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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0k1z | Problem:
A square with sides of length $1~\mathrm{cm}$ is given. There are many different ways to cut the square into four rectangles. Let $S$ be the sum of the four rectangles' perimeters. Describe all possible values of $S$ with justification. | [
"Solution:\n\nThe answer is $6 < S \\leq 10$. This can be shown by considering several cases, as shown in the 14 figures below.\n\n\nSquare 1\n\nSquare 8\n\nSquare 2\n\nSquare 9\n\nSquare 3\n... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 6 < S ≤ 10 | |
0hzl | Problem:
A circle is randomly chosen in a circle of radius $1$ in the sense that a point is randomly chosen for its center, then a radius is chosen at random so that the new circle is contained in the original circle. What is the probability that the new circle contains the center of the original circle? | [
"Solution:\nIf the center of the new circle is more than $1/2$ away from the center of the original circle then the new circle cannot possibly contain the center of the original one. Let $x$ be the distance between the centers (by symmetry this is all we need to consider), then for $0 \\leq x \\leq 1/2$ the probabi... | United States | Harvard-MIT Math Tournament | [
"Statistics > Probability > Counting Methods > Other",
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | proof and answer | 1 - ln 2 | |
0k00 | Problem:
There are 12 students in a classroom; 6 of them are Democrats and 6 of them are Republicans. Every hour the students are randomly separated into four groups of three for political debates. If a group contains students from both parties, the minority in the group will change his/her political alignment to that... | [
"Solution:\n\nWhen the party distribution is $6$-$6$, the situation can change (to $3$-$9$) only when a group of three contains three people from the same party, and the remaining three are distributed evenly across the other three groups (to be converted).\n\nTo compute the probability, we assume that the groups a... | United States | HMMT November 2017 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 341/54 | |
0f5b | Problem:
Show that given any $2m + 1$ different integers lying between $-(2m-1)$ and $2m-1$ (inclusive) we can always find three whose sum is zero. | [] | Soviet Union | 17th ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
002h | Paladino tiene $n$ tarjetas numeradas de $1$ a $n$ y las divide en dos grupos ($n > 1$). Una división es perfecta si por lo menos uno de los grupos contiene dos tarjetas tales que la suma de los números de esas tarjetas es igual al cuadrado de un número natural. ¿Cuál es el menor valor de $n$ para el cual todas las div... | [] | Argentina | XIV Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Español | proof and answer | 15 | |
0e7g | Problem:
1. Na sliki sta graf logaritemske funkcije $f$ s predpisom $f(x)=2 \log (x-2)+1$ in premica $p$.
a) Zapiši enačbo premice $p$.
b) Izračunaj ničlo logaritemske funkcije. Rezultat naj bo natančen. | [] | Slovenia | Državno tekmovanje | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | a) Requires the diagram to read off specific points. b) x = 2 + 10^{-1/2} | |
0e6i | Problem:
Koliko odstotkov od
$$
\frac{\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{3}}{\frac{1}{4} \cdot\left(1-\frac{1}{3}\right)}-\frac{1}{1+\frac{1}{2}}
$$
je
$$
\frac{1}{12} \cdot \frac{\left(0,5-\frac{1}{4}\right)^{2} \cdot 8}{0,008 \cdot\left(2+\frac{1}{2}\right)^{4}}
$$
? Rezultat zaokroži na stotine.... | [
"Solution:\n\nPoenostavimo prvi izraz\n$$\n\\frac{\\frac{1}{4}+\\frac{1}{8}}{\\frac{1}{4} \\cdot \\frac{2}{3}}-\\frac{1}{\\frac{3}{2}} = \\frac{\\frac{3}{8}}{\\frac{2}{12}}-\\frac{2}{3} = \\frac{9}{4}-\\frac{2}{3} = \\frac{19}{12}.\n$$\n\nPoenostavimo tudi drugi izraz\n$$\n\\frac{1}{12} \\cdot \\frac{\\left(\\frac{... | Slovenia | Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 8.42% | |
00j0 | Let $k$ be a circle with mid-point $M$. $T$ is a point on $k$ and $t$ the tangent of $k$ in $T$. $P$ is a point on $t$ with $P \neq T$ and $g$ a line containing $P$ with $g \neq t$. $g$ has the points $U$ and $V$ in common with $k$ ($U \neq V$), and $S$ is the mid-point of the arc $UV$ not containing $T$. $Q$ is the po... | [
"Let $R$ be the common point of $t$ and the tangent $s$ of $k$ in $S$. Since $S$ is the mid-point of the arc $UV$, $s$ is parallel to $g$ (and not to $t$). Since $MR$ is perpendicular to $TS$ and bisects $\\angle SRT$, $QP$ bisects $\\angle UPT$. Let $W$ be the common point of $PQ$ and $TS$. Because of the given sy... | Austria | AustriaMO2011 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0dj0 | Positive real numbers $a_1, a_2, \dots, a_n, b_1, b_2, \dots, b_n$ satisfy $a_1 \ge a_2 \ge \dots \ge a_n$ and $b_1b_2\dots b_k \ge a_1a_2\dots a_k$ for each $k = 1, 2, \dots, n$. Prove that
$$
b_1 + b_2 + \dots + b_n \ge a_1 + a_2 + \dots + a_n.
$$ | [
"Note that for each $k = 1, 2, \\dots, n$, we have\n$$\n\\frac{b_1}{a_1} + \\frac{b_2}{a_2} + \\dots + \\frac{b_k}{a_k} \\ge k \\sqrt[k]{\\frac{b_1}{a_1} \\frac{b_2}{a_2} \\dots \\frac{b_k}{a_k}} \\ge k,\n$$\nwhich means that\n$$\ns_k = \\frac{b_1 - a_1}{a_1} + \\frac{b_2 - a_2}{a_2} + \\cdots + \\frac{b_k - a_k}{a... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof only | null | |
04p9 | Kings and rooks are placed on an $8 \times 8$ board, so that none of the pieces is under attack. A king attacks neighbouring squares (there are eight of them, except when the king is placed on the edge of the board), and a rook attacks all squares in the same row or column as the square where the rook is placed. At mos... | [] | Croatia | Croatian Mathematical Society Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 8 | |
09wl | Given is the sequence of numbers $a_0, a_1, a_2, \dots, a_{2020}$ with $a_0 = 0$. Furthermore, the following holds for every $k = 1, 2, \dots, 2020$:
$$
a_k = \begin{cases} a_{k-1} \cdot k & \text{if } k \text{ is divisible by } 8, \\ a_{k-1} + k & \text{if } k \text{ is not divisible by } 8. \end{cases}
$$
What are th... | [
"02"
] | Netherlands | Second Round | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | final answer only | 02 | |
0cxn | Consider a triangle $ABC$ and a point $P$ in its interior. Lines $PA$, $PB$, $PC$ intersect $BC$, $CA$, $AB$ at $A'$, $B'$, $C'$, respectively. Prove that
$$
\frac{BA'}{BC} + \frac{CB'}{CA} + \frac{AC'}{AB} = \frac{3}{2}
$$
if and only if at least two of the triangles $PAB$, $PBC$, $PCA$ have the same area. | [
"\n\nLet $AB = c$, $BC = a$, $CA = b$, $BA' = x$, $CB' = y$, $AC' = z$. The condition in the problem $\\frac{x}{a} + \\frac{y}{b} + \\frac{z}{c} = \\frac{3}{2}$ is equivalent to\n$$\n\\begin{equation*}\nbcx + cay + abz = \\frac{3}{2}abc \\tag{1}\n\\end{equation*}\n$$\nFrom Ceva's Theorem,\n... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem"
] | English | proof only | null | |
05fz | Problem:
Soit $ABC$ un triangle avec $\widehat{BAC}=60^{\circ}$. Soient $D$ et $E$ les pieds des bissectrices issues respectivement des sommets $B$ et $C$. Soit $I$ le point d'intersection des droites $(BD)$ et $(CE)$. Montrer que les points $A$, $I$, $D$ et $E$ sont cocycliques. | [
"Solution:\n\n\n\nOn va faire une chasse aux angles pour démontrer que $\\widehat{IDA}+\\widehat{IEA}=180^{\\circ}$. Notons $x=\\widehat{ACE}=\\widehat{ECB}$ ainsi que $y=\\widehat{ABD}=\\widehat{DBC}$. Comme la somme des angles d'un triangle vaut $180^{\\circ}$ on a ainsi\n$$\n\\begin{alig... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"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 | |
0b28 | Problem:
In $\triangle ABC$, $AB = 20$, $BC = 21$, and $CA = 29$. Point $M$ is on side $AB$ with $\frac{AM}{MB} = \frac{3}{2}$, while point $N$ is on side $BC$ with $\frac{CN}{NB} = 2$. $P$ and $Q$ are points on side $AC$ such that the line $MP$ is parallel to $BC$ and the line $NQ$ is parallel to $AB$. Suppose that $... | [
"Solution:\n\nWe use similar triangles here. Note that triangles $ABC$, $AMP$, $QNC$ and $QRP$ are all similar right (by the Pythagorean theorem, since $20^{2} + 21^{2} = 29^{2}$) triangles by AA similarity and corresponding angle theorem. We see that $AP = \\frac{3}{5} \\cdot 29 = \\frac{87}{5}$ and $CQ = \\frac{2... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 224/15 | |
08sj | Consider a square consisting of 9 little squares of side length $1$ arranged in $3$ rows and $3$ columns as in the figure below. Let us number the little squares as indicated in the figure.
| ① | ② | ③ |
|----|----|----|
| ⑧ | | ④ |
| ⑦ | ⑥ | ⑤ |
Let $A$, $B$, $C$, $D$ be points chosen from the interior (not on th... | [
"We partition the intersection of the quadrilateral $ABCD$ and the union of squares $1$, $3$, $5$, $7$ into $8$ triangles $T_1, T_2, \\dots, T_8$ as indicated in the figure below. Then, $X$ equals the sum of the areas of the triangles $T_1, T_2, \\dots, T_8$. We also partition the intersection of $ABCD$ and the uni... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
03g1 | Some cities of country Graphland are connected with roads provided that
* (i) from each city we can reach any other city.
It turned out that every city $A$ can choose its favourite number $1 \le f(A) \le 2024$ which is an integer, such that the following condition holds:
* (ii) the favourite numbers of any two cities c... | [
"Answer: For each $m$ coprime with $2024$.\n\nLet us first assume that $(m, 2024) > 1$. Then the tourist, starting from the capital with a favourite number $r$, can reach only cities with numbers $km + r$ (mod $2024$), $k \\in \\mathbb{N}$. Since it is not a complete system of residues modulo $2024$, it means that ... | Bulgaria | 6 TST for BMO | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | All integers m between 1 and 2024 that are coprime to 2024 | |
07d2 | Find the largest natural number $n$ for which there exist $n$ positive integers such that none of them divides another one, but in any triplet of these numbers, one divides the sum of the other two. | [
"In our example, four numbers are $2$, $3$, $7$, $17$ and the fifth number $x$ is a number satisfying the following relations. Such $x$ exists due to the **Chinese Remainder Theorem**.\n$$\n3 \\mid 7 + x, \\quad 7 \\mid 17 + x, \\quad 17 \\mid 3 + x, \\quad 2 \\nmid x.\n$$\n(For instance, $x = 473$.) Note that $x$ ... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 6 | |
07x7 | Determine, with proof, the smallest positive integer $N$ for which the equation
$$
x^2 - y^2 = N
$$
has exactly 24 solutions $(x, y)$ with positive integers $x$ and $y$. | [
"Because $N > 0$, we need to have $x > y > 0$ for any solution. Since $x^2 - y^2 = (x + y)(x - y)$, each solution gives a factorisation of $N$ into two factors, say $N = a \\cdot b$, where $a = x + y > x - y = b$.\nIf we start with a factorisation $N = a \\cdot b$ with $a > b$, there is at most one solution $(x, y)... | Ireland | IRL_ABooklet_2024 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 10080 | |
0jmz | Problem:
Consider a $1 \times 1$ grid of squares. Let $A, B, C, D$ be the vertices of this square, and let $E$ be the midpoint of segment $CD$. Furthermore, let $F$ be the point on segment $BC$ satisfying $BF = 2\,CF$, and let $P$ be the intersection of lines $AF$ and $BE$. Find $\frac{AP}{PF}$. | [
"Solution:\n\nLet line $BE$ hit line $DA$ at $Q$. It's clear that triangles $AQP$ and $FBP$ are similar so\n$$\n\\frac{AP}{PF} = \\frac{AQ}{BF} = \\frac{2AD}{\\frac{2}{3}BC} = 3\n$$"
] | United States | HMMT November | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 3 | |
0cex | Let $ABC$ be an isosceles triangle with apex at $A$, and let $M, N, P$ be the midpoints of the sides $BC, CA, AB$, respectively. Let $Q$ and $R$ be points inside the segments $BM$ and $CM$, respectively, so that $\angle BAQ = \angle MAR$. The segment $NP$ crosses $AQ$ and $AR$ at $U$ and $V$, respectively. The point $S... | [
"\nWe first show that $AVMS$ is cyclic. Since the triangle $ABC$ is isosceles, $AM \\perp BC$, and since $NP$ is midline, $V$ is the midpoint of $AR$. Then $MV$ is the $M$-median of the right triangle $AMR$, so $VM = VA = VR$. Hence $V$ lies on the perpendicular bisector of $AM$. By hypothe... | Romania | Seventeenth Stars of Mathematics Competition | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
05py | Problem:
Soit $ABC$ un triangle, $O$ le centre de son cercle circonscrit. Soit $P$ un point sur $(AO)$ et $D, E, F$ les projections orthogonales de $P$ sur $(AB)$, $(BC)$ et $(CA)$. Soit $X$ et $Y$ les intersections des cercles circonscrits à $DEF$ et à $BCP$.
Montrer que $\widehat{BAX} = \widehat{YAC}$ | [
"Solution:\n\n\n\nÉtape 1 : Montrons que $(DF)$ est parallèle à $(BC)$.\n\nSoit $P_0$ le point diamétralement opposé à $A$ sur le cercle circonscrit à $ABC$. $(P_0C)$ est perpendiculaire à $(AC)$ donc parallèle à $(PF)$ et $(P_0B)$ est de même parallèle à $(PD)$.\n\nDonc, d'après le théorèm... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point ci... | null | proof only | null | |
0erq | The value of $\sqrt{2 \times 0 \times 1 \times 6}$ is
(A) 6 (B) 3 (C) 2 (D) 1 (E) 0 | [
"$2 \\times 0 \\times 1 \\times 6 = 0$, and $\\sqrt{0} = 0$"
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | E | |
04lo | We call a point $P$ inside a triangle $ABC$ marvelous if exactly $27$ rays can be drawn from it, intersecting the sides of $ABC$ such that the triangle is divided into $27$ smaller triangles of equal areas. Determine the total number of marvelous points inside a given triangle $ABC$.
(Naboj) | [
"Let $P$ be a marvelous point of $ABC$ and let (without loss of generality) the area of $ABC$ be $27$. We first note that among the $27$ rays, there must always be $PA$, $PB$ and $PC$; otherwise not all the shapes determined by the rays will be triangles.\n\nEach of the $27$ smaller triangles is contained within (o... | Croatia | Mathematical competitions in Croatia | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 325 | |
0asc | Problem:
Let $\log_{14} 16$ be equal to $a$. Express $\log_{8} 14$ in terms of $a$. | [
"Solution:\n\n$\\frac{4}{3a}$"
] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | 4/(3a) | |
0hy4 | Problem:
Edward, the author of this test, had to escape from prison to work in the grading room today. He stopped to rest at a place $1,875$ feet from the prison and was spotted by a guard with a crossbow. The guard fired an arrow with an initial velocity of $100\ \mathrm{ft} / \mathrm{s}$. At the same time, Edward st... | [
"Solution:\n\nWe use the formula for distance, $d = \\frac{1}{2} a t^{2} + v t + d_{0}$. Then after $t$ seconds, Edward is at location $1875 + \\frac{1}{2}(1)(t^{2})$ from the prison. After $t$ seconds, the arrow is at location $\\frac{1}{2}(-1)(t^{2}) + 100 t$ from the prison. When the arrow hits Edward, both obje... | United States | HMMT | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 75 ft/s | |
069c | Let $m_1 < m_2 < \dots < m_s$ be a sequence of $s \ge 2$ positive integers, none of which can be written as the sum of (two or more) distinct other numbers in the sequence. For every integer $r$ with $1 \le r < s$ prove that $r m_r + m_s \ge (r+1)(s-1)$. | [
"For $k, l$ with $0 \\le k \\le r$ and $k+1 \\le l \\le s$, we introduce the auxiliary value\n$$\nT(k, l) := m_l + \\sum_{i=1}^{k} m_i.\n$$\n\nWe claim that these $\\frac{1}{2}(r+1)(2s - r)$ auxiliary values are all pairwise distinct: Let us assume that $T(k, \\ell) = T(u, v)$. Without loss of generality $k \\le u$... | Greece | 21st Mediterranean Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | English | proof only | null | |
0e01 | Problem:
Za katero vrednost realnega števila $a$ je $1+\frac{1}{2+\frac{1}{3+\frac{1}{a}}}=0$? | [
"Solution:\n\nPoenostavimo verižni ulomek do oblike $\\frac{10a+3}{7a+2}$. Ta je enak nič, če je števec enak $0$. Tako je $10a+3=0$ oziroma $a=-\\frac{3}{10}$."
] | Slovenia | Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | -3/10 | |
0dmy | Problem:
На страницама $AB$, $AC$ и $BC$ троугла $ABC$ дате су, редом, тачке $M$, $X$ и $Y$ тако да је $AX = MX$ и $BY = MY$. Нека су $K$ и $L$, редом, средишта дужи $AY$ и $BX$, а $O$ центар описане кружнице троугла $ABC$. Ако су $O_{1}$ и $O_{2}$ тачке симетричне тачки $O$ у односу на $K$ и $L$, редом, доказати да т... | [
"Solution:\n\nПоставимо координатни систем са почетком у тачки $M$ и $x$-осом дуж праве $AB$. Нека тачке $X$ и $Y$ имају координате $(a, b)$ и $(c, d)$ редом. Због $AX = XM$ и $BY = YM$, координате тачака $A$ и $B$ су $(2a, 0)$ и $(2c, 0)$, а тачака $K$ и $L$ су $\\left(a + \\frac{c}{2}, \\frac{d}{2}\\right)$ и $\\... | Serbia | СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
013d | Problem:
The 25 member states of the European Union set up a committee with the following rules:
(1) the committee should meet daily;
(2) at each meeting, at least one member state should be represented;
(3) at any two different meetings, a different set of member states should be represented; and
(4) at the $n$'t... | [
"Solution:\n\nIf one member is always represented, rules 2 and 4 will be fulfilled. There are $2^{24}$ different subsets of the remaining 24 members, so there can be at least $2^{24}$ meetings. Rule 3 forbids complementary sets at two different meetings, so the maximal number of meetings cannot exceed $\\frac{1}{2}... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2^{24} = 16777216 | |
0l8a | Let $k$ be a real number such that the system
$$
\begin{aligned}
|25 + 20i - z| &= 5 \\
|z - 4 - k| &= |z - 3i - k|
\end{aligned}
$$
has exactly one complex solution $z$. The sum of all possible values of $k$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. Here $i ... | [] | United States | 2025 AIME I | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | 77 | |
0i56 | Problem:
A square and a regular hexagon are drawn with the same side length. If the area of the square is $\sqrt{3}$, what is the area of the hexagon? | [
"Solution:\nThe hexagon is composed of six equilateral triangles each of side length $\\sqrt[4]{3}$ (with base $b = \\sqrt[4]{3}$ and height $\\frac{\\sqrt{3}}{2} b$), so the total area is $\\frac{9}{2}$."
] | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | 9/2 | |
0h57 | Andriy and Olesia play such game. Firstly, Andriy chooses a chess king and places it on the chessboard. Then they move in turn by the rules of the king. However, it is not allowed to put the king on the field that Andriy began from or was already used. The loser is the one who cannot move. Who wins if both are trying t... | [
"Olesia always wins.\n\nFor every chessman the chessboard is divided into couples of squares that are connected by the move of the chosen chessman. Then the win strategy of Olesia is as follows: Andriy moves the chessman to the square of some couple (it also concerns the first Andriy's choice of placing the chessma... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round (Second Tour) | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Olesia always wins. | |
06z0 | Problem:
The sides of a triangle form an arithmetic progression. The altitudes also form an arithmetic progression. Show that the triangle must be equilateral. | [
"Solution:\n\nLet the sides be $a$, $a + d$, $a + 2d$ with $d \\geq 0$. Then the altitudes are $k / a$, $k / (a + d)$, $k / (a + 2d)$, where $k$ is twice the area. We claim that $k / a + k / (a + 2d) > 2k / (a + d)$ unless $d = 0$. This is equivalent to $(a + d)(a + 2d) + a(a + d) > 2a(a + 2d)$ or $2d^2 > 0$, which... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Jensen/smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
03j1 | Problem:
An integer is digitally divisible if
(a) none of its digits is zero;
(b) it is divisible by the sum of its digits (e.g., $322$ is digitally divisible).
Show that there are infinitely many digitally divisible integers. | [
"Solution:\n\nLet us construct infinitely many digitally divisible integers.\n\nConsider the integer $111\\ldots 1$ consisting of $k$ digits, all of which are $1$ (for any $k \\geq 1$). None of its digits is zero, so condition (a) is satisfied.\n\nThe sum of its digits is $k$ (since there are $k$ digits, each $1$).... | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0dyu | Problem:
Poenostavi izraz $x^{(x+1)^2} \cdot \left(x^{x-1}\right)^{x+1} : \frac{x^{x^2}}{x^{-2x}}$ in nato izračunaj vrednost izraza za $x=\sqrt{2}$. | [
"Solution:\n\nV izrazu $x^{(x+1)^2} = x^{x^2 + 2x + 1}$ uporabimo kvadriranje dvočlenika v eksponentu.\n\nV potenci $\\left(x^{x-1}\\right)^{x+1}$ uporabimo pravilo za potenciranje potenc, tako dobimo $\\left(x^{x-1}\\right)^{x+1} = x^{x^2 - 1}$.\n\nUlomek odpravimo z upoštevanjem pravila za deljenje potenc.\n\nUpo... | Slovenia | Državno tekmovanje | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | proof and answer | Simplified form: x^{x^2}; at x = sqrt(2): 2 | |
02by | Problem:
(a) Se $r=\sqrt{2}+\sqrt{3}$, mostre que $\sqrt{6}=\frac{r^{2}-5}{2}$.
(b) Se $s=\sqrt{215}+\sqrt{300}$, mostre que $s^{2}>1015$. | [
"Solution:\n\n(a) Como\n$$\nr^{2}=(\\sqrt{2}+\\sqrt{3})^{2}=(\\sqrt{2})^{2}+2(\\sqrt{2})(\\sqrt{3})+(\\sqrt{3})^{2}=2+2 \\sqrt{6}+3=5+2 \\sqrt{6}\n$$\nportanto $r^{2}-5=2 \\sqrt{6} \\Longrightarrow \\sqrt{6}=\\frac{r^{2}-5}{2}$.\n\n(b) Pelo mesmo argumento temos que\n$$\n\\begin{aligned}\ns^{2} & =(\\sqrt{215}+\\sq... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0i0c | Problem:
Reduce the number $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$. | [
"Solution:\n\nObserve that $\\left(\\sqrt[3]{2+\\sqrt{5}}+\\sqrt[3]{2-\\sqrt{5}}\\right)^{3} = (2+\\sqrt{5}) - 3\\sqrt[3]{2+\\sqrt{5}} - 3\\sqrt[3]{2-\\sqrt{5}} + (2-\\sqrt{5}) = 4 - 3\\left(\\sqrt[3]{2+\\sqrt{5}} + \\sqrt[3]{2-\\sqrt{5}}\\right)$.\n\nHence $\\sqrt[3]{2+\\sqrt{5}} + \\sqrt[3]{2-\\sqrt{5}}$ is a roo... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 1 | |
0ijq | Problem:
Ann and Anne are in bumper cars starting 50 meters apart. Each one approaches the other at a constant ground speed of $10~\mathrm{km}/\mathrm{hr}$. A fly starts at Ann, flies to Anne, then back to Ann, and so on, back and forth until it gets crushed when the two bumper cars collide. When going from Ann to Ann... | [
"Solution:\n\nSuppose that at a given instant the fly is at Ann and the two cars are $12d$ apart. Then, while each of the cars travels $4d$, the fly travels $8d$ and meets Anne. Then the fly turns around, and while each of the cars travels $d$, the fly travels $3d$ and meets Ann again. So, in this process described... | United States | Harvard-MIT Mathematics Tournament | [
"Math Word Problems"
] | null | final answer only | 55 | |
0hds | Prove that one can choose 7 pairwise distinct numbers from $1, 2, \ldots, 10000$, such that none of them is a perfect square and no sum of the several numbers out of the 7 chosen ones is a perfect square either. | [
"Consider the following 7 numbers: $2^1$, $2^3$, $\\ldots$, $2^{13} = 8192 < 10000$. It is clear that the sum of any subset of them is such that the highest power of $2$ that divides the sum is odd, thus, it is not a perfect square."
] | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
041j | Let $\odot I$ be the incircle of $\triangle ABC$ with $AB > AC$. $\odot I$ is tangent to $BC$ and $AD$ at $D$ and $E$, respectively. The tangent line $EP$ of $\odot I$ intersects the extended line of $BC$ at $P$. Segment $CF$ is parallel to $PE$ and intersects $AD$ at point $F$. Line $BF$ intersects $\odot I$ at points... | [
"Therefore, $\\angle IGP = \\angle IEP = 90^\\circ$, that is, $IG \\perp PG$. Hence, points $P$, $S$ and $T$ are collinear.\n\nLine $PST$ intersects $\\triangle ABC$. By Menelaus' Theorem, we have\n$$\n\\frac{AS}{SC} \\cdot \\frac{CP}{PB} \\cdot \\frac{BT}{TA} = 1.\n$$\nSince $AS = AT$, $CS = CD$ and $BT = BD$, we ... | China | China Southeastern Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, n... | English | proof only | null | |
0c1h | In $n$ transparent boxes there are red balls and blue balls. One needs to choose 50 boxes such that, together, they contain at least half of the red balls and at least half of the blue balls. Is such a choice possible irrespective on the number of balls and on the way they are distributed in the boxes, if:
a) $n = 100$... | [
"a) The answer is in the negative. If we have 100 boxes, and 25 of the boxes only contain one red ball each, while the other 75 boxes contain just one blue ball each, then one needs to choose at least 13 of the boxes containing red balls and at least 38 of the boxes containing blue balls, which means one would need... | Romania | 69th NMO Selection Tests for JBMO | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) No; b) Yes | |
05l0 | Problem:
Montrer qu'il existe une infinité d'entiers positifs $n$ tels que le plus grand facteur premier de $n^{4}+n^{2}+1$ soit égal au plus grand facteur premier de $(n+1)^{4}+(n+1)^{2}+1$. | [
"Solution:\n\nPour tout $n \\geq 1$, on note $p_{n}$ le plus grand diviseur premier de $n^{4}+n^{2}+1$, et $q_{n}$ le plus grand diviseur premier de $n^{2}+n+1$.\nOn a donc $p_{n}=q_{n^{2}}$.\n\nDe plus, de l'identité\n$$\nn^{4}+n^{2}+1=\\left(n^{2}+n+1\\right)\\left(n^{2}-n+1\\right)=\\left(n^{2}+n+1\\right)\\left... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0ap0 | Problem:
If $2A99561$ is equal to the product when $3 \times (523 + A)$ is multiplied by itself, find the digit $A$. | [
"Solution:\n\nWe are given that $2A99561 = [3 \\times (523 + A)]^2$, which is equivalent to $2A99561 = 9 \\times (523 + A)^2$.\n\nSince $(523 + A)^2$ is an integer, it follows that $2A99561$ is divisible by $9$.\n\nBy the rule on divisibility by $9$, after adding all the digits of $2A99561$, it suffices to find the... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 4 | |
0asr | Problem:
Let $\mathbb{R}^{\star}$ be the set of all real numbers, except $1$. Find all functions $f: \mathbb{R}^{\star} \rightarrow \mathbb{R}$ that satisfy the functional equation
$$
x + f(x) + 2 f\left(\frac{x+2009}{x-1}\right) = 2010
$$ | [
"Solution:\nLet $g(x) = \\frac{x+2009}{x-1}$. Then the given functional equation becomes\n$$\nx + f(x) + 2 f(g(x)) = 2010\n$$\nReplacing $x$ with $g(x)$ in $(1\\star)$, and after noting that $g(g(x)) = x$, we get\n$$\ng(x) + f(g(x)) + 2 f(x) = 2010\n$$\nEliminating $f(g(x))$ in $(1\\star)$ and $(2\\star)$, we obtai... | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = (x^2 + 2007x - 6028) / (3(x - 1)) | |
05dy | Problem:
Let $m > 1$ be an integer. A sequence $a_{1}, a_{2}, a_{3}, \ldots$ is defined by $a_{1} = a_{2} = 1$, $a_{3} = 4$, and for all $n \geq 4$,
$$
a_{n} = m\left(a_{n-1} + a_{n-2}\right) - a_{n-3}.
$$
Determine all integers $m$ such that every term of the sequence is a square. | [
"Solution:\nThe only such $m$ are $m = 2$ and $m = 10$.\n\nConsider an integer $m > 1$ for which the sequence defined in the problem statement contains only perfect squares. We shall first show that $m - 1$ is a power of $3$.\n\nSuppose that $m - 1$ is even. Then $a_{4} = 5m - 1$ should be divisible by $4$ and henc... | European Girls' Mathematical Olympiad (EGMO) | EGMO 2020 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | m = 2 or m = 10 | |
0fg0 | Problem:
Hallar la ecuación de la circunferencia que pasa por los afijos de las soluciones de la ecuación
$$
z^{3}+(-1+i) z^{2}+(1-i) z+i=0
$$ | [
"Solution:\nSe comprueba sin dificultad que $z=-i$ es una solución de la ecuación; como\n$$\nz^{3}+(-1+i) z^{2}+(1-i) z+i=(z+i)\\left(z^{2}-z+1\\right)\n$$\ny las raíces de $z^{2}-z+1$ son\n$$\n\\frac{1 \\pm i \\sqrt{3}}{2}\n$$\nLas tres raíces tienen módulo 1, y es inmediato que la circunferencia pedida es la circ... | Spain | OME 21 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | null | proof and answer | |z|=1 | |
0afi | Во $\triangle ABC$, $\overline{AC}=5 \text{ cm}$ и $\overline{BC}=9 \text{ cm}$. Од средината на страната $AB$ е издигната нормала која што страната $BC$ ја сече во точка $E$. Точката $E$ е поврзана со темето $A$. Пресметај го периметарот на $\triangle AEC$.
 | [
"$\\triangle ABE$ е рамнокрак, со основа $AB$, па $\\overline{AE} = \\overline{EB}$. Тогаш имаме\n$$L_{ACE} = \\overline{AC} + (\\overline{CE} + \\overline{EA}) = 5 + 9 = 14 \\text{ cm.}$$"
] | North Macedonia | Регионален натпревар по математика за основно образование | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | Macedonian, English | proof and answer | 14 cm | |
031q | Problem:
Some of the vertices of a convex $n$-gon are connected by segments such that any two of them have no a common interior point. Prove that for any $n$ points in general position (i.e., any three of them are not collinear) there is an one-to-one correspondence between the points and the vertices of the $n$-gon s... | [
"Solution:\n\nLet $A_{1} A_{2} \\ldots A_{n}$ be a convex $n$-gon. Denote by $\\mathcal{B}$ the set of the connected vertices and let $B_{1} B_{2} \\ldots B_{k}$ be its convex hull. We shall prove by induction on $n$ that there is a map with the desired properties that in addition sends two given adjacent vertices ... | Bulgaria | Team selection test for 44. IMO | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | null | proof only | null | |
0l2v | Problem:
Compute the unique ordered pair $(x, y)$ of real numbers satisfying the system of equations
$$
\frac{x}{\sqrt{x^{2}+y^{2}}}-\frac{1}{x}=7 \quad \text{ and } \quad \frac{y}{\sqrt{x^{2}+y^{2}}}+\frac{1}{y}=4 .
$$ | [
"Solution:\n\nConsider vectors\n$$\n\\binom{\\frac{x}{\\sqrt{x^{2}+y^{2}}}}{\\frac{y}{\\sqrt{x^{2}+y^{2}}}} \\text{ and } \\binom{-\\frac{1}{x}}{\\frac{1}{y}}.\n$$\nThey are orthogonal and add up to $\\binom{7}{4}$, which has length $\\sqrt{7^{2}+4^{2}}=\\sqrt{65}$. The first vector has length $1$, so by Pythagoras... | United States | HMMT February 2024 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | (-13/96, 13/40) | |
01zi | Does there exist a polynomial $p(x)$ with integer coefficients such that
$$
p(\sqrt{2}) = \sqrt{2} \quad \text{und} \quad p(2\sqrt{2}) = 2\sqrt{2} + 2?
$$ | [
"Suppose that such polynomial $p(x)$ with integer coefficients exists. It follows from the equality $p(\\sqrt{2}) = \\sqrt{2}$ that $p(-\\sqrt{2}) = -\\sqrt{2}$, i.e. the numbers $\\sqrt{2}$ and $-\\sqrt{2}$ are roots of the polynomial $p(x) - x$. According to Bezout's theorem, the polynomial $p(x) - x$ is divisibl... | Belarus | Belarus2022 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Algebraic Number Theory > Quadratic fields"
] | English | proof only | null | |
0l0l | Suppose $A$, $B$, and $C$ are points in the plane with $AB = 40$ and $AC = 42$, and let $x$ be the length of the line segment from $A$ to the midpoint of $BC$. Define a function $f$ by letting $f(x)$ be the area of $\triangle ABC$. Then the domain of $f$ is an open interval $(p, q)$, and the maximum value of $f(x)$ occ... | [
"By the Triangle Inequality, $BC$ is between $40 + 42 = 82$ and $42 - 40 = 2$. The corresponding bounding values of $x$ are $p = 42 - \\frac{40+42}{2} = 1$ and $q = 42 - \\frac{42-40}{2} = 41$.\n\nThe area of $\\triangle ABC$ is maximized when $\\angle BAC$ is a right angle, in which case the area is $r = \\frac{1}... | United States | 2024 AMC 12 B | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | MCQ | C | |
04qh | Let $m$ and $n$ be positive integers of different parity. Prove that
$$
\frac{3m^2 + 5mn}{3n^2 + mn}
$$
is not a positive integer.
(Ilko Brnetić) | [
"Let $d$ be the greatest common divisor of $m$ and $n$, i.e. $m = d m'$ and $n = d n'$, where $m'$ and $n'$ are relatively prime positive integers of different parity.\nNow we need to prove that\n$$\n\\frac{3d^2 m'^2 + 5d^2 m' n'}{3d^2 n'^2 + d m' n'} = \\frac{m'(3m' + 5n')}{n'(m' + 3n')}\n$$\nis not a positive int... | Croatia | Croatian Junior Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0280 | Problem:
Em um certo país há 21 cidades e o governo pretende construir $n$ estradas (todas de mão dupla), sendo que cada estrada liga exatamente 2 das cidades do país. Qual o menor valor de $n$ para que, independente de como as estradas sejam construídas, seja possível viajar entre quaisquer 2 cidades (passando, possi... | [
"Solution:\n\nDiremos que um grupo de cidades é conexo se é possível viajar de uma delas para qualquer outra por meio de estradas. Naturalmente grupos de cidades conexas distintos não podem possuir cidades em comum. Se o governo construir todas as estradas de um grupo de 20 cidades e deixar uma cidade isolada sem e... | Brazil | null | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 191 | |
0i2m | Problem:
Given that $7,999,999,999$ has at most two prime factors, find its largest prime factor. | [
"Solution:\n$7,999,999,999 = 8 \\cdot 10^{9} - 1 = 2000^{3} - 1 = (2000 - 1)\\left(2000^{2} + 2000 + 1\\right)$, so $\\left(2000^{2} + 2000 + 1\\right) = 4,002,001$ is its largest prime factor."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 4002001 | |
00wr | Problem:
Is it possible to put two tetrahedra of volume $\frac{1}{2}$ without intersection into a sphere with radius $1$? | [
"Solution:\n\nNo, it is not. Any tetrahedron that does not contain the centre of the sphere as an internal point has a height drawn to one of its faces less than or equal to the radius of the sphere. As each of the faces of the tetrahedron is contained in a circle with radius not greater than $1$, its area cannot e... | Baltic Way | Baltic Way | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | No | |
0cq4 | For every positive numbers $x$, $y$, $z$ prove the inequality
$$
\frac{x+1}{y+1} + \frac{y+1}{z+1} + \frac{z+1}{x+1} \le \frac{x}{y} + \frac{y}{z} + \frac{z}{x}.
$$
Даны положительные числа $x$, $y$, $z$. Докажите неравенство
$$
\frac{x+1}{y+1} + \frac{y+1}{z+1} + \frac{z+1}{x+1} \le \frac{x}{y} + \frac{y}{z} + \frac{... | [
"Заметим, что $\\frac{a+1}{b+1} - \\frac{a}{b} = \\frac{b-a}{b(b+1)}$ для любых положительных $a$ и $b$. Значит, после переноса всех членов в левую часть требуемое неравенство приобретает вид\n$$\n\\frac{y-x}{y(y+1)} + \\frac{z-y}{z(z+1)} + \\frac{x-z}{x(x+1)} \\le 0. \\quad (1)\n$$\nМожно считать, что $x$ — наибол... | Russia | Russian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English, Russian | proof only | null | |
0ji8 | Problem:
Let $ABC$ be a triangle and $D$ a point on $BC$ such that $AB = \sqrt{2}$, $AC = \sqrt{3}$, $\angle BAD = 30^{\circ}$, and $\angle CAD = 45^{\circ}$. Find $AD$. | [
"Solution:\n\n$\\boxed{\\dfrac{\\sqrt{6}}{2}}$ OR $\\dfrac{\\sqrt{3}}{\\sqrt{2}}$\n\nNote that $[BAD] + [CAD] = [ABC]$. If $\\alpha_1 = \\angle BAD$, $\\alpha_2 = \\angle CAD$, then we deduce\n$$\n\\frac{\\sin(\\alpha_1 + \\alpha_2)}{AD} = \\frac{\\sin \\alpha_1}{AC} + \\frac{\\sin \\alpha_2}{AB}\n$$\nupon division... | United States | HMMT November 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | sqrt(6)/2 | |
0i2a | Problem:
A balloon that blows up in the shape of a perfect cube is being blown up at a rate such that at time $t$ fortnights, it has surface area $6 t$ square furlongs. At how many cubic furlongs per fortnight is the air being pumped in when the surface area is $144$ square furlongs? | [
"Solution:\n\nThe surface area at time $t$ is $6 t$, so the volume is $t^{3/2}$. Hence the air is being pumped in at a rate of $\\frac{3}{2} \\sqrt{t}$. When the surface area is $144$, $t=24$, so the answer is $3 \\sqrt{6}$."
] | United States | Harvard-MIT Math Tournament | [
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | 3 sqrt 6 | |
097f | Problem:
Fie funcția continuă $f:(1 ;+\infty) \rightarrow \mathbb{R}$, $f(x)=\frac{x}{\sqrt[4]{(x-1)^{3}(x+1)^{5}}}$. Determinați primitivele $F:(1 ;+\infty) \rightarrow \mathbb{R}$ ale funcției $f$. | [
"Solution:\n\n$$\n\\begin{aligned}\n& \\int \\frac{x\\, dx}{\\sqrt[4]{(x-1)^{3}(x+1)^{5}}} = \\int \\frac{x\\, dx}{(x-1)(x+1)^{4} \\sqrt{\\frac{x+1}{x-1}}} = \\left|\\begin{array}{c}\nt = \\sqrt[4]{\\frac{x+1}{x-1}} \\\\\nx = \\frac{2}{t^{4}-1}+1 \\\\\ndx = -\\frac{8 t^{3}}{\\left(t^{4}-1\\right)^{2}} dt\n\\end{arr... | Moldova | Olimpiada Republicană la Matematică | [
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | proof and answer | F(x) = ln((√[4]{x+1} + √[4]{x-1}) / (√[4]{x+1} − √[4]{x-1})) − 2 arctan(√[4]{(x+1)/(x−1)}) − 2 √[4]{(x−1)/(x+1)} + C | |
0ga8 | 平面上有一個凸 $3n$ 邊形，其每個頂點上都有一台機器人，每台機器人都射出一道雷射光指向另一台機器人。你每次操作可以選取一台機器人，叫它順時鐘旋轉，直到它的雷射光指向一台新的機器人為止。當三台機器人 $A$、$B$、$C$，其中 $A$ 的雷射光射向 $B$，$B$ 的雷射光射向 $C$，而 $C$ 的雷射光射向 $A$ 時，我們稱這三台機器人構成一個三角形。試問：至少要多少次操作，才能保證平面上出現 $n$ 個三角形？
There's a convex $3n$-polygon on the plane with a robot on each of its vertices. Each robot fires a laser... | [
"答案：$\\frac{9n^2-7n}{2}$ 次。\n\n以下以 $EX$ 代表隨機變數 $X$ 的期望值。\n\n1. 首先，對於任兩點 $A$ 和 $B$，令 $N_{AB}$ 為要將 $A$ 的雷射指向 $B$ 所需的步數。假設我們以均勻分布在所有點中隨機選取相異的 $A$ 和 $B$，則 $EN_{AB} = \\frac{3n-2}{2}$（事實上，易知 $N_{AB}$ 為 $\\{0,1,\\dots,3n-2\\}$ 上的均勻分布。）\n\n2. 對於任三點 $A$、$B$ 和 $C$，令 $N_{ABC}$ 為要旋轉三點上的雷射，使得 $ABC$ 要成為三角形所需步數。假設我們以均勻分布在所有點中隨機選... | Taiwan | 二〇一六數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | (9n^2-7n)/2 | |
0dkh | Let $A_1, A_2, \dots, A_{20}$ be 20 distinct subsets of size 3 of the set $X = \{1, 2, \dots, 10\}$. We say that a subset $S$ of $X$ is a covering subset if for every $1 \le i \le 20$, it holds that $S \cap A_i \ne \emptyset$. What is the minimum possible value of $k$, such that there always exists a covering subset of... | [
"First, we select all the 3-element subsets of $X_1 = \\{1, 2, 3, 4, 5\\}$ and of $X_2 = \\{6, 7, 8, 9, 10\\}$. Then, the number of subsets will be $\\binom{5}{3} + \\binom{5}{3} = 20$. Suppose there is a covering set of size $k \\le 5$, then one of the two sets $X_1, X_2$ will have no more than 2 elements, which m... | Saudi Arabia | Saudi Booklet | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 6 | |
09e6 | a, b, c are integers and $a < b < c$. An integer $n$ is said to be *quadratical* if $a^2 + b^2 - c^2 = n$. Find number of quadratical numbers no less than 1 and no greater than 2014. | [
"Note that $(3n)^2 + (4n)^2 = (5n^2)$ from where we derive $(3n)^2 + (4n - 1)^2 - (5n - 1)^2 = 2n$ and $(3n + 2)^2 + (4n)^2 - (5n - 1)^2 = 2n + 3$.\n\na. Let's consider the case of quadratical number is odd. Set $a = 3n + 2$, $b = 4n$, $c = 5n + 1$ in the equality $(3n + 2)^2 + (4n)^2 - (5n - 1)^2 = 2n + 3$. If $n ... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 2014 | |
0213 | Problem:
Let $ABC$ be a scalene acute triangle. Let $B_{1}$ be the point on ray $[AC$ such that $|AB_{1}| = |BB_{1}|$. Let $C_{1}$ be the point on ray $[AB$ such that $|AC_{1}| = |CC_{1}|$. Let $B_{2}$ and $C_{2}$ be the points on line $BC$ such that $|AB_{2}| = |CB_{2}|$ and $|BC_{2}| = |AC_{2}|$. Prove that $B_{1}, C... | [
"Solution:\nBy construction, lines $B_{1}C_{2}$ and $B_{2}C_{1}$ bisect segments $[AB]$ and $[AC]$, respectively, so their intersection $O$ is the circumcentre of $ABC$. Hence $\\angle BOC = 2\\angle A$ and $\\angle CBO = \\angle OCB = 90^{\\circ} - \\angle A$. Now, by construction, $\\angle OC_{1}B = 90^{\\circ} -... | Benelux Mathematical Olympiad | Benelux 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... | null | proof only | null | |
096n | Problem:
În piramida patrulateră regulată $V A B C D$, înălțimea are lungimea $h$ și este diametru al unei sfere, iar $m(\angle A V B)=\varphi$. Determinați lungimea curbei obținute la intersecția sferei cu suprafața laterală a piramidei. | [
"Solution:\n\n\n\nCurba obținută la intersecția sferei cu suprafața laterală a piramidei constă din 4 arce congruente ale cercurilor, obținute la intersecția fețelor laterale cu sfera.\n\nDeterminăm lungimea arcului $N M$ de cerc cu centrul în punctul $I$, obținut la intersecția sferei cu p... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 4 h φ √(cos φ) / cos(φ/2) | |
0l7y | On $\triangle ABC$ points $A$, $D$, $E$, and $B$ lie in that order on side $\overline{AB}$ with $AD = 4$, $DE = 16$, and $EB = 8$. Points $A$, $F$, $G$, and $C$ lie in that order on side $\overline{AC}$ with $AF = 13$, $FG = 52$, and $GC = 26$. Let $M$ be the reflection of $D$ through $F$, and let $N$ be the reflection... | [] | United States | 2025 AIME I | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | final answer only | 588 | |
0jkw | Problem:
Let $f(x) = x^{2} - 2$, and let $f^{n}$ denote the function $f$ applied $n$ times. Compute the remainder when $f^{24}(18)$ is divided by $89$. | [
"Solution:\nLet $L_{n}$ denote the Lucas numbers given by $L_{0} = 2$, $L_{1} = 1$, and $L_{n+2} = L_{n+1} + L_{n}$. Note that $L_{n}^{2} - 2 = L_{2n}$ when $n$ is even (one can show this by induction, or explicitly using $L_{n} = \\left(\\frac{1+\\sqrt{5}}{2}\\right)^{n} + \\left(\\frac{1-\\sqrt{5}}{2}\\right)^{n}... | United States | HMMT November 2014 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 47 | |
09aj | Given are $2n$ people and it is known that their heights are all different. They have to stand in two rows, each with $n$ people. How many different positions are there, if the front row person is always shorter than the back row person? | [
"In the first seats of the rows, we can choose 2 people $\\frac{2n(2n-1)}{2}$ different ways. Then in the second seats of the rows, we can choose 2 people $\\frac{(2n-2)(2n-3)}{2}$ different ways. Continuing this, we have\n$$\n\\frac{2n(2n-1)}{2} \\cdot \\frac{(2n-2)(2n-3)}{2} \\cdots \\frac{2 \\cdot 1}{2} = \\frac... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | Mongolian | proof and answer | (2n)! / 2^n | |
03x1 | Given an integer $n \ge 3$, prove that there exists a set $S$ of $n$ distinct positive integers such that for any two distinct nonempty subsets $A$ and $B$ of $S$, the numbers
$$
\sum_{x \in A} x, \quad \sum_{x \in B} x
$$
are two coprime composite integers, and $\sum_{x \in X} x$ denotes the sum of all elements of a f... | [
"Let $f(X)$ be the average of elements of the finite number set $X$.\n\nFirst of all, make $n$ different primes $p_1, p_2, \\dots, p_n$ which are all bigger than $n$, and we prove that for any different nonempty subsets $A$, $B$ of the set $S_1 = \\{ \\prod (1/p_j) : 1 \\le j \\le n \\}$, $f(A) \\ne f(B)$ always ho... | China | Chinese Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
07bu | Quadrilateral $ABCD$ is both inscribed and circumscribed. Let $E$ be the intersection point of $AD$ and $BC$, $F$ the intersection point of $AB$ and $CD$, $S$ the intersection point of $AC$ and $BD$ and $O$ the circumcenter of quadrilateral $ABCD$. $E'$ and $F'$ are selected on $AB$ and $AD$ such that $\angle BEE' = \a... | [
"An inversion with center $O$ and constant $r^2$ ($r$ is the radius of the circumcircle of quadrilateral), takes the circle with diameter $OS$ to the line $EF$ and the circumcircle of the triangle $OAM$ to the line $AM$. Furthermore, since $XO$ is the bisector of $\\angle AXB$ and $OA = OB$, we get the quadrilatera... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > C... | null | proof only | null | |
0fqv | Problem:
Prueba que para todo $a, b, c > 0$ se cumple que
$$
\frac{a^{2}}{b^{3} c} - \frac{a}{b^{2}} \geq \frac{c}{b} - \frac{c^{2}}{a}
$$
¿En qué caso se cumple la igualdad? | [
"Solution:\n\nMultiplicando por $a b^{3} c$ toda la desigualdad para eliminar los denominadores, tendremos que\n$$\n\\begin{gathered}\na^{3} - a^{2} b c \\geq a c^{2} b^{2} - c^{3} b^{3} \\Longleftrightarrow a^{3} - a c^{2} b^{2} \\geq a^{2} b c - c^{3} b^{3} \\Longleftrightarrow \\\\\na^{2}(a - b c) \\geq c^{2} b^... | Spain | OME fase local | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | Equality holds if and only if a = b c. | |
0gv2 | For an integer $n$, let $\sigma(n)$ be the sum of all positive divisors of $n$. A sequence of positive integers $(a_i)_{i=0}^{\infty}$ is defined as follows: $a_0 = 1$ and for each $n > 1$, $a_n$ is the smallest integer greater than 1 such that
$$
\sigma(a_0 a_1 \cdots a_{n-1}) \mid \sigma(a_0 a_1 \cdots a_n).
$$
Deter... | [
"Answer: 36.\n\nWe will show that this sequence is increasing and consists of all the numbers of the form $p^{2^n}$, where $p$ is a prime and $n$ is a non-negative integer. Thus, all the terms satisfying the condition are $1, 2, 11, 23, \\dots, 2^{1024}, 11^{1024}, 23^{1024}, 2^{2048}, 2^{4096}$.\n\nLet $S$ be the ... | Turkey | Team Selection Test for IMO 2024 | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 36 | |
00s0 | We have $n$ students sitting at a round table. Initially each student is given one candy. At each step each student having candies either picks one of its candies and gives it to one of its neighbouring students, or distributes all of its candies to its neighbouring students in any way he wishes. A distribution of cand... | [
"The answer turns out to be $\\binom{2n-1}{n}$ if $n$ is odd and $\\binom{2n-1}{n} - 2\\binom{3n-1}{n}$ if $n$ is even.\n\nCase 1. Suppose $n$ is odd, say $n=2m+1$. In this case we will show that any distribution of candies is legal. Thus the number of legal distributions is indeed $\\binom{2n-1}{n}$.\n\nIn this ca... | Balkan Mathematical Olympiad | BMO 2017 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | If n is odd: C(2n−1, n). If n is even: C(2n−1, n) − 2·C(3n−1, n). | |
056e | Mari writes 8 prime numbers (not necessarily different) to her notebook, all smaller than 200. She then adds 1 to the first number, 2 to the second number and so on until adding 8 to the eighth number. She then finds the product of the eight sums. Find the largest power of two which can divide the product found. | [
"To have the product divisible by as great power of 2 as possible, each of the factors has to be divisible by the greatest power of 2 possible. By adding an even number to a prime number, the sum is divisible by 2 only if the original prime is even. As the only even prime number is 2, it has to be chosen to all eve... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 2^31 | |
07ys | Problem:
Determinare tutti gli interi positivi di tre cifre che sono uguali a 34 volte la somma delle loro cifre. | [
"Solution:\nGli interi richiesti sono 102, 204, 306, 408.\n\nIndichiamo infatti con $a, b, c$, rispettivamente, la cifra delle centinaia, delle decine, delle unità di un intero di 3 cifre. La condizione data si traduce allora in $100a + 10b + c = 34(a + b + c)$, da cui, con semplici passaggi algebrici, si ricava ch... | Italy | null | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 102, 204, 306, 408 | |
08oo | Problem:
Let $a$, $b$, $c$ be positive real numbers. Prove that
$$
\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>2
$$ | [
"Solution:\nStarting from the double expression on the left-hand side of given inequality, and applying twice the Arithmetic-Geometric mean inequality, we find that\n$$\n\\begin{aligned}\n2 \\frac{a}{b}+2 \\sqrt{\\frac{b}{c}}+2 \\sqrt[3]{\\frac{c}{a}} & =\\frac{a}{b}+\\left(\\frac{a}{b}+\\sqrt{\\frac{b}{c}}+\\sqrt{... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0atm | Problem:
Let $a$, $b$ and $c$ be positive integers such that $\frac{a \sqrt{2013}+b}{b \sqrt{2013}+c}$ is a rational number. Show that $\frac{a^{2}+b^{2}+c^{2}}{a+b+c}$ and $\frac{a^{3}-2 b^{3}+c^{3}}{a+b+c}$ are both integers. | [
"Solution:\nBy rationalizing the denominator, $\\frac{a \\sqrt{2013}+b}{b \\sqrt{2013}+c}=\\frac{2013 a b-b c+\\sqrt{2013}\\left(b^{2}-a c\\right)}{2013 b^{2}-c^{2}}$. Since this is rational, then $b^{2}-a c=0$. Consequently,\n$$\n\\begin{aligned}\na^{2}+b^{2}+c^{2} & =a^{2}+a c+c^{2}=(a+c)^{2}-a c=(a+c)^{2}-b^{2} ... | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Other",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0isx | Problem:
A root of unity is a complex number that is a solution to $z^{n}=1$ for some positive integer $n$. Determine the number of roots of unity that are also roots of $z^{2}+a z+b=0$ for some integers $a$ and $b$. | [
"Solution:\nAnswer: 8\n\nThe only real roots of unity are $1$ and $-1$.\n\nIf $\\zeta$ is a complex root of unity that is also a root of the equation $z^{2}+a z+b=0$, then its conjugate $\\bar{\\zeta}$ must also be a root. In this case, $|a|=|\\zeta+\\bar{\\zeta}| \\leq |\\zeta|+|\\bar{\\zeta}| = 2$ and $b=\\zeta \... | United States | 11th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 8 | |
0ha8 | For positive integers $m$ and $n$, compare the numbers $A = m^{525} + n^{525}$ and $B = (m+n)(m^2+n^2)(m^4+n^4)(m^8+n^8)\dots(m^{128}+n^{128})$. | [
"If $m = n$, then $A = 2m^{525}$, $B = 2m \\cdot 2m^2 \\cdot 2m^4 \\dots \\cdot 2m^{128} = 2^8 m^{255} > A = 2m^{525}$\n$\\Leftrightarrow m = n = 1$, while for $m = n > 1$ we get $A > B$.\n\nNow, without the loss of generality, $m > n$. Then we can do the transformations:\n$$\n\\begin{aligned}\nB \\le (m-n)B &= (m-... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | B > A only when both integers equal one; in all other cases A > B. | |
06jv | Let $ABCD$ be inscribed in a circle with centre $O$. Let $E$ be the intersection point of $AC$ and $BD$. $M$ and $N$ are the midpoints of the arcs $AB$ and $CD$ respectively (the arcs not containing any other vertices). Let $P$ be the intersection point of $EO$ and $MN$. Suppose $BC = 5$, $AC = 11$, $BD = 12$ and $AD =... | [
"We have $\\frac{MN}{NP} = \\frac{23}{13}$.\nLet $I$ and $J$ be the incentres of $\\triangle ADE$ and $\\triangle BCE$ respectively. Note that $I$ is the intersection point of $AN$ and $DM$, while $J$ is the intersection point of $BN$ and $CM$. Applying Pascal's theorem to $ANBDMC$, we find that $I, J, E$ are colli... | Hong Kong | Year 2016 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point ci... | null | proof and answer | 23/13 | |
093y | Problem:
Let $n$ be a positive integer. Prove that in a regular $6 n$-gon, we can draw $3 n$ diagonals with pairwise distinct ends and partition the drawn diagonals into $n$ triplets so that:
- the diagonals in each triplet intersect in one interior point of the polygon and
- all these $n$ intersection points are disti... | [
"Solution:\nFor $n=1$ take the main diagonals.\n\nFor $n=2$ we have the following construction:\n\n\n\nDenote the vertices $A_{1}, A_{2}, \\ldots, A_{12}$. We will show that the lines $A_{1} A_{4}$, $A_{2} A_{6}$ and $A_{3} A_{11}$ are concurrent. Let $X$ be the intersection point of $A_{2}... | Middle European Mathematical Olympiad (MEMO) | 15th Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
02h6 | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$
f(x)f(y) + f(x) = 2f(x) + xy
$$
for every real $x, y$. | [
"First, $f$ is bijective because, plugging $x = 1$, we get $f(f(y) + f(1)) = 2f(1) + x$; $x + 2f(1)$ spans all real numbers (so $f$ is surjective) and $f(x) = f(y) \\iff f(x) + f(1) = f(y) + f(1) \\iff f(f(x) + f(1)) = f(f(y) + f(1)) \\iff 2f(1) + x = 2f(1) + y \\iff x = y$ (so $f$ is injective).\nChoose $x \\neq 0... | Brazil | Brazilian Math Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | f(x) = x + 1 | |
07jv | $n > 10$ is a positive integer. $n$ lines are drawn on the plane such that no three are concurrent and no two are parallel. Among the finite regions formed by these lines, at least $\frac{n^2}{8} + 1$ of them were blacked. We call a triangle formed by three lines *good* if there is only one black region inside of it. P... | [
"First, we prove that for every line $\\ell$, there exist lines $d_1$, $d_2$ such that the triangle $\\ell d_1 d_2$ has exactly one black region inside it. On one of the two sides of $\\ell$, there is a black region. Without loss of generality, assume this region is above $\\ell$. To see this, take the closest poin... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
094p | Problem:
Let $A B C$ be an acute triangle with $A B < A C$. Let $J$ be the center of the $A$-excircle of $A B C$. Let $D$ be the projection of $J$ on line $B C$. The internal bisectors of angles $B D J$ and $J D C$ intersect lines $B J$ and $J C$ at $X$ and $Y$, respectively. Segments $X Y$ and $J D$ intersect at $P$.... | [
"Solution:\n\nLet $E$ and $F$ be the points symmetric to $D$ with respect to lines $J C$ and $B J$, respectively. Then $D$, $E$ and $F$ are the points of tangency of the $A$-excircle of $A B C$ with $B C$, $C A$ and $A B$, respectively. In particular, $A E = A F$.\n\nLet $Z$ be the point on ray $D P \\rightarrow$ s... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
093h | Problem:
Determine the smallest and the greatest possible values of the expression
$$
\left(\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}+\frac{1}{c^{2}+1}\right)\left(\frac{a^{2}}{a^{2}+1}+\frac{b^{2}}{b^{2}+1}+\frac{c^{2}}{c^{2}+1}\right)
$$
provided $a$, $b$, and $c$ are non-negative real numbers satisfying $ab+bc+ca=1$. | [
"Solution:\n\nLet us denote\n$$\nx=\\frac{a^{2}}{a^{2}+1}+\\frac{b^{2}}{b^{2}+1}+\\frac{c^{2}}{c^{2}+1}, \\quad y=\\frac{2 a b c}{(a+b)(b+c)(c+a)}\n$$\nto simplify notation.\nDenominators in $x$ can be manipulated using $ab+bc+ca=1$ as\n$$\na^{2}+1=a^{2}+ab+bc+ca=(a+b)(a+c)\n$$\nand similarly for $b^{2}+1$ and $c^{... | Middle European Mathematical Olympiad (MEMO) | MEMO Team Competition | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | minimum 27/16, maximum 2 | |
041v | Given an integer $n \ge 2$. Suppose $A_1, A_2, \dots, A_n$ are $n$ nonempty finite sets satisfying: $|A_i \Delta A_j| = |i-j|$ for all $i, j \in \{1, 2, \dots, n\}$.
Find the minimum value of $|A_1| + |A_2| + \dots + |A_n|$. (Here $|X|$ denotes the number of elements of a finite set $X$ and $X \Delta Y = \{a \mid a \in... | [
"For each positive integer $k$, we prove that the minimum value of $S_{2k}$ is $k^2 + 2$; the minimum value of $S_{2k+1}$ is $k(k+1)+2$.\nFirstly, define the sets $A_1, A_2, \\dots, A_{2k}, A_{2k+1}$ as follows:\n$$\nA_i = \\{i, i+1, \\dots, k\\}, \\quad i = 1, 2, \\dots, k; \\quad A_{k+1} = \\{k, k+1\\};\n$$\n$$\n... | China | China Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | ⌊n^2/4⌋ + 2 | |
09gp | 00, 01, 02, 03, ..., 97, 98, 99 дугаартай 100 ширхэг билетээс ямар ч $k$ ширхэг билетийг сонгон авахад, сонгосон билетүүд дотор аравтын орны цифрүүд нь хоорондоо ялгаатай, мөн нэгжийн орны цифрүүд нь хоорондоо ялгаатай байх 4 ширхэг билет заавал олддог байв. $k$-ийн боломжит хамгийн бага утгыг ол. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 31 | |
0aw6 | Problem:
An ant situated at point $A$ decides to walk 1 foot east, then $\frac{1}{2}$ foot northeast, then $\frac{1}{4}$ foot east, then $\frac{1}{8}$ foot northeast, then $\frac{1}{16}$ foot east and so on (that is, the ant travels alternately between east and northeast, and the distance travelled is decreased by hal... | [
"Solution:\n\nThe distance is the hypotenuse of a right triangle. The length of its base is\n$$\n\\begin{aligned}\n1+\\frac{1}{2 \\sqrt{2}}+\\frac{1}{4}+\\frac{1}{8 \\sqrt{2}}+\\frac{1}{16}+\\frac{1}{32 \\sqrt{2}}+\\cdots & =\\left(1+\\frac{1}{4}+\\frac{1}{16}+\\cdots\\right)+\\frac{1}{2 \\sqrt{2}}\\left(1+\\frac{1... | Philippines | 18th PMO National Stage Oral Phase | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2/3 * sqrt(5 + 2*sqrt(2)) | |
0j1x | Problem:
Let $x$ be a real number. Find the maximum value of $2^{x(1-x)}$. | [
"Solution:\n\nAnswer: $\\sqrt[4]{2}$\n\nConsider the function $2^{y}$. This is monotonically increasing, so to maximize $2^{y}$, you simply want to maximize $y$. Here, $y = x(1-x) = -x^{2} + x$ is a parabola opening downwards. The vertex of the parabola occurs at $x = \\frac{-1}{-2} = \\frac{1}{2}$, so the maximum ... | United States | Harvard-MIT November Tournament | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | sqrt[4]{2} | |
0jgm | Problem:
In triangle $ABC$, $\angle BAC = 60^\circ$. Let $\omega$ be a circle tangent to segment $AB$ at point $D$ and segment $AC$ at point $E$. Suppose $\omega$ intersects segment $BC$ at points $F$ and $G$ such that $F$ lies in between $B$ and $G$. Given that $AD = FG = 4$ and $BF = \frac{1}{2}$, find the length of... | [
"Solution:\n\nAnswer: $\\boxed{\\frac{16}{5}}$ Let $x = CG$. First, by power of a point, $BD = \\sqrt{BF(BF + FG)} = \\frac{3}{2}$, and $CE = \\sqrt{x(x + 4)}$. By the law of cosines, we have\n$$\n\\left(x + \\frac{9}{2}\\right)^2 = \\left(\\frac{11}{2}\\right)^2 + (4 + \\sqrt{x(x + 4)})^2 - \\frac{11}{2}(4 + \\sqr... | United States | HMMT November | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 16/5 | |
010t | Problem:
Find the smallest positive integer $k$ which is representable in the form $k = 19^{n} - 5^{m}$ for some positive integers $m$ and $n$. | [
"Solution:\nAnswer: 14.\nAssume that there are integers $n, m$ such that $k = 19^{n} - 5^{m}$ is a positive integer smaller than $19^{1} - 5^{1} = 14$. For obvious reasons, $n$ and $m$ must be positive.\n\nCase 1: Assume that $n$ is even. Then the last digit of $k$ is 6. Consequently, we have $19^{n} - 5^{m} = 6$. ... | Baltic Way | Baltic Way | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 14 | |
09zo | In a $3 \times 2$ rectangle, we put the numbers $1$ to $6$ in the squares in such a way that each number occurs exactly once. The *score* of such a distribution is determined as follows: for each two adjacent squares we compute the difference between their two numbers and we add up all these differences. In the example... | [] | Netherlands | Second Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 11 | |
0ime | Problem:
Bob the bomb-defuser has stumbled upon an active bomb. He opens it up, and finds the red and green wires conveniently located for him to cut. Being a seasoned member of the bomb-squad, Bob quickly determines that it is the green wire that he should cut, and puts his wirecutters on the green wire. But just bef... | [
"Solution:\n\nAnswer: $\\frac{23}{30}$. Suppose Bob makes $n$ independent decisions, with probabilities of switching $p_{1}, p_{2}, \\ldots, p_{n}$. Then in the expansion of the product\n$$\nP(x)=\\left(p_{1}+\\left(1-p_{1}\\right) x\\right)\\left(p_{2}+\\left(1-p_{2}\\right) x\\right) \\cdots\\left(p_{n}+\\left(1-... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof and answer | 23/30 | |
04r2 | Prove that for each integer number $n$, $n \ge 3$, the following $2n$-digit number
$$
\underbrace{1\dots12}_{n-1} \underbrace{8\dots8}_{n-2} 96
$$
is a perfect square. | [
"The number under consideration can be expressed as follows:\n$$\n\\begin{aligned}\n& (10^{2n-1} + 10^{2n-2} + \\dots + 10^{n+1}) + 2 \\cdot 10^n + 8 \\cdot (10^{n-1} + 10^{n-2} + \\dots + 10^2) + 96 \\\\\n&= 10^{n+1} \\cdot \\frac{10^{n-1}-1}{9} + 2 \\cdot 10^n + 8 \\cdot 10^2 \\cdot \\frac{10^{n-2}-1}{9} + 96 \\\... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof only | null | |
0iiy | Problem:
Two $18$-$24$-$30$ triangles in the plane share the same circumcircle as well as the same incircle. What's the area of the region common to both the triangles? | [
"Solution:\n\nNotice, first of all, that $18$-$24$-$30$ is $6$ times $3$-$4$-$5$, so the triangles are right. Thus, the midpoint of the hypotenuse of each is the center of their common circumcircle, and the inradius is $\\frac{1}{2}(18+24-30)=6$. Let one of the triangles be $ABC$, where $\\angle A < \\angle B < \\a... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | 132 | |
09h4 | $n$ girls are standing in a circle, each holding exactly 1 card. One girl gives her card to the girl on her left, who in turn gives 2 cards to the girl on her left. The girl who got the cards gives 1 card to the girl on her left. The girl who got the card gives 2 cards to the girl on her left. Continuing this way, each... | [
"Answer: $n = 2$, $n = 2^k + 1$, $n = 2^k + 2$, $k \\ge 1$.\nSolution omitted."
] | Mongolia | Mongolian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | n = 2, or n = 2^k + 1, or n = 2^k + 2 for k ≥ 1 | |
031s | Problem:
A set of at least three positive integers is called uniform if removing any of its elements the remaining set can be disjoint into two subsets with equal sums of elements. Find the minimal cardinality of a uniform set. | [
"Solution:\nLet $A=\\{a_{1}, a_{2}, \\ldots, a_{n}\\}$ be a uniform set. Set $S=a_{1}+a_{2}+\\cdots+a_{n}$. It follows from the given condition that $S-a_{i}$ is an even number for any $i=1,2, \\ldots, n$. Suppose that the number $S$ is even. Then all the numbers $a_{i}$ are even. Set $a_{i}=2 b_{i}, i=1,2, \\ldots... | Bulgaria | 52. Bulgarian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 7 | |
0ft5 | Problem:
Seien $n \in \mathbb{N}$ und $t_{1}, t_{2}, \ldots, t_{k}$ verschiedene positive Teiler von $n$. Eine Identität der Form $n = t_{1} + t_{2} + \ldots + t_{k}$ heißt Darstellung von $n$ als Summe verschiedener Teiler. Zwei solche Darstellungen gelten als gleich, wenn sie sich nur um die Reihenfolge der Summande... | [
"Solution:\n\nSo ein $M$ existiert nicht. Wir nennen eine Darstellung von $n$ als Summe verschiedener Teiler im Folgenden kurz Darstellung. Wir zeigen nun, dass gilt\n$$\na\\left(3 \\cdot 2^{n}\\right) \\geq n\n$$\nWegen $6 = 1 + 2 + 3$ ist $a(6) \\geq 1$ und es genügt zu zeigen, dass gilt $a\\left(3 \\cdot 2^{n+1}... | Switzerland | IMO - Selektion | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0aub | Problem:
Segment $CD$ is tangent to the circle with center $O$, at $D$. Point $A$ is in the interior of the circle, and segment $AC$ intersects the circle at $B$. If $OA = 2$, $AB = 4$, $BC = 3$, and $CD = 6$, find the length of segment $OC$.
 | [] | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2*sqrt(15) | |
0kmo | Problem:
Two unit squares $S_{1}$ and $S_{2}$ have horizontal and vertical sides. Let $x$ be the minimum distance between a point in $S_{1}$ and a point in $S_{2}$, and let $y$ be the maximum distance between a point in $S_{1}$ and a point in $S_{2}$. Given that $x=5$, the difference between the maximum and minimum po... | [
"Solution:\n\nConsider what must happen in order for the minimum distance to be exactly $5$. Let one square, say $S_{1}$, have vertices of $(0,0)$, $(0,1)$, $(1,0)$, and $(1,1)$. Further, assume WLOG that the center of $S_{2}$ is above the line $y=\\frac{1}{2}$ and to the right of the line $x=\\frac{1}{2}$, determi... | United States | HMMT November 2021 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | final answer only | 472 | |
05nt | Problem:
Soit $ABC$ un triangle. On note $P$ le symétrique de $B$ par rapport à $(AC)$ et $Q$ le symétrique de $C$ par rapport à $(AB)$.
Soit $T$ l'intersection entre $(PQ)$ et la tangente en $A$ au cercle circonscrit à $(APQ)$.
Montrer que le symétrique de $T$ par rapport à $A$ appartient à $(BC)$. | [
"Solution:\n\n\n\nSoient $B'$ et $C'$ les symétriques de $B$ et $C$ par rapport à $A$. Comme le triangle $AB'P$ est isocèle en $A$, on a $\\widehat{AB'P} = \\frac{1}{2}\\left(180^{\\circ} - \\widehat{B'AP}\\right) = \\frac{1}{2} \\widehat{PAB} = \\widehat{BAC}$, donc $(B'P) \\parallel (AC)$... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null |
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