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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0165 | Determine all integral solutions of the equation
$$
x^2y = (x + y)^2 + 1.
$$ | [
"We have $y > 0$. Also, $y \\mid x^2 + 1$, so we can write $x^2 + 1 = a y$, where $a > 0$. Hence follows\n$$\na(x^2y - (x+y)^2 - 1) - y(x^2a - (x+a)^2 - 1) = (a-y)(-x^2 + a y - 1) = 0,\n$$\nso $x$ satisfies\n$$\nx^2 a = (x+a)^2 + 1.\n$$\nThus, $x$ satisfies\n$$\nx^2 z = (x+z)^2 + 1 \\quad (*)\n$$\nfor $z = y$ and $... | Baltic Way | Baltic Way SHL | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (-1, 1), (-1, 2), (5, 2), (5, 13) | |
0ce7 | Let $u, v \in \mathbb{C}$ so that $|1 - u| = |1 - v|$ and $|1 - u^2| = |1 - v^2|$. Prove that $u = v$ or $u = \bar{v}$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof only | null | |
0b52 | Problem:
Fie $n$ un număr natural nenul. Este posibil ca orice număr raţional strict pozitiv să fie exprimat sub forma
$$
\frac{a^{n}+b^{n+2}}{c^{n+1}+d^{n+3}}
$$
unde $a, b, c, d$ sunt numere naturale nenule? | [] | Romania | TESTUL 1 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0b5s | Determine all polynomials $P \in \mathbb{Z}[X]$ such that there exists $k \in \mathbb{N}^*$, so that for all primes $q$, $P(q)$ has at most $k$ distinct prime divisors. | [
"Clearly a constant non-null polynomial $P$ fulfills the requirements, so we shall assume in the sequel $\\deg P \\ge 1$. Before further proceeding with the solution, we state the following \"folklore\" preliminaries.\n(1) **Dirichlet's Theorem.** The arithmetic progression $(an + b)_{n \\in \\mathbb{N}}$ with $a \... | Romania | Local Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | All polynomials of the form c·X^m with c a nonzero integer and m a nonnegative integer. | |
0gbt | 令 $\mathbb{R}$ 表示所有實數所成的集合。試求所有可能的函數 $f : \mathbb{R} \to \mathbb{R}$ 滿足:
對任意實數 $x, y$, 恆有 $f(f(x) + y) = f(x^2 - y) + 4(y - 2)(f(x) + 2)$. | [
"令 $y = \\frac{x^2 - f(x)}{2}$ 代入原式，得\n$$\nf\\left(\\frac{x^2 + f(x)}{2}\\right) = f\\left(\\frac{x^2 + f(x)}{2}\\right) + 4\\left(\\frac{x^2 - f(x)}{2} - 2\\right)(f(x) + 2).\n$$\n得知：$(x^2 - f(x) - 4)(f(x) + 2) = 0$；因此，對每一個實數 $x$，恆有\n$$\nf(x) = x^2 - 4 \\text{ 或 } f(x) = -2.\n$$\n\n更進一步地，不論上述哪一種情況下，$f(\\sqrt{2}) =... | Taiwan | 二〇一八數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = -2 and f(x) = x^2 - 4 | |
0fef | Problem:
En un tablero de ajedrez $10 \times 10$ se colocan $41$ torres. Probar que se pueden elegir al menos $5$ de ellas que no se coman entre sí. | [
"Solution:\n\nPues $41 = 4 \\times 10 + 1$, existe una fila $A$ que contiene al menos $5$ torres.\n\nQuedan $9$ filas con, al menos, $31$ torres. Una fila $B$, de entre estas $9$, contiene, al menos, $4$ torres.\n\nQuedan $8$ filas con, al menos, $21$ torres. Una fila $C$, de entre estas $8$, contiene, al menos, $3... | Spain | null | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof only | null | |
0cwx | Let $P_1(x)$ and $P_2(x)$ be monic quadratic polynomials (i.e., quadratic polynomials with leading coefficient $1$). Let points $A_1$ and $A_2$ be vertices of parabolas $y = P_1(x)$ and $y = P_2(x)$, respectively. By $m(g(x))$ denote the minimal value of the function $g(x)$. It happens that the differences $m(P_1(P_2(x... | [
"Let the given trinomials be $P_1(x) = (x - x_1)^2 + y_1$ and $P_2(x) = (x - x_2)^2 + y_2$, where $A_1(x_1; y_1)$ and $A_2(x_2; y_2)$ are the coordinates of the vertices of the parabolas. Then $m(P_1(x)) = y_1$, and $P_1(P_2(x)) = ((x - x_2)^2 + y_2 - x_1)^2 + y_1$. If $y_2 \\le x_1$, then the minimal value of the ... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | Russian | proof and answer | 45 degrees | |
0kun | Problem:
Let $ABC$ be an equilateral triangle of side length $15$. Let $A_b$ and $B_a$ be points on side $AB$, $A_c$ and $C_a$ be points on side $AC$, and $B_c$ and $C_b$ be points on side $BC$ such that $\triangle AA_bA_c$, $\triangle BB_cB_a$, and $\triangle CC_aC_b$ are equilateral triangles with side lengths $3$, ... | [
"Solution:\n\n\n\nLet $\\triangle XYZ$ be the triangle formed by lines $A_bA_c$, $B_aB_c$, and $C_aC_b$. Then, the desired circle is the incircle of $\\triangle XYZ$, which is equilateral. We have\n$$\n\\begin{aligned}\nYZ & = YA_c + A_cA_b + A_bZ \\\\\n & = A_cC_a + A_cA_b + A_bB_a \\\\\... | United States | HMMT November 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 3\sqrt{3} | |
04h4 | For a real number $a$, let $P_a$ be the parabola given by the equation $y = x^2 + a x + (2014 - a)$. Prove that all parabolae $P_a$ pass through the same point. | [
"Let us find a point $(x_0, y_0)$ that lies on all parabolae $P_a$ for any real $a$.\n\nThe equation of $P_a$ is $y = x^2 + a x + (2014 - a)$.\n\nLet us try to find $x_0$ and $y_0$ such that for all $a$, $y_0 = x_0^2 + a x_0 + (2014 - a)$.\n\nRewriting:\n$$\ny_0 = x_0^2 + a x_0 + 2014 - a\n$$\n$$\ny_0 = x_0^2 + 201... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
03xj | Given two moving points $A(x_1, y_1)$ and $B(x_2, y_2)$ on parabola curve $y^2 = 6x$ with $x_1 + x_2 = 4$ and $x_1 \neq x_2$, and the perpendicular bisector of segment $AB$ intersects $x$-axis at point $C$. Find the maximum area of $\triangle ABC$. | [
"Let the midpoint of $AB$ be $M(x_0, y_0)$. Then $x_0 = \\frac{x_1 + x_2}{2} = 2$ and $y_0 = \\frac{y_1 + y_2}{2}$. We have\n$$\nk_{AB} = \\frac{y_2 - y_1}{x_2 - x_1} = \\frac{\\frac{y_2}{6} - \\frac{y_1}{6}}{\\frac{y_2}{6} + \\frac{y_1}{6}} = \\frac{\\frac{6}{y_2} + \\frac{6}{y_1}}{\\frac{6}{y_2} + \\frac{6}{y_1}}... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Linear Algebra > Determinants",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 14/3 * sqrt(7) | |
0dub | Problem:
Za poljubno naravno število $n$ označimo s $p(n)$ produkt števk tega naravnega števila, zapisanega desetiško. Izračunaj vsoto
$$
p(1)+p(2)+\cdots+p(2001)
$$ | [
"Solution:\n\nIzračunamo $p(1)+p(2)+\\cdots+p(9)=45$.\n\nČe označimo s $k$ stotice in z $j$ desetice, potem je\n$$\n\\begin{aligned}\np(j 0)+p(j 1)+\\cdots+p(j 9) & =j \\cdot 45 \\\\\np(k j 0)+p(k j 1)+\\cdots+p(k j 9) & =k \\cdot j \\cdot 45 \\text{ in } \\\\\np(1 k j 0)+p(1 k j 1)+\\cdots+p(1 k j 9) & =k \\cdot j... | Slovenia | 45. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 184320 | |
0c84 | Let $n$ be a nonnegative integer and $M = \{n^3, n^3 + 1, n^3 + 2, \dots, n^3 + n\}$. Consider $A$ and $B$ two nonempty, disjoint subsets of $M$ such that the sum of elements of the set $A$ divides the sum of elements of the set $B$. Prove that the number of elements of the set $A$ divides the number of elements of the... | [
"Denote $A = \\{n^3 + n_1, n^3 + n_2, \\dots, n^3 + n_a\\}$, $B = \\{n^3 + m_1, n^3 + m_2, \\dots, n^3 + m_b\\}$ and $k \\in \\mathbb{N}$ such that $n^3 + m_1 + n^3 + m_2 + \\dots + n^3 + m_b = k(n^3 + n_1 + n^3 + n_2 + \\dots + n^3 + n_a)$. Then $n^3(ka - b) = m_1 + m_2 + \\dots + m_b - k(n_1 + n_2 + \\dots + n_a)... | Romania | SELECTION TESTS FOR THE 2019 JBMO | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
02ga | Providence Ave. has infinitely many traffic lights, all equally spaced and synchronized. The distance between any two consecutive ones is $1500$ m. The traffic lights stay green $1.5$ minute, red $1$ minute, then green again $1.5$ minute, and so on. Suppose that a car is passing through Providence Ave. at constant spee... | [
"Suppose that at instant $0$ the traffic lights turn green and the car passes through the first light at instant $t_0 \\ge 0$ (time is measured in seconds). The traffic lights will stay green between the time instants $150k + 90$ and $150(k+1)$, for each integer $k$. The car will pass through the lights at the inst... | Brazil | XXII OBM | [
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | All speeds of the form 20/k meters per second for positive integers k. | |
077l | Let $k$ be a positive integer. A sequence of integers $\langle a_i \rangle_{i=1}^\infty$ is called $k$-pop if the following holds: for every $n \in \mathbb{N}$, $a_n$ is equal to the number of distinct elements in the set $\{a_1, \dots, a_{n+k}\}$. Determine, as a function of $k$, how many $k$-pop sequences there are. | [
"The answer is $2^k$. Before we prove this, we will obtain a characterization of $k$-pop sequences. To that end, note that in a $k$-pop sequence, since the set $\\{a_1, \\dots, a_{n+k+1}\\}$ has either the same number of distinct elements as $\\{a_1, \\dots, a_{n+k}\\}$ or precisely one more, it follows that $a_{n+... | India | EGMO TST Day 2 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 2^k | |
0079 | In a school with 5 grades there are 250 girls and 250 boys. Each grade has 100 students. Teams of one girl and one boy from the same grade must be formed for a contest. At least 19 students in each grade are girls and at least 19 are boys. Find the greatest number of teams that can be formed with certainty. | [
"The answer is $126$. Let there be $a_i$ girls and $b_i$ boys in grade $i$, $1 \\le i \\le 5$. Consider a $2 \\times 5$ table with $a_1, \\dots, a_5$ in the first row and $b_1, \\dots, b_5$ in the second row. Mark the smaller of the numbers $a_i, b_i$ for each $i$. The number of teams that can be formed is the sum ... | Argentina | National Olympiad of Argentina | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 126 | |
02xs | Problem:
O triângulo $ABC$ é isósceles com $AB = BC$. A bissetriz do ângulo $\angle CAB$ encontra o lado $BC$ no ponto $D$. A diferença entre as medidas de dois ângulos internos do triângulo $ABD$ é $40^{\circ}$. Encontre os possíveis valores do ângulo $\angle ACB$.
 | [
"Solution:\n\nSeja $x = \\angle BAC = \\angle ABC$. Assim, $\\angle ACB = 180^{\\circ} - 2x$. Consequentemente, os ângulos internos do triângulo $ABD$ são $\\angle BAD = x/2$, $\\angle DBA = x$ e $\\angle ADB = 180^{\\circ} - 3x/2$. Consideraremos todos os casos para os quais dois ângulos podem diferir por $40^{\\c... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 4°, 20°, 40°, 68° | |
0epj | If $\frac{12}{x}$ is a natural number and $x$ is a natural number, how many possible values are there for $x$?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8 | [
"$x$ must be a divisor of $12$, and the possibilities are $1$, $2$, $3$, $4$, $6$, $12$, which is $6$ in all."
] | South Africa | South African Mathematics Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | C | |
033n | Problem:
In any cell of an $n \times n$ table a number is written such that all the rows are different. Prove that one can remove a column such that the rows in the new table are still different. | [
"Solution:\n\nConsider a graph with vertices the lines of the table and edges that join two vertices if the respective lines are different exactly in one position. Write on the edges the respective elements that are different.\n\nAssume that the given statement is not true. Then the graph has $n$ vertices and $n$ e... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
02ji | Problem:
Se $2\left(2^{2x}\right) = 4^{x} + 64$, então $x$ é igual a:
A) -2
B) -1
C) 1
D) 2
E) 3 | [
"Solution:\n\nSolução 1: Notamos que os termos do lado direito da equação dada podem ser escritos como potências de 2; de fato, $4^{x} = \\left(2^{2}\\right)^{x} = 2^{2x}$ e $64 = 2^{6}$. Desse modo, a equação se torna $2\\left(2^{2x}\\right) = 2^{2x} + 2^{6}$. Temos então $2\\left(2^{2x}\\right) - 2^{2x} = 2^{6}$,... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | MCQ | E | |
0gri | Let $O$ be the circumcenter of an acute triangle $ABC$. A line perpendicular to $AO$ intersects the line segments $[AC]$ and $[AB]$ at $D$ and $E$, respectively. Let $K$ be a point on $[BC]$ which is not on the line $AO$. The line $AK$ intersects the circumcircle of triangle $ADE$ again at $L$. Let $M$ be the point sym... | [
"\n\nLet $AM$ and $DE$ intersect at $N$, and $F$ be the foot of the perpendicular line from $O$ to $AC$. Since $AO \\perp DE$ and $\\angle BAO = 90^\\circ - \\angle ACB$ we get $\\angle ALD = \\angle AED = \\angle ACB$ and hence the points $K$, $C$, $D$, $L$ are concyclic. Using power equat... | Turkey | 23rd Junior Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneo... | English | proof only | null | |
091r | Problem:
Let $n \geqslant 3$ be an integer. At a MEMO-like competition, there are $3n$ participants, there are $n$ languages spoken, and each participant speaks exactly three different languages.
Prove that at least $\left\lceil\frac{2n}{9}\right\rceil$ of the spoken languages can be chosen in such a way that no parti... | [
"Solution:\n\nConsider the classifications of the set of $n$ available languages into easy, medium, and hard languages. There are $3^{n}$ possible classifications in total and we denote by $S$ the set of all possible classifications. For each classification $s \\in S$, let $A(s)$ be the number of easy languages and... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
00qv | Isaac and Jeremy play a game. Isaac tells Jeremy that he is thinking of some $2^n$ integers $k_1, \dots, k_{2^n}$, where $n$ is a positive integer. Jeremy asks questions of the form "is $k_i < k_j$?", to which Isaac answers truthfully. After $n2^{n-1}$ questions, Jeremy must state whether or not Isaac's numbers are all... | [
"We adopt the following notation. If $a = a_1, \\dots, a_{2^n}$ is a sequence of integers, then we will denote by $Q(a)_j$ the $j$th question that Jeremy asks, and $A(a)_j$ the $j$th answer that Isaac gives, when Isaac chooses the sequence $k_i = a_i$. $Q(a)_j$ is allowed to depend on $Q(a)_1, \\dots, Q(a)_{j-1}$ a... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0ene | Let $n \ge 3$ be an integer, and let $a_2, a_3, \dots, a_n$ be positive real numbers such that $a_2 a_3 \dots a_n = 1$. Prove that
$$
(1 + a_2)^2 (1 + a_3)^3 \dots (1 + a_n)^n > n^n.
$$ | [
"$$\n\\begin{aligned}\n(a_k + 1)^k &= \\left( a_k + \\frac{1}{k-1} + \\frac{1}{k-1} + \\dots + \\frac{1}{k-1} \\right)^k \\\\\n&\\ge k^k \\cdot a_k \\cdot \\frac{1}{(k-1)^{k-1}} \\\\\n&= \\frac{k^k}{(k-1)^{k-1}} \\cdot a_k\n\\end{aligned}\n$$\nThe inequality is strict unless $a_k = \\frac{1}{k-1}$. Multiplying anal... | South Africa | International Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
005b | En un triángulo $ABC$, $\vec{A} = 2\vec{C}$ y $2\vec{B} = \vec{A} + \vec{C}$. La bisectriz del ángulo $\vec{C}$ corta al lado $AB$ en $E$, y $F$ es el punto medio del segmento $AE$. La altura correspondiente al lado $BC$ es $AD$. La mediatriz del segmento $DF$ corta al lado $AC$ en $M$.
Demostrar que $AM = CM$. | [] | Argentina | XIIIª OLIMPÍADA de MAYO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | Español | proof only | null | |
0ia1 | Problem:
In a classroom, 34 students are seated in 5 rows of 7 chairs. The place at the center of the room is unoccupied. A teacher decides to reassign the seats such that each student will occupy a chair adjacent to his/her present one (i.e. move one desk forward, back, left or right). In how many ways can this reass... | [
"Solution:\n\nColor the chairs red and black in checkerboard fashion, with the center chair black. Then all 18 red chairs are initially occupied. Also notice that adjacent chairs have different colors. It follows that we need 18 black chairs to accommodate the reassignment, but there are only 17 of them. Thus, the ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 0 | |
0ky9 | Fix a function $f: \mathbb{N} \to \mathbb{N}$ and for any $m, n \in \mathbb{N}$ define
$$
\Delta(m, n) = \underbrace{f(f(\dots f(m)\dots))}_{f(n) \text{ times}} - \underbrace{f(f(\dots f(n)\dots))}_{f(m) \text{ times}}.
$$
Suppose $\Delta(m, n) \neq 0$ for any distinct $m, n \in \mathbb{N}$. Show that $\Delta$ is unbou... | [
"Suppose for the sake of contradiction that $|\\Delta(m, n)| \\le N$ for all $m, n$. Note that $f$ is injective, as\n$$\nf(m) = f(n) \\implies \\Delta(m, n) = 0 \\implies m = n,\n$$\nas desired.\n\nLet $G$ be the “arrow graph” of $f$, which is the directed graph with vertex set $\\mathbb{N}$ and edges $n \\to f(n)$... | United States | USA TST | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0dao | 1. Let $p$ be an odd prime number.
a. Show that $p$ divides $n \cdot 2^{n} + 1$ for infinitely many positive integers $n$.
b. Find all $n$ satisfying the condition above when $p = 3$. | [
"a. We will show that for all $k > 0$, the number $n = (p k + 1)(p - 1)$ satisfies $p \\mid n \\cdot 2^{n} + 1$. Indeed, by Fermat's Little Theorem, we have\n$$\n2^{p-1} \\equiv 1 \\pmod{p} \\text{ so } 2^{n} = 2^{(p k + 1)(p - 1)} \\equiv 1 \\pmod{p}.\n$$\nAnd then\n$$\nn \\cdot 2^{n} + 1 \\equiv (p k + 1)(p - 1) ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | Infinitely many n exist; for example, all n of the form (p k + 1)(p − 1). For p = 3, all positive integers congruent to 1 or 2 modulo 6. | |
03k2 | Problem:
Prove that the product of the first $n$ natural numbers is divisible by the sum of the first $n$ natural numbers if and only if $n+1$ is not an odd prime. | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
09kx | Michael, who runs at a speed three times faster than his walking pace, typically walks to school but occasionally needs to run to arrive on time. Today, it took him 12 minutes longer to walk to school compared to yesterday. How many additional minutes did he spend going to school today compared to yesterday? | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | English | proof and answer | 8 | |
0eqm | Find the value of $2 + 2 \times 3^2$ | [
"The value is $2 + 2 \\times 9 = 2 + 18 = 20$"
] | South Africa | South African Mathematics Olympiad Second Round | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 20 | |
0jyd | Problem:
Emilia wishes to create a basic solution with $7\%$ hydroxide $(\mathrm{OH})$ ions. She has three solutions of different bases available: $10\%$ rubidium hydroxide $(\mathrm{Rb}(\mathrm{OH}))$, $8\%$ cesium hydroxide $(\mathrm{Cs}(\mathrm{OH}))$, and $5\%$ francium hydroxide $(\mathrm{Fr}(\mathrm{OH}))$. (The... | [
"Solution:\n\nSuppose that Emilia uses $R$ liters of $\\mathrm{Rb}(\\mathrm{OH})$, $C$ liters of $\\mathrm{Cs}(\\mathrm{OH})$, and $F$ liters of $\\mathrm{Fr}(\\mathrm{OH})$, then we have\n$$\n\\frac{10\\% \\cdot R + 8\\% \\cdot C + 5\\% \\cdot F}{R + C + F} = 7\\% \\text{ and } \\frac{5\\% \\cdot F}{R + C + F} \\l... | United States | HMMT November 2017 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 1% | |
0aek | Кој од следните изрази е поголем: $A = \frac{1+a+a^2+...+a^{n-1}}{1+a+a^2+...+a^n}$,
$$
B = \frac{1+b+b^2+...+b^{n-1}}{1+b+b^2+...+b^n}, \text{ ако } a>b>0.
$$ | [
"Изразот $\\frac{1}{A}$ можеме да го запишеме во облик:\n$$\n\\frac{1}{A} = \\frac{1+a+a^2+...+a^n}{1+a+a^2+...+a^{n-1}} = 1 + \\frac{a^n}{1+a+a^2+...+a^{n-1}} = 1 + \\frac{1}{\\frac{1}{a} + \\frac{1}{a^{n-1}} + ... + \\frac{1}{a}}\n$$\nСлично и за $B$, изразот $\\frac{1}{B}$ можеме да го запишеме во облик:\n$$\n\\... | North Macedonia | Републички натпревар по математика за средно образование | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | Macedonian, English | proof and answer | B > A | |
052o | Let $a$ and $b$ be positive integers such that $b$ is divisible by $a$ and writing $a$ and $b$ one after another in this order gives $(a+b)^2$. Prove that $\frac{b}{a} = 6$. | [
"Let $n$ be the number of digits of $b$ and let $b = k a$. Then by the conditions of the problem, $10^n \\cdot a + k a = (a + k a)^2$, or\n$$\na = \\frac{10^n + k}{(k+1)^2} \\qquad (1)\n$$\nIf $k$ were odd, then the numerator on the r.h.s. of (1) would be odd and the denominator even, so $a$ could not be an integer... | Estonia | Open Contests | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Other"
] | English | proof only | null | |
01cf | Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying, for all $x \neq 0$ and all $y$,
$$
f(x + y^2) = f(x) + f(y)^2 + \frac{2f(xy)}{x}.
$$ | [
"Answer: $f(z) = 0$ or $f(z) = z^2$.\nReplacing $y$ by $-y$ gives us\n$$\nf(x) + f(y)^2 + \\frac{2f(xy)}{x} = f(x + y^2) = f(x) + f(-y)^2 + \\frac{2f(-xy)}{x},\n$$\nwhich implies that\n$$\nf(y)^2 + \\frac{2f(xy)}{x} = f(-y)^2 + \\frac{2f(-xy)}{x} \\quad (7)\n$$\nfor all $x \\neq 0$ and all $y$. Let $x = 1$ and comp... | Baltic Way | Baltic Way 2015 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(z) = 0 or f(z) = z^2 | |
0e4a | The lengths $a$, $b$ and $c$ of the sides of the triangle $ABC$ satisfy $c^2 = 2ab$ and $a^2 + c^2 = 3b^2$. The inner angles of the triangle $ABC$ measure
(A) $30^\circ$, $60^\circ$ and $90^\circ$.
(B) $45^\circ$, $60^\circ$ and $75^\circ$.
(C) $45^\circ$, $45^\circ$ and $90^\circ$.
(D) $60^\circ$, $60^\circ$ and $60^\... | [
"From $c^2 = 2ab$ and $a^2 + c^2 = 3b^2$ we get $a^2 + 2ab = 3b^2$ or $(a + b)^2 = 4b^2$. It follows that $(a + b - 2b)(a + b + 2b) = 0$. Since $a$ and $b$ are positive, the only possibility is that $a = b$. Then $c^2 = 2a^2 = a^2 + b^2$. Hence, we have a right isosceles triangle and the inner angles measure $45^\\... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | C | |
01o4 | The graphs of the functions $f(x) = ax^2 + bx + c$, $g(x) = \frac{cx + b}{cx + a}$, $h(x) = cx + b$ have the unique common point of the intersection, and the graph of $h(x)$ has no another common points with the graphs of $f(x)$ and $g(x)$. All $a, b, c$ are non-zero pairwise distinct real numbers.
Find all possible va... | [
"Since the graphs of $f(x)$ and $h(x)$ has exactly one common point, the equation $ax^2 + bx + c = cx + b$, i.e. the equation\n$$\nax^2 - (c-b)x + (c-b) = 0, \\quad (1)\n$$\nhas exactly one root. So, its discriminant is equal to $0$, i.e. $(c-b)^2-4a(c-b) = 0 \\Leftrightarrow (c-b)(c-b-4a) = 0$. By condition, $c \\... | Belarus | Belorusija 2012 | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | a = 3/11, b = -8/11, c = 4/11 | |
05ci | Call a positive integer $n$ interesting if both the sum of digits of $n$ and the sum of digits of $n+1$ are perfect squares, whereas $n$ and $n+1$ have the same number of digits. Find all positive integers $k$ for which there exists an interesting $k$-digit number. | [
"For any positive integer $a$, let $s(a)$ denote the sum of digits of $a$. Clearly $s(n + 1) = s(n) + 1$ unless the last digit of $n + 1$ is zero. The only two consecutive integers that are both perfect squares are $0$ and $1$, but $s(n) = 0$ is impossible for a positive $n$. The contradiction shows that the last d... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | all integers k ≥ 10 | |
0hj2 | Problem:
Let $x$ and $y$ be positive integers. Can $x^{2}+2y$ and $y^{2}+2x$ both be squares? | [
"Solution:\n\nThe answer is no. Because of the symmetry between $x$ and $y$, we can assume that $x \\geq y$. Then\n$$\nx^{2} < x^{2} + 2y \\leq x^{2} + 2x < x^{2} + 2x + 1 = (x+1)^{2}.\n$$\nThus $x^{2} + 2y$, which lies between two consecutive squares, cannot be a square."
] | United States | Berkeley Math Circle Monthly Contest 8 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | No | |
015e | Some computer is operating with real numbers. Develop an algorithm for calculating $ac - bd$ and $ad + bc$ from given real numbers $a$, $b$, $c$, $d$, using only three operations of multiplication. (The number of additions and subtractions can be arbitrary but no division is allowed.) | [
"Calculate $x = (a + b)(c + d)$, $y = ac$, $z = bd$. Then $ac - bd = y - z$ and $ad + bc = x - y - z$."
] | Baltic Way | Baltic Way SHL | [
"Discrete Mathematics > Algorithms",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0509 | Find all pairs $(n, k)$ of positive integers that satisfy the equality $n! + (n+1)! = k! + 120$. | [
"Note that for every $n$, $n! + (n+1)! = n! + n! \\cdot (n+1) = n! \\cdot (n+2) \\le (n+2)!$. Thus if $n! + (n+1)! = k! + 120$, then due to $120 = 5!$ we have $k! + 5! \\le (n+2)!$. This inequality in turn implies $k < n+2$ and $5 < n+2$. Hence $5 \\le n+1$, leading to $0 \\le (n+1)! - 5! = k! - n!$. Consequently, ... | Estonia | Selected Problems from Open Contests | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | (4, 4), (5, 6) | |
0e99 | Žan wrote a sequence of four positive real numbers. The first term in the sequence was the number $3$, and the last term was the number $9$. The first three terms formed a geometric sequence, and the last three terms formed an arithmetic sequence. Determine all four terms of Žan's sequence. | [
"Denote the second and the third term of Žan's sequence by $x$ and $y$. Then $3$, $x$, $y$ is a geometric sequence and $x^2 = 3y$. On the other hand, $x$, $y$, $9$ is an arithmetic sequence, so $2y = x + 9$. We may rewrite the second equation as $y = \\frac{x+9}{2}$. Plugging this into the first equation and rearra... | Slovenia | National Math Olympiad 2013 - First Round | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 3, 9/2, 27/4, 9 | |
0l3l | Problem:
Jasper and Rose are playing a game. Twenty-six $32$-ounce jugs are in a line, labeled Quart $A$ through Quart $Z$ from left to right. All twenty-six jugs are initially full. Jasper and Rose take turns making one of the following two moves:
- Remove a positive integer number of ounces from the leftmost nonempt... | [
"Solution:\n\nNotice that after any sequence of moves, the leftmost nonempty jug has at most as many ounces as the second leftmost nonempty jug.\n\nJasper's strategy for $31$ is as follows: as long as at least two jugs are nonempty, remove all but one ounce from the first jug. This will ensure Rose only ever gets t... | United States | HMMT November 2024 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 31 | |
04t7 | Let $p > 3$ be a prime. Find the number of ordered sextuples $(a, b, c, d, e, f)$ of positive integers, whose sum is $3p$, and all the fractions
$$
\frac{a+b}{c+d}, \quad \frac{b+c}{d+e}, \quad \frac{c+d}{e+f}, \quad \frac{d+e}{f+a}, \quad \frac{e+f}{a+b}
$$
are integers. | [
"Taking the product of the 1st, the 3rd and the 5th fractions reveals that their value has to be 1, that is\n$$\na + b = c + d = e + f = p. \\tag{1}\n$$\nthe form of the second and of the fourth fraction implies\n$$\nf + a \\mid d + e \\quad \\text{and} \\quad d + e \\mid b + c. \\tag{2}\n$$\nthat is first $f + a$ ... | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | p + 2 if p ≡ 1 (mod 3); p + 1 if p ≡ 2 (mod 3) | |
0dcf | Let sequences of real numbers $\left(x_{n}\right)$ and $\left(y_{n}\right)$ satisfy $x_{1}=y_{1}=1$ and
$$
x_{n+1}=\frac{x_{n}+2}{x_{n}+1} \quad \text{ and } \quad y_{n+1}=\frac{y_{n}^{2}+2}{2 y_{n}} \text{ for } n=1,2, \ldots .
$$
Prove that $y_{n+1}=x_{2^{n}}$ holds for $n=0,1,2, \ldots$
| [
"Note that, since $x_{1}=1$, $x_{n+1}=\\frac{x_{n}+2}{x_{n}+1}$ we can write\n$$\n\\begin{aligned}\nx_{n+1} & =\\frac{x_{n} \\cdot x_{1}+2}{x_{n}+x_{1}}=\\frac{\\frac{x_{n-1}+2}{x_{n-1}+1} \\cdot x_{1}+2}{\\frac{x_{n-1}+2}{x_{n-1}+1}+x_{1}}=\\frac{x_{n-1}\\left(x_{1}+2\\right)+2\\left(x_{1}+1\\right)}{x_{n-1}\\left... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
06p1 | The sequence $f(1), f(2), f(3), \ldots$ is defined by
$$
f(n)=\frac{1}{n}\left(\left\lfloor\frac{n}{1}\right\rfloor+\left\lfloor\frac{n}{2}\right\rfloor+\cdots+\left\lfloor\frac{n}{n}\right\rfloor\right),
$$
where $\lfloor x\rfloor$ denotes the integer part of $x$.
a. Prove that $f(n+1)>f(n)$ infinitely often.
b. Pro... | [
"Let $g(n)=n f(n)$ for $n \\geq 1$ and $g(0)=0$. We note that, for $k=1, \\ldots, n$,\n$$\n\\left\\lfloor\\frac{n}{k}\\right\\rfloor-\\left\\lfloor\\frac{n-1}{k}\\right\\rfloor=0\n$$\nif $k$ is not a divisor of $n$ and\n$$\n\\left\\lfloor\\frac{n}{k}\\right\\rfloor-\\left\\lfloor\\frac{n-1}{k}\\right\\rfloor=1\n$$\... | IMO | IMO 2006 Shortlisted Problems | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
03t2 | Let $S = \{(a, b) \mid 1 \le a, b \le 5,\ a, b \in \mathbb{Z}\}$. Let $T$ be the set of integer points in the plane such that for any point $P$ in $S$, there exists a different point $Q$ in $T$ such that $PQ$ does not contain integer points except $P$ and $Q$. Find the minimum value of $|T|$, where $|T|$ denotes the nu... | [
"We first prove that $|T| \\ne 1$.\nIf $|T| = 1$, let $T = \\{Q(x_0, y_0)\\}$. We may take point $P(x_1, y_1)$ in $S$ satisfying the conditions: (1) $(x_1, y_1) \\ne (x_0, y_0)$, (2) $x_1$ and $x_0$ have the same parity, $y_1$ and $y_0$ have the same parity. Then, the midpoint of $PQ$ is an integer, which is a cont... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2 | |
05ju | Problem:
On considère une grille carrée de $4 \times 4$ cases.
Deux cases distinctes ayant un côté commun sont dites voisines. Initialement, toutes les cases sont rouges. Par la suite, une case pourra éventuellement changer de couleur et être soit rouge soit bleue. Effectuer une opération sur la case $c$ signifie que l... | [
"Solution:\nRegroupons les cases de la grille en quatre tuiles, contenant chacune une case marquée, comme indiqué dans la figure suivante :\n\nIl est facile de vérifier qu'une opération quelconque change la couleur d'une et une seule de ces cases marquées et qu'une opération effectuée dans ... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 4, 6, 8, 10, 12 | |
0dk5 | Given an integer $n > 1$, let $1 = a_1 < a_2 < \dots < a_t = n - 1$ be all the positive integers that are less than $n$ and coprime to $n$. Determine all values of $n$ for which there is no index $i \in \{1, \dots, t-1\}$ satisfying $3 \mid a_i + a_{i+1}$. | [
"It is easy to check that for $n = 2, 4, 10$ the sequences are $(1), (1, 3), (1, 3, 7, 9)$ respectively. Thus these numbers are solution of the given problem. Considering $n \\geq 3$, if $n$ is odd then $\\text{gcd}(n, 2) = 1$ so clearly $a_1 = 1, a_2 = 2$ and their sum is divisible by $3$, which not satisfy. Hence... | Saudi Arabia | Saudi Arabia booklet 2024 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 2, 4, 10 | |
039f | Find the least number $m$ for which any five equilateral triangles with combined area $m$ can cover an equilateral triangle of area $1$. | [
"We prove that $m = 2$. First we show that $m \\ge 2$. It suffices for any $s \\in (0,1)$ to find five equilateral triangles with combined area greater than $2s$, which can not cover an equilateral triangle $\\triangle ABC = \\Delta$ of area $1$. Let $A_1B_1C_1$ be an equilateral triangle of area $(1+s)/2$ and vert... | Bulgaria | Bulgarian National Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 2 | |
05y9 | Problem:
Soit $ABC$ un triangle isocèle en $A$ dont tous les angles sont aigus, et soit $D$ un point situé sur le segment $[BC]$. Soit $\ell$ la droite parallèle à $(BC)$ passant par $A$, puis $X$ le point de $\ell$ pour lequel $(XD)$ est perpendiculaire à $(BC)$, et $\Gamma$ le cercle de centre $X$ passant par $D$. L... | [
"Solution:\n\n\n\nNotons $T$ le point d'intersection des bissectrices de $\\widehat{EYA}$ et $\\widehat{FZA}$, $U$ le point d'intersection des droites $(EY)$ et $(FZ)$, et $\\mathcal{C}$ le cercle de rayon $[AU]$. Puisque les angles $\\widehat{UEA}$ et $\\widehat{AFU}$ sont droits, $E$ et $... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
07oj | Determine the real solutions of the equation
$$
\sqrt{2x^2 + 2x + 3} + \sqrt{2x^2 + 2} = \sqrt{3x^2 + 2x - 1} + \sqrt{x^2 + 6}.
$$ | [
"Let $a$, $b$, $c$, $d$ be the quadratic polynomials defined by\n$$\n\\begin{aligned}\na(x) &= 2x^2 + 2x + 3 & b(x) &= 2x^2 + 2 \\\\\nc(x) &= 3x^2 + 2x - 1 & d(x) &= x^2 + 6.\\end{aligned}\n$$\nThe equation to be solved is writable in the form\n$$\n\\sqrt{a(x)} + \\sqrt{b(x)} = \\sqrt{c(x)} + \\sqrt{d(x)}.\n$$\n\nO... | Ireland | Irska 2014 | [
"Algebra > Equations and Inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | {-3, -2, 1, 2} | |
0k1u | Problem:
Consider the addition problem:
| | C | A | S | H |
| :---: | :---: | :---: | :---: | :---: |
| + | | | M | E |
| O | S | I | D | E |
where each letter represents a base-ten digit, and $C, M, O \neq 0$. (Distinct letters are allowed to represent the same digit) How many ways are there to assign values ... | [
"Solution:\n\nClearly, $C A S H$ and $M E$ cannot add up to $11000$ or more, so $O=1$ and $S=0$. By examining the units digit, we find that $H=0$. Then $C A S H+M E<9900+99<10000$, so there are no solutions."
] | United States | HMMT November 2018 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Logic"
] | null | proof and answer | 0 | |
0i54 | Problem:
How many four-digit numbers are there in which at least one digit occurs more than once? | [
"Solution:\n\nThere are $9000$ four-digit numbers altogether. If we consider how many four-digit numbers have all their digits distinct, there are $9$ choices for the first digit (since we exclude leading zeroes), and then $9$ remaining choices for the second digit, then $8$ for the third, and $7$ for the fourth, f... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 4464 | |
037c | Problem:
Let $a_{1}, a_{2}, \ldots$ be a sequence of real numbers less than $1$ and such that $a_{n+1}(a_{n}+2)=3$, $n \geq 1$. Prove that:
a) $-\frac{7}{2}<a_{n}<-2$
б) $a_{n}=-3$ for any $n$. | [
"Solution:\n\na) It follows from $a_{n}<1$ and $a_{n+1}=\\frac{3}{a_{n}+2}<1$ that $a_{n}<-2$. Thus, $a_{n+1}<-2$, and therefore $a_{n}+2=\\frac{3}{a_{n+1}}> -\\frac{3}{2}$, i.e. $a_{n}>-\\frac{7}{2}$.\n\nб) First solution. Set $b_{n}=a_{n}+3$. Then $b_{n+1}=\\frac{3 b_{n}}{b_{n}-1}$. It follows from a) that $-\\fr... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a_n = -3 for all n | |
0cf3 | Determine all positive integers $n$ satisfying the following condition: There exist pairwise distinct integers $a_1, \dots, a_n, b_1, \dots, b_n$ such that
$$
\prod_{i=1}^{n} (a_k^2 + a_i a_k + b_i) = \prod_{i=1}^{n} (b_k^2 + a_i b_k + b_i) = 0 \quad \text{for all } k = 1, \dots, n.
$$ | [
"The required integers are $n = 1$ and $n = 2$. In the former case, $(a_1, b_1) = (1, -2)$ is the unique pair of integers satisfying the conditions in the statement; and in the latter, only $(a_1, b_1, a_2, b_2) = (2, 0, -1, -2)$ and $(a_1, b_1, a_2, b_2) = (-1, -2, 2, 0)$ fit the bill. Verification is routine.\n\n... | Romania | Seventeenth Stars of Mathematics Competition | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | 1, 2 | |
0bv0 | Problem:
Se consideră șirul $\left(x_{n}\right)_{n \geq 1}$, în care $x_{1}=\frac{1}{3}$ și $x_{n+1}=(n+1) x_{n}+\frac{(n+1)(n+1)!}{3^{n+1}}, \forall n \in \mathbb{N}^{*}$.
Să se calculeze $\lim _{n \rightarrow \infty} n\left(\frac{x_{n}}{n!}-\frac{3}{4}\right)$. | [] | Romania | OLIMPIADA DE MATEMATICĂ ETAPA LOCALĂ | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 0 | |
0hxh | Problem:
In the game Pogémon, Pogé Balls contain Pichagus with a $50\%$ chance and otherwise contain nothing. The new Pogémon universe started with one Pichagu and has existed for five hours. Every hour, every Pichagu at the time opened a new Pogé Ball.
a) What is the expected number of Pichagus created in the last h... | [
"Solution:\n\na) Every hour, the expected number of Pichagus increases $50\\%$. Thus, the expected number of Pichagus after $4$ hours is $\\left(\\frac{3}{2}\\right)^{4}$. Each of these Pichagus has a $50\\%$ chance of creating a new Pichagu in the last hour, so the answer is $\\frac{1}{2}\\left(\\frac{3}{2}\\right... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | a) 81/32; b) 17091/512 | |
06jx | In $\triangle ABC$, let $AD$ be the angle bisector of $\angle BAC$, with $D$ on $BC$. The perpendicular from $B$ to $AD$ intersects the circumcircle of $\triangle ABD$ at $B$ and $E$. Prove that $E, A$ and the circumcentre $O$ of $\triangle ABC$ are collinear. | [
"By simple angle chasing, we find that\n$$\n\\begin{align*} \n\\angle BAE &= \\angle BAD + \\angle DAE = \\frac{A}{2} + \\angle DBE = \\frac{A}{2} + B - \\left(90^\\circ - \\frac{A}{2}\\right) \\\\ \n&= A + B - 90^\\circ = 90^\\circ - C = \\angle BAO. \n\\end{align*}\n$$\nAs $O$ and $E$ lie on the same side of $AB$... | Hong Kong | HKG TST | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
06uy | Let $O$ be the circumcentre, and $\Omega$ be the circumcircle of an acute-angled triangle $A B C$. Let $P$ be an arbitrary point on $\Omega$, distinct from $A, B, C$, and their antipodes in $\Omega$. Denote the circumcentres of the triangles $A O P, B O P$, and $C O P$ by $O_{A}, O_{B}$, and $O_{C}$, respectively. The ... | [
"Let the lines $\\ell_{B}$ and $\\ell_{C}$ meet at $L_{A}$; define the points $L_{B}$ and $L_{C}$ similarly. Note that the sidelines of the triangle $L_{A} L_{B} L_{C}$ are perpendicular to the corresponding sidelines of $A B C$. Points $O_{A}, O_{B}, O_{C}$ are located on the corresponding sidelines of $L_{A} L_{B... | IMO | IMO Shortlisted Problems | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0324 | Problem:
Find all real values of $a$ such that the system
$$
\left\lvert\,
\begin{aligned}
& \frac{a x+y}{y+1}+\frac{a y+x}{x+1}=a \\
& a x^{2}+a y^{2}=(a-2) x y-x
\end{aligned}
\right.
$$
has a unique solution. | [
"Solution:\nIf $x \\neq -1$ and $y \\neq -1$, we easily get that $y = a$. Plugging it in the second equation gives\n$$\na x^{2} - (a^{2} - 2a - 1)x + a^{3} = 0\n$$\nIf $a = 0$ the system has a unique solution $(0 ; 0)$. If $a \\neq 0$, we consider the following two cases.\n\nCase 1. $-1$ is a root of $(*)$. Then $(... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a = 0 or a = -1/3 | |
0i2y | Problem:
Let $a_{1}=3$ and define $a_{n+1} = \left(3 a_{n}^{2} + 1\right) / 2 - a_{n}$ for $n \geq 1$. If $n$ is a power of $3$, prove that $a_{n}$ is divisible by $n$. | [
"Solution:\n\nThe main trick is finding a closed-form expression for $a_{n}$, which requires some experimentation. We will show that $a_{n} = \\left(2^{2^{n}+1} + 1\\right) / 3$ for all $n$ by induction. It is easy to check that the formula holds for $n=1$. And if it holds for some $n$, then\n\n$$\n\\begin{aligned}... | United States | Berkeley Math Circle | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
04zr | There are 8 identical dice. The numbers $4$, $5$, $6$ are written on three faces of the dice, as shown in the figure, and the remaining faces carry the numbers $1$, $2$, $3$ so that the sum of the numbers written on each pair of opposite faces is $7$.
a) Show that using these dice, it is pos... | [
"a) Put together four dice as depicted in Fig. 7. These dice form the lower layer of the cube. On top of this, place another similar layer turned upside down. By the construction of the layer, the numbers on the faces touching each other within one layer coincide everywhere. As the second layer is turned upside dow... | Estonia | Selected Problems from the Final Round of National Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) Yes, it is possible. b) No, it is impossible. | |
09vw | Each of Kwik, Kwek, and Kwak is lying on two consecutive days of the week and is telling the truth on the other five days. No two of them are lying on the same day. Uncle Donald wants to know who of his nephews ate his sweets. The three nephews know all too well who did it. On Sunday, Kwik says that Kwek ate the sweets... | [
"C) It was Kwak."
] | Netherlands | First Round | [
"Discrete Mathematics > Logic"
] | English | MCQ | C | |
0ji5 | Problem:
Let $ABCD$ be a convex quadrilateral. Extend line $CD$ past $D$ to meet line $AB$ at $P$ and extend line $CB$ past $B$ to meet line $AD$ at $Q$. Suppose that line $AC$ bisects $\angle BAD$. If $AD=\frac{7}{4}$, $AP=\frac{21}{2}$, and $AB=\frac{14}{11}$, compute $AQ$. | [
"Solution:\n\nAnswer: $\\frac{42}{13}$\n\nWe prove the more general statement $\\frac{1}{AB}+\\frac{1}{AP}=\\frac{1}{AD}+\\frac{1}{AQ}$, from which the answer easily follows.\n\nDenote $\\angle BAC=\\angle CAD=\\gamma$, $\\angle BCA=\\alpha$, $\\angle ACD=\\beta$. Then we have that by the law of sines,\n$$\n\\frac{... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 42/13 | |
0c7g | We consider the sets $M = \{0, 1, 2, \dots, 2019\}$ and
$$
A = \left\{ x \in M \mid \frac{x^3 - x}{24} \in \mathbb{N} \right\} .
$$
a)
How many elements has the set $A$?
b)
Determine the smallest positive integer $n$, $n \ge 2$, which has the property that any subset with $n$ elements of the set $A$ contains two dist... | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | a) 1263; b) 26 | |
0i4v | Problem:
The Dyslexian alphabet consists of consonants and vowels. It so happens that a finite sequence of letters is a word in Dyslexian precisely if it alternates between consonants and vowels (it may begin with either). There are 4800 five-letter words in Dyslexian. How many letters are in the alphabet? | [
"Solution:\n\nSuppose there are $c$ consonants, $v$ vowels. Then there are $c \\cdot v \\cdot c \\cdot v \\cdot c + v \\cdot c \\cdot v \\cdot c \\cdot v = (c v)^2 (c+v)$ five-letter words. Thus, $c+v = 4800 / (c v)^2 = 3 \\cdot (40 / c v)^2$, so $c v$ is a divisor of $40$. If $c v \\leq 10$, we have $c+v \\geq 48$... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 12 | |
0kgm | The base-nine representation of the number $N$ is 27,006,000,052$_{nine}$. What is the remainder when $N$ is divided by 5?
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4 | [] | United States | AMC 12 A | [
"Number Theory > Modular Arithmetic"
] | null | MCQ | D | |
07hs | Consider a natural number $n$. Amin and Ali play a game, each player at his own turn chooses a number $i$ from the set $\{0, 1, 2, \dots, n\}$ that has been never chosen before, and chooses a non-zero rational number $a_i$. Ali takes the first move. When all the numbers $\{0, 1, 2, \dots, n\}$ get chosen, they form a p... | [
"We claim that Ali has a winning strategy for even $n$. If $n$ is even, Ali does the last move, assuming that he has to choose $a_k$. Choose a rational number $q$ which is not a root of $Q(x) = \\sum_{i \\neq k} a_i x^i$. Now Ali can choose $a_k = \\frac{-Q(q)}{q^k}$ and win the game.\n\nIf $n$ is odd we show that ... | Iran | 40th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | null | proof and answer | All even n | |
0d6u | Let $I_{a}$ be the excenter of triangle $A B C$ with respect to $A$. The line $A I_{a}$ intersects the circumcircle of triangle $A B C$ at $T$. Let $X$ be a point on segment $T I_{a}$ such that $X I_{a}^{2}=X A \cdot X T$. The perpendicular line from $X$ to $B C$ intersects $B C$ at $A^{\prime}$. Define $B^{\prime}$ an... | [
"Let $P$ be the foot of perpendicular from $I_{a}$ to $B C$, $D$ the midpoint of $B C$ and $S_{a}$ the reflection of $A$ with respect to $S$. The line through $I$ parallel to $B C$ cuts $A P$, $A A^{\\prime}$, $A V$ at $Q$, $R$, $L$, respectively.\nRedefine $A^{\\prime}$ as follow: Let $I$ be the incenter and $S$ t... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane ... | English | proof only | null | |
0e5s | Let $p$ be a polynomial of degree $2$ with at least one non-integer coefficient. Suppose that for every integer $n$, the value $p(n)$ is also an integer. Prove that all the coefficients of the polynomial $q(x) = p(x) - \frac{1}{2}x(x+1)$ are integers. | [
"Denote $P(x) = \\alpha x^2 + \\beta x + \\gamma$. Then $P(0) = \\gamma$, $P(1) - P(0) = (\\alpha + \\beta + \\gamma) - \\gamma = \\alpha + \\beta$ and $P(-1) - P(0) = (\\alpha - \\beta + \\gamma) - \\gamma = \\alpha - \\beta$ must be integers. We thus know that $\\alpha + \\beta, \\alpha - \\beta \\in \\mathbb{Z}$... | Slovenia | National Math Olympiad 2012 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
02ao | Problem:
Uma faixa retangular de cartolina tem $5~\mathrm{cm}$ por $1~\mathrm{cm}$. Corte a faixa com 4 cortes retilíneos de modo a poder montar um quadrado com as peças obtidas (não vale superposição das peças). | [
"Solution:\n\n"
] | Brazil | Desafios | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
07nw | We say that a doubly infinite sequence
$$
\dots, s_{-2}, s_{-1}, s_0, s_1, s_2, \dots
$$
is subaveraging if $s_n = (s_{n-1} + s_{n+1})/4$ for all integers $n$.
a. Find a subaveraging sequence in which all entries are different from each other. Prove that all entries are indeed distinct.
b. Show that if $(s_n)$ is a s... | [
"a. A sequence $(s_n)$ is subaveraging iff $s_{n+1} = 4s_n - s_{n-1}$ for all $n \\in \\mathbb{Z}$. This is equivalent to $s_{n-1} = 4s_n - s_{n+1}$ so that we can go backwards by the same recursion. The sequence $(s_n)$ is completely determined by $s_0$ and $s_1$.\n\nFor $s_0 = 0$ and $s_1 = 1$ we prove by inducti... | Ireland | Ireland | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0crn | An expression $\frac{a}{b} \cdot \frac{c}{d} \cdot \frac{e}{f}$ is written on the blackboard (numbers $a$, $b$, $c$, $d$, $e$, and $f$ are positive integers). If one increases $a$ by $1$, then the value of this expression increases by $3$. If one increases $c$ by $1$ in the initial expression, then its value increases ... | [
"**Первое решение.** Пусть значение исходного выражения равно $A$. Тогда в результате первой операции произведение примет значение $\\frac{a+1}{a} \\cdot A = A + 3$, откуда $A = 3a$. Значит, $A$ — натуральное число. Кроме того, из этого равенства следует, что оно делится на $3$. Аналогично доказывается, что число $... | Russia | XL Russian mathematical olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 60 | |
0a3g | Problem:
Speler Zero en speler One spelen een spel op een $n \times n$-bord $(n \geq 1)$. De kolommen van dit $n \times n$-bord zijn genummerd met tweemachten, dus we hebben kolom 1, kolom 2, kolom 4 tot en met kolom $2^{n-1}$. Om en om zetten de spelers hun eigen getal (dus Zero een 0 en One een 1) in één van de lege... | [
"Solution:\n\na.\nOplossing 1. Merk eerst op dat als $n=1$, het unieke vakje een 0 krijgt omdat speler Zero begint, dus in dit geval is het mogelijk.\nWe gaan nu bewijzen dat voor alle andere $n$ Zero kan voorkomen dat One wint. Omdat $4 \\mid 2^{k}$ voor alle $k \\geq 2$, bepalen alleen de eerste twee kolommen wie... | Netherlands | IMO-selectietoets | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | null | proof and answer | a) n = 1.
b) n ≡ 0, 1, 3 (mod 4). | |
0a4s | Problem:
Find all positive integers $n$ such that $n^{4} - n^{3} + 3n^{2} + 5$ is a perfect square. | [
"Solution:\nLet $f(n) = 4n^{4} - 4n^{3} + 12n^{2} + 20 = 4(n^{4} - n^{3} + 3n^{2} + 5)$ and note that $(n^{4} - n^{3} + 3n^{2} + 5)$ is a perfect square if and only if $f(n)$ is. First note that:\n$$\n(2n^{2} - n + 5)^{2} - f(n) = 9n^{2} - 10n + 5 = 4n^{2} + 5(n - 1)^{2} > 0.\n$$\nAlso note that\n$$\nf(n) - (2n^{2}... | New Zealand | NZMO Round One | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 2 | |
0ff9 | Problem:
Dado un pentágono regular, se dibujan sus cinco segmentos diagonales. Se pide determinar el número total de triángulos que aparecen construidos en la figura y clasificar este conjunto de triángulos en clases de triángulos iguales (directa o inversamente) entre sí. | [
"Solution:\n\n\n\nEn cada figura se ha dibujado en color gris un representante de cada clase de triángulos y en el centro el número de los que hay en esa clase."
] | Spain | Olimpiadas Matemáticas Españolas | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Rotation",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 35 triangles in total, partitioned into congruence classes: one class with 10 triangles and five classes with 5 triangles each. | |
0fc8 | Problem:
Un club tiene 25 miembros. Cada comité está formado por 5 miembros. Dos comités cualesquiera tienen como mucho un miembro en común. Prueba que el número de comités no puede ser superior a 30. | [
"Solution:\n\nSupongamos que haya 31 comités. Entonces hay, al menos, $31 \\times 5 = 155$ asientos en estos 31 comités. Como sólo hay 25 miembros, al menos, uno de ellos tendrá que ocupar, al menos, 7 de los 155 asientos. Consideremos este miembro $A$ y 7 de los comités donde se sienta. Hay, al menos, otros 28 asi... | Spain | null | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0ia8 | Problem:
For what value of $a > 1$ is
$$
\int_{a}^{a^{2}} \frac{1}{x} \log \frac{x-1}{32} \, dx
$$
minimum? | [
"Solution:\nLet $f(a) = \\int_{a}^{a^{2}} \\frac{1}{x} \\log \\frac{x-1}{32} \\, dx$. Then we want $\\frac{d f}{d a} = 0$; by the Fundamental Theorem of Calculus and the chain rule, this implies that\n\n$$\n2a \\left( \\frac{1}{a^{2}} \\log \\frac{a^{2}-1}{32} \\right) - \\frac{1}{a} \\log \\frac{a-1}{32} = \\frac{... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Calculus > Derivatives"
] | null | proof and answer | a = 3 | |
07d4 | a.
Prove that there doesn't exist a sequence of positive integers like $a_1, a_2, a_3, \dots$ such that for all $i, j \in \mathbb{Z}^+$ with $i < j$
$$
\text{gcd}(a_i + j, a_j + i) = 1.
$$
b.
Assume that $p$ is an odd prime number. Prove that there exists a sequence of positive integers like $a_1, a_2, a_3, \dots$ suc... | [
"a.\nIf $i, a_i, j$ and $a_j$ are all odd or are all even then $\\text{gcd}(a_i + j, a_j + i)$ will be an even number. So except for at most two values of $i$, the parity of $i, a_i$ is different. So there exists some even numbers $i, j$ such that $a_i, a_j$ are both odd which, again, we conclude that $\\text{gcd}(... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
016e | Determine all positive integers $n$ such that there exists a partition of the set $\{n, n+1, n+2, \dots, n+8\}$ such that the product of all elements of the first subset coincides with the product of all elements of the second subset. | [
"We will prove that there is no such $n$.\nAssume the contrary. Obviously, the members of $\\{n, n+1, n+2, \\dots, n+8\\}$ can have prime factors $p \\le 7$ only. Otherwise exactly one member contains this factor and hence only one product will have this factor.\nAmong the 9 numbers there are exactly 5 odd if $n$ i... | Baltic Way | Baltic Way SHL | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | No positive integer n satisfies the condition. | |
00is | Determine all integer solutions $(x, y, z)$ of the equation $x^4 + x^2 = 7^z y^2$. | [
"Clearly, we have solutions for any value of $z$, if we also have $x = y = 0$. We claim that there are no other solutions.\n\nIn order to show this, we first assume that $z$ is even. In this case, the number $7^z y^2$ is a perfect square, and therefore $x^4 + x^2 = x^2(x^2 + 1)$ must also be a perfect square. The o... | Austria | AustriaMO2011 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | (x, y, z) = (0, 0, z) for all integers z; no other integer solutions exist. | |
0d7t | In a school there are totally $n > 2$ classes and not all of them have the same numbers of students. It is given that each class has one head student. The students in each class wear hats of the same color and different classes have different hat colors. One day all the students of the school stand in a circle facing t... | [
"Suppose that there are $m$ students in total. We number the students standing in the circle by $1, 2, 3, \\ldots, m$, clockwise (starting with an arbitrary student). Denote by $a_{1} < a_{2} < a_{3} < \\cdots < a_{n}$ the positions of the $n$ head students obtained in this numbering.\n\nWe consider an $n \\times m... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
02gq | $ABCD$ is a rhombus. Take points $E, F, G, H$ on sides $AB, BC, CD, DA$ respectively so that $EF$ and $GH$ are tangent to the incircle of $ABCD$. Show that $EH$ and $FG$ are parallel. | [
"Let $O$ be the center of the incircle. We show first that $AE \\cdot CF = AO^2$. Let $\\angle AOE = \\theta$. Then if $OX$ is the perpendicular from $O$ to $AB$, we have $\\angle AOX = B/2$ and hence $\\angle XOE = \\theta - B/2$. If $EF$ touches the circle at $Y$, $\\angle EOY = \\angle XOE$, so $\\angle BEF = 2(... | Brazil | XXV OBM | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0doy | A family $L$ of $2006$ lines on the plane is given in such a way that it doesn't contain parallel lines and it doesn't contain three concurrent lines. We say that the line $l_1 \in L$ is *bounding* the line $l_2 \in L$, if all intersection points of the line $l_2$ with other lines from $L$ lie on the one side of the li... | [
"Assume the contrary, i.e. there aren't lines $l$ and $l'$, satisfying the conditions 1) and 2).\nLet's choose an arbitrary $l \\in L$. There are two points $a, b \\in l$ such that $a, b$ are intersection points of $l$ with lines from $L$, and all intersection points of the line $l$ with lines from $L$ lie on the s... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0k2u | Problem:
Michelle has a word with $2^{n}$ letters, where a word can consist of letters from any alphabet. Michelle performs a switcheroo on the word as follows: for each $k = 0, 1, \ldots, n-1$, she switches the first $2^{k}$ letters of the word with the next $2^{k}$ letters of the word. For example, for $n = 3$, Mich... | [
"Solution:\n\nLet $m(n)$ denote the number of switcheroos needed to take a word of length $2^{n}$ back to itself. Consider a word of length $2^{n}$ for some $n > 1$. After 2 switcheroos, one has separately performed a switcheroo on the first half of the word and on the second half of the word, while returning the (... | United States | HMMT February | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2^n | |
01eo | Prove that there are infinitely many positive integers $n$, which are not divisible by $10$ and such that $s(n^2) < s(n) - 5$ where $s(n)$ is the sum of digits of $n$. | [
"All integers of the form $499\\ldots99$ satisfy the condition. Indeed, if $n = 4\\underbrace{99\\ldots99}_{k} = 5 \\cdot 10^k - 1$ then\n$$\nn^2 = 25 \\cdot 10^{2k} - 10^{k+1} + 1 = 24 \\underbrace{99\\ldots9}_{k-1} \\underbrace{00\\ldots00}_{k} 1.\n$$\nIn such a case $s(n) = 4 + 9k$, but $s(n^2) = 7 + 9(k - 1) = ... | Baltic Way | Baltic Way shortlist | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0f45 | Problem:
Find 21 consecutive integers, each with a prime factor less than 17. | [] | Soviet Union | 15th ASU | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | The 21 consecutive integers from 20570 to 20590 inclusive. | |
00dy | A positive integer is *happy* if
* all of its digits are different and non-zero,
* one of its digits is equal to the sum of the others.
For example, $253$ is a happy number. How many happy numbers are there? | [
"The largest digit should be equal to the sum of the rest of the digits. We separate in cases according to the value of the largest digit.\n\nObserve that if the largest digit is $1$, then it is the only digit and the number is not happy. In the case that $2$ is the largest digit then either it is the only digit or... | Argentina | XXXIII Cono Sur Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 264 | |
031r | Problem:
Let $D$ be a point on the side $AC$ of $\triangle ABC$ such that $BD = CD$. A line parallel to $BD$ intersects the sides $BC$ and $AB$ at points $E$ and $F$, respectively. Set $G = AE \cap BD$. Prove that $\Varangle BCG = \Varangle BCF$. | [
"Solution:\n\nLet $H = AC \\cap EF$. Then $\\Varangle CDG = \\Varangle FHC$ and\n$$\n\\frac{CD}{DG} = \\frac{BD}{DG} = \\frac{FH}{HE} = \\frac{FH}{HC}\n$$\nIt follows that $\\triangle CDG \\sim \\triangle FHC$ which implies that $\\Varangle GCD = \\Varangle CFH$. Hence\n$$\n\\begin{aligned}\n\\Varangle BCG & = \\Va... | Bulgaria | 52. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
06oe | The sequence of real numbers $a_{0}, a_{1}, a_{2}, \ldots$ is defined recursively by
$$
a_{0}=-1, \quad \sum_{k=0}^{n} \frac{a_{n-k}}{k+1}=0 \quad \text{ for } \quad n \geq 1
$$
Show that $a_{n}>0$ for $n \geq 1$. | [
"The proof goes by induction. For $n=1$ the formula yields $a_{1}=1 / 2$. Take $n \\geq 1$, assume $a_{1}, \\ldots, a_{n}>0$ and write the recurrence formula for $n$ and $n+1$, respectively as\n$$\n\\sum_{k=0}^{n} \\frac{a_{k}}{n-k+1}=0 \\quad \\text{ and } \\quad \\sum_{k=0}^{n+1} \\frac{a_{k}}{n-k+2}=0 .\n$$\nSub... | IMO | IMO 2006 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0h9y | In the acute triangle $ABC$ there are altitudes $BP$ and $CQ$, point $T$ is the intersection of altitudes of $\triangle PAQ$. It turns out that $\angle CTB = 90^\circ$. Find the value of $\angle BAC$. | [
"The statement of the problem yields $\\angle BQC = \\angle BTC = \\angle BPC = 90^\\circ$, hence, points $Q, T, P$ lie on the circle with diameter $BC$, and in this exact order: $B, Q, T, P, C$ (Fig. 6), since $\\triangle ABC$ is acute. Then,\n$$\n\\angle QTP = 180^\\circ - \\angle ABP = 180^\\circ - (90^\\circ - ... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 45° | |
06oq | Let $ABC$ be a triangle with incentre $I$. A point $P$ in the interior of the triangle satisfies
$$
\angle PBA + \angle PCA = \angle PBC + \angle PCB.
$$
Show that $AP \geq AI$ and that equality holds if and only if $P$ coincides with $I$. | [
"Let $\\angle A = \\alpha$, $\\angle B = \\beta$, $\\angle C = \\gamma$. Since $\\angle PBA + \\angle PCA + \\angle PBC + \\angle PCB = \\beta + \\gamma$, the condition from the problem statement is equivalent to $\\angle PBC + \\angle PCB = (\\beta + \\gamma)/2$, i.e. $\\angle BPC = 90^\\circ + \\alpha/2$.\n\nOn t... | IMO | IMO 2006 Shortlisted Problems | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > ... | English | proof only | null | |
0jwg | Problem:
Let $P(x), Q(x)$ be nonconstant polynomials with real number coefficients. Prove that if
$$
\lfloor P(y)\rfloor=\lfloor Q(y)\rfloor
$$
for all real numbers $y$, then $P(x)=Q(x)$ for all real numbers $x$. | [
"Solution:\n\nBy the condition, we know that $|P(x)-Q(x)| \\leq 1$ for all $x$. This can only hold if $P(x)-Q(x)$ is a constant polynomial. Now take a constant $c$ such that $P(x)=Q(x)+c$. Without loss of generality, we can assume that $c \\geq 0$. Assume that $c>0$. By continuity, if $\\operatorname{deg} P=\\opera... | United States | February 2017 | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
03t7 | Let $k$ be a positive integer no less than $3$ and $\theta$ be a real number. Prove that, if both $\cos(k - 1)\theta$ and $\cos k\theta$ are rational numbers, then there exists a positive integer $n > k$, such that both $\cos(n - 1)\theta$ and $\cos n\theta$ are rational numbers. | [
"First we prove a lemma.\nLemma Let $\\alpha$ be a real number. If $\\cos \\alpha$ is rational, then $\\cos m\\alpha$ is rational for any positive integer $m$.\nWe prove by induction on $m$. By $\\cos 2\\alpha = 2\\cos^2\\alpha - 1$, we get that the lemma is true for $m = 2$.\nWe suppose that the lemma is true for... | China | China Western Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials"
] | English | proof only | null | |
0cll | Let $ABCDEF$ be a convex hexagon with $\angle A \equiv \angle C \equiv \angle E$ and $\angle B \equiv \angle D \equiv \angle F$, such that there is a point $P$ in its interior that is equidistant from the sides $AB$, $CD$ and $EF$. If $G_1 \neq G_2$ are the centroids of triangles $ACE$ and $BDF$, respectively, prove th... | [
"If $\\angle A = \\angle C = \\angle E = \\alpha$ and $\\angle B = \\angle D = \\angle F = \\beta$, then the sum of the measures of the angles of the hexagon $ABCDEF$ is $3\\alpha + 3\\beta = 720^\\circ$, so $\\alpha + \\beta = 240^\\circ$. Since $\\angle B + \\angle C = \\alpha + \\beta = 240^\\circ$, the lines $A... | Romania | 75th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geome... | English | proof only | null | |
0gsi | There are two boxes: one contains $29$ distinguishable balls and the other one is empty. In a move applied to a picked box we choose some collection of balls from this box and transfer them to the other box. Starting with the non-empty box we repeatedly apply these moves, picking boxes alternatively. What is the maxima... | [
"Answer: $2^{29}-2$.\n\nLet us show that the total number of moves cannot be $2^{29}-1$. If the total number of moves is $2^{29}-1$ then all non-empty subsets of a $29$ element set should be selected. The total number of subsets containing a given element is $2^{28}$, which is an even number. Therefore, at the end ... | Turkey | Turkish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2^29 - 2 | |
02dn | Five spheres of radius $r$ are inside a right circular cone. Four of the spheres lie on the base of the cone. Each touches two of the others and the sloping sides of the cone. The fifth sphere touches each of the other four and also the sloping sides of the cone. Find the volume of the cone. | [
"The left-hand diagram shows the four spheres on the base. Evidently $AC = 2r\\sqrt{2}$.\n\nThe right-hand diagram shows a vertical section through $A$, $C$ and the center $O$ of the top sphere. $P$ is the apex of the cone and $QR$ is a diameter of its base. Evidently $PQR$ is similar to $OAC$ and its sides are par... | Brazil | IV OBM | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | pi * r^3 * (2*sqrt(2) + 1)^3 / 3 | |
0aa7 | Problem:
Find all integers $n$ and $m$, $n > m > 2$, such that a regular $n$-sided polygon can be inscribed in a regular $m$-sided polygon so that all the vertices of the $n$-gon lie on the sides of the $m$-gon. | [
"Solution:\nIt works only for $n = 2m$, and for $m = 3$ and $n = 4$.\n\nTo begin with, let's see why it works for $n = 2m$. For a $2m$-gon we can choose two points on each side, symmetrically, so that the distance between the two of them is equal to the distance between two close points on adjacent sides.\n\nFor $n... | Nordic Mathematical Olympiad | The 31st Nordic Mathematical Contest | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | All pairs with n = 2m for m > 2, and the pair (n, m) = (4, 3). | |
0bv1 | Problem:
Se consideră $I_{n} = \int_{0}^{4} \sqrt[n]{x^{n} + (4-x)^{n}} \, dx$, $n \in \mathbb{N}$, $n \geq 2$. Să se calculeze $\lim_{n \rightarrow \infty} I_{n}$. | [] | Romania | Olimpiada de Matematică Etapa Locală | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Limits"
] | null | proof and answer | 12 | |
0l7o | Let $H$ be the orthocenter of acute triangle $ABC$, let $F$ be the foot of the altitude from $C$ to $AB$, and let $P$ be the reflection of $H$ across $BC$. Suppose that the circumcircle of triangle $AFP$ intersects line $BC$ at two distinct points $X$ and $Y$. Prove that $C$ is the midpoint of $XY$. | [] | United States | 16th United States of America Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
... | null | proof only | null |
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