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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
04p7 | How many different bracelets consisting of four black and four white beads arranged in a circle are there? Two bracelets are considered different if they cannot be turned over so that the beads are equally aligned on them. | [] | Croatia | Croatian Mathematical Society Competitions | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 8 | |
0lcp | Let $(a_n)$ be the sequence such that $a_1 = \frac{3}{2}$ and
$$
a_{n+1} = a_n - \frac{3n+2}{2n(n+1)(2n+1)},\ n \ge 1.
$$
Find the limit $\lim_{n \to +\infty} a_n$. | [] | Vietnam | MOCK TEST FOR VMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | ln 2 | |
0f8n | Problem:
7 boys each went to a shop 3 times. Each pair met at the shop. Show that 3 must have been in the shop at the same time. | [] | Soviet Union | 23rd ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
01gr | Consider $2n$ rays (half-lines) in the plane such that no two rays are parallel (the endpoints of the rays may coincide). Prove that there exists a line in the plane that does not pass through any of the endpoints and intersects with exactly $n$ rays. | [
"Let us choose any circle such that all of the endpoints of the rays are inside the circle. Let us also choose a tangent line on the circle that is not parallel to any of the rays. Let this tangent line be $l_0$ and let $l_\\alpha$ be a tangent line we get after rotating $l_0$ counterclockwise by angle $\\alpha$ wi... | Baltic Way | Baltic Way 2020 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0g8w | 試求所有正整數對 $(x, y)$, 滿足
$$
\sqrt[3]{7x^2 - 13xy + 7y^2} = |x - y| + 1.
$$ | [
"答案為 $x = y = 1$ 與 $\\{x, y\\} = \\{m^3 + m^2 - 2m - 1,\\ m^3 + 2m^2 - m - 1\\}$，其中 $m \\ge 2$。\n\n1. 若 $x = y$，則原式等價於 $x^{2/3} = 1$，故 $x = y = 1$。\n\n2. 若 $x > y$，令 $n = x - y$，則原式可改寫為\n$$\n\\sqrt[3]{7(y + n)^2 - 13(y + n)y + 7y^2} = n + 1.\n$$\n等號兩邊同時立方並化簡後, 我們有\n$$\ny^2 + yn = n^3 - 4n^2 + 3n + 1.\n$$\n為讓左式配方, 我... | Taiwan | 二〇一五數學奧林匹亞競賽第一階段選訓營 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | x = y = 1, or {x, y} = {m^3 + m^2 − 2m − 1, m^3 + 2m^2 − m − 1} for m ≥ 2 | |
0a7t | Problem:
Let $n \geq 2$ and let $x_{1}, x_{2}, \ldots, x_{n}$ be real numbers satisfying $x_{1}+x_{2}+\ldots+x_{n} \geq 0$ and $x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}=1$. Let $M=\max \{x_{1}, x_{2}, \ldots, x_{n}\}$. Show that
$$
M \geq \frac{1}{\sqrt{n(n-1)}}
$$
When does equality hold in (1)? | [
"Solution:\n\nDenote by $I$ the set of indices $i$ for which $x_{i} \\geq 0$, and by $J$ the set of indices $j$ for which $x_{j}<0$. Let us assume $M<\\frac{1}{\\sqrt{n(n-1)}}$. Then $I \\neq\\{1,2, \\ldots, n\\}$, since otherwise we would have $|x_{i}|=x_{i} \\leq \\frac{1}{\\sqrt{n(n-1)}}$ for every $i$, and $\\s... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 9 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | M ≥ 1/√(n(n−1)). Equality holds exactly when n−1 of the numbers equal 1/√(n(n−1)) and the remaining number equals (1−n)/√(n(n−1)). | |
0fmd | Problem:
Sea $ABCD$ un cuadrilátero convexo y $P$ un punto interior. Determina cuáles son las condiciones que deben cumplir el cuadrilátero y el punto $P$ para que los cuatro triángulos $PAB$, $PBC$, $PCD$ y $PDA$ tengan la misma área. | [
"Solution:\n\nConsideremos, primero, los triángulos $PCD$ y $PCB$. Tienen la base común $PC$ y alturas correspondientes $DX$ y $BY$. Si queremos que tengan la misma área, las alturas deben ser iguales. Por lo tanto, el punto $Q$ tiene que ser el punto medio de la diagonal $BD$. La recta $CP$ debe pasar por $Q$. Aná... | Spain | Spain | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0eix | Problem:
Na državno tekmovanje v računanju se lahko uvrsti največ 30 tekmovalcev. Na letošnjem državnem tekmovanju so tekmovalci reševali 4 naloge, pri čemer je $\frac{1}{3}$ tekmovalcev rešila natanko 3 naloge, $\frac{1}{4}$ tekmovalcev je rešila natanko 2 nalogi, $\frac{1}{6}$ tekmovalcev je rešila natanko 1 nalogo,... | [
"Solution:\n\nNajmanjši skupni večkratnik števil $3, 4, 6$ in $8$ je $24$. Število tekmovalcev na tekmovanju mora biti torej deljivo s $24$. Ker pa so vsi večkratniki števila $24$, razen števila $24$, večji od $30$, je bilo na tekmovanju $24$ tekmovalcev, od katerih so vse $4$ naloge rešili\n$$\n\\left(1-\\frac{1}{... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | C | |
0kbs | Problem:
Let $x$ and $y$ be non-negative real numbers that sum to $1$. Compute the number of ordered pairs $(a, b)$ with $a, b \in \{0,1,2,3,4\}$ such that the expression $x^{a} y^{b} + y^{a} x^{b}$ has maximum value $2^{1-a-b}$. | [
"Solution:\n\nLet $f(x, y) = x^{a} y^{b} + y^{a} x^{b}$. Observe that $2^{1-a-b}$ is merely the value of $f\\left(\\frac{1}{2}, \\frac{1}{2}\\right)$, so this value is always achievable.\n\nWe claim (call this result (*)) that if $(a, b)$ satisfies the condition, so does $(a+1, b+1)$. To see this, observe that if $... | United States | HMMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 17 | |
00a3 | Consider the points $O = (0,0)$, $A = (-2,0)$ and $B = (0,2)$ in the coordinate plane. Let $E$ and $F$ be the midpoints of $OA$ and $OB$ respectively. Rotate triangle $OEF$ clockwise about $O$ to reach a triangle $OE'F'$ and, for each rotated position, let $P = (x, y)$ be the intersection of lines $AE'$ and $BF'$. Find... | [
"Let $R$ be the clockwise $90^{\\circ}$ rotation about $O$. Apparently $R$ takes $A$ to $B$ and also $R(E') = F'$ for each rotated position $OE'F'$ of the initial right isosceles triangle $OEF$. Hence $R$ takes line $AE'$ to line $BF'$. The angle between a line and its image under any rotation equals the angle of r... | Argentina | Argentine National Olympiad 2015 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coo... | English | proof and answer | (1 + sqrt(3))/2 | |
0f7f | Problem:
$ABCDE$ is a convex pentagon with $\angle ABC = \angle ADE$ and $\angle AEC = \angle ADB$. Show that $\angle BAC = \angle DAE$. | [] | Soviet Union | 21st ASU | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates"
] | null | proof only | null | |
0e5j | Find all integers $a, b, c$ and $d$ that satisfy the equality
$$
a\sqrt{2} + b\sqrt{5} + c = d\sqrt{10}.
$$ | [
"One solution is straightforward: $a = b = c = d = 0$. We will prove that it is the only one. Suppose there is another solution $(a, b, c, d)$. We may suppose that the integers $a, b, c$ and $d$ are coprime, otherwise their greatest common divisor could be deleted from the equation (because not all numbers are equa... | Slovenia | National Math Olympiad 2012 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | a=0, b=0, c=0, d=0 | |
09ei | 4 problems are posed on a certain examination. 98% of students solved I problem, 90% solved II problem, 85% solved III problem. What is the least and the most percentage of students that solved all three problems? | [
"First we will prove a lemma which is a generalized form of the given problem. Let $|\\Omega|$ be the universal set.\n\n**Lemma:** If $a_1 \\leq |A| \\leq a_2$; $b_1 \\leq |B| \\leq b_2$ then the double inequality\n$$\n\\max\\{0, a_1 + a_2 - |\\Omega|\\} \\leq |A \\cap B| \\leq \\min\\{a_2, b_2\\}\n$$\nholds.\n\n**... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | proof and answer | least 73%, greatest 85% | |
0ctg | Find all pairs of distinct real $x$ and $y$ such that $x^{100} - y^{100} = 2^{99}(x - y)$ and $x^{200} - y^{200} = 2^{199}(x - y)$.
Найдите все пары различных действительных чисел $x$ и $y$ такие, что $x^{100} - y^{100} = 2^{99}(x - y)$ и $x^{200} - y^{200} = 2^{199}(x - y)$. | [
"$(x, y) = (2, 0)$ and $(x, y) = (0, 2)$.\n\nSet $x = 2a$, $y = 2b$. We have $a^{100} - b^{100} = a^{200} - b^{200} = a - b \\neq 0$, whence $a^{100} + b^{100} = 1$. The case $ab = 0$ is easy. Assume that $ab \\neq 0$; then $|a|, |b| < 1$. Since $a^{100} - a = b^{100} - b$, we have $ab > 0$. Now,\n$$\n1 = \\left| \... | Russia | Russian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English; Russian | proof and answer | (x, y) = (2, 0) and (x, y) = (0, 2) | |
0jd1 | Problem:
Generalization: Given a segment $AB$ and a point $M$ inside of it, construct circle $\omega_{l}$ centered at $O_{l}$ passing through $A$ and $M$ and $\omega_{r}$ centered at $O_{r}$ passing through $M$ and $B$ so that $O_{l}$ and $O_{r}$ are on the same side of $AB$ and $\angle A O_{l} M = \angle M O_{r} B = ... | [
"Solution:\n\nAs above, $N$ is on the same side of $AB$ as $O_{l}$ and $O_{r}$.\n\nFor the first part, $\\angle ANM = x$ because it spans the arc $AM$; hence $\\angle MND = 180^{\\circ} - x$. As $MNDB$ is cyclic, we have $\\angle MBD = x$.\n\nFor the second part, $\\angle ANB = \\angle ANM + \\angle MNB = x + x = 2... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0i8l | Problem:
The rational numbers $x$ and $y$, when written in lowest terms, have denominators $60$ and $70$, respectively. What is the smallest possible denominator of $x+y$? | [
"Solution:\n\nWrite $x + y = \\dfrac{a}{60} + \\dfrac{b}{70} = \\dfrac{7a + 6b}{420}$. Since $a$ is relatively prime to $60$ and $b$ is relatively prime to $70$, it follows that none of the primes $2, 3, 7$ can divide $7a + 6b$, so we won't be able to cancel any of these factors in the denominator. Thus, after redu... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 84 | |
03zx | Let $P$ be a point on the image of $y = x + \frac{2}{x}$ ($x > 0$). Through $P$ draw lines perpendicular to $y = x$ and $y$-axis with foot points $A, B$, respectively. Then the value of $\vec{PA} \cdot \vec{PB}$ is ______. | [
"Let $P(x_0, x_0 + \\frac{2}{x_0})$. The expression for line $PA$ is then\n$$\ny - \\left(x_0 + \\frac{2}{x_0}\\right) = -(x - x_0),\n$$\nor\n$$y = -x + 2x_0 + \\frac{2}{x_0}.$$\nFrom\n$$\n\\begin{cases} y = x, \\\\ y = -x + 2x_0 + \\frac{2}{x_0}, \\end{cases}\n$$\nwe get $A(x_0 + \\frac{1}{x_0}, x_0 + \\frac{1}{x_... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | -1 | |
0fsb | Problem:
Soit $n$ un nombre entier strictement positif. Soient $x_{1} \leq x_{2} \leq \ldots \leq x_{n}$ des nombres réels tels que $x_{1}+x_{2}+\ldots+x_{n}=0$ et $x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}=1$. Montrer que $x_{1} x_{n} \leq -1 / n$. | [
"Solution:\n\nOn commence par élever au carré la condition que la somme des $x_{i}$ est $0$ :\n$$\n\\underbrace{\\sum x_{i}^{2}}_{=1}+\\sum_{i \\neq j} 2 x_{i} x_{j}=\\left(\\sum x_{i}\\right)^{2}=0\n$$\net donc on conclut que\n$$\n\\sum_{i \\neq j} 2 x_{i} x_{j}=-1\n$$\nLe but est de faire apparaître seulement des... | Switzerland | null | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0gx2 | We know that at some natural $n$ the number $n^2 + 2008n$ written in decimal notation ends with 4. Find what digit is in the ten's place of the number. | [
"It's clear that the number $2000n$ does not influence the answer, which implies that the sought digits will be the same for numbers $A = n^2 + 2008n$ and $B = n^2 + 8n^2$. As number $(B+16)$ equals $(n+4)^2$ (being the square of the natural number) and ends in $0$, this number should end in $00$. Thus $B = \\overl... | Ukraine | Ukrajina 2008 | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 8 | |
06de | Find all integers $n$ satisfying all three conditions
$$
n \equiv 2 \pmod{3}, \quad n \equiv -1 \pmod{5} \quad \text{and} \quad n \equiv 3 \pmod{7}.
$$ | [
"The answer is any integer of the form $105k + 59$ where $k \\in \\mathbb{Z}$.\nSince $n \\equiv 2 \\equiv -1 \\pmod{3}$ and $n \\equiv -1 \\pmod{5}$, we have\n$$\nn \\equiv -1 \\pmod{15}.\n$$\nTesting $n = -1, 14, 29, 44, 59$, we see that $n = 59$ satisfies $n \\equiv 3 \\pmod{7}$. Therefore, $n = 59$ is one solut... | Hong Kong | IMO HK TST | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | n ≡ 59 (mod 105) | |
0bl7 | Let $n$ be a positive integer. Prove that the polynomial
$$(X^3 + X + 1)^n + 5X^2 + 30X + 5$$
is irreducible in $\mathbb{Z}[X]$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof only | null | |
0cvh | Let $BL$ be an internal angle bisector in a scalene triangle $ABC$, with $L \in AC$. The extension of the median through $B$ meets the circumcircle $\omega$ of $ABC$ at point $D$. A line $\ell$ through the circumcenter of the triangle $BDL$ is parallel to $AC$. Prove that $\omega$ is tangent to $\ell$. | [
"Пусть $M$ — середина отрезка $AC$, $S$ — вторая точка пересечения прямой $BL$ с окружностью $\\omega$, $N$ — середина дуги $ABC$ (см. рис. 8). Тогда $S$ — середина меньшей дуги $AC$ окружности $\\omega$, а точки $M, S, N$ лежат на серединном перпендикуляре к отрезку $AC$. Прямая $BN$ — внешняя биссектрица угла $AB... | Russia | Regional round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point ... | English; Russian | proof only | null | |
0j0r | Problem:
Express the following in closed form, as a function of $x$:
$\sin^{2}(x) + \sin^{2}(2x) \cos^{2}(x) + \sin^{2}(4x) \cos^{2}(2x) \cos^{2}(x) + \cdots + \sin^{2}\left(2^{2010} x\right) \cos^{2}\left(2^{2009} x\right) \cdots \cos^{2}(2x) \cos^{2}(x)$. | [
"Solution:\n\n$1 - \\dfrac{\\sin^{2}\\left(2^{2011} x\\right)}{4^{2011} \\sin^{2}(x)}$\n\nNote that\n\\[\n\\begin{aligned}\n& \\sin^{2}(x) + \\sin^{2}(2x) \\cos^{2}(x) + \\cdots + \\sin^{2}\\left(2^{2010} x\\right) \\cos^{2}\\left(2^{2009} x\\right) \\cdots \\cos^{2}(x) \\\\\n& \\quad = \\left(1 - \\cos^{2}(x)\\rig... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | final answer only | 1 - sin^2(2^{2011} x) / (4^{2011} sin^2(x)) | |
0hm1 | Problem:
There are three prisoners in a prison. A warden has 2 red and 3 green hats and he has decided to play the following game: He puts the prisoners in a row one behind the other and on the head of each prisoner he puts a hat. The first prisoner in the row can't see any of the hats, the second prisoner can see onl... | [
"Solution:\n\nIf the first two prisoners had red hats, the third one won't be silent (he would conclude that his hat is green). Hence, at least one of the first two prisoners has a green hat, and everybody knows that (because the third prisoner is silent). Thus if the first prisoner had red hat, the second one woul... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Logic"
] | null | proof and answer | green | |
0ibl | Problem:
Suppose the function $f(x)-f(2 x)$ has derivative $5$ at $x=1$ and derivative $7$ at $x=2$. Find the derivative of $f(x)-f(4 x)$ at $x=1$. | [
"Solution:\nLet $g(x)=f(x)-f(2 x)$. Then we want the derivative of\n$$\nf(x)-f(4 x)=(f(x)-f(2 x))+(f(2 x)-f(4 x))=g(x)+g(2 x)\n$$\nat $x=1$. This is $g'(x)+2 g'(2 x)$ at $x=1$, or $5+2 \\cdot 7=19$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Calculus > Derivatives"
] | null | final answer only | 19 | |
0la1 | Let $a > 2$ be a real number and $f_n(x) = a^{10} x^{n+10} + x^n + \dots + x + 1$ ($n = 1, 2, \dots$). Prove that for every positive integer $n$ the equation $f_n(x) = a$ has exactly a real root $x_n \in (0; +\infty)$. Prove that the sequence $(x_n)$ has a finite limit when $n \to +\infty$. | [
"For every $n$ we define $g_n(x) = f_n(x) - a$. Then $g_n(x)$ is a continuous and increasing function on $[0; +\\infty)$. We have $g_n(0) = 1 - a < 0$; $g_n(1) = a^{10} + n + 1 - a > 0$ so $g_n(x) = 0$ has the only root $x_n$ in $(0; +\\infty)$.\n\nTo prove the existence of the limit $\\lim_{n \\to \\infty} x_n$, w... | Vietnam | Vijetnam 2007 | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | English | proof only | null | |
01t0 | We say that a diagonal of a convex pentagon is *good* if it divides the pentagon into a triangle and a circumscribed quadrilateral.
Find the greatest number of good diagonals in a convex pentagon. (I. Gorodnin) | [
"Answer: 2.\n\nShow that any two intersecting diagonals of the pentagon cannot be good at the same time. Suppose, contrary to our claim, that there are two good intersecting diagonals. Without loss of generality, we assume that $AD$ and $BE$ are good diagonals of the pentagon $ABCDE$ (see Fig. 1). Then $BCDE$ and $... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 2 | |
036y | Problem:
Find the maximum of the function
$$
f(x) = \frac{\lg x \cdot \lg x^{2} + \lg x^{3} + 3}{\lg^{2} x + \lg x^{2} + 2}
$$
and the values of $x$, when it is attained. | [
"Solution:\nThe domain of $f(x)$ is $x > 0$. Setting $y = \\lg x$ gives\n$$\nF(y) = \\frac{2y^{2} + 3y + 3}{y^{2} + 2y + 2}\n$$\nSince the denominator is positive, the function $F(y)$ is defined for all real $y$.\nLet $M$ be the desired value of $f(x)$ (if it exists). Then for any real $y$ we have\n$$\n\\begin{gath... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Maximum value 2.5, attained only at x = 0.01 | |
0eow | A die has 20 identical equilateral triangular faces numbered from $1$ to $20$. If two such dice are rolled the most probable sum of the numbers showing on the top faces is
(A) $18$ (B) $19$ (C) $20$ (D) $21$ (E) $2$ | [
"It is easy to see that for $n = 1, 2, \\ldots, 20$ there are $n$ equally probable ways to obtain a total of $n+1$. (One die shows any number $x$ between $1$ and $n$, and the other die shows $n+1-x$.) In particular, a total of $21$ can be obtained with $20$ different throws. Beyond that, the number of possibilities... | South Africa | South African Mathematics Olympiad | [
"Statistics > Probability > Counting Methods > Other"
] | English | MCQ | D | |
02jc | Problem:
Se $x$, $y$ e $z$ são números inteiros positivos tais que $x y z = 240$, $x y + z = 46$ e $x + y z = 64$, qual é o valor de $x + y + z$?
A) 19
B) 20
C) 21
D) 24
E) 36 | [
"Solution:\n\nSolução 1:\nDe $x y z = 240$ segue que $x y = \\frac{240}{z}$; substituindo em $x y + z = 46$ obtemos $\\frac{240}{z} + z = 46$, ou seja, $z^{2} - 46z + 240 = 0$. As raízes desta equação são números cuja soma é $46$ e cujo produto é $240$, ou seja, as raízes são $6$ e $40$. Logo, $z = 6$ ou $z = 40$ (... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | B | |
06iq | Consider six points in the interior of a square of side length $3$. Prove that among the six points, there are two whose distance is less than $2$. | [
"Partition the square as shown. There are two congruent rectangles of size $1.3 \\times 1.5$ above, and three congruent rectangles of size $1.7 \\times 1$ below. By the pigeonhole principle, $2$ of the $6$ points must lie inside the same rectangle.\n\nIf there are $2$ points belonging to the same $1.3 \\times 1.5$ ... | Hong Kong | 1997-2023 IMO HK TST | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0k4y | Problem:
A permutation of $\{1,2, \ldots, 7\}$ is chosen uniformly at random. A partition of the permutation into contiguous blocks is correct if, when each block is sorted independently, the entire permutation becomes sorted. For example, the permutation $(3,4,2,1,6,5,7)$ can be partitioned correctly into the blocks $... | [
"Solution:\nLet $\\sigma$ be a permutation on $\\{1, \\ldots, n\\}$. Call $m \\in\\{1, \\ldots, n\\}$ a breakpoint of $\\sigma$ if $\\{\\sigma(1), \\ldots, \\sigma(m)\\}=\\{1, \\ldots, m\\}$. Notice that the maximum partition is into $k$ blocks, where $k$ is the number of breakpoints: if our breakpoints are $m_{1},... | United States | HMMT February 2018 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 151/105 | |
06tt | There are $n \geqslant 3$ islands in a city. Initially, the ferry company offers some routes between some pairs of islands so that it is impossible to divide the islands into two groups such that no two islands in different groups are connected by a ferry route.
After each year, the ferry company will close a ferry ro... | [
"Initially, we pick any pair of islands $A$ and $B$ which are connected by a ferry route and put $A$ in set $\\mathcal{A}$ and $B$ in set $\\mathcal{B}$. From the condition, without loss of generality there must be another island which is connected to $A$. We put such an island $C$ in set $\\mathcal{B}$. We say tha... | IMO | IMO 2016 Shortlisted Problems | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0868 | Problem:
a. Si hanno sette numeri interi positivi $a, b, c, d, e, f, g$ tali che i prodotti $ab, bc, cd, de, ef, fg, ga$ sono tutti cubi perfetti. Dimostrare che anche $a, b, c, d, e, f, g$ sono cubi perfetti.
b. Si hanno sei numeri interi positivi $a, b, c, d, e, f$ tali che i prodotti $ab, bc, cd, de, ef, fa$ sono ... | [
"Solution:\n\nSi noti innanzitutto che il prodotto e il quoziente (quando questo è un numero intero) di due cubi perfetti è ancora un cubo perfetto. La quantità\n$$\n\\frac{(ab)(cd)(ef)(ga)}{(bc)(de)(fg)} = a^{2}\n$$\nallora è un cubo perfetto. Ora, se $a^{2}$ è un cubo perfetto, anche $a$ è un cubo perfetto: difat... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Linear Algebra > Matrices"
] | null | proof and answer | a) Yes, all seven integers must be perfect cubes. b) No; for example, taking a=2, b=4, c=2, d=4, e=2, f=4 gives all adjacent products equal to 8 (a perfect cube) while the integers themselves are not perfect cubes. | |
0hmu | Problem:
Let $a$ and $b$ be positive real numbers. Prove that
$$
\sqrt{a^{2}-a b+b^{2}} \geq \frac{a+b}{2}
$$ | [
"Solution:\nSquaring both sides (which is OK since both sides are positive), it's equivalent to show that $4\\left(a^{2}-a b+b^{2}\\right) \\geq (a+b)^{2}$. But their difference is\n$$\n4\\left(a^{2}-a b+b^{2}\\right)-(a+b)^{2}=3 a^{2}-6 a b+3 b^{2}=3(a-b)^{2} \\geq 0.\n$$"
] | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0kc0 | Problem:
Let $\varphi(n)$ denote the number of positive integers less than or equal to $n$ which are relatively prime to $n$. Let $S$ be the set of positive integers $n$ such that $\frac{2 n}{\varphi(n)}$ is an integer. Compute the sum
$$
\sum_{n \in S} \frac{1}{n}
$$ | [
"Solution:\nLet $T_{n}$ be the set of prime factors of $n$. Then\n$$\n\\frac{2 n}{\\phi(n)}=2 \\prod_{p \\in T} \\frac{p}{p-1}\n$$\nWe can check that this is an integer for the following possible sets:\n$$\n\\varnothing,\\{2\\},\\{3\\},\\{2,3\\},\\{2,5\\},\\{2,3,7\\} .\n$$\nFor each set $T$, the sum of the reciproc... | United States | HMMT February 2020 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 10/3 | |
00oj | Let $ABCD$ be a trapezoid with parallel sides $AB$ and $CD$, with $\angle BAD = 90^\circ$ and with $AB + CD = BC$. Furthermore, let $M$ be the mid-point of $AD$.
Prove that $\angle CMB = 90^\circ$. | [
"We reflect the points $B$ and $C$ in $M$ and obtain the points $E$ and $F$, respectively. We clearly have $EC = BF = AB + AF = AB + CD = BC = EF$, therefore, the quadrilateral $BCEF$ is a rhombus. Since the diagonals in a rhombus are orthogonal, we get $BE \\perp CF$ and we obtain $\\angle BMC = 90^\\circ$ as desi... | Austria | Austrian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
07qx | Let
$$ f(n) = 4n^4 + 7n^2 + 3n + 6. $$
Prove that if $n$ is an integer, then $f(n)$ is not the cube of an integer. | [
"Suppose for the sake of contradiction that $n$ and $z$ are integers satisfying $f(n) = z^3$. Write $f(n) = 3(n^4 + n + 2) + n^4 - 7n^2$ and let $\\tau \\in \\{0, 1, 2\\}$ be the remainder of $n$ on division by 3.\n\nSuppose 3 divides $z$. Then 3 divides $n^4 + 7n^2 = n^2(n^2 + 7)$. But $r^2 + 7 \\in \\{7, 8, 11\\}... | Ireland | Ireland_2017 | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0hht | Find all functions $f: \mathbb{R} \to \mathbb{R}$, such that for any real $x$, $y$ holds the following:
$$
xf(x) + yf(xy) = xf(x + yf(y))
$$ | [
"Let $P(x, y)$ be the given assertion,\n$$\nP(0,1): f(0) = 0\n$$\nAssume that there exists $a \\neq 0$ such that $f(a) = 0$. Then $P(x, a)$:\n$$\nxf(x) + af(xa) = xf(x) \\Rightarrow \\forall x \\in \\mathbb{R} \\ f(xa) = 0 \\Rightarrow \\forall x \\in \\mathbb{R} \\ f(x) = 0\n$$\nAnd we found the first solution.\n\... | Ukraine | Problems from Ukrainian Authors | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | f(x) = 0 for all real x; f(x) = x for all real x | |
0b5o | Consider a convex quadrilateral $ABCD$ with
$$
AB = CB \quad \text{and} \quad \angle ABC + 2\angle CDA = \pi
$$
and let $E$ be the midpoint of $AC$. Show that $\angle CDE = \angle BDA$. | [
"Let point $X$ be lying on line $BE$ such that $\\angle CXE = \\angle CDE$ ($X$ is the (other than $C$) meeting point of the circumcircle of $\\triangle CDE$ and the line $BE$).\n\nTherefore the quadrilateral $DECX$ is cyclic, so $\\angle DXE = \\angle DCE = \\pi - \\angle CDA - \\angle CAD$. But $\\angle CDA = \\f... | Romania | Local Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinate... | English | proof only | null | |
04u4 | Paul is filling the cells of a rectangular table alternately with crosses and circles (he starts with a cross). When the table is filled in completely, he determines his score as $X - O$ where $X$ is the sum of squares of the numbers of crosses in all the rows and columns, and $O$ is the sum of squares of the numbers o... | [
"Let $n = 67$ and denote by $k = \\frac{1}{2}(n^2+1)$ the total number of crosses in the table. A row containing $a$ crosses and $n-a$ circles contributes $a^2-(n-a)^2 = 2n \\cdot a - n^2$ to the total score and thus all the $n$ rows combined contribute\n$$\n2n \\cdot k - n \\cdot n^2 = 2n \\cdot \\frac{n^2+1}{2} -... | Czech Republic | 67th Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 134 | |
0k86 | Problem:
A sequence of real numbers $a_{0}, a_{1}, \ldots, a_{9}$ with $a_{0}=0$, $a_{1}=1$, and $a_{2}>0$ satisfies
$$
a_{n+2} a_{n} a_{n-1}=a_{n+2}+a_{n}+a_{n-1}
$$
for all $1 \leq n \leq 7$, but cannot be extended to $a_{10}$. In other words, no values of $a_{10} \in \mathbb{R}$ satisfy
$$
a_{10} a_{8} a_{7}=a_{10}+... | [
"Solution:\nSay $a_{2}=a$. Then using the recursion equation, we have $a_{3}=-1$, $a_{4}=\\frac{a+1}{a-1}$, $a_{5}=\\frac{-a+1}{a+1}$, $a_{6}=-\\frac{1}{a}$, $a_{7}=-\\frac{2 a}{a^{2}-1}$, and $a_{8}=1$.\n\nNow we have $a_{10} a_{8} a_{7}=a_{10}+a_{8}+a_{7}$. No value of $a_{10}$ can satisfy this equation iff $a_{8... | United States | HMMT November 2019 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | sqrt(2) - 1 | |
0knn | Let $n \ge 4$ be an integer. Find all positive real solutions to the following system of $2n$ equations:
$$
\begin{aligned}
a_1 &= \frac{1}{a_{2n}} + \frac{1}{a_2}, & a_2 &= a_1 + a_3, \\
a_3 &= \frac{1}{a_2} + \frac{1}{a_4}, & a_4 &= a_3 + a_5, \\
a_5 &= \frac{1}{a_4} + \frac{1}{a_6}, & a_6 &= a_5 + a_7, \\
\vdots & &... | [] | United States | USA Junior MO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | The unique positive solution is a_1 = a_3 = ⋯ = a_{2n-1} = 1 and a_2 = a_4 = ⋯ = a_{2n} = 2. | |
0bvq | Find the largest subsets $A_1, A_2 \subset (0, \infty)$ such that:
$$
ab + cd \ge \sqrt{a^2 + b^2} + \sqrt{c^2 + d^2}, \quad \forall a, b, c, d \in A_1, \quad (1)
$$
$$
ab + cd \ge \sqrt{a^2 + c^2} + \sqrt{b^2 + d^2}, \quad \forall a, b, c, d \in A_2. \quad (2)
$$ | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | A1 = [sqrt(2), ∞) and A2 = [sqrt(2), ∞). | |
0hly | Problem:
A $3 \times 3 \times 3$ cube is made out of 27 subcubes. On every face shared by two subcubes, there is a door allowing you to move from one cube to the other. Is it possible to visit every subcube exactly once if
(a) You may start and end wherever you like
(b) You must start at the center subcube? | [
"Solution:\n(a) It is possible. Here is one of many possible routes.\n\nLevel 1\nLevel 2\nLevel 3\n\n(b) It is impossible. Color the subcubes black and white alternately as shown:\n\nLevel 1\n\nLevel 2\n\nLevel 3\nEv... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) Yes. b) No. | |
0gw2 | Find all triplets of real positive numbers $x$, $y$ and $z$ such that
$$
\begin{cases}
\sqrt{2x - \frac{2}{y}} + \sqrt{2y - \frac{2}{z}} + \sqrt{2z - \frac{2}{x}} = \sqrt{3(x + y + z)}, \\
x^2 + y^2 + z^2 = 6.
\end{cases}
$$ | [
"Відповідь: $x = y = z = \\sqrt{2}$. Із системи випливає, що\n$$\n\\sqrt{x - \\frac{1}{y}} + \\sqrt{y - \\frac{1}{z}} + \\sqrt{z - \\frac{1}{x}} = \\frac{1}{2} \\sqrt{x + y + z} \\cdot \\sqrt{x^2 + y^2 + z^2}.\n$$\nЗвідси за нерівністю Коші-Буняковського маємо:\n$$\n\\sqrt{x - \\frac{1}{y}} + \\sqrt{y - \\frac{1}{z... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | x = y = z = √2 | |
0hk6 | Problem:
A cube $3 \times 3 \times 3$ is made of cheese and consists of 27 small cubical cheese pieces arranged in the $3 \times 3 \times 3$ pattern. A mouse is eating the cheese in such a way that it starts at one of the corners and eats smaller pieces one by one. After he finishes one piece, he moves to the adjacent... | [
"Solution:\n\nColor the pieces of cheese alternatively in red and green such that corners are green and any two adjacent cubes are of different colors. We easily see that the mouse is moving always from the cube of one color to the cube of the other color. There are 14 green and 13 red cubes, the central cube being... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
0099 | There is a person standing in each square of a $2012 \times 2012$ checkerboard; each one can be a truth-teller, someone who always tells the truth, or a liar, someone who always lies. Each person states the same: "In my row, there are as many liars as in my column." Determine the minimum amount of truth-tellers that th... | [] | Argentina | XXIX Olimpíada Matemática Argentina National Round | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 92172 | |
08ve | Let $H$ be the orthocenter of an acute triangle $ABC$, and let $D$ be the intersection of the two lines $AH$, $BC$. Let $E$ be the point of intersection of the circumcircle to the triangle $ABD$ and the line $CH$, lying outside of the triangle $ABC$. And let $F$ be the point of intersection of the circumcircle to the t... | [
"Let $K$, $L$ be the feet of the perpendicular lines drawn from $B$ to the side $CA$ and from $C$ to the side $AB$, respectively. Since the line segment $AB$ is a diameter of the circumcircle to the triangle $ABD$, $\\angle AEB = 90^\\circ$. We see that the triangles $AEB$ and $ALE$ are similar, since they have the... | Japan | Japan Junior 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... | null | proof only | null | |
0imr | Problem:
The equation $x^{2}+2x=i$ has two complex solutions. Determine the product of their real parts. | [
"Solution:\n\nAnswer: $\\frac{1-\\sqrt{2}}{2}$. Complete the square by adding $1$ to each side. Then $(x+1)^{2}=1+i=e^{\\frac{i \\pi}{4}} \\sqrt{2}$, so $x+1= \\pm e^{\\frac{i \\pi}{8}} \\sqrt[4]{2}$. The desired product is then\n$$\n\\left(-1+\\cos \\left(\\frac{\\pi}{8}\\right) \\sqrt[4]{2}\\right)\\left(-1-\\cos... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (1-\sqrt{2})/2 | |
032r | Problem:
Let $AA_{1}$, $BB_{1}$ and $CC_{1}$ be the altitudes of an acute $\triangle ABC$ ($A_{1} \in BC$, $B_{1} \in CA$ and $C_{1} \in AB$). Denote by $O$ the circumcenter of $\triangle ABC$, and by $H_{1}$ the orthocenter of $\triangle A_{1}B_{1}C_{1}$. Prove that the midpoint of the segment $OH_{1}$ coincides with ... | [
"Solution:\nDenote by $H$ the orthocenter of $\\triangle ABC$, and by $G_{1}$ the centroid of $\\triangle A_{1}B_{1}C_{1}$. Let $O_{1}$ be the midpoint of the segment $OH$. It is well-known that $O_{1}$ is the circumcenter of $\\triangle A_{1}B_{1}C_{1}$, and $H$ is its incenter. Then $\\overrightarrow{H_{1}G_{1}} ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof only | null | |
0g47 | Problem:
Let $k$ be a circle centred at $O$ and let $X, A, Y$ be three points on $k$ in this order such that the tangent to the circumcircle of triangle $O X A$ through $X$ and the tangent to the circumcircle of $O A Y$ through $Y$ are parallel. Show that $\angle X A Y=120^{\circ}$ if $A$ lies on the minor arc $X Y$. | [
"Solution:\nLet $P$ be the intersection of $O X$ with the tangent through $Y$, and $Q$ any point on the tangent through $X$ such that $A$ and $Q$ are not on the same side of $O X$. Note that $\\angle O A Y=\\angle O Y A$, as $O Y=O A$. Also, by the tangent chord theorem, $\\angle O A Y=\\angle O Y P$. Similarly, $\... | Switzerland | Second round 2022 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04x2 | Given is a convex hexagon $ABCDEF$, such that $\angle A = \angle C = \angle E$ and $AB = BC$, $CD = DE$, $EF = FA$. Prove that the lines $AD$, $BE$ and $CF$ have a common point. | [
"Assume that the angle bisectors of the angles $\\angle B$ and $\\angle D$ intersect at $P$ (Fig. 1). We shall prove that the hexagon $ABCDEF$ has an inscribed circle, whose center is $P$. Then the conclusion follows from Brianchon's Theorem.\nThe equality $AB = BC$ implies that the triangles $ABP$ and $CBP$ are co... | Czech-Polish-Slovak Mathematical Match | Czech-Slovak-Polish Match | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | null | proof only | null | |
0j7t | Problem:
Sixteen wooden $C$s are placed in a $4$-by-$4$ grid, all with the same orientation, and each is to be colored either red or blue. A quadrant operation on the grid consists of choosing one of the four $2$-by-$2$ subgrids of $C$s found at the corners of the grid and moving each $C$ in the subgrid to the adjacen... | [
"Solution:\n\nAnswer: $1296$\n\nFor each quadrant, we have three distinct cases based on the number of $C$s in each color:\n- Case 1: all four the same color: $2$ configurations (all red or all blue)\n- Case 2: $3$ of one color, $1$ of the other: $2$ configurations (three red or three blue)\n- Case 3: $2$ of each c... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 1296 | |
0jf2 | Problem:
Trapezoid $ABCD$ is inscribed in the parabola $y = x^{2}$ such that $A = (a, a^{2})$, $B = (b, b^{2})$, $C = (-b, b^{2})$, and $D = (-a, a^{2})$ for some positive reals $a, b$ with $a > b$. If $AD + BC = AB + CD$, and $AB = \frac{3}{4}$, what is $a$? | [
"Solution:\n\n$t^{2} = (a - b)^{2} [1 + (a + b)^{2}] = (a - b)^{2} [1 + t^{2}]$. Thus $a = \\frac{t + \\frac{t}{\\sqrt{1 + t^{2}}}}{2} = \\frac{\\frac{3}{4} + \\frac{3}{5}}{2} = \\frac{27}{40}$."
] | United States | HMMT November 2013 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 27/40 | |
07zp | Problem:
La professoressa Scappavia insegna matematica in una scuola in cui si fanno 6 ore al giorno di lezione, dal lunedì al venerdì. Il suo orario settimanale prevede 18 ore di insegnamento ed ella, per ragioni personali, gradirebbe non insegnare mai nell'ultima ora di lezione. La commissione che fa l'orario conced... | [] | Italy | Italian Mathematical Olympiad - Febbraio Round | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | MCQ | D | |
07f9 | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that for any three real numbers $a, b, c$ that satisfy $a + f(b) + f(f(c)) = 0$, the following equality holds:
$$
f(a)^3 + b f(b)^2 + c^2 f(c) = 3abc.
$$ | [
"The answers are $f(x) = x$, $f(x) = -x$ and $f(x) = 0$.\n\nFirst, let's prove that $f(x)$ is injective at point $0$. Assume there exist two distinct real numbers $t_1$ and $t_2$ such that $f(t_1) = f(t_2) = 0$. Comparing $P(-f(b) - f(0), b, t_1)$ and $P(-f(b) - f(0), b, t_2)$ gives us\n$$\n(f(b) + f(0)) b t_1 = (f... | Iran | 37th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | f(x) = x; f(x) = -x; f(x) = 0 | |
02qq | Problem:
Os discos $A, B, C$ e $D$ representam polias de diâmetros $8, 4, 6$ e $2~\mathrm{cm}$, respectivamente, unidas por correias que se movimentam sem deslizar. Quando o disco $A$ dá uma volta completa no sentido horário, o que acontece com o disco $D$?
A) Dá 4 voltas no sentido horário... | [
"Solution:\n\nA figura mostra que os discos $A$ e $B$ giram no mesmo sentido, os discos $B$ e $C$ em sentidos opostos e os discos $C$ e $D$ no mesmo sentido.\n\n\n\nAssim, $D$ gira no sentido anti-horário. Lembramos que o perímetro $p$ de um círculo de raio $r$ é dado por $p = 2\\pi r$. Com... | Brazil | Brazilian Mathematical Olympiad | [
"Math Word Problems"
] | null | MCQ | D | |
004z | Sea $ABC$ un triángulo acutángulo, tal que $AB < AC$. Se traza una circunferencia con diámetro $AC$, y sobre ella un punto $P$ tal que $AP = AB$ y $P$ está en el semiplano determinado por $AC$ que no contiene a $B$. $BP$ corta a la circunferencia nuevamente en $Q$, y $AQ$ corta en $R$ a la recta perpendicular a $BC$ qu... | [] | Argentina | XVI Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Spanish | proof only | null | |
0fen | Problem:
En un triángulo acutángulo $ABC$ consideramos su ortocentro, $H$. Sean $A'$, $B'$ y $C'$ los simétricos de $H$ con respecto a los lados $BC$, $CA$ y $AB$, respectivamente. Probar que si los triángulos $ABC$ y $A'B'C'$ tienen un ángulo igual, entonces también tiene un lado igual. ¿Es cierto el recíproco? | [
"Solution:\n\nPor perpendicularidad de sus lados, $\\angle CAH = \\angle HBC$. Por simetría con respecto a $CB$, $\\angle CBA' = \\angle HBC$, por lo que $A'$ está sobre la circunferencia circunscrita a $ABC$, y análogamente $B'$ y $C'$.\n\nPor el teorema del seno, siendo $a$ el lado opuesto al ángulo $\\alpha$ y $... | Spain | null | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | The converse is false. | |
0b9f | Find all polynomials $P$, $Q$ with real coefficients, such that, for infinitely many positive integers $n$, $P(1)P(2)\dots P(n) = Q(n!)$. | [
"Let $P(x)$ and $Q(x)$ be polynomials with real coefficients such that for infinitely many positive integers $n$, $P(1)P(2)\\dots P(n) = Q(n!)$.\n\nLet $d$ be the degree of $P(x)$ and $e$ the degree of $Q(x)$.\n\nFor large $n$, $P(1)P(2)\\dots P(n)$ is a product of $n$ terms, each of degree $d$, so the degree of $P... | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | P(x) = x^d and Q(x) = x^d for some integer d ≥ 0 | |
06xt | Determine all positive, composite integers $n$ that satisfy the following property: if the positive divisors of $n$ are $1=d_{1}<d_{2}<\cdots<d_{k}=n$, then $d_{i}$ divides $d_{i+1}+d_{i+2}$ for every $1 \leqslant i \leqslant k-2$. | [
"Answer: $n=p^{r}$ is a prime power for some $r \\geqslant 2$.\n\nSolution 1. It is easy to see that such an $n=p^{r}$ with $r \\geqslant 2$ satisfies the condition as $d_{i}=p^{i-1}$ with $1 \\geqslant i \\geqslant k=r+1$ and clearly\n$$\np^{i-1} \\mid p^{i}+p^{i+1}\n$$\nNow, let us suppose that there is a positiv... | IMO | International Mathematical Olympiad Shortlist | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | n = p^r for a prime p and integer r ≥ 2 | |
0e0w | Problem:
Dokaži neenakost
$$
\frac{9}{4}<\log_{2} \pi+\log_{4} \pi<\frac{5}{2}
$$ | [
"Solution:\n\nKer velja\n$$\n\\log_{2} \\pi+\\log_{4} \\pi=\\frac{1}{\\log_{\\pi} 2}+\\frac{1}{\\log_{\\pi} 4}=\\frac{3}{2 \\cdot \\log_{\\pi} 2}=\\frac{3}{2} \\log_{2} \\pi\n$$\nje potrebno videti, da je $\\frac{9}{2}<3 \\log_{2} \\pi<5$.\n\nNeenakost $\\frac{9}{2}<3 \\log_{2} \\pi$ je enakovredna $3<\\log_{2} \\p... | Slovenia | Slovenian Secondary School Mathematical Competition | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof only | null | |
033o | Problem:
The points $P$ and $Q$ lie respectively on the diagonals $AC$ and $BD$ of a quadrilateral $ABCD$ and $\frac{AP}{AC} + \frac{BQ}{BD} = 1$. The line $PQ$ meets the sides $AD$ and $BC$ at points $M$ and $N$. Prove that the circumcircles of the triangles $AMP$, $BNQ$, $DMQ$ and $CNP$ are concurrent. | [
"Solution:\n\nLet $AC \\cap BD = O$ and $X$ be the second intersection point of the circumcircles of $\\triangle AOB$ and $\\triangle BOC$. Set $\\Varangle XBO = \\Varangle XCO = \\alpha$ and $\\Varangle XAO = \\Varangle XDO = \\beta$. Since $\\triangle AXC \\sim \\triangle DXB$, then $\\frac{XD}{XA} = \\frac{BD}{A... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jyt | Problem:
Sam spends his days walking around the following $2 \times 2$ grid of squares.
| 1 | 2 |
| :--- | :--- |
| 4 | 3 |
Say that two squares are adjacent if they share a side. He starts at the square labeled $1$ and every second walks to an adjacent square. How many paths can Sam take so that the sum of the numb... | [
"Solution:\n\nAnswer: $167$\n\nNote that on the first step, Sam can either step on $2$ or $4$. On the second step, Sam can either step on $1$ or $3$, regardless of whether he is on $2$ or $4$. Now, for example, say that Sam takes $8$ steps. His total sum will be $2+1+2+1+2+1+2+1+2a$, where $a$ is the number of time... | United States | February 2017 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 167 | |
0bsp | Consider the isosceles right triangle $ABC$, with $m(\widehat{BAC}) = 90^\circ$. Take now the point $D$ so that $BD \perp BC$ and $AD = BC$. Find the measure of the angle $\widehat{BAD}$. | [
"Case 1: $D$ and $A$ are on different sides of $BC$ (figure 1).\nDenote $\\{E\\} = AC \\cap DB$. Then $m(\\widehat{ABE}) = 45^\\circ$, therefore $[BA]$ is bisector and altitude in the triangle $BEC$. So, $[AE] = [AC]$. Construct $AM \\perp BE$, Then $[AM]$ is a midline in the triangle $EBC$, hence $AM = \\frac{1}{2... | Romania | 67th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 15° or 105° | |
02bf | Problem:
Na figura a seguir, o círculo de centro $B$ é tangente ao círculo de centro $A$ em $X$. O círculo de centro $C$ é tangente ao círculo de centro $A$ em $Y$. Além disto, os círculos de centros $B$ e $C$ também são tangentes. Se $A B=6, A C=5$ e $B C=9$, quanto mede $A X$ ?
 | [
"Solution:\nSejam $r_{a}, r_{b}$ e $r_{c}$ os raios dos círculos de centros $A, B$ e $C$, respectivamente. Se $Z$ é o ponto de tangência dos círculos de centros $B$ e $C$, os dados do problema nos permitem montar o seguinte sistema de equações:\n$$\n\\begin{aligned}\nA B & = A X - B X \\\\\n6 & = r_{a} - r_{b} \\\\... | Brazil | null | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 10 | |
03ge | Problem:
Let $n$ be a five digit number (whose first digit is non-zero) and let $m$ be the four digit number formed from $n$ by deleting its middle digit. Determine all $n$ such that $n / m$ is an integer. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All five-digit multiples of 1000: n = 1000·t for integers t from 10 to 99 (i.e., numbers of the form ab000 with a in 1..9 and b in 0..9). | |
08c5 | Problem:
Alberto, Barbara e Ciro si ritrovano un giorno per preparare dei ravioli per una cena di beneficenza a favore delle olimpiadi di matematica. Come prima cosa decidono di ripartire equamente le ore di lavoro fra la mattina e il pomeriggio, e ovviamente lavorano contemporaneamente e per la stessa quantità di tem... | [
"Solution:\n\nLa risposta è (D). Sia $h$ il numero di ore di lavoro durante la mattinata (e dunque anche durante il pomeriggio). Sappiamo che Alberto prepara $2 h \\cdot 90 = 180 h$ ravioli, mentre Barbara ne prepara $h \\cdot 110 + h \\cdot 70 = 180 h$. Sia ora $t_1$ il tempo che Ciro impiega a preparare i primi $... | Italy | Gara di Febbraio | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | D | |
0529 | Let $a$, $b$ and $c$ be real numbers for which $abc = 1$. Prove that
$$
\frac{1}{1+a^{2014}} + \frac{1}{1+b^{2014}} + \frac{1}{1+c^{2014}} > 1.
$$ | [
"Let $a^{2014} = u$, $b^{2014} = v$ and $c^{2014} = w$; then $abc = 1$ gives that $uvw = 1$. As the numerators of the l.h.s. of the inequality to be proven are positive, the inequality is equivalent to\n$$\n(1+v)(1+w) + (1+w)(1+u) + (1+u)(1+v) > (1+u)(1+v)(1+w).\n$$\n\nBy expanding, simplifying and using $uvw = 1$,... | Estonia | Final Round of National Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
01il | For sets $S$ and $T$ consisting of positive real numbers, define $S + T = \{s + t \mid s \in S, t \in T\}$ and $\frac{1}{S} = \{\frac{1}{s} \mid s \in S\}$. Define the sets $A_1, A_2, A_3, \dots$ recursively by $A_1 = \{1\}$ and
$$
A_n = \bigcup_{i=1}^{n-1} \left( (A_i + A_{n-i}) \cup \left( \frac{1}{\frac{1}{A_i} + \f... | [
"*Solution:* We start with a lemma.\n**Lemma.** For any $n$, $x \\in A_n$ implies $\\frac{1}{x} \\in A_n$.\n*Proof.* Induction. Case $n = 1$ is clear. Now, if $x = a_i + a_{n-i} \\in A_i + A_{n-i} \\subset A_n$, then $\\frac{1}{a_i} \\in A_i$ and $\\frac{1}{a_{n-i}} \\in A_{n-i}$, and thus\n$$\n\\frac{1}{x} \\in \\... | Baltic Way | Baltic Way 2023 Shortlist | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0c93 | For $n \in \mathbb{N}$, $n \ge 2$, consider $A$ a matrix $n \times n$ with complex entries, such that $A^2 = \text{tr}(A) \cdot A$. Prove that the matrices $ABA$ and $ACA$ commute, for any $n \times n$ matrices $B$ and $C$, with complex entries.
Mihai Opincaru | [
"Observe that $\\text{tr}(A) = 0$ implies $A^2 = O_{n \\times n}$ and\n$$\nABA \\cdot ACA = O_{n \\times n} = ACA \\cdot ABA,\n$$\n\nIf $\\text{tr}(A) \\neq 0$, we show that $\\text{rank}(A) = 1$. Let $r = \\text{rank}(A)$. Then we can find matrices $X \\in \\mathcal{M}_{n \\times r}(\\mathbb{C})$ and $Y \\in \\mat... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices"
] | English | proof only | null | |
01ox | For two positive integers $a$ and $b$ the number $\overline{a.b}$ is equal to the decimal fraction which we have if after the number $a$ we put the decimal point and then write the number $b$. For example, for $a = 20, b = 13$ we get $\overline{a.b} = 20.13$, and $\overline{b.a} = 13.2$.
Prove that there are infinite n... | [
"Show that if $n = 9k \\pm 3$, $k \\in \\mathbb{N}$, then the given equation has no natural solutions.\nLet the decimal representations of $a$ and $b$ consist of $m$ and $l$ digits respectively. Then the initial equation is equivalent to the equation\n$$\n\\left(a + \\frac{b}{10^l}\\right) \\left(b + \\frac{a}{10^m... | Belarus | BelarusMO 2013_s | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof only | null | |
04xo | For any positive integer $n$, let $\tau(n)$ denote the number of positive divisors of $n$ and $\varphi(n)$ the number of positive integers not greater than $n$ which are relatively prime to $n$. Find all positive integers $n$ for which one of the three numbers $n$, $\tau(n)$, and $\varphi(n)$ is the arithmetic mean of ... | [
"We have $\\tau(1) = \\varphi(1) = 1$, that is $n = 1$ satisfies the given condition. In the following, we assume $n > 1$. For such $n$, clearly $\\tau(n) \\le n$ and $\\varphi(n) < n$. This means $n$ cannot be the arithmetic mean of $\\tau(n)$ and $\\varphi(n)$. We are left with two cases.\n\n*Case 1:* $\\tau(n) =... | Czech-Polish-Slovak Mathematical Match | 12th Czech-Polish-Slovak Mathematics Competition | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | {1, 4, 6, 9} | |
0kp1 | Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$, respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$, as shown. Suppose that $AB = 2$, $O_1O_2 = 15$, $CD = 16$, and $ABO_1CDO_2$ is a convex hexagon... | [] | United States | AIME II | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 135 | |
011z | Problem:
What is the smallest positive odd integer having the same number of positive divisors as $360$? | [
"Solution:\nAn integer with the prime factorization $p_1^{r_1} \\cdot p_2^{r_2} \\cdot \\ldots \\cdot p_k^{r_k}$ (where $p_1, p_2, \\ldots, p_k$ are distinct primes) has precisely $(r_1+1) \\cdot (r_2+1) \\cdot \\ldots \\cdot (r_k+1)$ distinct positive divisors.\n\nSince $360 = 2^3 \\cdot 3^2 \\cdot 5$, it follows ... | Baltic Way | Baltic Way | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 3465 | |
0kpy | Problem:
In the Cartesian plane, let $A=(0,0)$, $B=(200,100)$, and $C=(30,330)$. Compute the number of ordered pairs $(x, y)$ of integers so that $\left(x+\frac{1}{2}, y+\frac{1}{2}\right)$ is in the interior of triangle $ABC$. | [
"Solution:\n\nWe use Pick's Theorem, which states that in a lattice polygon with $I$ lattice points in its interior and $B$ lattice points on its boundary, the area is $I + B/2 - 1$. Also, call a point center if it is of the form $\\left(x+\\frac{1}{2}, y+\\frac{1}{2}\\right)$ for integers $x$ and $y$.\n\nThe key o... | United States | HMMT February | [
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 31480 | |
066c | We consider the sequence of real numbers $(a_n)$, $n=1,2,3,...$
$$
a_1 = 2 \text{ and } a_n = \left(\frac{n+1}{n-1}\right) (a_1 + a_2 + \dots + a_{n-1}), \quad n \ge 2.
$$
Determine the term $a_{2013}$. | [
"We observe that:\n$$\na_1 = 2,\\ a_2 = \\frac{3}{2} \\cdot a_1 = 3 \\cdot 2,\\ a_3 = \\frac{4}{2} \\cdot (a_1 + a_2) = \\frac{4}{2} \\cdot 4 \\cdot 2 = 4 \\cdot 2^2,\n$$\n$$\na_4 = \\frac{5}{3} \\cdot (a_1 + a_2 + a_3) = \\frac{5}{3} \\cdot 24 = 5 \\cdot 2^3,\n$$\n$$\na_5 = \\frac{6}{4} \\cdot (a_1 + a_2 + a_3 + a... | Greece | Hellenic Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 2014 * 2^{2012} | |
0lb0 | a_0 = 1, a_1 = 3 \text{ and } a_{n+2} = 1 + \left\lfloor \frac{a_{n+1}^2}{a_n} \right\rfloor \text{ for all } n \ge 0.
Show that $a_{n+2} \cdot a_n - a_{n+1}^2 = 2^n$ for all integers $n$. | [] | Vietnam | IMO2011 Selection | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
09nv | Show that $\underbrace{66\dots6}_{61}\underbrace{11\dots1}_{61}$ is divisible by $61$.
(Nursoltan Khavalbolot) | [] | Mongolia | MMO2025 Round 3 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof only | null | |
0b44 | Problem:
The set $S = \{1, 2, \ldots, 2022\}$ is to be partitioned into $n$ disjoint subsets $S_1, S_2, \ldots, S_n$ such that for each $i \in \{1, 2, \ldots, n\}$, exactly one of the following statements is true:
(a) For all $x, y \in S_i$ with $x \neq y$, $\operatorname{gcd}(x, y) > 1$.
(b) For all $x, y \in S_i$ ... | [
"Solution:\n\nThe answer is $15$.\n\nNote that there are $14$ primes at most $\\sqrt{2022}$, starting with $2$ and ending with $43$. Thus, the following partition works for $15$ sets. Let $S_1 = \\{2, 4, \\ldots, 2022\\}$, the multiples of $2$ in $S$. Let $S_2 = \\{3, 9, 15, \\ldots, 2019\\}$, the remaining multipl... | Philippines | Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 15 | |
08w0 | For a real number $r$ denote by $[r]$ the greatest integer less than or equal to $r$. How many positive integers $n$ are there for which
$$
\lfloor \frac{1000000}{n} \rfloor - \lfloor \frac{1000000}{n+1} \rfloor = 1
$$
is satisfied? | [
"First, we prove the following Lemma.\n\n**Lemma.** If real numbers $x, y$ and an integer $k$ satisfy $k < x - y < k + 1$, then $[x] - [y] = k$ or $k + 1$ must hold.\n**Proof:** Since $0 \\le x - [x] < 1$ and $0 \\le y - [y] < 1$, we have $x - y - 1 < [x] - [y] < x - y + 1$. This, together with $k < x - y < k + 1$,... | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 1172 | |
08g5 | Problem:
Marina riempie le caselle di una griglia $4 \times 4$ scrivendo dentro ciascuna il numero $1$, il numero $2$ o il numero $3$. Quanti sono i modi di riempire la griglia tali che la somma di ogni riga e la somma di ogni colonna siano divisibili per $3$?
(A) $3^{8}-1$
(B) $3^{8}$
(C) $2 \cdot 3^{8}$
(D) $3^{9}$... | [
"Solution:\n\nLa risposta è (D). Iniziamo a riempire la sotto-tabella $3 \\times 3$ in alto a sinistra in un modo a piacere: per ognuna delle $9$ caselle abbiamo $3$ scelte, dunque in totale $3^{9}$ possibilità. Ora mostriamo che, per ognuna di queste, la scelta delle altre $7$ caselle risulta obbligata e sempre po... | Italy | Olimpiadi di Matematica - Febbraio | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | MCQ | D | |
0757 | Let $(a_0, a_1, a_2, ...)$ and $(b_0, b_1, b_2, ...)$ be two infinite sequences of integers such that
$$
(a_n - a_{n-1})(a_n - a_{n-2}) + (b_n - b_{n-1})(b_n - b_{n-2}) = 0,
$$
for all integers $n \ge 2$. Prove that there exists a positive integer $K$ such that
$$
a_{K+2011} = a_{K+(2011)^{2011}}.
$$ | [
"Consider points $P_j = (a_j, b_j)$ in the plane. The slope of the lines $P_nP_{n-1}$ and $P_nP_{n-2}$ are\n$$\nr = \\frac{b_n - b_{n-1}}{a_n - a_{n-1}}, \\quad s = \\frac{b_n - b_{n-2}}{a_n - a_{n-2}},\n$$\nrespectively. The given condition implies that $rs = -1$. Hence it follows that the lines $P_nP_{n-1}$ and $... | India | Indija TS | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0ekl | Problem:
Dan je izraz $\frac{x^{n-1}}{x^{n}-2 x^{n-1}}-\frac{x^{n}}{x^{n+1}-4 x^{n-1}}$. Kateri izraz je ekvivalenten izrazu za $x \neq 0$?
(A) $\frac{1}{(x-2)}$
(B) $\frac{2}{(x-2)(x+2)}$
(C) $\frac{1}{(x+2)}$
(D) $\frac{2 x}{(x-2)(x+2)}$
(E) $\frac{1-x}{(x-2)(x+2)}$ | [
"Solution:\n\nV imenovalcih ulomkov izpostavimo skupni faktor ter krajšamo, kar se da\n\n$\\frac{x^{n-1}}{x^{n}-2 x^{n-1}}-\\frac{x^{n}}{x^{n+1}-4 x^{n-1}} = \\frac{x^{n-1}}{x^{n-1}(x-2)}-\\frac{x^{n}}{x^{n-1}\\left(x^{2}-4\\right)} = \\frac{1}{(x-2)}-\\frac{1}{x^{-1}\\left(x^{2}-4\\right)}$.\n\nSeštejemo ulomka\n\... | Slovenia | 22. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | B | |
0c7i | A set of prime numbers is called *interesting* if the following holds: *the sum of any three distinct numbers from the set is also prime*.
Find the maximum number of elements an interesting set should have. | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 4 | |
07bi | $H$ is the foot of the altitude of vertex $A$ of triangle $ABC$ and $H'$ is the reflection of $H$ with respect to the midpoint of $BC$. If tangents to the circumcircle of triangle $ABC$ at points $B$ and $C$ intersect each other at $X$ and the perpendicular to $XH'$ at $H'$ intersects lines $AB$ and $AC$ at $Y$ and $Z$... | [
"Let $P$, $Q$ and $M$ be the feet of perpendicular lines from $X$ to $AB$, $AC$ and $BC$, respectively. Obviously, $M$ is the midpoint of $BC$. We have\n$$\n\\angle ZXY = \\angle ZXH' + \\angle H'XY = \\angle AQH' + \\angle APH' = \\angle PH'Q - \\angle A\n$$\nSince we know $\\angle BXC = 180^\\circ - \\angle 2A$, ... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | null | proof only | null | |
0csn | Все клетки квадратной таблицы $100 \times 100$ пронумерованы в некотором порядке числами от $1$ до $10000$. Петя закрашивает клетки по следующим правилам. Вначале он закрашивает $k$ клеток по своему усмотрению. Далее каждым ходом Петя может закрасить одну еще не закрашенную клетку с номером $a$, если для неё выполнено ... | [
"Докажем вначале следующее утверждение.\n**Лемма.** Для любых двух клеток $A$ и $B$ существует такая клетка $C$, закрасив которую, можно затем закрасить и $A$, и $B$ (возможно, $C$ совпадает с $A$ или с $B$.)\n\n**Доказательство.** Можно считать, что номер $a$ клетки $A$ меньше, чем номер $b$ клетки $B$. Пусть $D$ ... | Russia | XL Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 1 | |
07zs | Problem:
Quanti sono i numeri naturali che in base 10 si scrivono con 3 cifre e in base 2 si scrivono con 7 cifre? | [] | Italy | Italian Mathematical Olympiad - Febbraio Round | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 28 | |
0is9 | Problem:
Let $n$ be a positive integer and let $S$ be the set $\{1,2, \ldots, n\}$. Define a function $f: S \rightarrow S$ by
$$
f(x)= \begin{cases}2 x & \text{ if } 2 x \leq n, \\ 2 n-2 x+1 & \text{ otherwise. }\end{cases}
$$
Define $f^{2}(x)=f(f(x)), f^{3}(x)=f(f(f(x)))$, and so on. If $m$ is a positive integer satis... | [
"Solution:\nFirst note that\n$$\nf(x) \\equiv \\pm 2 x \\quad \\bmod 2 n+1\n$$\nIt follows that\n$$\nf^{p}(x) \\equiv \\pm 2^{p} x \\quad \\bmod 2 n+1\n$$\nThus if $f^{m}(1)=1$, $2^{m} \\equiv \\pm 1$ and so, for any $k \\in S$,\n$$\nf^{m}(k) \\equiv \\pm 2^{m} k \\equiv \\pm k \\quad \\bmod 2 n+1\n$$\nthat is, $f^... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Number Theory > Other",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof only | null | |
027e | Problem:
Consideremos o conjunto $A=\{1,2,3,4, \ldots, n\}$. Um subconjunto de $A$ é chamado hierárquico se satisfaz as seguintes duas propriedades:
- O subconjunto deve ter mais de um número.
- Há um número no subconjunto que coincide com a soma dos outros números do subconjunto.
Deseja-se dividir o conjunto $A$ em s... | [
"Solution:\n\na) Vamos supor que foi possível dividir o conjunto em $\\ell$ grupos. Em cada grupo, o maior coincide com a soma dos outros números do grupo. Então, a soma de todos os números do grupo seria duas vezes o maior número do grupo. Provamos assim que a soma dos números dentro de cada grupo é sempre um núme... | Brazil | null | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
07jm | Let $n \in \mathbb{N}$ be a positive integer. We call a function $f(x, y)$ a *friend* of $n$ if for at least one percent of positive integers $k$ such that $0 \le k \le n$ the equation $f(x, y) = k$ has a solution $(x_0, y_0)$ in positive integers such that $\frac{y_0}{x_0} \in [\frac{1}{100}, 100]$. Let $g(x, y)$ be a... | [
"First, note that given the positive coefficients, if $a x^m y^n$ is the highest degree term appearing in $g$, we have $a x^n y^m \\le g(x, y)$. Therefore, if $f(p, q) = k < n$ and $(p, q)$ are in the specified region, we have:\n$$\na p^n \\left(\\frac{p}{100}\\right)^m \\le a p^n q^m \\le n \\implies p \\le \\sqrt... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
08sv | Suppose there are 5 cards, each one of which has a distinct number from the set $\{2, 3, 4, 5, 6\}$ written on it. When these cards are placed in line from left to right randomly, what is the probability that for each $i$, $1 \leq i \leq 5$ the number written on the card placed on the $i$-th spot from the left is great... | [
"Let $N_i$ be the number written on the card placed on the $i$-th position from the left. If $N_i \\geq i$ is satisfied for all $i$ ($1 \\leq i \\leq 5$), then $N_5$ has 2 possibilities as it can either be 5 or 6. When $N_5$ is determined, $N_4$ has 2 possibilities as it can be one of the numbers 4, 5, 6 different ... | Japan | Japan Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Permutations"
] | English | proof and answer | 2/15 | |
05li | Problem:
Déterminer tous les couples d'entiers positifs ou nuls $(x, y)$ pour lesquels $x^{2}+y^{2}$ divise à la fois $x^{3}+y$ et $x+y^{3}$. | [
"Solution:\n\nTout d'abord, remarquons que les couples $(x, y) \\in \\{(0,0),(1,0),(0,1),(1,1)\\}$ sont solutions.\n\nOn se place maintenant dans le cas où $(x, y)$ est une solution éventuelle autre que celles-ci. De plus, $x$ et $y$ jouant des rôles symétriques, on suppose ici que $x \\leqslant y$, donc que $y \\g... | France | Olympiades Françaises de Mathématiques - Test de Janvier | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | [(0,0), (1,0), (0,1), (1,1)] | |
0j80 | Let $A$ be a set with $|A| = 225$, meaning that $A$ has $225$ elements. Suppose further that there are eleven subsets $A_1, \dots, A_{11}$ of $A$ such that $|A_i| = 45$ for $1 \le i \le 11$ and $|A_i \cap A_j| = 9$ for $1 \le i < j \le 11$. Prove that $|A_1 \cup A_2 \cup \dots \cup A_{11}| \ge 165$, and give an example... | [
"Let $S$ be the complement of $A_1 \\cup A_2 \\cup \\dots \\cup A_{11}$ in $A$; we wish to prove that $|S| \\le 60$. For $\\ell \\ge 0$, define\n$$\n\\theta(\\ell) = \\left(1 - \\frac{\\ell}{2}\\right) \\left(1 - \\frac{\\ell}{3}\\right) = 1 - \\frac{2}{3}\\ell + \\frac{1}{3}\\binom{\\ell}{2}.\n$$\nNote that $\\the... | United States | USAMO | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 165 | |
0kuh | Problem:
Suppose that point $D$ lies on side $BC$ of triangle $ABC$ such that $AD$ bisects $\angle BAC$, and let $\ell$ denote the line through $A$ perpendicular to $AD$. If the distances from $B$ and $C$ to $\ell$ are $5$ and $6$, respectively, compute $AD$. | [
"Solution:\n\n\n\nLet $\\ell$, the external angle bisector, intersect $BC$ at $X$. By the external angle bisector theorem, $AB : AC = XB : XC = 5 : 6$, so $BD : DC = 5 : 6$ by the angle bisector theorem. Then $AD$ is a weighted average of the distances from $B$ and $C$ to $\\ell$, namely\n$... | United States | HMMT November 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 60/11 | |
016u | Consider positive integers that can be expressed in the form $\binom{n}{k}$ where $n \ge 4$ and $2 \le k \le n-2$. Prove that every such integer has at least two distinct prime divisors. | [
"Assume that there exists a prime $p$ and positive integers $n, k, t$ such that\n$$\np^t = \\binom{n}{k} = \\frac{n}{k} \\cdot \\frac{n-1}{k-1} \\cdots \\frac{n-k+1}{1}\n$$\nDenote by $\\text{ord}(x)$ the exponent of $p$ in the prime decomposition of $x$. Let $m$ be the number from the set $\\{n-k+1, \\dots, n\\}$ ... | Baltic Way | BALTIC WAY | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0lfr | Problem:
Find all the functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $f(4x+3y) = f(3x+y) + f(x+2y)$ for all integers $x$ and $y$. | [
"Solution:\nPutting $x=0$ in the original equation\n$$\nf(4x+3y) = f(3x+y) + f(x+2y)\n$$\nwe get\n$$\nf(3y) = f(y) + f(2y)\n$$\nNext, (1) for $y=-2x$ gives us $f(-2x) = f(x) + f(-3x) = f(x) + f(-x) + f(-2x)$ (in view of (2)). It follows that\n$$\nf(-x) = -f(x)\n$$\nNow, let $x=2z-v$, $y=3v-z$ in (1). Then\n$$\nf(5z... | Zhautykov Olympiad | XVI International Zhautykov Olympiad in Mathematics | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | All functions defined by choosing integers a and b and setting f(n) = b·n for integers n not divisible by five, and f(n) = a·(n/5) for integers n divisible by five. | |
0ck7 | Let $n$ be a given positive integer. For a finite set $M$ of points in the plane, we say that distinct points $A, B \in M$ are connected if the line $AB$ contains exactly $n+1$ points in $M$.
Determine the smallest positive integer $m$ for which there exists a set $M$ of $m$ points in the plane with the property that ... | [
"Let $M = \\{A_1, A_2, \\dots, A_m\\}$ be a set of $m$ points with the given property and $A_1 \\in M$. Since $A_1$ is connected to other points, there is a line $d_0$ that contains exactly $n$ other points $A_2, \\dots, A_{n+1}$ from the set $M$. Since each of the points $A_1, A_2, \\dots, A_{n+1}$ is already conn... | Romania | 75th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | (n+1)(n+2)/2 | |
0b1n | Problem:
Let $P = (3^{1} + 1)(3^{2} + 1)(3^{3} + 1) \ldots (3^{2020} + 1)$. Find the largest value of the integer $n$ such that $2^{n}$ divides $P$. | [
"Solution:\n\nIf $k$ is even, then note that $3^{k} + 1 \\equiv 2 \\pmod{4}$ and so $2 \\mid\\mid 3^{k} + 1$, i.e., $4 \\nmid 3^{k} + 1$.\n\nOn the other hand, if $k$ is odd, note that $3^{k} + 1 \\equiv 4 \\pmod{8}$ so $4 \\mid\\mid 3^{k} + 1$, i.e., $4 \\mid 3^{k} + 1$ but $8 \\nmid 3^{k} + 1$.\n\nThus the greate... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 3030 |
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