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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0aew | Македонската математичка олимпијада се одржува во две соби означени со броеви $1$ и $2$. На почетокот сите натпреварувачи влегуваат во соба бр.$1$. Краен распоред на натпреварувачите по соби се добива на следниот начин: листа со имиња на неколку од натпреварувачите се чита на глас; кога едно име ќе биде прочитано, тој ... | [
"Ќе покажеме дека вкупниот број можни крајни рапореди е парен број па затоа не може да биде $2009$. Доволно е да се покаже дека постои листа имиња на некои од натпреварувачите со која сите натпреварувачи од соба бр.$1$ се префрлаат во соба бр.$2$. (ако ова важи тогаш за секој можен краен распоред и обратниот распор... | North Macedonia | XVI Македонска математичка олимпијада | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | Macedonian, English | proof only | null | |
00e5 | Initially we have a paper triangle $ABC$ such that $\angle BAC = 120^\circ$. In the first step, we draw the angle bisectors of the three angles of the triangle, which intersect in $I$, and then using a pair of scissors we cut along segments $AI$, $BI$ and $CI$, obtaining 3 triangles: $ABI$, $BCI$ and $CAI$. In the seco... | [
"Let $\\angle CAB = 2\\alpha$, $\\angle ABC = 2\\beta$ and $\\angle BCA = 2\\gamma$. Notice that $\\alpha+\\beta+\\gamma = 90^\\circ$, and if $I$ is the incenter of $\\triangle ABC$ we can compute $\\angle AIB$, $\\angle BIC$, $\\angle CIA$ in terms of these variables.\n\n\n**Fact 1:** The ... | Argentina | Rioplatense Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 32 | |
0ks2 | Problem:
Five cards labeled $1, 3, 5, 7, 9$ are laid in a row in that order, forming the five-digit number $13579$ when read from left to right. A swap consists of picking two distinct cards, and then swapping them. After three swaps, the cards form a new five-digit number $n$ when read from left to right. Compute the... | [
"Solution:\n\nFor a given card, let $p(n)$ denote the probability that it is in its original position after $n$ swaps. Then\n$$\np(n+1) = p(n) \\cdot \\frac{3}{5} + (1 - p(n)) \\cdot \\frac{1}{10},\n$$\nby casework on whether the card is in the correct position or not after $n$ swaps. In particular, $p(0) = 1$, $p(... | United States | HMMT February 2022 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 50308 | |
0cj2 | Consider the set
$$
A = \{(x, y, z) \mid x, y, z \in \mathbb{R}, 88(x+y+z) = 33(xy+xz+yz) = 24(x^2+y^2+z^2) \neq 0\}.
$$
$$
\text{Determine } \max_{(x,y,z) \in A} (\max\{x, y, z\} - \min\{x, y, z\}).
$$ | [] | Romania | 75th NMO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | 2*sqrt(3) | |
0kbk | Problem:
Let $\Gamma$ be a circle of radius $1$ centered at $O$. A circle $\Omega$ is said to be friendly if there exist distinct circles $\omega_{1}, \omega_{2}, \ldots, \omega_{2020}$, such that for all $1 \leq i \leq 2020$, $\omega_{i}$ is tangent to $\Gamma$, $\Omega$, and $\omega_{i+1}$. (Here, $\omega_{2021}=\om... | [
"Solution:\n\nLet $P$ satisfy $OP = x$. (For now, we focus on $f(P)$ and ignore the $A$ and $B$ from the problem statement.) The key idea is that if we invert at some point along $OP$ such that the images of $\\Gamma$ and $\\Omega$ are concentric, then $\\omega_{i}$ still exist. Suppose that this inversion fixes $\... | United States | HMMT February | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof and answer | 1000π/9 | |
0kp9 | Problem:
Compute the number of permutations $\pi$ of the set $\{1,2, \ldots, 10\}$ so that for all (not necessarily distinct) $m, n \in\{1,2, \ldots, 10\}$ where $m+n$ is prime, $\pi(m)+\pi(n)$ is prime. | [
"Solution:\nSince $\\pi$ sends pairs $(m, n)$ with $m+n$ prime to pairs $(m', n')$ with $m'+n'$ prime, and there are only finitely many such pairs, we conclude that if $m+n$ is composite, then so is $\\pi(m)+\\pi(n)$. Also note that $2 \\pi(1)=\\pi(1)+\\pi(1)$ is prime because $2=1+1$ is prime. Thus, $\\pi(1)=1$.\n... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 4 | |
08is | Problem:
Each side of the arbitrary triangle is divided into $2002$ congruent segments. After that each interior division point of the side is joined with the opposite vertex. Prove that the number of obtained regions of the triangle is divisible by $6$. | [] | JBMO | The first selection test for IMO 2003 and BMO 2003 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
08u6 | 30 students took a test consisting of three problems. Problems were worth 1, 2, 4 points respectively if solved correctly, and no partial credits were given for any of the problems. Suppose that for each of the problems there were 10 students answering it correctly. How many different possibilities were there for the s... | [
"1296\n\nBecause of the way the grading was done, it is the case that two students receive the same grades only if the problems they answer correctly are exactly the same. Designate problems as the problem I, II, and III, corresponding to the grades 1, 2 and 4 for a correct answer, respectively. The possibilities f... | Japan | Japan Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1296 | |
0d6d | Let $ABC$ be an acute, non-isosceles triangle, $AX$, $BY$, $CZ$ are the altitudes with $X$, $Y$, $Z$ belonging to $BC$, $CA$, $AB$ respectively. Respectively denote $(O_{1})$, $(O_{2})$, $(O_{3})$ as the circumcircles of triangles $AYZ$, $BZX$, $CXY$. Suppose that $(K)$ is a circle that is internally tangent to $\left(... | [
"Let $H$ be the orthocenter of triangle $ABC$.\n\n\n\nWe can see that $HA \\cdot HX = HB \\cdot HY = HC \\cdot HZ = k$. We consider the inversion with center $H$ and ratio $-k$ as the function $f$.\n\nIt is easy to see that\n$$\nf(A) = X, \\quad f(B) = Y, \\quad f(C) = Z\n$$\nso $f((O)) = (... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof only | null | |
01et | Prove, that for any $p, q \in \mathbb{N}$, such that $\sqrt{11} > \frac{p}{q}$, following inequality holds:
$$
\sqrt{11} - \frac{p}{q} > \frac{1}{2pq}.
$$ | [
"We can assume that $p$ and $q$ are coprime, and since both sides of first inequality are positive, we can change it to $11q^2 > p^2$. The same way we can change second inequality:\n$$\n11p^2q^2 > p^4 + p^2 + \\frac{1}{4}.\n$$\nTo see this one holds, we will prove stronger one:\n$$\n11p^2q^2 \\geq p^4 + 2p^2.\n$$\n... | Baltic Way | Baltic Way shortlist | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0ic2 | Problem:
NASA has proposed populating Mars with $2,004$ settlements. The only way to get from one settlement to another will be by a connecting tunnel. A bored bureaucrat draws on a map of Mars, randomly placing $N$ tunnels connecting the settlements in such a way that no two settlements have more than one tunnel conn... | [
"Solution:\n\nThe problem is equivalent, in general, to finding the least number of edges required so that a graph on $n$ vertices will be connected, i.e., one can reach any vertex from any other vertex by following the edges of the graph. (We are letting settlements be vertices and tunnels be edges, of course.) Th... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2005004 | |
0adm | Calculate the angles in the triangle $ABC$, if the angle between the altitude from $C$ and the bisector of the angle $ACB$ is $90^\circ$, and the angle between the bisectors of the exterior angles at the vertices $A$ and $B$ is $61^\circ$. | [
"Let us denote with $\\alpha, \\beta, \\gamma$ the angles in the triangle $ABC$ at the vertices $A, B, C$ correspondently and with $\\alpha_1, \\beta_1, \\gamma_1$ the correspondent exterior angles. Let $D$ be the base of the altitude from $C$, $E$ be the base of the bisector of the angle $ACB$ and $F$ be the inter... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | ∠A = 70°, ∠B = 52°, ∠C = 58° | |
0d7e | Let $ABC$ be an acute, non-isosceles triangle with the circumcircle ($O$). Denote $D, E$ as the midpoints of $AB, AC$ respectively. Two circles $(ABE)$ and $(ACD)$ intersect at $K$ differs from $A$. Suppose that the ray $AK$ intersects ($O$) at $L$. The line $LB$ meets $(ABE)$ at the second point $M$ and the line $LC$ ... | [
"1) Denote $G$ as the intersection of $BE, CD$, then $G$ is the centroid of triangle $ABC$. We assume that $AB < AC$ and the solution is similar to all other cases.\nSince $ABLC$ is cyclic, we have $180^{\\circ} - \\angle AKN = \\angle ACN = \\angle ABM = \\angle AKM$, hence\n$$\n\\angle AKM + \\angle AKN = 180^{\\... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumce... | English | proof only | null | |
0kiu | Problem:
Let $n$ be a positive integer. Alice writes $n$ real numbers $a_{1}, a_{2}, \ldots, a_{n}$ in a line (in that order). Every move, she picks one number and replaces it with the average of itself and its neighbors (that is, $a_{n}$ is not a neighbor of $a_{1}$, nor vice versa). A number changes sign if it chang... | [
"Solution:\n\nThe maximum number is $n-1$. We first prove the upper bound. For simplicity, color all negative numbers red, and all non-negative numbers blue. Let $X$ be the number of color changes among adjacent elements (i.e. pairs of adjacent elements with different colors). It is clear that the following two sta... | United States | HMIC 2021 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | n - 1 | |
0hl9 | Problem:
Calculate $\sum_{n=1}^{2001} n^{3}$. | [
"Solution:\n\n$\\sum_{n=1}^{2001} n^{3} = \\left(\\sum_{n=1}^{2001} n\\right)^{2} = \\left(\\frac{2001 \\cdot 2002}{2}\\right)^{2} = 4012013006001$."
] | United States | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 4012013006001 | |
01o8 | The lengths of some three sides of a quadrilateral are equal to $2$, $7$, and $11$.
Find the area of the quadrilateral if it has the greatest area among all quadrilaterals with the mentioned lengths of their sides. | [
"Answer: $30\\sqrt{3}$.\n\nIt is easy to show that if the area of the quadrilateral with three given sides is a maximum, then the quadrilateral is convex. Let $AB$, $BC$, $CD$ be given and the quadrilateral $ABCD$ have the maximal area. We have $S(ABCD) = S(ABD) + S(BCD)$ and $S(ABD) = 0.5 \\cdot AB \\cdot BD \\sin... | Belarus | Belorusija 2012 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 30√3 | |
0094 | Problem:
An integer number $n > 2$ is called *k-beta* if two different numbers can be chosen from the set $\{1, 2, 3, \dots, n\}$ so that their product is equal to $k$ times the sum of the other $n-2$ numbers. For each positive integer $k$, find all the *k-beta* numbers. | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | Let T = n(n+1)/2.
- For k ≥ 7: no k-beta numbers exist.
- For k = 6: the only k-beta number is n = 3 (e.g., choose 2 and 3).
- For k = 5: no k-beta numbers exist.
- For k = 4: the only k-beta number is n = 4 (e.g., choose 3 and 4).
- For k = 3: the only k-beta number is n = 6 (e.g., choose 5 and 6).
- For k = 2: the on... | |
0555 | Find all possibilities: how many acute angles can there be in a convex polygon? | [
"A square has 0 acute angles, a right-angled trapezium has 1 acute angle, an obtuse triangle has 2 acute angles, an acute triangle has 3.\n\n\nFig. 1\n\nFig. 2\n\nFig. 3\nLet us show that 4 or more acute angles is not possible. By moving alo... | Estonia | Open Contests | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 0, 1, 2, 3 | |
07ty | Let $P_1, P_2, \dots, P_{2021}$ be $2021$ points in the quarter plane $\{(x, y) \mid x \ge 0, y \ge 0\}$.
The centroid of these $2021$ points lies at the point $(1, 1)$.
Show that there are two distinct points $P_i, P_j$ such that the distance from $P_i$ to $P_j$ is no more than $\sqrt{2}/20$. | [
"The appearance of $\\sqrt{2}$ is a clue that this question will end up with a pigeon-hole principle on squares of a grid with spacing $1/20$.\nWe prove a more general result:\n\n**Theorem:** Let $k$ be a positive integer, let $n \\ge k(k+1)/2$ and let $h > 0$. Suppose that a set of points $(x_j, y_j)$ for $j = 1, ... | Ireland | IRL_ABooklet | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0etb | Let $PAB$ and $PBC$ be two similar right-angled triangles (in the same plane) with $\angle PAB = \angle PBC = 90^\circ$ such that $A$ and $C$ lie on opposite sides of the line $PB$. If $PC = AC$, calculate the ratio $\frac{PA}{AB}$. | [
"\nFigure 1\n\nWe may assume that $AB = 1$. Let $PA = x$, so that $PB = \\sqrt{x^2+1}$, by Pythagoras. From the similarity of triangles $PAB$ and $PBC$, we have $BC = \\frac{\\sqrt{x^2+1}}{x}$, so that $PC = x + \\frac{1}{x}$, again by Pythagoras. Thus $AC = PC = x + \\frac{1}{x}$.\nNow let... | South Africa | The South African Mathematical Olympiad, Third Round | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English, Afrikaans | proof and answer | sqrt(2) | |
04vv | Let $T$ be the centroid of a triangle $ABC$. Consider two isosceles right-angled triangles $BTK$ and $CTL$ so that $K$ lies in the half-plane $BTC$ and $L$ lies in the half-plane $CTA$. Finally, denote the centre of the side $BC$ as $D$ and the centre of $KL$ as $E$. Determine all the possible values of the ratio $\fra... | [
"We shall prove that the ratio has to be equal to $2\\sqrt{2}$.\n\nFirst, we shall observe that the triangles $BTC$ and $KTL$ (coloured turquoise and yellow in the diagram) are similar, since\n$$\n\\angle BTC = \\angle BTK + \\angle KTC = 45^\\circ + \\angle KTC = \\angle KTC + \\angle CTL ... | Czech Republic | First Round of the 73rd Czech and Slovak Mathematical Olympiad (take-home part) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 2√2 | |
0fyy | Problem:
Finde alle Funktionen $f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}$ so dass für alle positiven rationalen Zahlen $x, y$ gilt
$$
f\left(f(x)^{2} y\right)=x^{3} f(x y)
$$ | [
"Solution:\nMit $y=1$ folgt\n$$\nf\\left(f(x)^{2}\\right)=x^{3} f(x)\n$$\ninsbesondere ist $f$ injektiv. Ersetzt man hier $x$ durch $x y$, dann erhält man\n$$\nf\\left(f(x y)^{2}\\right)=x^{3} y^{3} f(x y)\n$$\nAndererseits kann man in der ursprünglichen Gleichung $y$ durch $f(y)^{2}$ ersetzen und erhält unter noch... | Switzerland | IMO Selektion | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | f(x) = 1/x for all positive rational x | |
0hze | Problem:
Let $A_{n}$ be the area outside a regular $n$-gon of side length $1$ but inside its circumscribed circle, let $B_{n}$ be the area inside the $n$-gon but outside its inscribed circle. Find the limit as $n$ tends to infinity of $\frac{A_{n}}{B_{n}}$. | [
"Solution:\nThe radius of the inscribed circle is $\\frac{1}{2} \\cot \\frac{\\pi}{n}$, the radius of the circumscribed circle is $\\frac{1}{2} \\csc \\frac{\\pi}{n}$, and the area of the $n$-gon is $\\frac{n}{4} \\cot \\frac{\\pi}{n}$. The diagram below should help you verify that these are correct.\n\n![](attache... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 2 | |
0h0v | The sum of a few consecutive integers (that are greater than $1$) is $2011$. Find all such numbers. | [
"Let $a_1$ be the first number and $a_n$ is the last one. We arrive at:\n$$\n\\frac{a_1 + a_n}{2} n = 2011 \\Leftrightarrow (a_1 + a_n)n = 2 \\cdot 2011.\n$$\nSince $2$ and $2011$ are prime, we have two options: $a_1 + a_n = 2$ and $n = 2011$ or $a_1 + a_n = 2011$ and $n = 2$. First case is clearly impossible, whil... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 1005 + 1006 | |
0fkc | Problem:
Calcular la suma $2\left[h\left(\frac{1}{2009}\right)+h\left(\frac{2}{2009}\right)+\ldots+h\left(\frac{2008}{2009}\right)\right]$, siendo
$$
h(t)=\frac{5}{5+25^{t}}, \quad t \in \mathbb{R}
$$ | [
"Solution:\nSe observa que la función $h$ es simétrica respecto al punto $\\left(\\frac{1}{2}, \\frac{1}{2}\\right)$. Por tanto, $h(1-t)=\\frac{5}{25^{1-t}+5}=\\frac{5 \\cdot 5^{t}}{25+5 \\cdot 25^{t}}=\\frac{25^{t}}{25^{t}+5}$, de donde $h(t)+h(1-t)=1$. La suma vale entonces $2 \\cdot 1004=2008$."
] | Spain | XLV Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2008 | |
03hg | Problem:
Let $ABCD$ be a rectangle with $BC = 3 AB$. Show that if $P, Q$ are the points on side $BC$ with $BP = PQ = QC$, then
$$
\angle DBC + \angle DPC = \angle DQC.
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
058l | Do there exist numbers $a$, $b$, $c$ that satisfy the equation
$$
2a(c-a) - b(2a+b) + c(2b-c) = 2020?
$$ | [
"Transforming the l.h.s. of the equation gives\n$$\n\\begin{aligned}\n2a(c-a) - b(2a+b) + c(2b-c) &= 2ac - 2a^2 - 2ab - b^2 + 2bc - c^2 \\\\\n&= -a^2 - (a+b-c)^2.\n\\end{aligned}\n$$\nThe equality $-a^2 - (a+b-c)^2 = 2020$ cannot be valid since all terms of its l.h.s. are non-positive whereas the r.h.s. is positive... | Estonia | Estonian Math Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | No | |
04gg | Let $a$ be a positive integer, and let $b$ and $c$ be integers such that the equation $a x^2 + b x + c = 0$ has two different solutions in the interval $\langle 0, \frac{1}{2} \rangle$. Prove that $a \ge 6$. | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
06yo | Let $\mathbb{Z}_{>0}$ denote the set of positive integers. Let $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ be a function satisfying the following property: for $m, n \in \mathbb{Z}_{>0}$, the equation
$$
f(m n)^2 = f\left(m^2\right) f(f(n)) f(m f(n))
$$
holds if and only if $m$ and $n$ are coprime.
For each positi... | [
"- From $P(1,1)$, we have $f(1)^2 = f(1) f(f(1))^2$, so $f(1) = f(f(1))^2$.\n- From $P(1, f(1))$, we have $f(f(1))^2 = f(1) f(f(f(1))) f(f(f(1)))$, so $f(f(f(1))) = 1$.\n- From $P(1, f(f(1)))$, we have $f(f(f(1)))^2 = f(1) f(f(f(f(1)))) f(f(f(f(1))))$, which simplifies to $1 = f(1)^3$, so $f(1) = 1$.\n- From $P(1, ... | IMO | IMO2024 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Dis... | English | proof and answer | For each positive integer n, the possible values of f(n) are exactly the positive integers whose set of prime divisors is the same as that of n; equivalently, f(n) can be any product of the primes dividing n with arbitrary positive exponents. | |
0cla | Andrei represents $2025$ as a sum of $40$ pairwise different positive integers. Find the lowest value that the largest of the $40$ numbers can achieve. | [
"Let $0 < a_1 < a_2 < a_3 < \\dots < a_{40}$ so that $a_1 + a_2 + a_3 + \\dots + a_{40} = 2025$. Then $a_2 \\ge a_1 + 1$, $a_3 \\ge a_2 + 1$, $a_4 \\ge a_3 + 1$, $\\dots$, $a_{40} \\ge a_{39} + 1$.\nConsequently, $a_{40} \\ge a_1 + 39 \\ge a_2 + 38 \\ge \\dots \\ge a_{38} + 2$.\nThen $40 \\cdot a_{40} \\ge (a_1+39)... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 71 | |
0ikf | Problem:
Let $n$ be an integer at least $5$. At most how many diagonals of a regular $n$-gon can be simultaneously drawn so that no two are parallel? Prove your answer. | [
"Solution:\n\nLet $O$ be the center of the $n$-gon. Let us consider two cases, based on the parity of $n$:\n\n- $n$ is odd. In this case, for each diagonal $d$, there is exactly one vertex $D$ of the $n$-gon, such that $d$ is perpendicular to line $O D$; and of course, for each vertex $D$, there is at least one dia... | United States | Harvard-MIT Mathematics Tournament, Team Round A | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | n | |
0l4i | Problem:
Amelia's mother proposes a game. "Pick two of the shapes below," she says to Amelia. (The shapes are an equilateral triangle, a parallelogram, an isosceles trapezoid, a kite, and an ellipse. These shapes are drawn to scale.) Amelia's mother continues: "I will draw those two shapes on a sheet of paper, in whate... | [
"Solution:\nAmelia should choose the parallelogram and the ellipse, which are then simultaneously bisected into congruent halves by the line through their centers. This works because these two shapes have half-turn rotational symmetry.\n\nEach of the other three shapes has only a finite number of lines cutting it i... | United States | 25th Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | parallelogram and ellipse | |
0288 | Problem:
Diogo recortou uma cruz de cartolina como mostrado abaixo. Nesta cruz, todos os lados têm comprimento igual a $1~\mathrm{cm}$, e todos os ângulos são retos. Fernanda desafiou Diogo a fazer dois cortes em linha reta nesta cruz, de modo a formar quatro peças que possam ser reencaixadas de modo a formar um quadr... | [
"Solution:\n\na) A área do quadrado obtido deve ser a mesma da cruz de cartolina. Como a cruz de cartolina pode ser dividida em cinco quadrados de lados iguais a um, sua área será igual a $5~\\mathrm{cm}^2$.\n\nb) Para que a área do quadrado obtido seja igual a $5~\\mathrm{cm}^2$, é necessário que seu lado seja igu... | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Area: 5 square centimeters; side length: square root of 5 centimeters. | |
020o | Problem:
An integer $n \geqslant 2$ having exactly $s$ positive divisors $1=d_{1}<d_{2}<\cdots<d_{s}=n$ is said to be good if there exists an integer $k$, with $2 \leqslant k \leqslant s$, such that $d_{k}>1+d_{1}+\cdots+d_{k-1}$. An integer $n \geqslant 2$ is said to be bad if it is not good.
a. Show that there are ... | [
"Solution:\n\na.\n\nSolution 1. We note that $n=2^{m}$ has $m+1$ divisors, $d_{k}=2^{k-1}$ for $1 \\leqslant k \\leqslant m+1$. Thus\n$$\n1+d_{1}+\\cdots+d_{k-1}=1+\\left(2^{k-1}-1\\right)=2^{k-1}=d_{k}\n$$\nfor each $k \\geqslant 2$, and hence each power of $2$ is a bad integer. This exhibits infinitely many bad i... | Benelux Mathematical Olympiad | 10th Benelux Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Other"
] | null | proof only | null | |
0f9j | Problem:
A quadratic polynomial $p(x)$ has positive real coefficients with sum $1$. Show that given any positive real numbers with product $1$, the product of their values under $p$ is at least $1$. | [] | Soviet Union | 24th ASU | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
0l7c | Problem:
Define $\operatorname{sgn}(x)$ to be $1$ when $x$ is positive, $-1$ when $x$ is negative, and $0$ when $x$ is $0$. Compute
$$
\sum_{n = 1}^{\infty}\frac{\operatorname{sgn}(\sin(2^{n}))}{2^{n}}.
$$
(The arguments to $\sin$ are in radians.) | [
"Solution:\nNote that each of the following is equivalent to the next.\n\n$\\cdot\\ \\operatorname{sgn}(\\sin (2^{n})) = +1$\n\n$\\cdot\\ 0 < 2^{n} \\bmod 2\\pi < \\pi$\n\n$\\cdot\\ 0 < \\frac{2^{n}}{\\pi} \\bmod 2 < 1$\n\nThe $n$th digit after the decimal point in the binary representation of $\\frac{1}{\\pi}$ is ... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 1 - 2/π | |
04gq | If $p$ and $p^2 + 8$ are prime numbers, prove that $p^3 + 4$ is also prime. | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0i2x | Problem:
Find all $x$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ such that $1-\sin^{4} x-\cos^{2} x=\frac{1}{16}$. | [
"Solution:\n$1-\\sin^{4} x-\\cos^{2} x = \\frac{1}{16} \\Rightarrow (16-16 \\cos^{2} x) - \\sin^{4} x - 1 = 0 \\Rightarrow 16 \\sin^{4} x - 16 \\sin^{2} x + 1 = 0$.\n\nUse the quadratic formula in $\\sin x$ to obtain $\\sin^{2} x = \\frac{1}{2} \\pm \\frac{\\sqrt{3}}{4}$.\n\nSince $\\cos 2x = 1 - 2 \\sin^{2} x = \\... | United States | Harvard-MIT Math Tournament | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | x = ±π/12, ±5π/12 | |
05uw | Problem:
Soit $a$ un réel strictement positif et $n \geqslant 1$ un entier. Montrer que
$$
\frac{a^{n}}{1+a+\ldots+a^{2 n}}<\frac{1}{2 n}
$$ | [
"Solution:\nPuisque la difficulté réside dans le dénominateur du membre de droite, et pour plus de confort, on peut chercher à montrer la relation inverse, à savoir :\n$$\n\\frac{1+a+\\ldots+a^{2 n}}{a^{n}}>2 n\n$$\nL'idée derrière la solution qui suit est \"d'homogénéiser\" le numérateur du membre de gauche, c'est... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0d9a | Find the number of ways one can put numbers $1$ or $2$ in each cell of a $8 \times 8$ chessboard in such a way that the sum of numbers in each column and in each row is an odd number (two ways are considered different if the number on some cell in the first way is different from the number on the cell at correspondent ... | [
"Consider the leftmost column and lowest row of table, we color all these cells. We can see that for all ways to put the number of the sub square $7 \\times 7$ that not colored, we can choose the number for the correspondent colored position at same column or row. Indeed, if sum of the $7$ numbers is odd, then we p... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 2^49 | |
04ku | Let $\triangle ABC$ be an acute-angled triangle such that $|AC| > |AB|$, and let $O$ be its circumcentre. The angle bisector of $\angle BAC$ meets the side $\overline{BC}$ at point $D$. The line through $B$ perpendicular to the line $AO$ intersects the line $AO$ at point $E$. Prove that points $A, B, D$ and $E$ all lie... | [] | Croatia | Mathematical competitions in Croatia | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"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 | |
03ie | Problem:
A walk consists of a sequence of steps of length $1$ taken in directions north, south, east or west. A walk is self-avoiding if it never passes through the same point twice. Let $f(n)$ denote the number of $n$-step self-avoiding walks which begin at the origin. Compute $f(1)$, $f(2)$, $f(3)$, $f(4)$, and show ... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | f(1) = 4, f(2) = 12, f(3) = 36, f(4) = 100; and for all n, 2^n < f(n) ≤ 4·3^{n−1}. | |
0jc9 | Problem:
You are repeatedly flipping a fair coin. What is the expected number of flips until the first time that your previous $2012$ flips are 'HTHT...HT'? | [
"Solution:\nAnswer: $\\left(2^{2014}-4\\right) / 3$\nLet $S$ be our string, and let $f(n)$ be the number of binary strings of length $n$ which do not contain $S$. Let $g(n)$ be the number of strings of length $n$ which contain $S$ but whose prefix of length $n-1$ does not contain $S$ (so it contains $S$ for the \"f... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (2^2014 - 4) / 3 | |
0eiq | Problem:
Poišči vse pare realnih števil $x$ in $y$, ki rešijo sistem enačb
$$
\begin{gathered}
\frac{y^{3}+15 x^{2}}{y^{4}-x^{3}}=\frac{y^{2}+15 x}{y^{3}-x^{2}} \\
\frac{1500 y^{3}+4 x^{2}}{9 y^{4}-4}=\frac{1500 y^{2}+4 x}{9 y^{3}-4}
\end{gathered}
$$ | [
"Solution:\n\nV obeh enačbah odpravimo ulomke in ju poenostavimo, da dobimo\n$$\n\\begin{aligned}\n-x^{2} y^{3}+15 x^{2} y^{3} & =-x^{3} y^{2}+15 x y^{4} \\\\\n-6000 y^{3}+36 x^{2} y^{3}-16 x^{2} & =-6000 y^{2}+36 x y^{4}-16 x\n\\end{aligned}\n$$\nPrvo enačbo preoblikujemo v $x^{2} y^{2}(x-y)=-15 x y^{3}(x-y)$ in n... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (0,1), (1,0), (-1/375,-1/375), (-5,1/3), (-15/2,1/2), (-5/2,1/6), (15,-1) | |
071f | Problem:
Pablo was trying to solve the following problem: find the sequence $x_0, x_1, x_2, \ldots, x_{2003}$ which satisfies $x_0 = 1$, $0 \leq x_i \leq 2 x_{i-1}$ for $1 \leq i \leq 2003$ and which maximises $S$. Unfortunately he could not remember the expression for $S$, but he knew that it had the form $S = \pm x_... | [
"Solution:\n\nFor any combination of signs the maximum is obtained by taking all $x_i$ as large as possible. Suppose we have a different set of $x_i$. Then for some $k$ we must have $x_k < 2 x_{k-1}$ and $x_i = 2 x_{i-1}$ for all $i > k$. Suppose $2 x^{k-1} - x^k = h > 0$. Then we can increase $x_k$ by $h$, $x_{k+1... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x_i = 2^i for i = 0, 1, ..., 2003 | |
05js | Problem:
Sur un échiquier $5$ sur $5$, on a placé sur des cases différentes $k$ cavaliers, de telle manière que chacun peut en prendre exactement $2$ autres.
Quelle est la plus grande valeur possible de $k$? | [
"Solution:\n\nLa configuration suivante donne un exemple avec $16$ cavaliers :\n\n\n\nOn montre maintenant que c'est optimal : quitte à échanger les couleurs, on peut supposer le centre noir. Notons $k$ le nombre de cavaliers sur une case noire, $l$ le nombre de cavaliers sur une case blanc... | France | OFM 2013-2014 Envoi 2 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 16 | |
0aed | Let $a$ and $b$ be integers such that $a = a^2 + b^2 - 8b - 2ab + 16$. Prove that $a$ is a square. | [
"$9a = a^2 + b^2 + 8a - 8b - 2ab + 16 = (a - b + 4)^2$ hence $9a$ is a square.\nNow it is obvious that $a$ is a square too."
] | North Macedonia | Junior Macedonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Macedonian, English | proof only | null | |
011b | Problem:
Find all positive integers $n$ such that $n$ is equal to $100$ times the number of positive divisors of $n$. | [
"Solution:\n\n$2000$ is the only such integer.\n\nLet $d(n)$ denote the number of positive divisors of $n$ and $p \\triangleright n$ denote the exponent of the prime $p$ in the canonical representation of $n$. Let $\\delta(n)=\\frac{n}{d(n)}$. Using this notation, the problem reformulates as follows: Find all posit... | Baltic Way | Baltic Way | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 2000 | |
0b2f | Problem:
In convex quadrilateral $A B C D$, $\angle C A B = \angle B C D$. $P$ lies on line $B C$ such that $A P = P C$, $Q$ lies on line $A P$ such that $A C$ and $D Q$ are parallel, $R$ is the point of intersection of lines $A B$ and $C D$, and $S$ is the point of intersection of lines $A C$ and $Q R$. Line $A D$ me... | [
"Solution:\n\nRefer to the figure shown below.\n\n\nBy angle-chasing (with directed angles), we have\n$$\n\\angle Q A R = \\angle P A B = \\angle C A B - \\angle C A P = \\angle B C D - \\angle P C A = \\angle A C D = \\angle Q D R\n$$\nThus quadrilateral $Q A D R$ is cyclic. Then,\n$$\n\\a... | Philippines | 23rd Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
095n | Problem:
În interiorul unui cerc mare sunt construite trei cercuri mici congruente, astfel încât fiecare cerc mic este tangent la cercul mare și la celelalte două cercuri mici. Dintr-un punct arbitrar $M$, situat pe cercul mare și diferit de punctele de tangență, este dusă câte o tangentă la fiecare cerc mic. Fie $l_{... | [
"Solution:\n\nLema 1. Dacă $K, L, N$ sunt punctele de tangență ale cercului mare cu cercurile mici, atunci triunghiul $K L N$ este echilateral.\n\n\n\nDemonstraţie. Deoarece $K O_{1} \\perp l$ și $K O \\perp l$ (unde $l$ este tangentă la cercul cu centrul $O_{1}$ în punctul $K$ ), atunci pu... | Moldova | A 62 - A OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04q0 | Denote by $\mathbb{N}$ the set of all positive integers. Find all functions $f: \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ that satisfy the following conditions:
* $f(a, b) + a + b = f(a, 1) + f(1, b) + ab$ holds for all positive integers $a$ and $b$.
* If any number among $a + b$ and $a + b - 1$ is divisible by prim... | [
"Plugging $(a, b) \\leftarrow (1, 1)$ into the first given condition, we get $f(1, 1) = 1$.\n\nOn the other hand, by plugging $(a, b) \\leftarrow (a, b+1)$ we get\n$$\nf(a, b + 1) + a + b + 1 = f(a, 1) + f(1, b + 1) + ab + a.\n$$\nUsing the first given condition in its original form, it follows that\n$$\nf(a, b + 1... | Croatia | Croatian Mathematical Society Competitions | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | f(a,b) = \binom{a+b}{2} = \frac{(a+b)(a+b-1)}{2} | |
07a6 | Two circles in the space are called *linking* if they intersect at two points or they are interlocked. Find a necessary and sufficient condition for four distinct points $A$, $B$, $A'$, $B'$ in the space such that every two different circles passing through $A$, $B$ and the other passing through $A'$, $B'$ respectively... | [
"The points should be on a circle (or line) and $A$, $B$ should separate $A'$, $B'$ on it. To prove necessity, first suppose that the points are not coplanar. Then there exist two parallel planes passing through $A$, $B$ and $A'$, $B'$ respectively. Any two circles in these planes are not linking. So the points sho... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | The four points must lie on a single circle or line, and the pair A, B must separate the pair A′, B′ on it. | |
0l5m | Problem:
Compute
$$
\frac{20 + \frac{1}{25 - \frac{1}{20}}}{25 + \frac{1}{20 - \frac{1}{25}}}.
$$ | [
"Solution:\nWe can use the fact that\n$$\nx + \\frac{1}{y - \\frac{1}{x}} = x + \\frac{x}{xy - 1} = \\frac{x^2 y}{xy - 1}.\n$$\nLetting $x = 20$, $y = 25$ and vice versa in the above expression, we get\n$$\n\\frac{x + \\frac{1}{y - \\frac{1}{x}}}{y + \\frac{1}{x - \\frac{1}{y}}} = \\frac{x^2 y}{x y^2} = \\frac{x}{y... | United States | HMMT February | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 4/5 | |
001r | Se tienen en el plano una línea quebrada cerrada y sin entrecruzamientos de $m$ lados y una línea quebrada cerrada y sin entrecruzamientos de $n$ lados. Estas dos líneas quebradas se intersectan en puntos interiores a sus lados (nunca en vértices). Se sabe que en total hay exactamente 102 puntos de intersección entre l... | [
"Cada línea quebrada cerrada de $k$ lados es un polígono de $k$ lados (posiblemente no convexo), y \"sin entrecruzamientos\" significa que no se cruza a sí misma.\n\nCada lado de la primera línea puede intersectar a cada lado de la segunda línea a lo sumo en un punto interior (pues no se permite que se crucen en lo... | Argentina | XIX Olimpíada Matemática Argentina | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | español | proof and answer | 23 | |
0b9h | Find all positive integers $r$ with the property that there exists positive prime numbers $p$ and $q$ so that $p^2 + pq + q^2 = r^2$. | [
"The given relation is equivalent to $(p+q)^2 = r^2 + pq$, which can be written $(p+q+r)(p+q-r) = pq$.\nThe divisors of $pq$ are $1$, $p$, $q$ and $pq$. Since $p+q > \\max\\{p,q\\}$, it follows that $p+q-r = 1$ and $p+q+r = pq$.\n\nAdding the last two equalities yields $2p+2q = pq+1$, that is $(p-2)(q-2) = 3$.\nThi... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 7 | |
03ka | Problem:
Let $y_{1}, y_{2}, y_{3}, \ldots$ be a sequence such that $y_{1}=1$ and, for $k>0$, is defined by the relationship:
$$
\begin{gathered}
y_{2k}= \begin{cases}2 y_{k} & \text{ if } k \text{ is even } \\
2 y_{k}+1 & \text{ if } k \text{ is odd }\end{cases} \\
y_{2k+1}= \begin{cases}2 y_{k} & \text{ if } k \text{ ... | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0czc | Let $f \in \mathbb{Z}[X], f = X^{2} + a X + b$, be a quadratic polynomial. Prove that $f$ has integer zeros if and only if for each positive integer $n$ there is an integer $u_{n}$ such that $n \mid f\left(u_{n}\right)$. | [
"If $f = (X - x_{1})(X - x_{2}),\\ x_{1}, x_{2} \\in \\mathbb{Z}$, then take $u_{n} = n + x_{1}$ and get $f\\left(u_{n}\\right) = n\\left(n + x_{1} - x_{2}\\right)$, hence $n \\mid f\\left(u_{n}\\right),\\ n \\geq 1$.\n\nConversely, assume that $f\\left(u_{n}\\right) = k_{n} \\cdot n$, for some integer $k_{n}$, $n ... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
08o0 | Problem:
A circle passing through the midpoint $M$ of the side $B C$ and the vertex $A$ of a triangle $A B C$ intersects the sides $A B$ and $A C$ for the second time at the points $P$ and $Q$, respectively. Show that if $\angle B A C=60^\circ$ then
$$
A P+A Q+P Q<A B+A C+\frac{1}{2} B C
$$ | [
"Solution:\nSince the quadrilateral $A P M Q$ is cyclic, we have $\\angle P M Q=180^\\circ-\\angle P A Q=180^\\circ-\\angle B A C=120^\\circ$. Therefore $\\angle P M B+\\angle Q M C=180^\\circ-\\angle P M Q=60^\\circ$.\nLet the point $B'$ be the symmetric of the point $B$ with respect to the line $P M$ and the poin... | JBMO | 17th Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0cwm | On the board 777 pairwise distinct *complex* numbers are written. It turned out that there are exactly 760 ways to choose a pair of numbers $a$ and $b$ written on the board so that
$$
a^2 + b^2 + 1 = 2ab.
$$
(Here pairs considered unordered, i.e., $(a, b)$ and $(b, a)$ is the same pair.) Prove that one can choose numb... | [
"Note that the condition $a^2 + b^2 + 1 = 2ab$ is equivalent to $(a-b)^2 = -1$ or $a-b = \\pm i$. Consider a graph whose vertices are the numbers written on the board, with an edge connecting two numbers if they differ by $i$. According to the problem's condition, this graph has exactly 760 edges.\n\nEach connected... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Algebra > Intermediate Algebra > Complex numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Russian | proof only | null | |
0l2s | What is the least value of $n$ such that $n!$ is a multiple of 2024?
(A) 11 (B) 21 (C) 22 (D) 23 (E) 253 | [
"Because the prime factorization of $2024$ is $2^3 \\cdot 11 \\cdot 23$, it follows that $n!$ is a multiple of $2024$ if and only if $n \\ge 23$. Therefore $23$ is the least value of $n$ such that $n!$ is a multiple of $2024$."
] | United States | AMC 10 A | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | MCQ | D | |
00eo | We say that an equilateral triangle on the plane is in *standard position* if one of its sides is horizontal and the opposite vertex lies above that side. We have $n$ equilateral triangles $S_1, S_2, \dots, S_n$, all of which are in standard position. For each triangle $S_i$, denote by $T_i$ its medial triangle. Let $S... | [
"We divide each triangle $S_i$ into four congruent triangles $T_i$, $A_i$, $B_i$, $C_i$.\nLet $A$ be the set of all points that belong to some $A_i$ but don't belong to any $T_i$. We define $B$ and $C$ analogously. It is clear then that $S = T \\cup A \\cup B \\cup C$.\nNext we will prove that $\\text{area}(A) \\le... | Argentina | Cono Sur Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0g9c | 令不等邊三角形 $\triangle ABC$ 的內切圓圓心為 $I$, 且該內切圓分別切 $CA, AB$ 邊於點 $E, F$. 設 $\triangle AEF$ 的外接圓在 $E$ 和 $F$ 的兩條切線交於點 $S$. 直線 $EF$ 與 $BC$ 交於點 $T$. 試證: 以 $ST$ 為直徑的圓垂直於 $\triangle BIC$ 的九點圓。
In a scalene triangle $ABC$ with incenter $I$, the incircle is tangent to sides $CA$ and $AB$ at points $E$ and $F$. The tangents to the c... | [
"解: 令 $D$ 為 $I$ 對 $BC$ 的垂足。令 $X, Y$ 分別為 $B, C$ 對 $CI, BI$ 的垂足。可證 $BIFX, CIEY$ 共圓, 因此可得 $X, Y$ 落在直線 $EF$ 上。將 $M$ 設為 $BC$ 的中點, $\\omega$ 為 $DMXY$ 的外接圓。原題等價於證明 $T$ 落在 $S$ 對 $\\omega$ 的極線上。\n\n設 $K$ 為 $AM$ 於 $EF$ 的交點, 根據 SL 2005 G6 可得 $K, I, D$ 共線。\n\n令 $N$ 為 $EF$ 的中點, $L$ 為 $KS$ 於 $BC$ 的交點。由\n$$\n-1 = (A, I; N, S) \\s... | Taiwan | 二〇一五數學奧林匹亞競賽第三階段選訓營 | [
"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 > Tangents",
"Geometry > Plane Geometry > Advanced Configurations... | null | proof only | null | |
03o6 | Problem:
The $n$ players of a hockey team gather to select their team captain. Initially, they stand in a circle, and each person votes for the person on their left.
The players will update their votes via a series of rounds. In one round, each player $a$ updates their vote, one at a time, according to the following ... | [
"Solution:\n\nInitially, all players are in a cycle. Note that once a player leaves the cycle, they cannot rejoin. Furthermore, a new cycle cannot be created. Hence, at any point in time, the graph corresponding to the votes will be a functional graph with a single cycle.\n\nWe will first prove that after $\\lfloor... | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Other"
] | null | proof only | null | |
07gq | Call a triple of numbers **nice** if one of them is the average of the other two. Assume that we have $2k + 1$ distinct numbers with $k^2$ nice triples. Prove that these numbers can be divided into two arithmetic progressions with equal ratios. | [
"Let the numbers be $a_1 < a_2 < \\dots < a_{2k+1}$ in increasing order. First notice that the numbers $a_1, a_2, \\dots, a_k, a_{k+1}, a_{k+2}, \\dots, a_{2k+1}$ can be the middle element of at most $0, 1, 2, \\dots, k-1, k, k-1, \\dots, 0$ nice triples, respectively. Therefore, in total we have at most $0 + 1 + 2... | Iran | 38th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0igg | Problem:
In how many ways can 4 purple balls and 4 green balls be placed into a $4 \times 4$ grid such that every row and column contains one purple ball and one green ball? Only one ball may be placed in each box, and rotations and reflections of a single configuration are considered different. | [
"Solution:\n\nThere are $4! = 24$ ways to place the four purple balls into the grid. Choose any purple ball, and place two green balls, one in its row and the other in its column. There are four boxes that do not yet lie in the same row or column as a green ball, and at least one of these contains a purple ball (ot... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 216 | |
01gq | Each vertex $v$ and each edge $e$ of a graph $G$ are assigned numbers $f(v) \in \{1, 2\}$ and $f(e) \in \{1, 2, 3\}$, respectively. Let $S(v)$ be the sum of numbers assigned to the edges incident to $v$ plus the number $f(v)$. We say that assignment $f$ is cool if $S(u) \neq S(v)$, for every pair of adjacent vertices i... | [
"Let $v_1, v_2, \\dots, v_n$ be any ordering of the vertices of $G$. Initially each vertex assigned number $1$, and each edge assigned number $2$. One may imagine that there is a chip lying on each vertex, while two chips are lying on each edge. We are going to refine this assignment so as to get a cool one by perf... | Baltic Way | Baltic Way 2020 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
044f | In $\triangle ABC$, $AB = 1$, $AC = 2$, $B - C = \frac{2\pi}{3}$. Then the area of $\triangle ABC$ is ______. | [
"By the law of sines, it follows that $\\frac{\\sin B}{\\sin C} = \\frac{AC}{AB} = 2$. Since $B - C = \\frac{2\\pi}{3}$, we have\n$$\n2 \\sin C = \\sin B = \\sin \\left( C + \\frac{2\\pi}{3} \\right) = -\\frac{1}{2} \\sin C + \\frac{\\sqrt{3}}{2} \\cos C,\n$$\nnamely, $\\frac{5}{2} \\sin C = \\frac{\\sqrt{3}}{2} \\... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 3√3/14 | |
0akh | Alice and Bob play the following game: They start with two non-empty piles of coins. Taking turns, with Alice playing first, each player chooses a pile with an even number of coins and moves half of this pile to the other pile. The game ends if a player cannot move, in which case the other player wins.
Determine all pa... | [
"By $\\nu_2(n)$ we denote the largest nonnegative integer $r$ such that $2^r \\mid n$. A position $(a,b)$ (i.e. two piles of sizes $a$ and $b$) is said to be $k$-happy if $\\nu_2(a) = \\nu_2(b) = k$ for some integer $k \\ge 0$, and $k$-unhappy if $\\min\\{\\nu_2(a), \\nu_2(b)\\} = k < \\max\\{\\nu_2(a), \\nu_2(b)\\... | North Macedonia | Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | Bob has a winning strategy exactly when both piles have the same highest power of two dividing their sizes and this common exponent is even. | |
07cx | a) A number $m$ is called **mirror-symmetry** if it is possible to divide the reverse decimal expansion of $m$ into some blocks such that the multiply of these blocks is equal to $m$. For instance, numbers $6$, $543$ and $21$ are such blocks for number $123456$, if the multiply of these $3$ numbers was equal to $123456... | [
"a) For any number $A$, let $\\overleftarrow{A}$ be the reverse decimal expansion of $A$. Assume that $A = \\overline{A_nA_{n-1}\\cdots A_1}$ is a **mirror-symmetry** number with $m$ digits, all from $\\{1, 2, 3\\}$, and $A_n, A_{n-1}, \\dots, A_1$ are blocks of $A$ with number of digits $m_n, \\dots, m_1$ such tha... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | a) Exactly the palindromic numbers whose digits are from the set {1, 2, 3}. b) Infinitely many exist; for example, every number of the form 315 times a power of ten is good. | |
0kqp | Problem:
Let triangle $A B C$ be an acute triangle with circumcircle $\Gamma$. Let $X$ and $Y$ be the midpoints of minor arcs $\widehat{A B}$ and $\widehat{A C}$ of $\Gamma$, respectively. If line $X Y$ is tangent to the incircle of triangle $A B C$ and the radius of $\Gamma$ is $R$, find, with proof, the value of $X ... | [
"Solution:\n\nNote that $X$ and $Y$ are the centers of circles $(A I B)$ and $(A I C)$, respectively, so we have $X Y$ perpendicularly bisects $A I$, where $I$ is the incenter. Since $X Y$ is tangent to the incircle, we have $A I$ has length twice the inradius. Thus, we get $\\angle A = 60^{\\circ}$. Thus, since $\... | United States | HMMT February 2022 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometr... | null | proof and answer | R√3 | |
02hh | Problem:
Uma loja de sabonetes realiza uma promoção com o anúncio "Compre um e leve outro pela metade do preço". Outra promoção que a loja poderia fazer oferecendo o mesmo desconto percentual é:
A) "Leve dois e pague um"
B) "Leve três e pague um"
C) "Leve três e pague dois"
D) "Leve quatro e pague três"
E) "Leve cinco... | [
"Solution:\n\n(D) Pela promoção, quem levar 2 unidades paga pelo preço de 1,5 unidade, logo quem levar 4 unidades paga pelo preço de 3 unidades, ou seja, leva quatro e paga três."
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | MCQ | D | |
061x | Problem:
Gegeben sind die positiven reellen Zahlen $a$ und $b$ und die natürliche Zahl $n$.
Man ermittle in Abhängigkeit von $a, b$ und $n$ das größte der $n+1$ Glieder in der Entwicklung von $(a+b)^n$. | [
"Solution:\n\nDas $k$-te Glied $G(k)$ in der Entwicklung von $(a+b)^n$ ist gegeben durch die Formel:\n$$\nG(k) = \\binom{n}{k-1} \\cdot a^{n-k+1} \\cdot b^{k-1},\n$$\nwobei $\\binom{n}{0} = 1$ und $1 \\leq k \\leq n+1$.\n\nDa es endlich viele Glieder gibt und jede endliche Zahlenmenge (mindestens) ein maximales Ele... | Germany | Auswahlwettbewerb zur IMO 2005 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | Let α = (n+1)b/(a+b). If α is not an integer, the unique largest term is the k-th term with k = ⌊α⌋ + 1, i.e., C(n, k−1) a^{n−k+1} b^{k−1}. If α is an integer m, then the m-th and (m+1)-th terms are equal and both maximal. | |
0kp0 | Problem:
Real numbers $x$ and $y$ satisfy the following equations:
$$
\begin{aligned}
x & = \log_{10}\left(10^{y-1}+1\right)-1 \\
y & = \log_{10}\left(10^{x}+1\right)-1
\end{aligned}
$$
Compute $10^{x-y}$. | [
"Solution:\n\nTaking 10 to the power of both sides in each equation, these equations become:\n$$\n\\begin{aligned}\n& 10^{x} = \\left(10^{y-1} + 1\\right) \\cdot 10^{-1} \\\\\n& 10^{y} = \\left(10^{x} + 1\\right) \\cdot 10^{-1}\n\\end{aligned}\n$$\nLet $a = 10^{x}$ and $b = 10^{y}$. Our equations become:\n$$\n\\beg... | United States | HMMT November 2022 | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 101/110 | |
06nd | In $\triangle ABC$, $AC = \sqrt{3} AB$ and $BC = 2$. $D$ is a point inside $\triangle ABC$ such that $\angle BDC = 90^\circ$, $\angle DAC = 18^\circ$ and $BD = 1$. Find $\angle DAB$. | [
"48°\n\n\n\nNote that $DC = \\sqrt{3}$, $\\angle DBC = 60^\\circ$ and $\\angle DCB = 30^\\circ$. In particular, we notice that $AB : AC = DB : DC$. Thus, the internal angle bisectors of $\\angle BAC$ and $\\angle BDC$ meet $BC$ at the same point $P$, and the external angle bisectors of $\\a... | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 48° | |
0azh | Problem:
The sum of the terms of an infinite geometric series is $2$ and the sum of the squares of the corresponding terms of this series is $6$. Find the sum of the cubes of the corresponding terms. | [
"Solution:\n\nLet $a$ be the first term and let $r \\in (-1,1)$ be the common ratio of such infinite geometric series. Then, $\\dfrac{a}{1-r} = 2$ \\hspace{2mm} (1) and $\\dfrac{a^{2}}{1-r^{2}} = 6$ \\hspace{2mm} (2).\n\nSquaring (1) gives $\\dfrac{a^{2}}{(1-r)^{2}} = 4$ and using (2) yields $\\dfrac{1-r}{1+r} = \\... | Philippines | 20th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 96/7 | |
099h | Let $x$, $y$, $z$ be positive real numbers such that $x + y + z = 1$. Prove that the following inequality holds:
$$
\frac{\sqrt{xyz}}{x^2 + y^2 + z^2 - x^3 - y^3 - z^3} \le \sqrt{\frac{xy}{(1-z)^2} + \frac{yz}{(1-x)^2} + \frac{zx}{(1-y)^2}}
$$
(proposed by N. Argilsan) | [
"$$\n\\left( \\frac{1}{x^2(1-x) + y^2(1-y) + z^2(1-z)} \\right)^2 \\le\n$$\n\n$$\n\\leq \\frac{1}{x(1-x)^2} + \\frac{1}{y(1-y)^2} + \\frac{1}{z(1-z)^2}.\n$$\nLet's consider $f(x) = \\frac{1}{x^2}$ function. Because $f''(x) = \\frac{6}{x^4} > 0$, hence apply Jensen's inequality and we choose $\\alpha = x$, $\\beta =... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
054j | Given positive integers $a$, $b$, $c$ and $d$ and
$$ (a+b)(a+c)(a+d)(b+c)(b+d)(c+d) = u, $$
$$ ab + ac + ad + bc + bd + cd = v, $$
prove that the product $uv$ is divisible by 3. | [
"If among numbers $a$, $b$, $c$, $d$ there are two that give either remainders $0$ and $0$ or remainders $1$ and $2$ modulo $3$, the sum of these two numbers is divisible by $3$. Hence $u$, as well as $uv$, is divisible by $3$.\n\nNow study the case where at most one among the numbers $a$, $b$, $c$, $d$ is divisibl... | Estonia | National Olympiad Final Round | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0j69 | Problem:
Let $P(x) = x^{4} + 2x^{3} - 13x^{2} - 14x + 24$ be a polynomial with roots $r_{1}, r_{2}, r_{3}, r_{4}$. Let $Q$ be the quartic polynomial with roots $r_{1}^{2}, r_{2}^{2}, r_{3}^{2}, r_{4}^{2}$, such that the coefficient of the $x^{4}$ term of $Q$ is $1$. Simplify the quotient $Q(x^{2}) / P(x)$, leaving your... | [
"Solution:\nAnswer: $x^{4} - 2x^{3} - 13x^{2} + 14x + 24$\n\nWe note that we must have\n$$\nQ(x) = (x - r_{1}^{2})(x - r_{2}^{2})(x - r_{3}^{2})(x - r_{4}^{2}) \\Rightarrow Q(x^{2}) = (x^{2} - r_{1}^{2})(x^{2} - r_{2}^{2})(x^{2} - r_{3}^{2})(x^{2} - r_{4}^{2})\n$$\nSince $P(x) = (x - r_{1})(x - r_{2})(x - r_{3})(x ... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | x^4 - 2x^3 - 13x^2 + 14x + 24 | |
06jy | Let the internal angle bisector of $\angle BAC$ of $\triangle ABC$ meet side $BC$ at $D$. Let $\Gamma$ be the circle through $A$ tangent to $BC$ at $D$. Suppose $\Gamma$ meets sides $AB$ and $AC$ at $E$ and $F$ again, respectively. Lines $BF$ and $CE$ meet $\Gamma$ again at $P$ and $Q$, respectively. Let $AP$ and $AQ$ ... | [
"Since $\\angle AFD = \\angle ADB$ and $\\angle DAF = \\angle BAD$, we have $\\angle ADF = \\angle ABD$. This implies $\\angle AEF = \\angle ADF = \\angle ABD$ so that $EF \\parallel BC$. It follows that $\\angle XBP = \\angle EFP = \\angle BAX$. Thus, $XB$ is a tangent to $(ABP)$. Hence, we have $XB^2 = XP \\cdot ... | Hong Kong | Pre-IMO 2017 Mock Exam | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0315 | Problem:
Find all values of $a$, for which the equation
$$
\sqrt{a x^{2}+a x+2}=a x+2
$$
has a unique root. | [
"Solution:\nIf $a x+2<0$, then the equation has no real roots. If $a x+2 \\geq 0$, it is equivalent to $a x^{2}+a x+2=(a x+2)^{2}$, i.e., $(a^{2}-a) x^{2}+3 a x+2=0$. The last equation has a unique real root in the following three cases.\n\nCase 1. The coefficient of $x^{2}$ vanishes and the respective linear equat... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a = -8 or a ≥ 1 | |
07a1 | Four metal pieces are joined to each other to form a quadrilateral in the space. The angle between them can vary freely. In a case that the quadrilateral is not planar, we mark one point of each piece such that the points lie in a plane. Prove that these four points are always coplanar as the quadrilateral varies.
![]... | [
"Let quadrilateral that these four pieces form in the space be $ABCD$ and we marked points $M, N, P$ and $Q$ on sides $AB, BC, CD$ and $DA$ respectively, so that the marked points are on a plane named $\\pi$. Let $a, b, c$ and $d$ be the distances from $A, B, C$ and $D$ to $\\pi$ respectively. So we have\n\n![](att... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof only | null | |
0c4w | Find all the functions $f: \mathbb{R} \to \mathbb{R}$ so that $f(x+y) = f(xy) + f(x) + f(y)$, for every $x, y \in \mathbb{R}$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) ≡ 0 | |
0e8d | Let $m$ and $n$ be coprime natural numbers of different parity. Prove that the numbers $2^{2m} + 2^{m+1} + 1$ and $2^{2n} + 2^{n+1} + 1$ are coprime as well. | [
"Denote $a = 2^{2n} + 2^{n+1} + 1$ and $b = 2^{2m} + 2^{m+1} + 1$. We notice that $a = (2^n + 1)^2$ and $b = (2^m + 1)^2$. Because we have $(2^{2n} - 1)^2 = (2^n + 1)^2(2^n - 1)^2 = a(2^n - 1)^2$ and $(2^{2m} - 1)^2 = b(2^m - 1)^2$, the greatest common divisor $D(a,b)$ divides $D((2^{2n} - 1)^2, (2^{2m} - 1)^2)$. W... | Slovenia | Selection Examinations for the IMO 2013 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
05d9 | Problem:
Let $A B C$ be a triangle with circumcentre $O$. The points $D, E$ and $F$ lie in the interiors of the sides $B C, C A$ and $A B$ respectively, such that $D E$ is perpendicular to $C O$ and $D F$ is perpendicular to $B O$. (By interior we mean, for example, that the point $D$ lies on the line $B C$ and $D$ is... | [
"Solution:\n\nLet $\\ell_{C}$ be the tangent at $C$ to the circumcircle of $\\triangle A B C$. As $C O \\perp \\ell_{C}$, the lines $D E$ and $\\ell_{C}$ are parallel. Now we find that\n$$\n\\angle C D E=\\angle\\left(B C, \\ell_{C}\\right)=\\angle B A C,\n$$\nhence the quadrilateral $B D E A$ is cyclic. Analogousl... | European Girls' Mathematical Olympiad (EGMO) | European Girls' Mathematical Olympiad 2012-Day 1 Solutions | [
"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 > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0gze | There are 16 consecutive positive integers written on the board. Andrew calculates their product and Olesya – their sum. Can it happen that in both numbers there coincide
a) three last digits,
b) four last digits? | [
"**Answer:** a) yes; b) no.\n\nIt's obvious that a number received by Andrew is divisible by $16$ and by $125$, because from $16$ consecutive numbers more than four are divisible by $2$ and at least $3$ are divisible by $5$. It also implies that three last digits in Andrew's number are $0$.\n\na) Let the numbers $a... | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a) yes; b) no | |
0g1a | Problem:
Finde alle Funktionen $f: \mathbb{R} \rightarrow \mathbb{R}$, sodass für alle $x, y \in \mathbb{R}$ gilt:
$$
f(x+y f(x))=f(x f(y))-x+f(y+f(x))
$$ | [
"Solution:\n\nWir setzen $x=y=0$ ein, um $f(f(0))=0$ zu erhalten. Mit $x=y=1$ wird die Gleichung zu $f(f(1))=1$. Mit diesen beiden Identitäten und einsetzen von $x=1$ und $y=0$ erhalten wir $f(1)=0$. Dann gilt auch $1=f(f(1))=f(0)$. Durch Einsetzen von $y=0$ wird die Originalgleichung zu $f(f(x))=x$. Setzen wir nun... | Switzerland | SMO Finalrunde | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 1 - x | |
01g5 | Let $p$ a prime number. Prove that $p^2 + p + 1$ is not a perfect cube. | [
"Assume that $p^2 + p + 1 = k^3$. Then we have that $p(p+1) = k^3 - 1 = (k-1)(k^2 + k + 1)$. Since $p$ is prime, either $p \\mid k-1$ or $p \\mid k^2 + k + 1$. If $p \\mid k-1$, then $p \\le k-1$ and $k^3 \\ge (p+1)^3 > p^2 + p + 1 = k^3$, which is impossible. Now consider the possibility that $p \\mid k^2 + k + 1$... | Baltic Way | Baltic Way 2019 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof only | null | |
0d1s | $ABCD$ is a cyclic quadrilateral and $\omega$ its circumcircle. The perpendicular line to $AC$ at $D$ intersects $AC$ at $E$ and $\omega$ at $F$. Denote by $\ell$ the perpendicular line to $BC$ at $F$. The perpendicular line to $\ell$ at $A$ intersects $\ell$ at $G$ and $\omega$ at $H$. Line $GE$ intersects $FH$ at $I$... | [
"To prove that $C$, $F$, $I$, and $J$ are concyclic, it is equivalent to prove that $\\angle CFI = \\angle CJI$ or $\\angle CFI + \\angle CJI = 180^{\\circ}$, depending on the configuration. We will present here the proof for one configuration. The proof for the other configuration is similar.\n\n![](attached_image... | Saudi Arabia | Selection tests for the Balkan Mathematical Olympiad 2013 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03we | Prove that
$$
-1 < \left( \sum_{k=1}^{n} \frac{k}{k^2 + 1} \right) - \ln n \le \frac{1}{2}, \quad n = 1, 2, \dots
$$ | [
"We first prove that\n$$\n\\frac{x}{1+x} < \\ln(1+x) < x, \\quad x > 0. \\qquad \\textcircled{1}\n$$\nLet\n$$\nh(x) = x - \\ln(1+x), \\\\\ng(x) = \\ln(1+x) - \\frac{x}{1+x}.\n$$\nThen, for $x > 0$,\n$$\nh'(x) = 1 - \\frac{1}{1+x} > 0, \\\\\ng'(x) = \\frac{1}{1+x} - \\frac{1}{(1+x)^2} = \\frac{x}{(1+x)^2} > 0.\n$$\n... | China | China Mathematical Competition (Complementary Test) | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof only | null | |
02wl | Problem:
Um hotel possui 5 quartos distintos, todos com camas individuais para até 2 pessoas. O hotel está sem outros hóspedes e 5 amigos querem passar a noite nele. De quantos modos os 5 amigos podem escolher seus quartos? | [
"Solution:\n\nAnalisando apenas a quantidade de pessoas por quarto, sem levar em consideração a ordem, as possíveis distribuições podem ser associadas às listas:\n$$\n(1,1,1,1,1),\\ (1,1,1,2) \\text{ ou } (2,2,1)\n$$\nAnalisaremos agora lista por lista o número de maneiras de distribuir os amigos.\n\ni) Na lista $(... | Brazil | Brazilian Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Permutations",
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 2220 | |
0c9f | Problem:
a) Scrieți numărul $2021$ ca sumă de puteri distincte cu baza $(-2)$.
b) Arătați că numărul $2021$ nu se poate scrie ca sumă de puteri distincte cu baza $(-3)$. | [] | Romania | Olimpiada Națională GAZETA MATEMATICĂ | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) 2021 = (-2)^12 + (-2)^11 + (-2)^5 + (-2)^2 + (-2)^0.
b) Impossible: any sum of distinct powers of −3 is congruent to 0 or 1 modulo 3, while 2021 is congruent to 2 modulo 3. | |
0ldc | A person wants to plant two different kinds of tree on a plot tabular grid size $m \times n$ (each square planted one tree). A planting way is called impressive if two following conditions are satisfied
i) The number of trees in each kind is equal.
ii) The difference between the number of two kinds of tree in each co... | [
"For convenience, consider this problem on a $m \\times n$ table and write $+1$ or $-1$ in each square to represent the trees.\n\na) For a $4 \\times 4$ table, the table below is satisfied.\n\n| | | | |\n|---|---|---|---|\n| A | A | A | B |\n| A | A | B | A |\n| A | B | B | B |\n| B | A | B | B |\n\nIt is c... | Vietnam | VMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) Tile the grid by repeating the given four by four pattern across the entire two thousand sixteen by two thousand sixteen grid; this yields an impressive planting. b) If an impressive planting exists, both m and n must be divisible by 4. | |
0835 | Problem:
In un parallelepipedo rettangolo $P$ la lunghezza della diagonale è $\sqrt{133}$ e la superficie totale è $228$. Sapendo che uno dei lati è medio proporzionale tra gli altri due, il volume di $P$ è
(A) $64$
(B) $125$
(C) $192$
(D) $216$
(E) $343$. | [
"Solution:\n\nLa risposta è (D). Se indichiamo con $x, y$ e $z$ le lunghezze dei lati in ordine crescente, le ipotesi si traducono come\n$$\n\\left\\{\n\\begin{array}{l}\nx y + x z + y z = \\frac{228}{2} = 114 \\\\\nx^{2} + y^{2} + z^{2} = 133 \\\\\nx z = y^{2}\n\\end{array}\n\\right.\n$$\n\nDalle prime due ottenia... | Italy | Progetto Olimpiadi di Matematica 2003 | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Surface Area",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
02s3 | Problem:
Um tetraedro regular é um sólido de quatro faces, sendo todas elas triângulos equiláteros de mesmo tamanho. A figura abaixo mostra um tetraedro regular.
O comprimento de qualquer aresta de um tetraedro regular é o mesmo. Por exemplo, no tetraedro acima, $\overline{AB} = \overline{AC}... | [
"Solution:\n\nComeçamos desenhando um cubo de lado 1:\n\n\nVamos traçar duas arestas do tetraedro que queremos inscrever neste cubo: uma será $\\overline{AC}$ e outra será $\\overline{EG}$:\n\n\nEm seguida, ligamos o ponto $A$ aos pontos $E$ e $G$, e ligamos o ponto... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0ch8 | Consider $A, B \in M_n(\mathbb{R})$, and the function $f: M_n(\mathbb{C}) \to M_n(\mathbb{C})$, defined by $f(Z) = AZ + B\bar{Z}$, $Z \in M_n(\mathbb{C})$, where $\bar{Z}$ is the matrix having as entries the conjugates of the entries of $Z$. Prove that the following are equivalent:
(1) $f$ is injective;
(2) $f$ is surj... | [
"For $Z \\in M_n(\\mathbb{C})$, there are $X, Y \\in M_n(\\mathbb{R})$ such that $Z = X + iY$, and $\\bar{Z} = X - iY$. Thus $f(Z) = (A+B)X + i(A-B)Y$.\n\n(1) $\\Rightarrow$ (3). Suppose that $A+B$ or $A-B$ are singular. In case $A+B$ is singular, $\\det(A+B) = 0$, so there is $C \\in M_{n,1}(\\mathbb{R}) \\setminu... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Algebra > Linear Algebra > Linear transformations"
] | English | proof only | null | |
06ml | Let $a$, $b$, $c$, $d$ be roots of the equation $x^4 + x + 1 = 0$. Let $a^5 + 2a + 1$, $b^5 + 2b + 1$, $c^5 + 2c + 1$, $d^5 + 2d + 1$ be roots of the equation $x^4 + px^3 + qx^2 + rx + s = 0$. Find the value of $p + 2q + 4r + 8s$. | [
"The answer is $30$.\nFirstly, since $a^4 + a + 1 = 0$, we have\n$$\na^5 + 2a + 1 = a(a^4 + a + 1) - a^2 + a + 1 = -a^2 + a + 1.\n$$\nLet $y = -x^2 + x + 1$. Then $x = \\frac{1 \\pm \\sqrt{-4y+5}}{2}$. Now,\n$$\n\\begin{aligned}\n0 &= x^4 + x + 1 \\\\\n &= \\left( \\frac{3 - 2y \\pm \\sqrt{-4y+5}}{2} \\right)^2 + ... | Hong Kong | IMO HK TST | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 30 | |
049l | Determine the largest possible quotient of a three-digit number and the sum of its digits. | [] | Croatia | CroatianCompetitions2011 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 100 | |
08i5 | Problem:
The real numbers $x_{1}, x_{2}, \ldots, x_{2003}$ satisfy the relations $x_{1} / 1 = x_{2} / 2 = x_{3} / 3 = \ldots = x_{2003} / 2003$ and
$$
\sqrt{1^{2} + 2^{2} + \ldots + 2003^{2}} + \sqrt{x_{1}^{2} + x_{2}^{2} + \ldots + x_{2003}^{2}} = \sqrt{(1 + x_{1})^{2} + (2 + x_{2})^{2} + \ldots + (2003 + x_{2003})^{... | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
07jh | Let $(a_n)$ be a sequence of positive real numbers such that for all $n > 2025$ we have
$$
a_n = \max_{1 \le i \le 2025} a_{n-i} - \min_{1 \le i \le 2025} a_{n-i}
$$
Prove that there is a positive integer $M$ such that $a_n < \frac{1}{1404}$, for all $n > M$. | [
"First, we prove that the sequence is bounded; divide the sequence into blocks of $2025$ terms. It can be easily seen that the maximum of these blocks is decreasing. If the infimum of the maximums of the blocks is zero, the statement follows. Therefore, assume this infimum is $M$. Due to the decreasing nature of th... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0eif | Problem:
Dokaži, da je za vsako naravno število $n$ število $7^{2018} + 9^{2020 n}$ deljivo s $5$. | [
"Solution:\n\nŠtevilo $9^{2020 n} = 9^{2 \\cdot 1010 n} = 81^{1010 n}$ ima pri deljenju s $5$ ostanek $1$, saj je njegova zadnja števka v desetiškem zapisu enaka $1$, ne glede na to, koliko je vrednost naravnega števila $n$.\n\nŠtevilo $7^{2018} = 7^{4 \\cdot 504 + 2} = 2401^{504} \\cdot 49$ pa ima pri deljenju s $... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Odbirno tekmovanje | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null |
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