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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
05x5 | Problem:
Anna et Elie jouent à un jeu. On leur donne à tous les deux le même ensemble $A$ composé d'un nombre fini d'entiers strictement positifs et distincts. Anna choisit un entier $a \in A$ secrètement. Si Elie choisit un entier $b$ (pas forcément dans $A$) et le donne à Anna, Anna lui donne le nombre de diviseurs ... | [
"Solution:\n\nNotons $P$ l'ensemble fini des nombres premiers divisant au moins un élément de $A$ et $n \\geqslant 1$ le plus grand entier tel qu'il existe $a \\in A$ et $p \\in P$ tels que $p^{n} \\mid a$. On peut remplacer $A$ par l'ensemble des entiers $m$ dont les facteurs premiers sont tous dans $P$ et tels qu... | France | Préparation Olympique Française de Mathématiques - Envoi 3: Arithmétique | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
07gv | Find all functions $f : \mathbb{N} \to \mathbb{N}$ such that for all positive integers $m$ and $n$
$$
f(n) + 1400m^2 \mid n^2 + f(f(m)).
$$ | [
"**Lemma.** *There are infinitely many positive integers $n$ such that*\n$$\nf(n) \\geq n^{\\frac{13}{10}}.\n$$\n*Proof.* Letting $(m,n) = (m,f(m))$ to obtain $f(f(m)) + 1400m^2 \\mid f(m)^2 + f(f(m))$ yielding $f(m) > m$. Plugging $n=1$ to obtain $f(1) + 1400m^2 \\mid 1 + f(f(m))$ yielding $f(f(m)) > m^2$. Assume ... | Iran | 38th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Other"
] | null | proof and answer | no such function | |
0114 | Problem:
We are given $1999$ coins. No two coins have the same weight. A machine is provided which allows us with one operation to determine, for any three coins, which one has the middle weight. Prove that the coin that is the $1000$-th by weight can be determined using no more than $1000000$ operations and that this... | [
"Solution:\n\nIt is possible to find the $1000$-th coin (i.e. the medium one among the $1999$ coins). First we exclude the lightest and heaviest coin—for this we use $1997$ weighings, putting the medium-weighted coin aside each time. Next we exclude the $2$-nd and $1998$-th coins using $1995$ weighings, etc. In tot... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0kie | Let $N$ be the positive integer $7777\ldots777$, a $313$-digit number where each digit is a $7$. Let $f(r)$ be the leading digit of the $r$th root of $N$. What is $f(2) + f(3) + f(4) + f(5) + f(6)$?
(A) 8 (B) 9 (C) 11 (D) 22 (E) 29 | [] | United States | 2021 AMC 10 B Fall | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | A | |
0bea | Find all real $x > 0$ and integer $n > 0$ so that $\lfloor x \rfloor + \left\{ \frac{1}{x} \right\} = 1.005 \cdot n$. | [
"The equation can be written $\\lfloor x \\rfloor + \\{1/x\\} = n + n/200$. Denote $q$, respectively $r$, the quotient and the remainder of the division of $n$ by $200$. Then $\\lfloor x \\rfloor + \\{1/x\\} = n + q + r/200$. The integer part of the left-hand member is $\\lfloor x \\rfloor$ and the integer part of ... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | proof and answer | x = 100/7, n = 14 | |
05vj | Problem:
On dit qu'un entier $n \geqslant 1$ est nippon s'il admet trois diviseurs $d_{1}, d_{2}$ et $d_{3}$ tels que $1 \leqslant d_{1}<d_{2}<d_{3}$ et $d_{1}+d_{2}+d_{3}=2022$.
Quel est le plus petit entier nippon? | [
"Solution:\n\nTout d'abord, on remarque que $2 \\times 2019$ est nippon, puisque les entiers $1$, $2$ et $2019$ figurent parmi ses diviseurs.\n\nSoit $n$ l'entier nippon minimal : il s'agit du PPCM des entiers $d_{1}, d_{2}$ et $d_{3}$. On pose alors $a_{k}=n / d_{k}$, de sorte que $1 \\leqslant a_{3}<a_{2}<a_{1}$,... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic",
"Algebra > Equations and Inequaliti... | null | proof and answer | 1344 | |
0c2g | Let $n$ be a positive integer. A set $A$ of positive integers is $n$-size complete if the set of all the remainders obtained when dividing an element of $A$ by an element of $A$ is $\{0, 1, 2, \dots, n\}$. For example, the set $\{3, 4, 5\}$ is a 4-size complete set.
Determine the minimum number of elements of a 100-siz... | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 51 | |
0043 | Hallar todos los enteros positivos $n$ tales que $[\sqrt{n}]-2$ divide a $n-4$ y $[\sqrt{n}]+2$ divide a $n+4$. | [] | Argentina | XVII OLIMPIADA MATEMATICA DEL CONO SUR | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Español | proof and answer | All n equal to 2, 36, or n = m^2 + 2m − 4 for integers m ≥ 3. | |
02lt | Let $A$ be one of the intersection points of two circles with centers $X$ and $Y$. The tangent lines to these circles passing through $A$ meet the circles again at $B$ and $C$. Let $P$ be the point in the plane such that $PXAY$ is a parallelogram. Prove that $P$ is the circumcenter of the triangle $ABC$. | [
"The tangent lines passing through $A$ are perpendicular to $AX$ and $AY$, so $AX \\perp AC$ and $AY \\perp AB$. Since $AXPY$ is a parallelogram, $PX$ is parallel to $AY$, so $PX \\perp AB$. Since $AB$ is a chord of a circle with center $X$, $PX$ is the perpendicular bisector of $AB$. Analogously, $PY$ is the perpe... | Brazil | XXXI Brazilian Math Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
00y2 | Problem:
Solve the system of equations:
$$
\left\{\begin{array}{l}
x^{5}=y+y^{5} \\
y^{5}=z+z^{5} \\
z^{5}=t+t^{5} \\
t^{5}=x+x^{5} .
\end{array}\right.
$$ | [
"Solution:\nAdding all four equations we get $x+y+z+t=0$. On the other hand, the numbers $x, y, z, t$ are simultaneously positive, negative or equal to zero. Thus, $x=y=z=t=0$ is the only solution."
] | Baltic Way | Baltic Way 1993 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | x = y = z = t = 0 | |
0bsj | The people of an ancient tribe used a language in which the words were formed with two letters only: $A$ and $B$. Researchers discovered that any two words of equal length differ in at least three positions. For instance, the words $ABBAA$ and $AAAAB$ differ in positions $2$, $3$ and $5$, that is, in three positions.
... | [
"If we denote by $C$ the set of all possible words of length $n$ (not necessarily from the language), we have $\\text{Card}(C) = 2^n$. If $x$ and $y$ are arbitrary words from $C$, let $d(x, y)$ be the number of positions at which the corresponding letters are different (the Hamming distance). Obviously, $d(x, x) = ... | Romania | 67th Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0glq | Let $H$ be the orthocenter of an acute triangle $ABC$. The circumcircle of $\triangle BCH$ intersects $AB$ and $AC$ again at points $A_1$ and $A_2$ respectively. Define points $B_1, B_2, C_1$ and $C_2$ analogously. Prove that the circumcenter of the triangle formed by lines $A_1A_2, B_1B_2$, and $C_1C_2$ is on the Eule... | [] | Thailand | T3MO 2017 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane G... | English | proof only | null | |
02k6 | Problem:
Iara possui $R\$ 50,00$ para comprar copos que custam $R\$ 2,50$ e pratos que custam $R\$ 7,00$. Ela quer comprar no mínimo 4 pratos e 6 copos. O que ela pode comprar? | [
"Solution:\n\nSejam $c$ e $p$ o número de copos e pratos que Iara pode comprar. Logo seu gasto é $2,5c + 7p$. Ela só tem $R\\$ 50,00$, logo $2,5c + 7p \\leq 50$ (I).\n\nAlém disso, ela quer comprar no mínimo 4 pratos e 6 copos, logo $p \\geq 4$ e $c \\geq 6$ (II).\n\nDevemos encontrar dois números inteiros $c$ e $p... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | She can buy 4 plates and 8 cups, or 5 plates and 6 cups. | |
0kkt | Problem:
There are $N$ lockers, labeled from $1$ to $N$, placed in clockwise order around a circular hallway. Initially, all lockers are open. Ansoon starts at the first locker and always moves clockwise. When she is at locker $n$ and there are more than $n$ open lockers, she keeps locker $n$ open and closes the next ... | [
"Solution:\n\nNote that in the first run-through, we will leave all lockers $2^{n}-1$ open. This is because after having locker $2^{n}-1$ open, we will close the next $2^{n}-1$ lockers and then start at locker $2^{n}-1+2^{n}-1+1=2^{n+1}-1$. Now we want $1$ to be the last locker that is open. We know that if $N<2046... | United States | HMMT November 2021 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 2046 | |
0kro | Problem:
Suppose $n \geq 3$ is a positive integer. Let $a_{1}<a_{2}<\cdots<a_{n}$ be an increasing sequence of positive real numbers, and let $a_{n+1}=a_{1}$. Prove that
$$
\sum_{k=1}^{n} \frac{a_{k}}{a_{k+1}} > \sum_{k=1}^{n} \frac{a_{k+1}}{a_{k}}.
$$ | [
"Solution:\nWe will use induction. The base case is $n=3$. In this case, we want to show that\n$$\n\\frac{a_{1}}{a_{2}}+\\frac{a_{2}}{a_{3}}+\\frac{a_{3}}{a_{1}}>\\frac{a_{2}}{a_{1}}+\\frac{a_{3}}{a_{2}}+\\frac{a_{1}}{a_{3}}.\n$$\nEquivalently, we want to show\n$$\n\\begin{aligned}\na_{1}^{2} a_{3}+a_{2}^{2} a_{1}+... | United States | HMMT February 2022 | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0ddj | Let $ABC$ be a triangle with $AB < AC$ inscribed in $(O)$. The tangent line at $A$ of $(O)$ cuts $BC$ at $D$. Take $H$ as the projection of $A$ on $OD$ and $E$, $F$ as projections of $H$ on $AB$, $AC$. Suppose that $EF$ cuts $(O)$ at $R$, $S$. Prove that $(HRS)$ is tangent to $OD$. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cw7 | In a kindergarden, a nurse took $n > 1$ congruent cardboard rectangles and gave them to $n$ kids, one per each. Each kid has cut its rectangle into congruent squares (the squares of different kids could be of different sizes). It turned out that the total number of the obtained squares is a prime number. Prove that all... | [
"Предположим противное: пусть прямоугольники имели размеры $k \\times l$, где $l < k$. Пусть $a_i$ — длина стороны квадрата у $i$-го ребёнка. Тогда каждая сторона прямоугольника составлена из нескольких отрезков длиной $a_i$, то есть $l = b_i a_i$ и $k = c_i a_i$, где $b_i$ и $c_i$ — натуральные числа. При этом у р... | Russia | Final round | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English; Russian | proof only | null | |
0694 | Let $\mathbb{R}_+ = (0, \infty)$. Determine all functions $f : \mathbb{R}_+ \to \mathbb{R}_+$ such that
$$
f(xf(y)) + yf(z) + zf(x) = xy + yz + zx,
$$
for all $x, y, z \in \mathbb{R}_+$. | [
"For $x = y = z = 1$ we have $f(f(1)) = 1$, and hence for $x = y = z = f(1)$ we have\n$$\n3f(f(1)) = 3f(1)^2 \\Rightarrow f(1)^2 = 1 \\Rightarrow f(1) = 1.\n$$\nFor $y = z = 1$, we have for every $x \\in \\mathbb{R}_+$:\n$$\n\\begin{aligned}\nf(f(x)) + f(x) &= 2x \\\\\nf(f(x)) &= 2x - f(x)\n\\end{aligned} \\qquad (... | Greece | SELECTION EXAMINATION 2019 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = x for all x > 0 | |
06ec | Given that $p, q$ and $r$ are positive real numbers, prove that
$$
\frac{1}{q+r} + \frac{1}{r+p} + \frac{1}{p+q} \ge \frac{9}{2(p+q+r)}
$$
Hence prove that if $m$ is a real number greater than $1$, then
$$
\frac{p^m}{q+r} + \frac{q^m}{r+p} + \frac{r^m}{p+q} \ge \frac{(p+q+r)^{m-1}}{2 \cdot 3^{m-2}}
$$ | [
"By the Cauchy-Schwarz inequality, we have\n$$\n((q+r) + (r+p) + (p+q)) \\left( \\frac{1}{q+r} + \\frac{1}{r+p} + \\frac{1}{p+q} \\right) \\ge (1+1+1)^2 = 9.\n$$\nDividing both sides by $2(p+q+r)$, we obtain\n$$\n\\frac{1}{q+r} + \\frac{1}{r+p} + \\frac{1}{p+q} \\ge \\frac{9}{2(p+q+r)}.\n$$\n\nNext, WLOG assume $p ... | Hong Kong | IMO HK TST | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0a2j | If you write the date 16 January 1091 with 8 digital digits in a row, it looks like this:
When you read this upside down, it reads as the exact same date. What is the first date in the future (22 June 2024 or later) for which it is also true that, written in 8 digital digits consecutively, the... | [] | Netherlands | Junior Mathematical Olympiad | [
"Math Word Problems"
] | null | final answer only | 10 December 2101 | |
01nf | Find all possible values of real number $a$ such that there exist a function $f : \mathbb{R} \to \mathbb{R}$, and real number $\alpha$ satisfying the equalities $f(\alpha) = 0$ and $f(f(x)) = x f(x) + a$ for all real $x$. | [
"Answer: $a = 0$.\nIndeed, if $a = 0$, then the function $f \\equiv 0$ satisfies the condition. Now let $a \\neq 0$. Suppose that $f(\\alpha) = 0$ for some $\\alpha$. We have $f(0) = f(f(\\alpha)) = \\alpha \\cdot f(\\alpha) + a = a$. Then $f(a) = f(f(0)) = 0 \\cdot f(0) + a = a$. Therefore, $a = f(a) = f(f(a)) = a... | Belarus | Belorusija 2012 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | 0 | |
08ap | Problem:
Dato il sistema
$$
\begin{cases}
x + y + z = 7 \\
x^{2} + y^{2} + z^{2} = 27 \\
xyz = 5
\end{cases}
$$
quante terne ordinate di numeri reali $(x, y, z)$ ne sono soluzione?
(A) 6
(B) 3
(C) 2
(D) 0
(E) Infinite. | [
"Solution:\n\nLa risposta è (B).\n\nPRIMA SOLUZIONE\n\nSiano $s = x + y + z = 7$, $q = xy + yz + zx$ e $p = xyz = 5$. Si può notare facilmente che $s^{2} = x^{2} + y^{2} + z^{2} + 2(xy + yz + zx) = 27 + 2q$, da cui $q = 11$. Consideriamo ora il polinomio $(t - x)(t - y)(t - z) = t^{3} - (x + y + z)t^{2} + (xy + yz ... | Italy | Progetto Olimpiadi della Matematica - Gara di Febbraio | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | MCQ | B | |
04s8 | Some objects are in each of four rooms. Let $n \ge 2$ be an integer. We move one $n$-th of objects from the first room to the second one. Then we move one $n$-th of (the new number of) objects from the second room to the third one. Then we move similarly objects from the third room to the fourth one and from the fourth... | [
"Let us compute backwards. Firstly we find the number of the objects in two rooms before the move. Let $a$ and $b$ be number of the objects in the rooms $A$ and $B$ before the move. This number after the move we denote by $a'$ and $b'$. By the conditions\n$$\na' = \\frac{n-1}{n}a, \\quad b' = b + \\frac{1}{n}a\n$$\... | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 5, attained when n = 3 | |
0ep2 | The number of squares that have $(-1; -1)$ as a vertex and at least one of the coordinate axes as an axis of symmetry is
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5 | [
"Answer E.\nIf we reflect $P(-1; -1)$ in the $x$-axis to map to $Q(-1; 1)$ we obtain 3 squares with the $x$-axis as a line of symmetry, as shown in the diagram. Two of the cases have $PQ$ as an edge/side and in one case $PQ$ is a diagonal of the square. Similarly, if we reflect $P$ in the $y$-axis to $R$, then ther... | South Africa | South African Mathematics Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | MCQ | E | |
075k | The circumcenter of the cyclic quadrilateral $ABCD$ is $O$. The second intersection point of the circles $ABO$ and $CDO$, other than $O$, is $P$, which lies in the interior of the triangle $DAO$. Choose a point $Q$ on the extension of $OP$ beyond $P$, and a point $R$ on the extension of $OP$ beyond $O$. Prove that $\an... | [
"Let $H$ be the radical center of the circles $ABCD$, $ABOP$ and $CDPO$. Then the radical axes of any two of these circles, i.e. the lines $AB$, $CD$ and $OP$, pass through $H$. Since $P$ lies on the shorter arcs $AO$ and $DO$, it follows that $H$ lies on the extension of $OP$ beyond $P$. The radical center satisfi... | India | Indija TS 2012 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03wd | Find all pairs $(p, q)$ of prime numbers such that $pq \mid 5^p + 5^q$. (Posed by Fu Yunhao) | [
"If $2 \\mid pq$, we suppose that $p = 2$ without loss of generality, and then $q \\mid 5^q + 25$. By Fermat's theorem we have $q \\mid 5^q - 5$, so $q \\mid 30$, where $(2, 3)$ and $(2, 5)$ are solutions [(2, 2) does not fit].\nIf $5 \\mid pq$, we suppose that $p = 5$ without loss of generality and then $5q \\mid ... | China | Chinese Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequaliti... | English | proof and answer | [[2, 3], [3, 2], [2, 5], [5, 2], [5, 5], [5, 313], [313, 5]] | |
05zi | Problem:
Antoine propose à Baptiste de jouer à l'"alphabet en folie" : Ils commencent par se mettre d'accord sur une lettre. Puis, à tour de rôle, chacun peut choisir de prononcer entre 1 et 2 lettres suivantes dans l'alphabet en partant de $A$. Celui qui prononce la lettre choisie a gagné. Si Antoine commence, pour q... | [
"Solution:\n\nMontrons que Antoine a une stratégie gagnante si, et seulement si, le rang de la lettre choisie dans l'alphabet n'est pas un multiple de $3$.\n\nColorons les lettres de l'alphabet en bleu-blanc-rouge dans l'ordre. Si la lettre d'arrivée est rouge (c'est à dire que son rang dans l'alphabet est un multi... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Antoine has a winning strategy if and only if the target letter’s position in the alphabet is not a multiple of three; he loses when it is a multiple of three. | |
068k | Let a square $ABCD$ of side length $8$ cm which is divided, with lines parallel to its sides, into $64$ small squares of side $1$ cm. Color $7$ small squares black, and the other $57$ small squares are white. Suppose that there is a positive integer $k$ such that no matter which $7$ squares are black, there is a rectan... | [
"We divide $ABCD$ into $8$ rectangles $4 \\times 2$. Since we color seven small squares with black color, from pigeonhole, there will be at least one $4 \\times 2$ rectangle that contains no black square and of course its area is $8\\,\\text{cm}^2$.\n\n\nFigure 1\n\nIn what follows we will ... | Greece | Hellenic Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 8 | |
04py | Determine the number of complex solutions of the equation
$$
z^{2019} = z + \bar{z}.
$$ | [] | Croatia | Croatian Mathematical Society Competitions | [
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | 2019 | |
06ag | Let $ABC$ be an acute angled triangle with $AB < AC < BC$, inscribed to the circle $\Gamma_1$ with center $O$. The circle $\Gamma_2$ with center $A$ and radius $AC$ intersects the line $BC$ at point $D$ and the circle $\Gamma_1$ at $E$. The circumcircle of the triangle $DEF$ (say, $\Gamma_3$) intersects the line $BC$ a... | [
"a) The triangle $ADC$ is isosceles ($AD = AC$ are radius of $\\Gamma_2$), so $\\angle D_1 = \\angle C$.\nThe angle $\\angle F_1$ is external of the cyclic $ACBF$, therefore $\\angle F_1 = \\angle C$.\nFrom the two equalities above we conclude that $\\angle D_1 = \\angle F_1$, thus\n$$\nBD = BF \\tag{1}.\n$$\n\n![]... | Greece | Selection examinations | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0b67 | Prove that every matrix $A \in \mathcal{M}_2(\mathbb{R})$ can be written in the form $A = X^3 + Y^3$, where $X, Y \in \mathcal{M}_2(\mathbb{R})$, $XY = YX$. | [] | Romania | Shortlisted Problems for the Romanian NMO | [
"Algebra > Linear Algebra > Matrices"
] | English | proof only | null | |
052c | The greatest common divisor of positive integers $a$, $b$, $c$ is $1$. It is known that $c$ divides $a + 2b$ and $a^2 - b^2$. Prove that $c$ also divides $a - b$. | [
"Let $d = \\gcd(a+b, c)$. Since $c \\mid a+2b$, also $d \\mid a+2b$. Now $(a+2b) - (a+b) = b$ and $2(a+b) - (a+2b) = a$ are divisible by $d$. Therefore $d$ is the common divisor of $a$, $b$, $c$ and due to our initial assumption of $a$, $b$, $c$ being relatively prime it has to be $1$. Hence $a+b$ and $c$ are also ... | Estonia | Open Contests | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
06ma | A 'palindrome' is a positive integer which reads the same from left to right as from right to left, such as $12321$ and $259952$. Someone wrote down a five-digit palindrome $m$ and then removed a digit of $m$ to obtain a four-digit positive integer $n$ (that does not start with $0$). How many possible values of $n$ are... | [
"**Answer:** $2358$\n\nNote that $m$ is of the form $ABCBA$ with $A$ nonzero, so $n$ is of the form $BCBA$, $ACBA$, $ABBA$, $ABCA$ or $ABCB$. The second, third and fourth cases can be combined, so we are down to three types of possible values of $n$:\n\n* Type I — Equal thousands digit and tens digit, with nonzero ... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | proof and answer | 2358 | |
0iqz | Problem:
Let $p(x)$ be the polynomial of degree $4$ with roots $1, 2, 3, 4$ and leading coefficient $1$. Let $q(x)$ be the polynomial of degree $4$ with roots $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ and leading coefficient $1$. Find $\lim_{x \rightarrow 1} \frac{p(x)}{q(x)}$. | [
"Solution:\n\nAnswer: $-24$\n\nConsider the polynomial $f(x) = x^4 q\\left(\\frac{1}{x}\\right)$—it has the same roots, $1, 2, 3, 4$, as $p(x)$. But this polynomial also has the same coefficients as $q(x)$, just in reverse order. Its leading coefficient is $q(0) = 1 \\cdot \\frac{1}{2} \\cdot \\frac{1}{3} \\cdot \\... | United States | 1st Annual Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | -24 | |
06gi | Evaluate $\frac{\sqrt{45+\sqrt{1}} + \sqrt{45+\sqrt{2}} + \dots + \sqrt{45+\sqrt{2024}}}{\sqrt{45-\sqrt{1}} + \sqrt{45-\sqrt{2}} + \dots + \sqrt{45-\sqrt{2024}}}$. | [
"The answer is $1 + \\sqrt{2}$.\nLet $a$ and $b$ be the numerator and the denominator respectively. For each $n = 1, 2, \\dots, 2024$, we have the identity\n$$\n\\sqrt{45 + \\sqrt{2025-n}} + \\sqrt{45 - \\sqrt{2025-n}} = \\sqrt{2} \\cdot \\sqrt{45 + \\sqrt{n}}.\n$$\nThis can be easily proved by squaring both sides ... | Hong Kong | IMO HK TST | [
"Algebra > Intermediate Algebra > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1 + sqrt(2) | |
09hg | Let $a_1 < a_2 < \dots$ be the positive divisors of a positive integer $a$ and let $b_1 < b_2 < \dots$ be the positive divisors of a positive integer $b$. Find all $a, b$ such that
$$
\begin{cases} a_{10} + b_{10} = a \\ a_{11} + b_{11} = b \end{cases}
$$ | [
"Answer: $a = 2^{10}$ and $b = 2^{11}$.\nIf $b = b_{11}$ then $a_{11} = 0$, which is impossible. Thus $b \\ge 2b_{11}$.\nLet $a = a_{10} \\cdot n$. Since $a > a_{10}$, we have $n \\ge 2$ and\n$$\nb_{10} = a - a_{10} = a_{10} \\cdot (x - 1) \\ge a_{10}.\n$$\nHence $2b_{10} \\ge b_{10} + a_{10} = a$. It follows that ... | Mongolia | Mongolian National Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | a = 2^10, b = 2^11 | |
06t0 | The points $P$ and $Q$ are chosen on the side $B C$ of an acute-angled triangle $A B C$ so that $\angle P A B = \angle A C B$ and $\angle Q A C = \angle C B A$. The points $M$ and $N$ are taken on the rays $A P$ and $A Q$, respectively, so that $A P = P M$ and $A Q = Q N$. Prove that the lines $B M$ and $C N$ intersect... | [
"Denote by $S$ the intersection point of the lines $B M$ and $C N$. Let moreover $\\beta = \\angle Q A C = \\angle C B A$ and $\\gamma = \\angle P A B = \\angle A C B$. From these equalities it follows that the triangles $A B P$ and $C A Q$ are similar (see Figure 1). Therefore we obtain\n$$\n\\frac{B P}{P M} = \\f... | IMO | 55th International Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0eeo | Let $ABC$ be a triangle such that $|AB| = 2|AC|$ and let $D$ be a point on the ray $CA$ such that $|CD| = 3|AC|$. Prove that the line from $C$ perpendicular to the line $BD$ bisects the segment $AB$. | [
"Denote the intersection of the line through $C$ perpendicular to $BD$ and the line $BD$ by $E$ and the intersection of the lines $CE$ and $AB$ by $M$.\n\nWe have $|AD| = 2|AC| = |AB|$, so the triangle $DBA$ is isosceles with the apex at $A$, and so $\\angle BDA = \\angle ABD$.\n\nThis implies $\\angle ACM = 90^\\c... | Slovenia | Slovenija 2016 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0l7t | A piecewise linear periodic function is defined by
$$
f(x) = \begin{cases} x & \text{if } x \in [-1, 1), \\ 2-x & \text{if } x \in [1, 3), \end{cases}
$$
and $f(x+4) = f(x)$ for all real numbers $x$. The graph of $f(x)$ has the sawtooth pattern depicted below.
The parabola $x = 34y^2$ intersec... | [] | United States | 2025 AIME I | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | 259 | |
07gy | Is it possible to assign numbers $1, 2, \ldots, 8$ to the vertices of a cube in a way that the assigned number of each vertex divides the sum of numbers assigned to its neighbours? (Note that every number should be used once). | [
"The answer is no. First, for every $1 \\leq i \\leq 8$ we define $N_i$ to be the set of numbers assigned to the vertices adjacent to the vertex with number $i$. By the problem's assumption, the sum of elements of $N_i$ is divisible by $i$. Since the sum of every three numbers less than $8$ is at most $7+6+5=18$, t... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | No | |
07h3 | For a positive integer $n$, let $\tau(n)$ and $\sigma(n)$ be the number of positive divisors of $n$ and the sum of positive divisors of $n$, respectively. Let $a$ and $b$ be positive integers such that $\sigma(a^n)$ divides $\sigma(b^n)$ for all $n \in \mathbb{N}$. Prove that each prime factor of $\tau(a)$ divides $\ta... | [
"$$\na = p_1^{\\alpha_1} p_2^{\\alpha_2} p_3^{\\alpha_3} \\dots p_n^{\\alpha_n} \\quad \\text{and} \\quad b = q_1^{\\beta_1} q_2^{\\beta_2} q_3^{\\beta_3} \\dots q_m^{\\beta_m}\n$$\nWe have $\\tau(a) = (\\alpha_1 + 1) \\dots (\\alpha_n + 1)$ and $\\tau(b) = (\\beta_1 + 1) \\dots (\\beta_m + 1)$ so it suffices to pr... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and ... | English | proof only | null | |
0jsx | Problem:
I have five different pairs of socks. Every day for five days, I pick two socks at random without replacement to wear for the day. Find the probability that I wear matching socks on both the third day and the fifth day. | [
"Solution:\n\nI get a matching pair on the third day with probability $\\frac{1}{9}$ because there is a $\\frac{1}{9}$ probability of the second sock matching the first. Given that I already removed a matching pair on the third day, I get a matching pair on the fifth day with probability $\\frac{1}{7}$. We multiply... | United States | HMMT November 2016 | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Permutations"
] | null | proof and answer | 1/63 | |
032y | Problem:
One chooses a point in the interior of $\triangle ABC$ with area $1$ and connects it with the vertices of the triangle. Then one chooses a point in the interior of one of the three new triangles and connects it with its vertices, etc. At any step one chooses a point in the interior of one of the triangles obt... | [
"Solution:\n\na) At each step the triangle from which we choose a point is divided into three new triangles, i.e., the number of triangles increases by two. Hence at the $n$-th step we have $2n+1$ triangles.\n\nb) We shall prove by induction on $n$ that removing any triangle, there is a pairing of the remaining tri... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof only | null | |
0jak | Determine which integers $n > 1$ have the property that there exists an infinite sequence $a_1, a_2, a_3, \dots$ of nonzero integers such that the equality
$$
a_k + 2a_{2k} + \dots + n a_{nk} = 0
$$
holds for every positive integer $k$. | [
"We will show that the sequence exists for all $n \\ge 3$.\n\nFor $n = 2$, the sequence cannot exist. If it existed, we would have $a_k = -2a_{2k}$ for all $k$, from which $a_1 = (-2)^r a_{2r}$ for all $r$ by induction. Then $a_1$ would have to be divisible by $2^r$ for all $r$, which is impossible for $a_1 \\ne 0$... | United States | USAMO | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Other"
] | null | proof and answer | All integers n ≥ 3 | |
0gvv | a) Prove that the equality $\{x\} + \{2y\} = \{y\} + \{2x\}$ for real numbers $x$ and $y$ implies $\{x\} = \{y\}$.
b) Do there exist integers $n \ge 3$ such that the equality $\{x\} + \{ny\} = \{y\} + \{nx\}$ for real numbers $x$ and $y$ implies $\{x\} = \{y\}$?
(Here $\{a\} = a - [a]$, where $[a]$ stands for the gre... | [
"a) Suppose that for some real numbers $x$ and $y$ the equality $\\{x\\} + \\{2y\\} = \\{y\\} + \\{2x\\}$ holds. Let $\\{x\\} = \\alpha$, $\\{y\\} = \\beta$, $\\alpha, \\beta \\in [0;1)$. It suffices for $\\alpha, \\beta \\in [0;1)$ to consider the equality $\\alpha + \\{2\\beta\\} = \\beta + \\{2\\alpha\\}$.\n\nIf... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | a) {x} = {y}. b) No, such integers do not exist. | |
0jnl | Problem:
For integers $a, b, c, d$, let $f(a, b, c, d)$ denote the number of ordered pairs of integers $(x, y) \in \{1,2,3,4,5\}^{2}$ such that $a x + b y$ and $c x + d y$ are both divisible by $5$. Find the sum of all possible values of $f(a, b, c, d)$. | [
"Solution:\n\nAnswer: $31$\n\nStandard linear algebra over the field $\\mathbb{F}_{5}$ (the integers modulo $5$). The dimension of the solution set is at least $0$ and at most $2$, and any intermediate value can also be attained. So the answer is $1 + 5 + 5^{2} = 31$."
] | United States | HMMT February | [
"Algebra > Linear Algebra > Vectors",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 31 | |
0doq | The incircle of $\triangle ABC$ with center $I$ touches the sides $AB$ and $AC$ at points $P$ and $Q$, respectively. $BI$ and $CI$ intersect $PQ$ at $K$ and $L$, respectively. Prove that circumcircle of $\triangle ILK$ touches the incircle of $\triangle ABC$ if and only if $AB + AC = 3BC$. | [
"Let $BC = a$, $AC = b$, $AB = c$ and $\\angle CAB = \\alpha$, $\\angle ABC = \\beta$, $\\angle BCA = \\gamma$. Let $D$ be the intersection point of $BL$ and $CK$. Note that $\\triangle PAQ$ is isosceles.\n$$\n\\angle BKL = \\angle APK - \\angle ABK = \\frac{\\pi - \\alpha}{2} - \\frac{\\beta}{2} = \\frac{\\alpha +... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0j00 | Problem:
How many 8-digit numbers begin with $1$, end with $3$, and have the property that each successive digit is either one more or two more than the previous digit, considering $0$ to be one more than $9$? | [
"Solution:\n\nGiven an 8-digit number $a$ that satisfies the conditions in the problem, let $a_{i}$ denote the difference between its $(i+1)$th and $i$th digit. Since $a_{i} \\in \\{1,2\\}$ for all $1 \\leq i \\leq 7$, we have $7 \\leq a_{1}+a_{2}+\\cdots+a_{7} \\leq 14$.\n\nThe difference between the last digit an... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | final answer only | 21 | |
05z0 | Problem:
Soit $n \geqslant 3$ un entier. Chaque ligne d'un tableau $(n-2) \times n$ contient les nombres de 1 à $n$ en un exemplaire chacun, on suppose de plus que dans chaque colonne tous les nombres sont différents. Montrer que l'on peut compléter le tableau en un tableau $n \times n$ de telle sorte que dans chaque ... | [
"Solution:\n\nNotons $a_{i}, b_{i}$ les deux nombres qu'il manque dans la colonne $i$. Chacun des nombres de 1 à $n$ apparaît en deux exemplaires parmi les $a_{i}, b_{i}$. Construisons un multigraphe dont les sommets sont les nombres de 1 à $n$, et pour chaque $i=1, \\ldots, n$, on ajoute une arête entre $a_{i}$ et... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
0h4q | In January, Petro used to buy from one to three toy cars every day. On February 1, he tried to make a rectangle of all his cars. When he arranged them into rows of 7 cars, one car remained. When he arranged the cars into rows of 10, there were 2 excessive cars. Can Petro arrange them into rows of 4 cars?
**Answer:** y... | [
"For some positive integers $n$, $k$ we have:\n$$\n7k+1=10n+2 \\text{ or } 7k=10n+1.\n$$\nTherefore, we need to find a number between 29 and 92 that ends with 1 and is divisible by 7. The minimal such number is 21, which is less than 30. The next one is 91. It's easy to check that no other number with these propert... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | English | proof and answer | yes | |
0fri | Sean $a$, $b$, $c$, $d$ números reales tales que
$$
a + b + c + d = 0 \quad \text{y} \quad a^2 + b^2 + c^2 + d^2 = 12.
$$
Halla el valor mínimo y el valor máximo que puede tomar el producto $abcd$, y determina para qué valores de $a$, $b$, $c$, $d$ se consiguen ese mínimo y ese máximo. | [
"De las condiciones del enunciado tenemos que no todos los números tienen el mismo signo. El producto $abcd$ tomará un valor positivo cuando dos números sean positivos y dos negativos, así que buscaremos el máximo suponiendo que $a$, $b > 0$ y $c$, $d < 0$. Notemos que\n$$\n2(ab + cd) \\leq a^2 + b^2 + c^2 + d^2 = ... | Spain | LVII Olimpiada Matemática Española Concurso Final Nacional | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Maximum abcd = 9, attained exactly when two numbers are square root of three and the other two are negative square root of three (any order). Minimum abcd = -3, attained exactly at permutations of (3, -1, -1, -1) (equivalently, (1, 1, 1, -3)). | |
0bpt | Problem:
a) Comparați numerele $A = \sqrt[3]{4} + \sqrt[3]{13}$ și $B = \sqrt[3]{9} + \sqrt[3]{8}$.
b) Demonstrați că $\log_{2} 3 + \log_{3} 4 \in (2, 3)$. | [] | Romania | Olimpiada Națională de Matematică - Etapa Locală | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | a) A < B.
b) log_2 3 + log_3 4 ∈ (2, 3). | |
0alm | Problem:
How many solutions has $\sin 2 \theta - \cos 2 \theta = \sqrt{6} / 2$ in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$?
(a) 1
(b) 2
(c) 3
(d) 4 | [] | Philippines | QUALIFYING STAGE | [
"Precalculus > Trigonometric functions"
] | null | MCQ | b | |
03do | Equilateral triangle of area $n^2$ is partitioned into $n^2$ small triangles with unit area with lines parallel to its sides. The vertices of all small triangles are called *knots*. Find the sum of the areas of all equilateral triangles with vertices knots as a polynomial of $n$ and write this polynomial as a product o... | [
"Answer: $\\frac{n(n+1)(n+2)(n+3)(n^2+3n+6)}{240}$.\n\nAny equilateral triangle $T$ under consideration can be embedded into an unique equilateral triangle $M$, such that the vertices of $T$ lie on the sides of $M$ and those sides are parallel to the sides of the bigger triangle. If $M$ has $k$ times bigger side th... | Bulgaria | Bulgaria 2022 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | n(n+1)(n+2)(n+3)(n^2+3n+6)/240 | |
05yc | Problem:
Soit $n \geqslant 1$ un entier. Bosphore a écrit $n$ fois le nombre $2$ au tableau. Il effectue ensuite, $n-1$ fois d'affilée, l'opération suivante : il choisit deux nombres écrits au tableau, qu'il appelle $a$ et $b$, puis les efface et écrit le nombre $\sqrt{(ab+1)/2}$ à la place. Enfin, il appelle $x$ le n... | [
"Solution:\n\nDans la suite, notons $x_{n}$ le plus petit réel $x$ auquel peut aboutir Bosphore en partant de $n$ entiers égaux à $2$, et posons $y_{n}=\\sqrt{(n+3)/n}$. Il s'agit de démontrer que $x_{n} \\geqslant y_{n}$, avec égalité une infinité de fois et inégalité stricte une infinité de fois. Une étude des pe... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
04ej | Prove that the number $\cos \frac{\pi}{2 \cdot 3^n}$ is irrational for every positive integer $n$.
(A real number is irrational if it cannot be expressed as a ratio of two integers.) | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0bd3 | In triangle $ABC$, the bisector $AD$ and the median $BE$ meet at $P$. The straight lines $AB$ and $CP$ meet at $F$. The parallel through $B$ at $CF$ meets the straight line $DF$ at $M$. Show that $DM = BF$.
Gheorghe Bumbăcea | [
"Ceva's Theorem implies\n$$\n\\frac{BF}{FA} \\cdot \\frac{AE}{EC} \\cdot \\frac{CD}{DB} = 1,\n$$\nhence $\\frac{BF}{FA} = \\frac{BD}{DC}$. The converse of Thales' Theorem yields $DF \\parallel AC$. From $\\triangle BFD \\sim \\triangle BAC$ follows $\\frac{BF}{BA} = \\frac{FD}{AC}$, whence $BF = \\frac{AB}{AC} \\cd... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0i18 | Problem:
Let $a_{1}, a_{2}, \ldots, a_{2000}$ be real numbers in the interval $[0,1]$. Find the maximum possible value of
$$
\sum_{1 \leq i<j \leq 2000}(j-i)\left|a_{j}-a_{i}\right|
$$ | [
"Solution:\nThe answer is $1,000,000,000$. First, note that the desired sum $S$ is convex as a function of each $a_{i}$. (Indeed, $\\left|a_{i}-a\\right|$ is convex for any real number $a$, and a sum of convex functions is itself convex.) Consequently, it attains a maximum at some point where each variable lies at ... | United States | Berkeley Math Circle Monthly Contest #1 | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1000000000 | |
0co4 | Given a nonisosceles acute-angled triangle $ABC$. Let $O$ be its circumcenter, let $AD$ be its altitude, and let $M$ be the midpoint of side $BC$. The lines passing through $O$, perpendicular to $AB$ and $AC$, intersect $AD$ at points $P$ and $Q$, respectively. Let $S$ be the circumcenter of triangle $OPQ$. Prove that ... | [
"Пусть для определённости $AB > AC$ (см. рис. 19). Обозначим через $L$ середину отрезка $AB$. Заметим, что $\\angle AOL = \\frac{1}{2} \\angle AOB = \\angle ACB$. Отсюда $\\angle BAO = 90^\\circ - \\angle AOL = 90^\\circ - \\angle ACB = \\angle CAD$.\n\nРис. 19\nСтороны треугольника $OPQ$ п... | Russia | Final round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chas... | English; Russian | proof only | null | |
0apg | Problem:
In how many ways can the letters of the word SPECIAL be permuted if the vowels are to appear in alphabetical order? | [
"Solution:\n\nWe first arrange the letters without restrictions. There are $7!$ such arrangements. There are $3!$ ways to arrange the vowels into three particular positions, but only one of these is where the vowels are arranged in alphabetical order. Thus, the desired number of arrangements is $7! \\div 3! = 4 \\c... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | final answer only | 840 | |
02gf | Initially, a calculator displays number $1$. An operation consists in pressing either key $\sin$ or key $\cos$, which calculates respectively the sine and cosine with the arguments in radians. After performing $2001$ operations, what is the greatest possible value that can be achieved? | [
"Obviously the value does not exceed $1$. We cannot get close to $1$ with $\\sin$. So the final press must be $\\cos$ and we want the previous value to be as close to $0$ as possible. We cannot get close to $0$ with $\\cos$, so the $2000$th press must be $\\sin$. Thus we want the previous value to be as close to $0... | Brazil | XXIII OBM | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | cos(sin(sin(…sin(1)…))) with 2000 applications of sin followed by cos | |
05ao | A number is called a *palindrome* if reversing its digits results in the exact same number.
Given an integer $n > 1$, how many $n$-digit natural numbers are there such that when added to the number obtained by reversing the order of its digits, the result is a palindrome? | [
"Let $\\overline{a_1a_2...a_n}$ be an $n$-digit number such that the sum of the number with the number obtained by reversing its digits is a palindrome. We consider two cases separately.\n\n* If the sum is an $n$-digit number, then obviously $a_1 + a_n < 10$. We show that no carry occurs when adding the numbers $\\... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | If n is even: 36 * 55^{(n-2)/2} + 8 * 9^{(n-2)/2}. If n is odd: 36 * 55^{(n-3)/2} * 5 + 8 * 9^{(n-3)/2}. | |
0257 | Problem:
Na figura, $O$ é o centro do semicírculo de diâmetro $PQ$, $R$ é um ponto sobre o semicírculo e $RM$ é perpendicular a $PQ$. Se a medida do arco $\widehat{PR}$ é o dobro da medida do arco $\overparen{RQ}$, qual é a razão entre $PM$ e $MQ$?
 | [
"Solution:\n\n$180^\\circ = 2 R\\widehat{O}Q + R\\widehat{O}Q = 3 R\\widehat{O}Q$\n\ndonde $R\\widehat{O}Q = 60^\\circ$. Mas, $OR = OQ$ é o raio do círculo, de modo que o triângulo $\\triangle ORQ$ é equilátero. Assim, sua altura $RM$ também é a mediana, ou seja, $OM = MQ$. Se $r$ é o raio do círculo, então $OM = M... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 3 | |
0fvf | Problem:
Die Ebene wird in Einheitsquadrate unterteilt. Jedes Feld soll mit einer von $n$ Farben gefärbt werden, sodass gilt: Können vier Felder mit einem L-Tetromino bedeckt werden, dann haben diese Felder vier verschiedene Farben (das L-Tetromino darf gedreht und gespiegelt werden). Bestimme den kleinsten Wert von $... | [
"Solution:\n\nJe zwei der sieben Felder innerhalb der Figur links in Abbildung 1 können gleichzeitig mit einem L-Tetromino bedeckt werden. Diese Felder haben also verschiedene Farben, insbesondere ist $n \\geq 7$. Nehme an, $n=7$ sei möglich und färbe die Felder im Gebiet wie in Abbildung 1. Die beiden Felder recht... | Switzerland | SMO Finalrunde | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 8 | |
06qg | Consider $2009$ cards, each having one gold side and one black side, lying in parallel on a long table. Initially all cards show their gold sides. Two players, standing by the same long side of the table, play a game with alternating moves. Each move consists of choosing a block of $50$ consecutive cards, the leftmost ... | [
"a.\nWe interpret a card showing black as the digit $0$ and a card showing gold as the digit $1$. Thus each position of the $2009$ cards, read from left to right, corresponds bijectively to a nonnegative integer written in binary notation of $2009$ digits, where leading zeros are allowed. Each move decreases this i... | IMO | IMO Problem Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | a. Yes. b. No; the second player wins. | |
01u6 | A positive integer is called *nice* if it is equal to the sum of the squares of its three distinct divisors. (A divisor may be equal to $1$ or to the number itself.)
a) Prove that any nice number is divisible by $3$.
b) Are there infinitely many nice numbers? | [
"a) Let $N$ be a nice number, i.e. $N = d_1^2 + d_2^2 + d_3^2$, where $d_1, d_2, d_3$ are distinct divisors of $N$. If some divisor of $N$ is divisible by $3$, then $N$ is divisible by $3$.\n\nSo we suppose that $d_1, d_2, d_3$ are not divisible by $3$. Then their squares are congruent to $1$ modulo $3$, i.e. $d_1^... | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | All nice numbers are divisible by 3; Yes, there are infinitely many nice numbers. | |
02of | Emerald wrote on the blackboard all the integers from $1$ to $2011$. Then she erased all the even numbers.
a. How many numbers were left on the board?
b. How many of the remaining numbers were written with only the digits $0$ and $1$? | [
"a. The erased numbers were $2 = 2 \\cdot 1$, $4 = 2 \\cdot 2$, $\\ldots$, $2010 = 2 \\cdot 1005$. So $2011 - 1005 = 1006$ numbers were left on the board.\n\nb. We can list the numbers: they are $1$, $11$, $101$, $111$, $1001$, $1011$, $1101$, $1111$, a total of $8$.\n\n**OR** we can argue that the number is of the... | Brazil | Brazilian Math Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | a) 1006; b) 8 | |
0aug | Problem:
Points $A$, $M$, $N$ and $B$ are collinear, in that order, and $AM = 4$, $MN = 2$, $NB = 3$. If point $C$ is not collinear with these four points, and $AC = 6$, prove that $CN$ bisects $\angle BCM$. | [
"Solution:\n\n\n\nSince $\\frac{CA}{AM} = \\frac{3}{2} = \\frac{BA}{AC}$ and $\\angle CAM = \\angle BAC$, then $\\triangle CAM \\sim \\triangle BAC$. Therefore,\n$$\n\\angle MCA = \\angle CBA.\n$$\nSince $AC = 6 = AN$, then $\\triangle CAN$ is isosceles. Therefore,\n$$\n\\angle ACN = \\angl... | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0432 | In convex quadrilateral $ABCD$, $\vec{BC} = 2\vec{AD}$. Point $P$ is on the plane of quadrilateral $ABCD$, satisfying $\vec{PA} + 2020\vec{PB} + 2020\vec{PC} = 2020\vec{PD} = \vec{0}$. Let $s$ and $t$ be the areas of quadrilateral $ABCD$ and $\triangle PAB$, respectively. Then the value of $\frac{t}{s}$ is ______. | [
"$$\n\\overrightarrow{PA} + \\overrightarrow{PC} = 2\\overrightarrow{PY}, \\quad \\overrightarrow{PB} + \\overrightarrow{PD} = 2\\overrightarrow{PX},\n$$\ncombining the given conditions, we know that $\\overrightarrow{PY} + 2020\\overrightarrow{PX} = \\overrightarrow{0}$. Hence, point $P$ lies on segment $XY$, and ... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 337/2021 | |
08a6 | Problem:
Consideriamo il polinomio $p(x)=\left(1+x^{3^{1}}\right)\left(1+x^{3^{2}}\right)\left(1+x^{3^{3}}\right)\left(1+x^{3^{4}}\right)\left(1+x^{3^{5}}\right)\left(1+x^{39}\right)$, e supponiamo di svolgere il prodotto, ottenendo quindi un'espressione del tipo $a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{402} x^{402}$, dove... | [
"Solution:\nLa risposta è $(\\mathbf{C})$. Osserviamo innanzitutto che i coefficienti non nulli sono al più $2^{6}=64$, a seconda che in ciascun fattore prendiamo il termine 1 o il termine in $x$. Tuttavia possiamo osservare che $x^{3} \\cdot x^{9} \\cdot x^{27}=x^{39}$, quindi ci sono termini nello svolgimento del... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | MCQ | C | |
0duc | Problem:
Naj bo $D$ nožišče višine na stranico $BC$ trikotnika $ABC$. Simetrala notranjega kota pri $C$ seka nasprotno stranico v točki $E$. Koliko meri kot $\Varangle EDB$, če je $\Varangle CEA = \frac{\pi}{4}$? | [
"Solution:\n\nOznačimo kote trikotnika z $\\alpha$, $\\beta$ in $\\gamma$ na običajen način. V trikotniku $AEC$ velja $\\alpha + \\frac{\\pi}{4} + \\frac{\\gamma}{2} = \\pi$, od koder z upoštevanjem $\\gamma = \\pi - \\alpha - \\beta$ izpeljemo $\\alpha = \\beta + \\frac{\\pi}{2}$.\n\nKer je $AD \\perp DB$, je $\\f... | Slovenia | 45. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | pi/4 | |
0e1o | Find all integers $n$, such that the equation $x^2 + nx + n + 5 = 0$ has only integer solutions. | [
"Let $x_1$ and $x_2$ be the integer solutions of this quadratic equation. We may assume that $x_1 \\le x_2$. By Viete's formulas we have $x_1 + x_2 = -n$ and $x_1x_2 = n + 5$. Adding both identities together we get $x_1x_2 + x_1 + x_2 = 5$, so\n\n$$(x_1 + 1)(x_2 + 1) = 6.$$\n\nSince $6 = 1 \\cdot 6 = 2 \\cdot 3 = (... | Slovenia | National Math Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | -5, -3, 7, 9 | |
0c8r | Let $k$ be a positive integer greater than $1$ and let $(a_n)_{n \ge 1}$ be a sequence of pairwise distinct positive integers. Show that the set $M = \{a_1^k, a_2^k, a_3^k, \dots\}$ does not contain an infinite arithmetic sequence. | [
"Assume, by way of contradiction, that $M$ contains the infinite arithmetic sequence $\\{b_1^k, b_2^k, b_3^k, \\dots\\}$, where $1 \\le b_1 < b_2 < b_3 < \\dots$.\nIf we denote by $r$ its common difference, we have $b_{n+1}^k - b_n^k = r$, for all $n \\ge 1$.\nObserve that\n$$\nr = b_{n+1}^k - b_n^k = (b_{n+1} - b_... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof only | null | |
00v4 | Let $ABC$ be an acute-angled triangle with $AB < AC$, orthocentre $H$, circumcircle $\Gamma$ and circumcentre $O$. Let $M$ be the midpoint of $BC$ and let $D$ be a point such that $ADOH$ is a parallelogram. Suppose that there exists a point $X$ on $\Gamma$ and on the opposite side of $DH$ to $A$ such that $\angle DXH +... | [
"Since $AH \\parallel DE$ we have:\n$$\n\\angle DEX = \\angle DEX = \\angle DEX = \\angle DEX\n$$\nSo $DXEH$ is cyclic - call this circle $\\omega$.\nIt's well-known that $AH = 2OM = ON$ so $NH = OA$ (which we'll use later) and $ON = AH = DO$ (as $ADOH$ is a parallelogram). This means $\\frac{OD}{ND} = \\frac{1}{2}... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous ... | English | proof only | null | |
0gi8 | 給定三角形 $ABC$。設 $BPCQ$ 為一平行四邊形 ($P$ 不在 $BC$ 上)。令 $U$ 為 $CA$ 與 $BP$ 的交點，$V$ 為 $AB$ 與 $CP$ 的交點，$X$ 為 $CA$ 與三角形 $ABQ$ 的外接圓異於 $A$ 的交點，$Y$ 為 $AB$ 與三角形 $ACQ$ 的外接圓異於 $A$ 的交點。證明 $\overline{BU} = \overline{CV}$ 若且唯若 $AQ, BX, CY$ 三線共點。
Given triangle $ABC$. Let $BPCQ$ be a parallelogram ($P$ is not on $... | [
"由根心定理，$AQ, BX, CY$ 共點若且唯若 $B, C, X, Y$ 共圓，透過 $\\angle BXC = \\angle BQA, \\angle BYC = \\angle AQC$，這又等價於 $QA$ 為 $\\angle BQC$ 的其中一條角平分線，也就是 $\\frac{\\sin \\angle BQA}{\\sin \\angle AQC} = \\pm 1$（注意到 $Q$ 不在 $BC$ 上$)$。\n\n另一方面，由角元西瓦定理，\n$$\n\\frac{\\sin \\angle BQA}{\\sin \\angle AQC} \\cdot \\frac{\\sin \\angle C... | Taiwan | IMO 2J, Mock Exam 1 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Misce... | Chinese; English | proof only | null | |
01nd | Determine the greatest possible value of $n$ that satisfies the following condition: for any choice of $n$ subsets $M_1, ..., M_n$ of the set $M = \{1, 2, ..., n\}$ satisfying the conditions
i) $i \in M_i$; and
ii) $i \in M_j \Leftrightarrow j \notin M_i$ for all $i \neq j$,
there exist $M_k$ and $M_l$ such that $M_k \... | [
"(Solution of A. Zhuk.) First, if $M = \\{1, 2, ..., 6\\}$, then due to the conditions i) and ii) we have $|M_1| + |M_2| + ... + |M_6| = 21$. It follows that $|M_k| \\ge 4$ for some index $k$. There is nothing to prove if $|M_k| = 6$. If $|M_k| = 5$, then there exists $l \\notin M_k$, and $M_k \\cup M_l = M$. If $|... | Belarus | Belorusija 2012 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 6 | |
0ada | The sides of the triangle are $a=3$, $b=4$ and $c=5$. Find if there is a point inside the triangle so that the distance to his sides is less than $1$. | [
"Let $ABC$ be the triangle such that $\\overline{BC}=a=3$, $\\overline{AC}=b=4$ and $\\overline{AB}=c=5$. Because $a^2 + b^2 = 3^2 + 4^2 = 5^2 = c^2$, we have that $ABC$ is a right triangle.\n\nLet $M$ be the point inside the triangle such that the distance to his sides is less than $1$. Let $K$, $L$ and $N$ be the... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | No, such a point does not exist. | |
06ac | The positive real numbers $\alpha, \beta, \gamma, \delta$ satisfy the equality:
$$
\alpha + \beta\gamma + \gamma\delta + \delta\beta + \frac{1}{\alpha\beta^2\gamma^2\delta^2} = 18.
$$
Find the maximal possible value of $\alpha$. | [
"Using the AM-GM inequality we get\n\n$$\n\\begin{align*}\n\\alpha + \\beta\\gamma + \\gamma\\delta + \\delta\\beta + \\frac{1}{\\alpha\\beta^2\\gamma^2\\delta^2} &= \\alpha + \\left( \\beta\\gamma + \\gamma\\delta + \\delta\\beta + \\frac{1}{\\alpha\\beta^2\\gamma^2\\delta^2} \\right) \\\\\n&\\ge \\alpha + 4\\sqrt... | Greece | 39th Hellenic Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 16 | |
042d | Find all positive integers $a$ such that $(2^n - n^2) \mid (a^n - n^a)$ for all positive integers $n \ge 5$. (posed by Yang Mingliang) | [
"First, we prove that $a$ is even. It follows from the given condition by choosing an even integer $n \\ge 6$.\n\nNext, we prove that $a$ has no odd prime factor. Suppose the contrary, let $p$ be an odd prime factor of $a$. If $p = 3$, let $n = 8$, then $2^n - n^2 = 192$ has a factor $3$, but $a^n - n^a$ is not div... | China | China Western Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | a = 2 or a = 4 | |
0j4h | Problem:
In how many ways can one fill a $4 \times 4$ grid with a $0$ or $1$ in each square such that the sum of the entries in each row, column, and long diagonal is even? | [
"Solution:\n\nAnswer: $256$\n\nFirst we name the elements of the square as follows:\n\n| $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ |\n| :--- | :--- | :--- | :--- |\n| $a_{21}$ | $a_{22}$ | $a_{23}$ | $a_{24}$ |\n| $a_{31}$ | $a_{32}$ | $a_{33}$ | $a_{34}$ |\n| $a_{41}$ | $a_{42}$ | $a_{43}$ | $a_{44}$ |\n\nWe claim... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | 256 | |
09yj | A positive integer $n$ is called *divisor primary* if for every positive divisor $d$ of $n$ at least one of the numbers $d-1$ and $d+1$ is prime. For example, $8$ is divisor primary, because its positive divisors $1$, $2$, $4$, and $8$ each differ by $1$ from a prime number ($2$, $3$, $5$, and $7$, respectively), while... | [
"Suppose $n$ is divisor primary. Then $n$ cannot have an odd divisor $d \\ge 5$. Indeed, for such a divisor, both $d-1$ and $d+1$ are even. Because $d-1 > 2$, these are both composite numbers and that would contradict the fact that $n$ is divisor primary. The odd divisors $1$ and $3$ can occur, because the integer ... | Netherlands | Final Round | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 96 | |
050l | a) Find all positive integers $n$, such that the sum of all integers from $1$ to $n + 1$ can be represented as the sum of $n$ consecutive integers.
b) Find all positive integers $n$, for which there exists an integer $a$, such that the sum of the integers from $a$ to $a + n$ is equal to the sum of the integers from $a... | [
"a)\nClearly the sum of the first two positive integers can be represented as the sum of one positive integer. Now, let $n \\ge 2$ and let us show that the sum of the $n + 1$ first positive integers cannot be represented as a sum of $n$ consecutive integers. Indeed, on one hand $1 + 2 + \\dots + n + (n+1) > 2 + 3 +... | Estonia | Estonian Math Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | a) n = 1. b) All positive integers n work; a valid choice is a = n^2. | |
0d0c | Find all positive integers $n$ such that $5^{n^2} + 7$ is divisible by $6$. | [
"We have $5^{n^2} + 7 = (6 - 1)^{n^2} + 7 \\equiv (-1)^{n^2} + 1 \\pmod{6}$. It follows that $5^{n^2} + 7$ is divisible by $6$ if and only if $(-1)^{n^2} + 1 = 0$. This is equivalent to $n^2$ being odd. Therefore, the possible values of $n$ are all odd positive integers."
] | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Number Theory > Modular Arithmetic"
] | English | proof and answer | All odd positive integers | |
020g | Problem:
An arithmetic progression is a set of the form $\{a, a+d, \ldots, a+k d\}$, where $a, d, k$ are positive integers and $k \geqslant 2$. Thus an arithmetic progression has at least three elements and the successive elements have difference $d$, called the common difference of the arithmetic progression.
Let $n$... | [
"Solution:\n\nThe maximum value is $n^{2}$, which is attained for the partition into $n$ arithmetic progressions $\\{1, n+1,2 n+1\\}, \\ldots,\\{n, 2 n, 3 n\\}$, each of difference $n$.\n\nSuppose indeed that the set has been partitioned into $N$ progressions, of respective lengths $\\ell_{i}$, and differences $d_{... | Benelux Mathematical Olympiad | BxMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | n^2 | |
0f51 | Problem:
$M$ is a point inside a regular tetrahedron. Show that we can find two vertices $A$, $B$ of the tetrahedron such that $\cos \angle AMB \leq -1/3$. | [] | Soviet Union | 16th ASU | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0jv7 | Problem:
Let $ABC$ be a triangle with $AB=5$, $BC=6$, and $AC=7$. Let its orthocenter be $H$ and the feet of the altitudes from $A$, $B$, $C$ to the opposite sides be $D$, $E$, $F$ respectively. Let the line $DF$ intersect the circumcircle of $AHF$ again at $X$. Find the length of $EX$. | [
"Solution:\nSince $\\angle AFH=\\angle AEH=90^\\circ$, $E$ is on the circumcircle of $AHF$. So $\\angle XEH=\\angle HFD=\\angle HBD$, which implies that $XE \\parallel BD$. Hence $\\frac{EX}{BD}=\\frac{EY}{YB}$. Let $DF$ and $BE$ intersect at $Y$. Note that $\\angle EDY=180^\\circ-\\angle BDF-\\angle CDE=180^\\circ... | United States | HMMT November 2016 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > D... | null | proof and answer | 190/49 | |
0jwn | Problem:
Five equally skilled tennis players named Allen, Bob, Catheryn, David, and Evan play in a round robin tournament, such that each pair of people play exactly once, and there are no ties. In each of the ten games, the two players both have a $50\%$ chance of winning, and the results of the games are independent... | [
"Solution:\n\nWe make the following claim: if there is a 5-cycle (a directed cycle involving 5 players) in the tournament, then there is a 4-cycle.\n\nProof: Assume that $A$ beats $B$, $B$ beats $C$, $C$ beats $D$, $D$ beats $E$ and $E$ beats $A$. If $A$ beats $C$ then $A, C, D, E$ forms a 4-cycle, and similar if $... | United States | HMMT November 2017 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 49/64 | |
097a | Problem:
Să se afle valorile reale $u$ și $v$ ce verifică egalitatea
$$
\left(u^{2020}-u^{2019}\right)+\left(v^{2020}-v^{2019}\right)=u \ln u+v \ln v .
$$ | [
"Solution:\nConform domeniului de valori ale egalității din enunț, trebuie să avem $u>0$ și $v>0$. Fixăm $u>0$ și $v>0$; fie funcția $f: \\square \\rightarrow \\square$, $f(x)=u^{x}+v^{x}$. Această funcție este derivabilă de două ori pe $\\square$. Calculând derivatele de ordin 1 și 2, obținem $f'(x)=u^{x} \\ln u+v... | Moldova | Olimpiada Republicană la Matematică | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | null | proof and answer | u = v = 1 | |
0bpo | a) Prove that $2^n + 3^n + 5^n + 8^n$ is not a perfect square for any positive integer $n$.
b) Find all positive integers $n$ so that $1^n + 4^n + 6^n + 7^n = 2^n + 3^n + 5^n + 8^n$. | [] | Romania | 67th NMO Shortlisted Problems | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | n = 1 and 2 | |
084h | Problem:
Sia $ABC$ un triangolo rettangolo in $A$, con $AB > AC$; sia $AH$ l'altezza relativa all'ipotenusa. Sulla retta $BC$ si prenda $D$ tale che $H$ sia punto medio di $BD$; sia poi $E$ il piede della perpendicolare condotta da $C$ ad $AD$. Dimostrare che $EH = AH$. | [
"Solution:\n\nIl triangolo $ABD$ è isoscele su $BD$, perché $AH$ è altezza e mediana. $AH$ è pertanto anche bisettrice dell'angolo $B \\widehat{A} D$, e quindi gli angoli $D \\widehat{A} H$, $B \\widehat{A} H$ sono uguali.\n\nGli angoli $A \\widehat{E} C$, $A \\widehat{H} C$ sono retti per costruzione; quindi $E$ e... | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08c1 | Problem:
Data una circonferenza $\omega$ di diametro $A B$ e $P$ un punto interno al segmento $A B$, sia $M$ il punto medio di $P B$. Siano $r, s$ due rette parallele passanti rispettivamente per $M, P$, non coincidenti con la retta $A B$ né ad essa ortogonali. Sia poi $H$ la proiezione ortogonale di $A$ su $s$ e sia ... | [
"Solution:\n\na.\nEssendo $A B$ un diametro di $\\omega$ l'angolo insistente sull'arco corrispondente è retto e dunque $\\widehat{A K B} = 90^{\\circ} = \\widehat{A H P}$; si conclude che $B K$ è parallelo a $r$ e a $s$. Allora si può applicare il teorema di Talete a queste tre parallele e alle trasversali $A B, A ... | Italy | Gara di Febbraio | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0603 | Problem:
On considère 51 entiers strictement positifs de somme 100 sur une ligne. Montrer que pour tout entier $1 \leqslant k < 100$, il existe des entiers consécutifs de somme $k$ ou $100-k$. | [
"Solution:\n\nOn reformule légèrement l'énoncé pour le rendre plus visuel : on se représente un cercle de périmètre 100 gradué (de 0 à 99 par exemple) sur lequel on a marqué en noir 51 graduations et en pointillés les 49 autres. Sans perte de généralité on marque en noir la graduation 0. Nos entiers positifs corres... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0ffk | Problem:
Sean $a$, $b$, $c$ enteros positivos, dos a dos primos entre sí. Demostrar que $2abc - ab - bc - ca$ es el mayor entero que no puede expresarse en la forma $x\,bc + y\,ca + z\,ab$, donde $x$, $y$, y $z$ son enteros no negativos. | [] | Spain | International Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
03uj | Let $n$ be a positive integer, set $A \subseteq \{1, 2, \dots, n\}$, and for every $a, b \in A$, $\text{lcm}(a, b) \le n$. Prove that
$$
|A| \le 1.9\sqrt{n} + 5.
$$ | [
"**Proof** For $a \\in (\\sqrt{n}, \\sqrt{2n}]$, $\\text{lcm}(a, a+1) = a(a+1) > n$, so $|A \\cap (\\sqrt{n}, \\sqrt{2n}]| \\le \\frac{1}{2}(\\sqrt{2}-1)\\sqrt{n} + 1$.\nFor $a \\in (\\sqrt{2n}, \\sqrt{3n}]$, we have\n$$\n\\text{lcm}(a, a+1) = a(a+1) > n,\n$$\n$$\n\\text{lcm}(a+1, a+2) = (a+1)(a+2) > n,\n$$\n$$\n\\... | China | China National Team Selection Test | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0jmm | Define a function $f : \mathbb{N} \to \mathbb{N}$ by $f(1) = 1$, $f(n + 1) = f(n) + 2^{f(n)}$ for every positive integer $n$. Prove that $f(1), f(2), \dots, f(3^{2013})$ leave distinct remainders when divided by $3^{2013}$.
(This problem was suggested by Evan O'Dorney.) | [
"We prove the following stronger statement: For any $k \\ge 0$ and $a \\ge 1$, the values $f(a), f(a+1), \\dots, f(a+3^k - 1)$ are all distinct modulo $3^k$; that is, these numbers form a complete set of residues modulo $3^k$. We will induct on $k$, the case $k=0$ being trivial.\n\nAssume the statement true for a g... | United States | IMO Team Selection Team Selection Test | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0hl5 | Problem:
A beaver walks from $(0,0)$ to $(4,4)$ in the plane, walking one unit in the positive $x$ direction or one unit in the positive $y$ direction at each step. Moreover, he never goes to a point $(x, y)$ with $y>x$. How many different paths can he walk? | [
"Solution:\n\n$C(4)=14$."
] | United States | null | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | null | final answer only | 14 | |
07po | Suppose a doubly infinite sequence of real numbers
$$
\dots, a_{-2}, a_{-1}, a_0, a_1, a_2, \dots
$$
has the following sub-Fibonacci property:
$$
a_{n+2} = \frac{a_n + a_{n+1}}{2}, \quad \text{for all integers } n.
$$
Show that if this sequence is bounded (i.e. if there exists a number $R$ such that $|a_n| \le R$ for a... | [
"For any $n \\in \\mathbb{Z}$ let $d_n = a_{n+1} - a_n$. Then, for $n \\in \\mathbb{Z}$,\n$$\n2d_{n+1} = 2a_{n+2} - 2a_{n+1} = (a_{n+1} + a_n) - 2a_{n+1} = a_n - a_{n+1} = -d_n.\n$$\nThis implies $d_n = (-2)^{-n}d_0$ for all $n \\in \\mathbb{Z}$.\nIf $d_0 = 0$, then $d_n = 0$ for all $n \\in \\mathbb{Z}$, hence $a_... | Ireland | Ireland | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0ird | Problem:
a. Write $1$ as a sum of $4$ distinct unit fractions.
b. Write $1$ as a sum of $5$ distinct unit fractions.
c. Show that, for any integer $k > 3$, $1$ can be decomposed into $k$ unit fractions. | [
"Solution:\n\na.\n$1 = \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{7} + \\frac{1}{42}$\n\nb.\n$1 = \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{7} + \\frac{1}{43} + \\frac{1}{43 \\cdot 42}$\n\nc.\nIf we can do it for $k$ fractions, simply replace the last one (say $\\frac{1}{n}$) with $\\frac{1}{n+1} + \\frac{1}{n(n+1)}$. T... | United States | Harvard-MIT November Tournament | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | a) 1 = 1/2 + 1/3 + 1/7 + 1/42
b) 1 = 1/2 + 1/3 + 1/7 + 1/43 + 1/(43*42)
c) For every integer k > 3, such a decomposition exists by repeatedly replacing a term 1/n with 1/(n+1) + 1/(n(n+1)) to increase the number of terms by one. | |
0kdq | Problem:
In quadrilateral $A B C D$, there exists a point $E$ on segment $A D$ such that $\frac{A E}{E D}=\frac{1}{9}$ and $\angle B E C$ is a right angle. Additionally, the area of triangle $C E D$ is 27 times more than the area of triangle $A E B$. If $\angle E B C=\angle E A B$, $\angle E C B=\angle E D C$, and $B ... | [
"Solution:\n\n\n\nExtend sides $A B$ and $C D$ to intersect at point $F$. The angle conditions yield $\\triangle B E C \\sim \\triangle A F D$, so $\\angle A F D=90^{\\circ}$. Therefore, since $\\angle B F C$ and $\\angle B E C$ are both right angles, quadrilateral $E B F C$ is cyclic and\n... | United States | HMMO 2020 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 320 | |
007x | Let $ABC$ be a triangle with $\angle C = 90^\circ$ and $AC = 1$. The median $AM$ intersects the incircle at points $P$ and $Q$ such that $AP = QM$. Find the length of $PQ$. | [
"One may assume $P$ between $A$ and $Q$. Let the incircle touch sides $BC$ and $CA$ at $U$ and $V$ respectively. By power of a point $AV^2 = AP \\cdot AQ$, $MU^2 = MQ \\cdot MP$. Also $AQ = MP$ as $AP = QM$, and so $AV^2 = MU^2$, $AV = MU$. On the other hand $CU = CV$ by equal tangents, hence $AC = AV + CV = MU + C... | Argentina | National Olympiad of Argentina | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasin... | English | proof and answer | √(2√5 − 4) |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.