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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0lg9 | Problem:
A convex hexagon $A B C D E F$ is inscribed in a circle with radius $R$. Diagonals $A D$ and $B E$, $B E$ and $C F$, $A D$ and $C F$ of the hexagon meet at points $M$, $N$, $K$ respectively. Let $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}, r_{6}$ be the inradii of the triangles $A B M$, $B C N$, $C D K$, $D E M$, $E F ... | [
"Solution:\nWe start with a lemma.\n\nLemma. Let $R$ be the circumradius of a quadrilateral $X Y Z T$, the diagonals of $X Y Z T$ meet at $U$, and $\\varphi=\\frac{1}{2} \\angle X U Y$. Then the radii $r_{1}$ and $r_{2}$ of the incentres of $X Y U$ and $Z T U$ satisfy\n$$\n\\frac{r_{1}+r_{2}}{R} \\leqslant 2 \\tan ... | Zhautykov Olympiad | IZhO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Jensen/smoothing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry >... | null | proof only | null | |
0hcv | It is known that nonzero real numbers $x$, $y$, $z$ satisfy the condition $xy + yz + zx = 0$. What value can the expression
$$
\frac{1}{x^2+2yz} + \frac{1}{y^2+2zx} + \frac{1}{z^2+2xy}
$$
be equal to? | [
"Since $xyz \\neq 0$, we can do the following transformation of the given expression:\n$$\n\\begin{aligned}\n\\frac{1}{x^2+2yz} + \\frac{1}{y^2+2zx} + \\frac{1}{z^2+2xy} &= \\frac{1}{x^2+2yz-xy-yz-zx} + \\frac{1}{y^2+2zx-xy-yz-zx} + \\frac{1}{z^2+2xy-xy-yz-zx} \\\\\n&= \\frac{1}{x^2+yz-xy-zx} + \\frac{1}{y^2+zx-xy-... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | 0 | |
0ak3 | Let $n > 1$ be a positive integer and $a_1, a_2, \dots, a_n$ be a sequence of $n$ positive integers. Let
$$
b_i = \left[ \frac{a_1 + a_2 + \dots + a_{i-1} + a_{i+1} + \dots + a_n}{n-1} \right], \quad 1 \leq i \leq n.
$$
Let $f$ be a mapping such that $f(a_1, a_2, \dots, a_n) = (b_1, b_2, \dots, b_n)$.
a) Let the funct... | [
"a.\nLet $n > 2$. We will show that for $m$ big enough, $g(m) = 1$.\nLet $a_1, a_2, \\dots, a_n$ be a sequence of positive integers. Then\n$$\nf(a_1, a_2, \\dots, a_n) = \\left( \\left[ \\frac{a_2 + a_3 + \\dots + a_n}{n-1} \\right], \\left[ \\frac{a_1 + a_3 + \\dots + a_n}{n-1} \\right], \\dots, \\left[ \\frac{a_1... | North Macedonia | Macedonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0g1g | Problem:
Sei $ABC$ ein Dreieck mit $AC > AB$. Sei $P$ der Schnittpunkt von $BC$ und der Tangente durch $A$ am Umkreis des Dreiecks $ABC$. Sei $Q$ der Punkt auf der Geraden $AC$, sodass $AQ = AB$ gilt und $A$ zwischen $C$ und $Q$ liegt. Seien $X$ respektive $Y$ die Mittelpunkte von $BQ$ respektive $AP$. Sei $R$ der Pun... | [
"Solution:\n\n\n\nSei $S$ der Schnittpunkt von $AP$ und $BQ$. Wegen dem Tangentenwinkelsatz gilt $\\angle SAB = \\angle ACB = \\angle QCB$ und nach Voraussetzung gilt $\\angle ABS = \\angle BQC$. Daraus folgt, dass die Dreiecke $ABS$ und $BQC$ ähnlich sind. Folglich sind die Nebenwinkel $\\... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00y1 | Problem:
Compute the sum of all positive integers whose digits form either a strictly increasing or a strictly decreasing sequence. | [
"Solution:\n\nDenote by $I$ and $D$ the sets of all positive integers with strictly increasing (respectively, decreasing) sequence of digits. Let $D_{0}, D_{1}, D_{2}$ and $D_{3}$ be the subsets of $D$ consisting of all numbers starting with $9$, not starting with $9$, ending in $0$ and not ending in $0$, respectiv... | Baltic Way | Baltic Way 1993 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 25617208995 | |
09fm | Find all positive integer solutions of the equation
$$
(x^2 - 1)^2 (y^2 - 1)^2 + 16x^2 y^2 = z^2.
$$ | [
"Let $a := x^2 - 1$ and $b := y^2 - 1$. Then the given equation becomes\n$$\na^2 b^2 + 16(a+1)(b+1) = z^2.\n$$\nWhen $a, b > 13$, it is not hard to show that\n$$\n(ab + 8)^2 < z^2 < (ab + 9)^2,\n$$\nand hence there is no solution to the equation. In the case of at least one of $a, b$ is not greater than $13$, or eq... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (x, y, z) = (1, n, 4n) or (n, 1, 4n), for any integer n ≥ 1 | |
0ccq | Let $M$ be the midpoint of the side $CD$ of the square $ABCD$. The perpendicular from $C$ onto $BM$ meets the lines $BM$ at $N$, and $AB$ at $E$. The line $BM$ intersects the line $AD$ in $P$. Let $F$ be the midpoint of the segment $BN$. Prove that:
a) the triangles $CBE$ and $BAP$ are congruent;
b) the segments $AN$... | [
"a) From $CB = BA$ and $\\angle CEB = \\angle BPA$ (they have the same complement $\\angle EBP$), follows that the right triangles $CBE$ and $BAP$ are congruent.\n\nb) Denote $T$ the common point of the lines $EC$ and $AP$. The segments $DM$ and $AT$ are midlines in the right triangles $APB$ and $BEC$, so $A$ is th... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | proof only | null | |
07yf | Problem:
Sia $\Gamma$ la circonferenza ex-inscritta al triangolo $ABC$ opposta al vertice $A$, ossia la circonferenza tangente a $BC$ e ai prolungamenti dei lati $AB$ e $AC$ dalla parte di $B$ e di $C$. Sia $D$ il centro di $\Gamma$ e siano $E$ ed $F$, rispettivamente, i punti di tangenza di $\Gamma$ con i prolungamen... | [
"Solution:\n\nLe rette $BD$ e $CD$ sono bisettrici degli angoli $\\widehat{EBC}$ e $\\widehat{BCF}$; detti $\\alpha, \\beta, \\gamma$ gli angoli interni del triangolo $ABC$ (rispettivamente in $A, B, C$), sappiamo perciò che $\\widehat{BCD} = \\frac{180^{\\circ} - \\gamma}{2}$ e $\\widehat{CBD} = \\frac{180^{\\circ... | Italy | null | [
"Geometry > Plane Geometry > Circles > Tangents",
"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 ... | null | proof only | null | |
01n0 | Determine the greatest possible value of the constant $c$ that satisfies the following condition: for any convex heptagon the sum of the lengths of all its diagonals is greater than $cP$, where $P$ is a perimeter of the heptagon. | [
"The sum of the diagonals $A_1A_3$, $A_3A_5$, $A_5A_7$, $A_7A_2$, $A_2A_4$, $A_4A_6$, $A_6A_1$ is obviously greater than the perimeter $P$ of the heptagon (see Fig. 1).\n\n\nFig. 1\n\nSimilarly, the sum of diagonals $A_1A_4$, $A_4A_7$, $A_7A_3$, $A_3A_6$, $A_6A_2$, $A_2A_5$, $A_5A_1$ is gre... | Belarus | 62nd Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 2 | |
0k8u | Problem:
Compute the sum of all positive real numbers $x \leq 5$ satisfying
$$
x=\frac{\left\lceil x^{2}\right\rceil+\lceil x\rceil \cdot\lfloor x\rfloor}{\lceil x\rceil+\lfloor x\rfloor} .
$$ | [
"Solution:\nNote that all integer $x$ work. If $x$ is not an integer then suppose $n < x < n+1$. Then $x = n + \\frac{k}{2n+1}$, where $n$ is an integer and $1 \\leq k \\leq 2n$ is also an integer, since the denominator of the fraction on the right hand side is $2n+1$. We now show that all $x$ of this form work.\n\... | United States | HMMT November 2019 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 85 | |
04ai | Let $(a_n)$ be a sequence defined by
$$
a_0 = 9 \quad \text{and} \quad a_{k+1} = 3a_k^4 + 4a_k^3 \quad \text{for all } k \ge 0.
$$
Prove that the decimal representation of $a_{11}$ ends with at least 2011 digits 9. | [
"We will prove by induction that for all $n \\ge 0$, $a_n \\equiv -1 \\pmod{10^{2^n}}$.\n\nBase case: $n = 0$. $a_0 = 9 \\equiv -1 \\pmod{10}$, since $9 + 1 = 10$.\n\nInductive step: Assume $a_n \\equiv -1 \\pmod{10^{2^n}}$.\n\nConsider $a_{n+1} = 3a_n^4 + 4a_n^3$.\n\nLet $m = 10^{2^n}$. Then $a_n \\equiv -1 \\pmod... | Croatia | CroatianCompetitions2011 | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0e53 | Find all integral solutions $x$ and $y$ of the equation
$$
3xy + 2x + y = 12.
$$ | [
"Rewrite the equation as $x(3y+2) = 12 - y$. Obviously, $3y+2 \\neq 0$ divides $12 - y$, so $3y+2$ divides $3(12 - y) + (3y+2) = 38$. Since $3y+2$ gives the remainder of $2$ when divided by $3$, there are four possibilities. The number $3y+2$ is equal to $-19, -1, 2$ or $38$, so $y$ is equal to $-7, -1, 0$ or $12$ ... | Slovenia | National Math Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (-1, -7), (-13, -1), (6, 0), (0, 12) | |
0izo | Problem:
How many sequences of ten binary digits are there in which neither two zeroes nor three ones ever appear in a row? | [
"Solution:\nLet $a_{n}$ be the number of binary sequences of length $n$ satisfying the conditions and ending in $0$, let $b_{n}$ be the number ending in $01$, and let $c_{n}$ be the number ending in $11$. From the legal sequences of length $2$ ($01$, $11$, $10$), we find that $a_{2}=b_{2}=c_{2}=1$. We now establish... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 28 | |
0im9 | Problem:
Solve for the positive integer(s) $n$ such that $\phi\left(n^{2}\right)=1000 \phi(n)$. | [
"Solution:\n\nAnswer: $1000$.\n\nThe unique solution is $n=1000$. For, $\\phi(p n)=p \\phi(n)$ for every prime $p$ dividing $n$, so that $\\phi\\left(n^{2}\\right)=n \\phi(n)$ for all positive integers $n$."
] | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof and answer | 1000 | |
00o2 | Let $ABCDEF$ be a regular hexagon with sidelength $s$. The points $P$ and $Q$ are on the diagonals $BD$ and $DF$, respectively, such that $BP = DQ = s$.
Prove that the three points $C$, $P$ and $Q$ are on a line. | [
"\n\nFigure 1: Problem 2\n\nSolution:\n\nOur strategy is to compute the angles $\\angle DCQ$ and $\\angle DCP$ to check that they are equal.\n\nThe interior angles of a regular hexagon equal $120^\\circ$. The triangle $DEF$ is isosceles and therefore, we get $\\angle DFE = \\angle EDF = 30^... | Austria | AUT_ABooklet_2023 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
02r5 | Problem:
Um número inteiro $n$ é simpático quando existem inteiros positivos $a, b$ e $c$ tais que $a < b < c$ e $n = a^{2} + b^{2} - c^{2}$. Por exemplo, os números 1 e 2 são simpáticos, pois $1 = 4^{2} + 7^{2} - 8^{2}$ e $2 = 5^{2} + 11^{2} - 12^{2}$.
a) Verifique que $(3x+1)^{2} + (4x+2)^{2} - (5x+2)^{2}$ é igual ... | [
"Solution:\n\na) Lembrando que $(a+b)^{2} = a^{2} + 2ab + b^{2}$, podemos simplificar a expressão $(3x+1)^{2} + (4x+2)^{2} - (5x+2)^{2}$ como segue:\n$$\n(3x+1)^{2} + (4x+2)^{2} - (5x+2)^{2} = 9x^{2} + 6x + 1 + 16x^{2} + 16x + 4 - 25x^{2} - 20x - 4\n$$\n$$\n= (9 + 16 - 25)x^{2} + (6 + 16 - 20)x + (1 + 4 - 4) = 2x +... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | m = 4, n = 3 | |
01y2 | In the Paircity there are $n \ge 2$ married couples numbered from $1$ to $n$. On the New Year's eve, each man congratulated two women from couples whose numbers differ by one. It turned out that for any two men from couples whose numbers differ by one, there is a woman who received congratulations from both of them.
P... | [
"Call a man *leftside* if he congratulated women from couples with numbers less than his own, otherwise call him *rightside*. Clearly, the man from the first couple is leftside and the man from the $n$-th couple is rightside. Hence, there is a number $i$, $1 \\le i < n - 1$, such that the man from the $i$-th couple... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
08mo | Problem:
Let $x, y$ be positive real numbers such that $x^{3}+y^{3} \leq x^{2}+y^{2}$. Find the greatest possible value of the product $x y$. | [
"Solution:\n\nWe have $(x+y)\\left(x^{2}+y^{2}\\right) \\geq (x+y)\\left(x^{3}+y^{3}\\right) \\geq \\left(x^{2}+y^{2}\\right)^{2}$, hence $x+y \\geq x^{2}+y^{2}$. Now $2(x+y) \\geq (1+1)\\left(x^{2}+y^{2}\\right) \\geq (x+y)^{2}$, thus $2 \\geq x+y$. Because $x+y \\geq 2 \\sqrt{x y}$, we will obtain $1 \\geq x y$. ... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 1 | |
0ldm | Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that
$$
f(xf(y) - f(x)) = 2f(x) + xy
$$
for all real numbers $x, y$. | [
"By taking $x = 1$ into (1), we get\n$$\nf(f(y) - f(1)) = y + 2f(1), \\quad \\forall y \\in \\mathbb{R}. \\quad (2)\n$$\nHence $f$ is bijection and so, there exists a unique real number $a$ such that $f(a) = 0$. Plugging $x = a$ into (1), we have\n$$\nf(af(y)) = ay, \\quad \\forall y \\in \\mathbb{R}. \\quad (3)\n$... | Vietnam | Vietnamese Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = 1 - x | |
0e77 | Find all prime numbers $p$ and $q$ such that $p^4 - q^6$ is a power of a prime number. (Numbers $7$ and $8$ are powers of prime numbers, but $6$ is not.) | [
"Write $p^4 - q^6 = r^n$ for some prime $r$ and some positive integer $n$. The expression $p^4 - q^6$ can be factored as $p^4 - q^6 = (p^2 - q^3)(p^2 + q^3)$. Since $p^4 - q^6 > 0$, the primes $p$ and $q$ cannot be equal and are therefore relatively prime. Let $d$ be the greatest common divisor of the numbers $p^2 ... | Slovenia | National Math Olympiad 2013 - Final Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | p=3, q=2 | |
0byn | Let $(a_n)_{n \ge 1}$ be an arithmetic sequence of positive integers and let $S_n = a_1^2 + a_2^2 + \dots + a_n^2$, $n \in \mathbb{N}^*$. Prove that:
a) if $p \ge 5$ is a prime number, then $p$ divides $S_p$;
b) $S_5$ is not a square. | [
"a) We have $a_n = a_1 + (n-1)r$, $n \\in \\mathbb{N}^*$, with $a_1 \\in \\mathbb{N}^*$, $r \\in \\mathbb{N}$. This yields\n$$\nS_p = p \\left( a_1^2 + a_1(p-1)r + \\frac{(p-1)(2p-1)}{6} r^2 \\right).\n$$\nSince $p \\ge 5$ is a prime number, we have $\\gcd(p, 2) = 1$, $\\gcd(p, 3) = 1$, hence $6 \\mid (p-1)(2p-1)$,... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Diophantine Equations > Infinite descent / root flipping"
] | English | proof only | null | |
02bc | Problem:
Todos os ângulos de um hexágono $A B C D E F$ são iguais. Mostre que $\mathrm{AB}-\mathrm{DE}=\mathrm{EF}-\mathrm{BC}=\mathrm{CD}-\mathrm{FA}$. | [
"Solution:\n\nProlonguemos os segmentos $AF$, $BC$ e $DE$ determinando os pontos de intersecção $X$, $Y$, $Z$, como mostrado na figura.\n\nComo a soma dos ângulos internos de um hexágono convexo é $180^{\\circ} \\times (6-2) = 720^{\\circ}$, cada ângulo interno deste hexágono mede $720^{\\circ} / 6 = 120^{\\circ}$.... | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0g8h | 令 $\mathbb{N}, \mathbb{Z}, \mathbb{Q}$ 分別代表所有正整數、整數及有理數所成集合。試求所有函數 $f : \mathbb{Q} \to \mathbb{Z}$, 滿足
$$
f\left(\frac{f(x)+a}{b}\right) = f\left(\frac{x+a}{b}\right)
$$
對於所有 $x \in \mathbb{Q}, a \in \mathbb{Z}$ 和 $b \in \mathbb{N}$ 都成立。
Let $\mathbb{N}, \mathbb{Z}, \mathbb{Q}$ denote the set of all positive integers,... | [
"$f = \\lfloor x \\rfloor$, $f = \\lceil x \\rceil$, 以及常數函數。\n\n我們首先驗證以上函數滿足題意。對於所有三元組 $(x, a, b) \\in \\mathbb{Q} \\times \\mathbb{Z} \\times \\mathbb{N}$, 令\n$$\nq = \\lfloor \\frac{x+a}{b} \\rfloor.\n$$\n易知 $bq \\le x + a < b(q+1) \\Rightarrow bq \\le \\lfloor x \\rfloor + a < b(q+1)$, 故\n$$\n\\lfloor \\frac{\\l... | Taiwan | 國際數學奧林匹亞競賽第二階段選訓營 獨立研究(三) | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All constant functions from the rationals to the integers, and the two functions f(x) equal to the floor of x and f(x) equal to the ceiling of x. | |
0ldb | Given a triangle $ABC$ with fixed vertices $B$, $C$; point $A$ moves such that triangle $ABC$ is acute. Let $D$ be the midpoint of $BC$ and $E$, $F$ be the projections of $D$ to $AB$, $AC$ respectively.
a) Let $O$ be the circumcenter of triangle $ABC$. $EF$ meets $AO$ and $BC$ at $M$, $N$ respectively. Prove that the ... | [
"a) Without loss of generality, suppose that $AB < AC$. Clearly, $N$ lies on the opposite ray of ray $BC$. Note that $AEDF$ is a cyclic quadrilateral and $\\angle OAC = 90^\\circ - \\angle ABC$, we have\n$$\n\\begin{align*}\n\\angle AMN &= \\angle MAE + \\angle MEA = 90^\\circ - \\angle ABC + \\angle ADF \\\\\n&= \... | Vietnam | VMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations... | English | proof only | null | |
03gs | Problem:
Let $f(n)$ be the sum of the first $n$ terms of the sequence
$$
0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, \ldots
$$
a) Give a formula for $f(n)$.
b) Prove that $f(s+t)-f(s-t)=s t$ where $s$ and $t$ are positive integers and $s>t$. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a) f(n) = ⌊n/2⌋·⌈n/2⌉ = ⌊n^2/4⌋.
b) f(s+t) − f(s−t) = s·t for positive integers s > t. | |
0459 | Given a prime number $p$ and an infinite set $A \subset \mathbb{Z}$. Show that one can always find a subset $B$ of $A$, $B$ contains $2p-2$ elements, and for any $p$ distinct elements of $B$, their arithmetic mean does not belong to $A$.
(Contributed by Fu Yunhao) | [
"Assume that, on the contrary, for some infinite set $A \\subset \\mathbb{Z}$, one cannot find a $(2p-2)$-element subset $B$ satisfying the problem conditions. Without loss of generality, assume that $A$ contains infinitely many positive integers; otherwise take $-A$ instead. Notice that the arithmetic mean of any ... | China | China National Team Selection Test | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
091q | Problem:
For an integer $n \geqslant 3$, let $\mathcal{M}$ be the set $\{(x, y) \mid x, y \in \mathbb{Z}, 1 \leqslant x \leqslant n, 1 \leqslant y \leqslant n\}$ of points in the plane. ($\mathbb{Z}$ is the set of integers.)
What is the maximum possible number of points in a subset $S \subseteq \mathcal{M}$ which does... | [
"Solution:\n\nWe will prove that the maximal cardinality of $S$ is $2 n-2$.\nThe set\n$$\nS=\\{1\\} \\times\\{2, \\ldots, n\\} \\cup\\{2, \\ldots, n\\} \\times\\{1\\}\n$$\nhas cardinality $2 n-2$ and it does not contain three distinct points that form a right triangle.\n\nWe will show that any subset $S \\subset \\... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2n - 2 | |
0he7 | How many distinct prime divisors does the number $11^8 + 11^7 - 132$ have? | [
"Consider the following transformation:\n$$\n11^8 + 11^7 - 132 = 11^6 \\cdot (11^2 + 11) - 132 = 11^6 \\cdot 132 - 132 = 132 \\cdot (11^6 - 1).\n$$\nSince $132 = 11 \\cdot 12 = 2^2 \\cdot 3 \\cdot 11$, it has three prime divisors: $2$, $3$, $11$.\n$$\n\\begin{aligned}\n11^6 - 1 &= (11^3 - 1)(11^3 + 1) = (11-1)(11^2... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 7 | |
01xn | The tangents to the circumcircle of the acute triangle $ABC$, passing through the vertices $B$ and $C$, meet at point $F$. The points $M$, $L$ and $N$ are the feet of the perpendiculars from the vertex $A$ to the lines $FB$, $FC$ and $BC$ respectively.
Prove the inequality $AM + AL \ge 2AN$. | [
"Note that $\\angle ABM = \\angle ACB$ by the property of the angle between the tangent to the circle at point $B$ and the chord $AB$. The right triangles $AMB$ and $ANC$ are similar since they have equal acute angles, therefore $\\frac{AM}{AN} = \\frac{AB}{AC}$, whence $AN = \\frac{AM \\cdot AC}{AB}$.\n\nBy analog... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0d2y | In acute triangle $A B C$, points $D$ and $E$ are the feet of the perpendiculars from $A$ to $B C$ and $B$ to $C A$, respectively. Segment $A D$ is a diameter of circle $\omega$. Circle $\omega$ intersects sides $A C$ and $A B$ at $F$ and $G$ (other than $A$), respectively. Segment $B E$ intersects segments $G D$ and $... | [
"We have $\\angle A F G=\\angle A D G$ since $A G D F$ is cyclic. On the other hand $\\angle D G A=\\angle A F D=90^\\circ$, since $A D$ is a diameter. We deduce that triangles $A G D$ and $A D B$ are similar, and therefore $\\angle A F G=\\angle C B A$.\n\nBecause $\\angle A E B=\\angle A D B=90^\\circ$, quadrilat... | Saudi Arabia | Selection tests for the Gulf Mathematical Olympiad 2013 | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08jb | Problem:
Let three congruent circles intersect in one point $M$ and $A_{1}$, $A_{2}$ and $A_{3}$ be the other intersection points for those circles. Prove that $M$ is the orthocenter for the triangle $A_{1} A_{2} A_{3}$.
Problem:
Trei cercuri egale au în comun un punct $M$ şi se intersectează câte două în puncte $A$... | [
"Solution:\n\nThe quadrilaterals $O_{3} M O_{2} A_{1}$, $O_{3} M O_{1} A_{2}$ and $O_{1} M O_{2} A_{3}$ are rhombuses. Therefore, $O_{2} A_{1} \\parallel M O_{3}$ and $M O_{3} \\parallel O_{1} A_{2}$, which imply $O_{2} A_{1} \\parallel O_{1} A_{2}$. Because $O_{2} A_{1} = O_{3} M = O_{1} A_{2}$, the quadrilateral ... | JBMO | 7th JBMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Dist... | null | proof only | null | |
0ksi | Problem:
Let $P(x) = x^{4} + a x^{3} + b x^{2} + x$ be a polynomial with four distinct roots that lie on a circle in the complex plane. Prove that $a b \neq 9$. | [
"Solution:\n\nIf either $a = 0$ the problem statement is clearly true. Thus, assume that $a \\neq 0$. Let the roots be $0, z_{1}, z_{2}, z_{3}$, and let the circle through these points be $C$. Note that we have\n$$\n\\begin{aligned}\n& \\frac{3}{z_{1} + z_{2} + z_{3}} = -\\frac{3}{a} \\\\\n& \\frac{\\frac{1}{z_{1}}... | United States | HMMT February 2022 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | null | proof only | null | |
06yt | Problem:
$O$ is the circumcenter of the triangle $ABC$. The lines $AO$, $BO$, $CO$ meet the opposite sides at $D$, $E$, $F$ respectively. Show that $1 / AD + 1 / BE + 1 / CF = 2 / AO$. | [
"Solution:\n\nProjecting onto the altitude from $A$, we have $AD \\cos (C-B) = AC \\sin C = 2R \\sin B \\sin C$, so $2R / AD = \\cos (C-B) / (\\sin B \\sin C)$.\n\nHence $2R / AD + 2R / BE + 2R / CF = \\cos (C-B) / (\\sin B \\sin C) + \\cos (A-C) / (\\sin C \\sin A) + \\cos (B-A) / (\\sin A \\sin B)$.\n\nSo $2R \\s... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0igb | Problem:
Write down a set $S$ of positive integers, all greater than $1$, such that for each $x \in S$, $x$ is a proper divisor of $(P / x) + 1$, where $P$ is the product of all the elements of $S$. Your score is $2n$, where $n = |S|$. | [] | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | null | |
0f97 | Problem:
One bird lives in each of $n$ bird-nests in a forest. The birds change nests, so that after the change there is again one bird in each nest. Also for any birds $A$, $B$, $C$, $D$ (not necessarily distinct), if the distance $\{AB\} < \{CD\}$ before the change, then $\{AB\} > \{CD\}$ after the change. Find all ... | [] | Soviet Union | 23rd ASU | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1 | |
0fx9 | Problem:
Sei $\mathbb{R}^{+}$ die Menge der positiven reellen Zahlen. Bestimme alle Funktionen $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$, sodass für alle $x, y>0$ gilt
$$
f(x+f(y))=f(x+y)+f(y)
$$ | [
"Solution:\n\nOffenbar ist $f(x)=2 x$ eine Lösung und wir zeigen, dass es die einzige ist. Wir zeigen zuerst, dass $f(z)>z$ gilt für alle $z>0$. Wäre $f(z)=z$, dann folgt mit $y=z$ sofort der Widerspruch $f(x+z)=f(x+z)+f(z)>f(x+z)$. Sei also $f(z)<z$ und setze $x=z-f(z)>0$ und $y=z$ in die Gleichung ein, dann folgt... | Switzerland | IMO Selektion 2008 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 2x for all positive real x | |
0as8 | Problem:
Let $a$, $b$, $c$ be three, not necessarily distinct, numbers chosen randomly from the set $\{3, 4, 5, 6, 7, 8\}$. Find the probability that $ab + c$ is even. | [
"Solution:\n\n(ans. $0.5$.\n\n$\\operatorname{Prob}(ab + c \\text{ is even}) = \\frac{4(3 \\cdot 3 \\cdot 3)}{6 \\cdot 6 \\cdot 6} = \\frac{4}{8}$. This is because $(ab + c$ even$)$ iff ($ab$ and $c$ even OR $ab$ and $c$ odd) iff ($a, b, c$ even OR $a$ odd and $b, c$ even OR $b$ odd and $a, c$ even OR $a, b, c$ odd... | Philippines | 13th Philippine Mathematical Olympiad | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 0.5 | |
012u | Problem:
Is it possible to select $1000$ points in a plane so that at least $6000$ distances between two of them are equal? | [
"Solution:\n\nLet's start with configuration of $4$ points and $5$ distances equal to $d$, like in this figure:\n$(\\alpha)$\n\n\n\nNow take $(\\alpha)$ and two copies of it obtainable by parallel shifts along vectors $\\vec{a}$ and $\\vec{b}$, $|\\vec{a}|=|\\vec{b}|=d$ and $\\angle(\\vec{a... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | Yes | |
076s | Find all functions $f : \mathbf{R} \to \mathbf{R}$ such that
$$
f(x^3 + f(y)) = x^2 f(x) + y,
$$
for all $x, y \in \mathbf{R}$. | [
"Let $f(0) = \\lambda$. Put $x = y = 0$ in (1) to get $f(\\lambda) = 0$. Taking $y = \\lambda$, we get $f(x^3) = x^2 f(x) + \\lambda$. Put $x = \\lambda, y = 0$ in (1) and we get\n$$\nf(\\lambda^3 + \\lambda) = \\lambda^2 f(\\lambda). \\qquad (2)\n$$\nBut $x = \\lambda, y = \\lambda$ gives $f(\\lambda^3) = \\lambda... | India | IND_TSExams | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | f(x) = x or f(x) = -x | |
0faz | Problem:
Does there exist a 4-digit integer which cannot be changed into a multiple of 1992 by changing 3 of its digits? | [
"Solution:\n\nThe only 4-digit multiples of $1992$ are: $1992$, $3984$, $5976$, $7968$, $9960$. All have first digit odd, second digit $9$, third digit $>5$ and last digit even, so it is easy to find a number which has all digits different from all of them."
] | Soviet Union | 1st CIS | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Other"
] | null | proof and answer | Yes | |
0072 | Hallar todos los números reales $x$ tales que
$$
\lfloor 2x \rfloor + \lfloor 3x \rfloor + \lfloor 7x \rfloor = 2008
$$ | [] | Argentina | Argentina 2009 | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | Spanish | proof and answer | [167 + 3/7, 167 + 1/2) | |
09fs | Let $a \ge 1$ and $b \ge 2$ be given positive integers. Show that there does not exist any non-constant polynomial $f(x)$ with integer coefficients such that $f(n^a)$ and $f(b^n)$ are relatively prime for every positive integer $n$.
(Battsengel B., Bayarmagnai G.) | [
"Suppose that such an $f$ exists.\nSince $f(b^a)$ and $f(b^b)$ are relatively prime, so are $f(0)$ and $b$. Since $f$ is non-constant polynomial, there exist a prime $p$ and a positive integer $m$ such that $p \\mid f(b^{am})$. It is clear that $b$ is relatively prime to $p$. We choose a positive integer $x_0$ such... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
0077 | Se considera un tablero de $a \times b$, con $a$ y $b$ enteros mayores o iguales que $2$. Inicialmente sus casillas están coloreadas de blanco y de negro como un tablero de ajedrez. La operación permitida consiste en elegir dos casillas con un lado común y recolorearlas de la siguiente manera: una casilla blanca pasa a... | [] | Argentina | Argentina 2009 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Modular Arithmetic"
] | Spanish | proof only | It is possible if and only if 3 divides a·b. | |
07eq | Find all functions $f : \mathbb{R} \to \mathbb{R}$ that for all $x, y \in \mathbb{R}$,
$$
f(x + y)f(x^2 - xy + y^2) = x^3 + y^3.
$$ | [
"Let $P(x, y)$ denote the assertion given in the statement of the problem. If there exists $a$ such that $f(a) = 0$, $P(a, 0)$ results in $a = 0$. Now\n$$\nP(x, x-y) \\rightarrow f(2x-y)f(x^2-xy+y^2) = (2x-y)(x^2-xy+y^2)\n$$\nBy dividing this equation by $P(x, y)$ (where $x + y \\neq 0$), it is obtained that\n$$\n\... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x or f(x) = -x | |
010l | Problem:
Let the numbers $\alpha, \beta$ satisfy $0<\alpha<\beta<\pi / 2$ and let $\gamma$ and $\delta$ be the numbers defined by the conditions:
(i) $0<\gamma<\pi / 2$, and $\tan \gamma$ is the arithmetic mean of $\tan \alpha$ and $\tan \beta$;
(ii) $0<\delta<\pi / 2$, and $\frac{1}{\cos \delta}$ is the arithmetic mea... | [
"Solution:\nLet $f(t)=\\sqrt{1+t^{2}}$. Since $f''(t)=\\left(1+t^{2}\\right)^{-3 / 2}>0$, the function $f(t)$ is strictly convex on $(0, \\infty)$. Consequently,\n$$\n\\begin{aligned}\n\\frac{1}{\\cos \\gamma} & =\\sqrt{1+\\tan ^{2} \\gamma}=f(\\tan \\gamma)=f\\left(\\frac{\\tan \\alpha+\\tan \\beta}{2}\\right)< \\... | Baltic Way | Baltic Way 1998 | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0jm3 | Problem:
Sammy has a wooden board, shaped as a rectangle with length $2^{2014}$ and height $3^{2014}$. The board is divided into a grid of unit squares. A termite starts at either the left or bottom edge of the rectangle, and walks along the gridlines by moving either to the right or upwards, until it reaches an edge ... | [
"Solution:\n\nAnswer: 4\n\nLet $R$ be the original rectangle and $R'$ the new rectangle which is different from $R$. We see that the perimeter of $R'$ depends on the possibilities for the side lengths of $R'$.\n\nWe will prove that the dividing line must have the following characterization: starting from the lower ... | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Number Theory > Diophantine Equa... | null | proof and answer | 4 | |
0eh9 | Problem:
Dan je izraz, v katerem je $x$ realno število in $x \notin \{-3, -1, 0, 1, 2, 3, 4\}$.
$$
\frac{x^{2}-4x}{5x-5} \cdot \left(\frac{x^{3}+1}{x^{2}+x}-1\right) \cdot \left(\left(1-\frac{3x-3}{x^{2}+x-6}\right) \cdot \left(\frac{6}{x+3}-\frac{1}{x-2}\right)^{-1}-1\right)^{-1}
$$
Poenostavi ga. | [
"Solution:\n\nRazstavljanje vsote kubov:\n\nIzračun $\\frac{x^{3}+1}{x^{2}+x}-1=\\frac{(x-1)^{2}}{x}$\n\nPreoblikovanje $1-\\frac{3x-3}{x^{2}+x-6} = \\frac{(x+1)(x-3)}{(x-2)(x+3)}$\n\nPreoblikovanje $\\frac{6}{x-3}-\\frac{1}{x-2} = \\frac{5(x-3)}{(x-2)(x+3)}$\n\nIzračun $\\left(1-\\frac{3x-3}{x^{2}+x-6}\\right) \\c... | Slovenia | Državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | x - 1 | |
0a2z | Exactly one of the following statements is true: which one? *Please note that the numbers $a$ and $b$ need not be integers.*
A) There do not exist $a > 0$ and $b > 0$ with $a \cdot b < \frac{a}{b} < a + b$.
B) There do not exist $a > 0$ and $b > 0$ with $a \cdot b < a + b < \frac{a}{b}$.
C) There do not exist $a > 0$ a... | [] | Netherlands | Dutch Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | C | |
00zc | Problem:
On an infinite checkerboard, two players alternately mark one unmarked cell. One of them uses $\times$, the other $o$. The first who fills a $2 \times 2$ square with his symbols wins. Can the player who starts always win? | [
"Solution:\n\nDivide the plane into dominoes in the way indicated by the thick lines in Figure 2. The second player can respond by marking the other cell of the same domino where the first player placed his mark. Since every $2 \\times 2$ square contains one whole domino, the first player cannot win.\n\n![](attache... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | No | |
0jiy | Problem:
Let $A B C D E F$ be a convex hexagon with the following properties.
a. $\overline{A C}$ and $\overline{A E}$ trisect $\angle B A F$.
b. $\overline{B E} \parallel \overline{C D}$ and $\overline{C F} \parallel \overline{D E}$.
c. $A B = 2 A C = 4 A E = 8 A F$.
Suppose that quadrilaterals $A C D E$ and $A D... | [
"Solution:\n\nFrom conditions (a) and (c), we know that triangles $A F E$, $A E C$ and $A C B$ are similar to one another, each being twice as large as the preceding one in each dimension. Let $\\overline{A E} \\cap \\overline{F C} = P$ and $\\overline{A C} \\cap \\overline{E B} = Q$. Then, since the quadrilaterals... | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 7295 | |
0age | Let $ABC$ be an acute triangle with orthocenter $H$, and let $M$ be the midpoint of $AC$. The point $C_1$ on $AB$ is such that $CC_1$ is an altitude of the triangle $ABC$. Let $H_1$ be the reflection of $H$ in $AB$. The orthogonal projections of $C_1$ onto the lines $AH_1$, $AC$ and $BC$ are $P$, $Q$ and $R$, respectiv... | [
"We first prove the following lemma.\n**Lemma.** Let $XYZT$ be a cyclic quadrilateral such that $XZ \\perp YT$. Let $XZ \\cap YT = O$, and let $V$ and $W$ be the orthogonal projections of $O$ to lines $XY$ and $ZT$, respectively. If $I$ lies in the middle of $YZ$, and $J$ lies in the middle of $XT$, then $IVJW$ is ... | North Macedonia | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry ... | English | proof only | null | |
08lh | Problem:
Consider $ABC$ an acute-angled triangle with $AB \neq AC$. Denote by $M$ the midpoint of $BC$, by $D, E$ the feet of the altitudes from $B, C$ respectively and let $P$ be the intersection point of the lines $DE$ and $BC$. The perpendicular from $M$ to $AC$ meets the perpendicular from $C$ to $BC$ at point $R$... | [
"Solution:\n\nLet $F$ be the foot of the altitude from $A$ and let $S$ be the intersection point of $AM$ and $RC$. As $PC$ is an altitude of the triangle $PRS$, the claim is equivalent to $RM \\perp PS$, since the latter implies that $M$ is the orthocenter of $PRS$. Due to $RM \\perp AC$, we need to prove that $AC ... | JBMO | 2008 Shortlist JBMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Pla... | null | proof only | null | |
063u | Problem:
Es seien $a_{1}, a_{2}, \ldots, a_{n}, k$ und $M$ positive ganze Zahlen mit den Eigenschaften
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}=k \quad \text{ und } \quad a_{1} a_{2} \ldots a_{n}=M .
$$
Man beweise: Für $M>1$ hat das Polynom
$$
P(x)=M(x+1)^{k}-\left(x+a_{1}\right)\left(x+a_{2}\right) ... | [
"Solution:\n\nWir zeigen $P(x)<0$ für alle $x>0$, also $M(x+1)^{k}<\\left(x+a_{1}\\right) \\ldots\\left(x+a_{n}\\right) \\Leftrightarrow$ $a_{1} a_{2} \\ldots a_{n}(x+1)^{\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\ldots+\\frac{1}{a_{n}}}<\\left(x+a_{1}\\right) \\ldots\\left(x+a_{n}\\right) \\Leftrightarrow \\prod_{i=1}^{n... | Germany | 2. Auswahlklausur | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
06mx | One edge of a triangular pyramid has length $6$ while every other edge has length $5$. Find the volume of the pyramid. | [
"Answer: $\\frac{5\\sqrt{39}}{2}$\n\nLet the vertices of the pyramid be $A$, $B$, $C$, $D$, where $BD = 6$ and all other edges have length $5$. Let also $M$ be the midpoint of $BD$ and $N$ be the foot of perpendicular from $A$ to the base $BCD$. Of course we have $BM = MD = 3$ and $CM = 4$. Note that\n$$\nAN^2 = AB... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | 5*sqrt(39)/2 | |
01kl | Fifteen red, blue and green points are marked on a plane. It is known that the sum of the distances between the red points and the blue points is $51$, the sum of the distances between the red points and the green points is $39$, the sum of the distances between the blue points and the green points is $1$.
How many poi... | [
"Let $n$, $k$, $m$ denote respectively the number of red, green, blue points. Then $m + n + k = 15$. From (*) in the solution of Problem 2, Category C, we have\n$$\n39m \\le 51k + n, \\quad (1)\n$$\n$$\n51k - n \\le 39m. \\quad (2)\n$$\nSubstitute for $k$ in (1) to obtain (after simple manipulations)\n$$\n18m + 10n... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Either red 8, green 3, blue 4; or red 13, green 1, blue 1. | |
00ab | Given two positive integers $a$ and $b$, a legal move is to choose a proper divisor of one of them and add it to either $a$ or $b$. Players $A$ and $B$ make legal moves in turns; $A$ plays first. The one who obtains a number $\ge 2015$ wins. Determine who wins if the game starts with:
$$
a) a=3, b=5; \qquad b) a=6, b=7... | [
"Player $B$ wins in part a); player $A$ wins in part b). Note that, by the rules, the player to move can always add $1$ to one of the current numbers.\n\nFor $a=6$, $b=7$ let $A$ start the game by adding $1$ to $6$, leading to the pair $7, 7$. If $a=3$, $b=5$, $A$'s first move is forced to be adding $1$ to one of t... | Argentina | Argentine National Olympiad 2015 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | a) Player B wins; b) Player A wins. | |
09io | Let $N \ge 3$ be an odd integer. For distinct divisors $n$ and $m$ of $N$ such that $n > m$, prove that
$$
\frac{n}{m} \ge 1 + \frac{2m+4}{N}.
$$ | [] | Mongolia | Mongolian Mathematical Olympiad Round 2 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
08xw | Let $\Gamma$ be the circumcircle of triangle $ABC$. Let $A'$ be the mid-point of arc $BC$ of the circle $\Gamma$ opposite to $A$, $B'$ be the mid-point of arc $CA$ opposite to $B$ and $C'$ be the mid-point of arc $AB$ opposite to $C$. If the area of the triangle $AB'C'$, $A'BC'$, $A'B'C'$ is $2$, $3$, $4$, respectively... | [
"$$\n\\frac{288}{35}\n$$\nFor a polygon $P$, let us denote by $S(P)$ its area.\nLet $I$ be the incenter of the circle $\\Gamma$. Since $AI$ is the bisector of the angle $\\angle CAB$, $I$ lies on the line segment $AA'$. Similarly, $I$ lies on the line segments $BB'$ and $CC'$. By using the theorem on angles at a po... | Japan | Japan 2015 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 288/35 | |
08qf | Problem:
Find all perfect squares $n$ such that if the positive integer $a \geqslant 15$ is some divisor of $n$ then $a+15$ is a prime power. | [
"Solution:\nWe call a positive integer $a$ \"nice\" if $a+15$ is a prime power.\nFrom the definition, the numbers $n=1, 4, 9$ satisfy the required property. Suppose that for some $t \\in \\mathbb{Z}^{+}$, the number $n=t^{2} \\geqslant 15$ also satisfies the required property. We have two cases:\n\n1. If $n$ is a p... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1, 4, 9, 16, 49, 64, 196 | |
0jmp | Problem:
For integers $m, n \geq 1$, let $A(n, m)$ be the number of sequences $(a_{1}, \cdots, a_{n m})$ of integers satisfying the following two properties:
(a) Each integer $k$ with $1 \leq k \leq n$ occurs exactly $m$ times in the sequence $(a_{1}, \cdots, a_{n m})$.
(b) If $i, j$, and $k$ are integers such that ... | [
"Solution:\n\nWe show that $A(n, m)$ is equal to the number of standard Young tableaux with $n$ rows and $m$ columns (i.e., fillings of an $n \\times m$ matrix with the numbers $1,2, \\ldots, n m$ so that numbers are increasing in each row and column).\n\nConsider the procedure where every time a $k$ appears in the... | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof only | null | |
0i19 | Problem:
Equilateral triangles $A B C_{1}$, $B C A_{1}$, $C A B_{1}$ are constructed outwards on the sides of a triangle $A B C$. Prove:
a. The centers of these three triangles form another equilateral triangle.
b. Segments $A A_{1}$, $B B_{1}$, $C C_{1}$ are concurrent.
c. Segments $A A_{1}$, $B B_{1}$, $C C_{1}$ ... | [
"Solution: (All angles will be directed modulo $\\pi$ for simplicity.)\n\na. Let the respective centers of triangles $A B C_{1}$, $B C A_{1}$, $C A B_{1}$ be denoted $O_{C}$, $O_{A}$, $O_{B}$. A rotation about $A$ by an angle of $\\pi / 6$, together with a scaling (about $A$) by a factor of $\\sqrt{3} / 3$, takes $... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Advanced Configurations > Napoleon and Fermat points",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0edx | The sum of every three consecutive terms of the sequence $a_1, a_2, a_3, a_4, \dots$ equals $2016$. It also holds that $a_{667} = 667$ and $a_{1004} = 1004$. What is the value of $a_{2016}$?
(A) 0
(B) 345
(C) 667
(D) 1004
(E) 2016 | [
"Since the sum of any three consecutive terms is the same we have $a_{n+3} = a_n$ for all $n$. This implies that $a_{667} = a_{667+3 \\cdot 449} = a_{2014}$ and $a_{1004} = a_{1004+3 \\cdot 337} = a_{2015}$. From $a_{2014} + a_{2015} + a_{2016} = 2016$ we conclude that $a_{2016} = 2016 - 667 - 1004 = 345$."
] | Slovenia | Slovenija 2016 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | MCQ | B | |
0f9g | Problem:
Given a point $X$ and $n$ vectors $\mathbf{x}_{\mathrm{i}}$ with sum zero in the plane. For each permutation of the vectors we form a set of $n$ points, by starting at $X$ and adding the vectors in order. For example, with the original ordering we get $\mathbf{X}_1$ such that $\mathbf{X}\mathbf{X}_1 = \mathbf{... | [] | Soviet Union | 24th ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | null | proof only | null | |
0kxh | Problem:
Quadrilateral $A B C D$ is inscribed in circle $\Gamma$. Segments $A C$ and $B D$ intersect at $E$. Circle $\gamma$ passes through $E$ and is tangent to $\Gamma$ at $A$. Suppose that the circumcircle of triangle $B C E$ is tangent to $\gamma$ at $E$ and is tangent to line $C D$ at $C$. Suppose that $\Gamma$ h... | [
"Solution:\n\nThe key observation is that $\\triangle A C D$ is equilateral. This is proven in two steps.\n- From tangency at $C$, we have\n$$\n\\angle D C A = \\angle D C E = \\angle E B C = \\angle D B C = \\angle D A C\n$$\nimplying that $C A = C D$.\n- Consider the common tangent of $\\gamma$ and $\\Gamma$ at $... | United States | HMMT February | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 9*sqrt(21)/7 | |
072d | Problem:
Let $\alpha$ and $\beta$ be positive integers such that
$$
\frac{43}{197}<\frac{\alpha}{\beta}<\frac{17}{77}
$$
Find the minimum possible value of $\beta$. | [
"Solution:\nWe have\n$$\n\\frac{77}{17}<\\frac{\\beta}{\\alpha}<\\frac{197}{43}\n$$\nThat is,\n$$\n4+\\frac{9}{17}<\\frac{\\beta}{\\alpha}<4+\\frac{25}{43}\n$$\nThus $4<\\frac{\\beta}{\\alpha}<5$. Since $\\alpha$ and $\\beta$ are positive integers, we may write $\\beta=4 \\alpha+x$, where $0<x<\\alpha$. Now we get\... | India | INMO | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 32 | |
0fe7 | Problem:
Calcula las soluciones reales de la ecuación:
$$
\sqrt[3]{1729-X}+\sqrt[3]{X}=19
$$ | [
"Solution:\nSi llamamos $a=\\sqrt[3]{1729-x}$ y $b=\\sqrt[3]{x}$, se tiene que $a$ y $b$ son raíces del sistema\n$$\n\\left.\\begin{array}{l}a+b=19 \\\\ a^{3}+b^{3}=1729\\end{array}\\right\\}\n$$\nPara resolver este sistema procedemos como sigue:\n$$\n19^{3}=(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)=1729+3 \\times 19 a b\n$... | Spain | null | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 1000, 729 | |
0176 | Let $n$ be a fixed positive integer. Does there exist an infinite subset $A$ of the set $\mathbb{N}$ of positive integers such that for every pairwise distinct $a_1, \dots, a_n \in A$ the numbers $a_1 + \dots + a_n$ and $a_1 \cdots a_n$ are coprime? | [
"For $n = 1$ the statement is obviously false. We assert that it is true for all $n > 1$.\n\nWe first consider the sequence $x_0, x_1, \\dots$ of positive integers which is recursively defined by $x_0 = n$ and $x_{k+1} = (x_0 + \\dots + x_k)! + 1$ for $k \\ge 0$. We claim that the set $A := \\{x_k \\mid k \\ge 1\\}... | Baltic Way | BALTIC WAY | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
09ao | Let $m \in \mathbb{N}$. $m^2 < a, b < m^2 + m$ and $a \neq b$. Find all the natural $c$, such that $c \mid ab$, $m^2 < c < m^2 + m$.
(proposed by D. Ganzorig) | [
"Let $d$ be a number such that $d \\mid ab$ and $d \\in (m^2, m^2 + m)$. Then $d \\mid (a-d)(b-d)$ and $|a-d| < m$, $|b-d| < m$. It follows that $|(a-d)(b-d)| < m^2 < d$. Hence\n$$\n(a-d)(b-d) = 0.\n$$\nWe have $d = a \\lor b$."
] | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | c = a or c = b | |
00fq | Determine the largest of all integers $n$ with the property that $n$ is divisible by all positive integers that are less than $\sqrt[3]{n}$. | [
"Observation from that $\\operatorname{lcm}(2,3,4,5,6,7)=420$ is divisible by every integer less than or equal to $T=[\\sqrt[3]{420}]$ and that $\\operatorname{lcm}(2,3,4,5,6,7,8)=840$ is not divisible by $9=[\\sqrt[3]{840}]$. One may guess $420$ is the required integer.\n\nLet $N$ be the required integer and suppo... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 420 | |
0byv | Consider $f : [a, b] \to [a, b]$ a differentiable function, such that $f'$ is continuous and positive. Show that there is a point $c \in (a, b)$ such that
$$
f(f(b)) - f(f(a)) = (f'(c))^2 (b - a).
$$ | [
"Obviously $f$ is increasing.\nUsing Lagrange's theorem on $[f(a), f(b)]$ we get a point $c_1 \\in (f(a), f(b))$ such that\n$$\nf(f(b)) - f(f(a)) = f'(c_1)(f(b) - f(a)).\n$$\nAnother use of the theorem for the interval $[a, b]$ gives a point $c_2 \\in (a, b)$ such that\n$$\nf(b) - f(a) = f'(c_2)(b - a).\n$$\nCollec... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | English | proof only | null | |
02yx | Problem:
No quadriculado da figura, dizemos que dois ou mais pontos nas intersecções das linhas são equilegais, em relação a um ponto fixo, quando suas distâncias são iguais a este. Por exemplo, $B$ e $C$ são equilegais em relação a $A$, mas $B$ e $D$ não são.
a) Marque, no quadriculado a s... | [
"Solution:\n\na) Supondo que a medida do lado de cada quadradinho seja $1~\\mathrm{cm}$, então a distância $d$ de $K$ a $J$ pode ser determinada pelo Teorema de Pitágoras: $d^{2} = 2^{2} + 1^{2}$, donde concluímos que $d = \\sqrt{5}~\\mathrm{cm}$. Todos os pontos das intersecções que estiverem à distância de $\\sqr... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | a) All grid intersection points at distance sqrt(5) from J (i.e., those with displacements (±2, ±1) or (±1, ±2) from J). b) A circle of radius sqrt(5) centered at J. c) A sphere of radius sqrt(5) centered at J. | |
0aa5 | Problem:
Let $n$ be a positive integer. Show that there exist positive integers $a$ and $b$ such that:
$$
\frac{a^{2}+a+1}{b^{2}+b+1}=n^{2}+n+1
$$ | [
"Solution:\nLet $P(x) = x^{2} + x + 1$. We have\n$$\nP(n) P(n+1) = (n^{2} + n + 1)(n^{2} + 3n + 3) = n^{4} + 4n^{3} + 7n^{2} + 6n + 3.\n$$\nAlso,\n$$\nP((n+1)^{2}) = (n+1)^{4} + (n+1)^{2} + 1 = n^{4} + 4n^{3} + 7n^{2} + 6n + 3.\n$$\nBy choosing $a = (n+1)^{2}$ and $b = n+1$, we get $P(a)/P(b) = P(n)$ as desired."
] | Nordic Mathematical Olympiad | The 31st Nordic Mathematical Contest | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0f61 | Problem:
Is $\ln 1.01$ greater or less than $2/201$? | [] | Soviet Union | 18th ASU | [
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Functions"
] | null | proof and answer | greater | |
0j9r | Problem:
Let $f(x) = x^{2} + a x + b$ and $g(x) = x^{2} + c x + d$ be two distinct real polynomials such that the $x$-coordinate of the vertex of $f$ is a root of $g$, the $x$-coordinate of the vertex of $g$ is a root of $f$, and both $f$ and $g$ have the same minimum value. If the graphs of the two polynomials inters... | [
"Solution:\n\nAnswer: $-8048$\n\nIt is clear, by symmetry, that $2012$ is equidistant from the vertices of the two quadratics. Then it is clear that reflecting $f$ about the line $x = 2012$ yields $g$ and vice versa. Thus the average of each pair of roots is $2012$. Thus the sum of the four roots of $f$ and $g$ is ... | United States | 15th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | -8048 | |
0igk | Problem:
Show that for $b$ even, there exists some $M$ such that for every $m, n > M$ with $m n$ even, an $m \times n$ rectangle is $(1, b)$-tileable. | [
"Solution:\nBy the diagram below, it is possible to tile a $(2 b+2) \\times (4 b+1)$ rectangle.\n\n\n\nSince we can already tile a $(2 b+2) \\times 2 b$ rectangle by above, and $2 b$ is relatively prime to $4 b+1$, this will allow us to tile any $(2 b+2) \\times n$ rectangle for $n$ suffici... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0b03 | Problem:
Determine the units digit of $(2!+2)(3!+3)(4!+2)(5!+3) \cdots (2018!+2)(2019!+3)$. | [
"Solution:\n\nIn mod $10$, the expression above becomes $(4)(9)(6)\\left(2^{1007}\\right)\\left(3^{1008}\\right) = 2^{1010} \\cdot 3^{1011}$ because starting from $5!$, the units digit will be alternating from $3$ and $2$. In mod $10$, $2^{1010} \\cdot 3^{1011}$ becomes $(4)(7) = 28$, which ends with $8$."
] | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 8 | |
076f | Problem:
From a set of 11 square integers, show that one can choose 6 numbers $a^{2}, b^{2}, c^{2}, d^{2}, e^{2}, f^{2}$ such that
$$
a^{2}+b^{2}+c^{2} \equiv d^{2}+e^{2}+f^{2} \quad(\bmod 12)
$$ | [
"Solution:\nThe first observation is that we can find 5 pairs of squares such that the two numbers in a pair have the same parity. We can see this as follows:\n\n| Odd numbers | Even numbers | Odd pairs | Even pairs | Total pairs |\n| :---: | :---: | :---: | :---: | :---: |\n| 0 | 11 | 0 | 5 | 5 |\n| 1 | 10 | 0 | 5... | India | INMO | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0k3a | Problem:
Let $ABC$ be a triangle with $AB = 20$, $BC = 10$, $CA = 15$. Let $I$ be the incenter of $ABC$, and let $BI$ meet $AC$ at $E$ and $CI$ meet $AB$ at $F$. Suppose that the circumcircles of $BIF$ and $CIE$ meet at a point $D$ different from $I$. Find the length of the tangent from $A$ to the circumcircle of $DEF$... | [
"Solution:\nLet $O = AI \\cap (AEF)$. We claim that $O$ is the circumcenter of $DEF$. Indeed, note that $\\angle EDF = \\angle ECI + \\angle FBI = \\frac{\\angle B + \\angle C}{2} = \\frac{\\angle EOF}{2}$, and $OE = OF$, so the claim is proven.\n\nNow note that the circumcircle of $DEF$ passes through the incenter... | United States | HMMT November 2018 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Isogon... | null | final answer only | 2*sqrt(30) | |
0kkn | Problem:
A semicircle with radius $2021$ has diameter $AB$ and center $O$. Points $C$ and $D$ lie on the semicircle such that $\angle AOC < \angle AOD = 90^\circ$. A circle of radius $r$ is inscribed in the sector bounded by $OA$ and $OC$ and is tangent to the semicircle at $E$. If $CD = CE$, compute $\lfloor r \rfloo... | [
"Solution:\n\nWe are given\n$$\nm \\angle EOC = m \\angle COD\n$$\nand\n$$\nm \\angle AOC + m \\angle COD = 2 m \\angle EOC + m \\angle COD = 90^\\circ.\n$$\nSo $m \\angle EOC = 30^\\circ$ and $m \\angle AOC = 60^\\circ$. Letting the radius of the semicircle be $R$, we have\n$$\n(R - r) \\sin \\angle AOC = r \\Righ... | United States | HMMT Spring 2021 Guts Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 673 | |
04k7 | Let $S$ be the centre of the circle $k$ with radius $1$. Vertices $A$ and $B$ of the square $ABCD$ belong to the circle $k$, and its side $CD$ passes through the point $S$. Find the length of the side of the square $ABCD$. | [] | Croatia | Mathematical competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2/sqrt(5) | |
05fr | Problem:
Trouver toutes les paires $(m,n)$ d'entiers strictement positifs telles que
$$
m! + n! = m^n
$$
Remarque. Pour tout entier $n$, on rappelle que $n!$ désigne l'entier $1 \times 2 \times 3 \times \cdots \times n$. | [
"Solution:\n\nOn vérifie facilement que les paires $(2,2)$ et $(2,3)$ sont solutions. On va montrer que ce sont les seules.\n\nOn commence par traiter le cas où $m$ ou $n$ vaut $1$ ou $2$.\n\nSi $m=1$, alors $m^n=1$ mais $m!+n!\\geqslant 2$, donc il n'y a pas de solution.\n\nSi $m=2$, l'équation devient $2 + n! = 2... | France | Préparation Olympique Française de Mathématiques | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | (2,2) and (2,3) | |
0abf | Maja went to the bookstore to buy two books. The price of the first book was $65\%$ and the price of the other $57.5\%$ from the money that Maja had with her. She needed additional $45$ denars to buy the two books. How much money did Maja have with her? | [
"Let Maja had $x$ denars with her. Then from the condition in the problem we have that\n$$\n\\frac{65}{100}x + \\frac{57.5}{100}x = 45 + x.\n$$\nSo we obtain that\n$$\n65x + 57.5x = 4500 + 100x.\n$$\n$$\n122.5x = 4500 + 100x\n$$\n$$\n22.5x = 4500\n$$\n$$\nx = 200.\n$$\nSo Maja had $200$ denars with her."
] | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 200 | |
0jut | Problem:
An infinite sequence of real numbers $a_{1}, a_{2}, \ldots$ satisfies the recurrence
$$
a_{n+3}=a_{n+2}-2 a_{n+1}+a_{n}
$$
for every positive integer $n$. Given that $a_{1}=a_{3}=1$ and $a_{98}=a_{99}$, compute $a_{1}+a_{2}+\cdots+a_{100}$.
Proposed by: Evan Chen | [
"Solution:\n\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} a_{k} & =a_{1}+a_{2}+a_{3}+\\sum_{k=1}^{n-3}\\left(a_{k}-2 a_{k+1}+a_{k+2}\\right) \\\\\n& =a_{1}+a_{2}+a_{3}+\\sum_{k=1}^{n-3} a_{k}-2 \\sum_{k=2}^{n-2} a_{k}+\\sum_{k=3}^{n-1} a_{k} \\\\\n& =2 a_{1}+a_{3}-a_{n-2}+a_{n-1}\n\\end{aligned}\n$$\n\nPutting $n=100$ gi... | United States | HMMT February 2016 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 3 | |
0hjy | Problem:
A toy slot machine accepts two kinds of coins: red and green. When a coin is inserted, the machine returns 5 coins of the other color. Laura starts with one green coin. Can it happen that after a while, she has the same number of coins of each color? | [
"Solution:\n\nNote that each time Laura uses the machine, the total number of coins she has increases by four, an even number. She begins with one coin, an odd number, so she will always have an odd number of coins and can never have the same number of each color."
] | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0auy | Problem:
Let $\square ABCD$ be a trapezoid with parallel sides $AB$ and $CD$ of lengths 6 units and 8 units, respectively. Let $E$ be the point of intersection of the extensions of the nonparallel sides of the trapezoid. If the area of $\triangle BEA$ is 60 square units, what is the area of $\triangle BAD$? | [
"Solution:\n\nNote that $\\triangle BEA \\sim \\triangle CED$ and $|EB| = \\frac{2 \\cdot 60}{6} = 20$. Thus, $|EC| = \\frac{8}{6} \\cdot 20 = \\frac{80}{3}$, and hence $|BC| = \\frac{80}{3} - 20 = \\frac{20}{3}$. Finally then,\n$$\n\\text{area of } \\triangle BAD = \\frac{1}{2} \\cdot 6 \\cdot \\frac{20}{3} = 20\n... | Philippines | 18th PMO National Stage Oral Phase | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 20 | |
0gk8 | If the polynomial
$$
f(x) = x^6 + a_1 x^5 + a_2 x^4 + a_3 x^3 + a_4 x^2 + a_5 x + 3
$$
with real coefficients possesses all negative roots, show that $f(2) \ge 27^2$. | [
"Let $a_6 = 3$ and $r_1, r_2, \\dots, r_6 < 0$ be all roots of $f(x)$.\nFor each $k = 1, 2, \\dots, 6$, we have\n$$\na_k = (-1)^k \\sum_{1 \\le j_1 < \\dots < j_k \\le 6} r_{j_1} r_{j_2} \\cdots r_{j_k} = \\sum_{1 \\le j_1 < \\dots < j_k \\le 6} |r_{j_1}| |r_{j_2}| \\cdots |r_{j_k}|\n$$\nUsing the AM-GM inequality,... | Thailand | Thai Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
07k1 | Initially, there are $n$ glasses numbered $1, 2, \ldots, n$, and the capacity of glass number $i$ is $i$ liters. A total of $n$ liters of water is distributed among these glasses such that the volume of water in each glass is an integer. In each step, we can pour water from one glass into another until the source glass... | [
"We call a glass that is not empty \"non-empty\" and a glass that has been emptied \"empty\". Each glass holds one liter of water, and all the water is in $n$ containers (glasses), each of which can be either non-empty or empty.\n\nSuppose in the initial state we have $f_1$ non-empty glasses, and we want to reach a... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0i1z | Problem:
Find $\log_{n}\left(\frac{1}{2}\right) \log_{n-1}\left(\frac{1}{3}\right) \cdots \log_{2}\left(\frac{1}{n}\right)$ in terms of $n$. | [
"Solution:\n\nUsing $\\log \\frac{1}{x} = -\\log x$ and $\\log_{b} a = \\frac{\\log a}{\\log b}$, we get that the product equals\n$$\n\\frac{(-\\log 2)(-\\log 3) \\cdots (-\\log n)}{\\log n \\cdots \\log 3 \\log 2} = (-1)^{n-1}.\n$$"
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | (-1)^{n-1} | |
00zy | Problem:
Given a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive integers in which every positive integer occurs exactly once. Prove that there exist integers $\ell$ and $m$, $1<\ell<m$, such that $a_{1}+a_{m}=2 a_{\ell}$. | [
"Solution:\n\nLet $\\ell$ be the least index such that $a_{\\ell}>a_{1}$. Since $2 a_{\\ell}-a_{1}$ is a positive integer larger than $a_{1}$, it occurs in the given sequence beyond $a_{\\ell}$. In other words, there exists an index $m>\\ell$ such that $a_{m}=2 a_{\\ell}-a_{1}$. This completes the proof."
] | Baltic Way | Baltic Way 1997 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
06iz | Determine all $n \times n$ tables of nonnegative integers with rows and columns labelled by $0, 1, 2, \dots, n-1$ such that for each $0 \le i, j \le n-1$, the number in the cell $(i, j)$ is the number of $i$'s of row $j$. One such table can be seen below.
| | column 0 | column 1 | column 2 | column 3 | column 4... | [
"There are only two such tables as given below.\n\n| 1 | 0 | 3 | 3 | 4 |\n|---|---|---|---|---|\n| 1 | 3 | 2 | 1 | 1 |\n| 0 | 1 | 0 | 1 | 0 |\n| 2 | 1 | 0 | 0 | 0 |\n| 1 | 0 | 0 | 0 | 0 |\n\n| 1 | 0 | 4 | 4 | 3 |\n|---|---|---|---|---|\n| 1 | 4 | 1 | 1 | 1 |\n| 0 | 0 | 0 | 0 | 1 |\n| 1 | 0 | 0 | 0 | 0 |\n| 2 | 1 | ... | Hong Kong | 1997-2023 IMO HK TST | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Only for n = 5, exactly two tables:
[[1,0,3,3,4],[1,3,2,1,1],[0,1,0,1,0],[2,1,0,0,0],[1,0,0,0,0]] and [[1,0,4,4,3],[1,4,1,1,1],[0,0,0,0,1],[1,0,0,0,0],[2,1,0,0,0]]. | |
022m | Problem:
Os pontos $E$ e $F$ estão nos lados $A D$ e $B C$, respectivamente, do quadrado $A B C D$. Sabendo que $B E=E F=F D=30$, encontre a área do quadrado.
 | [
"Solution:\n\nSejam $G$ e $H$ os pés das perpendiculares traçadas de $E$ e $F$ aos lados $B C$ e $A D$, respectivamente.\n\n\n\nComo $A B = C D$ e $B E = F D$, aplicando o Teorema de Pitágoras, temos:\n$$\n\\begin{aligned}\nA E & = \\sqrt{B E^{2} - A B^{2}} \\\\\n& = \\sqrt{F D^{2} - C D^{2... | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 810 | |
04v2 | Find all integers $n$ such that $2^n + n^2$ is a square of an integer. | [] | Czech Republic | Final Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | n = 0 and n = 6 | |
03r2 | For a given positive integer $n \geq 2$, suppose positive integers $a_i$ ($i=1, 2, \dots, n$) satisfy $a_1 < a_2 < \dots < a_n$ and $\sum_{i=1}^{n} \frac{1}{a_i} \leq 1$.
Prove that, for any real number $x$, the following inequality holds,
$$
\left[ \sum_{i=1}^{n} \frac{1}{a_i^2 + x^2} \right]^2 \leq \frac{1}{2} \cdot ... | [
"For $x^2 \\geq a_1(a_1-1)$, from $\\sum_{i=1}^{n} \\frac{1}{a_i} \\leq 1$ we have\n$$\n\\begin{aligned}\n\\left[ \\sum_{i=1}^{n} \\frac{1}{a_i^2 + x^2} \\right]^2 &\\leq \\left( \\sum_{i=1}^{n} \\frac{1}{2a_i |x|} \\right)^2 = \\frac{1}{4x^2} \\left( \\sum_{i=1}^{n} \\frac{1}{a_i} \\right)^2 \\\\\n&\\leq \\frac{1}... | China | China Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof only | null | |
0bda | Problem:
Az $\left(a_{n}\right)_{n \geq 1}$ sorozat növekvő és korlátos. Számítsd ki a
$$
\lim_{n \rightarrow \infty} \left(2 a_{n} - a_{1} - a_{2}\right) \left(2 a_{n} - a_{2} - a_{3}\right) \cdots \left(2 a_{n} - a_{n-2} - a_{n-1}\right) \left(2 a_{n} - a_{n-1} - a_{1}\right)
$$
határértéket! | [] | Romania | Matematika tantárgyverseny Megyei szakasz | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 0 | |
08fc | Problem:
Sia $ABC$ un triangolo acutangolo con $AB < AC$. Siano
- $D$ il piede della bisettrice dell'angolo in $A$,
- $E$ il punto del segmento $BC$ (diverso da $B$) tale che $AB = AE$,
- $F$ il punto del segmento $BC$ (diverso da $B$) tale che $BD = DF$,
- $G$ il punto del segmento $AC$ tale che $AB = AG$.
Dimostrare... | [
"Solution:\n\nPrimo metodo (angle chasing)\nPer le proprietà dell'angolo alla circonferenza (anche nel caso limite della tangente) la tesi è equivalente a dimostrare che $\\angle EGA = \\angle EFG$.\nIndichiamo con $\\theta$ l'ampiezza di $\\angle GBC$. Osserviamo che i punti $B, E, G$ appartengono ad una stessa ci... | Italy | XXXVIII Olimpiade Italiana di Matematica, Cesenatico | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Tr... | null | proof only | null | |
0akw | Let $p$ and $q$ be odd prime numbers and $a$ be a positive integer satisfying $p|a^q + 1$ and $q|a^p + 1$. Prove that $p|a+1$ or $q|a+1$. | [] | North Macedonia | Macedonian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
09f8 | Count the number of ways to fill unit squares of a $4 \times 4$ table with two colours, red and blue, such that no two rows and no two columns are painted the same. | [
"Let $X$ be the set of the painting of the $4 \\times 4$ table with two colors, red and blue, so that no two rows are painted the same. Since we can color the $1 \\times 4$ table with two colors in $2^4 = 16$ ways, $|X| = 16 \\cdot 15 \\cdot 14 \\cdot 13$.\n\nNow let $B_{ij} \\subset X$ be the set of the painting o... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 33864 | |
0bvg | Let $m$ and $n$ be two positive integers such that $n - 1 > m > 1$, let $S_n$ be the group of the permutations of degree $n$ and let $S$ be the subset of $S_n$ made by all permutations $\sigma$ with the property $\sigma(m) < \sigma(m+1) < \dots < \sigma(n)$. Show that $S$ contains a unique subgroup of $S_n$, which cont... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Algebra > Abstract Algebra > Group Theory",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English | proof only | null | |
02tp | Problem:
Sejam $a$ e $b$ dois dígitos diferentes de zero não necessariamente diferentes. O número de dois dígitos $\overline{a b}$ é chamado de curioso, se ele for um divisor do número $\overline{b a}$, que é formado pela troca da ordem dos dígitos de $\overline{a b}$. Ache todos os números curiosos.
Observação: O tra... | [
"Solution:\nO número de dois dígitos $\\overline{a b}$ pode ser escrito como $10 a + b$, assim como $\\overline{b a} = 10 b + a$. Se $10 a + b$ é divisor de $10 b + a$, temos $10 b + a = (10 a + b) k$, onde $k$ é um inteiro menor ou igual a 9 já que os dois números possuem dois dígitos. Segue que\n$$\n\\begin{align... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | {11, 22, 33, 44, 55, 66, 77, 88, 99} | |
0a87 | Problem:
The function $f$ is defined for non-negative integers and satisfies the condition
$$
f(n)= \begin{cases}f(f(n+11)), & \text{ if } n \leq 1999 \\ n-5, & \text{ if } n>1999\end{cases}
$$
Find all solutions of the equation $f(n)=1999$. | [
"Solution:\nIf $n \\geq 2005$, then $f(n)=n-5 \\geq 2000$, and the equation $f(n)=1999$ has no solutions. Let $1 \\leq k \\leq 4$. Then\n$$\n\\begin{gathered}\n2000-k=f(2005-k)=f(f(2010-k)) \\\\\n=f(1999-k)=f(f(2004-k))=f(1993-k)\n\\end{gathered}\n$$\nLet $k=1$. We obtain three solutions $1999=f(2004)=f(1998)=f(199... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 13 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | All n of the form n = 6k with k = 1, 2, ..., 334 (i.e., n = 6, 12, ..., 2004). |
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