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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0bm3 | Given non-negative real numbers $a$, $b$, $c$ such that $ab + bc + ca + 2abc = 1$, show that $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge 2$ and determine the cases of equality. | [
"The condition in the statement is equivalent to $\\frac{1}{1+a} + \\frac{1}{1+b} + \\frac{1}{1+c} = 2$. If $t$ is a non-negative real number, then $\\sqrt{t} \\ge \\frac{2t}{1+t} = 2 - \\frac{2}{1+t}$, and equality holds if and only if $t$ is either $0$ or $1$. Consequently,\n$$\n\\begin{aligned}\n\\sqrt{a} + \\sq... | Romania | 2015 Ninth STARS OF MATHEMATICS Competition | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | The minimum possible value of the sum of square roots is two, attained precisely when one of the numbers is zero and the other two are equal to one (up to permutation). | |
0if8 | Problem:
In triangle $ABC$ with altitude $AD$, $\angle BAC = 45^{\circ}$, $DB = 3$, and $CD = 2$. Find the area of triangle $ABC$. | [
"Solution:\n\nSuppose first that $D$ lies between $B$ and $C$. Let $ABC$ be inscribed in circle $\\omega$, and extend $AD$ to intersect $\\omega$ again at $E$. Note that $A$ subtends a quarter of the circle, so in particular, the chord through $C$ perpendicular to $BC$ and parallel to $AD$ has length $BC = 5$. Ther... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 15 | |
08q7 | Problem:
Let $P$ be a point in the interior of a triangle $A B C$. The lines $A P, B P$ and $C P$ intersect again the circumcircles of the triangles $P B C, P C A$, and $P A B$ at $D, E$ and $F$ respectively. Prove that $P$ is the orthocenter of the triangle $D E F$ if and only if $P$ is the incenter of the triangle $A... | [
"Solution:\nIf $P$ is the incenter of $A B C$, then $\\angle B P D = \\angle A B P + \\angle B A P = \\frac{\\hat{A} + \\hat{B}}{2}$, and $\\angle B D P = \\angle B C P = \\frac{\\hat{C}}{2}$. From triangle $B D P$, it follows that $\\angle P B D = 90^{\\circ}$, i.e. that $E B$ is one of the altitudes of the triang... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fc4 | Problem:
El triángulo $ABC$ es isósceles en $C$, y sea $\Gamma$ su circunferencia circunscrita. Sea $M$ el punto medio del arco $BC$ de $\Gamma$ que no contiene a $A$, y sea $N$ el punto donde la paralela a $AB$ por $M$ vuelve a cortar a $\Gamma$. Se sabe que $AN$ es paralela a $BC$. ¿Cuáles son las medidas de los áng... | [
"Solution:\n\nSi $AN$ es paralela a $BC$, entonces $ABCN$ es un trapecio con circunferencia circunscrita, y por lo tanto isósceles.\n\n\n\nSe tiene entonces que $\\angle ANC = \\angle BAN$. Pero $\\angle NAC = \\angle ACB$ y $\\angle ANM = \\angle ABC$ por ser $AN$ y $BC$ paralelas, y ser $... | Spain | null | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | Angle A = Angle B = 72°, Angle C = 36° | |
09g6 | Let $ABC$ be a scalene triangle. The midpoints of the sides $AB$, $BC$ and $CA$ are denoted $C_0$, $A_0$ and $B_0$ respectively. Let $l_A$, $l_B$ and $l_C$ denote the bisectors of the interior angles of $A$, $B$ and $C$ respectively. If $M$ is the intersection of the perpendicular from $C$ to $l_C$ and the perpendicula... | [
"Let $I$ denote the incenter of $ABC$, i.e. the intersection of the bisectors $l_A$, $l_B$ and $l_C$.\n\nWe denote by $P$ and $Q$ the bases of the perpendiculars from $A$ to $l_C$ and $C$ to $l_A$, respectively. Let $PQ$ intersect $AB$ and $BC$ at $R$ and $S$ respectively. Since $APQC$ is inscribed, we have $\\angl... | Mongolia | 2015 Mongolian IMO Team Selection Tests | [
"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",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
036f | Problem:
There are $40$ knights in a kingdom. Every morning they fight in pairs (everyone has exactly one enemy to fight with) and every evening they sit around a table (during the evening they do not change their seats). Find the least number of days such that:
a) the fights can be arranged in a way that every two k... | [] | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | a) 39; b) 20 | |
0gtk | In a triangle $ABC$ with $\hat{B} < \hat{C}$, let $K$ be the center of the excircle that is tangent to the side $[AC]$. The lines $AK$ and $BC$ intersect at $D$, and $E$ is the center of the circumcircle of $BKC$. Prove that
$$
\frac{1}{|KA|} = \frac{1}{|KD|} + \frac{1}{|KE|}.
$$ | [
"\nWe will show that $\\frac{|KA|}{|KD|} + \\frac{|KA|}{|KE|} = 1$. Let $\\angle A = 2\\alpha$, $\\angle B = 2\\beta$ and $\\angle C = 2\\theta$. Recall that $\\alpha + \\beta + \\theta = 90^\\circ$. By angle chasing we obtain that $\\angle ADC = \\theta - \\beta$, $\\angle ABK = \\angle KB... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof only | null | |
03p6 | 
From point $P$ outside a circle draw two tangents to the circle touching at points $A$ and $B$. Draw a secant line intersecting the circle at points $C$ and $D$, with $C$ between $P$ and $D$. Choose point $Q$ on the chord $CD$ such that $\angle DAQ = \angle PBC$. Prove that $\angle DBQ = \angl... | [
"Using $\\angle DAB = \\angle DCB$, $\\angle DAB = \\angle DAQ + \\angle QAB$, $\\angle DCB = \\angle PBC + \\angle BPQ$, and $\\angle DAQ = \\angle PBC$, we get\n\n$\\angle QAB = \\angle BPQ$, so points $P, A, Q, B$ share a common circle. Then $\\angle BQP = \\angle PAB$, that is, $\\angle DBQ + \\angle CDB = \\an... | China | China Mathematical Competition (Extra Test) | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0kxt | Problem:
Triangle $ABC$ with $\angle BAC > 90^{\circ}$ has $AB = 5$ and $AC = 7$. Points $D$ and $E$ lie on segment $BC$ such that $BD = DE = EC$. If $\angle BAC + \angle DAE = 180^{\circ}$, compute $BC$. | [
"Solution:\n\nLet $M$ be the midpoint of $BC$, and consider dilating about $M$ with ratio $-\\frac{1}{3}$. This takes $B$ to $E$, $C$ to $D$, and $A$ to some point $A'$ on $AM$ with $AM = 3A'M$. Then the angle condition implies $\\angle DAE + \\angle EA'D = 180^{\\circ}$, so $AD A' E$ is cyclic. Then by power of a ... | United States | HMMT February | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(111) | |
082g | Problem:
Dati due numeri $x$ e $y$ con $0 < x < 1$, $0 < y < 1$ quale delle seguenti affermazioni è sempre vera?
(A) $x + y < 1$
(B) $\frac{x}{y} < 1$
(C) $x + y > 1$
(D) $x^{2} + y^{2} > 1$
(E) $x < \frac{1}{y}$. | [
"Solution:\n\nLa risposta è $\\mathbf{(E)}$. Tale affermazione è sempre vera, poiché $x < 1$, mentre $\\frac{1}{y} > 1$. Per $x = \\frac{1}{2}$, $y = \\frac{1}{3}$ le ipotesi del testo sono verificate, ma (B), (C) e (D) sono false. Infine $x = y = \\frac{3}{4}$ fornisce un controesempio per (A)."
] | Italy | Progetto Olimpiadi di Matematica 2003 GARA di SECONDO LIVELLO BIENNIO | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | E | |
0fhm | Problem:
Dado un número natural $n>0$ y un número complejo $z = x + i y$ de módulo unidad, $x^2 + y^2 = 1$, se puede cumplir o no la igualdad
$$
\left(z + \frac{1}{z}\right)^n = 2^{n-1}\left(z^n + \frac{1}{z^n}\right)
$$
Fijado $n$, designaremos por $S(n)$ al subconjunto de complejos de módulo unidad para los que se c... | [
"Solution:\n\nComo $z = \\cos t + i \\operatorname{sen} t$, sustituyendo en la igualdad resulta la ecuación trigonométrica\n$$\n\\cos^n t = \\cos n t\n$$\nque para $n=2,3,4$ tiene las soluciones\n$$\n\\begin{array}{ll}\nn=2, & S(n)=\\{1,-1\\} \\\\\nn=3, & S(n)=\\{1,-1, i,-i\\} \\\\\nn=4, & S(n)=\\{1,-1, \\cos t + i... | Spain | OME 28 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials"
] | null | proof and answer | For unit-modulus z, the condition is equivalent to cos^n t = cos(nt) with z = cos t + i sin t.
- n = 2: S(2) = {1, −1}.
- n = 3: S(3) = {1, −1, i, −i}.
- n = 4: S(4) = {1, −1} together with the four points where tan t = ±√6 (so cos^2 t = 1/7); in total, S(4) has 6 elements.
- For n > 5: an upper bound on the size is |S... | |
0apu | Problem:
Given that $x+2$ and $x-3$ are factors of $p(x)=a x^{3}+a x^{2}+b x+12$, what is the remainder when $p(x)$ is divided by $x-1$? | [
"Solution:\nSince $x+2$ is a factor of $p(x)=a x^{3}+a x^{2}+b x+12$, Factor Theorem guarantees that\n$$\np(-2) = -8a + 4a - 2b + 12 = 0 \\quad \\text{or} \\quad 2a + b = 6\n$$\nSimilarly, we also have\n$$\np(3) = 27a + 9a + 3b + 12 = 0 \\quad \\text{or} \\quad 12a + b = -4\n$$\nSolving the system of equations invo... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 18 | |
04kt | A triangle $ABC$ is given. Circle $k$ touches $\overline{BC}$ from outside the triangle at point $K$, and the extensions of lines $\overline{AB}$ and $\overline{AC}$ over points $B$ and $C$ at points $L$ and $M$, respectively. The circle with diameter $\overline{BC}$ intersects segment $LM$ at points $P$ and $Q$ so tha... | [
"Denote $\\alpha = \\langle CAB$, $\\beta = \\langle ABC$, $\\gamma = \\langle BCA$.\n\n\nThe centre of circle $k$ is the intersection of the bisectors of angles $\\angle CBL$ and $\\angle BCM$.\nTherefore, it suffices to show that those bisectors intersect line $LM$ at points $P$ and $Q$ r... | Croatia | Mathematical competitions in Croatia | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hbz | Circle $k$ of radius $r$ is inscribed in $\triangle ABC$. Tangent lines of $k$, that are parallel to sides $AB$, $BC$ and $CA$, intersect other sides of $\triangle ABC$ at points $M, N$; $P, Q$ and $L, T$ ($P, T \in AB$, $L, N \in BC$ and $M, Q \in AC$). Denote by $r_1, r_2, r_3$ radii of circles inscribed in triangles... | [
"Since all these triangles are similar, we get\n$$\n\\frac{r_1 + r_2 + r_3}{r} = \\frac{p_1 + p_2 + p_3}{p}\n$$\nIt is not hard to see that (fig. 28)\n$$\n\\begin{aligned}\n2p_1 &= CM + CN + MN = \\\\\n&= CM + CN + MZ + ZN = \\\\\n&= CM + MX + CN + NY = 2p - (AX + AB + BY) = 2p - 2c.\n\\end{aligned}\n$$\nThus,\n$$\... | Ukraine | 59th Ukrainian National 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 > Distance chasing"
] | English | proof only | null | |
02hr | Problem:
1) Qual é o maior dos números?
(A) $2 \times 0 \times 2006$
(B) $2 \times 0+6$
(C) $2+0 \times 2006$
(D) $2 \times(0+6)$
(E) $2006 \times 0+0 \times 6$ | [
"Solution:\n\n(D) Lembre que se num produto um dos fatores é zero, então o produto também é zero. Temos:\n$2 \\times 0 \\times 2006=0$\n$2 \\times 0+6=0+6=6$\n$2+0 \\times 2006=2+0=2$\n$2 \\times(0+6)=2 \\times 6=12$\n$2006 \\times 0+0 \\times 6=0+0=0$\nLogo, o maior é $2 \\times(0+6)$."
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | D | |
0dof | Problem:
За природан број $n$, са $v_{2}(n)$ означавамо највећи цео број $k \geqslant 0$ такав да $2^{k} \mid n$. Претпоставимо да функција $f: \mathbb{N} \rightarrow \mathbb{N}$ задовољава услове:
(i) $f(x) \leqslant 3 x$ за све $x \in \mathbb{N}$;
(ii) $v_{2}(f(x)+f(y))=v_{2}(x+y)$ за све $x, y \in \mathbb{N}$.
Дока... | [
"Solution:\n\nЗаменом $x=y$ добијамо $v_{2}(f(x))=v_{2}(x)$. Ако је $v_{2}(a)=k>0$, посматрањем функције $g(x)=f\\left(2^{k} x\\right) / 2^{k}$ тврђење сводимо на случај непарног $a$. Зато надаље сматрамо да $2 \\nmid a$.\n\nПриметимо да, ако је $x \\not \\equiv y\\left(\\bmod 2^{k}\\right)$, онда је $f(x) \\not \\... | Serbia | 14. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / ... | null | proof only | null | |
00hb | A positive integer is called fancy if it can be expressed in the form
$$
2^{a_{1}} + 2^{a_{2}} + \cdots + 2^{a_{100}}
$$
where $a_{1}, a_{2}, \ldots, a_{100}$ are non-negative integers that are not necessarily distinct.
Find the smallest positive integer $n$ such that no multiple of $n$ is a fancy number. | [
"Let $k$ be any positive integer less than $2^{101}-1$. Then $k$ can be expressed in binary notation using at most 100 ones, and therefore there exists a positive integer $r$ and non-negative integers $a_{1}, a_{2}, \\ldots, a_{r}$ such that $r \\leq 100$ and $k = 2^{a_{1}} + \\cdots + 2^{a_{r}}$. Notice that for a... | Asia Pacific Mathematics Olympiad (APMO) | APMO 2016 | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 2^{101} - 1 | |
0ad1 | A $4 \times 4$ table is divided into 16 white unit square cells. Two cells are called neighbours if they share a common side. A move consists in choosing a cell and changing its color and the colors of its neighbours from white to black or from black to white. After exactly $n$ moves all the 16 cells were black. Find a... | [
"Figure covers at most 5 cells, so we need at least 4 steps to change color every cell. If we place figure 4 times so that center of figure lies in dark cell we consider that is possible to do that if $n=4$.\n\nFurthermore, applying operation two times on same cells we get that is possible to get table with every c... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | All even integers greater than or equal to four | |
0ebv | Problem:
Izračunaj naravni števili, katerih vsota je $168$, največji skupni delitelj pa $24$. Zapiši vse možne rešitve. | [
"Solution:\n\nNaj bosta iskani števili $a$ in $b$. Podano je:\n\n- $a + b = 168$\n- $\\gcd(a, b) = 24$\n\nKer je njun največji skupni delitelj $24$, lahko zapišemo $a = 24m$, $b = 24n$, kjer sta $m$ in $n$ tuji števili ($\\gcd(m, n) = 1$).\n\nTorej:\n$$\na + b = 24m + 24n = 24(m + n) = 168\n$$\n$$\nm + n = \\frac{1... | Slovenia | 15. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | (24, 144), (48, 120), (72, 96) | |
08ci | Problem:
Una gara di matematica consta di 90 domande a risposta multipla. Camilla ha risposto a tutte le domande: quale dei seguenti non può essere il punteggio totalizzato da Camilla, sapendo che una risposta corretta vale 5 punti e una risposta sbagliata vale -1 punto?
(A) -78
(B) 116
(C) 204
(D) 318
(E) 402 | [
"Solution:\n\nLa risposta è (B). Sia $r$ il numero di risposte corrette date da Camilla; per differenza, il numero di risposte sbagliate è $90 - r$, e il punteggio totale è $5r + (-1)(90 - r) = 6r - 90 = 6(r - 15)$. Ne segue che il punteggio di Camilla è multiplo di 6, e quindi non può essere uguale a $116 = 2^{2} ... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | B | |
03v4 | It is given the sequence $\{a_n\}$: $a_1 = 1$,
$$
a_{n+1} = 2a_n + n \cdot (1 + 2^n), \quad n = 1, 2, 3, \dots
$$
Find the general term $a_n$. | [
"Divide the recursion formula by $2^{n+1}$ throughout, we obtain\n$$\n\\frac{a_{n+1}}{2^{n+1}} = \\frac{a_n}{2^n} + \\frac{n}{2^{n+1}} + \\frac{n}{2},\n$$\nthat is,\n$$\n\\frac{a_{n+1}}{2^{n+1}} - \\frac{a_n}{2^n} = \\frac{n}{2^{n+1}} + \\frac{n}{2}.\n$$\nThen\n$$\n\\sum_{i=1}^n \\left( \\frac{a_{i+1}}{2^{i+1}} - \... | China | China Southeastern Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | a_n = 2^{n-2}(n^2 - n + 6) - n - 1 (n ≥ 2) | |
0lf1 | A fractional number $x$ is called ‘pretty’ if it has finite expression in base $b$ numeral system where $b$ is a positive integer in $[2; 2022]$. Prove that there exists finite positive integers $n \ge 4$ such that with every $m$ in $\left(\frac{2n}{3}, n\right)$, there is at least one pretty number among two numbers $... | [
"Call a positive integer $n$ 'good' if there exists $m$ that $m$ in the interval $\\left(\\frac{2n}{3}, n\\right)$ and $\\frac{m}{n-m}, \\frac{n-m}{m}$ is not pretty. Next, we will prove the following claims.\n\n**Claim 1.** If $n$ is good then any multiple of $n$ is also good.\n\n*Proof.* Consider a good number $n... | Vietnam | IMO Team Selection Test | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
06hq | There were $36$ participants in a party, some of whom shook hands with each other, such that any two participants shook hands with each other at most once. Each participant then recorded the number of handshakes made, and it was found that no two participants with the same number of handshakes made had shaken hands wit... | [] | Hong Kong | HONG KONG PRELIMINARY SELECTION CONTEST | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English; Chinese | proof and answer | 546 | |
0hi8 | Is it possible to write positive integers in the cells of the board $2022 \times 2022$ in such a way that the sum of the numbers in any rectangle $R$ is a perfect square if and only if $R$ is a square?
(Arsenii Nikolaiev) | [
"Yes, you can.\n\nNumber the columns from left to right and the rows from top to bottom with the numbers $1, 2, \\dots, 2022$. Write in the cell $(r, c)$, i.e. the one in the $r$-th row and $c$-th column, the number $x^{2(r+c)}$, where the number $x$ is selected later.\n\nConsider an arbitrary rectangle $R$, which ... | Ukraine | Problems from Ukrainian Authors | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0l1j | Problem:
Given that the 32-digit integer
$$
64312311692944269609355712372657
$$
is the product of 6 consecutive primes, compute the sum of these 6 primes. | [
"Solution:\nBecause the product is approximately $64 \\cdot 10^{30}$, we know the primes are all around $200000$. Say they are $200000 + x_{i}$ for $i = 1, \\ldots, 6$.\nBy expanding $\\prod_{i=1}^{6}\\left(200000 + x_{i}\\right)$ as a polynomial in $200000$, we see that\n$$\n31231 \\cdot 10^{25} = 200000^{5}\\left... | United States | HMMT February 2024 Guts Round | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | 1200974 | |
0dm4 | Each student in the class wrote down in his notebook a geometric progression consisting of different positive numbers. Let $A$ be the sum of the first terms of all these progressions, $B$ the sum of their second terms, and $C$ the sum of their third terms. Prove that the sequence $A$, $B$, $C$ is not an arithmetic prog... | [] | Saudi Arabia | Saudi Booklet | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
02eu | $P_0 = (1,0), P_1 = (1,1), P_2 = (0,1), P_3 = (0,0)$. $P_{n+4}$ is the midpoint of $P_n P_{n+1}$. $Q_n$ is the quadrilateral $P_n P_{n+1} P_{n+2} P_{n+3}$. $A_n$ is the interior of $Q_n$. Find $\bigcap_{n \ge 0} A_n$. | [
"Interpreting the points $P_k$ as vectors, we have $P_{n+4} = \\frac{P_n + P_{n+1}}{2}$, which is a linear homogeneous recursion. Its characteristic polynomial is $2x^4 - x - 1 = (x - 1)(2x^3 + 2x^2 + 2x + 1)$. So let $\\alpha, \\beta, \\gamma$ be the roots of $f(x) = 2x^3 + 2x^2 + 2x + 1$, so that $P_n = Q_0 + Q_1... | Brazil | XIII OBM | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | (3/7, 4/7) | |
0bm4 | Let $ABCD$ be a cyclic quadrangle, let the diagonals $AC$ and $BD$ cross at $O$, and let $I$ and $J$ be the incentres of the triangles $ABC$ and $ABD$, respectively. The line $IJ$ crosses the segments $OA$ and $OB$ at $M$ and $N$, respectively. Prove that the triangle $OMN$ is isosceles. | [
"\n\nWe show that $\\angle OMN \\equiv \\angle ONM$. To this end, let the line $IJ$ cross the segments $AD$ and $BC$ at $P$ and $Q$, respectively, and consider the position of $J$ relative to the line $AC$, to write $\\angle OMN \\equiv \\angle AJP \\pm \\angle JAM \\equiv \\angle AJP \\pm ... | Romania | 2015 Danube Mathematical Competition | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08du | Problem:
Sia $ABC$ un triangolo e sia $D$ il piede della bisettrice uscente dal vertice $A$. Sia $\omega$ la circonferenza tangente ad $AC$ in $A$ e passante per $D$. Sia $P$ la seconda intersezione di $\omega$ con la retta $BC$. Sapendo che $AC=54$, $AD=63$ e $CP=108$, trovare $AB$.
(A) 72
(B) $\frac{147}{2}$
(C) 98... | [
"Solution:\n\nLa risposta è $(\\mathbf{C})$. Dato che $CA$ è tangente a $\\omega$, abbiamo $CP=\\frac{AC^{2}}{CD}=\\frac{54^{2}}{108}=27$. Inoltre $\\angle DAC=\\angle CPA$, ma $AD$ è la bisettrice di $\\angle BAC$, quindi $\\angle BAD=\\angle CPA$. Di conseguenza la circonferenza circoscritta al triangolo $ABP$ è ... | Italy | Olimpiadi della Matematica | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | C | |
0dg8 | Do there exist positive integers $m$ and $n$ such that the decimal representation of $5^m$ starts with $2^n$ and the decimal representation of $2^m$ starts with $5^n$? | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Number Theory > Other"
] | English | proof and answer | No | |
0gxm | Consider the set $M = \{1,2,3,4,6,8,12,16,24,48\}$ and all of its four-element subsets. Denote by $n$ the number of these subsets such that the product of their elements is greater than $2009$ and by $m$ the number of these subsets such that the product of their elements is smaller than $2009$. Which number is bigger -... | [
"Let $M_1 = \\{a, b, c, d\\}$ be a four-element subset of $M$ such that $abcd < 2009$. Then the subset $M_2 = \\{\\frac{48}{a}, \\frac{48}{b}, \\frac{48}{c}, \\frac{48}{d}\\}$ is also a four-element subset of $M$ because all of its elements are distinct and belong to $M$ and $\\frac{48}{a} \\cdot \\frac{48}{b} \\cd... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | n > m | |
09ek | 148 points lie on a circle. One wants to join these points by line segments in such a way that no two of line segments intersect inside the circle. Prove that there exist 50 points such that any two of them are not connected. | [
"Let $F_n$ denote the set of configurations of $n$ points joined by line segments in such a way that no two of the line segments intersect inside the circle. If $f \\in F_n$ then the given $n$ points are said to be vertices of subconfiguration $f$. First we shall prove that it is possible to colour vertices of $f$ ... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0a9n | Problem:
Find the smallest positive integer $n$, such that there exist $n$ integers $x_{1}, x_{2}, \ldots, x_{n}$ (not necessarily different), with $1 \leq x_{k} \leq n, 1 \leq k \leq n$, and such that
$$
x_{1}+x_{2}+\cdots+x_{n}=\frac{n(n+1)}{2}, \quad \text{ and } \quad x_{1} x_{2} \cdots x_{n}=n!
$$
but $\left\{x_{1... | [
"Solution:\nIf it is possible to find a set of numbers as required for some $n=k$, then it will also be possible for $n=k+1$ (choose $x_{1}, \\ldots, x_{k}$ as for $n=k$, and let $x_{k+1}=k+1$). Thus we have to find a positive integer $n$ such that a set as required exists, and prove that such a set does not exist ... | Nordic Mathematical Olympiad | The 26th Nordic Mathematical Contest | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 9 | |
063r | Problem:
Es sei $ABCD$ ein konvexes Viereck. Die Diagonale $BD$ halbiere den Winkel $\measuredangle CBA$. Der Umkreis des Dreiecks $ABC$ schneide die Strecken $\overline{CD}$ und $\overline{DA}$ in den inneren Punkten $P$ bzw. $Q$. Die durch den Punkt $D$ verlaufende Parallele zur Geraden $AC$ schneide die Geraden $BA... | [
"Solution:\n\nSchritt 1: Wegen $\\measuredangle BRD = \\measuredangle BAC = \\measuredangle BPC = \\pi - \\measuredangle DPB$ ist $BPDR$ ein Sehnenviereck. Analog zeigt man, dass auch $BSDQ$ ein Sehnenviereck ist.\n\nSchritt 2: Es sei $X$ definiert als der Schnittpunkt von $BD$ mit dem Umkreis des Dreiecks $ABC$. D... | Germany | 1. Auswahlklausur | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0e18 | Find all functions $g: \mathbb{R} \to \mathbb{R}$, for which there exists a strictly increasing function $f: \mathbb{R} \to \mathbb{R}$, such that
$$
f(x + y) = f(x)g(y) + f(y)
$$
for all real $x$ and $y$. | [
"Inserting $y = 0$ into the equation, we get $f(x) = f(x)g(0) + f(0)$ or\n$$\nf(x)(1 - g(0)) = f(0).\n$$\nIf $g(0) \\neq 1$ then $f(x) = \\frac{f(0)}{1-g(0)}$ and $f$ is constant. This is not possible since $f$ is strictly increasing. We conclude that $g(0) = 1$ and $f(0) = 0$. From\n$$\nf(x)g(y) + f(y) = f(x + y) ... | Slovenia | Selection Examinations for the IMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | g(x) = a^x for all real x, where a > 0 | |
088m | Problem:
Per entrare nel castello di Burian bisogna usare una parola chiave che è costituita da almeno 6 caratteri; inoltre un carattere non si può mai ripetere due o più volte consecutivamente, e una coppia di caratteri consecutivi non può comparire in un altro punto della parola chiave. Sul tastierino funzionano orm... | [
"Solution:\n\nLa risposta è $\\mathbf{(C)}$. Consideriamo i primi due caratteri della parola chiave: poiché essi devono essere diversi, abbiamo $3 \\cdot 2 = 6$ possibilità di sceglierli. Senza perdere generalità, supponiamo che essi siano $NE$. Possiamo a questo punto fare una semplice analisi per casi, scoprendo ... | Italy | Olimpiadi di Matematica | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | MCQ | C | |
0e0l | Let $x$ be a real number such that $x + \frac{1}{x} + 1$ is a positive integer. Prove that $x^2 + \frac{1}{x^2} + 1$ is a positive integer divisible by $x + \frac{1}{x} + 1$. | [
"We have\n$$\nx^2 + \\frac{1}{x^2} + 1 = \\left(x + \\frac{1}{x}\\right)^2 - 1 = \\left(x + \\frac{1}{x} + 1\\right)\\left(x + \\frac{1}{x} - 1\\right).\n$$\nSince $x + \\frac{1}{x} + 1$ is a positive integer, $x + \\frac{1}{x} - 1$ is an integer and so is $x^2 + \\frac{1}{x^2} + 1$. Since $x^2 + \\frac{1}{x^2} + 1... | Slovenia | National Math Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
03g2 | The segment $BC$ is fixed in the plane. Let $A$ be a point, such that $\triangle ABC$ is acute. Let $BP$ ($P \in AC$) and $CQ$ ($Q \in AB$) be altitudes in $\triangle ABC$ and $O$ is the center of the circumscribed circle about $ABC$. The points $K$ and $L$ are symmetric to $O$ with respect to the lines $AB$ and $AC$ r... | [] | Bulgaria | 6 TST for BMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plan... | English | proof only | null | |
0jnk | Problem:
Find all integers $n$, not necessarily positive, for which there exist positive integers $a, b, c$ satisfying $a^{n}+b^{n}=c^{n}$. | [
"Solution:\nAnswer: $\\pm 1, \\pm 2$\nBy Fermat's Last Theorem, we know $n < 3$. Suppose $n \\leq -3$. Then $a^{n} + b^{n} = c^{n} \\Longrightarrow (b c)^{-n} + (a c)^{-n} = (a b)^{-n}$, but since $-n \\geq 3$, this is also impossible by Fermat's Last Theorem. As a result, $|n| < 3$.\n\nFurthermore, $n \\neq 0$, as... | United States | HMMT November 2015 | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | -2, -1, 1, 2 | |
07pe | The points $N$, $E$, $F$ and $M$ are on the sides $AB$, $BC$, $CD$ and $DA$, respectively, of a unit square $ABCD$ such that $|AN| = |BE| = |CF| = |DM| = x$ and $0 < x < 1$. The lines $AE$, $BF$, $CM$ and $DN$ form a quadrilateral $GHKL$. Express the area of $GHKL$ in terms of $x$. | [
"The right triangles $\\triangle ABE$ and $\\triangle BCF$ have legs of the same length, hence they are congruent. This implies that $\\angle GBE + \\angle BEG = 90^\\circ$ and so\n\n$\\angle EGB = 90^\\circ$. This shows that $\\triangle BGE$ is similar to $\\triangle BCF$. Using symmetry we see now that $GHKL$ is ... | Ireland | Ireland | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | (1 - x)^2 / (1 + x^2) | |
0fof | El conjunto $M$ está formado por números enteros de la forma $a^2 + 13b^2$, con $a$ y $b$ enteros distintos de cero.
• Demostrar que el producto de dos elementos cualesquiera de $M$ es un elemento de $M$.
• Determinar, razonadamente, si existen infinitos pares de enteros $(x, y)$ tales que $x + y$ no pertenece a $M$,... | [] | Spain | L Olimpiada Matemática Española | [
"Number Theory > Algebraic Number Theory > Quadratic fields",
"Number Theory > Algebraic Number Theory > Quadratic forms",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | Spanish | proof and answer | Yes. There are infinitely many pairs, for example all integer pairs with x + y equal to one. | |
04l2 | Let $x$ and $y$ be positive real numbers such that
$$
2^{x^2} = 16^y \quad \text{and} \quad \log_{\sqrt{2017}} x + \log_{\sqrt{2017}} y > 0.
$$
Prove that $y > \frac{1}{2}$. (Kristina Ana Škreb) | [] | Croatia | Mathematical competitions in Croatia | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0jc2 | Problem:
Let $a_{0} = -2$, $b_{0} = 1$, and for $n \geq 0$, let
$$
\begin{aligned}
& a_{n+1} = a_{n} + b_{n} + \sqrt{a_{n}^{2} + b_{n}^{2}} \\
& b_{n+1} = a_{n} + b_{n} - \sqrt{a_{n}^{2} + b_{n}^{2}}
\end{aligned}
$$
Find $a_{2012}$. | [
"Solution:\nAnswer: $2^{1006} \\sqrt{2^{2010} + 2} - 2^{2011}$ We have\n$$\n\\begin{gathered}\na_{n+1} + b_{n+1} = 2\\left(a_{n} + b_{n}\\right) \\\\\na_{n+1} b_{n+1} = \\left(a_{n} + b_{n}\\right)^{2} - a_{n}^{2} - b_{n}^{2} = 2 a_{n} b_{n}\n\\end{gathered}\n$$\nThus,\n$$\n\\begin{aligned}\na_{n} + b_{n} & = -2^{n... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 2^{1006} sqrt(2^{2010} + 2) - 2^{2011} | |
0bbl | Let $f : \mathbb{R} \to \mathbb{R}$ be a non-decreasing function and $F : \mathbb{R} \to \mathbb{R}$ a function having right and left finite derivatives at any point in $\mathbb{R}$ and $F(0) = 0$. Suppose that $\lim_{x \uparrow x_0} f(x) \le F'_s(x_0)$ and $\lim_{x \downarrow x_0} f(x) \ge F'_d(x_0)$, for any $x_0 \in... | [
"Consider the function $G : \\mathbb{R} \\to \\mathbb{R}$, given by $G(x) = F(x) - \\int_0^x f(t) \\, dt$. Let $x_0 \\in \\mathbb{R}$, $x \\ne x_0$. Then\n$$\nF(x) = \\frac{F(x) - F(x_0)}{x - x_0} \\cdot (x - x_0) + F(x_0).\n$$\nAs $F$ has right and left finite derivatives in $x_0$, the function $x \\mapsto \\frac{... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | proof only | null | |
0eva | Let $ABC$ be an isosceles triangle with $AC = BC$. Let $D$ be a point on a line $BA$ such that $A$ lies between $B$ and $D$. Let $O_1$ be the circumcircle of triangle $DAC$. $O_1$ meets $BC$ at point $E$. Let $F$ be the point on the line $BC$ such that $FD$ is tangent to circle $O_1$, and let $O_2$ be the circumcircle ... | [
"We first show that both $DB$ and $DE$ are tangent to circle $O$. Since $DFBG$ is concyclic, we have $\\angle FDG = \\angle GBE$. Since $FD$ is tangent to $O_1$, we have $\\angle DEG = \\angle FDG$. Hence $\\angle GBE = \\angle DEG$, which means that $DE$ is tangent to $O$. On the other hand, since $ACED$ is concyc... | South Korea | Korean Mathematical Olympiad Final Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0bf0 | The function $f: [a, b] \to \mathbb{R}$ is continuous and $f(y) \le \max\{f(x), f(z)\}$ for every $x, y, z \in [a, b]$ so that $x < y < z$. Denote $m = \min_{x \in [a, b]} f(x)$ and define the set $A = \{x \in [a, b] \mid f(x) = m\}$.
a) Prove that $A$ is a closed interval, possibly degenerated.
b) Prove that, if $a,... | [] | Romania | Shortlisted Problems for the 64th NMO | [
"Precalculus > Functions"
] | null | proof only | null | |
0hp8 | Problem:
There are infinitely many bowls arranged on the number line, one at each integer. Initially each bowl has one fruit in it. In a move, one may take any fruit and move it to an adjacent bowl (bowls may hold more than one fruit, or no fruits at all).
Is it possible that after 999 moves, every bowl still has exac... | [
"Solution:\n\nThe answer is no.\nAfter 999 moves, only finitely many bowls were touched (i.e. received or lost any fruit), so we can consider a large interval $[L, M]$ such that all touched bowls are in this interval.\n\nNow, consider the sum $S$ of the indices of all fruits inside this interval. By hypothesis, $S$... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
04gs | Let $a$, $b$ and $c$ be positive real numbers. Prove the inequality
$$
\frac{a^2}{a+b} + \frac{b^2}{b+c} \ge \frac{3a+2b-c}{4}.
$$
(Belarus 2010) | [
"$$\n\\frac{a^2}{a+b} + \\frac{a+b}{4} \\ge a \\quad \\text{and} \\quad \\frac{b^2}{b+c} + \\frac{b+c}{4} \\ge b.\n$$\nAdding these two inequalities we get\n$$\n\\left( \\frac{a^2}{a+b} + \\frac{a+b}{4} \\right) + \\left( \\frac{b^2}{b+c} + \\frac{b+c}{4} \\right) \\ge a+b = \\frac{3a+2b-c}{4} + \\frac{a+b}{4} + \\... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0846 | Problem:
Alberto e Barbara stanno salendo con una seggiovia. Alberto occupa il sedile n. 48 e Barbara il sedile n. 180. Nell'istante in cui Alberto incrocia il sedile n. 75 Barbara incrocia il sedile n. 169. Quanti sedili ci sono sulla seggiovia?
Si supponga che i sedili siano ugualmente distanziati e che procedano in... | [] | Italy | Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO TRIENNIO | [
"Number Theory > Modular Arithmetic"
] | null | MCQ | A | |
0e8s | What is the value of the product $x \cdot y$ if $3^x = a$ and $a^y = 81$?
(A) 4
(B) 3
(C) 12
(D) 0
(E) 1 | [
"Since $81 = a^y = (3^x)^y = 3^{x y}$, and $81 = 3^4$, we have $x y = 4$. The correct answer is **A**."
] | Slovenia | National Math Olympiad 2013 - First Round | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | MCQ | A | |
0fm1 | Problem:
Una sucesión $\left(a_{n}\right)_{n \geq 1}$ se define mediante la recurrencia
$$
a_{1}=1,\ a_{2}=5,\ a_{n}=\frac{a_{n-1}^{2}+4}{a_{n-2}}, \text{ para } n \geq 3
$$
Demostrar que todos los términos de la sucesión son números enteros y encontrar una fórmula explícita para $a_{n}$. | [
"Solution:\n\nObservamos a partir de la definición que $a_{k} a_{k-2}=a_{k-1}^{2}+4$ y $a_{k+1} a_{k-1}=a_{k}^{2}+4$. Restando a la segunda ecuación de la primera, resulta\n$$\na_{k+1} a_{k-1}-a_{k} a_{k-2}=a_{k}^{2}-a_{k-1}^{2} \\Leftrightarrow a_{k-1}^{2}+a_{k-1} a_{k+1}=a_{k}^{2}+a_{k} a_{k+2}\n$$\nque es equiva... | Spain | 48 aME | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | a_n = ((2+√2)/4)·(3−2√2)^n + ((2−√2)/4)·(3+2√2)^n | |
0jh6 | Problem:
Points $A$, $B$, $C$ lie on a circle $\omega$ such that $BC$ is a diameter. $AB$ is extended past $B$ to point $B'$, and $AC$ is extended past $C$ to point $C'$ such that line $B'C'$ is parallel to $BC$ and tangent to $\omega$ at point $D$. If $B'D = 4$ and $C'D = 6$, compute $BC$. | [
"Solution:\n\nLet $x = AB$ and $y = AC$, and define $t > 0$ such that $BB' = t x$ and $CC' = t y$. Then $10 = B'C' = (1 + t) \\sqrt{x^2 + y^2}$, $4^2 = t(1 + t) x^2$, and $6^2 = t(1 + t) y^2$ (by power of a point), so $52 = 4^2 + 6^2 = t(1 + t)(x^2 + y^2)$ gives $\\frac{13}{25} = \\frac{52}{10^2} = \\frac{t(1 + t)}... | United States | HMMT November | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 24/5 | |
0ae0 | Докажи дека равенката $x^n - a_1 x^{n-1} - a_2 x^{n-2} - \dots - a_{n-1} x - a_n = 0$, каде што $a_k \ge 0, 1 \le k \le n$, нема две различни позитивни решенија. | [
"Равенката за $x \\neq 0$ е еквивалентна со равенката $1 = \\frac{a_1}{x} + \\frac{a_2}{x^2} + \\dots + \\frac{a_n}{x^n}$. Ќе воведеме ознака $f(x) = \\frac{a_1}{x} + \\frac{a_2}{x^2} + \\dots + \\frac{a_n}{x^n}$ за $x \\in (0, +\\infty)$.\n\nНе е тешко да се види дека за $0 < x_1 < x_2$, $f(x_1) > f(x_2)$. Според ... | North Macedonia | Републички натпревар по математика за средно образование | [
"Algebra > Algebraic Expressions > Polynomials"
] | Macedonian, English | proof only | null | |
040j | Find all integers $k \ge 3$, with the following properties: There exist integers $m$ and $n$ satisfying $(m, k) = (n, k) = 1$ and $k \mid (m-1)(n-1)$ with $1 < m < k$, $1 < n < k$ and $m+n > k$. | [
"If $k$ has a factor of square number greater than $1$, let $t^2 \\mid k$, $t > 1$, then taking $m = n = k - \\frac{k}{t} + 1$, we see that such $k$ has the properties.\n\nIf $k$ has no factor of square number, if there are two primes $p_1, p_2$ such that $(p_1-2)(p_2-2) \\ge 4$ and $p_1 p_2 \\mid k$. Let $k = p_1 ... | China | China National Team Selection Test | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | All integers at least three except odd primes, twice an odd prime, and thirty. | |
0cvf | Let $ABC$ be an acute-angled triangle with $AC < BC$. A circle passes through $A$ and $B$, and crosses the segments $AC$ and $BC$ again at $A_1$ and $B_1$, respectively. The circumcircles of the triangles $ABC$ and $A_1B_1C$ meet again at $P$. The segments $AB_1$ and $BA_1$ cross at $S$. Let $Q$ and $R$ be the reflecti... | [
"Используя окружности $(PABC)$ и $(PA_1B_1C)$, получаем $\\angle PAB = 180^\\circ - \\angle PCB = \\angle PA_1B_1$ и аналогично $\\angle PBA = \\angle PB_1A_1$. Тем самым, $\\angle PAB \\sim \\angle PA_1B_1$. Из этого подобия вытекает, что $\\angle APAA_1 = \\angle APB \\pm \\angle BPA_1 = \\angle BPB_1$ и $PA/PA_1... | Russia | Final round | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurat... | English; Russian | proof only | null | |
02rd | Ana drew two distinct hexagons, *ABCDEF* and *PQRSTU*, each with all internal angles measuring $120^\circ$.
a. Given that $AB = CD = 5$, $BC = 8$ and $EF = 3$, find the perimeter of *ABCDEF*.
b. Given that $PQ = 3$, $QR = 4$, $RS = 5$ and $TU = 1$, find $ST + PU$. | [
"A hexagon with all internal angles measuring $120^\\circ$ can be inscribed in an equilateral triangle by extending three of its sides:\n\n\n\nIn both items, $x$, $y$, $z$ and $w$ are given. The equilateral triangle has sidelength $x + y + z$, so the remaining two sides of the hexagon have ... | Brazil | Brazilian Math Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | a) 41; b) 14 | |
0joj | Problem:
Find all the roots of the polynomial $x^{5}-5 x^{4}+11 x^{3}-13 x^{2}+9 x-3$. | [
"Solution:\nThe $x^{5}-5 x^{4}$ at the beginning of the polynomial motivates us to write it as $(x-1)^{5}+x^{3}-3 x^{2}+4 x-2$ and again the presence of the $x^{3}-3 x^{2}$ motivates writing the polynomial in the form $(x-1)^{5}+(x-1)^{3}+(x-1)$. Let $a$ and $b$ be the roots of the polynomial $x^{2}+x+1$. It's clea... | United States | HMMT November 2015 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | 1, (3 + i√3)/2, (3 - i√3)/2, (1 + i√3)/2, (1 - i√3)/2 | |
07qm | Let $a_1, a_2, \dots, a_m$ be positive integers, none of which is equal to $10$, such that $a_1 + a_2 + \dots + a_m = 10m$. Prove that
$$
(a_1 a_2 a_3 \cdots a_m)^{1/m} \le 3\sqrt{11}.
$$ | [] | Ireland | Irish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
039x | Find all values of the real parameter $a$ such that the inequality $\sqrt{x - x^2 - a} + \sqrt{6a - 2x - x^2} \le \sqrt{10a - 2x - 4x^2}$ has a unique solution. | [] | Bulgaria | Fall Mathematical Competition | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | a = 0 or a = 1/4 | |
04o1 | Let $a$, $b$ and $c$ be positive real numbers such that $a + b + c = 2$. Prove that
$$
\frac{(a-1)^2}{b} + \frac{(b-1)^2}{c} + \frac{(c-1)^2}{a} \ge \frac{1}{4} \left( \frac{a^2+b^2}{a+b} + \frac{b^2+c^2}{b+c} + \frac{c^2+a^2}{c+a} \right).
$$ | [
"By the Cauchy-Bunyakovsky-Schwarz inequality, we have\n$$\n\\frac{(a-1)^2}{b} + \\frac{(b-1)^2}{c} \\ge \\frac{(2-a-b)^2}{b+c} = \\frac{c^2}{b+c},\n$$\nand similar inequalities hold for all pairs. By adding the inequalities, we get\n$$\n\\frac{(a-1)^2}{b} + \\frac{(b-1)^2}{c} + \\frac{(c-1)^2}{a} \\ge \\frac{1}{2}... | Croatia | Croatian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0idm | Problem:
Let $A C E$ be a triangle with a point $B$ on segment $A C$ and a point $D$ on segment $C E$ such that $B D$ is parallel to $A E$. A point $Y$ is chosen on segment $A E$, and segment $C Y$ is drawn. Let $X$ be the intersection of $C Y$ and $B D$. If $C X=5, X Y=3$, what is the ratio of the area of trapezoid $... | [
"Solution:\n\n\n\nDraw the altitude from $C$ to $A E$, intersecting line $B D$ at $K$ and line $A E$ at $L$. Then $C K$ is the altitude of triangle $B C D$, so triangles $C K X$ and $C L Y$ are similar. Since $C Y / C X=8 / 5, C L / C K=8 / 5$. Also triangles $C K B$ and $C L A$ are similar... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 39/25 | |
00nl | Let $ABC$ be a triangle with circumcenter $U$ such that $\angle CBA = 60^\circ$ and $\angle CBU = 45^\circ$. Let $D$ be the point of intersection of the lines $BU$ and $AC$.
Prove that $AD = DU$. | [
"\n\nIn the isosceles triangle $AUB$, we have\n$$\n\\angle BAU = \\angle UBA = 60^\\circ - 45^\\circ = 15^\\circ,\n$$\nand therefore\n$$\n\\angle AUB = 180^\\circ - \\angle BAU - \\angle UBA = 150^\\circ.\n$$\nThe inscribed angle theorem implies\n$$\n\\angle BCA = \\frac{1}{2} \\angle BUA =... | Austria | Austrian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03my | Consider the following system of 10 equations in 10 real variables $v_1, \dots, v_{10}$:
$$
v_i = 1 + \frac{6 v_i^2}{v_1^2 + v_2^2 + \dots + v_{10}^2} \quad (i = 1, \dots, 10).
$$
Find all 10-tuples $(v_1, v_2, \dots, v_{10})$ that are solutions of this system. | [
"For a particular solution $(v_1, v_2, \\dots, v_{10})$, let $s = v_1^2 + v_2^2 + \\dots + v_{10}^2$.\nThen\n$$\nv_i = 1 + \\frac{6v_i^2}{s} \\quad \\Rightarrow \\quad 6v_i^2 - sv_i + s = 0.\n$$\nLet $a$ and $b$ be the roots of the quadratic $6x^2 - sx + s = 0$, so for each $i$, $v_i = a$ or $v_i = b$. We also have... | Canada | Kanada | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Either all ten entries are 8/5, or nine entries are 4/3 and one entry is 4 (in any order). | |
0832 | Problem:
Per quante coppie $(p, q)$ di numeri primi (positivi) il polinomio $x^{2}+p x+q$ ha due radici intere? NOTA: Si ricorda che 1 non è un numero primo.
(A) 0
(B) 1
(C) 2
(D) 4
(E) infinite. | [
"Solution:\n\nLa risposta è (B). Infatti siano $a$ e $b$ le due radici intere del polinomio. Si ha quindi che $(x-a)(x-b)=x^{2}-(a+b)x+ab=x^{2}+p x+q$ da cui:\n$$\na+b=-p, \\quad ab=q\n$$\nPoiché $q$ è un primo, dalla seconda relazione si ricava che $a, b= \\pm 1$ o $a, b= \\pm q$. Dovendo essere $a+b$ negativo si ... | Italy | Progetto Olimpiadi di Matematica 2003 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | MCQ | B | |
0kyx | Problem:
A polynomial $f \in \mathbb{Z}[x]$ is called splitty if and only if for every prime $p$, there exist polynomials $g_p, h_p \in \mathbb{Z}[x]$ with $\operatorname{deg} g_p, \operatorname{deg} h_p < \operatorname{deg} f$ and all coefficients of $f - g_p h_p$ are divisible by $p$. Compute the sum of all positive... | [
"Solution:\n\nWe claim that $x^4 + a x^2 + b$ is splitty if and only if either $b$ or $a^2 - 4b$ is a perfect square. (The latter means that the polynomial splits into $(x^2 - r)(x^2 - s)$.)\n\nAssuming the characterization, one can easily extract the answer. For $a = 16$ and $b = n$, one of $n$ and $64 - n$ has to... | United States | HMMT February 2024 | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Quadratic residues"... | null | proof and answer | 693 | |
0ikm | Problem:
Let $n$ be a positive integer, and let Pushover be a game played by two players, standing squarely facing each other, pushing each other, where the first person to lose balance loses. At the HMPT, $2^{n+1}$ competitors, numbered 1 through $2^{n+1}$ clockwise, stand in a circle. They are equals in Pushover: wh... | [
"Solution:\n\nAt any point during this competition, we shall say that the situation is living if both players 1 and $2^{n}$ are still in the running. A living situation is far if those two players are diametrically opposite each other, and near otherwise, in which case (as one can check inductively) they must be ju... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | (2^n - 1)/8^n | |
0lf2 | There are 4 identical fair dices. Denote $x_i$ ($1 \leq x_i \leq 6$) be the number of dots on a face appearing on the $i$-th dice $1 \leq i \leq 4$.
a) Find the number of possible tuples $(x_1, x_2, x_3, x_4)$.
b) Find the probability that there exists a number $x_j$ such that $x_j$ is equal to the sum of the remaining... | [
"a) Using the principle of multiplication, the answer is $6^4 = 1296$.\n\nb) Firstly, there are 4 ways to choose $i$ such that $x_i$ is equal to the sum of the remaining numbers. For each $i$, we choose $x_i$ first and using the star-bar problem, one can obtain $\\binom{x_i-1}{2}$ ways to choose the other 3 numbers... | Vietnam | Vietnamese Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | a) 1296; b) 5/81; c) 26/81 | |
012l | Problem:
Find all sequences $a_{0} \leqslant a_{1} \leqslant a_{2} \leqslant \ldots$ of real numbers such that
$$
a_{m^{2}+n^{2}}=a_{m}^{2}+a_{n}^{2}
$$
for all integers $m, n \geqslant 0$. | [
"Solution:\nDenoting $f(n)=a_{n}$ we have\n$$\nf\\left(m^{2}+n^{2}\\right)=f^{2}(m)+f^{2}(n) .\n$$\nSubstituting $m=n=0$ into (7) we get $f(0)=2 f^{2}(0)$, hence either $f(0)=\\frac{1}{2}$ or $f(0)=0$. We consider these cases separately.\n\n(1) If $f(0)=\\frac{1}{2}$ then substituting $m=1$ and $n=0$ into (7) we ob... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Exactly three sequences: (i) a_n = 0 for all n; (ii) a_n = 1/2 for all n; (iii) a_n = n for all n. | |
047d | Given an acute triangle $ABC$ with $AB < BC < CA$. Let $D$ be a moving point on side $BC$, and $E$ be a moving point on the minor arc $\widehat{BC}$ of the circumcircle of $ABC$, such that $\angle BAD = \angle BED$.
Let $F$ be the intersection point of the line through $D$ perpendicular to $AB$ with the extension of $A... | [
"*Proof*. Let $H$ be the orthocenter of triangle $ABD$. Since $DF \\perp AB$, the points $H, D, F$ are colinear, as shown:\n\n\n**Case 1:** When $\\angle ADB < 90^\\circ$:\n• $H$ lies inside triangle $ABD$\n• By orthocenter properties, $\\angle BAD$ and $\\angle BHD$ are supplementary\n• Gi... | China | 2024 CGMO | [
"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 | |
0bwj | Let $ABC$ be a triangle, with $m(\angle BAC) = 90^\circ$ and $AB = 2AC$. On the ray $CA$ we consider the point $D$ such that $CD = 3CA$. Let $E \in BC$ such that $DE = DB$. Show that $AE \perp BC$.
Adrian Bud | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0any | Problem:
Consider a function $f(x) = a x^{2} + b x + c$, $a > 0$ with two distinct roots a distance $p$ apart. By how much, in terms of $a, b, c$ should the function be translated downwards so that the distance between the roots becomes $2p$? | [] | Philippines | AREA STAGE | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | 3(b^2 - 4ac)/(4a) | |
0e2l | Problem:
Za katera naravna števila $n$ obstaja večkratnik števila 7, ki ima vsoto števk enako $n$? | [
"Solution:\n\nŠtevilo, katerega vsota števk je 1, je potenca števila 10 in ni večkratnik 7. Zato ne obstaja tak večkratnik števila 7, da bi bila vsota števk enaka 1.\n\nPoskusimo najti tak večkratnik, da bo vsota števk enaka 2. To število mora imeti dve števki enaki 1. Preverimo po vrsti nekaj takih naravnih števil... | Slovenia | 54. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | all natural numbers except 1 | |
0hxl | Problem:
Suppose $x$ satisfies $x^{3}+x^{2}+x+1=0$. What are all possible values of $x^{4}+2 x^{3}+2 x^{2}+2 x+1$? | [
"Solution:\n$x^{4}+2 x^{3}+2 x^{2}+2 x+1 = (x+1)\\left(x^{3}+x^{2}+x+1\\right) = 0$ is the only possible solution."
] | United States | null | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 0 | |
0fm4 | Let $a$, $b$, $c$ be three positive real numbers with $a \cdot b \cdot c = 1$. Prove that if $a + b + c > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$, then exactly one of the three numbers is greater than $1$. | [
"Let $a$, $b$, $c > 0$ and $a \\cdot b \\cdot c = 1$.\n\nSuppose $a + b + c > \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}$.\n\nNote that $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = \\frac{b c + a c + a b}{a b c} = b c + a c + a b$ since $a b c = 1$.\n\nSo the inequality becomes:\n\n$$\na + b + c > a b + b c + c a\... | Spain | Spanija 2012 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0d2e | Define Fibonacci sequence $\{F\}_{n=0}^{\infty}$ as $F_{0}=0$, $F_{1}=1$ and $F_{n+1}=F_{n}+ F_{n-1}$ for every integer $n>1$. Determine all quadruples $(a, b, c, n)$ of positive integers with $a<b<c$ such that each of $a, b, c, a+n, b+n, c+2n$ is a term of the Fibonacci sequence. | [
"Let $(a, b, c, n)$ be a quadruplet of positive integers with $a<b<c$ such that each of $a, b, c, a+n, b+n, c+2n$ is a term of the Fibonacci sequence, and let\n$$\nb+n=F_{k}\n$$\nfor some positive integer $k$. Because $b<b+n$ and $a+n<b+n$, we have $\\max \\{b, a+n\\} \\leq F_{k-1}$.\nAssume $\\min \\{b, a+n\\} \\l... | Saudi Arabia | Selection tests for the Balkan Mathematical Olympiad 2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | (1, 2, 3, 1) | |
0d24 | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ which satisfy for all $x, y \in \mathbb{R}$ the relation
$$
f(f(f(x)+y)+y)=x+y+f(y) .
$$ | [
"Plug in $y=0$. The functional equation becomes\n$$\nf(f(f(x)))=x+f(0),\n$$\nfor all $x \\in \\mathbb{R}$. Since the map $x \\mapsto x+f(0)$ is bijective, then so is $f$.\n\nPlug in $y=-x$. The functional equation becomes\n$$\nf(f(f(x)-x)-x)=f(-x),\n$$\nfor all $x \\in \\mathbb{R}$. By injectivity of $f$, we can ca... | Saudi Arabia | Selection tests for the Balkan Mathematical Olympiad 2013 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | f(x) = x | |
09cp | Бүхэл тоо $3 \le n$ ба $a_2 \cdot a_3 \dots a_n = 1$ байх зэрэг бодит тоо $a_2, a_3 \dots, a_n$ өгчээ. $(1+a_2)^2 (1+a_3)^3 \dots (1+a_n)^n > n^n$ гэж батал. | [
"The substitution $a_2 = \\frac{x_2}{x_1}$, $a_3 = \\frac{x_3}{x_2}$, ..., $a_n = \\frac{x_1}{x_{n-1}}$ transforms the original problem into the inequality\n$$\n(x_1 + x_2)^2 (x_2 + x_3)^3 \\cdots (x_{n-1} + x_1)^n > n^n x_1^2 x_2^3 \\cdots x_{n-1}^{n-1}, \\quad (*)\n$$\nfor all $x_1, \\dots, x_{n-1} > 0$. To prove... | Mongolia | ОУМО-53 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | Mongolian | proof only | null | |
065a | A pupil has $7$ pieces of paper. He chooses some of them and cuts each of them into seven pieces. In the sequel, he chooses some of the pieces and cuts each of them into seven pieces. He continues this procedure many times with the pieces he has in hands every time. Is it possible to have some time $2009$ pieces of pap... | [
"Let he choose at the beginning $\\alpha_1$ from the seven pieces and each of them into seven pieces. Then he will have totally $7 - \\alpha_1 + 7\\alpha_1 = 7 + 6\\alpha_1$ pieces of paper. Suppose that in the next step he chooses $\\alpha_2$ pieces of paper and cuts each of them into seven pieces. Then he will ha... | Greece | SELECTION EXAMINATION | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | No | |
0gt5 | Find all real numbers $a$ for which there are different real numbers $x, y, z$ such that
$$
\frac{x^3 + a}{y+z} = \frac{y^3 + a}{x+z} = \frac{z^3 + a}{x+y} = -3.
$$ | [
"Answer: $-2 < a < 0$ and $0 < a < 2$.\n\nLetting $c = x + y + z$, one sees that $x, y, z$ are the roots of the polynomial $P(t) = t^3 - 3t + a + 3c$. Therefore, Vieta's theorem implies $x + y + z = 0$, thus $c = 0$ and $P(t) = t^3 - 3t + a$. Consequently the polynomial $P(t) = t^3 - 3t + a$ has three distinct real... | Turkey | 30th Turkish Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | English | proof and answer | -2 < a < 0 or 0 < a < 2 | |
04fc | Let $N$ and $K$ be positive integers. A number of students is first divided into $N$ non-empty groups and then the same students are divided into $N + K$ non-empty groups. Prove that in the second distribution at least $K+1$ students are in a smaller group than in the first distribution. (Yugoslavia 1981) | [
"Let $S$ be the set of all students. For $s \\in S$ we denote by $a_s$ the number of students in the group of the student $s$ in the first distribution and by $b_s$ the number of students in the group of student $s$ in the second distribution.\nWe have\n$$\n\\sum_{s \\in S} \\frac{1}{a_s} = N, \\quad \\sum_{s \\in ... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0e1v | Let $a$ and $b$ be real numbers such that $|a| \neq |b|$ and $\frac{a+b}{a-b} + \frac{a-b}{a+b} = 6$.
Find the value of the expression $\frac{a^3 + b^3}{a^3 - b^3} + \frac{a^3 - b^3}{a^3 + b^3}$. | [
"The equality\n$$\n6 = \\frac{a+b}{a-b} + \\frac{a-b}{a+b} = \\frac{2a^2 + 2b^2}{a^2 - b^2}\n$$\nimplies $6a^2 - 6b^2 = 2a^2 + 2b^2$, or $4a^2 = 8b^2$. From here we get $a = \\pm b\\sqrt{2}$. Now, we can conclude that\n$$\n\\frac{a^3 + b^3}{a^3 - b^3} + \\frac{a^3 - b^3}{a^3 + b^3} = \\frac{2a^6 + 2b^6}{a^6 - b^6} ... | Slovenia | National Math Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 18/7 | |
02tk | Problem:
a) "Se um quadrilátero $ABCD$ é tal que $\angle ABC = \angle ADC = 90^\circ$, então $AB^2 - CD^2 = AD^2 - BC^2$."
b) "Se um quadrilátero $ABCD$ é tal que $\angle ACB = \angle ADC = 90^\circ$, então $AB^2 = BC^2 + CD^2 + AD^2$."
Prove esses resultados. | [
"Solution:\na) Como $\\angle ABC = \\angle ADC = 90^\\circ$, sabemos que os triângulos $\\triangle ABC$ e $\\triangle ADC$ são retângulos. Pelo Teorema de Pitágoras, temos\n$$\n\\begin{aligned}\nAB^2 + BC^2 & = AC^2 \\\\\nAD^2 + CD^2 & = AC^2\n\\end{aligned}\n$$\nPortanto,\n$$\n\\left(AB^2 - CD^2\\right) - \\left(A... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
04kd | Let $ABC$ be an acute-angled triangle. Let point $B'$ be the reflection of $B$ across the line $AC$, and point $C'$ the reflection of $C$ across the line $AB$. Circles circumscribed to triangles $ABB'$ and $ACC'$ intersect at points $A$ and $P$. Prove that the circumcentre of triangle $ABC$ lies on the line $AP$. (Russ... | [
"Denote $\\alpha = \\angle BAC$, $\\beta = \\angle CBA$ and $\\gamma = \\angle ACB$, and let $O$ be the circumcentre of triangle $ABC$.\n\nPoint $P$ is on the circle circumscribed to triangle $ACC'$, so $\\angle APC = \\angle AC'C$ since they are subtended by $\\overarc{AC}$. Because of the... | Croatia | Mathematical competitions in Croatia | [
"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"
] | null | proof only | null | |
0lef | Define the sequence $(a_n)$ as $a_1 = 1$, $a_{2n} = a_n$ and $a_{2n+1} = a_n + 1$ for all positive integers $n$.
a) Find all positive integers $n$ such that $a_{kn} = a_n$ for all integers $1 \le k \le n$.
b) Prove that there are infinitely many positive integers $m$ such that $a_{km} \ge a_m$ for all positive intege... | [
"a)\nWe prove by induction on $n$ that $a_n = s_2(n)$ for all $n \\ge 1$ where $s_2(n)$ is the sum of the digits of $n$ in binary representation.\n\nThe base case $n = 1$ is trivial. Now, we assume that $a_n = s_2(n)$ is proved for $n = 1, 2, \\dots, k$, and prove it for $n = k + 1$.\n\nWe shall prove the statement... | Vietnam | TST | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a) All n with n = 2 or n = 2^t − 1 for some integer t ≥ 1. b) Infinitely many choices exist; for example, any m = 2^p works. | |
0fpv | Encontrar cuántas soluciones enteras tiene la ecuación
$$
|5 - x_1 - x_2| + |5 + x_1 - x_2| + |5 + x_2 + x_3| + |5 + x_2 - x_3| = 20.
$$ | [
"Podemos reescribir la ecuación en la forma\n$$\n|y_1| + |y_2 - y_1| + |y_3 - y_2| + |20 - y_3| = 20, \\quad (1)\n$$\ndonde $y_1 = 5 - x_1 - x_2$, $y_2 = 10 - 2x_2$ y $y_3 = 15 - x_2 + x_3$, por tanto toda solución entera de la ecuación original da una solución entera de (1) con $y_2$ un número par. Recíprocamente,... | Spain | LII Olimpiada Matemática Española | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | Spanish | proof and answer | 891 | |
0d49 | Turki has divided a square into finitely many white and green rectangles, each with sides parallel to the sides of the square. Within each white rectangle, he writes down its width divided by its height. Within each green rectangle, he writes down its height divided by its width. Finally, he calculates $S$, the sum of ... | [
"Let $s$ be the sidelength of the original square, $m$ and $n$ the numbers of white and green rectangles $W_{1}, W_{2}, \\ldots, W_{m}$ and $G_{1}, G_{2}, \\ldots, G_{n}$, respectively, $a_{i}$ and $b_{i}$ the width and height of the white rectangle $W_{i}$, respectively, for $i=1,2, \\ldots, m$, and $c_{j}$ and $d... | Saudi Arabia | SAMC | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English, Arabic | proof and answer | 5/2 | |
0hra | Problem:
Given a triangle $ABC$, let $D$ be the point of the ray $BA$ such that $BD = BA + AC$. If $K$ and $M$ are points on the sides $BA$ and $BC$, respectively, such that the triangles $BDM$ and $BCK$ have the same areas, prove that $\angle BKM = \frac{1}{2} \angle BAC$. | [
"Solution:\n\nSince the area of the triangle $BDM$ is equal to $BD \\cdot BM \\cdot \\sin \\angle DBM$ and a similar formula holds for the area of $\\triangle BCK$, we immediately have that $BD \\cdot BM = BC \\cdot BK$. This implies that $\\frac{BD}{BK} = \\frac{BC}{BM}$, hence $KM \\parallel CD$. Thus $\\angle BK... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06ar | We consider a $100 \times 100$ square consisting of $10^4$ unit squares. We call tiles of type A the rectangular tiles $8 \times 1$ or $1 \times 8$ consisting of 8 unit squares. We call tiles of type B the rectangular tiles $4 \times 2$ or $2 \times 4$ consisting of 8 unit squares. Examine if it is possible to cover th... | [
"Suppose such a coverage is possible with $N$ tiles of type A and with $N$ tiles of type B. Calculating the area of the tiles, which is $8 = 2^3$ and of the square which is equal to $100^2 = 2^4 \\cdot 5^4$, we observe that we need $5^4 = 625$ of tiles of type A and 625 of tiles of type B, so $N = 625$.\n\nWe color... | Greece | Selection Examination | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | No; such a tiling is impossible | |
06uc | Sir Alex plays the following game on a row of 9 cells. Initially, all cells are empty. In each move, Sir Alex is allowed to perform exactly one of the following two operations:
(1) Choose any number of the form $2^{j}$, where $j$ is a non-negative integer, and put it into an empty cell.
(2) Choose two (not necessarily ... | [
"Solution 1. We will solve a more general problem, replacing the row of 9 cells with a row of $k$ cells, where $k$ is a positive integer. Denote by $m(n, k)$ the maximum possible number of moves Sir Alex can make starting with a row of $k$ empty cells, and ending with one cell containing the number $2^{n}$ and all ... | IMO | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > C... | English | proof and answer | 2 * sum_{j=0}^{8} C(n, j) - 1 | |
08mw | Problem:
We can change a natural number $n$ in three ways:
a) If the number $n$ has at least two digits, we erase the last digit and we subtract that digit from the remaining number (for example, from $123$ we get $12-3=9$);
b) If the last digit is different from $0$, we can change the order of the digits in the opp... | [
"Solution:\n\nThe answer is NO. We will prove that if the first number is divisible by $11$, then all the numbers which we can get from $n$ are divisible by $11$.\n\nWhen we use a), from the number $10a + b$, we will get the number $m = a - b = 11a - n$, so $11 \\mid m$ since $11 \\mid n$.\n\nIt's well-known that a... | JBMO | JBMO Shortlist | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
0b34 | Problem:
A positive integer is called lucky if it is divisible by $7$, and the sum of its digits is also divisible by $7$. Fix a positive integer $n$. Show that there exists some lucky integer $\ell$ such that $|n-\ell| \leq 70$. | [
"Solution:\n\nSuppose we have some lucky integer $n$. We will show that the gap between it and the next lucky integer is no more than $2 \\times 70$.\n\nIn one iteration, we increment $n \\rightarrow n+7$. Clearly the number we have is still divisible by $7$, so it will suffice for us to show that the digit-sum wil... | Philippines | 23rd Philippine Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
04qt | Suppose that every integer has been given one of the colors red, blue, green, yellow. Let $x$ and $y$ be odd integers such that $|x| \neq |y|$. Show that there are two integers of the same color whose difference has one of the following values: $x$, $y$, $x+y$, $x-y$. | [
"We denote colors by capital initial letters. Let us suppose that there exists a coloring $f : \\mathbb{Z} \\to \\{R, G, B, Y\\}$ such that for any $a \\in \\mathbb{Z}$ we have $f\\{a, a+x, a+y, a+x+y\\} = \\{R, G, B, Y\\}$. We now define a coloring of an integer lattice $g : \\mathbb{Z} \\times \\mathbb{Z} \\to \\... | Czech Republic | 6-th Czech-Slovak Match | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
08w9 | A positive integer whose 1's digit is not 0 is called a *palindromic number* if the number remains the same when its digits are read in reverse order. For example, the number $12321$ is a palindromic number, since the number obtained by reading its digits in reverse order is the same $12321$, while $1234$ is not since ... | [
"Let us call a palindromic number a p.d.number. We count p.d.numbers less than $2012$ by classifying them according to the number of digits.\n\n* A 4-digit number is a p.d.number if and only if its thousand's digit and one's digit coincide and also its hundred's digit and ten's digit coincide. Therefore, $2002$ is ... | Japan | Japan Junior Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Other"
] | English | proof and answer | 119 | |
0fcy | Problem:
Supongamos que tenemos un tablero con dieciséis casillas dispuestas en cuatro filas y cuatro columnas.
a) Prueba que se pueden colocar siete fichas, nunca dos en la misma casilla, de forma que al eliminar dos filas y dos columnas cualesquiera, siempre quede alguna ficha sin eliminar.
b) Prueba que si se col... | [
"Solution:\n\n(a) Una solución es:\n\n\n\n(b) Si se tienen 6 fichas en el tablero, alguna columna tendrá al menos dos fichas; eliminamos esa columna.\n\nQuedan, como máximo, 4 fichas, y exactamente tres columnas. Por el mismo procedimiento podemos ahora eliminar una columna de forma que nos... | Spain | null | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
025x | Problem:
A razão entre o número de homens e o de mulheres na cidade de Campo Verde é $\frac{2}{3}$. A idade média dos homens é 37 anos e a das mulheres é 42 anos. Qual é a idade média dos habitantes de Campo Verde? | [
"Solution:\n\nSe $H$ indica o número de homens e $M$ o de mulheres, então:\n$$\n\\frac{H}{M} = \\frac{2}{3} \\quad \\Rightarrow \\quad M = \\frac{3H}{2}\n$$\nA idade média da população é:\n$$\n\\frac{37H + 42M}{H + M} = \\frac{37H + 42 \\frac{3H}{2}}{H + \\frac{3H}{2}} = \\frac{100H}{\\frac{5H}{2}} = \\frac{100 \\t... | Brazil | Nível 2 | [
"Math Word Problems",
"Statistics > Mathematical Statistics"
] | null | final answer only | 40 | |
013w | Problem:
Let the medians of the triangle $A B C$ meet at $M$. Let $D$ and $E$ be different points on the line $B C$ such that $D C = C E = A B$, and let $P$ and $Q$ be points on the segments $B D$ and $B E$, respectively, such that $2 B P = P D$ and $2 B Q = Q E$. Determine $\angle P M Q$. | [
"Solution:\n\nDraw the parallelogram $A B C A'$, with $A A' \\parallel B C$. Then $M$ lies on $B A'$, and $B M = \\frac{1}{3} B A'$. So $M$ is on the homothetic image (centre $B$, dilation $1 / 3$) of the circle with centre $C$ and radius $A B$, which meets $B C$ at $D$ and $E$. The image meets $B C$ at $P$ and $Q$... | Baltic Way | Baltic Way 2005 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 90° | |
0gow | Show that
$$
\frac{1}{x + y^{20} + z^{11}} + \frac{1}{y + z^{20} + x^{11}} + \frac{1}{z + x^{20} + y^{11}} \le 1
$$
for all positive real numbers $x, y, z$ satisfying $xyz = 1$. | [
"By the Cauchy-Schwarz inequality we have\n$$\n\\frac{1}{a + b^{20} + c^{11}} \\le \\frac{a^{13} + b^{-6} + c^3}{(a^7 + b^7 + c^7)^2}\n$$\nfor all positive real numbers $a, b, c$. Summing up the inequalities for $(a, b, c) = (x, y, z), (y, z, x)$ and $(z, x, y)$ gives that it suffices to show that\n$$\nx^{13} + y^{... | Turkey | 19th Turkish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof only | null | |
0l6l | Problem:
Jerry places at most one rook in each cell of a $2025 \times 2025$ grid of cells. A rook attacks another rook if the two rooks are in the same row or column and there are no other rooks between them.
Determine, with proof, the maximum number of rooks Jerry can place on the grid such that no rook attacks 4 ot... | [
"Solution:\n\nThe answer is $2024 \\times 4 = 8096$. More generally, for an $n \\times n$ grid, the answer is $4n - 4$. Call a rook that attacks at most 3 other rooks good.\n\nWe use the following observation in both parts of the solution: a rook on the border of the grid must be good.\n\nLower Bound: Place rooks o... | United States | HMMT February 2025 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 8096 | |
06xe | Professor Oak is feeding his 100 Pokémon. Each Pokémon has a bowl whose capacity is a positive real number of kilograms. These capacities are known to Professor Oak. The total capacity of all the bowls is 100 kilograms. Professor Oak distributes 100 kilograms of food in such a way that each Pokémon receives a non-negat... | [
"Answer: The answer is $D=50$.\n\nSolution 1. First, consider the situation where 99 bowls have a capacity of 0.5 kilograms and the last bowl has a capacity of 50.5 kilograms. No matter how Professor Oak distributes the food, the dissatisfaction level of every Pokémon will be at least 0.5. This amounts to a total d... | IMO | International Mathematical Olympiad Shortlist | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 50 | |
0izf | Problem:
A function $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ is linear in each of the $x_{i}$ and $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)=\frac{1}{x_{1} x_{2} \cdots x_{n}}$ when $x_{i} \in\{3,4\}$ for all $i$. In terms of $n$, what is $f(5,5, \ldots, 5)$? | [
"Solution:\nAnswer: $\\frac{1}{6^{n}}$\n\nLet $f_{n}\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)$ denote the $n$-variable version of the function. We will prove that $f_{n}(5, \\ldots, 5)=\\frac{1}{6^{n}}$ by induction.\n\nThe base case was done in the two previous problems. Suppose we know that $f_{n-1}(5,5, \\ldot... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | final answer only | 1/6^n |
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