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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0f5k | Problem:
$a_n$ is the last digit of $[10^{n / 2}]$. Is the sequence $a_n$ periodic?
$b_n$ is the last digit of $[2^{n / 2}]$. Is the sequence $b_n$ periodic? | [] | Soviet Union | 17th ASU | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Modular Arithmetic"
] | null | proof only | a_n is not periodic; b_n is not periodic | |
0hif | What is the maximum number of points that can be placed on the plane so that there are exactly $2012$ straight lines, which pass through at least two of them?
(Danylo Mysak)
Fig. 24
**Fig. 25** | [
"It is easy to place $2012$ points for the condition of the problem being fulfilled: we will place $2011$ points on a straight line, and the last one — outside of it (fig. 24).\n\n*Proof.* Let's draw all the lines through every pair of points from the set $M$. We will find a line and a point from $M$ such that the ... | Ukraine | Problems from Ukrainian Authors | [
"Geometry > Plane Geometry > Combinatorial Geometry > Sylvester's theorem",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 2012 | |
0hp1 | Problem:
The Cannibal Club of California (CCC) had 30 members yesterday morning - but that was before their festive annual dinner! After the dinner, it turned out that among any six members of the club, there was a pair one of whom ate the other. Prove that at least six members of the CCC are now nested inside one ano... | [
"Solution:\n\nWe must assume that nobody was eaten by more than one person for the problem statement to make sense. To each cannibal we assign a numerical \"depth\" as follows: the depth of cannibal $C$ is the largest integer $n$ such that there exist cannibals $C_{1}, C_{2}, \\ldots, C_{n}=C$ such that $C_{i}$ ate... | United States | Berkeley Math Circle Take-Home Contest #2 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0h4c | Non-intersecting circles $\omega_1$ and $\omega_2$ of radii are inscribed into angle $BAC$ with
$B \in \omega_1$, $C \in \omega_2$, and the radius of $\omega_1$ is smaller than that of $\omega_2$. Let $K \neq B$ and $N \neq C$ be the points of intersection of $BC$ with $\omega_1$ and $\omega_2$ respectively. Let also ... | [
"Проведемо дотичну $AF$ до описаного кола трикутника $APB$. Тоді $\\angle PAF = \\angle PBA = \\angle PKB$, а тому $AF \\parallel BC$. Отже, $\\angle FAC = \\angle ACN = \\angle CMA$. З цього випливає, що пряма $AF$ дотикається до описаного кола трикутника $ACM$. Оскільки описані кола трикутників $АСМ$ і $АВР$ в то... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0fxh | Problem:
Ein Palindrom ist eine natürliche Zahl, die im Dezimalsystem vorwärts und rückwärts gelesen gleich gross ist (z.B. $1129211$ oder $7337$). Bestimme alle Paare $(m, n)$ natürlicher Zahlen, sodass
ein Palindrom ist. | [
"Solution:\n\nDies gilt genau dann, wenn $m \\leq 9$ oder $n \\leq 9$ gilt. Sei zuerst oBdA $n \\geq m$ und $m \\leq 9$. Mit Hilfe der schriftlichen Multiplikation aus der Grundschule erhält man dann\n\nalso tatsächlich ein Palindrom. Wir nehmen nun $n, m \\geq 10$ an und betrachten wieder ... | Switzerland | SMO Finalrunde | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | proof and answer | All (m, n) with min(m, n) ≤ 9 | |
03qm | In a planar rectangular coordinate system, a sequence of points $\{A_n\}$ on the positive half of the $y$-axis and a sequence of points $\{B_n\}$ on the curve $y = \sqrt{2x}$ ($x \ge 0$) satisfy the condition $|OA_n| = |OB_n| = \frac{1}{n}$. The $x$-intercept of line segment $A_nB_n$ is $a_n$, and the $x$-coordinate of... | [
"(1) According to the stated conditions we have $A_n(0, \\frac{1}{n})$, and $B_n(b_n, \\sqrt{2b_n})$ ($b_n > 0$). From $|OB_n| = \\frac{1}{n}$ we get\n$$\nb_n^2 + 2b_n = \\left(\\frac{1}{n}\\right)^2.\n$$\nThus,\n$$\nb_n = \\sqrt{\\left(\\frac{1}{n}\\right)^2 + 1} - 1, \\quad n \\in \\mathbb{N}.\n$$\n\nSince $2n^2b... | China | China Mathematical Competition (Extra Test) | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0fo7 | Let $n \ge 1$ be a positive integer and for all $t \in \mathbb{R}$ let
$$
P(t) = 1 + t + t^2 + \dots + t^{2n}
$$
If $x \in \mathbb{R}$, $P(x) \in \mathbb{Q}$ and $P(x^2) \in \mathbb{Q}$, then show that $x \in \mathbb{Q}$. | [
"It is easily to check that holds:\n\n(i) $P(t) > 0$ for all $t \\in \\mathbb{R}$.\n\n(ii) $P(t)P(-t) = P(t^2)$ for all $t \\in \\mathbb{R}$.\n\n(iii) $t = \\frac{P(t) + P(-t) - 2}{P(t) - P(-t)}$ for all $t \\neq 0$.\n\nLet $x \\in \\mathbb{R}$ such that $P(x) \\in \\mathbb{Q}$ and $P(x^2) \\in \\mathbb{Q}$. Then f... | Spain | International Mathematical Arhimede Contest | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | Spanish | proof only | null | |
09k9 | Find all positive integers $n$ for which checkers can be placed on the cells of an $n \times n$ chessboard in such a way that each cell has exactly two neighboring cells with checkers. Two cells are considered neighbors if they share a common side. | [
"Answer: $n$ even.\n\nLet's start by proving that any even number $n$ is a good number.\nFor the $n = 2k$ case we can construct a checkered pattern that all the cells of the frame contains checker and contains $2k - 4$ case in the middle. The figure below illustrates this construction:\n\nNow, let's show that for a... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | all even positive integers | |
0fjp | Problem:
Se representa por $\mathbb{Z}$ el conjunto de todos los enteros. Hallar todas las funciones $f: \mathbb{Z} \rightarrow \mathbb{Z}$ tales que, para cualesquiera $x, y$ enteros se cumple
$$
f(x+f(y))=f(x)-y
$$ | [
"Solution:\n\nPrimeramente observemos que $f(x+n f(y))=f(x)-n y$.\nPara $n=0$ es obvio, y por inducción, suponemos que para el entero $n \\geq 1$ se cumple\n$$\nf(x+(n-1) f(y))=f(x)-(n-1) y\n$$\nEntonces\n$$\n\\begin{aligned}\n& f(x+n f(y))=f(x+(n-1) f(y)+f(y))= \\\\\n& \\quad=f(x+(n-1) f(y))-y= \\\\\n& \\quad=f(x)... | Spain | Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | no such function exists | |
0dth | Find the minimum value of
$$
\frac{x_1^3 + \cdots + x_n^3}{x_1 + \cdots + x_n}
$$
where $x_1, x_2, \dots, x_n$ are distinct positive integers. | [
"The minimum value $\\frac{1}{2}n(n+1)$ is achieved by letting $x_k = k$ for $1 \\le k \\le n$. To prove the inequality, it suffices to prove that\n$$\nx_1^3 + \\cdots + x_n^3 \\ge (x_1 + \\cdots + x_n)^2,\n$$\nsince $x_1 + \\cdots + x_n \\ge 1 + 2 + \\cdots + n = n(n+1)/2$.\nWe may assume that $x_1 < x_2 < \\cdots... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | n(n+1)/2 | |
0cq9 | At some moment, three different points were marked in the plane. After that, at each minute three marked points are chosen (denote them by $A$, $B$, $C$), and a point $D$ symmetrical to $A$ with respect to the perpendicular bisector to $BC$ is marked.
A day later, it appeared that there exist three collinear marked poi... | [
"Предположим противное; тогда исходные три точки лежат на некоторой окружности $\\omega$. Докажем индукцией по количеству минут, что все отмеченные точки также лежат на $\\omega$. Действительно, изначально это верно. Пусть в некоторый момент по точкам $A$, $B$, $C$ строится точка $D$. Тогда серединный перпендикуляр... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English, Russian | proof only | null | |
0kj0 | In a particular game, each of 4 players rolls a standard 6-sided die. The winner is the player who rolls the highest number. If there is a tie for the highest roll, those involved in the tie will roll again and this process will continue until one player wins. Hugo is one of the players in this game. What is the probab... | [
"First observe that if $q$ players tie on the initial roll, the probability that any one of these $q$ players will ultimately win is $\\frac{1}{q}$. Let $N$ be the value of Hugo's first roll.\n\nConsider four cases based on the number of highest scoring rolls in the first round. The probability that Hugo will roll ... | United States | Fall 2021 AMC 10 B | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | MCQ | C | |
0gq0 | There are $n$ chests placed on the vertices of a regular $n$-gon and a bead. Alice and Bob play a game. At the beginning Alice hides the bead in one of the chests. Each move consists of three stages:
* Alice, if she wishes, can secretly move the bead from the located chest to any of two neighboring chests if the neighb... | [
"Let $f(n)$ be the minimal number of moves necessary for Bob to guarantee winning. We will show that if $n = 5$ Bob cannot win and\n$$\nf(n) = \\begin{cases} 2 & \\text{if } n = 3, 4, n \\ge 12 \\\\ 3 & \\text{if } 6 \\le n \\le 11 \\end{cases}\n$$\n\nLet the vertices be $v_1, v_2, \\dots, v_n$ in clockwise directi... | Turkey | Team Selection Test for JBMO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Bob cannot win for n = 5. Otherwise, the minimal guaranteed number of moves is f(n) = 2 for n = 3, 4 and for all n ≥ 12, and f(n) = 3 for 6 ≤ n ≤ 11. | |
0jge | Problem:
A 50-card deck consists of 4 cards labeled "$i$" for $i = 1, 2, \ldots, 12$ and 2 cards labeled "13". If Bob randomly chooses 2 cards from the deck without replacement, what is the probability that his 2 cards have the same label? | [
"Solution:\n\nAll pairs of distinct cards (where we distinguish cards even with the same label) are equally likely. There are $\\binom{2}{2} + 12 \\binom{4}{2} = 73$ pairs of cards with the same label and $\\binom{50}{2} = 100 \\cdot \\frac{49}{4} = 1225$ pairs of cards overall, so the desired probability is $\\fra... | United States | HMMT November 2013 | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 73/1225 | |
0fgw | Problem:
Se atribuye al matemático renacentista Leonardo da Pisa (más conocido como Fibonacci) la sucesión definida de la manera siguiente
$$
\begin{aligned}
& a_{1}=1 \\
& a_{2}=1 \\
& a_{i}=a_{i-1}+a_{i-2} \quad \text{ para } i>2
\end{aligned}
$$
Expresar $a_{2 n}$ en función solamente de los tres términos $a_{n-1}$... | [
"Solution:\n\nPrimera solución (combinatoria)\n\nLlamemos $f_{n}$ a la sucesión de Fibonacci propiamente dicha, es decir, la que cumple $f_{n+1} = f_{n} + f_{n-1}$, con $f_{1} = 1$ y $f_{2} = 1$, y pongamos $a_{n} = f_{n+1}$ para $n \\geq 1$. La sucesión $a_{n}$ cumple también la recurrencia $a_{n+1} = a_{n} + a_{n... | Spain | OME 24 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | a_{2n} = a_n (a_{n-1} + a_{n+1}) | |
0c2p | Given a non-negative integer $k$, show that there are infinitely many positive integers $n$ such that the product of any $n$ consecutive integers is divisible by $(n+k)^2 + 1$. | [
"*First solution.* Since the product of $n$ consecutive integers is divisible by $n!$, it is sufficient to show that there are infinitely many positive integers $n$ such that $n!$ is divisible by $(n+k)^2 + 1$.\nTo obtain infinitely many positive integers $n$ such that $n!$ is divisible by $(n+k)^2 + 1$, it is suff... | Romania | 69th NMO Selection Tests for BMO and IMO | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
04tr | Let $p$ be a circle with center $K$ passing through $M$, $q$ a semicircle with diameter $KM$ and $L$ a point inside the segment $KM$. A line through $L$ perpendicular to $KM$ intersects $q$ at point $Q$ and $p$ at points $P_1, P_2$ such that $P_1Q > P_2Q$. Line $MQ$ intersects $p$ for the second time at $R \ne M$. Prov... | [
"The circle containing semicircle $q$ is the image of $p$ in homothety with center $M$ and factor $1/2$, hence $Q$ is the midpoint of $RM$. Since triangles $MP_1Q$, $P_2RQ$ share the angle by $Q$, we have\n$$\n\\frac{S_1}{S_2} = \\frac{\\frac{1}{2} P_1 Q \\cdot M Q \\cdot \\sin \\angle P_1 Q M}{\\frac{1}{2} P_2 Q \... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
016n | Let $a, b, c, d$ be real numbers such that $a, b \le c, d$. Prove
$$
(a + b + c + d)^2 \ge 8(ac + bd).
$$
When does equality hold? | [
"We have\n$$\n\\begin{aligned}\nD &= (a+b+c+d)^2 - 8(ac+bd) \\\\\n&= (a+c)^2 - 4ac + (b+d)^2 - 4bd + 2[(a+c)(b+d) - 2ac - 2bd] \\\\\n&= (c-a)^2 + (d-b)^2 + 2[(d-a)(c-b) - (b-a)(d-c)].\n\\end{aligned}\n$$\nWithout loss of generality $a \\le b$ can be assumed. Then, if $c > d$, the first term in the expression is pos... | Baltic Way | BALTIC WAY | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | Equality holds if and only if a + d = b + c and either a = d or b = c. | |
005n | Se tienen $N$ segmentos cerrados en una recta. Se sabe que para cada $d$, $0 < d \le 1$, existen dos puntos en un segmento o en dos segmentos distintos que se encuentran a distancia $d$.
a) Demuestre que la suma de las longitudes de los segmentos es mayor o igual que $\frac{1}{N}$.
b) Demuestre, para cada $N$, que $\fr... | [] | Argentina | XVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | Spanish | proof only | null | |
02rf | In this problem, a *box* is a parallelepiped $P \in \mathbb{R}^3$. We define the *size* of a box $P$ as $a^s + b^s + c^s$, $a, b, c$ being its dimensions and $s$ a fixed integer.
Find all values of $s$ such that the following statement is true: if a box $P_1$ is inside box $P_0$ then the size of $P_1$ does not exceed s... | [] | Brazil | Brazilian Math Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | s = 0 and s = 1 | |
07mz | Find all integers $n$ for which $n^5 + n^4 + n^3 + n^2 + n + 1$ is divisible by 199. | [
"Note that $(n-1)(n^5 + n^4 + n^3 + n^2 + n + 1) = n^6 - 1 = (n^3 - 1)(n^3 + 1) = (n-1)(n+1)(n^2 - n + 1)(n^2 + n + 1)$, and so\n$$\nn^5 + n^4 + n^3 + n^2 + n + 1 = (n+1)(n^2 - n + 1)(n^2 + n + 1).\n$$\nBecause 199 is a prime number, this expression is divisible by 199 iff one of the factors $n+1$, $n^2-n+1$ or $n^... | Ireland | Ireland | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof and answer | n ≡ 92, 93, 106, 107, 198 (mod 199) | |
0a4q | Problem:
Show that for all positive integers $k$, there exists a positive integer $n$ such that $n2^{k} - 7$ is a perfect square. | [
"Solution:\n\nProof by induction on $k$.\n\nFor the base cases ($k \\leq 3$) we can simply choose $n = 2^{3 - k}$ to get $n2^{k} - 7 = 2^{3} - 7 = 1^{2}$.\n\nFor the inductive step let $k \\geq 3$ and assume there exist integers $a$ and $n$ such that\n$$\na^{2} = n2^{k} - 7.\n$$\nWe will now endeavour to find integ... | New Zealand | New Zealand Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
019n | A polynomial $P$ with integer coefficients satisfies
$$
P(x_1) = P(x_2) = \dots = P(x_k) = 54
$$
and
$$
P(y_1) = P(y_2) = \dots = P(y_n) = 2013
$$
for distinct integers $x_1, \dots, x_k; y_1, \dots, y_n$. Determine the maximal value of $kn$. | [
"Letting $Q(x) = P(x) - 54$, we see that $Q$ has $k$ zeroes at $x_1, \\dots, x_k$, while $Q(y_i) = 1959$ for $i = 1, \\dots, n$. We notice that $1959 = 3 \\cdot 653$, and an easy check shows that $653$ is a prime number. As\n$$\nQ(x) = \\prod_{j=1}^{k} (x - x_j)S(x),\n$$\nand $S(x)$ is a polynomial with integer coe... | Baltic Way | Baltic Way 2013 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 6 | |
0el7 | Problem:
V kraju Zmajski Vrh so se prebivalci odločili, da bodo uporabljali nov način merjenja dnevnega časa. Vsak dan so s poldnevom in polnočjo razdelili na dve enaki polovici. Namesto, da bi vsako polovico razdelili na 12 ur s po 60 minutami, so jo razdelili na 10 zmajskih ur s po 100 zmajskimi minutami. Župan Zmaj... | [
"Solution:\n\nČas $8.25$ v zmajskih urah in zmajskih minutah ustreza $8 \\frac{25}{100} = 8 \\frac{1}{4} = \\frac{33}{4}$ zmajskih ur. Ker $10$ zmajskih ur ustreza $12$ običajnim uram, $1$ zmajska ura ustreza $\\frac{12}{10}$ običajne ure. Torej $\\frac{33}{4}$ zmajskih ur ustreza $\\frac{33}{4} \\cdot \\frac{12}{1... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | E | |
0dv2 | Problem:
Naj bo $A'$ nožišče višine na stranico $BC$ ostrokotnega trikotnika $ABC$. Krožnica s premerom $AA'$ seka stranico $AB$ v točkah $A$ in $D$, stranico $AC$ pa v točkah $A$ in $E$. Dokaži, da leži središče očrtane krožnice trikotnika $ABC$ na nosilki višine na $DE$ trikotnika $ADE$. | [
"Solution:\n\nOznačimo z $A''$ nožišče višine iz $A$ trikotnika $ADE$. Središče trikotniku $ABC$ očrtane krožnice leži na premici $AA''$, če je $\\angle BAA'' = \\frac{\\pi}{2} - \\gamma$, kjer je $\\gamma = \\angle ACB$. Ker je $AA'' \\perp DE$ in $AB \\perp A'D$ ter je štirikotnik $ADA'E$ tetiven, je $\\angle BAA... | Slovenia | 46. matematično tekmovanje srednješolcev Slovenije | [
"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 only | null | |
0abp | a)
$$
1\frac{5}{8} + \left(1\frac{1}{2} + \left(\left(1\frac{1}{2} - \frac{1}{6}\right) \cdot \left(\frac{1}{3} + \frac{1}{4}\right) + 1\right)\right) : \frac{2}{3}
$$
b)
$$
0,6 : \frac{1\frac{1}{2} + 0,5 \cdot 2\frac{1}{2} - 0,25}{15 - 0,5}
$$ | [
"a)\n$$\n\\begin{aligned}\n1\\frac{5}{8} + \\left(1\\frac{1}{2} + \\left(\\left(1\\frac{1}{2} - \\frac{1}{6}\\right) \\cdot \\left(\\frac{1}{3} + \\frac{1}{4}\\right) + 1\\right)\\right) : \\frac{2}{3} &= 1\\frac{5}{8} + \\left(1\\frac{1}{2} + \\left(\\frac{2}{6} \\cdot \\frac{7}{12} + 1\\right)\\right) : \\frac{2}... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | final answer only | a) 5 2/3; b) 6 | |
0e38 | Let $x$, $y$ and $z$ be real numbers such that $0 \le x, y, z \le 1$. Prove that
$$
xyz + (1-x)(1-y)(1-z) \le 1.
$$
When does the equality hold? | [
"The inequality is equivalent to $xy + yz + zx \\le x + y + z$. Since $x$ is positive and $y \\le 1$ we have $xy \\le x$. A similar reasoning shows that $yz \\le y$ and $zx \\le z$. Hence, the inequality holds.\n\nThe equality holds if and only if $xy = x$, $yz = y$ and $zx = z$. If $x=0$, then the third equality i... | Slovenia | National Math Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Equality holds if and only if x=y=z=0 or x=y=z=1. | |
0icz | Problem:
Shown on your answer sheet is a $20 \times 20$ grid. Place as many queens as you can so that each of them attacks at most one other queen. (A queen is a chess piece that can move any number of squares horizontally, vertically, or diagonally.) It's not very hard to get 20 queens, so you get no points for that,... | [
"Solution:\n\nAn elementary argument shows there cannot be more than 26 queens: we cannot have more than 2 in a row or column (or else the middle queen would attack the other two), so if we had 27 queens, there would be at least 7 columns with more than one queen and thus at most 13 queens that are alone in their r... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0a54 | Problem:
A sequence of $A$s and $B$s is called antipalindromic if writing it backwards, then turning all the $A$s into $B$s and vice versa, produces the original sequence. For example $A B B A A B$ is antipalindromic. For any sequence of $A$s and $B$s we define the cost of the sequence to be the product of the position... | [
"Solution:\nFor each integer $0\\leq k\\leq 1009$ define a $k$-pal to be any sequence of 2020 $A$s and $B$s, where the first $k$ terms are $B$, the last $k$ terms are $B$, and the middle $(2020 - 2k)$ terms form an antipalindromic sequence.\n\nNow for any $k$, define $f(k)$ to be sum of the costs of all $k$-pals. N... | New Zealand | New Zealand Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof and answer | 2021^1010 | |
010u | Problem:
Does there exist a finite sequence of integers $c_{1}, \ldots, c_{n}$ such that all the numbers $a+c_{1}, \ldots, a+c_{n}$ are primes for more than one but not infinitely many different integers $a$? | [
"Solution:\n\nAnswer: yes.\nLet $n=5$ and consider the integers $0, 2, 8, 14, 26$. Adding $a=3$ or $a=5$ to all of these integers we get primes. Since the numbers $0, 2, 8, 14$ and $26$ have pairwise different remainders modulo $5$ then for any integer $a$ the numbers $a+0, a+2, a+8, a+14$ and $a+26$ have also pair... | Baltic Way | Baltic Way | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
028d | Problem:
O preço da gasolina - Em 1972 encher o tanque de gasolina de um carro pequeno custava $R\$ 29,90$, e em 1992, custava $\$ 149,70$ para encher o mesmo tanque. Qual dos valores abaixo melhor aproxima o percentual de aumento no preço da gasolina nesse período de 20 anos?
(a) $20\%$
(b) $125\%$
(d) $300\%$
(d) $4... | [
"Solution:\n\nO aumento do valor foi\n$$\n149,70 - 29,90 = 119,80 \\text{ reais }\n$$\nque corresponde a:\n$$\n\\frac{119,80}{29,90} \\times 100\\% = 400,66\\%\n$$\nA opção correta é (d)."
] | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | (d) | |
0g4r | Problem:
during the World Cup, there are $n$ different Panini stickers to collect. Marco's friends are trying to complete their collection, but nobody has a full set of stickers yet! A pair of his friends are said to be wholesome if their combined collection has at least one of each sticker. Marco knows the contents o... | [
"Solution:\n\nto show that at least $n$ different tables are necessary, suppose we had $n$ people, each of which are only missing a sticker, with each person missing a different sticker. Any two of these people are wholesome together, so they must be seated at different tables.\n\nNow, to show that $n$ is always su... | Switzerland | Second round 2023 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n | |
0l58 | Problem:
A parallelogram $P$ can be folded over a straight line so that the resulting shape is a regular pentagon with side length $1$. Compute the perimeter of $P$. | [
"Solution:\n\n\n\nIn regular pentagon $ABCDE$ (labeled clockwise), reflect $ABDE$ across $AB$ to obtain $ABD'E'$. Then, $CDE'D'$ is one such parallelogram $P$. The length of $CD'$ is\n$$\nCB + BD = 1 + 2\\cos \\angle CBD = 1 + 2\\cos (\\pi /5) = 1 + \\frac{\\sqrt{5} + 1}{2} = \\frac{\\sqrt{... | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 5 + sqrt(5) | |
038s | A circle is called *good colored* if the vertices of any equilateral triangle inscribed in this circle are colored in distinct colors. Let $k$ be a circle with radius $2$.
a) Is there a coloring of the points on $k$ and inside $k$ in three colors such that $k$ and any circle with radius at least $1$ that touches $k$ a... | [
"a) Assume that such a coloring exists and $a$, $b$ and $c$ are the colors. Let $O$ be the center of $k$ and consider an equilateral triangle $OBC$ with side $\\sqrt{3}$. Its circumcircle has radius $1$ and touches $k$. If the color of $O$ is $a$, then the colors of $B$ and $C$ are $b$ and $c$. This shows that the ... | Bulgaria | Spring Mathematical Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) No. b) Yes. | |
07lw | Let $ABCD$ be a square. The line segment $AB$ is divided internally at $H$ so that $|AB| \cdot |BH| = |AH|^2$. Let $E$ be the midpoint of $AD$ and $X$ be the midpoint of $AH$. Let $Y$ be the point on $EB$ such that $XY$ is perpendicular to $BE$. Prove that $|XY| = |XH|$. | [
"Let $ABCD$ have side length $2a$ and write $x = |AH|$. Then, by assumption\n$$\nx^2 = 2a(2a - x).\n$$\n\nBecause $|AB| = 2|EA|$, Pythagoras gives $|BE|^2 = |EA|^2 + |AB|^2 = 5|EA|^2$.\n\nObserve that $\\triangle BXY$ and $\\triangle BEA$ are similar. Hence, $\\frac{|BE|}{|EA|} = \\frac{|BX... | Ireland | Irska | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0i4c | Problem:
A continuous real function $f$ satisfies the identity $f(2x) = 3f(x)$ for all $x$. If $\int_{0}^{1} f(x) dx = 1$, what is $\int_{1}^{2} f(x) dx$? | [
"Solution:\nLet $S = \\int_{1}^{2} f(x) dx$. By setting $u = 2x$, we see that\n$$\n\\int_{1/2}^{1} f(x) dx = \\int_{1/2}^{1} \\frac{f(2x)}{3} dx = \\int_{1}^{2} \\frac{f(u)}{6} du = S/6.\n$$\nSimilarly, $\\int_{1/4}^{1/2} f(x) dx = S/36$, and in general\n$$\n\\int_{1/2^{n}}^{1/2^{n-1}} f(x) dx = S/6^{n}.\n$$\nAddin... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 5 | |
0829 | Problem:
In un torneo di pallacanestro 8 squadre sono divise in due gruppi di 4 squadre ciascuno. Al termine degli incontri preliminari, si disputano le semifinali, in cui la prima classificata del primo gruppo incontrerà la seconda classificata del secondo gruppo e la prima classificata del secondo gruppo incontrerà ... | [
"Solution:\n\nLa risposta è (D). La probabilità cercata si può calcolare come segue: la squadra $A$ raggiunge la semifinale se si classifica prima o seconda nel suo girone, e questo avviene con probabilità $\\frac{2}{4}=\\frac{1}{2}$. Nel caso che $A$ raggiunga la semifinale, solo un'altra squadra fra $B, C, D$ rag... | Italy | Progetto Olimpiadi di Matematica | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | MCQ | D | |
04vz | Find the largest set of positive integers whose sum is $2024$, and such that each number except the smallest one is a multiple of the sum of all the smaller numbers. (Patrik Bak) | [] | Czech Republic | First Round (take-home) | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | The largest set has size 1, namely {2024}. | |
08eu | Problem:
Sia $ABC$ un triangolo acutangolo, sia $M$ il punto medio di $BC$, e sia $H$ il piede dell'altezza uscente da $B$. Indichiamo con $Q$ il centro della circonferenza circoscritta al triangolo $ABM$, e con $X$ l'intersezione tra l'altezza $BH$ e l'asse di $BC$.
Dimostrare che i seguenti due fatti sono equivalent... | [
"Solution:\n\nSia $Y$ la proiezione di $X$ su $AB$. Dimostriamo che le circonferenze circoscritte ai triangoli $AMC$ e $AXH$ passano entrambe per $Y$. Questo è equivalente a dimostrare che $BY \\cdot BA = BM \\cdot BC$, e questo a sua volta è vero in quanto entrambi i prodotti sono uguali a $BX \\cdot BH$, dal mome... | Italy | XXXVII Olimpiade Italiana di Matematica | [
"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 | |
0300 | Problem:
Em uma turma existem 70 alunos, tais que:
I) 14 meninos passaram em Matemática;
II) 12 meninos passaram em Física;
III) 10 meninos e 16 meninas não passaram em Matemática nem em Física;
IV) 32 são meninos;
V) 10 passaram nas duas disciplinas;
VI) 22 passaram apenas em Matemática.
Quantas meninas passaram somen... | [
"Solution:\nPara resolver o problema, vamos utilizar o diagrama abaixo, no qual o retângulo superior representa as quantidades de meninos em cada caso e o inferior as quantidades de meninas; na circunferência da esquerda, a quantidade de alunos que passou em matemática, enquanto que na da direita, a quantidade que ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 4 | |
0cib | Let $(a_n)_{n \ge 1}$ and $(b_n)_{n \ge 1}$ be the sequences defined by the equations
$$
\prod_{k=1}^{n} (2k^2 + i) = a_n + ib_n, \quad n = 1, 2, \dots
$$
Prove that the sequence $\left(\frac{a_n}{b_n}\right)_{n \ge 1}$ is convergent and find its limit. | [] | Romania | 75th NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | 1 | |
0gmu | If $a$, $b$, $c$ are side lengths and $r$ is the radius of the incircle of a triangle, prove that
$$
\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \le \frac{1}{4r^2}.
$$ | [] | Turkey | XIII. National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
056a | Prove that for all positive real numbers $x, y, z$
$$
\frac{y^2 z}{x} + y^2 + z \geqslant \frac{9y^2 z}{x + y^2 + z}.
$$ | [
"By bringing all the terms to the same side and to the common denominator, we get an equivalent inequality\n$$\n\\frac{y^2z(x + y^2 + z) + xy^2(x + y^2 + z) + xz(x + y^2 + z) - 9xy^2z}{x(x + y^2 + z)} \\geq 0.\n$$\nSince $x$, $y$, and $z$ are positive, the denominator $x(x + y^2 + z)$ is positive as well. Therefore... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
08rk | $x$, $y$ and $z$ are distinct 2-digit positive integers. The first digit of $x$ is equal to the second digit of $y$, the first digit of $y$ is equal to the second digit of $z$, and the first digit of $z$ is equal to the second digit of $x$. How many positive integers can be the greatest common divisor of $x$, $y$ and $... | [
"We can write $x = 10a + b$, $y = 10b + c$, $z = 10c + a$ with positive integers $a, b, c$ less than $10$. Let $d$ be the greatest common divisor of $x$, $y$ and $z$. $11$ cannot divide $d$, since otherwise it contradicts the fact that $x$, $y$ and $z$ are distinct.\n\n$x + y + z = (10a + b) + (10b + c) + (10c + a)... | Japan | The 16th Japanese Mathematical Olympiad - The First Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 7 | |
0cvd | The cells of the $2 \times 2019$ table are to be filled with real numbers (one number in each cell) so that the following rules will be satisfied. The first row should contain $2019$ pairwise distinct real numbers; the second row should be a permutation of the first row. Each column should contain two distinct real num... | [
"**Оценка.** Докажем, что в первой строке таблицы, в которой числа расставлены по правилам, не менее трёх рациональных чисел (и, соответственно, не более $2016$ иррациональных чисел). Каждое из чисел, встречающихся в таблице, записано ровно в двух клетках, одна из которых находится в верхней строке, а другая — в ни... | Russia | Regional round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English; Russian | proof and answer | 2016 | |
06wo | Let $\omega$ be the circumcircle of a triangle $A B C$, and let $\Omega_{A}$ be its excircle which is tangent to the segment $B C$. Let $X$ and $Y$ be the intersection points of $\omega$ and $\Omega_{A}$. Let $P$ and $Q$ be the projections of $A$ onto the tangent lines to $\Omega_{A}$ at $X$ and $Y$, respectively. The ... | [
"Let $D$ be the point of tangency of $B C$ and $\\Omega_{A}$. Let $D'$ be the point such that $D D'$ is a diameter of $\\Omega_{A}$. Let $R'$ be (the unique) point such that $A R' \\perp B C$ and $R' D' \\parallel B C$. We shall prove that $R'$ coincides with $R$.\n\nLet $P X$ intersect $A B$ and $D' R'$ at $S$ and... | IMO | IMO 2021 Shortlisted Problems | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
0l3z | Problem:
Rishabh has $2024$ pairs of socks in a drawer. He draws socks from the drawer uniformly at random, without replacement, until he has drawn a pair of identical socks. Compute the expected number of unpaired socks he has drawn when he stops. | [
"Solution:\n\nWe solve for the expected number of total socks drawn and subtract two at the end.\nLet $E_{n}$ be the expected number of socks drawn for $n$ pairs of socks, so that $E_{1}=2$. Suppose there are $n$ pairs of socks, Rishabh continued to draw socks until the drawer was empty, and without loss of general... | United States | HMMT February 2024 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 4^{2024}/\binom{4048}{2024} - 2 | |
08hf | Problem:
Let $n \geq 1$ be a positive integer. For every $k = 1, 2, \ldots, n$ the functions $f_k: \mathbb{R} \rightarrow \mathbb{R}$, $f_k(x) = a_k x^2 + b_k x + c_k$ with $a_k \neq 0$ are given. Find the greatest possible number of parts of the rectangular plane $xOy$ which can be obtained by the intersection of the ... | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | n^2 + 1 | |
0a9m | Problem:
Given a triangle $ABC$, let $P$ lie on the circumcircle of the triangle and be the midpoint of the arc $BC$ which does not contain $A$. Draw a straight line $l$ through $P$ so that $l$ is parallel to $AB$. Denote by $k$ the circle which passes through $B$, and is tangent to $l$ at the point $P$. Let $Q$ be th... | [
"Solution:\n\nThere are three possibilities: $Q$ between $A$ and $B$, $Q = B$, and $B$ between $A$ and $Q$. If $Q = B$ we have that $\\angle ABP$ is right, and $AP$ is a diameter of the circumcircle. The triangles $ABP$ and $ACP$ are then congruent (they have $AP$ in common, $PB = PC$, and both have a right angle o... | Nordic Mathematical Olympiad | The 26th Nordic Mathematical Contest | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cjp | The circles $\Gamma_1$, $\Gamma_2$, and $\Gamma_3$ in the plane are pairwise externally tangent. Let $P_2$ be the point of tangency between the circles $\Gamma_1$ and $\Gamma_3$, and $P_1$ the point of tangency between the circles $\Gamma_2$ and $\Gamma_3$. Consider points $A$ and $B$ on the circle $\Gamma_3$ that are ... | [
"Let $\\{P_3\\} = \\Gamma_1 \\cap \\Gamma_2$ be the second point of intersection of the circles $\\Gamma_1$ and $\\Gamma_2$, and let $O_1, O_2, O_3$ be the centers of the circles $\\Gamma_1, \\Gamma_2$, and $\\Gamma_3$, respectively. Denote by $O_4$ the intersection point of the common tangents to the circles $\\Ga... | Romania | 75th NMO Selection Tests | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | proof only | null | |
0914 | Problem:
Let $ABC$ be an isosceles triangle with $|AC| = |BC|$. Its incircle touches $AB$ and $BC$ at $D$ and $E$, respectively. A line (different from $AE$) passes through $A$ and intersects the incircle at $F$ and $G$. The lines $EF$ and $EG$ intersect the line $AB$ at $K$ and $L$, respectively. Prove that $|DK| = |... | [
"Solution:\n\nIn view of symmetry, suppose that $AF < AG$, and, in addition, that $G$ is on the smaller arc $DE$ (for the other case see the last two sentences below).\nIf the incircle touches $AC$ at $J$, then $\\angle CAB = \\angle CJE = \\angle JDE = \\angle JFE$ (Fig. 1), hence $AJFK$ is a cyclic quadrilateral.... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geom... | null | proof only | null | |
07g5 | Find all positive integers $n$ such that
$$
d(n) \mid 2^{\sigma(n)} - 1.
$$
(Where $d(n)$ and $\sigma(n)$ are the total number and the sum of positive divisors of $n$.) | [
"We first prove the following lemma.\n**Lemma.** Let $L(n)$ be the least prime divisor of $n$, then $L(d(n)) \\le L(\\sigma(n))$, for all positive integers $n$.\n\n*Proof.* Let $n = p_1^{\\alpha_1} \\dots p_t^{\\alpha_t}$ for some distinct prime numbers $p_1, \\dots, p_t$. It follows that\n$$\nd(n) = (1 + \\alpha_1... | Iran | 38th Iranian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibi... | null | proof and answer | 1 | |
07co | Find all functions $f : \mathbb{R}^{+} \times \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ that satisfy the following conditions for all positive real numbers $x, y, z$
$$
\begin{gathered} f(f(x, y), z) = x^2 y^2 f(x, z), \\ f(x, 1 + f(x, y)) \geq x^2 + xyf(x, x). \end{gathered} \qquad (\rightarrow \text{p.57})
$$ | [
"* The function $g(x) = f(x, 1)$ is bijective.\nAssume that $a, b$ are two positive numbers with $f(a, 1) = f(b, 1)$.\nBy comparing $P(a, 1, 1)$, $P(b, 1, 1)$ we obtain\n$$\na^2 f(a, 1) = f(f(a, 1), 1) = f(f(b, 1), 1) = b^2 f(b, 1) \\implies a = b.\n$$\nSo $f(a, 1)$ is injective. Also\n$$\nP(1, y, 1) : f(f(1, y), 1... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | f(x, y) = x^2 y | |
01ce | Find all functions $f: \mathbf{R} \to \mathbf{R}$ satisfying, for all real numbers $x$ and $y$, the equation
$$
|x|f(y) + yf(x) = f(xy) + f(x^2) + f(f(y)).
$$ | [
"Answer: all functions $f(x) = c(|x| - x)$, where $c$ is a real number. Choosing $x = y = 0$, we find\n$$\nf(f(0)) = -2f(0).\n$$\nDenote $a = f(0)$, so that $f(a) = -2a$, and choose $y = 0$ in the initial equation:\n$$\na|x| = a + f(x^2) + f(a) = a + f(x^2) - 2a \\Rightarrow f(x^2) = a(|x| + 1).\n$$\nIn particular,... | Baltic Way | Baltic Way 2015 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All functions f(x) = c(|x| - x), where c is any real constant. | |
0e2z | Problem:
Simetrala diagonale $AC$ pravokotnika $ABCD$, v katerem je $|AB| > |BC|$, seka stranico $CD$ v točki $E$. Krožnica s središčem $E$ in polmerom $AE$ seka stranico $AB$ še v točki $F$. Naj bo $G$ pravokotna projekcija točke $C$ na premico $EF$. Pokaži, da točka $G$ leži na diagonali $BD$. | [
"Solution:\n\nOznačimo $\\angle FAE = \\alpha$. Ker točka $E$ leži na simetrali daljice $AC$, je enako oddaljena od $A$ in $C$, zato je središče krožnice, na kateri ležijo točke $A$, $C$ in $F$. Tako velja $|AE| = |CE| = |FE|$. Zato je $\\angle EFA = \\angle FAE = \\alpha$. Zaradi vzporednosti $AB$ in $CD$ sledi še... | Slovenia | 54. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof only | null | |
06e9 | Suppose there are $4n$ line segments of unit length inside a circle of radius $n$. Furthermore, a straight line $L$ is given. Prove that there exists a straight line $L'$ that is either parallel or perpendicular to $L$ and that $L'$ cuts at least two of the given line segments. | [
"Let $AB$ and $CD$ be the diameters of the circle which are parallel and perpendicular to $L$ respectively. Let $P_iQ_i$ be the projection of each segment on $AB$, and let $X_iY_i$ be the projection of each segment on $CD$. Note that $P_iQ_i + X_iY_i \\ge 1$. Therefore, we have\n$$\n\\sum_{j=1}^{4n} P_iQ_i + \\sum_... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0aqn | Problem:
The sum of the product and the sum of two integers is $95$. The difference between the product and the sum of these integers is $59$. Find the integers. | [
"Solution:\n\nLet the two integers be $x$ and $y$.\n\nLet $S = x + y$ (sum), $P = x y$ (product).\n\nWe are given:\n\n$P + S = 95$ \\quad (1)\n\n$P - S = 59$ \\quad (2)\n\nAdd (1) and (2):\n\n$P + S + P - S = 95 + 59$\n\n$2P = 154$\n\n$P = 77$\n\nNow substitute $P = 77$ into (1):\n\n$77 + S = 95$\n\n$S = 18$\n\nS... | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 7 and 11 | |
0ko8 | Problem:
Aerith thinks $(1.4)^{(1.4)^{(1.4)}}$ is well-defined, but Bob thinks it diverges. Who is right? | [
"Solution:\n\nBecause $(1.4)^2 = 1.96 < 2$, the tetrations of $1.4$ can never surpass $2$, so the expression does not diverge to infinity; Aerith is right."
] | United States | Berkeley Math Circle Monthly Contest 7 | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | Aerith is right. | |
0b2d | Problem:
Given triangle $ABC$, let $D$ be a point on side $AB$ and $E$ be a point on side $AC$. Let $F$ be the intersection of $BE$ and $CD$. If $\triangle DBF$ has an area of $4$, $\triangle BFC$ has an area of $6$, and $\triangle FCE$ has an area of $5$, find the area of quadrilateral $ADFE$. | [
"Solution:\n\nLet the area of quadrilateral $ADFE$ be $x$. By Menelaus' Theorem, $\\frac{AD}{DB} \\cdot \\frac{BF}{FE} \\cdot \\frac{EC}{CA} = 1$. Since $\\frac{AD}{DB} = \\frac{x+5}{10}$, $\\frac{BF}{FE} = \\frac{6}{5}$, and $\\frac{EC}{CA} = \\frac{11}{x+15}$, we have $\\frac{66(x+5)}{50(x+15)} = 1$, or $x = \\fr... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem"
] | null | proof and answer | 105/4 | |
0kvu | Problem:
Let $ABCD$ be a convex quadrilateral such that $\angle ABD = \angle BCD = 90^{\circ}$, and let $M$ be the midpoint of segment $BD$. Suppose that $CM = 2$ and $AM = 3$. Compute $AD$. | [
"Solution:\n\nSince triangle $BCD$ is a right triangle, we have $CM = BM = DM = 2$. With $AM = 3$ and $\\angle ABM = 90^{\\circ}$, we get $AB = \\sqrt{5}$. Now\n$$\nAD^{2} = AB^{2} + BD^{2} = 5 + 16 = 21\n$$\nso $AD = \\sqrt{21}$."
] | United States | HMMT February 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | sqrt(21) | |
05cv | Solve the system of equations
$$
\begin{cases}
x + y = z, \\
x^2 + y^2 = 4z, \\
x^3 + y^3 = 18z.
\end{cases}
$$ | [
"**Solution 1:** If $x \\neq 0$ or $y \\neq 0$, then from the second equation $z > 0$. Thus $z = 0$ can only hold if $x = y = 0$. The triple $(x, y, z) = (0, 0, 0)$ satisfies all equations. Now assume $z \\neq 0$.\nSquaring the first equation yields $x^2 + 2xy + y^2 = z^2$. Subtracting the second equation yields\n$... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | (x, y, z) = (0, 0, 0) or z = 6 with {x, y} = {3 + sqrt(3), 3 − sqrt(3)} | |
0e32 | Problem:
Naj bo $O$ središče ostrokotnemu trikotniku $ABC$ očrtane krožnice $\mathcal{K}$. Simetrala notranjega kota pri $A$ seka krožnico $\mathcal{K}$ še v točki $D$, simetrala notranjega kota pri $B$ pa seka krožnico $\mathcal{K}$ še v točki $E$. Označimo z $I$ središče trikotniku $ABC$ včrtane krožnice. Kolikšna j... | [
"Solution:\n\nKer je trikotnik $ABC$ ostrokotni, ležita točki $I$ in $O$ na istem bregu premice $ED$. Iz pogoja, da ležijo točke $D, E, I$ in $O$ na isti krožnici, zato sledi $\\angle DOE = \\angle DIE$. Označimo kote v trikotniku z $\\alpha, \\beta$ in $\\gamma$ in z njimi izrazimo kota $\\angle DOE$ in $\\angle D... | Slovenia | 54. matematično tekmovanje srednješolcev Slovenije | [
"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 | pi/3 | |
0dct | Let $n$ be a positive integer. Each number from $1, 2, 3, \ldots, 1000$ was painted one of $n$ colors. It turned out that every two distinct numbers, one of which is a divisor of the other one, have different colors. Find the smallest $n$ for which such situation is possible. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 10 | |
0khh | Problem:
A tournament among $2021$ ranked teams is played over $2020$ rounds. In each round, two teams are selected uniformly at random among all remaining teams to play against each other. The better ranked team always wins, and the worse ranked team is eliminated. Let $p$ be the probability that the second best ranke... | [
"Solution:\nIn any given round, the second-best team is only eliminated if it plays against the best team. If there are $k$ teams left and the second-best team has not been eliminated, the second-best team plays the best team with probability $\\frac{1}{\\binom{k}{2}}$, so the second-best team survives the round wi... | United States | HMMT Spring 2021 Guts Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 674 | |
08oe | Problem:
In a country with $n$ cities, all direct airlines are two-way. There are $r > 2014$ routes between pairs of different cities that include no more than one intermediate stop (the direction of each route matters). Find the least possible $n$ and the least possible $r$ for that value of $n$. | [
"Solution:\n\nDenote by $X_{1}, X_{2}, \\ldots, X_{n}$ the cities in the country and let $X_{i}$ be connected to exactly $m_{i}$ other cities by direct two-way airline. Then $X_{i}$ is a final destination of $m_{i}$ direct routes and an intermediate stop of $m_{i}(m_{i}-1)$ non-direct routes. Thus $r = m_{1}^{2} + ... | JBMO | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n = 14; r = 2016 | |
09b8 | Мянга есөн зуун хэдэн оны нэг өдөр Бат өөрийн төрсөн өдрөөрөө төрсөн оных нь цифрүүдийн нийлбэр өөрийнх нь настай яг таарч байгааг анзаарав. Мөн түүний ах Болд яг энэ өдөр төрсөн бөгөөд нас нь Батынхтай адил зүй тогтолтой байгааг мэджээ. Хэрэв тэд 99-ээс бага настай бол Болд Батаас хэдэн насаар ах вэ? | [
"Тухайн оныг $A$ гэвэл $1900 \\leq A < 2000$ болно. Батын төрсөн он нь $\\overline{18ab}$ эсвэл $\\overline{19ab}$ байна. Тэгвэл\n$$\nA = \\overline{18ab} + 1 + 8 + a + b = 1809 + 11a + 2b\n$$\nэсвэл\n$$\nA = \\overline{19ab} + 1 + 9 + a + b = 1910 + 11a + 2b\n$$\n\nБолдын хувьд $18cd$ эсвэл $19cd$ онд төрсөн байга... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | Mongolian | proof and answer | 9 | |
0gqy | Let $(a_n)_{n=1}^\infty$ be a sequence of integers such that $a_1 = -5$, $a_2 = -6$ and
$$
a_{n+1} = a_n + (a_1+1)(2a_2+1)(3a_3+1)\cdots((n-1)a_{n-1}+1)((n^2+n)a_n+2n+1)
$$
for all integers $n \ge 2$. Prove that if prime number $p$ divides $n a_n + 1$ for some positive integer $n$, then there exists an integer $m$ such... | [
"Define $b_n = (a_1 + 1)(2a_2 + 1) \\cdots ((n-1)a_{n-1} + 1)$ for $n = 2, 3, \\dots$ and let $b_1 = 1$. By using mathematical induction we will prove that for all positive values of $n$\n$$\nb_{n+1} = (b_1 + 2b_2 + \\dots + n b_n)^2 - 5 \\quad (1)\n$$\nFor $n=1$ we have $b_2 = a_1 + 1 = -4$ and hence $b_2 = -4 = b... | Turkey | Team Selection Test | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathe... | null | proof only | null | |
02en | Show that there are infinitely many positive integer solutions to $a^3 + 1990b^3 = c^4$. | [
"Take, for instance, $a = b = 1991k^4$ and $c = 1991k^3$ (there are lots of other solutions; can you find them all?)."
] | Brazil | XII OBM | [
"Number Theory > Diophantine Equations"
] | English | proof only | null | |
08lv | Problem:
Is it possible to arrange the numbers $1^{1}, 2^{2}, \ldots, 2008^{2008}$ one after the other, in such a way that the obtained number is a perfect square? (Explain your answer.) | [
"Solution:\nWe will use the following lemmas.\n\nLemma 1. If $x \\in \\mathbb{N}$, then $x^{2} \\equiv 0$ or $1 \\pmod{3}$.\n\nProof: Let $x \\in \\mathbb{N}$, then $x=3k$, $x=3k+1$ or $x=3k+2$, hence\n$$\n\\begin{aligned}\n& x^{2}=9k^{2} \\equiv 0 \\pmod{3} \\\\\n& x^{2}=9k^{2}+6k+1 \\equiv 1 \\pmod{3}, \\\\\n& x^... | JBMO | 2008 Shortlist JBMO | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | No | |
0386 | Problem:
Each side of a sheet of paper is a map of 5 countries. The countries on one of the maps are colored in 5 different colors. Prove that it is possible to color the countries on the other map in such a way that every two are colored in different colors and at least $20\%$ of the sheet is colored in the same color... | [
"Solution:\nDenote by $A_{1}, A_{2}, \\ldots, A_{5}$ and $B_{1}, B_{2}, \\ldots, B_{5}$ the countries on the respective sides of the sheet of paper. Let $S_{ij}$ be the area of the part of $A_{i}$ which belongs to the country $B_{j}$ on the other side of the sheet. (If $A_{i}$ and $B_{j}$ do not have a common area,... | Bulgaria | Spring Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0fw3 | Problem:
Sei $ABCD$ ein Trapez mit $AB \parallel CD$ und $AB > CD$. Die Punkte $K$ und $L$ liegen auf den Seiten $AB$ bzw. $CD$ mit $\dfrac{AK}{KB} = \dfrac{DL}{LC}$. Die Punkte $P$ und $Q$ liegen so auf der Strecke $KL$, dass gilt
$$
\angle APB = \angle BCD \quad \text{und} \quad \angle CQD = \angle ABC
$$
Zeige, das... | [
"Solution:\n\nWir bezeichnen die beiden gegebenen Winkelgrößen mit $\\alpha = \\angle ABC = \\angle CQD$ und $\\beta = \\angle BCD = \\angle APB$. Weil $AB \\parallel CD$ gilt $\\alpha + \\beta = 180^{\\circ}$. Ebenfalls weil $AB \\parallel CD$ folgt aus $\\dfrac{AK}{KB} = \\dfrac{DL}{LC}$, dass sich die Geraden $A... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0akg | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$
f(\max\{x, y\} + \min\{f(x), f(y)\}) = x + y \quad (1)
$$
for all $x, y \in \mathbb{R}$. | [
"Let the function $f$ is such that for every $x, y \\in \\mathbb{R}$ it is fulfilled the equation (1).\nIf in (1) we put $y = x$, we obtain\n$$\nf(x + f(x)) = 2x, \\text{ for every } x \\in \\mathbb{R}. \\quad (2)\n$$\nFurthermore, for $x = 0$ from (2) follows $f(f(0)) = 0$, and for $x = y = \\frac{f(0)}{2}$ from (... | North Macedonia | Macedonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x for all real x | |
0i8w | Problem:
Daniel and Scott are playing a game where a player wins as soon as he has two points more than his opponent. Both players start at par, and points are earned one at a time. If Daniel has a $60\%$ chance of winning each point, what is the probability that he will win the game? | [
"Solution:\n\nConsider the situation after two points. Daniel has a $9/25$ chance of winning, Scott, $4/25$, and there is a $12/25$ chance that the players will be tied. In the latter case, we revert to the original situation. In particular, after every two points, either the game returns to the original situation,... | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 9/13 | |
05fb | Problem:
Soit $\Omega$ et $\Gamma$ deux cercles sécants. On note $A$ une de leurs intersections. Soit $d$ une droite quelconque passant par le point $A$. On note $P$ et $Q$ les intersections respectives de la droite $d$ avec les cercles $\Omega$ et $\Gamma$ différentes de $A$.
Montrer qu'il existe un point indépendant... | [
"Solution:\n\nPar symétrie, on peut supposer que le rayon du cercle $\\Omega$ est supérieur au rayon du cercle $\\Gamma$.\nSoit $O_1$ le centre du cercle $\\Omega$ et $O_2$ le centre du cercle $\\Gamma$. Soit $B$ le point tel que le quadrilatère $AO_1BO_2$ soit un parallélogramme. On va montrer que le point $B$ app... | France | ENVOI 1 : GÉOMÉTRIE Corrigé | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0juj | Problem:
A square can be divided into four congruent figures as shown:
For how many $n$ with $1 \leq n \leq 100$ can a unit square be divided into $n$ congruent figures? | [
"Solution:\nWe can divide the square into congruent rectangles for all $n$, so the answer is $100$."
] | United States | HMMT November | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 100 | |
03oa | Problem:
A rectangle $R$ is divided into a set $S$ of finitely many smaller rectangles with sides parallel to the sides of $R$ such that no three rectangles in $S$ share a common corner. An ant is initially located at the bottom-left corner of $R$. In one operation, we can choose a rectangle $r \in S$ such that the ant... | [
"Solution:\nConsider the following version of the problem:\nA rectangle $R$ is divided into a set $S$ of finitely many smaller rectangles such that no three rectangles in $S$ share a common corner. For each $r \\in S$, draw two non-intersecting arcs inside $r$, connecting the pairs of adjacent corners of $r$ (there... | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0j8e | Problem:
Let $XYZ$ be a triangle with $\angle XYZ = 40^{\circ}$ and $\angle YZX = 60^{\circ}$. A circle $\Gamma$, centered at the point $I$, lies inside triangle $XYZ$ and is tangent to all three sides of the triangle. Let $A$ be the point of tangency of $\Gamma$ with $YZ$, and let ray $\overrightarrow{XI}$ intersect ... | [
"Solution:\n\nAnswer: $10^{\\circ}$\n\nLet $D$ be the foot of the perpendicular from $X$ to $YZ$. Since $I$ is the incenter and $A$ the point of tangency, $IA \\perp YZ$, so\n\n$$\nAI \\parallel XD \\Rightarrow \\angle AIB = \\angle DXB\n$$\n\nSince $I$ is the incenter,\n$$\n\\angle BXZ = \\frac{1}{2} \\angle YXZ =... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 10° | |
0cbg | Find all the pairs $(x, y)$ of real numbers fulfilling
$$
3 \cdot \left\{ \frac{3x+2}{3} \right\} + 4 \cdot \left\lfloor \frac{4y+3}{4} \right\rfloor = 4 \cdot \left\{ \frac{4y+3}{4} \right\} + 3 \cdot \left\lfloor \frac{3x+2}{3} \right\rfloor = 18.
$$ | [
"Since $0 \\le \\{a\\} < 1$ and $\\lfloor b \\rfloor$ is an integer, the equality $3\\{a\\} + 4\\lfloor b \\rfloor = 18$ is possible only when $\\{a\\} = \\frac{2}{3}$ and $\\lfloor b \\rfloor = 4$.\nSince $0 \\le \\{c\\} < 1$ and $\\lfloor d \\rfloor$ is an integer, the equality $4\\{c\\} + 3\\lfloor d \\rfloor = ... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (6, 13/4), (5, 4) | |
0cb0 | The positive integer $n$ is a perfect square. Find the quotient of the division of $2023$ by $n$, if the remainder is $223 - \frac{3}{2} \cdot n$. | [
"Denote by $c$ the quotient of the division. From the quotient-remainder theorem we obtain $2023 = n \\cdot c + 223 - \\frac{3}{2} \\cdot n$, thus $(2c-3)n = 3600$. (1)\n\nThe remainder $223 - \\frac{3}{2} \\cdot n$ is a positive integer, therefore $n$ is even and $0 \\le 223 - \\frac{3}{2} \\cdot n < n$, whence we... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 14 | |
0cz7 | Consider the sequence $x_{n} = 2^{n} - n$, $n = 0, 1, 2, \ldots$. Find all integers $m \geq 0$ such that $s_{m} = x_{0} + x_{1} + x_{2} + \ldots + x_{m}$ is a power of $2$. | [
"We have\n$$\ns_{m} = \\sum_{k=0}^{m} \\left(2^{k} - k\\right) = 2^{m+1} - 1 - \\frac{m(m+1)}{2}\n$$\nWe prove that for $m \\geq 3$, we have $2^{m} < s_{m} < 2^{m+1}$. This inequality is equivalent to\n$$\n2^{m} < 2^{m+1} - 1 - \\frac{m(m+1)}{2} < 2^{m+1}\n$$\nThe right inequality is obvious. The left inequality is... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | m = 0, 1, 2 | |
02e7 | A number is written in each square of a chessboard, so that each number not on the border is the mean of the 4 neighboring numbers. Show that if the largest number is $N$, then there is a number equal to $N$ in the border squares. | [
"Take the leftmost $N$. Suppose it is not in the border. Then it must be the mean of the 4 neighboring numbers. Hence each of the 4 neighboring numbers must also be $N$, but one of them is to the left of $N$. Contradiction."
] | Brazil | VIII OBM | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Logic"
] | English | proof only | null | |
08eg | Problem:
Sia $ABC$ un triangolo equilatero di lato unitario e sia $P$ un punto dalla parte opposta della retta $AB$ rispetto al punto $C$, tale che l'angolo $\widehat{APB}$ misuri $60^{\circ}$. Supponiamo che la bisettrice dell'angolo $\overline{APB}$ intersechi i segmenti $AB$ e $AC$ nei punti $X$ e $Y$, rispettivame... | [
"Solution:\n\nLa risposta è $\\mathbf{(A)}$.\n\n\n\nSia $O$ il centro del triangolo. Il punto $P$ giace per ipotesi sull'arco della circonferenza circoscritta ad $AOB$ esterno al triangolo. La bisettrice dell'angolo $A\\hat{P}B$ passa per il punto medio dell'arco $AB$ opposto a quello sui c... | Italy | Italian Mathematical Olympiad - February Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > A... | null | MCQ | (A) | |
0ipl | Problem:
Suppose we have an (infinite) cone $\mathcal{C}$ with apex $A$ and a plane $\pi$. The intersection of $\pi$ and $\mathcal{C}$ is an ellipse $\mathcal{E}$ with major axis $BC$, such that $B$ is closer to $A$ than $C$, and $BC = 4$, $AC = 5$, $AB = 3$. Suppose we inscribe a sphere in each part of $\mathcal{C}$ c... | [
"Solution:\nAnswer: $\\sqrt{\\frac{1}{3}}$\n\nIt can be seen that the points of tangency of the spheres with $\\mathcal{E}$ must lie on its major axis due to symmetry. Hence, we consider the two-dimensional cross-section with plane $ABC$. Then the two spheres become the incentre and the excentre of the triangle $AB... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | final answer only | 1/3 | |
0hch | Determine the smallest possible value of $x^6 + x^4y^2 + x^2y^4 + y^6$, given that the product of real numbers $x, y$ is $1$? | [
"Without loss of generality, let $x, y$ be positive. Then we can factor the expression as $(x^4 + y^4)(x^2 + y^2)$. Since $(x^2 - y^2)^2 = x^4 - 2x^2y^2 + y^4 \\ge 0$, then $x^4 + y^4 \\ge 2x^2y^2 = 2$, and similarly $x^2 + y^2 \\ge 2xy = 2$. Therefore, the smallest number is $4$, that can be obtained when $x = y =... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 4 | |
0166 | Determine all positive integers $n$, for which $2^{n+1} - n^2$ is a prime number. | [
"This occurs exactly if $n = 1$ or $n = 3$.\n\nTo see this, first note that if $n$ is even, then $2^{n+1} - n^2$ is a multiple of $4$ and hence in particular composite. Now let $n$ be odd. Writing $n = 2m - 1$ for some positive integer $m$, we find\n$$\n2^{n+1} - n^2 = (2^m)^2 - (2m-1)^2 = [2^m + (2m-1)] \\cdot [2^... | Baltic Way | Baltic Way SHL | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | n = 1 or n = 3 | |
0fhs | Problem:
Demostrar que si entre los infinitos términos de una progresión aritmética de números enteros hay un cuadrado perfecto, entonces infinitos términos de la progresión son cuadrados perfectos. | [
"Solution:\nBastará probar que a partir de un cuadrado perfecto podemos construir otro. Sea la progresión:\n$$\na^{2},\\ a^{2}+d,\\ a^{2}+2d,\\ \\ldots,\\ a^{2}+kd\\ \\ldots\n$$\nComo $(a+d)^{2} = a^{2} + 2ad + d^{2} = a^{2} + (2a + d)d$, basta tomar $k = 2a + d$ para obtener otro cuadrado en la progresión."
] | Spain | OME 30 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
00iv | Let $p$ and $q$ be real numbers such that the quadratic equation
$$
x^2 + px + q = 0
$$
has two real solutions $x_1$ and $x_2$.
The following two conditions hold:
(i) The numbers $x_1$ and $x_2$ differ by 1.
(ii) The numbers $p$ and $q$ differ by 1.
Show that $p, q, x_1$ and $x_2$ are integers. | [
"Without loss of generality, we assume $x_1 = x_2 + 1$. By Vieta's formulas, this implies $p = -(x_1 + x_2) = -2x_2 - 1$ and $q = x_1 x_2 = x_2^2 + x_2$.\nTherefore, it is enough to check that $x_2$ has to be an integer.\n\n**Case 1: $q = p - 1$**\nThis implies $x_2^2 + x_2 = -2x_2 - 1 - 1$ and therefore $x_2^2 + 3... | Austria | AustriaMO2011 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
00q7 | Let $m_1$, $m_2$, $m_3$, $n_1$, $n_2$ and $n_3$ be positive real numbers such that
$$
(m_1 - n_1)(m_2 - n_2)(m_3 - n_3) = m_1 m_2 m_3 - n_1 n_2 n_3.
$$
Prove that
$$
(m_1 + n_1)(m_2 + n_2)(m_3 + n_3) \geq 8m_1 m_2 m_3.
$$ | [
"Divide both sides of the given equality by $m_1 m_2 m_3$ and set $a = \\frac{n_1}{m_1}$, $b = \\frac{n_2}{m_2}$ and $c = \\frac{n_3}{m_3}$. The equality $(m_1 - n_1)(m_2 - n_2)(m_3 - n_3) = m_1 m_2 m_3 - n_1 n_2 n_3$ becomes\n$$\n(1 - a)(1 - b)(1 - c) = 1 - abc \\iff a + b + c = ab + bc + ca\n$$\nand we have to sh... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0gfh | 找出所有的實係數多項式 $P(x)$，使得對滿足 $2xyz = x + y + z$ 的非零實數，皆有
$$
\frac{P(x)}{yz} + \frac{P(y)}{zx} + \frac{P(z)}{xy} = P(x - y) + P(y - z) + P(z - x)
$$ | [
"定\n$$\nQ(x, y, z) = xP(x) + yP(y) + zP(z) - xyz[P(x - y) + P(y - z) + P(z - x)]\n$$\n則 $Q(x, y, z)$ 也是實係數多項式，且當 $xyz \\neq 0$ 時\n$$\n2xyz = x + y + z \\Rightarrow Q(x, y, z) = 0\n$$\n上面的性質可延伸到複數上面，即 $x, y, z$ 也可用複數帶入。當 $(x, y, z) = (t, -t, 0)$ 帶入得出 $P(t) = P(-t)$ 知 $P(x)$ 是偶函數。又帶入\n$$\n(x, y, z) = \\left(x, \\frac... | Taiwan | 國際奧林匹亞競賽第三次訓練營 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Intermediate Algebra > Complex numbers"
] | Chinese; English | proof and answer | P(x) = ax^2 + b for real constants a and b | |
0g5y | 給定正整數 $n$, 考慮 $n$ 維空間的所有整數點 (即每個座標都是整數的點)。當兩個整數點的直線距離為 $1$ 時, 我們稱它們互相相鄰。試問是否可能將其中一部分的整數點做標記, 使得對於每一個整數點, 在該點本身和它所有相鄰的點這 $(2n + 1)$ 個點中, 總是恰有一個被標記? | [
"可以！\n令 $x_1, \\dots, x_n$ 為整數點的座標, 將所有滿足\n$$\n(2n + 1) \\mid (x_1 + 2x_2 + \\dots + nx_n)\n$$\n的點標記, 即達成條件; 對於每個點, $x_1 + 2x_2 + \\dots + nx_n$ 可以唯一的表示為 $(2n+1)l \\pm k$, 其中 $l$ 為整數, $k = 0, 1, \\dots, n$. 當 $k=0$ 時即該點被標記, 否則即是沿著第 $k$ 個座標方向的兩個相鄰點之一被標記。"
] | Taiwan | 二〇一一數學奧林匹亞競賽第二階段選訓營 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | Yes | |
095x | Problem:
$M$ este o mulţime de 2018 numere naturale, nici unul dintre care nu se divide cu 2018. Să se arate, că există o submulţime a lui $M$, care are suma elementelor divizibilă cu 2018. | [
"Solution:\n\nFie $M = \\{a_{1}, a_{2}, \\ldots, a_{2018}\\}$. Se formează sumele\n$$\na_{1},\\ a_{1}+a_{2},\\ a_{1}+a_{2}+a_{3},\\ \\ldots,\\ a_{1}+a_{2}+a_{3}+\\ldots+a_{2018}\n$$\nDacă printre cele 2018 numere din (*) este unul divizibil cu 2018, termenii sumei respective formează mulţimea necesară.\n\nÎn caz co... | Moldova | A 62-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other"
] | null | proof only | null | |
0g6s | 設 $\triangle ABC$ 為一銳角三角形，點 $P$ 是 $BC$ 邊上的一點。已知線段 $PE$ 垂直於 $AC$，線段 $PD$ 垂直於 $AB$ 邊。試求：三角形 $PDE$ 面積最大，若且唯若 $P$ 為 $BC$ 中點。 | [
"首先注意到\n$$\n2\\Delta PDE \\text{ 的面積} = PD \\times PE \\times \\sin \\angle DPE = PD \\times PE \\times \\sin(\\angle B + \\angle C),\n$$\n因此 $\\triangle PDE$ 面積達到最大值若且唯若 $PD \\times PE$ 達到最大值。\n\n接著，令 $x, y$ 分別為線段 $PE, PD$ 長，$b, c$ 分別為線段 $AC, AB$ 長。令 $d$ 為三角形 $ABC$ 面積，則顯然 $xb + yc = 2d$。故 $y = (2d - xb)/c$。因此，\n$$... | Taiwan | 二〇一二數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04l7 | Find all polynomials $P$ with real coefficients such that
$$
P(x^2) + 2P(x) = (P(x))^2 + 2
$$
holds for every real number $x$. | [
"The original equation is equivalent to $P(x^2) - 1 = P^2(x) - 2P(x) + 1$, i.e.\n$$\nP(x^2) - 1 = (P(x) - 1)^2.\n$$\nIf we define $Q(x) := P(x) - 1$, the above equation becomes\n$$\nQ(x^2) = Q^2(x).\n$$\nIf $Q(x)$ is a constant polynomial, this implies that it has to be $0$ or $1$. On the other hand, if the degree ... | Croatia | Mathematical competitions in Croatia | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | P(x) = 1; P(x) = 2; and P(x) = x^n + 1 for any positive integer n | |
01x4 | The line $l$ passes through the vertex $D$ of the rectangle $ABCD$ and doesn't have any other common points with this rectangle. The point $M$ lies on $l$, and area of the triangle $MBD$ is twice the area of the triangle $MAD$.
Find all possible values of the ratio of the areas of the triangles $MCD$ and $MAD$. | [
"$S(MCD) : S(MAD) = 1$."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | 1 | |
0h6f | It is known that the arithmetic average of the numbers *a*, *b* is equal to the number *c*, so $c = \frac{1}{2}(a+b)$, and that the geometric average number of *a*, *c* is equal to the number *b*, so $b = \sqrt{ac}$. Is it necessary that numbers *a*, *b*, *c* are equal? | [
"Let's rewrite the equality $b^2 = ac$ using $c = \\frac{a+b}{2}$:\n\n$$\nb^2 = a \\cdot \\frac{a+b}{2} \\Leftrightarrow 2b^2 = ba + a^2 \\Leftrightarrow (b-a)(2b+a) = 0.\n$$\nNow let's denote, for example, $b=2$, which means $a=-4$ and $c=-1$, hence we receive three different numbers satisfying the conditions."
] | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | No; for example, a = -4, b = 2, c = -1. | |
0ane | Problem:
Point $P$ on side $BC$ of triangle $ABC$ satisfies
$$
|BP| : |PC| = 2 : 1
$$
Prove that the line $AP$ bisects the median of triangle $ABC$ drawn from vertex $C$. | [
"Solution:\n\nLet $Q$ be the midpoint of line segment $BP$. The conditions of the problem imply $|BQ| = |QP| = |PC| = \\frac{1}{3}|BC|$. Let $R$ be the midpoint of line segment $AB$. Then $RQ$ is a midline of $ABP$. Consequently, $RQ \\parallel AP$. Ray $AP$ bisects side $CQ$ of triangle $CRQ$ while being parallel ... | Philippines | 18th PMO Area Stage | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0ats | Problem:
Find all positive integers $m$ and $n$ so that for any $x$ and $y$ in the interval $[m, n]$, the value of $\frac{5}{x} + \frac{7}{y}$ will also be in $[m, n]$. | [] | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (m,n) = (1,12), (2,6), (3,4) | |
09j9 | Throwing an ankle-bone results in four positions – horse, camel, sheep and goat – with equal probability. A player throws four ankle-bones and receives the following number of points depending on the positions.
| Position | Score |
|----------------------------------|-------|
| Four horses ... | [] | Mongolia | Mongolian Mathematical Olympiad Round 2 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | 23 | |
0bu8 | Problem:
Fie $f:[0,1] \rightarrow [0,1]$ o funcție crescătoare și fie
$$
a_{n} = \int_{0}^{1} \frac{1 + (f(x))^{n}}{1 + (f(x))^{n+1}} \, \mathrm{d}x, \quad n \in \mathbb{N}^{*}
$$
Arătați că șirul $\left(a_{n}\right)_{n \in \mathbb{N}^{*}}$ este convergent și calculați limita sa. | [
"Solution:\n\nÎn mod evident, $a_{n} \\geq 1$, oricare ar fi $n \\in \\mathbb{N}^{*}$. Pe de altă parte,\n$$\n\\begin{aligned}\na_{n} - a_{n+1} &= \\int_{0}^{1} \\left( \\frac{1 + (f(x))^{n}}{1 + (f(x))^{n+1}} - \\frac{1 + (f(x))^{n+1}}{1 + (f(x))^{n+2}} \\right) \\, \\mathrm{d}x \\\\\n&= \\int_{0}^{1} \\frac{(f(x)... | Romania | Olimpiada Naţională de Matematică | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Functions"
] | null | proof and answer | 1 | |
07jq | Point $M$ is the midpoint of side $BC$ of triangle $ABC$. The line perpendicular to $AM$ at point $A$ intersects the circumcircle of $ABC$ for the second time at $K$. The altitudes $BE$ and $CF$ of triangle $ABC$ intersect line $AK$ at points $P$ and $Q$, respectively. Prove that the radical axis of the circumcircles o... | [
"**Claim 1.** *If $\\Xi$ is the $\\Xi$-point corresponding to vertex $A$, then $K\\Xi \\perp BC$.*\n\n*Proof*. Let $\\Xi'$ and $X$ be the reflections of $\\Xi$ with respect to $BC$ and $M$ respectively. According to the properties of the $\\Xi$-point, we know that these points lie on the circumcircle of $ABC$ and $... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle cha... | null | proof only | null |
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