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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
04hy | The area of the intersection of a smaller and larger square equals two thirds of the area of the smaller square, as well as one fifth of the area of their union. Determine the ratio of the sides of the smaller and larger square.
(Andrea Aglić-Aljinović) | [] | Croatia | Croatia Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | sqrt(3) | |
0hj7 | Problem:
Let $x$ and $y$ be integers such that $\frac{3x + 4y}{5}$ is an integer. Prove that $\frac{4x - 3y}{5}$ is an integer. | [
"Solution:\n\nThe basic strategy is to combine the facts that $x$, $y$, and $\\frac{3x + 4y}{5}$ are all integers. Here is one solution:\n$$\n2(x) + 1(y) - 2\\left(\\frac{3x + 4y}{5}\\right) = \\frac{5(2x + y) - 2(3x + 4y)}{5} = \\frac{10x + 5y - 6x - 8y}{5} = \\frac{4x - 3y}{5}\n$$\nSince the left side is clearly ... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof only | null | |
0iyz | Problem:
We write $\{a, b, c\}$ for the set of three different positive integers $a, b$, and $c$. By choosing some or all of the numbers $a, b$ and $c$, we can form seven nonempty subsets of $\{a, b, c\}$. We can then calculate the sum of the elements of each subset. For example, for the set $\{4,7,42\}$ we will find ... | [
"Solution:\n\nThe answer is five.\nFor example, the set $\\{2,3,5\\}$ has $2,3,5,5,7,8$, and $10$ as its sums, and the first five of those are prime. If you're worried about $5$ appearing twice in that list, then try $\\{2,3,11\\}$ which has $2,3,11,5,13,14$, and $16$ as its subsets' sums, so now we see five differ... | United States | Bay Area Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 5; for example {2,3,11} | |
0jag | Problem:
Alice is sitting in a teacup ride with infinitely many layers of spinning disks. The largest disk has radius $5$. Each succeeding disk has its center attached to a point on the circumference of the previous disk and has a radius equal to $2/3$ of the previous disk. Each disk spins around its center (relative ... | [
"Solution:\n\nAnswer: $18\\pi$\n\nSuppose the center of the largest teacup is at the origin in the complex plane, and let $z = \\frac{2}{3} e^{\\pi i t / 6}$. The center of the second disk is at $5 e^{\\pi i t / 6}$ at time $t$; that is, $\\frac{15}{2} z$. Then the center of the third disk relative to the center of... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Transformations > Inversion",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 18π | |
0cgq | Consider a positive integer $n$ and the set $A_n = \{1, 3, 5, \dots, 2n - 1\}$. For each pair $(a, b)$, where $a, b \in A_n$ we construct the concatenated number $m = \overline{ab}$, obtained by joining the numbers $a$ and $b$. For instance, for $19, 37 \in A_{30}$, the concatenated number is $m = 1937$.
a) What is th... | [
"a) We cannot obtain a perfect square by concatenating two elements from the set $A_{10} = \\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\\}$.\n\nThe elements $1$ and $21$ from the set $A_{11} = \\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21\\}$ yield the perfect square $121$. Hence, the answer is $n = 11$.\n\nb) By concatenating ... | Romania | 74th Romanian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a) n = 11; b) 7921 | |
090m | In the interior of an acute triangle $ABC$ that is not isosceles, there are three distinct points $A_1, B_1, C_1$ such that $AB_1 : CB_1 = AB : CB$ and $AC_1 : BC_1 = AC : BC$. Let $A_2$ be the point symmetric to $A_1$ with respect to the line $BC$, let $B_2$ be the point symmetric to $B_1$ with respect to the line $AC... | [
"For three distinct points $X_1, X_2, X_3$, when the line $X_1X_2$ is rotated counterclockwise by an angle $\\theta$ about $X_1$ to coincide with the line $X_1X_3$, this $\\theta$ is denoted by $\\angle X_2X_1X_3$. Note that differences of $180^\\circ$ are disregarded. First, we show the following lemma concerning ... | Japan | The 35th Japanese Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangent... | English | proof only | null | |
06mm | Let $ABCD$ be a cyclic quadrilateral and $E$ be the intersection of $AC$ and $BD$. $P$ and $Q$ are two points on $AC$ such that the points $A$, $E$, $Q$, $P$, $C$ lie on the same straight line in this order, and that $BP$ bisects $\angle ABC$ whereas $DQ$ bisects $\angle ADC$. If $AE = 4$, $EQ = 2$ and $QP = 3$, find t... | [
"Note that $\\sin \\angle DAB = \\sin \\angle DCB$ as the two angles are supplementary. Using $[XYZ]$ to denote the area of $XYZ$, we have\n$$\n\\frac{AE}{EC} = \\frac{[DAB]}{[DCB]} = \\frac{\\frac{1}{2} \\cdot AD \\cdot AB}{\\frac{1}{2} \\cdot DC \\cdot BC} = \\frac{AB}{BC} \\cdot \\frac{AD}{DC} \\quad (1)\n$$\nBy... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | 15 | |
0kd5 | Problem:
Let $ABCD$ be a tetrahedron such that its circumscribed sphere of radius $R$ and its inscribed sphere of radius $r$ are concentric. Given that $AB = AC = 1 \leq BC$ and $R = 4r$, find $BC^2$. | [
"Solution:\n\nLet $O$ be the common center of the two spheres. Projecting $O$ onto each face of the tetrahedron will divide it into three isosceles triangles. Unfolding the tetrahedron into its net, the reflection of any of these triangles about a side of the tetrahedron will coincide with another one of these tria... | United States | HMMT February 2020 | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | 1 + sqrt(7/15) | |
07ad | For nonnegative integers $m$ and $n$, the sequence $a(m, n)$ of real numbers is defined as follows: $a(0, 0)$ is equal to $2$, and for each natural number $n$, $a(0, n) = 1$ and $a(n, 0) = 2$. Also for $m, n \in \mathbb{N}$:
$$
a(m, n) = a(m - 1, n) + a(m, n - 1)
$$
Prove that for each natural number $k$, all roots of ... | [
"Let $Q_k(x) = \\sum_{i=0}^{k} a(i, 2k - 2i)x^i$. According to the recurrence relation for $a(m, n)$, we get\n$$\n\\begin{align*}\nP_k(x) &= xP_{k-1}(x) + Q_k(x) \\\\\nQ_k(x) &= xQ_{k-1}(x) + P_{k-1}(x).\n\\end{align*}\n$$\nSo $Q_k(x) = P_k(x) - xP_{k-1}(x)$ and therefore, $P_k(x) - xP_{k-1}(x) = x(P_{k-1}(x) - xP_... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0bn3 | A triangle $ABC$ with $AB = AC$ is obtuse at $A$. Let $M$ be the mirror image of $A$ across $C$. The perpendicular bisector of the line segment $AM$ meet the line $AB$ at point $P$. Given that lines $PM$ and $BC$ are perpendicular, prove that $APM$ is an equilateral triangle.
Marcel Neferu | [
"Lines $BC$ and $PM$ meet at $D$. Denote $x$ the measure of $\\angle ABC$. Then $\\angle MCD = \\angle ACB = \\angle ABC = x$ and $\\angle PMC = 90^\\circ - x$. The triangle $PAM$ is isosceles, since $PC$ is median as well as perpendicular bisector of the line segment $AM$, hence $\\angle PMC = \\angle PAC$. On the... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
0i5w | Problem:
given that $a, b, c$ are positive integers satisfying
$$
a+b+c=\operatorname{gcd}(a, b)+\operatorname{gcd}(b, c)+\operatorname{gcd}(c, a)+120
$$
determine the maximum possible value of $a$. | [
"Solution:\n\n240. Notice that $(a, b, c) = (240, 120, 120)$ achieves a value of 240. To see that this is maximal, first suppose that $a > b$. Notice that\n$$\na + b + c = \\operatorname{gcd}(a, b) + \\operatorname{gcd}(b, c) + \\operatorname{gcd}(c, a) + 120 \\leq \\operatorname{gcd}(a, b) + b + c + 120,\n$$\nor $... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 240 | |
0b1j | Problem:
Find the sum of the squares of the real roots of the equation $2x^{4} - 3x^{3} + 7x^{2} - 9x + 3 = 0$. | [
"Solution:\nBy synthetic division by $(x-1)$, we get\n$$\n\\begin{aligned}\n2x^{4} - 3x^{3} + 7x^{2} - 9x + 3 &= (x-1)\\left(2x^{3} - x^{2} + 6x - 3\\right) \\\\\n&= (x-1)(2x-1)\\left(x^{2} + 3\\right)\n\\end{aligned}\n$$\nHence, the only real roots of $2x^{4} - 3x^{3} + 7x^{2} - 9x + 3 = 0$ are $1$ and $\\frac{1}{... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 5/4 | |
0479 | Given an integer $n > 1$, let the real number $x > 1$ satisfy
$$
x^{101} - n x^{100} + n x - 1 = 0.
$$
Prove that for any real numbers $0 < a < b < 1$, there exists a positive integer $m$ such that
$$
a < \{x^m\} < b.
$$
Here $\{t\} = t - \lfloor t \rfloor$ denotes the fractional part of the real number $t$. | [
"*Proof.* We will sequentially prove the following conclusions:\n\n(1) The equation (5) has 99 roots with modulus equal to 1.\n\nClearly, $x = 1$ is a root of the equation (5). Consider the equation\n$$\n\\frac{x^{101} - n x^{100} + n x - 1}{x - 1} = 0,\n$$\ni.e.,\n$$\nf(x) = x^{100} - (n - 1) \\sum_{j=1}^{99} x^j ... | China | The 65th IMO China National Team Selection Test | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Intermediate Algebra > Complex numbers",
"Discrete Mathematics > Combinatorics > P... | English | proof only | null | |
017j | A lizard wants to walk from one corner to the diametrically opposite corner of a regular dodecahedron with edge length $1$. Prove that the lizard has to walk a distance of at least $4$.
(A regular dodecahedron is a Platonic solid consisting of twelve regular pentagons.) | [
"\nLet $\\theta$ be the angle between two diagonals in a regular pentagon. Then the angle at each vertex is $3\\theta$. Let $d$ denote the length of the diagonal in the pentagons. Then considerations of similar triangles give\n$$\n\\frac{d-1}{1} = \\frac{1}{d}\n$$\n$$\n\\text{so } d^2 = d +... | Baltic Way | BALTIC WAY | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
08kn | Problem:
Let $p$ be a prime number and let $a$ be an integer. Show that if $n^{2}-5$ is not divisible by $p$ for any integer $n$, there exist infinitely many integers $m$ so that $p$ divides $m^{5}+a$. | [
"Solution:\nWe start with a simple fact:\n\nLemma: If $b$ is an integer not divisible by $p$ then there is an integer $s$ so that $s b$ has the remainder $l$ when divided by $p$.\nFor a proof, just note that numbers $b, 2b, \\ldots, (p-1)b$ have distinct non-zero remainders when divided by $p$, and hence one of the... | JBMO | OJBM | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0c8t | In a country there are $n$ airports and $n$ air companies operating return flights. Each company operates an odd number of flights forming a closed route. Prove that a traveller can complete a closed route consisting of an odd number of flights operated by pairwise distinct companies. | [
"In graph-theoretic setting, the statement reads:\nConsider a collection of $n$ odd cycles, not necessarily distinct, all on the same vertex set of size $n$. Prove that at most one edge can be chosen from each of these cycles to form a collection that contains the edges of an odd cycle.\nCall a set of edges *rainbo... | Romania | Romanian Master of Mathematics | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
01oe | $N$ boys ($N \ge 3$), no two of them having the same height, are arranged along a circle. A boy in the given arrangement is said to be *middle* if he is taller than one of his neighbors and shorter than the other one.
Prove that if $N$ is odd number then there exists at least one middle boy in any arrangement of the bo... | [
"Consider arbitrary arrangement of the boys along the circle. We put\n\nthe signs \"+\" or \"-\" before any boy in accordance with the following rule: we move clockwise along the circle and put the sign \"+\" before the boy if he is taller than the previous boy and we put the sign \"-\" if ... | Belarus | 62nd Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
00p6 | Find all pairs of integers $(x, y)$, such that $x^3 = 2y^2 + 1$. | [
"Write $x^3 = (1 + y\\sqrt{-2})(1 - y\\sqrt{-2})$. The identity\n$$\n(1 - y\\sqrt{-2} - y^2)(1 + y\\sqrt{-2}) - y^2(1 - y\\sqrt{-2}) = 1\n$$\nshows that $1 + y\\sqrt{-2}$ and $1 - y\\sqrt{-2}$ are relatively prime in $\\mathbb{Z}[\\sqrt{-2}]$. Since $\\mathbb{Z}[\\sqrt{-2}]$ is a unique factorization domain,\n$$\n1... | Balkan Mathematical Olympiad | BMO 2010 Shortlist | [
"Number Theory > Algebraic Number Theory > Unique factorization",
"Number Theory > Algebraic Number Theory > Quadratic fields",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (1, 0) | |
077k | Problem:
Let $\mathbb{R}[x]$ be the set of all polynomials with real coefficients, and let $\operatorname{deg} P$ denote the degree of a nonzero polynomial $P$. Find all functions $f: \mathbb{R}[x] \rightarrow \mathbb{R}[x]$ satisfying the following conditions:
- $f$ maps the zero polynomial to itself,
- for any non-ze... | [
"Solution:\nWe have $f(p)=p$ for all $p \\in \\mathbb{R}[x]$, or $f(p)=-p$ for all $p \\in \\mathbb{R}[x]$. These clearly satisfy the given conditions.\n\nProof\nClaim 1 For all $p \\in \\mathbb{R}[x], f(f(p))=p$.\nProof. Using condition 3 on the polynomials $p$ and $f(p)$, we see that $p-f(f(p))$ has the same set ... | India | INMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof and answer | f(p) = p for all polynomials p, or f(p) = -p for all polynomials p | |
0f78 | Problem:
Given $n$ points can one build $n-1$ roads, so that each road joins two points, the shortest distance between any two points along the roads belongs to $\{1, 2, 3, \ldots, n(n-1)/2\}$, and given any element of $\{1, 2, 3, \ldots, n(n-1)/2\}$ one can find two points such that the shortest distance between them... | [] | Soviet Union | 20th ASU | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Possible exactly for n = 2, 3, and 4 (e.g., a path with edge lengths 1; 1,2; and 1,3,2, respectively). Impossible for n ≥ 5. | |
076g | Let $\triangle ABC$ be triangle in which $AB = AC$. Suppose the orthocentre of the triangle lies on the in-circle. Find the ratio $AB/BC$. | [
"Since the triangle is isosceles, the orthocentre lies on the perpendicular $AD$ from $A$ on to $BC$. Let it cut the in-circle at $H$. Now we are given that $H$ is the orthocentre of the triangle. Let $AB = AC = b$ and $BC = 2a$. Then $BD = a$. Observe that $b > a$ since $b$ is the hypotenuse and $a$ is a leg of a ... | India | IND_National | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles ... | null | proof and answer | 3/4 | |
0hak | 60 participants took part in the Olympiad. They were offered 8 tasks, evaluated from 0 to 7 points each. Prove, that in total there are 3 participants, whose results differ in not more than 1 point. Would the statement be true, if 58 took part in the Olympiad?
*Result of a participant at Olympiad is the total amount o... | [
"Minimal amount of points, that was possible to earn equals to $0$, maximum – to $56$. Consider such segments in points: $[0; 1]$, $[2; 3]$, $[4; 5]$, ..., $[54; 55]$ and $56$ points. If at least 3 students are in at least one of these segments, the statement is proved. If not, then for each of these segments there... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | Yes for sixty participants; no for fifty-eight participants. A counterexample for fifty-eight is two participants at each even total from zero to fifty-six. | |
05f6 | Problem:
Déterminer tous les triplets d'entiers $(x, y, z)$ vérifiant la propriété suivante :
$$
\operatorname{pgcd}(x, y, z)<\operatorname{pgcd}(x+y, y+z, z+x)
$$ | [
"Solution:\nNotons que puisque $\\operatorname{pgcd}(x, y, z)$ divise chacun des $x, y, z$, il divise également $\\operatorname{pgcd}(x+y, y+z, z+x)$.\nOn remarque que si le triplet $(x, y, z)$ est solution, alors les triplets $(k x, k y, k z)$ sont solutions pour tout $k \\in \\mathbb{N}^{*}$. On peut donc suppose... | France | ENVOi 3 : ARITHMÉTIQUE | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All triples that are positive integer multiples of a triple of odd integers; equivalently, those for which dividing each entry by their greatest common divisor yields three odd integers. | |
00nm | Let $p$ be a prime and let $m$ and $n$ be positive integers such that $p^2 + m^2 = n^2$.
Prove that $m > p$. | [
"We have $p^2 = n^2 - m^2 = (n - m)(n + m)$. Since $p$ is a prime, the number $p^2$ has the divisors $1$, $p$ and $p^2$. Since the two factors $n - m$ and $n + m$ are distinct, they cannot be both equal to $p$. Furthermore, $n - m$ is smaller than $n + m$, therefore, $n - m = 1$, i.e. $n = m + 1$.\n\nWe find\n$$\np... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
01bj | Is it possible to subdivide a convex $2014$-gon, by means of diagonals that do not intersect, into triangles, in such a way that each vertex be incident to an odd number of triangles? | [
"No, it is not possible. First, note that each triangulated polygon can be bicoloured: The triangles may each be assigned either of two colours, black and white, in such a way that the colours alternate (i.e. adjacent triangles have opposite colours). This is easily shown inductively, since a triangulated polygon c... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
01zv | Problem:
Find all polynomials $p(x)$ with real coefficients such that
$$
p(a+b-2 c)+p(b+c-2 a)+p(c+a-2 b)=3 p(a-b)+3 p(b-c)+3 p(c-a)
$$
for all $a, b, c \in \mathbb{R}$. | [
"Solution:\nFor $a=b=c$, we have $3 p(0)=9 p(0)$, hence $p(0)=0$. Now set $b=c=0$, then we have\n$$\np(a)+p(-2 a)+p(a)=3 p(a)+3 p(-a)\n$$\nfor all $a \\in \\mathbb{R}$. So we find a polynomial equation\n$$\np(-2 x)=p(x)+3 p(-x)\n$$\nNote that the zero polynomial is a solution to this equation. Now suppose that $p$ ... | Benelux Mathematical Olympiad | Benelux Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All real polynomials of the form p(x) = ax^2 + bx with real a, b. | |
08yd | Call a positive integer a good number if its every digit is a prime number. Find all the good numbers having three digits and with the property that their squares are good numbers of five digits. | [
"235\n\nLet $n$ be a three-digit good number, then we can write $n = 100a + 10b + c$, where $a, b, c$ are 1-digit primes. If $n^2$ is a good number with 5 digits, we must have $n^2 < 10^5$, from which it follows that we have $n < 320$. Since $10b + c \\ge 22$, we conclude that $a = 2$ must hold. When $c = 2, 3, 5, ... | Japan | 2019 Japan Mathematical Olympiad First Stage | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 235 | |
0ecu | During the first school hour the ratio of the number of boys and girls in the classroom was $3 : 4$. After 4 more girls entered the classroom and 4 boys left it before the second school hour, the ratio became $2 : 5$. How many more girls than boys were in the classroom during the second school hour?
(A) 5
(B) 6
(C) 7
... | [
"Denote the number of boys in the classroom during the first school hour by $f$ and the number of girls by $d$. Then $\\frac{f}{d} = \\frac{3}{4}$. During the second school hour there are $f - 4$ boys and $d + 4$ girls in the classroom, therefore the ratio equals $\\frac{f-4}{d+4} = \\frac{2}{5}$. From first equati... | Slovenia | National Math Olympiad 2015 – First Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | E | |
004m | Un polígono de doce lados, cuyos vértices pertenecen a una circunferencia $C$, tiene seis lados de longitud $2$ y seis lados de longitud $\sqrt{3}$. Calcule el radio de la circunferencia $C$. | [] | Argentina | XVI Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | Spanish | proof and answer | √13 | |
0cu5 | Let $\omega$ be the incircle of a triangle $ABC$ (with $AB < AC$). Let the excircle $\omega_A$ touch the side $BC$ at $A'$. A variable point $X$ is chosen on the segment $AA'$ so that the segment $A'X$ has no common points with $\omega$. The tangents to $\omega$ passing through $X$ meet the side $BC$ at $Y$ and $Z$. Pr... | [
"Let $\\omega$ touch $BC$ at $A''$. We assume that $A''Y < A''Z$. Let $\\omega'$ be the excircle of the triangle $XYZ$ touching the side $XZ$, and let $T$ be the meeting point of $\\omega$ and $AA'$ closest to $A$; the tangent at $T$ to $\\omega$ is parallel to $BC$. The homothety centered at $X$ mapping $\\omega$ ... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English; Russian | proof only | null | |
04st | Suppose that the real numbers $x, y, z$ satisfy equalities
$$
15(x + y + z) = 12(xy + yz + zx) = 10(x^2 + y^2 + z^2)
$$
and that at least one of them is different from zero.
a) Prove that $x + y + z = 4$.
b) Find the smallest interval $(a, b)$, which contains all three numbers from any triplet $(x, y, z)$ satisfying th... | [] | Czech Republic | Czech and Slovak Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | x + y + z = 4; the smallest interval is [2/3, 2]. | |
0ilv | Problem:
Find all nonconstant polynomials $P(x)$, with real coefficients and having only real zeros, such that $P(x+1) P\left(x^{2}-x+1\right)=P\left(x^{3}+1\right)$ for all real numbers $x$. | [
"Solution:\nAnswer: $\\left\\{P(x)=x^{k} \\mid k \\in \\mathbb{Z}^{+}\\right\\}$.\n\nNote that if $P(\\alpha)=0$, then by setting $x=\\alpha-1$ in the given equation, we find $0=P\\left(x^{3}+1\\right)=P\\left(\\alpha^{3}-3 \\alpha^{2}+3 \\alpha\\right)$. Because $P$ is nonconstant, it has at least one zero. Becaus... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | P(x) = x^k for any positive integer k | |
0h27 | Find all natural numbers $N$, that have two digits and are equal to the sum of their digits added to the cube of this sum. | [
"Let $n$ be the sum of the digits, then: $N = n + n^3$. The cube has to be less than $99$, that gives us four options: $1$, $2$, $3$, $4$. An easy check shows that the only possible answer is $30$."
] | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 30 | |
043v | Let $m$ be a real number. If the real and imaginary parts of complex $z = 1 + i + \frac{m}{1+i}$, with $i$ being the imaginary unit, are greater than zero, then the range of $m$ is ______. | [
"After calculation, we get $z = 1 + i + \\frac{(1-i)m}{2} = \\frac{2+m}{2} + \\frac{2-m}{2}i$.\n\nBy the condition, it follows that $\\frac{2+m}{2} > 0$, $\\frac{2-m}{2} > 0$, and then we find the solution is $-2 < m < 2$."
] | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | −2 < m < 2 | |
0gfg | 證明：對任意正實數 $a, b$ 及任意正整數 $n$, 不等式
$$
(a+b)^n - a^n - b^n \ge \frac{2^n-2}{2^{n-2}} \cdot ab(a+b)^{n-2}.
$$ | [] | Taiwan | 2022 數學奧林匹亞競賽第一階段培訓營, 獨立研究(三) | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Chinese; English | proof only | null | |
09ad | Let $ABCD$ be a tangential quadrilateral. Let $\omega$ be externally inscribed circle in $ABCD$, tangent to $AD$, $BC$ and $AB$. Denote by $X_{AB}$ the point of tangency of $\omega$ and the circle that passes through $A$ and $B$ and internally tangent to $\omega$. Let us define $X_{BC}$, $X_{CD}$ and $X_{DA}$, analogou... | [
"\n\nLet $I$ be the center of the incircle of $ABCD$, and $AB$ touches the incircle at $Q$ and $\\omega$ at $P$. It will be sufficient to show that $X_{AB}$, $P$, $I$ are collinear and $\\angle BX_{AB}P = \\angle AX_{AB}P$ (all passes through $I$). Let $(IP) \\cap \\omega = X$, $PP'$ and $Q... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Mongolian | proof only | null | |
01ii | Let $\mathbb{Z}^+$ be the set of positive integers. Find all strictly increasing functions $\mathbb{Z}^+ \to \mathbb{Z}^+$ with $f(1) = 1$ that satisfy the equation
$$
3 \cdot (f(1) + f(2) + \dots + f(n)) = f(n+1) + f(n+2) + \dots + f(2n)
$$
for all $n \in \mathbb{Z}^+$. | [
"The strictly increasing function $\\mathbb{Z}^+ \\to \\mathbb{Z}^+$ with $f(n) = 2n-1$ for all $n \\in \\mathbb{Z}^+$ satisfies $f(1) = 1$ and solves the functional equation, since $1+3+\\dots+(2n-1) = n^2$ and $(2n+1)+(2n+3)+\\dots+(4n-1) = (2n)^2-n^2 = 3n^2$ for all $n \\in \\mathbb{Z}^+$.\nWe claim that no othe... | Baltic Way | Baltic Way 2023 Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | f(n) = 2n - 1 | |
03b4 | Lame rook is chess piece that is allowed to make moves only in the neighboring by edge cell. For 100 moves lame rook visited all cells of a $10 \times 10$ chessboard exactly once and finally arriving in the initial cell. Prove that some two consecutive moves of the rook are left turns. | [] | Bulgaria | Selection test for 27. Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
05ys | Problem:
Soit $a \geqslant 2$ et $d \geqslant 2$ deux entiers premiers entre eux. On pose $x_{1}=1$; puis, pour tout entier $k \geqslant 1$, on pose $x_{k+1}=x_{k} / a$ si $a$ divise $x_{k}$, et $x_{k+1}=x_{k}+d$ sinon.
Trouver, en fonction de $a$ et de $d$, l'entier $\ell$ maximal pour lequel $a^{\ell}$ divise l'un d... | [
"Solution:\n\nPour tout entier $n \\geqslant 1$, soit $f(n)$ le plus petit entier tel que $a f(n) \\geqslant n$ et $a f(n) \\equiv n \\pmod{d}$, et soit $\\delta(n)=(a f(n)-n) / d$. Par construction, si l'on dispose d'un terme $x_{k}=n$ de la suite, les termes suivants de la suite seront $x_{k+1}=n+d, x_{k+2}=n+2 d... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Logarithmic functio... | null | proof and answer | ℓ = ⌊log_a d⌋ + 1 (equivalently, ℓ is the unique integer satisfying d < a^ℓ < a d) | |
0dau | Determine the maximal number of disjoint crosses (5 squares) which can be put inside $8 \times 8$ chessboard such that sides of a cross are parallel to sides of the chessboard. | [
"Let's note, that 4 corner cells can't be covered by crosses. So we assume that they are removed from the board and we have only 60 cells.\nFrom the first row at most two cells can be covered by crosses. So at least 4 cells will be not covered. The same argument works for last row, first column and last column. So ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 8 | |
09mt | Let $A = \{a_1, a_2, \dots, a_n\}$ be a set of positive integers with $n \ge 1$ elements. The set $A$ is called good if, for any two distinct subsets $X$ and $Y$ of $A$ (i.e., $X \neq Y$), the difference $S(X) - S(Y)$ is not divisible by $2^n$. Here, $S(X) = \sum_{a \in X} a$ denotes the sum of the elements in the subs... | [
"Answer: $2^{n(n-1)/2}$.\nLet $N = 2^n$ and $v_2(a) = s$ if and only if $2^s \\mid a$ and $2^{s+1} \\nmid a$, for a positive integer $a$. Let $v_2(A) = \\{v_2(a) \\mid a \\in A\\}$ for the set $A$.\n**Claim:** A set $A$ is good if and only if $v_2(A) = \\{0, 1, \\dots, n-1\\}$.\nFirst we show that sets of the form ... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof and answer | 2^{n(n-1)/2} | |
0kig | Let $N$ be the positive integer $7777\ldots777$, a $313$-digit number where each digit is a $7$. Let $f(r)$ be the leading digit of the $r$th root of $N$. What is $f(2) + f(3) + f(4) + f(5) + f(6)$?
(A) 8 (B) 9 (C) 11 (D) 22 (E) 29 | [
"Because $10^r$ is written as a $1$ followed by $r$ zeros, the $r$th root of any number smaller than $10^r$ when written as a decimal has only one digit to the left of the decimal point. Extending this reasoning, the leading digit of the $r$th root of $N$ is the same as the leading digit of the $r$th root of the in... | United States | Fall 2021 AMC 10 B | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | A | |
0il8 | Problem:
Compute the value of the infinite series
$$
\sum_{n=2}^{\infty} \frac{n^{4}+3 n^{2}+10 n+10}{2^{n} \cdot\left(n^{4}+4\right)}
$$ | [
"Solution:\nWe employ the difference of squares identity, uncovering the factorization of the denominator: $n^{4}+4=\\left(n^{2}+2\\right)^{2}-(2 n)^{2}=\\left(n^{2}-2 n+2\\right)\\left(n^{2}+2 n+2\\right)$. Now,\n$$\n\\begin{aligned}\n\\frac{n^{4}+3 n^{2}+10 n+10}{n^{4}+4} & =1+\\frac{3 n^{2}+10 n+6}{n^{4}+4} \\\\... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 11/10 | |
0dfz | Determine the smallest positive integer $a$ for which there exist a prime number $p$ and a positive integer $b \ge 2$ such that
$$
\frac{a^p - a}{p} = b^2.
$$ | [
"If $p=2$, our equation becomes $a(a-1) = 2b^2$, whose smallest solution in $\\mathbb{N}$ is $a=9$.\n\nNow let $p \\ge 3$. Since $a$ and $a^{p-1}-1$ are coprime and $a(a^{p-1}-1) = pb^2$, either $a$ or $a^{p-1}-1$ must be a square, and it is obviously not the latter; hence $a$ is a square. Assume that $a=4$. Then\n... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 9 | |
0dgr | Given $\triangle ABC$, $D$ is on $BC$ and $P$ is on $AD$. A line $\ell$ is passing through $D$ intersects $AB$, $PB$ at $M$, $E$ respectively, and intersects $AC$ extended and $PC$ extended at $F$, $N$ respectively. Let $DE = DF$. Prove that $DM = DN$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | English | proof only | null | |
03o9 | Problem:
Let $ABC$ be a triangle with circumcircle $\Gamma$ and $AB \neq AC$. Let $D$ and $E$ lie on the arc $BC$ of $\Gamma$ not containing $A$ such that $\angle BAE = \angle DAC$. Let the incenters of $BAE$ and $CAD$ be $X$ and $Y$ respectively, and let the external tangents of the incircles of $BAE$ and $CAD$ inters... | [
"Solution:\nLet $AX$ and $AY$ intersect $(ABC)$ again at $P$ and $Q$, let the inradii of $ABE$ and $ACD$ be $r_B$ and $r_C$, and let $(AXY)$ intersect $(ABC)$ again at $N$.\n\nFirst note that $\\angle BAP = \\frac{1}{2}\\angle BAE = \\frac{1}{2}\\angle CAD = \\angle QAC$, so $XP = BP = CQ = CY$. This thus implies t... | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chas... | null | proof only | null | |
08ic | Problem:
Let $x$ be a real number. Find the smallest value of the expression
$$\sqrt{x^{2}+2x+4}+\sqrt{x^{2}-\sqrt{3}\,x+1}.$$ | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | sqrt(5 + 2*sqrt(3)) | |
07gg | An $n \times n$ Latin square is given. We are allowed to do the following operation on an $n \times n$ table. We can choose a cell in the square and add the same integer to the number of all the cells in the union of the row and the column of the chosen cell. Prove that one can do a finite number of such operations to ... | [
"Denote the number of cell $(i, j)$ in the square by $a_{ij}$. For $i, j, r, k \\in \\{1, 2, \\dots, n\\}$, with $i \\neq j$ and $r \\neq k$, we define the operation $P_{ijrk}$ in this way: Add $1$ to all the cells in the union of row and column of cell $(i, j)$, add $1$ to all the cells in the union of row and col... | Iran | 38th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
022e | Problem:
Enchendo uma piscina - Uma piscina vazia foi abastecida de água por duas torneiras $A$ e $B$, ambas com vazão constante. Durante 4 horas, as duas torneiras ficaram abertas e encheram $50\%$ da piscina. Em seguida, a torneira $B$ foi fechada e durante 2 horas a torneira $A$ encheu $15\%$ do volume da piscina. ... | [
"Solution:\n\nComo as torneiras $A$ e $B$ despejam água na piscina com vazão constante, o volume de água despejado na piscina por cada torneira é proporcional ao tempo em que ela fica aberta. Assim, se durante 2 horas a torneira $A$ enche $15\\%$ do volume da piscina, então em 4 horas ela encherá $30\\%$ do volume ... | Brazil | Nível 3 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 7 hours | |
0h46 | $$
\begin{cases} (x+y)(1+xy) + (x-y)^2 = 2, \\ x^3 + y^3 = 1. \end{cases}
$$ | [
"Перше рівняння системи запишемо, з урахуванням рівності $x^3 + y^3 = 1$, у вигляді $(x + y)(1 + xy) + (x - y)^2 - (x + y)(x^2 - xy + y^2) - 1 = 0$, тобто $(x + y - 1)(1 - (x - y)^2) = 0$. Подальші міркування є очевидними.\n\nВідповідь: $(1; 0)$, $(0; 1)$."
] | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | (1, 0), (0, 1) | |
0bxp | Let $ABCD$ be a square. We take the points $E \in (AB)$, $N \in (CD)$ and $F, M \in (BC)$, such that the triangles $AMN$ and $DEF$ are equilateral. Prove that $PQ = FM$, where $\{P\} = AN \cap DE$ and $\{Q\} = AM \cap EF$.
 | [
"Because $\\triangle ABM \\equiv \\triangle ADN$ and $\\triangle DAE \\equiv \\triangle DCF$, we have $m(\\overline{BAM}) = m(\\overline{DAN}) = m(\\overline{ADE}) = m(\\overline{CDF}) = 15^\\circ$.\nThen $\\triangle ABM \\equiv \\triangle ADN \\equiv \\triangle DAE \\equiv \\triangle DCF$ and $\\triangle AMN \\equ... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
01ug | a) Given the ten digits from $0$ to $9$, prove that three numbers $A$, $B$, and $C$ can be formed by combining these digits, provided that each digit is used exactly once and $A+B=C$. Notice that $0$ may not be the first digit of any of the numbers.
b) Find all possible values of the sum of the digits of $C$. | [
"a)\nFor example, $765 + 324 = 1089$.\n\nb)\nSince any integer $Y$ is congruent modulo $9$ to the sum of its digits, we have\n$$\nA + B + C \\equiv S(A) + S(B) + S(C) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 0 = 45 \\equiv 0 \\pmod{9}.\n$$\nBy condition, we have $A + B = C$, so $2C \\equiv 0 \\pmod{9}$, whence it foll... | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 9 or 18 | |
050a | Consider the diagonals $A_1A_3$, $A_2A_4$, $A_3A_5$, $A_4A_6$, $A_5A_1$ and $A_6A_2$ of a convex hexagon $A_1A_2A_3A_4A_5A_6$. The hexagon whose vertices are the points of intersection of the diagonals is regular. Can we conclude that the hexagon $A_1A_2A_3A_4A_5A_6$ is also regular? | [
"We show that the hexagon $A_1A_2A_3A_4A_5A_6$ has all its side lengths equal and all its angles equal. As the internal hexagon is regular, the grey triangles in Fig. 2 all have two angles of equal size and so they are isosceles. Additionally, all these six isosceles triangles have their bases of equal lengths, thu... | Estonia | Selected Problems from Open Contests | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof and answer | Yes, the original hexagon must be regular. | |
0cvk | 24 students attend a mathematical circle. For any team consisting of 6 students, the teacher considers it to be either good or OK. For the tournament of mathematical battles, the teacher wants to partition all the students into 4 teams of 6 students each. May it happen that every such partition contains either exactly ... | [
"Приведём один из возможных примеров. Выделим трёх школьников. Будем называть сыгранными команды, в которых содержится 1 или 3 выделенных школьника, а остальные — несыгранными.\n\nВыделенные школьники могут либо оказаться в трёх разных командах, или тогда мы получим три сыгранные команды и одну несыгранную, либо ок... | Russia | Final round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | Yes | |
0bwb | Let $2^{-n_1} + 2^{-n_2} + \dots + 2^{-n_k} + \dots$, where $1 \le n_1 < n_2 < \dots < n_k < \dots$, be the binary expansion of $(\sqrt{5}-1)/2$. Prove that $n_k \le 2^{k-1}-2$ for all integers $k \ge 4$.
*Amer. Math. Monthly* | [
"We show that $n_{k+1} \\le 2n_k + 2$ for all indices $k$. Since $n_4 = 6 = 2^{4-1} - 2$, the conclusion follows inductively. (None of the first three exponents, $n_1 = 1$, $n_2 = 4$, $n_3 = 5$, satisfies the inequality in the statement.)\n\nWrite $\\alpha = (\\sqrt{5}-1)/2$, $m = 2^{n_k} \\sum_{j=1}^k 2^{-n_j}$ an... | Romania | Eleventh STARS OF MATHEMATICS Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Algebraic Number Theory > Algebraic numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
08io | Problem:
Let $a, b, c$ be positive numbers such that $a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = 3$. Prove that
$$
a + b + c \geq a b c + 2
$$
Problem:
Fie $a, b, c$ numere pozitive astfel ca $a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = 3$. Demonstrați că $a + b + c \geq a b c + 2$. | [
"Solution:\nWe can consider the case $a \\geq b \\geq c$ which implies $c \\leq 1$. The given inequality writes\n$$\na + b - 2 \\geq (a b - 1) c \\geq (a b - 1) c^{2} = (a b - 1) \\frac{3 - a^{2} b^{2}}{a^{2} + b^{2}}\n$$\nPut $x = \\sqrt{a b}$. From the inequality $3 a^{2} b^{2} \\geq a^{2} b^{2} + b^{2} c^{2} + c... | JBMO | 7th JBMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
04up | Let $k \le 2022$ be a positive integer. Alice and Bob play a game on a $2022 \times 2022$ board. Initially, all cells are white. Alice starts and the players alternate. In her turn, Alice can either color one white cell in red or pass her turn. In his turn, Bob can either color a $k \times k$ square of white cells in b... | [] | Czech Republic | Czech-Polish-Slovak Match | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Let n = 2022 and q = floor(n/k). Under optimal play Bob colors exactly floor(q^2/2) blocks, i.e., floor(q^2/2)·k^2 cells. Hence Bob never gets more than half the board. The outcome is a draw if and only if 2k divides 2022 (equivalently k ∈ {1, 3, 337, 1011}); for all other k, Alice has a winning strategy. Bob has no wi... | |
05ia | Problem:
Soient $x_{1}, x_{2}, \ldots, x_{5}$ des réels tels que
$$
\left|x_{2}-x_{1}\right|=2\left|x_{3}-x_{2}\right|=3\left|x_{4}-x_{3}\right|=4\left|x_{5}-x_{4}\right|=5\left|x_{1}-x_{5}\right| .
$$
Montrer que ces cinq réels sont égaux. | [
"Solution:\nNotons $\\alpha=\\left|x_{2}-x_{1}\\right|$ et montrons par l'absurde que $\\alpha=0$. D'après l'énoncé, il existe des signes $\\varepsilon_{1}, \\ldots, \\varepsilon_{5}$ tels que\n$$\nx_{2}-x_{1}=\\varepsilon_{1} \\alpha, \\quad x_{3}-x_{2}=\\varepsilon_{2} \\frac{\\alpha}{2}, \\quad x_{4}-x_{3}=\\var... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
0fxv | Problem:
Betrachte eine Menge $A$ von 2009 Punkten in der Ebene, von denen keine drei auf einer Geraden liegen. Ein Dreieck, dessen Eckpunkte alle in $A$ liegen, heißt internes Dreieck. Beweise, dass jeder Punkt aus $A$ im Innern einer geraden Anzahl interner Dreiecke enthalten ist. | [
"Solution:\n\nWir betrachten eine Menge $B$ aus 2008 Punkten und einen weiteren Punkt $P \\notin B$, sodass keine drei der Punkte aus $B \\cup\\{P\\}$ auf einer Geraden liegen. Wir zeigen, dass die Anzahl $\\alpha(P)$ der Dreiecke mit Eckpunkten in $B$, welche $P$ im Innern enthalten, gerade ist. Die Menge aller St... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
01n3 | A cubic trinomial $x^3 + px + q$ with integer coefficients $p$ and $q$ is said to be *irrational* if it has three pairwise distinct real irrational roots $\alpha_1$, $\alpha_2$, $\alpha_3$.
Find all irrational cubic trinomials for which the value of $|\alpha_1| + |\alpha_2| + |\alpha_3|$ is the minimal possible. | [
"First, for $R(x) = x^3 + px + q$ to have three distinct real roots it is necessary that $p < 0$ (because the derivative $R'(x) = 3x^2 + p$ cannot be nonnegative). Let now $p < 0$, then the equation $R'(x) = 0$ has two real roots $x_1 = -\\sqrt{-p/3}$, $x_2 = \\sqrt{-p/3}$. Now, the condition that $R(x)$ has three ... | Belarus | Belorusija 2012 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | English | proof and answer | x^3 - 3x - 1 and x^3 - 3x + 1 | |
03w8 | Find the integer solutions of the function $x^2 - 2xy + 126y^2 = 2009$. | [
"Suppose that the integers $x, y$ satisfy\n$$\nx^2 - 2xy + 126y^2 - 2009 = 0.\n$$\nLooking at this as a quadratic function of $x$,\n$$\n\\Delta = 4y^2 - 4 \\times (126y^2 - 2009) = 500(4^2 - y^2) + 36\n$$\nshould be a square number.\nIf $y^2 > 4^2$, then $\\Delta < 0$. So $y^2 < 4^2$, when $y^2 \\in \\{0, 1^2, 2^2,... | China | China Southeastern Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | (1, 4), (7, 4), (-1, -4), (-7, -4) | |
0a32 | Charlie has a thick book of $n$ pages in which, after opening, the first page on the right shows the page number $1$. Then all the pages are numbered continuously so that if you open the book just anywhere, you see an even page number on the left and an odd page number on the right. The very last page before the back c... | [] | Netherlands | Dutch Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1230 | |
07nh | $AB$ and $CD$ are two parallel line segments. $AD$ and $BC$ intersect at $P$. Prove that the circumcircles of the triangles $ABP$ and $CDP$ touch at $P$. | [
"Whether or not $P$ lies between the parallel lines gives two cases to distinguish. Let $S$ and $T$ be points on the tangent to the circumcircle of $\\triangle ABP$ at $P$ such that $S$ and $A$ are on the same side of the line $BC$ and $T$ and $B$ are on the same side of $AD$.\n\n**First Solution:**\nWe have $\\ang... | Ireland | Ireland | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0bfx | For every non-negative integer $n$, let $s_n$ be the sum of the digits in the decimal expansion of $2^n$. Is the sequence $(s_n)_{n \in \mathbb{N}}$ eventually increasing? | [
"Suppose, if possible, that the sequence is eventually increasing, say from some rank $n_0$ on. Fix a non-negative integer $m$ such that $6m \\ge n_0$ to write\n$$\n\\begin{cases} \ns_{6m+1} \\ge s_{6m} + 1, \\\\\ns_{6m+2} \\ge s_{6m+1} + 2, \\\\\ns_{6m+3} \\ge s_{6m+2} + 4, \\\\\ns_{6m+4} \\ge s_{6m+3} + 8, \\\\\n... | Romania | The Tenth IMAR Mathematical Competition | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | No, the sequence is not eventually increasing. | |
0b9w | Find all the functions $f : [-\frac{\pi}{2}; \frac{\pi}{2}] \to \mathbb{R}$ such that:
a. $f$ is differentiable on $[-\frac{\pi}{2}, \frac{\pi}{2}]$;
b. there exists an antiderivative $F$ of $f$ such that $F(x) + f'(x) \le 0, \forall x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ and $F(-\frac{\pi}{2}) = F(\frac{\pi}{2}) = 0$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Calculus > Differential Equations > ODEs",
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Trigonometric functions"
] | null | proof and answer | All solutions are f(x) = C · sin x for an arbitrary real constant C. | |
0au5 | Problem:
Suppose that the function $y = f(x)$ satisfies $1 - y = \frac{9 e^{x} + 2}{12 e^{x} + 3}$. If $m$ and $n$ are consecutive integers so that $m < \frac{1}{y} < n$ for all real $x$, find the value of $mn$. | [] | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 12 | |
0kt8 | Problem:
Kimothy starts in the bottom-left square of a $4 \times 4$ chessboard. In one step, he can move up, down, left, or right to an adjacent square. Kimothy takes $16$ steps and ends up where he started, visiting each square exactly once (except for his starting/ending square). How many paths could he have taken? | [
"Solution:\n\nThe problem is asking to count the number of cycles on the board that visit each square once. We first count the number of cycle shapes, then multiply by $2$ because each shape can be traversed in either direction. Each corner must contain an L-shaped turn, which simplifies the casework. In the end th... | United States | HMMT November 2022 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 12 | |
07dr | Let $P(z) = a_d z^d + a_{d-1} z^{d-1} + \dots + a_1 z + a_0$ be a polynomial with complex coefficients, the **reverse** of this polynomial is defined as
$$
P^*(z) = \overline{a_0}z^d + \overline{a_1}z^{d-1} + \dots + \overline{a_d}
$$
a. Prove that
$$
P^*(z) = z^d \overline{P\left(\frac{1}{z}\right)}
$$
b. Let $m$ be... | [
"a.\nWe have\n$$\nP^*(z) = \\bar{a}_0 z^d + \\bar{a}_1 z^{d-1} + \\cdots + \\bar{a}_d = z^d \\left( \\bar{a}_0 + \\bar{a}_1 \\frac{1}{z} + \\cdots + \\bar{a}_d \\left( \\frac{1}{z} \\right)^d \\right)\n$$\nso\n$$\nP^*(z) = z^d \\overline{\\left( a_0 + a_1 \\frac{1}{z} + \\cdots + a_d \\left( \\frac{1}{z} \\right)^d... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
02c0 | Problem:
Dois tipos de vela têm o mesmo comprimento mas são feitas de material diferente; uma queima completamente em 3 horas e a outra em 4 horas, ambas queimam com velocidade uniforme. A que horas as velas devem ser acesas de modo que às 16 horas o comprimento de uma seja o dobro do da outra?
(a) $1: 24$
(b) $1: 28... | [] | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | c | |
01t8 | Let $ABC$ be a triangle with $\angle C = 90^\circ$, and let $H$ be the foot of the altitude from $C$. A point $D$ is chosen inside the triangle $CBH$ so that $CH$ bisects $AD$. Let $P$ be the intersection point of the lines $BD$ and $CH$. Let $\omega$ be the semicircle with diameter $BD$ that meets the segment $CB$ at ... | [] | Belarus | 66th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane ... | English | proof only | null | |
0inz | Problem:
Consider the polynomial $P(x) = x^{3} + x^{2} - x + 2$. Determine all real numbers $r$ for which there exists a complex number $z$ not in the reals such that $P(z) = r$. | [
"Solution:\n\nAnswer: $r > 3$, $r < \\frac{49}{27}$. Because such roots to polynomial equations come in conjugate pairs, we seek the values $r$ such that $P(x) = r$ has just one real root $x$. Considering the shape of a cubic, we are interested in the boundary values $r$ such that $P(x) - r$ has a repeated zero. Th... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | All real r with r < 49/27 or r > 3 | |
02qx | Problem:
Um polígono convexo é elegante quando ele pode ser decomposto em triângulos equiláteros, quadrados ou ambos, todos com lados de mesmo comprimento. Abaixo, mostramos alguns polígonos elegantes, indicando para cada um deles uma decomposição e o número de lados.
4 lados
![](attached_ima... | [
"Solution:\n\na) Um exemplo de polígono elegante com oito lados aparece abaixo.\n\n\nb) Como um polígono elegante é convexo e é formado colocando lado a lado quadrados e triângulos equiláteros, seus ângulos são somas de parcelas iguais a $60^\\circ$ ou $90^\\circ$ que não ultrapassem $180^\... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Possible interior angles are 60, 90, 120, and 150 degrees; the number of sides is at most 12. | |
02nt | Problem:
Uma companhia de eletricidade instalou um poste num terreno plano. Para fixar bem o poste, foram presos cabos no poste, a uma altura de $1{,}4$ metros do solo e a $2$ metros de distância do poste, sendo que um dos cabos mede $2{,}5$ metros, conforme a figura.
Um professor de Matemática, após analisar estas... | [
"Solution:\n\nPara que o poste fique perpendicular ao solo, o ângulo em $A$ deve ser reto e, portanto, o triângulo $\\triangle ABC$ deve ser retângulo (ver figura). Nesse caso, os dados do problema dão que a hipotenusa mede $2{,}5\\ \\mathrm{m}$ e os catetos $1{,}4\\ \\mathrm{m}$ e $2\\ \\mathrm{m}$. Assim, pelo Te... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | Yes; the pole is not perpendicular to the ground. | |
01b7 | Given positive real numbers $a, b, c, d$ that satisfy equalities
$$
a^2 + d^2 - ad = b^2 + c^2 + bc \quad \text{and} \quad a^2 + b^2 = c^2 + d^2,
$$
find all possible values of the expression $\frac{ab+cd}{ad+bc}$. | [
"**Answer:** $\\frac{\\sqrt{3}}{2}$.\n\nLet $A_1BC_1$ be a triangle with $A_1B = b$, $BC_1 = c$ and $\\angle A_1BC_1 = 120^\\circ$, and $C_2DA_2$ be another triangle with $C_2D = d$, $DA_2 = a$ and $\\angle C_2DA_2 = 60^\\circ$. By the law of cosines and the assumption $a^2 + d^2 - ad = b^2 + c^2 + bc$, we have $A_... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | sqrt(3)/2 | |
0e0d | Let $ABC$ be an acute triangle and let $I$ be the incentre of the triangle $ABC$. The lines $AI$ and $BI$ meet the circumcircle of the triangle $ABC$ again at $A_1$ and $B_1$. The segments $A_1B_1$ and $CI$ meet at $C_1$. Let $S$ be the circumcentre of the triangle $IA_1C_1$. The lines $SC_1$ and $BI$ meet the segment ... | [
"Let us write $\\angle ACB = \\gamma$. Then $\\angle CBA = 2\\gamma$ and $\\angle BAC = \\pi - 3\\gamma$. If we draw a figure where $\\angle CBA = 2\\angle ACB$, we notice that $IA_1C_1$ is a right triangle. Let us prove this.\nSince $ABA_1B_1$ is a cyclic quadrilateral we have $\\angle C_1A_1I = \\angle B_1A_1A = ... | Slovenia | Selection Examinations for the IMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | ∠BAC = 2π/5, ∠CBA = 2π/5, ∠ACB = π/5 | |
01f7 | Find all functions $f: \mathbb{R} \to [-2019, 2019]$, for which
$$
2f(f(x)) + f(f(-x)) + 8f(x) + 4f(-x) + 4x = 0
$$
for all real numbers $x$. | [
"Let us define an auxiliary function $g: \\mathbb{R} \\to \\mathbb{R}$ by setting\n$$\ng(x) = f(f(x)) + 4f(x) + 4x\n$$\nfor all real numbers $x$. The given functional equation may be written as\n$$\n2g(x) = -g(-x),\n$$\nfor all real numbers $x$, so that\n$$\n4g(x) = -2g(-x) = -(-g(-(-x))) = g(x),\n$$\nfor all real ... | Baltic Way | Baltic Way 2019 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | No such function exists. | |
01ud | Every one of six pupils attends exactly two of four hobby groups. There are no pupils attending the same two hobby groups. Each hobby group is open every day. During some consecutive days, every one of these six pupils has attended one of her/his hobby group. It has been observed that each of these days each hobby grou... | [
"Answer: 24 days.\nNote that from four hobby groups one can form exactly $4 \\cdot 3/2 = 6$ different pairs. Since we have exactly six pupils, for each pair of the hobby group there exists exactly one pupil attending just these two hobby groups. By condition, each hobby group is attended by either one or two pupils... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 24 | |
0iaq | Problem:
Consider a $2003$-gon inscribed in a circle and a triangulation of it with diagonals intersecting only at vertices. What is the smallest possible number of obtuse triangles in the triangulation? | [
"Solution:\n\nBy induction, it follows easily that any triangulation of an $n$-gon inscribed in a circle has $n-2$ triangles. A triangle is obtuse unless it contains the center of the circle in its interior (in which case it is acute) or on one of its edges (in which case it is right). It is then clear that there a... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 1999 | |
05qs | Problem:
Soit $n \geqslant 1$ un entier. Pour tout sous-ensemble non vide $A$ de $\{1,2, \ldots, n\}$, on note $P(A)$ le produit de tous les éléments de $A$. Par exemple, pour $A=\{2,4,7\}$, on a $P(A)=56$. Déterminer la somme des $\frac{1}{P(A)}$ lorsque $A$ parcourt tous les sous-ensembles non vides de $\{1,2, \ldot... | [
"Solution:\n\nOn a l'identité $\\left(1+x_{1}\\right)\\left(1+x_{2}\\right) \\cdots\\left(1+x_{n}\\right)=1+\\sum x_{i}+\\sum_{i<j} x_{i} x_{j}+\\cdots$. En prenant $x_{i}=\\frac{1}{a_{i}}$, on obtient\n\n$$\n\\sum_{k \\geqslant 0} \\sum_{a_{1}<\\cdots<a_{k}} \\frac{1}{a_{1} a_{2} \\cdots a_{k}}=\\left(1+\\frac{1}{... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | n | |
087s | Problem:
Dimostrare che esistono infiniti numeri primi che dividono almeno un intero della forma $2^{n^{3}+1}-3^{n^{2}+1}+5^{n+1}$ con $n$ intero positivo. | [
"Solution:\n\nSupponiamo che l'insieme $S$ dei primi che dividono gli interi della forma $a_{n}=2^{n^{3}+1}-3^{n^{2}+1}+5^{n+1}$ sia finito. Troveremo una contraddizione esibendo un intero $n$ tale che $a_{n}$ possiede almeno un fattore primo che non appartiene ad $S$.\n\nSia $n$ il prodotto di tutti i numeri $p-1$... | Italy | Cesenatico | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
02f2 | Find all $n > 5$ for which it is possible to find a convex polyhedron with all $n$ faces congruent such that each face has another face parallel to it. | [
"Since for each face there is another parallel to it, $n$ must be even. For $n = 4k$, attach two congruent regular pyramids with $2k$ lateral faces by its bases. For $n = 4k + 2$, attach two congruent regular pyramids with $2k + 1$ lateral faces and twist one of them. Possible coordinates for the vertices of this s... | Brazil | XIV OBM | [
"Geometry > Solid Geometry > 3D Shapes"
] | English | proof and answer | All even integers greater than 5 | |
03vm | Find the least positive integer $n$ with the following property: Paint each vertex of a regular $n$-gon arbitrarily with one of three colors, say red, yellow and blue, there must exist four vertices of the same color that constitute the vertices of some isogonal trapezoid. | [
"We claim that the least positive integer $n$ is $17$.\n\nFirstly we prove that $n = 17$ has the required property. By contradiction, assume that we have a painting pattern with three colors for the regular $17$-gon such that any group of $4$ vertices of the same color cannot constitute an isogonal trapezoid.\n\nAs... | China | China Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 17 | |
05u0 | Problem:
Trouver tous les quadruplets d'entiers relatifs $(a, b, c, p)$ tels que $p$ soit un nombre premier et pour lesquels
$$
73 p^{2}+6=9 a^{2}+17 b^{2}+17 c^{2}
$$ | [
"Solution:\n\nL'égalité de l'énoncé met en jeu de nombreux carrés. La première chose à faire consiste donc à l'étudier modulo un nombre $n$ pour lequel il y a peu de résidus quadratiques. On étudie donc le cas $n=8$, car les carrés modulo 8 sont 0, 1 et 4.\n\nEn particulier, si $p$ est impair, l'équation devient $a... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | (±1, ±1, ±4, 2) and (±1, ±4, ±1, 2) | |
00fc | Determine all pairs $(h, s)$ of positive integers with the following property: If one draws $h$ horizontal lines and another $s$ lines which satisfy
(i) they are not horizontal,
(ii) no two of them are parallel,
(iii) no three of the $h+s$ lines are concurrent,
then the number of regions formed by these $h+s$ lines is ... | [
"Let $a_{h, s}$ be the number of regions formed by $h$ horizontal lines and $s$ other lines as described in the problem. Let $\\mathcal{F}_{h, s}$ be the union of the $h+s$ lines and pick any line $\\ell$. If it intersects the other lines in $n$ (distinct!) points then $\\ell$ is partitioned into $n-1$ line segment... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | (995, 1), (176, 10), (80, 21) | |
0fvr | Problem:
Drei gleich grosse Kreise $k_{1}$, $k_{2}$, $k_{3}$ schneiden sich nichttangential in einem Punkt $P$. Seien $A$ und $B$ die Mittelpunkte der Kreise $k_{1}$ und $k_{2}$. Sei $D$ bzw. $C$ der von $P$ verschiedene Schnittpunkt von $k_{3}$ mit $k_{1}$ bzw. $k_{2}$. Zeige, dass $A B C D$ ein Parallelogramm ist. | [
"Solution:\n\nSei $M$ der Mittelpunkt von $k_{3}$. Man betrachte das Viereck $P B C M$. Alle vier Seiten dieses Vierecks sind nach Voraussetzung gleich lang. Folglich ist $P B C M$ ein Rhombus und $B C$ ist parallel zu $P M$. Analog kann gezeigt werden, dass $A D$ parallel ist zu $P M$. Daraus folgt direkt, dass $A... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
06rf | Let $m$ be a positive integer and consider a checkerboard consisting of $m$ by $m$ unit squares. At the midpoints of some of these unit squares there is an ant. At time 0, each ant starts moving with speed 1 parallel to some edge of the checkerboard. When two ants moving in opposite directions meet, they both turn $90^... | [
"For $m=1$ the answer is clearly correct, so assume $m>1$. In the sequel, the word collision will be used to denote meeting of exactly two ants, moving in opposite directions.\n\nIf at the beginning we place an ant on the southwest corner square facing east and an ant on the southeast corner square facing west, the... | IMO | 52nd International Mathematical Olympiad 2011 Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 3m/2 - 1 | |
0dvb | Problem:
Nariši graf funkcije s predpisom $f(x) = -\sqrt{x^{2} - 6x + 9}$ in izračunaj ploščino lika, ki ga oklepa graf dane funkcije s koordinatnima osema. | [
"Solution:\n\nZapis $f(x) = -\\sqrt{(x-3)^{2}}$\n\nZapis $f(x) = -|x-3|$\n\nIzračunana ničla $(m)$, začetna vrednost $(n): m = 3, n = -3$\n\nNarisan graf funkcije $g(x) = -(x-3)$\n\nNarisan graf funkcije $f(x) = -|x-3|$\n\nIzračunana ploščina trikotnika $S = \\frac{9}{2}$"
] | Slovenia | 2. matematično tekmovanje dijakov srednjih tehniških in strokovnih šol | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 9/2 | |
03x3 | Suppose that the equation $\lg kx = 2\lg(x+1)$ has exactly one real root. Then the range of $k$ is ____. | [
"We have\n$$\nkx > 0, \\qquad \\textcircled{1}\n$$\n$$\nx + 1 > 0, \\qquad \\textcircled{2}\n$$\n$$\nkx = (x + 1)^2. \\qquad \\textcircled{3}\n$$\nThe expression ③ can be standardized as\n$$\nx^2 + (2 - k)x + 1 = 0. \\qquad \\textcircled{4}\n$$\nThe two roots of ④ are\n$$\nx_1, x_2 = \\frac{1}{2}[k - 2 \\pm \\sqrt{... | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | k < 0 or k = 4 | |
0eex | Problem:
Izbrati si moramo sedemmestno geslo, ki vsebuje vsaj eno črko in vsaj eno števko. Izbiramo lahko med znaki $A, B, C, G, J, M, R, Z$ in $3$.
a) Koliko je vseh možnih izbir za geslo, če se znaki ne smejo ponavljati?
b) Koliko je vseh možnih izbir za geslo, če se znaki lahko ponavljajo? | [
"Solution:\n\na.\nUpoštevanje, da je v geslu 6 črk in ena števka.\n\nUgotovitev, da je 7 možnih položajev za števko.\n\nIzračun števila možnih razporeditev črk v geslo $8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3$.\n\nIzračun števila možnih izbir za geslo $7 \\cdot 8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 ... | Slovenia | 16. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | a) 141120; b) 2685816 | |
0cnh | A balance and $100$ coins are given. Several coins are false. The number of false coins is greater than $0$ but less than $99$. All genuine coins have equal weights, and all false coins also have equal weights. The weight of a genuine coin is greater than the weight of a false one. It is permitted to perform a weighing... | [
"**Первое решение.** Заплатив монету перед первым взвешиванием, мы имеем $99$ монет, среди которых есть хотя бы одна настоящая. Положим на две чаши весов по одной монете. Если одна из них перевешивает, то она настоящая. Если же весы в равновесии, то либо обе взвешиваемые монеты настоящие, либо обе фальшивые. Заплат... | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms"
] | English; Russian | proof only | null | |
09pd | Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, and $CF$. Let $\omega$ be the circle with diameter $BC$, and suppose it intersects the segment $AD$ at point $K$ inside triangle $ABC$. On ray $KD$, let $L$ be a point such that $KA = KL$. Let lines $BL$ and $CL$ intersect the circle $\omega$ again at points $P$... | [
"\n\nSince $\\angle BEC = \\angle BFC = 90^\\circ$, points $E$ and $F$ lie on circle $\\omega$.\n\nWe observe that:\n$$\n\\angle LAF = \\angle DAB = \\angle FCB = \\angle FPL,\n$$\nso quadrilateral $AFLP$ is cyclic; denote its circumcircle by $\\omega_1$. Similarly, we get that $AQLE$ is cy... | Mongolia | MMO2025 Round 4 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0b09 | Problem:
For a positive integer $x$, let $f(x)$ be the last two digits of $x$. Find
$$
\sum_{n=1}^{2019} f\left(7^{7^{n}}\right).
$$ | [] | Philippines | Philippines Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | 50493 | |
03xa | Suppose that the line $l$: $y = kx + m$ ($k$, $m$ are integers) intercepts an ellipse $\frac{x^2}{16} + \frac{y^2}{12} = 1$ at two different points $A$, $B$, and intercepts the hyperbola $\frac{x^2}{4} - \frac{y^2}{12} = 1$ at two different points $C$, $D$. Can the line $l$ be such that $\vec{AC} + \vec{BD} = 0$? If ye... | [
"For $\\begin{cases} y = kx + m, \\\\ \\frac{x^2}{16} + \\frac{y^2}{12} = 1 \\end{cases}$ by eliminating $y$ and simplifying it, we get\n$$\n(3 + 4k^2)x^2 + 8kmx + 4m^2 - 48 = 0.\n$$\nDefine $A(x_1, y_1)$, $B(x_2, y_2)$. Then $x_1 + x_2 = -\\frac{8km}{3+4k^2}$.\n$$\n\\Delta_1 = (8km)^2 - 4(3 + 4k^2)(4m^2 - 48) > 0.... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 9 | |
0cha | Consider a continuous function $f : [0, 1] \to \mathbb{R}$ with $f(1) = 0$. Prove that the following limit exists and calculate its value:
$$
\lim_{t \to 1^+} \left( \frac{1}{1-t} \cdot \int_{0}^{1} x(f(tx) - f(x)) \, dx \right).
$$ | [
"Consider $g : [0, 1] \\to \\mathbb{R}$ defined by $g(x) = x f(x)$, for any $x \\in [0, 1]$. Being continuous, $g$ has an antiderivative $G : [0, 1] \\to \\mathbb{R}$ with $G(0) = 0$. Then $g(1) = f(1) = 0$ and\n$$\n\\int_{0}^{1} x f(x) \\, dx = \\int_{0}^{1} g(x) \\, dx = G(1),\n$$\nand\n$$\n\\int_{0}^{1} x f(tx) ... | Romania | 74th Romanian Mathematical Olympiad | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Calculus > Applications",
"Precalculus > Limits"
] | English | proof and answer | 2\int_{0}^{1} x f(x) \, dx | |
023f | Problem:
Guilherme escreveu um número em cada casa de um tabuleiro $8 \times 8$ de modo que a soma dos números das casas vizinhas de cada casa do tabuleiro é igual a 1. Calcule a soma de todos os números escritos por Guilherme.
Observação: duas casas são vizinhas se possuem um lado em comum. | [
"Solution:\n\nNumere as casas do tabuleiro conforme mostrado na figura 117.1.\n\nA soma dos números das casas marcadas com um mesmo número é igual a 1, porque elas são as vizinhas a uma determinada casa.\n\n| 1 | 2 | 1 | 7 | 8 | 7 | 8 | 9 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 2 | 1... | Brazil | Desafios | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 20 | |
0aj0 | Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that
$$
\frac{a+b+c+3}{4} \ge \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a}.
$$ | [
"Rewrite the left hand side of inequality in following way:\n$$\n\\frac{a+b+c+3}{4} = \\frac{a+b+c+3}{4\\sqrt{abc}} = \\frac{a+1}{4\\sqrt{abc}} + \\frac{b+1}{4\\sqrt{abc}} + \\frac{c+1}{4\\sqrt{abc}}\n$$\nRewrite denominators:\n$$\n\\frac{a+1}{4\\sqrt{abc}} + \\frac{b+1}{4\\sqrt{abc}} + \\frac{c+1}{4\\sqrt{abc}} = ... | North Macedonia | European Mathematical Cup | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof only | null | |
0ed9 | Let $ABCD$ be a cyclic quadrilateral and let $\mathcal{K}$ be its circumcircle. Denote the intersection of the lines $AB$ and $CD$ by $E$, so that $|AB| = |BE|$. Denote the intersection of the tangents to $\mathcal{K}$ from $B$ and $D$ by $F$, so that $AB$ and $DF$ are parallel. Show that the points $A, C$ and $F$ are ... | [
"Let us use Pascal's Theorem for the points $A, B, C$ and $D$ on the circle. Here we use the points $B$ and $D$ twice and denote their copies by $B_1$ and $D_1$. We have $AB \\cap CD = E$, $BB_1 \\cap DD_1 = F$ and the lines $AD_1$ and $B_1C$ either intersect or are parallel.\n\nIf $AD_1$ and $B_1C$ intersect, deno... | Slovenia | Slovenija 2016 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Ad... | null | proof only | null | |
0hdo | A palindromic number is a number whose digits stand symmetrically with respect to the center of the number's decimal notation, for instance, $7$, $1221$ and $57575$ are palindromic, while $1212$ and $3330$ are not. Prove that for any number of pairwise distinct palindromic numbers the sum of their reciprocals is not gr... | [
"Let an arbitrary natural number $n$ and consider all palindromic numbers that have exactly $n$ digits. We consider $n$ odd and $n$ even separately.\n\nWhen $n = 2m$, palindrome becomes $a_1a_2\\dots a_{m-1}a_m a_m a_{m-1}\\dots a_2a_1$, the only restriction on its digits is $a_1 \\neq 0$. Hence there are $9 \\cdot... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0bq8 | Problem:
Se consideră funcţia $f:(0, \infty) \rightarrow \mathbf{R}$ cu proprietatea că
$$
f(x y)=f(x)+f(y) \text{ pentru orice } x, y \in (0, \infty)
$$
a) Arătaţi că $f(1)=0$ şi $f\left(\frac{1}{x}\right)=-f(x)$ pentru orice $x \in (0, \infty)$;
b) Arătaţi că dacă 1 este singura soluţie a ecuaţiei $f(x)=0$, atunci f... | [
"Solution:\n\na.\nFolosim proprietatea dată:\n$$\nf(xy) = f(x) + f(y), \\quad \\forall x, y \\in (0, \\infty)\n$$\n\nPentru $x = 1$ și orice $y > 0$:\n$$\nf(1 \\cdot y) = f(1) + f(y) \\implies f(y) = f(1) + f(y)\n$$\nRezultă că $f(1) = 0$.\n\nPentru orice $x > 0$, alegem $y = \\frac{1}{x}$:\n$$\nf(x \\cdot \\frac{1... | Romania | Olimpiada Națională de Matematică - Etapa Locală | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof only | null | |
0bui | Problem:
Produsul a două numere este 540. Dacă primul număr s-ar mări cu 5, atunci produsul lor ar fi 600. Aflați numerele. | [
"Solution:\n\nNotăm $a$ și $b$ cele două numere. Avem $a \\cdot b = 540$, (1) și $(a + 5) \\cdot b = 600$, (2).\n\nDin (1) și (2) avem $a \\cdot b + 5 \\cdot b = 600 \\Leftrightarrow 540 + 5 \\cdot b = 600 \\Leftrightarrow 5 \\cdot b = 60 \\Leftrightarrow b = 12$.\n\nCum $a \\cdot 12 = 540 \\Leftrightarrow a = 45$.... | Romania | Olimpiada de Matematică | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 45 and 12 |
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