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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
07h8 | Let $a$, $b$, $c$, $d$ be four non-zero complex numbers such that
$$
2|a - b| \le |b|, \quad 2|b - c| \le |c|, \quad 2|c - d| \le |d|, \quad 2|d - a| \le |a|.
$$
Prove that
$$
\left| \frac{b}{a} + \frac{c}{b} + \frac{d}{c} + \frac{a}{d} \right| > \frac{7}{2}.
$$ | [
"$$\n\\left| \\frac{a}{b} - 1 \\right| \\le \\frac{1}{2}, \\quad \\left| \\frac{b}{c} - 1 \\right| \\le \\frac{1}{2}, \\quad \\left| \\frac{c}{d} - 1 \\right| \\le \\frac{1}{2}, \\quad \\left| \\frac{d}{a} - 1 \\right| \\le \\frac{1}{2}.\n$$\nPutting $(\\frac{a}{b}, \\frac{b}{c}, \\frac{c}{d}, \\frac{d}{a}) = (x, y... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0ajb | Let $ABC$ be an acute triangle and let $k$ be the circle circumscribed around it. The point $O$ in the interior of the triangle is such that $\overline{CE} = \overline{CF}$, where $E$ and $F$ are points on $k$ and $E$ lies on $AO$, and $F$ lies on $BO$. Prove that $O$ lies on the bisector of the angle at the vertex $C$... | [
"From $\\overline{CE} = \\overline{CF}$ it follows that $\\angle CAE = \\angle CBF$ (1), as inscribed angles subtending equal chords.\n\nLet us assume first that the triangle is isosceles. Then, from the fact that $O$ lies in the interior of $ABC$ and (1), it follows that $\\angle BAO = \\angle BAC - \\angle CAO = ... | North Macedonia | Macedonian Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0ekj | Problem:
Drugi največji delitelj nekega naravnega števila $n$ je 2022. Kateri je tretji največji delitelj tega naravnega števila $n$?
(A) 337
(B) 674
(C) 1011
(D) 1348
(E) 2021 | [
"Solution:\n\nDrugi največji delitelj naravnega števila $n$ je enak $\\frac{n}{p}$, kjer je $p$ najmanjše praštevilo, ki deli $n$. Torej je $n = 2022 p = 2 \\cdot 3 \\cdot 337 \\cdot p$. Od tod sledi, da je $n$ deljiv z $2$, torej je $p = 2$. Tretji največji delitelj števila $n$ je zato enak $2 \\cdot 337 \\cdot 2 ... | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | D | |
023u | Problem:
Dois jogadores se enfrentam em um jogo de combate com dados. O atacante lançará três dados e o defensor, dois. O atacante derrotará o defensor em apenas um lance de dados se, e somente se, as duas condições seguintes forem satisfeitas:
i) O maior dado do atacante for maior do que o maior dado do defensor.
ii)... | [
"Solution:\n\na) Para ganhar, precisamos tirar ao menos dois $6$. A probabilidade será igual a tirar:\n- três seis: $P_{1} = \\left(\\frac{1}{6}\\right)^{3}$; ou\n- dois seis e outro número qualquer menor que $6$: $P_{2} = 3 \\cdot \\left(\\frac{1}{6}\\right)^{2} \\cdot \\frac{5}{6} = \\frac{15}{6^{3}}$.\nPortanto,... | Brazil | null | [
"Statistics > Probability > Counting Methods > Permutations",
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | a) 2/27; b) 43/216 | |
0avv | Problem:
Pentagon $A B C D E$ is inscribed in a circle. Its diagonals $A C$ and $B D$ intersect at $F$. The bisectors of $\angle B A C$ and $\angle C D B$ intersect at $G$. Let $A G$ intersect $B D$ at $H$, let $D G$ intersect $A C$ at $I$, and let $E G$ intersect $A D$ at $J$. If $F H G I$ is cyclic and
$$
J A \cdot ... | [
"Solution:\n\nSince $\\angle B A C$ and $\\angle B D C$ subtend the same arc, we can let $\\alpha=\\angle B A G=\\angle G A C=\\angle C D G=\\angle G D B$. Since $\\angle B A G=\\angle B D G$, then $G$ is a point on the circumcircle.\nLet $x=\\angle F H I$ and $y=\\angle F I H$. Since $A H I D$ is cyclic $(\\angle ... | Philippines | 18th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03si | Suppose there are $8$ white balls and $2$ red balls in a packet. Each time one ball is drawn and replaced by a white one. Then the probability of drawing out all of the red balls just in the fourth draw is ______. | [
"The following three cases can satisfy the condition.\n\n| | 1st draw | 2nd draw | 3rd draw | 4th draw |\n|--------|----------|----------|----------|----------|\n| Case 1 | Red | White | White | Red |\n| Case 2 | White | Red | White | Red |\n| Case 3 | White | White | Re... | China | China Mathematical Competition | [
"Statistics > Probability > Counting Methods > Other"
] | English | final answer only | 0.0434 | |
016o | Albert, Ben and Carla are looking at the dust in the air, and Ben says that if there are $1000$ dust grains in a $10\text{cm} \times 10\text{cm} \times 10\text{cm}$ box, then no matter how they are situated, he can choose a point such that there are at least $10$ dust grains in a distance of at most $2$ cm from the poi... | [
"Carla is right. Take each dust grain and colour all points in a distance of at most $2$ cm and at least $1$ cm from the grain. Then we have coloured a volume of $1000 \\cdot \\frac{4}{3} \\cdot \\pi \\cdot (2^3 - 1^3) = \\frac{28000}{3}\\pi\\text{cm}^3 > 28000\\text{cm}^3$ counted with multiplicity. All the colour... | Baltic Way | BALTIC WAY | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Solid Geometry > Volume"
] | null | proof and answer | Carla | |
0ci6 | Let $BC$ be a fixed segment in the plane, and let $A$ be a variable point in the plane not on the line $BC$. Distinct points $X$ and $Y$ are chosen on the rays $\vec{CA}$ and $\vec{BA}$, respectively, such that $\angle CBX = \angle YCB = \angle BAC$. Assume that the tangents to the circumcircle of $ABC$ at $B$ and $C$ ... | [
"Let $X', Y', P'$, and $Q'$ be the reflections across $BC$ of $X, Y, P$, and $Q$, respectively. Then $\\Omega_1$ and $\\Omega_2$ are just the circles $PXX'P'$ and $QYY'Q'$, respectively.\nDenote $\\alpha = \\angle BAC = \\angle CBX = \\angle YCB$. Let $XY$ cross $BC$ at $W$; the case $XY \\parallel BC$ may be treat... | Romania | Romanian Master of Mathematics | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmon... | English | proof only | null | |
0245 | Problem:
Potências de $3-$ Se $3^{a}=2$, quanto vale $27^{2 a}$ ? | [
"Solution:\n\nTemos $27^{2 a} = (3^{3})^{2 a} = 3^{6 a} = (3^{a})^{6} = 2^{6} = 64$."
] | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | final answer only | 64 | |
06zt | Problem:
Show that any convex polygon of area $1$ is contained in some parallelogram of area $2$. | [
"Solution:\nLet the vertices $X$, $Y$ of the polygon be the two which are furthest apart. The polygon must lie between the lines through $X$ and $Y$ perpendicular to $XY$ (for if a vertex $Z$ lay outside the line through $Y$, then $ZY > XY$). Take two sides of a rectangle along these lines and the other two sides a... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals"
] | null | proof only | null | |
0cf6 | Let $a$, $b$ and $c$ be three real numbers such that $ab + bc + ca = 3$. Prove that
$$
\frac{a^4 + b^4 + (a+b)^4}{a^2 + b^2 + ab} + \frac{b^4 + c^4 + (b+c)^4}{b^2 + c^2 + bc} + \frac{c^4 + a^4 + (c+a)^4}{c^2 + a^2 + ca} \ge 18.
$$ | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0dvn | Problem:
Dani sta premici z enačbama $(1-a) x - 2 a y - 2 = 0$ in $-2 x + a y - 1 = 0$. Določi $a$ tako, da se bosta premici sekali na simetrali lihih kvadrantov. | [
"Solution:\n\nČe je $a = 0$, sta premici med seboj vzporedni (njuni enačbi sta tedaj $x - 2 = 0$ in $2 x + 1 = 0$), zato privzemimo, da $a \\neq 0$.\n\nIzrazimo $y = \\frac{(1-a)x - 2}{2a}$ iz prve in $y = \\frac{2x + 1}{a}$ iz druge enačbe.\n\nIzenačimo dobljeni desni strani:\n$$\n\\frac{(1-a)x - 2}{2a} = \\frac{2... | Slovenia | 3. matematično tekmovanje dijakov srednjih tehniških in strokovnih sol | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | a = 1 | |
0ekt | Problem:
Trapez $ABCD$ je včrtan krožnici $\mathcal{K}$. Nosilki stranic $AD$ in $BC$ se sekata v točki $M$, tangenti na krožnico $\mathcal{K}$ v točkah $B$ in $D$ pa se sekata v točki $N$. Dokaži, da sta daljici $MN$ in $AB$ vzporedni. | [
"Solution:\n\nNaj bo $S$ središče krožnice $\\mathcal{K}$ in $T$ presečišče premice $MS$ s stranico $AB$. Ker je trapez $ABCD$ tetiven, je enakokrak s krakoma $AD$ in $BC$. Označimo $\\alpha=\\angle BAD=\\angle CBA$. Zaradi simetrije lahko predpostavimo, da je $|AB|>|CD|$. Trikotnik $BMA$ je enakokrak z vrhom pri $... | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0411 | Let $m$ be a positive integer, $n = 2^m - 1$, and $P_n = \{1, 2, \dots, n\}$ be the set of $n$ points on the number axis. A grasshopper jumps between adjacent points on $P_n$. Find the maximal number of $m$ such that for any $x, y \in P_n$, the number of ways that a grasshopper jumping from $x$ to $y$ by 2012 steps is ... | [
"If $m \\ge 11$, then $n = 2^m - 1 > 2013$. Since there is only one way a grasshopper jumps from point $1$ to point $2013$ by $2012$ steps, we see that $m \\le 10$.\n\nIn the following, we show that the answer is $m = 10$. To show this, we will prove a stronger proposition by induction on $m$: for any $k \\ge n = 2... | China | China Southeastern Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 10 | |
08k8 | Problem:
Find all the integer solutions of the equation
$$
9 x^{2} y^{2} + 9 x y^{2} + 6 x^{2} y + 18 x y + x^{2} + 2 y^{2} + 5 x + 7 y + 6 = 0
$$ | [
"Solution:\nThe equation is equivalent to the following one\n$$\n\\begin{aligned}\n& \\left(9 y^{2} + 6 y + 1\\right) x^{2} + \\left(9 y^{2} + 18 y + 5\\right) x + 2 y^{2} + 7 y + 6 = 0 \\\\\n& \\Leftrightarrow (3 y + 1)^{2} \\left(x^{2} + x\\right) + 4(3 y + 1) x + 2 y^{2} + 7 y + 6 = 0\n\\end{aligned}\n$$\nTheref... | JBMO | OJBM | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (-2, 0), (-3, 0), (0, -2), (-1, 2) | |
0fiw | Problem:
Demuestra que no existe ninguna función $f: \mathbb{N} \rightarrow \mathbb{N}$ que cumpla
$$
f(f(n))=n+1
$$ | [
"Solution:\nSupongamos que exista $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ tal que $f(f(n))=n+1$.\nSe tiene que $f(0)=a \\in \\mathbb{N}$. Por el enunciado\n$$\nf(f(0))=1 ; \\quad f(f(0))=f(a)=1\n$$\ny del mismo modo,\n$$\nf(1)=a+1,\\ f(a+1)=2,\\ f(2)=a+2,\\ \\ldots\n$$\nSupongamos que $f(n-1)=a+n-1$; entonces $f(... | Spain | Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0huc | Problem:
In a bag are $n$ fair, six-sided dice whose faces are colored white and red in such a way that the total numbers of white and red sides are equal. Let $p$ be the probability that the same color comes up twice when taking one die randomly out of the bag and throwing it twice. Let $q$ be the probability that th... | [
"Solution:\n\nConsider the following procedure: Remove one die randomly from the bag, roll it, replace it in the bag, remove another die randomly from the bag, and roll it. It is clear that each roll is an independent and random choice of one of the $6 n$ sides of all the dice; hence the probability of getting the ... | United States | Berkeley Math Circle | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof only | null | |
0e8a | Problem:
Naj bo $\mathcal{K}_1$ krožnica s središčem $S_1$ in polmerom $r$. Naj bo $\mathcal{K}_2$ krožnica s središčem $S_2$ na krožnici $\mathcal{K}_1$ in polmerom $\frac{2}{3} r$. Presečišče premice $S_1 S_2$ s krožnico $\mathcal{K}_2$, ki leži zunaj kroga, omejenega s krožnico $\mathcal{K}_1$, označimo z $A$. Eno ... | [
"\n\nKer je štirikotnik $E S_2 C D$ tetiven, je $\\angle S_2 E D=\\angle S_2 C A=\\angle C A S_2$. Torej je trikotnik $E A D$ enakokrak z vrhom $D$. Od tod sledi $|A H|=|E H|$ oziroma $|A H|=\\frac{1}{2}|E A|=\\frac{1}{2}\\left(2 r+\\frac{2}{3} r\\right)=\\frac{4}{3} r$. Ker je $\\frac{4}{3... | Slovenia | 57. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > M... | null | proof only | null | |
083c | Problem:
Ad un pranzo sono state invitate $n$ persone, che siederanno attorno ad una tavola rotonda, i cui posti sono stati contrassegnati da 1 ad $n$ mediante opportuni cartellini segnaposto, distribuiti da un maestro cerimoniere.
Il cameriere ha deciso di servire le portate seguendo un procedimento originale: sceglie... | [
"Solution:\nÈ possibile sistemare i segnaposto nel modo voluto se e solo se $n$ è pari.\nSe $n=2k$ è pari, una possibile soluzione è la seguente: procedendo in senso orario lungo la tavola, il maestro cerimoniere piazza prima il segnaposto numero $2k$, poi tutti quelli pari in ordine crescente e quindi tutti i disp... | Italy | XIX Gara Nazionale di Matematica | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | n is even | |
08ao | Problem:
Quanti interi positivi sono una potenza di $4$ e si scrivono in base $3$ usando solo le cifre $0$ e $1$, lo $0$ quante volte si vuole (anche nessuna) e l'$1$ al più due volte?
(A) $4$
(B) $2$
(C) $1$
(D) $0$
(E) Infiniti. | [
"Solution:\n\nLa risposta è (B). Un numero che in base $3$ termina con zero è un multiplo di $3$; siccome nessuna potenza di $4$ è un multiplo di $3$, i numeri che cerchiamo, in base $3$, finiscono con $1$. Se usiamo esattamente una cifra $1$, l'unica possibilità è quindi il numero che si scrive come \"1\" in base ... | Italy | Progetto Olimpiadi della Matematica - Gara di Febbraio | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | MCQ | B | |
0eqi | The tens digit of the product $1 \times 2 \times 3 \times \cdots \times 98 \times 99$ is
(A) 0 (B) 1 (C) 2 (D) 4 (E) 9 | [] | South Africa | South African Mathematics Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | A | |
003w | Un conjunto de enteros positivos distintos se dice *especial* si para todo par de estos enteros, $a, b$, se verifica que $\frac{a+b}{a-b}$ es un número entero (no necesariamente positivo).
Encontrar un conjunto especial de 5 números, y determinar si existe un conjunto especial de 10 números. | [] | Argentina | Argentina 2006 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Español | proof and answer | One example is {6, 8, 9, 10, 12}. No special set of 10 numbers exists. | |
0fmb | Problem:
En un triángulo rectángulo de hipotenusa unidad y ángulos de $30^{\circ}, 60^{\circ}$ y $90^{\circ}$, se eligen 25 puntos cualesquiera. Demuestra que siempre habrá 9 de ellos que podrán cubrirse con un semicírculo de radio $\frac{3}{10}$. | [
"Solution:\n\nEste triángulo se puede descomponer en tres triángulos congruentes y semejantes al triángulo inicial.\n\n\n\nTenemos 3 triángulos y 25 puntos. En algún triángulo habrá al menos 9 puntos. La hipotenusa de cada uno de estos triángulos semejantes al inicial mide $\\frac{\\sqrt{3}... | Spain | Spain | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
05d2 | The angle bisectors of an acute triangle $ABC$ meet at point $I$. The line $AI$ meets the circumcircle of the triangle $ABC$ at point $D$ ($D \neq A$) and the side $BC$ at point $E$. The line $BI$ meets the circumcircle of the triangle $CDI$ at point $K$ whereas the line $CI$ meets the circumcircle of the triangle $BDI... | [
"Let $\\alpha = \\angle CAI = \\angle IAB$, $\\beta = \\angle ABI = \\angle IBC$, $\\gamma = \\angle BCI = \\angle ICA$. Then $\\alpha + \\beta + \\gamma = 90^\\circ$ and $\\angle CBD = \\angle CAD = \\alpha = \\angle DAB = \\angle DCB$, yielding\n$$\n\\angle KBD = \\angle IBD = \\alpha + \\beta = 90^\\circ - \\gam... | Estonia | Estonian 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 > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0gm3 | Given a circle with center $O$ and a point $A$ in the interior of this circle, find the geometric locus of the intersection of $[AB]$ with the inner bisection of $\angle AOB$, where $B$ is a point on the circle outside the line $OA$. | [] | Turkey | VIIth NATIONAL MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
04un | Let $ABC$ be a scalene triangle, $I$ its incenter and $k$ its circumcircle. Rays $BI, CI$ meet $k$ again at $S_b \neq B$, $S_c \neq C$, respectively. Prove that the tangent to $k$ at $A$, the line $S_bS_c$, and the line through $I$ parallel to $BC$ are concurrent. (Patrik Bak) | [] | Czech Republic | First Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Advanced Configu... | English | proof only | null | |
09kr | Denote by $k!!$ the product $k \times (k-2) \times \cdots \times 1$ for any odd integer $k \ge 1$. Show that $(2^m - 1)!! - 1$ is divisible by $2^m$ for any integer $m \ge 3$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof only | null | |
0kl4 | Problem:
Find the number of ways in which the letters in "HMMTHMMT" can be rearranged so that each letter is adjacent to another copy of the same letter. For example, "MMMMTTHH" satisfies this property, but "HHTMMMTM" does not. | [
"Solution:\nThe final string must consist of \"blocks\" of at least two consecutive repeated letters. For example, MMMMTTHH has a block of 4 M's, a block of 2 T's, and a block of 2 H's. Both H's must be in a block, both T's must be in a block, and all M's are either in the same block or in two blocks of 2. Therefor... | United States | HMMT November 2021 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 12 | |
043l | As shown in Fig. 11.1, in a plane rectangular coordinate system $xOy$, the left and right foci of the ellipse $\Gamma : \frac{x^2}{2} + y^2 = 1$ are $F_1, F_2$, respectively. Let $P$ be a point on $\Gamma$ in the first quadrant, and the extensions of $PF_1, PF_2$ intersect $\Gamma$ at points $Q_1, Q_2$, respectively. L... | [
"It is easy to find $F_1 = (-1, 0), F_2 = (1, 0)$.\nDenote $P(x_0, y_0)$, $Q_1(x_1, y_1)$, $Q_2(x_2, y_2)$. By the given condition, it follows that\n$$\nx_0, y_0 > 0, \\quad y_1 < 0, \\quad y_2 < 0.\n$$\nBy the definition of ellipse we get\n$$\n|PF_1| + |PF_2| = |Q_1F_1| + |Q_1F_2| = |Q_2F_1| + |Q_2F_2| = 2\\sqrt{2... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algeb... | null | proof and answer | 1/3 | |
0fmu | Deslizamos un cuadrado de 10 cm de lado por el plano $OXY$ de forma que los vértices de uno de sus lados estén siempre en contacto con los ejes de coordenadas, uno con el eje $OX$ y otro con el eje $OY$. Determina el lugar geométrico que en ese movimiento describen:
1. El punto medio del lado de contacto con los ejes.
... | [
"Sean $PQRS$ el cuadrado de lado 10 cm, $PQ$ el lado de apoyo, $M(m_1, m_2)$ el punto medio de dicho lado y $C(c_1, c_2)$ el centro del cuadrado tal y como muestra la figura donde, además, señalamos los puntos $A, B, D$ y $E$.\n\n\n\na) Caso del punto medio $M$.\n$$\nOM = PM = \\frac{1}{2}P... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | Spanish | proof and answer | 1) Midpoint M of the supporting side: circle x^2 + y^2 = 25.
2) Center C: moves along the angle bisectors as (±5λ, ±5λ) with λ in [1, √2]; in the first quadrant this is the segment on y = x from (5, 5) to (5√2, 5√2).
3) Vertices of the supporting side P and Q (first quadrant): P lies on the x-axis segment {(t, 0) : 0... | |
0eqo | Together, the two positive integers $a$ and $b$ have 9 digits and contain each of the digits $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$ exactly once. For which possible values of $a$ and $b$ is the fraction $a/b$ closest to $1$? | [
"If $a > b$, then $a$ has at least five digits, so $a \\ge 12345$, and $b$ has at most four digits, so $b \\le 9876$. In this case, we have\n$$\n\\frac{a}{b} \\ge \\frac{12345}{9876} > 1,\n$$\nso the value of $a/b$ that is closest to $1$ is\n$$\n\\frac{12345}{9876} = 1 + \\frac{2469}{9876}\n$$\nin this case.\n\nOn ... | South Africa | The South African Mathematical Olympiad Third Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a = 9876, b = 12345 | |
0c6t | Let $ABC$ be an acute triangle, and let $D, E, F$ be the feet of the altitudes from $A, B, C$, respectively. The lines $BC$ and $EF$ cross at $P$, and the line through $D$ and parallel to $EF$ crosses the lines $AC$ and $AB$ at $Q$ and $R$, respectively. Prove that the circle $PQR$ passes through the midpoint of the si... | [
"Let $M$ be the midpoint of the side $BC$. If $AB = AC$, then $P$ is the ideal point of the line $BC$, the points $Q$ and $R$ fall at $C$ and $B$, respectively, and the circle $PQR$ degenerates into the line $BC$ on which $M$ clearly lies.\n\n\n\nAssume henceforth that $AB \\neq AC$, say, $... | Romania | 70th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneo... | English | proof only | null | |
08hg | Problem:
The circle with the center $O$ is tangent to the sides $[AB]$, $[BC]$, $[CD]$ and $[DA]$ of the convex quadrilateral $ABCD$ at the points $M$, $N$, $K$ and $L$ respectively. The straight lines $MN$ and $AC$ are parallel and the straight line $MK$ intersects the line $LN$ at the point $P$. Prove that the point... | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ae8 | Определи ги комплексните броеви $z$ за кои
$$
|z| = \frac{1}{|z|} = |z - 1|.
$$ | [
"Јасно е дека равенките се определени за $z \\neq 0$. Од равенката $|z| = \\frac{1}{|z|}$, добиваме $|z|^2 = 1$, односно\n$$\n|z| = 1. \\qquad (1)\n$$\nОд претходната равенка и равенката $|z| = |z - 1|$ ја добиваме равенката\n$$\n|z - 1| = 1. \\qquad (2)\n$$\nАко комплексниот број $z$ го запишеме во алгебарски обли... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Algebra > Intermediate Algebra > Complex numbers"
] | Macedonian, English | proof and answer | {1/2 + i*sqrt(3)/2, 1/2 - i*sqrt(3)/2} | |
09oc | If $a, b, c$ are nonzero numbers such that $\sqrt[3]{abc}(a+b+c) = ab+bc+ca$
then prove that these numbers, written in some order, form a geometric progression.
(Otgonbayar Uuye) | [] | Mongolia | MMO2025 Round 3 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0it7 | Problem:
Compute $2009^{2} - 2008^{2}$. | [
"Solution:\nFactoring this product with difference of squares, we find it equals:\n$$\n(2009 + 2008)(2009 - 2008) = (4017)(1) = 4017\n$$"
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 4017 | |
04sy | A triangle $ABC$ is given every two sides of which differ in length by at least $d > 0$. Denote by $T$ its centroid, $I$ incentre and $\rho$ inradius. Prove that
$$
S_{AIT} + S_{BIT} + S_{CIT} \geq \frac{2}{3} \rho d,
$$
where $S_{XYZ}$ denotes the area of triangle $XYZ$. | [] | Czech Republic | Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof only | null | |
0aj3 | The contestants of this year's MMO are "well" distributed in $n$ columns (a distribution in columns is "well" if no two contestants in the same column are acquaintances), but the same cannot be obtained in less than $n$ columns. Show that there exist contestants $M_1, M_2, \dots, M_n$ for which the following hold:
(1) ... | [
"We will perform a rearrangement with respect to columns. First we move to the first column each contestant from the second column who doesn't have an acquaintance in the first column. **(1 point)** The new arrangement is “well”, and therefore at least one contestant remains in the second column. Now we move to the... | North Macedonia | Macedonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
097d | Problem:
Triunghiul ascuțitunghic isoscel $ABC$, $m(\angle B) = m(\angle C) = \alpha$, este baza prismei $ABC A_{1} B_{1} C_{1}$. Muchia laterală $A_{1}A$ este perpendiculară muchiei $AC$, iar $m\left(\angle A_{1}AB\right) = \beta < 90^{\circ}$. Determinați aria laterală a prismei, dacă $A_{1}A = BC = a$. | [
"Solution:\n\nȚinând cont de faptul că $A_{1}ACC_{1}$ este dreptunghi, $AC = AB = \\frac{a}{2 \\cos \\alpha}$, obținem $\\mathcal{A}_{A_{1}ACC_{1}} = \\frac{a^{2}}{2 \\cos \\alpha}$. $\\mathcal{A}_{A_{1}ABB_{1}} = \\frac{a^{2}}{2 \\cos \\alpha} \\sin \\beta$.\n\nConsiderăm dreapta $d \\parallel AB$, $C \\in d$. Fie... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"... | null | proof and answer | A_lat = (a^2 / (2 cos α)) (1 + sin β + sqrt(4 cos^2 α − cos^2 β)) | |
0053 | Alex y Bruno escriben, entre los dos, un número natural de 6 dígitos distintos. Cada uno, en su turno, escribe un dígito a la derecha del último dígito que escribió el otro. Empieza Alex con el primer dígito de la izquierda y termina Bruno con el último dígito de la derecha. (Está prohibido escribir un dígito que ya se... | [] | Argentina | XIIIª OLIMPÍADA de MAYO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization"
] | Español | proof and answer | Alex | |
0d32 | Find all pairs of positive integers $(a, b)$ such that $a^{2} + b^{2}$ divides both $a^{3} + 1$ and $b^{3} + 1$. | [
"We have\n$$\n0 \\equiv (a^{3} + 1) - (b^{3} + 1) \\equiv (a - b)(a^{2} + a b + b^{2}) \\equiv (a - b) a b \\pmod{a^{2} + b^{2}}.\n$$\nLet $d$ be a common divisor of $a$ and $a^{2} + b^{2}$. Then $d$ divides $a^{3} + 1$ and $a^{3}$, so it divides $1$. Hence $a$ and $a^{2} + b^{2}$ are coprime. In a similar way $b$ ... | Saudi Arabia | Selection tests for the Gulf Mathematical Olympiad 2013 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (1,1) | |
0g81 | 設 $ABCD$ 為凸四邊形, $BC$ 與 $AD$ 兩邊並不平行。假設 $BC$ 邊上有一點 $E$ 使得四邊形 $ABED$ 與四邊形 $AECD$ 都有內切圓。試證: $AD$ 邊上存在一點 $F$ 使得四邊形 $ABCF$ 與四邊形 $BCDF$ 都有內切圓的充要條件是 $AB$ 平行於 $CD$。 | [
"設 $\\omega_1, \\omega_2$ 分別為四邊形 $ABED$, $AECD$ 的內切圓, 點 $O_1, O_2$ 分別為 $\\omega_1, \\omega_2$ 的圓心。存在滿足題設中的一點 $F$ 的充分條件是如果 $\\omega_1, \\omega_2$ 也分別是四邊形 $ABCF$, $BCDF$ 的內切圓。\n\n自 $B$ 向 $\\omega_2$ 引異於 $BC$ 的切線, 並且自 $C$ 向 $\\omega_1$ 引異於 $BC$ 的切線, 令此兩條切線分別與 $AD$ 邊交於點 $F_1, F_2$。我們需要證明: $F_1 = F_2$ 的充要條件是 $AB \\paral... | Taiwan | 二〇一三數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0046 | Hay 390 monedas de oro distribuidas en 30 cofres: 13 monedas en cada cofre. Cada moneda pesa un número entero de gramos, mayor o igual que 1 y menor o igual que 13 y hay 13 monedas de cada peso.
Se sabe que si dos monedas están en un mismo cofre, la diferencia entre sus pesos es menor o igual que 4 gramos. Determinar c... | [] | Argentina | Argentina 2006 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | Español | proof and answer | 364 | |
03zd | Determine, with proof, whether there is any odd integer $n \ge 3$ and $n$ distinct prime numbers $p_1, p_2, \dots, p_n$, such that all $p_i + p_{i-1}$ ($i=1, 2, \dots, n$, and $p_{n-1} = p_1$) are perfect squares? | [
"Suppose that there exists odd integer $n \\ge 3$ and $n$ distinct prime numbers $p_1, p_2, \\dots, p_n$ satisfying the given condition.\nIf all $p_1, p_2, \\dots, p_n$ are odd, then all the sums $p_i + p_{i+1}$ are multiples of $4$, so the prime numbers $p_1, p_2, \\dots, p_n$ modulo $4$ appear to be $1$ and $3$ a... | China | 2011 China Western Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | English | proof and answer | No; such an odd number and primes do not exist. | |
09qv | Problem:
Zij $I$ het middelpunt van de ingeschreven cirkel van driehoek $ABC$. Een lijn door $I$ snijdt het inwendige van lijnstuk $AB$ in $M$ en het inwendige van lijnstuk $BC$ in $N$. We nemen aan dat $BMN$ een scherphoekige driehoek is. Laat nu $K$ en $L$ punten op lijnstuk $AC$ zijn zodat $\angle BMI = \angle ILA$... | [
"Solution:\n\nNoem $D$, $E$ en $F$ de voetpunten van $I$ op respectievelijk $BC$, $CA$ en $AB$. Er geldt dat $N$ tussen $C$ en $D$ ligt: als namelijk $N$ tussen $D$ en $B$ ligt, dan is $\\angle BNI$ groter dan $\\angle BDI = 90^{\\circ}$, maar gegeven is dat $\\triangle BMN$ scherphoekig is. Dus $N$ ligt tussen $C$... | Netherlands | Toets 6 juni 2012 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | null | proof only | null | |
04g7 | Let $a_1, b_1, c_1, a_2, b_2, c_2$ be positive real numbers such that $b_1^2 \le 4a_1c_1$ and $b_2^2 \le 4a_2c_2$. Prove that $4(a_1 + a_2 + 5)(c_1 + c_2 + 1) > (b_1 + b_2 + 2)^2$. (Macedonia 2013) | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0cmi | Prove that there exists a positive integer $n > 1$ such that the product of some $n$ consecutive positive integers equals the product of some $n + 100$ consecutive positive integers. | [
"For example, let $n = (1 \\cdot 2 \\cdot 3 \\cdots 101) - 101$. Then the product of the first $n + 100$ natural numbers equals the product of $n$ consecutive numbers starting from $102$ and ending at $n + 101$.\n\nIndeed, after cancellation, the equality\n$$\n1 \\cdot 2 \\cdot 3 \\cdots (n+100) = 102 \\cdot 103 \\... | Russia | Russian mathematical olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | English; Russian | proof only | null | |
03me | For any positive integers $n$ and $k$, let $L(n, k)$ be the least common multiple of the $k$ consecutive integers $n, n+1, \dots, n+k-1$. Show that for any integer $b$, there exist integers $n$ and $k$ such that $L(n, k) > b L(n + 1, k)$.
Soit $L(n, k)$ le plus petit commun multiple de la suite des $k$ entiers consécu... | [
"**I.** Let $p > b$ be prime, let $n = p^3$ and $k = p^2$. If $p^3 < i < p^3 + p^2$, then no power of $p$ greater than 1 divides $i$, while $p$ divides $p^3 + p$. It follows that $L(p^3, p^2) = p^2 L(p^3 + 1, p^2 - 1)$. A similar calculation shows that $L(p^3 + 1, p^2) = p L(p^3 + 1, p^2 - 1)$. Thus $L(p^3, p^2) = ... | Canada | Kanada 2012 | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English, French | proof only | null | |
0ksq | Problem:
Let $S=\{(x, y) \in \mathbb{Z}^2 \mid 0 \leq x \leq 11, 0 \leq y \leq 9\}$. Compute the number of sequences $(s_0, s_1, \ldots, s_n)$ of elements in $S$ (for any positive integer $n \geq 2$) that satisfy the following conditions:
- $s_0 = (0,0)$ and $s_1 = (1,0)$,
- $s_0, s_1, \ldots, s_n$ are distinct,
- for ... | [
"Solution:\nLet $a_n$ be the number of such possibilities where there are $n$ $90^{\\circ}$ turns. Note that $a_0 = 10$ and $a_1 = 11 \\cdot 9$.\n\nNow suppose $n = 2k$ with $k \\geq 1$. The path traced out by the $s_i$ is uniquely determined by a choice of $k+1$ nonnegative $x$-coordinates and $k$ positive $y$-coo... | United States | HMMT February 2022 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | final answer only | 646634 | |
0e95 | A rectangle has been divided into four parts by three line segments, as shown in the picture. After that, the four shapes obtained have been rearranged to form a square. What is the perimeter of this square?
 | [
"Let us use the notation suggested in the figure. By Pythagoras' theorem we have $y = \\sqrt{15^2 - 9^2} = \\sqrt{144} = 12$. The two right triangles on the left side are similar, so $\\frac{x}{5} = \\frac{y}{15}$, or $x = \\frac{y}{3} = 4$. Hence, the sides of the rectangle measure $9$ and $16$, and its area is $1... | Slovenia | National Math Olympiad 2013 - First Round | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 48 | |
0kfm | Problem:
A fair coin is flipped eight times in a row. Let $p$ be the probability that there is exactly one pair of consecutive flips that are both heads and exactly one pair of consecutive flips that are both tails. If $p=\frac{a}{b}$, where $a, b$ are relatively prime positive integers, compute $100 a+b$. | [
"Solution:\nSeparate the sequence of coin flips into alternating blocks of heads and tails. Of the blocks of heads, exactly one block has length $2$, and all other blocks have length $1$. The same statement applies to blocks of tails. Thus, if there are $k$ blocks in total, there are $k-2$ blocks of length $1$ and ... | United States | HMMO 2020 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 1028 | |
03r1 | The sequence $\{a_n\}$ satisfies $a_1 = a_2 = 1$ and
$$
a_{n+2} = \frac{1}{a_{n+1}} + a_n, \quad n = 1, 2, \dots
$$
Find $a_{2004}$. | [
"According to the assumption we have\n$$\na_{n+2} a_{n+1} - a_{n+1} a_n = 1.\n$$\nThus, $\\{a_{n+1} a_n\\}$ is an arithmetic progression with first term $1$ and common difference $1$. Hence\n$$\na_{n+1} a_n = n, \\quad n = 1, 2, \\dots\n$$\nSo $a_{n+2} = \\frac{n+1}{a_{n+1}} = \\frac{n+1}{n} = \\frac{n+1}{n} a_n, \... | China | China Western Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | (3*5*...*2003)/(2*4*...*2002) | |
0czv | Let $a$, $b$, $c$, $d$ be positive integers such that $a + b + c + d = 2011$. Prove that $2011$ is not a divisor of $a b - c d$. | [
"We have\n$$\n(a + c)(b + c) = a b + a c + b c + c^2 = (a + b + c + d) c + a b - c d = 2011 c + a b - c d\n$$\nBecause $2011$ is a prime, if $2011 \\mid a b - c d$, then $2011 \\mid a + c$ or $2011 \\mid b + c$. This is not possible since $0 < a + c < 2011$, and $0 < b + c < 2011$, a contradiction."
] | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0a2e | Joah has a number of large pots with marbles in them. At the beginning of the week, all pots contain a different positive number of marbles. On the first day of the week, he adds one marble to each pot. On the second day, he adds a marble to all pots whose number of marbles is divisible by $2$. On the third day, he add... | [] | Netherlands | Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | null | proof and answer | 2 | |
08ke | Problem:
Let $ABC$ be a triangle inscribed in a circle $K$. The tangent from $A$ to the circle meets the line $BC$ at point $P$. Let $M$ be the midpoint of the line segment $AP$ and let $R$ be the intersection point of the circle $K$ with the line $BM$. The line $PR$ meets again the circle $K$ at the point $S$. Prove t... | [
"Solution:\n\nFigure 2\nAssume that point $C$ lies on the line segment $BP$. By the Power of Point theorem we have $MA^{2} = MR \\cdot MB$ and so $MP^{2} = MR \\cdot MB$. The last equality implies that the triangles $MR$ and $MPB$ are similar. Hence $\\angle MPR = \\angle MBP$ and since $\\... | JBMO | OJBM | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0f9u | Problem:
What is the largest possible value of $|\ldots |a_1 - a_2| - a_3| - \ldots - a_{1990}|$, where $a_1, a_2, \ldots, a_{1990}$ is a permutation of $1, 2, 3, \ldots, 1990$? | [
"Solution:\nAnswer $1989$\n\nSince $|a - b| \\leq \\max(a, b)$, a trivial induction shows that the expression does not exceed $\\max(a_1, a_2, \\ldots, a_{1990}) = 1990$. But for integers, $|a - b|$ has the same parity as $a + b$, so a trivial induction shows that the expression has the same parity as $a_1 + a_2 + ... | Soviet Union | 24th ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 1989 | |
00l5 | Let $\alpha \in \mathbb{Q}^+$. Determine all functions $f: \mathbb{Q}^+ \to \mathbb{Q}^+$ such that
$$
f\left(\frac{x}{y} + y\right) = \frac{f(x)}{f(y)} + \alpha x
$$
holds for all $x, y \in \mathbb{Q}^+$.
Here, $\mathbb{Q}^+$ denotes the set of positive rational numbers. | [
"Setting $y = x$ and $y = 1$ yields\n$$\nf(x+1) = 1 + f(x) + \\alpha x \\qquad (1)\n$$\nand\n$$\nf(x+1) = \\frac{f(x)}{f(1)} + f(1) + \\alpha x \\qquad (2)\n$$\nrespectively. Equating (1) and (2) implies\n$$\nf(x)\\left(1 - \\frac{1}{f(1)}\\right) = f(1) - 1.\n$$\nAs $f$ cannot be constant due to (1), we obtain $f(... | Austria | National Competition | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | α = 2 and f(x) = x^2 for all positive rational x; no solutions exist for other α. | |
0j7r | Problem:
Find all real values of $x$ for which
$$
\frac{1}{\sqrt{x}+\sqrt{x-2}}+\frac{1}{\sqrt{x+2}+\sqrt{x}}=\frac{1}{4}
$$ | [
"Solution:\nWe note that\n$$\n\\begin{aligned}\n\\frac{1}{4} &= \\frac{1}{\\sqrt{x}+\\sqrt{x-2}}+\\frac{1}{\\sqrt{x+2}+\\sqrt{x}} \\\\\n&= \\frac{\\sqrt{x}-\\sqrt{x-2}}{(\\sqrt{x}+\\sqrt{x-2})(\\sqrt{x}-\\sqrt{x-2})} + \\frac{\\sqrt{x+2}-\\sqrt{x}}{(\\sqrt{x+2}+\\sqrt{x})(\\sqrt{x+2}-\\sqrt{x})} \\\\\n&= \\frac{\\s... | United States | Harvard-MIT November Tournament | [
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | 257/16 | |
0bha | Let $\triangle ABC$ be isosceles, with $AB = AC$, and $P$, $Q$ be points on the side $AC$ so that $m(\widehat{ABP}) = m(\widehat{PBQ}) = m(\widehat{QBC})$. If $[AD]$ is an altitude, $D \in BC$, $BP \cap AD = \{M\}$, $BQ \cap AD = \{N\}$ and $\triangle ABN$ is isosceles, prove that:
a) $M$ is the orthocenter of triangl... | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Mi... | null | proof and answer | M is the orthocenter of triangle ABC, and MN = (AB/AD)·(AB − AD). | |
0i8y | Problem:
There are 16 members on the Height-Measurement Matching Team. Each member was asked, "How many other people on the team - not counting yourself - are exactly the same height as you?" The answers included six 1's, six 2's, and three 3's. What was the sixteenth answer? (Assume that everyone answered truthfully.... | [
"Solution:\n\nFor anyone to have answered $3$, there must have been exactly $4$ people with the same height, and then each of them would have given the answer $3$. Thus, we need at least four $3$'s, so $3$ is the remaining answer. (More generally, a similar argument shows that the number of members answering $n$ mu... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | final answer only | 3 | |
0emp | Given a scalene triangle $ABC$ with $|AB| + |CA| = 2|BC|$, show that the line joining the incenter and the centroid of the triangle is parallel to $BC$. | [
"Adopting the usual notation, $2a = b + c$, hence $s = a + b + c = \\frac{3}{2}a$. The area of the triangle is $K = sr = \\frac{3}{2}ar = \\frac{1}{2}a h_a$, where $h_a$ is the altitude from $A$ onto $BC$. Therefore $h_a = 3r$, and so the incenter is a third of the way to the vertex. The centroid is also a third of... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0dwt | Problem:
Poišči najmanjše praštevilo $p$, za katerega ima število $p^{3}+2 p^{2}+p$ natanko 42 pozitivnih deliteljev. | [
"Solution:\n\nNajprej zapišemo $p^{3}+2 p^{2}+p = p(p+1)^{2}$. Ker $p$ in $p+1$ nimata skupnih deliteljev (razen 1), je vsak delitelj števila $p(p+1)^{2}$ enak bodisi 1-krat neki delitelj števila $(p+1)^{2}$ bodisi $p$-krat ta delitelj. Ker ima število $p(p+1)^{2}$ natanko 42 deliteljev, ima $(p+1)^{2}$ natanko 21 ... | Slovenia | 49. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 23 | |
0b6d | Let $A \in \mathcal{M}_3(\mathbb{C})$ be such that $\text{tr}(A^2) = \text{tr}(A^*)$. Show that there exist $\alpha, \beta \in \mathbb{C}$ such that the matrix $(A + \alpha I_3)^3 + \beta I_3$ is nilpotent. | [] | Romania | Shortlisted Problems for the Romanian NMO | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | English | proof only | null | |
00uu | Prove that there is a positive integer number $n$ such that the decimal representation of the number:
$$
\sum_{k=1}^{\lfloor \frac{n}{3} \rfloor} \binom{n}{3k} 8^k
$$
ends in 2023 digits 8. | [
"Let $f(n) = \\sum_{k=1}^{\\lfloor \\frac{n}{3} \\rfloor} \\binom{n}{3k} 8^k$ and $\\omega \\neq 1$ be a third root of the unity. Using the fact that for every integer $k \\geq 0$:\n$$\n1 + \\omega^k + \\omega^{2k} = \\begin{cases} 3, & \\text{if } 3 \\mid k \\\\ 0, & \\text{otherwise,} \\end{cases}\n$$\nwe get tha... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Intermediate A... | English | proof only | null | |
0ka6 | Problem:
Will stands at a point $P$ on the edge of a circular room with perfectly reflective walls. He shines two laser pointers into the room, forming angles of $n^{\circ}$ and $(n+1)^{\circ}$ with the tangent at $P$, where $n$ is a positive integer less than $90$. The lasers reflect off of the walls, illuminating th... | [
"Solution:\n\nNote that we want the path drawn out by the lasers to come back to $P$ in as few steps as possible. Observe that if a laser is fired with an angle of $n$ degrees from the tangent, then the number of points it creates on the circle is $\\frac{180}{\\operatorname{gcd}(180, n)}$. (Consider the regular po... | United States | HMMT November 2019 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 28 | |
0dh3 | A rectangle $R$ is partitioned into smaller rectangles whose sides are parallel with the sides of $R$. Let $B$ be the set of all boundary points of all the rectangles in the partition, including the boundary of $R$. Let $S$ be the set of all (closed) segments whose points belong to $B$. Let a maximal segment be a segme... | [
"Let a minor intersection be a point in $S$ where exactly three rectangles meet and let the number of minor intersections be $j$. Let side segments be segments corresponding to a side of a rectangle in the partition and let proper segments be segments into which intersection points cut up maximal segments.\nLet the... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F"
] | English | proof only | null | |
0g5e | $f(x_1, \dots, x_n)$ 為次數小於 $n$ 的整係數多項式, 證明滿足
$$
f(x_1, \dots, x_n) \equiv 0 \pmod{13}
$$
的有序 $n$-元組 $(x_1, \dots, x_n)$ 的個數必為 13 的倍數, 其中 $0 \le x_i \le 12$. | [
"解：以下同餘皆模 13. 我們先證明\n$$\n\\sum_{x=0}^{12} x^k \\equiv 0, \\text{對於 } 0 \\le k < 12.\n$$\n$k=0$ 的情形易證, 故設 $k>0$. 令 $g$ 是模 13 的原根; 故 $g, 2g, \\dots, 12g$ 是 $1, 2, \\dots, 12$ 的某個排列. 故\n$$\n\\sum_{x=0}^{12} x^k \\equiv \\sum_{x=0}^{12} (gx)^k = g^k \\sum_{x=0}^{12} x^k,\n$$\n因 $g^k \\ne 1$, 必有 $\\sum_{x=0}^{12} x^k = ... | Taiwan | 二〇一一數學奧林匹亞競賽第三階段選訓營 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n"
] | null | proof only | null | |
0lch | Let $F$ be the set of all functions $f: \mathbb{Z} \setminus \{0\} \to \mathbb{N}^*$ with the following property: If $a, b \in \mathbb{Z} \setminus \{0\}$ and $a$ is not divisible by $b$ then there exist integers $r, s$ such that $a = br + s$ and $f(s) < f(b)$.
Find all functions $f_0 \in F$ such that $\forall f \in F,... | [
"We will prove that the required function $f_0$ is $g(n) = \\lceil \\log_2 |n| \\rceil + 1$.\n\n(1) We prove that if $g(n) = \\lceil \\log_2 |n| \\rceil + 1$ then $g(n) \\in F$.\nConsider numbers $a, b \\in \\mathbb{Z} \\setminus \\{0\\}$ and $a$ is not divisible by $b$. Assume that $r', s'$ are integers such that ... | Vietnam | Vietnamese Mathematical Competitions | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | f0(n) = ⌈log2 |n|⌉ + 1 | |
0fwr | Problem:
Seien $m, n$ natürliche Zahlen. Betrachte ein quadratisches Punktgitter aus $(2m+1) \times (2n+1)$ Punkten in der Ebene. Eine Menge von Rechtecken heisst gut, falls folgendes gilt:
a. Für jedes der Rechtecke liegen die vier Eckpunkte auf Gitterpunkten und die Seiten parallel zu den Gitterlinien.
b. Keine zw... | [
"Solution:\n\nWir führen Koordinaten ein, sodass das Gitter genau aus den Punkten $(x, y)$ mit ganzzahligen Koordinaten $-m \\leq x \\leq m$ und $-n \\leq y \\leq n$ besteht. Wir bestimmen die grösstmögliche Anzahl Rechtecke in einer guten Menge, die ein festes Einheitsquadrat überdecken können. Aus Symmetriegründe... | Switzerland | IMO Selektion 2008 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | m n (m+1)(n+1) | |
08x9 | Let $n$ be a positive integer. Points $P_1, P_2, \dots, P_{4n}$ are placed in a plane in such a way that no 3 points among them lie on any straight line. Furthermore, for each $i = 1, 2, \dots, 4n$ if we rotate the half-line $P_i P_{i-1}$ starting at $P_i$ around the point $P_i$ by $90^\circ$ clockwise, then the half l... | [
"Let for $k = 1, 2, \\dots, n$ $A_k = P_{4k-3}$, $B_k = P_{4k-2}$, $C_k = P_{4k-1}$, $D_k = P_{4k}$. Also, let $A_{n+1} = A_1$ and $B_{n+1} = B_1$. From now on let us say that the directed line segments $A_i B_i, B_i C_i, C_i D_i, D_i A_{i+1}$ are leftward, downward, rightward, upward segments, respectively. Furthe... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | (2n-1)(n-1) | |
0ggr | 令 $\mathbb{R}$ 代表所有實數所成的集合。試確定所有單射函數 $f: \mathbb{R} \to \mathbb{R}$ 使得
$$
(f(a) - f(b))(f(b) - f(c))(f(c) - f(a)) = f(ab^2 + bc^2 + ca^2) - f(a^2b + b^2c + c^2a)
$$
對所有實數 $a, b, c$ 都成立。 | [
"$f(x) = \\alpha x + \\beta$ or $f(x) = \\alpha x^3 + \\beta$ where $\\alpha \\in \\{-1, 0, 1\\}$ and $\\beta \\in \\mathbb{R}$.\n\nIt is straightforward to check that above functions satisfy the equation. Now let $f(x)$ satisfy the equation, which we denote $E(a, b, c)$. Then clearly $f(x) + C$ also does; therefor... | Taiwan | 2022 數學奧林匹亞競賽第三階段選訓營, 獨立研究 (二) | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | Chinese; English | proof and answer | All injective solutions are f(x) = x + β, f(x) = -x + β, f(x) = x^3 + β, or f(x) = -x^3 + β, where β is any real constant. | |
04km | Determine all complex numbers $z$ for which the ratio of the imaginary part of the fifth power of $z$ to the fifth power of the imaginary part of $z$ is the smallest possible. (Revista de Matematică din Timișoara 1984) | [] | Croatia | Mathematical competitions in Croatia | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | The minimum value of Im(z^5) / (Im z)^5 is −4, attained for all nonzero complex numbers with arg z = π/4 + k·(π/2) for any integer k. | |
0h3i | Vandal Peter cut a rectangular head teacher's portrait along a straight line. After this he cut one of the pieces along a straight line, then he cut one of the new pieces etc. After he had made 100 cuts, the head teacher arrived and forced Peter to pay 2 kopecks for each triangular piece and 1 kopeck for each quadrangu... | [
"Очевидно, що отримані під час розрізання шматки є опуклими многокутнимами. Одним розрізанням загальна кількість вершин збільшується щонайбільше на 4. Тому після 100 розрізань многокутники матимуть не більше за 404 вершини. З іншого боку, після 100 розрізань утворився 101 многокутник. Нехай серед них $n$ трикутникі... | Ukraine | Ukrainian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | English | proof only | null | |
04g9 | The points $P$ and $Q$ lie on the side $\overline{AB}$ of the rectangle $ABCD$ such that $|AP| = |PQ| = |QB|$. The line $DQ$ meets the lines $AC$ and $CP$ at points $K$ and $L$ respectively, and the line $DB$ meets the lines $AC$ and $CP$ at points $N$ and $M$ respectively.
Determine the ratio of the areas of quadrilat... | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | 1/40 | |
0gp2 | Prove that
$$
\frac{(a+1)(b+2)}{(b+1)(b+5)} + \frac{(b+1)(c+2)}{(c+1)(c+5)} + \frac{(c+1)(a+2)}{(a+1)(a+5)} \ge \frac{3}{2}
$$
for all positive real numbers $a$, $b$, $c$ satisfying the condition $a^2 + b^2 + c^2 \ge 3$. | [
"Since $4(x+2)^2 - 3(x+1)(x+5) = (x-1)^2 \\ge 0$, we have $\\frac{x+2}{(x+1)(x+5)} \\ge \\frac{3}{4(x+2)}$. Therefore it suffices to show that\n$$\n\\frac{a+1}{b+2} + \\frac{b+1}{c+2} + \\frac{c+1}{a+2} \\ge 2\n$$\nfor all positive real numbers satisfying $a^2 + b^2 + c^2 \\ge 3$.\n\nThe Cauchy-Schwarz Inequality g... | Turkey | Team Selection Test for IMO 2011 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0ca0 | Problem:
Fie $a, b, c$ numere strict pozitive, astfel încât $a+b+c=1$. Arătaţi că
$$
\frac{1}{a b c}+\frac{4}{a^{2}+b^{2}+c^{2}} \geq \frac{13}{a b+b c+c a}
$$ | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
05wx | Problem:
Soit $ABC$ un triangle rectangle en $B$ avec $BC < BA$. Soit $D$ le point du segment $[AB]$ tel que $BD = BC$. La perpendiculaire à $(AC)$ passant par $D$ intersecte $(AC)$ en $E$. Soit $B'$ le symétrique de $B$ par rapport à $(CD)$. Montrer que $(EC)$ est la bissectrice de l'angle $\widehat{BEB'}$. | [
"Solution:\n\n\n\nOn remarque que le cercle de diamètre $[CD]$ apparaît assez naturellement. En effet, on a des angles droits $\\widehat{DB'C} = \\widehat{CBD} = \\widehat{DEC} = 90^{\\circ}$, les points $B$, $B'$, et $E$ sont sur le cercle de diamètre $[DC]$, autrement dit $C$, $B$, $D$, $... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0bcc | Problem:
Fie $ABC$ un triunghi având punctul $O$ ca centru al cercului său circumscris. Punctele $D$, $E$ şi $F$ se află respectiv pe laturile $BC$, $CA$ şi $AB$, astfel încât dreapta $DE$ este perpendiculară pe $CO$ iar dreapta $DF$ este perpendiculară pe $BO$. (De exemplu, punctul $D$ se află pe dreapta $BC$, fiind ... | [] | Romania | Olimpiada europeana de matematica a fetelor | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chas... | null | proof only | null | |
06fu | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that
(i) the set $\left\{ \frac{f(x)}{x} \mid x \ne 0 \right\}$ is finite,
(ii) $f(3x - 1 - f(x)) = 3(f(x) - 1 - 3x)$ for all $x \in \mathbb{R}$. | [
"The only solution is $f(x) = 3x$.\nWhen $f(x) = 3x$, we have\n$$\nf(3x - 1 - f(x)) = f(-1) = -3 = 3(-1) = 3(f(x) - 1 - 3x).\n$$\nAlso, $\\frac{f(x)}{x} = 3$ is a constant. So $f(x) = 3x$ is a solution.\n\nNow, we show that this is the only solution. Let $g(x) = 3x - 1 - f(x)$ for any $x \\in \\mathbb{R}$. If $g(x)... | Hong Kong | IMO HK TST | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | f(x) = 3x | |
0aef | Куќите во една улица се нумерирани од $1$ до $100$. Колку пати во броевите на куќите се јавува цифрата $7$? | [
"Броевите на куќите што ја содржат цифрата $7$ се: $7$, $17$, $27$, $37$, $47$, $57$, $67$, $70$, $71$, $72$, $73$, $74$, $75$, $76$, $77$, $78$, $79$, $87$, $97$. Во тие броеви таа се појавува вкупно $20$ пати (двапати ја има во бројот $77$)."
] | North Macedonia | Регионален натпревар по математика за основно образование | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | Macedonian, English | final answer only | 20 | |
09xt | We consider an integer $n > 1$ with the following property: for every positive divisor $d$ of $n$ we have that $d+1$ is a divisor of $n+1$. Prove that $n$ is a prime number. | [
"Suppose by contradiction that $n$ is not prime. Now consider the greatest divisor $d < n$ of $n$. Then we can write $n$ as $de$. Since $n$ is not prime, we have $d > 1$ and hence also $e < n$. Now $e$ must satisfy $e > 1$ and $e \\le d$ (because $d$ is the greatest divisor satisfying $d < n$). Now $d+1$ must be a ... | Netherlands | Dutch Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization"
] | English | proof only | null | |
089r | Problem:
Determinare tutte le terne di interi strettamente positivi $(a, b, c)$ tali che
- $a \leq b \leq c$;
- $\operatorname{MCD}(a, b, c)=1$;
- $a$ è divisore di $b+c$, $b$ è divisore di $c+a$ e $c$ è divisore di $a+b$. | [
"Solution:\n\nLe uniche terne di soluzioni sono $(1,1,1)$, $(1,1,2)$ e $(1,2,3)$.\n\nDimostriamo innanzitutto che $a, b, c$ sono a due a due coprimi (mostriamo solo che $\\operatorname{MCD}(a, b)=1$; per le altre coppie la dimostrazione è la stessa).\nSe $d$ è il massimo comun divisore tra $a$ e $b$, allora $d$ div... | Italy | Progetto Olimpiadi della Matematica | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (1,1,1), (1,1,2), (1,2,3) | |
0b1p | Problem:
Find the 2020th term of the following sequence:
$$
1, 1, 3, 1, 3, 5, 1, 3, 5, 7, 1, 3, 5, 7, 9, 1, 3, 5, 7, 9, 11, \ldots
$$ | [
"Solution:\nWe have that for each $n \\in \\mathbb{N}$, the $(1+2+\\cdots+n)$th term is $2n-1$. The first $n+1$ odd positive integers are then listed. Observe that the largest triangular number less than or equal to $2020$ is $\\frac{63 \\times 64}{2} = 2016$. Therefore, the 2020th term is $7$."
] | Philippines | 22nd Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 7 | |
0kr6 | Problem:
For each $i \in \{1, \ldots, 10\}$, $a_{i}$ is chosen independently and uniformly at random from $[0, i^{2}]$. Let $P$ be the probability that $a_{1} < a_{2} < \cdots < a_{10}$. Estimate $P$.
An estimate of $E$ will earn $\left\lfloor 20 \min \left(\frac{E}{P}, \frac{P}{E}\right)\right\rfloor$ points. | [
"Solution:\n\nThe probability that $a_{2} > a_{1}$ is $7/8$. The probability that $a_{3} > a_{2}$ is $7/9$. The probability that $a_{4} > a_{3}$ is $23/32$. The probability that $a_{5} > a_{4}$ is $17/25$. The probability that $a_{6} > a_{5}$ is $47/72$. The probability that $a_{7} > a_{6}$ is $31/49$. The probabil... | United States | HMMT November 2022 | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | null | |
0966 | Problem:
Rezolvaţi în $\mathbb{R}$ ecuaţia
$$
\sqrt{2 x^{5}+x^{4}+4 x^{3}+2 x^{2}+2 x+1}+\sqrt{17-2 x+34 x^{2}-4 x^{3}+17 x^{4}-2 x^{5}}=7 x^{2}-8 x+22
$$ | [
"Solution:\nVom aplica inegalitatea $\\frac{a+b}{2} \\leq \\sqrt{\\frac{a^{2}+b^{2}}{2}}$, care este adevărată pentru orice numere reale $a$ şi $b$.\nPe $DVA$ are loc inegalitatea\n$$\n\\frac{\\sqrt{2 x^{5}+x^{4}+4 x^{3}+2 x^{2}+2 x+1}+\\sqrt{17-2 x+34 x^{2}-4 x^{3}+17 x^{4}-2 x^{5}}}{2} \\leq \\sqrt{\\frac{18 x^{4... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 4 | |
0519 | Two circles $c$ and $c'$ with centers $O$ and $O'$ lie completely outside each other. Points $A$, $B$, and $C$ lie on the circle $c$ and points $A'$, $B'$, and $C'$ lie on the circle $c'$ so that segment $AB \parallel A'B'$, $BC \parallel B'C'$, and $\angle ABC = \angle A'B'C'$. The lines $AA'$, $BB'$, and $CC'$ are al... | [
"The triangles $ABP$ and $A'B'P$ are similar, because their corresponding sides are parallel (Fig. 1). Hence $\\frac{|AB|}{|A'B'|} = \\frac{|BP|}{|B'P|}$. Likewise the triangles $BCP$ and $B'C'P$ are similar, hence $\\frac{|BC|}{|B'C'|} = \\frac{|BP|}{|B'P|}$. Thus $\\frac{|AB|}{|A'B'|} = \\frac{|BC|}{|B'C'|}$, and... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
093n | Problem:
Let $a$, $b$ and $c$ be positive integers satisfying $a < b < c < a + b$. Prove that $c(a-1) + b$ does not divide $c(b-1) + a$. | [
"Solution:\n\nPut $A = c(a-1) + b$, $B = c(b-1) + a$ and suppose that $A$ is a divisor of $B$. Then $A$ is also a divisor of the number $C = bA - aB$. Since\n$$\nC = b(c(a-1) + b) - a(c(b-1) + a) = (b-a)(a+b-c) > 0\n$$\nit follows from $c > b-a > 0$ and $a-1 \\geq a+b-c > 0$ that\n$$\nA = c(a-1) + b > c(a-1) > (b-a... | Middle European Mathematical Olympiad (MEMO) | MEMO Team Competition | [
"Number Theory > Divisibility / Factorization",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0fhf | Problem:
Dado un número natural $n$, se designa por $s(n)$ la suma de las cifras del número $n$, expresado en el sistema de numeración binario, es decir, el número de cifras 1 que tiene. Determinar, para todo número natural $k$
$$
\sigma(k)=s(1)+s(2)+\cdots+s\left(2^{k}\right)
$$ | [
"Solution:\n\nEscribiendo los números $1, 2, \\ldots, 2^{k}$ en base 2 tenemos:\n$$\n\\begin{aligned}\n& 0=\\quad \\begin{array}{llllll}\n0 & 0 & \\ldots & 0 & 0 ; & k \\text{ ceros }\n\\end{array} \\\\\n& 1=\\quad \\begin{array}{llllll}\n0 & 0 & \\ldots & 0 & 1 ; & k-1 \\text{ ceros }\n\\end{array} \\\\\n& 2=\\qua... | Spain | OME 27 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | k·2^{k-1} + 1 | |
050q | a) Prove that for every real number $x$ the arithmetic mean of $\sqrt{1 + \sin x}$ and $\sqrt{1 - \sin x}$ is equal to one of the following: $\sin \frac{x}{2}$, $\cos \frac{x}{2}$, $-\sin \frac{x}{2}$, $-\cos \frac{x}{2}$.
b) Can one leave out one of the four numbers listed in part a) in such a way that the claim stil... | [
"a) Denote the arithmetic mean given in the problem by $A(x)$. As\n$$\n1 + \\sin x = \\sin^2 \\frac{x}{2} + \\cos^2 \\frac{x}{2} + 2 \\sin \\frac{x}{2} \\cos \\frac{x}{2} = \\left( \\sin \\frac{x}{2} + \\cos \\frac{x}{2} \\right)^2,\n$$\n\n$$\n1 - \\sin x = \\sin^2 \\frac{x}{2} + \\cos^2 \\frac{x}{2} - 2 \\sin \\fr... | Estonia | Estonian Math Competitions | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | No; none can be left out. | |
04lk | Determine the number of positive integers $c \le 1000000$, that can be expressed as $c = a^2 + 3b^2 - 4ab$ for some non-zero integers $a$ and $b$. | [] | Croatia | Mathematical competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 749998 | |
0g1z | Problem:
Seien $a$, $b$ und $c$ natürliche Zahlen. Finde den kleinsten Wert, den folgender Ausdruck an oxnehmen kann:
$$
\frac{a}{\operatorname{ggT}(a+b, a-c)}+\frac{b}{\operatorname{ggT}(b+c, b-a)}+\frac{c}{\operatorname{ggT}(c+a, c-b)}
$$ | [
"Solution:\n\nZuerst bemerken wir, dass\n$$\n\\operatorname{ggT}(a+b, a-c)=\\operatorname{ggT}(a+b-(a-c), a-c)=\\operatorname{ggT}(b+c, a-c) \\leq b+c\n$$\ngilt. Daraus folgt dann\n$$\n\\frac{a}{\\operatorname{ggT}(a+b, a-c)}+\\frac{b}{\\operatorname{ggT}(b+c, b-a)}+\\frac{c}{\\operatorname{ggT}(c+a, c-b)} \\geq \\... | Switzerland | SMO - Finalrunde | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 3/2 | |
044a | For real numbers $x_1, x_2, \dots, x_{60} \in [-1, 1]$, find the maximum of
$$
\sum_{i=1}^{60} x_i^2 (x_{i+1} - x_{i-1}),
$$
where $x_0 = x_{60}, x_{61} = x_1$. | [
"The maximum is $40$. First, notice that\n$$\n\\begin{align*}\n\\sum_{i=1}^{60} x_i^2 (x_{i+1} - x_{i-1}) &= \\sum_{i=1}^{60} x_i^2 x_{i+1} - \\sum_{i=1}^{60} x_i^2 x_{i-1} \\\\\n&= \\sum_{i=1}^{60} x_i^2 x_{i+1} - \\sum_{i=1}^{60} x_{i+1}^2 x_i \\\\\n&= \\sum_{i=1}^{60} x_i x_{i+1} (x_i - x_{i+1}).\n\\end{align*}\... | China | China National Team Selection Test | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 40 | |
0i35 | Let $a > b > c > d$ be positive integers and suppose
$$
ac + bd = (b + d + a - c)(b + d - a + c).
$$
Prove that $ab + cd$ is not prime. | [
"**First Solution.** For the sake of contradiction, assume that $ab + cd$ is prime. Note that\n$$\nab + cd = (a + d)c + (b - c)a = m \\cdot \\gcd(a + d, b - c)\n$$\nfor some positive integer $m$. Writing $g = \\gcd(a+d, b-c)$, we have\n$$\nm = \\frac{a+d}{g} \\cdot c + \\frac{b-c}{g} \\cdot a \\geq c+a > 1.\n$$\nTh... | United States | USA IMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Algebraic Number Theory > Unique factorization",
"Number Theory > Algebraic Number Theory > Quadratic fields",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > ... | English | proof only | null | |
0hx9 | Problem:
We wish to place exactly $100$ dominoes (of size $2 \times 1$ or $1 \times 2$) without overlapping on a $20 \times 20$ chessboard so that every $2 \times 2$ square contains at least two uncovered unit squares which lie in the same row or column. In how many ways can this be done? | [
"Solution:\n\nThe answer is $\\left(\\begin{array}{l}20 \\\\ 10\\end{array}\\right)^{2}$.\n\nGeneralizing the problem slightly, the answer is $\\left(\\begin{array}{c}m+n \\\\ n\\end{array}\\right)^{2}$ for a $2m \\times 2n$ rectangle. We provide a \"proof without words\" with the following bijection:\n\n![](attach... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | (20 choose 10)^2 | |
0ig9 | Problem:
In how many ways can 6 purple balls and 6 green balls be placed into a $4 \times 4$ grid of boxes such that every row and column contains two balls of one color and one ball of the other color? Only one ball may be placed in each box, and rotations and reflections of a single configuration are considered diffe... | [
"Solution:\nIn each row or column, exactly one box is left empty. There are $4! = 24$ ways to choose the empty spots. Once that has been done, there are 6 ways to choose which two rows have 2 purple balls each. Now, assume without loss of generality that boxes $(1,1)$, $(2,2)$, $(3,3)$, and $(4,4)$ are the empty on... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 5184 | |
0itc | Problem:
Let $C_{1}$ and $C_{2}$ be externally tangent circles with radius $2$ and $3$, respectively. Let $C_{3}$ be a circle internally tangent to both $C_{1}$ and $C_{2}$ at points $A$ and $B$, respectively. The tangents to $C_{3}$ at $A$ and $B$ meet at $T$, and $TA = 4$. Determine the radius of $C_{3}$. | [
"Solution:\n\nAnswer: $8$\n\nLet $D$ be the point of tangency between $C_{1}$ and $C_{2}$. We see that $T$ is the radical center of the three circles, and so it must lie on the radical axis of $C_{1}$ and $C_{2}$, which happens to be their common tangent $TD$. So $TD = 4$.\n\n\n\nWe have\n$... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 8 | |
0kfq | Problem:
While waiting for their next class on Killian Court, Alesha and Belinda both write the same sequence $S$ on a piece of paper, where $S$ is a 2020-term strictly increasing geometric sequence with an integer common ratio $r$. Every second, Alesha erases the two smallest terms on her paper and replaces them with... | [
"Solution:\n\nBecause we only care about when the ratio of $A$ to $B$ is an integer, the value of the first term in $S$ does not matter. Let the initial term in $S$ be $1$. Then, we can write $S$ as $1, r, r^{2}, \\ldots, r^{2019}$. Because all terms are in terms of $r$, we can write $A = r^{a}$ and $B = r^{b}$. We... | United States | HMMO 2020 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | final answer only | 2018 | |
0kn1 | Problem:
Compute the product of all positive integers $b \geq 2$ for which the base $b$ number $111111_{b}$ has exactly $b$ distinct prime divisors. | [
"Solution:\nNotice that this value, in base $b$, is\n$$\n\\frac{b^{6}-1}{b-1} = (b+1)\\left(b^{2}-b+1\\right)\\left(b^{2}+b+1\\right)\n$$\nThis means that, if $b$ satisfies the problem condition, $(b+1)\\left(b^{2}-b+1\\right)\\left(b^{2}+b+1\\right)$ has more than $p_{1} \\ldots p_{b}$, where $p_{i}$ is the $i$th ... | United States | HMMT Spring 2021 Guts Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 24 | |
00zv | Problem:
A rectangle can be divided into $n$ equal squares. The same rectangle can also be divided into $n+76$ equal squares. Find all possible values of $n$. | [
"Solution:\nLet $ab = n$ and $cd = n+76$, where $a, b$ and $c, d$ are the numbers of squares in each direction for the partitioning of the rectangle into $n$ and $n+76$ squares, respectively. Then $\\frac{a}{c} = \\frac{b}{d}$, or $ad = bc$. Denote $u = \\gcd(a, c)$ and $v = \\gcd(b, d)$, then there exist positive ... | Baltic Way | Baltic Way 1997 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 324 | |
0jrl | Problem:
A graph consists of 6 vertices. For each pair of vertices, a coin is flipped, and an edge connecting the two vertices is drawn if and only if the coin shows heads. Such a graph is good if, starting from any vertex $V$ connected to at least one other vertex, it is possible to draw a path starting and ending at ... | [
"Solution:\nFirst, we find the probability that all vertices have even degree. Arbitrarily number the vertices $1, 2, 3, 4, 5, 6$. Flip the coin for all the edges out of vertex $1$; this vertex ends up with even degree with probability $\\frac{1}{2}$. Next we flip for all the remaining edges out of vertex $2$; rega... | United States | HMMT November 2015 | [
"Discrete Mathematics > Graph Theory"
] | null | final answer only | 507/16384 |
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