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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
04s2 | **Find all triples of real numbers $x$, $y$ and $z$ for which**
$$
x(y^2 + 2z^2) = y(z^2 + 2x^2) = z(x^2 + 2y^2).
$$ | [
"If for example $x = 0$, we get a system $0 = y z^2 = 2 y^2 z$ which means that one of the unknowns $y$ and $z$ vanishes and the other can be arbitrary. The cases $y = 0$ or $z = 0$ are discussed similarly. Thus we have obtained three groups of solutions $(x, y, z)$ which are formed by triples $(t, 0, 0)$, $(0, t, ... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Other"
] | English | proof and answer | (t, 0, 0), (0, t, 0), (0, 0, t), (t, t, t), (4t, t, 2t), (2t, 4t, t), (t, 2t, 4t) for any real t | |
0jip | Problem:
There are 100 students who want to sign up for the class Introduction to Acting. There are three class sections for Introduction to Acting, each of which will fit exactly 20 students. The 100 students, including Alex and Zhu, are put in a lottery, and 60 of them are randomly selected to fill up the classes. W... | [
"Solution:\n\nAnswer: $\\frac{19}{165}$\n\nThere is a $\\frac{60}{100} = \\frac{3}{5}$ chance that Alex is in the class. If Alex is in the class, the probability that Zhu is in his section is $\\frac{19}{99}$. So the answer is $\\frac{3}{5} \\cdot \\frac{19}{99} = \\frac{19}{165}$."
] | United States | HMMT 2014 | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 19/165 | |
0ha2 | There are 30 cities in the country, some of them are connected by flights. The total amount of flights satisfies the following: if one does not consider any 26 cities with all flights that connect one of these cities and any other, then one can get from any of the 4 cities that is left to any other city of the four, ma... | [
"Let $A$ is a vertex with the smallest degree $k(A)$. If $k(A) < 27$, then there exist at least 3 vertices that are not connected with a chosen vertex. Then 3 of those vertices together with $A$\n\nmake a not connected graph, so it contradicts the condition. Thus, the smallest size of graph is $\\frac{1}{2} \\cdot ... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 405 | |
05bd | Point $D \notin \{B, C\}$ is chosen on segment $BC$ and point $A$ is chosen on the plane such that it does not lie on the line $BC$. Points $P$ and $Q$ are chosen such that $ABPD$ and $ACQD$ are parallelograms (with the vertices in this order). The circumcircles of triangles $BDP$ and $CDQ$ intersect in $R \neq D$. Pro... | [
"A reflection across the centre of the segment $BD$ sends the triangle $BDP$ into the triangle $DBA$. Thus the circumcircle of $BDP$ goes to the circumcircle of $DBA$. Analogously we see that a reflection across the centre of $CD$ sends the circumcircle of $CDQ$ into the circumcircle of $DCA$.\n\n![](attached_image... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
03g9 | Problem:
Determine which of the two numbers $\sqrt{c+1}-\sqrt{c}$, $\sqrt{c}-\sqrt{c-1}$ is greater for any $c \geq 1$. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | sqrt(c) - sqrt(c-1) is greater than sqrt(c+1) - sqrt(c) for all c >= 1 | |
0j5m | Problem:
Let $n$ be an odd positive integer, and suppose that $n$ people sit on a committee that is in the process of electing a president. The members sit in a circle, and every member votes for the person either to his/her immediate left, or to his/her immediate right. If one member wins more votes than all the othe... | [
"Solution:\n\nLet $x$ be the probability Hermia is elected if Lysander votes for her, and let $y$ be the probability that she wins if Lysander does not vote for her. We are trying to find $x$, and do so by first finding $y$.\n\nIf Lysander votes for Hermia with probability $\\frac{1}{2}$ then the probability that H... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | (2^n - 1) / (n * 2^{n-1}) | |
038v | Let $a_1 > \frac{1}{12}$ and $a_{n+1} = \sqrt{(n+2)a_n + 1}$ for $n \ge 1$. Prove that:
a) $a_n > n - \frac{2}{n}$;
b) the sequence $b_n = 2^n \left(\frac{a_n}{n} - 1\right)$, $n = 1, 2, \dots$, is convergent. | [
"a) We shall prove the statement by induction if $a_1 > \\frac{19}{243} \\left(\\frac{1}{12} > \\frac{19}{243}\\right)$. It is true for $n \\le 3$, since $a_2 > \\sqrt{3 \\cdot \\frac{19}{243} + 1} = \\frac{10}{9}$ and $a_3 > \\sqrt{4 \\cdot \\frac{10}{9} + 1} = \\frac{7}{3}$.\n\nAssume that $a_n > n - \\frac{2}{n}... | Bulgaria | Spring Mathematical Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
06b7 | Let $\alpha_1, \alpha_2, \dots, \alpha_\nu, \beta_1, \beta_2, \dots, \beta_\nu$ be $2\nu$ positive integers such that the $\nu + 1$ products
$$
\begin{align*}
\gamma_1 &= \alpha_1 \alpha_2 \alpha_3 \cdots \alpha_{\nu-1} \alpha_\nu, & \gamma_2 &= \beta_1 \alpha_2 \alpha_3 \cdots \alpha_{\nu-1} \alpha_\nu, \\
\gamma_3 &... | [] | Greece | Selection Examination | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | ν! | |
02vu | Problem:
Determine os números primos $p$ tais que a representação decimal da fração $\frac{1}{p}$ tenha período de tamanho 5.
Observação: Se a representação decimal de um número possui uma sequência de dígitos que se repete de forma periódica, o tamanho da menor sequência de dígitos que se repete é o tamanho da repres... | [
"Solution:\nSejam $q$ o tamanho da parte não periódica e $\\overline{abcde}$ o período da representação decimal de $1/p$. Assim, se representarmos os dígitos da parte não periódica por meio do símbolo $\\star$, temos\n$$\n\\begin{aligned}\n\\frac{1}{p} & = 0, \\star \\star \\ldots \\star \\text{ abcdeabcde } \\ldot... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 41, 271 | |
0i4w | Problem:
A path of length $n$ is a sequence of points $(x_{1}, y_{1}),(x_{2}, y_{2}), \ldots,(x_{n}, y_{n})$ with integer coordinates such that for all $i$ between $1$ and $n-1$ inclusive, either
(1) $x_{i+1}=x_{i}+1$ and $y_{i+1}=y_{i}$ (in which case we say the $i$th step is rightward) or
(2) $x_{i+1}=x_{i}$ and $y_{... | [
"Solution:\nThis is just the number of paths of length $10$. The $i$th step can be either upward or rightward, so there are $2^{10}=1024$ such paths."
] | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | final answer only | 1024 | |
02sl | Problem:
Sobre um quadro negro, Kiara escreveu 20 números inteiros, todos eles diferentes de zero. Em seguida, para cada par de números escritos por Kiara, Yndira escreveu sobre o mesmo quadro o respectivo produto entre eles (inclusive, se o resultado de algum produto já estava escrito, Yndira o repetiu).
Por exemplo... | [
"Solution:\n\na) A quantidade de números escritos por Yndira é igual ao número de pares que podem ser formados com os $20$ números escritos por Kiara.\n\nDe quantas maneiras podemos escolher um par de elementos em um conjunto contendo $20$ elementos? Para a escolha do primeiro elemento temos $20$ possibilidades. Pa... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | a) 210; b) 14 | |
0gvf | For which real numbers $x > 1$ there exists a triangle with sides of lengths $x^4 + x^3 + 2x^2 + x + 1$, $2x^3 + x^2 + 2x + 1$ and $x^4 - 1$? | [
"Дійсно, при всіх $x > 1$, як нескладно переконатися, мають місце нерівності\n$$\nx^4 + x^3 + 2x^2 + x + 1 > 2x^3 + x^2 + 2x + 1,\n$$\n$$\nx^4 + x^3 + 2x^2 + x + 1 > x^4 - 1,\n$$\n$$\nx^4 + x^3 + 2x^2 + x + 1 < (x^4 - 1) + (2x^3 + x^2 + 2x + 1).\n$$"
] | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | All real numbers greater than 1 | |
0eeg | Problem:
Poišči vse pare realnih števil $x$ in $y$, ki rešijo sistem enačb
$$
\begin{aligned}
& \log_{3} x^{2} + \log_{2} y^{3} = 1 \\
& \log_{9} x^{4} + \log_{4} y^{9} = 2
\end{aligned}
$$ | [
"Solution:\nS prehodom na novo osnovo sistem enačb preoblikujemo do\n$$\n\\begin{aligned}\n& \\frac{\\log x^{2}}{\\log 3} + \\frac{\\log y^{3}}{\\log 2} = 1 \\\\\n& \\frac{\\log x^{4}}{\\log 9} + \\frac{\\log y^{9}}{\\log 4} = 2\n\\end{aligned}\n$$\nOpazimo, da mora biti število $y$ nujno pozitivno, zato je $\\log ... | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | (x, y) = (1/sqrt(3), 2^(2/3)) or (x, y) = (-1/sqrt(3), 2^(2/3)) | |
0g5l | 令 $n \ge 3$. 請問什麼樣的凸 $n$ 邊形可以完整的分割成有限多個平行四邊形? (不只是哪些 $n$, 還包括形狀的描述。) | [
"令 $G$ 是一個可以被分割的凸 $n$ 邊形。對於 $G$ 上的任意一個邊 $a$, 先將它轉到水平的位置, 然後可以由 $a$ 一路向上找到一個序列的平行四邊形, 每個都有兩邊平行於 $a$ 而且依序相連。\n\n(P): 在 $G$ 上存在唯一的邊, 使得上述序列的最後一個平行四邊形的上方貼在這個邊上。\n\n原因是：若這個序列沒有碰到 $G$ 的邊界，或是碰到某個 $G$ 的邊但不是如上述平行相貼的，就可以繼續往上找到下一個平行四邊形；再加分割是使用有无限多個平行四邊形，所以性質 (P) 成立。若不是非唯一的（可以有其他不同的序列），則不可能是凸邊形。我們稱性質 (P) 所描述的邊為 $a$ 的平行對邊，並以 $O... | Taiwan | 二〇一一數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | null | |
0gag | 已知非常數的整係數多項式 $f(x)$ 滿足
$$
(x^3 + 4x^2 + 4x + 3)f(x) = (x^3 - 2x^2 + 2x - 1)f(x + 1).
$$
證明：對所有正整數 $n$ ($n \ge 8$), $f(n)$ 至少有五個不同的質因數。
Let $f(x)$ be the polynomial with integer coefficients ($f(x)$ is not constant) such that
$$
(x^3 + 4x^2 + 4x + 3)f(x) = (x^3 - 2x^2 + 2x - 1)f(x + 1).
$$
Prove that for each positive ... | [
"題設等價於\n$$\n(x + 3)(x^2 + x + 1)f(x) = (x - 1)(x^2 - x + 1)f(x + 1). \\quad (1)\n$$\n在 (1) 式中分別令\n$$\nx = -3, \\frac{-1 - \\sqrt{3}i}{2}, \\frac{-1 + \\sqrt{3}i}{2}, 1\n$$\n則\n$$\nf(-2) = f\\left(\\frac{1 - \\sqrt{3}i}{2}\\right) = f\\left(\\frac{1 + \\sqrt{3}i}{2}\\right) = f(1) = 0.\n$$\n在 (1) 式中令 $x = -2, 0$. 則 ... | Taiwan | 二〇一六數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Intermediate Algebra > Complex numbers",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, ... | null | proof only | null | |
0bfd | Parallels to the opposite sides are drawn from the vertices $C$ and $B$ of an equilateral triangle $ABC$. A variable straight line meets the two parallels at $M$ and $N$ respectively and the straight lines $BM$ and $CN$ meet at $O$. Prove that the sum $m(\angle MOA) + m(\angle MAC)$ is constant. | [] | Romania | Shortlisted Problems for the 64th NMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
0bbp | a) Show that, for every positive integer $n$, there exists uniquely determined positive integers $x_n, y_n$ such that $(1 + \sqrt{33})^n = x_n + y_n\sqrt{33}$.
b) Prove that if $x_n, y_n$ are defined as above and $p$ is a positive prime, then at least one of the numbers $y_{p-1}, y_p, y_{p+1}$ is divisible by $p$. | [
"a) The equality is obtained for $x_n = \\binom{n}{0} + 33\\binom{n}{2} + 33^2\\binom{n}{4} + \\dots$ and $y_n = \\binom{n}{1} + 33\\binom{n}{3} + 33^2\\binom{n}{5} + \\dots$.\nAlso, if $a + b\\sqrt{33} = c + d\\sqrt{33}$, $a, b, c, d \\in \\mathbb{N}$ and $b \\neq d$, then $\\sqrt{33} = \\frac{a-c}{d-b}$ is ration... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficien... | null | proof only | null | |
0l3y | Problem:
Let $ABTCD$ be a convex pentagon with area $22$ such that $AB = CD$ and the circumcircles of triangles $TAB$ and $TCD$ are internally tangent. Given that $\angle ATD = 90^{\circ}$, $\angle BTC = 120^{\circ}$, $BT = 4$, and $CT = 5$, compute the area of triangle $TAD$. | [
"Solution:\n\nPaste $\\triangle TCD$ outside the pentagon to get $\\triangle ABX \\cong \\triangle DCT$. From the tangent circles condition, we get\n$$\n\\begin{aligned}\n\\angle XBT & = 360^{\\circ} - \\angle XBA - \\angle ABT \\\\\n& = 360^{\\circ} - \\angle DCT - \\angle ABT \\\\\n& = 360^{\\circ} - 270^{\\circ}... | United States | HMMT February 2024 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 64(2 - sqrt(3)) | |
0iv0 | Problem:
Find the ratio of the radius of $\omega_{i+1}$ to the radius of $\omega_{i}$, where $\omega_{1}$ is the incircle of an isosceles triangle $T$ with base $10$ and height $12$, and $\omega_{i+1}$ is tangent to $\omega_{i}$ and both legs of the triangle for $i > 1$. | [
"Solution:\n\nLet $r$ be the radius of $\\omega_{i}$ and $r'$ the radius of $\\omega_{i+1}$. The centers of $\\omega_{i}$ and $\\omega_{i+1}$ both lie on the angle bisector from the vertex opposite the base, and each circle is tangent to both legs of the triangle.\n\nThe configuration is similar for each $\\omega_{... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | 4/9 | |
06yv | Problem:
In the triangle $ABC$, the midpoints of $AC$ and $AB$ are $M$ and $N$ respectively. $BM$ and $CN$ meet at $P$. Show that if it is possible to inscribe a circle in the quadrilateral $AMPN$ (touching every side), then $ABC$ is isosceles. | [
"Solution:\n\n\n\nIf the quadrilateral has an inscribed circle then $AM + PN = AN + PM$ (consider the tangents to the circle from $A, M, P, N$). But if $AB > AC$, then $BM > CN$ (see below). We have $AN = AB / 2$, $PM = BM / 3$, $AM = AC / 2$, $PN = CN / 3$, so it follows that $AM + PN < AN... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triang... | null | proof only | null | |
0hl1 | Problem:
12 points are placed around the circumference of a circle. How many ways are there to draw 6 non-intersecting chords joining these points in pairs? | [
"Solution:\n\n$C$ (number of chords) $= C(6) = 132$."
] | United States | null | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 132 | |
06gn | Find a function $f : \mathbb{R}^{\ge 0} \to \mathbb{R}^{\ge 0}$ satisfying $f(2x + 1) = 4f(x) + 9$ for all $x \ge 0$. ($\mathbb{R}^{\ge 0}$ is the set of nonnegative real numbers.) | [
"A possible function is $f(x) = 0$ for any $x \\in [0, 1)$ and $f(x) = 3(4^k - 1)$ for any $x \\in [2^k - 1, 2^{k+1} - 1)$ where $k \\in \\mathbb{Z}^+$.\nIt suffices to check the above function satisfies the conditions. It is clear that $f(x) \\ge 0$ for any $x \\ge 0$. If $x \\in [0, 1)$, then $2x + 1 \\in [1, 3)$... | Hong Kong | Year 2012 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | One example is f(x) = 0 for x in [0, 1) and f(x) = 3(4^k - 1) for x in [2^k - 1, 2^{k+1} - 1) where k is any positive integer. | |
08h1 | Problem:
Solve in nonnegative integers the equation $5^{t} + 3^{x} 4^{y} = z^{2}$. | [
"Solution:\nIf $x=0$ we have\n$$\nz^{2} - 2^{2y} = 5^{t} \\Longleftrightarrow (z + 2^{y})(z - 2^{y}) = 5^{t}\n$$\nPutting $z + 2^{y} = 5^{a}$ and $z - 2^{y} = 5^{b}$ with $a + b = t$ we get $5^{a} - 5^{b} = 2^{y+1}$. This gives us $b = 0$ and now we have $5^{t} - 1 = 2^{y+1}$. If $y \\geq 2$ then consideration by m... | JBMO | null | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (t, x, y, z) = (1, 0, 1, 3), (0, 1, 0, 2), (2, 2, 2, 13), (0, 1, 2, 7) | |
0jqp | Problem:
Let $AB$ be a segment of length $2$ with midpoint $M$. Consider the circle with center $O$ and radius $r$ that is externally tangent to the circles with diameters $AM$ and $BM$ and internally tangent to the circle with diameter $AB$. Determine the value of $r$. | [
"Solution:\n\nLet $X$ be the midpoint of segment $AM$. Note that $OM \\perp MX$ and that $MX = \\frac{1}{2}$ and $OX = \\frac{1}{2} + r$ and $OM = 1 - r$. Therefore by the Pythagorean theorem, we have\n$$\nOM^2 + MX^2 = OX^2 \\Longrightarrow (1 - r)^2 + \\frac{1}{2^2} = \\left(\\frac{1}{2} + r\\right)^2\n$$\nwhich ... | United States | HMMT November 2015 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | r = 1/3 | |
02lm | Let $S$ be a set of $6n$ points in a line. Choose arbitrarily $4n$ of these points and paint them blue; the other $2n$ points are painted green. Prove that there exists a line segment that contains exactly $3n$ points from $S$, $2n$ of them blue and $n$ of them green. | [
"Let $A_i$ be the segment with exactly $3n$ points from $S$, the leftmost being the $i$-th point from left to right of $S$, $i = 1, 2, \\dots, 3n + 1$. Define $f(i)$ as the number of blue points in $A_i$. We have to prove that $f(i) = 2n$ for some $i$.\n\nNotice that $|f(i + 1) - f(i)| \\le 1$ since $A_i$ and $A_{i... | Brazil | XXX OBM | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0cvr | One is given $n$ coins of pairwise distinct weights and $n$ scales, $n > 2$. On each weighing, it is permitted to put two coins onto the pans of one of the scales, check the result, and take the coins back from the pans. One of the scales (nobody knows which one) is possibly broken, and it provides random results (some... | [
"Докажем сначала, что за $2n - 1$ взвешивание можно найти самую тяжёлую монету. Более точно, мы докажем по индукции по $n$, что самую тяжёлую из $n \\ge 2$ данных монет можно определить за $2n - 1$ взвешивание, имея трое весов, одни из которых, возможно, испорчены.\n\nЕсли $n = 2$, то взвесим данные две монеты по о... | Russia | Final round | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms"
] | English; Russian | proof and answer | 2n - 1 | |
09fr | Determine all sequences of positive integers $\{a_n\}_{n=1}^{\infty}$ satisfying the following conditions for any positive integer $k$:
(i) $a_{2^{k+1}} = 2 \cdot a_{2^k}$,
(ii) The set $\{a_1, a_2, \dots, a_k\}$ is a complete set of residue classes modulo $k$. | [
"The two sequences $(1, 2, 3, \\ldots)$ and $(3, 2, 1, 4, 5, 6, \\ldots)$.\n\nFor a positive integer $n$, let $S_n := \\{a_1, \\dots, a_n\\}$.\n\nNow fix $n$ and let $M := \\max S_n$ and $m := \\min S_n$ and let $k := M - m$. Since all elements of $S_n$ are different by (ii), we have $k \\ge n-1$. If $k \\ge n$ the... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | null | proof and answer | (1, 2, 3, 4, 5, 6, ...) and (3, 2, 1, 4, 5, 6, ...) | |
0cqc | Some cells of a $100 \times 100$ desk contain a token. We call a cell *beautiful* if the total number of tokens in its neighbors is even (two cells are neighbors if they share a common side). Can it appear that there exists exactly one beautiful cell? (K. Knop)
В некоторых клетках доски 100 × 100 стоит по фишке. Назов... | [
"**Лемма.** Для любой клетки доски $X$ существует множество $S$, состоящее из чётного количества клеток и содержащее $X$, такое, что у каждой клетки доски чётное число соседей лежит в $S$.\n\n**Доказательство.** Раскрасим клетки доски в шахматном порядке; можно считать, что $X$ — чёрная. Для начала рассмотрим одну ... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English, Russian | proof and answer | No | |
017k | Find all quadruples of real numbers $(a, b, c, d)$ satisfying the system of equations
$$
\begin{cases}
(b+c+d)^{2010} = 3a \\
(a+c+d)^{2010} = 3b \\
(a+b+d)^{2010} = 3c \\
(a+b+c)^{2010} = 3d.
\end{cases}
$$ | [
"There are two solutions: $(0, 0, 0, 0)$ and $(\\frac{1}{3}, \\frac{1}{3}, \\frac{1}{3}, \\frac{1}{3})$.\nIf $(a, b, c, d)$ satisfies the equations, then we may as well assume $a \\le b \\le c \\le d$. These are non-negative because an even power of a real number is always non-negative. It follows that\n$$\nb+c+d \... | Baltic Way | BALTIC WAY | [
"Algebra > Equations and Inequalities"
] | null | proof and answer | (0, 0, 0, 0) and (1/3, 1/3, 1/3, 1/3) | |
0kvy | Problem:
Let $ABCD$ be a convex quadrilateral such that $\angle ABC = \angle BCD = \theta$ for some angle $\theta < 90^{\circ}$. Point $X$ lies inside the quadrilateral such that $\angle XAD = \angle XDA = 90^{\circ} - \theta$. Prove that $BX = XC$. | [
"Solution:\n\nLet lines $AB$ and $CD$ meet at $T$. Notice that\n$$\n\\begin{aligned}\n& \\angle ATD = 180^{\\circ} - \\angle ABC - \\angle DBC = 180^{\\circ} - 2\\theta \\\\\n& \\angle AXD = 180^{\\circ} - 2(90^{\\circ} - \\theta) = 2\\theta\n\\end{aligned}\n$$\nTherefore, $A$, $T$, $X$, an... | United States | HMMT February | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Eul... | null | proof only | null | |
0eek | Let $\mathcal{K}$ be a circle with centre $O$, and let $\mathcal{K}'$ be a circle that goes through the point $O$ with a radius that is greater than twice the radius of the circle $\mathcal{K}$. A common tangent of the circles $\mathcal{K}$ and $\mathcal{K}'$ touches the circle $\mathcal{K}$ at point $A$, and it touche... | [
"Let the tangent from $B$ to the circle $K$ (distinct from the tangent $AB$) touch the circle $K$ at $A'$. We show that the points $B$, $A'$ and $E$ are collinear.\n\nWe have $|AB| = |A'B|$ and $|AO| = |A'O|$, so the triangles $ABO$ and $A'BO$ have three equal sides and are therefore congruent. This implies $\\angl... | Slovenia | Slovenija 2016 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07d7 | Let $\omega$ be the circumcircle of isosceles triangle $ABC$ where $AB = AC$. Points $P$ and $Q$ lie on $\omega$ and $BC$ respectively such that $AP = AQ$. Lines $AP$ and $BC$ intersect at $R$. Prove that the tangents from $B$ and $C$ to the incircle of triangle $AQR$ (different from $BC$) are concurrent on $\omega$. | [
"Let $I, I_a$ be the incenter and $A$-excenter of triangle $AQR$. To prove the claim of the problem, it suffices to show that $\\widehat{BIC} = 90^\\circ + \\frac{1}{2}\\widehat{BAC}$.\n\n\n\nWe have\n$$\n\\begin{aligned}\n\\widehat{ARC} &= \\widehat{PCB} - \\widehat{RPC} = \\widehat{PCA} +... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Polar ... | null | proof only | null | |
04xc | Given an integer $n \ge 16$, consider the set
$$
G = \{(x, y) : x, y \in \{1, 2, \dots, n\}\}
$$
consisting of $n^2$ points in the plane. Let $A$ be any subset of $G$ containing at least $4n\sqrt{n}$ points. Prove that there are at least $n^2$ convex quadrangles with all their vertices in $A$ such that their diagonals ... | [
"Let $|A| = m \\ge 4n\\sqrt{n}$ and let $S$ be the set of all segments with endpoints in $A$. Clearly, $|S| = \\binom{m}{2}$. The coordinates of every midpoint of a segment from $S$ are integer multiples of $1/2$. In the convex hull of $G$ there are less than $4n^2$ such points so there exists a point $B$ which is ... | Czech-Polish-Slovak Mathematical Match | Cesko-Slovacko-Poljsko | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Quadrilaterals"
] | English | proof only | null | |
0j06 | Problem:
Suppose that $x$, $y$, and $z$ are non-negative real numbers such that $x + y + z = 1$. What is the maximum possible value of $x + y^{2} + z^{3}$? | [
"Solution:\nSince $0 \\leq y, z \\leq 1$, we have $y^{2} \\leq y$ and $z^{3} \\leq z$. Therefore $x + y^{2} + z^{3} \\leq x + y + z = 1$. We can get $x + y^{2} + z^{3} = 1$ by setting $(x, y, z) = (1, 0, 0)$."
] | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1 | |
08jd | Problem:
The positive real numbers $a$, $b$, $c$ satisfy the relation $a^{2} + b^{2} + c^{2} = 3 a b c$. Prove the inequality
$$
\frac{a}{b^{2} c^{2}} + \frac{b}{c^{2} a^{2}} + \frac{c}{a^{2} b^{2}} \geq \frac{9}{a + b + c}
$$ | [] | JBMO | The first selection test for JBMO 2003 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0ft7 | Problem:
Die Funktion $f:[0,1] \rightarrow \mathbb{R}$ habe die folgenden Eigenschaften:
a. $f(x) \geq 0$ für alle $x \in [0,1]$
b. $f(1) = 1$
c. $f(x+y) \geq f(x) + f(y)$ für alle $x, y, x+y \in [0,1]$
Beweise: $f(x) \leq 2x$ für alle $x \in [0,1]$ | [
"Solution:\n\nAus (a) und (c) folgt, dass $f$ monoton steigend ist, denn für $0 \\leq x \\leq y \\leq 1$ gilt $f(y) = f(x + (y - x)) \\geq f(x) + f(y - x) \\geq f(x) + 0 = f(x)$.\n\nAus (b) und (c) erhält man $2 f(1/2) \\leq f(1) = 1$, also $f(1/2) \\leq 1/2$. Genauso zeigt man $f(1/4) \\leq 1/4$ und mit Induktion ... | Switzerland | IMO - Selektion | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof only | null | |
02ak | Problem:
O número 119 tem a seguinte propriedade:
- a divisão por 2 deixa resto 1 ;
- a divisão por 3 deixa resto 2 ;
- a divisão por 4 deixa resto 3 ;
- a divisão por 5 deixa resto 4 ;
- a divisão por 6 deixa resto 5 .
Quantos inteiros positivos menores que 2007 satisfazem essa propriedade? | [
"Solution:\n\nInicialmente note que se $N$ dividido por $d$ deixa resto $r$, então somando a $N$ um múltiplo de $d$, o resto não se altera, isto é:\n$$\n\\frac{(N+\\text{ múltiplo de } d)}{d} \\text{ também deixa resto } r\n$$\nPor exemplo: $38$ dividido por $3$ deixa resto $2$, logo o resto da divisão de $(38+5 \\... | Brazil | null | [
"Number Theory > Modular Arithmetic",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | 33 | |
0ggq | 令 $\mathbb{R}^+$ 為全體正實數所成的集合。找出所有函數 $f: \mathbb{R}^+ \to \mathbb{R}^+$ 使得
$$
f(x + y^2 f(y)) = f(1 + y f(x)) f(x)
$$
對所有正實數 $x, y$ 恆成立。 | [
"令 $P(x, y)$ 表示將 $(x, y)$ 帶入原題條件。\n$$\nP(1, 1) \\Rightarrow f(1) = 1.\n$$\n$$\nP(1, y) \\Rightarrow f(1 + y^2 f(y)) = f(1 + y). \\qquad (1)\n$$\n$$\nP(x, 1) \\Rightarrow f(x + 1) = f(1 + f(x)) f(x). \\qquad (2)\n$$\n比較 $P(1 + x^2 f(x), y)$ 和 $P(1 + y^2 f(y), x)$，有\n$$\n\\begin{aligned}\nf(1 + x^2 f(x) + y^2 f(y)) &... | Taiwan | 2022 數學奧林匹亞競賽第二階段選訓營, 獨立研究 (二) | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | Chinese; English | proof and answer | f(x) = 1 for all x > 0; and f(x) = 1/x for all x > 0 | |
0434 | There are 10 cards, each of which has two numbers, numbered $1$, $2$, $3$, $4$, $5$, written on it, and the numbers on any two cards are not exactly identical. The 10 cards are placed in five boxes labelled $1$, $2$, $3$, $4$, $5$, and a card with $i$ and $j$ written on it can only be placed in box $i$ or $j$. One plac... | [
"Denote the card with $i, j$ written on it as $\\{i, j\\}$. It is easy to know that these 10 cards are exactly $\\{i, j\\}$ ($1 \\le i < j \\le 5$).\n\nConsider the “good” placements of the cards. There are 10 cards in the five boxes, so there are at least 3 cards in box 1. The only cards that will fit in box 1 are... | China | China Mathematical Competition | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 120 | |
0a5d | Problem:
Find all triples $(x,p,n)$ of non-negative integers such that $p$ is prime and
$$2x(x + 5) = p^{n} + 3(x - 1).$$ | [
"Solution:\nThe equation rearranges to be\n$$p^{n} = 2x(x + 5) - 3(x - 1) = 2x^{2} + 7x + 3 = (2x + 1)(x + 3).$$\nSince $x$ is a non-negative integer, both factors $(2x + 1)$ and $(x + 3)$ must be positive integers. Therefore both $(2x + 1)$ and $(x + 3)$ are both powers of $p$. Let\n$$2x + 1 = p^{a}$$\n$$x + 3 = p... | New Zealand | NZMO Round One | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis... | null | proof and answer | (2, 5, 2) and (0, 3, 1) | |
08se | How many pairs of integers $(a, b)$ satisfy $a^2 b^2 = 4a^5 + b^3$? | [
"If $a$ or $b$ is $0$, both must be $0$.\n\nConsider the case $a, b \\neq 0$. Let $g > 0$ be the G.C.D. of $a$ and $b$, and $a = g a'$, $b = g b'$. Substituting them into the given equation we get\n$$\nga'^2 (b'^2 - 4g a'^3) = b'^3.\n$$\nSince $a'$ divides $b'^3$ and $a'$ and $b'$ are relatively prime, $a' = \\pm 1... | Japan | Japan 2007 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 8 | |
06rc | The rows and columns of a $2^{n} \times 2^{n}$ table are numbered from $0$ to $2^{n}-1$. The cells of the table have been colored with the following property being satisfied: for each $0 \leq i, j \leq 2^{n}-1$, the $j$th cell in the $i$th row and the $(i+j)$th cell in the $j$th row have the same color. (The indices of... | [
"Throughout the solution we denote the cells of the table by coordinate pairs; $(i, j)$ refers to the $j$th cell in the $i$th row.\n\nConsider the directed graph, whose vertices are the cells of the board, and the edges are the arrows $(i, j) \\rightarrow (j, i+j)$ for all $0 \\leq i, j \\leq 2^{n}-1$. From each ve... | IMO | 51st IMO Shortlisted Problems | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic",
"... | English | proof only | null | |
0brk | Given a positive integer $n$, show that for no set of integers modulo $n$, whose size exceeds $1 + \sqrt{n+4}$, is it possible that the pairwise sums of unordered pairs be all distinct. | [
"Let $S \\subseteq \\mathbb{Z}/n\\mathbb{Z}$ be a set whose pairwise sums of unordered pairs are distinct, and consider the pairwise differences of *ordered* pairs.\nThe crucial observation is that if a difference $d \\neq 0$ occurs twice in $S$, then these two occurrences must be adjacent in a 3-term arithmetic pr... | Romania | 67th NMO Selection Tests for BMO and IMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | English | proof only | null | |
0dpu | For every positive integer $m$ prove the inequality $|\{\sqrt{m}\} - \frac{1}{2}| > \frac{1}{8(\sqrt{m}+1)}$. | [
"Let $[\\sqrt{m}] = k$. Then\n$$\n|\\{\\sqrt{m}\\} - \\frac{1}{2}| = \\frac{1}{2}|2k + 1 - 2\\sqrt{m}| = \\frac{1}{2} \\cdot \\frac{|4k^2 + 4k + 1 - 4m|}{2k + 1 + 2\\sqrt{m}}.\n$$\nThe numerator of the last fraction is an odd positive integer, and therefore at least $1$. Thus,\n$$\n|\\{\\sqrt{m}\\} - \\frac{1}{2}| ... | Silk Road Mathematics Competition | SILK ROAD MATHEMATICS COMPETITION XX | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0idr | Problem:
A triangle has side lengths $18$, $24$, and $30$. Find the area of the triangle whose vertices are the incenter, circumcenter, and centroid of the original triangle. | [
"Solution:\n\nThere are many solutions to this problem, which is straightforward. The given triangle is a right $3$-$4$-$5$ triangle, so the circumcenter is the midpoint of the hypotenuse. Coordinatizing for convenience, put the vertex at $(0,0)$ and the other vertices at $(0,18)$ and $(24,0)$. Then the circumcente... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 3 | |
0iq8 | Problem:
Evaluate the sum
$$
1 - 2 + 3 - 4 + \cdots + 2007 - 2008
$$ | [
"Solution:\nEvery odd integer term can be paired with the next even integer, and this pair sums to $-1$. There are $1004$ such pairs, so the total sum is $-1004$."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | -1004 | |
03q3 | The solution set of the inequality $|x|^3 - 2x^2 - 4|x| + 3 < 0$ is
________. | [
"Notice that $|x| = 3$ is a root of the equation $|x|^3 - 2x^2 - 4|x| + 3 = 0$. Then the original inequality can be rewritten as $(|x| - 3)(|x|^2 + |x| - 1) < 0$, that is\n$$\n(|x| - 3)\\left(|x| - \\frac{-1 + \\sqrt{5}}{2}\\right)\\left(|x| - \\frac{-1 - \\sqrt{5}}{2}\\right) < 0.\n$$\nSince $|x| - \\frac{-1 - \\s... | China | China Mathematical Competition (Shaanxi) | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | (-3, -(\sqrt{5} - 1)/2) \cup ((\sqrt{5} - 1)/2, 3) | |
0dz3 | Let the numbers $a$ and $b$ be such that $a^3 + b^3 = 13$ and $a^9 + b^9 = -299$. Find the value of $ab$, given that $ab$ is real. | [
"**First solution**\nSince $a^9 + b^9 = (a^3 + b^3)(a^6 - a^3b^3 + b^6)$, we have $a^6 - a^3b^3 + b^6 = -\\frac{299}{13} = -23$.\n\nSubtracting this equality from $a^6 + 2a^3b^3 + b^6 = (a^3 + b^3)^2 = 13^2 = 169$ we get $3a^3b^3 = 192$, which implies $ab = 4$ since $ab$ is a real number.\n\n\n**Second solution**\n... | Slovenia | Slovenija 2008 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 4 | |
0cbw | Find all positive integers $n$ so that there are exactly $1 + 2^{n+2}$ integers between $3^n$ and $3^{n+1}$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2 | |
0ewq | Problem:
Given a circle $c$ and two fixed points $A$, $B$ on it. $M$ is another point on $c$, and $K$ is the midpoint of $BM$. $P$ is the foot of the perpendicular from $K$ to $AM$.
a. Prove that $KP$ passes through a fixed point (as $M$ varies).
b. Find the locus of $P$. | [
"Solution:\n\na.\nTake $Y$ on the circle so that $\\angle ABY = 90^\\circ$. Then $AY$ is a diameter and so $\\angle AMY = 90^\\circ$. Take $X$ as the midpoint of $BY$. Then triangles $BXK$ and $BYM$ are similar, so $XK$ is parallel to $YM$. Hence $XK$ is perpendicular to $AM$, and so $P$ is the intersection of $XK$... | Soviet Union | 3rd ASU | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | a) KP always passes through X, where Y is the point on the circle such that angle ABY is a right angle and X is the midpoint of BY. b) The locus of P is the circle with diameter AX. | |
00jb | A postman has $n$ parcels of weights $1$, $2$, $3$, $4$, $\ldots$, $n$. He wants to divide the parcels into three groups of equal weight. Is this possible for
a. $n = 2011$
b. $n = 2012$? | [
"In the first case, the total weight\n$$\n1 + 2 + \\cdots + 2010 + 2011 = \\frac{2011 \\cdot 2012}{2}\n$$\nis not a multiple of $3$. Therefore, there is no solution in this case.\n\nIn the second case, we distribute the first $8$ parcels as follows: Parcels $1$, $2$, $3$, $6$ (of total weight $12$) are put into the... | Austria | Austrian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | null | proof and answer | a) No; b) Yes | |
0363 | Problem:
Two tangent circles with centers $O_{1}$ and $O_{2}$ are inscribed in a given angle. Prove that if a third circle with center on the segment $O_{1} O_{2}$ is inscribed in the angle and passes through one of the points $O_{1}$ and $O_{2}$ then it passes through the other one too. | [
"Solution:\nDenote the circles by $k_{1}(O_{1}, r)$, $k_{2}(O_{2}, R)$ and $k(O, x)$, where $r < x < R$. Let $A$, $B$ and $C$ be the feet of the perpendiculars from $O_{1}$, $O$ and $O_{2}$, respectively, to the arm $p$ of the given angle $O p q$. Let $l$ be the line through $O_{1}$ parallel to $p$ and let $l$ meet... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0k6n | Problem:
The polynomial $x^{3}-3 x^{2}+1$ has three real roots $r_{1}$, $r_{2}$, and $r_{3}$. Compute
$$
\sqrt[3]{3 r_{1}-2}+\sqrt[3]{3 r_{2}-2}+\sqrt[3]{3 r_{3}-2}
$$ | [
"Solution:\n\nLet $r$ be a root of the given polynomial. Then\n$$\nr^{3}-3 r^{2}+1=0 \\Longrightarrow r^{3}-3 r^{2}+3 r-1=3 r-2 \\Longrightarrow r-1=\\sqrt[3]{3 r-2}\n$$\nNow by Vieta the desired value is $r_{1}+r_{2}+r_{3}-3=3-3=0$."
] | United States | HMMT November 2019 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 0 | |
0el2 | Problem:
Na matematičnem tekmovanju so tekmovalci reševali 3 naloge. Prvo nalogo je pravilno rešilo $90 \%$ tekmovalcev, drugo $80 \%$ in tretjo $70 \%$ tekmovalcev. Najmanj koliko odstotkov tekmovalcev je pravilno rešilo vse tri naloge?
(A) 30
(B) 35
(C) 40
(D) 50
(E) 70 | [
"Solution:\n\nDenimo, da je na tekmovanju sodelovalo $n$ tekmovalcev. Če sta $A$ in $B$ dve množici tekmovalcev, potem je $n \\geq |A \\cup B| = |A| + |B| - |A \\cap B|$ oziroma $|A \\cap B| \\geq |A| + |B| - n$. Ker je prvo nalogo pravilno rešilo $\\frac{9}{10} n$ tekmovalcev, drugo nalogo pa $\\frac{8}{10} n$ tek... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | MCQ | C | |
0kd0 | Problem:
Two players play a game where they are each given 10 indistinguishable units that must be distributed across three locations. (Units cannot be split.) At each location, a player wins at that location if the number of units they placed there is at least 2 more than the units of the other player. If both player... | [
"Solution:\n\nBy stars and bars, the total number of distributions is $\\binom{12}{2}^2 = 66^2$.\n\nIf no locations are won, either both distributions are identical or the difference between the two is $(1,0,-1)$, in some order.\n\nThe first case has 66 possibilities.\n\nIf the difference is $(1,0,-1)$, we can cons... | United States | HMMO 2020 | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 1011 | |
01hn | Let $p$ be a polynomial with real coefficients and $n \ge 1$ be an integer. Prove that there exists a non-zero polynomial $q$ such that the coefficients of $p \cdot q$ vanish for each power that is not a multiple of $n$. | [
"Let $x$ be the variable. We want to find a real non-zero polynomial $q$ such that $p(x) \\cdot q(x) = s(x^n)$ for some real polynomial $s$. If $p$ is the zero polynomial then $p \\cdot q = 0$ for every polynomial $q$. It can therefore be assumed that $p \\neq 0$.\n\nLet $y = x^n$. Then $y^m$ is a real polynomial f... | Baltic Way | Baltic Way 2021 Shortlist | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
0l7l | Let $k$ and $d$ be positive integers. Prove that there exists a positive integer $N$ such that for every odd integer $n > N$, the digits in the base-$2n$ representation of $n^k$ are all greater than $d$. | [
"Let $k$ and $d$ be given positive integers.\n\nLet $n$ be an odd integer, and consider the base-$2n$ representation of $n^k$.\n\nRecall that in base-$b$, the digits of a number $x$ are the coefficients $a_i$ in the expansion:\n$$\nx = a_0 + a_1 b + a_2 b^2 + \\cdots + a_m b^m,\n$$\nwhere $0 \\leq a_i < b$.\n\nWe w... | United States | 16th United States of America Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
07ah | Do there exist natural numbers $a$, $b$ and $c$ such that $a^2 + b^2 + c^2$ is divisible by $2013(ab + bc + ca)$? | [
"**Lemma 1.** Let $A \\equiv 2 \\pmod{3}$ be a positive integer. Then there exists a prime number $p$ such that $p \\equiv 2 \\pmod{3}$ and $p^\\alpha \\mid A$ where $\\alpha$ is an odd integer.\n\n**Proof of lemma.** Assume to the contrary that there is not such a prime number $p$. Therefore, if $A = p_1^{\\alpha_... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | No, such natural numbers do not exist. | |
0bee | A function $f : (0, \infty) \to (0, \infty)$ is called *contractive* if, for every numbers $x, y \in (0, \infty)$, we have $\lim_{n \to \infty} (f^n(x) - f^n(y)) = 0$, where $f^n = f \circ f \circ \dots \circ f$. Prove
a) If $f : (0, \infty) \to (0, \infty)$ is contractive, continuous and has a fixed point (there is $... | [
"a) Suppose the contrary, that is $f$ has another fixed point $x_1 \\in (0, \\infty) \\setminus \\{x_0\\}$. Then $\\lim_{n \\to \\infty} (f^n(x_0) - f^n(x_1)) = x_0 - x_1 \\ne 0$, a contradiction. In this case, the continuity of $f$ (intermediate value property) implies $f(x) < x$, for all $x \\in (0, x_0)$ or $f(x... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
069b | If $\alpha$, $\beta$, $\gamma$ are positive real numbers such that $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = 3$, prove that
$$
\frac{\alpha + \beta}{\alpha^2 + \alpha\beta + \beta^2} + \frac{\beta + \gamma}{\beta^2 + \beta\gamma + \gamma^2} + \frac{\gamma + \alpha}{\gamma^2 + \gamma\alpha + \alpha^2} \le... | [
"By putting $x = \\frac{1}{\\alpha}$, $y = \\frac{1}{\\beta}$, $z = \\frac{1}{\\gamma}$, we have $x + y + z = 3$ and the inequality takes the form:\n$$\n\\frac{xy(x+y)}{x^2+xy+y^2} + \\frac{yz(y+z)}{y^2+yz+z^2} + \\frac{zx(z+x)}{z^2+zx+x^2} \\le 2.\n$$\nWe have $\\frac{xy}{x^2+xy+y^2} \\le \\frac{1}{3} \\Leftrighta... | Greece | SELECTION EXAMINATION 2019 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 2; equality when α = β = γ = 1. | |
02at | Problem:
O número
$$
A=(\sqrt{6}+\sqrt{2})(\sqrt{3}-2) \sqrt{\sqrt{3}+2}
$$
é igual a:
(a) $-\sqrt{3}$
(b) $-\sqrt{2}$
(c) -2
(d) 1
(d) 2 | [
"Solution:\nComo\n$$\n\\begin{aligned}\nA^{2} & =[(\\sqrt{6}+\\sqrt{2})(\\sqrt{3}-2) \\sqrt{\\sqrt{3}+2}]^{2} \\\\\n& =(\\sqrt{6}+\\sqrt{2})^{2}(\\sqrt{3}-2)^{2}(\\sqrt{\\sqrt{3}+2})^{2} \\\\\n& =(\\sqrt{6}+\\sqrt{2})^{2}(\\sqrt{3}-2)^{2}(\\sqrt{3}+2) \\\\\n& =(\\sqrt{6}+\\sqrt{2})^{2}(\\sqrt{3}-2)[(\\sqrt{3}-2)(\\... | Brazil | null | [
"Algebra > Intermediate Algebra > Other"
] | null | MCQ | c | |
0ct7 | Let $f(x)$ be a cubic polynomial. We say that a triple $(a, b, c)$ of distinct real numbers is a cycle if $f(a) = b$, $f(b) = c$, and $f(c) = a$. Assume that there exist eight cycles $(a_i, b_i, c_i)$, $i = 1, 2, \dots, 8$, containing 24 distinct real numbers. Prove that among eight sums of the form $a_i + b_i + c_i$, ... | [
"If four of the sums of the form $a_i + b_i + c_i$ are equal to $s$, then the corresponding 12 numbers of the form $a_i, b_i, c_i$ are roots of the polynomial $x + f(x) + f(f(x)) - s$, whose degree is 9.",
"Предположим противное; тогда у каких-то четырёх циклов $(a_i, b_i, c_i)$ суммы чисел одинаковы и равны неко... | Russia | Russian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English; Russian | proof only | null | |
0esg | If $x - y > x$, then which of the sentences MUST be true?
(A) $x > 0$ (B) $y < 0$ (C) $x > y$ (D) $y > 0$ (E) $x < 0$ | [
"If $x - y > x$ then $-y > 0$, so $y < 0$"
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | MCQ | B | |
0ao1 | Problem:
The polynomial function $p(x)$ has the form $x^{10} - 4x^{9} + \ldots + a x + k$ where $a, k \in \mathbb{R}$. If $p(x)$ has integral zeros, find the minimum possible positive value of $k$. | [] | Philippines | Area Stage | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 3 | |
0jty | Problem:
Let $A$ denote the set of all integers $n$ such that $1 \leq n \leq 10000$, and moreover the sum of the decimal digits of $n$ is $2$. Find the sum of the squares of the elements of $A$. | [
"Solution:\nFrom the given conditions, we want to calculate\n$$\n\\sum_{i=0}^{3} \\sum_{j=i}^{3}\\left(10^{i}+10^{j}\\right)^{2}\n$$\nBy observing the formula, we notice that each term is an exponent of $10$. $10^{6}$ shows up $7$ times, $10^{5}$ shows up $2$ times, $10^{4}$ shows up $9$ times, $10^{3}$ shows up $4... | United States | HMMT February 2016 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 7294927 | |
0ksx | Problem:
In a country with $n$ cities, there is a recurrent one-way flight between every pair of cities. Every flight has a constant price in the range $\$100, \$120, \$140, \$160, \$180$. A $\$N$ flight ticket gives unlimited access to flights which cost $\$N$, and tickets can be traded for tickets of lower prices. F... | [
"Solution:\n\nFor each city $C$, let $f_C(\\$N)$ be the maximum distance one can travel starting at $C$ with an $\\$N$ ticket. If Aerith cannot take 4 flights in a row, $f_C$ is a nondecreasing function that remains between $0$ and $3$, of which there are $\\binom{4+5-1}{5-1} = \\binom{8}{4} = 70$ such functions. H... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 57 | |
00mb | Es seien $ABC$ ein spitzwinkeliges Dreieck, $H$ sein Höhenschnittpunkt und $D$, $E$ und $F$ die Fußpunkte der Höhen durch $A$, $B$ bzw. $C$. Der Schnittpunkt von $DF$ mit der Höhe durch $B$ sei $P$. Die Normale auf $BC$ durch $P$ schneide die Seite $AB$ in $Q$. Der Schnittpunkt von $EQ$ mit der Höhe durch $A$ sei $N$.
... | [
"Siehe Abbildung 1. Es seien $\\beta = \\angle ABC$ und $\\gamma = \\angle ACB$. Wegen $\\angle AFH = \\angle AEH = 90^\\circ$ ist $AFHE$ ein Sehnenviereck, und es folgt aus der Parallelität von $DA$ und $PQ$ die Beziehung $\\angle FQP = \\angle FAH = \\angle FEH = \\angle FEP$. Daher ist auch $QFPE$ ein Sehnenvier... | Austria | 48. Österreichische Mathematik-Olympiade | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometr... | German | proof only | null | |
0643 | Problem:
Die reellen Zahlen $r_{1}, r_{2}, \ldots, r_{2019}$ erfüllen die Bedingungen $r_{1}+r_{2}+\ldots+r_{2019}=0$ sowie $r_{1}^{2}+r_{2}^{2}+\ldots+r_{2019}^{2}=1$. Es sei $a=\min \left(r_{1}, r_{2}, \ldots, r_{2019}\right)$ und $b=\max \left(r_{1}, r_{2}, \ldots, r_{2019}\right)$. Man beweise: $a b \leq \frac{-1}... | [
"Solution:\n\nWeil wegen (2) die $r_{i}$ nicht alle $0$ sein und wegen (1) nicht alle das gleiche Vorzeichen haben können, gilt $b>0$ und $a<0$. Mit $P=\\left\\{i: u_{i}>0\\right\\}$ und $N=\\left\\{i: u_{i} \\leq 0\\right\\}$ sowie $p=|P|$ und $n=|N|$ gilt $p+n=2019$ und aus (1) folgt\n\n$0=\\sum_{i=1}^{2019} u_{i... | Germany | Auswahlklausur | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
06fj | Let $x_1$ and $y_1$ be positive integers satisfying $x_1 + y_1 = 2^r$, where $r$ is a positive integer. If $x_1 < y_1$, let $x_2 = 2x_1$ and $y_2 = 2^r - x_2$. On the other hand, if $x_1 > y_1$, then we let $y_2 = 2y_1$ and $x_2 = 2^r - y_2$. Apply the same procedure to $(x_2, y_2)$, and so on. Show that the procedure ... | [
"Note that for any positive integers $x$ and $y$ satisfying $x + y = 2^r$, we have\n$$\nv_2(x) = v_2(2^r - y) = v_2(y),\n$$\nwhere $v_2(n)$ is the largest $m$ such that $2^m \\mid n$. Therefore, WLOG we may assume $x_k < y_k$ in some step. Since\n$$\nv_2(x_{k+1}) = v_2(2x_k) = v_2(x_k) + 1,\n$$\nthe sequence $\\{v_... | Hong Kong | Year 2008 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0a92 | Problem:
The squares of a $100 \times 100$ chessboard are painted with 100 different colours. Each square has only one colour and every colour is used exactly 100 times. Show that there exists a row or a column on the chessboard in which at least 10 colours are used. | [
"Solution:\n\nDenote by $R_{i}$ the number of colours used to colour the squares of the $i$'th row and let $C_{j}$ be the number of colours used to colour the squares of the $j$'th column. Let $r_{k}$ be the number of rows on which colour $k$ appears and let $c_{k}$ be the number of columns on which colour $k$ appe... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 20 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
05t2 | Problem:
On dit qu'une suite $\left(u_{n}\right)_{n \geqslant 0}$ d'entiers naturels non nuls est sicilienne si
$$
u_{n+1} \in\left\{u_{n} / 2, u_{n} / 3, 2 u_{n}+1, 3 u_{n}+1\right\}$$
pour tout $n \geqslant 0$. Démontrer que, pour tout entier $k \geqslant 1$, il existe une suite sicilienne $\left(u_{n}\right)_{n \ge... | [
"Solution:\n\nNous allons démontrer, pour tout entier $k \\geqslant 2$, la propriété $\\mathcal{P}_{k}$ suivante : il existe une suite sicilienne finie $\\left(u_{0}, \\ldots, u_{\\ell}\\right)$ telle que $u_{0}=k$ et $u_{\\ell}<k$.\n\nLa propriété $\\mathcal{P}_{k}$ est immédiate si $k$ est pair, puisqu'il suffit ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0j2t | Problem:
How many ordered pairs $(S, T)$ of subsets of $\{1,2,3,4,5,6,7,8,9,10\}$ are there whose union contains exactly three elements? | [
"Solution:\nLet the three elements in the union be $a$, $b$, and $c$. We know that $a$ can be only in $S$, only in $T$, or both, so there are 3 possibilities for placing it. (Recall that $S=\\{a\\}$, $T=\\{b, c\\}$ is different from $S=\\{b, c\\}$, $T=\\{a\\}$ because $S$ and $T$ are an ordered pair.) Likewise for ... | United States | Harvard-MIT November Tournament | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 3240 | |
0k26 | Problem:
Triangle $\triangle ABC$ has $AB = 21$, $BC = 55$, and $CA = 56$. There are two points $P$ in the plane of $\triangle ABC$ for which $\angle BAP = \angle CAP$ and $\angle BPC = 90^{\circ}$. Find the distance between them. | [
"Solution:\n\nLet $P_1$ and $P_2$ be the two possible points $P$, with $AP_1 < AP_2$. Both lie on the $\\angle A$-bisector and the circle $\\gamma$ with diameter $BC$. Let $D$ be the point where the $\\angle A$-bisector intersects $BC$, let $M$ be the midpoint of $BC$, and let $X$ be the foot of the perpendicular f... | United States | HMMT February 2018 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | 5*sqrt(409)/2 | |
0kmb | Problem:
Let $f(x) = x^{2} + x + 1$. Determine, with proof, all positive integers $n$ such that $f(k)$ divides $f(n)$ whenever $k$ is a positive divisor of $n$. | [
"Solution:\nThe answer is $n$ can be $1$, a prime that is $1 \\bmod 3$, or the square of any prime except $3$. It is easy to verify that all of these work.\n\nFirst note that $n$ must be $1 \\bmod 3$ since $1$ divides $n$ implies $f(1)$ divides $f(n)$.\n\nNext, suppose for sake of contradiction that $n = a b$, with... | United States | HMMT Spring 2021 Team Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | n equals 1, or a prime congruent to 1 modulo 3, or the square of any prime other than 3 | |
0b79 | Let $ABCD$ be a cyclic quadrilateral. The lines $AD$, $BC$ meet at $P$; $AB$, $CD$ at $Q$; and $AC$, $BD$ at $R$. The perpendicular bisectors of $AB$, respectively $BC$, meet $PR$ at $X$, respectively $QR$ at $Y$. Prove that $XY$ passes through $B$. | [
"All poles and polars are considered with respect to the given circumcircle of $ABCD$.\nTo start with, notice that line $q = PR$ is the polar of $Q$ and line $p = QR$ is the polar of $P$. As $Y$ lies on the polar of $P$, it follows that $P$ lies on the polar $y$ of $Y$. The pole of the line $OY$ is the point at inf... | Romania | Local Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | English | proof only | null | |
00bl | Let $\triangle ABC$ be a triangle with perimeter $100$ and incenter $I$. The parallel to $AB$ through $I$ divides the median through $A$ in ratio $7:3$, counted from $A$. Find the length of side $AB$. | [
"Let $AM$ be the median through $A$, $CJ$ the bisector through $C$, and let the parallel to $AB$ through $I$ intersect $AM$ at $P$. Set $AP:PM = \\lambda$; in our problem, $\\lambda = \\frac{7}{3}$.\n\nIf $N$ is the midpoint of $CL$ then $MN \\parallel AB$ as $M$ is the midpoint of $BC$. Hence $MN \\parallel IP$. N... | Argentina | Argentina_2018 | [
"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 and answer | 35 | |
0kym | Consider the following operation. Given a positive integer $n$, if $n$ is a multiple of $3$, then you replace $n$ by $\frac{n}{3}$. If $n$ is not a multiple of $3$, then you replace $n$ by $n+10$. Then continue this process. For example, beginning with $n=4$, this procedure gives $4 \rightarrow 14 \rightarrow 24 \right... | [
"**Answer (C):** The first several iterations give\n$$\n100 \\rightarrow 110 \\rightarrow 120 \\rightarrow 40 \\rightarrow 50 \\rightarrow 60 \\rightarrow 20 \\rightarrow 30 \\rightarrow 10 \\rightarrow 20 \\rightarrow 30 \\rightarrow 10 \\rightarrow \\dots\n$$\nThe values will then continue to cycle through $20 \\... | United States | AMC 10 A | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | null | MCQ | C | |
0huw | Problem:
Let $ABC$ be an acute triangle with orthocenter $H$, and let $M$ and $N$ denote the midpoints of $AB$ and $AC$. Rays $MH$ and $NH$ intersect the circumcircle of $ABC$ again at points $X$ and $Y$. Prove that the four points $M, N, X, Y$ lie on a circle. | [
"Solution:\nLet $\\Omega$ be the circumcircle of $\\triangle ABC$. Let $P$ and $Q$ be the reflections of $H$ across $M$ and $N$.\n\nSince $\\angle APB = \\angle AHB = 180^\\circ - \\angle C$, point $P$ lies on $\\Omega$; thus so does $Q$. Moreover, $MN \\parallel PQ$, so\n$$\n\\angle XMN = ... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05la | Problem:
Soit $n$ un entier strictement positif. On considère $2n+1$ entiers distincts compris au sens large entre $-2n+1$ et $2n-1$. Montrer qu'on peut en choisir 3 dont la somme soit nulle. | [
"Solution:\n\nOn raisonne par récurrence sur $n$.\n\nInitialisation : Pour $n=1$, on prend $2n+1=3$ entiers distincts parmi $-1, 0, 1$, donc ces trois entiers. Leur somme est bien nulle.\n\nHérédité : Supposons l'énoncé vrai au rang $n$ : si on prend $2n+1$ entiers distincts compris au sens large entre $-2n+1$ et $... | France | Envoi de combinatoire | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0b2o | Problem:
Find the smallest positive integer $n$ for which there are exactly 2323 positive integers less than or equal to $n$ that are divisible by 2 or 23, but not both. | [
"Solution:\nThe number of positive integers from $1$ to $n$ that are divisible by $2$ or $23$, but not both, is\n$$\nf(n) = \\left\\lfloor \\frac{n}{2} \\right\\rfloor + \\left\\lfloor \\frac{n}{23} \\right\\rfloor - 2\\left\\lfloor \\frac{n}{46} \\right\\rfloor\n$$\nWe need to find $f(n) = 2323$, which can be done... | Philippines | 23rd Philippine Mathematical Olympiad Qualifying Stage | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 4644 | |
00b2 | A positive integer is written in each box of a $4 \times 4$ board, so that the 16 numbers are all different. For every row and every column, the number written in one of its boxes equals the sum of the remaining three. Let $M$ be the greatest of the 16 numbers. Find the minimum possible value of $M$. | [
"For $i = 1, \\dots, 4$, let $a_i$ be the maximum number in column $i$, and let $b_1, b_2, \\dots, b_{12}$ be the remaining 12 numbers written on the board (different from $a_1, a_2, a_3, a_4$). Then, for every $i$, $a_i$ is the sum of the other three numbers in column $i$; therefore,\n$$\na_1 + a_2 + a_3 + a_4 = b... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 21 | |
0heb | Let $D$ be the point on the arc $AC$ of the circumcircle of the triangle $ABC$ ($AB < BC$), that doesn't contain point $B$. Let $X$ and $X'$ be any two points on the side $AC$ such that $\angle ABX = \angle CBX'$. Show that regardless of the choice of point $X$, the circumcircle of $\triangle DXX'$ passes through a fix... | [
"Let $W$ be the circumcircle of $\\triangle ABC$, let $w$ be the circumcircle of $\\triangle BXX'$, and let $v$ be the circumcircle of $\\triangle DXX'$. Let $TB$ be a tangent line to the circle $W$, where $T$ belongs to the line $AC$. Then $TB^2 = TA \\cdot TC$, as well as (Fig. 12)\n$$\n\\angle TBX = \\angle TBA ... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0leu | Given a convex polyhedron with 2022 faces. In 3 arbitrary faces, there are already numbers 26, 4 and 2022 (each face contains one number). One wants to fill in each other face a real number which is the arithmetic mean of every number in faces that share a common edge with that face. Prove that there is exactly one way... | [
"First, we will prove the following lemma:\n\n**Lemma.** Given a positive integer $n$. Prove that the system of linear equations with $n$ variables $(x_1, x_2, \\dots, x_n)$\n$$\n\\left\\{ \\begin{array}{l} a_{11}x_1 + \\cdots + a_{1n}x_n = b_1, \\\\ \\cdots \\\\ a_{n1}x_1 + \\cdots + a_{nn}x_n = b_n \\end{array} \... | Vietnam | TST | [
"Algebra > Linear Algebra > Matrices",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
06sv | Let $n \geqslant 2$ be an integer. Consider an $n \times n$ chessboard divided into $n^{2}$ unit squares. We call a configuration of $n$ rooks on this board happy if every row and every column contains exactly one rook. Find the greatest positive integer $k$ such that for every happy configuration of rooks, we can find... | [
"Answer. $\\lfloor\\sqrt{n-1}\\rfloor$.\n\nLet $\\ell$ be a positive integer. We will show that (i) if $n>\\ell^{2}$ then each happy configuration contains an empty $\\ell \\times \\ell$ square, but (ii) if $n \\leqslant \\ell^{2}$ then there exists a happy configuration not containing such a square. These two stat... | IMO | 55th International Mathematical Olympiad Shortlist | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | ⌊√(n−1)⌋ | |
023d | Problem:
Seja $x_{1}, x_{2}, \ldots, x_{n}$ uma sequência na qual cada termo é $0$, $1$ ou $-2$. Se
$$
\left\{
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=-5 \\
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=19
\end{array}
\right.
$$
determine $x_{1}^{5}+x_{2}^{5}+\cdots+x_{n}^{5}$. | [
"Solution:\n\nSejam $a$ a quantidade de termos iguais a $1$ e $b$ a quantidade de termos iguais a $-2$. Podemos escrever:\n$$\n\\left\\{\n\\begin{array}{l}\na \\cdot 1 + b \\cdot ( -2 ) = -5 \\\\\na \\cdot 1^{2} + b \\cdot ( -2 )^{2} = 19\n\\end{array}\n\\Longleftrightarrow\n\\left\\{\n\\begin{array}{l}\na - 2b = -... | Brazil | Nível 3 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | -125 | |
07p1 | Let $ABC$ be a triangle with greatest side $BC$ and let $D$ and $E$ be internal points on $AB$ and $AC$, respectively, such that $|AD| = |AE|$. Let $F$ and $G$ be internal points on $BC$ such that $|BD| = |BG|$ and $|CF| = |CE|$. Prove that $D$, $E$, $F$ and $G$ are concyclic. | [
"The relative position of $F$ and $G$ leads to different situations. If $F = G$ we have nothing to show. There remain two cases to consider: case (i) $F$ is between $C$ and $G$; case (ii) $G$ is between $C$ and $F$.\n\ncase (i)\n\ncase (ii)\nIn both cases, denote $\... | Ireland | Irska 2014 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07c6 | Points $X$ and $Y$ are located on sides $AB$ and $AC$ of triangle $ABC$ ($X, Y \neq A$) such that the reflection of line $BC$ with respect to $XY$ is tangent to the circumcircle of triangle $AXY$. If $O$ denotes the circumcenter of triangle $ABC$, prove that the circumcircle of triangle $AXY$ is tangent to the circumci... | [
"Let $\\omega$ be the circumcircle of triangle $AXY$. Since the reflection of line $BC$ with respect to $XY$ is tangent to $\\omega$, the reflection of $\\omega$ with respect to $XY$ is tangent to line $BC$ at a point that is denoted by $T$.\n\nLet $P$ be the second intersection point of circumcircles of triangles ... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
08d7 | Problem:
La pianta di un castello è realizzata in questo modo: si consideri una circonferenza lunga $2019$ metri con inscritto un poligono regolare di $2019$ vertici. Una volta numerati i vertici del poligono da $1$ a $2019$ in senso orario si traccino delle circonferenze di lunghezza $2019$ m centrate in ogni punto n... | [
"Solution:\n\nLa risposta è (B). Siano $\\Gamma$ la circonferenza iniziale e $C_{1}, C_{2}, \\ldots, C_{44}$ le circonferenze disegnate con centro nei vertici del poligono etichettati con quadrati perfetti; si noti che, dato che il raggio $R$ di ciascuna circonferenza $C_{i}$ è uguale al raggio di $\\Gamma$ (tutte ... | Italy | Progetto Olimpiadi della Matematica | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | MCQ | B | |
05oo | Problem:
Pour quel $n \geqslant 1$ peut-on remplir un échiquier $n \times n$ avec des pièces de la forme
sans qu'elles se recouvrent? (Les rotations sont autorisées). | [
"Solution:\nSupposons que nous avons réussi à recouvrir un échiquier $n \\times n$. Une pièce recouvre 4 cases. Donc clairement 4 divise $n^{2}$ donc $n$ est pair. Maintenant, chaque pièce recouvre soit trois cases noires et une case blanche de l'échiquier, soit l'inverse. Or, comme $n$ est pair, il y a autant de c... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Exactly when n is divisible by 4 | |
0g1x | Problem:
Déterminer toutes les paires $(f, g)$ de fonctions $f, g: \mathbb{R} \rightarrow \mathbb{R}$ telles que pour tous $x, y \in \mathbb{R}$
- $f(x) \geq 0$,
- $f(x+g(y))=f(x)+f(y)+2 y g(x)-f(y-g(y))$. | [
"Solution:\n\nSoit $(f, g)$ une solution du problème. Si $g \\equiv 0$, alors toutes les fonctions $f \\geq 0$ satisfont le problème. On obtient la première solution.\n\nSupposons maintenant qu'il existe $a$ tel que $g(a) \\neq 0$.\n\nSubstituer $x=y-g(y)$ donne $y g(y-g(y))=0$ pour tout $y$ et donc $g(y-g(y))=0$ p... | Switzerland | SMO-Selektion | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | Two families:
1) g(x) ≡ 0 for all x, and f is any function with f(x) ≥ 0 for all x.
2) g(x) = x for all x, and f(x) = x^2 + b x + c with c ≥ b^2/4. | |
0fcz | Problem:
Demostrad que si $-1 < x < 1$, $-1 < y < 1$,
$$
\left|\frac{x-y}{1-x y}\right| \leq \frac{|x|+|y|}{1+|x y|}
$$ | [
"Solution:\nSi $x, y$ tienen signos opuestos, se tiene $|x-y| = |x| + |y|$, $|1-x y| = 1 - x y = 1 + |x y|$.\nAsí pues, la desigualdad es, realmente, una igualdad.\nSi la desigualdad se cumple para un par de números $(x, y)$, se cumple para el par opuesto $(-x, -y)$. Por tanto, sin pérdida de generalidad, podemos s... | Spain | null | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
03xo | The minimum of $y = (a \cos^2 x - 3) \sin x$ is -3. Then the range of real number $a$ is ______. | [
"Let $\\sin x = t$. The expression is then changed to\n$$\ng(t) = (-a t^2 + a - 3)t,\n$$\nor\n$$\ng(t) = -a t^3 + (a-3)t.\n$$\nFrom $-a t^3 + (a-3)t \\ge -3$, we get\n$$\n-a t (t^2 - 1) - 3 (t - 1) \\ge 0,\n$$\n$$\n(t - 1)(-a t (t + 1) - 3) \\ge 0.\n$$\nSince $t - 1 \\le 0$, we have $-a t (t + 1) - 3 \\le 0$, or\n$... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | [-3/2, 12] | |
0dbz | Two circles with centers $A$ and $B$ intersect at points $M$ and $N$. Radii $A P$ and $B Q$ are parallel (on opposite sides of $A B$). If the common external tangents meet $A B$ at $D$, and $P Q$ meets $A B$ at $C$, prove that $C N D$ is a right angle. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0fpy | Sea $n \ge 2$ un número entero. Determinar el menor número real positivo $\gamma$ de modo que para cualesquiera números reales positivos $x_1, x_2, \dots, x_n$ y cualesquiera números reales $y_1, y_2, \dots, y_n$ con $0 \le y_1, y_2, \dots, y_n \le \frac{1}{2}$ que cumplan $x_1 + x_2 + \dots + x_n = y_1 + y_2 + \dots +... | [
"Sea $n \\ge 2$ un número entero. Determinar el menor número real positivo $\\gamma$ de modo que para cualesquiera números reales positivos $x_1, x_2, \\dots, x_n$ y cualesquiera números reales $y_1, y_2, \\dots, y_n$ con $0 \\le y_1, y_2, \\dots, y_n \\le \\frac{1}{2}$, verificando $x_1 + x_2 + \\dots + x_n = y_1 ... | Spain | LII Olimpiada Matemática Española | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | Spanish | proof and answer | gamma = (1/2) * (1/(n-1))^(n-1) | |
0fdp | Problem:
¿Cuántos números, comprendidos entre $1{,}000$ y $9{,}999$, verifican que la suma de sus cuatro dígitos es mayor o igual que el producto de los mismos?
¿Para cuántos de ellos se verifica la igualdad? | [
"Solution:\n\nSi el número tuviera algún cero entre sus cifras, entonces tendríamos la desigualdad estricta. Hay exactamente $9000 - 9^{4} = 2439$ números de este tipo, esto es, con una cifra igual a cero.\n\nConsideremos el número \"abcd\" escrito en su expresión decimal, y supondremos que no contiene ninguna cifr... | Spain | null | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 2502 numbers satisfy the inequality; 12 numbers satisfy the equality. | |
0c5r | Given a positive real number $a$, determine the minimum of the expression
$$
\left( \int_{0}^{1} f(x) \, dx \right)^{2} - (a + 1) \int_{0}^{1} x^{2a} f(x) \, dx
$$
may achieve, as $f$ runs through the class of all concave functions $f : [0, 1] \to \mathbb{R}$ such that $f(0) = 1$. | [
"The required minimum is $(2a - 1)/(8a + 4)$ and is achieved for $f: [0, 1] \\to \\mathbb{R}$, $f(x) = 1 - x$. The verification is routine and hence omitted.\n\nFix a concave function $f: [0, 1] \\to \\mathbb{R}$ such that $f(0) = 1$ to write\n$$x^a f(x) + 1 - x^a = x^a f(x) + (1 - x^a)f(0) \\le f(x^a \\cdot x + (1... | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (2a - 1)/(8a + 4) | |
0jp0 | Problem:
Let $ABC$ be an acute triangle with orthocenter $H$, circumcenter $O$, and incenter $I$. Prove that ray $AI$ bisects $\angle HAO$. | [
"Solution:\n\nWithout loss of generality, $AB < AC$. It follows that $\\angle BAH = 90^{\\circ} - \\angle B$, since the extension of $AH$ is perpendicular to $BC$. Moreover, we also have $\\angle AOC = 2\\angle B$; but since $OA = OC$, this implies $\\angle OAC = \\frac{1}{2}\\left(180^{\\circ} - \\angle AOC\\right... | 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 | |
0dgz | Let $A$ be one of two points of intersection of two unequal circles in the same plane. If two particles from $A$ move each on a circle in a clockwise direction with two uniform velocities until they return to point $A$ at the same instant. Prove that there is always a point in the plane that is equidistant from the two... | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | (0,1) | |
0efo | Problem:
Kateri od navedenih zapisov predstavlja definicijsko območje funkcije $f(x)=\frac{2x}{|x|-2}$?
(A) $\mathbb{R}$
(B) $(-\infty, 2) \cup (2, \infty)$
(C) $\mathbb{R}^{+}$
(D) $\mathbb{R} \setminus \{-2, 2\}$
(E) $(-2, 2)$ | [
"Solution:\n\nFunkcija ni definirana za $|x|-2=0$. Rešitvi enačbe sta $x_{1}=-2$, $x_{2}=2$. Definicijsko območje funkcije $f$ je $\\mathbb{R} \\setminus \\{-2, 2\\}$. Pravilen odgovor je (D)."
] | Slovenia | 17. tekmovanje dijakov srednjih tehniških in strokovnih šol v znanju matematike | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
0j7l | Problem:
Determine the number of quadratic polynomials $P(x) = p_{1} x^{2} + p_{2} x - p_{3}$, where $p_{1}, p_{2}, p_{3}$ are not necessarily distinct (positive) prime numbers less than $50$, whose roots are distinct rational numbers. | [
"Solution:\n\nThe existence of distinct rational roots means that the given quadratic splits into linear factors. Then, since $p_{1}, p_{3}$ are both prime, we get that the following are the only possible factorizations:\n- $(p_{1} x - p_{3})(x + 1) \\Rightarrow p_{2} = p_{1} - p_{3}$\n- $(p_{1} x + p_{3})(x - 1) \... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 31 |
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