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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0064 | Demuestre que no existen enteros positivos $x$ e $y$ tales que
$$
x^{2008} + 2008! = 21^y.
$$ | [] | Argentina | XXIII Olimpíada Iberoamericana de Matemática | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Spanish | proof only | null | |
01ib | Find all polynomials $P$ with integer coefficients such that the number $P(a) - P(b)$ is divisible by $a + b$ for all integers $a, b$, provided that $a + b \neq 0$ | [
"Answer: all polynomials whose every odd-degree term has zero coefficient.\n\nLet $P(x) = P_0(x) + P_1(x)$, where $P_0$ and $P_1$ are polynomials whose all non-zero terms have either even or odd degree, respectively.\nThen we can write $P_0(x) = Q(x^2)$, where polynomial $Q$ is obtained from polynomial $P_0$ by div... | Baltic Way | Baltic Way 2021 Shortlist | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All polynomials with integer coefficients whose odd-degree terms all have zero coefficients (i.e., even polynomials). | |
0jgx | Problem:
In triangle $ABC$, $\angle A = 45^{\circ}$ and $M$ is the midpoint of $\overline{BC}$. $\overline{AM}$ intersects the circumcircle of $ABC$ for the second time at $D$, and $AM = 2MD$. Find $\cos \angle AOD$, where $O$ is the circumcenter of $ABC$. | [
"Solution:\n\n$\\cos \\angle AOD = -\\frac{1}{8}$. $\\angle BAC = 45^{\\circ}$, so $\\angle BOC = 90^{\\circ}$. If the radius of the circumcircle is $r$, $BC = \\sqrt{2} r$, and $BM = CM = \\frac{\\sqrt{2}}{2} r$. By power of a point, $BM \\cdot CM = AM \\cdot DM$, so $AM = r$ and $DM = \\frac{1}{2} r$, and $AD = \... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > ... | null | proof and answer | -1/8 | |
0blv | Prove that if $a$, $b$, $c$ are the length of the sides of a triangle, then
$$
\sqrt{\frac{a}{-a+b+c}} + \sqrt{\frac{b}{a-b+c}} + \sqrt{\frac{c}{a+b-c}} \ge 3.
$$ | [
"Since $a$, $b$, $c$ are the length of the sides of a triangle, numbers $-a+b+c$, $a-b+c$ and $a+b-c$ are positive. The GM-HM inequality yields\n$$\n\\sqrt{\\frac{a}{-a+b+c}} = \\sqrt{1 \\cdot \\frac{a}{-a+b+c}} \\ge \\frac{2}{1+\\frac{-a+b+c}{a}} = \\frac{2a}{b+c}\n$$\nand the other two similar relations. So, it i... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
06gc | The incircle of a triangle $ABC$ touches the three sides $BC$, $CA$ and $AB$ at $D$, $E$ and $F$ respectively. The line segments $BE$ and $CF$ intersect the incircle at $Y$ and $Z$ respectively. Suppose the lines $FE$ and $BC$ intersect externally at $X$. Prove that the three points $X$, $Y$ and $Z$ are collinear. | [
"Let $X' = EF \\cap YZ$. Apply Pascal's theorem to the points $EEFFZY$ on the incircle. Then $C = EE \\cap FZ$, $X' = EF \\cap ZY$ and $B = FF \\cap YE$ are collinear. Hence, $X' = X$, and $X, Y, Z$ are collinear.\n\n"
] | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Pappus theorem",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof only | null | |
0722 | Problem:
Find all primes $p$ and $q$, and even numbers $n>2$, satisfying the equation
$$
p^{n}+p^{n-1}+\cdots+p+1=q^{2}+q+1
$$ | [
"Solution:\nObviously $p \\neq q$. We write this in the form\n$$\np\\left(p^{n-1}+p^{n-2}+\\cdots+1\\right)=q(q+1)\n$$\nIf $q \\leq p^{n / 2}-1$, then $q<p^{n / 2}$ and hence we see that $q^{2}<p^{n}$. Thus we obtain\n$$\nq^{2}+q<p^{n}+p^{n / 2}<p^{n}+p^{n-1}+\\cdots+p\n$$\nsince $n>2$. It follows that $q \\geq p^{... | India | INMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | p=2, q=5, n=4 | |
0d4d | Let $a, b$ be two nonnegative real numbers and $n$ a positive integer. Prove that
$$
\left(1-2^{-n}\right)\left|a^{2^{n}}-b^{2^{n}}\right| \geq \sqrt{a b}\left|a^{2^{n}-1}-b^{2^{n}-1}\right|
$$ | [
"First Solution. If $a = b$, the inequality is satisfied. If $a \\neq b$, we can assume without loss of generality that $a > b$. In this case, the inequality can be written\n$$\n\\left(2^{n}-1\\right)\\left(a^{2^{n}}-b^{2^{n}}\\right) \\geq 2^{n} \\sqrt{a b}\\left(a^{2^{n}-1}-b^{2^{n}-1}\\right),\n$$\nor equivalent... | Saudi Arabia | SAMC | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English, Arabic | proof only | null | |
0jmk | Problem:
Find the number of digits in the decimal representation of $2^{41}$. | [
"Solution:\nAnswer: $13$\n\nNoticing that $2^{10} = 1024 \\approx 1000$ allows for a good estimate. Alternatively, the number of decimal digits of $n$ is given by $\\left\\lfloor \\log_{10}(n) \\right\\rfloor + 1$. Using $\\log_{10}(2) \\approx 0.31$ also gives the correct answer. The exact value of $2^{41}$ is $21... | United States | HMMT November 2014 | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | 13 | |
03i8 | Problem:
Eve and Odette play a game on a $3 \times 3$ checkerboard, with black checkers and white checkers. The rules are as follows:
I. They play alternately.
II. A turn consists of placing one checker on an unoccupied square of the board.
III. In her turn, a player may select either a white checker or a black checke... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
09nd | Let $\alpha$, $\beta$, and $\gamma$ be three angles of a triangle. Prove that
$$
\cos(\alpha) \cos(3\alpha) + \cos(\beta) \cos(3\beta) + \cos(\gamma) \cos(3\gamma) + \frac{7}{4} \ge 2 \cos(\alpha) \cos(\beta) \cos(\gamma).
$$
(Otgonbayar Uuye) | [
"First recall that $2 \\prod \\cos \\alpha = 1 - \\sum \\cos^2 \\alpha$. Now using the identity $\\cos(3x) = 4 \\cos^3(x) - 3 \\cos(x)$, we see that\n$$\n\\begin{aligned}\n4 \\sum \\cos(\\alpha) \\cos(3\\alpha) - 8 \\prod \\cos(\\alpha) + 7 &= 16 \\sum \\cos^4(\\alpha) - 8 \\sum \\cos^2(\\alpha) + 3 \\\\\n&= \\sum ... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
00pg | Given an odd number $n > 1$, let $S = \{k : 1 \le k < n, (k,n) = 1\}$ and let $T = \{k : k \in S, (k+1,n) = 1\}$. For each $k \in S$, let $r_k$ be the remainder left by $(k^{|S|} - 1)/n$ upon division by $n$. Show that
$$
\prod_{k \in T} (r_k - r_{n-k}) \equiv |S|^{|T|} \pmod{n}.
$$ | [
"Since $n$ is odd, $|S| = \\phi(n) = n \\prod_{p|n} (1 - 1/p)$ is even. Given an element $k$ of $S$, write\n$$\nk^{|S|} \\equiv 1 + n r_k \\pmod{n^2} \\quad \\text{and} \\quad (n-k)^{|S|} \\equiv 1 + n r_{n-k} \\pmod{n^2},\n$$\nand notice that\n$$\n(n-k)^{|S|} \\equiv k^{|S|} - |S| \\cdot n \\cdot k^{|S|-1} \\pmod{... | Balkan Mathematical Olympiad | shortlistBMO 2011 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
07qy | Five teams play in a soccer competition where each team plays one match against each of the other four teams. A winning team gains 5 points and a losing team 0 points. For a 0-0 draw both teams gain 1 point, and for other draws (1-1, 2-2, etc.) both teams gain 2 points. At the end of the competition, we write down the ... | [
"Ten matches are played each one contributing either 2, 4 or 5 points. Hence the total number of points is between 20 and 50.\nIf the team scores are five consecutive integers, then the total number of points must be a multiple of 5. If the total number of points is 20, all teams will score 4 and if the total numbe... | Ireland | Ireland_2017 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 6 | |
00oy | Integers are written in the cells of a table $2010 \times 2010$. Adding $1$ to all the numbers in a row or in a column is called a *move*. We say that the table is *equilibrium* if one can obtain after finitely many moves a table in which all the numbers are equal.
a) Find the largest positive integer $n$, for which t... | [
"a) We shall prove that for a table $m \\times m$ ($m \\ge 2$) the answer is $n = 2m-2$; in particular $n = 4018$ for $m = 2010$.\n\nDenote by $a_{ij}$ the number written in the cell $(i,j)$ ($1 \\le i,j \\le m$). Let $a$, $b$, $c$ and $d$ be numbers written in cells, which centers form a rectangle with the sides p... | Balkan Mathematical Olympiad | BMO 2010 Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | a) 4018; b) 2^{4019} - 2^{2010} + 1 | |
0h2o | Solve the following equation:
$$
|[x]| = |[x]|,
$$
where $[a]$ stands for the greatest integer that does not exceed $a$. | [
"**Answer:** All non-negative reals and negative integers.\n\nConsider 3 cases.\n\n1) $x \\ge 0$. Then $|x| = x$, $[x] \\ge 0$, and so $|[x]| = [x] = [|x|]$, which means that any non-negative $x$ is a solution of our equation.\n\n2) $x$ is a negative integer. Then $[x] = x$, $[-x] = -x$, $|x| = -x$, and so $|[x]| =... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | All non-negative real numbers and negative integers | |
04za | Call a positive integer $n$ prime-prone if there exist at least three prime numbers from which we can get $n$ by removing the last digit. Prove that every two prime-prone positive integers differ from each other by at least $3$. (Juniors.) | [
"As the prime numbers under consideration have at least two digits, the last digit can be only $1$, $3$, $7$, or $9$. Thus $n$ is prime-prone if and only if, among numbers $10n + 1$, $10n + 3$, $10n + 7$, and $10n + 9$, at least three are primes.\n\nIf $n = 3k$, then $10n + 3 = 30k + 3$ and $10n + 9 = 30k + 9$ are ... | Estonia | Estonija 2010 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
01nr | Define $M_n = \{1, 2, \dots, n\}$, for any $n \in \mathbb{N}$. A collection of 3-element subsets of $M_n$ is said to be *fine* if for any coloring of elements of $M_n$ in two colors there is a subset of the collection all three elements of which are of the same color.
For any $n \ge 5$ find the minimal possible number ... | [
"We call any 3-element subset a *triple*; a triple is said to be *monochromatic* if all its three elements are the same color. Let $f(n)$ denote the minimal possible number of the triples in a fine collection of $M_n$.\n\nFirst, we have $f(5) = 10 = \\binom{5}{3}$ — the total number of triples in $M_5$.\nIndeed, if... | Belarus | Belorusija 2012 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | Minimal number f(n): f(5) = 10, f(6) = 10, and f(n) = 7 for all n ≥ 7. | |
0679 | Let $p$ prime and $m$ a positive integer. Determine all pairs $(p, m)$ satisfying the equation: $p(p+m)+p=(m+1)^3$. (A. Fellouris) | [
"The given equation is written\n\n$$\np(p+m+1) = (m+1)^3\n$$\nTherefore the prime $p$ is a divisor of $(m+1)^3$. Hence $p \\mid (m+1)$, which means that there exists positive integer $k$ such that $m+1 = kp$. Then, from (1) we get:\n$$\np(p+kp) = (kp)^3 \\Leftrightarrow k+1 = k^3p \\Rightarrow k^3 \\mid (k+1) \\Rig... | Greece | 31st Hellenic Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (2, 1) | |
08jr | Problem:
Se consideră numerele naturale nenule $x$ și $y$ astfel încât $3x + 4y$ și $4x + 3y$ sunt ambele pătrate perfecte. Să se arate că numerele $x$ și $y$ sunt ambele divizibile cu $7$. | [] | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
08ww | Suppose the least common multiple of three positive integers $x$, $y$, $z$ is $2100$. What is the minimum possible value that the sum $x + y + z$ can take? | [
"44\nSince $2100 = 2^2 \\cdot 3 \\cdot 5^2 \\cdot 7$, if we take $x = 5^2 = 25$, $y = 7$, $z = 2^2 \\cdot 3 = 12$, then the least common multiple of $x$, $y$, $z$ is $2100$ and we get $x + y + z = 44$.\n\nNow let us show that $44$ is the desired minimum value. Assume that the least common multiple of $x$, $y$, $z$ ... | Japan | Japan 2013 Initial Round | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 44 | |
0her | Given a triangle $ABC$, where $O$ is its circumcenter, $M$ is a midpoint of $BC$, and $W$ is a point of the second intersection of bisector of $C$ with the circumcircle. The line parallel to $BC$ that passes through $W$ intersects $AB$ at point $K$, so that $BK = BO$. Find the angle $WMB$. | [
"Let the line that passes through $W$ parallel to $AB$ intersect the line $BC$ at point $T$ (Fig. 6). Then, $KWTB$ is a parallelogram and:\n$$WT = BK = BO = WO.$$\nNotice that $WO \\perp AB$, since $\\triangle ABO$ is isosceles, and $CW$ is a bisector of $\\angle BCA$, thus $\\angle OWT = 90^\\circ$. It is also cle... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 45° | |
07i2 | Let $M$ be the midpoint of side $BC$ of acute-angled triangle $ABC$, and let $E$ and $F$ be the feet of the perpendiculars from $M$ to sides $AC$ and $AB$, respectively. Points $X$ and $Y$ are such that $\triangle CEY \sim \triangle XEC$ and $\triangle XBF \sim \triangle BYF$ (the corresponding vertices of the triangle... | [
"We begin the proof with a lemma.\n**Lemma 1.** Triangle $ABC$ and point $X'$ are given such that $\\triangle BX'A \\sim \\triangle AX'C$ and $X'$ doesn't lie on line $BC$. The reflection of $A$ with respect to $X'$ lies on the circumcircle of $ABC$.\n*Proof.* We denote by $O$ the circumcenter of $ABC$ and by $AB, ... | Iran | 40th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line,... | null | proof only | null | |
00jw | A square and an equilateral triangle are inscribed to a circle. The seven vertices form a convex heptagon $H$ that is inscribed to the circle. (As a special case $H$ can be a hexagon if a vertex of the square coincides with a vertex of the triangle.)
For which positions of the triangle relative to the square does $H$ ... | [
"The square divides the circle into four arcs. None of them can contain more than one of the triangle vertices, because the distance between two triangle vertices is equivalent to an inner angle of $120^{\\circ}$, whereas the distance between two vertices of the square is only equivalent to an inner angle of $90^{\... | Austria | AustriaMO2013 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | Maximum area: when a triangle vertex lies at the midpoint of the arc between two adjacent square vertices, equivalently when the side opposite that vertex is parallel to a side of the square. Minimum area: when one triangle vertex coincides with a vertex of the square (e.g., the triangle shares C or D with the square),... | |
0a36 | Let $n$ be a positive integer. A grasshopper stands on the number line at the number $1$ and may make either a jump of length $2$ or of length $3$ each time. Each time, the grasshopper must land on an integer from $1$ through $n$ where the grasshopper has not been before. The grasshopper would like to visit all integer... | [
"We distinguish different cases for $n$ based on the remainder of $n$ when dividing by $3$.\n\nIf $n = 3k$, then the grasshopper can jump as follows:\n$1; 3, 6, \\ldots, 3(k-1); 3(k-1)+2, 3(k-2)+2, \\ldots, 2; 4, 7, \\ldots, 3(k-1)+1; 3k$.\nThe grasshopper jumps over triples on the way out, over triples plus $2$ on... | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms"
] | null | proof only | null | |
077q | Let $r > 0$ be a real number. All the interior points of the disc $D(r)$ of radius $r$ are colored with one of two colors, red or blue.
1. If $r > \frac{\pi}{\sqrt{3}}$, show that we can find two points $A$ and $B$ in the interior of the disc such that the distance $AB = \pi$ and $A$ and $B$ have the same color.
2. D... | [] | India | EGMO TST Day 1 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 1) True for radius greater than pi divided by square root of three. 2) Yes, it also holds for radius greater than pi over two. | |
095y | Problem:
Să se arate, că pentru oricare număr natural $n$ există numerele naturale $x$ şi $y$ astfel, încât $2(n-x y)=x(x+1)+y(y+3)$. | [
"Solution:\n\nDin egalitate se obţine\n$$\n2 n = x^{2} + 2 x y + y^{2} + x + 3 y = (x + y)^{2} + (x + y) + 2 y = (x + y)(x + y + 1) + 2 y\n$$\nde unde $n = \\frac{(x + y)(x + y + 1)}{2} + y$. Fie $x + y = z$; atunci $n = \\frac{z(z + 1)}{2} + y$. Fracţia este un număr triunghiular; $\\frac{z(z + 1)}{2} = T_{z}$. Se... | Moldova | A 62-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0435 | In ellipse $\Gamma$, $A$ is an endpoint of the major axis, $B$ is an endpoint of the minor axis, and $F_1$, $F_2$ are the foci. If $\overrightarrow{AF_1} \cdot \overrightarrow{AF_2} + \overrightarrow{BF_1} \cdot \overrightarrow{BF_2} = 0$, then find the value of $\tan \angle ABF_1 \cdot \tan \angle ABF_2$. | [
"By symmetry, suppose the equation of $\\Gamma$ is $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$ ($a > b > 0$), and $A(a, 0)$, $B(0, b)$, $F_1(-c, 0)$, $F_2(c, 0)$, where $c = \\sqrt{a^2 - b^2}$.\nBy the given conditions, we know that\n$$\n\\begin{aligned}\n\\overrightarrow{AF_1} \\cdot \\overrightarrow{AF_2} + \\overr... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | -1/5 | |
0as0 | Problem:
Give three real roots of $\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1$. | [
"Solution:\nAny three numbers in $[5,10]$.\n\n$\\sqrt{x+3-4 \\sqrt{x-1}}+\\sqrt{x+8-6 \\sqrt{x-1}}=1$\n\ncan be written\n\n$\\sqrt{(x-1)-4 \\sqrt{x-1}+4}+\\sqrt{(x-1)-6 \\sqrt{x-1}+9}=\\sqrt{(\\sqrt{x-1}-2)^{2}}+\\sqrt{(\\sqrt{x-1}-3)^{2}}=1$\n\nhence $|\\sqrt{x-1}-2|+|\\sqrt{x-1}-3|=1$.\n\nThere are four cases.\n\... | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Other"
] | null | final answer only | Any three numbers in the interval [5, 10] | |
0cvn | Let $n \ge 2$ be a positive integer. Petya and Vasya play the following game. Petya chooses $2n$ (not necessarily distinct) nonnegative real numbers $x_1, x_2, \dots, x_{2n}$ whose sum equals $1$, and tells those numbers to Vasya. Vasya arranges those numbers in a circle by his own choice, computes the product in each ... | [
"Если Петя выберет числа $0, \\frac{1}{2}, \\frac{1}{4(n-1)}, \\frac{1}{4(n-1)}, \\dots, \\frac{1}{4(n-1)}$, то, как бы ни расставлял эти числа Вася, число $\\frac{1}{2}$ будет в одной паре с числом $\\frac{1}{4(n-1)}$. Значит, одно из произведений будет равно $\\frac{1}{8(n-1)}$, а остальные будут не больше него. ... | Russia | Regional round | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English; Russian | proof and answer | 1/(8(n-1)) | |
06fi | Let $H$ be an arbitrary point on the altitude $CP$ of an acute triangle $ABC$. The lines $AH$ and $BH$ intersect $BC$ and $AC$ in $M$ and $N$ respectively. Show that $\angle NPC = \angle MPC$. | [
"(Blanchet's theorem) Suppose the line passing through $C$ and parallel to $AB$ meets $PN$ and $PM$ at $X$ and $Y$ respectively. Since $PC \\perp XY$, the result is the same as $CX = CY$ (as this implies $\\triangle PCX \\cong \\triangle PCY$).\n\nBy Ceva's theorem, we have\n$$\n\\frac{AP}{PB} \\times \\frac{BM}{MC... | Hong Kong | Year 2008 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0dma | Problem:
Одредити све парове природних бројева $(x, n)$ који су решења једначине
$$
x^{3}+2x+1=2^{n}
$$ | [
"Solution:\n\nПровером се добија да је за $n \\leqslant 2$ једино решење пар $(1,2)$. Докажимо да за $n \\geqslant 3$ нема решења.\n\nБрој $x$ мора бити непаран, па је $x^{2}+2 \\equiv 3 \\pmod{8}$. Сада из $x\\left(x^{2}+2\\right) \\equiv -1 \\pmod{8}$ следи да је $x \\equiv 5 \\pmod{8}$. Шта више, како $3 \\mid x... | Serbia | СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Quadratic reciprocity"
] | null | proof and answer | (1, 2) | |
09kj | Let $ABC$ be an acute triangle, where the altitudes $BD$ and $CE$ are drawn. Let $L$ be the point on segment $BD$ such that $AD = DL$, and let $K$ be the point on segment $CE$ such that $AE = EK$. Denote $M$ as the midpoint of segment $KL$. The circumcircle of triangle $ABC$ intersects line $AL$ again at point $T$ and ... | [
"\nLet $P = (BS) \\cap (TC)$. Since $M$ is the midpoint of $LK$, it follows that $\\frac{\\sin \\angle LAM}{\\sin \\angle MAK} = \\frac{AK}{AL}$. Since $\\angle BAL = \\angle KAC = \\angle BAC - 45^\\circ$, $\\angle SBC = \\angle TCB$ or $BTSC$ is an isosceles trapezoid. Applying the sine t... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06mg | Find the period of the repetend of the fraction $\frac{39}{1428}$ using binary numbers, i.e. its binary decimal representation.
(Note: When a proper fraction is expressed as a decimal number (of any base), either the decimal number terminates after finite steps, or it is of the form
$$
0.b_1b_2\cdots b_s a_1a_2\cdots ... | [
"Note that $\\frac{39}{1428} = \\frac{13}{476}$ in lowest term. Let\n$$\n\\frac{13}{476} = 0 \\cdot b_1 b_2 \\cdots b_s \\overline{a_1 a_2 \\cdots a_k} \\qquad (1)\n$$\nin binary decimal representation, where $a_1 a_2 \\cdots a_k$ is its repetend and $k$ is its period. We can write $0.b_1 b_2 \\cdots b_s = \\frac{B... | Hong Kong | The Twenty-fifth Hong Kong (China) Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 24 | |
04f2 | Prove that any 2001-element subset of the set $\{1, 2, 3, \dots, 3000\}$ contains three elements such that each two of them are relatively prime. | [
"Let $S$ be a set of 2001 distinct positive integers from the set $\\{1, 2, 3, \\dots, 3000\\}$. Let us look at 500 sets\n$$\nK_j = \\{6j + i \\mid i = 1, 2, 3, 4, 5, 6\\}, \\quad j = 0, 1, 2, \\dots, 499.\n$$\nThere is a set with at least five elements of $S$ among them (since $4 \\cdot 500 < 2001$).\nIf three of ... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
041g | Find all nonempty sets $S$ of integers such that $3m-2n \in S$ for all (not necessarily distinct) $m, n \in S$. | [
"Call a set $S$ \"good\" if it satisfies the property as stated in the problem.\n\n(1) If $S$ has only one element, $S$ is \"good\".\n\n(2) Now we can assume that $S$ contains at least two elements. Let\n$$\nd = \\min\\{|m-n| : m, n \\in S, m \\neq n\\}.\n$$\nThen there is an integer $a$, such that $a+d, a+2d \\in ... | China | China Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | All such sets are exactly the following, for some integers a and positive integer d: (1) the singleton {a}; (2) the set {a + k d where k ranges over all integers not divisible by three}; (3) the full arithmetic progression {a + k d where k ranges over all integers}. | |
0e1l | A teacher invited a group of children to sit down at a round table. There were three times as many boys as there were girls. The teacher walked around the table and observed the pairs of children sitting next to each other. She noticed that the number of same-sex pairs was twice the number of boy-girl pairs. At least h... | [
"Let $x$ be the number of girls and $y$ the number of boys. Then $y = 3x$ and there are $4x$ children altogether.\n\nLet $a$ be the number of boy-girl pairs. Then the number of same-sex pairs is equal to $2a$ and there are $3a$ pairs altogether. The number of pairs equals the number of all children. Indeed, every c... | Slovenia | National Math Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 12 | |
00z9 | Problem:
Consider the functions $f$ defined on the set of integers such that
$$
f(x) = f\left(x^{2} + x + 1\right),
$$
for all integers $x$. Find
(a) all even functions,
(b) all odd functions of this kind. | [
"Solution:\n(a) For $f$ even, we have $f(x-1) = f\\left((x-1)^{2} + (x-1) + 1\\right) = f\\left(x^{2} - x + 1\\right) = f\\left((-x)^{2} - x + 1\\right) = f(-x) = f(x)$ for any $x \\in \\mathbb{Z}$. Hence $f$ has a constant value; any constant will do.\n\n(b) For $f$ odd, a similar computation yields $f(x-1) = -f(x... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (a) All constant functions on the integers. (b) Only the zero function. | |
01t3 | Points $K$ and $N$ are marked on the sides $AB$ and $AC$ of the triangle $ABC$, respectively, such that $KB = KN$. The bisector of the angle $ACB$ meets the circumcircle of $ABC$ at points $C$ and $R$. The perpendicular from $R$ on $AB$ intersects the segment $BN$ at $D$.
Prove that the points $A$, $K$, $D$, $N$ are co... | [
"Since $R$ is the midpoint of the arc $\\angle AB$, the diameter of the circumcircle passing through $R$ is perpendicular to the chord $AB$. So, $RD$ is a perpendicular bisector of the segment $AB$. Therefore, $AD = BD$ and $\\angle ABD = \\angle BAD$.\n\nBy condition, $KB = KN$, then $\\angle KBN = \\angle KNB$. H... | Belarus | 66th Belarusian Mathematical Olympiad | [
"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 | |
0c8p | Consider $a, b \in \mathbb{N}^*$. Define the sequence $(x_n)_{n \in \mathbb{N}}$ by $x_0 = 0, x_1 = 1$ and
$$
x_{n+2} = a x_{n+1} + b x_n, \quad \forall n \in \mathbb{N}
$$
and the matrix $A$, by
$$
A = \begin{bmatrix} a & b \\ 1 & 0 \end{bmatrix}.
$$
a) Prove that
$$
A^n = \begin{bmatrix} x_{n+1} & b x_n \\ x_n & b x... | [
"a) We shall use induction on $n$.\nAs $x_2 = a \\cdot 1 + b \\cdot 0 = a$, we have\n$$\nA = \\begin{bmatrix} a & b \\\\ 1 & 0 \\end{bmatrix} = \\begin{bmatrix} x_2 & b x_1 \\\\ x_1 & b x_0 \\end{bmatrix},\n$$\nso the property is true for $n = 1$.\nConsider now that the formula is true for some $n \\ge 1$. Recurren... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0a1h | There are $2023$ people sitting at a round table. Each person is either a knave or a knight. Knights always speak the truth and knaves always lie. The first person says: "There is at least one knave at this table." The person to the left says: "There is at least one knight at this table." The third one says: "There are... | [
"B) $506$"
] | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Logic"
] | English | MCQ | B | |
02ic | Problem:
Quais os valores de $x$ que satisfazem $\frac{1}{x-2}<4$?
(A) $x<\frac{9}{4}$
(B) $x>2$
(C) $2<x<\frac{9}{4}$
(D) $x<-2$
(E) $x<2$ ou $x>\frac{9}{4}$ | [
"Solution:\n\n$\\frac{1}{x-2}<4 \\Rightarrow \\frac{1}{x-2}-4<0 \\Rightarrow \\frac{1-4(x-2)}{x-2}=\\frac{9-4x}{x-2}<0$\n\n$1^\\circ$ caso : $9-4x>0$ e $x-2<0$ :\n$9-4x>0 \\Rightarrow x<\\frac{9}{4}$ e $x-2<0 \\Rightarrow x<2$.\nComo $2<\\frac{9}{4}$ a solução são todos os números $x$ menores que 2, isto é $x<2$.\n... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | (E) | |
043e | Suppose $f(x)$ is an odd function with domain $\mathbb{R}$. If $f(1) = 2$, $f(2) = 3$, then the value of $f(f(-1))$ is ______. | [
"By the given conditions we have $f(-1) = -f(1) = -2$. Therefore,\n$$\nf(f(-1)) = f(-2) = -f(2) = -3.\n$$"
] | China | China Mathematical Competition | [
"Precalculus > Functions"
] | null | final answer only | -3 | |
0erj | A and $B$ are opposite vertices of a regular hexagon. $C$ and $D$ are midpoints of two opposite sides. If the area of the hexagon is $126$, then $AB \times CD$ is
(A) $129$ (B) $132$ (C) $84\sqrt{3}$ (D) $168$ (E) $248$ | [
"By joining opposite vertices, the hexagon can be divided into six congruent equilateral triangles. If we form a rectangle around the hexagon by drawing lines through $A$ and $B$ parallel to $CD$, then the area of the rectangle is $AB \\times CD$. Next, the portion of the rectangle outside the hexagon is composed o... | South Africa | South African Mathematics Olympiad First Round | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | MCQ | D | |
051w | Ats and Pets both thought of two positive integers that do not exceed some positive integer $n$. If they both added the numbers they thought of, then both sums gave the same remainder when divided by $n$. But if both of them multiplied the numbers they thought of, then both products also gave equal remainders when divi... | [
"a) If Ats thought of numbers $1$ and $21$ and Pets of numbers $10$ and $12$, then both get the sum $22$ and the products will be $21$ and $120$, respectively, both of which give the remainder $21$ when divided by $99$.\n\nb) Let the numbers Ats chose be $a$ and $b$, the ones Pets chose $c$ and $d$. According to th... | Estonia | Final Round of National Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) No; for n = 99, for example, one pair can be 1 and 21 and the other 10 and 12. b) Yes; for n = 101 the pairs must be identical. | |
0ajc | A group of mathematicians is attending a conference. We say that a mathematician is $k$-content if he is in room with at least $k$ people he admires or if he is admired by at least $k$ other people in the room. It is known that when all participants are in the same room then they are all at least $3k+1$-content. Prove ... | [
"We will for simplicity and clarity of presentation use some basic graph theoretic terms, this is in no way essential.\nWe represent the situation by a directed graph (abbr. digraph) $G(V, E)$ where each vertex $v \\in V(G)$ represents a mathematician and each edge $e \\in E(G)$ represents an admiration relation. G... | North Macedonia | European Mathematical Cup | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
0bbw | Find all positive integers $n$ for which there exists a set $S \subset \mathbb{C}$ with $n$ elements, such that
i) each element of $S$ has modulus $1$,
ii) $\sum_{z \in S} z = 0$, and
iii) $z + w \neq 0$, for each $z, w \in S$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Circles"
] | null | proof and answer | All positive integers except 1, 2, and 4 | |
011e | Problem:
Determine all positive real numbers $x$ and $y$ satisfying the equation
$$
x + y + \frac{1}{x} + \frac{1}{y} + 4 = 2 \cdot (\sqrt{2x+1} + \sqrt{2y+1}).
$$ | [
"Solution:\n$x = y = 1 + \\sqrt{2}$.\n\nNote that\n$$\nx + \\frac{1}{x} + 2 - 2 \\sqrt{2x+1} = \\frac{x^2 + 2x + 1 - 2x \\sqrt{2x+1}}{x} = \\frac{1}{x}(x - \\sqrt{2x+1})^2.\n$$\nHence the original equation can be rewritten as\n$$\n\\frac{1}{x}(x - \\sqrt{2x+1})^2 + \\frac{1}{y}(y - \\sqrt{2y+1})^2 = 0.\n$$\nFor $x,... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | x = y = 1 + sqrt(2) | |
05di | Problem:
Let $n$ be a positive integer.
a. Prove that there exists a set $S$ of $6 n$ pairwise different positive integers, such that the least common multiple of any two elements of $S$ is no larger than $32 n^{2}$.
b. Prove that every set $T$ of $6 n$ pairwise different positive integers contains two elements the ... | [
"Solution:\n\na.\nLet the set $A$ consist of the $4 n$ integers $1, 2, \\ldots, 4 n$ and let the set $B$ consist of the $2 n$ even integers $4 n + 2, 4 n + 4, \\ldots, 8 n$. We claim that the $6 n$-element set $S = A \\cup B$ has the desired property.\n\nIndeed, the least common multiple of two (even) elements of $... | European Girls' Mathematical Olympiad (EGMO) | EGMO | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
00o3 | Alice and Bob play a game on a strip of $n \ge 3$ squares with two game pieces. At the beginning, Alice's piece is on the first square while Bob's piece is on the last square. The figure shows the starting position for a strip of $n = 7$ squares.
The players alternate. In each move, they adva... | [
"Bob wins for $n = 3k + 2$ with $k \\in \\mathbb{Z}_{\\ge 1}$, Alice wins for all other $n \\ge 3$.\n\nIt is easily checked that Alice wins for 3, 4, 6 and 7 squares while Bob wins for 5 or 8 squares. We conjecture that Bob wins for all $n$ of the form $3k + 2$ and prove it by induction.\n\nWe include the case $k =... | Austria | AUT_ABooklet_2023 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Bob can force a win exactly when n ≡ 2 (mod 3). Alice can force a win for all other n ≥ 3. | |
069l | Determine all pairs $(x, y)$ of positive integers satisfying the equation
$$
yx^y = y + 1
$$ | [
"Let $y = \\frac{p}{q}$, with $p, q \\in \\mathbb{N}^*$, $(p, q) = 1$. Then the equation is written in the form:\n$$\nx^q = \\frac{y+1}{y} = \\frac{\\frac{p}{q}+1}{\\frac{p}{q}} = \\frac{p+q}{p} \\Rightarrow x = \\sqrt[q]{\\frac{(p+q)^q}{p^q}}.\n$$\nTherefore the right hand side must be rational. Since $(p+q, p)=1$... | Greece | 36th Hellenic Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | (2, 1) | |
0epl | How many zeroes will there be in the answer if the following number is divided by $111$?
$1110222222003333333330004444444444440000$ | [
"$\\frac{1110222222003333333330004444444444440000}{111} = 100020020000300300300000400400400400000$, which contains $28$ zeroes."
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | 28 | |
0ija | Problem:
At time $0$, an ant is at $(1,0)$ and a spider is at $(-1,0)$. The ant starts walking counterclockwise along the unit circle, and the spider starts creeping to the right along the $x$-axis. It so happens that the ant's horizontal speed is always half the spider's. What will the shortest distance ever between ... | [
"Solution:\n\nPicture an instant in time where the ant and spider have $x$-coordinates $a$ and $s$, respectively. If $1 \\leq s \\leq 3$, then $a \\leq 0$, and the distance between the bugs is at least $1$. If $s > 3$, then, needless to say, the distance between the bugs is at least $2$. If $-1 \\leq s \\leq 1$, th... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | sqrt(7/8) | |
03pj | Let $a_1, a_2, \dots, a_{2n}$ be real numbers with $\sum_{i=1}^{2n-1} (a_{i+1} - a_i)^2 = 1$. Find the maximum value of $(a_{n+1} + a_{n+2} + \dots + a_{2n}) - (a_1 + a_2 + \dots + a_n)$. (posed by Leng Gangsong) | [
"First, for $n=1$, we have $(a_2 - a_1)^2 = 1$, $a_2 - a_1 = \\pm 1$. Then the maximum value of $a_2 - a_1$ is $1$.\n\nSecondly, for $n \\ge 2$, let $x_1 = a_1$, $x_{i+1} = a_{i+1} - a_i$, $i = 1, 2, \\dots, 2n-1$. Then $\\sum_{i=2}^{2n} x_i^2 = 1$, and $a_k = x_1 + \\dots + x_k$, $k = 1, 2, \\dots, 2n$. Using Cauc... | China | China Western Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | sqrt(n(2n^2+1)/3) | |
0g8m | 設一個三角形的三邊長分別為 $a, b, c$, 而 $a, b, c$ 三邊所對應的高分別為 $h_a, h_b, h_c$。
證明 $(\frac{a}{h_a})^2 + (\frac{b}{h_b})^2 + (\frac{c}{h_c})^2 \ge 4$.
In a triangle, let $a, b, c$ be the lengths of sides, and $h_a, h_b, h_c$ be the lengths of corresponding heights. Prove that $(\frac{a}{h_a})^2 + (\frac{b}{h_b})^2 + (\frac{c}{h_c})^2... | [
"將三角形的面積記為 $A$。易知 $ah_a = bh_b = ch_c = 2A$。故原命題等價於證明\n$$\na^4 + b^4 + c^4 \\ge 16A^2. \\quad (1)\n$$\n將(1)式的左邊減去右邊, 並由 Heron 公式得\n$$\n\\begin{aligned}\n& a^4 + b^4 + c^4 - 16A^2 \\\\ \n= & a^4 + b^4 + c^4 - (a+b+c)(a+b-c)(a-b+c)(-a+b+c) \\\\ \n= & 2a^4 + 2b^4 + 2c^4 - 2a^2b^2 - 2a^2c^2 - 2b^2c^2 \\\\ \n= & (a^2 - ... | Taiwan | 二〇一四數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
05s8 | Problem:
Soit $ABC$ un triangle, et soit $E$ et $F$ deux points appartenant respectivement aux droites $(AB)$ et $(AC)$, distincts de $A$, $B$ et $C$. Soit également $\Omega$ le cercle circonscrit à $ABC$, soit $O$ le centre de $\Omega$, et soit $\Gamma$ le cercle circonscrit à $AEF$. Enfin, soit $P$ le point d'inters... | [
"Solution:\n\nPuisque $(BC)$ et $(EF)$ sont deux droites d'intérêt notoire, on note $T$ leur point d'intersection, éventuellement rejeté à l'infini. Alors $Q$ appartient à la droite $(BC)$ si et seulement si, en angles de droites, on a $(BT, ET) = (ET, PT)$.\n\nOr, toujours en angles de droites, on sait déjà que\n$... | France | Préparation Olympique Française de Mathématiques | [
"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"
] | null | proof only | null | |
09x4 | Problem:
Een verzameling $S$ die bestaat uit 2019 (verschillende) positieve gehele getallen heeft de volgende eigenschap: het product van elke 100 elementen van $S$ is een deler van het product van de overige 1919 elementen. Wat is het maximale aantal priemgetallen dat $S$ kan bevatten? | [
"Solution:\nHet maximale aantal priemgetallen is 1819.\n\nWe beginnen met de constructie. Kies verschillende priemgetallen $p_{1}, p_{2}, \\ldots, p_{1819}$ en zij $P=p_{1} p_{2} \\cdots p_{1819}$. Neem\n$$\nS=\\left\\{p_{1}, p_{2}, \\ldots, p_{1819}, P, P \\cdot p_{1}, \\ldots, P \\cdot p_{199}\\right\\}\n$$\nVoor... | Netherlands | Selectietoets | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1819 | |
0apf | Problem:
It is given that $\triangle ABC \sim \triangle DEF$. If the area of $\triangle ABC$ is $\frac{3}{2}$ times that of $\triangle DEF$ and $AB = BC = AC = 2$, what is the perimeter of $\triangle DEF$? | [
"Solution:\n\n$2 \\sqrt{6}$\n\nWe first define a notation. Let $(XYZ)$ denote the area of $\\triangle XYZ$.\n\n$$\n\\begin{gathered}\n\\left(\\frac{AB}{DE}\\right)^2 = \\frac{(ABC)}{(DEF)} = \\frac{\\frac{3}{2}(DEF)}{(DEF)} = \\frac{3}{2} \\Longrightarrow \\frac{AB}{DE} = \\sqrt{\\frac{3}{2}} \\\\\nDE = \\sqrt{\\fr... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | final answer only | 2\sqrt{6} | |
07vk | Let $A$, $B$, $C$, $D$, $E$ be five points on a circle such that $|AB| = |CD|$ and $|BC| = |DE|$. The segments $AD$ and $BE$ intersect at $F$. Let $M$ denote the midpoint of segment $CD$. Prove that the circle of center $M$ and radius $ME$ passes through the midpoint of segment $AF$. | [
"**Solution 1.** Because the arcs $AB$ and $CD$, as well as $BC$ and $DE$ are equal, the cyclic quadrilaterals $ABCD$ and $BCDE$ are isosceles trapeziums with $BC \\parallel AD$ and $CD \\parallel BE$. Hence, $BFDC$ is a parallelogram. As a consequence we see that $|BF| = |CD| = |BA|$ and so $\\triangle AFB$ is iso... | Ireland | IRL_ABooklet_2023 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
01ct | Find all combinations of four integers $(a, b, c, d)$ satisfying the equations
$$
\begin{cases} -a^2 + b^2 + c^2 + d^2 = 1 \\ 3a + b + c + d = 1. \end{cases}
$$ | [
"**Answer:** The solutions are: $(0, 1, 0, 0)$, $(0, 0, 1, 0)$, $(0, 0, 0, 1)$ and $(1, 0, -1, -1)$, $(1, -1, 0, -1)$, $(1, -1, -1, 0)$.\n\nWe write the equations as:\n$$\n\\begin{aligned}\nb^2 + c^2 + d^2 &= 1 + a^2 \\\\\nb + c + d &= 1 - 3a\n\\end{aligned}\n$$\nWe now apply the root-mean square and the arithmetic... | Baltic Way | Baltic Way 2016 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1), (1, 0, -1, -1), (1, -1, 0, -1), (1, -1, -1, 0) | |
0hk0 | Problem:
Six distinct numbers are chosen from the list $1, 2, \ldots, 10$. Prove that their product is divisible by a perfect square greater than 1. | [
"Solution:\n\nIf all the odd numbers are chosen, then in particular $9$ is chosen and the product is divisible by $9$.\nIf not all the odd numbers are chosen, then at most $4$ odds and thus at least $2$ evens are chosen. Therefore the product is divisible by $4$.",
"Solution:\n\nAt most one of the numbers is $1$,... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
05gu | Problem:
Soit $n, m \geqslant 1$ des entiers, avec $m$ impair.
Prouver que $2^{m}-1$ et $2^{n}+1$ sont premiers entre eux. | [
"Solution:\n\nPar l'absurde : supposons qu'il existe un nombre premier $p$ qui divise à la fois $2^{m}-1$ et $2^{n}+1$.\nAlors, on a $2^{m} \\equiv 1 \\bmod [p]$ et $2^{n} \\equiv -1 \\bmod [p]$,\nd'où $2^{2n} \\equiv 1 \\bmod [p]$.\nSoit $\\omega$ l'ordre de $2$ modulo $p$. On sait qu'alors $\\omega$ divise $m$ et... | France | Olympiades Françaises de Mathématiques, Envoi No. 6 | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0dlr | Find all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying the following conditions:
(i) there is at most one number $a$ such that $f(a) = 0,$
(ii) $f(f(x+y)f(x-y) + 25xy + y^2) = 24yf(x) + xf(x+y)$ for all real $x, y$. | [
"First, setting $y = 0$ to give $f(f(x)^2) = x f(x)$.\n\nThen setting $x = 0$ to get $f(f(y)f(-y) + y^2) = 24y f(0)$. Now change $y$ by $-y$ then the LHS does not change, so we will get\n$$\n24y f(0) = -24y f(0), \\forall y \\implies f(0) = 0.\n$$\nFrom $f(f(y)f(-y) + y^2) = 0$, one can get $f(y)f(-y) = -y^2$.\n\nS... | Saudi Arabia | Saudi Booklet | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x for all real x; f(x) = -x for all real x | |
06od | A sequence of real numbers $a_{0}, a_{1}, a_{2}, \ldots$ is defined by the formula
$$
a_{i+1}=\left\lfloor a_{i}\right\rfloor \cdot\left\langle a_{i}\right\rangle \quad \text{ for } \quad i \geq 0
$$
here $a_{0}$ is an arbitrary real number, $\left\lfloor a_{i}\right\rfloor$ denotes the greatest integer not exceeding $... | [
"First note that if $a_{0} \\geq 0$, then all $a_{i} \\geq 0$. For $a_{i} \\geq 1$ we have (in view of $\\left\\langle a_{i}\\right\\rangle<1$ and $\\left\\lfloor a_{i}\\right\\rfloor>0$ )\n$$\n\\left\\lfloor a_{i+1}\\right\\rfloor \\leq a_{i+1}=\\left\\lfloor a_{i}\\right\\rfloor \\cdot\\left\\langle a_{i}\\right\... | IMO | IMO 2006 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
0bj0 | Let $f : [0, 1] \to \mathbb{R}$ be a derivable function, with continuous derivative, and let $s_n = \sum_{k=1}^{n} f\left(\frac{k}{n}\right)$.
Prove that the sequence $(s_{n+1} - s_n)_{n \in \mathbb{N}^*}$ converges to $\int_0^1 f(x) dx$. | [
"Using the mean value theorem, we obtain\n$$\n\\begin{align*}\ns_{n+1} - s_n &= \\sum_{k=1}^{n+1} f\\left(\\frac{k}{n+1}\\right) - \\sum_{k=1}^{n} f\\left(\\frac{k}{n}\\right) \\\\\n&= f(1) - \\sum_{k=1}^{n} \\left( f\\left(\\frac{k}{n}\\right) - f\\left(\\frac{k}{n+1}\\right) \\right) \\\\\n&= f(1) - \\frac{1}{n(n... | Romania | 65th Romanian Mathematical Olympiad | [
"Calculus > Differential Calculus > Applications",
"Calculus > Integral Calculus > Techniques > Single-variable"
] | null | proof only | null | |
0coh | 2009 weights are placed in a row. Each weight is of integer number of grams, and no weight is heavier than 1 kg. It turned out that each neighboring weights differ by exactly 1 g, while the total weight of all weights (in grams) is an even number. Prove that it is possible to partition all the weights into two sets wit... | [
"Ясно, что веса всех гирь, стоящих на нечётных местах, имеют одинаковую чётность, а веса всех остальных гирь — другую чётность. Так как общий вес чётен, то 1005 гирь на нечётных местах имеют чётные веса.\n\nПоставим первую гирьку на левую чашку весов (пусть ее вес равен $2a \\le 1000$ г), остальные 2008 гирь разобь... | Russia | Regional round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English; Russian | proof only | null | |
00oe | Determine the smallest constant $C$ such that the inequality
$$
(X + Y)^2 (X^2 + Y^2 + C) + (1 - XY)^2 \geq 0
$$
holds for all real numbers $X$ and $Y$.
For which values of $X$ and $Y$ does equality hold for this smallest constant $C$? | [
"The smallest constant is $C = -1$. Equality holds for $X = Y = \\frac{1}{\\sqrt{3}}$ or $X = Y = -\\frac{1}{\\sqrt{3}}$.\n\nWe first investigate the case $X = Y$. It is easily seen that the inequality becomes equivalent to\n$$\n(3X^2 - 1)^2 + 4(C + 1)X^2 \\geq 0\n$$\nwhich implies $C \\geq -1$ by setting $X^2 = \\... | Austria | Austrian Mathematical Olympiad | [
"Algebra > Equations and Inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | C = -1, with equality at (X, Y) = (1/√3, 1/√3) or (X, Y) = (−1/√3, −1/√3). | |
0axz | Problem:
Determine the area of the polygon formed by the ordered pairs $(x, y)$ where $x$ and $y$ are positive integers which satisfy the equation
$$
\frac{1}{x} + \frac{1}{y} = \frac{1}{13}
$$ | [] | Philippines | Philippine Mathematical Olympiad Area Stage | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 12096 | |
0ecb | Problem:
Reši enačbo $\frac{4^{\log_{3} x}}{6} = \frac{2^{\log_{3}(x+1)}}{3}$. | [] | Slovenia | 15. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | (3 + sqrt(21)) / 2 | |
04y8 | Let $ABC$ be a triangle. Line $\ell$ is parallel to $BC$ and it respectively intersects side $AB$ at point $D$, side $AC$ at point $E$, and the circumcircle of the triangle $ABC$ at points $F$ and $G$, where points $F, D, E, G$ lie in this order on $\ell$. The circumcircles of triangles $FEB$ and $DGC$ intersect at poi... | [
"Let $\\omega_B, \\omega_C$ be the circumcircles of triangles $FEB$ and $DGC$ respectively. Since $PQ$ is the radical axis of $\\omega_B$ and $\\omega_C$, it is sufficient to prove that the powers of $A$ with respect to $\\omega_B$ and $\\omega_C$ are equal. If we denote by $X$ the second intersection of $\\omega_B... | Czech-Polish-Slovak Mathematical Match | null | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
048x | Let $ABCD$ be a unit square. The unit circle $k$ has center $C$. Determine the radius of the circle $k_1$ which touches the circle $k$ and line segments $\overline{AB}$ and $\overline{AD}$. | [] | Croatia | Hrvatska 2011 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 3 - 2√2 | |
0i5s | Problem:
Let $n$ be an integer. Prove that
$$
\left\lfloor\frac{n}{2}\right\rfloor+\left\lfloor\frac{n+1}{2}\right\rfloor=n .
$$ | [
"Solution:\nSuppose $n=2m$ is even. Then $\\lfloor n/2\\rfloor=\\lfloor m\\rfloor=m$ and $\\lfloor(n+1)/2\\rfloor=\\lfloor m+1/2\\rfloor=m$, whose sum is $m+m=2m=n$.\n\nOtherwise $n=2m+1$ is odd. In this case $\\lfloor n/2\\rfloor=\\lfloor m+1/2\\rfloor=m$ and $\\lfloor(n+1)/2\\rfloor=\\lfloor m+1\\rfloor=m+1$, who... | United States | HMMT 2002 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0jmg | Problem:
Let $\mathcal{H}$ be a regular hexagon with side length one. Peter picks a point $P$ uniformly and at random within $\mathcal{H}$, then draws the largest circle with center $P$ that is contained in $\mathcal{H}$. What is the probability that the radius of this circle is less than $\frac{1}{2}$? | [
"Solution:\nAnswer: $\\frac{2 \\sqrt{3}-1}{3}$\n\nWe first cut the regular hexagon $\\mathcal{H}$ by segments connecting its center to each vertex into six different equilateral triangles with side lengths $1$. Therefore, each point inside $\\mathcal{H}$ is contained in some equilateral triangle. We first see that ... | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | (2√3 - 1)/3 | |
08n5 | Problem:
Point $D$ lies on the side $[BC]$ of $\triangle ABC$. The circumcenters of $\triangle ADC$ and $\triangle BAD$ are $O_{1}$ and $O_{2}$, respectively, and $O_{1}O_{2} \parallel AB$. The orthocenter of $\triangle ADC$ is $H$ and $AH = O_{1}O_{2}$. Find the angles of $\triangle ABC$ if $2m(\angle C) = 3m(\angle ... | [
"Solution:\n\n\nAs $AD$ is the radical axis of the circumcircles of $\\triangle ADC$ and $\\triangle BAD$, we have that $O_{1}O_{2} \\perp AD$, therefore $\\widehat{DAB} = 90^{\\circ}$. Let $F$ be the midpoint of $[CD]$ and $[CE]$ be a diameter of the circumcircle of $\\triangle ADC$. Then ... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneo... | null | proof and answer | Angle A = 105 degrees, Angle B = 30 degrees, Angle C = 45 degrees | |
07ux | Let $N$ be a positive integer, $L$ and $x_i$ ($1 \le i \le N$) real numbers such that
$$
0 \le x_1 < x_2 < x_3 < \dots < x_N \le L.
$$
What is the least value of $N$ which ensures that at least 10 successive numbers $x_i$ all lie within a distance 1 of each other (i.e. we want $x_i, \dots, x_{i+9}$ so that $x_{i+9} \le... | [
"Let $M = \\lfloor L \\rfloor$ be the least integer greater than or equal to $L$. We claim that the least value of $N$ is $9M + 1$.\n\nFirst, we show that $9M + 1$ points are enough. Partition $[0, L]$ into subintervals $I_1, \\dots, I_M$, where\n$$\n\\begin{aligned}\nI_k &= [k-1, k), \\quad 1 \\le k < M \\\\\nI_M ... | Ireland | IRL_ABooklet | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 9*ceil(L) + 1 | |
0b1f | Problem:
In square $A B C D$, $P$ and $Q$ are points on sides $C D$ and $B C$, respectively, such that $\angle A P Q = 90^{\circ}$. If $A P = 4$ and $P Q = 3$, find the area of $A B C D$. | [
"Solution:\n\nNote that triangles $A D P$ and $P C Q$ are similar, so $A D / P C = A P / P Q = 4 / 3$. Let $A D = 4x$ and $P C = 3x$. Since $A B C D$ is a square, $P D = x$. Applying Pythagorean theorem on triangle $A D P$, we have $x^{2} + (4x)^{2} = 16$, so that $x^{2} = 16 / 17$. Hence, the area of square $A B C... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 256/17 | |
0amn | Problem:
Let $DAN$ be a triangle whose vertices lie on a circle $C$. Let $AE$ be the angle bisector of $\angle DAN$ with $E$ on $C$. If $DA = 2$, $AN = 1$, $AE = 2.5$, and $AE$ intersects $DN$ at $I$, find $AI$. | [] | Philippines | Area Stage | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 4/5 | |
0ksw | Problem:
Find, with proof, all functions $f: \mathbb{R} \setminus \{0\} \to \mathbb{R}$ such that
$$
f(x)^2 - f(y) f(z) = x(x+y+z)(f(x) + f(y) + f(z))
$$
for all real $x, y, z$ such that $x y z = 1$. | [
"Solution:\nThe answer is either $f(x) = 0$ for all $x$ or $f(x) = x^2 - \\frac{1}{x}$ for all $x$. These can be checked to work.\n\nNow, I will prove that these are the only solutions. Let $P(x, y, z)$ be the assertion of the problem statement.\n\nLemma 1. $f(x) \\in \\{0, x^2 - \\frac{1}{x}\\}$ for all $x \\in \\... | United States | HMMT February 2022 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | Either f(x) = 0 for all nonzero real x, or f(x) = x^2 − 1/x for all nonzero real x. | |
0gn4 | Given a circle with diameter $AB$ and a point $Q$ on the circle different from $A$ and $B$, let $H$ be the foot of the perpendicular dropped from $Q$ to $AB$. If the circle with center at $Q$ and radius $QH$ intersects the circle with diameter $AB$ at $C$ and $D$, prove that $CD$ bisects $QH$. | [] | Turkey | Team Selection Examination for the International Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null | |
0i05 | Problem:
If 5 points are placed in the plane at lattice points (i.e. points $(x, y)$ where $x$ and $y$ are both integers) such that no three are collinear, then there are 10 triangles whose vertices are among these points. What is the minimum possible number of these triangles that have area greater than $1/2$? | [
"Solution:\n\nBy the pigeonhole principle, the 5 points cannot all be distinct modulo 2, so two of them must have a midpoint that is also a lattice point. This midpoint is not one of the 5 since no 3 are collinear. Pick's theorem states that the area of a polygon whose vertices are lattice points is $B/2 + I - 1$ w... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 4 | |
0ijb | Problem:
Let $C$ be the unit circle. Four distinct, smaller congruent circles $C_{1}, C_{2}, C_{3}, C_{4}$ are internally tangent to $C$ such that $C_{i}$ is externally tangent to $C_{i-1}$ and $C_{i+1}$ for $i=1, \ldots, 4$ where $C_{5}$ denotes $C_{1}$ and $C_{0}$ represents $C_{4}$. Compute the radius of $C_{1}$. | [
"Solution:\nLet $O$ and $O'$ be the centers of $C$ and $C_{1}$ respectively, and let $C_{1}$ be tangent to $C, C_{2}, C_{4}$ at $P, Q$, and $R$ respectively. Observe that $Q O R O'$ is a square and that $P, O'$, and $O$ are collinear. Thus, if $r$ is the desired radius, $1 = r + O O' = r + r \\sqrt{2}$, so that $r ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | sqrt(2) - 1 | |
0lb7 | Given a sequence of real numbers $(x_n)$
$$
x_1 = 3 \quad \text{and} \quad x_n = \frac{n+2}{3n}(x_{n-1} + 2), \quad \forall n \ge 2.
$$
Prove that the sequence has a finite limit as $n \to \infty$ and calculate this limit. | [
"For $n \\ge 1$, we have\n$$\nx_{n+1} - x_n = \\left(\\frac{n+3}{3(n+1)} - 1\\right)x_n + \\frac{2(n+3)}{3(n+1)} = \\frac{2}{3(n+1)}(n+3 - nx_n). \\quad (1)\n$$\nWe first prove that\n$$\nx_n > 1 + \\frac{3}{n} \\quad \\forall n \\ge 2. \\quad (2)\n$$\nThe proof proceeds by induction on $n$. For $n=2$ we have\n$$\nx... | Vietnam | Vietnam Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | Vietnamese | proof and answer | 1 | |
07ts | Using straight edge and compass only, show how to construct an equilateral triangle equal in area to a given triangle. | [
"**Analysis:** Let $ABC$ be the given triangle and let $2x$ be the length of the sides of the required equilateral triangle $XYZ$. Let $AD$ be an altitude of $ABC$, $h = |AD|$ and $a = |BC|$. The area of $ABC$ is equal to $ah/2$ and the area of $XYZ$ is $\\sqrt{3}x^2$. Then we must have $ah/2 = \\sqrt{3}x^2$, i.e.\... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | proof only | null | |
04ys | Find the minimal distance between two points, one of which is on the graph of function $y = e^x$ and the other on the graph of function $y = \ln x$. | [
"The graphs of functions $y = e^x$ and $y = \\ln x$ are symmetrical with respect to the line $y = x$ (see Fig. 18).\n\nHence, the distance between points on these graphs is minimal if and only if both points are at minimal distance from the line $y = x$.\n\nThe distance between the graph of the function $y = e^x$ a... | Estonia | Estonija 2010 | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications",
"Precalculus > Functions"
] | null | proof and answer | sqrt(2) | |
06zc | Problem:
$I$ is the incenter of the triangle $ABC$ and the incircle touches $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. $AD$ meets the incircle again at $P$. $M$ is the midpoint of $EF$. Show that $P$, $M$, $I$, $D$ are cyclic (or the points are collinear). | [
"\n\n$\\angle AEI = \\angle AME = 90^\\circ$, so $AEI$ and $AME$ are similar. Hence $AM / AE = AE / AI$ or $AM \\cdot AI = AE^2$. $AE$ is tangent to the incircle, so $AE^2 = AP \\cdot AD$. Hence $AM \\cdot AI = AP \\cdot AD$, so if $P$, $M$, $I$, $D$ are not collinear, then they are cyclic.... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06lf | Let $D$ be an arbitrary point inside $\triangle ABC$. Let $\Gamma$ be the circumcircle of $\triangle BCD$. The external angle bisector of $\angle ABC$ meets $\Gamma$ again at $E$. The external angle bisector of $\angle ACB$ meets $\Gamma$ again at $F$. The line $EF$ meets the extension of $AB$ and $AC$ at $P$ and $Q$ r... | [
"Let $BE$ and $CF$ intersect at the $A$-excentre $J$ of $\\triangle ABC$. We claim that $J$ is the desired fixed point.\n\nWe only consider the configuration as shown since the other cases are similar. Since $BE$ is the external angle bisector, we have $\\angle PBE = \\angle EBC$. This is equal to $\\angle EFJ$ as ... | 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"
] | null | proof only | null | |
0cui | The circle $\omega$ is circumscribed about an acute-angled triangle $ABC$. Points $D$ and $E$ are chosen on the sides $AB$ and $BC$, respectively, so that $AC \parallel DE$. Points $P$ and $Q$ are chosen on the smaller arc $AC$ of $\omega$ so that $DP \parallel EQ$. Rays $QA$ and $PC$ meet $DE$ at $X$ and $Y$, respecti... | [
"Since $ABCQ$ is cyclic and $AC \\parallel DE$, we have $\\angle BEX = \\angle BCA = \\angle BQA = \\angle BQX$. Therefore, $XBEQ$ is cyclic; similarly, $YBDP$ is also cyclic. Thus $\\angle XBEQ = \\angle XEQ = \\angle DEQ$ and $\\angle PBY = \\angle PDE$.\n\nBy the condition, $DP \\parallel EQ$. Therefore, $180^\\... | Russia | XLIII Russian mathematical olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
09oz | In triangle $ABC$, the internal angle bisectors $BE$ and $CF$ intersect at point $I$. Let $F \in AB$, $E \in AC$. Let $D$ be the foot of the perpendicular from $I$ to $BC$. If $\angle A = 60^\circ$, prove that
$$
S_{BCF} + S_{BCE} = 3S_{BIC},
$$
where $S_{XYZ}$ denotes the area of triangle $XYZ$.
(Bilegdemberel Bat-Amg... | [] | Mongolia | MMO2025 Round 3 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0ire | Problem:
Determine all real numbers $a$ such that the inequality $|x^{2}+2 a x+3 a| \leq 2$ has exactly one solution in $x$. | [
"Solution:\n\nAnswer: $1, 2$\n\nLet $f(x) = x^{2} + 2 a x + 3 a$. Note that $f(-3/2) = 9/4$, so the graph of $f$ is a parabola that goes through $(-3/2, 9/4)$. Then, the condition that $|x^{2} + 2 a x + 3 a| \\leq 2$ has exactly one solution means that the parabola has exactly one point in the strip $-2 \\leq y \\l... | United States | 11th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 1, 2 | |
02hs | Problem:
Determine o valor de $123456123456 \div 1000001 =$. | [
"Solution:\n\nÉ claro que com números tão grandes, a questão não pretende que se efetue a divisão. Para resolvê-la vamos usar alguns truques aritméticos:\n$$\n\\begin{aligned}\n& 123456123456 = 123456000000 + 123456 = 123456 \\times 1000000 + 123456 = \\\\\n& = 123456 \\times (1000000 + 1) = 123456 \\times 1000001\... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 123456 | |
086i | Problem:
Francesco e Andrea decidono di consultare l'oracolo matematico per sapere se hanno delle coppie $(x, y)$ di numeri (reali) fortunati. Per determinare la coppia (o le coppie) di numeri fortunati, l'oracolo chiede sia a Francesco che a Andrea il giorno $(g)$ e mese $(m)$ di nascita, dopodiché per ciascuno di lo... | [
"Solution:\n\nLa risposta è (D). Esaminiamo la situazione di Francesco: perché il sistema ammetta infinite soluzioni è necessario che la seconda equazione sia equivalente alla prima, ovvero che differiscano al più per una costante moltiplicativa. Il termine noto della seconda equazione è $362 = 181 \\times 2$ quind... | Italy | Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | MCQ | D | |
0fbo | Problem:
Probar que si el producto de $n$ números reales y positivos es igual a $1$, su suma es mayor o igual que $n$. | [
"Solution:\n\nPrimera solución\n\nConsideremos $a_{1} \\cdot a_{2} \\cdot \\ldots \\cdot a_{n}=1$ y $S=a_{1}+a_{2}+\\cdots+a_{n}$.\nSuponemos que no todos son $1$.\nSea $a_{r}$ el mayor y $a_{s}$ el menor de los elementos $a_{i}$, $a_{r} \\geq 1$ y $a_{s} \\leq 1$. Se consideran los $n$ números obtenidos al cambiar... | Spain | OME 12 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
0l1g | Problem:
It can be shown that there exists a unique polynomial $P$ in two variables such that for all positive integers $m$ and $n$,
$$
P(m, n)=\sum_{i=1}^{m} \sum_{j=1}^{n}(i+j)^{7} .
$$
Compute $P(3,-3)$. | [
"Solution:\nNote that for integers $m>0, n>1$,\n$$\nP(m, n)-P(m, n-1)=\\sum_{i=1}^{m}(i+n)^{7}\n$$\nFor any given positive integer $m$, both sides are a polynomial in $n$, so they must be equal as polynomials. In particular,\n$$\nP(3, x)-P(3, x-1)=\\sum_{i=1}^{3}(i+x)^{7}=(x+1)^{7}+(x+2)^{7}+(x+3)^{7}\n$$\nfor all ... | United States | HMMT February 2024 Guts Round | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | final answer only | -2445 | |
01vr | A point $O$ is chosen inside a triangle $ABC$ so that the lengths of segments $OA$, $OB$ and $OC$ are equal to $15$, $12$ and $20$, respectively. It is known that the feet of the perpendiculars from $O$ to the sides of the triangle $ABC$ are the vertices of an equilateral triangle.
Find the value of the angle $ABC$. | [
"Answer: $\\angle B = 90^\\circ$.\n\nLet $A_1$, $B_1$ and $C_1$ be the feet of the perpendiculars from $O$ to the sides $BC$, $CA$ and $AB$ respectively. Since $\\angle OA_1B = 90^\\circ = \\angle OC_1B$, the quadrilateral $BA_1OC_1$ is cyclic and $OB$ is the diameter of its circumcircle. From the sine law\n\nfor t... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 90° | |
0c0h | Given an integer $n \ge 2$, determine the integral part of the number
$$
\sum_{k=1}^{n-1} \frac{1}{\left(1 + \frac{1}{n}\right) \cdots \left(1 + \frac{k}{n}\right)} - \sum_{k=1}^{n-1} \left(1 - \frac{1}{n}\right) \cdots \left(1 - \frac{k}{n}\right).
$$ | [
"The required integral part is $0$. The difference of the two sums is positive, since positivity clearly holds termwise.\n\nTo show it less than $1$, let $u_k = n^k/k!$, $k = 0, 1, 2, \\dots$, and express the two sums in terms of the $u_k$. The first sum is $(u_{n+1} + \\cdots + u_{2n-1})/u_n$, and the second is $(... | Romania | 69th NMO Selection Tests for BMO and IMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 0 | |
099k | Let $a$, $b$, $c > 1$ positive integer numbers. Prove that there exists infinitely many $a$, $b$, $c$ such that $a|b^2 - 1$, $b|c^2 - 1$, $c|a^2 - 1$.
(proposed by B. Battsengel) | [
"Let $n$ be a positive integer number. $a = 2n + 1$, $b = 4n + 1$, $c = 4n$ satisfy the desired conditions."
] | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0jgn | Problem:
a.
Given a finite set $X$ of points in the plane, let $f_{X}(n)$ be the largest possible area of a polygon with at most $n$ vertices, all of which are points of $X$. Prove that if $m, n$ are integers with $m \geq n > 2$, then $f_{X}(m) + f_{X}(n) \geq f_{X}(m+1) + f_{X}(n-1)$.
b.
Let $P_{0}$ be a 1-by-2 rect... | [
"Solution:\n\n## 1 Convexity of the largest area of a polygon function\n\na.\nLet $V$ be a finite set of points in the plane. Let $f_{V}$ be a function that takes integers $\\geq 3$ as input, and outputs a polygon of largest area with vertices in $V$.\n\nLemma 1.1. $\\left[f_{V}(a+1)\\right] + \\left[f_{V}(a-1)\\ri... | United States | HMIC | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | sqrt(2)/2^1006 | |
0ess | One hundred glasses are arranged in a $10 \times 10$ array. Now we pick $a$ of the rows and pour blue liquid into all glasses in these rows, so that they are half full. The remaining rows are filled halfway with yellow liquid. Afterwards, we pick $b$ of the columns and fill them up with blue liquid. The remaining colum... | [
"The total number of glasses that are green at the end of the procedure is\n$$\na(10 - b) + b(10 - a) = 10a + 10b - 2ab = 2(5a + 5b - ab).\n$$\nWe immediately observe that this number is always even, so the number of green glasses cannot be $25$ (i.e., one quarter). Hence the answer to the second question is no.\n\... | South Africa | The South African Mathematical Olympiad Third Round | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a. Exactly half are green if and only if either a = 5 or b = 5 (the other can be any integer from 0 to 10). b. No, one quarter green is impossible. | |
00hz | Let $ABC$ be an acute triangle. Let $D$ be a point on side $AB$ and $E$ be a point on side $AC$ such that lines $BC$ and $DE$ are parallel. Let $X$ be an interior point of $BCED$. Suppose rays $DX$ and $EX$ meet side $BC$ at points $P$ and $Q$, respectively, such that both $P$ and $Q$ lie between $B$ and $C$. Suppose t... | [
"\nLet $\\ell$ be the radical axis of circles $BQX$ and $CPX$. Since $X$ and $Y$ are on $\\ell$, it is sufficient to show that $A$ is on $\\ell$. Let line $AX$ intersect segments $BC$ and $DE$ at $Z$ and $Z'$, respectively. Then it is sufficient to show that $Z$ is on $\\ell$. By $BC \\para... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
098m | Problem:
Șirul $\left(x_{n}\right)_{n=0}^{\infty}$ este definit prin:
$$
x_{0}=1,\ x_{n+1}=x_{n}-\frac{x_{n}^{2}}{2023},\ n \in \mathbf{N}
$$
Demonstrați că şirul este convergent și $\lim _{n \rightarrow \infty} x_{n}=0$. | [
"Solution:\n\nVom arăta prin inducție că $x_{n} \\in(0 ; 1], \\forall n \\in \\mathbf{N}$. Pentru $n=0$ este adevărat, deoarece $x_{0}=1 \\in(0 ; 1]$. Presupunem că $x_{m} \\in(0 ; 1]$ pentru un oarecare $m \\geq 0$. Atunci și $\\frac{x_{m}}{2023} \\in(0 ; 1]$ și obținem\n$$\nx_{m+1}=x_{m}-\\frac{x_{m}^{2}}{2023}=x... | Moldova | Moldova National Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 0 | |
0dpn | Given a positive integer $n$. Let us call by a *word* any sequence of $n$ letters of an alphabet. Define the distance $\rho(A, B)$ between the words $A = a_1a_2...a_n$ and $B = b_1b_2...b_n$ as the number of positions in which they differ (i.e. the number of indices $i$ for which $a_i \ne b_i$). We say that the word $C... | [
"Answer: $n+1$ for $n \\ne 2$ and $4$ for $n = 2$.\nFor $n \\ne 2$, take all words consisting only of letters $a$ and $b$, in which each letter $a$ to the left of each letter $b$, and for $n = 2$ the words $aa, ab, ba, bb$.\n\nLet us prove that more words can not be chosen. Consider two selected words $A$ and $B$, ... | Silk Road Mathematics Competition | SILK ROAD MATHEMATICS COMPETITION XVII | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n+1 for n ≠ 2; 4 for n = 2 | |
0hvh | Problem:
A bulldozer is touring Pascal's triangle. It starts at the top of the triangle, at $\left(\begin{array}{l}0 \\ 0\end{array}\right)=1$. Each move, it travels to an adjacent positive integer, but can never return to a spot it has already visited. Moreover, if it has visited two numbers $a>b$, it may not visit $a... | [
"Solution:\nThe main idea is to visit odd numbers!\nWe claim inductively that the first $2^{n}$ rows of Pascal's triangle satisfy the following properties:\n- The $2^{n}$th row contains only odd numbers.\n- The first $2^{n}$ rows contain $3^{n}$ odd numbers.\n- When taken modulo 2, there is 120 degree symmetry\n- T... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null |
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