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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0k22 | Problem:
Is the number
$$
\left(1+\frac{1}{2}\right)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{6}\right) \ldots\left(1+\frac{1}{2018}\right)
$$
greater than, less than, or equal to $50$? | [
"Solution:\nCall the expression $S$. Note that\n$$\n\\left(1+\\frac{1}{2}\\right)\\left(1+\\frac{1}{4}\\right)\\left(1+\\frac{1}{6}\\right) \\ldots\\left(1+\\frac{1}{2018}\\right)<\\left(1+\\frac{1}{1}\\right)\\left(1+\\frac{1}{3}\\right)\\left(1+\\frac{1}{5}\\right) \\ldots\\left(1+\\frac{1}{2017}\\right)\n$$\nMul... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | less than 50 | |
0dut | Problem:
Tine je zbiral znamke. Za rojstni dan je dobil nov album, v katerega bo lahko spravil veliko znamk. Iz hranilnika je vzel $2002$ tolarja in sklenil, da bo ves denar porabil za nakup znamk. Prijatelj mu je ponudil manjše znamke po $10$ tolarjev in večje po $28$ tolarjev. Tine se je odločil, da bo kupil čim več... | [
"Solution:\n\nDenimo, da bo Tine kupil $x$ znamk po $10$ tolarjev in $y$ znamk po $28$ tolarjev. Tedaj velja $10x + 28y = 2002$ oziroma $5x + 14y = 1001$, od tod pa $5x + 5y = 1001 - 9y$ oziroma $x + y = \\frac{1001 - 9y}{5}$. Vrednost vsote bo tem večja, čim manjši bo $y$. Ker je $y$ naravno število, lahko poskuša... | Slovenia | 46. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | 193 | |
0k38 | Problem:
Over all real numbers $x$ and $y$, find the minimum possible value of
$$
(x y)^{2}+(x+7)^{2}+(2 y+7)^{2}
$$ | [
"Solution 1: Rewrite the given expression as $\\left(x^{2}+4\\right)\\left(1+y^{2}\\right)+14(x+2 y)+94$. By Cauchy-Schwartz, this is at least $(x+2 y)^{2}+14(x+2 y)+94=(x+2 y+7)^{2}+45$. The minimum is $45$, attained when $x y=2$, $x+2 y=-7$.\n\n\nSolution 2: Let $z=2 y$, $s=x+z$, $p=x z$. We seek to minimize\n$$\... | United States | HMMT November 2018 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 45 | |
0kya | Problem:
For integers $2 \leq n \leq 100$, let $f(n)$ denote the unique integer $1 \leq a < n$ such that $101 a \equiv 1 \pmod{n}$. Estimate
$$
\sum_{n=2}^{100} \frac{f(n)}{n}
$$ | [] | United States | HMMT November | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 49 + H_100/101 ≈ 49.05136 | |
06vb | Let $n$ be a positive integer. Harry has $n$ coins lined up on his desk, each showing heads or tails. He repeatedly does the following operation: if there are $k$ coins showing heads and $k>0$, then he flips the $k^{\text{th}}$ coin over; otherwise he stops the process. (For example, the process starting with $T H T$ w... | [
"We represent the problem using a directed graph $G_{n}$ whose vertices are the length-$n$ strings of $H$'s and $T$'s. The graph features an edge from each string to its successor (except for $T T \\cdots T T$, which has no successor). We will also write $\\bar{H}=T$ and $\\bar{T}=H$.\n\nThe graph $G_{0}$ consists ... | IMO | IMO 2019 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Algebraic prope... | English | proof and answer | n(n+1)/4 | |
0bzj | Given a positive odd integer $n$, show that the arithmetic mean of the fractional parts $\{k^{2n}/p\}$, $k = 1, \dots, (p-1)/2$, is the same for infinitely many primes $p$. | [
"Notice that $\\{k^{2n}/p\\} = r_k/p$, where $r_k$ is the remainder $k^{2n}$ leaves upon division by $p$. Clearly, the $r_k$ are quadratic residues modulo $p$.\nIf $p$ is prime, and $p-1$ and $n$ are relatively prime, then the $r_k$, $k = 1, \\dots, (p-1)/2$, are pairwise distinct, since the $k^{2n}$, $k = 1, \\dot... | Romania | THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Other"
] | English | proof only | null | |
0brl | Find all the positive integers $p$ with the property that the sum of the first $p$ positive integers is a four-digit positive integer whose decomposition into prime factors is of the form $2^m3^n(m+n)$, where $m, n \in \mathbb{N}^*$. | [
"The number $m+n$ is prime and, obviously, $m+n \\ge 5$. If $m+n=5$, the largest value for $N = 2^m3^n(m+n)$ is $2^1 \\cdot 3^4 \\cdot 5 = 810$, which has only three digits.\nSuppose now that $m+n \\ge 11$. Then $N \\ge 2^{10} \\cdot 3 \\cdot 11 > 10000$, hence $N$ cannot have four digits.\nSo, $m+n=7$. In this cas... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 63 | |
0i7a | Problem:
Suppose $f$ is a differentiable real function such that $f(x) + f'(x) \leq 1$ for all $x$, and $f(0) = 0$. What is the largest possible value of $f(1)$? (Hint: consider the function $e^{x} f(x)$.) | [
"Solution:\n\n$1 - 1/e$\n\nLet $g(x) = e^{x} f(x)$; then $g'(x) = e^{x} (f(x) + f'(x)) \\leq e^{x}$. Integrating from $0$ to $1$, we have\n$$\ng(1) - g(0) = \\int_{0}^{1} g'(x) \\, dx \\leq \\int_{0}^{1} e^{x} \\, dx = e - 1.\n$$\nBut $g(1) - g(0) = e \\cdot f(1)$, so we get $f(1) \\leq (e - 1)/e$.\n\nThis maximum ... | United States | Harvard-MIT Math Tournament | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Equations > ODEs"
] | null | proof and answer | 1 - 1/e | |
07jy | In the triangle $ABC$ the point $M$ is the midpoint of $AB$, and the point $B'$ is the foot of the altitude from $B$ to $AC$. The circle ($CB'M$) intersects $BC$ again at $D$. The circles ($ABD$) and ($CB'M$) intersect again at $K$. The line parallel to $AB$ passing through $C$ intersects circle ($CB'M$) again at $L$. ... | [
"Since $CL \\parallel AB$ and $A$, $B$, $D$, $K$ are concyclic, we have\n$$\n180^\\circ - \\angle AKD = \\angle ABD = \\angle BCL = \\angle DKL\n$$\nThus, $\\angle AKD + \\angle DKL = 180^\\circ$. This implies that $A$, $K$, $L$ are collinear.\n\n\n\nNotice that $B'M$ is the median to the h... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
013v | Problem:
a. What is the smallest number of circles of radius $\sqrt{2}$ that are needed to cover a rectangle of size $6 \times 3$?
b. What is the smallest number of circles of radius $\sqrt{2}$ that are needed to cover a rectangle of size $5 \times 3$? | [
"Solution:\n\na. Consider the four corners and the two midpoints of the sides of length $6$. The distance between any two of these six points is $3$ or more, so one circle cannot cover two of these points, and at least six circles are needed.\n\nOn the other hand, one circle will cover a $2 \\times 2$ square, and i... | Baltic Way | Baltic Way 2005 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | a: 6; b: 5 | |
0dqu | Find all polynomials $P(x)$ with real coefficients such that
$$
P(a) \in \mathbb{Z} \text{ implies that } a \in \mathbb{Z}.
$$ | [
"Let $P(x) = a_n x^n + \\cdots + a_1 x + a_0$. Define $Q(x) = P(x+1) - P(x)$. Then $Q(x)$ is of degree $n-1$. We'll prove by contradiction that $|Q(x)| \\le 3$ for all $x$. This will imply that $n \\le 1$.\n\nAssume that $|Q(a)| > 3$ for some $a \\in \\mathbb{R}$. Then $|P(a+1) - P(a)| > 3$. Thus there are 3 intege... | Singapore | Singapore Mathematical Olympiad (SMO) 2011 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof and answer | All constant polynomials with non-integer value, and all linear polynomials of the form P(x) = (x + q)/p where p and q are integers with p nonzero. | |
02nk | Problem:
Dois irmãos - A diferença de idade entre dois irmãos é de três anos. Um ano atrás, a idade do pai desses irmãos era o dobro da soma das idades dos irmãos e, dentro de vinte anos, a idade do pai será a soma das idades desses dois filhos. Qual é a idade de cada um dos irmãos? | [
"Solution:\n\nSeja $x$ a idade do irmão mais novo e $x + 3$ a idade do irmão mais velho.\n\nSeja $p$ a idade do pai.\n\nUm ano atrás:\n- Idade do irmão mais novo: $x - 1$\n- Idade do irmão mais velho: $x + 2$\n- Idade do pai: $p - 1$\n\nA soma das idades dos irmãos, um ano atrás, era $(x - 1) + (x + 2) = 2x + 1$.\n... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | Younger sibling: 10 years; older sibling: 13 years | |
0afv | Во квадратна шема со димензии $3 \times 3$ Димитар може да ги запишува броевите $\frac{1}{2}$, $\frac{1}{3}$ и $\frac{1}{6}$. Дали може Димитар во секое квадратче да запише по еден од овие броеви така што збировите на броевите во трите редици, трите колони и двете дијагонали да се различни меѓу себе. | [
"Можни збирови на три броја од множеството $\\{\\frac{1}{2}, \\frac{1}{3}, \\frac{1}{6}\\}$ се\n$$\n\\frac{1}{2} + \\frac{1}{2} + \\frac{1}{2} = \\frac{3}{2}\n$$\n$$\n\\frac{1}{2} + \\frac{1}{2} + \\frac{1}{3} = \\frac{4}{3}\n$$\n$$\n\\frac{1}{2} + \\frac{1}{2} + \\frac{1}{6} = \\frac{7}{6}, \\quad \\frac{1}{3} + \... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Macedonian, English | proof and answer | No | |
01zn | Integers $a, b, c, d, e, f$ satisfy the system
$$
\begin{cases}
ace + 3adf - 3bcf + 3bde = 5, \\
acf - ade + bce + 3bdf = 2.
\end{cases}
$$
Find all possible values of the expression $abcde$. | [
"Answer: $0$.\n\nLet $x = ac + 3bd$, $y = ad - bc$. We have\n$$\n\\begin{align*}\n37 = 5^2 + 3 \\cdot 2^2 &= (ace + 3adf - 3bcf + 3bde)^2 + 3(acf - ade + bce + 3bdf)^2 \\\\\n&= (ex + 3fy)^2 + 3(fx - ey)^2 = (e^2 + 3f^2)(x^2 + 3y^2) \\\\\n&= (e^2 + 3f^2)((ac + 3bd)^2 + 3(ad - bc)^2) \\\\\n&= (e^2 + 3f^2)(c^2 + 3d^2)... | Belarus | SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO | [
"Number Theory > Algebraic Number Theory > Quadratic fields",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 0 | |
0g7o | 在 $999 \times 999$ 的棋盤中, 有些格子塗成白色, 剩下的格子塗成紅色。考慮三個格子所形成的序列 $(C_1, C_2, C_3)$, 其中 $C_1, C_2$ 在同一列、$C_2, C_3$ 在同一行, 且 $C_1, C_3$ 為白色, $C_2$ 為紅色。令滿足這些條件的序列總數為 $T$。求 $T$ 的最大可能值。 | [
"設第 $i$ 列與第 $j$ 行分別有 $a_i, b_j$ 個白色方格，並設 $R$ 是紅色方格所成的集合。對於坐標為 $(i, j)$ 的紅色方格，會有 $a_i b_j$ 個可行序列 $(C_1, C_2, C_3)$，其中 $C_2 = (i, j)$。因此\n$$\nT = \\sum_{(i,j) \\in R} a_i b_j.\n$$\n利用不等式 $2ab \\le a^2 + b^2$ 可得\n$$\nT \\le \\frac{1}{2} \\sum_{(i,j) \\in R} (a_i^2 + b_j^2) = \\frac{1}{2} \\sum_{i=1}^{n} (n-a_i)a_i^2 +... | Taiwan | 二〇一三數學奧林匹亞競賽第三階段選訓營, 獨立研究 (三) | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 4 * 999^4 / 27 | |
082j | Problem:
Un gioco è costituito da 10 lanci di un normale dado cubico con le facce numerate da 1 a 6. Alla fine si sommano i punteggi ottenuti, con la regola che se si ottiene 6 in un lancio i punti del lancio successivo vengono contati raddoppiati e che se si fa 6 all'ultimo lancio si ha diritto ad un (solo) tiro supp... | [
"Solution:\n\nLa risposta è (D). Infatti, supponiamo per assurdo che 114 sia un punteggio ammissibile. Indichiamo con $N$ il numero dei lanci con cui è stato ottenuto, $N$ può essere 10 o 11. Il primo punteggio fatto è minore o uguale a sei, dunque si è fatto almeno $114-6=108$ con $N-1$ lanci; notiamo che $108=12 ... | Italy | Progetto Olimpiadi di Matematica 2003 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | MCQ | D | |
0e78 | Find all prime numbers $p$, $q$ and $r$ such that $p + q^2 = r^4$. | [
"Rewrite the equations as $p = r^4 - q^2 = (r^2 - q)(r^2 + q)$. Since $p$ is prime, we have $r^2 - q = 1$ and $r^2 + q = p$. From the first equality we obtain $q = r^2 - 1 = (r-1)(r+1)$. Since $q$ is prime we get $r-1=1$. So, $r=2$ and $q=3$. Finally, from $r^2 + q = p$ we find $p=7$."
] | Slovenia | National Math Olympiad 2013 - Final Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | p = 7, q = 3, r = 2 | |
04u9 | Let $a \neq b$ be positive real numbers. Consider the equation
$$
\lfloor a x + b \rfloor = \lfloor b x + a \rfloor
$$
where $\lfloor y \rfloor$ denotes the largest integer not exceeding $y$. Prove that the set of real solutions $x$ to this equation contains an interval of length at least
$$
\frac{1}{\max\{a, b\}} \qqu... | [
"Consider linear functions $f(x) = a x + b$, $g(x) = b x + a$. Since $a, b$ are distinct and positive, their graphs are two distinct lines with positive slope. As $f(1) = g(1) = a + b$, point $P = [1, a + b]$ is the intersection of these lines (Fig. 3).\n\nWithout loss of generality, assume $b > a$ (i.e. the line d... | Czech Republic | 67th Czech and Slovak Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0la9 | Find all the positive integers $m$ such that there exist polynomials with real coefficients $P(x)$, $Q(x)$, and $R(x, y)$, satisfying the following condition:
$$
P(R(a, b)) = a \text{ and } Q(R(a, b)) = b
$$
for all real numbers $a, b$ such that $a^m - b^2 = 0$. | [] | Vietnam | Vijetnam 2008 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | m = 1 | |
03sz | Suppose $f(x) = x^3 + \log_2(x + \sqrt{x^2+1})$. For any $a, b \in \mathbb{R}$, to satisfy $f(a) + f(b) \ge 0$, the condition $a + b \ge 0$ is ( ).
(A) necessary and sufficient
(B) not necessary but sufficient
(C) necessary but not sufficient
(D) neither necessary nor sufficient | [
"Obviously $f(x) = x^3 + \\log_2(x + \\sqrt{x^2+1})$ is an odd function and is monotonically increasing. So, if $a+b \\ge 0$, i.e. $a \\ge -b$, we get $f(a) \\ge f(-b)$, $f(a) \\ge -f(b)$, and that means $f(a)+f(b) \\ge 0$.\n\nOn the other hand, if $f(a) + f(b) \\ge 0$, then $f(a) \\ge -f(b) = f(-b)$. So $a \\ge -b... | China | China Mathematical Competition | [
"Precalculus > Functions"
] | English | MCQ | A | |
0jv8 | Problem:
Let $ABCDE$ be a convex pentagon with $CD = DE$ and $\angle BCD = \angle DEA = 90^{\circ}$. Point $F$ lies on $AB$ such that $\frac{AF}{AE} = \frac{BF}{BC}$. Prove that $\angle FCE = \angle ADE$ and $\angle FEC = \angle BDC$. | [
"Solution:\n\nLet $\\omega$ denote the circumcircle of $\\triangle CDE$ and let $D_{1}$ be the point opposite to $D$. Let $DA$ meet $\\omega$ at $A_{1}$ and let $F' = A_{1}C \\cap AB$. If we let $\\alpha = \\angle DA_{1}C = \\angle DD_{1}C = \\angle ED_{1}D$ then\n$$\n\\frac{AF'}{AE} = \\frac{AF' \\cdot AD_{1}}{AA_... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02jo | Problem:
A pista de um autódromo tem $20~\mathrm{km}$ de comprimento e forma circular. Os pontos marcados na pista são: $A$, que é o ponto de partida, $B$ que dista $5~\mathrm{km}$ de $A$ no sentido do percurso, $C$ que dista $3~\mathrm{km}$ de $B$ no sentido do percurso, $D$ que dista $4~\mathrm{km}$ de $C$ no sentid... | [
"Solution:\n\n(C) Vamos marcar os 4 pontos a partir de $A$.\nComo o comprimento é de $20~\\mathrm{km}$, o comprimento de cada um dos 4 quadrantes é $5~\\mathrm{km}$. Podemos então marcar os pontos. Como $367 = 18 \\times 20 + 7$, o carro deu 18 voltas completas e percorreu mais $7~\\mathrm{km}$ a partir de $A$. Log... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic"
] | null | MCQ | C | |
0hz9 | Problem:
Matt has somewhere between 1000 and 2000 pieces of paper he's trying to divide into piles of the same size (but not all in one pile or piles of one sheet each). He tries $2,3,4,5,6,7$, and $8$ piles but ends up with one sheet left over each time. How many piles does he need? | [
"Solution:\n\nThe number of sheets will leave a remainder of $1$ when divided by the least common multiple of $2,3,4,5,6,7$, and $8$, which is $8 \\cdot 3 \\cdot 5 \\cdot 7 = 840$. Since the number of sheets is between $1000$ and $2000$, the only possibility is $1681$. The number of piles must be a divisor of $1681... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 41 | |
0bpd | Problem:
Fie $\mathcal{C}$ mulțimea funcțiilor $f:[0,1] \rightarrow \mathbb{R}$, de două ori derivabile pe $[0,1]$, care au cel puțin două zerouri, nu neapărat distincte, în $[0,1]$ și $\left|f''(x)\right| \leq 1$, oricare ar fi $x$ în $[0,1]$. Determinați valoarea maximă pe care o poate lua integrala
$$
\int_{0}^{1}|... | [
"Solution:\n\nDacă există $a \\in [0,1]$, astfel încât $f(a)=f'(a)=0$, considerăm $x \\in [0,1], x \\neq a$. Atunci $f(x)=\\frac{1}{2}(x-a)^{2} f''(\\theta)$, pentru un anumit $\\theta$ între $a$ și $x$. Deoarece $\\left|f''(t)\\right| \\leq 1$, oricare ar fi $t$ în $[0,1]$, rezultă că $|f(x)| \\leq \\frac{1}{2}(x-... | Romania | Olimpiada Naţională de Matematică, Etapa Naţională | [
"Calculus > Differential Calculus > Applications",
"Calculus > Integral Calculus > Applications"
] | null | proof and answer | Maximum value: 1/6. The maximizers are exactly f(x) = x^2/2, f(x) = (1 − x)^2/2, and their negatives. | |
02zi | Problem:
No trapézio $ABCD$, os lados $AB$ e $CD$ são paralelos. Sejam $M$ o ponto médio da diagonal $AC$, $N$ o ponto médio da diagonal $BD$ e $P$ o ponto médio do lado $AB$. Sabemos que $AB = 15\ \mathrm{cm}$, $CD = 24\ \mathrm{cm}$ e a altura do trapézio é $h = 14\ \mathrm{cm}$.
a) Calcule a medida do comprimento ... | [
"Solution:\n\n\n\na) Se $P$ é o ponto médio de $AB$ e $L$ o ponto médio de $AD$, segue que $ML$ é base média do triângulo $ACD$ e assim $LM = 12\\ \\mathrm{cm}$. Como $LN$ é base média do triângulo $ABD$, segue que $LN = 7,5\\ \\mathrm{cm}$. Portanto, $MN = LM - LN = 4,5\\ \\mathrm{cm}$\n\n... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | MN = 4.5 cm; Area of triangle MNP = 15.75 cm^2 | |
0crw | The real numbers $x$, $y$, and $z$ are chosen in such a way that $x^2 + y^2 = 1$, and the three numbers $x + yz$, $y + zx$, and $z + xy$ are rational. Prove that the number $xyz^2$ is also rational. | [
"Из условия следует, что число $(x + yz)(y + zx) = xy + (x^2 + y^2)z + xyz^2 = (xy + z) + xyz^2$ рационально. Поскольку число $xy + z$ также рационально по условию, то и число $xyz^2 = (x + yz)(y + zx) - (xy + z)$ также рационально."
] | Russia | XL Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
0jbj | Problem:
Let $\pi$ be a randomly chosen permutation of the numbers from $1$ through $2012$. Find the probability that $\pi(\pi(2012)) = 2012$. | [
"Solution:\nAnswer: $\\frac{1}{1006}$\n\nThere are two possibilities: either $\\pi(2012) = 2012$ or $\\pi(2012) = i$ and $\\pi(i) = 2012$ for $i \\neq 2012$.\n\nThe first case occurs with probability $\\frac{2011!}{2012!} = \\frac{1}{2012}$, since any permutation on the remaining $2011$ elements is possible.\n\nSim... | United States | HMMT November | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | final answer only | 1/1006 | |
0526 | Juku writes down all 20-digit numbers in which each of digits $3$, $4$, $5$ and $6$ appear five times in a row (in some order). Prove that it is possible to choose two of those numbers such that their difference is divisible by $207$. | [
"As $207 = 9 \\cdot 23$ and $9$ and $23$ are relatively prime, it suffices to find a difference that would be divisible by both $9$ and $23$.\n\nAll the $20$-digit numbers listed are divisible by $9$ because the sum of their digits is $90$. Therefore the difference of any two of them is also divisible by $9$.\n\nIt... | Estonia | Final Round of National Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
053c | In an acute triangle the feet of altitudes drawn from vertices $A$ and $B$ are $D$ and $E$, respectively. Let $M$ be the midpoint of side $AB$. Line $CM$ intersects the circumcircle of $CDE$ again in point $P$ and the circumcircle of $CAB$ again in point $Q$. Prove that
$$
|MP| = |MQ|.
$$ | [
"The orthocenter $H$ of the triangle $ABC$ is located on the circumcircle of the triangle $CDE$, because $\\angle HDC + \\angle HEC = 90^\\circ + 90^\\circ = 180^\\circ$ (see fig. 35). Let $\\alpha = \\angle BAC$; then also $\\angle CHE = 90^\\circ - \\angle ECH = \\alpha$. Therefore $\\angle MPE = 180^\\circ - \\a... | Estonia | IMO Team Selection Contest | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Mi... | null | proof only | null | |
0ces | Let $n$ be a positive integer and let $x$ and $y$ be positive divisors of $2n^2 - 1$. Prove that $x + y$ is not divisible by $2n + 1$. | [
"Suppose, if possible, that $x + y$ is divisible by $2n + 1$. Note that $2n + 1$ and $2n^2 - 1$ are relatively prime, as $(2n - 1)(2n + 1) - 2(2n^2 - 1) = 1$. Hence $2n + 1$ is coprime to $\\gcd(x, y)$, so $x/\\gcd(x, y)$ and $y/\\gcd(x, y)$ are coprime divisors of $2n^2 - 1$ whose sum is divisible by $2n + 1$.\n\n... | Romania | Twentieth IMAR Mathematical Competition | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials... | English | proof only | null | |
0iy9 | Problem:
There are 15 stones placed in a line. In how many ways can you mark 5 of these stones so that there are an odd number of stones between any two of the stones you marked? | [
"Solution:\n\nNumber the stones $1$ through $15$ in order. We note that the condition is equivalent to stipulating that the stones have either all odd numbers or all even numbers. There are $\\binom{8}{5}$ ways to choose $5$ odd-numbered stones, and $\\binom{7}{5}$ ways to choose all even-numbered stones, so the to... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 77 | |
00ic | We consider a group of trees in a nature reserve, all of which have a positive integral age. The average age is 41 years. After destruction of a tree with an age of 2010 years by lightning, the average age of the remaining trees is 40 years.
Determine the original number of trees in the group. What is the maximal numb... | [
"The original number of trees is denoted by $n$ and the sum of their ages by $s$.\nThen we have\n$$\ns = 41n.\n$$\nOn the other hand, we have\n$$\ns - 2010 = 40(n - 1).\n$$\nThis immediately yields\n$$\nn = 1970 \\quad \\text{and} \\quad s = 80770.\n$$\nAs $80770 = 2010 \\cdot 40 + 370 = 2010 \\cdot 39 + 2380$, the... | Austria | Austria 2010 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | Original number of trees: 1970; Maximal number of 2010-year-old trees: 39 | |
08qp | Problem:
Prove that there doesn't exist any prime $p$ such that every power of $p$ is a palindrome (palindrome is a number that is read the same from the left as it is from the right; in particular, number that ends in one or more zeros cannot be a palindrome). | [
"Solution:\nNote that by criterion for divisibility by $11$ and the definition of a palindrome we have that every palindrome that has even number of digits is divisible by $11$.\nSince $11^{5} = 161051$ is not a palindrome and since $11$ cannot divide $p^{k}$ for any prime other than $11$ we are now left to prove t... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
059n | There are 25 green, 20 brown and 15 orange chameleons in a zoo. Whenever exactly two chameleons of distinct colours meet, both change their colour to the third one. Otherwise, the chameleons do not change their colours. Is it possible that:
a. at some time instant, we have the same number of chameleons of each colour?... | [
"*Answer:* (a) No; (b) No.\n\nAfter each meet of two chameleons, the numbers of chameleons of two colours decrease by 1 and the number of chameleons of the remaining colour increases by 2. Thus the difference of the numbers of chameleons of each two colours is constant modulo 3. As the remainders modulo 3 of the nu... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | a. No; b. No. | |
0a4a | Problem:
Beschouw een rechthoekig bord van $m \times n$ vakjes met $m, n \ge 1$. De hoekpunten van de vakjes vormen een $(m+1) \times (n+1)$-grid. We noemen een driehoek met hoekpunten punten van het grid *laag* als die minstens één zijde heeft die parallel is met een zijde van het bord zo dat de hoogte van de driehoek... | [
"Solution:\nAls $m, n \\ge 2$ en minstens één van de twee is even, dan is het antwoord 0. Anders (minstens één van de twee is 1, of ze zijn beide oneven) is het antwoord 2. \nWe tekenen eerst een voorbeeld voor $n = 1$ en $m \\ge 1$ met twee bijzondere driehoeken, een voorbeeld voor $n = 2$ en $m \\ge 3$ met nul bi... | Netherlands | IMO-selectietoets I | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 0 if both side lengths are at least two and at least one is even; otherwise 2 | |
08l9 | Problem:
On a $5 \times 5$ board, $n$ white markers are positioned, each marker in a distinct $1 \times 1$ square. A smart child got an assignment to recolor in black as many markers as possible, in the following manner: a white marker is taken from the board; it is colored in black, and then put back on the board on a... | [
"Solution:\na) Position 20 white markers on the board such that the left-most column is empty. This positioning is good because the coloring can be realized column by column, starting with the second (from left), then the third, and so on, so that the white marker on position $(i, j)$ after the coloring is put on p... | JBMO | 2008 Shortlist JBMO | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0c5z | a) Prove that, if $x, y \ge 1$, then
$$
x + y - \frac{1}{x} - \frac{1}{y} \ge 2\sqrt{xy} - \frac{2}{\sqrt{xy}}.
$$
b) Prove that, if $a, b, c, d \ge 1$ and $abcd = 16$, then
$$
a + b + c + d - \frac{1}{a} - \frac{1}{b} - \frac{1}{c} - \frac{1}{d} \ge 6.
$$ | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0hkf | Problem:
Find all pairs $(m, n)$ of natural numbers such that $200 m + 6 n = 2006$. | [
"Solution:\nFirst we divide both sides of the equation by $2$ and get: $100 m + 3 n = 1003$.\n\nSince $m$ and $n$ are natural numbers we immediately get that $m \\leq 10$.\n\nSince $3 n$ is divisible by $3$ and $1003$ gives remainder $1$ upon division by $3$, we conclude that $100 m$ must also give the remainder $1... | United States | Berkeley Math Circle | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (1, 301), (4, 201), (7, 101), (10, 1) | |
0gah | 令 $n$ 為一正整數, 並在黑板上寫下 $1, 2, \dots, n$ 等數字。阿發和小李輪流從黑板上選擇一個數字, 規則如下:
(i) 你不能選之前被任何人選過的數字。
(ii) 如果你之前選過 $k$, 你不能選 $k-1$ 或 $k+1$。
(iii) 如果所有數字被選完則雙方平手；否則，先沒有數字可選的人輸。
假設阿發先選。試求所有小李有必勝法的正整數 $n$。
Let $n$ be a positive integer, and write down $1, 2, \dots, n$ on the blackboard. Alpha and Lee take turn choosing a number from t... | [
"除了 $n = 1, 2, 4, 6$ 外，小李都必勝。\n方便起見，令 $[n] = \\{1, 2, \\dots, n\\}$。我們先證明一個引理：\n引理：如果小李第一次選擇 $n$，且阿發已經選了他的第 $k$ 個數字 ($k \\ge 2$)，則小李必可以選擇他的第 $k$ 個數字。\n證明：令 $a_1 < a_2 < \\dots < a_k$ 為阿發所選的頭 $k$ 個數字；注意到 $a_k < n$。基於 $a_{i+1} - a_i > 1$，因此對於所有 $1 \\le i \\le k$ 必然存在 $b_i$ 使得 $a_i < b_i < a_{i+1}$，其中 $a_{k+1} := n$。又... | Taiwan | 二〇一六數學奧林匹亞競賽第二階段選訓營 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | All positive integers n except 1, 2, 4, 6 | |
08i0 | Problem:
In the triangle $ABC$ with semiperimeter $p$, the points $M$, $N$, and $P$ lie on the sides $BC$, $CA$, and $AB$ respectively. Show that
$$
p < AM + BN + CP < 3p.
$$ | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
05j1 | Problem:
On appelle diviseur propre d'un entier $n$ un diviseur positif de $n$ qui est différent de $1$ et de $n$.
Existe-t-il un entier $n$ dont le produit des diviseurs propres est égal à $2013$ ? | [
"Solution:\n\nLa décomposition en facteurs premiers de $2013$ est $3 \\times 11 \\times 61$. Ainsi, si le produit des diviseurs stricts d'un entier $n$ vaut $2013$, cela implique que $n$ est lui-même divisible par $3$, $11$ et $61$. Or, dans ce cas, il compte parmi ses diviseurs stricts au moins les nombres suivant... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
07pr | Let $f(t) = 2 + \cos t + \cos \sqrt{2} t$, for all real numbers $t$. Prove that $f$ is strictly positive on $(-\infty, \infty)$, and is not periodic. | [
"At any rate, $f$ is non-negative since the cosine takes no value smaller than $-1$, which means that $f$ takes no value smaller than $2 - 1 - 1 = 0$.\n\nSuppose $f(t) = 0$ for some real number $t$, so that\n$$\n0 = (1 + \\cos t) + (1 + \\cos \\sqrt{2} t).\n$$\nBut the expression on the right is a sum of two non-ne... | Ireland | Ireland | [
"Precalculus > Trigonometric functions",
"Precalculus > Functions"
] | null | proof only | null | |
037v | Problem:
Prove that if $a, b, c > 0$, then
$$
\frac{a b}{3 a + 4 b + 5 c} + \frac{b c}{3 b + 4 c + 5 a} + \frac{c a}{3 c + 4 a + 5 b} \leq \frac{a + b + c}{12}
$$ | [
"Solution:\nWe shall prove the following generalization of the given inequality: if $x, y \\geq 1$ and $a, b, c > 0$ then\n$$\n\\frac{a b}{x a + y b + 2 c} + \\frac{b c}{x b + y c + 2 a} + \\frac{c a}{x c + y a + 2 b} \\leq \\frac{a + b + c}{x + y + 2}\n$$\nThe given inequality is obtained for $x = \\frac{6}{5}$ an... | Bulgaria | Team selection test for 47. IMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
09k8 | Initially, all the cells of a $10 \times 10$ table contain the number $-1$. The following operation is allowed: select two consecutive rows and two consecutive columns, take the four numbers at their intersections, and reverse the sign of all other numbers in the selected rows and columns. Is it possible to transform a... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | No | |
0ghz | 有一等腰梯形 $ABCD$，其中 $AD$ 平行於 $BC$，並設 $\Omega$ 為其外接圓。令 $X$ 為 $D$ 關於 $BC$ 的對稱點，$Q$ 為 $\Omega$ 的弧 $BC$ (不含 $A$) 上一點，$P$ 為 $DQ$ 與 $BC$ 的交點。設點 $E$ 滿足 $EQ$ 平行於 $PX$ 且 $EQ$ 平分 $\angle BEC$。證明 $EQ$ 亦平分 $\angle AEP$。
Let $\Omega$ be the circumcircle of an isocles trapezoid $ABCD$, in which $AD$ is paral... | [
"令 $R$ 為 $AE$ 與 $BC$ 的交點。要證明原命題, 我們只需證明 $\\odot(EBC)$ 與 $\\odot(ERP)$ 相切於 $E$ (因為這告訴我們 $EP$, $EA$ 關於 $EB$, $EC$ 逆平行, 而後者關於 preserved $EQ$, $EQ$ 逆平行)。設 $AE$ 分別與 $\\Omega$, $\\odot(EBC)$ 交於 $S$, $T$, 則由 Reim 定理知 $P$, $Q$, $R$, $S$ 共圓。因為 $EQ$ 平分 $\\angle BEC$, $EQ$ 與 $\\odot(EBC)$ 的交點 $M$ 為 $\\odot(EBC)$ 上弧 $BC$ 的中點。由... | Taiwan | 2023 數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Chinese (Traditional) | proof only | null | |
0d4y | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying the following conditions
(a) $f(1)=1$
(b) $\forall (x, y) \in \mathbb{R}^2$, $f(x+y)=f(x)+f(y)$
(c) $\forall x \in \mathbb{R} \setminus \{0\}$, $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^2}$. | [
"Put $x=y=0$ into equation (b) to get $f(0)=0$. Put $y=-x$ into equation (b) to get $f(-x)=-f(x)$ and deduce that $f(x-y)=f(x)-f(y)$.\n\nLet $x \\in \\mathbb{R} \\setminus \\{0,1\\}$. We have\n$$\nf\\left(\\frac{1}{x-1}\\right)=\\frac{f(x-1)}{(x-1)^2}=\\frac{f(x)-1}{(x-1)^2}\n$$\nOn the other hand,\n$$\n\\begin{ali... | Saudi Arabia | SAMC 2015 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English, Arabic | proof and answer | f(x) = x for all real x | |
02gc | Let $C$ be a wooden cube. For each pair $(X, Y)$ of vertices of $C$, we cut $C$ through the plane orthogonal to $\overline{XY}$ and passing through its midpoint. Into how many pieces is the cube divided? | [
"Let's call the plane orthogonal to and passing through the midpoint of a segment $\\overline{AB}$ the \"medial plane\" of $\\overline{AB}$. There exist three different kinds of medial planes: the medial planes of the edges of $C$ (there are 3 different such planes); the medial planes of the diagonals of the faces ... | Brazil | XXII OBM | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof and answer | 96 | |
04fo | Find all integer solutions of the equation $x^4 + 68 = 4y^4$. (Tomislav Pejković) | [
"Let us rewrite the equation:\n$$x^4 + 68 = 4y^4$$\n$$x^4 - 4y^4 = -68$$\n$$ (x^2 - 2y^2)(x^2 + 2y^2) = -68 $$\n\nNow, $-68$ factors as $(-1) \\times 68$, $(-2) \\times 34$, $(-4) \\times 17$, $(-17) \\times 4$, $(-34) \\times 2$, $(-68) \\times 1$ and their negatives. We consider all pairs $(a, b)$ such that $a \\... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (x, y) = (4, 3), (4, -3), (-4, 3), (-4, -3) | |
0klr | Problem:
How many ways are there to place 31 knights in the cells of an $8 \times 8$ unit grid so that no two attack one another?
(A knight attacks another knight if the distance between the centers of their cells is exactly $\sqrt{5}$.) | [
"Solution:\nConsider coloring the squares of the chessboard so that 32 are black and 32 are white, and no two squares of the same color share a side. Then a knight in a square of one color only attacks squares of the opposite color. Any arrangement of knights in which all 31 are placed on the same color therefore w... | United States | HMMT November 2021 Team Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 68 | |
0d1h | Determine if there are polynomials $p(x)$ and $q(x)$ with real coefficients such that
$$
\frac{p(n)}{q(n)} = 1 + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{n!}
$$
for every positive integer $n$. | [
"Assume that there are polynomials $p, q \\in \\mathbb{R}[X]$ such that\n$$\n\\frac{p(n)}{q(n)} = 1 + \\frac{1}{2!} + \\dots + \\frac{1}{n!}, \\quad n \\ge 1. \\quad (1)\n$$\nThen\n$$\n\\frac{p(n+1)}{q(n+1)} - \\frac{p(n)}{q(n)} = \\frac{1}{(n+1)!}, \\quad n \\ge 1,\n$$\nso\n$$\n\\frac{p(n+1)q(n) - p(n)q(n+1)}{q(n)... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | No | |
0fak | Problem:
Show that if 15 numbers lie between 2 and 1992 and each pair is coprime, then at least one is prime. | [
"Solution:\n\nSuppose not. Then since $452 = 2025 > 1992$, each of the numbers must have a prime factor $\\leq 43$. But there are only 14 such primes: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43$. Hence\nthere must be two numbers with the same prime factor and they cannot be coprime. Contradiction."
] | Soviet Union | 1st CIS | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0cc4 | Let $n \in \mathbb{N}$, $n \ge 2$, and three sets of real numbers $A, B, C$, pairwise disjoint, each of them having $n$ elements.
Let $a$ be the number of triples $(x, y, z) \in A \times B \times C$ for which $x < y < z$ and $b$ be the number of triples $(x, y, z) \in A \times B \times C$ for which $x > y > z$. Prove t... | [
"Consider $y \\in B$, arbitrarily chosen. Denote by $A_y$ the number of pairs $(x, z) \\in A \\times C$ for which $x < y < z$, and by $B_y$ the number of pairs $(x, z) \\in A \\times C$ for which $x > y > z$.\n\nLet $A = \\{a_1 < a_2 < \\dots < a_n\\}$ and $C = \\{c_1 < c_2 < \\dots < c_n\\}$.\nOn the real axis, $y... | Romania | THE 73rd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD - THIRD SELECTION TEST | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0k8h | Problem:
Let $ABCD$ be an isosceles trapezoid, and let $E$ be the foot of the altitude from $A$ to line $BC$. Prove that line $DE$ passes through the centroid of $\triangle ABC$. | [
"Solution:\n\nLet $M$ be the midpoint of $BC$. Also, let $F$ be the foot from $D$ to $BC$. Then $AEFD$ is a rectangle. Define $G$ as the intersection of $AM$ and $DE$. Then $\\triangle AGD \\sim \\triangle EMG$, and we get\n\n$$\n\\frac{GM}{GA} = \\frac{ME}{DA} = \\frac{ME}{FE} = \\frac{1}{2}\n$$\n\nwhich implies $... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08vn | Let $n$ be an integer greater than or equal to $2$. Consider the ways of lining up $n^2$ numbers consisting of $n$ numbers each of $\{1, 2, \dots, n\}$ from left to right. Is there a way to line them up to satisfy the following condition?
**Condition:** For any integer $k$ with $1 \le k \le n^2 - 1$, the remainder obt... | [
"We first note that in order to show the existence of such an arrangement, it is enough to show the following:\n> It is possible to line up from left to right $n^2$ numbers consisting of $n$ numbers each of $\\{0, 1, \\dots, n-1\\}$ in such a way that for each $k$ with $1 \\le k \\le n^2-1$ the sum of the $k$ numbe... | Japan | Japan Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Other"
] | null | proof only | null | |
06yx | Problem:
Define the sequence $\{p\}_1, \{p\}_2, \{p\}_3, \ldots$ as follows. $\{p\}_1 = 2$, and $\{p\}_n$ is the largest prime divisor of $\{p\}_1 \{p\}_2 \ldots \{p\}_{n-1} + 1$. Prove that $5$ does not occur in the sequence. | [] | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
026i | Problem:
Diferença de quadrados - Se $(x+y)^2-(x-y)^2=20$, então $x y$ é igual a:
(a) 0
(b) 1
(c) 2
(d) 5
(e) 10 | [
"Solution:\n\nComo $(x+y)^2 = x^2 + 2 x y + y^2$ e $(x-y)^2 = x^2 - 2 x y + y^2$, temos:\n$$\n(x+y)^2 - (x-y)^2 = x^2 + 2 x y + y^2 - x^2 + 2 x y - y^2 = 4 x y = 20\n$$\nsegue-se que $x y = 5$. A opção correta é (d)."
] | Brazil | Nível 2 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | d | |
011w | Problem:
Let $f$ be a real-valued function defined on the positive integers satisfying the following condition: For all $n > 1$ there exists a prime divisor $p$ of $n$ such that
$$
f(n) = f\left(\frac{n}{p}\right) - f(p)
$$
Given that $f(2001) = 1$, what is the value of $f(2002)$? | [
"Solution:\nFor any prime $p$ we have $f(p) = f(1) - f(p)$ and thus $f(p) = \\frac{f(1)}{2}$. If $n$ is a product of two primes $p$ and $q$, then $f(n) = f(p) - f(q)$ or $f(n) = f(q) - f(p)$, so $f(n) = 0$. By the same reasoning we find that if $n$ is a product of three primes, then there is a prime $p$ such that\n... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 2 | |
0kpa | Problem:
A regular tetrahedron has a square shadow of area $16$ when projected onto a flat surface (light is shone perpendicular onto the plane). Compute the sidelength of the regular tetrahedron.
(For example, the shadow of a sphere with radius $1$ onto a flat surface is a disk of radius $1$.) | [
"Solution:\nImagine the shadow of the skeleton of the tetrahedron (i.e. make the entire tetrahedron translucent except for the edges). The diagonals of the square shadow must correspond to a pair of opposite edges of the tetrahedron. Both of these edges must be parallel to the plane—if they weren't, then edges corr... | United States | HMMT November 2022 | [
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | 4√2 | |
0hy7 | Problem:
The Houson Association of Mathematics Educators decides to hold a grand forum on mathematics education and invites a number of politicians from the United States to participate. Around lunch time the politicians decide to play a game. In this game, players can score 19 points for pegging the coordinator of th... | [
"Solution:\n\nAnswer: 1209. Attainable scores are positive integers that can be written in the form $8a + 9b + 19c$, where $a, b$, and $c$ are nonnegative integers. Consider attainable number of points modulo $8$.\n\nScores that are $0 \\pmod{8}$ can be obtained with $8a$ for positive $a$.\n\nScores that are $1 \\p... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | final answer only | 1209 | |
0hap | For positive numbers $x$, $y$, $z$ prove inequality:
$$
\frac{1}{3}(x^3 + y^3 + z^3) \geq xyz + \frac{2}{9}(x + y + z)(x - z)^2 .
$$ | [
"From known identity\n$$\nx^3 + y^3 + z^3 - 3xyz = (x + y + z) \\cdot \\frac{(x-y)^2 + (y-z)^2 + (z-x)^2}{2} .\n$$\n\nFrom inequality between $n$ arithmetic mean and root-means-square we have, that\n$$\n\\begin{aligned} (x - y)^2 + (y - z)^2 + (z - x)^2 &\\ge \\frac{1}{3}(|x - y| + |y - z| + |z - x|)^2 \\ge \\\\ &\... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0c3q | Problem:
Arătaţi că, dacă $n \geq 2$ este un număr întreg, atunci există matricele inversabile $A_{1}, A_{2}, \ldots, A_{n} \in \mathcal{M}_{2}(\mathbb{R})$, cu elementele nenule, aşa încât $A_{1}^{-1}+A_{2}^{-1}+\ldots+A_{n}^{-1}=\left(A_{1}+A_{2}+\ldots+A_{n}\right)^{-1}$. | [
"Solution:\nEgalitatea este echivalentă cu $\\left(A_{1}+A_{2}+\\ldots+A_{n}\\right)\\left(A_{1}^{-1}+A_{2}^{-1}+\\ldots+A_{n}^{-1}\\right)=I_{2}$.\n\nVom lua $A_{1}=A$, $A_{2}=\\ldots=A_{n}=B$. Cerinţa devine $I_{2}+(n-1) A B^{-1}+(n-1) B A^{-1}+(n-1)^{2} I_{2}=I_{2}$.\n\nNotând $A B^{-1}=X$, căutăm $X$ astfel înc... | Romania | Olimpiada Nationala de Matematica | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof only | null | |
0d5o | Let $ABC$ be a triangle, with $AB < AC$, $D$ the foot of the altitude from $A$, $M$ the midpoint of $BC$, and $B'$ the symmetric of $B$ with respect to $D$. The perpendicular line to $BC$ at $B'$ intersects $AC$ at point $P$. Prove that if $BP$ and $AM$ are perpendicular then triangle $ABC$ is right-angled. | [
"Let $E$ be the intersection point of $AD$ and $BP$. Because $AD$ is perpendicular to $BM$ and $BP$ is perpendicular to $AM$, the point $E$ is the orthocenter of triangle $ABM$ and therefore $ME$ is perpendicular to $AB$.\n\n\n\nBecause point $D$ is the midpoint of the segment $BB'$ and $DE... | Saudi Arabia | SAMC 2015 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English, Arabic | proof only | null | |
0at8 | Problem:
The incircle of a triangle has radius $4$, and the segments into which one side is divided by the point of contact with the incircle are of lengths $6$ and $8$. What is the perimeter of the triangle? | [
"Solution:\n\n$42$"
] | Philippines | Philippines Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | final answer only | 42 | |
05v3 | Problem:
Soit $m, n \geqslant 2$ des entiers tels que $\operatorname{PGCD}(m, n)=\operatorname{PGCD}(m, n-1)=1$. On définit la suite $\left(n_{k}\right)_{k \in \mathbb{N}}$ par $n_{0}=m$ et $n_{k+1}=n \cdot n_{k}+1$ pour $k \in \mathbb{N}$. Montrer que les entiers $n_{1}, \ldots, n_{m-1}$ ne peuvent pas tous être des ... | [
"Solution:\n\nTout d'abord, la suite $(n_{k})$ est une suite arithmético-géométrique. On peut donc établir une formule close pour son terme général, à savoir\n$$\nn_{k}=n^{k} m+\\frac{n^{k+1}-1}{n-1},\n$$\nque l'on peut aussi obtenir par récurrence à partir du calcul des premiers termes. De cette expression on dédu... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0gfr | 設 $n$ 為正整數, 且 $n$ 恰有 36 個不同的質因數。對於 $k = 1, 2, 3, 4, 5$, 令 $c_n$ 代表在區間 $\left[\frac{(k-1)n}{5}, \frac{kn}{5}\right]$ 中與 $n$ 互質的整數個數。證明：
$$
\sum_{1 \le i < j \le 5} (c_i - c_j)^2 \ge 2^{36}.
$$ | [] | Taiwan | 2022 數學奧林匹亞競賽第一階段培訓營, 獨立研究(三) | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Algebraic Expressions > Sequences and... | Chinese; English | proof only | null | |
01h1 | Consider the Euclidean plane, the points $A = (0,0)$ and $B = (1,0)$ and the open half-strip
$$
S = \{(x, y) : 0 < x < 1, y > 0\}
$$
with width 1 and vertices $A$ and $B$.
Find all functions $f: S \to S$ satisfying the following conditions for all $P, Q \in S$:
(i) $f(f(P)) = P$
(ii) If $P, Q, A$ are collinear, then $f... | [
"Fix any $0 < \\alpha < 90^\\circ$ and consider the ray of points $P \\in S$ with $\\angle BAP = \\alpha$. This ray (or the part of it which is in $S$) is mapped by (ii) under $f$ to a ray starting from $B$ with a certain angle $\\beta = \\beta(\\alpha)$, i.e., $\\angle f(P)BA = \\beta(\\alpha)$ for all such $P$. B... | Baltic Way | Baltic Way 2020 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | The unique function maps each point to the orthocenter of the triangle formed by that point and the two endpoints of the base segment. | |
0ele | Problem:
Poišči vsa realna števila $a \neq -1$, za katera je razmerje med rešitvama kvadratne enačbe $(2a+2)x^{2} + (2a+3)x + 1 = 0$ enako $1 : 3$. | [
"Solution:\nOpazimo, da lahko levo stran enačbe razstavimo in dobimo\n$$\n((2a+2)x+1)(x+1)=0\n$$\nRešitvi sta torej $x_{1} = -1$ in $x_{2} = -\\frac{1}{2a+2}$. Torej mora biti bodisi $x_{2} = -3$ bodisi $x_{2} = -\\frac{1}{3}$. V prvem primeru sledi $2a+2 = \\frac{1}{3}$, od koder dobimo $a = -\\frac{5}{6}$. V drug... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a = 1/2 or a = -5/6 | |
0cnz | Eight players participated in a chess tournament, and each pair of players have played exactly once. It appeared that if two players $A$ and $B$ played a draw then the resulting numbers of points of $A$ and $B$ are different. Find the greatest possible number of draws in this tournament. (Each win is worth $1$ point, e... | [
"Answer: $20$.\n\nWe will estimate the number $S$ — the sum of the numbers of draws for all $8$ chess players. This sum is exactly twice the number of draws in the tournament (since each draw is counted twice — for both players).\n\nWe will prove that $S \\le 41$ — then the number of drawn games in the tournament d... | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | 20 | |
08wt | Suppose you write down on a blackboard without repetition each of those positive integers which are less than or equal to $10^6$ and are divisible by $3$. How many $1$'s do you have to write on the blackboard? | [
"For $k = 0, 1, 2, 3, 4, 5$ let us denote by $A_k$ the set of all multiples of $3$ less than or equal to $10^6$ whose $10^k$'s digit is $1$, and define $N_k$ to be the number of elements in the set $A_k$. Then the desired answer for the problem is $\\sum_{k=0}^5 N_k$.\n\nThe set $A_5$ consists of multiples of $3$ l... | Japan | Japan Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | English | proof and answer | 199998 | |
0kvw | Problem:
Triangle $A B C$ has incircle $\omega$ and $A$-excircle $\omega_{A}$. Circle $\gamma_{B}$ passes through $B$ and is externally tangent to $\omega$ and $\omega_{A}$. Circle $\gamma_{C}$ passes through $C$ and is externally tangent to $\omega$ and $\omega_{A}$. If $\gamma_{B}$ intersects line $B C$ again at $D$... | [
"Solution:\n\nLet $\\overline{B C}$ touch the incircle at $X$ and the $A$-excircle at $Y$. Since $B X=C Y$, it suffices to show that $B P \\cdot B C=B X \\cdot B Y$ by symmetry.\n\n\n\nThe inversion centered at $B$ with radius $\\sqrt{B X \\cdot B Y}$\n- fixes lines $\\overline{A B}$ and $\... | United States | HMIC 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Ca... | null | proof only | null | |
0dlz | Determine all triples $(a, b, c)$ of integers such that
$$
a^3 + b^3 + c^3 = 25(abc + a^2b + b^2c + c^2a).
$$ | [
"We will show that there is no solution other than the trivial $a = b = c = 0$. For any other triple, we can write $a = dx$, $b = dy$, $c = dz$, where $\\gcd(x, y, z) = 1$ and $d$ is a positive integer. After dividing out $d$, we obtain the same equation but for $x$, $y$, and $z$. We will demonstrate that each of $... | Saudi Arabia | Saudi Booklet | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | a = b = c = 0 | |
0l4r | Problem:
Compute the number of ways to arrange the numbers $1$, $2$, $3$, $4$, $5$, $6$, and $7$ around a circle such that the product of every pair of adjacent numbers on the circle is at most $20$. (Rotations and reflections count as different arrangements.) | [
"Solution:\n\nFix the position of the number $7$. Note that the only numbers that can be next to $7$ are $1$ and $2$, so they must occupy the two slots adjacent to $7$.\n\nNow, the number $6$ can only be adjacent to $1$, $2$, and $3$. As $6$ can no longer be adjacent to both $1$ and $2$, we conclude that $6$ must b... | United States | HMMT February | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | proof and answer | 56 | |
061z | Problem:
Gegeben seien ein Kreis $K$ und eine Gerade $g$, die keinen gemeinsamen Punkt haben.
Ferner sei $\overline{AB}$ der Durchmesser von $K$, der orthogonal zu $g$ ist, wobei $B$ näher an $g$ liegt als $A$. Weiter sei ein beliebiger Punkt $C$, verschieden von $A$ und $B$, auf $K$ gegeben. Die Gerade $AC$ schneidet... | [
"Solution:\n\nWir bezeichnen den zweiten Schnittpunkt von $CF$ und $K$ mit $H$ sowie den Schnittpunkt von $AB$ und $g$ mit $X$. Wegen $AB \\perp g$ reicht es zu zeigen, dass $GH \\parallel g$. Dies ist genau dann der Fall, wenn $\\angle AGH = \\angle AFD$. Umfangswinkelsatz und Scheitelwinkel liefern $\\angle AGH =... | Germany | Auswahlwettbewerb zur IMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0dax | Let $a$, $b$, $c \geq 0$ satisfy $a + b + c = 1$. Prove that
$$
\frac{\sqrt{a}}{b + 1} + \frac{\sqrt{b}}{c + 1} + \frac{\sqrt{c}}{a + 1} > \frac{1}{2}(\sqrt{a} + \sqrt{b} + \sqrt{c}) .
$$ | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0fkt | Problem:
Dado un número natural $n$ mayor que $1$, hallar todos los pares de números enteros $a$ y $b$ tales que las dos ecuaciones $x^{n} + a x - 2008 = 0$ y $x^{n} + b x - 2009 = 0$ tengan, al menos, una raíz común real. | [
"Solution:\n\nRestando ambas ecuaciones tenemos que $(b-a)x = 1$. Luego, si estas ecuaciones van a tener una raíz común, tiene que ser $x = 1/(b-a)$. Notar que $a$ no puede ser igual a $b$.\n\nSustituyendo en una de las ecuaciones, tendremos que\n$$\n(b-a)^{n-1}(a - 2008(b-a)) = -1\n$$\ny que, por ser $a$ y $b$ ent... | Spain | XLV Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (2007, 2008) and ((-1)^n - 2008, (-1)^n - 2009) | |
00uw | Let $\mathbb{R}^+ = (0, \infty)$ be the set of all positive real numbers. Find all functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ and polynomials $g(x)$ with non-negative coefficients and $g(0) = 0$ that satisfy the equality:
$$
f(f(x) + g(y)) = f(x - y) + 2y
$$
for all positive real numbers $x > y$. | [
"Assume that $f : \\mathbb{R}^+ \\to \\mathbb{R}^+$ and the polynomial $g$ with non-negative coefficients and $g(0) = 0$ satisfy the conditions of the problem. For positive reals with $x > y$, we shall write $P(x, y)$ for the relation:\n$$\nf(f(x) + g(y)) = f(x - y) + 2y.\n$$\n1. Step 1. $f(x) \\ge x$. Assume that ... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Polynomials"
] | English | proof and answer | f(x) = x for all positive x; g(x) = x. | |
069z | Solve in the real numbers the inequality:
$$
\frac{(x+2)^4}{x^3} - \frac{(x+2)^2}{2x} \geq -\frac{x}{16}.
$$ | [
"For $x \\neq 0$ we have:\n$$\n\\begin{align*}\n\\frac{(x+2)^4}{x^3} - \\frac{(x+2)^2}{2x} &\\geq -\\frac{x}{16} \n\\Leftrightarrow \\frac{(x+2)^4}{x^3} - \\frac{(x+2)^2}{2x} + \\frac{x}{16} \\geq 0 \\\\\n&\\Leftrightarrow \\frac{16(x+2)^4 - 8(x+2)^2 x^2 + x^4}{16x^3} \\geq 0 \n\\Leftrightarrow \\frac{(4(x+2)^2 - x... | Greece | 37th Hellenic Mathematical Olympiad 2020 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | x > 0 or x = -4/3 or x = -4 | |
06aq | Let $ABC$ be a triangle with $AB > AC$, $AD$ its bisector, where $D$ is a point on the side $BC$ and $I$ its incenter. If $M$ is the midpoint of the segment $AD$ and $F$ is the point of intersection of the line $MB$ with the circumcircle of the triangle $BIC$, prove that: $AF \perp FC$.
(E. Psarras) | [
"Since $\\angle CII_a = \\angle BII_a = 90^\\circ$ the excenter $I_a$ belongs to the circumcircle of the triangle $BIC$. Moreover in the triangle $CDA$, $CI, CI_a$ are the internal and external bisector, respectively, and therefore the points $A, D$ are the conjugate harmonics of the points $I, I_a$. Since $M$ is t... | Greece | 40th Hellenic Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry >... | English | proof only | null | |
05zj | Problem:
Soit $ABC$ un triangle et $\Omega$ son cercle circonscrit. On note $A'$ le point diamétralement opposé à $A$ dans le cercle $\Omega$. Soit $I$ le centre du cercle inscrit au triangle $ABC$, $E$ et $F$ les points de contact du cercle inscrit avec les côtés $AC$ et $AB$ respectivement. Le cercle circonscrit au ... | [
"Solution:\n\n\n\nPuisque $F$ est le point de contact du cercle inscrit avec le côté $[AC]$, l'angle $\\widehat{IFA}$ est droit et le segment $[IA]$ est un diamètre du cercle circonscrit au triangle $AEF$. On en déduit que\n$$\n\\widehat{AXI} = 90^\\circ = \\widehat{AXA'}\n$$\noù on a utili... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - Envoi 5 : Pot Pourri | [
"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 | |
02if | Problem:
Um artesão começa a trabalhar às 8 h e produz 6 braceletes a cada vinte minutos; já seu auxiliar começa a trabalhar uma hora depois e produz 8 braceletes do mesmo tipo a cada meia hora. O artesão pára de trabalhar às $12~\mathrm{h}$, mas avisa ao seu auxiliar que este deverá continuar trabalhando até produzir... | [
"Solution:\n\nO artesão produz 6 braceletes a cada 20 minutos. Como 1 hora $=60$ minutos $=3 \\times 20$ minutos, o artesão produz $6 \\times 3=18$ braceletes em 1 hora. Como ele trabalhou $12$ horas $-8$ horas $=4$ horas, o número de braceletes feitos pelo artesão é $18 \\times 4=72$.\n\nO auxiliar produz 8 bracel... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
0jhz | Determine if there exists a (three-variable) polynomial $P(x, y, z)$ with integer coefficients satisfying the following property: A positive integer $n$ is *not* a perfect square if and only if there is a triple $(x, y, z)$ of positive integers such that $P(x, y, z) = n$.
(This problem was suggested by Gerhard Woeginge... | [
"The answer is *yes*. Suppose that $Q(x, y, z)$ is a polynomial with integer coefficients such that for all integers $x, y, z$, we have\n\n* $Q(x, y, z) \\ge 0$,\n* if $Q(x, y, z) = 0$ then $x$ is a non-square, and\n* for each positive non-square $x$, there exist $y, z \\in \\mathbb{Z}$ with $Q(x, y, z) = 0$.\n\nWe... | United States | TST | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | yes | |
067k | Prove that for the polynomial $P(x) = x^4 - x^3 - 3x^2 - x + 1$, there exist infinitely many positive integers $n$, for which $P(3^n)$ is composite. | [
"1. For $x = 3^{2n-1}$ we have:\n$$\n\\begin{aligned}\nP(3^{2n-1}) &= 81^{2n-1} - 27^{2n-1} - 3 \\cdot (-1)^{2n-1} - 3^{2n-1} + 1 \\\\\n&\\equiv 1^{2n-1} - 2^{2n-1} - 3 \\cdot 9^{2n-1} - 3^{2n-1} + 1 \\pmod{5} \\\\\n&\\equiv 5 - (2^{2n-1} + 3^{2n-1}) \\pmod{5}.\n\\end{aligned}\n$$\nFor odd exponents $m$, the number... | Greece | Mediterranean Mathematical Competition PETER O' HALLORAN MEMORIAL | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
03es | Given a rational number $q > 3$ such that $q^2 - 4$ is the square of a rational number. The row $\{a_i\}_{i=0}^{\infty}$ is defined as follows:
$$
a_0 = 2, \ a_1 = q, \ a_{i+1} = q a_i - a_{i-1}, \text{ for each } i = 1, 2, \dots
$$
Do there exist a natural number $n$ and nonzero integers $b_0, b_1, \dots, b_n$ such t... | [
"We will prove that such numbers do not exist.\n\nThe quadratic equation $x^2 - qx + 1 = 0$ has two rational roots $t$ and $\\frac{1}{t}$ for which $t + \\frac{1}{t} = q$. It easily follows by induction that $a_m = t^m + \\frac{1}{t^m}$. Let's assume that there exist numbers $b_0, b_1, \\dots, b_n$ satisfying the c... | Bulgaria | Bulgarian Winter Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
08dn | Problem:
Alberto e Barbara sono seduti, l'uno accanto all'altra, davanti a un tavolo su cui hanno disposto in fila, da sinistra verso destra, 15 cioccolatini. Alcuni dei cioccolatini sono al latte, gli altri al cioccolato fondente. A turno, iniziando da Alberto, giocano al seguente gioco: durante il proprio turno, cia... | [
"Solution:\n\nSupponiamo che, più in generale, un giocatore si trovi a dover effettuare una mossa con una fila di $n$ cioccolatini numerati da 1 a $n$; denoteremo con effettuare la mossa $k$, per il giocatore di turno, l'atto di mangiare i cioccolatini $\\{1, \\ldots, k\\}$. Diremo che la mossa $k$ è legale se $1 \... | Italy | XXXV Olimpiade Italiana di Matematica | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 8320 | |
0emn | On a $5 \times 5$ board, two players alternately mark numbers on empty cells. The first player always marks $1$'s, the second $0$'s. One number is marked per turn, until the board is filled. For each of the nine $3 \times 3$ squares the sum of the nine numbers on its cells is computed. Denote by $A$ the maximum of thes... | [
"First, notice that player two can always ensure that $A \\le 6$. The squares of the grid can be partially tiled with $2 \\times 1$ tiles, as shown below. Whenever a $1$ is placed inside a tile, player two can ensure that it also contains a $0$. As every $3 \\times 3$ square contains three complete tiles, it will h... | South Africa | South-Afrika 2011-2013 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 6 | |
0c36 | The point $T$ is taken on the edge $AD$ of the regular pyramid $VABCD$, with apex $V$, so that $\frac{AT}{TD} = \frac{1}{4}$, and the straight line $TS$, perpendicular on $VB$, is drawn ($S \in VB$). If the straight lines $TS$ and $AD$ are perpendicular, prove that the height of the pyramid equals the length of the sid... | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
04m6 | Let $\tau(n)$ be the number of positive divisors of $n$. Let $\tau_1(n)$ be the number of positive divisors of $n$ which give remainder $1$ when divided by $3$. Find all possible integral values of the fraction
$$
\frac{\tau(10n)}{\tau_1(10n)}.
$$
$(\text{IMO Shortlist 2016})$ | [
"Let $n = 3^a \\cdot b \\cdot m$, where $a$ is a non-negative integer, $b$ and $m$ positive integers such that all prime factors of $b$ give remainder $1$ when divided by $3$, and all prime factors of $m$ give remainder $2$.\nThen $\\tau(10n) = (a+1) \\cdot \\tau(b) \\cdot \\tau(10m)$.\nFor $\\tau_1(n)$, a prime fa... | Croatia | Croatian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | All even positive integers and all odd composite integers (equivalently, all integers greater than one except odd primes). | |
0f1z | Problem:
$S$ is a set of $1976$ points which form a regular $1976$-gon. $T$ is the set of all points which are the midpoint of at least one pair of points in $S$. What is the greatest number of points of $T$ which lie on a single circle? | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | 1976 | |
06nz | A two digit number $s$ is *special* if $s$ is the common two leading digits of the decimal expansion of $4^n$ and $5^n$, where $n$ is a certain positive integer. Given that there are two special numbers, find these two special numbers. | [
"The condition on $s$ means there exists a positive integer $n$ and nonnegative integers $a$ and $b$ such that\n$$\n10^a s \\le 5^n < 10^a (s+1) \\quad \\text{and} \\quad 10^b s \\le 4^n < 10^b (s+1).\n$$\nNote that $10^a s = 5^n$ and $10^b s = 4^n$ cannot hold simultaneously. Multiplying the second inequality and ... | Hong Kong | IMO HK TST | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | 21 and 46 | |
0j0s | Problem:
Suppose that there are real numbers $a, b, c \geq 1$ and that there are positive reals $x, y, z$ such that
$$
\begin{aligned}
a^{x}+b^{y}+c^{z} & =4 \\
x a^{x}+y b^{y}+z c^{z} & =6 \\
x^{2} a^{x}+y^{2} b^{y}+z^{2} c^{z} & =9
\end{aligned}
$$
What is the maximum possible value of $c$ ? | [
"Solution:\nAnswer: $\\sqrt[3]{4}$\n\nThe Cauchy-Schwarz inequality states that given 2 sequences of $n$ real numbers $x_{1}, x_{2}, \\ldots, x_{n}$ and $y_{1}, y_{2}, \\ldots, y_{n}$, then\n$$\n\\left(x_{1}^{2}+x_{2}^{2}+\\ldots+x_{n}^{2}\\right)\\left(y_{1}^{2}+y_{2}^{2}+\\ldots+y_{n}^{2}\\right) \\geq \\left(x_{... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | sqrt[3]{4} | |
08qe | Problem:
Find all triples $(a, b, c)$ of nonnegative integers that satisfy
$$
a! + 5^{b} = 7^{c}
$$ | [
"Solution:\nWe cannot have $c=0$ as $a!+5^{b} \\geqslant 2 > 1 = 7^{0}$.\n\nAssume first that $b=0$. So we are solving $a!+1=7^{c}$. If $a \\geqslant 7$, then $7 \\mid a!$ and so $7 \\nmid a!+1$. So $7 \\nmid 7^{c}$ which is impossible as $c \\neq 0$. Checking $a<7$ by hand, we find the solution $(a, b, c) = (3, 0,... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (3, 0, 1), (2, 1, 1), (4, 2, 2) | |
0133 | Problem:
Let $n$ be a positive integer such that the sum of all the positive divisors of $n$ (except $n$) plus the number of these divisors is equal to $n$. Prove that $n=2 m^{2}$ for some integer $m$. | [
"Solution:\nLet $t_{1}<t_{2}<\\cdots<t_{s}$ be all positive odd divisors of $n$, and let $2^{k}$ be the maximal power of $2$ that divides $n$. Then the full list of divisors of $n$ is the following:\n$$\nt_{1}, \\ldots, t_{s}, 2 t_{1}, \\ldots, 2 t_{s}, \\ldots, 2^{k} t_{1}, \\ldots, 2^{k} t_{s} .\n$$\nHence,\n$$\n... | Baltic Way | Baltic Way | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0d5c | Find the number of 6-tuples $(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6})$ of distinct positive integers satisfying the following two conditions:
a. $a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} = 30$;
b. We can write $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ on sides of a hexagon such that after a finite number of t... | [
"We label the vertex of the hexagon by $1,2,3,4,5,6$ and suppose that six numbers are written in the order $a, b, c, d, e, f$ on the edges $(1,2), (2,3), \\ldots, (6,1)$, respectively.\n\nWe first notice that by choosing a vertex of the hexagon and adding 1 to the two numbers written on two adjacent sides to the ve... | Saudi Arabia | SAMC 2015 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English, Arabic | proof and answer | 13680 | |
0dd6 | For a non-empty set $\mathcal{T}$ denote by $p(\mathcal{T})$ the product of all elements of $\mathcal{T}$. Does there exist a set $\mathcal{T}$ of 2021 elements such that for any $a \in \mathcal{T}$ one has that $p(\mathcal{T}) - a$ is an odd integer. Consider two cases:
1. All elements of $\mathcal{T}$ are irrational ... | [
"1. Consider the polynomial\n$$\nf(x) = x(x+2)(x+4)\\dots(x+4040) - x - (2m-1)\n$$\nfor some $m \\in \\mathbb{Z}^+$ then $\\lim_{x \\to +\\infty} f(x) = +\\infty$ and $f(0) = 2^{2020} \\cdot 2020! - (2m-1)$. Take $m$ big enough to get $f(0) < 0$ then by the continuous property of $f(x)$, there exists some real numb... | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Case 1: Yes, such a set exists with all elements irrational. Case 2: No, it is impossible if at least one element is rational. | |
09b6 | Let $t$, $k$, $m$ be positive integers and $t > \sqrt{km}$. Prove that
$$
\binom{2m}{0} + \binom{2m}{1} + \dots + \binom{2m}{m-t-1} < \frac{2^{2m}}{2k}
$$ | [
"**Lemma.** For all integers $0 \\le t, s \\le m$ such that $t + s \\le m$, $\\frac{\\binom{2m}{m-s}}{\\binom{2m}{m-t-s}} > \\frac{t^2}{m}$.\n\n**Proof of lemma.**\n$$\n\\begin{aligned}\nA &= \\frac{\\binom{2m}{m-s}}{\\binom{2m}{m-t-s}} = \\frac{(m-t-s)!(m+t+s)!}{(m-s)!(m+s)!} = \\\\\n&= \\left( \\frac{(m+t+1)(m+t+... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
07r8 | The sequence $a = (a_0, a_1, a_2, \dots)$ is defined by $a_0 = 0$, $a_1 = 2$ and
$$
a_{n+2} = 2a_{n+1} + 41a_n \quad \text{for all } n \ge 0.
$$
Prove that $a_{2016}$ is divisible by $2017$. | [
"The equation $x^2 - 2x - 41 = 0$ has roots $\\alpha = 1 + \\sqrt{42}$ and $\\beta = 1 - \\sqrt{42}$. Using $a_0 = 0$ and $a_1 = 2$, we obtain\n$$\na_n = \\frac{\\alpha^n - \\beta^n}{\\sqrt{42}} \\quad \\text{for all } n \\ge 0. \\qquad (2)\n$$\nNote that if $h$ and $k$ are positive integers, then\n$$\n\\alpha^{h(k... | Ireland | Ireland_2017 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > ... | English | proof only | null | |
0cof | Given a right-angled triangle $ABC$ with $\angle ACB = 90^\circ$. One has increased each of the lengths of its legs $AC$ and $BC$ by $1$ (and $\angle ACB$ remained right). Could it happen that the length of the hypothenuse $AB$ increased by more than $\sqrt{2}$? (S. Volchenkov)
Каждый катет прямоугольного треугольника... | [
"**Первое решение.** Пусть длины катетов исходного прямоугольного треугольника были равны $x$ и $y$. Тогда его гипотенуза имела длину $\\sqrt{x^2 + y^2}$, а после увеличения катетов стала $\\sqrt{(x+1)^2 + (y+1)^2}$. Предположим, что гипотенуза увеличится более, чем на $\\sqrt{2}$. Тогда\n\n$$\n\\begin{aligned}\n\\... | Russia | Regional round | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English; Russian | proof and answer | No | |
0bcd | Problem:
Fie $a, b \in \mathbb{R}$ şi $z \in \mathbb{C} \setminus \mathbb{R}$ astfel încât $|a-b| = |a+b-2z|$.
a) Să se arate că ecuaţia $|z-a|^{x} + |\bar{z}-b|^{x} = |a-b|^{x}$, cu necunoscuta $x \in \mathbb{R}$, are soluţie unică.
b) Să se rezolve inecuaţia $|z-a|^{x} + |\bar{z}-b|^{x} \leq |a-b|^{x}$, cu necunos... | [] | Romania | Olimpiada Naţională de Matematică Etapa Judeţeană şi a Municipiului Bucureşti | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | a) x = 2. b) x ≥ 2. | |
027l | Problem:
Em um torneio, quaisquer dois jogadores jogam entre si. Cada jogador obtém um ponto por vitória, $1/2$ por empate e $0$ ponto por derrota. Seja $S$ o conjunto das $10$ menores pontuações. Sabemos que cada jogador obteve metade da sua pontuação jogando contra jogadores de $S$.
a) Qual a soma das pontuações do... | [
"Solution:\n\na) Os jogadores de $S$, em partidas disputadas apenas entre si, obtiveram $\\frac{10(10-1)}{2}=45$ pontos. Como os que eles obtiveram jogando entre si correspondem a metade dos pontos que cada um obteve no torneio, podemos concluir que a soma dos pontos dos jogadores de $S$ é $45+45=90$.\n\nb) Sejam $... | Brazil | NÍVEL 3 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a) 90; b) 25 | |
06cw | a. Let $a$, $n$, $k$ be positive integers. Prove that $a^{4n+k}$ and $a^k$ have identical first (rightmost) digits in their decimal representations.
b. Find the first digit of the decimal representation of the number
$$
2^{1999} + 7^{1999} + 9^{1999}.
$$ | [
"a. By the Euler-Fermat theorem, since $\\varphi(10) = 4$ and $4n + k \\equiv k \\pmod{4}$, we have\n$$\na^{4n+k} \\equiv a^k \\pmod{10}\n$$\nfor any positive integers $a$, $n$, $k$. This means $a^{4n+k}$ and $a^k$ have the same rightmost digit.\n\nb. The rightmost digit is $0$.\nBy part (a), since $1999 \\equiv 3 ... | Hong Kong | HKG TST | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | 0 |
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