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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0c4a | Problem:
Considerăm un număr natural nenul $n$, un cerc de lungime $6n$ și $3n$ puncte care împart cercul în $3n$ arce mici, astfel încât $n$ dintre aceste arce au lungimea $1$, alte $n$ dintre aceste arce au lungimea $2$, iar cele $n$ arce rămase au lungimea $3$.
Arătați că printre punctele considerate există două c... | [
"Solution:\n\nPunctele considerate sunt o parte dintre vârfurile unui poligon cu $6n$ laturi, înscris în cercul dat.\n\nPresupunem contrariul. Atunci capetele fiecărui arc de lungime $1$ au ca puncte diametral opuse două puncte situate în interiorul unui arc de lungime $3$. Astfel, fiecărui arc de lungime $1$ i se ... | Romania | Olimpiada Naţională de Matematică Etapa Naţională | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0jyo | Problem:
Rebecca has twenty-four resistors, each with resistance $1$ ohm. Every minute, she chooses any two resistors with resistance of $a$ and $b$ ohms respectively, and combine them into one by one of the following methods:
- Connect them in series, which produces a resistor with resistance of $a+b$ ohms;
- Connect... | [
"Solution:\n\nThis is the same problem as in OEIS A153588. It is helpful to see (or guess) that neither the numerator or the denominator of the final resistance exceed the $(n+1)$-th Fibonacci number, which in this case is $F_{25}=75025$, using concepts on the line of continued fractions. So $75025^{2} \\approx 5.6... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Intermediate Algebra > Other",
"Number Theory > Other"
] | null | final answer only | null | |
005u | Sea $ABCD$ un paralelogramo de lados $AB$, $BC$, $CD$, $AD$, tal que $AB > AD$ y $\frac{AC}{BD} = 3$. Sea $r$ la recta simétrica de $AD$ con respecto a $AC$ y sea $s$ la recta simétrica de $BC$ con respecto a $BD$. Si $r$ y $s$ se cortan en $P$, calcular el valor de $\frac{PA}{PB}$. | [] | Argentina | Argentina 2008 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | Spanish | proof and answer | 9 | |
0dxc | Problem:
Naj bo $x$ tako realno število, da je $x+\frac{1}{x}$ celo število. Dokaži, da je tedaj za vsako naravno število $n$ tudi $x^{n}+\frac{1}{x^{n}}$ celo število. | [
"Solution:\n\nTrditev bomo dokazali z matematično indukcijo. V bazi indukcije preverimo primera $n=1$ in $n=2$. Števili\n$$\nx^{1}+\\frac{1}{x^{1}}=x+\\frac{1}{x}\n$$\nin\n$$\nx^{2}+\\frac{1}{x^{2}}=\\left(x+\\frac{1}{x}\\right)^{2}-2\n$$\nsta celi števili, saj je $x+\\frac{1}{x}$ po predpostavki naloge celo števil... | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0ejv | Problem:
Dan je polinom $p(x)=x^{6}+x^{5}+\ldots+x+1$. Dokaži, da polinom $p(x)$ deli polinom $p\left(x^{9}\right)$. | [
"Solution:\n\nPri dokazu bomo nekajkrat uporabiti razcep\n$$\na^{n}-1=(a-1)\\left(a^{n-1}+a^{n-2}+\\ldots+a+1\\right), \\quad n \\in \\mathbb{N}\n$$\nNajprej opazimo, da je $(x-1) p(x)=x^{7}-1$. Torej je\n$$\n\\begin{aligned}\n\\left(x^{9}-1\\right) p\\left(x^{9}\\right) & =\\left(x^{9}\\right)^{7}-1=x^{7 \\cdot 9}... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof only | null | |
0k7s | Problem:
Do there exist four points $P_{i} = (x_{i}, y_{i}) \in \mathbb{R}^{2}$ ($1 \leq i \leq 4$) on the plane such that:
- for all $i = 1,2,3,4$, the inequality $x_{i}^{4} + y_{i}^{4} \leq x_{i}^{3} + y_{i}^{3}$ holds, and
- for all $i \neq j$, the distance between $P_{i}$ and $P_{j}$ is greater than $1$? | [
"Solution:\n\nIn fact, there are! One might think that the region\n$$\n\\left\\{(x, y) \\in \\mathbb{R}^{2} \\mid x^{4} + y^{4} \\leq x^{3} + y^{3}\\right\\}\n$$\nis inside the ball defined by $x^{2} + y^{2} \\leq x + y$, which is a ball of radius $1 / \\sqrt{2}$. It turns out that it is not the case.\nWe claim tha... | United States | HMIC | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
03vg | Given positive integers $m, n \ge 2$, first select two different $a_i, a_j$ ($j > i$) in the integer set $A = \{a_1, a_2, \dots, a_n\}$ and take the difference $a_j - a_i$. Then arrange the $\binom{n}{2}$ differences in ascending order to form a new sequence, which we call 'derived sequence' and is denoted by $\bar{A}$... | [
"**Proof** For any integer $m \\ge 2$, if the remainder of $x$ divided by $m$ is $i$, $i \\in \\{0, 1, \\dots, m-1\\}$, then $x$ belongs to the residue class modulus $m$, $K_i$.\nSuppose in the set $A = \\{a_1, a_2, \\dots, a_n\\}$, the number of elements that belong to $K_i$ is $n_i$ ($i = 0, 1, 2, \\dots, m-1$), ... | China | China Southeastern Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
09yr | Consider an acute triangle $ABC$ with $|AB| > |CA| > |BC|$. The vertices $D$, $E$, and $F$ are the base points of the altitudes from $A$, $B$, and $C$, respectively. The line through $F$ parallel to $DE$ intersects $BC$ in $M$. The angular bisector of $\angle MFE$ intersects $DE$ in $N$. Prove that $F$ is the circumcen... | [
"Because of the requirement on the length, the configuration is fixed: $M$ lies on the ray $CB$ past $B$, and $N$ lies on the ray $ED$ past $D$. See the figure. Let $\\alpha = \\angle BAC$ and $\\beta = \\angle ABC$. Moreover, let $H$ be the orthocentre of the triangle (in other words: the intersection of $AD$, $BE... | Netherlands | IMO Team Selection Test 2 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0j9w | Problem:
Answer the following two questions and justify your answers:
(1) What is the last digit of the sum $1^{2012} + 2^{2012} + 3^{2012} + 4^{2012} + 5^{2012}$?
(2) What is the last digit of the sum $1^{2012} + 2^{2012} + 3^{2012} + 4^{2012} + \cdots + 2011^{2012} + 2012^{2012}$? | [
"Solution:\n\nThe final digit of a power of $k$ depends only on the final digit of $k$, so there are 10 cases to consider. These are easy to work out. For $k$ ending in 1, the final digits are $1, 1, 1, 1, \\ldots$ For $k$ ending in 2 they are $2, 4, 8, 6, 2, 4, 8, 6, \\ldots$, et cetera. In fact all 10 possible fi... | United States | Bay Area Mathematical Olympiad | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | Part (1): 9; Part (2): 0 | |
0hpe | Problem:
Let $n$ be an integer greater than $2$. Positive real numbers $x$ and $y$ satisfy $x^{n} = x + 1$ and $y^{n+1} = y^{3} + 1$. Prove that $x < y$. | [
"Solution:\nIt is clear from the given that $x^{n} > 1$ and $y^{n+1} > 1$; therefore $x > 1$, $y > 1$. From this we get\n$$\n0 < (y - 1)(y^{2} - 1) = y^{3} - y^{2} - y + 1 \\Rightarrow y^{2} + y < y^{3} + 1 = y^{n+1}\n$$\nand, dividing by $y$, we obtain $y + 1 < y^{n}$. Thus $y^{n} - y > 1 = x^{n} - x$. Now if $y \... | United States | Berkeley Math Circle Take-Home Contest #1 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0at6 | Problem:
Let $X$, $Y$, and $Z$ be points on the sides $BC$, $AC$, and $AB$ of $\triangle ABC$, respectively, such that $AX$, $BY$, and $CZ$ are concurrent at point $O$. The area of $\triangle BOC$ is $a$. If $BX : XC = 2 : 3$ and $CY : YA = 1 : 2$, what is the area of $\triangle AOC$? | [
"Solution:\n$3a$"
] | Philippines | Philippines Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates"
] | null | final answer only | 3a | |
0902 | Let us call a point in the $xy$-plane a good point if each of its coordinates is an integer from $1$ to $2000$. Let us also call polyline $ABCD$ a Z-shaped polyline if four points $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$, $D(x_4, y_4)$ satisfy all the following conditions:
* $A$, $B$, $C$, $D$ are good points.
* $x_... | [
"Let us call a good point on $x = 1$ or $x = 2000$ excluding $(1, 1)$ a special point. Consider Z-shaped polyline $ABCD$ and let $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$, $D(x_4, y_4)$. Since $1 \\le x_1 < x_2 \\le 2000$, $1 \\le x_3 < x_2 \\le 2000$ and $1 \\le x_3 < x_4 \\le 2000$, any special point on Z-shape... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | English | proof and answer | 1333 | |
02k3 | Problem:
Uma certa máquina tem um visor, onde aparece um número inteiro $x$, e duas teclas $A$ e $B$. Quando se aperta a tecla $A$ o número do visor é substituído por $2x+1$. Quando se aperta a tecla $B$ o número do visor é substituído por $3x-1$.
Se no visor está o número $5$, o maior número de dois algarismos que se... | [
"Solution:\n\n(D) O diagrama a seguir mostra os resultados que podem ser obtidos a partir do número $5$ apertando-se cada uma das duas teclas.\n\n"
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Algorithms"
] | null | MCQ | D | |
0a9a | Problem:
A point $P$ is chosen in an arbitrary triangle. Three lines are drawn through $P$ which are parallel to the sides of the triangle. The lines divide the triangle into three smaller triangles and three parallelograms. Let $f$ be the ratio between the total area of the three smaller triangles and the area of the... | [
"Solution:\n\nLet $ABC$ be the triangle and let the lines through $P$ parallel to its sides intersect the sides in the points $D, E; F, G$ and $H, I$. The triangles $ABC$, $DEP$, $PFG$ and $IPH$ are similar and $BD=IP$, $EC=PF$. If $BC=a$, $IP=a_1$, $DE=a_2$ and $PF=a_3$, then $a_1+a_2+a_3=a$. There is a positive $... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 23 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | The ratio is at least one third, with equality if and only if the point is the centroid. | |
0g62 | 令 $ABCDE$ 是一凸五邊形使得
$$
BC \parallel AE, AB = BC + AE, 且 \angle ABC = \angle CDE.
$$
令 $M$ 是 $CE$ 之中點, $O$ 為三角形 $BCD$ 外接圓之圓心。
已知 $\angle DMO = 90^\circ$, 試證 $2\angle BDA = \angle CDE$. | [
"在射線 $AE$ 上取一點 $T$ 使得 $AT = AB$; 則由 $BC \\parallel AE$, 得\n$$\n\\angle CBT = \\angle ATB = \\angle ABT,\n$$\n所以 $BT$ 是 $\\angle ABC$ 的角平分線。\n\n另一方面，我們有\n$$\nET = AT - AE = AB - AE = BC,\n$$\n因此四邊形 $BCTE$ 是一平行四邊形且 $M$ 是對角線 $CE$ 之中點也是對角線 $BT$ 之中點。\n\n其次, 令 $K$ 是 $D$ 對於 $M$ 的對稱點。則 $OM$ 垂直平分線段 $DK$, 因此 $OD = OK$, 即點 $K... | Taiwan | 二〇一一數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Construct... | null | proof only | null | |
09qz | Problem:
Laat $a$, $b$, $c$ en $d$ positieve reële getallen zijn. Bewijs dat
$$
\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b} \geq 0
$$ | [
"Solution:\nEr geldt\n$$\n\\frac{a-b}{b+c}=\\frac{a-b+b+c}{b+c}-1=\\frac{a+c}{b+c}-1\n$$\nDoor hetzelfde met de andere drie breuken te doen en daarna de vier keer $-1$ naar de andere kant te halen, krijgen we dat we moeten bewijzen:\n$$\n\\frac{a+c}{b+c}+\\frac{b+d}{c+d}+\\frac{c+a}{d+a}+\\frac{d+b}{a+b} \\geq 4\n$... | Netherlands | Toets 6 juni 2012 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
08xy | When five positive integers $a, b, c, d, e$ satisfy $a < b < c < d < e < a^2 < b^2 < c^2 < d^2 < e^2 < a^3 < b^3 < c^3 < d^3 < e^3$, determine the minimum possible value that the sum $a+b+c+d+e$ can take. | [
"From the given inequalities, it follows that $a+4 \\le e$ and $e^2+1 \\le a^3$ must hold. Therefore, we have\n$$(a+4)^2 \\le e^2 \\le a^3 - 1,$$\nfrom which it follows that $(a + 4)^2 \\le a^3 - 1$, i.e.,\n$$\n(a-4)(a^2+3a+4) \\geq 1 > 0.\n$$\nThis means that we must have $a > 4$. Consequently, we have\n$$\na+b+c+... | Japan | Japan 2015 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 35 | |
01v7 | The bisector of the angle $CAB$ of triangle $ABC$ intersects the side $CB$ at $L$. The point $D$ is the foot of the perpendicular from $C$ to $AL$ and the point $E$ is the foot of the perpendicular from $L$ to $AB$. The lines $CB$ and $DE$ meet at $F$.
Prove that $AF$ is an altitude of the triangle $ABC$. | [
"Let $F'$ be the foot of an altitude from $A$ in the triangle $ABC$.\n\n\n\nIt is enough to prove that the points $D$, $F'$ and $E$ are colinear, since it will lead to $F = F'$ which means that $AF$ is an altitude of the triangle $ABC$. Since $\\angle CDA = \\angle CF'A = 90^\\circ$, the qu... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
05x6 | Problem:
On dit qu'un nombre rationnel strictement positif $q$ est magnifique s'il existe quatre entiers strictement positifs $a, b, c, d$ tels que
$$
q = \frac{a^{2021} + b^{2023}}{c^{2022} + d^{2024}}
$$
Existe-t-il un rationnel strictement positif qui n'est pas magnifique? | [
"Solution:\n\nLa question posée est assez déroutante : il a l'air d'être dur de décider ou non si un nombre peut s'écrire de cette forme. On peut donc essayer de se fixer un rationnel strictement positif de la forme $\\frac{r}{s}$ avec $r, s$ des entiers strictement positifs, et chercher des bons $a, b, c, d$ pour ... | France | Préparation Olympique Française de Mathématiques - ENVOI 5 : Pot-POURRI | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | No; every positive rational number is magnificent. | |
0iid | Problem:
A sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive reals satisfies $a_{n+1}=\sqrt{\frac{1+a_{n}}{2}}$. Determine all $a_{1}$ such that $a_{i}=\frac{\sqrt{6}+\sqrt{2}}{4}$ for some positive integer $i$. | [
"Solution:\nClearly $a_{1}<1$, or else $1 \\leq a_{1} \\leq a_{2} \\leq a_{3} \\leq \\ldots$\nWe can therefore write $a_{1}=\\cos \\theta$ for some $0<\\theta<90^{\\circ}$.\nNote that $\\cos \\frac{\\theta}{2}=\\sqrt{\\frac{1+\\cos \\theta}{2}}$, and $\\cos 15^{\\circ}=\\frac{\\sqrt{6}+\\sqrt{2}}{4}$.\nHence, the p... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | a1 ∈ { (sqrt(6)+sqrt(2))/4, sqrt(3)/2, 1/2 } | |
0cpn | Consider $2011$ nonzero integer numbers $a_1, a_2, \dots, a_{2011}$. It appears that the sum of each of them with the product of all the others is negative. Let us partition these $2011$ numbers into two nonempty groups and find the product of the numbers in each group. Prove that the sum of these two products is also ... | [
"Предположим, что среди данных чисел четное количество отрицательных. Тогда среди них есть положительное число $a$, и произведение всех чисел, кроме $a$, положительно. Это противоречит условию.\n\nЗначит, среди данных чисел нечетное число отрицательных. Пусть $x_1, x_2, \\dots, x_k$ и $y_1, y_2, \\dots, y_m$ — две ... | Russia | Russian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English, Russian | proof only | null | |
03of | It is given that $f(x)$ is a function defined on $\mathbb{R}$, satisfying $f(1) = 1$, and for any $x \in \mathbb{R}$,
$$
f(x+5) \ge f(x)+5,
$$
and $f(x+1) \le f(x)+1$.
If $g(x) = f(x)+1-x$, then $g(2002) = \underline{\hspace{2cm}}$. | [
"We determine $f(2002)$ first. From the conditions given, we have\n$$\n\\begin{aligned}\nf(x)+5 &\\le f(x+5) \\le f(x+4)+1 \\\\\n&\\le f(x+3)+2 \\le f(x+2)+3 \\\\\n&\\le f(x+1)+4 \\le f(x)+5.\n\\end{aligned}\n$$\nThus the equality holds for all. So we have $f(x+1)=f(x)+1$.\n\nHence, from $f(1)=1$, we get $f(2)=2$, ... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | 1 | |
0cxp | Evaluate the sum
$$
1+2+3-4-5+6+7+8-9-10+\ldots-2010
$$
where each three consecutive signs $+$ are followed by two signs $-$. | [
"We can write the sum as follows\n$$\n\\begin{gathered}\n\\sum_{k=0}^{401}[5k+1+5k+2+5k+3-(5k+4)-(5k+5)] \\\\\n=\\sum_{k=0}^{401}(5k-3)=5\\sum_{k=1}^{401}k-3\\cdot 402 \\\\\n=5 \\cdot \\frac{401 \\cdot 402}{2}-3 \\cdot 402 \\\\\n=402 \\cdot\\left(5 \\cdot \\frac{401}{2}-3\\right)=201 \\cdot 1999=401799\n\\end{gathe... | Saudi Arabia | SAMC | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | final answer only | 401799 | |
0jh8 | Problem:
Evaluate $1 + 2(-4)^2 + (-4)^3 + 2(-4)^5 + (-4)^6$. | [
"Solution:\n\n$1 + 2(-4)^2 + (-4)^3 + 2(-4)^5 + (-4)^6 = 1 - 2 \\cdot 4^2 + 2 \\cdot 4^5 = 2049 - 32 = 2017$."
] | United States | HMMT November 2013 | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 2017 | |
0bzr | Find the prime numbers $a > b > c$ given that $a - b$, $b - c$ and $a - c$ are distinct primes. | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | a=7, b=5, c=2 | |
0516 | Let $k$ be a positive integer. Determine the largest number of snakes, consisting of four squares (see figure), which can be placed on a $(2k+1) \times (2k+1)$ chessboard so that the snakes neither overlap nor stick out across the edges of the chessboard. The snakes can be turned and reflected.
![](attached_image_1.pn... | [
"First show that $k^2$ snakes can be placed on a $(2k+1) \\times (2k+1)$ chessboard. Divide the chessboard into strips of width $2$ (one strip of width $1$ remains). On any strip we can place $k$ snakes, one after another; so on $k$ strips, it is possible to place $k^2$ snakes.\n\nIt remains to prove that one can n... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | k^2 | |
0d6f | Given a polynomial $P(x) = a_{n} x^{n} + a_{n-1} x^{n-1} + \cdots + a_{1} x + a_{0}$ of real coefficients. Suppose that $P(x)$ has $n$ real roots (not necessarily distinct), and there exists a positive integer $k$ such that $a_{k} = a_{k-1} = 0$. Prove that $P(x)$ has a real root of multiplicity $k+1$.
(Note: we call ... | [
"We will show that $a_{k} = a_{k-1} = a_{k-2} = \\cdots = a_{0} = 0$ by induction on $n$, the degree of $P(x)$.\n\nIn fact, we may assume that the leading coefficient of $P(x)$ is $1$. For $n = 1, 2$, the result follows immediately.\n\nAssume that the induction hypothesis is true for every $n < m$, we shall prove i... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof only | null | |
0bze | Show that the positive divisors of no integer greater than $1$ may be placed in the cells of a rectangular array so that the four conditions below be simultaneously fulfilled:
(a) each cell contains exactly one divisor;
(b) distinct cells contain distinct divisors;
(c) the sum of the divisors on each row is the same; a... | [
"Suppose, if possible, that the positive divisors of some integer $n > 1$ may be arranged as required in an $k \\times \\ell$ rectangular array, where $k$ is the number of rows, and $\\ell$ is the number of columns; clearly, $k$ and $\\ell$ must both be greater than $1$.\nLet $s$ be the common value of the row sums... | Romania | THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | English | proof only | null | |
071n | Problem:
Let $a, b, c$ be three real numbers such that $1 \geq a \geq b \geq c \geq 0$. Prove that if $\lambda$ is a root of the cubic equation $x^{3}+a x^{2}+b x+c=0$ (real or complex), then $|\lambda| \leq 1$. | [
"Solution:\nSince $\\lambda$ is a root of the equation $x^{3}+a x^{2}+b x+c=0$, we have\n$$\n\\lambda^{3} = -a \\lambda^{2} - b \\lambda - c\n$$\nThis implies that\n$$\n\\begin{aligned}\n\\lambda^{4} & = -a \\lambda^{3} - b \\lambda^{2} - c \\lambda \\\\\n& = (1-a) \\lambda^{3} + (a-b) \\lambda^{2} + (b-c) \\lambda... | India | INMO | [
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
0c0w | Find all integers $n$ such that
$$
2^{2018} + 2^{2022} + 2^{2023} + 2^{2024} + 2^{2026} + 2^n
$$
is a square. | [] | Romania | Shortlisted problems for the 69th NMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 2026 | |
0ccw | Let $n \ge 2$ be an integer, and let $f$ be a $4n$-variable polynomial with real coefficients, such that, for any $2n$ points $(x_1, y_1), \dots, (x_{2n}, y_{2n})$ in the Cartesian plane,
$$
f(x_1, y_1, \dots, x_{2n}, y_{2n}) = 0
$$
if and only if they form the vertices of a regular $2n$-gon in some order, or are all e... | [
"The smallest possible degree is $2n$. In what follows, we will frequently write $A_i = (x_i, y_i)$, and abbreviate $P(x_1, y_1, \\dots, x_{2n}, y_{2n})$ to $P(A_1, \\dots, A_{2n})$ or as a function of any $2n$ points.\n\nSuppose that $f$ is valid. First, we note a key property:\n\n**Claim (Sign of $f$).** $f$ atta... | Romania | THE Fifteenth ROMANIAN MASTER OF MATHEMATICS | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 2n | |
0f1o | Problem:
50 watches, all keeping perfect time, lie on a table. Show that there is a moment when the sum of the distances from the center of the table to the center of each dial equals the sum of the distances from the center of the table to the tip of each minute hand. | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0hqb | Problem:
Let $ABCD$ be a convex quadrilateral such that $\angle ABD = \angle ACD$. Prove that $ABCD$ can be inscribed in a circle. | [
"Solution:\n\nThere are many ways to structure the proof. The following method seems to have minimal logical difficulties.\nBecause points $A$, $B$, and $C$ are not collinear, we can draw the circumscribed circle $\\omega$ of $\\triangle ABC$. The arc $AC$ of $\\omega$, not containing $B$, is intercepted by inscrib... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0hft | Fedir and Mykhailo have three piles of stones: the first contains $100$ stones, the second $101$, the third $102$. They are playing a game with the following rules: they move in turns, in his turn a player chooses any two piles of stones, containing, say, $a$ and $b$ stones correspondingly, and takes from each of them ... | [
"Let's provide the winning strategy for Mykhailo. Suppose that before Fedir's move, the piles contained $(2n, 2n+1, 2n+2)$ stones for some integer $n > 1$. Then there are 3 possible moves.\n\nIf he takes first two piles, then, as $(2n, 2n+1) = 1$, after his move we will have the following numbers of stones: $(2n-1,... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | Mykhailo (the second player) wins. | |
0amc | Problem:
What is the units digit of $25^{2010} - 3^{2012}$?
(a) 8
(b) 6
(c) 2
(d) 4 | [] | Philippines | Qualifying Round | [
"Number Theory > Modular Arithmetic"
] | null | MCQ | d | |
04o0 | Prove that
$$
3abc + a + b + c \ge 2(ab + bc + ca)
$$
holds for all real numbers $a, b, c \ge 1$.
Determine all cases for which the equality is obtained. | [
"Let us denote $x = a-1$, $y = b-1$ and $z = c-1$. Then $x, y, z \\ge 0$, and the given inequality easily transforms into\n$$\n3xyz + xy + yz + zx \\ge 0,\n$$\nwhich is true since all addends on the left-hand side are non-negative.\n\nThe equality is obtained if and only if $xyz = xy = yz = zx = 0$, which is true i... | Croatia | Croatian Junior Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Equality holds if and only if at least two of the numbers are equal to one. | |
07j3 | Find all the functions $f : \mathbb{R} \to \mathbb{R}$ such that for all $x, y \in \mathbb{R}$ we have:
$$
f(yf(x) + f(x)f(y)) = xf(y) + f(xy)
$$ | [
"First Solution. Plugging $(x, y) = (1, yf(z) + f(z)f(y))$ yields\n$$\n\\begin{align*}\n& f(C(yf(z) + f(z)f(y) + f(yf(z) + f(z)f(y)))) \\\\\n&= f(C(yf(z) + zf(y) + f(y)f(z) + f(zy)) \\\\\n&= 2f(yf(z) + f(z)f(y)) = 2zf(y) + 2f(zy)\n\\end{align*}\n$$\n\n$$\nzf(y) + f(zy) = yf(z) + f(yz).\n$$\nHence, $f(y) = Cy$ for s... | Iran | 41th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0, f(x) = x, or f(x) = -2x | |
0gb0 | 2. 任選橢圓 $C: x^2 + 2y^2 = 2098$ 上的一個有理點 $P_0 = (x_p, y_p)$。我們將依以下方式遞迴決定 $P_1, P_2, \cdots$：對於所有 $i = 0, 1, \cdots,$
(1) 選取一個不在 $C$ 上的整點 $Q_i = (x_i, y_i)$，使得 $|x_i| < 50$ 且 $|y_i| < 50$。
(2) 連接 $\overline{P_iQ_i}$，並令其與 $C$ 的另一交點為 $P_{i+1}$。
試證：對於任何 $P_0$，我們都可以適當選取 $Q_0, Q_1, \cdots$，使得存在某個非負整數 $k$，讓 $\overline{OP_k} = 2... | [
"1. 易知 $C$ 上的所有整數點為 $(\\pm44, \\pm9)$，且 $44^2 + 9^2 = 2017$，故我們只要證明經過適當的操作後，某個 $P_k$ 是整點即可。\n\n2. 若 $P_0$ 是整點，由 1. 知取 $k=0$ 即可，故假設 $P_0 = (a/m, b/m)$ 不為整點，其中 $a, b, m \\in \\mathbb{Z}$ 且 $m > 0$。\n\n– 顯然存在整數 $s$ 和 $t$ 滿足 $|s - \\frac{a}{m}| \\le 1/2$ 及 $|t - \\frac{b}{m}| \\le 1/2$。又，$|s| < \\sqrt{2098} + 1 < 50$，同... | Taiwan | 二〇一七數學奧林匹亞競賽第三階段選訓營 | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0hn0 | Problem:
How many ways are there to color the five vertices of a regular 17-gon either red or blue, such that no two adjacent vertices of the polygon have the same color? | [
"Solution:\n\nThe answer is zero! Call the polygon $A_{1} A_{2} \\ldots A_{17}$. Suppose for contradiction such a coloring did exist.\nIf we color $A_{1}$ red, then $A_{2}$ must be blue. From here we find $A_{3}$ must be red, then $A_{4}$ must be blue; thus $A_{5}$ must be red, $A_{6}$ must be blue. Proceeding in t... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | 0 | |
0jbp | Problem:
Find all ordered triples $(a, b, c)$ of positive reals that satisfy: $\lfloor a\rfloor b c=3$, $a\lfloor b\rfloor c=4$, and $a b\lfloor c\rfloor=5$, where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$. | [
"Solution:\n\nAnswer: $\\left(\\frac{\\sqrt{30}}{3}, \\frac{\\sqrt{30}}{4}, \\frac{2 \\sqrt{30}}{5}\\right),\\left(\\frac{\\sqrt{30}}{3}, \\frac{\\sqrt{30}}{2}, \\frac{\\sqrt{30}}{5}\\right)$\n\nWrite $p=a b c$, $q=\\lfloor a\\rfloor\\lfloor b\\rfloor\\lfloor c\\rfloor$. Note that $q$ is an integer.\n\nMultiplying ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (sqrt(30)/3, sqrt(30)/4, 2sqrt(30)/5), (sqrt(30)/3, sqrt(30)/2, sqrt(30)/5) | |
0ezk | Problem:
Show that given $200$ integers you can always choose $100$ with sum a multiple of $100$. | [] | Soviet Union | 4th ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other"
] | null | proof only | null | |
04wq | There are 5 distinct points $A$, $B$, $C$, $D$, $E$ lying in this order on a circle with radius $r$ satisfying $AC = BD = CE = r$. There is a triangle having ortocentres of triangles $ACD$, $BCD$, $BCE$ as its vertices. Prove that this triangle is right-angled. | [
"In any obtuse triangle $XYZ$ with obtuse angle in $Z$ and ortocentre $W$, angles $XYZ$ and $XWZ$ are equal, as complementing the angle $YXW$ to $90$ degrees (fig. 1). Moreover, points $Y$ and $W$ lie in different half-planes determined by $XZ$.\n\n\nFig. 1\n\nLet $P$, $Q$, $R$ be ortocentr... | Czech-Polish-Slovak Mathematical Match | Cesko-Slovacko-Poljsko 2006 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03mh | Determine all polynomials $P(x)$ with real coefficients such that
$$
(x + 1)P(x - 1) - (x - 1)P(x)
$$
is a constant polynomial. | [
"The answer is $P(x)$ being any constant polynomial and $P(x) \\equiv kx^2 + kx + c$ for any (nonzero) constant $k$ and constant $c$.\nLet $\\Lambda$ be the expression $(x+1)P(x-1) - (x-1)P(x)$, i.e. the expression in the problem statement.\nSubstituting $x = -1$ into $\\Lambda$ yields $2P(-1)$ and substituting $x ... | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All real polynomials of the form P(x) = a x^2 + a x + c for real a and c (including constant polynomials when a = 0). | |
0260 | Problem:
Um serviço de vigilância vai ser instalado num parque na forma de uma rede de estações. As estações devem ser conectadas por linhas de telefone, de modo que qualquer uma das estações possa se comunicar com todas as outras, seja por uma conexão direta seja através de no máximo uma outra estação. Cada estação p... | [
"Solution:\n\nO exemplo mostra que podemos conectar pelo menos 7 estações dentro das condições propostas. Começamos com uma estação particular, e vamos pensar nela como se fosse a base da rede. Ela pode ser conectada a 1, 2 ou 3 estações conforme mostra o diagrama.\n\n\n\nAgora, as estações... | Brazil | Lista 6 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 10 | |
0bbs | Consider $n$ persons, each of them speaking at most 3 languages. From any 3 persons there are at least two which speak a common language.
i) For $n \le 8$, exhibit an example in which no language is spoken by more than two persons.
ii) For $n \ge 9$, prove that there exists a language which is spoken by at least thre... | [
"i) Split the 8 persons in two groups of 4. Set any pair of persons in each group to speak a different language for a total of $6 + 6 = 12$ languages, each spoken by 2 persons, each person speaking 3 languages.\n\nFor $n \\le 7$, just remove $8-n$ persons.\n\nii) Assume by contrary that each language is spoken by a... | Romania | 62nd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
05k0 | Problem:
Le sommet $B$ d'un angle $\widehat{A B C}$ se trouve à l'extérieur d'un cercle $\omega$ tandis que les demi-droites $[B A)$ et $[B C)$ le traversent. Soit $K$ un point d'intersection du cercle $\omega$ avec $[B A)$. La perpendiculaire à la bissectrice de l'angle $\widehat{A B C}$ passant par $K$ recoupe le cer... | [
"Solution:\n\nSoit $O$ le centre du cercle $\\omega$. Notons $\\Delta$ la bissectrice de $\\widehat{A B C}$, $\\Delta'$ la parallèle à $\\Delta$ passant par $O$ et $N$ le symétrique de $O$ par rapport à $\\Delta$.\nLa symétrie $s_{\\Delta'}$ par rapport à $\\Delta'$ envoie $O$ sur $O$ et $P... | France | OFM | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0hvx | Problem:
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(x y+z)=f(x) f(y)+f(z)
$$
for all real numbers $x$, $y$, and $z$. | [
"Solution:\nAnswer: $f(x)=0$ or $f(x)=x$. Both of these trivially satisfy the equation.\n\nWe plug in $x=y=z=0$ to get\n$$\n\\begin{array}{r}\nf(0)=f(0)^{2}+f(0) \\\\\n0=f(0)^{2} \\\\\n0=f(0)\n\\end{array}\n$$\nThen we plug in $x=y=1$, $z=0$ to get\n$$\n\\begin{aligned}\nf(1) & =f(1)^{2}+f(0) \\\\\n0 & =f(1)^{2}-f(... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | f(x) = 0 for all real x, or f(x) = x for all real x | |
0447 | For each positive integer $N$, let $\tau(N)$ be the number of positive factors of $N$; $\omega(N)$ be the number of distinct prime factors of $N$; $\Omega(N)$ be the number of prime factors (counts multiplicities) of $N$. Prove: for each positive integer $n$,
$$
\sum_{m=1}^{n} 5^{\omega(m)} \le \sum_{k=1}^{n} \lfloor \... | [
"First, note that $\\lfloor \\frac{n}{k} \\rfloor$ represents the number of multiples of $k$ among $1, 2, \\dots, n$. Hence,\n$$\n\\sum_{k=1}^{n} \\lfloor \\frac{n}{k} \\rfloor \\tau(k)^2 = \\sum_{k=1}^{n} \\sum_{1 \\le m \\le n,\\ k|m} \\tau(k)^2 = \\sum_{m=1}^{n} \\sum_{k|m} \\tau(k)^2.\n$$\nTo prove the problem ... | China | China National Team Selection Test | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
07vt | Suppose $ABC$ is an equilateral triangle whose circumcircle $\Gamma$ has radius $1$. Prove that if $P$ is within or on $\Gamma$, then the product $|PA| \cdot |PB| \cdot |PC|$ does not exceed $2$. Also determine the points $P$ for which the product is equal to $2$. | [
"We will use complex numbers and suppose that $\\Gamma$ is centred at the origin so that $\\Gamma$ is the set of complex numbers $w$ such that $|w| = 1$. Moreover, w.l.o.g., we can suppose that $A = 1$, $B = \\omega$, $C = \\omega^2$, where $\\omega = e^{2\\pi i/3}$ is a cube root of unity and $1 + \\omega + \\omeg... | Ireland | IRL_ABooklet_2023 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof and answer | Maximum product equals 2, attained exactly at the three points on the circumcircle diametrically opposite A, B, and C (equivalently, the images of A, B, and C under a sixty-degree rotation about the circumcenter). | |
0gtf | Let $(x, y, z)$ be a triple of positive real numbers satisfying
$$
xyz = 1 \quad \text{and} \quad \frac{y}{z}(y-x^2) + \frac{z}{x}(z-y^2) + \frac{x}{y}(x-z^2) = 0,
$$
and $t_1, t_2$ and $t_3$ be the smallest, the median and the largest of $x, y, z$, respectively. Find the smallest possible value of
$$
\frac{t_1 + t_3}{... | [
"Answer: $\\frac{5}{\\sqrt[5]{256}}$.\nLet $\\frac{x^2}{y} = a$, $\\frac{y^2}{z} = b$ and $\\frac{z^2}{x} = c$. The problem conditions in this new variables take the following form\n---\n\n$$\nabc = 1, \\quad a+b+c = ab+bc+ca.\n$$\nNow we readily get $(a-1)(b-1)(c-1) = 0$. Therefore, at least one of the numbers $a,... | Turkey | Team Selection Test | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 5/√[5]{256} | |
0gl8 | Let $p \ge 5$ be a positive prime number. Show that there exists an integer $t \in \{1, 2, ..., p\}$ such that the equation
$$
x^2 = y^{\frac{p-1}{2}} + t
$$
has no integer solutions. | [
"We consider three cases depending on the value of $p$.\n\n*Case 1.* $p \\equiv 1 \\pmod 4$.\n\nPicking $t = 2$ works. Since $x^2$ and $y^{\\frac{p-1}{2}}$ are perfect squares, their difference cannot be $2$.\n\n*Case 2.* $p \\equiv 3 \\pmod 4$ and $p > 7$.\n\nNote that $y^{\\frac{p-1}{2}} \\equiv -1, 0$ or $1 \\pm... | Thailand | The first T3MO | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0jfl | Let $n$ be a positive integer. There are $\frac{n(n+1)}{2}$ marks, each with a black side and a white side, arranged into an equilateral triangle, with the biggest row containing $n$ marks. Initially, each mark has the black side up. An *operation* is to choose a line parallel to one of the sides of the triangle, and f... | [
"**Solution** (By Warut Suksompong). The answer is $6\\lfloor\\frac{n}{4}\\rfloor + n - 4\\lfloor\\frac{n}{4}\\rfloor = n + 2\\lfloor\\frac{n}{4}\\rfloor$.\nFor $n=1$ the answer is clearly 1, since there is only one configuration other than the initial one, and that configuration takes 1 step to get to. From now on... | United States | USAMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Linear Algebra > Linear transformations"
] | null | proof and answer | n + 2 floor(n/4) | |
092d | Problem:
Determine all functions $f: \mathbb{R} \setminus \{0\} \to \mathbb{R} \setminus \{0\}$ such that
$$
f\left(x^{2} y f(x)\right)+f(1)=x^{2} f(x)+f(y)
$$
holds for all nonzero real numbers $x$ and $y$. | [
"Solution:\nLet $f$ be any function with the desired property and set $\\alpha=f(1)$.\n\nLemma. Let $x \\in \\mathbb{R}^{\\neq 0}$ be arbitrary and put $z=x^{2} f(x)$. Then $f(z)=z$, $f\\left(z^{2}\\right)=2 z-\\alpha$ and $z^{2}=3 z-2 \\alpha$.\n\nProof. Substituting $y=1$ and $y=z$ into the given functional equat... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 1/x^2 or f(x) = -1/x^2 for all nonzero real x | |
0afg | Дали постојат реални броеви *a*, *b*, *c*, *d* такви што условите:
a) равенката $ax^2 + bdx + c = 0$ има реални различни корени $x_1, x_2$
б) равенката $bx^2 + cdx + a = 0$ има реални различни корени $x_2, x_3$
в) равенката $cx^2 + adx + b = 0$ има реални различни корени $x_3, x_1$.
да важат истовремено. | [
"Нека претпоставиме дека такви броеви постојат. Тогаш равенките под а), б) и в) имаат по две различни решенија, секоја посебно, па според тоа $a \\neq 0, b \\neq 0$ и $c \\neq 0$. Од Виетовите формули имаме $x_1 x_2 = \\frac{c}{a}$, $x_2 x_3 = \\frac{a}{b}$ и $x_3 x_1 = \\frac{b}{c}$. Ако последните три равенства г... | North Macedonia | Републички натпревар по математика за средно образование | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | Macedonian, English | proof and answer | No | |
0k5k | Show that there is an absolute constant $c < 1$ with the following property: whenever $\mathcal{P}$ is a polygon with area 1 in the plane, one can translate it by a distance of $\frac{1}{100}$ in some direction to obtain a polygon $\mathcal{Q}$, for which the intersection of the interiors of $\mathcal{P}$ and $\mathcal... | [
"The following solution is due to Brian Lawrence. We will prove the result with the generality of any measurable set $\\mathcal{P}$ (rather than a polygon). For a vector $v$ in the plane, write $\\mathcal{P} + v$ for the translate of $\\mathcal{P}$ by $v$.\nSuppose $\\mathcal{P}$ is a polygon of area 1, and $\\vare... | United States | USA TSTST | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
05sd | Problem:
Soit $S=\{1, \ldots, n\}$, avec $n \geqslant 3$ un entier, et soit $k$ un entier strictement positif. On note $S^{k}$ l'ensemble des $k$-uplets d'éléments de $S$. Soit $f: S^{k} \rightarrow S$ telle que, si $x=\left(x_{1}, \ldots, x_{k}\right) \in S^{k}$ et $y=\left(y_{1}, \ldots, y_{k}\right) \in S^{k}$ avec... | [
"Solution:\n\nNous montrerons le résultat par récurrence sur $k$. Le cas $k=1$ est trivial, supposons donc le résultat vrai pour $k-1 \\geqslant 1$ et montrons-le pour $k$.\n\nSupposons l'existence de $k-1$ éléments $a_{2}, \\ldots, a_{k}$ de $S$ tels que la fonction $\\varphi: a \\in S \\mapsto f\\left(a, a_{2}, \... | France | Préparation Olympique Française de Mathématiques - ENVOI 4 : POT-POURRI | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0521 | A train departed from the station 12 minutes later than planned. If the train would not make any stops on the way and would travel at average speed equal to what would be the average speed between stops according to the timetable, then it would reach the destination exactly at the right time. But if the train would sto... | [
"Let the time we are looking for be $t$. The conditions of the problem imply that $t - 12 \\text{ min} = 1.4 \\cdot (t - 24 \\text{ min})$, from which $0.4t = 21.6 \\text{ min}$ and $t = 54 \\text{ min}$."
] | Estonia | Final Round of National Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 54 minutes | |
068d | Let $x$, $y$, $z$ be positive real numbers such that $x + y + z = 9xyz$. Prove that:
$$
\frac{x}{\sqrt{x^2 + 2yz + 2}} + \frac{y}{\sqrt{y^2 + 2zx + 2}} + \frac{z}{\sqrt{z^2 + 2xy + 2}} \geq 1.
$$
When does the equality hold? | [
"From the inequality $2yz \\le y^2 + z^2$ we have that $x^2 + 2yz + 2 \\le x^2 + y^2 + z^2 + 2$, so\n$$\n\\frac{x}{\\sqrt{x^2 + 2yz + 2}} \\ge \\frac{x}{\\sqrt{x^2 + y^2 + z^2 + 2}}\n$$\nWorking similarly and adding we have that\n$$\n\\frac{x}{\\sqrt{x^2 + 2yz + 2}} + \\frac{y}{\\sqrt{y^2 + 2zx + 2}} + \\frac{z}{\\... | Greece | Selection Examination | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Equality holds at x = y = z = 1/√3. | |
0h8s | Let $[x_1, x_2]$ and $[y_1, y_2, y_3]$ be the least common multiple of $x_1, x_2$ and $y_1, y_2, y_3$ respectively. For any positive integers $a, b, c, d$ let $A$ and $B$ be such that:
$$
A = [a, b, c] \cdot [a, b, d] \cdot [a, c, d] \cdot [b, c, d] \text{ and } B = [a, b] \cdot [a, c] \cdot [a, d] \cdot [b, c] \cdot [... | [
"Take prime $p$ that divides $abcd$. Without loss of generality, let the prime number be a factor of $a, b, c, d$ of degree $a_1 \\geq b_1 \\geq c_1 \\geq d_1$ respectively. We can find the biggest degree of $p$—$p_A$ and $p_B$, that divide $A$ and $B$ respectively.\n\n$$\np_A = 3a_1 + b_1, \\quad p_B = 3a_1 + 2b_1... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
08jj | Problem:
In the rectangular coordinate system every point with integer coordinates is called a lattice point. Let $P_n(n, n+5)$ be a lattice point and denote by $f(n)$ the number of lattice points on the open segment $\left(OP_n\right)$, where the point $O(0,0)$ is the coordinate system origin. Calculate the number $f... | [] | JBMO | The second selection test for JMBO 2003 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 1600 | |
0dsw | Let $x_1$, $x_2$, $x_3$, $y_1$, $y_2$, $y_3$ be real numbers in $[-1, 1]$. Find the maximum value of
$$
(x_1y_2 - x_2y_1)(x_2y_3 - x_3y_2)(x_3y_1 - x_1y_3)
$$ | [] | Singapore | Singapore International Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | 84*sqrt(3) - 144 | |
0g6y | 試求出所有由正整數集映至正整數集的函數對 $(f, g)$ 滿足
$$
f^{g(n)+1}(n) + g^{f(n)}(n) = f(n+1) - g(n+1) + 1
$$
對所有正整數 $n$ 皆成立。這裡定義 $f^1(n) = f(n)$, $f^{k+1}(n) = f(f^k(n))$。 | [
"唯一滿足題目敘述的函數對 $(f, g)$ 是 $f(n) = n, g(n) = 1$。\n由條件可知對所有正整數 $n$ 都有\n$$\nf(f^{g(n)}(n)) < f(n+1).\n$$\n將函數 $f$ 能取到的所有值依大小記為 $y_1 < y_2 < \\dots$ (這個序列的長度可能是有限或無限), 我們接下來要運用數學歸納法證明：\n$$\n(i)_n : f(x) = y_n \\text{ 若且唯若 } x = n,\n$$\n$$\n(ii)_n : y_n = n.\n$$\n\n$n$ 有 $f(x) = a = y_a$ 若且唯若 $x = a$. 注意到這也表示對於任意 $1 \\le... | Taiwan | 二〇一二數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | f(n) = n and g(n) = 1 for all positive integers n | |
04d2 | Determine all functions $f : \mathbb{R} \to \mathbb{R}$ such that for all real numbers $x$ and $y$ holds
$$
f(x^2 + f(y)) = (f(x) + y^2)^2.
$$ | [
"Let $a = f(0)$. Setting $x = y = 0$ in the given equation gives $f(a) = a^2$. Setting $y = 0$ gives\n$$\nf(x^2 + a) = (f(x))^2, \\quad \\forall x \\in \\mathbb{R}. \\qquad (\\circ)\n$$\nSetting $x = a$ gives $f(a^2 + a) = (f(a))^2 = a^4$.\nAssume that $a < 0$. Then there exists $b > 0$ such that $a = -b^2$. Settin... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = x^2 for all real x | |
0ibd | Problem:
Prove that
$$
\frac{\sigma(1)}{1}+\frac{\sigma(2)}{2}+\frac{\sigma(3)}{3}+\cdots+\frac{\sigma(n)}{n} \leq 2 n
$$
for every positive integer $n$. | [
"Solution:\nThis is similar to the previous solution. If $d$ is a divisor of $i$, then so is $i/d$, and $(i/d)/i = 1/d$. Summing over all $d$, we see that $\\sigma(i)/i$ is the sum of the reciprocals of the divisors of $i$, for each positive integer $i$. So, summing over all $i$ from $1$ to $n$, we get the value $1... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
084u | Problem:
Quanto vale $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$?
(A) $\sqrt[3]{9-4 \sqrt{5}}$
(B) 1
(C) $\frac{3}{2}$
(D) $\sqrt[3]{4}$
(E) $2 \sqrt[3]{2}$. | [
"Solution:\n\nLa risposta è (B). Con opportuni raccoglimenti, si scrive il cubo di $r=\\sqrt[3]{2+\\sqrt{5}}+\\sqrt[3]{2-\\sqrt{5}}$ come $r^{3}=4+3 \\sqrt[3]{2^{2}-5}(\\sqrt[3]{2+\\sqrt{5}}+\\sqrt[3]{2-\\sqrt{5}})=4-3(\\sqrt[3]{2+\\sqrt{5}}+\\sqrt[3]{2-\\sqrt{5}})=4-3 r$. Perciò $r$ verifica la condizione che $r^{... | Italy | Progetto Olimpiadi di Matematica 2006 GARA di SECONDO LIVELLO | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | B | |
03m2 | Let $A$, $B$, $P$ be three points on a circle. Prove that if $a$ and $b$ are the distances from $P$ to the tangents at $A$ and $B$ and $c$ is the distance from $P$ to the chord $AB$, then $c^2 = ab$. | [
"Let $r$ be the radius of the circle, and let $a'$ and $b'$ be the respective lengths of $PA$ and $PB$. Since $b' = 2r \\sin \\angle PAB = 2rc/a'$, $c = a'b'/(2r)$. Let $AC$ be the diameter of the circle and $H$ the foot of the perpendicular from $P$ to $AC$. The similarity of the triangles $ACP$ and $APH$ imply th... | Canada | Kanada 2010 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"... | null | proof only | null | |
0ajy | Let $x$, $y$, $z$ be positive integers such that $x \neq y \neq z \neq x$. Prove that $(x+y+z)(xy+yz+zx-2) \geq 9xyz$.
When does the equality hold? | [
"Since $x$, $y$, $z$ are distinct positive integers, the required inequality is symmetric and WLOG we can suppose that $x \\geq y+1 \\geq z+2$. We consider 2 possible cases:\n\n**Case 1.** $y \\geq z+2$. Since $x \\geq y+1 \\geq z+3$ it follows that\n$$\n(x-y)^2 \\geq 1, \\quad (y-z)^2 \\geq 4, \\quad (x-z)^2 \\geq... | North Macedonia | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Equality holds if and only if the three integers are consecutive: {x, y, z} = {k, k+1, k+2} for some positive integer k (in any order). | |
0i1x | Problem:
Let $J H I Z$ be a rectangle, and let $A$ and $C$ be points on sides $Z I$ and $Z J$, respectively. The perpendicular from $A$ to $C H$ intersects line $H I$ in $X$, and the perpendicular from $C$ to $A H$ intersects line $H J$ in $Y$. Prove that $X, Y$ and $Z$ are collinear (lie on the same line). | [
"Solution:\n\nObserve that $\\angle X A I = \\angle X H C = \\angle H C J$. Hence $\\triangle X A I \\sim \\triangle H C J$, and thus $X I / H J = A I / C J$. Likewise, $Y J / H I = C J / A I$. Putting these together yields $X I / H J = H I / Y J$, and hence\n$$\n\\frac{X I}{Z I} = \\frac{Z J}{Y J} \\Rightarrow \\t... | United States | 3rd Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
01rt | a) After division of a positive integer $n$ by two positive integers one has two remainders different from zero.
Is it possible for $n$ to be the sum of these two remainders?
b) After division of a positive integer $n$ by $29$, $39$, and $59$ one has three nonzero remainders such that their sum is equal to $n$.
Find a... | [
"a) Let $n = a q_1 + r_1 = b q_2 + r_2$, $a, b \\in \\mathbb{N}$, $a < b$, $r_1 \\neq 0$, $r_2 \\neq 0$. Suppose that $n = r_1 + r_2$. Then $n = a q_1 + r_1 = b q_2 + r_2 = r_1 + r_2$, ($q_1 \\ge 0$, $q_2 \\ge 0$) and $r_1 < a$, $r_2 < b$, so $b q_2 = r_1 < a \\le b$. The inequality $b q_2 < b$ holds only if $q_2 =... | Belarus | FINAL ROUND | [
"Number Theory > Modular Arithmetic"
] | English | proof and answer | a) No. b) 112. | |
0gxr | $$
\frac{AO^2}{BC} + \frac{BO^2}{CA} + \frac{CO^2}{AB} \ge \frac{AO + BO + CO}{\sqrt{3}}
$$
where $O$ is the point inside triangle $ABC$ such that $\angle AOB = \angle BOC = \angle COA = 120^\circ$. | [
"Denote the length of $AO$ by $x$, $BO$ by $y$ and $CO$ by $z$. From the cosine law we have: (fig.26)\n\n\nFig.26\n\n$$\nAB = \\sqrt{x^2 + xy + y^2}, \\quad BC = \\sqrt{y^2 + yz + z^2},\n$$\n$CA = \\sqrt{z^2 + zx + x^2}$ and so we can rewrite our inequality:\n$$\n\\frac{x^2}{\\sqrt{y^2 + yz... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
0b94 | On the exterior of a non-equilateral triangle $ABC$ consider the similar triangles (in this order) $ABM$, $BCN$ and $CAP$, such that the triangle $MNP$ is equilateral. Find the angles of the triangles $ABM$, $BCN$ and $CAP$.
Nicolae Bourbăcuţ | [
"The given similarity rewrites as\n$$\n\\frac{m-b}{a-b} = \\frac{n-c}{b-c} = \\frac{p-a}{c-a} = k,\n$$\nhence\n$$\nm = ka + (1-k)b,\n$$\n$$\nn = kb + (1-k)c,\n$$\n$$\np = kc + (1-k)a.\n$$\nSince the triangle $MNP$ is equilateral, we have\n$$\nm + \\varepsilon n + \\varepsilon^2 p = 0,\n$$\nwhere $\\varepsilon = \\c... | Romania | Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | English | proof and answer | 30°, 30°, 120° | |
0hug | Problem:
Let $A_{0}, A_{1}, \ldots, A_{n}$ be points in a plane such that
(i) $A_{0}A_{1} \leq \frac{1}{2}A_{1}A_{2} \leq \cdots \leq \frac{1}{2^{n-1}}A_{n-1}A_{n}$ and
(ii) $0 < \measuredangle A_{0}A_{1}A_{2} < \measuredangle A_{1}A_{2}A_{3} < \cdots < \measuredangle A_{n-2}A_{n-1}A_{n} < 180^{\circ}$,
where all these... | [
"Solution:\nSuppose that $A_{k}A_{k+1} \\cap A_{m}A_{m+1} \\neq \\emptyset$ for some $k, m > k+1$. Without loss of generality we may suppose that $k=0, m=n-1$ and that no two segments $A_{k}A_{k+1}$ and $A_{m}A_{m+1}$ intersect for $0 \\leq k < m-1 < n-1$ except for $k=0$, $m=n-1$. Also, shortening $A_{0}A_{1}$, we... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof only | null | |
0hwj | Problem:
Let $a$, $b$, and $c$ be positive real numbers such that
$$
a b c = 1
$$
Prove that for every positive integer $n$,
$$
a + b + c > \frac{1}{a} + \frac{1}{b} + \frac{1}{c} .
$$ | [
"Solution:\n\nUsing $a b c = 1$, we transform the given condition to\n$$\na^{n} + b^{n} + c^{n} > \\frac{1}{a^{n}} + \\frac{1}{b^{n}} + \\frac{1}{c^{n}} .\n$$\n$$\na + b + c > b c + c a + a b\n$$\nor\n$$\n- b c - c a - a b + a + b + c > 0 .\n$$\nWe then add $a b c - 1 (= 0)$ to the left side, getting\n$$\na b c - b... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0laa | Let $n > 3$ be a positive integer and let $T$ denote the set of the first $n$ positive integers. A subset $S$ of $T$ is called a *scattered set* if $S$ has the following property: There exists a positive integer $c$, not exceeding $\frac{n}{2}$ such that $|s_1 - s_2| \neq c$ for any two numbers $s_1, s_2$ in $S$. What ... | [] | Vietnam | Vijetnam 2008 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | floor(2n/3) | |
0czf | In the isosceles triangle $ABC$, with $AB = AC$, the angle bisector of $\widehat{B}$ intersects side $AC$ at $B'$. Suppose that $BB' + B'A = BC$. Find the angles of the triangle. | [
"On the side $BC$ take point $M$ such that $CM = AB'$. The angle bisector theorem implies\n$$\n\\frac{AB'}{B'C} = \\frac{AB}{BC} = \\frac{AC}{BC} \\text{, hence } \\frac{MC}{B'C} = \\frac{AC}{BC} \\text{.}\n$$\n\nIt follows $\\frac{MC}{AC} = \\frac{B'C}{BC}$, that is $\\triangle MCB' \\sim ... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | ∠B = ∠C = 40°, ∠A = 100° | |
0asi | Problem:
A figure consists of two overlapping circles that have radii $4$ and $6$. If the common region of the circles has area $2\pi$, what is the area of the entire figure? | [
"Solution:\n$50\\pi$"
] | Philippines | Philippines Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles"
] | null | final answer only | 50π | |
06xg | Let $\mathbb{R}_{>0}$ be the set of positive real numbers. Determine all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that
$$
x(f(x)+f(y)) \geqslant (f(f(x))+y) f(y)
$$
for every $x, y \in \mathbb{R}_{>0}$. | [
"Answer: All functions $f(x)=\\frac{c}{x}$ for some $c>0$.\n\nSolution 1. Let $f: \\mathbb{R}_{>0} \\rightarrow \\mathbb{R}_{>0}$ be a function that satisfies the inequality of the problem statement. We will write $f^{k}(x)=f(f(\\cdots f(x) \\cdots))$ for the composition of $f$ with itself $k$ times, with the conve... | IMO | International Mathematical Olympiad Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | f(x) = c/x for some c > 0 | |
05hk | Problem:
Soit $ABC$ un triangle isocèle en $A$, $\Gamma$ est un cercle tangent à $(AC)$ en $C$ à l'extérieur du triangle $ABC$. On note $\omega$ le cercle passant par $A$ et $B$ et tangent intérieurement à $\Gamma$ en $D$. $E$ est la seconde intersection de $(AD)$ et de $\Gamma$, montrer que $(BE)$ est tangente à $\Ga... | [
"Solution:\n\nOn note $I_{A}$ l'inversion de centre $A$ et de rayon $r = AB = AC$, ainsi, $B$ et $C$ sont fixes par $I_{A}$, la puissance de $A$ par rapport au cercle $\\omega$ est exactement le rayon au carré donc le cercle $\\omega$ est fixe par inversion. Cela montre que $D$ et $E$ sont échangés dans $I_{A}$. Ma... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0f3m | Problem:
$ABCD$ is a tetrahedron. Angles $ACB$ and $ADB$ are $90^\circ$. Let $k$ be the angle between the lines $AC$ and $BD$. Show that $\cos k < \dfrac{CD}{AB}$. | [] | Soviet Union | ASU | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0br4 | Find all pairs of positive integers $(x, y)$ so that
$$
x + y = \sqrt{x} + \sqrt{y} + \sqrt{xy}.
$$ | [
"The equality can be written $(x + y) - \\sqrt{x} = \\sqrt{xy} + \\sqrt{y}$. Squaring yields $x^2 + xy + y^2 + x - y = 2(2y + x)\\sqrt{x}$. Since $2y + x \\neq 0$, $\\sqrt{x}$ must be rational, therefore $x$ is a perfect square. In the same way, $y$ is a perfect square.\n\nDenote now $\\sqrt{x} = a$, $\\sqrt{y} = b... | Romania | 67th Romanian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | [(1, 4), (4, 1), (4, 4)] | |
02sm | Problem:
Um número inteiro positivo é chamado ziguezague, se satisfaz as seguintes três condições:
- Seus algarismos são não nulos e distintos.
- Não possui três algarismos consecutivos em ordem crescente.
- Não possui três algarismos consecutivos em ordem decrescente.
Por exemplo, $14385$ e $2917$ são ziguezague, mas... | [
"Solution:\n\na) O número $978563412$ é, claramente, um número ziguezague. Vamos mostrar que, se $N$ é um número ziguezague, então\n$$\nN \\leq 978563412\n$$\nA maior quantidade de algarismos que um número ziguezague pode possuir é igual a $9$. Portanto, podemos supor que $N$ é da forma\n$$\nN=\\overline{a b c d e ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) 978563412; b) 1260 | |
0h0g | Points $K$ and $M$ are chosen on the sides $AB$ and $BC$ of triangle $ABC$ in such a way, that $AK = KM = MC$. Let $N$ be the point of intersection of $AM$ and $CK$, $P$ – the feet of perpendicular from point $N$ to the line $KM$, and $Q$ is such point inside segment $KM$, that $MQ = KP$. Prove, that incircle of $KMB$ ... | [
"Let $I$ be incenter of $KMB$. Then, $KI$ and $MI$ are bisectors of $\\angle BKM$ and $\\angle BMK$ respectively. $AKM$ and $CMK$ are isosceles triangles, thus, $\\angle KAM = \\angle KMA = \\angle MKI = \\angle BKI$ and $\\angle MKC = \\angle MCK = \\angle KMI = \\angle BMI$, moreover, these angles are acute (Fig.... | Ukraine | 50th Mathematical Olympiad in Ukraine, Fourth Round (March 24, 2010) | [
"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",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | English | proof only | null | |
0ho8 | Problem:
Determine all integers $n$ for which $n^{2}+15$ is the square of an integer. | [
"Solution:\n\nWe may limit the search to nonnegative integers $n$, since $n^{2}+15 = (-n)^{2}+15$. Suppose there is a nonnegative integer $m$ such that\n$$\n\\begin{aligned}\nn^{2}+15 & = m^{2} \\\\\n15 & = m^{2} - n^{2} \\\\\n& = (m+n)(m-n) .\n\\end{aligned}\n$$\nNote that the factors $m+n$ and $m-n$ are not both ... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | -7, -1, 1, 7 | |
04g2 | Let $n$ be a positive integer and let $S$ be the sum of all integers from $1$ to $n$. Prove that $S + 1$ is not divisible by $3$. (Ilko Brnetić) | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | proof only | null | |
0fbm | Problem:
Calcular el límite
$$
\lim_{n \rightarrow \infty} \frac{1}{n}\left(\frac{1}{n^{k}}+\frac{2^{k}}{n^{k}}+\cdots+\frac{(n-1)^{k}}{n^{k}}+\frac{n^{k}}{n^{k}}\right)
$$
(Para el cálculo del límite se puede seguir el procedimiento de construcción de la integral). | [
"Solution:\n\nPara el cálculo del límite se puede seguir el procedimiento de construcción de la integral.\n$$\n\\lim_{n \\rightarrow \\infty} \\frac{1}{n}\\left(\\frac{1}{n^{k}}+\\frac{2^{k}}{n^{k}}+\\cdots+\\frac{(n-1)^{k}}{n^{k}}+\\frac{n^{k}}{n^{k}}\\right) = \\int_{0}^{1} x^{k} dx = \\frac{1}{k+1}\n$$"
] | Spain | OME 12 | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Limits"
] | null | proof and answer | 1/(k+1) | |
05wh | Problem:
Trouver toutes les triplets d'entiers positifs $(x, y, z)$ satisfaisant l'équation
$$
x! + 2^{y} = z!
$$ | [
"Solution:\nSoit $(x, y, z)$ un triplet solution.\nOn cherche dans un premier temps à réduire le nombre de valeurs que $x$ peut prendre. Pour ce faire, notons que $x! < z!$ donc $x < z$. En particulier, $x!$ divise $z!$ et $x!$, donc $x!$ divise $z! - x! = 2^{y}$. Supposons par l'absurde que $x \\geqslant 3$, dans ... | France | Préparation Olympique Française de Mathématiques - ENVOI 5 : Pot-POURRI | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (0, 0, 2), (1, 0, 2), (2, 2, 3) | |
04gl | In the interior of an acute triangle $ABC$ the point $P$ is chosen, so that
$$
\angle APB = \angle CBA + \angle ACB, \quad \angle BPC = \angle ACB + \angle BAC.
$$
Prove that
$$
\frac{|AC| \cdot |BP|}{|BC|} = \frac{|BC| \cdot |AP|}{|AB|}. \qquad (\text{Belarus 2013})
| [
"Let us denote $\\angle BAC = \\alpha$, $\\angle CBA = \\beta$ and $\\angle ACB = \\gamma$.\n\n\nFrom $\\angle APB = \\beta + \\gamma$ and $\\angle BPC = \\gamma + \\alpha$ we get\n$$\n\\angle CPA = 360^\\circ - \\angle APB - \\angle BPC = 360^\\circ - \\beta - \\gamma - \\alpha = \\alpha +... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0d4r | Aws plays a solitaire game on a fifty-two card deck: whenever two cards of the same color are adjacent, he can remove them. Aws wins the game if he removes all the cards. If Aws starts with the cards in a random order, what is the probability for him to win? | [
"Let us consider the positions $1,2, \\ldots, 52$, from left to right, of the fifty-two cards at the beginning of the game. Whenever Aws removes two adjacent cards, we subtract $2$ from the positions of all the cards on the right of these adjacent removed cards.\n\nWe notice that, at each operation, if a card is no... | Saudi Arabia | SAMC | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English, Arabic | proof and answer | binom(26,13)^2 / binom(52,26) | |
05we | Problem:
Soit $n \geqslant 2$ un entier. Anna a écrit au tableau $n$ entiers $a_{1}, a_{2}, \ldots, a_{n}$ deux à deux distincts. Elle remarque alors que, quelle que soit la manière de sélectionner $n-1$ de ces entiers, leur somme est divisible par $n$.
Démontrer que la somme de l'ensemble des $n$ entiers est divisibl... | [
"Solution:\n\nSoit $s$ la somme de tous les entiers. Sélectionner $n-1$ entiers revient à choisir l'entier, disons $a_{i}$, que l'on n'a pas sélectionné. La somme de nos $n-1$ entiers est alors égale à $s-a_{i}$. Ainsi, Anna a simplement remarqué que $a_{i} \\equiv s \\pmod{n}$.\nCeci étant valable pour tout $i$, o... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0hry | Problem:
Suppose that $S$ is a set of 2001 positive integers, and $n$ different subsets of $S$ are chosen so that their sums are pairwise relatively prime. Find the maximum possible value of $n$. (Here the "sum" of a finite set of numbers means the sum of its elements; the empty set has sum 0.) | [
"Solution:\nThe answer is $2^{2000}+1$. To see that we cannot do better than this, note that at least half of the $2^{2001}$ possible subsets of $S$ have even sums. Indeed, if all elements of $S$ are even, then all subsets have even sums; on the other hand, if there exists some odd $s \\in S$, we can divide the sub... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 2^{2000}+1 | |
0kei | Problem:
A positive integer $N$ is piquant if there exists a positive integer $m$ such that if $n_{i}$ denotes the number of digits in $m^{i}$ (in base 10), then $n_{1}+n_{2}+\cdots+n_{10}=N$. Let $p_{M}$ denote the fraction of the first $M$ positive integers that are piquant. Find $\lim _{M \rightarrow \infty} p_{M}$. | [
"Solution:\nFor notation, let $n_{i}(m)$ denote the number of digits of $m^{i}$ and $N(m)=n_{1}(m)+n_{2}(m)+\\cdots+n_{10}(m)$. Observe that $n_{i}(10 m)=n_{i}(m)+i$ so $N(10 m)=N(m)+55$. We will determine, for $k \\rightarrow \\infty$, how many of the integers from $N\\left(10^{k}\\right)$ to $N\\left(10^{k+1}\\ri... | United States | HMMT February 2020 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 32/55 | |
060x | Problem:
Aline et Elsa jouent au jeu suivant. Elles disposent de 100 pierres qu'elles séparent en deux piles (pas forcément de même taille) au début du jeu. Puis chacune à leur tour, en commençant par Aline, elles effectuent le mouvement suivant : elles choisissent une pile, puis un entier strictement positif inférieu... | [
"Solution:\n\nOn observe sur le cas à une seule pile en commençant par la fin que les solutions perdantes pour Aline sont les piles à $2^{n}-1$ pierres.\n\nRevenons au cas d'une configuration initiale générique $(a, b)$. Par définition, la position $(1,1)$ est perdante. $(a, b)$ est perdante pour la personne qui do... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | (50,50), (67,33), (33,67), (95,5), (5,95); equivalently, those splits with (pile1+1)/(pile2+1) a power of two. | |
047e | Given an isosceles triangle $ABC$ with $AB = AC$. Let a moving point $D$ satisfy $AD \parallel BC$ and $DB > DC$. Let a moving point $E$ be on the arc $\widearc{BC}$ of the circumcircle of $\triangle ABC$ that does not contain $A$, such that $EB < EC$. Let $F$ be a point on the extension of $BC$ such that $\angle DFE =... | [
"\n\n**Proof 1.** Let $AE$ and $DE$ intersect line $BC$ at points $P$ and $Q$, respectively. Since $AB = AC$ and $A$, $B$, $E$, $C$ are concyclic, we have $\\angle ABP = \\angle ACB = \\angle AEB$. Hence, $AP \\cdot AE = AB^2$.\n\nNote that $Q$ might be on the extension of $BC$, but $F$ can... | China | The 65th IMO China National Team Selection Test | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof only | null | |
08nd | Problem:
Do there exist prime numbers $p$ and $q$ such that $p^{2}(p^{3}-1)=q(q+1)$? | [
"Solution:\n\nWrite the given equation in the form\n$$\np^{2}(p-1)\\left(p^{2}+p+1\\right)=q(q+1)\n$$\nFirst observe that it must not be $p=q$, since in this case the left hand side of (9) is greater than its right hand side. Hence, since $p$ and $q$ are distinct primes, (9) immediately yields $p^{2} \\mid q+1$, th... | JBMO | JBMO Shortlist | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | No | |
0055 | Se tiene una cantidad finita de números positivos. Hay que distribuir los números en grupos, de modo que la razón entre dos números de un mismo grupo sea siempre distinta de $2007$.
Determine el mínimo número de grupos para los que esto puede lograrse. | [] | Argentina | XVI Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | 2 | |
02cq | Problem:
Produto de três números - No diagrama abaixo cada círculo representa um algarismo. Preencha o diagrama colocando em cada círculo um dos algarismos de 0 a 9, utilizando cada algarismo uma única vez.
$$
\text{○} \times \text{○○} \times \text{○○○} = \text{○○○○}
$$ | [
"Solution:\n\nSejam $a, b, c, \\ldots$ os números em cada círculo como indicado abaixo.\n\n$$(a) \\times (b)(c) \\times (d)(e)(f) = (g)(h)(i)(j)$$\n\nTemos que $a$, $c$ e $f$ não podem ser zero, pois $0 \\times x = 0$.\n\nMas, o produto dos três números é um número de 4 algarismos, assim, $a b d < 10$ e portanto os... | Brazil | Lista 6 | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | null | |
01dp | A finite sequence $d_{n-1}, d_{n-2}, \dots, d_1, d_0$ of digits is called a *stable final segment of length $n$* if it has the following property: If $m$ is any positive integer such that the last $n$ digits of $m$ are $d_{n-1}d_{n-2}\dots d_1d_0$ (in this order), then for every positive integer $k$ the last $n$ digits... | [
"Let $a = d_{n-1}\\dots d_1 d_0$ (where initial zeros are ignored if there are any). The sequence $d_{n-1}, d_{n-2}, \\dots, d_1, d_0$ is a stable final segment if and only if $a^k - a \\equiv 0 \\pmod{10^n}$ for all integers $k \\ge 2$. This again is equivalent to $a^2 - a \\equiv 0 \\pmod{10^n}$ because $a^2 - a ... | Baltic Way | Baltic Way 2016 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 4 | |
0bwi | Consider the cube $ABCDEFGH$ and the points $M$ – the midpoint of the side $EF$ and $S$ – the center of the face $BCGF$. Prove that the straight lines $AS$ and $BM$ are perpendicular. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof only | null | |
0fyz | Problem:
Finde alle Polynome $P \neq 0$ mit reellen Koeffizienten, welche die folgende Bedingung erfüllen:
$$
P(P(k)) = P(k)^2 \text{ für } k = 0, 1, 2, \ldots, (\operatorname{deg} P)^2
$$ | [
"## Lösung"
] | Switzerland | IMO Selektion | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All such polynomials are P(x) = 1, P(x) = m x with m ≠ 0, and P(x) = x^2. | |
0d1k | Let $a_1, a_2, \dots$ be a sequence of positive integers such that
$$
a_{n+1} = \begin{cases} \frac{a_n}{2} & \text{if } a_n \text{ is even,} \\ 2^r + \frac{a_n+1}{2} & \text{if } a_n \text{ is odd and } 2^{r-1} \le a_n < 2^r. \end{cases}
$$
Prove that no matter what the value of $a_1$ is, there exists an $N$ such that... | [
"Suppose $a_i$, when expressed in binary, is a number formed by appending some number of copies of $10$ at the beginning followed by a generic $k$-digit binary number. We show that continuing the recurrence from $a_i$ will eventually result in an $N$ for which $n > N$ implies $a_n = a_{n+2}$. We do so by strong ind... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null |
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