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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
004i | Dado un entero positivo $m$, se define la sucesión $\{a_n\}$ de la siguiente manera:
$$
a_1 = \frac{m}{2}, \quad a_{n+1} = a_n [a_n], \text{ si } n \ge 1.
$$
Determinar todos los valores de $m$ para los cuales $a_{2007}$ es el primer entero que aparece en la sucesión.
Nota: Para un número real $x$ se define $[x]$ como ... | [] | Argentina | XXII Olimpiada Iberoamericana de Matemática | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Spanish | proof and answer | All m with v2(m−1) = 2006; equivalently m = 1 + 2^2006(2k+1), i.e., m ≡ 1 + 2^2006 (mod 2^2007). | |
0jpq | Problem:
Let $a$, $b$, $c$ be positive integers such that $a^{2}-b c$ is a perfect square. Prove that the number $2a+b+c$ is not a prime number. | [
"Solution:\n\nSuppose that $a^{2}-b c=d^{2}$, so that $(a-d)(a+d)=b c$. Then, we can find positive integers $w, x, y, z$ such that $a-d=w x$, $a+d=y z$, $b=w y$, $c=x z$. Thus,\n$$\n2a+b+c=(a-d)+(a+d)+b+c=w x+w y+x z+y z=(w+z)(x+y)\n$$\nwhich is clearly not prime."
] | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0ctu | A circle $\omega$ is tangent to the sides of an angle $BAC$ at $B$ and $C$. A line $l$ intersects the segments $AB$ and $AC$ at $K$ and $L$, respectively. The circle $\omega$ meets $l$ at $P$ and $Q$. The points $S$ and $T$ are chosen on the segment $BC$ so that $KS \parallel AC$ and $LT \parallel AB$. Prove that the p... | [
"If $l \\nmid BC$, set $X = l \\cap BC$ (see Fig. 12). Then $\\frac{XB}{XT} = \\frac{XK}{XL} = \\frac{XS}{XC}$, so $XT \\cdot XS = XB \\cdot XC = XP \\cdot XQ$.\n\n\n\nSince $B$, $C$, $P$, and $Q$ lie on $\\omega$, we have $XB \\cdot XC = XP \\cdot XQ$. Thus, $XT \\cdot XS = XP \\cdot XQ$; ... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English; Russian | proof only | null | |
06b6 | A $7 \times 7$ square is divided into $49$ unit squares with sides parallel to those of $T$. We color each of the $49$ unit squares with two colors, red and blue, so that the following two conditions are simultaneously satisfied:
(i) There are exactly $4$ rows in each of which the blue squares are more than the red sq... | [
"In fact, if such a square, let $S$, existed, then each of the rows and columns of $T$ to which the sides of $S$ belong have at least $4$ squares of the same color (let with color blue in the figure $6$). This contradicts condition (ii). Therefore, for every coloring, each monochromatic square has side less or equa... | Greece | Selection Examination | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 3 | |
053y | Denote by $f^n(x)$ the result of applying the function $f$ $n$ times to $x$ (e.g. $f^1(x) = f(x)$, $f^2(x) = f(f(x))$, $f^3(x) = f(f(f(x)))$ etc). Find all functions from real numbers to real numbers which satisfy $f^d(x) = 2015 - x$ for all divisors $d$ of $2015$, which are greater than $1$, and for all real $x$. | [
"Since $5$ is a divisor of $2015$, we have for any real $z$:\n$$\n\\begin{aligned}\nf^{25}(z) &= f^5(f^5(f^5(f^5(z)))) = 2015 - f^5(f^5(f^5(f^5(z)))) \\\\\n&= 2015 - (2015 - f^5(f^5(f^5(z)))) = f^5(f^5(f^5(z))) = \\dots = f^5(z) = 2015 - z.\n\\end{aligned}\n$$\nSince $13$ is also a divisor of $2015$, we have for an... | Estonia | Estonian Math Competitions | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | f(x) = 2015 - x | |
0go7 | Determine the number of positive integers $n$ for which $(n + 15)(n + 2010)$ is a perfect square. | [
"Let $m^2 = n^2 + 2025n + 2010 \\cdot 15$ where $m$ is a positive integer. Then $(2n + 2m + 2025)(2n - 2m + 2025) = 1995^2$. Hence $n = (A + B - 4050)/4$ and $m = (A - B)/4$\n\nwhere $A$ and $B$ are integers such that $AB = 1995^2$, $A > B$ and $A + B > 4050$. Since $1995^2 \\equiv 1 \\pmod 4$, these formulas alway... | Turkey | 15th Junior Turkish Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadra... | English | proof and answer | 39 | |
0cs1 | A sphere $\omega$ passes through a vertex $S$ of a triangular pyramid $SABC$ and intersects the edges $SA$, $SB$, $AC$ at $A_1 \neq S$, $B_1 \neq S$, and $C_1 \neq S$, respectively. The sphere $\omega$ intersects the circumsphere $\Omega$ of $SABC$ by a circle lying in a plane parallel to $(ABC)$. The points $A_2$, $B_... | [
"**Первое решение.** Утверждение задачи эквивалентно равенству $SA_2 \\cdot SA = SB_2 \\cdot SB = SC_2 \\cdot SC$. Значит, ввиду равенств $AA_1 = SA_2$ и двух аналогичных, достаточно доказать, что $AA_1 \\cdot AS = BB_1 \\cdot BS = CC_1 \\cdot CS$.\nПусть $\\ell$ — прямая, проходящая через центры сфер $\\Omega$ и $... | Russia | XL Russian mathematical olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
073b | Let $X$ be the set of all bijective functions from the set $S = \{1, 2, 3, ..., n\}$ to itself. For each $f \in X$, define
$$
T_f(j) = \begin{cases} 1, & \text{if } f^{(12)}(j) = j, \\ 0, & \text{otherwise.} \end{cases}
$$
Determine
$$ \sum_{f \in X} \sum_{j=1}^{n} T_f(j). $$
(Here $f^{(k)}(x) = f(f^{(k-1)}(x))$ for $... | [
"Suppose $n \\ge 12$. The elements of $X$ are permutations of $\\{1, 2, 3, ..., n\\}$. If $j$ belongs to a $k$-cycle of $f$, where $k$ is a divisor of $12$, then $f^{(k)}(j) = j$. The number of divisors of $12$ is $6$. Let $m(j,k)$ be the number of elements $f$ of $T$ in which $j$ belongs to a $k$-cycle of $f$, $1 ... | India | Indija TS 2007 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | t(n) · n!, where t(n) is the number of positive divisors of 12 that are ≤ n (so t(n)=6 for n≥12; for n=1..11, t(n)=1,2,3,4,4,5,5,5,5,5,5 respectively). | |
0c2c | Lucia has a total of 2018 yellow, blue and green balls. There are 4 times more green balls than blue balls. In one *exchange* Lucia offers 13 yellow balls to her friend Cristina and receives 5 blue balls and 7 green balls. After a certain number of such *exchanges*, Lucia has no yellow balls, but she has 1271 green bal... | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 1313 | |
0250 | Problem:
Na figura abaixo, o triângulo $A B C$ é retângulo em $C$ e tanto $B C D E$ quanto $C A F G$ são quadrados. Se o produto das áreas dos triângulos $E A B$ e $B F A$ é $64$, determine a área do triângulo $A B C$.
 | [
"Solution:\n\nComo $D A$ é paralelo a $E B$, a área do triângulo $A E B$ é $\\frac{E B \\cdot B C}{2} = \\frac{B C^{2}}{2}$. Da mesma forma, a área do triângulo $A B F$ é $\\frac{A C^{2}}{2}$. Portanto,\n$$\n\\begin{aligned}\n\\left[ A B E \\right] \\cdot \\left[ A B F \\right] &= \\frac{B C^{2}}{2} \\cdot \\frac{A... | Brazil | null | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 8 | |
09pc | For the real number sequence $\{a_n\}_{n=1}^\infty$, we are given that $a_1 = 1$, $a_2 = 3$, and for $n \ge 1$,
$$
a_{n+2} = a_{n+1} + \frac{3a_{n+1}^2 - 1}{a_{n+1} - a_n}
$$
Prove that the all terms of the sequence $\{a_n\}$ are natural numbers and $a_{61} > 4^{61}$.
(Otgonbayar Uuye) | [] | Mongolia | MMO2025 Round 3 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
08iw | Problem:
The quadrilateral $ABCD$ is inscribed in the circle with center $O$, the points $M$ and $N$ are the middle points of the diagonals $[AC]$ and $[BD]$ respectively and $P$ is the intersection point of the diagonals. It is known that the points $O, M, N$ and $P$ are distinct. Prove that the points $O, M, B$ and ... | [] | JBMO | The first selection test for IMO 2003 and BMO 2003 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
000z | Una fila de hormigas marchan todas a la misma velocidad por un sendero rectilíneo. La distancia entre la primera y la última hormiga es de $15$ metros. La hormiga inspectora recorre la fila comenzando desde la última hormiga, y cuando alcanza a la primera hormiga, regresa hasta encontrar nuevamente a la última hormiga.... | [] | Argentina | XIX Olimpíada Matemática Argentina | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | español | final answer only | 32 | |
04nf | In how many ways can the letters $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $I$ be arranged so that all of the vowels and all of the consonants are ordered alphabetically? | [] | Croatia | Croatia_2018 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | final answer only | 84 | |
092e | Problem:
There are $n$ students standing in line in positions $1$ to $n$. While the teacher looks away, some students change their positions. When the teacher looks back, they are standing in line again. If a student who was initially in position $i$ is now in position $j$, we say the student moved for $|i-j|$ steps. ... | [
"Solution:\n\nLet us denote $x_{i}$ the place of student $i$ after switching places. Since $\\sum_{i=1}^{n} i - x_{i} = \\sum_{i=1}^{n} i - \\sum_{i=1}^{n} x_{i} = 0$ holds, the sum of summands $i - x_{i}$ which are negative is the same as the sum of absolute values of summands $i - x_{i}$ which are negative. There... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | ⌊n^2/2⌋ | |
05jf | Problem:
Prouver que pour tous réels strictement positifs $a, b, c$ tels que $a b c=1$, on a
$$
\frac{1}{1+a^{2}+(b+1)^{2}}+\frac{1}{1+b^{2}+(c+1)^{2}}+\frac{1}{1+c^{2}+(a+1)^{2}} \leqslant \frac{1}{2}
$$ | [
"Solution:\nSoit $a, b, c$ des réels strictement positifs.\nCompte tenu de l'inégalité bien connue $x^{2}+y^{2} \\geqslant 2 x y$ valable pour tous réels $x, y$, on a\n$$\n\\frac{1}{1+a^{2}+(b+1)^{2}}=\\frac{1}{2+a^{2}+b^{2}+2 b} \\leqslant \\frac{1}{2(1+a b+b)}\n$$\nDe même, et puisque $a b c=1$, on a\n$$\n\\frac{... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
07vd | Leonard Larsson's Language
The linguist Leonard Larsson made up a new language in which all words have exactly seven letters and only the twenty letters from $A$ to $T$ are used. Furthermore, any pair of distinct words differ in at least two places. (For instance, if the words $LEONARD$ and $LARSSON$ are part of this l... | [
"Let's generalise and consider exactly $n$ letters in a word, chosen from an alphabet of $m$ letters, with the restriction that any two distinct words must differ in at least two places. We claim that this allows for at most $m^{n-1}$ words, and that this bound can be obtained. Since $20^{7-1} = 64\\ 000\\ 000$ lie... | Ireland | IRL_ABooklet_2023 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
00vo | Let $n$ and $m$ be positive integers with $n \ge 2$. There are $n$ piles having $a_1, \dots, a_n$ stones such that for each $1 \le i \le n$ we have $m \le a_i \le m \cdot i$ (so $a_1 = m$). Aida and Bob play the following game: on each round, Bob picks two non-empty piles (if possible) and he removes a number of stones... | [
"**Answer:** $k = m$.\nLet the piles be labeled $1, 2, \\dots, n$ such that the pile labeled $i$ has $a_i$ stones.\nWe first provide a construction achieving $k \\ge m$. Let $a_1 = a_2 = \\dots = a_{n-1} = m$ and $a_n = m \\cdot n$. Then each stone removed from the pile $n$ can be matched to a stone removed from on... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | k = m | |
067r | Let $ABC$ be a scalene acute angled triangle with $AB < AC < BC$ and circumcenter $c(O, R)$. The ex-circle ($c_A$) corresponding to the vertex $A$, has center $I$ and is tangent to the sides $BC$, $AC$, $AB$ at $D$, $E$, $Z$, respectively. The line $AI$ intersects the circle $c(O, R)$ at $M$ and the circumcircle, say $... | [
"Since $AZ$ and $AE$ are tangents to the circle ($c_A$), we have $IZ \\perp AZ$ and $IE \\perp AE$. Hence, the circle ($c_2$) contains $I$ and $AI$ is a diameter. First we will prove that $AN$ passes through $O$, that is $KNO = KNA$ (*).\n\nMoreover we have $KNO = KMO$ and from the isosceles triangle $OKM$ we get t... | Greece | SELECTION EXAMINATION | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0885 | Problem:
Sia $n$ un intero positivo. Un treno ferma in $2 n$ stazioni, incluse quella iniziale e finale, numerate in ordine dalla prima alla $2 n$-esima. Si sa che in una certa carrozza, per ogni coppia di interi $i, j$ tali che $1 \leq i<j \leq 2 n$, è stato prenotato esattamente un posto per il tragitto tra la stazi... | [
"Solution:\n\nConsideriamo il tratto dalla stazione $n$-esima alla $n+1$-esima. In questo tratto nella carrozza in questione ci sono tutti i passeggeri saliti in una delle $n$ stazioni dalla prima all' $n$-esima e diretti verso una delle $n$ stazioni dalla $n+1$-esima alla $2 n$-esima. Il totale dei passeggeri pres... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | n^2 | |
0dwl | Problem:
Reši enačbo:
$$
\log \left(\frac{1}{x}\right) \cdot \log \left(\frac{4}{x}\right) = \frac{3}{4} \cdot (\log 4)^2
$$ | [
"Solution:\n\nNajprej uporabimo pravilo za logaritmiranje količnika:\n$$(\\log 1 - \\log x) \\cdot (\\log 4 - \\log x) = \\frac{3}{4}(\\log 4)^2.$$\nKer je $\\log 1 = 0$, dobimo:\n$$(-\\log x) \\cdot (\\log 4 - \\log x) = \\frac{3}{4}(\\log 4)^2.$$\nRazpremo oklepaj:\n$$-\\log x \\cdot \\log 4 + (\\log x)^2 = \\fra... | Slovenia | 4. državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 8, 1/2 | |
08d2 | Problem:
Annalisa, Bruna e Cecilia giocano a calcio: una di loro sta in porta e le altre in campo. Chi fa gol rimane in campo, mentre chi non ha segnato si scambia con il portiere. Sapendo che Annalisa è stata in campo per 12 turni e Bruna per 21 turni, mentre Cecilia è stata in porta 8 volte, chi ha cominciato in por... | [
"Solution:\n\nLa risposta è (A). Siano $a, b, c$ il numero di turni in cui rispettivamente Annalisa, Bruna e Cecilia sono in porta, mentre $A, B, C$ il numero di turni in cui sono in campo. Se $x$ è il numero totale di turni giocati, allora avremo $a = x - 12$, $b = x - 21$, $c = 8$ e $A = 12$, $B = 21$, $C = x - 8... | Italy | GARA di FEBBRAIO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | MCQ | A | |
0l9q | Solve the following system of equations
$$
\begin{cases}
\sqrt{x^2 - 2x + 6} \log_3 (6-y) = x \\
\sqrt{y^2 - 2y + 6} \log_3 (6-z) = y \\
\sqrt{z^2 - 2z + 6} \log_3 (6-x) = z.
\end{cases}
$$ | [
"Conditions for $x$, $y$, $z$ of the definition of the expressions in the equation are: $x, y, z < 6$.\nThe given system of equations is equivalent to the system:\n$$\n\\left\\{\n\\begin{aligned}\n\\log_3(6-y) &= \\frac{x}{\\sqrt{x^2-2x+6}} && (1) \\\\\n\\log_3(6-z) &= \\frac{y}{\\sqrt{y^2-2y+6}} && (2) \\\\\n\\log... | Vietnam | Vijetnam 2006 | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | x = y = z = 3 | |
06oy | Points $A_{1}$, $B_{1}$, $C_{1}$ are chosen on the sides $BC$, $CA$, $AB$ of a triangle $ABC$, respectively. The circumcircles of triangles $AB_{1}C_{1}$, $BC_{1}A_{1}$, $CA_{1}B_{1}$ intersect the circumcircle of triangle $ABC$ again at points $A_{2}$, $B_{2}$, $C_{2}$, respectively ($A_{2} \neq A$, $B_{2} \neq B$, $C... | [
"We will work with oriented angles between lines. For two straight lines $\\ell, m$ in the plane, $\\angle(\\ell, m)$ denotes the angle of counterclockwise rotation which transforms line $\\ell$ into a line parallel to $m$ (the choice of the rotation centre is irrelevant). This is a signed quantity; values differin... | IMO | IMO 2006 Shortlisted Problems | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | English | proof only | null | |
003a | Determinar si el número $2004$ se puede escribir como suma de los cuadrados de dos enteros positivos. ¿Y $2004$? | [] | Argentina | Argentina 2006 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | Español | proof and answer | No; 2004 cannot be written as the sum of two squares of positive integers because its factorization includes primes congruent to three modulo four with odd exponents. | |
0bc1 | Given a positive integer number $n$, determine the maximum number of edges a simple graph on $n$ vertices may have in order that it contain no cycles of even length. | [
"The required maximum is $\\lfloor 3(n-1)/2 \\rfloor$. It is achieved, for instance, by an $(n-1)/2$ arm wind-mill if $n$ is odd, and an $(n-2)/2$ arm wind-mill with an extra edge joined at the hub if $n$ is even.\n\nTo show that a simple graph on $n$ vertices with no cycles of even length has at most $3(n-1)/2$ ed... | Romania | 62nd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | floor(3(n-1)/2) | |
06u9 | Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of integers and $b_{0}, b_{1}, b_{2}, \ldots$ be a sequence of positive integers such that $a_{0}=0$, $a_{1}=1$, and
$$
a_{n+1}=\left\{\begin{array}{ll}
a_{n} b_{n}+a_{n-1}, & \text{ if } b_{n-1}=1 \\
a_{n} b_{n}-a_{n-1}, & \text{ if } b_{n-1}>1
\end{array} \quad \text{ f... | [
"The value of $b_{0}$ is irrelevant since $a_{0}=0$, so we may assume that $b_{0}=1$.\n\nLemma. We have $a_{n} \\geqslant 1$ for all $n \\geqslant 1$.\n\nProof. Let us suppose otherwise in order to obtain a contradiction. Let\n$$\n\\begin{equation*}\nn \\geqslant 1 \\text{ be the smallest integer with } a_{n} \\leq... | IMO | International Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0407 | Let $E$ be a given set with $n$ elements. Suppose that $A_1, A_2, \dots, A_k$ are $k$ distinct nonempty subsets of $E$, with either $A_i \cap A_j = \emptyset$, $A_i \subset A_j$, or $A_j \subset A_i$ for any $1 \le i < j \le k$. Find the maximum value of $k$. | [] | China | China Western Invitational Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 2n - 1 | |
04pl | The product of a certain number of distinct positive integers less than $1000$ is not divisible by $250$. At most how many numbers have been multiplied? | [] | Croatia | Croatian Mathematical Society Competitions | [
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | 802 | |
0bog | Let $m \ge n \ge 2$ be integers, and let $A \in \mathcal{M}_{m,n}(\mathbb{C})$ and $B \in \mathcal{M}_{n,m}(\mathbb{C})$, be two matrices such that there exist $k \in \mathbb{N}^*$, and $a_0, a_1, \dots, a_k \in \mathbb{C}$, such that $a_k(AB)^k + a_{k-1}(AB)^{k-1} + \dots + a_1(AB) + a_0I_m = O_m$, and $a_k(BA)^k + a_... | [
"Suppose that $a_0 \\ne 0$. Rewriting the given equality as\n$$\nAB \\left( -\\frac{a_k}{a_0}(AB)^{k-1} - \\frac{a_{k-1}}{a_0}(AB)^{k-2} - \\dots - \\frac{a_1}{a_0}I_m \\right) = I_m,\n$$\nwe find that $AB$ is invertible, hence $\\mathrm{rank}(AB) = m \\ge n$.\nOn the other hand, we have $m = \\mathrm{rank}(AB) \\l... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
015l | Characterise all non-negative real numbers $a$ and $b$ such that the equation
$$
\sqrt{x + 2\sqrt{a}} - \sqrt{x - 2\sqrt{a}} = 2b
$$
has at least one real solution. (All roots are assumed to be real.) | [
"Squaring the equation yields\n$$\nx + 2\\sqrt{a} + x - 2\\sqrt{a} - 2\\sqrt{x^2 - 4a} = 4b^2 \\Leftrightarrow \\sqrt{x^2 - 4a} = x - 2b^2.\n$$\nAfter squaring again, we obtain\n$$\nx^2 - 4a = x^2 - 4b^2x + 4b^4 \\Leftrightarrow x = b^2 + \\frac{a}{b^2}.\n$$\nThe equation therefore has the unique solution $x = b^2 ... | Baltic Way | Baltic Way SHL | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | All nonnegative pairs with either b > 0 and a >= b^4, or b = 0 and a = 0. | |
07ga | Find all non-constant polynomials $P(x)$ with integer coefficients such that for each $n$ the polynomial $P^n(x)$ only has integer roots.
(Where $P^n(x)$ means the $n^{\text{th}}$ fold composition of $P(x)$ with itself.) | [
"We shall prove that $P(x)$ only has one integer root. Assume to the contrary, that $P(x)$ at least has two different integer roots. We shall then prove the following lemma.\n**Lemma.** The polynomial $P^n(x)$ has at least $n+1$ distinct integer roots.\n*Proof.* We prove by induction on $n$. The base is true, assum... | Iran | 38th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | All such polynomials are exactly of the forms a x^d, x − r, and −x − r, with integers a, d ≥ 1, and r. | |
0hln | Problem:
If $x$ is a positive real number, find the smallest possible value of $2x + \frac{18}{x}$. | [
"Solution:\n\nThe answer is $12$. This is achieved when $x = 3$; to see it is optimal, note that\n$$\n2x + \\frac{18}{x} \\geq 12 \\Longleftrightarrow x^2 - 6x + 9 \\geq 0,\n$$\nwhich is obviously true since the left-hand side is $(x-3)^2 \\geq 0$.\n\nAlternatively, those who know the so-called AM-GM inequality may... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 12 | |
065p | Let $\mathbb{N}^* = \{1,2,3,...\}$ be the set of positive integers. Find all functions $f: \mathbb{N}^* \rightarrow \mathbb{N}^*$ such that
$$
f(f(m)^2 + 2f(n)^2) = m^2 + 2n^2, \text{ for all } m, n \in \mathbb{N}^*.
$$ | [
"First we prove that $f$ is injective. In fact, for any fixed $n$, if $f(m_1) = f(m_2)$, then: $m_1^2 + 2n^2 = f(f(m_1)^2 + 2f(n)^2) = f(f(m_2)^2 + 2f(n)^2) = m_2^2 + 2n^2$, whence $m_1^2 = m_2^2$ and $m_1 = m_2$.\n\nSince $f$ is injective we have:\n$$\nf(m)^2 + 2f(n)^2 = f(p)^2 + 2f(q)^2 \\Leftrightarrow m^2 + 2n^... | Greece | 26th Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | f(n) = n for all positive integers n | |
0f7t | Problem:
$ABCDEFG$ is a regular $7$-gon. Prove that $1 / AB = 1 / AC + 1 / AD$. | [] | Soviet Union | 21st ASU | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0lbd | Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying all of the following conditions:
1. $f$ is a bijective function.
2. $f$ is an increasing function.
3. $f(f(x)) = f(x) + 12x, \forall x \in \mathbb{R}$. | [] | Vietnam | Vietnam Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | Vietnamese | proof and answer | f(x) = 4x | |
0h1h | Given natural number $N$ we write two different nonzero digits from the right. It turns out, that new number is divisible by $N$. What is maximal value of $N$ can be? | [
"Assume that two digits form the number $\\overline{ab}$. Then we have $\\overline{Nab}:N$, or $100N+\\overline{ab}:N$, which implies $\\overline{ab}:N$. Since $\\overline{ab}$ is form by two distinct digits then $a \\le 98$, hence $N \\le 98$. The number $N=98$ satisfies all the requirements."
] | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | 98 | |
0cux | In a product of 3 positive integers each multiple was decreased by 3. Could it happen that the product increases by 2016 after this change?
В произведении трёх положительных целых чисел каждый сомножитель уменьшили на 3. Могло ли случиться, что произведение увеличилось на 2016 после этого изменения? | [
"Yes. An example is $1 \\cdot 1 \\cdot 676$.\n\nAfter the operation, we have $(-2) \\cdot (-2) \\cdot 673 = 4 \\cdot 673 = 2692 = 676 + 2016$.\n\n**Note.** The given example is unique. Here is how it can be found. Suppose two of the factors are $1$, and the third is $a$. Their product is $a$, and after decreasing e... | Russia | XLIII Russian mathematical olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English; Russian | proof and answer | Yes; for example 1, 1, and 676. | |
09on | For any real numbers $t_1, t_2, \dots, t_n \le 2$, show that the inequality
$$
\frac{t_1^3}{t_2^2 + 4} + \frac{t_2^3}{t_3^2 + 4} + \dots + \frac{t_{n-1}^3}{t_n^2 + 4} + \frac{t_n^3}{t_1^2 + 4} \le n
$$
holds. | [
"Let $s_i = t_i^2/4$. Since $t_i^3 \\le 2t_i^2 = 8s_i$, so for $0 \\le s_1, \\dots, s_n \\le 1$, it suffices to prove\n$$\n\\frac{s_1}{s_2+1} + \\frac{s_2}{s_3+1} + \\dots + \\frac{s_n}{s_1+1} \\le \\frac{n}{2}.\n$$\nNow for any $0 \\le x, y \\le 1$, we consider:\n$$\n(1+y)(1-y+x) = 2x + (1-y)(1+y-x) \\ge 2x\n$$\nw... | Mongolia | MMO2025 Round 4 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0gwm | Side $BC$ of the parallelogram $ABCD$ is extended beyond the point $C$ and point $K$ is placed on the extension so that $\triangle CDK$ is isosceles triangle with base $CK$. Side $DC$ is extended beyond the point $C$ and point $L$ is placed on the extension so that $CBL$ is isosceles triangle with base $CL$. Bisectors ... | [
"Since trapezium $ABKD$ is equilateral, points $A, B, K, D$ are on one circle. Thereafter, points $B, L, D, A$ reside on one circle too. Thus all the five points $B, L, D, A, K$ are on the same circle $w$ circumscribed about the triangle $\\triangle ALK$. Since trapeziums $ABKD$ and $BLDA$ share one of the diagonal... | Ukraine | Ukrajina 2008 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | a/(2 sin 2 alpha) | |
025k | Problem:
Um trapézio - A figura dada representa um trapézio $ABCD$ em que $AB$ é paralelo a $CD$ e as diagonais $AC$ e $BD$ cortam-se no ponto $P$. Se as áreas dos triângulos $\triangle APB$ e $\triangle CPD$ medem $4$ e $9\ \mathrm{cm}^2$, respectivamente, qual é a área do triângulo $\triangle PCB$?
![](attached_ima... | [
"Solution:\n\nOs triângulos $\\triangle APB$ e $\\triangle CPD$ são semelhantes, pois o ângulo $A\\widehat{P}B$ é igual ao ângulo $C\\widehat{P}D$ (opostos pelo vértice) e o ângulo $A\\widehat{B}D$ é igual ao ângulo $B\\widehat{D}C$ (alternos internos).\n\n\n\nComo a razão entre suas áreas ... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 6 cm^2 | |
01eu | Let $n \ge 3$ be an integer, such that $4n + 1$ is a prime number. Prove that $4n + 1$ divides $n^{2n} - 1$. | [
"Since $p := 4n + 1$ is a prime number, each non-zero remainder modulo $p$ possesses a unique multiplicative inverse. Since $-4 \\cdot n \\equiv 1 \\pmod{p}$, we have $n \\equiv (-4)^{-1} \\pmod{p}$, from which we deduce that $n \\equiv -(2^{-1})^2$. Consequently,\n$$\nn^{2n} - 1 \\equiv (-(2^{-1})^2)^{2n} - 1 \\eq... | Baltic Way | Baltic Way shortlist | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof only | null | |
045f | Let $a, b, c, p, q, r$ be positive integers with $p, q, r \ge 2$. Denote
$$
Q = \{(x, y, z) \in \mathbb{Z}^3 \mid 0 \le x \le a, 0 \le y \le b, 0 \le z \le c\}.
$$
Initially, some stones are put at each point of $Q$, with total $M$ stones. Then one can perform the following three types of operations repeatedly:
(1) rem... | [
"**Proof:** The smallest positive integer $M$ is $p^a q^b r^c$.\n\nFirst, if we initially put less than $p^a q^b r^c$ stones at the point $(a, b, c)$, then one cannot perform a sequence of operations to put a stone at $(0, 0, 0)$. The reason is as follows. For a stone $u$ at $(x, y, z)$, define its weight as $w(u) ... | China | 2022 China Team Selection Test for IMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | p^a q^b r^c | |
0aya | Problem:
A regular hexagon is inscribed in another regular hexagon such that each vertex of the inscribed hexagon divides a side of the original hexagon into two parts in the ratio $2:1$. Find the ratio of the area of the inscribed hexagon to the area of the larger hexagon. | [
"Solution:\n\nWithout loss of generality, we may assume that the original hexagon has side length $3$. Let $s$ then be the side length of the inscribed hexagon. Notice that there are six triangles with sides $1$, $2$, and $s$, and the angle between the sides of lengths $1$ and $2$ is $120^{\\circ}$, as shown below:... | Philippines | 20th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 7/9 | |
06hx | Let $a$, $b$, $c$ be distinct nonzero real numbers. If the equations $ax^3 + bx + c = 0$, $bx^3 + cx + a = 0$ and $cx^3 + ax + b = 0$ have a common root, prove that at least one of these equations has three real roots (not necessarily distinct). | [
"Let $t$ be the common root. Adding the three equations $at^3 + bt + c = 0$, $bt^3 + ct + a = 0$ and $ct^3 + at + b = 0$, we get $(a + b + c)(t^3 + t + 1) = 0$.\n\nIf $t^3 + t + 1 = 0$, then $at^3 + bt + c = 0$ becomes $(b - a)t + (c - a) = 0$. Similarly, $bt^3 + ct + a$ becomes $(c - b)t + (a - b) = 0$. As $a$, $b... | Hong Kong | The Seventeenth Hong Kong (China) Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof only | null | |
0itk | Problem:
Let $ABC$ be an equilateral triangle with side length $2$, and let $\Gamma$ be a circle with radius $\frac{1}{2}$ centered at the center of the equilateral triangle. Determine the length of the shortest path that starts somewhere on $\Gamma$, visits all three sides of $ABC$, and ends somewhere on $\Gamma$ (no... | [
"Solution:\n\nAnswer: $\\sqrt{\\frac{28}{3}}-1$\n\nSuppose that the path visits sides $AB$, $BC$, $CA$ in this order. Construct points $A'$, $B'$, $C'$ so that $C'$ is the reflection of $C$ across $AB$, $A'$ is the reflection of $A$ across $BC'$, and $B'$ is the reflection of $B$ across $A'C'$. Finally, let $\\Gamm... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(28/3) - 1 | |
0fw8 | Problem:
Bestimme die beiden kleinsten natürlichen Zahlen, die sich in der Form $7 m^{2}-11 n^{2}$ mit natürlichen Zahlen $m$ und $n$ schreiben lassen. | [
"Solution:\n\nNehme an, es gelte $7 m^{2}-11 n^{2}=c$ mit einer natürlichen Zahl $c$.\nWir betrachten die Gleichung modulo 7. Es muss also gelten $c \\equiv -11 n^{2} \\equiv 3 n^{2} \\pmod{7}$. Wegen $n^{2} \\equiv 0,1,2,4$ ist daher $c \\equiv 0,3,5,6 \\pmod{7}$. Analog erhält man modulo 11 die Gleichung $c \\equ... | Switzerland | IMO Selektion | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 7 and 13 | |
01st | a) Determine all functions $f: \mathbb{Z} \to \mathbb{Z}$ such that
$$
f(x - f(y)) = f(f(x)) - f(y) - 1
$$
for all integers $x$ and $y$.
b) The same question if
$$
f(x - f(y)) = f(f(x)) - f(y) - 2
$$
for all integers $x$ and $y$. | [
"**a)** $f(x) = -1$ or $f(x) = x + 1$.\n\n(Alternative solution by I. Voronovich)\nSetting $y = f(x)$ in the given equation\n$$\nf(x - f(y)) = f(f(x)) - f(y) - 1, \\qquad (1)\n$$\nwe obtain $f(x - f(f(x))) = -1$, i.e., there exists an integer $\\lambda$ such that $f(\\lambda) = -1$. Set $y = \\lambda$ in (1), then\... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | a) All solutions: f(x) = -1 for all integers x; and f(x) = x + 1 for all integers x.
b) All solutions: f(x) = -2 for all integers x; f(x) = x + 2 for all integers x; and, for any odd integer d, the function defined by f(x) = x + 2 for even x and f(x) = x + d for odd x. | |
00d2 | Sea $ABC$ un triángulo isósceles rectángulo con ángulo recto en $A$. Sean $E$ y $F$ puntos en $AB$ y $AC$ respectivamente tales que $\angle ECB = 30°$ y $\angle FBC = 15°$. Las rectas $CE$ y $BF$ se cortan en $P$ y la recta $AP$ corta al lado $BC$ en $D$. Calcular la medida del ángulo $FDC$. | [
"Demostraremos que $FD$ es perpendicular a $BC$.\n\nSea $D'$ en $BC$ tal que $FD'$ es perpendicular a $BC$ y sea $P'$ el punto de intersección de $AD'$ y $BF$.\n\nQueremos demostrar que $\\angle BCP' = 30°$, lo que implica que $P = P'$ y por lo tanto que $D = D'$.\n\nObservamos que $ABD'F$ es cíclico, ya que $\\ang... | Argentina | Nacional OMA | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Spanish | proof and answer | 90° | |
03at | Find all coprime positive integers $a$ and $b$ with the following property: There exists positive integer $N$ such that any positive integer, greater than $N$, has the form $\sum a^i b^j$, where $i$ and $j$ are non-negative integers and none of the summands divides any other. | [] | Bulgaria | Selection test for 27. Balkan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | a = 2 and b = 3 (up to order) | |
01dx | Let there be an operator $*$. Given an expression that includes this operator, one can make the following transformations:
1. An expression of the form $x * (y * z)$ can be rewritten as $((1 * x) * y) * z$;
2. An expression of the form $x * 1$ can be rewritten as $x$.
The transformations may be performed only on the en... | [
"\\begin{align*}\n1 &\\overset{(2)}{\\rightleftharpoons} 1 * 1 \\\\\n&\\overset{(2)}{\\rightleftharpoons} (1 * 1) * 1 \\\\\n&\\overset{(2)}{\\rightleftharpoons} ((1 * 1) * 1) * 1 \\\\\n&\\overset{(1)}{\\rightleftharpoons} 1 * (1 * 1) \\\\\n&\\overset{(2)}{\\rightleftharpoons} (1 * (1 * 1)) * 1 \\\\\n&\\overset{(2)}... | Baltic Way | Baltic Way shortlist | [
"Discrete Mathematics > Logic"
] | English | proof and answer | n = 1, 2, 3, 4 | |
08u3 | For an acute triangle $ABC$ satisfying $AB \neq AC$, denote by $H$ the foot of the perpendicular line segment drawn from the point $A$ to the side $BC$. Take points $P$ and $Q$ in such a way that the 3 points $A$, $B$, $P$ and the 3 points $A$, $C$, $Q$ lie on a straight line in the given order, respectively. If the 4 ... | [
"Without loss of generality we may assume that $AB < AC$ is satisfied. Let us write, for the sake of simplicity, $\\angle B$ for $\\angle ABC$ and $\\angle C$ for $\\angle BCA$.\n\nUsing the fact that the points $B$, $P$, $Q$, $C$ lie on the circumference of the same circle, we get\n$$\n\\angle AQH - \\angle APH = ... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0aai | Problem:
Let $ABC$ be an acute-angled triangle with circumscribed circle $k$ and centre of the circumscribed circle $O$. A line through $O$ intersects the sides $AB$ and $AC$ at $D$ and $E$. Denote by $B'$ and $C'$ the reflections of $B$ and $C$ over $O$, respectively. Prove that the circumscribed circles of $ODC'$ an... | [
"Solution:\n\nLet $P$ be the intersection of the circles $k$ and the circumscribed circle of triangle $ADE^{1}$. Let $C_{1}$ be the second intersection of the circumscribed circle of $\\triangle DOP$ with $k$. We will prove that $C_{1}=C'$, i.e. the reflection of $C$ over $O$. We know that $|OC_{1}|=|OP|$, and henc... | Nordic Mathematical Olympiad | Nordic Mathematical Contest | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler ... | null | proof only | null | |
03jw | Problem:
A particle can travel at speeds up to $2$ metres per second along the $x$-axis, and up to $1$ metre per second elsewhere in the plane. Provide a labelled sketch of the region which can be reached within one second by the particle starting at the origin. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0g5t | 令 $n$ 為某個大於 3 的整數且 $S = \{1, 2, \dots, n\}$。令 $A_1, A_2, \dots, A_n$ 為 $S$ 的子集，其中 $|A_i| \ge 2$ 對所有 $i$。假設對於任意兩個元素的子集 $S' \subseteq S$，都存在唯一的 $i$ 使得 $S' \subseteq A_i$，試證：$A_i \cap A_j \ne \emptyset$ 對於任意 $1 \le i < j \le n$。 | [
"由題意得\n$$\n\\sum_{i=1}^{n} \\binom{|A_i|}{2} = \\binom{n}{2}. \\qquad (1)\n$$\n令 $d_i = |\\{k \\mid i \\in A_k\\}|$。明顯地，\n$$\n\\sum_{i=1}^{n} d_i = \\sum_{k=1}^{n} |A_k|. \\qquad (2)\n$$\n由「存在與唯一」性質，得到\n$$\n\\sum_{i=1}^{n} \\binom{d_i}{2} = \\sum_{1 \\le i < j \\le n} |A_i \\cap A_j|.\n$$\n再加上 $|A_i \\cap A_j| \\le... | Taiwan | 二〇一一數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0iil | Problem:
Compute
$$
\sum_{n_{60}=0}^{2} \sum_{n_{59}=0}^{n_{60}} \cdots \sum_{n_{2}=0}^{n_{3}} \sum_{n_{1}=0}^{n_{2}} \sum_{n_{0}=0}^{n_{1}} 1
$$ | [
"Solution:\nThe given sum counts the number of non-decreasing 61-tuples of integers $\\left(n_{0}, \\ldots, n_{60}\\right)$ from the set $\\{0,1,2\\}$. Such 61-tuples are in one-to-one correspondence with strictly increasing 61-tuples of integers $\\left(m_{0}, \\ldots, m_{60}\\right)$ from the set $\\{0,1,2, \\ldo... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 1953 | |
0hyg | Problem:
Let $C$ be a circle in the $x y$-plane with center on the $y$-axis and passing through $A=(0, a)$ and $B=(0, b)$ with $0<a<b$. Let $P$ be any other point on the circle, let $Q$ be the intersection of the line through $P$ and $A$ with the $x$-axis, and let $O=(0,0)$. Prove that $\angle B Q P=\angle B O P$. | [
"Solution:\n\nWe make use of the fact that an angle inscribed in a circle has measure equal to one-half of the arc subtended. Since the $x$- and $y$-axes meet in a right angle, the circle $C_{1}$ through $B$, $O$, and $Q$ has $Q B$ as a diameter. Also, $\\angle A P B$ is a right angle, since $A B$ is the diameter o... | United States | 1st Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
051t | Real numbers $x_1$, $x_2$, $x_3$, $x_4$ in $[0; 1]$ are such that the product
$$
K = |x_1 - x_2| \cdot |x_1 - x_3| \cdot |x_1 - x_4| \cdot |x_2 - x_3| \cdot |x_2 - x_4| \cdot |x_3 - x_4|
$$
is as large as possible. Prove that $\frac{1}{27} > K > \frac{4}{243}$. | [
"If some two numbers among $x_1$, $x_2$, $x_3$, $x_4$ are equal then $K = 0$ which is not maximal. Thus assume w.l.o.g. that $x_1 > x_2 > x_3 > x_4$. Applying AM-GM for $x_1 - x_2$, $x_2 - x_3$ and $x_3 - x_4$ gives\n$$\n\\sqrt[3]{(x_1 - x_2)(x_2 - x_3)(x_3 - x_4)} \\le \\frac{(x_1 - x_2) + (x_2 - x_3) + (x_3 - x_4... | Estonia | Final Round of National Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
08pn | Problem:
Let $a$, $b$, $c$ be positive numbers such that $ab + bc + ca = 3$. Prove that
$$
\frac{a}{\sqrt{a^{3}+5}} + \frac{b}{\sqrt{b^{3}+5}} + \frac{c}{\sqrt{c^{3}+5}} \leq \frac{\sqrt{6}}{2}
$$ | [
"Solution:\nFrom the AM-GM inequality we have\n$$\na^{3} + a^{3} + 1 \\geq 3a^{2} \\Rightarrow 2(a^{3} + 5) \\geq 3(a^{2} + 3)\n$$\nUsing the condition $ab + bc + ca = 3$, we get\n$$\n(a^{3} + 5) \\geq 3(a^{2} + ab + bc + ca) = 3(c + a)(a + b)\n$$\ntherefore\n$$\n\\frac{a}{\\sqrt{a^{3} + 5}} \\leq \\sqrt{\\frac{2a^... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
05u3 | Problem:
Pour s'entraîner en prévision du dernier test POFM de l'année, Jean-Baptiste et Marie-Odile ont collecté 100 problèmes de mathématiques, et s'attellent désormais à la confection d'un programme de révisions. Pendant les 100 jours qui les séparent du test POFM, chacun devra traiter un problème par jour. On note... | [
"Solution:\n\nTout d'abord, il existe $100!$ manières de choisir l'ordre dans lequel Jean-Baptiste traitera les problèmes. Ces $100!$ manières jouent toutes des rôles symétriques. Sans perte de généralité, on numérote les problèmes de $1$ à $100$, dans l'ordre où Jean-Baptiste les traite, et il s'agit de démontrer ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof only | null | |
05gz | Problem:
Soit $ABCD$ un quadrilatère cyclique. Un cercle passant par $A$ et $B$ coupe $[AC]$ et $[BD]$ en $E$ et $F$. Les droites $(AF)$ et $(BE)$ coupent $[BC]$ et $[AD]$ en $P$ et $Q$ respectivement. Montrer que $(PQ)$ est parallèle à $(CD)$. | [
"\n\n$\\widehat{AQB} = \\widehat{AQE} = \\pi - \\widehat{QEA} - \\widehat{EAQ} = \\widehat{AEB} - \\widehat{CAD}$ et de même, $\\widehat{APB} = \\widehat{AFB} - \\widehat{CBD}$. Or, $\\widehat{AEB} = \\widehat{AFB}$ et $\\widehat{CAD} = \\widehat{CBD}$, donc $\\widehat{AQB} = \\widehat{APB}... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hh5 | Consider a complete graph with 4046 vertices with edges colored in some colors. We call a graph $k$-good graph if all the vertices of the graph can be divided into 2023 pairs in such a way that among the colors of the 2023 of the edges connecting the vertices in the pairs are exactly $k$ distinct colors. Is it possible... | [
"Number the vertices of the graph with numbers from 1 to 4046. Consider the following graph: all edges between vertices of different parity are colored 1 and all other edges are colored in their unique colors. Consider any partition of the vertices into 2023 into pairs. Suppose that in $x$ of these pairs the vertic... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Yes | |
07tq | Through a point *P* on the hypotenuse $AB$ of the right-angled triangle $ABC$ lines are drawn parallel to the other two sides. These parallels meet $BC$ and $AC$ at $D$ and $E$, respectively. Prove $|BP| \cdot |AP| = |BD| \cdot |CD| + |AE| \cdot |CE|$. | [
"Because *PE* is parallel to $BC$ and *PD* is parallel to $AC$, the two right angled triangles $APE$ and $PBD$ are similar. This implies that there exists a positive number $s$ (the similarity factor) such that\n$$\n|AP| = s|BP|, \\quad |AE| = s|DP| \\quad \\text{and} \\quad |EP| = s|BD|.\n$$\n\n![](attached_image_... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cvm | Let $BH$ be an altitude in an acute-angled triangle $ABC$. Let $M$ and $N$ be the midpoints of $AH$ and $CH$, respectively. Let $BB'$ be a diameter in the circumcircle of $\triangle BMN$. Prove that $AB' = CB'$. | [
"Let $O$ be the center of the circle $\\Omega$; that is, $O$ is the midpoint of the diameter $BB'$. Denote by $H'$ and $P$ the projections of the points $B'$ and $O$, respectively, onto the line $AC$ (see Fig. 9). Since $O$ lies on the perpendicular bisector of $MN$, we get that $P$ is the midpoint of $MN$. Since $... | Russia | Regional round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English; Russian | proof only | null | |
09zt | Alicia writes down $a$ distinct integers on a piece of paper and Britt writes down $b$ distinct integers on another piece of paper. Alicia wrote down at least one integer that Britt did not write down, and Britt wrote at least one integer down that Alicia did not write down. Vera counts the number of distinct integers ... | [
"We first note that there is a useful relation between $a$, $b$, $d$, and $v$. The total number of integers on the two pieces of paper is $a + b$, the number of integers on Alicia's piece of paper plus the number of integers on Britt's piece of paper. This, however, also equals $v + d$: the total number of distinct... | Netherlands | Second Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | proof and answer | a) Example exists: Alicia writes 1 through 2022 and Britt writes 1012 through 3033; then d = 1011 and v = 3033 and the equality holds. b) Yes: Alicia {1, 2, 3} and Britt {3, 4, 5}; then a = b = 3, d = 1, v = 5 and the equality holds. c) Impossible: the relation forces a = d while the conditions require a > d, a contrad... | |
0dvj | Problem:
Metka stoji $60~\mathrm{m}$ vzhodno in $80~\mathrm{m}$ južno od kraja, kjer stoji Tine. Oba sta enako oddaljena od lipe v mestnem parku, ki je naravnost vzhodno od kraja, kjer je Tine. Sočasno se vsak s svojega kraja odpravita naravnost proti lipi. Koliko metrov poti bo vsak izmed njiju prehodil do njunega sr... | [
"Solution:\n\nKer sta Metka in Tine enako oddaljena od lipe, velja $60 + x = \\sqrt{80^{2} + x^{2}}$. Če obe strani enačbe kvadriramo in nato enačbo uredimo, dobimo $120x = 2800$, od tod pa izrazimo $x = \\frac{70}{3}~\\mathrm{m}$. Vsak izmed njiju bo do srečanja prehodil $83\\,\\frac{1}{3}~\\mathrm{m}$."
] | Slovenia | 47. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 83 1/3 m | |
0ckv | Determine all functions $f: \mathbb{C} \to \mathbb{C}$ such that $|wf(z) + zf(w)| = 2|zw|$, for any $z, w \in \mathbb{C}$. | [
"For $z = 1$ and $w = 0$ we obtain $f(0) = 0$. For $w = z \\in \\mathbb{C}^*$ we obtain $|f(z)| = |z|$; this equality is also verified for $z = 0$, so $|f(z)| = |z|$, for any $z \\in \\mathbb{C}$. We have $|f(1)| = 1$ and, for $w = 1$, we obtain $2|z| = |f(z) + zf(1)| \\le |f(z)| + |zf(1)| = 2|z|$ for any $z \\in \... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | All solutions are f(z) = c z for all z, where c is a complex constant with |c| = 1. | |
0gon | Let $I$ be the incenter and $AD$ be a diameter of the circumcircle of a triangle $ABC$. If the point $E$ on the ray $BA$ and the point $F$ on the ray $CA$ satisfy the condition
$$
BE = CF = \frac{AB + BC + CA}{2},
$$
show that the lines $EF$ and $DI$ are perpendicular. | [
"Let $a = BC$, $b = CA$, $c = AB$, and $u = (a + b + c)/2$. Let $P$ be the point where the line drawn parallel to $AB$ through $I$ meets $BD$. Similarly define $Q$. Then $IP = u - b = AF$, $IQ = u - c = AE$ and $\\angle PIQ = \\angle BAC = \\angle FAE$. Hence the triangles $PIQ$ and $FAE$ are congruent.\n\nLet $P'$... | Turkey | Team Selection Test for IMO 2011 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0azl | Problem:
A square is inscribed in a circle, and a rectangle is inscribed in the square. Another circle is circumscribed about the rectangle, and a smaller circle is tangent to three sides of the rectangle, as shown below. The shaded area between the two larger circles is eight times the area of the smallest circle, whi... | [] | Philippines | 21st PMO Area Stage | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | 1/2 | |
0giq | 找出所有 $p$ 為質數的正整數數對 $(a, p)$，使得 $p^a + a^4$ 為完全平方數。
Determine all pairs $(a, p)$ of positive integers with $p$ prime such that $p^a + a^4$ is a perfect square. | [
"Let $p^a + a^4 = b^2$ for some positive integer $b$. Then we have\n$$\np^a = b^2 - a^4 = (b + a^2)(b - a^2).\n$$\nHence both $b + a^2$ and $b - a^2$ are powers of $p$.\nLet $b - a^2 = p^x$ for some integer $x$. Then $b + a^2 = p^{a-x}$ and $a - x > x$. Therefore we have\n$$\n2a^2 = (b + a^2) - (b - a^2) = p^{a-x} ... | Taiwan | IMO 1J, Mock Exam 2 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Chinese; English | proof and answer | [(1, 3), (2, 3), (6, 3), (9, 3)] | |
0ksm | Problem:
The number $0$ is written on a blackboard. Every minute, Aerith simultaneously replaces every $0$ with a $1$ and every $1$ with a $10$. For example, if the current number were $1100$, in one minute it would be $101011$. She eventually gets tired and leaves, leaving some number $N$ written on the board. If $9 ... | [
"Solution:\n\nLet $a_{n}$ be the number after $n$ minutes.\nAfter 2 minutes, we note that $a_{2}=10$ is $a_{1}=1$ followed by $a_{0}=0$, so waiting $n$ more minutes gives that $a_{n+2}$ is $a_{n+1}$ followed by $a_{n}$. Thus, $a_{n}$ has $F_{n+1}$ digits, so\n$$\na_{n+2}=a_{n+1} \\cdot 10^{F_{n+1}}+a_{n}\n$$\nWe no... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0gqb | In a convex quadrilateral $ABCD$, the diagonals intersect at $E$, $BE = \sqrt{2} \cdot ED$ and $\angle BEC = 45^\circ$. Let $F$ be the foot of the perpendicular from $A$ to $BC$ and $P$ be the second intersection point of the circumcircle of triangle $BFD$ and the line segment $[DC]$. Find $\angle APD$. | [
"Let $Q$ be the foot of the perpendicular from $B$ to the line $AC$. Then we obtain $BQ = EQ = ED$ which means that $\\angle AQD = 22.5^\\circ$. Since the points $B, F, Q, A$ are concyclic, we have $CF \\cdot CB = CQ \\cdot CA$. Since the points $B, F, P, D$ are also concyclic we have $CF \\cdot CB = CP \\cdot CD$ ... | Turkey | Team Selection Test for JBMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 22.5° | |
02pe | Problem:
Possuímos 32 pedras, todas com pesos diferentes. Descreva um processo para mostrar que podemos encontrar as duas pedras mais pesadas com 35 pesagens em uma balança de pratos. | [
"Solution:\n\nDividimos as pedras em 16 pares, pesamos cada par e pegamos as 16 mais pesadas. Repetimos o processo com as 16 pedras obtendo 8 pedras com oito pesagens a mais, 4 pedras com quatro pesagens, 2 pedras com 2 pesagens e a pedra mais pesada com a última pesagem.\n\nAté este momento foram usadas $16+8+4+2+... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
0bd6 | Let $P$ and $Q$ be two real polynomials so that $\lfloor P(x) \rfloor = \lfloor Q(x) \rfloor$, for every $x \in \mathbb{R}$. Prove that $P = Q$. | [
"Let $P(x)$ and $Q(x)$ be real polynomials such that $\\lfloor P(x) \\rfloor = \\lfloor Q(x) \\rfloor$ for all $x \\in \\mathbb{R}$.\n\nSuppose $P(x) \\ne Q(x)$. Then $P(x) - Q(x)$ is a nonzero polynomial, so $P(x) - Q(x)$ is not identically zero. Thus, there exists $x_0 \\in \\mathbb{R}$ such that $P(x_0) \\ne Q(x... | Romania | Shortlisted Problems for the 64th NMO | [
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
068v | Find all positive integers $x$, $y$, $z$ with $z$ odd, which satisfy the equation
$$
2018^x = 100^y + 1918^z
$$ | [
"We write $2018 = 2 \\cdot 1009$, $100 = 2^2 \\cdot 5^2$ and $1918 = 2 \\cdot 7 \\cdot 137$. Then, the equation has the form\n$$\n2^x \\cdot 1009^x = 2^{2y} \\cdot 5^{2y} + 2^z \\cdot 7^z \\cdot 137^z, \\quad (1)\n$$\n\nConsider the following cases:\n\n* If $2y < z$, then the power of $2$ that divides the right han... | Greece | Selection Examination | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (1, 1, 1) | |
0c2f | Let $n$ be a positive integer and let $C$ be a circle whose circumference is equal to $6n$. The circle $C$ is divided by $3n$ points into $3n$ small arcs such that $n$ of them have the lengths equal with $1$, other $n$ have the lengths equal with $2$ and the remaining $n$ arcs have the lengths equal with $3$. Prove tha... | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof only | null | |
0d2i | Define a regular $n$-pointed star to be a union of $n$ line segments $P_{1}P_{2}$, $P_{2}P_{3}$, $\ldots$, $P_{n}P_{1}$ such that
- the points $P_{1}, P_{2}, \ldots, P_{n}$ are coplanar and no three of them are collinear;
- each of the $n$ line segments intersects at least one of the other line segments at a point othe... | [
"Because all angles are congruent and all line segments are congruent, the regular $n$-pointed star is cyclic. Indeed, consider any four consecutive vertices $P_{i}, P_{i+1}, P_{i+2}, P_{i+3}$, for some $i \\in \\{1,2, \\ldots, n\\}$. They form an isosceles trapezoid. So they are cocyclic. By induction, all vertice... | Saudi Arabia | Selection tests for the Gulf Mathematical Olympiad 2013 | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common div... | English | proof and answer | 61, 77, 93, 99, 122, 124, 154, 186, 198 | |
04tc | Let $ABC$ be an acute triangle with altitudes $BD$, $CE$. Given that $AE \cdot AD = BE \cdot CD$, what is the smallest possible measure of $\angle BAC$? (Patrik Bak) | [
"We denote $\\angle BAC$ by $\\alpha$ and express lengths $AE$, $AD$, $BE$, $CD$ in terms of $\\cos \\alpha$ and the side lengths $b = AC$, $c = AB$ of triangle $ABC$. The condition rewrites as\n\n$$\nb \\cos \\alpha \\cdot c \\cos \\alpha = (c - b \\cos \\alpha)(b - c \\cos \\alpha),\n$$\nwhich simplifies to $bc =... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 60° | |
0lg2 | Problem:
Let $A_{n}$ be the set of partitions of the sequence $(1,2, \ldots, n)$ into several subsequences such that every two neighbouring terms of each subsequence have different parity, and let $B_{n}$ be the set of partitions of the sequence $(1,2, \ldots, n)$ into several subsequences such that all the terms of ea... | [
"Solution:\nFor each partition $\\pi$ of $\\{1,2, \\ldots, n\\}$, with the elements within each block written in ascending order, denote by $k(\\pi)$ the number of blocks of $\\pi$ ending in an even number and by $\\ell(\\pi)$ the number of blocks of $\\pi$ ending in an odd number.\nAlso denote $f_{n}(x, y)=\\sum_{... | Zhautykov Olympiad | XI International Zhautykov Olympiad in Sciences | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof only | null | |
0cuo | Let $\omega$ and $\Omega$ be the incircle and the circumcircle of an equilateral triangle $ABC$, respectively. Points $P$ and $Q$ are chosen on the sides $AC$ and $AB$, respectively, so that the segment $PQ$ contains the center of the triangle $ABC$. Circles $\Omega_b$ and $\Omega_c$ have diameters $BP$ and $CQ$, respe... | [
"Let $O$ be the center of $ABC$, and $B_1$ and $C_1$ be the midpoints of the lesser arcs $AC$ and $AB$ of $\\Omega$, respectively. Then the triangles $OPB_1$ and $OQC_1$ are isosceles (see Fig. 8). Next, the rays $B_1P$ and $C_1Q$ meet at a point $B' = C'$ on $\\Omega$ which is one of the common points of $\\Omega_... | Russia | XLIII Russian mathematical olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
... | English; Russian | proof only | null | |
0img | Problem:
In triangle $ABC$, $\angle ABC$ is obtuse. Point $D$ lies on side $AC$ such that $\angle ABD$ is right, and point $E$ lies on side $AC$ between $A$ and $D$ such that $BD$ bisects $\angle EBC$. Find $CE$, given that $AC = 35$, $BC = 7$, and $BE = 5$. | [
"Solution:\n\nAnswer: 10. Reflect $A$ and $E$ over $BD$ to $A'$ and $E'$ respectively. Note that the angle conditions show that $A'$ and $E'$ lie on $AB$ and $BC$ respectively. $B$ is the midpoint of segment $AA'$, and $CE' = BC - BE' = 2$. Menelaus' theorem now gives\n$$\n\\frac{CD}{DA} \\cdot \\frac{AA'}{A'B} \\c... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 10 | |
00gw | Prove that for any positive integer $k$, there exists an arithmetic sequence
$$
\frac{a_{1}}{b_{1}}, \quad \frac{a_{2}}{b_{2}}, \quad \ldots, \quad \frac{a_{k}}{b_{k}}
$$
of rational numbers, where $a_{i}, b_{i}$ are relatively prime positive integers for each $i=1,2, \ldots, k$, such that the positive integers $a_{1},... | [
"For $k=1$, there is nothing to prove. Henceforth assume $k \\geq 2$.\nLet $p_{1}, p_{2}, \\ldots, p_{k}$ be $k$ distinct primes such that\n$$\nk < p_{k} < \\cdots < p_{2} < p_{1}\n$$\nand let $N = p_{1} p_{2} \\cdots p_{k}$. By Chinese Remainder Theorem, there exists a positive integer $x$ satisfying\n$$\nx \\equi... | Asia Pacific Mathematics Olympiad (APMO) | XXI Asian Pacific Mathematics Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
00e1 | A positive integer $n > 1$, whose positive divisors are
$$
1 = d_1 < d_2 < \dots < d_k = n,
$$
is called *sureño* if all of the numbers $d_2 - d_1, d_3 - d_2, \dots, d_k - d_{k-1}$ are divisors of $n$.
a. Find a positive integer that is not sureño and has exactly 2022 positive divisors that are sureño.
b. Prove that ... | [
"We will prove that the number $2^{2022}7^k$ is not sureño and has exactly 2022 sureño divisors (for every positive integer $k$).\n\nEvery power of 2 is sureño. Indeed, the divisors of $2^k$ are $2^j$ for $j = 0, \\dots, k$ and $d_j - d_{j-1} = 2^j - 2^{j-1} = 2^{j-1}$ is a divisor of $2^k$. Therefore we have that ... | Argentina | XXXIII Cono Sur Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Other"
] | English | proof and answer | n = 2^{2022} · 7^k for any positive integer k (e.g., 2^{2022} · 7 for part a) | |
00xz | Problem:
Suppose two functions $f(x)$ and $g(x)$ are defined for all $x$ such that $2 < x < 4$ and satisfy $2 < f(x) < 4$, $2 < g(x) < 4$, $f(g(x)) = g(f(x)) = x$ and $f(x) \cdot g(x) = x^{2}$ for all such values of $x$. Prove that $f(3) = g(3)$. | [
"Solution:\n\nLet $h(x) = \\frac{f(x)}{x}$. Then we have $g(x) = \\frac{x^{2}}{f(x)} = \\frac{x}{h(x)}$ and $g(f(x)) = \\frac{f(x)}{h(f(x))} = x$ which yields $h(f(x)) = \\frac{f(x)}{x} = h(x)$. Using induction we easily get $h\\left(f^{(k)}(x)\\right) = h(x)$ for any natural number $k$ where $f^{(k)}(x)$ denotes $... | Baltic Way | Baltic Way 1993 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
06d8 | Find all positive integers $n$ for which the number $n^8 - 2^n$ is not divisible by 72. | [
"$n$ can be any positive integer of the form $4k - 2$ where $k \\in \\mathbb{Z}^+$. When $n = 4k - 2$, we have\n$$\nn^8 - n^2 = n^2(n^6 - 1) = 4(2k - 1)^2(n^6 - 1).\n$$\nSince both $2k-1$ and $n^6-1$ are odd, this number is not divisible by 8, and hence not divisible by 72.\n\nWe now claim that for any other $n$, $... | Hong Kong | IMO HK TST | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All positive integers n with n ≡ 2 (mod 4), i.e., n = 4k − 2 for k ∈ Z^+. | |
00gn | A regular ($5 \times 5$)-array of lights is defective, so that toggling the switch for one light causes each adjacent light in the same row and in the same column as well as the light itself to change state, from on to off, or from off to on. Initially all the lights are switched off. After a certain number of toggles,... | [
"We assign the following first labels to the 25 positions of the lights:\n| 1 | 1 | 0 | 1 | 1 |\n| :--- | :--- | :--- | :--- | :--- |\n| 0 | 0 | 0 | 0 | 0 |\n| 1 | 1 | 0 | 1 | 1 |\n| 0 | 0 | 0 | 0 | 0 |\n| 1 | 1 | 0 | 1 | 1 |\n\nFor each on-off combination of lights in the array, define its first value to be the su... | Asia Pacific Mathematics Olympiad (APMO) | XIX Asian Pacific Mathematics Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | The center (row 3, column 3) and the four positions (row 2, column 2), (row 2, column 4), (row 4, column 2), (row 4, column 4). | |
0keq | Problem:
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x \neq 0$,
$$
f(x)+2 f\left(\frac{x-1}{x}\right)=3 x .
$$
(Note that the function must be defined for all $x \in \mathbb{R}$, including $x=0$.) | [
"Solution:\nPlugging $\\frac{x-1}{x}$ in for $x$ in the functional equation gives\n$$\nf\\left(\\frac{x-1}{x}\\right)+2 f\\left(\\frac{\\frac{x-1}{x}-1}{\\frac{x-1}{x}}\\right)=f\\left(\\frac{x-1}{x}\\right)+2 f\\left(\\frac{1}{1-x}\\right)=\\frac{3(x-1)}{x} .\n$$\nNow plugging $\\frac{1}{1-x}$ in for $x$ gives\n$$... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = (-x^3 + 3x^2 + 2) / (3x(1 - x)) for x ≠ 0, 1; f(0) = c; f(1) = 3 − 2c, where c is any real number. | |
0lcq | Let $(x_n)$, $(y_n)$ be two positive sequences defined by $x_1 = 1$, $y_1 = \sqrt{3}$ and
$$
\begin{cases} x_{n+1}y_{n+1} - x_n = 0 \\ x_{n+1}^2 + y_n = 2 \end{cases}
$$
for all positive integers $n$. Prove that these sequences are convergent and find their limits. | [
"We can see that $x_1 = 1 = 2 \\sin \\frac{\\pi}{6}$, $y_1 = \\sqrt{3} = 2 \\cos \\frac{\\pi}{6}$.\nSo we will prove, by induction, that for all positive integers $n$, we have\n$$\nx_n = 2 \\sin \\frac{\\pi}{3 \\cdot 2^n}, \\quad y_n = 2 \\cos \\frac{\\pi}{3 \\cdot 2^n}. \\qquad (1)\n$$\nIndeed, for $n=1$ the state... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | lim x_n = 0, lim y_n = 2 | |
06vm | Let $A B C D E$ be a convex pentagon with $C D = D E$ and $\angle E D C \neq 2 \cdot \angle A D B$. Suppose that a point $P$ is located in the interior of the pentagon such that $A P = A E$ and $B P = B C$. Prove that $P$ lies on the diagonal $C E$ if and only if $\operatorname{area}(B C D) + \operatorname{area}(A D E)... | [
"\nFor the equivalence with the collinearity condition, let $F$ denote the foot of the perpendicular from $P^{\\prime}$ to $A B$, so that $F$ is the midpoint of $P P^{\\prime}$. We have that $P$ lies on $C E$ if and only if $F$ lies on $M N$, which occurs if and only if we have the equality... | IMO | IMO 2019 Shortlisted Problems | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
04it | Let $I$ be the incentre of the triangle $ABC$. If $|AI| = |BC|$ and $\angle ACB = 2\angle BAC$, determine the angles of the triangle $ABC$. (Brazil 2011) | [] | Croatia | Croatia 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 and answer | Angles are 40°, 60°, and 80° | |
0h2y | A positive integer is written on a blackboard. Each minute Andrew looks at his clock and adds the number of minutes the clock is showing (an integer between $0$ and $59$) to the number written on the blackboard.
a) Prove that at some moment a composite number will be written on the blackboard.
b) Will necessarily a n... | [
"a) У першому, третьому, п'ятому і т.д. записаних числах буде чергуватись парність (адже сума двох послідовних кількостей хвилин непарна). Отже, серед них будуть парні числа. При цьому лише перше з них може дорівнювати $2$ (отримане як $2 + 0$ або $1 + 1$), наступні парні числа будуть більшими за $2$, а тому складе... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Other"
] | English | proof and answer | No | |
0cpk | A grid square $2010 \times 2010$ is decomposed into «corners» (i.e. L-shaped figures consisting of 3 cells each). Prove that it is possible to mark one cell in each corner so that each row and each column would contain the same number of marked cells.
Клетчатый квадрат $2010 \times 2010$ разрезан на трёхклеточные угол... | [
"Обозначим $n = 2010/3 = 670$. Назовём линией строку или столбец; пронумеруем строки сверху вниз, а столбцы — слева направо. Заметим, что для выполнения условий задачи достаточно, чтобы при любом $k$ в левых $k$ столбцах и в верхних $k$ строках содержалось бы по $kn$ отмеченных клеток.\n\nСкажем, что уголок смотрит... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English, Russian | proof only | null | |
0bbx | Let $A_0A_1A_2$ be a non-equilateral triangle. The incircle of the triangle $A_0A_1A_2$ touches the side $A_iA_{i+1}$ at the point $T_{i+2}$ (indices are reduced modulo 3). Let $X_i$ be the perpendicular foot dropped from the point $T_i$ onto the line $T_{i+1}T_{i+2}$. Show that the lines $A_iX_i$ are concurrent at a p... | [
"The lines $A_iA_{i+1}$ and $X_iX_{i+1}$ are parallel, for they are both antiparallel to the line $T_iT_{i+1}$. Hence the triangles $A_0A_1A_2$ and $X_0X_1X_2$ are homologous: the three lines $A_iX_i$ are concurrent at the homology centre which lies on the homology line. The latter passes through the incentres of t... | Romania | 62nd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Advanced Conf... | null | proof only | null | |
07dh | The point $P$ is chosen in the interior of the isosceles trapezoid $ABCD$, where $AB \parallel CD$, such that $\angle APB > \angle ADC$ and $\angle DPC > \angle ABC$. Prove that $AB + CD > AD + BC$. | [
"Since $ABCD$ is an isosceles trapezoid, we know that $\\angle ADC = \\angle ABx$ and $\\angle ABC = \\angle DCy$. (Take a look at the figure below.)\n\n\n\nConsider a circle tangent to $AD$, $BC$ at $A$, $B$ respectively. We know that $P$ is in the interior of this circle because $\\angle ... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
058f | How many positive integers, where the only allowed digits are $0$ and $1$, are less than $1111100100$? | [
"The given number has $11$ digits. There are $2^{11} - 1 = 2047$ positive integers with at most $11$ digits each of which is either $0$ or $1$. We solve the problem by subtracting the number of positive integers that are not less than the given number.\nThe $11$-digit numbers larger than the given number are all of... | Estonia | Estonian Math Competitions | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Other",
"Number Theory > Other"
] | English | final answer only | 2019 | |
0jai | Problem:
Two infinite rows of evenly-spaced dots are aligned as in the figure below. Arrows point from every dot in the top row to some dot in the lower row in such a way that:
- No two arrows point at the same dot.
- No arrow can extend right or left by more than $1006$ positions.
Show that at most $2012$ dots in the... | [
"Solution:\n\nCall dots in the lower line that lie at the endpoints of arrows \"target dots\" and those that are not, \"missed dots\". If an arrangement has $2013$ or more missed dots, pick a contiguous set $S$ of dots in the lower line that includes exactly $2013$ missed dots and $t$ target dots.\n\nConsider the s... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0dhv | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that
$$
2(y + 1)f(x)f(y - 1) = 2yf(xy) - f(2x)
$$
for all $x, y \in \mathbb{R}$. | [
"Put $x = 0, y = -1$: $0 = -2f(0) - f(0)$ thus $f(0) = 0$.\n\nPut $y = 1$: $2f(x) = f(2x)$, for all $x$ so the given condition can rewrite as\n$$\n(y + 1)f(x)f(y - 1) = yf(xy) - f(x), \\forall x, y\n$$\nContinue to put $y = 0$ then $f(x)f(-1) = -f(x)$. If $f(-1) \\neq -1$ then $f(x) = 0$ for all $x$, which is satis... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = 0 for all real x; f(x) = x for all real x | |
07r7 | Suppose $A$, $B$, and $C$ are the angles in an acute-angled triangle. Prove that
$$
\frac{\sin 2A + \sin 2B + \sin 2C}{\cos A + \cos B + \cos C} \le \sqrt{3}.
$$ | [
"Since $\\cos x$ is decreasing on $[0, \\pi/2]$ and $\\sin x$ is increasing on this interval, by Chebyshev's inequality,\n$$\n\\frac{\\sin 2A + \\sin 2B + \\sin 2C}{\\cos A + \\cos B + \\cos C} = \\frac{2}{\\cos A + \\cos B + \\cos C} \\le \\frac{2}{3} (\\sin A + \\sin B + \\sin C).\n$$\nBut, just as in the solutio... | Ireland | Ireland_2017 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
0fz6 | Problem:
Eine ganze Zahl $m$ ist eine echte Potenz, falls es positive ganze Zahlen $a$ und $n$ gibt, sodass $n>1$ und $m=a^{n}$.
a. Zeige, dass es 2012 verschiedene positive ganze Zahlen gibt, sodass sich keine nichtleere Teilmenge davon zu einer echten Potenz aufsummiert.
b. Zeige, dass es 2012 verschiedene positiv... | [
"Solution:\n\na. Sei $p$ eine Primzahl, welche grösser als $2012^{2}$ ist, dann erfüllen die Zahlen $p, 2p, \\ldots, 2012p$ die Bedingung. Sei $B$ nun eine nicht leere Teilmenge der Zahlen, dann teilt $p$ die Summe $\\sum_{b \\in B} b$, aber $\\sum_{b \\in B} b < 2012 \\cdot 2012p < p^{2}$ und somit nicht durch $p^... | Switzerland | IMO Selektion | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
07xl | Suppose $a$, $b$, $c$ are real numbers such that $a + b + c = 1$ and $a^2 + b^2 + c^2 = 1$. Prove that $a^3 + b^3 + c^3 \ge \frac{5}{9}$. | [
"In the solution to Problem 7 we have seen that $a + b + c = 1$ and $a^2 + b^2 + c^2 = 1$ imply $a^3 + b^3 + c^3 = 1 + 3abc$. Hence, $a^3 + b^3 + c^3 \\ge \\frac{5}{9}$ iff $abc \\ge -\\frac{4}{27}$. Several proofs of this inequality can be found in the solution to Problem 11."
] | Ireland | IRL_ABooklet_2025 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null |
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