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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
03hn | Problem:
$A$, $B$, $C$, $D$ are four "consecutive" points on the circumference of a circle and $P$, $Q$, $R$, $S$ are points on the circumference which are respectively the midpoints of the $\operatorname{arcs}\ AB$, $BC$, $CD$, $DA$. Prove that $PR$ is perpendicular to $QS$. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous"
] | null | proof only | null | |
06fn | Let $\theta_1, \theta_2, \dots, \theta_{2008}$ be real numbers. Find the maximum value of
$$ \sin \theta_1 \cos \theta_2 + \sin \theta_2 \cos \theta_3 + \dots + \sin \theta_{2007} \cos \theta_{2008} + \sin \theta_{2008} \cos \theta_1. $$ | [
"The maximum value is $1004$.\nNote that $\\sin \\theta_j \\cos \\theta_{j+1} \\le \\frac{1}{2}(\\sin^2 \\theta_j + \\cos^2 \\theta_{j+1})$ for any $j$, where the indices are taken modulo $2008$. It follows that\n$$\n\\sum_{j=1}^{2008} \\sin \\theta_j \\cos \\theta_{j+1} \\le \\frac{1}{2} \\sum_{j=1}^{2008} (\\sin^... | Hong Kong | null | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1004 | |
03mv | Let $ABC$ be an acute-angled triangle with circumcenter $O$. Let $\Gamma$ be a circle with centre on the altitude from $A$ in $ABC$, passing through vertex $A$ and points $P$ and $Q$ on sides $AB$ and $AC$. Assume that $BP \cdot CQ = AP \cdot AQ$. Prove that $\Gamma$ is tangent to the circumcircle of triangle $BOC$. | [
"Let $\\omega$ be the circumcircle of $BOC$. Let $M$ be the point diametrically opposite to $O$ on $\\omega$ and let the line $AM$ intersect $\\omega$ at $M$ and $K$. Since $O$ is the circumcenter of $ABC$, it follows that $OB = OC$ and therefore that $O$ is the midpoint of the arc $\\widehat{BOC}$ of $\\omega$. Si... | Canada | Kanada 2015 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
09b9 | Яг зургаан натурал тоон хуваагчтай ба тэдгээрийн нийлбэр 2010 байх бүх натурал $n$ тоог ол. | [
"$n$ тоо яг 6 хуваагчтай гэдгээс $n = p^5$ эсвэл $n = p^2 \\cdot q$ ($p, q$-нь анхны тоо) хэлбэртэй байна.\n\na. $n = p^5$ байг. Тэгвэл\n$$\n1 + p + p^2 + p^3 + p^4 = 2010\n$$\nболно. $p = 2, p = 3$ үед 2010-аас бага, $p = 5$ үед 2010-аас их тоо гарна. Иймд $n = p^5$ байх боломжгүй.\n\nb. $n = p^2 \\cdot q$ байг. Т... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | Mongolian | proof and answer | no such natural numbers | |
0gs3 | Let $ABC$ be a triangle satisfying $|AB| = |AC|$. Let $D$ be a point on smaller arc $AC$ of circumcircle $\omega$ of $ABC$. Let $E$ be the symmetric point of $B$ with respect to the line $AD$. The line $BE$ intersects $\omega$ at $F$. The tangent line of $\omega$ at $F$ intersects the line $AC$ at $K$ and the lines $DF... | [
"Let $M$ be a point on $AD$ such that $D \\in [AM]$. Since $\\angle EDA = \\angle ADB = \\angle ACB = \\angle ABC = \\angle CDM$, we conclude that the points $C$, $D$, $E$ are collinear.\n\n\n\nUsing Pascal theorem for the points $D$, $C$, $A$, $B$, $F$, $F$, we find that the points $E = DC... | Turkey | Team Selection Test for EGMO 2019 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0gco | 令 $Q_{>0}$ 表示所有正有理數所成之集合。試求所有函數 $f: Q_{>0} \to Q_{>0}$ 滿足
$$
f(x^2 f(y)^2) = f(x)^2 f(y), \text{ 對所有 } x, y \in Q_{>0} \text{ 均成立.}
$$ | [
"$f(x) = 1$ for all $x \\in Q_{>0}$.\nTake any $a, b \\in Q_{>0}$. By substituting $x = f(a)$, $y = b$ and $x = f(b)$, $y = a$ into the assumption of the problem we get\n$$\n(f(f(a)))^2 f(b) = f(f(a)^2 f(b)^2) = f(f(b))^2 f(a),\n$$\nwhich yields\n$$\n\\frac{f(f(a))^2}{f(a)} = \\frac{f(f(b))^2}{f(b)} \\text{ for all... | Taiwan | 二〇一九數學奧林匹亞競賽第二階段選訓營, 獨立研究(二) | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | f(x) = 1 for all positive rational x | |
0i70 | Problem:
Equilateral triangle $A B C$ of side length $2$ is drawn. Three squares external to the triangle, $A B D E$, $B C F G$, and $C A H I$, are drawn. What is the area of the smallest triangle that contains these squares? | [
"Solution:\n\nThe equilateral triangle with sides lying on lines $D G$, $E H$, and $F I$ has minimal area. (The only other reasonable candidate is the triangle with sides along $D E$, $F G$, $H I$, but a quick sketch shows that it is larger.) Let $J$, $K$, and $L$ be the vertices of this triangle closest to $D$, $H... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 12 + 7√3 | |
03xd | Let $n$ be an integer greater than two, and let $A_1, A_2, \dots, A_{2n}$ be pairwise distinct nonempty subsets of $\{1, 2, \dots, n\}$.
Determine the maximum value of $\sum_{i=1}^{2n} \frac{|A_i \cap A_{i+1}|}{|A_i| \cdot |A_{i+1}|}$.
(Here, we set $A_{2n+1} = A_1$. For a set $X$, let $|X|$ denote the number of elemen... | [
"The answer is $n$.\n\nWe consider each summand $s_i = \\frac{|A_i \\cap A_{i+1}|}{|A_i| \\cdot |A_{i+1}|}$.\nIf $A_i \\cap A_{i+1}$ is the empty set, then $s_i = 0$.\nIf $A_i \\cap A_{i+1}$ is nonempty, because $A_i \\neq A_{i+1}$, at least one of $A_i$ and $A_{i+1}$ has more than one element, that is, $\\max\\{|A... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n | |
0jqd | Problem:
Kelvin the Frog is trying to hop across a river. The river has 10 lilypads on it, and he must hop on them in a specific order (the order is unknown to Kelvin). If Kelvin hops to the wrong lilypad at any point, he will be thrown back to the wrong side of the river and will have to start over. Assuming Kelvin i... | [
"Solution:\n\nKelvin needs (at most) $i(10-i)$ hops to determine the $i$th lilypad he should jump to, then an additional 11 hops to actually get across the river. Thus he requires $$\\sum_{i=1}^{10} i(10-i) + 11 = 176$$ hops to guarantee success."
] | United States | HMMT November 2015 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms"
] | null | final answer only | 176 | |
0341 | Problem:
The diagonals $AC$ and $BD$ of a cyclic quadrilateral $ABCD$ with circumcenter $I$ intersect at a point $E$. If the midpoints of segments $AD$, $BC$ and $IE$ are collinear, prove that $AB = CD$. | [
"Solution:\n\nDenote the midpoints of $AD$ and $BC$ by $M$ and $N$, respectively. Without loss of generality we may assume that $I$ lies in the interior of $AMNB$ and $E$ lies in the interior of $MDCN$. Then we have\n$$\n\\begin{aligned}\nS_{MIN} & = S_{AMNB} - S_{ABI} - S_{AMI} - S_{BNI} \\\\\n& = S_{AMNB} - S_{AB... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0jwr | Problem:
Consider the graph in 3-space of
$$
0 = x y z (x + y)(y + z)(z + x)(x - y)(y - z)(z - x)
$$
This graph divides 3-space into $N$ connected regions. What is $N$? | [
"Solution:\n\nNote that reflecting for each choice of sign for $x$, $y$, $z$, we get new regions. Therefore, we can restrict to the case where $x, y, z > 0$. In this case, the sign of the expression only depends on $(x - y)(y - z)(z - x)$. It is easy to see that for this expression, every one of the $3! = 6$ orderi... | United States | February 2017 | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 48 | |
07p6 | Let $k \ge 3$ be an integer and let functions $f_0, f_1, \dots, f_k$ be defined for positive integers such that
$$
\begin{align*}
f_j(1) &= 1 && \text{for } 0 \le j \le k, \\
f_0(n+1) &= \sum_{i=0}^{k} f_i(n) && \text{for } n \ge 1 \text{ and} \\
f_{j+1}(n+1) &= f_j(n+1) + f_j(n) && \text{for } 0 \le j < k \text{ and }... | [
"With $m = n + k + 1$ and $j = k + 1 - i$, we obtain for $n \\ge 1$\n$$\nf_k(n + k + 1) = \\sum_{i=0}^{k} \\binom{k+1}{i} f_k(n + i).\n$$\nThis shows that $a_i = \\binom{k+1}{i}$ are the integers that solve the problem.",
"Define $r = \\sqrt[k+1]{2}$, so that $r^{k+1} = 2$ and let\n$$\nF = 1 + r + r^2 + \\dots + ... | Ireland | Irska 2014 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | a_i = \binom{k+1}{i} for 0 \le i \le k | |
0h0h | Find all functions such that $f: \mathbb{Z} \to \mathbb{Z}$ that satisfy the following two conditions:
1) $f(x + f(x + 2y)) = f(2x) + f(2y)$ for all integers $x, y$;
2) $f(0) = 2$. | [
"**Answer:** $f(k) = k + 2$.\n\nSubstitute $x = 0$ and $y = 0 \\Rightarrow$\n$$\nf(f(2y)) = f(2y) + 2, \\qquad (1)\n$$\n$$\nf(x + f(x)) = f(2x) + 2, \\qquad (2)\n$$\nLet us prove by induction, that $f(2n) = 2n + 2$. Base $n = 0$ is already proved. Let us suppose that we have $f(2n) = 2n + 2$ and prove that $f(2n + ... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010) | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | f(k) = k + 2 | |
05ep | Problem:
Soit $ABCD$ un quadrilatère et $M, N, P, Q$ les milieux respectifs de $[AB]$, $[BC]$, $[CD]$, $[DA]$.
Montrer que le quadrilatère $MNPQ$ est un parallélogramme.
 | [
"Solution:\n\nPuisque les points $M$ et $N$ sont les milieux respectifs des côtés $[AB]$ et $[BC]$, les droites $(MN)$ et $(AC)$ sont parallèles.\n\nPuisque les points $P$ et $Q$ sont les milieux respectifs des côtés $[CD]$ et $[DA]$, les droites $(PQ)$ et $(AC)$ sont parallèles.\n\nDonc les droites $(PQ)$ et $(MN)... | France | ENVOI 1 : GÉOMÉTRIE Corrigé | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
01v2 | Find all pairs of positive integers $(m, n)$ such that
$$
9^m - 7^m = 2^n.
$$ | [
"**Answer:** $(m, n) = (1, 1)$, $(m, n) = (2, 5)$.\nIt is easy to see that the pair $(m, n) = (1, 1)$ satisfies the condition. For $m > 1$ consider the equation\n$$\na^m - b^m = (a - b)(a^{m-1} + a^{m-2}b + a^{m-3}b^2 + \\dots + a^{m-1}b^m).\n$$\n\nFor $a = 9$, $b = 7$, the first factor is $2$ and the second is gre... | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (m, n) = (1, 1) and (2, 5) | |
0aw3 | Problem:
Let $n$ be a positive integer greater than $1$. If $2n$ is divided by $3$, the remainder is $2$. If $3n$ is divided by $4$, the remainder is $3$. If $4n$ is divided by $5$, the remainder is $4$. If $5n$ is divided by $6$, the remainder is $5$. What is the least possible value of $n$? | [
"Solution:\n\nWe have $2n \\equiv 2 \\pmod{3}$, $3n \\equiv 3 \\pmod{4}$, $4n \\equiv 4 \\pmod{5}$, and $5n \\equiv 5 \\pmod{6}$. This is equivalent to saying that $n \\equiv 1 \\pmod{3,4,5,6}$. The smallest such $n$ is one more than the $\\mathrm{LCM}$ of $3,4,5,$ and $6$, which is $61$."
] | Philippines | 18th PMO National Stage Oral Phase | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | final answer only | 61 | |
0648 | Problem:
Entscheiden Sie, ob es eine Menge $M$ positiver ganzer Zahlen gibt, die folgende Eigenschaft hat: Für jede positive rationale Zahl $r<1$ existiert genau eine endliche Teilmenge $S$ von $M$, sodass $\sum_{s \in S} 1 / s = r$ gilt, das heißt die Summe der Kehrwerte aller Elemente von $S$ ist gleich $r$. | [
"Solution:\n\nEine solche Menge existiert nicht, was wir mithilfe eines Widerspruchsbeweises zeigen werden. Angenommen, $S$ hat die angegebene Eigenschaft. Offenbar ist $M$ dann unendlich, und wir dürfen ohne Beschränkung der Allgemeinheit annehmen, dass $1 \\notin M$. Wir bezeichnen die Elemente von $M$ der Größe ... | Germany | 2. Auswahlklausur | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | No, such a set does not exist. | |
0h21 | Let $P(x)$ be a polynomial with integer coefficients. Given that for some integer $a$ there exists $n$ such that $\underbrace{P(P(...P(a)...))}_{n} = a$, prove that $P(P(a)) = a$. | [
"Define a sequence $(a_i)$ in the following way: $a_0 = a$, $a_i = P(a_{i-1})$ for all natural $i$. Consider minimal number $m$, for which there exists $j < m$, such that $a_j = a_m$. From the condition of the problem it follows that this number does exist. Observe, that $a_0, a_1, ..., a_{m-1}$ are distinct.\nWe s... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
01uj | Given a convex $2n$-gon $H$ with pairwise parallel opposite sides.
a) Prove that there exists a pair of the opposite sides of $H$ which possesses the following property: there exists a straight line that is perpendicular to these sides and intersects each of them.
b) Are there any values of $n$ such that for a convex... | [
"b) there are no such $n$.\n\na) A diagonal of the convex $2n$-gon is said to be main if there are $n-1$ vertices of this polygon between the vertices connected by this diagonal. Suppose that there exist a convex $2n$-gon $A_0A_1...A_{2n-1}$ such that its opposite sides are parallel ($A_iA_{i+1} \\parallel A_{i+n}A... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof and answer | a) Such a pair always exists. b) No values of n. | |
0818 | Problem:
Siano $A$ e $B$ due punti nello spazio. L'insieme dei punti dello spazio la cui distanza da $B$ è doppia di quella da $A$ è
(A) un piano
(B) una circonferenza
(C) una superficie sferica
(D) la superficie laterale di un cono
(E) la superficie laterale di un cilindro. | [] | Italy | Progetto Olimpiadi di Matematica 2000 GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Solid Geometry > Other 3D problems"
] | null | MCQ | C | |
0b8o | Let $ABC$ be a triangle with $AB = AC$ and $\angle BAC = 40^\circ$. The points $S$ and $T$ lie on the sides $AB$ and $BC$, respectively, such that $\angle BAT = \angle BCS = 10^\circ$. The straight lines $AT$ and $CS$ meet at point $P$.
Show that $BT = 2PT$. | [
"Triangle $ABC$ is isosceles, so $\\angle ABC = \\angle ACB = 70^\\circ$.\n\nNotice that $\\angle TAC = 40^\\circ - 10^\\circ = 30^\\circ$ and $\\angle ACS = 70^\\circ - 10^\\circ = 60^\\circ$, hence $\\angle APC = 90^\\circ$.\n\nTriangles $ABT$ and $BSC$ are similar, whence $\\frac{BS}{BC} = \\frac{BT}{AB}$.\n\nAl... | Romania | Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
09z5 | For real numbers $x$ and $y$ we define $M(x, y)$ to be the maximum of the three numbers $xy$, $(x-1)(y-1)$, and $x+y-2xy$. Determine the smallest possible value of $M(x, y)$ where $x$ and $y$ range over all real numbers satisfying $0 \le x, y \le 1$. | [
"We will show that the minimum value is $\\frac{4}{9}$. This value can be attained by taking $x = y = \\frac{2}{3}$. Then we have $xy = \\frac{4}{9}$, $(x-1)(y-1) = \\frac{1}{9}$, and $x+y-2xy = \\frac{4}{9}$, and the maximum is indeed $\\frac{4}{9}$.\n\nNow we will prove that $M(x, y) \\ge \\frac{4}{9}$ for all $x... | Netherlands | IMO Team Selection Test 1 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 4/9 | |
033d | Problem:
In a group of $n$ tourists, among every three of them there are at least two that are not familiar. For any partition of the group into two groups, there are at least two familiar tourists in some of the groups. Prove that there is a tourist who is familiar with at most $\frac{2n}{5}$ tourists. | [
"Solution:\n\nConsider a graph $G$ with $n$ vertices corresponding to the tourists and two vertices connected when they are familiar.\n\nThe first condition of the problem means that there is no triangle in $G$.\n\nThe second condition of the problem means that there is a cycle of odd length in the graph. Indeed, i... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0alf | Problem:
The 6-digit number $739ABC$ is divisible by $7$, $8$, and $9$. What values can $A$, $B$, and $C$ take? | [
"Solution:\n\nSince $\\gcd(7,8,9) = 1$, then $739ABC$ is divisible by $7$, $8$, and $9$ if and only if it is divisible by $7 \\cdot 8 \\cdot 9 = 504$. Note that the only integers between $739000$ and $739999$ which are divisible by $504$ are $739368$ and $739872$. So, $(A, B, C) \\in \\{(3,6,8), (8,7,2)\\}$."
] | Philippines | 18th PMO Area Stage | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | A, B, C can be (3,6,8) or (8,7,2). | |
03hj | Problem:
Simplify
$$
\left(\frac{1 \cdot 2 \cdot 4 + 2 \cdot 4 \cdot 8 + \cdots + n \cdot 2n \cdot 4n}{1 \cdot 3 \cdot 9 + 2 \cdot 6 \cdot 18 + \cdots + n \cdot 3n \cdot 9n}\right)^{1/3}
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 2/3 | |
0bdl | Let $n$ be a positive integer and let $x_1, \dots, x_n$ be positive real numbers. Show that
$$
\begin{aligned}
\min (x_1, 1/x_1 + x_2, \dots, 1/x_{n-1} + x_n, 1/x_n) &\le 2 \cos \left(\frac{\pi}{n+2}\right) \\
&\le \max (x_1, 1/x_1 + x_2, \dots, 1/x_{n-1} + x_n, 1/x_n).
\end{aligned}
$$ | [
"Only the first inequality will be proved; the second is dealt with similarly. Suppose, if possible, that each of the $n+1$ positive real numbers $x_1, 1/x_1 + x_2, \\dots, 1/x_{n-1} + x_n, 1/x_n$ is greater than $2 \\cos \\alpha$, where $\\alpha = \\pi/(n+2)$, and show recursively that $x_k > \\sin((k+1)\\alpha) /... | Romania | 64th NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials"
] | null | proof only | null | |
055f | Find the smallest real constant $C$ such that for any positive real numbers $a_1, a_2, a_3, a_4$ and $a_5$ (not necessarily distinct), one can always choose distinct subscripts $i, j, k$ and $l$ such that $\left|\frac{a_i}{a_j} - \frac{a_k}{a_l}\right| \le C$. | [
"See IMO 2016 shortlist, problem A2."
] | Estonia | IMO Team Selection Contest I | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 1/2 | |
0e7z | Let $D$ be the midpoint of the side $BC$, $E$ the midpoint of the side $CA$ and $T$ the center of mass of a triangle $ABC$. Let the lines $AT$, $BT$ and $CT$ also intersect the circumcircle of the triangle $ABC$ at the points $P$, $Q$ and $R$, respectively. Suppose $\angle ACB = \angle RQP$. Prove that $DCET$ is a cycl... | [
"The quadrilateral $RBCQ$ is cyclic, so $\\angle RQB = \\angle RCB$, and the assumptions of the problem imply $\\angle ECT = \\angle ACR = \\angle BQP$. The quadrilateral $QABP$ is also cyclic, so $\\angle BQP = \\angle BAP$. Since $E$ and $D$ are the midpoints of the two sides, the lines $AB$ and $DE$ are parallel... | Slovenia | National Math Olympiad 2013 - Final Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06lm | In $\triangle ABC$, let $D$ be a point on side $BC$. Suppose the incircle $\omega_1$ of $\triangle ABD$ touches sides $AB$ and $AD$ at $E, F$ respectively, and the incircle $\omega_2$ of $\triangle ACD$ touches sides $AD$ and $AC$ at $F, G$ respectively. Suppose the segment $EG$ intersects $\omega_1$ and $\omega_2$ aga... | [
"Let $O$ be the circumcentre of $\\triangle FPQ$. First of all,\n$$\n\\angle OFP = 90^\\circ - \\angle PQF = \\angle FQG - 90^\\circ = \\frac{\\angle FAG}{2} = \\angle GEF = \\angle AFP\n$$\nwhere the second last equality follows from the fact that $AE = AF = AG$. This shows that $O$ lies on $AD$.\n\nNext, Let $I_1... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | null | proof only | null | |
0dap | A set $S$ is called neighboring if it has the following two properties:
i) $S$ has exactly 4 elements,
ii) for every element $x \in S$ at least one of the $x-1$ or $x+1$ belongs to $S$.
Find the number of all neighboring subsets of the set $\{1,2, \ldots, n\}$. | [
"Let $x$ be the smallest index and $y$ be the largest index of the neighboring set. Then our set consists of elements\n$$\na_{x}, a_{x+1}, a_{y-1}, a_{y}\n$$\nSo the number of neighboring sets is equal to the number of pairs $(x, y)$ where $y-x \\geq 3$. The number of such sets is equal to\n$$\n1+2+\\cdots+(n-3)=\\... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | ((n-3)(n-2))/2 | |
0hrl | Problem:
Suppose that $b$ and $c$ are real numbers such that the equation
$$
x^{2}+b x+c=0
$$
has two different solutions $x_{1}, x_{2}$. Suppose that
a. The (positive) difference between $x_{1}$ and $x_{2}$ is $1$;
b. The (positive) difference between $b$ and $c$ is also $1$.
Find all possible values of $b$ and $c... | [
"Solution:\n\nGiven the two solutions $x_{1}$ and $x_{2}$, we know that the equation factors:\n$$\nx^{2}+b x+c=\\left(x-x_{1}\\right)\\left(x-x_{2}\\right)=0.\n$$\nFrom this we derive that\n$$\nb=-x_{1}-x_{2} \\text{ and } c=x_{1} x_{2}.\n$$\nIn condition (a), we switch the labels on $x_{1}$ and $x_{2}$, if necessa... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (b, c) = (-1, 0), (5, 6), (1, 0), (3, 2) | |
06ao | Let $AΒΓ$ be a triangle and $M$, $N$ the midpoints of $AB$, $AΓ$, respectively. The points $Δ$ and $E$ lie on the segment $BN$, such that $ΓΔ \parallel ME$ and $BΔ < BE$. Prove that:
$$
BΔ = 2 \cdot EN.
$$ | [
"Since $ME \\parallel ΓΔ$ it follows that $ΓΔE = M\\hat{E}Δ$, and so\n$$\n180^\\circ - ΓΔE = 180^\\circ - M\\hat{E}Δ \\Rightarrow BΔΓ = M\\hat{E}N \\quad (1)\n$$\nSince the points $M$, $N$ are the midpoints of the sides $AB$, $AΓ$, respectively, we conclude that\n$$\nMN \\parallel BΓ, \\quad MN = \\frac{BΓ}{2}. \\q... | Greece | 40th Hellenic Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
0cwl | The sport competition consists of 25 contests; in each contest there is exactly one winner who receives a gold medal. 25 athletes participate in this competition, each of them participates in all 25 contests. There are 25 sports experts. Each of 25 experts is to make his *prediction* how many gold medals each athlete w... | [
"**Ответ.** 24.\n\n**Решение.** *Upper bound.* We will show that $k \\le 24$, i.e., that any expert could be incompetent. If this expert believes that all athletes will receive one medal each, we can refute them with the result $(25, 0, 0, ..., 0)$. Otherwise, the expert believes that several (at least one) athlete... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Russian | proof and answer | 24 | |
0l6r | Problem:
Let $\triangle ABC$ be an equilateral triangle with side length $6$. Let $P$ be a point inside triangle $\triangle ABC$ such that $\angle BPC = 120^\circ$. The circle with diameter $\overline{AP}$ meets the circumcircle of $\triangle ABC$ again at $X \neq A$. Given that $AX = 5$, compute $XP$. | [
"Solution:\n\nLet $A'$ be the antipode of $A$. As $\\angle AXA' = 90^\\circ$, we have $X$, $P$, and $A'$ are collinear. As $\\angle BPC = 120^\\circ$ and $\\angle BA'C = 180^\\circ - \\angle BAC = 120^\\circ$, it follows that $P$ lies on the circle with center $A'$ passing through $B$ and $... | United States | HMMT February 2025 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | sqrt(23) - 2*sqrt(3) | |
0ki3 | Distinct lines $\ell$ and $m$ lie in the $xy$-plane. They intersect at the origin. Point $P(-1, 4)$ is reflected about line $\ell$ to point $P'$, and then $P'$ is reflected about line $m$ to point $P''$. The equation of line $\ell$ is $5x - y = 0$, and the coordinates of $P''$ are (4, 1). What is the equation of line $... | [
"The reflection through $\\ell$ followed by the reflection through $m$ is equivalent to a rotation about the intersection of the two lines by twice the angle formed by the two lines. As the result of the two reflections in the present case, $P$ was rotated by $90^\\circ$ clockwise around the origin to $P''(4, 1)$. ... | United States | Fall 2021 AMC 10 B | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | MCQ | D | |
0djv | Let $ABC$ be an acute triangle with $AB < AC$. Let $I$ be its incenter and $\Gamma$ be its circumcircle. Let $M$ be the midpoint of $BC$, $K$ the midpoint of arc $BC$ not containing $A$, $L$ the midpoint of arc $BC$ containing $A$ and $J$ the reflection of $I$ with respect to the line $KL$. The line $LJ$ cuts $\Gamma$ ... | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0cwy | Petya chose 100 pairwise distinct positive real numbers, each less than $1$, and arranged these numbers on a circle. Then he performs the following moves. By one move he takes some three consecutive numbers $a, b, c$ (in this order) and replaces the middle number $b$ by $a - b + c$. Find the greatest possible $k$ such ... | [
"Estimate. We will show that the number of integers never exceeds $50$.\nWe will track the differences between each number and the next one in clockwise order. If three consecutive numbers were $a$, $b$, and $c$, their differences were $a-b$ and $b-c$. After applying the operation to $b$, the numbers become $a$, $a... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Russian | proof and answer | 50 | |
0bhz | a) Give an example of matrices $A$ and $B$ from $M_2(\mathbb{R})$, such that
$$
A^2 + B^2 = \begin{pmatrix} 2 & 3 \\ 3 & 2 \end{pmatrix}.
$$
b) Let $A$ and $B$ be matrices from $M_2(\mathbb{R})$, such that $A^2 + B^2 = \begin{pmatrix} 2 & 3 \\ 3 & 2 \end{pmatrix}$. Prove that $AB \neq BA$. | [
"a) An example is $A = \\frac{\\sqrt{3}}{2} \\begin{pmatrix} 1 & 1 \\\\ 1 & 1 \\end{pmatrix}$ and $B = \\begin{pmatrix} 0 & 1 \\\\ -1 & 0 \\end{pmatrix}$.\n\nb) If the matrices $A$ and $B$ commute, then $A^2 + B^2 = (A + iB)(A - iB)$, hence\n$$\n\\begin{aligned}\n|\\det(A + iB)|^2 &= \\det(A + iB) \\det(A - iB) = \... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | One example is A = (sqrt(3)/2) * [[1, 1], [1, 1]] and B = [[0, 1], [-1, 0]]. Moreover, for any A and B with A^2 + B^2 = [[2, 3], [3, 2]], we must have AB ≠ BA. | |
071q | Problem:
Show that the equation
$$
x^{2}+y^{2}+z^{2}=(x-y)(y-z)(z-x)
$$
has infinitely many solutions in integers $x, y, z$. | [
"Solution:\nWe seek solutions $(x, y, z)$ which are in arithmetic progression. Let us put $y-x=z-y=d>0$ so that the equation reduces to the form\n$$\n3 y^{2}+2 d^{2}=2 d^{3}\n$$\nThus we get $3 y^{2}=2(d-1) d^{2}$. We conclude that $2(d-1)$ is 3 times a square. This is satisfied if $d-1=6 n^{2}$ for some $n$. Thus ... | India | INMO | [
"Number Theory > Diophantine Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0hlp | Problem:
Suppose $a, b, c$ are positive integers such that
$$
\begin{aligned}
& b=a^{2}-a \\
& c=b^{2}-b \\
& a=c^{2}-c
\end{aligned}
$$
Prove that $a=b=c=2$. | [
"Solution:\nIf $a=1$, we get $b=0$ which is impossible. So it is enough to show that $a$ cannot be greater than $2$. If $a>2$, we have\n$$\nb=a^{2}-a=a(a-1)>a(2-1)=a.\n$$\nSo $b>a$; in particular $b>2$, so applying the same logic to the second equation we get $c>b$. Lastly, we have $c>2$ so applying the same logic ... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a = b = c = 2 | |
052j | Prove that for any integer $n \ge 3$ we have $(2n)! < n^{2n}$. | [
"For $n = 3$ the claim holds: $(2n)! = 6! = 720$ and $n^{2n} = 3^6 = 729$.\n\nSuppose $n \\ge 4$. Divide the numbers $2, 3, \\dots, 2n-2$ into pairs $(k, 2n-k)$ with $2 \\le k \\le n-1$, leaving $n$ alone. For each pair we have\n$$\nk(2n-k) = (n - (n-k))(n + (n-k)) = n^2 - (n-k)^2 < n^2.\n$$\nHence $2 \\cdot 3 \\cd... | Estonia | Open Contests | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0d0l | The median $AM$, $M \in BC$, of the acute-angled triangle $ABC$ intersects the circumcircle of triangle at $D$. Let $E$ be the symmetric point of $A$ with respect to $M$. Prove that $BC$ is the common tangent of the circumcircles of triangles $BDE$ and $CDE$. | [
"\n\n$ABEC$ is a parallelogram. We have $\\overline{CBD} \\equiv \\overline{CAD} \\equiv \\overline{AEB}$, so $BC$ is tangent to the circumcircle of triangle $BDE$. Similarly, from $\\overline{BCD} \\equiv \\overline{BAD} \\equiv \\overline{AEC}$, it follows that $BC$ is tangent to the circ... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0fej | Problem:
Dada una semicircunferencia de diámetro $A B=2 R$, se considera una cuerda $C D$ de longitud fija $c$. Sea $E$ la intersección de $A C$ con $B D$ y $F$ la intersección de $A D$ con $B C$.
Probar que el segmento $E F$ tiene longitud constante y dirección constante al variar la cuerda $C D$ sobre la semicircunf... | [
"Solution:\n\nComo los triángulos $E F C$ y $E D F$ son rectángulos, el cuadrilátero $E D F C$ es inscriptible y $E F$ es el diámetro. Llamemos $r$ al radio del circuncírculo de $E C D$, por el teorema de los senos en $E C D$:\n$$\nE F = 2 r = \\frac{c}{\\operatorname{sen} E}\n$$\nPongamos $\\alpha = \\angle B O D$... | Spain | null | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscella... | null | proof only | null | |
0cuk | In a product of 7 positive integers each multiple was decreased by 3. Could it happen that the product becomes exactly 13 times larger than the initial one?
В произведении 7 положительных целых чисел каждый сомножитель уменьшили на 3. Могло ли случиться так, что произведение стало ровно в 13 раз больше исходного? | [
"Yes.\nAn example is $1 \\cdot 1 \\cdot 1 \\cdot 1 \\cdot 2 \\cdot 16 = 32$ (there are other examples).\n\nAfter decreasing each factor by $3$, the numbers become $-2, -2, -2, -2, -1, 13$. Their product is $(-2)^4 \\cdot (-1) \\cdot 13 = 16 \\cdot (-1) \\cdot 13 = -208$.\n\nBut the Russian solution gives a more pre... | Russia | XLIII Russian mathematical olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English; Russian | proof and answer | Yes; for example, 1, 1, 1, 1, 1, 2, 16. | |
0fkf | Problem:
Prueba que para cualesquiera números reales $a, b$ tales que $0<a, b<1$, se cumple la desigualdad siguiente:
$$
\sqrt{a b^{2}+a^{2} b}+\sqrt{(1-a)(1-b)^{2}+(1-a)^{2}(1-b)}<\sqrt{2}
$$ | [
"Solution:\n\nSe verifica que $\\sqrt{x}<\\sqrt[3]{x}$ para todo $x \\in(0,1)$. Teniendo en cuenta que $0<\\frac{a+b}{2}<1$, utilizando la desigualdad anterior y aplicando la desigualdad entre las medias aritmética y geométrica, se tiene:\n$$\n\\sqrt{a b\\left(\\frac{a+b}{2}\\right)}<\\sqrt[3]{a b\\left(\\frac{a+b}... | Spain | XLIV Olimpiada Matemática Española | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
08a1 | Problem:
Sia $ABCD$ un quadrato all'interno del quale vengono tracciati due segmenti che dividono l'angolo in $A$ in tre angoli uguali e il quadrato in due triangoli uguali e un quadrilatero. Qual è il rapporto tra l'area del quadrilatero e quella di uno dei due triangoli?
(A) $\sqrt{3}-\frac{1}{\sqrt{3}}$
(B) $2 \sq... | [
"Solution:\n\nLa risposta è (E). Siano $AT$ e $AS$ i due segmenti tracciati, con $T$ sul lato $BC$, $S$ sul lato $CD$; l'angolo in $A$ è retto e lo abbiamo diviso in tre angoli uguali, dunque $B\\widehat{A}T = S\\widehat{A}D = 30^\\circ$. I triangoli rettangoli $ABT$ e $ASD$ sono perciò due metà di un triangolo equ... | Italy | Progetto Olimpiadi della Matematica | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | E | |
0b9p | Find all real numbers $x$, $y$, $z$, $t \in [0, \infty)$ so that
$$
x + y + z \le t, \quad x^2 + y^2 + z^2 \ge t \text{ and } x^3 + y^3 + z^3 \le t.
$$ | [
"Adding $x + y + z \\le t$, $-2x^2 - 2y^2 - 2z^2 \\le -2t$ and $x^3 + y^3 + z^3 \\le t$, one gets $x(1-x)^2 + y(1-y)^2 + z(1-z)^2 \\le 0$.\nSince $x$, $y$, $z \\in [0, \\infty)$, it follows $x$, $y$, $z \\in \\{0, 1\\}$.\nIf $x = y = z = 0$, then $t = 0$.\nIf exactly two of the numbers $x$, $y$, $z$ are nil, then $... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (x,y,z) ∈ {0,1}^3 and t = x + y + z; explicitly, (x,y,z,t) ∈ {(0,0,0,0), (1,0,0,1), (0,1,0,1), (0,0,1,1), (1,1,0,2), (1,0,1,2), (0,1,1,2), (1,1,1,3)} | |
01m4 | Point $M$ is the midpoint of the side $BC$ of an acute-angled triangle $ABC$, $N$ and $K$ are the feet of its altitudes $AN$ and $CK$, $H$ is the orthocenter of $ABC$. The bisector of the angle $ACB$ meets the segment $AH$ at point $T$.
Find the length of the circumradius of the triangle $NBK$ if $CT \parallel MH$, $TH... | [] | Belarus | 61st Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof and answer | 5 | |
0jay | Problem:
Let $Q$ be the product of the sizes of all the non-empty subsets of $\{1,2, \ldots, 2012\}$, and let $M= \log_{2}\left(\log_{2}(Q)\right)$. Give lower and upper bounds $L$ and $U$ for $M$. If $0<L \leq M \leq U$, then your score will be $\min \left(23,\left\lfloor\frac{23}{3(U-L)}\right\rfloor\right)$. Otherwi... | [
"Solution:\nAnswer: 2015.318180... In this solution, all logarithms will be taken in base 2. It is clear that $\\log (Q)=\\sum_{k=1}^{2012}\\binom{2012}{k} \\log (k)$. By paring $k$ with $2012-k$, we get $\\sum_{k=1}^{2011} 0.5 * \\log (k(2012-k))\\binom{2012}{k}+\\log (2012)$, which is between $0.5 * \\log (2012) ... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 2014 ≤ M ≤ 2016 | |
00pm | Let $ABC$ be a triangle with circumcircle $c$ and circumcenter $O$, and let $D$ be a point on the side $BC$ different from the vertices and the midpoint of $BC$. Let $K$ be the point where the circumcircle $c_1$ of the triangle $BOD$ intersects $c$ for the second time and let $Z$ be the point where $c_1$ meets the line... | [
"Since the quadrilateral $ODCE$ is cyclic, we have $\\angle O_1 = \\angle C$. (The angles are as shown in the figure.)\n\nSince the quadrilateral $ODBZ$ is cyclic, we have $\\angle O_2 = \\angle B$. Adding these we obtain $\\angle EOUZ = \\angle B + \\angle C = 180^\\circ - \\angle A$. Ther... | Balkan Mathematical Olympiad | Balkan 2012 shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous >... | English | proof only | null | |
0e9z | Let $m$ and $n$ be positive integers such that $5m + n$ divides $5n + m$. Prove that $m$ divides $n$. | [
"There exists a positive integer $k$ such that $5n + m = k(5m + n)$, or $(5 - k)n = (5k - 1)m$. The right-hand side is strictly positive, so the left-hand side must be strictly positive as well, which implies $5 - k > 0$ and so $k \\in \\{1, 2, 3, 4\\}$. If $k = 1$, then $4n = 4m$, so $n = m$. If $k = 2$, then $3n ... | Slovenia | National Math Olympiad in Slovenia | [
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
0bht | Two circles $\gamma_1$ and $\gamma_2$ meet at two points; let $A$ be one of these points. The tangent to $\gamma_1$ at $A$ meets again $\gamma_2$ at $B$, the tangent to $\gamma_2$ at $A$ meets again $\gamma_1$ at $C$, and the line $BC$ meets again $\gamma_1$ and $\gamma_2$ at $D_1$ and $D_2$, respectively. Let $E_1$ an... | [
"To begin, apply Menelaus' theorem to triangles $ABD_2$, $ACD_1$, $ABC$, to write\n$$\n\\frac{NB}{NA} \\cdot \\frac{CD_2}{CB} \\cdot \\frac{E_2A}{E_2D_2} = 1, \\quad \\frac{MA}{MC} \\cdot \\frac{E_1D_1}{E_1A} \\cdot \\frac{BC}{BD_1} = 1, \\quad \\frac{MC}{MA} \\cdot \\frac{NA}{NB} \\cdot \\frac{PB}{PC} = 1,\n$$\nso... | Romania | THE 2014 DANUBE MATHEMATICAL COMPETITION | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cwj | Let $n$ be a positive integer. Positive integers $1, 2, \dots, n$ are written in a row in some order. For any two neighboring numbers their GCD is written on the paper. Find the greatest possible number of distinct numbers among all $n-1$ numbers written on the paper. | [
"**Ответ.** $\\lfloor n/2 \\rfloor$.\n\n**Решение.** *Upper bound.* Assume one of the written numbers is greater than $\\lfloor n/2 \\rfloor$, say, $\\text{GCD}(a, b) = d > \\lfloor n/2 \\rfloor$. Then the larger of the numbers $a, b$ must be at least $2d$, which exceeds $n$ - a contradiction. Therefore, each writt... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | Russian | proof and answer | ⌊n/2⌋ | |
02hm | Problem:
Qual é o maior fator primo de $2006$? | [
"Solution:\n\nA decomposição de $2006$ em fatores primos é $2006 = 2 \\times 17 \\times 59$. Logo, o maior fator primo de $2006$ é $59$."
] | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 59 | |
0jm4 | Problem:
Let $ABCD$ be a quadrilateral inscribed in a circle with diameter $\overline{AD}$. If $AB=5$, $AC=6$, and $BD=7$, find $CD$. | [
"Solution:\n\n$AD^{2} = AB^{2} + BD^{2} = AC^{2} + CD^{2}$, so\n$$\nCD = \\sqrt{AB^{2} + BD^{2} - AC^{2}} = \\sqrt{38}.\n$$"
] | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | sqrt(38) | |
012t | Problem:
A lattice point in the plane is a point whose coordinates are both integral. The centroid of four points $\left(x_{i}, y_{i}\right), i=1,2,3,4$, is the point $\left(\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}, \frac{y_{1}+y_{2}+y_{3}+y_{4}}{4}\right)$. Let $n$ be the largest natural number with the following property: T... | [
"Solution:\nTo prove $n \\geq 12$, we have to show that there are 12 lattice points $\\left(x_{i}, y_{i}\\right)$, $i=1,2, \\ldots, 12$, such that no four determine a lattice point centroid. This is guaranteed if we just choose the points such that $x_{i} \\equiv 0(\\bmod 4)$ for $i=1, \\ldots, 6$, $x_{i} \\equiv 1... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 12 | |
071i | Problem:
The sequences $a_0, a_1, a_2, \ldots$ and $b_0, b_1, b_2, \ldots$ are defined by $a_0 = 1$, $b_0 = 4$, $a_{n+1} = a_n^{2001} + b_n$, $b_{n+1} = b_n^{2001} + a_n$. Show that no member of either sequence is divisible by $2003$. | [
"Solution:\n\n$2003$ is prime, so $a^{2002} = 1 \\bmod 2003$ for any $a$ not divisible by $2003$. Thus $a_{n+1} = a_n^{-1} + b_n \\bmod 2003$, $b_{n+1} = b_n^{-1} + a_n \\bmod 2003$. Put $c_n = a_n b_n$. Then $c_{n+1} = c_n + 1 / c_n + 2 = (c_n + 1)^2 / c_n \\bmod 2003$. So if $c_n \\neq 0 \\bmod 2003$, then $c_{n+... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
04v1 | Let $ABC$ be a triangle with $AB < AC$ and circumcenter $O$. The angle bisector of $\angle BAC$ meets the side $BC$ at $D$. The line through $D$ perpendicular to $BC$ meets the segment $AO$ at $X$. Furthermore, let $Y$ be the midpoint of segment $AD$. Prove that points $B, C, X, Y$ lie on a single circle. (Karl Czakler... | [] | Czech Republic | Czech-Polish-Slovak Match | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06bg | Let a sequence of real numbers $a_0, a_1, a_2, \dots$ satisfies the condition
$$
\sum_{n=0}^{m} a_n \cdot (-1)^n \cdot \binom{m}{n} = 0
$$
for all sufficiently large values of $m$. Show that there exists a polynomial $P$ such that $a_n = P(n)$ for all $n \ge 0$. | [
"For any function $f(x)$ and any integer $m \\ge 0$, define\n$$\n\\Delta_m f(x) = \\sum_{n=0}^{m} f(x+n) (-1)^n \\binom{m}{n}.\n$$\n**Claim.** Let $Q(x)$ be a polynomial of degree $d$. Then $\\Delta_m Q(x) = 0$ for all $m \\ge d$.\n*Proof.* This is a standard result about finite differences of polynomials. Indeed, ... | Hong Kong | 1997-2023 IMO HK TST | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
01g4 | Chessboard has a form of $n \times n$ torus, where $n = 2^{2k} + 1$. Prove that $n$ queens in the cells $(i, 2^k \cdot i)$ ($1 \le i \le n$, multiplication modulo $n$) do not attack each other.
 | [
"It is clear that each vertical line contains 1 queen. Since $2^k$ is invertible modulo $2^{2k} + 1$, each horizontal line contains 1 queen. Each diagonal line also contains 1 queen because $2^k \\pm 1$ is invertible modulo $2^{2k} + 1$."
] | Baltic Way | Baltic Way 2019 | [
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof only | null | |
0e8o | If at the cinema three box offices are open, the visitors have to wait 15 min to buy a ticket. By how many minutes is the waiting time reduced if two more box offices are opened?
(A) 3
(B) 5
(C) 6
(D) 7
(E) 10 | [
"With 3 box offices open the waiting time is 15 minutes. If 1 box office is open the waiting time is 3 times longer, i.e. 45 minutes. If 5 box offices are open the waiting time is $\\frac{45}{5} = 9$ minutes. So, if two additional box offices are opened the waiting time is reduced by $15 - 9 = 6$ minutes. The corre... | Slovenia | National Math Olympiad 2013 - First Round | [
"Math Word Problems"
] | null | MCQ | C | |
02p8 | For each positive integer $N$ with $2k$ digits, let $\text{odd}(N)$ be the $k$-digit number obtained by writing the digits of odd order of $N$ and $\text{even}(N)$ be the $k$-digit number obtained by writing the digits of even order of $N$. For example, $\text{odd}(249035) = 405$ and $\text{even}(249035) = 293$. Prove ... | [
"We will prove by induction that $\\text{odd}(N) \\cdot \\text{even}(N) < N$ for all positive integers $N$ with $2k$ digits.\n\nIf $N = 10a + b$, $a, b \\in \\{0, 1, 2, \\dots, 9\\}$, $a \\neq 0$, $N = 10a + b > a \\cdot b + b \\ge a \\cdot b = \\text{even}(N) \\cdot \\text{odd}(N)$.\n\nNow suppose that $N$ has $2k... | Brazil | Brazilian Math Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0l4k | Problem:
Sasha wants to bake 6 cookies in his $8$ inch $\times$ $8$ inch square baking sheet. With a cookie cutter, he cuts out from the dough six circular shapes, each exactly $3$ inches in diameter. Can he place these six dough shapes on the baking sheet without the shapes touching each other? If yes, show us how. I... | [
"Solution:\n\nYes, but just barely! The centers of the cookies must lie within the middle $5 \\times 5$ square. Divide this square in half one way and in thirds the other way, creating a grid of six $\\frac{5}{2} \\times \\frac{5}{3}$ rectangles, and place cookie centers at alternate intersections of this grid as s... | United States | 25th Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | Yes. Place the six cookie centers inside the middle 5×5 square of the sheet at alternate intersections of a grid formed by dividing one direction into halves and the other into thirds; the nearest center distance is sqrt(325)/6 > 3, so the cookies do not touch. | |
0jx2 | Problem:
Sean is a biologist, and is looking at a string of length $66$ composed of the letters $A, T, C, G$. A substring of a string is a contiguous sequence of letters in the string. For example, the string $A G T C$ has $10$ substrings: $A, G, T, C, A G, G T, T C, A G T, G T C, A G T C$. What is the maximum number ... | [
"Solution:\n\nLet's consider the number of distinct substrings of length $\\ell$. On one hand, there are obviously at most $4^{\\ell}$ distinct substrings. On the other hand, there are $67-\\ell$ substrings of length $\\ell$ in a length $66$ string. Therefore, the number of distinct substrings is at most\n$$\n\\sum... | United States | February 2017 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2100 | |
0c9h | Problem:
Se consideră triunghiul ascuţitunghic $ABC$. Fie $O$ centrul cercului circumscris acestuia şi $D$ piciorul înălţimii din $A$. Ştiind că $OD \parallel AB$, arătaţi că $\sin 2B = \operatorname{ctg} C$. | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geom... | null | proof only | null | |
0e49 | What is the size of the angle $x$ in degrees, if $2 \cos 10^\circ + \sin 100^\circ + \sin 1000^\circ + \sin 10000^\circ = \sin x$ and $-90^\circ \le x \le 90^\circ$ holds?
(A) -80
(B) -10
(C) 0
(D) 10
(E) 80 | [
"Since $2\\cos 10^\\circ + \\sin 100^\\circ + \\sin 1000^\\circ + \\sin 10000^\\circ = 2\\cos 10^\\circ + \\sin 80^\\circ - \\sin 80^\\circ - \\sin 80^\\circ = \\sin 80^\\circ$, we have $x = 80^\\circ$."
] | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | MCQ | E | |
0dhj | Prove that for any natural number $n > 1$ then $2^n - 1 \nmid 3^n - 1$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof only | null | |
0d6a | Let $n \geq 3$ and $x_{1}, x_{2}, \ldots, x_{n}$ be $n$ distinct integers. Prove that
$$
\left(x_{1}-x_{2}\right)^{2}+\left(x_{2}-x_{3}\right)^{2}+\ldots+\left(x_{n}-x_{1}\right)^{2} \geq 4 n-6 .
$$ | [
"We prove by induction on $n$.\n\nWhen $n=3$, we can assume that $x_{1}>x_{2}>x_{3}$ (only when $n=3$) and the LHS $\\geq 1^{2}+1^{2}+2^{2}=6$.\n\nMoving from $n$ to $n+1$, the change in LHS is\n$$\nC_{n+1}=\\left(x_{n}-x_{n+1}\\right)^{2}+\\left(x_{n+1}-x_{1}\\right)^{2}-\\left(x_{n}-x_{1}\\right)^{2} .\n$$\nWe ne... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0hc8 | Let $m$ and $n$ be positive integers, neither is divisible by $6$. A rectangle $m \times n$ is filled with $2 \times 2$ and $3 \times 3$ squares. Show that such rectangle can be filled with squares of one of the types: $2 \times 2$ or $3 \times 3$.
(Edvard Turkevych)
**Fig. 12** | [
"Consider the case when one of the numbers is not divisible by two and another one is not divisible by three. Without loss of generality, let $m$ be not divisible by $2$, and let $n$ be not divisible by $3$, and let $m$ be the width and $n$ be the height of the rectangle.\n\nWe will color in black all the rows of t... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
08ad | Problem:
Trovare tutte le coppie $(a, b)$ di numeri interi positivi tali che $a+1$ sia un divisore di $b-1$ e $b$ sia un divisore di $a^{2}+a+2$. | [
"Solution:\n\nConsideriamo $b$; dal momento che $a+1$ divide $b-1$ possiamo scrivere $b = k(a+1) + 1$, con $k$ naturale. Riformuliamo la seconda condizione come $(k(a+1)+1) h = a^{2} + a + 2$, ovvero $k h(a+1) + h = a(a+1) + 2$ per qualche $h \\in \\mathbb{N}$.\n\nSupponiamo anzitutto che $k$ sia positivo; se $k h ... | Italy | Progetto Olimpiadi della Matematica - Gara di Febbraio | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All pairs are (a, 1) with a any positive integer, and (2k, 2k^2 + k + 1) with k any positive integer. | |
01km | There is a heap of $25201$ stones. Nick and Mary play the following game. They, in turn, remove the stones from the heap. Per move it is allowed to remove either exactly $m$ or exactly $n$ stones. The player wins if he removes the last stone. If the last stone can not be removed by any of the players, then the result o... | [
"If Nick fixes the number $n$ and defines himself as the first player, then Mary wins if she sets $m = 11 - n$. Indeed, if Nick removes $n$ ($m$) stones, then Mary removes $m$ ($n$) stones. So exactly $11 = m + n$ stones are removed from the heap after each pair of moves (Nick - Mary). Since $25201 = 2291 \\cdot 11... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof only | null | |
0e57 | A triangle $ABC$ is given with the points $D$ on the side $AB$, $E$ on the side $BC$ and $F$ on the side $AC$, such that the line $CD$ is perpendicular to the side $AB$, the line $DE$ is perpendicular to the side $BC$ and the line $DF$ is perpendicular to the side $AC$. Prove that the points $A$, $B$, $E$ and $F$ lie o... | [
"Denote $\\angle BAC = \\alpha$. The point $D$ lies on the segment $AB$, so the angles $\\angle BAC$ and $\\angle CBA$ cannot be obtuse. If the point $D$ coincides with $A$ or with $B$, the statement is obviously true. For the remainder of the solution we assume that $D$ lies in the interior of the segment $AB$.\n\... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0327 | Problem:
The dragon Spas has one head. His family tree consists of Spas, the Spas parents, their parents, etc. It is known that if a dragon has $n$ heads, then his mother has $3 n$ heads and his father has $3 n+1$ heads. A positive integer is called good if it can be written in a unique way as a sum of the numbers of ... | [
"Solution:\n\nWe shall say that the order of Spas is $1$, the order of his parents is $2$, the order of their parents is $3$, etc. Write the number of the heads of any dragon in ternary base: Spas has $1$ head, his mother $10$ heads, his father $11$ heads, etc. It follows by induction on the order of the dragons th... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Other"
] | null | proof and answer | 2003 is a good number; there are 404 good numbers less than 2003. | |
0ifa | Problem:
The sides of a regular hexagon are trisected, resulting in 18 points, including vertices. These points, starting with a vertex, are numbered clockwise as $A_{1}, A_{2}, \ldots, A_{18}$. The line segment $A_{k} A_{k+4}$ is drawn for $k=1,4,7,10,13,16$, where indices are taken modulo 18. These segments define a... | [
"Solution:\n\nLet us assume all sides are of side length $3$. Consider the triangle $A_{1} A_{4} A_{5}$. Let $P$ be the point of intersection of $A_{1} A_{5}$ with $A_{4} A_{8}$. This is a vertex of the inner hexagon. Then $\\angle A_{4} A_{1} A_{5} = \\angle A_{5} A_{4} P$, by symmetry. It follows that $A_{1} A_{4... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 9/13 | |
0fqe | Problem:
Para cada número de cuatro cifras $a b c d$, denotamos por $S$ al número $S = \overline{a b c d} - \overline{d c b a}$. Demuestra que $S$ es múltiplo de $37$ si y sólo si las dos cifras centrales del número $a b c d$ son iguales. | [
"Solution:\n\nEscribimos el número como $\\overline{a b c d}$ como $1000 a + 100 b + 10 c + d$ y el número $\\overline{d c b a}$ como $1000 d + 100 c + 10 b + a$. Por tanto,\n$$\n\\begin{array}{r}\nS = \\overline{a b c d} - \\overline{d c b a} = 1000 a + 100 b + 10 c + d - (1000 d + 100 c + 10 b + a) = \\\\\n999(a ... | Spain | OME fase local | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0c1a | Prove that the number
$$
\sqrt[n]{\sqrt{2018} + \sqrt{2017}} + \sqrt[n]{\sqrt{2018} - \sqrt{2017}}
$$
is not rational for $n \ge 2$. | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Number Theory > Algebraic Number Theory > Quadratic fields",
"Algebra > Intermediate Algebra > Other"
] | null | proof only | null | |
03lk | Problem:
Let $(a, b, c)$ be a Pythagorean triple, i.e., a triplet of positive integers with $a^{2}+b^{2}=c^{2}$.
a) Prove that $(c / a+c / b)^{2}>8$.
b) Prove that there does not exist any integer $n$ for which we can find a Pythagorean triple $(a, b, c)$ satisfying $(c / a+c / b)^{2}=n$. | [
"Solution:\nLet $(a, b, c)$ be a Pythagorean triple. View $a, b$ as lengths of the legs of a right angled triangle with hypotenuse of length $c$; let $\\theta$ be the angle determined by the sides with lengths $a$ and $c$. Then\n$$\n\\begin{aligned}\n\\left(\\frac{c}{a}+\\frac{c}{b}\\right)^{2} & =\\left(\\frac{1}{... | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequal... | null | proof only | null | |
04dr | A regular 2013-gon $A_1A_2...A_{2013}$ is given. Let $S$ be the intersection of the segments $A_1A_4$ and $A_3A_5$. Determine the angle $\angle A_3SA_4$. | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 180 − 540/2013 degrees | |
03t5 | $AB$ is a diameter of the circle $O$, the point $C$ lies on the extended line $AB$ produced. A line passing through $C$ intersects with the circle $O$ at points $D$ and $E$. $OF$ is a diameter of the circumcircle $O_1$ of $\triangle BOD$. Join $CF$ and its extension, which intersects the circle $O_1$ at $G$. Prove that... | [
"\n\nThus, $G$, $A$, $C$, $D$ are concyclic. Hence,\n$$\n\\angle AGC = \\angle ADC, \\qquad ①\n$$\n$$\n\\angle AGC = \\angle AGO + \\angle OGF = \\angle AGO + \\frac{\\pi}{2}, \\qquad ②\n$$\n$$\n\\angle ADC = \\angle ADB + \\angle BDC = \\angle BDC + \\frac{\\pi}{2}. \\qquad ③\n$$\nCombinin... | China | China Western Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0kan | Problem:
Initially, all the squares of an $8 \times 8$ grid are white. You start by choosing one of the squares and coloring it gray. After that, you may color additional squares gray one at a time, but you may only color a square gray if it has exactly 1 or 3 gray neighbors at that moment (where a neighbor is a squar... | [
"Solution:\n\nIt is not possible. Let $L(t)$ denote the length of the boundary of the gray region after $t$ squares have been colored gray. We have $L(1) = 4$ since the perimeter of the first square colored is $4$. After $t \\geq 1$ squares have been colored, if we add a square that has exactly one gray neighbor, t... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | It is not possible. | |
0jpk | Problem:
For an integer $n$, let $f(n)$ denote the number of pairs $(x, y)$ of integers such that $x^{2}+x y+y^{2}=n$. Compute the sum
$$
\sum_{n=1}^{10^{6}} n f(n)
$$
Write your answer in the form $a \cdot 10^{b}$, where $b$ is an integer and $1 \leq a<10$ is a decimal number.
If your answer is written in this form, y... | [
"Solution:\nAnswer: $1.813759629294 \\cdot 10^{12}$\n\nRewrite the sum as\n$$\n\\sum_{x^{2}+x y+y^{2} \\leq 10^{6}}\\left(x^{2}+x y+y^{2}\\right),\n$$\nwhere the sum is over all pairs $(x, y)$ of integers with $x^{2}+x y+y^{2} \\leq 10^{6}$. We can find a crude upper bound for this sum by noting that\n$$\nx^{2}+x y... | United States | HMMT February | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | null | final answer only | 1.813759629294 * 10^12 | |
003k | Los números $1$, $2$, $3$, ..., $n^2$ se colocan en las casillas de una cuadrícula de $n \times n$, en algún orden, un número por casilla. Una ficha se encuentra inicialmente en la casilla con el número $n^2$. En cada paso, la ficha puede avanzar a cualquiera de las casillas que comparten un lado con la casilla donde s... | [] | Argentina | 21° OLIMPIADA IBEROAMERICANA DE MATEMÁTICA | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | Español | proof and answer | Minimum: N = n^2 if n is even, and N = n^2 + 1 if n is odd (with the trivial case n = 1 giving N = 0). Maximum: N = n^3 − n. | |
0fgs | Problem:
Sean $x_{1}, x_{2}, \ldots, x_{n}$, números reales que cumplen $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$. Demostrar que para cada entero $k \geq 2$ existen enteros no todos nulos $a_{1}, a_{2}, \ldots, a_{n}$, tales que $\left|a_{i}\right| \leq k-1$ para todo $i$ y
$$
\left|a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n}... | [] | Spain | International Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
04nk | A building consists of the ground floor and $100$ additional floors. The lift in the building has only two buttons, $A$ and $B$. By pressing $A$ the lift rises $7$ floors, and by pressing $B$ the lift goes down $9$ floors. Is it possible to reach every floor from any other floor using this lift? | [] | Croatia | Croatia_2018 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | Yes | |
0jpz | Problem:
Let $ABC$ be a triangle and $P$ a point inside it. Rays $BP$ and $CP$ meet $AC$ and $AB$ at $Y$ and $X$, respectively. Prove that if $AP$ bisects $BC$ then $XY \parallel BC$. | [
"Solution:\n\nLet $Q$ be the reflection of $P$ across $M$ (with $M$ the midpoint of $BC$). Accordingly, $BPCQ$ is a parallelogram.\n\n\n\nFrom this, we see that $\\triangle AXP \\sim \\triangle ABQ$ and $\\triangle AYP \\sim \\triangle ACQ$, and thus we deduce\n\n$$\n\\frac{AX}{AB} = \\frac... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0b6v | Consider a triangle $ABC$ and a point $M$ in its interior. Denote by $D, E, F$ the projections of $M$ onto the sides $BC, CA$ respectively $AB$ and by $r$ the radius of its incircle. Prove that if
$$
\frac{BC}{MD} + \frac{CA}{ME} + \frac{AB}{MF} = \frac{AB + BC + CA}{r},
$$
then the straight lines $AD$, $BE$ and $CF$ a... | [] | Romania | Shortlisted Problems for the Romanian NMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
00h4 | Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$, and let $P$ be a point on the extension of $AC$ such that $PB$ and $PD$ are tangent to $\omega$. The tangent at $C$ intersects $PD$ at $Q$ and the line $AD$ at $R$. Let $E$ be the second point of intersection between $AQ$ and $\omega$. Prove that $B$, $E$, $... | [
"To show $B$, $E$, $R$ are collinear, it is equivalent to show the lines $AD$, $BE$, $CQ$ are concurrent. Let $CQ$ intersect $AD$ at $R$ and $BE$ intersect $AD$ at $R'$. We shall show $RD / RA = R'D / R'A$ so that $R = R'$.\n\nSince $\\triangle PAD$ is similar to $\\triangle PDC$ and $\\triangle PAB$ is similar to ... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0cz4 | Prove that for any positive real numbers $a, b, c$,
$$
2\left(a^{3}+b^{3}+c^{3}+a b c\right) \geq (a+b)(b+c)(c+a).
$$ | [
"The inequality is equivalent to\n$2 a^{3} + 2 b^{3} + 2 c^{3} + 2 a b c \\geq a^{2}(b+c) + b^{2}(c+a) + c^{2}(a+b) + 2 a b c$,\nhence\n$$\n2 a^{3} + 2 b^{3} + 2 c^{3} \\geq a b(a+b) + b c(b+c) + c a(c+a) \\tag{1}\n$$\nFor any positive real numbers $x$ and $y$ we have\n$$\nx^{3} + y^{3} \\geq x y(x+y)\n$$\nIndeed, ... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0daz | Find all functions $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ that for all real numbers $x, y, z$ satisfies the equation
$$
f(f(x, z), f(z, y)) = f(x, y) + z.
$$ | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | null | |
03v0 | Let $A = [-2, 4)$, $B = \{x \mid x^2 - a x - 4 \leq 0\}$. If $B \subseteq A$, then the range of real $a$ is ( ). | [
"$x^2 - a x - 4 = 0$ has two roots:\n$$\nx_1 = \\frac{a}{2} - \\sqrt{4 + \\frac{a^2}{4}}, \\quad x_2 = \\frac{a}{2} + \\sqrt{4 + \\frac{a^2}{4}}\n$$\nWe have $B \\subseteq A \\Leftrightarrow x_1 \\geq -2$ and $x_2 < 4$. This means that\n$$\n\\frac{a}{2} - \\sqrt{4 + \\frac{a^2}{4}} \\geq -2, \\quad \\frac{a}{2} + \... | China | China Mathematical Competition | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | [0, 3) | |
0dij | Denote by $\mathbb{N}$ the set of positive integers. Find all functions $f: \mathbb{N} \to \mathbb{N}$ such that $x - y$ divides $x^{f(x)} - y^{f(y)}$ for every two coprime integers $x$ and $y$. | [
"(Base on the solution of Hadi Alaithan, IMO 2023's team member)\nNote that for $x, y$ are coprime positive integers, we have\n$$\nx - y \\mid x^{f(x)} - y^{f(y)} \\Leftrightarrow x - y \\mid x^{|f(x) - f(y)|} - 1.\n$$\nNow we state the following lemma:\n**Lemma (Zsigmondy's Theorem):** for $a > b \\ge 1$ be coprim... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Other"
] | English | proof and answer | All functions with f(1) arbitrary in N and f(n)=c for all n≥2, where c is an arbitrary positive integer. | |
03bo | Points $M$, $N$ and $P$ lie on the sides $BC$, $CA$, $AB$ of triangle $ABC$. Triangles $CNM$, $APN$ and $BMP$ are acute and let $H_C$, $H_A$ and $H_B$ be their respective orthocenters. Prove that if the three lines $AH_A$, $BH_B$ and $CH_C$ are concurrent then $MH_A$, $NH_B$ and $PH_C$ are also concurrent. | [
"Denote by $A_1$, $B_1$ and $C_1$ the projections of $A$, $B$ and $C$ on $NP$, $PM$ and $MN$, respectively. It follows from the condition of the problem (Carnot's theorem) that $(NC_1^2 - MC_1^2) + (MB_1^2 - PB_1^2) + (PA_1^2 - NA_1^2) = 0$. Hence $0 = (CN^2 - CM^2) + (BM^2 - BP^2) + (AP^2 - AN^2) = (CN^2 - AN^2) +... | Bulgaria | Team selection test for the 54th IMO | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0cr0 | A safe contains $n$ boxes numbered from $1$ to $n$. Each box initially contained a card with the number of this box on it. Then Basil rearranged the cards so that the $i$th box now contains a card with number $a_i$. Pete may switch any two cards with numbers $x$ and $y$; for such an operation he should pay $2|x - y|$ r... | [
"Let $(b_1, \\dots, b_n)$ be an arbitrary arrangement of the cards (here $b_i$ is the number on the card in the $i$th box). Define its cost as $|b_1 - 1| + |b_2 - 2| + \\dots + |b_n - n|$.\n\n**Lemma.** For any arrangement $(b_1, \\dots, b_n)$ in which not all cards are in their places, it is possible to swap two c... | Russia | XL Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0l03 | Equilateral $\triangle ABC$ with side length $14$ is rotated about its center by angle $\theta$, where $0 < \theta \le 60^\circ$, to form $\triangle DEF$. See the figure. The area of hexagon $ADBECF$ is $91\sqrt{3}$. What is $\tan \theta$?
(A) $\frac{3}{5}$
(B) $\frac{5\sqrt{3}}{11}$
(C) $\fra... | [
"**Answer (B):** Let $O$ be the center of $\\triangle ABC$ and $\\triangle DEF$, let $P$ be the foot of the perpendicular from $D$ to $\\overline{AB}$, and let $M$ be the midpoint of $\\overline{AB}$.\n\nThe condition $\\theta \\le 60^\\circ$ implies that $P$ lies on $\\overline{AM}$ (as op... | United States | 2024 AMC 12 B | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angl... | null | MCQ | B | |
0ap2 | Problem:
Let $k$ be a positive integer. A positive integer $n$ is said to be a $k$-flip if the digits of $n$ are reversed in order when it is multiplied by $k$. For example, $1089$ is a $9$-flip because $1089 \times 9 = 9801$, and $21978$ is a $4$-flip because $21978 \times 4 = 87912$. Explain why there is no $7$-flip... | [
"Solution:\n\nSuppose, by way of contradiction, that the number $A \\ldots Z$ is a $7$-flip. Then $A \\ldots Z \\times 7 = Z \\ldots A$. So that there will be no carry-over in the multiplication of the last digit, $A$ should be $1$. This will imply two contradicting statements: (1) $Z \\geq 7$ and (2) the product $... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Other"
] | null | proof only | null | |
0d4u | Find all positive integers $n$ such that
$$
3^{n} + 4^{n} + \cdots + (n+2)^{n} = (n+3)^{n}.
$$ | [
"Notice first that\n$$\n3^{2} + 4^{2} = 5^{2}, \\quad \\text{and} \\quad 3^{3} + 4^{3} + 5^{3} = 6^{3}.\n$$\nHowever\n$$\n3^{4} + 4^{4} + 5^{4} + 6^{4} = 2258 < 2401 = 7^{4}\n$$\nAssume that\n$$\n3^{n} + 4^{n} + \\cdots + (n+2)^{n} < (n+3)^{n}\n$$\nfor some $n \\geq 4$. We have\n$$\n\\begin{aligned}\n3^{n+1} + 4^{n... | Saudi Arabia | SAMC | [
"Algebra > Intermediate Algebra > Exponential functions",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English, Arabic | proof and answer | n = 2, 3 | |
0kp3 | Problem:
Vijay chooses three distinct integers $a, b, c$ from the set $\{1,2,3,4,5,6,7,8,9,10,11\}$. If $k$ is the minimum value taken on by the polynomial $a(x-b)(x-c)$ over all real numbers $x$, and $l$ is the minimum value taken on by the polynomial $a(x-b)(x+c)$ over all real numbers $x$, compute the maximum possi... | [
"Solution:\n\nQuadratics are minimized at the average of their roots, so\n$$\n\\begin{aligned}\n& k = a\\left(\\frac{b+c}{2} - b\\right)\\left(\\frac{b+c}{2} - c\\right) \\\\\n& l = a\\left(\\frac{b-c}{2}\\right)\\left(\\frac{c-b}{2}\\right) = -\\frac{a(b-c)^2}{4}, \\text{ and } \\\\\n& l = a\\left(\\frac{b-c}{2} -... | United States | HMMT November 2022 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | 990 | |
0etq | *South African Magical Flights* (SAMF) operates flights between South African airports. If there is a flight from airport $A$ to airport $B$, there will also be a flight from $B$ to $A$.
The SAMF headquarters are located in Kimberley. Every airport that is served by SAMF can be reached from Kimberley in precisely one ... | [
"We can depict the situation as a connected graph without cycles, i.e., a tree. The vertices represent the airports and there is an edge between two vertices if and only if there is a direct flight possible between the two airports represented by these two vertices. The leaves of the tree (vertices (other than the ... | South Africa | The South African Mathematical Olympiad Third Round | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0gu3 | Show that if $m$ and $n$ are relatively prime positive integers, then
$$
\frac{n^4 + m}{m^2 + n^2} \quad \text{and} \quad \frac{n^4 - m}{m^2 - n^2}
$$
can not be integers simultaneously. | [
"Assume the contrary. The sum of the numbers is\n$$\n\\frac{2n^2m(n^2m-1)}{(m^2+n^2)(m^2-n^2)}\n$$\nSince $m$ and $n$ are relatively prime, we have\n$$\n\\gcd(n^2m, m^2+n^2) = \\gcd(n^2m, m^2-n^2) = 1.\n$$\nThen we obtain $(m^2+n^2)(m^2-n^2) \\mid 2(n^2m-1)$. Note that $m, n \\ge 1$ and we can not have $m=n=1$, and... | Turkey | 31st Junior Turkish Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0bna | Let $\gamma$, $\gamma_0$, $\gamma_1$, $\gamma_2$ be coplanar circles such that $\gamma_i$ is internally tangent to $\gamma$ at $A_i$, and $\gamma_i$ and $\gamma_{i+1}$ are externally tangent at $B_{i+2}$, $i = 0, 1, 2$ (indices are reduced modulo $3$). The tangent at $B_i$, common to $\gamma_{i-1}$ and $\gamma_{i+1}$, ... | [
"Let $\\gamma_i$ cross the lines $A_iC_{i+1}$ and $A_iC_{i+2}$ again at $X_{i+2}$ and $Y_{i+1}$, respectively.\n\nSince the homothety centred at $A_i$, transforming $\\gamma_i$ into $\\gamma$, sends $X_{i+2}$ to $C_{i+1}$ and $Y_{i+1}$ to $C_{i+2}$, the lines $X_{i+2}Y_{i+1}$ and $C_{i+1}C_{i+2}$ are parallel, so $... | Romania | 2015 Ninth STARS OF MATHEMATICS Competition | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods >... | English | proof only | null |
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