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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0i5g | Problem:
Suppose $a, b, c, d$ are real numbers such that
$$
|a-b|+|c-d|=99 ; \quad|a-c|+|b-d|=1
$$
Determine all possible values of $|a-d|+|b-c|$. | [
"Solution:\n99 If $w \\geq x \\geq y \\geq z$ are four arbitrary real numbers, then $|w-z|+|x-y|=|w-y|+|x-z|=w+x-y-z \\geq w-x+y-z=|w-x|+|y-z|$. Thus, in our case, two of the three numbers $|a-b|+|c-d|,|a-c|+|b-d|,|a-d|+|b-c|$ are equal, and the third one is less than or equal to these two. Since we have a 99 and a... | United States | Harvard-MIT Math Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 99 | |
05i8 | Problem:
Soit $a, b, c, d$ des entiers naturels tels que $0<|a d-b c|<\min (c, d)$.
Prouver que pour tous entiers $x, y>1$ premiers entre eux, le nombre $x^{a}+y^{b}$ n'est pas divisible par $x^{c}+y^{d}$. | [
"Solution:\n\nPar l'absurde : on suppose que les entiers $x, y>1$ sont premiers entre eux et que $s=x^{c}+y^{d}$ divise $x^{a}+y^{b}$. On a alors:\n$$\nx^{c}=-y^{d} \\quad \\bmod s \\quad \\text { et } \\quad x^{a}=-y^{b} \\quad \\bmod s .\n$$\nD'où\n$$\nx^{a d}=(-1)^{d} y^{b d} \\bmod s \\quad \\text { et } \\quad... | France | Olympiades Françaises de Mathématiques - Épreuve en temps limité de Janvier | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0edy | Problem:
V zaporedju $a_{1}, a_{2}, a_{3}, a_{4}, \ldots$ je vsota vsakih treh zaporednih členov enaka 2016. Velja še $a_{667}=667$ in $a_{1004}=1004$. Koliko je $a_{2016}$?
(A) 0
(B) 345
(C) 667
(D) 1004
(E) 2016 | [
"Solution:\n\nKer je vsota vsakih treh zaporednih členov zaporedja enaka, velja $a_{n+3}=a_{n}$ za vsak $n$. Od tod sledi $a_{667}=a_{667+3 \\cdot 449}=a_{2014}$ in $a_{1004}=a_{1004+3 \\cdot 337}=a_{2015}$. Ker je $a_{2014}+a_{2015}+a_{2016}=2016$, je $a_{2016}=2016-667-1004=345$."
] | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | B | |
0do4 | Problem:
За задат природан број $k$, нека је $n_{k}$ најмањи природан број такав да постоји коначан скуп $A$ целих бројева са следећим особинама:
- за свако $a \in A$ постоје $x, y \in A$ (не обавезно различити) такви да
$$
n_{k} \mid a-x-y
$$
- не постоји подскуп $B$ скупа $A$ за који важи $|B| \leqslant k$ и $n_{k} ... | [
"Solution:\n\nСа $F_{i}$ означавамо Фибоначијеве бројеве: $F_{1}=F_{2}=1, F_{i+1}=F_{i}+F_{i-1}$. Посматрајмо скуп $A=\\{a_{1}, a_{2}, \\ldots, a_{k+1}\\}$ са $n=F_{k+2}+3$, где су\n$$\na_{i}=(-1)^{k-1-i} F_{i+1} \\text{ за } i=1,2, \\ldots, k-1, \\quad a_{k}=F_{k}+1 \\quad \\text{и} \\quad a_{k+1}=F_{k}+2\n$$\nУ с... | Serbia | 12. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0187 | For any real number $a$ we define a sequence $x_0, x_1, \dots$ such that $x_0 = a$ and $x_{i+1} = 3x_i - x_i^3$ for all $i \ge 0$. Determine the number of reals $a$ for which $x_{2011} = x_0$. | [
"If $|x_i| > 2$ then $|x_{i+1}| = |x_i| \\cdot |3 - x_i^2| > |x_i|$ it follows that the sequence $(|x_i|)$ is strictly increasing therefore the sequence $(|x_i|)$ cannot be periodic. It is thus enough to consider the case when $|a| \\le 2$.\nDenote $x_i = 2 \\sin \\alpha$, where $-\\frac{\\pi}{2} \\le \\alpha \\le ... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 3^{2011} | |
0kku | Problem:
After the Guts round ends, HMMT organizers will collect all answers submitted to all 66 questions (including this one) during the individual rounds and the guts round. Estimate $N$, the smallest positive integer that no one will have submitted at any point during the tournament.
An estimate of $E$ will receive... | [
"Solution:\nThe correct answer was 139."
] | United States | HMMT Spring 2021 Guts Round | [
"Math Word Problems"
] | null | final answer only | 139 | |
0bx4 | Two children, Alex and Cristi, play several times a game, in which the winner receives $x$ points, and the loser $y$ points ($x$ and $y$ are nonnegative integers, with $x > y$, and in any game one of the children is the winner and the other is the loser). The final score is $147$ to $123$, in Alex's favour. Cristi has ... | [
"Denote by $a$ the number of games won by Alex. Then $a x + 6 y = 147$ and $6 x + a y = 123$.\nSubtracting the above equalities, we obtain $a x + 6 y - 6 x - a y = 24$, or $(a - 6)(x - y) = 24$.\n\n$a - 6$ and $x - y$ are positive integers, because $a$, $x$ and $y$ are nonnegative integers, with $x > y$ and $a > 6$... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | x = 13, y = 5 | |
08td | How many positive integers of $2009$ or less digits can be represented in the form $a^{2009} + b^{2009}$ using integers $a$ and $b$? | [
"By symmetry, it is sufficient to determine the number of those positive integers having digits less than or equal to $2009$ that can be represented in the form $a^{2009} + b^{2009}$ by using a pair of integers $a$ and $b$ with the additional hypothesis $a \\ge b$.\n\nNote also that the requirement that $a^{2009} +... | Japan | Japan Junior Mathematical Olympiad First Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 99 | |
02i5 | Problem:
Qual o maior número de 6 algarismos que se pode encontrar suprimindo-se 9 algarismos do número $778157260669103$ sem mudar a ordem dos algarismos?
(A) $778152$
(B) $781569$
(C) $879103$
(D) $986103$
(E) $987776$ | [
"Solution:\n\nSolução 1. Para que seja o maior possível, o número deve começar com o maior algarismo. Para termos 6 algarismos sem mudar a ordem, o maior é $8$ depois $7$, faltam agora $4$ algarismos para completar o número, escolhemos $9103$. Logo, o número é $879103$ ($77$-$8793$).\n\nSolução 2. As opções D e E n... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms"
] | null | MCQ | C | |
0jmt | Problem:
Two circles are said to be orthogonal if they intersect in two points, and their tangents at either point of intersection are perpendicular. Two circles $\omega_{1}$ and $\omega_{2}$ with radii $10$ and $13$, respectively, are externally tangent at point $P$. Another circle $\omega_{3}$ with radius $2 \sqrt{2... | [
"Solution:\n\nLet $\\omega_{i}$ have center $O_{i}$ and radius $r_{i}$. Since $\\omega_{3}$ is orthogonal to $\\omega_{1}$, $\\omega_{2}$, $\\omega_{4}$, it has equal power $r_{3}^{2}$ to each of them. Thus $O_{3}$ is the radical center of $\\omega_{1}$, $\\omega_{2}$, $\\omega_{4}$, which is equidistant to the thr... | United States | HMMT 2014 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | 92/61 | |
0d6j | Let $x$, $y$, $z$ be positive real numbers satisfy the condition $x^{2} + y^{2} + z^{2} = 2(xy + yz + zx)$. Prove that
$$
x + y + z + \frac{1}{2xyz} \geq 4
$$ | [
"Without loss of generality, assume that $z = \\min \\{x, y, z\\}$. From the condition $x^{2} + y^{2} + z^{2} = 2(xy + yz + zx)$, we get\n$$\n(x + y)^{2} - 2z(x + y) + z^{2} = 4xy\n$$\nor\n$$\n(x + y - z)^{2} = 4xy\n$$\nUsing the AM-GM inequality, we have\n$$\n\\frac{x + y - z}{2} + \\frac{x + y - z}{2} + 2z + \\fr... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0b3i | Problem:
What is the largest multiple of $7$ less than $10,\!000$ which can be expressed as the sum of squares of three consecutive numbers? | [
"Solution:\n\nLet the number be expressed as $a^{2} + (a+1)^{2} + (a+2)^{2}$, where $a$ is an integer. It may be checked that this expression is a multiple of $7$ if and only if the remainder when $a$ is divided by $7$ is $1$ or $4$. \n\nIn the former case, the largest possible value of $a$ that places the value of... | Philippines | 24th Philippine Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | 8750 | |
00sq | Let $AD$, $BE$, and $CF$ denote the altitudes of triangle $\triangle ABC$. Points $E'$ and $F'$ are the reflections of $E$ and $F$ over $AD$, respectively. The lines $BF'$ and $CE'$ intersect at $X$, while the lines $BE'$ and $CF'$ intersect at the point $Y$. Prove that if $H$ is the orthocenter of $\triangle ABC$, the... | [
"We will prove that the desired point of concurrency is the midpoint of $BC$. Assume that $\\triangle ABC$ is acute. Let $(ABC)^5$ intersect $(AEF)$ at the point $Y'$; we will prove that $Y = Y'$.\n\nFigure 7: G7\nUsing the fact that $H$ is the incenter of $\\triangle DEF$ we get that $D$, ... | Balkan Mathematical Olympiad | BMO 2019 Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ai0 | Prove that there are infinitely many positive integers which can't be expressed as $a^{d(a)} + b^{d(b)}$ where $a$ and $b$ are positive integers.
For positive integer $a$ expression $d(a)$ denotes the number of positive divisors of $a$. | [
"If $a$ is a square of an integer, any its power is also square of an integer.\nIf $a$ is not a perfect square, number of its positive divisors is even. We can prove this by pairing divisors of $a$ as $d$ and $\\frac{a}{d}$. A divisor $d$ won't be paired with itself because that would imply $a = d^2$.\nThis proves ... | North Macedonia | European Mathematical Cup | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof only | null | |
00si | Anna and Bob play a game on the set of all points of the form $(m, n)$ where $m, n$ are integers with $|m|, |n| \leq 2019$. Let us call the lines $x = \pm 2019$ and $y = \pm 2019$ the *boundary lines* of the game. The points of these lines are called the *boundary points*. The *neighbours* of point $(m, n)$ are the poi... | [
"Anna does not have a winning strategy. We will provide a winning strategy for Bob. It is enough to describe his strategy for the deletions on the line $y = 2019$.\n\nBob starts by deleting $(0, 2019)$ and $(-1, 2019)$. Once Anna completes her step, he deletes the next two available points on the left if Anna decre... | Balkan Mathematical Olympiad | BMO 2019 Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Anna does not have a winning strategy. | |
0ixw | Problem:
In how many ways can you rearrange the letters of "HMMTHMMT" such that the consecutive substring "HMMT" does not appear? | [
"Solution:\n\nThere are $8!/(4!2!2!) = 420$ ways to order the letters. If the permuted letters contain \"HMMT\", there are $5 \\cdot 4!/2! = 60$ ways to order the other letters, so we subtract these. However, we have subtracted \"HMMTHMMT\" twice, so we add it back once to obtain $361$ possibilities."
] | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 361 | |
0fpn | Sean $m \ge 1$ un entero positivo, $a$ y $b$ enteros positivos distintos mayores estrictamente que $m^2$ y menores estrictamente que $m^2 + m$. Hallar todos los enteros $d$, que dividen al producto $ab$ y cumplen $m^2 < d < m^2 + m$. | [
"Sea $d$ un entero positivo que divida a $ab$ y tal que $d \\in (m^2, m^2+m)$. Entonces $d$ divide a $(a-d)(b-d) = ab-da-db+d^2$. Como que $|a-d| < m$ y $|b-d| < m$, deducimos que $|(a-d)(b-d)| < m^2 < d$ lo que implica que $(a-d)(b-d) = 0$. Así $d = a$ o $d = b$."
] | Spain | LII Olimpiada Matemática Española | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Spanish | proof and answer | d equals a or b | |
07bb | Find all polynomials $P(x)$ and $Q(x)$ with rational coefficients such that
$$
P(x)^3 + Q(x)^3 = x^{12} + 1.
$$ | [
"We have $x^{12} + 1 = (x^4 + 1)(x^8 - x^4 + 1)$. It is easy to check that these two factors are irreducible in $\\mathbb{Z}[x]$ (or equivalently in $\\mathbb{Q}[x]$). On the other hand, $P^3 + Q^3 = (P+Q)(P^2-PQ+Q^2)$. Therefore, our goal is to find $P$ and $Q$ such that $(P+Q)(P^2-PQ+Q^2) = (x^4+1)(x^8-x^4+1)$. I... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | null | proof and answer | The only solutions are P(x) = x^4, Q(x) = 1 and P(x) = 1, Q(x) = x^4. | |
0c3n | Problem:
Fie $n \in \mathbb{N}$, $n \geq 2$ şi numerele $a_{1}, a_{2}, \ldots, a_{n} \in (1, \infty)$. Demonstraţi că funcţia $f:[0, \infty) \rightarrow \mathbb{R}$, definită prin relaţia
$$
f(x)=\left(a_{1} a_{2} \ldots a_{n}\right)^{x}-a_{1}^{x}-a_{2}^{x}-\ldots-a_{n}^{x}
$$
pentru orice $x \in [0, \infty)$, este str... | [
"Solution:\nVom realiza demonstraţia prin inducţie matematică.\nPentru $n=2$, fie $a_{1}, a_{2} \\in (1, \\infty)$. Avem\n$$\nf(x)=\\left(a_{1} a_{2}\\right)^{x}-a_{1}^{x}-a_{2}^{x}=\\left(a_{1}^{x}-1\\right)\\left(a_{2}^{x}-1\\right)-1\n$$\nDeoarece funcţiile $f_{1}, f_{2}:[0, \\infty) \\rightarrow \\mathbb{R}$, d... | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Intermediate Algebra > Exponential functions",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0csx | The treasurer of Math republic chose a number $\alpha > 2$ and issued coins with values of 1 rouble and of $\alpha^k$ roubles for all positive integer $k$. It turns out that all the values of coins (except for 1) are irrational. May it happen that for any positive integer $n$, one may take several coins whose values su... | [
"Ответ. Могло.\n\nПокажем, что математики могли выбрать число $\\alpha = \\frac{-1 + \\sqrt{29}}{2}$; это число является корнем уравнения $\\alpha^2 + \\alpha = 7$. Ясно, что $\\alpha > 2$. Нетрудно видеть, что при натуральных $m$ мы имеем $(2\\alpha)^m = a_m + b_m\\sqrt{29}$, где $a_m$ и $b_m$ — целые числа, причё... | Russia | XL Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Algebraic Number Theory > Algebraic numbers"
] | null | proof and answer | Yes | |
0fas | Problem:
If you have an algorithm for finding all the real zeros of any cubic polynomial, how do you find the real solutions to $\{x\} = \{p(y)\}$, $\{y\} = \{p(x)\}$, where $p$ is a cubic polynomial? | [
"Solution:\nLet $p(x) \\equiv a x^3 + b x^2 + c x + d$. Finding the solutions with $x = y$ is obvious, just solve the cubic $a x^3 + b x^2 + (c - 1)x + d = 0$.\n\nFor $x \\neq y$, we have $x - y = a(y^3 - x^3) + b(y^2 - x^2) + c(y - x)$.\n\nDividing by $y - x$ gives $a(x^2 + x y + y^2) + b(x + y) + c + 1 = 0$.\n\nP... | Soviet Union | 1st CIS | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0jjd | Problem:
Let $f: \mathbb{N} \rightarrow \mathbb{N}$ be a function satisfying the following conditions:
(a) $f(1)=1$.
(b) $f(a) \leq f(b)$ whenever $a$ and $b$ are positive integers with $a \leq b$.
(c) $f(2 a)=f(a)+1$ for all positive integers $a$.
How many possible values can the 2014-tuple $(f(1), f(2), \ldots, f(2... | [
"Solution:\n\nAnswer: $1007$\n\nNote that $f(2014)=f(1007)+1$, so there must be exactly one index $1008 \\leq i \\leq 2014$ such that $f(i)=f(i-1)+1$, and for all $1008 \\leq j \\leq 2014$, $j \\neq i$ we must have $f(j)=f(j-1)$. We first claim that each value of $i$ corresponds to exactly one 2014-tuple $(f(1), \\... | United States | HMMT 2014 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 1007 | |
085o | Problem:
È data una circonferenza di diametro $A B$ e centro $O$. Sia $C$ un punto sulla circonferenza (diverso da $A$ e da $B$), e si tracci la retta $r$ parallela ad $A C$ per $O$. Sia $D$ l'intersezione di $r$ con la circonferenza dalla parte opposta di $C$ rispetto ad $A B$.
i) Dimostrare che $D O$ è bisettrice d... | [
"Solution:\n\nAbbiamo $A \\widehat\\{C\\} D = C \\widehat\\{D\\} O$, perché alterni interni rispetto alle parallele $A C$ e $D O$; inoltre $A \\widehat\\{C\\} D = A \\widehat\\{B\\} D$, dato che insistono sullo stesso arco di circonferenza. Il triangolo $D O B$ è formato da due raggi, e quindi isoscele; da ciò si r... | Italy | Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
09cx | $ABC$ гурвалжны $\angle BAC = 90^\circ$ болно. $A$ оройтоос $BC$ төмлө татсан өндрийн суурь $D$ бөгөөд $ABD$, $ACD$ гурвалжинд багтсан тойргийн төвүүд харгалзан $I_1$, $I_2$ болог. $I_1$ ба $I_2$ цэгүүдээс $AD$ хэрчимд татсан перпендикулярийн сууриуд харгалзан $M$ ба $K$ бөгөөд $I_1 M + I_2 K = \frac{1}{4} BC$ ба $ABC$... | [
"$m = BD$, $n = CD$, $I_1 M = d_1$, $I_2 K = d_2$, $AD = h$ болог.\n$$\n\\begin{align*}\nBN &= BP \\text{ ба } AP = AM \\text{ байх нь}\\ \\text{ойломжтой.}\n\\end{align*}\n$$\n$$\n\\text{Иймд } AB = AM + BN \\text{ ба}\n$$\n$$\nc = h - d_1 + m - d_1. \\text{ Адилаар}\n$$\n$$\nb = h - d_2 + n - d_2. \\text{ Эндээс}... | Mongolia | ММО-48 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | Mongolian | proof and answer | ∠B = 30°, ∠C = 60°, or ∠B = 60°, ∠C = 30° (with ∠A = 90°). | |
0jat | Problem:
Mac is trying to fill 2012 barrels with apple cider. He starts with 0 energy. Every minute, he may rest, gaining 1 energy, or if he has $n$ energy, he may expend $k$ energy $(0 \leq k \leq n)$ to fill up to $n(k+1)$ barrels with cider. What is the minimal number of minutes he needs to fill all the barrels? | [
"Solution:\n\nAnswer: 46\n\nFirst, suppose that Mac fills barrels during two consecutive minutes. Let his energy immediately before doing so be $n$, and the energy spent in the next two minutes be $k_{1}, k_{2}$, respectively. It is not difficult to check that he can fill at least as many barrels by spending $k_{1}... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 46 | |
0i6p | Problem:
Determine the positive value of $a$ such that the parabola $y = x^{2} + 1$ bisects the area of the rectangle with vertices $(0,0)$, $(a, 0)$, $(0, a^{2} + 1)$, and $(a, a^{2} + 1)$. | [
"Solution:\n$\\sqrt{3}$\n\nThe area of the rectangle is $a^{3} + a$. The portion under the parabola has area $\\int_{0}^{a} x^{2} + 1\\, dx = a^{3} / 3 + a$. Thus we wish to solve the equation $a^{3} + a = 2\\left(a^{3} / 3 + a\\right)$; dividing by $a$ and rearranging gives $a^{2} / 3 = 1$, so $a = \\sqrt{3}$."
] | United States | Harvard-MIT Math Tournament | [
"Calculus > Integral Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | proof and answer | sqrt(3) | |
018h | Aino and Väinö start to play the game GCD($m, n$) where $m$ and $n$ are positive integers. In the beginning there are two piles of stones on the table, one with $m$ stones, another with $n$ stones. The one whose turn it is, takes away a number of stones from one of the piles. This number is a multiple of the number of ... | [
"Choose $\\alpha = (1 + \\sqrt{5})/2$, so that $\\alpha^2 = \\alpha + 1$ holds. We prove by induction on the sum $m+n$ that if $m > \\alpha n$, then Aino has a winning strategy in GCD($m, n$), otherwise if $\\alpha n \\ge m > n$, then Väinö has.\n\n1) If $n \\mid m$, then Aino can remove all of the stones from the ... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | (1 + sqrt(5)) / 2 | |
076p | Let $n$ be an odd natural number. We consider an $n$ by $n$ grid which is made up of $n^2$ unit squares and $2n(n+1)$ edges. We colour each of these edges either red or blue. If there are at most $n^2$ red edges, then show that there exists a unit square at least three of whose edges are blue. | [
"Suppose on the contrary that each unit square has at least two red edges. Each red edge is part of at most two unit squares. Therefore\n$$\n2n^2 \\le \\sum_{\\text{unit squares}} \\text{(red edges of the square)} = \\sum_{\\text{red edges}} \\text{(unit squares containing the edge)} \\le 2n^2.\n$$\nThis implies th... | India | IND_TSExams | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
06ix | Let $ABCD$ be a square of side length $1234$. $E$ is a point on $CD$ such that $CEFG$ is a square of side length $567$ with $F$, $G$ outside $ABCD$. The circumcircle of $\triangle ACF$ meets $BC$ again at $H$. Find $CH$. | [
"Note that $BD$ is the perpendicular bisector of $AC$, while $EG$ is the perpendicular bisector of $CF$. Thus the intersection of $BD$ and $EG$, which we denote by $O$, is the circumcentre of $\\triangle ACF$.\n\nAs $\\angle OBG = \\angle OGB = 45^\\circ$, $\\triangle OBG$ is isosceles. Since $OH = OC$, we have $BH... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 667 | |
0k69 | Problem:
In the Year $0$ of Cambridge there is one squirrel and one rabbit. Both animals multiply in numbers quickly. In particular, if there are $m$ squirrels and $n$ rabbits in Year $k$, then there will be $2 m + 2019$ squirrels and $4 n - 2$ rabbits in Year $k+1$. What is the first year in which there will be stric... | [
"Solution:\n\nIn year $k$, the number of squirrels is\n$$\n2(2(\\cdots(2 \\cdot 1 + 2019) + 2019) + \\cdots) + 2019 = 2^{k} + 2019 \\cdot \\left(2^{k-1} + 2^{k-2} + \\cdots + 1\\right) = 2020 \\cdot 2^{k} - 2019\n$$\nand the number of rabbits is\n$$\n4(4(\\cdots(4 \\cdot 1 - 2) - 2) - \\cdots) - 2 = 4^{k} - 2 \\cdo... | United States | HMMT February 2019 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | 13 | |
00j7 | Two circles $k_1$ and $k_2$ with radii $r_1$ and $r_2$ are externally tangent in $Q$. The other end-points of the diameter through $Q$ are named $P$ on $k_1$ and $R$ on $k_2$. We choose two points $A$ and $B$, one on each of the arcs $PQ$ on $k_1$. (PBQA is convex.) Furthermore, $C$ is the second common point of the li... | [
"A homothety with center $Q$ and ratio $-r_2/r_1$ maps $k_1$ onto $k_2$.\n\nThis homothety maps $A$ to $C$, $B$ to $D$, and $P$ to $R$. It therefore follows that $PB = PU$ and $RD = RV$ are parallel, as are $PA = PV$ and $RC = RU$. $PURV$ must therefore be a parallelogram (no two of these p... | Austria | AustriaMO2011 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
06n1 | A special calculator contains a red button, which counts the number of even digits of an integer. For instance, when the screen shows $2022$, pressing the red button gives $4$ since all $4$ digits of $2022$ are even. Someone inputs a positive integer $n$ into the calculator and keeps pressing the red button until $0$ i... | [
"Answer: $2 \\times 10^{19}$\nSuppose the sequence of numbers shown on the calculator screen is\n$$\nn \\to p \\to q \\to r \\to 0\n$$\nwith $p$, $q$, $r$ nonzero. Note that $r$ is at least $1$ and so $q$ consists of at least one even digit, which means $q \\geq 2$. Hence $p$ consists of at least two even digits an... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2 × 10^19 | |
0kao | Problem:
In triangle $\triangle A B C$, we have marked points $A_{1}$ on side $B C$, $B_{1}$ on side $A C$, and $C_{1}$ on side $A B$ so that $A A_{1}$ is an altitude, $B B_{1}$ is a median, and $C C_{1}$ is an angle bisector. It is known that $\triangle A_{1} B_{1} C_{1}$ is equilateral. Prove that $\triangle A B C$ ... | [
"Solution:\n\nLet equilateral $\\triangle A_{1} B_{1} C_{1}$ have sides of length $s$. Since $\\triangle A A_{1} C$ is right with midpoint $B_{1}$ on the hypotenuse $A C$,\n$$\nB_{1} A = B_{1} C = B_{1} A_{1} = s = B_{1} C_{1},\n$$\nso points $A$, $C_{1}$, $A_{1}$, and $C$ lie on a circle centered at $B_{1}$. There... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
08jy | Problem:
Two circles $k_{1}$ and $k_{2}$ intersect at points $A$ and $B$. A circle $k_{3}$ centered at $A$ meets $k_{1}$ at $M$ and $P$ and $k_{2}$ at $N$ and $Q$, such that $N$ and $Q$ are on different sides of $MP$ and $AB > AM$.
Prove that the angles $\angle MBQ$ and $\angle NBP$ are equal. | [
"Solution:\n\nAs $AM = AP$, we have\n$$\n\\angle MBA = \\frac{1}{2} \\operatorname{arc} AM = \\frac{1}{2} \\operatorname{arc} AP = \\angle ABP\n$$\nand likewise\n$$\n\\angle QBA = \\frac{1}{2} \\operatorname{arc} AQ = \\frac{1}{2} \\operatorname{arc} AN = \\angle ABN\n$$\nSumming these equalities yields $\\angle MB... | JBMO | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0844 | Problem:
Sia $P$ un punto interno ad un triangolo $ABC$. Le rette $AP$, $BP$ e $CP$ intersecano i lati di $ABC$ in $A'$, $B'$ e $C'$ rispettivamente. Ponendo
$$
x = \frac{AP}{PA'}, \quad y = \frac{BP}{PB'}, \quad z = \frac{CP}{PC'}
$$
dimostrare che $xyz = x + y + z + 2$.
 | [
"Solution:\n\nSi verifica facilmente che, comunque scelti tre numeri reali non nulli $a, b, c$, e ponendo $x = \\frac{a + b}{c}$, $y = \\frac{b + c}{a}$ e $z = \\frac{a + c}{b}$, la relazione da dimostrare diventa un'identità algebrica. Per trovare $a, b, c$ si proceda come segue: siano $D$ ed $E$ le intersezioni d... | Italy | XX Olimpiade Italiana di Matematica | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
07hk | Suppose that $n$ is a positive integer. Consider a regular $2n$-gon such that one of its largest diagonals is parallel to the $x$-axis. Find the smallest integer $d$ such that there is a polynomial $P(x)$ of degree $d$ whose graph intersects all sides of the polygon on points other than its vertices. | [
"First of all, we show that $d$ should be at least $n$. The vertices of the polygon are on $n+1$ different vertical lines and between any two such lines the polynomial should intersect two edges of the polygon, one above and one below the $x$-axis. So by the *intermediate value theorem* the polynomial should have a... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | n | |
057n | Given is an $m \times m$ table with $2n$ distinct unit squares marked with a ring ($2n \le m^2$). Juku wishes to connect these $2n$ rings into pairs using $n$ (possibly curved) lines in a way that meets the following conditions:
(1) Each line begins from some ring and ends in some other ring;
(2) Every two unit squares... | [
"Color the unit squares black and white in such a way that unit squares with a common side are of different color. Each line goes from a black square to a white square and vice versa; thus whenever the endpoints of a line are in squares of equal color, the line visits an odd number of squares, and otherwise, the li... | Estonia | Open Contests | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0gfa | 從集合 $S=\{1,2,3, ..., 2024\}$ 中，取出 1000 個數而造出具有 1000 個元素的子集 $T$，而 $T$ 的最小元素是 $k$，求 $k$ 的期望值。 | [
"設 $M$ 是所求的期望值，則有：\n$$\n\\begin{aligned}\n\\binom{2024}{1000} M &= 1 \\cdot \\binom{2023}{999} + 2 \\cdot \\binom{2022}{999} + 3 \\cdot \\binom{2021}{999} + \\cdots + 1025 \\cdot \\binom{999}{999} \\\\\n&= \\sum_{a+b=1025} \\binom{a}{1} \\binom{b+999}{999} \\\\\n&= \\binom{1025+999+1}{1+999+1} \\\\\n&= \\binom{2025... | Taiwan | 國際奧林匹亞競賽第三次訓練營 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | Chinese; English | proof and answer | 2025/1001 | |
0j8k | Problem:
Let $n$ be an even positive integer. Prove that $\varphi(n) \leq \frac{n}{2}$. | [
"Solution:\n\nAgain, let $A_{n}$ be the set of all positive integers $x \\leq n$ such that $\\operatorname{gcd}(n, x)=1$. Since $n$ is even, no element of $A_{n}$ may be even, and, by definition, every element of $A_{n}$ must be at most $n$. It follows that $\\varphi(n)$, the number of elements of $A_{n}$, must be ... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
02hq | Problem:
Em 1998, a população do Canadá era de 30,3 milhões. Qual das opções abaixo representa a população do Canadá em 1998?
A) 30300000
B) 303000000
C) 30300
D) 303000
E) 30300000000 | [
"Solution:\n\nTemos que 1 milhão $= 1000000$. Logo, 30,3 milhões $= 30,3 \\times 1000000 = 30300000$"
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | A | |
03df | Let $n \ge 4$ be an integer and $x_1, x_2, \dots, x_n, x_{n+1}, x_{n+2}$ are reals, such that $x_{n+1} = x_1$ and $x_{n+2} = x_2$. Given that there exists a positive real $a$, such that $x_i^2 = a + x_{i+1}x_{i+2}$ for all $i = 1, 2, \dots, n$. Prove that at least two of the numbers $x_1, x_2, \dots, x_n$ are negative. | [
"Notice that $x_i^2 x_{i+1} = a x_{i+1} + x_{i+1}^2 x_{i+2}$ holds for all $i = 1, 2, \\dots, n$. After summing these equations we get $\\sum_{i=1}^n x_i^2 x_{i+1} = a \\sum_{i=1}^n x_i + \\sum_{i=1}^n x_i^2 x_{i+1}$ and therefore $\\sum_{i=1}^n x_i = 0$. Since $a > 0$ at least one of the numbers $x_i$ is not equal... | Bulgaria | Bulgaria 2022 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
02jk | Problem:
Sabendo-se que $0,333\ldots=\frac{1}{3}$, qual é a fração irredutível equivalente a $0,1333\ldots$ ?
A) $\frac{1}{13}$
B) $\frac{1}{15}$
C) $\frac{1}{30}$
D) $\frac{2}{15}$
E) $\frac{1333}{10000}$ | [
"Solution:\n\nSolução 1 - Usando o dado da questão temos: $0,1333\\ldots=\\frac{1,333\\ldots}{10}=\\frac{1+0,333\\ldots}{10}=\\frac{1}{10}\\left(1+\\frac{1}{3}\\right)=\\frac{1}{10} \\times \\frac{4}{3}=\\frac{2}{15}$.\n\n\nSolução 2 - Usando a regra que fornece a geratriz de uma dízima periódica, temos: $0,1333\\l... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | D | |
0c4l | At a dinner there are $5$ people. Among them there are $7$ pairs of acquaintances (if $A$ knows $B$, then $B$ knows $A$). Prove that there exists a group of $3$ people who know each other. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 70th NMO | [
"Discrete Mathematics > Graph Theory > Turán's theorem"
] | English | proof only | null | |
07b2 | A **partial sum** of $n$ real numbers $a_1, a_2, ..., a_n$ is the sum of some of them; that is, $\epsilon_1 a_1 + \epsilon_2 a_2 + ... + \epsilon_n a_n$, where for each $1 \le i \le n$, $\epsilon_i$ is either 0 or 1 and at least one of them is nonzero. Now, having these partial sums, we want to find the numbers.
Years... | [
"a) Let $a_1 \\le a_2 \\le \\dots \\le a_n$ be the stolen numbers. Since all the partial sums are positive, all the stolen numbers must be positive. Obviously $a_1$ is the smallest number among the partial sums. Suppose that numbers $a_1, a_2, \\dots, a_i$ have been determined. Omit all the partial sums of $a_1, a_... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
0g2f | Problem:
Sei $ABC$ ein Dreieck und sei $\gamma = \angle ACB$, sodass $\gamma / 2 < \angle BAC$ und $\gamma / 2 < \angle CBA$ gilt. Sei $D$ der Punkt auf der Strecke $BC$, sodass $\angle BAD = \gamma / 2$ gilt. Sei $E$ der Punkt auf der Strecke $CA$, sodass $\angle EBA = \gamma / 2$ gilt. Ausserdem sei $F$ der Schnittp... | [
"Solution:\n\nWegen $\\angle FAD = \\angle BAD = \\gamma / 2 = \\angle FCD$ liegen die Punkte $A, F, D$ und $C$ auf einem Kreis, das heisst $AFDC$ ist ein Sehnenviereck. Mit dem Peripheriewinkelsatz folgt nun $\\angle ADF = \\angle ACF = \\gamma / 2 = \\angle FAD$. Hieraus folgt wiederum, dass das Dreieck $AFD$ gle... | Switzerland | SMO - Vorrunde | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0g1s | Problem:
Seien $a$ und $b$ natürliche Zahlen, sodass
$$
\frac{3 a^{2}+b}{3 a b+a}
$$
eine ganze Zahl ist. Bestimme alle Werte, die obiger Ausdruck annehmen kann. | [
"Solution:\n\n$3 a b+a=a(3 b+1)$ teilt $3 a^{2}+b$, also ist insbesondere $a$ ein Teiler von $3 a^{2}+b$. Nun teilt $a$ ebenfalls $3 a^{2}+b-(3 a) a=b$. Somit können wir $b=a s$ schreiben. Einsetzen in den Ausdruck liefert $3 a^{2} s+a \\mid 3 a^{2}+a s$ und nach kürzen mit $a$ bekommen wir $3 a s+1 \\mid 3 a+s$. D... | Switzerland | SMO - Vorrunde | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1 | |
0j35 | Problem:
For each integer $x$ with $1 \leq x \leq 10$, a point is randomly placed at either $(x, 1)$ or $(x,-1)$ with equal probability. What is the expected area of the convex hull of these points? Note: the convex hull of a finite set is the smallest convex polygon containing it. | [
"Solution:\n\nLet $n=10$. Given a random variable $X$, let $\\mathbb{E}(X)$ denote its expected value. If all points are collinear, then the convex hull has area zero. This happens with probability $\\frac{2}{2^{n}}$ (either all points are at $y=1$ or all points are at $y=-1$). Otherwise, the points form a trapezoi... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Expected values",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | null | proof and answer | 1793/128 | |
028a | Problem:
Seja $S(n)$ a soma dos dígitos de um inteiro $n$. Por exemplo, $S(327)=3+2+7=12$. Encontre o valor de
$$
A=S(1)-S(2)+S(3)-S(4)+\ldots-S(2016)+S(2017)
$$ | [
"Solution:\nSe $m$ é par, o número $m+1$ possui os mesmos dígitos que $m$ com exceção do dígito das unidades, que é uma unidade maior. Portanto, $S(m+1)-S(m)=1$. Isso nos permite agrupar os termos da sequência em pares com diferença igual a 1 :\n$$\n\\begin{array}{r}\nS(1)-S(2)+S(3)-S(4)+\\ldots-S(2016)+S(2017)= \\... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 1009 | |
0g46 | Problem:
Let $ABC$ be a triangle and let $P$ be a point in the interior of the side $BC$. Let $I_{1}$ and $I_{2}$ be the incenters of the triangles $APB$ and $APC$, respectively. Let $X$ be the closest point to $A$ on the line $AP$ such that $XI_{1}$ is perpendicular to $XI_{2}$. Prove that the distance $AX$ is indepe... | [
"Solution:\n\nLet $\\omega_{1}$ and $\\omega_{2}$ be the two incircles, centered at $I_{1}$ and $I_{2}$ respectively. We first introduce the points $R_{1}$ and $S_{1}$ on $\\omega_{1}$ such that $XR_{1}$ and $XS_{1}$ are tangent to $\\omega_{1}$, with the condition that $S_{1}$ is on the line $AP$. Similarly, we in... | Switzerland | Final round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0j5a | Problem:
Find all irrational numbers $x$ such that $x^{3}-17 x$ and $x^{2}+4 x$ are both rational numbers. | [
"Solution:\nAnswer: $-2 \\pm \\sqrt{5}$\nFrom $x^{2}+4 x \\in \\mathbb{Q}$, we deduce that $(x+2)^{2}=x^{2}+4 x+4$ is also rational, and hence $x=-2 \\pm \\sqrt{y}$, where $y$ is rational. Then $x^{3}-17 x=(26-6 y) \\pm (y-5) \\sqrt{y}$, which forces $y$ to be $5$. Hence $x=-2 \\pm \\sqrt{5}$. It is easy to check t... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | -2 + sqrt(5) and -2 - sqrt(5) | |
0kx5 | Problem:
Let $ABCD$ be a square of side length $5$. A circle passing through $A$ is tangent to segment $CD$ at $T$ and meets $AB$ and $AD$ again at $X \neq A$ and $Y \neq A$, respectively. Given that $XY = 6$, compute $AT$. | [
"Solution:\n\nLet $O$ be the center of the circle, and let $Z$ be the foot from $O$ to $AD$. Since $XY$ is a diameter, $OT = ZD = 3$, so $AZ = 2$. Then $OZ = \\sqrt{5}$ and $AT = \\sqrt{OZ^2 + 25} = \\sqrt{30}$."
] | United States | HMMT November | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(30) | |
0kmv | A two-digit positive integer is said to be *cuddly* if it is equal to the sum of its nonzero tens digit and the square of its units digit. How many two-digit positive integers are cuddly?
(A) 0 (B) 1 (C) 2 (D) 3 (E) 4 | [
"Solution:\n\nThe value of a two-digit cuddly number $\\underline{ab}$ is $10a + b$. Therefore $10a + b = a + b^2$, so $9a = b(b - 1)$. It follows that $9$ divides $b(b - 1)$. If $3 \\mid b$ and $3 \\mid (b - 1)$, then $3 \\mid 1$, which is a contradiction. Thus either $9 \\mid b$ or $9 \\mid (b - 1)$. Furthermore,... | United States | AMC 10 A | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | MCQ | B | |
02ul | Problem:
No desenho abaixo, os pontos $E$ e $F$ pertencem aos lados $AB$ e $BD$ do triângulo $\triangle ABD$ de modo que $AE = AC$ e $CD = FD$. Se $\angle ABD = 60^{\circ}$, determine a medida do ângulo $\angle ECF$.
 | [
"Solution:\nSejam $2\\alpha = \\angle EAC$ e $2\\beta = \\angle FDC$. Como os triângulos $\\triangle EAC$ e $\\triangle FDC$ são isósceles, segue que $\\angle ACE = \\angle AEC = 90^{\\circ} - \\alpha$ e $\\angle DCF = \\angle CFD = 90^{\\circ} - \\beta$. Consequentemente, $\\angle ECF = \\alpha + \\beta$.\n\nAnali... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 60° | |
055y | Find all integer pairs $(a, b)$ for which $(2a^2 + b)^3 = b^3 a$. | [
"If $b = 0$, then according to the equation $2a^2 + b = 0$ from which $a = 0$.\n\nAssume now that $b \\neq 0$. As $b^3$ and $(2a^2 + b)^3$ are perfect cubes, their ratio $a$ is the cube of a rational number; as it is an integer, it is the cube of an integer $c$. By taking cubic root from each side of the equation w... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | (0, 0), (8, 128), (27, 729), (-1, -1) | |
058b | Let $n$ be a fixed positive integer. Find all triples $(a, b, c)$ of integers satisfying the following system of equations:
$$
\begin{cases}
a^{n+3} + b^{n+2}c + c^{n+1}a^2 + a^n b^3 = 0 \\
b^{n+3} + c^{n+2}a + a^{n+1}b^2 + b^n c^3 = 0 \\
c^{n+3} + a^{n+2}b + b^{n+1}c^2 + c^n a^3 = 0
\end{cases}
$$ | [
"If $a = 0$ then the first equation implies $b^{n+2}c = 0$. Hence $b = 0$ or $c = 0$; w.l.o.g., $b = 0$. Then the last equation reduces to $c^{n+3} = 0$ which implies $c = 0$. Thus if $a = 0$ then $a = b = c = 0$. Analogously we can prove that if $b = 0$ or $c = 0$ then $a = b = c = 0$. The triple $(0,0,0)$ satisfi... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Infinite descent / root flipping"
] | English | proof and answer | a = b = c = 0 | |
069o | Determine all positive integers which are equal to $13$ times the sum of their digits. | [
"Let $\\kappa$ be the number of digits of the integer $A$ which is equal to $13$ times the sum of its digits. The least possible $A$ is $10^{\\kappa-1}$, while the maximal possible sum of the digits is $9\\kappa$. Therefore we need to have:\n$$\n10^{\\kappa-1} \\leq 13 \\cdot 9\\kappa = 117\\kappa. \\qquad (1)\n$$\... | Greece | 36th Hellenic Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | 117, 156, 195 | |
02q5 | Problem:
De quantas formas é possível colorir as 12 arestas de um cubo de branco ou de preto? Duas colorações são iguais quando é possível obter uma a partir da outra por uma rotação. | [] | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 218 | |
09rc | Problem:
In trapezium $A B C D$ is $A B \| C D$. Zij $M$ het midden van diagonaal $A C$. Neem aan dat driehoeken $A B M$ en $A C D$ dezelfde oppervlakte hebben. Bewijs dat $D M \| B C$. | [
"Solution:\n\nOmdat $M$ het midden van $A C$ is, is de oppervlakte van driehoek $A B M$ gelijk aan de oppervlakte van driehoek $B C M$. Dus de oppervlakte van driehoek $A B C$ is twee keer zo groot als de oppervlakte van driehoek $A B M$ en daarmee ook twee keer zo groot als de oppervlakte van driehoek $A C D$. De ... | Netherlands | Selectietoets | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0i2l | Problem:
$P$ is a polynomial. When $P$ is divided by $x-1$, the remainder is $-4$. When $P$ is divided by $x-2$, the remainder is $-1$. When $P$ is divided by $x-3$, the remainder is $4$. Determine the remainder when $P$ is divided by $x^{3}-6x^{2}+11x-6$. | [
"Solution:\n\nThe remainder polynomial is simply the order two polynomial that goes through the points $(1, -4)$, $(2, -1)$, and $(3, 4)$: $x^{2} - 5$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | x^2 - 5 | |
0l0w | Problem:
Suppose that $a$ and $b$ are positive integers such that $\operatorname{gcd}\left(a^{3}-b^{3},(a-b)^{3}\right)$ is not divisible by any perfect square except 1. Given that $1 \leq a-b \leq 50$, compute the number of possible values of $a-b$ across all such $a, b$. | [
"Solution:\nClaim 1. Let $a$ and $b$ be positive integers. Then, $\\operatorname{gcd}\\left(a^{3}-b^{3},(a-b)^{3}\\right)$ is squarefree if and only if $\\operatorname{gcd}(a, b)=1$, $a-b$ is squarefree, and $a-b$ is not divisible by 3.\n\nProof. If $\\operatorname{gcd}(a, b)=d>1$, then $g$ is divisible by $d^{3}$,... | United States | HMMT November 2024 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 23 | |
01og | $N$ boys ($N \ge 3$), no two of them having the same height, are arranged along a circle. A boy in the given arrangement is said to be *middle* if he is taller than one of his neighbors and shorter than the other one.
Find all possible numbers of middle boys in the arrangement. | [
"Answer: any integer number from $0$ to $N-2$ of the same parity as $N$.\n\nConsider arbitrary arrangement of the boys along the circle. We say that a boy in the given arrangement is *tall* if he is taller than both of his neighbors, and a boy is *short* if he is shorter than both of his neighbors.\n\nThe numbers o... | Belarus | Belorusija 2012 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | Any integer from 0 to N−2 with the same parity as N. | |
0gvn | Which minimal quantity of sides of even lengths can have a polygon on a squared paper built of $2005$ dominoes $1 \times 2$? (Each domino covers two adjacent unit squares of the paper, and the boundary of the polygon is a connected closed polygonal line that does not touch and does not cross itself.) | [
"Розфарбуємо дошку як шахівницю, а разом з нею і многокутник. Проведемо в усіх клітинках дошки, які входять до складу многокутника, діагоналі, тим самим розбивши кожну з цих клітинок на чотири рівнобедрених прямокутних трикутнички з одиничною гіпотенузою. Оскільки многокутник утворено з додоміно, то кількість чорни... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2 | |
0dkf | If $a, b, c \ge 0$ with $a + b + c = 3$. Prove that
$$
(ab + c)(ac + b) \le 4.
$$ | [] | Saudi Arabia | Saudi Booklet | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0l6s | Problem:
In an $11 \times 11$ grid of cells, each pair of edge-adjacent cells is connected by a door. Karthik wants to walk a path in this grid. He can start in any cell, but he must end in the same cell he started in, and he cannot go through any door more than once (not even in opposite directions). Compute the maxim... | [
"Solution:\n\nThis is simply asking for the longest circuit in the adjacency graph of this grid. Note that this grid has $4 \\cdot 9 = 36$ cells of odd degree, 9 along each side. If we color the cells with checkerboard colors so that the corners are black, then 20 of these 36 cells are whit... | United States | HMMT February | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 200 | |
098p | Problem:
Calculați: $\int_{0}^{\pi} \sqrt{1+\cos (4046 x)}\, d x$. | [
"Solution:\nUtilizând periodicitatea funcției $f(t)=\\cos t$, obținem\n$$\n\\begin{gathered}\n\\int_{0}^{\\pi} \\sqrt{1+\\cos (4046 x)}\\, d x = \\sqrt{2} \\int_{0}^{\\pi} |\\cos (2023 x)|\\, d x =\n\\left|\\begin{array}{c}\nt = 2023 x \\\\\nd x = \\frac{1}{2023} d t \\\\\nx = 0 \\Rightarrow t = 0 \\\\\nx = \\pi \\... | Moldova | Olimpiada Republicană la Matematică | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Trigonometric functions"
] | null | proof and answer | 2√2 | |
0jck | Problem:
In the game of rock-paper-scissors-lizard-Spock, rock defeats scissors and lizard, paper defeats rock and Spock, scissors defeats paper and lizard, lizard defeats paper and Spock, and Spock defeats rock and scissors, as shown in the below diagram. As before, if two players choose the same move, then there is ... | [
"Solution:\n\nLet the three players be $A$, $B$, $C$. Our answer will simply be the sum of the probability that $A$ beats both $B$ and $C$, the probability that $B$ beats both $C$ and $A$, and the probability that $C$ beats $A$ and $B$, because these events are all mutually exclusive. By symmetry, these three proba... | United States | HMMT November 2012 | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 12/25 | |
0f08 | Problem:
Find the largest integer $n$ such that $4^{27} + 4^{1000} + 4^{n}$ is a square. | [] | Soviet Union | ASU | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1972 | |
01oa | Determine the greatest possible value of the area of a quadrilateral $ABCD$ if the length of broken line $ABDC$ is equal to $L$. | [
"Answer: $S(ABCD) = L^2/8$.\nLet the area of $ABCD$ be a maximum for some $AB = x$, $BD = y$, $CD = z$, $x + y + z = L$. Since $S(ABCD) = S(ABD) + S(DBC) = \\frac{1}{2} AB \\cdot BD \\sin \\angle ABD + \\frac{1}{2} BD \\cdot DC \\sin \\angle BDC$, we see that the area of the quadrilateral with fixed values of $x$, ... | Belarus | Belorusija 2012 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | L^2/8 | |
016w | Find out whether or not there exist two disjoint infinite sets $A$ and $B$ in the plane satisfying the following conditions:
(i) No three points in $A \cup B$ are collinear and the distance of any pair of points in $A \cup B$ is at least $1$.
(ii) There is a point of $A$ in any triangle whose vertices are in $B$ and th... | [
"We first observe that for some set $S$ of five points in $A$, the convex hull of $S$ contains no further points of $A$. For let $S_1$ be a set of five points in $A$, and let $P, Q \\in S_1$ be such that $S_1$ is in the half plane determined by the line $PQ$. We may suppose that $S_1$ is on the left hand side as on... | Baltic Way | BALTIC WAY | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | No; such sets do not exist. | |
0ipt | Problem:
If $p$ and $q$ are positive integers and $\frac{2008}{2009} < \frac{p}{q} < \frac{2009}{2010}$, what is the minimum value of $p$? | [
"Solution:\nAnswer: $4017$\n\nBy multiplying out the fraction inequalities, we find that $2008q + 1 \\leq 2009p$ and $2010p + \\leq 2009q$. Adding $2009$ times the first inequality to $2008$ times the second, we find that $2008 \\cdot 2009q + 4017 \\leq 2008 \\cdot 2009q + p$, or $p \\geq 4017$. This minimum is att... | United States | 1st Annual Harvard-MIT November Tournament | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | 4017 | |
01q6 | Let $O$ be the circumcenter of an acute-angled triangle $ABC$. Let $AH$ be the altitude of this triangle, $M$, $N$, $P$, $Q$ be the midpoints of the segments $AB$, $AC$, $BH$, $CH$, respectively.
Let $\omega_1$ and $\omega_2$ be the circumcircles of the triangles $AMN$ and $POQ$.
Prove that one of the intersection poi... | [
"Let $M$, $N$ and $R$ be the midpoints of the sides $AB$, $AC$, $BC$, respectively. Let $X$ denote the foot of the perpendicular from $O$ on $AH$. We claim that $X$ is the point of intersection of $\\omega_1$ and $\\omega_2$.\n\n\n\nSince $O$ is the center of the circumcircle of the triangl... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
0hpl | Problem:
Cheryl chooses a word in this problem and tells its first letter to Aerith and its last letter to Bob. The following conversation ensues over a series of emails:
Aerith: "I don't know her word, do you?"
Bob: "No, in fact, I don't know if we can ever figure out what her word is without having more information."... | [
"Solution:\nWe will process their conversation message by message.\n\"I don't know her word, do you?\"\nAerith would know Cheryl's word if the first letter was unique, so Aerith does not have any of $b$ (Bob), $k$ (know), $m$ (more), $p$ (problem), $s$ (series), $y$ (you).\n\n\"No, in fact, I don't know if we can e... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Logic"
] | null | proof and answer | then | |
0138 | Problem:
There are 2003 pieces of candy on a table. Two players alternately make moves. A move consists of eating one candy or half of the candies on the table (the "lesser half" if there is an odd number of candies); at least one candy must be eaten at each move. The loser is the one who eats the last candy. Which pl... | [
"Solution:\n\nLet us prove inductively that for $2n$ pieces of candy the first player has a winning strategy. For $n=1$ it is obvious. Suppose it is true for $2n$ pieces, and let's consider $2n+2$ pieces. If for $2n+1$ pieces the second is the winner, then the first eats 1 piece and becomes the second in the game s... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | second player | |
0g3w | Problem:
Let $\mathbb{R}_{>0}$ denote the set of positive real numbers. Find all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that
$$
x+f(y f(x)+1)=x f(x+y)+y f(y f(x))
$$
for all positive real numbers $x$ and $y$. | [
"Solution:\nLet $f$ be a solution to the FE. By plugging $y=\\frac{x}{f(x)}$, we obtain\n$$\nx+f(x+1)=x f\\left(x+\\frac{x}{f(x)}\\right)+x \\Longleftrightarrow f(x+1)=x f\\left(x+\\frac{x}{f(x)}\\right)\n$$\nPlug in $x=1$ in (1) to get\n$$\nf(2)=f\\left(1+\\frac{1}{f(1)}\\right)\n$$\nNote that if $f$ is injective,... | Switzerland | IMO Selection | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | f(x) = 1/x for all x > 0 | |
0gcr | 給定一個由若干個正整數所成的集合 $S$，試證下列兩個敍述至少有一成立：
(1) 存在 $S$ 中的相異有限子集合 $F$ 與 $G$ 使得
$$
\sum_{x \in F} \frac{1}{x} = \sum_{x \in G} \frac{1}{x};
$$
(2) 存在一正有理數 $r < 1$，使得對 $S$ 的任一個有限子集合 $F$，都有
$$
\sum_{x \in F} \frac{1}{x} \neq r.
$$ | [
"Argue indirectly. Agree, as usual, that the empty sum is $0$ to consider rationals in $[0, 1)$; adjoining $0$ causes no harm, since $\\sum_{x \\in F} 1/x = 0$ for no nonempty finite subset $F$ of $S$. For every rational $r$ in $[0, 1)$, let $F_r$ be the unique finite subset of $S$ such that\n$$\n\\sum_{x \\in F_r}... | Taiwan | 二〇一九數學奧林匹亞競賽第二階段選訓營,模擬競賽(一) | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
0hll | Problem:
Alice picks an odd integer $n$ and writes the fraction
$$
\frac{2 n+2}{3 n+2}
$$
Show that this fraction is already in lowest terms. (For example, if $n=5$ this is the fraction $\frac{12}{17}$.) | [
"Solution:\n\nLet $A=2 n+2$ and $B=3 n+2$. Now notice that\n$$\n3 A-2 B=3(2 n+2)-2(3 n+2)=2.\n$$\nSo if some integer $d \\geq 1$ divides both $A$ and $B$, it also divides $3 A-2 B=2$. Hence $d$ must be $1$ or $2$.\n\nBut since $n$ was odd, the number $3 n+2$ is odd, and so we can't have $d=2$. Thus the only common ... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
03ip | Problem:
In the diagram, $OB_{i}$ is parallel and equal in length to $A_{i}A_{i+1}$ for $i=1,2,3$ and $4$ ($A_{5}=A_{1}$). Show that the area of $B_{1}B_{2}B_{3}B_{4}$ is twice that of $A_{1}A_{2}A_{3}A_{4}$.
 | [
"Solution:\n\nLet $O$ be the origin. Let $A_{i}$ have position vector $\\vec{a}_{i}$ for $i=1,2,3,4$.\n\nSince $OB_{i}$ is parallel and equal in length to $A_{i}A_{i+1}$, we have:\n\n$$\n\\vec{OB}_{i} = \\vec{b}_{i} = \\vec{a}_{i+1} - \\vec{a}_{i}\n$$\n\nSo $B_{i}$ has position vector $\\vec{b}_{i} = \\vec{a}_{i+1}... | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
012b | Problem:
A set $S$ of four distinct points is given in the plane. It is known that for any point $X \in S$ the remaining points can be denoted by $Y, Z$ and $W$ so that
$$
|X Y| = |X Z| + |X W|.
$$
Prove that all the four points lie on a line. | [
"Solution:\nLet $S = \\{A, B, C, D\\}$ and let $AB$ be the longest of the six segments formed by these four points (if there are several longest segments, choose any of them). If we choose $X = A$ then we must also choose $Y = B$. Indeed, if we would, for example, choose $Y = C$, we should have $|AC| = |AB| + |AD|$... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
00se | Let $a, b, c$ be real numbers such that $0 \le a \le b \le c$. Prove that if
$$
a + b + c = ab + bc + ca > 0,
$$
then $\sqrt{bc}(a + 1) \ge 2$. When does the equality hold? | [
"Let $a + b + c = ab + bc + ca = k$. Since $(a + b + c)^2 \\ge 3(ab + bc + ca)$, we get that $k^2 \\ge 3k$. Since $k > 0$, we obtain that $k \\ge 3$.\nWe have $bc \\ge ca \\ge ab$, so from the above relation we deduce that $bc \\ge 1$.\nBy AM-GM, $b + c \\ge 2\\sqrt{bc}$ and consequently $b + c \\ge 2$. The equalit... | Balkan Mathematical Olympiad | BMO 2019 Shortlist | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds if and only if a = b = c = 1 or a = 0 and b = c = 2. | |
00qe | Let $ABC$ be an isosceles triangle, ($AB = AC$). Let $D$ and $E$ be two points on the side $BC$ such that $D \in BE$, $E \in DC$ and $2\angle DAE = \angle BAC$. Prove that we can construct a triangle $XYZ$ such that $XY = BD$, $YZ = DE$ and $ZX = EC$. Find $\angle BAC + \angle YXZ$. | [
"Let $\\omega$ be the circle of center $A$ and radius $AB$. Let $A'$ be a point on the circle $\\omega$, lying on the minor arc $\\widehat{BC}$, such that $\\angle BAD = \\angle A'AD$. Since $2\\angle DAE = \\angle A$, it is easy to see that $\\angle CAE = \\angle A'AE$.\n\n\n\nWe deduce th... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry... | English | proof and answer | 180° | |
0ht0 | Problem:
Determine the greatest real number $a$ such that the inequality
$$
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2} \geq a\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{5}\right)
$$
holds for every five real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$. | [
"Solution:\n\nNote that\n$$\nx_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2} = \\left(x_{1}^{2}+\\frac{x_{2}^{2}}{3}\\right) + \\left(\\frac{2 x_{2}^{2}}{3}+\\frac{x_{3}^{2}}{2}\\right) + \\left(\\frac{x_{3}^{2}}{2}+\\frac{2 x_{4}^{2}}{3}\\right) + \\left(\\frac{x_{4}^{2}}{3}+x_{5}^{2}\\right) .\n$$\n\nNow applyi... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 2/√3 | |
03lt | Problem:
$ABCD$ is a convex quadrilateral in which $AB$ is the longest side. Points $M$ and $N$ are located on sides $AB$ and $BC$ respectively, so that each of the segments $AN$ and $CM$ divides the quadrilateral into two parts of equal area. Prove that the segment $MN$ bisects the diagonal $BD$. | [
"Solution:\n\nSince $[MADC] = \\frac{1}{2}[ABCD] = [NADC]$, it follows that $[ANC] = [AMC]$, so that $MN \\parallel AC$. Let $m$ be a line through $D$ parallel to $AC$ and $MN$ and let $BA$ produced meet $m$ at $P$ and $BC$ produced meet $m$ at $Q$. Then\n$$\n[MPC] = [MAC] + [CAP] = [MAC] + [CAD] = [MADC] = [BMC]\n... | Canada | 40th Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
08rm | Tomohiro and Akinori read mathematical books as follows. Akinori reads 2 pages a day. Tomohiro reads 3 pages a day. However, each of the two stop reading of that day if he reaches the end of a chapter.
There is a mathematical book which consists of 10 chapters and 120 pages. Find the smallest value of the difference b... | [
"It is clear that the smallest value exists. Let $B$ be a book with $n_i$ pages for the $i$-th chapter which attains the smallest value.\n\nWe first prove that none of $n_1, n_2, \\dots, n_{10}$ are equivalent to $3$ or $5$ modulo $6$. Assume that $n_1 \\equiv 3, 5 \\pmod{6}$. Checking the parity, we can assume wit... | Japan | The 16th Japanese Mathematical Olympiad - The First Round | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 14 | |
0k9y | Problem:
In triangle $ABC$ with $AB < AC$, let $H$ be the orthocenter and $O$ be the circumcenter. Given that the midpoint of $OH$ lies on $BC$, $BC = 1$, and the perimeter of $ABC$ is $6$, find the area of $ABC$. | [
"Solution:\n\nLet $A'B'C'$ be the medial triangle of $ABC$, where $A'$ is the midpoint of $BC$ and so on. Notice that the midpoint of $OH$, which is the nine-point-center $N$ of triangle $ABC$, is also the circumcenter of $A'B'C'$ (since the midpoints of the sides of $ABC$ are on the nine-point circle). Thus, if $N... | United States | HMMT February 2019 February 16, 2019 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geom... | null | proof and answer | 6/7 | |
0ft0 | Problem:
Sei $P$ ein Polynom vom Grad $n$, sodass gilt
$$
P(k)=\frac{k}{k+1} \quad \text{ für } \quad k=0,1,2, \ldots, n
$$
Finde $P(n+1)$. | [
"Solution:\n\nBetrachte das Polynom $Q(x) = (x+1) P(x) - x$. $Q$ hat Grad $n+1$ und nach Voraussetzung die $n+1$ Nullstellen $k = 0, 1, \\ldots, n$. Es gibt also eine Konstante $a$ mit\n$$\nQ(x) = a x(x-1)(x-2) \\cdots (x-n)\n$$\nAußerdem ist $Q(-1) = ((-1)+1) P(-1) - (-1) = 1$. Einsetzen von $x = -1$ in obige Form... | Switzerland | IMO - Selektion | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | P(n+1) = 1 if n is odd, and P(n+1) = n/(n+2) if n is even. | |
02pr | Problem:
O quadrado de $13$ é $169$, que tem como algarismo das dezenas o número $6$. O quadrado de outro número tem como algarismo das dezenas o número $7$. Quais são os possíveis valores para o algarismo das unidades desse quadrado? | [
"Solution:\n\nSuponhamos que o número é $10a + b$, com $b$ um algarismo. Quando elevamos ao quadrado obtemos\n$$\n(10a + b)^2 = 100a^2 + 20ab + b^2\n$$\nque tem três parcelas: $100a^2$, $20ab$ e $b^2$.\nA primeira parcela termina em $00$, enquanto a segunda termina em um número par seguido por zero. Assim, para o a... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 6 | |
02ui | Problem:
A calculadora científica de João possui uma tecla especial que transforma qualquer número $x$ escrito na tela e que seja diferente de $1$ no número $\frac{1}{1-x}$.
a) O que acontece se o número $2$ estiver escrito na tela e apertarmos a tecla especial três vezes?
b) O que acontece se o número $2$ estiver e... | [
"Solution:\n\na) Após apertarmos a tecla três vezes, obtemos:\n$$\n2 \\xrightarrow{1^{a}} \\frac{1}{1-2} = -1 \\xrightarrow{2^{a}} \\frac{1}{1-(-1)} = \\frac{1}{2} \\xrightarrow{3^{a}} \\frac{1}{1-1/2} = 2\n$$\n\nb) Em virtude do item anterior, a cada três toques na tecla especial, tudo se passa como se o número $2... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | a) 2; b) -1; c) 1/2 | |
00oq | Let $\alpha$ and $\beta$ be real numbers with $\beta \neq 0$. Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$
f(\alpha f(x) + f(y)) = \beta x + f(y)
$$
holds for all real $x$ and $y$. | [
"The function $f$ is injective using the variable $x$ (on the left $x$ only occurs as $f(x)$, on the right $x$ is free with a non-vanishing factor, so substituting $x = a$ and $x = b$ with $f(a) = f(b)$ gives the desired conclusion).\n\nWe set $x = 0$ and remove the outer $f$ due to the injectivity and obtain $f(y)... | Austria | Austrian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | All solutions are of the form f(x) = x + C with parameters satisfying α = β and (α + 1)C = 0. Equivalently: (i) If α = β (and β ≠ 0) then f(x) = x; (ii) If α = β = −1 then f(x) = x + C for any real C. | |
00u9 | A hare and a tortoise run in the same direction, at constant but different speeds, around the base of a tall square tower. They start together at the same vertex, and the run ends when both return to the initial vertex simultaneously for the first time. Suppose the hare runs with speed $1$, and the tortoise with speed ... | [
"Suppose that $x = \\frac{p}{q}$ where $p, q$ are positive integers with $p < q$ and $\\gcd(p, q) = 1$. Suppose that the hare takes $p$ minutes for a full turn about the tower. Then the tortoise takes $q$ minutes for a full turn. They will meet again at the same vertex $pq$ minutes when the hare will make $q$ full ... | Balkan Mathematical Olympiad | BMO 2022 shortlist | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Other"
] | English | proof and answer | 1/8, 1/7, 1/5, 3/23 | |
03oe | As shown in the diagram, in $\triangle ABC$, $\angle A = 60^\circ$, $AB > AC$, point $O$ is a circumcenter and $H$ is the intersection point of two altitudes $BE$ and $CF$. Points $M$ and $N$ are on the line segments $BH$ and $HF$ respectively, and satisfy $BM = CN$. Determine the value of $\frac{MH + NH}{OH}$.
![](at... | [
"We take $BK = CH$ on $BE$ and join $OB$, $OC$ and $OK$.\nFrom the property of the circumcenter of a triangle, we know that $\\angle BOC = 2\\angle A = 120^\\circ$. From the property of the orthocenter of a triangle, we get $\\angle BHC = 180^\\circ - \\angle A = 120^\\circ$. So $\\angle BOC = \\angle BHC$. Then fo... | China | China Mathematical Competition (Extra Test) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / C... | English | proof and answer | sqrt(3) | |
04d1 | Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all real numbers $x$ and $y$ holds
$$
f(y + f(x)) - f(x + f(y)) = f(x - y)(f(x + y) - 1).
$$ | [
"Setting $x = y$ gives $0 = f(0)(f(2x) - 1)$, $\\forall x \\in \\mathbb{R}$.\n\nFirst case: $f(0) \\ne 0$\nWe have $f(2x) - 1 = 0$, $\\forall x \\in \\mathbb{R}$, so $f(x) = 1$, $\\forall x \\in \\mathbb{R}$. We check that this function is a solution.\n\nSecond case: $f(0) = 0$\nSetting $y = 0$ gives $f(f(x)) - f(x... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 1 for all real x; f(x) = 0 for all real x | |
0ed5 | Let
$$
N = 2^{15} \cdot 2015.
$$
How many divisors of $N^2$ are strictly smaller than $N$ and do not divide $N$? | [
"A number with prime factorization $p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_k^{\\alpha_k}$ has\n$$\n\\tau(p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_k^{\\alpha_k}) = (\\alpha_1 + 1)(\\alpha_2 + 1) \\dots (\\alpha_k + 1)\n$$\ndivisors. The number $N$ can be factored as $N = 2^{15} \\cdot 5 \\cdot 13 \\cdot 31$, so\n$... | Slovenia | Slovenija 2016 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 291 | |
01zb | Let $ABC$ be an isosceles triangle with the base $BC$. Points $X$, $Y$ and $Z$ are chosen on the sides $BC$, $AC$ and $AB$, respectively, such that $\triangle ABC \sim \triangle YXZ$. Let $W$ be the reflection of $X$ with respect to the midpoint of the segment $BC$.
Prove that the points $X$, $Y$, $Z$ and $W$ are cocyc... | [
"Denote by $O$ the center of the circumcircle of the triangle $XYZ$. Then\n$$\n\\angle YOZ = 2\\angle YXZ = 2\\angle ABC = 180^\\circ - \\angle BAC = 180^\\circ - \\angle ZAY,\n$$\nwhence the quadrilateral $ZAYO$ is cyclic. The chords $OY$ and $OZ$ are equal so $\\angle ZAO = \\angle OAY$. Thus $O$ lies on the bise... | Belarus | Belarus2022 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0evk | Let $U$ be a set of $m$ triangles. Prove that there exists a subset $W$ of $U$ satisfying the following conditions.
(i) The number of triangles in $W$ is at least $0.45m^{4/5}$.
(ii) There exist no 6 distinct points $A, B, C, D, E$, and $F$ such that $W$ contains 6 triangles $ABC, BCD, CDE, DEF, EFA$, and $FAB$. | [
"Let $U'$ be a subset of $U$ by choosing each triangle with the probability $p$ independently at random. Then the expected number of triangles in $U'$ is $mp$.\n\nA sequence of 6 distinct points $(x_1, x_2, \\dots, x_6)$ is called a *bad configuration* if all 6 triangles $x_1x_2x_3, x_2x_3x_4, \\dots, x_4x_5x_6, x_... | South Korea | Korean Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0ftm | Problem:
Sei $ABC$ ein beliebiges Dreieck und $D, E, F$ die Seitenmitten von $BC, CA, AB$. Die Schwerlinien $AD, BE$ und $CF$ schneiden sich im Schwerpunkt $S$. Mindestens zwei der Vierecke
$$
AFSE, \quad BDSF, \quad CESD
$$
seien Sehnenvierecke. Zeige, dass das Dreieck $ABC$ gleichseitig ist. | [
"Solution:\n\nWir können aus Symmetriegründen annehmen, dass $AFSE$ und $BDSF$ Sehnenvierecke sind.\n\nBeachte, dass $AB$ parallel ist zu $ED$, analog für die anderen Seiten. In den folgenden Gleichungen bedeutet (*) Gleichheit von Stufenwinkeln und (**) Peripheriewinkelsatz im Kreis. Es gilt\n$$\n\\varphi = \\Vara... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneo... | null | proof only | null | |
09yf | Problem:
Bepaal alle positieve gehele getallen $n$ met de volgende eigenschap: voor ieder drietal $(a, b, c)$ van positieve reële getallen is er een drietal $(k, \ell, m)$ van niet-negatieve gehele getallen zodat dat $a n^{k}$, $b n^{\ell}$ en $c n^{m}$ de lengtes van de zijden van een (niet-gedegenereerde) driehoek v... | [
"Solution:\n\nHet is duidelijk dat $n=1$ niet voldoet, want niet elke drie positieve reële getallen $a, b$ en $c$ zijn de lengtes van een driehoek. We bewijzen nu eerst dat $n \\geq 5$ niet voldoet door het drietal $(1,2,3)$ te bekijken. Stel dat er $k, \\ell, m$ bestaan zodat $n^{k}, 2 n^{\\ell}$ en $3 n^{m}$ de l... | Netherlands | IMO-selectietoets II | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 2, 3, 4 | |
05e1 | Problem:
We are given an acute triangle $A B C$. Let $D$ be the point on its circumcircle such that $A D$ is a diameter. Suppose that points $K$ and $L$ lie on segments $A B$ and $A C$, respectively, and that $D K$ and $D L$ are tangent to circle $A K L$.
Show that line $K L$ passes through the orthocentre of $A B C$.... | [
"\nFigure 1: Diagram to solution 1\n\nLet $M$ be the midpoint of $K L$. We will prove that $M$ is the orthocentre of $A B C$. Since $D K$ and $D L$ are tangent to the same circle, $|D K|=|D L|$ and hence $D M \\perp K L$. The theorem of Thales in circle $A B C$ also gives $D B \\perp B A$ a... | European Girls' Mathematical Olympiad (EGMO) | EGMO 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations... | null | proof only | null | |
07tp | Prove that $\sqrt[3]{\frac{x^6 + 1}{2}} \le \frac{3x^2 - 4x + 3}{2}$, for all real $x$, with equality iff $x = 1$. | [
"Note that for all real $x$\n$$\nx^6 + 1 = (x^2 + 1)(x^4 - x^2 + 1) = (x^2 + 1)(x^2 + \\sqrt{3}x + 1)(x^2 - \\sqrt{3}x + 1),\n$$\nis a product of three positive numbers, since $x^2 \\pm \\sqrt{3}x + 1 = (x \\pm \\frac{\\sqrt{3}}{2})^2 + \\frac{1}{4}$.\nWe use AGM for three positive variables: $\\sqrt[3]{ABC} \\le \... | Ireland | IRL_ABooklet | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0eht | Problem:
Naj bo $a=256$, $b$ pa zmnožek vseh pozitivnih deliteljev števila $a$. Katera od naslednjih enakosti je pravilna?
(A) $b=a^{4}$
(B) $b=a^{9}$
(C) $b^{2}=a^{7}$
(D) $b^{2}=a^{9}$
(E) $b^{3}=a^{10}$ | [
"Solution:\n\nKer je $a=2^{8}$, je $b=1 \\cdot 2 \\cdot 2^{2} \\cdot \\ldots \\cdot 2^{8}=2^{0+1+2+\\ldots+8}=2^{36}$. Torej velja $b^{2}=2^{72}=a^{9}$. Pravilen odgovor je (D)."
] | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | D | |
05y0 | Problem:
Alice a disposé 200 boîtes dans son salon. Chaque boîte contient un papier sur lequel elle a écrit un entier naturel non nul; les 200 entiers ne sont pas nécessairement distincts. Chaque minute, et tant que c'est possible, Alice effectue une action de la forme suivante : elle choisit trois boîtes, contenant d... | [
"Solution:\n\nCi-dessous, on dira qu'un entier change si Alice remplace $c$ par un entier $k \\times c$. Elle peint en bleu les entiers qui ne changent qu'un nombre fini de fois, et en rouge ceux qui changent un nombre infini de fois. Par construction, le nombre initialement minimal ne pourra jamais changer, donc i... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null |
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