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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
02n7 | Problem:
Idades múltiplas - Quando Isabel nasceu, sua mãe estava fazendo aniversário de 20 anos. Se Isabel e sua mãe viverem mais 100 anos, quantas vezes terão sido múltiplas as idades das duas? | [
"Solution:\n\nQuando Isabel tem $a$ anos, sua mãe tem $20+a$ anos. Se $a$ é divisor de $20+a$, então $(20+a)/a = (20/a) + 1$ é um número inteiro e, consequentemente, $20/a$ também é inteiro. Então, $a$ é um divisor de $20$ e, portanto, $a$ pode ser $1$, $2$, $4$, $5$, $10$ ou $20$. Assim, temos um total de $6$ veze... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 6 | |
02n9 | Problem:
Ana tem um cubo de $10~\mathrm{cm}$ de lado. Ela cortou o cubo em cubinhos de $1~\mathrm{cm}$ de lado e, com esses cubinhos, ela brinca de formar outros blocos retangulares, mas sem que sobrem cubinhos. Por exemplo, ela formou um bloco de $10 \times 20 \times 5$. No total, quantos blocos diferentes ela pode c... | [] | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 19 | |
0jxw | Problem:
On a $3 \times 3$ chessboard, each square contains a Chinese knight with $\frac{1}{2}$ probability. What is the probability that there are two Chinese knights that can attack each other? (In Chinese chess, a Chinese knight can attack any piece which is two squares away from it in a particular direction and on... | [
"Solution:\n\nSuppose the $3 \\times 3$ square is\n\n| A | B | C |\n| :---: | :---: | :---: |\n| D | E | F |\n| G | H | I |\n\nWe count the number of ways a board could have two knights attack each other using PIE. First notice that in any setup with two knights attack each other, the center square must be empty. A... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 79/256 | |
0gfd | 證明不等式
$$
2(a^3 + b^3 + c^3) + abc + ab + bc + ca \ge 2(a^2b + b^2c + c^2a) + a^2 + b^2 + c^2 + a + b + c
$$
對任意非負實數 $a$, $b$, $c$ 恆成立。 | [] | Taiwan | 2022 數學奧林匹亞競賽第一階段培訓營, 獨立研究(一) | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | Chinese; English | proof only | null | |
0bg5 | Consider $ABCD$ a rectangle of center $O$ with $AB \ne BC$. The perpendicular dropped from $O$ to $BD$ intersects lines $AB$ and $BC$ in $E$ and $F$. Let $M$ and $N$ be the midpoints of segments $[CD]$ and $[AD]$. Prove that $FM \perp EN$. | [
"Let $P$ be the midpoint of $[BC]$ and $Q$ be the intersection point of $EO$ and $CD$. As $[PM]$ is the midsegment of the triangle $BCD$, it results that $PM \\parallel BD$ and $OQ \\perp BD$, so $QF \\perp PM$. But $PC \\perp MQ$, so $F$ is the orthocenter of the triangle $MPQ$, which leads to $PQ \\perp MF$. Sinc... | Romania | The Danube Mathematical Competition | [
"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 | |
0jx0 | Problem:
Let $a$ and $b$ be complex numbers satisfying the two equations
$$
\begin{aligned}
a^{3}-3 a b^{2} & =36 \\
b^{3}-3 b a^{2} & =28 i
\end{aligned}
$$
Let $M$ be the maximum possible magnitude of $a$. Find all $a$ such that $|a|=M$. | [
"Solution:\nNotice that\n$$\n\\begin{aligned}\n(a-b i)^{3} & =a^{3}-3 a^{2} b i-3 a b^{2}+b^{3} i \\\\\n& =\\left(a^{3}-3 a b^{2}\\right)+\\left(b^{3}-3 b a^{2}\\right) i \\\\\n& =36+i(28 i) \\\\\n& =8\n\\end{aligned}\n$$\nso that $a-b i=2+i$. Additionally\n$$\n\\begin{aligned}\n(a+b i)^{3} & =a^{3}+3 a^{2} b i-3 a... | United States | February 2017 | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof and answer | a ∈ {3, -3/2 + (3√3/2)i, -3/2 - (3√3/2)i} | |
0jar | Problem:
Alice and Bob are playing a game of Token Tag, played on an $8 \times 8$ chessboard. At the beginning of the game, Bob places a token for each player on the board. After this, in every round, Alice moves her token, then Bob moves his token. If at any point in a round the two tokens are on the same square, Ali... | [
"Solution:\n\na. Color the checkerboard in the standard way so that half of the squares are black and the other half are white. Bob's winning strategy is to place the two coins on the same color, so that Alice must always move her coin on to a square with the opposite color as the square containing Bob's coin.\n\nb... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
00mi | Let $\alpha$ be an arbitrary positive real number. Determine for this number $\alpha$ the greatest real number $C$ such that the inequality
$$
\left(1 + \frac{\alpha}{x^2}\right) \left(1 + \frac{\alpha}{y^2}\right) \left(1 + \frac{\alpha}{z^2}\right) \geq C \cdot \left(\frac{x}{z} + \frac{z}{x} + 2\right)
$$
is valid f... | [
"By replacing $\\alpha$ by $xy + yz + zx$ and clearing fractions we get the equivalent inequality\n$$\n(x^2 + xy + xz + yz)(y^2 + yx + yz + xz)(z^2 + zx + zy + xy) \\geq Cxy^2z(x^2 + z^2 + 2xz).\n$$\nThis inequality is homogeneous of degree 6, thus no further constraint has to be considered. As each of the three fa... | Austria | 49th Austrian Mathematical Olympiad, National Competition (Final Round, part 1) | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | C = 16; equality when x = y = z = sqrt(alpha/3). | |
0dbf | Let $P(x)$ be a polynomial of degree $n \geq 2$ with rational coefficients such that $P(x)$ has $n$ pairwise different real roots forming an arithmetic progression. Prove that among the roots of $P(x)$ there are two that are also the roots of some polynomial of degree 2 with rational coefficients. | [
"Denote $x_{1} < x_{2} < \\ldots < x_{n}$ as the roots of $P(x)$. Let $d = x_{n} - x_{n-1} = \\cdots = x_{2} - x_{1} > 0$. Since $P(x)$ has rational coefficients then by applying Vieta's theorem, we have\n$$\n\\sum_{i=1}^{n} x_{i} \\in \\mathbb{Q} \\text{ and } \\sum_{1 \\leq i < j \\leq n} x_{i} x_{j} \\in \\mathb... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
06fe | Show that the product of three consecutive integers is the sum of four integer cubes. | [
"Note that the product of three consecutive integers is divisible by $6$. Let the product be $6k$ for some $k \\in \\mathbb{Z}$. Then we easily check that\n$$\n(k + 1)^3 + (k - 1)^3 + (-k)^3 + (-k)^3 = 6k.\n$$\nThis completes the proof."
] | Hong Kong | Year 2008 | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
03qe | Let $EFGH$, $ADC$ and $E_1F_1G_1H_1$ be three convex quadrilaterals, satisfying:
(a) Points $E$, $F$, $G$ and $H$ lie on sides $AB$, $BC$, $CD$ and $DA$, respectively, and $\frac{AE}{EB} \cdot \frac{BF}{FC} \cdot \frac{CG}{GD} \cdot \frac{DH}{HA} = 1$;
(b) points $A$, $B$, $C$ and $D$ lie on sides $H_1E_1$, $E_1F_1$, $... | [
"(1) If $EF \\parallel AC$, then $\\frac{BE}{EA} = \\frac{BF}{FC}$.\nSo $\\frac{DH}{HA} = \\frac{DG}{GC}$ using condition (a). Then $HG \\parallel AC$, giving $E_1F_1 \\parallel AC \\parallel H_1G_1$. That\n\nmeans $\\frac{F_1C}{CG_1} = \\frac{E_1A}{AH_1} = \\lambda$.\n\n(2) If $EF$ is not ... | China | China Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Quadrilaterals"
] | English | proof and answer | lambda | |
0kw5 | Problem:
Let $ABC$ be a triangle with $\angle BAC > 90^{\circ}$. Let $D$ be the foot of the perpendicular from $A$ to side $BC$. Let $M$ and $N$ be the midpoints of segments $BC$ and $BD$, respectively. Suppose that $AC = 2$, $\angle BAN = \angle MAC$, and $AB \cdot BC = AM$. Compute the distance from $B$ to line $AM$... | [
"Solution:\n\n\n\nExtend $AM$ to meet the circumcircle of $\\triangle ABC$ at $X$. Then, we have $\\triangle ABM \\sim \\triangle CXM$, which implies that $\\frac{CX}{CM} = \\frac{AB}{AM}$. Using the condition $AB \\cdot BC = AM$, we get that $CX = \\frac{1}{2}$.\n\nNow, the key observation... | United States | HMMT February 2023 | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(285) / 38 | |
06nj | There are 20 students in a class, numbered $1$ to $20$. They have taken a test with $n$ questions. After the test, the teacher tried to pick some students to draft the solutions, and as such it is necessary that the picked students together can solve all questions. The teacher found that this can be only done by choosi... | [
"Answer: $18$\n\nFor convenience we shall call 'a student whose class number is even' an even student. We make the following observations.\n\n(1) Every question is solved by at least one even student (as the teacher could choose all even students).\n\n(2) Similarly, every question is solved by at least one of stude... | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 18 | |
0l0n | Problem:
Compute $\sqrt[4]{5508^{3}+5625^{3}+5742^{3}}$, given that it is an integer. | [
"Solution:\nLet $a=5625=75^{2}$ and $b=117$. Then we have\n$$\n5508^{3}+5625^{3}+5742^{3}=(a-b)^{3}+a^{3}+(a+b)^{3}=3 a^{3}+6 a b^{2}=3 a\\left(a^{2}+2 b^{2}\\right)\n$$\nWe have $3 a=3^{3} \\cdot 5^{4}$, so $a^{2}+2 b^{2}=3^{4} \\cdot\\left(625^{2}+2 \\cdot 19^{2}\\right)$ should be 3 times a fourth power. This me... | United States | HMMT February 2024 Guts Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 855 | |
0gm2 | Find all functions $f : \mathbb{Q}^+ \to \mathbb{Q}^+$ such that
$$
f(xf(x) + f(y)) = (f(x))^2 + y
$$
for all positive rationals $x, y$. | [] | Thailand | The 14th Thailand Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | f(x) = x for all positive rationals x | |
0aae | Problem:
Let $A$, $B$, $C$ and $D$ be points on the circle $\omega$ such that $ABCD$ is a convex quadrilateral. Suppose that $AB$ and $CD$ intersect at a point $E$ such that $A$ is between $B$ and $E$ and that $BD$ and $AC$ intersect at a point $F$. Let $X \neq D$ be the point on $\omega$ such that $DX$ and $EF$ are p... | [
"Solution:\n\nIt can be difficult to find out what to do, but the key is to show that $AYFE$ is cyclic. The motivation for this is that we want to show that $\\angle BAY = \\angle BAX$. But $\\angle BAX = \\angle BDX$ since $BXDA$ is cyclic, and $\\angle BDX = \\angle FDX = \\angle DFE$ since $DX$ and $EF$ are para... | Nordic Mathematical Olympiad | The 35th Nordic Mathematical Contest | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0c9v | Problem:
În triunghiul ascuţitunghic $ABC$ se notează cu $O$ centrul cercului circumscris şi cu $D$ piciorul înălţimii din $A$.
Fie $M, N, P, Q$ mijloacele segmentelor $AB, AC, BD$, respectiv $CD$. Arătaţi că unul dintre punctele de intersectie ale cercurilor circumscrise triunghiurilor $AMN$ şi $POQ$ este situat pe î... | [] | Romania | Al treilea test de selecţie pentru OBMJ Bucureşti | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscella... | null | proof only | null | |
0b7n | Let $\mathcal{L}$ be a finite collection of lines in the plane in general position (no two lines in $\mathcal{L}$ are parallel and no three are concurrent). Consider the open circular discs inscribed in the triangles enclosed by each triple of lines in $\mathcal{L}$. Determine the number of such discs intersected by no... | [
"The complement of the union of all lines in $\\mathcal{L}$ is the disjoint union of\n$$\n\\begin{pmatrix} |\\mathcal{L}| \\\\ 0 \\end{pmatrix} + \\begin{pmatrix} |\\mathcal{L}| \\\\ 1 \\end{pmatrix} + \\begin{pmatrix} |\\mathcal{L}| \\\\ 2 \\end{pmatrix}\n$$\nopen convex sets called *rooms*, of which exactly\n$$\n... | Romania | NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | C(|L|-1, 2) | |
0hgo | Does there exist a trinomial $f(x) = ax^2 + bx + c$ with integer coefficients such that $a$ isn't divisible by $2022$ and all numbers $f(1), f(2), \ldots, f(2022)$ give different remainders under the division by $2022$? | [
"Consider the following trinomial:\n$$\nf(x) = 1011x^2 + 1012x = 1011x(x + 1) + x.\n$$\nThe first term is always divisible by $2022$, so $f(x)$ gives remainder $x$. Proof completed."
] | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
08rj | We can make a cube by connecting every two centroids on neighboring faces of a regular octahedron. How many times is the cube as large as the octahedron? | [
"Consider a regular octahedron $A$-$BCDE$-$F$. Denote by $\\pi_1$, $\\pi_2$ and $\\pi_3$ the planes passing through the diagonals $BD$ and $CE$, the diagonals $CE$ and $AF$, and the diagonals $BD$ and $EF$, respectively.\nBy symmetry, $\\pi_1$, $\\pi_2$ and $\\pi_3$ divide the octahedron into 8 equal parts. The oct... | Japan | The 4th Japanese Junior Mathematical Olympiad | [
"Geometry > Solid Geometry > Volume"
] | English | proof and answer | 2/9 | |
0500 | Find all pairs $(m, n)$ of positive integers for which the $m \times n$ grid contains exactly $225$ rectangles whose side lengths are odd and whose edges lie on the lines of the grid. | [
"The $m \\times n$ grid is formed by $m+1$ horizontal and $n+1$ vertical lines. Number the horizontal lines with numbers from $1$ to $m+1$ and the vertical lines with numbers from $1$ to $n+1$. Rectangles with odd side lengths arise if and only if two horizontal lines with different parity and two vertical lines wi... | Estonia | Selected Problems from Open Contests | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | (1,29), (29,1), (5,9), (9,5) | |
0652 | Circles $(K_1, r_1)$ and $(K_2, r_2)$ are tangent and also they are tangent to the sides of the rectangular $AB\Gamma\Delta$, with $AB = a$ and $B\Gamma = b$.
(i) Express the sides $a$ and $b$ with respect to the radius $r_1$ and $r_2$.
(ii) If the common internal tangent of the two circles passes through $\Delta$ comp... | [
"(i) We draw the line segment $K_1H \\perp K_2Z$.\nIn the triangle $K_1HK_2$ we have $K_1K_2 = r_1 + r_2$ and $K_2H = r_2 - r_1$ and hence from Pythagorean theorem we find $K_1H = 2\\sqrt{r_1r_2}$. From the rectangular $K_1EZH$ we have $EZ = 2\\sqrt{r_1r_2}$. Moreover we have\n$$\nAB = AE+EZ+ZB \\Leftrightarrow a =... | Greece | Selection Examination for Juniors | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | (i) a = (sqrt(r1) + sqrt(r2))^2, b = 2*r2. (ii) r1/r2 = (3 - sqrt(5))/2 and ΔK = ((1 + sqrt(5))/2)*r2. | |
08wo | Determine all the positive integers $n$ for which the product of all positive factors equals $24^{240}$. | [
"If a positive integer $l$ is a factor of a positive integer $n$, then $n/l$ is also a factor of $n$ and vice versa. Hence, if $\\{a_1, a_2, \\dots, a_r\\}$ is a listing of all the positive factors of $n$, then so is $\\{n/a_1, n/a_2, \\dots, n/a_r\\}$. So, if we denote by $N$ the product of all the positive factor... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | n = 2^15 · 3^5 | |
06c9 | $I$ is the incentre of $\triangle ABC$. $P, Q, R$ are the circumcentres of the triangles $IBC, ICA$ and $IAB$ respectively. Prove that the points $A, R, B, P, C, Q$ are concyclic. | [
"Since $\\angle BIC = 90^\\circ + \\frac{A}{2}$, the circumcentre $P$ of $\\triangle BIC$ must lie on the different side of $BC$ as $A$ and $I$. As\n$$\n\\angle BPC = 360^\\circ - 2\\angle BIC = 360^\\circ - 2\\left(90^\\circ + \\frac{A}{2}\\right) = 180^\\circ - A,\n$$\nthe point $P$ lies on $(ABC)$. By symmetry, ... | Hong Kong | HKG TST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
02xf | Problem:
Os 2.019 armários dos 2.019 alunos de uma escola são numerados com os quadrados dos 2.019 primeiros naturais positivos, ou seja, o primeiro armário tem o número $1^{2}=1$, o segundo armário tem o número $2^{2}=4$, o terceiro armário tem o número $3^{2}=9$, e assim até o último armário que tem o número $2.019^... | [
"Solution:\n\na) Vamos dividir em grupos pela quantidade de algarismos:\nI) 1 algarismo: 3 armários $\\left(1^{2}, 2^{2}, 3^{2}\\right)$;\nII) 2 algarismos: $9-3=6$ armários ( $4^{2}$ a $9^{2}$ );\nIII) 3 algarismos: $31-9=22$ armários $\\left(10^{2}\\right.$ a $\\left.31^{2}\\right)$;\nIV) 4 algarismos: $99-31=68$... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | final answer only | a) 358; b) 0 | |
0g1b | Problem:
Sei $n \geq 3$ eine natürliche Zahl. Wie viele Diagonalen eines regulären $n$-Ecks kann man maximal einzeichnen, sodass falls sich zwei Diagonalen im Innern schneiden, sie senkrecht aufeinander stehen? | [
"Solution:\n\nWir unterscheiden zwei Fälle:\n\nn ungerade:\nWir zeigen zuerst, dass es keine zwei Diagonalen gibt, die rechtwinklig aufeinander stehen. Da das regelmäßige $n$-Eck in einen Kreis einbeschrieben werden kann, ist aufgrund des Peripheriesatzes und der Symmetrie der Winkel von jedem Punkt zu jeder (nicht... | Switzerland | IMO-Selektion | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | If n is odd, the maximum is n−3; if n is even, the maximum is n−2. | |
0anv | Problem:
Find the range of
$$
f(x) = \frac{(x-a)(x-b)}{(c-a)(c-b)} + \frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-c)(b-a)}
$$
where $a$, $b$, $c$ are distinct real numbers.
(a) all real numbers
(b) $\{1\}$
(c) $[-a-b-c, +\infty)$
(d) $\{a+b+c\}$ | [] | Philippines | QUALIFYING STAGE | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | null | MCQ | b | |
0ggv | 平面上有 2022 個相異格子點。令 $I$ 為這些點中，有幾對點之間的距離恰為 1 的點對數量。求 $I$ 的最大可能值。
註：格子點為 $x$ 座標與 $y$ 座標皆為整數的點。 | [
"最大可能的 $I$ 是 3954；此最大值會在 $45 \\times 45$ 的點陣扣除三個角落時達到。\n\n對於任何 2022 個相異格子點，假設其共有 $n$ 個 $x$ 座標 $x_1, x_2, \\dots, x_n$，且 $x$ 座標為 $x_i$ 的點共有 $a_i$ 個。注意到 $x = x_i$ 直排上的點，最多只能造出 $a_i - 1$ 個長度為 1 且垂直 $x$ 軸的線段，故垂直 $x$ 軸且長度為 1 的線段總量 $\\le \\sum_{i=1}^n (a_i - 1) = 2022 - n$。\n\n同理，若這些點共有 $m$ 個不同的 $y$ 座標，則垂直 $y$ 軸且長度為 1 的線... | Taiwan | 2022 數學奧林匹亞競賽第二階段選訓營, 獨立研究 (二) | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | Chinese; English | proof and answer | 3954 | |
0gj4 | 設 $a_1, a_2, \dots, a_{123}$ 為滿足下列條件的正整數：
• $a_1, a_2, \dots, a_{123}$ 是 $1, 2, \dots, 123$ 的一種排列
• $|a_1 - a_2|, |a_2 - a_3|, \dots, |a_{122} - a_{123}|$ 是 $1, 2, \dots, 122$ 的一種排列
證明 $\max(a_1, a_{123}) \ge 32$。
Let $a_1, a_2, \dots, a_{123}$ be positive integers such that
* $a_1, a_2, \dots, a_{123}$ is a permutati... | [
"考慮以下的一般命題：\n設 $N$ 為正整數, 且 $a_1, a_2, \\dots, a_{2N-1}$ 為滿足下列條件的正整數：\n• $a_1, a_2, \\dots, a_{2N-1}$ 是 $1, 2, \\dots, 2N-1$ 的一種排列\n• $|a_1 - a_2|, |a_2 - a_3|, \\dots, |a_{2N-2} - a_{2N-1}|$ 是 $1, 2, \\dots, 2N-2$ 的一種排列\n則 $a_1 + a_{2N-1} \\ge N + 1$, 推得 $\\max(a_1, a_{2N-1}) \\ge \\lfloor \\frac{N+1}{2} \\rfloor$。... | Taiwan | IMO 3J, Independent Study 1 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Chinese; English | proof only | null | |
03xn | Let $A = \{a_1, a_2, \ldots, a_{2010}\}$ and $B = \{b_1, b_2, \ldots, b_{2010}\}$ be two sets of complex numbers, such that the equality
$$
\sum_{1 \le i < j \le 2010} (a_i + a_j)^n = \sum_{1 \le i < j \le 2010} (b_i + b_j)^n
$$
holds for every $n = 1, 2, \ldots, 2010$. Prove that $A = B$. | [
"**Proof** Let $S_k = \\sum_{i=1}^{2010} a_i^k$ and $\\tilde{S}_k = \\sum_{i=1}^{2010} b_i^k$. We first show by induction that $S_k = \\tilde{S}_k$ for $k = 1, 2, \\ldots, 2010$.\n\nSetting $n = 1$ in the given equality, we have $2009S_1 = 2009\\tilde{S}_1$, and hence $S_1 = \\tilde{S}_1$. Assume that $S_j = \\tild... | China | China National Team Selection Test | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0e84 | Problem:
V splošni obliki zapiši predpise za vse polinome četrte stopnje, ki imajo ničle $-2, 0, 1$ (ena izmed njih je dvojna) in njihovi grafi potekajo skozi točko $M(-1, 5)$. | [] | Slovenia | Državno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | The three polynomials are:
1) -(5/2)x^4 -(5/2)x^3 + 5x^2
2) (5/2)x^4 + (15/2)x^3 - 10x
3) -(5/4)x^4 + (15/4)x^2 - (5/2)x | |
0f3d | Problem:
Given $1980$ vectors in the plane. The sum of every $1979$ vectors is a multiple of the other vector. Not all the vectors are multiples of each other. Show that the sum of all the vectors is zero. | [
"Solution:\n\nLet the vectors be $\\mathbf{x}_i$ and their sum $\\mathbf{s}$. Then we have $\\mathbf{s} - \\mathbf{x}_i = n_i \\mathbf{x}_i$ for some scalar $n_i$. Hence $(n_i + 1) \\mathbf{x}_i = \\mathbf{s}$. If $\\mathbf{s}$ is nonzero, then it follows that every vector is a multiple of $\\mathbf{s}$ and hence a... | Soviet Union | ASU | [
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
00md | Determine all polynomials $P(x) \in \mathbb{R}[x]$ satisfying the following two conditions:
(a) $P(2017) = 2016$ and
(b) $(P(x)+1)^2 = P(x^2+1)$ for all real numbers $x$. | [
"Letting $Q(x) := P(x) + 1$ we get the two new conditions $Q(2017) = 2017$ and $Q(x^2+1) = Q(x)^2 + 1$, $x \\in \\mathbb{R}$.\n\nWe now define the sequence $\\langle x_n \\rangle_{n \\ge 0}$ recursively by $x_0 = 2017$ and $x_{n+1} = x_n^2 + 1$, $n \\ge 0$. A straightforward induction yields $Q(x_n) = x_n$, $n \\ge... | Austria | 48th Austrian Mathematical Olympiad National Competition (Final Round, part 1) | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials"
] | English | proof and answer | P(x) = x - 1 | |
0alz | Problem:
Find all ordered pairs $(x, y)$ satisfying the system
$$
\begin{aligned}
x^{2} + 4y^{2} - x y &= 10 \\
2x - 4y + 3x y &= 11
\end{aligned}
$$ | [] | Philippines | Area Stage | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | (3, 1), (-2, -3/2) | |
08pg | Problem:
Find all positive integers $n$ such that the number $A_{n} = \frac{2^{4n+2} + 1}{65}$ is
a) an integer;
b) a prime. | [
"Solution:\n\na) Note that $65 = 5 \\cdot 13$.\nObviously, $5 = 2^{2} + 1$ is a divisor of $\\left(2^{2}\\right)^{2n+1} + 1 = 2^{4n+2} + 1$ for any positive integer $n$. Since $2^{12} \\equiv 1 \\pmod{13}$, if $n \\equiv r \\pmod{3}$, then $2^{4n+2} + 1 \\equiv 2^{4r+2} + 1 \\pmod{13}$. Now, $2^{4 \\cdot 0 + 2} + 1... | JBMO | Junior Balkan Mathematics Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | a) A_n is an integer if and only if n ≡ 1 (mod 3).
b) There is no n for which A_n is prime. | |
03yl | If $\min_{x \in \mathbb{R}} \frac{a x^2 + b}{\sqrt{x^2 + 1}} = 3$, find
(1) the range of $b$; | [
"Denote $f(x) = \\frac{a x^2 + b}{\\sqrt{x^2+1}}$. It is easy to see that $a > 0$. By $f(0) = b$, we see that $b \\ge 3$.\n\na. If $b - 2a \\ge 0$,\n$$\nf(x) = \\frac{a x^2 + b}{\\sqrt{x^2 + 1}} = a \\sqrt{x^2 + 1} + \\frac{b-a}{\\sqrt{x^2 + 1}} \\ge 2\\sqrt{a(b-a)} = 3,\n$$\nequality holds if $a \\sqrt{x^2+1} = \\... | China | China Southeastern Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | [3, +infinity) | |
0ag7 | Let $n$ be a natural number. Prove that $13|7^n + n^5$ if and only if $13|n^7 \cdot 7^n + 1$. | [
"Let $13|7^n + n^5$. Clearly $13$ does not divide $n$, so by Little Fermat's theorem we have that $n^{12} \\equiv 1 \\pmod{13}$, i.e. $13|n^{12} - 1$.\n\nNow we have $13|7^n + n^5$, i.e. $13|n^7(7^n + n^5)$ or $13|n^7 7^n + 1 + n^{12} - 1$, and because of $13|n^{12} - 1$ we get $13|n^7 \\cdot 7^n + 1$.\n\nNow let $... | North Macedonia | null | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
07du | 100 points lie on the perimeter of a circle and they are arbitrary labeled by $1, 2, \ldots, 100$. We call a triangle **clockwise** if the increasing order of its vertices is in a clockwise fashion. Prove that it is not possible to have exactly $2017$ clockwise triangles. | [
"Consider an operation in which we replace the labels of two consecutive points. We will focus on the parity of the total number of clockwise triangles.\n\nLet $A, B$ be two consecutive points. Consider a clockwise triangle that none of points $A, B$ are its vertices or exactly one of them is a vertex of this trian... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof only | null | |
03qo | Assume that $M = \{(x, y) \mid x^2 + 2y^2 = 3\}$, and $N = \{(x, y) \mid y = mx + b\}$. If $M \cap N \neq \emptyset$ for all $m \in \mathbb{R}$, then $b$ takes values from ( ).
(A) $\left[-\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}\right]$
(B) $\left(-\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}\right)$
(C) $\left(-\frac{\sqrt{6... | [
"For any $m \\in \\mathbb{R}$ we have $M \\cap N \\neq \\emptyset$, which means point $(0, b)$ is on or in the ellipsoid $\\frac{x^2}{3} + \\frac{2y^2}{3} = 1$. Therefore\n$$\n\\frac{2b^2}{3} \\le 1, \\text{ or } -\\frac{\\sqrt{6}}{2} \\le b \\le \\frac{\\sqrt{6}}{2}.\n$$\nAnswer: A."
] | China | China Mathematical Competition (Hainan) | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | MCQ | A | |
0bvn | Let $P_1P_2\dots P_n$ be a regular $n$-gon. A frog situated at a vertex $P_k$ ($1 \le k \le n$) can jump to one of the vertices $P_{k+2}$ or $P_{k-3}$, the indexes being considered modulo $n$. Determine the set of positive integers $n \ge 3$ for which the frog can make $n$ jumps such that it visits all the vertices of ... | [
"If $n$ has the given property, denote by $a$ the number of jumps of type $P_k \\mapsto P_{k+2}$, and let $b$ be the number of jumps of type $P_k \\mapsto P_{k-3}$. Then $a+b = n$ and $n \\mid 2a-3b$, which leads to $n \\mid 5a$ and $n \\mid 5b$. As $0 \\le a, b \\le n$, we must have either $a = 0$, $b = n$, or $b ... | Romania | Eleventh STARS OF MATHEMATICS Competition | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All integers n at least 3 for which either n is not divisible by 6 or n is divisible by 5; equivalently, all n except multiples of 6 that are not multiples of 5. | |
0j8y | Problem:
Let $ABCDEF$ be a regular hexagon of area $1$. Let $M$ be the midpoint of $DE$. Let $X$ be the intersection of $AC$ and $BM$, let $Y$ be the intersection of $BF$ and $AM$, and let $Z$ be the intersection of $AC$ and $BF$. If $[P]$ denotes the area of polygon $P$ for any polygon $P$ in the plane, evaluate $[BX... | [
"Solution:\n\nLet $O$ be the center of the hexagon. The desired area is $[ABCDEF] - [ACDM] - [BFEM]$. Note that $[ADM] = [ADE]/2 = [ODE] = [ABC]$, where the last equation holds because $\\sin 60^{\\circ} = \\sin 120^{\\circ}$. Thus, $[ACDM] = [ACD] + [ADM] = [ACD] + [ABC] = [ABCD]$, but the area of $ABCD$ is half t... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 0 | |
0fj3 | Problem:
Consideramos el conjunto $N=\{1,2,3, \ldots\}$ de los números naturales y la aplicación $f$ : $N \rightarrow N$ que cumple las dos siguientes condiciones:
a) $f(f(n))=n$ para todo $n \in N$.
b) $f(f(n)+1)=\left\{\begin{array}{cc}n-1, & \text{ si } n \text{ es par;} \\ n+3, & \text{ si } n \text{ es impar. }\en... | [
"Solution:\nPor la condición a), es obvio que $f$ es biyectiva. Y, como por la condición b) se observa que nunca es $f(f(n)+1)=2$, forzosamente tendrá que ser 2 la imagen del único elemento que no es de la forma $f(n)+1$, o sea de 1: $f(1)=2$. Y de nuevo por a), $f(2)=1$.\nVamos a probar, por inducción sobre $n$, q... | Spain | Viernes 19 de enero de 2001 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | f(n) = { n+1 if n is odd; n-1 if n is even } | |
0j20 | Problem:
An ant starts at the point $(1,0)$. Each minute, it walks from its current position to one of the four adjacent lattice points until it reaches a point $(x, y)$ with $|x|+|y| \geq 2$. What is the probability that the ant ends at the point $(1,1)$? | [
"Solution:\n\nFrom the starting point of $(1,0)$, there is a $\\frac{1}{4}$ chance we will go directly to $(1,1)$, a $\\frac{1}{2}$ chance we will end at $(2,0)$ or $(1,-1)$, and a $\\frac{1}{4}$ chance we will go to $(0,0)$. Thus, if $p$ is the probability that we will reach $(1,1)$ from $(0,0)$, then the desired ... | United States | Harvard-MIT November Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 7/24 | |
0dps | Prove that for any positive integer $m$ there exists a positive integer $n$, such that any $n$ different points on a plane can be partitioned into $m$ non-empty sets, *convex hulls* of which would share a common point.
*Convex hull* of a finite set $X$ of points on a plane is a set of points that lie inside or on the ... | [
"Let's remind **Helly's theorem**: if in a finite set of convex sets of points on a plane each three intersect, then all of them intersect.\n\nLet's prove that $n = 9m$ satisfies the problem statement. Let $X$ be an arbitrary set of $9m$ different points on a plane, and $Y$ — the set of subsets of $X$ of size $6m +... | Silk Road Mathematics Competition | XIX Silk Road Mathematical Competition | [
"Geometry > Plane Geometry > Combinatorial Geometry > Helly's theorem",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
09xk | A frog jumps around on the grid points in the plane, from one grid point to another. The frog starts at the point $(0,0)$. Then it makes, successively, a jump of one step horizontally, a jump of $2$ steps vertically, a jump of $3$ steps horizontally, a jump of $4$ steps vertically, et cetera. Determine all $n > 0$ such... | [
"(a) When the frog has finished $n$ jumps, it has made $1+2+3+\\cdots+n = \\frac{1}{2}n(n+1)$ steps in total. To get back at $0$, the frog must make the same number of steps to the left and the right. Thus, the total number of steps must be even. This means that $\\frac{1}{2}n(n+1)$ is even, and hence $n(n+1)$ is a... | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | All n with n ≡ 0 or 7 modulo 8 | |
09jj | Find all positive integer pairs $(n, a)$ such that the quotient
$$
\frac{a^{n+1} + 3^{n+1} + 1}{a^n + 3^n + 1}
$$
is an integer. | [] | Mongolia | Mongolian Mathematical Olympiad Round 2 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (n, a) = (1, 9) and (1, 22) | |
072u | Let $ABC$ be an equilateral triangle, and $D, E, F$ be points on $BC, CA, AB$ respectively. Let $\angle BAD = \alpha$, $\angle CBE = \beta$, $\angle ACF = \gamma$. Suppose $\alpha + \beta + \gamma \ge 120^\circ$. Prove that the union $\Sigma$ of triangular regions $BAD, CBE, ACF$ covers the triangle $ABC$. | [
"Suppose some point $P$ does not lie in any of the triangles $BAD, CBE, ACF$. Then $\\angle BAD < \\angle BAP$, $\\angle CBE < \\angle CBP$ and $\\angle ACF < \\angle ACP$. Thus\n$$\n\\alpha + \\beta + \\gamma < \\angle BAP + \\angle CBP + \\angle ACP.\n$$\nIf $P$ is the circumcentre of $ABC$, then $\\angle BAP = \... | India | Indija TS 2006 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ej1 | Problem:
Pravilna, pokončna, štiristrana prizma ima površino $128~\mathrm{m}^2$ in višino $6~\mathrm{m}$. Koliko meri njena telesna diagonala?
(A) $D \doteq 8,23~\mathrm{m}$
(B) $D \doteq 8,24~\mathrm{m}$
(C) $D \doteq 7,22~\mathrm{m}$
(D) $D \doteq 8,25~\mathrm{m}$
(E) $D \doteq 7,21~\mathrm{m}$ | [
"Solution:\n\nS pomočjo obrazca za izračun površine pravilne štiristrane prizme $P=2 a^{2}+4 a v$, dobimo kvadratno enačbo $128=2 a^{2}+24 a$. Rešitvi kvadratne enačbe sta $a=4$ in $a=-16$. Ugotovimo, da je pravilna rešitev $a=4$. Osnovni rob prizme meri $a=4~\\mathrm{m}$. Z obrazcem $D=\\sqrt{2 a^{2}+v^{2}}$ izrač... | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | null | MCQ | D | |
017c | Prove that there exists an integer which is divisible by $2010$ and whose decimal notation contains exactly two different digits. | [
"Let $\\varphi$ be Euler's totient function. The Fermat-Euler theorem states that if $a$ and $n$ are relatively prime, then\n$$\na^{\\varphi(n)} \\equiv 1 \\pmod{n}.\n$$\nNow $201 = 3 \\times 67$ is relatively prime to $10 = 2 \\times 5$ and we therefore have that $201$ divides $10^{\\varphi(201)} - 1$. It follows ... | Baltic Way | BALTIC WAY | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0k1e | Problem:
Consider an unusual biased coin, with probability $p$ of landing heads, probability $q \leq p$ of landing tails, and probability $\frac{1}{6}$ of landing on its side (i.e. on neither face). It is known that if this coin is flipped twice, the likelihood that both flips will have the same result is $\frac{1}{2}... | [
"Solution:\n\nThe probability that both flips are the same is $p^{2} + q^{2} + \\frac{1}{36}$. For this to be $\\frac{1}{2}$, we must have\n$$\np^{2} + q^{2} + \\frac{1}{36} = p^{2} + \\left(\\frac{5}{6} - p\\right)^{2} + \\frac{1}{36} = \\frac{1}{2}\n$$\nUsing the quadratic formula, $p = \\frac{2}{3}$ or $\\frac{1... | United States | HMMT November 2018 | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 2/3 | |
0bh8 | Find the infimum and the supremum of the set
$$
\{\sin 1 + \sin 3 + \dots + \sin(2n + 1) \mid n \in \mathbb{N}\}.
$$ | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | infimum = 0, supremum = 1/sin(1) | |
0jnd | Problem:
A number $n$ is bad if there exists some integer $c$ for which $x^{x} \equiv c \pmod{n}$ has no integer solutions for $x$. Find the number of bad integers between $2$ and $42$ inclusive. | [
"Solution:\nCall a number good if it is not bad. We claim all good numbers are products of distinct primes, none of which are equivalent to $1$ modulo another.\n\nWe first show that all such numbers are good. Consider $n = p_{1} p_{2} \\ldots p_{k}$, and let $x$ be a number satisfying $x \\equiv c \\pmod{p_{1} p_{2... | United States | HMMT November 2015 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / F... | null | proof and answer | 25 | |
0ab4 | In a school there are $32$ classes. Six of them have $29$ students each, eight of them have $30$ students each, two of them have $31$ students each and the rest have $32$ students each. How many students are there in the school? | [
"The number of classes with $32$ students each is $32 - 6 - 8 - 2 = 16$.\n\nThe total number of students is $6 \\cdot 29 + 8 \\cdot 30 + 2 \\cdot 31 + 16 \\cdot 32 = 174 + 240 + 62 + 512 = 988$."
] | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 988 | |
07cj | Let $a$, $b$ and $c$ be positive real numbers with $c \ge b \ge a$. Prove that
$$
\frac{(c-a)^2}{6c} \le \frac{a+b+c}{3} - \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}
$$ | [
"**First solution.** Just to simplify the calculations, define $S = a + c$ and $P = ac$. In this case, the RHS of the inequality becomes\n$$\nRHS = \\frac{a+b+c}{3} - \\frac{3}{\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}} = \\frac{S+b}{3} - \\frac{3Pb}{Sb+P}.\n$$\nThe goal is to minimize the above phrase as a functi... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
09qh | Problem:
Alle positieve gehele getallen worden rood of groen gekleurd, zodat aan de volgende voorwaarden wordt voldaan:
- Er zijn zowel rode als groene getallen.
- De som van drie (niet noodzakelijk verschillende) rode getallen is rood.
- De som van drie (niet noodzakelijk verschillende) groene getallen is groen.
Vind... | [
"Solution:\n\nStel dat we het getal $k$ rood kleuren. We bewijzen met inductie ver $n$ dat dan ook $(2n+1)k$ rood is voor alle $n \\geq 0$. Voor $n=0$ is dit natuurlijk waar. Stel dat $(2n-1)k$ rood is voor zekere $n$, dan is ook, $k+k+(2n-1)k = (2n+1)k$ rood. Dit voltooit de inductie. Analoog geldt dat als $k$ gro... | Netherlands | Dutch TST | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Other"
] | null | proof and answer | Exactly two colorings: either all odd integers are red and all even integers are green, or all odd integers are green and all even integers are red. | |
01b8 | Does there exist a positive integer $m$ and a polynomial $P(x)$ with real coefficients for which $x^m + x + 2 = P(P(x))$ for all real $x$? | [
"**Answer:** No.\n\nThe case $m = 1$ is trivial, since in that case the only possibility is $P(x) = Ax + B$ for some integers $A$ and $B$, but by comparing the coefficients we see that $A^2 = 2$, a contradiction.\n\nNow let $m > 1$. It is well-known that $a - b \\mid P(a) - P(b)$ (this is seen directly by writing o... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | null | proof only | null | |
0ii2 | Problem:
Find the largest positive integer $n$ such that $1! + 2! + 3! + \cdots + n!$ is a perfect square. Prove that your answer is correct. | [
"Solution:\n\nClearly $1! + 2! + 3! = 9$ works. For $n \\geq 4$, we have\n$$\n1! + 2! + 3! + \\cdots + n! \\equiv 1! + 2! + 3! + 4! \\equiv 3 \\pmod{5}\n$$\nbut there are no squares congruent to $3$ modulo $5$."
] | United States | Harvard-MIT Mathematics Tournament, Team Round B | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 3 | |
0bnb | Let $a$, $b$, $c > 0$ be real numbers so that $a + b + c = 1$. Prove that
$$
\frac{bc + a + 1}{a^2 + 1} + \frac{ca + b + 1}{b^2 + 1} + \frac{ab + c + 1}{c^2 + 1} \le \frac{39}{10}.
$$ | [
"(Solution by Lenca Cuturela) Since $a + b + c = 1$, we have $a - a^2 = ab + ca$, so\n$$\n\\frac{bc + a + 1}{a^2 + 1} = 1 + \\frac{bc + a - a^2}{a^2 + 1} = 1 + \\frac{ab + bc + ca}{a^2 + 1},\n$$\nand the similar relations. The inequality to be proven is equivalent with\n$$\n\\frac{1}{a^2 + 1} + \\frac{1}{b^2 + 1} +... | Romania | 66th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
09jr | Let $M$ be the midpoint of the side $CD$ of the trapezoid $ABCD$. Let segment $BM$ intersect diagonal $AC$ at point $E$. On the segment $AE$, a point $L$ (distinct from $C$) is chosen such that $EC = EL$. Line $BL$ intersects the base $AD$ at point $K$. Prove that $AK = KD$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | proof only | null | |
0jpj | Problem:
Let $N$ denote the sum of the decimal digits of $\binom{1000}{100}$. Estimate the value of $N$. If your answer is a positive integer $A$ written fully in decimal notation (for example, 521495223), your score will be the greatest integer not exceeding $25 \cdot (0.99)^{|A-N|}$. Otherwise, your score will be zer... | [
"Solution:\nAnswer: 621 http://www.wolframalpha.com/input/?i=sum+of+digits+of+nCr (1000,100)\nTo see this, one can estimate there are about 150 digits, and we expect the digits to be roughly random, for $150 \\cdot 4.5 \\approx 675$, which is already very close to the actual answer. The actual number of digits is 1... | United States | HMMT February | [
"Number Theory > Other"
] | null | final answer only | 621 | |
05i6 | Problem:
Trouver toutes les fonctions $f$ de $\mathbb{R}$ dans $\mathbb{R}$ telles que $f(a b)=f(a+b)$ pour tous nombres irrationnels $a$ et $b$. | [
"Solution:\nIl est clair que les fonctions constantes conviennent. Montrons que ce sont les seules. Soit $u$ un nombre irrationnel positif. Alors $\\sqrt{u}$ est irrationnel. En prenant $(a, b) = (\\sqrt{u}, -\\sqrt{u})$, il vient $f(-u) = f(0)$. En prenant $(a, b) = (-\\sqrt{u}, -\\sqrt{u})$, il vient $f(u) = f(-2... | France | Olympiades Françaises de Mathématiques - Épreuve en temps limité de Janvier | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All constant functions | |
00f0 | Let $S$ be a set consisting of $m$ pairs $(a, b)$ of positive integers with the property that $1 \leq a < b \leq n$. Show that there are at least
$$
4m \frac{\left(m-\frac{n^{2}}{4}\right)}{3n}
$$
triples $(a, b, c)$ such that $(a, b), (a, c)$, and $(b, c)$ belong to $S$. | [
"Call a triple $(a, b, c)$ good if and only if $(a, b), (a, c)$, and $(b, c)$ all belong to $S$. For $i$ in $\\{1, 2, \\ldots, n\\}$, let $d_{i}$ be the number of pairs in $S$ that contain $i$, and let $D_{i}$ be the set of numbers paired with $i$ in $S$ (so $|D_{i}| = d_{i}$). Consider a pair $(i, j) \\in S$. Our ... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1989 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0dyz | In an acute triangle $ABC$ we have $|AC| > |AB|$. Let $D$ and $E$ be two points on the sides $AC$ and $AB$ such that $|CD| = |BE|$. Denote the intersection of the segments $BD$ and $CE$ by $F$ and let $G$ be a point on the segment $AC$ such that the line $GF$ is parallel to the bisector of the angle $\angle BAC$. Prove... | [
"Denote the intersection of lines $AB$ and $FG$ by $H$ and let $J$ be the point where the bisector of the angle $\\angle BAC$ meets the segment $BC$. Since the line $FG$ is parallel to $AJ$, we have $\\angle HGA = \\angle FGD = \\angle BAJ = \\angle AHG$, so the triangle $GAH$ is isosceles and $|AH| = |AG|$. We kno... | Slovenia | Slovenija 2008 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0k65 | Problem:
A convex 2019-gon $A_{1} A_{2} \ldots A_{2019}$ is cut into smaller pieces along its 2019 diagonals of the form $A_{i} A_{i+3}$ for $1 \leq i \leq 2019$, where $A_{2020}=A_{1}$, $A_{2021}=A_{2}$, and $A_{2022}=A_{3}$. What is the least possible number of resulting pieces? | [
"Solution:\nEach time we draw in a diagonal, we create one new region, plus one new region for each intersection on that diagonal. So, the number of regions will be\n$$\n1+ (\\text{number of diagonals}) + (\\text{number of intersections})\n$$\nwhere (number of intersections) counts an intersection of three diagonal... | United States | HMMT November 2019 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 5049 | |
0l68 | Problem:
There exists a unique circle that is both tangent to the parabola $y = x^{2}$ at two points and tangent to the curve $x = \sqrt{\frac{y^{3}}{1 - y}}$. Compute the radius of this circle.
Proposed by: Karthik Venkata Vedula | [
"Solution:\n\n\n\nWe can square both sides of the second curve to get $x^{2} = \\frac{y^{3}}{1 - y}$, which further rearranges to\n$$\n\\frac{x^{2}}{(x^{2} + y^{2})^{2}} = \\frac{y}{x^{2} + y^{2}}.\n$$\nThis relation implies that curves $y = x^{2}$ and $x^{2} = \\frac{y^{3}}{1 - y}$ map to ... | United States | HMMT February | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | final answer only | sqrt(5)/2 | |
0dvk | Problem:
Naj bo $X$ presečišče diagonal konveksnega štirikotnika $ABCD$. Dokaži, da se trikotnikoma $ABX$ in $CDX$ očrtani krožnici dotikata natanko tedaj, ko je $AB \parallel CD$. | [
"Solution:\n\nOznačimo $\\angle BAX = \\alpha$ in $\\angle DCX = \\beta$. Potem je $\\angle BO_1X = 2\\alpha$ in $\\angle DO_2X = 2\\beta$, od koder izračunamo\n$$\n\\angle O_1XB = \\frac{\\pi}{2} - \\alpha \\text{ in } \\quad \\angle O_2XD = \\frac{\\pi}{2} - \\beta,\n$$\nsaj je $O_1BX$ enakokraki trikotnik z vrho... | Slovenia | 47. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ig6 | Problem:
In how many ways can $8$ people be arranged in a line if Alice and Bob must be next to each other, and Carol must be somewhere behind Dan? | [
"Solution:\n\nLet us place Alice and Bob as a single person; there are then $7! = 5040$ different arrangements. Alice can be in front of Bob or vice versa, multiplying the number of possibilities by $2$, but Carol is behind Dan in exactly half of those, so that the answer is just $5040$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 5040 | |
08l3 | Problem:
Find all triples $(x, y, z)$ of real numbers that satisfy the system
$$
\left\{\begin{array}{l}
x + y + z = 2008 \\
x^{2} + y^{2} + z^{2} = 6024^{2} \\
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{2008}
\end{array}\right.
$$ | [
"Solution:\nThe last equation implies $x y z = 2008(x y + y z + z x)$, therefore\n$$\nx y z - 2008(x y + y z + z x) + 2008^{2}(x + y + z) - 2008^{3} = 0.\n$$\n$(x - 2008)(y - 2008)(z - 2008) = 0$.\nThus one of the variables equals $2008$. Let this be $x$. Then the first equation implies $y = -z$. From the second on... | JBMO | 2008 Shortlist JBMO | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | All permutations of (2008, 4016, -4016) | |
01el | Heights $BB_1$ and $CC_1$ of acute triangle $ABC$ intersect in point $H$. $B_2$ and $C_2$ are points on segments $BH$ and $CH$ respectively such that $BB_2 = B_1H$ and $CC_2 = C_1H$. Circumcircle of the triangle $B_2HC_2$ intersects circumcircle of triangle $ABC$ in points $D$ and $E$. Prove that triangle $DEH$ is righ... | [
"Despite of the logical symmetry of the picture the right angle in triangle $\\triangle DEH$ is not $H$ but either $D$ or $E$.\nDenote by $w$ the circumcircle of the triangle $B_2HC_2$. Midperpendicular to the segment $C_2H$ is also the midperpendicular to $CC_1$ therefore it passes through the midpoint $X$ of side... | Baltic Way | Baltic Way shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0g1n | Problem:
Soit $ABC$ un triangle avec $AB = AC \neq BC$ et $I$ le centre de son cercle inscrit. La droite $BI$ coupe $AC$ en $D$, et la perpendiculaire à $AC$ passant par $D$ coupe $AI$ en $E$. Montrer que la réflexion de $I$ par rapport à la droite $AC$ est sur le cercle circonscrit au triangle $BDE$. | [
"Solution:\n\nOn note $I'$ le symétrique de $I$ par rapport à $AC$, $J$ et $F$ les symétriques respectifs de $I$ et $C$ par $D$, $M$ le milieu de $II'$, $A'$ le milieu de $BC$, $S$ l'intersection de $AI$ et $I'C$, $K$ l'intersection de $II'$ avec la médiatrice de $BI$ et $\\Gamma$ le cercle de centre $E$ passant pa... | Switzerland | IMO-Selektion | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0g57 | 已知 $a, b, c, d$ 為非負實數, 試求滿足下列方程組的解 $(a, b, c, d)$:
$$
a^2(b+c)(b+c+d) = \sqrt{b+c}\sqrt[3]{b+c+d}
$$
$$
b^2(c+d)(c+d+a) = \sqrt{c+d}\sqrt[3]{c+d+a}
$$
$$
c^2(d+a)(d+a+b) = \sqrt{d+a}\sqrt[3]{d+a+b}
$$
$$
d^2(a+b)(a+b+c) = \sqrt{a+b}\sqrt[3]{a+b+c}
$$ | [
"若 $a, b, c, d$ 其中有一項為 $0$, 容易推得 $a = b = c = d = 0$ 為一解, 因此以下不妨設 $a, b, c, d > 0$。\n將原式整理後得到\n$$\na^{12}(b+c)^3(b+c+d)^4 = 1\n$$\n$$\nb^{12}(c+d)^3(c+d+a)^4 = 1\n$$\n$$\nc^{12}(d+a)^3(d+a+b)^4 = 1\n$$\n$$\nd^{12}(a+b)^3(a+b+c)^4 = 1\n$$\n考慮 $x$ 為 $16$ 個數字 $\\left\\{\\frac{x}{y}\\mid x, y \\in a, b, c, d\\right\\}$... | Taiwan | 二〇一一數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | Solutions are (0, 0, 0, 0) and (648^{-1/19}, 648^{-1/19}, 648^{-1/19}, 648^{-1/19}). | |
0dk0 | Find the maximum value of $k$ such that: for all $a, b, c$ are sidelengths of some triangle, then
$$
k \cdot \frac{3(a^2 + b^2 + c^2)}{(a + b + c)^2} + \sqrt{\frac{ab + bc + ca}{a^2 + b^2 + c^2}} \leq k + 1.
$$ | [
"The given inequality can be written as\n$$\n2k \\cdot \\frac{(a^2 + b^2 + c^2) - (ab + bc + ca)}{(a + b + c)^2} \\leq \\frac{\\sqrt{a^2 + b^2 + c^2} - \\sqrt{ab + bc + ca}}{\\sqrt{a^2 + b^2 + c^2}}\n$$\n$$\n2k \\cdot \\frac{(a^2 + b^2 + c^2) - (ab + bc + ca)}{(a + b + c)^2} \\leq \\frac{(a^2 + b^2 + c^2) - (ab + b... | Saudi Arabia | Saudi Arabia booklet 2024 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 2 - sqrt(2) | |
0f7p | Problem:
Find all sets of 6 weights such that for each of $n = 1, 2, 3, \ldots, 63$, there is a subset of weights weighing $n$. | [] | Soviet Union | 21st ASU | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 1, 2, 4, 8, 16, 32 | |
0bot | a) Find all positive integers $a$ for which
$$
\frac{1}{4} < \frac{1}{a+1} + \frac{1}{a+2} + \frac{1}{a+3} < \frac{1}{3}.
$$
b) Prove that for any integer $p \ge 2$ there exist $p$ consecutive positive integers $a_1, a_2, \dots, a_p$ such that
$$
\frac{1}{p+1} < \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_p} < ... | [
"a) Set $S = \\frac{1}{a+1} + \\frac{1}{a+2} + \\frac{1}{a+3}$ and notice that $\\frac{3}{a+3} < S < \\frac{3}{a+1}$ to infer from $\\frac{1}{4} < S < \\frac{1}{3}$ that $\\frac{1}{4} < \\frac{3}{a+1}$ and $\\frac{3}{a+3} < \\frac{1}{3}$. Consequently $6 < a < 11$, so $a \\in \\{7, 8, 9, 10\\}$. It is easy to check... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | a ∈ {8, 9, 10}; for part (b), one valid choice is a_i = p^2 + i for i = 1, 2, ..., p | |
05yo | Problem:
Soit $A, B, C, D$ quatre points situés dans cet ordre sur un cercle $\omega$. Soit $P$ et $Q$ deux points situés sur la droite $(A B)$, de sorte que les points $Q, A, B, P$ soient alignés dans cet ordre, que le cercle circonscrit à $A D Q$ soit tangent à la droite $(A C)$, et que le cercle circonscrit à $B C ... | [
"Solution:\n\n\n\nSoit $X$ le point d'intersection éventuel dont on doit démontrer l'existence : nous allons en chercher des caractérisations alternatives. Pour ce faire, et au vu des nombreuses relations de tangence et de cocyclicité dont on dispose, une chasse aux angles s'impose.\n\nPar ... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian ... | null | proof only | null | |
0jxx | Problem:
Let $\ldots, a_{-1}, a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of positive integers satisfying the following relations: $a_{n}=0$ for $n<0$, $a_{0}=1$, and for $n \geq 1$,
$$
a_{n}=a_{n-1}+2(n-1) a_{n-2}+9(n-1)(n-2) a_{n-3}+8(n-1)(n-2)(n-3) a_{n-4} .
$$
Compute
$$
\sum_{n \geq 0} \frac{10^{n} a_{n}}{n!}
$$ | [
"Solution:\n\nLet $y=\\sum_{n \\geq 0} \\frac{x^{n} a_{n}}{n!}$. Then $y^{\\prime}=\\left(1+2 x+9 x^{2}+8 x^{3}\\right) y$ by definition. So $y=C \\exp \\left(x+x^{2}+3 x^{3}+2 x^{4}\\right)$. Take $x=0$ to get $C=1$. Take $x=10$ to get the answer."
] | United States | February 2017 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | e^{23110} | |
0krr | A straight river that is 264 meters wide flows from west to east at a rate of 14 meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at 80 meters per minute, and Sherry swims at 60 meters per minute. ... | [
"Because the two women cross the river in the same amount of time, the north-south components of their velocities are the same value $y$. The east-west components of Melanie's and Sherry's velocities must be values $-x$ and $x$, respectively, because the two women meet halfway between their starting points. But Mel... | United States | 2022 AIME I | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 550 | |
0b2w | Problem:
Let $\triangle ABC$ be an isosceles triangle with a right angle at $A$, and suppose that the diameter of its circumcircle $\Omega$ is $40$. Let $D$ and $E$ be points on the arc $BC$ not containing $A$ such that $D$ lies between $B$ and $E$, and $AD$ and $AE$ trisect $\angle BAC$. Let $I_{1}$ and $I_{2}$ be th... | [
"Solution:\n\nLet $O$ be the center of $\\Omega$. Note that $\\angle OBD = \\angle CBD = \\frac{\\pi}{3}$, so $\\triangle OBD$ is an equilateral triangle. Thus, $BD = BO = \\frac{40}{2} = 20$. This implies that $EC = DE = BD = 20$.\n\nClearly, $I_{1}$ and $I_{2}$ lie on the segments $AD$ and $AE$ respectively. It i... | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 20 | |
0dts | Two rhombi *ABCD* and *AXYZ* are external to each other with a common vertex *A*. The vertices of the rhombi are labelled clockwise. If $\angle DAX = \angle BAZ$, prove that the centres of the rhombi and the midpoint of the segment *BZ* form an isosceles triangle. | [
"Let the centres be $P$, $Q$ and the midpoint of the segment $BZ$ be $R$ as shown in the figure. Since $\\angle DAX = \\angle BAZ$, we have $DZ = BX$. Since $P$ and $R$ are the midpoints of $XZ$ and $BZ$, respectively, $PR = BX/2$. Similarly $QR = DZ/2$. Therefore $QR = PR$ and it follows that $\\triangle PQR$ is i... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0jjl | Problem:
Find the area of a triangle with side lengths $14$, $48$, and $50$. | [
"Solution:\n\nNote that this is a multiple of the $7$-$24$-$25$ right triangle. The area is therefore\n$$\n\\frac{14 \\times 48}{2} = 336\n$$"
] | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | 336 | |
08o1 | Problem:
Let $P$ and $Q$ be the midpoints of the sides $BC$ and $CD$, respectively, of a rectangle $ABCD$. Let $K$ and $M$ be the points of intersection of the line $PD$ with $QB$ and $QA$, respectively, and let $N$ be the point of intersection of the lines $PA$ and $QB$.
Let $X, Y, Z$ be the midpoints of the line segm... | [
"Solution:\nLet $R$ be the midpoint of the side $AD$. Then the lines $BR$ and $PD$ are parallel. Since $\\angle MAN = \\angle QAP = \\angle QBR = \\angle QKM$, the points $A, N, K, M$ are concyclic.\n\n\n\nLet $\\ell_{4}$ be the line passing through the midpoint $W$ of the line segment $MK$... | JBMO | 17th Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0htx | Problem:
Given a triangle $ABC$, we draw three circles with respective diameters $AB$, $BC$, and $CA$. Prove that there exists a point that is inside all three circles. | [
"Solution:\n\nWe claim that the center $I$ of the triangle's inscribed circle is such a point. To see this, note that\n$$\n\\angle BIC = 180 - \\angle ICB + \\angle CBI = 180 - \\frac{\\angle C + \\angle B}{2} = 180 - \\frac{180 - A}{2} = 90 + \\frac{A}{2} > 90.\n$$\nThus $\\triangle BIC$ is obtuse and if we drop a... | United States | Berkeley Math Circle | [
"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 | |
0cus | Some cells of a $200 \times 200$ checkered square contain red or blue tokens — one per cell; the other cells are empty. We say that a token *sees* another token if they are situated either in one row or in one column. Assume that each red token sees exactly five blue tokens (and, perhaps, some red tokens), and each blu... | [
"Answer. $3800$ tokens.\n\n*Example.* For each side of the square, choose a $5 \\times 190$ rectangle adjacent to this side at distance $5$ from the two neighboring sides. Put red tokens into two such rectangles of different orientations and blue tokens into the other two.\n\n*Bound.* Say that a line (i.e., row or ... | Russia | XLIII Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | 3800 | |
0j3m | Problem:
Let $O$ be the point $(0,0)$. Let $A$, $B$, $C$ be three points in the plane such that $A O = 15$, $B O = 15$, and $C O = 7$, and such that the area of triangle $A B C$ is maximal. What is the length of the shortest side of $A B C$? | [
"Solution:\nAnswer: $20$\n\nWe claim that $O$ should be the orthocenter of the triangle $A B C$. If $O$ is not on an altitude of $\\triangle A B C$, suppose (without loss of generality) that $\\overline{A O}$ is not perpendicular to $\\overline{B C}$. We can rotate $A$ around $O$, leaving $B$ and $C$ fixed, to make... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Misc... | null | proof and answer | 20 | |
0bo4 | a) Show that the last two digits of $1038^2$ are 4.
b) Show that there are infinitely many perfect squares whose last three digits are 4.
c) Prove that there is no perfect square whose last four digits are 4. | [
"a) $1038^2 = 1077444$.\n\nb) By squaring a number which ends in 038 we get a number ending in 444, as shows the diagram:\n$$\n\\begin{array}{cccccc}\n\\dots & \\dots & \\dots & 0 & 3 & 8 \\\\\n\\dots & \\dots & \\dots & 0 & 3 & 8 \\\\\\hline\n\\dots & \\dots & \\dots & 3 & 0 & 4 \\\\\n\\dots & \\dots & \\dots & 1 ... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
05g5 | Problem:
Les diagonales du quadrilatère convexe $ABCD$ sont perpendiculaires et se rencontrent en $O$. La perpendiculaire à $(AB)$ passant par $O$ rencontre $(AB)$ en $M$ et $(CD)$ en $M'$. La perpendiculaire à $(BC)$ passant par $O$ rencontre $(BC)$ en $N$ et $(DA)$ en $N'$. La perpendiculaire à $(CD)$ passant par $O... | [
"Solution:\n\n\n\nCommençons par prouver que le quadrilatère $MNPQ$ est inscriptible :\nOn note que le quadrilatère $BMON$ a deux angles droits en $M$ et en $N$, donc il est inscrit dans le cercle de diamètre $[OB]$. De même, les quadrilatères $CNOP$, $DPOQ$ et $AQOM$ sont inscrits respecti... | France | Olympiades Françaises de Mathématiques, Envoi No. 6 | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
078n | All the squares of a $2024 \times 2024$ board are coloured white. In one move, Mohit can select one row or column whose every square is white, choose exactly $1000$ squares in this row or column, and colour all of them red. Find the maximum number of squares that Mohit can colour red in a finite number of moves. | [
"Let $n = 2024$ and $k = 1000$. We claim that the maximum number of squares that can be coloured in this way is $k(2n - k)$, which evaluates to $3048000$.\n\nIndeed, call a row/column *bad* if it has at least one red square. After the first move, there are exactly $k+1$ bad rows and columns: if a row was picked, th... | India | INMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 3048000 | |
03n8 | Problem:
Consider an arrangement of tokens in the plane, not necessarily at distinct points. We are allowed to apply a sequence of moves of the following kind: Select a pair of tokens at points $A$ and $B$ and move both of them to the midpoint of $A$ and $B$.
We say that an arrangement of $n$ tokens is collapsible if... | [
"Solution:\n\nFor a given positive integer $n$, consider an arrangement of $n$ tokens in the plane, where the tokens are at points $A_{1}, A_{2}, \\ldots, A_{n}$. Let $G$ be the centroid of the $n$ points, so as vectors (after an arbitrary choice of origin),\n$$\n\\vec{G}=\\frac{\\vec{A}_{1}+\\vec{A}_{2}+\\cdots+\\... | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
0hai | Let $a, b, x, y$ be non-negative numbers, $k > 0$ and for these numbers the following inequalities are true:
$$
a + kx \le 1; \quad a + ky \le 1; \quad b + \frac{x}{k} \le 1; \quad b + \frac{y}{k} \le 1.
$$
Prove that the following equation is true:
$$
(a + x)^2 + (b + y)^2 \le 2.
$$ | [
"(Edward Turkevich)\n**Answer:** the equation is true when $a = b, x = y = 1 - a, k = 1$.\n\nFrom $a + kx \\le 1$: $x \\le \\frac{1-a}{k}$ and from $b + \\frac{x}{k} \\le 1$: $x \\le k(1-b)$.\nSo, $x^2 \\le (1-a)(1-b)$ and $x \\le \\frac{1}{2}(2-a-b)$ (based on the inequality of arithmetic and geometric means of $1... | Ukraine | The Problems of Ukrainian Authors | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
01k3 | Integers from $1$ to $80$ are written on a blackboard. Tom and Jerry play the following game. They, in turn (Tom is the first), erase these numbers. Per move it is allowed to erase exactly $1$, or exactly $5$, or exactly $8$ numbers. The player wins if he erases the last number.
Who wins if both of the players play to ... | [
"It is easy to see that if either $1$ or $5$ numbers remain on the blackboard, then the player who must move wins, but if $2$ or $4$ numbers remain, then this player loses. So we will solve the problem moving backward. We write all numbers from $1$ to $80$ and mark them with \"+\" or \"-\". If $k$ numbers remain on... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Jerry | |
0839 | Problem:
Un museo ha la pianta quadrata ed è suddiviso in $n^{2}$ stanze quadrate tutte uguali (con $n>1$ ). Ogni coppia di stanze adiacenti (cioè con un muro in comune) comunica mediante una porta (come mostrato nell'esempio a fianco per $n=4$ ). Il guardiano notturno vuole organizzare il suo giro di ispezione in modo... | [
"Solution:\nSe $n$ è pari si riesce a costruire il percorso per qualsiasi $k$. Infatti si riesce ad esibire un percorso chiuso che passa una volta da ciascuna stanza, e facendo $k$ giri si ottiene il tragitto del guardiano.\nSupponiamo per comodità che il museo sia orientato lungo i punti cardinali. Un modo per cos... | Italy | XIX GARA NAZIONALE DI MATEMATICA | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | If the grid side length is even, it is possible for every positive required per-room count. If the grid side length is odd, it is possible only when the required per-room count equals one. | |
0f9k | Problem:
The line joining the midpoints of two opposite sides of a convex quadrilateral makes equal angles with the diagonals. Show that the diagonals are equal. | [
"Solution:\n\n\nLet $L$, $M$, $N$ be the midpoints of $BC$, $CD$, $DA$. Assume that $NL$ makes equal angles with $AC$ and $BD$, so $\\angle NLM = \\angle BEL = \\angle AFN = \\angle LNM$, so $LM = MN$ and hence $BD = AC$."
] | Soviet Union | 24th ASU | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00cv | Fede debe elegir 50 números enteros distintos, desde $1$ hasta $100$ inclusive de modo que su suma sea igual a $2900$. Determinar cuál es la menor cantidad de números pares que puede haber entre los 50 números que elija Fede. | [
"Calculamos la suma de los 50 impares entre $1$ y $100$:\n$$\n1+3+5+\\ldots+99=(1+2+\\ldots+100)-(2+4+\\ldots+100)=50\\cdot101-50\\cdot51=2500.\n$$\nFaltan $400$ para llegar a $2900$. Ahora cambiamos los menores enteros impares por los mayores enteros pares y lo hacemos en grupos de $2$ porque $400$ es par. Comenza... | Argentina | Nacional OMA | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Spanish | proof and answer | 6 | |
0h39 | Let $S(a)$ denote the sum of decimal digits of a positive integer $a$. Positive integer $n$ is such that $S(n) = 406$, $S(2012n) = 2012$. Find all possible values of $S(2011n)$. | [
"Зауважимо, $2012n$ є сумою двох чисел $1000n$, одного числа $10n$ та двох чисел $n$. Якщо не відбувається перенесень у наступний розряд під час додавання цих чисел в стовпчик, то сума цифр утвореного числа буде в 5 разів більшою за суму цифр числа $n$. Під час перенесення однієї одиниці в старший розряд сума цифр ... | Ukraine | Ukrainian Mathematical Olympiad | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 1609, 1615, 1624 | |
0ig4 | Problem:
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a smooth function such that $f'(x)^2 = f(x) f''(x)$ for all $x$. Suppose $f(0) = 1$ and $f^{(4)}(0) = 9$. Find all possible values of $f'(0)$. | [
"Solution: $\\pm \\sqrt{3}$\nLet $f'(0) = a$. Then the equation gives $f''(0) = a^2$. Differentiating the given equation gives\n$$\n2 f'(x) f''(x) = f(x) f'''(x) + f'(x) f''(x)\n$$\nor $f'(x) f''(x) = f(x) f'''(x)$. Differentiating once more gives\n$$\nf'(x) f'''(x) + f''(x)^2 = f(x) f^{(4)}(x) + f'(x) f'''(x)\n$$\... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Equations > ODEs",
"Calculus > Differential Calculus > Derivatives"
] | null | proof and answer | ± sqrt(3) | |
03gy | Problem:
Consider all line segments of length $4$ with one end-point on the line $y = x$ and the other end-point on the line $y = 2x$. Find the equation of the locus of the midpoints of these line segments. | [
"Solution:\nLet the end-points of the segment be $A(a, a)$ on $y = x$ and $B(b, 2b)$ on $y = 2x$.\n\nThe length of $AB$ is $4$:\n$$\nAB = \\sqrt{(b - a)^2 + (2b - a)^2} = 4\n$$\nSo,\n$$\n(b - a)^2 + (2b - a)^2 = 16\n$$\nExpanding:\n$$\n(b - a)^2 + (2b - a)^2 = (b - a)^2 + (4b^2 - 4ab + a^2) = (b - a)^2 + 4b^2 - 4ab... | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 25x^2 + 13y^2 - 36xy = 4 | |
0jtz | Problem:
Determine the number of triples $0 \leq k, m, n \leq 100$ of integers such that
$$
2^{m} n - 2^{n} m = 2^{k}
$$ | [
"Solution:\nFirst consider when $n \\geq m$, so let $n = m + d$ where $d \\geq 0$. Then we have\n$$\n2^{m}(m + d - 2^{d} m) = 2^{m}(m(1 - 2^{d}) + d)\n$$\nwhich is non-positive unless $m = 0$. So our first set of solutions is $m = 0, n = 2^{j}$.\n\nNow, we can assume that $m > n$, so let $m = n + d$ where $d > 0$. ... | United States | HMMT February | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 22 | |
0ebi | Problem:
Za koliko naravnih števil $n$, $n \leq 2015$, ulomek $\frac{3 n-1}{2 n^{2}+1}$ ni okrajšan? | [
"Solution:\n\nDenimo, da ulomek $\\frac{3 n-1}{2 n^{2}+1}$ ni okrajšan. Potem obstaja naravno število $a$, različno od $1$, ki deli $3 n-1$ in $2 n^{2}+1$. Sledi, da $a$ deli tudi $3\\left(2 n^{2}+1\\right)-2 n(3 n-1)=2 n+3$ in zato tudi $3(2 n+3)-2(3 n-1)=11$. Ker je $11$ praštevilo, sledi $a=11$, torej $11$ deli ... | Slovenia | 59. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | 183 | |
09iq | Let $ABC$ be a triangle. Let $D$ be a point on $AB$ and let $E$ be a point on $AC$ such that $BCED$ is cyclic. Let $F$ be the intersection of the segment $BE$ and the circumcircle of $ADC$ and let $G$ be the intersection of $CD$ and the circumcircle of $ABE$. Suppose that the segments $BG$ and $CF$ meets at $S$. Show t... | [
"Since the quadrilaterals $ADFC$, $DBCE$, $AEGB$ are cyclic, we have\n$$\n\\angle ADC = \\angle AFC = \\angle ADC = \\angle AEB = \\angle AGB.\n$$\nBecause $\\angle AEF = \\angle AFC$, we have $\\triangle AEF \\sim \\triangle ACF$. Hence $AF^2 = AE \\cdot AC$. In the same way, we have $AG^2 = AD \\cdot AB$. It foll... | Mongolia | Mongolian Mathematical Olympiad Round 3 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0doe | Problem:
У трапезу $ABCD$ чији унутрашњи углови нису прави, дијагонале $AC$ и $BD$ секу се у тачки $E$. Нека су $P$ и $Q$ редом подножја нормала из темена $A$ и $B$ на праве $BC$ и $AD$. Описане кружнице троуглова $CEQ$ и $DEP$ секу се у тачки $F \neq E$. Доказати да се праве $AP$, $BQ$ и $EF$ секу у једној тачки или ... | [
"Solution:\n\nСлучај када је $AD \\parallel BC$ је једноставан. Наиме, тада је $E$ средиште дијагонале $AC$, те је $EA = EC = EP$, а слично важи и $EB = ED = EQ$. Одатле следи да су кругови $CEQ$ и $DEP$ симетрични у односу на симетралу дужи $CP$ и $DQ$, те је $EF \\perp CP$, тј. $EF \\parallel AP \\parallel BQ$.\n... | Serbia | 14. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Pappus theorem"
] | null | proof only | null |
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