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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
07gf | Find all finite sets $\{a_1, a_2, \dots, a_k\}$ of integers for which there is an injective function $f : \mathbb{R} \to \mathbb{R}$ such that for all positive integer $n$
$$
\{x \in \mathbb{R} : f^n(x) - x\} = \{n + a_1, n + a_2, \dots, n + a_k\}.
$$
(Here $f^n(x)$ is the function that obtained from composing $f$ for ... | [
"Let us denote the set $\\{a_1, \\dots, a_k\\}$ by $A$ and $\\{n+a_1, \\dots, n+a_k\\}$ by $A+n$. Assume that the statement of the problem is true for the set $A$.\n**Lemma.** All the elements of $A$ are non-negative.\n\n*Proof.* Assume that there is a positive integer $m$ such that $-m \\in A$. This means that the... | Iran | 38th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | {0} | |
0evn | Let $n$ be a positive integer. There are $n$ boxes $B_1, B_2, \dots, B_n$ each of which contains some balls. One can perform the following moves.
Choose positive integers $i$ and $j$ with $1 \le i \le j \le n$, and add exactly one ball to each of the boxes $B_i, B_{i+1}, \dots, B_j$.
For positive integers $x_1, x_2, \d... | [
"The answer is $\\left\\lceil \\frac{2n+2}{3} \\right\\rceil$.\n\nFor $i = 0, 1, \\dots, n$, let $y_i = x_i - x_{i+1}$ where $x_0 = x_{n+1} = 0$. Note that $y_0 + y_1 + \\dots + y_n = 0$. The move in the problem is equivalent to the following.\nChoose non-negative integers $i$ and $j$ with $0 \\le i < j \\le n$, an... | South Korea | The 36th KOREAN MATHEMATICAL OLYMPIAD Final Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | ceil((2n+2)/3) | |
0jsh | Problem:
The very hungry caterpillar lives on the number line. For each non-zero integer $i$, a fruit sits on the point with coordinate $i$. The caterpillar moves back and forth; whenever he reaches a point with food, he eats the food, increasing his weight by one pound, and turns around. The caterpillar moves at a sp... | [
"Solution:\n\nOn the $n$th straight path, the caterpillar travels $n$ units before hitting food and his weight is $n-1$. Then his speed is $2^{1-n}$. Then right before he turns around for the $n$th time, he has traveled a total time of $\\sum_{i=1}^{n} \\frac{i}{2^{1-i}}=\\frac{1}{2} \\sum_{i=1}^{n} i \\cdot 2^{i}$... | United States | HMMT November | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 9217 | |
0fr4 | Problem:
Sea $ABC$ un triángulo acutángulo. Sea $D$ el pie de la altura correspondiente al lado $BC$; $M$ el punto medio del lado $BC$ y $F$ el punto de corte de la bisectriz interior del ángulo $\angle BAC$ con el lado $BC$. Determinar todos los triángulos para los cuales $F$ es el punto medio del segmento $DM$. | [
"Solution:\n\nLa condición $DF = FM$ se puede reescribir como\n$$\n2BF = BM + BD\n$$\nSi llamamos $a$, $b$ y $c$ a las longitudes de los lados $BC$, $CA$ y $AB$, respectivamente, y aplicamos el teorema de la bisectriz tendremos que\n$$\n\\frac{BF}{c} = \\frac{FC}{b} = \\frac{a}{b+c}\n$$\nPor consiguiente,\n$$\nBF =... | Spain | FASE LOCAL DE LA OLIMPIADA MATEMÁTICA ESPAÑOLA. | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof and answer | Either b = c with a < sqrt(2)·b, or a = (b + c)/sqrt(2) with c/3 < b < c. | |
0fa8 | Problem:
$n$ numbers are written on a blackboard. Someone then repeatedly erases two numbers and writes half their arithmetic mean instead, until only a single number remains. If all the original numbers were $1$, show that the final number is not less than $1/n$. | [
"Solution:\n\nPut $c = (a + b)/4$. We have $1/c = 4/(a + b) \\leq 1/a + 1/b$, so each move does not increase the sum of the reciprocals of the numbers. If the final number is $k$, then the final sum of reciprocals is $1/k$. The initial sum is $n$, so $1/k \\leq n$, or $k \\geq 1/n$."
] | Soviet Union | 25th ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0efb | Problem:
Za kateri $x$ bo imel izraz $\frac{x^{3}-2 x^{2}-9 x+18}{x^{2}-9}$ vrednost 0?
(A) 9
(B) 3
(C) -3
(D) 0
(E) 2 | [
"Solution:\n$\\frac{x^{3}-2 x^{2}-9 x+18}{x^{2}-9}=\\frac{x^{2}(x-2)-9(x-2)}{x^{2}-9}=\\frac{(x-2)\\left(x^{2}-9\\right)}{x^{2}-9}=x-2$\n\nin vrednost je enaka 0 za $x=2$. Pravilen odgovor je (E)."
] | Slovenia | 17. tekmovanje dijakov srednjih tehniških in strokovnih šol v znanju matematike | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | E | |
0af5 | Од средините на страните на остроаголниот триаголник **ДАВС** се повлечени нормали кон соседните страни. Докажи дека шестоаголникот што тие го формираат има два пати помала плоштина од плоштината на **ДАВС**.
 | [
"Нека $O$ е центарот на опишаната кружница околу триаголникот $\\Delta ABC$. Нека $A_1, B_1, C_1$ се средините на страните $BC, CA, AB$ соодветно. Точката $O$ ја поврзуваме со $A_1, B_1, C_1$ и добиваме три паралелограми $C_1OB_1K, A_1OC_1L$ и $B_1OA_1M$, чијашто вкупна плоштина е еднаква на плоштината на шестоагол... | North Macedonia | Републички натпревар по математика за основно образование | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | Macedonian, English | proof only | null | |
07t2 | Two circles, centres $O_1$ and $O_2$, intersect at $A$ and $B$. Let $O_1C_1$ and $O_2C_2$ be parallel radii of these circles such that $C_1$ and $C_2$ are on the same side of $O_1O_2$. Prove that the circumcircles of triangles $C_1AC_2$ and $C_1BC_2$ have the same radius. | [
"Let $D_1$ and $D_2$ be points on the circles such that $C_1D_1$ and $C_2D_2$ are diameters. We want to show that the reflection in the line $C_1C_2$ of the circumcircle of $\\triangle C_1BC_2$ is the circumcircle of $\\triangle C_1AC_2$. Equivalently we can show that the reflection of $B$ in the line $C_1C_2$ is o... | Ireland | IRL_ABooklet_2020 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02yg | Problem:
Os inteiros positivos $x$ e $y$ são tais que o número $x^{2019}+x+y^{2}$ é divisível por $x y$.
a) Dê um exemplo de tais inteiros $x$ e $y$, com $x>y$.
b) Verifique que, necessariamente, $x$ é um quadrado perfeito. | [
"Solution:\na) Basta escolher $x=4$ e $y=2$, pois $4^{2019}+4+2^{2}=8 \\cdot (2^{4035}+1)$ é divisível por $4 \\cdot 2=8$.\n\nb) Seja $d=mdc(x, y)$. Assim, $x=d m$ e $y=d n$, com $mdc(m, n)=1$. Daí,\n$$\n\\frac{x^{2019}+x+y^{2}}{x y}=\\frac{d^{2018} m^{2019}+m+d n^{2}}{d m n}\n$$\né um inteiro. Como $d$ divide $d^{... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | (x, y) = (4, 2); necessarily, x is a perfect square. | |
0dd9 | Suppose that $a, b, c, d$ are positive real numbers satisfying $(a+c)(b+d) = ac+bd$. Find the smallest possible value of
$$
\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}
$$ | [
"First of all, apply the AM-GM inequality as follows:\n$$\n\\left(\\frac{a}{b} + \\frac{c}{d}\\right) + \\left(\\frac{b}{c} + \\frac{d}{a}\\right) \\ge 2\\sqrt{\\frac{ac}{bd}} + 2\\sqrt{\\frac{bd}{ac}} = \\frac{2(ac + bd)}{\\sqrt{abcd}}\n$$\nContinuing to apply the AM-GM inequality, then\n$$\n\\left(\\frac{a}{b} + ... | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 8 | |
0kwp | Problem:
In the following figure—not drawn to scale!—$E$ is the midpoint of $BC$, triangle $FEC$ has area $7$, and quadrilateral $DBEG$ has area $27$. Triangles $ADG$ and $GEF$ have the same area, $x$. Find $x$.
 | [
"Solution:\n\nThe answer is $x=8$.\n\nUse the notation $[\\cdot]$ to denote the area of a polygon. Draw $GB$; notice that triangles $GBE$ and $GEC$ have equal bases and altitudes, so $[GBE]=[GEC]=x+7$. Since $[ABE]=27+x$, we have $[GDB]=20-x$.\n\nLikewise, if we draw $AC$, we see that $[ABE]=[AEC]=27+x$, so $[AGC]=... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 8 | |
0gfb | 設 $n$ 為正整數。試決定所有函數 $f : \mathbb{R} \to \mathbb{R}$，滿足
$$
(f(x))^n f(x + y) = (f(x))^{n+1} + x^n f(y)
$$
對所有的實數 $x, y$ 均成立。 | [] | Taiwan | 2022 數學奧林匹亞競賽第一階段培訓營, 國際競賽實作(一) | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | Chinese; English | proof and answer | All solutions are: (1) f(x) = 0 for all real x; (2) f(x) = x for all real x; and additionally (3) f(x) = -x for all real x when n is even. | |
0f7m | Problem:
Some lattice points in the plane are marked. $S$ is a set of non-zero vectors. If you take any one of the marked points $P$ and place each vector in $S$ with its beginning at $P$, then more vectors will have their ends on marked points than not. Show that there are an infinite number of points. | [] | Soviet Union | 21st ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Prealgebra / Basic Algebra > Other",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
00gj | Let $S$ be a set of 9 distinct integers all of whose prime factors are at most $3$. Prove that $S$ contains 3 distinct integers such that their product is a perfect cube. | [
"Without loss of generality, we may assume that $S$ contains only positive integers. Let\n$$\nS = \\{2^{a_i} 3^{b_i} \\mid a_i, b_i \\in \\mathbb{Z},\\ a_i, b_i \\geq 0,\\ 1 \\leq i \\leq 9\\}.\n$$\nIt suffices to show that there are $1 \\leq i_1, i_2, i_3 \\leq 9$ such that\n$$\n\\begin{equation*}\na_{i_1} + a_{i_... | Asia Pacific Mathematics Olympiad (APMO) | XIX Asian Pacific Mathematics Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0bwy | Determine all positive integers $k$ and $n$ for which $2^k + 10n^2 + n^4$ is the square of an integer. | [
"The required integers are $k = 3$ and $n = 2$. It is readily checked that these integers satisfy the condition in the statement.\n\nTo show that there are no other such, write $2^k + 10n^2 + n^4 = N^2$, where $N$ is a positive integer, so $(N - n^2 - 5)(N + n^2 + 5) = 2^k - 25$.\n\nThe latter shows that $N \\ge n^... | Romania | THE Tenth STARS OF MATHEMATICS COMPETITION | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | k = 3, n = 2 | |
07gu | Natural numbers are placed in the cells of an infinite grid, such that the number in each cell is equal to the number of its adjacent cells having the same number (two cells of the grid are adjacent if they have a common vertex). Find the maximum value of distinct numbers this infinite grid can have. | [
"The answer is $5$. For the example consider the following table (see the cover photo of the booklet).\n$$\na_{ij} = \\begin{cases} 2 & i \\equiv 1 \\pmod{8} \\\\ 5 & i \\equiv 4, 5 \\pmod{8} \\\\ 4 & i \\equiv 0 \\pmod{8} \\text{ or } i \\equiv 7 \\pmod{8} \\text{ and } j \\not\\equiv 2, 5 \\pmod{6} \\\\ 1 & i \\e... | Iran | 38th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 5 | |
065l | If the nonnegative real numbers $x$, $y$ and $z$ have sum $2$, prove that:
$$
x^2y^2 + y^2z^2 + z^2x^2 + xyz \le 1.
$$
For which values of $x$, $y$ and $z$ the equality is valid? | [
"We will use the well-known inequality $2\\alpha\\beta \\le \\alpha^2 + \\beta^2$, which is valid for all $\\alpha, \\beta \\in \\mathbb{R}$. The equality holds for $\\alpha = \\beta$. Thus we have\n\n$$\n\\begin{align*}\nx^2 y^2 + y^2 z^2 + z^2 x^2 + xyz &= \\frac{1}{2}(2x^2 y^2 + 2y^2 z^2 + 2z^2 x^2 + 2xyz) \\\\\... | Greece | 26th Hellenic Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | (1, 1, 0) or (1, 0, 1) or (0, 1, 1) | |
05g3 | Problem:
On pose 23 allumettes sur une table et 2 joueurs jouent à un jeu : chacun, à son tour, retire entre 1 et 4 allumettes (inclus). Celui qui prend la dernière gagne. Existe-t-il une stratégie gagnante pour l'un des deux?
Une stratégie gagnante est une manière de jouer qui permet à l'un des deux de gagner peu im... | [
"Solution:\n\nLe $1^{\text{er}}$ joueur possède une stratégie gagnante. La voici :\n\n1. $J_1$ prend 3 allumettes : il en reste un multiple de 5.\n\n2. $J_2$ en retire $r \\in \\{1,2,3,4\\}$.\n\n3. $J_1$ en enlève $5 - r$ : il en reste un multiple de 5.\n\nEt ainsi de suite. Donc à chaque fois que $J_2$ joue, il y ... | France | ENVOi 3 : ARITHMÉTIQUE | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Yes — the first player. Take three first, then always take enough matches so that your move plus the opponent’s previous move totals five. | |
0fwq | Problem:
Bestimme alle Funktionen $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$, sodass für alle $x, y>0$ gilt:
$$
f(x y) \leq \frac{x f(y)+y f(x)}{2}
$$ | [
"Solution:\n\nSei $a=f(1)>0$. Setze $y=1$, dann folgt $f(x) \\leq \\frac{1}{2}(x a+f(x))$, also $f(x) \\leq a x$.\n\nSetze nun $y=\\frac{1}{x}$, unter Verwendung der eben gezeigten Abschätzung folgt weiter\n$$\na \\leq \\frac{1}{2}\\left(x f\\left(\\frac{1}{x}\\right)+\\frac{1}{x} f(x)\\right) \\leq \\frac{1}{2}\\l... | Switzerland | SMO Finalrunde | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All functions of the form f(x) = a x for x positive, where a is any positive constant. | |
03ro | Assume that positive numbers $a$, $b$, $c$, $x$, $y$, $z$ satisfy $cy + bz = a$; $az + cx = b$ and $bx + ay = c$. Find the minimum value of the function $f(x, y, z) = \frac{x^2}{1+x} + \frac{y^2}{1+y} + \frac{z^2}{1+z}$. | [
"By assumption, $b(az + cx - b) + c(bx + ay - c) - a(cy + bz - a) = 0$, i.e. $2bcx + a^2 - b^2 - c^2 = 0$, we get $x = \\frac{b^2 + c^2 - a^2}{2bc}$. For the similar reason, $y = \\frac{a^2 + c^2 - b^2}{2ac}$ and $z = \\frac{a^2 + b^2 - c^2}{2ab}$.\n\nSince $a$, $b$, $c$, $x$, $y$, $z$ are positive, by the above th... | China | China Mathematical Competition (Extra Test) | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 1/2 | |
01cs | Prove that
$$
\sqrt{\frac{1}{3x+1}} + \sqrt{\frac{x}{x+3}} \ge 1
$$
holds for all $x > 0$. For which values of $x$ is there an equality? | [
"As both sides of the inequality are positive by definition, one may equivalently replace both sides by their squares. The left side minus the right side then becomes\n$$\n\\begin{align*}\n& 0 \\le \\frac{1}{3x+1} + \\frac{x}{x+3} + 2\\sqrt{\\frac{x}{(3x+1)(x+3)}} - 1 \\\\\n&= \\frac{-8x}{(3x+1)(x+3)} + 2\\sqrt{\\f... | Baltic Way | Baltic Way 2016 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x = 1 | |
0ia9 | Problem:
Let $a$, $b$, $c$ be the three roots of $p(x) = x^{3} + x^{2} - 333 x - 1001$. Find $a^{3} + b^{3} + c^{3}$. | [
"Solution:\n\nWe know that $x^{3} + x^{2} - 333 x - 1001 = (x - a)(x - b)(x - c) = x^{3} - (a + b + c) x^{2} + (ab + bc + ca) x - abc$.\n\nAlso, $(a + b + c)^{3} - 3(a + b + c)(ab + bc + ca) + 3abc = a^{3} + b^{3} + c^{3}$.\n\nThus,\n$$\na^{3} + b^{3} + c^{3} = (-1)^{3} - 3(-1)(-333) + 3 \\cdot 1001 = 2003.\n$$"
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 2003 | |
0f8z | Problem:
Find the least possible value of $(x + y)(y + z)$ for positive reals satisfying $(x + y + z)xyz = 1$. | [] | Soviet Union | 23rd ASU | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 2 | |
07i1 | Prove that for each natural number $n$ one can find complex numbers $\omega_1, \omega_2, \dots, \omega_n$ on the unit circle such that
$$
|\sum \omega_j| = |\sum \omega_j^2| = n - 1.
$$ | [
"Change the ordering of $\\omega_i$ so that they are indexed by their arguments.\n\n**Lemma 1.** After a rotation, one can assume that all $\\omega_i$ are lying on the upper half plane and $\\omega_1 = 1$.\nWe multiply the $\\omega_i$ by $\\text{cis}(\\theta)$ so that $\\sum \\omega_i$ becomes imaginary. So we have... | Iran | 40th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Intermediate Alg... | null | proof only | null | |
08o6 | Problem:
For any real number $a$, let $\lfloor a\rfloor$ denote the greatest integer not exceeding $a$. In positive real numbers solve the following equation
$$
n+\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor=2014
$$ | [
"Solution:\nObviously $n$ must be a positive integer. Now note that $44^2=1936<2014<2025=45^2$ and $12^3<1900<2014<13^3$.\nIf $n<1950$ then $2014=n+\\lfloor\\sqrt{n}\\rfloor+\\lfloor\\sqrt[3]{n}\\rfloor<1950+44+12=2006$, a contradiction!\nSo $n \\geq 1950$. Also if $n>2000$ then $2014=n+\\lfloor\\sqrt{n}\\rfloor+\\... | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1958 | |
0jzp | Problem:
Let $ABCD$ be a convex quadrilateral with $AC = 7$ and $BD = 17$. Let $M, P, N, Q$ be the midpoints of sides $AB$, $BC$, $CD$, $DA$ respectively. Compute $MN^{2} + PQ^{2}$. | [
"Solution:\n\n$MPNQ$ is a parallelogram whose side lengths are $3.5$ and $8.5$ so the sum of squares of its diagonals is $\\frac{7^{2} + 17^{2}}{2} = 169$"
] | United States | February 2017 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof and answer | 169 | |
0es5 | Andile and Zandre play a game on a $2017 \times 2017$ board. At the beginning, Andile declares some of the squares *forbidden*, meaning that nothing may be placed on such a square. After that, they take turns to place coins on the board, with Zandre placing the first coin. It is not allowed to place a coin on a forbidd... | [
"The minimum number is $2017$. For example, Andile can achieve a win by declaring all squares of the last row forbidden, so that $2016$ rows remain. After that, there will be exactly $2016$ moves possible, no matter how the two play, since placing a coin always eliminates exactly one row and one column from further... | South Africa | The South African Mathematical Olympiad Third Round | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 2017 | |
07w9 | A positive integer is *totally square* if the sum of its digits (written in base 10) is a square number. For example, $13$ is totally square because $1 + 3 = 2^2$, but $16$ is not totally square. Show that there are infinitely many positive integers that are not the sum of two totally square integers. | [
"The squares modulo $9$ are $\\{0, 1, 4, 7\\}$. Any integer is congruent modulo $9$ to the sum of its digits (as powers of $10$ are congruent to $1$ modulo $9$). Therefore any totally square integer is congruent modulo $9$ to one of $\\{0, 1, 4, 7\\}$. Taking all combinations of pairs, the sum of two totally square... | Ireland | IRL_ABooklet_2023 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0h60 | There is a white square $8 \times 8$. In one move Dmitry can choose a totally white square $2 \times 2$ and paint in black color any two cells of this square, located on the diagonal. What is the maximum number of cells according to the following rules Dmitry will paint?
Fig. 31 | [
"**Answer:** 42.\n\nFirst show how to achieve the required number of colorings. In each box $4 \\times 4$ will painting in such sequence. First select all four squares $2 \\times 2$, on which the $4 \\times 4$ square is split and paint diagonals, as shown in Fig. 31 (black squares). After this there is a completely... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 42 | |
0iop | Problem:
Let $S$ denote the set of all triples $(i, j, k)$ of positive integers where $i + j + k = 17$. Compute
$$
\sum_{(i, j, k) \in S} i j k.
$$ | [
"Solution:\nAnswer: $11628 = \\binom{19}{5}$. We view choosing five objects from a row of $19$ objects in an unusual way. First, remove two of the chosen objects, the second and fourth, which are not adjacent nor at either end, forming three nonempty groups of consecutive objects. We then have $i, j$, and $k$ choic... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof and answer | 11628 | |
0e3n | Problem:
Oskrbnik planinske koče je več let spremljal, koliko časa porabijo planinci za pot od vznožja v dolini do planinske koče tik pod vrhom gore. Pohodniki so sami zapisovali porabljen čas, oskrbnik pa je podatke zbral in uredil frekvenčno tabelo, kjer je zapisal relativne frekvence.
| porabljen čas v minutah | r... | [
"Solution:\n\nVsota relativnih frekvenc mora biti $100\\%$. Tako izračunamo $f_{6} = 6,5$.\n\nZa grupirane podatke se aritmetična sredina izračuna z obrazcem\n$$\n\\bar{x} = \\frac{f_{1} \\cdot x_{1} + f_{2} \\cdot x_{2} + f_{3} \\cdot x_{3} + \\ldots + f_{k} \\cdot x_{k}}{N}\n$$\nVrednosti $x$ so sredine razredov ... | Slovenia | 10. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Statistics > Mathematical Statistics"
] | null | final answer only | 6.5% and 131.4 minutes | |
038f | Problem:
For any positive integer $n$ denote by $f(n)$ the smallest positive integer $m$ such that the sum $1+2+\cdots+m$ is divisible by $n$. Find all $n$ such that $f(n)=n-1$. | [
"Solution:\nIf $f(n)=n-1$, then $n$ divides the sum $1+2+\\cdots+(n-1)=\\frac{(n-1)n}{2}$, which implies that $n$ is odd.\n\nThe numbers $n=p^{s}$, where $p>2$ is a prime number and $s \\geq 1$, are solutions. Indeed, if $k \\in \\mathbb{N}$ and $k<p^{s}-1$, then the sum $1+2+\\cdots+k=\\frac{k(k+1)}{2}$ is not div... | Bulgaria | Spring Mathematical Competition | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All n of the form p^s where p is an odd prime and s is a positive integer. | |
0im1 | Problem:
Convex quadrilateral $ABCD$ has right angles $\angle A$ and $\angle C$ and is such that $AB = BC$ and $AD = CD$. The diagonals $AC$ and $BD$ intersect at point $M$. Points $P$ and $Q$ lie on the circumcircle of triangle $AMB$ and segment $CD$, respectively, such that points $P$, $M$, and $Q$ are collinear. Su... | [
"Solution:\n\nAnswer: $36$. Note that $m \\angle QPB = m \\angle MPB = m \\angle MAB = m \\angle CAB = \\angle BCA = \\angle CDB$. Thus, $MP \\cdot MQ = MB \\cdot MD$. On the other hand, segment $CM$ is an altitude of right triangle $BCD$, so $MB \\cdot MD = MC^{2} = 36$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 36 | |
0kv4 | Problem:
Let $ABCD$ be a square of side length $10$. Point $E$ is on ray $\overrightarrow{AB}$ such that $AE = 17$, and point $F$ is on ray $\overrightarrow{AD}$ such that $AF = 14$. The line through $B$ parallel to $CE$ and the line through $D$ parallel to $CF$ meet at $P$. Compute the area of quadrilateral $AEPF$. | [
"Solution:\n\n\n\nFrom $BP \\parallel CE$, we get that $[BPE] = [BPC]$. From $DP \\parallel CF$, we get that $[DPF] = [DPC]$. Thus,\n$$\n\\begin{aligned}\n[AEPF] & = [BACP] + [BPE] + [DPF] \\\\\n& = [BACP] + [BPC] + [DPC] \\\\\n& = [ABCD] \\\\\n& = 10^2 = 100.\n\\end{aligned}\n$$"
] | United States | HMMT November 2023 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 100 | |
05zu | Problem:
Soit $n \geqslant 3$ un entier. On colorie $2n$ sommets d'un $4n+1$-gone régulier. Montrer qu'il existe trois sommets coloriés qui forment un triangle isocèle. | [
"Solution:\n\nOn procède par l'absurde en supposant qu'aucun triplet de sommets coloriés ne forme de triangle isocèle.\n\nDans un premier temps, fixons un sommet colorié $O$ et notons $OP_{1}, \\ldots, P_{2n}, P_{-2n}, \\ldots, P_{-1}$ le reste du polygone. Puisque le triangle $OP_{i}P_{-i}$ est isocèle pour $1 \\l... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0drg | Let $n \ge 3$ be an integer. Prove that there exist positive integers $\ge 2$, $a_1, a_2, \dots, a_n$, such that $a_1 a_2 \cdots \hat{a}_i \cdots a_n \equiv 1 \pmod{a_i}$, for $i = 1, \dots, n$. Here $\hat{a}_i$ means the term $a_i$ is omitted. | [
"Let $a_1 = 2$, $a_2 = 3$. For $i = 3, \\dots, n-1$, let $a_i = a_1 a_2 \\cdots a_{i-1} + 1$. Let $a_n = a_1 a_2 \\cdots a_{n-1} - 1$. Clearly, $a_1 a_2 \\cdots a_{n-1} \\equiv 1 \\pmod{a_n}$. Also $a_{i+1} \\equiv a_{i+2} \\equiv \\cdots \\equiv a_{n-1} \\equiv 1 \\pmod{a_i}$. For $i = 1, \\dots, n-1$, we have\n$$... | Singapore | Singapur 2015 | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0jiu | Problem:
Fix a positive real number $c>1$ and positive integer $n$. Initially, a blackboard contains the numbers $1, c, \ldots, c^{n-1}$. Every minute, Bob chooses two numbers $a, b$ on the board and replaces them with $c a + c^{2} b$. Prove that after $n-1$ minutes, the blackboard contains a single number no less tha... | [
"Solution:\n\nBy a simple reverse induction, we can show that at any instant, any number on the board takes the form $\\sum_{\\alpha \\in A} c^{r_{\\alpha}} c^{\\alpha}$ for a certain $A \\subseteq \\{0, \\ldots, n-1\\}$ for non-negative integer weights $r_{\\alpha}$ satisfying $\\sum_{\\alpha \\in A} \\phi^{-r_{\\... | United States | HMMT 2014 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0876 | Problem:
Un quadrato $ABCD$ di lato $1$ è inscritto in una circonferenza $\gamma$. Si costruiscano i simmetrici degli archi $\overparen{AB}$, $\overparen{BC}$, $\overparen{CD}$, $\overparen{DA}$ di $\gamma$ rispetto ai lati $AB$, $BC$, $CD$, $DA$ rispettivamente. Indichiamo con $L, M, N, O$ i punti medi degli archi co... | [
"Solution:\n\nLa risposta è (E). Il raggio della circonferenza circoscritta al quadrato è $\\frac{\\sqrt{2}}{2}$, essendo metà della diagonale del quadrato stesso. Anche $LMNO$ è un quadrato, essendo la figura simmetrica per rotazioni di $90$ gradi. Ora, i vertici del quadrato unitario dividono la circonferenza in ... | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | E | |
0j19 | Problem:
Call a positive integer in base $10$ $k$-good if we can split it into two integers $y$ and $z$, such that $y$ is all digits on the left and $z$ is all digits on the right, and such that $y = k \cdot z$. For example, $2010$ is $2$-good because we can split it into $20$ and $10$ and $20 = 2 \cdot 10$. $20010$ is... | [
"Solution:\nWe wish to find integers $a$, $z$ such that $48z \\cdot 10^{a} + z = z\\left(48 \\cdot 10^{a} + 1\\right)$ is a perfect square, where $z < 10^{a}$. This would prove that there exists a $48$-good perfect square because we are pulling off the last $a$ digits of the number and get two integers $48z$ and $z... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null | |
0cct | Let $ABCDEF$ be a regular hexagon of side length $2$. Through the vertices and the sides' midpoints we construct parallels to the sides, which divide the hexagon into $24$ congruent equilateral triangles, whose vertices are called *nodes*. A *sheet* is any (non-degenerate) equilateral triangle whose vertices are nodes.... | [
"a) Consider the regular hexagon $ABCDEF$, centered at $O$, and let $T$, $U$, $V$, $X$, $Y$, $Z$ be the midpoints of the sides $AB$, $BC$, $CD$, $DE$, $EF$, $FA$, respectively.\nNotice that the nodes situated on the sides of the hexagon cannot have trios, so a node that admits a trio is either $O$, or situated at d... | Romania | THE 73rd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD - SECOND SELECTION TEST | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | a) 11√3/4. b) The center node has trios that cover the whole hexagon; a node at distance one from the center (e.g., R) does not. c) 96. | |
03en | Find all natural numbers $k$ for which there exist natural numbers $x, y$ such that the number $\frac{x^k y}{y^2 - x^2}$ is prime.
(Konstantin Delchev, Stanislav Harizanov) | [
"Let $d = (x, y)$ be the greatest common divisor of the numbers $x$ and $y$. Then, $x = d x_1$, $y = d y_1$, where $x_1, y_1$ are coprime naturals. The expression in the condition can be rewritten as\n$$\n\\frac{d^{k-1}}{y_1^2 - x_1^2} x_1^k y_1 = p,\n$$\nwhere we want $p$ to be prime. But $(x_1^k, y_1^2 - x_1^2) =... | Bulgaria | Bulgarian Winter Tournament | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 2, 4 | |
0229 | Problem:
Quatro números inteiros - Se quatro inteiros positivos distintos $m, n, p$ e $q$ satisfazem a equação
$$
(7-m)(7-n)(7-p)(7-q)=4
$$
então a soma $m+n+p+q$ é igual a:
(a) 10
(b) 21
(c) 24
(d) 26
(e) 28 | [
"Solution:\n\nComo $m, n, p$ e $q$ são inteiros, então $7-m$, $7-n$, $7-p$ e $7-q$ também são inteiros. Agora,\n$$\n4 = (-1) \\times (-2) \\times 1 \\times 2\n$$\né a única decomposição de $4$ em um produto de números inteiros distintos. Segue que\n$$\n(7-m)+(7-n)+(7-p)+(7-q)=(-1)+(-2)+1+2\n$$\ne daí obtemos $m+n+p... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | (e) | |
0h6l | In every cell of the table $5 \times 9$ either $0$ or $1$ is written. Then the sums of the numbers for every row and column are calculated. What is the maximum amount of different values that can be among these $14$ numbers?
 | [
"The sums vary from $0$ to $9$, thus there exist $10$ different values for the sum. Let us show that all $10$ values cannot be present simultaneously. Let us suppose that there is a row or a column with the sum $0$, thus all the cells of that row or column are zeroes. If it is a column, then the maximum value for t... | Ukraine | UkraineMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 9 | |
0bjw | Let $A = \{1, 3, 3^2, 3^3, \dots, 3^{2014}\}$. We obtain a *partition* of $A$ if $A$ is written as a disjoint union of nonempty subsets.
a) Prove that there is no partition of $A$ such that the product of elements in each subset is a square.
b) Prove that there exists a partition of $A$ such that the sum of elements ... | [
"a) Assume that such a partition exists. Then the product of all elements of $A$ must be a square as well. But this equals $3^{1+2+3+\\dots+2014} = 3^{2015 \\cdot 1007}$, obviously not a square.\n\nb) Observe that $3^{2n} + 3^{2n+1} = (3^n \\cdot 2)^2$, hence a possible partition is\n$$\nA = \\{1,3\\} \\cup \\{3^2,... | Romania | 65th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0b0h | Problem:
A musical performer has three different outfits. In how many ways can she dress up for seven different performances such that each outfit is worn at least once? (Assume that outfits can be washed and dried between performances.) | [
"Solution:\n\nLet the three outfits be $A$, $B$, and $C$. For each performance, the performer can choose any of the three outfits, so there are $3^7$ total ways to assign outfits to the seven performances.\n\nHowever, we require that each outfit is worn at least once. We use the principle of Inclusion-Exclusion.\n\... | Philippines | Philippines Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 1806 | |
0gcy | 考慮數列 $a_0, a_1, a_2, \dots$, 其中 $a_n = 2^n + 2^{\lfloor n/2 \rfloor}$。證明數列中有無窮多項可寫成兩個以上不同項的和。也有無窮多項不能寫成這樣的和。 | [
"Call a nonnegative integer *representable* if it equals the sum of several (possibly 0 or 1) distinct terms of the sequence. We say that two nonnegative integers $b$ and $c$ are *equivalent* (written as $b \\sim c$) if they are either both representable or both non-representable.\n\nOne can easily compute (or chec... | Taiwan | 二〇一九數學奧林匹亞競賽第二階段選訓營, 獨立研究 (三) | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0fqf | Problem:
Un conjunto de números enteros $T$ es orensano si existen enteros $a < b < c$ tales que $a$ y $c$ pertenecen a $T$ y $b$ no pertenece a $T$. Hallar el número de subconjuntos $T$ de $\{1,2, \ldots, 2019\}$ que son orensanos. | [
"Solution:\n\nEl número de subconjuntos de $\\{1,2, \\cdots, 2019\\}$ es $2^{2019}$ como es bien conocido. Contemos ahora el número de estos conjuntos que NO tienen la propiedad pedida. Claramente, el conjunto vacío, y los subconjuntos de $\\{1,2, \\ldots, 2019\\}$ con un único elemento no tienen la propiedad pedid... | Spain | LV Olimpiada matemática Española | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 2^2019 - 2039191 | |
0cel | Let $ABC$ be an acute triangle, and let $D$, $E$, $F$ be the feet of its altitudes from $A$, $B$, $C$, respectively. The lines $AB$ and $DE$ cross at $K$ and the lines $AC$ and $DF$ cross at $L$. Let $M$ be the midpoint of the side $BC$ and let the line $AM$ cross the circle $ABC$ again at $N$. Finally, the parallel th... | [
"\nWe will show that $PM$ and $PN$ are the tangents from $P$ to $\\omega$; the conclusion then follows at once.\n\nWe first prove that $PM$ is the tangent of $\\omega$ at $M$. The tangent of $\\omega$ at $M$ is parallel to the tangent of $\\Gamma$ at $A$, which is in turn parallel to $EF$. ... | Romania | Nineteenth IMAR Mathematical Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
... | English | proof only | null | |
0c2w | Determine the positive integers $n \ge 3$ such that, for every integer $m \ge 0$, there exist integers $a_1, a_2, \dots, a_n$ such that $a_1 + a_2 + \dots + a_n = 0$ and $a_1a_2 + a_2a_3 + \dots + a_{n-1}a_n + a_na_1 = -m$. | [
"Any $n \\ge 5$ has the desired property: one can choose $a_1 = 1 - m$, $a_2 = a_3 = \\dots = a_{n-3} = 0$, $a_{n-2} = -1$, $a_{n-1} = m$, $a_n = 0$.\n\nNumbers $n = 3$ and $n = 4$ do not have the property.\n\nFor $n = 3$, $-2m = 2a_1a_2 + 2a_2a_3 + 2a_3a_1 = (a_1 + a_2 + a_3)^2 - a_1^2 - a_2^2 - a_3^2$ comes to $a... | Romania | 69th NMO Selection Tests for JBMO | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | all integers n with n >= 5 | |
00g9 | Determine all finite nonempty sets $S$ of positive integers satisfying
$$
\frac{i+j}{(i, j)} \quad \text{is an element of } S \text{ for all } i, j \text{ in } S,
$$
where $(i, j)$ is the greatest common divisor of $i$ and $j$. | [
"Let $k \\in S$. Then $\\frac{k+k}{(k, k)}=2$ is in $S$ as well.\n\nSuppose for the sake of contradiction that there is an odd number in $S$, and let $k$ be the largest such odd number. Since $(k, 2)=1$, $\\frac{k+2}{(k, 2)}=k+2>k$ is in $S$ as well, a contradiction. Hence $S$ has no odd numbers.\n\nNow suppose tha... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | {2} | |
01sd | A sequence is called *N-nice* if it consists of a finite number of the consecutive positive integers and the sum of all its terms is equal to $N$.
Find the total number of all $4030^k$-nice sequences, where $k$ is a positive integer. (*N*-nice sequence may consist of a single term.) | [
"Let $P(N)$ be the number of $N$-nice sequences, $D(N)$ be the number of odd divisors of $N$. Then $P(N) = D(N)$ (see the solution of Problem C.7). Since $D(4030^k) = D(2^k \\cdot 5^k \\cdot 13^k \\cdot 31^k) = (k+1)^3$, we see that the number of the required $4030^k$-nice sequences is equal to $(k+1)^3$."
] | Belarus | FINAL ROUND | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | (k+1)^3 | |
0efl | Problem:
V kvadrat s stranico dolžine $a$ je včrtana krožnica s središčem v $O$. Štiri manjše krožnice s središči $O_{1}$, $O_{2}$, $O_{3}$ in $O_{4}$ se dotikajo večje krožnice in po dveh stranic kvadrata (glej sliko). Koliko je ploščina kvadrata $O_{1} O_{2} O_{3} O_{4}$?
(A) $3 a^{2}(3-2 \sqrt{2})$
(B) $4 a^{2}(3-... | [
"Solution:\n\nOznačimo oglišča osnovnega kvadrata z $A_{1}$, $A_{2}$, $A_{3}$ in $A_{4}$, tako da je za vsak $i$ oglišče $A_{i}$ najbližje točki $O_{i}$. Polmer večje krožnice je $R=\\frac{a}{2}$, polmer manjših krožnic pa označimo z $r$. Tedaj je $|A_{1} O_{1}|=r \\sqrt{2}$, saj je to diagonala majhnega kvadratka ... | Slovenia | Slovenian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | B | |
0ch3 | Let $n$ be a positive integer and let $a$ and $b$ be positive integers congruent to $1$ modulo $4$. Prove that there exists a positive integer $k$ such that at least one of the numbers $a^k - b$ and $b^k - a$ is divisible by $2^n$. | [
"**Lemma.** For any integer $p \\ge 2$, if $c-1 \\equiv 2^p \\pmod{2^{p+1}}$, then $c^2 - 1 \\equiv 2^{p+1} \\pmod{2^{p+2}}$.\n*Proof.* Write $c^2 - 1 = (c-1)(c+1)$. By hypothesis, $c-1$ is divisible by $2^p$. As $c+1$ leaves remainder 2 upon division by $2^p$, it follows that $2^{p+1}$ is the highest power of 2 di... | Romania | 74th NMO Selection Tests for BMO and IMO | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Abstract Algebra > Group Theory"
] | English | proof only | null | |
07dk | Let $x$, $y$ be distinct real numbers such that $x^4 - y^4 = x - y$. Prove that
$$
\frac{x - y}{x^6 - y^6} \le \frac{4}{3}(x + y)
$$ | [
"It is deduced, from the state of the problem that\n$$\n(x^2 + y^2)(x + y) = 1\n$$\nWithout loss of generality, assume that $x > y$. Therefore we have\n$$\n\\begin{align*}\n\\frac{x-y}{x^6-y^6} &\\le \\frac{4}{3}(x+y) \\\\\n\\Leftrightarrow \\quad &x^6 - y^6 \\ge \\frac{3}{4}(x-y) \\\\\n\\Leftrightarrow \\quad &x^6... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Other"
] | null | proof only | null | |
051v | In a $2n \times 2n$ grid exactly half of the squares have been coloured black and the other half are white. In one step one can take some $2 \times 2$ square in this grid and reflect its four squares w.r.t. the horizontal or vertical central axis. Which positive integers $n$ make it possible to get from any initial con... | [
"In case of $n = 1$ it is not possible to get the chessboard-pattern if $2 \\times 2$ the initial configuration is like in fig. 19, because adjacent same-coloured squares are same-coloured also after reflecting.\n\n\n\nLet us now show that for any $n \\ge 2$ we can start from any initial co... | Estonia | Final Round of National Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | n ≥ 2 | |
00ez | Let $A_{1}$, $A_{2}$, $A_{3}$ be three points in the plane, and for convenience, let $A_{4}=A_{1}$, $A_{5}=A_{2}$. For $n=1,2$, and $3$, suppose that $B_{n}$ is the midpoint of $A_{n} A_{n+1}$, and suppose that $C_{n}$ is the midpoint of $A_{n} B_{n}$. Suppose that $A_{n} C_{n+1}$ and $B_{n} A_{n+2}$ meet at $D_{n}$, a... | [
"Let $G$ be the centroid of triangle $A B C$, and also the intersection point of $A_{1} B_{2}$, $A_{2} B_{3}$, and $A_{3} B_{1}$.\n\nBy Menelao's theorem on triangle $B_{1} A_{2} A_{3}$ and line $A_{1} D_{1} C_{2}$,\n$$\n\\frac{A_{1} B_{1}}{A_{1} A_{2}} \\cdot \\frac{D_{1} A_{3}}{D_{1} B_{1... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1989 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | proof and answer | 25/49 | |
053i | Non-equilateral triangle $ABC$ has a $60°$ angle at vertex $A$. Let the angle bisector drawn from vertex $A$ intersect the opposite side at point $D$, and let $Q$ and $R$ be the feet of the altitudes drawn from vertices $B$ and $C$, respectively. Prove that lines $AD$, $BQ$ and $CR$ intersect in three distinct points t... | [
"If line $AD$ passed through the point of intersection of lines $BQ$ and $CR$, the line segment $AD$ would be an altitude of triangle $ABC$. As $AD$ is also the angle bisector, triangle $ABC$ would be isosceles with $AB = AC$. As $\\angle BAC = 60°$, triangle $ABC$ would be equilateral, contradicting the assumption... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hgx | Circles $w_1$ and $w_2$ intersect at the points $P$ and $Q$. Suppose $AB$ and $CD$ are parallel diameters of the circles $w_1$ and $w_2$ respectively. Moreover, none of the points $A, B, C, D$ coincides with $P$ or $Q$, and the points are located on the circles in the following order: $A, B, P, Q$ on $w_1$ and $C, D, P... | [
"Let $O_1$ be the center of circle $w_1$ and $X' = BP \\cap AQ$ as on Fig. 26. Since $AP \\perp BP$ and $BQ \\perp AQ$, then $AP$ and $BQ$ are heights of the triangle $ABX'$, and $X$ is the orthocenter. Then $X'X$ is also a height in the triangle, therefore $X'X \\perp AB$ and $\\angle PX'X = 90^\\circ - \\angle AB... | Ukraine | Problems from Ukrainian Authors | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Ge... | English | proof only | null | |
04fi | Prove that there are infinitely many positive integers $n$ such that $2^n - 8$ is divisible by $n$. (Kristina Ana Škreb) | [
"We will prove that for all positive integers of the form $n = 3p$ (where $p > 3$ is a prime number)\n$$\nn \\mid 2^n - 8.\n$$\nBy Fermat's little theorem we have $2^p \\equiv 2 \\pmod{p}$, so it follows that\n$$\n2^{3p} - 8 = (2^p)^3 - 8 \\equiv 2^3 - 8 = 0 \\pmod{p}. \\quad (9)\n$$\nAnalogously, since $3p$ is an ... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
00o1 | Determine whether there exists a real number $r$ such that the equation
$$
x^3 - 2023x^2 - 2023x + r = 0
$$
has three different rational solutions. | [
"Let $N = 2023$. We assume that the equation $x^3 - Nx^2 - Nx + r = 0$ has three rational solutions $\\frac{a}{k}$, $\\frac{b}{k}$, $\\frac{c}{k}$, where $a, b, c$ are integers and $k$ is a positive integer with $\\gcd(a, b, c, k) = 1$. According to Vieta we have $\\frac{a}{k} + \\frac{b}{k} + \\frac{c}{k} = N$ and... | Austria | AUT_ABooklet_2023 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | No, such an r does not exist. | |
0cqf | A convex hexagon $ABCDEF$ is given. It appears that $\angle FAE = \angle BDC$, and that each of the quadrilaterals $ABDF$ and $ACDE$ is cyclic. Prove that the lines $BF$ and $CE$ are parallel.
Дан выпуклый шестиугольник $ABCDEF$. Известно, что углы $\angle FAE$ и $\angle BDC$ равны, а четырехугольники $ABDF$ и $ACDE$ ... | [
"Пусть $K$ — точка пересечения отрезков $AE$ и $BF$. Поскольку четырёхугольники $ABDF$ и $ACDE$ вписанные, мы имеем $\\angle AFB = \\angle ADB$ и $\\angle ADC = \\angle AEC$. Отсюда и из условия задачи получаем $\\angle AKB = \\angle AFB + \\angle FAE = \\angle ADB + \\angle BDC = \\angle ADC = \\angle AEC$. Итак, ... | Russia | Russian mathematical olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
0gq4 | Show that for all positive real numbers $a$, $b$, $c$ satisfying $a + b + c = 1$ the following inequality is held:
$$
\frac{a^4 + 5b^4}{a(a + 2b)} + \frac{b^4 + 5c^4}{b(b + 2c)} + \frac{c^4 + 5a^4}{c(c + 2a)} \ge 1 - ab - bc - ca
$$ | [
"Let $f(a, b, c) = \\frac{a^4 + 5b^4}{a(a + 2b)} + \\frac{b^4 + 5c^4}{b(b + 2c)} + \\frac{c^4 + 5a^4}{c(c + 2a)}$. Since $a + b + c = 1$ we will prove that $f(a, b, c) \\ge a^2 + b^2 + c^2 + ab + bc + ac$.\n\nBy Cauchy-Schwarz inequality for positive $x_1, \\dots, x_n$\n$$\n(x_1 + \\dots + x_n) \\left( \\frac{a_1^2... | Turkey | Team Selection Test for JBMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0iwb | Problem:
In terms of $n$, what is the minimum number of edges a finite graph with chromatic number $n$ could have? Prove your answer. | [
"Solution:\n\nThe minimum number of edges is $\\frac{n(n-1)}{2}$.\n\nWe prove this claim by induction. It holds for $n=1$.\n\nNow assume the claim holds for $n$, and consider a graph of chromatic number $n+1$. This graph must have at least one vertex of degree $n$, or else, by part a), it could be colored with only... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | n(n-1)/2 | |
0gix | 如果整數數列 $a_0, a_1, a_2, \dots$ 符合 $a_0 = 0, a_1 = 1$, 且對於所有正整數 $n$ 有
$$
(a_{n+1} - 3a_n + 2a_{n-1})(a_{n+1} - 4a_n + 3a_{n-1}) = 0,
$$
則我們稱它為卡哇伊數列。一個整數如果屬於某個卡哇伊數列，則我們稱這個整數為卡哇伊。
如果連續兩個正整數 $m$ 及 $m+1$ 皆卡哇伊 (不一定要屬於相同的卡哇伊數列)，證明 3 整除 $m$ 且 $m/3$ 也卡哇伊。
A sequence of integers $a_0, a_1, a_2, \dots$ is called *kawaii*, if $a_0... | [
"We start by rewriting the condition in the problem as:\n$$\na_{n+1} = 3a_n - 2a_{n-1}, \\text{ or } a_{n+1} = 4a_n - 3a_{n-1}.\n$$\nWe have $a_{n+1} \\equiv a_n \\pmod{2}$ and $a_{n+1} = a_{n-1} \\pmod{3}$ for all $n \\ge 1$.\nNow, since $a_0 = 0$ and $a_1 = 1$, we have that $a_n \\equiv 0, 1 \\pmod{3}$ for all $n... | Taiwan | IMO 3J, Mock Exam 2 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | Chinese; English | proof only | null | |
0eji | Problem:
Zaporedje $\{a_{n}\}_{n \in \mathbb{N}}$ je podano s prvim členom $a_{1}=3$ in rekurzivno zvezo $(3-a_{n+1})(6+a_{n})=18$ za vse $n \geq 1$. Dokaži, da za vsako naravno število $n$ velja
$$
\sum_{k=1}^{n} \frac{1}{a_{k}}=\frac{2^{n+1}}{3}-\frac{n+2}{3}
$$ | [
"Solution:\n\nZa vsak $n \\geq 1$ iz rekurzivne zveze izrazimo $a_{n+1}=3-\\frac{18}{6+a_{n}}=\\frac{3 a_{n}}{6+a_{n}}$. Od tod sledi\n$$\n\\frac{1}{a_{n+1}}=\\frac{a_{n}+6}{3 a_{n}}=\\frac{1}{3}+\\frac{2}{a_{n}}\n$$\nTo zvezo uporabimo večkrat zapored, da izpeljemo\n$$\n\\begin{aligned}\n\\frac{1}{a_{n+1}} & =\\fr... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0lf5 | Given three functions
$$
P(x) = (x^2 - 1)^{2023}, Q(x) = (2x + 1)^{14}, R(x) = \left(2x + 1 + \frac{2}{x}\right)^{34}.
$$
Initially, we pick a set $S$ containing two of these functions, and we perform some operations on it. Allowed operations include:
* Take two functions $p, q \in S$ and add one of $p+q, p-q$, or $pq$... | [
"First of all, from $P(x), Q(x)$, after all of the operations, we get a polynomial variable $x$, in which $R(x)$ is not a polynomial, so it is not possible to have\n$$\nP, Q \\to R.\n$$\nNext, we will show that $R, P \\to Q$ cannot be obtained. Indeed,\n$$\nP'(x) = 4046x(x^2-1)^{2022}, R'(x) = 34\\left(2-\\frac{2}{... | Vietnam | Team selection tests | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Abstract Algebra > Ring Theory"
] | English | proof only | null | |
0gru | Let $x$, $y$, $z$ be positive real numbers such that
$$
\sqrt{x}, \sqrt{y}, \sqrt{z} \text{ are sides of a triangle and } \frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 5.
$$
Prove that
$$
\frac{x(y^2 - 2z^2)}{z} + \frac{y(z^2 - 2x^2)}{x} + \frac{z(x^2 - 2y^2)}{y} \ge 0.
$$ | [
"Since $\\sqrt{x}$, $\\sqrt{y}$, $\\sqrt{z}$ are sides of a triangle, we have\n$$\n(\\sqrt{x} + \\sqrt{y} + \\sqrt{z})(\\sqrt{x} + \\sqrt{y} - \\sqrt{z})(\\sqrt{y} + \\sqrt{z} - \\sqrt{x})(\\sqrt{z} + \\sqrt{x} - \\sqrt{y}) = 2(xy + yz + zx) - x^2 - y^2 - z^2 \\ge 0.\n$$\nWe also have\n$$\n5xy = \\left( \\frac{x}{y... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0jza | Problem:
Let $ABCD$ be a quadrilateral with an inscribed circle $\omega$ and let $P$ be the intersection of its diagonals $AC$ and $BD$. Let $R_{1}, R_{2}, R_{3}, R_{4}$ be the circumradii of triangles $APB$, $BPC$, $CPD$, $DPA$ respectively. If $R_{1}=31$ and $R_{2}=24$ and $R_{3}=12$, find $R_{4}$. | [
"Solution:\n\nNote that $\\angle APB = 180^{\\circ} - \\angle BPC = \\angle CPD = 180^{\\circ} - \\angle DPA$ so $\\sin APB = \\sin BPC = \\sin CPD = \\sin DPA$. Now let $\\omega$ touch sides $AB$, $BC$, $CD$, $DA$ at $E$, $F$, $G$, $H$ respectively. Then $AB + CD = AE + BF + CG + DH = BC + DA$ so\n$$\n\\frac{AB}{\... | United States | February 2017 | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 19 | |
0c74 | Let $ABC$ be a triangle, in which $m(\angle ABC) = 45^\circ$ and $m(\angle BAC) > 90^\circ$. Let $O$ be the midpoint of the side $[BC]$. We consider the point $M \in (AC)$ such that $m(\angle COM) = m(\angle CAB)$. The perpendicular through $M$ on $AC$ intersects the line $AB$ at the point $P$.
a) Find the measure of ... | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | a) ∠BCP = 45°.
b) If ∠BAC = 105°, then PB = 2·MO. | |
0jvo | Problem:
For $i = 0, 1, \ldots, 5$ let $l_{i}$ be the ray on the Cartesian plane starting at the origin, at angle $\theta = i \frac{\pi}{3}$ counterclockwise from the positive $x$-axis. For each $i$, point $P_{i}$ is chosen uniformly at random from the intersection of $l_{i}$ with the unit disk. Consider the convex hu... | [
"Solution:\n\nA vertex $P_{i}$ is part of the convex hull if and only if it is not contained in the triangle formed by the origin and the two adjacent vertices. Let the probability that a given vertex is contained in the aforementioned triangle be $p$. By linearity of expectation, our answer is simply $6(1-p)$.\n\n... | United States | HMMT February 2016 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 2 + 4 ln 2 | |
06to | For every positive integer $n$ with prime factorization $n=\prod_{i=1}^{k} p_{i}^{\alpha_{i}}$, define
$$
\mho(n)=\sum_{i: p_{i}>10^{100}} \alpha_{i}
$$
That is, $\mho(n)$ is the number of prime factors of $n$ greater than $10^{100}$, counted with multiplicity.
Find all strictly increasing functions $f: \mathbb{Z} \rig... | [
"Answer. $f(x)=a x+b$, where $b$ is an arbitrary integer, and $a$ is an arbitrary positive integer with $\\mho(a)=0$.\n\nA straightforward check shows that all the functions listed in the answer satisfy the problem condition. It remains to show the converse.\n\nAssume that $f$ is a function satisfying the problem c... | IMO | 56th International Mathematical Olympiad Shortlisted Problems | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Functional Equations > Inject... | English | proof and answer | All functions of the form f(x) = a x + b, where b is any integer and a is a positive integer whose prime factors are all at most ten to the one hundred. | |
0g88 | 令 $a_i > 0, i = 1, 2, \dots, n, \sum_{i=1}^{n} a_i = 1$.
試證：對任意正整數 $k$,
$$
(a_1^k + \frac{1}{a_1^k})(a_2^k + \frac{1}{a_2^k})\cdots(a_n^k + \frac{1}{a_n^k}) \ge (n^k + \frac{1}{n^k})^n.
$$
Let $a_i > 0, i = 1, 2, \dots, n, \sum_{i=1}^{n} a_i = 1$.
Prove that for any positive integer $k$,
$$
(a_1^k + \frac{1}{a_1^k})(a... | [
"$$\n\\begin{aligned}\na_1^k + \\frac{1}{a_1^k} &= a_1^k + \\frac{1}{n^{2k} a_1^k} + \\cdots + \\frac{1}{n^{2k} a_n^k} \\\\\n&\\ge (n^{2k} + 1) \\left( \\frac{a_1^k}{(n^{2k} a_1^{2k})^{n^{2k}}} \\right)^{\\frac{1}{n^{2k}+1}} \\\\\n&= (n^k + \\frac{1}{n^k}) \\left( \\frac{1}{n a_1} \\right)^{\\frac{k(n^{2k}-1)}{n^{2... | Taiwan | 國際數學奧林匹亞競賽第二階段選訓營 獨立研究(三) | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
03pe | Suppose $\frac{3}{2} \le x \le 5$. Prove that $2\sqrt{x+1} + \sqrt{2x-3} + \sqrt{15-3x} < 2\sqrt{19}$. | [
"By Cauchy's inequality, we have\n$$\n\\begin{aligned}\n& 2\\sqrt{x+1} + \\sqrt{2x-3} + \\sqrt{15-3x} \\\\\n&= \\sqrt{x+1} + \\sqrt{x+1} + \\sqrt{2x-3} + \\sqrt{15-3x} \\\\\n&\\le \\sqrt{[(x+1) + (x+1) + (2x-3) + (15-3x)](1^2 + 1^2 + 1^2 + 1^2)} \\\\\n&= 2\\sqrt{x+14} \\le 2\\sqrt{19},\n\\end{aligned}\n$$\nand the ... | China | China Mathematical Competition (Shaanxi) | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0hwm | Problem:
Let $k$ be a positive integer. Prove that there exist POSITIVE integers $a_{0}, \ldots, a_{k}$ such that for all integers $x \geq 0$,
$$
x^{k}=a_{0}\left(\begin{array}{l}
x \\
k
\end{array}\right)+a_{1}\left(\begin{array}{c}
x+1 \\
k
\end{array}\right)+\cdots+a_{k-1}\left(\begin{array}{c}
x+k-1 \\
k
\end{array... | [
"Solution:\nGiven $k$ and $x$, let $A$ be the set of all sets of the form\n$$\n\\left\\{m_{1}+\\frac{1}{k}, m_{2}+\\frac{2}{k}, \\ldots, m_{k}+\\frac{k}{k}\\right\\}\n$$\nwhere the $m_{i}$ are integers, $1 \\leq m_{i} \\leq x$. Clearly, $A$ consists of $x^{k}$ sets, each of which has $k$ distinct elements. Now we c... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0cpa | Consider nine quadratic polynomials $x^2 + a_1x + b_1$, $x^2 + a_2x + b_2$, ..., $x^2 + a_9x + b_9$ such that the sequences $a_1, a_2, ..., a_9$ and $b_1, b_2, ..., b_9$ are both arithmetical progressions, and the sum of all nine polynomials is a polynomial having at least one real root. Denote by $k$ the number of ori... | [
"Обозначим $P_i(x) = x^2 + a_i x + b_i$, $P(x) = P_1(x) + \\dots + P_9(x)$. Заметим, что $P_i(x) + P_{10-i}(x) = 2x^2 + (a_i + a_{10-i})x + (b_i + b_{10-i}) = 2P_5(x)$. Значит, $P(x) = 9P_5(x)$, и условие равносильно тому, что $P_5(x)$ имеет хотя бы один корень.\n\nОбозначим через $x_0$ какой-нибудь из его корней. ... | Russia | Russian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English, Russian | proof and answer | 4 | |
011p | Problem:
A set of 8 problems was prepared for an examination. Each student was given 3 of them. No two students received more than one common problem. What is the largest possible number of students? | [
"Solution:\nAnswer: 8.\nDenote the problems by $A, B, C, D, E, F, G, H$, then 8 possible problem sets are $ABC$, $ADE$, $AFG$, $BDG$, $BFH$, $CDH$, $CEF$, $EGH$. Hence, there could be 8 students.\n\nSuppose that some problem (e.g., $A$) was given to 4 students. Then each of these 4 students should receive 2 differe... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 8 | |
0cx5 | Find all non-empty sets $S$ of nonzero real numbers such that
a) $S$ has at most 5 elements;
b) If $x$ is in $S$, then so are $1-x$ and $\frac{1}{x}$. | [
"Let $x \\in S$. Then\n$$\n1-x, \\frac{1}{x}, \\frac{1}{1-x}, 1-\\frac{1}{x}, 1-\\frac{1}{1-x} \\in S.\n$$\n(1) Since $|S| \\leq 5$, it follows that at least two of these numbers are equal. Considering all 15 possible cases it follows $x=-1$ or $x=\\frac{1}{2}$. We get\n$$\nS=\\left\\{-1, \\frac{1}{2}, 2\\right\\}\... | Saudi Arabia | SAMC | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English | proof and answer | {-1, 1/2, 2} | |
0h5b | Find all functions $f: \mathbb{R} \to \mathbb{R}$ that for all real $x, y$ fulfill the equality:
$$
f(f(x) - y^2) = f(x^2) + y^2 f(y) - 2f(xy).
$$ | [
"Substitute $y = 1$ and $y = -1$:\n$$\nf(f(x)-1) = f(x^2) + f(1) - 2f(x) \\text{ and } f(f(x)-1) = f(x^2) + f(-1) - 2f(-x). \\ (*)\n$$\nCombining both equalities, get $f(1) - 2f(x) = f(-1) - 2f(-x)$. Hence, if $x = 1$: $f(1) = f(-1)$, so $f(x) = f(-x)$, the function $f$ is even.\n\nIf $x = y = 1$, $f(f(1) - 1) = 0$... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round (Second Tour) | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | f(x) = 0 for all x, and f(x) = x^2 for all x | |
0itt | Let $m$ and $n$ be positive integers. Mr. Fat has a set $S$ containing every rectangular tile with integer side lengths and area a power of $2$. Mr. Fat also has a rectangle $R$ with dimensions $2^m \times 2^n$ and a $1 \times 1$ square removed from one of the corners. Mr. Fat wants to choose $m + n$ rectangles from $S... | [
"**Solution 1.** We call each of the rectangles in $S$ a tile, and the tile with area $2^0 = 1$ the unit tile. We may assume without loss of generality that the missing $1 \\times 1$ square in $R$ is the top-left corner. Suppose Mr. Fat walks on the path of squares starting from the top right corner square and goin... | United States | Team Selection Test 2009 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
05dp | Problem:
Determine all integers $n \geq 2$ for which there exist integers $x_{1}, x_{2}, \ldots, x_{n-1}$ satisfying the condition that if $0 < i < n$, $0 < j < n$, $i \neq j$ and $n$ divides $2i + j$, then $x_{i} < x_{j}$. | [
"Solution:\n\nSuppose that $n$ has one of these forms. For an integer $i$, let $x_{i}$ be the largest integer such that $2^{x_{i}}$ divides $i$. Now assume that $0 < i < n$, $0 < j < n$, $i \\neq j$, $n$ divides $2i + j$ and $x_{i} \\geq x_{j}$. Then the highest power of $2$ dividing $2i + j$ is $2^{x_{j}}$ and the... | European Girls' Mathematical Olympiad (EGMO) | European Girls' Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | All integers of the form 2^k with k ≥ 1 and 3·2^k with k ≥ 0. | |
0let | Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that
$$
f\left(\frac{f(x)}{x} + y\right) = 1 + f(y), \forall x, y > 0.
$$ | [
"Firstly, we will prove that $\\frac{f(x)}{x}$ is a constant.\nAssume that there exists $a, b \\in (0, +\\infty)$ such that $\\frac{f(a)}{a} \\neq \\frac{f(b)}{b}$.\nWithout loss of generality, we assume that $\\frac{f(a)}{a} < \\frac{f(b)}{b}$. By plugging $x = a, x = b$ into the relation, we get\n$$\nf\\left(y + ... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | f(x) = x for all x > 0 | |
0czx | Let $\left(F_{n}\right)_{n \geq 0}$ be the sequence of Fibonacci numbers: $F_{0}=0$, $F_{1}=1$ and $F_{n+2}=F_{n+1}+F_{n}$, for every $n \geq 0$. Prove that for any prime $p \geq 3$, $p$ divides $F_{2p}-F_{p}$. | [
"Observe that\n$$\n\\begin{gathered}\nF_{2p} = \\frac{1}{\\sqrt{5}} \\left[ \\left( \\frac{1+\\sqrt{5}}{2} \\right)^{2p} - \\left( \\frac{1-\\sqrt{5}}{2} \\right)^{2p} \\right] \\\\\n= \\frac{1}{\\sqrt{5}} \\left[ \\left( \\frac{1+\\sqrt{5}}{2} \\right)^p - \\left( \\frac{1-\\sqrt{5}}{2} \\right)^p \\right] \\left[... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
010y | Problem:
Let $a$, $b$, $c$ and $d$ be prime numbers such that $a > 3b > 6c > 12d$ and $a^{2} - b^{2} + c^{2} - d^{2} = 1749$. Determine all possible values of $a^{2} + b^{2} + c^{2} + d^{2}$. | [
"Solution:\nThe only possible value is $1999$.\n\nSince $a^{2} - b^{2} + c^{2} - d^{2}$ is odd, one of the primes $a$, $b$, $c$ and $d$ must be $2$, and in view of $a > 3b > 6c > 12d$ we must have $d = 2$. Now\n$$\n1749 = a^{2} - b^{2} + c^{2} - d^{2} > 9b^{2} - b^{2} + 4d^{2} - d^{2} = 8b^{2} - 12,\n$$\nimplying $... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1999 | |
0grx | There are $2019$ bags: each bag contains $2019$ beads with total weight $1$ kg. In each bag the beads are numbered by $1, 2, \ldots, 2019$. A proper collection is a collection of differently numbered beads containing at most one bead from each bag and having total weight not less than $1$ kg. Find the maximal possible ... | [
"A good collection is a collection of differently numbered beads containing exactly one bead from each bag. Let $(\\sigma_1, \\sigma_2, \\dots, \\sigma_{2019})$ be a good collection, where for each $i = 1, 2, \\dots, 2019$, $\\sigma_i$ is a number of the bead taken from $i$-th bag. We say that two good collections ... | Turkey | Team Selection Test for IMO 2019 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 2018! | |
096j | Problem:
Arătați că toate soluțiile ecuației $\left(\frac{1+i z}{1-i z}\right)^{2019}=\frac{1+\sqrt{3} i}{1-\sqrt{3} i}$ sunt reale și determinați aceste soluții. | [
"Solution:\nObservăm că $\\left|\\frac{1+i z}{1-i z}\\right|^{2019}=\\left|\\frac{1+\\sqrt{3} i}{1-\\sqrt{3} i}\\right|=1$.\nAtunci $|1+i z|=|1-i z| \\Leftrightarrow|1+i z|^{2}=|1-i z|^{2} \\Leftrightarrow(1+i z)(1-i \\bar{z})=(1-i z)(1+i \\bar{z}) \\Leftrightarrow$\n$$\n1-i \\bar{z}+i z+z \\bar{z}=1+i \\bar{z}-i z... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | All solutions are real: z = tan(π/6057 + kπ/2019), for integers k with −1009 ≤ k ≤ 1009. | |
0fnl | Problem:
Es posible disponer sobre una circunferencia los números $0, 1, 2, \ldots, 9$ de tal manera que la suma de tres números sucesivos cualesquiera sea, como mucho
a) $13$,
b) $14$,
c) $15$? | [] | Spain | L Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | a) No; b) No; c) Yes | |
04he | Let $\alpha$ be a real number. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$f(x + \alpha + f(y)) = f(f(x)) + f(\alpha) + y,$$
for all $x, y \in \mathbb{R}$. | [
"Letting $x = y = -\\alpha$ we get\n$$\nf(f(-\\alpha)) = f(f(-\\alpha)) + f(\\alpha) - \\alpha \\implies f(\\alpha) = \\alpha.\n$$\nLetting $x = -\\alpha, y = \\alpha$ we get\n$$\n\\alpha = f(f(\\alpha)) = f(f(-\\alpha)) + \\alpha + \\alpha \\implies f(f(-\\alpha)) = -\\alpha.\n$$\nNow let us denote $f(-\\alpha) = ... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x | |
089p | Problem:
In una variante del gioco della battaglia navale Anna posiziona una portaerei (che possiamo pensare come rettangolino $5 \times 1$) in una griglia $10 \times 10$, indifferentemente in verticale o in orizzontale, senza farla vedere a Jacopo. Jacopo prova a colpire la portaerei, dicendole volta per volta le coo... | [
"Solution:\n\nLa risposta è 20. 20 colpi sono necessari: se Jacopo spara 19 colpi o meno, allora esiste almeno una riga o almeno una colonna bersagliata da un solo colpo. Tale colpo divide quella riga (o colonna) in due spazi, uno dei quali dev'essere lungo almeno 5 quadretti, e dunque potrebbe nascondere la portae... | Italy | Progetto Olimpiadi della Matematica | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 20 | |
03zl | Given $n$ points $P_1, P_2, \dots, P_n$ on a plane, let $M$ be any point on segment $AB$ on the plane. Denote by $|P_iM|$ the distance between $P_i$ and $M$, $i = 1, 2, 3, \dots, n$. Prove that
$$
\sum_{i=1}^n |P_iM| \le \max\left\{\sum_{i=1}^n |P_iA|, \sum_{i=1}^n |P_iB|\right\}.
$$
(posed by Jin Mengwei) | [
"Let $O$ be the origin. Then we have $\\overrightarrow{OM} = t \\overrightarrow{OA} + (1-t) \\overrightarrow{OB}$, $t \\in (0, 1)$.\n$$\n\\begin{aligned}\n|P_iM| &= |\\overrightarrow{OM} - \\overrightarrow{OP_i}| \\\\\n&= |t \\overrightarrow{OA} + (1-t) \\overrightarrow{OB} - t \\overrightarrow{OP_i} - (1-t) \\over... | China | China Southeastern Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
082y | Problem:
Un tastierino numerico quadrato di lato $4$ ha $16$ tasti numerati da $1$ a $16$. L'agente $007$ deve premere due tasti contemporaneamente per penetrare nella base nemica, e se sbaglia farà suonare l'allarme. Tuttavia sa soltanto che i due tasti non sono contigui (cioè non hanno un lato o un vertice in comune... | [
"Solution:\n\nLa risposta è (B). La probabilità richiesta è l'inverso del numero delle possibili paia di tasti non contigui. Per contare tali paia distinguiamo vari casi:\n\na) uno dei tasti è tra i quattro centrali: allora l'altro deve essere sul bordo, e vanno esclusi 5 tasti contigui.\nPossibilità: $4 \\cdot 7 =... | Italy | Progetto Olimpiadi di Matematica 2003 | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | MCQ | B | |
0dng | Problem:
Правилан $n$-тоугао подељен је на троуглове помоћу $n-3$ дијагонале од којих никоје две немају заједничких унутрашњих тачака. Колико највише међу овим троугловима може бити међусобно неподударних?
(Душан Ђукић) | [
"Solution:\n\nОдговор је $\\left[\\frac{3 n-7}{4}\\right]$ за $n>3$, односно 1 за $n=3$.\n\nТроуглове са по две, једном или ниједном страницом која је уједно страница $n$-тоугла ( $n>3$ ) зовемо редом ушиима, танким и дебелим троугловима. Нека у подели има $a$ дебелих троуглова, $b$ танких и $c$ ушију. Број страниц... | Serbia | 8. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | For n > 3, the maximum is floor((3n − 7)/4); for n = 3, it is 1. | |
06qd | Let $a$, $b$, $c$ be positive real numbers such that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = a + b + c$. Prove that
$$
\frac{1}{(2a + b + c)^2} + \frac{1}{(2b + c + a)^2} + \frac{1}{(2c + a + b)^2} \leq \frac{3}{16}
$$ | [
"For positive real numbers $x$, $y$, $z$, from the arithmetic-geometric-mean inequality,\n$$\n2x + y + z = (x + y) + (x + z) \\geq 2 \\sqrt{(x + y)(x + z)}\n$$\nwe obtain\n$$\n\\frac{1}{(2x + y + z)^2} \\leq \\frac{1}{4(x + y)(x + z)}\n$$\nApplying this to the left-hand side terms of the inequality to prove, we get... | IMO | IMO Problem Shortlist | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
0h4x | A right trapezoid is given with the following property: a square can be inscribed into it such that all its vertices lie on different edges of the trapezoid and none of them coincide with any vertex of the trapezoid. Construct this square with a ruler and a compass.
(Mariya Rozhkova) | [
"**Analysis.** Let $ABCD$ be our trapezoid with right angles $A$ and $B$. Let $EFGH$ be the required square centered at $O$, and suppose $E \\in AB$, $F \\in BC$. In the quadrilateral $EBFO$ two opposite angles are right, hence, it's cyclic. This implies that $\\angle EFO = \\angle EBO = 45^\\circ$ as they intercep... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | null | |
078e | A sleeping rabbit lies in the interior of a convex $2024$-gon. A hunter picks three vertices of the polygon and he lays a trap which covers the interior and the boundary of the triangular region determined by them. Determine the minimum number of times he needs to do this to guarantee that the rabbit will be trapped. | [
"Let the $2024$-gon be $A_0A_1 \\cdots A_{2023}$. We claim that the answer is $2022$ which is achieved by picking the triangles $\\triangle A_0A_iA_{i+1}$ for $i = 1, 2, \\dots, 2022$.\n\nFor any $0 \\le i \\le 2023$, we claim that the entirety of $\\angle A_iA_{i+1}A_{i+2}$ is covered by triangles with one of thei... | India | IMO TST | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2022 | |
070k | Problem:
Given $1997$ points inside a circle of radius $1$, one of them the center of the circle. For each point take the distance to the closest (distinct) point. Show that the sum of the squares of the resulting distances is at most $9$. | [
"Solution:\n\nLet the points be $P_i$ for $i = 1, 2, \\ldots, 1997$. Take $P_1$ to be the center of the given unit circle. Let $x_i$ be the distance from $P_i$ to the closest of the other $1996$ points. Let $C_i$ be the circle centered at $P_i$ with radius $x_i / 2$. Then $C_i$ and $C_j$ cannot overlap by more than... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0k1v | Problem:
Let $n$ be a positive integer. There are $n$ islands, and you want to build some number of bridges so that there is a path between any two islands. (Each bridge joins a pair of islands, and can be traveled in both directions.) At least how many bridges must you build? | [
"Solution:\n\nYou need to build at least $n-1$ bridges. Imagine you start at one island and are only building bridges that start from islands you can already reach. Then, each new bridge connects you to only one new island, so to be connected to all of them, you need $n-1$ bridges. Conversely, $n-1$ bridges is enou... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | n-1 | |
0i3s | Let $A$ be a finite set of positive integers. Prove that there exists a finite set $B$ of positive integers such that $A \subseteq B$ and
$$
\prod_{x \in B} x = \sum_{x \in B} x^2.
$$ | [
"For any finite set $S$ of positive integers, let\n$$\nD(S) = \\prod_{x \\in S} x - \\sum_{x \\in S} x^2.\n$$\nIf $D(A) = 0$, then we take $B = A$.\n\nIf $D(A) < 0$, then let $m = \\max A$. Write $A'_k = A \\cup \\{m+1, m+2, \\dots, m+k\\}$. Then there is a positive integer $k$ such that\n$$\n\\begin{aligned}\n-D(A... | United States | USA IMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
09cv | $a, b, c > 0$ бол
$$
\frac{a+b+3c}{3a+3b+2c} + \frac{a+3b+c}{3a+2b+3c} + \frac{3a+b+c}{2a+3b+3c} \ge \frac{15}{8}
$$
тэгшитгэл биш биелэхийг батал. | [
"$x = 2a + 3b + 3c$, $y = 3a + 2b + 3c$, $z = 3a + 3b + 2c$ гэсэн орлуулга хийе. Эндээс $a, b, c > 0$ тул $x, y, z > 0$ ба\n$$\na = \\frac{-5x + 3y + 3z}{8}, \\quad b = \\frac{3x - 5y + 3z}{8}, \\quad c = \\frac{3x + 3y - 5z}{8}\n$$\nболохыг хялбархан олж болно. Эндээс\n$$\na + b + 3c = \\frac{7x + 7y - 9z}{8};\n$$... | Mongolia | ММО-48 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | Mongolian | proof only | null | |
00yj | Problem:
Prove that any irreducible fraction $\frac{p}{q}$, where $p$ and $q$ are positive integers and $q$ is odd, is equal to a fraction $\frac{n}{2^{k}-1}$ for some positive integers $n$ and $k$. | [
"Solution:\n\nSince the number of congruence classes modulo $q$ is finite, there exist two non-negative integers $i$ and $j$ with $i>j$ which satisfy $2^{i} \\equiv 2^{j} \\pmod{q}$. Hence, $q$ divides the number $2^{i}-2^{j}=2^{j}\\left(2^{i-j}-1\\right)$. Since $q$ is odd, $q$ has to divide $2^{i-j}-1$. Now it su... | Baltic Way | Baltic Way | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0ja3 | Problem:
You are given an unlimited supply of red, blue, and yellow cards to form a hand. Each card has a point value and your score is the sum of the point values of those cards. The point values are as follows: the value of each red card is $1$, the value of each blue card is equal to twice the number of red cards, ... | [
"Solution:\n\nAnswer: $168$\n\nIf there are $B$ blue cards, then each red card contributes $1 + 2B$ points (one for itself and two for each blue card) and each yellow card contributes $3B$ points. Thus, if $B > 1$, it is optimal to change all red cards to yellow cards. When $B = 0$, the maximum number of points is ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 168 |
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