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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0dba | Let Pascal triangle be an equilateral triangular array of numbers, consisting of $2019$ rows and except for the numbers in the bottom row, each number is equal to the sum of two numbers immediately below it. How many ways to assign each of numbers $a_{0}, a_{1}, \ldots, a_{2018}$ (from left to right) in the bottom row ... | [
"First, by induction, one can show that\n$$\nS = \\binom{n}{0} a_{0} + \\binom{n}{1} a_{1} + \\cdots + \\binom{n}{n} a_{n}\n$$\nif the Pascal triangle consists of $n$ rows.\n\nNote that for any odd prime $p$, we also have:\n\n**Claim 1.** $\\binom{2p}{p} \\equiv 2 \\pmod{p}$.\nIndeed,\n$$\n\\begin{aligned}\n& \\bin... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 2^{2016} | |
09bb | Ангийн салаа тус бүр 5, 7, 9 сурагчтай. Салаадын хооронд зохиогдсон тэмцээнд түрүүлсэн салааг тортоор урамшуулах байв. Гэхдээ тортыг тэмцээнээс өмнө хэсгүүдэд хувааж (хэсгүүд нь хоорондоо заавал тэнцүү байх албагүй) тавих хэрэгтэй ба аль ч салааг түрүүлэхэд тортоо дахин хуваалгүйгээр тэр салааны хүүхдүүдэд яг тэнцүү ху... | [
"**VII-B1.** (Н.Аргилсан) $A = n^4 - 4n^3 + 22n^2 - 36n + 18 = (n^2 - 2n)^2 + 18(n^2 - 2n) + 18$ гэсэн хувиргая. $n^2 - 2n = x$ гэвэл $A = x^2 + 18x + 18 = y^2$ болно ($y \\in \\mathbb{N}$). Эндээс $(x+9)^2 - 63 = y^2$ гэсэн тэгшитгэл үүснэ.\n$$\n\\begin{aligned}\n(x + 9)^2 - y^2 &= 63 \\Rightarrow \\\\\n&\\Rightar... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Mongolian | proof and answer | 19 | |
0ah7 | Let $A$, $B$, $C$ be points lying on a circle $\Gamma$ with the center $O$ and assume that $\angle ABC > 90^\circ$. Let $D$ be the point of intersection of the line $AB$ and the line perpendicular to $AC$ at $C$. Let $l$ be the line through $D$ and perpendicular to $AO$. Let $E$ be the point of intersection of $l$ and ... | [
"Let $l \\cap AO = \\{K\\}$, and $G$ be the other end point of the diameter of $\\Gamma$ through $A$. Then $D$, $C$, $G$ are collinear. Moreover, $E$ is the orthocenter of triangle $ADG$. Therefore $GE \\perp AD$ and $G$, $E$, $B$ are collinear.\n\nAs $\\angle CDF = \\angle GDK = \\angle GAC = \\angle GFC$, $FG$ is... | North Macedonia | 29-th Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
060h | Problem:
Déterminer tous les triplets $(p, q, r)$ de nombres premiers tels que $p+q^{2}=r^{4}$. | [
"Solution:\n\nNotons que l'équation se réécrit $p=(r^{2})^{2}-q^{2}=(r^{2}-q)(r^{2}+q)$. Comme $r^{2}+q$ est strictement positif, et $p$ aussi, $r^{2}-q$ aussi. En particulier, d'après l'équation précédente, on a que $r^{2}-q=1$ et $r^{2}+q=p$.\n\nSi $r$ et $q$ sont impairs, alors $r^{2}-q$ est pair, ce qui est con... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (7,3,2) | |
0gcp | 設三角形 $ABC$ 的垂心為 $H$, 外接圓為 $\Gamma$。取點 $P$ 為 $\Gamma$ 上異於 $A$, $B$, $C$ 的一點, 並令 $M$ 為線段 $HP$ 的中點。分別在直線 $BC$, $CA$, $AB$ 上取點 $D$, $E$, $F$ 使得 $AP \parallel HD$, $BP \parallel HE$, $CP \parallel HF$。證明: $D$, $E$, $F$, $M$ 共線。 | [
"(∠ 代表有向角。)\n\n\n\n令 $A'$, $P'$ 分別為 $A$, $P$ 關於 $\\Gamma$ 的對徑點, $M_a$, $M'$ 分別為 $\\overline{HA'}$, $\\overline{HP'}$ 的中點, $\\Omega$ 為 $\\triangle ABC$ 的九點圓, 則 $M$, $M_a$, $M'$ 位於 $\\Omega$ 上。設 $M'H$ 交 $\\Omega$ 另\n\n一點於 $X$, $D'$ 為 $AH$ 與 $BC$ 的交點, 則\n$$\n\\angle XD'D = \\angle XD'M_a = \\a... | Taiwan | 二〇一九數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous... | null | proof only | null | |
06hz | Let $S$ be a proper subset of $\mathbb{R}$ (i.e. $S \neq \mathbb{R}$) having at least two elements. Suppose there exists a function $f: \mathbb{R} \to \mathbb{R}$ satisfying the following conditions:
(i) $f(a + x + y) + f(f(a)) + f(x) + f(y) = x + y$; and
(ii) $f(axy) + f(a) + f(x)f(y) = xy$
for any real numbers $a \no... | [
"The only solution is $f(a) = 0$ for $a \\notin S$ and $f(x) = x$ for $x \\in S$.\nLabel the equations as follows.\n$$\nf(a + x + y) + f(f(a)) + f(x) + f(y) = x + y \\quad (1)\n$$\n$$\nf(axy) + f(a) + f(x)f(y) = xy \\tag{2}\n$$\n\n**Case 1.** $0 \\notin S$\nPutting $a = 0$ in (2), we obtain\n$$\nf(x)f(y) = xy - 2f(... | Hong Kong | IMO HK TST | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = x for x in S, and f(t) = 0 for t not in S | |
04m4 | Let $\overline{AD}$ be the altitude of an acute-angled triangle $ABC$. On the line $AD$ there are distinct points $E$ and $F$ such that $|DE| = |DF|$ and the point $E$ is inside the triangle $ABC$. The circumcircle of the triangle $BEF$ meets segments $\overline{BC}$ and $\overline{AB}$ again at points $K$ and $M$, res... | [
"Notice that the circumcentre of triangle $BEF$ is on the segment $\\overline{BC}$. Therefore, segment $\\overline{BK}$ is a diameter of the circumcircle of triangle $BEF$, so $\\angle BMK = 90^\\circ$, and analogously $\\angle LNC = 90^\\circ$.\n\n\n\nLet lines $AD$ and $KM$ intersect at $... | Croatia | Croatian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09po | Problem:
Je hebt 2007 kaarten. Op elke kaart is een positief geheel getal kleiner dan 2008 geschreven. Als je een aantal (minstens 1) van deze kaarten neemt, is de som van de getallen op de kaarten niet deelbaar door 2008. Bewijs dat op elke kaart hetzelfde getal staat. | [
"Solution:\n\nNoem de getallen op de kaarten $a_1, a_2, \\ldots, a_{2007}$, waarbij $1 \\leq a_i \\leq 2007$ voor alle $i$.\n\nStel dat er twee kaarten zijn met verschillende getallen, zeg $a_1 \\neq a_2$.\n\nNeem nu de som $S = a_1 + a_2 + \\cdots + a_{2007}$. Omdat alle $a_i < 2008$, geldt $S < 2007 \\times 2007 ... | Netherlands | TOETS TRAININGSKAMP | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0jz9 | Problem:
Kelvin the Frog was bored in math class one day, so he wrote all ordered triples $(a, b, c)$ of positive integers such that $a b c = 2310$ on a sheet of paper. Find the sum of all the integers he wrote down. In other words, compute
$$
\sum_{\substack{a b c = 2310 \\ a, b, c \in \mathbb{N}}} (a + b + c)
$$
wher... | [
"Solution:\nNote that $2310 = 2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11$. The given sum clearly equals $3 \\sum_{a b c = 2310} a$ by symmetry. The inner sum can be rewritten as\n$$\n\\sum_{a \\mid 2310} a \\cdot \\tau\\left(\\frac{2310}{a}\\right)\n$$\nas for any fixed $a$, there are $\\tau\\left(\\frac{2310}{a}\\righ... | United States | February 2017 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 49140 | |
0hbt | Let the triangle **ABC** be such that $2AC = AB$ and $\angle A = 2\angle B$. Let **AL** be its bisector and let **M** be the midpoint of **AB**. It turns out that $CL = ML$. Show that $\angle B = 30^\circ$.
(Danylo Khilko)
**Fig. 4** | [
"Since **AL** is a bisector, then $\\angle CAL = \\angle LAB = \\angle CBA$ (Fig. 4). Then $\\triangle ALB$ is isosceles, so **LM** is its altitude and a median. Thus $\\angle LMA = 90^\\circ$. Consider the triangles **AML** and **ALC**. Let **C**' be the projection of **L** on **AC**. Then right triangles **AML** ... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | 30° | |
0f1s | Problem:
$n$ real numbers are written around a circle. One of the numbers is $1$ and the sum of the numbers is $0$. Show that there are two adjacent numbers whose difference is at least $n/4$. Show that there is a number which differs from the arithmetic mean of its two neighbours by at least $8/n^2$. Improve this res... | [] | Soviet Union | ASU | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | Adjacent difference bound: at least n/4. Deviation from neighbor average: at least 8/n^2; can be improved to k/n^2 for some k > 8. For n = 30, one can take k = 1800/113, and there exists an example of 30 numbers with all deviations at most 2/113. | |
01xp | The angle $ADC$ of the parallelogram $ABCD$ equals $40^\circ$. The point $K$ is given such that the segments $AK$ and $BC$ intersect, $AK = BC$ and $\angle BAK = 80^\circ$. The point $L$ is given such that the segments $CL$ and $AD$ intersect, $CL = AB$ and $\angle BCL = 80^\circ$.
Find the angles of the triangle $BKL$... | [
"All angles equal $60^\\circ$."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 60°, 60°, 60° | |
0b33 | Problem:
Find the sum of all $k$ for which $x^{5} + k x^{4} - 6 x^{3} - 15 x^{2} - 8 k^{3} x - 12 k + 21$ leaves a remainder of $23$ when divided by $x + k$.
(a) $-1$
(b) $-\frac{3}{4}$
(c) $\frac{5}{8}$
(d) $\frac{3}{4}$ | [] | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | MCQ | b | |
04wb | Find the largest integer $n$ with the following property: Whenever five distinct points are given in a plane such that some two of them lie inside the triangle formed by the remaining three points, then some three of these five points can be denoted by $X, Y, Z$ such that $n^\circ < |\angle XYZ| \le 180^\circ$. | [] | Czech Republic | National Round | [
"Geometry > Plane Geometry > Advanced Configurations > Napoleon and Fermat points",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | 119 | |
0e6v | Let $ABC$ be a triangle. Let $D$ denote the intersection point of the bisector of the angle $\angle BAC$ and the side $BC$, and let $E$ denote the intersection point of the bisector of the angle $\angle CBA$ and the side $AC$. Suppose $|CD| = |CE|$. Prove that the triangle $ABC$ is isosceles. | [
"Denote by $F$ the intersection point of the lines $AD$ and $BE$. Then the line $CF$ is the bisector of the angle $\\angle ACB$. The triangles $CFD$ and $CFE$ coincide in two sides and the angle between them, hence they are congruent. The angles $\\angle CDF$ and $\\angle FEC$ are thus equal. The triangles $ADC$ an... | Slovenia | National Math Olympiad 2012 | [
"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 | |
0d9i | A chessboard has $64$ cells painted black and white in the usual way. A bishop path is a sequence of distinct cells such that two consecutive cells have exactly one common point. At least how many bishop paths can the set of all white cells be divided into? | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 2 | |
09eb | Diagonals of the convex quadrilateral $ABCD$ inscribed in a circle intersect at point $E$. The line passing $E$ perpendicular to the side $AD$ intersects the side $BC$ at point $N$. The line passing $E$ perpendicular to the side $BC$ intersects the side $AD$ at point $M$. Prove that $2MN = AB + CD$ implies that $ABCD$ ... | [
"Let denote $\\angle BAD = \\alpha$. Then $\\angle BCA = BAD = \\alpha$.\n\nSince $\\angle BCA = 90^\\circ - \\alpha = \\angle AEM \\Rightarrow \\angle MED = \\alpha$ and we conclude that $\\triangle EMD$ is. Similarly, it is easy to show $AM = EM = MD$. In other words, point $M$ is the midpoint of $AD$. Similarly,... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof only | null | |
0ewf | Problem:
Given a fixed regular pentagon $ABCDE$ with side $1$. Let $M$ be an arbitrary point inside or on it. Let the distance from $M$ to the closest vertex be $\mathbf{r}_1$, to the next closest be $\mathbf{r}_2$ and so on, so that the distances from $M$ to the five vertices satisfy $\mathbf{r}_1 \leq \mathbf{r}_2 \l... | [
"Solution:\nLet $X$ be the midpoint of $AB$ and $O$ the center of $ABCDE$. Suppose $M$ lies inside $AXO$. Then $ME = \\mathbf{r}_3$. So we maximise $\\mathbf{r}_3$ by taking $M$ at $X$, with distance $1.5590$, and we minimise $\\mathbf{r}_3$ by taking $M$ as the intersection of $AO$ and $EB$ with distance $0.8090$.... | Soviet Union | 2nd ASU | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | a) The five midpoints of the diagonals. b) The five midpoints of the sides. | |
0izb | Problem:
Circles $\omega_{1}$ and $\omega_{2}$ intersect at points $A$ and $B$. Segment $PQ$ is tangent to $\omega_{1}$ at $P$ and to $\omega_{2}$ at $Q$, and $A$ is closer to $PQ$ than $B$. Point $X$ is on $\omega_{1}$ such that $PX \parallel QB$, and point $Y$ is on $\omega_{2}$ such that $QY \parallel PB$. Given th... | [
"\n\nLet $C$ be the fourth vertex of parallelogram $APCQ$. The midpoint $M$ of $\\overline{PQ}$ is the intersection of the diagonals of this parallelogram. Because $M$ has equal power with respect to the two circles $\\omega_{1}$ and $\\omega_{2}$, it lies on $\\overleftrightarrow{AB}$, the... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 2 - sqrt(3) | |
0cp4 | A company consists of $7$ persons, and each $6$ of them can sit at a round table so that each two neighbors would be familiar to each other. Prove that all $7$ persons can sit at a round table with the same property to be satisfied. (S. Volchenkov)
В компании из семи человек любые шесть могут сесть за круглый стол так... | [
"Заметим, что у каждого в компании не менее трёх знакомых.\n\nДействительно, если бы некто $X$ был знаком менее, чем с тремя, то, исключив из компании одного из его знакомых, мы получили бы шестёрку людей, в которой у $X$ не более одного знакомого, т. е. посадить их за круглый стол невозможно. Более того, если бы у... | Russia | Regional round | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English; Russian | proof only | null | |
0674 | Let $a_1, a_2, \dots, a_n$, and $b_1, b_2, \dots, b_n$, $2n$ real numbers. Prove that there exists an integer $k$ with $1 \le k \le n$, such that
$$
\sum_{i=1}^{n} |a_i - a_k| \le \sum_{i=1}^{n} |b_i - a_k|.
$$ | [] | Greece | 17th Mediterranean Mathematical Competition | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0bcg | Problem:
Fie $a, b \in \mathbb{C}$. Să se arate că $|a z + b \bar{z}| \leq 1$, pentru orice $z \in \mathbb{C}$, cu $|z| = 1$, dacă şi numai dacă $|a| + |b| \leq 1$. | [] | Romania | Olimpiada Naţională de Matematică Etapa Judeţeană şi a Municipiului Bucureşti | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
02ys | Problem:
Um conjunto contém 4 números. As seis somas de dois elementos desse conjunto são 189, 320, 287, 264, $x$ e $y$. Encontre o maior valor possível para $x+y$. | [
"Solution:\n\nSejam $a, b, c$ e $d$ os quatro números do conjunto. Temos dois casos a considerar:\n\nI) $x = a + b$ e $y = c + d$ (somas sem parcelas em comum). Então $a + c$, $a + d$, $b + c$ e $b + d$ são, em alguma ordem, os números 189, 320, 287 e 264. Adicionando essas quatro somas, obtemos $a + b + c + d = 53... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 761 | |
0etg | Consider $16$ points arranged as shown, with horizontal and vertical distances of $1$ between consecutive rows and columns. In how many ways can one choose four of these points such that the distance between every two of those four points is strictly greater than $2$?
 | [
"Label the sixteen points $A$, $B$, $C$, $\\ldots$, $P$, as shown below in Figure 1:\n\n\n\nA selection of four points satisfying the condition that the distance between every two of the four points is greater than $2$, will be called a valid selection.\n\nLet us first consider a valid sele... | South Africa | The South African Mathematical Olympiad Third Round | [
"Discrete Mathematics > Combinatorics",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 7 | |
0d16 | Show that for every positive integers $n \ge 3$ there are distinct positive integers $a_1, a_2, \dots, a_n$ with
$$
a_1! a_2! \dots a_{n-1}! = a_n!
$$ | [
"For $n = 3$ we have $3! \\cdot 5! = 6!$.\n\nAssume that $a_1! a_2! \\dots a_{k-1}! = a_k!$. Since $a_k!(a_k! - 1)! = (a_k!)!$, it follows that\n$$\na_1! a_2! \\dots a_{k-1}! (a_k! - 1)! = a_k! (a_k! - 1)! = (a_k!)!\n$$\nand obviously $a_k! > a_k! - 1 > a_{k-1}!$. We are done by induction."
] | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0k1a | Problem:
Let $n$ and $m$ be positive integers which are at most $10^{10}$. Let $R$ be the rectangle with corners at $(0,0)$, $(n, 0)$, $(n, m)$, $(0, m)$ in the coordinate plane. A simple non-self-intersecting quadrilateral with vertices at integer coordinates is called far-reaching if each of its vertices lie on or i... | [
"Solution:\n\nLet $g = \\gcd(n, m)$, with $n = g \\cdot a$ and $m = g \\cdot b$. Note that the number of points on the diagonal of $R$ connecting $(0,0)$ and $(n, m)$ is $g+1$.\n\nWe construct two far-reaching quadrilaterals and show that at least one of them has small area.\n\nFor our first quadrilateral, let $(x_... | United States | HMMT February | [
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0k5f | For an integer $n > 0$, denote by $\mathcal{F}(n)$ the set of integers $m > 0$ for which the polynomial $p(x) = x^2 + mx + n$ has an integer root.
a. Let $S$ denote the set of integers $n > 0$ for which $\mathcal{F}(n)$ contains two consecutive integers. Show that $S$ is infinite but
$$
\sum_{n \in S} \frac{1}{n} \le ... | [
"We prove the following.\n**Claim.** The set $S$ is given explicitly by $S = \\{x(x+1)y(y+1) \\mid x, y > 0\\}$.\n*Proof.* Note that $m, m+1 \\in \\mathcal{F}(n)$ if and only if there exist integers $q > p \\ge 0$ such that\n$$\nm^2 - 4n = p^2 \\\\\n(m+1)^2 - 4n = q^2.\n$$\nSubtraction gives $2m + 1 = q^2 - p^2$, s... | United States | USA TSTST | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
... | null | proof only | null | |
0dgb | Let $ABC$ be an acute-angled triangle. Point $P$ is such that $AP = AB$ and $PB \parallel AC$. Point $Q$ is such that $AQ = AC$ and $CQ \parallel AB$. Segments $CP$ and $BQ$ meet at point $X$. Prove that the circumcenter of triangle $ABC$ lies on the circumcircle of triangle $PXQ$. | [
"Let $D$ be the vertex of parallelogram $ABDC$. Then $APDC$ and $AQDB$ are isosceles trapezoids. Therefore the perpendicular bisectors to segments $PD$ and $QD$ coincide with the perpendicular bisectors to $AC$ and $AB$ respectively, the circumcenter $O$ of triangle $ABC$ is also the circumcenter of $DPQ$ and $\\an... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
01v9 | Let $ABCD$ be a cyclic quadrilateral with the circumcircle $\omega$. The points $B_1$ and $D_1$ are symmetric to $A$ with respect to the midpoints of $BC$ and $CD$. The circumcircle of the triangle $CB_1D_1$ intersects $\omega$ at $C$ and $G$.
Prove that $AG$ is the diameter of $\omega$.
(A. Voidelevich) | [
"Since $B_1$ is symmetric to $A$ with respect to the midpoint of $BC$, $ABB_1C$ is a parallelogram, so $BB_1 \\parallel AC$ and $BB_1 = AC$. Likewise $ACD_1D$ is a parallelogram, $DD_1 \\parallel AC$ and $DD_1 = AC$. Therefore the triangle $CD_1B_1$ is obtained from the triangle $ADB$ by the transfer along the line... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0aok | Problem:
If $\sqrt{4+x} + \sqrt{10-x} = 6$, find the product $\sqrt{4+x} \sqrt{10-x}$.
(a) 13
(b) 7
(c) 17
(d) 11 | [] | Philippines | Qualifying Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | d | |
03io | Problem:
Eleven theatrical groups participated in a festival. Each day, some of the groups were scheduled to perform while the remaining groups joined the general audience. At the conclusion of the festival, each group had seen, during its days off, at least one performance of every other group. At least how many days... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 6 | |
0h2i | One corner cell is removed from a $2011 \times 2011$ square. Is it possible to cut the obtained figure along the lines of the grid into less than $121$ squares? | [
"We will write $a \\to x$, if an $a \\times a$ square without one corner cell can be cut along the lines of the grid into $x$ squares.\n\nEvidently, $2 \\to 3$. Also note that $7 \\to 8$ and $9 \\to 9$ (fig. 24 and 25). We will show that if $a \\to x$ and $b \\to y$, then $ab \\to x+y$. Indeed, a square with the si... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Yes | |
07ci | Let $f, g : \mathbb{R}^+ \to \mathbb{R}^+$ be two functions such that for all positive real numbers $x$ and $y$
$$
f(x + g(y))^2 = f(x^2) + y^2.
$$
Prove that the range of $g$ is not bounded from above. | [
"Let $P(x, y)$ be the assertion\n$$\nf(x + g(y))^2 = f(x^2) + y^2\n$$\nIf $g(f(t)) < t$ for some $t > 0$, then $P(t - g(f(t)), f(t))$ implies\n$$\nf(t)^2 = f((t - g(t))^2) + f(t)^2 \\implies f((t - g(t))^2) = 0\n$$\nwhich is impossible.\nSo $g(f(x)) \\ge x$, for all positive real numbers $x$. Hence the solution is ... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
01sn | Prove that
$$
\left(a^2 + \frac{c^2}{b^2}\right) \left(b^2 + \frac{a^2}{c^2}\right) \left(c^2 + \frac{b^2}{a^2}\right) \ge \frac{abc}{\sqrt{3}} \left(a + \frac{1}{a}\right) \left(b + \frac{1}{b}\right) \left(c + \frac{1}{c}\right)
$$
for all positive real $a, b, c$. | [
"(Solution by S. Shaban, A. Vasileuski) By the Cauchy-Bunyakovsky-Schwarz inequality,\n$$\n\\sqrt{a^2 + \\frac{c^2}{b^2}} \\sqrt{b^2 + \\frac{a^2}{c^2}} \\geq \\left( ab + \\frac{a}{b} \\right) = a \\left( b + \\frac{1}{b} \\right),\n$$\n$$\n\\sqrt{b^2 + \\frac{a^2}{c^2}} \\sqrt{c^2 + \\frac{b^2}{a^2}} \\geq \\left... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
07d6 | Let $A_1, A_2, \ldots, A_k$ be subsets of $\{1, 2, 3, \ldots, n\}$ such that for all $1 \le i, j \le k : A_i \cap A_j \ne \emptyset$. Prove that there are $n$ distinct positive integers $x_1, x_2, \ldots, x_n$ such that for each $1 \le j \le k$
$$
\operatorname{lcm}_{i \in A_j} \{x_i\} > \operatorname{lcm}_{i \notin A_... | [
"Let $p_1, \\dots, p_k$ be distinct prime numbers and for all $1 \\le i \\le n$ define $x_i$ to be $x_i = \\prod_{j \\in A_j} p_j$. We prove that these $x_i$'s satisfy the problem's conditions.\n\n$$\n1 \\le i, j \\le k \\implies A_i \\cap A_j \\ne \\emptyset \\implies \\exists l : l \\in A_i, l \\in A_j \\implies ... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0kse | Problem:
Let $A B C D E F$ be a regular hexagon and let point $O$ be the center of the hexagon. How many ways can you color these seven points either red or blue such that there doesn't exist any equilateral triangle with vertices of all the same color? | [
"Solution:\n\nWithout loss of generality, let $O$ be blue. Then we can't have any two adjacent blues on the perimeter of $A B C D E F$. However, because of the two larger equilateral triangles $A C E$ and $B D F$, we need at least two blues to keep us from having an all red equilateral triangle. We can't have three... | United States | HMMT November 2022 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 6 | |
02nf | Problem:
Soma e inverte - Usando somente as duas operações "+1=\text{somar 1}" e " $-i=\text{menos o inverso}$", podemos formar várias sequências a partir de um número inicial. Por exemplo, iniciando com o número $3$, podemos formar a sequência
$$
3 \xrightarrow{+1} 4 \xrightarrow{+1} 5 \xrightarrow{-i}-\frac{1}{5} \x... | [
"Solution:\n\nComo $0$ não é o inverso de número algum, qualquer sequência que comece e termine em $0$ deve ser dada por\n$$\n0 \\xrightarrow{+1} 1 \\longrightarrow \\cdots \\longrightarrow -1 \\xrightarrow{+1} 0.\n$$\nUma sequência dessas é a seguinte.\n$$\n\\begin{aligned}\n& 0 \\xrightarrow{+1} 1 \\xrightarrow{+... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 0 → +1 → 1 → −i → −1 → +1 → 0 | |
0j1l | Problem:
Estimate the sum of all the prime numbers less than $1,000,000$. If the correct answer is $X$ and you write down $A$, your team will receive $\min \left(\left\lfloor\frac{25 X}{A}\right\rfloor,\left\lfloor\frac{25 A}{X}\right\rfloor\right)$ points, where $\lfloor x\rfloor$ is the largest integer less than or ... | [
"Solution:\n\nAnswer: $37,550,402,023$\n\nA decent approximation to the sum of all the primes can be obtained with the following two facts. First, there are approximately $\\frac{n}{\\ln n}$ primes less than $n$ and second, the $n^{\\text{th}}$ prime is approximately $n \\ln n$. We'll approximate $\\ln 1000000$ as ... | United States | Harvard-MIT November Tournament | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 37,550,402,023 | |
01bd | Let $n \ge 4$ be an even integer. In how many ways can one select four different positive integers $k$, $1 \le k \le n$, so that the sum of two of the chosen numbers equals the sum of the other two? | [
"**Answer:**\n$$\n\\frac{(n-2)(2n^2 - 5n)}{24}\n$$\n\nSolution:\nLetting $a$ be the smallest and $b$ the largest of the chosen numbers, the sum in the problem has to be $a+b$. So given $a$ and $b$, the other numbers $c$ and $d$ have to satisfy $a < c, d < b$ and $c+d = a+b$. For the smaller of $c, d$, say $c$, one ... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | (n-2)(2n^2 - 5n)/24 | |
0gta | There are $n$ students and $29$ boards. Each student writes one of the numbers $1$, $2$ or $3$ to each board (a student can write different numbers to different boards). It is observed that any two students wrote different numbers to at least one board and any three students wrote the same number to at least one board.... | [
"Answer: $3^{28}$.\nLet us numerate boards by $1$, $2$, $\\ldots$, $29$ and for each student define a vector $(a_1, a_2, \\ldots, a_{29})$, where $a_i$ is a number written by this student to the board number $i$. Since any two students wrote different numbers to at least one board, we get a set $V$ of $n$ different... | Turkey | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 3^28 | |
0kfc | Problem:
Altitudes $B E$ and $C F$ of acute triangle $A B C$ intersect at $H$. Suppose that the altitudes of triangle $E H F$ concur on line $B C$. If $A B = 3$ and $A C = 4$, then $B C^{2} = \frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Compute $100 a + b$. | [
"Solution:\n\n\n\nLet $P$ be the orthocenter of $\\triangle E H F$. Then $E H \\perp F P$ and $E H \\perp A C$, so $F P$ is parallel to $A C$. Similarly, $E P$ is parallel to $A B$. Using similar triangles gives\n$$\n1 = \\frac{B P}{B C} + \\frac{C P}{B C} = \\frac{A E}{A C} + \\frac{A F}{A... | United States | HMMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 33725 | |
0l09 | Problem:
Let $ABC$ be a triangle with $\angle BAC = 90^\circ$. Let $D$, $E$, and $F$ be the feet of the altitude, angle bisector, and median from $A$ to $BC$, respectively. If $DE = 3$ and $EF = 5$, compute the length of $BC$. | [
"\nSince $F$ is the circumcenter of $\\triangle ABC$, we have that $AE$ bisects $\\angle DAF$. So by the angle bisector theorem, we can set $AD = 3x$ and $AF = 5x$. Applying Pythagorean theorem to $\\triangle ADE$ then gives\n$$\n(3x)^2 + (5+3)^2 = (5x)^2 \\Longrightarrow x = 2\n$$\nSo $AF ... | United States | HMMT February 2024 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 20 | |
03u1 | Let $O$ be an interior point of the triangle $ABC$. Prove that there exist positive integers $p, q$ and $r$, such that
$$
|p \cdot \vec{OA} + q \cdot \vec{OB} + r \cdot \vec{OC}| < \frac{1}{2007}.
$$ | [
"It is well-known that there are positive real numbers $\\beta, \\gamma$ such that\n$$\n\\vec{OA} + \\beta \\vec{OB} + \\gamma \\vec{OC} = \\vec{0}.\n$$\nSo for positive integer $k$, we have\n$$\nk \\vec{OA} + k\\beta \\vec{OB} + k\\gamma \\vec{OC} = \\vec{0}.\n$$\nLet $m(k) = [k\\beta]$, $n(k) = [k\\gamma]$, where... | China | China Western Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
0h4n | Find all integers $n$ that have more than $\frac{n}{2}$ divisors. | [
"**Answer:** $n \\in \\{1; 2; 3; 4; 6\\}$.\n\nClearly a number cannot have divisors greater than $\\frac{n}{2}$, besides $n$ itself. Therefore, in order to have more than $\\frac{n}{2}$ divisors it must be divisible by all numbers from $1$ to $\\frac{n}{2}$ and by $n$. Denote by $m$ the integer number which equals ... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 1, 2, 3, 4, 6 | |
0h0c | Let $\omega$ be a circumscribed circle of $ABC$. Chord $AD$ is a bisector of triangle $ABC$ and meets $BC$ at $L$, chord $DK$ is perpendicular to $AC$ and meets it at $M$. Find the ratio $\frac{AM}{MC}$, if $\frac{BL}{LC} = \frac{1}{2}$.
Fig.15 | [
"Using the property of our bisector, we get $\\frac{BL}{AB} = \\frac{LC}{AC} \\Rightarrow \\frac{AC}{AB} = \\frac{LC}{BL}$, or $\\frac{AC}{AB} = 2$, hence, $AC = 2AB$ (Fig.15).\n\n$$\n\\angle BAD = \\angle DAC \\ (AD \\text{ is a bisector of } \\angle BAC), \\text{ thus } BD = DC. \\text{ Consider } S \\text{ - mid... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010) | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 3 | |
06fq | Let $\triangle ABC$ be a right-angled triangle with $\angle C = 90^\circ$. $CD$ is the altitude from $C$ to $AB$, with $D$ on $AB$. $\omega$ is the circumcircle of $\triangle BCD$. $\omega_1$ is a circle situated in $\triangle ACD$, which is tangent to the segments $AD$ and $AC$ at $M$ and $N$ respectively, and is also... | [
"(i) This is an immediate consequence of Casey's theorem. We provide an elementary proof as follows assuming the result in part (ii) (which will be proved later).\nLet $a = BC$, $b = CA$ and $c = AB$. By part (ii), we have $BM = BC = a$. Then we have\n$$\n\\begin{aligned}\nAN &= AM = c - a, \\\\\nCN &= b - AN = a +... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0hnl | Problem:
A circle with center at $O$ has radius $1$. Points $P$ and $Q$ outside the circle are placed such that $PQ$ passes through $O$. Tangent lines to the circle through $P$ hit the circle at $P_{1}$ and $P_{2}$, and tangent lines to the circle through $Q$ hit the circle at $Q_{1}$ and $Q_{2}$. If $\angle P_{1} P P_... | [
"Solution:\n$$(45-30)^{\\circ} = \\frac{\\pi}{12}.$$"
] | United States | null | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | π/12 | |
0bi1 | For each positive integer $n$ the function $f_n : [0, n] \to \mathbb{R}$ is defined by $f_n(x) = \arctg(\lfloor x \rfloor)$. Prove that $f_n$ is a Riemann integrable function
and find
$$
\lim_{n \to \infty} \frac{1}{n} \int_{0}^{n} f_{n}(x) dx.
$$ | [
"The function $f_n$ is locally constant, hence Riemann integrable.\n\nNext, we have\n$$\n\\int_{0}^{n} f_{n}(x) dx = \\sum_{i=0}^{n-1} \\int_{i}^{i+1} f_{n}(i) dx = \\sum_{i=0}^{n-1} \\arctan i.\n$$\nApplying Stolz-Cesàro theorem, we obtain\n$$\n\\lim_{n \\to \\infty} \\frac{\\arctan 1 + \\arctan 2 + \\dots + \\arc... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | π/2 | |
0kw8 | Problem:
The Fibonacci numbers are defined recursively by $F_{0}=0$, $F_{1}=1$, and $F_{i}=F_{i-1}+F_{i-2}$ for $i \geq 2$. Given 30 wooden blocks of weights $\sqrt[3]{F_{2}}, \sqrt[3]{F_{3}}, \ldots, \sqrt[3]{F_{31}}$, estimate the number of ways to paint each block either red or blue such that the total weight of th... | [
"Solution:\n\nTo get within an order of magnitude, one approach is to let $X_{n}$ be a random variable which takes the value $\\pm \\sqrt[3]{F_{n}}$, with the sign chosen uniformly at random. We want the probability that $S=\\sum_{i=2}^{31} X_{i}$ is in $[-1,1]$. We can attempt to approximate the distribution of $S... | United States | HMMT February 2023 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 4064598 | |
0ibr | Problem:
Two positive rational numbers $x$ and $y$, when written in lowest terms, have the property that the sum of their numerators is $9$ and the sum of their denominators is $10$. What is the largest possible value of $x + y$? | [
"Solution:\n\nFor fixed denominators $a < b$ (with sum $10$), we maximize the sum of the fractions by giving the smaller denominator as large a numerator as possible: $8/a + 1/b$. Then, if $a \\geq 2$, this quantity is at most $8/2 + 1/1 = 5$, which is clearly smaller than the sum we get by setting $a = 1$, namely ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 73/9 | |
049b | Let $a$, $b$, $c$ be different real numbers none of which is zero. Consider quadratic equations:
$$
ax^2 + bx + c = 0, \quad bx^2 + cx + a = 0, \quad cx^2 + ax + b = 0.
$$
If $\frac{c}{a}$ is a root of the first equation, prove that all three of them have a common root. What is the product of the other roots of those e... | [] | Croatia | CroatianCompetitions2011 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 1 | |
0ar2 | Problem:
What is the remainder when $(0! + 1! + 2! + \cdots + 2011!)^2$ is divided by $10$? | [
"Solution:\n\nFirst, observe the units digit of $n!$ for $n \\geq 5$. For $n \\geq 5$, $n!$ is divisible by $10$, so its units digit is $0$.\n\nTherefore, $0! + 1! + 2! + \\cdots + 2011!$ has the same units digit as $0! + 1! + 2! + 3! + 4!$.\n\nCompute:\n\n$0! = 1$\n\n$1! = 1$\n\n$2! = 2$\n\n$3! = 6$\n\n$4! = 24$\n... | Philippines | 13th Philippine Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | final answer only | 6 | |
05yf | Problem:
Soit $a_{1}, a_{2}, \ldots$ des nombres réels strictement positifs tels que
$$
a_{n+1}^{2}+a_{n} a_{n+2} \leqslant a_{n}+a_{n+2}
$$
pour tout entier $n \geqslant 1$. Démontrer que $a_{2023} \leqslant 1$. | [
"Solution:\n\nL'inégalité donnée dans l'énoncé et celle qu'il nous est demandé de démontrer suggèrent de s'intéresser aux termes de la forme $b_{n}=a_{n}-1$, car alors l'équation de l'énoncé se réécrit comme\n$$\nb_{n+1}\\left(b_{n+1}+2\\right)+b_{n} b_{n+2} \\leqslant 0,\n$$\net il s'agit de démontrer que $b_{2023... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
049j | Find all pairs $(x, y)$ of integers such that $x^2(y - 1) + y^2(x - 1) = 1$. | [
"The given equation is equivalent to\n$$\n\\begin{aligned}\nx^2y - x^2 + xy^2 - y^2 &= 1 \\\\\nxy(x + y) - (x^2 + y^2) &= 1 \\\\\nxy(x + y) - ((x + y)^2 - 2xy) &= 1\n\\end{aligned}\n$$\nIf we put $u = x + y$ and $v = xy$ we get the equation:\n$$\nuv - (u^2 - 2v) = 1\n$$\nfrom which follows $uv + 2v = u^2 + 1$, that... | Croatia | CroatianCompetitions2011 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | [(1, 2), (2, 1), (2, -5), (-5, 2)] | |
01g8 | Let $a_0 > 0$ be a real number, and let
$$
a_n = \frac{a_{n-1}}{\sqrt{1 + 2020 \cdot a_{n-1}^2}}, \quad \text{for } n = 1, 2, \dots, 2020.
$$
Show that $a_{2020} < \frac{1}{2020}$. | [
"Let $b_n = \\frac{1}{a_n^2}$. Then $b_0 = \\frac{1}{a_0^2}$ and\n$$\nb_n = \\frac{1 + 2020 \\cdot a_{n-1}^2}{a_{n-1}^2} = b_{n-1} \\left( 1 + 2020 \\cdot \\frac{1}{b_{n-1}} \\right) = b_{n-1} + 2020.\n$$\nHence $b_{2020} = b_0 + 2020^2 = \\frac{1}{a_0^2} + 2020^2$ and $a_{2020}^2 = \\frac{1}{\\frac{1}{a_0^2} + 202... | Baltic Way | Baltic Way 2020 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
0861 | Problem:
Indicando con $x_{1}, x_{2}, x_{3}$ e $x_{4}$ le soluzioni dell'equazione $x^{4}-2 x^{3}-7 x^{2}-2 x+1=0$, quanto vale $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}$?
(A) 1
(B) $\frac{1}{2}$
(C) 2
(D) 4
(E) 7. | [
"Solution:\n\nLa risposta è (C). È possibile determinare esplicitamente i valori delle soluzioni dell'equazione. Si tratta infatti di una equazione reciproca di prima specie (equazioni in cui il coefficiente di $x^{k}$ è uguale a quello di $x^{n-k}$, dove $n$ è il grado dell'equazione e $k=0,1, \\ldots, n$) ed hann... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | MCQ | C | |
0erw | For how many positive three-digit numbers is the hundreds digit smaller than the units digit? | [
"360\n\nSystematic counting soon reveals a pattern.\n\nIn the 900s there are clearly no cases.\n\nIn the 800s there are 10: 809, 819, 829, ..., 899.\n\nIn the 700s there are 20: 708, 718, 728, ..., 798\n709, 719, 729, ..., 799.\n\nIn the 600s there are 30: 607, 617, 627, ..., 697\n608, 618, 628, ..., 698\n609, 619,... | South Africa | South African Mathematics Olympiad Second Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | final answer only | 360 | |
0hq8 | Problem:
Let $ABC$ be a triangle, and let $M$ and $N$ be the respective midpoints of $AB$ and $AC$. Suppose that
$$
\frac{CM}{AC} = \frac{\sqrt{3}}{2}
$$
Prove that
$$
\frac{BN}{AB} = \frac{\sqrt{3}}{2}
$$ | [
"Solution:\nLet $L$ be the midpoint of $BC$. Let $BC = 2a$, $AC = 2b$, and $AB = 2c$. Applying the parallelogram law to $CLMN$ gives\n$$\nCM^{2} + c^{2} = a^{2} + b^{2} + a^{2} + b^{2}\n$$\nor\n$$\nCM^{2} = 2a^{2} + 2b^{2} - c^{2}.\n$$\nSquaring both sides of the given equation and substituting yields\n$$\n\\frac{2... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0d9s | Find all functions $f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+}$ satisfying $f(1)=2$, $f(2) \neq 4$, and
$$
\max \{f(m)+f(n), m+n\} \mid \min \{2m+2n, f(m+n)+1\}
$$
for all $m, n \in \mathbb{Z}^{+}$. | [
"In given condition, substitute $m=n$, we have\n$$\n\\max \\{2 f(n), 2 n\\} \\mid \\min \\{4 n, f(2 n)+1\\},\n$$\nthus $4 n \\geq 2 f(n)$ or $f(n) \\leq 2 n$.\nContinue substitute $m=1$, we have\n$$\n\\max \\{2+f(n), 1+n\\} \\mid \\min \\{2+2 n, f(1+n)+1\\} \\quad (*)\n$$\nThus $f(n+1)+1 \\geq f(n)+2$ or $f(n+1) \\... | Saudi Arabia | Team selection tests for IMO 2018 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | f(n) = n + 1 for all n in positive integers | |
03hh | Problem:
Let
$$
f(x) = a_{0} + a_{1} x + a_{2} x^{2} + \cdots + a_{n} x^{n}
$$
be a polynomial with coefficients satisfying the conditions:
$$
0 \leq a_{i} \leq a_{0}, \quad i = 1, 2, \ldots, n
$$
Let $b_{0}, b_{1}, \ldots, b_{2n}$ be the coefficients of the polynomial
$$
\begin{aligned}
(f(x))^{2} &= \left(a_{0} + a_{... | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
03sk | Let $2006$ be expressed as the sum of five positive integers $x_1, x_2, x_3, x_4, x_5$, and $S = \sum_{1 \le i < j \le 5} x_i x_j$. We ask:
(1) What value of $x_1, x_2, x_3, x_4, x_5$ will make $S$ the maximum?
(2) Further, if $|x_i - x_j| \le 2$ for any $1 \le i, j \le 5$, then what value of $x_1, x_2, x_3, x_4, x_5$ ... | [
"(1) Obviously the number of the values of $S$ is finite, so the maximum and minimum exist. Suppose $x_1 + x_2 + x_3 + x_4 + x_5 = 2006$ such that $S = \\sum_{1 \\le i < j \\le 5} x_i x_j$ reaches the maximum, we must have\n$$\n|x_i - x_j| \\le 1, \\quad (1 \\le i, j \\le 5).\n$$\nOtherwise, assume that this does n... | China | China Mathematical Competition | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | Maximum: {402, 401, 401, 401, 401}. Minimum under |xi − xj| ≤ 2: {402, 402, 402, 400, 400}. | |
0hbr | In expression $1\ 2\ 3\ 4\ 5\ 6\ 7\ 8\ 9$ you should arrange parentheses and four arithmetic symbols $+, -, \times, :$ between some of the digits to obtain the largest possible number.
(Bogdan Rublyov) | [
"It is clear that the largest number you can get if you use only the actions of multiplication and addition. We will show that there is actually no sense to put any characters and is the largest nine-digit number $123456789$.\n\nFirst remove one \"+\" sign, that is write $\\overline{XY}$ instead of expression $X + ... | Ukraine | Ukrainian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | English | proof and answer | 123456789 | |
0j5j | Problem:
Let $n$ be an odd positive integer, and suppose that $n$ people sit on a committee that is in the process of electing a president. The members sit in a circle, and every member votes for the person either to his/her immediate left, or to his/her immediate right. If one member wins more votes than all the othe... | [
"Solution:\n\nLet $x$ be the probability Hermia is elected if Lysander votes for her, and let $y$ be the probability that she wins if Lysander does not vote for her. We are trying to find $x$, and do so by first finding $y$.\n\nIf Lysander votes for Hermia with probability $\\frac{1}{2}$ then the probability that H... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | (2^n - 1) / (n * 2^(n-1)) | |
0iap | Problem:
For how many integers $n$, for $1 \leq n \leq 1000$, is the number $\frac{1}{2}\binom{2 n}{n}$ even? | [
"Solution:\n\n990\n\nIn fact, the expression $\\binom{2 n}{n}$ is always even, and it is not a multiple of four if and only if $n$ is a power of 2, and there are 10 powers of 2 between 1 and 1000.\n\nLet $f(N)$ denote the number of factors of 2 in $N$. Thus,\n\n$$\nf(n!)=\\left\\lfloor\\frac{n}{2}\\right\\rfloor+\\... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | final answer only | 990 | |
0ji1 | Let $n$ be a positive integer. Consider a triangular array of nonnegative integers as follows:
Call such a triangular array *stable* if for every $0 \le i < j < k \le n$ we have
$$
a_{i,j} + a_{j,k} \le a_{i,k} \le a_{i,j} + a_{j,k} + 1.
$$
For $s_1, \dots, s_n$ any nondecreasing sequence of... | [
"We begin by proving a lemma on the stability of triangular arrays.\n\n**Lemma 3.** Let $T$ be a stable triangular array with row sums $s_1, s_2, \\dots, s_n$. Then there is a unique index $i$ such $a_{i,n}$ in $T$ can be increased by one to get a new stable triangular array $T^+$ with row sums $s_1, s_2, \\dots, s... | United States | IMO Team Selection Team Selection Test | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
0641 | Problem:
Es sei $ABC$ ein Dreieck. Der dem Punkt $A$ gegenüberliegende Ankreis $\omega$ berühre die Strecke $\overline{BC}$ sowie die Strahlen $AC$ und $AB$ in den Punkten $D$, $E$ bzw. $F$. Der Umkreis des Dreiecks $AEF$ schneide die Gerade $BC$ in den Punkten $P$ und $Q$. Schließlich sei $M$ der Mittelpunkt der Stre... | [
"Solution:\n\nEs sei $J$ der Mittelpunkt des Ankreises $\\omega$. Dann gilt $JE \\perp AE$ und $JF \\perp AF$, der Thaleskreis $\\Omega$ von $\\overline{AJ}$ verläuft also durch $E$ und $F$, und damit auch durch $P$ und $Q$. Der Strahl $AD$ schneide $\\Omega$ und $\\omega$ erneut in $N$ bzw. $T$.\n\n![](attached_im... | Germany | 1. Auswahlklausur | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion"
] | null | proof only | null | |
06my | Let $f(n) = \left| \binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \dots + \binom{n}{3\left\lfloor \frac{n}{3} \right\rfloor} \right| - \frac{2^n}{3}$, where $[x]$ is the greatest integer not exceeding $x$. Find $f(1) + f(2) + \dots + f(2021)$. | [
"The answer is 898.\n\nIndeed, let $\\omega = e^{\\frac{2\\pi i}{3}}$ be a cube root of unity. By the binomial theorem, we have\n$$\n(1 + 1)^n = \\binom{n}{0} + \\binom{n}{1} + \\dots + \\binom{n}{n},\n$$\n$$\n(1 + \\omega)^n = \\binom{n}{0} + \\binom{n}{1}\\omega + \\dots + \\binom{n}{n}\\omega^n,\n$$\n$$\n(1 + \\... | Hong Kong | IMO HK TST | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 898 | |
03i1 | Problem:
$N$ is an integer whose representation in base $b$ is $777$. Find the smallest positive integer $b$ for which $N$ is the fourth power of an integer. | [
"Solution:\n\nThe number $N$ written as $777$ in base $b$ means:\n$$\nN = 7b^2 + 7b + 7 = 7(b^2 + b + 1)\n$$\nWe want $N$ to be a perfect fourth power, i.e., $N = k^4$ for some integer $k$.\nSo,\n$$\n7(b^2 + b + 1) = k^4\n$$\nSince $7$ is prime, $7$ must divide $k^4$, so $k$ is divisible by $7$.\nLet $k = 7m$, then... | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 18 | |
09tu | Problem:
Zij $ABC$ een scherphoekige driehoek en zij $D$ het voetpunt van de hoogtelijn uit $A$. Op $AD$ liggen verschillende punten $E$ en $F$ zodat $|AE| = |BE|$ en $|AF| = |CF|$. Een punt $T \neq D$ voldoet aan $\angle BTE = \angle CTF = 90^\circ$. Toon aan dat $|TA|^2 = |TB| \cdot |TC|$.
... | [
"Solution:\n\nOplossing I. Zij $M$ het midden van $AB$ en zij $N$ het midden van $AC$. Uit de gegevens volgt dat $E$ op de middelloodlijn van $AB$ ligt, dus $\\angle BME = 90^\\circ$. Omdat ook $\\angle BTE = 90^\\circ$ en $\\angle BDE = 90^\\circ$, is $BDTEM$ een koordenvijfhoek wegens Thales. Zo ook is $CFDTN$ ee... | Netherlands | IMO-selectietoets II | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dxb | Problem:
Naj bosta $F_{1}$ in $F_{2}$ gorišči elipse $4 x^{2}+9 y^{2}=36$ in $K$ taka točka na tej elipsi, da je $\left|K F_{1}\right| : \left|K F_{2}\right|=2: 1$. Izračunaj ploščino trikotnika $K F_{1} F_{2}$. | [
"\n\nEnačbo elipse delimo s $36$, da dobimo $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$. Dana elipsa je v središčni legi, njena velika polos je enaka $a=3$, njena mala polos pa $b=2$. Ker točka $K$ leži na elipsi, velja $\\left|K F_{1}\\right|+\\left|K F_{2}\\right|=2 a=6$. Ker pa je $\\left|K F_... | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 4 | |
0ea4 | Exactly $p-1$ distinct positive integers are written on the blackboard for some prime number $p$. The number $p$ is among these $p-1$ numbers. For any pair of the numbers the absolute value of their difference is also on the board. Prove that all the numbers on the blackboard are divisible by $p$. | [
"Denote the numbers on the board by $a_1, a_2, \\dots, a_{p-1}$.\nWithout loss of generality we may assume that $a_1 < a_2 < \\dots < a_{p-1}$. Then the numbers $a_2 - a_1 < a_3 - a_1 < \\dots < a_{p-1} - a_1$ are also written on the blackboard. There are $p-2$ of them and they are all distinct. This is only possib... | Slovenia | National Math Olympiad in Slovenia | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0jj2 | Problem:
How many integers $n$ in the set $\{4,9,14,19, \ldots, 2014\}$ have the property that the sum of the decimal digits of $n$ is even? | [
"Solution:\nAnswer: 201\nWe know that 2014 does not qualify the property. So, we'll consider $\\{4,9,14, \\ldots, 2009\\}$ instead. Now, we partition this set into 2 sets: $\\{4,14,24, \\ldots, 2004\\}$ and $\\{9,19,29, \\ldots, 2009\\}$.\n\nFor each, the first and second set are basically $x4$ and $x9$, where $x =... | United States | HMMT November 2014 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 201 | |
0k43 | Problem:
A bag contains nine blue marbles, ten ugly marbles, and one special marble. Ryan picks marbles randomly from this bag with replacement until he draws the special marble. He notices that none of the marbles he drew were ugly. Given this information, what is the expected value of the number of total marbles he d... | [
"Solution:\nThe probability of drawing $k$ marbles is the probability of drawing $k-1$ blue marbles and then the special marble, which is $p_{k} = \\left(\\frac{9}{20}\\right)^{k-1} \\times \\frac{1}{20}$. The probability of drawing no ugly marbles is therefore $\\sum_{k=1}^{\\infty} p_{k} = \\frac{1}{11}$.\n\nThen... | United States | HMMT February 2018 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 20/11 | |
0ff0 | Problem:
¿Existe un conjunto infinito de números naturales que NO se pueden representar en la forma
$$
n^{2}+p
$$
siendo $n$ natural y $p$ primo? Razónese la contestación. | [
"Solution:\nLa respuesta es afirmativa. Vamos a demostrar que hay un número no acotado de cuadrados que no se puede expresar de esa manera.\nSea $m^{2}=n^{2}+p$. Entonces $m^{2}-n^{2}=p \\Leftrightarrow (m+n)(m-n)=p$, y al ser $p$ primo, necesariamente\n$$\nm-n=1,\\ m+n=p\n$$\nDe aquí que\n$$\nm=\\frac{p+1}{2}\n$$\... | Spain | TANDA III | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
06k0 | a. Does there exist $5$ circles in the plane such that each circle passes through exactly $3$ centres of other circles?
b. Does there exist $6$ circles in the plane such that each circle passes through exactly $3$ centres of other circles? | [
"a. No. Let the centres be $A$, $B$, $C$, $D$, $E$. WLOG assume $A$ is the centre of $(BCD)$. One of the following must occur.\n\n* If $B$ is the centre of $(ACD)$, then $AC = AD = AB = BC = BD$. So both $\\triangle ABC$ and $\\triangle ABD$ are equilateral triangles. Note that at least one of $A$, $B$ must lie on ... | Hong Kong | HKG TST | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Transformations > Translation"
] | null | proof and answer | a. No; b. Yes | |
0fft | Problem:
Sean $a, b, c$ y $d$ enteros impares tales que $0<a<b<c<d$ y $a d = b c$. Demostrar que si $a + d = 2^{k}$ y $b + c = 2^{m}$ para ciertos enteros $k$ y $m$, entonces $a = 1$. | [] | Spain | International 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"
] | null | proof only | null | |
0iia | Problem:
Find the prime factorization of
$$
2006^{2} \cdot 2262 - 669^{2} \cdot 3599 + 1593^{2} \cdot 1337
$$
(No proof is necessary.) | [
"Solution:\nUpon observing that $2262 = 669 + 1593$, $3599 = 1593 + 2006$, and $1337 = 2006 - 669$, we are inspired to write $a = 2006$, $b = 669$, $c = -1593$. The expression in question then rewrites as $a^{2}(b-c) + b^{2}(c-a) + c^{2}(a-b)$. But, by experimenting in the general case (e.g. setting $a = b$), we fi... | United States | Harvard-MIT Mathematics Tournament, Team Round B | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 2 * 3 * 7 * 13 * 29 * 59 * 61 * 191 | |
0kmr | Problem:
Let $O$ and $A$ be two points in the plane with $OA = 30$, and let $\Gamma$ be a circle with center $O$ and radius $r$. Suppose that there exist two points $B$ and $C$ on $\Gamma$ with $\angle ABC = 90^{\circ}$ and $AB = BC$. Compute the minimum possible value of $\lfloor r \rfloor$. | [
"Solution:\n\nLet $f_1$ denote a $45^{\\circ}$ counterclockwise rotation about point $A$ followed by a dilation centered at $A$ with scale factor $1/\\sqrt{2}$. Similarly, let $f_2$ denote a $45^{\\circ}$ clockwise rotation about point $A$ followed by a dilation centered at $A$ with scale factor $1/\\sqrt{2}$. For ... | United States | HMMT Spring 2021 | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | 12 | |
0626 | Problem:
Gegeben sei ein Dreieck $ABC$ mit $\overline{AB} + \overline{BC} = 3 \overline{AC}$. Sein Inkreis habe den Mittelpunkt $I$ und berühre die Seiten $AB$ in $D$ bzw. $BC$ in $E$. Weiter seien $K$ und $L$ die Spiegelpunkte von $D$ bzw. $E$ bezüglich $I$.
Man beweise, dass das Viereck $ACKL$ ein Sehnenviereck ist. | [
"Solution:\n\nZu den in der Aufgabenstellung genannten Punkten bezeichnen wir den Berührpunkt des Inkreises mit $AC$ mit $F$, den Mittelpunkt von $AC$ mit $S$, den Schnittpunkt von $w_{\\beta}$ und der Mittelsenkrechten $m_{AC}$ mit $P$, den Spiegelpunkt von $B$ an $I$ mit $R$, den Mittelpunkt von $BI$ mit $T$ und ... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotat... | null | proof only | null | |
044b | In a plane rectangular coordinate system $xOy$, given parabola $\Gamma: y^2 = 2px$ ($p > 0$), a line with inclination angle $\frac{\pi}{4}$ intersects $\Gamma$ at point $P(3, 2)$ and another point $Q$. Then the area of $\triangle OPQ$ is ______. | [
"Since point $P(3, 2)$ is on $\\Gamma$, we have $2p = \\frac{4}{3}$.\nThe slope of the mentioned line is $1$ and it passes through point $P(3, 2)$. Therefore, its equation is $y = x - 1$, and it passes through point $A(1, 0)$ on the $x$-axis. Substitute $x = y + 1$ into $y^2 = \\frac{4}{3}x$, eliminating $x$ and ar... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | 4/3 | |
09ic | Find all positive integer solutions to the equation $4^n - 2^n + 1 = 3^m$. | [] | Mongolia | Mongolian Mathematical Olympiad Round 2 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (n, m) = (1, 1) | |
0eg8 | Problem:
Določi vrednosti parametrov $a$ in $b$ tako, da bo premica z enačbo $y=3 x+5$ tangenta na graf funkcije $f$, $f(x)=a x^{2}+b x$, v točki z absciso $-1$. | [
"Solution:\n\nIzračunamo ordinato dotikališča $y=3(-1)+5=2$.\n\nUpoštevamo koordinate dotikališča v $f(x)=a x^{2}+b x$ in dobimo enačbo $a-b=2$.\n\nOdvajamo funkcijo $f$ in dobimo $f'(x)=2 a x+b$.\n\nUpoštevamo dotikališče $D(-1,2)$ in smerni koeficient premice $k=3$ ter dobimo enačbo $-2 a+b=3$.\n\nRešimo sistem e... | Slovenia | 17. tekmovanje dijakov srednjih tehniških in strokovnih šol v znanju matematike | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a = -5, b = -7 | |
00l3 | Determine all positive integers $k$ and $n$ satisfying the equation
$$k^2 - 2016 = 3^n$$ | [
"We immediately see that $n = 1$ does not lead to a solution, while $n = 2$ yields the solution $(k, n) = (45, 2)$.\n\nWe show that there is no solution with $n \\ge 3$. In that case $3^n$ is divisible by $9$ and thus $k^2$ is divisible by $9$ which implies that $k = 3l$ for some positive integer $l$. After divisio... | Austria | Regional Competition | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof and answer | (45, 2) | |
0ai5 | Two tangents are drawn from a point $M$ to circle $k$, that touch it at points $G$ and $H$. If $O$ is the center of $k$ and $K$ is the orthocenter of the triangle $MGH$, prove that $\angle GMH = \angle OGK$.
Од точка $M$ кон кружница $k$ се повлечени две тангенти со допирни точки $G$ и $H$. Ако $O$ е центарот на $k$ и... | [
"Let us notice that $K$ must lie on $OM$. From $HK \\perp GM$ and $OG \\perp GM$, it follows that $HK \\parallel OG$. Analogously $OH \\parallel GK$. From $\\overline{OG} = \\overline{OH}$, it follows that $OHKG$ is a rhombus. Let us notice that $O$, $H$, $M$ and $G$ lie on the circle with diameter $OM$. Hence $\\a... | North Macedonia | Macedonian Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
08q2 | Problem:
We have a group of $n$ kids. For each pair of kids, at least one has sent a message to the other one. For each kid $A$, among the kids to whom $A$ has sent a message, exactly $25\%$ have sent a message to $A$. How many possible two-digit values of $n$ are there? | [
"Solution:\n\nIf the number of pairs of kids with two-way communication is $k$, then by the given condition the total number of messages is $4k + 4k = 8k$. Thus the number of pairs of kids is $\\frac{n(n-1)}{2} = 7k$. This is possible only if $n \\equiv 0,1 \\pmod{7}$.\n\n- In order to obtain $n = 7m + 1$, arrange ... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 26 | |
04ig | Let $\triangle ABC$ be an acute triangle with $|AC| > |AB|$. Let $N$ be the foot of the altitude from point $A$ to side $\overline{BC}$. Let $P$ be a point on the extension of $\overline{AB}$ over point $B$, and let $Q$ be a point on the extension of $\overline{AC}$ over point $C$ such that $BPQC$ is a cyclic quadrilat... | [
"The sum of opposite angles in a cyclic quadrilateral is $180^\\circ$, therefore $\\angle CQP = 180^\\circ - \\angle PBC$ and $\\angle QPB = 180^\\circ - \\angle BCQ$, from which it follows that $\\angle AQP = \\angle CBA$ and $\\angle QPA = \\angle ACB$.\nLet $O$ be the centre of the circle circumscribed to $APQ$.... | Croatia | Croatia Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0ack | Find all prime numbers $p, q, r$, such that $\frac{p}{q} - \frac{4}{r+1} = 1$. | [
"$(p, q, r) \\in \\{(3, 2, 7), (5, 3, 5), (7, 3, 2)\\}$\n\nWe can rewrite in the form\n$$\n\\frac{pr + p - 4q}{q(r + 1)} = 1 \\Rightarrow pr + p - 4q = qr + q \\Rightarrow r(p - q) = 5q - p.\n$$\n\nHence $p \\neq q$\n$$\nr = \\frac{5q-p}{p-q} = \\frac{4q+q-p}{p-q} \\quad \\text{i.e.} \\quad r = \\frac{4q}{p-q} - 1.... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | (3, 2, 7), (5, 3, 5), (7, 3, 2) | |
00t7 | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that
$$
f(x^2 + y) \ge \left(\frac{1}{x} + 1\right) f(y)
$$
holds for all $x \in \mathbb{R} \setminus \{0\}$ and all $y \in \mathbb{R}$. | [
"We will show that $f(x) = 0$ for all $x \\in \\mathbb{R}$ which obviously satisfies the equation.\n\nFor $x = -1$ and $y = t + 1$ we get $f(t) \\geq 0$ for every $t \\in \\mathbb{R}$.\n\nFor $x = \\frac{1}{n}$, we get that\n$$\nf\\left(y + \\frac{1}{n^2}\\right) \\geq (n + 1)f(y).\n$$\nTherefore\n$$\nf\\left(y + \... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 0 for all real x | |
086k | Problem:
Sia $ABC$ un triangolo con tutti gli angoli maggiori di $45^{\circ}$ e minori di $90^{\circ}$.
a. Dimostrare che si possono disporre all'interno di $ABC$ tre quadrati con le seguenti proprietà:
- i tre quadrati hanno tutti lo stesso lato;
- i tre quadrati hanno un vertice comune $K$ interno al triangolo;
- d... | [
"Solution:\n\nPer dimostrare il punto (a) lavoriamo al contrario, disponendo arbitrariamente dei quadrati di lato fissato ed osservando quali sono le caratteristiche del triangolo $ABC$ generato da tale disposizione. Per l'esattezza (con riferimento alla figura) scegliamo tre angoli\n$$\n\\left\\{\\begin{array}{l}\... | Italy | Cesenatico | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions a... | null | proof only | null | |
0h45 | Find all pairs of positive rational numbers $x$ and $y$ such that both $x + y^{-1}$ and $y + x^{-1}$ are integers. | [
"Нехай $x = \\frac{m}{n}$, $y = \\frac{p}{q}$, де $m, n, p, q$ — натуральні числа, причому $(m; n) = (p; q) = 1$. Оскільки $x + y^{-1} = \\frac{m}{n} + \\frac{q}{p} = \\frac{mp + nq}{np}$ має бути цілим, то $np \\mid mp + nq$. Аналогічно, $y + x^{-1} = \\frac{p}{q} + \\frac{n}{m} = \\frac{pm + nq}{qm}$ має бути ціл... | Ukraine | Ukrainian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof and answer | (x, y) ∈ {(1, 1), (1/2, 2), (2, 1/2)} | |
06w1 | A magician intends to perform the following trick. She announces a positive integer $n$, along with $2n$ real numbers $x_{1} < \ldots < x_{2n}$, to the audience. A member of the audience then secretly chooses a polynomial $P(x)$ of degree $n$ with real coefficients, computes the $2n$ values $P(x_{1}), \ldots, P(x_{2n})... | [
"Answer: No, she cannot.\n\nLet $x_{1} < x_{2} < \\ldots < x_{2n}$ be real numbers chosen by the magician. We will construct two distinct polynomials $P(x)$ and $Q(x)$, each of degree $n$, such that the member of audience will write down the same sequence for both polynomials. This will mean that the magician canno... | IMO | IMO 2020 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Linear Algebra > Matrices"
] | null | proof and answer | No, she cannot. | |
0blu | Let $(R, +, \cdot)$ be a unit ring such that for all $x \in R$ one can find $e_1, e_2 \in R$, such that $e_1^2 = e_1$, $e_2^2 = e_2$ and $x = e_1e_2$. Show that:
a) $1$ is the only invertible element in $R$;
b) $x^2 = x$, for all $x \in R$. | [
"a) Let $x$ be an invertible element of $R$, let $e_1$ and $e_2$ be idempotent elements of $R$ such that $x = e_1e_2$, and notice that $1 - e_1 = (1 - e_1) \\cdot 1 = (1 - e_1)e_1e_2x^{-1} = (e_1 - e_1^2)e_2x^{-1} = 0$, so $e_1 = 1$ and $x^2 = e_2^2 = e_2 = x$. Since $x$ is invertible, it follows that $x = 1$.\n\nb... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Abstract Algebra > Ring Theory"
] | null | proof only | null | |
07vo | The Mosers' spindle is a seven point configuration in the plane in which **ABC**, **BCD**, **AEF** and **EFG** are equilateral triangles with side length $1$. The points $A$, $B$, $D$, $G$, $F$ form a convex pentagon with $|DG| = 1$.
Determine which of the angles, $\angle BAF$ or $\angle BDG$,... | [
"Connect $A$ to $D$ and $G$. Because $\\triangle ABC$ and $\\triangle BCD$ are equilateral, $\\angle BAD = \\angle BDA = 30^\\circ$ and $|AD| = \\sqrt{3}$. Similarly, $\\angle GAF = 30^\\circ$ and $|AG| = \\sqrt{3}$. We now have $\\angle BDG = \\angle BDA + \\angle ADG = 30^\\circ + \\angle ADG$\n![](attached_image... | Ireland | IRL_ABooklet_2023 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | ∠BDG > ∠BAF | |
054t | Around each vertex of a regular hexagon of side length $\sqrt{3}$ in a plane, one draws a circle of radius $1$ with centre at that vertex and paints the region inside the circle blue. Find the area of the part of the plane that is painted blue. | [
"The circles drawn around two neighbouring vertices of the hexagon intersect, since $2 \\cdot 1 > \\sqrt{3}$. Hence every two neighbouring circles have a common region of the shape of a lens. Since a regular hexagon can be put together from six equilateral triangles, the circumradius of the hexagon equals $\\sqrt{3... | Estonia | National Olympiad Final Round | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 4π + 3√3 | |
0hk9 | Problem:
Solve
$$
2 \sqrt{1+x \sqrt{1+(x+1) \sqrt{1+(x+2) \sqrt{1+(x+3)(x+5)}}}}=x
$$ | [
"Solution:\n\nAs the left hand side is nonnegative, we see that any solution will have $x \\geq 0$. For such $x$ we have\n$$\n\\sqrt{1+(x+3)(x+5)} = \\sqrt{x^2+8x+16} = \\sqrt{(x+4)^2} = |x+4| = x+4.\n$$\nProceeding similarly we get\n$$\n\\begin{aligned}\n& 2 \\sqrt{1+x \\sqrt{1+(x+1) \\sqrt{1+(x+2) \\sqrt{1+(x+3)(... | United States | Berkeley Math Circle | [
"Algebra > Intermediate Algebra > Other",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | no real solutions | |
0koc | Find the remainder when
$$
\binom{3}{2} + \binom{4}{2} + \dots + \binom{40}{2}
$$
is divided by $1000$. | [] | United States | AIME II | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 659 | |
05i2 | Problem:
Soit $ABC$ un triangle, $\Gamma$ son cercle circonscrit et $\omega$ le cercle de même centre que $\Gamma$ et tangent à la droite $(BC)$. Les tangentes au cercle $\omega$ passant par $A$ coupent $(BC)$ en un point $X$ du côté de $B$ et en un point $Y$ du côté de $C$. La tangente au cercle $\Gamma$ en $B$ et la... | [
"Solution:\n\nAfin d'éviter d'avoir à séparer les différents cas, en fonction de la position du point $Y$ par rapport au segment $[BC]$, nous allons utiliser les angles de droite orientés : $(AB, CD)$ désigne l'angle (relatif) dont il faut tourner la droite $(AB)$ pour qu'elle soit parallèle à la droite $(CD)$.\n\n... | France | ENVOI 1 : GÉOMÉTRIE Corrigé | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0fsc | Problem:
Zwei Kreise schneiden sich in den beiden Punkten $M$ und $N$. Sei $A$ ein weiterer Punkt auf dem ersten Kreis, verschieden von $M$ und $N$. Die Geraden $A M$ und $A N$ schneiden den zweiten Kreis nochmals in den Punkten $B$ und $C$. Zeige, dass die Tangente an den ersten Kreis im Punkt $A$ parallel zur Gerade... | [
"Solution:\n\nSei $P$ ein Punkt auf der Tangente an den ersten Kreis durch $A$, sodass $P$ und $N$ auf verschiedenen Seiten der Geraden $A M$ liegen. Nach dem Tangentenwinkelsatz gilt $\\Varangle P A M = \\Varangle A N M$. Da $M N C B$ ein Sehnenviereck ist, gilt ausserdem $\\Varangle C B M = 180^{\\circ} - \\Varan... | Switzerland | IMO - Selektion | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jgy | Two incongruent triangles $ABC$ and $XYZ$ are called a pair of *pals* if they satisfy the following conditions:
(a) the two triangles have the same area;
(b) let $M$ and $W$ be the respective midpoints of sides $BC$ and $YZ$. The two sets of lengths $\{AB, AM, AC\}$ and $\{XY, XW, XZ\}$ are identical 3-element sets of ... | [
"The answer is *yes*.\nWe start with the following observations.\n\n**Lemma 2.** The following statement and its converse are both true: If $q, r, s$ are three distinct positive real numbers such that $q, r, 2s$ are side lengths of a triangle, then there is an unique triangle $PQR$ with $PQ = q, PR = r$, and $PS = ... | United States | IMO Team Selection Test | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | null | proof and answer | Yes | |
0ci8 | A polynomial $P$ with integer coefficients is *square-free* if it is not expressible in the form $P = Q^2R$, where $Q$ and $R$ are polynomials with integer coefficients and $Q$ is not constant. For a positive integer $n$, let $P_n$ be the set of polynomials of the form
$$
1 + a_1x + a_2x^2 + \dots + a_nx^n
$$
with $a_1... | [
"Clearly, $|P_n| = 2^n$. Alternatively, but equivalently, we prove that less than $\\frac{1}{100} \\cdot 2^n$ of the polynomials in $P_n$ are not square-free for all but finitely many $n$. Throughout the solution $n$ is always assumed to be sufficiently large to allow room for as large integers $r \\le n$ as the di... | Romania | Romanian Master of Mathematics | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Intermediate Algebra > Complex numbers",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
... | English | proof only | null |
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