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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0k2n | Problem:
What is the 3-digit number formed by the $9998^{\text{th}}$ through $10000^{\text{th}}$ digits after the decimal point in the decimal expansion of $\frac{1}{998}$?
Note: Make sure your answer has exactly three digits, so please include any leading zeroes if necessary. | [
"Solution:\nAnswer: 042\n\nNote that $\\frac{1}{998} + \\frac{1}{2} = \\frac{250}{499}$ repeats every 498 digits because 499 is prime, so $\\frac{1}{998}$ does as well (after the first 498 block). Now we need to find $38^{\\text{th}}$ to $40^{\\text{th}}$ digits. We expand this as a geometric series\n$$\n\\frac{1}{... | United States | HMMT November 2018 | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | final answer only | 042 | |
08vf | Let $a, b, c, d, e, f, g$ be 7 distinct positive integers less than or equal to 7. Determine all the prime numbers which can be represented in the form
$$
a \times b \times c \times d + e \times f \times g.
$$ | [
"Let $A = \\{a, b, c, d\\}$, $B = \\{e, f, g\\}$. Suppose the number $X = abcd + efg$ is a prime. Then, we see that the numbers $2, 4, 6$ must belong to the same set $A$ or $B$, because, otherwise, both $abcd$ and $efg$ become even, and therefore $X$ must be even, and since $X \\ne 2$, $X$ cannot be a prime. Also, ... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 179 | |
02d7 | (i) $ABCD$ is a square with side $1$. $M$ is the midpoint of $AB$, and $N$ is the midpoint of $BC$. The lines $CM$ and $DN$ meet at $I$. Find the area of the triangle $CIN$.
(ii) The midpoints of the sides $AB$, $BC$, $CD$, $DA$ of the parallelogram $ABCD$ are $M$, $N$, $P$, $Q$ respectively. Each midpoint is joined t... | [
"(i) $CIN$ is similar to $DCN$, so area $CIN = \\left(\\frac{CN}{DN}\\right)^2$ area $DCN = \\left(\\frac{\\frac{1}{2}}{\\sqrt{5/2}}\\right)^2 \\cdot \\frac{1}{4} =$\n\n$\\frac{1}{20}$\n\n\n\n(ii) If we stretch the plane parallel to one of sides of the square then all areas are increased by... | Brazil | I OBM | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry... | English | proof and answer | (i) 1/20; (ii) 2/5 of the area of the parallelogram; (iii) 1/3 of the area of ABC; (iv) No, it is not true for all convex quadrilaterals. | |
0hig | $n \ge 4$ real numbers are placed on the circle. It is known that for any four consecutive numbers $a, b, c, d$ that go around the circle in this specified order, the condition $a + d = b + c$ holds. For which $n$ can we conclude that all the numbers are equal? | [
"We denote the numbers in the circle by $a_1, a_2, \\ldots, a_n$. Choose the largest among them (or any of them, if there are several such numbers). Without loss of generality, let this number be $a_1$. From the condition of the problem we have that\n$$\na_1 + a_2 = a_n + a_3, \\quad a_1 + a_n = a_2 + a_{n-1} \\Rig... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | all odd n | |
0b98 | Given a positive integer number $n$, determine the minimum of
$$
\max \left\{ \frac{x_1}{1+x_1}, \frac{x_2}{1+x_1+x_2}, \dots, \frac{x_n}{1+x_1+x_2+\dots+x_n} \right\},
$$
as $x_1, x_2, \dots, x_n$ run through all non-negative real numbers which add up to 1. | [
"Let\n$$\nf(x_1, x_2, \\dots, x_n) = \\max \\left\\{ \\frac{x_k}{1 + x_1 + x_2 + \\dots + x_k} : k = 1, 2, \\dots, n \\right\\},\n$$\nwhere $x_1 \\ge 0, x_2 \\ge 0, \\dots, x_n \\ge 0$ and $x_1 + x_2 + \\dots + x_n = 1$, and notice that\n$$\n\\frac{a_1}{1+a_1} = \\frac{a_2}{1+a_1+a_2} = \\dots = \\frac{a_n}{1+a_1+a... | Romania | NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | 1 - 2^{-1/n} | |
0422 | As shown in Fig. 7.1, $\odot O_1$ and $\odot O_2$ are tangent externally at point $T$. The quadrilateral $ABCD$ is inscribed in $\odot O_1$. The lines $DA$ and $CB$ are tangent to $\odot O_2$ at points $E$ and $F$, respectively. $BN$, the bisector of $\angle ABF$, intersects the segment $EF$ at point $N$. Line $FT$ int... | [
"Let $P$ be the intersection of line $AM$ with $EF$. Join $AT$, $BM$, $BP$, $BT$, $CM$, $CT$, $ET$, $TP$. As shown in Fig. 7.2.\n\nAs $BF$ is tangent to $\\odot O_2$ at $F$, we have $\\angle BFT = \\angle FET$.\n\nAs $\\odot O_1$ is tangent to $\\odot O_2$ at $T$, we have $\\angle MBT = \\angle FET$.\n\nHence, $\\a... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: c... | English | proof only | null | |
0dqg | Suppose that $a_1, \dots, a_{15}$ are prime numbers forming an arithmetic progression with common difference $d > 0$. If $a_1 > 15$, prove that $d > 30,000$. | [
"Lemma: Suppose $p$ is prime and $a_1, \\dots, a_p$ are primes forming an A.P. with common difference $d$. If $a_1 > p$, we claim that $p \\mid d$.\n\n*Proof:* Since $p$ is prime and every $a_i$ is a prime $> p$, $p$ does not divide $a_i$ for any $i$. By the pigeonhole principle, there exist $1 \\le i < j \\le p$ s... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0bv2 | Problem:
Fie $ABCD$ un patrulater convex. Punctele $M$, $N$, $P$ şi $Q$ împart segmentele $[AB]$, $[BC]$, $[CD]$ şi respectiv $[DA]$ în acelaşi raport. Demonstraţi că dreptele $AC$, $BD$, $MP$ şi $NQ$ sunt concurente dacă şi numai dacă $ABCD$ este paralelogram. | [] | Romania | OLIMPIADA DE MATEMATICĂ ETAPA LOCALĂ | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
075z | Problem:
Find all positive integers $m$, $n$, and primes $p \geq 5$ such that
$$
m\left(4 m^{2}+m+12\right)=3\left(p^{n}-1\right)
$$ | [
"Solution:\nRewriting the given equation we have\n$$\n4 m^{3}+m^{2}+12 m+3=3 p^{n}\n$$\nThe left hand side equals $(4 m+1)\\left(m^{2}+3\\right)$.\nSuppose that $\\left(4 m+1, m^{2}+3\\right)=1$. Then $\\left(4 m+1, m^{2}+3\\right)=\\left(3 p^{n}, 1\\right),\\left(3, p^{n}\\right),\\left(p^{n}, 3\\right)$ or $\\lef... | India | INMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization >... | null | proof and answer | (m, n, p) = (12, 4, 7) | |
02i0 | Problem:
Uma fábrica embala 8 latas de palmito em caixas de papelão cúbicas de $20~\mathrm{cm}$ de lado. Estas caixas são colocadas, sem deixar espaços vazios, em caixotes de madeira de $80~\mathrm{cm}$ de largura por $120~\mathrm{cm}$ de comprimento por $60~\mathrm{cm}$ de altura. Qual o número máximo de latas de pal... | [
"Solution:\n\nEm cada caixote de madeira de dimensões $a \\times b \\times c$ cabem, empilhados regularmente, $\\frac{a}{l} \\times \\frac{b}{l} \\times \\frac{c}{l}$ cubos de lado $l$. No nosso caso, $a=60$, $b=80$, $c=120$ e $l=20$. Como $60$, $80$ e $120$ são múltiplos de $20$, podemos encher o caixote sem deixa... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems"
] | null | MCQ | A | |
087p | Problem:
Nel paese di Cuccagna si gioca al seguente solitario. Si parte da una stringa finita di zeri e uni, e sono concesse le mosse seguenti:
(i) cancellare due uni consecutivi;
(ii) cancellare tre zeri consecutivi;
(iii) se all'interno della stringa c'è la sottostringa 01, si può sostituire questa sottostringa con ... | [
"Solution:\n\nDimostro che l'unica stringa di dieci cifre partendo dalla quale non è possibile vincere il solitario è 1111100000 (cinque uni seguiti da cinque zeri).\nData una stringa considero queste due quantità:\n- il numero degli uni,\n- il peso totale degli zeri, dove uno zero alla cui destra compaiano $i$ uni... | Italy | Cesenatico | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 1 | |
07sf | Alice has written down an integer $x$ in decimal notation. Bob takes the first (left-most) digit of $x$ (which is not zero) and moves it to the far right, shifting all the other digits one space to the left. Alice remarks that the result of this operation is $3x$. Find all possible values of $x$. | [
"Let the number of digits of $x$ be $d$, the leading digit of $x$ be $1 \\le a \\le 9$ and the remaining $d-1$ digits be $1 \\le b \\le 10^d - 1$. There is no solution with $d=1$, so we can assume $d > 1$. Then the equations to solve are:\n$$\n\\begin{align*}\nx &= 10^{d-1}a + b \\\\\n3x &= 10b + a\n\\end{align*}\n... | Ireland | IRL_ABooklet_2020 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n"
] | null | proof and answer | All integers consisting of c repetitions of 142857 or c repetitions of 285714 for any positive integer c; equivalently x = ((10^{6c} − 1)/(10^6 − 1)) · 142857 · a with c ≥ 1 and a ∈ {1, 2}. | |
005q | Sean $ABC$ un triángulo escaleno y $r$ la bisectriz externa del ángulo $ABC$. Se consideran $P$ y $Q$ los pies de las perpendiculares a la recta $r$ que pasan por $A$ y $C$, respectivamente. Las rectas $CP$ y $AB$ se intersectan en $M$ y las rectas $AQ$ y $BC$ se intersectan en $N$. Demuestre que las rectas $AC$, $MN$ ... | [] | Argentina | XXIII Olimpíada Iberoamericana de Matemática | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > ... | Spanish | proof only | null | |
07sc | In triangle *ABC*, the length of the altitude through *A* is equal to $\frac{15\sqrt{3}}{14}$, $\angle BAC = 120^\circ$ and $|BC| = 7$. Find the lengths of the other two sides of triangle *ABC*. | [
"If we let $a = |BC|$, $b = |CA|$ and $c = |AB|$ and use that $\\cos(120^\\circ) = -1/2$, the Cosine Rule gives $49 = a^2 = b^2 + c^2 + bc$. Moreover, the area of triangle $ABC$ is equal to $\\frac{7}{2} \\cdot \\frac{15\\sqrt{3}}{14} = \\frac{15\\sqrt{3}}{4}$ and this is equal to $\\frac{bc\\sin(120^\\circ)}{2} = ... | Ireland | IRL_ABooklet_2020 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 3 and 5 | |
054m | Prove that in every triangle there is a median whose length squared is at least $\sqrt{3}$ times the area of the triangle. | [
"Assume w.l.o.g. that $BC$ is the shortest side of the triangle. Then the least angle of the triangle is by vertex $A$. Denote $a = BC$, $b = CA$, $c = AB$, $\\alpha = \\angle BAC$ and let $m$ be the length of the median drawn from vertex $A$. By assumptions made at the beginning of the solution, $\\alpha \\le 60^\... | Estonia | National Olympiad Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > ... | English | proof only | null | |
0gva | Find all pairs of positive integers $m$ and $n$ such that
$$
\sqrt{m} + \frac{2005}{\sqrt{n}} = 2006.
$$ | [
"$$\n\\sqrt{n} = \\frac{2006^2 n + 2005^2 - m n}{2 \\cdot 2005 \\cdot 2006},\n$$\nso $\\sqrt{n} \\in \\mathbb{Q}$, and therefore, as is known, the number $n$ is the square of a natural number. It also follows that $m = k^2$, $k \\in \\mathbb{N}$, and moreover $k \\mid 2005$. By checking all natural divisors of $200... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | (m, n) ∈ {(1, 1), (2576025, 25), (4004001, 160801), (4020025, 4020025)} | |
00io | Two dissections of a square into three rectangles are considered essentially different if one cannot be switched to the other by simple rearrangement of the pieces.
How many essentially different dissections of the $2010 \times 2010$ square into three rectangles with integer side lengths exist such that the area of one... | [
"There are two distinct possibilities for dissections that we must consider. The rectangles can either be in the form of three \"strips\" (i.e. all with one side of length $2010$) or there can be one such strip, with the other two rectangles resulting from a cut at right angles to the first cut.\n\nWe first conside... | Austria | Austria 2010 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 1678 | |
0k1w | Problem:
A square and an equilateral triangle have the property that the area of each is the perimeter of the other. What is the area of the square? | [
"Solution:\nAssuming the square has side length $x$, it has area $x^{2}$, so the equilateral triangle has side length $x^{2} / 3$. The area of the equilateral triangle is then given in two ways by\n$$\n4x = \\frac{\\sqrt{3}}{4} \\left(\\frac{x^{2}}{3}\\right)^{2}.\n$$\nSolving gives $x^{3} = \\frac{144}{\\sqrt{3}} ... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 12 * 4^(1/3) | |
03v6 | Suppose that $a_k > 0$, $k = 1, 2, \dots, 2008$.
Prove that if and only if $\sum_{k=1}^{2008} a_k > 1$, there is a sequence $\{x_n\}$ satisfying
(1) $0 = x_0 < x_n < x_{n+1}$, $n = 1, 2, 3, \dots$;
(2) $\lim_{n \to \infty} x_n$ exists;
(3) $x_n - x_{n-1} = \sum_{k=1}^{2008} a_k x_{n+k} - \sum_{k=0}^{2007} a_{k+1} x_{n+... | [
"Proof of necessity: Assume that there exists $\\{x_n\\}$ satisfying (1)–(3). Notice that the expression in (3) can be written as\n$$\nx_n - x_{n-1} = \\sum_{k=1}^{2008} a_k (x_{n+k} - x_{n+k-1}), \\quad n \\in \\mathbb{N}.\n$$\nAs $x_0 = 0$, we then have\n$$\n\\begin{align*}\nx_n &= \\sum_{l=1}^{n} (x_l - x_{l-1})... | China | China Mathematical Competition (Complementary Test) | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | English | proof only | null | |
09gk | Given a natural number $k$ and a sequence $\{a_n\}_{n \ge 1}$ defined as $a_1 = 2 + 2016$, $a_2 = 2^2 + 2016$, $a_3 = 2^{2^2} + 2016$, ..., $a_n = 2^{2^n} + 2016$, .... If there exist two members of the sequence that are multiples of $k$, then prove that there are infinitely many members of the sequence that are multip... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
03fw | Find all functions $f : \mathbb{R}^+ \to \mathbb{R}^+$, such that
$$
f(af(b) + a)(f(bf(a)) + a) = 1
$$
for any positive reals $a, b$.
(Lyuben Lichev) | [] | Bulgaria | 5 Bulgarian National Olympiad - Final Round | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = 1/x for all x > 0 | |
04nv | A box holds $n$ balls, some of which are white, and some are black. Find $n$ if the probability that by drawing two balls we draw exactly one white and one black is $\frac{1}{2}$, and we know that there are 2018 more balls of one colour than the other. | [] | Croatia | Croatia_2018 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 2018^2 | |
0fcx | Problem:
Sean $x_{1}, x_{2}$ las raíces del polinomio $P(x)=3 x^{2}+3 m x+m^{2}-1$, siendo $m$ un número real. Probar que $P\left(x_{1}^{3}\right)=P\left(x_{2}^{3}\right)$. | [
"Solution:\nTenemos $x_{1}+x_{2}=-m, \\quad x_{1} \\cdot x_{2}=\\frac{m^{2}-1}{3}$ y\n$$\n\\begin{aligned}\nP\\left(x_{1}^{3}\\right)-P\\left(x_{2}^{3}\\right) & =3 x_{1}^{6}+3 m x_{1}^{3}+m^{2}-1-\\left(3 x_{2}^{6}+3 m x_{2}^{3}+m^{2}-1\\right) \\\\\n& =3\\left(x_{1}^{6}-x_{2}^{6}\\right)+3 m\\left(x_{1}^{3}-x_{2}... | Spain | null | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0gry | Let $S$ be a set having $n$ elements and $A_1, A_2, \dots, A_n$ be subsets of $S$ such that the union of any three of them is equal to $S$ and the union of any two of them is not equal to $S$. Find the maximal possible value of $n$. | [
"The answer is $64$. Assume that $n \\geq 65$. Then there are at least $\\binom{65}{2} = 2080$ unordered pairs of subsets $A_i, A_j$. By conditions, to each unordered pair $A_i, A_j$ we can correspond an element $c(i, j) \\in S$ such that $c(i, j) \\notin A_i \\cup A_j$. Since $2080 > 2019$, for some $A_k, A_l$ and... | Turkey | Team Selection Test for EGMO 2019 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 64 | |
070b | Problem:
$n = k^2 - k + 1$, where $k$ is a prime plus one. Show that we can color some squares of an $n \times n$ board black so that each row and column has exactly $k$ black squares, but there is no rectangle with sides parallel to the sides of the board which has its four corner squares black. | [
"Solution:\n\nWe can regard the rows as lines and the columns as points. Black squares denote incidence. So line 3 contains point 4 iff square $(3,4)$ is black. The condition about rectangles then means that there is at most one line through two distinct points.\n\nSuppose we take the points to be $(a, b, c)$, wher... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Abstract Algebra > Field Theory"
] | null | proof only | null | |
0ihq | Problem:
Larry can swim from Harvard to MIT (with the current of the Charles River) in 40 minutes, or back (against the current) in 45 minutes. How long does it take him to row from Harvard to MIT, if he rows the return trip in 15 minutes? (Assume that the speed of the current and Larry's swimming and rowing speeds re... | [
"Solution:\n\nLet the distance between Harvard and MIT be $1$, and let $c$, $s$, $r$ denote the speeds of the current and Larry's swimming and rowing, respectively. Then we are given\n$$\ns + c = \\frac{1}{40} = \\frac{9}{360}, \\quad s - c = \\frac{1}{45} = \\frac{8}{360}, \\quad r - c = \\frac{1}{15} = \\frac{24}... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 14:24 | |
0ifo | Problem:
Let $ABCD$ be a rectangle with area $1$, and let $E$ lie on side $CD$. What is the area of the triangle formed by the centroids of triangles $ABE$, $BCE$, and $ADE$? | [
"Solution:\nLet the centroids of $ABE$, $BCE$, and $ADE$ be denoted by $X$, $Y$, and $Z$, respectively. Let $d(P, QR)$ denote the distance from $P$ to line $QR$. Since the centroid lies two-thirds of the distance from each vertex to the midpoint of the opposite edge, $d(X, AB) = d(Y, CD) = d(Z, CD) = \\frac{1}{3} B... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 1/9 | |
03p0 | Assume that $n$ is a given positive integer. Find all of the integer groups $(a_1, a_2, \dots, a_n)$ satisfying the conditions:
(1) $a_1 + a_2 + \dots + a_n \ge n^2$;
(2) $a_1^2 + a_2^2 + \dots + a_n^2 \le n^3 + 1$. | [
"Suppose $(a_1, a_2, \\dots, a_n)$ is an integer group which satisfies the conditions. Then by Cauchy's inequality we have\n$$\na_1^2 + \\dots + a_n^2 \\ge \\frac{1}{n} (a_1 + \\dots + a_n)^2 \\ge n^3. \\qquad \\textcircled{1}\n$$\nCombining $a_1^2 + \\dots + a_n^2 \\le n^3 + 1$, we see that it can only be $a_1^2 +... | China | China Western Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (n, n, ..., n) | |
09zd | Kira has $3$ blocks with the letter $A$, $3$ blocks with the letter $B$, and $3$ blocks with the letter $C$. She puts these $9$ blocks in a sequence. She wants to have as many distinct distances between blocks with the same letter as possible. For example, in the sequence $ABCAABCBC$ the blocks with the letter $A$ have... | [
"We will show that the maximum number of distinct distances is $7$. First we prove that the number of distinct distances cannot be more than $7$, then we will show that there is a sequence of blocks with $7$ distances.\n\nThe possible distances between two blocks in the sequence are the numbers $1$ to $8$. Therefor... | Netherlands | Final Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 7 | |
07yz | Problem:
Dato un numero primo $p$, determinare tutte le coppie ordinate di numeri naturali $(m, n)$ che verificano l'equazione:
$$
\frac{1}{m} + \frac{1}{n} = \frac{1}{p}
$$ | [
"SOLUZIONE"
] | Italy | Progetto Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (m, n) = (p + 1, p + p^2), (2p, 2p), (p + p^2, p + 1). Equivalently, (m, n) = (p + d, p + p^2/d) where d ∈ {1, p, p^2}. | |
0kc2 | Problem:
For a real number $r$, the quadratics $x^{2}+(r-1)x+6$ and $x^{2}+(2r+1)x+22$ have a common real root. The sum of the possible values of $r$ can be expressed as $\frac{a}{b}$, where $a, b$ are relatively prime positive integers. Compute $100a+b$. | [
"Solution:\nLet the common root be $s$. Then,\n$$\ns^{2}+(r-1)s+6 = s^{2}+(2r+1)s+22\n$$\nand $s = -\\frac{16}{r+2}$. Substituting this into $s^{2}+(r-1)s+6=0$ yields\n$$\n\\frac{256}{(r+2)^{2}} - \\frac{16(r-1)}{r+2} + 6 = 0\n$$\nAfter multiplying both sides by $(r+2)^{2}$, the equation becomes\n$$\n256 - 16(r-1)(... | United States | HMMO 2020 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 405 | |
03af | The incircle of $\triangle ABC$ has center $I$ and touches the sides $BC$, $AC$ and $AB$ at points $A_1$, $B_1$ and $C_1$, respectively. An arbitrary line $\ell$ through $I$ is given and the points $A'$, $B'$ and $C'$ are symmetric to $A_1$, $B_1$ and $C_1$, respectively, with respect to $\ell$. Prove that the lines $A... | [
"Denote by $d_c(X)$ the distance from the point $X$ to the line $AB$ and analogously for the lines $BC$ and $CA$. It is not difficult to see that the Sine version of the Ceva theorem implies that the equality\n$$\n\\frac{d_b(A')}{d_c(A')} \\cdot \\frac{d_c(B')}{d_a(B')} \\cdot \\frac{d_a(C')}{d_b(C')} = 1\n$$\nis n... | Bulgaria | 58. National mathematical olympiad Final round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Tria... | English | proof only | null | |
08c0 | Problem:
a. Dimostrare che esistono infinite terne $(x, y, z)$ di interi positivi tali che $x^{2}+y^{2}+z^{2}$ sia un quadrato perfetto.
b. Dimostrare che esistono infinite terne $(x, y, z)$ di interi positivi tali che $x^{2}+y^{2}+z^{2}$ sia un quadrato perfetto e con la proprietà che il massimo comun divisore dei t... | [
"Solution:\n\na. Osserviamo dapprima che $x=1, y=2, z=2$ è una soluzione, in quanto $1^{2}+2^{2}+2^{2}=3^{2}$. Ne segue allora che, per ogni intero positivo $n$, la terna $x=n, y=2 n, z=2 n$ soddisfa la richiesta del problema, in quanto\n$$\nn^{2}+(2 n)^{2}+(2 n)^{2}=9 n^{2}=(3 n)^{2}.\n$$\n\nb. Per ogni intero pos... | Italy | Gara di Febbraio | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0j5u | An integer is assigned to each vertex of a regular pentagon so that the sum of the five integers is $2011$. A turn of a solitaire game consists of subtracting an integer $m$ from each of the integers at two neighboring vertices and adding $2m$ to the opposite vertex, which is not adjacent to either of the first two ver... | [
"Let $a_1, a_2, a_3, a_4$, and $a_5$ represent the integers at vertices $v_1$ to $v_5$ (in order around the pentagon) at the start of the game. We will first show that the game can be won at only one of the vertices. Observe that the quantity $a_1 + 2a_2 + 3a_3 + 4a_4 \\pmod{5}$ is an invariant of the game. For ins... | United States | USAMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0b4y | Problem:
Encontre uma maneira de se escrever todos os algarismos de $1$ a $9$ em sequência e sem repetição, de forma que os números determinados por quaisquer dois algarismos consecutivos da sequência sejam divisíveis por $7$ ou $13$. | [] | Portuguese Language Countries Olympiad (OMCPLP) | 9a Olimpíada de Matemática da Comunidade dos Países de Língua Portuguesa | [] | Português | null | null | |
04mh | Let $a$, $b$ and $c$ be different positive real numbers such that $(a+b-c)(b+c-a)(c+a-b) \neq 0$. Prove that at least one of the numbers
$$
\frac{a+b}{a+b-c}, \quad \frac{b+c}{b+c-a}, \quad \frac{c+a}{c+a-b}
$$
lies in the interval $(1, 2)$, and that at least one of these numbers does not lie in that interval. | [] | Croatia | Croatia_2018 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0f46 | Problem:
$AB$ is a diameter of the circle $C$. $M$ and $N$ are any two points on the circle. The chord $MA'$ is perpendicular to the line $NA$ and the chord $MB'$ is perpendicular to the line $NB$. Show that $AA'$ and $BB'$ are parallel. | [] | Soviet Union | 15th ASU | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
06kf | Find the first digit after the decimal point of the number $\frac{1}{1009} + \frac{1}{1010} + \dots + \frac{1}{2016}$. | [
"The answer is 6.\nLet $a_n = \\frac{1}{n+1} + \\frac{1}{n+2} + \\dots + \\frac{1}{2n}$. Firstly, we find that\n$$\na_{n+1} - a_n = \\frac{1}{2n+1} + \\frac{1}{2n+2} - \\frac{1}{n+1} = \\frac{1}{(2n+1)(2n+2)} > 0.\n$$\nThis shows the sequence is strictly increasing. Thus, we easily find that\n$$\na_{1008} > a_3 = \... | Hong Kong | HKG TST | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 6 | |
0gyc | Let be given positive numbers $a$, $b$, $c$. Prove that
$$
\frac{a}{3a^2 + b^2 + 2ac} + \frac{b}{3b^2 + c^2 + 2ab} + \frac{c}{3c^2 + a^2 + 2bc} \le \frac{3}{2(a+b+c)}
$$ | [
"Using for the first denominator such inequality $3a^2 + b^2 + 2ac = 2a^2 + a^2 + b^2 + 2ac \\ge 2a^2 + 2ab + 2ac = 2a(a+b+c)$ and analogously for the rest two denominators. Then the left-hand side becomes in the form of:\n$$\n\\begin{aligned}\n\\frac{a}{3a^2 + b^2 + 2ac} + \\frac{b}{3b^2 + c^2 + 2ab} + \\frac{c}{3... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
000w | Determinar el menor entero positivo $k$ de modo que la ecuación
$$
2002x + 273y = 200201 + k
$$
tenga soluciones enteras, y para ese valor de $k$, hallar la cantidad de soluciones $(x, y)$ con $x, y$ enteros positivos que tiene la ecuación. | [] | Argentina | XIX Olimpíada Matemática Argentina | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | español | proof and answer | k = 90; number of positive solutions = 33 | |
05to | Problem:
Soit $C$ un entier naturel non nul. Trouver toutes les fonctions $f: \mathbb{N}^{*} \rightarrow \mathbb{N}^{*}$ telles que, pour tous les entiers $a$ et $b$ de somme $a+b \geqslant C$, l'entier $a+f(b)$ divise $a^{2}+b f(a)$. | [
"Solution:\n\nTout d'abord, toute fonction linéaire strictement croissante est solution. En effet, pour tout entier $k \\geqslant 1$, l'entier $a+k b$ divise bien $a^{2}+b \\times(k a)=a(a+k b)$. Réciproquement, montrons que toute solution est une fonction linéaire (qui sera strictement croissante, puisque à valeur... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All functions of the form f(n) = k n for a positive integer k | |
07t8 | Show that there exists a convex hexagon in the plane such that the distance between every pair of vertices is an integer. | [
"Let $P_1$, $P_2$, $P_3$ be the vertices of an equilateral triangle with sides of length $1$ inscribed in a circle. Obtain points $Q_1$, $Q_2$, $Q_3$ by rotating the triangle $P_1P_2P_3$ so that $Q_i$ lies on the minor arc $P_iP_{i+1}$. Let $|P_1Q_1| = a$ and $|Q_1P_2| = b$. Applying Ptolemy's theorem to the cyclic... | Ireland | IRL_ABooklet_2020 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geomet... | null | proof only | null | |
03wt | Let $a$, $b$ be positive real numbers with $b - a > 2$. Prove that for any two distinct integers $m$, $n$ in the interval $[a, b)$, there is a nonempty set $S$ consisting of some integers in the interval $[ab, (a+1)(b+1))$, such that $\frac{\prod x}{mn}$ is a square of a rational number. (Posed by Yu Hongbing) | [
"We first prove the following lemma:\n\nLemma. Let $u$ be an integer with $a \\le u < u+1 < b$. Then there are two distinct integers $x$, $y$ in the interval $[ab, (a+1)(b+1))$, such that $\\frac{xy}{u(u+1)}$ is a square of an integer.\n\nProof of lemma. Let $v$ be the smallest integer not less than $\\frac{ab}{u}$... | China | China National Team Selection Test | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Other"
] | English | proof only | null | |
05if | Problem:
Soit $ABC$ un triangle et $\omega$ un cercle qui passe par $A$ et $B$ et recoupe les côtés $[BC]$ et $[AC]$ respectivement en $D$ et $E$. Les points $K$ et $L$ sont respectivement les centres du cercle inscrit dans $DAC$ et du cercle inscrit dans $BEC$. Soit $N$ l'intersection des droites $(EL)$ et $(DK)$.
P... | [
"Solution:\n\n\n\nSolution 1. On va montrer l'égalité entre les angles de droites ($KL, KN$) et $(LN, KL)$. Notons d'abord que $C, K, L$ sont alignés (sur la bissectrice intérieure en $C$ du triangle $ABC$).\n$$\n\\begin{aligned}\n(KL, KN) & = (CK, CD) + (DC, DN) = \\frac{1}{2}((\\overright... | France | Test EGMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscel... | null | proof only | null | |
0hh3 | Let $P(x)$, $Q(x)$ and $R(x)$ be polynomials with integer coefficients, such that the equality holds: $P(x) = Q(x)R(x)$. We denote by $a$ and $b$ the maximum of the absolute values of coefficients of the polynomials $P(x)$ and $Q(x)$ respectively. Does the condition $b \le 2023a$ always hold? | [
"**Answer:** No, not necessarily.\n\nAs an example, consider the following two polynomials:\n\n$$\nQ(x) = 1 + 2x + 3x^2 + 4x^3 + \\cdots + 2023x^{2022} + 2024x^{2023} + 2023x^{2024} + \\cdots + x^{4046}, \\quad R(x) = x - 1,\n$$\n$$\n\\text{then } P(x) = Q(x)(x-1) = -1 - x - x^2 - \\cdots - x^{2023} + x^{2024} + x^... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof and answer | No | |
0fkr | Problem:
Dada una circunferencia y dos puntos $P$ y $Q$ en su interior, inscribir un triángulo rectángulo cuyos catetos pasen por $P$ y $Q$. ¿Para qué posiciones de $P$ y $Q$ el problema no tiene solución? | [
"Solution:\n\n\n\nBasta trazar la circunferencia de diámetro $PQ$, las eventuales intersecciones con la circunferencia inicial de centro $O$ nos proporcionan dos puntos $A$ y $B$ que unidos con $P$ y $Q$ definen las soluciones.\n\nNo existe solución cuando ambas circunferencias no se cortan... | Spain | FASE LOCAL DE LA XLIV OME | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | No solution exactly when OM + PQ/2 < r, where r is the radius of the given circle and M is the midpoint of PQ. | |
0ieo | Problem:
Three noncollinear points and a line $\ell$ are given in the plane. Suppose no two of the points lie on a line parallel to $\ell$ (or $\ell$ itself). There are exactly $n$ lines perpendicular to $\ell$ with the following property: the three circles with centers at the given points and tangent to the line all c... | [
"Solution:\nThe condition for the line is that each of the three points lies at an equal distance from the line as from some fixed point; in other words, the line is the directrix of a parabola containing the three points. Three noncollinear points in the coordinate plane determine a quadratic polynomial in $x$ unl... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 1 | |
0cw6 | The altitudes of an acute-angled triangle $ABC$ meet at $H$. The points $B_1$ and $C_1$ are chosen on the segments $BH$ and $CH$, respectively, so that $B_1C_1 \parallel BC$. It appeared that the center of the circumcircle $\omega$ of $\triangle B_1HC_1$ lies on the line $BC$. Prove that $\omega$ is tangent to the circ... | [
"Обозначим через $\\Gamma'$ окружность, описанную около треугольника $BHC$ (см. рис. 1). На касательной к $\\Gamma'$ в точке $H$ отметим точку $X$, лежащую внутри угла $BCH$. Тогда $\\angle BHX = \\angle BCH = \\angle B_1C_1H$ (последнее равенство следует из того, что $BC \\parallel B_1C_1$). Значит, окружность $\\... | Russia | Regional round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
0boj | Find all non constant polynomial functions $f: [0, 1] \to \mathbb{R}^*$, having rational coefficients, for which the following property holds: for every $x$ in $[0, 1]$, there exist two polynomial functions $g_x, h_x: [0, 1] \to \mathbb{R}$, with rational coefficients, such that $h_x(x) \neq 0$ and
$$
\int_0^x \frac{1}... | [
"The required functions are $f(x) = a(x-b)^n$, $0 \\le x \\le 1$, where $n$ is an integer greater than $1$, and $a$ and $b$ are rational numbers, $a \\neq 0$ and $b \\notin [0, 1]$. Clearly, these functions satisfy the conditions in the statement.\n\nSince the closed unit interval $[0, 1]$ is uncountable, and there... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | All such polynomials are f(x) = a(x - b)^n with n an integer greater than 1, a and b rational, a not zero, and b not in the unit interval. | |
033j | Problem:
Let $p(x)$ and $q(x)$ be polynomials with $m \geq 2$ non-zero coefficients. If $\frac{p(x)}{q(x)}$ is not a constant function, find the least possible number of the non-zero coefficients of the polynomial $f(u, v)=p(u) q(v)-p(v) q(u)$. | [
"Solution:\nConsidering the polynomials $p(x)=x^{m-1}+x^{m-2}+\\cdots+x+1$ and $q(x)=x^{m-1}+x^{m-2}+\\cdots+x+a$, $a \\neq 1$, shows that the desired minimal number does not exceed $2m-2$ (one has that $f(u, v)=(a-1)\\left(u^{m-1}+u^{m-2}+\\cdots+u\\right)+(1-a)\\left(v^{m-1}+v^{m-2}+\\cdots+v\\right)$). We shall ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 2m-2 | |
0ewy | Problem:
How many (algebraically) different expressions can we obtain by placing parentheses in $a_1 / a_2 / \ldots / a_n$? | [
"Solution:\nAnswer $2^{n - 2}$. $a_1$ must be in the numerator, and $a_2$ must be in the denominator, but the other symbols can be in either. This is easily proved by induction."
] | Soviet Union | 4th ASU | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2^{n-2} | |
0117 | Problem:
Two positive integers are written on the blackboard. Initially, one of them is $2000$ and the other is smaller than $2000$. If the arithmetic mean $m$ of the two numbers on the blackboard is an integer, the following operation is allowed: one of the two numbers is erased and replaced by $m$. Prove that this o... | [
"Solution:\n\nEach time the operation is performed, the difference between the two numbers on the blackboard will become one half of its previous value (regardless of which number was erased). The mean value of two integers is an integer if and only if their difference is an even number. Suppose the initial numbers... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Maximum number of operations: 10; example initial pair: 2000 and 976 | |
0drr | Let $a_1, a_2, \dots, a_9$ be a sequence of numbers satisfying $0 < p \le a_i \le q$ for each $i = 1, 2, \dots, 9$. Prove that
$$
\frac{a_1}{a_9} + \frac{a_2}{a_8} + \dots + \frac{a_9}{a_1} \le 1 + \frac{4(p^2 + q^2)}{pq}.
$$ | [
"Note that for any positive numbers $a, b$ between $p$ and $q$, we may assume (due to symmetry) that $0 < p \\le a \\le b \\le q$. We have\n$$\n\\frac{a}{b} + \\frac{b}{a} \\le \\frac{p}{q} + \\frac{q}{p} \\iff \\frac{a^2 + b^2}{ab} \\le \\frac{p^2 + q^2}{pq} \\\\\n\\iff p^2 ab + q^2 ab - a^2 pq - b^2 pq = (bq - ap... | Singapore | Singapur | [
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
0hgk | You are given $n \ge 4$ positive real numbers. It turned out that their $\frac{n(n-1)}{2}$ pairwise products form an arithmetic progression in some order. Prove that all of these numbers are equal.
(Anton Trygub) | [
"If some two products are equal, then all products are equal, and all numbers are equal. If some two numbers are equal, then some products are equal, so all numbers are equal. Now consider 4 largest numbers $a < b < c < d$. The largest two products are $cd, bd$. Then the difference of the progression is $cd - bd$. ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
03ei | Let $n \in \mathbb{N}$ and $\mathcal{A}$ be a nonempty family of nonempty subsets of $\{1, 2, \dots, n\}$ with the following property – if $A \in \mathcal{A}$ and $A \subset B \subseteq \{1, 2, \dots, n\}$, then $B \in \mathcal{A}$. Prove that the function
$$
f(x) := \sum_{A \in \mathcal{A}} x^{|A|} (1-x)^{n-|A|}
$$
is... | [
"Let $0 < p < q < 1$. Notice that $p^* = \\frac{q-p}{1-p} \\in (0, 1)$. We construct the sets $X$ and $Y$ as follows: For each element $i \\in \\{1, 2, \\dots, n\\}$ we put $i$ in $X$ with probability $p$ (independently of each other), and for each element $j \\in \\{1, 2, \\dots, n\\} \\setminus X$ we put $j$ in $... | Bulgaria | 2 Bulgarian Winter Tournament | [
"Discrete Mathematics > Combinatorics > Expected values"
] | English | proof only | null | |
09nm | Uchral, together with a few friends, bought some books and some notebooks from a store. The price of one book is $A$ times more expensive than one notebook. Uchral and her friends decided to divide the purchased books and notebooks such that the monetary value each person receives is equal.
Uchral received $\frac{1}{1... | [] | Mongolia | MMO2025 Round 3 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | {24, 9, 4, 3/2} | |
03ym | Let $a_1, a_2, \dots, a_n$ ($n \ge 3$) be real numbers. Prove that
$$
\sum_{i=1}^{n} a_i^2 - \sum_{i=1}^{n} a_i a_{i+1} \le \left[ \frac{n}{2} \right] (M-m)^2,
$$
where $a_{n+1} = a_1$, $M = \max_{1 \le i \le n} a_i$, $m = \min_{1 \le i \le n} a_i$. $[x]$ is the largest integer not exceeding $x$. | [
"If $n = 2k$ ($k$ is a positive integer), then\n$$\n2 \\left( \\sum_{i=1}^{n} a_i^2 - \\sum_{i=1}^{n} a_i a_{i+1} \\right) = \\sum_{i=1}^{n} (a_i - a_{i+1})^2 \\le n(M-m)^2,\n$$\ntherefore,\n$$\n\\sum_{i=1}^{n} a_i^2 - \\sum_{i=1}^{n} a_i a_{i+1} \\le \\frac{n}{2} (M-m)^2 = \\left[ \\frac{n}{2} \\right] (M-m)^2.\n$... | China | China Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
01t6 | Let $A$, $B$, $C$ denote the intersection points of the diagonals $A_1A_4$ and $A_2A_5$, $A_1A_6$ and $A_2A_7$, $A_1A_9$ and $A_2A_{10}$ of the regular decagon $A_1A_2...A_9A_{10}$, respectively.
Find the angles of the triangle $ABC$.
(Folklore) | [
"(Solution by A. Goloubitskaya.) Let $\\Gamma$ be the circumcircle of the given regular hexagon. It is evident that $B$ is the center of $\\Gamma$. Since all sides of the regular hexagon are equal, we have $2\\alpha = \\text{arc } A_1A_2 = \\text{arc } A_2A_3 = \\dots = \\text{arc } A_9A_{10} = \\text{arc } A_{10}A... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | ∠ABC = 90°, ∠BCA = 36°, ∠CAB = 54° | |
08dl | Problem:
Sia $ABC$ un triangolo acutangolo. Siano $D$ il piede della bisettrice interna da $A$ ed $M$ il punto medio di $AD$. Sia inoltre $X$ un punto sul segmento $BM$ tale che $\angle MXA = \angle DAC$.
Dimostrare che $AX$ è perpendicolare a $XC$. | [
"Solution:\n\nDato che per ipotesi vale $\\angle MXA = \\angle DAC = \\angle MAB$, i due triangoli $MXA$ e $MAB$ sono simili. Dunque\n$$\n\\frac{XM}{AM} = \\frac{AM}{BM}\n$$\nSia $H$ il piede della perpendicolare condotta da $A$ su $BC$. D'ora in poi assumiamo che $H$ sia fra $B$ e $D$. L'altro caso, ovvero $D$ fra... | Italy | XXXV Olimpiade Italiana di Matematica | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler lin... | null | proof only | null | |
08ar | Problem:
Un numero naturale si dice palindromo se è uguale al numero che si ottiene leggendo le cifre della sua scrittura in base dieci da destra verso sinistra (ad esempio, 68386 e 44 sono palindromi, 220 non lo è). Sappiamo che il numero naturale $x$ e il numero $x+312$ sono entrambi palindromi; $x$ ha quattro cifre... | [
"Solution:\n\nLa risposta è $(\\mathbf{E})$. Se $x$ è un numero di 4 cifre (dunque minore di 10000), $x+312$, che deve averne 5, è compreso fra 10000 e 10311 (estremi inclusi). Questo significa che le prime due cifre di $x+312$ sono 1 e 0; poiché si tratta di un numero palindromo, deve dunque essere della forma $10... | Italy | Progetto Olimpiadi della Matematica - GARA di FEBBRAIO | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | E | |
0gvc | We are given $n$ ($n \ge 3$) points in the plane not all lying on the same line. For an arbitrary point $M$ we define $f(M)$ to be the sum of the distances from these $n$ points to $A$. It is known that there exists a point $M_1$ such that for every point $M$ of the plane the inequality $f(M_1) \le f(M)$ holds. Let $M_... | [
"It is easy to prove that if a point $M$ is the midpoint of a segment $BC$, then for any point $A$ the inequality $AM \\le \\frac{1}{2}(AB + AC)$ holds. Suppose that the points $M_1$ and $M_2$ mentioned in the problem statement do not coincide. Let $M_3$ be the midpoint of the segment $M_1M_2$. Then $A_kM_3 \\le \\... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0dx9 | Problem:
Za katere kote $\alpha$ z lastnostjo $0<\alpha<2 \pi$ velja neenakost
$$
\frac{5 \sin \alpha-2}{\sin \alpha} \geq 2 \sin \alpha ?
$$ | [
"Solution:\n\n1. Očitno mora biti $\\alpha \\neq \\pi$, saj sicer ulomek na levi strani neenakosti ne bi bil dobro definiran. Obravnavajmo dva primera.\n\nČe je $0<\\alpha<\\pi$, tedaj je $\\sin \\alpha>0$. Torej če neenakost pomožimo s $\\sin \\alpha$, se neenačaj ohrani in dobimo $5 \\sin \\alpha-2 \\geq 2 \\sin ... | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | [π/6, 5π/6] ∪ (π, 2π) | |
0ky8 | Let $m$ and $n$ be fixed positive integers. Tsvety and Freyja play a game on an infinite grid of unit square cells. Tsvety has secretly written a real number inside of each cell so that the sum of the numbers within every rectangle of size either $m \times n$ or $n \times m$ is zero. Freyja wants to learn all of these ... | [
"The answer is the following:\n* If $\\gcd(m, n) > 1$, then Freyja cannot win.\n* If $\\gcd(m, n) = 1$, then Freyja can win in a minimum of $(m-1)^2 + (n-1)^2$ questions.\n\nFirst, we dispose of the case where $\\gcd(m, n) > 1$. Write $d = \\gcd(m, n)$. The idea is that any labeling where each $1 \\times d$ rectang... | United States | USA TST | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Linear Algebra > Vectors",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd... | null | proof and answer | If gcd(m, n) > 1, no finite number of questions suffices. If gcd(m, n) = 1, the minimum number is (m − 1)^2 + (n − 1)^2. | |
0il2 | Problem:
Solve for all complex numbers $z$ such that $z^{4} + 4z^{2} + 6 = z$. | [
"Solution:\n\nRewrite the given equation as $\\left(z^{2} + 2\\right)^{2} + 2 = z$. Observe that a solution to $z^{2} + 2 = z$ is a solution of the quartic by substitution of the left hand side into itself. This gives $z = \\frac{1 \\pm i \\sqrt{7}}{2}$. But now, we know that $z^{2} - z + 2$ divides into $\\left(z^... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | z = (1 ± i√7)/2, z = (-1 ± i√11)/2 | |
045c | Fix an equilateral triangle $ABC$ with side length $1$. We call $(\triangle DEF, \triangle XYZ)$ a good triangle pair if the points $D$, $E$, and $F$ lie in the interior of segments $BC$, $CA$, and $AB$, respectively, the points $X$, $Y$, and $Z$ lie on the lines $BC$, $CA$, and $AB$, respectively, and they satisfy the... | [
"(1) First, consider the rotation of $90^\\circ$ (clockwise or counterclockwise), centered at an arbitrary point on the plane. Then the images $X'$, $Y'$, and $Z'$ of the points $X$, $Y$, and $Z$, respectively, satisfy\n$$\nX'Y' \\parallel DE, \\quad Y'Z' \\parallel EF, \\quad Z'X' \\parallel FD.\n$$\nIn this case,... | China | Chinese Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Napoleon and Fermat points",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine... | English | proof and answer | (97√2 + 40√3)/15 | |
033v | Problem:
Prove that among any $2n+1$ irrational numbers there are $n+1$ numbers such that the sum of any $2,3, \ldots, n+1$ of them is an irrational number. | [
"Solution:\n\nLet the given numbers be $a_{1}, a_{2}, \\ldots, a_{2n+1}$. Choose first 1 and then choose at any step (if it is possible) a number that is not a linear combination with rational coefficients of the already chosen numbers. We may assume that the chosen numbers are $a_{0}=1, a_{1}, a_{2}, \\ldots, a_{k... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Linear Algebra > Vectors",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
01c4 | Let $m$ and $n$ be positive integers, assumed relatively prime. Determine all possible values of
$$
\text{gcd}(2^m - 2^n, 2^{m^2+mn+n^2} - 1).
$$ | [
"We may assume $m \\ge n$. It is well known that\n$$\n\\text{gcd}(2^p - 1, 2^q - 1) = 2^{\\text{gcd}(p,q)} - 1,\n$$\nso that\n$$\n\\begin{aligned}\n\\text{gcd}(2^m - 2^n, 2^{m^2+mn+n^2} - 1) &= \\text{gcd}(2^{m-n} - 1, 2^{m^2+mn+n^2} - 1) \\\\\n&= 2^{\\text{gcd}(m-n, m^2+mn+n^2)} - 1.\n\\end{aligned}\n$$\nNext, con... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 1 and 7 | |
0ex3 | Problem:
Reduce each of the first billion natural numbers (billion $= 10^9$) to a single digit by taking its digit sum repeatedly. Do we get more 1s than 2s? | [
"Taking digit sums repeatedly gives the remainder after dividing the number by $9$, or $9$ if the number is exactly divisible by $9$. $10^9 - 1 = 9n$, and for any $r \\geq 0$ the nine consecutive numbers $9r + 1$, $9r + 2$, ..., $9r + 9$ include just one number giving remainder $1$ and one number giving remainder $... | Soviet Union | 4th ASU | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | Yes; there is exactly one more 1 than 2. | |
0g72 | 設 $a$, $b$, $c$ 是正實數且滿足 $\min\{a+b, b+c, c+a\} > \sqrt{2}$ 與 $a^2 + b^2 + c^2 = 3$。
試證：
$$
\frac{a}{(b+c-a)^2} + \frac{b}{(c+a-b)^2} + \frac{c}{(a+b-c)^2} \ge \frac{3}{(abc)^2}
$$ | [
"由 $b+c > \\sqrt{2}$ 推得 $b^2 + c^2 > 1$，故 $a^2 = 3 - (b^2 + c^2) < 2$，即 $a < \\sqrt{2} < b+c$。\n故我們有 $b+c-a > 0$，同樣地有 $c+a-b > 0$ 與 $a+b-c > 0$。\n我們將用 HÖLDER 不等式：\n$$\n\\frac{x_1^{p+1}}{y_1^p} + \\frac{x_2^{p+1}}{y_2^p} + \\cdots + \\frac{x_n^{p+1}}{y_n^p} \\ge \\frac{(x_1 + x_2 + \\cdots + x_n)^{p+1}}{(y_1 + y_2 +... | Taiwan | 二〇一二數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof only | null | |
0fws | Problem:
Ein Weg in der Ebene führt vom Punkt $(0,0)$ zum Punkt $(6,6)$, wobei man in jedem Schritt entweder um $1$ nach rechts oder um $1$ nach oben gehen kann. Wieviele Wege gibt es, die weder den Punkt $(2,2)$ noch den Punkt $(4,4)$ enthalten? | [
"Solution:\n\nFür $a, b \\geq 0$ ist allgemein die Anzahl Wege vom Punkt $(x, y)$ zum Punkt $(x+a, y+b)$, bei denen man in jedem Schritt um $1$ nach rechts oder nach oben geht, gleich $\\binom{a+b}{a} = \\binom{a+b}{b}$. Denn ein solcher Weg ist dadurch festgelegt, dass man in der Folge der $a+b$ nötigen Schritte d... | Switzerland | Vorrundenprüfung | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 300 | |
0i2r | Problem:
A circle has two parallel chords of length $x$ that are $x$ units apart. If the part of the circle included between the chords has area $2 + \pi$, find $x$. | [
"Solution:\n\nLet $C$ be the area of the circle, $S$ be the area of the square two of whose edges are the chords, and $A$ be the area of the part of the circle included between the chords. The radius of the circle is $\\frac{\\sqrt{2}}{2} x$, so $C = \\frac{\\pi}{2} x^{2}$, and $S = x^{2}$. Then the area $A$ is the... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals"
] | null | proof and answer | 2 | |
09fd | Find all maps $f: \mathbb{Z}_{\ge 1} \to \mathbb{Z}_{\ge 1}$ such that for any positive integers $a$ and $b$, the number
$$
af(a)^2 + bf(b)^2 + 3ab(f(a) + f(b))
$$
is a perfect cube. | [
"For any prime $p$ let us set $a = b = p$. Hence $2pf(p)(f(p) + p)$ is a perfect cube, we have $p \\mid f(p)$. Let $f(p) = kp$, $k \\in \\mathbb{Z}_{\\ge 1}$. Setting $a = p$, $b = 1, 2$ we get that\n$$\n\\begin{aligned}\npf(p)^2 + f(1)^2 + 3p(f(p) + f(1)) &= m^3 \\\\\npf(p)^2 + 2f(2)^2 + 6p(f(p) + f(2)) &= n^3.\n\... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | f(n) = n for all positive integers n | |
07cg | Three non-planar rings are located in the space.
a) Is it always possible to find a circle that passes through all three circles?
b) Is it always possible to find a square that passes through all three circles? | [
"a) No! As a counterexample, consider three small equal circles on the circumference of a sphere. Each of these circles divides the circumference of the sphere into two parts. Call the smaller part for each circle its \"inner domain\", and assume that the inner domains of these circles are mutually disjoint. Now, i... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | a) No. b) Yes. | |
0hve | Problem:
Let $ABC$ be a triangle with circumcircle $\Gamma$, whose incircle touches $BC$, $CA$, $AB$ at $D$, $E$, $F$. We draw a circle tangent to segment $BC$ at $D$ and to minor $\operatorname{arc} \widehat{BC}$ of $\Gamma$ at the point $A_{1}$. Define $B_{1}$ and $C_{1}$ in a similar way. Prove that lines $A_{1}D$,... | [
"Solution:\n\nBy the so-called \"shooting lemma\" (which is proved by taking homothety at $A_{1}$) we find that line $A_{1}D$ passes through the arc midpoint of $\\widehat{BAC}$ of $\\Gamma$; denote this arc midpoint by $X$. Define $Y$ and $Z$ similarly, so that $Y$ lies on line $B_{1}E$ and $Z$ lies on line $C_{1}... | United States | Berkeley Math Circle | [
"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 | |
04g5 | Determine all complex numbers $z$ such that $\frac{\bar{z}}{z} + \frac{z}{\bar{z}}$ is a positive integer. | [
"Let $z = re^{i\\theta}$, where $r > 0$ and $\\theta \\in \\mathbb{R}$. Then $\\bar{z} = re^{-i\\theta}$.\n\nWe have:\n\\[\n\\frac{\\bar{z}}{z} + \\frac{z}{\\bar{z}} = \\frac{re^{-i\\theta}}{re^{i\\theta}} + \\frac{re^{i\\theta}}{re^{-i\\theta}} = e^{-2i\\theta} + e^{2i\\theta} = 2\\cos(2\\theta)\n\\]\n\nWe are tol... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | All nonzero complex numbers with argument equal to any integer multiple of pi, or equal to plus or minus pi over six plus any integer multiple of pi; equivalently, z ≠ 0 with arg z ∈ {kπ, π/6 + kπ, −π/6 + kπ} for integer k. | |
0beb | Suppose $A$ is a singular matrix of order $n$ with complex entries, all of which having absolute value equal to $1$.
a. Let $n = 3$. Show that two lines or two columns of the matrix $A$ are proportional.
b. Find, with proof, if the above claim holds for $n = 4$. | [
"a. By suitable multiplication on each row and column, the matrix $A$ can be written as $k \\begin{pmatrix} 1 & 1 & 1 \\\\ 1 & a & b \\\\ 1 & c & d \\end{pmatrix}$, where $a, b, c, d, k$ are complex numbers of absolute value $1$.\n\nThe relation $\\det(A) = 0$ gives $(a-1)(d-1) = (b-1)(c-1)$. Take the complex conju... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | The claim holds for size three: some two rows or two columns are proportional. It fails for size four: there exists a singular four by four unimodular-entry matrix with no proportional pair of rows or columns. | |
03wk | Let $H$ be the orthocenter of an acute triangle $ABC$, and $D$ be the midpoint of the side $BC$. A line passing through the point $H$ meets the sides $AB, AC$ at the points $F, E$ respectively, such that $AE = AF$. The ray $DH$ meets the circumcircle of $\triangle ABC$ at the point $P$.
Prove that $P, A, E, F$ are con... | [
"On the ray $HD$, mark a point $M$ such that $HD = DM$. Join the segments $BM, CM, BH$ and $CH$. As $D$ is the midpoint of $BC$, the quadrilateral $BHCM$ is a parallelogram, and so\n$$\n\\angle BMC = \\angle BHC = 180^\\circ - \\angle BAC.\n$$\nHence,\n$$\n\\angle BMC + \\angle BAC = 180^\\circ,\n$$\nand the point ... | China | China Western 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"
] | English | proof only | null | |
08sq | Put 6 points $A$, $B$, $C$, $D$, $E$, $F$ on the circumference of a circle in this order in such a way that the following conditions are satisfied: arc $AB$ and arc $BC$, arc $CD$ and arc $DE$, and arc $EF$ and arc $FA$ have same lengths, respectively. Determine the value of the angle $\angle BFD$ if $\angle ACE = 68^\... | [
"From the well-known property of a quadrilateral inscribed in a circle, we have $\\angle ACE + \\angle AFE = 180^\\circ$, and therefore, $\\angle AFE = 180^\\circ - 68^\\circ = 112^\\circ$.\n\nSince the arcs $CD$ and $DE$ have the same lengths, we have from the theorem on inscribed angles that $\\angle CFD = \\angl... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 56° | |
09t1 | Problem:
Zij $n$ een even positief geheel getal. Een rijtje van $n$ reële getallen noemen we volledig als voor elke gehele $m$ met $1 \leq m \leq n$ geldt dat de som van de eerste $m$ termen of de som van de laatste $m$ termen van het rijtje geheel is. Bepaal het minimale aantal gehele getallen in een volledig rijtje ... | [
"Solution:\n\nWe bewijzen dat het minimale aantal gehele getallen in een volledig rijtje gelijk aan $2$ is.\n\nBekijk eerst het geval $n=2$. Noem $a_{1}$ en $a_{2}$ de getallen in het rijtje. Dan is $a_{1}$ of $a_{2}$ geheel, zeg zonder verlies van algemeenheid $a_{1}$. Verder is $a_{1}+a_{2}$ geheel, maar dan is o... | Netherlands | Selectietoets | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 2 | |
0i8c | Problem:
The graph of $x^{4}=x^{2} y^{2}$ is a union of $n$ different lines. What is the value of $n$? | [
"Solution:\nThe equation $x^{4}-x^{2} y^{2}=0$ factors as $x^{2}(x+y)(x-y)=0$, so its graph is the union of the three lines $x=0$, $x+y=0$, and $x-y=0$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 3 | |
0b65 | Let $m, p$ be integers larger than $2$. Find the least positive integer $n$, such that every $n$-element subset of $\{1, 2, 3, \dots, pm\}$ contains two numbers of sum divisible by $p$. | [] | Romania | Shortlisted Problems for the Romanian NMO | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | If p is odd: n = m*(p−1)/2 + 2. If p is even: n = m*(p/2 − 1) + 3. | |
06a9 | Find all pairs of nonzero integers $(x, y)$ such that the integer $x^2 + y^2$ is a common divisor of the integers $x^5 + y$ and $y^5 + x$. | [
"We have that $x^2 + y^2 \\mid x(x^5 + y)$ as well as $x^2 + y^2 \\mid y(y^5 + x)$ and hence\n$$\nx^2 + y^2 \\mid x(x^5 + y) + y(y^5 + x) \\Rightarrow x^2 + y^2 \\mid x^6 + y^6 + 2xy. \\quad (1)\n$$\nFrom the well-known identity for the sum of cubes we have:\n$$\nx^2 + y^2 \\mid (x^2)^3 + (y^2)^3 \\qquad (2),\n$$\n... | Greece | 39th Hellenic Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | [(1, 1), (-1, -1), (1, -1), (-1, 1)] | |
02kl | Problem:
Dados $a$ e $b$ números reais seja $a \diamond b = a^{2} - a b + b^{2}$. Quanto vale $1 \diamond 0$?
A) 1
B) 0
C) 2
D) -2
E) -1 | [
"Solution:\n\nFazendo $a = 1$ e $b = 0$ em $a \\diamond b = a^{2} - a b + b^{2}$ obtemos:\n$$\n1 \\diamond 0 = 1^{2} - 1 \\times 0 + 0^{2} = 1\n$$"
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | MCQ | A | |
0jys | Problem:
Let $G$ be a weighted bipartite graph $A \cup B$, with $|A|=|B|=n$. In other words, each edge in the graph is assigned a positive integer value, called its weight. Also, define the weight of a perfect matching in $G$ to be the sum of the weights of the edges in the matching.
Let $G'$ be the graph with vertex ... | [
"Solution:\n\nLet $m$ denote the minimum weight of a matching in $G$. Let $G''$ be the (multi)graph formed by taking the union of all minimum weight perfect matchings, but keeping edges multiple times. Note that $G''$ is regular. Now, assume that some perfect matching in $G'$ (equivalently $G''$) has weight $M > m$... | United States | HMIC | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0iya | Problem:
A line in the plane is called strange if it passes through $(a, 0)$ and $(0, 10-a)$ for some $a$ in the interval $[0, 10]$. A point in the plane is called charming if it lies in the first quadrant and also lies below some strange line. What is the area of the set of all charming points? | [
"Solution:\nThe strange lines form an envelope (set of tangent lines) of a curve $f(x)$, and we first find the equation for $f$ on $[0, 10]$. Assuming the derivative $f'$ is continuous, the point of tangency of the line $\\ell$ through $(a, 0)$ and $(0, b)$ to $f$ is the limit of the intersection points of this lin... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 50/3 | |
0dfs | Let positive numbers be written along the circle, such that all of them are less than $1$. Prove that one can split the circle to $3$ parts such that for each two arcs the sums of numbers written on them differs by at most $1$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0imx | Problem:
Three real numbers $x$, $y$, and $z$ are such that $(x+4)/2 = (y+9)/(z-3) = (x+5)/(z-5)$. Determine the value of $x/y$. | [
"Solution:\n\nAnswer: $1/2$. Because the first and third fractions are equal, adding their numerators and denominators produces another fraction equal to the others: $((x+4)+(x+5))/(2+(z-5)) = (2x+9)/(z-3)$. Then $y+9 = 2x+9$, etc."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 1/2 | |
09oy | Let $\alpha = \sqrt{60} + \sqrt{61}$.
(1) Prove that $\alpha$ is irrational.
(2) Find an example of polynomial with integer coefficients for which $\alpha$ is a root.
(Batbayasgalan Balkhuu) | [] | Mongolia | MMO2025 Round 2 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Algebraic Number Theory > Algebraic numbers"
] | English | proof and answer | alpha is irrational; an example polynomial with integer coefficients having alpha as a root is x^4 - 242 x^2 + 1. | |
061v | Problem:
Gegeben sind die sechs reellen Zahlen $a, b, c$ und $x, y, z$ so, dass
a) $0 < b - c < a < b + c$
b) $a x + b y + c z = 0$
Man ermittle (mit Begründung!) das Vorzeichen von $a y z + b x z + c x y$. | [
"Solution:\n\nAus $0 < b - c < a < b + c$ folgt, dass $a$, $b$ und $c$ positiv sind. Außerdem kann man mit Strecken dieser Längen ein Dreieck konstruieren. Daraus folgt, dass die Zahlen $-a + b + c$, $a - b + c$ und $a + b - c$ positiv sind.\n\nAus $a x + b y + c z = 0$ folgt $x = -\\frac{b y + c z}{a}$.\n\nEs ergi... | Germany | Auswahlwettbewerb zur IMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof and answer | ≤ 0 | |
06oa | In $\triangle ABC$, $\angle BAC = 150^\circ$ and $BC = 74$. $D$ is a point on $BC$ such that $BD = 14$ and $\angle ADB = 60^\circ$. Find the area of $\triangle ABC$. | [
"Let $b$, $c$ and $d$ be the lengths of $AB$, $AC$ and $AD$ respectively.\nThe area of $\\triangle ABC$ is $\\frac{1}{2}bc\\sin 150^\\circ$. Alternatively, using $BC$ as base, the height of the triangle is $d\\sin 60^\\circ$ and so the area is also equal to $\\frac{1}{2}(74)d\\sin 60^\\circ$. This gives\n$$\n\\frac... | Hong Kong | IMO Preliminary Selection Contest — Hong Kong | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 111√3 | |
0ifm | Problem:
The Red Sox play the Yankees in a best-of-seven series that ends as soon as one team wins four games. Suppose that the probability that the Red Sox win Game $n$ is $\frac{n-1}{6}$. What is the probability that the Red Sox will win the series? | [
"Solution:\n\nNote that if we imagine that the series always continues to seven games even after one team has won four, this will never change the winner of the series. Notice also that the probability that the Red Sox will win Game $n$ is precisely the probability that the Yankees will win Game $8-n$. Therefore, t... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 1/2 | |
0lcz | Let $a$, $b$ and $c$ be non-negative real numbers. Prove that
$$
3(a^2 + b^2 + c^2) \ge (a+b+c)(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) + (a-b)^2 + (b-c)^2 + (c-a)^2 \ge (a+b+c)^2.
$$ | [
"It is well-known that $\\sqrt{ab} + \\sqrt{bc} + \\sqrt{ca} - a - b - c \\le 0$, then we have\n$$\n\\begin{aligned}\n& (a+b+c)(\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ca}) + (a-b)^2 + (b-c)^2 + (c-a)^2 \\\\\n= & (a+b+c)(\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ca}) + 3(a^2+b^2+c^2) - (a+b+c)^2 \\\\\n= & (a+b+c)(\\sqrt{ab}+\\sqrt{bc}+\\s... | Vietnam | VMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
067n | Show that in every triangle there exist some vertex such that with the sides which are concurrent in that vertex, and with any inner cevian passing with this vertex, it is possible to construct a triangle. | [
"We suppose that $\\hat{A} \\le \\hat{B} \\le \\hat{C}$. Let $AM$ be an internal cevian of the triangle passing through $A$. Since $AM\\hat{B} > \\hat{C} \\ge \\hat{B}$, it follows that $AM > AB$. Hence it is enough to prove that: $AB < AC + AM$.\n\nLet $\\hat{C} \\le 90^\\circ$. We will prove that $c < b + h_a$, w... | Greece | Mediterranean Mathematical Competition PETER O' HALLORAN MEMORIAL | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
01ip | Let $n$ be a positive integer. Each cell of an $n \times n$ table is coloured in one of $k$ colours where every colour is used at least once. Two colours $A$ and $B$ are said to touch each other, if there exists a cell coloured in $A$ sharing a side with a cell coloured in $B$. The table is coloured in such a way that ... | [
"$k = 2n-1$ when $n \\neq 2$ and $k = 4$ when $n = 2$.\n$k = 2n - 1$ is possible by colouring diagonally as shown in the figure below and when $n = 2$, $k = 4$ is possible by colouring each cell in a unique colour.\n\n\nWe consider the graph, where each node represents a colour and two node... | Baltic Way | Baltic Way 2023 Shortlist | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory"
] | English | proof and answer | k = 2n − 1 for n ≠ 2, and k = 4 for n = 2 | |
085a | Problem:
Si consideri il piano tassellato con triangoli equilateri, e sia $F_{0}$ uno qualsiasi di essi. Si costruisce una sequenza di figure sempre più grandi in questo modo: $F_{1}$ è il poligono che si ottiene aggiungendo ad $F_{0}$ la cornice formata da tutti i triangoli della tassellazione che toccano $F_{0}$ (pe... | [
"Solution:\n\nLa risposta è (B). Per $n \\geq 1$ la figura $F_{n}$ è un esagono avente tre lati lunghi e tre lati corti, alternati fra loro. Ciascuno di quelli lunghi è composto da $n+1$ lati di triangoli, mentre ciascuno di quelli corti è composto da $n$ lati di triangoli. Detto $x_{n}$ il numero dei triangoli del... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | B | |
0gxd | Solve the system of equations:
$$
\begin{cases}
\frac{9}{2(x+y)} = \frac{1}{x} + \frac{1}{y}, \\
\sqrt{x^2 - 2} = \sqrt{3 - y^2}.
\end{cases}
$$ | [
"Consider the first equations of the system. It is easy to see that $xy \\neq 0$, and we\n$$\n\\text{get } \\frac{9}{2(x+y)} = \\frac{x+y}{xy} \\Leftrightarrow 9xy = 2x^2 + 4xy + 2y^2 \\Leftrightarrow 2y^2 - 5xy + 2x^2 = 0\n$$\n\n$y = \\frac{5x \\pm \\sqrt{25x^2 - 16x^2}}{4} = \\frac{5x \\pm 3x}{4}$. It follows tha... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Algebra > Intermediate Algebra > Other",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | [(2, 1), (-2, -1)] | |
00cr | Los números naturales desde $1$ hasta $300$ inclusive se ubican alrededor de una circunferencia. Decimos que un tal ordenamiento es *alternado* si cada número es menor que sus dos vecinos o es mayor que sus dos vecinos. A un par de números vecinos lo llamaremos *par bueno* si al quitar ese par de la circunferencia, los... | [] | Argentina | Nacional OMA 2019 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | 150 | |
0ihi | Problem:
All the chairs in a classroom are arranged in a square $n \times n$ array (in other words, $n$ columns and $n$ rows), and every chair is occupied by a student. The teacher decides to rearrange the students according to the following two rules:
a. Every student must move to a new chair.
b. A student can only... | [
"Solution:\n\nIf $n$ is even, there are many ways to do this. One is simply to exchange adjacent students in each row: the student in an odd-numbered chair $k$ exchanges places with the student in chair $k+1$. In other words, exchange students in chairs 1 and 2, those in chairs 3 and 4, and so on. Since there are a... | United States | Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
015h | Let $n$ be a positive integer. Consider sums
$$
S_n = \sum_{m=0}^{n} (-1)^{j_m} m^2,
$$
where $j_m \in \{0, 1\}$. Show that it is always possible to choose the numbers $j_m$ in such a way that $0 \le S_n \le 4$. | [
"Since\n$$\n(n + 2)^2 - (n + 1)^2 - n^2 + (n - 1)^2 = 4,\n$$\nwe can always choose 8 consecutive indices such that the corresponding terms in the sum add up to zero. As the first term is a zero anyway, it suffices to note that $1^2 = 1$, $2^2 - 1^2 = 3$, $3^2 - 2^2 - 1^2 = 4$, $4^2 - 3^2 - 2^2 - 1^2 = 2$, $5^2 - 4^... | Baltic Way | Baltic Way SHL | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0djz | Find all non-constant polynomials $P(x)$ with real coefficients that satisfy
$$
P(x^3 - 7x) = P(x - 7)P(x - 8)P(x - 3)
$$
for all $x \in \mathbb{R}$. | [
"Let $n = \\deg P(x) \\ge 1$ and $a \\ne 0$ be the leading coefficient of $P$, then comparing the leading coefficients of both sides to get $a = \\pm 1$. Note that if $P(x)$ satisfies then so does $-P(x)$; without loss of generality, we assume that $a = 1$.\n\nFirst, let solve the problem when $n = 1$, consider $P(... | Saudi Arabia | Saudi Arabia booklet 2024 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | P(x) = ±(x + 6)^n for any positive integer n |
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