id stringlengths 4 4 | problem_markdown stringlengths 36 3.59k | solutions_markdown listlengths 0 10 | images images listlengths 0 15 | country stringclasses 58
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|---|---|---|---|---|---|---|---|---|---|
0hwc | Problem:
A lattice point is a point in the coordinate plane both of whose coordinates are integers. In $\triangle ABC$, all three vertices are lattice points and the area of the triangle is $1/2$. Prove that the orthocenter of $\triangle ABC$ is also a lattice point. | [
"Solution:\n\nLet us position our coordinate system so that $A$ is the origin. Let $B = (a, b)$ and $C = (c, d)$. Then the formula for the area of a triangle with given vertex coordinates gives $\\frac{1}{2} = \\frac{1}{2}|ad - bc|$, i.e. $ad - bc = \\pm 1$.\n\nLet $H = (x, y)$ be the orthocenter of $\\triangle ABC... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0iaz | Problem:
An infinite table of nonnegative integers is constructed as follows: in the top row, some number is $1$ and all other numbers are $0$'s; in each subsequent row, every number is the sum of some two of the three closest numbers in the preceding row. An example of such a table is shown below.
$$
\begin{array}{c... | [
"Solution:\n\nWe use induction on $n$. It is clear that any number in row $1$ is at most $1 = 2^{0}$. Now, if every number in row $n$ is at most $2^{n-1}$, then every number in row $n+1$ is the sum of two numbers in row $n$ and so is at most $2^{n-1} + 2^{n-1} = 2^{n}$. This gives the induction step, and the result... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
06td | Let $\mathcal{V}$ be a finite set of points in the plane. We say that $\mathcal{V}$ is balanced if for any two distinct points $A, B \in \mathcal{V}$, there exists a point $C \in \mathcal{V}$ such that $AC = BC$. We say that $\mathcal{V}$ is center-free if for any distinct points $A, B, C \in \mathcal{V}$, there does n... | [
"Part (a).\nAssume that $n$ is odd. Consider a regular $n$-gon. Label the vertices of the $n$-gon as $A_{1}, A_{2}, \\ldots, A_{n}$ in counter-clockwise order, and set $\\mathcal{V} = \\{A_{1}, \\ldots, A_{n}\\}$. We check that $\\mathcal{V}$ is balanced. For any two distinct vertices $A_{i}$ and $A_{j}$, let $k \\... | IMO | 56th International Mathematical Olympiad Shortlisted Problems | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Circles",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory >... | English | proof and answer | a) For every integer n at least three, there exists a balanced set of n points. b) Balanced, center-free sets exist exactly for odd n at least three; they do not exist for even n. | |
0fx5 | Problem:
Ein $8 \times 11$-Rechteck wird irgendwie in 21 Gebiete zerlegt, wobei jedes dieser Gebiete zusammenhängend ist und aus Einheitsquadraten besteht, deren Kanten parallel zu den Kanten des Rechtecks liegen. Beweise, dass mindestens zwei dieser Gebiete bis auf Rotationen und Spiegelungen dieselbe Form haben. | [
"Solution:\n\nNehme an, dies sei nicht der Fall. Das Rechteck hat Fläche $88$. Die Anzahl zusammenhängender Teile aus $1$, $2$, $3$ respektive $4$ Einheitsquadraten ist gleich $1$, $1$, $2$ respektive $5$, wie man sich leicht überlegt. Wenn nun keine zwei Teile in der Zerlegung dieselbe Form haben, dann ist die Ges... | Switzerland | SMO Finalrunde | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
046m | A nonempty set $A$ of integers is called a “beautiful set” if for any $a \in A$ and $k \in \{1, 2, \dots, 2023\}$, the set
$$
\{b \in A \mid \lfloor \frac{b}{3^k} \rfloor = \lfloor \frac{a}{3^k} \rfloor \}
$$
has exactly $2^k$ elements.
Prove that: if the intersection of an integer set $S$ and any beautiful set is not ... | [
"For a positive integer $n$, a non-empty set of integers $A$ is called an “order $n$ strong beautiful set” if $A \\subset \\{0, 1, \\dots, 3^n - 1\\}$ and for any $a \\in A$ and $1 \\le k \\le n$, we have $\\#\\{b \\in A \\mid \\lfloor 3^{-k}b \\rfloor = \\lfloor 3^{-k}a \\rfloor \\} = 2^k$. It is clear that a 2023... | China | 2023 Chinese IMO National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0l95 | Prove that there exist an integer $m \ge 2002$ and $m$ distinct positive integers $a_1, a_2, \dots, a_m$ such that the number
$$
\prod_{i=1}^{m} a_i^2 - 4 \sum_{i=1}^{m} a_i^2
$$
is a perfect square. | [] | Vietnam | CONTEST FOR THE SELECTION OF VIETNAMESE INTERNATIONAL MATHEMATICAL OLYMPIAD TEAM | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0dw2 | Problem:
Poišči vse celoštevilske rešitve enačbe $\sqrt{x} + \sqrt{y} = \sqrt{2004}$. | [
"Solution:\n\nKvadrirajmo enačbo $\\sqrt{x} = \\sqrt{2004} - \\sqrt{y}$ in izrazimo $2 \\sqrt{y \\cdot 2004} = 2004 + y - x$. Od tod sledi, da mora biti $2 \\sqrt{y \\cdot 2004} = 4 \\sqrt{y \\cdot 501}$ celo število, zato mora biti $y = 501 \\cdot k^{2}$, kjer je $k$ nenegativno celo število. Če sedaj $y$ vstavimo... | Slovenia | 48. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | [(2004, 0), (501, 501), (0, 2004)] | |
06jq | Let $a$, $b$ and $c$ be positive real numbers satisfying $abc = 1$. Determine the smallest possible value of $\frac{a^3+8}{a^3(b+c)} + \frac{b^3+8}{b^3(c+a)} + \frac{c^3+8}{c^3(a+b)}$. | [
"The minimum value is $\\frac{27}{2}$.\nLet $x = \\frac{1}{a}$, $y = \\frac{1}{b}$ and $z = \\frac{1}{c}$. Then we have $xyz = 1$. By the AM-GM inequality, we have\n$$\na^3 + 2 = a^3 + 1 + 1 \\ge 3\\sqrt[3]{(a^3)(1)(1)} = 3a.\n$$\nTherefore,\n$$\n\\frac{a^3 + 8}{a^3(b+c)} \\ge \\frac{3a + 6}{a^3(b+c)} = \\frac{3a^2... | Hong Kong | Year 2016 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | 27/2 | |
070i | Problem:
Let $o(n)$ be the number of $2n$-tuples $(a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n})$ such that each $a_{i}, b_{j} = 0$ or $1$ and $a_{1} b_{1} + a_{2} b_{2} + \ldots + a_{n} b_{n}$ is odd. Similarly, let $e(n)$ be the number for which the sum is even. Show that $\dfrac{o(n)}{e(n)} = \dfrac{2^{n... | [
"Solution:\nWe prove by induction that $o(n) = 2^{2n-1} - 2^{n-1}$. For $n = 1$, this reads $o(1) = 2^{1} - 2^{0} = 1$, which is obviously true—the only such $2$-tuple is $(1, 1)$. Suppose it is true for $n$.\n\nIf $(a_{1}, a_{2}, \\ldots, a_{n}, b_{1}, b_{2}, \\ldots, b_{n})$ gives an odd sum, then we can take $(a... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | (2^n - 1)/(2^n + 1) | |
04ge | Let $A$ be the number of all six-digit numbers whose product of digits equals $105$, and let $B$ be the number of all six-digit numbers whose product of digits equals $147$. Determine the ratio $A : B$. (Kristina Ana Škreb) | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 2:1 | |
0848 | Problem:
Siano dati 13 numeri reali $a_{1}, a_{2}, \ldots, a_{13}$, tutti diversi da zero, di cui almeno tre positivi, e sia $n$ il numero di $a_{i}$ negativi. Sapendo che, tra tutti i possibili prodotti $a_{i} a_{j}$, esattamente 22 risultano negativi, quanto vale $n$? ($a_{i} a_{j}$ e $a_{j} a_{i}$ contano come un s... | [] | Italy | UNIONE MATEMATICA ITALIANA Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO BIENNIO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | A | |
0cbb | For a positive integer $n$, denote $s(n)$ the sum of its odd digits. For instance, $s(512418) = 5 + 1 + 1 = 7$ and $s(82624) = 0$.
Compute $s(1) + s(2) + s(3) + \dots + s(2023)$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 16064 | |
01q3 | Let $I$ be the incenter of a triangle $ABC$. The circle passing through $I$ and centered at $A$ meets the circumference of the triangle $ABC$ at points $M$ and $N$.
Prove that the line $MN$ touches the incircle of the triangle $ABC$. | [
"Let $\\angle BAC = 2\\alpha$, $\\angle ABC = 2\\beta$, $\\angle BCA = 2\\gamma$. Since $AM = AN$, we obtain $\\angle AMN = \\angle ANM$. We have $\\angle AMN = \\angle ACN$ and $\\angle MNC = \\angle MAC$ (as inscribed angles). Further,\n$$\n\\begin{aligned}\n\\angle AKN &= \\angle MAK + \\angle AMK = \\\\\n&= \\a... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscri... | English | proof only | null | |
0aw1 | Problem:
Suppose that Ethan has four red chips and two white chips. He selects three chips at random and places them in Urn 1, while the remaining chips are placed in Urn 2. He then lets his brother Josh draw one chip from each urn at random. What is the probability that the chips drawn by Josh are both red? | [
"Solution:\nThe possibilities are\n\n| Urn 1 | Urn 2 | Number of Ways |\n| :--------------------------: | :--------------------------: | :-------------------------------------: |\n| $1\\ \\mathrm{R},\\ 2\\ \\mathrm{W}$ | $3\\ \\mathrm{R}$ ... | Philippines | 18th PMO National Stage Oral Phase | [
"Statistics > Probability > Counting Methods > Combinations",
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 2/5 | |
0hjh | Problem:
Find all integer solutions to $x y = 2003(x + y)$ | [
"Solution:\nWe get\n$$\n\\begin{gathered}\n2003^2 - 2003x - 2003y - x y = 2003^2 \\\\\n(2003 - x)(2003 - y) = 2003^2\n\\end{gathered}\n$$\nSo $(2003 - x)$ is a divisor of $2003^2$. As $2003$ is a prime number, we conclude that $2003 - x$ is $-2003^2$, $-2003$, $-1$, $1$, $2003$, or $2003^2$, and $x$. Then for $x, y... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All integer solutions (x, y) are (4014012, 2004), (2004, 4014012), (4006, 4006), (2002, -4010006), (-4010006, 2002), (0, 0). | |
092r | Problem:
A class of high school students wrote a test. Every question was graded as either 1 point for a correct answer or 0 points otherwise. It is known that each question was answered correctly by at least one student and the students did not all achieve the same total score.
Prove that there was a question on the t... | [
"Solution:\nLet $n$ be the number of the students in the class and $a$ their average score. Denote by $P$ and $S$ the set of all problems, and the set of all students respectively, and let $S(p)$ be the non-empty set of students who solved problem $p \\in P$. For any student $s$, let $\\operatorname{sc}(s)$ be the ... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0bw5 | Show that, for any two distinct elements of the set $\{20^{17}, 2^{107}, 7^{102}\}$, neither their sum, nor their difference, is a perfect square. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
08zj | Let us call a positive integer a *good number* if the digit $2$ appears more frequently than the digit $3$, and a *bad number* if the digit $3$ appears more frequently than the digit $2$. For example, $2023$ is a good number because the digit $2$ appears twice and the digit $3$ once, and $123$ is neither a good number ... | [
"$22$\n\nFor a positive integer $k$, let us define the changed number of $k$ as the number obtained by switching each digit $2$ in $k$ to $3$, and each digit $3$ in $k$ to $2$. By this definition, if $l$ is the changed number of $k$, then the changed number of $l$ is $k$ itself. For a good number $m \\le 1999$ and ... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 22 | |
0er2 | Mrs. Habana turns $45$ years old in $2016$ and her son $17$. In which year will Mrs. Habana turn double her son's age?
(A) $2033$ (B) $2044$ (C) $2031$ (D) $2040$ (E) $2027$ | [
"Mrs Habana will be $45 + x$ years old and her son will be $17 + x$ old where $x$ is the number of years from $2016$. Therefore $45 + x = 2(17 + x)$, giving $45 + x = 34 + 2x$, so $x = 45 - 34 = 11$. The year will then be $2016 + 11 = 2027$."
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | E | |
0kwc | Let $n \ge 3$ be an integer and let $K_n$ be the complete graph on $n$ vertices. Each edge of $K_n$ is colored either red, green, or blue. Let $A$ denote the number of triangles in $K_n$ with all edges of the same color, and let $B$ denote the number of triangles in $K_n$ with all edges of different colors. Prove that
... | [
"* each monochromatic triangle has a charge of +6,\n* each bichromatic triangle has a charge of 0, and\n* each trichromatic triangle has a charge of -3.\nSince each vee contributes to exactly one triangle, we obtain that the total charge is $6A - 3B$.\n\n\\begin{align*}\n& 2 \\left[ \\binom{a}{2} + \\binom{b}{2} + ... | United States | USA TSTST | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0k7x | Problem:
Prove that if $x, y, z$ are positive real numbers, then
$$
x^{2}+2 y^{2}+3 z^{2}>x y+3 y z+z x .
$$ | [
"Solution:\nWe note the identity\n$$\n2\\left(x^{2}+2 y^{2}+3 z^{2}\\right)-2(x y+3 y z+z x)=(x-y)^{2}+(x-z)^{2}+3(y-z)^{2}+2 z^{2} \\geq 2 z^{2}>0\n$$\nsince all squares of real numbers are nonnegative."
] | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0ev6 | Given an one-to-one correspondence $f : \{1, 2, \dots, n\} \to \{1, 2, \dots, n\}$ for some positive integer $n$, define four sets $A$, $B$, $C$, $D$ as follows:
$$
A = \{i \mid i > f(i)\}
$$
$$
B = \{(i, j) \mid i < j \le f(j) < f(i) \text{ or } f(j) < f(i) < i < j\}
$$
$$
C = \{(i, j) \mid i < j \le f(i) < f(j) \text... | [
"We use an induction by the number $|D|$: Firstly, if $|D| = 0$, $f$ is identity and $A = B = C = \\emptyset$. So it holds for $|D| = 0$.\n\nNext, assume that it holds that for all bijections with $|D| < k$ for some positive integer $k$. Given $f$ with $|D_f| = k > 0$, there exists $i$ such that $f(i) > f(i+1)$. De... | South Korea | The 26th Korean Mathematical Olympiad Final Round | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
021f | Problem:
Let $ABC$ be a triangle with incentre $I$ and circumcircle $\Omega$. Let $D, E, F$ be the midpoints of the arcs $\overline{BC}, \overline{CA}, \overline{AB}$ of $\Omega$ not containing $A, B, C$, respectively. Let $D'$ be the point of $\Omega$ diametrically opposite to $D$. Show that $I, D'$, and the midpoint ... | [
"Solution:\n\nBy definition of $D$, $E$ and $F$, we know that $ID, IE$ and $IF$ are the angle bisectors of $ABC$. Using angles in $\\Omega$, we can compute \n$$\n\\overline{EFI} = \\overline{EFC} = \\overline{EBC} = \\frac{\\overline{ABC}}{2} = 90^{\\circ} - \\frac{\\overline{ACB}}{2} - \\... | Benelux Mathematical Olympiad | 17th Benelux Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0l4x | Problem:
A plane $\mathcal{P}$ intersects a rectangular prism at a hexagon which has side lengths $45$, $66$, $63$, $55$, $54$, and $77$, in that order. Compute the distance from the center of the rectangular prism to $\mathcal{P}$. | [
"Solution 1: Translate $\\mathcal{P}$ so that it contains the center. The intersection of the translated plane with the rectangular prism is a centrally symmetric hexagon. Let its side lengths be $a$, $b$, $c$, $a$, $b$, and $c$, in that order. Then, for some $t_{a}$, $t_{b}$, and $t_{c}$, the side lengths of the h... | United States | HMMT February 2025 | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | sqrt(95/24) | |
04u1 | Find the largest possible size of a set $M$ of integers with the following property: Among any three distinct numbers from $M$, there exist two numbers whose sum is a power of $2$ with non-negative integer exponent. | [
"The set $\\{-1, 3, 5, -2, 6, 10\\}$ attests that $M$ can have $6$ elements: The sum of any two numbers from the triplet $(-1, 3, 5)$ is a power of two and the same is true for triplet $(-2, 6, 10)$. For the sake of contradiction, assume that some set $M$ has more than $6$ elements.\n\nClearly, $M$ can't contain th... | Czech Republic | 67th Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | 6 | |
01ve | The sum of several positive numbers from $(0, 1]$ is equal to $S$. It is known that one with the guarantee can divide all given numbers into two groups such that the sum of numbers in the first group does not exceed $1$ and the sum of numbers in the second group does not exceed $5$.
Find the maximum possible value of $... | [
"Answer: $5.5$.\n\nWe first show that $S \\le 5.5$ (obviously, $S \\le 6$). Suppose that $S > 5.5$. Then we may write $S = 5.5 + 19x$, where $0 < x \\le \\frac{1}{38}$. It can occur that we have $11$ numbers, $10$ of which are equal to $0.5 + 2x$ and the last one is equal to $0.5 - x$. It is clear that it is imposs... | Belarus | Selection and Training Session | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 5.5 | |
08nm | Problem:
Solve the following equation for $x, y, z \in \mathbb{N}$
$$
\left(1+\frac{x}{y+z}\right)^{2}+\left(1+\frac{y}{z+x}\right)^{2}+\left(1+\frac{z}{x+y}\right)^{2}=\frac{27}{4}
$$ | [
"Solution:\nCall $a=1+\\frac{x}{y+z}$, $b=1+\\frac{y}{z+x}$, $c=1+\\frac{z}{x+y}$ to get\n$$\na^{2}+b^{2}+c^{2}=\\frac{27}{4}\n$$\nSince it is also true that\n$$\n\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=2\n$$\nthe quadratic-harmonic means inequality implies\n$$\n\\frac{3}{2}=\\sqrt{\\frac{a^{2}+b^{2}+c^{2}}{3}} \\ge... | JBMO | JBMO Shortlist | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | All triples with x = y = z ∈ N | |
0gqt | Let $ABC$ be a triangle with circumcircle $\omega$ and let $\omega_A$ be a circle drawn outside $ABC$ and tangent to side $BC$ at $A_1$ and tangent to $\omega$ at $A_2$. Let the circles $\omega_B$ and $\omega_C$ and the points $B_1, B_2, C_1, C_2$ are defined similarly. Prove that if the lines $AA_1, BB_1, CC_1$ are co... | [
"Let $A_1A_2$ intersect $\\omega$ at $A_3$. Consider the common tangent of circles $\\omega$ and $\\omega_A$ passing through $A_2$ and let $U$ be a point on this common tangent which is at the same side of the line $A_1A_2$ with respect to $B$. Then by tangency, we have $\\angle A_1A_2U = \\angle BA_1A_2$ and $\\an... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0e0h | Problem:
Poišči vsa soda naravna števila $n$, za katera velja
$$
-53 < \frac{2009}{53-n} < 53-n
$$ | [
"Solution:\n\nČe je $n < 53$, je $\\frac{2009}{53-n} > 0 > -53$. Neenakost $\\frac{2009}{53-n} < 53-n$ je enakovredna $2009 < (53-n)^2$ oziroma $\\sqrt{2009} < 53-n$. Ker je $\\sqrt{2009} > 44$, sledi $53-n > 44$ oziroma $9 > n$. Možne vrednosti $n$ so $2, 4, 6$ in $8$, pri vseh pa je $53-n \\geq 45 > \\sqrt{2009}$... | Slovenia | Slovenian Secondary School Mathematical Competition | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 2, 4, 6, 8, 92, 94, 96 | |
0ej8 | Problem:
Za katere vrednosti parametra $a \in \mathbb{R}$ je funkcija $f(x)=\log \left(x^{2}+(a+4) x+9\right)$ definirana na množici vseh realnih števil?
(A) $a<2$
(B) $a>-10$
(C) $-2<a<10$
(D) $-10<a<2$
(E) $a=-4$ | [
"Solution:\n\nUgotovimo, da bo definicijsko območje dane funkcije $f$ množica vseh realnih števil ob pogoju $x^{2}+(a+4) x+9>0$. To pa bo izpolnjeno, če bo $D<0$. Poenostavimo levo stran neenačbe:\n$$\nD=b^{2}-4 a c=(a+4)^{2}-4 \\cdot 1 \\cdot 9=a^{2}+8 a-20=(a+10)(a-2)\n$$\nRešimo kvadratno neenačbo $(a+10)(a-2)<0... | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | D | |
06er | In triangle $ABC$, the altitude, angle bisector and median from $C$ divide the angle $\angle C$ into four equal angles. Find $\angle B$. | [
"$\\angle B$ can be $22.5^\\circ$ or $67.5^\\circ$.\n\nWLOG assume $AC < BC$. Let $D$ be the foot of altitude from $C$, let $E$ be the foot of internal angle bisector from $C$, and let $F$ be the midpoint of $AB$. Let $P$ be the projection of $E$ on $BC$, and let $DP$ meet $AC$ at $Q$.\n\nFirstly, since $\\angle AC... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 22.5° or 67.5° | |
0lfc | There are $n \ge 2$ classes organized $m \ge 1$ learning groups for students. Every class has students participating in at least one group. Every group has exactly $a$ classes that the students in this group participate in. For any two groups, there are no more than $b$ classes with students participating in both group... | [
"a) If $m = 1$, then the class will have 8 group to attend, this is contradiction. If $m \\ge 3$, consider any 3 groups $X_1, X_2, X_3$ then\n$$\nn \\ge |X_1 \\cup X_2 \\cup X_3| \\ge |X_1| + |X_2| + |X_3| - |X_1 \\cap X_2| - |X_2 \\cap X_3| - |X_1 \\cap X_3| \\ge 4 + 4 + 4 - 1 - 1 - 1 = 9, \\text{ (contradiction).... | Vietnam | Vietnamese Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal... | English | proof and answer | a) m = 2; b) n ≥ 20; c) minimum n = 16 | |
0jtr | Problem:
Find the number of ordered pairs of integers $(a, b)$ such that $a, b$ are divisors of $720$ but $ab$ is not. | [
"Solution:\nFirst consider the case $a, b > 0$. We have $720 = 2^4 \\cdot 3^2 \\cdot 5$, so the number of divisors of $720$ is $5 \\times 3 \\times 2 = 30$. We consider the number of ways to select an ordered pair $(a, b)$ such that $a, b, ab$ all divide $720$. Using the balls and urns method on each of the prime f... | United States | HMMT February 2016 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 2520 | |
03no | Problem:
At a dinner party there are $N$ hosts and $N$ guests, seated around a circular table, where $N \geq 4$. A pair of two guests will chat with one another if either there is at most one person seated between them or if there are exactly two people between them, at least one of whom is a host. Prove that no matte... | [
"Solution:\n\nLet a run refer to a maximal group of consecutive dinner party guests all of whom are the same type (host or guest). Suppose that there are exactly $k$ runs of hosts and $k$ runs of guests. Let $G_{i}$ and $H_{i}$ denote the number of runs of guests and hosts, respectively, of length exactly $i$. Furt... | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
01ot | Let $A_1$, $B_1$, $C_1$ be midpoints of sides $BC$, $CA$, $AB$ of an acute-angled triangle $ABC$ respectively. Let $C_1L$ and $C_1K$ be bisectrices of $\triangle ACC_1$ and $\triangle BCC_1$ respectively. The line $C_1B_1$ intersects the line $LK$ at $A_2$, and the line $C_1A_1$ intersects the line $LK$ at $B_2$.
Prove... | [
"First show that the line $LK$ is parallel to the side $AB$. Indeed, since $C_1L$ and $C_1K$ are the bisectrices of the angles $AC_1C$ and $BC_1C$ respectively we have\n$$\n\\frac{AL}{LC} = \\frac{AC_1}{CC_1} = [AC_1 = BC_1] = \\frac{C_1B}{CC_1} = \\frac{BK}{KC}.\n$$\n\n\nSo $LK \\parallel ... | Belarus | BelarusMO 2013_s | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0kwh | Problem:
Let $ABCD$ be a square, and let $M$ be the midpoint of side $BC$. Points $P$ and $Q$ lie on segment $AM$ such that $\angle BPD = \angle BQD = 135^{\circ}$. Given that $AP < AQ$, compute $\frac{AQ}{AP}$. | [
"Solution:\n\nNotice that $\\angle BPD = 135^{\\circ} = 180^{\\circ} - \\frac{\\angle BAD}{2}$ and $P$ lying on the opposite side of $BD$ as $C$ means that $P$ lies on the circle with center $C$ through $B$ and $D$. Similarly, $Q$ lies on the circle with center $A$ through $B$ and $D$.\n\nLet the side length of the... | United States | HMMT February | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(5) | |
04am | $$
a_0 = 1, \quad a_{n+1} = \begin{cases} \frac{a_n}{2}, & \text{if } a_n \text{ is even,} \\ a_n + d, & \text{if } a_n \text{ is odd.} \end{cases}
$$
Determine all $d$ such that $a_n = 1$ for some $n > 0$. (Italy 2005) | [
"First notice that for even $d$ the sequence is of the form $a_n = 1 + nd$, i.e. all the terms of the sequence are odd and the sequence is monotonically increasing. Therefore $a_n \\neq 1$ for $n > 0$.\n\nLet $d$ be an arbitrary odd number. We can easily show by induction that $a_n < d$ if $a_n$ is odd, and $a_n < ... | Croatia | CroatianCompetitions2011 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | all odd positive integers | |
07ou | a. Let $a_0$, $a_1$, $a_2$ be real numbers and consider the polynomial
$$
P(x) = a_0 + a_1 x + a_2 x^2.
$$
Assume that $P(-1)$, $P(0)$ and $P(1)$ are integers.
Prove that $P(n)$ is an integer for all integers $n$.
b. Let $a_0$, $a_1$, $a_2$, $a_3$ be real numbers and consider the polynomial
$$
Q(x) = a_0 + a_1 x + a_2... | [
"a. By assumption $P(0) = a_0 \\in \\mathbb{Z}$, $P(-1) = a_0 - a_1 + a_2 \\in \\mathbb{Z}$ and $P(1) = a_0 + a_1 + a_2 \\in \\mathbb{Z}$. This implies that $a_1 + a_2$ and $a_2 - a_1$ are integers, hence also $2a_1$ and $2a_2$. From $P(n) = a_0 + a_1 n + a_2 n^2 = a_0 + a_1(n + n^2) + (a_2 - a_1)n^2$ we see now th... | Ireland | Irska 2014 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
0egd | Problem:
a. V domu za starejše občane so praznovali rojstni dan najstarejše oskrbovanke. Pripravili so 15 litrov napitka iz domačega hruškovega soka, razredčenega z vodo, tako da je bilo v napitku $20 \%$ vode. Ker je bil še vedno presladek, so dolili še 5 litrov vode. Izračunaj delež naravnega soka v $\%$ v dobljenem... | [
"Solution:\n\na. V 15 litrih soka je $20 \\%$ od 15 litrov, torej 3 litri vode. Naravnega soka je torej 12 litrov v 20 litrih napitka, kar pomeni, da je delež naravnega soka $60 \\%$.\n\nb. Ugotovimo, da je dvoposteljnih in troposteljnih sob skupaj $141-70=71$. Zapišemo $D+T=71$. Skupaj je v dvoposteljnih in tropos... | Slovenia | 18. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Odbirno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | a) 60%; b) 43 two-bed rooms and 28 three-bed rooms | |
0kdr | Problem:
For odd primes $p$, let $f(p)$ denote the smallest positive integer $a$ for which there does not exist an integer $n$ satisfying $p \mid n^{2}-a$. Estimate $N$, the sum of $f(p)^{2}$ over the first $10^{5}$ odd primes $p$.
An estimate of $E>0$ will receive $\left\lfloor 22 \min (N / E, E / N)^{3}\right\rfloo... | [
"Solution:\n\nOf course, there is no such thing as a uniform random prime. More rigorously, for any $n$, the joint distributions of $\\left(\\frac{p_{1}}{q}\\right), \\ldots, \\left(\\frac{p_{n}}{q}\\right)$ where $q$ is a uniform random prime less than $N$ converges in distribution to $n$ independent coin flips be... | United States | HMMT February 2020 | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity"
] | null | final answer only | 22 * 10^5 | |
0bbr | A square with side length $\ell$ is contained in a unit square whose centre is not interior to the former. Show that $\ell \le 1/2$. | [
"**Proof from The Book.** Notice that there is a line through the centre of the unit square separating the square with side length $\\ell$ and a standard quarter of the unit square (i.e., a square with side length $1/2$ and one vertex at the centre of the unit square). To conclude, apply the celebrated theorem of E... | Romania | 62nd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0l5q | Problem:
Regular hexagon $ABCDEF$ has side length $2$. Circle $\omega$ lies inside the hexagon and is tangent to segments $\overline{AB}$ and $\overline{AF}$. There exist two perpendicular lines tangent to $\omega$ that pass through $C$ and $E$, respectively. Given that these two lines do not intersect on line $AD$, c... | [
"Solution:\n\n\nLet $O$ be the center of $\\omega$, and let the two tangent lines intersect at $P$. Note that $O$ lies on the external angle bisector of $\\angle CPE$ because the tangents are symmetric about line $PO$. Additionally, $O$ lies on the perpendicular bisector of $CE$ by symmetry... | United States | HMMT February | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"... | null | final answer only | (3√3 - 3)/2 | |
0bip | Let $A_0A_1A_2$ be a triangle and let $O$ be its circumcenter. The lines $OA_k$ and $A_{k+1}A_{k+2}$ meet at $B_k$, and the tangent of the circumcircle $A_0A_1A_2$ at $A_k$ meets the line $B_{k+1}B_{k+2}$ at $C_k$, $k = 0, 1, 2$, indices being reduced modulo $3$. Show that the points $C_0, C_1, C_2$ are collinear. | [
"\nTo this end, let the parallel through $O$ to the tangent $A'_{k+1}A'_{k+2}$ meet the tangents $A'_k A'_{k+1}$ and $A'_k A'_{k+2}$ at $X_k$ and $Y_k$, respectively, and notice that:\n\n(1) the angles $OX_k B_k$ and $OA_{k+2} B_k$ are equal, since the points $A_{k+2}$, $B_k$, $O$, $X_k$ al... | Romania | 65th NMO Selection Tests for BMO and IMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Concurrency a... | null | proof only | null | |
0h29 | There are 10 piles of stones, with $3$, $4$, $5$, $\ldots$, $12$ stones respectively. At one step one can pick three piles and add $1$ stone to the first pile, $2$ stones to the second pile, $3$ stones to the third pile, or pick any three piles and take $1$ stone out of first pile, $2$ stones out of second pile, $3$ st... | [
"After each operation the total number of stones changes by a number which is divisible by $3$. At the end the total number is $2011 \\cdot 10$ which is not divisible by $3$. However, at the starting moment the total number $75$ is divisible by $3$, which provides a contradiction."
] | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | No | |
006k | Un cuadrado de $2n \times 2n$ se cubre, sin salirse del cuadrado, sin huecos ni superposiciones, con rectángulos de $1 \times 2$ y piezas como las de la figura (que cubren exactamente 4 cuadrados de $1 \times 1$). Las figuras se pueden girar o dar vueltas. Demuestre que en el recubrimiento hay al menos $n + 1$ rectángu... | [] | Argentina | XVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof only | null | |
057z | Let $D$ be the foot of the altitude drawn to the hypotenuse $AB$ of a right triangle $ABC$. The inradii of the triangles $ABC$, $CAD$ and $CBD$ are $r$, $r_1$ and $r_2$, respectively. Prove that $CD = r + r_1 + r_2$. | [
"Let $a = |BC|$, $b = |CA|$, $c = |AB|$ and $h = |CD|$; then $ab = ch = (a + b + c)r$.\n\nNote that the triangles *ABC*, *ACD* and *CBD* are similar by two equal angles (Fig. 28). As the similarity ratio of triangles *ACD* and *ABC* is $\\frac{b}{c}$ and that of triangles *CBD* and *ABC* is $\\frac{a}{c}$, we have ... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ian | Problem:
At a certain college, there are 10 clubs and some number of students. For any two different students, there is some club such that exactly one of the two belongs to that club. For any three different students, there is some club such that either exactly one or all three belong to that club. What is the larges... | [
"Solution:\n\nLet $C$ be the set of clubs; each student then corresponds to a subset of $C$ (the clubs to which that student belongs). The two-student condition implies that these subsets must be all distinct. Now (assuming there is more than one student) some student belongs to a nonempty set $S$ of clubs. For eve... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 513 | |
03u2 | Find all functions $f: Q^+ \to Q^+$ such that
$$
f(x) + f(y) + 2xyf(xy) = \frac{f(xy)}{f(x+y)}, \qquad \textcircled{1}
$$
Where $Q^+ = \{q \mid q \text{ is a positive rational number}\}$. | [
"(1) Prove that $f(1) = 1$.\nPut $y = 1$ in \\textcircled{1}, and write $f(1) = a$. Then\n$$\nf(x) + a + 2x f(x) = \\frac{f(x)}{f(x+1)}\n$$\nThus\n$$\nf(x+1) = \\frac{f(x)}{(1 + 2x)f(x) + a} \\qquad \\textcircled{2}\n$$\nHence\n$$\nf(2) = \\frac{a}{4a} = \\frac{1}{4},\n$$\n$$\nf(3) = \\frac{1}{\\frac{1}{4} + a} = \... | China | China National Team Selection Test | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 1/x^2 for all positive rational x | |
08vm | Suppose a magician can perform each of the following 3 kinds of tricks any number of times:
* Trick A: Change 1 orange and 1 grape into 2 apples.
* Trick B: Change 1 grape and 1 apple into 3 oranges
* Trick C: Change 1 apple and 1 orange into 4 grapes.
Initially, there were 2011 each of apples, grapes and oranges, and... | [
"The number of apples decreases by $1$ after $1$ trick of type $B$ or $C$ is performed, and increases by $2$ after $1$ trick of type $A$ is performed. The number of grapes decreases by $1$ after $1$ trick of type $A$ or $B$ is performed, and increases by $4$ after $1$ trick of type $C$ is performed. Suppose the mag... | Japan | Japan Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 2024 | |
05yw | Problem:
Soient $x$, $y$, $z$ des réels strictement positifs tels que $x y + y z + z x = 1$. Montrer que
$$
2\left(x^{2}+y^{2}+z^{2}\right)+\frac{4}{3}\left(\frac{1}{x^{2}+1}+\frac{1}{y^{2}+1}+\frac{1}{z^{2}+1}\right) \geqslant 5
$$ | [
"Solution:\n\nDans cet exercice, on aimerait bien appliquer l'inégalité arithmético-géométrique de façon à obtenir une minoration du type $\\left(x^{2}+1\\right)+\\frac{1}{x^{2}+1} \\geqslant 2$. Mais si on s'y prend mal, on n'arrive pas à conclure. En effet, on aurait pu s'en douter en étudiant le cas d'égalité de... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOi 2 : Algèbre | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
066t | Determine the greatest possible value of $M$ for which:
$$
\frac{x}{1+\frac{yz}{x}} + \frac{y}{1+\frac{zx}{y}} + \frac{z}{1+\frac{xy}{z}} \ge M,
$$
for all real numbers $x, y, z > 0$ satisfying the equation: $xy + yz + zx = 1$. | [
"The inequality is equivalent to\n$$\n\\frac{x^2}{x+yz} + \\frac{y^2}{y+zx} + \\frac{z^2}{z+xy} \\ge M. \\quad (1)\n$$\nSince $x, y, z > 0$, from Cauchy-Schwarz inequality we find\n$$\n\\begin{aligned}\n& \\left( \\frac{x^2}{x+yz} + \\frac{y^2}{y+zx} + \\frac{z^2}{z+xy} \\right) (x+yz+y+zx+z+xy) \\ge (x+y+z)^2 \\\\... | Greece | Selection Examination A | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 3/(√3+1) | |
09c9 | $f(x), g(y)$ нь комплекс коэф. бүхий $n$ зэргийн унитар олон гишүүнтүүд ба $f(x) - g(y) = \prod_{j=1}^{n} (a_j x + b_j y + c_j)$, $a_j, b_j, c_j \in \mathbb{C}$, $j = \overline{1, n}$ байг. $\exists a, b, c \in \mathbb{C}$, $f(x) = (x+a)^n + c$, $g(y) = (y+b)^n + c$ гэж батал. | [
"$\\Pi_{j=1}^{n} a_j = 1$ -д хувааж\n$$\nf(x) - g(y) = \\Pi_{j=1}^{n} (x - \\alpha_j y + \\beta_j)\n$$\nгэж бичвэл\n$$\n\\Pi_{j=1}^{n} (x - \\alpha_j y) = x^n - y^n = \\Pi_{j=1}^{n} (x - \\varepsilon^j y), \\varepsilon \\in U'_{n}\n$$\nболохыг төвөггүй ажиглаж болно. Шаардлагатай гэж үзвэл дахин дугаарлах замаар $\... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | Mongolian | proof only | null | |
0jjx | Problem:
If you flip a fair coin 1000 times, what is the expected value of the product of the number of heads and the number of tails? | [
"Solution:\nAnswer: 249750\nWe solve the problem for $n$ coins. We want to find\n$$\nE(n) = \\sum_{k=0}^{n} \\frac{1}{2^{n}} \\binom{n}{k} k(n-k)\n$$\nWe present three methods for evaluating this sum.\n\nMethod 1:\nDiscard the terms $k=0$, $k=n$. Since $\\binom{n}{k} k(n-k) = n(n-1) \\binom{n-2}{k-1}$ by the factor... | United States | HMMT November 2014 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | final answer only | 249750 | |
0c8j | The diagonals of the faces of the rectangular parallelepiped $ABCD'A'B'C'D'$ satisfy the relation:
$$
CA^2 = \frac{2 \cdot B'A^2 \cdot D'A^2}{B'A^2 + D'A^2}.
$$
Prove that $AC' \le \sqrt{3} \cdot AA'$. | [
"Denote $AB = a$, $AD = b$, $AA' = c$. By Pythagorean Theorem, we have $a^2 + b^2 = \\frac{2(b^2 + c^2)(c^2 + a^2)}{b^2 + c^2 + c^2 + a^2}$, therefore $a^4 + b^4 = 2c^4$.\nWe also have $4c^4 = 2(a^4 + b^4) \\ge (a^2 + b^2)^2$ thus $2c^2 \\ge a^2 + b^2$.\nIn the end, $C'A = \\sqrt{a^2 + b^2 + c^2} \\le \\sqrt{3c^2} ... | Romania | Romanian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
087u | Problem:
Nella classe di Sergio, dopo la correzione dell'ultimo compito di matematica, al quale tutti gli alunni erano stati presenti, la media aritmetica delle insufficienze è risultata $4,6$, mentre la media aritmetica delle sufficienze è risultata $7,1$. Sapendo che il professore ha dato soltanto voti interi, quant... | [
"Solution:\n\nLa risposta è (C). Sia $a$ la somma dei voti insufficienti, $b$ la somma di quelli sufficienti, $m$ il numero di insufficienze nella classe, $n$ il numero di sufficienze nella classe. La media delle insufficienze è $\\frac{a}{m}$ e quella delle sufficienze $\\frac{b}{n}$. Possiamo considerare $a, b, m... | Italy | UNIONE MATEMATICA ITALIANA SCUOLA NORMALE SUPERIORE DI PISA Progetto Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization"
] | null | MCQ | C | |
0h2l | There are $n \ge 3$ burrows on a straight line. Mouse Jerry is hiding in one of these burrows. Cat Tom has a possibility to put his paw into one of the burrows and to catch Jerry if he is hiding in this burrow. After every Tom's attempt Jerry necessarily runs to the neighboring (left or right) burrow. Can Tom always ca... | [
"Enumerate the burrows from left to right with numbers from $1$ to $n$ and define the \"distance\" between the burrows $i$ and $j$ by $i - j$ (it can be negative).\n\nFirst Tom checks all the burrows from $1$ to $n$ one after another. If at the moment when Tom checks the first burrow Jerry is in a burrow with an od... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Yes, Tom can always catch Jerry. | |
08kl | Problem:
Let $p$ be an odd prime. Prove that $p$ divides the integer
$$
\frac{2^{p!}-1}{2^{k}-1}
$$
for all integers $k=1,2, \ldots, p$. | [
"Solution:\nAt first, note that $\\frac{2^{p!}-1}{2^{k}-1}$ is indeed an integer.\n\nWe start with the case $k=p$. Since $p \\mid 2^{p}-2$, then $p / 2^{p}-1$ and so it suffices to prove that $p \\mid 2^{p!}-1$. This is obvious as $p \\mid 2^{p-1}-1$ and $\\left(2^{p-1}-1\\right) \\mid 2^{p!}-1$.\n\nIf $k=1,2, \\ld... | JBMO | OJBM | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0hw7 | Problem:
Show that if $n$ is a positive integer for which $2+2 \sqrt{1+12 n^{2}}$ is an integer, then it is a perfect square. | [
"Solution:\nLet the value of this expression be $2k$, say. Then we may conclude that\n$$\n12 n^{2} + 1 = (k-1)^{2} \\Longrightarrow 12 n^{2} = k(k-2).\n$$\nObviously $k$ is even and hence $\\gcd(k, k-2) = 2$. So we may write this factorization as $2x^{2}$ and $6y^{2}$ in some order.\nIf $k = 2x^{2}$ then $2k = 4x^{... | United States | Berkeley Math Circle: Monthly Contest 8 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
01jf | Let $\triangle ABC$ be a triangle with centroid $G$. Let $D$, $E$ and $F$ be the circumcenters of $\triangle BCG$, $\triangle CAG$ and $\triangle ABG$, respectively. The point $X$ is defined as the intersection of the perpendiculars from $E$ to $AB$ and $F$ to $AC$. Prove that $DX$ bisects the segment $EF$. | [
"The two parts may be completed independently, and in the three solutions below we demonstrate different approaches to both parts, though one can create valid solutions combining either first part with either second part.\nLet $\\omega_B$, $\\omega_C$ denote the circumcircles of triangles $ABG$ and $ACG$ respective... | Baltic Way | Baltic Way 2023 Shortlist | [
"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",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
08i2 | Problem:
A given rectangular table has at least one column and at least one line. It is fully completed by the first positive integers, written consecutively from left to right and beginning with the first line. It is known that the number $170$ is written on the middle line, and in the same column with him on the las... | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 371 or 477 | |
0l0p | Two straight pipes (circular cylinders), with radii $1$ and $\frac{1}{4}$, lie parallel and in contact on a flat floor. The figure below shows a head-on view. What is the sum of the possible radii of a third parallel pipe lying on the same floor and in contact with both?
 | [
"**Answer (C):** There are two possible positions for the third pipe—either nestled in the gap between the pipes or outside. See the figure below.\n\n\n\nConsider the blown-up figure below. In this diagram, $A$ is the center of the circle of radius $1$, $B$ is the center of the circle of ra... | United States | AMC 10 B | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 10/9 | |
0dzz | Problem:
Izrazi $(A-B)^{-1}$ z $x$, če je $A=\frac{3 x^{-\frac{1}{3}}}{x^{\frac{2}{3}}-2 x^{-\frac{1}{3}}}$ in $B=\frac{x^{\frac{1}{3}}}{x^{\frac{4}{3}}-x^{\frac{1}{3}}}$. Izraz poenostavi. | [
"Solution:\n\nPoenostavimo izraz $A$, tako da ulomek razširimo z $x^{\\frac{1}{3}}$. Dobimo $A=\\frac{3}{x-2}$. Izraz $B$ pa razširimo z $x^{-\\frac{1}{3}}$. Dobimo $B=\\frac{1}{x-1}$. Izračunamo razliko $A-B=\\frac{3}{x-2}-\\frac{1}{x-1}=\\frac{2x-1}{(x-2)(x-1)}$. Zapišemo še $(A-B)^{-1}=\\frac{(x-2)(x-1)}{2x-1}$.... | Slovenia | Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Intermediate Algebra > Other"
] | null | final answer only | (x-2)(x-1)/(2x-1) | |
0l5x | Four unit squares form a $2 \times 2$ grid. Each of the 12 unit line segments forming the sides of the squares is colored either red or blue in such a way that each unit square has 2 red sides and 2 blue sides. One example is shown below (red is solid, blue is dashed). Find the number of such colorings.
![](attached_im... | [] | United States | AIME II | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 82 | |
068f | Let $a, b, c$ positive real numbers such that $a + b + c = 1$. Prove that:
$$
(a+1)\sqrt{2a(1-a)} + (b+1)\sqrt{2b(1-b)} + (c+1)\sqrt{2c(1-c)} \ge 8(ab+bc+ca).
$$
When does equality hold? | [
"Since $a + b + c = 1$, we have $a + 1 = 2a + b + c$ and $1 - a = b + c$. Hence we have the equivalent inequality\n$$\n(2a + b + c)\\sqrt{2a(b + c)} + (2b + c + a)\\sqrt{2b(c + a)} + (2c + a + b)\\sqrt{2c(a + b)} \\ge 8(ab + bc + ca)\n$$\nFrom the inequality of arithmetic and geometric mean we get: $2a + b + c \\ge... | Greece | SELECTION EXAMINATION | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds if and only if a = b = c = 1/3. | |
0g54 | Problem:
Finde alle ganzzahligen Werte, die der Ausdruck
$$
\frac{p q+p^{p}+q^{q}}{p+q}
$$
annehmen kann, wobei $p$ und $q$ Primzahlen sind. | [
"Antwort: Der einzige ganzzahlige Wert ist 3 .\n\nLösung: Falls $p$ und $q$ beide ungerade sind, dann ist der Zähler ungerade und der Nenner gerade. Weil eine gerade Zahl nie eine ungerade Zahl teilt, ergibt dies keinen ganzzahligen Wert. Wir können also annehmen, dass mindestens eine unserer Primzahlen gerade ist,... | Switzerland | Zweite Runde 2023 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 3 | |
0cxm | Find all integers $n$ for which $n(n+2010)$ is a perfect square. | [
"Let $n(n+2010) = k^2$ for some integer $k$.\n\nWe can write:\n$$\nn(n+2010) = k^2\n$$\n$$\nn^2 + 2010n - k^2 = 0\n$$\nThis is a quadratic in $n$:\n$$\nn^2 + 2010n - k^2 = 0\n$$\nThe discriminant must be a perfect square for $n$ to be integer:\n$$\n\\Delta = 2010^2 + 4k^2 = m^2\n$$\nfor some integer $m$.\n\nSo:\n$$... | Saudi Arabia | SAMC | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All integers n of the form n = -1005 ± (d + 2010^2/d)/4, where d ranges over positive divisors of 2010^2 such that d + 2010^2/d is divisible by 4. | |
0980 | Problem:
Într-o grădină zoologică locuiesc hameleoni de trei culori: $x$ hameleoni de culoare sură, $2022$ hameleoni de culoare albă și $100$ hameleoni de culoare roz. Hameleonii se pot întâlni între ei, însă doar câte doi. Dacă se întâlnesc $2$ hameleoni de aceeași culoare, atunci ei nu își schimbă culoarea. Dacă se ... | [
"Solution:\n\nPresupunem că într-un moment concret de timp în grădina zoologică sunt: $a$ hameleoni suri, $b$ hameleoni albi și $c$ hameleoni roz, adică avem tripletul $(a, b, c)$. După momentul întâlnirii a oricăror doi hameleoni, acest triplet se schimbă în unul dintre următoarele trei:\n\na) $(a-1, b-1, c+2)$, s... | Moldova | Olimpiada Republicană la Matematică | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | All integers x that are multiples of 3 | |
08ok | Problem:
Find all nonnegative integers $x, y, z$ such that
$$
2013^{x} + 2014^{y} = 2015^{z}
$$ | [
"Solution:\nClearly, $y > 0$, and $z > 0$. If $x = 0$ and $y = 1$, then $z = 1$ and $(x, y, z) = (0, 1, 1)$ is a solution.\n\nIf $x = 0$ and $y \\geq 2$, then modulo 4 we have $1 + 0 \\equiv (-1)^{z}$, hence $z$ is even ($z = 2z_{1}$ for some integer $z_{1}$). Then $2^{y} 1007^{y} = (2015^{z_{1}} - 1)(2015^{z_{1}} ... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (0, 1, 1) | |
07qa | Point $D$ is on side $AC$ of triangle $ABC$ such that $\angle ABC = \angle DAC = 30^\circ$ and $\angle ADB = 45^\circ$. Prove that $|BD| = |DC|$.
 | [
"Let $O$ be the circumcentre of $\\triangle ABC$ and $E$ the point where the line $AD$ meets the circumcircle again.\n\n\n\nBecause central angles are twice inscribed angles subtended by the same arc, we obtain $\\angle COA = 2\\angle CBA = 60^\\circ$ and $\\angle EOC = 2\\angle EAC = 60^\\... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordi... | null | proof only | null | |
01u0 | Let $E$ be the intersection point of the segments $AB$ and $CD$. Points $X$ and $Y$ are marked in the plane such that the quadrilaterals $AECX$ and $BEDY$ are parallelograms.
Prove that the lines $AD$, $BC$, and $XY$ are either pairwise parallel or concurrent.
(M. Karpuk, A. Voidelevich) | [
"Let the segment $XY$ meet the segments $AB$ and $CD$ at $F$ and $G$, respectively. Since $AX \\parallel DC$ and $DC \\parallel YB$, the triangles $AXF$ and $BYF$ are similar. Therefore we have $AF/FB = AX/YB$. In the same way we conclude that $DG/GC = DY/XC$.\n\n\n\nSuppose that the lines ... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0hbb | Are there integers $a < b < c < d$ such that
$$
\frac{a}{a} + \frac{a}{b} + \frac{a}{c} + \frac{a}{d} = \frac{b}{a} + \frac{b}{b} + \frac{b}{c} + \frac{b}{d} ?
$$ | [
"**Answer.** Yes, for example $a = -28$, $b = -14$, $c = -7$ and $d = 4$.\n\nClearly, all numbers cannot be positive. Let's take $a = -4$, $b = -2$, $c = -1$ and find the corresponding $d$:\n$$\n\\begin{aligned}\n\\frac{-4}{-4} + \\frac{-4}{-2} + \\frac{-4}{-1} + \\frac{-4}{d} &= \\frac{-2}{-4} + \\frac{-2}{-2} + \... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | a = -28, b = -14, c = -7, d = 4 | |
03l6 | Problem:
Suppose that the real numbers $a_{1}, a_{2}, \ldots, a_{100}$ satisfy
$$
\begin{gathered}
a_{1} \geq a_{2} \geq \cdots \geq a_{100} \geq 0 \\
a_{1}+a_{2} \leq 100 \\
a_{3}+a_{4}+\cdots+a_{100} \leq 100
\end{gathered}
$$
Determine the maximum possible value of $a_{1}^{2}+a_{2}^{2}+\cdots+a_{100}^{2}$, and find ... | [
"Solution:\nWe have $a_{1}+a_{2}+\\cdots+a_{100} \\leq 200$, so\n$$\n\\begin{aligned}\na_{1}^{2}+a_{2}^{2}+\\cdots+a_{100}^{2} & \\leq (100-a_{2})^{2}+a_{2}^{2}+a_{3}^{2}+\\cdots+a_{100}^{2} \\\\\n& = 100^{2}-200 a_{2}+2 a_{2}^{2}+a_{3}^{2}+\\cdots+a_{100}^{2} \\\\\n& \\leq 100^{2}-(a_{1}+a_{2}+\\cdots+a_{100}) a_{... | Canada | Canadian Mathematics Olympiad | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Maximum value: 10000. Achieved by the sequences (100, 0, 0, ..., 0) and (50, 50, 50, 50, 0, 0, ..., 0). | |
061u | Problem:
Im Dreieck $A B C$ ist $A D$ ($D \in B C$) eine Seitenhalbierende, $E$ ein Punkt auf $A C$ und $F$ der Schnittpunkt von $B E$ mit $A D$.
Man beweise: Falls $\frac{B F}{F E}=\frac{B C}{A B}+1$, dann ist $B E$ eine Winkelhalbierende. | [
"Solution:\n\nSeien $B'$, $C'$, $E'$ die Projektionen von $B$, $C$ und $E$ auf die Gerade $A D$. Die rechtwinkligen Dreiecke $\\triangle D B B'$ und $\\triangle D C C'$ sind kongruent, da $D$ der Mittelpunkt von $B C$ ist und die spitzen Winkel bei $D$ gleich groß sind. Daraus folgt, dass $B B' = C C'$.\nIm Dreieck... | Germany | Auswahlwettbewerb zur IMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
075r | For a positive integer $n$, a cubic polynomial $p(x)$ is said to be $n$-good if there exist $n$ distinct integers $a_1, a_2, \dots, a_n$ such that all the roots of the polynomial $p(x) + a_i = 0$ are integers for $1 \le i \le n$. Given a positive integer $n$ prove that there exists an $n$-good cubic polynomial. | [
"Let $f(x) = x^3 - mx^2 + nx$, $k$ an integer, and $a, b, c$ the roots of $f(x) + k = 0$. Then\n$$\n4m^2 - 12n = 3(a-c)^2 + (a - 2b + c)^2.\n$$\nSince the equation $x^2 + 3y^2 = 1$ has a rational solution, it has infinitely many rational solutions. Therefore one can find $D$ such that $4D$ can be expressed as $3s^2... | India | Indija TS 2013 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | null | proof only | null | |
0988 | Problem:
Fie $a, b, c$ numere reale pozitive distincte. Demonstrați, că ecuația
$$
(a+b+c) x^{2}+2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right) x+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0
$$
are două soluţii reale distincte. | [
"Solution:\n\nFie $\\Delta$ discriminantul ecuației date. Atunci $\\Delta_{1}=\\frac{\\Delta}{4}=\\left(\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}\\right)^{2}-(a+b+c)\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right)=$\n$$\n\\begin{aligned}\n& =\\frac{a^{2}}{b^{2}}+\\frac{b^{2}}{c^{2}}+\\frac{c^{2}}{a^{2}}+2 \\frac{... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
02ir | Problem:
Seja $n=9867$. Se você calculasse $n^{3}-n^{2}$, encontraria um número cujo algarismo das unidades é:
A) 0
B) 2
C) 4
D) 6
E) 8 | [
"Solution:\n\nSolução 1: O algarismo final de $9867^{3}$ é o mesmo que o de $7^{3}=343$, isto é, 3; o algarismo final de $9867^{2}$ é o mesmo que o de $7^{2}=49$, isto é, 9. Se de um número terminado em 3 subtraímos outro terminado em 9, o algarismo final do resultado é 4.\n\nComentário: Observe que:\nalgarismo das... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | C | |
0g4l | Problem:
Let $D$ be the set of real numbers excluding $-1$. Find all functions $f: D \rightarrow D$ such that for all $x, y \in D$ satisfying $x \neq 0$ and $y \neq -x$, the equality
$$
(f(f(x)) + y) f\left(\frac{y}{x}\right) + f(f(y)) = x
$$
holds. | [
"Solution:\nPlugging in $x = y \\neq 0$ yields $f(f(x)) = \\alpha x$ for some constant $\\alpha$. This holds for all $x \\neq 0, -1$. If $\\alpha$ was any real number different from $0$ or $1$, we would get $f(f(-1/\\alpha)) = -1$, which is a contradiction. Hence, only $\\alpha = 0$ or $\\alpha = 1$ are possible. I... | Switzerland | Swiss Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = (1 - x) / (1 + x) for all real x ≠ -1 | |
0ej4 | Problem:
Lucijana je iz množice števil $\{1,2,3,4,5,6,7,8,9\}$ izbrala 3 različna števila. Z njimi je zapisala največje možno trimestno število in najmanjše možno trimestno število. Dobljeni števili je seštela in dobila število 545. Koliko je vsota 3 števil, ki jih je izbrala Lucijana?
(A) 6
(B) 7
(C) 9
(D) 11
(E) 13 | [
"Solution:\n\nNaj bodo $a, b$ in $c$ števila, ki jih je izbrala Lucijana, pri čemer je $a > b > c$. Največje možno število, ki ga z njimi lahko zapiše, je tedaj $\\overline{abc}$, najmanjše možno število pa $\\overline{cba}$. Za njuno vsoto velja, da je $100a + 10b + c + 100c + 10b + a = 101(a + c) + 20b = 545$. Če... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | B | |
0g37 | Problem:
Sei $ABC$ ein Dreieck mit $AB > AC$. Die Winkelhalbierenden bei $B$ und $C$ treffen sich im Punkt $I$ innerhalb des Dreiecks $ABC$. Der Umkreis des Dreiecks $BIC$ schneidet $AB$ ein zweites Mal in $X$ und $AC$ ein zweites Mal in $Y$. Zeige, dass $CX$ parallel zu $BY$ ist. | [
"Solution:\n\nWir zeigen, dass $\\angle ACX = \\angle AYB$. Dies ist ausreichend, um zu zeigen, dass die Linien parallel sind. Wir nehmen für unseren Beweis an, dass $X$ zwischen $A$ und $B$, $Y$ nicht zwischen $A$ und $C$ liegt.\n\nDa $CI$ die Winkelhalbierende von $\\angle ACB$ ist, haben wir $\\angle ACI = \\ang... | Switzerland | Vorrunde | [
"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 | |
0au6 | Problem:
The product of the two roots of $\sqrt{2014}\, x^{\log_{2014} x} = x^{2014}$ is an integer. Find its units digit. | [] | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Number Theory > Modular Arithmetic"
] | null | final answer only | 6 | |
052d | Nonzero integers $a$, $b$ and $c$ satisfy $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$. Prove that among $a$, $b$, $c$ there are two integers which have a common divisor larger than 1. | [
"Multiplying the given equation by $abc$ we get $bc + ca + ab = 0$.\n\nIf $a$, $b$, $c$ were all odd, then $bc$, $ca$ and $ab$ were also odd and their sum could not be $0$.\n\nIf one of the numbers $a$, $b$, $c$ was even and the others were odd, then two of the numbers $bc$, $ca$ and $ab$ were even and one odd, whi... | Estonia | Open Contests | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
02fd | A triangle has semi-perimeter $s$, circumradius $R$ and inradius $r$. Show that it is right-angled iff $2R = s - r$. | [
"Let $a \\ge b \\ge c$ be the sides of the triangle. If the triangle is right-angled, then\n$r = \\frac{b+c-a}{2}$, $2R = a$ and $s - r = \\frac{a+b+c}{2} - \\frac{b+c-a}{2} = a = 2R$.\n\nIf $2R = s - r$, consider\n$$\nE = (2R-a)(2R-b)(2R-c) = 8R^3 - 4R^2(a+b+c) + 2R(ab+bc+ca) - abc\n$$\nLet $\\Delta$ be the area o... | Brazil | XVI OBM | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0e8z | Find all natural numbers $n$ of the form $n = \overline{23ab16c}$, such that all their digits are different and they are divisible by $9$ and $11$. Here, $a$, $b$ and $c$ are digits. | [
"The number $n$ is divisible by $99$. We may write\n$$\n\\begin{aligned}\nn &= \\overline{23ab1} \\cdot 100 + 60 + c = \\overline{23ab1} \\cdot 99 + \\overline{23ab1} + 60 + c = \\\\\n&= \\overline{23ab1} \\cdot 99 + \\overline{23a} \\cdot 100 + 10b + 1 + 60 + c = \\\\\n&= (\\overline{23ab1} \\cdot 99 + \\overline{... | Slovenia | National Math Olympiad 2013 - First Round | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 2379168 and 2389167 | |
0kkc | At noon on a certain day, Minneapolis is $N$ degrees warmer than St. Louis. At 4:00 the temperature in Minneapolis has fallen by $5$ degrees while the temperature in St. Louis has risen by $3$ degrees, at which time the temperatures in the two cities differ by $2$ degrees. What is the product of all possible values of ... | [] | United States | AMC 12 B | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | C | |
0hr2 | Problem:
Determine all integers $x$ such that the product $x(x+1)(x+2)$ is the square of an integer. | [
"Solution:\nThe answers are $x=0$, $x=-1$, and $x=-2$, for all of which the product is $0=0^{2}$.\n\nAssume that there is another solution. We can immediately rule out the case $x<-2$, as then the product is negative. So $x$ is a positive integer. Note that the greatest common divisor of $x$ and $x+1$ is $1$, as tw... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | x = -2, -1, 0 | |
03jb | Problem:
A chord $ST$ of constant length slides around a semicircle with diameter $AB$. $M$ is the mid-point of $ST$ and $P$ is the foot of the perpendicular from $S$ to $AB$. Prove that angle $SPM$ is constant for all positions of $ST$. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0euq | A prime $p$ is called *nice prime* if there exists sequences of positive integers $(n_1, n_2, \dots, n_k)$ satisfying the following conditions for infinitely many positive integers $k$, but there does not exist such for $k = 1$.
1. For $i = 1, 2, \dots, k$, $n_i \ge \frac{i+1}{2}$.
2. For $i = 1, 2, \dots, k$, $p^{n_i}... | [
"Let us denote by $a|b$ when a positive integer $a$ divides $b$. We write $p^c \\nmid n$ when a positive integer $n$ is a multiple of $p^c$, but not of $p^{c+1}$ for a prime $p$.\n\n**Lemma 1.** For a given prime $q$ and a positive integer $a$ such that $(a, q) = 1$, let $s$ be the smallest positive integer $x$ sat... | South Korea | 23rd Korean Mathematical Olympiad Final Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof only | null | |
0df7 | Point $M$ on side $AB$ of quadrilateral $ABCD$ is such that quadrilaterals $AMCD$ and $BMDC$ are circumscribed around circles centered at $O_1$ and $O_2$ respectively. Line $O_1O_2$ cuts an isosceles triangle with vertex $M$ from angle $CMD$. Prove that $ABCD$ is a cyclic quadrilateral. | [
"If $AB \\parallel CD$ then the incircles of $AMCD$ and $BMDC$ have equal radii; now the problem conditions imply that the whole picture is symmetric about the perpendicular from $M$ to $O_1O_2$, and hence $ABCD$ is an isosceles trapezoid (or a rectangle). The conclusion in this case is true.\n\n![](attached_image_... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point ... | English | proof only | null | |
0ekq | Problem:
Naj bo $p$ praštevilo in $n \leq p^{p}$ naravno število. Dokaži, da vsaj eno od števil $n+p^{p}$ in $n \cdot p^{p}$ ni popoln kvadrat. | [
"Solution:\n\nDenimo, da sta obe števili popolna kvadrata, torej $n+p^{p}=a^{2}$ in $n \\cdot p^{p}=b^{2}$, kjer sta $a$ in $b$ naravni števili.\n\nČe je $p=2$, tedaj je $n \\leq 2^{2}=4$. Toda nobeno od števil $1+2^{2}$, $2+2^{2}$, $3+2^{2}$ in $4+2^{2}$ ni popoln kvadrat. Torej mora biti $p$ liho praštevilo.\n\nP... | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0egk | Problem:
Kaj dobimo po racionalizaciji izraza $\frac{4+2 \sqrt{6}}{(1+\sqrt{2})+\sqrt{3}}$?
(A) $1-\sqrt{2}-\sqrt{3}$
(B) $1-\sqrt{2}+\sqrt{3}$
(C) $1+\sqrt{2}+\sqrt{3}$
(D) $-1+\sqrt{2}+\sqrt{3}$
(E) $-1+\sqrt{2}-\sqrt{3}$ | [
"Solution:\nŠtevec in imenovalec pomnožimo z izrazom $(1+\\sqrt{2})-\\sqrt{3}$ in dobimo $\\frac{4+2 \\sqrt{6}}{(1+\\sqrt{2})+\\sqrt{3}} \\cdot \\frac{(1+\\sqrt{2})-\\sqrt{3}}{(1+\\sqrt{2})-\\sqrt{3}}$.\n\n$= \\frac{(4+2\\sqrt{6})((1+\\sqrt{2})-\\sqrt{3})}{((1+\\sqrt{2})+\\sqrt{3})((1+\\sqrt{2})-\\sqrt{3})}$\n\nIme... | Slovenia | Državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | MCQ | D | |
092m | Problem:
Let $ABC$ be an acute-angled triangle with $\Varangle BAC > 45^{\circ}$ and with circumcentre $O$. The point $P$ lies in its interior such that the points $A, P, O, B$ lie on a circle and $BP$ is perpendicular to $CP$. The point $Q$ lies on the segment $BP$ such that $AQ$ is parallel to $PO$.
Prove that $\Vara... | [
"Solution:\nSince $\\Varangle BAC > 45^{\\circ}$, we have $\\Varangle BOC > 90^{\\circ}$, and the points $A, P, O, B$ lie on a circle in this order.\nInstead of the equality $\\Varangle QCB = \\Varangle PCO$, we show the equivalent statement $\\Varangle OCB = \\Varangle PCQ$.\nWe know that $\\Varangle OCB = 90^{\\c... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06nv | A point $P$ lies inside an equilateral triangle $ABC$ such that $AP = 15$ and $BP = 8$. Find the maximum possible value of the sum of areas of triangles $ABP$ and $BCP$. | [
"Rotate $\\triangle PBC$ about point $B$ by $60^\\circ$ in the anticlockwise direction, so that $BA$ is the image of $BC$ after rotation. Let $Q$ be the image of $P$ after rotation. Then\n$$\n[ABP] + [BCP] = [PAQB] = [PQB] + [QPA].\n$$\n\n\n\nNow, $\\triangle PQB$ is an equilateral triangle... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 60 + 16√3 | |
0cpi | Let $C_0$ and $B_0$ be respectively the midpoints of the sides $AB$ and $AC$ of a non-isosceles acute-angled triangle $ABC$. Denote by $O$ and $H$ its circumcenter and its orthocenter, respectively. The lines $BH$ and $OC_0$ intersect at point $P$, while the lines $CH$ and $OB_0$ intersect at point $Q$. Given that the ... | [
"Первое решение. Пусть $BB'$ и $CC'$ — высоты треугольника. Так как $OB_0$ и $OC_0$ — серединные перпендикуляры к сторонам $AC$ и $AB$, то отрезки $B'B_0$ и $C'C_0$ равны высотам ромба $OPHQ$, значит, $B'B_0 = C'C_0$. Но эти отрезки являются проекциями отрезка $OH$ на прямые $AB$ и $AC$; значит, $OH$ составляет рав... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distan... | English, Russian | proof only | null | |
01m8 | a) Does there exist a function $f: \mathbb{R} \to \mathbb{R}$, such that for any real $x$ the following equality holds $f(\sin x) + f(\cos x) = 1$?
b) The same question for $f(\sin x) + f(\cos x) = \sin x$.
(I. Gorodnin) | [] | Belarus | 61st Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | a) Yes. For example, f(x) = 1/2 for all real x works. b) No such function exists. | |
06y6 | Let $n$ be a positive integer. A class of $n$ students run $n$ races, in each of which they are ranked with no draws. A student is eligible for a rating $(a, b)$ for positive integers $a$ and $b$ if they come in the top $b$ places in at least $a$ of the races. Their final score is the maximum possible value of $a-b$ ac... | [
"The answer can be achieved by the students finishing in the same order in every race. To show that this is the maximum, we will apply a series of modifications to the results of the races, each of which does not decrease the total score, such that after $k$ such modifications the first $k$ positions are the same i... | IMO | IMO2024 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | n(n-1)/2 | |
09rw | Problem:
Vind alle functies $f: \mathbb{R} \setminus \{0\} \rightarrow \mathbb{R}$ waarvoor geldt:
$$
x f(x y)+f(-y)=x f(x)
$$
voor alle reële $x, y$ ongelijk aan $0$. | [
"Solution:\n\nVul in $x=1$, dat geeft $f(y)+f(-y)=f(1)$, oftewel $f(-y)=f(1)-f(y)$ voor alle $y$.\n\nVul nu in $y=-1$, dat geeft $x f(-x)+f(1)=x f(x)$. Als we hierin $f(-x)=f(1)-f(x)$ invullen, krijgen we $x(f(1)-f(x))+f(1)=x f(x)$, dus $x f(1)+f(1)=2 x f(x)$.\n\nWe zien dat $f$ van de vorm $f(x)=c+\\frac{c}{x}$ is... | Netherlands | Selectietoets | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = c + c/x for any real constant c, defined on nonzero real numbers | |
06xk | Let $n$ be a positive integer. We arrange $1+2+\cdots+n$ circles in a triangle with $n$ rows, such that the $i^{\text{th}}$ row contains exactly $i$ circles. The following figure shows the case $n=6$.
In this triangle, a ninja-path is a sequence of circles obtained by repeatedly going from a c... | [
"Answer: The maximum value is $k=1+\\left\\lfloor\\log_{2} n\\right\\rfloor$.\n\nSolution 1. Write $N=\\left\\lfloor\\log_{2} n\\right\\rfloor$ so that we have $2^{N} \\leqslant n \\leqslant 2^{N+1}-1$.\nWe first provide a construction where every ninja-path passes through at most $N+1$ red circles. For the row $i=... | IMO | International Mathematical Olympiad Shortlist | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions >... | null | proof and answer | 1 + floor(log_2 n) | |
03ni | Problem:
Let $S$ be a set of $n \geq 3$ positive real numbers. Show that the largest possible number of distinct integer powers of three that can be written as the sum of three distinct elements of $S$ is $n-2$. | [
"Solution:\n\nWe will show by induction that for all $n \\geq 3$, it holds that at most $n-2$ powers of three are sums of three distinct elements of $S$ for any set $S$ of positive real numbers with $|S|=n$. This is trivially true when $n=3$.\n\nLet $n \\geq 4$ and consider the largest element $x \\in S$. The sum o... | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | n - 2 | |
0h1v | On the diagonals $AC$ and $BD$ of the cyclic quadrilateral $ABCD$ consider points $X$ and $Y$ such that $ABXY$ is a parallelogram. Prove that the circumradii of $BXD$ and $CYA$ are equal. | [
"Since $ABCD$ is cyclic, then $\\angle ABD = \\angle ACD$. Moreover, $\\angle ABD = \\angle ABY = \\angle ABY$. Hence, $CXYD$ is cyclic because $\\angle XCD = \\angle XYB$ (fig. 12).\n\n\n\nThis implies that $\\angle XCY = \\angle XDY$, from what it follows that $\\sin \\angle ACY = \\sin \... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03pi | Let the three sides of a triangle be integers $l$, $m$, $n$, respectively, satisfying $l > m > n$ and $\{\frac{3^l}{10^4}\} = \{\frac{3^m}{10^4}\} = \{\frac{3^n}{10^4}\}$, where $\{x\} = x - [x]$ and $[x]$ denotes the integral part of the number $x$. Find the minimum perimeter of such a triangle. | [
"Since\n$$\n\\frac{3^l}{10^4} - \\left\\lfloor \\frac{3^l}{10^4} \\right\\rfloor = \\frac{3^m}{10^4} - \\left\\lfloor \\frac{3^m}{10^4} \\right\\rfloor = \\frac{3^n}{10^4} - \\left\\lfloor \\frac{3^n}{10^4} \\right\\rfloor,\n$$\nwe have\n$$\n\\begin{align}\n3^l &\\equiv 3^m \\equiv 3^n \\pmod{10^4} \\\\\n&\\iff \\b... | China | China Mathematical Competition (Extra Test) | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | English | proof and answer | 3003 |
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