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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
07n4 | Given $n$ points $a_1, a_2, \dots, a_n$ in the complex plane, determine a point the sum of whose squared distances from the points and the real axis is a minimum. | [
"Let $z$ stand for an arbitrary complex number and denote by $\\operatorname{Im}(z)$ its imaginary part. Then we want to minimize the expression\n$$\nf(z) = |\\operatorname{Im}(z)|^2 + \\sum_{k=1}^{n} |z - a_k|^2.\n$$\nIf $m$ denotes the centroid of the given points, so that $m = \\frac{1}{n} \\sum_{k=1}^{n} a_k$, ... | Ireland | Ireland | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | If m = a + i b is the centroid of the given points, the minimizing point is z* = a + i · n b/(n + 1). | |
0czk | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
2 f(x) = f(x + y) + f(x + 2y)
$$
for all $x \in \mathbb{R}$ and for all $y \geq 0$. | [
"Without loss of generality, we may assume that $f(0) = 0$. Let $y > 0$ and $n$ a positive integer. For $x = n y$ we get\n$$\n2 f(n y) = f((n + 1) y) + f((n + 2) y).\n$$\nThe sequence $a_n = f(n y), n = 0, 1, 2, \\ldots$, satisfies the second order linear recursive relation\n$$\na_{n+2} = -a_{n+1} + 2 a_n\n$$\nwith... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | All constant functions | |
04k6 | Let $a$ and $b$ be integers of different parity. Prove that there exists an integer $c$ such that the numbers $ab + c$, $a + c$ and $b + c$ are squares of integers. | [
"For arbitrary $a$ and $b$ of different parity define\n$$\nc = \\frac{1 + a^2 + b^2 - 2a - 2b - 2ab}{4}.\n$$\nNote that $c$ is an integer, since $a$ and $b$ are of different parity (for example, if $a$ is even and $b$ is odd, then $1 - 2b + b^2 = (1 - b)^2$ and $a^2 - 2a - 2ab$ are divisible by $4$).\nThen\n$$\n\\b... | Croatia | Mathematical competitions in Croatia | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0epr | If $4350$ is written as a product of its prime factors, then the largest prime factor is
(A) $5$ (B) $13$ (C) $19$ (D) $29$ (E) $31$ | [
"Clearly $4350$ is divisible by $5$, since the last two digits are $50$. It is also divisible by $3$, since the sum of the digits is divisible by $3$. Dividing by $150$ gives a quotient of $29$, which is prime, so the prime factorization is $4350 = 2 \\times 3 \\times 5^2 \\times 29$, and the largest prime factor i... | South Africa | South African Mathematics Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | MCQ | D | |
0bhh | Let $A$ and $B$ be two matrices from $M_3(C)$, such that $A^2 = AB + BA$. Prove that the matrix $AB - BA$ is singular. | [
"Let $A, B \\in M_3(\\mathbb{C})$ such that $A^2 = AB + BA$.\n\nLet $X = AB - BA$.\n\nWe want to prove that $X$ is singular, i.e., $\\det(X) = 0$.\n\nLet us compute $\\operatorname{Tr}(X)$:\n\n$\\operatorname{Tr}(AB - BA) = \\operatorname{Tr}(AB) - \\operatorname{Tr}(BA) = 0$,\n\nsince $\\operatorname{Tr}(AB) = \\o... | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof only | null | |
0k0z | Problem:
Let $m > 1$ be a fixed positive integer. For a nonempty string of base-ten digits $S$, let $c(S)$ be the number of ways to split $S$ into contiguous nonempty strings of digits such that the base-ten number represented by each string is divisible by $m$. These strings are allowed to have leading zeroes.
In term... | [
"Solution:\nAnswer: $0$ and $2^{n}$ for all nonnegative integer $n$\n\nFirst, we note that $c(1) = 0$ and $c(00 \\ldots 0) = 2^{n-1}$ if there are $n$ zeroes in the string. Now we show that these are the only possibilities. Note that a split can be added if and only if the string before this split (ignoring all oth... | United States | HMIC 2018 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 0 and 2^n for all nonnegative integers n | |
0afj | Сашка имала 3 корпи со јаболка. Во корпите имало 12, 14 и 22 јаболка. Дозволено е Сашка да избере две корпи, од трите корпи кои ги имала, и да префрлува јаболка од едната во другата корпа. Притоа, мора да префрли од една во друга корпа што ги избрала онолку јаболка колку што има во корпата во која ги додава(префла) јаб... | [
"Најпрвин Сашка ги избрала корпите со 22 и 14 јаболка. Од корпата во која има 22 јаболка префрлила 14 јаболка во корпата со 14 јаболка и после првото префрлување, во корпите има 8, 28 и 12 јаболка.\n\nВториот пат ги избрала корпите со 28 и 12 јаболка. Од корпата со 28 јаболка префрлила 12 јаболка во корпата со 12 ј... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Macedonian, English | final answer only | One way: first choose the baskets with twenty two and fourteen, and move fourteen from the larger to the smaller to get eight, twenty eight, and twelve. Next choose the baskets with twenty eight and twelve, and move twelve from the larger to the smaller to get sixteen, twenty four, and eight. Finally choose the baskets... | |
0co7 | Two distinct real numbers $a$ and $b$ are chosen in such a way that the equation
$$
(x^2 + 20a x + 10b)(x^2 + 20b x + 10a) = 0
$$
has no real roots. Prove that the number $20(b-a)$ is not an integer.
Различные действительные числа $a$ и $b$ таковы, что уравнение
$$
(x^2 + 20a x + 10b)(x^2 + 20b x + 10a) = 0
$$
не имее... | [
"10.5. См. решение задачи 9.5."
] | Russia | Final round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English; Russian | proof only | null | |
0l0v | Let $\triangle ABC$ be a triangle with integer side lengths and the property that $\angle B = 2\angle A$. What is the least possible perimeter of such a triangle?
(A) 13 (B) 14 (C) 15 (D) 16 (E) 17 | [
"Let $a$, $b$, and $c$ be the lengths of the sides opposite vertices $A$, $B$, and $C$, respectively. Note that $a < b$. Applying the Law of Sines in $\\triangle ABC$, together with the identities $\\sin B = \\sin(2A) = 2 \\sin A \\cos A$ and\n$$\n\\sin C = \\sin(\\pi - 3A) = \\sin(3A) = (\\sin A)(-1 + 4\\cos^2 A),... | United States | 2024 AMC 12 B | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | C | |
0kpx | Problem:
Let $P(x)$ be a quadratic polynomial with real coefficients. Suppose that $P(1)=20$, $P(-1)=22$, and $P(P(0))=400$. Compute the largest possible value of $P(10)$. | [
"Solution:\nLet $P(x) = a x^{2} + b x + c$. The given equations give us:\n$$\n\\begin{aligned}\n& a + b + c = 20 \\\\\n& a - b + c = 22\n\\end{aligned}\n$$\nHence $b = -1$, $a + c = 21$, and so the final equation gives us $a c^{2} = 400$. Substituting $a = 21 - c$ and solving the cubic in $c$, we get $c = -4, 5, 20... | United States | HMMT November 2022 | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 2486 | |
01qs | Find all pairs $(a; b)$ of integers $a$ and $b$ satisfying the equality $(b^2 + 7(a - b))^2 = a^3 b$. | [
"Answer: $a = b = t$, $t \\in \\mathbb{Z}$, $(a; b) = (-18; -2)$, $(a; b) = (0; 7)$, $(a; b) = (12; 3)$.\n\nWe rearrange the initial equation $(b^2 + 7(a-b))^2 = a^3 b$:\n$$\n\\begin{align*}\n(b^2 + 7(a-b))^2 &= a^3 b \\Leftrightarrow b^4 + 14b^2(a-b) + 49(a-b)^2 = a^3 b \\Leftrightarrow \\\\\n&\\qquad a^3 b - b^4 ... | Belarus | Final Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | All pairs with a equal to b for any integer, plus the pairs (-18, -2), (12, 3), and (0, 7). | |
0387 | Problem:
Let $n \geq 3$ be a positive integer and $M$ be the set of the first $n$ prime numbers. For every nonempty subset $X$ of $M$ denote by $P(X)$ the product of the elements of $X$. Let $N$ be a set of fractions of the form $\frac{P(A)}{P(B)}$, where $A \subset M$, $B \subset M$, $A \cap B = \varnothing$ such that... | [
"Solution:\nConsider the following three element sets\n$$\n\\begin{aligned}\nN_{1} & = \\left\\{ \\frac{p_{3} p_{4} \\ldots p_{n-1}}{p_{1}}, \\frac{p_{2} p_{3} \\ldots p_{n-1}}{p_{1}}, \\frac{p_{2} p_{3} \\ldots p_{n-1} p_{n}}{p_{1}} \\right\\} \\\\\nN_{2} & = \\left\\{ \\frac{p_{1} p_{4} \\ldots p_{n-1}}{p_{2}}, \... | Bulgaria | Team selection test for 47. IMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | Maximum size equals 3n−2 for n in {3, 4, 5, 6}, and equals 3n−3 for all n at least 7. | |
075n | Show that there exist infinitely many pairs $(a, b)$ of positive integers with the property that $a + b$ divides $ab + 1$, $a - b$ divides $ab - 1$, $b > 1$ and $a > b\sqrt{3} - 1$. | [
"Observe $b^2 - 1 = b(a+b) - (ab+1)$ and $b^2 - 1 = b(a-b) - (ab-1)$. Hence $a+b$ and $a-b$ both divide $b^2 - 1$. Thus $\\text{lcm}(a+b, a-b)$ divides $b^2 - 1$. Since both are positive, $\\text{lcm}(a+b, a-b) \\le b^2 - 1$.\nLet $d = \\text{gcd}(a, b)$. Then $d|ab$ and $d|a+b|ab+1$. Hence $d|1$ showing $d=1$. If ... | India | Indija TS 2012 | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis... | English | proof only | null | |
0f2k | Problem:
Two players play a game by moving a piece on an $n \times n$ chessboard. The piece is initially in a corner square. Each player may move the piece to any adjacent square (which shares a side with its current square), except that the piece may never occupy the same square twice. The first player who is unable ... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | Starting at a corner: for even board size the first player has a winning strategy; for odd board size the second player has a winning strategy. If the initial square is adjacent to the corner, the outcome reverses: for even board size the second player wins, and for odd board size the first player wins. | |
029t | Problem:
Soma dos terminados em 9 - A soma $S_{n}=9+19+29+39+\cdots+a_{n}$ denota a soma dos primeiros $n$ números naturais terminados em $9$. Qual é o menor valor de $n$ para que $S_{n}$ seja maior do que $10^{5}$? | [
"Solution:\n\nNote que $S_{n}$ é a soma dos $n$ primeiros termos de uma progressão aritmética cujo primeiro termo é $a_{1}=9$ e a razão é $r=10$. Substituindo esses dados na fórmula $a_{n}=a_{1}+(n-1) r$ obtemos $a_{n}=9+10(n-1)$. Por outro lado, note que:\n\n$$\n\\begin{aligned}\n& 9=9+0 \\cdot 10 \\\\\n& 19=9+1 \... | Brazil | Lista 2 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 142 | |
0k4h | Problem:
Let $ABC$ be an equilateral triangle of side length $1$. For a real number $0 < x < 0.5$, let $A_1$ and $A_2$ be the points on side $BC$ such that $A_1B = A_2C = x$, and let $T_A = \triangle AA_1A_2$. Construct triangles $T_B = \triangle BB_1B_2$ and $T_C = \triangle CC_1C_2$ similarly.
There exist positive ... | [
"Solution:\n\nNotice that the given expression is defined and continuous not only on $0 < x < 0.5$, but also on $0 \\leq x \\leq 0.5$. Let $f(x)$ be the function representing the area of the (possibly degenerate) hexagon for $x \\in [0, 0.5]$. Since $f(x)$ is equal to the given expression over $(0, 0.5)$, we can co... | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | (8, 2) | |
0l14 | What is the number of ordered triples $(a, b, c)$ of positive integers, with $a \le b \le c \le 9$, such that there exists a (non-degenerate) triangle $\triangle ABC$ with an integer inradius for which $a, b$, and $c$ are the lengths of the altitudes from $A$ to $\overline{BC}$, $B$ to $\overline{AC}$, and $C$ to $\ove... | [
"Let $x, y$, and $z$ be the lengths of $\\overline{BC}$, $\\overline{AC}$, and $\\overline{AB}$, respectively. Let $r$ be the inradius of $\\triangle ABC$. Then\n$$\n\\text{Area}(\\triangle ABC) = \\frac{1}{2}xa = \\frac{1}{2}yb = \\frac{1}{2}zc = \\frac{1}{2}(x + y + z)r.\n$$\nTherefore $x = \\frac{2\\text{Area}(\... | United States | 2024 AMC 12 B | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | B | |
03t6 | Let $X$ be a set of 56 elements. Find the least positive integer $n$ such that for any 15 subsets of $X$, if the union of every 7 sets of these subsets contains at least $n$ elements, then there exist 3 of the 15 subsets whose intersection is nonempty. (posed by Leng Gangsong) | [
"Suppose there exist 15 subsets of $X$, the union of every 7 of these 15 subsets contains at least 41 elements, and there exists no 3 of the 15 subsets whose intersection is nonempty.\nSince every element of $X$ can only belong 2 of the 15 subsets, we can suppose that every element of $X$ belongs to exactly 2 of th... | China | China Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 41 | |
0d9u | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f\left(2 x^{3}+f(y)\right)=y+2 x^{2} f(x),
$$
for all real numbers $x, y$. | [
"Denote ($*$) as the given condition. Replacing $x = y = -\\frac{1}{2}$ in ($*$), we get\n$$\nf\\left(2\\left(f\\left(-\\frac{1}{2}\\right)\\right)^{2}-\\frac{1}{2}\\right)=0\n$$\nIf there exists some $a \\neq 0$ such that $f(a)=0$. Substituting $x=a$ in ($*$), we get $f(y)=0$ for any $y$, which is obviously a solu... | Saudi Arabia | Team selection tests for BMO 2018 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | f(x) = 0; f(x) = x; f(x) = -x | |
0ggp | 設 $\mathbb{N}$ 代表所有正整數的集合，$\mathbb{N}^2$ 代表正整數數對的集合。在所有從 $\mathbb{N}^2$ 到 $\mathbb{N}$ 的函數中，對於所有非負整數 $k$，我們遞迴定義 $k$ 級好函數如下：
(i) 函數 $f(a,b) = a$、函數 $f(a,b) = b$ 都是 0 級好函數。
(ii) 若 $f(a,b), g(a,b)$ 分別為 $p$ 級、$q$ 級好函數，則 $\text{gcd}(f(a,b), g(a,b))$ 為 $(p+q)$ 級好函數，且 $f(a,b)g(a,b)$ 為 $(p+q+1)$ 級好函數。
證明：若對於某個正整數 $n \ge 3$，$f(... | [
"Let $N = \\binom{n}{3}$. Let's first show that the statement holds for $t \\le \\binom{N+3}{2}$ and all good functions of order at most $N+1$. This can be proved by induction on $N$. It is trivial when $N=0$. Now suppose that the statement holds for all smaller $N$, then for any good function with $N$ multiplicati... | Taiwan | 2022 數學奧林匹亞競賽第三階段選訓營, 獨立研究(一) | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | Chinese; English | proof only | null | |
0djw | Mohammed has a chess board, colored white and black in the usual way and mentioned a point, which lies strictly inside one of the 64 cells. Ahmed can draw any closed broken line without self-intersections and ask Mohammed whether his point is inside or outside the broken line. How many questions does Ahmed need to ask ... | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 2 | |
01qi | $$
f(x) = \begin{cases} x + \frac{1}{2}, & \text{if } x < \frac{1}{2}, \\ x^2, & \text{if } x \ge \frac{1}{2}. \end{cases}
$$
Let $a$ and $b$ be two real numbers such that $0 < a < b < 1$. We define the sequences $a_n$ and $b_n$ by $a_0 = a, b_0 = b$, and
$$
a_n = f(a_{n-1}),\ b_n = f(b_{n-1}) \text{ for } n > 0.
$$
Sh... | [
"2. See IMO-2014 Shortlist, Problem A2."
] | Belarus | SELECTION and TRAINING SESSION | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0hls | Problem:
The country of Squareland is shaped like a square and is divided into $64$ congruent square cities. We want to divide Squareland into states and assign to each state a capital city so that the following rules are satisfied:
a. Every city lies entirely within one state.
b. Given any two states, the numbers o... | [
"Solution:\n\nIn the diagram below, no city shares a corner with any two of the cities marked X. Therefore the nine X's are\n\nin nine different states. The diagram at right shows that nine states are also sufficient ( $*$ denotes capital).\n"
] | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 9 | |
038o | In acute $\triangle ABC$ denote by $M$ and $N$ the midpoints of the altitudes $BB_1$ and $CC_1$, respectively, $P = AM \cap CC_1$ and $Q = AN \cap BB_1$. Prove that:
a) the points $M, N, P$ and $Q$ are concyclic;
b) if the points $B, C, P$ and $Q$ are concyclic then $\triangle ABC$ is isosceles. | [
"a) Since $\\triangle ACC_1 \\sim \\triangle ABB_1$ and $AN$ and $AM$ are medians in these triangles we have\n$$\n\\asymp ANC_1 = \\asymp AMB_1 \\Rightarrow \\asymp QNB = \\asymp PMQ,\n$$\ni.e. the points $M, N, P$ and $Q$ are concyclic.\n\nb) If the points $B, C, P$ and $Q$ are concyclic then $\\asymp QCP = \\asym... | Bulgaria | Winter Mathematical Competition | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
05d6 | Problem:
There are $2017$ lines in a plane such that no $3$ of them go through the same point. Turbo the snail can slide along the lines in the following fashion: she initially moves on one of the lines and continues moving on a given line until she reaches an intersection of $2$ lines. At the intersection, she follow... | [
"Solution:\n\nWe show that this is not possible.\nThe lines divide the plane into disjoint regions. We claim that there exists an alternating $2$-coloring of these regions, that is each region can be colored in black or white, such that if two regions share a line segment, they have a different color. We show this ... | European Girls' Mathematical Olympiad (EGMO) | null | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | No | |
06do | Find the greatest real $K$ such that for every positive $u$, $v$ and $w$ with $u^2 > 4vw$, the inequality
$$
(u^2 - 4vw)^2 > K(2v^2 - uw)(2w^2 - uv)
$$
holds. Justify your claim. | [
"The greatest $K$ is $16$.\n\nWe first prove the inequality when $K = 16$. Note that\n$$\n\\begin{aligned}\nu^2 - 4vw &= u^2 + 2vw - 6vw \\\\\n&\\ge u^2 + 2vw - 3(v^2 + w^2) \\\\\n&= u^2 + (v+w)^2 - 4(v^2 + w^2) \\\\\n&\\ge 2u(v+w) - 4(v^2 + w^2) \\\\\n&= 2(uw - 2v^2) + (uv - 2w^2).\n\\end{aligned}\n$$\nIf $uw - 2v... | Hong Kong | CHKMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 16 | |
01p2 | Let $\mathbb{Z}$ and $\mathbb{Q}$ be the sets of integers and rationals respectively.
a) Does there exist a partition of $\mathbb{Z}$ into three non-empty subsets $A, B, C$ such that the sets $A+B$, $B+C$, $C+A$ are disjoint?
b) Does there exist a partition of $\mathbb{Q}$ into three non-empty subsets $A, B, C$ such ... | [] | Belarus | BelarusMO 2013_s | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Modular Arithmetic",
"Algebra > Abstract Algebra > Other"
] | null | proof and answer | a) Yes. For example, take the three residue classes modulo three.
b) No. | |
06bz | Five numbers $1$, $2$, $3$, $4$, $5$ are written on a blackboard. A student may erase any two of the numbers $a$ and $b$ on the board and write the numbers $a+b$ and $ab$ replacing them. If this operation is repeatedly performed, can the numbers $21$, $27$, $64$, $180$, $540$ ever appear on the board at the same time? | [
"No. We consider the numbers modulo $3$.\n* If $3 \\mid a, b$, then $3 \\mid a + b$ and $3 \\mid ab$. The number of multiples of $3$ remains unchanged.\n* If $3 \\mid a$ and $3 \\nmid b$, then $3 \\nmid a + b$ and $3 \\mid ab$. The number of multiples of $3$ remains unchanged.\n* If $a \\equiv b \\equiv \\pm 1 \\pm... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | No | |
04dz | Let $ABC$ be a right triangle with height $\overline{CN}$. If $|AC| = |BN| = 1$, determine the length of the hypotenuse $\overline{AB}$. | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | (1 + sqrt(5)) / 2 | |
0ibu | Problem:
Find the largest integer $n$ such that $3^{512}-1$ is divisible by $2^{n}$. | [
"Solution:\nWrite\n$$\n\\begin{aligned}\n3^{512}-1 & =\\left(3^{256}+1\\right)\\left(3^{256}-1\\right)=\\left(3^{256}+1\\right)\\left(3^{128}+1\\right)\\left(3^{128}-1\\right) \\\\\n& =\\cdots=\\left(3^{256}+1\\right)\\left(3^{128}+1\\right) \\cdots(3+1)(3-1)\n\\end{aligned}\n$$\nNow each factor $3^{2^{k}}+1$, $k \... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | 11 | |
05bk | How many integers $k$ are there for which $1 \le k \le n$ and the sum $1 + 2 + \dots + k$ is divisible by $n$, if
a) $n = 2023^{2024}$;
b) $n = 2024^{2023}$? | [
"Note that $1+2+\\dots+k = \\frac{k(k+1)}{2}$.\n\na) Since $n$ is odd, the number $\\frac{k(k+1)}{2}$ is divisible by $n$ if and only if $k(k+1)$ is divisible by $n$.\n\nSince $2023 = 7 \\cdot 17^2$, the number $n$ can be expressed as $p_1^{\\alpha_1} p_2^{\\alpha_2}$, where $p_1 = 7$, $p_2 = 17$ and $\\alpha_1, \\... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | a) 4; b) 3 | |
01ub | Find all possible values of a real number $a$ such that there exist pairwise distinct nonzero real numbers $x, y, z$ satisfying the equalities
$$
x - \frac{y}{z} - \frac{z}{y} = y - \frac{z}{x} - \frac{x}{z} = z - \frac{x}{y} - \frac{y}{x} = a.
$$ | [
"Answer: $a = 1$.\n\nLet $A = x - \\frac{y}{z} - \\frac{z}{y}$, $B = y - \\frac{z}{x} - \\frac{x}{z}$, and $C = z - \\frac{x}{y} - \\frac{y}{x}$. By condition, it follows that\n$$\n0 = A - B = (x - y) + \\frac{x - y}{z} - \\frac{z(x - y)}{xy} = (x - y) \\left( 1 + \\frac{1}{z} - \\frac{z}{xy} \\right) = \\frac{(x -... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Other"
] | English | proof and answer | 1 | |
08id | Problem:
In the space a geometrical configuration, which includes $n$ ($n \geq 3$) distinct points, is given. A point $A$ of this configuration has the following properties: if $A$ is excluded from the configuration, then among the remaining points there are no colinear points; after the elimination of $A$ from the co... | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 9, 22, 30 | |
07oq | In a triangle $ABC$ the internal bisector of the angle $\angle ABC$ meets the external bisector of $\angle BCA$ at $D$. The circumcircle of the triangle $ABC$ cuts $BD$ at $E$. Prove that $E$ is the circumcentre of the triangle $ACD$. | [
"Because $BD$ bisects $\\angle ABC$, we have $|EA| = |EC|$.\n\n\n\nLet $I$ be the incentre of $\\triangle ABC$, then $\\angle DCI = 90^\\circ$.\n\nWe have $\\angle EIC = \\angle ICB + \\angle IBC = \\angle ACI + \\angle ABE = \\angle ACI + \\angle ACE = \\angle ECI$.\n\nThus $\\angle ECD = ... | Ireland | Irska 2014 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a4k | Problem:
A positive integer is called sparkly if it has exactly 9 digits, and for any $n$ between $1$ and $9$ (inclusive), the $n^{\text{th}}$ digit is a positive multiple of $n$. How many positive integers are sparkly? | [
"Solution:\nFor each $n = 1, 2, \\ldots, 9$ there are $\\lfloor 9 / n \\rfloor$ different possibilities for the $n^{\\mathrm{th}}$ digit. For example, there are $\\lfloor 9 / 2 \\rfloor = 4$ possible choices for the second digit (these being $2, 4, 6$ and $8$). Therefore the answer is\n$$\n\\left\\lfloor \\frac{9}{... | New Zealand | New Zealand Mathematical Olympiad | [
"Discrete Mathematics > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 216 | |
0clu | Let $ABC$ be an acute triangle with $AB < AC$ and let $H$ and $O$ be its orthocentre and circumcentre, respectively. Let $\Gamma$ be the circle $BOC$. The line $AO$ and the circle of radius $AO$ centred at $A$ cross $\Gamma$ again at $A'$ and $F$, respectively. Prove that $\Gamma$, the circle on diameter $AA'$ and the ... | [
"Consider the composition $\\iota$ of an inversion centered at $A$ and the reflection in the bisector $AM$ that swaps $B$ and $C$. Then $\\iota$ swaps the circle $\\Omega$ with the line $BC$, hence it swaps $O$ with the reflection $L$ of $A$ in $BC$. Hence $\\iota(\\Gamma)$ is the circle $\\Gamma^* = (BCL)$,\n\ni.e... | Romania | Seventeenth ROMANIAN MASTER OF MATHEMATICS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Quadrilaterals > ... | English | proof only | null | |
0cwk | Find the least $k$ satisfying the following condition: for any polynomial $f(x)$ of degree $100$ with real coefficients there exists a polynomial $g(x)$ of degree not greater than $k$ with real coefficients such that the graphs $y = f(x)$ and $y = g(x)$ have exactly $100$ common points. | [
"**Answer.** $98$.\n\nSet $n = 100$.\n\n1. We will show how to construct a polynomial $g$ of degree at most $n-2$ for a given polynomial $f$ of degree $n$. Multiplying $f$ by a nonzero constant doesn't change the condition (we can multiply $g$ by the same constant), so we assume the leading coefficient of $f$ is $1... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Russian | proof and answer | 98 | |
003j | Sea $a$ un número real tal que $\frac{1}{a} = a - [a]$. Demostrar que $a$ es irracional. | [] | Argentina | Argentina 2006 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | Español | proof only | null | |
06dp | Let $ABCDEF$ be a regular hexagon of side length $1$, and $O$ be the centre of the hexagon. In addition to the sides of the hexagon, line segments are drawn from $O$ to each vertex, making a total of twelve unit line segments. Find the number of paths of length $2003$ along these line segments that start at $O$ and ter... | [
"The answer is $\\frac{7 - \\sqrt{7}}{14}(1 + \\sqrt{7})^{2003} + \\frac{7 + \\sqrt{7}}{14}(1 - \\sqrt{7})^{2003}$.\n\nFor each integer $n \\ge 0$, let $a_n$ be the number of paths of length $n$ starting at $O$ and terminating at $O$. Also, let $b_n$ be the number of paths of length $n$ starting at $O$ and terminat... | Hong Kong | CHKMO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | (7 - sqrt(7))/14 * (1 + sqrt(7))^2003 + (7 + sqrt(7))/14 * (1 - sqrt(7))^2003 | |
0bqa | Problem:
În triunghiul oarecare $ABC$, se consideră $M$ şi $N$ mijloacele segmentelor $BC$, respectiv $AM$, punctul $D$ simetricul punctului $C$ faţă de $A$, $BN \cap AC = \{S\}$, $DM \cap AB = \{T\}$ și punctul $P$ mijlocul segmentului $[SC]$.
a) Demonstrați că $AC = 3PC$.
b) Demonstrați că dreptele $ST$ și $BC$ su... | [] | Romania | OLIMPIADA DE MATEMATICĂ - ETAPA LOCALĂ | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof and answer | 4 cm^2 | |
0kz9 | There are real numbers $x$, $y$, $h$, and $k$ that satisfy the system of equations
$$
\begin{aligned}
x^2 + y^2 - 6x - 8y &= h \\
x^2 + y^2 - 10x + 4y &= k.
\end{aligned}
$$
What is the minimum possible value of $h + k$?
(A) -54 (B) -46 (C) -34 (D) -16 (E) 16 | [
"**Answer (C):** Adding the two equations and then completing the squares gives\n$$\n2(x - 4)^2 + 2(y - 1)^2 = h + k + 32 + 2.\n$$\nTo ensure a real solution, it follows that $h + k$ is at least $-34$. This solution can be obtained by setting $x = 4$ and $y = 1$, in which case $h = 4^2 + 1^2 - 6 \\cdot 4 - 8 \\cdot... | United States | 2024 AMC 12 B | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | C | |
05vy | Problem:
Soit $A, B, C$ et $D$ quatre points situés sur un cercle $\Omega$. Soit $E$ et $F$ les points d'intersection des demi-droites $[B A)$ et $[B C)$ avec la tangente à $\Omega$ en $D$. Soit $T$ un point, situé à l'intérieur du triangle $A B C$, tel que $(T E)$ soit parallèle à $(C D)$ et que $(T F)$ soit parallèl... | [
"Solution:\n\nSoit $P$ et $Q$ les points d'intersection des droites $(T E)$ et $(T F)$ avec $(A C)$. Puisque $(E D)$ est tangente à $\\Omega$, on sait que\n$$\n(P A, P E)=(C A, C D)=(D A, D E)\n$$\ndonc que les points $A, D, E$ et $P$ appartiennent à un même cercle $\\alpha$. On démontre de même que les points $C, ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07yc | Problem:
Sia $ABCD$ un quadrilatero convesso tale che $\widehat{CAB} = \widehat{CDA}$ e $\widehat{BCA} = \widehat{ACD}$. Detto $M$ il punto medio di $AB$ si dimostri che $\widehat{BCM} = \widehat{DBA}$. | [
"Solution:\n\nChiamiamo per comodità $\\widehat{BCM} = \\alpha$. Consideriamo i triangoli $\\triangle ADC$ e $\\triangle BAC$: essi hanno due angoli congruenti, e quindi anche il terzo è congruente per differenza ($\\widehat{CAD} = \\widehat{CBA}$). Per il primo criterio di similitudine sono dunque simili; in parti... | Italy | null | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
01ei | A convex quadrilateral $ABCD$ is right-angled at $A$ and fulfils $BC + CD = 1$. Determine its greatest possible area. | [
"Answer: $\\frac{1}{8}(1 + \\sqrt{2})$.\nReflect the quadrilateral in line $AB$, and reflect the resulting pentagon again in line $AD$; see Figure 1. This produces an octagon of fixed perimeter\n$$\n4(BC + CD) = 4\n$$\nand area four times that of $ABCD$. The maximal area is obtained for a regular octagon of edge le... | Baltic Way | Baltic Way shortlist | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | 1/8 * (1 + sqrt(2)) | |
0fyo | Problem:
In einem Dorf mit mindestens einem Einwohner gibt es mehrere Vereine. Jeder Einwohner des Dorfes ist Mitglied in mindestens $k$ Vereinen und je zwei verschiedene Vereine haben höchstens ein gemeinsames Mitglied. Zeige dass mindestens $k$ dieser Vereine dieselbe Anzahl Mitglieder haben. | [
"Solution:\n\nWir können annehmen, dass jeder Verein mindestens ein Mitglied besitzt, denn das Entfernen aller leeren Vereine ändert nichts an den Voraussetzungen. Wähle unter allen Vereinen einen mit maximaler Mitgliederzahl $m$. Dann ist jeder trip dieser $m$ Mitglieder noch in mindestens $k-1$ anderen Vereinen u... | Switzerland | SMO Finalrunde | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0kto | Problem:
Suppose $P(x)$ is a monic polynomial of degree $2023$ such that
$$
P(k) = k^{2023} P\left(1 - \frac{1}{k}\right)
$$
for every positive integer $1 \leq k \leq 2023$. Then $P(-1) = \frac{a}{b}$, where $a$ and $b$ are relatively prime integers. Compute the unique integer $0 \leq n < 2027$ such that $b n - a$ is ... | [
"Solution:\n\nLet $n = 2023$. If $P(x) = x^{n} + a_{n-1} x^{n-1} + \\cdots + a_{0}$, then let\n$$\nR(x) = x^{n} P\\left(1 - \\frac{1}{x}\\right) = (x-1)^{n} + a_{n-1}(x-1)^{n} x + \\cdots + a_{0} x^{n}\n$$\nThen, note that $Q(x) = P(x) - R(x)$ is a polynomial of degree at most $n$, and it has roots $1, 2, \\ldots, ... | United States | HMMT February 2022 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | 406 | |
04iz | Determine the largest positive integer $n$ such that
$$
n + 5 \mid n^4 + 1395.
$$ | [
"Let $d = n + 5$. Then $d \\mid n^4 + 1395$.\n\nWe have $n = d - 5$, so\n$$\nn^4 + 1395 = (d - 5)^4 + 1395.\n$$\nExpand $(d - 5)^4$:\n$$\n(d - 5)^4 = d^4 - 4d^3 \\cdot 5 + 6d^2 \\cdot 25 - 4d \\cdot 125 + 625 = d^4 - 20d^3 + 150d^2 - 500d + 625.\n$$\nSo\n$$\nn^4 + 1395 = d^4 - 20d^3 + 150d^2 - 500d + 625 + 1395 = d... | Croatia | Croatia Mathematical Competitions | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 2015 | |
08py | Problem:
Let $a, b, c, d$ be positive real numbers such that $a b c d = 1$. Prove the inequality
$$
\frac{1}{a^{3}+b+c+d}+\frac{1}{a+b^{3}+c+d}+\frac{1}{a+b+c^{3}+d}+\frac{1}{a+b+c+d^{3}} \leqslant \frac{a+b+c+d}{4}
$$ | [
"Solution:\nFrom the Cauchy-Schwarz Inequality, we obtain\n$$\n(a+b+c+d)^{2} \\leqslant (a^{3}+b+c+d)\\left(\\frac{1}{a}+b+c+d\\right)\n$$\nUsing this, together with the other three analogous inequalities, we get\n$$\n\\begin{aligned}\n\\frac{1}{a^{3}+b+c+d}+\\frac{1}{a+b^{3}+c+d}+\\frac{1}{a+b+c^{3}+d} & +\\frac{1... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0jwx | Problem:
On a $3 \times 3$ chessboard, each square contains a knight with $\frac{1}{2}$ probability. What is the probability that there are two knights that can attack each other? (In chess, a knight can attack any piece which is two squares away from it in a particular direction and one square away in a perpendicular... | [
"Solution:\n\nNotice that a knight on the center square cannot attack any other square on the chessboard, so whether it contains a knight or not is irrelevant.\n\nFor ease of reference, we label the other eight squares as follows:\n\n| 0 | 5 | 2 |\n| :---: | :---: | :---: |\n| 3 | X | 7 |\n| 6 | 1 | 4 |\n\nNotice t... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 209/256 | |
0gme | Let $n \ge 2$ be an integer and $(a_1, a_2, \dots, a_n)$ be a permutation of $1, 2, \dots, n$. For each $1 \le k \le n$, $a_k$ apples are placed at the point $k$ on the real axis. Children named $A, B, C$ are assigned points $x_A, x_B, x_C \in \{1, 2, \dots, n\}$, respectively. For each $k \in \{1, 2, \dots, n\}$, chil... | [] | Turkey | X. NATIONAL MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | All integers n ≥ 2 | |
09ho | If $m$ and $n$ are positive integers, can the number $m^4 + 2mn + n^2 - 2021$ be the product of three or more consecutive integers? | [
"Answer: No.\nSuppose that $N = m^4 + 2mn + n^2 - 2021$ is the product of three or more consecutive integers. Then $N$ and $m^4 - m^2 = m[(m-1)m(m+1)]$ are divisible by $3$. Thus $(m+n)^2 \\equiv N - (m^4 - m^2) + 2021 \\equiv 2 \\pmod{3}$, which is a contradiction, since $2$ is not a quadratic residue modulo $3$."... | Mongolia | Round 3 | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | No | |
0b7y | Let $A$, $B$, $C \in M_n(\mathbb{R})$ be such that $ABC = 0_n$ and $\operatorname{rank} B = 1$. Prove that $AB = O_n$ or $BC = O_n$. | [
"Let $K \\in \\mathcal{M}_{n,1}(\\mathbb{C})$ and $L \\in \\mathcal{M}_{1,n}(\\mathbb{C})$ be such that $KL = B$.\nThen $O_n = (AK)(LB)$ with $AK \\in \\mathcal{M}_{n,1}(\\mathbb{C})$, $LB \\in \\mathcal{M}_{1,n}(\\mathbb{C})$.\nLet $AK = {}^t(a_1, a_2, \\dots, a_n)$ (where ${}^t X$ denotes the transpose of $X$), a... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Vectors"
] | English | proof only | null | |
02xm | Problem:
No Calendário Jupiteriano, os meses são Julius, Uranius, Plutônius, Ílius, Terrius, Eráclitus e Raley. Os meses que começam com consoantes possuem 17 dias e os meses que começam com vogais têm 19 dias. O ano começa em Július e segue a sequência mencionada, anteriormente, encerrando-se em Raley. Assim, como no... | [
"Solution:\n\na) Se ele nasceu em 11 de Plutônio de 1.999, viveu $17-11=6$ dias neste mês, 19 dias em Ílius, 17 dias em Terrius, 19 dias em Eraclitus, 17 dias em Raley, 17 dias em Julius e 5 em Uranius, ou seja, Diógenes completará 100 dias em 5 de Uranius de 2.000.\n\nb) Cada ano neste calendário possui $17 \\cdot... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Other"
] | null | proof and answer | a) 5 de Uranius de 2000. b) segunda. | |
039w | Let $\alpha$ and $\beta$, $\alpha \neq \beta$, be acute angles such that
$$
(\cos^2 \alpha + \cos^2 \beta)(1 + \tan \alpha \tan \beta) = 2.
$$
Prove that $\alpha + \beta = 90^\circ$. | [
"Let $x = \\tan \\alpha$, $y = \\tan \\beta$.\n\nWe have $\\cos^2 \\alpha = \\dfrac{1}{1 + x^2}$ and $\\cos^2 \\beta = \\dfrac{1}{1 + y^2}$.\n\nSo,\n$$\n\\left( \\dfrac{1}{1 + x^2} + \\dfrac{1}{1 + y^2} \\right)(1 + x y) = 2.\n$$\n\nMultiply both sides by $(1 + x^2)(1 + y^2)$:\n$$\n\\left[ (1 + y^2) + (1 + x^2) \\r... | Bulgaria | Fall Mathematical Competition | [
"Precalculus > Trigonometric functions"
] | English | proof only | null | |
0h3e | 2012 people stand in a line. Each of them is either a knight (and always tells truth) or a knave (and always lies). Each of the people announced: «There are more knaves on my left than knights on my right». How many knaves are in the line? | [
"Нехай у шерензі стоять $b$ брехунів та $r$ лицарів. Оскільки найлівіший з лицарів сказав правду, маємо $b > r - 1$. Оскільки найправіший з брехунів збрехав, то маємо $b - 1 \\le r$. Отже, $r = b$ або $r = b - 1$. Другий випадок неможливий, оскільки загальна кількість людей парна. Тому $r = b = 1006$.\n\n*Відповідь... | Ukraine | Ukrainian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Logic"
] | English | proof and answer | 1006 | |
0bx8 | Determine all integers $n \ge 2$ such that $a + \sqrt{2}$ and $a^n + \sqrt{2}$ are both rational for some real number $a$ depending on $n$. | [
"There is only one such $n$, namely, $n = 2$, in which case we may take $a = 1/2 - \\sqrt{2}$. Verifications are routine and hence omitted.\n\nTo rule out the case $n \\ge 3$, let $a$ be a real number such that $a + \\sqrt{2}$ is rational. For $a^n + \\sqrt{2}$ to be rational, it is necessary and sufficient that $a... | Romania | THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factoriz... | English | proof and answer | n = 2 | |
0fnb | Sea $Q_0$ el conjunto de los números racionales mayores que cero. Sea $f: Q_0 \to \mathbb{R}$ una función que satisface las tres siguientes condiciones:
(i) $f(x)f(y) \geq f(xy)$ para todos los $x, y \in Q_0$;
(ii) $f(x + y) \geq f(x) + f(y)$ para todos los $x, y \in Q_0$;
(iii) existe un número racional $a > 1$ tal qu... | [
"Demostrar que $f(x) = x$ para todo $x \\in Q_0$.\n\nDemostramos sucesivamente las siguientes propiedades adicionales que ha de tener cualquier función $f$ que satisfaga las condiciones del enunciado:\n\n(1) $f(1) \\ge 1$. En efecto, tomando $x = 1, y = a$ en (i) y usando (iii), obtenemos que $a = f(a) \\le f(1)f(a... | Spain | Olimpiada Internacional de Matemáticas | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Spanish | proof only | null | |
00a6 | A $+1$ or $-1$ is written at each vertex of a regular $n$-gonal prism so that the product of numbers on each face is $-1$. For which $n \geq 3$ is this possible? | [
"Call *vertical* the $n$ edges that join corresponding vertices of the two bases. A vertical edge is *odd* if it has different number at its endpoints and *even* otherwise. Take two adjacent vertical edges $e_1$ and $e_2$. They determine a lateral face $F$ which is a rectangle with opposite sides $e_1$ and $e_2$. T... | Argentina | Argentine National Olympiad 2015 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | All n that are multiples of 4 | |
0ag8 | The real numbers $a, b, c, d$ satisfy simultaneously the equations
$$
abc - d = 1, \quad bcd - a = 2, \quad cda - b = 3, \quad dab - c = -6.
$$
Prove that $a + b + c + d \neq 0$. | [
"Suppose that $a + b + c + d = 0$. Then\n$$\nabc + bcd + cda + dab = 0 \\quad (1)\n$$\nIf $abcd = 0$, then one of the numbers, say $d$, must be $0$. In this case $abc = 0$, and so at least one of the numbers $a, b, c$ will be equal to $0$, making one of the given equations impossible. Hence $abcd \\neq 0$ and, from... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof only | null | |
01os | Prove that $\left(C_{2n}^{n+1}\right)^2 \le C_{n^2+n}^{n+1}$ for any $n \in \mathbb{N}$. | [
"Note that\n$$\n(t + k)^2 \\le (k + 1)(t^2 + k) \\quad (*)\n$$\nfor all integers $k$ and $t$, $0 \\le k < t$. Indeed, we have\n$$\n(*) \\Leftrightarrow t^2+2tk+k^2 \\le kt^2+t^2+k^2+k \\Leftrightarrow kt^2-2tk+k \\ge 0 \\Leftrightarrow k(t-1)^2 \\ge 0.\n$$\nIt is obvious that the last inequality holds for $0 \\le k... | Belarus | BelarusMO 2013_s | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
077p | Let $S$ be a finite set of positive integers. Assume that there are precisely 2023 ordered pairs $(x, y)$ in $S \times S$ so that the product $xy$ is a perfect square. Prove that one can find at least four distinct elements in $S$ so that none of their pairwise products is a perfect square.
*Note:* As an example, if $... | [
"Let $S$ be a finite set of positive integers such that there are exactly 2023 ordered pairs $(x, y) \\in S \\times S$ with $xy$ a perfect square.\n\nLet us analyze the structure of $S$.\n\nFor any $x, y \\in S$, $xy$ is a perfect square if and only if the product of their squarefree parts is a square. For each $x ... | India | INMO_2023 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Algebraic Number Theory > Quadratic forms",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
02lp | Emerald writes $2009^2$ integers in a $2009 \times 2009$ table, one number in each entry of the table. She sums all the numbers in each row and in each column, obtaining 4018 sums. She notices that all sums are distinct. Is it possible that all such sums are perfect squares? | [
"Yes, it is. For instance, consider the table\n\n| $(3^1 - 2)^2$ | $3(2 \\cdot 3^1 - 1)$ | $0$ | $0$ | $\\ldots$ | $0$ |\n|---|---|---|---|---|---|\n| $0$ | $(3^2 - 2)^2$ | $3(2 \\cdot 3^2 - 1)$ | $0$ | $\\ldots$ | $0$ |\n| $0$ | $0$ | $(3^3 - 2)^2$ | $3(2 \\cdot 3^3 - 1)$ | $\\ldots$ | $0$ |\n| $\\ldots$ | $\\ldot... | Brazil | XXXI Brazilian Math Olympiad | [
"Discrete Mathematics > Other",
"Number Theory > Other"
] | English | proof and answer | Yes | |
04ch | Real numbers $x$, $y$ and $z$ are consecutive terms of an arithmetic sequence. If the numbers $\cos^2 x$, $\cos^2 y$ and $\cos^2 z$ are mutually different and consecutive terms of some arithmetic sequence, determine all possible values of $y$. | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Sequences and Series"
] | English | proof and answer | y = π/4 + k·(π/2), for any integer k | |
0j5i | Problem:
For a positive integer $n$, let $p(n)$ denote the product of the positive integer factors of $n$. Determine the number of factors $n$ of $2310$ for which $p(n)$ is a perfect square. | [
"Solution:\n\nAnswer: $27$\n\nNote that $2310 = 2 \\times 3 \\times 5 \\times 7 \\times 11$. In general, we see that if $n$ has $d(n)$ positive integer factors, then $p(n) = n^{\\frac{d}{2}}$ since we can pair factors $(d, \\frac{n}{d})$ which multiply to $n$. As a result, $p(n)$ is a square if and only if $n$ is a... | United States | Harvard-MIT November Tournament | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 27 | |
09ka | Let $ABCD$ be a quadrilateral with $\angle DAB = \angle ABC$. The circumcircle of triangle $ABC$ intersects segment $AD$ at point $K$ and segment $CD$ at point $L$. Segment $AL$ intersects segment $CK$ at point $P$. If $\angle ADB = \angle PDC$, prove that $CP = CL$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof only | null | |
023t | Problem:
Vladimir escolheu três algarismos $a, b$ e $c$ tais que $a > b > c > 0$ e com eles formou os números $abc$, $cba$ e $cab$. Note que $abc$ não é o produto de $a, b$ e $c$, mas sim o número de algarismos $a, b$ e $c$. Por exemplo, se $a = 1$, $b = 2$ e $c = 3$, $abc$ será o número 123.
Depois de escolher estes t... | [
"Solution:\nComo $a > c$, o número $abc$ é maior que os números $cba$ e $cab$, e então devemos ter $abc = cba + cab$. Assim,\n$$\n100 \\cdot a + 10 \\cdot b + c = (100 \\cdot c + 10 \\cdot b + a) + (100 \\cdot c + 10 \\cdot a + b)\n$$\ne então $89a = 199c + b$.\nNote que $10 > a = (199 \\cdot c + b)/89 > 2c$, e que... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 954, 459, 495 | |
0cc6 | Let $ABC$ be an acute-angled triangle with $BC > AB$, such that the points $A$, $H$, $I$ and $C$ are concyclic (where $H$ is the orthocenter and $I$ is the incenter of triangle $ABC$). The line $AC$ intersects the circumcircle of triangle $BHC$ at point $T$, and the line $BC$ intersects the circumcircle of triangle $AH... | [
"Let lines $CH$ and $PT$ meet at $U$ and denote by $A'$, $B'$, $C'$ the feet of the altitudes of the triangle $ABC$.\n\n$AHIC$ is a cyclic quadrilateral, so $\\angle CHI = \\angle CAI = \\frac{\\angle A}{2}$. Since $PT \\parallel HI$, it follows that $\\angle HUT = \\angle CHI = \\frac{\\angle A}{2}$.\nHence, $\\an... | Romania | THE 73rd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD - SECOND SELECTION TEST | [
"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"
] | null | proof and answer | A = 80°, B = 60°, C = 40° | |
0am0 | Problem:
Find all real numbers $a$ and $b$ so that for all real numbers $x$,
$$
2 \cos^2\left(x+\frac{b}{2}\right) - 2 \sin\left(a x - \frac{\pi}{2}\right) \cos\left(a x - \frac{\pi}{2}\right) = 1
$$ | [] | Philippines | 18th PMO Area Stage | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | All solutions are a = 1 with b = π/2 + 2πk, or a = −1 with b = −π/2 + 2πk, for any integer k. | |
0dd1 | Each cell of an $8 \times 8$ board is either black or white. It is known that among any nine cells forming a $3 \times 3$ square there are evenly many white ones. What is the minimum possible number of black cells? | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 4 | |
0aw2 | Problem:
How many different integral solutions $(x, y)$ does $3|x|+5|y|=100$ have? | [
"Solution:\n\nFrom $3|x|+5|y|=100$, we have $3|x|=100-5|y|=5(20-|y|)$. This implies that $20-|y|$ must be divisible by $3$ and it must be nonnegative. Thus, $y= \\pm 2, \\pm 5, \\pm 8, \\pm 11, \\pm 14, \\pm 17, \\pm 20$. Since each possible value of $y$ has two possible $x$ values except for $y= \\pm 20$, then the... | Philippines | 18th PMO National Stage Oral Phase | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 26 | |
0le8 | A sequence $(x_n)$ is defined as follows $x_1 = 1$ and $x_{n+1} = x_n + 3\sqrt{x_n} + \frac{n}{\sqrt{x_n}}$ for all positive integers $n$.
a) Prove that $\lim_{n \to +\infty} \frac{n}{x_n} = 0$.
b) Find the limit $\lim_{n \to +\infty} \frac{n^2}{x_n}$. | [
"a. We will prove $x_n \\ge n^2$ for all positive integers by induction. The statement is obvious in case $n=1$. Assume that $x_n \\ge n^2$, hence\n$$\nx_{n+1} > x_n + 3\\sqrt{x_n} \\ge x_n + 3n \\ge (n+1)^2.\n$$\nTherefore, the statement is proved. Thus\n$$\n0 < \\frac{n}{x_n} \\geq \\frac{1}{n}.\n$$\nBy the Squee... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | a) 0; b) 4/9 | |
06nl | We say that a real number is 'almost an integer' if it differs from an integer by at most $0.1$. For example, $2023$, $-2023.9$ and $2023.0822$ are almost integers. Show that among any $10$ real numbers, there exist two different real numbers whose difference is almost an integer. | [
"Let $x_1, x_2, \\dots, x_{10}$ be the real numbers. For each $j$, we write $x_j = a_j + b_j$ where $a_j$ is the largest integer not exceeding $x_j$. Note that we must have $0 \\le b_j < 1$. WLOG we may assume $b_1 \\le b_2 \\le \\dots \\le b_{10}$.\n\n* If $b_{j+1} - b_j \\le 0.1$ for some $j = 1, 2, \\dots, 9$, t... | Hong Kong | Hong Kong Team Selection Test 1 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
02j5 | Problem:
Ana, Bento e Lucas participam de um concurso que consta de 20 perguntas com a seguinte regra:
- cada resposta certa ganha 5 pontos,
- cada resposta errada perde 3 pontos,
- cada resposta em branco perde 2 pontos.
Veja os resultados na tabela a seguir:
| | Número de respostas certas | Número de respostas er... | [
"Solution:\n\nO número de pontos de cada um deles é:\n\nAna: $5 \\times 12 + (-3) \\times 4 + (-2) \\times 4 = 60 - 12 - 8 = 40$\n\nBento: $5 \\times 13 + (-3) \\times 7 + (-2) \\times 0 = 65 - 21 = 44$\n\nLucas: $5 \\times 12 + (-3) \\times 3 + (-2) \\times 5 = 60 - 9 - 10 = 41$\n\nLogo, Bento foi o mais bem class... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | E | |
0gii | 考慮多項式函數 $P(x) = x^2 - 1$。設實數 $a$ 滿足
$$P(P(P(a))) = 2024.$$
將 $a^2$ 之值寫成 $m+\sqrt{n}$，其中 $m, n$ 為正整數且 $n$ 不被大於 1 的平方數整除，則 $m+n$ 之值為⑥⑦。
Consider the polynomial $P(x) = x^2 - 1$. There is a real number $a$ that satisfies
$$P(P(P(a))) = 2024.$$
Write $a^2$ as $m + \sqrt{n}$, where $m$ and $n$ are integers and $n$ is squar... | [
"由 $P(P(P(a))) = 2024$，得\n$$\nP(P(a))^2 - 1 = 2024 \\Rightarrow P(P(a)) = \\pm\\sqrt{1+2024} = \\pm45.\n$$\n但負不合，因為 $P$ 的值域為 $[-1, \\infty)$。再操作一次可得\n$$\nP(a)^2 - 1 = 45 \\Rightarrow (a^2 - 1)^2 = 46 \\Rightarrow a^2 = 1 + \\sqrt{46}.\n$$\n故 $m+n=1+46=47$."
] | Taiwan | APMO Taiwan Preliminary Round 1 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | Chinese; English | final answer only | 47 | |
02vo | Problem:
Cada um dos números $1,2,3, \ldots, 25$ é arranjado em uma das casas de um tabuleiro $5 \times 5$. Em cada linha, eles aparecem em ordem crescente, da esquerda para a direita. Encontre os valores máximo e mínimo possíveis para as somas dos números da terceira coluna. | [
"Solution:\n\nPodemos numerar as linhas e colunas do tabuleiro com os números de 1 a 5, de cima para baixo e da esquerda para a direita. Além disso, podemos denotar o número escrito na linha de número $i$ e na coluna de número $j$ por $a_{ij}$, como indicado na figura abaixo:\n\n| $a_{11}$ | $a_{12}$ | $a_{13}$ | $... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | minimum 45, maximum 85 | |
0apq | Problem:
Find the largest integer $n$ such that
$$
\frac{n^{2007}+n^{2006}+\cdots+n^{2}+n+1}{n+2007}
$$
is an integer. | [
"Solution:\nLet $f(n) = n^{2007} + n^{2006} + \\cdots + n^{2} + n + 1$. As a geometric series, we have\n$$\nf(n) = \\frac{n^{2008} - 1}{n - 1}\n$$\nUsing the Division Algorithm, we can write\n$$\nf(n) = \\frac{n^{2008} - 1}{n - 1} = (n + 2007) g(n) + R\n$$\nwhere $g(n)$ is an integer-valued function, and $R = f(-20... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | (2007^{2008} - 1)/2008 - 2007 | |
0gyd | Let a given triangle $A_1A_2A_3$ and a point $X$ which doesn't lie on each of lines $A_1A_2$, $A_1A_3$ and $A_2A_3$ be given. Denote by $l_i$ line which passes through the point $A_i$ and is normal to the chord line $XA_i$, $i=1,2,3$. Let are the points $B_1 = l_2 \cap l_3$, $B_2 = l_3 \cap l_1$, $B_3 = l_1 \cap l_2$, ... | [
"Since $A_3H_2 \\parallel A_1X$ as the lines which are normal to the $l_i$, analogously $A_1H_2 \\parallel A_3X$ then the quadrangle $A_1XA_3H_2$ is a parallelogram. Then we have that $A_1XA_2H_3$ and $A_3XA_2H_1$ are the parallelogram too. Therefore the $A_3H_2$ and $A_2H_3$ are the equal and parallel segments. Th... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
08kq | Problem:
Let $ABCD$ be a trapezoid with $AB \parallel CD$ and $AB > CD$. It is given that the sum of the angles at the base $AB$ is equal to $90^{\circ}$. Prove that the distance between the midpoints of the parallel sides is equal to $\frac{1}{2}(AB - CD)$. | [
"Solution:\n\nLet $E$ be the intersection point of $AD$ and $BC$. The point $E$ is on the line through the midpoints $K$ and $F$ of the parallel sides. Let $G$ and $H$ be the points on $AB$, such that $EG$ and $CH$ are parallel to $AE$. Then the triangle $CHG$ is right-angled with the right angle at $C$ and the med... | JBMO | JBMO Shortlist | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
01kt | Circles $\Gamma_1$ and $\Gamma_2$ touch each other externally at point $M_3$. They touch a circle $\Gamma_3$ internally at points $M_1$ and $M_2$ respectively. Let $S$ be the circumcenter of the triangle $M_1M_2M_3$. Prove that the line $SM_1$ touches $\Gamma_3$.
(M. Karpuk) | [
"Let $O_1, O_2, O_3$ be respectively the centers of the circles $\\Gamma_1, \\Gamma_2, \\Gamma_3$, and $R_1, R_2, R_3$ be their radii.\nLet $\\Gamma$ be the circumscribed circle of the triangle $O_1O_2O_3$ (see Fig. 1), and let $\\Gamma$ touch the lines $O_3O_1, O_3O_2, O_1O_2$ at points $K_1, K_2, K_3$ respectivel... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0je9 | Find all triples $(x, y, z)$ of positive integers such that $x \le y \le z$ and
$$
x^3(y^3 + z^3) = 2012(xyz + 2).
$$ | [
"The solution is $(x, y, z) = (2, 251, 252)$.\nBy the given, $x$ divides $2 \\cdot 2012 = 2^3 \\cdot 503$. Moreover, $x$ cannot be a multiple of $503$, by considering the exponent of $503$ in both terms of the equation. Similarly, by considering the powers of $2$ on both sides of the equation, we conclude that $x$ ... | United States | IMO Team Selection Test | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | (2, 251, 252) | |
0c5d | Let $A_1 A_2 \dots A_n$ be a regular polygon with circumcircle $C$. For each direction $d$, denote $B_k$ the second intersection with $C$ of the line of direction $d$, passing through $A_k$, $k = 1, 2, \dots, n$. Prove that the sums
$$
S = \sum_{k=1}^{n} \overrightarrow{A_k B_k}, \quad \sigma = \sum_{k=1}^{n} |\overrig... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Linear Algebra > Vectors"
] | English | proof only | null | |
0dw6 | Problem:
Naj bo $n$ naravno število, ki je enako vsoti svojih od $n$ manjših deliteljev. Tako število je denimo število $28$. Kolikšna je vsota recipročnih vrednosti vseh njegovih deliteljev? | [
"Solution:\n\nNaj bodo $d_{1}, d_{2}, \\ldots, d_{k}$ delitelji števila $n$ in naj velja $1 = d_{1} < d_{2} < \\ldots < d_{k} = n$. Potem je $d_{i} \\cdot d_{k+1-i} = n$, zato je\n$$\n\\frac{1}{d_{1}} + \\frac{1}{d_{2}} + \\ldots + \\frac{1}{d_{k}} = \\frac{1}{d_{1}} + \\frac{1}{d_{2}} + \\ldots + \\frac{1}{d_{k}} ... | Slovenia | 48. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 2 | |
0cey | A *partition* of a set $S$ is a set of pairwise disjoint subsets of $S$ whose union is $S$. Let $P$ be a partition of the set $\{1, 2, ..., 2024\}$ into 2-element sets, satisfying the following condition: For every set $\{a, b\}$ in $P$, either $|a - b| = 1$ or $|a - b| = 506$. Assume that $\{1518, 1519\}$ belongs to $... | [
"Assume the square in the upper-left corner is white.\nBy the condition in statement, a set $\\{a, b\\}$ in $P$ is a horizontal domino if $|a-b| = 1$ and a vertical domino if $|a - b| = 506$, with the possible exceptions $\\{506, 507\\}$, $\\{1012, 1013\\}$ and $\\{1518, 1519\\}$. As this latter is an all-black mem... | Romania | Eighteenth STARS OF MATHEMATICS Competition | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 506 | |
009e | Alan plays a solitaire game in his computer. To begin, Alan chooses a positive integer $n$.
Then he chooses a positive odd divisor of $n$.
If the chosen divisor is $d=1$, the computer replaces $n$ with $n+1$.
If the chosen divisor is a number $d$ greater than 1, the computer replaces $n$ with $n/d$.
Alan continues the ... | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | No | |
03ln | Problem:
Let's say that an ordered triple of positive integers $(a, b, c)$ is $n$-powerful if $a \leq b \leq c$, $\operatorname{gcd}(a, b, c)=1$, and $a^{n}+b^{n}+c^{n}$ is divisible by $a+b+c$. For example, $(1,2,2)$ is 5-powerful.
a) Determine all ordered triples (if any) which are $n$-powerful for all $n \geq 1$.
... | [
"Solution:\n\nLet $T_{n}=a^{n}+b^{n}+c^{n}$ and consider the polynomial\n$$\nP(x)=(x-a)(x-b)(x-c)=x^{3}-(a+b+c) x^{2}+(a b+a c+b c) x-a b c .\n$$\nSince $P(a)=0$, we get $a^{3}=(a+b+c) a^{2}-(a b+a c+b c) a+a b c$ and multiplying both sides by $a^{n-3}$ we obtain $a^{n}=(a+b+c) a^{n-1}-(a b+a c+b c) a^{n-2}+(a b c)... | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra ... | null | proof and answer | a) (1, 1, 1) and (1, 1, 4). b) No such triples exist. | |
0dqz | There are $2012$ distinct points in the plane each of which is to be coloured using one of $n$ colours so that the number of points of each colour are distinct. A set of $n$ points is said to be multi-coloured if their colours are distinct. Determine $n$ that maximizes the number of multi-coloured sets. | [
"Let $m_1 < m_2 < \\dots < m_n$ be the number of points of each colour. We call $m_1, m_2, \\dots, m_n$ the colour distribution. Then $m_1 + \\dots + m_n = 2012$ and the number of multi-coloured sets is $M = m_1 m_2 \\dots m_n$. We have the following observations.\n\n(i) $m_1 > 1$. For if $m_1 = 1$, then $m_1 m_2 \... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Number Theory > Divisibility / Fact... | English | proof and answer | 61 | |
09k0 | Let $ABC$ be an acute triangle with $AB > AC$. Consider point $N$ inside the triangle such that $\angle ANB = 180^\circ - \angle BAC$, $\angle NBA = 180^\circ - 2\angle ACB$. The angle bisector of $\angle NBA$ intersects the angle bisector of $\angle NAC$ at point $P$. The line passing through $N$ and parallel to line ... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | proof only | null | |
04qz | On the table there are $k$ heaps of $1, 2, \dots, k$ stones, where $k \ge 3$. In the first step, we choose any three of the heaps on the table, merge them into a single new heap, and remove 1 stone (throw it away from the table) from this new heap. In the second step, we again merge some three of the heaps together int... | [
"After $i$ steps, there will be $k - 2i$ heaps left on the table; thus if a single heap is to remain in the end, the number $k$ must be odd and the total number of steps has to be $\\frac{1}{2}(k-1)$. Let us distinguish two cases, according as the remainder of $k$ upon division by 4 is 1 or 3.\n\nThe case of $k = 4... | Czech Republic | Czech-Slovak-Polish Match | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diop... | English | proof and answer | 161 | |
0a7h | Problem:
Let $n>1$ be an integer and let $a_{1}, a_{2}, \ldots, a_{n}$ be $n$ different integers. Show that the polynomial
$$
f(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \cdot \ldots \cdot\left(x-a_{n}\right)-1
$$
is not divisible by any polynomial with integer coefficients and of degree greater than zero but less th... | [
"Solution:\nSuppose $g(x)$ is a polynomial of degree $m$, where $1 \\leq m < n$, with integer coefficients and leading coefficient $1$, such that\n$$\nf(x) = g(x) h(x)\n$$\nwhere $h(x)$ is a polynomial. Let\n$$\n\\begin{aligned}\n& g(x) = x^{m} + b_{m-1} x^{m-1} + \\cdots + b_{1} x + b_{0} \\\\\n& h(x) = x^{n-m} + ... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 6 | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | null | proof only | null | |
08pm | Problem:
Let $a, b, c, d$ and $x, y, z, t$ be real numbers such that
$$
0 \leq a, b, c, d \leq 1, \quad x, y, z, t \geq 1 \text{ and } a+b+c+d+x+y+z+t=8
$$
Prove that
$$
a^{2}+b^{2}+c^{2}+d^{2}+x^{2}+y^{2}+z^{2}+t^{2} \leq 28
$$ | [
"Solution:\nWe observe that if $u \\leq v$ then by replacing $(u, v)$ with $(u-\\varepsilon, v+\\varepsilon)$, where $\\varepsilon>0$, the sum of squares increases. Indeed,\n$$\n(u-\\varepsilon)^{2}+(v+\\varepsilon)^{2}-u^{2}-v^{2}=2 \\varepsilon(v-u)+2 \\varepsilon^{2}>0\n$$\nThen, denoting\n$$\nE(a, b, c, d, x, y... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0l7k | Let $\mathbb{Z}$ be the set of integers, and let $f: \mathbb{Z} \to \mathbb{Z}$ be a function. Prove that there are infinitely many integers $c$ such that the function $g: \mathbb{Z} \to \mathbb{Z}$ defined by $g(x) = f(x) + c x$ is not bijective.
*Note*: A function $g: \mathbb{Z} \to \mathbb{Z}$ is *bijective* if for... | [] | United States | 16th United States of America Junior Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
00l0 | Let $I$ be the incenter of triangle $ABC$ and let $k$ be a circle through the points $A$ and $B$. This circle intersects
* the line $AI$ in points $A$ and $P$,
* the line $BI$ in points $B$ and $Q$,
* the line $AC$ in points $A$ and $R$ and
* the line $BC$ in points $B$ and $S$,
with none of the points $A, B, P, Q, R$ ... | [
"We define angles $\\alpha = \\angle BAC$ and $\\beta = \\angle CBA$ as usual, cf. Figure 4. Since points $A$,\n\nFigure 4: Problem 5\n$B, S$ and $R$ lie on a common circle, we have $\\angle BSR = 180^\\circ - \\alpha$, and therefore $\\angle RSC = \\alpha$. Similarly, $\\angle CRS = \\beta... | Austria | Austrian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
03bc | Let $O$ be an interior point of an acute-angled $\triangle ABC$. Denote by $A_1$, $B_1$ and $C_1$ its projection on the sides $BC$, $AC$ and $AB$. Let $P$ be the intersecting point of the lines through $A$ and $B$, orthogonal to $B_1C_1$ and $A_1C_1$, respectively. If $H$ is the projection of $P$ on $AB$, prove that th... | [
"Since $\\not\\asymp PBC_1 = \\not\\asymp OC_1A_1$ and $\\not\\asymp OC_1A_1 = \\not\\asymp OBA_1$ (why?), then $\\not\\asymp PBA = \\not\\asymp OBC$. Analogously $\\not\\asymp PAB = \\not\\asymp OAC$. Denote by $M$ and $N$ the projections of $P$ on $BC$ and $AC$. Since\n$$\nBC_1 \\cdot BH = BO \\cdot BP \\cos \\an... | Bulgaria | Bulgaria | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
00nb | The pages of a notebook are numbered consecutively such that the first sheet contains the numbers $1$ and $2$, the second sheet contains the numbers $3$ and $4$, and so on. One sheet is torn out of the notebook. The page numbers on the remaining sheets are added. The resulting sum equals $2021$.
a. How many pages can ... | [
"There is exactly one solution. The notebook had $64$ pages and the sheet with the page numbers $29$ and $30$ is ripped out.\n\nLet $b > 0$ be the number of sheets. The number of pages will be $2b$. We are looking for a number $2b$ such that\n$$\n1 + 2 + \\cdots + (2b - 1) + 2b = \\frac{(2b) \\cdot (2b + 1)}{2} > 2... | Austria | Austrian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | a) 64 pages. b) The torn sheet had pages 29 and 30. | |
092p | Problem:
Let $\mathbb{R}$ denote the set of real numbers. Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(x) f(y) = x f(f(y-x)) + x f(2x) + f\left(x^{2}\right)
$$
holds for all real numbers $x$ and $y$. | [
"Solution:\nWe set $x=0$ and get $f(0) f(y) = f(0)$, so $f(0) = 0$ or $f(y) = 1$ for all $y$. The latter leads to a contradiction.\nWe set $y = x + z$ and get\n$$\nf(x) f(x+z) = x f(f(z)) + x f(2x) + f\\left(x^{2}\\right)\n$$\nfor all $x$ and $z$. Setting $z=0$ yields\n$$\nf(x)^{2} = x f(2x) + f\\left(x^{2}\\right)... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0 for all real x; f(x) = 3x for all real x | |
0l0y | What is the value of
$$
\tan^2 \frac{\pi}{16} \cdot \tan^2 \frac{3\pi}{16} + \tan^2 \frac{\pi}{16} \cdot \tan^2 \frac{5\pi}{16} + \tan^2 \frac{3\pi}{16} \cdot \tan^2 \frac{7\pi}{16} + \tan^2 \frac{5\pi}{16} \cdot \tan^2 \frac{7\pi}{16} ?
$$ | [
"**Answer (B):** Let $\\theta = \\frac{\\pi}{16}$, and let $c = \\cos \\theta$, $s = \\sin \\theta$, and $t = \\tan \\theta$. Then\n$$\ne^{8i\\theta} = e^{\\frac{\\pi}{2}i} = (c + is)^8 = 0 + 1i.\n$$\nExpanding the binomial and taking the real part gives\n$$\n0 = 1c^8 - 28c^6s^2 + 70c^4s^4 - 28c^2s^6 + 1s^8.\n$$\nD... | United States | AMC 12 A | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | 68 | |
03us | It is given the set $S = \{1, 2, 3, \dots, 3n\}$, where $n$ is a positive integer. $T$ is a subset of $S$ such that: for any $x, y, z \in T$ (where $x, y, z$ can be the same), $x+y+z \notin T$. Find the maximum value of the number of elements in such set. | [
"Set $T_0 = \\{n+1, n+2, \\dots, 3n\\}$, where $|T_0| = 2n$ and the sum of any three elements in $T_0$ is larger than $3n$, that is to say, the sum does not belong to $T_0$. Thus, $\\max |T| \\ge 2n$.\n\nOn the other hand, construct a sequence of sets\n$$\nA_0 = \\{n, 2n, 3n\\}, \\quad A_k = \\{k, 2n-k, 2n+k\\}, \\... | China | China Southeastern Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2n | |
07x2 | Jimmy the grasshopper is sitting on a corner square of a rectangular $9 \times n$ board, where $n \ge 8$. To describe the hops Jimmy can make, we label the squares of the board by pairs of positive integers $(a, b)$, where $1 \le a \le 9$ and $1 \le b \le n$.
Jimmy can hop in both directions between squares $(a, b)$ an... | [
"Let $g = \\gcd(45, n)$, then the answer will be $9n/g$.\n\n$$\n\\begin{align*}\nV &= (2, 7) \\quad \\text{i.e. hopping from } (a, b) \\text{ to } (a + 2, b + 7) \\\\\nV^{-1} &= (-2, -7) \\quad \\text{i.e. hopping from } (a, b) \\text{ to } (a - 2, b - 7) \\\\\nH &= (7, 2) \\quad \\text{i.e. hopping from } (a, b) \... | Ireland | IRL_ABooklet_2024 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 9n / gcd(45, n) | |
0bal | Let $u : [a, b] \to \mathbb{R}$ be a continuous function which has at each point $x \in (a, b]$ a finite left derivative, denoted by $u'_s(x)$. Prove that the function $u$ is increasing if and only if $u'_s(x) \ge 0$, for all $x \in (a, b]$. | [
"We shall use the following results which can be viewed as generalizations of the theorems of Rolle and Lagrange.\n\n**Lemma 1.** If a function $u : [\\alpha, \\beta] \\to \\mathbb{R}$ has left and right finite derivatives at any point in $[\\alpha, \\beta]$ and $u(\\alpha) = u(\\beta)$, then there is $c \\in [\\al... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | null | proof only | null | |
032t | Problem:
Let $m \geq 3$ and $n \geq 2$ be integers. Prove that in a group of $N = m n - n + 1$ people such that there are two familiar people among any $m$, there is a person who is familiar with $n$ people. Does the statement remain true if $N < m n - n + 1$? | [
"Solution:\n\nConsider a group with maximal number of people such that any two of them are not familiar. It is clear that if there are $l$ people in this group, then $l \\leq m-1$. Moreover, the maximality of $l$ implies that any of the other $N-l$ people is familiar to at least one of the $l$ people in the group. ... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | No |
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