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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0620 | Problem:
Gegeben seien zwei positive ganze Zahlen $n$ und $k$. In der Ebene liegen $n$ Kreise ($n \geq 2$), so dass jeder Kreis jeden anderen zweimal schneidet und alle diese Schnittpunkte paarweise verschieden sind.
Jeder Schnittpunkt wird mit einer von $n$ Farben so gefärbt, dass jede Farbe wenigstens einmal verwend... | [
"Solution:\n\nDie Antwort lautet: $2 \\leq k \\leq n \\leq 3$ oder $3 \\leq k \\leq n$.\nOffensichtlich gilt $k \\leq n$ nach Aufgabenstellung sowie $k \\geq 2$, weil für $k=1$ alle Punkte dieselbe Farbe hätten, während die Anzahl $n$ der Farben $\\geq 2$ sein soll. Wir nummerieren die Kreise und die Farben von 1 b... | Germany | Auswahlwettbewerb zur IMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2 ≤ k ≤ n ≤ 3 or 3 ≤ k ≤ n | |
0iko | Problem:
Given a convex $n$-gon, $n \geq 4$, at most how many diagonals can be drawn such that each drawn diagonal intersects every other drawn diagonal either in the interior of the $n$-gon or at a vertex? Prove your answer. | [
"Solution:\n\nFirst of all, assume without loss of generality that the $n$-gon is regular (this has no effect as far as diagonal intersection is concerned). Also, treat $n=4$ as a special case; obviously the answer is 2 here.\n\nIf $n$ is odd, simply draw $n$ diagonals, connecting each vertex to the ones $(n-1) / 2... | United States | Harvard-MIT Mathematics Tournament, Team Round A | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Maximum equals n for all n ≥ 5, and equals 2 for n = 4. | |
0l0m | Problem:
There exists a unique line tangent to the graph of $y = x^{4} - 20 x^{3} + 24 x^{2} - 20 x + 25$ at two distinct points. Compute the product of the $x$-coordinates of the two tangency points. | [
"Solution:\nIf $f(x)$ is tangent to the $x$-axis at $(c, 0)$, then $f(x)$ will be divisible by $(x-c)^{2}$. Thus, if $f(x)$ is tangent at the $x$-axis at $c_{1}$ and $c_{2}$, then $f(x) = P(x) (x-c_{1})^{2} (x-c_{2})^{2}$ for some polynomial $P(x)$. By adding $mx + b$, we see that $f(x)$ is tangent to $y = mx + b$ ... | United States | HMMT November 2024 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | -38 | |
07j2 | Let $(a_n)_{n \in \mathbb{N}}$ be a sequence of natural numbers such that every prime number greater than $2024$ has a multiple in this sequence. Prove that the number of prime divisors of the elements of the sequence $b_n = a_1 a_2 \dots a_n - 1$ is infinite. | [
"Assume to the contrary, let $P(b_n)$ be finite and\n$$\nP(b_n) = \\{2, p_1, \\dots, p_n, q_1, \\dots, q_s\\}\n$$\nwhere $p_i \\equiv 1 \\pmod 4$, $q_i \\equiv -1 \\pmod 4$. Thus, any element of $b_n$ can be written in the form\n$$\nb_n = 2^a p_1^{\\alpha_1} \\cdots p_n^{\\alpha_n} q_1^{\\beta_1} \\cdots q_s^{\\bet... | Iran | 41th Iranian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof only | null | |
0km9 | Problem:
Can $x^{2020}-8$ be written as the product of two nonconstant polynomials with integer coefficients? | [
"Solution:\n\nThe answer is no. Indeed, suppose we had $x^{2020}-8 = f \\cdot g$. Since every complex root of $f$ has absolute value $\\sqrt[2020]{8}$, and the product of these roots is an integer (by Vieta's formula), we conclude that\n$$\n8^{\\frac{\\operatorname{deg} f}{2020}} = 2^{\\frac{3 \\operatorname{deg} f... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | No | |
02lv | Prove that there are no positive integers $x$ and $y$ such that $x^3 + y^3 = 2^{2009}$. | [
"One can verify that $x^3 \\equiv 0, 1, -1 \\pmod{7}$. Since $2^{2009} = (2^3)^{669} \\cdot 2^2 \\equiv 1^{669} \\cdot 4 \\equiv 4 \\pmod{7}$ and $x^3 + y^3 \\equiv -2, -1, 0, 1, 2 \\pmod{7}$, it is not possible that $x^3 + y^3 \\equiv 2^{2009} \\pmod{7}$, so the equation $x^3 + y^3 = 2^{2009}$ has no solutions.\n\... | Brazil | XXXI Brazilian Math Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0baj | Find all positive integers $a, b$ for which there exists sets $A, B$ of positive integers so that $A \cap B = \emptyset$, $A \cup B = \mathbb{N}^*$ and $aA = bB$ (if $x$ is a number and $M$ is a set of numbers, $xM = \{xm \mid m \in M\}$). | [
"We can assume that $1 \\in A$. Then $a \\in bB$, hence there exists $p \\in B$ such that $a = pb$. Moreover, $p \\ge 2$, because $1 \\in A$.\n\nEvery pair $(pb, b)$, with $b \\in \\mathbb{N}^*$, is a solution: we use the partition\n\n$$\nA = \\{p^{2n}q \\mid n \\in \\mathbb{N}, q \\in \\mathbb{N}^*, p \\nmid q\\}\... | Romania | 2011 CLOCK-TOWER SCHOOL JUNIORS COMPETITION | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | All pairs (a, b) such that one divides the other with quotient at least 2; equivalently, a = p b or b = p a for some integer p ≥ 2. | |
08f6 | Problem:
Lucio compra un tavolo a forma di esagono regolare e una tovaglia rettangolare di area $1024~\mathrm{cm}^2$ che ha lato minore pari al lato dell'esagono e copre esattamente la porzione di tavolo compresa fra due lati opposti. Quanto vale l'area del tavolo, espressa in $\mathrm{cm}^2$? | [
"Solution:\n\nLa risposta è 1536.\n\n\n\nSia $A B C D E F$ un esagono regolare che rappresenti il tavolo; la tovaglia di area $1024~\\mathrm{cm}^2$ copre esattamente il rettangolo $A B D E$. Detto $O$ il centro dell'esagono e detti $H, K$ i punti di intersezione fra i segmenti $B D$ e $O C$... | Italy | Italian Mathematical Olympiad - February Round | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof and answer | 1536 | |
03d1 | In each of the cells of an $n \times m$ table is written one of the letters $a$, $b$, or $c$. The corresponding letter in every two rows, respectively every two columns of the table, coincide in at most $k \ge 0$ positions. Determine $m$, $n$ and $k$, if
$$
\frac{2mn + 6k}{3(m + n)} \ge k + 1.
$$ | [
"Denote the table rows by $u_1, u_2, \\dots, u_n$. Let $d(u_i, u_j)$, $i \\ne j$ be the number of positions, in which the rows $u_i$ and $u_j$ differ. According to the statement $d(u_i, u_j) \\ge m - k$. Then:\n$$\n(1) \\qquad S = \\sum_{1 \\le i < j \\le n} d(u_i, u_j) \\ge \\frac{n(n-1)}{2}(m-k).\n$$\n\nConsider ... | Bulgaria | Bulgaria 2022 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | m = 3, n = 3, k = 0 | |
00r9 | Prove that among any $20$ consecutive positive integers there exists a number $d$ such that for each positive integer $n$ we have the inequality
$$
n\sqrt{d}\{n\sqrt{d}\} > \frac{5}{2}
$$
where $\{x\}$ denotes the fractional part of the real number $x$. | [
"Among the given numbers there is a number of the form $20k + 15 = 5(4k + 3)$. We shall prove that $d = 5(4k + 3)$ satisfies the statement's condition. Since $d \\equiv -1 \\pmod 4$, it follows that $d$ is not a perfect square, and thus for any $n \\in \\mathbb{N}$ there exists $a \\in \\mathbb{N}$ such that $a + 1... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
02th | Problem:
Arnaldo e Bernaldo decidem jogar um jogo que possui um número limitado de jogadas. Arnaldo escreve o número 1 no quadro em sua primeira jogada. Em seguida, Bernaldo escreve 2 ou 4 no quadro. Depois disso, Arnaldo escreve 3 ou 9 no quadro. Os dois continuam jogando alternadamente mantendo a regra de que na jog... | [
"Solution:\n\na) Veja que 15 é divisível por 3, então independente da última jogada, 15 ou $15^{2}$, o resto na divisão por 3 não será alterado. Vejamos a jogada de número 14. Veja que 14 deixa resto 2 na divisão por 3 enquanto que $14^{2}=196$ deixa resto 1 na divisão por 3. Como 14 é par, quem fará tal jogada é B... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | null | proof and answer | a) Bernaldo can force a win when the game ends on the fifteenth move. b) Arnaldo can force a win when the game ends on the seventh move; his winning strategy is: if Bernaldo writes two on his second move, Arnaldo writes twenty‑five on his fifth move; if Bernaldo writes four on his second move, Arnaldo writes five on hi... | |
0gsu | Let $k, n$ be positive integers with $k \ge n!$ Prove that
$$
\phi(k) \ge (n-1)!
$$ | [
"For the solution we will show that if\n$$\n\\phi(k) < (n-1)! \\text{ then } k < n!\n$$\nLet $k = q_1^{\\alpha_1} \\dots q_s^{\\alpha_s}$ with $q_1 < q_2 < \\dots < q_s$. It suffices to show that\n$$\n\\frac{\\phi(k)}{k} \\geq \\frac{1}{n}.\n$$\nSince\n$$\n\\frac{\\phi(k)}{k} = (1 - \\frac{1}{q_1}) \\dots (1 - \\fr... | Turkey | 30th Turkish Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0a4z | Problem:
Let $ABCD$ be a square and let $X$ be any point on side $BC$ between $B$ and $C$. Let $Y$ be the point on line $CD$ such that $BX = YD$ and $D$ is between $C$ and $Y$. Prove that the midpoint of $XY$ lies on diagonal $BD$. | [
"Solution:\n\nConstruct point $E$ on diagonal $BD$ so that $EX$ is parallel to $CD$. So $\\angle BXE = \\angle BCD = 90^\\circ$. Also $\\angle XBE = 45^\\circ$ because $E$ is on diagonal $BD$. Therefore triangle $\\triangle BEX$ is an isosceles right-angled triangle. Hence\n$$EX = BX = YD.$... | New Zealand | NZMO Round One | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0axf | Problem:
Let $P$ be a point inside the isosceles trapezoid $ABCD$ where $AD$ is one of the bases, and let $PA$, $PB$, $PC$, and $PD$ bisect angles $A$, $B$, $C$, and $D$ respectively. If $PA = 3$ and $\angle APD = 120^\circ$, find the area of trapezoid $ABCD$. | [] | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 6√3 | |
0kvi | Problem:
Compute the number of ways to color the vertices of a regular heptagon red, green, or blue (with rotations and reflections distinct) such that no isosceles triangle whose vertices are vertices of the heptagon has all three vertices the same color. | [
"Solution:\n\nNumber the vertices $1$ through $7$ in order. Then, the only way to have three vertices of a regular heptagon that do not form an isosceles triangle is if they are vertices $1,2,4$, rotated or reflected. Thus, it is impossible to have four vertices in the heptagon of one color because it is impossible... | United States | HMMT November 2023 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 294 | |
09or | A group of people needs to be transported from point $A$ to point $B$ by a bus that can carry only half of them. The bus first departs with half of the people while the others start walking. After dropping off the passengers partway, the bus returns, picks up the rest, and continues to $B$. The walking group and the bu... | [] | Mongolia | MMO2025 Round 2 | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | English | proof and answer | 3:5 | |
0kjn | Problem:
Let $f: \mathbb{N} \rightarrow \mathbb{N}$ be a strictly increasing function such that $f(1)=1$ and $f(2 n) f(2 n+1)=9 f(n)^2+3 f(n)$ for all $n \in \mathbb{N}$. Compute $f(137)$. | [
"Solution:\nPlugging in $n=1$ gives $f(2) f(3)=12$, therefore $(f(2), f(3))=(2,6)$ or $(3,4)$. However, the former implies\n$$\nf(4) f(5) \\geq (6+1)(6+2)>42=9 \\cdot 2^2+3 \\cdot 2\n$$\nwhich is impossible; therefore $f(2)=3$ and $f(3)=4$. We now show by induction with step size 2 that $f(2 n)=3 f(n)$ and $f(2 n+1... | United States | HMMT Spring 2021 Guts Round | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 2215 | |
009k | Let $ABC$ be a right triangle. It is known that there are points $D$ on side $AC$ and $E$ on side $BC$ such that $AB = AD = BE$ and $BD \perp DE$. Find $\frac{AB}{BC}$ and $\frac{BC}{CA}$. | [
"Denote $BC = a$, $CA = b$, $AB = c$. The assumptions imply $c \\le a$, $c \\le b$. First we prove that $b + c = 2a$, without using the condition that $ABC$ is a right triangle.\n\nLet $F$ be the midpoint of $BE$. By $BD \\perp DE$ triangle $BED$ is right at $D$, so $DF$ is the median to its hypotenuse $BE$. Hence\... | Argentina | NATIONAL XXX OMA | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | AB/BC = 3/4, BC/CA = 4/5 | |
0kgh | Problem:
A circle contains the points $(0,11)$ and $(0,-11)$ on its circumference and contains all points $(x, y)$ with $x^{2}+y^{2}<1$ in its interior. Compute the largest possible radius of the circle. | [
"Solution:\n\nSuch a circle will be centered at $(t, 0)$ for some $t$; without loss of generality, let $t>0$. Our conditions are that\n$$\nt^{2}+11^{2}=r^{2}\n$$\nand\n$$\nr \\geq t+1\n$$\nSo, $t^{2} \\leq (r-1)^{2}$, which means\n$$\n(r-1)^{2}+11^{2} \\geq r^{2} \\Longrightarrow 122 \\geq 2r\n$$\nso our answer is ... | United States | HMMT Spring 2021 | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 61 | |
0hl6 | Problem:
After walking so much that his feet get really tired, the beaver staggers so that, at each step, his coordinates change by either $(+1,+1)$ or $(+1,-1)$. Now he walks from $(0,0)$ to $(8,0)$ without ever going below the $x$-axis. How many such paths are there? | [
"Solution: $C(4)=14$."
] | United States | null | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 14 | |
0dqa | Find the sum of all the 5-digit integers which are not multiples of 11 and whose digits are 1, 3, 4, 7, 9. | [
"First note that an integer is divisible by 11 if and only if the alternating sum of the digits is divisible by 11. In our case, these are the integers where $1$, $4$ and $7$ are at the odd positions. Let $S$ be the sum of all the 5-digit integers formed by $1$, $3$, $4$, $7$, $9$ and let $T$ be the sum of those wh... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 5842368 | |
0dbg | Prove that there are no positive integers $x$, $y$, $z$ such that
$$
(3x + 4y)(4x + 5y) = 7^{z}
$$ | [
"From the condition, we can see that both numbers $3x + 4y$ and $4x + 5y$ are powers of $7$, so such must be also their division. However\n$$\n1 < \\frac{4x + 5y}{3x + 4y} < 2\n$$\nand can't be power of $7$."
] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
05vi | Problem:
Maena et Théodore jouent à un jeu. Ils jouent sur une grille carrée formée de $99 \times 99$ cases. On considère que deux cases sont adjacentes si elles ont un sommet ou un côté en commun.
Initialement, Maéna numérote les cases de la grille de 1 à $99^{2}$, de façon arbitraire. Théodore place alors un jeton ... | [
"Solution:\n\nTout d'abord, Théodore peut toujours se débrouiller pour effectuer au moins trois mouvements. Pour ce faire, il lui suffit de sélectionner un carré de taille $2 \\times 2$ à l'intérieur du carré de taille $99 \\times 99$, puis d'en parcourir les quatre cases, qui sont nécessairement adjacentes puisqu'... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 3 | |
04is | For any two positive real numbers $x$ and $y$ prove
$$
\log^2(xy) \ge \log(x^2)\log(y^2).
$$ | [] | Croatia | Croatia Mathematical Competitions | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0ipe | Problem:
Evaluate the integral $\int_{0}^{1} \ln x \ln (1-x) d x$. | [
"Solution:\nAnswer: $2-\\frac{\\pi^{2}}{6}$\n\nWe have the MacLaurin expansion $\\ln (1-x) = -x - \\frac{x^{2}}{2} - \\frac{x^{3}}{3} - \\cdots$. So\n$$\n\\int_{0}^{1} \\ln x \\ln (1-x) d x = -\\int_{0}^{1} \\ln x \\sum_{n=1}^{\\infty} \\frac{x^{n}}{n} d x = -\\sum_{n=1}^{\\infty} \\frac{1}{n} \\int_{0}^{1} x^{n} \... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2 - π^2/6 | |
00pp | Let $P$ and $Q$ be points inside a triangle $ABC$ such that $\angle PAC = \angle QAB$ and $\angle PBC = \angle QBA$. Let $D$ and $E$ be the feet of the perpendiculars from $P$ to the lines $BC$ and $AC$, and $F$ be the foot of perpendicular from $Q$ to the line $AB$. Let $M$ be the intersection of the lines $DE$ and $A... | [
"Let $G$ be the foot of the perpendicular from $P$ to the line $AB$, and $H$ and $I$ be the feet of the perpendiculars from $Q$ to the lines $CB$ and $CA$, respectively. Observe that we also have $\\angle PCA = \\angle QCB$ by the trigonometric form of Ceva's Theorem.\n\nThe quadrilaterals $AEPG$ and $AFQI$ are sim... | Balkan Mathematical Olympiad | Balkan 2012 shortlist | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
09rz | Problem:
Laat $a$, $b$ en $c$ rationale getallen zijn waarvoor $a+b c$, $b+a c$ en $a+b$ allemaal ongelijk aan $0$ zijn en waarvoor geldt dat
$$
\frac{1}{a+b c}+\frac{1}{b+a c}=\frac{1}{a+b}
$$
Bewijs dat $\sqrt{(c-3)(c+1)}$ rationaal is. | [
"Solution:\n\nOplossing I. Er geldt\n$$\n\\frac{1}{a+b c}+\\frac{1}{b+a c}=\\frac{(b+a c)+(a+b c)}{(a+b c)(b+a c)}=\\frac{(a+b)+(a+b) c}{a b+a^{2} c+b^{2} c+a b c^{2}}\n$$\nUit de gegeven gelijkheid volgt dus\n$$\n(a+b)((a+b)+(a+b) c)=a b+a^{2} c+b^{2} c+a b c^{2}\n$$\noftewel\n$$\n\\begin{aligned}\n(a+b)^{2} & =a ... | Netherlands | IMO-selectietoets | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0dql | Let $OT$ be a diameter of a circle. Let $A$ and $B$ be two distinct points on the circle both on the same side of $OT$, and let $C$ be the intersection of the tangents to the circle at $A$ and $B$. The tangent to the circle at $T$ meet the lines $OA$, $OB$ and $OC$ at $A'$, $B'$ and $C'$ respectively. Prove that $C'$ i... | [
"\n\nConsider the inversion in the circle $\\omega$ centred at $O$ with radius $OT$. The circle $\\alpha$ with diameter $OT$ is inverted into the tangent line $\\alpha'$ to $\\alpha$ at $T$. Thus $A'$ and $B'$ are the inverses of $A$ and $B$ respectively. The circle $\\beta$ centred at $C$ ... | Singapore | Singapore International Mathematical Olympiad Committee National Team Selection Test | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof only | null | |
050v | Jüri wishes to draw $n$ circles and any number of lines on the plane such that all the lines meet at one point, and for every two circles there exist two lines that touch both of these circles.
a)
Is it possible for Jüri to solve this problem for any $n \ge 2$?
b)
For which natural numbers $n$ is it possible to solve... | [
"a.\nJüri can draw two lines and draw any number of circles such that they touch both of the lines.\n\n\nFig. 8\n\nb.\nAssume that Jüri has solved the problem for some $n$, where $n > 1$.\n\nLet $O$ be the intersection point of all the lines. Look at any circle $c$. From the premises of the... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | a) Yes, it is possible for any number at least two.
b) It is possible exactly for n ≤ 4. | |
0hx3 | Problem:
Let $n$ be a positive integer which is divisible by $5$ and which can be written as the sum of two (not necessarily distinct) squares. Prove that $n$ can be written as the sum of two squares one of which is greater than or equal to four times the other. | [
"Solution:\n\nLet $n = a^2 + b^2$, where $a$ and $b$ are nonnegative integers. Suppose that each of the squares $a^2$ and $b^2$ is less than $4$ times the other, so\n$$\na < 2b \\text{ and } b < 2a.\n$$\nSince $a^2 + b^2$ is divisible by $5$, so is $a^2 + b^2 - 5b^2 = a^2 - 4b^2 = (a + 2b)(a - 2b)$. Thus $5$ divide... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | null | proof only | null | |
07u9 | Let $n \ge 3$ be an integer and let $(p_1, p_2, p_3, \dots, p_n)$ be a permutation of $\{1, 2, 3, \dots, n\}$. For this permutation we say that $p_t$ is a *turning point* if $2 \le t \le n-1$ and
$$
(p_t - p_{t-1})(p_t - p_{t+1}) > 0.
$$
For example, for $n=8$, the permutation $(2, 4, 6, 7, 5, 1, 3, 8)$ has two turning... | [
"**Solution 1.** We claim that $q(n) = 2^n - 4$ and that this is a perfect square only when $n = 3$.\nIf there is a unique turning point, then it is either a minimum or maximum. We count the number of permutations where the turning point is a maximum, and so $q(n)$ is double this number. As we are interested in the... | Ireland | IRL_ABooklet | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 3 | |
0h8i | Determine if there exist non-integer $x, y$, so that for any integer $a, b$, both $x+y$ and $ax+by$ are integers. | [
"Suppose such numbers exist. Then $ax+by = a(x+y) + (b-a)y$. Hence $(b-a)y$ is an integer. Let $b=2, a=1$, then $y$ is an integer – contradiction."
] | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | No such numbers exist. | |
092a | Problem:
Let $ABCD$ be a cyclic quadrilateral. Let $E$ be the intersection of lines parallel to $AC$ and $BD$ passing through points $B$ and $A$, respectively. The lines $EC$ and $ED$ intersect the circumcircle of $AEB$ again at $F$ and $G$, respectively. Prove that points $C$, $D$, $F$, and $G$ lie on a circle.
![](a... | [
"Solution:\nThe solution uses directed angles. It suffices to show $\\angle GDC = \\angle GFC$, which is done as follows\n$$\n\\begin{aligned}\n\\angle GDC & = \\angle EDC = \\angle EDB + \\angle BDC \\\\\n& = \\angle DEA + \\angle BAC = \\angle GEA + \\angle ABE \\\\\n& = \\angle GBA + \\angle ABE = \\angle GBE = ... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04b4 | How many perfect squares are there between $4^9$ and $9^4$, excluding those two numbers? | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | final answer only | 430 | |
07rb | Given a point $P$ between the legs of an angle with vertex $A$. Show, with proof, how to construct a line through $P$ that intersects the legs of the angle at points $B$ and $C$ so that $|PB| = |PC|$. | [
"First, draw the circle centre $P$ through $A$ and let $D$ be the second intersection point of this circle with the line $AP$. Then construct the two lines that pass through $D$ and are parallel to the legs of the given angle. They intersect the legs of the angle at $B$ and $C$.\n\n\n\nBy c... | Ireland | Ireland_2017 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof only | null | |
0fpk | Determinar el mayor número de alfiles que se pueden colocar en un tablero de ajedrez de $8 \times 8$, tal que no haya dos alfiles en la misma casilla y cada alfil sea amenazado como máximo por uno de los otros alfiles.
Nota. Un alfil amenaza a otro si ambos se encuentran en dos casillas diferentes de una misma diagona... | [] | Spain | XXXI Olimpiada Iberoamericana de Matemáticas | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | Spanish | proof and answer | 20 | |
09y0 | Three wheels are pushed together so they don't slip if we turn them. The circumferences of the wheels are $14$, $10$, and $6$ cm, respectively. On each wheel an arrow is drawn, pointing downwards. Someone turns the big wheel and the other wheels turn with it. This stops at the first moment all arrows point downwards ag... | [
"5"
] | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Modular Arithmetic"
] | English | final answer only | 5 | |
04qm | In a rectangle with dimensions $2 \times 3$ there is a polyline of length 36, which can have self-intersections. Show that there exists a line parallel to two sides of the rectangle, which intersects the other two sides in their interior points and intersects the polyline in fewer than 10 points.
![](attached_image_1.... | [
"Consider an arbitrary line segment of the polyline and denote by $d$ its length and by $x$ and $y$ the lengths of its perpendicular projections on the sides of lengths 2 and 3, respectively. Cauchy-Schwarz inequality gives us\n$$\n(2x + 3y)^2 \\leq (2^2 + 3^2)(x^2 + y^2) = 13d^2,\n$$\nwhich means $2x + 3y \\leq d ... | Czech Republic | null | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0jhe | Problem:
Find the number of integers $n$ such that
$$
1+\left\lfloor\frac{100 n}{101}\right\rfloor=\left\lceil\frac{99 n}{100}\right\rceil
$$ | [
"Solution:\nConsider $f(n)=\\left\\lceil\\frac{99 n}{100}\\right\\rceil-\\left\\lfloor\\frac{100 n}{101}\\right\\rfloor$. Note that $f(n+10100)=\\left\\lceil\\frac{99 n}{100}+99 \\cdot 101\\right\\rceil-\\left\\lfloor\\frac{100 n}{101}+100^{2}\\right\\rfloor=f(n)+99 \\cdot 101-100^{2}=f(n)-1$. Thus, for each residu... | United States | HMMT | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 10100 | |
09c1 | $1, 2, 3, \ldots, 100$ тоонуудыг дараалсан 2 тоо хөрш нүдэнд (ерөнхий талтай нүд) бичигдсэн байхаар $10 \times 10$ хүснэгтэд байрлуулав. Тэгвэл ядаж 2 ширхэг бүхэл тооны квадрат агуулах мөр эсвэл багана олдохыг батал. | [
"\n\nЭсрэгээс $1^2$, $2^2$, $\\ldots$, $10^2$ тоонуудыг аль ч 2 нь нэг мөр эсвэл баганад оршихгүй байхаар байрлуулж чадсан гэе. $10 \\times 10$ хүснэгтийн мөр ба багануудыг зурагт үзүүлсэн байдлаар дугаарлая. Тэгээд нүд бүрт $(x, y)$ тоог харгалзуулья. $x$-мөрийн дугаар, $y$-баганы дугаар.\... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | Mongolian | proof only | null | |
0275 | Problem:
a) Qual o valor de
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\cdots+\frac{1}{\sqrt{99}+\sqrt{100}} ?
$$
b) Se $x=\sqrt{1+2 \cdot \sqrt{1+3 \cdot \sqrt{1+4 \cdot \sqrt{1+5 \cdot \sqrt{\ldots}}}}}$ é um número real, qua... | [
"Solution:\na) Perceba que\n$$\n\\begin{aligned}\n\\frac{1}{\\sqrt{n}+\\sqrt{n+1}} \\cdot \\frac{\\sqrt{n}-\\sqrt{n+1}}{\\sqrt{n}-\\sqrt{n+1}} & =\\frac{\\sqrt{n}-\\sqrt{n+1}}{(\\sqrt{n})^{2}-(\\sqrt{n+1})^{2}} \\\\\n& =\\frac{\\sqrt{n}-\\sqrt{n+1}}{n-(n-1)} \\\\\n& =-\\sqrt{n}+\\sqrt{n+1}\n\\end{aligned}\n$$\nPort... | Brazil | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | a) 9; b) 3 | |
0b7x | Suppose $a \in [-2, \infty)$, $r \in [0, \infty)$ and let $n$ be a positive integer.
Show that
$$
r^{2n} + a r^n + 1 \geq (1 - r)^{2n}.
$$ | [
"If $r = 0$, the relation is obvious.\nOtherwise, dividing by $r^{2n}$, one gets the same inequality with $r$ replaced by $\\frac{1}{r}$, so one can assume $r \\in (0, 1]$.\nSince $r^{2n} + a r^n + 1 \\geq r^{2n} - 2 r^n + 1 = (1 - r^n)^2$ and $1 - r^n \\geq 0$, it is enough to prove that $1 - r^n \\geq (1 - r)^n$.... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0c1t | Consider the natural numbers $a \neq 0$, $b = 2a + 1000$, $c = a + 1$ and $d = 2a + 1002$.
a) Show that $\frac{a}{b} < \frac{c}{d}$.
b) For $a = 9$, find the least natural number $n$ such that
$$
\frac{a+n}{b+n} > \frac{c+n}{d+n}.
$$ | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 1001 | |
0b3h | Problem:
How many ways are there to arrange the first ten positive integers such that the multiples of $2$ appear in increasing order, and the multiples of $3$ appear in decreasing order?
(a) $720$
(b) $2160$
(c) $5040$
(d) $6480$ | [] | Philippines | 24th Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | MCQ | d | |
0bck | Problem:
Se consideră funcţia $f: \mathbb{R} \rightarrow \mathbb{R}$,
$$
f(x)= \begin{cases} a x, & x \in \mathbb{Q} \\ b x, & x \in \mathbb{R} \setminus \mathbb{Q} \end{cases}
$$
unde $a$ şi $b$ sunt două numere reale nenule.
Să se arate că $f$ este injectivă dacă şi numai dacă $f$ este surjectivă. | [] | Romania | Olimpiada Naţională de Matematică Etapa Judeţeană şi a Municipiului Bucureşti | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof only | null | |
0i4g | Problem:
This question forms a three question multiple choice test. After each question, there are 4 choices, each preceded by a letter. Please write down your answer as the ordered triple (letter of the answer of Question $\# 1$, letter of the answer of Question $\# 2$, letter of the answer of Question \#3). If you f... | [
"Solution:\n\n(A, C, D).\n\nQuestion 2: In order for the answer to be consistent with the condition, \"If the answer to Question 2 were Letter (D),\" the answer to this question actually must be \"Letter (D).\" The letter of this answer is (C).\n\nQuestion 1: If a fourth question had an answer with letter (C), then... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Logic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | null | MCQ | (A, C, D) | |
0acb | Find all the real functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$
f(x^2 - y^2) = (x - y)[f(x) + f(y)]
$$ | [
"If $x = y$, we have\n$$\nf(x^2 - y^2) = (x - y)[f(x) + f(y)], \\quad f(0) = 0 \\cdot [2f(x)] = 0.\n$$\nThen $f(0) = 0$.\nIf $y = -x$, we have\n$$\n\\begin{aligned}\nf[x^2 - (-x)^2] &= [x - (-x)][f(x) + f(-x)], \\quad f(0) = 2x[f(x) + f(-x)], \\\\\n0 &= -2x[f(x) + f(-x)].\n\\end{aligned}\n$$\nFrom the last equation... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = k x for any real constant k | |
009b | In a shop there are two classes of packages: $11$-kg and $12$-kg ones. The total weight of all the packages is $5940$ kg. It is known that there are packages of $12$ kg, but the amount of packages of each kind is unknown. Show that these packages can be divided in $11$ groups with the same weight. | [] | Argentina | XXIX Olimpíada Matemática Argentina National Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0hr8 | Problem:
Aerith and Bob take turns picking a nonnegative integer, each time changing exactly one digit from the other's last number. The first person to pick a number that (s)he picked before loses. If Aerith goes first, and both play optimally, who wins?
(Note: There are no leading zeroes, except in the number $0$ i... | [
"Solution:\n\nBob wins. One winning strategy for Bob is: each time Aerith picks an even number, add one, and each time Aerith picks an odd number, subtract one. This only changes the last digit, since there are no carryovers in either case.\n\nBob would only get into a situation where he repeated an even number if ... | United States | Berkeley Math Circle: Monthly Contest 8 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Bob wins | |
0g0b | Problem:
Trouver tous les triplets d'entiers naturels $(a, b, c)$ tels que pour tout entier naturel $n$ qui n'a pas de diviseur premier plus petit que $2015$
$$
n+c \mid a^{n}+b^{n}+n
$$ | [
"Solution:\n\nLemme - Il existe une infinité de nombres premiers $p$ tel que $p \\equiv 2\\pmod{3}$.\n\nDémonstration : Par l'absurde, soient $p_{1}, \\ldots, p_{n}$ les uniques premiers congrus à $2$ modulo $3$.\nSi $n$ est pair, alors $p_{1} p_{2} \\ldots p_{n}+1 \\equiv 2\\pmod{3}$ et donc ce nombre possède au m... | Switzerland | IMO-Selektion | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | (1,1,2) | |
040o | Let $f(x)$ be an odd function on $\mathbb{R}$, and $f(x) = x^2$ for $x \ge 0$. Suppose for any $x \in [a, a+2]$, $f(x+a) \ge 2f(x)$. Then the range of real number $a$ is ______. | [
"According to the given condition, we have\n$$\nf(x) = \\begin{cases} x^2 & (x \\ge 0), \\\\ -x^2 & (x < 0). \\end{cases}\n$$\nSo $2f(x) = f(\\sqrt{2}x)$. Therefore, the original inequality is equivalent to $f(x+a) \\ge f(\\sqrt{2}x)$.\n\nAs $f(x)$ is increasing over $\\mathbb{R}$, then $x + a \\ge \\sqrt{2}x$, i.e... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | [√2, +∞) | |
0154 | Problem:
The polynomial $P$ has integer coefficients and $P(x)=5$ for five different integers $x$. Show that there is no integer $x$ such that $-6 \leq P(x) \leq 4$ or $6 \leq P(x) \leq 16$. | [
"Solution:\n\nAssume $P\\left(x_{k}\\right)=5$ for different integers $x_{1}, x_{2}, \\ldots, x_{5}$. Then\n$$\nP(x)-5=\\prod_{k=1}^{5}\\left(x-x_{k}\\right) Q(x)\n$$\nwhere $Q$ is a polynomial with integral coefficients. Assume $n$ satisfies the condition in the problem. Then $|n-5| \\leq 11$. If $P\\left(x_{0}\\r... | Baltic Way | Baltic Way 2008 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0d78 | Given four numbers $x, y, z, t$, let $(a, b, c, d)$ be a permutation of $(x, y, z, t)$ and set $x_{1}=|a-b|$, $y_{1}=|b-c|$, $z_{1}=|c-d|$, and $t_{1}=|d-a|$. From $x_{1}, y_{1}, z_{1}, t_{1}$, form in the same fashion the numbers $x_{2}, y_{2}, z_{2}, t_{2}$, and so on. It is known that $x_{n}=x$, $y_{n}=y$, $z_{n}=z$... | [
"First, consider 4 sequences $\\left(x_{n}\\right),\\left(y_{n}\\right),\\left(z_{n}\\right),\\left(t_{n}\\right)$ with $x_{1}=|a-b|$, $y_{1}=|b-c|$, $z_{1}=|c-d|$, $t_{1}=|d-a|$, $(a, b, c, d)$ is permutation of $(x, y, z, t)$. And\n$$\n\\left\\{\n\\begin{array}{l}\nx_{n+1}=\\left|a_{n}-b_{n}\\right| \\\\\ny_{n+1}... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | All permutations of (0, 0, 0, 0), (a, a, 0, 0) with a > 0, and (2a, a, a, 0) with a > 0. | |
0dq0 | In a language, an alphabet with 25 letters is used; *words* are exactly all sequences of (not necessarily different) letters of length $17$. Two ends of a paper strip are glued so that the strip forms a ring; the strip bears a sequence of $5^{18}$ letters. Say that a word is *singular* if one can cut out a piece bearin... | [
"Let the alphabet consist of letters $a_1, a_2, \\dots, a_{25}$. By a *piece* we always mean a piece of the strip containing exactly $17$ consecutive letters; different pieces may contain the same word. Say that a piece is *singular* if the word it contains is such.\n\nWe start with constructing an example containi... | Silk Road Mathematics Competition | XXI SILK ROAD MATHEMATICAL COMPETITION | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2*5^17 | |
0gw3 | a) Prove that for any rational number $\alpha \in (0; 1)$ there exists an infinite set of real numbers that satisfy the equation $\{x[x\{x\}]\} = \alpha$ and any two of them have the same fractional part.
b) Prove that for any rational number $\alpha \in (0; 1)$ there exists an infinite set of real numbers that satisf... | [
"a) Let $\\alpha = \\frac{p}{q}$, where $p, q \\in \\mathbb{N}$, $p < q$. Consider $y = pq + \\frac{1}{q}$. Then all $x = y + m$, where $m = q^n$, $n \\in \\mathbb{N}$, $n \\ge 2$, have equal fractional parts and satisfy the given equation. Indeed, then we have:\n$$\n\\{x\\} = \\{y\\}, \\quad m\\{y\\} \\in \\mathbb... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof only | null | |
0a0y | Find all positive integers $n$ for which there exist $n$ distinct positive integers $a_1, a_2, \dots, a_n$, none of them greater than $n^2$, such that
$$
\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n} = 1.
$$ | [
"The answer is that the given property holds for all $n \\ne 2$. For $n = 1$, the set $\\{1\\}$ satisfies (3). For $n = 2$, note that no set satisfies (3); if $a_1$ or $a_2$ equals $1$, then $\\frac{1}{a_1} + \\frac{1}{a_2} > 1$, if $a_1$ and $a_2$ are both at least two, then $\\frac{1}{a_1} + \\frac{1}{a_2} \\le \... | Netherlands | IMO Team Selection Test 1 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof and answer | All positive integers except 2 | |
09pi | Problem:
In een groep van scholieren spreken 50 scholieren Duits, 50 scholieren Frans en 50 scholieren Spaans. Sommige scholieren spreken meer dan één taal.
Bewijs dat de scholieren in 5 groepen verdeeld kunnen worden zodat in elke groep precies 10 scholieren Duits spreken, 10 Frans en 10 Spaans. | [
"Solution:\n\nScholieren die geen enkele taal spreken kunnen we buiten beschouwing laten, want we kunnen ze willekeurig over de groepen verdelen.\nWe onderscheiden zeven typen scholieren, al naar gelang de talen die ze spreken: DFS, FS, SD, DF, D, F en S, waarbij bijvoorbeeld een S-scholier alleen Spaans spreekt.\n... | Netherlands | UITWERKINGEN TOETS TRAININGSKAMP | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0h97 | The cube is cut into $2010$ smaller cubes. $2008$ of them are unit cubes (cubes of side $1$) and edges of another $2$ cubes are assigned integer numbers (different from $1$). Find the volume of the big cube.
(Serhiy Zhydkov) | [
"Let's denote the edge of the big cube as $a$ and the edges of two not unit as $b$ and $c$.\nThen we have an equation:\n$$\nb^3 + c^3 + 2008 = a^3.\n$$\nBecause $12^3 = 1728 < 2008$, then $a > 12$, so $a \\ge 13$.\nNow let's find an upper bound of $a$. It is clear that $a \\ge b+c$,\n$$\nb^3 + 3b^2c + 3bc^2 + c^3 \... | Ukraine | The Problems of Ukrainian Authors | [
"Geometry > Solid Geometry > 3D Shapes",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 2197 | |
0jfw | Problem:
Compute the value of $1^{25} + 2^{24} + 3^{23} + \ldots + 24^{2} + 25^{1}$. If your answer is $A$ and the correct answer is $C$, then your score on this problem will be $\left\lfloor 25 \min \left(\left(\frac{A}{C}\right)^{2}, \left(\frac{C}{A}\right)^{2}\right)\right\rfloor$. | [
"Solution:\n\nAnswer: 66071772829247409\n\nThe sum is extremely unimodal, so we want to approximate it using its largest term. Taking logs of each term, we see that the max occurs when $(26-n) \\log n$ peaks, and taking derivatives gives\n$$\nx + x \\log x = 26\n$$\nFrom here it's easy to see that the answer is aro... | United States | HMMT 2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | 66071772829247409 | |
09wx | Problem:
Vind alle functies $f: \mathbb{Z} \rightarrow \mathbb{Z}$ die voldoen aan
$$
f(-f(x)-f(y))=1-x-y
$$
voor alle $x, y \in \mathbb{Z}$. | [
"Solution:\n\nOplossing I. Invullen van $x=y=1$ geeft $f(-2 f(1))=-1$. Invullen van $x=n, y=1$ geeft $f(-f(n)-f(1))=-n$. Nu kiezen we $x=-f(n)-f(1)$ en $y=-2 f(1)$, dat geeft\n$$\nf(-f(-f(n)-f(1))-f(-2 f(1)))=1-(-f(n)-f(1))-(-2 f(1))\n$$\nwaarbij we de linkerkant verder kunnen uitrekenen als $f(--n--1)=f(n+1)$ en d... | Netherlands | IMO-selectietoets III | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x - 1 | |
02y6 | Problem:
Um triminó é um retângulo $3 \times 1$ e um monominó é um único quadrado $1 \times 1$. Quais são as possíveis posições de um monominó na cobertura de um tabuleiro $8 \times 8$ usando 21 triminós e 1 monominó?
 | [
"Solution:\n\nPinte os quadradinhos do tabuleiro $8 \\times 8$ com as cores 1, 2 e 3 como indicado nos tabuleiros a seguir.\n\n| 1 | 2 | 3 | 1 | 2 | 3 | 1 | 2 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| 3 | 1 | 2 | 3 | 1 | 2 | 3 | 1 |\n| 2 | 3 | 1 | 2 | 3 | 1 | 2 | 3 |\n| 1 | 2 | 3 | 1 | 2 | 3 |... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | (3,3), (3,6), (6,3), (6,6) | |
07sj | Let $a, b, c > 0$. Prove that
$$
\frac{2}{(1+a)^2} + \frac{2}{(1+b)^2} + \frac{2}{(1+c)^2} \ge \frac{1}{1+ab} + \frac{1}{1+bc} + \frac{1}{1+ca}
$$
with equality if and only if $a = b = c = 1$. | [
"We first prove that if $a, b > 0$, then\n$$\n\\frac{1}{(1+a)^2} + \\frac{1}{(1+b)^2} \\ge \\frac{1}{1+ab}\n$$\nwith equality if and only if $a = b = 1$. To prove this we multiply through by $(1+a)^2(1+b)^2(1+ab)$ and simplify to obtain the equivalent inequality $a^3b+ab^3-a^2b^2-2ab+1 \\ge 0$. This can be written ... | Ireland | IRL_ABooklet_2020 | [
"Algebra > Equations and Inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
04j5 | Let $ABCD$ be a convex quadrilateral such that $\angle BAD = 50^\circ$, $\angle ADB = 80^\circ$ and $\angle ACB = 40^\circ$ holds. If $\angle DBC = 30^\circ + \angle BDC$, determine $\angle BDC$. | [
"Let $ABCD$ be a convex quadrilateral with the given angles. Let $\\angle BDC = x$.\n\nGiven:\n$\\angle BAD = 50^\\circ$\n$\\angle ADB = 80^\\circ$\n$\\angle ACB = 40^\\circ$\n$\\angle DBC = 30^\\circ + x$\n\nLet us denote $E = AB \\cap CD$ (if needed), but first, let's focus on triangle $ABD$.\n\nIn $\\triangle AB... | Croatia | Croatia Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 40° | |
01ek | Let $\omega$ be a circle and $A$ a point outside of $\omega$. Draw the tangents from $A$ to $\omega$ and call the points of tangency $X$ and $Y$. Let $B$ and $C$ be points on the segments $AX$ and $AY$, respectively, such that the perimeter of $\triangle ABC$ is equal to the length of the segment $AX$. Let $D$ be the r... | [
"Let $B'$ be the reflection of $A$ through $B$. Since the perimeter of $\\triangle ABC$ equals the length of the segment $AX$, $AB$ is less than half of $AX$ and, therefore, $B'$ lies on the segment $AX$.\nLet the point $C'$ lie on $AY$ such that $B'C'$ touches $\\omega$ in the point $Z$. Let $C''$ be the midpoint ... | Baltic Way | Baltic Way shortlist | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof only | null | |
0bfe | Let $a, b \in \mathbb{N}$, $a, b > 2$. Prove that $2^a + 1$ cannot be divisible by $2^b - 1$. | [] | Romania | Shortlisted Problems for the 64th NMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null | |
0hf7 | For which values of $n > 1$ is there a set of pairwise distinct natural numbers $(a_1, a_2, ..., a_n)$ such that the following condition is satisfied:
$$
(a_1+1)!+(a_2+1)!+...+(a_n+1)! : a_1!+a_2!+...+a_n!
$$ | [
"$a_1, a_2, ..., a_n$ Without the loss of generality, let us assume that $a_1 < a_2 < ... < a_n$. Then $(a_i + 1)! \\le a_i!(a_n + 1)$, and equality is reached only when $i = n$.\n\nLet $(a_1 + 1)! + (a_2 + 1)! + ... + (a_n + 1)! = N(a_1! + a_2! + ... + a_n!)$ for some natural number $N$. Then the condition above y... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | no values of n greater than 1 | |
02sw | Problem:
Uma pulga, que está no ponto $A$ de uma reta, pula exatamente $1~\mathrm{m}$ de cada vez, sem nunca sair dessa reta.
a) Se a pulga quer chegar no ponto $B$ localizado sobre a reta, a uma distância de $5~\mathrm{m}$ à direita de $A$, com exatamente 7 pulos, de quantas maneiras ela pode fazer isso?
b) Se a pu... | [
"Solution:\n\na) Para chegar em $B$, a pulga deve dar exatamente um passo para a esquerda, e seis para a direita, em qualquer ordem. Como esse passo para a esquerda pode ser dado em qualquer momento, há 7 momentos possíveis para dá-lo! Logo, são 7 maneiras distintas da pulga chegar em $B$ com 7 passos.\n\nb) Para c... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) 7; b) 36; c) No | |
03o5 | Problem:
Initially, three non-collinear points, $A$, $B$, and $C$, are marked on the plane. You have a pencil and a double-edged ruler of width 1. Using them, you may perform the following operations:
- Mark an arbitrary point in the plane.
- Mark an arbitrary point on an already drawn line.
- If two points $P_1$ and ... | [
"Solution:\n\nClaim 1. It is possible to draw internal/external angle bisectors.\nProof. Let $A$, $B$, $C$ be marked. To bisect $\\angle A B C$, draw the parallel line to $A B$ unit 1 away from it on the opposite side as $C$, and draw the parallel line to $B C$ unit 1 away from it on the opposite side as $A$. Let t... | Canada | CMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometr... | null | proof only | null | |
07jz | We call a positive integer *n lucky* if it has at least two prime divisors and can be written in the form
$$
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
$$
where $p_1, \dots, p_k$ are distinct prime divisors of $n$. (Note that $n$ may have other prime divisors than $p_1, \dots, p_k$.) Prove that for every prime number... | [
"We claim that there exist $n$ with three prime factors. Choose a prime $p_1$. Using quadratic reciprocity and Dirichlet's theorem on arithmetic progressions, choose a prime $p_2$ such that\n$$\n\\left( \\frac{p_1}{p_2} \\right) = \\left( \\frac{p_2}{p_1} \\right) = 1\n$$\nand choose $p_3$ such that\n$$\n\\left(\\f... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null | |
0jrq | Problem:
If $a$ and $b$ satisfy the equations $a + \frac{1}{b} = 4$ and $\frac{1}{a} + b = \frac{16}{15}$, determine the product of all possible values of $a b$. | [
"Solution:\n\nWe multiply $a + \\frac{1}{b} = 4$ and $\\frac{1}{a} + b = \\frac{16}{15}$ to get\n$$\n\\left(a + \\frac{1}{b}\\right) \\left(\\frac{1}{a} + b\\right) = 4 \\cdot \\frac{16}{15} = \\frac{64}{15}.\n$$\nExpanding the left side:\n$$\na \\cdot \\frac{1}{a} + a b + \\frac{1}{b} \\cdot \\frac{1}{a} + \\frac{... | United States | HMMT November 2016 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 1 | |
0db2 | Let $n$ be a positive integer and $p > n + 1$ a prime. Prove that $p$ divides the following sum
$$
S = 1^{n} + 2^{n} + \ldots + (p-1)^{n}.
$$ | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof only | null | |
049i | If $\cos \gamma = 2 \sin \alpha \sin \beta - 1$, prove that the triangle with angles $\alpha, \beta, \gamma$ is isosceles. | [] | Croatia | Hrvatska 2011 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
0d3m | Let $\Gamma$ be a circle with center $O$ and $A E$ be a diameter. Point $D$ lies on segment $O E$ and point $B$ is the midpoint of one of the $\operatorname{arcs} \overparen{A E}$ of $\Gamma$. Construct point $C$ such that $A B C D$ is a parallelogram. Lines $E B$ and $C D$ meet at $F$. Line $O F$ meets the minor $\ope... | [
"Because $D C$ and $A B$ are parallel, $\\angle C D E = \\angle B A O = 45^\\circ$. Because $B C$ and $A E$ are parallel, $\\angle C B E = \\angle O E B = 45^\\circ$. Hence quadrilateral $B C E D$ is cyclic and therefore $\\angle B E C = \\angle B D C$.\n\nOn the other hand, $\\angle D F B = \\angle D E F + \\angle... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English, Arabic | proof only | null | |
0j7z | Problem:
Let $a \star b = a b + a + b$ for all integers $a$ and $b$. Evaluate $1 \star (2 \star (3 \star (4 \star \ldots (99 \star 100) \ldots)))$. | [
"Solution:\n\nAnswer: $101! - 1$\n\nWe will first show that $\\star$ is both commutative and associative.\n\n- **Commutativity:** $a \\star b = a b + a + b = b a + b + a = b \\star a$\n\n- **Associativity:**\n \\[\n a \\star (b \\star c) = a (b c + b + c) + a + b c + b + c = a b c + a b + a c + a + b c + b + c\n ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 101! - 1 | |
0hs8 | Problem:
$n$ boxes initially contain $1, 2, \ldots, n$ marbles respectively ($n \geq 1$). Charlotte first adds a marble to each box. Then she adds a marble to each box in which the number of marbles is divisible by $2$, then a marble to each box in which the number of marbles is divisible by $3$, and so on, until she ... | [
"Solution:\n\nThe answer is: all $n$ such that $n+1$ is prime.\n\nLemma 1. If box $A$ is to the left of box $B$ (we can arrange the boxes so that they initially contain $1, \\ldots, n$ marbles respectively from left to right), then at no stage of the process can box $A$ have more marbles than box $B$.\n\nProof. At ... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | All n such that n+1 is prime | |
069h | Find all pairs $(x, n)$ of positive integers satisfying the equation:
$$
3 \cdot 2^x + 4 = n^2.
$$ | [
"If $x = 1$, then we have no solutions, while for $x = 2$, we have the solution $(x, n) = (2, 4)$.\n\nNow we suppose that $x \\ge 3$. The left part of the equation is even and so $n$ is even, say $n = 2k$. Then the equation can be written as: $3 \\cdot 2^x + 4 = 4k^2 \\Leftrightarrow 3 \\cdot 2^{x-2} = k^2 - 1$, an... | Greece | SELECTION EXAMINATION | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (x, n) = (2, 4), (5, 10), (6, 14) | |
05da | Problem:
Let $n$ be a positive integer. Find the greatest possible integer $m$, in terms of $n$, with the following property: a table with $m$ rows and $n$ columns can be filled with real numbers in such a manner that for any two different rows $[a_{1}, a_{2}, \ldots, a_{n}]$ and $[b_{1}, b_{2}, \ldots, b_{n}]$ the fol... | [
"Solution:\nThe largest possible $m$ is equal to $2^{n}$.\n\nIn order to see that the value $2^{n}$ can be indeed achieved, consider all binary vectors of length $n$ as rows of the table. We now proceed with proving that this is the maximum value.\n\nLet $[a_{k}^{i}]$ be a feasible table, where $i=1, \\ldots, m$ an... | European Girls' Mathematical Olympiad (EGMO) | European Girls' Mathematical Olympiad 2012-Day 1 Solutions | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2^n | |
0l6c | Problem:
Ethan initially writes some numbers on a blackboard, each of which is either a $3$ or a $5$. He then repeatedly picks two numbers and replaces them with their sum, difference, product, or quotient (if the divisor is nonzero). Let $f(n)$ denote the minimum number of numbers Ethan must initially write for him t... | [
"Solution:\n\nThe following expressions represent optimal ways for Ethan to make each of the 8 given numbers.\n\n$$164 = 3(5(5 + 5) + 3) + 5$$\n\n$$187 = 3(3 + 5)(3 + 5) - 5$$\n\n$$191 = 5 \\cdot 5(3 + 5) - 3 \\cdot 3$$\n\n$$224 = (3 + 5)(5 \\cdot 5 + 3)$$\n\n$$255 = 5 \\cdot 5(5 + 5) + 5$$\n\n$$286 = (3 + 5 + 5)(5... | United States | HMMT February | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | final answer only | 6, 6, 6, 5, 5, 6, 6, 7 | |
0ga7 | 已知 $x, y$ 為滿足 $x + y = 1$ 的正實數。試證：
$$
\frac{x}{x^2 + y^3} + \frac{y}{x^3 + y^2} \le 2 \left( \frac{x}{x+y^2} + \frac{y}{x^2 + y} \right).
$$
Let $x, y$ be positive real numbers such that $x + y = 1$. Prove that
$$
\frac{x}{x^2 + y^3} + \frac{y}{x^3 + y^2} \le 2 \left( \frac{x}{x+y^2} + \frac{y}{x^2 + y} \right).
$$ | [
"令 $t = xy$，則\n$$\nx^2 + y^2 = 1 - 2t,\n$$\n$$\nx^3 + y^3 = 1 - 3t,\n$$\n$$\nx^4 + y^4 = 1 - 4t + t^2\n$$\n$$\nx^5 + y^5 = 1 - 5t + 5t^2.\n$$\n因為 $x^2 + y = x + y^2$，所以原不等式等價於\n$$\n\\begin{aligned}\n& \\frac{x}{x^2 + y^3} + \\frac{y}{x^3 + y^2} \\le \\frac{2}{x+y^2} \\\\\n\\Leftrightarrow & \\frac{x^4 + y^4 + xy}{(... | Taiwan | 二〇一六數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
06aa | Let $ABC$ be a triangle with $AB < AC < BC$. On the ray $BC$ we consider a point $D$ such that $BD = BA$ and on the ray $CB$ we consider a point $E$ such that $CE = CA$. Let $K$ be the circumcenter of the triangle $ADE$ and let $F$, $G$ the intersections of the lines $AD$, $KC$ and $AE$, $KB$, respectively. Prove that ... | [
"\nLet $M$ the midpoint of $AD$. Then the points $B$, $G$, $K$, $M$ belong to the perpendicular bisector of $AD$, that is they are collinear. Moreover, since $GD = GA$ the circle $c_3$ contains $A$.\n\nSimilarly, if $N$ is the midpoint of $AE$, the points $G$, $F$, $K$, $N$ belong to the pe... | Greece | 39th Hellenic Mathematical Olympiad | [
"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 > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneo... | null | proof only | null | |
0bri | For $n \in \mathbb{N}$, consider the system $(S_n) := \begin{cases} x^2 + n y^2 = z^2 \\ n x^2 + y^2 = t^2 \end{cases}$, where $x, y, z, t \in \mathbb{N}$. If $M_1 = \{ n \in \mathbb{N} \mid \text{system } (S_n) \text{ has infinitely many solutions} \}$, and $M_2 = \{ n \in \mathbb{N} \mid \text{system } (S_n) \text{ h... | [
"a) Notice that $x = 1, y = 3, z = 8, t = 4$ is a solution to the system $(S_7)$, and so is $(k, 3k, 8k, 4k)$, for any $k \\in \\mathbb{N}$, hence $7 \\in M_1$.\n\nIf $(x, y, z, t)$ is a solution to the system $(S_{10})$, it would follow that $11(x^2 + y^2) = z^2 + t^2$. From $11 \\mid z^2 + t^2$ we get $11 \\mid z... | Romania | 67th NMO Selection Tests for JBMO | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
02uk | Problem:
Em um grupo de 200 pessoas, apenas $1\%$ é mulher. Determine o número de homens que devem abandonar o grupo para que $98\%$ das pessoas restantes sejam do sexo masculino. | [
"Solution:\n\nO número de mulheres é $200 \\cdot \\frac{1}{100} = 2$. Para que tal número represente $2\\% = 100\\% - 98\\%$ da nova quantidade total de pessoas $x$, devemos ter $2 = x \\cdot \\frac{2}{100}$, ou seja, $x = 100$. Assim, devem sair $198 - 98 = 100$ pessoas do sexo masculino do grupo."
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 100 | |
0hzb | Problem:
Let $C$ be a circle with two diameters intersecting at an angle of $30$ degrees. A circle $S$ is tangent to both diameters and to $C$, and has radius $1$. Find the largest possible radius of $C$. | [
"Solution:\n\nFor $C$ to be as large as possible we want $S$ to be as small as possible. It is not hard to see that this happens in the situation shown below. Then the radius of $C$ is $1 + \\csc 15 = \\mathbf{1} + \\sqrt{\\mathbf{2}} + \\sqrt{\\mathbf{6}}$. The computation of $\\sin 15$ can be done via the half an... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 1 + sqrt(2) + sqrt(6) | |
0eac | Problem:
Brez uporabe računala izračunaj vrednost izraza $\sqrt{|2 \sqrt{5}-6|}-\sqrt{2 \sqrt{5}+6}$. | [] | Slovenia | 14. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | final answer only | -2 | |
09jh | Let $n \ge 1$. Let $a_1, \dots, a_{2n+2}$ be a sequence of pairwise distinct integers. Prove that $a_{2n+2} - a_1$ is divisible by $2n+1$ if we have $|a_i - a_j| \le n$ whenever $|i-j| \le n$. | [
"Let $a_k$ with $1 \\le k \\le 2n+2$ denote the minimum. Translating the sequence by a constant, we may assume that $a_k = 0$. Moreover, reversing the order of the sequence if necessary, we may assume that $1 \\le k \\le n+1$. By the minimality of $a_k$, we have $a_{k+1}, \\dots, a_{k+n} \\ge 1$ and from the distan... | Mongolia | Mongolian Mathematical Olympiad Round 3 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
00vg | Prove that the inequality
$$
\left(\frac{a^2 + b^2}{a + b}\right)^3 + \left(\frac{b^2 + c^2}{b + c}\right)^3 + \left(\frac{c^2 + a^2}{c + a}\right)^3 \ge a^3 + b^3 + c^3
$$
holds for all $a, b, c > 0$. | [
"The desired inequality holds if (and only if) the following one does\n$$\n2 \\left( \\frac{a^2 + b^2}{a+b} \\right)^3 \\geq a^3 + b^3 \\quad (*)_{a,b}\n$$\nfor all $a, b > 0$. Indeed, $(*)_{a,b}$ is implied by the problem statement by setting $b = c$. Conversely, we recover the problem statement by summing $(*)_{a... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlisted Problems | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
06s3 | Call admissible a set $A$ of integers that has the following property:
If $x, y \in A$ (possibly $x = y$) then $x^{2} + k x y + y^{2} \in A$ for every integer $k$.
Determine all pairs $m, n$ of nonzero integers such that the only admissible set containing both $m$ and $n$ is the set of all integers. | [
"A pair of integers $m, n$ fulfills the condition if and only if $\\gcd(m, n) = 1$. Suppose that $\\gcd(m, n) = d > 1$. The set\n$$\nA = \\{ \\ldots, -2d, -d, 0, d, 2d, \\ldots \\}\n$$\nis admissible, because if $d$ divides $x$ and $y$ then it divides $x^{2} + k x y + y^{2}$ for every integer $k$. Also $m, n \\in A... | IMO | 53rd International Mathematical Olympiad Shortlisted Problems with Solutions | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | gcd(m, n) = 1 | |
0144 | Problem:
Consider the sequence $a_{k}$ defined by $a_{1}=1$, $a_{2}=\frac{1}{2}$,
$$
a_{k+2}=a_{k}+\frac{1}{2} a_{k+1}+\frac{1}{4 a_{k} a_{k+1}} \quad \text{for } k \geq 1
$$
Prove that
$$
\frac{1}{a_{1} a_{3}}+\frac{1}{a_{2} a_{4}}+\frac{1}{a_{3} a_{5}}+\cdots+\frac{1}{a_{98} a_{100}}<4
$$ | [
"Solution:\nNote that\n$$\n\\frac{1}{a_{k} a_{k+2}}<\\frac{2}{a_{k} a_{k+1}}-\\frac{2}{a_{k+1} a_{k+2}}\n$$\nbecause this inequality is equivalent to the inequality\n$$\na_{k+2}>a_{k}+\\frac{1}{2} a_{k+1}\n$$\nwhich is evident for the given sequence. Now we have\n$$\n\\begin{aligned}\n\\frac{1}{a_{1} a_{3}}+\\frac{... | Baltic Way | Baltic Way 2005 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
02ap | Problem:
5. Calculadora diferente - Uma fábrica produziu uma calculadora original que efetua duas operações:
- a adição usual $+$
- a operação $\circledast$
Sabemos que para todo número natural $a$ tem-se:
$$
\text{(i)}\ a \circledast a = a \quad \text{e (ii)}\ a \circledast 0 = 2a
$$
e, para quaisquer quatro naturai... | [
"Solution:\n\nPara calcular $(2+3) \\circledast (0+3)$ utilizaremos a propriedade (iii), e temos:\n$$\n(2+3) \\circledast (0+3) = (2 \\circledast 0) + (3 \\circledast 3)\n$$\nAgora, por (ii) temos $2 \\circledast 0 = 2 \\times 2 = 4$, e por (i) temos $3 \\circledast 3 = 3$. Portanto,\n$$\n(2+3) \\circledast (0+3) =... | Brazil | null | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 7; 2000 | |
08dh | Problem:
Nel cassetto di Alice ci sono 30 calzini di 5 colori: 6 bianchi, 6 gialli, 6 rossi, 6 verdi e 6 azzurri. Il fratellino birichino prende 10 buste nere e inserisce in ogni busta tre calzini (presi dal cassetto) di tre colori diversi. Ora Alice deve andare a Cesenatico e dovrà avere in valigia almeno tre paia di... | [
"Solution:\n\nLa risposta è $\\mathbf{(C)}$. Osserviamo che, se Alice prende 3 buste, non può essere sicura di avere le 3 paia di calzini che le servono: ad esempio la prima busta potrebbe contenere un calzino bianco, uno giallo e uno rosso, la seconda uno bianco, uno giallo e uno verde e la terza uno bianco, uno g... | Italy | Gara di Febbraio | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | MCQ | C | |
0a30 | Milou has 100 long envelopes of different sizes. Each envelope has a width equal to one of the integers $21, \ldots, 30$ and a height equal to one of the integers $11, \ldots, 20$ and each combination occurs exactly once. Milou wants to organise the envelopes into piles. An envelope may only be placed on top of another... | [] | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 10 | |
0i7q | Problem:
Show that every consistent 2-configuration of order 4 on a finite set $A$ has a subset that is a consistent 2-configuration of order 2. | [
"Solution:\nFirst, assume the 2-configuration has just one cell. We claim there exists a sequence $a_{0}, a_{1}, \\ldots, a_{n}$ of elements of $A$ (not necessarily all distinct) such that the list\n$$\n\\{a_{0}, a_{1}\\}, \\{a_{1}, a_{2}\\}, \\ldots, \\{a_{n-1}, a_{n}\\}, \\{a_{n}, a_{0}\\}\n$$\ncontains each elem... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0675 | Let $\triangle ABC$ be a triangle and let $M$ the middle of the side $BC$. Externally of the triangle we consider parallelogram $BCDE$, such that $BE \parallel AM$ and $BE = AM / 2$. Prove that the line $EM$ passes from the middle point of the segment $AD$.
 | [
"We extend $AM$ till it meets $ED$ at point $N$. Then $BMNE$ and $MCDN$ are parallelograms and hence $EN = BM = MC = ND$. Hence $N$ is the middle of $ED$. Moreover we observe that $\\frac{AM}{MN} = 2$ and $M$ lie on the median of the triangle $EAD$. Hence $M$ is the centroid of the triangle $AED$. Therefore the lin... | Greece | 31st Hellenic Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
06y8 | On a board with $2024$ rows and $2023$ columns, Turbo the snail tries to move from the first row to the last row. On each attempt, he chooses to start on any cell in the first row, then moves one step at a time to an adjacent cell sharing a common side. He wins if he reaches any cell in the last row. However, there are... | [
"First we demonstrate that there is no winning strategy if Turbo has $2$ attempts.\nSuppose that $(2, i)$ is the first cell in the second row that Turbo reaches on his first attempt. There can be a monster in this cell, in which case Turbo must return to the first row immediately, and he cannot have reached any oth... | IMO | IMO2024 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 3 | |
007u | Given several integers, it is allowed to replace two of them by their nonnegative difference. The operation is repeated until only one number remains. If the initial numbers are $1, 2, \ldots, 2010$, what can be the last number remaining? | [
"The operation replaces $a$ and $b$ by $|b-a|$ which is even if $a$ and $b$ have the same parity and odd otherwise. So the number $N$ of odd numbers either remains unchanged or decreases by $2$ after each step. Initially $N$ is odd ($N = 1005$), so the last number will be odd, and clearly between $1$ and $2010$. Co... | Argentina | National Olympiad of Argentina | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | All odd integers from 1 to 2009 inclusive | |
04r3 | In a group of 21 different integers, a sum of arbitrary eleven ones is greater than a sum of the remaining ten numbers.
a) Prove that every considered number is greater than 100.
b) Find all such groups of 21 different integers containing number 101. | [
"a. Let the numbers be $a_1 < a_2 < a_3 < \\dots < a_{21}$. Since they are integers, for every $i \\in \\{1, 2, \\dots, 20\\}$ it holds that $a_{i+1} - a_i \\ge 1$, and therefore $a_{i+10} - a_i \\ge 10$ for every $i \\in \\{1, 2, \\dots, 11\\}$.\n\nThe problem condition is fulfilled if and only if the sum\n$$\na_1... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) Every number in the set is greater than 100.
b) Exactly those sets consisting of 101 together with any 20 consecutive integers all greater than 101; in such sets the sum of the 11 smallest exceeds the sum of the 10 largest by 1. | |
0hoz | Problem:
In this fragment of a computer keyboard, the keys are congruent squares touching along their edges, and each letter refers to the point at the center of the corresponding key. Prove that triangles $Q A Z$ and $E S Z$ have the same area.
 | [
"Solution:\n\nLet us use measuring units in which the side length of each key is $1$. We express the area of quadrilateral $Q A Z E$ in two ways:\n\na. By dividing into triangles $Q A Z$ and $Q Z E$. Since $\\triangle Q Z E$ has base $Q E = 2$ and height $2$, we get\n$$\n\\text{Area } Q A Z E = \\text{Area } Q A Z ... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
07nc | Prove for all positive real numbers $a$, $b$, $c$, $d$ that
$$
\frac{a^2}{b+c+d} + \frac{b^2}{a+c+d} + \frac{c^2}{a+b+d} \ge \frac{4a+4b+4c-3d}{9}.
$$ | [
"For arbitrary real numbers $x$, $y$ we have $9x^2 - 6xy + y^2 = (3x - y)^2 \\ge 0$. Hence, if $y > 0$, $\\frac{x^2}{y} \\ge \\frac{6x-y}{9}$ with equality iff $y = 3x$. Hence,\n$$\n\\frac{a^2}{b+c+d} + \\frac{b^2}{a+c+d} + \\frac{c^2}{a+b+d} \\\\\n\\ge \\frac{6a-b-c-d}{9} + \\frac{6b-a-c-d}{9} + \\frac{6c-a-b-d}{9... | Ireland | Ireland | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0h95 | A sequence $(x_n)$ satisfies the following conditions: $x_1 = a$, $x_{n+1} = \frac{1}{2}\left(x_n - \frac{1}{x_n}\right)$, $n \in \mathbb{N}$. Prove that there exists a number $a$ such that the sequence $(x_n)$ has exactly 2018 pairwise distinct elements.
(If one of the elements of the sequence equals 0, then the seque... | [
"Let us denote $x_1 = a = \\operatorname{ctg} \\alpha$. Then\n$$\nx_2 = \\frac{1}{2}\\left(x_1 - \\frac{1}{x_1}\\right) = \\frac{1}{2}(\\operatorname{ctg} \\alpha - \\operatorname{tg} \\alpha) = \\frac{1}{2} \\cdot \\frac{\\cos^2 \\alpha - \\sin^2 \\alpha}{\\sin \\alpha \\cdot \\cos \\alpha} = \\frac{\\cos 2\\alpha... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
064i | Problem:
Für eine feste positive ganze Zahl $m$ sei $A$ eine Teilmenge von $\{0,1,2, \ldots, 5^{m}\}$, die aus $4m+1$ Elementen besteht.
Beweisen Sie, dass es in $A$ stets drei Zahlen $a, b, c$ gibt, für die $a < b < c$ und $c + 2a > 3b$ gilt. | [
"Solution:\n\nWir nehmen an, dass es $4m+2$ Elemente $x_{0} < x_{1} < \\ldots < x_{4m+1}$ aus $\\{0,1,2, \\ldots, 5^{m}\\}$ gibt, für welche die Behauptung nicht erfüllt ist. Dann gilt insbesondere $x_{4m+1} + 2x_{i} \\leq 3x_{i+1}$ für alle $i = 0, 1, \\ldots, 4m-1$. Umformen ergibt $x_{4m+1} - x_{i} \\geq \\frac{... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade 2022 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null |
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