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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0g3g | Problem:
Let $ABC$ be an acute triangle with $AB = AC$ and let $D$ be a point on the side $BC$. The circle with centre $D$ passing through $C$ intersects the circumcircle of $ABD$ in $P$ and $Q$, where $Q$ is the point closer to $B$. The line $BQ$ intersects $AD$ in $X$ and $AC$ in $Y$. Prove that $PDXY$ is cyclic. | [
"Solution:\n\nWe first claim that $P \\in AC$. Indeed, let $P'$ be the second intersection between the circle centered at $D$ and $AC$. Then\n$$\n\\angle ABD = \\angle ABC = \\angle ACB = \\angle P'CD = 180^\\circ - \\angle AP'D\n$$\nso that $ABDP'$ is cyclic. This implies that $P = P'$, in particular $P \\in AC$.\... | Switzerland | Final round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0h37 | Three boys picked nuts. After discovering that $420$ nuts are picked in total, the boys decided to share them evenly. First, one of the boys gave each of two others one fourth of the nuts he picked and a nut. Then another boy gave each of two others one fourth of the nuts he collected (those he picked and those he obta... | [
"З рівностей\n$$\nx_3 - \\frac{1}{4}x_3 - \\frac{1}{4}x_3 - 2 = 140,\n$$\n$$\nx_2 + \\frac{1}{4}x_3 + 1 = 140.\n$$\n$$\nx_1 + \\frac{1}{4}x_3 + 1 = 140\n$$\nзнайдемо кількість горіхів у кожного з хлопчиків на передостанньому етапі. Аналогічно відновлюємо весь «ланцюжок»:\n$$\n(140, 140, 140) \\leftarrow (68, 68, 28... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | first boy: 68, second boy: 122, third boy: 230 | |
0dv5 | Problem:
V škofjeloški grajski kleti 7 palčkov hrani svoj zaklad. Zaklad je za 12 vrati, vsaka vrata pa so zaklenjena z 12 ključavnicami. Vse ključavnice so različne. Vsak palček ima ključe za nekaj ključavnic. Katerikoli 3 palčki imajo skupaj ključe za vse ključavnice. Dokaži, da imajo palčki skupaj vsaj 333 (ne nujn... | [
"Solution:\n\nZagotovo obstajajo 4 palčki, od katerih ima vsak vsaj 48 ključev (sicer bi lahko izbrali 3, ki bi skupaj imeli manj kot $3 \\cdot 48=144$ ključev in ne bi mogli odpreti vseh ključavnic). Preostali 3 palčki imajo skupaj vsaj 144 ključev. Torej imamo 4 palčke z vsaj 48 ključi in trojico z vsaj 144 ključ... | Slovenia | 46. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0cug | Initially $n > 1$ positive integers are written on the board. On each minute, a new number that is the sum of squares of all already written numbers appears on the board. (For example, if initial numbers were $1$, $2$, $2$, then on the first minute the number $1^2 + 2^2 + 2^2$ appears.) Prove that the $100$th new numbe... | [
"Let $S_i$ be the number appearing on the board on the $i$th minute. Then $S_{i+1} = S_i(S_i + 1)$, so $S_{i+1}$ contains all prime divisors of $S_i$ plus at least one more.\n\nLet $S_1, \\dots, S_{100}$ be the numbers that were written on the board in the first $100$ minutes. Suppose that before writing the number... | Russia | XLIII Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English; Russian | proof only | null | |
0bsw | Let $m$ be a positive integer. Consider a $4m \times 4m$ array of square unit cells. Two different cells are *related* to each other if they are in either the same row or in the same column. No cell is related to itself. Some cells are colored blue, such that every cell is related to at least two blue cells. Determine ... | [
"The required minimum is $6m$ and is achieved by a diagonal string of $m \\times 4$ blocks of the form below (bullets mark centers of blue cells):\n\n\nIn particular, this configuration shows that the required minimum does not exceed $6m$.\n\nWe now show that any configuration of blue cells... | Romania | 2016 European Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 6m | |
06v1 | Let $n \geqslant 2018$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ be pairwise distinct positive integers not exceeding $5n$. Suppose that the sequence
$$
\frac{a_{1}}{b_{1}}, \frac{a_{2}}{b_{2}}, \ldots, \frac{a_{n}}{b_{n}}
$$
forms an arithmetic progression. Prove that the terms ... | [
"Suppose that (1) is an arithmetic progression with nonzero difference. Let the difference be $\\Delta=\\frac{c}{d}$, where $d>0$ and $c, d$ are coprime.\nWe will show that too many denominators $b_{i}$ should be divisible by $d$. To this end, for any $1 \\leqslant i \\leqslant n$ and any prime divisor $p$ of $d$, ... | IMO | IMO Shortlisted Problems | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Com... | null | proof only | null | |
0g02 | Problem:
Gegeben seien ein Kreis $k$ und zwei Punkte $A$ und $B$ ausserhalb des Kreises. Gib an, wie man mit Zirkel und Lineal einen Kreis $\ell$ konstruieren kann, sodass $A$ und $B$ auf $\ell$ liegen und sich $k$ und $\ell$ berühren. | [
"Solution:\n\nAls Erstes konstruieren wir die Mittelsenkrechte $m$ der Strecke $A B$. Sei $O$ der Mittelpunkt von $k$. Falls $O$ auf $m$ liegt, dann wählen wir einen der beiden Schnittpunkte von $m$ mit $k$ und nennen ihn $P$. Wir können $P$ immer so wählen, dass $A, B$ und $P$ nicht auf einer Geraden liegen. Konst... | Switzerland | SMO - Finalrunde | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
00tj | Let $ABC$ be an acute triangle with $AC > AB$ and circumcircle $\Gamma$. The tangent from $A$ to $\Gamma$ intersects $BC$ at $T$. Let $M$ be the midpoint of $BC$ and let $R$ be the reflection of $A$ in $B$. Let $S$ be a point so that $SABT$ is a parallelogram and finally let $P$ be a point on line $SB$ such that $MP$ i... | [
"Let $N$ be the midpoint of $BS$ which, as $SABT$ is a parallelogram, is also the midpoint of $TA$. Using $ST \\parallel AB \\parallel MP$ we get:\n$$\n\\frac{NB}{BP} = \\frac{1}{2} \\cdot \\frac{SB}{BP} = \\frac{TB}{2 \\cdot BM} = \\frac{TB}{BC}\n$$\nwhich shows that $TA \\parallel CP$.\n\n![](attached_image_1.png... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07hn | Suppose that $3 \le n$ is a natural number. Find the maximum value of the natural number $k$, for which, there are real numbers $a_1, a_2, \dots, a_n \in [0, 1)$ (not necessarily distinct) such that for every natural number $j$ satisfying $j \le k$, there exists a sum of a certain number of $a_i$'s equal to $j$. | [
"We claim that the answer is $n-2$. First, note that since the sum of a number of variables (at least 2) is one, and the rest of the numbers are less than one, the sum of all the variables is less than $n-1$. Suppose the numbers are $a_i = 1-b_i$.\n\n**First solution.** Using induction, we will prove that for each ... | Iran | 40th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | n - 2 | |
0021 | Sea $a \geq 4$ un entero positivo. Determinar el menor valor de $n \geq 5$, tal que $a$ se puede representar de la forma
$$
\sigma = \frac{x_1^2 + x_2^2 + \dots + x_n^2}{x_1 x_2 \dots x_n}
$$
para una elección adecuada de los $n$ enteros positivos $x_1, x_2, \dots, x_n$. | [] | Argentina | XX OLIIMPÍADA MATEMÁTICA ARGENTINA | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | español | proof and answer | The minimal n is 5 for a = 4 or a = 5, and n = a for all a ≥ 6. | |
01x3 | Let $n$ be a given positive integer. Sisyphus performs a sequence of turns on a board consisting of $(n+1)$ squares in a row, numbered from $0$ to $n$ from left to right. Initially, $n$ stones are put into square $0$, and the other squares are empty. At every turn, Sisyphus chooses any nonempty square, say with $k$ sto... | [] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
0le5 | Consider the following polynomial
$$
f(x) = x^2 - \alpha x + 1
$$
with $\alpha \in \mathbb{R}$.
a) For $\alpha = \frac{\sqrt{15}}{2}$, express $f(x)$ as the quotient of two polynomials with non-negative coefficients.
b) Find all values of $\alpha$ such that $f(x)$ can be written as the quotient of two polynomials wit... | [
"a. We consider the following transformation\n$$\n\\left(x^2 - \\frac{\\sqrt{15}}{2}x + 1\\right) \\left(x^2 + \\frac{\\sqrt{15}}{2}x + 1\\right) = x^4 - \\frac{7}{4}x^2 + 1,\n$$\n$$\n\\left(x^{4} - \\frac{7}{4}x^{2} + 1\\right) \\left(x^{4} + \\frac{7}{4}x^{2} + 1\\right) = x^{8} - \\frac{17}{16}x^{4} + 1,\n$$\n$$... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | a) f(x) = (x^16 + (223/256)x^8 + 1) / [(x^2 + (sqrt(15)/2)x + 1)(x^4 + (7/4)x^2 + 1)(x^8 + (17/16)x^4 + 1)]. b) All real alpha with alpha < 2. | |
0hqp | Problem:
Prove that there exists an infinite sequence of $a_{1}, a_{2}, \ldots$ positive integers such that the following condition holds: $\operatorname{gcd}\left(a_{m}, a_{n}\right)=1$ if and only if $|m-n|=1$. | [
"Solution:\nEnumerate the primes $p_{1}, q_{1}, p_{2}, q_{2}, \\ldots$ and define\n$$\na_{n}=p_{n} q_{n} \\cdot \\begin{cases}\\prod_{k=1}^{n-2} p_{k} & n \\text{ even } \\\\ \\prod_{k=1}^{n-2} q_{k} & n \\text{ odd. }\\end{cases}\n$$\nThis works by construction. The idea is that you just take every pair $i<j$ you ... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
06id | Let $ABC$ and $PQR$ be two triangles. If $\cos A = \sin P$, $\cos B = \sin Q$ and $\cos C = \sin R$, what is the largest angle (in degrees) among the six interior angles of the two triangles? (1 mark)
設 $ABC$ 和 $PQR$ 為三角形。若 $\cos A = \sin P$、$\cos B = \sin Q$ 且 $\cos C = \sin R$，則兩個三角形六個內角中最大的一個（以「度」為單位）是多少？ (1分) | [
"Let $A$, $B$, $C$ be the angles of triangle $ABC$, and $P$, $Q$, $R$ be the angles of triangle $PQR$.\n\nGiven:\n$$\n\\cos A = \\sin P,\\ \\cos B = \\sin Q,\\ \\cos C = \\sin R\n$$\n\nRecall that $\\sin x = \\cos(90^\\circ - x)$, so:\n$$\n\\cos A = \\sin P = \\cos(90^\\circ - P) \\implies A = 90^\\circ - P \\text{... | Hong Kong | HONG KONG PRELIMINARY SELECTION CONTEST | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English; Chinese | final answer only | 90° | |
0ex9 | Problem:
a. Each of $x_{1}, \ldots, x_{n}$ is $-1$, $0$ or $1$. What is the minimal possible value of the sum of all $x_{i}x_{j}$ with $1 \leq i < j \leq n$?
b. Is the answer the same if the $x_{i}$ are real numbers satisfying $0 \leq |x_{i}| \leq 1$ for $1 \leq i \leq n$? | [
"Solution:\n\na. Answer: $-\\left[ n / 2 \\right]$.\n\nLet $A = (x_{1} + \\ldots + x_{n})^{2}$, $B = x_{1}^{2} + \\ldots + x_{n}^{2}$. Then we must minimize $A - B$. For $n$ even, we separately minimize $A$ and maximize $B$ by taking half the $x$'s to be $+1$ and half to be $-1$. For $n$ odd we can take $[n / 2]$ $... | Soviet Union | 5th ASU | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | -floor(n/2) | |
0hti | Problem:
Suppose $a, b, c$ are rational numbers such that
$$
\begin{aligned}
& \left(a^{2}+1\right)^{3}=b+1 \\
& \left(b^{2}+1\right)^{3}=c+1 \\
& \left(c^{2}+1\right)^{3}=a+1
\end{aligned}
$$
Prove that $a=b=c=0$. | [
"Solution:\nWe have that $b=\\left(a^{2}+1\\right)^{3}-1$, $c=\\left(b^{2}+1\\right)^{3}-1$, and $a=\\left(c^{2}+1\\right)^{3}-1$. By direct substitution we derive that $a$ satisfies the following polynomial equation of degree 216:\n$$\n\\left(\\left(\\left(\\left(\\left(a^{2}+1\\right)^{3}-1\\right)^{2}+1\\right)^... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | null | proof only | null | |
0c5m | Let $r$ be a rational number consider integers $a_1, a_2, \dots, a_6, b_1, b_2, \dots, b_6$ such that $1 \le b_1 < b_2 < \dots < b_6 \le 11$ and
$$
r = \frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} = \frac{a_4}{b_4} = \frac{a_5}{b_5} = \frac{a_6}{b_6}.
$$
Prove that $r$ is an integer. | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof only | null | |
09bc | $ n^4 - 4n^3 + 22n^2 - 36n + 18 $ нь натурал тооны квадрат болох бүх эерэг бүхэл $ n $ тоонуудыг ол. | [
"**VII-B2.** (Н.Дайвий-Од) Тортоо дугуй хэлбэртэй гэж үзье.\nбайдлаар хуваалт хийсэн гэе. Эхлээд нэг дугуйг 5 тэнцүү, дараа нь өөр нэг дугуйг 9 тэнцүү сегментээр хуваая.\nДараа нь тэдгээрийг давхцуулан тавихдаа нэг нэг радиус давх-\nцаж байхаар байрлуулъя. Энэ тохиолдолд 5 + 7 + 9 - 2 = 19\nхэсэгт хуваагдаж байгаа ... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
... | Mongolian | proof and answer | n = 1, 3 | |
04av | Let $ABC$ be a triangle with centroid $T$ and circumcenter $O$ such that $OT \perp AT$. Let $A'$ be the other intersection of the line $AT$ and the circumcircle of the triangle $ABC$. Let $D$ be the intersection of the lines $BA'$ and $AC$, and let $E$ be the intersection of the lines $CA'$ and $AB$. Prove that the cir... | [
"Let $A_1, B_1$ and $C_1$ be the midpoints of the sides $\\overline{BC}, \\overline{CA}$ and $\\overline{AB}$ respectively. Let $k$ be the circumcircle of the triangle $ABC$.\n\n\n\nNotice that from $OT \\perp AA'$ it follows that $OT$ is the bisector of the chord $AA'$ of the circle $k$ so... | Croatia | CroatianCompetitions2011 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
08w7 | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that
$$
f(f(x+y)f(x-y)) = x^2 - y f(y)
$$
for all $x, y \in \mathbb{R}$. | [] | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x for all real x | |
0kv6 | Problem:
Let $x$, $y$, and $N$ be real numbers, with $y$ nonzero, such that the sets $\{(x+y)^2, (x-y)^2, x y, x / y\}$ and $\{4, 12.8, 28.8, N\}$ are equal. Compute the sum of the possible values of $N$. | [
"Solution:\nFirst, suppose that $x$ and $y$ were of different signs. Then $x y < 0$ and $x / y < 0$, but the set has at most one negative value, a contradiction. Hence, $x$ and $y$ have the same sign; without loss of generality, we say $x$ and $y$ are both positive.\n\nLet $(s, d) := (x+y, x-y)$. Then the set given... | United States | HMMT February 2023 | [
"Algebra > Intermediate Algebra > Other",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 85.2 | |
0kr4 | Problem:
Estimate $A$, the number of unordered triples of integers $(a, b, c)$ so that there exists a nondegenerate triangle with side lengths $a$, $b$, and $c$ fitting inside a $100 \times 100$ square. An estimate of $E$ earns $\max (0,\lfloor 20-|A-E| / 1000\rfloor)$ points. | [
"Solution:\n\nLet's first count the number of such triangles with perimeter equal to $p$. By Stars and Bars, there are $\\binom{p}{2} \\approx \\frac{p^{2}}{2}$ ordered triples of positive integers that sum to $p$. Additionally, note that only about a quarter of them satisfy the triangle inequality, we have only $\... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > ... | null | final answer only | 187500 | |
07ly | The sum of a certain number of consecutive positive integers is equal to $2010$.
Find these integers. | [
"Let $a$ be the smallest and $b$ be the largest integer in the sum, that is we wish to find all possible positive integers $a \\leq b$ such that $\\sum_{k=a}^{b} k = 2010$. We have\n$$\n\\begin{aligned}\n\\sum_{k=a}^{b} k &= \\sum_{k=0}^{b-a} a + k = (b-a+1)a + \\sum_{k=0}^{b-a} k \\\\\n&= (b-a+1)a + \\frac{(b-a)(b... | Ireland | Irish Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | [4..63], [91..110], [127..141], [162..173], [400..404], [501..504], [669..671], [2010] | |
0c96 | Consider a real number $a \ge 1$. The sequence $(x_n)_{n \ge 1}$ is given by $x_1 = a$ and $x_{n+1} = 1 + \log_2 x_n$, for any $n \in \mathbb{N}^*$. Determine values of $a$ such that all terms of the sequence are rational numbers. | [
"For $a = 1$ or $a = 2$ we get the constant sequence $1$ or $2$ respectively.\nWe shall prove that these are the only values satisfying the problem.\nTo see this, let $x_n = \\frac{k}{l}$ and $x_{n+1} = \\frac{p}{q}$, with $k, l$ coprime positive integers, $p, q$ coprime positive integers with $q \\ge 2$. Then\n$$\... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | a = 1 or a = 2 | |
081l | Problem:
Un puzzle da 1000 pezzi può essere montato incastrando i pezzi uno dopo l'altro, in modo da inserire ciascun nuovo pezzo nella porzione di puzzle già composta, oppure costruendo diversi gruppi di pezzi e poi unendo questi tra di loro. Ogni unione (di due singoli pezzi, o di due gruppi, o di un pezzo a un grup... | [
"Solution:\n\nLa risposta è 999. Dimostriamo per induzione che per costruire un nucleo di $n$ pezzi sono necessarie $n-1$ mosse, comunque si proceda. L'affermazione è chiaramente vera per un puzzle costituito da un solo pezzo. Supponiamo che questa affermazione sia vera per tutti i nuclei con meno di $n$ pezzi. L'u... | Italy | Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 999 | |
00aj | Given 100 infinitely large boxes with markers in them, the following procedure is carried out. At step 1 one adds one marker in every box. At step 2 one marker is added in every box containing an even number of markers. At step 3 one marker is added in every box in which the number of markers is divisible by 3, and so ... | [
"The answer is *no*. Regardless of the initial distribution all boxes will contain the same number of markers after finitely many steps. Moreover this is true for any number of boxes.\n\nDenote by $x_n$ the number of markers in a certain box before step $n$, $n = 1, 2, \\dots$. Suppose that $x_n = n$ for some $n$. ... | Argentina | Argentine National Olympiad 2016 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | no | |
02ob | Problem:
Qual é a soma? Se $x + |x| + y = 5$ e $x + |y| - y = 6$, qual é o valor da soma $x + y$?
(a) $-1$
(b) $11$
(c) $\frac{9}{5}$
(d) $1$
(e) $-11$ | [
"Solution:\n\n1º Caso: Se $x \\leq 0$, então $|x| = -x$ e, pela primeira equação, temos $x + (-x) + y = 5$, ou seja, $y = 5$. Substituindo esse valor na segunda equação, obtemos $x = 6$, o que não é possível, pois estamos supondo $x \\leq 0$. Logo, não há solução nesse caso $x \\leq 0$.\n\n2º Caso: Se $y \\geq 0$, ... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | c | |
02ub | Problem:
Em um torneio de xadrez, todos os jogadores enfrentaram todos os outros exatamente uma vez. Em cada partida, o jogador ganha 1 ponto se vencer, $1/2$ se empatar e 0 ponto se perder. Ao final do torneio, um repórter somou as pontuações de todos os jogadores e obteve 190 pontos. Nesse tipo de torneio, o vencedo... | [
"Solution:\n\na) Seja $J$ o número de jogadores. Cada partida vale no total 1 ponto, seja $1+0=1$ ou $1/2+1/2=1$. Então a pontuação total é igual ao número de partidas. Como cada um dos $J$ jogadores enfrenta cada um dos outros $J-1$ jogadores, poderíamos pensar que o total de jogos seria $J(J-1)$ embates. Entretan... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | a) 20 players. b) André did not win. | |
0ahf | On a board there are $n$ nails each two connected by a string. Each string is colored in one of $n$ given distinct colors. For each three distinct colors, there exist three nails connected with strings in these three colors. Can $n$ be
a) $6$?
b) $7$? | [
"a. The answer is no.\n\nSuppose it is possible. Consider some color, say blue. Each blue string is the side of $4$ triangles formed with vertices on the given points. As there exist $\\binom{5}{2} = \\frac{5 \\cdot 4}{2} = 10$ pairs of colors other than blue, and for any such pair of colors together with the blue ... | North Macedonia | XVI-th Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) no; b) yes | |
03if | Problem:
A gambling student tosses a fair coin and scores one point for each head that turns up and two points for each tail. Prove that the probability of the student scoring exactly $n$ points is
$$
\frac{1}{3}\left[2+\left(-\frac{1}{2}\right)^{n}\right].
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof and answer | (1/3)[2 + (-1/2)^n] | |
08g0 | Problem:
In un sacchetto ci sono delle biglie di vari colori. Si sa che tutte le biglie tranne 6 sono gialle, tutte le biglie tranne 7 sono rosse, tutte le biglie tranne 10 sono blu. Inoltre, c'è almeno una biglia blu e potrebbero esserci anche biglie di colori diversi da giallo, rosso e blu. Quante biglie contiene il... | [
"Solution:\n\nLa risposta è (B). Chiamiamo $n$ il numero totale di biglie presenti nel sacchetto. Le biglie gialle sono allora $n-6$, quelle rosse sono $n-7$, e quelle blu sono $n-10$. Dato che potrebbero anche esserci biglie di altri colori, vale la seguente disuguaglianza: $n-6+n-7+n-10 \\leq n$, che implica $n \... | Italy | Italian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | B | |
09at | Let $(2m + 1, 2n + 1) = 1$ for natural numbers $m$ and $n$. Find
$$
(2^{2m+1} + 2^{m+1} + 1, 2^{2n+1} + 2^{n+1} + 1).
$$
Here $(a, b)$ denotes the greatest common denominator of $a$ and $b$. | [
"Let $d = (2^{2m+1} + 2^{m+1} + 1, 2^{2n+1} + 2^{n+1} + 1)$. It is well known that\n$$\n(2^k - 1, 2^n - 1) = 2^{(k,n)} = 1.\n$$\nSince\n$$\n(2^{2a+1} + 2^{a+1} + 1)(2^{2a+1} - 2^{a+1} + 1) = (2^{2a+1} + 1)^2 - (2^{a+1})^2 = 2^{4a+2} + 1,\n$$\n$d \\mid (2^{4a+2} + 1, 2^{4b+2} + 1)$. This implies\n$$\nd \\mid (2^{8a+... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | 5 if m and n are both divisible by 4; otherwise 1 | |
0alv | Problem:
Find the sum of the digits of the integer $10^{1001} - 9$.
(a) 9010
(b) 9001
(c) 9100
(d) 9009 | [] | Philippines | Qualifying Round | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | b | |
05xx | Problem:
Montrer qu'il existe une infinité de couples $(m, n)$ d'entiers strictement positifs distincts tels que $m!n!$ soit un carré parfait. | [
"Solution:\n\nOn aurait envie de prendre $m = n$ pour avoir $m!n! = (n!)^2$ et avoir un carré parfait. Mais l'énoncé force $m \\neq n$. Malgré cela, on voit déjà un moyen de faire apparaître naturellement des carrés parfaits.\n\nSupposons sans perte de généralité $m > n$ (comme $m!n! = n!m!$, quitte à remplacer $(m... | France | Préparation Olympique Française de Mathématiques - Envoi 3: Arithmétique | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
09wq | Problem:
Voor een positief getal $n$ schrijven we $d(n)$ voor het aantal positieve delers van $n$. Bepaal alle positieve gehele getallen $k$ waarvoor er positieve gehele getallen $a$ en $b$ bestaan met de eigenschap
$$
k = d(a) = d(b) = d(2a + 3b)
$$ | [
"Solution:\nVoor $i \\geq 0$ kiezen we $a = 2 \\cdot 5^{i}$ en $b = 3 \\cdot 5^{i}$. Dan hebben $a$ en $b$ elk $2(i+1)$ delers. Verder is $2a + 3b = 4 \\cdot 5^{i} + 9 \\cdot 5^{i} = 13 \\cdot 5^{i}$ en dat heeft ook $2(i+1)$ delers. Dus dit voldoet met $k = 2(i+1)$. We zien dat alle even waarden van $k$ voldoen.\n... | Netherlands | IMO-selectietoets III | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | all even positive integers | |
07i3 | Find all functions $f : \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that for all $0 < x, y, z$ the following numbers are the side lengths of a triangle:
$$
x + f(y),\ f(f(y)) + z,\ f(f(z)) + f(x)
$$
and for every positive number $a$ there exists $0 < b$ such that $f(b) < a$. | [
"Let $(x, y, z) \\rightarrow (f(y), y, f(y))$. Then the triangle inequalities give:\n$$\nf(f(f(y))) < 3f(y)\n$$\nIf $z$ is in the range of $f$, we obtain $f(f(z)) < 3z$. Now let $z$ be in the range of $f$.\n\nLet $(x, y, z) \\rightarrow (f(y), y, z)$. Then:\n$$\n2f(y) < 2f(f(y)) + f(f(z)) + z < 2f(f(y)) + 4z \\\\\n... | Iran | 40th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof and answer | f(x) = x for all x > 0 | |
0cfh | Solve the equation
$$
((((((((n - \frac{1}{2}) \cdot 2 - \frac{2}{3}) \cdot 3 - \frac{3}{4}) \cdot 4 - \frac{4}{5}) \cdot \dots) \cdot 2023 - \frac{2023}{2024}) \cdot 2024 = 1.
$$ | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof and answer | 1 | |
05sb | Problem:
Soit $\left(a_{n}\right)_{n \geqslant 0}$ une suite de réels. On suppose que $a_{n}=\left|a_{n+1}-a_{n+2}\right|$ pour tout entier naturel $n$. De plus, $a_{0}$ et $a_{1}$ sont strictement positifs et distincts. Montrer que la suite $\left(a_{n}\right)_{n \geqslant 0}$ n'est pas bornée. | [
"Solution:\n\nIl est clair que la suite $\\left(a_{n}\\right)$ est à termes positifs.\n\nSoit $i$ tel que $a_{i}<a_{j}$ pour $j<i$. Supposons par l'absurde que $i \\geqslant 4$. Alors $a_{i-2}=\\left|a_{i}-a_{i-1}\\right|=a_{i-1}-a_{i}<a_{i-1}$ donc $a_{i-3}=\\left|a_{i-2}-a_{i-1}\\right|=a_{i-1}-a_{i-2}=a_{i}$, ce... | France | Préparation Olympique Française de Mathématiques - ENVOI 4 : POT-POURRI | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0j83 | Problem:
Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep a running tally of the sum of the results of all rolls made. A player wins if, after he rolls, the number on the running tally is a multiple of $7$. Play continues until either player wins, or else indefinitely. I... | [
"Solution:\n\n$\\boxed{\\dfrac{5}{11}}$\n\nFor $1 \\leq k \\leq 6$, let $x_k$ be the probability that the current player, say $A$, will win when the number on the tally at the beginning of his turn is $k$ modulo $7$. The probability that the total is $l$ modulo $7$ after his roll is $\\frac{1}{6}$ for each $l \\not... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 5/11 | |
0axr | Problem:
Let $x, y, z$ be positive integers such that
$$
\begin{aligned}
& (x+y)(y+z)=2016 \\
& (x+y)(z+x)=1080
\end{aligned}
$$
Determine the smallest possible value for $x+y+z$. | [
"Solution:\nNote that $2016=2^{5} \\times 3^{2} \\times 7$ and $1080=2^{3} \\times 3^{3} \\times 5$. Moreover\n$$\nx+y+z=\\frac{1}{2}((x+y)+(y+z)+(z+x))\n$$\nSince $x+y$ is a common factor for both $2016$ and $1080$, and we want $x+y+z$ to be as small as possible, then we try to find the largest possible factor for... | Philippines | Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 61 | |
0ar6 | Problem:
Show that $\sqrt[n]{2}-1 \leq \sqrt{\frac{2}{n(n-1)}}$ for all positive integers $n \geq 2$. | [
"Solution:\n\nLet $x_n = \\sqrt[n]{2} - 1 \\geq 0$. Then $2 = (1 + x_n)^n \\geq 1 + n x_n + \\frac{n(n-1)}{2} x_n^2 \\geq 1 + \\frac{n(n-1)}{2} x_n^2$.\n\nThus, $\\frac{n(n-1)}{2} x_n^2 \\leq 2 - 1$ and the desired inequality follows."
] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof only | null | |
02vx | Problem:
Uma partição do Conjunto dos Números Naturais é uma coleção de conjuntos $A_{1}, A_{2}, \ldots, A_{k}$ de modo que cada número natural pertença a exatamente um deles. Veja que em qualquer partição do Conjunto dos Números Naturais pelo menos um desses conjuntos é infinito, pois caso contrário o Conjunto dos Nú... | [
"Solution:\n\na) Veja que $x$ possui infinitos múltiplos no conjunto dos números naturais que estarão divididos entre os conjuntos da partição. Se cada conjunto tivesse apenas uma quantidade finita de múltiplos de $x$, então o número total de múltiplos de $x$ entre os naturais, por ser uma união desses conjuntos, s... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Logic",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof only | null | |
01rs | Does there exist a function $f$, $f: \mathbb{R} \to \mathbb{R}$, such that
$$
\begin{cases}
\{f(x)\} \sin^2 x + \{x\} \cos f(x) \cos x = f(x), \\
f(f(x)) = f(x),
\end{cases}
$$
for all real $x$.
(Here $\{y\}$ stands for the fractional part of $y$.) | [
"Assume that there exists a function $f(x)$ satisfying the problem condition:\n$$\n\\begin{cases}\n\\{f(x)\\} \\sin^2 x + \\{x\\} \\cos f(x) \\cos x = f(x), \\\\\n f(f(x)) = f(x),\n\\end{cases}\n$$\nfor all real $x$.\nReplacing $x$ by $f(x)$ in the first equality, we obtain\n$$\n\\{f(f(x))\\} \\sin^2 f(x) + \\{f(x)... | Belarus | FINAL ROUND | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | No | |
056s | The polynomial $x^3 + px + q$, where $p$ and $q$ are real numbers and at least one of them is non-zero, has a real root $a$ that satisfies
$$
a^2 \le -\frac{4}{3}p.
$$
Prove that this polynomial has a real root different from $a$. | [
"The assumption $a^3 + pa + q = 0$ implies $q = -a(a^2 + p)$, whence $x^3 + px + q = (x-a)(x^2 + ax + a^2 + p)$. The discriminant of $x^2 + ax + a^2 + p$ is $D = a^2 - 4(a^2 + p) = -(3a^2 + 4p)$; the assumption $a^2 \\le -\\frac{4}{3}p$ implies $D \\ge 0$. Hence there are real numbers $b$ and $c$ such that $x^2 + a... | Estonia | Final Round of National Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0fwi | Problem:
Bestimme alle natürlichen Zahlen $n$, für die genau eine ganze Zahl $a$ mit $0 < a < n!$ existiert, sodass gilt
$$
n! \mid a^{n} + 1
$$ | [
"Solution:\n\nOffensichtlich ist $n = 2$ eine Lösung. Für $n \\geq 4$ ist $n!$ durch $4$ teilbar, und aus $n! \\mid a^{n} + 1$ folgt daher $a^{n} \\equiv 3 \\pmod{4}$. Wegen $a^{2} \\not\\equiv 3 \\pmod{4}$ ist $n$ also ungerade.\n\nFür jede ungerade natürliche Zahl $n$ ist $a = n! - 1$ eine Lösung, denn es gilt $a... | Switzerland | IMO Selektion | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory ... | null | proof and answer | All prime numbers | |
0ceb | Let $G = \{f : [0, 1] \to [0, 1] \mid f \text{ is one to one and continuous}\}$, and $\circ$ be the function composition; $(G, \circ)$ is a group.
a) Give an example of an infinite subgroup $H$ of $G$, which contains nonincreasing functions and $H \neq G$.
b) Let $H$ be a finite subgroup of $G$. Prove that $H$ has at... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
096u | Problem:
Să se afle toate perechile $(x, y)$ de numere naturale, care satisfac ecuaţia
$$
x^{2}-6 x y+8 y^{2}+5 y-5=0
$$ | [
"Solution:\nEcuaţia se ordonează ca o ecuaţie de gradul 2 în raport cu necunoscuta $x$ :\n$$\nx^{2}-6 y \\cdot x+\\left(8 y^{2}+5 y-5\\right)=0\n$$\nDiscriminantul ei, $\\Delta=4 y^{2}-20 y+20$ trebuie să fie un pătrat perfect. Fie $4 y^{2}-20 y+20=k^{2}$; pentru $k$ sunt suficiente valorile naturale. Urmează $(2 y... | Moldova | A 63-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (2,1), (4,1), (11,4), (13,4) | |
0j7v | Problem:
Let $ABC$ be a triangle such that $AB = 7$, and let the angle bisector of $\angle BAC$ intersect line $BC$ at $D$. If there exist points $E$ and $F$ on sides $AC$ and $BC$, respectively, such that lines $AD$ and $EF$ are parallel and divide triangle $ABC$ into three parts of equal area, determine the number o... | [
"Solution:\n\nAnswer: 13\n\n\n\nNote that such $E, F$ exist if and only if\n$$\n\\frac{[ADC]}{[ADB]} = 2\n$$\n([] denotes area.) Since $AD$ is the angle bisector, and the ratio of areas of triangles with equal height is the ratio of their bases,\n$$\n\\frac{AC}{AB} = \\frac{DC}{DB} = \\frac... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof and answer | 13 | |
07nm | The orthocentre $H$ of triangle $ABC$ is reflected in each of the three sides of the triangle, giving points $D$, $E$ and $F$.
Prove that $H$ is the incentre of triangle $DEF$. | [
"Construct the circumcircle of $\\triangle ABC$. Let $D'$, $E'$ and $F'$ denote the points where the altitudes meet the circumcircle. Let $K$ be the intersection of $AD'$ with $BC$ and denote the orthocentre of $\\triangle ABC$ by $H$. Then $\\angle BAD' = 90^\\circ - \\angle ABC = \\angle BCF$ and $\\angle BAD' = ... | Ireland | Ireland | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jr4 | Problem:
Consider an $8 \times 8$ grid of squares. A rook is placed in the lower left corner, and every minute it moves to a square in the same row or column with equal probability (the rook must move; i.e. it cannot stay in the same square). What is the expected number of minutes until the rook reaches the upper righ... | [
"Solution:\n\nLet the expected number of minutes it will take the rook to reach the upper right corner from the top or right edges be $E_{e}$, and let the expected number of minutes it will take the rook to reach the upper right corner from any other square be $E_{c}$. Note that this is justified because the expect... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 70 | |
08qt | Problem:
Find all pairs $(p, q)$ of prime numbers such that
$$
1+\frac{p^{q}-q^{p}}{p+q}
$$
is a prime number. | [
"Solution:\nIt is clear that $p \\neq q$. We set\n$$\n1+\\frac{p^{q}-q^{p}}{p+q}=r\n$$\nand we have that\n$$\np^{q}-q^{p}=(r-1)(p+q)\n$$\nFrom Fermat's Little Theorem we have\n$$\np^{q}-q^{p} \\equiv -q \\quad(\\bmod p)\n$$\nSince we also have that\n$$\n(r-1)(p+q) \\equiv -r q - q \\quad(\\bmod p)\n$$\nfrom (3) we ... | JBMO | JBMO | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | (2,5) | |
03rs | Determine if there exists a convex polyhedron such that
(1) it has 12 edges, 6 faces and 8 vertices;
(2) it has 4 faces with each pair of them sharing a common edge of the polyhedron. | [
"The answer is yes, as shown in the figure.\n\n"
] | China | China Girls' Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F"
] | English | final answer only | Yes | |
0eiz | Problem:
V katerih točkah na krivulji, podani $z$ enačbo $f(x)=x^{3}-2 x^{2}+3$, tangenta $z$ abscisno osjo oklepa kot $135^{\circ}$?
(A) $T_{1}(1,-4)$ in $T_{2}(-2,0)$.
(B) $T_{1}(1,-4)$ in $T_{2}(2,0)$.
(C) $T_{1}(1,0)$ in $T_{2}(-1,4)$.
(D) $T_{1}(1,2)$ in $T_{2}\left(\frac{1}{3}, \frac{28}{9}\right)$
(E) $T_{1}(1... | [
"Solution:\n\nVrednost odvoda funkcije $f$ v iskanih točkah mora biti enaka tangensu naklonskega kota tangente $v$ teh točkah $f'(x)=\\tan 135^{\\circ}=-1$. Rešimo enačbo $3 x^{2}-4 x=-1$. Enačbo uredimo in izračunamo abscisi iskanih točk $x_{1}=1$ in $x_{2}=\\frac{1}{3}$. Izračunamo funkcijski vrednosti $f\\left(x... | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | null | MCQ | E | |
01cv | Is it true that for any real numbers $a$, $b$, $c$ and $d$ satisfying $a^2 + b^2 + (a-b)^2 = c^2 + d^2 + (c-d)^2$ also the equality
$$
a^3 + b^3 + (a-b)^3 = c^3 + d^3 + (c-d)^3
$$
$$
a^4 + b^4 + (a-b)^4 = c^4 + d^4 + (c-d)^4
$$
holds? | [
"a) No, for example, if $a = b = 7$, $c = 8$ and $d = 3$ then\n$$\n7^2 + 7^2 + 0^2 = 98 = 8^2 + 3^2 + 5^2,\n$$\nbut\n$$\n7^3 + 7^3 + 0^3 = 686 \\neq 664 = 8^3 + 3^3 + 5^3.\n$$\n\nb) Yes, because\n$$\n(a^2 + b^2 + (a-b)^2)^2 = 2(a^4 + b^4 + (a-b)^4)\n$$\n(this is verified by simple algebra)."
] | Baltic Way | Baltic Way 2016 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Part (a): No. Part (b): Yes. | |
019w | Let $k$ and $n$ be two integers satisfying $1 \le k < n$. Consider $kn + 1$ rooks placed on an $n \times n$-chessboard. Prove that among them one may find $k + 1$ rooks no two of which attack each other. | [
"Let us first of all consider the case $k=1$. Now $n \\ge 2$, there are $n+1$ rooks on an $n \\times n$-chessboard, and we are to prove that some pair of them does not attack each other. Observe that the box principle tells us that there has to be some column $C$ containing at least two rooks. As $C$ consists of $n... | Baltic Way | Baltic Way 2013 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
06cy | Let $ABC$ be an acute triangle. Suppose a circle $\Gamma_1$, with centre $O_1$, touches the sides $BC$ produced at $E$, $AC$ produced at $G$, and $AB$ at $C'$. Suppose also that another circle $\Gamma_2$, with centre $O_2$, touches the sides $AB$ produced at $H$, $BC$ produced at $F$, and $AC$ at $B'$. Let the extensio... | [
"Let $EC'$ meet $FB'$ at $D$. Note that $EC'$ is parallel to the internal angle bisector of $\\angle CBA$, which is $BO_2$. Therefore, $ED \\perp PF$. Similarly, $FD \\perp PE$. This implies $D$ is the orthocentre of $\\triangle PEF$, and hence $PD \\perp EF$. It suffices to show $P, A, D$ are collinear, since this... | Hong Kong | HKG TST | [
"Geometry > Plane Geometry > Circles > Tangents",
"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 > Miscellaneous > Angle chasing"
... | null | proof only | null | |
08n6 | Problem:
Inside the square $A B C D$, the equilateral triangle $\triangle A B E$ is constructed. Let $M$ be an interior point of the triangle $\triangle A B E$ such that $M B=\sqrt{2}$, $M C=\sqrt{6}$, $M D=\sqrt{5}$ and $M E=\sqrt{3}$. Find the area of the square $A B C D$. | [
"Solution:\n\nLet $K, F, H, Z$ be the projections of point $M$ on the sides of the square.\nThen by Pythagorean Theorem we can prove that $M A^2 + M C^2 = M B^2 + M D^2$.\nFrom the given condition we obtain $M A = 1$.\nWith center $A$ and angle $60^{\\circ}$, we rotate $\\triangle A M E$, so we construct the triang... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 3 + sqrt(6) | |
0afb | Во равенството $25!=15\ 511 \times 10\ 043\ 330\ у85\ 984\ z00\ 000$ определи ги цифрите $x,y$ и $z$ за да тоа е точно. | [
"По дефиниција $25!=1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdots 25$. Ако овој број го разложиме на прости множители (направиме негова канонична факторизација), се добива:\n$$\n25! = 2^{22} \\cdot 3^{10} \\cdot 5^6 \\cdot 7^3 \\cdot 11^2 \\cdot 13 \\cdot 17 \\cdot 23 = 10^6 \\cdot 2^{16} \\cdot 3^{10} \\cdot 7^3 \... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Macedonian, English | proof and answer | x=2, y=9, z=0 | |
0je5 | Problem:
Define the sequence of positive integers $\{a_n\}$ as follows. Let $a_1=1$, $a_2=3$, and for each $n>2$, let $a_n$ be the result of expressing $a_{n-1}$ in base $n-1$, then reading the resulting numeral in base $n$, then adding $2$ (in base $n$). For example, $a_2=3_{10}=11_2$, so $a_3=11_3+2_3=6_{10}$. Expre... | [
"Solution:\n\nAnswer: 23097\n\nWe claim that for nonnegative integers $m$ and for $0 \\leq n < 3 \\cdot 2^m$, $a_{3 \\cdot 2^m + n} = (3 \\cdot 2^m + n)(m+2) + 2n$. We will prove this by induction; the base case for $a_3 = 6$ (when $m=0$, $n=0$) is given in the problem statement.\n\nNow, suppose that this is true f... | United States | HMMT 2013 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | final answer only | 23097 | |
078p | Let $x_1, x_2, \dots, x_{2024}$ be non-negative real numbers such that $x_1 \le x_2 \le \dots \le x_{2024}$, and $x_1^3 + x_2^3 + \dots + x_{2024}^3 = 2024$. Prove that
$$
\sum_{1 \le i < j \le 2024} (-1)^{i+j} x_i^2 x_j \ge -1012.
$$ | [
"We want that\n$$\n\\sum_{1 \\le i < j \\le 2024} (-1)^{i+j} x_i^2 x_j \\ge -1012.\n$$\nNow, observe that the LHS is\n$$\n- \\left( \\sum_{i=1}^{1012} x_{2i-1}^2 x_{2i} \\right) + \\sum_{i=1}^{1012} \\left( (x_{2i}^2 - x_{2i-1}^2) \\left( \\sum_{j<i \\le 1012} (x_{2j} - x_{2j-1}) \\right) \\right).\n$$\nBut, $x_{2i... | India | IMO TST | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
094b | Problem:
Let $\mathbb{R}$ denote the set of all real numbers. For each pair $(\alpha, \beta)$ of nonnegative real numbers subject to $\alpha+\beta \geq 2$, determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying
$$
f(x) f(y) \leq f(x y)+\alpha x+\beta y
$$
for all real numbers $x$ and $y$. | [
"Solution:\nWe know $f(x) f(y) \\leq f(x y)+\\alpha x+\\beta y$ and by exchanging $x$ and $y$ we get $f(x) f(y) \\leq f(x y)+\\beta x+\\alpha y$. Combining the two we get\n$$\nf(x) f(y) \\leq f(x y)+\\gamma x+\\gamma y\n$$\nwhere $\\gamma=\\frac{\\alpha+\\beta}{2}$. Notice that $\\gamma \\geq 1$.\n\nSetting $x=y=-1... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | The only case with a solution is when the parameters both equal one, in which case the unique function is f of x equals x plus one. For any other parameter pair there is no function. | |
06yc | Let $n$ be a positive integer. Given an $n \times n$ board, the unit cell in the top left corner is initially coloured black, and the other cells are coloured white. We then apply a series of colouring operations to the board. In each operation, we choose a $2 \times 2$ square with exactly one cell coloured black and w... | [
"Now we prove that if such a colouring is possible for $n$ then $n$ must be a power of 2. Suppose it is possible to colour an $n \\times n$ board where $n>1$. Identify the top left corner of the board by $(0,0)$ and the bottom right corner by $(n, n)$. Whenever an operation takes place in a $2 \\times 2$ square cen... | IMO | IMO2024 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | n is a power of 2 | |
0kj5 | For real numbers $x$, let
$$P(x) = 1 + \cos(x) + i \sin(x) - \cos(2x) - i \sin(2x) + \cos(3x) + i \sin(3x),$$
where $i = \sqrt{-1}$. For how many values of $x$ with $0 \le x < 2\pi$ does $P(x) = 0$?
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4 | [] | United States | AMC 12 B | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | MCQ | A | |
0g9j | 令 $c$ 爲正整數。令 $a_1 = c$, 並遞迴定義
$$
a_{n+1} = a_n^3 - 4c \times a_n^2 + 5c^2 \times a_n + c.
$$
證明: 對於所有正整數 $n \ge 2$, 存在質數 $p$ 整除 $a_n$, 但對於任何 $i < n$, $p$ 都不整除 $a_i$.
Let $c \ge 1$ be an integer. Define a sequence of positive integers by $a_1 = c$ and
$$
a_{n+1} = a_n^3 - 4c \times a_n^2 + 5c^2 \times a_n + c
$$
for al... | [
"令 $x_0 = 0$ 且 $x_n = a_n/c$。易見 $x_1 = 1$, $x_2 = 2c^2 + 1$ 且\n$$\nx_{n+1} = c^2(x_n^3 - 4x_n^2 + 5x_n) + 1. \\quad (1)\n$$\n很明顯的, $x_n$ 為遞增數列。要證明原題, 我們僅需改為證明對數列 $x_n$ 成立即可。\n\n以下先證明三個引理：\n\n(1) 引理一：若 $i = j \\pmod m$, 則 $x_i = x_j \\pmod{x_m}$.\n此引理等價於 $x_{i+m} = x_i \\pmod{x_m}$。固定 $m$, 此時對 $i = 0$ 顯然成立; 而若 $x_{i... | Taiwan | 二〇一五數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | null | proof only | null | |
0jn4 | Problem:
Evan's analog clock displays the time $12:13$; the number of seconds is not shown. After 10 seconds elapse, it is still $12:13$. What is the expected number of seconds until $12:14$? | [
"Solution:\n\nAt first, the time is uniformly distributed between $12:13:00$ and $12:13:50$. After 10 seconds, the time is uniformly distributed between $12:13:10$ and $12:14:00$. Thus, it takes on average 25 seconds to reach $12:14$ (:00)."
] | United States | HMMT February 2015 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 25 | |
012s | Problem:
Let $\mathbb{Q}_{+}$ be the set of positive rational numbers. Find all functions $f: \mathbb{Q}_{+} \rightarrow \mathbb{Q}_{+}$ which for all $x \in \mathbb{Q}_{+}$ fulfil
(1) $f\left(\frac{1}{x}\right)=f(x)$
(2) $\left(1+\frac{1}{x}\right) f(x)=f(x+1)$ | [
"Solution:\n\nSet $g(x)=\\frac{f(x)}{f(1)}$. Function $g$ fulfils (1), (2) and $g(1)=1$. First we prove that if $g$ exists then it is unique. We prove that $g$ is uniquely defined on $x=\\frac{p}{q}$ by induction on $\\max (p, q)$. If $\\max (p, q)=1$ then $x=1$ and $g(1)=1$. If $p=q$ then $x=1$ and $g(x)$ is uniqu... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All solutions are f(p/q) = a · p · q for coprime integers p, q and a in positive rationals. | |
01p0 | Three of six segments (three sides and three medians of a triangle) are painted red, and three others are painted blue.
Can one construct a triangle using the segments of the same color as its sides? | [
"Answer: yes, one can.\n\nLet $G$ be a gravicenter of the triangle $ABC$, and $A_1$, $B_1$, $C_1$ be the midpoints of the sides $BC$, $AC$, $AB$ respectively. Denote the sides and the medians of the triangle $ABC$ in the following way: $AB = c$, $BC = a$, $CA = b$, $AA_1 = d$, $BB_1 = e$, $CC_1 = f$.\n\nSuppose tha... | Belarus | BelarusMO 2013_s | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof and answer | yes | |
08aa | Problem:
Un cavallo è posto in una casella d'angolo di una scacchiera $3 \times 3$. Una mossa consiste nello spostare il cavallo in una casella raggiungibile mediante due passi in orizzontale seguiti da un passo in verticale, o due passi in verticale seguiti da un passo in orizzontale. In quanti modi è possibile spost... | [
"Solution:\n\nLa risposta è 992. Osserviamo che il cavallo si muoverà sempre su caselle adiacenti al perimetro della scacchiera (tutte tranne quella centrale) e che da ogni casella sono possibili due mosse: una che porta il cavallo avanti di 3 caselle sul perimetro in senso orario, e una che lo sposta di 3 caselle ... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 992 | |
0imh | Problem:
Let $x$, $y$, $n$ be positive integers with $n > 1$. How many ordered triples $(x, y, n)$ of solutions are there to the equation $x^{n} - y^{n} = 2^{100}$? | [
"Solution:\n\nAnswer: 49. Break all possible values of $n$ into the four cases: $n = 2$, $n = 4$, $n > 4$ and $n$ odd. By Fermat's theorem, no solutions exist for the $n = 4$ case because we may write $y^{4} + (2^{25})^{4} = x^{4}$.\n\nWe show that for $n$ odd, no solutions exist to the more general equation $x^{n}... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Pol... | null | proof and answer | 49 | |
04n3 | An $n \times n$ square board is given, where $n$ is an odd positive integer. Each of the $2n(n+1)$ unit segments delimiting the unit squares is coloured either red or blue. It is known that there are no more than $n^2$ red unit segments.
Prove that there is a unit square on the board whose border comprises at least thr... | [
"Assume the contrary, i.e. that there is no square bordered by three or four blue segments. Then each square is bordered by at least two red segments.\n\nNow we count the pairs $(P, r)$, where $P$ is a unit square, and $r$ is a red segment adjacent to $P$. We will count them in two different ways.\n\nSince every sq... | Croatia | Croatia_2018 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
03xk | Prove that equation $2x^3 + 5x - 2 = 0$ has exactly one real root (denoted as $r$), and there is a unique strictly increasing sequence $\{a_n\}$ such that $\frac{2}{5} = r^{a_1} + r^{a_2} + r^{a_3} + \dots$. | [
"Let $f(x) = 2x^3 + 5x - 2$. Then we have $f'(x) = 6x^2 + 5 > 0$, which means $f(x)$ is strictly increasing. Furthermore, $f(0) = -2 < 0$, $f(\\frac{1}{2}) = \\frac{3}{4} > 0$. Therefore, $f(x)$ has a unique real root $r \\in (0, \\frac{1}{2})$. From $2r^3 + 5r - 2 = 0$, we have\n$$\n\\frac{2}{5} = \\frac{r}{1 - r^... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | The unique sequence is a_n = 3n − 2 for n ≥ 1; the real root r is the unique solution in the interval between zero and one half of the equation 2x^3 + 5x − 2 = 0. | |
0c4o | The differentiable function $f : (0, \infty) \to \mathbb{R}$ is such that the limit $\lim_{x \to \infty} f'(x)$ exists and $x(f(x+1) - f(x)) = f(x)$, $\forall x > 0$. Prove that $f(x) = ax$, $\forall x \in (0, \infty)$, for some real number $a$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof only | null | |
09s7 | Problem:
Gegeven zijn positieve gehele getallen $r$ en $k$ en een oneindige rij positieve gehele getallen $a_{1} \leq a_{2} \leq \ldots$ zodat $\frac{r}{a_{r}}=k+1$. Bewijs dat er een $t$ is met $\frac{t}{a_{t}}=k$. | [
"Solution:\n\nWe bewijzen dit uit het ongerijmde. Stel dat zo'n $t$ niet bestaat. Als $a_{k}=1$, dan zou $\\frac{k}{a_{k}}=k$, tegenspraak met onze aanname. Dus $a_{k} \\geq 2$. We bewijzen nu met inductie naar $i$ dat $a_{i k} \\geq i+1$. De inductiebasis hebben we zojuist gedaan. Stel nu dat voor zekere $i \\geq ... | Netherlands | Selectietoets | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0fo4 | Problem:
Sean $B$ y $C$ dos puntos fijos de una circunferencia de centro $O$, que no sean diametralmente opuestos. Sea $A$ un punto variable sobre la circunferencia, distinto de $B$ y $C$, y que no pertenece a la mediatriz de $BC$. Sean $H$, el ortocentro del triángulo $ABC$; y $M$ y $N$ los puntos medios de los segmen... | [
"Solution:\nEmpezaremos considerando el caso en que $\\triangle ABC$ es acutángulo. En primer lugar, denotaremos por $A'$ el punto diametralmente opuesto a $A$ con lo que los triángulos $ACA'$ y $ABA'$ son rectángulos. Los segmentos $HB$ y $CA'$ son paralelos por ser perpendiculares a $AC$. Igualmente, $HC$ y $BA'$... | Spain | L Olimpiada matemática Española (Concurso Final) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a55 | Problem:
Let $\triangle ABC$ be an acute triangle with $AB > AC$. Let $P$ be the foot of the altitude from $C$ to $AB$ and let $Q$ be the foot of the altitude from $B$ to $AC$. Let $X$ be the intersection of $PQ$ and $BC$. Let the intersection of the circumcircles of triangle $\triangle AXC$ and triangle $\triangle PQC... | [
"Solution:\nLet $Z$ be the point where $PY$ intersects $AX$. The problem asks us to prove that $AZ = ZX$.\n\n\n\nSince $\\angle BPC = \\angle BQC = 90^\\circ$ we conclude that $BPQC$ is a cyclic quadrilateral. Hence $BPYQC$ is a cyclic pentagon. A quick angle chase gives us:\n\n$$\n\\angle ... | New Zealand | NZMO Round One | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | null | proof only | null | |
0bup | Problem:
Fie $A \in M_{2}(\mathbb{C})$ și $A^{2}-3 A+5 I_{2}=O_{2}$.
a) Aflați inversa matricei $A$.
b) Calculați $\operatorname{det}\left(A^{2}-I_{2}\right)+\operatorname{det}\left(A^{2}+A\right)-\operatorname{det}\left(A^{2}+2 I_{2}\right)$. | [] | Romania | OLIMPIADA DE MATEMATICĂ ETAPA LOCALĂ | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof and answer | A^{-1} = (3 I_2 - A)/5; the value equals 45 | |
0cfm | The path from Little Red Riding Hood to Grandma has ups, downs and flat portions. When walking, LRRH goes up two times slower and goes down two times faster than on flat ground. When riding her bike, she goes up three times slower and goes down three times faster than on flat ground.
LRRH noticed that, when walking, t... | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 60% | |
0188 | Let $f: \mathbb{Z} \to \mathbb{Z}$ be a function such that, for all integers $x$ and $y$, the following holds:
$$
f(f(x) - y) = f(y) - f(f(x)).
$$
Show that $f$ is bounded, ie. that there is a $C$ such that
$$
-C < f(x) < C
$$
for all $x$. | [
"First, setting $y = f(x)$ one obtains $f(0) = 0$. Secondly $y = 0$ yields $f(f(x)) = 0$ for all $x$, thus\n$$\nf(f(x) - y) = f(y).\n$$\nSetting $x = 0$ yields $f(-y) = f(y)$, and finally $y := -z$ yields\n$$\nf(f(x) + z) = f(-z) = f(z).\n$$\nIf $f(x) = 0$ for all $x$, then $f$ is obviously bounded. If on the other... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof only | null | |
018o | There are $2011$ people in a city. For some period of time every day a group of at least $4$ people went to a restaurant to have dinner. No group of $3$ people went together to more than one dinner. Prove that there exists a group of $24$ people such that at every dinner there was a person not belonging to this group. | [
"We can assume that at every dinner there were *exactly* $4$ people (just remove the surplus people from every dinner, which does not affect the condition that no group of $3$ people went together to two different dinners, and can only make the task of finding a suitable $24$-people group harder).\n\nConsider a gro... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
07rw | Let $p$, $q$ be real numbers. Show there exists $1 \le x \le 4$ such that
$$
|px + q + \frac{8}{x}| \ge 1
$$ | [
"We prove a stronger statement, that $|px + q + \\frac{8}{x}| \\ge 1$ for one of $x \\in \\{1, 2, 4\\}$. Let us write:\n$$\nf(x) = px + q + \\frac{8}{x}\n$$\nWe compare $f(2)$ to the linear interpolation of $f(1)$ and $f(4)$, which is:\n$$\n\\begin{aligned}\n\\frac{2}{3}f(1) + \\frac{1}{3}f(4) &= \\left(\\frac{2}{3... | Ireland | Irish | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
08oa | Problem:
Let $a$, $b$, $c$ be positive real numbers. Prove that
$$
\left(\left(3 a^{2}+1\right)^{2}+2\left(1+\frac{3}{b}\right)^{2}\right)\left(\left(3 b^{2}+1\right)^{2}+2\left(1+\frac{3}{c}\right)^{2}\right)\left(\left(3 c^{2}+1\right)^{2}+2\left(1+\frac{3}{a}\right)^{2}\right) \geq 48^{3}
$$
When does equality hold? | [
"Solution:\nLet $x$ be a positive real number. By AM-GM we have $\\frac{1+x+x+x}{4} \\geq x^{\\frac{3}{4}}$, or equivalently $1+3 x \\geq 4 x^{\\frac{3}{4}}$. Using this inequality we obtain:\n$$\n\\left(3 a^{2}+1\\right)^{2} \\geq 16 a^{3} \\text{ and } 2\\left(1+\\frac{3}{b}\\right)^{2} \\geq 32 b^{-\\frac{3}{2}}... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Equality holds when a = b = c = 1. | |
0919 | Problem:
All positive divisors of a positive integer $N$ are written on a blackboard. Two players $A$ and $B$ play the following game taking alternate moves. In the first move, the player $A$ erases $N$. If the last erased number is $d$, then the next player erases either a divisor of $d$ or a multiple of $d$. The pla... | [
"Solution:\n\nLet $N = p_1^{a_1} p_2^{a_2} \\ldots p_k^{a_k}$ be the prime factorization of $N$. In an arbitrary move the players write down a divisor of $N$, which we can represent as a sequence $(b_1, b_2, \\ldots, b_k)$, where $b_i \\leq a_i$ (such a sequence represents the number $p_1^{b_1} p_2^{b_2} \\ldots p_... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All positive integers that are perfect squares | |
0hq0 | Problem:
Find all the values of $m$ for which the zeros of $2x^{2} - m x - 8$ differ by $m-1$. | [
"Solution:\n6, $-\\frac{10}{3}$."
] | United States | null | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 6, -10/3 | |
01o9 | Pirate Bob has 14 silver, 15 gold, and 16 platinum coins, and Pirate Bill has 16 silver, 15 gold, and 14 platinum coins. From time to time they exchange their coins using the following rule: one of the pirates gives to the other pirate two coins of the same metal and instead of them gets two coins from the other two me... | [
"Let $(S, G, P)$ be the set of gold, silver and platinum coins of Bill at some moment. The initial set is $(16, 15, 14)$. Note that Bob and Bill have together $30$ gold, $30$ silver and $30$ platinum coins. So, $S \\le 30$, $G \\le 30$, and $P \\le 30$ at any moment. By condition, $S + G + P = 16 + 15 + 14 = 45$ at... | Belarus | Belorusija 2012 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 17, 20, 23, 26, 29 | |
07qv | Does there exist an even positive integer $n$ for which $n + 1$ is divisible by 5 and the two numbers $2^n + n$ and $2^n - 1$ are co-prime? | [
"Because $(2^n + n) - (2^n - 1) = n + 1$, we have $\\text{gcd}(2^n + n, 2^n - 1) = \\text{gcd}(n + 1, 2^n - 1)$.\nFrom $2^2 \\equiv 4 \\pmod{5}$, $2^3 \\equiv 3 \\pmod{5}$ and Fermat's Little Theorem we see that $2^n \\equiv 1 \\pmod{5}$ iff $n$ is divisible by 4. Hence, when $n \\equiv -1 \\pmod{5}$ and $n \\equiv... | Ireland | Ireland_2017 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof and answer | 34 | |
0kj9 | Right triangle $ABC$ has side lengths $BC = 6$, $AC = 8$, and $AB = 10$. A circle centered at $O$ is tangent to line $BC$ at $B$ and passes through $A$. A circle centered at $P$ is tangent to line $AC$ at $A$ and passes through $B$. What is $OP$?
(A) $\frac{23}{8}$ (B) $\frac{29}{10}$ (C) $\frac{35}{12}$ (D) $\frac{73}... | [
"**Answer (C):** More generally, let $a = BC$, $b = AC$, and $c = AB$ where $a < b$; then $c = \\sqrt{a^2 + b^2}$. Because $AB$ is a chord of both circles, their centers $O$ and $P$ must lie on the perpendicular bisector of $AB$. Letting $M$ be the midpoint of $AB$, observe that $\\triangle OMB$ is a right triangle... | United States | AMC 12 B | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | C | |
0iwi | Problem:
$$
\sum_{n=2009}^{\infty} \frac{1}{\binom{n}{2009}}
$$
Note that $\binom{n}{k}$ is defined as $\frac{n!}{k!(n-k)!}$. | [
"Solution:\n\nAnswer: $\\frac{2009}{2008}$\n\nObserve that\n$$\n\\begin{aligned}\n\\frac{k+1}{k}\\left(\\frac{1}{\\binom{n-1}{k}}-\\frac{1}{\\binom{n}{k}}\\right) & = \\frac{k+1}{k} \\frac{\\binom{n}{k}-\\binom{n-1}{k}}{\\binom{n}{k}\\binom{n-1}{k}} \\\\\n& = \\frac{k+1}{k} \\frac{\\binom{n-1}{k-1}}{\\binom{n}{k}\\... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | final answer only | 2009/2008 | |
09fe | Let $a$, $b$, $c$ and $d$ be non-negative real numbers satisfying $a + b + c + d = 2$. Prove that
$$
(a^2 + b^2)(b^2 + c^2)(c^2 + d^2)(d^2 + a^2) \le 1.
$$ | [
"Since $a$, $b$, $c$, $d \\ge 0$, we see that\n$$\n(a+b)^2 \\ge a^2 + b^2, \\quad (b+c)^2 \\ge b^2 + c^2, \\quad (c+d)^2 \\ge c^2 + d^2, \\quad (d+a)^2 \\ge d^2 + a^2.\n$$\n\nTherefore it is enough to prove that $(a+b)(b+c)(c+d)(d+a) \\le 1$. However, by the Cauchy-Schwartz inequality, we have\n$$\n(a+b)(b+c)(c+d)(... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
02kj | Problem:
A festa de aniversário de André tem menos do que 120 convidados. Para o jantar, ele pode dividir os convidados em mesas completas de 6 pessoas ou em mesas completas de 7 pessoas. Nos dois casos são necessárias mais do que 10 mesas e todos os convidados ficam em alguma mesa. Quantos são os convidados? | [
"Solution:\n\nComo podemos repartir o total de convidados em mesas de 6 ou 7, o número de convidados é um múltiplo de 6 e de 7. Como o menor múltiplo comum de 6 e 7 é $42$, podemos ter $42, 84, 126, \\ldots$ convidados. Como são menos do que $120$ convidados, só podemos ter $42$ ou $84$ convidados. Por outro lado, ... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | 84 | |
0k5w | Problem:
Let $d$ be a real number such that every non-degenerate quadrilateral has at least two interior angles with measure less than $d$ degrees. What is the minimum possible value for $d$? | [
"Solution:\n\nThe sum of the internal angles of a quadrilateral is $360^{\\circ}$. To find the minimum $d$, we note the limiting case where three of the angles have measure $d$ and the remaining angle has measure approaching zero. Hence, $d \\geq 360^{\\circ} / 3 = 120$. It is not difficult to see that for any $0 <... | United States | HMMT February 2019 February 16, 2019 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 120 | |
04b6 | One of the numbers $x^2$ and $(1-x)^2$ is smaller, and the other is greater than $1$.
Prove that $0 < x^2 - x < 2$. | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0d5p | Let $ABC$ be a triangle, $\Gamma$ its circumcircle, $I$ its incenter, and $\omega$ a tangent circle to the line $AI$ at $I$ and to the side $BC$. Prove that the circles $\Gamma$ and $\omega$ are tangent. | [
"Let $M$ be the midpoint of arc $BC$ not containing $A$, $E$ the tangent point of $BC$ to $\\omega$, $F$ the second intersection point of $EM$ with $\\omega$. Remember that $M$ is the circumcenter of triangle $BCI$ and therefore $MI = MB$.\n\n\n\nBecause $MI$ is tangent to $\\omega$, we hav... | Saudi Arabia | SAMC 2015 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English, Arabic | proof only | null | |
0ax7 | Problem:
A circle with radius $5$ is tangent to the $x$-axis, the $y$-axis, and the line $4x - 3y + 10 = 0$. Find its center. | [
"Solution:\nThe $x$ and $y$-intercepts of the given line are $-5/2$ and $10/3$, respectively. This means that the line and the coordinate axes determine a circle on the second quadrant, and so the center is at $(-5, 5)$"
] | Philippines | Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | final answer only | (-5, 5) | |
0cpz | Let $P(x)$ be a quadratic polynomial with a unit leading coefficient. Given that the polynomials $P(x)$ and $P(P(P(x)))$ have a common root, prove that $P(0)P(1) = 0$. (A. Khrabrov)
Квадратный трёхчлен $P(x)$ с единичным старшим коэффициентом таков, что многочлены $P(x)$ и $P(P(P(x)))$ имеют общий корень. Докажите, чт... | [
"Пусть $t$ — общий корень данных многочленов. Тогда $0 = P(P(P(t))) = P(P(0))$. Пусть $P(x) = x^2 + a x + b$; тогда $P(0) = b$, $P(1) = a + b + 1$, а значит, $0 = P(P(0)) = P(b) = a b + b^2 + b = b(a + b + 1) = P(0) \\cdot P(1)$, что и требовалось доказать."
] | Russia | Russian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English, Russian | proof only | null | |
0jpr | Problem:
Alice Czarina is bored and is playing a game with a pile of rocks. The pile initially contains $2015$ rocks. At each round, if the pile has $N$ rocks, she removes $k$ of them, where $1 \leq k \leq N$, with each possible $k$ having equal probability. Alice Czarina continues until there are no more rocks in the... | [
"Solution:\n\nAnswer: $-501$\n\nWe claim that\n$$\np = \\frac{1}{5} \\frac{6}{10} \\frac{11}{15} \\frac{16}{20} \\cdots \\frac{2006}{2010} \\frac{2011}{2015}.\n$$\nLet $p_n$ be the probability that, starting with $n$ rocks, the number of rocks left after each round is a multiple of $5$. Indeed, using recursions we ... | United States | HMMT February 2015 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | -501 | |
0ax0 | Problem:
The edges of a square are to be colored either red, blue, yellow, pink, or black. Each side of the square can only have one color, but a color may color many sides. How many different ways are there to color the square if two ways that can be obtained from each other by rotation are identical? | [] | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 165 | |
0d0y | Triangle $ABC$ is inscribed in circle $\omega$. Line $\ell$ is tangent to $\omega$ at $A$. Points $B_1$ and $C_1$ lie on $\ell$ such that rays $CA$ and $BA$ bisect $\widehat{BCB_1}$ and $\widehat{CBC_1}$, respectively. Segments $BB_1$ and $CC_1$ intersect at $P$. The line through $P$ parallel to segment $BC$ intersects... | [
"Assume that $P$ is the midpoint of segment $B_2C_2$; that is, $B_2P = C_2P$. We will show that $AB = AC$.\n\nSet $\\widehat{ABC} = B$, $\\widehat{BCA} = C$, and $\\widehat{CAB} = A$. Extend segment $AP$ through $P$ to meet segment $BC$ at $A_3$. Then it is clear $A_3$ is the midpoint of side $BC$ or $BA_3 = A_3C$.... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | English | proof only | null | |
088h | Problem:
Quanti sono i numeri interi positivi di 10 cifre $abcdefghij$, con tutte le cifre diverse e che verificano le condizioni $a+j=b+i=c+h=d+g=e+f=9$?
Nota: un numero non può iniziare con $0$.
(A) 3456
(B) 3528
(C) 3645
(D) 3840
(E) 5040. | [
"Solution:\n\nLa risposta è (A). Chiamiamo, come nel testo, $abcdefghij$ le 10 cifre del numero.\n\nPer i numeri della forma richiesta, fissare le prime 5 cifre $a, b, c, d, e$ determina univocamente tutto il numero per la condizione imposta (dato che possiamo ricavare $f=9-e$, $g=9-d$, $h=9-c$, $i=9-b$, $j=9-a$).\... | Italy | Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | MCQ | A | |
062s | Problem:
Gegeben sei ein Sehnenviereck $ABCD$, dessen Diagonalen $AC$ und $BD$ sich im Punkt $E$ schneiden und dessen Seiten $AD$ und $BC$ auf Geraden liegen, die sich im Punkt $F$ schneiden. Die Mittelpunkte der Strecken $AB$ und $CD$ seien mit $G$ bzw. $H$ bezeichnet. Man beweise, dass die Gerade $EF$ in $E$ den Kre... | [
"Solution:\n\nEine zentrische Streckung mit Zentrum $E$ und Streckfaktor $2$ bildet $G$ auf $G'$ und $H$ auf $H'$ ab. Dann ist\n\n(1) $\\Varangle GHE = \\Varangle G'H'E$.\n\nWegen $\\Varangle ABF = 180^{\\circ} - \\Varangle CBA = \\Varangle ADC$ und $\\Varangle FAB = 180^{\\circ} - \\Varangle BAD = \\Varangle DCB$ ... | Germany | IMO-Auswahlklausur | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09sx | Problem:
Gegeven zijn cirkels $\Gamma_{1}$ met middelpunt $A$ en $\Gamma_{2}$ met middelpunt $B$, waarbij $A$ op $\Gamma_{2}$ ligt. Op $\Gamma_{2}$ ligt verder een variabel punt $P$, niet op $A B$. Een lijn door $P$ die $\Gamma_{1}$ raakt in $S$, snijdt $\Gamma_{2}$ nogmaals in $Q$, waarbij $P$ en $Q$ aan dezelfde kan... | [
"Solution:\n\nOplossing I. Punt $P$ ligt buiten $\\Gamma_{1}$, omdat er anders geen raaklijn $P S$ aan $\\Gamma_{1}$ bestaat. Aangezien $P$ en $Q$ aan dezelfde kant van $A B$ liggen, ligt $S$ op het deel van $\\Gamma_{1}$ aan diezelfde kant van $A B$ dat buiten $\\Gamma_{2}$ ligt. (In het extreme geval dat $P$ op $... | Netherlands | IMO-selectietoets III | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | null | proof only | null |
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