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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0kx0 | Let $ABC$ be a scalene triangle and let $P$ and $Q$ be two distinct points in its interior. Suppose that the angle bisectors of $\angle PAQ$, $\angle PBQ$, and $\angle PCQ$ are the altitudes of triangle $ABC$. Prove that the midpoint of $\overline{PQ}$ lies on the Euler line of $ABC$. | [
"Let $H$ be the orthocenter of $ABC$, and construct $P'$ using the following claim.\n\n**Claim** — There is a point $P'$ for which\n$$\n\\angle APH + \\angle AP'H = \\angle BPH + \\angle BP'H = \\angle CPH + \\angle CP'H = 0.\n$$\n\n*Proof*. After inversion at $H$, this is equivalent to the fact that $P$'s image ha... | United States | USA TSTST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geom... | null | proof only | null | |
0h8n | Cut right triangle with one angle $30^\circ$ into three pairwise distinct isosceles triangles so that each has a non acute angle. | [
"We can cut $\\triangle ABC$ with $\\angle BAC = 30^\\circ$ into triangles as shown in Fig. 25.\n\n\n\nHere $BC = CD$ and $\\triangle DBC$ is isosceles with right angle. Then $BN = MD$ and $\\triangle DBM$ is isosceles with angle $\\angle DMB = 150^\\circ$. Thus $\\angle DMA = 30^\\circ$, $... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
000s | Un supermercado vende Rocacola y Rocajugo en botellas con tapita. Además, cambia $15$ tapitas de Rocacola por una botella de Rocajugo llena, y cambia $20$ tapitas de Rocajugo por una botella de Rocacola llena.
Lucas tiene $511$ tapitas que va cambiando en el supermercado por botellas llenas. Después de haber el conten... | [] | Argentina | XII Olimpíada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | español | proof and answer | 163 and 462 | |
07oo | Show that from eight consecutive integers one can always pick four whose squares have the same sum as the squares of the remaining four. | [
"Let the given eight consecutive integers be $n, n+1, \\dots, n+7$. Experimenting with the case $n=0$ leads to the identity\n$$\nn^2 + (n+3)^2 + (n+5)^2 + (n+6)^2 = (n+1)^2 + (n+2)^2 + (n+4)^2 + (n+7)^2.\n$$\nAlternatively, one may observe that\n$$\nn^2 + (n+3)^2 - (n+1)^2 - (n+2)^2 = 4.\n$$\nSubtracting from it th... | Ireland | Irska 2014 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
05z2 | Problem:
Soit $n \geqslant 2$. Montrer qu'il existe des entiers $a, b$ tels que, pour tout entier $m$, le nombre $m^{3}+a m+b$ ne soit pas un multiple de $n$. | [
"Solution:\n\n$m^{3}+a m+b$ est un multiple de $n$ si et seulement si $m^{3}+a m \\equiv -b \\pmod{n}$. On remarque alors qu'il suffit de trouver $a$ tel que $m^{3}+a m$ ne prend pas toutes les valeurs possibles modulo $n$ (et de prendre pour $-b$ une des valeurs qui $n$ n'est pas prise). Comme $m^{3}+a m$ modulo $... | France | Préparation Olympique Française de Mathématiques - Envoi 3: Arithmétique | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0drf | There are 30 children, $a_1, a_2, \dots, a_{30}$, seated clockwise in a circle on the floor. The teacher walks behind the children in the clockwise direction with a box of 1000 candies. She drops a candy behind the first child $a_1$. She then skips one child and drops a candy behind the third child, $a_3$. Now she skip... | [
"When the $k^{\\text{th}}$ candy is dropped, the teacher has skipped $1+2+\\dots+(k-1) = \\dfrac{k(k-1)}{2}$ children. Then it is dropped behind $a_i$, where $i = k + \\dfrac{k(k-1)}{2} = \\dfrac{k(k+1)}{2}$ (mod 30). Thus $2i = k(k+1)$ (mod 60). From here it is clear that the $i^{\\text{th}}$ and $j^{\\text{th}}$ ... | Singapore | Singapur 2015 | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 18 | |
0553 | Let $n$ and $m$ be positive integers. What is the biggest number of points that can be marked in the vertices of the squares of the $n \times m$ grid in such a way that no three of the marked points lie in the vertices of any right-angled triangle? | [
"All vertices of squares lie on $n+1$ horizontal and $m+1$ vertical lines. Suppose that at least $n+m+1$ points are marked in the grid. Because $m > 0$, the number of marked points is greater than $n+1$. Hence by the pigeonhole principle, at least one horizontal line contains more than one marked point. Hence at mo... | Estonia | Open Contests | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n + m | |
0gy1 | Find all positive integer numbers $n$ such that the number $3\pi$ is a period of the function $f(x) = \cos nx \cdot \sin \frac{2009x}{n^2}$. | [
"$3\\pi$ is a period of the function $f(x)$ if and only if $f(x+3\\pi) = f(x)$ for arbitrary real number $x$. Putting $x=0$ we obtain that $f(0)=0$, since $f(3\\pi) = \\cos 3\\pi n \\cdot \\sin \\frac{3 \\cdot 2009\\pi}{n^2}$ we have $0 = \\cos 3\\pi n \\cdot \\sin \\frac{3 \\cdot 2009\\pi}{n^2} \\Leftrightarrow 0 ... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Precalculus > Trigonometric functions",
"Precalculus > Functions"
] | English | proof and answer | 1, 7 | |
0c4x | Let $A$, $B \in \mathcal{M}_n(\mathbb{C})$ and $c \in \mathbb{C}^*$ so that $AB - BA = c \cdot (A - B)$. Prove that the matrices $A$ and $B$ have the same eigenvalues. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Linear transformations"
] | English | proof only | null | |
060t | Problem:
Soit $k$ un entier premier avec $n$ vérifiant $1 \leqslant k < n$. Augustin colorie les entiers de $\{1,2, \ldots, n-1\}$ avec autant de couleurs qu'il le souhaite. Cependant, si $j$ est un entier vérifiant $1 \leqslant j \leqslant n-1$, les entiers $j$ et $n-j$ sont de la même couleur. De plus, si $i$ est un... | [
"Solution:\n\nPour $k=1$, on voit que $n$ a la même couleur que $n-1$, puis que $n-2$ a la même couleur que $n-1$, etc. et tous les entiers ont la même couleur.\n\nPour $k=2$, $n$ est impair et on voit que $n, n-2, \\ldots, 1$ ont la même couleur, puis que $n-1$ a la même couleur que $1$, et que $n-1, n-3, \\ldots,... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof only | null | |
04yn | A unit L-shape consists of three unit squares as shown in the picture. Prove that for any positive integer $k$ it is possible to cut a similar L-shape with $k$ times larger side lengths into unit L-shapes.
 | [
"Let the L-shape be placed so that the two longer sides meet at the top left corner. Starting from the top left we place on it $k$ unit L-shapes diagonally with the same orientation as the large L-shape (Fig. 12). The rest consists of two equal staircase-like parts; it is enough to show that one of them, e.g. the l... | Estonia | Estonija 2010 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
014t | Problem:
For a positive integer $n$, let $S(n)$ denote the sum of its digits. Find the largest possible value of the expression $\frac{S(n)}{S(16 n)}$. | [
"Solution:\n\nIt is obvious that $S(a b) \\leq S(a) S(b)$ for all positive integers $a$ and $b$. From here we get\n$$\nS(n) = S(n \\cdot 10000) = S(16 n \\cdot 625) \\leq S(16 n) \\cdot 13;\n$$\nso we get $\\frac{S(n)}{S(16 n)} \\leq 13$.\n\nFor $n = 625$ we have an equality. So the largest value is $13$."
] | Baltic Way | Baltic Way 2008 | [
"Number Theory > Other"
] | null | proof and answer | 13 | |
0hs2 | Problem:
Prove that every positive real number $y$ satisfies
$$
2 y \geq 3-\frac{1}{y^{2}}
$$
When does equality occur? | [
"Solution:\nSince $y$ is positive, we can multiply both sides by $y$ and rearrange to get the equivalent inequality\n$$\n2 y^{3}-3 y^{2}+1 \\geq 0\n$$\nThis factors as\n$$\n(2 y+1)(y-1)^{2} \\geq 0.\n$$\nThe first factor is positive since $y>0$, and the second is nonnegative since it is a real number squared. Thus,... | United States | Berkeley Math Circle: Monthly Contest 8 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | y=1 | |
02x1 | Problem:
Ana multiplica dois números inteiros positivos cuja diferença é $202$, mas comete um erro e obtém um número $1000$ unidades menor que o correto. Ao dividir o resultado de Ana pelo menor dos números que deveria multiplicar, o quociente é $288$ e o resto é $67$. Quais os dois números que Ana multiplicou? | [
"Solution:\n\nSejam $x$ e $x+202$ os inteiros que Ana multiplica. A partir do resultado da divisão de Ana, podemos escrever:\n$$\nx(x+202)-1000=288x+67\n$$\nDaí,\n$$\nx^2-86x-1067=0\n$$\nResolvendo a equação anterior, encontramos como raízes $x=-11$ e $x=97$. Como $x$ é um inteiro positivo, devemos ter $x=97$. O ou... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 97 and 299 | |
0dil | Let $ABC$ be an acute triangle with $M$ is the midpoint of $BC$ and $\angle BAM < 90^\circ$, $\angle AMB = 60^\circ$. On ray $MA$, take the point $N$ such that $BN = AC$. Prove that $BC = 2AN$ and the orthocenters and circumcenters of two triangles $BMN$, $AMC$ form the four vertices of an isosceles trapezoid. | [
"On $MN$, take the point $D$ such that $ND = AM$. Then, $AN = DM$. According to the sine theorem then\n$$\n\\frac{BM}{\\sin \\angle BNM} = \\frac{BN}{\\sin \\angle BMN} = \\frac{AC}{\\sin \\angle AMC} = \\frac{MC}{\\sin \\angle MAC},\n$$\n\nso $\\angle BNM = \\angle MAC < 90^\\circ$, which implies that $\\triangle ... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > D... | English | proof only | null | |
09ti | Problem:
We hebben 1000 ballen in 40 verschillende kleuren, waarbij er van elke kleur precies 25 ballen zijn. Bepaal de kleinste waarde van $n$ met de volgende eigenschap: als je de 1000 ballen willekeurig in een cirkel legt, zijn er altijd $n$ ballen naast elkaar te vinden waarbij minstens 20 verschillende kleuren vo... | [
"Solution:\n\nOplossing I. Bekijk de cirkel van ballen waarbij de 25 ballen van één kleur steeds allemaal naast elkaar liggen. Om nu 20 verschillende kleuren te hebben, moet je minstens 18 van deze groepen nemen plus nog een bal aan de ene kant daarvan en een bal aan de andere kant. Totaal heb je dus minimaal $18 \... | Netherlands | Selectietoets | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 452 | |
07xw | $$
\begin{aligned}
a + b + c &= p \\
a^2 + b^2 + c^2 &= q
\end{aligned}
\qquad
\begin{aligned}
bc - x - 2ya &= 0 \\
ca - x - 2yb &= 0 \\
ab - x - 2yc &= 0,
\end{aligned}
$$
where $p^2 < 3q$. Determine the possible values of $abc$. | [
"Label the given equations as follows:\n$$\na + b + c = p \\qquad (23)\n$$\n$$\na^2 + b^2 + c^2 = q \\qquad (24)\n$$\n$$\nbc - x - 2ya = 0 \\qquad (25)\n$$\n$$\nca - x - 2yb = 0 \\qquad (26)\n$$\n$$\nab - x - 2yc = 0. \\qquad (27)\n$$\n\nAdding $(25) \\cdot (c - b)$, $(26) \\cdot (a - c)$ and $(27) \\cdot (b - a)$,... | Ireland | IRL_ABooklet_2025 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (5 p^3 - 9 p q ± (p^2 - 3 q) sqrt(2(3 q - p^2)))/54 | |
0kjw | For $n$ a positive integer, let $R(n)$ be the sum of remainders when $n$ is divided by $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, and $10$. For example, $R(15) = 1 + 0 + 3 + 0 + 3 + 1 + 7 + 6 + 5 = 26$. How many two-digit positive integers $n$ satisfy $R(n) = R(n + 1)$?
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4 | [] | United States | AMC 12 B | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | MCQ | C | |
0jog | Problem:
Let $n$ be a three-digit integer with nonzero digits, not all of which are the same. Define $f(n)$ to be the greatest common divisor of the six integers formed by any permutation of $n$'s digits. For example, $f(123) = 3$, because $\operatorname{gcd}(123, 132, 213, 231, 312, 321) = 3$. Let the maximum possible... | [
"Solution:\nLet $n = \\overline{abc}$, and assume without loss of generality that $a \\geq b \\geq c$. We have $k \\mid 100a + 10b + c$ and $k \\mid 100a + 10c + b$, so $k \\mid 9(b - c)$. Analogously, $k \\mid 9(a - c)$ and $k \\mid 9(a - b)$. Note that if $9 \\mid n$, then $9$ also divides any permutation of $n$'... | United States | HMMT November 2015 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 5994 | |
0buy | Problem:
a) Câte progresii aritmetice de numere naturale există cu primul termen $1$ şi care conţin numărul $45001$?
b) Arătaţi că nu există progresii aritmetice neconstante de numere naturale cu toţi termenii pătrate perfecte. | [
"Solution:\na) Fie $(a_n)$ o progresie aritmetică cu primul termen $1$ şi care conţine numărul $45001$. Atunci $45001 = 1 + (n-1) \\cdot r \\Rightarrow (n-1) \\cdot r = 45000$, deci $r$ este divizor al lui $45000$.\n\n$45000 = 2^{3} \\cdot 3^{2} \\cdot 5^{4}$. Deci $r = 2^{i} \\cdot 3^{j} \\cdot 5^{k}$, $i = \\over... | Romania | Olimpiada de Matematică | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) 60; b) No non-constant progression exists. | |
07k2 | In the scalene triangle *ABC*, the in-circle touches side *BC* at the point *D*. The angle bisectors of $\angle ABC$ and $\angle ACB$ meet the circumcircle of the triangle at the points $T_B$ and $T_C$, respectively. Let $AX_B$ and $AX_C$ be diameters in the circumcircles of triangles $ACD$ and $ABD$, respectively. Pro... | [
"Let $N$ and $Y$ be the midpoint of the arc $\\widearc{BAC}$ and the intersection of the lines $ID, T_bN$, respectively. Now, since $\\angle YT_bA = \\frac{\\widearc{AN}}{2} = \\frac{B-C}{2} = \\angle AIY$, therefore $AYT_bI$ is cyclic, so $\\angle AYD = \\angle AT_bI = \\angle C$. Thus, $AYCD$ is cyclic and $AYX_b... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0i3q | Problem:
A point on a circle inscribed in a square is $1$ and $2$ units from the two closest sides of the square. Find the area of the square. | [
"Solution:\n\nCall the point in question $A$, the center of the circle $O$, and its radius $r$. Consider a right triangle $BOA$ with hypotenuse $OA$: $OA$ has length $r$, and $BO$ and $BA$ have lengths $r-1$ and $r-2$. By the Pythagorean theorem,\n$$(r-1)^2 + (r-2)^2 = r^2$$\nwhich gives\n$$r^2 - 6r + 5 = 0$$\nso $... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 100 | |
087w | Problem:
Gabriele, l'amante dei cubi, ha comprato un magnifico pezzo da collezione: un cubo interamente composto di cioccolato, avente gli spigoli lunghi $10~\mathrm{cm}$. Purtroppo, avendo perso una scommessa con due suoi amici, dovrà cedere due terzi del volume del blocco di cioccolato. Gabriele ha deciso di prender... | [
"Solution:\n\nLa risposta è (D). Il blocco di cioccolato ceduto ha la stessa superficie del cubo originale, pari a $6 \\cdot 10^{2}~\\mathrm{cm}^{2}$. Infatti, quando si taglia via il cubo più piccolo, 3 delle sue facce sono esattamente uguali ai 3 quadrati che si formano sul blocco di cioccolato ceduto."
] | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > 3D Shapes"
] | null | MCQ | D | |
0eyj | Problem:
The streets in a city are on a rectangular grid with $m$ east-west streets and $n$ north-south streets. It is known that a car will leave some (unknown) junction and move along the streets at an unknown and possibly variable speed, eventually returning to its starting point without ever moving along the same ... | [] | Soviet Union | 2nd ASU | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F"
] | null | proof and answer | (m - 1)(n - 1) | |
0bpv | Problem:
a. Demonstraţi că $\frac{2a}{ab+1} \leq \sqrt{\frac{a}{b}}$ pentru orice $a, b \in (0, \infty)$.
b. Dacă $a, b, c > 0$ astfel încât $abc = 1$, demonstraţi că
$$
\frac{2a}{ab+1} + \frac{2b}{bc+1} + \frac{2c}{ac+1} \leq a \sqrt{c} + b \sqrt{a} + c \sqrt{b}
$$ | [] | Romania | Olimpiada Națională de Matematică - Etapa Locală | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0ikl | Problem:
A circle of radius $t$ is tangent to the hypotenuse, the incircle, and one leg of an isosceles right triangle with inradius $r=1+\sin \frac{\pi}{8}$. Find $r t$. | [
"Solution:\nThe distance between the point of tangency of the two circles and the nearest vertex of the triangle is seen to be both $r\\left(\\csc \\frac{\\pi}{8}-1\\right)$ and $t\\left(\\csc \\frac{\\pi}{8}+1\\right)$, so\n$$\n\\begin{aligned}\nr t & =\\frac{r^{2}\\left(\\csc \\frac{\\pi}{8}-1\\right)}{\\csc \\fr... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance cha... | null | proof and answer | (2+\sqrt{2})/4 | |
041i | Given positive integers $u, v$, the sequence $\{a_n\}$ is defined as: $a_1 = u + v$, and for $m \ge 1$,
$$
\begin{cases} a_{2m} = a_m + u, \\ a_{2m+1} = a_m + v. \end{cases}
$$
Denote $S_m = a_1 + a_2 + \dots + a_m$ ($m = 1, 2, \dots$). Prove that there are infinite terms in sequence $\{S_n\}$ that are square numbers. | [
"For positive integer $n$, we have\n$$\n\\begin{align*} \nS_{2^{n+1}-1} &= a_1 + (a_2 + a_3) + (a_4 + a_5) + \\dots + (a_{2^{n+1}-2} + a_{2^{n+1}-1}) \\\\\n&= u + v + (a_1 + u + a_1 + v) + (a_2 + u + a_2 + v) + \\dots + \\\\\n& \\qquad (a_{2^{n-1}} + u + a_{2^{n-1}} + v) \\\\\n&= 2^n (u + v) + 2S_{2^{n-1}}. \n\\end... | China | China Mathematical Competition (Complementary Test) | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
04yy | Let $x$, $y$ and $z$ be positive integers satisfying $\text{gcd}(x, y, z) = 1$. Prove that if $(y^2 - x^2) - (z^2 - y^2) = ((y - x) - (z - y))^2$, then $x$ and $z$ are perfect squares. | [
"Remove the parentheses, collect the terms and divide both sides by $2$ to get $x^2 + y^2 + z^2 - 2xy - 2yz + xz = 0$. This equality can be written as $(x - y + z)^2 = xz$. Hence $xz$ is a square of an integer. If $x$ and $z$ have a common divisor $d$, then $xz$ is divisible by $d^2$, and by the previous equality $... | Estonia | Estonija 2010 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0bxl | Let $A$ be a point outside the circle $\mathcal{C}$. The tangents from $A$ touch the circle at $B$ and $C$. Let $P$ be an arbitrary point on $AC$ produced, $Q$ the projection of $C$ onto $PB$ and $E$ the second intersection point of the circumcircle of $ABP$ with the circle $\mathcal{C}$. Prove that $\angle PEQ = 2\ang... | [
"Let $\\{D\\} = (BP \\cap \\mathcal{C})$ and $\\{M\\} = DE \\cap AC$.\n\nThen $\\angle BDE = \\angle BCE = 180^\\circ - \\angle ACB - \\angle PCE = 180^\\circ - \\angle ABC - \\angle CBE = \\angle APE$, hence $\\Delta MPD \\sim \\Delta MEP$ (AA), which leads to $\\angle MPD \\equiv \\angle MEP$ and $MP^2 = MD \\cdo... | Romania | THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ihr | Problem:
Vernonia High School has 85 seniors, each of whom plays on at least one of the school's three varsity sports teams: football, baseball, and lacrosse. It so happens that 74 are on the football team; 26 are on the baseball team; 17 are on both the football and lacrosse teams; 18 are on both the baseball and foo... | [
"Solution:\n\nSuppose that $n$ seniors play all three sports and that $2n$ are on the lacrosse team. Then, by the principle of inclusion-exclusion,\n$$\n85 = (74 + 26 + 2n) - (17 + 18 + 13) + n = 100 + 2n - 48 + n = 52 + 3n.\n$$\nIt is easily seen that $n = 11$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 11 | |
07n6 | $n \ge 3$ people seated at a round table play a game as follows. Initially each player is dealt a green or orange card. Game play then takes place in a series of rounds. In each round,
(i) Each player notes the colours of the cards held by the two players to his/her immediate right. If these are different, the player r... | [
"(a) Denote the configuration of cards after round $k \\ge 0$ by a binary sequence $a_1^{(k)}, a_2^{(k)}, a_3^{(k)}, \\dots, a_n^{(k)}$, where the subscripts are interpreted modulo $n$ (here $k=0$ corresponds to the initial cards dealt to the players). The outcome of each round may then be computed recursively by\n... | Ireland | Ireland | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
09i8 | There is a country, where the price of every fruit is an integer prime. Here 6 melons, 5 oranges, 5 apples and 3 pineapples cost 130 coins. Also, 3 melons, 3 oranges, 5 apples and 6 pineapples cost 130 coins. What is the price of each fruit? | [] | Mongolia | Mongolian Mathematical Olympiad Round 1 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | final answer only | Melon = 11, Orange = 3, Apple = 2, Pineapple = 13 | |
07gp | Given circles $\omega_1$ and $\omega_2$ with two intersection points $A$ and $B$. Points $X$ on $\omega_1$ and $Y$ on $\omega_2$ are chosen such that $XY$ is tangent to both circles and $XY$ is closer to $B$ than $A$. If $C$ and $D$ are the reflections of $X$ and $Y$ with respect to $B$, prove that $\angle CAD < 90^\ci... | [
"Suppose that $AB$ intersects $XY$ and $CD$ at $M$ and $N$, respectively. By comparing the power of $M$ with respect to $\\omega_1$ and $\\omega_2$ we have\n$$\nMX^2 = MB \\cdot MA = MY^2 \\implies MX = MY\n$$\n\n$\\angle CAD < 90^\\circ$ it suffices to prove that $A$ is outside the circle with diagonal $CD$; or eq... | Iran | 38th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Rotation",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0apw | Problem:
In an arithmetic sequence, the third, fifth and eleventh terms are distinct and form a geometric sequence. If the fourth term of the arithmetic sequence is $6$, what is its $2007$th term? | [
"Solution:\n\n6015\nLet $a$ and $d$ be the first term and the common difference of the given arithmetic sequence. Since there are distinct terms of the sequence, it follows that $d \\neq 0$. Since the third, fifth and eleventh terms form a geometric sequence, we have\n$$\n\\frac{a+4d}{a+2d} = \\frac{a+10d}{a+4d} \\... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | final answer only | 6015 | |
0csu | Pete and Bazil have a deck of $4026$ cards, $2013$ of which contain a digit $1$, while $2013$ others contain a digit $2$. Initially, Bazil uses all these cards to make a $4026$-digit number. After that, Pete makes several moves, each consisting in switching two cards and paying $1$ rouble to Bazil. The process stops wh... | [
"Рассмотрим $4026$-значное число $A$, состоящее из $2013$ единиц и $2013$ двоек. Пусть в этом числе в нечётных разрядах стоит $k$ единиц и $\\ell = 2013 - k$ двоек, тогда в чётных разрядах будет $k$ двоек и $\\ell$ единиц (здесь $k$ может принимать любое целое значение от $0$ до $2013$). Разность сумм цифр в нечётн... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 5 | |
0em4 | A little boy Mak has drawn a rectangle of size $20 \times 15$, and divided it by straight lines into unit squares. How many squares (of any size) are there in Mak's picture? | [
"Consider a square of side length $d$, and the position of its bottom-right unit square. It has $d-1$ squares above, and $d-1$ squares to the left of it. Hence there are $(21-d)(16-d)$ possibilities for the bottom-right corner, resulting in $(21-d)(16-d)$ squares of size $d$ appearing in the grid. The total number ... | South Africa | South-Afrika 2011-2013 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 1840 | |
0847 | Problem:
Determinare il più piccolo intero $n$ con la seguente proprietà: dati comunque $n$ interi $a_{1}, \ldots, a_{n}$, ne esistono due distinti tali che la loro somma o la loro differenza è divisibile per $9$.
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7. | [] | Italy | UNIONE MATEMATICA ITALIANA Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO TRIENNIO | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic"
] | null | MCQ | D | |
01uc | Prove that
$$
\frac{xyzt}{(x+y)(z+t)} \le \frac{(x+z)^2(y+t)^2}{4(x+y+z+t)^2}
$$
for all positive numbers $x$, $y$, $z$, $t$. | [
"The required inequality\n$$\n\\frac{xyzt}{(x+y)(z+t)} \\le \\frac{(x+z)^2(y+t)^2}{4(x+y+z+t)^2}\n$$\nis equivalent to the inequality\n$$\n\\frac{(x+y)(z+t)}{xyzt} \\ge \\frac{4(x+y+z+t)^2}{(x+z)^2(y+t)^2}\n$$\n(1)\nNote that\n$$\n\\frac{2(x+y+z+t)}{(x+z)(y+t)} = \\frac{2}{x+z} + \\frac{2}{y+t}\n$$\nBy A.M. - G.M. ... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0hfw | On the sides $AB$ and $BC$ of the square $ABCD$ we marked points $K$ and $L$ correspondingly so that $KB = LC$. Segments $AL$ and $CK$ intersect at point $P$. Prove that lines $DP$ and $KL$ are perpendicular. | [
"By construction $\\Delta ADK = \\Delta BAL$. Then $\\angle BAL + \\angle ALB = \\angle BAL + \\angle AKD = 90^\\circ$, and $\\Delta AMK$ has a right angle $M$, so $AL \\perp DK$ (fig. 14).\n\nSimilarly $DL \\perp CK$. Then lines $AL$ and $CK$, which intersect at point $P$, contain two altitudes of the $\\Delta DKL... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0fhq | Problem:
Demostrar que todo número primo $p$ distinto de 2 y de 5 tiene infinitos múltiplos escritos sólo con unos (es decir, de la forma 111...1). | [
"Solution:\nSi consideramos las clases de restos módulo $p$, entre las potencias $10^{0}, 10^{1}, 10^{2}, \\ldots, 10^{p-1}$, tiene que haber al menos 2 que pertenezcan a la misma clase. La clase 0 no puede aparecer pues $p$ es distinto de 2 y de 5. Sean esas potencias $10^{a}$ y $10^{b}$ y supongamos $b>a$. $10^{a... | Spain | OME 29 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
003m | Se han elegido varios segmentos en una recta (con posibles superposiciones). Si a cada segmento se le colorea de azul sus dos terceras partes de la izquierda, el conjunto de los puntos azules es un segmento de longitud $31$. Si a cada segmento se le colorea de rojo sus dos terceras partes de la derecha, el conjunto de ... | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | Español | proof and answer | 24 | |
0bd7 | Given an integer $n \ge 2$, let $a_n$, $b_n$, $c_n$ be integer numbers such that $(\sqrt[3]{2}-1)^n = a_n + b_n\sqrt[3]{2} + c_n\sqrt[3]{4}$. Show that $c_n \equiv 1 \pmod{3}$ if and only if $n \equiv 2 \pmod{3}$. | [
"The binomial expansion of $(\\sqrt[3]{2}-1)^n$ yields\n$$\nc_n = \\sum_{k \\equiv 2 \\pmod{3}} (-1)^{n-k} \\cdot 2^{(k-2)/3} \\binom{n}{k} \\equiv (-1)^n \\sum_{k \\equiv 2 \\pmod{3}} \\binom{n}{k} \\pmod{3}.\n$$\nSince\n$$\n\\sum_{k \\equiv 2 \\pmod{3}} \\binom{n}{k} = \\frac{1}{3}((1+1)^n + \\varepsilon(1+\\vare... | Romania | 64th NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomia... | null | proof only | null | |
08ek | Problem:
Ambra costruisce una sequenza di numeri, a partire da numero reale positivo $x_{0}$, nel seguente modo: dato l'$n$-esimo termine $x_{n}$, il termine successivo è $x_{n+1} = \{1 - \frac{1}{x_{n}}\}$, dove $\{\alpha\}$ rappresenta la parte frazionaria di $\alpha$, cioè la differenza fra $\alpha$ e il massimo nu... | [
"Solution:\n\na. Se possiamo scrivere $x_{n}$ nella forma $x_{n} = \\frac{p_{n}}{q_{n}}$ con $p_{n}$ e $q_{n}$ interi coprimi, allora anche $x_{n+1}$ potrà essere scritto nella stessa forma. Infatti se definiamo il numero intero $k$ come $k = -\\left\\lfloor 1 - \\frac{1}{x_{n}} \\right\\rfloor$ (dove $\\lfloor x \... | Italy | Italian Mathematical Olympiad, February Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
06tf | Let $S$ be a nonempty set of positive integers. We say that a positive integer $n$ is clean if it has a unique representation as a sum of an odd number of distinct elements from $S$. Prove that there exist infinitely many positive integers that are not clean.
(U.S.A.) | [
"Define an odd (respectively, even) representation of $n$ to be a representation of $n$ as a sum of an odd (respectively, even) number of distinct elements of $S$. Let $\\mathbb{Z}_{>0}$ denote the set of all positive integers.\nSuppose, to the contrary, that there exist only finitely many positive integers that ar... | IMO | 56th International Mathematical Olympiad Shortlisted Problems | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0d7d | Find all integers $n$ such that there exists a polynomial $P(x)$ with integer coefficients satisfying
$$
P\left(\sqrt[3]{n^{2}}+\sqrt[3]{n}\right)=2016 n+20 \sqrt[3]{n^{2}}+16 \sqrt[3]{n} .
$$ | [
"First, we prove 2 lemmas as the problem in Level 4's Test. Then turn to the problem, since $P\\left(\\sqrt[3]{n^{2}}+\\sqrt[3]{n}\\right)=2016 n+20 \\sqrt[3]{n^{2}}+16 \\sqrt[3]{n}$, then by the second lemma, we obtain $n-1 \\mid 20-16=4$.\nThat is, $n \\in\\{5,3,2,0,-1,-3\\}$, and we can easily check that these n... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | n ∈ {5, 3, 2, 0, -1, -3} | |
020b | Problem:
Find all positive integers $n>1$ with the following property:
for each two positive divisors $k, \ell<n$ of $n$, at least one of the numbers $2 k-\ell$ and $2 \ell-k$ is a (not necessarily positive) divisor of $n$ as well. | [
"Solution:\nIf $n$ is prime, then $n$ has the desired property: if $k, \\ell<n$ are positive divisors of a prime $n$, we have $k=\\ell=1$, in which case $2 k-\\ell=1$ is a divisor of $n$ as well.\n\nAssume now that a composite number $n$ has the desired property. Let $p$ be its smallest prime divisor and let $m=n /... | Benelux Mathematical Olympiad | Benelux Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All prime numbers, and 6, 9, 15 | |
0fti | Problem:
Sei $ABC$ ein gleichseitiges Dreieck und $P$ ein Punkt in dessen Innern. $X$, $Y$ und $Z$ seien die Fusspunkte der Lote von $P$ auf die Seiten $BC$, $CA$ und $AB$. Zeige, dass die Summe der Flächen der Dreiecke $BXP$, $CYP$ und $AZP$ nicht von $P$ abhängt. | [
"Solution:\n\nSchraffiere die Dreiecke $BXP$, $CYP$ und $AZP$. Zeichne die drei gleichseitigen Dreiecke, die $P$ als Ecke haben und deren gegenüberliegende Seite auf einer der Seiten von $ABC$ liegt. Durch diese drei gleichseitigen Dreiecke wird $ABC$ in drei Dreiecke und drei Parallelogramme zerlegt, die offenbar ... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
04p0 | Find all real numbers $x$ such that
$$
\left\lfloor \frac{x^2 + 1}{x + 2} \right\rfloor + \left\lfloor \frac{x - 1}{2} \right\rfloor = \frac{x(3x + 1)}{2(x + 2)}
$$
For a real number $t$, $\lfloor t \rfloor$ denotes the largest integer not greater than $t$. For example, if $t = 3.14$, then $\lfloor t \rfloor = 3$. | [
"Note that\n$$\n\\frac{x^2 + 1}{x + 2} + \\frac{x - 1}{2} = \\frac{x(3x + 1)}{2(x + 2)}.\n$$\nSince $\\lfloor t \\rfloor \\le t$ holds for all real numbers $t$, and the equality is attained if and only if $t$ is an integer, it follows that both\n$$\n\\frac{x^2 + 1}{x + 2} \\quad \\text{and} \\quad \\frac{x - 1}{2}\... | Croatia | Croatian Mathematical Society Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | x ∈ { -7, -3, -1, 3 } | |
0fw2 | Problem:
Betrachte einen Würfel mit Kantenlänge $2a$. In jedem Eckpunkt, jedem Kantenmittelpunkt und jedem Flächenmittelpunkt befindet sich eine Stadt. Zwei Städte sind durch eine Strasse miteinander verbunden, falls ihr Abstand $a$ beträgt. Gibt es eine Reiseroute, die durch jede Stadt genau einmal führt? | [
"Solution:\n\nFärbe alle Städte in den Eck- und Flächenmittelpunkten rot und alle Städte in den Kantenmittelpunkten blau. Jede Strasse verbindet dann eine rote und eine blaue Stadt. Längs jeder Reiseroute alternieren daher die Farben der besuchten Städte. Daraus folgt, dass sich die Anzahl blauer und roter Städte l... | Switzerland | Vorrundenprüfung | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof only | null | |
0a7l | Problem:
A hexagon is inscribed in a circle of radius $r$. Two of the sides of the hexagon have length $1$, two have length $2$ and two have length $3$. Show that $r$ satisfies the equation
$$
2 r^{3} - 7 r - 3 = 0
$$ | [
"Solution:\n(See Figure 5.) We join the vertices of the hexagon to the center $O$ of its circumcircle. We denote by $\\alpha$ the central angles corresponding to the chords of length $1$, by $\\beta$ those corresponding to the chords of length $2$, and by $\\gamma$ those corresponding to the chords of length $3$. C... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 7 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 2 r^3 - 7 r - 3 = 0 | |
0ko4 | Problem:
Positive integers $a$, $b$, and $c$ are all powers of $k$ for some positive integer $k$. It is known that the equation $a x^{2} - b x + c = 0$ has exactly one real solution $r$, and this value $r$ is less than $100$. Compute the maximum possible value of $r$.
Proposed by: Akash Das | [
"Solution:\n\nNote that for there to be exactly one solution, the discriminant must be $0$, so $b^{2} - 4 a c = 0$. Thus, $b$ is even, so $k = 2$. Since $r = \\frac{b}{2a}$, then $r$ is also a power of $2$, and the largest power of $2$ less than $100$ is $64$. This is achieved by $(x - 64)^{2} = x^{2} - 128 x + 409... | United States | HMMT February 2022 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 64 | |
0irf | Problem:
Let $a$, $b$, and $c$ be positive real numbers satisfying $a^{b} > b^{a}$ and $b^{c} > c^{b}$. Does it follow that $a^{c} > c^{a}$? | [
"Solution:\n\nYes. We have\n$$\n\\left(a^{c}\\right)^{b} = \\left(a^{b}\\right)^{c} > \\left(b^{a}\\right)^{c} = \\left(b^{c}\\right)^{a} > \\left(c^{b}\\right)^{a} = \\left(c^{a}\\right)^{b} ;\n$$\nthe desired inequality follows by taking the $b$th root."
] | United States | Berkeley Math Circle Monthly Contest 2 | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | Yes | |
0e08 | Let $(a_n)$ be a non-constant arithmetic sequence with the initial term $a_1 = 1$. The terms $a_2$, $a_5$, $a_{11}$ form a geometric sequence. Find the sum of the first 2009 terms of the sequence $(a_n)$. | [
"Let $d$ be the difference of the sequence $(a_n)$. Then $a_2 = 1 + d$, $a_5 = 1 + 4d$ and $a_{11} = 1 + 10d$. Since $a_2$, $a_5$ and $a_{11}$ form a geometric progression, we have $(1 + 4d)^2 = (1 + d)(1 + 10d)$ or $6d^2 = 3d$. Since the sequence is not constant, we conclude that $d = \\frac{1}{2}$ and the sum of ... | Slovenia | National Math Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 1010527 | |
0fxg | Problem:
Betrachte $n$ Kinder, von denen keine zwei gleich gross sind. Wie viele Möglichkeiten gibt es, diese Kinder in eine Reihe zu stellen, sodass jedes Kind ausser dem grössten einen Nachbarn besitzt, der grösser ist als es. | [
"Solution:\n\nWir gehen vom grössten Kind aus. Das zweitgrösste muss links oder rechts daneben stehen, sonst wäre(n) sein(e) Nachbar(n) kleiner als es. Das drittgrösste muss wiederum direkt neben diesem Pärchen stehen, das viertgrösste direkt neben dieser Dreiergruppe usw. Insgesamt gibt ab dem zweitgrössten Kind a... | Switzerland | Vorrundenprüfung | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 2^{n-1} | |
01bf | Two players play a game on an $N \times N$ board. The players alternately mark a cell in such a way that there is never a diagonal on the board containing two marked cells. For which $N > 0$ does the starting player have a winning strategy? | [
"**Answer:** For $N$ is odd.\n\nWe call the starting player $A$ and the other player $B$.\n\nIf $N$ is even $B$ has a winning strategy by symmetry. Every time $A$ places a mark on cell $(a, b)$, $B$ marks the cell $(N-a+1, b)$. Since $N$ is even $N-a+1 \\neq a$ for any $a$. Further every cell in a diagonal going th... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | All odd positive integers N | |
06ct | Let $O$ be the circumcentre of $\triangle ABC$. Suppose $AB > AC > BC$. Let $D$ be a point on the minor arc $BC$ such that $AD$ is not a diameter of the circumcircle of $\triangle ABC$. Let $E$ and $F$ be points on $AD$ such that $AB \perp OE$ and $AC \perp OF$. Let $P$ be the intersection of $BE$ and $CF$. If $PB = PC... | [
"Note that $OE$ is the perpendicular bisector of $AB$. Let $\\angle EBA = \\angle EAB = x$. Similarly, let $\\angle FCA = \\angle FAC = y$. Then we have\n$$\n\\angle BPC = 180^\\circ - \\angle PBC - \\angle PCB = 2x + 2y = 2 \\angle BAC = \\angle BOC.\n$$\nThis implies $O$, $B$, $C$, $P$ are concyclic. Applying Pto... | Hong Kong | CHKMO | [
"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 and answer | 30° | |
0jfo | Problem:
On a game show, Merble will be presented with a series of 2013 marbles, each of which is either red or blue on the outside. Each time he sees a marble, he can either keep it or pass, but cannot return to a previous marble; he receives 3 points for keeping a red marble, loses 2 points for keeping a blue marble... | [
"Solution:\n\nThe probability that Merble fails is $\\frac{1}{2^{2012}}$.\n\nFirst, we note that if all the marbles are red or all are blue, then it is impossible for Merble to win; we claim that he can guarantee himself a win in every other case. In particular, his strategy should be to keep the first red and firs... | United States | HMMT | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 1/2^2012 | |
0041 | Sean $AM$ y $AN$ las rectas tangentes a una circunferencia $\Gamma$ trazadas desde un punto $A$ ($M$ y $N$ pertenecen a la circunferencia). Una recta por $A$ corta a $\Gamma$ en $B$ y $C$ con $B$ entre $A$ y $C$, y $\frac{AB}{BC} = \frac{2}{3}$.
Si $P$ es el punto de intersección de $AB$ y $MN$, calcular $\frac{AP}{PC}... | [] | Argentina | Argentina 2006 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | Español | proof and answer | 4/3 | |
0f6k | Problem:
Is there an integer $n$ such that the sum of the (decimal) digits of $n$ is $1000$ and the sum of the squares of the digits is $1000^2$? | [] | Soviet Union | 19th ASU | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof only | null | |
0eak | Consider a regular $n$-gon where $n > 1$ is an odd integer. At most how many vertices can be coloured red so that the centre of the $n$-gon does not lie inside a polygon, determined by the red vertices? | [
"The greatest number of vertices that can be painted red is $\\frac{n+1}{2}$. If we paint $\\frac{n+1}{2}$ consecutive vertices of the $n$-gon, then the condition is satisfied.\n\nNow, assume that we have painted at least $\\frac{n+1}{2}+1$ vertices. Let us denote the vertices of the $n$-gon in order using positive... | Slovenia | National Math Olympiad in Slovenia | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | (n+1)/2 | |
0bw4 | If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 1$, show that
$$
\frac{bc}{2 - a^2} + \frac{ca}{2 - b^2} + \frac{ab}{2 - c^2} \le \frac{3}{5}.
$$ | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0cmw | In a country, some pairs of cities are connected by the roads, which do not intersect outside the cities. At each city, there is a sign showing the least length of the route beginning at this city and passing through all other cities (a route can pass through some cities several times; it should not necessarily return ... | [
"Рассмотрим самый короткий маршрут $\\ell$, проходящий по всем городам. Пусть он начинается в городе $A$, заканчивается в городе $B$, а его длина равна $N$. Тогда числа на табличках в городах $A$ и $B$ равны $N$, а все остальные не меньше $N$. Пусть $C$ — один из оставшихся городов. Он лежит на данном маршруте, поэ... | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof only | null | |
0j0k | Problem:
In the country of Francisca, there are $2010$ cities, some of which are connected by roads. Between any two cities, there is a unique path which runs along the roads and which does not pass through any city twice. What is the maximum possible number of cities in Francisca which have at least $3$ roads running... | [
"Solution:\n\nAnswer: $1004$\n\nThe restrictions on how roads connect cities directly imply that the graph of the cities of Francisca with the roads as edges is a tree. Therefore the sum of the degrees of all the vertices is $2009 \\cdot 2 = 4018$.\n\nSuppose that $b$ vertices have degree $\\geq 3$. The other $2010... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 1004 | |
0c1k | Find all the positive integers $\overline{abcd}$ knowing that the following conditions hold:
a) $a^2 + b^2 + c^2 + d^2$ is divisible by $4$;
b) the remainder of $\overline{abcd}$ when divided by $c$ is $7$. | [] | Romania | 69th Romanian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | All four-digit numbers with c = 9, and a, b, d ∈ {1,3,5,7,9}, such that a + b + d ≡ 7 (mod 9). There are no solutions with c = 8. | |
0l20 | Problem:
Compute the number of ways to shade exactly 4 distinct cells of a $4 \times 4$ grid such that no two shaded cells share one or more vertices. | [
"Solution:\n\nAssign coordinates to the cells of the grid so that the bottom-left, bottom-right, and top-right corners are $(0,0)$, $(3,0)$, and $(3,3)$ respectively.\n\nObserve that for each quadrant of the grid, all four cells of that quadrant share a vertex. Thus, any valid coloring must have exactly one shaded ... | United States | HMMT November 2024 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 79 | |
000t | En cada casilla del tablero de $4 \times 4$ debe haber un número natural de $1$ a $16$ inclusive, sin repetir, de modo que la suma de los cuatro números de cada una de las cuatro filas, la suma de los cuatro números de cada una de las cuatro columnas y la suma de los cuatro números de cada una de las dos diagonales sea... | [] | Argentina | XIX Olimpíada Matemática Argentina | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | español | proof and answer | Completed grid:
4 5 7 14
6 13 3 15
11 12 9 1
10 2 16 8 | |
0jdx | Problem:
Find all triples of positive integers $(x, y, z)$ such that $x^{2} + y - z = 100$ and $x + y^{2} - z = 124$. | [
"Solution:\n\nAnswer: $(12, 13, 57)$\n\nCancel $z$ to get $24 = (y - x)(y + x - 1)$. Since $x, y$ are positive, we have $y + x - 1 \\geq 1 + 1 - 1 > 0$, so $0 < y - x < y + x - 1$. But $y - x$ and $y + x - 1$ have opposite parity, so $(y - x, y + x - 1) \\in \\{(1, 24), (3, 8)\\}$ yields $(y, x) \\in \\{(13, 12), (... | United States | HMMT November 2013 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (12, 13, 57) | |
0j4z | Problem:
Let $ABCD$ be a cyclic quadrilateral, and suppose that $BC = CD = 2$. Let $I$ be the incenter of triangle $ABD$. If $AI = 2$ as well, find the minimum value of the length of diagonal $BD$. | [
"Solution:\n\nAnswer: $2\\sqrt{3}$\n\nLet $T$ be the point where the incircle intersects $AD$, and let $r$ be the inradius and $R$ be the circumradius of $\\triangle ABD$. Since $BC = CD = 2$, $C$ is on the midpoint of arc $BD$ on the opposite side of $BD$ as $A$, and hence on the angle bisector of $A$. Thus $A$, $... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"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 ... | null | proof and answer | 2√3 | |
0iy8 | Problem:
How many perfect squares divide $2^{3} \cdot 3^{5} \cdot 5^{7} \cdot 7^{9}$? | [
"Solution:\n\nThe number of such perfect squares is $2 \\cdot 3 \\cdot 4 \\cdot 5$, since the exponent of each prime can be any nonnegative even number less than the given exponent."
] | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | final answer only | 120 | |
05zy | Problem:
On considère un $2022$-gone régulier de côté $1$. Les points sont numérotés $A_{1}, \ldots, A_{2022}$ dans un certain ordre. Au départ, Elie part du point $A_{1}$, puis saute de point en point vers le point $A_{2}$, puis de $A_{2}$ vers $A_{3}$ etc... chaque fois en prenant l'arc le plus court. Lorsqu'il atte... | [
"Solution:\n\nLe problème demande de trouver la valeur maximale d'une somme de longueurs, il contient nécessairement deux parties. Dans un premier temps, on montre que le chemin parcouru par Elie est forcément de longueur inférieure à $2\\left(N^{2}-N+1\\right)$, cette partie s'appelle l'analyse. Dans un second tem... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2042222 | |
01db | Let $M$ and $N$ be the midpoints of the sides $AC$ and $AB$, respectively, of an acute triangle $ABC$. Let $\omega_B$ be the circle centered at $M$ passing through $B$, and let $\omega_C$ be the circle centered at $N$ passing through $C$. Let the point $D$ be such that $ABCD$ is an isosceles trapezoid with $AD$ paralle... | [
"Let $E$ be such that $ABEC$ is a parallelogram with $AB \\parallel CE$ and $AC \\parallel BE$, and let $\\omega$ be the circumcircle of $ABC$ with center $O$.\nIt is known that the radical axis of two circles is perpendicular to the line connecting the two centers. Since $BE \\perp MO$ and $CE \\perp NO$, this mea... | Baltic Way | Baltic Way 2016 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance cha... | null | proof only | null | |
0lbg | Cho trước một số nguyên tố $p$ không nhỏ hơn $17$. Chứng minh rằng $t=3$ là số nguyên dương lớn nhất có tính chất: Với mọi bộ số nguyên $a, b, c, d$ sao cho $abc$ không chia hết cho $p$, nhưng $a+b+c$ chia hết cho $p$ thì luôn tồn tại $x, y, z \in \{0, 1, ..., \lfloor \frac{p}{t} \rfloor - 1\}$ để $ax+by+cz+d$ chia hết... | [] | Vietnam | Kì thi chọn học sinh vào Đội tuyển Quốc gia Dự thi IMO | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | Vietnamese | proof and answer | 3 | |
02ca | Problem:
Cinco bolas iguais estão se movendo na mesma direção ao longo de uma reta fixa, mantendo uma certa distância de uma para outra. Na mesma direção, mas no sentido oposto, outras cinco bolas se movem de encontro às primeiras. As velocidades de todas as bolas são iguais. Quando duas bolas colidem, voltam na mesma... | [
"Solution:\n\nUma solução clara para o problema seria fazer todo o percurso das bolas, mas adotaremos outra estratégia.\nImagine que quando há a colisão de duas bolas, ao invés de gerar a volta das mesmas, uma bola se transforma na outra, como se não houvesse a colisão. Chamaríamos a esse processo de transmutação.\... | Brazil | Desafios | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 25 | |
0kpf | Problem:
Aerith has 5 coins, all with heads facing up. She wants to flip them so that they all have tails facing up.
a) If she must flip exactly three coins at a time (from heads to tails or vice versa), is this possible?
b) What if she must flip exactly two coins at a time?
For each part, either show a way it can ... | [
"Solution:\n\na) This is possible. One solution is $\\text{HHHHH} \\rightarrow \\text{TTTHH} \\rightarrow \\text{THHTH} \\rightarrow \\text{TTTTT}$.\n\nb) This is not possible. Since you are always flipping two coins at a time, the number of heads will always go down by two, go up by two, or stay the same. Since th... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) Possible. b) Impossible. | |
05qn | Problem:
On dit qu'un entier $k$ est olympique s'il existe quatre entiers $a, b, c$ et $d$, tous premiers avec $k$, tels que $k$ divise $a^{4}+b^{4}+c^{4}+d^{4}$. Soit $n$ un entier quelconque.
Montrer que $n^{2}-2$ est olympique. | [
"Solution:\n\nSoit $p_{1}^{\\alpha_{1}} \\cdots p_{\\ell}^{\\alpha_{\\ell}}$ la décomposition de $n^{2}-2$ en produits de facteurs premiers. Grâce au théorème Chinois, il nous suffit en fait de montrer que chacun des entiers $p_{i}^{\\alpha_{i}}$ est olympique.\n\nSupposons tout d'abord que $p_{i}=2$. Puisque $n^{2... | France | Préparation Olympique Française de Mathématiques | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Res... | null | proof only | null | |
0dtj | Seven triangles of area $7$ lie inside a square of area $27$. Prove that among the $7$ triangles there are $2$ that intersect in a region of area not less than $1$. | [
"Let the triangular regions be $P_i$, $i = 1, \\dots, 7$, with $A(P_i) = 7$ and suppose that $A(P_i \\cap P_j) < 1$ if $i \\neq j$. Then\n$$\n\\begin{align*}\nA(P_1 \\cup P_2) &= A(P_1) + A(P_2) - A(P_1 \\cap P_2) > 14 - 1 = 13 \\\\\nA(P_1 \\cup P_2 \\cup P_3) &= A(P_1 \\cup P_2) + A(P_3) - A((P_1 \\cup P_2) \\cap ... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0bis | Given a positive integer $n$ and an increasing real-valued function $f$ on the closed unit interval $[0, 1]$, determine the maximum value the sum
$$
\sum_{k=1}^{n} f\left(\left|x_k - \frac{2k-1}{2n}\right|\right)
$$
may achieve subject to $0 \le x_1 \le \dots \le x_n \le 1$. | [
"Let $a_k = (2k-1)/(2n)$, $k = 1, \\dots, n$. The required maximum is $\\sum_{k=1}^n f(a_k)$ and is achieved, for instance, at $x_1 = \\dots = x_n = 0$ or at $x_1 = \\dots = x_n = 1$.\n\nTo show that $\\sum_{k=1}^n f(a_k)$ is an upper bound for the sum under consideration subject to the given constraint, fix an $n$... | Romania | 65th NMO Selection Tests for BMO and IMO | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | \sum_{k=1}^{n} f\left(\frac{2k-1}{2n}\right) | |
0ena | Let $n$ be a positive integer. A train stops at $2n$ stations, including the first and last ones, numbered in order from the first to the $2n^{th}$. It is known that on a certain car, for each pair of integers $i, j$ such that $1 \le i < j \le 2n$, exactly one seat has been reserved for the trip from the $i^{th}$ stati... | [
"When the train leaves station $k$, the passengers on board the train are those that are travelling from station $i$ to $j$, with $i \\le k < j$. Hence there are $k(2n-k)$ people on board the train at that point, with a maximum of $n^2$ between stations $n$ and $n+1$.\n\nMoreover, at station $k$ there are $k-1$ pas... | South Africa | South-Afrika 2011-2013 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | n^2 | |
0528 | Let a finite decimal fraction be given. Juku starts appending digits to this fraction in such a way that each new digit equals the remainder of the sum of all digits existing so far in division by $10$. (For instance, if the initial fraction is $27.35$ then the digits added to the end are $7$, $4$, $8$ etc.)
Prove that... | [
"It suffices to show that the infinite decimal fraction is periodic. For that, note that each new digit except for the first digit is congruent to twice the previous digit modulo $10$. Indeed, let $a_1, \\dots, a_{k-1}, a_k$ be the existing at some time moment digits where $a_k$ is already added by Juku. Then the n... | Estonia | Final Round of National Olympiad | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0biz | Let $a, b, c$ be real numbers such that:
$$
|a - b| \ge |c|, \quad |b - c| \ge |a|, \quad |c - a| \ge |b|.
$$
Prove that one of the numbers $a, b, c$ equals the sum of the other two. | [
"Squaring the first inequality gives $(a-b)^2 \\ge c^2$, hence $(a-b+c)(b+c-a) \\ge 0$. Multiplying the latter with the other two similar inequalities implies $(a+b-c)^2(b+c-a)^2(c+a-b)^2 \\le 0$, hence one of $a, b, c$ is the sum of the other two."
] | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
05yi | Problem:
Nimatha et Thanima jouent à un jeu sur un échiquier $8 \times 8$. Tour par tour en commençant par Nimatha, chaque joueur choisit une case qui n'a pas encore été choisie et la colorie dans sa couleur (rouge pour Nimatha, bleu pour Thanima). Montrez que Thanima peut toujours faire que Nimatha ne puisse colorier... | [
"Solution:\n\nOn considère le coloriage suivant :\n\n| 1 | 1 | 2 | 2 | 3 | 3 | 4 | 4 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 5 | 6 | 6 | 7 | 7 | 8 | 8 | 5 |\n| 9 | 9 | 10 | 10 | 11 | 11 | 12 | 12 |\n| 13 | 14 | 14 | 15 | 15 | 16 | 16 | 13 |\n| 17 | 17 | 18 | 18 | 19 | 19 | 20 | 20 |\... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
020y | Problem:
Pebbles are placed on a $2021 \times 2021$ board in such a way that each square contains at most one pebble. The pebble set of a square of the board is the collection of all pebbles which are in the same row or column as this square. Determine the least number of pebbles that can be placed on the board in suc... | [
"Solution:\n\nLet $N \\geqslant 1$ be a positive integer. We claim that the least number of pebbles that can be placed on a $(2N+1) \\times (2N+1)$ chessboard in such a way that no two squares of the board have the same pebble set is $3N+1$. The problem has $N=1010$, so at least $3031$ pebbles are needed.\n\nWe beg... | Benelux Mathematical Olympiad | 13th Benelux Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 3031 | |
01ef | Let $ABCD$ be a trapezoid with $AD \parallel BC$ and $\angle ADC = \angle BAD$, and let $\ell$ be a line not intersecting the linesegments $AC$ or $BD$. Assume that $\ell$ intersects the lines $AC, AD, BC, BD$ and $CD$ in the points $P, Q, R, S$ and $T$ respectively. Show that the three circles $\odot(DQT), \odot(BRQ)$... | [
"\n\nNote first that $ABCD$ is concyclic since it is an isosceles trapezoid. We define $X$ to be the intersection of $\\odot(DQT)$ and $\\odot(ABCD)$, and it now suffices to prove that $BRQX$ and $BPSX$ are cyclic quadrilaterals. We have that\n$$\n\\begin{aligned}\n\\angle BXQ &= \\angle BX... | Baltic Way | Baltic Way shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point"
] | English | proof only | null | |
04e7 | All points $(x, y)$ with integer coordinates where $1 \le x \le 200$ and $1 \le y \le 100$ are marked in the coordinate system, 20000 points in total. How many lines of length $\sqrt{5}$ whose ends are marked points are there? (Mea Bombardelli) | [
"A line segment of length $\\sqrt{5}$ between two integer points must have a vector $\\vec{v} = (a, b)$ with $a^2 + b^2 = 5$. The integer solutions are $(\\pm 1, \\pm 2)$ and $(\\pm 2, \\pm 1)$.\n\nSo, the possible displacements are $(1,2)$, $(1,-2)$, $(-1,2)$, $(-1,-2)$, $(2,1)$, $(2,-1)$, $(-2,1)$, $(-2,-1)$.\n\n... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 78208 | |
0huk | Problem:
Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that $f(f(x)+y+1)=x+f(y)+1$ for every two integers $x$ and $y$. | [
"Solution:\n\nSubstituting $x=0$ we get $f(f(0)+y+1)=f(y)+1$.\n\nNow we have $x+f(y)+1=f(f(x)+y+1)=f[(f(x)+1)+y]=f[f(f(0)+x+1)+(y-1)+1]=f(y-1)+(f(0)+x+1)+1$ from where we conclude that\n\n$$\nf(y)=f(y-1)+f(0)+1 .\n$$\n\nUsing induction we obtain $f(n)=f(0)+n(f(0)+1)$ for all integers $n$.\n\nIf we set $y=0$ in (1) ... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | f(n) = n for all integers n; and f(n) = -n - 2 for all integers n | |
0eff | Problem:
Na koliko različnih načinov lahko sedejo v vrsto z 8 sedeži v gledališču 4 pari, če želi vsak posamezen par sedeti skupaj?
(A) $4!$
(B) $2 \cdot 4!$
(C) $24$
(D) $384$
(E) $256$ | [
"Solution:\n\nNačinov razporeditev 4 parov v vrsto je $4! = 24$. Če želijo pari sedeti skupaj, se lahko posedejo na $4! \\cdot 2^4 = 384$ načinov. Pravilen odgovor je (D)."
] | Slovenia | 17. tekmovanje dijakov srednjih tehniških in strokovnih šol v znanju matematike | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | MCQ | D | |
0ia7 | Problem:
You have infinitely many boxes, and you randomly put 3 balls into them. The boxes are labeled $1,2, \ldots$. Each ball has probability $1 / 2^{n}$ of being put into box $n$. The balls are placed independently of each other. What is the probability that some box will contain at least 2 balls? | [
"Solution:\n$$\n\\frac{5}{7}\n$$\nNotice that the answer is the sum of the probabilities that boxes $1,2, \\ldots$, respectively, contain at least 2 balls, since those events are mutually exclusive. For box $n$, the probability of having at least 2 balls is\n$$\n3\\left[\\left(1 / 2^{n}\\right)^{2}\\left(1-1 / 2^{n... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 5/7 | |
0kqa | Problem:
Let $ABC$ be a triangle with $AB=13$, $BC=14$, and $CA=15$. Pick points $Q$ and $R$ on $AC$ and $AB$ such that $\angle CBQ=\angle BCR=90^{\circ}$. There exist two points $P_{1} \neq P_{2}$ in the plane of $ABC$ such that $\triangle P_{1}QR$, $\triangle P_{2}QR$, and $\triangle ABC$ are similar (with vertices i... | [
"Solution:\nLet $T$ be the foot of the $A$-altitude of $ABC$. Recall that $BT=5$ and $CT=9$.\nLet $T'$ be the foot of the $P$-altitude of $PQR$. Since $T'$ is the midpoint of the possibilities for $P$, the answer is\n$$\n\\sum_{P} d(P, BC)=2 d\\left(T', BC\\right)\n$$\nSince $T'$ splits $QR$ in a $5:9$ ratio, we ha... | United States | HMMT November 2022 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | 48 | |
0idn | Problem:
Farmer John is grazing his cows at the origin. There is a river that runs east to west 50 feet north of the origin. The barn is 100 feet to the south and 80 feet to the east of the origin. Farmer John leads his cows to the river to take a swim, then the cows leave the river from the same place they entered and... | [
"Solution:\nSuppose we move the barn to its reflection across the river's edge. Then paths from the origin to the river and then to the old barn location correspond to paths from the origin to the river and then to the new barn location, by reflecting the second part of the path across the river, and corresponding ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 40*sqrt(29) | |
08zz | Find the number of tuples $(a_1, a_2, \dots, a_{2100})$ of integers between $0$ and $2099$ inclusive that satisfy the following condition:
There exists a tuple of integers $(b_1, b_2, \dots, b_{2100})$ such that for every integer $i$ from $1$ to $2100$,
$$
a_i \equiv \sum_{\substack{\gcd(j-i, 2100)=1 \\ 1 \le j \le 210... | [
"$$\n\\boxed{\\frac{2100^{210}}{2^{164} \\cdot 3^{30}}}\n$$\nLet the function $\\gcd(a, b)$ denote the greatest common divisor of $|a|$ and $|b|$, with the definition that the greatest common divisor of $0$ and a non-negative integer $x$ is $x$. Let us also denote that, for conditions $P_1, P_2, \\dots, P_k$, the n... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > Möbius inversion",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 2100^{210}/(2^{164}*3^{30}) | |
0l31 | Problem:
Let $ABCD$ be a convex quadrilateral with area $202$, $AB=4$, and $\angle A=\angle B=90^{\circ}$ such that there is exactly one point $E$ on line $CD$ satisfying $\angle AEB=90^{\circ}$. Compute the perimeter of $ABCD$. | [
"Solution:\n\n\n\nThe locus of point $E$ such that $\\angle AEB=90^{\\circ}$ is the circle $\\omega$ with diameter $AB$. Thus, if there exists unique point $E$, the circle $\\omega$ must intersect line $CD$ at exactly one point and hence line $CD$ must be tangent to $\\omega$.\n\nNow, since... | United States | HMMT November | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 206 | |
0752 | Consider two polynomials $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ and $Q(x) = b_n x^n + b_{n-1} x^{n-1} + \dots + b_1 x + b_0$ with integer coefficients such that $a_n - b_n$ is a prime, $a_{n-1} = b_{n-1}$ and $a_n b_0 - a_{n-1} b_n \neq 0$. Suppose there exists a rational number $r$ such that $P(r) = ... | [
"Let $r = u/v$ where $\\gcd(u, v) = 1$. Then we get\n\n$$\n\\begin{align*}\na_n u^n + a_{n-1} u^{n-1} v + \\dots + a_1 u v^{n-1} + a_0 v^n &= 0, \\\\\nb_n u^n + b_{n-1} u^{n-1} v + \\dots + b_1 u v^{n-1} + b_0 v^n &= 0.\n\\end{align*}\n$$\n\nSubtraction gives\n$$\n(a_n - b_n)u^n + (a_{n-2} - b_{n-2})u^{n-2}v^2 + \\... | India | Indija mo 2011 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0gk7 | Does there exist a function $f: \mathbb{Z} \to \mathbb{Z}$ such that
$$
f(f(n) - 2n) = 2f(n) + n
$$
for all integers $n$? | [
"There are many functions $f$ that satisfy the given condition. One of them is given by the following definition:\n$$\nf(n) = \\begin{cases} n & \\text{if } n \\ge 0 \\\\ -3n & \\text{if } n < 0 \\end{cases}\n$$\nwhich can be verified in the following 3 cases.\n\n$n > 0$: $f(f(n) - 2n) = f(-n) = 3n$ and $2f(n) + n ... | Thailand | Thai Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | Yes. For example, f(n) = n for n >= 0 and f(n) = -3n for n < 0. | |
0kzt | Integers $a$, $b$, and $c$ satisfy $ab + c = 100$, $bc + a = 87$, and $ca + b = 60$. What is $ab + bc + ca$? | [
"Notice that the difference between $100$ and $87$ is $13$, a prime number. This fact will help to simplify the problem. Subtract the second equation from the first to get\n$$\n\\begin{aligned}\n13 &= (ab + c) - (bc + a) \\\\\n&= ab - bc - a + c \\\\\n&= b(a - c) - (a - c) \\\\\n&= (b - 1)(a - c).\n\\end{aligned}\n... | United States | AMC 12 A | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 276 | |
09m6 | Let $m \ge 2$ be a positive integer and $p \ge 5$ be a prime number. Consider the sequence $a_n$ defined by $a_n = (p-2)^n + p^n - m$ for $n \ge 1$, and let $q$ be the smallest prime that divides a member of the sequence. Determine all pairs $(m, p)$ such that $q \ge m$.
(Bayarmagnai Gombodorj) | [
"Answer: $p \\ge 5$ prime, $(m, p) = (2, p), (3, p)$.\nLet $p \\ge 5$ be a prime number.\n\nFirst, consider the case where $m \\ge 2$ is even. If $m$ is even, consider $a_1 = 2(p-1)-m$. Since $a_1$ is even, we have $q = 2$. Therefore, $q \\ge m$ is satisfied only for $m = 2$. For $m \\ge 4$, the condition is not sa... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | (m, p) = (2, p) or (3, p) for any prime p at least five | |
0c1d | Let $ABC$ be a triangle, and let $E$ and $F$ be two arbitrary points on the sides $AB$ and $AC$, respectively. The circumcircle of triangle $AEF$ meets the circumcircle of triangle $ABC$ again at point $M$. Let $D$ be the reflection of point $M$ across the line $EF$ and let $O$ be the circumcenter of triangle $ABC$. Pr... | [
"Then $\\angle AEF = \\angle AMF < \\angle AMC = \\angle ABC$, therefore the line $EF$ does meet the line $BC$ at a point $G$ such that $B$ is between $C$ and $G$. It is known that the circumcircles of triangles $ABC$, $AEF$, $EBG$ and $FCG$ have a common point, Miquel's point of the complete quadrilateral $BCFEAG$... | Romania | 69th NMO Selection Tests for JBMO | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cycli... | null | proof only | null | |
0agp | Let $p$ be a positive integer, $p > 1$. Find the number of $m \times n$ matrices with entries in the set $\{1,2,3,..., p\}$ and such that the sum of elements on each row and each column is not divisible by $p$. | [
"Denote by $X$ the set of all matrices with entries in the set $\\{1,2,3,..., p\\}$ and let $A_i$, respectively $B_j$, the set of matrices in which the sum of elements in the row $i$, respectively on the column $j$, be divisible by $p$. Then we have to obtain the cardinality $N$ of the set\n$$\nX \\setminus \\left(... | North Macedonia | Mediterranean Mathematics Competition | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | p^{mn-m-n} ((p-1)^{m+n} + (-1)^{m+n} (p-1)) | |
0j12 | Problem:
Let $p$ be a monic cubic polynomial such that $p(0)=1$ and such that all the zeros of $p'(x)$ are also zeros of $p(x)$. Find $p$. Note: monic means that the leading coefficient is 1. | [
"Solution:\n\n$(x+1)^3$\n\nA root of a polynomial $p$ will be a double root if and only if it is also a root of $p'$. Let $a$ and $b$ be the roots of $p'$. Since $a$ and $b$ are also roots of $p$, they are double roots of $p$. But $p$ can have only three roots, so $a = b$ and $a$ becomes a double root of $p'$. This... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (x+1)^3 | |
07yr | Problem:
Per ogni intero $n \geq 2$, determinare:
a. il più grande numero reale $c_{n}$ tale che
$$
\frac{1}{1+a_{1}}+\frac{1}{1+a_{2}}+\ldots+\frac{1}{1+a_{n}} \geq c_{n}
$$
per ogni scelta dei numeri reali positivi $a_{1}, a_{2}, \ldots, a_{n}$ tali che $a_{1} \cdot a_{2} \cdot \ldots \cdot a_{n}=1$;
b. il più gra... | [
"Solution:\n\n(a) $c_{n}=1$ per ogni $n \\geq 2$.\nDimostriamo innanzitutto che per ogni $n \\geq 2$ si ha\n$$\n\\frac{1}{1+a_{1}}+\\frac{1}{1+a_{2}}+\\ldots+\\frac{1}{1+a_{n}} \\geq 1\n$$\nPer $n=2$, ponendo $a_{1}=a$ si ha $a_{2}=\\frac{1}{a}$ e l'espressione a sinistra della disuguaglianza si riduce a\n$$\n\\fra... | Italy | null | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | c_n = 1 for all n >= 2; d_2 = 2/3 and d_n = 1 for all n >= 3 | |
0hjb | Problem:
Find the number of multiples of $3$ which have six digits, none of which is greater than $5$. | [
"Solution:\n\nThe first digit can be any number from $1$ to $5$, making $5$ possibilities. Each of the succeeding digits, from the ten-thousands digit to the tens digit, can be any of the six digits from $0$ to $5$.\n\nFinally, we claim that there are exactly two possibilities for the last digit. Given the first fi... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Discrete Mathematics > Other",
"Number Theory > Other"
] | null | proof and answer | 12960 |
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