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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0asd | Problem:
Find the values of $a$ and $b$ such that $a x^{4} + b x^{2} + 1$ is divisible by $x^{2} - x - 2$. | [
"Solution:\n$a = \\frac{1}{4},\\ b = -\\frac{5}{4}$"
] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a = 1/4, b = -5/4 | |
0im2 | Problem:
Define $x \star y = \frac{\sqrt{x^{2} + 3 x y + y^{2} - 2 x - 2 y + 4}}{x y + 4}$. Compute
$$
((\cdots((2007 \star 2006) \star 2005) \star \cdots) \star 1).
$$ | [
"Solution:\n\nAnswer: $\\frac{\\sqrt{15}}{9}$.\n\nNote that\n$$\nx \\star 2 = \\frac{\\sqrt{x^{2} + 6x + 4 - 2x - 4 + 4}}{2x + 4} = \\frac{\\sqrt{(x+2)^{2}}}{2(x+2)} = \\frac{1}{2}\n$$\nfor $x > -2$.\n\nBecause $x \\star y > 0$ if $x, y > 0$, we need only compute\n$$\n\\frac{1}{2} \\star 1 = \\frac{\\sqrt{\\frac{1}... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | proof and answer | sqrt(15)/9 | |
0dok | Observe that the fraction $1/7 = 0.\overline{142857}$ is a pure periodical decimal with period $6 = 7 - 1$, and in one period one has $142 + 857 = 999$. For $n = 1, 2, \dots$, find a sufficient and necessary condition that the fraction $1/(2n+1)$ has the same properties as above and find two such fractions other than $... | [
"Suppose $1/(2n+1) = \\dot{a}_1 a_2 \\cdots a_n b_1 b_2 \\cdots \\dot{b}_n$, $a_i + b_i = 9$, $i = 1, 2, \\dots, n$, and $2n$ is its period. Since\n$$\n1 + \\frac{1}{10^{2n}} + \\frac{1}{10^{4n}} + \\dots = \\frac{1}{1 - \\frac{1}{10^{2n}}} = \\frac{10^{2n}}{10^{2n} - 1}\n$$\nwe have\n$$\n\\begin{align*}\n(1) \\qua... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | Necessary and sufficient condition: 2n+1 divides 10^n + 1 and 2n+1 does not divide 10^i + 1 for any i with 1 ≤ i < n. Examples: 1/17 and 1/19. | |
0kb8 | Let $ABC$ be a triangle with incenter $I$. Points $K$ and $L$ are chosen on segment $BC$ such that the incircles of $\triangle ABK$ and $\triangle ABL$ are tangent at $P$, and the incircles of $\triangle ACK$ and $\triangle ACL$ are tangent at $Q$. Prove that $IP = IQ$. | [
"Let $I_B, J_B, I_C, J_C$ be the incenters of $\\triangle ABK$, $\\triangle ABL$, $\\triangle ACK$, $\\triangle ACL$ respectively.\n\n\n\nWe begin with the following claim which does not depend on the existence of tangency points $P$ and $Q$.\n**Claim** — Lines $BC$, $I_BJ_C$, $J_BI_CI_C$ m... | United States | USA TSTST | [
"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",
... | null | proof only | null | |
06lb | Let $\triangle ABC$ be an acute triangle with incentre $I$ and orthocentre $H$. $AI$ meets the circumcircle of $\triangle ABC$ again at $M$. Suppose the length $IM$ is exactly the circumradius of $\triangle ABC$. Show that $AH \ge AI$. | [
"Let $O$ be the circumcentre of $\\triangle ABC$. Since $BM = CM = IM = OM$ (where $OM$ is the circumradius of $\\triangle ABC$), $B$, $C$, $I$, $O$ are concyclic. Therefore, we have $\\angle BIC = \\angle BOC$. This yields $90^\\circ + \\frac{A}{2} = 2A$, and hence $A = 60^\\circ$. Now\n$$\n\\angle BHC = 180^\\cir... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric In... | null | proof only | null | |
08l1 | Problem:
If for the real numbers $x$, $y$, $z$, $k$ the following conditions are valid, $x \neq y \neq z \neq x$ and
$$
x^{3} + y^{3} + k(x^{2} + y^{2}) = y^{3} + z^{3} + k(y^{2} + z^{2}) = z^{3} + x^{3} + k(z^{2} + x^{2}) = 2008,
$$
find the product $x y z$. | [
"Solution:\n$x^{3} + y^{3} + k(x^{2} + y^{2}) = y^{3} + z^{3} + k(y^{2} + z^{2}) \\Rightarrow x^{2} + xz + z^{2} = -k(x + z)$ \\:(1)\nand $y^{3} + z^{3} + k(y^{2} + z^{2}) = z^{3} + x^{3} + k(z^{2} + x^{2}) \\Rightarrow y^{2} + yx + x^{2} = -k(y + x)$ \\:(2)\n\n- From (1) $-(2) \\Rightarrow x + y + z = -k$ \\:(*)\n... | JBMO | 2008 Shortlist JBMO | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1004 | |
04r6 | There are two touching circles, $k_1(S_1, r_1)$ and $k_2(S_2, r_2)$ in a rectangle $ABCD$ with $|AB| = 9$, $|BC| = 8$. Moreover, $k_1$ touches $AD$ and $CD$, while $k_2$ touches $AB$ and $BC$.
a) Prove $r_1 + r_2 = 5$.
b) What is the least and what is the greatest possible area of $AS_1S_2$? | [
"a) Let $M$ and $N$ be intersections of the line through $S_1$ parallel to $AD$. Analogously, let $K$ and $L$ be intersections of the line through $S_2$ parallel to $AB$. Let $P$ be the intersection of $KL$ and $MN$ (see Fig. 1). The Pythagoras theorem for $S_1PS_2$ gives\n$$\n(r_1 + r_2)^2 = (8 - r_1 - r_2)^2 + (9... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | r1 + r2 = 5; minimum area = 14; maximum area = 31/2 | |
06r7 | Let $A_{1} A_{2} \ldots A_{n}$ be a convex polygon. Point $P$ inside this polygon is chosen so that its projections $P_{1}, \ldots, P_{n}$ onto lines $A_{1} A_{2}, \ldots, A_{n} A_{1}$ respectively lie on the sides of the polygon. Prove that for arbitrary points $X_{1}, \ldots, X_{n}$ on sides $A_{1} A_{2}, \ldots, A_{... | [
"Denote $P_{n+1}=P_{1}, X_{n+1}=X_{1}, A_{n+1}=A_{1}$.\n\nLemma. Let point $Q$ lies inside $A_{1} A_{2} \\ldots A_{n}$. Then it is contained in at least one of the circumcircles of triangles $X_{1} A_{2} X_{2}, \\ldots, X_{n} A_{1} X_{1}$.\n\nProof. If $Q$ lies in one of the triangles $X_{1} A_{2} X_{2}, \\ldots, X... | IMO | 51st IMO Shortlisted Problems | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0alg | Problem:
Find the sum of
$$
\frac{1}{2 \times 5}+\frac{1}{5 \times 8}+\frac{1}{8 \times 11}+\cdots+\frac{1}{2009 \times 2012}
$$
(a) $\frac{335}{2012}$
(b) $\frac{545}{2012}$
(c) $\frac{865}{2012}$
(d) $\frac{1005}{2012}$ | [] | Philippines | QUALIFYING STAGE | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | a | |
01k2 | Prove that $|a\sqrt{3} - b\sqrt{5}| > \frac{4}{4a + 5b}$ for any naturals $a$ and $b$.
(V. Bernik) | [
"Multiplying the left-hand side of the required inequality by $a\\sqrt{3} + b\\sqrt{5}$, we obtain $|a\\sqrt{3} - b\\sqrt{5}|(a\\sqrt{3} + b\\sqrt{5}) = |3a^2 - 5b^2|$. If $|3a^2 - 5b^2| = 1$ for some integers $a$ and $b$, then either $3a^2 - 5b^2 = 1$ or $3a^2 - 5b^2 = -1$. In the first case we have $3a^2 \\equiv ... | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
02hd | Given positive integers $a$, $c$ and integer $b$, prove that there exists a positive integer $x$ such that
$$
a^x + x \equiv b \pmod{c},
$$
that is, there exists a positive integer $x$ such that $c$ is a divisor of $a^x + x - b$. | [
"Let $\\ell$ be the length of the cycle of the sequence $a, a^2, a^3, \\dots$ mod $c$. Thus\n$$\na^{\\ell + \\ell k} \\equiv a^\\ell \\pmod{c} \\quad (*)\n$$\nfor all positive integer $k$ and $\\ell$ large enough.\n\nLet $d = \\gcd(\\ell, c)$. The multiples of $\\ell$ mod $c$ are the multiples of $d$. Let's prove t... | Brazil | Brazil | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0bzg | Given an integer $n \ge 2$, determine the maximum value the sum
$$
a_1/a_2 + a_2/a_3 + \dots + a_{n-1}/a_n
$$
may achieve, and the points at which the maximum is achieved, as $a_1, \dots, a_n$ run over all of positive real numbers subject to
$$
a_k \ge a_1 + \dots + a_{k-1}, \quad k = 2, \dots, n.
$$ | [
"*First solution.* The required maximum is $n/2$ and is achieved if and only if $a_k = 2^{k-2}a_1$, $k = 2, \\dots, n$, and $a_1$ is any positive real number.\nTo prove this, let $A_k = a_1 + \\dots + a_k$, $k = 0, \\dots, n-1$, where empty sums are zero, and refer to the condition in the statement, $A_k \\le a_{k+... | Romania | THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Maximum value: n/2. Equality holds if and only if a_k = 2^{k-2} a_1 for k = 2, ..., n, with any a_1 > 0. | |
0b7m | Let $ABC$ be a scalene triangle. The tangents to the nine-point circle at the foot of the perpendicular dropped from $A$ on the line $BC$ and at the midpoint of the side $BC$ meet at the point $A'$; the points $B'$ and $C'$ are defined similarly. Prove that the lines $AA'$, $BB'$ and $CC'$ are concurrent.
Gazeta Matem... | [
"The tangent at $A$ to the circumcircle $ABC$ meets the line $BC$ at the point $A''$; the points $B''$ and $C''$ are defined similarly. The points $A''$, $B''$ and $C''$ are collinear on Lemoine's line. We shall prove that the lines $AA'$, $BB'$ and $CC'$ are the polars of the points $A''$, $B''$ and $C''$, respect... | Romania | NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry >... | English | proof only | null | |
0is7 | Problem:
The function $f$ satisfies
$$
f(x)+f(2x+y)+5xy = f(3x-y)+2x^{2}+1
$$
for all real numbers $x, y$. Determine the value of $f(10)$. | [
"Solution:\n\nSetting $x=10$ and $y=5$ gives $f(10)+f(25)+250=f(25)+200+1$, from which we get $f(10)=-49$."
] | United States | 11th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | -49 | |
0j8g | Problem:
Determine the number of angles $\theta$ between $0$ and $2\pi$, other than integer multiples of $\pi/2$, such that the quantities $\sin \theta$, $\cos \theta$, and $\tan \theta$ form a geometric sequence in some order. | [
"Solution:\nIf $\\sin \\theta$, $\\cos \\theta$, and $\\tan \\theta$ are in a geometric progression, then the product of two must equal the square of the third. Using this criterion, we have 3 cases.\n\n- Case 1: $\\sin \\theta \\cdot \\tan \\theta = \\cos^{2} \\theta$. This implies that $\\left(\\sin^{2} \\theta\\... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem"
] | null | proof and answer | 4 | |
04ta | Find all positive integers $n$ with the following property: Numbers $1, 2, \ldots, n$ can be split into three disjoint non-empty subsets with mutually different sizes such that, for any pair of subsets, the subset with fewer elements has larger sum of its elements.
(A size of a subset is the number of its elements.) | [
"We first exclude small values of $n$.\nThe three subsets have to have in total at least $1 + 2 + 3 = 6$ elements, hence $n \\ge 6$.\nFor $n=6$, the smallest subset (in size) contains a single number and its sum is therefore at most $6$. The sum of the remaining numbers is at least $15$, hence the sum of at least o... | Czech Republic | 66th Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | n = 9 or n ≥ 12 | |
0ilr | Problem:
Find all pairs $(n, k)$ of positive integers such that
$$
sigma(n) \phi(n) = \frac{n^{2}}{k}
$$ | [
"Solution:\nAnswer: $(1, 1)$.\n\nIt is clear that for a given integer $n$, there is at most one integer $k$ for which the equation holds. For $n=1$ this is $k=1$. But, for $n>1$, problem 1 asserts that $\\sigma(n) \\phi(n) \\leq n^{2}-1 < n^{2}$, so that $k \\geq 2$. We now claim that $2 > \\frac{n^{2}}{\\sigma(n) ... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebrai... | null | proof and answer | (1, 1) | |
0idg | Problem:
A committee of 5 is to be chosen from a group of 9 people. How many ways can it be chosen, if Bill and Karl must serve together or not at all, and Alice and Jane refuse to serve with each other? | [
"Solution:\nIf Bill and Karl are on the committee, there are $\\binom{7}{3} = 35$ ways for the other group members to be chosen. However, if Alice and Jane are on the committee with Bill and Karl, there are $\\binom{5}{1} = 5$ ways for the last member to be chosen, yielding 5 unacceptable committees.\n\nIf Bill and... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 41 | |
0gm5 | Let $f(x,y)$ and $g(x,y)$ be real-valued functions defined for all $x, y \in \{1, 2, ..., 2000\}$. If one can find two 1000-element subsets $X, Y$ of $1, 2, ..., 2000$ such that $f(x, y) = g(x, y)$ whenever $x \notin X$ and $y \notin Y$, find the maximal number of ordered pairs $(x, y)$ at which $f$ and $g$ can take di... | [] | Turkey | VIIth NATIONAL MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 3000000 | |
0e1n | Find all prime numbers $p$, $q$ and $r$ such that $p > q > r$ and the numbers $p-q$, $p-r$ and $q-r$ are also prime. | [
"We have $r \\ge 2$, so $p$ and $q$ must be odd primes. Thus, $p-q$ is even and equal to $2$. We get $p-q=2$. The numbers $p-r=q-r+2$ and $q-r$ are prime and differ by $2$, so they have the same parity. We conclude that both must be odd. Since $q$ and $q-r$ are odd, $r$ must be even. Thus, $r=2$.\n\nThe numbers $q$... | Slovenia | National Math Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | p=7, q=5, r=2 | |
0ajj | Нека $ABC$ е остроаголен триаголник и нека $H$ е неговиот ортоцентар. Точката $G$ припаѓа на рамнината на триаголникот при што $ABGH$ е паралелограм. Точката $I$ припаѓа на правата $GH$ така што правата $AC$ ја полови отсечката $HI$. Правата $AC$ ја сече опишаната кружница околу триаголникот $GCI$ по вторпат во точката... | [
"Бидејќи $HG \\parallel AB$ и $BG \\parallel AH$, добиваме дека $BG \\perp BC$ и $CH \\perp GH$. Според тоа, четириаголникот $BGCH$ е тетивен. Бидејќи $H$ е ортоцентар на триаголникот $ABC$, добиваме дека $\\angle HAC = 90^\\circ - \\angle ACB = \\angle CBH$. Бидејќи $BGCH$ и $CGJI$ се тетивни четириаголници, добив... | North Macedonia | IMO Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
0gf0 | 對於所有正整數 $n$, 令 $V_n = \lfloor 2^n \sqrt{2020} \rfloor + \lfloor 2^n \sqrt{2021} \rfloor$。試證數列 $V_1, V_2, \dots$ 中有無窮多個奇數, 也有無窮多個偶數。
註. $\lfloor x \rfloor$ 代表不超過實數 $x$ 的最大整數。
For each positive integer $n$, define $V_n = \lfloor 2^n \sqrt{2020} \rfloor + \lfloor 2^n \sqrt{2021} \rfloor$. Prove that, in the sequence $V_1... | [
"令 $a_n = 1_{\\{\\{2^n\\sqrt{2020}\\}\\} > \\frac{1}{2}}$, $b_n = 1_{\\{\\{2^n\\sqrt{2021}\\}\\} > \\frac{1}{2}}$。假設在某項 $N$ 之後 $V_n$ 的奇偶性皆相同, 則 $V_{n+1} - 2V_n = a_n + b_n$ 的奇偶性也必然相同。這表示 $a_n = b_n \\forall n \\ge N$, 或 $a_n = 1 - b_n \\forall n \\ge N$。換言之, 在二進位下, $2^N\\sqrt{2020}$ 與 $2^N\\sqrt{2021}$ 要不小數部分相同, 要不... | Taiwan | 2021 數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Other"
] | null | proof only | null | |
04vt | Suppose that a point $P$ lying in the interior of a convex quadrilateral $ABCD$ satisfies
$$
\angle PAD = \angle ADP = \angle CBP = \angle PCB = \angle CPD.
$$
Let $O$ be the circumcentre of the triangle $CPD$. Prove that $OA = OB$. | [
"From the given equalities, one sees that $PC \\parallel AD$ and $PD \\parallel BC$. Since the lines $PC$ and $PD$ are distinct (we know that $\\angle CPD \\neq 0$), the lines $AD$ and $BC$ are not parallel, so they intersect at a unique point $X$ such that $PCXD$ is a rhombus.\n\nNow, note that the quadrilateral $... | Czech Republic | Final Round of the 73rd Czech and Slovak Mathematical Olympiad (March 17–20, 2024) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
07ny | Suppose $a$, $b$, and $c$ are real numbers, with $a \neq b$. Prove that the minimum value of the quadratic polynomial
$$
p(x) = 3x^2 + 2c x - (c(a+b) + (a^2 + ab + b^2)), \quad -\infty < x < \infty,
$$
is negative. Prove also that $p(a) + p(b)$ is positive. | [
"$$\n\\begin{align*}\n3p(x) &= (3x)^2 + 2c(3x) - (3c(a+b) + 3(a^2 + ab + b^2)) \\\\\n&= (3x + c)^2 - (c^2 + 3c(a+b) + 3(a^2 + ab + b^2)) \\\\\n&= (3x + c)^2 - \\left( \\left(c + \\frac{3}{2}(a+b)\\right)^2 - \\frac{9}{4}(a+b)^2 + 3(a^2 + ab + b^2) \\right) \\\\\n&= (3x + c)^2 - \\left( c + \\frac{3}{2}(a+b) \\right... | Ireland | Ireland | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0fdh | Problem:
Sea $P$ un punto del lado $BC$ de un triángulo $ABC$. La paralela por $P$ a $AB$ corta al lado $AC$ en el punto $Q$ y la paralela por $P$ a $AC$ corta al lado $AB$ en el punto $R$. La razón entre las áreas de los triángulos $RBP$ y $QPC$ es $k^2$.
Determínese la razón entre las áreas de los triángulos $ARQ$ y ... | [
"Solution:\n\nLos triángulo $RBP$ y $QPC$ son semejantes, de razón $k$. El cuadrilátero $ARPQ$ es un paralelogramo, y $PQ = RA$.\nSi $BR = x$, entonces\n$$\n\\begin{gathered}\nPQ = RA = kx ; \\quad BA = (1+k)x \\\\\n\\text{Área } RBP = S = \\frac{BR \\cdot PX}{2} \\\\\nPX = \\frac{x h}{2}\n... | Spain | null | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | k/(1+k)^2 | |
0je3 | Problem:
Find the sum of $\frac{1}{n}$ over all positive integers $n$ with the property that the decimal representation of $\frac{1}{n}$ terminates. | [
"Solution:\nThe decimal representation of $\\frac{1}{n}$ terminates if and only if $n = 2^{i} 5^{j}$ for some nonnegative integers $i, j$, so our desired sum is\n$$\n\\sum_{i \\geq 0} \\sum_{j \\geq 0} 2^{-i} 5^{-j} = \\sum_{i \\geq 0} 2^{-i} \\sum_{j \\geq 0} 5^{-j} = \\left(1 - 2^{-1}\\right)^{-1} \\left(1 - 5^{-... | United States | HMMT November 2013 | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 5/2 | |
029z | Problem:
Um quadriculado - O retângulo quadriculado na figura é feito de 31 segmentos de $0{,}5~\mathrm{cm}$, e compreende 12 quadrados. Rosa desenhou numa folha retangular de $21~\mathrm{cm}$ por $29{,}7~\mathrm{cm}$ quadriculada com quadrados de lado $0{,}5~\mathrm{cm}$, um grande retângulo quadriculado feito com 19... | [
"Solution:\n\nSejam $m$ e $n$ respectivamente, o número de segmentos de $0{,}5~\\mathrm{cm}$ sobre dois lados consecutivos do retângulo. Sabemos que o número total de segmentos de $0{,}5~\\mathrm{cm}$ na divisão do retângulo em $m \\times n$ quadrados de lado $0{,}5~\\mathrm{cm}$ é: $m(n+1)+n(m+1)$ (prove isso). As... | Brazil | Nível 2 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 798, 936, or 966 | |
0dup | Problem:
Ali obstaja funkcija $f: \mathbb{N} \rightarrow \mathbb{N}$, da bo
$$
f(f(2002))=17, \quad f(m n)=f(m) f(n) \quad \text{ in } \quad f(n) \leq n
$$
za vsaka $m, n \in \mathbb{N}$ ? | [
"Solution:\nKer je $2002=2 \\cdot 7 \\cdot 11 \\cdot 13$, z upoštevanjem zveze $f(x y)=f(x) f(y)$ izpeljemo\n$$\nf(f(2002))=f(f(2)) \\cdot f(f(7)) \\cdot f(f(11)) \\cdot f(f(13))=17\n$$\nKer je $f(x) \\leq x$, je $f(f(x)) \\leq f(x) \\leq x$, in zato $f(f(2)) \\leq 2, f(f(7)) \\leq 7, f(f(11)) \\leq 11$ in $f(f(13)... | Slovenia | 46. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | No, such a function does not exist. | |
01jg | In an acute triangle $\triangle ABC$ with $|AB| \neq |AC|$, the perpendicular bisectors of sides $AC$ and $AB$ intersect segment $BC$ at points $D$ and $E$, respectively. The tangents to $\odot(ABC)$ at the points $B$ and $C$ intersect $\odot(ABD)$ and $\odot(ACE)$ at points $Y$ and $Z$, respectively. Suppose that line... | [
"Let $O$ be the circumcentre of $\\odot(ABC)$. We proceed in several steps.\n\n**Step 1:** Point $O$ lies on $\\odot(ABD)$ and $\\odot(ACE)$.\n*Proof.* Note that since $|AD| = |DC|$, we have $\\angle CDA = 180^\\circ - 2\\angle ACB$, therefore $\\angle ADB = 2\\angle ACB = \\angle AOB$, which means that quadrilater... | Baltic Way | Baltic Way 2023 Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Colline... | English | proof only | null | |
0it8 | Problem:
Find $p$ so that $\lim_{x \rightarrow \infty} x^{p}(\sqrt[3]{x+1}+\sqrt[3]{x-1}-2 \sqrt[3]{x})$ is some non-zero real number. | [
"Solution:\nAnswer: $\\frac{5}{3}$\n\nMake the substitution $t=\\frac{1}{x}$. Then the limit equals to\n$$\n\\lim_{t \\rightarrow 0} t^{-p}\\left(\\sqrt[3]{\\frac{1}{t}+1}+\\sqrt[3]{\\frac{1}{t}-1}-2 \\sqrt[3]{\\frac{1}{t}}\\right)=\\lim_{t \\rightarrow 0} t^{-p-\\frac{1}{3}}(\\sqrt[3]{1+t}+\\sqrt[3]{1-t}-2)\n$$\nW... | United States | Harvard-MIT Mathematics Tournament | [
"Precalculus > Limits",
"Calculus > Differential Calculus > Other"
] | null | proof and answer | 5/3 | |
0b9j | Prove that if $n$ and $p$ are integers, $1 < p < n$, then the number
$$
(2p+1)n^3 + 6n(1^2 + 2^2 + 3^2 + \dots + p^2)
$$
can be written as the sum of $2p + 1$ different perfect cubes. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0jzz | Problem:
An ordered pair of sets $(A, B)$ is good if $A$ is not a subset of $B$ and $B$ is not a subset of $A$. How many ordered pairs of subsets of $\{1,2, \ldots, 2017\}$ are good? | [
"Solution:\n\nFirstly, there are $4^{2017}$ possible pairs of subsets, as each of the $2017$ elements can be in neither subset, in $A$ only, in $B$ only, or in both.\n\nNow let us count the number of pairs of subsets for which $A$ is a subset of $B$. Under these conditions, each of the $2017$ elements could be in n... | United States | February 2017 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 4^2017 - 2*3^2017 + 2^2017 | |
0do1 | Problem:
У равни је дато $n$ правих међу којима никоје две нису паралелне и никоје три се не секу у једној тачки. Под пресечним тачкама сматрамо све тачке у којима се секу неке две од ових правих.
a. Доказати да међу датим правим постоји једна са чије се сваке стране налази бар по
$$
\left\lfloor\frac{(n-1)(n-2)}{10}... | [
"Solution:\n\nЗа сваку од датих правих, пресечних тачака ван ње има\n$$\nm=\\frac{(n-1)(n-2)}{2}\n$$\nПретпоставимо да, за сваку праву, с једне њене стране има не више од $k$ пресечних тачака, где је $k \\leqslant \\frac{n}{2}$. Пребројмо на два начина тројке $(p, A, B)$, где су $A$ и $B$ пресечне тачке са различит... | Serbia | 12. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | n ≤ 7 | |
00zf | Problem:
Four heaps contain $38$, $45$, $61$, and $70$ matches respectively. Two players take turns choosing any two of the heaps and take some non-zero number of matches from one heap and some non-zero number of matches from the other heap. The player who cannot make a move, loses. Which one of the players has a winn... | [
"Solution:\n\nThe first player wins by making moves so that the opponent must face positions of the form $(a, a, a, b)$, where $a \\leq b$."
] | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | First player | |
0cbr | A diagonal line of a (not necessarily convex) polygon with at least four sides is any line through two non-adjacent vertices of that polygon. Determine all polygons with at least four sides satisfying the following condition: The reflexion of each vertex in each diagonal line lies inside or on the boundary of the polyg... | [
"Begin by noticing that $K$ is convex: Otherwise, the convex hull $\\hat{K}$ of $K$ would have a side $ab$ whose line of support is a diagonal line of $K$ ($a$ and $b$ are, of course, non-adjacent vertices of $K$ and there might virtually be other vertices or even sides of $K$ along the line segment $ab$). The refl... | Romania | THE 73rd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS - FIRST SELECTION TEST | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Transformations",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | Rhombi (convex quadrilaterals with all sides equal). | |
0jdz | Problem:
Tim and Allen are playing a match of tenus. In a match of tenus, the two players play a series of games, each of which is won by one of the two players. The match ends when one player has won exactly two more games than the other player, at which point the player who has won more games wins the match. In odd-... | [
"Solution:\n\nLet the answer be $E$. If Tim wins the first game and Allen wins the second game or vice versa, which occurs with probability $(3/4)^2 + (1/4)^2 = 5/8$, the expected number of additional games is just $E$, so the expected total number of games is $E + 2$. If, on the other hand, one of Tim and Allen wi... | United States | HMMT 2013 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | 16/3 | |
05q0 | Problem:
Prouver qu'il existe une infinité d'entiers $n$ tels que $2^{2^{n}+1}+1$ est divisible par $n$, mais $2^{n}+1$ ne l'est pas. | [
"Solution:\n\nPour $m \\geqslant 1$, posons $a_{m}=2^{3^{m}}+1$. On a\n$$\na_{m}=\\left(2^{3^{m-1}}+1\\right)\\left(\\left(2^{3^{m-1}}\\right)^{2}-2^{3^{m-1}}+1\\right)=a_{m-1}\\left(\\left(2^{3^{m-1}}\\right)^{2}-2^{3^{m-1}}+1\\right)\n$$\nNotons que pour $a \\in \\mathbb{N}^{*}$, si $p \\mid a+1$ et $p \\mid a^{2... | France | OCympiades Françaises de Mathématiques | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof only | null | |
0gm1 | Find the minimum value of $\frac{a^3 + b^3 + c^3}{abc}$ when $a$, $b$, and $c$ are sides of a right triangle. | [
"Assume without loss of generality that $a^2 + b^2 = c^2$. By the AM-GM inequality,\n$$\nc^2 = a^2 + b^2 \\ge 2ab.\n$$\nThat is,\n$$\nc^3 \\ge 2abc. \\quad (1)\n$$\nNext, we will show that $a^3 + b^3 \\ge \\sqrt{2}abc$. Consider $a^3 + b^3 = (a+b)(a^2 + b^2) - ab(a+b)$ and\n$$\n\\begin{align*}\n(a+b)(a^2 + b^2) &= ... | Thailand | The 14th Thailand Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | 2 + sqrt(2) | |
0f6s | Problem:
The quadratic $x^{2} + a x + b + 1$ has roots which are positive integers. Show that $(a^{2} + b^{2})$ is composite. | [
"Solution:\n\nLet the roots be $c$, $d$, so $c + d = -a$, $cd = b+1$. Hence $a^{2} + b^{2} = (c^{2} + 1)(d^{2} + 1)$."
] | Soviet Union | 20th ASU | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0af1 | Даден е $\triangle ABC$ во кој должините на страните се последователни природни броеви. Тежишната линија од темето $A$ е нормална на симетралата на аголот кај темето $B$. Да се пресмета периметарот на триаголникот $\triangle ABC$.
 | [
"Нека $D$ и $E$ се пресечните точки на тежишната линија од $A$ и симетралата на аголот од $B$, со страните $BC$ и $AC$ соодветно.\nОд $\\triangle ABD$ следува $\\overline{AB} = \\overline{BD}$, бидејќи $BE$ е симетрала на $\\angle B$, а ја сече $AD$ под прав агол. Значи $\\overline{BC} = 2 \\cdot \\overline{AB}$. Б... | North Macedonia | Републички натпревар по математика за основно образование | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Macedonian, English | proof and answer | 9 | |
07sk | Let $AB$ be a chord of a circle $\Omega$ with centre $M$, and $P$ a point on the segment $AB$. Circles $\Omega_A$ and $\Omega_B$ pass through $P$ and are tangent to $\Omega$ at $A$ and $B$, respectively. Let $Q$ be the second intersection point of $\Omega_A$ and $\Omega_B$. Prove that $\angle PQM$ is a right angle. | [
"Let $T$ be the intersection point of the tangents to $\\Omega$ at $A$ and $B$. Because of the right angles at $A$ and $B$, the points $A$, $M$, $B$, $T$ all lie on the circle with diameter $TM$. Because $|TA| = |TB|$ and these lines are tangents to $\\Omega_A$ and $\\Omega_B$, respectively, $T$ must be on the radi... | Ireland | IRL_ABooklet_2020 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00ud | Prove that there exists a real number $c < 3/4$ such that for each sequence $\{x_i\}_{i=1}^\infty$ satisfying $0 \le x_i \le 1$ for all $i$, there are infinitely many pairs $(m, n)$ with $m > n$ such that
$$
|x_m - x_n| \le \frac{c}{m}.
$$ | [
"Assume on the contrary that for every $c < 3/4$ there exists a sequence $\\{x_i\\}_{i=1}^\\infty$ such that there are finitely many such $(m, n)$ pairs. So, there exists a positive integer $N$ such that $|x_m - x_n| > \\frac{c}{m}$ for all $m > n > N$. Consider $x_{N+1}, x_{N+2}, \\dots, x_{N+K}$ where $K$ is a su... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0goq | Let $D$ be a point on the side $[BC]$ of the triangle $ABC$ different from the vertices and $E$ be the midpoint of $[CD]$. The line perpendicular to $BC$ at $E$ intersects the side $[AC]$ at point $F$ satisfying $AF \cdot BC = AC \cdot EC$. Let $G$ be the second point where the circumcircle of the triangle $ADC$ inters... | [
"We will show that $EF$ is a common tangent of the circumcircles of $AGF$ and $BGE$. Note that it is enough to show that $\\angle GBE = \\angle GEF$ and $\\angle GAF = \\angle GFE$.\n\nLet $H$ be the point of intersection of $AB$ and the line passing through $F$ parallel to $BC$.\n\nThen $AHF \\sim ABC$ and $\\frac... | Turkey | 19th Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0cec | The sum of seven consecutive integers equals $2023^n$. Find the positive integers $n$ such that at least one summand is a perfect square. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | all odd positive integers | |
0ldz | A farmer has 2 rectangle lands of size $120 \text{ m} \times 100 \text{ m}$.
a. On the first land, there are 9 circle gardens of diameter $5 \text{ m}$. Prove that regardless to the position of gardens, he always can builds a rectangle garden of size $25 \text{ m} \times 35 \text{ m}$.
b. On the second land, he build... | [
"a. Consider the rectangle $ABCD$ with $AB = CD = 120$ and $AD = BC = 100$. Divide it into 10 subrectangles of size $30 \\times 40$ as follows.\n\n\n\nConsider 9 centers of the given gardens. By the pigeonhole principle, there is some subrectangle that does not contain any point among them.... | Vietnam | VN IMO Booklet | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0ddb | Let $p$ be an odd prime and let $N = \frac{1}{4}(p^3 - p) - 1$. The numbers $1, 2, \dots, N$ are painted arbitrarily in two colors, red and blue. For any positive integer $n \le N$, denote by $r(n)$ the fraction of integers in $\{1, 2, \dots, n\}$ that are red (number of red numbers divided by $n$). Prove that there ex... | [
"Denote by $R(n)$ the number of red numbers in $\\{1, 2, \\dots, n\\}$, i.e., $R(n) = n r(n)$. Similarly, denote by $B(n)$ and $b(n) = B(n)/n$ the number and proportion of blue numbers in $\\{1, 2, \\dots, n\\}$, respectively. Notice that $B(n) + R(n) = n$ and $b(n) + r(n) = 1$. Therefore, the statement of the prob... | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
06sk | Let $\omega$ be the circumcircle of a triangle $A B C$. Denote by $M$ and $N$ the midpoints of the sides $A B$ and $A C$, respectively, and denote by $T$ the midpoint of the $\operatorname{arc} B C$ of $\omega$ not containing $A$. The circumcircles of the triangles $A M T$ and $A N T$ intersect the perpendicular bisect... | [
"Let $O$ be the center of $\\omega$, thus $O = M Y \\cap N X$. Let $\\ell$ be the perpendicular bisector of $A T$ (it also passes through $O$). Denote by $r$ the operation of reflection about $\\ell$. Since $A T$ is the angle bisector of $\\angle B A C$, the line $r(A B)$ is parallel to $A C$. Since $O M \\perp A B... | IMO | International Mathematical Olympiad Shortlisted Problems | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
03s5 | Let $m$ and $n$ be positive integers with $m > n \ge 2$. Set $S = \{1, 2, \dots, m\}$, and $T = \{a_1, a_2, \dots, a_n\}$ is a subset of $S$ such that every element in $S$ is divisible by at most one element in $T$. Prove that
$$
\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n} < \frac{m+n}{m}.
$$ | [
"For every $i$ with $1 \\le i \\le n$, we define set\n$$\nS_i = \\{b \\mid b \\text{ is an element in } S \\text{ and is divisible by } a_i\\}.\n$$\nThere are $\\lfloor \\frac{m}{a_i} \\rfloor$ elements in $S_i$. Since every element in $S$ is not divisible by any two distinct elements in $T$, it follows that $S_i \... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
09gb | Let say a 52-digit number $\overline{a_1a_2\ldots a_{52}}$ is "zig-zag" if
$$
a_1 \le a_2 \ge a_3 \le a_4 \ge \dots \le a_{50} \ge a_{51} \le a_{52}
$$
holds. Find the number of all zig-zag numbers composed of the digits $1$, $2$ and $3$. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Linear Algebra > Matrices"
] | English | proof and answer | 2342398945624433584 | |
00on | Let $\mathcal{ABC}$ be an acute triangle with orthocenter $H$. The circumcircle of the triangle $\mathcal{BHC}$ intersects $\mathcal{AC}$ a second time in point $P$ and $\mathcal{AB}$ a second time in point $Q$.
Prove that $H$ is the circumcenter of the triangle $\mathcal{APQ}$. | [
"\nFigure 2: Problem 6\n\nLet $H_a$ be the foot of the altitude on $BC$. With the angle sum in triangle $AH_aC$, we get\n$$\n\\angle HAC = 90^\\circ - \\angle BCA.\n$$\nLet $H_b$ be the foot of the altitude on $AC$. With the angle sum in triangle $CH_bB$, we get\n$$\n\\angle CBH = 90^\\circ... | Austria | Austrian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0jse | Problem:
A right triangle has side lengths $a$, $b$, and $\sqrt{2016}$ in some order, where $a$ and $b$ are positive integers. Determine the smallest possible perimeter of the triangle. | [
"Solution:\nThere are no integer solutions to $a^{2}+b^{2}=2016$ due to the presence of the prime $7$ on the right-hand side (by Fermat's Christmas Theorem). Assuming $a<b$, the minimal solution $(a, b)=(3,45)$ which gives the answer above."
] | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Algebraic Number Theory > Quadratic forms",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 48 + sqrt(2016) | |
0kec | Problem:
Find a formula for the sum of the squares of the numbers in the $n$th row of Pascal's triangle (i.e. the numbers $\left(\begin{array}{l}n \\ 0\end{array}\right),\left(\begin{array}{l}n \\ 1\end{array}\right), \ldots,\left(\begin{array}{l}n \\ n\end{array}\right)$ ). | [
"Solution:\nWe claim that the sum of the squares of the numbers in row $n$ is\n$$\n\\left(\\begin{array}{l}\nn \\\\\n0\n\\end{array}\\right)^2+\\left(\\begin{array}{c}\nn \\\\\n1\n\\end{array}\\right)^2+\\cdots+\\left(\\begin{array}{l}\nn \\\\\nn\n\\end{array}\\right)^2=\\left(\\begin{array}{c}\n2 n \\\\\nn\n\\end{... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | (2n choose n) | |
02tz | Problem:
É dado um quadrado $3 \times 3$ com números escritos em cada casinha $1 \times 1$. As jogadas permitidas são escolher uma linha, uma coluna ou uma diagonal e somar ou subtrair 1 dos três números que estiverem nela. Prove que não é possível começar com os números na configuração da esquerda e chegar aos número... | [
"Solution:\n\nSeja $S$ a soma de todos os números nas casinhas do quadrado. Ao somarmos 1 em três casinhas, trocamos $S$ por $S+3$ e, ao subtrairmos 1 em três casinhas, trocamos $S$ por $S-3$. Como estamos sempre somando ou subtraindo 3, o resto da soma de todos os números na divisão por 3 não se altera. Na primeir... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
01gh | Find all polynomial functions $P(x, y)$ with real coefficients which for all real numbers $x$ and $y$ satisfy
$$
P(x + y, x - y) = 2P(x, y).
$$ | [
"Answer: $P(x, y) = (a + b)x^2 + axy + by^2$, where $a$ and $b$ are real numbers.\nFrom the given equation,\n$$\nP(2x, 2y) = P((x+y) + (x-y), (x+y) - (x-y)) = 2P(x+y, x-y) = 4P(x, y). \\quad (1)\n$$\nLet $m, n$ be arbitrary non-negative integers. Consider term $c_{m,n}x^m y^n$ of the polynomial $P(x, y)$. Two polyn... | Baltic Way | Baltic Way 2020 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | P(x, y) = (a + b)x^2 + axy + by^2 for arbitrary real a, b. | |
0k9d | Problem:
Prove that if the line joining the circumcenter $O$ and the incenter $I$ is parallel to side $BC$ of an acute triangle, then $\cos B + \cos C = 1$. | [
"Solution:\nWe prove the following more general result, called Carnot's theorem: in a triangle $ABC$ we have\n$$\n\\cos A + \\cos B + \\cos C = 1 + \\frac{r}{R}\n$$\nwhere $r$ and $R$ are the inradius and circumradius of $\\triangle ABC$.\n\nThe altitude from vertex $C$ divides side $AB$ into two segments (one of w... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
036p | Problem:
Given a triangle $ABC$, let $BL$, $L \in AC$, be the bisector of $\angle ABC$ and $AH$, $H \in BC$, the altitude to $BC$. Prove that $\angle AHL = \angle ALB$ if and only if $\angle BAC = \angle ACB + 90^\circ$. | [] | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
0hon | Problem:
Find all real numbers $x$ for which $\tan (x / 2)$ is defined and greater than $\sin (x)$. | [
"Solution:\nWe claim that the answer is $x \\in \\left(k \\pi - \\frac{\\pi}{2},\\ k \\pi\\right)$ for $k \\in \\mathbb{Z}$.\n\nThe statement is never true for $x / 2$ a multiple of $\\frac{\\pi}{2}$ because $\\tan (x / 2)$ is either undefined or equal to $0 = \\sin x$. Thus $x / 2$ is not a multiple of $\\frac{\\p... | United States | Berkeley Math Circle: Monthly Contest 8 | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | x ∈ (kπ − π/2, kπ) for all integers k | |
0fnv | Halle todos los polinomios $P(x)$ con coeficientes reales tales que $P(2014) = 1$ y, para algún entero $c$, se cumple que
$$
x(P(x) - c) = (x - 2014)P(x)
$$ | [] | Spain | XXIX Olimpiada Iberoamericana de Matemáticas | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | Spanish | proof and answer | P(x) = x/2014 | |
0fs1 | Problem:
Soit $ABC$ un triangle aigu. Soient $M_{A}$, $M_{B}$ et $M_{C}$ les milieux respectifs des côtés $BC$, $CA$ et $AB$. Soient $M_{A}^{\prime}$, $M_{B}^{\prime}$ et $M_{C}^{\prime}$ les milieux respectifs des arcs mineurs $BC$, $CA$ et $AB$ sur le cercle circonscrit au triangle $ABC$. Soit $P_{A}$ l'intersection... | [
"Solution:\n\nOn montre d'abord que $AP_{A}$ est la bissectrice de l'angle $\\angle BAC$. Pour cela, on peut d'abord remarquer que $M_{B}^{\\prime}M_{B}$ et $M_{C}^{\\prime}M_{C}$ sont les médiatrices de $AC$ et $AB$, respectivement, s'intersectant ainsi en $O$ le centre du cercle circonscrit. Mais $\\triangle OM_{... | Switzerland | null | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometr... | null | proof only | null | |
01h6 | An integer valued function $f(n)$ is multiplicative (that means $f(ab) = f(a)f(b)$ for all coprime $a$ and $b$) and satisfies the equation
$$
f(m + k - 3) = f(m) + f(k) - f(3) \quad \text{for all primes } m \text{ and } k.
$$
Find $f(11)$. | [
"Answer: 1 or 11.\nNote that $f(1) = 1$ and the following equalities hold:\n$$\n\\begin{aligned}\nf(1) &= f(2 + 2 - 3) = f(2) + f(2) - f(3), \\\\\nf(7) &= f(5 + 5 - 3) = f(5) + f(5) - f(3), \\\\\nf(10) &= f(2)f(5) = f(11 + 2 - 3) = f(11) + f(2) - f(3), \\\\\nf(11) &= f(7 + 7 - 3) = f(7) + f(7) - f(3), \\\\\nf(15) &... | Baltic Way | Baltic Way 2020 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Other"
] | null | proof and answer | 1 or 11 | |
0lao | Given a positive integer $n$. Given a square table of sizes $3 \times 3$. One uses $n$ colors to paint the squares of the table in such a way, that each square is painted by one color. Two ways of painting is considered to be similar if they can be obtained from each other by rotating the table around its center. How m... | [
"Mark the squares of the table as shown below.\nIn the sequel, the squares that lie symmetric with respect to square $x$ are called symmetric; a square that is marked with odd number is called odd square; a square that is marked with even number is called even square.\nLet $s$ denote the number to be found.\nWe hav... | Vietnam | Vietnamese Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof and answer | n^3(n^6 + n^2 + 2)/4 | |
05l5 | Problem:
Soit $a, b, c, d > 0$ des réels tels que $a b c d = 1$.
Prouver que
$$
\frac{1}{a+b+2} + \frac{1}{b+c+2} + \frac{1}{c+d+2} + \frac{1}{d+a+2} \leqslant 1.
$$ | [
"Solution:\n\nIl est bien connu que, si $x, y > 0$ sont des réels, on a $x + y \\geqslant 2 \\sqrt{x y}$.\n\nPar suite, $\\frac{1}{a+b+2} \\leqslant \\frac{1}{2(\\sqrt{a b} + 1)}$ et $\\frac{1}{c+d+2} \\leqslant \\frac{1}{2(\\sqrt{c d} + 1)}$, d'où\n$$\n\\begin{aligned}\n\\frac{1}{a+b+2} + \\frac{1}{c+d+2} & \\leqs... | France | Olympiades Françaises de Mathématiques - Test de Février | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0fpu | Dos circunferencias $C$ y $C'$ son secantes en dos puntos $P$ y $Q$. La recta que une los centros corta a $C$ en $R$ y a $C'$ en $R'$, la que une $P$ y $R'$ corta a $C$ en $X \neq P$ y la que une $P$ y $R$ corta a $C'$ en $X' \neq P$. Si los tres puntos $X$, $Q$, $X'$ están alineados se pide:
1. hallar el ángulo $\angl... | [
"(i) Sean $F$ y $F'$ los puntos diametralmente opuestos a $R$ y $R'$ en $C$ y $C'$, respectivamente. Por el Teorema del ángulo inscrito se tiene que $\\angle PFQ = \\angle PXQ = \\alpha$ que, por simetría, es el doble de $\\angle PFR$, luego $\\angle PFR = \\alpha/2$. Como el triángulo $PFR$ es rectángulo en $P$ (a... | Spain | LII Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | Spanish | proof and answer | (i) ∠XPX′ = π/3 (sixty degrees). (ii) (d + r − r′)(d − r + r′) = rr′. | |
07fs | In acute-angled triangle $\triangle ABC$, altitudes $BE, CF$ meet at $H$. A perpendicular line is drawn from $H$ to $EF$ and intersects arc $BC$ of the circumcircle of $\triangle ABC$ (the one that doesn't contain $A$) at $K$. If $AK, BC$ meet at $P$, prove that $PK = PH$. | [
"Without loss of generality, let's assume that $AC \\ge AB$. Let $D$ be the second intersection point of line $AH$ and the circumcircle of $\\triangle ABC$ and let $O$ be circumcenter of $\\triangle ABC$. Let the lines $OD$ and $BC$ meet at $J$. Then we have $\\angle JDH = \\angle JHD$ since $BC$ is perpendicular b... | Iran | 37th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
064x | Determine the integer $n$ for which $A - B$, where $A = \sqrt{n^2 + 24}$ and $B = \sqrt{n^2 - 9}$, is an integer. | [
"It is necessary to have $n^2 - 9 \\ge 0 \\Leftrightarrow n \\ge 3$ or $n \\le -3$.\nLet $A - B = \\sqrt{n^2 + 24} - \\sqrt{n^2 - 9} = d \\in \\mathbb{Z}$. Then $d > 0$ and\n$$\n\\begin{aligned}\n\\sqrt{n^2 + 24} &= \\sqrt{n^2 - 9} + d \\\\\n\\Rightarrow n^2 + 24 &= n^2 - 9 + d^2 + 2d\\sqrt{n^2 - 9} \\\\\n\\Leftrig... | Greece | 24th Hellenic Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | n = -5 or n = 5 | |
0bcn | Problem:
Se consideră o funcţie monotonă $f: \mathbb{R} \rightarrow \mathbb{R}$.
a) Demonstraţi că $f$ are limite laterale în fiecare punct $x_{0} \in \mathbb{R}$.
b) Definim funcția $g: \mathbb{R} \rightarrow \mathbb{R}$, $g(x)=\lim_{t>x} f(t)$, i.e. $g(x)$ este limita la stânga în punctul $x$. Arătaţi că dacă func... | [] | Romania | Olimpiada Naţională de Matematică | [
"Precalculus > Functions",
"Precalculus > Limits"
] | null | proof only | null | |
0gn5 | Let $A_1$, $B_1$ and $C_1$ be the feet of the altitudes belonging to the vertices $A$, $B$ and $C$, respectively, of an acute triangle $ABC$, and let $O_A$, $O_B$ and $O_C$ be the incenters of the triangles $AB_1C_1$, $BC_1A_1$ and $CA_1B_1$, respectively. Let $T_A$, $T_B$ and $T_C$ be the points of tangency of the inc... | [
"Let $r$ be the inradius and $I$ the incenter of the triangle $ABC$.\nThe triangle $ABC$ is similar to the triangle $AB_1C_1$ with scale factor $\\cos \\angle A$.\nLet $X$ and $Y$ be the points of tangency of the incircle of the triangle $AB_1C_1$ with the line segments $AC_1$ and $AB_1$, respectively. As $AX/AT_B ... | Turkey | 14th Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | English | proof only | null | |
0jew | Problem:
Let $ABCD$ be a parallelogram with $AB = 8$, $AD = 11$, and $\angle BAD = 60^{\circ}$. Let $X$ be on segment $CD$ with $CX / XD = 1 / 3$ and $Y$ be on segment $AD$ with $AY / YD = 1 / 2$. Let $Z$ be on segment $AB$ such that $AX$, $BY$, and $DZ$ are concurrent. Determine the area of triangle $XYZ$. | [
"Solution:\n\nAnswer: $\\dfrac{19 \\sqrt{3}}{2}$\n\nLet $AX$ and $BD$ meet at $P$. We have $DP / PB = DX / AB = 3 / 4$. Now, applying Ceva's Theorem in triangle $ABD$, we see that\n$$\n\\frac{AZ}{ZB} = \\frac{DP}{PB} \\cdot \\frac{AY}{YD} = \\frac{3}{4} \\cdot \\frac{1}{2} = \\frac{3}{8}\n$$\nNow,\n$$\n\\frac{[AYZ]... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof and answer | 19√3/2 | |
00d8 | En el pizarrón hay dibujado un polígono de ocho lados. Mili debe escribir un número entero entre $1$ y $16$, sin repeticiones, en cada uno de sus lados y en cada uno de sus vértices. A continuación, para cada lado, Mili calcula la suma de los números escritos en sus dos vértices más el número escrito en ese lado. Obtie... | [
"Notamos $v$ a la suma de los números en los vértices del octógono y $a$ a la suma de los números en sus lados. Entonces $v + a = 1 + 2 + 3 + \\dots + 15 + 16 = 136$. Además, $2v + a = 8S$, de donde $2v + (136 - v) = 8S$ y tenemos que $v + 136 = 8S$. Sabemos que\n$$\n\\begin{aligned}\n1 + 2 + \\dots + 8 \\leq v \\l... | Argentina | Nacional OMA | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | All integers from 22 to 29 inclusive | |
05o6 | Problem:
On définit une suite ainsi:
$$
\left\{\begin{array}{l}
u_{0}=15,\ u_{1}=57 \\
u_{n}=u_{n-1}+u_{n-2}\ \text{ pour tout } n \geq 2
\end{array}\right.
$$
Trouver le plus grand entier $k$ tel que $3^{k} \mid u_{2017}$. | [
"Solution:\n\nSoit $k$ l'entier cherché. Les premiers termes de la suite sont $15$, $57$, $72$, $129$, $201$, $330$, $541$, ... apparemment tous divisibles par $3$. En effet, $u_{0}$ et $u_{1}$ sont multiples de $3$, donc $u_{2}=u_{0}+u_{1}$ aussi. De même, $u_{3}$ est divisible par $3$. Plus généralement, si $u_{n... | France | OCympiades Françaises de Mathématiques | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | null | proof and answer | 1 | |
0cks | Let $f : [0, \infty) \to [0, \infty)$ be a continuous bijective function, such that
$$
\lim_{x \to \infty} \frac{f^{-1}(f(x)/x)}{x} = 1.
$$
a) Show that $\lim_{x \to \infty} \frac{f(x)}{x} = \infty$ and $\lim_{x \to \infty} \frac{f^{-1}(ax)}{f^{-1}(x)} = 1$, for any $a > 0$.
b) Give an example of a function $f$ that ... | [
"a) Since $f$ is continuous and bijective, we deduce that $f(0) = 0$ and $f$ is increasing, with $\\lim_{x \\to \\infty} f(x) = \\infty$. Then $f^{-1} : [0, \\infty) \\to [0, \\infty)$ is also increasing, with $f^{-1}(0) = 0$ and $\\lim_{x \\to \\infty} f^{-1}(x) = \\infty$. The limit from the hypothesis ensures th... | Romania | 75th Romanian Mathematical Olympiad | [
"Precalculus > Limits",
"Precalculus > Functions"
] | English | proof and answer | f(x) = e^x - 1 | |
0gsf | In a right triangle $ABC$ with $\angle ACB = 90^\circ$ let $D$ be the foot of the altitude from $C$. Let $E$ and $F$ be the reflections of $D$ with respect to $AC$ and $BC$, respectively. Let $O_1$ and $O_2$ be the circumcenters of the triangles $ECB$ and $FCA$, respectively. Prove that
$$
2|O_1O_2| = |AB|.
$$ | [
"Let $\\angle ABC = \\alpha$ and $\\angle BAC = 90 - \\alpha = \\beta$. $\\Rightarrow \\angle BCD = \\angle EAC = \\beta$ and $\\angle ACD = \\angle ACE = \\alpha$. Let $K$ be the second intersection point of the circumcircle of the triangle $ECB$ and the line $AB$. $\\Rightarrow \\angle AKE = \\angle KEA = \\beta$... | Turkey | Team Selection Test for IMO 2019 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry ... | English | proof only | null | |
0kjk | What is the average number of pairs of consecutive integers in a randomly selected subset of $5$ distinct integers chosen from the set $\{1, 2, 3, \ldots, 30\}$? (For example, the set $\{1, 17, 18, 19, 30\}$ has $2$ pairs of consecutive integers.)
(A) $\frac{2}{3}$ \quad (B) $\frac{29}{36}$ \quad (C) $\frac{5}{6}$ \qu... | [] | United States | AMC 12 B | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | MCQ | A | |
0cpl | Let positive integers $d$ and $d'$ ($d' > d$) be two divisors of positive integer $n$. Prove that $d' > d + \frac{d^2}{n}$.
Натуральные числа $d$ и $d'$, $d' > d$ -- делители натурального числа $n$. Докажите, что $d' > d + \frac{d^2}{n}$. | [
"Поскольку числа $f = n/d$ и $f' = n/d'$ целые, а $f > f'$, имеем $f - f' \\ge 1$, или\n$$\n1 \\le \\frac{n}{d} - \\frac{n}{d'} = \\frac{(d' - d)n}{dd'} < \\frac{(d' - d)n}{d^2}.\n$$\nДомножая на $\\frac{d^2}{n}$, получаем $d' - d > \\frac{d^2}{n}$, что и требовалось доказать."
] | Russia | Russian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization"
] | English, Russian | proof only | null | |
09zk | There are $25$ guests at a party, one of which is Medan. Among the other guests, there are $12$ that each shook hands with exactly $18$ people. The other $12$ each shook hands with exactly $6$ people.
With how many guests did Medan shake hands?
A) $0$
B) $6$
C) $12$
D) $18$
E) $24$ | [] | Netherlands | Junior Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | MCQ | C | |
0i0d | Problem:
Let $\frac{1}{1-x-x^{2}-x^{3}}=\sum_{i=0}^{\infty} a_{n} x^{n}$, for what positive integers $n$ does $a_{n-1}=n^{2}$? | [
"Solution:\nMultiplying both sides by $1-x-x^{2}-x^{3}$ the right hand side becomes $a_{0}+(a_{1}-a_{0}) x+(a_{2}-a_{1}-a_{0}) x^{2}+\\ldots$, and setting coefficients of $x^{n}$ equal to each other we find that $a_{0}=1$, $a_{1}=1$, $a_{2}=2$, and $a_{n}=a_{n-1}+a_{n-2}+a_{n-3}$ for $n \\geq 3$. Thus the sequence ... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Generating functions"
] | null | proof and answer | 1, 9 | |
0jk2 | Problem:
Seven little children sit in a circle. The teacher distributes pieces of candy to the children in such a way that the following conditions hold.
- Every little child gets at least one piece of candy.
- No two little children have the same number of pieces of candy.
- The numbers of candy pieces given to any t... | [
"Solution:\n\nAnswer: $44$\n\nAn optimal arrangement is $2$-$6$-$3$-$9$-$12$-$4$-$8$.\n\nNote that at least two prime factors must appear. In addition, any prime factor that appears must appear in at least two non-prime powers unless it is not used as a common factor between any two adjacent little children. Thus w... | United States | HMMT November 2014 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 44 | |
05yn | Problem:
Soit $n \in \mathbb{N}^{\star}$. Montrer que si $2n+1$ et $3n+1$ sont des carrés parfaits, alors $5n+3$ n'est pas premier. | [
"Solution:\n\nSupposons que $2n+1$ et $3n+1$ sont des carrés parfaits : on dispose de $a, b$ deux entiers strictement positifs ($2n+1, 3n+1 > 0$) tels que $2n+1 = a^{2}$ et $3n+1 = b^{2}$. Ici, il s'agit de remarquer que $5n+3 = 4 \\times (2n+1) - (3n+1) = 4a^{2} - b^{2}$.\n\nPour trouver cette identité, on essaie ... | France | Préparation Olympique Française de Mathématiques - Envoi 3: Arithmétique | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
016i | Problem:
A $n$-trønder walk is a walk starting at $(0,0)$, ending at $(2 n, 0)$ with no self intersection and not leaving the first quadrant, where every step is one of the vectors $(1,1)$, $(1,-1)$ or $(-1,1)$.
(The figure shows the possible 2-trønder walks.)
Find the number of $n$-trønder ... | [] | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | Catalan number C_n = binomial(2n, n) / (n + 1) | |
0gf1 | 給定任意正整數 $k$, 證明存在質數 $p$ 及相異整數 $a_1, a_2, ..., a_{k+3} \in \{1, 2, ..., p-1\}$, 使得對於所有 $i = 1, 2, ..., k$, $p$ 都整除 $a_i a_{i+1} a_{i+2} a_{i+3} - i$。
Given a positive integer $k$, show that there exists a prime $p$ and such that one can choose distinct integers $a_1, a_2, \dots, a_{k+3} \in \{1, 2, \dots, p-1\}$ such t... | [
"First, construct distinct positive rational numbers $r_1, r_2, \\dots, r_{k+3}$ such that $r_i r_{i+1} r_{i+2} r_{i+3} = i$ for all $i = 1, 2, \\dots, k$. Since $r_{i+4} = \\frac{i+1}{i} r_i$, $\\{r_{4n+r}\\}_n$ are strictly increasing sequences for $r = 1, 2, 3, 4$. Choose $r_1, r_2, r_3$ to be distinct primes la... | Taiwan | 2021 數學奧林匹亞競賽第一階段選訓營 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
07lh | Suppose $a$, $b$, $c$ are complex numbers such that $a + b + c = 0$. Prove that
$$
2(a - b)^2(b - c)^2(c - a)^2 = (a^2 + b^2 + c^2)^3 - 54a^2b^2c^2.
$$ | [
"Let $ab + bc + ca = -p$ and $abc = q$. From $a + b + c = 0$ we obtain\n$$\n0 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) = a^2 + b^2 + c^2 - 2p, \\text{ i.e.,}\n$$\n$$\na^2 + b^2 + c^2 = 2p,\n$$\nand $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0$, i.e.,\n$$\na^3 + b^3 + c^3 = 3q... | Ireland | Irska | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Linear Algebra > Determinants"
] | English | proof only | null | |
0h2v | In the coordinate plane $xOy$, depict the set of all points such that $|y - [x]| = 2 - y - [x]$, where $[x]$ is the integer part of $x$, i.e. the largest integer not exceeding $x$. | [
"Вихідне співвідношення рівносильне системі\n$$\n\\begin{cases}\ny = 1, \\\\\n[x] = 1; \\\\\ny \\le 2 - [x].\n\\end{cases}\n$$\nСлід зобразити множину точок $\\{(x; y) : 1 \\le x < 2, y < 1\\} \\cup \\{(x; 1) : x < 2\\}$."
] | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
005m | En cada casilla de un tablero de $a$ filas y $b$ columnas está escrito un $0$ o un $1$ de modo que se verifican las siguientes condiciones.
* Si una fila y una columna se intersecan en una casilla con $0$ entonces contienen el mismo número de ceros.
* Si una fila y una columna se intersecan en una casilla con $1$ enton... | [] | Argentina | XVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | All pairs (a, b) with a ≤ b such that b = a or b = 2a. | |
0joy | Problem:
Let $k$ be a positive integer. Prove that there exist integers $x$ and $y$, neither of which is divisible by $3$, such that $x^{2} + 2y^{2} = 3^{k}$. | [
"Solution:\nFor the first several values of $k$ it is straightforward to find solutions $x_{k}$ and $y_{k}$ satisfying $x_{k}^{2} + 2y_{k}^{2} = 3^{k}$, leading to this table of solutions.\n\n| $k$ | $x_{k}$ | $y_{k}$ |\n| ---: | ---: | ---: |\n| 1 | 1 | 1 |\n| 2 | 1 | 2 |\n| 3 | 5 | 1 |\n| 4 | 7 | 4 |\n| 5 | 1 | 1... | United States | BAMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Algebraic Number Theory > Quadratic fields"
] | null | proof only | null | |
06nu | Let $n$ be a positive integer. Prove that there exists a positive integer $m$ such that
$$
3^n + 5^n - 1
$$
is divisible by $7^n$. | [
"We show that $m = 7^{n-1}$ works. By the lifting the exponent lemma, we have\n$$\nv_7(3^m + 4^m) = v_7(3^m + 4^m) + v_7(m) = 1 + (n-1) = n.\n$$\nThis implies $7^m \\mid 3^m + 4^m$, and hence $3^m \\equiv -4^m \\pmod{7^n}$. Similarly, $5^m \\equiv -2^m \\pmod{7^n}$. It follows that\n$$\n3^m + 5^m - 1 \\equiv -4^m -... | Hong Kong | Hong Kong Team Selection Test 2 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof only | null | |
0k8b | Problem:
Let $N$ be the number of sequences of positive integers $(a_{1}, a_{2}, a_{3}, \ldots, a_{15})$ for which the polynomials
$$
x^{2}-a_{i} x+a_{i+1}
$$
each have an integer root for every $1 \leq i \leq 15$, setting $a_{16}=a_{1}$. Estimate $N$. | [
"Solution:\nWe note that $a_{i+1}=x(a_{i}-x)$ for some positive integer $x$, so $a_{i+1} \\geq a_{i}-1$. So, the only way $a_{i}$ can decrease is decreasing by 1.\n\nAs it cannot decrease that quickly, we will make the assumption that if $a_{i} \\geq 10$, $a_{i+1}=a_{i}-1$, as otherwise it will increase at least ab... | United States | HMMT November 2019 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Algorithms"
] | null | proof and answer | 1409 | |
0jlr | Let $\mathbb{N}$ be the set of positive integers. Find all functions $f : \mathbb{N} \to \mathbb{N}$ that satisfy the equation
$$
f^{abc-a}(abc) + f^{abc-b}(abc) + f^{abc-c}(abc) = a + b + c
$$
for all $a, b, c \ge 2$.
(Here $f^1(n) = f(n)$ and $f^k(n) = f(f^{k-1}(n))$ for every integer $k$ greater than 1.) | [
"Clearly if $f(n) = n - 1$ for $n > 1$, the desired identity will be satisfied. In fact one can easily check that the value of $f(2)$ is also irrelevant, so any $f$ such that $f(n) = n - 1$ for all $n > 2$ will work. We show that these are the only such functions.\nPlug in $a = b = c$ to obtain $f^{a^3-a}(a^3) = a$... | United States | IMO Team Selection Team Selection Test | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All functions f with f(n) = n − 1 for all n > 2; the value of f(2) is arbitrary. | |
03sv | Solve the following system of equations.
$$
\begin{cases} x - y + z - w = 2, \\ x^2 - y^2 + z^2 - w^2 = 6, \\ x^3 - y^3 + z^3 - w^3 = 20, \\ x^4 - y^4 + z^4 - w^4 = 66. \end{cases}
$$ | [
"Let $p = x + z$, $q = xz$. The second to fourth equations of the system become\n$$\n\\begin{aligned}\np^2 &= x^2 + z^2 + 2q, \\\\\np^3 &= x^3 + z^3 + 3pq, \\\\\np^4 &= x^4 + z^4 + 4p^2q - 2q^2.\\end{aligned}\n$$\nSimilarly, let $s = y + w$, $t = yw$. The second to fourth equations of the system become\n$$\ns^2 = y... | China | China Mathematical Competition (Extra Test) | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | Four solutions: (x=3, y=2, z=1, w=0); (x=3, y=0, z=1, w=2); (x=1, y=2, z=3, w=0); (x=1, y=0, z=3, w=2). | |
0fm8 | Problem:
Sea $ABC$ un triángulo acutángulo con $\hat{A}=45^{\circ}$, y sea $P$ el pie de la altura por $B$. Trazamos la circunferencia de centro $P$ que pasa por $C$ y que vuelve a cortar a $AC$ en el punto $X$ y a la altura $PB$ en el punto $Y$. Sean $r$ y $s$ las rectas perpendiculares a la recta $AY$ por $P$ y $X$,... | [
"Solution:\n\nPor construcción es $PX=PY=PC$. Los triángulos $PAY$ y $PCB$, rectángulos en $P$, son iguales ya que $AP=PB$ (el triángulo rectángulo $APB$ es isósceles) y $PY=PC$. Por tanto los ángulos $\\alpha$ y $\\beta$ son iguales.\n\nEl triángulo rectángulo $PYQ$ es semejante a los anteriores, de manera que el ... | Spain | Spain | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
06um | Call a rational number short if it has finitely many digits in its decimal expansion. For a positive integer $m$, we say that a positive integer $t$ is $m$-tastic if there exists a number $c \in \{1,2,3, \ldots, 2017\}$ such that $\frac{10^{t}-1}{c \cdot m}$ is short, and such that $\frac{10^{k}-1}{c \cdot m}$ is not s... | [
"If $m=2^{a} \\cdot 5^{b} \\cdot s$, with $\\operatorname{gcd}(s, 10)=1$, then $\\frac{10^{t}-1}{m}$ is short if and only if $s$ divides $10^{t}-1$. So we may (and will) suppose without loss of generality that $\\operatorname{gcd}(m, 10)=1$. Define\n$$\nC=\\{1 \\leqslant c \\leqslant 2017: \\operatorname{gcd}(c, 10... | IMO | International Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 807 | |
074f | Let $a$, $b$ be two distinct odd positive integers. Define a sequence $(a_n)_{n \ge 1}$ by
$$
a_1 = a,\quad a_2 = b,\quad a_n \text{ is the largest odd divisor of } a_{n-1} + a_{n-2}, \text{ for all } n \ge 3.
$$
Show that there exists a natural number $N$ such that $a_n = \text{gcd}(a, b)$ for all $n \ge N$. | [
"Each $a_n$ is odd for $n \\ge 1$. Hence $a_{n-1} + a_{n-2}$ is even for $n \\ge 3$. This implies $a_n \\le (a_{n-1} + a_{n-2})/2$, for all $n \\ge 3$. An easy induction shows that $a_n \\ge \\max\\{a, b\\}$, for all $n$. Thus the number of pairs of the form $\\{a_{n-1}, a_{n-2}\\}$ is finite. It follows that $a_n$... | India | Indija TS 2009 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
04u5 | In a certain club, some pairs of members are friends. Given $k \ge 3$, we say that a club is *k*-good if every group of *k* members can be seated around a round table such that every two neighbors are friends. Prove that if a club is 6-*good* then it is 7-*good*.
(Josef Tkadlec) | [
"Consider a 6-good club and denote some seven of its members by $A, \\dots, G$. It suffices to show that $A, \\dots, G$ can be seated around a table as required. Consider only friendships among $A, \\dots, G$. First, we show that every member has at least three friends.\n\nWithout loss of generality consider $G$. B... | Czech Republic | 67th Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0j8h | Problem:
Let $f:[0,1) \rightarrow \mathbb{R}$ be a function that satisfies the following condition: if
$$
x=\sum_{n=1}^{\infty} \frac{a_{n}}{10^{n}}= . a_{1} a_{2} a_{3} \ldots
$$
is the decimal expansion of $x$ and there does not exist a positive integer $k$ such that $a_{n}=9$ for all $n \geq k$, then
$$
f(x)=\sum_{n... | [
"Solution:\nAnswer: 0\n\nNote that $\\frac{1}{3}=\\sum_{n=1}^{\\infty} \\frac{3}{10^{n}}$.\n\nClearly $f$ is an increasing function. Also for any integer $n \\geq 1$, we see from decimal expansions that\n$$\nf\\left(\\frac{1}{3} \\pm \\frac{1}{10^{n}}\\right)-f\\left(\\frac{1}{3}\\right)= \\pm \\frac{1}{10^{2 n}}.\... | United States | Harvard-MIT Mathematics Tournament | [
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Functions"
] | null | proof and answer | 0 | |
02o2 | Problem:
Qual é a menor fração? Quantas frações da forma $\frac{n}{n+1}$ são menores do que $\frac{7}{9}$, sabendo que $n$ é um número inteiro positivo?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5 | [
"Solution:\n\nAs frações da forma $\\frac{n}{n+1}$, com $n$ inteiro positivo, são\n\n\n\nObserve que temos $\\frac{1}{2}<\\frac{2}{3}<\\frac{3}{4}<\\frac{4}{5}<\\ldots$, ou seja, essa sequência de frações é crescente. Para comparar cada uma dessas frações com $\\frac{7}{9}$, precisamos igua... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | c | |
0g6n | 已知 $n$ 為一個三位數之奇數, 且其所有因數 (包含 1 與 $n$) 的末位數字之和為 33. 試求 $n$ 之值。 | [
"答 $n = 27^2 = 729$.\n由 $n$ 為奇數知：其每個因數皆為奇數。顯然，其每個因數的個位數都是奇數。\n又由於所有因數的末位數字之和為 33，是一個奇數。故 $n$ 有奇數個因數。從而，$n$ 為平方數。\n因為質數只有 2 個因數, 平方後只有 3 個因數, 其末位數字之和小於 27, 故滿足條件的三位數只能是合數的平方, 共有如下四個:\n$$\n27^2, 25^2, 21^2, 15^2.\n$$\n逐一驗證知：$27^2 = 729$ 為所求。"
] | Taiwan | 二〇一二數學奧林匹亞競賽第一階段選訓營 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 729 | |
01fb | Magician puts on the $3 \times 100$ board cards with numbers from $1$ to $300$ in a "snake-like" line so that consecutive numbers are side-to-side adjacent (either horizontally or vertically, not diagonally). The numbers are written on the bottom sides of the cards, the upper sides are empty. After that the magician tu... | [
"**Answer:** $k = 2$.\n\nExample. Put $300$ in the low left corner and $101$ in the cell above it.\n\nEstimation. One open card does not determine the snake uniquely, because all the cells of $3 \\times 100$ board can be considered as one cyclic path: let the rows of the board be denoted by letters *a*, *b*, *c* an... | Baltic Way | Baltic Way 2019 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Other"
] | English | proof and answer | 2 | |
0du8 | Problem:
Tina je na pet listkov zapisala po eno naravno število, a ni hotela izdati, katera števila je zapisala. Prebrisani Žan jo je prepričal, da mu je povedala vse vsote po dveh števil. Zvedel je, da so bile vsote 17, 20, 28, $14, 42, 36, 28, 39, 25$ in 31. Katera števila je zapisala Tina? | [
"Solution:\n\nOznačimo števila z $a, b, c, d$ in $e$. Predpostavimo lahko, da velja $a \\leq b \\leq c \\leq d \\leq e$. Uredimo vsote po velikosti: $14, 17, 20, 25, 28, 28, 31, 36, 39, 42$. Gotovo je $a+b=14$, $a+c=17$, $d+e=42$ in $c+e=39$. Če seštejemo vse vsote, se vsako od števil pojavi v tej vsoti natanko šti... | Slovenia | 45. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 3, 11, 14, 17, 25 | |
0071 | En el pizarrón está indicada una multiplicación de 26 números enteros positivos, o sea, 26 enteros separados por signos $\times$. Lucía cambia dos de los signos $\times$ por signos $+$ y calcula el resultado de la nueva expresión. Repite este procedimiento para cada posible elección de dos signos $\times$ en la expresi... | [] | Argentina | Argentina 2009 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | Spanish | proof and answer | 13, 14, 15 | |
09of | Let $n \ge 2$. Determine the number of ways to color the cells of a $4 \times n$ grid such that every $2 \times 2$ subgrid contains exactly one colored cell.
(Batbayasgalan Balkhuu) | [] | Mongolia | MMO2025 Round 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | 2^{\lfloor n/2 \rfloor} + 2^{\lceil n/2 \rceil} + 4 | |
0br7 | Two circles, $\omega_1$ and $\omega_2$, centred at $O_1$ and $O_2$, respectively, meet at points $A$ and $B$. A line through $B$ meets $\omega_1$ again at $C$, and $\omega_2$ again at $D$. The tangents to $\omega_1$ and $\omega_2$ at $C$ and $D$, respectively, meet at $E$, and the line $AE$ meets the circle $\omega$ th... | [
"Begin by noticing that the lines $CO_1$ and $DO_2$ meet at a point $P$ on $\\omega$, since $\\angle(PO_1, PO_2) = \\angle(O_1C, CB) + \\angle(BD, DO_2) = \\angle(CB, BO_1) + \\angle(O_2B, BD) = \\angle(O_2B, BO_1) = \\angle(O_1A, AO_2)$. In what follows, we consider the case where $O_1$ and $O_2$ lie on the segmen... | Romania | 67th NMO Selection Tests for BMO and IMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null |
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