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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0av1 | Problem:
How many real numbers $x$ satisfy the equation
$$
\left(\left|x^{2}-12 x+20\right|^{\log x^{2}}\right)^{-1+\log x}=\left|x^{2}-12 x+20\right|^{1+\log (1 / x)} ?
$$ | [
"Solution:\nLet $A = |x^2 - 12x + 20|$.\n\nThe equation is:\n$$\n\\left(A^{\\log x^2}\\right)^{-1 + \\log x} = A^{1 + \\log(1/x)}\n$$\n\nFirst, $A \\geq 0$ and $x \\neq 0$ (since $\\log x^2$ is defined for $x \\neq 0$).\n\nAlso, $\\log x$ is defined for $x > 0$.\n\nSo, $x > 0$.\n\nNow, $\\log(1/x) = -\\log x$, so $... | Philippines | 19th Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 5 | |
04kg | Let $x$ and $y$ be distinct real numbers such that
$$
x + 4 = (y - 2)^2 \quad \text{and} \quad y + 4 = (x - 2)^2.
$$
Determine $x^2 + y^2$. | [
"Let us write the given equations:\n$$\nx + 4 = (y - 2)^2 \\tag{1}\n$$\n$$\ny + 4 = (x - 2)^2 \\tag{2}\n$$\nExpand the right sides:\nFrom (1):\n$$\nx + 4 = y^2 - 4y + 4\n$$\nSo\n$$\nx = y^2 - 4y\n$$\nFrom (2):\n$$\ny + 4 = x^2 - 4x + 4\n$$\nSo\n$$\ny = x^2 - 4x\n$$\nNow substitute $x$ from above into the expression... | Croatia | Mathematical competitions in Croatia | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | null | proof and answer | 15 | |
0l38 | Problem:
The vertices of a cube are labeled with the integers $1$ through $8$, with each used exactly once. Let $s$ be the maximum sum of the labels of two edge-adjacent vertices. Compute the minimum possible value of $s$ over all such labelings. | [
"Solution:\n\nThe answer must be at least $11$, because the label $8$ is adjacent to three vertices, one of which has label at least $3$.\n\nTo show $11$ is achievable, note that the following labelling achieves $s=11$:\n\n\n\nThus the answer is $11$."
] | United States | HMMT November 2024 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 11 | |
031h | Problem:
Let $M$ be the centroid of $\triangle ABC$ with $\Varangle AMB = 2 \Varangle ACB$. Prove that:
a) $AB^{4} = AC^{4} + BC^{4} - AC^{2} \cdot BC^{2}$;
b) $\Varangle ACB \geq 60^{\circ}$. | [
"Solution:\na) We shall use the standard notation for the elements of $\\triangle ABC$. The Cosine theorem for $\\triangle ABM$ gives\n$$\n\\cot \\Varangle AMB = \\frac{AM^{2} + BM^{2} - AB^{2}}{2 AM \\cdot MB \\sin \\Varangle AMB} = \\frac{a^{2} + b^{2} - 5c^{2}}{12 S_{ABC}}.\n$$\nSince $\\cot 2\\gamma = \\frac{2 ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
08p9 | Problem:
A splitting of a planar polygon is a finite set of triangles whose interiors are pairwise disjoint, and whose union is the polygon in question. Given an integer $n \geq 3$, determine the largest integer $m$ such that no planar $n$-gon splits into less than $m$ triangles. | [
"Solution:\nThe required maximum is $\\lceil n / 3 \\rceil$, the least integer greater than or equal to $n / 3$. To describe a planar $n$-gon splitting into this many triangles, write $n = 3m - r$, where $m$ is a positive integer and $r = 0, 1, 2$, and consider $m$ coplanar equilateral triangles $A_{3i} A_{3i+1} A_... | JBMO | Junior Balkan Mathematics Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | ⌈n/3⌉ | |
09c7 | a, b - 1-ээс их натурал тоонууд ба $a \mid b^3 - 1$, $b \mid a - 1$ бол $a = b^{3/2} + 1$ эсвэл $a = b^2 + b + 1$ гэж батал. | [
"$$\nb^3 - 1 = ak \\quad (k \\ge 1) \\quad q \\equiv 1(q) \\Rightarrow k \\equiv -1(b)\n$$\n---\n$$\nk = lb - 1 \\ (l \\ge 1) \\quad a = \\frac{b^3 - 1}{lb - 1} \\Rightarrow l < b^2\n$$\nba\n$$\nlb - 1|b^2 - l,\\ lb - 1|b - l^2\n$$\nгэж гарна.\n$$\na \\neq b^{3/2} + 1 \\text{ гээ } (l \\neq b^{1/2}).\n$$\nХэрэв $1 ... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Mongolian | proof only | null | |
02yv | Problem:
No triângulo retângulo isósceles $A O B$, os pontos $P, Q$ e $S$ são escolhidos sobre os lados $O B, O A$ e $A B$, respectivamente, de modo que $P Q R S$ é um quadrado. Se os comprimentos de $O P$ e $O Q$ são $a$ e $b$, respectivamente, e a área do quadrado $P Q R S$ é $2/5$ da área do triângulo $A O B$, dete... | [
"Solution:\n\nSeja $C$ o pé da perpendicular do ponto $S$ ao segmento $O B$. Os triângulos $S P C$ e $P Q O$ possuem os mesmos ângulos, pois\n$$\n\\begin{aligned}\n\\angle C P S & = \\angle 180^\\circ - \\angle S P Q - \\angle O P Q \\\\\n& = 90^\\circ - \\angle O P Q \\\\\n& = \\angle P Q O\n\\end{aligned}\n$$\nCo... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 2 | |
09ag | A circle touches $BC$ side of triangle $ABC$ at the point $M$ and intersects sides $AB$ and $AC$ at $D$ and $E$ respectively. If $DE \parallel BC$, prove that $EM = MD$. | [
"Since $BC$ is tangent to the circle, we have $\\angle EMC = \\angle EAM = \\angle EDM$ and $\\angle DMB = \\angle DAM = \\angle DEM$. From $DE \\parallel BC$, we have $\\angle EDM = \\angle DMB$. So $AM$ bisects $\\angle CAB$."
] | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Mongolian | proof only | null | |
05e6 | Problem:
We call a positive integer $n$ peculiar if, for any positive divisor $d$ of $n$, the integer $d(d+1)$ divides $n(n+1)$. Prove that for any four different peculiar positive integers $A, B, C$ and $D$, the following holds:
$$
\operatorname{gcd}(A, B, C, D)=1
$$
Here $\operatorname{gcd}(A, B, C, D)$ is the large... | [
"Solution:\n\nObserve that $n=1$ is peculiar and that every prime is peculiar. Consider $\\frac{n}{d}$, where $d>1$ is a divisor of $n$. Then $\\frac{n}{d}\\left(\\frac{n}{d}+1\\right)$ divides $n(n+1)$, equivalent to $n+d$ dividing $d^{2}(n+1)$. Since $n \\equiv -d \\pmod{n+d}$, we obtain that $n+d$ divides $d^{3}... | European Girls' Mathematical Olympiad (EGMO) | EGMO 2024 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
03hq | Problem:
Let $k$ be a positive integer. Find all polynomials
$$
P(x) = a_{0} + a_{1} x + \cdots + a_{n} x^{n}
$$
where the $a_{i}$ are real, which satisfy the equation
$$
P(P(x)) = [P(x)]^{k}
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | All solutions are: (i) P(x) ≡ c where c is a real constant satisfying c = c^k (that is, for k = 1 any real c; for k ≥ 2, c = 0 or 1, and also c = −1 when k is odd), and (ii) the nonconstant polynomial P(x) = x^k. | |
0kgv | Problem:
Acute triangle $ABC$ has circumcircle $\Gamma$. Let $M$ be the midpoint of $BC$. Points $P$ and $Q$ lie on $\Gamma$ so that $\angle APM = 90^{\circ}$ and $Q \neq A$ lies on line $AM$. Segments $PQ$ and $BC$ intersect at $S$. Suppose that $BS = 1$, $CS = 3$, $PQ = 8 \sqrt{\frac{7}{37}}$, and the radius of $\Ga... | [
"Solution:\n\nLet $A'$ be the $A$-antipode in $\\Gamma$, let $O$ be the center of $\\Gamma$, and let $T := AA' \\cap BC$. Note that $A'$ lies on line $PM$. The key observation is that $T$ is the reflection of $S$ about $M$; this follows by the Butterfly Theorem on chords $\\overline{PA'}$ and $\\overline{AQ}$.\n\nL... | United States | HMMT Spring 2021 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 3703 | |
0d19 | Let $p$ be a prime. At any vertex of a regular polygon with $p$ sides it is written an integer. For any vertex of the polygon we compute the difference between the sum of the integers written at his neighbors and his number. After that we delete all the initial integers and replace them by the new obtained integers. Pr... | [
"Let $A_0A_1...A_{p-1}$ be the regular polygon and let $(a_0, a_1, ..., a_{p-1})$ be the integers written at its vertices, $a_k$ at $A_k$ for $k = 0, 1, ..., p-1$. Consider the polynomial with integer coefficients\n$$\nH(x) = a_0 + a_1x + ... + a_{p-1}x^{p-1}.\n$$\nApplying the transformation\n$$\na_0 \\to a_{p-1} ... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Linear Algebra > Matrices",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
00cf | Sea $n$ un entero positivo. Se tienen $n$ bolillas numeradas del 1 al $n$ y tres cajas de diferentes colores.
Hallar el menor $n$ tal que para toda ubicación de las $n$ bolillas en las tres cajas siempre haya en una misma caja dos bolillas tales que la diferencia de los números escritos en ellas (el mayor menos el meno... | [] | Argentina | Nacional OMA 2019 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | 27 | |
0kxn | Problem:
Six standard fair six-sided dice are rolled and arranged in a row at random. Compute the expected number of dice showing the same number as the sixth die in the row. | [
"Solution:\n\nFor each $i = 1, 2, \\ldots, 6$, let $X_{i}$ denote the indicator variable of whether the $i$-th die shows the same number as the sixth die. Clearly, $X_{6} = 1$ always. For all other $i$, $X_{i}$ is $1$ with probability $\\frac{1}{6}$ and $0$ otherwise, so $\\mathbb{E}\\left[X_{i}\\right] = \\frac{1}... | United States | HMMT November 2023 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 11/6 | |
07ut | At the start of a game, a positive integer $M$ is fixed and you are given boxes $B_i$ for each $i \in \mathbb{N}$, all are empty. You can adjust the number of marbles in the boxes by making a series of moves. The allowable moves are as follows:
**Move A:** Add a marble to $B_1$ and to $B_2$.
**Move B:** Add a marble to... | [
"We denote by $m_{i,j}$ the number of marbles in Box $B_i$ after $j$ moves, and we define the associated sum $s_j$ by\n$$\ns_j := \\sum_{i=1}^{\\infty} m_{i,j} 2^{i-1}.\n$$\nWe have $s_0 = 0$ and each $s_j - s_{j-1}$ is non-negative. In fact, $s_j - s_{j-1}$ equals:\n$$\n\\left. \\begin{array}{c} 3 \\\\ 1+2^{M-1} \... | Ireland | IRL_ABooklet | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | You can win the game if and only if M is odd. | |
00ly | Wie viele Lösungen hat die Gleichung
$$
\lfloor \frac{x}{20} \rfloor = \lfloor \frac{x}{17} \rfloor
$$
über der Menge der positiven ganzen Zahlen?
Dabei bezeichnet $\lfloor a \rfloor$ die größte ganze Zahl, die kleiner oder gleich $a$ ist.
(Karl Czakler) | [
"Es sei\n$$\n\\lfloor \\frac{x}{20} \\rfloor = \\lfloor \\frac{x}{17} \\rfloor = n.\n$$\nDann gilt $20n \\leq x < 20n + 20$ und $17n \\leq x < 17n + 17$. Daher muss für alle möglichen Lösungen $x$\ngelten\n$$\n20n \\leq x < 17n + 17. \\tag{1}\n$$\nFür den Wert $n$ folgt $20n < 17n + 17$ also $n \\in \\{0, 1, 2, 3, ... | Austria | 48. Österreichische Mathematik-Olympiade Landeswettbewerb für Anfängerinnen und Anfänger | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | German | proof and answer | 56 | |
08p7 | Problem:
The natural numbers from $1$ to $50$ are written down on the blackboard. At least how many of them should be deleted, in order that the sum of any two of the remaining numbers is not a prime? | [
"Solution:\n\nNotice that if the odd, respectively even, numbers are all deleted, then the sum of any two remaining numbers is even and exceeds $2$, so it is certainly not a prime. We prove that $25$ is the minimal number of deleted numbers. To this end, we group the positive integers from $1$ to $50$ in $25$ pairs... | JBMO | Junior Balkan Mathematics Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | null | proof and answer | 25 | |
0ghe | 令 $N$ 為正整數。怪怪國有 $N$ 座城堡，其中每對城堡之間至多只有一條道路。每條道路上有至多 4 名守衛。為了節省人事開銷，怪怪國王頒布以下命令：
(1) 如果三座城堡之間都互有道路，則其中任何一條道路都不能有 4 名守衛；
(2) 如果四座城堡之間都互有道路，則從其中任何一座城堡出發，通往另外三座城堡的三條道路不能全部都站有 3 名守衛。
證明：在此命令下，怪怪國在道路上的守衛總數不超過 $N^2$。
註：只證出守衛總數不超過 $cN^2$，其中 $c > 1$ 與 $N$ 無關，將依 $c$ 值來給分。
Let $N$ be a positive integer. Kingdom Wierdo has $N$ cast... | [
"以城堡為點集 $V$，道路為邊集 $E$，守衛數量為邊權重做 weighted graph。我們將對這個圖做一系列的操作，使得國王命令持續被滿足，且守衛數量不減。\n定義 $W(ab)$ 為 $ab$ 邊上的權重，而 $W(a) = \\sum_{ab \\in E} W(ab)$ 為點 $a$ 所有連邊的權重總和，此外，對任何不相連的兩點 $a$ 和 $b$，定義它們兩點等價，若且唯若對於任何其他 $c \\in V$，$ac$ 連邊且 $W(ac) = t$ 若且唯若 $bc$ 連邊且 $W(bc) = t$。\n現在考慮以下操作：若有 $a$、$b$ 兩點不相鄰也不等價，假設 $W(a) \\ge W(b)$，則對於... | Taiwan | 2023 數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | Chinese (Traditional) | proof only | null | |
0axu | Problem:
For how many primes $p < 50$ is $p^{4} + 5p^{3} + 4$ divisible by $5$? | [
"Solution:\n\nClearly, the expression is not divisible by $5$ when $p = 5$. For any prime $p$ other than $5$, note that $p^{4} \\equiv 1 \\pmod{5}$, so $p^{4} + 5p^{3} + 4 \\equiv 0 \\pmod{5}$. Hence, the problem reduces to counting the number of primes less than $50$ other than $5$. There are $13$ such primes."
] | Philippines | Philippine Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | final answer only | 14 | |
051o | A $(2k+1) \times (2k+1)$ table, where $k$ is a positive integer, contains one real number in each entry, where these numbers are pairwise different. After each row, one writes the median of the row, i.e., the number occurring in this row such that the row contains the same amount of numbers less than it and greater tha... | [
"Each row contains $k$ numbers less than the median and $k$ numbers greater than the median. Thus $k+1$ numbers in each row do not exceed the median of that row. In rows whose median does not exceed $m$, these $k+1$ numbers do not exceed $m$ either. There are $k+1$ such rows. Consequently, there are at least $(k+1)... | Estonia | Final Round of National Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0inb | Problem:
Let $a$ and $b$ be integer solutions to $17 a + 6 b = 13$. What is the smallest possible positive value for $a - b$? | [
"Solution:\n\nFirst group as $17(a-b) + 23b = 13$. Taking this equation modulo $23$, we get $-6(a-b) \\equiv -10 \\pmod{23}$. Since $-4$ is an inverse of $-6$ modulo $23$, then we multiply to get $(a-b) \\equiv 17 \\pmod{23}$. Therefore, the smallest possible positive value for $(a-b)$ is $17$. This can be satisfie... | United States | $10^{\text {th }}$ Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | final answer only | 17 | |
0ggy | 試決定滿足下述性質的所有正整數 $n \ge 3$：對每一個各邊長皆為 1 的凸 $n$ 邊形，都可以在其圍成的區域中放入一個邊長為 1 的正三角形。
(註：一個凸多邊形所圍成的區域，係指其內部及其邊。) | [
"All odd $n \\ge 3$.\n\nFirst we show that for every even $n \\ge 4$ there exists an $n$-polygon violating the required statement. Consider a regular $k$-gon $A_0A_1...A_{k-1}$ with side length 1. Let $B_1, B_2, ..., B_{n/2-1}$ be the points symmetric to $A_1, A_2, ..., A_{n/2-1}$ with respect to the line $A_0A_{n/... | Taiwan | 2022 數學奧林匹亞競賽第三階段選訓營, 獨立研究 (二) | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | Chinese; English | proof and answer | All odd integers n ≥ 3 | |
0ivn | Problem:
Find all solutions to $x^{4} + 2x^{3} + 2x^{2} + 2x + 1 = 0$ (including non-real solutions). | [
"Solution:\nAnswer: $-1, i, -i$\n\nWe can factor the polynomial as $(x+1)^{2}(x^{2}+1)$."
] | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | -1, i, -i | |
0hvu | Problem:
Given $a > b > c > 0$, prove that
$$
a^{4} b + b^{4} c + c^{4} a > a b^{4} + b c^{4} + c a^{4}.
$$ | [
"Solution:\nObserve that equality holds whenever $a = b$, $b = c$, or $a = c$. Knowing this, it is not difficult to factor the difference between the two sides:\n$$\n\\begin{aligned}\n& a^{4} b + b^{4} c + c^{4} a - a b^{4} - b c^{4} - c a^{4} \\\\\n& = (a^{4} b - a b^{4}) + (a c^{4} - b c^{4}) + (-a^{4} c + b^{4} ... | United States | Berkeley Math Circle Monthly Contest | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0l6o | Problem:
A pond has $2025$ lily pads arranged in a circle. Two frogs, Alice and Bob, begin on different lily pads. A frog jump is a jump which travels $2$, $3$, or $5$ positions clockwise. Alice and Bob each make a series of frog jumps, and each frog ends on the same lily pad that it started from. Given that each lily... | [] | United States | TST2025 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof only | null | |
0gtg | A school with $2023$ pupils organized either a museum tour or a nature tour every day during the summer holidays. No pupil participated in the same type of tour twice, and all tours were attended by different numbers of pupils. If no two pupils participated in two different tours together, find the maximal possible val... | [
"Answer: $77$.\n\nFirst of all, let us give an example for $77$ tours. Let us take $26 \\times 77 = 2002$ school pupils and divide them into groups $A$ and $B$ consisting of $26 \\times 51 = 1326$ and $26 \\times 26 = 676$ pupils, respectively. To each pupil from $A$ we assign a different pair from the set\n$$\nA =... | Turkey | 30th Turkish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 77 | |
08qh | Problem:
Alice and Bob play the following game: starting with the number $2$ written on a blackboard, each player in turn changes the current number $n$ to a number $n + p$, where $p$ is a prime divisor of $n$. Alice goes first and the players alternate in turn. The game is lost by the one who is forced to write a num... | [
"Solution:\n\nWe prove that Alice wins the game. For argument's sake, suppose that Bob can win by proper play regardless of what Alice does on each of her moves. Note that Alice can force the line $2 \\rightarrow \\mathbf{4} \\rightarrow 6 \\rightarrow \\mathbf{8} \\rightarrow 10 \\rightarrow \\mathbf{12}$ at the b... | JBMO | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | Alice | |
0ki2 | A bug starts at a vertex of a grid made up of equilateral triangles of side length $1$. At each step the bug moves in one of the $6$ possible directions along the grid lines randomly and independently with equal probability. What is the probability that after $5$ moves the bug never will have been more than $1$ unit aw... | [] | United States | AMC 12 B | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | MCQ | 13/108 | |
077e | Problem:
A stromino is a $3 \times 1$ rectangle. Show that a $5 \times 5$ board divided into twenty-five $1 \times 1$ squares cannot be covered by 16 strominos such that each stromino covers exactly three unit squares of the board and every unit square is covered by either one or two strominos. (A stromino can be plac... | [
"Solution:\n\nSuppose on the contrary that it is possible to cover the board with 16 strominos such that each unit square is covered by either one or two strominos. If there are $k$ squares that are covered by exactly one stromino then $2(25-k)+k=16 \\times 3=48$ and hence $k=2$. Thus there are exactly two squares ... | India | INMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0fln | Problem:
En un triángulo llamaremos $O$ al circuncentro, $I$ al incentro y $r$ al radio de la circunferencia inscrita. Si la mediatriz del segmento $O I$ corta a la circunferencia circunscrita en $L$, y $L I$ vuelve a cortarla en $M$, demuestra que $I M=2 r$. | [
"Solution:\n\nPor el Teorema de Euler, $(O I)^2=R^2-2 r R$. Sean $T$ y $Q$ los puntos de corte de la recta $O I$ con la circunferencia circunscrita. Entonces tenemos\n$$\nI L \\cdot I M=I T \\cdot I Q\n$$\nPor simetría, $I L=O L=R$. Por otra parte, $I T=O I+O T=O I+R$, y también tenemos $I ... | Spain | XLVII Olimpiada Matemática Española Primera Fase | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06v6 | Let $x_{1}, x_{2}, \ldots, x_{n}$ be different real numbers. Prove that
$$
\sum_{1 \leqslant i \leqslant n} \prod_{j \neq i} \frac{1-x_{i} x_{j}}{x_{i}-x_{j}}= \begin{cases}0, & \text{ if } n \text{ is even } \\ 1, & \text{ if } n \text{ is odd }\end{cases}
$$ | [
"Solution 1 (Lagrange interpolation). Since both sides of the identity are rational functions, it suffices to prove it when all $x_{i} \\notin\\{ \\pm 1\\}$. Define\n$$\nf(t)=\\prod_{i=1}^{n}\\left(1-x_{i} t\\right)\n$$\nand note that\n$$\nf\\left(x_{i}\\right)=\\left(1-x_{i}^{2}\\right) \\prod_{j \\neq i} 1-x_{i} ... | IMO | IMO 2019 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | English | proof only | null | |
0aii | Let $a$, $b$, $c$ be real numbers for which $a + b + c = 4$ and $a, b, c > 1$. Prove that
$$
\frac{1}{a-1} + \frac{1}{b-1} + \frac{1}{c-1} \ge 8 \left( \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \right).
$$
Нека $a$, $b$, $c$ се реални броеви за кои $a + b + c = 4$ и $a, b, c > 1$. Докажи дека
$$
\frac{1}{a-1} + \f... | [
"Since it holds that $\\frac{1}{a-1} - \\frac{8}{b+c} = \\frac{1}{a-1} - \\frac{8}{4-a} = \\frac{12-9a}{(a-1)(4-a)} = \\frac{3(4-3a)}{(a-1)(4-a)}$ the given inequality is equivalent to\n$$\n3 \\left( \\frac{4-3a}{(a-1)(4-a)} + \\frac{4-3b}{(b-1)(4-b)} + \\frac{4-3c}{(c-1)(4-c)} \\right) \\ge 0.\n$$\nWithout loss of... | North Macedonia | Macedonian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
0d37 | We call a positive integer good if it doesn't have a zero digit and the sum of the squares of its digits is a perfect square. For example, $122$ and $34$ are good and $304$ and $12$ are not good. Prove that there exists a $n$-digit good number for every positive integer $n$. | [
"We prove this by induction on $n$.\n\nFor $n=1,2,3$, the numbers $1$, $34$, and $122$ are good numbers and all their odd digits are less than $5$.\n\nAssume that $a_{n}$ is an $n$-digit good number and all its odd digits are less than $5$.\n\nIf $a_{n}$ has an even digit $2m$, remove this digit and replace it by $... | Saudi Arabia | Selection tests for the Balkan Mathematical Olympiad 2013 | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
06ee | Show that there exist infinitely many squarefree positive integers $n$ that divide $2005^n - 1$. (An integer is squarefree if it contains no factors of the form $d^2$, $d > 1$.) | [
"Firstly, note that $2005 \\equiv 1 \\pmod{3}$. Therefore, $3 \\mid 2005 - 1$. Suppose we have chosen distinct primes $p_1, p_2, \\dots, p_k$ such that $p_1 p_2 \\cdots p_k \\mid 2005^{p_1 p_2 \\cdots p_{k-1}} - 1$, where the exponent of $2005$ is $1$ when $k=1$. Then\n$$\n2005^{p_1 p_2 \\cdots p_k} = (2005^{p_1 p_... | Hong Kong | CHKMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof only | null | |
0hik | Are there any 10 numbers, not all of which are the same, each of which is equal to the square of the sum of all the other numbers? | [
"Suppose that such numbers exist. Since they are equal to some squares, each of these numbers is nonnegative. Let's denote the sum of all these 10 numbers by $S$. Let's pick one of these numbers and denote it by $a$, then $S \\ge a$, and also\n$$\na = (S - a)^2 \\Rightarrow a^2 - (2S + 1)a + S^2 = 0.\n$$\nIf there ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | No | |
0kew | Problem:
A sphere is centered at a point with integer coordinates and passes through the three points $(2,0,0)$, $(0,4,0)$, $(0,0,6)$, but not the origin $(0,0,0)$. If $r$ is the smallest possible radius of the sphere, compute $r^{2}$. | [
"Solution: Let $(x, y, z)$ be the center of the sphere. By the given condition, we have\n$$\n(x-2)^{2} + y^{2} + z^{2} = x^{2} + (y-4)^{2} + z^{2} = x^{2} + y^{2} + (z-6)^{2}.\n$$\nSubtracting $x^{2} + y^{2} + z^{2}$ yields\n$$\nx^{2} - (x-2)^{2} = y^{2} - (y-4)^{2} = z^{2} - (z-6)^{2},\n$$\nor\n$$\n4(x-1) = 8(y-2)... | United States | HMMO | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 51 | |
03r9 | Let $p$ and $q$ be two coprime positive integers, and let $n$ be a nonnegative integer. Determine the number of integers that can be written in the form $ip + jq$, where $i$ and $j$ are nonnegative integers with $i + j \le n$. (posed by Li Weigu) | [
"Define a set\n$$\nS(p, q, n) = \\{ip + jq \\mid i \\text{ and } j \\text{ are nonnegative integers with } i + j \\le n\\}\n$$\nLet $s_n = |S(p, q, n)|$, where $|X|$ denotes the number of elements in set $X$. The answer of the problem is\n$$\ns_n = \\begin{cases} \\dfrac{(n+1)(n+2)}{2}, & \\text{if } n < r, \\\\ \\... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof and answer | Let r = max{p, q}. The number is s_n = ((n+1)(n+2))/2 for n < r, and s_n = (r(2n - r + 3))/2 for n ≥ r. | |
09hw | For real numbers $a$, $b$, $c$, prove that
$$
a^2 + b^2 + c^2 + \frac{2ab}{1 + |a - b|} \geq \frac{2bc}{1 + |b + c|} + \frac{2ca}{1 + |c + a|}.
$$ | [
"$$\nA = a^2 + b^2 + c^2 + \\frac{2ab}{1 + |a - b|} + \\frac{2bc}{1 + |b - c|} + \\frac{2ca}{1 + |c - a|} \\geq 0\n$$\nfor all real numbers $a$, $b$, $c$. Let\n$$\nx = |b - c|, \\quad y = |c - a|, \\quad z = |a - b|.\n$$\nAs $(ab) \\cdot (bc) \\cdot (ca) = (abc)^2 \\geq 0$, we may assume that $ab \\geq 0$. Since $x... | Mongolia | Round 3 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
02yy | Problem:
José quer preencher as casas de um tabuleiro $2 \times n$ com zeros e uns de modo que dois números vizinhos iguais, em uma mesma linha, impeçam que se preencha também com números iguais as casas correspondentes da outra linha. Por exemplo, no desenho abaixo, os valores de $A$ e $B$ não podem ser iguais.
| 0 ... | [
"Solution:\n\na) Temos os preenchimentos:\n\n| 0 | 0 | 1 |\n| :--- | :--- | :--- |\n| 1 | 0 | 0 |\n| 0 | 0 | 0 |\n| :--- | :--- | :--- |\n| 1 | 0 | 1 |\n| 0 | 0 | 1 |\n| :--- | :--- | :--- |\n| 1 | 0 | 1 |\n\nb) A princípio, existem 2 escolhas possíveis para $A$ e outras duas possíveis para $B$. Apenas a escolha $A... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | a) Bottom row can be 1 0 0 or 1 0 1. b) 3. c) 4*3^2019. | |
079a | We call the permutation $\pi$ of $\{1,2, ..., n\}$ *consistent* if the set $\{\pi(k) - k \mid k = 1,2, ..., n\}$ has 2 members. Prove that the total number of *consistent* permutations is $\sigma(n) - \tau(n)$, where $\sigma(n)$ is the sum of the positive divisors of $n$ and $\tau(n)$ is the number of positive divisors... | [
"Since $\\sum_{k=1}^{n}(\\pi(k) - k) = 0$, so from the two members of $\\{\\pi(k) - k \\mid k = 1, 2, ..., n\\}$, one should be positive and the other one negative. Let $a$ be the positive and $-b$ be the negative members. Also let $A = \\{k \\mid \\pi(k) - k = a\\}$ and $B = \\{k \\mid \\pi(k) - k = -b\\}$. First ... | Iran | 27th Iranian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | σ(n) - τ(n) | |
0aix | Let $a$, $b$, $c$ be positive real numbers for which $abc = 1$. Prove that
$$
a^2 b + b^2 c + c^2 a \geq \sqrt{(a+b+c)(ab+bc+ca)}.
$$ | [
"$$\n\\begin{align*}\n(a^2 b + b^2 c + c^2 a)^2 &\\ge 3 (a^2 b \\cdot b^2 c + b^2 c \\cdot c^2 a + c^2 a \\cdot a^2 b) = 3abc (b^2 a + c^2 b + a^2 c) = 3(b^2 a + c^2 b + a^2 c). \\quad &(\\text{2 points}) \\\\\n(a^2 b + b^2 c + c^2 a)^2 &= (a^2 b + b^2 c + c^2 a) \\cdot (a^2 b + b^2 c + c^2 a) \\ge 3\\sqrt[3]{a^3 b... | North Macedonia | Macedonian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
04eb | If $60^a = 3$, $60^b = 5$ and $x = \frac{1-a-b}{2(1-b)}$, prove that $12^x$ is a positive integer. | [
"Let us first find $a$ and $b$.\n\nSince $60^a = 3$, take logarithms:\n$$\n60^a = 3 \\implies a = \\frac{\\log 3}{\\log 60}\n$$\nSimilarly,\n$$\n60^b = 5 \\implies b = \\frac{\\log 5}{\\log 60}\n$$\nNow, compute $1 - a - b$:\n$$\n1 - a - b = 1 - \\frac{\\log 3}{\\log 60} - \\frac{\\log 5}{\\log 60} = 1 - \\frac{\\l... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof only | null | |
03qc | Let $\theta$ be an acute angle such that the equation $x^2 + 4x\cos\theta + c\cot\theta = 0$ involving variable $x$ has multiple roots. Then the measure of $\theta$ in radians is ( ).
(A) $\frac{\pi}{6}$
(B) $\frac{\pi}{12}$ or $\frac{5\pi}{12}$
(C) $\frac{\pi}{6}$ or $\frac{5\pi}{12}$
(D) $\frac{\pi}{12}$ | [
"Since the equation $x^2 + 4x\\cos\\theta + c\\cot\\theta = 0$ has multiple roots, we have\n$$\n\\Delta = 16\\cos^2\\theta - 4\\cot\\theta = 0,\n$$\nor\n$$\n4\\cot\\theta(2\\sin 2\\theta - 1) = 0.\n$$\n\nIt follows that\n$$\n2\\theta = \\frac{\\pi}{6} \\text{ or } 2\\theta = \\frac{5\\pi}{6}.\n$$\nThus $\\theta = \... | China | China Mathematical Competition (Hainan) | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | MCQ | B | |
0h8v | Find all functions $f: [0, +\infty) \to [0, +\infty)$, which for all not-negative $x, y$ satisfy equality:
$$
f(f(x) + f(y)) = xyf(x + y).
$$ | [
"For $y=0$ from the condition we have, that $f(f(x)+f(0))=0$ for any $x$, and there is such $a$ that $f(a)=0$. Thus, for $x=a$, $f(f(0))=0$. Then, for $x=y=0$ it follows, that $f(2f(0))=0$, and for $x=y=f(0)$ we have, that\n$$\nf(f(f(0)) + f(f(0))) = f^2(0)f(2f(0)) = 0,\n$$\ni.e. $f(0)=0$ and $f(f(x))=0$ for any $x... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | f(x) = 0 for all x ≥ 0 | |
08ux | For a positive integer $n$, denote by $S(n)$ the sum of the digits of $n$. Determine the number of positive integers $n$ less than or equal to $999$ for which $S(9n) = 27$ is valid. | [
"First of all let us determine the number of possible ways of forming the triplets $(a, b, c)$ of integers satisfying the condition $n \\ge a \\ge b \\ge c \\ge 1$ for a given positive integer $n$. If we let $a' = a + 2$, $b' = b + 1$, $c' = c$, then we see that a triplet $(a, b, c)$ satisfying the condition will c... | Japan | Japan Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Other"
] | null | proof and answer | 165 | |
07jp | In a $n \times n$ table, consider the main diagonal and the cells below. We call these cells a triangular grid of length $n$.
We want to place a real number in each cell of a triangular table of length $n$ such that for each cell $(i, j)$, the sum of the numbers in all cells of its row $i$ and all cells of its column ... | [
"Consider the top-left cell, it follows that the sum of the numbers in the first row is zero. Considering the cells in the first row and summing them up, it follows that the sum of the numbers in the table is zero. Analogously, it follows that the sum of the numbers in the first column is zero. Using this and writi... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 1 | |
02fz | Let $N = \{0, 1, 2, 3, \dots\}$. Find all functions $f: N \to N$ which satisfy $f(2f(n)) = n + 1998$ for all $n$. | [
"First of all, $f$ is injective: indeed, $f(x) = f(y) \\implies f(2f(x)) = f(2f(y)) \\iff x + 1998 = y + 1998 \\iff x = y$.\n\nMoreover, $f$ takes all integer values bigger than $1997$: if $m \\ge 1998$ then $m = 1998 + k$ for some $k \\ge 0$ and $f(2f(k)) = k+1998 = m$.\n\nNotice that the values of $m$ such that $... | Brazil | XX OBM | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | no function exists | |
00km | Let $a$, $b$, $c$, $d$ be positive numbers. Prove that
$$
(a^2 + b^2 + c^2 + d^2)^2 \geq (a+b)(b+c)(c+d)(d+a).
$$
When does equality hold? | [
"By the inequality between the arithmetic and the geometric mean, we have\n$$\n(a+b)(b+c)(c+d)(d+a) \\leq \\left( \\frac{(a+b)+(b+c)+(c+d)+(d+a)}{4} \\right)^4 = 2^4 \\left( \\frac{a+b+c+d}{4} \\right)^4.\n$$\nBy the inequality between the quadratic and the arithmetic mean, we have\n$$\n2^4 \\left( \\frac{a+b+c+d}{... | Austria | Austrian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | a = b = c = d | |
0eqx | Two sequences of real numbers are defined as follows:
$$
u_1 = 0, \quad u_{n+1} = \frac{1}{2}(u_n + v_n)$$
$$v_1 = 1, \quad v_{n+1} = \frac{1}{4}(u_n + 3v_n)$$
Find the value of $v_{2016} - u_{2016}$. | [
"$u_1 = 0$\n$v_1 = 1$\n$$\nu_{n+1} = \\frac{1}{2}(u_n + v_n)$$\n$$v_{n+1} = \\frac{1}{4}(u_n + 3v_n)$$\n$$\n\\begin{aligned}\nv_{2016} - u_{2016} &= \\frac{1}{4}(u_{2015} + 3v_{2015}) - \\frac{1}{2}(u_{2015} + v_{2015}) \\\\\n&= \\frac{1}{4}v_{2015} - \\frac{1}{4}u_{2015} \\\\\n&= \\frac{1}{4}(v_{2015} - u_{2015}) ... | South Africa | South African Mathematics Olympiad Third Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | final answer only | 1/4^{2015} | |
0a4f | Problem:
Gegeven is $\triangle ABC$ met omgeschreven cirkel $\Gamma$. Zij $M$ het midden van de boog $BC$ van $\Gamma$ waar $A$ niet op ligt. Het punt $N$ op $\Gamma$ is de antipode van $A$. De lijn door $B$ loodrecht op $AM$ snijdt $AM$ in het punt $D$ en snijdt $\Gamma$ een tweede keer in het punt $P \neq B$. De lij... | [
"Solution:\n\n\n\nOplossing I. We noteren de helft van de hoek bij $A$ als $\\alpha = \\frac{1}{2} \\angle BAC = \\angle BAM = \\angle MAC$, want $M$ is het midden van boog $BC$. Dan merken we op dat $\\angle ABP = \\angle ABD = 90^\\circ - \\angle DAB = 90^\\circ - \\alpha$ en dat $\\angle... | Netherlands | Maarttoets | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point"
] | null | proof only | null | |
07wg | For $n \ge 3$, a *special n-triangle* is a triangle with $n$ distinct numbers on each side such that the sum of the numbers on a side is the same for all sides. For instance, because $41 + 23 + 43 = 43 + 17 + 47 = 47 + 19 + 41$, the following is a special 3-triangle:
$$
\begin{array}{c c c}
& 41 & \\
23 & 19 & \\
43 & ... | [
"For each $N > 1$, let $A_N$ denote the set of all positive integers that are not multiples of $N$, and let $a_{N,1} < a_{N,2} < a_{N,3} < \\dots$ be the elements of $A_N$ in increasing order. The set $A_N$ consists of blocks of $N-1$ consecutive integers followed by a skipped integer. If $N \\equiv 1 \\pmod 3$ the... | Ireland | IRL_ABooklet_2023 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
050z | An equilateral triangle with side length $3$ is divided into $9$ equilateral triangles with side length $1$. An integer from $1$ to $10$ is written into every point that is a vertex of a small triangle (colored vertices on the figure), such that all numbers are written exactly once. For every small triangle, the sum of... | [
"Any three small triangles, from which no two have common vertices, take up nine of the ten numbers written into the vertices of the small triangles. So, the sum of the numbers inside those small triangles is $55 - a$, where $a$ is the number at the last vertex. Now it is sufficient to prove that we can choose the ... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0fhc | Problem:
Sean $a$ y $b$ enteros diferentes de $0$, $1$ y $-1$, y consideremos la matriz
$$
\left(
\begin{array}{ccccc}
a+b & a+b^{2} & a+b^{3} & \ldots & a+b^{m} \\
a^{2}+b & a^{2}+b^{2} & a^{2}+b^{3} & \ldots & a^{2}+b^{m} \\
a^{3}+b & a^{3}+b^{2} & a^{3}+b^{3} & \ldots & a^{3}+b^{m} \\
\vdots & \vdots & \vdots & & \v... | [
"Solution:\nMultiplicando la primera fila por $\\lambda$, la segunda por $1-\\lambda$, y sumando, resulta para un elemento cualquiera la relación\n$$\n\\lambda\\left(a+b^{k}\\right)+(1-\\lambda)\\left(a^{2}+b^{k}\\right)=\\lambda a+a^{2}-\\lambda a^{2}+b^{k}\n$$\nPor lo tanto, para que éste sea un elemento cualquie... | Spain | OME 27 | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | A minimal generating set is S consisting of the first two rows. For each i ≥ 3, the i-th row equals λ·(first row) + (1 − λ)·(second row), where λ = −a(1 + a + a^2 + … + a^{i−3}) (equivalently, λ = a(a^{i−2} − 1)/(1 − a)). For n = 3, λ = −a. Rows 1 and 2 are included in S. | |
0chw | Let $ABC$ be a triangle. An arbitrary circle which passes through the points $B$ and $C$ intersects the sides $AC$ and $AB$ for the second time in $D$ and $E$, respectively. The line $BD$ intersects the circumcircle of triangle $AEC$ at $P$ and $Q$, and the line $CE$ intersects the circumcircle of the triangle $ABD$ at... | [
"a) Denote by $X$ the intersection of the lines $CE$ and $BD$. Writing the power of $X$ with respect to all three circles, we obtain:\n$$\nXR \\cdot XS = XB \\cdot XD = XE \\cdot XC = XP \\cdot XQ.\n$$\nSince $\\{X\\} = PQ \\cap RS$, we infer that the points $P, Q, R$ and $S$ are concyclic.\n\nb) The triangles $ADP... | Romania | 74th NMO Selection Tests for JBMO | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0dlk | Find all positive integers $n < 2027$ that satisfy the following conditions:
(i) For every positive divisor $d$ of $n$, numbers $1^d, 2^d, \dots, 2026^d$ all have distinct remainders when divided by $2027$.
(ii) $\tau(n)^2 \mid n$ with $\tau(n)$ is the number of positive divisors of $n$. | [
"Let call the numbers satisfying the given condition as \"good\" number. Consider some good number $n$. One can see that if $n$ is even, then $d = 2$ is a divisor of $2$, however, then $1^2$ and $2026^2$ both divide $2027$ with remainder $1$, which does not satisfy. Therefore, $n$ is odd. We see that $n = 1$ satisf... | Saudi Arabia | Saudi Booklet | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theo... | null | proof and answer | 1, 9, 625, 2025 | |
078y | A finite set $S$ of positive integers is called *cardinal* if $S$ contains the integer $|S|$, where $|S|$ denotes the number of distinct elements in $S$. Let $f$ be a function from the set of positive integers to itself, such that for any cardinal set $S$, the set $f(S)$ is also cardinal. Here $f(S)$ denotes the set of... | [
"**Solution 1.** The possible values are 1, 2, and 2024.\n\n**Construction.** The function $f(x) = 1$ for all $x \\in \\mathbb{N}$ works. Also, $f(x) = 1$ for all $x \\neq 2024$ and $f(2024) = 2$, works. Finally, $f(x) = x$ for all $x \\in \\mathbb{N}$ works as well.\nIt remains to show these are the only possible ... | India | INMO | [
"Discrete Mathematics > Combinatorics > Functional equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 1, 2, 2024 | |
058k | Let $n$ be a positive integer. Find the largest number of knights that can be placed on a board of size $n \times n$ in such a way that no two knights attack each other. A knight attacks precisely the squares that are located either horizontally by one square and vertically by two squares away or horizontally by two sq... | [
"Clearly one can place only $1$ knight on an $1 \\times 1$ board, which is $\\lfloor \\frac{1^2}{2} \\rfloor$, and at most $4$ knights on an $2 \\times 2$ board. In the rest, assume $n \\ge 3$.\n\nIf a set of unit squares is divided into pairs in such a way that a knight on one square of any pair attacks the other ... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | floor(n^2/2) | |
0b9o | Find the real numbers $x$ and $y$ such that
$$
(x^2 - x + 1)(3y^2 - 2y + 3) - 2 = 0.
$$ | [
"From $x^2 - x + 1 = (x - \\frac{1}{2})^2 + \\frac{3}{4} \\ge \\frac{3}{4}$, $3y^2 - 2y + 3 = 3(y - \\frac{1}{3})^2 + \\frac{8}{3} \\ge \\frac{8}{3}$\nwe derive that $(x^2 - x + 1)(3y^2 - 2y + 3) \\ge \\frac{3}{4} \\cdot \\frac{8}{3} = 2$, for any real numbers $x$ and $y$. Equality holds for $(3y - 1)^2 = 0$ and $(... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | x = 1/2, y = 1/3 | |
0f4g | Problem:
The reals $a$ and $b$ are such that $a \cos x + b \cos 3x > 1$ has no real solutions. Show that $|b| \leq 1$. | [] | Soviet Union | 15th ASU | [
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0hda | Square $ABCD$ of size $2019 \times 2019$ is divided by two lines into four rectangles with integer side lengths, and some of the rectangles might be squares. Turns out, the area of the rectangle containing vertex $A$ equals the perimeter of the rectangle containing vertex $B$. What is the area of the smallest of the fo... | [
"Let us denote the sides of the rectangle containing vertex $A$ as $a$ and $b$ (Fig. 21). Then the sides of the rectangle containing vertex $B$ are equal to $a$ and $2019-b$. The equality of the area of the first rectangle and perimeter of the second yields:\n$$\nab = 2 \\cdot (a + 2019 - b) \\Rightarrow ab - 2a + ... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 16 | |
06l2 | Let $\Gamma_1$ and $\Gamma_2$ be two circles with different radii, with $\Gamma_1$ the smaller one. The two circles meet at distinct points $A$ and $B$. $C$ and $D$ are two points on the circles $\Gamma_1$ and $\Gamma_2$ respectively, and such that $A$ is the midpoint of the segment $CD$. $CB$ is extended to meet the c... | [] | Hong Kong | HKG TST | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
000i | La sucesión de números reales $a_1, a_2, \dots$ se define como:
$$
a_1 = 56 \quad y \quad a_{n+1} = a_n - \frac{1}{a_n} \quad \text{para cada entero } n \ge 1.
$$
Demuestre que existe un entero $k$, $1 \le k \le 2002$, tal que $a_k < 0$. | [] | Argentina | XVII Olimpíada Iberoamericana de Matemática | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | español | proof only | null | |
0g4u | Problem:
Sei $ABC$ ein Dreieck, welches $2 \cdot \angle CBA = 3 \cdot \angle ACB$ erfüllt. Die Punkte $D$ und $E$ liegen auf der Seite $AC$, sodass $BD$ und $BE$ den Winkel $\angle CBA$ in drei gleich große Winkel unterteilen und sodass $D$ zwischen $A$ und $E$ liegt. Sei $F$ außerdem der Schnittpunkt von $AB$ und der... | [
"Solution:\n\nDie Bedingung $2 \\cdot \\angle CBA = 3 \\cdot \\angle ACB$ in der Aufgabe bedeutet genau, dass $\\angle DCF = \\angle FCB = \\angle EBD = \\angle DBF$ gilt. Aus $\\angle DCF = \\angle DBF$ können wir folgern, dass $FDCB$ ein Sehnenviereck ist. Daher gilt $\\angle FDB = \\angle FCB = \\angle EBD$, was... | Switzerland | Zweite Runde 2023 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08ak | Problem:
Su una circonferenza di centro $A$ e raggio $R$ vengono presi nell'ordine quattro punti distinti $B, C, G, H$ in modo tale che $G$ giaccia sul prolungamento della mediana del triangolo $A B C$ condotta da $B$, e $H$ giaccia sul prolungamento dell'altezza di $A B C$ condotta da $B$. Detta $X$ l'intersezione fr... | [] | Italy | Olimpiade Italiana di Matematica | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0k90 | Problem:
Let $n$, $k$, $r$ be positive integers. Suppose we have a collection of sets $S_{1}, S_{2}, \ldots, S_{r}$, where each $S_{i}$ is a subset of $\{1,2, \ldots, n\}$ consisting of one or more consecutive integers. We say that such a collection is a $k$-fold perfect cover of $\{1,2, \ldots, n\}$ if each element o... | [
"Solution:\n\nWe argue by contradiction. Fix $n$. Consider the smallest collection of $S_{i}$'s (smallest as in least number of sets) that is a $k$-fold perfect cover for some $k$ and can't be partitioned as desired. Choose the $S_{i}$ of the form $\\{1,2, \\ldots, m\\}$ where $m$ is as small as possible. If $m=n$,... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0i1d | Let $\{a_n\}_{n \ge 0}$ be a sequence of real numbers such that $a_{n+1} \ge a_n^2 + \frac{1}{5}$ for all $n \ge 0$. Prove that $\sqrt{a_{n+5}} \ge a_{n-5}^2$ for all $n \ge 5$. | [
"It suffices to prove that $a_{n+5} \\ge a_n^2$ for $n \\ge 0$, for then we would have $\\sqrt{a_{n+5}} \\ge a_n$ and $a_n \\ge a_{n-5}^2$ for all $n \\ge 5$, which implies the desired result. Adding the inequalities\n$$\n\\begin{align*}\na_{n+5} &\\ge a_{n+4}^2 + \\frac{1}{5}, \\\\\na_{n+4} &\\ge a_{n+3}^2 + \\fra... | United States | USA IMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
00xn | Problem:
There are 13 cities in a certain kingdom. Between some pairs of cities two-way direct bus, train or plane connections are established. What is the least possible number of connections to be established in order that choosing any two means of transportation one can go from any city to any other without using t... | [
"Solution:\n\nAn example for 18 connections is shown in Figure 1 (where single, double and dashed lines denote the three different kinds of transportation). On the other hand, a connected graph with 13 vertices has at least 12 edges, so the total number of connections for any two kinds of vehicle is at least 12. Th... | Baltic Way | Baltic Way 1993 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 18 | |
0cfp | Let $n \ge 3$ be an integer. A 3-element subset $A$ of the set $M = \{1, 2, 3, \dots, n\}$ will be called interesting if there exists a set $B$ of positive integers so that:
(1) if $a \in A, b \in A, a \neq b$, then $(a+b) \in B$;
(2) if $x, y, z \in B, x < y < z$, then $2 \cdot y = x + z$.
Denote $A_n$ the number of t... | [] | Romania | 74th NMO Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | A_7 = 9 and A_91 = 2025 (> 2024) | |
0hnc | Problem:
Find the largest number $n$ having the following properties:
(a) No two digits of $n$ are equal.
(b) The number formed by reversing the digits of $n$ is divisible by 8. | [
"Solution:\nBy condition (a), the number $n$ cannot have more than 10 digits. Write $m$ for the number formed by reversing the digits of $n$.\n\nThe first digit of $n$ is the last digit of $m$, and as such must be even, and thus at most 8. Assume that the first digit is 8.\n\nThen the second digit of $n$ is the ten... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 8697543210 | |
0dzl | Anja has a number of $1 \times 1$ square tiles $\Box$, while Bojan has the L-shaped tiles $\Box\Box$. They take turns putting their tiles onto a rectangular board, one tile at a time. Anja wins if Bojan cannot fit another one of his tiles onto the board when his turn comes despite there being squares left uncovered. Ot... | [
"(a) A $6 \\times 9$ board has $54$ squares. If Bojan wants to win, they have to cover the entire board, because Anja can always fit in another one of her tiles as long as there are empty squares left. After Bojan and Anja each put $13$ tiles onto the board there will be two squares left uncovered. Regardless of wh... | Slovenia | Slovenija 2008 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0b2g | Problem:
In a convex polygon, the number of diagonals is $23$ times the number of its sides. How many sides does it have?
(a) $46$
(b) $49$
(c) $66$
(d) $69$ | [] | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | b | |
05al | A positive integer $n$ is given. There are $n$ computers in a network, numbered with natural numbers $1, 2, \dots, n$. The computers are connected with one-way communication lines such that information can be sent from any computer to all other computers either directly or through other computers in the network. Initia... | [
"We will show that $2n - 1$ initiations are necessary regardless of the network of lines. Consider an arbitrary sequence of initiations, at the end of which all computers have output the numbers of all computers. Let $A$ be the computer with the latest time of the initiation of the first procedure. Since the proced... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Algorithms"
] | English | proof and answer | 2n - 1 | |
0hto | Problem:
The tower function of twos, $T(n)$, is defined by $T(1)=2$ and $T(n+1)=2^{T(n)}$ for $n \geq 1$. Prove that $T(n)-T(n-1)$ is divisible by $n!$ for $n \geq 2$. | [
"Solution:\n\nLet $U(n) = T(n) - T(n-1)$. Note that\n$$\nU(n) = T(n) - T(n-1) = 2^{T(n-1)} - 2^{T(n-2)} = 2^{T(n-2)}\\left(2^{T(n-1)-T(n-2)} - 1\\right) = T(n-1)\\left(2^{U(n-1)} - 1\\right)\n$$\n\nWe first prove two lemmas, the second a refinement of the first.\n\nLemma 1. $U(n-1) \\mid U(n)$.\n\nProof. By inducti... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic Expr... | null | proof only | null | |
080d | Problem:
Una scatola contiene 3 palline bianche e 2 palline nere. Marco estrae una pallina e la rimette nella scatola aggiungendo un'altra pallina dello stesso colore. A questo punto egli estrae una nuova pallina dalla scatola. Qual è la probabilità che quest'ultima sia bianca?
(A) $\frac{1}{2}$
(B) $\frac{7}{12}$
(C... | [] | Italy | Progetto Olimpiadi di Matematica 2000 GARA di SECONDO LIVELLO | [
"Statistics > Probability > Counting Methods > Other"
] | null | MCQ | C | |
09ve | Problem:
In koordenvierhoek $A B C D$ is $E$ het snijpunt van de diagonalen. Een lijn door $E$, ongelijk aan $A C$ of $B D$, snijdt $A B$ in $P$ en $B C$ in $Q$. De cirkel die raakt aan $P Q$ in $E$ en verder door $D$ gaat, snijdt de omgeschreven cirkel van $A B C D$ nogmaals in punt $R$. Bewijs dat $B, P, R$ en $Q$ o... | [
"Solution:\n\nVanwege koordenvierhoek $D B C R$ geldt $\\angle Q C R = 180^{\\circ} - \\angle B C R = \\angle B D R$. Omdat $E Q$ raakt aan de cirkel door $E, D$ en $R$ geldt $\\angle B D R = \\angle E D R = \\angle Q E R$, dus $\\angle Q C R = \\angle Q E R$. Dit betekent dat $E R Q C$ een koordenvierhoek is. Anal... | Netherlands | IMO-selectietoets III | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0duy | Problem:
Na enotski krožnici s središčem v koordinatnem izhodišču izberemo krožni lok s krajiščema $A$ in $B$, ki leži v prvem kvadrantu. Naj bo $p_{1}$ ploščina lika pod krožnim lokom in nad abcisno osjo, $p_{2}$ pa naj bo ploščina lika levo od krožnega loka in desno od ordinatne osi (glej sliko). Dokaži, da je vsota... | [
"Solution:\n\nOznačimo točke, kot kaže slika. Ploščina krožnega izseka, ki pripada loku $\\widehat{A B}$, je enaka $p_{\\widehat{A B}}=\\frac{\\beta-\\alpha}{2}$, kjer smo kota $\\alpha$ in $\\beta$ merili v radianih. Sledi\n$$\n\\begin{aligned}\n& p_{1}=p_{\\widehat{A B}}+p_{O B B^{\\prime \\prime}}-p_{O A A^{\\pr... | Slovenia | 46. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
00rc | Let $a$, $b$, $c$ be positive real numbers. Prove that
$$
\sqrt{a^3 b + a^3 c} + \sqrt{b^3 c + b^3 a} + \sqrt{c^3 a + c^3 b} \geq \frac{4}{3}(ab + bc + ca)
$$ | [
"W.L.O.G. $a \\ge b \\ge c$.\n$$\na \\ge b \\ge c \\Rightarrow ab \\ge ac \\ge bc \\Rightarrow ab + ac \\ge ab + bc \\ge ac + bc \\Rightarrow \\sqrt{ab+ac} \\ge \\sqrt{bc+ba} \\ge \\sqrt{ac+bc}\n$$\n$$\n\\sqrt{a^3 b + a^3 c} + \\sqrt{b^3 c + b^3 a} + \\sqrt{c^3 a + c^3 b} = a\\sqrt{ab+ac} + b\\sqrt{bc+ba} + c\\sqrt... | Balkan Mathematical Olympiad | BMO 2016 Short List Final | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0emc | $B_1$ and $C_1$ are marked on the bisector of angle $A$ in triangle $ABC$ so that $BB_1 \perp AB$, $CC_1 \perp AC$. Let $M$ be the midpoint of $B_1C_1$. Prove that $MB = MC$. | [
"Drop a perpendicular from $C_1$ onto $AB$ meeting $AB$ at $C_2$. Similarly define $B_2$ such that $B_1B_2 \\perp AC$. Let $M_c$ be on $AC$ such that $MM_c \\perp AC$ and $M_b$ on $AB$ such that $MM_b \\perp AB$. Without loss of generality, assume that the points on the angle bisector of $\\angle A$ are in the orde... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0092 | Let $ABCD$ be a convex quadrilateral. Let $P$ and $Q$ be points on the sides $AB$ and $AD$, respectively, such that $area(ABQ) = area(ADP) = \frac{1}{3} area(ABCD)$.
$PQ$ and the diagonal $AC$ meet at the point $R$.
Calculate the ratio $\frac{AR}{RC}$. | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof and answer | 1:2 | |
0fyu | Problem:
Die Gerade $g$ schneide den Kreis $k$ in den Punkten $A$ und $B$. Die Mittelsenkrechte der Strecke $A B$ schneide $k$ noch einmal in $C$ und $D$. Sei nun $P$ ein weiterer Punkt auf $g$, der ausserhalb von $k$ liegt. Die Parallelen zu $C A$ und $C B$ durch $P$ schneiden die Geraden $C B$ und $C A$ in den Punkt... | [
"Solution:\n\nWegen $C A \\parallel P X$ und $C B \\parallel Y B$ kriegen wir ähnliche Dreiecke $\\triangle X P B \\equiv \\triangle C A B \\equiv Y A P$. Daher können wir Kreise $k_{1}$ und $k_{2}$ einführen mit Mittelpunkt $X$ bzw $Y$ die durch die Punkte $P$ und $B$, bzw. durch $P$ und $A$ gehen. Sei nun $H$ der... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09mk | Let $O$ be the center of the circumcircle of an acute-angled triangle $ABC$. Let $CD$ be an altitude of the triangle, and let $M$ be a point on the segment $CD$. Let $E$ and $F$ be feet of perpendiculars from $M$ to $BC$ and $AC$, respectively. Let $G$ be the point symmetric to $C$ with respect to the line $EF$. Prove ... | [] | Mongolia | Mongolian 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 > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Confi... | English | proof only | null | |
091y | Problem:
Let $ABC$ be an isosceles triangle with $AC = BC$. Let $N$ be a point inside the triangle such that $2 \angle ANB = 180^{\circ} + \angle ACB$. Let $D$ be the intersection of the line $BN$ and the line parallel to $AN$ that passes through $C$. Let $P$ be the intersection of the angle bisectors of the angles $C... | [
"Solution:\n\nSince $AC = BC$, there is a circle $k$ such that the lines $AC$ and $BC$ are the tangents to $k$ at the points $A$ and $B$. The condition defining the point $N$ implies that the point $N$ lies on the circle $k$.\n\nBy the tangent-chord theorem, we have $\\angle BAN = \\angle DBC$ and $\\angle CAN = \\... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point ... | null | proof only | null | |
0ldn | Given 2017 positive real numbers $a_1, a_2, \dots, a_{2017}$. For each positive integer $n > 2017$, let
$$
a_n = \max\{a_{i_1} a_{i_2} a_{i_3} \mid i_1 + i_2 + i_3 = n, 1 \le i_1 \le i_2 \le i_3 \le n-1\}.
$$
Prove there exists some positive integers $m \le 2017$ and $N > 4m$ such that $a_n a_{n-4m} = a_{n-2m}^2$ for e... | [
"For every $n > 0$, let $b_n = \\ln a_n$. We can reduce the problem to: Given 2017 reals $b_1, b_2, \\dots, b_{2017}$. For every $n > 2017$, let\n$$\nb_n = \\max\\{b_{i_1} + b_{i_2} + b_{i_3} \\mid i_1 + i_2 + i_3 = n, 1 \\le i_1 \\le i_2 \\le i_3 \\le n-1\\}.\n$$\nProve there exist positive integers $m \\le 2017$ ... | Vietnam | Vietnamese Team Selection Test for IMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0c53 | Let $A_1, A_2, \dots, A_m$ be $n \times n$ matrices with real elements, where $m, n \in \mathbb{N}^*$, and $b_1, b_2, \dots, b_m$ be real numbers. Consider the set
$$
\mathcal{B} = \{B \in \mathcal{M}_n(\mathbb{R}) \mid \operatorname{tr}(BA_i^T) = b_i, \forall i = \overline{1, m}\}.
$$
Suppose that the matrix $C = (c_{... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Algebra > Linear Algebra > Vectors"
] | English | proof only | null | |
0i9n | Problem:
Let $ABCD$ be a square, and let $E$ be an internal point on side $AD$. Let $F$ be the foot of the perpendicular from $B$ to $CE$. Suppose $G$ is a point such that $BG = FG$, and the line through $G$ parallel to $BC$ passes through the midpoint of $EF$. Prove that $AC < 2 \cdot FG$. | [
"Solution:\n\nFirst note that, for given $E, F$, there is only one point $G$ with the required properties: since $BG = FG$, $G$ must lie on the perpendicular bisector of $BF$, and by the second condition, $G$ lies on the line through the midpoint of $EF$ parallel to $BC$; $G$ must thus be the unique intersection of... | United States | 5th Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mea... | null | proof only | null | |
08wb | Let $a, b, c, d, e, f, g, h, i$ be the distinct integers lying in between $1$ and $9$ (both $1$ and $9$ inclusive). Let $N$ be the maximum of the three numbers $a \times b \times c$, $d \times e \times f$ and $g \times h \times i$. Determine the minimum value the number $N$ can take. | [
"Let us write $p = abc$, $q = def$ and $r = ghi$.\nWe first show that it is possible to make the maximum of the three numbers $p, q, r$ to be no more than $72$. Indeed, if we consider the following, we see that this is possible:\n$$\n1 \\times 8 \\times 9 = 72, \\quad 2 \\times 5 \\times 7 = 70, \\quad 3 \\times 4 ... | Japan | Japan Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 72 | |
05as | On the base $BC$ of the isosceles triangle $ABC$ points $D$ and $E$ are chosen such that $BD = EC = 7$ cm. A line perpendicular to side $AB$ and passing through point $A$ intersects base $BC$ at point $F$. An altitude of triangle $ACE$ is $EG$, a median of triangle $AEF$ is $AH$. It is known that $\angle DAF = \frac{1}... | [
"Since triangle $BAC$ is isosceles (Fig. 25), its base angles are equal; denote $\\angle ABC = \\angle BCA = \\beta$. Since the sum of the interior angles of triangle $BAC$ is $180^\\circ$, we have $\\angle BAC = 180^\\circ - 2\\beta$. Then $\\angle DAF = \\frac{180^\\circ - 2\\beta}{2} = 90^\\circ - \\beta$ becaus... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 3.5 cm | |
07le | Consider a rectangular billiards table $a$ meters wide and $b$ meters long with pockets at its corners, where $a$ and $b$ are both positive integers. A ball is placed at the lower left corner of the table and shot at a $45$ degree angle. It travels without friction until it lands in one of the pockets. Every time it hi... | [
"Following the hint, we observe that the ball lands in a pocket as soon as the line segment reaches a point $(c, c)$ where $c$ is divisible by both $a$ and $b$; clearly $c$ is the least common multiple of $a$ and $b$. The ball touches a vertical rail $c$ times and a horizontal rail $c$ times. If $c$ is even/odd, th... | Ireland | Irska | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and prod... | English | proof and answer | 672084 | |
05lq | Problem:
On considère $n^{2}+2n+1$ points dans un carré de côté $n$. Montrer que trois d'entre eux sont les sommets d'un triangle (éventuellement dégénéré) d'aire au plus $1/2$. | [
"Solution:\n\nOn raisonne sur l'enveloppe convexe des $(n+1)^{2}$ points : si elle contient beaucoup de points, il y aura 3 sommets voisins de cette enveloppe assez proches pour faire un triangle d'aire petite. Si elle n'en contient pas beaucoup, on la triangule et on regarde les nombreux points dans les petits tri... | France | Envoi de combinatoire | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0fjb | Problem:
Nueve personas han celebrado cuatro reuniones diferentes sentados alrededor de una mesa circular. ¿Han podido hacerlo sin que existan dos de esas personas que se hayan sentado una junto a la otra en más de una reunión? Razona la respuesta. | [
"Solution:\nLa respuesta es sí, pueden celebrar las cuatro reuniones de modo que al final cada persona haya estado sentada junto a otras dos diferentes cada vez. Para demostrarlo, consideramos las siguientes cuatro formas de ordenar los números del $1$ al $9$, que representan cuatro maneras de sentarse alrededor de... | Spain | Viernes 19 de enero de 2001 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Yes | |
05wm | Problem:
Comptez le nombre de réarrangements $a_{1}, a_{2}, \ldots, a_{2023}$ de la séquence $1,2, \ldots, 2023$ telle que $a_{k}>k$ pour exactement une valeur de $k$. | [
"Solution:\n\nÀ un réarrangement valide, on peut lui associer un sous-ensemble de $1, \\ldots, 2023$ de cardinal au moins $2$ : l'ensemble des $k$ tels que $a_{k} \\neq k$.\n\nOn peut montrer qu'à un sous-ensemble de cardinal au moins $2$ de $1, \\ldots, 2023$, on peut associer un unique réarrangement valide qui lu... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof and answer | 2^2023 - 2024 | |
0coq | Can the 4 incenters of the 4 faces of some tetrahedron be coplanar?
Могут ли 4 центра вписанных в грани тетраэдра окружностей лежать в одной плоскости? | [
"**Ответ.** Не могут.\n\nПусть $I_A, I_B, I_C, I_D$ — центры вписанных окружностей треугольников $BCD, ACD, ABD, ABC$ соответственно. Предположим, что они лежат в одной плоскости. Тогда либо они образуют выпуклый четырёхугольник, либо одна из этих точек лежит в треугольнике, образованном тремя другими.\n\nСлучай 1.... | Russia | Final round | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English; Russian | proof only | null | |
0h59 | Circles $w_1$ and $w_2$ with centers $O_1$ and $O_2$ respectively intersect in points $A$ and $B$. A straight line $O_1O_2$ intersects $w_1$ in a point $Q$, that is not inside $w_2$, and $w_2$ in a point $X$, that is inside $w_1$. Around the triangle $O_1AX$ a circle $w_3$ is circumscribed and intersects $w_1$ for the ... | [
"a) Let $\\angle AO_1X = \\alpha$, then $\\angle ATX = \\alpha$, because they are subtended by the same arc of the circle $w_3$ (Fig. 24), moreover $\\angle AO_1X = \\angle ATB$, so $\\angle ATX = \\angle ATB$, therefore $T$, $X$, $B$ are collinear.\n\nb) Let $K_1 = XH \\cap TQ$, $\\angle O_2XH = \\alpha$, $\\angle... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round (Second Tour) | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0fjl | Problem:
¿Cuáles son las posibles áreas de un hexágono convexo con todos los ángulos iguales y cuyos lados miden $1,2,3,4,5$ y $6$, en algún orden? | [
"Solution:\n\nLa idea es prolongar los lados para formar un triángulo equilátero.\n\n\n\nTenemos $a+b+c+d+e+f=21$ y $\\ell=a+b+c=c+d+e=e+f+a$ de donde sale $3 \\ell=21+a+c+e$, y por tanto,\n$$\n\\ell=7+\\frac{a+c+e}{3}\n$$\nEl valor más pequeño de $a+c+e$ es $6$ y el más grande $15$, así qu... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 65*sqrt(3)/4 and 67*sqrt(3)/4 | |
0ayi | Problem:
A Vitas word is a string of letters that satisfies the following conditions:
- It consists of only the letters $B$, $L$, $R$.
- It begins with a $B$ and ends in an $L$.
- No two consecutive letters are the same.
How many Vitas words are there with $11$ letters? | [
"Solution:\nLet $a_n$ be the number of $n$-letter Vitas words that start with $B$ and end with $L$.\n\nLet us generalize and define:\n- $a_n$: number of $n$-letter words starting with $B$ and ending with $L$.\n- $b_n$: number of $n$-letter words starting with $B$ and ending with $R$.\n\nWe do not consider words end... | Philippines | 21st PMO Area Stage | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 341 | |
0bay | Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that its 2-fold composition is equal to the floor function, i.e. $f(f(x)) = \lfloor x \rfloor$, for any real number $x$. Prove that there exist distinct real numbers $a$ and $b$ such that $|f(a) - f(b)| \ge |a - b|$. | [
"We claim that $f(n) \\in \\mathbb{Z}$, for any integer $n$. Indeed, write $f(f(f(x))) = f([x]) = [f(x)]$ to derive that $f(n) = [f(n)]$ for any integer $n$, implying $f(n) \\in \\mathbb{Z}$.\n\nSuppose that for all $a, b \\in \\mathbb{Z}$ we have $|f(a) - f(b)| < |a - b|$. Then $|f(n + 1) - f(n)| < 1$, for any int... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0hq6 | Problem:
There are 1000 cities in the country of Euleria, and some pairs of cities are linked by dirt roads. It is possible to get from any city to any other city by traveling along these roads. Prove that the government of Euleria may pave some of the roads so that every city will have an odd number of paved roads le... | [
"Solution:\n\nCall a city \"even\" or \"odd\" according to whether the number of paved roads coming out of it is even or odd. Note the following.\n\nLemma: No matter which roads are paved, there will be an even (possibly zero) number of even cities.\n\nTo see why this is true, let $d_{i}$ be the number of paved roa... | United States | null | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
00bt | The pentagon $ABCDE$, with sides $AB$, $BC$, $CD$, $DE$ and $EA$, satisfies the following conditions:
$$
\bullet \angle ABC = \angle BCD = \angle CDE = 90^\circ
$$
$\bullet$ $CD$ is longer than $AB$.
$\bullet$ $AB = 28$, $BC = 15$, $DE = 10$ and $EA = 13$.
Calculate the area of the pentagon. | [
"Extend $BA$ and $DE$ so that they meet at point $P$.\n\n$AB = 28$, $BC = 15$, $DE = 10$ and $EA = 13$.\nThe quadrilateral $PBCD$ has three right angles, so the fourth is also a right angle, hence it is a rectangle. Then $PD = BC = 15$, therefore $PE = PD - DE = 15 - 10 = 5$. Now we apply P... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof and answer | 570 | |
083x | Problem:
Calcolare l'area dell'intersezione di tre cerchi aventi come rispettivi diametri i tre lati di un triangolo rettangolo isoscele con i cateti di lunghezza unitaria.
(A) $\frac{\pi-2}{8}$
(B) $\pi-3$
(C) $\frac{2 \pi-5}{8}$
(D) $\frac{\pi-1}{16}$
(E) $\frac{2 \pi-3}{16}$. | [] | Italy | Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO BIENNIO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | A | |
05v4 | Problem:
Soient $m$, $n$ et $x$ des entiers strictement positifs. Montrer que
$$
\sum_{i=1}^{n} \min \left(\left\lfloor\frac{x}{i}\right\rfloor, m\right)=\sum_{i=1}^{m} \min \left(\left\lfloor\frac{x}{i}\right\rfloor, n\right)
$$ | [
"Solution:\nQuitte à échanger $m$ et $n$, on peut supposer que $m \\leqslant n$. On procède alors par récurrence sur $n$ à $m$ fixé.\n\nInitialisation : Si $m=n$, l'égalité est triviale.\n\nHérédité : Supposons l'égalité vraie pour un certain $n \\geqslant m$ et cherchons à montrer que l'égalité est vraie pour $n+1... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null |
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