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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0e89 | Problem:
Denimo, da obstajata taki točki $D$ na stranici $AB$ in $E$ na stranici $AC$ trikotnika $ABC$, da je $|AE| = |ED| = |DB|$ in $|AD| = |DC| = |CB|$. Določi velikosti kotov trikotnika $ABC$. | [
"Solution:\n\n\n\nOznačimo $\\angle EBD = \\alpha$. Potem je $\\angle DEB = \\alpha$, torej je $\\angle EDA = 2\\alpha$ in $\\angle DAE = 2\\alpha$. Sledi $\\angle DEC = 4\\alpha$ oziroma $\\angle BEC = 3\\alpha$. Hkrati je $\\angle ACD = \\angle DAC = 2\\alpha$, torej je $\\angle BDC = 4\\... | Slovenia | 57. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | Angle at A is 36 degrees, angle at B is 72 degrees, angle at C is 72 degrees. | |
08oj | Problem:
Prove there are no integers $a$ and $b$ satisfying the following conditions:
i) $16 a - 9 b$ is a prime number
ii) $a b$ is a perfect square
iii) $a + b$ is a perfect square | [
"Solution:\nSuppose $a$ and $b$ be integers satisfying the given conditions. Let $p$ be a prime number, $n$ and $m$ be integers. Then we can write the conditions as follows:\n$$\n\\begin{aligned}\n& 16 a - 9 b = p \\\\\n& a b = n^2 \\\\\n& a + b = m^2\n\\end{aligned}\n$$\nMoreover, let $d = \\gcd(a, b)$ and $a = d ... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equa... | null | proof only | null | |
00xe | Problem:
Is it possible to draw a hexagon with vertices in the knots of an integer lattice so that the squares of the lengths of the sides are six consecutive positive integers? | [
"Solution:\n\nThe sum of any six consecutive positive integers is odd. On the other hand, the sum of the squares of the lengths of the sides of the hexagon is equal to the sum of the squares of their projections onto the two axes. But this number has the same parity as the sum of the projections themselves, the lat... | Baltic Way | Baltic Way 1992 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | No | |
01dl | Let $a_1, \dots, a_{2016}$ and $b_1, \dots, b_{2016}$ be two reorderings of the numbers $1, \dots, 2016$. Prove that
$$
2017 \mid a_i b_i - a_j b_j
$$
for some distinct indices $i$ and $j$. | [
"The number $2017$ is prime. Clearly, all $a_n b_n \\neq 0 \\pmod{2017}$. If all $a_n b_n$ were non-congruent modulo $2017$, then\n$$\na_1 b_1, a_2 b_2, \\dots, a_{2016} b_{2016}\n$$\nwould be another reordering of $1, \\dots, 2016$ modulo $2017$, and so\n$$\n\\prod_{n=1}^{2016} a_n b_n \\equiv 2016! \\equiv -1 \\p... | Baltic Way | Baltic Way 2016 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null | |
078f | Let $ABC$ be an acute-angled triangle with $AB < AC$, incenter $I$, and let $M$ be the midpoint of major arc $BAC$. Suppose the perpendicular line from $A$ to segment $BC$ meets lines $BI$, $CI$, and $MI$ at points $P$, $Q$, and $K$ respectively. Prove that the $A$-median line in $\triangle AIK$ passes through the circ... | [
"**Solution A**\nObserve that $\\angle PIQ = 90 - \\frac{\\angle A}{2}$, $\\angle IQP = 90 - \\frac{\\angle C}{2}$ and $\\angle IPQ = 90 - \\frac{\\angle B}{2}$. Thus, if $\\triangle DEF$ is the orthic triangle of $\\triangle IPQ$, then it is similar to $\\triangle ABC$.\nLet $\\ell$ be the $I$ midline in $\\triang... | India | IMO TST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Homothet... | null | proof only | null | |
044m | Function $f(x)$ with domain $\mathbb{R}$ satisfies: when $x \in [0, 1)$, $f(x) = 2^x - x$, and for any real number $x$, there is $f(x) + f(x+1) = 1$. Denote $a = \log_2 3$, and then the value of expression $f(a) + f(2a) + f(3a)$ is ______. | [
"By the conditions, we know that $f(x+n) = 1 - f(x)$ when $n$ is odd and $f(x+n) = f(x)$ when $n$ is even.\nNote that $a = \\log_2 3 \\in [1, 2)$, $2a = \\log_2 9 \\in [3, 4)$, $3a = \\log_2 27 \\in [4, 5)$. Therefore,\n$$\n\\begin{aligned}\nf(a) + f(2a) + f(3a) &= 1 - f(a - 1) + 1 - f(2a - 3) + f(3a - 4) \\\\\n&= ... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | 17/16 | |
071v | Problem:
1. For a convex hexagon $A B C D E F$ in which each pair of opposite sides is unequal, consider the following six statements:
$$
\begin{array}{ll}
\text{(a$_1$) } A B \text{ is parallel to } D E ; & (a_2)\ A E = B D \\
(b_1)\ B C \text{ is parallel to } E F ; & (b_2)\ B F = C E \\
\text{(c$_1$) } C D \text{ i... | [
"Solution:\n\n(a) Suppose all the six statements are true. Then $A B D E$, $B C E F$, $C D F A$ are isosceles trapeziums; if $K, L, M, P, Q, R$ are the mid-points of $A B$, $B C$, $C D$, $D E$, $E F$, $F A$ respectively, then we see that $K P \\perp A B, E D$; $L Q \\perp B C, E F$ and $M R \\perp C D, F A$.\n\n![]... | India | INMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0b4s | Problem:
There are three novel series Peter wishes to read. Each consists of $4$ volumes that must be read in order, but not necessarily one after the other. Let $N$ be the number of ways Peter can finish reading all the volumes. Find the sum of the digits of $N$. (Assume that he must finish a volume before reading a ... | [] | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Statistics > Probability > Counting Methods > Permutations",
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 18 | |
09pn | Problem:
Vier punten $P$, $Q$, $R$ en $S$ liggen in deze volgorde op een cirkel, zodat $\angle P S R = 90^\circ$. Zij $H$ en $K$ de voetpunten van de loodlijnen uit $Q$ op respectievelijk $P R$ en $P S$. Zij $T$ het snijpunt van $H K$ en $Q S$. Bewijs dat $|S T| = |T Q|$. | [] | Netherlands | TOETS TRAININGSKAMP | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
05ch | The midpoint of the hypotenuse $AB$ of a right-angled triangle $ABC$ is $M$. A point $D$ lies on the side $BC$ such that the circumcircle of the triangle $ACD$ intersects the line $DM$ at some point $K$ between the points $D$ and $M$. Let $L$ be the reflection of the point $K$ from the point $M$. The circumcircles of t... | [
"The right angle $DCA$ subtends the chord $AD$ of the circumcircle of triangle $ACD$ (Fig. 26), therefore $AD$ is a diameter of this circle. Since point $K$ lies on the same circle, $\\angle AKD = 90^\\circ$. From supplementary angles we get $\\angle MKA = 180^\\circ - \\angle AKD = 90^\\circ$.\n\nFrom the conditio... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 90° | |
0dzk | Let $K$ be a subset of the set $\mathbb{N}$ of all natural numbers with the property that for any two numbers $a$ and $b$ in $K$ either $a$ divides $b$ or $b$ divides $a$. Prove that every $c \in K$ is greater than the sum of all those numbers in $K$ that are smaller than $c$. | [
"Denote the elements of the set $K$ by $a_1, a_2, a_3, \\dots$, so that $a_1 < a_2 < a_3 < \\dots$. We will prove the claim by induction. Obviously, the sum of all numbers less than $a_2$ is equal to $a_1$ and is therefore less than $a_2$.\n\nNow, assume that for a positive integer $n$ we have $a_n > a_{n-1} + a_{n... | Slovenia | Slovenija 2008 | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
02hp | Problem:
Encontre o produto: $\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right) \ldots\left(1-\frac{1}{225}\right)$.
(A) $\frac{10}{125}$
(B) $\frac{5}{9}$
(C) $\frac{3}{5}$
(D) $\frac{8}{15}$
(E) $\frac{1}{120}$ | [
"Solution:\n\n(D) Cada um dos fatores é da forma \"diferença de quadrados\", isto é $a^{2}-b^{2}$, onde $a=1$\n\n$\\left(1-\\frac{1}{4}\\right)\\left(1-\\frac{1}{9}\\right)\\left(1-\\frac{1}{16}\\right) \\ldots\\left(1-\\frac{1}{225}\\right)=\\left(1-\\frac{1}{2^{2}}\\right)\\left(1-\\frac{1}{3^{2}}\\right)\\left(1... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | D | |
0d91 | Prove that there are infinitely many positive integers $n$ such that $n$ divides $2017^{2017^{n}-1}-1$ but $n$ does not divide $2017^{n}-1$. | [
"We will prove that if $a>1$ is a positive integer such that $a-1$ is not a power of $2$, then there are infinitely many positive integers $n$ such that $n$ divides $a^{a^{n}-1}-1$ but $n$ does not divide $a^{n}-1$. Therefore, the given problem follows directly as an application with $a=2017$.\n\nIn fact, there is ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
00s3 | Let $ABC$ be an acute triangle and $D$ a variable point on side $AC$. Point $E$ is on $BD$ such that $BE = \frac{BC^2 - CD \cdot CA}{BD}$. As $D$ varies on side $AC$ prove that the circumcircle of $ADE$ passes through a fixed point other than $A$. | [
"Let the circumcircle of triangle $CED$ intersect $BC$ at point $G$. From power of point we have\n$$\nBG \\cdot BC = BE \\cdot BD\n$$\nCombining (1) with the problem statement we get\n$$\n\\frac{BG \\cdot BC}{BD} = BE = \\frac{BC^2 - CD \\cdot CA}{BD}\n$$\nand from here we get\n$$\nCD \\cdot CA = BC(BC - BG) = BC \... | Balkan Mathematical Olympiad | BMO 2017 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centro... | English | proof only | null | |
07tt | Prove that
$$
\frac{3x^2 - 2x + 3}{4} \le \sqrt[3]{\frac{x^6 + 1}{2}} \le 2x^2 - 3x + 2,
$$
for all non-negative real $x$, with equality iff $x = 1$. | [
"These inequalities are surely true when $x = 0$, so assume $x > 0$ and let $y = x + \\frac{1}{x}$, so that $y \\ge 2$, and\n$$\ny^3 - 3y = x^3 + 3x + \\frac{3}{x} + \\frac{1}{x^3} - 3\\left(x + \\frac{1}{x}\\right) = x^3 + \\frac{1}{x^3} = \\frac{x^6 + 1}{x^3}.\n$$\nAlso\n$$\n\\frac{3x^2 - 2x + 3}{4} = \\frac{x}{4... | Ireland | IRL_ABooklet | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
04vy | Let $k$ be a circumcircle of an acute triangle $ABC$. Consider a point $P$ on the shorter arc $BC$ of the circle $k$. Denote by $O$ the intersection of segments $AP$ and $BC$. Let $O_1$ and $O_2$ be the circumcentres of triangles $BPQ$ and $CPQ$, respectively. Prove that if the line $O_1O_2$ passes through some vertex ... | [
"Circles with centers $O_1$ and $O_2$ have a common chord $PQ$. The line $O_1O_2$ therefore intersects the segment $PQ$ in its midpoint since it is the perpendicular bisector of $PQ$. Thus, the line $O_1O_2$ cannot pass through the vertex $A$, since it lies on the line $PQ$, but not inside the segment $PQ$. Let us ... | Czech Republic | Second Round of the 73rd Czech and Slovak Mathematical Olympiad (January 16th, 2024) | [
"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"
] | English | proof only | null | |
0f35 | Problem:
A grasshopper hops about in the first quadrant $ (x, y \geqslant 0) $. From $ (x, y) $ it can hop to $ (x + 1, y - 1) $ or to $ (x - 5, y + 7) $, but it can never leave the first quadrant. Find the set of points $ (x, y) $ from which it can never get further than a distance $1000$ from the origin. | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | All points (x, y) with x ≥ 0, y ≥ 0, and x + y < 5. | |
03f9 | Given is a convex 2024-gon $A_1A_2\dots A_{2024}$ and 1000 points inside it, so that no three points are collinear. Some pairs of the points are connected with segments so that the interior of the polygon is divided into triangles. Every point is assigned one number among $\{1, -1, 2, -2\}$, so that the sum of the numb... | [
"Clearly, if there are two adjacent points $A_i$ and $A_{i+1}$ with opposite numbers, the problem is solved. Without restriction, let $A_1 = 1$ and consider all segments $A_iA_{i+1}$ for $i = 1, 2, \\dots, 1012$. Since $A_{1013} = -1$, among the considered segments there is an odd number whose ends are one positive... | Bulgaria | Bulgarian Spring Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | English | proof only | null | |
0kfg | Problem:
Let $\Gamma_{1}$ and $\Gamma_{2}$ be concentric circles with radii $1$ and $2$, respectively. Four points are chosen on the circumference of $\Gamma_{2}$ independently and uniformly at random, and are then connected to form a convex quadrilateral. What is the probability that the perimeter of this quadrilater... | [
"Solution:\n\n\n\nDefine a triplet as three points on $\\Gamma_{2}$ that form the vertices of an equilateral triangle. Note that due to the radii being $1$ and $2$, the sides of a triplet are all tangent to $\\Gamma_{1}$. Rather than choosing four points on $\\Gamma_{2}$ uniformly at random... | United States | HMMT February 2020 | [
"Geometry > Plane Geometry > Circles > Tangents"
] | null | final answer only | 22/27 | |
0di9 | Consider an acute-angled triangle $ABC$, with $AC > AB$, and let $\Gamma$ be its circumcircle. Let $E$ and $F$ be the midpoints of the sides $AC$ and $AB$, respectively. The circumcircle of the triangle $CEF$ and $\Gamma$ meet at $X$ and $C$, with $X \neq C$. The line $BX$ and the tangent to $\Gamma$ through $A$ meet a... | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0kkd | Problem:
Aerith and Bob play a game where they start with a row of 20 squares, and take turns crossing out one of the squares. The game ends when there are two squares left. Aerith wins if the two remaining squares are next to each other, and Bob wins if they are not next to each other.
a) If Aerith goes first, who h... | [
"Solution:\n\na) Bob has a winning strategy. Since he goes second, he will get to take the last turn. If none of the last three squares remaining are together, he wins no matter what. If they are all in a row, he chooses the middle one, and if they are two together and one separate, he chooses one of the two togeth... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) Bob has a winning strategy. b) Aerith has a winning strategy. | |
05kw | Problem:
Soit $k \geqslant 1$ un entier. À chaque client, un opérateur téléphonique propose $k$ numéros pour lesquels la communication est gratuite (si une personne $A$ choisit le numéro de $B$, alors les appels de $A$ vers $B$ et de $B$ vers $A$ sont gratuits). On considère un groupe de $n$ personnes.
1) Si $n \geqs... | [
"Solution:\n\nOn dira pour simplifier qu'une personne $A$ choisit une personne $B$ si le numéro de $B$ fait partie des numéros choisis par $A$ vers lesquels la communication est gratuite.\n\n1)\nComme chaque personne peut choisir au plus $k$ autres personnes, il y a au plus $kn$ paires de personnes qui peuvent comm... | France | Olympiades Françaises de Mathématiques - Test de Janvier | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
00b8 | Let $P$ be a point in the exterior of a circumference $\Gamma$, and let $PA$ be one of the tangents from $P$ to $\Gamma$. The line $l$ passes through $P$ and intersects $\Gamma$ in $B$ and $C$, with $B$ between $P$ and $C$. Let $D$ be the point symmetric to $B$ with respect to $P$. Let $\omega_1$ and $\omega_2$ be the ... | [
"\nNote that $E\\hat{B}P = E\\hat{A}P = E\\hat{A}D = E\\hat{F}D$; then, $l \\parallel DF$. As a consequence, if $K$ is the second intersection point of the line $BC$ with $\\omega_1$, it follows that $CFDK$ is an isosceles trapezoid and so, $CF = KD$.\n\nOn the other hand, since $K\\hat{D}A... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
05nx | Problem:
Soit $n \geq 1$ un entier. Un groupe de $2n$ personnes se réunit. Chacune de ces personnes possède au moins $n$ amies dans ce groupe (en particulier, si $A$ est amie avec $B$ alors $B$ est amie avec $A$, et on n'est pas ami avec soi-même). Prouver que l'on peut disposer ces $2n$ personnes autour d'une table r... | [
"Solution:\n\nConsidérons une disposition arbitraire des invités :\nAB.... $A$.\nou $A, B$ désignent des personnes différentes, mais les deux $A$ représentent la même personne (on est autour d'une table ronde).\nDans ce qui suit, on considérera la disposition comme une suite de personnes, ordonnée de gauche à droit... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Other"
] | null | proof only | null | |
0cos | A polynomial $P(x)$ of degree $n \ge 3$ has $n$ real roots $x_1 < x_2 < \dots < x_n$ such that $x_2 - x_1 < x_3 - x_2 < \dots < x_n - x_{n-1}$. Prove that the maximum of the function $y = |P(x)|$ on the segment $[x_1, x_n]$ is attained at a point belonging to the segment $[x_{n-1}, x_n]$.
Многочлен $P(x)$ степени $n \... | [
"Заметим, что максимум функции $|P(x)|$ не может достигаться в точке $x_i$, ибо $|P(x_i)| = 0$. Рассмотрим произвольную точку $a \\in (x_i, x_{i+1})$ при $i < n-1$; положим $t = a - x_i$, $b = x_n - t$. Заметим, что $b \\in (x_{n-1}, x_n)$, поскольку $x_n > b > x_n - (x_{i+1} - x_i) > x_n - (x_n - x_{n-1}) = x_{n-1... | Russia | Final round | [
"Algebra > Algebraic Expressions > Polynomials"
] | English; Russian | proof only | null | |
0fz3 | Problem:
Seien $a$, $b$, $c$ positive reelle Zahlen mit $a b c \geq 1$. Beweise die Ungleichung
$$
\frac{a^{4}-1}{a b^{3}+a b c+a c^{3}}+\frac{b^{4}-1}{b c^{3}+a b c+b a^{3}}+\frac{c^{4}-1}{c a^{3}+a b c+c b^{3}} \geq 0
$$ | [
"Solution:\n\nLösung 1.\n$$\n\\begin{aligned}\n\\sum_{cyc} \\frac{a^{4}-1}{a b^{3}+a b c+a c^{3}} & =\\sum_{cyc} \\frac{\\frac{1}{9} a^{3}+\\frac{4}{9} a^{3}+\\frac{4}{9} a^{3}-\\frac{1}{a}}{b^{3}+b c+c^{3}} \\\\\n& \\geq \\sum_{cyc} \\frac{\\frac{1}{9}\\left(a^{3}+4 b^{3}+4 c^{3}\\right)-\\frac{1}{a}}{b^{3}+b c+c^... | Switzerland | IMO Selektion | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof only | null | |
08ka | Problem:
A triangle with area $2003$ is divided into non-overlapping small triangles. The number of all the vertices of all those triangles is $2005$. Show that at most one of the smaller triangles has area less or equal to $1$. | [
"Solution:\nSince all the vertices are $2005$, and the vertices of the big triangle are among them, it follows that the number of the small triangles is at least $2003$. So, it follows that at least one of the small triangles has area at most $1$."
] | JBMO | OJBM | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
06hj | $\triangle ABC$ is acute with $AB = 13$ and $BC = 7$. $D$ and $E$ are points on $AB$ and $AC$ respectively such that $BD = BC$ and $\angle DEB = \angle CEB$. Find the product of all possible values of the length of $AE$. | [
"Obviously one possible position of $E$ arises from the case when $BE \\perp CD$ (or equivalently, $ED = EC$). It is denoted by $E_1$ in the figure, in which case $E_1DBC$ is a kite. In particular, $BE_1$ bisects $\\angle CBA$, and hence $\\frac{AE_1}{E_1C} = \\frac{AB}{BC} = \\frac{13}{7}$. With $b$ denoting the l... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 507/10 | |
0033 | La lotería de Binarilandia sortea un número de 100 dígitos 0 y 1 (puede empezar con 0). Un número será premiado si coincide en al menos 51 posiciones con el número sorteado. Determinar la menor cantidad de números que se deben jugar para tener la certeza de que al menos uno de ellos será premiado. | [] | Argentina | Argentina 2006 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Español | proof and answer | 4 | |
08zf | On a $5 \times 5$ square board, a number of tiles consisting of 4 squares, as shown in the figure, are placed along the grid. Note that the placed tiles may be rotated or flipped over. Furthermore, the placed tiles may be overlapped, but must not protrude beyond the board. Assume that every square of the board is cover... | [
"Consider the tile placement obtained by superposing the two given arrangements below. In this placement, each square of the board should be covered by at most two tiles. Furthermore, with the exception of the central square, each of the 24 remaining squares are covered by at least one tile.\n\n![](attached_image_2... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 24 | |
0354 | Problem:
A circle through the vertex $A$ of $\triangle ABC$, $AB \neq AC$, meets the sides $AB$ and $AC$ at points $M$ and $N$, respectively, and the side $BC$ at points $P$ and $Q$, where $Q$ lies between $B$ and $P$. Find $\Varangle BAC$, if $MP \parallel AC$, $NQ \parallel AB$ and $\frac{BP}{CQ}=\frac{AB}{AC}$. | [
"Solution:\nSet $BC = a$, $CA = b$ and $AB = c$. Then $BA \\cdot BM = BP \\cdot BQ$, and\n$$\n\\frac{BM}{c} = \\frac{BP}{a}\n$$\nHence $BQ = \\frac{c^2}{a}$ and analogously $CP = \\frac{b^2}{a}$. Then $BP = \\frac{a^2 - b^2}{a}$, $CQ = \\frac{a^2 - c^2}{a}$ and the condition $\\frac{BP}{CQ} = \\frac{AB}{AC}$ become... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 120° | |
0hm3 | Problem:
Find the value of the infinite continued fraction
$$
1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \frac{1}{\ldots}}}}
$$ | [
"Solution:\nLet the value of the fraction be $x$. Then we get\n$$\nx = 1 + \\frac{1}{x + 1} \\Longrightarrow x(x + 1) = (x + 1) + 1 \\Longrightarrow x^{2} = 2\n$$\nThus, since $x$ is positive, we must have $x = \\sqrt{2}$."
] | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | sqrt(2) | |
0f4l | Problem:
Cucumber River has parallel banks a distance $1$ meter apart. It has some islands with total perimeter $8$ meters. It is claimed that it is always possible to cross the river (starting from an arbitrary point) by boat in at most $3$ meters. Is the claim always true for any arrangement of islands? [Neglect the... | [] | Soviet Union | 16th ASU | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | Yes | |
0f6f | Problem:
The points $A$, $B$, $C$, $D$, $E$, $F$ are equally spaced on the circumference of a circle (in that order) and $AF$ is a diameter. The center is $O$. $OC$ and $OD$ meet $BE$ at $M$ and $N$ respectively. Show that $MN + CD = OA$. | [] | Soviet Union | 19th ASU | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0aao | If $\mathbb{N}$ is the set of positive integers, find all functions $f: \mathbb{N} \to \mathbb{N}$ such that
(i) For any positive integer $a > 2021^{2021}$ it holds that $f(a) \le a$.
(ii) $\frac{a^2 f(b) + b^2 f(a)}{f(a) + f(b)}$ is a positive integer for all $a, b \in \mathbb{N}$. | [] | North Macedonia | Fourth Memorial Mathematical Contest | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | All constant functions f(n) = c for any fixed positive integer c, and the identity function f(n) = n for all positive integers n. | |
09qr | Problem:
Bewijs dat er geen oneindige rij priemgetallen $p_{0}, p_{1}, p_{2}, \ldots$ bestaat met de eigenschap dat voor alle positieve gehele $k$ geldt:
$$
p_{k}=2 p_{k-1}+1 \quad \text{of} \quad p_{k}=2 p_{k-1}-1
$$ | [
"Solution:\n\nStel dat er zo'n oneindige rij bestaat. Door eventueel de eerste twee elementen weg te laten, kunnen we zorgen dat het eerste priemgetal in de rij minstens $5$ is. We nemen dus zonder verlies van algemeenheid aan dat $p_{0} \\geq 5$. Dan weten we dat $p_{0} \\not\\equiv 0 \\bmod 3$.\n\nStel dat $p_{0}... | Netherlands | Dutch TST | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0bcf | Problem:
Fie $n$ un număr întreg strict pozitiv. Determinaţi cel mai mare număr întreg $m$ cu proprietatea că un tablou cu $m$ linii şi $n$ coloane poate fi umplut cu numere reale în aşa fel încât pentru oricare două linii diferite $[a_{1}, a_{2}, \ldots, a_{n}]$ si $[b_{1}, b_{2}, \ldots, b_{n}]$ următoarea relaţie e... | [] | Romania | Olimpiada europeana de matematica a fetelor | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2^n | |
00v7 | Let $f: \pi \to \mathbb{R}$ be a function from the Euclidean plane to the real numbers such that
$$
f(A) + f(B) + f(C) = f(O) + f(G) + f(H)
$$
for any acute triangle $ABC$ with circumcenter $O$, centroid $G$, and orthocenter $H$. Prove that $f$ is constant. | [
"\nLet $G_1 \\neq G_2$ be arbitrary points and let $d$ be the perpendicular bisector of $G_1G_2$. We shall construct two congruent triangles, symmetric with respect to $d$ and with centroids at $G_1$ and $G_2$. Choose an acute triangle $A_1BC$ with centroid $G_1$, and rotate it around $G_1$... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Meth... | English | proof only | null | |
0aqb | Problem:
Two regular polygons with the same number of sides have sides $48~\mathrm{cm}$ and $55~\mathrm{cm}$ in length. What is the length of one side of another regular polygon with the same number of sides whose area is equal to the sum of the areas of the given polygons? | [
"Solution:\n\nLet the number of sides be $n$.\n\nLet $a_1 = 48~\\mathrm{cm}$ and $a_2 = 55~\\mathrm{cm}$ be the side lengths of the two polygons.\n\nThe area $A$ of a regular polygon with $n$ sides and side length $a$ is:\n$$\nA = \\frac{n a^2}{4} \\cot \\frac{\\pi}{n}\n$$\n\nLet $a_3$ be the side length of the req... | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | 73 cm | |
0gnz | Show that
$$
\frac{(b+c)(a^4 - b^2c^2)}{ab + 2bc + ca} + \frac{(c+a)(b^4 - c^2a^2)}{bc + 2ca + ab} + \frac{(a+b)(c^4 - a^2b^2)}{ca + 2ab + bc} \geq 0
$$
for all positive real numbers $a, b, c$. | [
"$$\n\\begin{align*}\n\\sum_{\\text{cyc}} \\frac{(b+c)(a^4 - b^2c^2)}{ab+2bc+ca} &\\ge \\frac{1}{2} \\sum_{\\text{cyc}} (a^3 + abc - b^2c - bc^2) \\\\\n&\\ge \\frac{1}{2} (a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b)) \\ge 0\n\\end{align*}\n$$\n\nby Schur's Inequality if $\\frac{(b+c)(a^4 - b^2c^2)}{ab + 2bc + ca} \\ge ... | Turkey | 17th Turkish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0g4t | Problem:
Soit $n$ un entier strictement positif. Roger a un jardin carré de dimensions $(2 n+1) \times (2 n+1)$, et il y place des barrières pour diviser son jardin en plusieurs parcelles rectangulaires. Une fois terminé, il aura ainsi formé exactement deux rectangles horizontaux $k \times 1$ et deux rectangles vertic... | [
"Solution:\n\nConsidérons les 4 plus grandes parcelles que Roger va délimiter. Nous allons prouver qu'elles forment la bordure du jardin.\nSi on considère une parcelle verticale $1 \\times 2k$, on remarque qu'un de ses petits côtés (horizontaux) doit toucher le bord du jardin, autrement il y aurait un espace de hau... | Switzerland | Deuxième tour 2023 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2^n | |
0ca8 | Problem:
Se consideră $n$ unghiuri în jurul unui punct având măsurile în grade exprimate prin $n$ numere prime distincte. Știind că unghiurile formate de bisectoarele oricăror două unghiuri adiacente dintre cele $n$ unghiuri date inițial au măsurile în grade exprimate prin numere prime, să se determine valorile posibi... | [] | Romania | Olimpiada Națională GAZETA MATEMATICĂ | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 12 | |
0l28 | Problem:
The largest known prime number as of October 2024 is $2^{136279841}-1$. It happens to be an example of a prime number of the form $2x^{2}-1$. Estimate the number of positive integers $x \leq 10^{6}$ such that $2x^{2}-1$ is prime.
Submit a positive integer $E$. If the correct answer is $A$, you will receive $... | [
"Solution:\n\nIf $x \\leq 10^{6}$, then $2x^{2}-1 < 2 \\cdot 10^{12}$. The density of prime numbers up to $2 \\cdot 10^{12}$ is roughly\n$$\n\\frac{1}{\\ln \\left(2 \\cdot 10^{12}\\right)-1} \\approx 0.0366\n$$\nHowever, $2x^{2}-1$ can never be divisible by $2$, $3$, or $5$. Only $\\frac{1}{2} \\cdot \\frac{2}{3} \... | United States | HMMT November 2024 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Other"
] | null | final answer only | 139313 | |
04ht | Determine positive integer $n$ such that the sum of his two smallest divisors is $6$ and the sum of his two largest divisors is $1122$. | [
"**1.1.** Let us denote by $2x_1, 2x_2, \\dots, 2x_{10}$ the number of coins that the knights sitting in chairs 1, 2, ..., 10 had in the beginning, respectively. We have to determine $2x_8$.\nWe have a system of equations: $x_{10}+x_2 = 22, x_1+x_3 = 24, x_2+x_4 = 26, x_3+x_5 = 28, \\dots, x_8+x_{10} = 38, x_9+x_1 ... | Croatia | First round – City competition | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 935 | |
0fj7 | Problem:
El encargado del faro de Finisterre ha recibido la comunicación de que va a haber un corte del suministro eléctrico y debe hacer funcionar el faro con ayuda del generador alimentado con gasóleo. Ese generador consume $6$ litros de gasóleo cada hora y medio litro más cada vez que hay que ponerlo en marcha (ini... | [
"Solution:\n\nLas mejor manera de ahorrar combustible es hacer paradas de $10$ minutos (más largas no están permitidas). Con cada parada ahorramos $1$ litro de gasóleo, pero gastamos medio litro en volver a poner en marcha el generador. Es obvio que, en principio, la mejor estrategia de ahorro consiste en intercala... | Spain | Viernes 19 de enero de 2001 | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 47.5 liters | |
0hfn | In the triangle $ABC$ the median $BM$ is equal to the half of the side $BC$. Prove, that
$$
\angle ABM = \angle BCA + \angle BAC.
$$ | [
"Mark on the extension of median $BM$ by point $M$ such point $D$, that $BM = MD$ (fig. 11). Then triangles $ABM$, $CDM$ are equal by two sides and the angle between them, and also $BD = 2BM = BC$. Then triangle $BDC$ is isosceles, and\n$$\n\\begin{align*}\n\\angle ABM &= \\angle CDM = \\angle BCD = \\angle BCA + \... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0bpg | Problem:
Fie $x, y, z > 0$. Arătaţi că
$$
\frac{x^{3}}{z^{3}+x^{2} y} + \frac{y^{3}}{x^{3}+y^{2} z} + \frac{z^{3}}{y^{3}+z^{2} x} \geq \frac{3}{2}
$$ | [
"Solution:\n\nCu inegalitatea mediilor avem $x^{2} y \\leq \\frac{x^{3}+x^{3}+y^{3}}{3}$, de unde rezultă că\n$$\n\\frac{x^{3}}{z^{3}+x^{2} y} \\geq \\frac{x^{3}}{z^{3}+\\frac{x^{3}+x^{3}+y^{3}}{3}} = \\frac{3 x^{3}}{2 x^{3}+y^{3}+3 z^{3}}\n$$\nşi analoagele.\n\nSumând şi notând $x^{3}=a$, $y^{3}=b$, $z^{3}=c$, est... | Romania | Primul test de selecţie pentru OBMJ | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
02mz | Problem:
Um quadrilátero especial - Os ângulos $A \widehat{B} C$ e $C \widehat{D} A$ do quadrilátero $A B C D$ da figura são retos e os quatro lados do quadrilátero medem números inteiros que são todos distintos. Se $A D=7$ e $B C=11$, quanto medem os lados $A B$ e $C D$ ?
 | [] | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | AB = 3 and CD = 9, or AB = 17 and CD = 19 | |
05wg | Problem:
Martin a écrit le couple d'entiers $(1011,1012)$ au tableau. Puis, chaque minute, si le couple $(a, b)$ est écrit au tableau, il l'efface et le remplace, selon son choix, par l'un des couples $(b, a)$, $(b+1, a-1)$ ou $(b-2, a+2)$, en s'imposant uniquement de n'écrire que des couples dont les deux nombres son... | [
"Solution:\n\nNulle opération ne change la somme des entiers écrits au tableau, donc Martin ne peut écrire que des couples de la forme $(a, 2013-a)$, avec $0 \\leqslant a \\leqslant 2013$.\n\nEn outre, si Martin part d'un couple $(a, b)$ pour lequel $a \\geqslant 1$, il peut écrire successivement les couples $(b+1,... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | (a, 2013 - a) for 0 ≤ a ≤ 2013 | |
0jo0 | Problem:
Each cell of a $2 \times 5$ grid of unit squares is to be colored white or black. Compute the number of such colorings for which no $2 \times 2$ square is a single color. | [
"Solution:\n\nLet $a_{n}$ denote the number of ways to color a $2 \\times n$ grid subject only to the given constraint, and $b_{n}$ denote the number of ways to color a $2 \\times n$ grid subject to the given constraint, but with the added restriction that the first column cannot be colored black-black.\n\nConsider... | United States | HMMT November 2015 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 634 | |
09iy | Find the minimum area of a right triangle with inradius $1$. | [] | Mongolia | Mongolian Mathematical Olympiad Round 1 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 3 + 2 sqrt(2) | |
058v | There are sticks of length $1$ with a number $1$, $2$ or $3$ written on each of them. There is an unlimited supply of sticks with every number. Two triangles consisting of three sticks are considered different if neither of the triangles can be composed from sticks of the other triangle.
a. How many different triangle... | [
"a.\nThere are $3$ triangles having the same number on each side ($111$, $222$, $333$). There are $6$ triangles having one number on two sides and another number on the third side ($112$, $113$, $221$, $223$, $331$, $332$). Only $1$ triangle has a different number on every side ($123$). Thus there are $10$ differen... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | a: 10, b: 26 | |
0c9z | Problem:
Se consideră un patrulater convex $\mathcal{P}$ şi un punct $X$ în interiorul lui $\mathcal{P}$. Fie $M, N, P, Q$ proiecţiile lui $X$ pe laturile lui $\mathcal{P}$. Se ştie că $M, N, P, Q$ se află pe un cerc de centru $L$. Arătaţi că $L$ şi mijloacele $J, K$ ale diagonalelor lui $\mathcal{P}$ sunt pe o dreaptă... | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01iw | (a) Let $n = 8$. Determine whether there exists a labelling of an $8 \times 8$ grid such that the following condition is fulfilled: The difference of any two transversal products is always divisible by $65$.
(b) Let $n = 10$. Determine whether there exists a labelling of a $10 \times 10$ grid such that the following c... | [
"See solution of Problem C 3."
] | Baltic Way | Baltic Way 2023 Shortlist | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Yes for n = 8 (mod 65), and yes for n = 10 (mod 101). | |
0ilk | Problem:
Compute:
$$
\lim_{x \rightarrow 0} \frac{x^{2}}{1-\cos(x)}
$$ | [
"Solution:\nSince $\\sin^{2}(x) = 1 - \\cos^{2}(x)$, we multiply the numerator and denominator by $1 + \\cos(x)$ and use the fact that $x / \\sin(x) \\rightarrow 1$, obtaining\n$$\n\\lim_{x \\rightarrow 0} \\frac{x^{2}}{1-\\cos(x)} = \\lim_{x \\rightarrow 0} \\frac{x^{2}(1+\\cos(x))}{1-\\cos^{2}(x)} = \\lim_{x \\ri... | United States | Harvard-MIT Mathematics Tournament | [
"Precalculus > Limits",
"Precalculus > Trigonometric functions"
] | null | final answer only | 2 | |
068h | Let $AB\Gamma$ be an acute angled triangle inscribed in the circle $c(O, R)$ (with $AB < A\Gamma < B\Gamma$) and let $\Delta, E, Z$ be the touching points of the incircle of the triangle with the sides $B\Gamma$, $A\Gamma$, $AB$, respectively. The circumcircle of the triangle $AEZ$ (say, $(c_1)$) intersects the circle ... | [
"From the inscribed in the circle $c_1$ quadrilateral $AA'IZ$ we have:\n$$\nA\\hat{A}'I = A\\hat{Z}I = 90^\\circ = T A\\hat{A}'I \\quad . (\\alpha)\n$$\nFrom the inscribed quadrilateral $\\Gamma\\Delta IE$ (since $\\Gamma I$ bisector), we have\n$$\n\\hat{\\Delta}_1 = \\frac{\\hat{\\Gamma}}{2} \\qquad (1)\n$$\nFrom ... | Greece | SELECTION EXAMINATION | [
"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 > Circles > Radical axis ... | English | proof only | null | |
07g4 | Consider an acute scalene triangle $ABC$ with circumcircle $\Gamma$. The external angle bisector of $\angle BAC$ meets $BC$ at $X$. Lines $\ell_b$ and $\ell_c$ are tangent lines from $B$ and $C$ to $\Gamma$. A line passes through $X$ and intersects $\ell_b$ and $\ell_c$ at points $Y$ and $Z$, such that $X, Y$ and $Z$ l... | [
"Let $NY$ and $NZ$ intersect $BC$ at $E$ and $F$, respectively. Note that since quadrilaterals $ABYN$ and $ACZN$ are cyclic,\n$$\n\\angle NZY + \\angle NZY = \\angle NAB + \\angle NAC = \\angle A \\implies \\angle ZNY = 180^{\\circ} - \\angle A,\n$$\nhence the pentagon $ACZNE$ is cyclic. Similarly one can show that... | Iran | 38th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triang... | null | proof only | null | |
0750 | Let $D$, $E$, $F$ be points on the sides $BC$, $CA$, $AB$ respectively of a triangle $ABC$ such that $BD = CE = AF$ and $\angle BDF = \angle CED = \angle AFE$. Prove that $ABC$ is equilateral. | [
"\n\nConsider the triangle $BDF$, $CED$ and $AFE$ with $BD$, $CE$ and $AF$ as bases. The sides $DF$, $ED$ and $FE$ make equal angles $\\theta$ with the bases of respective triangles. If $B \\geq C \\geq A$, then it is easy to see that $FD \\geq DE \\geq EF$. Now using the triangle $FDE$, we... | India | Indija mo 2011 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
05w6 | Problem:
Soit $\left(a_{n}\right)_{n \geqslant 1}$ une suite d'entiers naturels non nuls telle que $a_{n+2 m}$ divise $a_{n}+a_{n+m}$ pour tous les entiers $m \geqslant 1$ et $n \geqslant 1$. Démontrer que cette suite est ultimement périodique, c'est-à-dire qu'il existe deux entiers $N \geqslant 1$ et $d \geqslant 1$ ... | [
"Solution:\n\nSupposons qu'il existe un entier $k \\geqslant 9$ tel que $a_{i}<a_{k}$ pour tout $i \\leqslant k-1$. Dans ces conditions, et pour tout entier $i<k / 2$, on sait que $a_{k-2 i}+a_{k-i}$ est un multiple de $a_{k}$ et que $a_{k-2 i}+a_{k-i}<2 a_{k}$, de sorte que $a_{k-2 i}+a_{k-i}=a_{k}$. En choisissan... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0las | Let $a$, $b$, $c$ be positive numbers satisfying:
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \le 16(a + b + c).
$$
Show that
$$
\frac{1}{(a + b + \sqrt{2(a + c)})^3} + \frac{1}{(b + c + \sqrt{2(b + a)})^3} + \frac{1}{(c + a + \sqrt{2(c + b)})^3} \le \frac{8}{9}.
$$
When does the equality occur? | [] | Vietnam | Vietnamese Team Selection for IMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | Equality occurs at a = b = c = 1/4, and the maximum value is 8/9. | |
04id | Let $O$ be the circumcentre of the acute triangle $ABC$ and let $N$ be the foot of the altitude from point $A$. Prove that $\angle BAN = \angle CAO$. (Hong Kong) | [] | Croatia | Croatia Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
03ua | Find the smallest positive integer $n \ge 9$ satisfying that for any group of integers $a_1, a_2, \dots, a_n$, there always exist $a_{i_1}, a_{i_2}, \dots, a_{i_9}$ ($1 \le i_1 < i_2 < \dots < i_9 \le n$) and $b_i \in \{4, 7\}$ ($i = 1, 2, \dots, 9$) such that $b_1 a_{i_1} + b_2 a_{i_2} + \dots + b_9 a_{i_9}$ is a mult... | [
"Let $a_1 = a_2 = 1$, $a_3 = a_4 = 3$, $a_5 = \\dots = a_{12} = 0$.\nIt is easy to check that any 9 integers of them will not meet the requirement. So $n \\ge 13$. We will prove that $n = 13$.\n\nWe only need to prove the following statement:\nGiven a group of $m$ integers $a_1, a_2, \\dots, a_m$, if there are not ... | China | China Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 13 | |
05kv | Problem:
On prend 1008 entiers distincts compris (au sens large) entre 1 et 2014.
a. Montrer qu'il existe trois entiers $a, b, c$ tels que le pgcd de $a$ et $b$ divise $c$ (le pgcd ou Plus Grand Diviseur Commun de deux entiers est le plus grand entier naturel qui les divise tous les deux).
b. Montrer qu'il existe tr... | [
"Solution:\n\na.\nGroupons les entiers de 1 à 2014 en paires d'entiers consécutifs : $(1, 2), (3, 4),\\ldots, (2013, 2014)$. Il y en a 1007. Par le principe des tiroirs, si on choisit 1008 entiers, il y en aura deux d'une même paire, $a$ et $b$, avec $b=a+1$. Leur pgcd vaut 1 (puisqu'il divise leur différence), don... | France | Envoi de combinatoire | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof only | null | |
01wi | Six-digit number $\overline{abcdef}$ is divisible by $3367$.
Prove that the sum $bcdefa + \overline{fabcd}$ is divisible by $3367$ as well.
(As usual, by $\overline{xyz}$ we denote an integer number, which decimal representation consists of digits $x, y, \dots, z$ in that order) | [
"Let $N = \\overline{abcdef}$ be the original six-digit number, which is divisible by $3367$.\n\nLet us write $N$ in terms of its digits:\n$$\nN = 100000a + 10000b + 1000c + 100d + 10e + f\n$$\n\nNow, consider $bcdefa$ and $\\overline{fabcd}$:\n\n$bcdefa$ is the number formed by shifting the first digit $a$ to the ... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof only | null | |
03cs | Find all pairs of real numbers $(x, y)$ such that
$$
\left| \begin{aligned} & \sqrt{x^2 - 2x + y^2 + 4y + 5} + \sqrt{x^2 - 8x + y^2 - 4y + 20} = 5 \\ & 16x^2 + 9y^2 = 68. \end{aligned} \right.
$$ | [
"Consider orthogonal coordinate system and the points $A(1, -2)$ and $B(4, 2)$. Since\n$$\n\\begin{aligned} \\sqrt{x^2 - 2x + y^2 + 4y + 5} &= \\sqrt{(x-1)^2 + (y+2)^2} \\\\ \\sqrt{x^2 - 8x + y^2 - 4y + 20} &= \\sqrt{(x-4)^2 + (y-2)^2}, \\end{aligned}\n$$\nwe conclude that the solutions of the first equation are al... | Bulgaria | Bulgaria 2022 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | (2, -2/3) | |
04bu | The figure consisting of five unit squares is called a “plus”. In how many ways can we place a *plus* on a *board* of the same shape that contains $5 \cdot 5^2$ unit squares, so that the *plus* covers exactly five unit squares of the *board*?
 | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Other"
] | English | proof and answer | 73 | |
050r | In an acute triangle $ABC$, a point $P$ is chosen such that all points symmetrical to $P$ with respect to the sides of $ABC$ lie on the circumcircle of $ABC$. Prove that $P$ is the orthocenter of $ABC$. | [
"Let $A'$, $B'$, $C'$ be points symmetric to the point $P$ with respect to the sides $BC$, $CA$, $AB$ (Fig. 17).\n\nThen $|C'A| = |PA| = |B'A|$, giving that the arcs $AC'$ and $AB'$ of the circumcircle of the triangle $ABC$ are equal. Since $A$ and $C'$ are on the same half-plane from the line $BB'$, and $C$ on the... | Estonia | Estonian Math Competitions | [
"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 | |
0f9x | Problem:
Find all integers $a$, $b$, $c$, $d$ such that $ab - 2cd = 3$, $ac + bd = 1$. | [
"Solution:\n\n$11 = (ab - 2cd)^2 + 2(ac + bd)^2 = (a^2 + 2d^2)(b^2 + 2c^2)$, so we must have either (1) $a^2 + 2d^2 = 1$, $b^2 + 2c^2 = 11$, or (2) $a^2 + 2d^2 = 11$, $b^2 + 2c^2 = 1$.\n\n(1) gives $a = \\pm 1$, $d = 0$, $b = \\pm 3$, $c = \\pm 1$. If $a = 1$ and $d = 0$, then $ac + bd = 1$ implies $c = 1$, and $ab... | Soviet Union | 25th ASU | [
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | null | proof and answer | [(1, 3, 1, 0), (-1, -3, -1, 0), (3, 1, 0, 1), (-3, -1, 0, -1)] | |
0jxj | Problem:
Rebecca has four resistors, each with resistance $1$ ohm. Every minute, she chooses any two resistors with resistance of $a$ and $b$ ohms respectively, and combine them into one by one of the following methods:
- Connect them in series, which produces a resistor with resistance of $a+b$ ohms;
- Connect them i... | [
"Solution:\n\nLet $R_{n}$ be the set of all possible resistances using exactly $n$ $1$-ohm circuit segments (without shorting any of them), then we get\n$$\nR_{n} = \\bigcup_{i=1}^{n-1} \\left( \\{ a+b \\mid a \\in R_{i},\\ b \\in R_{n-i} \\} \\cup \\left\\{ \\left. \\frac{a b}{a+b} \\ \\right|\\ a \\in R_{i},\\ b ... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 15 | |
03o1 | Problem:
Let $ABC$ be a triangle with incenter $I$. Suppose the reflection of $AB$ across $CI$ and the reflection of $AC$ across $BI$ intersect at a point $X$. Prove that $XI$ is perpendicular to $BC$.
(The incenter is the point where the three angle bisectors meet.) | [
"Solution:\n\nSuppose the reflection of $AC$ across $BI$ intersects $BC$ at $E$. Define $F$ similarly for the reflection of $AB$ across $CI$. Also suppose $CI$ intersects $AB$ at $M$ and $BI$ intersects $AC$ at $N$. Since $CA$ and $CF = BC$ are reflections across $CI$, and so are $MA$ and $MF = XM$, we have that $A... | Canada | CMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Transformations"
] | null | proof only | null | |
0e5w | Problem:
Za celi števili $x$ in $y$ velja $x + x y + y^{2} = 1$ in $y(5 + x) \geq 0$. Katerim celim številom je lahko enaka vrednost izraza $x - y$? | [
"Solution:\n\nIz enakosti sledi $x(1 + y) = 1 - y^{2} = (1 + y)(1 - y)$. Če je $y = -1$, je enakost izpolnjena, iz neenakosti pa potem sledi $-(5 + x) \\geq 0$ oziroma $x \\leq -5$. Tako je $x - y = x + 1 \\leq -4$.\n\nČe pa je $y \\neq -1$, potem lahko iz enakosti pokrajšamo $(1 + y)$, da dobimo $x = 1 - y$. Sledn... | Slovenia | 56. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All integers less than or equal to -4, together with -3, -1, and 1 | |
0fdm | Problem:
Encontrar todas las funciones $f: \Re \rightarrow \Re$ tales que
$$
x^{2} \cdot f(x)+f(1-x)=2 x-x^{4}
$$ | [
"Solution:\nSustituimos $x$ por $1-x$ y obtenemos\n$$\n(1-x)^{2} \\cdot f(1-x)+f(x)=2(1-x)-(1-x)^{4}\n$$\nAl eliminar $f(1-x)$ entre esta ecuación y la dada, resulta $f(x)=1-x^{2}$.\nSe comprueba que $f(x)=1-x^{2}$ satisface la ecuación propuesta.\nLa solución buscada es, por tanto, $f(x)=1-x^{2}$."
] | Spain | null | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 1 - x^2 | |
0a96 | Problem:
A line through a point $A$ intersects a circle in two points, $B$ and $C$, in such a way that $B$ lies between $A$ and $C$. From the point $A$ draw the two tangents to the circle, meeting the circle at points $S$ and $T$. Let $P$ be the intersection of the lines $S T$ and $A C$. Show that $A P / P C = 2 \cdot ... | [
"Solution:\nFirst we show that if we fix the points $A$, $B$ and $C$ but vary the circle, then the point $P$ stays fixed. To that end, suppose we have two different circles through $B$ and $C$. Draw the tangents from $A$ to one circle, meeting the circle at points $S_1$ and $T_1$, and the tangents to the other circ... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 21 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0dlc | Let $ABC$ be a triangle with incentre $I$ such that $AB < AC < BC$. The second intersections of $AI$, $BI$ and $CI$ with the circumcircle of triangle $ABC$ are $M_A$, $M_B$ and $M_C$, respectively. Lines $AI$ and $BC$ intersect at $D$ and lines $BM_C$ and $CM_B$ intersect at $X$. Suppose the circumcircles of triangles ... | [
"Let $O$ be the circumcentre of triangle $ABC$. First we note from standard properties of the Miquel point $S$, we have:\n$$\n\\bullet \\ \\triangle SM_C M_B \\sim \\triangle SBC \\sim \\triangle SPQ; (*)\n$$\n* I and S are inverses with respect to circle $ABC$;\n* $\\angle OSX = 90^\\circ$.\n![](attached_image_1.p... | Saudi Arabia | Saudi Booklet | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformation... | null | proof only | null | |
04bn | Ante wrote all integers from $1$ to $40$ in a row without spacings, thus obtaining the multi-digit number $12345\ldots383940$. Then he decided to erase $60$ digits of this number. What is the largest number Ante can get by doing this? | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms"
] | English | proof and answer | 99967383940 | |
0grr | Let $a_0, a_1, \dots, a_{100}$ and $b_1, b_2, \dots, b_{100}$ be two real sequences such that for all $n = 0, 1, \dots, 99$
$$
a_{n+1} = \frac{a_n}{2}, \quad b_{n+1} = \frac{1}{2} - a_n \quad \text{or} \quad a_{n+1} = 2a_n^2, \quad b_{n+1} = a_n
$$
holds. If $a_{100} \le a_0$, find the maximal value of $b_1 + b_2 + \do... | [
"The answer is $50$. If $a_0 < 0$ then it is clear that $a_{100} > a_0$. If $a_0 = 0$, then $b_n \\in \\{0, 1/2\\}$ and hence $S = b_1 + b_2 + \\dots + b_{100} \\le 50$. Assume that $a_0 > 0$. In this case all terms of the sequence $(a_n)$ are positive. We have\n$$\n(a_n - \\frac{a_{n-1}}{2})(a_n - 2a_{n-1}^2) = 0 ... | Turkey | Team Selection Test | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | 50 | |
0bty | All the 16 squares of a $4 \times 4$ array are white. Define a *move* by selecting a rectangle $1 \times 3$ or $3 \times 1$ and switching the colors of each of its squares from white to black or from black to white. Is it possible that all squares turn black after a sequence of moves? | [
"<table><tr><td>1</td><td>2</td><td>3</td><td>1</td></tr><tr><td>2</td><td>3</td><td>1</td><td>2</td></tr><tr><td>3</td><td>1</td><td>2</td><td>3</td></tr><tr><td>1</td><td>2</td><td>3</td><td>1</td></tr></table>\nto observe that a move will change colors in one square of each number. As initially there are six squ... | Romania | 67th NMO Selection Tests for JBMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | No | |
0e1e | Find integers $a$ and $b$, such that $\sqrt{2010 + 2\sqrt{2009}}$ is a solution of the quadratic equation $x^2 + a x + b = 0$. Prove that for such $a$ and $b$ the number $\sqrt{2010 - 2\sqrt{2009}}$ is not the solution of the given equation. | [
"We notice that\n$$\n\\sqrt{2010 + 2\\sqrt{2009}} = \\sqrt{1 + 2\\sqrt{2009} + 2009} = \\sqrt{(1 + \\sqrt{2009})^2} = 1 + \\sqrt{2009}.\n$$\nSimilarly, $\\sqrt{2010 - 2\\sqrt{2009}} = \\sqrt{2009} - 1$. Since $1 + \\sqrt{2009}$ is the solution of the quadratic equation $x^2 + a x + b = 0$, we have $2010 + 2\\sqrt{2... | Slovenia | National Math Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | a = -2, b = -2008 | |
0bkq | Find all four digit numbers $abcd$ which are equal to the sum of all the positive integers starting with $ab$ and finishing with $cd$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1353, 1863, 3388, 3591 | |
07tw | The function $g : [0, \infty) \to [0, \infty)$ satisfies the functional equation:
$$
g(g(x)) = \frac{3x}{x+3}, \text{ for all } x \ge 0.
$$
You are also told that $g(x) = \frac{x+1}{2}$ for $2 \le x \le 3$.
a. Find $g(2021)$.
b. Find $g(1/2021)$. | [
"For part (a), we determine $g(y)$ for any $y \\ge 6$. Applying the functional equation we have:\n$$\ng(g(y)) = \\frac{3y}{y+3}.\n$$\nAlso, as $2 < \\frac{3y}{y+3} < 3$ we can apply $g$ to each side and apply the second condition to get:\n$$\ng(g(g(y))) = g\\left(\\frac{3y}{y+3}\\right) = \\frac{1}{2}\\left(\\frac{... | Ireland | IRL_ABooklet | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | g(2021) = 24261/4057, g(1/2021) = 6/12127 | |
0484 | Given a positive integer $k$, let integers $a, b$ (not both zero) satisfy $0 \le a < k$, $0 \le b < k+1$. Define the sequence $\{T_n\}_{n \ge k}$ as follows:
$$
T_k = a, \quad T_{k+1} = b,
$$
and for $n \ge k + 2$,
$$
T_n \equiv T_{n-1} + T_{n-2} \pmod{n} \quad \text{with} \quad 0 \le T_n < n.
$$
Concatenating the deci... | [
"**Proof:** First, we prove that $\\{T_n\\}_{n=k}^{\\infty}$ is unbounded. If not, let $M = \\max_{n \\ge k}\\{T_n\\}$, then for any $n > 2M$, from $T_{n-1} \\le M$ and $T_{n-2} \\le M$, we have $T_n = T_{n-1} + T_{n-2}$. By the boundedness of $T_n$, the only possibility is $T_{n-1} = T_{n-2} = 0$, which inductivel... | China | China-TST-2025A | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | English | proof only | null | |
00aa | Alex has thought of a number $N$ in $S = \{1, 2, ..., 1001\}$, and Bibi has to find it via the following procedure. She gives Alex a list of subsets of $S$, Alex reads it and tells Bibi how many subsets in her list contain $N$. If Bibi wishes she can repeat the same with a second list, and then with a third one, but no... | [
"The least number of subsets is 28. Suppose that Bibi has 3 lists 1, 2, 3 which enable her to find $N$ with certainty. Let the lists contain $a_1, a_2, a_3$ subsets respectively. For list $i=1,2,3$ Alex announces the number $x_i$ of subsets in the list that contain $N$, and the ordered triple $x_1, x_2, x_3$ is the... | Argentina | Argentine National Olympiad 2015 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 28 | |
00ir | Each brick of a set has 5 holes in a horizontal row. We can either place pins into individual holes or brackets into two neighboring holes. No hole is allowed to remain empty. We place $n$ such bricks in a row in order to create patterns running from left to right, in which no two brackets are allowed to follow another... | [
"Since 3 pins (P) or 2 brackets (B) may not lie in a row, they may not do so on an individual brick. This means that there are only three different types of brick, which we name A (PBPP), B (PPBP) and C (BPB). Naming the number of possible patterns of $n$ bricks with a brick A at the end $a_n$, and analogously $b_n... | Austria | AustriaMO2011 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | F_{n+3} | |
0k58 | Problem:
Evan has a simple graph with $v$ vertices and $e$ edges. Show that he can delete at least $\frac{e-v+1}{2}$ edges so that each vertex still has at least half of its original degree. | [
"Solution:\n\nFix $v$. We use strong induction on the number of edges $e$. If $e \\leq v-1$, the result trivially holds by removing 0 edges. Now take $e > v-1$ and assume the result has been shown for all smaller values of $e$. Consider a graph $G$ with $v$ vertices and $e$ edges.\n\nSuppose $G$ contains a cycle $C... | United States | HMMT February | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0cor | Let $f(x) = bx^2 + cx + a$, $g(x) = cx^2 + ax + b$, where $a, b, c$ are integers. Given that $f(1234) = g(1234)$, determine whether the equality $f(1) = 2009$ can hold.
Целые числа $a, b, c$ таковы, что значения квадратных трёхчленов $bx^2 + cx + a$ и $cx^2 + ax + b$ при $x = 1234$ совпадают. Может ли первый трёхчлен ... | [
"**Первое решение.** Подставляя $x = 1234$ в оба трёхчлена и приравнивая их, получаем $1234^2 \\cdot b + 1234 \\cdot c + a = 1234^2 \\cdot c + 1234 \\cdot a + b$, или, после переноса всех членов в левую часть, $(1234^2-1)b+(1234-1234^2)c+(1-1234)a = 0$. Разделив последнее равенство на 1233, имеем $1235b - 1234c - a... | Russia | Regional round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization"
] | English; Russian | proof and answer | No | |
0087 | A rectangular sheet of grid paper with dimensions $59 \times 133$ must be divided into a maximum number of pieces by making two cuts. Before cutting it is allowed to fold the sheet along grid lines as many times as desired. The folded sheet is placed on the table and then cut twice along grid lines that are visible aft... | [
"Let $l$ and $m$ be grid lines in the initial rectangle that were cut by the first cutting. There was a folding that made them coincide, hence $l \\parallel m$ (a folding cannot make perpendicular lines coincide). Also it follows that $l$ and $m$ are separated by at least one line of folding which is intact. So the... | Argentina | Mathematical Olympiad Rioplatense | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2010 | |
0jn1 | Problem:
A thief starts at the point $x=0$ on a number line, and runs either left or right at a constant rate. One hour later, a policeman who moves twice as fast as the thief appears at the point $x=0$. However, the policeman does not know which direction the thief went and cannot see the thief. Show that the policema... | [
"Solution:\nAssume for simplicity that the rate at which the thief moves is $1$ unit per hour. The police should drive left for an hour, then drive right for four hours.\n\nThis works for the following reason: if the thief runs left, he will be caught in one hour by the left-moving police at the position $x=-2$. If... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Algorithms",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof only | null | |
0fzi | Problem:
Gegeben sind $2 n$ Chips, die in einer Reihe liegen. In einem Zug kann man zwei benachbarte Chips vertauschen. Wieviele Züge muss man machen, damit jeder Chip einmal am Anfang und einmal am Ende der Reihe war? | [
"Solution:\n\nMan muss insgesamt $3 n^{2}-2 n$ Züge machen.\n\nUntere Schranke:\nWir zählen, wie oft alle Chips zusammen um eins verschoben werden müssen.\nJeder Chip wird sicher von seiner Anfangsposition bis zum näheren Rand, vom einen Rand zum anderen Rand und vom anderen Rand zu seiner Endposition verschoben. E... | Switzerland | IMO-Selektionsprüfung | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 3n^2 - 2n | |
00tw | There are 100 positive integer numbers written on a board. At each step, Alex composes 50 fractions using each number written on the board exactly once, brings these fractions to their irreducible form, and then replaces the 100 numbers on the board with the new numerators and denominators to create 100 new numbers. Fi... | [
"Equivalently, we have a graph on 100 vertices and a positive integer written on each vertex. At each step we pick a perfect matching (i.e. a set of disjoint edges covering all vertices) and for each edge of the matching we divide the numbers in its endpoints with their highest common divisor.\n\nIf initially the n... | Balkan Mathematical Olympiad | BMO 2022 shortlist | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 99 | |
0kx7 | Problem:
Let $A$, $E$, $H$, $L$, $T$, and $V$ be chosen independently and at random from the set $\left\{0, \frac{1}{2}, 1\right\}$. Compute the probability that $\lfloor T \cdot H \cdot E\rfloor = L \cdot A \cdot V \cdot A$. | [
"Solution:\n\nThere are $3^{3} - 2^{3} = 19$ ways to choose $L$, $A$, and $V$ such that $L \\cdot A \\cdot V \\cdot A = 0$, since at least one of $\\{L, A, V\\}$ must be $0$, and $3^{3} - 1 = 26$ ways to choose $T$, $H$, and $E$ such that $\\lfloor T \\cdot H \\cdot E\\rfloor = 0$, since at least one of $\\{T, H, E... | United States | HMMT February 2023 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 55/81 | |
0hwd | Problem:
Suppose there are 100 prisoners, each of whom is given a number between 1 and 100. There is also a room with 100 boxes, labeled 1 to 100, and 100 pieces of paper with the numbers 1 to 100 on them. Each piece of paper is randomly placed inside one of the 100 boxes.
One at a time, each prisoner is allowed to e... | [
"Solution:\n\nThey can use the following strategy: each prisoner opens the box labeled with his own number. He then looks at the number on the paper inside and opens the box labeled with that number next. He continues doing so until he finds his own number or he has opened 50 boxes. For instance, prisoner 2 might o... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
05gk | Problem:
Déterminer tous les réels $t$ pour lesquels le polynôme
$$
P(x)=x^{3}+3 t x^{2}+(4 t-1) x+t
$$
possède deux racines réelles dont la différence est égale à 1. | [
"Solution:\nSi $t$ est un réel fixé, on pose $P_{t}(x)=x^{3}+3 t x^{2}+(4 t-1) x+t$.\nLa première chose à remarquer est que, pour tout $t$, on a $P_{t}(-1)=0$. Par suite, -1 est une racine du polynôme $P_{t}$, ce qui assure que $P_{t}(x)$ est factorisable par $x+1$. On trouve ainsi que\n$$\nP_{t}(x)=(x+1)\\left(x^{... | France | null | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 10/9, 6/5, 0 | |
0giw | If $x$ is a three-digit integer and $x^2$ divided by $2024$ leaves a remainder of $1$, then the smallest possible $x$ is $\underline{\underline{\stackrel{\bigcirc}{10}\ \stackrel{\bigcirc}{11}\ \stackrel{\bigcirc}{12}}}$. | [
"答案：$461$\n\n$2024 = 8 \\cdot 11 \\cdot 23$, so $x^2 \\equiv 1 \\pmod{2024}$ iff $x \\equiv 1, 10 \\pmod{11}$ and $x$ is odd. Thus, $x \\equiv 1, 45, -1, -45 \\pmod{2 \\cdot 11 \\cdot 23}$. The smallest 3 digits one is $506 - 45 = 461$."
] | Taiwan | APMO Taiwan Preliminary Round 1 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | Chinese; English | final answer only | 461 | |
0iro | Problem:
The polynomial $a x^{2}-b x+c$ has two distinct roots $p$ and $q$, with $a$, $b$, and $c$ positive integers and with $0<p, q<1$. Find the minimum possible value of $a$. | [
"Solution:\nLet $x$ and $y$ be the roots. Then:\n\n$$\n\\begin{gathered}\n\\frac{b}{a}=x+y<2 \\Rightarrow b<2a \\\\\n\\frac{c}{a}=x y<1 \\Rightarrow c<a \\Rightarrow a>1 \\\\\nb^{2}>4 a c>4 c^{2} \\Rightarrow b>2c\n\\end{gathered}\n$$\n\nEvaluated at $1$, the polynomial must be greater than $0$, so $a+c>b$. Then:\n... | United States | 1st Annual Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 5 | |
0ezv | Problem:
You are given three positive integers. A move consists of replacing $m \leq n$ by $2m$, $n - m$. Show that you can always make a series of moves which results in one of the integers becoming zero. [For example, if you start with $4, 5, 10$, then you could get $8, 5, 6$, then $3, 10, 6$, then $6, 7, 6$, then $... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0hvz | Problem:
Let $c > 0$ be a positive real number. We define the sequence $(x_n)$ by $x_0 = 0$ and
$$
x_{n+1} = x_n^2 + c
$$
for each $n \geq 0$. For which values of $c$ is it true that $|x_n| < 2016$ for all $n$? | [
"Solution:\nThe answer is $c \\leq \\frac{1}{4}$.\n\nFirst, we show that $c \\leq \\frac{1}{4}$ all work. Clearly it suffices to prove the result when $c = \\frac{1}{4}$. In that case, the sequence is defined by $x_{n+1} = x_n^2 + \\frac{1}{4}$. We claim that $x_n \\leq \\frac{1}{2}$ for all $n$. Indeed, this follo... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | c ≤ 1/4 | |
082h | Problem:
Nel paese di $\mathrm{Oz}$, oltre alle persone normali (che possono mentire oppure dire la verità), vivono cavalieri (che dicono sempre la verità) e furfanti (che mentono sempre). Una strana legge impone che in ogni matrimonio i coniugi siano o entrambi normali, oppure uno cavaliere e l'altro furfante. Arriva... | [
"Solution:\n\nLa risposta è $\\mathbf{( E )}$. Se il signor Bianchi non fosse normale, sarebbe un furfante o un cavaliere. Nel primo caso, la moglie sarebbe un cavaliere e avrebbe mentito affermando che il marito è un cavaliere, questo è assurdo. Nel secondo caso (signor Bianchi cavaliere) la moglie sarebbe un furf... | Italy | Progetto Olimpiadi di Matematica 2003 GARA di SECONDO LIVELLO TRIENNIO | [
"Discrete Mathematics > Logic"
] | null | MCQ | E | |
0iv5 | Problem:
You start with a number. Every second, you can add or subtract any number of the form $n!$ to your current number to get a new number. In how many ways can you get from $0$ to $100$ in $4$ seconds? ($n!$ is defined as $n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$, so $1! = 1$, $2! = 2$, $3! = ... | [
"Solution:\n\nTo get to $100$, you have to use one number which is at least $5! = 120$, because $24 \\times 4 = 96$, which is less than $100$. If you use $6! = 720$ or anything larger, you need to get back from $720$ to $100$ (or further) in three seconds. Since $3 \\cdot 5! < 620$, there is no way to do this in $3... | United States | 2nd Annual Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 36 | |
05qj | Problem:
Adalbert et Babette jouent aux dominos sur une grille rectangulaire de 2 cases de hauteur et 2018 cases de largeur. Adalbert commence en posant un domino de taille $1 \times 2$ en position horizontale, de façon à ce qu'il recouvre exactement deux cases de la grille. Puis Babette joue de même un domino $1 \tim... | [
"Solution:\n\nDivisons la grille en 1009 blocs consécutifs de taille $2 \\times 2$. Adalbert commence par poser un domino dans un de ces blocs. Il est alors assuré de pouvoir le compléter plus tard par un autre domino. Babette doit alors jouer dans un autre bloc (et elle ne peut pas jouer à cheval entre deux blocs,... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 5 : Pot-POURRI | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0hnu | Problem:
Define the digitlength of a positive integer to be the total number of letters used in spelling its digits. For example, since "two zero one one" has a total of 13 letters, the digitlength of 2011 is 13. We begin at any positive integer and repeatedly take the digitlength. Show that after some number of steps... | [
"Solution:\n\nWe first claim that if a number has at least two digits, then the digitlength is less than the number itself. To see this, note that any digit has at most five letters; thus a 2-digit number, which is at least $10$, has digitlength at most $5 \\cdot 2 = 10$; a 3-digit number, which is at least $100$, ... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null |
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