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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0734 | Problem:
Let $n$ be a natural number such that $n = a^{2} + b^{2} + c^{2}$, for some natural numbers $a, b, c$. Prove that
$$
9 n = (p_{1} a + q_{1} b + r_{1} c)^{2} + (p_{2} a + q_{2} b + r_{2} c)^{2} + (p_{3} a + q_{3} b + r_{3} c)^{2}
$$
where $p_{j}$'s, $q_{j}$'s, $r_{j}$'s are all nonzero integers. Further, if $3$... | [
"Solution:\nIt can be easily seen that\n$$\n9 n = (2 b + 2 c - a)^{2} + (2 c + 2 a - b)^{2} + (2 a + 2 b - c)^{2}\n$$\nThus we can take $p_{1} = p_{2} = p_{3} = 2$, $q_{1} = q_{2} = q_{3} = 2$ and $r_{1} = r_{2} = r_{3} = -1$. Suppose $3$ does not divide $\\gcd(a, b, c)$. Then $3$ does divide at least one of $a, b,... | India | INMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
09qj | Problem:
In de scherphoekige driehoek $A B C$ is $\angle C$ groter dan $\angle A$. Zij $E$ zodat $A E$ een middellijn is van de omgeschreven cirkel $\Gamma$ van $\triangle A B C$. Zij $K$ het snijpunt van $A C$ en de raaklijn in $B$ aan $\Gamma$. Zij $L$ het voetpunt van de loodlijn vanuit $K$ op $A E$ en zij $D$ het ... | [
"Solution:\n\nOplossing I. Door de voorwaarden in de opgave ligt de configuratie vast. Omdat $\\angle L A D=\\angle E A B=\\angle E C B$ vanwege de omtrekshoekstelling op koorde $E B$ van $\\Gamma$, geldt\n$$\n\\begin{gathered}\n\\angle B D K=\\angle A D L=180^\\circ-\\angle D L A-\\angle L A D=90^\\circ-\\angle L ... | Netherlands | Dutch TST | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a38 | Problem:
Voor een positief geheel getal $n$ definiëren we $\alpha(n)$ als het gemiddelde van alle positieve delers van $n$, en $\beta(n)$ als het gemiddelde van alle positieve gehele getallen $k \leq n$ zodat $\operatorname{ggd}(k, n)=1$.
Vind alle positieve gehele getallen $n$ waarvoor geldt dat $\alpha(n)=\beta(n)$. | [
"Solution:\nAntwoord: $n=1$ en $n=6$.\nWe merken eerst op dat $n=1$ voldoet.\nWe bewijzen nu dat $\\beta(n)=\\frac{n}{2}$ voor $n \\geq 2$. Er geldt namelijk $\\operatorname{ggd}(k, n)=\\operatorname{ggd}(k-n, n)=\\operatorname{ggd}(n-k, n)$, dus $\\operatorname{ggd}(k, n)=1$ dan en slechts dan als $\\operatorname{... | Netherlands | IMO-selectietoets | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | n = 1 or n = 6 | |
0fc9 | Problem:
Sean $x, y, z$ reales positivos tales que $x+y+z=3$. Halla el valor máximo alcanzado por
$$
\sqrt{x}+\sqrt{2 y+2}+\sqrt{3 z+6}
$$
¿Para qué valores de $x, y, z$ se alcanza dicho máximo? | [
"Solution:\n\nConsideremos los vectores $(\\sqrt{x}, \\sqrt{y+1}, \\sqrt{z+2})$ y $(\\sqrt{1}, \\sqrt{2}, \\sqrt{3})$, cuyas coordenadas son todas reales y positivas, cuyos módulos respectivos son $\\sqrt{x+y+z+3}=\\sqrt{6}$ y $\\sqrt{1+2+3}=\\sqrt{6}$, y cuyo producto escalar es la expresión cuyo máximo se pide ha... | Spain | null | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | Maximum value: 6; attained at x = y = z = 1. | |
02nx | Arnold and Bernold play the following game in a $m \times n$ board: Arnold chooses one of its cells and places a knight on it. Then Bernold and Arnold move the knight alternately, with the condition that the knight visits a cell at most once. The player who is unable to move the knight loses. Determine, in terms of $m$... | [
"Suppose, without loss of generality, $m \\le n$. If $m = 2$, Arnold has winning strategy if and only if $n$ is not a multiple of $4$; for $m \\ge 3$, Arnold has winning strategy if and only if $m$ and $n$ are both odd.\n\nSuppose $m = 2$. If $4$ does not divide $n$, Arnold can win placing the knight on the first c... | Brazil | Brazilian Math Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | Assuming the smaller side is at most the larger: if the smaller side equals two, then the first player wins if and only if the larger side is not divisible by four; if the smaller side is at least three, then the first player wins if and only if both side lengths are odd. Otherwise the second player wins. | |
09bs | $ABC$ гурвалжинд $AH$ өндөр, $AL$ биссектрис татав. $AL$ шулуун руу $BP$ ба $CQ$ өндөр буулгав. $M$ нь $BC$-ийн дундаж цэг бол $HPMQ$ цэгүүд нэг тойрог дээр оршино гэж батал. | [
"\n\n$(CQ) \\cap (AB) = B_1$ гэе. $AL$ биссектрис гэдгээс $CQ = QB_1$ буюу $MQ$ нь $\\triangle B_1CB$-ын дундаж шугам болж $MQ \\parallel AB$ болно. Эндээс $\\angle LQM = \\angle LAB$.\n\n$\\angle APB = \\angle AHB = 90^\\circ \\Rightarrow H, P, A, B$ цэгүүд нэг тойрог дээр оршино гэдгээс $... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | Mongolian | proof only | null | |
0leo | Let $D$ be the intersection of the two tangent lines of $(O)$ at $B$ and $C$. The circle passing through $A$ and tangent to $BC$ at $B$ intersects the median passing $A$ of the triangle $ABC$ at $G$. Lines $BG$, $CG$ intersect $CD$, $BD$ at $E, F$ respectively.
a) The line passing through the midpoints of $BE$ and $CF... | [
"a) Let $I$, $X$ and $Y$ be the midpoints of $BC$, $BE$ and $CF$, respectively. Let $IX$, $IY$ intersect $AC$, $AB$ at $S$, $T$ respectively. Since $IB$ is tangent to $(ABG)$, we have\n$$\nIB^2 = IG \\cdot IA = IC^2,\n$$\nso $IC$ is tangent to $(AGC)$ as well. Since $I$, $Y$ are midpoints of $BC$, $CF$, we have $IY... | Vietnam | VMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometr... | English | proof only | null | |
0d5h | Let $\left(a_{n}\right)_{n \geq 0}$ be a sequence of positive integers such that $a_{n}^{2}$ divides $a_{n-1} a_{n+1}$, for all $n \geq 1$. Prove that if there exists an integer $k \geq 2$ such that $a_{k}$ and $a_{1}$ are relatively prime, then $a_{1}$ divides $a_{0}$. | [
"Assume, for the sake of contradiction, that there exists an integer $k \\geq 2$ such that $a_{k}$ and $a_{1}$ are relatively prime and that $a_{1}$ does not divide $a_{0}$. We deduce that there exists a prime number $p$ such that $0 \\leq v_{p}\\left(a_{0}\\right)<v_{p}\\left(a_{1}\\right)$, where $v_{p}(a)$ is th... | Saudi Arabia | SAMC 2015 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English, Arabic | proof only | null | |
01ba | We are to paint $n$ seats in a row, each red or green. We call painting odd, if each monochromatic sequence is of odd length. By monochromatic sequence we mean a sequence of seats in one color, which is bounded by seats of the other color or a wall. Count how many ways of odd painting are there. | [
"**Answer:** $2f_n$, where $f_n$ is the $n$-th element of the Fibonacci sequence.\n\nLet $g_k$, $r_k$ be the numbers of possible odd paintings of $k$ seats such that the first seat is painted green or red respectively. Obviously $g_k = r_k$ for any $k$. Note that $g_k = r_{k-1} + g_{k-2} = g_{k-1} + g_{k-2}$ as $r_... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 2 f_n | |
04z1 | Find all positive integers $n$ such that $1 + 2^2 + 3^3 + 4^n$ is a perfect square. | [
"Let $1 + 2^2 + 3^3 + 4^n = x^2$. This implies $32 = x^2 - 4^n$ or, equivalently, $2^5 = (x - 2^n)(x + 2^n)$. As the l.h.s. is a power of 2, the factors in the r.h.s. are of the form $x - 2^n = 2^a$ and $x + 2^n = 2^{5-a}$ where $a$ is 0, 1 or 2. Subtracting the first of the two equalities from the second gives $2^... | Estonia | Estonija 2010 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1 | |
09i0 | A word is *palindromic* if it reads the same from the left and the right. What is the maximum length of a word written in two letters that cannot be partitioned into four or fewer palindromic subwords?
Here, for example, the word *abbaa* can be partitioned into two palindromic subwords as $(abba)(a)$ or three palindrom... | [] | Mongolia | Round 2 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 9 | |
0epo | Steve was sorting $1\,000$ eggs into sizes. He got paid $20$ cents for each egg that he sorted. For each egg that he broke while sorting he did not get paid and had to pay his employer $R1$. Steve was paid $R176$. How many eggs did Steve break? | [
"Suppose Steve breaks $n$ eggs, so $1\\,000 - n$ eggs are unbroken. He receives $R0.20 \\times (1\\,000 - n)$, but has to repay $R1 \\times n$. Thus $0.2(1\\,000 - n) - n = 176$, giving $200 - 1.2n = 176$, so $1.2n = 200 - 176 = 24$ and $n = 20$."
] | South Africa | South African Mathematics Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | English | proof and answer | 20 | |
00no | Let $ABC$ be an isosceles triangle with $AC = BC$ and circumcircle $k$. The point $D$ lies on the shorter arc of $k$ over the chord $BC$ and is different from $B$ and $C$. Let $E$ denote the intersection of $CD$ and $AB$.
Prove that the line through $B$ and $C$ is a tangent of the circumcircle of the triangle $BDE$. | [
"We denote the center of the circumcircle of the triangle $BDE$ by $M$ and $\\angle BAC = \\angle CBA$ by $\\alpha$. Since the quadrilateral $ABDC$ is cyclic, we obtain $\\angle BDE = \\alpha$. By the inscribed angle theorem, $\\angle BME = 2\\alpha$ and thus $\\angle EBM = \\angle MEB = 90^\\circ - \\alpha$. There... | Austria | Austrian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
07bv | Polynomial $p(x) \in \mathbb{R}[x]$ is of odd degree $m$ greater than one. Also $f : \mathbb{R} \to \mathbb{Z}$ is a function such that for each real number $x$, we have $p(f(x)) = f(p(x))$.
a) Prove that the range of function $f$ is a finite set.
b) If $f$ is a non-constant function, prove that the equation $p(x) = ... | [
"First we claim that $p : \\mathcal{R}(f) \\to \\mathcal{R}(f)$ is a surjective function, where by $\\mathcal{R}(f)$ we mean the range of the function $f$.\n$$\ny \\in \\mathcal{R}(f) \\Rightarrow \\exists x_0 \\in \\mathbb{R}; f(x_0) = y \\Rightarrow p(y) = p(f(x_0)) = f(p(x_0)) \\Rightarrow p(y) \\in \\mathcal{R}... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange"
] | null | proof only | null | |
089y | Problem:
In quali basi $b>6$ la scrittura 5654 rappresenta una potenza di un numero primo? | [
"Solution:\n\nLa scrittura 5654 in base $b$ rappresenta il numero $N=5 b^{3}+6 b^{2}+5 b+4=(b+1)\\left(5 b^{2}+b+4\\right)$. Se $b$ è dispari, $b+1$ è pari; viceversa, se $b$ è pari, $5 b^{2}+b+4$ è pari. In ogni caso, $N$ è pari, e quindi è una potenza di 2.\n\nNe segue che $b+1$ e $5 b^{2}+b+4$ sono entrambe pote... | Italy | Cesenatico | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 7 | |
09wf | On a $4 \times 4$ board, there are $16$ grass hoppers, each on its own square. At a certain time, each grass hopper jumps to an adjacent square: to the square above, below, left, or right of its current square, but not diagonally and not leaving the board.
What is the maximum number of squares that can be empty after t... | [
"C) $10$"
] | Netherlands | First Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | MCQ | C | |
0hpi | Problem:
Let $f(n)$ be the number of digits of a positive integer $n$ (in base 10). Prove that
$$
f\left(2^{n}\right)+f\left(5^{n}\right)=n+1
$$ | [
"Solution:\nLet $f\\left(2^{n}\\right)=x$. Then since the smallest number with $x$ digits is $10^{x-1}$ and the largest is $10^{x}-1$, we have\n$$\n10^{x-1} \\leq 2^{n}<10^{x}\n$$\nHowever, a power of $2$ (other than $1$) cannot also be a power of $10$, so the inequality is strict:\n$$\n10^{x-1}<2^{n}<10^{x}.\n$$\n... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0cuy | Some pairs of cities in a country are connected with one-directional direct flights (between any two cities, there is at most one flight).
We say that a city *A* is *accessible* for a city *B* if one may reach the city *A* starting at *B* (perhaps with zero flights or more than one flight in a chain). Assume that for a... | [
"Choose the city with a maximal number of cities accessible from it.\n\nFirst solution. Number all the cities in the country as $A_1, A_2, \\dots, A_n$. By the condition, there exists a city $B_2$ for which both $A_1$ and $A_2$ are accessible. Next, the cities $A_3$ and $B_2$ are accessible for some city $B_3$. Sin... | Russia | XLIII Russian mathematical olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Other"
] | English; Russian | proof only | null | |
0hh4 | You are given $n \ge 2$ distinct positive integers. For each pair $a < b$ of these numbers, consider the difference $b - a$. For each of these differences, Vlada writes down the maximum power of two by which this difference is divisible. What is the largest possible number of distinct numbers that Vlada could write?
(O... | [
"We will prove that there are at most $n-1$ different degrees with induction by $n$. The base case for $n=2$ is obvious, let's prove the transition. Let the statement be proved for $k \\le n-1$, let us prove it for $n$ numbers. Suppose that all numbers are divisible by the same power of $2$: $2^k$. If $k > 0$, then... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n - 1 | |
02v5 | Problem:
Para determinar a quantidade de divisores positivos de um número, basta fatorá-lo como potências de primos distintos e multiplicar os sucessores dos expoentes. Por exemplo, $2016=2^{5} \cdot 3^{2} \cdot 5^{1}$ possui $(5+1)(2+1)(1+1)=36$ divisores positivos. Considere o número $n=2^{7} \cdot 3^{4}$.
a) Deter... | [
"Solution:\n\na) A partir da fatoração de $n$, podemos determinar a fatoração de $n^{2}$ :\n$$\n\\begin{aligned}\nn^{2} & =\\left(2^{7} \\cdot 3^{4}\\right)^{2} \\\\\n& =2^{14} \\cdot 3^{8}\n\\end{aligned}\n$$\nEntão o número $n^{2}$ possui $(14+1)(8+1)=15 \\cdot 9=135$ divisores positivos.\n\nb) Note que todos os ... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) 135; b) 67; c) 28 | |
0jfr | Problem:
How many positive integers $k$ are there such that
$$
\frac{k}{2013}(a+b)=\operatorname{lcm}(a, b)
$$
has a solution in positive integers $(a, b)$? | [
"Solution:\nFirst, we can let $h=\\operatorname{gcd}(a, b)$ so that $(a, b)=(h A, h B)$ where $\\operatorname{gcd}(A, B)=1$. Making these substitutions yields\n$$\n\\frac{k}{2013}(h A+h B)=h A B,\n$$\nso\n$$\nk=\\frac{2013 A B}{A+B}.\n$$\nBecause $A$ and $B$ are relatively prime, $A+B$ shares no common factors with... | United States | HMMT 2013 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | final answer only | 1006 | |
04pa | In a bathroom of size $6\ \mathrm{m} \times 6\ \mathrm{m}$, one corner is occupied by a rectangular bathtub of size $2\ \mathrm{m} \times 1.5\ \mathrm{m}$. What is the radius of the largest possible circular carpet that can be spread on the bathroom floor? (Petar Bakić) | [] | Croatia | Croatian Mathematical Society Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | (13 - 6√2)/2 | |
0e3g | Problem:
Učiteljica je Mateju izročila štiri liste papirja, na vsakem je bila zapisana neničelna števka. Matej je liste postavil v vrsto in tako oblikoval štirimestno število. Ko je dva lista med sabo zamenjal, ne da bi ju pri tem obrnil ali zavrtel, je oblikoval še eno štirimestno število. Ali je lahko oblikoval štev... | [
"Solution:\n\nOdgovor je da. Če je katera izmed števk soda, lahko Matej oblikuje dve sodi števili, saj sodo števko postavi na mesto enic, med sabo pa zamenja dva izmed ostalih listov. Če je med štirimi števkami število $5$, Matej postopa podobno. Petico postavi na mesto enic, med sabo pa nato zamenja dva izmed osta... | Slovenia | 54. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | Yes | |
0cck | A natural number $n$ is interesting if it can be written as $n = \lfloor \frac{1}{a} \rfloor + \lfloor \frac{1}{b} \rfloor + \lfloor \frac{1}{c} \rfloor$, where $a, b$ and $c$ are positive real numbers, such that $a + b + c = 1$.
Determine all the interesting numbers. ($[x]$ denotes the floor of the real number $x$.) | [
"Without loss of generality, we may choose $a \\le b \\le c$.\n\nThe positive real numbers $a, b, c$ are such that $a + b + c = 1$, thus $a, b, c \\in (0, 1)$. From $\\frac{1}{a}, \\frac{1}{b}, \\frac{1}{c} \\in (1, \\infty)$ follows that $[\\frac{1}{a}], [\\frac{1}{b}], [\\frac{1}{c}] \\in \\mathbb{N}^*$, therefor... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | All natural numbers n ≥ 7 | |
0471 | Given a positive integer $n$, a large equilateral triangle of side length $3n$ is divided into $9n^2$ small equilateral triangles of side length 1. Each small triangle is colored in one of three colors: red, yellow, or blue, such that there are $3n^2$ triangles of each color. A trapezoid consisting of 3 small triangles... | [
"The maximum number of multicolored trapezoids is $18n^2 - 9n$.\n\nUse the numbers 1, 2, and 3 to represent the three colors. First, consider a small equilateral triangle of side length 2 formed by 4 small equilateral triangles. Let the middle small equilateral triangle be labeled $a$, and the other three small equ... | China | The 65th IMO China National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 18n^2 - 9n | |
0iqe | Problem:
What is the units digit of $7^{2009}$? | [
"Solution:\nAnswer: 7\nNote that the units digits of $7^{1}, 7^{2}, 7^{3}, 7^{4}, 7^{5}, 7^{6}, \\ldots$ follows the pattern $7, 9, 3, 1, 7, 9, 3, 1, \\ldots$. The 2009th term in this sequence should be 7.\n\n\nAlternate method:\nNote that the units digit of $7^{4}$ is equal to 1, so the units digit of $(7^{4})^{50... | United States | 1st Annual Harvard-MIT November Tournament | [
"Number Theory > Modular Arithmetic",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | final answer only | 7 | |
0dif | Let $n \ge 3$ be an integer and let $x_1, x_2, \dots, x_n$ be real numbers in the interval $[0, 1]$. Let $s = x_1 + x_2 + \dots + x_n$, with $s \ge 3$. Prove that there exist integers $i$ and $j$ with $1 \le i < j \le n$ such that
$$
2^{j-i} x_i x_j > 2^{s-3}.
$$ | [
"Let $1 \\le a < b \\le n$ be such that $2^{b-a} x_a x_b$ is maximal. This choice of $a$ and $b$ implies that\n$$\nx_{a+t} \\le 2^t x_a, \\forall t = 1-a, 2-a, \\dots, b-a-1,\n$$\nand similarly\n$$\nx_{b-t} \\le 2^t x_b, \\forall t = b-n, b-n+1, \\dots, b-a+1.\n$$\nNow, suppose that $x_a \\in (\\frac{1}{2^{u+1}}, \... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
034q | Problem:
Solve the inequality
$$
\log_{a}\left(x^{2}-x-2\right)>\log_{a}\left(3+2x-x^{2}\right)
$$
if it is known that $x=a+1$ is a solution. | [
"Solution:\nThe inequality is defined for $x^{2}-x-2>0$ and $3+2x-x^{2}>0$, whence $x \\in (2,3)$.\n\nSince $x=a+1$ is a solution, we have $a \\in (1,2)$.\n\nThen the inequality is equivalent to\n$$\nx^{2}-x-2 > 3+2x-x^{2} \\Longleftrightarrow (x+1)(2x-5)>0\n$$\nand therefore $x \\in \\left(\\frac{5}{2}, 3\\right)$... | Bulgaria | Bulgarian Mathematical Competitions | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (5/2, 3) | |
0g09 | Problem:
Seien $a$, $b$, $c$ reelle Zahlen, sodass gilt:
$$
\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = 1
$$
Bestimme alle Werte, welche folgender Ausdruck annehmen kann:
$$
\frac{a^{2}}{b+c} + \frac{b^{2}}{c+a} + \frac{c^{2}}{a+b}
$$ | [
"Solution:\n$$\n\\begin{aligned}\n& a + b + c = a + b + c \\\\\n& \\overbrace{\\left(\\frac{a}{b+c} + \\frac{b}{a+c} + \\frac{c}{a+b}\\right)}^{1} \\cdot (a + b + c) = a + b + c \\\\\n& \\frac{a^{2}}{b+c} + \\frac{a(b+c)}{b+c} + \\frac{b^{2}}{a+c} + \\frac{b(a+c)}{a+c} + \\frac{c^{2}}{a+b} + \\frac{c(a+b)}{a+b} = a... | Switzerland | SMO - Finalrunde | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 0 | |
06dw | In a conference, there are $2002$ representatives from $100$ countries. The number of representatives from each country is at least $1$ and at most $45$. They are seated in rows with each row consisting of $45$ seats. It is required that the representatives from the same country must be seated in the same row. What is ... | [
"At least $86$ rows are needed.\nSuppose there are $86$ (type A) countries with $23$ representatives each, $10$ countries with $2$ representatives each, and $4$ countries with $1$ representative each. Then there are $86 + 10 + 4 = 100$ countries and $86 \\times 23 + 10 \\times 2 + 4 \\times 1 = 2002$ representative... | Hong Kong | IMO HK TST | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | 86 | |
04fv | Andrija and Boris have 2014 cards labelled with numbers from 1 to 2014. Andrija has all the cards with even, and Boris all the cards with odd numbers. Andrija arranged his cards in a circle clockwise, from 2 to 2014 respectively, and put them upside down, so that the numbers on the cards can not be seen. Boris knows th... | [
"The largest number of points for which Boris can be certain to gain is $503$.\n\nWe claim that if Boris arranges his cards clockwise, but in the reverse order, from $2013$ to $1$, then regardless of Andrija's arrangement of cards, Boris will gain $503$ points.\n\n\n\nFor such Boris' arrang... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 503 | |
0gs5 | Let $ABC$ be a triangle with $|AB| > |AC|$. Let $D$ be the foot of the altitude drawn from $A$ to $BC$, $K$ be the intersection of $AD$ and the internal bisector of the angle at $B$, $M$ be the foot of the perpendicular drawn from $B$ to $CK$ and $N$ be the intersection of $BM$ and $AK$. Let $T$ be the intersection of ... | [
"Let $R$ and $S$ be the reflections of $C$ in the lines $BK$ and $BN$, respectively. It is easy to see the following: $C, N, R$ are collinear, $A, B, R$ are collinear and $C, M, S$ are collinear.\n\n\n\nBy symmetry, $\\angle BRK = \\angle BSK = \\angle KCB$. But, $\\angle KCB = \\angle BNK$... | Turkey | Team Selection Test for IMO 2019 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03ak | The excircles of $\triangle ABC$ touch the sides $AB$, $BC$ and $CA$ at points $M$, $N$ and $P$, respectively. Let $I$ and $O$ be the incenter and the circumcenter of $\triangle ABC$. Prove that if $AMNP$ is a cyclic quadrilateral, then:
a) the points $M$, $P$ and $I$ are collinear;
b) the points $I$, $O$ and $N$ are... | [
"By Carnot's theorem, the perpendiculars from the points $M$, $N$ and $P$ to the lines $AB$, $BC$ and $CA$, respectively, have a common point $X$. Then the quadrilateral $AMXP$ is cyclic. Now it is easy to see that $X = N$ and hence $AN$ is a diameter of the circumcircle of $AMNP$.\n\na) Denote by $I_B$ and $I_C$ t... | Bulgaria | Team selection test for 50. IMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Pappus theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler l... | English | proof only | null | |
023x | Problem:
Escolhi quatro frações entre $1/2$, $1/4$, $1/6$, $1/10$ e $1/12$ cuja soma é $1$. Quais foram as frações que eu não escolhi? | [
""
] | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 1/10 | |
0hv2 | Problem:
How many multiples of $7$ between $10^{6}$ and $10^{9}$ are perfect squares? | [
"Solution:\n\n$\\left[\\sqrt{\\frac{10^{9}}{7^{2}}}\\right] - \\left[\\sqrt{\\frac{10^{6}}{7^{2}}}\\right] = 4517 - 142 = 4375$."
] | United States | null | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | final answer only | 4375 | |
0aow | Problem:
Consider the function $f$ defined by
$$
f(x) = 1 + \frac{2}{x}
$$
Find the roots of the equation
$$
(\underbrace{f \circ f \circ \cdots \circ f}_{10 \text{ times}})(x) = x
$$
where "o" denotes composition of functions. | [
"Solution:\n$-1$ and $2$\n\nLet $f^{(n)}(x) = (\\underbrace{f \\circ f \\circ \\cdots \\circ f}_{n \\text{ times}})(x)$. For allowed values of $x$, note that $f^{(n)}(x)$ is of the form\n$$\nf^{(n)}(x) = \\frac{a_n x + b_n}{c_n x + d_n}\n$$\nwhere $a_n, b_n, c_n, d_n \\in \\mathbb{Z}$ for all integers $n \\geq 1$. ... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | [-1, 2] | |
0jzv | Problem:
Reimu has a wooden cube. In each step, she creates a new polyhedron from the previous one by cutting off a pyramid from each vertex of the polyhedron along a plane through the trisection point on each adjacent edge that is closer to the vertex. For example, the polyhedron after the first step has six octagona... | [
"Solution:\n\nNotice that the number of vertices and edges triple with each step. We always have 3 edges meeting at one vertex, and slicing off a pyramid doesn't change this (we make new vertices from which one edge from the previous step and two of the pyramid edges emanate). So at each step we replace the sliced-... | United States | HMMT November 2017 | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F"
] | null | final answer only | 974 | |
0dr5 | The quadrilateral $ABCD$ is inscribed in a circle which has diameter $BD$. Points $A'$ and $B'$ are symmetric to $A$ and $B$ with respect to the line $BD$ and $AC$ respectively. If the lines $A'C$, $BD$ intersect at $P$ and $AC$, $B'D$ intersect at $Q$, prove that $PQ$ is perpendicular to $AC$. | [
"Let $AC$ intersect $BD$ at $R$. Then $\\angle BAR = \\angle BAC = \\angle BA'P = \\angle BAP$. That is $AB$ bisects $\\angle PAR$. As $\\angle BAD = 90^\\circ$, we also have $AD$ is the external bisector of $\\angle PAR$. By the angle bisector theorem, we have\n$$\n\\frac{BR}{BP} = \\frac{DR}{DP} = \\frac{AR}{AP} ... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0k87 | Problem:
For positive integers $a$ and $b$ such that $a$ is coprime to $b$, define $\operatorname{ord}_{b}(a)$ as the least positive integer $k$ such that $b \mid a^{k}-1$, and define $\varphi(a)$ to be the number of positive integers less than or equal to $a$ which are coprime to $a$. Find the least positive integer $... | [
"Solution:\nThe maximum order of an element modulo $n$ is the Carmichael function, denoted $\\lambda(n)$. The following properties of the Carmichael function are established:\n- For primes $p>2$ and positive integers $k$, $\\lambda\\left(p^{k}\\right)=(p-1) p^{k-1}$.\n- For a positive integer $k$,\n$$\n\\lambda\\le... | United States | HMMT February 2019 | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | 240 | |
030r | Problem:
Determinați cel mai mare număr natural $n$ pentru care este adevărată afirmația:
Există $n$ numere naturale nenule distincte $x_{1}, x_{2}, \ldots, x_{n}$ cu proprietatea că oricare ar fi numerele $a_{1}, a_{2}, \ldots, a_{n} \in\{-1,0,1\}$, nu toate nule, numărul $n^{3}$ nu divide numărul $a_{1} x_{1}+a_{2} x... | [
"Solution:\nPentru $n=9$ alegem $x_{1}=2^{0}, x_{2}=2^{1}, \\ldots, x_{9}=2^{8}$. Oricare ar fi numerele $a_{1}, a_{2}, \\ldots, a_{n} \\in\\{-1,0,1\\}$, avem:\n$$\n\\left|a_{1} x_{1}+a_{2} x_{2}+\\ldots+a_{n} x_{n}\\right| \\leq 1+2+\\ldots+2^{8}=2^{9}-1<9^{3}\n$$\nDacă $9^{3}$ divide $a_{1} x_{1}+a_{2} x_{2}+\\ld... | Brazil | Al doilea baraj de selecție pentru OBMJ | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 9 | |
0djg | Let $ABCD$ be a cyclic quadrilateral. Take $F \in AB$, $E \in CD$ such that $(FCD)$ is tangent to $AB$ and $(EAB)$ is tangent to $CD$. Denote $G = AE \cap DF$, $H = BE \cap CF$. Prove that $EF$ is the perpendicular bisector of $GH$. | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellan... | English | proof only | null | |
04i5 | Let $I$ be the incentre of the acute triangle $ABC$ and let $|AC| > |BC|$. The angle bisector and the altitude from vertex $C$ close an angle of $10^\circ$. If $\angle AIB = 120^\circ$, determine the angles of the triangle $ABC$. (Ilko Brnetić) | [
"**1.5.** It suffices to show that $\\triangleq ECA = \\triangleq HGA$.\nSince the quadrilateral $ACBE$ is cyclic, we have $\\triangleq ECA = \\triangleq EBA$, so it suffices to show that $ABGH$ is a cyclic quadrilateral.\nFrom triangle $DEH$ we have $\\triangleq DHE = 180^\\circ - \\triangleq EDH - \\triangleq HED... | Croatia | First round – City competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | English | proof and answer | ∠A = 50°, ∠B = 70°, ∠C = 60° | |
04ui | Let $ABCD$ be a trapezoid ($AB \parallel CD$) and denote $P = BC \cap AD$. Let $k_1, k_2$ be circles with diameters $BC, AD$, respectively, and denote $P = BC \cap AD$. Prove that the tangents from $P$ to $k_1$ form the same angle as the tangents from $P$ to $k_2$. (Patrik Bak) | [] | Czech Republic | First Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | proof only | null | |
0gr5 | For each positive integer $n$ let $d(n)$ be the number of prime divisors of $n$. Show that for each positive integer $n$ there are positive integers $k, m$ satisfying $k - m = n$ and $d(k) - d(m) = 1$. | [
"Let $p$ be the smallest prime not dividing $n$. Then all prime divisors of $p-1$ divide $n$ and $d((p-1)n) = d(n)$. Now note that $k = pn$ and $m = (p-1)n$ satisfy the conditions. Indeed, $k - m = n$ and $d(k) - d(m) = d(pn) - d((p-1)n) = (d(n) + 1) - d(n) = 1$."
] | Turkey | 25th Turkish Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0f12 | Problem:
Each side of a convex hexagon is longer than $1$. Is there always a diagonal longer than $2$? If each of the main diagonals of a hexagon is longer than $2$, is there always a side longer than $1$? | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0ee7 | Problem:
Katera izmed naštetih kvadratnih enačb ima rešitvi $1-\sqrt{2}$ in $\frac{1}{\sqrt{2}-1}$?
(A) $x^{2}+2x-1=0$
(B) $x^{2}+\sqrt{2}x-1=0$
(C) $x^{2}+x-1=0$
(D) $x^{2}-2x-1=0$
(E) $x^{2}-x+2=0$ | [
"Solution:\nDrugo ničlo racionaliziramo $\\frac{1}{\\sqrt{2}-1}=\\frac{\\sqrt{2}+1}{2-1}=\\sqrt{2}+1$. Predpisani rešitvi ima torej kvadratna enačba $(x-(1-\\sqrt{2}))(x-(\\sqrt{2}+1))=0$. Ko levo stran zmnožimo, dobimo $x^{2}-2x-1=0$. Pravilen odgovor je (D)."
] | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | D | |
0346 | Problem:
A square $n \times n$ ($n \geq 2$) is divided into $n^{2}$ unit squares colored in black or white such that the squares at the four corners of any rectangle (containing at least four squares) have not the same color. Find the maximum possible value of $n$. | [
"Solution:\n\nWe shall show that the desired value is equal to $4$.\n\nLet $(i, j)$ be the unit square in the $i$-th row and the $j$-th column of a $4 \\times 4$ square. It is easy to check that if the squares $(1,1), (1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,4)$ are white and the others are black, then the cond... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 4 | |
0h8j | Determine if there exist noninteger $x$, $y$ such that for any integer $a$, $b$ that are either both odd or both even, numbers $x + y$ and $a x + b y$ are integers? | [
"Let $x = y = \\frac{1}{2}$, then $x + y = 1$, $a x + b y = \\frac{1}{2}(a + b) \\in \\mathbb{Z}$."
] | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic"
] | English | proof and answer | Yes; for example x = 1/2 and y = 1/2. | |
0j87 | Problem:
To survive the coming Cambridge winter, Chim Tu doesn't wear one T-shirt, but instead wears up to FOUR T-shirts, all in different colors. An outfit consists of three or more T-shirts, put on one on top of the other in some order, such that two outfits are distinct if the sets of T-shirts used are different or... | [
"Solution:\n\nWe note that there are 4 choices for Chim Tu's innermost T-shirt, 3 choices for the next, and 2 choices for the next. At this point, he has exactly 1 T-shirt left, and 2 choices: either he puts that one on as well or he discards it. Thus, he has a total of $4 \\times 3 \\times 2 \\times 2 = 48$ outfit... | United States | Harvard-MIT November Tournament | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | proof and answer | 144 | |
00gb | Let a set $S$ of 2004 points in the plane be given, no three of which are collinear. Let $\mathcal{L}$ denote the set of all lines (extended indefinitely in both directions) determined by pairs of points from the set. Show that it is possible to colour the points of $S$ with at most two colours, such that for any point... | [
"Choose any point $p$ from $S$ and color it, say, blue. Let $n(q, r)$ be the number of lines from $\\mathcal{L}$ that separates $q$ and $r$. Then color any other point $q$ blue if $n(p, q)$ is odd and red if $n(p, q)$ is even.\n\nNow it remains to show that $q$ and $r$ have the same color if and only if $n(q, r)$ i... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
05pk | Problem:
On considère un quadrilatère convexe $ABCD$ dont les diagonales sont perpendiculaires et se coupent en $E$. Soit $P \in [AD]$ un point tel que $P \neq A$ et $PE = EC$. On suppose que le cercle circonscrit du triangle $BCD$ recoupe $[AD]$ en un point $Q$ différent de $A$. On note $R$ le deuxième point d'inters... | [
"Solution:\n\nIntroduisons $F$, le point d'intersection du cercle circonscrit du triangle $ARD$ et du segment $[DE]$. La puissance de $E$ par rapport au cercle passant par $A, R, F, D$ est $ER \\cdot EA = EF \\cdot ED$. Par ailleurs, la puissance de $E$ par rapport au cercle passant par $A, R, P$ est\n$$\nER \\cdot... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Misce... | null | proof only | null | |
08ba | Problem:
Per quante quaterne $(a, b, c, d)$ di numeri interi non negativi le tre espressioni $a^{2}-c^{2}$, $b^{2}-d^{2}$ e $a b+b c+c d+d a$ sono tutte uguali a $1024$?
(A) $0$
(B) $1$
(C) $4$
(D) $9$
(E) $11$ | [
"Solution:\n\nLa risposta è (B). Scriviamo le tre equazioni nella forma $(a+c)(a-c)=1024$, $(b+d)(b-d)=1024$ e $(a+c)(b+d)=1024$. Si osservi che $a+c$ e $a-c$ non possono essere nulli, dal momento che $(a+c)(a-c)=1024$. Confrontando prima e terza equazione otteniamo allora $b+d=\\frac{1024}{a+c}=a-c$, che sostituit... | Italy | Progetto Olimpiadi della Matematica | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | B | |
0k72 | Problem:
Annie has a permutation $\left(a_{1}, a_{2}, \ldots, a_{2019}\right)$ of $S=\{1,2, \ldots, 2019\}$, and Yannick wants to guess her permutation. With each guess Yannick gives Annie an $n$-tuple $(y_{1}, y_{2}, \ldots, y_{2019})$ of integers in $S$, and then Annie gives the number of indices $i \in S$ such that... | [
"Solution:\n\nPart (a) uses the idea that for $x \\neq y$, the guess $(x, y, y, \\ldots, y)$ returns $0$ if the first number is $y$, $2$ if the first number is $x$, and $1$ if the number is something else. If he tests the pairs $(x, y)=(1,2), \\ldots,(2017,2018)$, he will get a $0$ or $2$ (and therefore find out wh... | United States | HMIC | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
06r5 | $n \geq 4$ players participated in a tennis tournament. Any two players have played exactly one game, and there was no tie game. We call a company of four players bad if one player was defeated by the other three players, and each of these three players won a game and lost another game among themselves. Suppose that th... | [
"For any tournament $T$ satisfying the problem condition, denote by $S(T)$ sum under consideration, namely\n$$\nS(T)=\\sum_{i=1}^{n}\\left(w_{i}-\\ell_{i}\\right)^{3}\n$$\nFirst, we show that the statement holds if a tournament $T$ has only 4 players. Actually, let $A=\\left(a_{1}, a_{2}, a_{3}, a_{4}\\right)$ be t... | IMO | 51st IMO Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof only | null | |
0im8 | Problem:
A candy company makes 5 colors of jellybeans, which come in equal proportions. If I grab a random sample of 5 jellybeans, what is the probability that I get exactly 2 distinct colors? | [
"Solution:\nThere are $\\binom{5}{2} = 10$ possible pairs of colors. Each pair of colors contributes $2^{5} - 2 = 30$ sequences of beans that use both colors. Thus, the answer is $10 \\cdot 30 / 5^{5} = 12 / 125$.\n\nAnswer: $\\dfrac{12}{125}$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 12/125 | |
0krd | Problem:
Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ from the integers to the integers satisfying
$$
f(m+f(n))-f(m)=n
$$
for all integers $m, n \in \mathbb{Z}$. | [
"Solution:\nAdding $f(m)$ to both sides, we get $f(m+f(n))=n+f(m)$. Swapping $m$ and $n$ gives $m+f(n)=f(n+f(m))=f(f(m+f(n)))$. By fixing $n$ and varying $m$, we can get $m+f(n)$ to be any integer $x$. Thus, $x=f(f(x))$.\nPlugging in $1$ for $m$ and $f(x)$ for $n$ to the original equation then gives $f(x)+f(1)=f(1+... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(n) = n or f(n) = -n | |
04gm | Prove that
$$
\log_2 \left(1 + \frac{1}{1}\right) + \log_2 \left(1 + \frac{1}{2}\right) + \dots + \log_2 \left(1 + \frac{1}{k}\right) + \dots + \log_2 \left(1 + \frac{1}{2014}\right) < 11.
$$ | [
"Let us denote the sum by $S$:\n$$\nS = \\log_2 \\left(1 + \\frac{1}{1}\\right) + \\log_2 \\left(1 + \\frac{1}{2}\\right) + \\dots + \\log_2 \\left(1 + \\frac{1}{2014}\\right)\n$$\n\nWe have:\n$$\n\\log_2 \\left(1 + \\frac{1}{k}\\right) = \\log_2 \\left(\\frac{k+1}{k}\\right) = \\log_2 (k+1) - \\log_2 k\n$$\n\nTher... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof only | null | |
0ew8 | Problem:
Given $a_0$, $a_1$, ..., $a_n$, satisfying $a_0 = a_n = 0$, and $a_{k - 1} - 2a_k + a_{k + 1} \geq 0$ for $k = 1, 2, ..., n-1$. Prove that all the numbers are negative or zero. | [
"Solution:\n\nThe essential point is that if we plot the values $a_r$ against $r$, then the curve formed by joining the points is cup shaped. Its two endpoints are on the axis, so the other points cannot be above it. There are many ways of turning this insight into a formal proof. Barry Paul's was neater than mine:... | Soviet Union | 2nd ASU | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
03pa | A new sequence is obtained from the sequence of the positive integers $\{1, 2, 3, \ldots\}$ by deleting all the perfect squares. Then the $2\ 003$rd term of the new sequence is ( ).
(A) $2\ 046$
(B) $2\ 047$
(C) $2\ 048$
(D) $2\ 049$ | [
"Since $\\sqrt{2046} = \\sqrt{2047} = \\sqrt{2048} = \\sqrt{2049} = 45$, and $2003 + 45 = 2048$, Answer: C."
] | China | China Mathematical Competition (Shaanxi) | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | C | |
0jwm | Problem:
Let $\mathbb{N}$ denote the natural numbers. Compute the number of functions $f: \mathbb{N} \rightarrow \{0,1, \ldots, 16\}$ such that
$$
f(x+17)=f(x) \quad \text{and} \quad f\left(x^{2}\right) \equiv f(x)^{2}+15 \pmod{17}
$$
for all integers $x \geq 1$. | [
"Solution:\nBy plugging in $x=0$, we get that $f(0)$ can be either $-1,2$. As $f(0)$ is unrelated to all other values, we need to remember to multiply our answer by $2$ at the end. Similarly, $f(1)=-1$ or $2$.\n\nConsider the graph $x \\rightarrow x^{2}$. It is a binary tree rooted at $-1$, and there is an edge $-1... | United States | February 2017 | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 12066 | |
090t | Consideramos la sucesión de números enteros $\{f(n)\}_{n=1}^{\infty}$ definida por:
* $f(1) = 1$.
* Si $n$ es par, $f(n) = f(n/2)$.
* Si $n > 1$ es impar y $f(n-1)$ es impar, entonces $f(n) = f(n-1) - 1$.
* Si $n > 1$ es impar y $f(n-1)$ es par, entonces $f(n) = f(n-1) + 1$
a) Calcula $f(2^{2020} - 1)$.
b) Demuestra qu... | [
"En primer lugar, hacemos notar que la sucesión está bien definida: para cada $n \\in \\mathbb{N}$, $n > 1$, el valor de $f(n)$ está perfectamente determinado a partir de los valores de $f(r)$ con $r < n$.\nConsideramos la sucesión $g(n)$ dada por $g(n) = 0$ si la expresión binaria de $n$ tiene un número par de uno... | Mexico | LVI Olimpiada Matemática Española (Concurso Final) | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | Spanish | proof and answer | 0 | |
00ce | Se tiene un tablero de $7 \times 7$. Julián colorea 29 casillas de negro. Luego, Pilar debe colocar sobre el tablero un codo que tapa exactamente tres casillas como las de la figura (orientado de cualquier manera). Si las tres casillas que tapa el codo son negras, gana Pilar.
Determinar si Jul... | [] | Argentina | Nacional OMA 2019 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | No; with 29 black squares, an all-black L-shaped triomino is unavoidable, so Pilar can always win. | |
0255 | Problem:
Um número menor do que $200$ é formado por três algarismos diferentes e o dobro desse número também tem todos os algarismos diferentes. Ainda, o número e seu dobro não têm algarismos em comum. Qual é esse número? Quantas soluções têm esse problema? | [
"Solution:\n\nInicialmente note que o dobro de um número inteiro é par, logo ele termina em $0, 2, 4, 6$ ou $8$. No entanto, o número procurado não pode terminar em $0$, pois nesse caso o seu dobro também terminaria em $0$, e ambos teriam o algarismo $0$ em comum. Portanto, temos os casos a seguir.\n\n![](attached_... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | The numbers are 152, 182, 192, 143, 153, 134, 154, 164, 135, 145, 185, 176, 186, 138, 148, 178, 139, 179. There are 18 solutions. | |
0bqx | Find the smallest positive integer of the form $n = \overbrace{aa\dots a}^{k \text{ times}} + a(a-2)^2$, $k > 1$, divisible by $2016$. | [] | Romania | 67th NMO Shortlisted Problems | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 77777777952 | |
0ely | In a convex quadrilateral $ABCD$ the sides $AB$, $BC$, $CD$ are equal and $AC = BD = AD$. Find the angles of $ABCD$. | [
"Notice that $\\triangle ABC \\equiv \\triangle BCD$ (s, s, s) and $\\triangle ABD \\equiv \\triangle DCA$ (s, s, s).\nLet $\\alpha = \\angle CBD = \\angle BCA = \\angle CAB = \\angle CDB$\nand $\\beta = \\angle ACD = \\angle ADC = \\angle BAD = \\angle ABD$.\n\n\n\nIt can then be seen that... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | angle A = 72°, angle B = 108°, angle C = 108°, angle D = 72° | |
0ien | Problem:
Triangle $ABC$ has $AB=1$, $BC=\sqrt{7}$, and $CA=\sqrt{3}$. Let $\ell_1$ be the line through $A$ perpendicular to $AB$, $\ell_2$ the line through $B$ perpendicular to $AC$, and $P$ the point of intersection of $\ell_1$ and $\ell_2$. Find $PC$. | [
"Solution:\nBy the Law of Cosines, $\\angle BAC = \\cos^{-1} \\left(\\frac{3+1-7}{2 \\sqrt{3}}\\right) = \\cos^{-1}\\left(-\\frac{\\sqrt{3}}{2}\\right) = 150^\\circ$.\n\nIf we let $Q$ be the intersection of $\\ell_2$ and $AC$, we notice that $\\angle QBA = 90^\\circ - \\angle QAB = 90^\\circ - 30^\\circ = 60^\\circ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 3 | |
007e | There are given $2k$ boxes ($k \ge 2$) with $2k-1$ pebbles in each one. A legal move is to choose $2k-2$ boxes and remove one pebble from each one of them. Players $A$ and $B$ make moves alternately; $A$ goes first. A player wins if a move of his empties two boxes. Determine which player has a winning strategy. | [
"The second player $B$ has a winning strategy.\n\nEach move does not affect (ignores) exactly two boxes $i, j$; then we denote it by $m_{i,j}$. Let $A$'s first move be $m_{1,2}$. Then $B$ divides the remaining boxes arbitrarily into $k-1$ pairs $\\{3,4\\}, \\{5,6\\}, \\dots, \\{2k-1,2k\\}$, and his first $k-1$ move... | Argentina | Mathematical Olympiad Rioplatense | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | Player B | |
0jnm | Problem:
Let $a$ and $b$ be positive real numbers. Determine the minimum possible value of
$$
\sqrt{a^{2}+b^{2}}+\sqrt{(a-1)^{2}+b^{2}}+\sqrt{a^{2}+(b-1)^{2}}+\sqrt{(a-1)^{2}+(b-1)^{2}}
$$ | [
"Solution:\nAnswer: $2 \\sqrt{2}$\nLet $ABCD$ be a square with $A=(0,0)$, $B=(1,0)$, $C=(1,1)$, $D=(0,1)$, and $P$ be a point in the same plane as $ABCD$. Then the desired expression is equivalent to $AP + BP + CP + DP$. By the triangle inequality, $AP + CP \\geq AC$ and $BP + DP \\geq BD$, so the minimum possible ... | United States | HMMT November 2015 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 2 sqrt(2) | |
02r7 | Let $N = 2012$. Find all $N^2$-uples of real numbers $a_{i,j}$, $1 \le i, j \le N$, such that the following limit is convergent:
$$
\lim_{x \to +\infty} \sum_{1 \le i,j \le N} \left(a_{i,j} \sqrt[j]{x+i}\right)
$$ | [] | Brazil | Brazilian Math Olympiad | [
"Precalculus > Limits"
] | null | proof and answer | Exactly those arrays satisfying sum_{i=1}^N a_{i,j} = 0 for every j = 1, 2, ..., N. In this case the limit exists and equals sum_{i=1}^N a_{i,1} · i. | |
0ctk | Let $L$ be a point on the side $AC$ of a triangle $ABC$ such that $BL$ is the bisector of the angle $ABC$. Let $M$ be a point on the segment $CL$. The tangent through $B$ to the circumcircle $\Omega$ of $ABC$ meets the ray $CA$ at $P$. Tangents through $B$ and $M$ to the circimcircle $\Gamma$ of $\triangle BLM$ meet at... | [
"By angle chasing we get $\\angle BPM = 180^\\circ - 2\\angle BLP = \\angle BQM$ (see Fig. 9). So the points $B, M, P$, and $Q$ are concyclic, and $\\angle QPM = \\angle QBM = \\angle BLP$.",
"Так как $BL$ — биссектриса $\\angle ABC$, имеем $\\angle ABL = \\angle LBC$. Поскольку $PB$ — касательная к $\\Omega$, им... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
0eg9 | Problem:
Reši enačbo
$$
\log^{2}\left(\sqrt{3^{x}}-1\right)=1
$$ | [
"Solution:\nIz dane enačbe dobimo enačbi $\\log \\left(\\sqrt{3^{x}}-1\\right)=1$ in $\\log \\left(\\sqrt{3^{x}}-1\\right)=-1$.\n\nPo preoblikovanju prve enačbe dobimo $\\sqrt{3^{x}}-1=10$, ki jo preoblikujemo do enačbe $3^{x}=121$. Rešitev prve enačbe je $x=\\log_{3} 121 \\doteq 4,37$.\n\nPri drugi enačbi dobimo $... | Slovenia | 17. tekmovanje dijakov srednjih tehniških in strokovnih šol v znanju matematike | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | x = log_3 121 or x = log_3(121/100) | |
06ti | Let $ABC$ be a triangle inscribed into a circle $\Omega$ with center $O$. A circle $\Gamma$ with center $A$ meets the side $BC$ at points $D$ and $E$ such that $D$ lies between $B$ and $E$. Moreover, let $F$ and $G$ be the common points of $\Gamma$ and $\Omega$. We assume that $F$ lies on the arc $AB$ of $\Omega$ not c... | [
"It suffices to prove that the lines $FK$ and $GL$ are symmetric about $AO$. Now the segments $AF$ and $AG$, being chords of $\\Omega$ with the same length, are clearly symmetric with respect to $AO$. Hence it is enough to show\n$$\n\\begin{equation*}\n\\angle KFA = \\angle AGL . \\tag{1}\n\\end{equation*}\n$$\nLet... | IMO | 56th International Mathematical Olympiad Shortlisted Problems | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0654 | For a given positive integer $n > 2$, let $C_1$, $C_2$, $C_3$ be the boundaries of three convex $n$-gons in the plane such that the sets $C_1 \cap C_2$, $C_2 \cap C_3$, $C_3 \cap C_1$ are finite. Find the maximum number of points of the set $C_1 \cap C_2 \cap C_3$. | [
"Let us first observe that, if a line intersects a convex $n$-gon at finitely many points, then the number of such points is at most $2$. Therefore any two of the $n$-gons may intersect in at most $2n$ points. Choose two of the $n$-gons, $C_1$, $C_2$, and say that their intersection points are $p_1, p_2, \\dots, p_... | Greece | 24th Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | floor(3n/2) | |
06kz | A circle is circumscribed around an isosceles triangle whose two base angles are equal to $x^\circ$. Two points are chosen independently and randomly on the circle, and a chord is drawn between them. The probability that the chord intersects the triangle is $\frac{14}{25}$. Find the sum of the largest and smallest poss... | [
"The answer is $120$.\n\nThe probability that the chord does not intersect the triangle is $\\frac{11}{25}$. The only way this can happen is when the two points are chosen on the same arc between two of the triangle vertices. The probability that a point is chosen on one of the arcs opposite to one of the base angl... | Hong Kong | HKG TST | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 120 | |
05r6 | Problem:
On définit une suite $\left(a_{n}\right)_{n \geqslant 0}$ ainsi : on choisit $a_{0}, a_{1} \in \mathbb{N}^{*}$, et pour tout entier $n \geqslant 0$, on pose
$$
a_{n+2}=\left\lfloor\frac{2 a_{n}}{a_{n+1}}\right\rfloor+\left\lfloor\frac{2 a_{n+1}}{a_{n}}\right\rfloor,
$$
où $\lfloor x\rfloor$ désigne le plus gr... | [
"Solution:\n\nPour tous $a, b \\in \\mathbb{N}^{*}$, on note $g(a, b):=\\left\\lfloor\\frac{2 a}{b}\\right\\rfloor+\\left\\lfloor\\frac{2 b}{a}\\right\\rfloor$. Il est bien connu que $\\left(\\frac{\\sqrt{a}}{\\sqrt{b}}-\\frac{\\sqrt{b}}{\\sqrt{a}}\\right)^{2}=\\frac{a}{b}+\\frac{b}{a}-2$, or un carré est toujours ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 5 : Pot-POURRI | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0337 | Problem:
A plane bisects the volume of the tetrahedron $A B C D$ and meets the edges $A B$ and $C D$ respectively at points $M$ and $N$ such that $\frac{A M}{B M}= \frac{C N}{D N} \neq 1$. Prove that the plane passes through the midpoints of the edges $A C$ and $B D$. | [
"Solution:\nLet a plane $\\pi$ bisect the volume of $A B C D$ and meet the edges $A B$, $B C$, $C D$ and $D A$ at points $M$, $Q$, $N$ and $P$, respectively. Set $x=\\frac{A M}{B M}$, $y=\\frac{C N}{D N}$, $z=\\frac{A P}{D P}$ and $t=\\frac{C Q}{B Q}$. If $T=\\pi \\cap A C$ (we assume $T=\\infty$ if $\\pi \\paralle... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem"
] | null | proof only | null | |
0f5s | Problem:
The center of a coin radius $r$ traces out a polygon with perimeter $p$ which has an incircle radius $R > r$. What is the area of the figure traced out by the coin? | [] | Soviet Union | 18th ASU | [
"Geometry > Plane Geometry > Combinatorial Geometry > Minkowski's theorem",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 2pr | |
0jtt | Problem:
Let $P R O B L E M Z$ be a regular octagon inscribed in a circle of unit radius. Diagonals $M R$, $O Z$ meet at $I$. Compute $L I$. | [
"Solution:\n\nIf $W$ is the center of the circle then $I$ is the incenter of $\\triangle R W Z$. Moreover, $P R I Z$ is a rhombus. It follows that $P I$ is twice the inradius of a $1-1-\\sqrt{2}$ triangle, hence the answer of $2-\\sqrt{2}$. So $L I=\\sqrt{2}$.\n\nAlternatively, one can show (note, really) that the ... | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | sqrt(2) | |
01g9 | We call a positive integer $N$ splendid, if
$$
N = (a - b)^2 + (b - c)^2 + (c - a)^2
$$
for some integers $a$, $b$ and $c$. If $M$ and $N$ are splendid positive integers, is the sum $M + N$ or the product $MN$ also necessarily splendid? How about the product $2MN$? | [
"Answer: no to the first question, yes to the second.\nThe number $2$ is splendid since\n$$\n(1 - 0)^2 + (0 - 1)^2 + (1 - 1)^2 = 1^2 + 1^2 + 0^2 = 2.\n$$\nThe number $4$ is not splendid. Namely, if we had $4 = (a-b)^2 + (b-c)^2 + (c-a)^2$, at least one of these squares would have to be $2^2$ and the other two would... | Baltic Way | Baltic Way 2020 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Other"
] | null | proof and answer | Sum M+N: no; Product MN: no; Product 2MN: yes. | |
098c | Problem:
Fie expresia $E=\frac{3(x+y)(z+t)}{8} \cdot\left(\frac{1}{x y}+\frac{1}{z t}\right)$, unde $x, y, z, t \in[1 ; 3]$. Determinați cea mai mare valoare și cea mai mică valoare posibile ale expresiei $E$. | [
"Solution:\n\nExpresia $\\frac{8 E}{3}=(x+y)(z+t)\\left(\\frac{1}{x y}+\\frac{1}{z t}\\right)=\\frac{(x z+y t+y z+x t)(z t+x y)}{x y z t}=$\n\n$=\\frac{x z^{2} t+x^{2} y z+y z t^{2}+x y^{2} t+y z^{2} t+x y^{2} z+x z t^{2}+x^{2} y t}{x y z t}=$\n\n$=\\frac{z}{y}+\\frac{x}{t}+\\frac{t}{x}+\\frac{y}{z}+\\frac{z}{x}+\\... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | minimum 3, maximum 5 | |
08qy | Problem:
Find all pairs $(a, b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of $5$. | [
"Solution:\nThe condition is symmetric so we can assume that $b \\leq a$.\n\nThe first case is when $a = b$. In this case, $a! + a = 5^{m}$ for some positive integer $m$. We can rewrite this as $a \\cdot ((a-1)! + 1) = 5^{m}$. This means that $a = 5^{k}$ for some integer $k \\geq 0$. It is clear that $k$ cannot be ... | JBMO | Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (1,4), (4,1), (5,5) | |
00p8 | Given an integer number $n \ge 2$, determine the minimum value the sum
$$
\sum_{i=1}^{n} x_i^2 \left( 1 + \frac{x_i^{n-2}}{x_1 \cdots x_{i-1} x_{i+1} \cdots x_n} \right)
$$
may achieve, when $x_1, x_2, \dots, x_n$ run through the positive real numbers subject to
$$
\sum_{i=1}^{n} \frac{1}{x_i + 1} = 1.
$$ | [
"The required minimum is $n^2(n-1)$ and is achieved if and only if the $x_i$ are all equal to $n-1$.\nWrite\n$$\n\\sum_{i=1}^{n} x_i^2 \\left( 1 + \\frac{x_i^{n-2}}{x_1 \\cdots x_{i-1} x_{i+1} \\cdots x_n} \\right) = \\sum_{i=1}^{n} x_i^2 + \\sum_{i=1}^{n} \\frac{x_i^n}{x_1 \\cdots x_{i-1} x_{i+1} \\cdots x_n}\n$$\... | Balkan Mathematical Olympiad | shortlistBMO 2011 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | n^2(n-1) | |
0k0v | Problem:
A positive integer is called primer if it has a prime number of distinct prime factors. Find the smallest primer number. | [
"Solution:\nAnswer: $6$\nA primer number must have at least two distinct prime factors, and $6$ will work."
] | United States | HMMT November 2018 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 6 | |
0h8d | In the triangle $ABC$ $AB > AC$. A tangent to the circumcircle of the triangle $ABC$ has been drawn through the point $A$. This tangent intersects the line $BC$ at the point $P$. In continuation of the side $BA$ after the point $A$ point $Q$ was selected such that $AQ = AC$. Let $X$ and $Y$ be the midpoints of the segm... | [
"Consider the point $M$, which is the midpoint of the segment $PQ$, then $MX$ and $MY$ are the mid-segments of $\\triangle CPQ$ and $\\triangle APQ$ respectively. It follows that $MX = y$ and $MY = x$, and also $MX \\parallel CP$ and $MY \\parallel AQ$. This parallel gives that $\\angle XMY = \\angle PBQ = \\beta$ ... | Ukraine | UkraineMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00pw | Prove that the polynomial
$$
P(x) = (x^2 - 8x + 25)(x^2 - 16x + 100) \cdots (x^2 - 8n x + 25 n^2) + 1, \ n \in \mathbb{N}^*,
$$
cannot be written as the product of two polynomials with integer coefficients of degree greater or equal to 1. | [
"The polynomial $P(x)$ can be written in the form\n$$\nP(x) = [(x-4)^2 + 3^2][(x-8)^2 + 6^2] \\cdots [(x-4n)^2 + 9n^2] + 1\n$$\nLet we can write $P(x)$ in the form $P(x) = Q(x)R(x)$, where\n$$\nQ(x) = a_0 + a_1 x + \\dots + a_k x^k, \\quad 1 \\le k < 2n, \\ a_0, a_1, \\dots, a_k \\in \\mathbb{Z}, \\ a_k \\ne 0\n$$\... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0l1x | Problem:
Points $K, A, L, C, I, T, E$ are such that triangles $C A T$ and $E L K$ are equilateral, share a center $I$, and points $E, L, K$ lie on sides $\overline{C A}$, $\overline{A T}$, $\overline{T C}$ respectively. If the area of triangle $C A T$ is double the area of triangle $E L K$ and $C I=2$, compute the min... | [
"Solution:\n\n\n\nFirst, compute that triangle $C A T$ has side length $2 \\sqrt{3}$ and area $3 \\sqrt{3}$. Since triangle $E L K$ has half the area of $C A T$, the triangles $C E K, A E L, T L K$ must each have $\\frac{1}{6}$ the area of $C A T$. Now by sine area formula, we have\n$$\n\\f... | United States | HMMT November 2024 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Ge... | null | proof and answer | sqrt(3) - 1 | |
03kd | Problem:
Evaluate the sum
$$
\sum_{n=1}^{1994} (-1)^n \frac{n^2 + n + 1}{n!}
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | final answer only | 1/1993! + 1/1994! - 1 | |
02ww | Problem:
João possui um único pedaço de corrente com $n$ elos, cada um pesando $1~\mathrm{g}$. Cada vez que um elo de algum pedaço de corrente é quebrado, obtemos 3 pedaços. Por exemplo, se um pedaço possui 9 elos e quebramos o quarto, passaremos a ter pedaços com as seguintes quantidades de elos: 3, 1 e 5. Um elo queb... | [
"Solution:\na) Basta quebrarmos o primeiro e o sexto elos obtendo os pedaços de comprimentos: 1, 4, 1 e 2. Para realizarmos qualquer peso de $1~\\mathrm{g}$ a $8~\\mathrm{g}$, considere as somas de pedaços:\n$$\n\\begin{aligned}\n& 1=1 \\\\\n& 2=2 \\\\\n& 3=1+2 \\\\\n& 4=4 \\\\\n& 5=1+4 \\\\\n& 6=2+4 \\\\\n& 7=1+2+... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
05l8 | Problem:
Soit $ABC$ un triangle dont les angles sont aigus, et tel que $AB \neq AC$. On note $D$ le pied de la bissectrice de $\widehat{BAC}$. Le point $E$ (resp. $F$) désigne le pied de la hauteur issue de $B$ (resp. de $C$). Le cercle circonscrit au triangle $DBF$ rencontre le cercle circonscrit au triangle $DCE$ en... | [
"Solution:\n\n\n\nLemme 1. Les cercles $AEF$, $BDF$ et $CDE$ sont concourants en $M$.\nCela découle immédiatement du théorème de Miquel mais rappelons tout de même la démonstration:\n$$(MF, ME) = (MF, MD) + (MD, ME) = (BF, BD) + (CD, CE) = (AB, CD) + (CD, AC) = (AB, AC)$$\ndonc $A, E, F, M$... | France | Olympiades Françaises de Mathématiques - Test de Janvier | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
02ii | Problem:
Se $x > 0$, $y > 0$, $x > y$ e $z \neq 0$, então a única opção errada é:
(A) $x + z > y + z$
(B) $x - z > y - z$
(C) $x z > y z$
(D) $\frac{x}{z^{2}} > \frac{y}{z^{2}}$
(E) $x z^{2} > y z^{2}$ | [
"Solution:\nNessa questão usaremos as propriedades das desigualdades.\nPodemos somar o mesmo número a ambos os membros de uma desigualdade sem alterar o sinal, temos: $x > y \\Rightarrow \\left\\{\\begin{array}{l} x + z > y + z \\text{ (somando $z$ a ambos os membros)} \\\\ x - z > y - z \\text{ (somando $-z$ a amb... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | C | |
0jds | Problem:
For how many integers $1 \leq k \leq 2013$ does the decimal representation of $k^{k}$ end with a $1$? | [
"Solution:\n\nAnswer: 202\n\nWe claim that this is only possible if $k$ has a units digit of $1$. Clearly, it is true in these cases. Additionally, $k^{k}$ cannot have a units digit of $1$ when $k$ has a units digit of $2, 4, 5, 6$, or $8$. If $k$ has a units digit of $3$ or $7$, then $k^{k}$ has a units digit of $... | United States | HMMT 2013 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | final answer only | 202 | |
0gca | 設三角形 $ABC$ 的內心為 $I$，直線 $l$ 是線段 $AI$ 的中垂線。設點 $P$ 落在三角形 $ABC$ 的外接圓上，且令直線 $AP$ 與 $l$ 交於點 $Q$。點 $R$ 位於 $l$ 上，滿足 $\angle IPR = 90^\circ$。設直線 $IQ$ 與三角形 $ABC$ 中平行於 $BC$ 邊的中位線交於點 $M$。
$$
\text{試證: } \angle AMR = 90^\circ.
$$
(註: 三角形中，任兩邊中點的連線，稱為中位線。) | [
"令 $A'$ 為 $A$ 關於 $M$ 的對稱點，$I'$ 為 $I$ 關於 $P$ 的對稱點。則 $A'$ 位於 $BC$ 上且 $R$ 為 $\\triangle AII'$ 的外心，因此 $\\angle AMR$ 為直角若且唯若 $A, I, A', I'$ 共圓。\n\n\n以下證明 $\\triangle AIP \\sim \\triangle IA'P$（$\\sim$ 代表正向相似）：取 $J$ 使得 $AIA'J$ 為平行四邊形，取 $P'$ 使得 $\\triangle AIP' \\sim \\triangle JIA'$，則\n$$\n\\tria... | Taiwan | 二〇一八數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations... | null | proof only | null | |
06ek | Let $M$ be a subset of $\{1, 2, \dots, 2006\}$ with the following property: For any three elements $x, y$ and $z$ ($x < y < z$) of $M$, $x + y$ does not divide $z$. Determine the largest possible size of $M$. Justify your claim. | [
"The answer is $1004$.\n\nWhen $M = \\{1003, 1004, \\dots, 2006\\}$, the sum of any two elements in $M$ is at least $2007$, which is larger than the largest element in $M$. Therefore, $x + y$ does not divide $z$ for any $x, y, z \\in M$. This gives a possible case for $|M| = 1004$.\n\nNext, consider any subset $M$ ... | Hong Kong | CHKMO | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1004 | |
074m | Let $P_1(x) = a x^2 - b x - c$, $P_2(x) = b x^2 - c x - a$, $P_3(x) = c x^2 - a x - b$ be three quadratic polynomials where $a$, $b$, $c$ are non-zero real numbers. Suppose there exists a real number $\alpha$ such that $P_1(\alpha) = P_2(\alpha) = P_3(\alpha)$. Prove that $a = b = c$. | [
"We have three relations:\n$$\n\\begin{aligned}\na \\alpha^2 - b \\alpha - c &= \\lambda, \\\\\nb \\alpha^2 - c \\alpha - a &= \\lambda, \\\\\nc \\alpha^2 - a \\alpha - b &= \\lambda,\n\\end{aligned}\n$$\nwhere $\\lambda$ is the common value. Eliminating $\\alpha^2$ from these, taking these equations pairwise, we g... | India | Indija mo 2011 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0jhx | Problem:
In a game, there are three indistinguishable boxes; one box contains two red balls, one contains two blue balls, and the last contains one ball of each color. To play, Raj first predicts whether he will draw two balls of the same color or two of different colors. Then, he picks a box, draws a ball at random, ... | [
"Solution:\n\nLet us call the box with two red balls box 1, the box with one of each color box 2, and the box with two blue balls box 3. Without loss of generality, assume that the first ball that Raj draws is red. If Raj picked box 1, then he would have picked a red ball with probability $1$, and if Raj picked box... | United States | HMMT 2013 | [
"Statistics > Probability > Counting Methods > Other",
"Math Word Problems"
] | null | proof and answer | 5/6 | |
0f8m | Problem:
Find all positive integers $n$ satisfying $$(1 + 1 / n)^{n + 1} = (1 + 1 / 1998)^{1998}.$$ | [
"Solution:\n\nAnswer: no solutions.\n\nWe have $$(1 + 1 / n)^{n + 1} > e > (1 + 1 / n)^{n}.$$"
] | Soviet Union | 22nd ASU | [
"Precalculus > Limits"
] | null | proof and answer | no solutions | |
07tc | Let $a$, $b$, $c \geq 0$ be real numbers with $a + b + c = 1$. Show that
$$
a + b + c \leq \sqrt{a(1+b)} + \sqrt{b(1+c)} + \sqrt{c(1+a)} \leq 2.
$$ | [
"We consider first the left hand inequality. As $a + b + c = 1$ we have:\n$$\na \\leq 1 \\leq 1 + b, \\text{ so } a = \\sqrt{a^2} \\leq \\sqrt{a(1+b)}\n$$\nApplying this similarly to the other square roots and adding quickly gives:\n$$\n1 = a + b + c \\leq \\sqrt{a(1+b)} + \\sqrt{b(1+c)} + \\sqrt{c(1+a)}.\n$$\nA di... | Ireland | IRL_ABooklet_2020 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0jq8 | Problem:
There is a colony consisting of $100$ cells. Every minute, a cell dies with probability $\frac{1}{3}$; otherwise it splits into two identical copies. What is the probability that the colony never goes extinct? | [
"Solution:\n\nThe answer is $1-\\left(\\frac{1}{2}\\right)^{100}$.\n\nLet $p$ be the probability that a colony consisting of just one cell will survive. Then\n$$\np = \\frac{1}{3} \\cdot 0 + \\frac{2}{3}\\left(1-(1-p)^{2}\\right)\n$$\nowing to the fact that when the cell splits in two, the probability both of them ... | United States | Berkeley Math Circle | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 1 - (1/2)^{100} | |
0fzk | Problem:
Bestimme alle Polynome $P$ mit reellen Koeffizienten, sodass für alle $x \in \mathbb{R}$ gilt:
$$
(x+2014) P(x) = x P(x+1)
$$ | [
"Solution:\nSetzt man $x=0$ sieht man, dass $P(0)=0$ gelten muss. Setzt man nun $x=-1$ folgt $P(-1)=0$ usw. Somit sehen wir, dass $0, -1, -2, \\ldots, -2013$ alles Nullstellen von $P$ sind. Wir zeigen nun, dass der Grad $n$ von $P$ genau $2014$ sein muss. Dazu schreiben wir $P(X) = a_{n} X^{n} + a_{n-1} X^{n-1} + \... | Switzerland | IMO-Selektionsprüfung | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | P(x) = c · x(x+1)(x+2)⋯(x+2013) for any real constant c | |
03ri | If one side of square $ABCD$ is on the line $y = 2x - 17$, and the other two vertices lie on parabola $y = x^2$. Then the minimum area of the square is ________. | [
"Assume that $AB$ is on the line $y = 2x - 17$ and the coordinates of the other two vertices on the parabola are $C(x_1, y_1)$ and $D(x_2, y_2)$. Then $CD$ is on a line $L$ whose equation is $y = 2x + b$. Combining this with the equation of the parabola, we get $x^2 = 2x + b \\Rightarrow x_1, x_2 = 1 \\pm \\sqrt{b+... | China | China Mathematical Competition (Jiangxi) | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | final answer only | 80 |
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