S$ is equal to $1 \\pmod{n}$. The sum of any nonempty subset $T$ of $S$ is equal to $\\#T \\pmod{d}$. Since $d \\ge n$, the only possibility for this to hold is if $d=n$ and $\\#T=n$, i.e., $T=S$. This proves that $d \... | Singapore | Singapur 2015 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | n-2 | |
0972 | Problem:
Determinați toate numerele întregi $n$ pentru care numărul $A=\sqrt[3]{n+\sqrt[3]{n-1}}$ este rațional. | [
"Solution:\n\nFie $A \\in \\mathbb{Q}$. Atunci $\\exists x \\in \\mathbb{Q}: x^{3} = n + \\sqrt[3]{n-1}$. Dar $\\sqrt[3]{n-1} = y \\in \\mathbb{Q}$, deci $y^{3} = n-1$. $n = y^{3} + 1$. Deci $x^{3} = y^{3} + y + 1$.\n\nDacă $n > 1$, atunci $y > 0$. Prin urmare $y^{3} < x^{3} = y^{3} + y + 1 < y^{3} + 3y^{2} + 3y + ... | Moldova | Olimpiada Republicană la Matematică, A doua zi | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Intermediate Algebra > Other",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | n = 0 or n = 1 | |
0htw | Problem:
Let $a$, $b$, $c$ be positive real numbers. Assume that
$$
\frac{a^{19}}{b^{19}} + \frac{b^{19}}{c^{19}} + \frac{c^{19}}{a^{19}} \leq \frac{a^{19}}{c^{19}} + \frac{b^{19}}{a^{19}} + \frac{c^{19}}{b^{19}}
$$
Prove that
$$
\frac{a^{20}}{b^{20}} + \frac{b^{20}}{c^{20}} + \frac{c^{20}}{a^{20}} \leq \frac{a^{20}}{c... | [
"Solution:\nIf we multiply the first equation by $(a b c)^{19}$ then it can be rewritten as\n$$\n\\left(a^{19} - b^{19}\\right)\\left(b^{19} - c^{19}\\right)\\left(c^{19} - a^{19}\\right) \\leq 0.\n$$\nSimilarly, the desired equation is equivalent to\n$$\n\\left(a^{20} - b^{20}\\right)\\left(b^{20} - c^{20}\\right)... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0fay | Problem:
Given an infinite sheet of square ruled paper. Some of the squares contain a piece. A move consists of a piece jumping over a piece on a neighbouring square (which shares a side) onto an empty square and removing the piece jumped over. Initially, there are no pieces except in an $m \times n$ rectangle ($m, n ... | [
"Solution:\n\n2 if $mn$ is a multiple of 3, 1 otherwise\n\nObviously $1 \\times 2$ and $2 \\times 2$ can be reduced to 1. Obviously $3 \\times 2$ can be reduced to 2. Note that pieces on the four $X$ squares can be reduced to a single $X$ provided that the square $Y$ is empty (call this the L move):\n\n. . . . . . ... | Soviet Union | 1st CIS | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 2 if mn is a multiple of 3, 1 otherwise | |
0egn | Problem:
Daljica $AB$ je dolga $20~\mathrm{cm}$, točka $O$ pa je njeno razpolovišče. Krožnica $\mathcal{K}$ se od zunaj dotika krožnic s premeroma $AO$ in $BO$ ter od znotraj dotika krožnice s premerom $AB$ (glej sliko). Koliko centimetrov je polmer krožnice $\mathcal{K}$?
(A) $\frac{5}{2}$
(B) $\frac{10}{3}$
(C) $\f... | [
"Solution:\n\nOznačimo razpolovišče daljice $AO$ s $P$, središče krožnice $\\mathcal{K}$ pa z $R$. Polmer krožnice $\\mathcal{K}$ označimo z $r$. Zaradi simetrije je premica $RO$ pravokotna na premico $AB$. Torej po Pitagorovem izreku velja $|PO|^{2} + |OR|^{2} = |PR|^{2}$. Merjeno v centimetrih je $|PO| = 5$, $|OR... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | B | |
04eh | Let $ABCDA'B'C'D'$ be a cube of edge length $1$. Points $P$, $Q$, $R$ and $S$ are given on the edges $\overline{AB}$, $\overline{AD}$, $\overline{C'D'}$ and $\overline{B'C'}$, respectively, so that $PQRS$ is a square whose centre is in the centre of the cube. What is the length of the side of the square? | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | 3*sqrt(2)/4 | |
0l6t | Problem:
Let $\mathcal{S}$ be the set of all nonconstant monic polynomials $P$ with integer coefficients satisfying $P\left(\sqrt{3} + \sqrt{2}\right) = P\left(\sqrt{3} - \sqrt{2}\right)$. If $Q$ is an element of $\mathcal{S}$ with minimal degree, compute the only possible value of $Q(10) - Q(0)$. | [
"Solution:\nFirst, note that the polynomial $x^{4} - 10x^{2} + 1$ has both $\\sqrt{3} + \\sqrt{2}$ and $\\sqrt{3} - \\sqrt{2}$ as roots. It suffices to check whether a polynomial of degree at most 3 belongs in $\\mathcal{S}$. Suppose $f(x) = a x^{3} + b x^{2} + c x + d \\in \\mathcal{S}$. We compute\n\n$$(\\sqrt{3}... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Algebraic Number Theory > Algebraic numbers"
] | null | proof and answer | 890 | |
0gd1 | 設 $n$ 為大於 3 的正整數。房間中有 $n$ 個人，其中有些人之間存在敵對關係（敵對關係是雙向的）。假設我們知道這群人同時滿足以下兩個性質：
a. 任意的 4 個人中，必存在兩人不互相敵對。
b. 對於任何正整數 $m \ge 1$，如果我們能找到其中 $m$ 個人，他們之間互相都不敵對，則在剩下的 $n-m$ 個人當中，必存在 3 個人，他們之間任兩人都互相敵對（註：自己不會敵對自己。）
試求 $n$ 的最小可能值。 | [
"答. $n$ 的最小可能值為 $7$。\n\n要構造 $n=7$ 的例子，只要將所有人編號 $1$ 到 $n$ 後，讓編號 $i$、$i+1$、$i+2$（mod $n$）互相敵對。易檢查這群人滿足題目條件。故僅須證明 $n=4, 5, 6$ 都是不可能的。\n\n(i) $n=4$：令此四人為 $A$ 到 $D$。由 (a) 知必存在某兩人 $AB$ 互不敵對；但由 (b)，考慮 $m=1$ 並扣除 $D$，則 $ABC$ 必須互相都敵對，故矛盾。\n\n(ii) $n=5$：令此五人為 $A$ 到 $E$，並且不失一般性假設其中 $A$ 敵對最多的人。令 $d$ 為 $A$ 敵對的人數。\n\na. $d=4$：注意到... | Taiwan | 二〇一九數學奧林匹亞競賽第二階段選訓營, 獨立研究(二) | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 7 | |
0awh | Problem:
Suppose that $x$ and $y$ are nonzero real numbers such that $\left(x+\frac{1}{y}\right)\left(y+\frac{1}{x}\right)=7$. Find the value of $\left(x^{2}+\frac{1}{y^{2}}\right)\left(y^{2}+\frac{1}{x^{2}}\right)$. | [] | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 25 | |
0btv | Let $n$ be an integer greater than $2$ and consider the set $A = \{2^n - 1, 3^n - 1, \dots, (n-1)^n - 1\}$. Given that $n$ does not divide any element of $A$, prove that $n$ is a square-free number. Does it necessarily follow that $n$ is a prime number?
Marius Bocanu | [
"Suppose not and write $n = p a$ for a prime $p$ and a number $a > 1$ with $p \\mid a$. Notice that $(a+1)^n - 1 = a((a+1)^{n-1} + (a+1)^{n-2} + \\dots + 1)$ and $a+1 \\equiv 1 \\pmod{p}$ to infer that $n$ divides $(a+1)^n - 1$, a contradiction.\n\nFurther, $n$ needs not be a prime number; take for example $n = 15 ... | Romania | 67th NMO Selection Tests for JBMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | n is square-free; not necessarily prime (for example, n = 15). | |
02da | Given a triangle $ABC$ and a point $P_0$ on the side $AB$. Construct points $P_i$, $Q_i$, $R_i$ as follows: $Q_i$ is the foot of the perpendicular from $P_i$ to $BC$, $R_i$ is the foot of the perpendicular from $Q_i$ to $AC$ and $P_i$ is the foot of the perpendicular from $R_{i-1}$ to $AB$. Show that the points $P_i$ c... | [
"It is clear from the diagram that\n$$\nQ_n Q_{n+1} = P_n P_{n+1} \\cos \\angle B\n$$\n$$\nR_n R_{n+1} = Q_n Q_{n+1} \\cos \\angle C\n$$\n$$\nP_n P_{n+1} = R_{n-1} R_n \\cos \\angle A\n$$\nHence $P_n P_{n+1} = k^n P_0 P_1$, where $|k| = |\\cos A \\cos B \\cos C| < 1$. If we take the direction $A$ to $B$ as positive... | Brazil | II OBM | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
04je | Determine the coefficient of $x^9$ in the polynomial $(1 + x^3 + x^6)^{10}$. | [] | Croatia | Croatia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | final answer only | 210 | |
04mj | Let $\triangle ABC$ be a triangle such that $\angle CAB = 2\angle ABC$. A point $D$ is given in the interior of the triangle $\triangle ABC$, such that $|AD| = |BD|$ and $|CD| = |AC|$. Prove that $\angle ACB = 3\angle DCB$. | [
"Let $E$ be the intersection of the bisector of the segment $\\overline{AB}$ with the segment $\\overline{BC}$.\nDenote $\\beta = \\angle ABC$ and notice that $\\angle ACB = 180^\\circ - 3\\beta$.\nSince $E$ lies on the bisector of the segment $\\overline{AB}$, we have $|AE| = |BE|$. Therefore, $\\angle BAE = \\ang... | Croatia | Croatia_2018 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0ef1 | Problem:
Premica poteka skozi presečišče premic $11 x+3 y-7=0, 12 x+y-19=0$ in razpolovišče daljice s krajiščema $A(3,-2)$ in $B(-1,6)$. Zapiši enačbo te premice $v$ vseh treh oblikah in jih poimenuj. Premico nariši.
 | [
"Solution:\n\nPo reševanju sistema enačb s katerokoli metodo dobimo presečišče $P(2,-5)$. Določimo razpolovišče daljice $AB: S(1,2)$. Izračunamo smerni koeficient $k=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{2+5}{1-2}=-7$.\n\nVstavimo podatke v enačbo premice npr.: $y-y_{1}=k\\left(x-x_{1}\\right)$ in po ureditvi dob... | Slovenia | 17. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Odbirno tekmovanje | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | Explicit (slope–intercept) form: y = -7x + 9; Implicit (general) form: 7x + y - 9 = 0; Intercept form: x/(9/7) + y/9 = 1 | |
0ato | Problem:
If $p$ is a real constant such that the roots of the equation $x^{3}-6 p x^{2}+5 p x+88=0$ form an arithmetic sequence, find $p$. | [
"Solution:\nLet the roots be $b-d$, $b$, and $b+d$. From Vieta's formulas,\n$$\n\\begin{aligned}\n-88 &= (b-d) b (b+d) = b \\left(b^{2} - d^{2}\\right) \\\\\n5p &= (b-d) b + b (b+d) + (b+d)(b-d) = 3b^{2} - d^{2} \\\\\n6p &= (b-d) + b + (b+d) = 3b\n\\end{aligned}\n$$\nFrom (3), $b = 2p$. Using this on (1) and (2) yi... | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 2 | |
0hqt | Problem:
A positive integer is written in each cell of an $8 \times 8$ table so that each entry is the arithmetic mean of some two of its neighbors. Find the maximum number of distinct integers that may appear in the table. | [
"Solution:\n\nFirst consider the minimum number $m$ in the table. If it appears in some cell $A$, two neighboring cells $B, C$ must also contain $m$ because there is no other way for $m$ to be the arithmetic mean of two numbers in the table. Since $B$ cannot neighbor $C$, it must have another neighbor $D$ in additi... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 58 | |
0ktp | Let $ABC$ be a triangle with centroid $G$. Points $R$ and $S$ are chosen on rays $GB$ and $GC$, respectively, such that
$$
\angle ABS = \angle ACR = 180^\circ - \angle BGC.
$$
Prove that $\angle RAS + \angle BAC = \angle BGC$. | [
"**Solution 1 using power of a point** From the given condition that $\\angle ACR = \\angle CGM$, we get that\n$$\nMA^2 = MC^2 = MG \\cdot MR \\Rightarrow \\angle RAC = \\angle MGA.\n$$\nAnalogously,\n$$\n\\angle BAS = \\angle AGN.\n$$\nHence,\n$$\n\\angle RAS + \\angle BAC = \\angle RAC + \\angle BAS = \\angle MGA... | United States | USA TSTST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
00hd | The country Dreamland consists of 2016 cities. The airline Starways wants to establish some one-way flights between pairs of cities in such a way that each city has exactly one flight out of it. Find the smallest positive integer $k$ such that no matter how Starways establishes its flights, the cities can always be par... | [
"The flights established by Starways yield a directed graph $G$ on 2016 vertices in which each vertex has out-degree equal to 1.\n\nWe first show that we need at least 57 groups. For this, suppose that $G$ has a directed cycle of length 57. Then, for any two cities in the cycle, one is reachable from the other usin... | Asia Pacific Mathematics Olympiad (APMO) | APMO 2016 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 57 | |
0jlk | Let $ABC$ be a triangle with orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $AHC$ with the internal bisector of the angle $\angle BAC$. Let $X$ be the circumcenter of triangle $APB$ and $Y$ the orthocenter of triangle $APC$. Prove that the length of segment $XY$ is equal to the c... | [
"It is well-known that the reflection $H'$ of the orthocenter $H$ in the line $AC$ lies on the circumcircle of triangle $ABC$. Hence, the circumcenter of triangle $CAH'$ coincides with the circumcenter of triangle $ABC$. But since $H'$ is the reflection of $H$ in the line $AC$, the triangles $ACH$ and $CAH'$ are sy... | United States | USAMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0gn0 | Find all positive integers $n$ for which every coefficient of the polynomial
$$
P_n(x) = (x^2 + x + 1)^n - (x^2 + x)^n - (x^2 + 1)^n - (x + 1)^n + x^{2n} + x^n + 1
$$
is divisible by $7$. | [
"Using the fact that\n$$\nQ(x)^{7^m} \\equiv Q(x^{7^m}) \\pmod{7}\n$$\nfor all polynomials $Q(x)$ with integer coefficients and for all positive integers $m$, we see that the integers $n = 7^k$ and $n = 7^k + 7^l$ with $0 \\le k \\le l$ satisfy the condition of the problem.\n\nNow we will show that if $n$ is not of... | Turkey | 14th Turkish Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | All positive integers n of the form n = 7^k or n = 7^k + 7^l with 0 ≤ k ≤ l. | |
0ban | Consider an isosceles trapezoid $ABCD$ with perpendicular diagonals. The parallel from the intersection point of the diagonals meets the non-parallel sides $[BC]$ and $[AD]$ at points $P$ and $R$ respectively. Point $Q$ is the mirror image of $P$ across the midpoint of $[BC]$. Show that
a) $QR = AD$;
b) $QR \perp AD$... | [
"Let the diagonals $AC$ and $BD$ meet at point $O$ and let $M$ be the midpoint of the line segment $[BC]$.\n\na) Since $OM$ joins the midpoints of two sides of the triangle $PQR$, $MO \\parallel RQ$ and $OM = \\frac{RQ}{2}$. On the other hand, $[OM]$ is a median of the right-angled triangle $BOC$, hence $OM = \\fra... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane G... | null | proof only | null | |
0h02 | A positive integer $n$ are given. Positive numbers $x_0, x_1, \dots, x_n$ such that $x_0 x_1 \dots x_n = 1$. Find all positive $\gamma$ such that inequality
$$
x_0^\gamma + x_1^\gamma + \dots + x_n^\gamma \ge \frac{1}{x_0} + \frac{1}{x_1} + \dots + \frac{1}{x_n}
$$
holds for any set of numbers $x_0, x_1, \dots, x_n$. | [
"**Answer:** $\\gamma \\ge n$.\n\nAt first we will show that for $0 < \\gamma < n$ there exists a set $x_0, x_1, \\dots, x_n$, for which the inequality from the statement of the problem is not held.\n\nLet $x_0 = x^{-n}$, $x_1 = x_2 = \\dots = x_n = x$ for some $x > 0$. Then we have\n$$\n\\begin{gathered}\nx_0^\\ga... | Ukraine | The Problems of Ukrainian Authors | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | γ ≥ n | |
0h2k | We call a natural number a *twin* if it has two natural divisors whose difference is equal to $2$. Determine whether there are more twin numbers or the numbers that are not twin among the first $20112012$ natural numbers. | [
"Let $N$ be the number of twin numbers that do not exceed $M$. Then $N \\ge N_3 + N_4 - N_{12}$, since all the numbers from $M_3$ and $M_4$ are twin, as they have divisors $1, 3$ and $2, 4$ respectively, and $M_2$ consists of all the numbers that belong to both $M_3$ and $M_4$. But there are also twin numbers that ... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | proof and answer | There are more twin numbers than non-twin numbers among the first 20112012 natural numbers. | |
0glv | Fourteen schools participate in the second Tha Sala Mathematics Talent competition, with each school sending 14 students. The students take tests in 14 rooms, with 14 students in a room such that every room does not contain students from the same school.
Among the students there are 15 students who also participated i... | [] | Thailand | The 14th Thailand Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 13 | |
011v | Problem:
Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of positive real numbers satisfying $i \cdot a_{i}^{2} \geqslant (i+1) \cdot a_{i-1} a_{i+1}$ for $i=1,2, \ldots$ Furthermore, let $x$ and $y$ be positive reals, and let $b_{i}=x a_{i}+y a_{i-1}$ for $i=1,2, \ldots$ Prove that the inequality $i \cdot b_{i}^{2} >... | [
"Solution:\n\nLet $i \\geqslant 2$. We are given the inequalities\n$$\n(i-1) \\cdot a_{i-1}^{2} \\geqslant i \\cdot a_{i} a_{i-2}\n$$\nand\n$$\ni \\cdot a_{i}^{2} \\geqslant (i+1) \\cdot a_{i+1} a_{i-1} .\n$$\nMultiplying both sides of (6) by $x^{2}$, we obtain\n$$\ni \\cdot x^{2} \\cdot a_{i}^{2} \\geqslant (i+1) ... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | proof only | null | |
0c58 | Show that if $a, b, c \in (0, \infty)$, then
$$
\frac{a^2b}{a^6 + b^2 + 4ac + 2} + \frac{b^2c}{b^6 + c^2 + 4ba + 2} + \frac{c^2a}{c^6 + a^2 + 4cb + 2} \le \frac{a+b+c}{8}.
$$ | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0eo8 | Let $O$ be the centre of a two-dimensional coordinate system, and let $A_1, A_2, \dots, A_n$ be points in the first quadrant and $B_1, B_2, \dots, B_m$ points in the second quadrant. We associate numbers $a_1, a_2, \dots, a_n$ to the points $A_1, A_2, \dots, A_n$ and numbers $b_1, b_2, \dots, b_m$ to the points $B_1, B... | [
"Consider first the case that one of the areas is zero, e.g. $\\text{area}(OA_1B_1) = 0$. Then either $a_1 = 0$ or $b_1 = 0$. If $a_1 = 0$, then $\\text{area}(OA_1B_k) = a_1b_k = 0$ for all $k$, which means that all $B_k$ lie on a straight line through $O$ and $A_1$. Likewise, if $b_1 = 0$, then all $A_j$ lie on a ... | South Africa | The South African Mathematical Olympiad Third Round | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0813 | Problem:
Un fiume è attraversato da due ponti $TS$ e $VM$; le due rive $TV$ e $SM$ sono due archi di circonferenza concentrici; i due ponti $TS$ e $VM$ sono allineati con il centro (si veda la figura). Una persona vuole arrivare in $V$ partendo da $T$ scegliendo il percorso più breve tra i due possibili:
(1) seguire ... | [
"Solution:\n\nLa risposta è (C). Misurando l'angolo $\\alpha$ in radianti, la lunghezza del primo percorso è $R \\alpha$, mentre la lunghezza del secondo è $2(R-r)+r \\alpha$. Perciò il primo percorso è quello più corto se e solo se\n$$\nR \\alpha < 2(R-r) + r \\alpha,\n$$\nciaè se e solo se\n$$\n(R-r)(\\alpha-2) <... | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | MCQ | C | |
0cuj | Let $O$ and $I$ be respectively the circumcenter and the incenter of a scalene triangle $ABC$. Let $B'$ be the point symmetric to $B$ with respect to the line $OI$; we assume that $B'$ lies inside the angle $ABI$. Prove that the tangents to the circumcircle of the triangle $BIB'$ at points $B'$ and $I$ meet on the line... | [
"Let $BI \\cap (ABC) = \\{B, S\\}$, and $SB' \\cap CA = P$ (see Fig. 14).\n\nThen $\\angle ATS = \\angle IBB' = \\angle IB'B$. Thus $SB' \\cdot ST = SA^2 = SI^2$, whence the circle $\\gamma = (TIB')$ is tangent to $SI$. This implies $\\angle ITB' = 2\\phi$, so $TA$ is the angle bisector of $\\angle B'TI$. Hence, de... | Russia | XLIII Russian mathematical olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English; Russian | proof only | null | |
0bb2 | The convex quadrilateral $ABCD$ has $\angle BCD = \angle ADC \ge 90^\circ$. The bisectors of the angles $\angle BAD$ and $\angle ABC$ meet at a point $M$, placed on the line $CD$. Prove that $M$ is the midpoint of the segment $[CD]$. | [
"Case I: $AD$ and $BC$ have a common point $E$. Then $M$ is the incenter of the triangle $ABE$, hence $(EM)$ is the bisector of the angle $\\angle AEB$.\nSince $\\angle ECD = \\angle EDC$, the triangle $EDC$ is isosceles with base $[DC]$. Therefore $[EM]$ is a median in triangle $EDC$, so $M$ is the midpoint of the... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"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 | |
0565 | The mediant of two rational numbers $u$ and $v$ is $x = \frac{a+c}{b+d}$, where $\frac{a}{b}$ and $\frac{c}{d}$ are the reduced fractions of $u$ and $v$ respectively. Prove that for any two distinct positive rational numbers $u$ and $x$, there exist infinitely many positive rational numbers $v$, such that $x$ is the me... | [
"Let $u = \\frac{a}{b}$ and $x = \\frac{c}{d}$ be the reduced fractions of $u$ and $x$. We are looking for rational numbers $v$ such that $v = \\frac{mc-a}{md-b}$, where $m$ is large enough integer such that $mc-a$ and $md-b$ are both positive. According to the definition, $x$ is the mediant of $u$ and $v$ as soon ... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0eb0 | Let $a$, $b$ and $c$ be positive integers such that $a^2 + b^2 + c^2$ is divisible by $7$. Prove that $a^4 + b^4 + c^4$ is also divisible by $7$. | [
"When divided by $7$ a perfect square can give the remainder $0$, $1$, $2$ or $4$. For $a^2 + b^2 + c^2$ to be divisible by $7$, the numbers $a^2$, $b^2$ and $c^2$ must either all give the remainder $0$, or they must give three different remainders, $1$, $2$ and $4$.\n\nIn the first case the numbers $a$, $b$, $c$ a... | Slovenia | National Math Olympiad in Slovenia | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof only | null | |
0hu5 | Problem:
Without a calculator, find a factor $85^{9}-21^{9}+6^{9}$ that is between 2000 and 3000. | [
"Solution:\nWe know that $85^{9}-21^{9}$ has $85-21=64$ as a factor, and $6^{9}$ also has $64$ as a factor, so the sum is divisible by $64$.\n\nSimilarly, $-21^{9}+6^{9}$ is divisible by $-21+6=-15$, which means it is divisible by $5$. Since $85^{9}$ is also divisible by $5$, the whole sum is divisible by $5$.\n\nF... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 2240 | |
00uo | Let $O$ and $H$ be the circumcenter and orthocenter of a scalene triangle $ABC$, respectively. Let $D$ be the intersection point of the lines $AH$ and $BC$. Suppose the line $OH$ meets the side $BC$ at $X$. Let $P$ and $Q$ be the second intersection points of the circumcircles of $\triangle BDH$ and $\triangle CDH$ wit... | [
"Let $M$ and $N$ be the midpoints of the sides $AB$ and $AC$, respectively, and $E$ and $F$ be the feet of altitudes drawn from the vertices $B$ and $C$ to the corresponding sides.\nFirst we claim that $N$ lies on the line $PH$. Let $B'$ be diametrically opposite to the vertex $B$ concerning the circumcircle of $\\... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Geometry > Plane Geometry > Transformations > Inversion",
"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 > Distanc... | null | proof only | null | |
0ip8 | Problem:
How many different values can $\angle ABC$ take, where $A, B, C$ are distinct vertices of a cube? | [
"Solution:\nAnswer: 5. In a unit cube, there are 3 types of triangles, with side lengths $(1, 1, \\sqrt{2})$, $(1, \\sqrt{2}, \\sqrt{3})$ and $(\\sqrt{2}, \\sqrt{2}, \\sqrt{2})$. Together they generate 5 different angle values."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 5 | |
03wj | Let $x$, $y$, $z$ be positive numbers, and $\sqrt{a} = x(y-z)^2$, $\sqrt{b} = y(z-x)^2$, $\sqrt{c} = z(x-y)^2$. Prove that $a^2 + b^2 + c^2 \ge 2(ab + bc + ca)$. (Posed by Tang Lihua) | [
"$$\n\\begin{aligned}\n\\sqrt{b} + \\sqrt{c} - \\sqrt{a} &= -(y+z)(z-x)(x-y), \\\\\n\\sqrt{c} + \\sqrt{a} - \\sqrt{b} &= -(z+x)(x-y)(y-z), \\\\\n\\sqrt{a} + \\sqrt{b} - \\sqrt{c} &= -(x+y)(y-z)(z-x),\n\\end{aligned}\n$$\nso\n$$\n\\begin{aligned}\n& (\\sqrt{b} + \\sqrt{c} - \\sqrt{a})(\\sqrt{c} + \\sqrt{a} - \\sqrt{... | China | China Southeastern Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0778 | Problem:
Let $AB$ be a diameter of a circle $\Gamma$ and let $C$ be a point on $\Gamma$ different from $A$ and $B$. Let $D$ be the foot of perpendicular from $C$ onto $AB$. Let $K$ be a point of the segment $CD$ such that $AC$ is equal to the semiperimeter of the triangle $ADK$. Show that the excircle of triangle $ADK... | [
"Solution:\n\nDraw another diameter $PQ \\perp AB$. Let $E$ be the point at which the excircle $\\Gamma_1$ touches the line $AD$. Join $QE$ and extend it to meet $\\Gamma$ in $L$. Draw the diameter $EN$ of $\\Gamma_1$ and draw $QS \\perp NE$ (extended). See the figure. We also observe that $DE = EM = EN / 2$.\n\n![... | India | Indian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0iig | Problem:
Let $ABC$ be a triangle with $AB = 2$, $CA = 3$, $BC = 4$. Let $D$ be the point diametrically opposite $A$ on the circumcircle of $ABC$, and let $E$ lie on line $AD$ such that $D$ is the midpoint of $\overline{AE}$. Line $l$ passes through $E$ perpendicular to $\overline{AE}$, and $F$ and $G$ are the intersec... | [
"Solution:\n\nUsing Heron's formula we arrive at $[ABC] = \\frac{3 \\sqrt{15}}{4}$. Now invoking the relation $[ABC] = \\frac{abc}{4R}$ where $R$ is the circumradius of $ABC$, we compute $R^2 = \\left(\\frac{2 \\cdot 3}{[ABC]^2}\\right) = \\frac{64}{15}$. Now observe that $\\angle ABD$ is right, so that $BDEF$ is a... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 1024/45 | |
0dbi | Each point of the plane has some color. It is known that on every straight line there are points in at most two different colors. What is the maximum possible number of colors present on this plane? | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2 | |
0djt | Let $x$, $y$ and $z$ be positive reals such that $xyz = 1$. Find the largest possible value of the constant $C$ such that the inequality
$$
\left(\frac{x}{1+x}\right)^2 + \left(\frac{y}{1+y}\right)^2 + \left(\frac{z}{1+z}\right)^2 \ge C
$$
must hold under the described condition. | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | 3/4 | |
00fn | Let $F$ be the set of all $n$-tuples $(A_{1}, A_{2}, \ldots, A_{n})$ where each $A_{i}$, $i=1,2, \ldots, n$ is a subset of $\{1,2, \ldots, 1998\}$. Let $|A|$ denote the number of elements of the set $A$. Find the number
$$
\sum_{(A_{1}, A_{2}, \ldots, A_{n})} |A_{1} \cup A_{2} \cup \ldots \cup A_{n}|.
$$ | [
"Let $M$ be a subset of the set $\\{1,2, \\ldots, 1998\\}$ and let $|M|=k$. Then the set $M$ can be obtained as the union of $t$ sets $A_{1}, A_{2}, \\ldots, A_{t}$ in $(2^{t}-1)^{k}$ different ways since each element $x \\in M$ can belong to $2^{t}-1$ nonempty families of subsets $A_{1}, A_{2}, \\ldots, A_{t}$.\n\... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof and answer | 1998 (2^n − 1) 2^{1997 n} | |
00d4 | Sean $a$ y $b$ números enteros positivos tales que $\frac{5a^4 + a^2}{b^4 + 3b^2 + 4}$ es un número entero. Demostrar que $a$ no es primo. | [
"Si $b$ es par, entonces $b^2$ y $b^4$ son divisibles por $4$, por lo tanto $b^4+3b^2+4$ es divisible por $4$. Si $b$ es impar, entonces $b^2 \\equiv 1 \\pmod{4}$, $b^4 \\equiv 1 \\pmod{4}$ y $3b^2 \\equiv 3 \\pmod{4}$. Así que $b^4+3b^2+4 \\equiv 1+3+4 \\equiv 0 \\pmod{4}$. Luego el denominador de la fracción es d... | Argentina | Nacional OMA | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | Spanish | proof only | null | |
0gq9 | Find all positive integers $n$ satisfying $2n + 7 \mid n! - 1$. | [
"The answer is $1$, $5$ and $8$.\n\nChecking by hand for $n = 1, 2, \\ldots, 6$, we see that $1$ and $5$ work. For $n \\ge 7$, $2n + 7$ should be a prime number. Because, otherwise there exists a prime divisor of $2n + 7$ which is less than or equal to $n$ since $2n + 7$ is odd, but it divides $n!$.\n\nNow let $2n ... | Turkey | Team Selection Test for JBMO | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 1, 5, 8 | |
0kzs | In a race among 5 snails, there is at most one tie, but that tie can involve any number of snails. For example, the result of the race might be that Dazzler is first; Abby, Cyrus, and Elroy are tied for second; and Bruna is fifth. How many different results of the race are possible?
(A) 180 (B) 361 (C) 420 (D) 431 (E) ... | [
"If there are no ties, then there are $5!$ possible race results. Suppose $k$ of the 5 snails are tied, where $2 \\le k \\le 5$. There are $\\binom{5}{k}$ ways to choose the snails that are tied, and then considering those snails as a group, there are $6-k$ entrants and therefore $(6-k)!$ orders of finish. The numb... | United States | AMC 10 B | [
"Statistics > Probability > Counting Methods > Permutations",
"Statistics > Probability > Counting Methods > Combinations"
] | null | MCQ | D | |
06kb | Let $f(x)$ be a monic cubic polynomial with $f(0) = -64$ and all roots of $f(x)$ are nonnegative real numbers. What is the largest possible value of $f(-1)$? (A polynomial is *monic* if its leading coefficient is 1.) | [
"The largest possible value of $f(-1)$ is $-125$.\n\nLet $f(x) = (x-a)(x-b)(x-c)$ where $a, b, c \\ge 0$ and $abc = -f(0) = 64$. Then we have\n$$\n\\begin{aligned}\nf(-1) &= - (1+a)(1+b)(1+c) \\\\\n&= -1 - (a+b+c) - (ab+bc+ca) - abc \\\\\n&\\le -1 - 3\\sqrt[3]{abc} - 3\\sqrt[3]{a^2b^2c^2} - abc \\\\\n&= -125.\n\\en... | Hong Kong | HKG TST | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | -125 | |
02zs | Problem:
Em uma loja de chocolates, existem caixas com $8$, $9$ e $10$ chocolates. Observe que algumas quantidades de chocolates não podem ser compradas exatamente como, por exemplo, $12$ chocolates.
a) Encontre outra quantidade de chocolates que não pode ser comprada.
b) Verifique que todo número maior que $56$ pod... | [
"Solution:\n\na) Não é possível comprarmos $15$ chocolates, pois $15 > 10$ e a soma das quantidades de quaisquer duas caixas é maior que $15$.\n\nb) Inicialmente note que os números de $57$ a $64$ podem ser escritos na forma $8x + 9y$:\n\n| $x$ | $y$ | $8x + 9y$ |\n| :---: | :---: | :---: |\n| 9 | 1 | 57 |\n| 5 | 2... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 31 | |
039k | Let $f(x)$ be a monic polynomial of even degree with integer coefficients. It is known that there exist infinitely many integers $x$ for which $f(x)$ is a perfect square. Prove that there exists a polynomial $g(x)$ with integer coefficients such that $f(x) = g^2(x)$. | [
"Let $n = 2k$ and $f(x) = x^{2k} + a_{2k-1}x^{2k-1} + \\dots + a_1x + a_0$, where $a_i$ are integers. First we prove that $f(x)$ can be written in the form\n$$\nf(x) = (x^k + b_{k-1}x^{k-1} + \\dots + b_1x + b_0)^2 + r(x),\n$$\nwhere $b_0, b_1, \\dots, b_{k-1}$ are rational numbers and $r(x)$ is a polynomial with r... | Bulgaria | Bulgarian National Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
065q | In the plane are given $\nu$ different points such that any three of them are not collinear. We color these points red, green and black. In the sequel we consider all line segments with ends these $\nu$ points and we correspond to each of them an “algebraic value” according to the following rules:
1) If at least one of... | [
"From the three rules for the determination of the algebraic value of the line segments we have the following table:\n\n\nLet now that we have $\\kappa$ red, $\\pi$ green and $\\mu$ black points. Then it is clear that $\\kappa + \\pi + \\mu = \\nu$.\n\nThe $\\kappa$ red points determine $\\... | Greece | SELECTION EXAMINATION | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | -⌊ν/2⌋ | |
072z | Let $ABC$ be a triangle with $AB = AC$, and let $\Gamma$ be its circumcircle. Suppose the incircle $\gamma$ of $ABC$ moves (slides) on $BC$ in the direction of $B$. Prove that when $\gamma$ touches $\Gamma$ internally, it also touches the altitude through $A$. | [
"Let $\\gamma'$ be the position of $\\gamma$, when it touches $\\Gamma$ internally, and let $K$ be its centre. Let $O$ be the circumcentre and $I$ be the in-centre of $ABC$. Since $AB = AC$, both of these lie on the altitude $AD$. If $T$ is the point of contact of $\\Gamma$ and $\\gamma'$, then $T, K, O$ are collin... | India | Indija TS 2007 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof only | null | |
0jb1 | Problem:
Laura won the local math olympiad and was awarded a "magical" ruler. With it, she can draw (as usual) lines in the plane, and she can also measure segments and replicate them anywhere in the plane. She can also divide a segment into as many equal parts as she wishes; for instance, she can divide any segment i... | [
"Solution:\n\nLaura should extend the line $A M$ beyond $M$. Measure $A M$ and find the point $P$ on the extension of $A M$ beyond $M$ such that $A M = M P$. Vertical angles $\\angle C M P = \\angle D M A$, $C M = M D$ and $A M = M P$ so $\\triangle P M C$ is congruent to $\\triangle A M D$ by $\\mathrm{SAS}$.\n\nB... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
06aw | Determine all pairs $(k, n)$ of positive integers satisfying the equation
$$
1! + 2! + \cdots + k! = 1 + 2 + \cdots + n.
$$ | [
"We first compute the entries of the following matrix\n\n| $k$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n|-----|---|---|---|---|---|----|-----|------|-------|--------|\n| $k!$ | 1 | 2 | 6 | 24 | 120 | 720 | 5040 | 40320 | 362880 | 3628800 |\n| $1!+2!+\\cdots+k!$ | 1 | 3 | 9 | 33 | 153 | 873 | 5913 | 46233 | 40911... | Greece | Selection Examination | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | (1, 1), (2, 2), (5, 17) | |
06dc | Let $a$, $b$, $c$ be positive real numbers. Prove that
$$
(a+b)^2 + (a+b+4c)^2 \ge \frac{100abc}{a+b+c}.
$$ | [
"By the AM-GM inequality, we have\n$$\n(a + b + 4c)^2 \\geq (2\\sqrt{(a+b)(4c)})^2 = 16(a+b)c\n$$\nand\n$$\n\\frac{100abc}{a+b+c} \\leq \\frac{100c}{a+b+c} \\left(\\frac{a+b}{2}\\right)^2 = \\frac{25(a+b)^2c}{a+b+c}.\n$$\nTherefore, it suffices to prove\n$$\nd^2 + 16cd \\geq \\frac{25cd^2}{c+d}\n$$\nwhere $d = a + ... | Hong Kong | IMO HK TST | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0amo | Problem:
A sequence of numbers is defined using the relation
$$
a_{n} = -a_{n-1} + 6 a_{n-2}
$$
where $a_{1} = 2$, $a_{2} = 1$. Find $a_{100} + 3 a_{99}$. | [] | Philippines | AREA STAGE | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 7*2^98 | |
02mf | Problem:
Duas partículas percorrem um caminho circular de $120~\mathrm{m}$ de comprimento. A velocidade de uma delas é $2~\mathrm{m}/\mathrm{s}$ maior do que a da outra e ela completa cada volta num tempo que é 3 segundos inferior ao da outra. Qual é a velocidade de cada partícula? | [
"Solution:\n\nDenotemos as partículas por $A$ e $B$ e seja $v$ a velocidade da partícula $B$. Supondo que $A$ seja a mais rápida, temos que $v+2$ é a velocidade de $A$. Assim, o tempo que $B$ demora para dar uma volta é $120 / v$ e o tempo que $A$ demora é $120 /(v+2)$. Como esse tempo é três segundos inferior ao d... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 8 m/s and 10 m/s | |
09v1 | We are given a triangle with an additional two points on each side. So in total, there are nine points (see figure).
We want to choose three of the nine points that are not on one line. For example, we could choose (1) the three vertices of the triangle, or (2) the left vertex and the two ad... | [] | Netherlands | Junior Mathematical Olympiad, September 2019 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | final answer only | 72 | |
0b0i | Problem:
In $\triangle PMO$, $PM = 6\sqrt{3}$, $PO = 12\sqrt{3}$, and $S$ is a point on $MO$ such that $PS$ is the angle bisector of $\angle MPO$. Let $T$ be the reflection of $S$ across $PM$. If $PO$ is parallel to $MT$, find the length of $OT$. | [] | Philippines | Philippines Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 2√183 |
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