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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0af0 | The Macedonian mathematical Olympiad is held in two rooms labeled with the numbers $1$ and $2$. In the beginning all the contestants are in room $1$. The final schedule is obtained in the following manner: a list of names of some of the contestants is read; when a contestant's name is read, he and all of his friends ch... | [
"We'll prove that the total number of possible schedules is an even number so it can't be equal to $2009$. It is enough to prove that there exists a list of names of some of the contestants such that all the contestants in room $1$ go to room $2$ (if this is possible then for every possible final schedule the rever... | North Macedonia | Sixteenth Macedonian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | Macedonian, English | proof only | null | |
0gvy | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$
f(f(x) - y^2) = (x - y)^2 \cdot f(x + y)
$$
for any real numbers $x$ and $y$. | [
"Відповідь: $f(x) = 0$, $x \\in \\mathbb{R}$; $f(x) = x^2$, $x \\in \\mathbb{R}$.\n\nЗ вихідного співвідношення отримуємо, що\n$$\n(x - y)^2 f(x + y) = (x + y)^2 f(x - y).\n$$\nПокладемо $x = \\frac{t+1}{2}$, $y = \\frac{t-1}{2}$ і одержимо, що $1^2 \\cdot f(t) = t^2 \\cdot f(1)$ при всіх $t \\in \\mathbb{R}$. Звід... | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 0 for all real x; f(x) = x^2 for all real x | |
04hl | In an acute triangle $ABC$ such that $|AC| < |BC|$, points $M$ and $N$ are respectively the feet of altitudes from vertices $A$ and $B$. The circumcircle of the triangle $ABC$ with the centre $O$ and the circumcircle of the triangle $MNC$ with the centre $S$ intersect in points $C$ and $D$. If the point $P$ is the midp... | [
"Firstly note that the circumcircle of the triangle $MNC$ passes through the ortho-centre $H$ of the triangle $ABC$. The segment $CH$ is the diameter of that circle. Since the segment $CD$ is the common chord of circumcircles of triangles $ABC$ and $MNC$, the line $CD$ is perpendicular to the line $SO$ through thei... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0b32 | Problem:
Determine the set of all polynomials $P(x)$ with real coefficients such that the set $\{P(n) \mid n \in \mathbb{Z}\}$ contains all integers, except possibly finitely many of them. | [
"Solution:\nWe claim that the only such polynomials are of the form $P(x)=\\frac{1}{i}(x+j)$ for some integers $i \\neq 0, j$.\nLet $\\mathcal{R}$ be the set $\\{P(n) \\mid n \\in \\mathbb{Z}\\}$.\n\nWithout loss of generality, we may assume that the leading coefficient of $P(x)$ is positive; otherwise we can consi... | Philippines | 23rd Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | P(x) = (x + j)/i for integers i ≠ 0 and j | |
02v6 | Problem:
Seja $n \geq 3$ um inteiro positivo. Sobre uma reta, são marcados os $n$ pontos $P_{1}, P_{2}, P_{3}, \ldots, P_{n}$, nessa ordem e igualmente espaçados entre si. Em seguida, cada um dos pontos deve ser pintado de azul ou de vermelho de modo que não existam três pontos $P_{x}, P_{\frac{x+y}{2}}$ e $P_{y}$ pin... | [
"Solution:\n\na. Marcaremos com $A$ e $V$ as cores dos pontos em ordem. Considere a seguinte forma de colorir os 8 pontos:\n$$\n\\begin{array}{cccccccc}\nP_{1} & P_{2} & P_{3} & P_{4} & P_{5} & P_{6} & P_{7} & P_{8} \\\\\nA & V & V & A & A & V & V & A .\n\\end{array}\n$$\nA condição dada requer que não existam pont... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Algebraic Number Theory > Combinatorial number theory: van der Waerden's theorem"
] | null | proof only | null | |
0dno | Problem:
Нека је $n$ природан број већи од 1. Доказати да постоји природан број $m$ већи од $n^{n}$ такав да је
$$
\frac{n^{m}-m^{n}}{n+m}
$$
природан број. | [
"Solution:\n\nЗа почетак приметимо да за $m>n \\geqslant 3$ важи $n^{m}>m^{n}$, те је $\\frac{n^{m}-m^{n}}{m+n}>0$. Заиста, функција $f(x)=\\frac{\\ln x}{x}$ је опадајућа за $x>e$ јер је $f^{\\prime}(x)=\\frac{1-\\ln x}{x^{2}}<0$, па је $\\frac{\\ln n}{n}>\\frac{\\ln m}{m}$, тј. $m \\ln n>n \\ln m$, и одатле $n^{m}... | Serbia | 10. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof only | null | |
0469 | A party is attended by $n$ people. Assume that there are at most $n$ pairs of friendships among them, and any two people shake hands at the party if and only if they have a common friend at the party.
Suppose $m$ is a positive integer satisfying $m \ge 3$ and $n \le m^3$. Prove that there exists a person $A$ such that ... | [
"*Proof.* We represent individuals as vertices, friendships as edges, and use a graph to depict the problem.\nThe given graph contains a connected subgraph $G = (V, E)$ satisfying $|E| \\le |V| \\le n$. We denote the degree of vertex $v$ as $d(v)$ and the number of individuals with whom $v$ has shaken hands as $\\b... | China | 2023 Chinese IMO National Team Selection Test | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
00k1 | We call a convex pentagon in the Euclidean plane "special" if either all of its sides are of equal length or all of its interior angles are equal. We call it "very special" if either all of its sides are of equal length and two of its interior angles equal or if all of its interior angles are equal and two of its sides... | [
"Let the pentagon have the vertices $A$, $B$, $C$, $D$ and $E$ in this order. We first assume that all sides are of equal length and two angles equal.\n\n\nIf the two equal angles are adjacent, we can place them at $A$ and $B$ without loss of generality. In this case, $EABC$ is an equilater... | Austria | AustriaMO2013 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
09vv | In an acute triangle $ABC$ the foot of the altitude from $A$ is called $D$. Let $D_1$ and $D_2$ be reflections of $D$ in $AB$ and $AC$, respectively. The intersection of $BC$ and the line through $D_1$ parallel to $AB$, is called $E_1$. The intersection of $BC$ and the line through $D_2$ parallel to $AC$, is called $E_... | [
"Let $K$ be the midpoint of $DD_1$, and let $L$ be the midpoint of $DD_2$. Then $K$ lies on $AB$ and $L$ lies on $AC$. Because $\\angle AKD = 90^\\circ = \\angle ALD$, the quadrilateral $AKDL$ is cyclic. Hence, $\\angle DLK = \\angle DAK = \\angle DAB = 90^\\circ - \\angle ABC$. Moreover, $KL$ is a midsegment in tr... | Netherlands | BxMO Team Selection Test, March 2020 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
00fp | Let $ABC'$ be a triangle and $D$ the foot of the altitude from $A$. Let $E$ and $F$ be on a line passing through $D$ such that $AE$ is perpendicular to $BE$, $AF$ is perpendicular to $CF$, and $E$ and $F$ are different from $D$. Let $M$ and $N$ be the midpoints of the line segments $BC$ and $EF$, respectively. Prove th... | [
"Let $P$ be such that $ADMP$ is a rectangle. Choose points $Q$ and $R$ on the line $AP$ such that $Q B D A$ and $A D C R$ are rectangles. Points $Q$, $B$ and $D$ lie on the circle of diameter $AB$, hence $ADEQ$ is a cyclic quadrilateral. Similarly, $R$, $C$ and $D$ lie on the circle of diameter $AC$, hence $ADFR$ i... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0etc | Determine the smallest integer $k > 1$ such that there exist $k$ distinct primes whose squares sum to a power of $2$. | [
"For $p_1^2 + p_2^2$ (where $p_1$ and $p_2$ are two distinct primes) to be equal to $2^n$ (where $n \\ge 2$), we must have both $p_1$ and $p_2$ odd. This would give $p_1^2 + p_2^2 \\equiv 2 \\pmod 4$, while $2^n \\equiv 0 \\pmod 4$, a contradiction.\n\nFor $p_1^2 + p_2^2 + p_3^2$ (where $p_1, p_2$ and $p_3$ are thr... | South Africa | The South African Mathematical Olympiad, Third Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English, Afrikaans | proof and answer | 5 | |
044w | Positive real numbers $x, y$ satisfy the following condition: there exist $a \in [0, x], b \in [0, y]$ such that
$$
a^2 + y^2 = 2, \quad b^2 + x^2 = 1, \quad ax + by = 1.
$$
Then the maximum of $x+y$ is ______. | [
"In a plane rectangular coordinate system $xOy$, for positive real number pairs $(x, y)$ that satisfy the condition, take points $L(x, 0)$, $M(x, y)$, $N(0, y)$, and then quadrilateral $OLMN$ is a rectangle. Points $P, Q$ are on sides $LM, MN$, respectively, as shown in Fig. 8.1.\nSince $a^2 + y^2 = 2$, $b^2 + x^2 ... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | final answer only | sqrt(5) | |
03ev | We will call a natural number $m$ remarkable if there exist integers $a$, $b$, $c$, for which $m = a^3 + 2b^3 + 4c^3 - 6abc$. Prove that there exists a natural number $n < 2024$ such that for infinitely many prime numbers $p$, the number $np$ is remarkable. | [
"Lemma. Let $p$ be a prime number and $a$, $b$, $c \\in \\mathbb{Z}/p\\mathbb{Z}$. Then there exist $x$, $y$, $z \\in \\mathbb{Z}$ such that $|x|$, $|y|$, $|z| < \\sqrt[3]{p}$, $(x, y, z) \\neq (0, 0, 0)$ and $ax + by + cz \\equiv 0 \\pmod{p}$.\n\n*Proof.* Consider the set $M := \\{(x, y, z) : x, y, z \\in \\{0, 1,... | Bulgaria | Bulgarian Winter Tournament | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | English | proof only | null | |
00uj | Once upon a time there are $n$ pairs of princes and princesses who are in love with each other. One day a witch comes along and turns all the princes into frogs; the frogs can be distinguished by sight but the princesses cannot tell which frog corresponds to which prince. The witch tells the princesses that if any of t... | [
"We claim that the princesses can guarantee saving $k$ princes and no more. To see that they can save $k$ princes, have each princess kiss the first $k$ frogs - clearly each of the first $k$ frogs will be saved.\n\nNow we will show by induction on $k$ that the princesses cannot guarantee saving more than $k$ prince... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | k | |
0hq3 | Problem:
Alex and Bob have $30$ matches. Alex picks up somewhere between one and six matches (inclusive), then Bob picks up somewhere between one and six matches, and so on. The player who picks up the last match wins. How many matches should Alex pick up at the beginning to guarantee that he will be able to win? | [
"Solution:\n\n$2$."
] | United States | null | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 2 | |
0lax | On the plane given a circle $(O)$ and a point $A$ lying outside the circle. Through $A$, draw the tangents to $(O)$; let $B$ and $C$ be the tangent points. Consider a point $P$ moving on the opposite ray to ray $BA$ and a point $Q$ moving on the opposite ray to ray $CA$, such that the line $PQ$ is tangent to $(O)$. The... | [] | Vietnam | IMO2011 Selection | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > C... | English | proof only | null | |
09mf | Consider $2 \times 60$ grids where each cell contains a distinct number from $1$ to $120$. For each grid, calculate the sum of the elements in each row and each column, resulting in $62$ sums. A grid is called an *even grid* if all $62$ sums are even numbers, and an *odd grid* if all $62$ sums are odd numbers. Which ty... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Odd grids occur more frequently. | |
04r5 | There are two touching circles, $k_1(S_1, r_1)$ and $k_2(S_2, r_2)$ in a rectangle $ABCD$ with $|AB| = 9$, $|BC| = 8$. Moreover, $k_1$ touches $AD$ and $CD$, while $k_2$ touches $AB$ and $BC$.
a) Prove $r_1 + r_2 = 5$.
b) What is the least and what is the greatest possible area of $AS_1S_2$? | [
"a) Let $M$ and $N$ be intersections of the line through $S_1$ parallel to $AD$. Analogously, let $K$ and $L$ be intersections of the line through $S_2$ parallel to $AB$. Let $P$ be the intersection of $KL$ and $MN$ (see Fig. 1). The Pythagoras theorem for $S_1PS_2$ gives\n$$\n\\begin{align*}\n(r_1 + r_2)^2 &= (8 -... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | r1 + r2 = 5; minimum area = 14, maximum area = 31/2 | |
078d | Let $r > 0$ be a real number. We call a monic polynomial with complex coefficients $r$-good if all of its roots have absolute value at most $r$. We call a monic polynomial with complex coefficients *primordial* if all of its coefficients have absolute value at most 1.
a. Prove that any 1-good polynomial has a primordi... | [
"First, we show that if all roots of $Q$ have absolute value at most $1$, then $Q$ has a primordial multiple. We use induction on $\\deg Q$.\n\nIf $Q$ is linear, then it is clearly primordial, so we are done.\n\nNow assume $\\deg Q > 1$, and let $a$ be a root of $Q$, and write $Q(x) = (x - a)Q_1(x)$. By induction h... | India | IMO TST | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
0dq8 | In an isosceles triangle $ABC$, $AB = AC$. Let $E$ be a point on $AC$ and extend $AB$ to a point $D$ such that $BD = EC$. Let $F$ be the intersection of $BC$ and $DE$ and let the circumcircle of $ABC$ intersect the circumcircle of $BDF$ at $G$. Prove that $DE$ is perpendicular to $FG$. | [
"\n\nNote that the points $B$, $D$, $G$, $F$ are concyclic and the points $A$, $B$, $G$, $C$ are also concyclic. Also since $\\angle EDG = \\angle FDG = \\angle FBG = \\angle CBG = \\angle GAC = \\angle GAE$, the points $A$, $D$, $G$, $E$ are concyclic. Since $\\angle GEC = \\angle ADG = \\... | Singapore | Singapore International Mathematical Olympiad Committee National Team Selection Test | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point"
] | null | proof only | null | |
08tp | Determine all non-negative real-valued functions $f$ defined for non-negative real numbers, which satisfy
$$
f(x^2) + f(y) = f(x^2 + y + x f(4y))
$$
for every pair of non-negative real numbers $x$ and $y$. | [
"First of all, let us show that the function $f$ is monotone increasing in the wide sense, namely it satisfies the property that $a < b \\implies f(a) \\le f(b)$. So, suppose $a < b$. Then we can choose a number $t > 0$ such that $b = t^2 + a + t f(4a)$. Substituting $x = t$, $y = a$ into the given functional equat... | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Functional equatio... | null | proof and answer | f(x) = sqrt(x) and f(x) ≡ 0 | |
09tm | Problem:
Zij $n \geq 0$ een geheel getal. Een rij $a_{0}, a_{1}, a_{2}, \ldots$ van gehele getallen wordt als volgt gedefinieerd: er geldt $a_{0}=n$ en voor $k \geq 1$ is $a_{k}$ het kleinste gehele getal groter dan $a_{k-1}$ waarvoor $a_{k}+a_{k-1}$ het kwadraat van een geheel getal is. Bewijs dat er precies $\lfloor... | [
"Solution:\n\nZij $m=\\lfloor\\sqrt{2 n}\\rfloor$. We bewijzen eerst dat de rij kwadraten $a_{0}+a_{1}, a_{1}+a_{2}$, ... precies de rij $(m+1)^{2},(m+2)^{2}, \\ldots$ is. Vervolgens laten we zien dat de verschillen $a_{i}-a_{i-1}$ een rij opeenvolgende even getallen en een rij opeenvolgende oneven getallen vormen.... | Netherlands | IMO-selectietoets | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Other"
] | null | proof and answer | ⌊√(2n)⌋ | |
013u | Problem:
Let $ABCD$ be a convex quadrilateral such that $BC = AD$. Let $M$ and $N$ be the midpoints of $AB$ and $CD$, respectively. The lines $AD$ and $BC$ meet the line $MN$ at $P$ and $Q$, respectively. Prove that $CQ = DP$. | [
"Solution:\n\nLet $A'$, $B'$, $C'$, $D'$ be the feet of the perpendiculars from $A$, $B$, $C$, $D$, respectively, onto the line $MN$. Then\n$$\nAA' = BB' \\quad \\text{and} \\quad CC' = DD'.\n$$\nDenote by $X$, $Y$ the feet of the perpendiculars from $C$, $D$ onto the lines $BB'$, $AA'$, respectively. We infer from... | Baltic Way | Baltic Way 2005 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cuf | Let $P(x)$ be a polynomial of degree $n \ge 2$ with nonnegative coefficients. Let $a$, $b$, and $c$ be the side lengths of some acute-angled triangle. Prove that the numbers $\sqrt[n]{P(a)}$, $\sqrt[n]{P(b)}$, and $\sqrt[n]{P(c)}$ are also the side lengths of some acute-angled triangle. | [
"Without loss of generality, let $a \\ge b \\ge c$; these three positive numbers are the side lengths of an acute-angled triangle if and only if $a^2 < b^2 + c^2$. Since the coefficients of $P(x)$ are nonnegative, we have $P(a) \\ge P(b) \\ge P(c) > 0$; thus, we need to check that $\\sqrt[n]{P(a)}^2 < \\sqrt[n]{P(b... | Russia | XLIII Russian mathematical olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Algebraic Expressions > Polynomials"
] | English; Russian | proof only | null | |
0aos | Problem:
Let $P$ be a point on the diagonal $AC$ of the square $ABCD$. If $AP$ is one-fourth of the length of one side of the square and the area of the quadrilateral $ABPD$ is 1 square unit, find the area of $ABCD$. | [
"Solution:\n$4\\sqrt{2}$ square units\n\nLet $s$ be the length of one side of the square $ABCD$. Let $(ABP)$ and $(ABPD)$ denote the areas of $\\triangle ABP$ and quadrilateral $ABPD$, respectively. Then $(ABP) = \\frac{1}{2}(ABPD) = \\frac{1}{2}$.\n\nSince the diagonals of a square are perpendicular to each other ... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 4√2 | |
00yc | Problem:
There are $n$ lines $(n>2)$ given in the plane. No two of the lines are parallel and no three of them intersect at one point. Every point of intersection of these lines is labelled with a natural number between $1$ and $n-1$. Prove that, if and only if $n$ is even, it is possible to assign the labels in such ... | [
"Solution:\n\nSuppose we have assigned the labels in the required manner. When a point has label $1$ then there can be no more occurrences of label $1$ on the two lines that intersect at that point. Therefore the number of intersection points labelled with $1$ has to be exactly $\\frac{n}{2}$, and so $n$ must be ev... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof only | null | |
09y2 | We consider a triangle $ABC$ and a point $D$ on the extended line segment $AB$ on the side of $B$. The point $E$ lies on side $AC$ such that the angles $\angle DBC$ and $\angle DEC$ are equal. The intersection of $DE$ and $BC$ is $F$. Suppose that $|BF| = 2$, $|BD| = 3$, $|AE| = 4$, and $|AB| = 5$. (Attention: the pict... | [
"a.\nBecause angles $\\angle AEC$ and $\\angle ABD$ are straight, we have\n$$\n\\angle ABC = 180^{\\circ} - \\angle DBC = 180^{\\circ} - \\angle DEC = \\angle AED.\n$$\nBecause angle $A$ occurs in both triangles, triangles $\\triangle ABC$ and $\\triangle AED$ have two equal angles, and hence the triangles are simi... | Netherlands | Dutch Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 8 | |
084y | Problem:
Una piramide a base quadrata ha il lato di base lungo $\sqrt{3}$ e tutti gli spigoli delle facce laterali sono lunghi $\sqrt{2}$. Quanti gradi misura l'angolo fra due spigoli non appartenenti alla stessa faccia laterale? | [
"Solution:\n\nLa risposta è $120$. L'angolo richiesto è angolo al vertice del triangolo isoscele che ha per base la diagonale della base e per lati obliqui due spigoli delle facce laterali. La base misura $\\sqrt{6}$ e il rapporto tra il lato obliquo e metà base è $\\frac{\\sqrt{3}}{2}$: questo è il seno di metà de... | Italy | Progetto Olimpiadi di Matematica 2006 GARA di SECONDO LIVELLO | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 120° | |
0c0t | Find the pairs of integers $(a, b)$ such that $a^2 + 2b^2 + 2a + 1$ is a divisor of $2ab$. | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | All integer pairs (a, b) with either a = 0 and any integer b, or b = 0 and any integer a except a = −1; together with the six nontrivial pairs (−1, 1), (−1, −1), (−3, 1), (−3, 2), (−3, −1), (−3, −2). | |
0d2k | $ABC$ is a triangle, $H$ its orthocenter, $I$ its incenter, $O$ its circumcenter and $\omega$ its circumcircle. Line $CI$ intersects circle $\omega$ at point $D$ different from $C$. Assume that $AB = ID$ and $AH = OH$. Find the angles of triangle $ABC$. | [
"We already know that $DA = DB = DI$. Because $DI = AB$, triangle $ADB$ is equilateral. But $\\angle ACB + \\angle BDA = 180^\\circ$, we deduce that $\\angle ACB = 120^\\circ$.\n\nBecause $\\angle BOA = 2 \\angle BDA = 120^\\circ$, we deduce that $\\angle OAB = 30^\\circ$. On the other hand, we have $\\angle CAH = ... | Saudi Arabia | Selection tests for the Gulf Mathematical Olympiad 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscella... | English | proof and answer | ∠A = 15°, ∠B = 45°, ∠C = 120° | |
02v9 | Problem:
Os números inteiros do conjunto $\{1,2, \ldots, 20\}$ serão pintados com duas cores, branco e preto, de modo que ambas as cores sejam usadas. Além disso, o produto dos números de uma cor não deve possuir fatores primos em comum com o produto dos números da outra cor. De quantos modos isso pode ser feito? | [
"Solution:\n\nIndependente das cores escolhidas para os outros números, temos duas opções de escolha para a cor do número $1$. Temos também duas opções para a cor do número $2$ e, uma vez que ela tenha sido escolhida, todos os números do conjunto $\\{4,6,8,10,12,14,16,18,20\\}$ devem possuir a mesma cor do número $... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 62 | |
0949 | Problem:
Let $\mathbb{N}$ denote the set of positive integers. Determine all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that $f(1) \leq f(2) \leq f(3) \leq \ldots$ and the numbers $f(n)+n+1$ and $f(f(n))-f(n)$ are both perfect squares for every positive integer $n$. | [
"Solution:\nFor a positive integer $n$, denote by $b_{n}$ the number $\\sqrt{f(n)+n+1}$. Note that we have $f(n)+n+1 > f(n-1)+(n-1)+1$ for all $n > 1$. In other words, the sequence $\\left(b_{n}\\right)_{n}$ is a strictly increasing sequence of positive integers. Since $b_{1} \\geq 2$, we have $b_{n} \\geq n+1$, or... | Middle European Mathematical Olympiad (MEMO) | MEMO Team Competition | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | f(n) = n^2 + n for all positive integers n | |
0ldg | Given a real number $a$ and consider the sequence $(u_n)$ defined by
$$
u_1 = a, \quad u_{n+1} = \frac{1}{2} + \sqrt{\frac{2n+3}{n+1}u_n + \frac{1}{4}}, \quad \forall n \in \mathbb{N}^*.$$
1.
If $a = 5$, prove that $(u_n)$ has a finite limit and finds that limit.
2.
Find all the values of $a$ such that the sequence $(... | [
"We will solve directly part 2), from which to deduce the result of part 1). It can be seen that the sequence $(u_n)$ defines if and only if $u_2$ defines. Since $u_2 = \\frac{1}{2} + \\sqrt{\\frac{5}{2}a + \\frac{1}{4}}$, then $u_2$ defines if and only if\n$$\na \\geq -\\frac{1}{10}.\n$$\nWe will prove that the se... | Vietnam | Vietnamese Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | For a = 5, the sequence converges to 3. For all a ≥ -1/10, the sequence is well-defined and converges to 3. | |
0f0o | Problem:
$p(x) = ax^2 + bx + c$ is a real quadratic such that $|p(x)| \leq 1$ for all $|x| \leq 1$. Prove that $|cx^2 + bx + a| \leq 2$ for $|x| \leq 1$. | [] | Soviet Union | ASU | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0czg | Let $n$ be a positive integer. Prove that at least one of the integers
$$
\left[2^{n} \cdot \sqrt{2}\right],\left[2^{n+1} \cdot \sqrt{2}\right], \ldots,\left[2^{2 n} \cdot \sqrt{2}\right]
$$
is even, where $[a]$ denotes the integer part of $a$. | [
"Assume by contradiction that all integers are odd. Then, we have\n$$\n2a-1 < 2^{n} \\sqrt{2} < 2a\n$$\nfor some positive integer $a$. Multiplying inequalities (1) by $2$, we get $4a-2 < 2^{n+1} \\sqrt{2} < 4a$. Since the integer $\\left[2^{n+1} \\sqrt{2}\\right]$ is odd, it follows\n$$\n4a-1 < 2^{n+1} \\sqrt{2} < ... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
00ti | Let $\triangle ABC$ be a right-angled triangle with $\angle BAC = 90^\circ$. Let the height from $A$ cut its side $BC$ at $D$. Let $I, I_B, I_C$ be the incenters of triangles $ABC, ABD, ACD$ respectively. Let also $E_B, E_C$ be the excenters of $ABC$ with respect to vertices $B$ and $C$ respectively. If $K$ is the poin... | [
"Since $\\angle E_CBI = 90^\\circ = ICE_B$, we conclude that $E_CBCE_B$ is cyclic. Moreover, we have that\n$$\n\\angle BAI_B = \\frac{1}{2}\\angle BAD = \\frac{1}{2}\\hat{C},\n$$\nso $AI_B \\perp CI$. Similarly $AI_C \\perp BI$. Therefore is the orthocenter of triangle $AI_BIC$. It follows that\n$$\n\\angle II_B I_... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0h1i | Let $\tau(n)$ be the number of divisors of a natural number $n$. Prove that there exists infinitely many natural numbers $N$ such that $(\tau(N)+\tau(N+1)+1) \equiv 3 \pmod{4}$. | [
"So, finally, we have that every number $N = 3n^3$ with $n = 24m+1$ satisfies the problem condition, which evidently means, that there are infinitely many such numbers."
] | Ukraine | Problems of Ukrainian Authors | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
05ik | Problem:
Soit $k > 1$ un entier. Une fonction $f: \mathbb{N}^* \rightarrow \mathbb{N}^*$ est dite $k$-tastrophique lorsque pour tout entier $n > 0$, on a $f_k(n) = n^k$ où $f_k$ est la $k$-ième itérée de $f$ :
$$
f_k(n) = \underbrace{f \circ f \circ \cdots \circ f}_{k \text{ fois }}(n)
$$
Pour quels $k$ existe-t-il un... | [
"Solution:\n\nOn va prouver que, pour tout entier $k \\geq 1$, il existe une fonction $k$-tastrophique.\n\nClairement, la fonction $f: n \\longmapsto n$ est 1-tastrophique. On suppose donc que $k \\geq 2$.\n\nOn procède maintenant de la façon suivante. On pose $f(1) = 1$, et si $n$ est le plus petit entier pour leq... | France | Olympiades Françaises de Mathématiques - Test de Sélection | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | for all integers k at least two | |
0if3 | Problem:
Let $ABCD$ be a regular tetrahedron with side length $2$. The plane parallel to edges $AB$ and $CD$ and lying halfway between them cuts $ABCD$ into two pieces. Find the surface area of one of these pieces. | [
"Solution:\n\n$1 + 2\\sqrt{3}$\n\nThe plane intersects each face of the tetrahedron in a midline of the face; by symmetry it follows that the intersection of the plane with the tetrahedron is a square of side length $1$. The surface area of each piece is half the total surface area of the tetrahedron plus the area ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > 3D Shapes"
] | null | proof and answer | 1 + 2√3 | |
0k57 | Problem:
In a game, $N$ people are in a room. Each of them simultaneously writes down an integer between $0$ and $100$ inclusive. A person wins the game if their number is exactly two-thirds of the average of all the numbers written down. There can be multiple winners or no winners in this game. Let $m$ be the maximum... | [
"Solution:\n\nSince the average of the numbers is at most $100$, the winning number is an integer which is at most two-thirds of $100$, or at most $66$. This is achieved in a room with $34$ people, in which $33$ people pick $100$ and one person picks $66$, so the average number is $99$.\n\nFurthermore, this cannot ... | United States | HMMT February 2018 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 34 | |
0grl | In a non-isosceles acute triangle $ABC$, $D$ is the midpoint of the edge $[BC]$. The points $E$ and $F$ lie on $[AC]$ and $[AB]$, respectively, and the circumcircles of $CDE$ and $AEF$ intersect at $P$ on $[AD]$. The angle bisector of $P$ in triangle $EFP$ intersects the line $EF$ at $Q$. Prove that the tangent line to... | [
"Since $C$, $D$, $P$, $F$ are cyclic, we have $\\angle DPE = 180^\\circ - \\angle C$.\n\n\n\nAs $A$, $E$, $P$, $F$ are cyclic, we also get $\\angle FPE = 180^\\circ - \\angle A$. Therefore, we have $\\angle FPD = 360^\\circ - (180^\\circ - \\angle A + 180^\\circ - \\angle C) = 180^\\circ - ... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
07rn | Find all functions $f(x) = a x^2 + b x + c$, with $a \neq 0$, such that
$$
f(f(1)) = f(f(0)) = f(f(-1)).
$$ | [
"**Solution 1.** First note that $f(0) = c$, $f(1) = a + b + c$, $f(-1) = a - b + c$. Second, note that if $f(\\alpha) = f(\\beta)$, then $a(\\alpha^2 - \\beta^2) + b(\\alpha - \\beta) = 0$, so $(\\alpha - \\beta)(a(\\alpha + \\beta) + b) = 0$. Third, a key observation is that for fixed $k$, there are at most two d... | Ireland | Irish | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | a(x^2 - 1/2), a(x^2 - x - 1) + 1/2, a(x^2 + x - 1) - 1/2, for any a ≠ 0 | |
0hpg | Problem:
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ which satisfy
$$
f(x+y)=f(x-y)+4 x y
$$
for all real numbers $x$ and $y$. | [
"Solution:\nThe answer is the functions $f(x)=x^{2}+c$, where $c$ is a constant. It is easy to check that all such functions work, since\n$$\n(x+y)^{2}+c=(x-y)^{2}+c+4 x y\n$$\nis trivially true.\n\nNow, we prove these are the only functions. Put $x=y=\\frac{1}{2} a$ to obtain\n$$\nf(a)=f(0)+a^{2}\n$$\nfor all real... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | f(x) = x^2 + c for some real constant c | |
04pm | Let $\triangle ABC$ be a triangle such that $\angle ABC = 2\angle BCA$, and the angle bisector of $\angle BAC$ intersects the side $\overline{BC}$ at point $D$ so that $|AB| = |CD|$. Find $\angle CAB$. (Estonia 2002) | [] | Croatia | Croatian Mathematical Society Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 72° | |
09kn | Show that the sum or the product of $a, b, c$ is equal to $1$ if $a, b, c$ are real numbers satisfying
$$
(ab - c)(bc - a) + (bc - a)(ca - b) + (ca - b)(ab - c) = 4abc.
$$ | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0hqx | Problem:
Find all polynomials $f$ that satisfy the equation
$$
\frac{f(9 x)}{f(3 x)} = \frac{243 x - 729}{x - 81}
$$
for infinitely many values of $x$. | [
"Solution:\nWe have\n$$\n(x-81) f(9 x) = (243 x - 729) f(3 x)\n$$\nfor infinitely many values of $x$. Since both sides of this equation are polynomials, they must then be equal for all $x$.\n\nPlugging in $x=3$, we get $f(27)=0$. Plugging in $x=9$ then gives $f(81)=0$. Plugging in $x=27$ then gives $f(243)=0$. Thus... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = a x^2 (x - 27)(x - 81)(x - 243) for real constants a | |
0ioa | Problem:
Three positive reals $x$, $y$, and $z$ are such that
$$
\begin{aligned}
x^{2}+2(y-1)(z-1) & =85 \\
y^{2}+2(z-1)(x-1) & =84 \\
z^{2}+2(x-1)(y-1) & =89
\end{aligned}
$$
Compute $x+y+z$. | [
"Solution:\n\nAdd the three equations to obtain\n$$\nx^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x-4 x-4 y-4 z+6=258\n$$\nwhich rewrites as $(x+y+z-2)^{2}=256$. Evidently, $x+y+z=2 \\pm 16$. Since $x$, $y$, and $z$ are positive, $x+y+z>0$ so $x+y+z=2+16=18$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 18 | |
0h1z | Let $ABCD$ be a cyclic quadruple. Let us denote the midpoints of $AB$, $BC$, $CD$ and $DA$ by $M$, $L$, $N$ and $K$ respectively. It is known, that $\angle BMN = \angle MNC$. Prove that:
a) $\angle DKL = \angle CLK$;
b) $ABCD$ has a pair of parallel sides. | [
"a) Using the properties of inscribed angles we get $KM \\perp BD$, $KN \\perp AC$, $\\angle ABD = \\angle ACD \\Rightarrow \\angle AMK = \\angle ABD = \\angle ACD = \\angle KND$. Thus $\\angle KMN = \\pi - \\angle AMK - \\angle BMN = \\pi - \\angle KND - \\angle MNC = \\angle KNM$, hence $\\square KMN$ is isoscele... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
021e | Problem:
Let $N \geq 2$ be a natural number. At a mathematical olympiad training camp, the same $N$ courses are organised every day. Each student takes exactly one of the $N$ courses each day. At the end of the camp, every student has taken each course exactly once, and any two students took the same course on at least... | [
"Solution:\nThe largest number of students at the camp is $(N - 1)!$. Since each student takes exactly one course each day and, at the end, has taken each course exactly once, the schedule of a student can be represented by a permutation of the set of the $N$ courses.\n\nTo show that $(N - 1)!$ is possible, we can,... | Benelux Mathematical Olympiad | 17th Benelux Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | (N - 1)! | |
0j7x | Problem:
Let $a$ and $b$ be positive real numbers. Define two sequences of real numbers $\{a_{n}\}$ and $\{b_{n}\}$ for all positive integers $n$ by $(a+b i)^{n}=a_{n}+b_{n} i$. Prove that
$$
\frac{\left|a_{n+1}\right|+\left|b_{n+1}\right|}{\left|a_{n}\right|+\left|b_{n}\right|} \geq \frac{a^{2}+b^{2}}{a+b}
$$
for all ... | [
"Solution:\nLet $z=a+b i$. It is easy to see that what we are asked to show is equivalent to\n$$\n\\frac{\\left|z^{n+1}+\\bar{z}^{n+1}\\right|+\\left|z^{n+1}-\\bar{z}^{n+1}\\right|}{\\left|z^{n}+\\bar{z}^{n}\\right|+\\left|z^{n}-\\bar{z}^{n}\\right|} \\geq \\frac{2 z \\bar{z}}{|z+\\bar{z}|+|z-\\bar{z}|}\n$$\nCross-... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | proof only | null | |
08a2 | Problem:
Quante sono le coppie ordinate $(A, B)$ di sottoinsiemi di $\{1,2,3,4,5\}$ tali che l'intersezione tra $A$ e $B$ abbia esattamente un elemento?
(A) 80
(B) 280
(C) 1280
(D) 751
(E) 405 | [
"Solution:\n\nLa risposta è $\\mathbf{(E)}$. Le coppie richieste si possono costruire nella maniera seguente: scegliamo innanzitutto l'elemento comune tra $A$ e $B$ (cinque possibilità), e per ogni elemento non appartenente all'intersezione decidiamo se esso stia in $A$, in $B$ o in nessuno dei due. Questo ci porta... | Italy | Progetto Olimpiadi della Matematica | [
"Discrete Mathematics > Combinatorics"
] | null | MCQ | E | |
03yq | Find all positive integers $n$ such that equation $\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$ has exactly 2011 positive integer solutions $(x, y)$ with $x \le y$. | [
"From the given equation, we have $xy - nx - ny = 0 \\Rightarrow (x-n)(y-n) = n^2$. Then, besides $x = y = 2n$, for any $x-n$ equal to a proper divisor of $n^2$, we will get a positive integer solution $(x, y)$ satisfying the required condition. Therefore, $n^2$ should have exactly 2010 proper divisors that are les... | China | China Girls' Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | n = p^{2010} for any prime p | |
0f8l | Problem:
There are 21 towns. Each airline runs direct flights between every pair of towns in a group of five. What is the minimum number of airlines needed to ensure that at least one airline runs direct flights between every pair of towns? | [
"Solution:\n\nAnswer: 21.\n\nThere are 210 pairs of towns. Each airline serves 10 pairs, so we certainly need at least 21 airlines. The following arrangement shows that 21 is possible:\n\n| 1 | 2 | 3 | 4 | 5 |\n|---|---|---|---|---|\n| 1 | 6 | 7 | 8 | 9 |\n| 1 | 10 | 11 | 12 | 13 |\n| 1 | 14 | 15 | 16 | 17 |\n| 1 |... | Soviet Union | 22nd ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 21 | |
01w5 | Two circles $\Omega$ and $\Gamma$ are internally tangent at the point $B$. The chord $AC$ of $\Gamma$ is tangent to $\Omega$ at the point $L$, and the segments $AB$ and $BC$ intersect $\Omega$ at the points $M$ and $N$. Let $M_1$ and $N_1$ be the reflections of $M$ and $N$ about the line $BL$; and let $M_2$ and $N_2$ b... | [
"By the Archimedes's lemma, $BL$ is the bisector of $\\angle ABC$, so $M_1 \\in BC$ and $N_1 \\in AB$. Denote $MM_1 \\cap LB = T$ and $MM_2 \\cap LA = P$. Since $\\angle MTL = \\angle MPL = 90^\\circ$, the quadrilateral $MTLP$ is cyclic with the diameter $ML$. Hence\n$$\n\\angle MM_1M_2 = \\angle MTP = \\angle MLP ... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane ... | English | proof only | null | |
0kwg | Problem:
Given a positive integer $N$ (written in base 10), define its integer substrings to be integers that are equal to strings of one or more consecutive digits from $N$, including $N$ itself. For example, the integer substrings of 3208 are $3,2,0,8,32,20,320,208$, and 3208. (The substring 08 is omitted from this l... | [
"Solution:\nThe answer is $88,888,888$.\n\nIn our solution, we'll make use of the well-known fact that an integer is divisible by 9 if and only if the sum of its digits (in base 10) is divisible by 9. It was permissible to use this fact without proof on the contest, but for the sake of completeness, a proof can be ... | United States | Bay Area Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 88,888,888 | |
071e | Problem:
$C$ is a point on the semicircle with diameter $AB$. $D$ is a point on the arc $BC$. $M$, $P$, $N$ are the midpoints of $AC$, $CD$ and $BD$. The circumcenters of $ACP$ and $BDP$ are $O$, $O'$. Show that $MN$ and $OO'$ are parallel. | [
"\nLet the center of the circle be $X$ and the radius $r$. Let $\\angle AXM = \\theta$, $\\angle BXN = \\varphi$. Note that $O$ is the intersection of $XM$ and the perpendicular to $CD$ at $Q$, the midpoint of $CP$. We have $XM = r \\cos \\theta$. Let $CD$ and $XM$ meet at $Y$. Then $\\angl... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscella... | null | proof only | null | |
07h4 | Given an acute-angled triangle $ABC$ with an altitude $AD$ and orthocenter $H$. Let $E$ be the reflection of $H$ with respect to $A$. A point $X$ lies on the circumcircle of the triangle $BDE$ such that $DX \parallel AC$, and similarly a point $Y$ lies on the circumcircle of the triangle $CDE$ such that $DY \parallel A... | [
"Denote by $\\Gamma$ the circumcircle of the triangle $ABC$. Letting $F$ and $G$ be the reflections of $H$ with respect to $AC$ and $AB$, respectively. It is well-known that $F$ and $G$ lie on $\\Gamma$. Also $EF$ is parallel to $AC$ since $A$ is the midpoint of the segment $EH$. Therefore $\\angle EFB = 90^\\circ$... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0awd | Problem:
Find the integer which is closest to the value of $\frac{1}{\sqrt[6]{5^{6}+1}-\sqrt[6]{5^{6}-1}}$. | [] | Philippines | 19th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 9375 | |
075a | Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(x+y)f(x-y) = (f(x)+f(y))^2 - 4x^2f(y), \quad (1)
$$
for all $x, y \in \mathbb{R}$, where $\mathbb{R}$ denotes the set of all real numbers. | [
"Put $x = y = 0$; we get $f(0)^2 = 4f(0)^2$ and hence $f(0) = 0$.\n\n$$\nf(y)(f(y) - f(-y)) = 0.\n$$\n\nWe may conclude that either $f(y) = 0$ or $f(y) = f(-y)$ for each $y \\in \\mathbb{R}$. Replacing $y$ by $-y$, we may also conclude that $f(-y)(f(-y) - f(y)) = 0$. If $f(y) = 0$ and $f(-y) \\neq 0$ for some $y$, ... | India | Indija mo 2011 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 0 for all real x; f(x) = x^2 for all real x | |
0et5 | Let $S$ be a square with sides of length $2$ and $R$ be a rhombus with sides of length $2$ and angles measuring $60^\circ$ and $120^\circ$. These quadrilaterals are arranged to have the same centre and the diagonals of the rhombus are parallel to the sides of the square. Calculate the area of the region on which the fi... | [
"Let $S$ be the square $ABCD$, with centre $O$. Let $R$ be the rhombus $EFGH$, with $EG$ the short diagonal, and $O$ the midpoint of $EG$. Since the diagonals of $R$ bisect the angles of $R$, we have that $\\angle OFE = 30^\\circ$, so that $\\sin 30^\\circ = \\frac{1}{2}$ forces $EG$ to have length $2$. We may ther... | South Africa | The South African Mathematical Olympiad Third Round | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | 4 - 2/sqrt(3) | |
0558 | a) Let $a$ and $b$ be arbitrary positive integers of equal parity. Can we always find noninteger numbers $x$ and $y$ such that $x + y$ and $ax + by$ are integers?
b) The same question when $a$ and $b$ have different parities. | [
"a) By taking $x = y = \\frac{1}{2}$ we have that $x + y = 1$ is an integer and so is $ax + by = \\frac{1}{2}(a + b)$, since $a + b$ is even by the assumption.\n\nb) We notice that $ax + by = a(x + y) + (b - a)y$. Assume that $x + y$ and $ax + by$ are integers. Since $a(x + y)$ is an integer, since it is a product ... | Estonia | Open Contests | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Other"
] | English | proof only | null | |
06zi | Problem:
$f$ is a function defined on all reals in the interval $[0,1]$ and satisfies $f(0) = 0$, $f(x/3) = f(x)/2$, $f(1-x) = 1 - f(x)$. Find $f(18/1991)$. | [] | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | final answer only | 5/128 | |
0jbx | Problem:
It has recently been discovered that the right triangle with vertices $(0,0)$, $(0,2012)$, and $(2012,0)$ is a giant pond that is home to many frogs. Frogs have the special ability that, if they are at a lattice point $(x, y)$, they can hop to any of the three lattice points $(x+1, y+1)$, $(x-2, y+1)$, and $(... | [
"Solution:\n\n\n\nWe transform the triangle as follows: map each lattice point $(x, y)$ to the point\n$$\nx(1,0) + y\\left(\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right) = \\left(x + \\frac{y}{2}, \\frac{y \\sqrt{3}}{2}\\right).\n$$\n\nThis transforms the right triangle into an equilateral tria... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
03pf | Suppose $A$, $B$, $C$ are three non-collinear points corresponding to complex numbers $z_0 = ai$, $z_1 = \frac{1}{2} + bi$, $z_2 = 1 + ci$ ($a$, $b$ and $c$ being real numbers), respectively. Prove that the curve
$$
z = z_0 \cos^4 t + 2z_1 \cos^2 t \cdot \sin^2 t + z_2 \sin^4 t \ (t \in \mathbb{R})
$$
shares a single c... | [
"Let $z = x + yi$ ($x, y \\in \\mathbb{R}$), then\n$$\nx + yi = a\\cos^4 t \\cdot i + 2\\left(\\frac{1}{2} + bi\\right)\\cos^2 t \\cdot \\sin^2 t + (1 + ci)\\sin^4 t.\n$$\nSeparating real and imaginary parts, we get\n$$\n\\begin{aligned}\nx &= \\cos^2 t \\cdot \\sin^2 t + \\sin^4 t = \\sin^2 t, \\\\\ny &= a(1-x)^2 ... | China | China Mathematical Competition (Shaanxi) | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | 1/2 + ((a+c+2b)/4)i | |
0f22 | Problem:
$P$ is a polygon. Its sides do not intersect except at its vertices, and no three vertices lie on a line. The pair of sides $AB$, $PQ$ is called special if (1) $AB$ and $PQ$ do not share a vertex and (2) either the line $AB$ intersects the segment $PQ$ or the line $PQ$ intersects the segment $AB$. Show that t... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
03i7 | Problem:
The sides $AD$ and $BC$ of a convex quadrilateral $ABCD$ are extended to meet at $E$. Let $H$ and $G$ be the midpoints of $BD$ and $AC$, respectively. Find the ratio of the area of the triangle $EHG$ to that of the quadrilateral $ABCD$. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 1/4 | |
0b3e | Problem:
How many positive integers $n < 2022$ are there for which the sum of the odd positive divisors of $n$ is $24$?
(a) $7$
(b) $8$
(c) $14$
(d) $15$ | [] | Philippines | 24th Philippine Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | d | |
0etk | Let $n \ge 3$ be an integer, and consider a set of $n$ points in three-dimensional space such that:
(i) every two distinct points are connected by a string which is either red, green, blue, or yellow;
(ii) for every three distinct points, if the three strings between them are not all of the same colour, then they are o... | [
"Let us fix a notation: Number the points $1, 2, \\dots, n$; The colour of the string between points $i$ and $j$ is denoted by $c(i, j)$, so that $c(i, j) \\in \\{\\text{B}, \\text{G}, \\text{R}, \\text{Y}\\}$ for all $1 \\le i, j \\le n$, $i \\ne j$, and where $\\text{B} = \\text{Blue}$, $\\text{G} = \\text{Green}... | South Africa | The South African Mathematical Olympiad Third Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 9 | |
0hwv | Problem:
Determine if there exist positive integers $a, b, m, n$ such that $a \neq b$, $m \geq 2$, $n \geq 2$, and
$$
\underbrace{a^{a^{\cdots a}}}_{m} = \underbrace{b^{b^{\cdots b}}}_{n}.
$$ | [
"Solution:\n\nThere do not exist such positive integers. Assume for a contradiction that there do, however. We may assume $m > n$, so that $b > a$. Since we have equality between a power of $a$ and a power of $b$, $b$ is a rational power of $a$. We write $b = a^{x}$, where we know that $x \\geq 1$ is a rational num... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Algebra > Intermediate Algebra > Exponential functions",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | No; such integers do not exist. | |
07af | We call the three variable polynomial $P$ cyclic if $P(x, y, z) = P(y, z, x)$. Prove that cyclic three variable polynomials $P_1$, $P_2$, $P_3$ and $P_4$ exist such that for each cyclic three variable polynomial $P$, there exists a four variable polynomial $Q$ such that
$$
P(x, y, z) = Q(P_1(x, y, z), P_2(x, y, z), P_3... | [
"Let\n$$\nQ(x, y, z) = P(x, y, z) + P(y, x, z)\n$$\n$$\nR(x, y, z) = P(x, y, z) - P(y, x, z).\n$$\nBy these definitions it is obvious that $Q$ is a symmetric polynomial and $R$ is an antisymmetric one. Note that $R(x, x, z) = 0$, so $R$ is divisible by $x - y$. Similarly, $y - z$ and $z - x$ also divide $R$. Theref... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English | proof only | null | |
09j3 | Let $p$ be an odd prime. Find all possible remainders modulo $p^2$ of sums of the form
$$
a^{p-1} + a^{p-2}b + \dots + ab^{p-2} + b^{p-1}
$$
for integers $a$ and $b$. | [] | Mongolia | Mongolian Mathematical Olympiad Round 2 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof and answer | {0, p} ∪ {1 + k·p : k = 0, 1, ..., p − 1} | |
0jzs | Problem:
Prove that there are infinitely many pairs of positive integers $(m, n)$ such that
$$
\frac{m+1}{n}+\frac{n+1}{m}
$$
is an integer. | [
"Solution:\nIn fact there are infinitely many pairs $(m, n)$ for which\n$$\n\\frac{m+1}{n}+\\frac{n+1}{m}=3 .\n$$\nTo see this, note that $(2,3)$ is a solution which gives $3$. Thereafter, we observe that the equation writes as\n$$\n3 m n = m^{2} + n^{2} + m + n \\quad \\text{or} \\quad m^{2} + (1-3 n) m + \\left(n... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof only | null | |
0kvr | Problem:
Compute the number of positive four-digit multiples of $11$ whose sum of digits (in base ten) is divisible by $11$. | [
"Solution:\nLet an arbitrary such number be $\\overline{a b c d}$. Then, we desire $11 \\mid a+b+c+d$ and $11 \\mid a-b+c-d$, where the latter comes from the well-known divisibility trick for $11$. Sums and differences of multiples of $11$ must also be multiples of $11$, so this is equivalent to desiring $11 \\mid ... | United States | HMMT November | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 72 | |
0dwr | Problem:
Za katera praštevila $p$ in $q$ je število $(p+1)^{q}$ popolni kvadrat? | [
"Solution:\n\nČe je $q=2$, je $(p+1)^{2}$ popolni kvadrat za vsako praštevilo $p$.\n\nSicer pa je $q$ liho, torej je $q=2k+1$ za neko naravno število $k$. Tedaj je $(p+1)^{q} = (p+1)^{2k+1} = (p+1)(p+1)^{2k}$, zato mora biti število $(p+1)$ popolni kvadrat, torej $p+1 = n^{2}$, od koder sledi $p = (n-1)(n+1)$. Ker ... | Slovenia | 49. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All prime pairs (p, q) with q = 2 (any prime p), or p = 3 (any prime q). | |
0b0c | Problem:
A function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies
$$
f(x)+f(y)=f(x) f(y)+1-\frac{4}{x y}
$$
for all nonzero real numbers $x$ and $y$. Given that $f(1)>0$, find the value of $f(4)$. | [
"Solution:\nSetting $x=y=1$ in the functional equation and letting $c=f(1)>0$, we have $2c=c^{2}-3$ or $(c+1)(c-3)=0$. Thus $c=f(1)=3$ and setting $y=1$ yields\n$$\nf(x)+3=3 f(x)+1-\\frac{4}{x} \\Longrightarrow f(x)=1+\\frac{2}{x}\n$$\nfor all nonzero reals $x$. Hence, $f(4)=\\frac{3}{2}$."
] | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 3/2 | |
0487 | Given an odd positive integer $n$, let $m = \frac{n+1}{2}$. Let positive integers $a_1, a_2, \dots, a_m$ be pairwise incongruent modulo $n$, and positive integers $b_1, b_2, \dots, b_m$ be pairwise incongruent modulo $n$. Prove that the set
$$
C = \{a_i + b_j \text{ (least nonnegative residue modulo } n) \mid i, j \in ... | [
"Let $X = \\{0, \\dots, n-1\\}$, $A = \\{a_1, \\dots, a_m\\}$, $B = \\{b_1, \\dots, b_m\\}$, and $S = X \\setminus C$. We need to show that $|S| < \\sqrt{n} + \\frac{1}{2}$.\n\nConsider the bipartite graph $G := (X \\sqcup X, E)$, where the edge set $E := \\{(x, y) : x + y \\in S\\}$. In particular, the restriction... | China | 2025 International Mathematical Olympiad China National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | English | proof only | null | |
0bl5 | Consider the real sequences $(a_n)_{n \ge 1}$, $(b_n)_{n \ge 1}$, $(c_n)_{n \ge 1}$. Prove that if the sequence
$$
p_n(x) = (x - a_n)(x - b_n)(x - c_n)
$$
converges for infinitely many values of $x$, then it converges for every $x \in \mathbb{R}$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
09j0 | We say a polynomial of degree three with integer coefficients is *good* if it has three real roots and all its roots are irrational numbers between $0$ and $1$.
(i) Is there a good polynomial with leading coefficient equal to $10$?
(ii) Is there a good polynomial with leading coefficient equal to $13$? | [
"Answer: (i) No, (ii) Yes.\n\na. (i)\nLet $a$, $b$, $c$ and $d > 0$ be integers and suppose that\n$$\nF(x) = a + bx + cx^2 + dx^3 = d(x - \\alpha)(x - \\beta)(x - \\gamma)\n$$\nis a good polynomial with $0 < \\alpha, \\beta, \\gamma < 1$. Let $Q(x) = x(1-x)(2x-1)$. Then it is easy to see that $|Q(x)| \\le \\frac{1}... | Mongolia | Mongolian Mathematical Olympiad Round 3 | [
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | (i) No, (ii) Yes | |
061l | Problem:
Man beweise, dass es keine positive ganze Zahl $n$ mit der folgenden Eigenschaft gibt: Für $k=1,2, \ldots, 9$ ist die - in dezimaler Schreibweise - linke Ziffer von $(n+k)!$ gleich $k$. | [
"Solution:\n\nWir nehmen an, dass es eine Zahl $n$ mit der verlangten Eigenschaft gibt. Dann kann keine der Fakultäten eine Zehnerpotenz sein, weil ab $3!$ alle Fakultäten durch $3$ teilbar sind und die ersten Fakultäten offensichtlich nicht die verlangte Eigenschaft haben. Es kann auch keine der Zahlen $n+2, \\ldo... | Germany | Auswahlwettbewerb zur IMO 2002 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0cy1 | Let $\left(a_{n}\right)_{n \geq 0}$ and $\left(b_{n}\right)_{n \geq 0}$ be sequences defined by $a_{n+2}= a_{n+1}+a_{n}$, $n=0,1, \ldots$, $a_{0}=1$, $a_{1}=2$, and $b_{n+2}=b_{n+1}+b_{n}$, $n=0,1, \ldots$, $b_{0}=2$, $b_{1}=1$. How many integers do the sequences have in common? | [
"We have $a_{2}=3$, $a_{3}=5$, $a_{4}=8$, $\\ldots$ and $b_{2}=3$, $b_{3}=4$, $b_{4}=7$, $\\ldots$ It follows $a_{0}=b_{0}$, $a_{1}=b_{1}$, $a_{2}=b_{2}$, and $b_{3}<a_{3}<b_{4}<a_{4}<b_{5}$. We prove by induction of step 2 that for $m \\geq 3$ we have $b_{m}<a_{m}<b_{m+1}$. The basis cases $m=3$, $m=4$ are verifie... | Saudi Arabia | SAMC | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 3 | |
01n6 | Let $m$, $n$, $k$ be pairwise relatively prime positive integers greater than $3$.
Find the minimal possible number of points on the plane with the following property: there are $x$ of them which are the vertices of a regular $x$-gon for $x = m$, $x = n$, $x = k$. | [
"Answer: $m + n + k - 4$."
] | Belarus | Belorusija 2012 | [
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | m + n + k - 4 | |
039p | (Peter Boyvalenkov) Find the least positive integer which cannot be written in the form $x^3 - x^2y + y^2 + x - y$, where $x$ and $y$ are positive integers. | [
"Let $F(x, y) = x^3 - x^2y + y^2 + x - y$. Note that $F(1, 1) = 1$ and $F(1, 2) = 2$. We shall prove that the equation $F(x, y) = 3$ has no solution in positive integers. Write this equation in the form\n$$\ny^2 - (1 + x^2)y + x^3 + x - 3 = 0.\n$$\nIts discriminant with respect to $y$ equals\n$$\nD = (1 + x^2)^2 - ... | Bulgaria | Spring Mathematical Tournament | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | 3 | |
0egr | Problem:
Tonček je izpisal tretjo potenco izraza $2 x^{3} y - 3 x y^{2}$. Ugotovil je, da imata potenci z osnovama $x$ in $y$ v enem izmed členov enaka eksponenta. Kolikšen je koeficient v tem členu?
(A) 36
(B) -36
(C) 18
(D) -54
(E) 54 | [
"Solution:\n\n$\\left(2 x^{3} y - 3 x y^{2}\\right)^{3} = \\left(2 x^{3} y\\right)^{3} - 3\\left(2 x^{3} y\\right)^{2} \\cdot \\left(3 x y^{2}\\right) + 3\\left(2 x^{3} y\\right) \\cdot \\left(3 x y^{2}\\right)^{2} - \\left(3 x y^{2}\\right)^{3} = 8 x^{9} y^{3} - 36 x^{7} y^{4} + 54 x^{5} y^{5} - 27 x^{3} y^{6}$.\n... | Slovenia | Državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | E | |
0j0v | Problem:
Danielle Bellatrix Robinson is organizing a poker tournament with 9 people. The tournament will have 4 rounds, and in each round the 9 players are split into 3 groups of 3. During the tournament, each player plays every other player exactly once. How many different ways can Danielle divide the 9 people into t... | [
"Solution:\n\nAnswer: $20160$\n\nWe first split the 9 people up arbitrarily into groups of 3. There are $$\\frac{\\binom{9}{3}\\binom{6}{3}\\binom{3}{3}}{3!} = 280$$ ways of doing this. Without loss of generality, label the people 1 through 9 so that the first round groups are $\\{1,2,3\\}$, $\\{4,5,6\\}$, and $\\{... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 20160 | |
06ey | Let $D$ be a point on the side $BC$ of triangle $ABC$ such that $AB + BD = AC + CD$. The line segment $AD$ cuts the incircle of triangle $ABC$ at $X$ and $Y$ with $X$ closer to $A$. Let $E$ be the point of contact of the incircle of triangle $ABC$ on the side $BC$. Show that
(i) $EY$ is perpendicular to $AD$,
(ii) $XD$... | [
"(i) Note that $AB + BD$ is the semiperimeter of $\\triangle ABC$, and so $D$ is the contact point of the $A$-excircle of $\\triangle ABC$ and $BC$. Since $A$ is a centre of homothety between the incircle and the $A$-excircle, $X$ and $D$ are corresponding points under this homothety. Therefore, the tangent at $X$ ... | Hong Kong | CHKMO | [
"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 only | null | |
0eko | Problem:
Hkrati vržemo 3 poštene igralne kocke različnih barv. V koliko primerih lahko dobimo vsoto pik 10?
(A) 30
(B) 27
(C) 10
(D) 6
(E) 36 | [
"Solution:\n\nVsoto $10$ pik dobimo v šestih možnih izidih, in sicer, v izidih $\\{1,3,6\\}$, $\\{1,4,5\\}$ in $\\{2,3,5\\}$ vsakič $3! = 6$, v izidih $\\{2,2,6\\}$, $\\{3,3,4\\}$ in $\\{4,4,2\\}$ pa vsakič $\\frac{3!}{2!} = 3$. Če seštejemo vse možnosti $6 \\cdot 3 + 3 \\cdot 3$ dobimo $27$ možnosti. Pravilen je o... | Slovenia | 22. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Statistics > Probability > Counting Methods > Permutations",
"Statistics > Probability > Counting Methods > Combinations"
] | null | MCQ | B | |
0ict | Problem:
A horse stands at the corner of a chessboard, a white square. With each jump, the horse can move either two squares horizontally and one vertically or two vertically and one horizontally (like a knight moves). The horse earns two carrots every time it lands on a black square, but it must pay a carrot in rent t... | [
"Solution:\nThe horse must alternate white and black squares, and it ends on the same square where it started. Thus it lands on the same number of black squares ($b$) as white squares ($w$). Thus, its net earnings will be $2b - (b + w) = b - w = 0$ carrots, regardless of its path."
] | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 0 | |
07wo | Let $\mathbb{Z}_+ = \{1, 2, 3, 4, \dots\}$ be the set of all positive integers. Determine all functions $f : \mathbb{Z}_+ \to \mathbb{Z}_+$ that satisfy
$f(mn) + 1 = f(m) + f(n)$ for all positive integers $m$ and $n$;
$f(2024) = 1;$
$f(n) = 1$ for all positive $n \equiv 22 \pmod{23}$. | [
"Our main tool will be the observation that\n$$\nf(mn) = 1 \\quad \\text{implies} \\quad f(m) = f(n) = 1. \\qquad (6)\n$$\nThis is true because the functional equation $f(mn) + 1 = f(m) + f(n)$ then leads to $2 = f(m) + f(n)$ which is only possible when both positive integers, $f(m)$ and $f(n)$, are equal to 1. In ... | Ireland | IRL_ABooklet_2024 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | f(n) = 1 for all positive integers n | |
09hl | Let $x_1 \le x_2 \le \dots \le x_{2n-1}$ be real numbers and $A$ be their arithmetic mean. Show that
$$
2 \sum_{i=1}^{2n-1} (x_i - A)^2 \ge \sum_{i=1}^{2n-1} (x_i - x_n)^2.
$$ | [
"Replacing $x_i$ by $x_i - x_n$ we can assume that $x_n = 0$. Since the left hand side of the inequality is $2(\\sum x_i^2) - 2(2n-1)A^2$, it is enough to show that $\\sum x_i^2 \\ge 2(2n-1)A^2$. By the Cauchy-Schwarz inequality, we have\n$$\n\\frac{2n-1}{n-1} \\left( \\frac{\\sum_{i=1}^{n-1} x_i^2}{n-1} \\right) \... | Mongolia | Mongolian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
01lc | Given a $20 \times 20$ table with one of two signs "+" or "-" in any of its cells. Per move one can replace the signs in all cells of some row (or of some column) by the opposite signs. At the beginning there are $8$ minuses in the table (all other signs are pluses). After some moves the table with exactly $50$ minuses... | [
"than $10$ (since $0 \\le x \\le 10$, $0 \\le y \\le 10$). It follows that exactly one number, namely $72$, may be presented as the product of two positive integers no greater than $10$ ($72 = 8 \\cdot 9$, e.g., $x = 1, y = 2$). It corresponds to the case when $7$ initial minuses are changed. This means that exactl... | Belarus | Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0a5s | Problem:
Find all real numbers $x$ such that $-1 < x \leq 2$ and
$$
\sqrt{2 - x} + \sqrt{2 + 2x} = \sqrt{\frac{x^{4} + 1}{x^{2} + 1}} + \frac{x + 3}{x + 1}.
$$ | [
"Solution:\nNotice that we are solving for $x$ in the domain $-1 < x \\leq 2$. Using Cauchy-Schwarz Inequality on the left hand side, one has\n$$\n\\sqrt{2 - x} + \\sqrt{2 + 2x} = \\sqrt{\\frac{1}{2} \\cdot (4 - 2x) + \\sqrt{2 + 2x}} \\leq \\sqrt{\\left(\\frac{1}{2} + 1\\right)(2 + 2x + 4 - 2x)} = 3.\n$$\nUsing the... | New Zealand | New Zealand Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x = 1 | |
0293 | Problem:
Existem 1999 cidades e 4000 estradas em um certo país (cada estrada conecta 2 cidades). Prove que existe um caminho fechado passando através de não mais que 20 cidades. | [
"Solution:\n\nInicialmente destrua as cidades, juntamente com suas estradas incidentes, se delas partem menos que 3 estradas. Como $2 \\cdot 1999 < 4000$, eventualmente irão sobrar estradas e cidades após esse processo. Assim, podemos assumir que de todas as cidades restantes partem pelo menos 3 estradas. Escolha u... | Brazil | null | [
"Discrete Mathematics > Graph Theory"
] | null | proof only | null | |
0gdr | 設不等邊三角形 $ABC$ 垂心為 $H$, $AH$ 交外接圓 $\Omega$ 於另一點 $P$, $BH, CH$ 分別和 $AC, AB$ 交於 $E, F$。設 $Q, R$ 分別為 $PE, PF$ 與 $\Omega$ 的另一個交點, $Y$ 在 $\Omega$ 上使得 $AY, QR, EF$ 共點, 證明 $PY$ 平分 $EF$。
Let point $H$ be the orthocenter of a scalene triangle $ABC$. Line $AH$ intersects with the circumcircle $\Omega$ of triangle $ABC$ again at ... | [
"考慮 $ACRQPY$, 由帕斯卡定理知 $E = AC \\cap QP$, $CR \\cap PY$, $QR \\cap AY$ 共點, 類似的有 $F = AB \\cap RP$, $BQ \\cap PY$, $QR \\cap AY$ 共點, 因此 $BQ, CR, PY, EF$ 共於一點 $M$。\n\n在 $\\Omega$ 上取一點 $B' \\neq B$ 使得 $AB = AB'$, 則\n$$\n\\angle AFE = \\angle BCA = \\angle BB'A = \\angle ABB' \\Rightarrow BB' \\parallel EF\n$$\n並且\n$$\n... | Taiwan | 2020 Taiwan IMO 1J | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
02z4 | Problem:
Analisando os números naturais de 4 algarismos:
a) Quantos deles têm todos os algarismos diferentes?
b) Quantos têm o algarismo 1 exatamente uma vez e todos os algarismos diferentes?
c) Quantos têm o algarismo 1? | [
"Solution:\n\na) Para escolher a unidade do milhar, temos 9 possibilidades, já que não podemos utilizar o zero, pois não seria um número de quatro algarismos; para escolher o algarismo da centena, temos 9 possibilidades, pois já utilizamos um dos algarismos; para a dezena, são 8 possibilidades; e para a unidade são... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | a) 4536; b) 1848; c) 3168 | |
0dgi | Marwan has chosen 8 cells of the chessboard $8 \times 8$ such that no two lie on the same line or in the same row (we call it general configuration). On each step Hamza chooses 8 cells in general configuration and puts coins on them. Then Marwan shows all coins that are out of cells chosen by Marwan. If Marwan shows ev... | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 3 | |
0f29 | Problem:
You are given a set of scales and a set of $n$ different weights. $R$ represents the state in which the right pan is heavier, $L$ represents the state in which the left pan is heavier and $B$ represents the state in which the pans balance. Show that given any $n$-letter string of $R$s and $L$s you can put the... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0dzs | Problem:
Dana je funkcija $f(x) = a + b c^{x}$. Določi realna števila $a$, $b$, $c$, če je $f(0) = 5$, $f(1) = 14$ in $f(2) = 50$. | [
"Solution:\n\nVstavimo dane podatke $f(0) = 5$, $f(1) = 14$ in $f(2) = 50$ v predpis funkcije $f(x)$. Dobimo sistem enačb:\n$$\n5 = a + b,\n$$\n$$\n14 = a + b c,\n$$\n$$\n50 = a + b c^{2}.\n$$\nUporabimo zamenjalni način reševanja in dobimo rešitvi za $c$: $c_{1} = 4$ in $c_{2} = 1$. Rešitev $c_{2}$ odpade zaradi n... | Slovenia | Državno tekmovanje | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | a = 2, b = 3, c = 4 | |
02c3 | Problem:
Os dois quadrados - As medidas em centímetros dos lados de cada um dos dois quadrados são números inteiros. Se o menor quadrado tivesse $2001~\mathrm{cm}^2$ a mais de área, os dois quadrados seriam iguais. Quanto pode medir o lado do maior quadrado?
 | [
"Solution:\n\nSe $a$ é a medida do lado do quadrado maior e $b$ a medida do lado do quadrado menor, então pelo enunciado temos\n$$\na^2 = b^2 + 2001\n$$\nLogo:\n$$\n2001 = a^2 - b^2 = (a+b)(a-b)\n$$\nComo $a$ e $b$ são números inteiros, temos que $a+b$ e $a-b$ são divisores de $2001$. Mas, $2001 = 3 \\times 23 \\ti... | Brazil | Nível 2 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 1001, 335, 55, 49 | |
0b7l | a) There exists a unique sequence of positive integers $a_1, a_2, a_3, \dots$ such that
$$
n = \sum_{d|n} a_d, \quad \text{for all } n \in \mathbb{N}^*.
$$
b) There exists a unique sequence of positive integers $b_1, b_2, b_3, \dots$ such that
$$
n = \prod_{d|n} b_d, \quad \text{for all } n \in \mathbb{N}^*.
$$ | [
"a) Euler's totient $\\varphi$ provides the desired sequence, since $\\sum_{d|n} \\varphi(d) = n$. Indeed, consider the fractions $\\frac{1}{n}, \\frac{2}{n}, \\dots, \\frac{n}{n}$ expressed in lowest terms. For each divisor $d$ of $n$, the fractions with denominator equal to $d$ are precisely those having the nume... | Romania | NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > Möbius inversion",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | a) a_n equals Euler’s totient function: a_n = φ(n). b) b_1 = 1; if n is a positive power of a single prime p then b_n = p; if n has at least two distinct prime factors then b_n = 1. | |
0awe | Problem:
If $g(x) = \frac{x-2}{x}$ and $f(-g(-x)) = \frac{x-2}{2x-6}$, find $f(3)$. | [
"Solution:\n\nNote that $-\\frac{-x-2}{-x} = 3 \\Longrightarrow x = -\\frac{1}{2}$.\n\nHence $f(3) = \\frac{-\\frac{1}{2} - 2}{2\\left(-\\frac{1}{2}\\right) - 6} = \\frac{5}{14}$."
] | Philippines | Philippine Mathematical Olympiad | [
"Precalculus > Functions"
] | null | final answer only | 5/14 |
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