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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
093r | Problem:
Find all positive integers $n$ for which there exist positive integers $x_{1}, x_{2}, \ldots, x_{n}$ such that
$$
\frac{1}{x_{1}^{2}}+\frac{2}{x_{2}^{2}}+\frac{4}{x_{3}^{2}}+\cdots+\frac{2^{n-1}}{x_{n}^{2}}=1
$$ | [
"Solution:\n- $n=1$ :\nHere, $x_{1}:=1$ provides a solution, since\n$$\n\\frac{1}{1^{2}}=1\n$$\n\n- $n=2$ :\nHere, no solution exists. Indeed, $x_{1}=1$ or $x_{2}=1$ yields $\\frac{1}{x_{1}^{2}}+\\frac{2}{x_{2}^{2}}>1$, while $x_{1}, x_{2} \\geq 2$ leads to\n$$\n\\frac{1}{x_{1}^{2}}+\\frac{2}{x_{2}^{2}} \\leq \\fra... | Middle European Mathematical Olympiad (MEMO) | 14th Middle European Mathematical Olympiad 2020 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | All positive integers except 2 | |
02j8 | Problem:
Um ônibus, um trem e um avião partem no mesmo horário da cidade $A$ para a cidade $B$. Se eu tomar o ônibus cuja velocidade média é $100~\mathrm{km}/\mathrm{h}$, chegarei à cidade $B$ às 20 horas. Se eu tomar o trem, cuja velocidade média é $300~\mathrm{km}/\mathrm{h}$, chegarei à cidade $B$ às 14 horas. Qual... | [
"Solution:\n\nSeja $d$ a distância entre as duas cidades e $h$ o horário de partida comum do ônibus, do trem e do avião. Como, distância $=$ velocidade $\\times$ tempo, temos:\n$$\nd = 100 \\times (20 - h)\n$$\ne\n$$\nd = 300 \\times (14 - h)\n$$\nLogo,\n$$\n100 \\times (20 - h) = 300 \\times (14 - h)\n$$\nDonde\n$... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 12:00 | |
0fsq | Problem:
Beweise, dass die Gleichung
$$
14 x^{2}+15 y^{2}=7^{2000}
$$
keine ganzzahlige Lösungen $(x, y)$ besitzt. | [
"Solution:\n\nBetrachte die Gleichung modulo $3$. Wegen $7 \\equiv 1\\ (\\bmod\\ 3)$ lautet sie\n$$\n2 x^{2} \\equiv 1 \\quad(\\bmod\\ 3)\n$$\nQuadrate sind aber $\\equiv 0,1\\ (\\bmod\\ 3)$, also nimmt die linke Seite nur die Werte $0$ und $2$ an, Widerspruch. Es gibt keine ganzzahligen Lösungen."
] | Switzerland | IMO - Selektion | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0j1t | Let $ABC$ be a triangle with $\angle A = 90^\circ$. Points $D$ and $E$ lie on sides $AC$ and $AB$, respectively, such that $\angle ABD = \angle DBC$ and $\angle ACE = \angle ECB$. Segments $BD$ and $CE$ meet at $I$. Determine whether or not it is possible for segments $AB, AC, BI, ID, CI, IE$ to all have integer length... | [
"The answer is *no*, it is not possible for segments $AB$, $BC$, $BI$, $ID$, $CI$, $IE$ to all have integer lengths.\n\nSuppose on the contrary that these segments do have integer side lengths. Set $\\alpha = \\angle ABD = \\angle DBC$ and $\\beta = \\angle ACE = \\angle ECB$. Note that $I$ is the incenter of trian... | United States | USAMO 2010 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | No | |
04n0 | There are two non-intersecting circles, of radii $r_1$ and $r_2$. The distance between the points of tangency of the inner common tangent to these circles is $12$, while the distance between the points of tangency of the outer common tangent to these circles is $16$. Determine the product $r_1 r_2$.
Inner tangent (is ... | [] | Croatia | Croatia_2018 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 28 | |
04zu | Each side of a convex quadrangle $ABCD$ is a diameter of a circle. All four circles pass through the same point $O$, different from the vertices of the quadrangle, and no two circles have common points other than those mentioned. Prove that $ABCD$ is a rhombus. | [
"Since $AB$, $BC$, $CD$, and $DA$ are diameters (Fig. 9), $AOB$, $BOC$, $COD$, and $DOA$ are right angles. Hence $AOC$ and $BOD$ are straight angles, i.e., the diagonals of the quadrangle meet at $O$. The circles drawn on the opposite sides $AB$ and $CD$ cannot have common points besides $O$, since otherwise one ci... | Estonia | Selected Problems from the Final Round of National Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ib0 | Problem:
A floor is tiled with equilateral triangles of side length $1$, as shown. If you drop a needle of length $2$ somewhere on the floor, what is the largest number of triangles it could end up intersecting? (Only count the triangles whose interiors are met by the needle - touching along edges or at corners doesn't... | [
"Solution:\nLet $L$ be the union of all the lines of the tiling. Imagine walking from one end of the needle to the other. We enter a new triangle precisely when we cross one of the lines of the tiling. Therefore, the problem is equivalent to maximizing the number of times the needle crosses $L$.\n\nNow, the lines o... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 8 | |
0i6b | Problem:
Give the set of all positive integers $n$ such that $\varphi(n) = 2002^{2} - 1$. | [
"Solution:\nThe empty set, $\\varnothing$. If $m$ is relatively prime to $n$ and $m < n$, then $n - m$ must likewise be relatively prime to $n$, and these are distinct for $n > 2$ since $n / 2, n$ are not relatively prime. Therefore, for all $n > 2$, $\\varphi(n)$ must be even. $2002^{2} - 1$ is odd, and $\\varphi(... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof and answer | empty set | |
02tw | Problem:
Na figura abaixo, $ABCD$ é um quadrado e os pontos $K$, $L$ e $M$ estão sobre os lados $AB$, $BC$ e $CD$ de modo que $\triangle KLM$ é um triângulo isósceles retângulo em $L$. Prove que $AL$ e $DK$ são perpendiculares.
| [
"Solution:\n\nSejam $\\angle MLC = \\alpha$ e $\\angle BAL = \\beta$. Como $\\angle KLM = \\angle KBL = \\angle LCM = 90^{\\circ}$, segue que $\\angle KLB = \\angle LMC = 90^{\\circ} - \\alpha$. Além disso, como $KL = LM$, os triângulos retângulos $\\triangle LMC$ e $\\triangle BLK$ são congruentes. De $BL = CM$ e ... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0kms | Let $M$ be the least common multiple of all the integers $10$ through $30$, inclusive. Let $N$ be the least common multiple of $32$, $33$, $34$, $35$, $36$, $37$, $38$, $39$, and $40$. What is the value of $\frac{N}{M}$?
(A) 1 (B) 2 (C) 37 (D) 74 (E) 2886 | [] | United States | AMC 12 A | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | MCQ | null | |
0kd9 | Problem:
Let $P_{1} P_{2} P_{3} P_{4}$ be a tetrahedron in $\mathbb{R}^{3}$ and let $O$ be a point equidistant from each of its vertices. Suppose there exists a point $H$ such that for each $i$, the line $P_{i} H$ is perpendicular to the plane through the other three vertices. Line $P_{1} H$ intersects the plane throu... | [
"Solution:\n\nNote that $A$ is the orthocenter of triangle $P_{2} P_{3} P_{4}$ by projecting the givens about $H$.\n\nNext, let $P_{2} A$ intersect $P_{3} P_{4}$ at $C$ and $\\left(P_{2} P_{3} P_{4}\\right)$ at $D$. Then $A D=2 A C$.\n\nNote that $H$ is the orthocenter of triangle $P_{1} P_{2} C$, with $A$ being th... | United States | HMIC | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
04nh | Let $n$ be a positive integer. A sequence of $2n$ real numbers is good if for each positive integer $1 \le m \le 2n$ the sum of first $m$ or the sum of last $m$ terms of the sequence is an integer. Determine the least possible number of integers in a good sequence. (The Netherlands 2017) | [] | Croatia | Croatia_2018 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | English | proof and answer | 2 | |
0eld | Problem:
Določi natančno zgornjo mejo zaporedja s splošnim členom $a_{n} = n \left(\frac{20}{23}\right)^{n}$ za vse $n \in \mathbb{N}$. | [
"Solution:\n\nPoglejmo, kdaj zaporedje narašča. Neenakost $a_{n+1} > a_{n}$ se z upoštevanjem formule za splošni člen zaporedja glasi\n$$(n+1)\\left(\\frac{20}{23}\\right)^{n+1} > n\\left(\\frac{20}{23}\\right)^{n}$$\nin je ekvivalentna neenakosti\n$$(n+1) \\cdot \\frac{20}{23} > n.$$\nNeenakost poenostavimo do\n$$... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 7*(20/23)^7 | |
07w0 | Suppose *ABC* is an isosceles triangle with $b = c$. Show that $4bc > a^2$. Give an example of a right-angled triangle with hypotenuse $a$ in which this inequality fails. | [
"By the triangle inequality, $a < b + c$, hence $a < 2b$, using $b = c$. Therefore $a^2 < 4bc$, as required.\n\nIf $x > 1$, and $a = x^2 + 1$, $b = 2x$, $c = x^2 - 1$, then\n$$\nb^2 + c^2 = 4x^2 + (x^4 - 2x^2 + 1) = x^4 + 2x^2 + 1 = a^2,\n$$\nand so $a$, $b$, $c$ are side lengths of a right-angled triangle with hyp... | Ireland | IRL_ABooklet_2023 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Number Theory > Diophantine Equations > Pythagorean triples"
] | English | proof and answer | Example counterexample: a right triangle with sides a=41, b=40, c=9 (hypotenuse 41), for which 41^2 > 4·40·9. | |
08ws | How many possible ways of writing down a sequence of positive integers are there satisfying the following conditions?
Conditions: You start out with writing down the number $2012$ and end up with writing down the number $1$, and after writing down a number $n$ you follow with writing an integer less than or equal to $\... | [
"Let $m$ be a positive integer and denote by $a_m$ the number of possible ways of writing down a sequence of positive integers satisfying the following, which we call the conditions $C_m$:\n$C_m$: Start with writing down the number $m$, and end up with writing down the number $1$, and after writing down a number $n... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 201 | |
0172 | Let $r$ be a positive integer. The following game is being played on a rectangular board divided into $20 \times 12$ unit squares. One is allowed to move a piece from a square to another, if the distance between (the centres of) these squares is $\sqrt{r}$. The goal is to find a sequence of moves leading from the botto... | [
"a) If $r$ is even, then $a + b$ is even for any solution of the Diophantine equation $a^2 + b^2 = r$, so that the parity of the sum of the coordinates is preserved under the moves. If $3$ divides $r$, then $a \\equiv b \\equiv 0 \\pmod{3}$ for all the solutions of this equation, as $c^2 \\equiv 0 \\pmod{3}$ or $c^... | Baltic Way | BALTIC WAY | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | a) Impossible when the distance parameter is divisible by two or by three. b) Possible for seventy-three (an explicit path exists). c) No solution for ninety-seven. | |
09bi | Нэгэн компани аймшигт амьтан үржүүлдэг. Хэрэв уг амьтан өнөөдөр $a$-ширхэг гар, $b$-ширхэг хөл, $c$-ширхэг толгойтой байсан бол маргааш нь харгалзан $b+c-a$, $c+a-b$, $a+b-c$ ширхэг гар, хөл, толгойтой болох бөгөөд энэ хувьсал өдөр болгон үргэлжлэн явагддаг. Тэгвэл анх төрөхдөө зөвхөн тэнцүү тооны гар, хөл, толгойтой т... | [
"Бүгд тэнцүү $a = b = c$ үед мөнхөд амьдрах нь илэрхий. Иймд $a \\neq b$ гэж үзэж чадна.\n$$\na + b + c = (a + b - c) + (b + c - a) + (c + a - b)\n$$\n-гэдгээс уг аймшигт амьтны гар, хөл, толгойн нийлбэр үргэлж тогтмол байна. $a \\neq b \\Rightarrow |a-b| \\geq 1$ ба II дахь өдөр энэ ялгавар\n$$\n|c + a - b - (c + ... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | Mongolian | proof only | null | |
0k4o | Problem:
To dissect a polygon means to divide it into several regions by cutting along finitely many line segments. For example, the diagram below shows a dissection of a hexagon into two triangles and two quadrilaterals:
An integer-ratio right triangle is a right triangle whose side length... | [
"Solution:\n\nAbbreviate integer-ratio right triangle by IRRT.\n\nThe square $(n=4)$ has such a decomposition. For example, a $12 \\times 12$ square can be cut into twenty-four 3-4-5 triangles as shown below:\n\n\n\nNow we show that $n=4$ is the only solution. The proof is by contradiction.... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisen... | null | proof and answer | n = 4 | |
001a | Lado un ángulo recto $X \hat{A}Y$ de vértice $A$ y una semicircunferencia $\Gamma$ interior a este ángulo con centro en el lado $AX$ y tangente al lado $AY$ en $A$, construir una tangente a $\Gamma$ tal que el triángulo que se recorta del ángulo $X \hat{A}Y$ sea de área mínima. | [] | Argentina | XII Olimpíada Matemática Rioplatense | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | español | proof and answer | The optimal tangent is the one that makes an angle of thirty degrees with the side containing the center (equivalently sixty degrees with the other side). | |
0g6w | 令 $S$ 是 $[0, 1]$ 區間中所有有理數的集合。給定一個無限長的實數數列
$$
\{x_1, x_2, \dots, x_k, \dots\},
$$
如果存在一個由 $S$ 映至實數的函數 $H(x)$ 滿足：
(i) $H$ 在 $[0, \frac{1}{2}]$ 遞增。即: 對任意有理數 $0 \le a \le b \le \frac{1}{2}$, 有 $H(a) \le H(b)$.
(ii) 對任意兩個整數 $0 < p, 0 \le q \le p$, 都有
$$
H(\frac{q}{p}) = \frac{\sum_{k=1}^{q} x_{p+1-k} - \sum_{k=1}^{q} x_k}{p... | [
"所有超乎想像的數列是：$x_k = a(k \\log(k) - (k-1) \\log(k-1)) + c$, 其中 $a$ 是非負實數, $c$ 是任意實數。\n\n首先易知若 $\\{x_1, x_2, \\cdots, x_k, \\cdots\\}$ 是超乎想像的, 則對任意實數 $c$ 及正實數 $r$, $\\{\\frac{x_1}{r} + c, \\frac{x_2}{r} + c, \\cdots, \\frac{x_k}{r} + c, \\cdots\\}$ 也是超乎想像的。\n因此不妨假設 $x_1 = 0$. 令 $f(p) = \\sum_{i=1}^{p} x_i$, 則易知 $H(\\f... | Taiwan | 二〇一二數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | All such sequences are exactly those of the form x_k = a (k log k − (k − 1) log(k − 1)) + c, where a is any nonnegative real number and c is any real number. | |
0dtb | Find all positive integers $k$ such that there exist positive integers $a$, $b$ such that
$$
a^2 + 4 = (k^2 - 4)b^2.
$$ | [
"Suppose $k, a, b$ satisfy the equation. Rewrite the equation as\n$$\na = \\sqrt{(k^2 - 4)b^2 - 4} = \\sqrt{(kb)^2 - 4(b^2 + 1)} \\quad (1)\n$$\nConsider the quadratic equation\n$$\nx^2 - kbx + (b^2 + 1) = 0 \\quad (2)\n$$\nIts solutions are\n$$\nc = \\frac{kb \\pm \\sqrt{(kb)^2 - 4(b^2 + 1)}}{2} = \\frac{kb \\pm a... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | k = 3 | |
0a3o | Consider the sequence $y_0, y_1, \dots$ such that $y_0 = -\frac{1}{4}$ and $y_1 = 0$, and furthermore
$$
y_{n+1} + y_{n-1} = 4y_n + 1
$$
for all $n \ge 1$. Prove that for all $n \ge 0$ the expression $2y_{2n} + \frac{3}{2}$ is
a) a positive integer, and
b) the square of an integer. | [
"We substitute $x_n = 4y_n + 2$. Then the equation becomes homogeneous:\n$$\nx_{n+1} + x_{n-1} = 4y_{n+1} + 2 + 4y_{n-1} + 2 = 4(4y_n + 1) + 4 = 16y_n + 8 = 4x_n,\n$$\nwith initial conditions $x_0 = 4(-\\frac{1}{4}) + 2 = 1$ and $x_1 = 4 \\cdot 0 + 2 = 2$. So all the numbers in the sequence $(x_i)$ are integers and... | Netherlands | IMO Team Selection Test 1 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
03tf | How many integers $a$ satisfy the condition: for each $a$, the equation $x^3 = a x + a + 1$ with respect to $x$ has roots which are even and $|x| < 1000$. | [
"Let $x_0 = 2n$, where $n$ is an integer and $|2n| < 1000$, then $|n| \\le 499$. So we can choose at most $2 \\times 499 + 1 = 999$ numbers, that is, $n \\in \\{-499, -498, \\dots, 0, 1, \\dots, 499\\}$. Substituting $x_0 = 2n$ into the equation, we get $a = \\frac{8n^3 - 1}{2n + 1}$.\n\nSet $f(n) = \\frac{8n^3 - 1... | China | China Southeastern Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 999 | |
09d9 | $2^{2012}$ тоог 4 бүхэл тооны квадратуудын нийлбэрт хичнээн аргаар задалж болох вэ? | [
"Өгөгдсөн болого нь $x^2 + y^2 + z^2 + t^2 = 2^{2012}$ (1) тэгшитгэлийн бүх бүхэл $(x, y, z, t)$ шийдийн тоог олохтой адил юм. Энэ нь $x, y, z, t \\ge 0$ байх шийдийн тооноос хялбархан гарах нь ойлгомжтой. $a = 2k - 1, k = 1, 2, \\dots$ сондгой натурал тооны квадратыг 8-д хуваахад 1 үлдэнэ. Үнэхээр\n$a^2 = (2k - 1)... | Mongolia | ММО-48 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Mongolian | proof and answer | 18 | |
01dd | Let $ABCD$ be a parallelogram such that $\angle BAD = 60^\circ$. Let $K$ and $L$ be the midpoints of $BC$ and $CD$, respectively. Assuming that $ABKL$ is a cyclic quadrilateral, find $\angle ABD$. | [
"**Answer:** $75^\\circ$.\n\nLet $\\angle BAL = \\alpha$. Observe that\n$$\n\\angle ADB = \\angle CBD = \\angle CKL = \\angle BAL = \\alpha.\n$$\nThe second equality holds since $KL$ is a mid-segment of $BCD$, and the third equality holds since $ABKL$ is inscribed.\n\nLet $P$ be the intersection point of $BD$ and $... | Baltic Way | Baltic Way 2016 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 75° | |
00ag | Let $a_1, a_2, ..., a_{100}$ be a permutation of $1, 2, ..., 100$. For each triple $\{a_i, a_{i+1}, a_{i+2}\}$ of consecutive numbers, $1 \le i \le 98$, the middle number in the triple is marked. For instance if $a_i = 7, a_{i+1} = 99, a_{i+2} = 22$ then $a_{i+2} = 22$ is marked. Let $S$ be the sum of all marked number... | [
"The desired minimum is $33 \\cdot 34 = 1122$. More generally, for $n = 3k + 1$ instead of $100$ the answer is $S_{\\min} = 2(1 + \\dots + k) = k(k+1)$. For clarity we state separately a fact used later in a proof of the lower bound $S(\\alpha) \\ge 2(1 + \\dots + k)$ for each permutation $\\alpha$ of $1, 2, ..., 3... | Argentina | Argentine National Olympiad 2016 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 1122 | |
03r4 | Given an acute triangle $ABC$ with $O$ as its circumcenter. Line $AO$ and side $BC$ meet at $D$. Points $E$ and $F$ are on sides $AB$ and $AC$ respectively, such that points $A$, $E$, $D$ and $F$ are on a circle. Prove that the length of the projection of line segment $EF$ on side $BC$ does not depend on the positions ... | [
"Let $M$ and $N$ be the feet of the perpendiculars from $D$ to lines $AB$ and $AC$ respectively. Let $E_0$, $F_0$, $M_0$ and $N_0$ be the feet of the perpendiculars from $E$, $F$, $M$ and $N$ to side $BC$ respectively.\n\n\n\nIt suffices to show that $E_0F_0 = M_0N_0$. Without loss of gener... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Mi... | English | proof only | null | |
0h5w | a) Andrii and Olesia both received the set of cards, on which all integer numbers from $1$ till $2015$ are written. After that Olesia leaves herself some amount of cards (but not all) from her set, and the rest she puts aside. Andrii does the same. There are $2015^2$ points on the coordinate plane, and coordinates are ... | [
"a) Without loss of generality, we will consider that Olesia didn't choose the card with the number $a$. Then the point $(a, a)$ cannot be painted. Olesia doesn't paint it because the first coordinate is $a$, Andrii doesn't paint it because the second coordinate is $a$.\n\nb) If there is one number, for instance $1... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) It is impossible to color all points; for any number not chosen by one player the point with both coordinates equal to that number remains uncolored. b) Olesia and Andrii must ensure that no number is omitted by both of them, i.e., their discarded sets are disjoint so that every number is chosen by at least one of t... | |
09eq | Denote $S_a = \frac{a}{1} + \frac{a^2}{2} + \dots + \frac{a^{p-1}}{p-1}$ for any whole number $a$. Prove that if for any whole number $n$ the relation $S_{2n} + S_{2n-1} - 2S_n - S_2 = \frac{m}{k}$, (where $m, k$ are relatively prime) holds then $m$ is divisible by $p$. | [
"If rational numbers $\\frac{p_1}{q_1}, \\frac{p_2}{q_2}$ satisfy the conditions $q_1 \\cdot q_2 \\neq (\\text{mod } p)$ and $p_1q_2 - p_2q_1 \\equiv 0 \\pmod{p}$, then we write $\\frac{p_1}{q_1} \\equiv \\frac{p_2}{q_2} \\pmod{p}$. Then our task is to show $S_{2n} + S_{2n-1} - 2S_n - S_2 \\equiv 0 \\pmod{p}$.\n\n$... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof only | null | |
0frq | Sea *P* un punto en el plano. Demuestra que es posible trazar tres semirrectas con origen en *P* con la siguiente propiedad: para toda circunferencia de radio $r$ que contiene a $P$ en su interior, si $P_1$, $P_2$ y $P_3$ son los puntos de corte de las semirrectas con la circunferencia, entonces
$$
|PP_1| + |PP_2| + |P... | [
"**Solución 1:** Trazamos tres semirrectas con origen en $P$ de manera que el ángulo formado por dos cualesquiera de las tres sea de $120^\\circ$. Imaginemos el círculo dividido en seis regiones mediante tres rectas que pasen por su centro y sean paralelas a las tres semirrectas. Observemos que, por simetría respec... | Spain | LVIII Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Powe... | Spanish | proof only | null | |
0az3 | Problem:
Find the minimum value of the expression
$$
\sqrt{(x-1)^2 + (y+1)^2} + \sqrt{(x+3)^2 + (y-2)^2}.
$$ | [] | Philippines | 21st PMO Area Stage | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 5 | |
0ev7 | Let $x$, $y$, $z$ be positive real numbers satisfying $x + y + z = 1$. Prove that
$$
\frac{(1 + xy + yz + zx)(1 + 3x^3 + 3y^3 + 3z^3)}{9(x + y)(y + z)(z + x)} \ge \left( \frac{x\sqrt{1+x}}{\sqrt[4]{3+9x^2}} + \frac{y\sqrt{1+y}}{\sqrt[4]{3+9y^2}} + \frac{z\sqrt{1+z}}{\sqrt[4]{3+9z^2}} \right)^2
$$ | [
"By using $x + y + z = 1$, we have\n$$\n1+xy+yz+zx = (x+y+z)^2+xy+yz+zx = (x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y).\n$$\nWith this equation and Cauchy-Schwarz inequality, we can deduce that\n$$\n\\begin{aligned}\n(LHS) &= \\frac{1}{9} \\left( \\frac{1}{1-x} + \\frac{1}{1-y} + \\frac{1}{1-z} \\right) \\left( (x+3x^3) + (y+3... | South Korea | Korean Mathematical Olympiad Final Round | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0da7 | Let $p$ be a prime number of the form $9k+1$. Show that there exists an integer $n$ such that $p \mid n^{3}-3n+1$. | [
"Note that the existence of an govern integer as described in the problem can be equivalently stated as follows: the polynomial $x^{3}-3x+1$ has a root in $\\mathbb{F}_{p}$. Following the classical method for solving cubic equations, we set $x = w + \\frac{1}{w}$. Then\n$$\nx^{3}-3x+1 = \\left(w + \\frac{1}{w}\\rig... | Saudi Arabia | Team selection tests for GMO 2018 | [
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | English | proof only | null | |
08v4 | Suppose 4 sides $AB$, $BC$, $CD$, $DA$ of a quadrilateral $ABCD$ are tangent to a circle with its center $O$, and conditions
$$
OA = 5,\ OB = 6,\ OC = 7,\ OD = 8
$$
are satisfied. Let $M$, $N$ be the midpoint of the line segments $AC$, $BD$ respectively. Find the value of $OM : ON$. Here for a line segment $XY$ its len... | [
"Let $P$, $Q$, $R$, $S$ be the points of tangency of the given circle to the sides $DA$, $AB$, $BC$, $CD$, respectively. Let, also $A'$, $B'$, $C'$, $D'$ be the midpoints of the line segments $PQ$, $QR$, $RS$, $SP$, respectively. Note that $A'$, $B'$, $C'$, $D'$ lie on the line segments $OA$, $OB$, $OC$, $OD$, resp... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneou... | null | proof and answer | 35 : 48 | |
0bgo | Problem:
a) Fie $f:[0, \infty) \rightarrow[0, \infty)$ o funcție derivabilă și convexă. Arătați că dacă $f(x) \leq x$, oricare ar fi $x \geq 0$, atunci $f'(x) \leq 1$, oricare ar fi $x \geq 0$.
b) Determinați funcțiile $f:[0, \infty) \rightarrow[0, \infty)$ derivabile și convexe care au proprietatea că $f(0)=0$ și $f'... | [
"Solution:\na) Presupunem contrariul. Există $a \\geq 0$ cu $f'(a)>1$, deci, cum $\\lim _{x \\searrow a} \\frac{f(x)-f(a)}{x-a}>1$, există $b>a$ cu $\\frac{f(b)-f(a)}{b-a}>1$.\nPentru orice $x>b$, din convexitatea funcției $f$ rezultă $\\frac{f(x)-f(b)}{x-b} \\geq \\frac{f(b)-f(a)}{b-a}=m>1$.\nAtunci $f(x) \\geq m ... | Romania | Olimpiada Naţională de Matematică Etapa Naţională | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | null | proof and answer | f(x) = x for all x ≥ 0 | |
01xh | Find all non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients satisfying the equality $P(Q(x)) = P(x)Q(x) - P(x)$. | [
"Answer: $P(x) = a x^2 - a(b+1)x + a b$, $Q(x) = x^2 - (b+1)x + 2b$, where $a, b \\in \\mathbb{R}$ and $a \\ne 0$.\n\nDenote the degrees of polynomials $P$ and $Q$ by $m$ and $n$, respectively. In the equality\n$$\nP(Q(x)) = P(x)Q(x) - P(x) \\quad (1)\n$$\ncompare the degrees of both sides:\n$$\nm n = m + n \\iff (... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All such pairs are P(x) = a x^2 − a(b+1) x + a b and Q(x) = x^2 − (b+1) x + 2 b, where a and b are real and a ≠ 0. | |
0247 | Problem:
Na figura abaixo, encontre o valor de
$$
\angle G A B+\angle A B C+\angle B C D+\angle C D E+\angle D E F+\angle E F G+\angle F G A
$$
 | [
"Solution:\nConsidere o ponto $O$ no interior da figura, como indicado no desenho abaixo.\n\nEntão\n$$\n\\begin{array}{r}\n\\angle G A B+\\angle A B C+\\angle B C D+\\angle C D E+\\angle D E F+\\angle E F G+\\angle F G A+ \\\\\n(\\angle G A O+\\angle O A B)+(\\angle A B O+\\angle O B C)+(\\... | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 540° | |
0bnc | Consider triangle $ABC$ inscribed in circle $\omega$, and an interior point $P$. Lines $AP, BP$ and $CP$ intersect the circle $\omega$ for the second time at points $D, E, F$, respectively. Let $A', B', C'$ be the reflections of $A, B, C$ in the lines $EF, FD, DE$ respectively. Show that triangle $A'B'C'$ is similar to... | [
"We shall prove that $\\triangle PAB \\sim \\triangle PA'B'$, and similarly $\\triangle PBC \\sim \\triangle PB'C'$ and $\\triangle PCA \\sim \\triangle PC'A'$; obviously, these three triangle similarities are enough in order to prove that $\\triangle ABC \\sim \\triangle A'B'C'$, either from angle equalities or si... | Romania | 66th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles"
] | null | proof only | null | |
0gel | 證明：若三個非零相異複數 $\alpha_1, \alpha_2, \alpha_3$ 在複數平面上不共線，且滿足 $\alpha_1+\alpha_2+\alpha_3 = 0$，則
$$
\sum_{i=1}^{3} \left( \frac{|\alpha_{i+1} - \alpha_{i+2}|}{\sqrt{|\alpha_i|}} \left( \frac{1}{\sqrt{|\alpha_{i+1}|}} + \frac{1}{\sqrt{|\alpha_{i+2}|}} - \frac{2}{\sqrt{|\alpha_i|}} \right) \right) \le 0 \quad (*)
$$
必成立。其中 $... | [
"考慮以此三點為頂點的三角形 $A_1A_2A_3$, 其中 $A_1(\\alpha_1), A_2(\\alpha_2), A_3(\\alpha_3)$。依條件我們知道其重心 $O$ 在原點; 再令 $s_i = |\\alpha_{i+1} - \\alpha_{i+2}|$ 為 $\\overline{A_{i+1}A_{i+2}}$ 的邊長。因此, $m_i = \\frac{3}{2}|\\alpha_i|$ 為 $\\overline{A_{i+1}A_{i+2}}$ 上的中線長。則(*)等價於 $\\sum_{i=1}^3 \\frac{s_i}{\\sqrt{m_i}} \\left( \\frac{1}... | Taiwan | 2021 數學奧林匹亞競賽第二階段選訓營, 獨立研究 (一) | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof and answer | The inequality holds for all such triples, and equality holds if and only if the three points form an equilateral triangle with centroid at the origin (equivalently, up to ordering, alpha_{i+1}/alpha_i = e^{2πi/3}). | |
03kq | Problem:
How many pairs of positive integers $x, y$ are there, with $x \leq y$, and such that $\operatorname{gcd}(x, y)=5!$ and $\operatorname{lcm}(x, y)=50!$.
Note. $\operatorname{gcd}(x, y)$ denotes the greatest common divisor of $x$ and $y$, $\operatorname{lcm}(x, y)$ denotes the least common multiple of $x$ and $... | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 16384 | |
0feo | Problem:
Se da un triángulo rectángulo isósceles $A B C$, con el ángulo recto en $C$, y los catetos de longitud $2$. Un arco de círculo $l$ con centro $A$ divide al triángulo en dos partes de la misma área, mientras que el arco de círculo $m$ con centro en $B$ es tangente al arco $l$ en un punto de la hipotenusa $A B$... | [
"Solution:\n\nSe $r$ el radio del arco $l$. El área del sector determinado así en el triángulo es $1/8$ del área del círculo. Por lo tanto,\n$$\n\\frac{1}{8} \\pi r^{2} = 1 \\Rightarrow r = 2 \\sqrt{\\frac{2}{\\pi}}\n$$\nEl radio del círculo $m$ es\n$$\nr_{1} = |A B| - r = 2 \\sqrt{2} - 2 \\sqrt{\\frac{2}{\\pi}} = ... | Spain | TANDA I | [
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | 2 sqrt(pi) - pi | |
0kl9 | Problem:
Let $S=\{1,2, \ldots, 9\}$. Compute the number of functions $f: S \rightarrow S$ such that, for all $s \in S$, $f(f(f(s)))=s$ and $f(s)-s$ is not divisible by $3$. | [
"Solution:\nSince $f(f(f(s)))=s$ for all $s \\in S$, each cycle in the cycle decomposition of $f$ must have length $1$ or $3$. Also, since $f(s) \\not\\equiv s \\pmod{3}$ for all $s \\in S$, each cycle cannot contain two elements $a, b$ such that $a=b \\pmod{3}$. Hence each cycle has exactly three elements, one fro... | United States | HMMT Spring 2021 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | final answer only | 288 | |
0fy8 | Problem:
Sei $g$ eine Gerade in der Ebene. Die Kreise $k_{1}$ und $k_{2}$ liegen auf derselben Seite von $g$ und berühren $g$ in den Punkten $A$ respektive $B$. Ein weiterer Kreis $k_{3}$ berühre $k_{1}$ in $D$ und $k_{2}$ in $C$. Beweise dass gilt:
(a) Das Viereck $A B C D$ ist ein Sehnenviereck.
(b) Die Geraden $B C... | [
"Solution:\n\n(a) Seien $M_{1}, M_{2}$ und $M_{3}$ die Mittelpunkte von $k_{1}, k_{2}$ und $k_{3}$. Die Strecken $A M_{1}$ und $B M_{2}$ stehen dann senkrecht auf $g$ und die Punkte $M_{1}, D$ und $M_{3}$ liegen auf einer Geraden. Weiter sei $\\angle D A B=\\alpha$ und $\\angle C B A=\\beta$. Durch Winkeljagd erhäl... | Switzerland | Vorrundenprüfung | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0exy | Problem:
Prove that you can always draw a circle radius $A/P$ inside a convex polygon with area $A$ and perimeter $P$. | [
"Solution:\n\nDraw a rectangle width $A/P$ on the inside of each side. The rectangles at each vertex must overlap since the angle at the vertex is less than $180$. The total area of the rectangles is $A$, so the area covered must be less than $A$. Hence we can find a point not in any of the rectangles. But this poi... | Soviet Union | 6th ASU | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof only | null | |
07rf | Find all pairs $(t, x)$ of real numbers that satisfy
$$
t^3 - 3t^2 + 3t - x = 0 \quad \text{and}
$$
$$
27(x-1)^4 + (1-x^2)^3 = 0.
$$ | [
"With $s = t - 1$ we have $s^3 = t^3 - 3t^2 + 3t - 1$, and the first equation can be written as $x = s^3 + 1$. From the second equation we get $27(s^3)^4 = (x^2 - 1)^3$, hence, by taking cube roots, $3s^4 = x^2 - 1$, i.e. $x^2 = 3s^4 + 1$. Comparing with the square of $x = s^3 + 1$ we obtain\n$$\n3s^4 + 1 = s^6 + 2... | Ireland | Ireland_2017 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | (-1, -7), (1, 1), (2, 2) | |
05zx | Problem:
- À quelle condition existe-t-il $n+1$ entiers (pas forcément distincts) tels que, pour tout choix de $n$ entiers parmi les $n+1$, leur somme est une puissance de $p$ ?
- À quelle condition existe-t-il $n+1$ entiers strictement positifs (pas forcément distincts) tels que, pour tout choix de $n$ entiers parmi... | [
"Solution:\n\nEssayons d'analyser l'énoncé. Posons $x_{1}, \\ldots, x_{n+1}$ les entiers tels que pour tout choix de $n$ entiers parmi les $n+1$ entiers, alors leur somme est une puissance de $p$. Il existe donc $\\alpha_{1}, \\ldots, \\alpha_{n+1}$ des entiers positifs tels que $x_{2}+\\cdots+x_{n+1}=p^{\\alpha_{1... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | First part: existence iff gcd(n, p−1) = 1. Second part: existence iff n is a power of p. | |
01om | Three points $A$, $B$, $C$, are marked on the hyperbola $y = 1/x$ so that the triangle $ABC$ is equilateral.
Find all possible values of the product of the sum of abscissae and the sum of ordinates of the vertices of $ABC$. | [
"Answer 9.\nLet $A(a; 1/a)$, $B(b; 1/b)$, $C(c; 1/c)$ be the marked points. Then the required product is equal to\n$$\nT = (a + b + c) \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right).\n$$\n\nSince any vertical and any horizontal line meets the hyperbola $y = 1/x$ at most at one ... | Belarus | BelarusMO 2013_s | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 9 | |
0kuw | Problem:
Suppose rectangle $F O L K$ and square $L O R E$ are on the plane such that $R L=12$ and $R K=11$. Compute the product of all possible areas of triangle $R K L$. | [
"Solution:\n\nThere are two possible configurations, as shown below.\n\n\n\nIf $R L=12$, the side length of the square is $6 \\sqrt{2}$. Now\n$$\n121=R K^{2}=R E^{2}+E K^{2}=(6 \\sqrt{2})^{2}+E K^{2}\n$$\nso $E K=7$. Then the possible values of $L K$ are $6 \\sqrt{2} \\pm 7$. Note that the ... | United States | HMMT November | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 414 | |
086e | Problem:
Determinare tutte le funzioni $f$, definite sull'insieme $\mathbb{Z}$ dei numeri interi relativi e a valori nell'insieme $\mathbb{R}$ dei numeri reali, che soddisfano simultaneamente le seguenti proprietà:
- per ogni coppia di interi $(m, n)$ con $m<n$ si ha $f(m)<f(n)$;
- per ogni coppia di interi $(m, n)$ e... | [
"Solution:\n\nÈ immediato verificare che le funzioni del tipo $f(n) = (n - n_{0}) a$ con $n_{0}$ intero e $a$ reale positivo soddisfano le ipotesi: se $m < n$ allora $(m - n_{0}) a < (n - n_{0}) a$ e $f(m) - f(n) = (m - n_{0}) a - (n - n_{0}) a = m a - n a = [(m - n + n_{0}) - n_{0}] a = f(k)$ con $k = m - n + n_{0... | Italy | Cesenatico | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | All functions of the form f(n) = a(n − n0), where n0 is an integer and a is a positive real number. | |
0h7b | Find all positive integers $n$, such that $11^n - 1$ is divisible by $10^n - 1$. | [
"Since $10^n - 1 = 9 \\cdot (10^{n-1} + \\dots + 10 + 1)$ we obtain that $11^n - 1$ is divisible by $9$. Considering the residues modulo $9$: $11^n - 1 \\equiv 2^n - 1 \\pmod{9}$ we get $n = 6k$. But then $10^{6k} - 1$ is divisible by $10^6 - 1$. Hence it is divisible by $10^3 + 1$ and $10 + 1 = 11$ which is not po... | Ukraine | UkraineMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | no positive integers | |
05cf | Let $ABC$ be a triangle and $P$ a point inside it. Let $A'$, $B'$, $C'$ be the reflections of $A$, $B$, $C$ from point $P$, respectively. Find the ratio of the areas of the hexagon $AB'CA'BC'$ and the triangle $ABC$. | [
"Denote the area of the figure $K$ by $S_K$. Note (Fig. 25) that\n$$\nS_{AB'CA'BC'} = S_{APB'} + S_{B'PC} + S_{CPA'} + S_{A'PB} + S_{BPC'} + S_{C'PA}.\n$$\nFurthermore, $S_{APB'} = S_{APB}$, because triangles $APB'$ and $APB$ have the same altitude drawn from vertex $A$ and the corresponding bases are equal. Simila... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 2 | |
00wi | Problem:
The midpoint of each side of a convex pentagon is connected by a segment with the intersection point of the medians of the triangle formed by the remaining three vertices of the pentagon. Prove that all five such segments intersect at one point. | [
"Solution:\n\nLet $A$, $B$, $C$, $D$ and $E$ be the vertices of the pentagon (in order), and take any point $O$ as origin. Let $M$ be the intersection point of the medians of the triangle $CDE$, and let $N$ be the midpoint of the segment $AB$. We have\n$$\n\\overline{OM} = \\frac{1}{3}(\\overline{OC} + \\overline{O... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0i42 | Problem:
Let $A B C$ be a right triangle with right angle at $B$. Let $A C D E$ be a square drawn exterior to triangle $A B C$. If $M$ is the center of this square, find the measure of $\angle M B C$. | [
"Solution:\n\n\nNote that triangle $M C A$ is a right isosceles triangle with $\\angle A C M=90^{\\circ}$ and $\\angle M A C=45^{\\circ}$. Since $\\angle A B C=90^{\\circ}$, there is a circle $k$ with diameter $A C$ which also passes through points $B$ and $C$. As inscribed angles, $\\angle... | United States | Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | 45° | |
0irk | Problem:
Find the sum of all primes $p$ for which there exists a prime $q$ such that $p^{2}+p q+q^{2}$ is a square. | [
"Solution:\n83 and 5 both work, because $3^{2}+3 \\cdot 5+5^{2}=49$. Now, say $p^{2}+p q+q^{2}=k^{2}$, for a positive integer $k$. Then $(p+q)^{2}-k^{2}=p q$, or:\n$$\n(p+q+k)(p+q-k)=p q\n$$\nSince $p+q+k$ is a divisor of $p q$, and it is greater than $p$ and $q$, $p+q+k=p q$. Then $p+q-k=1$. So:\n$$\n2 p+2 q=p q+1... | United States | 1st Annual Harvard-MIT November Tournament | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 8 | |
098x | Problem:
Numerele reale $x, y, z$ satisfac relația $x+y+z=a$, unde $a$ este un număr real fixat. Determinați valoarea maximală posibilă a sumei $S=xy+yz+zx$. Pentru care valori $x, y, z$ această valoare maximală se atinge? | [
"Solution:\n\n$x+y+z=a \\Rightarrow z=a-x-y \\quad \\Rightarrow \\quad S=xy+y(a-x-y)+x(a-x-y)=$\n$=xy+ay-xy-y^{2}+ax-x^{2}-xy=-x^{2}+(a-y)x+y(a-y)$.\nConsiderăm $S$ ca funcție de gradul doi în $x$ cu coeficientul superior $-1$, termenul liber $y(a-y)$ și coeficientul lui $x$ egal cu $a-y$. Deci $S(x)=-x^{2}+(a-y)x+... | Moldova | OLIMPIADA REPUBLICANĂ LA MATEMATICĂ | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Maximum S is a^2/3, attained when x = y = z = a/3. | |
0inc | Problem:
Find the real number $\alpha$ such that the curve $f(x)=e^{x}$ is tangent to the curve $g(x)=\alpha x^{2}$. | [
"Solution:\n$\\boxed{e^{2}/4}$. Suppose tangency occurs at $x=x_{0}$. Then $e^{x_{0}}=\\alpha x_{0}^{2}$ and $f'(x_{0})=2 \\alpha x_{0}$. On the other hand, $f'(x)=f(x)$, so $\\alpha x_{0}^{2}=2 \\alpha x_{0}$. Clearly, $\\alpha=0$ and $x_{0}=0$ are impossible, so it must be that $x_{0}=2$. Then $\\alpha=e^{x_{0}}/... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | null | proof and answer | e^2/4 | |
0eca | Problem:
Naj bo $ABCD$ štirikotnik, za katerega velja $\Varangle BAC = \Varangle ACB = 20^\circ$, $\Varangle DCA = 30^\circ$ in $\Varangle CAD = 40^\circ$. Določi velikost kota $\Varangle CBD$.
 | [
"Solution:\n\n1. način. Opazimo da je trikotnik $ABC$ enakokrak z vrhom pri $B$. Naj bo $E$ taka točka na premici $CD$, da je $\\Varangle CAE = 30^\\circ$, tako da bo tudi trikotnik $ACE$ enakokrak z vrhom pri $E$. Ker je $\\Varangle CAD = 40^\\circ$, točka $E$ leži med $C$ in $D$. Potem je štirikotnik $ABCE$ delto... | Slovenia | 59. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof and answer | 80° | |
02y1 | Problem:
Liu e Lia brincam no quadro da sala de aula. Um deles escreve dois números naturais positivos e o outro tem que fazer transformações MULTISSÔMICAS até transformar o menor no maior. Transformação MULTISSÔMICA é trocar um número $a = m + n$ por $m \cdot n$, por exemplo, podemos trocar $10$ por $2 \cdot 8 = 16$.... | [
"Solution:\n\na)\n$$\n6 = 2 + 4 \\rightarrow 2 \\cdot 4 = 8 \\rightarrow 8 = 3 + 5 \\rightarrow 3 \\cdot 5 = 15\n$$\n\nb) Qualquer natural positivo $n$ pode ser escrito como $n = (n-1) + 1$, consequentemente, pode ser transformado em $(n-1)$. Começando com $5$, temos:\n$$\n\\begin{aligned}\n5 = 2 + 3 &\\rightarrow ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 13 | |
0djy | A rhombus of side length 1 and internal angle 60° is called a diamond. Suppose that a regular hexagon of side length $n$ is dissected into diamonds. Prove that each main diagonal of the hexagon halves exactly $n$ diamonds. | [
"Divide the hexagon into $6n^2$ triangular cells (in each dissection every diamond will consist of two of them) and consider a half $H$ of the hexagon (on one side on a fixed main diagonal $d$).\n\n"
] | Saudi Arabia | Saudi Arabia booklet 2024 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | English | proof only | null | |
0il6 | Problem:
Four spheres, each of radius $r$, lie inside a regular tetrahedron with side length $1$ such that each sphere is tangent to three faces of the tetrahedron and to the other three spheres. Find $r$. | [
"Solution:\n\nLet $O$ be the center of the sphere that is tangent to the faces $ABC$, $ABD$, and $BCD$. Let $P, Q$ be the feet of the perpendiculars from $O$ to $ABC$ and $ABD$ respectively. Let $R$ be the foot of the perpendicular from $P$ to $AB$. Then, $OPRQ$ is a quadrilateral such that $\\angle P$, $\\angle Q$... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | (sqrt(6) - 1)/10 | |
0ht4 | Problem:
Given five vertices of a regular heptagon, construct the two remaining vertices using straightedge alone. | [
"Solution:\nLet $A, B, C, D$, and $E$ be the known vertices and $F$ and $G$ the unknown vertices. The arrangement of $A, B, C, D$, and $E$ depends on the relative positions of $F$ and $G$ as shown in the diagram.\n\nBy connecting the intersections of $B C$ with $D E$ and $B D$ with $C E$ on... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0ip2 | Problem:
Find the minimum of $x^{2}-2x$ over all real numbers $x$. | [
"Solution:\nWrite $x^{2}-2x = x^{2}-2x+1-1 = (x-1)^{2}-1$. Since $(x-1)^{2} \\geq 0$, it is clear that the minimum is $-1$.\n\nAlternate method: The graph of $y = x^{2}-2x$ is a parabola that opens up. Therefore, the minimum occurs at its vertex, which is at $\\frac{-b}{2a} = \\frac{-(-2)}{2} = 1$. But $1^{2}-2 \\c... | United States | 1st Annual Harvard-MIT November Tournament | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | -1 | |
03xv | The number of integral points (i.e., the points whose $x$- and $y$-coordinates are both integers) within the area (not including the boundary) enclosed by the right branch of hyperbola $x^2 - y^2 = 1$ and line $x = 100$ is ______. | [
"By symmetry, we only need to consider the part of the area above the $x$-axis. Suppose line $y = k$ intercepts the right branch of the hyperbola and line $x = 100$ at points $A_k$ and $B_k$ ($k = 1, 2, \\dots, 99$), respectively. Then the number of integral points within the segment $A_k B_k$ is $99 - k$. Therefor... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
... | English | final answer only | 9800 | |
05ji | Problem:
Montrer que tout polyèdre a deux faces qui ont le même nombre de sommets. | [
"Solution:\n\nSoit $n$ le nombre de faces du polyèdre, et soit $F$ une face fixée : les arêtes de $F$ séparent $F$ de faces toutes différentes, donc le nombre d'arêtes de $F$ est inférieur ou égal au nombre de faces autres que $F$. De plus, $F$ a au moins 3 arêtes. Chaque face a autant d'arêtes que de sommet, donc ... | France | OFM 2013-2014 Envoi 2 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof only | null | |
01ff | Several points are given in the plane. A child wants to draw $k$ (closed) discs in such a manner, that for any two points $A, B$ ($A \neq B$) there exists a disc that contains only one of these points. What is the minimum $k$, such that for any initial configuration of 2019 points it is possible to draw the $k$ discs w... | [
"Answer: $k = 1010 = \\lfloor n/2 \\rfloor$, where $n$ is the number of points.\n\nWe say that discs separates two points if it contains only one of them.\n\nEstimation. Consider a set of $n$ points on a circle. Then any disc can separate at most two consecutive pairs, therefore we need at least $\\lfloor n/2 \\rfl... | Baltic Way | Baltic Way 2019 | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1010 | |
008o | Let $n \in \mathbb{N}$ be such that $n^3 + 1$ is divisible by $56$, and $d_1^6 + d_2^6 + \dots + d_k^6$ is divisible by $112$, where $d_1, d_2, \dots, d_k$ are all the positive divisors of $n$. Which is the smallest amount of divisors that $n$ can have? | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 112 | |
0h4f | There are 2012 cities in the land of Olympia. Some of the cities are connected by direct two-way routes (each route connects exactly two cities and no two routes connect the same pair of cities). It is known that no city is directly connected to more than 8 other cities. Prove that it is possible to close at most 2012 ... | [
"Let $G$ be the graph of the problem. Partition the set $V_G$ of its 2012 vertices into three subsets $A$, $B$, and $C$ so that the sum $S$ of the numbers of edges in the subgraphs $G(A)$, $G(B)$, and $G(C)$ is minimal (clearly, this is possible). Suppose that in one of the subsets (let it be $A$), there is a verte... | Ukraine | Ukrainian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0gyg | On the infinite checked paper $2006$ sides of $1 \times 1$ squares are painted black, all others are painted white. It is allowed to choose a square and to paint all its sides into opposite color. It is known that it is possible to make all square sides white in such manner. Whether is it enough
a) $260000$ repainting ... | [
"a) We look at the projections of painted black squares onto coordinate axis. Denote the lengths of these projections by $a$ and $b$. In such case area of all squares with marked sides does not exceed $ab$, and the number of black sides of squares is at least $2(a+b) \\le 2006$ (it is easy to see that there should ... | Ukraine | The Problems of Ukrainian Authors | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | a) yes; b) no | |
0fy2 | Problem:
Bestimme alle Lösungen in natürlichen Zahlen der Gleichung
$$
a b + b c + c a = 2(a + b + c)
$$ | [
"Solution:\nWir können aus Symmetriegründen $a \\leq b \\leq c$ annehmen. Sei zuerst $a \\geq 2$. Dann erhält man für die linke Seite die Abschätzung\n$$\na b + b c + c a \\geq 2a + 2b + 2c\n$$\nnach Voraussetzung gilt hier aber Gleichheit, also muss $a = b = c = 2$ sein. Dies ist tatsächlich eine Lösung. Sei im Fo... | Switzerland | Vorrundenprüfung | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All permutations of (2, 2, 2) and (1, 2, 4). | |
0b99 | Given a polynomial $f(x)$ with rational coefficients, of degree $d \ge 2$, we define the sequence of sets $f^0(\mathbb{Q}), f^1(\mathbb{Q}), \dots$, by $f^0(\mathbb{Q}) = \mathbb{Q}$ and then $f^{n+1}(\mathbb{Q}) = f(f^n(\mathbb{Q}))$ for $n \ge 0$.
(Given a set $S$, we write $f(S)$ for the set $\{f(x) \mid x \in S\}$.... | [
"For any function $f$, denote its $n$-th iterate $f^n$. Take $d = \\deg f \\ge 2$. One can write $f(x) = \\frac{1}{N}(a x^d + g(x))$ for some $N \\in \\mathbb{Z}_+^*$, $a \\in \\mathbb{Z}^*$, and some $g \\in \\mathbb{Z}[x]$, with $\\deg g \\le d-1$, $g(x) = \\sum_{i=0}^{d-1} a_i x^i$, $a_i \\in \\mathbb{Z}$, for a... | Romania | Local Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
06pp | Let $b$, $n > 1$ be integers. Suppose that for each $k > 1$ there exists an integer $a_{k}$ such that $b - a_{k}^{n}$ is divisible by $k$. Prove that $b = A^{n}$ for some integer $A$. | [
"Let the prime factorization of $b$ be $b = p_{1}^{\\alpha_{1}} \\ldots p_{s}^{\\alpha_{s}}$, where $p_{1}, \\ldots, p_{s}$ are distinct primes. Our goal is to show that all exponents $\\alpha_{i}$ are divisible by $n$, then we can set $A = p_{1}^{\\alpha_{1} / n} \\ldots p_{s}^{\\alpha_{s} / n}$.\n\nApply the cond... | IMO | 48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
08kk | Problem:
Find all natural numbers $n$ such that $5^{n} + 12^{n}$ is a perfect square. | [
"Solution:\nBy checking the cases $n=1,2,3$ we get the solution $n=2$ and $13^{2}=5^{2}+12^{2}$.\n\nIf $n=2k+1$ is odd, we consider the equation modulo $5$ and we obtain\n$$\n\\begin{aligned}\nx^{2} & \\equiv 5^{2k+1}+12^{2k+1} \\pmod{5} \\equiv 2^{2k} \\cdot 2 \\pmod{5} \\\\\n& \\equiv (-1)^{k} \\cdot 2 \\pmod{5} ... | JBMO | OJBM | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 2 | |
07fl | $\omega$ is a circle with diameter $AB$. Points $C, D$ lie on $\omega$ such that $C, D$ are on different sides of $AB$. A line passing through $C$ and parallel to $AD$ cuts $AB$ at $F$, and a line passing through $D$ and parallel to $AC$ cuts $AB$ at $E$. $A, B, C$ and $D$ are in a way that $E, F$ are inside $\omega$. ... | [
"Let $K$ be the intersection point of $CD$, $AX$ and $L$ be the intersection point of $CD$, $AY$.\n\n\n\nThen, since $DAEX$ is a cyclic quadrilateral, we have\n$$\n\\angle KDX = \\angle BAC = \\angle AED = \\angle AXD \\implies DK = KX = AK.\n$$\nSimilarly, $AL = LY = LC$. By Thales's theor... | Iran | 37th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03lx | Problem:
Define
$$
f(x, y, z) = \frac{(x y + y z + z x)(x + y + z)}{(x + y)(x + z)(y + z)}.
$$
Determine the set of real numbers $r$ for which there exists a triplet $(x, y, z)$ of positive real numbers satisfying $f(x, y, z) = r$. | [
"Solution:\nWe prove that $1 < f(x, y, z) \\leq \\frac{9}{8}$, and that $f(x, y, z)$ can take on any value within the range $\\left(1, \\frac{9}{8}\\right]$.\nThe expression for $f(x, y, z)$ can be simplified to\n$$\nf(x, y, z) = 1 + \\frac{x y z}{(x + y)(x + z)(y + z)}.\n$$\nSince $x, y, z$ are positive, we get $1... | Canada | CANADIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (1, 9/8] | |
0aoh | Problem:
If $f$ is a real-valued function, defined for all nonzero real numbers, such that $f(a) + \frac{1}{f(b)} = f\left(\frac{1}{a}\right) + f(b)$, find all possible values of $f(1) - f(-1)$.
(a) $\{-2, 2\}$
(c) $\{1, 2\}$
(b) $\{0, -1, 1\}$
(d) $\{0, -2, 2\}$ | [] | Philippines | QUALIFYING STAGE | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | MCQ | (d) | |
05ni | Problem:
On note $S(k)$ la somme des chiffres d'un nombre entier $k$. On dit qu'un entier $a$ est d'ordre $n$ s'il existe une suite d'entiers $a_{0}, a_{1}, \ldots, a_{n}$ tels que $a_{n}=a$ et $a_{i+1}=a_{i}-S\left(a_{i}\right)$ pour tout $i=0,1, \ldots, n-1$. Montrer que pour tout entier $n \geqslant 1$ il existe un... | [
"Solution:\n\nOn pose $F(x)=x-S(x)$, et on considère la suite définie par $u_{0}=1$ et $u_{i+1}=10^{u_{i}}$ pour $i \\geqslant 0$. Soit $n \\geqslant 1$ fixé, et considérons un entier $N>u_{n+1}$. Nous allons montrer que $X=F^{(n)}\\left(10^{N}-5 N\\right)$ (on itère $n$ fois $F$) est d'ordre $n$ mais pas $n+1$. Po... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES - ENVOI No. 3 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebr... | null | proof only | null | |
0jk6 | Problem:
Let $ABC$ be an equilateral triangle of side length $6$ inscribed in a circle $\omega$. Let $A_1, A_2$ be the points (distinct from $A$) where the lines through $A$ passing through the two trisection points of $BC$ meet $\omega$. Define $B_1, B_2, C_1, C_2$ similarly. Given that $A_1, A_2, B_1, B_2, C_1, C_2$... | [
"Solution:\n\nAnswer: $\\frac{846 \\sqrt{3}}{49}$\n\nLet $A'$ be the point on $BC$ such that $2BA' = A'C$. By law of cosines on triangle $AA'B$, we find that $AA' = 2\\sqrt{7}$. By power of a point, $A'A_1 = \\frac{2 \\times 4}{2\\sqrt{7}} = \\frac{4}{\\sqrt{7}}$. Using side length ratios, $A_1A_2 = 2 \\frac{AA_1}{... | United States | HMMT 2014 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 846*sqrt(3)/49 | |
0cxv | Consider the arithmetic sequence $8, 21, 34, 47, \ldots$.
a) Prove that this sequence contains infinitely many integers written only with digit $9$.
b) How many such integers less than $2010^{2010}$ are in the sequence? | [
"(a) We are looking for integers $m$ such that\n$$\n10^{m}-1=13 n+8 \\text{ for some positive integer } n\n$$\nThe last relation is equivalent to $10^{m}-9 \\equiv 0 \\pmod{13}$.\nWe will prove that $\\operatorname{ord}_{13}(10)=6$, that is, the smallest positive integer $s$ such that $10^{s} \\equiv 1 \\pmod{13}$ ... | Saudi Arabia | SAMC | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof and answer | 1106 | |
0jcd | Problem:
Five points are chosen on a sphere of radius $1$. What is the maximum possible volume of their convex hull? | [
"Solution:\n\nAnswer: $\\sqrt{3}/2$\n\nLet the points be $A$, $B$, $C$, $X$, $Y$ so that $X$ and $Y$ are on opposite sides of the plane defined by triangle $ABC$. The volume is $1/3$ the product of the area of $ABC$ and the sum of the distances from $X$ and $Y$ to the plane defined by $ABC$.\n\nThe area of $ABC$ is... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | sqrt(3)/2 | |
0id0 | Problem:
A binary string of length $n$ is a sequence of $n$ digits, each of which is $0$ or $1$. The distance between two binary strings of the same length is the number of positions in which they disagree; for example, the distance between the strings $01101011$ and $00101110$ is $3$ since they differ in the second, s... | [
"Solution:\nThe maximum possible number of such strings is $20$. An example of a set attaining this bound is\n\n| 00000000 | 00110101 |\n| :--- | :--- |\n| 11001010 | 10011110 |\n| 11100001 | 01101011 |\n| 11010100 | 01100110 |\n| 10111001 | 10010011 |\n| 01111100 | 11001101 |\n| 00111010 | 10101100 |\n| 01010111 |... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | final answer only | 20 | |
0cqj | A collection of $n$ real numbers is written on the blackboard. It appears that the square of each written number is greater than the product of any two other written numbers. Find the greatest possible value of $n$. (I. Bogdanov)
На доске написаны несколько чисел. Известно, что квадрат любого записанного числа больше ... | [
"$n = 3$.\n\nSuppose there are at least four numbers, and let $a$ be the number with the smallest absolute value. Among the remaining numbers, at least two have the same sign (both nonnegative or both nonpositive). Denote them by $b$ and $c$; then $bc = |bc| \\geq |a|^2 = a^2$, which contradicts the condition.\n\nI... | Russia | Russian mathematical olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English; Russian | proof and answer | 3 | |
0hxp | Problem:
Let $S$ be the locus of all points $(x, y)$ in the first quadrant such that $\frac{x}{t}+\frac{y}{1-t}=1$ for some $t$ with $0<t<1$. Find the area of $S$. | [
"Solution:\n\nSolving for $t$ in the given equation, we get $t^{2}+(y-x-1) t+x=0$. Using the quadratic equation, we get\n$$\nt=\\frac{(x+1-y) \\pm \\sqrt{(y-x-1)^{2}-4 x}}{2}.\n$$\nFor all valid combinations of $(x, y)$, $t$ is positive and less than $1$ (this is easy to see by inspection). All valid combinations o... | United States | HMMT | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 1/6 | |
050u | An equilateral triangle with side length $3$ is divided into $9$ equilateral triangles with side length $1$. An integer from $1$ to $10$ is written into every point that is a vertex of a small triangle (colored vertices on the figure), such that all numbers are written exactly once. For every small triangle, the sum of... | [
"In a triangle, which has $10$ at one vertex, the sum is at least $13$. If $10$ is not at one of the vertices of the large triangle, the number of triangles with sum greater than $12$ is at least $3$ and the problem is solved. If $10$ is at the vertex of the large triangle, then look, where is the number $9$. If $9... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0207 | Problem:
Let $\triangle ABC$ be a triangle with circumcircle $\Gamma$, and let $I$ be the center of the incircle of $\triangle ABC$. The lines $AI$, $BI$ and $CI$ intersect $\Gamma$ in $D \neq A$, $E \neq B$ and $F \neq C$. The tangent lines to $\Gamma$ in $F$, $D$ and $E$ intersect the lines $AI$, $BI$ and $CI$ in $R$... | [
"Solution:\nWe first prove that $|DB| = |DI|$. (This may also be claimed by referring to the lemma that $D$ is the centre of the circumcircle of $BIC I_{a}$.) By the constant angle theorem and the fact that $AD$ and $BE$ are angle bisectors of triangle $ABC$, we see that\n$$\n\\angle DBI = \\angle DBC + \\angle CBI... | Benelux Mathematical Olympiad | 5th Benelux Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0c1l | The rhombus $ABCD$ has $m(\widehat{BAD}) = 30^\circ$, and the bisector of the angle $\angle ADB$ intersects the side $[AB]$ in the point $E$. Find $m(\widehat{DEC})$. | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordi... | null | proof and answer | 45° | |
07hq | A polynomial $S(x) \in \mathbb{R}[x]$ is "simple" if it is divisible by $x$ but not by $x^2$. For the polynomial $P(x) \in \mathbb{R}[x]$, we know that there exists a simple polynomial $Q(x)$ such that $P(Q(x)) - Q(2x)$ is divisible by $x^2$. Prove that there exists a simple polynomial $R(x)$ such that $P(R(x)) - R(2x)... | [
"We prove this statement by induction. The base is clear, by the induction hypothesis assumes that\n$$\nx^{n-1}|P(R_{n-1}(x)) - R_{n-1}(2x)\n$$\nif it is also divisible by $x^n$ then we are done. Otherwise, set $R_n(x) = R_{n-1}(x) + a x^{n-1}$. Then $R_n(x) \\equiv R_{n-1}(x) \\pmod{x^{n-1}}$ and so we have\n$$\nx... | Iran | 40th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
04qi | Denote by $\mathbb{N}$ the set of all positive integers. Let $f: \mathbb{N} \to \mathbb{N}$ be a *multiplicative* function such that $f(4) = 4$ and
$$
f(m^2 + n^2) = f(m^2) + f(n^2)
$$
holds for all positive integers $m$ and $n$. Prove that $f(m^2) = m^2$ holds for all positive integers $m$.
(A function is called *mult... | [
"We easily get $f(1) = 1$, $f(2) = f(1) + f(1) = 2$, $f(5) = f(4) + f(1) = 5$, $f(20) = f(4)f(5) = 20$ and $f(16) = f(20) - f(4) = 16$.\nWe prove the claim by complete mathematical induction, assuming it holds for all positive integers $m \\le n$ ($P(m)$), and performing the induction step for $n + 1$:\n\n1) If $n$... | Croatia | Croatian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Other"
] | English | proof only | null | |
00kv | We consider the following operation applied to a positive integer: The integer is represented in an arbitrary base $b \ge 2$, in which it has exactly two digits and in which both digits are different from $0$. Then the two digits are swapped and the result in base $b$ is the new number.
Is it possible to transform ever... | [
"We show that each number $> 10$ can be transformed to a smaller number. In that way, we will eventually reach a number $\\le 10$.\n\nIf the number $n = 2k + 1$ is odd, we choose base $b = k$ with $n = (21)_k$. Swapping the two digits, we obtain the new number $(12)_k = k + 2$. Since $k \\ge 5$, the choice of $b = ... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Yes | |
07cl | $k$, $n$ are two arbitrary positive integers. Prove that there exists at least $(k-1)(n-k+1)$ positive integers that can be produced by $n$ number of $k$'s and using only $+$, $-$, $\times$, $\div$ operations and adding parentheses between them, but cannot be produced using $n-1$ number of $k$'s. | [
"Consider numbers of the form $k^m(k^i + j)$ for $i \\in \\{1, \\dots, n-k+1\\}$, $j \\in \\{1, \\dots, k-1\\}$ and $m \\ge 0$.\nNote that for such $i$ and $j$ in these intervals, we have $i + j \\le n$. Based on $n - (i + j)$ is an odd number or an even number, we have two cases.\n\n* If $n - (i + j)$ is odd, ther... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0dwi | Problem:
Osnovna ploskev pokončne prizme je deltoid, ki ima krajšo diagonalo dolgo $e$. Notranja kota deltoida z vrhoma v krajiščih daljše diagonale merita $90^{\circ}$ in $60^{\circ}$. Višina prizme je enaka daljši diagonali deltoida. Izrazi prost notino prizme z $e$. Rezultat naj bo točen. | [
"Solution:\n\nProstornina prizme je $V = \\frac{e \\cdot f}{2} \\cdot f$. Daljša diagonala je razdeljena na dela $f_1$ in $f_2$. Oba dela izrazimo z dolžino $e$ krajše diagonale. Pravokotni trikotnik $DBC$ je enakokrak, zato je $f_2 = \\frac{e}{2}$. Trikotnik $BDA$ je enakostraničen, zato je $f_1 = \\frac{e \\sqrt{... | Slovenia | 4. državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | V = \frac{e^3(2+\sqrt{3})}{4} | |
0c5s | Find all functions $f : \mathbb{R} \to (0, \infty)$, such that
$$
2^{-x-y} \le \frac{f(x)f(y)}{(x^2+1)(y^2+1)} \le \frac{f(x+y)}{(x+y)^2+1},
$$
for all $x, y \in \mathbb{R}$. | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = (x^2 + 1) 2^{-x} | |
0616 | Problem:
Zwei Sehnen $AC$ und $BD$ eines Kreises mit Mittelpunkt $M$ schneiden sich in $P$. Die Umkreise der Dreiecke $PBC$ und $PDA$ haben ihre Mittelpunkte in $E$ bzw. $F$ und schneiden sich ein zweites Mal in $Q$.
Man beweise, dass $\overline{MF} = \overline{QE}$ gilt. | [] | Germany | Auswahlwettbewerb zur IMO 2000 | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03lg | Problem:
Let $A, B, C, D$ be four points on a circle (occurring in clockwise order), with $A B < A D$ and $B C > C D$. Let the bisector of angle $B A D$ meet the circle at $X$ and the bisector of angle $B C D$ meet the circle at $Y$. Consider the hexagon formed by these six points on the circle. If four of the six side... | [
"Solution:\nWe're given that $A B < A D$. Since $C Y$ bisects $\\measuredangle B C D$, $B Y = Y D$, so $Y$ lies between $D$ and $A$ on the circle, as in the diagram above, and $D Y > Y A$, $D Y > A B$. Similar reasoning confirms that $X$ lies between $B$ and $C$ and $B X > X C$, $B X > C D$. So if $A B X C D Y$ has... | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
07vw | The triangle $ABC$ has circumcentre $O$ and circumcircle $\Gamma$. Let $AI$ be a diameter of $\Gamma$. The ray $AI$ extends to intersect the circumcircle $\omega$ of $\angle BOC$ for the second time at a point $P$.
Let $AD$ and $IQ$ be perpendicular to $BC$, with $D$ and $Q$ on $BC$. Let $M$ be the midpoint of $BC$.
1... | [
"a.\nWe denote the angles of $\\triangle ABC$ by $\\angle A$, $\\angle B$, $\\angle C$. Since $AI$ is a diameter, we have $\\angle ACI = \\angle ABI = 90^\\circ$. Also, $\\angle AIC = \\angle B$ (both standing on arc $AC$), and $\\angle CAI = \\angle CBI$ (both standing on arc $CI$). Therefore,\n\n![](attached_imag... | Ireland | IRL_ABooklet_2023 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > ... | English | proof only | null | |
0bua | Problem:
Se dau numerele $a, b, c \in \mathbb{C}$. Calculați, scriind rezultatul sub formă de produs:
a) $\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3}\end{array}\right|$
b) $\left|\begin{array}{ccc}a+b & b+c & c+a \\ a-b & b-c & c-a \\ a^{2}-b^{2} & b^{2}-c^{2} & c^{2}-a^{2}\end{array}\rig... | [] | Romania | Olimpiada Națională de Matematică | [
"Algebra > Linear Algebra > Determinants",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | a) (a+b+c)(a-b)(b-c)(c-a)
b) 2(a+b+c)(a-b)(b-c)(c-a) | |
0djh | Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that $f(0) = 0$ and
$$
2f\left(-\frac{1}{2}xy + f(x + y)\right) = x f(y) + y f(x)
$$
for all $x, y \in \mathbb{R}$. Prove that $f(2) = f(-2) = 0$. | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof only | null | |
03ft | Is it true that for any positive integer $n > 1$, there exists an infinite arithmetic progression $M_n$ of positive integers, such that for any $m \in M_n$, the number $n^m - 1$ is not a perfect power (a positive integer is a perfect power if it is of the form $a^b$ for positive integers $a, b > 1$)? | [
"(Victor Kostadinov) The answer is yes. Fix a positive integer $n$ and two large distinct primes $p, q > n$. Let $d_p = \\text{ord}_p(n)$, $d_q = \\text{ord}_q(n)$, $c_p = \\nu_p(n^{d_p} - 1)$, $c_q = \\nu_q(n^{d_q} - 1)$ and let $c = \\nu_p(d_q)$, $d = \\nu_q(d_p)$. Pick two sufficiently large constants $a, b$, an... | Bulgaria | Bulgarian National Olympiad - Final Round | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | Yes | |
0e7a | Find all functions $f: \mathbb{R} \setminus \{-1\} \to \mathbb{R}$ such that
$$
f(x) + f(y) = (x + y + 2)f(x)f(y)
$$
for all $x, y \in \mathbb{R} \setminus \{-1\}$. | [
"Plug in $x = y = 0$ to find $f(0) = f(0)^2$. So, either $f(0) = 0$ or $f(0) = 1$.\n\nIf $f(0) = 0$, then plug in $y = 0$ to see that $f(x) = 0$ for all $x$. Obviously, this function satisfies the conditions of the problem.\n\nIf, on the other hand, $f(0) = 1$, plug in $y = 0$ to see that $f(x) = \\frac{1}{x+1}$. I... | Slovenia | National Math Olympiad 2013 - Final Round | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 0 for all x, or f(x) = 1/(x+1) for all x ≠ −1 | |
0fsj | Problem:
Ein Quadrat ist in Rechtecke zerlegt, deren Seiten parallel zu den Quadratseiten liegen. Für jedes dieser Rechtecke wird das Verhältnis seiner kürzeren Seite zu seiner längeren gebildet. Zeige, dass die Summe dieser Verhältnisse mindestens 1 beträgt. | [
"Solution:\n\nDas Quadrat habe Seitenlänge $s$ und sei in die Rechtecke $R_{i}$ mit Seitenlängen $a_{i} \\leq b_{i}$ zerlegt. Die Summe der Flächen aller Rechtecke $R_{i}$ ist gleich der Fläche des Quadrates, also gleich $s^{2}$. Dies ergibt\n$$\ns^{2} = \\sum_{i} a_{i} \\cdot b_{i} = \\sum_{i} \\frac{a_{i}}{b_{i}}... | Switzerland | IMO - Selektion | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0kir | Problem:
Let $ABC$ be a right triangle with $\angle A = 90^{\circ}$. A circle $\omega$ centered on $BC$ is tangent to $AB$ at $D$ and $AC$ at $E$. Let $F$ and $G$ be the intersections of $\omega$ and $BC$ so that $F$ lies between $B$ and $G$. If lines $DG$ and $EF$ intersect at $X$, show that $AX = AD$. | [
"Solution:\n\nIn all solutions, let $O$ be the center of $\\omega$. Then $\\angle DOE = 90^{\\circ}$, so $\\angle DFE = 45^{\\circ}$, so $\\angle DXE = 135^{\\circ}$. Let $\\Gamma$ be the circle centered at $A$ with radius $AD = AE$, and let $X' = \\overrightarrow{AX} \\cap \\Gamma$. Then $\\angle DXE = \\angle DX'... | United States | HMMT Spring 2021 Team Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Ad... | null | proof only | null |
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