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|---|---|---|---|---|---|---|---|---|---|
06dg | a. Using the characteristic equation method or otherwise, find a formula in terms of $n$ for the sequence $a_0 = 1$, $a_1 = 4$ and $a_{n+2} = 5a_{n+1} - 6a_n$ for $n = 0, 1, 2, \dots$
b. Using the characteristic equation method or otherwise, find a formula in terms of $n$ for the sequence $b_0 = 3$, $b_1 = 7$ and $b_{... | [
"a.\nThe answer is $a_n = 2 \\cdot 3^n - 2^n$.\nThe characteristic equation is $\\lambda^2 - 5\\lambda + 6 = 0$. The roots are $\\lambda = 3, 2$. Therefore, we have\n$$\na_n = A \\cdot 3^n + B \\cdot 2^n\n$$\nfor some constants $A$ and $B$. Putting $n = 0$ and $n = 1$, we need to solve\n$$\n\\begin{cases} A + B = 1... | Hong Kong | IMO HK TST | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | a_n = 2·3^n − 2^n; b_n = (n + 6)·2^{n−1} | |
00xd | Problem:
Find an infinite non-constant arithmetic progression of positive integers such that each term is neither a sum of two squares, nor a sum of two cubes (of positive integers). | [
"Solution:\nFor any natural number $n$, we have $n^{2} \\equiv 0$ or $n^{2} \\equiv 1 \\pmod{4}$ and $n^{3} \\equiv 0$ or $n^{3} \\equiv \\pm 1 \\pmod{9}$. Thus $\\{36n + 3 \\mid n = 1, 2, \\ldots\\}$ is a progression with the required property."
] | Baltic Way | Baltic Way 1992 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | {36n + 3 | n = 1, 2, ...} | |
0ka7 | Problem:
In a rectangular box $A B C D E F G H$ with edge lengths $A B = A D = 6$ and $A E = 49$, a plane slices through point $A$ and intersects edges $B F$, $F G$, $G H$, $H D$ at points $P$, $Q$, $R$, $S$ respectively. Given that $A P = A S$ and $P Q = Q R = R S$, find the area of pentagon $A P Q R S$. | [
"Solution:\n\nLet $A D$ be the positive $x$-axis, $A B$ be the positive $y$-axis, and $A E$ be the positive $z$-axis, with $A$ the origin. The plane, which passes through the origin, has equation $k_{1} x + k_{2} y = z$ for some undetermined parameters $k_{1}, k_{2}$. Because $A P = A S$ and $A B = A D$, we get $P ... | United States | HMMT February 2019 February 16, 2019 | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | 141*sqrt(11)/2 | |
0ado | Реши ја равенката
$$
(x^2 - x + 1)(4y^2 + 6y + 4)(4z^2 - 12z + 25) = 21.
$$
во множеството реални броеви. | [
"Имаме\n$$\nx^2 - x + 1 = \\left(x - \\frac{1}{2}\\right)^2 + \\frac{3}{4} \\ge \\frac{3}{4}, \\qquad (1)\n$$\n$$\n4y^2 + 6y + 4 = 4\\left(y + \\frac{3}{4}\\right)^2 + \\frac{7}{4} \\ge \\frac{7}{4} \\qquad (2)\n$$\nИ\n$$\n4z^2 - 12z + 25 = 4\\left(z - \\frac{3}{2}\\right)^2 + 16 \\ge 16. \\qquad (3)\n$$\nОд (1),(2... | North Macedonia | Републички натпревар по математика за средно образование | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | Macedonian, English | proof and answer | x = 1/2, y = -3/4, z = 3/2 | |
0gv9 | The sum of positive real numbers $a$, $b$, $c$ equals $1$. Prove the inequality
$$
\sqrt{\frac{1}{a} - 1}\sqrt{\frac{1}{b} - 1} + \sqrt{\frac{1}{b} - 1}\sqrt{\frac{1}{c} - 1} + \sqrt{\frac{1}{c} - 1}\sqrt{\frac{1}{a} - 1} \ge 6.
$$ | [
"$$\n\\frac{1}{3} \\left( \\sqrt{\\frac{(b+c)(c+a)}{ab}} + \\sqrt{\\frac{(c+a)(a+b)}{bc}} + \\sqrt{\\frac{(a+b)(b+c)}{ca}} \\right) \\geq 2.\n$$\nЗа нерівністю Коші\n$$\n\\frac{1}{3} \\left( \\sqrt{\\frac{(b+c)(c+a)}{ab}} + \\sqrt{\\frac{(c+a)(a+b)}{bc}} + \\sqrt{\\frac{(a+b)(b+c)}{ca}} \\right) \\geq \\sqrt[3]{\\f... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0kii | Problem:
Almondine has a bag with $N$ balls, each of which is red, white, or blue. If Almondine picks three balls from the bag without replacement, the probability that she picks one ball of each color is larger than 23 percent. Compute the largest possible value of $\left\lfloor\frac{N}{3}\right\rfloor$. | [
"Solution:\n\nIf $k = \\left\\lfloor \\frac{N}{3} \\right\\rfloor$, then the maximum possible probability is $\\frac{6k^{3}}{(3k)(3k-1)(3k-2)}$, with equality when there are $k$ balls of each of the three colors.\n\nGoing from $3k \\rightarrow 3k+1$ replaces $\\frac{k}{3k-2} \\rightarrow \\frac{k+1}{3k+1}$, which i... | United States | HMMT Spring 2021 Guts Round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 29 | |
093u | Problem:
Let $ABC$ be an acute triangle and $D$ an interior point of segment $BC$. Points $E$ and $F$ lie in the half-plane determined by the line $BC$ containing $A$ such that $DE$ is perpendicular to $BE$ and $DE$ is tangent to the circumcircle of $ACD$, while $DF$ is perpendicular to $CF$ and $DF$ is tangent to the... | [
"Solution:\n\nDenote by $T$ the intersection point of $BE$ and $CF$. Clearly, $D, E, F, T$ are concyclic because of the right angle $DET$ and $TFD$.\n\n\n\nThe tangent line $DE$ gives $\\angle ADE = \\angle ACD$. Similarly $\\angle FDA = \\angle DBA$, therefore $\\angle FDE = 180^{\\circ} -... | Middle European Mathematical Olympiad (MEMO) | 15th Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Triangles > Triangle trigonomet... | null | proof only | null | |
05n3 | Problem:
Soit $ABC$ un triangle rectangle en $C$. La bissectrice de $\widehat{BAC}$ coupe $[BC]$ en $P$, et celle de $\widehat{ABC}$ coupe $[AC]$ en $Q$. Soient $M$ et $N$ sur $[AB]$ tels que $(MP)$ et $(NQ)$ soient perpendiculaires à $(AB)$. Combien vaut l'angle $\widehat{MCN}$ ?
 | [
"Solution:\n\nComme les angles $\\widehat{AMP}$ et $\\widehat{ACP}$ sont droits, les points $A, C, P$ et $M$ sont cocycliques sur le cercle de diamètre $[AP]$. De même, les points $B, C, Q$ et $N$ sont cocycliques sur le cercle de diamètre $[BQ]$. On peut donc faire une chasse aux angles :\n\n$$\n\\begin{aligned}\n... | France | French Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 45° | |
0ix5 | Trapezoid $ABCD$, with $AB \parallel CD$, is inscribed in circle $\omega$ and point $G$ lies inside triangle $BCD$. Rays $AG$ and $BG$ meet $\omega$ again at points $P$ and $Q$, respectively. Let the line through $G$ parallel to line $AB$ intersect segment $BD$ and $BC$ at points $R$ and $S$, respectively. Prove that q... | [
"**Solution 1.** First, we prove the “if” direction by assuming that ray $BG$ bisects $\\angle CBD$; that is, we assume that $\\widehat{DQ} = \\widehat{CQ}$. It is easy to see that $ABCD$ is an isosceles trapezoid with $AD = BC$. In particular, $\\widehat{AD} = \\widehat{BC}$ and $\\widehat{AC} = \\widehat{BD}$. Be... | United States | USAMO 2009 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | null | proof only | null | |
0gdw | 在坐標平面上有 $N$ 個銳角三角形, 它們的三頂點皆為整數點, 面積皆為 $2^{2020}$, 且兩兩不全等。試求 $N$ 的最大值。
註：$(x, y)$ 為整數點, 若且唯若 $x$ 與 $y$ 皆為整數。
There are $N$ acute triangles on the plane. Their vertices are all integer points, their areas are all equal to $2^{2020}$, but no two of them are congruent. Find the maximum possible value of $N$.
Note: $(x, y)... | [
"我們先觀察出以下性質。\n\n**引理一：** 所有整點三角形若面積是整數，則必有 1 或 3 條邊的中點是整點。\n*Proof.* 由 Pick 定理知，面積整數的三角形落在邊上的整點必為偶數個。假設在三角形的某邊 $AB$ 上有整點 $P$。設中點為 $M$，則 $P$ 對 $M$ 的對稱點 $P'$ 也是整點（$A + B = P + P'$），所以三頂點和三中點的整點數總和必須是偶數，故有奇數個中點是整點，得證。□\n\n**引理二：** 沒有整點正三角形。\n*Proof.* 整點三角形的面積必為有理數。但假設存在整點正三角形且邊長為 $a$，那麼其面積為 $\\frac{\\sqrt{3}}{4}a^2$，且... | Taiwan | 2020 Taiwan IMO 2J | [
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2^{2020} - 1 | |
074n | Problem:
Find all natural numbers $n > 1$ such that $n^{2}$ does not divide $(n-2)!$. | [
"Solution:\nSuppose $n = p q r$, where $p < q$ are primes and $r > 1$. Then $p \\geq 2$, $q \\geq 3$ and $r \\geq 2$, not necessarily a prime. Thus we have\n$$\n\\begin{aligned}\n& n-2 \\geq n-p = p q r - p \\geq 5p > p \\\\\n& n-2 \\geq n-q = q(p r - 1) \\geq 3q > q \\\\\n& n-2 \\geq n-p r = p r(q-1) \\geq 2 p r >... | India | INMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All primes; all numbers equal to twice a prime; and the numbers 8 and 9. | |
054h | Let $ABC$ be a scalene triangle with median $AM$. Let $K$ be the point of tangency of the incircle of triangle $ABC$ with the side $BC$. Prove that if the length of the side $BC$ is the arithmetic mean of the lengths of the sides $AB$ and $AC$ then the bisector of the angle $BAC$ passes through the midpoint of the line... | [
"Let $N$ be the intersection point of the bisector of angle $BAC$ and side $BC$; it suffices to prove that $KN = MN$ (Fig. 19). The bisector property implies $\\frac{NC}{NB} = \\frac{AC}{AB}$. Substituting $NC = BC - NB$ gives\n$$\nNB = \\frac{BC}{1 + \\frac{AC}{AB}} = \\frac{AB \\cdot BC}{AB + AC}.\n$$\nAs $AB + A... | Estonia | National Olympiad Final Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
02yn | Problem:
Os três inteiros positivos $a, b$ e $c$ satisfazem
$$
4^{a} \cdot 5^{b} \cdot 6^{c}=8^{8} \cdot 9^{9} \cdot 10^{10}
$$
Determine o valor de $a+b+c$. | [
"Solution:\n$$\n\\begin{aligned}\n4^{a} \\cdot 5^{b} \\cdot 6^{c} & =8^{8} \\cdot 9^{9} \\cdot 10^{10} \\\\\n2^{2a} \\cdot 5^{b} \\cdot (2 \\cdot 3)^{c} & = (2^{3})^{8} \\cdot (3^{2})^{9} \\cdot (2 \\cdot 5)^{10} \\\\\n2^{2a} \\cdot 5^{b} \\cdot 2^{c} \\cdot 3^{c} & = 2^{24} \\cdot 3^{18} \\cdot 2^{10} \\cdot 5^{10... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 36 | |
0375 | Problem:
Let $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ be a geometric progression with $a_{1}=3-2 a$ and ratio $q=\frac{3-2 a}{a-2}$, where $a \neq \frac{3}{2}, 2$ is a real number. Set $S_{n}=\sum_{i=1}^{n} a_{i}$, $n \geq 1$. Prove that if the sequence $\left\{S_{n}\right\}_{n=1}^{\infty}$ is convergent and its limit is ... | [
"Solution:\nSince $S_{n}=a_{1} \\cdot \\frac{1-q^{n}}{1-q}$, the sequence $\\left\\{S_{n}\\right\\}_{n=1}^{\\infty}$ converges if and only if $|q|<1$. Therefore $-1<\\frac{3-2 a}{a-2}<1$, whence $a \\in\\left(1, \\frac{5}{3}\\right) \\backslash\\left\\{\\frac{3}{2}\\right\\}$. In this case\n$$\nS=\\lim _{n \\righta... | Bulgaria | Spring Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0ade | Solve the system
$$
\begin{cases}
x^{\log y} + \sqrt{y^{\log x}} = 110 \\
xy = 1000
\end{cases}
$$ | [
"By definition we have $x, y > 0$. If $A = x^{\\log y}$, $B = y^{\\log x}$ then $\\log A = \\log x \\cdot \\log y = \\log B$, from where $A = B$, and we have $x^{\\log y} = y^{\\log x} = t$. Then the first equation of the system is $t + \\sqrt{t} = 110$, and with $\\sqrt{t} = z$, we have $t, z \\ge 0$ from where we... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | [(10, 100), (100, 10)] | |
0b64 | Let $a, b, c \in (0, 1)$ and $x, y, z \in (0, \infty)$ be such that $a = (bc)^x$, $b = (ca)^y$, $c = (ab)^z$. Show that
$$
\sum \frac{1}{x+y+2} \le 1.
$$ | [] | Romania | Shortlisted Problems for the Romanian NMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof only | null | |
04d7 | Let $ABCD$ be a cyclic quadrilateral such that $|AD| = |BD|$ and let $M$ be the intersection of its diagonals. Let $N$ be the second intersection of the diagonal $AC$ with the circle passing through $B$, $M$ and the incentre of the triangle $BCM$. Prove that $|AN| \cdot |NC| = |CD| \cdot |BN|$. | [
"We denote $\\angle BAD = \\angle DBA = \\alpha$ in the isosceles triangle $ABD$. Then $\\angle ACB = \\angle ADB = 180^\\circ - 2\\alpha$.\nLet $I$ be the incentre of the triangle $BCM$. We have\n$$\n\\begin{aligned}\n\\angle MIB &= 180^\\circ - \\frac{1}{2} (\\angle BMC + \\angle CBM) = 180^\\circ - \\frac{1}{2} ... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
07aq | Points $A_1$ and $A_2$ are the intersection points of the incircle and the angle bisector of vertex $A$ with side $BC$ of triangle $ABC$, respectively. Points $B_1$, $B_2$, $C_1$ and $C_2$ are defined similarly. Suppose that the perpendicular from $A_1$ to line $B_2C_2$ intersects the angle bisector of vertex $A$ in $A... | [
"We start by proving a simple lemma.\n\n**Lemma 1.** Let $X'$ be the reflection of the circumcenter of triangle $XYZ$ with respect to side $YZ$. Then $A'$ and $X$ are symmetric with respect to the center of the nine-point circle of triangle $XYZ$.\n\n*Proof.* Let $O$ and $H$ be the circumcenter and orthocenter of t... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > A... | English | proof only | null | |
0ixb | Problem:
Let $a$, $b$, and $c$ be the 3 roots of $x^{3}-x+1=0$. Find $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$. | [
"Solution:\nWe can substitute $x = y - 1$ to obtain a polynomial having roots $a+1$, $b+1$, $c+1$, namely,\n$$(y-1)^{3} - (y-1) + 1 = y^{3} - 3y^{2} + 2y + 1.$$\nThe sum of the reciprocals of the roots of this polynomial is, by Viete's formulas,\n$$\n\\frac{2}{-1} = -2.\n$$"
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | -2 | |
0c52 | Let $x, y, z$ be complex numbers so that $x \neq z$ and $\frac{x-y}{x-z} \notin \mathbb{R}$. Prove that
$$ \min(|2x - y - z|, |2y - x - z|, |2z - x - y|) \le \frac{1}{\sqrt{3}}(|x - y| + |y - z| + |z - x|). $$ | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof only | null | |
0hnx | Problem:
What is the maximum number of squares on an $8 \times 8$ chessboard on which pieces may be placed so that no two of these squares touch horizontally, vertically, or diagonally? | [
"Solution:\n\nThe answer is $16$.\n\nDivide the chessboard into sixteen $2 \\times 2$ regions (see figure). Notice that no two pieces can lie in the same region. So there are at most $16$ pieces. By placing a piece in the lower left corner of each region, we see that $16$ is achievable."
] | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 16 | |
0fjq | Problem:
¿Existe alguna potencia de 2, que al escribirla en el sistema decimal tenga todos sus dígitos distintos de cero y sea posible reordenar los mismos para formar con ellos otra potencia de 2 distinta? Justificar la respuesta. | [
"Solution:\n\nSupongamos que exista tal potencia de $2$, es decir, que haya dos potencias de $2$ cuyas expresiones decimales sólo difieran en el orden de colocación de los dígitos. Claramente ninguna de las dos potencias es divisible por $3$ y ambas dejan el mismo resto cuando se dividen por $9$. Esto último se deb... | Spain | Olimpiada Matemática Española | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof only | null | |
03cu | We will refer to a permutation $\sigma$ of the numbers from 1 to 10 as *bad*, if there exist three indices $i, j, k$, such that $1 \le i < j < k \le 10$, but $\sigma(j) < \sigma(k) < \sigma(i)$, and as *good*, otherwise. Determine the number of the good permutations. | [
"We will derive a closed formula for the good permutations of the numbers from 1 to $n$. Consider a good permutation $\\sigma$, where $j = \\sigma^{-1}(1)$ is an arbitrary index inbetween 1 and $n$. Then, the first $j-1$ positions of $\\sigma$ contain the numbers $2, 3, \\ldots, j$, forming a good permutation of $j... | Bulgaria | Bulgaria 2022 | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 16796 | |
0ckk | Find the prime numbers $a, b, c, d$, with $a \le b$ and $c \le d$, satisfying
(1) $a + b = c + d + 1$;
(2) $a^2 + b^2 + c^2 + d^2 = 3543$. | [
"Since $c + d + 1 = a + b$, the number $a + b + c + d$ is odd. Therefore either three of the numbers $a, b, c, d$ are even (and, being primes are equal to $2$), or exactly one of them $a, b, c, d$ is even (therefore equal to $2$).\n\nIf three of the numbers $a, b, c, d$ are equal to $2$, condition (2) implies that ... | Romania | 75th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | a=2, b=43, c=3, d=41 | |
01fv | Let $ABC$ be a triangle and $H$ its orthocenter. Point $D$ lies on segment $AC$ and $E$ is its foot onto the line $BC$. Prove that $EH \perp BD$ iff $BD$ bisects $AE$. | [
"Let $BD \\cap AH = X$. Then $XH \\perp BE$, so $EH \\perp BD \\iff H$ is orthocenter $BXE \\iff BH \\perp EX \\iff EX \\parallel AC \\iff AXED$ is parallelogram $\\iff BD$ bisects $AE$."
] | Baltic Way | Baltic Way 2019 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08tn | 2 positive integers each with 3 digits are given. Suppose the one's digit and the ten's digit are both 9 for both of these numbers. Write down all possible numbers that can appear as the thousand's digit of the product of these 2 numbers. | [
"$[8, 9]$.\nWe can represent 2 three-digit numbers of the problem in the form $100m - 1$, $100n - 1$ with some pairs of integers $m, n$ satisfying $2 \\le m, n \\le 10$. Then, the product of the 2 given numbers can be written $10000mn - 100(m + n) + 1 = (mn - 1)10000 + (100 - m - n)100 + 1$, and therefore the value... | Japan | Japan Junior Mathematical Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | [8, 9] | |
0el8 | Problem:
V pravilni 8-kotnik je včrtan kvadrat z oglišči v ogliščih pravilnega 8-kotnika (glej sliko). Ploščina kvadrata je enaka $2~\mathrm{cm}^2$. Koliko kvadratnih centimetrov je ploščina pravilnega 8-kotnika?
(A) $1+\sqrt{2}$
(B) $\frac{5}{2}$
(C) $2 \sqrt{2}$
(D) 3
(E) $3 \sqrt{2}-1$
 | [
"Solution:\n\nKvadrat s ploščino $2~\\mathrm{cm}^2$ ima stranico dolgo $\\sqrt{2}~\\mathrm{cm}$ in diagonalo dolgo $2~\\mathrm{cm}$. Če povežemo oglišča 8-kotnika z njegovim središčem, dobimo torej 8 enakokrakih trikotnikov s kraki dolžine $r=1~\\mathrm{cm}$ in notranjim kotom med kraki enakim $\\alpha=45^\\circ$. ... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | C | |
0i97 | Let $\triangle ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E$, respectively. Rays $BA$ and $ED$ intersect at $F$ while lines $BD$ and $CF$ intersect at $M$. Prove that $MF = MC$ if and only if $MB \cdot MD = MC^2$. | [
"Extend segment $DM$ through $M$ to $G$ such that $FG \\parallel CD$.\n\nThen $MF = MC$ if and only if quadrilateral $CDFG$ is a parallelogram, or, $FD \\parallel CG$. Hence $MC = MF$ if and only if $\\angle GCD = \\angle FDA$, that is, $\\angle FDA + \\angle CGF = 180^\\circ$.\n\nBecause q... | United States | USA IMO 2003 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0j9z | Problem:
For what positive integers $n$ do there exist functions $f, g:\{1,2, \ldots, n\} \rightarrow \{1,2, \ldots, n\}$ such that for each $1 \leq i \leq n$, either $f(g(i))=i$ or $g(f(i))=i$, but not both? | [
"Solution:\n\nAnswer: $n$ even\n\nWe claim that this is possible for all even $n$. First, a construction: set $f(2m-1) = f(2m) = 2m-1$ and $g(2m-1) = g(2m) = 2m$ for $m = 1, \\ldots, \\frac{n}{2}$. It is easy to verify that this solution works.\n\nNow, we show that this is impossible for odd $n$. Without loss of ge... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Functional equations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | n even | |
0e4z | The circles $K_1$ and $K_2$ with the centres $O_1$ and $O_2$ intersect at the points $A$ and $B$, so that $\angle O_1AO_2 > \frac{\pi}{2}$. The line $O_1B$ intersects the circle $K_2$ again at $C$, the line $O_2B$ intersects the circle $K_1$ again at $D$. The line through the point $B$ parallel to the line $CD$ interse... | [] | Slovenia | Selection Examinations for the IMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
049d | In the quadrilateral $ABCD$ the following holds:
$$
\angle ABC = \angle ADC = 90^\circ, \quad |AB| = |BC|, \quad |CD| + |DA| = m.
$$
Determine the area of $ABCD$ in terms of $m$. | [] | Croatia | Hrvatska 2011 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | m^2/4 | |
09qe | Problem:
Gegeven is een vierkant $ABCD$ met omgeschreven cirkel $\Gamma_{1}$. Zij $P$ een punt op boog $AC$ waar ook $B$ op ligt. Een cirkel $\Gamma_{2}$ raakt inwendig aan $\Gamma_{1}$ in $P$ en raakt daarnaast diagonaal $AC$ in $Q$. Zij $R$ een punt op $\Gamma_{2}$ zodat de lijn $DR$ raakt aan $\Gamma_{2}$. Bewijs d... | [
"Solution:\n\nOplossing I. Zij $M$ het snijpunt van $AC$ en $BD$ (oftewel het middelpunt van $\\Gamma_{1}$) en zij $N$ het middelpunt van $\\Gamma_{2}$. We gaan allereerst bewijzen dat $P$, $Q$ en $D$ op een lijn liggen. Als $P = B$, dan is $Q = M$ en is het triviaal. Zo niet, definieer dan $S$ als het snijpunt van... | Netherlands | toets | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals"
] | null | proof only | null | |
0epy | Suppose that $a$ is an integer, and that $n! + a$ divides $(2n)!$ for infinitely many positive integers $n$. Prove that $a = 0$. | [
"$$(2n)! = \\binom{2n}{n} \\cdot n!^2 \\equiv \\binom{2n}{n} \\cdot (-a)^2 \\mod (n! + a),$$\nso if $n! + a$ divides $(2n)!$, then it also divides $a^2 \\binom{2n}{n}$. We will show that when $n$ is large, $n! + a$ is greater than $a^2 \\binom{2n}{n}$ and therefore does not divide it (unless $a = 0$). Assume in the... | South Africa | South African Mathematics Olympiad | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 0 | |
0kbe | Problem:
Chelsea goes to La Verde's at MIT and buys 100 coconuts, each weighing 4 pounds, and 100 honeydews, each weighing 5 pounds. She wants to distribute them among $n$ bags, so that each bag contains at most 13 pounds of fruit. What is the minimum $n$ for which this is possible? | [
"Solution:\n\nThe answer is $n=75$, given by 50 bags containing one honeydew and two coconuts (13 pounds), and 25 bags containing two honeydews (10 pounds).\n\nTo show that this is optimal, assign each coconut 1 point and each honeydew 2 points, so that 300 points worth of fruit are bought in total. Then, we claim ... | United States | HMMO 2020 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 75 | |
0ldp | Let $ABC$ be an acute triangle inscribed in the circle $(O)$ and $I$ is the circumcenter of triangle $OBC$. Point $G$ belongs to the arc $BC$ (not contains $O$) of $(I)$. The circle $(ABG)$ intersects $AC$ at $E$ and circle $ACG$ intersects $AB$ at $F$ (points $E, F$ differ from $A$).
1. Denote $K$ as the intersection... | [
"1)\nWe have\n$$\n\\begin{aligned}\n\\angle EGF &= \\angle BGE + \\angle CGF - \\angle EGF \\\\\n&= 360^{\\circ} - 2\\angle BAC - (180^{\\circ} - 2\\angle BAC) = 180^{\\circ}\n\\end{aligned}\n$$\nthen three points $E, G, F$ are collinear. Since $\\angle ABK + \\angle ACK = \\angle AGE + \\angle AGF = 180^{\\circ}$ ... | Vietnam | Vietnamese Mathematical Olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurat... | null | proof only | null | |
05pr | Problem:
Un constructeur de jouets crée au moins un jouet par jour. Il n'est pas capable de créer plus de 725 jouets par an. Pour tout $n \in \mathbb{N}$, montrer qu'il existe des jours consécutifs pendant lesquels il a créé exactement $n$ jouets. | [
"Solution:\n\nSoit $x_{i} \\in \\mathbb{N}$ le nombre de jouets qu'il a créés pendant le $i$-ème jour. Si\n$$\nx_{1} + x_{2} + \\ldots + x_{n} < 2n\n$$\nétudions les nombres suivants :\n$$\n1 \\leqslant x_{1} < x_{1} + x_{2} < \\ldots < x_{1} + x_{2} + \\ldots + x_{n} < 2n.\n$$\nOn a deux cas. Soit ils représentent... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0kwy | Problem:
Elbert and Yaiza each draw 10 cards from a 20-card deck with cards numbered $1,2,3, \ldots, 20$. Then, starting with the player with the card numbered $1$, the players take turns placing down the lowest-numbered card from their hand that is greater than every card previously placed. When a player cannot place... | [
"Solution:\n\nPut each card in order and label them based on if Elbert or Yaiza got them. We will get a string of $E$'s and $Y$'s like $EEYYYE\\ldots$, and consider the \"blocks\" of consecutive letters. It is not hard to see that only the first card of each block is played, and the number of cards played is exactl... | United States | HMMT February 2023 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 324 | |
0dvu | Problem:
Zapiši enačbo polinoma 3. stopnje (lahko tudi v razstavljeni obliki), katerega graf poteka skozi točke $A(4,-5)$, $B(-1,0)$, $C(0,5)$ in $D(5,0)$. Skiciraj graf polinoma. | [
"Solution:\n\nTočki $B$ in $D$ sta ničli polinoma, zato lahko zapišemo $y = a(x+1)(x-5)(x-x_3)$. Polinom seka ordinatno os v točki $C(0,5)$, torej je $5 = a \\cdot 1 \\cdot (-5) \\cdot (-x_3)$ oziroma $5 a x_3 = 5$, od koder dobimo $x_3 = \\frac{1}{a}$, saj je $a \\neq 0$.\n\nKončno upoštevamo, da gre polinom skozi... | Slovenia | 3. matematično tekmovanje dijakov srednjih tehniških in strokovnih sol | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | null | final answer only | y = 1/2 (x+1)(x-5)(x-2) | |
0gr1 | Find all triples $(a, b, c)$ of positive integers such that $3^a + 3^b + 3^c$ is a perfect square. | [
"Let $3^a + 3^b + 3^c = m^2$. Since $m$ is odd we have $m^2 \\equiv 1 \\pmod{8}$. For each positive integer $k$ we have $3^k \\equiv 1$ or $3^k \\equiv 3 \\pmod{8}$. Therefore, $3^a \\equiv 3^b \\equiv 3^c \\equiv 1 \\pmod{8}$ and we get that $a$, $b$ and $c$ are odd numbers. Let $a \\leq b \\leq c$. Then $3^a (3^{... | Turkey | 22nd Junior Turkish Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | (a, b, c) = (2k - 1, 2k - 1, 2k - 1) for any positive integer k | |
0jnw | Problem:
Find the least positive integer $N > 1$ satisfying the following two properties:
- There exists a positive integer $a$ such that $N = a(2a - 1)$.
- The sum $1 + 2 + \cdots + (N - 1)$ is divisible by $k$ for every integer $1 \leq k \leq 10$. | [
"Solution:\nAnswer: 2016\n\nThe second condition implies that $16$ divides $a(2a - 1)(2a^{2} - a - 1)$, which shows that $a \\equiv 0$ or $1$ modulo $16$. The case $a = 1$ would contradict the triviality-avoiding condition $N > 1$. $a$ cannot be $16$, because $7$ does not divide $a(2a - 1)(2a^{2} - a - 1)$. $a$ can... | United States | HMMT February | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2016 | |
0fs4 | Problem:
We are given $n$ distinct rectangles in the plane. Prove that between the $4 n$ interior right angles formed by these rectangles at least $4 \sqrt{n}$ are distinct. | [
"Solution:\n\nFirst of all, let's make the whole picture easier to handle: We can split the rectangles into groups such that in each group, all sides of rectangles are parallel or perpendicular to each other. We also choose these groups to be maximal, in particular: For any two rectangles of different groups, they ... | Switzerland | null | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
067c | We consider an $n \times n$ chessboard, where $n$ is an even positive integer. On the board we put all numbers $1,2,3, \dots, n^2$, one at each square. Let $S_1$ be the sum of the numbers lying at the white squares and let $S_2$ be the sum of the numbers lying on the blank squares. Find all the numbers $n$ for which it... | [
"The given relation is equivalent to: $S_1 = \\frac{39}{103}(S_1 + S_2)$. Since\n$$\nS_1 + S_2 = 1 + 2 + \\dots + n^2 = \\frac{n^2(n^2+1)}{2}\n$$\nand $S_1$ is a natural number we conclude that $103 \\mid \\left\\lfloor \\frac{n^2(n^2+1)}{2} \\right\\rfloor$. Since $103$ is a prime number of the form $4\\kappa+3$, ... | Greece | 31st Hellenic Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | All even n divisible by 103, i.e., n = 206k for positive integers k. | |
0krv | Problem:
Determine all complex numbers $w$ such that
$$
10|w|^{2}=2|w+2|^{2}+\left|w^{2}+1
ight|^{2}+20 .
$$ | [
"Solution:\nLet $w = a + b i$, where $a$ and $b$ are real. The equation becomes\n$$\n\\begin{aligned}\n10\\left(a^{2}+b^{2}\\right) & = 2\\left[(a+2)^{2}+b^{2}\\right] + \\left(a^{2}-b^{2}+1\\right)^{2} + (2 a b)^{2} + 20 \\\\\n& = 2 a^{2} + 8 a + 8 + 2 b^{2} + \\left(a^{4} - 2 a^{2} b^{2} + b^{4}\\right) + 2\\left... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | w = -1 ± 2i | |
0a9b | Problem:
On a faded piece of paper it is possible, with some effort, to discern the following:
$$
\left(x^{2}+x+a\right)\left(x^{15}-\ldots\right)=x^{17}+x^{13}+x^{5}-90 x^{4}+x-90
$$
Some parts have got lost, partly the constant term of the first factor of the left side, partly the main part of the other factor. It wo... | [
"Solution:\nWe denote the polynomial $x^{2}+x+a$ by $P_{a}(x)$, the polynomial forming the other factor of the left side by $Q(x)$ and the polynomial on the right side by $R(x)$. The polynomials are integer valued for every integer $x$.\n\nFor $x=0$ we get $P_{a}(0)=a$ and $R(0)=-90$, so $a$ is a divisor of $90=2 \... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 23 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 2 | |
0ccd | Find all twice differentiable functions $f : \mathbb{R} \to (0, \infty)$ with the property $3(f'(x))^2 \le 2f(x)f''(x)$, for every $x \in \mathbb{R}$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Equations > ODEs"
] | null | proof and answer | f(x) ≡ c for some constant c > 0 | |
0cjj | Determine the four-digit natural numbers $\overline{abcd}$, for which there exists a prime number $p$, such that $\overline{cd} - \overline{ab} = p + 2$ and $\sqrt{\overline{ab} - 4} + \sqrt{\overline{cd}} = p^2$. | [
"Since $\\sqrt{\\overline{ab} - 4} + \\sqrt{\\overline{cd}} < 10 + 10 = 20$, we obtain $p = 2$ or $p = 3$.\nIf $p = 2$, then $\\overline{cd} = \\overline{ab} + 4$ and $\\sqrt{\\overline{ab} - 4} + \\sqrt{\\overline{ab} + 4} = 4$, therefore $\\overline{ab} = 5$, false.\nIf $p = 3$, then $\\overline{cd} = \\overline{... | Romania | 75th Romanian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 2025 | |
0l5e | Problem:
A subset $S$ of $\{1,2,3,\ldots ,2025\}$ is called balanced if for all elements $a$ and $b$ both in $S$, there exists an element $c$ in $S$ such that $2025$ divides $a + b - 2c$. Compute the number of nonempty balanced subsets. | [
"Solution:\nWe work mod $2025$, so the condition becomes that for any $a$, $b \\in S$, we have $\\frac{a + b}{2} \\in S$.\n\nFirst, we prove that $S$ must be an arithmetic sequence. Observe that if $S$ is balanced, then so is the shift $S + k = \\{s + k \\mid s \\in S\\}$ for all $k$, so we can assume $0 \\in S$. L... | United States | HMMT February | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | final answer only | 3751 | |
0f7o | Problem:
The set $S_0 = \{1, 2!, 4!, 8!, 16!, \ldots\}$. The set $S_{n+1}$ consists of all finite sums of distinct elements of $S_n$. Show that there is a positive integer not in $S_{1987}$. | [] | Soviet Union | 21st ASU | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0cgm | Let $ABCDE$ be an inscriptible pentagon for which $AB = BC = CD$ and the centroid of the pentagon coincides with the center of the circumscribed circle. Show that the pentagon $ABCDE$ is regular. | [
"We consider an orthonormal coordinate system centred in $O$, where $O$ is the centre of the circumcircle $C$ of the pentagon $ABCDE$, with the unit length equal to the radius of the circle $C$ and with the real axis being the perpendicular bisector of the segment $BC$. Let $z_X$ denote the complex number represent... | Romania | 74th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06m3 | Mandy needs to wake up early for attending a mathematics contest. She has set an alarm in her smartphone every 15 minutes since 5:30 am. If an alarm is not pressed off by her or her mother (or anybody else), it will ring for a while, stop for a while, then will ring again 9 minutes later as the first ring, and so on (e... | [
"If each ring-tone lasts for exactly $3$ minutes, the alarm will be ringing at $5:30$–$5:33$, $5:39$–$5:42$, $5:45$–$5:51$ and $5:54$–$6:30$. It will ring for $3 + 3 + 6 + 36 = 48$ minutes in total. Therefore, we must have $x > 3$.\n\nLet $x = 3 + y$ where $0 < y < 3$. In addition to the above periods, the alarm wi... | Hong Kong | HongKong 2022-23 IMO Selection Tests | [
"Math Word Problems"
] | English | proof and answer | 11/3 | |
0d7u | Let $P \in \mathbb{Q}[x]$ be a polynomial of degree $2016$ whose leading coefficient is $1$. A positive integer $m$ is "nice" if there exists some positive integer $n$ such that $m = n^{3} + 3n + 1$. Suppose that there exist infinitely many positive integers $n$ such that $P(n)$ are nice. Prove that there exists an ari... | [
"For convenience, denote $3d = 2016$.\n\nLemma. There exist a polynomial $Q(x) \\in \\mathbb{Q}[x]$ such that\n$$\n\\lim_{x \\rightarrow +\\infty} [\\sqrt[3]{P(x)} - Q(x)] = 0\n$$\nProof. Suppose that $S(x) \\in \\mathbb{Q}[x]$ is a polynomial such that $\\operatorname{deg}(P(x) - S^{3}(x))$ is minimized (if there ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization"
] | English | proof only | null | |
0h2h | Let $a, b, c, d$ be distinct natural numbers such that $ab + cd$ is divisible by $ac + bd$. Prove that $ac + bd$ is a composite number. | [
"To the contrary, assume that $ac + bd$ is prime. Then $ac + bd \\mid ab + cd \\Rightarrow ac + bd \\mid ac + bd + ab + cd = (a + d)(b + c)$, and so $ac + bd \\mid a + d$ or $ac + bd \\mid b + c$. But this is impossible because for distinct $a, b, c, d$ we have that $ac + bd > a + d$ and $ac + bd > b + c$. This con... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
020i | Problem:
Let $n$ be a positive integer. Suppose that its positive divisors can be partitioned into pairs (i.e. can be split in groups of two) in such a way that the sum of each pair is a prime number. Prove that these prime numbers are distinct and that none of these are a divisor of $n$. | [
"Solution:\nLet $d_{1}$ and $d_{2}$ be positive divisors of $n$ that form a pair as given in the problem. If $d_{1}$ and $d_{2}$ have a non-trivial prime divisor $p$ in common, then $p \\mid d_{1}+d_{2}$ and $p \\leqslant d_{1} < d_{1}+d_{2}$, so $d_{1}+d_{2}$ cannot be prime. Hence $\\gcd(d_{1}, d_{2}) = 1$, which... | Benelux Mathematical Olympiad | 8th Benelux Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
06zg | Problem:
The number $1$ or the number $-1$ is assigned to each vertex of a cube. Then each face is given the product of its four vertices. What are the possible totals for the resulting $14$ numbers? | [
"Solution:\n\nIf every vertex is $1$, we get $14$ and that is clearly the highest possible total. The lowest possible total cannot be lower than $-14$, but we cannot even achieve that because if all the vertices are $-1$, then all the faces are $1$.\n\nIf we change a vertex, then we also change three faces. If the ... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | -10, -6, -2, 2, 6, 14 | |
0bzl | Prove that $\frac{m}{n} + \frac{n}{m} \neq 1$, for any positive integers $m, n$. | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0fq6 | Problem:
¿De cuántas maneras se puede escribir $111$ como suma de tres números enteros en progresión geométrica? | [
"Solution:\nSean $a$, $a r$, $a r^{2}$ números enteros que satisfacen la ecuación\n$$\na + a r + a r^{2} = 111\n$$\nsiendo $r$ un número racional.\nHabida cuenta de que $1 + r + r^{2} = \\left(r + \\frac{1}{2}\\right)^{2} + \\frac{3}{4} > 0$, existen números naturales $p$ y $q$, primos entre sí, tales que\n$$\n1 + ... | Spain | null | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greate... | null | proof and answer | 17 | |
07sq | Prove that if $x, y, z$ are non-negative numbers, then
$$
x^3 + y^3 + z^3 + 3xyz \geq x^2(y + z) + y^2(z + x) + z^2(x + y).
$$
Determine the cases of equality. | [
"This inequality is known as Schur's inequality. By symmetry, we may assume\n$0 \\le x \\le y \\le z$ and introduce $t = y - x$ and $s = z - y$, which satisfy $t, s \\ge 0$.\nUsing $y = t + x$ and $z = y + s = t + s + x$ the desired inequality can be\nrestated as follows.\n$$\n\\begin{align*}\n0 &\\le \\sum_{\\text... | Ireland | IRL_ABooklet_2020 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof and answer | Equality holds precisely when either two numbers are equal and the third is zero, or all three numbers are equal. | |
0d3a | The excircle $\omega_{B}$ of triangle $ABC$ opposite $B$ touches side $AC$, rays $BA$ and $BC$ at $B_{1}$, $C_{1}$ and $A_{1}$, respectively. Point $D$ lies on major arc $\widetilde{A_{1}C_{1}}$ of $\omega_{B}$. Rays $DA_{1}$ and $C_{1}B_{1}$ meet at $E$. Lines $AB_{1}$ and $BE$ meet at $F$. Prove that line $FD$ is tan... | [
"Let $F'$ be the intersection point of the tangents to circle $\\omega_{B}$ at $B_{1}$ and $D$.\n\n\n\nLet $G$ be the intersection point of lines $B_{1}A_{1}$ and $DC_{1}$.\n\nConsider the six cyclic points $C_{1}, C_{1}, B_{1}, A_{1}, A_{1}, D$. Because the tangent lines to $\\omega_{B}$ a... | Saudi Arabia | Selection tests for the Balkan Mathematical Olympiad 2013 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | English | proof only | null | |
08a9 | Problem:
Qual è l'esponente del primo $2$ nella fattorizzazione del numero
$$
(5-1)\left(5^{5}-1\right) \ldots\left(5^{5 . .^{5}}-1\right)
$$
dove in ogni fattore compare ad esponente un "5" in più che nel precedente e nell'ultimo ne compaiono, come esponenti, $2014$? | [
"Solution:\n\nLa risposta è $4030$. Se $d$ è un intero dispari, l'espressione $5^{d}-1$ si può fattorizzare come $(5-1)\\left(5^{d-1}+5^{d-2}+\\ldots+1\\right)$ dove il secondo fattore contiene $d$ addendi dispari, ovvero un numero dispari di addendi dispari, e dunque è dispari. Un numero della forma $5^{d}-1$ con ... | Italy | Progetto Olimpiadi della Matematica - Gara di Febbraio | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 4030 | |
0jsv | Problem:
Sherry is waiting for a train. Every minute, there is a $75\%$ chance that a train will arrive. However, she is engrossed in her game of sudoku, so even if a train arrives she has a $75\%$ chance of not noticing it (and hence missing the train). What is the probability that Sherry catches the train in the nex... | [
"Solution:\n\nDuring any given minute, the probability that Sherry doesn't catch the train is $\\frac{1}{4} + \\left(\\frac{3}{4}\\right)^2 = \\frac{13}{16}$. The desired probability is thus one minus the probability that she doesn't catch the train for the next five minutes: $1 - \\left(\\frac{13}{16}\\right)^5$."... | United States | HMMT February | [
"Statistics > Probability > Counting Methods > Other",
"Math Word Problems"
] | null | final answer only | 1 - (13/16)^5 | |
01v6 | Given the hyperbola $y = \frac{1}{x}$ and four circles $S_1, S_2, S_3, S_4$. The circle $S_1$ intersects the hyperbola at points $Z_1, Z_2, Z_3, Z_4$; $S_2$ intersects the hyperbola at points $Z_3, Z_4, Z_5, Z_6$; $S_3$ intersects the hyperbola at points $Z_5, Z_6, Z_7, Z_8$; $S_4$ intersects the hyperbola at points $Z... | [
"Answer: $\\sqrt{R_1^2 + R_3^2 - R_2^2}$.\n\nFirst we prove the following\n\n**Lemma.** Let $x_1, x_2, x_3, x_4$ be abscissae of the intersection points of a circle with the hyperbola $y = \\frac{1}{x}$. Then\n$$\nR^2 = \\frac{1}{4}(x_1^2 + x_2^2 + x_3^2 + x_4^2 + \\frac{1}{x_1^2} + \\frac{1}{x_2^2} + \\frac{1}{x_3... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Circles",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | sqrt(R1^2 + R3^2 - R2^2) | |
03g6 | Let $n \ge 4$ be an integer number and $S_n = \{1, 2, 3, \dots, 2^n\}$. Two sets $A, B$ are given, $A \subset S_n, B \subset S_n \setminus S_{n-1}$, such that $|A| = n+1, |B| = 2$. Is it possible $ab-1$ be a perfect cube for any $a \in A, b \in B$?
(Dragomir Grozev) | [
"Answer: NO. Let us argue by contradiction. Arrange the numbers in $A$ and $B$ in increasing order $1 \\le a_1 < a_2 < \\dots < a_{n+1} \\le 2^n$ and $2^{n-1} < b_1 < b_2 \\le 2^n$. Apparently, there exists an index $i \\le n$ such that $a_i < a_{i+1} \\le 2a_i$. We denote:\n$$\na_i b_1 - 1 = q_1^3,\\quad a_{i+1} b... | Bulgaria | 6 TST for BMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | NO | |
0gal | 令 $n \ge 5$ 為一與 $6$ 互質的正整數。我們將一個正 $n$ 邊形的 $n$ 個頂點，各塗上三種顏色中的一種，使得被塗上每種顏色的頂點數都是奇數。試證：我們必可從這 $n$ 個頂點中找出三個點，它們的顏色都不相同，且此三點的連線構成等腰三角形。 | [
"令 $a_k$ 為所有等腰三角形中，三頂點恰包含 $k$ 種顏色的三角形個數，則題目等價於證明 $a_3 \\ge 1$。\n\n我們採取歸謬證法。假設 $a_3 = 0$。考慮集合\n$$\nX = \\{(\\Delta, E) : \\Delta \\text{是等腰三角形}, E \\text{是 } \\Delta \\text{的一邊}, E \\text{的兩端點不同色}\\}.\n$$\n讓我們用兩種不同方式計算 $X$ 中的元素個數：\n\n– 首先，對於每個三角形：\n\n* 只有一色的三角形必沒有兩端點異色的邊中。\n* 只有兩色的三角形則恰有 $2$ 條邊的兩端點異色。\n* 由假設，不存在三色的三... | Taiwan | 二〇一七數學奧林匹亞競賽第三階段選訓營 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0d3f | Let $a_{1}, a_{2}, \ldots, a_{2 n}$ be positive real numbers such that $a_{i}+a_{n+i}=1$, for all $i=1, \cdots, n$. Prove that there exist two different integers $1 \leq j, k \leq 2 n$ for which
$$
\sqrt{a_{j}^{2}-a_{k}^{2}}<\frac{1}{\sqrt{n}+\sqrt{n-1}}
$$ | [
"Assume, without loss of generality, that\n$$\n0<a_{1} \\leq a_{2} \\leq \\cdots \\leq a_{n} \\leq \\frac{1}{2} \\leq a_{2 n} \\leq a_{2 n-1} \\leq \\cdots \\leq a_{n+1}<1,\n$$\nand let\n$$\nK=\\min \\left\\{\\left|a_{i}^{2}-a_{j}^{2}\\right|, \\text{ with } i, j=1, \\ldots, 2 n, \\text{ and } i \\neq j\\right\\} .... | Saudi Arabia | SAMC | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English, Arabic | proof only | null | |
0avz | Problem:
An urn contains five red chips numbered $1$ to $5$, five blue chips numbered $1$ to $5$, and five white chips numbered $1$ to $5$. Two chips are drawn from this urn without replacement. What is the probability that they have either the same color or the same number? | [
"Solution:\n\nThe required probability is\n$$\n\\frac{3 \\cdot \\binom{5}{2} + 5 \\cdot \\binom{3}{2}}{\\binom{15}{2}} = \\frac{30 + 15}{105} = \\frac{3}{7}\n$$"
] | Philippines | 18th PMO National Stage Oral Phase | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 3/7 | |
04ho | Let $a$, $b$, $c$ be distinct positive integers and let $r$, $s$, $t$ be positive integers such that:
$$
ab + 1 = r^2, \quad ac + 1 = s^2, \quad bc + 1 = t^2.$$
Prove that it is not possible that all three fractions $\frac{rs}{t}$, $\frac{rt}{s}$ and $\frac{st}{r}$ are positive integers. (Miljen Mikić) | [
"Assume on the contrary that $\\frac{rs}{t}$, $\\frac{rt}{s}$, $\\frac{st}{r}$ are positive integers.\nWithout loss of generality we may assume that $a < b < c$. Since $\\frac{rs}{t}$ is a positive integer, $\\frac{r^2 s^2}{t^2}$ is also a positive integer, and hence\n$$\n\\frac{r^2 s^2}{t^2} = \\frac{a^2 bc + ab +... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0cj7 | Let $ABCD$ be a square with side length $1$.
a) Prove that there is no point $M$ in the plane of the square such that the lengths of the segments $MA$, $MB$, $MC$, and $MD$ are natural numbers.
b) Find the points $M$ in space such that the lengths of the segments $MA$, $MB$, $MC$, and $MD$ are natural numbers. | [] | Romania | 75th NMO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Solid Geometry > Other 3D problems",
"Number Theory > Other"
] | English | proof and answer | a) No such point exists in the plane.
b) Exactly the points on the line perpendicular to the square through its center, with coordinates (one half, one half, plus or minus the square root of k squared minus one half), where k is a positive integer; for these points all four distances equal k. | |
07bw | For positive real numbers $a$, $b$, $c$ and $d$ such that $\sum_{cyc} \frac{1}{ab} = 1$, prove that
$$
abcd + 16 \geq 8\sqrt{(a+c)\left(\frac{1}{a} + \frac{1}{c}\right)} + 8\sqrt{(b+d)\left(\frac{1}{b} + \frac{1}{d}\right)}.$$ | [
"First note that $(a+c)\\left(\\frac{1}{a}+\\frac{1}{c}\\right) = \\frac{a}{c} + \\frac{c}{a} + 2 = \\left(\\sqrt{\\frac{a}{c}} + \\sqrt{\\frac{c}{a}}\\right)^2$ and so\n$$\n\\sqrt{(a+c)\\left(\\frac{1}{a}+\\frac{1}{c}\\right)} = \\sqrt{\\frac{a}{c}} + \\sqrt{\\frac{c}{a}}\n$$\nSimilarly $\\sqrt{(b+d)\\left(\\frac{... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0cva | Each point $A$ in the plane is assigned a real number $f(A)$. It is known that $f(M) = f(A) + f(B) + f(C)$ whenever $M$ is the centroid of a triangle $ABC$. Prove that $f(A) = 0$ for every point $A$. | [
"Возьмём произвольную точку $M$ плоскости и докажем, что $f(M) = 0$. Для этого рассмотрим произвольный треугольник $ABC$, для которого точка $M$ является точкой пересечения медиан. Обозначим через $D$, $E$ и $F$ точки пересечения медиан треугольников $BCM$, $CAM$ и $ABM$, соответственно (см. рис. 17).\n\nЗаметим, ч... | Russia | Final round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English; Russian | proof only | null | |
0gaf | Let $\triangle ABC$ be an acute triangle, and let $M$ be the midpoint of $AC$. A circle $\omega$ passing through $B$ and $M$ meets the sides $AB$ and $BC$ again at $P$ and $Q$, respectively. Let $T$ be the point such that the quadrilateral $BPTQ$ is a parallelogram. Suppose that $T$ lies on the circumcircle of the tria... | [
"$BT/BM = \\sqrt{2}$。\n\n令 $S$ 為平行四邊形 $BPTM$ 的中心, 並令點 $B' \\neq B$ 是在射線 $BM$ 上滿足 $BM = MB'$ 的點, 如圖 1 所示。易知 $ABCB'$ 為平行四邊形。\n\n\n\n圖 1\n\n所以, $\\angle ABB' = \\angle PQM$, 以及 $\\angle BB'A = \\angle B'BC = \\angle MPQ$, 由此得 $\\triangle ABB' \\sim \\triangle MQP$。於是 $AM$ 與 $MS$ 分別是這兩個相似三角形對應邊... | Taiwan | 二〇一六數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(2) | |
04pf | The base of a right square pyramid has sides of length $12$, and the height of the pyramid is of length $8$. Find the surface area of the pyramid. (Nikola Adžaga) | [] | Croatia | Croatian Mathematical Society Competitions | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > 3D Shapes"
] | English | final answer only | 384 | |
0bk2 | Prove that three discs of radius $1$ cannot cover entirely a square surface of side $2$, but they can cover more than $99.75\%$ of it. | [
"Denote $ABCD$ the square and $S_1$, $S_2$, $S_3$ the discs.\n\nSuppose that $S_1$, $S_2$ and $S_3$ cover the whole square. Since there are three discs and the square has four vertices, one of the discs must cover two vertices of the square, say $S_1$ covers $A$, $B$. Then $[AB]$ is a diameter for $S_1$, so $S_1$ c... | Romania | 65th Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0fra | Problem:
Determinar todos los números de cuatro cifras $n=\overline{a b c d}$ tales que al insertar un dígito 0 en cualquier posición se obtiene un múltiplo de 7. | [
"Solution:\n\nComenzamos observando que el número que resulta de insertar un 0 al final de $n$ es $10 n$, que al ser múltiplo de 7 obliga a que $n$ también lo sea. De hecho, son múltiplos de 7 los siguientes cinco números:\n$$\n\\begin{aligned}\nn=\\overline{a b c d} & =1000 a+100 b+10 c+d \\\\\nx=\\overline{a 0 b ... | Spain | Spain | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 7000, 7007, 7070, 7077, 7700, 7707, 7770, 7777 | |
0ac8 | Let $n$ be a positive integer. Does $(2n-3)(2n-1)(2n+1)(2n+3)+16$ equal the square of a positive integer? Does the number $2005 \times 2007 \times 2009 \times 2011 + 16$ equal the square of a positive integer? | [
"Because $(2n-3)(2n+3) = 4n^2 - 9$ and $(2n-1)(2n+1) = 4n^2 - 1$, we have\n$$(2n-3)(2n-1)(2n+1)(2n+3) + 16 = (4n^2 - 9)(4n^2 - 1) + 16.$$\nExpanding:\n$$(4n^2 - 9)(4n^2 - 1) = 16n^4 - 4n^2 - 36n^2 + 9 = 16n^4 - 40n^2 + 9.$$\nSo,\n$$(2n-3)(2n-1)(2n+1)(2n+3) + 16 = 16n^4 - 40n^2 + 9 + 16 = 16n^4 - 40n^2 + 25.$$\nNoti... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | Yes; for all positive integers n it equals (4n^2 − 5)^2. For 2005 × 2007 × 2009 × 2011 + 16, yes; it equals (4 × 1004^2 − 5)^2. | |
0kny | Problem:
Compute $\sqrt{2022^{2}-12^{6}}$. | [
"Solution:\n\n$$\n\\begin{aligned}\n2022^{2}-12^{6} & =\\left(2022-12^{3}\\right)\\left(2022+12^{3}\\right) \\\\\n& =294 \\cdot 3750 \\\\\n& =\\left(2 \\cdot 3 \\cdot 7^{2}\\right)\\left(2 \\cdot 3 \\cdot 5^{4}\\right),\n\\end{aligned}\n$$\nso the answer is $2 \\cdot 3 \\cdot 5^{2} \\cdot 7=1050$."
] | United States | HMMT November 2022 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 1050 | |
0gpo | Show that for all positive real numbers $x$, $y$ and $z$ the following inequality is held:
$$
\frac{x(2x - y)}{y(2z + x)} + \frac{y(2y - z)}{z(2x + y)} + \frac{z(2z - x)}{x(2y + z)} \ge 1
$$ | [
"The first term $\\frac{x(2x-y)}{y(2z+x)} + 1 = \\frac{2(x^2+yz)}{y(2z+x)}$. The similar transformation of two other terms yields:\n$$\nf(x, y, z) = \\frac{x^2 + yz}{y(2z+x)} + \\frac{y^2 + zx}{z(2x+y)} + \\frac{z^2 + xy}{x(2y+z)} \\ge 2\n$$\nBy Cauchy-Schwarz inequality for positive $x_1, \\dots, x_n$\n$$\n(x_1 + ... | Turkey | 20th Turkish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
036m | Problem:
Let $O$ be the circumcenter of a triangle $ABC$ with $AC = BC$. The line $AO$ meets the side $BC$ at $D$. If the lengths of $BD$ and $CD$ are integers, and $AO - CD$ is a prime number, find these three numbers. | [] | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | proof and answer | BD = 8, CD = 4, AO − CD = 2 | |
04r4 | Find all pairs of primes $p$, $q$ for which there exists a positive integer $a$ such that
$$
\frac{pq}{p+q} = \frac{a^2+1}{a+1}.
$$ | [
"First, we will deal with the case when the wanted primes $p$ and $q$ are distinct. Then, the numbers $pq$ and $p+q$ are relatively prime: the product $pq$ is divisible by two primes only (namely $p$ and $q$), while the sum $p+q$ is divisible by neither of these primes.\nWe will look for a positive integer $r$ whic... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | p = q = 2 and p = q = 5 | |
0035 | Mauro escribió la lista de los números de $12$ dígitos con cada dígito igual a $0$ ó $1$ tales que la suma de los dígitos en las posiciones pares es igual a la suma de los dígitos en las posiciones impares. Determinar cuántos números tiene la lista de Mauro.
ACLARACIÓN: Todos los números de la lista tienen el primer d... | [] | Argentina | Argentina 2006 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | Español | proof and answer | 462 | |
0k7q | Problem:
Reimu and Sanae play a game using 4 fair coins. Initially both sides of each coin are white. Starting with Reimu, they take turns to color one of the white sides either red or green. After all sides are colored, the 4 coins are tossed. If there are more red sides showing up, then Reimu wins, and if there are ... | [
"Solution:\n\nClearly Reimu will always color a side red and Sanae will always color a side green, because their situation is never worse off when a side of a coin changes to their own color. Since the number of red-only coins is always equal to the number of green-only coins, no matter how Reimu and Sanae color th... | United States | HMMT February 2019 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 5/16 | |
0bn6 | The matrices $A, B \in \mathcal{M}_2(\mathbb{R})$ have the property $(A - B)^2 = O_2$.
a) Show that $\det(A^2 - B^2) = (\det(A) - \det(B))^2$.
b) Show that $\det(AB - BA) = 0$ if and only if $\det(A) = \det(B)$. | [
"a) From $(A - B)^2 = O_2$ follows $\\det(A - B) = 0$ and $\\mathrm{Tr}(A - B) = 0$, hence $\\mathrm{Tr}(A) = \\mathrm{Tr}(B) =: a$.\nDenote $b = \\det(A) - \\det(B)$. Then\n$$\n\\begin{cases}\nA^2 - aA + \\det(A)I_2 = O_2 \\\\\nB^2 - aB + \\det(B)I_2 = O_2\n\\end{cases}\n$$\nThis shows that $\\det(A^2 - B^2) = \\d... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof only | null | |
02t1 | Problem:
A partir do meio-dia, João faz, a cada 80 minutos, uma marca na posição do ponteiro das horas do seu relógio.
a) Depois de quanto tempo não será mais necessário fazer novas marcas no relógio?
b) Qual a soma dos ângulos internos do polígono formado pelas marcas? | [
"Solution:\na) O ponteiro das horas concluirá uma volta completa após $12 \\cdot 60 = 720$ minutos e ao longo dela nenhuma marca será repetida. Como $720$ é múltiplo de $80$, durante esse período são feitas exatamente $\\frac{12 \\cdot 60}{80} = 9$ marcas no relógio e, além disso, os dois ponteiros voltam às suas p... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | a) 720 minutes; b) 1260 degrees | |
0eia | Problem:
Naj bo $b=\sqrt[2019]{\sqrt[2018]{\ldots \sqrt[4]{\sqrt[3]{\sqrt[2]{a}}}}},\ a>0$. Kateri izraz je enakovreden $\sqrt[2020]{\sqrt[2019]{\sqrt[2018]{\ldots \sqrt[4]{\sqrt[3]{a^{404}}}}}}$ ?
(A) $\sqrt[2020]{b}$
(B) $\sqrt[5]{b}$
(C) $\sqrt[5]{b^{2}}$
(D) $b^{404}$
(E) $b^{202}$ | [
"Solution:\n\nDrugi izraz je zelo podoben izrazu $b$, namesto $\\sqrt[2]{a}$ pa imamo $a^{404}$. Lahko pa $a^{404}$ pišemo kot $\\sqrt[2]{a^{808}}$. Vse skupaj pa lahko preoblikujemo v $\\sqrt[2020]{b^{808}}=\\sqrt[5]{b^{2}}$."
] | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Intermediate Algebra > Other"
] | null | MCQ | C | |
03jr | Problem:
There are $5$ monkeys and $5$ ladders and at the top of each ladder there is a banana. A number of ropes connect the ladders, each rope connects two ladders. No two ropes are attached to the same rung of the same ladder. Each monkey starts at the bottom of a different ladder. The monkeys climb up the ladders ... | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
02mn | Problem:
Encontre todos os números de oito algarismos formados somente com os algarismos $1$, $2$, $3$ e $4$, cada um deles duas vezes, tais que:
a. exista um único algarismo entre os dois algarismos $1$;
b. existam dois algarismos entre os dois algarismos $2$;
c. existam três algarismos entre os dois algarismos $3$;... | [] | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 41312432, 23421314 | |
0dis | Find all monic integer polynomials $P(x)$ of second degree such that there exist some integer polynomial $Q(x)$ satisfying $P(x)Q(x)$ have all coefficients in the set $\{-1, 0, 1\}$. | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | All and only the following monic quadratics: x^2 + 2x + 1, x^2 + x + 1, x^2 + 1, x^2 − x + 1, x^2 − 2x + 1, x^2 + x, x^2, x^2 − x, x^2 + x − 1, x^2 − 1, x^2 − x − 1. | |
0der | Let $(a_n)$ be the integer sequence which is defined by $a_1 = 1$ and
$$
a_{n+1} = a_n^2 + n \cdot a_n, \forall n \ge 1.
$$
Let $S$ be the set of all primes $p$ such that there exists an index $i$ such that $p|a_i$. Prove that the set $S$ is an infinite set and it is not equal to the set of all primes. | [
"First, we shall show that $3 \\notin S$ by proving that\n$$\na_{3k-2} \\equiv 2 \\pmod{3},\\ a_{3k-1} \\equiv 2 \\pmod{3}\\ \\text{ and }\\ a_{3k} \\equiv 2 \\pmod{3}.\n$$\nSince $a_1 = 1$, $a_2 = 2$, $a_3 = 2^2 + 2 \\cdot 2 = 8 \\equiv 2 \\pmod{3}$ so the claim is true for $k = 1$.\nSuppose that the claim holds f... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0672 | Determine all the pairs of real numbers $(x, y)$, which are solutions of the system:
$$
\begin{cases}
(x^2 + y^2)^2 - xy(x+y)^2 = 19 \\
|x-y| = 1
\end{cases}
$$ | [
"Performing some manipulation into the first equation we find:\n$$\n\\left\\{ \\begin{array}{l} (x^2 + y^2)^2 - xy(x+y)^2 = 19 \\\\ |x-y| = 1 \\end{array} \\right\\} \\Leftrightarrow \\left\\{ \\begin{array}{l} x^4 + 2x^2y^2 + y^4 - x^3y - 2x^2y^2 - xy^3 = 19 \\\\ |x-y| = 1 \\end{array} \\right\\} \\Leftrightarrow ... | Greece | SELECTION EXAMINATION | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | [(3,2), (-2,-3), (-3,-2), (2,3)] | |
09nw | Determine the number of ways to color cells of a $4 \times 5$ grid such that every $2 \times 2$ subgrid contains exactly one colored cell. | [] | Mongolia | MMO2025 Round 2 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 6 | |
0ehk | Problem:
Taja in Lili igrata igro, pri kateri je na mizi postavljenih 10 kroglic, oštevilčenih z naravnimi števili od $1$ do $10$. V prvi potezi igre Taja izbere naravno število $n$, nato pa dekleti izmenjaje z mize jemljeta vsaka po eno kroglico, dokler kroglic ne zmanjka. Prvo kroglico z mize vzame Lili, zadnjo pa T... | [
"Solution:\n\nPokazali bomo, da ima zmagovito strategijo Lili. Vsota števil vseh kroglic na mizi je enaka $\\frac{10 \\cdot 11}{2} = 55$. Torej bo na koncu vsota števil ene od deklet večja kot $\\frac{55}{2} = 27 \\frac{1}{2}$, vsota števil druge pa manjša od $27 \\frac{1}{2}$, hkrati pa bosta obe vsoti od števila ... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije Državno tekmovanje | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Lili | |
002p | Determinar si existen enteros positivos $x$, $y$, $z$ tales que el producto $(x + y)(y + z)(z + x)$ sea igual a:
a) $6767$
b) $7676$
c) $6776$
En cada caso, si la respuesta es afirmativa, hallar todas las ternas ordenadas $(x, y, z)$ que satisfacen la condición. | [] | Argentina | XIV Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Español | proof and answer | a) No solutions.
b) No solutions.
c) All solutions are the permutations of (15, 7, 7). | |
0gd4 | 令 $\mathbb{R}$ 表示所有實數所成的集合。試求所有的函數 $f : (0, \infty) \to \mathbb{R}$ 滿足
$$ \left(x + \frac{1}{x}\right) f(y) = f(xy) + f\left(\frac{y}{x}\right), \text{ 對所有 } x, y > 0 \text{ 均成立。} $$ | [
"答案是 $f(x) = C_1x + \\frac{C_2}{x}$ with arbitrary constants $C_1$ and $C_2$.\n\n**Solution 1.** Fix a real number $a > 1$, and take a new variable $t$. For the values $f(t), f(t^2), f(at)$ and $f(a^2t^2)$, the relation of the assumption provides a system of linear equations:\n$$\nx = y = t : \\left(t + \\frac{1}{t... | Taiwan | 二〇一九數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | f(x) = C1*x + C2/x | |
0cln | Determine all natural numbers $n$ such that $2^n - n^2 + 1$ is a perfect square. | [
"We check by direct computation that $n = 2, 3, 4$ are solutions. We will prove that these are the only ones. We may assume $n \\ge 5$ after checking all smaller cases. We distinguish two cases.\n\n(I) If $n$ is even, write $n = 2t$ (then $t \\ge 3$). One can prove that\n$$\n(2^t - 1)^2 < 2^n - n^2 + 1 < (2^t)^2 = ... | Romania | 75th NMO Selection Tests | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Eq... | English | proof and answer | n = 2, 3, 4 | |
00oi | For each prime number $p$, determine the number of residue classes modulo $p$ which can be represented as $a^2 + b^2$ modulo $p$, where $a$ and $b$ are arbitrary integers. | [
"All $p$ residue classes.\n\nWith $a^2 + 0^2$ we first obtain all quadratic residue classes.\nSince not all residue classes are quadratic residues, there is a quadratic residue class $a^2$ that is followed by a quadratic non-residue class, so that $n = a^2 + 1$ is not a quadratic residue and therefore of course $n ... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | p | |
0db6 | Find the largest prime number that divides both $2^{63}-1$ and $2^{91}-1$. | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 127 | |
09in | 21 points divide a circle into 21 arcs of equal length. How many sets of 7 points, where no two points have 3 unit or 7 unit arc distance, are there? | [
"Answer: 126.\nLet $1, 2, 3, \\ldots, 21$ be points on a circle in clockwise order and let us use the following notation.\n$$\n\\begin{aligned}\na_1 &= 1, & a_2 &= 8, & a_3 &= 15 \\\\\nb_1 &= 4, & b_2 &= 11, & b_3 &= 18 \\\\\nc_1 &= 7, & c_2 &= 14, & c_3 &= 21 \\\\\nd_1 &= 10, & d_2 &= 17, & d_3 &= 3 \\\\\ne_1 &= 1... | Mongolia | Mongolian Mathematical Olympiad Round 3 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 126 | |
0hzz | Problem:
Let $r$ be the radius of the inscribed circle of triangle $ABC$. Take a point $D$ on side $BC$, and let $r_{1}$ and $r_{2}$ be the inradii of triangles $ABD$ and $ACD$. Prove that $r$, $r_{1}$, and $r_{2}$ can always be the side lengths of a triangle. | [
"Solution:\nWe must show that $r$, $r_{1}$, and $r_{2}$ satisfy the triangle inequality, i.e. that the sum of any two of them exceeds the third. Clearly $r$ is the largest of the three, so we need only verify that $r_{1} + r_{2} > r$.\n\nLet $K$ and $s$ be the area and semiperimeter of triangle $ABC$. Similarly def... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
0klf | Problem:
Find the number of 10-digit numbers $\overline{a_{1} a_{2} \cdots a_{10}}$ which are multiples of 11 such that the digits are non-increasing from left to right, i.e. $a_{i} \geq a_{i+1}$ for each $1 \leq i \leq 9$. | [
"Solution:\nIt is well known that $\\overline{a_{1} a_{2} \\cdots a_{10}}$ is divisible by 11 if and only if $S=a_{1}-a_{2}+a_{3}-\\cdots-a_{10}$ is. By the non-increasing condition, we deduce that\n$$\nS=\\left(a_{1}-a_{2}\\right)+\\left(a_{3}-a_{4}\\right)+\\cdots+\\left(a_{9}-a_{10}\\right) \\geq 0\n$$\nAlso,\n$... | United States | HMMT November 2021 Team Round | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2001 | |
0kw7 | Problem:
The number $2027$ is prime. For $i = 1, 2, \ldots, 2026$, let $p_{i}$ be the smallest prime number such that $p_{i} \equiv i \pmod{2027}$. Estimate $\max \left(p_{1}, \ldots, p_{2026}\right)$.
Submit a positive integer $E$. If the correct answer is $A$, you will receive $\left\lfloor 25 \min \left((E / A)^{8... | [
"Solution:\n\nIn this solution, all logs are in base $e$. Let $p_{1}, p_{2}, \\ldots$ be the primes in sorted order. Let $q_{i} = p_{i} \\bmod 2027$. Since the residues of primes modulo $2027$ should be uniformly distributed, we can make the probabilistic approximation that the $q_{i}$ are random variables uniforml... | United States | HMMT February 2023 | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Expected values",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 117448 | |
074j | Let $ABCDEF$ be a convex hexagon in which the diagonals $AD$, $BE$, $CF$ are concurrent at $O$. Suppose the area of triangle $OAF$ is the geometric mean of those of $OAB$ and $OEF$; and the area of triangle $OBC$ is the geometric mean of those of $OAB$ and $OCD$. Prove that the area of triangle $OED$ is the geometric m... | [
"Let $OA = a$, $OB = b$, $OC = c$, $OD = d$, $OE = e$, $OF = f$, $[OAB] = x$, $[OCD] = y$, $[OEF] = z$, $[ODE] = u$, $[OFA] = v$ and $[OBC] = w$. We are given that $v^2 = zx$, $w^2 = xy$ and we have to prove that $u^2 = yz$. Since $\\angle AOB = \\angle DOE$, we have\n\n$$\n\\frac{u}{x} = \\frac{\\frac{1}{2} de \\s... | India | Indija mo 2011 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof only | null |
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