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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
047n | Prove that the real-coefficient polynomial in $x, y, z$,
$$
x^{4}(x - y)(x - z) + y^{4}(y - z)(y - x) + z^{4}(z - x)(z - y),
$$
cannot be expressed as a finite sum of squares of real-coefficient polynomials in $x, y, z$. | [
"Denote the given polynomial by $F(x, y, z)$. By contradiction, assume there exist real-coefficient polynomials $f_1(x, y, z)$, $f_2(x, y, z)$, ..., $f_m(x, y, z)$ satisfying\n$$\nF(x, y, z) = \\sum_{i=1}^{m} f_i(x, y, z)^2. \\qquad (1)\n$$\nFirst, all $f_i$ must have degree at most 3. If some $f_i$ had degree $d >... | China | China-TST-2025A | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
08yv | Let $n$ be a positive integer. Determine all integers $1 \le k \le 2n^2$ satisfying the following condition.
> There is a $2n \times 2n$ board. When $k$ distinct cells are chosen and colored black, while the other cells are colored white, the minimum number of the $2 \times 2$ squares which have both black and white c... | [
"In this answer, the rows are numbered from the top and, the columns are numbered from the left. We denote the cell in the $i$-th row, the $j$-th column by $(i, j)$, $2 \\times 2$ square consisting of $(i, j)$, $(i+1, j)$, $(i, j+1)$, $(i+1, j+1)$ by $[i, j]$. Also, a set of cells is said to be mixed, if it has bot... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | All integers k with n^2 − n + 1 ≤ k ≤ n^2, and all multiples of 2n with n^2 + 1 ≤ k ≤ 2n^2. | |
06gb | Given that $a$ and $b$ are natural numbers, when $a^2 + b^2$ is divided by $a+b$, the quotient is $Q$ and the remainder is $R$, where $R < a+b$. Find all possible pairs of numbers $a$ and $b$ such that $Q^2 + R = 2010$. | [
"We are given that\n$$\na^2 + b^2 = Q(a + b) + R.\n$$\nSince $R < a + b$, we have\n$$\n(a+b)^2 \\le 2(a^2+b^2) < 2(Q+1)(a+b),\n$$\nwhich implies $a + b < 2(Q + 1)$, and hence $R < 2(Q + 1)$. Thus, we have\n$$\nQ^2 \\le Q^2 + R \\le Q^2 + 2Q + 1 = (Q + 1)^2.\n$$\nSince $Q^2 + R = 2010$, we must have $Q = \\lfloor\\s... | Hong Kong | IMO HK TST | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | [[53, 31], [31, 53]] | |
0fgu | Problem:
Sobre una circunferencia se eligen $n>3$ puntos y se numeran de 1 a $n$ en cualquier orden. Diremos que dos puntos no consecutivos $a$ y $b$ están relacionados si en uno de los dos arcos de extremos $a$ y $b$, todos los puntos están marcados con números menores que las marcas de $a$ y $b$.
Demostrar que el nú... | [
"Solution:\n\n\n\nEjemplo de configuración\n\nVamos a unir con un segmento dos puntos relacionados. Veamos que en una configuración de $n$ puntos hay un segmento más que en una configuración de $n-1$ puntos, lo que proporciona el resultado que se quiere probar junto con el hecho de que en u... | Spain | OME 24 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0lcm | Let $ABC$ be an acute and non-isosceles triangle with $\angle A = 45^\circ$. The altitudes $AD, BE, CF$ of triangle $ABC$ are concurrent at orthocenter $H$. The line $EF$ meets the line $BC$ at $P$. Point $I$ is the midpoint of segment $BC$ and $IF$ cuts $PH$ at $Q$.
1. Prove that $\angle IQH = \angle AIE$.
2. Let $K$ ... | [
"1. Without loss of generality, we can assume that $AB < AC$, then $B$ will lie between points $P, C$. The other case can be proved similarly.\nFirst, we will prove that the line $PH$ is perpendicular to $AI$.\nLet $U, V$ be the midpoints of two segments $AH, IH$, respectively, then $UV \\parallel AI$.\nIt is easy ... | Vietnam | Vietnamese Mathematical Competitions | [
"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 > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Co... | null | proof only | null | |
0j1n | Problem:
Call a positive integer almost-square if it can be written as $a \cdot b$, where $a$ and $b$ are integers and $a \leq b \leq \frac{4}{3} a$. How many almost-square positive integers are less than or equal to $1000000$? Your score will be equal to $25-65 \frac{|A-C|}{\min (A, C)}$. | [
"Solution:\n\nAnswer: $130348$\n\nTo get a good estimate for the number of almost-square integers, note that any number of the form $a \\cdot b$, with $b \\leq \\frac{4}{3} a$, will be by definition almost-square. Let's assume that it's relatively unlikely that a number is almost-square in more than one way. Then t... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 130348 | |
03d5 | For a triangle $ABC$ it is known that $AB = c$, $BC = a$, $AC = b$ and $\angle BAC = 80^\circ$. Point $M$ from the side $AB$ is such that $\angle AMC = 70^\circ$. If $AM + AC = BC$ prove that $a^2 = b(b+c)$. | [
"Choose a point $M_1$ on the ray $CA$ such that $AM_1 = AM$ and $CM_1 = BC$. It follows that triangle $M_1MA$ is isosceles and $\\angle AM_1M = \\angle AMM_1 = \\frac{1}{2} \\angle BAC = 40^\\circ$.\n\nTriangles $CM_1M$ and $CBM$ are congruent ($CM_1 = CB$, $CM = CB$, $\\angle CMM_1 = \\angle CMB = 110^\\circ$). Th... | Bulgaria | Bulgaria 2022 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0fon | Problem:
Sobre la gráfica de una función polinómica con coeficientes enteros, se eligen dos puntos con coordenadas enteras. Probar que si la distancia entre ellos es un número entero, entonces el segmento que los une es paralelo al eje de abscisas. | [
"Solution:\n\nSea el polinomio $f(x)=\\sum_{i=0}^{n} a_{i} x^{i},\\ a_{i} \\in \\mathbb{Z}$ y sean $A(c, f(c))$ y $B(d, f(d))$ dos puntos con coordenadas enteras. Entonces\n$$\nf(c)-f(d)=\\sum_{i=1}^{n} a_{i}\\left(c^{i}-d^{i}\\right)\n$$\nTodos los sumandos de esta suma son divisibles por $c-d$, así que\n$$\nf(c)-... | Spain | LI Olimpiada matemática Española (Concurso Final) | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0fty | Problem:
Sei $ABCD$ ein Parallelogramm. Die Punkte $P$ und $Q$ liegen im Innern von $ABCD$ auf der Diagonalen $AC$, dabei gilt $|AP| = |CQ| < \frac{1}{2}|AC|$. Die Gerade $BP$ schneidet $AD$ im Punkt $E$, die Gerade $BQ$ schneidet $CD$ in $F$. Zeige, dass $EF$ parallel zur Diagonalen $AC$ ist. | [
"Solution:\n\nBei Punkt $P$ können wir den ersten Strahlensatz anwenden:\n$$\n\\frac{|PE|}{|PB|} = \\frac{|PA|}{|PC|}\n$$\nEbenso bei Punkt $Q$:\n$$\n\\frac{|QF|}{|QB|} = \\frac{|QC|}{|QA|}\n$$\nNach Voraussetzung ist $|QC| = |PA|$ und $|QA| = |PC|$. Daher gilt\n$$\n\\frac{|PE|}{|PB|} = \\frac{|QC|}{|QA|} = \\frac{... | Switzerland | Vorrundenprüfung | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0brj | Let $x$ and $y$ be real nonzero numbers, such that $x^3 + y^3 + 3x^2y^2 = x^3y^3$. Determine the set of the possible values of $E = \frac{1}{x} + \frac{1}{y}$. | [
"Rewrite the given condition successively $(x + y)^3 - 3xy(x + y) = x^3y^3 - 3x^2y^2$, i.e. $(x + y)^3 - (xy)^3 = 3xy(x + y) - 3x^2y^2$, or $(x + y - xy)(x^2 + 2xy + y^2 + x^2y + xy^2 + x^2y^2) = 3xy(x + y - xy)$. We either have $x + y = xy$, which leads to $E = 1$ (obtained for $x = y = 2$), or $x^2 + 2xy + y^2 + ... | Romania | 67th NMO Selection Tests for JBMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | {-2, 1} | |
09y8 | Problem:
Vind alle viertallen $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$ van reële getallen zodat de volgende zes gelijkheden gelden:
$$
\begin{aligned}
& x_{1}+x_{2}=x_{3}^{2}+x_{4}^{2}+6 x_{3} x_{4}, \\
& x_{1}+x_{3}=x_{2}^{2}+x_{4}^{2}+6 x_{2} x_{4}, \\
& x_{1}+x_{4}=x_{2}^{2}+x_{3}^{2}+6 x_{2} x_{3}, \\
& x_{2}+x_{... | [
"Solution:\n\nOplossing I. De eerste twee vergelijkingen van elkaar afhalen geeft $x_{2}-x_{3}=x_{3}^{2}-x_{2}^{2}+6 x_{4}\\left(x_{3}-x_{2}\\right)$, wat we kunnen ontbinden als $0=\\left(x_{3}-x_{2}\\right)\\left(x_{3}+x_{2}+1+6 x_{4}\\right)$. We zien dat $x_{2}=x_{3}$ of $x_{2}+x_{3}+1+6 x_{4}=0$. Analoog geldt... | Netherlands | IMO-selectietoets | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | All permutations of the following quadruples: (0, 0, 0, 0); (1/4, 1/4, 1/4, 1/4); (-1/4, -1/4, -1/4, 3/4); (-1/2, -1/2, -1/2, 5/2). | |
09e3 | Let $x$, $y$, $z$ be arbitrary real numbers. Prove that $(x - y)^2 + (y - z)^2 + (z - x)^2$ and $(x - y)(y - z)(z - x)$ have the same sign. | [
"It is possible to prove that for odd natural number $n > 1$ and for arbitrary real numbers $a$, $b$, $c$ which satisfy the condition $a + b + c = 0$, the numbers $abc$ and $a^n + b^n + c^n$ have the same sign. It is obvious that after setting $n = 5$, $a = x - y$, $b = y - z$, $c = z - x$ follows required statemen... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
01f3 | For all non-negative real numbers $x$, $y$, $z$ with $x \ge y$, prove inequality
$$
\frac{x^3 - y^3 + z^3 + 1}{6} \ge (x - y) \sqrt{xyz}.
$$ | [
"From AM-GM inequality we have $\\sqrt{xyz} \\le \\frac{xy+z}{2}$. Hence it suffices to prove that\n$$\n\\frac{x^3 - y^3 + z^3 + 1}{3} \\ge (x - y)(xy + z),\n$$\nwhich is equivalent to\n$$\n\\frac{(x - y)^3 + z^3 + 1}{3} \\ge z(x - y).\n$$\nThis inequality follows directly from AM-GM inequality applied to the numbe... | Baltic Way | Baltic Way 2019 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
09pl | Problem:
Zij $\lfloor x\rfloor$ het grootste gehele getal kleiner dan of gelijk aan $x$. Laat $n \in \mathbb{N}, n \geq 7$ gegeven zijn.
Bewijs dat $\binom{n}{7}-\left\lfloor\frac{n}{7}\right\rfloor$ deelbaar is door 7. | [
"Solution:\n\nSchrijf $n=7k+\\ell$ voor gehele $k$ en $\\ell$ met $0 \\leq \\ell \\leq 6$, dan geldt $\\left\\lfloor\\frac{n}{7}\\right\\rfloor=\\left\\lfloor k+\\frac{\\ell}{7}\\right\\rfloor=k$.\nTe bewijzen: $\\binom{n}{7} \\equiv k \\pmod{7}$.\n\nBewijs: Het uitschrijven van de binomiaalcoëfficiënt geeft\n\n$$\... | Netherlands | UITWERKINGEN TOETS TRAININGSKAMP | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
0840 | Problem:
Da un sacchetto della tombola, contenente i numeri da $1$ a $90$, estraiamo contemporaneamente due numeri. Qual è la probabilità che la somma faccia $56$?
(A) $\frac{3}{445}$
(B) $\frac{14}{2025}$
(C) $\frac{1}{150}$
(D) $\frac{11}{1620}$
(E) $\frac{11}{1602}$. | [] | Italy | UNIONE MATEMATICA ITALIANA Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO TRIENNIO | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | MCQ | A | |
09s1 | Problem:
Zij $\triangle ABC$ een driehoek met $|AC| = 2|AB|$ en zij $O$ het middelpunt van de omgeschreven cirkel. Zij $D$ het snijpunt van de bissectrice van $\angle A$ met $BC$. Zij $E$ de loodrechte projectie van $O$ op $AD$ en zij $F \neq D$ het punt op $AD$ waarvoor $|CD| = |CF|$. Bewijs dat $\angle EBF = \angle ... | [
"Solution:\n\nOplossing I. Als $E = F$ zijn hoeken $EBF$ en $ECF$ beide $0$, dus aan elkaar gelijk. We nemen dus verder aan dat $E \\neq F$. Zij $G$ het tweede snijpunt van $AD$ en de omgeschreven cirkel van $\\triangle ABC$. Er zijn meerdere configuraties mogelijk. We bekijken de configuratie waarbij de punten op ... | Netherlands | IMO-selectietoets | [
"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... | null | proof only | null | |
04z9 | Let $a$ be a fixed real number. Find all real numbers $b$ such that, for every real number $x$, at least one of the numbers $x^2 + ax + b$ and $x^2 - ax + b$ is non-negative. | [
"Note that $x^2 + ax + b$ and $x^2 - ax + b$ sum up to $2x^2 + 2b$. If $b \\ge 0$, then it is non-negative for arbitrary real number $x$, implying that at least one of the numbers added was non-negative. If $b < 0$, then taking $x = 0$ turns both summands negative."
] | Estonia | Estonija 2010 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | b ≥ 0 | |
07vi | Suppose that $a$, $b$, $c$ are positive real numbers and $a + b + c = 3$. Prove that
$$
\frac{a+b}{c+2} + \frac{b+c}{a+2} + \frac{c+a}{b+2} \ge 2
$$
and determine when equality holds. | [
"**Solution 1.** Adding $3$ to each side, we have to show\n$$\n\\begin{aligned}\n5 &\\le \\frac{a+b}{c+2} + 1 + \\frac{b+c}{a+2} + 1 + \\frac{c+a}{b+2} + 1 \\\\\n&= (a+b+c+2) \\left( \\frac{1}{a+2} + \\frac{1}{b+2} + \\frac{1}{c+2} \\right) \\\\\n&= 5 \\left( \\frac{1}{a+2} + \\frac{1}{b+2} + \\frac{1}{c+2} \\right... | Ireland | IRL_ABooklet_2023 | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | Equality holds if and only if a = b = c = 1. | |
070m | Problem:
The incircle of the triangle $ABC$ touches $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. $AD$ meets the circle again at $Q$. Show that the line $EQ$ passes through the midpoint of $AF$ iff $AC = BC$. | [
"Solution:\n\n\n\n$\\angle AQ M = \\angle EQ D$ (opposite angle) $= \\angle EDC$ ($CD$ tangent to circle $EQD$) $= \\left(180^\\circ - \\angle C\\right) / 2 = \\angle A / 2 + \\angle B / 2$ $(*)$.\n\n$MF^2 = MQ \\cdot ME$ ($MF$ tangent to circle $FQE$). So $AM = AF$ is equivalent to $AM^2 =... | 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 > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0brg | Given a positive integer $k$ and an integer $a \equiv 3 \pmod{8}$, show that $a^m + a + 2$ is divisible by $2^k$ for some positive integer $m$. | [
"Proceed by induction on $k$. Since $a \\equiv 3 \\pmod{8}$, $m = 1$ works for $k = 1, 2, 3$, so let $k \\ge 3$ and let $m$ be a positive integer such that $a^m + a + 2$ is divisible by $2^k$.\n\nIf $(a^m + a + 2)/2^k$ is even, then $a^m + a + 2$ is clearly divisible by $2^{k+1}$.\nIf $(a^m + a + 2)/2^k$ is odd, we... | Romania | 67th NMO Selection Tests for BMO and IMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof only | null | |
0am6 | Problem:
Given that $\tan \alpha + \cot \alpha = 4$, find $\sqrt{\sec^{2} \alpha + \csc^{2} \alpha - \frac{1}{2} \sec \alpha \csc \alpha}$. | [] | Philippines | AREA STAGE | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | sqrt(14) | |
00r8 | For an arbitrary positive integer $s$, we denote by $v_2(s)$ the exponent of the biggest power of $2$ which divides $s$, i.e. $2^{v_2(s)}$ is a divisor of $s$, but $2^{v_2(s)+1}$ is not. Show that for every positive integer $m$ the following equality holds:
$$
v_2\left(\prod_{n=1}^{2^m} \binom{2n}{n}\right) = m2^{m-1} ... | [
"First we will prove the following auxiliary lemma:\n\n**Lemma.** For $n$ positive integer, the number of ones in its binary representation is exactly $v_2\\left(\\binom{2n}{n}\\right)$.\n\n**Proof of the Lemma.** Let $n_{(2)} = \\overline{b_k b_{k-1} \\dots b_1 b_0}$ be the binary representation of $n$. Then $n = ... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof only | null | |
00p4 | Let $c(O, R)$ be a circle with diameter $AB$ and $C$ a point on it different than $A$ and $B$ such that $\angle AOC > 90^\circ$. On the radius $OC$ we consider the point $K$ and the circle ($c_1$) with center $K$ and radius $KC = R_1$. We draw the tangents $AD$ and $AE$ from $A$ to the circle ($c_1$). Prove that the st... | [
"Let the lines $DE$ and $CA$ meet at point $L$. We will prove that the line $BK$ passes through $L$ (see figure 1).\nThe circle $c(O, R)$ is homothetic to the circle $c_1(K, R_1)$ with respect to homothety with center $A$ and ratio $m = \\frac{R}{R_1}$, say $H(A, \\frac{R}{R_1})$. The extension of $CD$ meets the ci... | Balkan Mathematical Olympiad | BMO 2010 Shortlist | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0iul | Problem:
Compute
$$
\sum_{k=1}^{2009} k\left(\left\lfloor\frac{2009}{k}\right\rfloor-\left\lfloor\frac{2008}{k}\right\rfloor\right)
$$ | [
"Solution:\n\nThe summand is equal to $k$ if $k$ divides $2009$ and $0$ otherwise. Thus the sum is equal to the sum of the divisors of $2009$, or $2394$."
] | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 2394 | |
0e0m | A positive integer only contains the digits $0$, $3$ and $7$. Prove that this integer is not a perfect square. | [
"The last digit of a perfect square can be $0$, $1$, $4$, $5$, $6$ or $9$. Denote the given number by $n$. If $n$ is to be a perfect square it has to end in $0$.\n\nAssume that the last $k$ digits of $n$ are equal to $0$ and the digit just to the left of them is not $0$. If $k$ is odd, $k = 2m - 1$. Then $\\frac{n}... | Slovenia | National Math Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
0dos | Find all pairs of positive integers $(m, n)$ for which it is possible to paint each unit square of an $m \times n$ chessboard either white or black in such a way that, for any unit square of the board, the number of unit squares which are painted the same color as that square and which have at least one common vertex w... | [
"We shall call a painting satisfying the condition of the problem a *good* painting. We shall also call squares sharing a vertex *neighbors*. Note that every square is a neighbor of itself.\n\nIf $mn$ is even, then there is a good painting: Suppose that the number of rows is even. Then we paint 1st and 2nd rows whi... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | All pairs with mn even (i.e., at least one of m or n is even). | |
0at2 | Problem:
There are 2008 blue, 2009 red, and 2010 yellow chips on a table. At each step, one chooses two chips of different colors, and recolor both of them using the third color. Can all the chips be of the same color after some steps? Prove your answer. | [
"Solution:\n\nAfter some steps, suppose that there are $a$ blue, $b$ red, and $c$ yellow chips on the table. We denote this scenario by the ordered triple $(a, b, c)$. Then the next step produces $(a-1, b-1, c+2)$, $(a+2, b-1, c-1)$, or $(a-1, b+2, c-1)$. One crucial observation on these three possibilities is the ... | Philippines | Philippines Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
0dkg | Let $a, b, c$ be nonnegative numbers such that $a \ge b \ge c$. Prove that
$$
a^3 + b^3 + c^3 - 3abc \ge \frac{9}{2}(a-b)(b^2-c^2).
$$ | [
"Write the inequality as\n$$\n(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \\ge \\frac{9}{2}(a - b)(b - c)(b + c).\n$$\nSince $a + b + c \\ge \\frac{3}{2}(b + c)$, it suffices to show that\n$$\na^2 + b^2 + c^2 - ab - bc - ca \\ge 3(a - b)(b - c).\n$$\nThis is equivalent to the obvious inequality $(a - 2b + c)^2 \\ge... | Saudi Arabia | Saudi Booklet | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0ask | Problem:
Both roots of the quadratic equation $x^{2} - 30x + 13k = 0$ are prime numbers. What is the largest possible value of $k$? | [
"17"
] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | final answer only | 17 | |
0ddo | Let $ABC$ be an acute, non-isosceles triangle with altitude $AD$ ($D \in BC$), $M$ is the midpoint of $AD$ and $O$ is the circumcenter. Line $AO$ meets $BC$ at $K$ and circle of center $K$, radius $KA$ cuts $AB$, $AC$ at $E$, $F$ respectively. Prove that $AO$ bisects $EF$. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry >... | null | proof only | null | |
0k7c | Problem:
Let $S$ be the set of all positive factors of $6000$. What is the probability that a random quadruple $(a, b, c, d) \in S^{4}$ satisfies
$$
\operatorname{lcm}(\operatorname{gcd}(a, b), \operatorname{gcd}(c, d)) = \operatorname{gcd}(\operatorname{lcm}(a, b), \operatorname{lcm}(c, d))?
$$ | [
"Solution:\nFor each prime factor, let the greatest power that divides $a, b, c, d$ be $p, q, r, s$. WLOG assume that $p \\leq q$ and $r \\leq s$, and further WLOG assume that $p \\leq r$. Then we need $r = \\min(q, s)$. If $q = r$ then we have $p \\leq q = r \\leq s$, and if $r = s$ then we have $p \\leq r = s \\l... | United States | HMMT February 2019 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 41/512 | |
0bpq | Problem:
a) Arătaţi că numărul $\log_{2015} 2016$ este număr iraţional;
b) Comparaţi numerele $\log_{5} 6$ şi $\log_{6} 7$;
c) Calculaţi $E=\lg^{3} 5+\lg^{3} 20+\lg 8 \cdot \lg (0,25)$. | [
"Solution:\n\na. Presupunem că numărul $\\log_{2015} 2016$ este raţional. Atunci există numerele $a, b \\in \\mathbb{N}^{*}$ astfel încât $\\log_{2015} 2016=\\frac{a}{b}$. Obținem $2016= 2015^{\\frac{a}{b}} \\Leftrightarrow 2016^{b}=2015^{a}$, relaţie imposibilă deoarece un număr e par şi celălalt impar.\n\nb. Apli... | Romania | Olimpiada de Matematică | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (a) log_{2015} 2016 is irrational; (b) log_{5} 6 > log_{6} 7; (c) E = 2 | |
0f07 | Problem:
Given 50 segments on a line show that you can always find either 8 segments which are disjoint or 8 segments with a common point. | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Combinatorial Geometry > Helly's theorem"
] | null | proof only | null | |
0k75 | Problem:
On floor $0$ of a weird-looking building, you enter an elevator that only has one button. You press the button twice and end up on floor $1$. Thereafter, every time you press the button, you go up by one floor with probability $\frac{X}{Y}$, where $X$ is your current floor, and $Y$ is the total number of times... | [
"Solution:\nBy induction, we can determine that after $n$ total button presses, your current floor is uniformly distributed from $1$ to $n-1$: the base case $n=2$ is trivial to check, and for the $n+1$th press, the probability that you are now on floor $i$ is $\\frac{1}{n-1}\\left(1-\\frac{i}{n}\\right)+\\frac{1}{n... | United States | HMMT February 2019 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 97/3 | |
08fk | Problem:
Sia $ABC$ un triangolo isoscele con $AB = AC$. L'altezza uscente da $A$ misura $15$ mentre l'altezza uscente da $B$ misura $24$. Quanto vale l'area di $ABC$?
(A) $180$
(B) $300$
(C) $240 \sqrt{2}$
(D) $200 \sqrt{3}$
(E) $320$ | [
"Solution:\n\nLa risposta è (B). Sia $M$ il punto medio del lato $BC$, che è anche il piede dell'altezza uscente da $A$ (in quanto $ABC$ è isoscele su base $BC$), e sia $P$ il piede dell'altezza uscente da $B$. Dal momento che l'altezza uscente da $A$ è più corta di quella uscente da $B$, l'angolo in $A$ risulta ot... | Italy | Italian Mathematical Olympiad - February Round | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | B | |
08y1 | Suppose you want to distribute nine distinct integers greater than or equal to $1$ and less than or equal to $9$ into nine square boxes of a $3 \times 3$ grid in such a way that the sums of the numbers on each row and each column are all multiples of $3$. How many possible ways of distributing numbers are there to atta... | [
"Let us replace every number $n$ distributed into a box in the grid by $p = 0, 1, 2$ which satisfies the identity $n \\equiv p \\pmod 3$. The condition that the sum of the numbers on each row and on each column must be a multiple of $3$ does not change after the replacement of the numbers $n$ by $p$. Among the trip... | Japan | Japan 2015 | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 5184 | |
0hrj | Problem:
Find all pairs of integers $x, y$ such that $2 x^{2} - 6 x y + 3 y^{2} = -1$. | [
"Solution:\n\nWe first characterize all solutions to the equation $a^{2} - 3 b^{2} = 1$ in nonnegative integers. This is a Pell equation, and the method used is standard. Suppose we have $(a + b \\sqrt{3})(a - b \\sqrt{3}) = a^{2} - 3 b^{2} = 1$ with $a > 1$; then we can set $a' = 2a - 3b$, $b' = 2b - a$. Then $(a ... | United States | Berkeley Math Circle | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Algebraic Number Theory > Quadratic fields"
] | null | proof and answer | For any nonnegative integer n, let A_n = (2 + √3)^n + (2 − √3)^n and B_n = ((2 + √3)^n − (2 − √3)^n)/√3. Then all integer solutions are given by x = ± A_n/2 and y = x ± B_n/2, with the two signs chosen independently. | |
00zs | Problem:
Five distinct points $A, B, C, D$ and $E$ lie on a line with
$$
|A B|=|B C|=|C D|=|D E|.
$$
The point $F$ lies outside the line. Let $G$ be the circumcentre of triangle $A D F$ and $H$ be the circumcentre of triangle $B E F$. Show that lines $G H$ and $F C$ are perpendicular.
 | [
"Solution:\n\nLet $O, H', G'$ be the circumcentres of the triangles $B D F$, $B C F$ and $C D F$, respectively (see Figure 6). Then $O, G$ and $G'$ lie on the perpendicular bisector of the segment $D F$, while $O, H$ and $H'$ lie on the perpendicular bisector of the segment $B F$. Moreover, $G$ and $H'$ lie on the ... | Baltic Way | Baltic Way 1997 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coord... | null | proof only | null | |
0fv4 | Problem:
Finde alle Tripel $(p, q, r)$ von Primzahlen, sodass auch die drei Differenzen
$$
|p-q|, \quad|q-r|, \quad|r-p|
$$
alle Primzahlen sind. | [
"Solution:\nWegen der Symmetrie des Problems können wir $p<q<r$ annehmen. Ist $p>2$, dann ist $r-p$ grösser als 2 und gerade, also keine Primzahl, Widerspruch. Ist $r>q+2$, dann folgt mit demselben Argument, dass $r-q$ nicht prim sein kann. Es gilt also $p=2$ und $r=q+2$. Nach Voraussetzung ist auch $q-p=q-2$ prim.... | Switzerland | Vorrundenprüfung | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (2, 5, 7) | |
0eb8 | Find all pairs of real numbers $x$ and $y$ which satisfy the equations
$$
x + \frac{1}{y-x} = 1,
$$
$$
y + \frac{1}{x-y} = 2.
$$ | [
"We add up the equations and get $x + y = 3$. From this we express $y = 3 - x$ and insert it in the first equation to get $x + \\frac{1}{3-2x} = 1$. We rearrange to get $2x^2 - 5x + 2 = 0$. We can factor the left side of this equation $2(x-2)(x - \\frac{1}{2}) = 0$. It follows $x = 2$ or $x = \\frac{1}{2}$. In the ... | Slovenia | National Math Olympiad 2015 – Final Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (2, 1) and (1/2, 5/2) | |
01ci | For $x \geq \frac{1}{2}$, what is the largest possible value of the expression
$$
\frac{x^4 - x^2}{x^6 + 16x^3 - 1}?
$$ | [
"Answer: $\\frac{1}{15}$.\n\nNote that, if $\\frac{1}{2} \\le x < 1$, then $x^4 - x^2$ is negative, while $x^6 + 16x^3 - 1$ is positive. Therefore, in this interval, the expression takes on only negative values. We can then consider only $x \\ge 1$.\n\nDenoting $t = x - \\frac{1}{x}$, hence $x^3 - \\frac{1}{x^3} = ... | Baltic Way | Baltic Way 2015 Shortlisted Problems | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | 1/15 | |
0boi | Let $k$ be a positive integer congruent to $1$ modulo $4$ which is not a perfect square, and let $a = (1 + \sqrt{k})/2$. Show that
$$
\{\lfloor a^2 n \rfloor - \lfloor a \lfloor an \rfloor \rfloor : n = 1, 2, 3, \dots\} = \{1, \dots, \lfloor a \rfloor\}.
$$ | [
"Let $a_n = an - \\lfloor an \\rfloor$, $n = 1, 2, 3, \\dots$. Since $a^2 = a + (k-1)/4$, it follows that $\\lfloor a^2 n \\rfloor = \\lfloor an \\rfloor + n(k-1)/4$, and $(a-1)\\lfloor an \\rfloor = (a-1)(an - a_n) = n(k-1)/4 - (a-1)a_n$, so, adding $\\lfloor an \\rfloor$ to each side, $a\\lfloor an \\rfloor = \\l... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
00u5 | Let $ABC$ be a triangle with $AB < AC$ and let $D$ be the other intersection point of the angle bisector of $A$ with the circumcircle of triangle $ABC$. Let $E$ and $F$ be points on the sides $AB$ and $AC$ respectively, such that $AE = AF$ and let $P$ be the point of intersection of $AD$ and $EF$. Let $M$ be the midpoi... | [
"Let $X$ be the other point of intersection of the circumcircles of the triangles $AEF$ and $ABC$. We have\n$$\n\\angle EXF = \\angle EAF = \\angle BAC = \\angle BXC\n$$\nand\n$$\n\\angle XFE = \\angle XAB = \\angle XCB,\n$$\nso the triangles $BXC$ and $EXF$ are similar. Since $P$ is the midpoint of the segment $EF... | Balkan Mathematical Olympiad | BMO 2022 shortlist | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
07ep | $n$ is a positive integer. Consider all $2^n$ sequences of numbers $0$ and $1$ with length $n$. At first, some of these numbers are marked. Two sequences are called *neighbors* if they have the same value in all their $n$ digits except for one. In each step, a non-marked sequence that has at least two marked neighbors ... | [
"First, we make an example. Assume these vertices\n$$\n\\begin{array}{ccccccccccc}\nu & u & m & m & \\dots & m & m & m & m \\\\\nm & m & u & u & \\dots & m & m & m & m \\\\\n\\vdots & \\vdots & \\vdots & \\vdots & \\ddots & \\vdots & \\vdots & \\vdots & \\vdots & m \\\\\nm & m & m & m & \\dots & u & u & w & m \\\\\... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | ⌊n/2⌋ + 1 | |
034a | Problem:
Consider the inequality $|x^{2}-5x+6| \leq x+a$, where $a$ is a real parameter.
a) Solve the inequality for $a=0$.
b) Find the values of $a$ for which the inequality has exactly three integer solutions. | [
"Solution:\n\na) We consider two cases. If $x \\in (-\\infty, 2] \\cup [3, \\infty)$, then the inequality becomes $x^{2}-6x+6 \\leq 0$, whence $x \\in [3-\\sqrt{3}, 3+\\sqrt{3}]$. Therefore the solutions of the inequality are $x \\in [3-\\sqrt{3}, 2] \\cup [3, 3+\\sqrt{3}]$.\n\nIf $x \\in (2,3)$, then the inequalit... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) x in [3 - sqrt(3), 3 + sqrt(3)]. b) a in [-2, 1). | |
05pm | Problem:
Soient $n \geqslant 2$ un entier et $X$ un ensemble à $n$ éléments. Montrer que le nombre de fonctions $f: X \rightarrow X$ telles que $f \circ f$ soit une fonction constante est égale à
$$
n \sum_{i=1}^{n-1}\binom{n-1}{i} i^{n-i-1}
$$ | [
"Solution:\n\nSoit $f: X \\rightarrow X$ telle que $f \\circ f = c$ pour un $c \\in X$. D'abord on a $f(c) = c$ parce que\n$$\nf \\circ f(c) = c \\Rightarrow f \\circ f \\circ f(c) = f(c),\n$$\net selon la définition $f \\circ f \\circ f(c) = f \\circ f(f(c)) = c$, ce qui nous donne $f(f(c)) = c$ donc $f(c) = c$. P... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Functional equations"
] | null | proof and answer | n \sum_{i=1}^{n-1} \binom{n-1}{i} i^{n-i-1} | |
0233 | Problem:
A figura a seguir mostra um segmento $AB$, seu ponto médio $C$ e as semicircunferências de diâmetros $AB$ e $AC$. Uma circunferência de centro $P$ é tangente às duas semicircunferências e também ao segmento $AB$. Sendo $AB=8~\mathrm{cm}$, e $O$, o ponto médio de $AC$, pergunta-se:
a)... | [
"Solution:\na) Seja $x$ o raio da circunferência de centro $P$. Traçamos $OP$, que passa pelo ponto de tangência $D$; $CP$, que passa pelo ponto de tangência $F$; e $PE$, perpendicular a $AB$ (e raio da circunferência destacada na figura). Temos $CP=4-x$ e $OP=2+x$. O perímetro do triângulo $OCP$ é\n$$\n\\begin{ali... | Brazil | null | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | Perimeter = 8 cm; radius = 16/9 cm | |
0i67 | Problem:
Consider a square of side length $1$. Draw four lines that each connect a midpoint of a side with a corner not on that side, such that each midpoint and each corner is touched by only one line. Find the area of the region completely bounded by these lines. | [
"Solution:\n\nIn unit square $ABCD$, denote by $E, F, G, H$ the respective midpoints of sides $AB, BC, CD, DA$. Let $I$ be the intersection of $AF$ and $DE$, let $J$ be the intersection of $BG$ and $AF$, let $K$ be the intersection of $CH$ and $BG$, and let $L$ be the intersection of $DE$ and $CH$. We want to find ... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 1/5 | |
0a18 | For the second round of the Mathematical Olympiad, 999 students are invited. Melanie creates invitation letters in order of participant number: $1, 2, 3, \ldots$. For some values of $n \geq 100$, she notices the following: the number of participant numbers from $1$ up to and including $n$ ending in a $5$ is exactly equ... | [
"9"
] | Netherlands | Dutch Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 9 | |
0dw5 | Problem:
Poišči vse celoštevilske rešitve enačbe $a^{b}=a b+2$. | [
"Solution:\n\nČe je $b<0$, število $a^{b}$ ni celo, če je $|a|>1$. Pri $a=1$ dobimo enačbo $1=b+2$ in od tod $b=-1$. Pri $a=-1$ dobimo enačbo $(-1)^{b}=-b+2$, ki pa nima rešitve: ker je $b<0$, je $-b+2 \\geq 3$, medtem ko je $(-1)^{b}= \\pm 1$.\n\nČe je $b=0$, dana enačba ni rešljiva.\n\nČe pa je $b>0$, število $a$... | Slovenia | 48. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | (-1, 3), (1, -1), (2, 3) | |
0acl | Solve the equation $\sqrt{1+3\sin^3 x} = 3-\sqrt{1-\cos^4 x}$. | [
"The given equation is equivalent to $\\sqrt{1+3\\sin^3 x} + \\sqrt{1-\\cos^4 x} = 3$.\n\nWe may make these assumptions: $\\sin x \\le 1$, i.e. $\\sin^3 x \\le 1$, from where $1+3\\sin^3 x \\le 4$ and $\\cos x \\ge 0$, i.e. $-\\cos^4 x \\le 0$, i.e. $1-\\cos^4 x \\le 1$. Then for the left side of the equation we ha... | North Macedonia | Macedonian Mathematical Competitions | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | x = \frac{\pi}{2} + 2k\pi,\ k \in \mathbb{Z} | |
0dzp | Problem:
Igralca imata kup enakih žetonov, s katerega izmenično jemljeta po enega in ga postavljata na poljubno prazno polje kvadratne tabele velikosti $2008 \times 2008$. Zmaga tisti, ki prvi postavi žeton tako, da skupaj s tremi drugimi tvori oglišča enakokrakega trapeza, ki ni pravokotnik, in katerega osnovnici sta... | [
"Solution:\n\nPoimenujmo enakokrak trapez, ki ni pravokotnik, in ima osnovnici vzporedni enemu izmed robov tabele, pravilen trapez. Zmaga drugi igralec in to ne glede na to kako igra prvi. Drugi igralec po vsaki potezi prvega preveri, ali lahko s postavitvijo svojega žetona tvori pravilen trapez. Če tega ne more na... | Slovenia | 52. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | second player | |
082i | Problem:
Da un punto $S$ esterno ad una circonferenza $\gamma$ di raggio 1 si tracci una retta tangente a $\gamma$ e si indichi con $T$ il punto di tangenza. Al variare di un punto $P$ su $\gamma$, il baricentro del triangolo $P S T$ descrive una curva $\gamma^{\prime}$. Qual è il rapporto tra la lunghezza di $\gamma^... | [
"Solution:\n\nLa risposta è (C). Indichiamo con $M$ il punto medio del segmento $T S$ (esso ovviamente resta fermo al variare di $P$ su $\\gamma$). Il baricentro $G$ del triangolo $P S T$ giace sicuramente sulla mediana $P M$, inoltre $M G=\\frac{1}{3} M P$ perché il baricentro divide la mediana in due parti, e la ... | Italy | Progetto Olimpiadi di Matematica 2003 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Ca... | null | MCQ | C | |
03jk | Problem:
A house is in the shape of a triangle, perimeter $P$ metres and area $A$ square metres. The garden consists of all the land within 5 metres of the house. How much land do the garden and house together occupy? | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | A + 5P + 25π | |
0hwy | Problem:
Let $x$ and $y$ be real numbers, and define a sequence $a_{0}, a_{1}, a_{2}, \ldots$ by
$$
a_{n}=\sum_{k=0}^{n} x^{k} y^{n-k}
$$
Suppose that $a_{m}, a_{m+1}, a_{m+2}, a_{m+3}$ are integers for some $m \geq 0$. Prove that $a_{n}$ is an integer for all $n \geq 0$. | [
"Solution:\nBy cancellation of terms we see that\n$$\na_{n+1}-x a_{n}=y^{n+1} \\quad \\text{ and } \\quad a_{n+1}-y a_{n}=x^{n+1}\n$$\nIn particular, $a_{n+2}-x a_{n+1}=y^{n+2}=y \\cdot y^{n+1}=y\\left(a_{n+1}-x a_{n}\\right)$, which we can write as\n$$\na_{n+2}=(x+y) a_{n+1}-x y a_{n} .\n$$\nWe let $s=x+y$ and $t=... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Linear Algebra > Determinants"
] | null | proof only | null | |
029h | Problem:
Duas tangentes são desenhadas de um ponto $A$ a um círculo de centro $O$, tocando-o em $B$ e $C$. Seja $H$ o ortocentro do triângulo $ABC$, sabendo que $\angle BAC = 40^\circ$, encontre o valor do ângulo $\angle HCO$.
 | [
"Solution:\n\nComo $AC$ é tangente ao círculo, temos $\\angle ACO = 90^\\circ$. Assim\n$$\n\\begin{aligned}\n\\angle HCO & = 90^\\circ - \\angle ACF \\\\\n& = \\angle CAF \\\\\n& = 40^\\circ\n\\end{aligned}\n$$"
] | Brazil | null | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 40° | |
0bpu | Problem:
Fie $a \in \mathbb{R}$. Demonstrați că dacă $a^{18} \in \mathbb{Q}$ și $a^{11} \in \mathbb{Q}$, atunci $a \in \mathbb{Q}$. | [] | Romania | OLIMPIADA DE MATEMATICĂ - ETAPA LOCALĂ | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
06qr | Let $ABCD$ be a circumscribed quadrilateral. Let $g$ be a line through $A$ which meets the segment $BC$ in $M$ and the line $CD$ in $N$. Denote by $I_{1}, I_{2}$, and $I_{3}$ the incenters of $\triangle ABM$, $\triangle MNC$, and $\triangle NDA$, respectively. Show that the orthocenter of $\triangle I_{1} I_{2} I_{3}$ ... | [
"Let $k_{1}, k_{2}$ and $k_{3}$ be the incircles of triangles $ABM$, $MNC$, and $NDA$, respectively (see Figure 1). We shall show that the tangent $h$ from $C$ to $k_{1}$ which is different from $CB$ is also tangent to $k_{3}$.\n\n\nFigure 1\n\nTo this end, let $X$ denote the point of inter... | IMO | IMO Problem Shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscel... | English | proof only | null | |
03xc | Let $a, b, c \in \{0, 1, 2, \dots, 9\}$. The quadratic equation $ax^2 + bx + c = 0$ has a rational root. Prove that the three-digit number $abc$ is not a prime number. | [
"If $\\overline{abc} = p$ is a prime number, and the roots of the equation $f(x) = ax^2 + bx + c = 0$ are rational numbers, then $b^2 - 4ac$ is a square number, $x_1, x_2$ are negative, and $f(x) = a(x - x_1)(x - x_2)$.\n\nSo $p = f(10) = a(10 - x_1)(10 - x_2)$, and we get $4ap = (20a - 2ax_1)(20a - 2ax_2)$.\n\nSin... | China | China Southeastern Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
0is4 | Problem:
Let $a$, $b$ be constants such that
$$
\lim_{x \rightarrow 1} \frac{(\ln (2-x))^{2}}{x^{2}+a x+b}=1.
$$
Determine the pair $(a, b)$. | [
"Solution:\n\nAnswer: $(-2,1)$\n\nWhen $x=1$, the numerator is $0$, so the denominator must be zero as well, so $1+a+b=0$.\n\nUsing l'Hôpital's rule, we must have\n$$\n1=\\lim_{x \\rightarrow 1} \\frac{(\\ln (2-x))^{2}}{x^{2}+a x+b}=\\lim_{x \\rightarrow 1} \\frac{2 \\ln (2-x)}{(x-2)(2 x+a)}\n$$\nand by the same ar... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Calculus > Applications",
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Limits"
] | null | proof and answer | (-2, 1) | |
01t2 | A point $A_1$ is marked inside an acute non-isosceles triangle $ABC$ such that $\angle A_1AB = \angle A_1BC$ and $\angle A_1AC = \angle A_1CB$.
Points $B_1$ and $C_1$ are defined in the same way. Let $G$ be the gravity center of the triangle $ABC$.
Prove that the points $A_1, B_1, C_1, G$ are concyclic. | [
"There is nothing to prove if some two of four points $A_1$, $B_1$, $C_1$, $G$ coincide. So, we can assume that all these points are pairwise distinct.\n\nLet $\\Gamma$ and $\\Gamma(A_1)$ denote the circumcircles of the triangles $ABC$ and $AA_1B$, respectively. Let $M_A$ be the intersection point of the lines $AA_... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0ida | Problem:
There exists a positive real number $x$ such that $\cos \left(\tan^{-1}(x)\right)=x$. Find the value of $x^{2}$. | [
"Solution:\nDraw a right triangle with legs $1$, $x$; then the angle $\\theta$ opposite $x$ is $\\tan^{-1} x$, and we can compute $\\cos (\\theta)=1 / \\sqrt{x^{2}+1}$. Thus, we only need to solve $x=1 / \\sqrt{x^{2}+1}$. This is equivalent to $x \\sqrt{x^{2}+1}=1$. Square both sides to get $x^{4}+x^{2}=1 \\Rightar... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | (-1 + sqrt(5)) / 2 | |
0hwi | Problem:
In triangle $ABC$, $\angle A = 60^\circ$. Let $E$ and $F$ be points on the extensions of $AB$ and $AC$ such that $BE = CF = BC$. The circumcircle of $ACE$ intersects $EF$ in $K$ (different from $E$). Prove that $K$ lies on the bisector of $\angle BAC$. | [
"Solution:\n\nLet the bisector of $\\angle BAC$ intersect the circumcircle of $\\triangle ACE$ at $K'$. The arcs, and hence the chords, $K'C$ and $K'E$ are equal; since $CB = BE$ is given, we have $\\triangle K'BC \\cong \\triangle K'BE$ and so $K'$ is on the bisector of $\\angle CBE$. This shows that $K'$ is the e... | United States | Berkeley Math Circle Monthly Contest 8 | [
"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 | |
06bc | Show that for any triangle $ABC$ with area $S$ and circumradius $R$,
$$
(a) \quad \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \le \frac{3\sqrt{3}}{4S} + \left(\frac{1}{a} - \frac{1}{b}\right)^2 + \left(\frac{1}{b} - \frac{1}{c}\right)^2 + \left(\frac{1}{c} - \frac{1}{a}\right)^2,
$$
$$
(b) \quad \tan \frac{A}{2} + \t... | [
"(a) We have\n$$\n\\begin{aligned}\n& \\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{c^2} - \\left(\\frac{1}{a} - \\frac{1}{b}\\right)^2 - \\left(\\frac{1}{b} - \\frac{1}{c}\\right)^2 - \\left(\\frac{1}{c} - \\frac{1}{a}\\right)^2 \\\\\n&= 2\\left(\\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca}\\right) - \\left(\\frac{1... | Hong Kong | 1997-2023 IMO HK TST | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > ... | null | proof only | null | |
0cx3 | Let $A_{1} A_{2} \ldots A_{2010}$ be a regular $2010$-gon. Find the number of obtuse triangles whose vertices are among $A_{1}, A_{2}, \ldots, A_{2010}$. | [
"We will solve the problem for a regular $n$-gon $A_{1} A_{2} \\ldots A_{n}$, $n \\geq 3$.\n\nSolution 1.\nLet $P(n)$ be the desired number of obtuse triangles and let $P_{1}(n)$ be the number of obtuse angles $A_{1} A_{i} A_{j}$, where $1 < i < j \\leq n$. Clearly $P(n) = n \\cdot P_{1}(n)$. Any of the considered ... | Saudi Arabia | SAMC | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 1012047060 | |
07nq | Triangle $ABC$ is inscribed in a circle $\Gamma$. The points $X, Y$ and $Z$ are the midpoints of those arcs $BC, CA$ and $AB$, respectively, on $\Gamma$ which do not contain the third point of the triangle. The intersection of the triangles $ABC$ and $XYZ$ form a hexagon $DEFGHK$. Prove that the diagonals $DG, EH$ and ... | [
"Join $AX$, $BY$ and $CZ$. These lines are the bisectors of the angles of $\\triangle ABC$ and intersect at its incentre $I$. Label the vertices of the hexagon as shown below and join $KI$ and $ZA$.\n\n\n\nWe have $\\angle KZI = \\angle XZC = \\angle XAC = \\angle KAI$, hence $AZKI$ is a cy... | Ireland | Ireland | [
"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 > Concurrency ... | null | proof only | null | |
0l6j | Problem:
A semicircle is inscribed in another semicircle if the smaller semicircle's diameter is a chord of the larger semicircle, and the smaller semicircle's arc is tangent to the diameter of the larger semicircle.
Semicircle $S_{1}$ is inscribed in a semicircle $S_{2}$, which is inscribed in another semicircle $S_{3... | [
"Solution:\n\nLet $P$, $Q$, and $R$ be the midpoints of the diameters (i.e., the center of the circular arcs) of $S_{3}$, $S_{2}$, and $S_{1}$, respectively. Observe that if one fixes $S_{3}$, the location of $S_{2}$ is uniquely determined by the angle between the diameters of $S_{2}$ and $... | United States | HMMT February 2025 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 20 | |
08rl | There are ten red cards, numbered $1$, $2$, $\ldots$, $10$, and ten blue cards, also numbered $1$, $2$, $\ldots$, $10$. How many ways are there to choose three from these twenty cards so that the sum of the numbers on the cards chosen is $16$ or less? | [
"Write $1$, $2$, $\\ldots$, $9$ or $10$ on the back of the cards so that each card has two numbers which add up to $11$.\n\nCall a set of three cards *good* if the sum of the numbers on their faces is $16$ or less, and call it *bad* if the sum of the numbers on their backs is $16$ or less.\n\nSince the sum of their... | Japan | The 4th Japanese Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 570 | |
0kjm | Problem:
Side $\overline{AB}$ of $\triangle ABC$ is the diameter of a semicircle, as shown below. If $AB=3+\sqrt{3}$, $BC=3\sqrt{2}$, and $AC=2\sqrt{3}$, then the area of the shaded region can be written as $\frac{a+(b+c\sqrt{d})\pi}{e}$, where $a, b, c, d, e$ are integers, $e$ is positive, $d$ is square-free, and $\o... | [
"Solution:\n\nDrop an altitude to point $D$ on $\\overline{AB}$ from $C$ and let $x=AD$. Solving for $x$, we find\n$$\n\\begin{aligned}\n12-x^{2}=18-(3+\\sqrt{3}-x)^{2} &\\Rightarrow 12=18-9-6\\sqrt{3}-3+2(3+\\sqrt{3})x-x^{2} \\\\\n&\\Rightarrow 6+6\\sqrt{3}=(6+2\\sqrt{3})x \\\\\n&\\Rightarrow x=\\sqrt{3}\n\\end{al... | United States | HMMT November 2021 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 147938 | |
04t8 | On the unit square $ABCD$ is given point $E$ on $CD$ in such a way, that $|\angle BAE| = 60^\circ$. Further let $X$ be an arbitrary inner point of the segment $AE$. Finally let $Y$ be the intersection of a line, perpendicular to $BX$ and containing $X$, with the line $BC$. What is the least possible length of $BY$? (Mi... | [
"Let us consider the Thales circle over $BY$, which is circumscribed to $BYX$. This circle contains $X$ and touches $AB$ in $B$. Of all such circles, the one which touches $AE$ (and it has to be in $X$) obviously has the least diameter (let us call the circle $k$). This circle is thus inscribed to the equilateral t... | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 2√3/3 | |
0dxe | Problem:
Dana je kocka $A B C D E F G H$. Na stranici $G H$ leži točka $K$, tako da velja $|G K|:|K H|=1:3$, na stranici $H E$ leži točka $L$, tako da velja $|H L|:|L E|=1:1$, na stranici $C G$ pa leži točka $M$. Kakšno mora biti razmerje $|C M|:|M G|$, da se bosta premici $A K$ in $L M$ sekali? | [
"Solution:\n\n\n\nNaj bo $P$ presečišče premic $A K$ in $L M$. Označimo vektorje $\\vec{a}=\\overrightarrow{A B}$, $\\vec{b}=\\overrightarrow{A D}$ in $\\vec{c}=\\overrightarrow{A E}$. Naj bo $\\overrightarrow{A P}=\\alpha \\overrightarrow{A K}$, $\\overrightarrow{L P}=\\beta \\overrightarr... | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | 2:1 | |
06di | Prove that if $a, b, c$ are positive real numbers, then
$$
\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \ge \frac{3}{2}.
$$ | [
"(Nesbitt's inequality) We have\n$$\n\\begin{align*}\n& \\frac{a}{b+c} + \\frac{b}{c+a} + \\frac{c}{a+b} \\ge \\frac{3}{2} \\\\\n\\Leftrightarrow & \\quad \\frac{a+b+c}{b+c} + \\frac{a+b+c}{c+a} + \\frac{a+b+c}{a+b} \\ge \\frac{9}{2} \\\\\n\\Leftrightarrow & \\quad (a+b+c) \\left( \\frac{1}{b+c} + \\frac{1}{c+a} + ... | Hong Kong | IMO HK TST | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0b4h | Problem:
Suppose that $p$ is a prime number which divides infinitely many numbers of the form $10^{n!} + 2023$ where $n$ is a positive integer. What is the sum of all possible values of $p$? | [] | Philippines | 25th Philippine Mathematical Olympiad Area Stage | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 34 | |
012q | Problem:
Let $P$ be a set of $n \geqslant 3$ points in the plane, no three of which are on a line. How many possibilities are there to choose a set $T$ of $\left(\begin{array}{c}n-1 \\ 2\end{array}\right)$ triangles, whose vertices are all in $P$, such that each triangle in $T$ has a side that is not a side of any oth... | [
"Solution:\n\nFor a fixed point $x \\in P$, let $T_{x}$ be the set of all triangles with vertices in $P$ which have $x$ as a vertex. Clearly, $\\left|T_{x}\\right|=\\left(\\begin{array}{c}n-1 \\\\ 2\\end{array}\\right)$, and each triangle in $T_{x}$ has a side which is not a side of any other triangle in $T_{x}$. F... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 1 if n = 3; n if n ≥ 4 | |
0ao6 | Problem:
Find the least possible value of $|x-1| + |x-3| + |x-5|$.
(a) 2
(b) 4
(c) 1
(d) 3 | [
"Solution:\nLet us consider the function $f(x) = |x-1| + |x-3| + |x-5|$.\n\nThe minimum value of the sum of absolute values $|x-a| + |x-b| + |x-c|$ occurs when $x$ is the median of $a, b, c$.\n\nHere, the numbers are $1, 3, 5$, so the median is $3$.\n\nLet us check $x = 3$:\n\n$$\n|3-1| + |3-3| + |3-5| = 2 + 0 + 2 ... | Philippines | Qualifying Round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | b | |
0eyu | Problem:
Given a pentagon with equal sides.
(a) Prove that there is a point $X$ on the longest diagonal such that every side subtends an angle at most $90$ degrees at $X$.
(b) Prove that the five circles with diameter one of the pentagon's sides do not cover the pentagon. | [] | Soviet Union | 3rd ASU | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09r6 | Problem:
Gegeven is een driehoek $A B C$. De bissectrice van $\angle C A B$ snijdt $B C$ in $L$. Op het inwendige van zijden $A C$ en $A B$ liggen respectievelijk de punten $M$ en $N$, zodat $A L$, $B M$ en $C N$ door één punt gaan en zodat $\angle A M N=\angle A L B$. Bewijs dat $\angle N M L=90^{\circ}$. | [
"Solution:\n\nNoem $T$ het snijpunt van $M N$ en $B C$. Merk op dat omdat $\\angle A C B=\\angle A L B-\\angle L A C=\\angle A M N-\\angle L A C<\\angle A M N$, geldt dat $T$ aan dezelfde kant van $C$ ligt als $B$ (en aan dezelfde kant van $M$ als $N$). Omdat $\\angle A M T=\\angle A M N=\\angle A L B=\\angle A L T... | Netherlands | Dutch TST | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ehi | Problem:
Računsko določi vse možne pare naravnih števil $a$ in $b$, $a > b$, tako da bo razlika kvadratov teh dveh števil enaka $60$. | [
"Solution:\n\nZapis $a^2 - b^2 = 60$\n\nRazstavljanje razlike kvadratov:\n$$a^2 - b^2 = (a - b)(a + b) = 60$$\n\nZapis vseh možnih razcepov števila $60$ na produkt dveh naravnih števil:\n$$1 \\cdot 60,\\ 2 \\cdot 30,\\ 3 \\cdot 20,\\ 4 \\cdot 15,\\ 5 \\cdot 12,\\ 6 \\cdot 10$$\n\nZapis vsaj enega sistema enačb:\n$$... | Slovenia | Državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | [(16, 14), (8, 2)] | |
0j7f | Problem:
In a plane, we are given line $l$, two points $A$ and $B$ neither of which lies on line $l$, and the reflection $A_{1}$ of point $A$ across line $l$. Using only a straightedge, construct the reflection $B_{1}$ of point $B$ across line $l$. Prove that your construction works.
Note: "Using only a straightedge"... | [
"Solution:\n\nWe are given a line, a pair of distinct points $A$ and $A_{1}$ that are reflections of each other across that line and a third point $B$ waiting to be reflected across the line. We can assume that the point labeled $A$ is on the same side of the line as $B$.\n\nGiven line $l$ and two points $A$ and $B... | United States | 13th Bay Area Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0bax | Let $a$ and $b$ be two complex numbers. Prove that the following statements are equivalent:
1) The absolute values of the roots of the equation $x^2 - a x + b = 0$ are respectively equal to the absolute values of the roots of the equation $x^2 - b x + a = 0$.
2) $a^3 = b^3$ or $b = \bar{a}$. | [
"Let $|x_1| = |x_3|$, $|x_2| = |x_4|$ (1) and notice that $|a| = |x_3 x_4| = |x_1 x_2| = |b|$ to derive that $|x_1 + x_2| = |x_3 + x_4|$ (2). The relations (1) and (2) show that there exists a number $k \\in \\mathbb{C}$ such that $x_2 = k x_1$, $x_4 = k x_3$ or $x_2 = k x_1$, $x_4 = \\bar{k} x_3$.\n\nIn the first ... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | null | proof only | null | |
03wr | There are given 12 red points on a circle. Find the minimum of $n$, such that there exist $n$ triangles, whose vertices are red points, satisfying every chord with red endpoints being a side of one triangle. (Posed by Tao Pingsheng) | [
"Let the set of 12 red points be $A = \\{A_1, A_2, \\dots, A_{12}\\}$. From $A_1$, one can get 11 chords with red points, but every triangle with vertex $A_1$ has two such chords, and so the 11 chords should be in at least 6 triangles with $A_1$ being an endpoint. The same applies to $A_i$ ($i = 2, 3, \\dots, 12$);... | China | China Southeastern Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 24 | |
06da | For any positive integer $n$, determine (with proof) if the polynomial
$$
f(x) = (x^2 + x)^{2n} + 1
$$
can be factored into a product of two non-constant polynomials with integer coefficients. | [
"No. Suppose on the contrary that $f(x)$ is reducible. Then so is\n$$\n4^{2n} f(x) = (4x^2 + 4x)^{2n} + 4^{2n} = ((2x + 1)^2 - 1)^{2n} + 4^{2n}.\n$$\nLet $y = 2x + 1$. Then we have $x = \\frac{y-1}{2}$, and hence\n$$\ng(y) = (y^2 - 1)^{2n} + 4^{2n} \\quad (1)\n$$\nis reducible over $\\mathbb{Q}$. By Gauss's lemma, ... | Hong Kong | IMO HK TST | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | It cannot be factored; for every positive integer n the polynomial is irreducible over the integers. | |
0g0q | Problem:
Bei 22 Mathematikwettbewerben werden jeweils 5 Preise verteilt. Nachdem alle Wettbewerbe durchgeführt sind, bemerken die Organisatoren, dass es für jede Kombination von zwei Wettbewerben genau einen gemeinsamen Preisträger gibt. Zeige, dass ein Teilnehmer bei allen Wettbewerben einen Preis gewonnen hat. | [
"Solution:\n\nWir betrachten einen Wettbewerb. In jedem anderen Wettbewerb muss einer der fünf Preisträger auch auf der Preisträgerliste stehen. Da es noch 21 andere Wettbewerbe gibt, muss mindestens einer der fünf Gewinner noch in mindestens fünf anderen Wettbewerben auch unter den besten fünf Schülern sein. Das h... | Switzerland | SMO - Vorrunde | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
04ft | Let $x_1, x_2, \dots, x_{100}$ be real numbers such that
$$
\begin{align*}
|2x_k - x_{k+1}| &= x_{k+2} \quad \text{for all } k \in \{1, 2, \dots, 98\}, \\
|2x_{99} - x_{100}| &= x_1, \\
|2x_{100} - x_1| &= x_2.
\end{align*}
$$
Prove that $x_1 = x_2 = \dots = x_{100}$.
(Tonći Kokan) | [
"Squaring the given equations we get\n$$\n\\begin{gathered}\n4x_k^2 - 4x_kx_{k+1} + x_{k+1}^2 = x_{k+2}^2 \\quad \\text{for } k \\in \\{1, 2, \\dots, 98\\}, \\\\\n4x_{99}^2 - 4x_{99}x_{100} + x_{100}^2 = x_1^2, \\quad 4x_{100}^2 - 4x_{100}x_1 + x_1^2 = x_2^2.\n\\end{gathered}\n$$\nAdding these equations we get\n$$\... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
024x | Problem:
Produtos consecutivos - Divida os números $2$, $3$, $5$, $7$, $11$, $13$ e $17$ em dois grupos de tal forma que, multiplicando todos os números de um grupo e todos do outro, encontremos números consecutivos. | [
"Solution:\n\nSolução 1: Como os produtos são números consecutivos, podemos denotá-los por $p$ e $p+1$. Temos, então,\n$$\np^2 + p = p(p+1) = 2 \\times 3 \\times 5 \\times 7 \\times 11 \\times 13 \\times 17 = 510510\n$$\nResolvendo a equação $p^2 + p - 510510 = 0$, encontramos uma única raiz positiva, $p = 714$. As... | Brazil | Nível 2 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | Group 1: {2, 3, 7, 17}; Group 2: {5, 11, 13}, yielding products 714 and 715 | |
0aro | Problem:
Find the domain of the function $f(x)=\frac{1}{\left[\left|x^{2}-x-2\right|\right]}$. | [
"Solution:\n$x$ must not satisfy $0 \\leq x^{2}-x-2<1 \\Rightarrow x$ must not satisfy $x^{2}-x-2=(x-2)(x+1) \\geq 0 \\Leftrightarrow x \\in(-\\infty,-1] \\cup[2,+\\infty)$ AND must not satisfy $x^{2}-x-2 \\leq 1 \\Leftrightarrow x^{2}-x-3<0 \\Leftrightarrow x \\in\\left(\\frac{1-\\sqrt{13}}{2}, \\frac{1+\\sqrt{13}... | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | (-∞, (1 - √13)/2] ∪ [ (1 - √5)/2, (1 + √5)/2 ] ∪ [ (1 + √13)/2, ∞ ) | |
082q | Problem:
Un dodecaedro è un solido regolare con 12 facce pentagonali. Una diagonale di un solido è un segmento che ha per estremi due vertici del solido che non appartengono ad una stessa faccia. Quante sono le diagonali del dodecaedro? | [
"Solution:\n\nLa risposta è $100$. È necessario contare le coppie (non ordinate) di vertici non appartenenti ad una stessa faccia. In ogni vertice si incontrano $3$ facce e in ciascuna di esse ci sono $2$ vertici che non sono su una faccia che contiene anche il vertice iniziale (totale $6$) più $2$ in comune con un... | Italy | Progetto Olimpiadi di Matematica 2003 | [
"Geometry > Solid Geometry > 3D Shapes",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 100 | |
0fbd | Problem:
En el plano se consideran los dos puntos $P(8,2)$ y $Q(5,11)$. Un móvil se desplaza de $P$ a $Q$ según un camino que ha de cumplir las condiciones siguientes: El móvil parte de $P$ y llega a un punto del eje $x$, a lo largo del cual recorre un segmento de longitud $1$; después se separa de este eje y se dirig... | [
"Solution:\n\nComo los trayectos a lo largo de cada eje son de longitud constante, sólo debemos preocuparnos de minimizar el resto del trayecto. Para ello se halla $P'$ simétrico de $P$ respecto del eje $OX$ y se transforma en $P''$ por el vector $(-1,0)$.\n\nAnálogamente, $Q'$ es el simétrico de $Q$ respecto del e... | Spain | OME 10 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 3 + sqrt(265) | |
0bzk | Let $\mathcal{F}$ be the set of all continuous functions $f: [0, 1] \to \mathbb{R}$, satisfying
$$ \max_{0 \le x \le 1} |f(x)| = 1, \text{ and let } I: \mathcal{F} \to \mathbb{R}, $$
$$
I(f) = \int_{0}^{1} f(x) \, dx - f(0) + f(1).
$$ | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Calculus > Integral Calculus > Applications",
"Precalculus > Functions"
] | null | proof and answer | The supremum of I(f) is 3 and the infimum is −3; neither is attained. The set of attainable values is the open interval (−3, 3). | |
027f | Problem:
No seguinte papel, foi desenhado um quadriculado de $4 \times 6$ e depois traçada a diagonal de $A$ a $B$.
Observe que a diagonal $A B$ intersecta o quadriculado em 9 pontos:
Se o quadriculado fosse de tamanho $12 \times 17$, em quantos pontos a diagonal $A B... | [
"Solution:\nVamos enumerar as linhas verticais e as linhas horizontais do quadriculado do seguinte modo:\n\nA diagonal $A B$ intersecta cada linha vertical em exatamente um ponto. Com isso, contaríamos 18 pontos de interseção. Também $A B$ intersecta cada linha horizontal em exatamente um p... | Brazil | null | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 29 | |
05th | Problem:
Les cases d'un échiquier $8 \times 8$ sont blanches. Un coup consiste à échanger les couleurs des cases d'un rectangle $3 \times 1$ ou $1 \times 3$ (les cases blanches deviennent noires et les cases noires deviennent blanches). Est-il possible d'aboutir en un nombre fini de coups à la configuration où toutes ... | [
"Solution:\n\nL'énoncé présente une suite finie d'opérations et le problème demande s'il est possible de partir de la situation initiale pour arriver à une certaine situation finale. Une première idée dans ce cas est de chercher un invariant.\n\nOn peut éventuellement essayer à la main de voir s'il est possible d'a... | France | Envoi 5: Pot Pourri | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0a9d | Problem:
There are 32 competitors in a tournament. No two of them are equal in playing strength, and in a one against one match the better one always wins. Show that the gold, silver, and bronze medal winners can be found in 39 matches. | [
"Solution:\n\nTo determine the gold medalist, we organize 16 pairs and matches, then 8 matches of the winners, 4 matches of the winners, 2 and finally one match, 31 matches altogether.\n\nNow the silver medal winner has at some point lost to number 1; as there were 5 rounds, there are 5 candidates. Let $C_{i}$ be t... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 23 | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
0a7q | Problem:
Determine all positive integers $n < 200$, such that $n^{2} + (n+1)^{2}$ is the square of an integer. | [
"Solution:\n\nWe determine the integral solutions of\n$$\nn^{2} + (n+1)^{2} = (n+p)^{2}, \\quad p \\geq 2\n$$\nThe root formula for quadratic equations yields\n$$\nn = p - 1 + \\sqrt{2p(p-1)} \\geq 2(p-1)\n$$\nBecause $n < 200$, we have $p \\leq 100$. Moreover, the number $2p(p-1)$ has to be the square of an intege... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 8 | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 3, 20, 119 | |
0475 | Let $I$ be the incenter of a triangle $ABC$. Write $L$, $M$, and $N$ for the midpoints of $AI$, $AC$, and $CI$, respectively. Assume that there is a point $D$ in the interior of the segment $AM$ such that $BD = BC$. The incircle of $\triangle ABD$ touches $AD$ and $BD$ at $E$ and $F$, respectively. Let $J$ be the circu... | [
"\n\n**Proof.** Let $K$ be the midpoint of $ID$, then $LN$ passes through $K$. Next, we prove that both $EF$ and $PQ$ pass through $K$.\nLet $X$ and $Y$ be the reflections of $D$ across $E$ and $F$, respectively. Let $Z$ be the projection of $I$ onto $AC$. Then, we have\n$$\nCX = 2DE + CD =... | China | 2024 CMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions an... | English | proof only | null | |
00uy | Let $n \ge 2$ and $S = \{1, 2, \dots, n^2\}$. For any function $f : S \to S$ let $\text{Fix}(f) = \{x \in S \mid f(x) = x\}$. Find the possible values of the expression
$$
|\text{Fix}(f)| + |\text{Im}(f)| + \max_{k \in S} |f^{-1}(k)|
$$
as $f$ ranges over all functions $f : S \to S$. | [
"We show that the answer is all values from $2n$ to $2n^2 + 1$. Assume $f$ has $k \\in \\{0, 1, \\dots, n^2\\}$ fixed points. Then say $|\\text{Im}(f)| = p$. Also let $s = \\max_{k \\in S} |f^{-1}(k)|$.\n\nUpper Bound: From the definitions of $s$ and $p$, we get $sp \\ge n^2$. We also have the bound $s \\le n^2 - p... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | All integers m with 2n <= m <= 2n^2 + 1 | |
062v | Problem:
Zwei Kreise $\Gamma$ und $\Gamma'$ mögen sich in den beiden voneinander verschiedenen Punkten $A$ und $B$ schneiden. Eine Gerade durch $B$ schneide $\Gamma$ und $\Gamma'$ so in $C$ bzw. $D$, dass $B$ zwischen $C$ und $D$ liege. Eine weitere Gerade durch $B$ schneide $\Gamma$ und $\Gamma'$ derart in $E$ bzw. $... | [
"Solution:\n\n(1) Die Dreiecke $A C D$ und $A E F$ sind gleichsinnig kongruent: Wir arbeiten mit orientierten Winkeln modulo $\\pi$. Es ist $\\varangle A D C = \\varangle A D B = \\varangle A F B = \\varangle A F E$ und $\\varangle D C A = \\varangle B C A = \\varangle B E A = \\varangle F E A$ (für das mittlere Gl... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers... | null | proof only | null | |
07vb | Let $ABCD$ be a square and let $\Gamma$ denote the circle with diameter $CD$. A tangent line is drawn to the circle $\Gamma$ from $B$, meeting the circle $\Gamma$ at $E$ and intersecting the segment $AD$ at $K$.
Prove that $|AD| = 4|KD|$. | [
"Let $O$ be the midpoint of $CD$, the centre of circle $\\Gamma$. Let $F$ be the point where line $CE$ meets $AD$, and let $CE$ meet $OB$ at $M$.\n\nAngle $\\angle DEC$ is standing on the diameter $CD$ of $\\Gamma$, hence is a right angle. Therefore, triangle $DEF$ has a right angle at $E$.... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08ue | Suppose there are $2010$ islands, and $2009$ bridges connecting some pair of these islands. Any pair of islands are either connected by one of the bridges or they are not. Also, any bridge connects two different islands. Furthermore, one can reach any island from any other island by crossing certain number of bridges. ... | [
"Throughout the discussion of this problem, we exclude the consideration of bridges both ends of which belong to the same island.\nFor a positive integer $n$ greater than or equal to $2$, we say that a collection of islands and bridges has the property $T(n)$ if the following three conditions (i), (ii), (iii) are s... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0daj | Let $ABC$ be an acute, non-isosceles triangle with $I$ as its incenter. Denote $D, E$ as the points of tangency of $(I)$ on $AB, AC$, respectively. The median segments with respect to vertex $A$ of triangles $ABE$ and $ACD$ meet $(I)$ at $P, Q$, respectively. Take points $M, N$ on the line $DE$ such that $AM \perp BE$ ... | [
"1) Since $AM$ passes through the midpoint of the segment $BE$ and $AP \\parallel BE$, we can see that\n$$\n(AP, AM, AB, AC) = -1 \\text{ or } (MP, MA, MD, ME) = -1.\n$$\nSuppose that $AM$ cuts $(I)$ at the second point $T$ then $MDTE$ is a harmonic quadrilateral, then $(Mx, MT, MD, ME) = -1$ with $Mx$ as the tange... | Saudi Arabia | Team selection tests for BMO 2018 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
... | English | proof only | null | |
0jp3 | Problem:
Neo has an infinite supply of red pills and blue pills. When he takes a red pill, his weight will double, and when he takes a blue pill, he will lose one pound. If Neo originally weighs one pound, what is the minimum number of pills he must take to make his weight 2015 pounds? | [
"Solution:\n\nAnswer: 13\nSuppose instead Neo started at a weight of 2015 pounds, instead had green pills, which halve his weight, and purple pills, which increase his weight by a pound, and he wished to reduce his weight to one pound. It is clear that, if Neo were able to find such a sequence of pills in the case ... | United States | HMMT November 2015 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 13 |
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