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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
02fi | Show that the $n$th root of a rational (for $n$ a positive integer) cannot be a root of the polynomial $x^5 - x^4 - 4x^3 + 4x^2 + 2$. | [
"First we prove that $P(x) = x^5 - x^4 - 4x^3 + 4x^2 + 2$ is irreducible in $\\mathbb{Z}$. Its possible rational roots are $\\pm 1, \\pm 2$. Since $P(x) = x^2(x-1)(x-2)(x+2)+2$ it is clear that $\\pm 2$ and $1$ are not roots and $P(-1) \\neq 0$ too. So if $P(x)$ is reducible then it must be written as a product of ... | Brazil | XVII OBM | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Abstract Algebra > Field Theory"
] | English | proof only | null | |
08kx | Problem:
Let $S$ be a point inside $\varangle p O q$, and let $k$ be a circle which contains $S$ and touches the legs $O p$ and $O q$ in points $P$ and $Q$ respectively. Straight line $s$ parallel to $O p$ from $S$ intersects $O q$ in a point $R$. Let $T$ be the point of intersection of the ray $P S$ and circumscribed ... | [
"Solution:\nLet $\\varangle O P S=\\varphi_{1}$ and $\\varangle O Q S=\\varphi_{2}$. We have that $\\varangle O P S=\\varangle P Q S=\\varphi_{1}$ and $\\varangle O Q S=\\varangle Q P S=\\varphi_{2}$ (tangents to circle $k$).\n\nBecause $R S \\| O P$ we have $\\varangle O P S=\\varangle R S T=\\varphi_{1}$ and $\\v... | JBMO | 2007 Shortlist JBMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dcb | Let be given a positive integer $n \geq 3$. Consider integers $a_{1}, a_{2}, \ldots, a_{n} > 1$ with the product equals to $A$ such that: for each $k \in \{1,2, \ldots, n\}$ then the remainder when $\frac{A}{a_{k}}$ divided by $a_{k}$ are all equal to $r$. Prove that $r \leq n-2$. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | proof only | null | |
0h4z | Winnie-the-Pooh and Piglet play the following game. There is a 15-inch-long stick. By his first move, Piglet breaks it into two pieces, then the players in turn break one of the existing pieces into two. The rules are that the resulting pieces must have integer length (in inches) and can't be 1-inch-long. The player wh... | [
"$$\n15 = 3+3+3+3+3 = 3+3+3+2+2+2 = 3+2+2+2+2+2+2.\n$$\nThis implies that in order to win Piglet needs to ensure the existence of two 3-inch and one 2-inch pieces (this will make the first and the third outcomes impossible). So, by his first move he must break the stick into 5 and 10 inches. The first piece will be... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Piglet | |
04kj | If $x$, $y$, $z$ and $w$ are real numbers such that
$$
x^2 + y^2 + z^2 + w^2 + x + 3y + 5z + 7w = 4,
$$
determine the largest possible value of $x + y + z + w$. | [
"By completing the squares of sums, the given equation can be written as\n$$\n\\left(x + \\frac{1}{2}\\right)^2 + \\left(y + \\frac{3}{2}\\right)^2 + \\left(z + \\frac{5}{2}\\right)^2 + \\left(w + \\frac{7}{2}\\right)^2 = 25.\n$$\nBy the inequality of arithmetic and quadratic means, we have\n$$\n\\frac{\\left(x + \... | Croatia | Mathematical competitions in Croatia | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 2 | |
0krp | Problem:
Let $S \subset \mathbb{N}$ be a set of positive integers whose product is $2021$ times its sum.
a) Given that $S$ has five (or more) elements, what is its minimum possible sum?
b) Given that $S$ has exactly four elements, what is its minimum possible sum?
c) Given that $S$ has exactly three elements, what ... | [
"Solution:\n\n$2021$ is prime factorized as $43 \\cdot 47$. There must therefore be at least one multiple of $43$ and at least one multiple of $47$.\n\na) The answer is $105 = 47 + 43 + 7 + 5 + 3 \\cdot 2 \\cdot 43 + 47$ is already too much, so any smaller answer would have to involve both $43$ and $47$. If $T$ is ... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | a) 105; b) 112; c) 240 | |
041r | Find the number of polynomials $f(x) = a x^3 + b x$ that satisfy the following conditions:
(1) $a, b \in \{1, 2, \dots, 2013\}$;
(2) the difference of any two numbers among $f(1)$, $f(2)$, $\dots$, $f(2013)$ is not a multiple of $2013$. | [
"2013 is factorized as $2013 = 3 \\times 11 \\times 61$. Let $p_1 = 3$, $p_2 = 11$, $p_3 = 61$. We denote by $a_i$ the residue of $a$ modulo $p_i$, by $b_i$ the residue of $b$ modulo $p_i$ ($i = 1, 2, 3$), $a, b \\in \\{1, 2, \\dots, 2013\\}$. By the Chinese Remainder Theorem, we have a bijection of $(a, b)$ with $... | China | China Girls' Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof and answer | 7200 | |
0byp | Let $n \ge 2$ be an integer and $A, B \in \mathcal{M}_n(\mathbb{C})$. If $(AB)^3 = O_n$, do we have always $(BA)^3 = O_n$? Justify. | [
"We will prove that the result is valid for $n \\le 3$ and not true for $n \\ge 4$. Matrices $C = AB$ and $D = BA$ have the same trace and both determinants are zero. If $C^2 = O_n$, then $D^3 = BC^2A = O_n$.\n\nFor $n=2$, matrices $C$ and $D$ have the same characteristic polynomial $P_C(X) = P_D(X) = X^2 - aX$, wi... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | English | proof only | null | |
0a72 | Problem:
Two concentric spheres have radii $r$ and $R$, $r < R$. We try to select points $A$, $B$ and $C$ on the surface of the larger sphere such that all sides of the triangle $A B C$ would be tangent to the surface of the smaller sphere. Show that the points can be selected if and only if $R \leq 2 r$. | [
"Solution:\n\nAssume $A$, $B$, and $C$ lie on the surface $\\Gamma$ of a sphere of radius $R$ and center $O$, and $A B$, $B C$, and $C A$ touch the surface $\\gamma$ of a sphere of radius $r$ and center $O$. The circumscribed and inscribed circles of $A B C$ then are intersections of the plane $A B C$ with $\\Gamma... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 2 | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0g4p | Problem:
Let $G$ be a graph whose vertices are the integers. Assume that any two integers are connected by a finite path in $G$. For two integers $x$ and $y$, we denote by $d(x, y)$ the length of the shortest path from $x$ to $y$, where the length of a path is the number of edges in it. Assume that $d(x, y) \mid x-y$ f... | [
"Solution:\nThe possible sets are $\\{0,1\\},\\{0,1,2\\},\\{0,1,2,3\\}$ and $\\mathbb{Z}_{\\geq 0}$.\nSince $d$ is defined as a distance, we also have the triangle inequality, stating that\n$$\nd(x, y) \\leq d(x, z)+d(z, y)\n$$\nfor all $x, y, z \\in \\mathbb{Z}$. Now note that for all $x \\in \\mathbb{Z}$ we must ... | Switzerland | Switzerland Selection Solution | [
"Discrete Mathematics > Graph Theory",
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | {0,1}, {0,1,2}, {0,1,2,3}, and Z_{>=0} | |
0l72 | Let $A$ be the set of positive integer divisors of $2025$. Let $B$ be a randomly selected subset of $A$. The probability that $B$ is a nonempty set with the property that the least common multiple of its elements is $2025$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | [] | United States | AIME II | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | final answer only | 237 | |
076l | Suppose $\alpha, \beta$ are two positive rational numbers. Assume for some positive integers $m, n$, it is known that $\alpha^{1/n} + \beta^{1/m}$ is a rational number. Prove that each of $\alpha^{1/n}$ and $\beta^{1/m}$ is a rational number. | [
"Let $l = \\text{lcm}(m, n)$ and write $n = l/u, m = l/v$. Then\n$$\n\\alpha^{1/n} = (\\alpha^u)^{1/l}, \\quad \\beta^{1/m} = (\\beta^v)^{1/l}.\n$$\nNote that $\\alpha^u$ and $\\beta^v$ are also rationals. Thus it is sufficient to consider the case $n = m$. Let us write $a = \\alpha^{1/n}$, $b = \\beta^{1/m}$ and $... | India | IND_TSExams | [
"Algebra > Abstract Algebra > Field Theory",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof only | null | |
01ut | Let $A = 2^7(7^{14} + 1) + 2^6 \cdot 7^{11} \cdot 10^2 + 2^6 \cdot 7^7 \cdot 10^4 + 2^4 \cdot 7^3 \cdot 10^6$.
Prove that the number $A$ ends with 14 zeroes. | [
"Answer: 14 zeroes.\n\n*First solution.* Derive the maximum power of 2 as a common factor:\n$$\nA = 2^7 \\cdot (7^{14} + 1 + 2 \\cdot 5^2 \\cdot 7^{11} + 2^3 \\cdot 5^4 \\cdot 7^7 + 2^3 \\cdot 5^6 \\cdot 7^3).\n$$\nNote that $50 = 7^2 + 1$ and represent 50 in this way in all terms:\n$$\n2^{-7}A = 7^{14} + (7^2 + 1)... | Belarus | Selection and Training Session | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0b25 | Problem:
For a positive integer $n$, denote by $\varphi(n)$ the number of positive integers $k \leq n$ relatively prime to $n$. How many positive integers $n$ less than or equal to $100$ are divisible by $\varphi(n)$? | [
"Solution:\n\nWe claim that any such integer $n$ must be either equal to $1$ or of the form $2^{a} 3^{b}$, where $a \\geq 1$ and $b \\geq 0$.\n\nFirst, we note that if $n > 1$, it must be even. This is because if $n$ admits a prime factorization $n = \\prod_{i=1}^{k} p_{i}^{r_{i}}$ over distinct primes $p_{i}$, the... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 16 | |
0cd7 | On a blackboard there are written, one after another, all positive integers from $1$ to $30000$, forming the following sequence of digits:
$123456789101112\ldots30000$.
Find the number of occurrences of $2023$ in this sequence. | [
"The sequence $2023$ appears $13$ times in the writing of $2023$, $12023$, $22023$ and of $20230$, $20231$, $20232$, $\\ldots$, $20239$. Moreover, $2023$ can appear by connecting the end of a number to the beginning of the next number. The writing $202|3$\n\nappears only once, in $3202|3203$, and the writing $20|23... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Discrete Mathematics > Other"
] | null | proof and answer | 25 | |
0hg6 | You are given 5 distinct positive integers. Can their arithmetic mean be:
a) exactly 3 times larger than their largest common divisor;
b) exactly 2 times larger than their largest common divisor; | [
"a) It's enough to provide an example of such 5 integers. One example is the set $1, 2, 3, 4, 5$, whose arithmetic mean is $3$, and the largest common divisor is $1$.\n\nb) Suppose that such numbers $a_1, a_2, a_3, a_4$ and $a_5$ exist, let $d$ be their largest common divisor, then these 5 integers can be rewritten... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | a) Yes; for example, 1, 2, 3, 4, 5. b) No. | |
0ah1 | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that:
$$
f(x + y f(x)) = f(f(x)) + x f(y)
$$ | [
"If we choose $x = 0$, then\n$$\nf(y f(0)) = f(f(0)).\n$$\nIf $f(0) \\ne 0$, then for $y = \\frac{1}{f(0)}$, we get\n$$\nf(t) = f(f(0)) = c,\n$$\ni.e. $f$ is a constant function. But then, substituting in (1), the equation gets the form\n$$\nc = c + c x, \\quad x \\in \\mathbb{R}.\n$$\nAs $x$ is arbitrary, we get t... | North Macedonia | XVIII-th Macedonian mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = 0 for all real x, and f(x) = x for all real x | |
03ed | Given an obtuse isosceles triangle $ABC$ with $CA = CB$ and circumcenter $O$. The point $P$ on $AB$ is such that $AP < \frac{AB}{2}$ and $Q$ on $AB$ is such that $BQ = AP$. The circle with diameter $CQ$ meets $(ABC)$ at $E$ and the lines $CE$, $AB$ meet at $F$. If $N$ is the midpoint of $CP$ and $ON$, $AB$ meet at $D$,... | [
"Let $T$ be the midpoint of $CQ$ (it is anyway the center of the important circle with diameter $CQ$). Then $OC$ is the perpendicular bisector of $AB$ (as $AC = BC$), triangle $ONT$ is isosceles by symmetry (as $P$ and $Q$ are symmetric with respect to the midpoint $M$ of $AB$ and hence with respect to $CO$, by the... | Bulgaria | Autumn tournament | [
"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 only | null | |
0iva | Problem:
Suppose $N$ is a 6-digit number having base-10 representation $\underline{a}\ \underline{b}\ \underline{c}\ \underline{d}\ \underline{e}\ \underline{f}$. If $N$ is $6/7$ of the number having base-10 representation $\underline{d}\ \underline{e}\ \underline{f}\ \underline{a}\ \underline{b}\ \underline{c}$, find... | [
"Solution:\n\nWe have $7(a b c d e f)_{10} = 6( d e f a b c )_{10}$, so $699400 a + 69940 b + 6994 c = 599300 d + 59930 e + 5993 f$. We can factor this equation as $6994(100 a + 10 b + c) = 5993(100 d + 10 e + f)$, which yields $538(a b c)_{10} = 461(d e f)_{10}$. Since $\\operatorname{gcd}(538, 461) = 1$, we must ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 461538 | |
0dkc | Let $ABC$ be an acute, non isosceles triangle with circumcircle $(O)$ and circumradius $R$. Denote $M$, $N$, $P$ as midpoints of segments $BC$, $CA$, $AB$ respectively. Let $(\omega)$ be the circle passes through $A$, $O$ and tangent to $OM$. Circle $(\omega)$ meets $AB$, $AC$ at $E$, $F$. Denote $I$ as midpoint of seg... | [
"Let circle $(\\omega)$ cuts $(O)$ again at $D$ and denote $O_1$ as center of $(\\omega)$. Since $AD$ is the radical axis of $(\\omega)$, $(O)$ then $OO_1 \\perp AD$. But $OO_1 \\perp OM$ since $OM$ is tangent to $(\\omega)$ leads to $AD \\parallel OM$ and then $AD \\perp BC$.\n\nThis means that $AD$, $AO$ are isog... | Saudi Arabia | Saudi Arabia booklet 2024 | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
... | English | proof only | null | |
06ld | Suppose there are $2019$ distinct points in a plane and the distances between pairs of them attain $k$ different values. Prove that $k$ is at least $44$. | [
"Suppose $k$ is less than $44$. Choose a point $P$ on the boundary of the convex hull of this set of $2019$ points.\nBy the pigeonhole principle, since $\\left\\lfloor \\frac{2018}{43} \\right\\rfloor > 45$, there is a circle centered at $P$ on which at least $45$ of the other points lie. Moreover, these $45$ point... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0cx0 | On the boundary of triangle $ABC$ points $D_1, D_2, E_1, E_2, F_1, F_2$ are chosen so that when going around the perimeter, the points are encountered in the following order: $A, F_1, F_2, B, D_1, D_2, C, E_1, E_2$. Given that $AD_1 = AD_2 = BE_1 = BE_2 = CF_1 = CF_2$, prove that two triangles formed by the triples of ... | [
"Let's begin with the following useful lemma.\n\n**Lemma.** Let points $F$ and $E$ be chosen on sides $AB$ and $AC$ of parallelogram $ABKC$ respectively, such that $BE = CF$. Then point $K$ is equidistant from lines $BE$ and $CF$ (see Fig. 1).\n\n\n\nРис. 1\n\n**Proof.** Since $BK \\paralle... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | Russian | proof only | null | |
0c0z | Determine whether there exist non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients satisfying
$$
P(x)^{10} + P(x)^9 = Q(x)^{21} + Q(x)^{20}.
$$ | [
"*First solution.* The answer is in the negative. Comparing the degrees of both sides in $(*)$ we get $\\deg P = 21n$ and $\\deg Q = 10n$ for some positive integer $n$. Take the derivative of $(*)$ to obtain\n$$\nP'P^8(10P + 9) = Q'Q^{19}(21Q + 20). \\quad (**)\n$$\nSince $\\gcd(10P + 9, P) = \\gcd(10P + 9, P + 1) ... | Romania | Eleventh Romanian Master of Mathematics | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | No | |
08w4 | A circle $X$ is inscribed in a quadrilateral $PQRS$. Construct 4 circles $A, B, C, D$, each of which is tangent to 3 of the 4 straight lines determined by the sides of the quadrilateral, as shown in the figure below. Suppose that the radii of the circles $A, B, C, X$ are $2$, $1$, $4$, $3$, respectively. Determine the ... | [
"Let us consider the situation, as indicated in the figure below, where $2$ circles and $2$ straight lines which are outer tangent to both of the circles and $1$ inner common tangent line are involved.\n\n\n\nLet $T$ and $U$ be the points of intersection of the common inner tangent line and... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 8 | |
0ka9 | Problem:
Let $p$ be a real number between $0$ and $1$. Jocelin has a coin that lands heads with probability $p$ and tails with probability $1-p$; she also has a number written on a blackboard. Each minute, she flips the coin, and if it lands heads, she replaces the number $x$ on the blackboard with $3x+1$; if it lands... | [
"Solution:\n\nIf the blackboard has the value $x$ written on it, then the expected value of the value after one flip is\n$$\nf(x) = p(3x + 1) + (1-p) \\frac{x}{2}\n$$\nBecause this expression is linear, we can say the same even if we only know the blackboard's initial expected value is $x$. Therefore, if the blackb... | United States | HMMT November 2019 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 1/5 | |
04xi | Let $a$, $b$, $c$ be positive real numbers satisfying $a^2 < bc$. Prove that
$$
b^3 + ac^2 > ab(a + c)
$$ | [
"Adding three AM-GM inequalities\n$$\n4a^3b + b^3c + 2c^3a \\geq 7a^2bc,\n$$\n$$\n4b^3c + c^3a + 2a^3b \\geq 7b^2ca,\n$$\n$$\n4c^3a + a^3b + 2b^3c \\geq 7c^2ab\n$$\nwe get\n$$\na^3b + b^3c + c^3a \\geq a^2bc + b^2ca + c^2ab \\quad (1)\n$$\nThe assumption $a^2 < bc$ implies $-a^3b > -b^2ca$ and this together with (1... | Czech-Polish-Slovak Mathematical Match | 11-th Czech-Slovak-Polish Match, 2011 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials"
] | English | proof only | null | |
001s | Sean $P_1, P_2, ..., P_n$, progresiones aritméticas infinitas de números enteros positivos, de diferencias $d_1, d_2, ..., d_n$, respectivamente. Demostrar que si todo número entero positivo figura en por lo menos una de las $n$ progresiones entonces una de las diferencias $d_i$ divide al mínimo común múltiplo de las r... | [] | Argentina | XIX Olimpíada Matemática Argentina | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | español | proof only | null | |
048a | Find all four-digit perfect squares of the form $\overline{aabb}$. | [] | Croatia | CroatianCompetitions2011 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 7744 | |
0f6h | Problem:
An $r \times s \times t$ cuboid is divided into $rst$ unit cubes. Three faces of the cuboid, having a common vertex, are colored. As a result exactly half the unit cubes have at least one face colored. What is the total number of unit cubes? | [] | Soviet Union | 19th ASU | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | [120, 144, 168, 180, 240] | |
02fj | $X$ has $n$ elements. $\mathcal{F}$ is a family of subsets of $X$ each with three elements, such that any two of the subsets have at most one element in common. Show that there is a subset of $X$ with at least $\lfloor\sqrt{2n}\rfloor$ members which does not contain any members of $\mathcal{F}$. | [
"Let $Y$ be a maximal subset of $X$ in the sense that if one adjoins another element from $X$ in $Y$ then it will contain a subset from $\\mathcal{F}$. Define $f: X \\setminus Y \\to \\binom{Y}{2}$, where $\\binom{Y}{2}$ is the family of 2-subsets from $Y$, with $f(x) = A$ if $A \\cup \\{x\\}$ is one of the sets fr... | Brazil | XVII OBM | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
06x4 | Let $ABCD E$ be a convex pentagon such that $BC = DE$. Assume there is a point $T$ inside $ABCDE$ with $TB = TD$, $TC = TE$ and $\angle TBA = \angle AET$. Let lines $CD$ and $CT$ intersect line $AB$ at points $P$ and $Q$, respectively, and let lines $CD$ and $DT$ intersect line $AE$ at points $R$ and $S$, respectively.... | [
"By the conditions we have $BC = DE$, $CT = ET$ and $TB = TD$, so the triangles $TBC$ and $TDE$ are congruent, in particular $\\angle BTC = \\angle DTE$.\n\nIn triangles $TBQ$ and $TES$ we have $\\angle TBQ = \\angle SET$ and $\\angle QTB = 180^\\circ - \\angle BTC = 180^\\circ - \\angle DTE = \\angle ETS$, so thes... | IMO | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ctd | Four distinct integers are written on the board. It is known that the absolute value of each of them is greater than $10^6$, and there is no positive integer greater than $1$ which divides all these numbers. Pete wrote on a paper six pairwise sums of the numbers from the board. Then he partitioned the obtained six numb... | [
"Yes.\n\nTwo possible examples are\n$$\n(N^2 - 3N + 1,\\ N^2 - N + 1,\\ -3N^2 + 3N - 1,\\ N^2 + N - 1)\n$$\nand\n$$\n(N^3 - N^2 + 1,\\ N^3 - 3N^2 + 2N - 1,\\ -N^3 + N^2 - 2N + 1,\\ -N^3 + N^2 + 2N - 1),\n$$\nwhere $N$ is an integer greater than $10^6$.\n\nFirst solution. For example, the numbers\n$$\nx = N^2 - 3N +... | Russia | Russian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English; Russian | proof and answer | Yes | |
0gz9 | Prove, that the equation $a^3 + b^3 + c^3 = a^2 + b^2 + c^2$ has infinite number of solutions in the integer numbers, if the biggest common divisor of numbers $a, b, c$ is 1. | [
"Lets choose such values: $b = 1 + x$, $c = 1 - x$ and put it in desired equation:\n$$\na^3 + 2(1 + 3x^2) = 2(1 + x^2) + a^2 \\Leftrightarrow a^2 - a^3 = 4x^2.\n$$\nWe put $1 - a = 4p^2$, then\n$$\na^2(1 - a) = (4p^2 - 1)^2 4p^2 = (2x)^2.\n$$\nWe put in a number $x = p(4p^2 - 1)$. Then three integer numbers $(a, b,... | Ukraine | The Problems of Ukrainian Authors | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
02vq | Problem:
Em uma festa, existem 25 crianças. Sabemos que quaisquer duas delas possuem pelo menos um de seus avós em comum (avô ou avó). Explique por que pelo menos 17 crianças possuem ou um mesmo avô ou uma mesma avó nessa família. | [
"Solution:\n\nPara cada criança, associe duas etiquetas, cada uma com o nome de um de seus avós. Se uma pessoa é o avô ou a avó de todas as crianças, então claramente a afirmação do enunciado é verdadeira. Tratemos então do caso em que nem todas as crianças são netas de uma mesma pessoa. Suponha que uma criança ten... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0dfx | Find all positive integers $n$ such that the products of its digits is equal to $n^2 - 10n - 22$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 12 | |
0f9a | Problem:
Show that $\{x\}^4 > \{x\} - 1/2$ for all real $x$. | [
"Solution:\n$\\{x\\}^4 - \\{x\\} + 1/2 = (\\{x\\}^2 - 1/2)^2 + (\\{x\\} - 1/2)^2 \\geq 0$. We could only have equality if $\\{x\\}^2 = \\{x\\} = 1/2$, which is impossible, so the inequality is strict."
] | Soviet Union | 24th ASU | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0hu7 | Problem:
Let $A B C D E F$ be a convex hexagon such that the quadrilaterals $A B D E$ and $A C D F$ are parallelograms. Prove that $B C E F$ is also a parallelogram. | [
"Solution:\n\nIt is well known that a quadrilateral is a parallelogram if and only if the two diagonals bisect each other, that is, have a common midpoint. Since $A B D E$ is a parallelogram, the midpoints of $A D$ and $B E$ coincide; since $A C D F$ is a parallelogram, the midpoints of $A D$ and $C F$ coincide. Co... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof only | null | |
0hh9 | In the quadrilateral $ABCD$ with $\angle ABC = \angle CDA = 90^\circ$. Let us denote $P = AC \cap BD$, $Q = AB \cap CD$, $R = AD \cap BC$. Let $l$ be the midline of the triangle $PQR$ parallel to $QR$. Prove that the circumcircle of the triangle formed by the lines $AB, AD$ and $l$ is tangent to the circumcircle of the... | [
"Let $l$ intersect $AB, AD, BC, CD$ at the points $M, N, K, L$ and let $E = AC \\cap QR$. Since $C$ is the orthocenter $AQR$, $AC \\perp QR$, and therefore $l$ is the perpendicular bisector of the segment $PE$ (fig. 15). Since $BC$ and $BA$ are the internal and external bisectors of the angle $DBE$, the points $K$ ... | Ukraine | 62nd Ukrainian National 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 > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
09mh | Prove that the equation $a! + b! = c^4 + 2024$ has a unique positive integer solution $(a, b, c)$ satisfying $a \le b$ and find the solution. | [
"Answer: $(a, b, c) = (5, 8, 14)$.\nWe prove that this is the only solution.\nConsider the equation modulo $16$. We have $c^4 + 2024 \\equiv 8, 9 \\pmod{16}$ on the right-hand side, since $x^4 \\equiv 0, 1 \\pmod{16}$. Now we consider the left-hand side.\nFirst, we have $b \\ge 7$, otherwise $2 \\cdot 6! = 1440 < 2... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | (5, 8, 14) | |
0dp5 | A (non-oriented) graph (without loops) with $2n$ vertices and $2n(n-1)$ edges is given, $n > 1$. Prove that some vertices and edges of the graph can be coloured in red in such way that each red edge connects red vertices and each red vertex belongs to exactly $n$ red edges. | [
"Consider a regular $2n$-gon (inscribed in a unit circle), which we label $n$ segments from the set of all sides and all diagonals.\nThe problem is equivalent to the following one: prove that some of unlabelled segments (including endpoints) can be coloured in red in such way that each red vertex belongs to exactly... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof only | null | |
0e8q | When a factory modernized its equipment, the productivity grew by $25\%$. Before, the management decided to fire $20\%$ of the employees. By how many $\%$ has the number of final products in this factory changed after both actions?
(A) It has decreased by $5\%$.
(B) It has decreased by $2.5\%$.
(C) It has decreased by ... | [
"Denote the number of products produced by the factory before the changes by $x$. Then they produced $\\frac{125}{100} \\cdot x$ products after the modernization of the equipment. After firing the employees the number has dropped to $\\frac{80}{100} \\cdot \\frac{125}{100} \\cdot x = x$ products. Hence, the number ... | Slovenia | National Math Olympiad 2013 - First Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
0htq | Problem:
Weighted coins numbered $2, 3, 4, \ldots, 2015$ are tossed. The coin numbered $i$ comes up heads with probability $1/(2i^2)$. What is the probability that an odd number of coins come up heads? | [
"Solution:\n\nLet $P_n$ be the probability that an odd number out of the coins whose numbers are at most $n$ ($1 \\leq n \\leq 2015$) come up heads. For $2 \\leq n \\leq 2015$, there are two ways for this to happen: coin $n$ is tails and an odd number of the preceding coins are heads, or coin $n$ is heads and an ev... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | final answer only | 1007/4030 | |
0120 | Problem:
Let $n \geqslant 2$ be a positive integer. Find whether there exist $n$ pairwise nonintersecting nonempty subsets of $\{1,2,3, \ldots\}$ such that each positive integer can be expressed in a unique way as a sum of at most $n$ integers, all from different subsets. | [
"Solution:\nAnswer: yes.\nLet $A_{1}$ be the set of positive integers whose only non-zero digits may be the $1$-st, the $(n+1)$-st, the $(2n+1)$-st etc. from the end; $A_{2}$ be the set of positive integers whose only non-zero digits may be the $2$-nd, the $(n+2)$-nd, the $(2n+2)$-nd etc. from the end, and so on. T... | Baltic Way | Baltic Way | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0hso | Problem:
If $\frac{1}{9}$ of $60$ is $5$, what is $\frac{1}{20}$ of $80$? | [
"Solution: In base 15, 6."
] | United States | null | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 4 | |
05zb | Problem:
Soit $ABC$ un triangle et $I$ le centre du cercle inscrit. On note $D$ et $E$ les pieds des bissectrices issues de $B$ et $C$. Soit $X$ l'intersection des symétriques de $(AB)$ et $(AC)$ par rapport à $(CE)$ et $(BD)$. Montrer que les droites $(XI)$ et $(BC)$ sont perpendiculaires. | [
"\n\nNotons $Y$ et $Z$ les intersections respectives de $(DX)$ et $(EX)$ avec la droite $(BC)$. On va montrer que le triangle $ZXY$ est isocèle en $X$ et que $I$ est le centre de son cercle inscrit. Si on arrive à montrer ces propriétés, la conclusion de l'exercice suivrait. En effet, la dr... | France | Préparation Olympique Française de Mathématiques - ENVOI 5 : Pot-POURRI | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Isogon... | null | proof only | null | |
0b36 | Problem:
A certain country wishes to interconnect $2021$ cities with flight routes, which are always two-way, in the following manner:
- There is a way to travel between any two cities either via a direct flight or via a sequence of connecting flights.
- For every pair $(A, B)$ of cities that are connected by a direct ... | [
"Solution:\nMore generally, consider the graph $G$ with $n$ vertices representing $n$ cities, two of them being connected by an edge if there is a two-way flight between them. We will prove that if $G$ is connected and every edge in $G$ belongs to a triangle, then $G$ must have at least $\\lfloor (3n-2)/2 \\rfloor$... | Philippines | 23rd Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 3030 | |
0231 | Problem:
$O$ triângulo acutângulo $ABC$ de ortocentro $H$ é tal que $AB = 48$ e $HC = 14$. O ponto médio do lado $AB$ é $M$ e o ponto médio do segmento $HC$ é $N$.
a. Mostre que o ângulo $MEN$ é reto. | [
"Solution:\n\nInicialmente observe que $ME$ é mediana relativa à hipotenusa do triângulo $AEB$. Portanto, $ME = AM = MB = 24$. Desse fato segue que o triângulo $BME$ é isósceles. Então $M\\hat{E}B = M\\hat{B}E = \\beta$.\n\nAnalogamente, como $N$ é o ponto médio da hipotenusa do triângulo $HEC$, temos $EN = HN = NC... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0g8a | 在凸六邊形 $ABCDEF$ 中, $AB \parallel DE$, $BC \parallel EF$, $CD \parallel FA$, 以及
$$
AB + DE = BC + EF = CD + FA.
$$
將邊 $AB$, $BC$, $DE$, $EF$ 的中點分別記為 $A_1, B_1, D_1, E_1$。設點 $O$ 為線段 $A_1D_1$ 及 $B_1E_1$ 的交點。證明 $\angle D_1OE_1 = \frac{1}{2}\angle DEF$。
In a convex hexagon $ABCDEF$, $AB \parallel DE$, $BC \parallel EF$, $CD... | [
"定義 $\\alpha = \\pi - \\angle FAB = \\pi - \\angle CDE$, $\\beta = \\pi - \\angle ABC = \\pi - \\angle DEF$, $\\gamma = \\pi - \\angle BCD = \\pi - \\angle EFA$。明顯有 $\\alpha + \\beta + \\gamma = \\pi$，且三者中至少兩者為銳角。令 $O'$ 為 $B_1E_1$ 及 $C_1F_1$ 的交點，$O''$ 為 $C_1F_1$ 及 $A_1D_1$ 的交點（$C_1, F_1$ 分別是 $CD$ 與 $FA$ 的中點）。再令 $\\... | Taiwan | 二〇一四年國際數學奧林匹亞競賽第二階段選訓營 獨立研究(二) | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Pla... | null | proof only | null | |
068y | In the blackboard are written some positive integers. We define the following movements:
(α) Every two successive numbers, say $n, n+1$, can be deleted by writing the number $n-2$.
(β) Every two numbers having difference 4, say $k, k+4$, can be deleted by writing the number $k-1$.
In the case we cannot apply any of ... | [
"**The answer is that that maximal possible value of $c$ is $-3$.**\n\nIn fact, this value is obtained if we start with the numbers $1,2,3,4,5$, and following the process:\n$$\n1, 2, 3, 4, 5 \\xrightarrow{\\alpha} 0, 1, 4, 5 \\xrightarrow{\\alpha} 0, 1, 2 \\xrightarrow{\\alpha} -1, 0 \\xrightarrow{\\beta} -3. \\qua... | Greece | SELECTION EXAMINATION | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | -3 | |
02r1 | Problem:
Escreva os algarismos de $0$ até $9$ em uma linha, na ordem que você escolher. Na linha debaixo junte os vizinhos, formando nove números novos, e some esses números como no exemplo:
| 2 | | 1 | | 3 | | 7 | | 4 | | 9 | | 5 | | 8 | | 0 | | 6 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- ... | [
"Solution:\n\nPara qualquer disposição dos algarismos, a soma dos vizinhos \"juntados\" terá sempre nove parcelas, sem repetição de algarismos nas unidades ou nas dezenas. O único algarismo que não aparece nas unidades é o primeiro e o único que não aparece nas dezenas é o último. Para que a soma seja máxima, o alg... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | B | |
0607 | Problem:
Soient $\mathrm{C}_{1}, \mathrm{C}_{2}, \ldots \mathrm{C}_{\mathrm{n}}$ des cercles de même rayon disposés dans le plan de sorte qu'ils ne soient jamais tangents 2 à 2 et qu'il existe toujours un chemin passant par les cercles pour aller d'un point de l'un d'entre eux à un autre (autrement dit, les cercles so... | [
"Solution:\n\nDans un premier temps, on reformule le problème en termes de graphes. On note $C$ l'ensemble des centres des cercles et $S$ l'ensemble des points d'intersections qui constituent les sommets d'un graphe $G$. On relie par une arête tout couple $(c, s) \\in C \\times S$ si $s$ appartient au cercle de cen... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
06v4 | Let $n \geqslant 3$ be a positive integer and let $(a_{1}, a_{2}, \ldots, a_{n})$ be a strictly increasing sequence of $n$ positive real numbers with sum equal to $2$. Let $X$ be a subset of $\{1,2, \ldots, n\}$ such that the value of
$$
\left|1-\sum_{i \in X} a_{i}\right|
$$
is minimised. Prove that there exists a str... | [
"Solution 1. Without loss of generality, assume $\\sum_{i \\in X} a_{i} \\leqslant 1$, and we may assume strict inequality as otherwise $b_{i}=a_{i}$ works. Also, $X$ clearly cannot be empty.\nIf $n \\in X$, add $\\Delta$ to $a_{n}$, producing a sequence of $c_{i}$ with $\\sum_{i \\in X} c_{i}=\\sum_{i \\in X^{c}} ... | IMO | IMO 2019 Shortlisted Problems | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
006u | Fede elige $2008$ enteros positivos tales que la multiplicación de esos $2008$ números termine en $75$, y los escribe en el pizarrón. A continuación Iván, sin ver los números de Fede, elige un entero positivo $k$ menor que $2008$. Si en la lista de Fede hay $k$ números tales que la multiplicación de esos $k$ números te... | [] | Argentina | Argentina 2009 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | All even integers k with 4 ≤ k ≤ 2006 | |
0h9c | The teacher wrote digits $123\ldots9123\ldots9123\ldots$ on the board until a $2018$-digit number was formed. After that Andriy and Olesya played a game as follows. Alternately (after Andriy begins) they cross out $2$ digits as follows: either the first two digits of the number remaining after the previous move, or the... | [
"Since we are only interested in divisibility by $3$, we can change all the digits as follows: $1, 4, 7 \\to 1$, $2, 5, 8 \\to 2$ and $3, 6, 9 \\to 3$, which will give us the equivalent problem. Then, after $2014 \\div 2 = 1007$ moves, a $4$-digit number will be left on the board, and Olesya will be making the last... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | Olesya | |
0456 | Consider all sequences of real numbers $x_0, x_1, x_2, \dots, x_{100}$ satisfying the following conditions:
(1) $x_0 = 0$;
(2) For any integer $i$, $1 \le i \le 100$, $1 \le x_i - x_{i-1} \le 2$ holds.
Find the largest positive integer $k \le 100$, such that
$$
x_k + x_{k+1} + \dots + x_{100} \ge x_0 + x_1 + \dots + x_... | [
"The answer is $67$.\n\nOn one hand, if $x_i = 2i$, $1 \\le i \\le 34$, $x_{34+j} = x_{34} + j = 68 + j$, $1 \\le j \\le 66$, then the sequence satisfies the conditions of the problem. We have\n$$\n\\sum_{j=68}^{100} x_j - \\sum_{i=0}^{67} x_i = \\sum_{j=1}^{33} (x_{67+j} - x_{34+j}) - \\sum_{i=1}^{34} x_i = 33^2 -... | China | 2022 CGMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 67 | |
05yr | Problem:
Est-il possible de trouver un bloc de 1000 nombres entiers strictement positifs consécutifs qui contient exactement 5 nombres premiers? | [
"Solution:\n\nA première vue, il semble difficile de garantir exactement 5 nombre premier dans un bloc de 1000 entiers consécutifs : aucun théorème d'arithmétique élémentaire permet de s'assurer d'avoir des nombres premiers rapprochés, mais pas d'autres nombres premier entre eux. On peut donc essayer de se demander... | France | Préparation Olympique Française de Mathématiques - ENVOI 5 : Pot-POURRI | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | null | proof and answer | Yes | |
02g6 | Given a triangle $ABC$, explain how to construct with ruler and compass a triangle $A'B'C'$ of minimum area such that $C' \in AC$, $A' \in AB$, $B' \in BC$ and $\angle B'A'C' = \angle BAC$, $\angle A'C'B' = \angle ACB$. | [
"Let $\\angle A$, $\\angle B$ and $\\angle C$ be the angles of $\\triangle ABC$ and $\\angle A'$, $\\angle B'$ and $\\angle C'$ be the angles of $\\triangle A'B'C'$. Thus $\\angle A = \\angle A'$ and $\\angle C = \\angle C'$.\n\nThe circumcircle of $\\triangle AA'C'$ meets the circumcircle of $\\triangle CC'B'$ at ... | Brazil | XXI OBM | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geomet... | English | proof only | null | |
00bq | Lucía writes the integer numbers from $1$ to $27$, in some order, around a circumference. Then, she calculates the sum of each pair of adjacent numbers, thus obtaining $27$ sums. We call $A$ the largest of these sums and $B$ the smallest. Find the minimum possible value of $A - B$.
Show how Lucía can write the numbers... | [
"It is easy to see that $A - B \\neq 0$. Indeed, we have $A - B = 0$ if and only if the $27$ sums are all equal; however, if $x, y, z$ are three consecutive numbers on the circumference, the sums $x+y$ and $y+z$ are different, since $x \\neq z$.\n\nWe will now prove that it is not possible that $A - B = 1$. If this... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2 | |
0imd | Problem:
For $a$ a positive real number, let $x_{1}, x_{2}, x_{3}$ be the roots of the equation $x^{3}-a x^{2}+a x-a=0$. Determine the smallest possible value of $x_{1}^{3}+x_{2}^{3}+x_{3}^{3}-3 x_{1} x_{2} x_{3}$. | [
"Solution:\n\nAnswer: $-4$. Note that $x_{1}+x_{2}+x_{3}=x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a$. Then\n$$\n\\begin{aligned}\n& x_{1}^{3}+x_{2}^{3}+x_{3}^{3}-3 x_{1} x_{2} x_{3}=\\left(x_{1}+x_{2}+x_{3}\\right)\\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-\\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}\\right)\\right) \\\\\n& \\q... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | -4 | |
0732 | Find all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying the equation
$$
f(x+y) + f(x)f(y) = (1+y)f(x) + (1+x)f(y) + f(xy), \quad (1)
$$
for all $x, y \in \mathbb{R}$. | [
"Taking $x = y = 0$ in (1), we obtain $f(0)^2 = 2f(0)$ and hence $f(0) = 0$ or $2$. If $f(0) = 2$, then $y = 0$ in (1) gives that $f(x) = 2 + x$ for all $x$. Putting this in (1), we get $2xy = 0$ for all $x, y$. This contradiction shows that there are no solutions with $f(0) = 2$. Thus $f(0) = 0$.\nPutting $x = y =... | India | Indija TS 2007 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) ≡ 0; f(x) = x^2 + x; f(x) = 3x | |
0j8v | Problem:
Let $ABC$ be a triangle with $AB=13$, $BC=14$, and $CA=15$. Let $D$ be the foot of the altitude from $A$ to $BC$. The inscribed circles of triangles $ABD$ and $ACD$ are tangent to $AD$ at $P$ and $Q$, respectively, and are tangent to $BC$ at $X$ and $Y$, respectively. Let $PX$ and $QY$ meet at $Z$. Determine ... | [
"Solution:\n\nAnswer: $\\frac{25}{4}$\n\nFirst, note that $AD=12$, $BD=5$, $CD=9$.\n\nBy equal tangents, we get that $PD=DX$, so $PDX$ is isosceles. Because $D$ is a right angle, we get that $\\angle PXD=45^{\\circ}$. Similarly, $\\angle XYZ=45^{\\circ}$, so $XYZ$ is an isosceles right triangle with hypotenuse $XY$... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | null | proof and answer | 25/4 | |
0kae | Problem:
How many ways are there to arrange the numbers $21, 22, 33, 35$ in a row such that any two adjacent numbers are relatively prime? | [
"Solution:\n\n$21$ cannot be adjacent to $33$ or $35$, so it must be on one end bordering $22$. $33$ cannot be adjacent to $21$ or $22$, so it must be on the other end bordering $35$. Thus, there are only $2$ orderings: $21, 22, 35, 33$, and $33, 35, 22, 21$."
] | United States | HMMT February 2019 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 2 | |
0cyy | Find positive integers $a_{1} < a_{2} < \ldots < a_{2010}$ such that
$$
a_{1}(1!)^{2010} + a_{2}(2!)^{2010} + \ldots + a_{2010}(2010!)^{2010} = (2011!)^{2010}.
$$ | [
"Consider\n$a_{1} = 2^{2010},\\ a_{2} = 3^{2010} - 1,\\ a_{3} = 4^{2010} - 1,\\ \\ldots,\\ a_{2010} = 2011^{2010} - 1$\nand get\n$$\n\\begin{gathered}\na_{1}(1!)^{2010} + a_{2}(2!)^{2010} + \\ldots + a_{2010}(2010!)^{2010} \\\\\n= (2!)^{2010} + (3!)^{2010} - (2!)^{2010} + (4!)^{2010} - (3!)^{2010} + \\ldots + \\\\\... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof and answer | a1 = 2^{2010}, a2 = 3^{2010} − 1, a3 = 4^{2010} − 1, ..., a2010 = 2011^{2010} − 1 | |
0jrs | Problem:
DeAndre Jordan shoots free throws that are worth 1 point each. He makes $40\%$ of his shots. If he takes two shots find the probability that he scores at least 1 point. | [
"Solution:\n\nWe want to find the probability of making at least one shot. The probability he makes no shots is $\\left(\\frac{3}{5}\\right)^{2}$, so the probability of making at least one is $1-\\left(\\frac{3}{5}\\right)^{2}=\\frac{16}{25}$."
] | United States | HMMT November 2016 | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 16/25 | |
004a | Para cada permutación $(x_1, x_2, \dots, x_{99})$ de $\{1, 2, \dots, 99\}$, sea
$$
L = |x_1 - x_2\sqrt{3}| + |x_2 - x_3\sqrt{3}| + \dots + |x_{98} - x_{99}\sqrt{3}| + |x_{99} - x_1\sqrt{3}|.
$$
Determinar el valor máximo de $L$, y para cuántas permutaciones de $\{1, 2, \dots, 99\}$ se alcanza este valor. | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | Español | proof and answer | Maximum L = 918 + 3618·sqrt(3). The number of permutations attaining it is 99 · C(36, 27) · 36! · 36! · 27!. | |
0a89 | Problem:
The infinite integer plane $\mathbb{Z} \times \mathbb{Z} = \mathbb{Z}^{2}$ consists of all number pairs $(x, y)$, where $x$ and $y$ are integers. Let $a$ and $b$ be non-negative integers. We call any move from a point $(x, y)$ to any of the points $(x \pm a, y \pm b)$ or $(x \pm b, y \pm a)$ a $(a, b)$-knight... | [
"Solution:\n\nIf the greatest common divisor of $a$ and $b$ is $d$, only points whose coordinates are multiples of $d$ can be reached by a sequence of $(a, b)$-knight moves starting from the origin. So $d=1$ is a necessary condition for the possibility of reaching every point in the integer plane. In any $(a, b)$-k... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 13 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | All nonnegative integer pairs with gcd(a, b) = 1 and a + b odd. | |
001f | Consideramos los números naturales $n$ de tres cifras, todas ellas distintas de cero. Diremos que un número $n$ es bueno si el número $n+1$ es múltiplo del número de dos cifras que se obtiene al suprimirle a $n$ la primera cifra de la izquierda (es decir, al suprimirle la cifra de las centenas). Por ejemplo, $123$ NO e... | [] | Argentina | XIX Olimpíada Matemática Argentina | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | español | proof and answer | 267, 343, 889, 917, 953 | |
06we | Let $n \geqslant 3$ be an integer. An integer $m \geqslant n+1$ is called $n$-colourful if, given infinitely many marbles in each of $n$ colours $C_{1}, C_{2}, \ldots, C_{n}$, it is possible to place $m$ of them around a circle so that in any group of $n+1$ consecutive marbles there is at least one marble of colour $C_... | [
"Answer: $m_{\\text{max}} = n^{2} - n - 1$.\n\nFirst suppose that there are $n(n-1)-1$ marbles. Then for one of the colours, say blue, there are at most $n-2$ marbles, which partition the non-blue marbles into at most $n-2$ groups with at least $(n-1)^{2} > n(n-2)$ marbles in total. Thus one of these groups contain... | IMO | IMO 2021 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n^2 - n - 1 | |
04ol | Points $P_1$, $P_2$ and $P_3$ are located on the side $AB$ of the triangle $ABC$ so that
$$
|AP_1| = |P_1P_2| = |P_2P_3| = |P_3B| = \frac{1}{4}|AB|.
$$
Parallels with the side $BC$ are drawn through those points and divide the triangle $ABC$ into four parts. The area of the part between the parallels through $P_2$ and ... | [] | Croatia | Croatian Mathematical Society Competitions | [
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | final answer only | 16 | |
02qr | Problem:
Uma folha retangular de $20~\mathrm{cm}$ por $30~\mathrm{cm}$ foi cortada ao longo das linhas tracejadas $AC$ e $BD$ em quatro pedaços: dois triângulos iguais e dois polígonos iguais de cinco lados cada um, como na Figura I. Os segmentos $AC$ e $BD$ têm o mesmo comprimento e se encontram no centro do retângul... | [
"Solution:\n\na) Vamos representar a folha original pelo retângulo $PQRS$ na figura abaixo. Seja $M$ o ponto onde os segmentos $AC$ e $BD$ se encontram. Como o centro do retângulo é o centro de simetria da figura, concluímos que $AM = MC = \\frac{1}{2} AC$. Por outro lado, sabemos que $AC = BD$, donde $AM = BM = CM... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | a) AB = 20 cm. b) Area of each triangular piece = 100 cm²; area of each five‑sided piece = 200 cm². c) Area of the rectangular hole = 200 cm². | |
03zz | It is known that each term of sequence $\{a_n\}$ is a non-zero real number, and for any positive integer $n$ holds the equation
$$
(a_1 + a_2 + \cdots + a_n)^2 = a_1^3 + a_2^3 + \cdots + a_n^3.
$$
(1) When $n = 3$, find out all the sequences consisting of three terms $a_1, a_2, a_3$.
(2) Does there exist an infinite ... | [
"(1) When $n = 1$, we have $a_1^2 = a_1^3$. Since $a_1 \\neq 0$, we get $a_1 = 1$.\nWhen $n = 2$, we have $(1+a_2)^2 = 1+a_2^3$. Since $a_2 \\neq 0$, we get $a_2 = 2$ or $a_2 = -1$.\nWhen $n = 3$, we have $(1+a_2+a_3)^2 = 1+a_2^3+a_3^3$. For $a_2 = 2$, we get $a_3 = 3$ or $a_3 = -2$; for $a_2 = -1$, we get $a_3 = 1... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | Part (1): The three-term sequences are (1, 2, 3), (1, 2, −2), and (1, −1, 1).
Part (2): Yes, it exists. One such sequence is given by a_n = n for 1 ≤ n ≤ 2012, and a_n = 2012(−1)^n for n ≥ 2013. | |
053v | Three workers must do a work completely. At first, one of them works as long as the other two would work together to complete one half of the work. Then another worker works as long as the other two would work together to complete one half of the work. Finally the third worker works as long as the other two would work ... | [
"Let the contributions of the first, second and third worker per a time unit be $x, y, z$, measured as percentages of the whole work. One half of the work would be done by the second and third worker together within $\\frac{1}{2(y+z)}$ time units, by the third and first worker together within $\\frac{1}{2(z+x)}$ ti... | Estonia | Estonian Math Competitions | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | proof and answer | 2.5 | |
0cyl | Let $a$, $b$, $c$ be positive real numbers. Prove that
$$
8(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \leq 9\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)
$$ | [
"The inequality is equivalent to\n$$\n8(a+b+c)(ab+bc+ca) \\leq 9(a+b)(b+c)(c+a)\n$$\nUsing the identity\n$$\n(a+b+c)(ab+bc+ca) = (a+b)(b+c)(c+a) + abc\n$$\nwe get\n$$\n8(a+b)(b+c)(c+a) + 8abc \\leq 9(a+b)(b+c)(c+a),\n$$\nhence $8abc \\leq (a+b)(b+c)(c+a)$. Applying AM-GM inequality, it follows\n$$\n2\\sqrt{ab} \\le... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
02g0 | Two mathematicians, lost in Berlin, arrived on the corner of Barbarossa street with Martin Luther street and need to arrive on the corner of Meininger street with Martin Luther street. Unfortunately they don't know which direction to go along Martin Luther Street to reach Meininger Street nor how far it is, so they mus... | [
"Since the mathematicians didn't know which side and at what distance their destination was, they should adopt the following strategy: walk $a_1$ blocks on one side (say, left), then go back to the starting point and walk $a_2$ blocks to the right, then go back again and walk $a_3$ blocks to the left, and so on, wi... | Brazil | XX OBM | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 9 | |
0arl | Problem:
How many times does the graph of $y= \pm \sqrt{\frac{x-1}{x+1}}$ cross the $x$-axis, the $y$-axis or the line $y=x$? | [
"Solution:\nTwice, at $(1,0)$ and in the third quadrant.\n\nFirst $x \\geq 1$ or $x<-1$. Solving $x$ in terms of $y$, we get $x=\\frac{1+y^{2}}{1-y^{2}}$. The graph of this equation has asymptotes $y=1$ and $y=-1$. For $x \\geq 1$ in this graph, $-1<y<1$ and this portion of the graph does not intersect by $y=x$. Fo... | Philippines | 13th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 2 | |
03ci | For a set $M$ of positive integers with $n$ elements, where $n$ is odd, a nonempty subset $T$ of $M$ is called good, if the product of the elements of $T$ is divisible by the sum of the elements of $M$, but not divisible by its square. If $M$ is good, find the maximum possible number of the good subsets of $M$? | [
"If $A \\cup B = M$ and $A \\cap B = \\emptyset$, then at most one of the sets $A$ and $B$ is good (otherwise $M$ is not good). Therefore the number of the good sets does not exceed half of the number of all subsets, i.e. $2^{n-1}$.\n\nWe will prove that the above estimate is sharp. Let $n = 2k + 1$ and $p$ be a pr... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 2^{n-1} | |
01ar | Let $X = \{x_0, \dots, x_{n-1}\}$ be an $n$-element set of real numbers such that $0 < |x_0| \le \dots \le |x_{n-1}|$. Prove that the sums of elements of all subsets of $X$ are $2^n$ consecutive members of an arithmetic sequence in some order if and only if
$$
|x_0| : \dots : |x_{n-1}| = 2^0 : \dots : 2^{n-1}.
$$ | [
"Let $d > 0$ be the difference of the arithmetic sequence of sums of elements of subsets. We prove by induction on $n$ that $|x_i| = d \\cdot 2^i$ for $0 \\le i < n$. The claim holds trivially for $n = 1$. Assume now that the claim holds for $n-1$ numbers.\nLet $s$ be the smallest among the $2^n$ sums and $s'$ be t... | Baltic Way | Baltic Way 2013 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0bix | Let $M$ be the set of palindromic numbers of the form $5n + 4$, where $n \ge 0$ is an integer. (A positive integer is a *palindromic number* if it remains the same when its digits are reversed. For instance, the numbers 7, 191, 23532, 3770773 are palindromic numbers.)
a) If we write the elements of $M$ in increasing or... | [
"a) The last (and hence, the first) digit of a number from $M$ equals 4 or 9. A direct count shows that $M$ contains 2 one digit numbers, 2 two digit numbers, 20 three digit and 20 four digit numbers, hence the 50th number is the 6th five digit number, that is, 40504.\n\nb) The greatest number in $M$ has the maximu... | Romania | 65th Romanian Mathematical Olympiad | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) 40504. b) Greatest: the palindrome starting and ending with 4 and having 2006 ones in between (i.e., 4 followed by 2006 ones and then 4). Smallest: the palindrome starting with 98, followed by 222 nines, and ending with 89 (i.e., 98 then 222 copies of 9 then 89). | |
04b1 | Find at least one integer $a$ such that for the polynomial $P(x) = x^5 + a x$ the statement
“if $n \mid P(k) - P(l)$ then $n \mid k - l$, for all $k, l \in \mathbb{Z}$”
holds only for finitely many positive integers $n$, including $n = 95$. | [
"One such number is $a = -95^4$.\nFor that $a$ we have $P(95) = P(0) = 0$, so $n$ divides $P(95) - P(0)$ for every positive integer $n$, and $n$ does not divide $95 - 0$ if $n$ is not a divisor of $95$. Therefore the statement is valid for only finitely many positive integers $n$.\n\nLet us show that the statement ... | Croatia | CroatianCompetitions2011 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | -95^4 | |
0e3p | Find all natural numbers $m$ and $n$, such that the sum of their greatest common divisor and their least common multiple equals $101$. | [
"Let $d$ denote the greatest common divisor of $m$ and $n$. Then $m = d m_1$ and $n = d n_1$, where $m_1$ and $n_1$ are coprime. The least common multiple of $m$ and $n$ is $d m_1 n_1$. We have\n$$\n101 = d + d m_1 n_1 = d(1 + m_1 n_1).\n$$\nSince $1 + m_1 n_1 \\ge 2$ and $101$ is prime, we can only have $d = 1$ an... | Slovenia | National Math Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | [(1, 100), (4, 25), (25, 4), (100, 1)] | |
0b8g | Let $a < c < b$ be real numbers and let $f : [a, b] \to \mathbb{R}$ be a function continuous at $c$. Show that if $f$ is the derivative of a function $F_a$ on $[a, c)$, and also the derivative of a function $F_b$ on $(c, b]$, then $f$ is the derivative of some function $F$ on the entire interval $[a, b]$. | [
"The function $f$ has primitives on the interval $[a, b]$ if and only if the function $g = f + 1 - f(c)$ has primitives on this interval. So, we can assume that $f(c) = 1$. Since $f$ is continuous at $c$, there exists $[\\alpha, \\beta] \\subset [a, b]$ such that $c \\in (\\alpha, \\beta)$ and $0 < f(x) < 2$, for e... | Romania | Romanian Mathematical Olympiad | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | English | proof only | null | |
03mw | Problem:
Let $p$ be a prime number for which $\frac{p-1}{2}$ is also prime, and let $a$, $b$, $c$ be integers not divisible by $p$. Prove that there are at most $1+\sqrt{2 p}$ positive integers $n$ such that $n<p$ and $p$ divides $a^{n}+b^{n}+c^{n}$. | [
"Solution:\nFirst suppose $b \\equiv \\pm a \\pmod{p}$ and $c \\equiv \\pm b \\pmod{p}$. Then, for any $n$, we have $a^{n}+b^{n}+c^{n} \\equiv \\pm a^{n}$ or $\\pm 3 a^{n} \\pmod{p}$. We are given that $p \\neq 3$ (since $\\frac{3-1}{2}$ is not prime) and $p \\nmid a$, so it follows that $a^{n}+b^{n}+c^{n} \\not\\e... | Canada | CMO | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0kv2 | Problem:
Compute the number of positive integers $n \leq 1000$ such that $\operatorname{lcm}(n, 9)$ is a perfect square. (Recall that lcm denotes the least common multiple.) | [
"Solution:\nSuppose $n=3^{a} m$, where $3 \\nmid m$. Then\n$$\n\\operatorname{lcm}(n, 9)=3^{\\max (a, 2)} m\n$$\nIn order for this to be a square, we require $m$ to be a square, and $a$ to either be even or $1$. This means $n$ is either a square (if $a$ is even) or of the form $3 k^{2}$ where $3 \\nmid k$ (if $a=1$... | United States | HMMT February | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | final answer only | 43 | |
08mi | Problem:
Consider a triangle $A B C$ with $\angle A C B=90^\circ$. Let $F$ be the foot of the altitude from $C$. Circle $\omega$ touches the line segment $F B$ at point $P$, the altitude $C F$ at point $Q$ and the circumcircle of $A B C$ at point $R$. Prove that points $A, Q, R$ are collinear and $A P=A C$.
![](attac... | [
"Solution:\n\nLet $M$ be the midpoint of $A B$ and let $N$ be the center of $\\omega$. Then $M$ is the circumcenter of triangle $A B C$, so points $M, N$ and $R$ are collinear. From $Q N \\parallel A M$ we get $\\angle A M R=\\angle Q N R$. Besides that, triangles $A M R$ and $Q N R$ are isosceles, therefore $\\ang... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | null | proof only | null | |
09ls | A number whose digits first increase and then decrease is called a *mountain-shaped number*. For example, $1310$ is mountain-shaped, while $1331$, $1234$, $4321$, and $3412$ are not mountain-shaped. How many four-digit mountain-shaped numbers are there in total? | [
"Let the four-digit number be $abcd$, where $a$, $b$, $c$, $d$ are its digits and $a \\neq 0$.\n\nA four-digit mountain-shaped number must satisfy:\n- The digits first increase, then decrease.\n- That is, there exists an index $k$ ($2 \\leq k \\leq 4$) such that $a < b < \\cdots < x$ (up to position $k$), then $x >... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 630 | |
03pp | Let $n$ be a given positive integer. Find the least positive integer $u_n$, such that for any positive integer $d$, the number of integers divisible by $d$ in every $u_n$ consecutive positive odd numbers is not less than the number of integers divisible by $d$ in $1, 3, 5, \dots, 2n-1$. (posed by Chen Yonggao) | [
"(1) $u_n \\ge 2n-1$. As $u_1 = 1$, we only need to consider $n \\ge 2$. Since the number of integers divisible by $2n-1$ in $1, 3, 5, \\dots, 2n-1$ is $1$ and that in $2(n+1)-1, 2(n+2)-1, \\dots, 2(n+2n-2)-1$ is $0$, then\n\n$u_n \\ge 2n - 1$.\n\n(2) $u_n \\le 2n-1$. We only need to consider the case when $2 \\nmi... | China | China Western Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | 2n-1 | |
08l8 | Problem:
Consider an integer $n \geq 4$ and a sequence of real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$. An operation consists in eliminating all numbers not having the rank of the form $4k+3$, thus leaving only the numbers $x_{3}, x_{7}, x_{11}, \ldots$ (for example, the sequence $4,5,9,3,6,6,1,8$ produces the seq... | [
"Solution:\nAfter the first operation 256 numbers remain; after the second one, 64 are left, then 16, next 4 and ultimately only one number.\n\nNotice that the 256 numbers left after the first operation are $3,7, \\ldots, 1023$, hence they are in arithmetical progression of common difference 4. Successively, the 64... | JBMO | 2008 Shortlist JBMO | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 683 | |
04oa | Let $k$ be a circle centred at $O$. Let $\overline{AB}$ be a chord of that circle and $M$ its midpoint. Tangents on $k$ at points $A$ and $B$ intersect at $T$. The line $\ell$ goes through $T$, intersects the shorter arc $\overarc{AB}$ at the point $C$ and the longer arc $\overarc{AB}$ at the point $D$, so that $|BC| =... | [
"Since $C$ and $D$ are on $k$, the power of the point $T$ with respect to $k$ equals $|TB|^2 = |TC| \\cdot |TD|$.\n\n\n\nFurthermore, since the right-angled triangles $TBM$ and $TOB$ are similar, we have $|TB|^2 = |TM| \\cdot |TO|$. Therefore, $|TC| \\cdot |TD| = |TM| \\cdot |TO|$, i.e. the... | Croatia | Croatian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
06me | At a party there are $1234$ participants, and each of them has shaken hands with exactly $137$ other participants. It is known that no three participants have shaken hands with each other. Furthermore, for any two participants $A$ and $B$ who have not shaken hands with each other, there must be exactly $k$ other partic... | [
"Answer: $17$\n\nConsider any participant $x$. He has shaken hands with $137$ other participants, say $y_1, y_2, \\dots, y_{137}$ (collectively known as Group Y participants). Also, there are $1234 - 1 - 137 = 1096$ participants who have not shaken hands with $x$; let's call them $z_1, z_2, \\dots, z_{1096}$ (colle... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 17 | |
0bng | Let $P$ be a point in the interior of the triangle $ABC$. The lines $AP$, $BP$, $CP$ intersect $BC$, $AC$, $AB$ at $A_1$, $B_1$, $C_1$, respectively. Given that
$$
s(PBA_1) + s(PCB_1) + s(PAC_1) = \frac{1}{2}s(ABC),
$$
(where $s(XYZ)$ denotes the area of $XYZ$) prove that $P$ lies on one of the medians of triangle $ABC... | [
"We have $\\frac{s(PBA_1)}{s(ABA_1)} = \\frac{PA_1}{AA_1} = \\frac{s(BPC)}{s(BAC)}$.\nDenote $s(BPC) = s_a$, etc.; it follows that $\\frac{s(PBA_1)}{s_c + s(PBA_1)} = \\frac{s_a}{s}$, where $s = s(ABC)$, hence $s(PBA_1) = \\frac{s_a s_c}{s - s_a} = \\frac{s_a s_c}{s_b + s_c}$.\nThe given equality becomes $\\frac{s_... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
07e6 | Find all polynomials $P(x, y)$ with real coefficients such that
$$
P(x, 2yz) + P(y, 2xz) + P(z, 2xy) = P(x + y + z, xy + xz + yz).
$$ | [
"Every polynomial in the form of $P(x, y) = Ax^2 + 2Ay + Bx$ for some $A, B \\in \\mathbb{R}$.\n\nLet $Q(a, b) = P(a, b^2)$ and define $Q_i(x, y)$ to be the homogeneous polynomial which consists of $i$-th degree coefficients of $Q(x, y)$. It is directly implied that\n$$\nQ_i(a, \\sqrt{2bc}) + Q_i(b, \\sqrt{2ac}) + ... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | P(x, y) = A x^2 + B x + 2 A y for real constants A, B | |
03k8 | Problem:
A number of schools took part in a tennis tournament. No two players from the same school played against each other. Every two players from different schools played exactly one match against each other. A match between two boys or between two girls was called a single and that between a boy and a girl was call... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 3 | |
0093 | Let there be a table with 100 columns and an unknown amount of rows. Starting by the first column, the first natural numbers are written in order, one number per cell, without skipping either numbers or cells, as shown in the picture. It is known that the number $38$ is written in the first column and $107$ is written ... | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | Rows: 69. Row 1, column 100: 6832. | |
03o0 | Problem:
A country with $n$ cities has some two-way roads connecting certain pairs of cities. Someone notices that if the country is split into two parts in any way, then there would be at most $k n$ roads between the two parts (where $k$ is a fixed positive integer). What is the largest integer $m$ (in terms of $n$ an... | [
"Solution:\nThe answer is $m = \\left\\lceil \\frac{n}{4k} \\right\\rceil$.\n\nCall a collection of cities independent if no two cities in the collection are joined by a road. Let $r$ and $k$ be integers such that $n = 4kq + r$ where $1 \\leq r \\leq 4k$.\n\nFirst we show that $m \\leq \\left\\lceil \\frac{n}{4k} \... | Canada | CMO 2023 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | m = ceil(n/(4k)) | |
05pu | Problem:
Soit $ABC$ un triangle dont les trois angles sont aigus. Soit $D \in [BC]$ tel que $\widehat{BAC} = \widehat{ADB}$. Soit $H$ le pied de la hauteur issue de $B$ dans $ABC$. La perpendiculaire à $(BC)$ passant par $H$ coupe $(AD)$ en $K$. On suppose que $K$ est à l'intérieur du triangle $ABC$. Soit $M$ le milie... | [
"Solution:\n\nNotons que $\\widehat{ABH} = 90^{\\circ} - \\widehat{BAC} = 90^{\\circ} - \\widehat{ADB} = \\widehat{AKH}$. Les points $A, B, K, H$ sont donc cocycliques.\n\nSoit $T$ le pied de la hauteur issue de $A$ dans $ABC$. Alors $\\widehat{BTA} = \\widehat{BHA} = 90^{\\circ}$ donc $A, B, H$ et $T$ sont cocycli... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0b7c | Given two real numbers $a$, $b$ such that $b - a^2 > 0$, describe all matrices $A \in M_2(\mathbb{R})$ such that $\det(A^2 - 2aA + bI_2) = 0$. | [
"The number $c = \\sqrt{b - a^2}$ is real. We have\n$$\nA^2 - 2aA + bI_2 = (A - (a + ic)I_2)(A - (a - ic)I_2),\n$$\nwhence the characteristic polynomial of $A$ has as a root $a + ic$, $a - ic$, or both. Since the polynomial has real coefficients, both are roots, hence eigenvalues. Therefore the polynomial is $x^2 -... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | All A ∈ M2(R) with trace(A) = 2a and det(A) = b; equivalently, all matrices of the form [[a + x, y], [(a^2 − x^2 − b)/y, a − x]] with x ∈ R and y ∈ R\{0}. | |
0g19 | Problem:
Sei $n \geq 1$ eine natürliche Zahl und seien $x_{1}, \ldots, x_{n}$ strikt positive reelle Zahlen. Zeige, dass man $a_{1}, \ldots, a_{n} \in \{-1,1\}$ wählen kann, sodass
$$
\sum_{i=1}^{n} a_{i} x_{i}^{2} \geq \left(\sum_{i=1}^{n} a_{i} x_{i}\right)^{2}
$$ | [
"Solution:\n\nOBdA können wir wie folgt sortieren: $x_{1} \\geq x_{2} \\geq \\cdots \\geq x_{n}$. Dann wählen wir $a_{i} = (-1)^{i+1}$. Für $1 \\leq i, j \\leq n$ gilt also $a_{i} a_{j} = (-1)^{i+j}$. Damit erhalten wir\n$$\n\\left(\\sum_{i=1}^{n} a_{i} x_{i}\\right)^{2} = \\sum_{i=1}^{n} x_{i}^{2} + \\sum_{i, j=1}... | Switzerland | IMO-Selektion | [
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof only | null | |
0at4 | Problem:
In $\triangle ABC$ with $BC = 24$, one of the trisectors of $\angle A$ is a median, while the other trisector is an altitude. What is the area of $\triangle ABC$? | [
"Solution:\n\n$32 \\sqrt{3}$"
] | Philippines | Philippines Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 32\sqrt{3} | |
000o | En la pantalla de la computadora hay inicialmente escritos dos $1$. El programa *insertar* hace que al apretar la tecla *Enter* se inserte entre cada par de números la suma de esos números.
En el primer paso se inserta un número y obtenemos $1$-$2$-$1$; en el segundo paso se insertan dos números y tenemos $1$-$3$-$2$-$... | [] | Argentina | XIX Olimpíada Matemática Argentina | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | español | proof and answer | 3^25 + 1 | |
042f | Let $n \ge 3$ be integer. Suppose that $\alpha, \beta, \gamma \in (0, 1)$ and $a_k, b_k, c_k \ge 0$ $(k = 1, 2, \dots, n)$ satisfy $\sum_{k=1}^n (k+\alpha)a_k \le \alpha$, $\sum_{k=1}^n (k+\beta)b_k \le \beta$ and $\sum_{k=1}^n (k+\gamma)c_k \le \gamma$. Find the minimum of $\lambda$ such that $\sum_{k=1}^n (k+\lambda)... | [
"Let $a_1 = \\frac{\\alpha}{1+\\alpha}$, $b_1 = \\frac{\\beta}{1+\\beta}$, $c_1 = \\frac{\\gamma}{1+\\gamma}$, $a_i, b_i, c_i = 0$ $(i = 2, 3, \\dots, n)$. We see that all conditions are satisfied. So, we must have\n$$\n(1 + \\lambda) \\frac{\\alpha}{1 + \\alpha} \\cdot \\frac{\\beta}{1 + \\beta} \\cdot \\frac{\\ga... | China | China Southeastern Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | alpha*beta*gamma / ((1+alpha)*(1+beta)*(1+gamma) - alpha*beta*gamma) |
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