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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
09rd | Problem:
Vind alle gehele getallen $n$ waarvoor $\frac{4 n-2}{n+5}$ het kwadraat van een rationaal getal is. (Een rationaal getal is een getal in $\mathbb{Q}$.) | [
"Solution:\n\nStel dat $\\frac{4 n-2}{n+5}$ het kwadraat van een rationaal getal is. Dan kunnen we het dus schrijven als $\\frac{p^{2}}{q^{2}}$ met $p, q$ niet-negatieve gehele getallen, $q \\neq 0$ en $\\operatorname{ggd}(p, q)=1$. Vanwege de voorwaarde $\\operatorname{ggd}(p, q)=1$, is $\\frac{p^{2}}{q^{2}}$ de m... | Netherlands | MO-selectietoets | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | n = 13 | |
08i3 | Problem:
Prove that for every positive integer $n$ the number $a = (2n+1)^5 - 2n - 1$ is divisible by $240$. | [
"Solution:\n\nLet $a = (2n+1)^5 - 2n - 1$.\n\nFirst, expand $(2n+1)^5$ using the binomial theorem:\n\n$$(2n+1)^5 = \\sum_{k=0}^5 \\binom{5}{k} (2n)^k (1)^{5-k} = 1 + 5 \\cdot 2n + 10 \\cdot (2n)^2 + 10 \\cdot (2n)^3 + 5 \\cdot (2n)^4 + (2n)^5$$\n\nCalculate each term:\n\n$1$\n\n$5 \\cdot 2n = 10n$\n\n$10 \\cdot (2n... | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0h5a | Prove the inequality for positive $a$, $b$, $c$, $d$:
$$
\left(\frac{a+b}{2c}\right)^2 + \left(\frac{b+c}{2d}\right)^2 + \left(\frac{c+d}{2a}\right)^2 + \left(\frac{d+a}{2b}\right)^2 \ge 4.
$$ | [
"Consecutively use the inequalities of means:\n$$\n\\begin{aligned}\n& \\left(\\frac{a+b}{2c}\\right)^2 + \\left(\\frac{b+c}{2d}\\right)^2 + \\left(\\frac{c+d}{2a}\\right)^2 + \\left(\\frac{d+a}{2b}\\right)^2 \n\\ge \\frac{ab}{c^2} + \\frac{bc}{d^2} + \\frac{cd}{a^2} + \\frac{da}{b^2} \n\\ge \\\\\n& \\ge 2\\sqrt{ab... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
02ju | Problem:
Camila e Lara têm, cada uma, um tabuleiro $4 \times 4$, inicialmente ambos em branco. Com estes tabuleiros elas fazem uma brincadeira do seguinte modo:
- Camila, escondida de Lara, pinta algumas casas de seu tabuleiro, de preto;
- Ainda em seu tabuleiro, Camila escreve em cada casa o número de casas vizinhas ... | [
"Solution:\n\nNotamos primeiro que se uma casa tem o algarismo $0$, então nenhuma das casas vizinhas pode estar pintada. Logo as casas marcadas com um $\\mathrm{X}$ na figura ao lado não foram pintadas:\n\nConsideremos agora a casa do canto superior direito, na qual aparece o número $1$. Ela tem $3$ vizinhas, e já ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | MCQ | B | |
0604 | Problem:
Montrer qu'il existe une infinité d'entiers $n$ tels que l'écriture de $n^{2}$ en base 4 ne contient que des 1 et des 2. | [
"Solution:\n\nOn montre un résultat un peu plus fort, qui se prête mieux à un raisonnement par récurrence : il existe une infinité d'entiers $n$ tels que l'écriture de $n^{2}$ en base 4 ne contient que des 1 et des 2, et tels que le premier et le dernier chiffre de l'écriture en base 4 de $n^{2}$ vaut 1. Un premier... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - Envoi 5 : Pot Pourri | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0dj5 | An acute scalene triangle $ABC$ with $\angle A = 45^\circ$ is inscribed in a circle ($O$). Let $AD$ be the altitude and $H$ be the orthocenter of $\triangle ABC$. Points $R$ and $S$ different from $A$ are taken on the rays $AB$ and $AC$ respectively so that $CA = CR$ and $BA = BS$. Points $X, Y$ and $K$ are the midpoin... | [
"Let $Z$, $M$ be the midpoints of $BR$, $BC$ respectively, then $ZY$ is the median of the triangle $BCR$ so $ZY = \\frac{BC}{2}$ and $ZY \\parallel BC$.\n\n\n\nIt is easy to see that $AH = 2OM = BC$ since triangle $OBC$ is right-angled at $O$. Hence $AK = \\frac{BC}{2} = ZY$. Let $L$ be the... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous >... | English | proof only | null | |
06yb | Let $N$ be a positive integer and let $a_{1}, a_{2}, \ldots$ be an infinite sequence of positive integers. Suppose that, for each $n > N$, $a_{n}$ is equal to the number of times $a_{n-1}$ appears in the list $a_{1}, a_{2}, \ldots, a_{n-1}$.
Prove that at least one of the sequences $a_{1}, a_{3}, a_{5}, \ldots$ and $a_... | [
"Let $M > \\max \\left(a_{1}, \\ldots, a_{N}\\right)$. We first prove that some integer appears infinitely many times. If not, then the sequence contains arbitrarily large integers. The first time each integer larger than $M$ appears, it is followed by a $1$. So $1$ appears infinitely many times, which is a contrad... | IMO | IMO2024 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
097y | Problem:
Să se rezolve în mulțimea numerelor reale ecuația
$$
\cos \frac{\pi}{x} = -x^{6} - 4 x^{5} + 2 x^{4} + 12 x^{3} - 9 x^{2} - 1
$$ | [
"Solution:\nAvem\n$$\n\\begin{gathered}\n1 \\geq -\\cos \\frac{\\pi}{x} = x^{6} + 4 x^{5} - 2 x^{4} - 12 x^{3} + 9 x^{2} + 1 = x^{2}\\left(x^{4} + 4 x^{3} - 2 x^{2} - 12 x + 9\\right) + 1 = \\\\\n= x^{2}\\left[\\left(x^{2} + 2 x\\right)^{2} - 6 x^{2} - 12 x + 9\\right] + 1 = x^{2}\\left[\\left(x^{2} + 2 x\\right)^{... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1 | |
0iqn | Problem:
Two mathematicians, Kelly and Jason, play a cooperative game. The computer selects some secret positive integer $n < 60$ (both Kelly and Jason know that $n < 60$, but they don't know what the value of $n$ is). The computer tells Kelly the unit digit of $n$, and it tells Jason the number of divisors of $n$. Th... | [
"Solution:\n\nThe only way in which Kelly can know that $n$ is divisible by at least two different primes is if she is given $0$ as the unit digit of $n$, since if she received anything else, then there is some number with that unit digit and not divisible by two primes (i.e., $1, 2, 3, 4, 5, 16, 7, 8, 9$). Then, a... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Logic",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 10 | |
04uf | Find the smallest positive integer $n$ such that for any coloring of numbers $1, 2, 3, \ldots, n$ by three colors there exist two numbers with the same color whose difference is a square of an integer. (Vojtech Bálint, Michal Rolínek, Josef Tkadlec) | [
"The answer is $n = 29$.\n\nFirst, for the sake of contradiction, assume that numbers $1, 2, \\ldots, 29$ can be colored by colors $A, B, C$ such that no two numbers with the same color differ by a square. Let $f(i)$ be the color of number $i$ for $i \\in \\{1, 2, \\ldots, 29\\}$.\n\nSince $9, 16$, and $25$ are squ... | Czech Republic | 67th Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 29 | |
01nq | Determine the greatest positive integer $k$ that satisfies the following property: The set of positive integers can be partitioned into $k$ subsets $A_1, \dots, A_k$ such that for all integers $n \ge 15$ and all $i \in \{1, 2, \dots, k\}$ there exist two distinct elements of $A_i$ whose sum is $n$. | [
"3. See IMO-2011 Shortlist, Problem C4."
] | Belarus | Belorusija 2012 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 2 | |
0ea2 | Problem:
Dokaži, da je za poljubno liho naravno število $n$ izraz $(n+2)^3 + 3(n+2)^2 - n - 5$ deljiv z $48$. | [
"Solution:\n\nNaj bo $n$ liho naravno število.\n\nRazpišimo izraz:\n\n$$(n+2)^3 + 3(n+2)^2 - n - 5$$\n\nNajprej razvijemo $(n+2)^3$ in $(n+2)^2$:\n\n$(n+2)^3 = n^3 + 3n^2 \\cdot 2 + 3n \\cdot 2^2 + 2^3 = n^3 + 6n^2 + 12n + 8$\n\n$3(n+2)^2 = 3(n^2 + 4n + 4) = 3n^2 + 12n + 12$\n\nSkupaj:\n\n$n^3 + 6n^2 + 12n + 8 + 3n... | Slovenia | 14. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
09zh | A box contains red, white and blue balls. The number of red balls is an even number and the total number of balls in the box is less than $100$. The number of white and blue balls together is exactly $4$ times the number of red balls. The number of red and blue balls together is exactly $6$ times the number of white ba... | [] | Netherlands | First Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | D | |
03eu | Maria and Bilyana play the following game. Maria has 2024, and Bilyana has 2023 fair coins. Coins are tossed randomly - the probability of each individual coin being heads after the toss is $\frac{1}{2}$. Maria wins if there are strictly more heads among her coins than Bilyana's, otherwise Bilyana wins. What is the pro... | [
"Let $p$ be the probability that Maria has more heads than Bilyana after tossing the first 2023 of Maria's coins. Then, for reasons of symmetry, the probability that Maria has fewer heads than Bilyana is also $p$, and therefore the probability that Maria and Bilyana have thrown an equal number of heads is $1 - 2p$.... | Bulgaria | Bulgarian Winter Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | English | proof and answer | 1/2 | |
08u4 | A positive integer with 2 digits was given. When it was multiplied by $7$ it became a $3$-digit number. When that $3$-digit number was multiplied again by $7$, it remained as a $3$-digit number. How many possibilities are there for the original $2$-digit number? | [
"Let $n$ be the given integer, then from the first condition we have $100 \\le 7n < 1000$, from which we get $15 \\le n < 142$. From the second condition, we have $100 \\le 7^2n < 1000$, which yields $3 \\le n \\le 20$. Therefore, we get $15 \\le n \\le 20$, and any integer $n$ satisfying the last inequalities sati... | Japan | Japan Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 6 | |
0biu | Let $\varepsilon$ be a positive real number. A positive integer will be called $\varepsilon$-squarish if it is the product of two integers $a$ and $b$ such that $1 < a < b < (1 + \varepsilon)a$. Prove that there are infinitely many occurrences of six consecutive $\varepsilon$-squarish integers. | [
"If $N$ is a large enough positive integer, then $N^2 - 1 = (N - 1)(N + 1)$ and $N^2 - 4 = (N - 2)(N + 2)$ are both $\\varepsilon$-squarish. Next, if $k$ is a large enough positive integer and $N = (k-1)(k+2) = k^2 + k - 2$, then $N^2 = (k-1)^2(k+2)^2$, $N^2 - 2 = (k^2 - 2)(k^2 + 2k - 1)$ and $N^2 - 5 = (k^2 - k - ... | Romania | THE 2014 Twelfth IMAR MATHEMATICAL COMPETITION | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
041t | There are $n$ ($n \ge 2$) coins in a row. If one of the coins is head, select an odd number of consecutive coins (or even 1 coin) with the one in head on the leftmost, and then flip all the selected coins upside down simultaneously. This is a *move*. No move is allowed if all $n$ coins are tails. Suppose $n$ coins are ... | [
"The answer is possible.\n\nFor any configuration of the coins, we define a corresponding 01-sequence $c_1c_2...c_n$ of length $n$ as follows: $c_i = 1$, if the $i$-th coin from the left is head, otherwise, $c_i = 0$. It is easy to see that the status of the $n$ coins has a one-to-one correspondence to such 01-sequ... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | possible | |
04m1 | A *crazy bishop* is an unorthodox chess piece which turns towards one of the four diagonally contiguous cells and attacks all cells directly in front, to the left, and to the right of it (just like a regular chess bishop which cannot see behind his back). We say that two cells of a board are *diagonally contiguous* if ... | [
"Imagine the figure of a crazy bishop like a dot from which we draw 3 arrows in directions of cells which are attacked by this crazy bishop. By the term *diagonal* we consider all the diagonals of the board, not just the main ones. The four corner squares are also considered diagonals (containing only one square).\... | Croatia | Croatian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 20 | |
0l27 | Problem:
A grid is called groovy if each cell of the grid is labeled with the smallest positive integer that does not appear below it in the same column or to the left of it in the same row. Compute the sum of the entries of a groovy $14 \times 14$ grid whose bottom left entry is $1$. | [
"Solution:\nThe following diagram is the entire $16 \\times 16$ groovy grid computed out. However, one will not need to write out every single entry to obtain the answer.\n\n| 16 | 15 | 14 | 13 | 12 | 11 | 10 | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :-... | United States | HMMT November 2024 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 1638 | |
0fwo | Problem:
Sei $A = (a_{1}, a_{2}, \ldots, a_{n})$ eine Folge ganzer Zahlen. Der Nachfolger von $A$ ist die Folge $A' = (a_{1}', a_{2}', \ldots, a_{n}')$ mit
$$
a_{k}' = \left|\{i < k \mid a_{i} < a_{k}\}\right| - \left|\{i > k \mid a_{i} > a_{k}\}\right|
$$
Sei $A_{0}$ eine endliche Folge ganzer Zahlen und für $k \geq ... | [
"Solution:\n\nWir nennen ein Paar $(i, j)$ von Indizes mit $i < j$ konstant bzw. steigend, falls $a_{i} = a_{j}$ bzw. $a_{i} < a_{j}$ gilt. Es gilt folgendes:\n\n(i) Ist $(i, j)$ steigend, dann bleibt es auch steigend. Denn gilt $a_{i} < a_{j}$, dann ist jedes Folgeglied rechts von $j$, das grösser als $a_{j}$ ist,... | Switzerland | IMO Selektion 2008 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
06sd | Let $n$ be a positive integer. Find the smallest integer $k$ with the following property: Given any real numbers $a_{1}, \ldots, a_{d}$ such that $a_{1}+a_{2}+\cdots+a_{d}=n$ and $0 \leqslant a_{i} \leqslant 1$ for $i=1,2, \ldots, d$, it is possible to partition these numbers into $k$ groups (some of which may be empty... | [
"Answer. $k=2n-1$.\n\nSolution 1. If $d=2n-1$ and $a_{1}=\\cdots=a_{2n-1}=n/(2n-1)$, then each group in such a partition can contain at most one number, since $2n/(2n-1)>1$. Therefore $k \\geqslant 2n-1$. It remains to show that a suitable partition into $2n-1$ groups always exists.\nWe proceed by induction on $d$.... | IMO | International Mathematical Olympiad Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 2n-1 | |
0g4o | Problem:
Let $n$ be a positive integer. Kimiko starts with $n$ piles of pebbles each containing a single pebble. She can take an equal number of pebbles from two existing piles and combine the removed pebbles to create a new pile. Determine, in terms of $n$, the smallest number of nonempty piles Kimiko can end up with... | [
"Solution:\n\nIf $n$ is a power of $2$ then there may be only one pile remaining; otherwise, there will be at least two piles remaining, but this can be attained. It is clear why you can reach one pile if $n$ is a power of $2$: the first $n/2$ piles can each receive pebbles from two piles with a pebble, the next $n... | Switzerland | Swiss Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | One if and only if the initial number of piles is a power of two; otherwise two. | |
0h5c | An acute-angled triangle $ABC$ has the side $BC > AB$, and the bisectrix $BL = AB$. On the segment $BL$ there exists point $M$, for which $\angle AML = \angle BCA$. Prove that $AM = LC$.
Fig. 16 | [
"On the segment $BC$ put a point such as $BD = BL$ (fig. 16). Then $\\Delta ABL = \\Delta BLD$, mark $\\angle LAB = \\angle BLA = \\angle BLD = \\angle BDL$. On the segment $BD$ choose a point $K$, such that $\\angle ALM = \\angle DLK$. Then $\\Delta ALM = \\Delta DLK$ because of $LD = AL$, but then $\\angle AML = ... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round (Second Tour) | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates"
] | English | proof only | null | |
0h42 | On the sides of an acute triangle $ABC$ three triangles $A'BC$, $AB'C$, $ABC'$ are constructed outside with $\angle ABC' = \angle A'BC = \angle B'AC = 30^\circ$ and $\angle BAC' = \angle AB'C = \angle A'CB = 90^\circ$. Prove that $A'C' \perp B'M$, where $M$ is the midpoint of $BC$. | [
"**Лема.** На сторонах гострокутного трикутника як на основах побудовано рівнобедрені трикутники $АХВ$, $BYC$, $CZA$, для яких $\\angle АХВ = \\angle ВҮС = 120^\\circ$, $\\angle СZA = 60^\\circ$. Тоді $XY \\perp BZ$.\n\n**Доведення.** Див. задачу 4 для 8 класу. $\\Box$\n\nПерейдемо до розв'язування задачі. Нехай $X... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
073k | Let $P(x) = x^2 + a x + b$ be a real polynomial with $a < 2$. Suppose $P(P(x)) = 0$ has four distinct real roots and the sum of some two of them is $\le -1$. Prove that $P(x+y) \ge P(x) + P(y)$ for all non-negative real numbers $x, y$. | [
"We have to consider two possibilities: $x_1 + x_2 \\le -1$ or $x_1 + x_3 \\le -1$. (Both the roots are from the same equation or each root coming from a different equation.) Suppose $x_1 + x_2 \\le -1$. Then $-a = x_1 + x_2$ shows that $a > 0$. Since the equations $P(x) = \\alpha$ and $P(x) = \\beta$ have distinct... | India | Indija TS 2008 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0a8y | Problem:
The circle $\mathcal{C}_1$ is inside the circle $\mathcal{C}_2$, and the circles touch each other at $A$. A line through $A$ intersects $\mathcal{C}_1$ also at $B$ and $\mathcal{C}_2$ also at $C$. The tangent to $\mathcal{C}_1$ at $B$ intersects $\mathcal{C}_2$ at $D$ and $E$. The tangents of $\mathcal{C}_1$ ... | [
"Solution:\n\n(See Figure 15.) Draw the tangent $CH$ to $\\mathcal{C}_2$ at $C$. By the theorem of the angle between a tangent and chord, the angles $ABH$ and $ACH$ both equal the angle at $A$ between $BA$ and the common tangent of the circles at $A$. But this means that the angles $ABH$ and $ACH$ are equal, and $C... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 19 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00vf | Let $f: \mathbb{R}^+ \to \mathbb{R}^+$ be a non-constant function. Prove that there exist $a, b \in \mathbb{R}^+$ such that
$$
f(a) + f(b) > 2f(\sqrt{ab}).
$$ | [
"Assume the contrary that for all $a, b \\in \\mathbb{R}^+$ we have that $f(a) + f(b) \\le 2f(\\sqrt{ab})$. Let $P(a, b)$ denote that assertion. First, $P(a, \\frac{1}{a})$ gives us $f(a) + f(\\frac{1}{a}) \\le 2f(1)$, and as $f$ is positive, we obtain that it is bounded. We will show by mathematical induction that... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlisted Problems | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof only | null | |
089e | Problem:
La quantità di inchiostro usato per comporre un testo per le Olimpiadi della Matematica segue una strana legge: negli anni dispari aumenta del $50\%$ rispetto all'anno precedente, negli anni pari diminuisce di un sesto (sempre rispetto all'anno precedente). Tra quanti anni per la prima volta sarà almeno il tr... | [
"Solution:\n\nLa risposta è $\\mathbf{( E )}$. Certamente la quantità di inchiostro supererà il triplo del 2012 per la prima volta in un anno dispari, giacché negli anni pari diminuisce. Se sono passati $2k+1$ anni, quindi, essa sarà diventata\n$$\n\\frac{3}{2} \\cdot \\left(\\frac{5}{6} \\cdot \\frac{3}{2}\\right)... | Italy | UNIONE MATEMATICA ITALIANA SCUOLA NORMALE SUPERIORE DI PISA GARA di SECONDO LIVELLO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | MCQ | E | |
0f6a | Problem:
$ABC$ is an acute angled triangle. The midpoints of $BC$, $CA$ and $AB$ are $D$, $E$, $F$ respectively. Perpendiculars are drawn from $D$ to $AB$ and $CA$, from $E$ to $BC$ and $AB$, and from $F$ to $CA$ and $BC$. The perpendiculars form a hexagon. Show that its area is half the area of the triangle. | [] | Soviet Union | 19th ASU | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | half the area of the triangle | |
0208 | Problem:
a) Find all positive integers $g$ with the following property: for each odd prime number $p$ there exists a positive integer $n$ such that $p$ divides the two integers
$$
g^{n}-n \quad \text{ and } \quad g^{n+1}-(n+1)
$$
b) Find all positive integers $g$ with the following property: for each odd prime number... | [
"Solution:\n\na.\nLet $g$ be a positive integer with the given property. So for each odd prime number $p$ there exists a positive integer $n$ such that $p \\mid g^{n}-n$ and $p \\mid g^{n+1}-(n+1)$.\nIf $g$ has an odd prime factor $p$, then from $p \\mid g^{n}-n$ it follows that $p \\mid n$, while from $p \\mid g^{... | Benelux Mathematical Olympiad | 5th Benelux Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | a) g = 2; b) g = 4 | |
014b | Problem:
162 pluses and 144 minuses are placed in a $30 \times 30$ table in such a way that each row and each column contains at most 17 signs. (No cell contains more than one sign.) For every plus we count the number of minuses in its row and for every minus we count the number of pluses in its column. Find the maxim... | [
"Solution:\n\nIn the statement of the problem there are two kinds of numbers: \"horizontal\" (that has been counted for pluses) and \"vertical\" (for minuses). We will show that the sum of numbers of each type reaches its maximum on the same configuration.\n\nWe restrict our attention to the horizontal numbers only... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2592 | |
035t | Problem:
The extensions of the sides $AB$ and $CD$ of a convex quadrilateral $ABCD$ meet at point $P$ and the extensions of the sides $BC$ and $AD$ meet at point $Q$. The point $O$ from the interior of the quadrilateral is such that $\Varangle BOP = \Varangle DOQ$. Prove that $\Varangle AOB + \Varangle COD = 180^\circ... | [
"Solution:\n\nThe Sine theorem for the triangles $ODQ$ and $AOQ$ gives\n$$\n\\frac{\\sin \\varphi}{\\sin \\beta} = \\frac{QD}{OD}\n$$\nand\n$$\n\\frac{\\sin (\\varphi + \\Varangle AOD)}{\\sin \\beta} = \\frac{AQ}{OA}\n$$\nwhence\n$$\n\\frac{\\sin (\\varphi + \\Varangle AOD)}{\\sin \\varphi} = \\frac{AQ}{OA} \\cdot ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05pc | Problem:
Soit $a \in [0 ; 1]$. On définit la suite $\left(x_{n}\right)$ par
$x_{0} = a$ et $x_{n+1} = 1 - \left|1 - 2 x_{n}\right|$, pour tout $n \geq 0$.
Prouver que la suite $\left(x_{n}\right)$ est périodique à partir d'un certain rang si et seulement si $a$ est un nombre rationnel.
(Note: on dit que $\left(x_{n}... | [
"Solution:\n\nOn note tout d'abord que si $x_{n} \\in [0 ; 1]$ alors $-1 \\leq 1 - 2 x_{n} \\leq 1$, d'où $0 \\leq x_{n+1} \\leq 1$. Puisque $x_{0} \\in [0 ; 1]$, on déduit ainsi par récurrence que $x_{n} \\in [0 ; 1]$ pour tout $n \\geq 0$.\n\n- Supposons que $a$ soit un nombre rationnel. Une récurrence immédiate ... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0avt | Problem:
Let $f(x)$ be a polynomial function of degree $2016$ whose $2016$ zeroes have a sum of $S$. Find the sum of the $2016$ zeroes of $f(2x-3)$ in terms of $S$. | [
"Solution:\nLet $r_{1}, r_{2}, \\cdots, r_{2016}$ be the zeroes of $f(x)$. We can then write $f$ as\n$$\nf(x) = c\\left(x - r_{1}\\right)\\left(x - r_{2}\\right) \\cdots \\left(x - r_{2016}\\right)\n$$\nwhere $\\sum_{i=1}^{2016} r_{i} = S$. Thus\n$$\nf(2x-3) = c\\left(2x-3 - r_{1}\\right)\\left(2x-3 - r_{2}\\right)... | Philippines | 18th PMO National Stage Oral Phase | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | S/2 + 3024 | |
0jng | Problem:
A group of friends, numbered $1,2,3, \ldots, 16$, take turns picking random numbers. Person 1 picks a number uniformly (at random) in $[0, 1]$, then person 2 picks a number uniformly (at random) in $[0, 2]$, and so on, with person $k$ picking a number uniformly (at random) in $[0, k]$. What is the probability ... | [
"Solution:\nAnswer: $\\frac{17^{15}}{16!^{2}}$\n\nSolution 1 (intuitive sketch). If person $i$ picks $a_{i}$, this is basically a continuous version of Catalan paths (always $y \\leq x$) from $(0,0)$ to $(17,17)$, with \"up-right corners\" at the $\\left(i, a_{i}\\right)$. A cyclic shifts argument shows that \"$\\f... | United States | HMMT February 2015 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | null | proof and answer | 17^{15} / 16!^{2} | |
0e2c | Let $a$, $b$ and $c$ be non-zero digits. Let $p$ be a prime number which divides the three-digit numbers $abc$ and $cba$. Show that $p$ divides at least one of the numbers $a+b+c$, $a-b+c$ and $a-c$. | [
"Since the prime number $p$ divides the numbers $\\overline{abc} = 100a + 10b + c$ and $\\overline{cba} = 100c + 10b + a$, it must also divide their difference\n$$\n\\overline{abc} - \\overline{cba} = 100(a-c) + (c-a) = 99(a-c).\n$$\nIf $p$ divides $a-c$ we are done. If not, then $p$ divides $99$, so $p=3$ or $p=11... | Slovenia | National Math Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
03lm | Problem:
Let $ABC$ be a triangle with circumradius $R$, perimeter $P$ and area $K$. Determine the maximum value of $K P / R^{3}$. | [
"Solution:\nSince similar triangles give the same value of $K P / R^{3}$, we can fix $R=1$ and maximize $K P$ over all triangles inscribed in the unit circle. Fix points $A$ and $B$ on the unit circle. The locus of points $C$ with a given perimeter $P$ is an ellipse that meets the circle in at most four points. The... | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Jensen/smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry ... | null | proof and answer | 27/4 | |
01pd | Find all pairs $(m, n)$ of positive integers $m$ and $n$ such that for all polynomial $P(x)$ with real coefficients and $\deg P(x) = m$ there exists a polynomial $Q(x)$ with real coefficients and $\deg Q(x) = n$ such that $Q(P(x))$ is divisible by $Q(x)$.
(A. Mirotin, S. Mazanik, I. Voronovich) | [
"Answer: all $(m, n)$ with odd $m$ and arbitrary $n$; or $(m, n)$ with even $m$ and even $n$.\n(Solution of A. Zhuk.)\n\n1. Let $m$ be odd. Show that any positive $n$ is appropriate.\nIndeed, if $P(x) \\equiv x$, then $Q(P(x)) \\not\\equiv Q(x)$ for any $Q \\in \\mathbb{R}[x]$.\nIf $P(x) \\not\\equiv x$, then $P(x)... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | English | proof and answer | All pairs with m odd and any positive n; or with m even and n even. | |
096x | Problem:
Rezolvaţi în $\mathbb{R}$ ecuaţia $2020^{x^{2}-2x} + \frac{x^{2}-2x}{2020^{x}} = 1$. | [
"Solution:\n1) Dacă $x^{2}-2x < 0$, atunci $2020^{x^{2}-2x} < 2020^{0} = 1$ și $\\frac{x^{2}-2x}{2020^{x}} < 0$. Prin urmare $2020^{x^{2}-2x} + \\frac{x^{2}-2x}{2020^{x}} < 1$, deci ecuația nu are soluții în acest caz.\n\n2) Dacă $x^{2}-2x > 0$, atunci $2020^{x^{2}-2x} > 2020^{0} = 1$ și $\\frac{x^{2}-2x}{2020^{x}}... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | {0, 2} | |
0fw9 | Problem:
Gegeben sei ein spitzwinkliges Dreieck $ABC$ und die Punkte $D$, $E$ und $F$ seien die Höhenfusspunkte der Höhen durch $A$, $B$ bzw. $C$. Sei $S$ der Schnittpunkt der Geraden $EF$ mit der Rechtwinkligen zu $AC$ durch $D$. Beweise, dass das Dreieck $DES$ gleichschenklig ist. | [
"Solution:\n\nSei $\\beta=\\angle ABC$. Wir zeigen, dass $\\triangle DES$ gleichschenklig ist, indem wir unabhängig voneinander (i) $\\angle ESD=90^{\\circ}-\\beta$ und (ii) $\\angle EDS=90^{\\circ}-\\beta$ beweisen (siehe Abbildung 1). Dazu benutzen wir die beiden Sehnenvierecke $BCEF$ (Thaleskreis über $BC$) und ... | Switzerland | Vorrundenprüfung | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dep | By $\text{rad}(x)$ we denote the product of all distinct prime factors of a positive integer $n$. Given $a \in \mathbb{N}$, a sequence $(a_n)$ is defined by $a_0 = a$ and $a_{n+1} = a_n + \text{rad}(a_n)$ for all $n \ge 0$. Prove that there exists an index $n$ for which $\frac{a_n}{\text{rad}(a_n)} = 2022$. | [
"*Solution.* To prove that $MP + NQ = BR$, by adding $RC$ to both sides it is equivalent to prove $CQ + MP = BC$. We start by defining point $T$ on the segment $CQ$ such that $BC = CT$, then it is sufficient to prove that $QT = PM$, since triangles $BTC$ and $NRC$ are isosceles. One can get that $BT$ is parallel to... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | English | proof only | null | |
03vw | Let $n$ be a positive integer, and $f(n)$ denote the number of $n$-digit integers $\overline{a_1a_2\cdots a_n}$ (called wave number) that satisfy the following conditions:
(i) $a_i \in \{1, 2, 3, 4\}$, and $a_i \neq a_{i+1}$, $i = 1, 2, \dots$;
(ii) When $n \ge 3$, the numbers $a_i - a_{i+1}$ and $a_{i+1} - a_{i+2}$ ha... | [
"(1) When $n \\ge 2$, if $a_1 < a_2 < \\overline{a_1a_2\\cdots a_n}$ is classified as A class. The number of $\\overline{a_1a_2\\cdots a_n}$ is denoted by $g(n)$. If $a_1 > a_2$, then $\\overline{a_1a_2\\cdots a_n}$ is classified as B class. By symmetry, the number of such $\\overline{a_1a_2\\cdots a_n}$ is also $g... | China | China Southeastern Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | f(10) = 8008; f(2008) ≡ 10 (mod 13) | |
09o6 | Prove that the polynomial $P(X) = X^5 + 61X + 2025$ is not the product of two non-constant polynomials with integer coefficients.
(Otgonbayar Uuye) | [
"Suppose, for contradiction, that $P(X)$ can be written as the product of two non-constant polynomials with integer coefficients. Since $P(X)$ is of degree $5$, the only possible degrees for the factors are $(1,4)$ or $(2,3)$.\n\nFirst, check if $P(X)$ has an integer root. If $a$ is an integer root, then $a^5 + 61a... | Mongolia | MMO2025 Round 2 | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0hff | There are $2022$ points on the circle, one of which is painted black, and the other $2021$ — white. In one move Hedgehog is allowed to do one of the following operations:
* repaint in the opposite color two consecutive points of the same color;
* repaint in the opposite colors two points of different colors, between wh... | [
"Let's enumerate the points in clockwise order (starting, for example, from the black one), by $1, 2, \\ldots, 2022$. Note that from the very beginning, there was one more black point in the positions with odd numbers than the black points in the positions with even numbers. After the first operation, we either add... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | No | |
0h07 | Do there exist pairwise distinct natural numbers $a_1, a_2, \dots, a_k$, greater than $1$, for which
$$
a_1 + a_2 + \dots + a_k = 2010 \cdot \left( \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_k} \right)
$$
a) if $k=2$;
b) if $k=12$. | [
"**Answer:** a) any pair of divisors of the number $2010$ with product $2010$, for example $a_1 = 2$, $a_2 = 1005$;\nb) $2$, $3$, $5$, $6$, $10$, $15$, $134$, $201$, $339$, $402$, $670$, $1005$.\n\nThe main idea is: if $n$ is not a square of an integer and $d$ is its divisor, $1 < d < n$, then $d' = \\frac{n}{d}$ i... | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | a) Yes: any two distinct divisors of 2010 whose product is 2010, for example 2 and 1005. b) Yes: for example 2, 3, 5, 6, 10, 15, 134, 201, 335, 402, 670, 1005. | |
0kth | Problem:
Call an ordered pair $(a, b)$ of positive integers fantastic if and only if $a, b \leq 10^{4}$ and
$$
\operatorname{gcd}(a \cdot n!-1, a \cdot(n+1)!+b)>1
$$
for infinitely many positive integers $n$. Find the sum of $a+b$ across all fantastic pairs $(a, b)$. | [
"Solution:\n\nWe first prove the following lemma, which will be useful later.\n\nLemma: Let $p$ be a prime and $1 \\leq n \\leq p-1$ be an integer. Then, $n!(p-1-n)!\\equiv(-1)^{n-1}(\\bmod p)$.\n\nProof. Write\n$$\n\\begin{aligned}\nn!(p-n-1)! & =(1 \\cdot 2 \\cdots n)((p-n-1) \\cdots 2 \\cdot 1) \\\\\n& \\equiv(-... | United States | HMMT November 2022 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 5183 | |
01op | Two parallel chords $AB$ and $CD$ are constructed on the hyperbola $y = 1/x$. The lines $AC$ and $BD$ meet ordinate axis $Oy$ at points $A_1$ and $D_1$ respectively, and meet abscissae axis $Ox$ at points $C_1$ and $B_1$ respectively.
Prove that the areas of the triangles $A_1OC_1$ and $D_1OB_1$ are equal. | [
"Let $A_1(0; a_1)$, $B_1(b_1; 0)$, $C_1(c_1; 0)$, $D_1(0; d_1)$, $A(a; 1/a)$, $B(b; 1/b)$, $C(c; 1/c)$, $D(d; 1/d)$ be the marked points. Since any vertical and any horizontal line meets the hyperbola $y = 1/x$ at most at one point we see that the numbers $a, b, c, d$ are pairwise distinct.\n\n![](attached_image_1.... | Belarus | BelarusMO 2013_s | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0135 | Problem:
A sequence $\left(a_{n}\right)$ is defined as follows: $a_{1}=\sqrt{2}$, $a_{2}=2$, and $a_{n+1}=a_{n} a_{n-1}^{2}$ for $n \geq 2$. Prove that for every $n \geq 1$ we have
$$
\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right)<(2+\sqrt{2}) a_{1} a_{2} \cdots a_{n} .
$$ | [
"Solution:\n\nFirst we prove inductively that for $n \\geq 1$, $a_{n}=2^{2^{n-2}}$. We have $a_{1}=2^{2^{-1}}$, $a_{2}=2^{2^{0}}$ and\n$$\na_{n+1}=2^{2^{n-2}} \\cdot\\left(2^{2^{n-3}}\\right)^{2}=2^{2^{n-2}} \\cdot 2^{2^{n-2}}=2^{2^{n-1}} .\n$$\nSince $1+a_{1}=1+\\sqrt{2}$, we must prove, that\n$$\n\\left(1+a_{2}\\... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof only | null | |
01yd | Let $\Omega$ be a circle, $S$ be a point outside $\Omega$ and $SA, SB$ be tangents from $S$ to $\Omega$. Arbitrary point $K$ is chosen on the segment $AB$. Points $U, W, R$ and $T$ are chosen on $\Omega$ such that $UW \cap RT = K$ and the points $W, T$ lie on the same half-plane with respect to the line $AB$. The line ... | [
"We will use multiple times the following well-known\n**Lemma.** Let the chords $AB$ and $CD$ of a circle intersect at the point $K$ then\n$$ \\frac{AC \\cdot AD}{BC \\cdot BD} = \\frac{CK}{DK} $$\n**Proof of lemma.** This equality directly follows from the sine laws for the triangles $ACK, ADK$ and $BCD$.\n\nThe q... | Belarus | SELECTION and TRAINING SESSION | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
0djj | An academic club of $2023$ members wants to organize some seminars. In each seminar, each member of the club will present exactly one of three subjects: Math, Physics, and Chemistry. It is given that for any two members, there exists some seminar where they do not present the same subject. Find the smallest possible va... | [
"Let $k$ be the number of seminars the club can organize. In the $i$-th seminar with $1 \\le i \\le k$, denote $A_i$, $B_i$, $C_i$ as the collection of participants presenting Math, Physics, Chemistry respectively. Then, for each $i$, we have $A_i$, $B_i$, $C_i$ are the partition of the members of the club. Replace... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 7 | |
0k70 | Problem:
2019 students are voting on the distribution of $N$ items. For each item, each student submits a vote on who should receive that item, and the person with the most votes receives the item (in case of a tie, no one gets the item). Suppose that no student votes for the same person twice. Compute the maximum pos... | [
"Solution:\n\nTo get an item, a student must receive at least 2 votes on that item. Since each student receives at most 2019 votes, the number of items one student can receive does not exceed $\\frac{2019}{2} = 1009.5$. So, the answer is at most 1009. This occurs when $N = 2018$ and item $i$ was voted to student $1... | United States | HMMT November 2019 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 1009 | |
0dnr | Problem:
У $\triangle ABC$ ($AB \neq AC$) уписана је кружница, чији је центар тачка $I$, и она додирује страницу $BC$ у тачки $D$. Нека је тачка $M$ средиште дужи $BC$. Доказати да се нормале из тачака $M$ и $D$ на праве $AI$ и $MI$, редом, секу на правој која садржи висину $\triangle ABC$ из темена $A$.
(Душан Ђукић... | [
"Solution:\n\nОзначимо са $\\gamma$ уписани круг $\\triangle ABC$, а са $\\gamma_{a}$ и $I_{a}$ приписани круг наспрам $A$ и његов центар. Тачка $E$ симетрична тачки $D$ у односу на $M$ је додирна тачка $\\gamma_{a}$ са страницом $BC$. Нормала $\\ell_{1}$ из $D$ на $MI$ је радикална оса круга $\\gamma$ и круга $\\o... | Serbia | 10. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"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 > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasin... | null | proof only | null | |
0f75 | Problem:
How many polynomials $p(x)$ have all coefficients $0$, $1$, $2$ or $3$ and take the value $n$ at $x = 2$? | [] | Soviet Union | 20th ASU | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | floor(n/2) + 1 | |
07lf | A square $ABCD$ contains a point $P$ such that
$$
|PA| = 3, \quad |PB| = 7 \quad \text{and} \quad |PD| = 5.
$$
Find the area of the square. | [
"Proceed as in the hint and let $P'$ denote the rotated position of $P$. Thus $|AP'| = 3$, $|BP'| = 5$ and $\\angle PAP' = 90^\\circ$. Since $PAP'$ is an isosceles right triangle with $|AP'| = |AP| = 3$, then $|PP'| = 3\\sqrt{2}$.\n\n\n\nThe area of the triangle $PBP'$ is, by Heron's formul... | Ireland | Irska | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 58 | |
0co0 | A teacher has $40$ pencils of $4$ colors (there are $10$ pencils of each color). He distributes them to $10$ kids so that each kid has $4$ pencils. Find the least value of $n$ such that the class leader can always find a group of $n$ kids having altogether the pencils of all $4$ colors.
Имеется $40$ карандашей четырех... | [
"**Ответ.** $3$ ребят.\n\nПокажем, что всегда можно выбрать трёх ребят так, чтобы у них нашлись карандаши всех цветов. Так как карандашей каждого цвета $10$, а каждому досталось по $4$ карандаша, то кому-то достались карандаши по крайней мере двух различных цветов. Осталось добавить к нему двух ребят, у которых ест... | Russia | Final round | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | 3 | |
0gi6 | $m$ 為正整數, 且實數 $a_1, a_2, \dots, a_m$ 滿足
$$
\frac{1}{m} \sum_{i=1}^{m} a_i = 1,
$$
$$
\frac{1}{m} \sum_{i=1}^{m} a_i^2 = 11,
$$
$$
\frac{1}{m} \sum_{i=1}^{m} a_i^3 = 1,
$$
$$
\frac{1}{m} \sum_{i=1}^{m} a_i^4 = 131.
$$
試證明 $m$ 必為 7 的倍數。 | [
"考慮多項式 $f(x) = (x-3)^2(x+4)^2 = x^4 + 2x^3 - 23x^2 - 24x + 144$, 則我們有\n$$\n0 \\le \\frac{1}{m} \\sum_{i=1}^{m} f(a_i) = 131 + 2 - 253 - 24 + 144 = 0,\n$$\n因此 $a_i$ 必須是 3 或 -4。假設其中有 $p$ 個 3 與 $q$ 個 -4, 則我們有\n$$\n\\begin{aligned}\n\\frac{3p - 4q}{m} &= \\frac{1}{m} \\sum_{i=1}^{m} a_i = 1, \\\\\n\\frac{9p + 16q}{m} &... | Taiwan | 2023 年台灣數學奧林匹亞考試 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | Chinese (Traditional) | proof only | null | |
05h9 | Problem:
Soient $x > 1$ et $y$ des entiers vérifiant $2 x^{2} - 1 = y^{15}$. Montrer que $x$ est divisible par $5$. | [
"Solution:\n\nL'entier $y$ est clairement impair et strictement plus grand que $1$. On factorise l'équation sous la forme\n$$\nx^{2} = \\left(\\frac{y^{5} + 1}{2}\\right)\\left(y^{10} - y^{5} + 1\\right)\n$$\nRemarquons que\n$$\ny^{10} - y^{5} + 1 \\equiv 3 \\pmod{y^{5} + 1}\n$$\net que donc $\\operatorname{pgcd}(y... | France | Envoi 1 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0204 | Problem:
In triangle $A B C$ the midpoint of $B C$ is called $M$. Let $P$ be a variable interior point of the triangle such that $\angle C P M = \angle P A B$. Let $\Gamma$ be the circumcircle of triangle $A B P$. The line $M P$ intersects $\Gamma$ a second time in $Q$. Define $R$ as the reflection of $P$ in the tange... | [
"Solution:\n\nWe claim $|Q R| = |B C|$, which will clearly imply that quantity $|Q R|$ is independent from the position of $P$ inside triangle $\\triangle A B C$ (and independent from the position of $A$).\nThis equality will follow from the equality between triangles $\\triangle B P C$ and $\\triangle R B Q$. This... | Benelux Mathematical Olympiad | 4th Benelux Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
07kt | Suppose $p \ge 4$. Determine the largest constant $q$ such that, for all $a, b > 0$,
$$
\frac{1}{a} + \frac{1}{b} + \frac{p}{a+b} \ge \frac{q}{\sqrt{ab}}
$$ | [
"Multiplying through by $\\sqrt{ab}$ we see that we require the largest number $q$ so that\n$$\n\\frac{a+b}{\\sqrt{ab}} + \\frac{\\sqrt{ab}}{a+b} p \\ge q, \\quad \\forall a, b > 0.\n$$\nTo deal with this, let\n$$\ns = \\frac{a+b}{\\sqrt{ab}},\n$$\nso that the inequality becomes\n$$\ns + \\frac{p}{s} \\ge q.\n$$\nT... | Ireland | Irska | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | 2\sqrt{p} | |
08sh | Determine the maximum possible value for the least common multiple of 4 distinct single digit positive integers. | [
"Possible prime factors for a single digit positive integer are $2$, $3$, $5$, $7$, and since $2^4 = 16$, $3^3 = 27$, $5^2 = 25$, $7^2 = 49$, are all bigger than $10$, orders of $2$, $3$, $5$, $7$ that can appear in a prime factorization of a single digit positive integer would be less than or equal to $3$, $2$, $1... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 2520 | |
01fo | A circle with center $O$ passes through the vertices $B$ and $C$ of the triangle $ABC$ and intersects for the second time segments $AB$ and $AC$ in points $C_1$ and $B_1$ correspondingly. The lines $AO$, $BB_1$ and $CC_1$ are concurrent. Prove that triangle $ABC$ is isosceles. | [
"Segments $BB_1$ and $CC_1$ intersect inside triangle $ABC$, point $A$ is outside the circle, therefore ray $AO$ lies inside angle $BAC$. Assume that triangle $ABC$ is not isosceles. Then $\\angle BAO \\neq \\angle OAC$. Without loss of generality $\\angle BAO < \\angle OAC$. Let $B'_1$ and $C'$ be points symmetric... | Baltic Way | Baltic Way 2019 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | English | proof only | null | |
041o | In $\triangle ABC$, point $B_2$ is the reflection of the center of B-excircle with respect to the midpoint of side $AC$, and point $C_2$ is the reflection of the center of C-excircle with respect to the midpoint of side $AB$. The A-excircle touches side $BC$ at point $D$. Prove that $AD \perp B_2C_2$. (posed by Bian Ho... | [
"Let $A_1$ be the center of A-excircle. By the properties about centers of ex-circles, the following sets of three points are collinear: $\\{B_1, A, C_1\\}$, $\\{A_1, C, B_1\\}$ and $\\{C_1, B, A_1\\}$, and $A_1A \\perp B_1C_1$.\n\n\nFig. 3. 1\n\nOn the plane, choose point $P$ such that $\\... | China | China Western Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometr... | English | proof only | null | |
01da | In triangle $ABC$, the points $D$ and $E$ are the intersections of the angular bisectors from $C$ and $B$ with the sides $AB$ and $AC$. Points $F$ and $G$ on the extensions of $AB$ and $AC$ beyond $B$ and $C$ satisfy $BF = CG = BC$. Prove $FG \parallel DE$. | [
"From $BC = BF$ follows $\\triangle BFC = \\triangle BCF$. From $\\triangle BFC + \\triangle BCF = \\triangle ABC$ then follows $\\triangle BFC = \\triangle ABE$. Hence $FC \\parallel BE$. Analogously $GB \\parallel CD$. Therefore\n$$\n\\frac{AF}{AG} = \\frac{AF \\ AC \\ AB}{AC \\ AB \\ AG} = \\frac{AB \\ AC \\ AD}... | Baltic Way | Baltic Way 2016 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
059g | On the side $AB$ of a triangle $ABC$, let $D$ be a point such that $\angle BDC = \angle ACB$. Let $K$ be the midpoint of $CD$ and let $E$ be the intersection of lines $BK$ and $AC$. Given that $\angle BKD = 2\angle BCD$, find $\angle AEB$. | [
"Answer: $90^\\circ$.\n\nDenote $\\angle CAB = \\alpha$ and $\\angle BCA = \\gamma$. Then $\\angle BDC = \\gamma$ and triangles *CBD* and *ABC* will be similar due to having two equal angles (see figure below).\n\nTherefore $\\angle BCD = \\alpha$ and $\\angle BKD = 2\\alpha$. But then $\\angle KBC = 2\\alpha - \\a... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 90° | |
0czl | The quadrilateral $ABCD$ has $AD = DC = CB < AB$ and $AB \parallel CD$. Points $E$ and $F$ lie on the sides $CD$ and $BC$ such that $\widehat{ADE} = \widehat{AEF}$. Prove that:
a. $4CF \leq CB$.
b. If $4CF = CB$, then $AE$ is the angle bisector of $\widehat{DAF}$. | [
"a.\n$\\widehat{FEC} = 180^\\circ - \\widehat{AEF} - \\widehat{DEA}$\n$$\n= 180^\\circ - \\widehat{ADE} - \\widehat{DEA} = \\widehat{DAE}\n$$\nFrom $AD = DC = CB < AB$ and $AB \\parallel CD$ it follows that $\\widehat{ADC} = \\widehat{DCB}$, hence triangles $ADE$ and $ECF$ are similar. This yields\n$$\n\\frac{AD}{E... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof only | null | |
0eb7 | The sides and one of the diagonals of a rhombus measure $60$ cm. Inside the rhombus there are $9$ points. Do there exist two of the points which are not more than $30$ cm apart? Justify your answer. | [
"The answer is yes. The shorter diagonal divides the rhombus into two equilateral triangles with the sides of length $60$ cm. Each of the two triangles can be further divided into $4$ equilateral triangles with the sides of length $30$ cm. We have thus divided the rhombus into $8$ equilateral triangles with the sid... | Slovenia | National Math Olympiad in Slovenia | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | Yes | |
0ign | Problem:
Let $m \circ n = \dfrac{m+n}{mn+4}$. Compute $((\cdots((2005 \circ 2004) \circ 2003) \circ \cdots \circ 1) \circ 0)$. | [
"Solution:\nNote that $m \\circ 2 = \\dfrac{m+2}{2m+4} = \\dfrac{1}{2}$, so the quantity we wish to find is just $\\left(\\dfrac{1}{2} \\circ 1\\right) \\circ 0 = \\dfrac{1}{3} \\circ 0 = \\dfrac{1}{12}$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | final answer only | 1/12 | |
00y6 | Problem:
The inscribed circle of the triangle $A_{1} A_{2} A_{3}$ touches the sides $A_{2} A_{3}$, $A_{3} A_{1}$ and $A_{1} A_{2}$ at points $S_{1}$, $S_{2}$, $S_{3}$, respectively. Let $O_{1}$, $O_{2}$, $O_{3}$ be the centres of the inscribed circles of triangles $A_{1} S_{2} S_{3}$, $A_{2} S_{3} S_{1}$ and $A_{3} S_... | [
"Solution:\n\nWe shall prove that the lines $S_{1} O_{1}$, $S_{2} O_{2}$, $S_{3} O_{3}$ are the bisectors of the angles of the triangle $S_{1} S_{2} S_{3}$. Let $O$ and $r$ be the centre and radius of the inscribed circle $C$ of the triangle $A_{1} A_{2} A_{3}$. Further, let $P_{1}$ and $H_{1}$ be the points where ... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hwh | Problem:
Let $A X Y Z B$ be a convex pentagon inscribed in a semicircle with diameter $\overline{A B}$, and let $K$ be the foot of the altitude from $Y$ to $\overline{A B}$. Let $O$ denote the midpoint of $\overline{A B}$ and $L$ the intersection of $\overline{X Z}$ with $\overline{Y O}$. Select a point $M$ on line $K... | [
"Solution:\n\nExtend the semicircle to a circle $\\Gamma$. Let line $L K$ meet $\\Gamma$ again at two points $P$ and $Q$. Let $W$ be the point on ray $O M$ such that $O W \\cdot O M = O A \\cdot O B$. So points $P, Q, W, O$ are concyclic, say on $\\gamma$.\n\nNow, $L$ is the radical center of $\\gamma, \\Gamma$, an... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry... | null | proof only | null | |
0l3e | Problem:
In triangle $ABC$, points $M$ and $N$ are the midpoints of $AB$ and $AC$, respectively, and points $P$ and $Q$ trisect $BC$. Given that $A$, $M$, $N$, $P$, and $Q$ lie on a circle and $BC = 1$, compute the area of triangle $ABC$. | [
"Solution:\n\nNote that $MP \\parallel AQ$, so $AMPQ$ is an isosceles trapezoid. In particular, we have $AM = MB = BP = PQ = \\frac{1}{3}$, so $AB = \\frac{2}{3}$. Thus $ABC$ is isosceles with base $1$ and legs $\\frac{2}{3}$, and the height from $A$ to $BC$ is $\\frac{\\sqrt{7}}{6}$, so the area is $\\frac{\\sqrt{... | United States | HMMT February 2024 Guts Round | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | √7/12 | |
080p | Problem:
Determinare tutte le coppie ordinate $(m, n)$ di interi positivi che soddisfano l'equazione
$$
\frac{1}{m}+\frac{1}{n}-\frac{1}{mn}=\frac{2}{5}
$$ | [
"Solution:\nConsideriamo dapprima le coppie $(m, n)$ con $m \\leq n$. Abbiamo\n$$\n\\frac{2}{5}=\\frac{1}{m}+\\frac{1}{n}-\\frac{1}{mn} \\leq \\frac{1}{m}+\\frac{1}{m}-\\frac{1}{mn}<\\frac{2}{m}\n$$\nda cui $m<5$. Inoltre\n$$\n\\frac{2}{5}=\\frac{1}{m}+\\frac{1}{n}-\\frac{1}{mn}>\\frac{1}{m}\n$$\nda cui $m>\\frac{5... | Italy | Progetto Olimpiadi di Matematica | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (3,10), (4,5), (10,3), (5,4) | |
0h90 | Andriy has counted the sum of the squares of positive integers from $1$ to $2016$, Vitaliy has counted the sum of the squares of natural numbers from $1$ to $2016$. Yuriy has added Andriy's and Vitaliy's numbers, multiplied by $3$ and added $2016$. What number has Yuriy got?
(Andriy Anikushin) | [
"Let's write down the number that Yuriy had got:\n$$\n\\begin{aligned}\n& 3 \\cdot (1^2 + 2^2 + \\dots + 2016^2) + 3 \\cdot (1 + 2 + \\dots + 2016) + 2016 = \\\\\n& = (3 \\cdot 1^2 + 3 \\cdot 1 + 1) + (3 \\cdot 2^2 + 3 \\cdot 2 + 1) + \\dots + (3 \\cdot 2016^2 + 3 \\cdot 2016 + 1) = \\\\\n& = (1^3 + 3 \\cdot 1^2 + ... | Ukraine | The Problems of Ukrainian Authors | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | final answer only | 2017^3 - 1 | |
0jr6 | Problem:
Find any quadruple of positive integers $(a, b, c, d)$ satisfying $a^{3} + b^{4} + c^{5} = d^{11}$ and $a b c < 10^{5}$. | [
"Solution:\nIt's easy to guess that there are solutions such that $a, b, c, d$ are in the form of $n^{x}$, where $n$ is a rather small number. After a few attempts, we can see that we obtain simple equations when $n=2$ or $n=3$: for $n=2$, the equation becomes in the form of $2^{t} + 2^{t} + 2^{t+1} = 2^{t+2}$ for ... | United States | HMMT November 2015 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | final answer only | (128, 32, 16, 4) | |
0kv7 | Problem:
An integer $n$ is chosen uniformly at random from the set $\{1,2,3, \ldots, 2023!\}$. Compute the probability that
$$
\gcd\left(n^{n}+50, n+1\right)=1
$$ | [
"Solution:\nIf $n$ is even, we need $\\gcd(n+1,51)=1$. If $n$ is odd, we need $\\gcd(n+1,49)=1$. Thus, the answer is\n$$\n\\frac{1}{2}\\left(\\frac{\\varphi(49)}{49}+\\frac{\\varphi(51)}{51}\\right)=\\frac{265}{357}\n$$"
] | United States | HMMT November 2023 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof and answer | 265/357 | |
0gqq | Determine the smallest value of $(a+5)^2 + (b-2)^2 + (c-9)^2$ for all real numbers $a, b, c$ satisfying $a^2 + b^2 + c^2 - ab - bc - ca = 3$. | [
"Note that $(a+5)^2 + (b-2)^2 + (c-9)^2 \\ge 72$ since\n$$\n(a+5)^2 + (b-2)^2 + (c-9)^2 + 4 \\cdot 3 - 84 = (a+5)^2 + (b-2)^2 + (c-9)^2 + 4(a^2+b^2+c^2-ab-bc-ca) - 84 \\\\\n= (a - 2b + 3)^2 + (b - 2c + 4)^2 + (c - 2a - 1)^2 \\ge 0\n$$\nThe equality is held at $(a, b, c) = (1, 2, 3)$."
] | Turkey | Team Selection Test | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 72 | |
0dsm | One hundred balls labelled $1$ to $100$ are to be put into two identical boxes so that each box contains at least one ball and the greatest common divisor of the product of the labels of all the balls in one box and the product of the labels of all the balls in the other box is $1$. Determine the number of ways that th... | [
"First let's assume the two boxes are different. The balls labelled $1$ and $2$ can be put into either box. All balls with even labels must then be put into the same box that the ball labelled $2$ goes. Thus any ball with label which has a common factor greater than $1$ with any one of those even labels must be put... | Singapore | Singapore Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 2047 | |
0k6c | Problem:
Bob is coloring lattice points in the coordinate plane. Find the number of ways Bob can color five points in $\{(x, y) \mid 1 \leq x, y \leq 5\}$ blue such that the distance between any two blue points is not an integer. | [
"Solution:\n\nWe can see that no two blue points can have the same $x$ or $y$ coordinate. The blue points then must make a permutation of $1,2,3,4,5$ that avoid the pattern of $3$-$4$-$5$ triangles. It is not hard to use complementary counting to get the answer from here.\n\nThere are $8$ possible pairs of points t... | United States | HMMT February 2019 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 80 | |
063i | Problem:
Man bestimme die kleinste positive ganze Zahl $n$, für die es ein Polynom
$$
P(X)=a_{2 n} X^{2 n}+a_{2 n-1} X^{2 n-1}+\ldots+a_{1} X+a_{0}
$$
mit reellen Koeffizienten gibt, das die beiden folgenden Eigenschaften aufweist:
- Für $i=0,1, \ldots, 2 n$ gilt $2014 \leq a_{i} \leq 2015$.
- Es gibt eine reelle Zahl... | [
"Solution:\n\nEs ist $n=2014$. Zunächst definiere für eine spezielle Wahl der Koeffizienten $a_{k}$, nämlich 2014 für gerades $k$ und 2015 für ungerades $k$, das Polynom $Q_{n}(X)=2014 X^{2 n}+2015 X^{2 n-1}+2014 X^{2 n-2}+\\ldots+2015 X+2014$. Es ist $Q_{2014}(-1)=0$, die gesuchte Zahl ist also höchstens 2014. Es ... | Germany | 1. Auswahlklausur 2014/2015 | [
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | 2014 | |
04d8 | The incircle of a scalene triangle $ABC$ touches the sides $\overline{AB}$ and $\overline{CA}$ at the points $M$ and $N$ respectively. The excircles opposite to vertices $B$ and $C$ touch the line $BC$ at points $P$ and $Q$ respectively. Prove that the quadrilateral $MNPQ$ is cyclic if and only if $\angle CAB = 90^\cir... | [
"Let $|BC| = a$, $|CA| = b$ and $|AB| = c$ be the lengths of the sides of the triangle $ABC$ and let $s = \\frac{1}{2}(a+b+c)$ be its semiperimeter. Without loss of generality we may assume $b < c$.\nLet $S$ be the intersection of lines $BC$ and $MN$, and let $X$ be the point where the incircle of the triangle $ABC... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Eule... | English | proof only | null | |
0amr | Problem:
How many positive integer pairs $x, y$ satisfy $\sqrt{x} + \sqrt{y} = \sqrt{600}$?
(a) 8
(b) 7
(c) 6
(d) 5 | [] | Philippines | Qualifying Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | d | |
0ehv | Problem:
Kolikšen je prvi člen rekurzivno podanega zaporedja s formulo $a_{n}=2 a_{n-1}+1$, če je peti člen enak $7$?
(A) $2$
(B) $15$
(C) $-\frac{7}{2}$
(D) $-\frac{1}{2}$
(E) $-5$ | [
"Solution:\nZapišemo $a_{5}=2 a_{4}+1$ in izračunamo $a_{4}=\\frac{a_{5}-1}{2}=\\frac{7-1}{2}=3$.\nPostopoma računamo vse člene do prvega člena:\n$a_{3}=\\frac{a_{4}-1}{2}=1$\n$a_{2}=\\frac{a_{3}-1}{2}=0$\n$a_{1}=\\frac{a_{2}-1}{2}=-\\frac{1}{2}$."
] | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | MCQ | D | |
08x8 | Let $ABC$ be an acute triangle and let $H$ be its orthocenter. Suppose that a circle going through the points $B$, $C$ and the circle having the line segment $AH$ as its diameter intersect at two distinct points $X$ and $Y$. Let $D$ be the foot of the perpendicular line drawn from $A$ to the line $BC$, and $K$ be the f... | [
"If $AB = AC$ holds, the center of a circle going through the points $B$, $C$ and the center of the circle having $AH$ as its diameter both lie on the perpendicular bisector of the side $BC$ of the triangle. Therefore, the line $XY$ and the perpendicular bisector of $BC$ intersect perpendicularly and the point of t... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter... | English | proof only | null | |
0ain | A convex quadrilateral $ABCD$ is given. Let $E$ be the intersection of $AB$ and $CD$, $F$ be the intersection of $AD$ and $BC$, and $G$ be the intersection of $AC$ and $EF$. Prove that the following two statements are equivalent:
(i) $BD$ and $EF$ are parallel
(ii) $G$ is the midpoint of the segment $\overline{EF}$
Да... | [
"We draw a line $l$ through $E$ which is parallel to $BC$. Let $H$ be the intersection of $l$ and $AG$. Now we have that $G$ is the intersection of the diagonals in the trapezoid $EHFC$.\n\n(i) $\\Rightarrow$ (ii): Let the lines $BD$ and $EF$ be parallel. Then, from Thales' theorem for parallel segments we have the... | North Macedonia | Macedonian Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0l7m | Let $m$ and $n$ be positive integers, and let $\mathcal{R}$ be a $2m \times 2n$ grid of unit squares.
A *domino* is a $1 \times 2$ or $2 \times 1$ rectangle. A subset $S$ of grid squares in $\mathcal{R}$ is *domino-tileable* if dominoes can be placed to cover every square of $S$ exactly once with no domino extending ou... | [] | United States | 16th United States of America Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | C(m+n, m)^2 | |
02x7 | Problem:
Na figura a seguir, $ABC$ é um triângulo equilátero, $D$, $E$ e $F$ são seus pontos médios e $P$ é o seu centro. Qual a fração que a área sombreada representa do total do triângulo $ABC$?
 | [
"Solution:\n\nOs pontos $I$, $G$ e $H$ são as interseções dos segmentos $DE$, $EF$ e $DF$ com os segmentos $AP$, $BP$ e $CP$. Pela simetria da figura, as áreas dos quadriláteros $EIPG$, $DIPH$ e $FGPH$ medem o mesmo valor $x\\ \\mathrm{cm}^2$. Além disso, pelo mesmo argumento, as áreas dos triângulos $AEI$ e $ADI$ ... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | 5/24 | |
0d6q | Let $ABC$ be a non-isosceles triangle inscribed in a circle $(O)$ and $BE$, $CF$ are two angle bisectors intersecting at $I$ with $E$ belonging to segment $AC$ and $F$ belonging to segment $AB$. Suppose that $BE$, $CF$ intersect $(O)$ at $M$, $N$ respectively. The line $d_{1}$ passes through $M$ and is perpendicular to... | [
"1) Denote the projections of $I$ on $AC$, $AB$ are $X$, $Y$ respectively.\nNote that $B$, $O$, $P$ are collinear because $\\angle BMP = 90^\\circ$. Because $O$, $H$ are the midpoints of $PB$, $PM$ respectively, we have $OH \\parallel BM$ and $OH = \\frac{1}{2} BM$.\nSimilarly, $OK \\parallel CN$ and $OK = \\frac{1... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry... | English | proof only | null | |
0fqn | Problem:
Calcular todos los pares de enteros $\left(x, y\right)$ tales que
$$
3^{4} 2^{3}\left(x^{2}+y^{2}\right)=x^{3} y^{3}
$$ | [
"Solution:\nNótese en primer lugar que si $x y=0$, entonces $x^{2}+y^{2}=0$, de donde resulta la solución $x=y=0$. Nótese también que si $x y<0$, entonces $x^{2}+y^{2}<0$, absurdo, luego $x, y$ tienen ambos el mismo signo. Como cambiar simultáneamente de signo a $x$ y a $y$ no altera la ecuación, podemos asumir a p... | Spain | LV Olimpiada matemática Española | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots >... | null | proof and answer | (x, y) = (-6, -6), (0, 0), (6, 6) | |
01r3 | Points $X$, $Y$ and $Z$ are marked on the sides $AD$, $AB$ and $BC$ of a parallelogram $ABCD$, respectively. It is known that $AX = CZ$.
a) Prove that at least one of the inequalities holds
$$
XY + YZ \geq AC \text{ or } XY + YZ \geq BD.
$$
b) Is it true that $XY + YZ \geq \frac{AC + BD}{2}$? | [
"b) It is not true.\n\na) Without loss of generality we assume that $\\alpha = \\angle BAD \\le 90^\\circ$. In this case we have\n$$\n\\begin{aligned}\nBD &= \\sqrt{AB^2 + AD^2 - 2AB \\cdot AD \\cos \\alpha} \\le \\\\\n&\\le \\sqrt{AB^2 + AD^2 + 2AB \\cdot AD \\cos \\alpha} = [AD = BC] = \\\\\n&= \\sqrt{AB^2 + BC^2... | Belarus | Final Round | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructio... | English | proof and answer | No | |
05qm | Problem:
Trouver toutes les valeurs possibles pour $f(2018)$ où $f$ est une fonction de $\mathbb{Q}$ dans $\mathbb{R}_{+}$ vérifiant les trois conditions suivantes:
- $f(2)=1 / 2$ ;
- pour tout rationnel $x$, si $f(x) \leqslant 1$, alors $f(x+1) \leqslant 1$;
- $f(x y)=f(x) f(y)$ pour tous rationnels $x$ et $y$. | [
"Solution:\nTout d'abord, l'égalité $1 / 2=f(2)=f(1 \\times 2)=f(1) f(2)=f(1) / 2$ montre que $f(1)=1$, et donc que $f(n) \\leqslant 1$ pour tout entier naturel non nul $n$.\n\nNotons de plus que pour tous $x, y$ avec $y \\neq 0$ on a $f(y) \\neq 0$, et $f(x / y)=f(x) / f(y)$ puisque $f(x / y) f(y)=f((x / y) y)=f(x... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1/2 | |
02qg | Problem:
Um número é enquadrado quando, ao ser somado com o número obtido invertendo a ordem de seus algarismos, o resultado é um quadrado perfeito. Por exemplo, $164$ e $461$ são enquadrados, pois $164+461 = 625=25^{2}$. Quantos são os números enquadrados entre $10$ e $100$?
A) 5
B) 6
C) 8
D) 9
E) 10 | [
"Solution:\nSeja $n$ um número enquadrado entre $10$ e $100$, $a$ seu algarismo das dezenas e $b$ seu algarismo das unidades; notamos que $1 \\leq a \\leq 9$ e $0 \\leq b \\leq 9$. Então $n=10a+b$ e o número obtido invertendo-se os algarismos de $n$ é $10b+a$. Como $n$ é enquadrado temos que $(10a+b)+(10b+a)=11a+11... | Brazil | Nível 2 | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | MCQ | C | |
09w7 | Let $ABC$ be an acute triangle and let $P$ be the intersection of the tangents in $B$ and $C$ to the circumcircle of $\triangle ABC$. The line through $A$ perpendicular to $AB$ and the line through $C$ perpendicular to $AC$ intersect in a point $X$. The line through $A$ perpendicular to $AC$ and the line through $B$ pe... | [
"Let $M$ be the circumcentre of $\\triangle ABC$ and let $\\alpha = \\angle BAC$. We first show that $\\triangle BYA \\sim \\triangle BMP$ and then that $\\triangle YBM \\sim \\triangle ABP$.\n\nBy the inscribed angle theorem we have $\\angle BMC = 2\\angle BAC = 2\\alpha$. Quadrilateral $PBMC$ is a kite with axis ... | Netherlands | IMO Team Selection Test 2, June 2020 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point ... | English | proof only | null | |
09ap | Let $A$, $C$ be interior points of square $XOBD$, such that $\angle AXC = \angle ABC = 45^\circ$. Prove that,
$$
S_{\triangle AXO} + S_{\triangle ABC} + S_{\triangle CXD} = S_{\triangle ACX} + S_{\triangle AOB} + S_{\triangle CBD}.
$$ | [
"Let $M$ and $N$ be exterior points of $XOBD$, such that $\\triangle XOA = \\triangle XDM$ and $\\triangle AOB = \\triangle NDB$. Then we have $S_{\\triangle ACX} = S_{\\triangle MCX}$, $S_{\\triangle ABC} = S_{\\triangle BCN}$. Since $MD = DN$ and $M$, $D$, $N$ are collinear, we have $S_{\\triangle MDC} = S_{\\tri... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0b0n | Problem:
For the upcoming semester, 100 math majors can take up to two out of five math electives. Suppose 22 will not take any math elective in the coming semester. Also,
- 7 will take Algebraic Number Theory and Galois Theory
- 12 will take Galois Theory and Hyperbolic Geometry
- 3 will take Hyperbolic Geometry and ... | [
"Solution:\n\nWe solve this problem by eliminating those math majors who do not fit the necessary criteria of taking exactly one of Galois Theory, Hyperbolic Geometry, or Topology. The 22 math majors who are not taking any math electives are immediately excluded. Also eliminated are the 7 taking Algebraic Number Th... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 17 | |
0bju | Show that there exists a multiple of $2013$ which ends in $2014$. | [
"Consider the numbers $a_1, a_2, \\dots, a_{2014}$, where $a_k$ is the $4k$-digit number whose digits are $k$ groups of the form $2014$, $1 \\le k \\le 2014$. Since there are only $2013$ possible remainders after the division by $2013$, two different $a$s must give the same remainder in this division, hence there e... | Romania | 65th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0j31 | Problem:
George has two coins, one of which is fair and the other of which always comes up heads. Jacob takes one of them at random and flips it twice. Given that it came up heads both times, what is the probability that it is the coin that always comes up heads? | [
"Solution:\n\nAnswer: $\\frac{4}{5}$.\n\nIn general, $P(A|B) = \\frac{P(A \\cap B)}{P(B)}$, where $P(A|B)$ is the probability of $A$ given $B$ and $P(A \\cap B)$ is the probability of $A$ and $B$ (See http://en.wikipedia.org/wiki/Conditional_probability for more information).\n\nIf $A$ is the event of selecting the... | United States | Harvard-MIT November Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 4/5 | |
09g7 | Let $n$ and $a$ be positive integers satisfying $a^n \equiv 1 \pmod{n}$. Show that there exists a positive integer $s$ such that $a^s + s \equiv 0 \pmod{n}$. | [
"We prove by induction on $n$. For $n=1$, there is nothing to prove. Suppose $n \\ge 2$ and let $d := \\gcd(\\phi(n), n)$.\n\nClearly $a$ and $n$ are relatively prime, thus we have $a^d \\equiv 1 \\pmod{n}$ by Euler's theorem. Since $d \\le \\phi(n) < n$ and $a^d \\equiv 1 \\pmod{d}$, there exists $m$ such that $a^... | Mongolia | 2015 Mongolian IMO Team Selection Tests | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0hm9 | Problem:
Prove that any triangle can be dissected into five isosceles triangles. (The pieces must not intersect except at their boundaries, and they must cover the given triangle.) | [
"Solution:\n\nWe present one of many possible solutions.\n\nFirst, draw the height to the longest side (or one of the longest sides), as shown in Figure 1. This height will always lie inside the triangle and will divide it into two right triangles. We will cut one of these right triangles into two isosceles pieces ... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
055m | Find all positive integers $n$ such that $n!$ is not divisible by $n^2$. | [
"For any prime $p$, $p^2 \\nmid p!$ as prime $p$ occurs only once in the prime factorization of $p!$. Additionally, $4^2 = 16$ does not divide $4! = 24$. We will show that $n^2 \\mid n!$ for all other positive integers $n$. Let $p$ be a prime factor of $n$, and $k$ the exponent of $p$ in the prime factorization of ... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | All prime numbers and 4 | |
09ji | A red ball of radius $7$, two green balls of radius $3$ and a yellow ball of radius no more than $2$ are placed on the floor so that each pair is tangent to each other. What is the radius of the yellow ball? | [] | Mongolia | Mongolian Mathematical Olympiad Round 2 | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 21*(sqrt(353) - 17)/32 | |
07vx | Find all functions $f : \mathbb{R} \to \mathbb{R}$ with the property that
$$
f(x)f(y) = (xy - 1)^2 f\left(\frac{x+y-1}{xy-1}\right)
$$
for all real numbers $x, y$ with $xy \neq 1$. | [
"Letting $x = 0, y = 1$ we get $f(1)f(0) = f(0)$, so $f(0) = 0$ or $f(1) = 1$.\n\n**Case 1:** $f(0) = 0$\nLetting $y = 0$ and $x = 1 - t$ we get $f(t) = 0$ in this case for all $t$. Indeed, this is a valid solution for our equation.\n\n**Case 2:** $f(0) \\neq 0$\nIn this case we have $f(1) = 1$. We first show that ... | Ireland | IRL_ABooklet_2023 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | All solutions are f(x) = 0 for all real x and f(x) = x^2 − x + 1 for all real x. |
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