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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
07jw | Let $ABC$ be an acute-angled scalene triangle. The internal angle bisector of vertex $A$ intersects the opposite side $BC$ at $E$ and the minor arc $BC$ of the circumcircle of $\triangle ABC$ at $M$. Suppose $D \neq M$ is a point on the minor arc $BC$ such that $ED = EM$. Also, let $P$ be a point on the line segment $A... | [
"From $O$, we draw a perpendicular to $AM$ which intersects $AD$ and $BC$ at $P'$ and $X$ respectively. Since $O$ and $E$ are both on the perpendicular bisector of $MD$, then $OE \\perp MD$. Since $E$ is the orthocenter of $\\triangle OXM$ (this implies $XE \\perp OM$, $ME \\perp OX$, and $OE \\perp XM$), and we ha... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angl... | null | proof only | null | |
0l44 | Problem:
Let $\zeta = \cos \frac{2\pi}{13} + i \sin \frac{2\pi}{13}$. Suppose $a > b > c > d$ are positive integers satisfying
$$
\left|\zeta^{a} + \zeta^{b} + \zeta^{c} + \zeta^{d}\right| = \sqrt{3}
$$
Compute the smallest possible value of $1000a + 100b + 10c + d$. | [
"Solution:\nWe may as well take $d = 1$ and shift the other variables down by $d$ to get $\\left|\\zeta^{a'} + \\zeta^{b'} + \\zeta^{c'} + 1\\right| = \\sqrt{3}$. Multiplying by its conjugate gives\n$$\n\\left(\\zeta^{a'} + \\zeta^{b'} + \\zeta^{c'} + 1\\right)\\left(\\zeta^{-a'} + \\zeta^{-b'} + \\zeta^{-c'} + 1\\... | United States | HMMT February 2024 | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 7521 | |
09h2 | Let $G$ be a graph with at least two vertices. Denote by $n(u)$ the sum of all degrees of vertices adjacent to a vertex $u$ of $G$. If $G$ does not contain cycles of length $3$ and $4$ then there are two vertices $u$ and $v$ of $G$ such that $n(u) = n(v)$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0fxb | Problem:
Sei $P$ ein reguläres Sechseck. Für einen Punkt $A$ seien $d_{1} \leq d_{2} \leq \ldots \leq d_{6}$ die Abstände von $A$ zu den sechs Eckpunkten von $P$, der Grösse nach geordnet. Finde den geometrischen Ort aller Punkte $A$ im Innern oder auf dem Rand von $P$, sodass
a) $d_{3}$ den kleinstmöglichen Wert ann... | [
"Solution:\n\nSei $M$ der Mittelpunkt von $P$. Betrachte alle Diagonalen von $P$, die $M$ enthalten, und alle Mittelsenkrechten durch die Seiten von $P$. Diese Geraden sind genau die Mittelsenkrechten auf den Verbindungsstrecken aller Paare von Eckpunkten von $P$. Aus Symmetriegründen können wir annehmen, dass $A$ ... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | a) The six points obtained by shrinking the hexagon by factor one half about its center, i.e., the midpoints of the segments from the center to each vertex. b) The center of the hexagon. | |
0brc | Two positive integers $x$ and $y$ are such that $\frac{2010}{2011} < \frac{x}{y} < \frac{2011}{2012}$. Find the smallest possible value of the sum $x + y$.
Neculai Stanciu | [
"The fraction $\\frac{x}{y}$ is subunitary, hence $x < y$, that is $x = y - d$, where $d$ is a positive integer. The given relation can be written $\\frac{2011-1}{2011} < \\frac{y-d}{y} < \\frac{2012-1}{2012}$ or $1 - \\frac{1}{2011} < 1 - \\frac{d}{y} < 1 - \\frac{1}{2012}$, whence $\\frac{1}{2011} > \\frac{d}{y} ... | Romania | 67th Romanian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 8044 | |
0es3 | $$
\sqrt{5050^2 - 4950^2} =
$$
(A) 10 (B) $10^2$ (C) $10^3$ (D) $10^4$ (E) $10^5$ | [
"$$\n\\sqrt{5050^2 - 4950^2} = \\sqrt{(5050 + 4950)(5050 - 4950)} = \\sqrt{10000 \\times 100} = \\sqrt{10^6} = 10^3.\n$$"
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | C | |
0gzg | Cube with side length $11$ is partitioned into unit cubes. The external slice of cubes is colored with yellow, the next slice (all cubes that share a vertex with the external slice) is colored with blue. Slice of cubes that touches the previous slice is colored with yellow and so on. Find the number of yellow and blue ... | [
"Obviously, there are $6$ slices of cubes. It is easy to see that the external yellow slice has height $11$, the next blue - $9$, the next yellow - $7$, the next blue - $5$, the next yellow - $3$ and the last one (6th slice) blue has height $1$, e.g. it's a unit cube. Further, we can calculate the number of cubes i... | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof and answer | yellow: 846, blue: 485 | |
0152 | Problem:
Let $M$ be a point on $B C$ and $N$ be a point on $A B$ such that $A M$ and $C N$ are angle bisectors of the triangle $A B C$. Given that
$$
\frac{\angle B N M}{\angle M N C}=\frac{\angle B M N}{\angle N M A}
$$
prove that the triangle $A B C$ is isosceles.
 | [
"Solution:\nLet $O$ and $I$ be the incentres of $A B C$ and $N B M$, respectively; denote angles as in the figure. We get\n$$\n\\alpha+\\beta=\\varepsilon+\\varphi, \\quad \\gamma+\\delta=2 \\alpha+2 \\beta, \\quad \\gamma=k \\cdot \\varepsilon, \\quad \\delta=k \\cdot \\varphi\n$$\nFrom here we get $k=2$. Therefor... | Baltic Way | Baltic Way 2008 | [
"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 | |
0axm | Problem:
Circles $\mathcal{C}_1$ and $\mathcal{C}_2$ with centers at $C_1$ and $C_2$, respectively, intersect at two distinct points $A$ and $B$. Points $P$ and $Q$ are varying points on $\mathcal{C}_1$ and $\mathcal{C}_2$, respectively, such that $P, Q$ and $B$ are collinear and $B$ is always between $P$ and $Q$. Let... | [
"Solution:\n\nLet $C$ be the intersection of the circle through $C_1, C_2$ and $A$, and the bisector of $\\angle C_1 A C_2$. It suffices to show that $C S = C A$. Let $\\angle C_1 A C_2 = C_1 B C_2 = 2 \\alpha$ (fixed), $\\angle C_1 P B = \\angle C_1 B P = 2 \\beta$ and $\\angle C_2 Q B = \\angle C_2 B Q = 2 \\gamm... | Philippines | 19th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Construct... | null | proof only | null | |
0e9c | Is it possible to line up the numbers $1, 2, 3, \ldots, 2013$ so that the arithmetic mean of any two of the numbers is never located between them? | [
"Let us show that the statement is true for any $n \\in \\mathbb{N}$. We claim that the numbers $1, 2, \\dots, n$ can be lined up so that for any pair their arithmetic mean does not lie somewhere in between.\n\nFirst, we will show that this is true for $n = 2^m$ for all $m \\in \\mathbb{N}$. We will use induction o... | Slovenia | National Math Olympiad in Slovenia | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Yes | |
0kaj | Problem:
Given that $A$, $B$ are nonzero base-10 digits such that $A \cdot \overline{A B} + B = \overline{B B}$, find $\overline{A B}$. | [
"Solution:\n\nWe know $A \\cdot \\overline{A B}$ ends in $0$. Since neither is $0$, they must be $2$, $5$ in some order. We easily find that $A = 2$, $B = 5$ works while the opposite doesn't, so $\\overline{A B} = 25$."
] | United States | HMMT February 2019 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 25 | |
07p8 | Given an $8 \times 8$ chess board, in how many ways can we select $56$ squares on the board while satisfying both of the following requirements:
1. All black squares are selected.
2. Exactly seven squares are selected in each column and in each row. | [
"Switching the definition of selection and non-selection, we see that an equivalent formulation is to select eight squares, with one in each row and in each column, and no black squares selected. We assume that the non-black squares are white.\n\nOrient the chessboard so that the square in row $1$, column $1$ is wh... | Ireland | Irska 2014 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 576 | |
08hw | Problem:
Prove that for all positive numbers $x, y, z$ such that $x + y + z = 1$ the following inequality holds
$$
\frac{x^{2}}{1 + y} + \frac{y^{2}}{1 + z} + \frac{z^{2}}{1 + x} \leq 1
$$ | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0898 | Problem:
Delle tre radici $a, b, c$ del polinomio $2x^{3}-7x^{2}-2x+12$ sappiamo che $4a=3(b+c)$. Quanto vale $a+bc$?
(A) $-\frac{5}{2}$
(B) $1-\sqrt{3}$
(C) $0$
(D) $\frac{7}{4}$
(E) $2+\sqrt{5}$. | [
"Solution:\n\nLa risposta è (A). Si ha $a+b+c=\\frac{7}{2}$ e $abc=-6$. Quindi $4a+4b+4c=14$ e, coi dati del testo, $3(b+c)+4b+4c=14$, da cui $b+c=2$. Quindi $a=\\frac{3}{2}$ e $bc=-4$, quindi $a+bc=\\frac{3}{2}-4=-\\frac{5}{2}$."
] | Italy | UNIONE MATEMATICA ITALIANA SCUOLA NORMALE SUPERIORE DI PISA GARA di SECONDO LIVELLO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | MCQ | A | |
0cev | Let $ABC$ be a triangle. A circle through $B$ and $C$ crosses the sides $AB$ and $AC$ at $P$ and $Q$, respectively. Points $X$ and $Y$ on segments $BQ$ and $CP$, respectively, satisfy $\angle ABY = \angle AXP$ and $\angle ACX = \angle AYQ$. Prove that $XY$ and $BC$ are parallel.
Andrei Chiriță
![](attached_image_1.png... | [
"Let $BY$ and $CX$ cross at $S$ and let circles $APX$ and $AQY$ cross again at $T$. We first prove that $A, S, T$ are collinear. Invert from $A$ with power $AP \\cdot AB = AQ \\cdot AC$. As $\\angle ABY = \\angle AXP$, the circle $APX$ is mapped to line $BY$. Similarly, the circle $AQY$ is mapped to line $CX$, so t... | Romania | Twentieth IMAR Mathematical Competition | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Concurrency and Collinearity > Pappus theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
047i | For an integer $n \ge 4$, if a sequence of real numbers $x_1, x_2, \dots, x_n$ satisfies the following three equalities:
$$
\begin{aligned}
x_1 + x_2 + \dots + x_n &= n, \\
x_1^2 + x_2^2 + \dots + x_n^2 &= 2n, \\
x_1^3 + x_2^3 + \dots + x_n^3 &= 3n,
\end{aligned}
$$
then we call $(x_1, x_2, \dots, x_n)$ a *regular* $n$... | [
"(1)\n\n**Solution 1**\nWe attempt to use the idea of a local inequality to prove that a number $a$ lies between the minimum and maximum of a regular $n$-tuple. To this end, we consider an indeterminate $b$ and consider the polynomial\n$$\nf(x) = (x - a)(x - b)^2 = x^3 - (a + 2b)x^2 + (2ab + b^2)x - ab^2,\n$$\nand ... | China | 2024 CMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Other"
] | English | proof and answer | C = sqrt(5), and one valid choice is λ = 1/6. | |
0e3b | Problem:
Prezračevalne naprave v lokalu čistijo zrak. Pretok zraka v odvisnosti od časa se za prvo napravo spreminja po formuli $f(t)=2^{t}$, za drugo pa $f(t)=2^{t+3}$. V lokalu imajo štiri naprave prvega tipa in eno drugega tipa. $S$ koliko napravami za prezračevanje s pretokom $f(t)=2^{t+2}$ bi lahko zamenjali obst... | [
"Solution:\n\nNastavimo enačbe $4 \\cdot 2^{t} + 2^{t+3} = a \\cdot 2^{t+2}$. Enačbo uredimo, tako da izpostavimo skupni faktor $2^{t}(4 + 8 - 4a) = 0$. Ugotovimo, da je $2^{t} \\neq 0$. Tako je $4 + 8 - 4a = 0$. Izračunamo $a = 3$."
] | Slovenia | 10. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | 3 | |
0a7u | Problem:
Show that there exist infinitely many mutually non-congruent triangles $T$, satisfying
(i) The side lengths of $T$ are consecutive integers.
(ii) The area of $T$ is an integer. | [
"Solution:\nLet $n \\geq 3$, and let $n-1, n, n+1$ be the side lengths of the triangle. The semiperimeter of the triangle then equals $\\frac{3n}{2}$. By Heron's formula, the area of the triangle is\n$$\n\\begin{gathered}\nT = \\sqrt{\\frac{3n}{2} \\cdot \\left(\\frac{3n}{2} - n + 1\\right)\\left(\\frac{3n}{2} - n\... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 9 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0eof | Pyramids with square bases were built by stacking magnetic balls. A pyramid with a $1 \times 1$ base contains one ball, a pyramid with a $2 \times 2$ base contains five balls, a pyramid with a $3 \times 3$ base contains 14 balls, a pyramid with a $4 \times 4$ base contains 30 balls. How many balls are there in a pyrami... | [
"The $n$-th pyramid is made by putting the $(n - 1)$-st pyramid on an $n \\times n$ square of balls. Thus the number of balls in the $n$-th pyramid is $1^2 + 2^2 + \\ldots + n^2$. The number of balls in the 10th pyramid is therefore $1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 = 385$."
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | final answer only | 385 | |
0iui | Problem:
How many ordered quadruples $(a, b, c, d)$ of four distinct numbers chosen from the set $\{1,2,3, \ldots, 9\}$ satisfy $b<a$, $b<c$, and $d<c$? | [
"Solution:\nGiven any 4 elements $p<q<r<s$ of $\\{1,2, \\ldots, 9\\}$, there are 5 ways of rearranging them to satisfy the inequality: prqs, psqr, qspr, qrps, and rspq. This gives a total of $\\binom{9}{4} \\cdot 5=630$ quadruples."
] | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 630 | |
0f4o | Problem:
$\mathbf{x}$ is a positive integer. Put $\mathbf{a} = \mathbf{x}^{1 / 12}$, $\mathbf{b} = \mathbf{x}^{1 / 4}$, $\mathbf{c} = \mathbf{x}^{1 / 6}$. Show that $2^{\mathrm{a}} + 2^{\mathrm{b}} \geq 2^{1 + \mathrm{c}}$. | [
"Solution:\n\nPut $\\mathbf{x} = \\mathbf{r}^{12}$. Since $\\mathbf{x}$ is a positive integer, we have $\\mathbf{r} \\geq 1$. We have to show that $\\dfrac{2^{\\mathrm{r}} + 2^{\\mathrm{r}^{3}}}{2} \\geq 2^{\\mathrm{r}^{2}}$. But this follows immediately from AM/GM."
] | Soviet Union | 16th ASU | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof only | null | |
0b0j | Problem:
How many positive divisors do $50,400$ and $567,000$ have in common? | [] | Philippines | Philippines Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 72 | |
0ef0 | Let $n \ge 3$ be a positive integer. We wish to inscribe into each square of the $n \times n$ array one of the numbers $1$, $2$ or $3$, so that any three squares which can be covered using the following shape [ ] [ ] [ ] (which can also be rotated) will contain three different numbers. In how many different ways can th... | [
"First notice that choosing two of the numbers in adjacent squares completely determines the remaining numbers in that row or column.\nIf we can inscribe the numbers as required then any permutation of the labels $1$, $2$ and $3$ will also give us a correct solution. We can put the numbers $1$, $2$ and $3$ in $6$ d... | Slovenia | Slovenija 2016 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 12 | |
0deh | Let $ABC$ be an acute triangle with $AB < AC$ and inscribed in the circle $(O)$. Denote $I$ as the incenter of $ABC$ and $D, E$ as the intersections of $AI$ with $BC, (O)$ respectively. Take a point $K$ on $BC$ such that $\angle AIK = 90^\circ$ and $KA, KE$ meet $(O)$ again at $M, N$ respectively. The rays $ND, NI$ mee... | [
"First, we know that $EI = EB = EC$ then $E$ is the circumcenter of triangle $BIC$. Thus $KI$ is the tangent line of circle $(BIC)$. Hence,\n$$\nKI^2 = KB \\cdot KC = KE \\cdot KN.\n$$\nBy combining with $KIE$ is right triangle, we get $IN \\perp KE$.\n\nSo $PE$ is the diameter of circle $(... | 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 > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Advanc... | null | proof only | null | |
09ib | Let $ABC$ be a triangle with $AB = 2BC$. Points $P$ and $Q$ are chosen on side the $AB$ to satisfy $AP : PQ : QB = 1 : 2 : 3$. A point $R$ is chosen on side the $AC$ to satisfy $AR : RC = 1 : 2$. Show that $\angle PRQ = 90^\circ$. | [] | Mongolia | Mongolian Mathematical Olympiad Round 1 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0ipw | Problem:
Solve the equation
$$
\sqrt{x+\sqrt{4 x+\sqrt{16 x+\sqrt{\ldots+\sqrt{4^{2008} x+3}}}}}-\sqrt{x}=1
$$
Express your answer as a reduced fraction with the numerator and denominator written in their prime factorization. | [
"Solution:\nAnswer: $\\frac{1}{2^{4016}}$\n\nRewrite the equation to get\n$$\n\\sqrt{x+\\sqrt{4 x+\\sqrt{16 x+\\sqrt{\\ldots+\\sqrt{4^{2008} x+3}}}}}=\\sqrt{x}+1\n$$\nSquaring both sides yields\n$$\n\\sqrt{4 x+\\sqrt{\\ldots+\\sqrt{4^{2008} x+3}}}=2 \\sqrt{x}+1\n$$\nSquaring again yields\n$$\n\\sqrt{16 x+\\sqrt{\\l... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | 1/2^{4016} | |
03rb | Let $f$: $\mathbb{R} \rightarrow \mathbb{R}$ be a function such that $f(0) = 1$ and for any $x$, $y \in \mathbb{R}$, $f(xy+1) = f(x)f(y) - f(y) - x + 2$ holds. Then $f(x) = \underline{\hspace{2cm}}$. | [
"Since for any $x$, $y \\in \\mathbb{R}$, $f(xy+1) = f(x)f(y) - f(y) - x + 2$, we have\n$$\nf(yx + 1) = f(y)f(x) - f(x) - y + 2.\n$$\nThus,\n$$\nf(x)f(y) - f(y) - x + 2 = f(y)f(x) - f(x) - y + 2,\n$$\nthat is,\n\nPut $y = 0$, we obtain $f(x) = x + 1$."
] | China | China Mathematical Competition (Hainan) | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = x + 1 | |
06wa | Let $n \geqslant 2$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers such that $a_{1}+a_{2}+\cdots+a_{n}=1$. Prove that
$$
\sum_{k=1}^{n} \frac{a_{k}}{1-a_{k}}\left(a_{1}+a_{2}+\cdots+a_{k-1}\right)^{2}<\frac{1}{3} .
$$ | [
"For all $k \\leqslant n$, let\n$$\ns_{k}=a_{1}+a_{2}+\\cdots+a_{k} \\quad \\text{ and } \\quad b_{k}=\\frac{a_{k} s_{k-1}^{2}}{1-a_{k}},\n$$\nwith the convention that $s_{0}=0$. Note that $b_{k}$ is exactly a summand in the sum we need to estimate. We shall prove the inequality\n$$\n\\begin{equation*}\nb_{k}<\\fra... | IMO | IMO 2021 Shortlisted Problems | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
0ba0 | Given a triangle $ABC$, let $D$ be the midpoint of the side $AC$ and let $M$ be the point that divides the segment $BD$ in the ratio $1/2$; that is, $MB/MD = 1/2$. The rays $AM$ and $CM$ meet the sides $BC$ and $AB$ at points $E$ and $F$, respectively. Assume the two rays are perpendicular: $AM \perp CM$. Show that the... | [
"Denote by $a$, $b$, $c$ the sidelengths, and by $m_a$, $m_b$, $m_c$ the lengths of the medians of the triangle $ABC$. Since $MD$ is median in the right-angled triangle $AMC$, it follows that $2m_b/3 = MD = AD = CD = b/2$, so $m_b = 3b/4$, whence $(3b/4)^2 = m_b^2 = (a^2 + c^2)/2 - b^2/4$; that is, $13b^2 = 8(a^2 +... | Romania | 62nd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Tri... | null | proof only | null | |
0bar | Determine all positive integers $n$ for which there exists a polynomial $f(x)$ with real coefficients, with the following properties:
1) for each integer $k$, the number $f(k)$ is an integer if and only if $n$ does not divide $k$;
2) the degree of $f$ is less than $n$. | [
"We will show that such polynomial exists if and only if $n = 1$ or $n$ is a power of a prime.\n\nWe will use two known facts stated in LEMMATA 1 and 2.\nLEMMA 1. If $p^a$ is a power of a prime and $k$ is an integer, then\n$$\n\\frac{(k-1)(k-2)\\cdots(k-p^a+1)}{(p^a-1)!}\n$$\nis divisible by $p$ if and only if $k$ ... | Romania | 2011 Fourth ROMANIAN MASTER OF MATHEMATICS | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Prime n... | null | proof and answer | n = 1 or n is a power of a prime | |
024i | Problem:
Uma ilha possui 50 clubes. Cada habitante da ilha é sócio de 1 ou 2 clubes. Cada clube tem no máximo 55 sócios e para cada par de clubes existe um habitante da ilha que é sócio dos dois clubes. Encontre todas as possibilidades para as quantidades possíveis de habitantes da ilha. Justifique sua resposta. | [
"Solution:\n\nExistem $\\binom{50}{2} = 1225$ pares de clubes e para cada um deles devemos ter um habitante que pertence a ambos. Denotemos esses habitantes por $H_{ij}$ com $i, j \\in \\{1,2, \\ldots, 50\\}$ e $i < j$. Se dois deles são iguais, um habitante estaria em três clubes e isso seria uma contradição. Port... | Brazil | null | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | [1225, 1525] | |
0163 | For which $n \ge 2$ is it possible to find $n$ pairwise non-similar triangles $A_1, A_2, \dots, A_n$ so that each one of them can be divided into $n$ triangles, similar to $A_1, A_2, \dots, A_n$? | [
"We can construct such triangles $A_i$ with angles $\\alpha$, $i\\alpha$, $(2n - i)\\alpha$, where $\\alpha = \\frac{\\pi}{2n+1}$. These triangles are non-similar to each other as the biggest angle of each triangle is different. Triangles $A_i$ and $A_{i+1}$ can be joined into a bigger triangle, as the largest angl... | Baltic Way | Baltic Way SHL | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | All integers n greater than or equal to 2 | |
04a9 | On the sides of a square $4n$ points are given: the four vertices of the square and another $n-1$ points on each side. Determine the number of all (non-degenerated) triangles with vertices at the given points. | [] | Croatia | CroatianCompetitions2011 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 2 n^2 (5 n - 3) | |
0gat | $\{a_n\}_{n \ge 0}$ 為公差 $d$ 的無窮整數數列, 其中首項滿足 $1 \le a_0 \le d$. 記此數列為 $S_0$, 我們以遞迴的方式定義一序列的新數列; 數列 $S_{n+1}$ 為 $S_n$ 經由底下兩步操作後得到:
步驟 1. 記 $S_n$ 的首項為 $b_n$. 將首項移除, 並把數列剩下的每一項往前挪。
步驟 2. 從新的首項開始算起, 將該數列的前 $b_n$ 項加 1.
試證：存在常數 $c$ 使得 $b_n = [ca_n]$ 對 $n \ge 0$ 恆成立。此處 $[\cdot]$ 表示高斯符號。 | [
"底下我們以歸納法證明\n$$\nc = \\frac{1 + \\sqrt{1 + \\frac{4}{d}}}{2},\n$$\n它是方程式 $c^2 - c - 1/d = 0$ 的正根。\n\n首先我們證明 $n = 0$ 命題成立。依定義 $b_0 = a_0$, 而\n$$\n\\begin{aligned}\n[ca_0] &= \\left[ \\frac{1 + \\sqrt{1 + 4/d}}{2} a_0 \\right] = a_0 + \\left[ \\frac{-1 + \\sqrt{1 + 4/d}}{2} a_0 \\right] \\\\\n&= a_0 + \\left[ \\frac{... | Taiwan | 二〇一七數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
06jp | Let $p$ be a prime number greater than $5$. Suppose there is an integer $k$ satisfying that $k^2 + 5$ is divisible by $p$. Prove that there are positive integers $m$ and $n$ such that $p^2 = m^2 + 5n^2$. | [
"Consider all numbers of the form $a + b k$ where $a, b$ are integers with $0 \\le a, b < \\sqrt{p}$. Since there are more than $(\\sqrt{p})^2 = p$ numbers, by the pigeonhole principle, two of them must be congruent modulo $p$. Their difference, which is of the form $c + d k$ with $|c|, |d| < \\sqrt{p}$ and $(c, d)... | Hong Kong | Year 2016 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Algebraic Number Theory > Quadratic forms",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
05jk | Problem:
Soit $ABC$ un triangle. On note $O$ le centre de son cercle circonscrit. Soient $D$, $E$, $F$ des points situés sur $[BC]$, $[CA]$ et $[AB]$ respectivement. On suppose que $FB = FD$ et $ED = EC$. Le cercle de centre $F$ et de rayon $FB$ et le cercle de centre $E$ et de rayon $EC$ se recoupent en $G$. Montrer ... | [
"Solution:\n\nLes triangles $BFD$ et $CED$ étant isocèles, on a $(DE, DF) = (DE, DC) + (DB, DF) = (CD, CE) + (BF, BD) = (BC, AC) + (AB, BC) = (AB, AC)$.\n\nComme $EG = ED$ et $FG = FD$, les points $E$ et $F$ appartiennent à la médiatrice de $[GD]$ donc le quadrilatère $FGE D$ est symétrique par rapport à $(FE)$. On... | France | null | [
"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 chasing"
] | null | proof only | null | |
026g | Problem:
O diretor da escola decidiu tirar uma foto dos formandos de 2008. Ele colocou os alunos em filas paralelas, todas com o mesmo número de alunos, mas essa disposição era muito larga para o campo de visão de sua máquina fotográfica. Para resolver esse problema, o diretor reparou que bastava tirar um aluno por fi... | [
"Solution:\n\nAs figuras a seguir representam a situação do problema, onde em preto estão representados os alunos que foram inicialmente retirados e em cinza os alunos retirados na segunda vez.\n\n\n\nSejam $n$ e $m$ o número de filas (linhas horizontais) e de colunas da formação inicial, r... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 24 | |
0j6q | Problem:
Consider the $8 \times 8 \times 8$ Rubik's cube below. Each face is painted with a different color, and it is possible to turn any layer, as you can with smaller Rubik's cubes. Let $X$ denote the move that turns the shaded layer shown (indicated by arrows going from the top to the right of the cube) clockwise... | [
"Solution:\n\nThere are two \"bands\" of individual unit cubes (called cubies) that are moved by $M$. Of those, only the cubies exactly three units from an edge of the cube, such as the cubie originally at the intersection of the shaded bands, can ever move out of a single plane of rotation. All the other cubes eit... | United States | 13th Bay Area Mathematical Olympiad | [
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof and answer | 28 | |
0f8r | Problem:
An insect is on a square ceiling side $1$. The insect can jump to the midpoint of the segment joining it to any of the four corners of the ceiling. Show that in $8$ jumps it can get to within $1/100$ of any chosen point on the ceiling. | [] | Soviet Union | 23rd ASU | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
07oh | Each of the four positive integers $N, N+1, N+2, N+3$ has exactly six positive divisors. There are exactly 20 different positive integers which are exact divisors of at least one of the integers. One of these is 27. Find all possible values of $N$.
(Both 1 and $m$ are counted as divisors of the integer $m$.) | [
"Each of the four integers has 1 as a divisor. Moreover, 2 is a common divisor of either $N$ and $N+2$ or $N+1$ and $N+3$. Hence, these four integers can have at most $6 \\times 4 - 3 - 1 = 20$ different divisors altogether. Because there are exactly 20 different positive divisors, only one of the four integers can... | Ireland | Irska 2014 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 242 | |
02dt | Show that the maximum number of spheres of radius $1$ that can be placed touching a fixed sphere of radius $1$ so that no pair of spheres has an interior point in common is between $12$ and $14$. | [
"We can place $6$ spheres with their centers coplanar with the fixed sphere. Then we can place $3$ more above and $3$ more below as shown above. Thus $12$ can be achieved.\n\n\n\nTake $O$ to be the center of the fixed sphere. Another sphere touching it blocks off a conical solid angle as sh... | Brazil | V OBM | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Surface Area"
] | English | proof only | null | |
070d | Problem:
An equilateral triangle of side $n$ is divided into $n^{2}$ equilateral triangles of side 1 by lines parallel to the sides. Initially, all the sides of all the small triangles are painted blue. Three coins $A$, $B$, $C$ are placed at vertices of the small triangles. Each coin in turn is moved a distance 1 alo... | [
"Solution:\n\n\nWe use induction. It is obvious for $n=1$ and $2$ - see diagram above. Note that $A$, $B$, $C$ start and end at vertices of the large triangle.\n\n\nNow assume that for $n$ we can find a solution with $A$, $B$, $C$ starting and ending at the vertices... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
00vr | Let $\triangle ABC$ be a triangle with $\angle BAC = 2\angle CBA$ and $AB < AC$. A point $P$ lies on segment $AC$ such that $PC = AB + AP$. Let $O$ be the circumcentre of triangle $\triangle ABP$, and let the line through $O$ parallel to $AB$ intersect $BP$ at $Q$. Show that $AQ$ passes through the midpoint of segment ... | [
"Let $M$ be the midpoint of $BC$ and let $D$ be the reflection of $C$ in $P$. From the length condition in the statement, we have $DA = DP - AP = DP - (PC - AB) = AB$. This together with $PM$ being the midline in $\\triangle CDB$ implies $\\angle MPC = \\angle BDA = \\frac{\\angle BAC}{2} = \\angle MBA$. So $ABMP$ ... | Balkan Mathematical Olympiad | 42nd Balkan 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 > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geom... | English | proof only | null | |
0edj | Three friends, Andrej, Blaž and Cene, were playing badminton. For each game, two of them were playing one against the other, and the third was free. After each game, the winner of the game played against the one that was free in the last game. Andrej played $17$ games and Blaž played $23$ games. At least how many games... | [
"Let $n$ be the number of games played by Cene. Then the total number of games was $\\frac{17+23+n}{2}$, which implies that $n$ is even. The total number of games was at least $23$ since Blaž has played this many of them. Since Cene was free for at most one game in a row he had to have played at least $11$ games, b... | Slovenia | Slovenija 2016 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | MCQ | D | |
01vg | Some cells of a checkered plane are marked so that the figure $A$ formed by marked cells satisfies the following two conditions:
1) any cell of the figure $A$ has exactly two adjacent cells of $A$; and
2) the figure $A$ can be divided into isosceles trapezoids of area $2$ with vertices at the grid nodes (and acute angl... | [
"It is clear that it is enough to prove the statement of the problem for *connected* figures $A$, since any disconnected figure is divided into several connected parts, each of which (as we prove) has a number of cells divided by $8$. Let's color some cells of the plane into four colors as shown in the figure.\n\nW... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0ktq | Problem:
Suppose $a$ and $b$ are positive integers such that $a^{b} = 2^{2023}$. Compute the smallest possible value of $b^{a}$. | [
"Solution:\nBy taking $a = 2^{2023}$ and $b = 1$, we get $b^{a} = 1$, which is clearly the minimum."
] | United States | HMMT February 2023 | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 1 | |
0ctw | Let $N > 1$ be an integer, and let $d_1 < \dots < d_s$ be all its positive divisors (thus, $d_1 = 1$ and $d_s = N$). Find all possible values of $N$ for which
$$
\gcd(d_1, d_2) + \gcd(d_2, d_3) + \dots + \gcd(d_{s-1}, d_s) = N - 2. \quad (\text{A. Kuznetsov})
$$
Саша выбрал натуральное число $N > 1$ и выписал в строку... | [
"$N = 3$.\n\nNotice that $\\gcd(d_i, d_{i+1}) \\le d_{i+1} - d_i$; thus, denoting $r_i = (d_{i+1} - d_i) - \\gcd(d_i, d_{i+1}) \\ge 0$ we get $r_1 + \\dots + r_{s-1} = 1$, so $r_k = 1$ for some $k$, and $r_i = 0$ for $i \\ne k$; this yields that $d_{k+1} - d_k = 2$ and that $d_k, d_{k+1}$ are odd. If $k \\ne s/2$, ... | Russia | Russian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English; Russian | proof and answer | N = 3 | |
0a6e | Problem:
A shop sells golf balls, golf clubs and golf hats. Golf balls can be purchased at a rate of 25 cents for two balls. Golf hats cost $1 each. Golf clubs cost $10 each. At this shop, Ross purchased 100 items for a total cost of exactly $100 (Ross purchased at least one of each type of item). How many golf hats di... | [
"Solution:\nLet the number of pairs of balls, clubs, and hats purchased be $b$, $c$, $h$ respectively. Then, we must have\n\n$$\n\\text{cost:}\\qquad \\frac{1}{4} b + h + 10c = 100\n$$\n$$\n\\text{quantity:}\\qquad 2b + h + c = 100\n$$\n\nSubtracting these yields $\\frac{7}{4} b - 9c = 0$ and thus\n$$\n7b = 36c.\n$... | New Zealand | NZMO Round One | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | final answer only | 21 | |
0agh | Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = 3$. Prove the inequality
$$
\frac{a^3+2}{b+2} + \frac{b^3+2}{c+2} + \frac{c^3+2}{a+2} \ge 3.
$$ | [
"From the inequality between the arithmetic and geometric means we have\n$$\n\\frac{a^3 + 2}{b + 2} = \\frac{a^3 + 1 + 1}{b + 2} \\ge \\frac{3\\sqrt[3]{a^3 \\cdot 1 \\cdot 1}}{b + 2} = \\frac{3a}{b + 2}.\n$$\nAnalogously we get the equations\n$$\n\\frac{b^3 + 2}{c + 2} \\ge \\frac{3b}{c + 2} \\quad \\text{and} \\qu... | North Macedonia | Macedonian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
03tu | Suppose $a$, $b$, $c$ are real numbers, with $a + b + c = 3$. Prove that
$$
\frac{1}{5a^2 - 4a + 11} + \frac{1}{5b^2 - 4b + 11} + \frac{1}{5c^2 - 4c + 11} \le \frac{1}{4}.
$$ | [
"$$\n\\text{If } a < \\frac{9}{5}, \\text{ then}\n$$\n$$\n\\frac{1}{5a^2 - 4a + 11} \\le \\frac{1}{24}(3-a). \\quad \\textcircled{1}\n$$\nIn fact,\n$$\n\\begin{align*}\n\\textcircled{1} &\\Leftrightarrow (3-a)(5a^2-4a+11) \\ge 24 \\\\\n&\\Leftrightarrow 5a^3 - 19a^2 + 23a - 9 \\le 0 \\\\\n&\\Leftrightarrow (a-1)^2(... | China | China Western Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0ehj | Problem:
Jure in Miha igrata igro z dvema posodama s kroglicami, v kateri poteze izvajata izmenično. Na začetku igre je v beli posodi $m$ kroglic, v črni pa $n$ kroglic. V vsaki potezi igralec bodisi odstrani eno kroglico iz ene od posod ali pa prestavi eno kroglico iz bele v črno posodo. Zmaga tisti igralec, ki odstr... | [
"Solution:\n\nPokazali bomo, da če sta obe števili $m$ in $n$ sodi, ima zmagovito strategijo Miha, če pa je vsaj eno od števil $m$ in $n$ liho, ima zmagovito strategijo Jure.\n\nDenimo najprej, da sta $m$ in $n$ sodi. Tedaj lahko Miha vsakič ponovi Juretovo potezo, saj bo tako po vsaki njegovi potezi število krogli... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Miha wins if both starting counts are even; otherwise Jure wins. | |
0e98 | Let $ABCDEF$ be a regular hexagon, let $P$ be the midpoint of the side $AB$ and let $R$ be the midpoint of the side $EF$, as shown in the figure. What is the ratio of the area of the quadrilateral $APRF$ to the area of the quadrilateral $BCDP$?
 | [
"Denote the length of the side of the regular hexagon by $a$. The quadrilateral $APRF$ is a trapezium with the altitude $\\frac{1}{2} \\cdot \\frac{a\\sqrt{3}}{2} = \\frac{a\\sqrt{3}}{4}$, so its area equals\n$$\nP_{APRF} = \\frac{a\\sqrt{3}}{4} \\cdot \\frac{|PR|+|AF|}{2} = \\frac{a\\sqrt{3}(\\frac{3}{2}a+a)}{8} =... | Slovenia | National Math Olympiad 2013 - First Round | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 5/8 | |
0ebw | We write numbers $1, 2, \dots, 2014$ in some order and denote them with $a_1, a_2, \dots, a_{2014}$. We then rewrite them in some new order and denote them with $b_1, b_2, \dots, b_{2014}$. Find the largest positive integer $k$ such that, regardless of the chosen orders, the number $3^k$ will certainly divide the numbe... | [
"Let's prove that the answer is $n = 672$.\n\nWe can check that the number $a_i^2 - b_i^2$ is divisible by $3$ if and only if the numbers $a_i$ and $b_i$ are either both divisible by $3$ or both co-prime to $3$. Among the numbers $1, 2, \\dots, 2014$ there are exactly $671$ of them divisible by $3$, $672$ of them h... | Slovenia | Selection Examinations for the IMO 2015 | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 672 | |
0knk | An equilateral triangle $\triangle \Delta$ of side length $L > 0$ is given. Suppose that $n$ equilateral triangles with side length $1$ and with non-overlapping interiors are drawn inside $\triangle \Delta$, such that each unit equilateral triangle has sides parallel to $\triangle \Delta$, but with opposite orientation... | [] | United States | USAJMO | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0k3h | Problem:
I have two cents and Bill has $n$ cents. Bill wants to buy some pencils, which come in two different packages. One package of pencils costs 6 cents for 7 pencils, and the other package of pencils costs a dime for a dozen pencils (i.e. 10 cents for 12 pencils). Bill notes that he can spend all $n$ of his cents... | [
"Solution:\n\nSuppose that Bill buys $a$ packages of 7 and $b$ packages of 12 in the first scenario and $c$ packages of 7 and $d$ packages of 12 in the second scenario. Then we have the following system:\n$$\n\\begin{aligned}\n& 6a + 10b = n \\\\\n& 6c + 10d = n + 2 \\\\\n& 7a + 12b > 7c + 12d .\n\\end{aligned}\n$$... | United States | HMMT November 2018 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 100 | |
0fye | Problem:
Die Punkte $X, Y, Z$ liegen in dieser Reihenfolge auf einer Geraden mit $|X Y| \neq |Y Z|$. Sei $k_{1}$ bzw. $k_{2}$ der Kreis mit Durchmesser $X Y$ bzw. $Y Z$. Die Punkte $A_{1}$ und $B_{1}$ bzw. $A_{2}$ und $B_{2}$ liegen auf $k_{1}$ bzw. $k_{2}$, sodass
$$
\angle A_{1} Y A_{2} = \angle B_{1} Y B_{2} = 90^{... | [
"Solution:\n\nEs gilt $\\angle A_{2} Z Y = 90^{\\circ} - \\angle A_{2} Y Z = \\angle A_{1} Y X$ und analog $\\angle A_{1} X Y = A_{2} Y Z$. Also ist $\\triangle A_{1} X Y \\sim \\triangle A_{2} Y Z$. Definiere $P$ als Schnittpunkt von $A_{1} A_{2}$ und $X Y$. Dann gilt $\\triangle P A_{1} Y \\sim P A_{2} Z$ und wir... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
00u0 | A cube of side length $2021$ is given. In how many ways can we place a $1 \times 1 \times 1$ cubelet on the border of this cube in such a way that the newly formed solid can be completely filled using $k \times 1 \times 1$, $1 \times k \times 1$ and $1 \times 1 \times k$ cuboids, for some $k \in \mathbb{N} \setminus \{... | [
"Suppose that for some $k > 1$ and some placed cubelet there is a valid filling. In each unit cubelet (of the original cube) with coordinates $(x, y, z)$ where $0 \\le x, y, z \\le 2020$, we assign the complex number $\\omega^{x+y+z}$ where $\\omega = e^{\\frac{2\\pi i}{k}}$. We also assign the number $\\omega^{a+b... | Balkan Mathematical Olympiad | BMO 2022 shortlist | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic E... | English | proof and answer | 13612182 | |
0285 | Problem:
Pedrinho está brincando de fazer arranjos com palitos. Ele dispõe seus palitos formando triângulos equiláteros, como mostra a figura abaixo:
Pedrinho quer pintar cada palito de seu arranjo de tal forma que cada triângulo tenha seus lados pintados de exatamente duas cores diferentes. ... | [
"Solution:\n\nComeçamos pintando os palitos de dentro do hexágono. Cada um desses palitos pode ser pintado com 3 cores distintas. Temos então $3^{6}$ maneiras de pintar estes palitos interiores.\n\nNotamos agora que cada um dos seis triângulos formados podem ser completados de duas maneiras diferentes, independente... | Brazil | null | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 46656 | |
09yb | Problem:
Bewijs dat voor elk positief geheel getal $n$ er positieve gehele getallen $a$ en $b$ bestaan met
$$
n \mid 4 a^{2}+9 b^{2}-1
$$ | [
"Solution:\n\nVoor $n=1$ voldoet elke keuze van $a$ en $b$. Stel nu dat $n>1$ en zij $p$ een priemdeler van $n$. Zij $k$ het aantal factoren $p$ in $n$. We geven een voorwaarde voor $a$ en $b$ modulo $p^{k}$ die garandeert dat $p^{k} \\mid 4 a^{2}+9 b^{2}-1$. Door dit voor elke priemdeler van $n$ te doen, krijgen w... | Netherlands | IMO-selectietoets II | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0eiu | Problem:
Dan je izraz $Z=5 a^{-x}\left(1-a^{-x}\right)^{-1}-3 a^{-x}\left(1+a^{-x}\right)^{-1}-2 a^{x}\left(a^{2 x}-1\right)^{-1}$, kjer je $a^{x} \neq 0,1,-1$.
a) Poenostavi izraz $Z$.
b) Izračunaj vrednost izraza $Z$ za $a=9^{b+c} \cdot 3^{2 b+c}: 27^{\frac{4}{3} b+c+\frac{1}{3}}$ in $x=1$. | [
"Solution:\n\nPoenostavimo vsak člen posebej. Prvi člen preoblikujemo v\n$$\n5 a^{-x}\\left(1-a^{-x}\\right)^{-1}=\\frac{5}{a^{x}}\\left(\\frac{a^{x}-1}{a^{x}}\\right)^{-1}=\\frac{5}{a^{x}-1}\n$$\nPodobno drugi člen preoblikujemo v\n$$\n3 a^{-x}\\left(1+a^{-x}\\right)^{-1}=\\frac{3}{a^{x}+1}\n$$\nImenovalec tretjeg... | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | Z = 8 / (a^(2x) − 1); for a = 9^(b+c) · 3^(2b+c) : 27^((4/3)b + c + 1/3) and x = 1, Z = −9 | |
0lfm | For each polynomial $P(x)$, define
$$
P_1(x) = P(x), \forall x \in \mathbb{R},
$$
$$
P_2(x) = P(P_1(x)), \forall x \in \mathbb{R},
$$
$$
P_{2024}(x) = P(P_{2023}(x)), \forall x \in \mathbb{R}.
$$
Let $a > 2$ be a real number. Is there a polynomial $P$ with real coefficients such that for all $t \in (-a, a)$, the equati... | [
"We will present two solutions for this problem.\n\n**First solution:**\nWe will prove that the polynomial $P(x) = \\frac{2}{a}x^2 - a$ satisfies the requirements of the problem. Specifically, we will prove the equation\n$$\nP_{2024}(x) = t \\quad (*)\n$$\nthere are exactly $2^{2024}$ distinct solutions in the inte... | Vietnam | Vietnamese MO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
018x | The incircle of a triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. Let $G$ be a point on the incircle such that $FG$ is a diameter. The lines $EG$ and $FD$ intersect at $H$. Prove that $CH \parallel AB$. | [
"We work in the opposite direction. Suppose that $H'$ is the point where $DF$ intersects the line through $C$ parallel to $AB$. We need to show that $H' = H$. For this purpose it suffices to prove that $E$, $G$, $H'$ are collinear, which reduces to showing that if $G' \\neq E$ is the common point of $EH'$ and the i... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"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",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jo9 | Problem:
Is there a polynomial $P(n)$ with integer coefficients such that $P(2)=4$ and $P(P(2))=7$? Prove your answer. | [
"Solution:\n\nThe answer is no. Let $P(n) = c_n x^n + \\cdots + c_0$. We are given that $P(2) = 4$ and $P(4) = 7$. The first equation implies that $c_0$ is even while the second implies that $c_0$ is odd, which is a contradiction."
] | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | No | |
0gr9 | Show that for each pair of positive integers $(a, b)$ there is a positive integer $n$ such that $n^2 + an + b$ has at least $2018$ distinct prime divisors. | [
"By induction over $k$ we will prove that for each pair of positive integers $(a, b)$ there is a positive integer $n$ such that $n^2 + an + b$ has at least $k$ distinct prime divisors. The case $k = 1$ is obvious. For given $(a, b)$ suppose that for some $n_k$, $n_k^2 + an_k + b$ has $k$ distinct prime divisors $p_... | Turkey | Team Selection Test | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0kol | Problem:
Compute the smallest positive integer $n$ for which there are at least two odd primes $p$ such that
$$
\sum_{k=1}^{n}(-1)^{\nu_{p}(k!)}<0
$$
Note: for a prime $p$ and a positive integer $m$, $\nu_{p}(m)$ is the exponent of the largest power of $p$ that divides $m$; for example, $\nu_{3}(18)=2$.
Proposed by: K... | [
"Solution:\n\nSay $n$ is $p$-good if $\\sum_{k=1}^{n}(-1)^{\\nu_{p}(k!)}<0$, where $p$ is an odd prime.\n\nClaim. $n$ is $p$-good iff\n$$\nn+1=\\sum_{i=0}^{k} a_{i} p^{2 i+1}\n$$\nwhere $a_{i}$ is an even integer with $\\left|a_{i}\\right|<p$.\n\nThe proof of this claim will be deferred to the end of the solution a... | United States | HMMT February 2022 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 229 | |
05s2 | Problem:
Soit $n$ un entier impair, et soit $S$ un ensemble de $n$ points du plan à coordonnées entières. On considère une permutation $f: S \rightarrow S$ qui satisfait la propriété suivante :
Pour toute paire de points $A$ et $B$ appartenant à $S$, la distance entre $f(A)$ et $f(B)$ est supérieure ou égale à la dist... | [
"Solution:\n\nPuisque $S$ est de cardinal impair, $f$ admet une orbite de cardinal impair. Sans perte de généralité, on peut donc supposer que $S$ est égal à cette orbite, et même que $S=\\{P_1, \\ldots, P_n\\}$ avec $f(P_i)=P_{i+1}$ pour tout $i \\leqslant n$, en posant $P_{n+1}=P_1$.\n\nSi $f$ n'a aucun point fix... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Number Theory > Other"
] | null | proof only | null | |
015g | Find all positive integers $n$ with the following property. Every function $f: \mathbf{R} \to \mathbf{R}$ such that
$$
f\left(\frac{a+b}{2}\right) = \frac{f(a)+f(b)}{2} \quad \text{for every } a, b \in \mathbf{R},
$$
satisfies also the condition
$$
f\left(\frac{x_1 + x_2 + \dots + x_n}{n}\right) = \frac{f(x_1) + f(x_2)... | [
"Suppose that $n$ has the property in the problem. Then for every function $f$, if $()$ holds then\n$$\n\\begin{aligned}\nf\\left(\\frac{x_1 + x_2 + \\dots + x_{2n}}{2n}\\right) &= f\\left(\\frac{1}{n}\\left(x_1 + x_2 + \\frac{x_3 + x_4}{2} + \\dots + \\frac{x_{2n-1} + x_{2n}}{2}\\right)\\right) \\\\\n&= \\frac{1}{... | Baltic Way | Baltic Way SHL | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | all positive integers | |
0d9j | Here are $2017$ points on the plane, no three of them are collinear. Some pairs of the points are connected by $n$ segments. Find the smallest value of $n$ so that there always exists two disjoint segments in any case. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | 2018 | |
0j8t | Determine whether or not there exist two different sets $A$, $B$, each consisting of at most $2011^2$ positive integers, such that every $x$ with $0 < x < 1$ satisfies the following inequality:
$$
\left| \sum_{a \in A} x^a - \sum_{b \in B} x^b \right| < (1-x)^{2011}.
$$ | [
"The answer is yes. We will show that a pair of such sets exists. Rewrite the desired inequality as\n$$\n\\left| \\sum_{a \\in A} (1-y)^a - \\sum_{b \\in B} (1-y)^b \\right| < y^{2011} \\quad (34)\n$$\nfor every $y$ such that $0 < y < 1$.\n\n*Step 1:* First, we show that there exist two different sets $A'$, $B'$ of... | United States | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Yes | |
0hir | Problem:
At a party with 100 people, everyone is either a knight, who always tells the truth, or a knave, who always lies. Each person says they shook hands with a different number of knights at the party, from 0 to 99. Each pair of people shook hands at most once, and everyone knows whether each other person is a kni... | [
"Solution:\n\nCall the person who said they shook hands with $i$ people person $i$ for each $i$ from $0$ to $99$. Now, if person $99$ is telling the truth, person $99$ must have shaken hands with everyone else, and all the other people must be knights. But then person $0$ would be lying, since they said they shook ... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1 | |
024f | Problem:
Quantos números inteiros entre $10$ e $999$ têm a soma de seus algarismos igual a $9$? | [
"Solution:\n\nVamos dividir em dois casos: números de dois algarismos e números de três algarismos.\n\nNo caso de números de dois algarismos, temos $18, 27, 36, 45, 54, 63, 72, 81$ e $90$, num total de $9$ números.\n\nDa mesma maneira, listamos os números de três algarismos, como segue:\n$$\n\\begin{aligned}\n108, ... | Brazil | Nível 2 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 54 | |
0jub | Problem:
Find a positive integer $N$ and $a_{1}, a_{2}, \ldots, a_{N}$, where $a_{k}=1$ or $a_{k}=-1$ for each $k=1,2, \ldots, N$, such that
$$
a_{1} \cdot 1^{3}+a_{2} \cdot 2^{3}+a_{3} \cdot 3^{3}+\cdots+a_{N} \cdot N^{3}=20162016,
$$
or show that this is impossible. | [
"Solution:\nIt is possible, as long as the sum $S$ desired is a multiple of $48$, with $N = S / 6$, which in this case is $3360336$, and the $a_{k}$ repeats the 8-term pattern $-1, 1, 1, -1, 1, -1, -1, 1$.\n\nUse the observation that if $f(x)$ is a degree-$k$ polynomial, then for any constant $h$, the difference $f... | United States | BAMO-12 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | It is possible. Take N = 3360336 and let a_k repeat the eight-term pattern −1, 1, 1, −1, 1, −1, −1, 1 for k = 1, 2, …, N; then the signed sum of cubes equals 20162016. | |
07js | Let $x_1 < x_2 < \dots < x_n$ be real numbers. Define the function $f : \mathbb{R} \rightarrow \mathbb{R}$
$$
f(x) = \left| \frac{(x-x_2)(x-x_3)\cdots(x-x_n)}{(x_1-x_2)(x_1-x_3)\cdots(x_1-x_n)} \right| + \cdots + \left| \frac{(x-x_1)(x-x_2)\cdots(x-x_{n-1})}{(x_n-x_1)(x_n-x_2)\cdots(x_n-x_{n-1})} \right|
$$
Prove that ... | [
"If for some $i$, $x_i = x_{i+1}$ then we have nothing to do. Assume now, that $x_i \\neq x_{i+1}$, $0 \\le i \\le n+1$. Consider the polynomial $P(x) = Q_1(x)^2 + \\dots + Q_n(x)^2 - 1$. It at most has $2(n-1)$ roots and it is easy to verify that $x_1, \\dots, x_n$ are roots. Hence, there is $i$, $1 \\le i \\le n-... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0cjg | At the robotics workshop, the registered students are grouped into teams made of three boys and a girl. One day two boys and a girl are missing, so the present students were regrouped into teams made by one girl and four boys.
How many students are registered at the robotics workshop? | [
"Initially, the students are grouped $(G, B, B, B), (G, B, B, B), (G, B, B, B), \\dots, (G, B, B, B)$.\nWhen two girls and a boy are missing, the situation (replacing a boy, if the missing boy is not in the same team as one of the missing girls) becomes:\n$(G, B, B), (G, B, B, B), (G, B, B, B), \\dots, (G, B, B, B)... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 28 | |
07eu | Two circle $\omega_1$ and $\omega_2$ intersect each other at points $P$ and $Q$. An arbitrary line passes from $P$ and intersects $\omega_1$ and $\omega_2$ respectively at points $A$ and $B$. A line parallel to $AB$ cuts $\omega_1$ at $D$ and $F$, and cuts $\omega_2$ at $C$ and $E$ in a way that $E$ and $F$ lie between... | [
"i) Notice that\n$$\n\\angle QEB = \\angle QPB = \\angle QDA \\implies \\angle QEX = \\angle QDX\n$$\nSo quadrilateral $XDEQ$ is cyclic and similarly quadrilateral $YCFQ$ is cyclic. Let $S$ be the second intersection point of these two circles. It's clear that $\\angle QSX = \\angle QEB = \\angle QCY$ so $X, S, Y$ ... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0hlb | Problem:
How many divisors does $2013^{13}$ have? (As usual, we count $1$ and $2013^{13}$ itself as divisors, but not negative integers.) | [
"Solution:\n\nThe prime factorization of $2013$ is $3 \\cdot 11 \\cdot 61$, so\n$$\n2013^{13} = 3^{13} \\cdot 11^{13} \\cdot 61^{13}.\n$$\nA divisor of this number is found by choosing $0$ to $13$ of the factors $3$ (there are $14$ possible choices), $0$ to $13$ of the factors $11$ ($14$ choices), and $0$ to $13$ o... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 2744 | |
032g | Problem:
Let $ABCD$ be a cyclic quadrilateral. Denote by $I$ and $J$ the incenters of $\triangle ABD$ and $\triangle BCD$. Prove that $ABCD$ is a circumscribed quadrilateral if and only if the points $A$, $I$, $J$ and $C$ are either collinear or concyclic. | [
"Solution:\nIt is easy to see that if the points $A$, $I$, $J$ and $C$ are collinear, then $AB = AD$ and $BC = CD$. Hence $ABCD$ is a circumscribed quadrilateral.\n\nSuppose that the points $A$, $I$, $J$ and $C$ are concyclic. Since $\\angle AIC > \\angle AIB$ or $\\angle AIC > \\angle AID$, it follows that $\\angl... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Pla... | null | proof only | null | |
04af | Circles $k_1$ and $k_2$, with radii $r, R$ ($r < R$) touch each other internally in $A$. Let $p$ be a line parallel to their common tangent and let $B$ be an intersection of the line $p$ with the circle $k_1$, and $C$ an intersection of $p$ with $k_2$, so that the points $B$ and $C$ lie on the same side of the line thr... | [] | Croatia | Hrvatska 2011 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | sqrt(rR) | |
0es4 | $p$, $q$ and $r$ represent the numbers $2$, $3$, $4$ in some order. What is the greatest possible value of $p^q \times r$? | [
"$3^4 \\times 2 = 81 \\times 2 = 162$"
] | South Africa | South African Mathematics Olympiad Second Round | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 162 | |
00t2 | Determine all positive integers $n$ such that $\frac{a^2 + n^2}{b^2 - n^2}$ is a positive integer for some positive integers $a$ and $b$. | [
"The required numbers are all even positive integers alone. Indeed, if $n$ is even, then let $a = n^2/2 - 1$ and $b = n^2/2 + 1$, to check that $\\frac{a^2 + n^2}{b^2 - n^2} = \\frac{n^4/4 + 1}{n^4/4 + 1} = 1$.\n\nSuppose now that such $a$ and $b$ exist for some positive odd integer $n$. Notice that we may and will... | Balkan Mathematical Olympiad | BMO Short List | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | all even positive integers | |
0hts | Problem:
Let $n$ be a positive integer which also divides $2^{n}-1$. Show that $n=1$. | [
"Solution:\n\nAssume not and let $p$ be the smallest prime divisor of $n$. We have $p \\mid 2^{n}-1$, and also $p \\mid 2^{p-1}-1$ by Fermat's little theorem. By using the classical fact that $\\gcd\\left(2^{x}-1,2^{y}-1\\right)=2^{\\gcd(x, y)}-1$, we conclude $p$ divides $2^{\\gcd(p-1, n)}-1$.\n\nBut since $p$ is ... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | n = 1 | |
0i40 | Problem:
Two circles have centers that are $d$ units apart, and each has diameter $\sqrt{d}$. For any $d$, let $A(d)$ be the area of the smallest circle that contains both of these circles. Find $\lim_{d \rightarrow \infty} \frac{A(d)}{d^{2}}$. | [
"Solution:\n\nThis equals $\\lim_{d \\rightarrow \\infty} \\frac{\\pi\\left(\\frac{d+\\sqrt{d}}{2}\\right)^{2}}{d^{2}} = \\frac{\\pi}{4}$."
] | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | π/4 | |
06n6 | Let $A_1A_2\dots A_n$ be a convex polygon where $n \ge 4$. For each $i = 1, 2, \dots, n$, let $f(i)$ be the maximum of the angles $\angle A_iA_jA_{i+1}$ where $A_j$ can be any vertex different from $A_i$ and $A_{i+1}$ (with $A_{n+1} = A_1$). For each $n$, find the smallest possible value of $f(1) + f(2) + \dots + f(n)$... | [
"For each $k = 2, 3, \\dots, n-1$, define $S_k = \\sum_{i=1}^{n} \\angle A_i A_{i+1} A_{i+1}$ where the indices are taken modulo $n$. Note that\n$$\nS_2 + S_3 + \\dots + S_{n-1} = \\angle A_1 + \\angle A_2 + \\dots + \\angle A_n = (n-2) \\cdot 180^\\circ \\quad (1)\n$$\nbecause each angle $\\angle A_i A_j A_k$ with... | Hong Kong | The 26th Hong Kong (China) Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 180° | |
0l8x | A positive integer $N_0$ is written on the board. Two persons, $A$ and $B$, play the following game: $A$ erases $N_0$ and writes a number $N_1 \in \{N_0 - 1; [N_0/3]\}$. Then $B$ erases $N_1$ and writes a number $N_2 \in \{N_1 - 1; [N_1/3]\}$. At his turn, $A$ erases $N_2$ and so on. They continue to play until on the ... | [] | Vietnam | CONTEST FOR THE SELECTION OF VIETNAMESE INTERNATIONAL MATHEMATICAL OLYMPIAD TEAM | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | 1: B; 2: A; 3: B | |
0cei | Let $g : \mathbb{R} \to \mathbb{R}$ be a continuous function. Prove that:
a) if the inequality $g\left(\int_0^1 f(t) dt\right) \le \int_0^1 g(f(t)) dt$ holds for every integrable function $f : [0, 1] \to \mathbb{R}$, then $g$ is convex;
b) if the inequality $g\left(\int_0^1 f(t) dt\right) \le \int_0^1 g(f(t)) dt$ hol... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
0guo | For the points $A$, $B$, $K$, $L$, $X$ located on the circle $\Gamma$ in the given order the lengths of arcs $\widehat{BK}$ and $\widehat{KL}$ are equal. The circle which passes through $A$ and is tangent to $BK$ at $B$ intersects the line segment $KX$ at points $P$ and $Q$. The second intersection point of the circle ... | [
"The oriented angle equalities $\\angle QXA = \\angle KBA = \\angle BPA$ and $\\angle XQA = \\angle PBA$ imply the oriented similarity $AXQ \\sim APB$, and the oriented angle equalities $\\angle XTA = \\angle LBA$ and $\\angle TXA = \\angle BLA$ imply the oriented similarity $AXT \\sim ALB$. If we put a C-coordinat... | Turkey | Team Selection Test for IMO 2023 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilate... | English | proof only | null | |
0h1e | Using the numbers $1, 2, \ldots, 20$ (each number once) as denominators and numerators, construct $10$ fractions with integer sum. | [
"$$\n\\frac{17}{3} + \\frac{13}{2} + \\frac{11}{6} + \\frac{19}{1} + \\frac{14}{7} + \\frac{18}{9} + \\frac{20}{10} + \\frac{16}{8} + \\frac{15}{5} + \\frac{12}{4} = 14 + 19 + 2 + 2 + 2 + 2 + 3 + 3 = 47.\n$$"
] | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | null | |
0ba9 | Given a prime number $p$ congruent to $1$ modulo $5$ such that $2p + 1$ is also prime, show that there exists a matrix of zeros and ones containing exactly $4p$ (respectively, $4p + 2$) ones no submatrix of which contains exactly $2p$ (respectively, $2p + 1$) ones. | [
"Let $p = 5q + 1$, $q \\ge 2$, and write $4p = 5(4q + 1) - 1$. Form a $5$-by-$(4q + 1)$ matrix consisting of ones only except for a single entry. Such a matrix has exactly $4p$ ones. A submatrix comprising $r$ rows, $1 \\le r \\le 5$, and $s$ columns, $1 \\le s \\le 4q + 1$, contains $rs$ or $rs-1$ ones. In the for... | Romania | 62nd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0fqr | In a triangle $ABC$, the internal bisector of $\angle A$ meets the side $BC$ at $D$. The lines through $D$ tangents to the circumcircles of triangles $\triangle ABD$ and $\triangle ACD$ meet the lines $AC$ and $AB$ at points $E$ and $F$, respectively. Lines $BE$ and $CF$ intersect at $G$. Prove that $\angle EDG = \angl... | [
"We have $\\angle ADE = \\angle B$ and $\\angle ADF = \\angle C$. So $AFDE$ is a cyclic\n\n\n\nquadrilateral and\n$$\n\\angle AFE = \\angle ADE = \\angle B \\text{ and } \\angle AEF = \\angle ADF = \\angle C,\n$$\nHence, $EF$ is parallel to $BC$ and $DE = DF$, so the triangle $FDE$ is isosc... | Spain | SPANISH MATHEMATICAL OLYMPIAD (FINAL ROUND) | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0g7e | 已知 $m \ge 0$, $f(x) = x^2 + \sqrt{mx} + m + 1$。試證: 對任意正實數 $x_1, x_2, \dots, x_n$,
$$
f(\sqrt[n]{x_1 \cdots x_n}) \le \sqrt[n]{f(x_1)f(x_2)\cdots f(x_n)},
$$
上述等號成立的充要條件為 $x_1 = x_2 = \cdots = x_n$。 | [
"因 $\\Delta = (\\sqrt{m})^2 - 4(m+1) < 0$, 所以 $f(x) > 0$.\n利用數學歸納法證明：當 $n = 2^k$, 題設成立。\n\na. 當 $k=1$ 時, $n=2$. 因\n$$\n\\sqrt{m}x_1x_2(\\sqrt{x_1}-\\sqrt{x_2})^2 + (m+1)(x_1-x_2)^2 + \\sqrt{m}(m+1)(\\sqrt{x_1}-\\sqrt{x_2})^2 \\ge 0,\n$$\n所以\n$$\n(x_1x_2 + \\sqrt{mx_1x_2} + m + 1)^2 \\le (x_1^2 + \\sqrt{m}x_1 + m + ... | Taiwan | 二〇一三數學奧林匹亞競賽第三階段選訓營, 獨立研究(一) | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0dy3 | Let $ABC$ be an acute triangle. Denote its orthocentre by $H$ and let $A'$, $B'$ and $C'$ be the feet of the altitudes from $A$, $B$ and $C$. Let $P$ be the midpoint of $AH$, let $Q$ be the intersection of lines $B'P$ and $AB$ and denote the intersection of segments $A'C'$ and $BB'$ by $R$. Prove that the line $QR$ is ... | [
"The midpoint of the hypotenuse is also the circumcentre, so $P$ is the circumcentre of the triangle $AHB'$ and $|AP| = |PH| = |PB'|$. The triangle $HPB'$ is isosceles with the apex at $P$. Let $QB'B = \\alpha$. Then $\\alpha = \\angle PB'H = \\angle B'HP = \\angle BHA'$. Since $\\angle HA'B + \\angle BC'H = \\frac... | Slovenia | Slovenija 2008 | [
"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 | |
03qu | For a given real number $a$ and a positive integer $n$, prove that:
(1) there exists exactly one sequence of real numbers $x_0, x_1, \dots, x_n, x_{n+1}$, such that
$$
\begin{cases}
x_0 = x_{n+1} = 0, \\
\frac{1}{2}(x_{i+1} + x_{i-1}) = x_i + x_i^3 - a^3, \quad i = 1, 2, \dots, n;
\end{cases}
$$
(2) the sequence $x_0... | [
"(1) Proof of existence: From $x_{i+1} = 2x_i + 2x_i^3 - 2a^3 - x_{i-1}$, $i = 1, 2, \\dots, n$ and $x_0 = 0$, we get that $x_i$ is a polynomial of $x_1$ with degree $3^{i-1}$ and real coefficients, for $1 \\le i \\le n+1$. Specifically, $x_{n+1}$ is a polynomial of $x_1$ with degree $3^n$ and real coefficients. As... | China | China Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | English | proof only | null | |
0lg1 | Problem:
Determine the maximum integer $n$ with the property that for each positive integer $k \leq \frac{n}{2}$ there exist two positive divisors of $n$ with difference $k$. | [
"Solution:\nIf there exists a positive integer $p \\leq \\lfloor n / 6\\rfloor$ such that $p \\nmid n$, then we have $\\lfloor n / 2\\rfloor>\\lfloor n / 6\\rfloor$, and taking $k=\\lfloor n / 2\\rfloor-p \\geq 2$ and two positive divisors $d, d+k$ of $n$, we need $d+(\\lfloor n / 2\\rfloor-p)$ to divide $n$. But $... | Zhautykov Olympiad | XI International Zhautykov Olympiad in Sciences | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | 24 | |
02qu | Problem:
A figura mostra um dodecágono regular decomposto em seis triângulos equiláteros, seis quadrados e um hexágono regular, todos com lados de mesma medida.
a) Se cada triângulo tem área igual a $1~\mathrm{cm}^2$ qual é a área do hexágono?
b) A figura abaixo foi obtida retirando doze tr... | [
"Solution:\n\na) A figura abaixo mostra que o hexágono pode ser decomposto em seis triângulos iguais aos triângulos que fazem parte do dodecágono. Como cada um desses triângulos tem área $1~\\mathrm{cm}^2$, segue que o hexágono tem área $6~\\mathrm{cm}^2$.\n\n\nb) $1^{a}$ solução: A figura ... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | a) 6 cm^2; b) 6 cm^2; c) 6 cm^2 | |
0kpd | Find the three-digit positive integer $abc$ whose representation in base nine is $\underline{b}\ \underline{c}\ \underline{a}$ base $9$, where $a$, $b$, and $c$ are (not necessarily distinct) digits. | [] | United States | 2022 AIME I | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 227 | |
06wg | The kingdom of Anisotropy consists of $n$ cities. For every two cities there exists exactly one direct one-way road between them. We say that a path from $X$ to $Y$ is a sequence of roads such that one can move from $X$ to $Y$ along this sequence without returning to an already visited city. A collection of paths is ca... | [
"We write $X \\rightarrow Y$ (or $Y \\leftarrow X$) if the road between $X$ and $Y$ goes from $X$ to $Y$. Notice that, if there is any route moving from $X$ to $Y$ (possibly passing through some cities more than once), then there is a path from $X$ to $Y$ consisting of some roads in the route. Indeed, any cycle in ... | IMO | IMO 2021 Shortlisted Problems | [
"Discrete Mathematics > Graph Theory > Menger's theorem / max-flow, min-cut",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0e16 | There were 64 contestants at a chess tournament. Every pair played a game that ended either with one of them winning or in a draw. If a game ended in a draw, then each of the remaining 62 contestants won against at least one of these two contestants. There were at least two games ending in a draw at the tournament. Sho... | [
"First, we show that for each contestant at most one game ended in a draw. Assume, to the contrary, that the contestant $A$ tied against $B$ and $C$. The game between $A$ and $B$ was tied, so $C$ won against at least one of them. The game between $A$ and $C$ was tied, so $C$ won against $B$. On the other hand, the ... | Slovenia | Selection Examinations for the IMO | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
02nz | Let $ABCD$ be a quadrilateral with $\angle ABC \neq 90^\circ$. Let $M$ and $N$ be the midpoints of $AD$ and $CD$, respectively. Prove that the lines perpendicular to $BC$ passing through $M$ and perpendicular to $AB$ passing through $N$ and $BD$ are concurrent if and only if the diagonals $BD$ and $AC$ are perpendicula... | [
"Consider a homothety with center on $D$ that takes $M$ to $A$ and $N$ to $C$. So the perpendicular lines are mapped to the altitudes of the triangle $ABC$ relative to $A$ and $C$, and the intersection $P$ of the perpendicular lines is mapped to the orthocenter $H$ of triangle $ABC$.\n\n\n\... | Brazil | Brazilian Math Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals"
] | null | proof only | null | |
0ia6 | Problem:
Prove that any 2-configuration containing $e$ elements is $m$-separable for some $m \leq \frac{1}{2} + \sqrt{2e + \frac{1}{4}}$. | [
"Solution:\nSuppose $m$ is the minimum integer for which the given configuration $C$ on set $A$ is $m$-separable, and fix a corresponding labeling of the elements of $A$. Let $A_{i}$ be the set of all elements with the label $i$. Then, for any $i, j$ with $1 \\leq i < j \\leq m$, there must exist $a_{i} \\in A_{i},... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0cci | Let $ABC$ be a triangle. The moving points $M$ on the half line $BC$, $N$ on the half line $CA$, and $P$ on the half line $AB$, start simultaneously from vertices $B$, $C$, and $A$, respectively, and move with constant speeds $v_1, v_2, v_3 > 0$, expressed using the same unit.
a) Knowing that there are three distinct ... | [
"Let $a, b, c \\in \\mathbb{C}$ be the affixes of the vertices of triangle $ABC$. For $t \\ge 0$, we have the following expressions for the affixes of the points $M(t)$, $N(t)$, and $P(t)$:\n$$\n\\left\\{ \\begin{array}{l} m(t) = b \\cdot (1 - v_1 \\cdot t) + c \\cdot v_1 \\cdot t; \\\\ n(t) = c \\cdot (1 - v_2 \\c... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety"... | null | proof only | null |
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