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values | competition stringlengths 3 108 ⌀ | topics_flat listlengths 0 12 | language stringclasses 18
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values | final_answer stringlengths 1 1.22k ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0kmt | The product of the lengths of the two congruent sides of an obtuse isosceles triangle is equal to the product of the base and twice the triangle's height to the base. What is the measure, in degrees, of the vertex angle of this triangle?
(A) 105
(B) 120
(C) 135
(D) 150
(E) 165 | [] | United States | AMC 12 B | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | D | |
0dcc | Let $n$ be a positive integer and let $a_{1}, a_{2}, \ldots, a_{n}$ be any real numbers. Prove that there exists $m, k \in \{1,2, \ldots, n\}$ such that
$$
\left|\sum_{i=1}^{m} a_{i}-\sum_{i=m+1}^{n} a_{i}\right| \leq\left|a_{k}\right| .
$$ | [
"Denote\n$$\nS_{x}=\\left|\\sum_{i=1}^{x} a_{i}-\\sum_{i=x+1}^{n} a_{i}\\right| .\n$$\nObviously $S_{0}=-S_{n}$ and $S_{x+1}-S_{x}=2 a_{x+1}$. We may assume that $S_{n} \\geq 0$ and consider $m$ for which $S_{m} \\leq 0$ and $S_{m+1} \\geq 0$. Then\n$$\n\\left|S_{m}\\right|+\\left|S_{m+1}\\right|=\\left|S_{m}-S_{m+... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
00kd | (a) For which triangles with sides of length $a$, $b$ and $c$ do the inequalities $a^2 + b^2 > c^2$, $b^2 + c^2 > a^2$ and $a^2 + c^2 > b^2$ hold (along with the usual triangle inequalities $a+b>c$, $b+c>a$ and $c+a>b$)?
(b) For which triangles with sides of length $a$, $b$ and $c$ do the inequalities $a^n + b^n > c^n... | [
"(a) By the cosine theorem, we have $c^2 = a^2 + b^2 - 2ab \\cos \\gamma$. Since $\\cos \\gamma > 0 \\iff \\gamma < 90^\\circ$, we see that $a^2 + b^2 > c^2$ is equivalent to $\\gamma < 90^\\circ$. Since the analogous results hold for the other two inequalities, we see that these all hold exactly for acute angled t... | Austria | Austria 2014 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | English | proof and answer | (a) Exactly the acute triangles. (b) Exactly the isosceles triangles with the two largest sides equal (b = c) and the third side not exceeding them (a ≤ b), equivalently the apex angle at the unequal side is at most sixty degrees. | |
07oy | Three different nonzero real numbers $a$, $b$, $c$ satisfy the equations
$$
a + \frac{2}{b} = b + \frac{2}{c} = c + \frac{2}{a} = p
$$
where $p$ is a real number. Prove that $abc + 2p = 0$. | [
"We eliminate variables to get an equation in $b$ and $p$. First,\n$$\na = p - \\frac{2}{b} = \\frac{bp - 2}{b}, \\quad \\text{so that} \\quad \\frac{1}{a} = \\frac{b}{bp - 2}. \\quad \\text{Then}\n$$\n$$\nc = p - \\frac{2}{a} = p - \\frac{2b}{bp-2} = \\frac{bp^2 - 2p - 2b}{bp-2}, \\quad \\text{so that} \\quad \\fr... | Ireland | Irska 2014 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof only | null | |
00wu | Problem:
Consider two points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ on the graph of the function $y=\frac{1}{x}$ such that $0<x_{1}<x_{2}$ and $|A B|=2 \cdot|O A|$ ($O$ is the reference point, i.e., $O(0,0)$). Let $C$ be the midpoint of the segment $A B$. Prove that the angle between the $x$-axi... | [
"Solution:\n\nWe have $A\\left(x_{1}, \\frac{1}{x_{1}}\\right)$, $B\\left(x_{2}, \\frac{1}{x_{2}}\\right)$ and $C\\left(\\frac{x_{1}+x_{2}}{2}, \\frac{1}{2 x_{1}}+\\frac{1}{2 x_{2}}\\right)$. Computing the coordinates of $\\bar{v}=|O C| \\cdot \\overline{A C}+|A C| \\cdot \\overline{O C}$ we find that the vector $\... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jxi | Problem:
For positive integers $a$ and $N$, let $r(a, N) \in \{0,1, \ldots, N-1\}$ denote the remainder of $a$ when divided by $N$. Determine the number of positive integers $n \leq 1000000$ for which
$$
r(n, 1000)>r(n, 1001)
$$ | [
"Solution:\nAnswer: 499500\nNote that $0 \\leq r(n, 1000) \\leq 999$ and $0 \\leq r(n, 1001) \\leq 1000$. Consider the $\\binom{1000}{2}=499500$ ways to choose pairs $(i, j)$ such that $i>j$. By the Chinese Remainder Theorem, there is exactly one $n$ such that $1 \\leq n \\leq 1000 \\cdot 1001$ such that $n \\equiv... | United States | February 2017 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 499500 | |
0ivg | Problem:
Let $f(x) = x^{2} + 2x + 1$. Let $g(x) = f(f(\cdots f(x)))$, where there are $2009$ $f$'s in the expression for $g(x)$. Then $g(x)$ can be written as
$$
g(x) = x^{2^{2009}} + a_{2^{2009}-1} x^{2^{2009}-1} + \cdots + a_{1} x + a_{0}
$$
where the $a_{i}$ are constants. Compute $a_{2^{2009}-1}$. | [
"Solution:\n$2^{2009} f(x) = (x+1)^{2}$, so $f\\left(x^{n} + c x^{n-1} + \\ldots\\right) = \\left(x^{n} + c x^{n-1} + \\ldots + 1\\right)^{2} = x^{2n} + 2c x^{2n-1} + \\ldots$. Applying the preceding formula repeatedly shows us that the coefficient of the term of second highest degree in the polynomial doubles each... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 2^{2009} | |
058g | Prove that $a^{2020} + 10a^{1010} + 1001$ is prime for no integers $a$. | [
"Note that $1001 = 11 \\cdot 91$. If $11 \\mid a$ then also $11 \\mid a^{2020} + 10a^{1010} + 1001$. Otherwise, consider $a^5$ modulo $11$. A case study shows that if $a$ is congruent to $1$, $3$, $4$, $5$, or $9$, then $a^5$ is congruent to $1$, and in all other cases, $a^5$ is congruent to $-1$. Hence $a^{10} \\e... | Estonia | Estonian Math Competitions | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0h1y | Two circles touch each other externally at point $C$. Consider two diameters $A_1A_2$, $B_1B_2$ of the same direction. Circle with the center on the common internal tangent passes through the point of intersection of $A_1B_2$, $A_2B_1$, and meets these lines at points $M$, $N$. Prove that $MN$ is perpendicular to $A_1A... | [
"Since point $C$ is a center of homothety that transforms one circle into another, then $C = A_1B_2 \\cap A_2B_1$, $A_1B_2 \\perp A_2B_1$ (fig. 21).\n\nLet $D = MN \\cap B_1B_2$. Then, $\\angle DB_2C = \\angle B_2A_1A_2 = \\angle A_2CO = \\angle CND$. Therefore $DB_2NC$ is cyclic and $\\angle B_2DN = \\angle B_2CA_... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08kw | Problem:
Let the inscribed circle of the triangle $\triangle ABC$ touch side $BC$ at $M$, side $CA$ at $N$ and side $AB$ at $P$. Let $D$ be a point from $[NP]$ such that $\frac{DP}{DN}=\frac{BD}{CD}$. Show that $DM \perp PN$. | [
"Solution:\n\nFrom $AP = AN$ it follows that $\\angle ANP = \\angle APN$ or $\\angle NPB = \\angle PNC$ (both obtuse). Hence the triangles $BDP$ and $CND$ are similar (SSA) and $\\angle CDN = \\angle BDP$ and $\\frac{CD}{BD} = \\frac{CN}{BP} = \\frac{CM}{BM}$. So $DM$ is a bisector of the angle $BDC$, from where $N... | JBMO | 2007 Shortlist JBMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0e6x | Let $ABCD$ be a cyclic quadrilateral with longest side $AB$. Let the bisectors of the angles $\angle DCB$ and $\angle ADC$ additionally intersect the circumscribed circle of the quadrilateral $ABCD$ in points $E$ and $F$. Call $G$ the intersection point of the lines $CE$ and $DF$, and $H$ the intersection point of the ... | [
"Because the points $C$, $D$, $E$ and $F$ are concyclic, the angles $\\angle FDC$ and $\\angle FEC$ are equal. Because the points $A$, $E$, $F$ and $D$ are concyclic, the angles $\\angle ADF$ and $\\angle HEF$ are equal. Hence the angles $\\angle HEF$ and $\\angle FEG$ are equal. We can show similarly that the angl... | Slovenia | National Math Olympiad 2012 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05zs | Problem:
Emile a créé un exercice pour Benoît. Il lui annonce qu'il a choisi secrètement un polynôme $P$ unitaire de degré $2023$ à coefficients entiers, c'est-à-dire de la forme
$$
P(X) = X^{2023} + a_{2022} X^{2022} + a_{2021} X^{2021} + \ldots + a_{1} X + a_{0}
$$
ou $a_{0}, a_{1}, \ldots, a_{2022}$ sont des entier... | [
"Solution:\n\nMontrons que le $k$ minimal est $k = 2023$. En effet, supposons d'abord que $k < 2023$ et qu'Emile choisisse $P$ et $n_{1}, \\ldots, n_{k}$. Posons\n$$\nQ = P + (X - n_{1})(X - n_{2}) \\ldots (X - n_{k})\n$$\nqui est un polynôme à coefficients entiers, toujours unitaire de degré $2023$, et différent d... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 2 : AlgèBre | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | 2023 | |
0f5y | Problem:
Show that $\dfrac{(a + b)^2}{2} + \dfrac{a + b}{4} \geq a\sqrt{b} + b\sqrt{a}$ for all positive $a$ and $b$. | [
"Solution:\nBy AM/GM $\\sqrt{ab} \\leq \\dfrac{a + b}{2}$, so $\\sqrt{2} (a + b) + \\sqrt{ab} \\leq a + b$. Hence $\\sqrt{2a + 2b} \\geq \\sqrt{a} + \\sqrt{b}$ $\\quad(*)$.\n\nBy AM/GM $(a + b) \\geq 2\\sqrt{ab}$ and $2(a + b) + 1 \\geq 2\\sqrt{2a + 2b}$. Multiplying, $(a + b)(2a + 2b + 1) \\geq 4\\sqrt{ab}\\sqrt{2... | Soviet Union | 18th ASU | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
00hi | Consider a $2018 \times 2019$ board with integers in each unit square. Two unit squares are said to be neighbours if they share a common edge. In each turn, you choose some unit squares. Then for each chosen unit square the average of all its neighbours is calculated. Finally, after these calculations are done, the num... | [
"Let $n$ be a positive integer relatively prime to $2$ and $3$. We may study the whole process modulo $n$ by replacing divisions by $2,3,4$ with multiplications by the corresponding inverses modulo $n$. If at some point the original process makes all the numbers equal, then the process modulo $n$ will also have all... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | No | |
0fup | Problem:
Sei $n \geq 1$ eine natürliche Zahl. Ein reguläres $4 n$-Eck sei irgendwie in endlich viele Parallelogramme zerlegt.
a. Beweise, dass mindestens eines der Parallelogramme in der Zerlegung ein Rechteck ist.
b. Bestimme die Summe der Flächen aller Rechtecke in der Zerlegung. | [
"Solution:\n\nWir führen erst einige Bezeichnungen ein. Eine endliche Zerlegung des $4 n$-Ecks in Parallelogramme nennen wir im Folgenden kurz Zerlegung. Für eine Zerlegung nennen wir jeden Eckpunkt eines Parallelogramms einen Knoten. Je nachdem, ob ein Knoten auf dem Rand oder im Innern des $4 n$-Ecks liegt, heiss... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | n | |
0dk4 | Find the smallest positive number $m$ such that there exists unique function $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that
$$
f(x + f(y)) = f(x) + \frac{mx^5 f(y)}{f(x)^2} + f(y^2), \forall x, y > 0.
$$ | [
"If there exist $y_0$ such that $f(y_0) < y_0^2$ then taking $x = y_0^2 - f(y_0)$ to get\n$$\nf(x + f(y_0)) = f(y_0^2).\n$$\nFrom this, one can get\n$$\nf(x) + \\frac{mx^5 f(y_0)}{f(x)^2} = 0,\n$$\nwhich is contradiction. So $f(y) \\ge y^2$ for $y > 0$. Put $x = y = 1$ then\n$$\nf(1 + f(1)) = 2f(1) + \\frac{m}{f(1)... | Saudi Arabia | Saudi Arabia booklet 2024 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | 2 | |
09q5 | Problem:
Zij $N$ het aantal geordende vijftallen $(a_{1}, a_{2}, a_{3}, a_{4}, a_{5})$ van positieve gehele getallen waarvoor geldt
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}+\frac{1}{a_{5}}=1.
$$
(Bij geordende vijftallen doet de volgorde er toe, dus $(2,3,15,15,30)$ en $(15,2,15,3,30)$ zijn v... | [
"Solution:\n\nOplossing I. Bekijk een ongeordend vijftal dat voldoet en stel dat het uit de verschillende getallen $b_{1}, \\ldots, b_{k}$ bestaat (met $k \\leq 5$), waarbij $b_{i}$ precies $t_{i}$ keer voorkomt. Er geldt dus $t_{1}+\\cdots+t_{k}=5$. Nu geldt dat dit vijftal op $\\frac{5!}{t_{1}!\\cdots t_{k}!}$ ma... | Netherlands | Dutch TST | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities... | null | proof only | odd | |
0ebu | Problem:
Določi $m \in \mathbb{R}$ tako, da bo najmanjša vrednost funkcije $f(x) = (m-1)x^{2} + (1-3m)x + 2m + 1$ enaka $-1$. | [] | Slovenia | 15. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 3 | |
09qw | Problem:
Bestaan er kwadratische polynomen $P(x)$ en $Q(x)$ met reële coëfficiënten zodat het polynoom $P(Q(x))$ als nulpunten precies $x=2, x=3, x=5$ en $x=7$ heeft? | [
"Solution:\n\nStel dat zulke polynomen bestaan en schrijf $Q(x) = a x^{2} + b x + c$. Als we $2$, $3$, $5$ en $7$ in $Q$ stoppen, moeten er precies de (hoogstens) twee nulpunten van $P$ uitkomen. Omdat er niet meer dan twee keer dezelfde waarde uit $Q(x)$ kan komen (want $Q$ is kwadratisch) krijgen we dus twee vers... | Netherlands | Dutch TST | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0k32 | Problem:
A polygon in the plane (with no self-intersections) is called equitable if every line passing through the origin divides the polygon into two (possibly disconnected) regions of equal area.
Does there exist an equitable polygon which is not centrally symmetric about the origin?
(A polygon is centrally symmetric... | [
"Solution:\nConsider the polygon with vertices\n$$\n\\begin{aligned}\n& A(1,0),\\ B(0,1),\\ C(0,5),\\ D(5,0),\\ E(7,0) \\\\\n& F(0,-7),\\ G(-7,0),\\ H(-\\sqrt{84},\\ 0),\\ I(0,\\ \\sqrt{84}),\\ J(0,6) \\\\\n& K(-6,0),\\ L(-5,0),\\ M(0,-5),\\ N(0,-1)\n\\end{aligned}\n$$\nNotice that its intersection with each of the... | United States | HMIC 2018 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Yes | |
0h0i | Point $P$ is chosen inside isosceles triangle $ABC$ with base $BC$ and $\angle BAC < 90^\circ$ in such a way, that $\angle BPC = 2\angle BAC$. Let $K$ be the feet of perpendicular from $A$ to the line which contains the bisector of adjacent angle of $\angle BPC$. Prove, that $BP + PC = 2AK$. | [
"We first show that point $K$ is always inside triangle $\\triangle ABC$. Consider the case, when $P = P_1$ lies on the side $AB$ and compute some angles. Let $\\angle BAC = \\alpha$ (Fig.13), then $\\angle BP_1C = 2\\alpha$, $\\angle ABC = \\frac{\\pi}{2} - \\frac{\\alpha}{2}$, $\\angle BCP_1 = \\frac{\\pi}{2} - \... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010) | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0hqw | Problem:
Let $ABCD$ be a rectangle. Let $E$ be the foot of perpendicular from $A$ to $BD$. Let $F$ be an arbitrary point of the diagonal $BD$ between $D$ and $E$. Let $G$ be the intersection of the line $CF$ with the perpendicular from $B$ to $AF$. Let $H$ be the intersection of the line $BC$ with the perpendicular fr... | [
"Solution:\n\nLet $X$ be the intersection of lines $BG$ and $AE$. Since $BX \\perp AF$ and $AX \\perp BF$, $X$ is the orthocenter of $\\triangle ABF$; hence $FX \\perp AB \\Rightarrow FX \\parallel BC \\parallel AD$. It follows:\n$$\n\\frac{EF}{ED} = \\frac{FX}{DA} = \\frac{FX}{BC} = \\frac{GF}{GC}\n$$\n(the first ... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ge3 | 有 110 個品種的天竺鼠各 110 隻，排成一個 110 × 110 的方陣。試求最大的正整數 $n$，使得不論天竺鼠們如何排列，我們都可以找到一直列或一橫條的 110 隻天竺鼠包含至少 $n$ 個不同品種。
There are 110 guinea pigs for each of the 110 species, arranging as a 110 × 110 array. Find the maximum integer $n$ such that, no matter how the guinea pigs align, we can always find a column or a row of 110 guine... | [
"令 $c_i$ 與 $r_i$ 分別為包含第 $i$ 個品種的列數與行數。基於這些列與行的交叉點必須包含全部 110 個 $i$ 品種天竺鼠, 我們有 $c_i \\times d_i \\ge 110$, 從而 $c_i + d_i \\ge 11 + 10 = 21$。這表示 $\\sum_i (c_i + d_i) \\ge 110 \\times 21 = 2310$。又橫行與直列共有 220 條, 依據鴿籠原理, 至少有一條包含 $\\left[\\frac{2310}{220}\\right]+1=11$ 個不同品種。最後, 考慮讓 $(x, y)$ 位置是一隻 $(11\\left[\\frac{x}{11}... | Taiwan | 2021 數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 11 | |
0cg2 | The integers from $1$ to $49$ are written in a $7 \times 7$ table, such that for any $k \in \{1, 2, ..., 7\}$, the product of the numbers in the $k$-th row equals the product of the numbers in the $(8-k)$-th row.
a) Prove that there exists a row such that the sum of the numbers written on it is a prime number. | [
"a) Let $p_1, p_2, ..., p_7$ be the products of the elements of the rows $1, 2, ..., 7$. We have $p_1 = p_7$, $p_2 = p_6$ and $p_3 = p_5$, so $49! = (p_1 p_2 p_3)^2 \\cdot p_4$, (1).\n\nNote that $49! = 2^{46} \\cdot 3^{22} \\cdot 5^{10} \\cdot 7^8 \\cdot 11^4 \\cdot 13^3 \\cdot 17^2 \\cdot 19^2 \\cdot 23^2 \\cdot ... | Romania | 74th NMO Selection Tests for JBMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
07al | Find all coprime natural numbers $a$ and $b$ for which $b.a = \frac{a}{b}$ (Note that if $a = 92$ and $b = 13$, then $b.a = 13.92$). | [
"Let $k$ be the number of digits of $a$. Therefore, $10^{k-1} \\le a < 10^k$. We have $\\frac{a}{b} = b.a = b + \\frac{a}{10^k}$ and consequently $\\frac{a-b^2}{b} = \\frac{a}{10^k}$. Since $(a,b) = 1$, $(a-b^2,b) = 1$; therefore, $\\frac{a-b^2}{b}$ is the irreducible form of $\\frac{a}{10^k}$. Thus, for some natur... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Decimal... | English | proof and answer | a=5, b=2 | |
0jsr | Problem:
Let $S$ be the set of all positive integers whose prime factorizations only contain powers of the primes $2$ and $2017$ (that is, $1$, powers of $2$, and powers of $2017$ are thus contained in $S$). Compute
$$
\sum_{s \in S} \frac{1}{s}.
$$ | [
"Solution:\nSince every $s$ can be written as $2^{i} \\cdot 2017^{j}$ for non-negative integers $i$ and $j$, the given sum can be written as\n$$\n\\left(\\sum_{i=0}^{\\infty} \\frac{1}{2^{i}}\\right)\\left(\\sum_{j=0}^{\\infty} \\frac{1}{2017^{j}}\\right).\n$$\nWe can easily find the sum of these geometric series s... | United States | HMMT November | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 2017/1008 | |
0frp | Sea $ABC$ un triángulo, con $AB < AC$, y sea $\Gamma$ su circuncírculo. Sean $D$, $E$ y $F$ los puntos de tangencia del incírculo con $BC$, $CA$ y $AB$, respectivamente. Sea $R$ el punto de $EF$ tal que $DR$ es una altura del triángulo $DEF$ y sea $S$ el punto de corte de la bisectriz exterior del ángulo $\angle BAC$ c... | [
"Sea $X$ el segundo punto de corte de $AR$ con $\\Gamma$. Por comodidad, pongamos $\\beta = \\angle ABC$ y $\\gamma = \\angle BCA$. Se tiene entonces que\n$$\n\\frac{BX}{XC} = \\frac{\\sin \\angle BAX}{\\sin \\angle CAX} = \\frac{FR}{RE} = \\frac{FD \\sin(\\gamma/2)}{DE \\sin(\\beta/2)} = \\frac{BD \\sin \\beta \\s... | Spain | LVIII Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate M... | Spanish | proof only | null | |
0jd3 | Problem:
Jarris the triangle is playing in the $(x, y)$ plane. Let his maximum $y$ coordinate be $k$. Given that he has side lengths $6$, $8$, and $10$ and that no part of him is below the $x$-axis, find the minimum possible value of $k$. | [
"Solution:\n\nBy playing around, we find that Jarris should have his hypotenuse flat on the $x$-axis. The desired minimum value of $k$ is then the length of the altitude to the hypotenuse. Thus, by computing the area of the triangle in two ways, $\\frac{1}{2} \\cdot 10 \\cdot k = \\frac{1}{2} \\cdot 6 \\cdot 8$ and... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 24/5 | |
0dzy | Maja can write positive integers onto a blackboard according to two rules. For every number $n$ that is already on the board, she can write down $3n+13$. Whenever one of the numbers on the board is a perfect square, she can also write down its square root.
a. Using only this two rules, can she get the number $55$ assu... | [
"a. With $256$ written on the board, Maja can get $55$ by taking the square root of $256$, obtaining $16$, and then writing down $3 \\cdot 16 + 13 = 61$ and $3 \\cdot 61 + 13 = 196$. Now, she can write down the square root of $196$, which is $14$, and finally she can get $55$ as $3 \\cdot 14 + 13 = 55$. (Note: ther... | Slovenia | Slovenija 2008 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | a: yes; b: no | |
01m6 | Find all functions $f : \mathbb{R} \to \mathbb{R}$, $g : \mathbb{R} \to \mathbb{R}$ satisfying the following equality $f(f(x+y)) = x f(y) + g(x)$ for all real $x$ and $y$. (I. Gorodnin) | [
"Answer: $f(x) = a$, $g(x) = a - a x$ for all $a \\in \\mathbb{R}$.\n\nLet $a = f(0)$, $b = g(0)$. Note that due to the symmetry the given equation implies\n$$\nx f(y) + g(x) = y f(x) + g(y). \\tag{1}\n$$\nSet $y = 0$ in (1), then we have $g(x) = b - a x$. Now set $y = 1$ in (1), then $f(x) = (f(1) - a)x + a = a + ... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = a, g(x) = a - a x for any real a | |
05hv | Problem:
Soit $ABC$ un triangle isocèle en $A$, $D$ le pied de la hauteur issue de $A$ et $M$ un point intérieur à $ADC$ tel que $\widehat{AMB}$ est obtus et $\widehat{DBM} + \widehat{DAM} = \widehat{MCB}$. Les droites $(CM)$ et $(AD)$ se coupent en $P$, et les droites $(BM)$ et $(AD)$ se coupent en $Q$. Soit $S$ un p... | [
"Solution:\n\n\n\nNotons $\\alpha = \\widehat{BAD}$, $\\beta = \\widehat{CBM}$, $\\gamma = \\widehat{MCB}$ et $\\theta = \\widehat{DAM}$. Par hypothèse, on a $\\beta + \\theta = \\gamma$, $\\widehat{SQB} = \\widehat{DPC} = \\frac{\\pi}{2} - \\gamma$ et $\\widehat{MRQ} = 2\\theta$.\n\nOn a $... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0fll | Problem:
Consideremos un alfabeto de $n$ letras, con el que formaremos palabras. Diremos que una palabra contiene un palíndromo si un trozo de esa palabra, de más de una letra, se lee igual al derecho que al revés. Por ejemplo, la palabra OLIMPIADAS contiene el palíndromo ADA. Siendo $k$ un entero mayor que 2, determi... | [
"Solution:\n\nObservemos que una palabra contiene un palíndromo de longitud impar si y sólo si contiene un palíndromo de longitud $3$. Por tanto, sólo hay que contar las palabras que no contengan un palíndromo de longitud $3$.\n\nPodemos enumerar todas las palabras pedidas, de la siguiente manera: para la primera l... | Spain | XLVII Olimpiada Matemática Española Primera Fase | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | n^2 (n-1)^{k-2} | |
05f5 | Problem:
Soit $A$, $B$, $C$ et $D$ quatre points sur un cercle dans cet ordre. Soit $U$ le point d'intersection des droites $(AB)$ et $(CD)$, et $V$ le point d'intersection des droites $(BC)$ et $(DA)$. Soit $K$ le point d'intersection de la bissectrice issue de $U$ dans le triangle $AUC$ et de la bissectrice issue de... | [
"Solution:\n\nLe point $L$ est le centre du cercle circonscrit au triangle $UKV$. Pour montrer que les points $U$, $L$ et $V$ sont alignés, il suffit de montrer que $\\widehat{ULV} = 180^{\\circ}$. Par le théorème de l'angle au centre, $\\widehat{UKV} = \\frac{1}{2} \\widehat{ULV}$. Il suffit donc de montrer que $\... | France | ENVOI 1 : GÉOMÉTRIE Corrigé | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0l9z | Let $b$ be a positive real number. Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying
$$
f(x+y) = f(x) \cdot 3^{b^y} + f(y)^{-1} + b^x (3^{b^y} + f(y)^{-1} - b^y) \quad \forall x, y \in \mathbb{R}.
$$ | [
"The given equation system is equivalent to the following equation system\n$$\nf(x+y) + b^{x+y} = (f(x)+b^x)3^{b^x+f(y)-1} \\quad \\forall x,y \\in \\mathbb{R} \\quad (1)\n$$\nLet $g(x) = f(x) + b^x$. Then $(1) \\Leftrightarrow g(x+y) = g(x)3^{g(y)-1} \\quad \\forall x, y \\in \\mathbb{R} \\quad (2)$\n\nSubstitute ... | Vietnam | Vijetnam 2007 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | f(x) = -b^x and f(x) = 1 - b^x | |
087b | Problem:
16 coni stradali sono messi in linea retta a distanza di 10 metri uno dall'altro. Si vuole dipingere sulla strada una linea continua che vada dal primo all'ultimo cono. Sapendo che per dipingere 100 metri di linea continua sono necessari 6 litri di vernice, quanti litri di vernice sono necessari per completar... | [
"Solution:\n\nLa risposta è (B). 16 coni stradali individuano 15 segmenti lunghi 10 metri, per un totale di 150 metri di linea, e per dipingerla è necessaria una volta e mezza la vernice rispetto ai 6 litri dei 100 metri, ossia 9 litri."
] | Italy | Olimpiadi di Matematica | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | (B) | |
0hxa | Problem:
Let $k$ be a positive integer, and let $(a_{1}, a_{2}, \ldots, a_{2k})$ and $(b_{1}, b_{2}, \ldots, b_{2k})$ be two sequences of real numbers such that $\frac{1}{2} \leq a_{1} \leq \cdots \leq a_{2k}$ and $\frac{1}{2} \leq b_{1} \leq \cdots \leq b_{2k}$. Let $M$ and $m$ be the maximum and minimum respectively ... | [
"Solution:\nLet $c_{i}=b_{\\sigma(i)}$ where $\\sigma$ is a permutation of the numbers $1$ through $2k$ that will vary. We begin with $\\sigma$ being the numbers $1$ through $2k$ written in order,\n$$\n12 \\cdots 2k,\n$$\nso $c_{i}=b_{i}$ for each $i$. We then switch $1$ with $2$, $1$ with $3$, $1$ with $4$, and so... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
03rq | Let $S = \{1, 2, \dots, 2005\}$. If there is at least one prime number in any subset of $S$ consisting of $n$ pairwise coprime numbers, find the minimum value of $n$. | [
"First we prove $n \\ge 16$. In fact, let\n$$\nA_0 = \\{1, 2^2, 3^2, 5^2, \\dots, 41^2, 43^2\\},\n$$\nwhere the members in $A_0$, other than $1$, are the squares of prime numbers not greater than $43$. Then $A_0 \\subseteq S$, $|A_0| = 15$ and the numbers in $A_0$ are pairwise coprime but $A_0$ contains no prime nu... | China | China Western Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 16 | |
03va | Suppose that a matrix of nonnegative entries,
$$
P = \begin{bmatrix} x_{11} & x_{12} & x_{13} & x_{14} & x_{15} & x_{16} & x_{17} & x_{18} & x_{19} \\ x_{21} & x_{22} & x_{23} & x_{24} & x_{25} & x_{26} & x_{27} & x_{28} & x_{29} \\ x_{31} & x_{32} & x_{33} & x_{34} & x_{35} & x_{36} & x_{37} & x_{38} & x_{39} \end{bma... | [
"Proof of (i). Assume that it is not true. There is a column in $S$ which contains no $u_i$. We may say that $u_i \\neq x_{i2}$, $i = 1, 2, 3$. By property (1), we have $u_i < x_{i2}$, $i = 1, 2, 3$. On the other hand, let $k = 2$ in (i). Then by property (O), there exists $i_0 \\in \\{1, 2, 3\\}$ such that $x_{i_0... | China | China Mathematical Competition (Complementary Test) | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Linear Algebra > Matrices"
] | English | proof only | null | |
0dp0 | The incircle $\omega$ of the triangle $ABC$ touches the side $BC$ at point $K$. Let's draw a circle passing through $B$, $C$ and touching $\omega$ at $S$. Prove that the line $SK$ passes through the center of the excircle touching the side $BC$ of the triangle $ABC$. | [
"The lengths of the sides of $ABC$ denote by $a, b, c$. Let $I$ be the center of the incircle, $r$ its radius, $M$ the midpoint of $BC$. Denote by $r_a$ the radius of the excircle touching $BC$. Let $L$ be the point of tangency of excircle with $BC$. $\\omega_0$ be the circle with center $O$, passing through $B$ an... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"... | English | proof only | null | |
0aql | Problem:
A circle has radius $4$ units, and a point $P$ is situated outside the circle. A line through $P$ intersects the circle at points $A$ and $B$. If $PA = 4$ units and $PB = 6$ units, how far is $P$ from the center of the circle? | [] | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | proof and answer | 2√10 | |
089z | Problem:
Sia $x$ il numero di zeri con cui termina $2000!$ quando è scritto in base $5$, e $y$ il numero di zeri con cui termina $2013!$ quando è scritto in base $10$. Calcolare $x-y$. (Ricordiamo che il numero $n!$, per $n$ intero positivo, è il prodotto di tutti gli interi positivi minori o uguali a $n$.)
(A) $-2$
... | [
"Solution:\n\nLa risposta è (A). Sappiamo che $y$ è il massimo esponente tale che $10^{y}$ divida $2013!$. Dunque $y$ è il minimo tra il numero di fattori $2$ e fattori $5$ che intervengono nella fattorizzazione di $2013!$. Siccome il numero di fattori due è chiaramente più grande del numero dei fattori $5$, $y$ è ... | Italy | Progetto Olimpiadi della Matematica | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | A | |
0f4t | Problem:
Let $\{a\}$ denote the difference between $a$ and the nearest integer. For example $\{3.8\} = 0.2$, $\{-5.4\} = 0.4$. Show that
$$
\{a\}\{2a\}\{3a\}\ldots\{na\} \geq \frac{\{a\}^n}{2^n}
$$ | [] | Soviet Union | 16th ASU | [
"Number Theory > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
04ka | Determine the digits $a$ and $b$ if the number $\overline{a2017b}$ is divisible by $72$. | [] | Croatia | Mathematical competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | final answer only | a=2, b=6 | |
0hmw | Problem:
With two properly chosen weights and a balance scale, it is possible to determine the weight of an unknown object known to weigh an integer number of pounds from $1$ to $n$. Find the largest possible value of $n$.
Remark. The balance scale tells whether the weights placed on each side are equal and, if not, ... | [
"Solution:\n\nLet $a$ and $b$ be the known weights. The balance scale allows one to compare the unknown weight with four known weights: $a$, $b$, $a+b$, and $a-b$ (the last of these is gotten by balancing $x+b$ on one side with $a$ on the other). After the comparisons are done, there are at most $4$ values that $x$... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 9 | |
0ee2 | Problem:
Kvadratni enačbi $x^{2} + a x + 2 = 0$ in $x^{2} + 2 x + a = 0$, pri čemer je $a$ realno število, imata realne rešitve. Vsota kvadratov rešitev prve enačbe je enaka vsoti kvadratov rešitev druge enačbe. Koliko je $a$?
(A) $-4$
(B) $-2$
(C) $0$
(D) $4$
(E) Nič od naštetega. | [
"Solution:\n\nEnačba $x^{2} + a x + 2 = 0$ ima diskriminanto $D = a^{2} - 8 \\geq 0$ in rešitvi $x_{1,2} = \\frac{-a \\pm \\sqrt{a^{2} - 8}}{2}$, za kateri velja $x_{1}^{2} + x_{2}^{2} = a^{2} - 4$.\n\nEnačba $x^{2} + 2 x + a = 0$ ima diskriminanto $D = 4 - 4a \\geq 0$ in rešitvi $x_{1,2} = \\frac{-2 \\pm \\sqrt{4 ... | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | (A) | |
0abj | Which digit has to be put in the place of the star in the number $2008*$ in order to be divisible with $6$? | [
"In order $2008*$ to be divisible with $6$ it has to be divisible with $2$ and $3$. In order to be divisible with $2$ it has to end on $0$, $2$, $4$, $6$ or $8$. From the numbers $20080$, $20082$, $20084$, $20086$ and $20088$ only $20082$ and $20088$ are divisible with $3$. So on the place of the $*$ we can put $2$... | North Macedonia | Macedonian Mathematical Competitions | [
"Number Theory > Divisibility / Factorization"
] | null | final answer only | 2 or 8 | |
0a42 | Determine all positive integers $n$ for which all prime factors of $2^n - 1$ are at most 7. | [
"We note that $2 \\nmid 2^n - 1$ for all $n > 1$, so we are looking for all $n$ such that 3, 5 and 7 are the only divisors, or $2^n - 1 = 3^a 5^b 7^c$. The answers are $n = 1, 2, 3, 4, 6$, which we easily check with outcomes 1, 3, 7, $3 \\cdot 5$, $3^2 \\cdot 7$.\n\nWe compute $2^2 \\equiv 1 \\pmod{3}$, $2^4 \\equi... | Netherlands | BxMO/EGMO Team Selection Test | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | n = 1, 2, 3, 4, 6 | |
0evy | Problem:
Given a rectangle $ABCD$ with $AC$ length $e$ and four circles centers $A$, $B$, $C$, $D$ and radii $a$, $b$, $c$, $d$ respectively, satisfying $a + c = b + d < e$. Prove you can inscribe a circle inside the quadrilateral whose sides are the two outer common tangents to the circles center $A$ and $C$, and the ... | [
"Solution:\nLet $O$ be the center of the rectangle. Let $r = (a + c)/2 = (b + d)/2$. The required circle has center $O$, radius $r$. Let an outer common tangent touch the circle center $A$ at $W$, and the circle center $C$ at $X$. Let $P$ be the midpoint of $WX$, then $OP$ is parallel to $AW$ and $CX$ and has lengt... | Soviet Union | 1st ASU | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0ir0 | Problem:
Let $(x, y)$ be a pair of real numbers satisfying
$$
56 x + 33 y = \frac{-y}{x^{2} + y^{2}}, \quad \text{ and } \quad 33 x - 56 y = \frac{x}{x^{2} + y^{2}}
$$
Determine the value of $|x| + |y|$. | [
"Solution:\nAnswer: $\\frac{11}{65}$ Observe that\n$$\n\\frac{1}{x + y i} = \\frac{x - y i}{x^{2} + y^{2}} = 33 x - 56 y + (56 x + 33 y) i = (33 + 56 i)(x + y i)\n$$\nSo\n$$\n(x + y i)^{2} = \\frac{1}{33 + 56 i} = \\frac{1}{(7 + 4 i)^{2}} = \\left(\\frac{7 - 4 i}{65}\\right)^{2}\n$$\nIt follows that $(x, y) = \\pm\... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | 11/65 | |
064r | Let $a$ be positive real number such that $a^3 = 6(a+1)$. Prove that the equation $x^2 + ax + a^2 - 6 = 0$ has no real solution. | [
"In order to have $x^2 + ax + a^2 - 6 \\neq 0$, for all $x \\in \\mathbb{R}$, it is enough the discriminant $\\Delta = 3(8-a^2)$ to be less than $0$, that is $\\Delta = 3(8-a^2) < 0$.\n\nIn fact, if we suppose that $\\Delta = 3(8-a^2) \\geq 0$, then\n$$\na^2 \\leq 8 \\Rightarrow a \\leq 2\\sqrt{2} \\Rightarrow \\fr... | Greece | 11th Junior Balkan Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0goj | Prove that
$$
1 \le \frac{(x+y)(x^3+y^3)}{(x^2+y^2)^2} \le \frac{9}{8}
$$
for all positive real numbers $x$ and $y$. | [
"The Cauchy-Schwarz inequality implies that $(x+y)(x^3+y^3) \\ge (x^2+y^2)^2$. Therefore,\n$$\n1 \\le \\frac{(x+y)(x^3+y^3)}{(x^2+y^2)^2}.\n$$\nAs $0 \\le ((x-y)^2 - 2xy)^2$, we have\n$$\n4xy(x-y)^2 \\le (x-y)^4 + 4x^2y^2.\n$$\n$$\n\\Rightarrow 8x^3y + 8y^3x \\le x^4 + 18x^2y^2 + y^4. \\Rightarrow 8x^3y + 8y^3x + 8... | Turkey | 16th Junior Turkish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0fji | Problem:
Probar que para cualquier primo $p$ distinto de $2$ y $5$ existe un múltiplo de $p$ cuyas cifras son todas nueves. Por ejemplo, si $p=13$, es $999999=13 \cdot 76923$. | [
"Solution:\n\nSea $a_{i}$ el número compuesto por $i$ nueves $a_{i}=\\overbrace{99 \\cdots 9}^{i}$.\n\nSupongamos que existe $p$ tal que $p \\nmid a_{i}$ $\\forall i \\in \\mathbb{N}$, para probar por contradicción el enunciado. Considérense en dicho caso los números $\\{a_{1}, a_{2}, \\ldots, a_{p}\\}$. En este co... | Spain | Olimpiada Matemática Española | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0di3 | In each square of a garden shaped like $2022 \times 2022$ board, there is initially a tree of height $0$. A gardener and a lumberjack alternate turns playing the following game, with the gardener taking the first turn:
* The gardener chooses a square in the garden. Each tree on that square and all the surrounding squa... | [
"The answer is $K = 5 \\cdot \\frac{2022^2}{9} = 2271380$. In general, for a $3N \\times 3N$ board, $K = 5N^2$.\n\nWe solve the problem for a general $3N \\times 3N$ board. First, we prove that the lumberjack has a strategy to ensure there are never more than $5N^2$ majestic trees. Giving the squares of the board c... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 2271380 | |
07g1 | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that for all real numbers $x$ and $y$
$$
f(y - f(x)) = f(x) - 2x + f(f(y)).
$$ | [
"1. Plugging $y = x + f(x)$ to obtain $f(f(x + f(x))) = 2x$. Hence, the function is *surjective* and there is a real number $r$ such that $f(r) = 0$. Plug $x = r$ to obtain\n$$\nf(y) = -2r + f(f(y)).\n$$\nFor each real number $z$ there is a real number $y$ such that $f(y) = z$, therefore\n$$z = -2r + f(z). \\text{ ... | Iran | 38th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = x | |
0aaz | Stefan, Filip and Nikola went on an excursion. They took with them $222$ denars. Stefan spent $\frac{1}{3}$ of his money, Filip spent $\frac{1}{5}$ of his and Nikola spent $\frac{7}{15}$ of his. On the end of the excursion they had equal amount of money. How much money did each of them take for the excursion? | [
"Let Stefan took with him $x$ denars, Filip $y$ and Nikola $z$. We have that $x + y + z = 222$.\n\nOn the end of the excursion Stefan has left $\\frac{2}{3}x$, Filip $\\frac{4}{5}y$ and Nikola $\\frac{8}{15}z$ denars.\n\nFrom the condition in the problem we have $\\frac{2}{3}x = \\frac{4}{5}y = \\frac{8}{15}z$, so ... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | Stefan 72 denars, Filip 60 denars, Nikola 90 denars | |
06q6 | Let $k$ and $n$ be integers with $0 \leq k \leq n-2$. Consider a set $L$ of $n$ lines in the plane such that no two of them are parallel and no three have a common point. Denote by $I$ the set of intersection points of lines in $L$. Let $O$ be a point in the plane not lying on any line of $L$.
A point $X \in I$ is colo... | [
"There are at least $\\frac{1}{2}(k+1)(k+2)$ points in the intersection set $I$ in view of the condition $n \\geq k+2$.\nFor each point $P \\in I$, define its order as the number of lines that intersect the open line segment $O P$. By definition, $P$ is red if its order is at most $k$. Note that there is always at ... | IMO | 49th International Mathematical Olympiad Spain | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0b31 | Problem:
How many ordered pairs of positive integers $(x, y)$ satisfy $20x + 21y = 2021$?
(a) 4
(b) 5
(c) 6
(d) infinitely many | [] | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | MCQ | b | |
0c7q | Let $APD$ be an acute-angled triangle and let $B \in (AP)$, $C \in (PD)$ be two points. The diagonals of the quadrilateral $ABCD$ meet at the point $Q$. Denote by $H_1$ and $H_2$ the orthocenters of the triangles $APD$ and $BPC$ respectively. The circumcircles of the triangles $ABQ$ and $CDQ$ meet again at the point $X... | [
"Let $AA'$, $DD'$ be altitudes in the triangle $\\triangle APD$ and let $BB'$, $CC'$ be altitudes in the triangle $\\triangle BPC$.\nWe have $\\angle XAC = \\angle XBD$ and $\\angle XCA = \\angle XDB$, and thus $\\triangle XAC \\sim \\triangle XBD$. We prove that the line $H_1H_2$ passes through the point $X$ if an... | Romania | The DANUBE Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous ... | English | proof only | null | |
05uc | Problem:
Pierre et Clara jouent au jeu suivant. Tout d'abord, Clara choisit un entier $c$. Puis Pierre choisit un nombre premier $p \geqslant c$ et écrit deux entiers $a$ et $b$ au tableau. Clara se permet alors d'effectuer les opérations suivantes : elle choisit un des deux nombres écrits au tableau, disons $n$, l'ef... | [
"Solution:\n\nSoit $p$ un nombre premier, arbitrairement grand, que Pierre est susceptible de choisir, et soit $f: \\mathbb{Z} / p \\mathbb{Z} \\to \\mathbb{Z} / p \\mathbb{Z}$ la fonction définie par $f(x) = x^{2} + 1$.\n\nLorsque Clara remplace un entier $n$ par un entier $m$, l'une des deux égalités $n = f(m)$ o... | France | Préparation Olympique Française de Mathématiques - Test du 14 et du 21 Février 2021 | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Discrete Mathematics > Graph Theory"
... | null | proof only | null | |
036x | Problem:
Let $A A'$, $B B'$ and $C C'$ be the angular bisectors of a triangle $A B C$ with incenter $I$. The segments $C I$ and $A' B'$ meet at $D$ and the midpoints of the segments $A I$ and $B I$ are denoted by $M$ and $N$, respectively.
a) If $a = B C$, $b = A C$ and $c = A B$, find the ratio $C D : D I$.
b) If $... | [
"Solution:\n\na)\nWe have $C A' = \\frac{a b}{b + c}$ and $\\frac{A I}{I A'} = \\frac{A C}{C A'} = \\frac{b + c}{a}$. By the Menelaus theorem for $\\triangle A I C$ and the line $B' A'$ we have\n$$\n\\frac{C D}{D I} \\cdot \\frac{I A'}{A' A} \\cdot \\frac{A B'}{B' C} = 1 \\Rightarrow \\frac{C D}{D I} = \\frac{a + b... | Bulgaria | Spring Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Misc... | null | proof and answer | CD:DI = (a + b + c) / c; D is the incenter of triangle KLC | |
07yn | Problem:
Due maghi si esibiscono nel seguente numero. All'inizio il primo mago rinchiude il secondo mago in una cabina dove non possa né vedere né sentire nulla. Per iniziare il gioco, il primo mago invita Daniele, un membro del pubblico, a porre su ogni casella di una scacchiera $n \times n$, a propria discrezione, u... | [
"Solution:\n\nRisolveremo un problema leggermente più generale: supponiamo che la scacchiera possa avere un numero qualunque $N$ di caselle, non necessariamente un quadrato. Mostreremo che i maghi hanno una strategia se e solo se $N$ è potenza di due. Nel caso particolare $N = n^{2}$, segue che $n$ deve essere pote... | Italy | null | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | Exactly when n is a power of two. | |
0i3e | Problem:
Equilateral triangle $A B C$ with side length $1$ is drawn. A square is drawn such that its vertex at $A$ is opposite to its vertex at the midpoint of $B C$. Find the area enclosed within the intersection of the insides of the triangle and square. Hint: $\sin 75 = \frac{\sqrt{2}(\sqrt{3}+1)}{4}$. | [
"Solution:\n\nLet $D$ be the midpoint of $B C$, $F \\neq A$ be the point of intersection of the square and triangle lying on $A C$, $b$ be the length of $F C$, $x$ be the side length of the triangle, and $y$ be the length of $A D$.\n\nBy the law of sines on triangle $C D F$, we have\n$$\n\\frac{2 \\sin 75}{x} = \\f... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof and answer | 3/(4(√3+1)) | |
04ap | At a party among each group of four people there are three that all know each other or three that don't know each other. Acquaintances are mutual. Prove that all the participants at that party can be separated into two rooms so that in one room everybody knows everyone, and in the other room nobody knows anyone. | [
"People and acquaintances can be represented by a complete graph in two colours, where the vertices represent people, blue edges connect people who know each other, and red edges connect people who don't know each other. We call a set with $m$ vertices a blue (red) $m$-gon if all of the vertices are connected by a ... | Croatia | CroatianCompetitions2011 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0cfj | Prove that there are no nonnegative real numbers $x$ such that $[nx]$ is an odd natural number for any positive integer $n$. | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
0ao2 | Problem:
Find the equation of a circle, in the form $(x-h)^2 + (y-k)^2 = r^2$, inscribed in a triangle whose vertex are located at the points $(-2,1)$, $(2,5)$, $(5,2)$. | [
"Solution:\nLet the vertices of the triangle be $A(-2,1)$, $B(2,5)$, $C(5,2)$.\n\nLet the equation of the inscribed circle be $(x-h)^2 + (y-k)^2 = r^2$.\n\nThe center $(h,k)$ of the incircle is the intersection point of the angle bisectors, i.e., the incenter.\n\nLet $a$, $b$, $c$ be the lengths of the sides opposi... | Philippines | AREA STAGE | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | (x-2)^2 + (y-3)^2 = 2 | |
0kzx | How many different remainders can result when the 100th power of an integer is divided by 125?
(A) 1 (B) 2 (C) 5 (D) 25 (E) 125 | [
"Write $N = 5k + r$ for $r = 0, 1, 2, 3$, or $4$. If $r = 0$, then $N = 5k$ and $N^{100}$ is divisible by $125$, so the remainder is $0$. If $r = 1, 2, 3$, or $4$, then $N^2 = 25k^2 + 10rk + r^2 = 5m + 1$ for some integer $m$. Now use the Binomial Theorem:\n$$\n\\begin{aligned} N^{100} = (N^2)^{50} = (5m \\pm 1)^{5... | United States | AMC 10 B | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | B | |
0gai | 設 $ABCD$ 為凸四邊形, 點 $P, Q, R, S$ 分別在邊 $AB, BC, CD, DA$ 邊上。直線 $PR$ 與 $QS$ 交於 $O$ 點。設四個四邊形 $APOS, BQOP, CROQ, DSOR$ 都有內切圓。證明：直線 $AC, PQ, RS$ 共點或是兩兩互相平行。
Let $ABCD$ be a convex quadrilateral, and let $P, Q, R, S$ be points on the sides of $AB, BC, CD, and DA$, respectively. Let the line segments $PR$ and $QS$ meet at $O$. ... | [
"對於 $\\triangle ABC$ 與直線 $PQ$、以及 $\\triangle ACD$ 與直線 $RS$ 使用孟氏定理 (Melelaus' theorem)，可知 $AC$ 分別與 $PQ$、$RS$ 有相同交點 (此交點可能在無窮遠處) 的充要條件是：\n$$\n\\frac{AP}{PB} \\cdot \\frac{BQ}{QC} \\cdot \\frac{CR}{RD} \\cdot \\frac{DS}{SA} = 1. \\qquad (1)\n$$\n所以我們把目標放在證明 (1) 式。\n\n先來證下面的結果。\n\nLemma 1. 設四邊形 $EFGH$ 有內切圓, $M$ 為內切圓圓心。... | Taiwan | 二〇一六數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geome... | null | proof only | null | |
09qy | Problem:
Zij $ABC$ een driehoek en zij $X$ een punt binnen de driehoek. De lijnen $XA$, $XB$ en $XC$ snijden de omgeschreven cirkel van $\triangle ABC$ nogmaals in respectievelijk $P, Q$ en $R$. Zij $U$ een punt op de halfrechte $XP$ (d.w.z. op de lijn $XP$ zodat $P$ en $U$ aan dezelfde kant van $X$ liggen). De lijn d... | [
"Solution:\n\nOplossing I. Omdat $AB$ en $UV$ evenwijdig zijn, geldt $\\triangle ABX \\sim \\triangle UVX$ (hh). Zo ook is $\\triangle ACX \\sim \\triangle UWX$. Uit deze gelijkvormigheden verkrijgen we de verhoudingen\n$$\n\\frac{|XA|}{|XB|} = \\frac{|XU|}{|XV|} \\quad \\text{en} \\quad \\frac{|XA|}{|XC|} = \\frac... | Netherlands | Dutch TST | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0ige | Problem:
For how many integers $n$ between $1$ and $2005$, inclusive, is $2 \cdot 6 \cdot 10 \cdots (4n-2)$ divisible by $n!$? | [
"Solution:\nNote that\n$$\n\\begin{aligned}\n2 \\cdot 6 \\cdot 10 \\cdots (4n-2) &= 2^n \\cdot 1 \\cdot 3 \\cdot 5 \\cdots (2n-1) \\\\\n&= 2^n \\cdot \\frac{1 \\cdot 2 \\cdot 3 \\cdots 2n}{2 \\cdot 4 \\cdot 6 \\cdots 2n} \\\\\n&= \\frac{1 \\cdot 2 \\cdot 3 \\cdots 2n}{1 \\cdot 2 \\cdot 3 \\cdots n}\n\\end{aligned}\... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 2005 | |
0ib9 | Problem:
Let $n$ be a positive odd integer. Prove that
$$
\left\lfloor\log _{2} n\right\rfloor+\left\lfloor\log _{2}(n / 3)\right\rfloor+\left\lfloor\log _{2}(n / 5)\right\rfloor+\left\lfloor\log _{2}(n / 7)\right\rfloor+\cdots+\left\lfloor\log _{2}(n / n)\right\rfloor=(n-1) / 2 .
$$ | [
"Solution:\n\nNote that $\\left\\lfloor\\log _{2} k\\right\\rfloor$ is the cardinality of the set $\\left\\{2,4,8, \\ldots, 2^{\\left\\lfloor\\log _{2} k\\right\\rfloor}\\right\\}$, i.e., the number of powers of $2$ that are even and are at most $k$. Then $\\left\\lfloor\\log _{2}(n / k)\\right\\rfloor$ is the numb... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0aiz | A domino is a $2 \times 1$ or $1 \times 2$ tile. Determine in how many exactly $n^2$ dominoes can be placed without overlapping on a $2n \times 2n$ chessboard so that every $2 \times 2$ square contains at least two uncovered unit squares which lie in the same row or column. | [
"The answer is $\\binom{2n}{n}^2$.\n\nDivide the chessboard into $2 \\times 2$ squares. There are exactly $n^2$ such squares on the chessboard. Each of these squares can have at most two unit squares covered by the dominos. As the dominos cover exactly $2n^2$ squares, each of them must have exactly two unit squares... | North Macedonia | Girls European Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | (2n choose n)^2 | |
0bde | Let $\gamma$ be a circle, and let $P$ be a point in its plane, not situated on $\gamma$. Two variable lines $\ell$ and $\ell'$ through $P$ meet $\gamma$ at $X$ and $Y$, and $X'$ and $Y'$, respectively. Show that the line through the centres of the circles $PXY'$ and $PX'Y$ passes through a fixed point. | [
"Let the circles $PXY'$ and $PX'Y$ meet again at $Q$. A suitable inversion of pole $P$ sends the circles $PXY'$ and $PX'Y$ onto the lines $XY'$ and $X'Y$, respectively, while leaving $\\gamma$ invariant. The image of $Q$ under this inversion is the point $R$ where the lines $XY'$ and $X'Y$ meet. Upon inversion, the... | Romania | 64th NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0d3b | A positive proper divisor is a positive divisor of a number, excluding itself. For positive integers $n \geq 2$, let $f(n)$ denote the number that is one more than the largest proper divisor of $n$. Determine all positive integers $n$ such that $f(f(n))=2$. | [
"Let $n \\geq 2$ such that $f(f(n))=2$. The largest proper divisor of $f(n)$ is 1 if and only if $f(n)$ is a prime number.\n\nIf $f(n)=2$, this is equivalent to $n$ being a prime number.\n\nIf $f(n)=p \\geq 3$, then $p-1$, the largest proper divisor of $n$ is an even number. This means that $n$ is even and therefor... | Saudi Arabia | SAMC | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English, Arabic | proof and answer | All primes n, and all n of the form n = 2(p − 1) where p is a prime. | |
0egc | Problem:
Iz 17 vžigalic sestavimo mrežo (glej sliko). Najmanj koliko vžigalic moramo odstraniti, da preostale vžigalice ne bodo tvorile nobenega kvadrata?
 | [
"Solution:\n\nVžigalice na sliki tvorijo 6 majhnih kvadratov s stranicami dolgimi 1 vžigalico ter 2 večja kvadrata s stranicami dolgimi 2 vžigalici. Osenčimo nekatere majhne kvadrate in oštevilčimo nekatere vžigalice, kot to prikazuje slika.\n\n\n\nKer moramo iz vsakega osenčenega kvadrata ... | Slovenia | Slovenian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 4 | |
03nm | Problem:
There are $19,998$ people on a social media platform, where any pair of them may or may not be friends. For any group of $9,999$ people, there are at least $9,999$ pairs of them that are friends. What is the least number of friendships, that is, the least number of pairs of people that are friends, that must ... | [
"Solution:\n\nIt is $5 \\cdot 9999 = 49995$. One possible construction is as follows: have the $19,998$ people form $3,333$ groups of $6$ people, and within each group every pair of people are friends. Now, for any group of $9,999$ people, say that there are $x_{1}, x_{2}, \\ldots, x_{3333}$ people in each of the $... | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 49995 | |
0bnt | Consider a triangle $ABC$, such that $\angle B = 90^\circ$. Denote by $I$ the in-center and let $F$, $D$ and $E$ be the points where the incircle touches sides $[AB]$, $[BC]$, and $[AC]$ respectively. If $CI \cap EF = \{M\}$ and $DM \cap AB = \{N\}$, show that:
a) $AI = ND$;
b) $FM = \frac{EI \cdot EM}{EC}$. | [
"a. Triangle $AFE$ is isosceles with $AE = AF$, and $AI \\perp FE$, hence $\\angle AEF = 90^\\circ - \\angle A/2$. In the same way from the isosceles triangle $CDE$ we get $\\angle DEC = 90^\\circ - \\angle C/2$. As a consequence $\\overline{MED} = 180^\\circ - \\overline{AEF} - \\overline{DEC} = 180^\\circ - (180^... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0iyb | Fix a prime number $p > 5$. Let $a, b, c$ be integers no two of which have their difference divisible by $p$. Let $i, j, k$ be nonnegative integers such that $i + j + k$ is divisible by $p-1$. Suppose that for all integers $x$, the quantity
$$
(x-a)(x-b)(x-c)[(x-a)^i(x-b)^j(x-c)^k - 1]
$$
is divisible by $p$. Prove tha... | [
"We first prove that $k$ is congruent to one of $1, 0, -1$ modulo $(p-1)$. We rephrase the hypothesis in terms of modular arithmetic: If $x \\not\\equiv a, b, c \\pmod p$, then\n$$\n(x - a)^i (x - b)^j \\equiv (x - c)^{-k} \\pmod{p}.\n$$\nBy Fermat's little theorem, there is no harm in shifting $i, j, k$ by multipl... | United States | Team Selection Test 2009 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Abstract Algebra > Field Theory"
] | null | proof only | null | |
0jhg | Problem:
A right triangle has area $5$ and a hypotenuse of length $5$. Find its perimeter. | [
"Solution:\n\nLet $x$ and $y$ denote the legs of the triangle. Then $\\frac{1}{2} x y = 5$, so $x y = 10$. Also, by the Pythagorean theorem, $x^2 + y^2 = 5^2 = 25$.\n\nWe want to find the perimeter: $x + y + 5$.\n\nNote that $(x + y)^2 = x^2 + 2 x y + y^2 = (x^2 + y^2) + 2 x y = 25 + 20 = 45$, so $x + y = \\sqrt{45... | United States | HMMT November 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 5 + 3√5 | |
0ai2 | 2014 lines are given in a plane, arranged in three groups of pairwise parallel lines. What is the greatest possible number of triangles formed by the lines (each side from such a triangle lies on one of the lines)?
Во рамнина, 2014 прави се распоредени во три групи заемно паралелни прави. Кој е најголемиот можен број ... | [
"Let $a \\ge b \\ge c$ be the numbers of the lines in the three groups for which the greatest possible number of triangles is attained. Then $a+b+c=2014$, and the greatest possible number of triangles is $abc$ (when no three lines have a common point). We will show that $a \\le c+1$. Let us suppose the opposite, i.... | North Macedonia | Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 302561952 | |
06e2 | A positive integer $n$ is said to be *good* if there exist positive integers $a$ and $b$ such that $a+b=n$ and $ab$ is divisible by $n$. For instance, $9$ is good since $3+6=9$ and $3 \times 6$ is divisible by $9$. Furthermore, $n$ is said to be *very good* if we can choose $a$ and $b$ to be distinct. Hence $9$ is very... | [
"a. Suppose there exist $a, b \\in \\mathbb{Z}^+$ such that $a+b = 2003$ and $2003 \\mid ab$. Since $2003$ is a prime, we must have $2003 \\mid a$ or $2003 \\mid b$. WLOG assume $2003 \\mid a$. Then we have $a+b \\ge 2003+1 > 2003$, which is a contradiction. Therefore, $2003$ is not good.\n\nb. No. Suppose there ex... | Hong Kong | IMO HK TST | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) 2003 is not good. b) 2002 is not good. c) Exactly the positive integers of the form 4 times a product of distinct odd primes or 8 times a product of distinct odd primes (including the cases 4 and 8) are good but not very good. | |
0i94 | Problem:
A certain lottery has tickets labeled with the numbers $1,2,3, \ldots, 1000$. The lottery is run as follows: First, a ticket is drawn at random. If the number on the ticket is odd, the drawing ends; if it is even, another ticket is randomly drawn (without replacement). If this new ticket has an odd number, the... | [
"Solution:\n$1 / 501$\nNotice that the outcome is the same as if the lottery instead draws all the tickets, in random order, and awards a prize to the holder of the odd ticket drawn earliest and each even ticket drawn before it. Thus, the probability of your winning is the probability that, in a random ordering of ... | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Permutations",
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 1/501 | |
07ht | If $d(n)$ is the number of positive divisors of $n$, prove that there exists a natural number $n$ such that:
$$
\forall i \in \mathbb{N},\ i \le 1402 : \frac{d(n)}{d(n \pm i)} > 1401.
$$ | [
"Let $n = p_1^{\\alpha_1} \\cdots p_s^{\\alpha_s}$. We define $\\Omega(n) = \\alpha_1 + \\cdots + \\alpha_s$.\n**Lemma 1.** $d(n) \\le 2^{\\Omega(n)}$\n*Proof.*\n$$\n\\forall 1 \\le i \\le s : 2^{\\alpha_i} \\ge \\alpha_i + 1 \\implies 2^{\\Omega(n)} \\ge d(n)\n$$\n\nWe denote prime numbers less than $1402$ as $q_1... | Iran | 40th Iranian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0ipd | Problem:
Let $ABC$ be an equilateral triangle with side length $2$, and let $\Gamma$ be a circle with radius $\frac{1}{2}$ centered at the center of the equilateral triangle. Determine the length of the shortest path that starts somewhere on $\Gamma$, visits all three sides of $ABC$, and ends somewhere on $\Gamma$ (no... | [] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geome... | null | proof and answer | sqrt(4/3) - 1 | |
02rk | Determine if there exist positive integers $n, a_1, a_2, \dots, a_{2012}$ all greater than 1, such that
$$
n^2 = a_1^2 + a_2^3 + a_3^5 + \dots + a_i^{p_i} + \dots + a_{2012}^{p_{2012}},
$$
where $p_i$ is the *i*-th prime. | [
"Just pick any odd $a_2$, any even $a_3, \\dots, a_{2012}$. Then $a_2^3 + \\dots + a_{2012}^{p_{2012}} = 2N+1$ is odd. Then pick $a_1 = N$ and $n = N+1$, and $n^2 - a_1^2 = (N+1)^2 - N^2 = 2N+1$."
] | Brazil | Brazilian Math Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | Yes | |
0dsb | Find all functions $f: \mathbb{Z}^+ \to \mathbb{Z}^+$ such that $f(k+1) > f(f(k))$ for all $k \ge 1$, where $\mathbb{Z}^+$ is the set of positive integers. | [
"We shall show that $f(k) = k$ for all $k$. It suffices to show that $f$ is a strictly increasing function for if $f$ is a strictly increasing function, then for an arbitrary $k$,\n$$\nf(k+1) > f(f(k)) \\Rightarrow k+1 > f(k) \\Rightarrow f(k) = k.\n$$\nConsider the set of integers $\\{f(1), f(2), \\dots\\}$. By th... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(n) = n | |
00pu | Some squares of an $n \times n$ chessboard have been marked ($n \in \mathbb{N}^*$). Prove that if the number of marked squares is at least $n(\sqrt{n} + \frac{1}{2})$, then there exists a rectangle whose vertices are centers of marked squares. | [
"For each $i = 1, \\dots, n$, define $S_i$ as the set of $j$ such that the square $(i, j)$ is marked. Suppose by contradiction that there is no rectangle with vertices in centers of marked squares. Then $|S_i \\cap S_j| \\le 1$, $\\forall i \\ne j$, and $|S_1| + |S_2| + \\dots + |S_n| \\ge n(\\sqrt{n} + \\frac{1}{2... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
023v | Problem:
Um número não divisível por 5 - Determine quais números naturais $n$ entre 2001 e 2007, tornam o número $1^{n} + 2^{n} + 3^{n} + 4^{n}$ não divisível por 5. | [
"Solution:\n\n2004"
] | Brazil | Desafios | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | 2004 | |
0k8m | Problem:
Let $ABC$ be a triangle and $P$ a point inside it. Lines $AP$, $BP$, $CP$ meet the opposite sides at $D$, $E$, $F$. Assume that the three quadrilaterals $PDCE$, $PEAF$, $PFBD$ are all bicentric. Prove that triangle $ABC$ is equilateral.
(A quadrilateral is bicentric if it can be inscribed inside a circle, an... | [
"Solution:\n\nFirst, from just the fact the three quadrilaterals are cyclic, we contend that $P$ is the orthocenter of $\\triangle ABC$. By power of a point, we have $AF \\cdot AB = AP \\cdot AD = AE \\cdot AC$, hence quadrilateral $BFEC$ is cyclic as well.\n$$\n\\measuredangle AEP = \\measuredangle AFP = \\measure... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Pla... | null | proof only | null | |
0kl5 | Problem:
Aerith and Bob play rounds of pool. At some point Bob had won more rounds than Aerith, but now Aerith has won $85\%$ of their rounds. Show that at some point, Aerith had won exactly $75\%$ of their rounds. | [
"Solution:\n\nAt any time, let $a$ and $b$ represent the respective number of rounds won by Aerith and Bob. Now, after $n$ games, $a-3b = n(85\\% - 3 \\cdot 15\\%) > 0$. However, at some point in the past $b$ was greater than $a$, so $a-3b$ was negative. Consider the first subsequent moment when $a-3b$ was non-nega... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
052v | On the board there are numbers $1$, $2$, $3$, $4$, $5$ and $6$. In every step Juku deletes some two numbers $a$ and $b$ on the board and writes $ab + a + b$ on the board instead. He repeats such steps until there is only one number on the board. Find all possibilities what could be the last number on the board. | [
"Since $(a+1)(b+1) = c+1$, where $c$ is a number that would be written on the board instead of $a$ and $b$, the product of the numbers that are greater by $1$ than the numbers on the board does not change in the process. In the beginning the product is $2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7$ or $5040$. Con... | Estonia | Open Contests | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 5039 | |
0d8x | Let $ABCD$ be a cyclic quadrilateral and triangles $ACD$, $BCD$ are acute. Suppose that the lines $AB$ and $CD$ meet at $S$. Denote by $E$ the intersection of $AC$, $BD$. The circles $(ADE)$ and $(BCE)$ meet again at $F$.
1. Prove that $SF \perp EF$.
2. The point $G$ is taken outside of the quadrilateral $ABCD$ such ... | [
"1) First, notice that $\\angle AFB = \\angle AFE + \\angle BFE = \\angle ADB + \\angle ACB = \\angle AOB$, then $A, F, O, B$ are concyclic. Similarly, $D, F, O, C$ are also concyclic.\n\nBy considering three radical axes of three circles $(O)$, $(ABOF)$, $(CDFO)$, we can see that three lines $AB$, $CD$, $FO$ are c... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscrib... | English | proof only | null | |
07gr | A pair of natural numbers $(a, b)$ is called *square-maker* if $ab + 1$ is a perfect square. Find all natural numbers $n$ such that $\{1, 2, 3, \dots, 2n\}$ can be partitioned into square-maker pairs. | [
"The answer is $n = 2t$, for some positive integer $t$. To give an example note that $x(x+2)+1=(x+1)^2$. Assume that $n = 2t$ and consider the partition\n$$\n(1, 3), (2, 4), \\dots, (4t-2, 4t).\n$$\nWe want to prove that it is impossible for $n = 2k+1$. Note that if $a = 4t+2$ and $ab+1 = n^2$, by looking at both s... | Iran | 38th Iranian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n = 2t for positive integers t | |
0fcw | Problem:
Sean $a, b, c$ números reales no nulos y $a \neq b$. Probar que si las ecuaciones $x^{2}+a x+b c=0$ y $x^{2}+b x+c a=0$ tienen una raíz común, entonces las restantes raíces verifican la ecuación $x^{2}+c x+a b=0$. | [
"Solution:\n\nSean $x_{1}, x_{2}$ las raíces de la ecuación $x^{2}+a x+b c=0$ y $x_{1}, x_{3}$ las de $x^{2}+b x+c a=0$.\nLa solución común $x_{1}$ verifica la ecuación $x^{2}+a x+b c-\\left(x^{2}+b x+c a\\right)=0$, de donde resulta $(a-b)_{x_{1}}=(a-b)_{c}$ y $x_{1}=c$.\nSe sigue que $x_{2}=b$ y $x_{3}=a$, con lo... | Spain | null | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof only | null | |
0d4s | Let $ABC$ be an acute triangle with $\angle A < \angle B \leq \angle C$, and $O$ its circumcenter. The perpendicular bisector of side $AB$ intersects side $AC$ at $D$. The perpendicular bisector of side $AC$ intersects side $AB$ at $E$. Express the angles of triangle $DEO$ in terms of the angles of triangle $ABC$. | [
"Let $M$ be the midpoint of $AC$. Because $\\measuredangle COM = \\frac{1}{2} \\measuredangle COA = \\measuredangle CBE$, the quadrilateral $BCOE$ is cyclic.\n\n\n\nLet $N$ be the midpoint of $AB$. Because $\\measuredangle NOB = \\frac{1}{2} \\measuredangle AOB = \\measuredangle DCB$, the q... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English, Arabic | proof and answer | ∠ODE = 90° − ∠ACB, ∠DEO = 90° − ∠CBA, ∠EOD = ∠CBA + ∠ACB | |
0g58 | 給定正偶數 $n > 2$, 已知實數 $x_1, x_2, \dots, x_n$ 滿足
$$
\sum_{1 \le i < j \le n} (x_i - x_j)^2 = 1.
$$
試求 $\sum_{i=1}^{n} (x_i - x_{i+1})^2$ 的最大值, 此處 $x_{n+1} = x_1$. | [
"最大值為 $\\frac{4}{n}$。\n首先在 $x_1 = x_3 = \\cdots = x_{n-1}, x_2 = \\cdots = x_n$ 時可檢驗此值被達成。\n因此只須證明 $(x_1 - x_2)^2 + \\cdots + (x_{n-1} - x_n)^2 + (x_n - x_1)^2 \\le \\frac{4}{n}$ 即可。\n可同時將所有 $x_i$ 增加或減少一個常數, 使得 $x_1 + \\cdots + x_n = 0$, 此時原條件等價於 $n(x_1^2 + \\cdots + x_n^2) = 1$。\n由於對任意實數 $a, b$ 有\n$$\n(a-b)^2 \\le... | Taiwan | 二〇一一數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 4/n | |
019m | Let $0 \le a, b, c \le 1$ be distinct real numbers. Determine the minimum value of
$$
\frac{1}{|a-b|^3} + \frac{1}{|b-c|^3} + \frac{1}{|c-a|^3}.
$$ | [
"The answer is $17$ and is obtained when $a = 1$, $b = \\frac{1}{2}$, $c = 0$ (or some permutation of this).\n\nWithout loss of generality, let $a \\ge b \\ge c$ and $b = \\frac{a+c}{2} + t$ where $-\\frac{a+c}{2} < t < \\frac{a+c}{2}$. Then\n$$\n\\frac{1}{|a-b|^3} + \\frac{1}{|b-c|^3} + \\frac{1}{|c-a|^3} = \\frac... | Baltic Way | Baltic Way 2013 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 17 | |
0dj7 | Find all positive integers $n \ge 2$ for which there exist $n$ real numbers
$$
a_1 < a_2 < \dots < a_n
$$
and a real number $r > 0$ such that the $\frac{n(n-1)}{2}$ differences $a_j - a_i$ for $1 \le i < j \le n$ are equal, in some order, to the numbers
$$
r^1, r^2, \dots, r^{\frac{n(n-1)}{2}}.
$$ | [
"The answer are $n \\in \\{2, 3, 4\\}$. We first show a solution for each $n \\in \\{2, 3, 4\\}$. We will later show the impossibility of finding such a solution for $n \\ge 5$.\n\nFor $n = 2$, take for example $(a_1, a_2) = (1, 3)$ and $r = 2$.\n\nFor $n = 3$, take the root $r > 1$ of $x^3 - x - 1 = 0$ (the golden... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | {2, 3, 4} | |
01uo | Find all $a \in \mathbb{R}_{\ge 0}$ such that $f(a) = 0$ for any function $f: \mathbb{R}_{\ge 0} \to \mathbb{R}_{\ge 0}$ satisfying the equality
$$
f(f(x) + f(y)) = y f(1 + y f(x))
$$
for all nonnegative real numbers $x$ and $y$. | [
"Answer : $a \\in \\{0\\} \\cup [1, \\infty)$.\nLet function $f: \\mathbb{R}_{\\ge 0} \\to \\mathbb{R}_{\\ge 0}$ satisfy the equality\n$$\nf(f(x) + f(y)) = y f(1 + y f(x)) \\tag{1}\n$$\nfor all nonnegative real numbers $x$ and $y$. Set $f(0) = c$ and $f(1) = b$.\nFor $y = 0$ from equality (1) we get\n$$\nf(f(x) + c... | Belarus | Selection and Training Session | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | {0} ∪ [1, ∞) | |
0c5o | Given an integer $k \ge 2$, determine all positive integers $n_1, n_2, \dots, n_k$ satisfying
$$
n_2 \mid 2^{n_1-1}, \quad n_3 \mid 2^{n_2-1}, \quad \dots, \quad n_k \mid 2^{n_{k-1}-1}, \quad n_1 \mid 2^{n_k-1}.
$$ | [
"The required numbers are $n_1 = n_2 = \\dots = n_k = 1$.\n\nFor every integer $m > 1$, let $p(m)$ denote the least prime divisor of $m$. We show that, if $m$ and $\\ell$ are integers greater than $1$, and $m \\mid 2^{\\ell} - 1$, then $p(m) < p(\\ell)$. Since $p(m)$ is odd, $p(m) \\mid 2^{p(m)-1} - 1$, and since $... | Romania | 70th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | n1 = n2 = ... = nk = 1 | |
0cdv | Consider a tetrahedron $ABCD$ and the points $M, N$ on the edges $AC$ and $BD$, respectively. Prove that for any point $P$ of the segment $MN$, $P \neq M$, $P \neq N$, there exists a unique pair of points $(X, Y)$, with $X$ and $Y$ on the edges $AB$ and $CD$, respectively, such that the points $X, P$ and $Y$ are collin... | [
"Since $P \\neq M$ and $P \\neq N$, if $X, Y, P$ are collinear, we have $X \\notin \\{A, B\\}$ and $Y \\notin \\{C, D\\}$. Indeed, if $X = A$, then $XY \\subset (ACD)$ and thus $P \\in (ACD)$, false. The other situations are analogous.\n\n2023 ROMANIAN MATHEMATICAL OLYMPIAD – FINAL ROUND\n\n**Existence:** Since $P ... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null |
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