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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0aye | Problem:
How many ordered quadruples $(a, b, c, d)$ of positive odd integers are there that satisfy the equation $a + b + c + 2d = 15$? | [
"Solution:\nUsing the substitution $(a, b, c, d) = (2a_1 + 1, 2b_1 + 1, 2c_1 + 1, 2d_1 + 1)$ where $a_1, b_1, c_1, d_1$ are nonnegative integers, the problem is equivalent to finding the number $N$ of nonnegative integer solutions of $2a_1 + 1 + 2b_1 + 1 + 2c_1 + 1 + 4d_1 + 2 = 15$ or $a_1 + b_1 + c_1 + 2d_1 = 5$. ... | Philippines | 20th Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 34 | |
05ix | Problem:
On dit qu'un nombre à 9 chiffres est intéressant si chaque chiffre de 1 à 9 y apparaît une unique fois, que les chiffres de 1 à 5 y apparaissent dans l'ordre mais pas les chiffres de 1 à 6, par exemple 189236457.
Combien y a-t-il de nombres intéressants? | [
"Solution:\n\nPour construire un nombre intéressant, on peut placer d'abord les chiffres de 1 à 5, puis intercaler le 6 quelque part, puis le 7, le 8 et le 9. L'ordre des chiffres de 1 à 5 est imposé. Ensuite, le 6 peut être placé n'importe où sauf après le 5 car les chiffres de 1 à 6 ne sont pas dans l'ordre. Il y... | France | OFM 2013-2014 Envoi 2 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2520 | |
0ij3 | Problem:
Find the area in the first quadrant bounded by the hyperbola $x^{2}-y^{2}=1$, the $x$-axis, and the line $3x=4y$. | [
"Solution:\n\nConvert to polar coordinates: the hyperbola becomes\n$$\n1 = r^{2} (\\cos^{2} \\theta - \\sin^{2} \\theta) = r^{2} \\cos(2\\theta)\n$$\nso, letting $\\alpha := \\arctan(3/4)$, the area is\n$$\nS := \\int_{0}^{\\alpha} \\frac{r^{2}}{2} d\\theta = \\frac{1}{2} \\int_{0}^{\\alpha} \\sec(2\\theta) d\\thet... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | ln(7)/4 | |
0jwq | Problem:
Find the sum of all real numbers $x$ for which
$$
\lfloor\lfloor\cdots\lfloor\lfloor\lfloor x\rfloor+x\rfloor+x\rfloor \cdots\rfloor+x\rfloor=2017 \text{ and }\{\cdots\{\{\{x\}+x\}+x\} \cdots\}+x\} = \frac{1}{2017}
$$
where there are $2017$ $x$'s in both equations. ($\lfloor x\rfloor$ is the integer part of $x... | [
"Solution:\nThe two equations are equivalent to $2017\\lfloor x\\rfloor=2017$ and $\\{2017 x\\}=\\frac{1}{2017}$, respectively. The first equation reduces to $\\lfloor x\\rfloor=1$, so we must have $x=1+r$ for some real $r$ satisfying $0 \\leq r<1$. From the second equation, we deduce that $\\{2017 x\\}=\\{2017+201... | United States | HMMT November | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 3025 1/2017 | |
0dph | Let $P(n)$ be the number of ways of partitioning a positive integer number $n$ as a sum of powers of $2$, where the order of summands is not important. For example, $P(5) = 4$, since $5 = 4+1 = 2+2+1 = 2+1+1+1 = 1+1+1+1+1$. Prove that for each positive integer $n$ the following identity holds:
$$
P(n) + (-1)^{a_1}P(n-1... | [] | Silk Road Mathematics Competition | XV Silk Road Mathematics Competition | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0dh6 | Find all positive integers $n$ such that
$$
3^{n-1} + 5^{n-1} \mid 3^n + 5^n.
$$ | [
"Let $d = 3^{n-1} + 5^{n-1}$. We want $d \\mid 3^n + 5^n$.\n\nNote that:\n$$\n3^n + 5^n = 3 \\cdot 3^{n-1} + 5 \\cdot 5^{n-1} = 3d - 2 \\cdot 3^{n-1} + 5d - 4 \\cdot 5^{n-1} = (3+5)d - 2 \\cdot 3^{n-1} - 4 \\cdot 5^{n-1} = 8d - (2 \\cdot 3^{n-1} + 4 \\cdot 5^{n-1})\n$$\n\nSo $d \\mid 3^n + 5^n$ if and only if $d \\... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 1 | |
07ua | Let $f : \mathbb{R} \to \mathbb{R}$ be the fractional part function, defined by $f(x) = x - \lfloor x \rfloor$, where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$. For instance, $f(5.4) = 0.4$ and $f(-5.4) = 0.6$.
Let $g : \mathbb{R} \to \mathbb{R}$ be defined by $g(x) = f(1 - 2022x)$. Find 202... | [
"The fractional part $f(x)$ of a number $x$ is characterised by\n$$\nf(x) \\in [0, 1) \\text{ and } x - f(x) \\in \\mathbb{Z}.\n$$\nThe equation $g(x) = x$, i.e. $f(1 - 2022x) = x$, is therefore equivalent to\n$$\nx \\in [0, 1) \\text{ and } (1 - 2022x) - x = 1 - 2023x \\in \\mathbb{Z}.\n$$\nIf $x = \\frac{m}{n}$ i... | Ireland | IRL_ABooklet | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | x = m/2023 for m = 0, 1, ..., 2022 | |
0bas | a) Let $z_1, z_2, z_3, z_4$ be distinct complex numbers of zero sum, having equal absolute values. Prove that the points of affixes $z_1, z_2, z_3, z_4$ are the vertices of a rectangle.
b) Let $x, y, z, t$ be real numbers such that $\sin x + \sin y + \sin z + \sin t = 0$ and $\cos x + \cos y + \cos z + \cos t = 0$. Pr... | [
"a) The equality $z_1 + z_2 + z_3 + z_4 = 0$ implies $\\bar{z}_1 + \\bar{z}_2 + \\bar{z}_3 + \\bar{z}_4 = 0$ and furthermore $\\frac{1}{z_1} + \\frac{1}{z_2} + \\frac{1}{z_3} + \\frac{1}{z_4} = 0$ (1), for $|z_1| = |z_2| = |z_3| = |z_4| \\ne 0$.\nSuppose $z_1 + z_2 = -z_3 - z_4 \\ne 0$. The relation (1) gives $z_1 ... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
0dp6 | Determine all polynomials $P(x)$ with real coefficients such that for any rational (number) $r$ the equation $P(x) = r$ has a rational solution. | [
"Answer: $P(x) = a x + b$, where $a \\neq 0$, $b$ are rational numbers.\n\nObviously, $P(x)$ is not constant, so $\\deg P > 0$.\n\nFirst of all we prove that the coefficients of $P(x)$ are rational numbers. Let $\\deg P = n$, consider $n+1$ distinct rational numbers $r_1, r_2, \\dots, r_{n+1}$ and rationals $x_1, x... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | English | proof and answer | P(x) = a x + b, where a and b are rational and a ≠ 0 | |
0bb1 | Let $VABC$ be a regular pyramid with the base $ABC$ having center $O$. Denote $I$ and $H$ the incenter and the orthocenter of the triangle $VBC$ and suppose that $AH = 3OI$. Find the measure of the angle made by a lateral edge of the pyramid with the plane of the base.
Mircea Fianu | [
"Denote $M$ the midpoint of the edge $[BC]$. Since triangle $VBC$ is isosceles with $VB = VC$, the points $V$, $H$, $I$ and $M$ are collinear.\n\nFrom $AC \\perp OB$ and $AC \\perp OV$ follows $AC \\perp (BOV)$, hence $VB \\perp AC$, which, together with $VB \\perp CH$ leads to $VB \\perp (ACH)$, whence $AH \\perp ... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"... | null | proof and answer | 30° | |
0ehy | Problem:
V konveksnem štirikotniku so zunanji koti v razmerju $7: 4: 6: 1$. V kakšnem razmerju so notranji koti tega štirikotnika pri istih ogliččih?
(A) $2: 5: 3: 8$
(B) $8: 3: 5: 2$
(C) $1: 6: 4: 7$
(D) $7: 4: 6: 1$
(E) $11: 14: 12: 17$ | [
"Solution:\n\nVemo, da je vsota zunanjih kotov v konveksnem večkotniku enaka $360^{\\circ}$. Zunanji koti torej merijo zapovrstjo $140^{\\circ}$, $80^{\\circ}$, $120^{\\circ}$ in $20^{\\circ}$. Notranji koti so zato v razmerju $40^{\\circ}: 100^{\\circ}: 60^{\\circ}: 160^{\\circ}$. Razmerje okrajšamo in dobimo $2: ... | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | A | |
0bdo | Find all pairs of real numbers $(a, b)$ such that the equality $|ax + by| + |bx + ay| = 2|x| + 2|y|$ holds for all reals $x$ and $y$. | [
"Set $x = y = 1$ to get $|a + b| = 2$, so $a + b \\in \\{-2, 2\\}$. For $x = 1$, $y = -1$ we get $|a - b| = 2$, hence $a - b \\in \\{-2, 2\\}$. Combining the two above leads to the pairs $(a, b) \\in \\{(2, 0), (0, 2), (-2, 0), (0, -2)\\}$, all of them verifying the given equality for any reals $x$ and $y$."
] | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (2, 0), (0, 2), (-2, 0), (0, -2) | |
0jd9 | Problem:
Arpon chooses a positive real number $k$. For each positive integer $n$, he places a marker at the point $(n, n k)$ in the $(x, y)$ plane. Suppose that two markers whose $x$ coordinates differ by $4$ have distance $31$. What is the distance between the markers at $(7,7 k)$ and $(19,19 k)$? | [
"Solution:\n\nThe difference of the $x$-coordinates of the markers is $12 = 3 \\cdot 4$. Thus, by similar triangles (where we draw right triangles whose legs are parallel to the axes and whose hypotenuses lie along the line $y = k x$), the distance between the markers is $3 \\cdot 31 = 93$."
] | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 93 | |
0gn1 | Show that
$$
\frac{27}{4}(x+y)(y+z)(z+x) \geq (\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^2 \geq 6\sqrt{3}
$$
for any positive real numbers $x$, $y$, $z$ with $xy + yz + zx = 1$. | [] | Turkey | Team Selection Examination for the International Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
086n | Problem:
Francesca e Giorgia fanno il seguente gioco. Su un tavolo ci sono inizialmente alcune colonne di monete. Ogni colonna contiene un certo numero di monete, che può eventualmente variare da colonna a colonna. A turno, ogni giocatrice fa una e una sola delle seguenti possibili mosse:
- sceglie una colonna contene... | [
"Solution:\n\nVerifichiamo innanzitutto che il gioco ha termine dopo un numero finito di mosse. Supponiamo dapprima che inizialmente ci sia una sola colonna di monete, con $2^{\\alpha} d$ monete dove $d$ è un numero dispari, e ragioniamo per induzione su $\\alpha$. Se $\\alpha=0$ il gioco si conclude dopo una sola ... | Italy | Cesenatico | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | a) Francesca.
b) Let A be the number of columns whose size is divisible by 4, B the number of columns whose size is divisible by 2 but not by 4, and δ = 1 if there is at least one column with an odd number of coins (otherwise δ = 0). Francesca has a winning strategy if and only if δ + A + δAB is odd. | |
0h08 | Cut a square into triangles in such a way, that each triangle has exactly 3 neighbor triangles (two triangles are neighbors if they share a line segment; if they share only one point they aren't). | [
"\nFig.04"
] | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0296 | Problem:
Num triângulo $ABC$, tomamos pontos $X, Y$ sobre os lados $AB, BC$, respectivamente. Se $AY$ e $CX$ se intersectam em $Z$ e
$$
AY = YC \text{ e } AB = ZC
$$
verifique que
$$
\angle CZY = \angle ABY
$$ | [
"Solution:\n\nSejam $l$ a mediatriz do segmento $AC$ e $B'$ o simétrico de $B$ com respeito a $l$. Então, como $A$ é o simétrico de $C$, os triângulos $ABY$ e $CB'Y$ são congruentes, donde\n$$\n\\begin{aligned}\n\\angle CB'Y & = \\angle ABY \\\\\nCB' & = AB\n\\end{aligned}\n$$\nAlém disso, ... | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
01xz | The numbers $S_1 = 2^2$, $S_2 = 2^4$, $\dots$, $S_n = 2^{2n}$ are given. A rectangle $OABC$ is constructed on the Cartesian plane according to these numbers. For this, starting from the point $O$ the points $A_1, A_2, \dots, A_n$ are consistently marked along the axis $Ox$, and the points $C_1, C_2, \dots, C_n$ are con... | [
"Answer: \n\na) The maximal possible value of area equals $\\frac{1}{3}(4^n - 1)(4^n + n - 1)$ and it is attained at sets $(1, 1, \\dots, 1, 2^{2n})$ and $(2^2, 2^4, \\dots, 2^{2n-2}, 1)$.\n\nb) The minimal possible value equals $4(2^n - 1)^2$ and it is attained in two cases: either the sets $x_i$ and $y_i$ equal $... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | a) Maximal area: (1/3)(4^n − 1)(4^n + n − 1). Achieved by (x_1, …, x_n) = (1, 1, …, 1, 2^{2n}) and (y_1, …, y_n) = (2^2, 2^4, …, 2^{2n−2}, 1).
b) Minimal area: 4(2^n − 1)^2. Achieved either by (x_1, …, x_n) = (2^2, 2^3, …, 2^{n+1}) and (y_1, …, y_n) = (2^0, 2^1, …, 2^{n−1}), or by both sets equal to (2^1, 2^2, …, 2^n)... | |
0bx6 | a) Determine all 4-tuples $(x_0, x_1, x_2, x_3)$ of pairwise distinct integers such that each $x_k$ is coprime to $x_{k+1}$ (indices are reduced modulo 4), and the cyclic sum $x_0/x_1 + x_1/x_2 + x_2/x_3 + x_3/x_0$ is an integer.
b) Show that there are infinitely many 5-tuples $(x_0, x_1, x_2, x_3, x_4)$ of pairwise d... | [
"a) The only 4-tuples satisfying the required conditions are\n$$\n(1, 2, -1, -2) \\quad \\text{and} \\quad (1, -2, -1, 2)\n$$\nalong with their cyclic permutations; the cyclic sum in question is $-3$ for the former and its cyclic permutations, and $3$ for the latter and its cyclic permutations.\nThe condition that ... | Romania | THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a) The only 4-tuples are (1, 2, -1, -2) and (1, -2, -1, 2), together with their cyclic permutations; the cyclic sum equals −3 for the former and 3 for the latter.
b) For every integer k greater than 1, the 5-tuple (1, −k, k+1, −1, −k^2 − k) satisfies the conditions; the cyclic sum equals −k^2 − 2k − 2. | |
034r | Problem:
Find all values of the real parameter $a$ such that the equation
$$
lg (a x+1)=\lg (x-1)+\lg (2-x)
$$
has exactly one solution. | [
"Solution:\nThe equation is equivalent to $x \\in (1,2)$ and $a x + 1 = (x-1)(2-x)$, which can be written as $x^2 + (a-3)x + 3 = 0$. Therefore we have to find the values of $a$ such that the equation\n$$\nf(x) = x^2 + (a-3)x + 3 = 0\n$$\nhas exactly one root in the interval $(1,2)$. This is possible exactly in the ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (-1, -\frac{1}{2}] \cup \{3 - 2\sqrt{3}\} | |
015n | Consider $n \ge 4$ persons at a party. Suppose that whenever $4 \le r \le n$, it is not possible to arrange any $r$ of these persons in a circle such that each of them knows the two others he stands next to. Prove that on the whole there are at most $\frac{3}{2}(n-1)$ acquaintances among these persons. | [
"It suffices to show that every graph $\\mathcal{G} = (V, E)$ with $|V| = n \\ge 1$ and $|E| > \\frac{3}{2}(n-1)$ contains for some $r \\in \\{4, 5, \\dots, n\\}$ a cycle $C_r$ possessing exactly $r$ vertices. If this was not the case, take a counterexample $\\mathcal{G} = (V, E)$ with $|V| + |E|$ minimum. Clearly,... | Baltic Way | Baltic Way SHL | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0cjf | The non-negative integers $a$, $b$, $c$ are such that the numbers
$$
m = \frac{5a + 6b + 7c + 6}{4a + 3b + 2c + 3} \quad \text{and} \quad n = \frac{a + 2b + 3c + 5}{3a + b + 2c + 5}
$$
are both integers.
a. Prove that $m \ge 2$.
b. Find $m$ and $n$. | [
"a. If $m \\le 1$, then $5a + 6b + 7c + 6 \\le 4a + 3b + 2c + 3$, that is $a + 3b + 5c + 3 \\le 0$ – impossible. So $m \\ge 2$.\n\nb. If $n \\ge 2$, then $a + 2b + 3c + 5 \\ge 6a + 2b + 4c + 10$, whence $0 \\ge 5a + c + 5$, false. Since $n > 0$, $n$ must be equal to 1.\nNow $n = 1$ yields $a + 2b + 3c + 5 = 3a + b ... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | m = 2, n = 1 | |
0jau | Problem:
$ABCD$ is a rectangle with $AB = 20$ and $BC = 3$. A circle with radius $5$, centered at the midpoint of $DC$, meets the rectangle at four points: $W$, $X$, $Y$, and $Z$. Find the area of quadrilateral $WXYZ$. | [
"Solution:\n\nAnswer: $27$\n\nSuppose that $X$ and $Y$ are located on $AB$ with $X$ closer to $A$ than $B$. Let $O$ be the center of the circle, and let $P$ be the midpoint of $AB$. We have $OP \\perp AB$ so $OPX$ and $OPY$ are right triangles with right angles at $P$. Because $OX = OY = 5$ and $OP = 3$, we have $X... | United States | HMMT November 2012 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 27 | |
09he | Let $a_1 < a_2 < \dots < a_{53}$ be positive integers such that the sum of any 27 integers is greater than the sum of the remaining 26 integers.
a) Find the minimum value of $a_1$.
b) Find all possible values of $a_2, \dots, a_{53}$, when $a_1$ is at the minimum value. | [
"Clearly, the condition on the sequence is equivalent to the condition\n$$\na_1 + a_2 + \\dots + a_{27} > a_{28} + \\dots + a_{53}. \\quad (*)\n$$\nWe have $a_{i+26} \\ge a_{i+25}+1 \\ge \\dots \\ge a_i+26$, for $1 \\le i \\le 27$. It implies that $a_{i+26}-a_i \\ge 26$. Thus from (*) and the above inequality, we h... | Mongolia | Mongolian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a1 = 677; and when a1 = 677, the remaining terms are consecutive: a2, a3, ..., a53 = n, n+1, ..., n+51 with n ≥ 678. | |
0a45 | Problem:
Voor een getal van vijf cijfers $n = abcde$ definiëren we de verdraaide som van $n$ als de waarde
$$bcdea + cdeab + deabc + eabcd.$$
De verdraaide som van $20253$ is bijvoorbeeld
$$05232 + 25320 + 53202 + 32025 = 113079.$$
Laat $m$ en $n$ twee vijfcijferige getallen zijn met dezelfde verdraaide som. Bewijs da... | [
"Solution:\n\nWe definiëren $S(n)$ als de som van de cijfers van $n$ en $T(n)$ als de verdraaide som van $n$. Als we de vijf termen van $T(n) + n$ bekijken, dan zien we dat elk cijfer van $n$ precies één keer op elk van de plekken (tienduizendtal, duizendtal, honderdtal, tiental, eenheid) staat. Dit betekent dat\n$... | Netherlands | Maarttoets | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0c5g | Let $n \in \mathbb{N}$, $n \ge 2$, and let $a_1, a_2, \ldots, a_n$, and $b_1, b_2, \ldots, b_n$ be positive real numbers such that $a_1 + a_2 + \ldots + a_n = b_1 + b_2 + \ldots + b_n = S$.
a) Prove that $\sum_{k=1}^{n} \frac{a_k^2}{a_k + b_k} \ge \frac{S}{2}$.
b) Prove that $\sum_{k=1}^{n} \frac{a_k^2}{a_k + b_k} = ... | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof only | null | |
08l0 | Problem:
Let $a$, $b$ be two co-prime positive integers. A number is called good if it can be written in the form $a x + b y$ for non-negative integers $x$, $y$. Define the function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ as $f(n) = n - n_{a} - n_{b}$, where $s_{t}$ represents the remainder of $s$ upon division by $t$. ... | [
"Solution:\nIf $n$ is good then $n = a x + b y$. Also $n_{a} = (b y)_{a}$ and $n_{b} = (a x)_{b}$ so\n$$\nf(n) = a x - (a x)_{b} + b y - (b y)_{a} = b y' + a x'\n$$\nis also good, thus the sequence contains only good numbers which are non-negative.\n\nNow we have to prove that if the sequence contains only non-nega... | JBMO | 2007 Shortlist JBMO | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0085 | The numbers $1, 2, \dots, 2010$ are written in a row. Two players take turns in writing $+$ or $\times$ between two consecutive numbers for as long as this is possible. The first player wins if the obtained algebraic sum is divisible by $3$; otherwise the second player wins. Find a winning strategy for one of the playe... | [
"The first player $A$ has a winning strategy.\n\nOne may reduce $1, 2, \\dots, 2010$ modulo $3$, so assume that the initial numbers are $1, 2, 0, \\dots, 1, 2, 0$ (670 triples $1, 2, 0$).\n\nLet $A$ divide the $2009$ gaps between numbers into one group of $2$ (a double), corresponding to the initial $1, 2, 0$, and ... | Argentina | National Olympiad of Argentina | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Other"
] | English | proof only | null | |
03s9 | There are $n$ new students. Suppose that there are two students who know each other in every three students and there are two students who do not know each other in every four students. Find the maximum value of $n$. (posed by Tang Lihua) | [
"The maximum value of $n$ is $8$.\nWhen $n=8$, the example shown in the diagram satisfies the requirements, where $A_1, A_2, \\dots, A_8$ represent $8$ students. The line segment between $A_i$ and $A_j$ means $A_i$ and $A_j$ know each other.\n\n\nNext, if $n$ students satisfy the conditions... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 8 | |
0dzj | Two players share a pile of coins, alternating in taking one coin off the pile and placing it on any empty square of a $2008 \times 2008$ chessboard. The player who puts the coin onto the board in such a way that together with three other coins already on the board it forms a rectangle with sides parallel to the edges ... | [
"The second player has the winning strategy. He or she can arrange the columns into 1004 pairs with columns 1 and 2 forming the first pair, columns 3 and 4 forming the second pair, and so on, with columns 2007 and 2008 forming the final pair. Whenever the first player puts a coin into one of the columns, the second... | Slovenia | Slovenija 2008 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | the second player | |
0g3k | Problem:
Let $\mathbb{N}$ be the set of positive integers. Let $f: \mathbb{N} \rightarrow \mathbb{N}$ be a function such that for every $n \in \mathbb{N}$
$$
f(n)-n<2021 \text{ and } \underbrace{f(f(\cdots f(f}_{f(n)}(n)) \cdots))=n
$$
Prove that $f(n)=n$ for infinitely many $n \in \mathbb{N}$. | [
"Solution:\nWe first prove that $f$ is surjective. This follows directly from the second condition since for all $a \\in \\mathbb{N}$ we have\n$$\nb=\\underbrace{f(f(\\cdots f(f}_{f(a)-1}(a)) \\cdots)) \\Longrightarrow f(b)=a\n$$\nNow pick any prime $p>2021$ and $n$ with $f(n)=p$. Now apply $f$ to the second condit... | Switzerland | Final round | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0cfr | The irrational numbers $x, y$ are such that the numbers $x^2 + 2024y$, $y^2 + 2024x$ and $xy$ are rational. Find $x + y$.
Mihaela Berindeanu | [] | Romania | 74th NMO Shortlisted Problems | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | 2024 | |
0h47 | Solve the equation $x + \sqrt{1-x} + 1 = \sqrt{x} + 3\sqrt{x - x^2}$. | [
"Областю допустимих значень рівняння є відрізок $[0; 1]$. Запишемо рівняння у вигляді\n$$\n(\\sqrt{x})^2 + \\sqrt{1-x} + (\\sqrt{x})^2 + (\\sqrt{1-x})^2 - \\sqrt{x} - 3\\sqrt{x}\\sqrt{1-x} = 0.\n$$\nРозкладаючи ліву частину рівняння на множники, одержуємо:\n$$\n(\\sqrt{x} - \\sqrt{1-x})(2\\sqrt{x} - \\sqrt{1-x} - 1... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Other"
] | English | proof and answer | 1/2, 16/25 | |
08f9 | Problem:
Sia $ABC$ un triangolo, sia $r$ la bisettrice interna dell'angolo acuto $\widehat{BAC}$ e siano $K$ la proiezione di $B$ su $r$, $L$ la proiezione di $K$ su $AB$ e $D$ il simmetrico di $B$ rispetto ad $L$. Chiamiamo infine $H$ il piede dell'altezza del triangolo $ABC$ uscente da $B$. Dimostrare che:
a. $BH =... | [
"Solution:\n\na.\nOsserviamo intanto che il quadrilatero $AHKB$ è inscrivibile in una circonferenza, in quanto gli angoli $\\widehat{AHB}$ e $\\widehat{AKB}$ sono uguali (in particolare, retti). Gli angoli alla circonferenza $\\widehat{KBH}$ e $\\widehat{KAH}$ insistono sul medesimo arco e sono quindi uguali. D'alt... | Italy | Italian Mathematical Olympiad - February Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangle... | null | proof only | null | |
0dg3 | Let $f(x) = \frac{9^x}{9^x+3}$. Evaluate the sum
$$
\sum_{k=0}^{2021} f\left(\frac{k}{2021}\right).
$$ | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 1011 | |
0kzq | Integers $a$ and $b$ are randomly chosen without replacement from the set of integers with absolute value not exceeding $10$. What is the probability that the polynomial $x^3 + a x^2 + b x + 6$ has $3$ distinct integer roots?
(A) $\frac{1}{240}$ (B) $\frac{1}{221}$ (C) $\frac{1}{105}$ (D) $\frac{1}{84}$ (E) $\... | [
"Let $r$, $s$, and $t$ be the roots of $x^3 + a x^2 + b x + 6$. Then\n$$\nx^3 + a x^2 + b x + 6 = (x - r)(x - s)(x - t) = x^3 - (r + s + t)x^2 + (rs + st + tr)x - rst,\n$$\nso $rst = -6$, $r + s + t = -a$, and $rs + st + tr = b$. The only triples of distinct integers that satisfy the first of these three equations ... | United States | 2024 AMC 12 B | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | C | |
0bgf | Problem:
Fie $f: \mathbb{R} \rightarrow \mathbb{R}$ o funcţie arbitrară şi $g: \mathbb{R} \rightarrow \mathbb{R}$ o funcţie de gradul al doilea, având proprietatea:
pentru orice numere reale $m$ şi $n$, ecuaţia $f(x)=m x+n$ are soluţii dacă şi numai dacă ecuaţia $g(x)=m x+n$ are soluţii.
Arătaţi că funcţiile $f$ şi $g$... | [
"Solution:\nObservăm că, dacă graficul funcţiei $h: \\mathbb{R} \\rightarrow \\mathbb{R}$, $h(x)=m x+n$ este tangent la graficul funcţiei $g: \\mathbb{R} \\rightarrow \\mathbb{R}$, $g(x)=a x^{2}+b x+c$, atunci ecuaţia $a x^{2}+b x+c=m x+n$, $a, b, c, m, n \\in \\mathbb{R}$, $a \\neq 0$ are discriminantul nul, deci ... | Romania | Olimpiada Naţională de Matematică Etapa Naţională | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0g1j | Problem:
Sei $n$ eine natürliche Zahl, sodass es genau 2017 verschiedene Paare natürlicher Zahlen $(a, b)$ gibt, welche die Gleichung
$$
\frac{1}{a}+\frac{1}{b}=\frac{1}{n}
$$
erfüllen. Zeige, dass $n$ eine Quadratzahl ist. | [
"Solution:\n\nBemerke, dass $a$ und $b$ beide grösser als $n$ sind. Wir multiplizieren die Gleichung mit $n a b$ und bekommen $b n + a n = a b$. Nach Addition von $n^{2}$ subtrahieren wir $b n + a n$ und können faktorisieren: $(a-n)(b-n) = n^{2}$. Da $a$ und $b$ grösser als $n$ sind, bekommen wir eine Bijektion von... | Switzerland | SMO Finalrunde | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0b39 | Problem:
In right triangle $A B C$, $\angle A C B = 90^{\circ}$ and $\tan A > \sqrt{2}$. $M$ is the midpoint of $A B$, $P$ is the foot of the altitude from $C$, and $N$ is the midpoint of $C P$. Line $A B$ meets the circumcircle of $C N B$ again at $Q$. $R$ lies on line $B C$ such that $Q R$ and $C P$ are parallel, $S... | [
"Solution:\n\nRefer to the figure shown below.\n\n\n\nSince $C P$ is the $C$-altitude of $A B C$, triangles $A C P$, $A B C$ and $C B P$ are similar, so $A P = \\frac{A C^{2}}{A B}$ and $B P = \\frac{B C^{2}}{A B}$. As $C N Q B$ is cyclic, we have $\\angle P N Q = \\angle C B P = \\angle A ... | Philippines | 23rd Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"... | null | proof only | null | |
0993 | Problem:
Numerele naturale nenule $n$ și $m$ satisfac relația $n < m$. Arătați că
$$
S = \frac{1}{n} + \frac{1}{n+1} + \cdots + \frac{1}{m}
$$
nu este număr natural. | [
"Solution:\n\nNotăm $M = \\{ n, n+1, n+2, \\ldots, m \\}$. Deoarece $m > n$, în mulțimea $M$ sunt numere pare (din două numere consecutive unul este par). Fie $p \\in M$ numărul care se împarte la cea mai mare putere a lui $2$, să o notăm cu $2^{\\alpha}$, $\\alpha \\in \\mathbb{N}^*$. Deci oricare ar fi numărul $x... | Moldova | Olimpiada Republicană la Matematică | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
00w1 | Find all polynomials $P(x)$ with integer coefficients such that there exists a positive integer $N$ such that for all positive integers $n > N$ we have $P(n) > 0$ and
$$
n + P(n) \mid n^{P(n)} + P(n)^n.
$$ | [
"The solutions are $P(n) = 1$ and $P(n) = 2^s n^k - n$ for $0 \\le s \\le 3$ and $k \\ge 1$, excluding $P(n) = 0$.\n\nFirstly, we denote $Q(n) = P(n) + n$. We then have $Q(n) > n$ for all $n > N$. Note that the condition can be rewritten as\n$$\nn^{Q(n)-n} + (Q(n) - n)^n \\equiv 0 \\pmod{Q(n)}\n$$\n$$\nn^{Q(n)-n} +... | Balkan Mathematical Olympiad | 42nd Balkan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operat... | English | proof and answer | P(x) = 1 or P(x) = 2^s x^k − x for integers s with 0 ≤ s ≤ 3 and k ≥ 1, excluding the case s = 0, k = 1 (i.e., excluding P(x) = 0). | |
0cdd | Let $ABCD$ be a convex quadrilateral and let $P$ be a point inside such that $\angle APB + \angle CPD = \angle APD + \angle BPC$, $\angle PAD + \angle PCD = \angle PAB + \angle PCB$ and $\angle PDC + \angle PBC = \angle PDA + \angle PBA$. Prove that the quadrilateral $ABCD$ is circumscriptible.
Flavian Georgescu | [
"Notice that $\\angle PDC + \\angle PBC = \\angle PDA + \\angle PBA$ implies $\\angle PDA + \\angle PBA = (B + D)/2$. In the same manner, $\\angle PAB + \\angle PCB = (A + C)/2$. Add the last equalities to obtain $180^\\circ = (A + B + C + D)/2 = (\\angle PBA + \\angle PAB) + \\angle PDA + \\angle PCB = (180^\\circ... | Romania | THE Sixteenth STARS OF MATHEMATICS Competition | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometr... | null | proof only | null | |
07qg | Ben considers year $N$ lucky for a person if $N$ is divisible by his/her age (as a whole number of years) on 1st January of that year, e.g. $2016$ is lucky for someone whose age is $2$ or $3$ or $4$, but not for someone whose age is $5$. Ben was surprised to discover that both $2015$ and $2016$ were lucky for himself a... | [
"Let $A$, $B$, and $C$ be the ages of Amy, Ben, and Chris as whole numbers on 1st January $2016$, so that\n$$\n30 \\le B + 15 \\le C \\le 99, \\quad 15 \\le A + 15 \\le B \\le 89, \\quad 0 \\le A \\le 79.\n$$\nSince $B$ is a factor of $2016$, it must equal one of the following numbers: $16$, $18$, $21$, $24$, $28$,... | Ireland | Irish Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 41 years and 1 day | |
0817 | Problem:
Nel registrare le dichiarazioni dei tre imputati ad un processo, il cancelliere è stato piuttosto trascurato, e dal verbale risulta quanto segue:
Carlo: il colpevole è ... ario.
Dario: il colpevole è Dario.
Mario: il colpevole è ....ario.
Sapendo che il colpevole ha mentito e almeno uno degli innocenti ha det... | [
"Solution:\n\nLa risposta è (D). Poiché il colpevole ha mentito non può essere Dario, che si è autoaccusato. Poiché uno degli innocenti ha detto la verità, il colpevole finisce per ario, e dunque è Mario. Perciò Mario ha mentito, e dunque ha accusato Dario."
] | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Discrete Mathematics > Logic"
] | null | MCQ | D | |
0g6m | 令 $N$ 表示所有正整數所成的集合。試求所有的函數 $f : N \to N$ 使得對任意正整數 $n$,
$$
\frac{1}{f(1)f(2)} + \frac{1}{f(2)f(3)} + \dots + \frac{1}{f(n)f(n+1)} = \frac{f(f(n))}{f(n+1)}.
$$ | [
"由於\n$$\n\\frac{1}{1 \\cdot 2} + \\frac{1}{2 \\cdot 3} + \\dots + \\frac{1}{n(n+1)} = \\frac{n}{n+1},\n$$\n猜測 $f(n) = n$ 是滿足題設之唯一解。\n將 $n = 1$ 帶入，得 $f(f(1))f(1) = 1$，故 $f(1) = 1$。\n題設之 $n$ 以 $n+1$ 代入，得\n$$\n\\frac{f(f(n))}{f(n+1)} + \\frac{1}{f(n+1)f(n+2)} = \\frac{f(f(n+1))}{f(n+2)}.\n$$\n上式等價於\n$$\nf(f(n))f(n+2) ... | Taiwan | 二〇一二數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | f(n) = n for all n in N | |
0049 | Sea $n > 1$ un entero impar. Sean $P_0$ y $P_1$ dos vértices consecutivos de un polígono regular de $n$ lados. Para cada $k \ge 2$, se define $P_{k}$ como el vértice del polígono dado que se encuentra en la mediatriz de $P_{k-1}$ y $P_{k-2}$. Determine para qué valores de $n$ la sucesión $P_0, P_1, P_2, \dots$ recorre ... | [] | Argentina | 21º OLIMPIADA IBEROAMERICANA DE MATEMÁTICA | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | Español | proof and answer | All odd n that are powers of 3, i.e., n = 3^t with t ≥ 1 | |
0lc2 | Find all functions $f : \mathbb{R} \to \mathbb{R}$ that satisfy $f(0) = 0$, $f(1) = 2013$ and
$$
(x - y) (f(f(x)^2) - f(f(y)^2)) = (f(x) - f(y)) (f(x)^2 - f(y)^2)
$$
for all real numbers $x, y$. | [
"Given $x \\neq 0$ and $y = 0$, we get\n$$\nx f(f(x)^2) = f(x)^3\n$$\nwhich implies\n$$\nf(f(x)^2) = \\frac{f(x)^3}{x}, \\forall x \\neq 0.\n$$\nSubstituting into the original equation, for all $x, y \\neq 0$\n$$\n(x - y) \\left[ \\frac{f(x)^3}{x} - \\frac{f(y)^3}{y} \\right] = (f(x) - f(y))(f(x)^2 - f(y)^2) \\quad... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 2013x for all real x | |
0c68 | Let $ABC$ be an acute triangle such that $AB < AC$. Let $I$ be the incentre of the triangle $ABC$, and let the incircle touch the side $BC$ at $D$. The line $AD$ crosses the circle $ABC$ again at $E$. Let $M$ be the midpoint of the side $BC$, and let $N$ be the midpoint of the circular arc $BAC$. The line $EN$ crosses ... | [
"The internal bisectrix $AI$ and the perpendicular bisectrix $MN$ of the side $BC$ cross at the midpoint $K$ of the arc $BEC$. It is a fact that the circle $BIC$ is centred at $K$.\n\nLet now the line $MP$ cross the circle $BIC$ again at $Q$, to infer that the arcs $BL$ and $CQ$ of this circle have equal angular sp... | Romania | SELECTION TESTS FOR THE 2019 BMO AND IMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0a49 | Problem:
Zij $n \ge 2$ een geheel getal, en laat $z_1, \dots, z_n$ positieve gehele getallen zijn die voldoen aan:
- $z_j \le j$ voor $j = 1, \dots, n$;
- $z_1 + \dots + z_n$ is even.
Bewijs dat er $s_1, \dots, s_n \in \{-1, 1\}$ bestaan zodat:
\[
s_1 z_1 + s_2 z_2 + \dots + s_n z_n = 0.
\] | [
"Solution:\n\nOplossing I. We bewijzen dit met tweestapsinductie naar $n$. Voor $n = 2$ volgt uit de voorwaarden dat $z_1 = z_2 = 1$, en dan geldt dus dat $z_1 - z_2 = 0$. Voor $n = 3$ volgt uit de voorwaarde dat $(z_1, z_2, z_3)$ gelijk is aan $(1, 1, 2)$, $(1, 2, 1)$ of $(1, 2, 3)$. Daarvoor kiezen we respectieve... | Netherlands | Maarttoets | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0hty | Problem:
Show that $n$ divides $\varphi\left(a^{n}-1\right)$ for any integers $a$ and $n$, where $\varphi$ is Euler's totient function. | [
"Solution:\nLet $N = a^{n} - 1$. Then $\\gcd(a, N) = 1$ and the order of $a \\pmod{N}$ is exactly equal to $n$. But $a^{\\varphi(N)} \\equiv 1 \\pmod{N}$ too. Thus $n$ divides $\\varphi(N)$."
] | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof only | null | |
03z1 | Let $a_1, a_2, \dots, a_n$ ($n \ge 4$) be positive real numbers with $a_1 < a_2 < \dots < a_n$. For any positive real number $r$, the number of ternary groups $(i, j, k)$ satisfying $\frac{a_j - a_i}{a_k - a_j} = r$ ($1 \le i < j < k \le n$) is denoted as $f_n(r)$. Prove $f_n(r) < \frac{n^2}{4}$. | [
"Given $j$ ($1 < j < n$), the number of ternary groups $(i, j, k)$ satisfying $1 \\le i < j < k \\le n$ and\n$$\n\\frac{a_j - a_i}{a_k - a_j} = r \\qquad \\textcircled{1}\n$$\nis denoted as $g_j(r)$. For fixed $i, j$ with $i < j$, there is at most one $k$ satisfying **①**; so there are $j-1$ ways to choose $i$, whi... | China | China Mathematical Competition (Complementary Test) | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Other"
] | English | proof only | null | |
03sl | Suppose $f(x) = x^2 + a$. Define $f^1(x) = f(x)$, $f^n(x) = f(f^{n-1}(x))$, $n = 2, 3, \dots$, and let $M = \{a \in \mathbb{R} \mid |f^n(0)| \le 2$, for any $n \in \mathbb{N}\}$. Prove that $M = [-2, \frac{1}{4}]$. | [
"(a) When $a < -2$, then $|f^1(0)| = |a| > 2$, therefore $a \\notin M$.\n\n(b) When $-2 \\le a < 0$, we have $|f^1(0)| = |a| \\le 2$. Assume that $|f^{k-1}(0)| \\le |a| \\le 2$ for $k \\ge 2$. Since $a^2 \\le -2a$ for $-2 \\le a < 0$, we get that\n$$\n-2 \\le -|a| = a \\le f^k(0) = (f^{k-1}(0))^2 + a \\le a^2 + a \... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0h1d | For nonnegative real numbers $a, b, c$, prove that
$$
(a+1)^2 + (b+1)^2 + (c+1)^2 \ge 3(1 + a\sqrt{b} + b\sqrt{c} + c\sqrt{a}).
$$ | [
"By AM-GM:\n$$\n\\begin{cases} a + a^2 + b \\ge 3a\\sqrt{b}, \\\\ b + b^2 + c \\ge 3b\\sqrt{c}, \\\\ c + c^2 + a \\ge 3c\\sqrt{a}. \\end{cases}\n$$\nAdding them up we get what is required."
] | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0gra | For every positive integer $n$, let $\sigma(n)$ denote the number of positive divisors of $n$. Determine all positive integers $k$ such that there exist positive integers $a$ and $b$ satisfying
$$
k = \sigma(a) = \sigma(b) = \sigma(2a + 3b).
$$ | [
"Answer: All positive even integers.\nFor $k = 2m$ where $m \\in \\mathbb{Z}^+$, $a = 2 \\cdot 5^{m-1}$ and $b = 3 \\cdot 5^{m-1}$ satisfy the required equalities.\n\nNow suppose that for some odd positive integer $k$ there exist such $a$ and $b$. Since $\\sigma(n)$ is odd if and only if $n$ is a perfect square, th... | Turkey | 23rd Junior Turkish Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All positive even integers. | |
0kah | Problem:
An equiangular hexagon has side lengths $1,2,3,4,5,6$ in some order. Find the nonnegative difference between the largest and the smallest possible area of this hexagon. | [
"Solution:\n\nExtending three sides of the equiangular hexagon gives an equilateral triangle. Thus, if the sides are $a, b, c, d, e, f$, in order, then $a+b+c = a+f+e \\Rightarrow b+c = e+f \\Rightarrow f-c = b-e$. By a symmetric argument, we see that $d-a = f-c = b-e$ holds, which means that they must be separated... | United States | HMMT February 2019 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | sqrt(3)/2 | |
0jg7 | Problem:
Spencer is making burritos, each of which consists of one wrap and one filling. He has enough filling for up to four beef burritos and three chicken burritos. However, he only has five wraps for the burritos; in how many orders can he make exactly five burritos? | [
"Solution:\n\nSpencer's burrito-making can include either $3$, $2$, or $1$ chicken burrito; consequently, he has $\\binom{5}{3} + \\binom{5}{2} + \\binom{5}{1} = 25$ orders in which he can make burritos."
] | United States | HMMT 2013 | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 25 | |
0jqt | Problem:
Suppose $(a_{1}, a_{2}, a_{3}, a_{4})$ is a 4-term sequence of real numbers satisfying the following two conditions:
- $a_{3} = a_{2} + a_{1}$ and $a_{4} = a_{3} + a_{2}$;
- there exist real numbers $a, b, c$ such that
$$
a n^{2} + b n + c = \cos(a_{n})
$$
for all $n \in \{1,2,3,4\}$.
Compute the maximum poss... | [
"Solution:\n\nAnswer: $-9 + 3 \\sqrt{13}$\n\nLet $f(n) = \\cos a_{n}$ and $m = 1$. The second (\"quadratic interpolation\") condition on $f(m), f(m+1), f(m+2), f(m+3)$ is equivalent to having a vanishing third finite difference\n$$\nf(m+3) - 3 f(m+2) + 3 f(m+1) - f(m) = 0\n$$\nThis is equivalent to\n$$\n\\begin{ali... | United States | HMMT February 2015 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | -9 + 3 sqrt(13) | |
05q9 | Problem:
Les entiers $1, 2, \ldots, 2018$ sont écrits au tableau. On effectue alors 2017 opérations comme suit : choisir deux nombres $a$ et $b$, les effacer, et écrire $a + b + 2ab$ à la place. À la fin, il ne reste qu'un seul entier sur le tableau.
Quelles sont les valeurs que son chiffre des unités peut prendre? | [
"Solution:\n\nPuisqu'il y a 1009 nombres impairs entre 1 et 2018, on sait que la somme des entiers initialement inscrits au tableau est impaire. En outre, l'un de ces entiers est congru à $2 \\pmod{5}$. Une récurrence immédiate permet alors de montrer que, après chaque opération, la somme des entiers écrits sur le ... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | 7 | |
01bh | What is the least number of cells that must be marked on an $n \times n$ board such that there is no series of diagonal cells of length $> \frac{n}{2}$ without a mark? | [
"For any $n$ it is possible to set $n$ marks on the board and get the desired property, if they are simply put on every field on row $\\lfloor \\frac{n}{2} \\rfloor$. We now show that $n$ is also the minimum amount of marks needed.\n\nIf $n$ is odd there are $2n$ series of diagonal cells with length $> \\frac{n}{2}... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n | |
061y | Problem:
Sei $E=\{-1,0,1\}$ und $M$ eine Menge von Gitterpunkten der Ebene. Die Punkte aus $M$ sind so durch ein Streckennetz $(S)$ miteinander verbunden, dass man von jedem Punkt aus $M$ zu jedem anderen Punkt aus $M$ gelangen kann ohne $(S)$ zu verlassen. Dabei können die Strecken des Streckennetzes außer den Gitter... | [
"Solution:\n\nTeil a): Aus Symmetriegründen reicht es zunächst das günstigste Streckennetz für die Punkte $O(0,0)$, $A(1,0)$ und $C(0,1)$ zu finden. Durch Spiegelung an $O$ erhält man dann das gewünschte Netz für die Punkte aus $M$. Durch eine Drehung der Ebene $xOy$ um $O$ mit Drehwinkel $60^{\\circ}$ (Fig. 1) wir... | Germany | Auswahlwettbewerb zur IMO 2005 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | a) sqrt(2) + sqrt(6); b) 4 + 2*sqrt(3) | |
0012 | En la casa de Gabriel son muy metódicos. Todos los días hábiles la mamá sale en su moto a la misma hora, a la misma velocidad y por el mismo camino a buscar a Gabriel al colegio. Llega al colegio exactamente a las 12 horas y de inmediato regresa a su casa con Gabriel, por el mismo camino y a la misma velocidad. Por sup... | [] | Argentina | XIX Olimpíada Matemática Argentina | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | español | proof and answer | 12 | |
0b8n | A strip of breadth $\ell$ is the set of those points in the plane, situated between or on two parallel lines at distance $\ell$ one from the other.
Consider a finite set $S$ of $n \geq 3$ (distinct) points in the plane, with the property that any three points of $S$ can be covered by a strip of breadth 1. Prove the set... | [
"All solutions encountered in the competition (including the Official one) make use of the following\n\n**LEMMA.** A triangle can be covered by a strip of breadth 1 if and only if (at least) one of its altitudes is at most 1 long.\n\n*Proof.* The other implication being trivial, consider a triangle $ABC$ covered by... | Romania | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chas... | English | proof only | null | |
0h5h | Several people gathered at a party. Each two of them who don't know each other have exactly two common acquaintances (if a person $A$ knows a person $B$, then necessarily the person $B$ knows the person $A$). Mykhailyk and Vitalyk know each other, but have no common acquaintances.
a) Prove that Mykhailyk and Vitalyk h... | [
"a) Denote Mykhailyk and Vitalyk by $A$ and $B$ respectively. Let $M(A)$ and $M(B)$ be the sets of acquaintances of $A$ and $B$ respectively (see Fig. 3).\n\nThere is no person in $M(A)$ that knows $B$, since in this case $A$ and $B$ would have a common acquaintance. So every person $X$ from $M(A)$ has exactly two ... | Ukraine | Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory"
] | English | proof and answer | a) Mykhailyk and Vitalyk have the same number of acquaintances. b) Yes, such a situation can occur with exactly 6 people. | |
02mc | Problem:
Duas tabelas - As linhas da primeira tabela dada são todas progressões aritméticas de uma mesma razão e as colunas dessa tabela são todas progressões aritméticas de uma mesma razão. Na segunda tabela dada foi utilizada a mesma lei de formação, mas alguém apagou alguns números deixando apenas três. Qual é o nú... | [
"Solution:\n\nVemos que na primeira tabela cada linha é uma progressão aritmética de razão $3$ e cada coluna é uma progressão aritmética de razão $7$.\n\n| 5 | 8 | 11 | 14 | 17 |\n|:---:|:---:|:---:|:---:|:---:|\n| 12 | 15 | 18 | 21 | 24 |\n| 19 | 22 | 25 | 28 | 31 |\n| 26 | 29 | 32 | 35 | 38 |\n| 33 | 36 | 39 | ... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | final answer only | 102 | |
06c3 | $PQRS$ is a cyclic quadrilateral with $\angle PSR = 90^\circ$. $H$, $K$ are the feet of the perpendiculars from $Q$ to $PR$, $PS$ (suitably extended if necessary) respectively. Show that $HK$ bisects $QS$. | [
"Let $HK$ meet $QS$ at $A$. Since $\\angle PHQ = \\angle PKQ = 90^\\circ$, the points $P$, $Q$, $H$, $K$ are concyclic. As both $QK$ and $RS$ are perpendicular to $PS$, they are parallel. Therefore, we have\n$$\n\\angle QKA = \\angle QKH = \\angle QPH = \\angle QPR = \\angle QSR = \\angle SQK = \\angle AQK.\n$$\nTh... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03e2 | A quadruplet of distinct positive integers $(a, b, c, d)$ is called $k$-good if the following conditions hold:
1. Among $a, b, c, d$, no three form an arithmetic progression.
2. Among $a + b, a + c, a + d, b + c, b + d, c + d$, there are $k$ of them, forming an arithmetic progression.
a) Find a 4-good quadruplet.
b) ... | [
"a) The quadruple $(7, 6, 4, 3)$ is nice because there are no three numbers in it to form an arithmetic progression, but of the six numbers\n$$\n7 + 6 = 13,\\ 7 + 4 = 11,\\ 7 + 3 = 10,\\ 6 + 4 = 10,\\ 6 + 3 = 9,\\ 4 + 3 = 7\n$$\nthe numbers $7, 9, 11, 13$ form an arithmetic progression.\n\nb) Without restriction le... | Bulgaria | Autumn tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | English | proof and answer | Example: (7, 6, 4, 3). The maximal k is 4. | |
0ftb | Problem:
Gegeben sind 24 Punkte im Raum. Je drei dieser Punkte spannen eine Ebene auf, und es ist bekannt, dass die 24 Punkte auf diese Weise genau 2002 verschiedene Ebenen aufspannen. Beweise, dass eine dieser Ebenen mindestens 6 der Punkte enthält. | [
"Solution:\n\nNehme an, keine der Ebenen enthalte 6 oder mehr Punkte. Sei $m$ die Anzahl Ebenen mit genau 4 Punkten und $n$ die Anzahl Ebenen mit genau 5 Punkten. Wir zählen jetzt die Anzahl Ebenen auf zwei Arten:\n\na. Nach Vorraussetzung sind es genau 2002.\n\nb. Wir zählen die Ebenen via Punktetripel. Es gibt $\... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof only | null | |
0don | Let $0 < a < b < 1$ be real numbers and
$$
g(x) = \begin{cases} x + 1 - a, & \text{if } 0 < x < a, \\ b - a, & \text{if } x = a, \\ x - a, & \text{if } a < x < b, \\ 1 - a, & \text{if } x = b, \\ x - a, & \text{if } b < x < 1. \end{cases}
$$
Assume that for some positive integer $n$ there are $n+1$ real numbers $0 < x_... | [
"For real numbers $r$ and $s$ we call them equivalent (and write $r \\equiv s$), if $r - s \\in \\mathbb{Z}$. Then,\n$$\ng(x) \\equiv \\begin{cases} x - a, & \\text{if } x \\not\\equiv a, b, \\\\ b - a, & \\text{if } x \\equiv a, \\\\ -a, & \\text{if } x \\equiv b. \\end{cases}\n$$\nNote that $g$ is a bijection on ... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0afk | Нека $a, b, c$ се позитивни реални броеви за кои важи $ab + bc + ca = 1/3$. Да се докаже неравенството
$$
\frac{a}{a^2 - bc + 1} + \frac{b}{b^2 - ca + 1} + \frac{c}{c^2 - ab + 1} \ge \frac{1}{a + b + c}
$$ | [
"Да забележиме дека именителите во левиот израз се позитивни.\nСо примена на *Неравенството на Коши - Буњаковски* имаме\n$$\n\\begin{aligned}\n\\frac{a}{a^2 - bc + 1} + \\frac{b}{b^2 - ca + 1} + \\frac{c}{c^2 - ab + 1} &= \\frac{a^2}{a^3 - abc + a} + \\frac{b^2}{b^3 - abc + b} + \\frac{c^2}{c^3 - abc + c} \\\\ &\\g... | North Macedonia | XVI Македонска математичка олимпијада | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Macedonian, English | proof only | null | |
04l9 | A triangle $ABC$ is given, with altitudes $AD$, $BE$ and $CF$, and orthocentre $H$. The segments $EF$ and $AD$ intersect at point $G$. The segment $AK$ is the diameter of the circumcircle of triangle $ABC$ and it intersects $BC$ at point $M$. Prove that the lines $GM$ and $HK$ are parallel. (Go Geometry) | [
"\nSince $BFC$ is a right-angled triangle, we have $\\angle BCF = 90^\\circ - \\beta$. As $CDH$ is also a right-angled triangle, we get $\\angle DHC = \\beta$.\nIt follows that $\\angle AHF = \\beta$ (because the angles $\\angle DHC$ and $\\angle AHF$ are vertically opposite).\nAs $ABKC$ is... | Croatia | Mathematical competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dpl | In the triangle $ABC$, points $H, L, K$ are chosen on the sides $AB, BC, AC$, respectively, so that $CH \perp AB$, $HL \parallel AC$, $HK \parallel BC$. In the triangle $HBL$, let $P$ and $Q$ be feets of altitudes from $H$ and $B$, respectively. Prove that in the triangle $AKH$, the feets of altitudes from $A$ and $H$ ... | [
"Suppose $PQ$ intersects $AK$ and $HK$ at $R$ and $S$, respectively. From the fact that $P$ and $Q$ lie on a circle with diameter $HB$ and from $HL \\parallel AC$, it follows that $\\angle LPQ = \\angle BHQ = \\angle BAR$, i.e. the points $A, R, P$ and $B$ lie on the same circle. Therefore, $CP \\cdot CB = CR \\cdo... | Silk Road Mathematics Competition | SILK ROAD MATHEMATICS COMPETITION XVII | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0e2x | Problem:
Naj bo $H$ višinska točka ostrokotnega trikotnika $ABC$ in $D$ točka v notranjosti trikotnika $ABH$. Vzporednica k premici $AH$ skozi točko $D$ seka stranici $BC$ in $AB$ v točkah $K$ in $L$. Vzporednica k premici $BH$ skozi točko $D$ seka daljici $AC$ in $AB$ v točkah $M$ in $N$.
Dokaži, da točke $C$, $D$ i... | [
"Solution:\n\nOznačimo kot $\\angle ABC$ z $\\beta$. Ker je premica $KL$ vzporedna z višino na stranico $BC$, je torej pravokotna na $BC$. Zato je trikotnik $KLB$ pravokotni in velja $\\angle KLB = \\frac{\\pi}{2} - \\beta$. Torej je $\\angle KLN = \\frac{\\pi}{2} - \\beta$.\n\nKer je štirikotnik $KLMN$ tetiven, ve... | Slovenia | 54. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
076z | A polynomial $p(x)$ with real coefficients is called a square if and only if it is not a constant and there exists a polynomial $q(x)$ with real coefficients such that $p(x) = q(x)^2$. Suppose that $f(x)$ and $g(x)$ are non-constant polynomials with real coefficients such that neither of them is a square, but $f(g(x))$... | [
"We can easily extend the definition of a square polynomial to polynomials with complex coefficients. In all the arguments below we consider polynomials with complex coefficients.\n\n**Lemma:** If $p(x)$ is a square and $\\alpha$ is a non-zero complex number then $p(x) - \\alpha$ is not a square.\n\n**Proof of Lemm... | India | India_2017 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0dfw | Let incircle of triangle $ABC$ has center $I$ and touches sides $BC$, $AC$ and $AB$ at points $D$, $E$, $F$ respectively. Let $J_1, J_2, J_3$ be the ex-centres opposite $A$, $B$, $C$ respectively. Let $J_2F$ and $J_3E$ intersect at $P$, $J_3D$ and $J_1F$ intersect at $Q$, $J_1E$ and $J_2D$ intersect at $R$. Show that $... | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0110 | Problem:
For all positive real numbers $x$ and $y$ let
$$
f(x, y) = \min \left(x, \frac{y}{x^{2} + y^{2}}\right).
$$
Show that there exist $x_{0}$ and $y_{0}$ such that $f(x, y) \leqslant f\left(x_{0}, y_{0}\right)$ for all positive $x$ and $y$, and find $f\left(x_{0}, y_{0}\right)$. | [
"Solution:\nAnswer: the maximum value is $f\\left(\\frac{1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}}\\right) = \\frac{1}{\\sqrt{2}}$.\n\nWe shall make use of the inequality $x^{2} + y^{2} \\geqslant 2 x y$. If $x \\leqslant \\frac{y}{x^{2} + y^{2}}$, then\n$$\nx \\leqslant \\frac{y}{x^{2} + y^{2}} \\leqslant \\frac{y}{2 x ... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1/sqrt(2) | |
0lds | An investor has two rectangular lands, each of size $120 \times 100$.
a. On the first land, she wants to build a house with a rectangular base of size $25 \times 35$ and nine circular flower pots with diameter $5$ outside the house. Prove that for all positions of the flower pots, the remaining land is still sufficien... | [
"a.\nConsider the rectangle $ABCD$ where $AB = CD = 120$ and $AD = BC = 100$. Divide the rectangle into $10$ sub-rectangles of size $30 \\times 40$ as shown below. Consider $9$ centers of the flower pots. By the pigeonhole principle, it is clear that there exists a sub-rectangle that does not contain any center.\n\... | Vietnam | VMO | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof only | null | |
0k7e | Problem:
In acute $\triangle ABC$ with centroid $G$, $AB = 22$ and $AC = 19$. Let $E$ and $F$ be the feet of the altitudes from $B$ and $C$ to $AC$ and $AB$ respectively. Let $G'$ be the reflection of $G$ over $BC$. If $E$, $F$, $G$, and $G'$ lie on a circle, compute $BC$. | [
"Solution:\n\nNote that $B$, $C$, $E$, $F$ lie on a circle. Moreover, since $BC$ bisects $GG'$, the center of the circle that goes through $E$, $F$, $G$, $G'$ must lie on $BC$. Therefore, $B$, $C$, $E$, $F$, $G$, $G'$ lie on a circle. Specifically, the center of this circle is $M$, the midpoint of $BC$, as $ME = MF... | United States | HMMT November 2019 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof and answer | 13 | |
0j36 | Problem:
Evaluate
$$
\sum_{k=1}^{2010} \cos^{2}(k)
$$ | [
"Solution:\nAnswer: $1005+\\dfrac{\\sin (4021)-\\sin (1)}{4 \\sin (1)}$\n\nWe use the identity $\\cos^{2}(k) = \\dfrac{1+\\cos(2k)}{2}$. Then our expression evaluates to\n$$\n\\sum_{k=1}^{2010} \\cos^{2}(k) = \\sum_{k=1}^{2010} \\frac{1+\\cos(2k)}{2} = \\frac{1}{2} \\sum_{k=1}^{2010} 1 + \\frac{1}{2} \\sum_{k=1}^{2... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1005 + (sin(4021) - sin(1)) / (4 sin(1)) | |
025r | Problem:
Você sabe repartir um quadrado em 7 quadrados menores? | [
"Solution:\n\n"
] | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
07rc | Show that, for all $x, y, z, w$,
$$
(x - w)(y - z) + (y - w)(z - x) + (z - w)(x - y) = 0,
$$
and
$$
\sin(x - w)\sin(y - z) + \sin(y - w)\sin(z - x) + \sin(z - w)\sin(x - y) = 0.
$$ | [
"The first identity follows since\n$$\n\\begin{aligned}\n(x - w)(y - z) + (y - w)(z - x) + (z - w)(x - y) \n&= (xy - xz - wy + wz) + (yz - yx - wz + wx) + (zx - zy - wx + wy) \\\\\n&= (xy + wx + yz + wx + zx + wy) - (xx + wy + yx + wz + zy + wx) \\\\\n&= 0.\n\\end{aligned}\n$$\n\nThe second because in the first pla... | Ireland | Ireland_2017 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0la0 | Let $ABCD$ be a trapezium with the bottom edge $BC$ ($BC \parallel AC$ and $BC > AD$) and inscribed in the circle $(O)$ ($O$ is the center of $(O)$). Let $P$ be a point moving on the line $BC$ outside the segment $BC$ such that $PA$ doesn't touch the circle $(O)$. The circle with the diameter $PD$ intersects $(O)$ in $... | [
"Let $A'$ be the image of the point $A$ by the reflection through the point $O$. We prove that $N$, $M$, $A'$ are collinear and therefore the line $MN$ passes the fixed point $A'$. First, $DE$ is the radical axis of the circle $(O)$ and the circle ($\\gamma_1$) with the diameter $PD$.\n\nSince $\\angle PNA' = 90^\\... | Vietnam | Vijetnam 2007 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03ib | Problem:
It is known in Euclidean geometry that the sum of the angles of a triangle is constant. Prove, however, that the sum of the dihedral angles of a tetrahedron is not constant.
Note. (i) A tetrahedron is a triangular pyramid, and (ii) a dihedral angle is the interior angle between a pair of faces.
![](attached_... | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof only | null | |
06yk | Determine all sequences $a_{1}, a_{2}, \ldots$ of positive integers such that, for any pair of positive integers $m \leqslant n$, the arithmetic and geometric means
$$
\frac{a_{m}+a_{m+1}+\cdots+a_{n}}{n-m+1} \quad \text{ and } \quad \left(a_{m} a_{m+1} \cdots a_{n}\right)^{\frac{1}{n-m+1}}
$$
are both integers. | [
"Solution 1. Without loss of generality, we may divide all elements of $\\mathcal{S}$ by any common factor, after which they cannot all be even. As $a \\nmid b+2 c$ for $a$ even and $b$ odd, the elements of $\\mathcal{S}$ are all odd.\nWe now divide into three cases:\nCase 1: $|\\mathcal{S}|=1$.\nThe set $\\mathcal... | IMO | IMO2024 Shortlisted Problems | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | All constant sequences of positive integers: a_m = c for all m, for any fixed positive integer c. | |
0bfb | Find all differentiable functions $f: [0, \infty) \to \mathbb{R}$, $f(0) = 0$, so that:
i) $f'$ is strictly positive and increasing;
ii) $\int_0^1 (f'(x))^2 dx \ge f(x+f(x))-f(x)$, for every $x \in [0, \infty)$. | [] | Romania | Shortlisted Problems for the 64th NMO | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Functions"
] | null | proof and answer | none | |
0i2v | In triangle $ABC$, $\angle B = 2\angle C$. Let $P$ and $Q$ be points on the perpendicular bisector of segment $BC$ such that rays $AP$ and $AQ$ trisect $\angle A$. Prove that $PQ < AB$ if and only if $\angle B$ is obtuse. | [
"Let $ABQ_1P$ be a rhombus and let segments $AQ_1$ and $BP$ intersect at $Q_2$. Let $X$ be a point on ray $Q_2Q_1$. Because lines $PQ_2$ and $Q_1Q_2$ are perpendicular, $PX$ increases as $X$ moves away from $Q_2$. Note that $AQ_1$ bisects angle $BAP$, that is, $Q$ is on the ray $Q_2Q_1$. Because $\\angle ABC > \\an... | United States | USA IMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0hhu | For a natural number $n \ge 2$, consider an $n \times n$ board. Let $n^2$ points denote the centers of each of the $1 \times 1$ squares on this board. What is the largest number of these points that can be marked in such a way that no three marked points form the vertices of a right triangle?
*(Mykhailo Shtandenko)* | [
"We will show that there is an example where the marked number of points satisfies the conditions of the problem. For this, we will denote the centers of all the cells in the first row and the first column, except for the center of the cell at the intersection of the first row and the first column (see Fig. 3). By ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2n-2 | |
0e2v | Problem:
Diagonala $AC$ konveksnega štirikotnika $ABCD$ je simetrala kota $\angle DCB$. Naj bo $E$ presečišče stranice $AB$ in trikotniku $ACD$ očrtane krožnice in naj bo $F$ presečišče stranice $AD$ in trikotniku $ABC$ očrtane krožnice. Dokaži, da se daljice $AC$, $DE$ in $BF$ sekajo v eni točki. | [
"Solution:\n\nNaj bo $G$ presečišče daljic $AC$ in $BF$. Točke $A, B, C$ in $F$ ležijo na isti krožnici, zato je $\\angle AFB = \\angle ACB$. Diagonala $AC$ razpolavlja kot $\\angle DCB$, zato je $\\angle ACB = \\angle DCA$. Torej velja $\\angle AFG = \\angle DCA$, od koder sledi $\\angle GFD + \\angle DCG = \\pi$.... | Slovenia | 54. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0h3u | Let $a, b, c \in [0, 1]$. Prove that
$$
\frac{1}{1+a+b} + \frac{1}{1+b+c} + \frac{1}{1+c+a} + a+b+c \le 3 + \frac{1}{3}(ab+bc+ca).
$$ | [
"Доведемо допоміжну нерівність\n$$\n\\frac{1}{1+a+b} \\le 1 - \\frac{a+b}{2} + \\frac{ab}{3}.\n$$\nЦя нерівність рівносильна таким нерівностям:\n$$\n6 \\le (1+a+b)(6-3(a+b)+2ab),\n$$\n$$\n6 \\le 6 + 6(a+b) - 3(a+b) - 3(a+b)^2 + 2ab + 2ab(a+b),\n$$\n$$\n2a^2b + 2ab^2 - 3a^2 - 3b^2 - 4ab + 3a + 3b \\ge 0,\n$$\n$$\n2a... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0a9l | Problem:
The real numbers $a$, $b$, $c$ are such that $a^{2} + b^{2} = 2c^{2}$, and also such that $a \neq b$, $c \neq -a$, $c \neq -b$. Show that
$$
\frac{(a+b+2c)\left(2a^{2}-b^{2}-c^{2}\right)}{(a-b)(a+c)(b+c)}
$$
is an integer. | [
"Solution:\nLet us first note that\n$$\n\\frac{a+b+2c}{(a+c)(b+c)} = \\frac{(a+c)+(b+c)}{(a+c)(b+c)} = \\frac{1}{a+c} + \\frac{1}{b+c}\n$$\nFurther we have\n$$\n2a^{2} - b^{2} - c^{2} = 2a^{2} - (2c^{2} - a^{2}) - c^{2} = 3a^{2} - 3c^{2} = 3(a+c)(a-c)\n$$\nand\n$$\n2a^{2} - b^{2} - c^{2} = 2(2c^{2} - b^{2}) - b^{2}... | Nordic Mathematical Olympiad | The 26th Nordic Mathematical Contest | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0jzg | Problem:
Let $r$ be a positive integer. Show that if a graph $G$ has no cycles of length at most $2r$, then it has at most $|V|^{2016}$ cycles of length exactly $2016r$, where $|V|$ denotes the number of vertices in the graph $G$. | [
"Solution:\n\nThe key idea is that there is at most $1$ path of length $r$ between any pair of vertices, or else you get a cycle of length $\\leq 2r$. Now, start at any vertex ($|V|$ choices) and walk $2015$ times. There's at most $|V|^{2016}$ ways to do this by the previous argument. Now you have to go from the en... | United States | February 2017 | [
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
08zb | In isosceles right triangle $ABC$ with $\angle BAC = 90^\circ$ and $AB = AC = 7$, let $D$, $E$ and $F$ be points on sides $BC$, $CA$ and $AB$ respectively. Given that $\angle EDF = 90^\circ$, $DE = 5$, $DF = 3$, find the length of $BD$. | [
"Let $P$ be a point on line $AB$ such that $PD \\perp BC$, then we have $\\angle CDE = \\angle PDF$ since $\\angle CDP = \\angle EDF = 90^\\circ$, and $\\angle DCE = \\angle DPF = 45^\\circ$, which yields $\\triangle PDF \\sim \\triangle CDE$ with a similarity ratio of $3:5$. In addition, we have $BD = DP$ since $\... | Japan | Japan 2022 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 21√2/8 | |
01f0 | Determine all integers which can be expressed in the form
$$
x_1x_2 + x_2x_3 + \dots + x_{k-1}x_k,
$$
where $x_1, x_2, \dots, x_k$ are positive integers whose sum equals $2019$. | [
"First, easy induction shows that $\\sum_{i=1}^{k-1} x_i x_{i+1} \\ge (\\sum_{i=1}^k x_i) - 1$.\n\nBase case $k=2$: $x_1x_2 - (x_1+x_2-1) = (x_1-1)(x_2-1) \\ge 0$. Assuming the inequality holds for a certain $k$ and adding the obvious $x_kx_{k+1} \\ge x_{k+1}$, we get the claim for $k+1$.\n\nGiven the conditions of... | Baltic Way | Baltic Way 2019 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expr... | English | proof and answer | All integers from 2018 to 1019090 inclusive. | |
07zx | Problem:
Siano $\mathcal{C}$, $\mathcal{C}_1$, $\mathcal{C}_2$, tre circonferenze di raggi rispettivamente $r$, $r_1$, $r_2$ con $0 < r_1 < r_2 < r$. Le circonferenze $\mathcal{C}_1$ e $\mathcal{C}_2$ sono tangenti internamente a $\mathcal{C}$ in due punti distinti $A$ e $B$ e si intersecano fra loro in due punti dist... | [
"Solution:\n\nSupponiamo dapprima che il segmento $AB$ passi per uno dei punti di intersezione fra $\\mathcal{C}_1$ e $\\mathcal{C}_2$, e sia $C$ tale punto. Siano inoltre $O$, $O_1$, $O_2$ rispettivamente i centri delle circonferenze $\\mathcal{C}$, $\\mathcal{C}_1$, $\\mathcal{C}_2$. I triangoli $OAB$, $O_1AC$ e ... | Italy | XV Gara Nazionale di Matematica | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
02ku | Problem:
Seqüência triangular - Qual é o $21^{\text{\textordmasculine}}$ termo da seqüência
$$
1 ; 2+3 ; 4+5+6 ; 7+8+9+10 ; 11+12+13+14+15 ; \ldots ?
$$ | [
"Solution:\n\nObserve que o $21^{\\circ}$ termo é a soma de 21 números consecutivos. Tomando a primeira parcela de cada termo, isto é, $1, 2, 4, 7, 11, 16, \\ldots$, temos que\n$$\n\\begin{aligned}\n2 & = 1 + 1 \\\\\n4 & = 2 + 1 + 1 \\\\\n7 & = 3 + 2 + 1 + 1 \\\\\n11 & = 4 + 3 + 2 + 1 + 1 \\\\\n16 & = 5 + 4 + 3 + 2... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 4641 | |
0d1l | Given an odd number $n \ge 3$, let
$$
S = \{k : 1 \le k \le n, \gcd(k, n) = 1\}
$$
and let
$$
T = \{k : k \in S, \gcd(k + 1, n) = 1\}.
$$
For each $k \in S$, let $r_k$ be the remainder left by $\frac{k|S| - 1}{n}$ upon division by $n$. Show that
$$
\prod_{k \in T} (r_k - r_{n-k}) \equiv |S|^{|T|} \pmod{n}.
$$ | [
"Since $n$ is odd,\n$$\n|S| = \\phi(n) = n \\prod_{p|n} (1 - 1/p)\n$$\nis even. Given an element $k$ of $S$, write\n$k^{|S|} \\equiv 1+nr_k \\pmod{n^2}$ and $(n-k)^{|S|} \\equiv 1+nr_{n-k} \\pmod{n^2}$, and notice that\n$$\n(n-k)^{|S|} \\equiv k^{|S|} \\cdot n \\cdot k^{|S|-1} \\pmod{n^2},\n$$\nto get\n$$\nr_k - r_... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0860 | Problem:
Una banda di ladri vuole aprire la cassaforte di una banca. Un basista ha fatto ubriacare il direttore della banca ed è riuscito a sapere che:
a. la combinazione è formata da 5 cifre da 0 a 9;
b. la combinazione è un numero pari;
c. esattamente una delle 5 cifre della combinazione è dispari;
d. nella comb... | [
"Solution:\n\nLa risposta è (B). Dal fatto che il numero sia pari segue che deve essere pari l'ultima cifra. Conviene distinguere a seconda se questa sia quella doppia o no.\n\nNel primo caso l'altra cifra doppia può occupare 3 diverse posizioni (non la 4) e per ciascuna di queste la cifra dispari può occupare una ... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | MCQ | B | |
07qd | The sides of triangle $ABC$ have lengths $a, b, c$ which satisfy
$$
\frac{a^6}{2b^3 + 2c^3} + \frac{b^6 + 2b^3c^3 + c^6}{2a^3} + \frac{2a^3b^3 + 2a^3c^3}{a^3 + b^3 + c^3} = a^3 + b^3 + c^3.
$$
Determine if one of the internal angles of triangle $ABC$ can be a right angle or an obtuse angle, or if all internal angles mu... | [
"If we use the abbreviation $u = a^3$ and $v = b^3 + c^3$, the given equation can be written as\n$$\n\\frac{u^2}{2v} + \\frac{v^2}{2u} + \\frac{2uv}{u+v} = u+v.\n$$\nThis is the same as\n$$\n\\frac{u^3 + v^3}{2uv} = u + v - \\frac{2uv}{u+v} = \\frac{u^2 + v^2}{u+v},\n$$\nor $(u+v)(u^3+v^3) = 2uv(u^2+v^2)$, which si... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
08mx | Problem:
In a group of $n$ people, each one had a different ball. They performed a sequence of swaps; in each swap, two people swapped the ball they had at that moment. Each pair of people performed at least one swap. In the end each person had the ball he/she had at the start. Find the least possible number of swaps,... | [
"Solution:\n\nWe will denote the people by $A, B, C, \\ldots$ and their initial balls by the corresponding small letters. Thus the initial state is $A a, B b, C c, D d, E e(, F f)$. A swap is denoted by the (capital) letters of the people involved.\n\na) Five people form 10 pairs, so at least 10 swaps are necessary... | JBMO | JBMO Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) 10; b) 16 | |
0147 | Problem:
A rectangular array has $n$ rows and six columns, where $n > 2$. In each cell there is written either $0$ or $1$. All rows in the array are different from each other. For each pair of rows $(x_{1}, x_{2}, \ldots, x_{6})$ and $(y_{1}, y_{2}, \ldots, y_{6})$, the row $(x_{1} y_{1}, x_{2} y_{2}, \ldots, x_{6} y_{... | [
"Solution:\nClearly there must be rows with some zeroes. Consider the case when there is a row with just one zero; we can assume it is $(0,1,1,1,1,1)$. Then for each row $(1, x_{2}, x_{3}, x_{4}, x_{5}, x_{6})$ there is also a row $(0, x_{2}, x_{3}, x_{4}, x_{5}, x_{6})$; the conclusion follows.\n\nConsider the cas... | Baltic Way | Baltic Way 2005 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0bpk | Problem:
Rezolvaţi în mulţimea numerelor naturale ecuaţia $21^{x} + 4^{y} = z^{2}$. | [
"Solution:\nAvem $(z - 2^{y})(z + 2^{y}) = 21^{x}$ şi, notând $d = (z - 2^{y}, z + 2^{y})$, rezultă că $d \\mid (z + 2^{y}) - (z - 2^{y})$, deci $d \\mid 2^{y+1}$. Întrucât $d \\mid z + 2^{y}$ şi $z + 2^{y} \\mid 21^{x}$, rezultă $d \\mid (2^{y+1}, 21^{x})$, deci $d = 1$.\n\nCa urmare, avem de analizat două cazuri:... | Romania | Primul test de selecţie pentru OBMJ | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (1, 1, 5) |
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