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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0kxd | Problem:
Let $\Omega$ be a sphere of radius $4$ and $\Gamma$ be a sphere of radius $2$. Suppose that the center of $\Gamma$ lies on the surface of $\Omega$. The intersection of the surfaces of $\Omega$ and $\Gamma$ is a circle. Compute this circle's circumference. | [
"Solution:\n\nTake a cross-section of a plane through the centers of $\\Omega$ and $\\Gamma$, call them $O_1$ and $O_2$, respectively. The resulting figure is two circles, one of radius $4$ and center $O_1$, and the other with radius $2$ and center $O_2$ on the circle of radius $4$. Let these two circles intersect ... | United States | HMMT February 2023 | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles"
] | null | final answer only | π√15 | |
0gqd | Between any two cities of a country consisting of $2013$ cities one-way flights are organized so that there is at least one departure from each city. Determine the maximal possible value of $k$ such that no matter how these flights are arranged there are $k$ cities reachable from any city of a country by using at most ... | [
"The answer is $k = 3$.\n\nWe will write $A \\to B$ if flight is from $A$ to $B$. The flight arrangement where $A \\to B \\to C \\to A$ and all other flights incident to $A, B, C$ are directed to $A, B, C$ shows that $k \\le 3$.\n\nIf a city $A$ is reachable from any other city by using at most two flights we will ... | Turkey | Team Selection Test for JBMO | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 3 | |
075y | Problem:
Let $\Gamma_{1}$ and $\Gamma_{2}$ be two circles touching each other externally at $R$. Let $l_{1}$ be a line which is tangent to $\Gamma_{2}$ at $P$ and passing through the center $O_{1}$ of $\Gamma_{1}$. Similarly, let $l_{2}$ be a line which is tangent to $\Gamma_{2}$ at $Q$ and passing through the center $... | [
"Solution:\nSuppose that $P$ and $Q$ lie on the opposite sides of line joining $O_{1}$ and $O_{2}$. By symmetry we may assume that the configuration is as shown in the figure below. Then we have $K P > K O_{1} > K Q$ since $K O_{1}$ is the hypotenuse of triangle $K Q O_{1}$. This is a contradiction to the given ass... | India | INMO | [
"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 > Angle chasing"
... | null | proof only | null | |
04ow | Borna wants to colour each of the numbers $2$, $3$, $\ldots$, $32$ in one of $k$ colours, in such a way that none of the numbers is of the same colour as one of its multiples. Find the smallest positive integer $k$ for which that is possible. | [] | Croatia | Croatian Mathematical Society Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 5 | |
0b8h | Determine all pairs of positive integers $a$ and $b$ for which $a^6 \ge 5^{b+1}$ and $b^6 \ge 5^{a+1}$. | [
"We will prove the inequality $5^{n+1} \\ge n^6$ holds for all positive integers $n$, except $n = 3$ and $n = 4$.\n\nIndeed, this is readily verified for $n = 1$, $n = 2$ and $n = 5$. When the thesis is true for $n \\ge 5$, then $5^{n+2} = 5 \\cdot 5^{n+1} \\ge 5n^6$. It is then sufficient to show that $5n^6 \\ge (... | Romania | Local Mathematical Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | (3,3), (4,4), (5,5) | |
02qh | Problem:
O número $abcde$ tem cinco algarismos distintos e diferentes de zero, cada um deles representado por uma das letras $a, b, c, d, e$. Multiplicando-se este número por $4$ obtém-se um número de cinco algarismos $edcba$. Qual o valor de $a+b+c+d+e$?
A) 22
B) 23
C) 24
D) 25
E) 27 | [
"Solution:\n\nA multiplicação pode ser esquematizada como\n\n\n\nA solução é baseada nas seguintes observações:\n\n- O algarismo $a$ só pode ser $1$ ou $2$, pois, se fosse $a \\geq 3$, então $4a$ seria um número de $2$ algarismos e portanto o número $edcba$ teria $6$ algarismos. Mas $a$ não... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Modular Arithmetic"
] | null | MCQ | E | |
0fv9 | Problem:
Eine Raumstation besteht aus 25 Kammern, und je zwei Kammern sind mit einem Tunnel verbunden. Es gibt insgesamt 50 Haupttunnel, die in beide Richtungen benutzt werden können, die restlichen sind alle Einbahntunnel. Eine Gruppe von vier Kammern heisst verbunden, falls man von jeder dieser Kammern in jede ander... | [
"Solution:\n\nGibt es in einer Vierergruppe eine Kammer, von der drei Einwegtunnel wegführen, so ist diese Gruppe sicher nicht verbunden. Wir nennen eine solche Kammer isoliert in der Gruppe. Nummeriere die Kammern und nehme an, von der $k$-ten Kammer führen $a_{k}$ Einwegtunnel weg. Insgesamt gibt es $\\binom{25}{... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 9650 | |
0ji3 | Problem:
How many of the first $1000$ positive integers can be written as the sum of finitely many distinct numbers from the sequence $3^{0}, 3^{1}, 3^{2}, \ldots$? | [
"Solution:\n\nAnswer: $105$\n\nWe want to find which integers have only $0$'s and $1$'s in their base $3$ representation. Note that $1000_{10} = 1101001_{3}$. We can construct a bijection from all such numbers to the binary strings, by mapping $x_{3} \\leftrightarrow x_{2}$. Since $1101001_{2} = 105_{10}$, we concl... | United States | HMMT November 2013 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 105 | |
01ty | Points $A_1, B_1, C_1$ are the midpoints of the sides $BC$, $AC$, $AB$ of the triangle $ABC$ ($AB \ge AC \ge BC$), respectively. Points $C_2$ and $C_3$ are marked on $AB$ so that the perimeter of the triangle $BA_1C_2$ is equal to the perimeter of the quadrilateral $ACA_1C_2$ and the perimeter of the triangle $AB_1C_3$... | [
"By condition, the perimeter $P(AB_2C_1)$ of the triangle $AB_2C_1$ is equal to the perimeter $P(BCB_2C_1)$ of the quadrilateral $BCB_2C_1$, i.e., (see the Fig.)\n\n\n\n$$P(AB_2C_1) = AB_2 + B_2C_1 + C_1A = BC + CB_2 + B_2C_1 + C_1B = P(BCB_2C_1),$$\nso, taking into account $C_1A = C_1B$, w... | Belarus | 66th Belarusian Mathematical Olympiad | [
"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"
] | English | proof only | null | |
00ao | Find the angles of a convex quadrilateral $ABCD$ such that $\hat{ABD} = 29^\circ$, $\hat{ADB} = 41^\circ$, $\hat{ACB} = 82^\circ$ and $\hat{ACD} = 58^\circ$. | [
"We have $\\hat{BAD} = 180^\\circ - (29^\\circ + 41^\\circ) = 110^\\circ$, $\\hat{BCD} = 82^\\circ + 58^\\circ = 140^\\circ$. Consider the circumcircle $\\gamma$ of triangle $BCD$. Since $\\hat{BAD} + \\hat{BCD} > 180^\\circ$, point $A$ is interior to $\\gamma$.\n\nExtend $CA$ beyond $A$ to meet $\\gamma$ at $E$. B... | Argentina | Argentine National Olympiad 2016 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof and answer | Angle A = 110°, Angle B = 49°, Angle C = 140°, Angle D = 61° | |
0dh2 | Let $n$ be a natural number. Find the number of permutations of the set $\{1, 2, \dots, n\}$ such that for each $i = 1, 2, \dots, n$, the first $i$ numbers in the permutation are not larger than $i + 1$. For example, there are 4 such permutations for $n = 3$: $\{1, 2, 3\}$, $\{2, 1, 3\}$, $\{1, 3, 2\}$ and $\{2, 3, 1\}... | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 2^{n-1} | |
00zz | Problem:
Twelve cards lie in a row. The cards are of three kinds: with both sides white, both sides black, or with a white and a black side. Initially, nine of the twelve cards have a black side up. The cards $1$-$6$ are turned, and subsequently four of the twelve cards have a black side up. Now cards $4$-$9$ are turn... | [
"Solution:\n\nAnswer: there are $9$ cards with one black and one white side and $3$ cards with both sides white.\n\nDivide the cards into four types according to the table below.\n\n| Type | Initially up | Initially down |\n| :---: | :---: | :---: |\n| $A$ | black | white |\n| $B$ | white | black |\n| $C$ | white |... | Baltic Way | Baltic Way 1997 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | one black and one white: 9; both white: 3; both black: 0 | |
0l0k | Points $P$ and $Q$ are chosen uniformly and independently at random on sides $\overline{AB}$ and $\overline{AC}$, respectively, of equilateral triangle $\triangle ABC$. Which of the following intervals contains the probability that the area of $\triangle APQ$ is less than half the area of $\triangle ABC$?
(A) $[\frac{... | [
"Without loss of generality let $AB = AC = BC = 1$; then the area of $\\triangle ABC$ is $\\frac{1}{4}\\sqrt{3}$. Let $x = AP$ and $y = AQ$. Then the area of $\\triangle APQ$ is\n$$\n\\frac{1}{2}xy \\cdot \\sin 60^\\circ = \\frac{1}{4}\\sqrt{3} \\cdot xy.\n$$\nThe probability that the area of $\\triangle APQ$ is le... | United States | AMC 12 A | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | D | |
09k2 | A positive integer $k$ is called *nice* if there exist positive integers $a_1, a_2, \dots, a_{2021}$ such that
(1) $a_i - a_j$ is not divisible by $2023$ for any pair $1 \le i < j \le 2021$,
(2) for any index $i$ there exists an index $j$ such that $a_i - k a_j$ is divisible by $2023$.
Find the largest good number $k$ ... | [
"Answer: $92$.\nSet $n = 2023 = 7 \\times 17^2$ and denote by $\\mathbb{Z}/n\\mathbb{Z}$ the set of all residues modulo $n$. For any natural number $k$, denote by $k: \\mathbb{Z}/n\\mathbb{Z} \\to \\mathbb{Z}/n\\mathbb{Z}$ the map defined by $a$ (mod $n$) $\\mapsto k a$ (mod $n$).\nAssume that $k$ is nice and $a_1,... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 92 | |
077u | Problem:
Let $N \ge 3$ be an integer. In the country of Sibyl, there are $N^2$ towns arranged as the vertices of an $N \times N$ grid, with each pair of towns corresponding to an adjacent pair of vertices on the grid connected by a road. Several automated drones are each given the instruction to traverse a rectangular... | [] | India | EGMO TST Day 1 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | If N is even, the minimum number is N−1. If N is odd, the minimum number is N. | |
0hiz | Problem:
Determine the number of ways to fill a $3 \times 3$ grid with $0$'s and $1$'s such that each row and column has an even sum. | [
"Solution:\n\n| $a_{1}$ | $a_{2}$ | $a_{3}$ |\n| :--- | :--- | :--- |\n| $b_{1}$ | $b_{2}$ | $b_{3}$ |\n| $c_{1}$ | $c_{2}$ | $c_{3}$ |\n\nObserve that the condition is equivalent to saying that every entry in the table has the same parity as the sum of the entries in its row (resp. column).\n\nWe claim that upon f... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 16 | |
07c9 | $p \neq 13$ is a prime number in the form of $8k + 5$ for some natural number $k$, and $39$ is a non-residue modulo $p$. Prove that equation $x_1^4 + x_2^4 + x_3^4 + x_4^4 \equiv 0 \pmod{p}$ has a solution in the set of integers such that $p \nmid x_1x_2x_3x_4$. | [
"Note that\n$$\n\\sum_{1 \\le x_i \\le p-1} \\left(1 - \\left(x_1^4 + x_2^4 + x_3^4 + x_4^4\\right)^{p-1}\\right) \\stackrel{p}{=} \n\\qquad (p-1)^4 - (p-1)^4 \\overbrace{\\left(\\sum_{\\substack{\\frac{p-1}{4}|\\theta_i; \\\\ \\sum \\theta_i = p-1}}^A \\begin{pmatrix} p-1 \\\\ \\theta_1, \\theta_2, \\theta_3, \\th... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof only | null | |
0akf | Determine the largest integer $N$, for which there exists a $6 \times N$ table $T$ that has the following properties:
(i) Every column contains the numbers $1,2,3,4,5,6$ in some ordering
(ii) For any two columns $i \neq j$, there exists a row $r$ such that $T(r,i)=T(r,j)$,
(iii) For any two columns $i \neq j$, there ex... | [] | North Macedonia | Mediterranean Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English | proof and answer | 120 | |
08qo | Problem:
Are there any positive integers $m$ and $n$ satisfying the equation
$$
m^{3}=9 n^{4}+170 n^{2}+289 ?
$$ | [
"Solution:\nWe will prove that the answer is no. Note that\n$$\nm^{3}=9 n^{4}+170 n^{2}+289=(9 n^{2}+17)(n^{2}+17)\n$$\nIf $n$ is odd then $m$ is even, therefore $8 \\mid m^{3}$. However,\n$$\n9 n^{4}+170 n^{2}+289 \\equiv 9+170+289 \\equiv 4 \\pmod{8}\n$$\nwhich leads to a contradiction. If $n$ is a multiple of $1... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots >... | null | proof only | null | |
08xv | Consider the following operation you can perform on any positive integer $n$:
Remove the lowest digit of $n$ and add to the resulting number 4 times the removed digit.
For example, if $n = 1234$ we get $123+16 = 139$, and if $n = 7$ then we get $0+28 = 28$.
If we start with $25^{2015}$ and keep on applying the operatio... | [
"For integers $k$, $\\ell$ and a positive integer $m$, let us write $k \\equiv \\ell \\pmod m$ to mean that $k - \\ell$ is divisible by $m$. For a non-negative integer $n$, let $a_n$ be the integer obtained by applying the operation stated for the problem $n$ times starting with the initial number $a_0 = 25^{2015}$... | Japan | Japan 2015 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 4 | |
08h3 | Problem:
Let $n$ ($n \geq 1$) be an integer. Consider the equation
$$
2 \cdot \left\lfloor \frac{1}{2x} \right\rfloor - n + 1 = (n+1)(1 - n x)
$$
where $x$ is the unknown real variable.
a. Solve the equation for $n=8$.
b. Prove that there exists an integer $n$ for which the equation has at least 2021 solutions. (For... | [
"Solution:\n\nLet $k = \\left\\lfloor \\frac{1}{2x} \\right\\rfloor$, $k \\in \\mathbb{Z}$.\n\na. For $n=8$, the equation becomes\n$$\nk = \\left\\lfloor \\frac{1}{2x} \\right\\rfloor = 8 - 36x \\Rightarrow x \\neq 0 \\text{ and } x = \\frac{8 - k}{36}\n$$\nSince $x \\neq 0$, we have $k \\neq 8$, and the last relat... | JBMO | null | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) x = 5/36 and x = 1/9. b) For example, any integer n with n ≥ 2021^2 (equivalently, with square root of n plus one greater than 2021) yields at least 2021 solutions. | |
0hzt | Problem:
A regular tetrahedron has two vertices on the body diagonal of a cube with side length $12$. The other two vertices lie on one of the face diagonals not intersecting that body diagonal. Find the side length of the tetrahedron. | [
"Solution:\nLet $ABCD$ be a tetrahedron of side $s$. We want to find the distance between two of its opposite sides. Let $E$ be the midpoint of $AD$, $F$ the midpoint of $BC$. Then $AE = s/2$, $AF = s\\sqrt{3}/2$, and $\\angle AEF = 90^{\\circ}$. So the distance between the two opposite sides is $EF = \\sqrt{AF^2 -... | United States | Harvard-MIT Math Tournament | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 4√3 | |
0bw7 | Find all functions $f : \mathbb{R} \to \mathbb{R}$ which fulfill
$$
f(x^2f(x) + f(y)) = f(f(x^3)) + y, \text{ for every } x, y \in \mathbb{R}.
$$ | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | f(x) = x for all real x, and f(x) = -x for all real x | |
04j4 | Let $ABCD$ be a tetrahedron such that $\angle BAC = \angle CAD = \angle DAB = 90^\circ$, $|AD| = 2\sqrt{2}$ and $|AB| = |AC| = 3$ holds. Determine the radius of the inscribed sphere of the tetrahedron. (Mea Bombardelli) | [] | Croatia | Croatia Mathematical Competitions | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Surface Area"
] | null | proof and answer | 3(√2 - 1)/2 | |
0bnm | Show that if $k \ge 1$ and $n \ge 2$ are integers such that there exists $A, B \in \mathcal{M}_n(\mathbb{Z})$ with the property that $A^3 = O_n$ and $A^k B + BA = I_n$, then $k = 1$ and $n$ is even. Conversely, show that if $n \ge 2$ is even, then there exists $A, B \in \mathcal{M}_n(\mathbb{Z})$ such that $A^3 = O_n$ ... | [
"If $k \\ge 3$, then $BA = I_n$ (since $A^k = O_n$), so $A$ is invertible, which contradicts $A^3 = O_n$.\n\nIf $k = 2$, then relation $A^2B + BA = I_n$, multiplied at the left by $A$ and then at the right by $A^2$, leads to $ABA = A$ and $A^2BA^2 = A^2$. The last equality can be written $A(ABA)A = A^2$, whence $A^... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
0d4a | $$
\frac{1}{n}=0.a_{1}a_{2}\ldots
$$
Suppose that $n=a_{1}+a_{2}+\cdots$. Determine all possible values of $n$. | [
"Let $n \\geq 2$ be such an integer. Because $n=a_{1}+a_{2}+\\cdots$, there exists a positive integer $k>0$ such that $a_{k} \\neq 0$ and $a_{i}=0$ for all integers $i>k$.\n\nBecause $\\frac{10^{k}}{n}=\\overline{a_{1}a_{2}\\cdots a_{k}}$ is an integer and $\\frac{10^{k-1}}{n}=\\overline{a_{1}a_{2}\\cdots a_{k-1},a... | Saudi Arabia | SAMC | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English, Arabic | proof and answer | 8 | |
0kgj | Problem:
Let $ABCD$ be a parallelogram. Let $E$ be the midpoint of $AB$ and $F$ be the midpoint of $CD$. Points $P$ and $Q$ are on segments $EF$ and $CF$, respectively, such that $A$, $P$, and $Q$ are collinear. Given that $EP = 5$, $PF = 3$, and $QF = 12$, find $CQ$. | [
"Solution:\n\nTriangles $PFQ$ and $PEA$ are similar, so $AE = FQ \\cdot \\frac{PE}{PF} = 12 \\cdot \\frac{5}{3} = 20$. Now, $CQ = CF - QF = 20 - 12 = 8$."
] | United States | HMMT November 2021 Team Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | 8 | |
08ga | Problem:
Per esattamente quattro valori interi di $n$ compresi fra 1 e 10, estremi inclusi, l'espressione $n^{9}+3^{n+1} n^{6}+3^{3 n}$ rappresenta un numero primo. Quanto vale la somma di questi quattro valori di $n$ ?
(A) 12
(B) 16
(C) 18
(D) 22
(E) 25 | [
"Solution:\n\nLa risposta è $\\mathbf{( B )}$. Se $n=3 k$ con $k$ intero, chiaramente $n^{9}+3^{n+1} n^{6}+3^{3 n}$ è divisibile per $3$ e quindi per $n=3,6,9$ l'espressione non rappresenta un numero primo.\n\nSupponiamo invece che valga $n=3 k+1$ con $k$ intero. Allora\n$$\nn^{9}+3^{n+1} n^{6}+3^{3 n}=(3 k+1)^{9}+... | Italy | Italian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | B | |
085q | Problem:
La rappresentazione in base 2 di un numero $a$ è $1110000100111010101110100001$. Qual è la settima cifra da sinistra della rappresentazione di $a$ in base 8?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6. | [
"Solution:\nLa risposta è (D). Sia lo sviluppo in base 2 di un generico intero $a$\n$$\n\\ldots + 2^{8} a_{8} + 2^{7} a_{7} + 2^{6} a_{6} + 2^{5} a_{5} + 2^{4} a_{4} + 2^{3} a_{3} + 2^{2} a_{2} + 2^{1} a_{1} + a_{0}\n$$\nallora, se raggruppiamo i termini a tre a tre a partire da destra raccogliendo un'opportuna pot... | Italy | Olimpiadi di Matematica | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | D | |
056k | There are distinct points $O$, $A$, $B$, $K_1$, $\ldots$, $K_n$, $L_1$, $\ldots$, $L_n$ on a plane such that no three points are collinear. The open line segments $K_1L_1$, $\ldots$, $K_nL_n$ are coloured red, other points on the plane are left uncoloured. An *allowed path* from point $O$ to point $X$ is a polygonal ch... | [
"A path $OX_1\\ldots X_{k-1}A$ is *suitable* if it is a shortest allowed path from $O$ to $A$ and $X_1A \\le X_1B$. Similarly, a path $OY_1\\ldots Y_{k-1}B$ is *suitable* if it is the shortest allowed path from $O$ to $B$ and $Y_1B \\le Y_1A$. Let us show that for $n = 2$ it is possible to choose points $A$, $B$, $... | Estonia | IMO Team Selection Contest | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | 2 | |
0efi | Problem:
Letalska družba potniku za prtljago ne zaračuna dodatnih stroškov, če masa prtljage ne preseže določene dovoljene mase. Vsak dodaten kilogram prtljage pa mora potnik doplačati. Gospod in gospa Kotnik imata skupno prtljago, zato se jima ne bi zaračunalo dodatnih stroškov, če masa skupne prtljage ne bi presegla... | [
"Solution:\n\nOznačimo največjo dovoljeno težo za enega potnika z $d$. Gospod in gospa Kotnik imata skupaj $60-2d$ dodatnih kilogramov prtljage, za katere sta plačala 11 evrov. Torej sta za vsak dodaten kilogram plačala $\\frac{11}{60-2d}$ evrov.\n\nGospod Novak ima $60-d$ dodatnih kilogramov prtljage, za katere je... | Slovenia | Slovenian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
00he | Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that
$$
(z+1) f(x+y) = f(x f(z) + y) + f(y f(z) + x)
$$
for all positive real numbers $x$, $y$, $z$. | [
"The identity function $f(x) = x$ clearly satisfies the functional equation. Now, let $f$ be a function satisfying the functional equation. Plugging $x = y = 1$ into (3) we get $2 f(f(z) + 1) = (z + 1) f(2)$ for all $z \\in \\mathbb{R}^{+}$. Hence, $f$ is not bounded above.\n\nLemma. Let $a, b, c$ be positive real ... | Asia Pacific Mathematics Olympiad (APMO) | APMO 2016 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x for all positive real x | |
00yn | Problem:
Find all triples $(x, y, z)$ of positive integers satisfying the system of equations
$$
\left\{\begin{array}{l}
x^{2}=2(y+z) \\
x^{6}=y^{6}+z^{6}+31\left(y^{2}+z^{2}\right)
\end{array}\right.
$$ | [
"Solution:\nFrom the first equation it follows that $x$ is even. The second equation implies $x>y$ and $x>z$. Hence $4x > 2(y+z) = x^{2}$, and therefore $x=2$ and $y+z=2$, so $y=z=1$. It is easy to check that the triple $(2,1,1)$ satisfies the given system of equations."
] | Baltic Way | Baltic Way | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (2, 1, 1) | |
0hyd | Problem:
Suppose $f(x)$ is a rational function such that $3 f\left(\frac{1}{x}\right)+\frac{2 f(x)}{x}=x^{2}$ for $x \neq 0$. Find $f(-2)$. | [
"Solution:\nLet $x=\\frac{-1}{2}$. Then\n$$\n\\begin{aligned}\n& 3 f(-2)+\\frac{2 f\\left(\\frac{-1}{2}\\right)}{\\frac{-1}{2}}=\\frac{1}{4} \\\\\n& \\Rightarrow 3 f(-2)-4 f\\left(\\frac{-1}{2}\\right)= \\frac{1}{4}\n\\end{aligned}\n$$\nLet $x=-2$. Then\n$$\n\\begin{aligned}\n& 3 f\\left(\\frac{-1}{2}\\right)+\\fra... | United States | HMMT 1998 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 67/20 | |
0c7x | Solve in the set of real numbers the equation
$$
\left[ x + \frac{1}{x} \right] = \left[ x^2 + \frac{1}{x^2} \right],
$$
where $[a]$ represents the integer part of the real number $a$. | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | x in ((sqrt(5) - 1)/2, (sqrt(5) + 1)/2) | |
0bd4 | Let $A$ be a non-invertible square matrix of order $n$ with real entries, $n \ge 2$, and let $A^*$ be the adjoint of $A$. Prove that $\text{tr}(A^*) \ne -1$ if and only if the matrix $I_n + A^*$ is invertible. | [
"As $A$ is non-singular, we get $\\text{rank}(A) \\le n - 1$. Distinguish two cases:\n\ni) $\\text{rank}(A) \\le n - 2$. Then $A^* = O_n$ and the conclusion follows immediately.\n\nii) $\\text{rank}(A) = n-1$. Then $AA^* = O_n$ and by Sylvester's inequality $0 \\ge \\text{rank}(A) + \\text{rank}(A^*) - n$, that is ... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof only | null | |
0khs | Problem:
Tetrahedron $A B C D$ has side lengths $A B=6$, $B D=6 \sqrt{2}$, $B C=10$, $A C=8$, $C D=10$, and $A D=6$. The distance from vertex $A$ to face $B C D$ can be written as $\frac{a \sqrt{b}}{c}$, where $a, b, c$ are positive integers, $b$ is square-free, and $\operatorname{gcd}(a, c)=1$. Find $100 a+10 b+c$. | [
"Solution:\n\nFirst, we see that faces $A B D$, $A B C$, and $A C D$ are all right triangles. Now, $A B D$ can be visualized as the base, and it can be seen that side $A C$ is then the height of the tetrahedron, as $A C$ should be perpendicular to both $A B$ and $A D$. Therefore, the area of the base is $\\frac{6^{... | United States | HMMT November 2021 | [
"Geometry > Solid Geometry > Volume"
] | null | final answer only | 2851 | |
0b49 | Problem:
Suppose $PQRS$ is a convex quadrilateral with $\angle SPQ = \angle PQR = 120^\circ$, $SP - QR = 36$, $RS = 84$, and $QR$ is a positive even integer. Let $T$ be the intersection of lines $SP$ and $QR$. What is the largest possible perimeter of $\triangle PQT$? | [] | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 54 + 12√366 | |
0ivu | Problem:
For how many integer values of $b$ does there exist a polynomial function with integer coefficients such that $f(2)=2010$ and $f(b)=8$? | [
"Solution:\n\nWe can take $f(x) = -\\frac{2002}{d}(x-b) + 2010$ for all divisors $d$ of $-2002$.\n\nTo see that we can't get any others, note that $b-2$ must divide $f(b)-f(2)$, so $b-2$ divides $-2002$ (this is because $b-2$ divides $b^{n}-2^{n}$ and hence any sum of numbers of the form $b^{n}-2^{n}$)."
] | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 32 | |
0c8f | Solve for $x \in \mathbb{R}$:
$$
2^{x+1} + \log_2(1 + \sqrt{x}) = 4^x + 1.
$$ | [
"The equation can be written $\\log_2(1 + \\sqrt{x}) = (2^x - 1)^2$.\nThe function $f : [0, \\infty) \\to [0, \\infty)$, defined by $f(x) = \\log_2(1 + \\sqrt{x})$, is one to one and onto and its inverse $f^{-1} : [0, \\infty) \\to [0, \\infty)$ is given by $f^{-1}(x) = (2^x - 1)^2$.\nAs $f$ is increasing strictly,... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof and answer | 0, 1 | |
074p | Let $ABC$ be a triangle in which $BC < AC$. Let $M$ be the mid-point of $AB$; $AP$ be the altitude from $A$ on to $BC$; and $BQ$ be the altitude from $B$ on to $AC$. Suppose $QP$ produced meets $AB$ (extended) in $T$. If $H$ is the ortho-centre of $ABC$, prove that $TH$ is perpendicular to $CM$. | [
"Complete the parallelogram $ADBC$. Join $CD$, $CH$ and $HD$. Let $S$ and $L$ be the midpoints of $CH$ and $HD$ respectively. Observe that $CPHQ$ is a cyclic quadrilateral and $CH$ is a diameter of the circumscribing circle. Thus $S$ is the centre of a circle $\\Gamma_1$ passing through $H, P, Q, C$. Similarly, $L$... | India | Indija TS 2010 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneo... | null | proof only | null | |
0ah3 | We define a grid of type $(n_1, n_2, ..., n_m)$, $n_1 \ge n_2 \ge ... \ge n_m$ in the following way: we put $n_1$ squares horizontally, one next to each other, then we put $n_2$ squares horizontally under the first $n_1$ squares. We repeat the procedure until we get a grid of squares having $n_1$ squares in the first r... | [
"We define a function $f$ on the $m$-tuples $(n_1, n_2, ..., n_m)$ in the following way: $f(n_1, n_2, ..., n_m) = 0$ for the $m$-tuples for which it does not hold that:\n$$\nn_1 \\ge n_2 \\ge \\dots \\ge n_m \\ge 0. \\tag{1}\n$$\nIf $n_1 \\ge n_2 \\ge \\dots \\ge n_m \\ge 0$, then it holds that\n$$\nf(n_1, n_2, \\d... | North Macedonia | XVIII-th Macedonian mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 168 | |
08mt | Problem:
Let $x_{1}, x_{2}, \ldots, x_{n}$ be real numbers satisfying $\sum_{k=1}^{n-1} \min \left(x_{k}, x_{k+1}\right)=\min \left(x_{1}, x_{n}\right)$.
Prove that $\sum_{k=2}^{n-1} x_{k} \geq 0$. | [
"Solution:\nCase I. If $\\min \\left(x_{1}, x_{n}\\right)=x_{1}$, we know that $x_{k} \\geq \\min \\left(x_{k}, x_{k+1}\\right)$ for all $k \\in\\{1,2,3, \\ldots, n-1\\}$. So $x_{1}+x_{2}+\\ldots+x_{n-1} \\geq \\sum_{k=1}^{n-1} \\min \\left(x_{k}, x_{k+1}\\right)=\\min \\left(x_{1}, x_{n}\\right)=x_{1}$, hence $\\s... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
014s | Problem:
Determine all polynomials $p(x)$ with real coefficients such that
$$
p\left((x+1)^3\right) = (p(x)+1)^3
$$
and
$$
p(0) = 0
$$ | [
"Solution:\nConsider the sequence defined by\n$$\n\\left\\{\\begin{array}{l}\na_0 = 0 \\\\\na_{n+1} = (a_n + 1)^3\n\\end{array}\\right.\n$$\nIt follows inductively that $p(a_n) = a_n$. Since the polynomials $p$ and $x$ agree on infinitely many points, they must be equal, so $p(x) = x$."
] | Baltic Way | Baltic Way 2008 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | p(x) = x | |
0i50 | Problem:
Consider the two hands of an analog clock, each of which moves with constant angular velocity. Certain positions of these hands are possible (e.g. the hour hand halfway between the 5 and 6 and the minute hand exactly at the 6), while others are impossible (e.g. the hour hand exactly at the 5 and the minute ha... | [
"Solution:\n\n143 We can look at the twelve-hour cycle beginning at midnight and ending just before noon, since during this time, the clock goes through each possible position exactly once. The minute hand has twelve times the angular velocity of the hour hand, so if the hour hand has made $t$ revolutions from its ... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Other"
] | null | final answer only | 143 | |
0kkl | Problem:
Let $ABCD$ be a unit square. A circle with radius $\frac{32}{49}$ passes through point $D$ and is tangent to side $AB$ at point $E$. Then $DE=\frac{m}{n}$, where $m, n$ are positive integers and $\operatorname{gcd}(m, n)=1$. Find $100m+n$. | [
"Solution:\n\nLet $O$ be the center of the circle and let $F$ be the intersection of lines $OE$ and $CD$. Also let $r=32/49$ and $x=DF$. Then we know\n$$\nx^2+(1-r)^2=DF^2+OF^2=DO^2=r^2\n$$\nwhich implies that $x^2+1-2r=0$, or $1+x^2=2r$. Now,\n$$\nDE=\\sqrt{DF^2+EF^2}=\\sqrt{1+x^2}=\\sqrt{... | United States | HMMT November 2021 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 807 | |
0fm5 | On a right-angled triangle with hypotenuse length $1$, $25$ points are drawn. Prove that at least $9$ of them can always be covered by a $\frac{3}{10}$ radius half-circle. | [] | Spain | Spanija 2012 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | English | proof only | null | |
0i7b | Problem:
$A$ and $B$ are two points on a circle with center $O$, and $C$ lies outside the circle, on ray $A B$. Given that $A B = 24$, $B C = 28$, $O A = 15$, find $O C$. | [
"Solution:\n\nLet $M$ be the midpoint of $A B$; then $\\triangle O M B$ is a right triangle with $O B = 15$, $M B = 12$, so $O M = 9$.\n\nNow $\\triangle O M C$ is a right triangle with $O M = 9$, $M C = 40$, so $O C = 41$."
] | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 41 | |
04pg | A quadruple $(a, b, c, d)$ of positive integers is called *green* if
$$
b = a^2 + 1, \quad c = b^2 + 1, \quad d = c^2 + 1
$$
holds, and $D(a) + D(b) + D(c) + D(d)$ is odd, where $D(k)$ denotes the number of positive divisors of $k$.
How many green quadruples having elements less than 1\ 000\ 000 are there? | [
"Note that $D(k)$ is odd if and only if $k$ is a perfect square.\nFrom $m^2 < m^2 + 1 < m^2 + 2m + 1 = (m+1)^2$ it follows that $m^2 + 1$ is not a perfect square for any positive integer $m$. Hence $b, c$ and $d$ of any green quadruple are not perfect squares, i.e. $D(b), D(c)$ and $D(d)$ are even.\nTherefore, $D(a... | Croatia | Croatian Mathematical Society Competitions | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 2 | |
07h2 | Find all functions $f : \mathbb{R} \to A$ such that
i. $x \in f(y)$ if and only if $y \in f(x)$.
ii. $|x - y| > 2$ if and only if $f(x) \cap f(y) = \emptyset$.
iii. For all real numbers $0 \le r \le 1$; $f(r) = [r^2 - 1, r^2 + 1]$. | [
"We shall firstly prove following lemmas;\n\n**Lemma 1.** For any real number $r$, the length of the interval $f(r)$ is at most 2.\n*Proof.* If $x, y \\in f(r)$ we have $r \\in f(x) \\cap f(y)$ hence we have $|x - y| \\le 2$.\n\n**Lemma 2.** If the interval $I$ has length 2, then there is a real number $r$ such tha... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Let n = floor(x). Define g(x) by:
- if n is even: g(x) = (x - n)^2 + n;
- if n is odd: g(x) = sqrt(x - n) + n.
Then the unique solution is f(x) = [g(x) - 1, g(x) + 1]. | |
04sm | Find all integers $k \ge 2$ for which there exists $k$-element set $M$ of positive integers such that the product of all numbers in $M$ is divisible by the sum of any two (different) numbers from $M$. | [] | Czech Republic | Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | all integers k ≥ 2 | |
0008 | Sea $ABC$ un triángulo con $\hat{C} = 60^\circ$. El punto $P$ es el simétrico de $A$ respecto del punto de tangencia de la circunferencia inscrita con el lado $BC$. Demostrar que si la mediatriz del segmento $CP$ corta a la recta que contiene a la bisectriz del ángulo $\hat{B}$ en el punto $Q$, entonces el triángulo $C... | [] | Argentina | XI Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | español | proof only | null | |
0jvj | Problem:
Let $q(x) = q^{1}(x) = 2x^{2} + 2x - 1$, and let $q^{n}(x) = q(q^{n-1}(x))$ for $n > 1$. How many negative real roots does $q^{2016}(x)$ have? | [
"Solution:\nDefine $g(x) = 2x^{2} - 1$, so that $q(x) = -\\frac{1}{2} + g\\left(x + \\frac{1}{2}\\right)$. Thus\n$$\nq^{N}(x) = 0 \\Longleftrightarrow \\frac{1}{2} = g^{N}\\left(x + \\frac{1}{2}\\right)\n$$\nwhere $N = 2016$.\nBut, viewed as function $g : [-1, 1] \\to [-1, 1]$ we have that $g(x) = \\cos(2 \\arccos(... | United States | HMMT February 2016 | [
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials"
] | null | proof and answer | (1/3)(2^2017 + 1) | |
08np | Problem:
In a circle of diameter $1$ consider $65$ points no three of which are collinear. Prove that there exist $3$ among these points which form a triangle with area less than or equal to $\frac{1}{72}$. | [
"Solution:\n\nLemma: If a triangle $ABC$ lies in a rectangle $KLMN$ with sides $KL = a$ and $LM = b$, then the area of the triangle is less than or equal to $\\frac{ab}{2}$.\n\nProof of the lemma: Without any loss of generality, assume that among the distances of $A, B, C$ from $KL$, that of $A$ is between the othe... | JBMO | JBMO Shortlist | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0kjq | Each of the 12 edges of a cube is labeled $0$ or $1$. Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the 6 faces of the cube equal to $2$?
(A) 8 (B... | [] | United States | AMC 10 A | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | MCQ | E | |
0134 | Problem:
Let $x$, $y$ and $z$ be positive real numbers such that $x y z = 1$. Prove that
$$
(1+x)(1+y)(1+z) \geq 2\left(1+\sqrt[3]{\frac{y}{x}}+\sqrt[3]{\frac{z}{y}}+\sqrt[3]{\frac{x}{z}}\right)
$$ | [
"Solution:\nPut $a = b x$, $b = c y$ and $c = a z$. The given inequality then takes the form\n$$\n\\begin{aligned}\n\\left(1+\\frac{a}{b}\\right)\\left(1+\\frac{b}{c}\\right)\\left(1+\\frac{c}{a}\\right) & \\geq 2\\left(1+\\sqrt[3]{\\frac{b^{2}}{a c}}+\\sqrt[3]{\\frac{c^{2}}{a b}}+\\sqrt[3]{\\frac{a^{2}}{b c}}\\rig... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
05ez | Problem:
Trouver tous les nombres entiers $z \in \mathbb{Z}$ tels que
$$
2^{z}+2=r^{2}
$$
où $r \in \mathbb{Q}$ est un nombre rationnel.
Un nombre rationnel est un nombre qui s'écrit sous la forme $\frac{\mathrm{a}}{\mathrm{b}}$ avec $\mathrm{a}, \mathrm{b}$ des entiers et $\mathrm{b} \neq 0$. | [
"Solution:\n\nNotons que si $(z, r)$ est un couple solution, $(z,-r)$ est également un couple solution. On peut donc supposer pour la suite que $r \\geqslant 0$. Etant donné que $2^{z}+2>0$, on a même $r>0$. On pose $r=\\frac{a}{b}$, avec $a$ et $b$ des entiers strictement positifs et premiers entre eux.\n\nSi $z \... | France | ENVOi 3 : ARITHMÉTIQUE | [
"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 | {-2, 1} | |
0eti | Let $a$, $b$, and $c$ be nonzero integers. Show that there exists an integer $k$ such that
$$
\gcd(a + kb, c) = \gcd(a, b, c).
$$
(Note: 'gcd' stands for 'greatest common divisor') | [
"We may assume that $a$, $b$, $c$ are all positive, since if $a$, $b$, $c$ are all positive, and $\\gcd(a + kb, c) = \\gcd(a, b, c)$ for some integer $k$, then we immediately have $\\gcd(-a + (-k)b, \\pm c) = \\gcd(-a + k(-b), \\pm c) = \\gcd(a + (-k)(-b), \\pm c) = \\gcd(\\pm a, \\pm b, \\pm c)$. Moreover, if at l... | South Africa | The South African Mathematical Olympiad Third Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
02zn | Problem:
A Sequência de Fibonacci é definida recursivamente por $F_{n+2}=F_{n+1}+F_{n}$ para $n \in \mathbb{Z}$ e $F_{1}=F_{2}=1$. Determine o valor de:
$$
\left(1-\frac{F_{2}^{2}}{F_{3}^{2}}\right)\left(1-\frac{F_{3}^{2}}{F_{4}^{2}}\right) \cdot \ldots \cdot\left(1-\frac{F_{2019}^{2}}{F_{2020}^{2}}\right)
$$ | [
"Solution:\n\nPodemos reescrever um termo genérico do produto dado como\n$$\n\\begin{aligned}\n1-\\frac{F_{i}^{2}}{F_{i+1}^{2}} & =\\frac{F_{i+1}^{2}-F_{i}^{2}}{F_{i+1}^{2}} \\\\\n& =\\frac{\\left(F_{i+1}-F_{i}\\right)\\left(F_{i+1}+F_{i}\\right)}{F_{i+1}^{2}} \\\\\n& =\\frac{F_{i-1} F_{i+2}}{F_{i+1}^{2}}\n\\end{al... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | F_{2021} / (2 F_{2019} F_{2020}) | |
0215 | Problem:
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
(x-y)(f(x)+f(y)) \leqslant f\left(x^{2}-y^{2}\right) \quad \text{ for all } x, y \in \mathbb{R}
$$ | [
"Solution:\nClearly, $f(x)=c x$ is a solution for each $c \\in \\mathbb{R}$ since $(x-y)(c x+c y)=c\\left(x^{2}-y^{2}\\right)$. To show that there are no other solutions, we observe that\n\n(1) $x=y: \\quad 0 \\leqslant f(0)$;\n\n$x=1, y=0: \\quad f(0)+f(1) \\leqslant f(1) \\Rightarrow f(0) \\leqslant 0$, whence $f... | Benelux Mathematical Olympiad | 15th Benelux Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All functions of the form f(x) = c x for some real constant c | |
0gem | 費氏數列 $F_0, F_1, F_2, \dots$ 滿足 $F_0 = 0, F_1 = 1$, 且 $F_{n+1} = F_n + F_{n-1}$ 對所有 $n \ge 1$ 都成立。給定 $n \ge 2$, 假設整數集 $S$ 滿足：對於所有 $k = 2, 3, \dots, n$, 存在 $x, y \in S$ 使得 $x - y = F_k$。試求 $S$ 元素個數的最小可能值。
The Fibonacci numbers $F_0, F_1, F_2, \dots$ are defined inductively by $F_0 = 0, F_1 = 1$, and $F_{n+1} = F_n + F_{... | [
"下界為 $|S| \\ge d+1$, 其中 $d$ 為大於等於 $\\frac{n}{2}$ 的最小整數。\n\n1. 估計：讓我們以 $S$ 的元素為點作圖 $G$, 其連邊方式為：對每一個 $1 \\le k \\le d$, 找到一組 $x, y \\in S$ 滿足 $|x - y| = F_{2k-1}$, 就把 $x$ 與 $y$ 連邊；如果有不只一組 $(x, y)$ 滿足 $|x - y| = F_{2k-1}$, 就只取其中一組連邊。同時定義邊 $(x, y)$ 的長度為 $|x - y|$。\n\n以下證明 $G$ 中沒有環。假設 $G$ 中有環 $(x_1, x_2, \\dots, x_\\ell... | Taiwan | 2021 數學奧林匹亞競賽第二階段選訓營, 獨立研究 (一) | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | ceil(n/2) + 1 | |
0gfj | 設 $a, b$ 為實數, 並考慮函數 $f(x) = ax + b + \frac{9}{x}$。證明: 在區間 $[1, 9]$ 中存在一數 $x_0$ 滿足 $|f(x_0)| \ge 2$。 | [] | Taiwan | 2022 數學奧林匹亞競賽第二階段培訓營, 獨立研究(一) | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | Chinese; English | proof only | null | |
0bvz | a) Let $n \in \{2, 3, \dots, 2015\}$ and $A \in \mathcal{M}_n(\mathbb{Q})$ such that $A^{2017} = I_n$. Show that $A = I_n$.
b) Is the previous affirmation true for every positive integer $n$? | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | English | proof and answer | a) A = I_n.
b) No. The statement fails starting at n = 2016; for example, the 2016×2016 companion matrix of the cyclotomic polynomial Φ_2017 satisfies A^{2017} = I but A ≠ I. For any n ≥ 2016, block-diagonal extensions give further counterexamples. | |
086u | Problem:
Quanti interi $n$ sono tali che $\sqrt{n}$ differisce da $\sqrt{101}$ per meno di 1?
(A) 19
(B) 21
(C) 40
(D) 41
(E) 42. | [
"Solution:\n\nLa risposta è (D). Indichiamo con $\\mathcal{A}$ l'insieme degli interi $n$ che soddisfano la proprietà richiesta. Osserviamo innanzitutto che: $\\sqrt{101}-\\sqrt{81}>\\sqrt{100}-\\sqrt{81}=1$, cioè $81 \\notin \\mathcal{A}$.\n\nSe $n$ è un intero positivo, la differenza\n$$\n\\sqrt{n+1}-\\sqrt{n}=\\... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | D | |
0bxs | Let $G$ be a group and let $m$ and $n$ be relatively prime positive integers. Show that, if the functions $f: G \to G$, $f(x) = x^{m+1}$, and $g: G \to G$, $g(x) = x^{n+1}$, are both surjective endomorphisms, then $G$ is commutative.
Liviu Vlaicu | [
"Since $f$ is a morphism, $(xy)^{m+1} = x^{m+1}y^{m+1}$, so $(yx)^m = x^m y^m$ for all $x$ and all $y$ in $G$. Further, $y^{m+1}x^{m+1} = (yx)^{m+1} = (yx)^m(yx) = (x^m y^m)(yx) = x^m y^{m+1}x$, so $y^{m+1}x^m = x^m y^{m+1}$. Since $f$ is surjective, the latter shows that $x^m$ is in the center of $G$. Similarly, $... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Abstract Algebra > Group Theory"
] | English | proof only | null | |
0hhi | Let $k$ and $n$ be arbitrary natural numbers that satisfy the condition $3 \le k \le n$. Prove that among any $n$ pairwise distinct real numbers, there are either $k$ numbers with a positive sum or $(k-1)$ numbers with a negative sum. | [
"If there are no positive numbers in the given set, then there is at most one number that is equal to zero, and all other numbers are negative. Therefore, there are $(n-1)$ negative numbers in the set. Hence, we can take any set with $k-1 \\ge 2$ numbers as the desired subset.\n\nOtherwise, if there is at least one... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0c6w | Let $(a_n)_{n \in \mathbb{N}^*}$ be a sequence such that
$$
2(a_1 + a_2 + \dots + a_n) = n a_{n+1}, \quad \text{for all } n \ge 1.
$$
a) Prove that $(a_n)_{n \in \mathbb{N}^*}$ is an arithmetic sequence.
b) If $[a_1] + [a_2] + \dots + [a_n] = [a_1 + a_2 + \dots + a_n]$, for all $n \in \mathbb{N}^*$, prove that the ter... | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0e4f | Problem:
Določi vse funkcije $f: \mathbb{R} \rightarrow \mathbb{R}$, za katere velja $f(x+y)=f(x-y)+2 f(y) \cos x$ za vsa realna števila $x$ in $y$. | [
"Solution:\n\nV funkcijsko enačbo vstavimo $x=0$ in dobimo $f(-y)=-f(y)$ za vsako realno število $y$. Funkcija $f$ je liha.\n\nVstavimo še $x=\\frac{\\pi}{2}$ in izpeljemo $f\\left(y+\\frac{\\pi}{2}\\right)=f\\left(\\frac{\\pi}{2}-y\\right)$ za vsa realna števila $y$. Z upoštevanjem lihosti dobimo $f\\left(y+\\frac... | Slovenia | 55. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = a sin x for any real constant a | |
0igh | Problem:
Show that for $b$ even, there exists some $M$ such that for every $n > M$, a $2b \times n$ rectangle is $(1, b)$-tileable. | [
"Solution:\n\nRecall from above that we can tile a $2 \\times 2b$ rectangle. Four columns of a $(b+1) \\times 2b$ rectangle can be tiled as shown below, and repeating this $\\frac{b}{2}$ times tiles the entire rectangle. Since any integer at least $b$ can be written as a positive linear combination of $2$ and $b+1$... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Other",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0ar8 | Problem:
Find the last 2 nonzero digits of $16!$ | [] | Philippines | 13th Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | final answer only | 88 | |
01e1 | Required are all functions $f$ mapping non-negative reals to non-negative reals, fulfilling the identity
$$
f(x_1^2 + \cdots + x_n^2) = f(x_1)^2 + \cdots + f(x_n)^2
$$
for any choice of numbers $x_1, \dots, x_n$. | [
"Answer: the functions $f(x) = 0$ and $f(x) = x$.\nA first observation is that\n$$\nf(1) = f(1^2) = f(1)^2,\n$$\nso that $f(1)$ is either 0 or 1.\nAssume first that $f(1) = 0$. For each positive integer $n$, we find\n$$\nf(n) = f(n \\cdot 1^2) = n f(1)^2 = 0.\n$$\nGiven an arbitrary $x$, find $y$ so that $x^2 + y^2... | Baltic Way | Baltic Way shortlist | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 0 and f(x) = x | |
0a9x | Problem:
Let $ABC$ be a triangle and $\Gamma$ the circle with diameter $AB$. The bisectors of $\angle BAC$ and $\angle ABC$ intersect $\Gamma$ (also) at $D$ and $E$, respectively. The incircle of $ABC$ meets $BC$ and $AC$ at $F$ and $G$, respectively. Prove that $D, E, F$ and $G$ are collinear. | [
"Solution:\nLet the line $ED$ meet $AC$ at $G'$ and $BC$ at $F'$. $AD$ and $BE$ intersect at $I$, the incenter of $ABC$. As angles subtending the same arc $\\widehat{BD}$, $\\angle DAB = \\angle DEB = \\angle G' EI$. But $\\angle DAB = \\angle CAD = \\angle G' AI$. This means that $E, A, I$ and $G'$ are concyclic, ... | Nordic Mathematical Olympiad | The 29th Nordic Mathematical Contest | [
"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... | null | proof only | null | |
0d3t | Let $S=\{(a, b) \mid a, b=1,2,3,4$, and $a \neq b\}$, and consider all nonzero polynomials $p(X, Y)$ with integer coefficients such that $p(a, b)=0$ for every element $(a, b)$ in $S$.
(a) What is the minimal degree of such polynomial $p(X, Y)$?
(b) Determine all such polynomials $p(X, Y)$ with minimal degree. | [
"(a) Let $p(X, Y)$ be such a polynomial. There are two possible cases:\n\nCase 1. If $p(1, Y)=p(2, Y)=p(3, Y)=p(4, Y)=0$. In this case\n$$\n(X-1)(X-2)(X-3)(X-4)\n$$\nis a factor of $p(X, Y)$ and therefore $\\deg p(X, Y) \\geq 4$.\n\nCase 2. If there exists $a_{0} \\in \\{1,2,3,4\\}$ such that $p\\left(a_{0}, Y\\rig... | Saudi Arabia | SAMC | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English, Arabic | proof and answer | (a) 3. (b) Exactly the integer multiples of (X+Y−5)(X^2+Y^2−5X−5Y+10). | |
015d | Let $a$, $b$, $c$ be positive numbers such that $a$, $b$, $c > 1$ and
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 2.
$$
Prove that $\sqrt{a+b+c} \ge \sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1}$. | [
"From Cauchy-Schwartz we get\n$$\n(a+b+c) \\left( \\frac{a-1}{a} + \\frac{b-1}{b} + \\frac{c-1}{c} \\right) \\geq \\left( \\sqrt{a-1} + \\sqrt{b-1} + \\sqrt{c-1} \\right)^2.\n$$\nAs\n$$\n\\frac{a-1}{a} + \\frac{b-1}{b} + \\frac{c-1}{c} = 3 - \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) = 1,\n$$\nwe g... | Baltic Way | Baltic Way SHL | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
090p | Alice plays a game using a board with 20 rows and 25 columns. Initially, no number is written in any of the $20 \times 25$ cells. The game proceeds in several turns, and on the $n$-th turn, the following operation is performed:
Choose a positive integer $k$ and $k$ empty cells $A_1, A_2, \dots, A_k$ such that for every... | [
"$20!^3$\n\nLet $k$ and $l$ be positive integers. We denote by $(k, l)$ the cell in the $k$-th column from the left and the $l$-th row from the bottom. For the cell $(k, l)$, we define its *height* as $k + l - 1$. Note that the possible values for the height of a cell range from 1 to 44. For each integer $1 \\le i ... | Japan | The 35th Japanese Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | 20!^3 | |
071z | Problem:
Do there exist three distinct positive real numbers $a, b, c$ such that the numbers $a, b, c, b+c-a, c+a-b, a+b-c$ and $a+b+c$ form a 7-term arithmetic progression in some order? | [
"Solution:\n\nWe show that the answer is NO. Suppose, if possible, let $a, b, c$ be three distinct positive real numbers such that $a, b, c, b+c-a, c+a-b, a+b-c$ and $a+b+c$ form a 7-term arithmetic progression in some order. We may assume that $a < b < c$. Then there are only two cases we need to check:\n\n(I) $a+... | India | INMO | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | No | |
0ci1 | Consider a square $ABCD$ and the points $E$ on the side $CD$, $M$ on the diagonal $AC$ and $P$ on the side $BC$, such that $\angle BAE = \angle AEM = \angle AMP$. Prove that:
a) the triangle $AMP$ is isosceles;
b) $EM = DE + PB$. | [
"a) We denote $\\angle BAE = \\angle AEM = \\angle AMP = \\alpha$.\nWe have, in turn $\\angle DAE = 90^\\circ - \\alpha$, $\\angle EAC = \\alpha - 45^\\circ$, $\\angle DEA = \\angle EAB = \\alpha$, $\\angle CEM = 180^\\circ - 2\\alpha$.\nFrom the triangle $CPM$ follows that $\\angle PMC = 180^\\circ - \\angle AMP =... | Romania | 74th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
04cj | Find all solutions of the equation $m! + 2 = n^2$, where $m$ and $n$ are positive integers. | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (m, n) = (2, 2) | |
0iwk | Problem:
How many hits does "3.1415" get on Google? Quotes are for clarity only, and not part of the search phrase. Also note that Google does not search substrings, so a webpage with $3.14159$ on it will not match $3.1415$. If $A$ is your answer, and $S$ is the correct answer, then you will get $\max \left(25 - | \ln... | [
"Solution:\n\nAnswer: 422000"
] | United States | Harvard-MIT November Tournament | [
"Math Word Problems"
] | null | final answer only | 422000 | |
0l22 | Problem:
Ash and Gary independently come up with their own lineups of 15 fire, grass, and water monsters. Then, the first monster of both lineups will fight, with fire beating grass, grass beating water, and water beating fire. The defeated monster is then substituted with the next one from their team's lineup; if the... | [
"Solution:\n\nFirst, we show Ash cannot do better. Notice there is a $\\frac{2^{15}}{3^{15}}$ chance that Gary's $i$-th monster ties or defeats Ash's $i$-th monster for each $i$. If this is the case, Ash cannot win, as Ash's $i$-th monster will always be defeated by Gary's $i$-th monster, if not sooner. Thus, Ash w... | United States | HMMT February 2024 Guts Round | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 1 - 2^15/3^15 | |
0gvj | Can the number $1$ be represented as a sum of $2005$ different terms, each one of the form $\frac{1}{3n-1}$ with a positive integer $n$? | [
"Якщо доданки даної суми звести до спільного знаменника, то за mod $3$ чисельник матиме остачу $1$, а знаменник — остачу $2$. Таким чином, така сума не може дорівнювати $1$."
] | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | No | |
0bgt | Problem:
Fie $(A, +, \cdot)$ un inel cu proprietatea că $x = 0$ este unica soluție a ecuației $x^{2} = 0$, $x \in A$. Fie $B = \{ a \in A \mid a^{2} = 1 \}$. Să se arate că:
a. $ab - ba = bab - a$, oricare ar fi $a \in A$ și $b \in B$.
b. $(B, \cdot)$ este grup. | [] | Romania | Olimpiada Naţională de Matematică, Etapa judeţeană şi a municipiului Bucureşti | [
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
00cp | Se tiene una progresión aritmética de 7 términos en la que todos los términos son números primos diferentes. Determinar el menor valor posible del último término de una progresión.
ACLARACIÓN: En una progresión aritmética de diferencia $d$ cada término es igual al anterior más $d$. | [] | Argentina | Nacional OMA 2019 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | Spanish | proof and answer | 1307 | |
0igx | Problem:
Find a real, irreducible quartic polynomial with leading coefficient 1 whose roots are all twelfth roots of unity. | [
"Solution:\n\nAll twelfth roots of unity are roots of\n$$\n\\begin{aligned}\nx^{12}-1 & =\\left(x^{6}-1\\right)\\left(x^{6}+1\\right) \\\\\n& =\\left(x^{3}-1\\right)\\left(x^{3}+1\\right)\\left(x^{6}+1\\right) \\\\\n& =(x-1)\\left(x^{2}+x+1\\right)(x+1)\\left(x^{2}-x+1\\right)\\left(x^{2}+1\\right)\\left(x^{4}-x^{2... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | x^4 - x^2 + 1 | |
0alq | Problem:
If $\frac{\log x}{\log y} = 500$, what is the value of $\frac{\log (y / x)}{\log y}$?
(a) -498
(b) -501
(c) -502
(d) -499 | [
"Solution:\nLet $\\frac{\\log x}{\\log y} = 500$.\nSo, $\\log x = 500 \\log y$.\n\nNow,\n\\begin{align*}\n\\frac{\\log (y / x)}{\\log y} &= \\frac{\\log y - \\log x}{\\log y} \\\\\n&= \\frac{\\log y - 500 \\log y}{\\log y} \\\\\n&= \\frac{(1 - 500) \\log y}{\\log y} \\\\\n&= 1 - 500 \\\\\n&= -499\n\\end{align*}\n\n... | Philippines | Qualifying Round | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | MCQ | d | |
079z | Let $\omega$ be the circumcircle of an acute triangle $ABC$. Let $D$ be the midpoint of arc $\widearc{BAC}$ in $\omega$ and $I$ be the incenter of triangle $ABC$. Suppose $DI$ intersects $BC$ at $E$ and $\omega$ at $F$ for the second time. Suppose the parallel line to $AI$ from $E$ meets $AF$ at $P$. Prove that $PE$ is... | [
"Let $T$ be the point of intersection of the perpendicular bisector of $BC$ and circle $\\omega$, so $TD$ is a diagonal of $\\omega$ and $\\angle DFT = 90^\\circ$. Since $D$ is the midpoint of arc $\\widearc{BAC}$, $FD$ is the angle bisector of $\\angle BFC$. Therefore $(JEBC) = -1$ where $J$ is the intersection po... | Iran | Iranian 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 > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometr... | English | proof only | null | |
0c63 | If $S$ is a $10^4$-digit binary string consisting of zeroes and ones, and $k \le 10^4$ is a positive integer, a $k$-block of $S$ is any substring consisting of $k$ consecutive digits; two $k$-blocks, $a_1a_2\dots a_k$ and $b_1b_2\dots b_k$, are of the same type if $a_i = b_i$, $i = 1, \dots, k$. Consider all $10^4$-dig... | [
"The required maximum is 504 and is achieved for a string (to be described below) containing all 7 possible 3-block types different from 000.\n\nTo prove this, let $f(k)$ be the maximum number of $k$-digit strings whose 3-blocks are of at most 7 types. Clearly, $f(1) = 2$, $f(2) = 4$ and $f(3) = 7$.\n\nWe will show... | Romania | SELECTION TESTS FOR THE 2019 BMO AND IMO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 504 | |
098g | Problem:
Fie numerele $a, b, c \in \mathbb{N}^{*}$, astfel încât $\frac{a \sqrt{2023}+b}{a+c \sqrt{2023}}$ este un număr rațional. Arătați că $\left(a^{2}+b^{2}+c^{2}\right) \vdots(a+b+c)$. | [
"Solution:\n\nFie $\\frac{a \\sqrt{2023}+b}{a+c \\sqrt{2023}} \\in \\mathbb{Q}$. Atunci $\\frac{a \\sqrt{2023}+b}{a+c \\sqrt{2023}}=\\frac{m}{n}$, $m, n \\in \\mathbb{N}^{*}$. Obținem relaţia $n a \\sqrt{2023}+n b=m a+m c \\sqrt{2023} \\Leftrightarrow(n a-m c) \\sqrt{2023}=m a-n b$, dar $a, b, c, m, n \\in \\mathbb... | Moldova | Olimpiada Republicană la Matematică | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0duv | Problem:
Od 25 učencev so pri pisni nalogi štirje dobili oceno 5, pet oceno 4 in pet oceno 2. Koliko učencev je dobilo oceno 1 in koliko učencev oceno 3, če je bila povprečna ocena natanko 3? Zapiši odgovor. | [
"Solution:\n\nZapis enačbe\n$$\n\\frac{4 \\cdot 5 + 5 \\cdot 4 + 2 \\cdot 5 + x + 3y}{25} = 3\n$$\n\n$x + y = 11$\n\nReševanje sistema - postopek\n\nRešitvi: $x = 4$\n\n$y = 7$\n\nOdgovor: Oceno 3 je dobilo 7 učencev, oceno 1 pa 4 učenci."
] | Slovenia | 2. matematično tekmovanje dijakov srednjih tehniških in strokovnih šol | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | Grade 1: 4; Grade 3: 7 | |
04p6 | Point $P$ is the midpoint of the segment $\overline{AB}$ of length $2$. Let $T$ be the point at which the tangent from $A$ touches the circle with diameter $\overline{PB}$. Find the length of the segment $\overline{PT}$.
(Petar Bakić) | [] | Croatia | Croatian Mathematical Society Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | 1/√3 | |
0bz7 | Let $a, b, c, d \in [0, 1]$. Prove that
$$
\frac{a}{1+b} + \frac{b}{1+c} + \frac{c}{1+d} + \frac{d}{1+a} + abcd \le 3.
$$ | [
"We notice that\n$$\n\\frac{a}{1+b} + \\frac{b}{1+c} + \\frac{c}{1+d} + \\frac{d}{1+a} + abcd \\le \\frac{a}{1+abcd} + \\frac{b}{1+abcd} + \\frac{c}{1+abcd} + \\frac{d}{1+abcd} + abcd = \\frac{a+b+c+d}{1+abcd} + abcd.\n$$\nUsing repeatedly the inequality $x + y \\le 1 + xy$, $\\forall x, y \\in [0, 1]$ (equivalent ... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
05pv | Problem:
Si $n \in \mathbb{N}^{*}$, on note $d(n)$ son nombre de diviseurs. Quels sont les entiers strictement positifs tels que $d(n)^{3}=4 n$ ? | [
"Solution:\n\nSoit $f(n):=\\frac{d(n)^{3}}{n}$, l'énoncé revient à trouver les antécédents de $4$ par $f$. On remarque que $f(a b)=f(a) f(b)$ si $a$ et $b$ sont premiers entre eux. Ceci nous incite à décomposer $n$ en produit de facteurs premiers :\n\n$$\nn=p_{1}^{a_{1}} \\times p_{2}^{a_{2}} \\times \\cdots \\time... | France | OCympiades Françaises de Mathématiques | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2, 128, 2000 | |
0kq4 | Problem:
A string consisting of letters $A$, $C$, $G$, and $U$ is untranslatable if and only if it has no $AUG$ as a consecutive substring. For example, $ACUGG$ is untranslatable.
Let $a_{n}$ denote the number of untranslatable strings of length $n$. It is given that there exists a unique triple of real numbers $(x, ... | [
"Solution:\n\nIf a sequence is untranslatable, the first $n-1$ letters must form an untranslatable sequence as well. Therefore, we can count $a_{n}$ by\n- Append any letter to an untranslatable sequence of length $n-1$, so $4 a_{n-1}$ ways.\n- Then, subtract with the case when the sequence ends with $AUG$. There ar... | United States | HMMT November 2022 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | (4, 0, -1) | |
02x9 | Problem:
Uma máquina de cortar grama mais eficiente está sendo desenvolvida. Para isso, em um vértice de um quadrado de grama, de lado $m$, prende-se a ponta de uma haste metálica de comprimento $p$ e na outra ponta da haste prende-se um triângulo equilátero, por um de seus vértices, de lado $l$, sendo $p+l<m$. O triâ... | [
"Solution:\n\nQuando o triângulo gira ao redor de um de seus vértices, seu alcance é limitado por uma circunferência de raio $l$. Assim, o raio do maior arco da área de corte é $(p+l)$, enquanto que o raio do menor arco da área de corte é $(p-l)$. Portanto, a área de corte é\n$$\n\\frac{\\pi(p+l)^2-\\pi(p-l)^2}{4}=... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | π p l | |
0bu5 | Problem:
Fie numerele reale pozitive $a, b, c$, astfel încât
$$
\frac{a}{b+c+1}+\frac{b}{a+c+1}+\frac{c}{a+b+1} \leq 1
$$
Demonstraţi că:
$$
\frac{1}{b+c+1}+\frac{1}{a+c+1}+\frac{1}{a+b+1} \geq 1
$$ | [
"Solution:\n\nDin inegalitatea dintre media aritmetică şi media armonică deducem că\n$$\n\\left(\\sum (b+c+1)\\right) \\sum \\frac{1}{b+c+1} \\geq 9\n$$\nde unde\n$$\n\\left(a+b+c+\\frac{3}{2}\\right) \\sum \\frac{1}{b+c+1} \\geq \\frac{9}{2}\n$$\nAvem însă\n$$\n\\left(a+b+c+\\frac{3}{2}\\right) \\cdot \\frac{1}{b+... | Romania | Olimpiada Naţională de Matematică, Etapa Judeţeană şi a Municipiului Bucureşti | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0ezr | Problem:
A switch has two inputs $1$, $2$ and two outputs $1$, $2$. It either connects $1$ to $1$ and $2$ to $2$, or $1$ to $2$ and $2$ to $1$. If you have three inputs $1$, $2$, $3$ and three outputs $1$, $2$, $3$, then you can use three switches, the first across $1$ and $2$, then the second across $2$ and $3$, and f... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | null | proof and answer | 6 | |
0ey4 | Problem:
Four lighthouses are arbitrarily placed in the plane. Each has a stationary lamp which illuminates an angle of $90$ degrees. Prove that the lamps can be rotated so that at least one lamp is visible from every point of the plane. | [
"Solution:\n\nTake a north direction, arbitrary except that no points are aligned north-south or east-west. Take the two most northerly points. Point the lamp for the more easterly of these two in the direction SW (so that it covers directions S to W). Point the lamp for the other in the direction SE. For the other... | Soviet Union | 1st ASU | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
0flh | Problem:
Cada número racional se pinta de un color, usando sólo dos colores, blanco y rojo. Se dice que una tal coloración es sanferminera cuando para cada dos números racionales $x, y$, con $x \neq y$, si se cumple una de las tres condiciones siguientes:
a) $x y=1$,
b) $x+y=0$,
c) $x+y=1$,
entonces $x$ e $y$ están pin... | [
"Solution:\nSi una coloración es sanferminera, podemos hallar otra coloración sanferminera intercambiando simultáneamente el color de cada racional, de rojo a blanco y de blanco a rojo; si en la coloración inicial dos racionales tienen distinto color, también lo tendrán en la resultante. Hallemos entonces el número... | Spain | XLVII Olimpiada Matemática Española, Fase nacional (Pamplona) | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2 | |
0ahc | If $a$, $b$, $c$, $d$ are positive real numbers such that $abcd = 1$ then prove that the inequality
$$
\frac{1}{bc+cd+da-1} + \frac{1}{ab+cd+da-1} + \frac{1}{ab+bc+da-1} + \frac{1}{ab+bc+cd-1} \le 2
$$
holds. | [
"By multiplying $1+bc+cd+da$ and $1+ab$ together we get\n$$\n(1+bc+cd+da)(1+ab) = 1+bc+cd+da+ab+ab^2c+abcd+a^2bd = \\\\\n= 2+ab+bc+cd+da+\\frac{b}{d}+\\frac{a}{c}\n$$\nFrom the inequality between the arithmetic and the geometric mean for the positive numbers $\\frac{b}{d}$ and $\\frac{a}{c}$, and from the equality ... | North Macedonia | 19-th Macedonian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0fei | Problem:
Hallad las cuatro últimas cifras de $3^{2004}$. | [
"Solution:\n\nTenemos que $3^{2}=9=10-1$. Gracias a ello, la fórmula del binomio de Newton nos permite simplificar los cálculos:\n$$\n\\begin{aligned}\n& 3^{2004}=(10-1)^{2002} \\equiv -\\left(\\begin{array}{c}\n1002 \\\\\n3\n\\end{array}\\right) 10^{3}+\\left(\\begin{array}{c}\n1002 \\\\\n2\n\\end{array}\\right) 1... | Spain | null | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 0081 |
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