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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0i51 | Problem:
Count how many 8-digit numbers there are that contain exactly four nines as digits. | [
"Solution:\n\nThere are $\\binom{8}{4} \\cdot 9^{4}$ sequences of 8 numbers with exactly four nines. A sequence of digits of length 8 is not an 8-digit number, however, if and only if the first digit is zero. There are $\\binom{7}{4} 9^{3}$-digit sequences that are not 8-digit numbers. The answer is thus $\\binom{8... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 433755 | |
08eq | Problem:
Date due frazioni $a / b$ e $c / d$, definiamo la loro somma pirata come
$$
\frac{a}{b} \diamond \frac{c}{d}=\frac{a+c}{b+d}
$$
dove si intende che le due frazioni iniziali sono ridotte ai minimi termini (cioè semplificate il più possibile), ed anche il risultato viene poi ridotto ai minimi termini. Così, p... | [
"Solution:\n\nPoniamo $M_{n}=1 / 2$ e $m_{n}=1 /(n-1)$. Dimostriamo che $M_{n}$ è il massimo valore possibile per l'ultima frazione, mentre $m_{n}$ è il minimo valore possibile per l'ultima frazione. La dimostrazione consiste di due parti: una costruzione che mostra che tali valori sono effettivamente realizzabili,... | Italy | XXXVII Olimpiade Italiana di Matematica | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Maximum = 1/2, Minimum = 1/(n-1) | |
0jh3 | Problem:
Let $a_{1}, a_{2}, \ldots, a_{2013}$ be real numbers satisfying the following conditions:
- $a_{1}=a_{2013}=0$
- $|a_{i}-a_{i+1}|<1$, for $1 \leq i \leq 2012$.
- $\sum_{i=1}^{2013} a_{i}=0$.
Find the greatest possible value of the sum $\sum_{i=1}^{m} a_{i}$, where $m$ ($1 \leq m \leq 2013$) is allowed to vary,... | [
"Solution:\nThe answer is that the sum can come arbitrarily close to $503^{2}=253009$, but (due to an unfortunate oversight in the problem statement) cannot equal it.\n\nSuppose that the sequence $\\{a_{i}\\}$ is fixed, and consider the $m$ that yields the largest value of the sum $\\sum_{i=1}^{m} a_{i}$. Clearly, ... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 253009 (supremum; maximum not attained) | |
076a | Problem:
Let $ABC$ be a right-angled triangle with $\angle B = 90^{\circ}$. Let $BD$ be the altitude from $B$ onto $AC$. Let $P$, $Q$ and $I$ be the incentres of triangles $ABD$, $CBD$ and $ABC$ respectively. Show that the circumcentre of the triangle $PIQ$ lies on the hypotenuse $AC$. | [
"Solution:\n\nWe begin with the following lemma:\n\nLemma: Let $XYZ$ be a triangle with $\\angle XYZ = 90 + \\alpha$. Construct an isosceles triangle $XEZ$, externally on the side $XZ$, with base angle $\\alpha$. Then $E$ is the circumcentre of $\\triangle XYZ$.\n\nProof of the Lemma: Draw $ED \\perp XZ$. Then $DE$... | India | INMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0dtr | In the triangle $ABC$, $\angle B > 90^\circ$, the incircle touches the sides $BC$ and $CA$ at $D$ and $E$, respectively. The lines $ED$ and $AB$ intersect at $P$. The incircle of the triangle $AEP$ touches the sides $PE$ and $AP$ at $D_1$ and $E_1$, respectively. The lines $E_1D_1$ and $AE$ intersect at $P_1$. Suppose ... | [
"Let $\\angle EPB = x$. Since $P, C, E, B$ are concyclic, $x = \\angle EPB = \\angle BCA$ and therefore $\\triangle AEP \\sim \\triangle ABC$. Thus the points $(A, B, C, D, E, P)$ correspond to the points $(A, E, P, D_1, E_1, P_1)$ in 2 similar configurations associated with triangles $ABC, AEP$. Therefore $\\angle... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0bj9 | We will say that a positive integer $n$ is subject to an *interesting* change if it is multiplied by $2$ and the result is increased by $4$, a *special* change if it is multiplied by $3$ and the result is increased by $9$ and an *awesome* change if it is multiplied by $4$ and the result is increased by $16$.
a) Show t... | [
"a) Before the last change the number must be $(2020 - 16) : 4 = 501$; the previous number must be $(501 - 9) : 3 = 164$ and the required number is $(164 - 4) : 2 = 80$.\n\nb) An interesting change produces a multiple of $2$, a special change gives a multiple of $3$ and an awesome change yields a multiple of $4$. S... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a) 80; b) 332 | |
0k3n | Problem:
How many ways are there for Nick to travel from $(0,0)$ to $(16,16)$ in the coordinate plane by moving one unit in the positive $x$ or $y$ direction at a time, such that Nick changes direction an odd number of times? | [
"Solution:\n\nThis condition is equivalent to the first and last step being in different directions, as if you switch directions an odd number of times, you must end in a different direction than you started. If the first step is in the $x$ direction and the last step is in the $y$ direction, it suffices to count t... | United States | HMMT February 2018 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2 * C(30, 15) | |
0123 | Problem:
Let $p$ and $q$ be two different primes. Prove that
$$
\left\lfloor\frac{p}{q}\right\rfloor+\left\lfloor\frac{2 p}{q}\right\rfloor+\left\lfloor\frac{3 p}{q}\right\rfloor+\ldots+\left\lfloor\frac{(q-1) p}{q}\right\rfloor=\frac{1}{2}(p-1)(q-1) .
$$
(Here $\lfloor x\rfloor$ denotes the largest integer not greate... | [
"Solution:\nThe line $y=\\frac{p}{q} x$ contains the diagonal of the rectangle with vertices $(0,0)$, $(q, 0)$, $(q, p)$ and $(0, p)$ and passes through no points with integer coordinates in the interior of that rectangle. For $k=1,2, \\ldots, q-1$ the summand $\\left\\lfloor\\frac{k p}{q}\\right\\rfloor$ counts th... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0hhj | $BE$ and $CF$ are the altitudes of the acute scalene $\triangle ABC$, $O$ is its circumcenter and $M$ is the midpoint of the side $BC$. If the point, which is symmetric to $M$ with respect to $O$, lies on the line $EF$, find all possible values of the ratio $\frac{AM}{AO}$.
Fig. 21 | [
"Let $H$ be the orthocenter of $\\triangle ABC$, $A'$ be the antipode of $A$ in $(ABC)$, $S$ be the point, which is symmetric to $M$ with respect to $O$ (fig. 21), and $T$ be the intersection of $BC$ and $EF$. It's well-known that in this construction $H$ is the orthocenter of $\\triangle AMT$, $BHCA'$ is parallelo... | Ukraine | Problems from Ukrainian Authors | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > ... | English | proof and answer | sqrt(2) | |
08a4 | Problem:
Le facce di due tetraedri regolari identici vengono colorate di rosso, bianco, verde, blu; i colori sono scelti casualmente, ma le quattro facce di ciascun tetraedro debbono essere tutte di colori diversi. Qual è la probabilità che dopo la colorazione i due tetraedri siano indistinguibili?
(A) $\frac{1}{4!}$
(... | [
"Solution:\nLa risposta è (D). Poggiamo entrambi i tetraedri sulla faccia rossa e ruotiamoli fino ad avere davanti a noi la faccia bianca. A questo punto ci sono due casi possibili, quella a destra può essere verde o blu. Di qualunque colore sia quella del primo tetraedro, la probabilità che nel secondo sia del med... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | MCQ | D | |
009n | Find the least $n$ with the following property: In each $n$-term sequence of positive integers with sum $2013$ there are several consecutive terms with sum $31$. | [
"The least $n$ in question is $1022$. An example that $n = 1021$ is not enough: arrange in a row $32$ blocks $1, 1, \\ldots, 1, 32$, with $30$ ones in each, then add $29$ ones. This gives a sequence of length $32 \\cdot 31 + 29 = 1021$ and sum $62 \\cdot 32 + 29 = 2013$. No consecutive terms in it have sum $31$.\n\... | Argentina | NATIONAL XXX OMA | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1022 | |
06sl | In a triangle $A B C$, let $D$ and $E$ be the feet of the angle bisectors of angles $A$ and $B$, respectively. A rhombus is inscribed into the quadrilateral $A E D B$ (all vertices of the rhombus lie on different sides of $A E D B$). Let $\varphi$ be the non-obtuse angle of the rhombus. Prove that $\varphi \leqslant \m... | [
"Let $K, L, M$, and $N$ be the vertices of the rhombus lying on the sides $A E, E D, D B$, and $B A$, respectively. Denote by $d(X, Y Z)$ the distance from a point $X$ to a line $Y Z$. Since $D$ and $E$ are the feet of the bisectors, we have $d(D, A B)=d(D, A C), d(E, A B)=d(E, B C)$, and $d(D, B C)=d(E, A C)=0$, w... | IMO | International Mathematical Olympiad Shortlisted Problems | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
097z | Problem:
Fie dat triunghiul $ABC$ și $BC = a$. Considerăm un punct $M$ pe semidreapta $[CA)$, astfel încât $\angle MBC = \angle BAC$. Găsiți cea mai mică distanță posibilă dintre centrele $O_1$ și $O_2$ ale cercurilor circumscrise triunghiurilor $ABC$ și $ABM$. | [
"Solution:\n\nSunt posibile 2 cazuri:\n1. $\\angle ABC > \\angle BAC$. Atunci $M \\in [AC]$.\n2. $\\angle ABC \\leq \\angle BAC$. Atunci $M$ se află pe prelungirea dreptei $(AC)$ după punctul $A$ și $M \\notin [AC]$.\n\nConsiderăm cazul 1. $\\triangle ABC$ este asemenea cu $\\triangle BMC$ (deoarece $\\angle C$ est... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | a/2 | |
0e5c | A red box contains twelve balls numbered from $1$ to $12$. Jan moves some of the balls into the green box. He then realizes that for any two balls from the green box the following is true: if these two balls are numbered $a$ and $b$, then the ball numbered $|a-b|$ is in the red box. At most how many balls has Jan moved... | [
"Jan can move $6$ balls. If he moves all odd-numbered balls, then the difference of the numbers on any two of them is an even number, which is clearly not written on any of the balls in the green box.\n\nNow, assume that Jan moved at least $7$ balls to the green box and denote the numbers on these balls by $a_1 < a... | Slovenia | National Math Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 6 | |
0j52 | Problem:
Let $f:[0,1] \rightarrow [0,1]$ be a continuous function such that $f(f(x)) = 1$ for all $x \in [0,1]$. Determine the set of possible values of $\int_{0}^{1} f(x) \, dx$. | [
"Solution:\n\nAnswer: $\\left(\\frac{3}{4}, 1\\right]$\n\nSince the maximum value of $f$ is $1$, $\\int_{0}^{1} f(x) \\, dx \\leq 1$.\n\nBy our condition $f(f(x)) = 1$, $f$ is $1$ at any point within the range of $f$. Clearly, $1$ is in the range of $f$, so $f(1) = 1$. Now $f(x)$ is continuous on a closed interval ... | United States | Harvard-MIT Mathematics Tournament | [
"Precalculus > Functions",
"Calculus > Integral Calculus > Applications"
] | null | proof and answer | (3/4, 1] | |
017z | The numbers from $1$ to $2010$ are partitioned into three subsets $A$, $B$, and $C$, each containing exactly $670$ elements. Prove that there exist three numbers, one in each set, such that one is the sum of the other two. | [
"Assume no such numbers exist. Without loss of generality, assume $1, 2, \\dots, n-1 \\in A$, but $n \\in B$. Consider an $x \\in C$, and suppose that $x - 1 \\notin A$.\n\nIf $x - 1 \\in B$, the triple $(1, x - 1, x)$ satisfies the condition. So suppose $x - 1 \\in C$. Then $x - n \\in A$ would yield a triple $(x ... | Baltic Way | BALTIC WAY | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0krt | Problem:
Let $p$ be a prime and let $\mathbb{F}_p$ be the set of integers modulo $p$. Call a function $f: \mathbb{F}_p^2 \rightarrow \mathbb{F}_p$ quasiperiodic if there exist $a, b \in \mathbb{F}_p$, not both zero, so that $f(x+a, y+b)=f(x, y)$ for all $x, y \in \mathbb{F}_p$. Find, with proof, the number of function... | [
"Solution:\n\nEvery function $\\mathbb{F}_p^2 \\rightarrow \\mathbb{F}_p$ can be written uniquely as a polynomial $\\sum_{i=0}^{p-1} \\sum_{j=0}^{p-1} a_{ij} x^i y^j$. We claim that a function can be written as a sum of quasiperiodic functions if and only if $a_{ij}=0$ for all $i+j \\geq p$. The only if direction f... | United States | HMIC | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Linear Algebra > Linear transformations"
] | null | proof and answer | p^{p(p+1)/2} | |
04zt | Find all quadruples $(w, x, y, z)$ of positive integers such that $w^x + w^y = w^z$. | [
"Consider the following cases.\nIf $w = 1$, then no solution can exist, since the l.h.s. of the equality equals $2$ while the r.h.s. equals $1$.\nIf $w \\ge 2$, then $x < z$ and $y < z$, i.e., $x \\le z - 1$ and $y \\le z - 1$. Thus $w^x + w^y \\le w^{z-1} + w^{z-1} = 2 \\cdot w^{z-1} \\le w \\cdot w^{z-1} = w^z$. ... | Estonia | Selected Problems from the Final Round of National Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | (w, x, y, z) = (2, n, n, n+1) for any positive integer n | |
0kzp | Problem:
$$
\frac{2+3+\cdots+100}{1}+\frac{3+4+\cdots+100}{1+2}+\cdots+\frac{100}{1+2+\cdots+99} .
$$ | [
"Solution:\n\nLet $A$ denote the sum. We have\n\n$$\n\\begin{aligned}\nA+99 & =(1+2+\\cdots+100)\\left(\\frac{1}{1}+\\frac{1}{1+2}+\\cdots+\\frac{1}{1+2+\\cdots+99}\\right) \\\\\n& =5050 \\sum_{k=1}^{99} \\frac{2}{k(k+1)} \\\\\n& =10100 \\cdot \\sum_{k=1}^{99}\\left(\\frac{1}{k}-\\frac{1}{k+1}\\right) \\\\\n& =1010... | United States | HMMT November 2024 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 9900 | |
01r6 | Point $L$ is marked on the side $AB$ of a triangle $ABC$. The incircle of the triangle $ABC$ meets the segment $CL$ at points $P$ and $Q$.
Is it possible that the equalities $CP = PQ = QL$ hold if $CL$ is
a) the median?
b) the bisector?
c) the altitude?
d) the segment joining vertex $C$ with the point $L$ of tangency ... | [
"a) yes, it is possible; this is valid for the triangles with $AC = 5x$, $AB = 10x$, $BC = 13x$, where $x$ is any positive real number;\n\nb) it is impossible;\n\nc) it is impossible;\n\nd) yes, it is possible; this is valid for the triangles with $AC = 3x$, $AB = 4x$, $BC = 5x$, where $x$ is any positive real numb... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | proof and answer | a) Yes; for triangles with AC = 5x, AB = 10x, BC = 13x. b) No; impossible. c) No; impossible. d) Yes; for triangles with AC = 3x, AB = 4x, BC = 5x. | |
06u8 | An integer $n \geqslant 3$ is given. We call an $n$-tuple of real numbers $(x_{1}, x_{2}, \ldots, x_{n})$ Shiny if for each permutation $y_{1}, y_{2}, \ldots, y_{n}$ of these numbers we have
$$
\sum_{i=1}^{n-1} y_{i} y_{i+1} = y_{1} y_{2} + y_{2} y_{3} + y_{3} y_{4} + \cdots + y_{n-1} y_{n} \geqslant -1.
$$
Find the la... | [
"Answer: $K = - (n-1)/2$.\n\nSolution 1. First of all, we show that we may not take a larger constant $K$. Let $t$ be a positive number, and take $x_{2} = x_{3} = \\cdots = t$ and $x_{1} = -1/(2t)$. Then, every product $x_{i} x_{j}$ ($i \\neq j$) is equal to either $t^{2}$ or $-1/2$. Hence, for every permutation $y... | IMO | International Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values"
] | English | proof and answer | -(n-1)/2 | |
06q8 | Let $n$ be a positive integer and let $p$ be a prime number. Prove that if $a, b, c$ are integers (not necessarily positive) satisfying the equations
$$
a^{n}+p b=b^{n}+p c=c^{n}+p a,
$$
then $a=b=c$. | [
"If two of $a, b, c$ are equal, it is immediate that all the three are equal. So we may assume that $a \\neq b \\neq c \\neq a$. Subtracting the equations we get $a^{n}-b^{n}=-p(b-c)$ and two cyclic copies of this equation, which upon multiplication yield\n$$\n\\frac{a^{n}-b^{n}}{a-b} \\cdot \\frac{b^{n}-c^{n}}{b-c... | IMO | 49th International Mathematical Olympiad Spain | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis... | English | proof only | null | |
0k01 | Problem:
Evaluate the sum
$$
\sum_{k=1}^{\infty}\left(\prod_{i=1}^{k} \frac{P_{i}-1}{P_{i+1}}\right)=\frac{1}{3}+\frac{1}{3} \cdot \frac{2}{5}+\frac{1}{3} \cdot \frac{2}{5} \cdot \frac{4}{7}+\frac{1}{3} \cdot \frac{2}{5} \cdot \frac{4}{7} \cdot \frac{6}{11}+\ldots
$$
where $P_{n}$ denotes the $n^{\text{th}}$ prime numb... | [
"Solution:\nRewrite the given sum as\n$$\n\\begin{gathered}\n\\frac{1}{3}+\\frac{2}{3} \\cdot \\frac{1}{5}+\\frac{2}{3} \\cdot \\frac{4}{5} \\cdot \\frac{1}{7}+\\frac{2}{3} \\cdot \\frac{4}{5} \\cdot \\frac{6}{7}+\\cdot \\frac{1}{11}+\\ldots \\\\\n=\\frac{1}{3}+\\left(1-\\frac{1}{3}\\right) \\frac{1}{5}+\\left(1-\\... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1 | |
04i7 | Does there exist a positive integer $n$ such that $n^2 + 2n + 2015$ is a perfect square? | [] | Croatia | First round – City competition | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | No | |
0hvl | Problem:
For which $n \geq 3$ can an $n \times n$ board with all four corners removed be tiled with "L" and "J" tetris pieces? | [
"Solution:\n\nWe claim that $n$ works iff $n \\equiv 2 \\pmod{4}$.\n\nThe board has $n^{2}-4$ squares. If $n$ is odd, this is odd, so the board wouldn't be coverable with tetris tiles (which have 4 cells each).\n\nThus $n$ is even. Alternately color the rows of the board black and white. Each of the tetris pieces m... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | All n ≥ 3 with n ≡ 2 (mod 4). | |
06xu | Let $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ be $2n$ positive integers such that the $n+1$ products
$$
\begin{gathered}
a_{1} a_{2} a_{3} \cdots a_{n}, \\
b_{1} a_{2} a_{3} \cdots a_{n}, \\
b_{1} b_{2} a_{3} \cdots a_{n}, \\
\vdots \\
b_{1} b_{2} b_{3} \cdots b_{n}
\end{gathered}
$$
form a strictly in... | [
"Answer: The smallest common difference is $n!$.\n\nSolution 1. The condition in the problem is equivalent to\n$$\nD=\\left(b_{1}-a_{1}\\right) a_{2} a_{3} \\cdots a_{n}=b_{1}\\left(b_{2}-a_{2}\\right) a_{3} a_{4} \\cdots a_{n}=\\cdots=b_{1} b_{2} \\cdots b_{n-1}\\left(b_{n}-a_{n}\\right),\n$$\nwhere $D$ is the com... | IMO | International Mathematical Olympiad Shortlist | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | n! | |
0l0a | Problem:
A lame king is a chess piece that can move from a cell to any cell that shares at least one vertex with it, except for the cells in the same column as the current cell.
A lame king is placed in the top-left cell of a $7 \times 7$ grid. Compute the maximum number of cells it can visit without visiting the same... | [
"Solution:\n\nColor the columns all-black and all-white, alternating by column. Each move the lame king takes will switch the color it's on. Assuming the king starts on a black cell, there are 28 black and 21 white cells, so it can visit at most $22+21=43$ cells in total, which is easily achievable:\n\n![](attached... | United States | HMMT February 2024 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 43 | |
05oy | Problem:
Soit $a_{0} < a_{1} < a_{2} < \ldots$ une suite infinie d'entiers strictement positifs. Prouver qu'il existe un unique entier $n \geqslant 1$ tel que
$$
a_{n} < \frac{a_{0} + a_{1} + a_{2} + \cdots + a_{n}}{n} \leqslant a_{n+1}.
$$ | [
"Solution:\n\nPour $n \\geqslant 1$, on pose $d_{n} = a_{0} + a_{1} + \\cdots + a_{n} - n a_{n}$.\nClairement, l'inégalité de gauche de l'encadrement souhaité est équivalente à $d_{n} > 0$.\nD'autre part, on a $d_{n+1} = a_{0} + \\cdots + a_{n} - n a_{n+1}$, donc l'inégalité de droite est équivalente à $d_{n+1} \\l... | France | Olympiades Françaises de Mathématiques - Envoi 2 (Algèbre) | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
05gx | Problem:
Trouver tous les entiers strictement positifs $n$ tels que $2^{n-1} n+1$ soit un carré parfait. | [
"Solution:\n\nOn veut résoudre $2^{n-1} n+1=m^{2}$, c'est-à-dire $2^{n-1} n=(m-1)(m+1)$. Puisque $n=1$ n'est pas solution, on a $n \\geq 2$, et donc $m$ est nécessairement impair, et $m-1$ et $m+1$ sont pairs (en particulier $n \\geq 3$ ). On pose $k=\\frac{m-1}{2}$. Il suffit alors de résoudre $2^{n-3} n=k(k+1)$. ... | France | Envoi 1 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | n = 5 | |
08vt | Suppose both the triangles $ABC$ and $PQR$ are right isosceles triangles with $\angle ACB = \angle PQR = 90^\circ$ and $AB = 5$, $PR = 3$ are satisfied. Suppose further that the 3 points $A, P, Q$ lie on a straight line in the written order, and the same is true for the 3 points $B, Q, R$, and also for the 3 points $R,... | [
"Since $\\angle ACB = \\angle AQB = 90^\\circ$, the 4 points $A$, $B$, $Q$, $C$ lie on a same circle. Since the angles $\\angle AQC$ and $\\angle ABC$ are subtended by the same arc of this circle at the points $Q$ and $C$ respectively, we have $\\angle AQB = \\angle ABC = 45^\\circ$. Therefore, $\\angle RQC = 90^\\... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 41/8 | |
01yv | The lines $a_1, a_2, b_1, b_2, c_1, c_2$ passing, respectively, through the points $A_1, A_2, B_1, B_2, C_1, C_2$ rotate uniformly and with the same angular velocity about the corresponding points. At an arbitrary moment $t$ by $A(t)$ denote the intersection point of the lines $a_1$ and $a_2$. Points $B(t)$ and $C(t)$ ... | [
"Since the lines rotate uniformly the angles between the lines $a_1(t_1), a_1(t_2)$ and between the lines $a_2(t_1), a_2(t_2)$ are equal for any moments $t_1, t_2$. Hence the points $A_1, A_2, A(t_1), A(t_2)$ lie on the circle, i.e. all points $A(t)$ lie on the same circle. Moreover if the lines rotated by $\\phi$ ... | Belarus | SELECTION and TRAINING SESSION | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0cm1 | Problem:
For an integer $n \geq 5$, two players play the following game on a regular $n$-gon. Initially, three consecutive vertices are chosen, and one counter is placed on each. A move consists of one player sliding one counter along any number of edges to another vertex of the $n$-gon without jumping over another co... | [
"Solution:\n\nWe shall prove that the first player wins if and only if the exponent of $2$ in the prime decomposition of $n-3$ is odd.\n\nSince the game is identical for both players, has finitely many possible states and always terminates, we can label the possible states Wins or Losses according as whether a play... | Romanian Master of Mathematics (RMM) | The 7th Romanian Master of Mathematics Competition | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | The first player wins if and only if v2(n − 3) is odd. | |
08zl | Let $c$ be a non-negative integer. Find all sequences of positive integers $\{a_n\}_{n \ge 1}$ such that for any positive integer $n$, the following condition holds:
There are exactly $a_n$ positive integers $i$ satisfying $a_i \le a_{n+1} + c$. | [
"Assume that $a_{n+1} \\ge a_{n+2}$ holds for some positive integers $n$. Since any positive integer $i$ with $a_i \\le a_{n+2} + c$ must also satisfy $a_i \\le a_{n+1} + c$, it follows that $a_n \\ge a_{n+1}$. Hence, if $a_n < a_{n+1}$ holds for some positive integer $n$, then $a_{n+1} < a_{n+2}$ follows, and by i... | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | a_n = n + c + 1 for all n ≥ 1 | |
0dgt | Prove that for $n \ge 1$ the following inequality holds
$$
1 + \frac{5}{6n - 5} \le 6^{1/n} \le 1 + \frac{5}{n}.
$$ | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
02ux | Problem:
Seja $D$ um ponto no lado $AB$ do triângulo $ABC$ de modo que $CD = AC$, como indica a figura abaixo. O incírculo do triângulo $ABC$ é tangente aos lados $AC$ e $AB$ nos pontos $E$ e $F$, respectivamente. Sejam $I$ o incentro do triângulo $BCD$ e $P$ o ponto de encontro dos segmentos $AI$ e $EF$. Além disso, ... | [
"Solution:\n\na) Como $AF = AE$, segue que $\\angle AFE = \\angle AEF = \\frac{180^{\\circ} - \\angle CAB}{2} = 90^{\\circ} - \\frac{\\angle CAB}{2}$. Além disso, dado que $AC = CD$, temos $\\angle CAD = \\angle CDA$. Sabendo que $DI$ é bissetriz de $\\angle CDB$, temos\n$$\n\\begin{aligned}\n\\angle IDB & = \\frac... | Brazil | Brazilian 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 > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | null | proof only | null | |
0cq5 | Do there exist three coprime positive integer numbers such that a square of each of them is divisible by a sum of the other two?
Существуют ли три взаимно простых в совокупности натуральных числа, квадрат каждого из которых делится на сумму двух оставшихся? | [
"Ответ. Не существуют.\n\nПервое решение. Предположим противное: пусть нашлись такие числа $a, b, c$. Заметим, что числа $a+b, b+c, c+a$ попарно взаимно просты. В самом деле, пусть, скажем, числа $a+b, b+c$ делятся на некоторое простое $p$. Поскольку $c^2: (a+b), a^2: (b+c)$, то числа $c$ и $a$ также делятся на $p$... | Russia | Russian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequaliti... | English, Russian | proof and answer | No | |
0dv6 | Problem:
Ali lahko pokrijemo šahovnico velikosti $15 \times 15$ s $55$ dominami velikosti $4 \times 1$ tako, da ostanejo središčno polje in vsa polja ob ogliščih šahovnice nepokrita? (Domine morajo v celoti ležati na šahovnici in se ne smejo prekrivati.) | [
"Solution:\n\nPobarvajmo šahovnico, kot kaže slika. Vsaka domina velikosti $4 \\times 1$ pokrije po $2$ polji vsake barve. Ker število belih polj ni enako številu osenčenih polj, šahovnice ne moremo pokriti na predpisani način."
] | Slovenia | 46. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0g3a | Problem:
Falls $p \geq 5$ eine Primzahl ist, sei $q$ die kleinste Primzahl sodass $q>p$ und sei $n$ die Anzahl der positiven Teiler von $p+q$ (1 und $p+q$ inklusive).
a) Zeige, dass egal welche Primzahl $p$ gewählt wurde, die Zahl $n$ grösser oder gleich 4 ist.
b) Finde den kleinstmöglichen Wert $m$, den $n$ annehme... | [
"Solution:\n\nDa $p \\geq 5$, sind die beiden Primzahlen $p, q$ beide ungerade und somit $p+q$ gerade, also $p+q$ durch $1, 2, \\frac{p+q}{2}$ und $p+q$ teilbar. Zusätzlich ist $p+q>4$ und somit sind die genannten Teiler alle unterschiedlich. Dies beweist Teil a).\n\nfür Teil $\\mathrm{b}_{1}$ ) können wir kleine F... | Switzerland | Vorrunde | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | proof and answer | 6 | |
03rg | Six points are chosen on the sides of an equilateral triangle $ABC$: $A_1, A_2$ on $BC$, $B_1, B_2$ on $CA$, and $C_1, C_2$ on $AB$. These points are the vertices of a convex hexagon $A_1A_2B_1B_2C_1C_2$ with sides of equal length. Prove that the lines $A_1B_2$, $B_1C_2$ and $C_1A_2$ are concurrent. (proposed by Romani... | [
"Assume $A_1A_2 = d$, and $AB = a$. Construct an equilateral triangle $A_0B_0C_0$ with side of length $a-d$. Points $A', B', C'$ are chosen on the sides of triangle $A_0B_0C_0$ such that $A'C_0 = A_2C$, $B'A_0 = B_2A$, and $C'B_0 = C_2B$.\n\nTherefore,\n$$\nA'B_0 = a-d-A'C_0 = BC-A_1A_2-A_2C = BA_1\n$$\n\n$$\nB'C_0... | China | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0dqi | A student divides an integer $m$ by a positive integer $n$, where $n \le 100$, and claims that
$$ \frac{m}{n} = 0.167a_1a_2\dots $$
Show the student must be wrong. | [
"We have\n$$\n0.167 \\le \\frac{m}{n} < 0.168 \\Rightarrow 167n \\le 1000m < 168n.\n$$\nMultiply by $6$, we get\n$$\n1002n \\le 6000m < 1008n \\Rightarrow 6000m - 1000n < 8n \\le 800.\n$$\nBut $6000m - 1000n \\ge 2n > 0$. Thus $6000m - 1000n \\ge 1000$ since it is a multiple of $1000$. We thus get a contradiction."... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0hf0 | What is the size of the largest set of numbers we can choose among $1$, $2$, ..., $2n$ in such a way that any two of them have a common divisor greater than $1$? | [
"If we choose all even numbers, they follow required condition and there are exactly $n$ of them.\n\nNow let's consider next pairs of integers: $(1, 2)$, $(3, 4)$, ..., $(2n-1, 2n)$. In each pair, numbers are relatively prime and we cannot choose both of them to our set. Therefore, it is impossible to choose more t... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | n | |
05ox | Problem:
Soit $[AB]$ le diamètre d'un demi-cercle sur lequel on prend deux points $C$ et $D$. Soit $S$ l'intersection de $(AC)$ et $(BD)$ et $T$ le pied de la perpendiculaire à $[AB]$ issue de $S$.
Montrer que $(ST)$ est la bissectrice de l'angle $\widehat{CTD}$. | [
"Solution:\n\n\n\nOn s'en sort par chasse aux angles. Les points $A, D, S, T$ sont cocycliques (sur le cercle de diamètre $[AS]$) et de même les points $B, C, S, T$. Donc $\\widehat{STD} = \\widehat{SAD}$, et $\\widehat{STC} = \\widehat{SBC}$. Comme $\\widehat{SAD} = \\widehat{CAD} = \\wide... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | null | proof only | null | |
0j2r | Problem:
How many different numbers are obtainable from five 5s by first concatenating some of the 5s, then multiplying them together? For example, we could do $5 \cdot 55 \cdot 55,555 \cdot 55$, or $55555$, but not $5 \cdot 5$ or $2525$. | [
"Solution:\nAnswer: $7$\n\nIf we do $55555$, then we're done.\n\nNote that $5$, $55$, $555$, and $5555$ all have completely distinguishable prime factorizations. This means that if we are given a product of them, we can obtain the individual terms. The number of $5555$'s is the exponent of $101$, the number of $555... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 7 | |
002c | Determinar los enteros positivos $n$ tales que el conjunto de todos los divisores positivos de $30^n$ se puede dividir en grupos de tres de modo que el producto de los tres números de cada grupo sea el mismo. | [] | Argentina | XX Olimpiada Matemática Argentina | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | español | proof and answer | n ≡ 2 (mod 6) | |
0jns | Problem:
Find the maximum possible value of $H \cdot M \cdot M \cdot T$ over all ordered triples $(H, M, T)$ of integers such that $H \cdot M \cdot M \cdot T = H + M + M + T$. | [
"Solution:\nIf any of $H, M, T$ are zero, the product is $0$. We can do better (examples below), so we may now restrict attention to the case when $H, M, T \\neq 0$.\n\nWhen $M \\in \\{-2, -1, 1, 2\\}$, a little casework gives all the possible $(H, M, T) = (2, 1, 4), (4, 1, 2), (-1, -2, 1), (1, -2, -1)$.\n\n- If $M... | United States | HMMT February | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 8 | |
0dcp | Let $ABC$ be an acute, non-isosceles triangle with $O$, $H$ as circumcenter and orthocenter, respectively. Prove that the nine-point circles of $AHO$, $BHO$, $CHO$ have two common points. | [
"Let $M$, $N$, $P$ be the midpoints of $BC$, $CA$, $AB$ and $D$ be the projection of $A$ on $BC$. Denote $(E)$ as the Euler circle of $\\triangle ABC$; then $M$, $N$, $P$, $D$ lie on $(E)$ and $E$ is the midpoint of $OH$. Similarly, let $H_a$, $O_a$ be the midpoints of $AH$, $AO$ and $K$ be the projection of $A$ on... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | English | proof only | null | |
03gc | Problem:
Find the sum of $1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \cdots + (n-1)(n-1)! + n \cdot n!$, where $n! = n(n-1)(n-2) \cdots 2 \cdot 1$. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | (n+1)! - 1 | |
0cdw | Let $P \in \mathbb{C}[X]$ be a polynomial with degree at least $2$ and $n \ge 2$ be an integer. Prove that, if $A \in \mathcal{M}_n(\mathbb{C})$ commutes with every matrix $B \in \mathcal{M}_n(\mathbb{C})$ having the property $P(B) = O_n$, then there exists $a \in \mathbb{C}$ such that $A = aI_n$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Linear transformations"
] | null | proof only | null | |
032b | Problem:
Find all $k > 0$ such that there is a function $f : [0,1] \times [0,1] \to [0,1]$ satisfying the following conditions:
a) $f(f(x, y), z) = f(x, f(y, z))$;
b) $f(x, y) = f(y, x)$;
c) $f(x, 1) = x$;
d) $f(z x, z y) = z^{k} f(x, y)$,
for any $x, y, z \in [0,1]$. | [
"Solution:\nNow let $x \\leq y \\leq z$, $x, y, z \\in (0,1)$. Then using (a), we get that $f\\left(x y^{k-1}, z\\right) = f\\left(x, y z^{k-1}\\right)$. Hence it follows from the above that\n$$\n\\{x y^{k-1} z^{k-1}, x^{k-1} y^{(k-1)^2} z\\} \\cap \\{x y^{k-1} z^{(k-1)^2}, x^{k-1} y z^{k-1}\\} \\neq \\varnothing\n... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | k = 1 or k = 2 | |
02t3 | Problem:
Pedro decidiu levar todos os seus filhos, meninos e meninas, para tomar sorvete na sorveteria Sorvete Matemático. Na sorveteria, há 12 sabores diferentes de sorvete e cada criança pediu um combo com 3 bolas de sorvete. Depois de sair da sorveteria, Pedro percebeu que, no total, foram pedidas exatamente duas b... | [
"Solution:\n\na) Seja $n$ o número de filhos de Pedro. No total, foram pedidas $3 n$ bolas de sorvete. Como cada um dos 12 sabores foi pedido duas vezes, temos $3 n = 2 \\cdot 12$, ou seja, $n = 8$. Portanto, Pedro possui 8 filhos.\n\nb) Sejam $x$ o número de meninos e $y$ o número de meninas. Pelo item anterior, s... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | a) 8; b) 6 | |
01n7 | Find all triples $(x; n; p)$ of positive integers $x$, $n$, and primes $p$, such that $2x^3 + x^2 + 10x + 5 = 2 \cdot p^n$. | [
"Answer: $(x; n; p) = (1; 2; 3)$, $(x; n; p) = (3, 2, 7)$.\nIt is easy to see that $2x^3 + x^2 + 10x + 5 = (x^2 + 5)(2x + 1)$, so the initial equality can be rewritten as\n$$\n(x^2 + 5)(2x + 1) = 2 \\cdot p^n. \\quad (1)\n$$\nSince $(2x + 1)$ is odd for all natural $x$, we have $p \\neq 2$ and $2x + 1 = p^k$, $x^2 ... | Belarus | Belorusija 2012 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | [[1, 2, 3], [3, 2, 7]] | |
0kgn | What is the sum of all possible values of $t$ between $0$ and $360$ such that the triangle in the coordinate plane whose vertices are $(\cos 40^\circ, \sin 40^\circ)$, $(\cos 60^\circ, \sin 60^\circ)$, and $(\cos t^\circ, \sin t^\circ)$ is isosceles?
(A) 100
(B) 150
(C) 330
(D) 360
(E) 380 | [] | United States | AMC 12 B | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | MCQ | E | |
018e | In an urn there are $100$ balls each coloured either blue, red or green. If you draw (without repetitions) two balls randomly from the urn, the probability of getting two balls of different colour is $58\%$, and the probability of getting a blue and a green ball is $8\%$. How many red balls are there among the $100$ ba... | [
"Let $b$, $r$ and $g$ be the number of blue, red and green balls, respectively. We know that $b + r + g = 100$.\n\nIf you draw two balls randomly from the urn, the probability of getting two balls of different colour is $58\\%$, and the probability of getting a blue and a green ball is $8\\%$ and hence the probabil... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 55 | |
00ib | Prove that the inequality
$$
\frac{(x-y)^7 + (y-z)^7 + (z-x)^7 - (x-y)(y-z)(z-x)((x-y)^4 + (y-z)^4 + (z-x)^4)}{(x-y)^5 + (y-z)^5 + (z-x)^5} \ge 3
$$
holds for all pairwise different integers $x$, $y$, $z$. When does equality hold? | [
"Since\n$$\n\\begin{aligned}\n(x - y)^7 - (x - y)(y - z)(z - x)(x - y)^4 &= (x - y)^5((x - y)^2 - (y - z)(z - x)) \\\\\n&= (x - y)^5(x^2 + y^2 + z^2 - xy - yz - zx),\n\\end{aligned}\n$$\nwe can write\n$$\n\\sum_{cyclic} (x - y)^7 - (x - y)(y - z)(z - x) \\cdot \\sum_{cyclic} (x - y)^4 = \\left( \\sum_{cyclic} (x - ... | Austria | Austria 2010 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | Equality holds when x, y, z are consecutive integers in any order, i.e., (x, y, z) = (m, m+1, m+2) up to permutation. | |
02o4 | For each finite subset $F$ of the space $R^3$, define $V_r(F)$ as the union of the open spheres with center on each point of $F$ and radius $r$. Prove that, for $0 < r < R$,
$$
\mathrm{vol}(V_R(F)) \le \frac{R^3}{r^3} \mathrm{vol}(V_r(F)).
$$ | [
"Let $F = \\{P_1, P_2, \\dots, P_n\\}$ and let $\\alpha_{ij}$ be the perpendicular plane bisector of $P_i$ and $P_j$ for $i \\neq j$. Those planes define $n$ convex regions $R_1, R_2, \\dots, R_n$, where $R_i$ is the intersection of the half-spaces determined by $\\alpha_{ij}$ that contain $P_i$. Finally, let $A_r(... | Brazil | Brazilian Math Olympiad | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof only | null | |
0hqv | Problem:
Let $a \leq b \leq c \leq d$ be real numbers such that
$$
a+b+c+d=0 \quad \text{ and } \quad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=0
$$
Prove that $a+d=0$. | [
"Solution:\n\nLet us rearrange the first equation to read\n$$\na+b=-(c+d)\n$$\nand the second to give\n$$\n\\begin{aligned}\n\\frac{1}{a}+\\frac{1}{b} & =-\\frac{1}{c}-\\frac{1}{d} \\\\\n\\frac{a+b}{a b} & =-\\frac{c+d}{c d}\n\\end{aligned}\n$$\nIf $a+b \\neq 0$, then we can divide (1) by (2) to yield $a b=c d$. So... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0ay2 | Problem:
Suppose each of five sticks is broken into a long part and a short part. The ten parts are arranged into five pairs and glued back together so that again there are five sticks. What is the probability that each long part will be paired with a short part? | [
"Solution:\nThe 5 short pieces and 5 long pieces can be lined up in a row in $\\frac{10!}{5!5!}$ ways. Consider each of the 5 pairs of consecutive pieces as defining the reconstructed sticks. Each of those pairs could combine a short piece $(S)$ and a long piece $(L)$ in two ways: $SL$ or $LS$. Therefore, the numbe... | Philippines | Philippine Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Combinations",
"Statistics > Probability > Counting Methods > Permutations"
] | null | proof and answer | 8/63 | |
0elx | Divide a regular hexagon into exactly 8 identical shapes. | [
"The following image is a possible dissection of a hexagon into 8 identical shapes:\n"
] | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | final answer only | null | |
05w0 | Problem:
Une permutation des entiers $1$ à $2022$ est une suite $\sigma = (\sigma_{1}, \ldots, \sigma_{2022})$ telle que chaque élément de l'ensemble $\{1, \ldots, 2022\}$ soit égal à exactement un terme $\sigma_{i}$. Quelle est la plus petite valeur possible que peut prendre la somme
$$
\left\lfloor\frac{a_{1}}{1}\ri... | [
"Solution:\n\nDe manière générale, notons $\\mathcal{S}(a_{1}, a_{2}, \\ldots, a_{n})$ la somme\n$$\n\\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{a_{2}}{2}\\right\\rfloor+\\ldots+\\left\\lfloor\\frac{a_{n}}{n}\\right\\rfloor,\n$$\net $\\mathcal{S}_{n}$ la plus petite valeur que peut prendre $\... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 11 | |
0f4j | Problem:
$P$ is a polygon with $2n + 1$ sides. A new polygon is derived by taking as its vertices the midpoints of the sides of $P$. This process is repeated. Show that we must eventually reach a polygon which is homothetic to $P$. | [
"Solution:\n\nLet $P$ be a polygon with $2n + 1$ sides. Let the vertices of $P$ be $A_1, A_2, \\ldots, A_{2n+1}$ in order. The process consists of forming a new polygon $P'$ whose vertices are the midpoints $M_1, M_2, \\ldots, M_{2n+1}$ of the sides $A_1A_2, A_2A_3, \\ldots, A_{2n+1}A_1$ respectively. This process ... | Soviet Union | 16th ASU | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Linear transformations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof only | null | |
0j0x | Problem:
Compute the remainder when
$$
\sum_{k=1}^{30303} k^{k}
$$
is divided by $101$. | [
"Solution:\nThe main idea is the following lemma:\n\nLemma. For any non-negative integer $n$ and prime $p$, $$\\sum_{k=n+1}^{n+p^{2}-p} k^{k} \\equiv 1 \\pmod{p}.$$ \n\nProof. Note that $a^{b}$ depends only on the value of $a \\pmod{p}$ and the value of $b \\pmod{p-1}$. Since $p$ and $p-1$ are relatively prime, the... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | 29 | |
03fg | Let $d \ge 3$ be a positive integer. The binary strings of length $d$ are split into $2^{d-1}$ pairs, such that the strings in each pair differ in exactly one position. Show that there exists an alternating cycle of length at most $2d-2$, i.e. at most $2d-2$ binary strings that can be arranged on a circle so that any p... | [
"Consider a graph $G$ with vertices the binary vectors of length $d$ and edges connecting the pairs of vectors that differ in exactly one position. Let $\\mathcal{M}$ be a complete $d$-dimensional pairing. We will prove that for each vertex $\\mathbf{x} \\in \\{0,1\\}^d$ of $G$ we can find an alternating cycle of l... | Bulgaria | 3 Bulgarian Spring Tournament | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
00yu | Problem:
Prove that there exists a number $\alpha$ such that for any triangle $A B C$ the inequality
$$
\max \left(h_{A}, h_{B}, h_{C}\right) \leq \alpha \cdot \min \left(m_{A}, m_{B}, m_{C}\right)
$$
holds, where $h_{A}, h_{B}, h_{C}$ denote the lengths of the altitudes and $m_{A}, m_{B}, m_{C}$ denote the lengths of... | [
"Solution:\n\nLet $h = \\max \\left(h_{A}, h_{B}, h_{C}\\right)$ and $m = \\min \\left(m_{A}, m_{B}, m_{C}\\right)$. If the longest height and the shortest median are drawn from the same vertex, then obviously $h \\leq m$.\n\nNow let the longest height and shortest median be $A D$ and $B E$, respectively, with $|A ... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 2 | |
00e2 | The numbers $1, 2, 3, \dots, 170$ are written on a board. We want to color each number with one of the $k$ colors $C_1, C_2, \dots, C_k$, so that the following condition is fulfilled: for each $i$ with $1 \le i < k$, the sum of all the numbers colored $C_i$ divides the sum of all the numbers colored $C_{i+1}$. Determin... | [
"Let $S_i$ be the cardinality of the set of the numbers colored $C_i$. We begin by bounding the number of colors for which $S_i = 1$. Let's say there are $\\ell$ of those. If $a_j$ are the numbers with such colors and we have $a_1 < a_2 < \\dots < a_\\ell$, then we must have $a_1 \\mid a_2 \\mid \\dots \\mid a_\\el... | Argentina | XXXIII Cono Sur Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | 89 | |
0hiq | Problem:
Fifty counters are on a table. Two players alternate taking away 1, 2, 3, 4, or 5 of them. Whoever picks up the last counter is the loser. Who has a winning strategy, the first player or the second? | [
"Solution:\n\nNote that if you make a turn and there is 1 counter left, you have won since the other player must pick up that counter.\n\nIf you make a turn and there are 7 counters left, you can win: if your opponent picks up 1, 2, 3, 4, or 5 of them, you can respectively take $5, 4, 3, 2$, or $1$ of them to leave... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | first player | |
0kvm | Problem:
Let $P_{1}(x), P_{2}(x), \ldots, P_{k}(x)$ be monic polynomials of degree $13$ with integer coefficients. Suppose there are pairwise distinct positive integers $n_{1}, n_{2}, \ldots, n_{k}$ for which, for all positive integers $i$ and $j$ less than or equal to $k$, the statement "$n_{i}$ divides $P_{j}(m)$ for... | [
"Solution:\nWe first consider which integers can divide a polynomial $P_{i}(x)$ for all $x$. Assume that $c \\mid P_{i}(x)$ for all $x$. Then, $c$ must also divide the finite difference $Q(x) = Q_{i}(x+1) - Q_{i}(x)$. Since $Q_{i}(x)$ is degree $13$ and monic, the leading term of $Q(x)$ is the leading term of $(x+1... | United States | HMMT February 2023 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 144 | |
054b | Find all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying the equality $f(2^x + 2y) = 2^y f(f(x))f(y)$ for every $x, y \in \mathbb{R}$. | [
"Substituting $y = -2^{x-1}$ into the original equation gives\n$$\nf(0) = \\frac{1}{2^{2^{x-1}}} f(f(x)) f(-2^{x-1}). \\quad (1)\n$$\nSo, if $f(-2^x) = 0$ for at least one $x$ then also $f(0) = 0$. Then taking $x = 0$ and arbitrary $y$ in the original identity gives $f(1+2y) = 0$, i.e., $f \\equiv 0$.\n\nAssume in ... | Estonia | Estonian Math Competitions | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | f(x) = 0 for all x; f(x) = 2^x for all x | |
05yp | Problem:
Au moins $n\left(\sqrt{n}+\frac{1}{2}\right)$ carrés d'un échiquier $n \times n$ ont été marqués. Montrer qu'il existe quatre cases marquées qui forment les coins d'un rectangle. | [
"Solution:\n\nOn essaye d'utiliser le principe des tiroirs. S'il y a deux cases marquées aux coordonnées $(i, j)$ et $(i, k)$, on met une chaussette dans le tiroir étiqueté $(j, k)$. Il y a $\\frac{n(n-1)}{2}$ tiroirs possibles. S'il y a deux chaussettes dans le même tiroir $(j, k)$, c'est qu'il existe $i, i'$ tels... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0evi | Find the maximum value of $(x^2 - yz)(y^2 - zx)(z^2 - xy)$ given that $x, y, z$ are real numbers satisfying $x^2 + y^2 + z^2 = 1$. | [
"Let $f(x, y, z) = (x^2 - yz)(y^2 - zx)(z^2 - xy)$. Considering $f(x, 0, z) = -x^3z^3$, it is easily seen that maximum of $f$ is positive. Since $f$ is symmetric in $x, y, z$ and $f(x, y, z) = f(-x, -y, -z)$, we may assume that $x \\ge y \\ge z$, $x + y + z \\ge 0$ and consequently $x^2 - yz > 0$. If $f$ takes its ... | South Korea | Korean Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | 1/8 | |
0jyz | Problem:
Mary has a sequence $m_{2}, m_{3}, m_{4}, \ldots$, such that for each $b \geq 2$, $m_{b}$ is the least positive integer $m$ for which none of the base-$b$ logarithms $\log_{b}(m), \log_{b}(m+1), \ldots, \log_{b}(m+2017)$ are integers. Find the largest number in her sequence. | [
"Solution:\n\nIt is not difficult to see that for all of the logarithms to be non-integers, they must lie strictly between $n$ and $n+1$ for some integer $n$. Therefore, we require $b^{n+1} - b^{n} > 2018$, and so $m_{b} = b^{n} + 1$ where $n$ is the smallest integer that satisfies the inequality. In particular, th... | United States | HMMT November 2017 | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 2188 | |
08j7 | Problem:
If $m$ is a number from the set $\{1,2,3,4\}$ and each point of the plane is painted in red or blue, prove that in the plane there exists at least an equilateral triangle with the vertices of the same colour and with length side $m$.
Problem:
Fie mulțimea $M=\{1,2,3,4\}$. Fiecare punct al planului este colo... | [
"Solution:\n\nSuppose that in the plane there does not exist an equilateral triangle with the vertices of the same colour and length side $m=1,2,3,4$.\n\nFirst assertion: we shall prove that in the plane there does not exist a segment with the length $2$ such that the ends and the midpoint of this segment have the ... | JBMO | 7th JBMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0lfo | Let $P(x) \in \mathbb{Z}[x]$ be a polynomial. Determine all polynomials $Q(x) \in \mathbb{Z}[x]$, such that for every positive integer $n$, there exists a polynomial $R_n(x) \in \mathbb{Z}[x]$ satisfies
$$
Q(x)^{2n} - 1 = R_n(x) (P(x)^{2n} - 1).
$$ | [
"By fixing $x \\in \\mathbb{Z}$, we obtain that $Q(x)^{2n} - 1$ is a multiple of $P(x)^{2n} - 1$ for all positive integer $n$. We now prove the following lemma.\n\n*Lemma.* If $a$ and $b$ are two integers larger than $1$ such that $a^n - 1 \\mid b^n - 1$ for all positive integer $n$ then $b = a^k$ for some positive... | Vietnam | Team selection tests | [
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | Q(x) = ±P(x)^k for some positive integer k, or Q(x) ≡ ±1. | |
0gwq | Prove inequality $\frac{a^2}{bc(a^2+b^2)} + \frac{b^2}{ca(b^2+c^2)} + \frac{c^2}{ab(c^2+a^2)} \ge \frac{9}{2}$ for random positive numbers $a, b, c$ obeying equality $ab + bc + ca = 1$. | [
"Let's consider the secondary inequality $\\frac{a^3}{a^2+b^2} \\ge a - \\frac{b}{2}$ which is proved by means of rearrangement: $2a^3 \\ge (a^2+b^2)(2a-b) = 2a^3 + 2ab^2 - a^2b - b^3 \\Leftrightarrow b(a-b)^2 \\ge 0$.\n\n$$\n\\frac{a^2}{bc(a^2+b^2)} + \\frac{b^2}{ca(b^2+c^2)} + \\frac{c^2}{ab(c^2+a^2)} = \\frac{1}... | Ukraine | Ukrajina 2008 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0gbi | 三角形 $ABC$ 不是正三角形，$I$ 為其內心，$I_A$ 為角 $A$ 內的旁心，$I'_A$ 為 $I_A$ 對直線 $BC$ 的對稱點，且 $\ell_A$ 為直線 $AI'_A$ 對 $AI$ 的對稱線。依此類推，可定義 $I_B, I'_B$ 與 $\ell_B$。令 $P$ 為 $\ell_A$ 與 $\ell_B$ 的交點。
a. 證明：若 $O$ 為三角形 $ABC$ 的外心，則 $P$ 會落在直線 $OI$ 上。
b. 設由 $P$ 對三角形 $ABC$ 的內切圓所引的某一條切線，交 $ABC$ 的外接圓於 $X, Y$ 兩點。證明 $\angle XIY = 120^\circ$。 | [
"(a) 令 $A'$ 為 $A$ 對 $BC$ 的對稱點，$M$ 為直線 $AI$ 與 $ABC$ 的外接圓 $\\Gamma$ 的另一個交點。由於 $\\triangle ABA'$ 與 $\\triangle AOC$ 都是等腰三角形，且 $\\angle ABA' = 2\\angle ABC = \\angle AOC$，它們互為相似三角形。同理 $\\triangle ABI_A$ 與 $\\triangle AIC$ 也相似。所以有\n$$\n\\frac{AA'}{AI_A} = \\frac{AA'}{AB} \\cdot \\frac{AB}{AI_A} = \\frac{AC}{AO} \\cdot \... | Taiwan | 二〇一七數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homot... | null | proof only | null | |
08mr | Problem:
Let $a$, $b$, $c$ be positive real numbers with $abc=1$. Prove the inequality:
$$
\frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1}+\frac{2 b^{2}+\frac{1}{b}}{c+\frac{1}{b}+1}+\frac{2 c^{2}+\frac{1}{c}}{a+\frac{1}{c}+1} \geq 3
$$ | [
"Solution:\nBy $AM-GM$ we have $2x^{2}+\\frac{1}{x}=x^{2}+x^{2}+\\frac{1}{x} \\geq 3 \\sqrt[3]{\\frac{x^{4}}{x}}=3x$ for all $x>0$, so we have:\n\n$\\sum_{\\text{cyc}} \\frac{2 a^{2}+\\frac{1}{a}}{b+\\frac{1}{a}+1} \\geq \\sum_{cyc} \\frac{3a}{1+b+bc}=3\\left(\\sum_{cyc} \\frac{a^{2}}{1+a+ab}\\right) \\geq \\frac{3... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof only | null | |
0aj6 | Circles $k_1$ and $k_2$ intersect in points $A$ and $B$, such that $k_1$ passes through the center $O$ of the circle $k_2$. The line $p$ intersects $k_1$ in points $K$ and $O$ and $k_2$ in points $L$ and $M$, such that the point $L$ is between $K$ and $O$. The point $P$ is orthogonal projection of the point $L$ to the ... | [
"Let the point $C$ be the midpoint of the line segment $AB$. We have to prove $MC \\parallel KP$.\nLet us introduce angle $\\alpha = \\angle BKA$. Notice that\n$$\n\\begin{aligned}\n\\angle BLA &= 180^\\circ - \\angle BMA = 180^\\circ - \\frac{1}{2}\\angle BOA = \\\\\n&= 180^\\circ - \\frac{1}{2}(180^\\circ - \\ang... | North Macedonia | European Mathematical Cup | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
02as | Problem:
Um quadrilátero especial - Dois lados consecutivos de um quadrilátero medem $10~\mathrm{cm}$ e $15~\mathrm{cm}$. Se cada diagonal divide o quadrilátero em duas regiões de mesma área, calcule o seu perímetro. | [
"Solution:\n\nSe cada diagonal divide o quadrilátero em duas regiões de mesma área temos:\n$$\n\\text{Área}(\\triangle ABD) = \\text{Área}(\\triangle BCD) \\text{ e Área}(\\triangle ABC) = \\text{Área}(\\triangle ACD).\n$$\nMas,\n$$\n\\begin{aligned}\n\\text{Área}(\\triangle ABD) & = X + W \\\\\n\\text{Área}(\\tria... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof and answer | 50 cm | |
0a3j | Problem:
Vind alle paren $(a, b)$ van positieve gehele getallen zodat $f(x)=x$ de enige functie $f: \mathbb{R} \rightarrow \mathbb{R}$ is die voldoet aan
$$
f^{a}(x) f^{b}(y)+f^{b}(x) f^{a}(y)=2 x y
$$
voor alle $x, y \in \mathbb{R}$.
Hierin staat $f^{n}(x)$ voor het $n$ keer toepassen van $f$ op $x$, dus $f^{1}(x)=f(... | [
"Solution:\n\nWe gaan bewijzen dat precies alle paren $(a, b)$ met $\\operatorname{ggd}(a, b)=1$ en met $a+b$ oneven voldoen.\nNeem eerst aan dat $\\operatorname{ggd}(a, b)=n \\neq 1$. Bekijk de functie\n$$\ng(x)=\\left\\{\\begin{array}{lll}\nx+1 & \\text{ als }\\lfloor x\\rfloor \\not \\equiv 0 & \\bmod n \\\\\nx+... | Netherlands | Maarttoets | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All pairs of positive integers that are coprime and whose sum is odd. | |
0hcm | Determine if there exists a connected figure $F$ that consists of $1\times 1$ squares and satisfies the following conditions:
* it consists of $s$ $1\times 1$ squares and is not a rectangle;
* any $k\times s$ rectangle, where $k \ge s$ can be split into figures that coincide with $F$?
A figure that consists of $1\times... | [
"Yes, such figure exists.\n\nConsider an L-shaped tromino that consists of three $1\\times 1$ cells. $3\\times 2$, $2\\times 3$ rectangles can be easily split into such figures, as well as $9\\times 5$ rectangle (Fig. 16). Now let every $1\\times 1$ square be a $2\\times 3$ rectangle. Then the tromino became a new ... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Yes | |
0bt2 | Let $f : \mathbb{R} \to \mathbb{R}$ be an increasing function, and let $a$ be a real number. Prove that $f$ is continuous at $a$ if and only if there exists a sequence $(a_n)_{n \ge 1}$ of positive real numbers such that
$$
\int_a^{a+a_n} f(x) \, dx + \int_a^{a-a_n} f(x) \, dx \le \frac{a_n}{n}, \quad n = 1, 2, 3, \dot... | [
"Let $F : \\mathbb{R} \\to \\mathbb{R}$, $F(x) = \\int_a^x f(t) \\, dt$. If $f$ is continuous at $a$, then $F$ is differentiable at $a$ and $F'(a) = f(a)$. For every positive integer $n$, there exists a positive real number $\\delta_n$ such that $|F(x)/(x-a) - f(a)| \\le 1/(2n)$ for all $x$ in the $\\delta_n$-neigh... | Romania | 67th Romanian Mathematical Olympiad | [
"Calculus > Integral Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Limits"
] | English | proof only | null | |
0k5e | Let $ABC$ be an acute triangle with incenter $I$, circumcenter $O$, and circumcircle $\Gamma$. Let $M$ be the midpoint of $\overline{AB}$. Ray $AI$ meets $\overline{BC}$ at $D$. Denote by $\omega$ and $\gamma$ the circumcircles of $\triangle BIC$ and $\triangle BAD$, respectively. Line $MO$ meets $\omega$ at $X$ and $Y... | [
"Henceforth assume $\\angle A \\neq 60^\\circ$; we prove the concurrence. Let $L$ denote the center of $\\omega$, which is the midpoint of minor arc $BC$.\n\n**Claim.** Let $K$ be the point on $\\omega$ such that $\\overline{KL} \\parallel \\overline{AB}$ and $\\overline{KC} \\parallel \\overline{AL}$. Then $\\over... | United States | USA TSTST | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",... | null | proof only | null | |
02hg | Problem:
Qual é o maior dos números?
(A) $1000+0,01$
(B) $1000 \times 0,01$
(C) $1000 / 0,01$
(D) $0,01 / 1000$
(E) $1000-0,01$ | [
"Solution:\n\nTemos: $1000+0,01=1000,01$ ; $1000 \\times 0,01=1000 \\times \\frac{1}{100}=10$;\n$$\n\\frac{1000}{0,01}=\\frac{1000}{\\frac{1}{100}}=1000 \\times 100=100000\n$$\n$\\frac{0,01}{1000}$ é o inverso de $\\frac{1000}{0,01}$, logo, de (C) temos que $\\frac{1000}{0,01}=\\frac{1}{100000}=0,00001$. Agora, $10... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | C | |
08xu | How many non-square numbers are there among the positive factors of the number $6000$? | [
"[34]\nSince $6000 = 2^4 3^1 5^3$, positive factors of $6000$ can be written as $2^a 3^b 5^c$ where $a = 0, 1, 2, 3, 4$; $b = 0, 1$; $c = 0, 1, 2, 3$. Therefore, there are altogether $5 \\cdot 2 \\cdot 4 = 40$ positive factors of $6000$. Among them, square numbers arise only when all of $a, b, c$ are even, and this... | Japan | Japan 2015 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | final answer only | 34 | |
0crx | A positive integer $n$ is good if every its positive divisor increased by $1$ is a divisor of $n+1$. Find all good positive integers. | [
"$1$ and all odd prime numbers.\n\nClearly, $n = 1$ satisfies the condition. All odd primes also satisfy it: if $n = p$, then its divisors, increased by $1$, are $2$ and $p+1$; both divide $p+1$.\n\nOn the other hand, any number $n$ satisfying the condition has $1$ as a divisor; thus, $n+1$ is divisible by $1+1$, i... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 1 and all odd prime numbers | |
0i5y | Problem:
Find the number of sequences $a_{1}, a_{2}, \ldots, a_{10}$ of positive integers with the property that $a_{n+2}=a_{n+1}+a_{n}$ for $n=1,2, \ldots, 8$, and $a_{10}=2002$. | [
"Solution:\nLet $a_{1}=a$, $a_{2}=b$; we successively compute $a_{3}=a+b$, $a_{4}=a+2b$, $\\ldots$, $a_{10}=21a+34b$. The equation $2002=21a+34b$ has three positive integer solutions, namely $(84,7)$, $(50,28)$, $(16,49)$, and each of these gives a unique sequence."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | final answer only | 3 | |
07p4 | Suppose $u, v$ are distinct roots of the cubic polynomial
$$
p(z) = z^3 - 5z^2 + 6z - 1.
$$
Prove that $u, v$, and $u/v$ are positive irrational real numbers. | [
"$$\np(0) = -1,\\ p(1) = 1,\\ p(2) = p(3) = -1,\\ p(4) = 7,\n$$\nthe roots of $p$ are positive real numbers that belong to the union of the intervals $(0, 1)$, $(1, 2)$ and $(3, 4)$. Hence the roots are distinct, and neither of them is an integer. But since the coefficients of $p$ are integers, neither can they be ... | Ireland | Irska 2014 | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss’s Lemma"
] | null | proof only | null | |
03e3 | The points $A_1$, $B_1$, $C_1$ are chosen on the sides $BC$, $CA$, $AB$ of a triangle $ABC$ so that $BA_1 = BC_1$ and $CA_1 = CB_1$. The lines $C_1A_1$ and $A_1B_1$ meet the line through $A$, parallel to $BC$, at $P$, $Q$. Let the circumcircles of the triangles $APC_1$ and $AQB_1$ meet at $R$. Given that $R$ lies on $A... | [
"Observe that $(A_1, C_1, R, B_1)$ are concyclic. Let $I$ be the incenter of triangle $ABC$. Then $BI$ and $CI$ perpendicularly bisect $C_1A_1$ and $A_1B_1$ respectively. Hence, $I$ is the circumcenter of $\\triangle A_1B_1C_1$. Let $\\angle ACB = 2\\gamma$. $\\angle A_1IB_1 = 2\\angle B_1C_1A_1 = 2\\angle A_1RB_1 ... | Bulgaria | Autumn tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
008k | There are 100 metal balls which look exactly the same; 50 of them are radioactive. There are also three radiation detectors. For any group of balls, each detector is supposed to establish if there are radioactive balls in it or not. But it is known that one detector always gives the right answer, another one always giv... | [] | Argentina | XXIX Olimpíada Matemática Argentina National Round | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Algorithms"
] | English | proof only | null | |
03v3 | In $\triangle ABC$, $AB = AC$, the inscribed circle $I$ touches $BC$, $CA$, $AB$ at points $D$, $E$ and $F$ respectively. $P$ is a point on arc $\widehat{EF}$ (not containing $D$). Line $BP$ intersects the circle $I$ at another point $Q$, and lines $EP$, $EQ$ meet line $BC$ at $M$, $N$ respectively. Prove that
(1) $P$... | [
"**Proof**\n\n(1) From the given condition, $EF \\parallel BC$, so\n$$\n\\begin{aligned}\n\\angle ABC &= \\angle AFE = \\angle AFP + \\angle PFE \\\\\n&= \\angle PEF + \\angle PFE = 180^\\circ - \\angle FPE,\n\\end{aligned}\n$$\nand thus $P$, $F$, $B$, $M$ are concyclic.\n\n\n(2) By the sine law, $EF \\parallel BC$... | China | China Western Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
02u7 | Problem:
Certo matemático adora pensar em problemas e cozinhar bolos. Após cozinhar seus bolos, ele os corta em pedaços iguais. As três figuras a seguir mostram bolos circulares de mesmo raio em que os dois primeiros foram cortados em 3 e 4 pedaços iguais, respectivamente. Ele deseja cortar o terceiro bolo, mas a únic... | [
"Solution:\n\nO compasso servirá para transportar distâncias. O primeiro passo é marcar um ponto $A$ de referência na lateral do terceiro bolo. Usando os comprimentos de arcos do primeiro bolo e o compasso, pode-se marcar um ponto $C$ tal que o arco $AC$ meça $\\frac{1}{3}$ do perímetro de sua circunferência. Em se... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
00mg | Find all pairs $(a, b)$ of non-negative integers such that
$$
2017^a = b^6 - 32b + 1.
$$ | [
"Answer: The two solutions are $(0, 0)$ and $(0, 2)$.\nSince $2017^a$ is always odd, $b$ must be even, so $b = 2c$, $c$ integer. Therefore, $2017^a = 64(c^6 - c) + 1$ and thus $2017^a \\equiv 1 \\pmod{64}$. But we find $2017 \\equiv 33 \\pmod{64}$ and $2017^2 \\equiv (1+32)^2 = 1+2\\cdot32+32^2 \\equiv 1 \\pmod{64}... | Austria | 48th Austrian Mathematical Olympiad National Competition (Final Round, part 1) | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (0, 0) and (0, 2) | |
0het | Let $ABCDEF$ be an inscribed hexagon, such that $AB = BC$, $CD = DE$ and $EF = FA$. Show that the lines $AD$, $BE$ and $CF$ intersect at exactly one point. | [
"Equality $AB = BC$ implies that $\\angle AEB = \\angle BEC$, hence $EB$ is a bisector of $\\triangle ACE$, similarly, one can conclude the same about $AD$ and $CF$ (Fig. 7). Therefore, they intersect at one point.\n\n\nFig. 7"
] | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07kf | Air Michael and Air Patrick operate direct flights connecting Belfast, Cork, Dublin, Galway, Limerick and Waterford. For each pair of cities exactly one of the airlines operates the route (in both directions) connecting the cities. Prove that there are four cities for which one of the airlines operates a round trip. (N... | [
"The problem can be reinterpreted in graph theory terminology as the following: If $K_6$ is edge-coloured with two colours then there exists a monochromatic $C_4$ (cycle of length 4). As $K_6$ has 15 edges one of the colours, say red, must appear at least 8 times. Suppose the other colour is blue. Consider all the ... | Ireland | Irish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
03jy | Problem:
Let $C$ be a circle and $P$ a given point in the plane. Each line through $P$ which intersects $C$ determines a chord of $C$. Show that the midpoints of these chords lie on a circle. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0b20 | Problem:
A fixed point of a function $f$ is a value of $x$ for which $f(x)=x$. Let $f$ be the quadratic function defined by $f(x)=x^{2}-c x+c$ where $c \in \mathbb{R}$. Find, in interval notation, the set consisting of all values of $c$ for which $f \circ f$ has four distinct fixed points. | [
"Solution:\n\nFirst, observe that both $x=c$ and $x=1$ are fixed points of $f$, and thus fixed points of $f \\circ f$. Indeed, a fixed point of $f$ is a value of $x$ such that $f(x)-x=0$, and we have $f(x)-x = x^{2}-(c+1)x+c = (x-1)(x-c)$. Moreover, $(f \\circ f)(x)-x$ is a quartic polynomial in $x$, and its roots ... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (-∞, -1) ∪ (3, ∞) | |
0i8o | Problem:
A lattice point is a point $(x, y)$ with both $x$ and $y$ integers. Find, with proof, the smallest $n$ such that every set of $n$ lattice points contains three points that are the vertices of a triangle with integer area. (The triangle may be degenerate, in other words, the three points may lie on a straight l... | [
"Solution:\nClearly, $n=4$ is too small, for the 4 points could be vertices of a unit square, and all the possible triangles will have areas of $1/2$. We will show that $n=5$ works. Every lattice point falls into one of four parity classes: (even, even), (even, odd), (odd, even), and (odd, odd). For example the poi... | United States | 5th Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 5 | |
0b5w | Determine all integer numbers $n \ge 3$ such that the regular $n$-gon can be decomposed into isosceles triangles by noncrossing diagonals. | [
"The required numbers are of the form $n = 2^r(2^s+1)$, where $r$ and $s$ are nonnegative integer numbers which do not vanish simultaneously. Clearly, any such $n$ works.\n\nTo establish the converse, let $K$ be a regular $n$-gon, $n \\ge 4$, which can be decomposed into isosceles triangles by noncrossing diagonals... | Romania | 2010 DANUBE MATHEMATICAL COMPETITION | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | All n of the form n = 2^r(2^s + 1), where r and s are nonnegative integers not both zero. | |
016g | Problem:
Let $d(k)$ denote the number of positive divisors of a positive integer $k$. Prove that there exist infinitely many positive integers $M$ that cannot be written as
$$
M = \left( \frac{2 \sqrt{n}}{d(n)} \right)^2
$$
for any positive integer $n$. | [] | Baltic Way | Baltic Way | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0jl5 | Problem:
In rectangle $ABCD$ with area $1$, point $M$ is selected on $\overline{AB}$ and points $X, Y$ are selected on $\overline{CD}$ such that $AX < AY$. Suppose that $AM = BM$. Given that the area of triangle $MXY$ is $\frac{1}{2014}$, compute the area of trapezoid $AXYB$. | [
"Solution:\n\nAnswer: $\\frac{1}{2} + \\frac{1}{2014}$ OR $\\frac{504}{1007}$\n\nNotice that $[AMX] + [BYM] = \\frac{1}{2}[ABCD] = \\frac{1}{2}$.\n\nThus,\n$$\n[AXYB] = [AMX] + [BYM] + [MXY] = \\frac{1}{2} + \\frac{1}{2014} = \\frac{504}{1007}\n$$"
] | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof and answer | 504/1007 | |
0ks6 | Problem:
A triple of positive integers $(a, b, c)$ is tasty if $\operatorname{lcm}(a, b, c) \mid a+b+c-1$ and $a<b<c$. Find the sum of $a+b+c$ across all tasty triples. | [
"Solution:\nThe condition implies $c \\mid b+a-1$. WLOG assume $c>b>a$; since $b+a-1<2c$ we must have $b+a-1=c$. Substituting into $b \\mid a+c-1$ and $a \\mid c+b-1$ gives\n$$\n\\begin{aligned}\n& b \\mid 2a-2 \\\\\n& a \\mid 2b-2\n\\end{aligned}\n$$\nSince $2a-2<2b$ we must either have $a=1$ (implying $a=b$, bad)... | United States | HMMT November 2022 | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 44 |
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