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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0e7w | Problem:
Reši enačbo $25^{\sqrt{x}} - 124 \cdot 5^{\sqrt{x}} = 125$. | [] | Slovenia | Državno tekmovanje | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 9 | |
02q1 | Problem:
Dez pontos são dados no plano e não existem três colineares. Quatro segmentos distintos ligando pares destes pontos são escolhidos ao acaso, mas todos com a mesma probabilidade. Qual é a probabilidade de três dos segmentos escolhidos formarem um triângulo? | [
"Solution:\n\nO número de possíveis segmentos entre os $10$ pontos é $\\binom{10}{2} = 45$ e o número de formas de escolher $4$ desses segmentos é $\\binom{45}{4}$.\n\nJá o número de formas de escolher $4$ segmentos de tal modo que três deles formem um triângulo é igual ao número de maneiras de escolher três vértic... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 16/473 | |
0aqh | Problem:
Determine all functions $f:(0,+\infty) \rightarrow \mathbb{R}$ such that $f(2009)=1$ and
$$
f(x) f(y)+f\left(\frac{2009}{x}\right) f\left(\frac{2009}{y}\right)=2 f(x y)
$$
for all positive real numbers $x$ and $y$. | [] | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 1 for all positive x | |
0e7e | Problem:
Največja števka števila $13$ je $3$, največja števka števila $77$ je $7$. Za posamezne neničelne števke ugotovite, kolikokrat nastopajo kot največje števke v naravnih številih od $1$ do $99$, in sestavite ustrezno frekvenčno tabelo. To frekvenčno porazdelitev prikažite s histogramom in s frekvenčnim poligonom... | [] | Slovenia | Državno tekmovanje | [
"Statistics > Mathematical Statistics"
] | null | proof and answer | For digit d from 1 to 9, the frequency is 2d + 1. Thus the table is: 1→3, 2→5, 3→7, 4→9, 5→11, 6→13, 7→15, 8→17, 9→19. The histogram has unit-width classes with these heights; the frequency polygon connects the midpoints of the tops of adjacent bars and is closed to the axis at the ends. | |
0in1 | Problem:
Compute
$$
\int_{1}^{2} \frac{9x+4}{x^{5}+3x^{2}+x} \, dx
$$
(No, your TI-89 doesn't know how to do this one. Yes, the end is near.) | [
"Solution:\nAnswer: $\\ln \\frac{80}{23}$. We break the given integral into two pieces:\n$$\n\\int_{1}^{2} \\frac{9x+4}{x^{5}+3x^{2}+x} \\, dx = 5 \\int_{1}^{2} \\frac{x^{4}+3x+1}{x^{5}+3x^{2}+x} \\, dx - \\int_{1}^{2} \\frac{5x^{4}+6x+1}{x^{5}+3x^{2}+x} \\, dx\n$$\nThese two new integrals are easily computed; for,... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | final answer only | ln(80/23) | |
0jks | Problem:
Let $ABC$ be a triangle with circumcenter $O$, incenter $I$, $\angle B = 45^{\circ}$, and $OI \parallel BC$. Find $\cos \angle C$. | [
"Solution:\n\nAnswer: $1 - \\frac{\\sqrt{2}}{2}$\n\nLet $M$ be the midpoint of $BC$, and $D$ the foot of the perpendicular from $I$ to $BC$. Because $OI \\parallel BC$, we have $OM = ID$. Since $\\angle BOC = 2\\angle A$, the length of $OM$ is $OA \\cos \\angle BOM = OA \\cos A = R \\cos A$, and the length of $ID$ ... | United States | HMMT 2014 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 1 - sqrt(2)/2 | |
0ak2 | Find all positive integers $n$ such that $n$ has number of ciphers which is the same as the number of its different prime divisors and the sum of the different prime divisors is equal to the sum of their powers. | [
"Let $n = p_1^{a_1}p_2^{a_2}\\ldots p_k^{a_k}$. From the condition of the problem\n$$\np_1 + p_2 + \\ldots + p_k = a_1 + a_2 + \\ldots + a_k.\n$$\nWe discuss the number of ciphers of the number $n$. If $n$ has 4 ciphers, then he has 4 different prime divisors. Then $n \\ge 2^{14} \\cdot 3 \\cdot 5 \\cdot 7 > 10^4$ ... | North Macedonia | Junior Macedonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 4, 48, 72 | |
0c9n | Problem:
Fie $f:[a, b] \rightarrow \mathbb{R}$ o funcție cu proprietatea lui Darboux astfel ca $f(a) \cdot f(b)<0$. Arătați că există $\alpha, \beta$ astfel ca $a<\alpha<\beta<b$ și $f(\alpha)+f(\beta)=f(\alpha) \cdot f(\beta)$. | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ | [
"Precalculus > Functions"
] | null | proof only | null | |
0f6v | Problem:
Find all positive integers $a$, $b$, $c$ so that $a^2 + b = c$ and $a$ has $n > 1$ decimal digits all the same, $b$ has $n$ decimal digits all the same, and $c$ has $2n$ decimal digits all the same. | [] | Soviet Union | 20th ASU | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | The only solutions occur for two-digit repeated numbers: (a, b, c) = (33, 22, 1111), (66, 88, 4444), (88, 33, 7777). | |
0ef3 | Problem:
Dana je pravilna šeststrana piramida z dolžino osnovnega roba $a=10~\mathrm{cm}$ in dolžino stranskega roba $s=13~\mathrm{cm}$.
a) Natančno izračunaj površino piramide.
b) Natančno izračunaj ploščino osnega preseka piramide, ki nastane, če piramido presekamo z ravnino, ki poteka skozi vrh piramide in dve og... | [
"Solution:\n\na)\nS pomočjo Pitagorovega izreka izračunamo višino stranske ploskve piramide $v_{s}=12~\\mathrm{cm}$. S pomočjo formule za izračun ploščine osnovne ploskve $S_{op}=\\frac{6 a^{2} \\sqrt{3}}{4}=\\frac{6 \\cdot 10^{2} \\sqrt{3}}{4}=150 \\sqrt{3}\\ \\mathrm{cm}^{2}$ in ploščine plašča $S_{pl}=6 \\cdot \... | Slovenia | 17. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Odbirno tekmovanje | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > Other 3D problems"
] | null | final answer only | a) 150√3 + 360; b) 10√69 | |
0210 | Problem:
A sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive integers satisfies $a_{1}>5$ and $a_{n+1}=5+6+\cdots+a_{n}$ for all positive integers $n$. Determine all prime numbers $p$ such that, regardless of the value of $a_{1}$, this sequence must contain a multiple of $p$. | [
"Solution:\nWe claim that the only prime number of which the sequence must contain a multiple is $p=2$. To prove this, we begin by noting that\n$$\na_{n+1}=\\frac{a_{n}\\left(a_{n}+1\\right)}{2}-10=\\frac{\\left(a_{n}-4\\right)\\left(a_{n}+5\\right)}{2}\n$$\nLet $p>2$ be an odd prime, and choose $a_{1} \\equiv -4 \... | Benelux Mathematical Olympiad | 13th Benelux Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 2 | |
0kju | Let $m \ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1 \le a_i \le m$ for all $i$ such that $m$ divides $a_1 + a_2 + a_3 + a_4$. There is a polynomial $q(x) = c_3x^3 + c_2x^2 + c_1x + c_0$ such that $D(m) = q(m)$ for all odd integers $m \ge 5$... | [] | United States | AMC 12 A | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | MCQ | E | |
0gvx | A toy factory hired three painters to paint $2006$ cubic bricks. A painter needs $5$ minutes to paint a face of a brick. Each face has to be painted by just one of the painters. It is acceptable that different painters paint different faces of the same brick; however they cannot do it at the same time. Find the minimal... | [
"Відповідь: $20060$ секунд. Час, необхідний для розфарбування, не може бути меншим за $\\frac{2006 \\cdot 6 \\cdot 5}{3} = 20060$ секунд (зрозуміло, що час розфарбування буде найменшим, якщо вдасться так розподілити роботу, щоб кожні $5$ секунд фарбувалася максимальна кількість граней). Цього можна досягти, наприкл... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 20060 seconds | |
0fha | Problem:
Se consideran $n$ puntos del plano de forma que no existan dos parejas equidistantes. Por cada punto se traza el segmento que le une al más próximo. Demostrar que ningún punto está unido a más de cinco puntos. | [
"Solution:\n\nSea $A_{0}$ uno cualquiera de los $n$ puntos. Los demás serán representados por $A_{1}, A_{2}$, etc. Consideremos el triángulo $A_{0} A_{1} A_{2}$. Podemos suponer que $A_{1} A_{2}$ es el mayor lado (ya que uno de los otros dos es el más pequeño); eso significa que el ángulo del triángulo con vértice ... | Spain | OME 26 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
028k | Problem:
Desenhe duas circunferências de mesmo centro, uma de raio $1~\mathrm{cm}$ e a outra de raio $3~\mathrm{cm}$. Na região exterior à circunferência de raio $1~\mathrm{cm}$ e interior à de raio $3~\mathrm{cm}$, desenhe circunferências que sejam simultaneamente tangentes às duas circunferências, como mostrado na f... | [
"Solution:\n\n(a) Como as circunferências de raios $1~\\mathrm{cm}$ e $3~\\mathrm{cm}$ são concêntricas, as outras circunferências mostradas na figura devem ter raio igual a $1~\\mathrm{cm}$.\n\n(b) Os centros das 3 circunferências de raio $1~\\mathrm{cm}$ mostradas na figura formam um triângulo equilátero de lado ... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | radius 1 cm; maximum 6 | |
0gyr | On the coordinate space $XOY$ draw the set of points, whose coordinates satisfy the following equality:
$$
\sqrt{1-x^2} + \sqrt{1-y^2} = 2 - x^2 - y^2.
$$ | [
"Since $0 \\leq (1 - x^2) \\leq 1$, $0 \\leq (1 - y^2) \\leq 1$, we get that\n$$\n\\sqrt{1-x^2} + \\sqrt{1-y^2} \\geq (1-x^2) + (1-y^2), \\text{ and so the equality holds if } \n\\begin{cases}\n\\sqrt{1-x^2} = 1-x^2 \\\\\n\\sqrt{1-y^2} = 1-y^2\n\\end{cases} \\Rightarrow \n\\begin{cases}\nx^2 \\in \\{0;1\\} \\\\\ny^... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | All points with x and y each in {−1, 0, 1}, i.e., the nine points formed by all pairs of −1, 0, and 1. | |
0fmv | En una sala de baile hay 15 chicos y 15 chicas dispuestos en dos filas paralelas de manera que se formarán 15 parejas de baile. Sucede que la diferencia de altura entre el chico y la chica de cada pareja no supera los 10 cm. Demostrar que si colocamos los mismos chicos y chicas en dos filas paralelas en orden creciente... | [
"Sean $P_1, P_2, \\dots, P_{15}$ las quince parejas iniciales. Ordenemos ahora los chicos por alturas $a_1 \\le a_2 \\le \\dots \\le a_{15}$ y también las chicas $b_1 \\le b_2 \\le \\dots \\le b_{15}$. Supongamos que una de las parejas tuviese una diferencia de alturas superior a 10 cm, digamos $a_k - b_k > 10$. En... | Spain | Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Spanish | proof only | null | |
0hb2 | Consider the family of parabolas of the form $y = kx^2 + (k-n)x + (k+n)$ on a plane, where $k, n$ are any integer numbers. Is there a point with integer coordinates that doesn't belong to any parabola? | [
"Let $x=1$, then the second coordinate is $y = k + (k-n) + (k+n) = 3k$, which is divisible by three. Therefore, $(1; 1)$ doesn't belong to any parabola."
] | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | (1, 1) | |
030h | Problem:
Em uma reunião de matemáticos, Carlos diz a Frederico: $O$ dobro do produto dos dois dígitos do número de matemáticos na reunião é exatamente a nossa quantidade. Qual a quantidade mínima de matemáticos que deve se juntar a nós para que nossa quantidade seja um número primo? Ajude Frederico a resolver o proble... | [
"Solution:\n\nVamos representar a quantidade de matemáticos por $\\overline{a b}$. Temos que:\n$$\n\\begin{aligned}\n2 a b & = \\overline{a b} \\\\\n2 a b & = 10 a + b \\\\\n2 a b - b & = 10 a \\\\\nb(2 a - 1) & = 10 a \\\\\nb & = \\frac{10 a}{2 a - 1}\n\\end{aligned}\n$$\nAnalisando os possíveis valores de $a$, le... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 1 | |
0bj5 | The cube $ABCD'A'B'C'D'$ has edges of length $a$. Take the points $E \in (AB)$ and $F \in (BC)$ so that $AE + CF = EF$.
a) Find the measure of the angle of the planes $(D'DE)$ and $(D'DF)$.
b) Compute the distance from $D'$ to the straight line $EF$.
 | [
"a) Extend the segment $BA$ with $AH \\equiv FC$. Then $\\triangle DAH \\equiv \\triangle DCF$ (SAS), whence $\\overline{HDA} \\equiv \\overline{FDC}$, so $m(\\overline{HDF}) = 90^\\circ$. Then $[DH] \\equiv [DF]$, leads to $\\triangle DHE \\equiv \\triangle DFE$ (SSS) and from here $m(\\overline{FDE}) = m(\\overli... | Romania | 65th Romanian Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | angle = 45 degrees; distance = a*sqrt(2) | |
04h8 | Let $ABCD$ be a convex quadrilateral such that $\angle BAD = 90^\circ$, $\angle BAC = 2\angle BDC$ and $\angle DBA + \angle DCB = 180^\circ$. Find $\angle DBA$. | [
"Denote $\\angle BDC = x$ and $\\angle DBA = y$. Then $\\angle BAC = 2x$. Let $P$ and $Q$ be the intersections of the angle bisector of $\\angle BAC$ with segment $\\overline{BC}$ and ray $DC$.\n\n\n\nSince $\\angle BAQ = \\angle BDQ = x$, the quadrilateral $ABQD$ is cyclic. Given that $\\a... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 45° | |
0kas | Problem:
How many ways can one tile a $2 \times 8$ board with $1 \times 1$ and $2 \times 2$ tiles? Rotations and reflections of the same configuration are considered distinct. | [
"Solution:\n\nLet $f(n)$ denote the number of ways to fill a $2 \\times n$ board. One can fill the leftmost column with two $1 \\times 1$ tiles, leaving $f(n-1)$ ways, or one can fill the leftmost two columns with one $2 \\times 2$ tile, leaving $f(n-2)$ ways. Therefore, $f(n) = f(n-1) + f(n-2)$. One can also direc... | United States | HMMT February 2019 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 34 | |
07ai | Let $X$, $Y$, $X'$ and $Y'$ be four points on a line $l$ with the same order as written. Consider arcs $\omega_1$ and $\omega_2$ with endpoints $\{X, Y\}$ and arcs $\omega'_1$ and $\omega'_2$ with endpoints $\{X', Y'\}$ on one side of $l$ such that $\omega_1$ is tangent to $\omega'_1$ and $\omega_2$ is tangent to $\ome... | [
"Without loss of generality, we can suppose that $\\omega_1$ is shorter than $\\omega_2$. Let $l'$ be the common external tangent of $\\omega'_1$ and $\\omega_2$ and let it intersect $l$ at $O$. Let $OA$ be tangent to $\\omega_1$ at $A$, $B := \\omega_1 \\cap \\omega'_1$, $C := \\omega'_1 \\cap l'$, $D := l' \\cap ... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | English | proof only | null | |
07u2 | An isosceles triangle $ABC$ is inscribed in a circle with $\angle ACB = 90^\circ$ and $EF$ is a chord of the circle such that neither $E$ nor $F$ coincide with $C$. Lines $CE$ and $CF$ meet $AB$ at $D$ and $G$ respectively. Prove $|CE| \cdot |DG| = |EF| \cdot |CG|$. | [
"As $ABC$ is isosceles and $\\angle ACB = 90^\\circ$ then it follows that\n$$\n\\angle CAB = \\angle CBA = 45^\\circ\n$$\nNow quadrilateral $ABEC$ is cyclic, so\n$$\n\\angle BEC = 180^\\circ - \\angle BAC = 135^\\circ\n$$\nTherefore by subtraction from $180^\\circ$, we know $\\angle BED = 45^\\circ$.\nLet us define... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analyti... | null | proof only | null | |
05c0 | The letters $M$, $A$, $T$, $E$, $I$ and $K$ are assigned the numbers $1$, $2$, $3$, $4$, $5$ and $6$ in a certain order so that different letters correspond to different numbers. The sum of the numbers corresponding to the letters of the word $MATEMAATIK$ (taking into account repetitions) is $42$ and the sum of the num... | [
"In the words $MATEMAATIK$, $KEEMIK$ and $IT$, all the given letters occur exactly $3$ times. Since the sum of the numbers corresponding to the letters $M$, $A$, $T$, $E$, $I$ and $K$ is $1+2+3+4+5+6=21$, the sum of the numbers corresponding to the letters in these three words is $3 \\cdot 21 = 63$. Consequently, t... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | final answer only | 8 | |
06pa | Let $c>2$, and let $a(1), a(2), \ldots$ be a sequence of nonnegative real numbers such that
$$
a(m+n) \leq 2 a(m)+2 a(n) \quad \text{ for all } m, n \geq 1, \tag{1}
$$
and
$$
a\left(2^{k}\right) \leq \frac{1}{(k+1)^{c}} \quad \text{ for all } k \geq 0 \tag{2}
$$
Prove that the sequence $a(n)$ is bounded. | [
"For convenience, define $a(0)=0$; then condition (1) persists for all pairs of nonnegative indices.\n\nLemma 1. For arbitrary nonnegative indices $n_{1}, \\ldots, n_{k}$, we have\n$$\na\\left(\\sum_{i=1}^{k} n_{i}\\right) \\leq \\sum_{i=1}^{k} 2^{i} a\\left(n_{i}\\right) \\tag{3}\n$$\nand\n$$\na\\left(\\sum_{i=1}^... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0c1p | Let $AD$, $BE$, $CF$ be the altitudes of the triangle $ABC$ and let $K$, $L$, $M$ be the orthocenters of triangles $AEF$, $BFD$, $CDE$, respectively. Denote by $G_1$ and $G_2$ the barycenters of triangles $DEF$, $KLM$, respectively. Prove that $HG_1 = G_1G_2$, where $H$ is the orthocenter of $ABC$.
Severius Moldoveanu | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"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 only | null | |
05dz | Problem:
Let $ABC$ be a triangle with an obtuse angle at $A$. Let $E$ and $F$ be the intersections of the external bisector of angle $A$ with the altitudes of $ABC$ through $B$ and $C$ respectively. Let $M$ and $N$ be the points on the segments $EC$ and $FB$ respectively such that $\angle EMA = \angle BCA$ and $\angle ... | [
"Solution:\nThe first solution is based on the main Lemma. We present this Lemma with two different proofs.\nLemma: Let $ABC$ be an acute triangle with $AB = BC$. Let $P$ be any point on $AC$. Line passing through $P$ perpendicular to $AB$, intersects ray $BC$ in point $T$. If the line $AT$ intersects the circumscr... | European Girls' Mathematical Olympiad (EGMO) | EGMO 2021 | [
"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 | |
0bj2 | Construct outside the square $ABCD$ the rhombus $BCMN$, with $\angle BCM$ an obtuse angle. The straight lines $BM$ and $AN$ meet at $P$. Prove that $DM \perp CP$ and triangle $DPM$ is right and isosceles. | [
"Isosceles triangles $CDM$ and $CBM$ yield $\\overline{CMD} \\equiv \\overline{CDM}$ and $\\overline{CMB} \\equiv \\overline{CBM}$. In triangle $MBD$, the sum of the angles is $180^\\circ = m(\\overline{DMB}) + m(\\overline{BDM}) + m(\\overline{DBM}) = m(\\overline{DMB}) + 45^\\circ + m(\\overline{CDM}) + 45^\\circ... | Romania | 65th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0fmt | Supongamos que el excírculo del triángulo $ABC$ opuesto al vértice $A$ es tangente al lado $BC$ en el punto $A_1$. Análogamente, se definen los puntos $B_1$ en $CA$ y $C_1$ en $AB$, utilizando los excírculos opuestos a $B$ y $C$ respectivamente. Supongamos que el circuncentro del triángulo $A_1B_1C_1$ pertenece a la ci... | [
"Sea $\\Gamma$ la circunferencia circunscrita a $ABC$. Usamos en la resolución el siguiente **Lema**. Para todo triángulo $ABC$, y con las definiciones de puntos dadas en el enunciado, las mediatrices de $BC$ y $B_1C_1$ se cortan sobre $\\Gamma$, más concretamente en el punto medio $M$ del arco $BC$ que contiene a ... | Spain | Olimpiada Internacional de Matemáticas | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | Spanish | proof only | null | |
06z3 | Problem:
Points $A_1, A_2, \ldots, A_n$ are equally spaced on the side $BC$ of the triangle $ABC$ (so that $BA_1 = A_1A_2 = \ldots = A_{n-1}A_n = A_nC$). Similarly, points $B_1, B_2, \ldots, B_n$ are equally spaced on the side $CA$, and points $C_1, C_2, \ldots, C_n$ are equally spaced on the side $AB$. Show that
$$(A... | [
"Solution:\n\nUsing the cosine formula, $AA_k^2 = AB^2 + k^2 BC^2/(n+1)^2 - 2k AB \\cdot BC/(n+1) \\cos B$. So\n$$\n\\sum AA_k^2 = n AB^2 + BC^2/(n+1)^2 (1^2 + 2^2 + \\ldots + n^2) - 2 AB \\cdot BC \\cos B (1 + 2 + \\ldots + n)/(n+1).\n$$\nSimilarly for the other two sides.\n\nThus the total sum is\n$$\nn(AB^2 + BC... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0jt3 | Problem:
Given that
$$
\begin{aligned}
a+b & =23 \\
b+c & =25 \\
c+a & =30
\end{aligned}
$$
determine (with proof) the value of $a b c$. | [
"Solution:\nWe add all three equations to obtain\n$$\n2(a+b+c)=78\n$$\nso $a+b+c=39$. Therefore,\n$$\n\\begin{aligned}\n& a=39-25=14 \\\\\n& b=39-30=9 \\\\\n& c=39-23=16 .\n\\end{aligned}\n$$\nThus $a b c=14 \\cdot 9 \\cdot 16=2016$."
] | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 2016 | |
08po | Problem:
Let $A$ be a set of positive integers with the following properties:
(a) If $n$ is an element of $A$ then $n \leqslant 2018$.
(b) If $S$ is a subset of $A$ with $|S|=3$ then there are two elements $n, m$ of $S$ with $|n-m| \geqslant \sqrt{n}+\sqrt{m}$.
What is the maximum number of elements that $A$ can have? | [
"Solution:\nAssuming $n>m$ we have\n$$\n\\begin{aligned}\n|n-m| \\geqslant \\sqrt{n}+\\sqrt{m} & \\Leftrightarrow (\\sqrt{n}-\\sqrt{m})(\\sqrt{n}+\\sqrt{m}) \\geqslant \\sqrt{n}+\\sqrt{m} \\\\\n& \\Leftrightarrow \\sqrt{n} \\geqslant \\sqrt{m}+1 .\n\\end{aligned}\n$$\nLet $A_{k}=\\{k^{2}, k^{2}+1, \\ldots,(k+1)^{2}... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 88 | |
059q | A function $f: \mathbb{R} \to \mathbb{R}$ satisfies for all real $x$ and $y$ the equation $2f(x)(f(y))^2 + y^2f(-x|y|) = f(xy^2)$. Find all possible values of $f(1)$. | [
"Denote $f(1) = a$ and $f(-1) = b$. Choosing $x = y = 1$ and $x = y = -1$ yields the system of equations\n$$\n\\begin{cases} 2a^3 + b = a \\\\ 2b^3 + a = b \\end{cases}\n$$\nAdding the equations and cancelling equal terms, we get $2b^3 = -2a^3$. Therefore $b = -a$. Substituting this into the first equation yields $... | Estonia | Estonian Math Competitions | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | -1, 0, 1 | |
0i99 | Problem:
In a certain country, there are 100 senators, each of whom has 4 aides. These senators and aides serve on various committees. A committee may consist either of 5 senators, of 4 senators and 4 aides, or of 2 senators and 12 aides. Every senator serves on 5 committees, and every aide serves on 3 committees. How ... | [
"Solution:\nIf each senator gets a point for every committee on which she serves, and every aide gets $1/4$ point for every committee on which he serves, then the 100 senators get 500 points altogether, and the 400 aides get 300 points altogether, for a total of 800 points. On the other hand, each committee contrib... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 160 | |
0fdb | Problem:
Sea $M$ un punto interior del segmento $A B$. Se construyen cuadrados $A M C D$ y $B E H M$ en el mismo lado de $A B$. Si $N$ es el segundo punto de intersección de las circunferencias circunscritas a dichos cuadrados, probar que:
1. los puntos $B, N$ y $C$ están alineados.
2. el punto $H$ es el ortocentro de... | [
"Solution:\n\n\n\n1. Tenemos $\\angle M N B=45^{\\circ}$.\nSi es la figura 1, se verifica que $\\angle C N M+\\angle M N B=135^{\\circ}+45^{\\circ}=180^{\\circ}$; si la figura 2, $\\angle M N C=\\angle M N B\\left(=45^{\\circ}\\right)$.\nEn ambas situaciones se concluye que $B, N$ y $C$ est... | Spain | null | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
070f | Problem:
$I$ is the incenter of the triangle $ABC$. A circle with center $I$ meets the side $BC$ at $D$ and $P$, with $D$ nearer to $B$. Similarly, it meets the side $CA$ at $E$ and $Q$, with $E$ nearer to $C$, and it meets $AB$ at $F$ and $R$, with $F$ nearer to $A$. The lines $EF$ and $QR$ meet at $S$, the lines $FD... | [
"Solution:\n\n\n\n$D$ and $P$ are the reflections of $Q$ and $E$ respectively in the line $CI$. Hence $PQ$ and $DE$ meet at a point on $CI$. So $U$ lies on $CI$. So $\\angle PIU = 1/2 \\angle PIE = \\angle PDE$ ($I$ is center of circle through $D$, $P$, $E$) $= \\angle PDU$ (same angle). He... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilate... | null | proof only | null | |
00z3 | Problem:
The real numbers $a$, $b$ and $c$ satisfy the inequalities $|a| \geq |b+c|$, $|b| \geq |c+a|$ and $|c| \geq |a+b|$. Prove that $a+b+c=0$. | [
"Solution:\n\nSquaring both sides of the given inequalities we get\n$$\n\\left\\{\n\\begin{array}{l}\na^{2} \\geq (b+c)^{2} \\\\\nb^{2} \\geq (c+a)^{2} \\\\\nc^{2} \\geq (a+b)^{2}\n\\end{array}\n\\right.\n$$\nAdding these three inequalities and rearranging, we get $(a+b+c)^{2} \\leq 0$. Clearly equality must hold, ... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0apr | Problem:
Let $P$ be a point outside a circle, and let the two tangent lines through $P$ touch the circle at $A$ and $B$. Let $C$ be a point on the minor arc $A B$, and let $\overrightarrow{P C}$ intersect the circle again at another point $D$. Let $L$ be the line that passes through $B$ and is parallel to $\overline{P ... | [
"Solution:\nRefer to Figure 9. We only need to see four pairs of similar triangles.\n$$\n\\begin{aligned}\n\\triangle A E B \\sim \\triangle A B C & \\Longrightarrow \\frac{B E}{A B}=\\frac{B C}{A C} \\\\\n\\triangle A F B \\sim \\triangle A B D & \\Longrightarrow \\frac{B F}{A B}=\\frac{B D}{A D} \\\\\n\\triangle ... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0byd | Prove the inequality
$$
sin \frac{\pi}{4n} \ge \frac{\sqrt{2}}{2n},
$$
where $n$ is a positive integer. | [
"If $a = \\cos \\frac{\\pi}{2n} + i \\sin \\frac{\\pi}{2n}$, then $|a| = 1$ and $a^n = i$. We obtain from here that\n$$\ni - 1 = a^n - 1 = (a - 1)(a^{n-1} + a^{n-2} + \\dots + 1),\n$$\nand thus $\\sqrt{2} \\le |a - 1| \\cdot n$.\nOn the other hand, $|a - 1| = |-2 \\sin^2 \\frac{\\pi}{4n} + 2i \\sin \\frac{\\pi}{4n}... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof only | null | |
00eh | A number is said to be an *almost palindrome* if it is possible to place a nonzero digit to its left so a palindrome is obtained, that is, a number that reads the same from left to right as from right to left. For instance, $2023$ is an almost palindrome, because we can place the digit $3$ to its left to obtain the num... | [
"A six-digit almost palindrome can be written as $abcbad$, where $a, d \\neq 0$. This number is a multiple of $9$ if and only if the sum $a + b + c + b + a + d$ is a multiple of $9$. Notice that if the values of $a, b$ and $c$ are fixed, then the remainder of $d$ when divided by $9$ is determined. We know that $d$ ... | Argentina | Rioplatense Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | 900 | |
0esy | Determine all positive integers $a$ for which $a^a$ is divisible by $20^{19}$. | [
"If $20^{19} \\mid a^a$, then $10 \\mid a$, hence $a \\in \\{10, 20, 30, \\dots\\}$. Clearly, $a = 10$ is not possible ($10^{10}$ is smaller than $20^{19}$), while $a = 20$ is possible ($20^{20} = 20 \\cdot 20^{19}$). Also, all $a = 10k$, $k \\ge 4$, are possible (for $q = 5^{10k-19} \\cdot 2^{10k-38} \\cdot k^{10k... | South Africa | The South African Mathematical Olympiad, Third Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All positive integers a with a = 20 or a = 10k for integer k ≥ 4. | |
0k93 | Problem:
A point $P$ is chosen uniformly at random inside a square of side length $2$. If $P_{1}, P_{2}, P_{3}$, and $P_{4}$ are the reflections of $P$ over each of the four sides of the square, find the expected value of the area of quadrilateral $P_{1} P_{2} P_{3} P_{4}$. | [
"Solution:\n\nLet $ABCD$ denote the square defined in the problem. We see that if $P_{1}$ is the reflection of $P$ over $\\overline{AB}$, then the area of $P_{1}AB$ is the same as the area of $PAB$. Furthermore, if $P_{4}$ is the reflection of $P$ over $\\overline{DA}$, $P_{1}$, $A$, and $P_{4}$ are collinear, as $... | United States | HMMT November 2019 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 8 | |
0igy | Problem:
Three distinct lines are drawn in the plane. Suppose there exist exactly $n$ circles in the plane tangent to all three lines. Find all possible values of $n$. | [
"Solution:\n\nIf the three lines form a triangle, then there are 4 circles, namely the incircle and the three excircles.\n\nIf the three lines concur or are all parallel, then there are 0 circles.\n\nIf two lines are parallel and the third is not, then there are 2 circles lying between the two parallel lines, one o... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 0, 2, 4 | |
09j2 | Let $ABC$ be a triangle and let $D$ be a point on $BC$, let $E$ be a point on $AC$, let $F$ be a point on $AB$ such that $\angle BFD = \angle AFE$ and $\angle AEF = \angle CED$. Let $M$ and $N$ be points on the line $EF$ such that $EN = FD$ and $FM = ED$, where the points $M, F, E, N$ are positioned in this order. The ... | [] | Mongolia | Mongolian Mathematical Olympiad Round 2 | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0e4j | Yesterday in the playground the ratio of boys to girls was $3 : 2$. Today, the number of boys is the square of the number of girls and there are $6$ boys and $7$ girls less than yesterday. How many children were there in the playground yesterday?
(A) 12
(B) 13
(C) 15
(D) 25
(E) 30 | [
"Yesterday there were $3t$ boys and $2t$ girls in the playground. Today, there are $3t-6$ boys and $2t-7$ girls. From $3t-6 = (2t-7)^2$ we get $4t^2-31t+55 = 0$, so $t_{1,2} = \\frac{31\\pm\\sqrt{81}}{8}$. The only possible solution is $t = 5$, since $t$ is an integer. Yesterday there were $3t = 15$ boys and $2t = ... | Slovenia | National Math Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | D | |
0cu9 | A circle centered at $I$ is inscribed into a quadrilateral $ABCD$. $BA$ and $CD$ meet at $P$, and rays $AD$ and $BC$ meet at $Q$. Given that $P$ lies on the circle $(AIC)$, prove that $Q$ also lies on this circle. | [
"Prove that $\\angle DAI = \\angle BAI = \\angle DCI = \\angle BCI$ (see Fig. 6).\n\nТак как четырёхгольник $AICP$ вписанный, то $\\angle PCI = 180^\\circ - \\angle PAI = \\angle BAI$; иначе говоря, $\\angle DCI = \\angle BAI$ (см. рис. 6). Центр $I$ вписанной окружности четырёхгольника лежит на биссектрисах его уг... | Russia | XLIII Russian mathematical olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
04l3 | Let $a$ and $b$ be positive integers of different parity. Prove that $(a+3b)(5a+7b)$ cannot be a perfect square. | [
"By considering the given expression modulo $8$, we get\n$$\n(a + 3b)(5a + 7b) \\equiv 5a^2 + 22ab + 21b^2 \\equiv 5a^2 - 10ab + 5b^2 \\equiv 5(a - b)^2 \\pmod{8}.\n$$\nLet us assume the contrary, that $(a + 3b)(5a + 7b) = k^2$ for some positive integer $k$.\nThen\n$$\nk^2 \\equiv 5(a - b)^2 \\pmod{8}.\n$$\nSince $... | Croatia | Mathematical competitions in Croatia | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof only | null | |
0jab | Problem:
Points $X$ and $Y$ are inside a unit square. The score of a vertex of the square is the minimum distance from that vertex to $X$ or $Y$. What is the minimum possible sum of the scores of the vertices of the square? | [
"Solution:\n\nAnswer: $\\frac{\\sqrt{6}+\\sqrt{2}}{2}$\n\nLet the square be $ABCD$. First, suppose that all four vertices are closer to $X$ than $Y$. Then, by the triangle inequality, the sum of the scores is $AX + BX + CX + DX \\geq AB + CD = 2$.\n\nSimilarly, suppose exactly two vertices are closer to $X$ than $Y... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Advanced Configurations > Napoleon and Fermat points",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane... | null | proof and answer | (sqrt(6) + sqrt(2)) / 2 | |
07nf | A given list $n_1, n_2, \dots, n_{2011}$ of positive integers has the property that $n_i n_{i+1}$ is different from $n_j n_{j+1}$ whenever $i, j$ are distinct integers less than $2011$. Find the minimum number of distinct integers that must be in any such list. | [
"Suppose a list $L = (n_i)_{i=1}^N$ is made up only of numbers selected from the (possibly much shorter) list of distinct numbers $(m_i)_{i=1}^k$, and that the products $n_i n_{i+1}$, $1 \\le i \\le 2010$, are all distinct. We write down the indices of the numbers $m_i$ occurring in $L$ as another list $M$ also of ... | Ireland | Ireland | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 63 | |
0h6a | In the trapezoid $ABCD$ with perpendicular diagonals, points $P$, $N$, $Q$, $M$ are the middles of sides $AB$, $BC$, $CD$, $DA$ respectively. On the base $CD$ there is a point $L$ (different from the point $Q$) for which the angle $MLN$ is straight. Find the angle $LPA$.
(Bogdan Rublyov) | [
"By Varignon's Theorem quadrangle $MPNQ$ is a parallelogram and its sides are parallel to the diagonals of trapezoid $ABCD$, so it is a rectangle (Fig. 38). Denote $O$ the intersection point of $AC$ and $BD$. Then from the properties of rectangular $\\triangle MNL$ we get that: $OL = OM = ON \\Rightarrow OL = OP = ... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 90° | |
0f2j | Problem:
An $n$-gon area $A$ is inscribed in a circle radius $R$. We take a point on each side of the polygon to form another $n$-gon. Show that it has perimeter at least $2A/R$. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0kag | Problem:
What is the smallest positive integer with remainder $2, 3, 4$ when divided by $3, 5, 7$ respectively? | [
"Solution:\nWe note that if we double the number then it leaves a remainder of $1$ when divided by all of $3, 5$, and $7$. The smallest even number satisfying this is $3 \\cdot 5 \\cdot 7 + 1 = 106$, so the smallest possible number is $106 / 2 = 53$."
] | United States | HMMT February 2019 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | final answer only | 53 | |
0fdr | Problem:
Sean $n_{1}$, $n_{2}$ dos números naturales. Demuestra que la suma $\sqrt{n_{1}} + \sqrt[3]{n_{2}}$ es un número entero o un número irracional. | [
"Solution:\nSi se pone $x = \\sqrt{n_{1}} + \\sqrt[3]{n_{2}}$, se obtiene\n$$\n\\begin{aligned}\nx - \\sqrt{n_{1}} & = \\sqrt[3]{n_{2}} \\Longrightarrow n_{2} = (x - \\sqrt{n_{1}})^{3} \\Longrightarrow (3x^{2} + n_{1}) \\sqrt{n_{1}} = x^{3} + 3 n_{1} x - n_{2} \\\\\n(3x^{2} + n_{1})^{2} n_{1} & = (x^{3} + 3 n_{1} x... | Spain | XLVII Olimpiada Matemática Española Primera Fase | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0ilz | Problem:
Points $A$, $B$, and $C$ lie in that order on line $\ell$, such that $AB = 3$ and $BC = 2$. Point $H$ is such that $CH$ is perpendicular to $\ell$. Determine the length $CH$ such that $\angle AHB$ is as large as possible. | [
"Solution:\n\nLet $\\omega$ denote the circumcircle of triangle $ABH$. Since $AB$ is fixed, the smaller the radius of $\\omega$, the bigger the angle $AHB$. If $\\omega$ crosses the line $CH$ in more than one point, then there exists a smaller circle that goes through $A$ and $B$ that crosses $CH$ at a point $H'$. ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | sqrt(10) | |
00m3 | Auf einer Kette sind $2016$ Perlen im Kreis angeordnet, von denen jede eine der Farben schwarz, blau oder grün hat. In jedem Schritt wird gleichzeitig jede Perle durch eine neue Perle ersetzt, wobei sich die Farbe der neuen Perle wie folgt bestimmt: Falls die beiden ursprünglichen Nachbarn dieselbe Farbe hatten, hat di... | [
"a. Da $2016$ durch $4$ teilbar ist, kann man abwechselnd zwei schwarze und zwei grüne Perlen nehmen. Im ersten Schritt werden dann bereits alle durch blaue Perlen ersetzt.\n\nb. Wenn wir der Farbe Blau die Zahl $0$ zuordnen, der Farbe Grün die Zahl $1$ und der Farbe Schwarz die Zahl $2$, gilt in jedem Schritt, das... | Austria | 48. Österreichische Mathematik-Olympiade | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | German | proof only | null | |
0dhd | Consider the lattice in the plane, from which we may cut rectangles, but only by making cuts along the lines of the lattice. Prove that for any integer $m > 12$ one may cut a rectangle of area greater than $m$ such, that from that rectangle one can't cut a rectangle of area $m$. The Yellow Fish | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0kem | Problem:
Regular hexagon $P_{1} P_{2} P_{3} P_{4} P_{5} P_{6}$ has side length $2$. For $1 \leq i \leq 6$, let $C_{i}$ be a unit circle centered at $P_{i}$ and $\ell_{i}$ be one of the internal common tangents of $C_{i}$ and $C_{i+2}$, where $C_{7}=C_{1}$ and $C_{8}=C_{2}$. Assume that the lines $\{\ell_{1}, \ell_{2},... | [
"Solution:\n\nThe only way for the lines $\\ell_{i}$ to bound a regular hexagon $H$ is if they are rotationally symmetric around the center $O$ of the original hexagon. (A quick way to see this is to note that the angle between the two internal common tangents of $C_{i}$ and $C_{i+2}$ cannot be a multiple of $60^{\... | United States | HMMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 1603 | |
07ba | A local supermarket is responsible for the distribution of $100$ supply boxes. Each box is ought to contain $10$ kilograms of rice and $30$ eggs. It is known that a total of $1000$ kilograms of rice and $3000$ eggs are in these boxes, but in some of them the amount of either item is more or less than the amount require... | [
"The answer is $99$. First consider the initial condition that all of eggs are in just one of boxes. In each step, we can transfer eggs to at most one new box and so we need at least $99$ steps. We claim that $99$ steps is always enough. For this end, we call a box containing exactly $30$ eggs and $10$ kilograms of... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 99 | |
08y4 | Suppose positive integers $a$, $b$, $c$ satisfy the following four conditions:
$\bullet$ $\text{gcd}\{a, b, c\} = 1$.
$\text{gcd}(a, b + c) > 1$.
$\text{gcd}(b, c + a) > 1$.
$\text{gcd}(c, a + b) > 1$.
Here we denote by $\text{gcd}\{ \}$ the greatest common divisor of the numbers in $\{ \}$.
Determine the minimum possi... | [
"$\\boxed{30}$\n\nLet $g_1 = \\text{gcd}\\{a, b+c\\}$, $g_2 = \\text{gcd}\\{b, c+a\\}$, $g_3 = \\text{gcd}\\{c, a+b\\}$. If there exists a prime $p$ which divides both $g_1$ and $g_2$, then $p$ must divide both $a$ and $b$. Furthermore, since $p$ divides $b+c$ as well, $c$ must be divisible by $p$, but this contrad... | Japan | Japan 2015 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 30 | |
0cjs | Let $n \ge 2$ be a natural number and let $a_1, a_2, \dots, a_n$ be nonzero integers such that $a_1 + a_2 + \dots + a_n = a_1 a_2 \dots a_n$.
Prove that the number $(a_1^2 - 1)(a_2^2 - 1)\dots(a_n^2 - 1)$ is a perfect square. | [
"For $n=2$, we have $a_1+a_2 = a_1 a_2$, which rewrites as $(a_1-1)(a_2-1) = 1$. We get $a_1 = 2, a_2 = 2$, so $(a_1^2 - 1)(a_2^2 - 1) = 9$.\n\n$$\n|a_1 a_2 \\dots a_n| = |a_1 + a_2 + \\dots + a_n| \\le |a_1| + |a_2| + \\dots + |a_n|.\n$$\nDividing by $|a_1 a_2 \\dots a_n| > 0$ we obtain\n$$\n1 \\le \\frac{|a_1|}{|... | Romania | 75th NMO Selection Tests | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof only | null | |
09mn | For $n \ge 1$, let $a_n = 1^1 + 2^2 + \dots + n^n$. Show that the sequence $\{a_n\}$ contains infinitely many terms that leave a remainder of $32$ when divided by $2024$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof only | null | |
001p | En una caja fuerte hay $128$ bolsas con oro, todas con el mismo aspecto, pero todas de distinto peso. El tesorero quiere determinar las dos bolsas más pesadas y para ello dispone de una balanza de dos platos. La única operación permitida es colocar una bolsa en cada plato y de este modo establecer cuál de las dos es má... | [] | Argentina | XIX Olimpíada Matemática Argentina | [
"Discrete Mathematics > Algorithms"
] | español | proof and answer | Yes | |
0kna | Problem:
An up-right path between two lattice points $P$ and $Q$ is a path from $P$ to $Q$ that takes steps of length 1 unit either up or to the right.
How many up-right paths from $(0,0)$ to $(7,7)$, when drawn in the plane with the line $y = x - 2.021$, enclose exactly one bounded region below that line? | [
"Solution:\n\nWe will make use of a sort of bijection which is typically used to prove the closed form for the Catalan numbers. We will count these paths with complementary counting. Since both the starting and ending points are above the line $x - 2.021$, any path which traverses below this line (and hence include... | United States | HMMT Spring 2021 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | null | proof and answer | 637 | |
08ce | Problem:
Un trapezio rettangolo con base maggiore $AB$ e base minore $CD$ è circoscritto ad una circonferenza di raggio $10$. Si sa che il lato obliquo $BC$ misura $24$. Qual è la distanza tra i punti medi di $BC$ e $AD$?
(A) 21
(B) $\frac{13}{2} \sqrt{11}$
(C) $\frac{33}{5} \sqrt{11}$
(D) 22
(E) 23 | [
"Solution:\n\nLa risposta è (D). Per il Teorema di Talete la distanza tra i punti medi dei lati obliqui è data dalla semisomma delle basi. Poiché il trapezio è circoscritto ad una circonferenza, la somma delle basi è pari alla somma dei lati obliqui. Poiché il trapezio è rettangolo, il lato obliquo più corto è pari... | Italy | Gara di Febbraio | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | MCQ | D | |
0ght | 給定首一實係數多項式 $P_1(x), \dots, P_n(x)$。對於任意實數 $y$，定義集合
$$
S_y = \{z \in \mathbb{R} \mid \text{存在某個 } i \in \{1, \dots, n\} \text{ 使得 } y = P_i(z)\}.
$$
證明：若對於任意兩個相異實數 $y_1, y_2$，集合 $S_{y_1}, S_{y_2}$ 的元素個數相等，則 $P_1, \dots, P_n$ 有相同的次數。
Given some monic polynomials $P_1, ..., P_n$ with real coefficients, for any real numbe... | [
"Without loss of generality, we can assume that $P_1, \\dots, P_n$ are pairwise distinct. We will first show that $|S_y| = n$ for any $y \\in \\mathbb{R}$ and $\\deg P_i$ is odd for any $i \\in [n]$. To see this, let $a$ be the number of polynomials with odd degrees, and $b$ be the number of those with even degrees... | Taiwan | 2023 數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | Chinese (Traditional) | proof only | null | |
09b3 | Let $ABC$ be a scalene triangle. The inscribed circle of $ABC$ touches the sides $BC$, $CA$ and $AB$ at the points $A_1$, $B_1$ and $C_1$ respectively. Let $I$ be the center of the inscribed circle, $O$ be the center of the circumscribed circle and $OI \cap BC = D$. The perpendicular line from $A_1$ to $B_1C_1$ interse... | [
"Let us consider a circle with center at $J = IA_1 \\cap B_1C_1$ and touches $BC$ ($\\omega$).\nIt suffices to prove that $E$ is on that circle. $G$ is midpoint of $BC$, $F$ is midpoint of $BC$. Let $S_1$ be a homothety, which maps circumcircle onto incircle and with center $R \\in OI$. Let $S_2$ be a homothety, wh... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Pappus theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Pla... | null | proof only | null | |
0kld | Problem:
The 2020 members of the society of game theorists are holding the annual election for their leadership board. All members are initially on the board, and are ranked based on their qualifications. They start off by voting on whether to keep the board the same size; if they fail to get a strict majority, the me... | [
"Solution:\n\nNumber members #1 to #2020, with #1 being the member with the highest rank. We claim that only if there are $2^{n}-1$ members still on the board for some $n \\in \\mathbb{N}$, they all remain on the board.\n\nProof. We use induction on $n$.\n\nBase case: $n=1$.\nIf there were $2^{1}-1=1$ persons on th... | United States | Berkeley Math Circle: Monthly Contest 3 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 1023 | |
06d4 | Let $G$ be a connected simple graph which has $2p$ vertices (where $p$ is a positive integer), but contains no triangles. Prove that the number $\#E$ of its edges satisfies $\#E \leq p^2$. | [
"We prove that any simple graph with $2n$ vertices and without triangles has at most $n^2$ edges. We prove this by induction. When $n=1$, clearly there is at most $n^2=1$ edge between 2 vertices. Assume there are at most $k^2$ edges when there are $2k$ vertices. Consider a simple graph with $2(k+1)$ vertices and wi... | Hong Kong | HKG TST | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
085x | Problem:
Priscilla è stata incaricata di preparare la scenografia per la recita della sua scuola. Ha bisogno di una falce di luna, e ha a disposizione un cerchio di cartone di raggio $r$ in cui ritagliarla; allora punta il compasso sul bordo del cerchio, disegna un arco di circonferenza di raggio $r \sqrt{2}$ e taglia... | [
"Solution:\n\nLa risposta è $(\\mathbf{A})$. Sia $O$ il centro del cerchio di cartone, $O'$ il punto in cui Priscilla punta il compasso, $A$ e $B$ le intersezioni dell'arco di circonferenza disegnato da Priscilla con il bordo del cerchio. Allora $A O' B$ è un triangolo rettangolo isoscele (lo possiamo giustificare ... | Italy | Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | MCQ | A | |
0gi7 | 令 $n$ 與 $k$ 為正整數。令 $A$ 為由平面上 $2n$ 個相異點所構成的集合，其中任三點不共線。$A$ 中的若干對點之間有連線，使得平面上有 $n^2 + k$ 條相異線段。試證：平面上至少有 $\frac{4}{3}k^{3/2}$ 個相異三角形，其頂點皆屬於 $A$，且其三邊皆為上述所連線段。 | [
"將問題視為圖 $G = (V, E)$。對於每個點 $v \\in V$, 令 $d(v)$ 為該點度數。對於每條邊 $e = uv \\in E$, 令 $s(e) = d(u) + d(v)$, 也就是其兩端點的度數和。我們拆解為以下三步驟：\n\n1. **引理一:** 對於每一條邊 $e \\in E$, 至少有 $s(e) - 2n$ 個三角形以其為邊。\n\n*Proof.* 令 $e$ 的兩端點為 $u$ 和 $v$, 則扣除 $uv$ 外, $A$ 和 $B$ 總計還要對其餘 $2n-2$ 個點連出 $s(e)-2$ 條邊, 故由排容原理知至少有 $(s(e)-2) - (2n-2) = s(e) - 2n... | Taiwan | 2023 年台灣數學奧林匹亞考試 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | Chinese (Traditional) | proof only | null | |
0atz | Problem:
Let $a$, $p$, and $q$ be positive integers with $p \leq q$. Prove that if one of the numbers $a^{p}$ and $a^{q}$ is divisible by $p$, then the other number must also be divisible by $p$. | [
"Solution:\nSuppose that $p \\mid a^{p}$. Since $p \\leq q$, it follows that $a^{p} \\mid a^{q}$, which implies that $p \\mid a^{q}$.\n\nNow, suppose that $p \\mid a^{q}$, and, on the contrary, $p \\nmid a^{p}$. Then there is a prime number $r$ and a positive integer $n$ such that $r^{n} \\mid p$ (which implies tha... | Philippines | 15th Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
01l0 | Bob has the collection of $n$ coins. Each of them weights an integer number of grams, and their total weight is equal to $300$ grams. Let $k, l, m$ be positive integers such that $k + l + m = 300$.
Find the smallest possible number of $n$ for which Bob always (independently of coin's weights and for all possible $k, l,... | [
"Answer: $201$.\n\nLet $a_1 \\le a_2 \\le \\dots \\le a_n$ be the weights of the coins in Bob's collection. If there exists an $i$ such that $a_i \\ge 101$, and $k = l = m = 100$, then, obviously, Bob cannot partition his collection into three equally weighted groups. It is easy to see that if $n \\le 200$, then th... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 201 | |
036r | Problem:
Are there exist two triangles whose angles (in some order) form an arithmetic progression with nonzero common difference. | [
"Solution:\n\n1. We shall prove that if the angles of two triangles form an arithmetic progression with difference $d$, then $d=0$.\nLet $\\alpha$ be the least of all six angles and let the progression be $\\alpha, \\alpha+d, \\ldots, \\alpha+5d$. Thus,\n$$\n3\\alpha + (k_{1} + k_{2})d = 3\\alpha + (k_{3} + k_{4} +... | Bulgaria | Team selection test for 23. BMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | No | |
0byr | Determine the smallest radius a circle passing through exactly three lattice points may have. | [
"The required minimum is $5\\sqrt{2}/6$: the circle of radius $5\\sqrt{2}/6$, centered at $(5/6, 5/6)$ passes through exactly three lattice points, namely, $(0, 0)$, $(1, 2)$ and $(2, 1)$.\n\nNext, consider a circle $\\omega$ of radius at most $5\\sqrt{2}/6$ passing through exactly three lattice points. We may and ... | Romania | THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 5*sqrt(2)/6 | |
02s9 | Problem:
Os cinco amigos Arnaldo, Bernaldo, Cernaldo, Dernaldo e Ernaldo disputaram entre si cinco corridas. Em cada corrida, o ganhador recebeu cinco pontos, o segundo colocado quatro pontos e assim sucessivamente até o último colocado, que recebeu apenas um ponto. Para obter a pontuação final de cada corredor, foram... | [
"Solution:\n\na) Como Arnaldo e Bernaldo obtiveram, respectivamente, 19 pontos e 16 pontos como pontuação final, o total de pontos por eles obtidos foi igual a 35. Como nenhum deles ganhou sequer uma das corridas, então, em cada uma das corridas, a soma dos pontos por eles obtidos foi, no máximo, igual a $4+3=7$. C... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | a) In every race, Arnaldo and Bernaldo finished second and third in some order. b) Cernaldo scored 15 points. c) Cernaldo won 2 races, Dernaldo won 2 races, Ernaldo won 1 race, and Arnaldo and Bernaldo won none. | |
0gdi | 設三角形 $ABC$ 的內心為 $I$，垂心為 $H$。平面上有一點 $K$ 滿足
$$
AH + AK = BH + BK = CH + CK.
$$
證明：$H$、$I$、$K$ 三點共線。 | [
"Let $\\Omega$ denote the circumcircle of $\\triangle ABC$. Recall that the reflection of $H$ across $\\overline{BC}$, which we denote by $X$, is a point of $\\Omega$. Define $Y, Z$ similarly. Thus the circumcircle $\\omega_A$ of $\\triangle HYZ$ is centered at $A$. Define $\\omega_B, \\omega_C$ similarly.\n\nThen,... | 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 > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cso | Число $x$ таково, что среди четырёх чисел $x - \sqrt{2}$, $x - 1/x$, $x + 1/x$, $x^2 + 2\sqrt{2}$ ровно одно не является целым. Найдите все такие $x$. | [
"Обозначим $a = x - \\sqrt{2}$, $b = x - 1/x$, $c = x + 1/x$, $d = x^2 + 2\\sqrt{2}$. Заметим, что $b$ и $c$ не могут одновременно быть целыми. Действительно, тогда число $b+c = 2x$ также целое, значит, $x$ рационально, поэтому как $a$, так и $d$ не будут целыми как суммы рационального и иррационального чисел. Итак... | Russia | XL Russian mathematical olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | sqrt(2) - 1 | |
0bhc | Let $a \in (0, 1)$. Find all functions $f : \mathbb{R} \to (0, \infty)$ so that $f(x) \le a^x$ and $f(x + y + z) \le f(x)f(y)f(z)$, for every $x, y, z \in \mathbb{R}$. | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = a^x | |
04uq | In a sequence of $71$ nonzero real numbers, each number (apart from the first one and the last one) is one less than the product of its two neighbors. Prove that the first and the last number are equal.
(Josef Tkadlec) | [] | Czech Republic | Final Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0aac | Problem:
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying that for every $x \in \mathbb{R}$,
$$
f(x(1+|x|)) \leq x \leq f(x)(1+|f(x)|)
$$ | [
"Solution:\nLet $g(x) = x(1 + |x|)$. We want that $f(g(x)) \\leq x \\leq g(f(x))$ for all $x$. The solution only uses that $g$ is strictly increasing and surjective with an inverse which is also strictly increasing.\n\nNotice that the inverse function to such $g$ gives a solution $f = g^{-1}$ since $x \\leq x \\leq... | Nordic Mathematical Olympiad | The 35th Nordic Mathematical Contest | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = { (-1 + sqrt(1 + 4x)) / 2 for x >= 0; (1 - sqrt(1 - 4x)) / 2 for x <= 0 } | |
0b41 | Problem:
In $\triangle ABC$, let $D$ be the point on side $BC$ such that $AB + BD = DC + CA$. The line $AD$ intersects the circumcircle of $\triangle ABC$ again at point $X \neq A$. Prove that one of the common tangents of the circumcircles of $\triangle BDX$ and $\triangle CDX$ is parallel to $BC$. | [
"Solution:\n\nRefer to the figure shown below:\n\n\nLet $V$ and $W$ be the midpoints of $\\operatorname{arcs} BD$ and $CD$ respectively. We claim that $VW$ is the desired common tangent. To prove this, let $E$ and $F$ be the orthogonal projections of $V$ and $W$ onto $BC$. Note that $E$ and... | Philippines | Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kjy | A quadratic polynomial $p(x)$ with real coefficients and leading coefficient $1$ is called *disrespectful* if the equation $p(p(x)) = 0$ is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial $\tilde{p}(x)$ for which the sum of the roots is maxim... | [
"Suppose $p(x) = (x - r)(x - s)$. Observe that $p(x)$ must have (two) real roots in order for $p(p(x))$ to have any roots at all. More specifically, if $y$ is a root of $p(p(x))$, then $p(y) = r$ or $p(y) = s$. That is, the equations\n$$\n(x - r)(x - s) - r = 0 \\quad \\text{and} \\quad (x - r)(x - s) - s = 0\n$$\n... | United States | AMC 10 A | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | (A) | |
09py | Problem:
Laat $a$, $b$ en $c$ positieve reële getallen zijn die voldoen aan $a+b+c \geq a b c$. Bewijs dat
$$
a^{2}+b^{2}+c^{2} \geq \sqrt{3} a b c
$$ | [
"Solution:\nWe weten dat $a^{2}+b^{2}+c^{2} \\geq a b+b c+c a$ en dus $3\\left(a^{2}+b^{2}+c^{2}\\right) \\geq(a+b+c)^{2}$. Wegens de ongelijkheid van het meetkundig en rekenkundig gemiddelde geldt $a+b+c \\geq 3(a b c)^{\\frac{1}{3}}$ en anderzijds is gegeven dat $a+b+c \\geq a b c$. Dus we hebben twee ongelijkhed... | Netherlands | Dutch TST | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0l4e | Problem:
Let $ABC$ be a triangle. Let $X$ be the point on side $\overline{AB}$ such that $\angle BXC = 60^{\circ}$. Let $P$ be the point on segment $\overline{CX}$ such that $BP \perp AC$. Given that $AB = 6$, $AC = 7$, and $BP = 4$, compute $CP$. | [
"Solution:\n\nConstruct parallelogram $BPCQ$. We have $CQ = 4$, $\\angle ACQ = 90^{\\circ}$, and $\\angle ABQ = 120^{\\circ}$. Thus, $AQ = \\sqrt{AC^{2} + CQ^{2}} = \\sqrt{65}$, so if $x = CP = BQ$, then by Law of Cosine, $x^{2} + 6x + 6^{2} = 65$. Solving this gives the answer $x = \\sqrt{38} - 3$.\n\n![](attached... | United States | HMMT February 2024 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | sqrt(38) - 3 | |
072a | Problem:
Let $S$ denote the set of all 6-tuples $(a, b, c, d, e, f)$ of positive integers such that $a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=f^{2}$. Consider the set
$$
T=\{a b c d e f:(a, b, c, d, e, f) \in S\}
$$
Find the greatest common divisor of all the members of $T$. | [
"Solution:\nWe show that the required gcd is 24. Consider an element $(a, b, c, d, e, f) \\in S$. We have\n$$\na^{2}+b^{2}+c^{2}+d^{2}+e^{2}=f^{2}\n$$\nWe first observe that not all $a, b, c, d, e$ can be odd. Otherwise, we have $a^{2} \\equiv b^{2} \\equiv c^{2} \\equiv d^{2} \\equiv e^{2} \\equiv 1$ $(\\bmod 8)$ ... | India | INMO 2004 | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 24 | |
085y | Problem:
Alberto, Barbara, Chiara e Davide mescolano un mazzo di 40 carte e poi ne distribuiscono 10 a testa. Alberto guarda le sue ed esclama: "Che strano, non ho nessuna carta di picche". Sapendo questa informazione, qual è la probabilità che anche Barbara non abbia nessuna carta di picche? (nota: le carte di picche... | [
"Solution:\n\nLa risposta è $(\\mathbf{D})$. Il punto di partenza è la definizione di probabilità come\n$$\n\\frac{\\text{casi favorevoli}}{\\text{casi possibili}}.\n$$\nIgnoriamo le carte di Alberto; la situazione si semplifica: ora sappiamo che ci sono 30 carte, di cui 10 di picche, e vogliamo determinare la prob... | Italy | Olimpiadi di Matematica | [
"Statistics > Probability > Counting Methods > Combinations",
"Statistics > Probability > Counting Methods > Permutations"
] | null | MCQ | D | |
06gl | Find all reals $x$, $y$ and $z$ such that
$$
\begin{cases}
(x + 1)yz = 27, \\
(y + 1)zx = 12, \\
(z + 1)xy = 12.
\end{cases}
$$ | [
"The solutions are $(x, y, z) = (2, -3, -3)$, $\\left(\\frac{8 \\mp 2\\sqrt{10}}{3}, 1 \\pm \\sqrt{10}, 1 \\pm \\sqrt{10}\\right)$.\n\nClearly, $x, y, z \\neq 0$. Since\n$$\n(y + 1)zx = 12 = (z + 1)xy,\n$$\nwe have $zx = xy$. This implies $y = z$. By the first equation $(x + 1)y^2 = 27$, we obtain\n$$\nx = \\frac{2... | Hong Kong | Year 2012 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (2, -3, -3); ((8 - 2√10)/3, 1 + √10, 1 + √10); ((8 + 2√10)/3, 1 - √10, 1 - √10) | |
0742 | Suppose the altitudes of a triangle are all integers whose sum does not exceed $20$ and its in-radius is also an integer. Find all possible values for the area of the triangle. | [
"We begin with the standard expression:\n$$\n\\frac{1}{r} = \\frac{1}{h_a} + \\frac{1}{h_b} + \\frac{1}{h_c},\n$$\nwhere $r$ is the inradius and $h_a, h_b, h_c$ are the altitudes of the given triangle, in standard notation.\n\nThe arithmetic mean - harmonic mean inequality gives\n$$\n\\frac{1}{r} \\ge \\frac{9}{h_a... | India | Indija TS 2009 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Prealgebra / Basic Algebra > ... | null | proof and answer | 3√3, 12√3, 100/√15 | |
0a2u | Caitlin took a long walk each day in October. Only on the 16 days when rain was expected, she took an umbrella with her. Of the 31 days of October, the rain forecast was correct on exactly 21 days. Fortunately, on the days when it rained, Caitlin always had her umbrella with her.
On how many days did it *not* rain?
A) ... | [] | Netherlands | Dutch Mathematical Olympiad | [
"Math Word Problems"
] | null | MCQ | E | |
0kbo | Problem:
The number $3003$ is the only number known to appear eight times in Pascal's triangle, at positions
$$
\binom{3003}{1},\binom{3003}{3002},\binom{a}{2},\binom{a}{a-2},\binom{15}{b},\binom{15}{15-b},\binom{14}{6},\binom{14}{8}
$$
Compute $a+b(15-b)$. | [
"Solution:\nWe first solve for $a$. Note that $3003=3 \\cdot 7 \\cdot 11 \\cdot 13$. We have $3003=\\binom{a}{2}=\\frac{a(a-1)}{2} \\approx \\frac{a^{2}}{2}$. This means we can estimate $a \\approx \\sqrt{3003 \\cdot 2}$, so $a$ is a little less than $80$. Furthermore, $11 \\mid 2 \\cdot 3003=a(a-1)$, meaning one o... | United States | HMMO 2020 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 128 | |
0g1r | Problem:
Seien $m, n \geq 2$ natürliche Zahlen. Wir haben vier Farben und wollen jedes Feld eines $m \times n$ Rechtecks mit einer davon einfärben, sodass in jedem $2 \times 2$ Quadrat alle vier Farben vorkommen. Wie viele verschiedene Möglichkeiten gibt es dafür?
Bemerkung: Wir zählen zwei Möglichkeiten als verschie... | [
"Solution:\n\nSeien zuerst $m, n>2$. Wir ersetzen die Farben der Einfachheit halber durch Zahlen und betrachten drei benachbarte Felder einer Zeile, die alle unterschiedlich gefärbt wurden (ObdA mit 1, 2 und 3). Es folgt der Reihe nach:\n\n\n\nAlso sind die gesamten Spalten, welche die drei... | Switzerland | SMO - Vorrunde | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 6·2^m + 6·2^n − 24 | |
02bt | Problem:
Quantos números inteiros e positivos satisfazem a dupla inequação $2000<\sqrt{n(n-1)}<2005$?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5 | [
"Solution:\n\nA opção correta é (e).\nComo os números que aparecem são todos positivos, podemos elevá-los ao quadrado mantendo o sentido das desigualdades, obtendo\n$$\n2000 \\times 2000 = 2000^{2} < n(n-1) < 2005^{2} = 2005 \\times 2005\n$$\nObserve que $n$ e $n-1$ são inteiros consecutivos, portanto, as únicas op... | Brazil | Nível 2 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | e | |
00rv | Consider integers $m \ge 2$ and $n \ge 1$. Show that there is a polynomial $P(x)$ of degree equal to $n$ with integer coefficients such that $P(0), P(1), \dots, P(n)$ are all perfect powers of $m$. | [
"Let $a_0, a_1, \\dots, a_n$ be integers to be chosen later, and consider the polynomial $P(x) = \\frac{1}{n!} Q(x)$ where\n$$\nQ(x) = \\sum_{k=0}^{n} (-1)^{n-k} \\binom{n}{k} a_k \\prod_{\\substack{0 \\le i \\le n \\\\ i \\ne k}} (x-i).\n$$\nObserve that for $l \\in \\{0, 1, \\dots, n\\}$ we have\n$$\n\\begin{alig... | Balkan Mathematical Olympiad | BMO 2017 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof only | null | |
08h4 | Problem:
For any set $A=\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\}$ of five distinct positive integers denote by $S_{A}$ the sum of its elements, and denote by $T_{A}$ the number of triples $(i, j, k)$ with $1 \leqslant i<j<k \leqslant 5$ for which $x_{i}+x_{j}+x_{k}$ divides $S_{A}$.
Find the largest possible value of $T_{... | [
"Solution:\nWe will prove that the maximum value that $T_{A}$ can attain is $4$. Let $A=\\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\\}$ be a set of five positive integers such that $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. Call a triple $(i, j, k)$ with $1 \\leqslant i<j<k \\leqslant 5$ good if $x_{i}+x_{j}+x_{k}$ divides $S_{A}$.... | JBMO | null | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 4 | |
0h4i | **Can real numbers $x$, $y$, $z$ satisfy**
$$
\frac{1}{(x-y)(x+y)} + \frac{1}{(y-z)(y+z)} + \frac{1}{(z-x)(z+x)} = 0?
$$ | [
"Denote $a = x^2 - y^2$, $b = y^2 - z^2$, then $-a - b = z^2 - x^2$ and the equation can be rewritten as:\n$$\n\\frac{1}{a} + \\frac{1}{b} = \\frac{1}{a+b} \\Leftrightarrow (a+b)^2 = ab \\Leftrightarrow a^2 - ab + b^2 = 0 \\Leftrightarrow \\left(a - \\frac{b}{2}\\right)^2 + \\frac{3b^2}{4} = 0.\n$$\nThe last equali... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | No | |
02ls | Let $q = 2p+1$, $p, q > 0$ primes. Prove that there exists a multiple of $q$ whose digits sum in decimal base is positive and at most 3. | [
"First notice that $10$ is a multiple of $5 = 2 \\cdot 2 + 1$, so we are done if $q$ and $10$ are not coprime. If $q$ and $10$ are coprime, since $q$ is prime, by Fermat's theorem $q \\mid 10^{q-1} - 1 \\Leftrightarrow q \\mid 10^{2p} - 1 \\Leftrightarrow q \\mid 10^p - 1$ or $q \\mid 10^p + 1$. If $q \\mid 10^p + ... | Brazil | XXXI Brazilian Math Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0ijz | Problem:
$$
\begin{array}{rlrr}
A=(0,428) & B=(9,85) & C=(42,865) & D=(192,875) \\
E=(193,219) & F=(204,108) & G=(292,219) & H=(316,378) \\
I=(375,688) & J=(597,498) & K=(679,766) & L=(739,641) \\
& M=(772,307) & & N=(793,0) \\
&
\end{array}
$$
A fly starts at $A$, visits all the other points, and comes back to $A$ in... | [
"Solution:\n\nThis problem is an instance of the Traveling Salesman Problem, which is NP-hard. There is an obvious algorithm in $O(n!)$ time (where $n$ is the number of points), but faster algorithms exist. Nonetheless, the best strategy for solving this problem is probably to draw the points and exercise your geom... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Algorithms",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | final answer only | null | |
01yh | Two lines pass through the point $F(0; \frac{1}{4})$ on the Cartesian plane. These lines are perpendicular to each other and intersect the parabola $y = x^2$ at the points $A$, $B$, $C$ and $D$ (these points are mentioned in the $x$-coordinate increasing order). The difference of projections of the segments $AD$ and $B... | [
"Answer: $\\frac{m^2}{2}$.\nDenote the abscissas of the points $A$, $B$, $C$ and $D$ by $a$, $b$, $c$ and $d$ respectively. It is easy to see that the perpendicularity of the lines $AC$ and $BD$ is equivalent to the equality $(a+c)(b+d) = -1$ and the condition about the difference of projections is equivalent to th... | Belarus | Belarus2022 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals"
] | English | proof and answer | m^2/2 | |
0dtm | Alice and Bob play a game. Bob starts by picking a set $S$ consisting of $M$ vectors of length $n$ with entries either $0$ or $1$. Alice picks a sequence of numbers $y_1 \le y_2 \le \cdots \le y_n$ from the interval $[0, 1]$, and a choice of real numbers $x_1, \dots, x_n \in \mathbb{R}$. Bob wins if he can pick a vecto... | [
"The answer is $M = n + 1$.\n\n$|S| \\ge n + 1$, since if Alice selects\n$$\n(y_1, y_2, \\dots, y_n) = \\underbrace{(0, 0, \\dots, 0)}_{k\\ 0's}, 1, 1, \\dots, 1)\n$$\n$$\n(x_1, x_2, \\dots, x_n) = \\underbrace{(-1, -1, \\dots, -1)}_{k\\ (-1)'s}, 1, 1, \\dots, 1),\n$$\nwhere $0 \\le k \\le n$, then Bob can only win... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | n + 1 |
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