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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0i0q | Problem:
An $n \times n$ matrix of integers is called "golden" if, for every row and every column, their union contains all of the numbers $1,2,3, \ldots, 2 n-1$. Find all golden matrices (of all sizes). | [
"Solution:\n\nThe $1 \\times 1$ matrix whose only entry is $1$ is certainly golden; we claim there are no others. To see this, note that each of the numbers $1,2, \\ldots, 2 n-1$ must appear at least $n$ times in the matrix. (Proof: Suppose some number $k$ appears less than $n$ times. Because there are $n$ columns,... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | Only the 1x1 matrix with the single entry 1. | |
054i | Call a tuple $(a_1, \dots, a_n)$ of real numbers *stable* if the sums $a_1 + a_2 + \dots + a_k$, as well as the sums $a_k + a_{k+1} + \dots + a_n$, where in both cases $0 < k \le n$, are either all negative or all non-negative.
For instance, the tuple $(3, -1, 2)$ is stable, since:
$$
\begin{array}{lclcl}
3 & \ge & 0, ... | [
"Consider terms whose absolute value is minimal in the tuple. If there exists a negative such element, denote it $a_i$, then the sum of $a_i$ and its any neighbour is non-negative. Thus $a_i$ is neither the first nor the last in the tuple because of stability of the tuple. But then both $a_{i-1} + a_i$ and $a_i + a... | Estonia | National Olympiad Final Round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0epp | Eskom announces that there is a 60% chance of Stage 1 load shedding for a specific week.
In a specific suburb Stage 1 load shedding is from 10:00 to 12:30 on a Monday, Wednesday, Friday and Sunday. What is the probability that in that specific suburb there will be load shedding at a moment in the week, without knowing ... | [
"There are $7 \\times 24 = 168$ hours in a week and there are $4 \\times 2.5 = 10$ hours during which load shedding may occur. Since there is a 60% chance of load shedding, the probability that there is load shedding at a particular moment during the week is\n$$\n\\frac{4 \\times 2.5}{7 \\times 24} \\times 60\\% = ... | South Africa | South African Mathematics Olympiad | [
"Statistics > Probability > Counting Methods > Other",
"Math Word Problems"
] | English | final answer only | 1/28 | |
09sc | Problem:
Gegeven is een gelijkzijdige driehoek $A B C$. Op de lijn door $B$ evenwijdig aan $A C$ ligt een punt $D$, zodat $D$ en $C$ aan dezelfde kant van lijn $A B$ liggen. De middelloodlijn van $C D$ snijdt de lijn $A B$ in $E$. Bewijs dat driehoek $C D E$ gelijkzijdig is. | [
"Solution:\n\nWe bekijken in alle oplossingen de configuratie waarin $E$ tussen $A$ en $B$ ligt. Het geval dat $B$ tussen $A$ en $E$ ligt gaat analoog. (En vanwege de voorwaarde dat $D$ en $C$ aan dezelfde kant van $A B$ liggen, kan $A$ niet tussen $B$ en $E$ liggen, dus hebben we zo alle gevallen gehad.)\n\nOmdat ... | Netherlands | IMO-selectietoets II | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
06ws | Let $a_{1}, a_{2}, a_{3}, \ldots$ be an infinite sequence of positive integers such that $a_{n+2m}$ divides $a_{n} + a_{n+m}$ for all positive integers $n$ and $m$. Prove that this sequence is eventually periodic, i.e. there exist positive integers $N$ and $d$ such that $a_{n} = a_{n+d}$ for all $n > N$. | [
"We will make repeated use of the following simple observation:\n\nLemma 1. If a positive integer $d$ divides $a_{n}$ and $a_{n-m}$ for some $m$ and $n > 2m$, it also divides $a_{n-2m}$. If $d$ divides $a_{n}$ and $a_{n-2m}$, it also divides $a_{n-m}$.\n\nProof. Both parts are obvious since $a_{n}$ divides $a_{n-2m... | IMO | IMO 2021 Shortlisted Problems | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
09z3 | Given a positive integer $c$, we construct a sequence of fractions $a_1, a_2, a_3, \dots$ as follows:
* $a_1 = \frac{c}{c+1}$;
* to get $a_n$, we take $a_{n-1}$ (in its most simplified form, with both the numerator and denominator chosen to be positive) and we add 2 to the numerator and 3 to the denominator. Then we si... | [
"a.\nThe sequence starts as follows.\n$$\n\\begin{aligned}\na_1 &= \\frac{10}{11}, & a_2 &= \\frac{12}{14} = \\frac{6}{7}, & a_3 &= \\frac{8}{10} = \\frac{4}{5}, & a_4 &= \\frac{6}{8} = \\frac{3}{4}, \\\\\na_5 &= \\frac{5}{7}, & a_6 &= \\frac{7}{10}, & a_7 &= \\frac{9}{13}\n\\end{aligned}\n$$\nIt seems that the las... | Netherlands | Final Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | a) 4; b) Yes (for example at n = 97); c) 7 and 27 | |
0blf | a) Show that the equation $p + x + y + z = xyz$ has positive integer solutions for every positive prime $p$.
b) Find the positive integer solutions of the equation $3 + x + y + z = xyz$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | a) For any prime p, for example (x, y, z) = (2, p+3, 1) and all its permutations are solutions.
b) All positive integer solutions are the permutations of (1, 2, 6). | |
06b9 | Problem:
Let $k$ be a positive integer. Find the maximal length $L$ of a sequence $a_1, \dots, a_L$ containing positive integers that satisfies the following two properties:
* Each term of the sequence is smaller or equal to $2^k$.
* There are no consecutive terms $a_i, a_{i+1}, \dots, a_j$ (with $1 \le i < j \le L$) ... | [] | Greece | Selection Examination | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 2k + 1 | |
0b07 | Problem:
PQR Entertainment wishes to divide their popular idol group PRIME, which consists of seven members, into three sub-units - PRIME-P, PRIME-Q, and PRIME-R - with each of these sub-units consisting of either two or three members. In how many different ways can they do this, if each member must belong to exactly ... | [
"Solution:\n\nNote that the only way to do this is to divide PRIME into two sub-units of two members and one sub-unit of three members. There are $\\binom{7}{3}\\binom{4}{2}\\binom{2}{2}=210$ ways to do this. Since there are 3 ways to choose which unit will have three members, our final answer is $210 \\cdot 3=630$... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | final answer only | 630 | |
08ix | Problem:
Prove that the equation
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{abc} = \frac{12}{a + b + c}
$$
has many solutions $(a, b, c)$ in strictly positive integers. | [] | JBMO | The first selection test for IMO 2003 and BMO 2003 | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof only | null | |
0d6z | There are totally $16$ teams participating in a football tournament; each team playing with every other exactly $1$ time. In each match, the winner gains $3$ points, the loser gains $0$ point and each team gains $1$ point for the tie match. Suppose that at the end of the tournament, each team gains the same number of p... | [
"See the solution to Problem 1 in the test of level 4. $\\square$"
] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0hrh | Problem:
Denote by $f(n)$ the integer obtained by reversing the digits of a positive integer $n$. Find the greatest integer that is certain to divide $n^{4}-f(n)^{4}$ regardless of the choice of $n$. | [
"Solution:\n\nThe answer is $99$.\n\nLet $x = \\overline{d_{n-1} \\ldots d_{2} d_{1} d_{0}}$, i.e. $x = 10^{n-1} d_{n-1} + \\cdots + 10^{2} d_{2} + 10 d_{1} + d_{0}$. Then $y = f(x) = 10^{n-1} d_{0} + 10^{n-2} d_{1} + \\cdots + 10 d_{n-2} + d_{n-1}$.\n\nLet us show separately that $9 \\mid x^{4} - y^{4}$ and that $... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 99 | |
03p7 | Put numbers $1, 2, 3, 4, 5, 6, 7$ and $8$ at the vertices of a cube, such that the sum of any three numbers on any face is not less than $10$. Find the minimum sum of the four numbers on a face. (posed by Qiu Zonghu) | [
"Suppose that the four numbers on a face of the cube are $a_1, a_2, a_3, a_4$ such that their sum reaches the minimum and $a_1 < a_2 < a_3 < a_4$. Since the maximum sum of any three numbers less than $5$ is $9$, we have $a_4 \\ge 6$, and then $a_1 + a_2 + a_3 + a_4 \\ge 16$.\n\n\n\nAs seen ... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 16 | |
07pa | A circle is drawn through three vertices $A$, $B$, $C$ of a parallelogram $ABCD$ intersecting the side $CD$ internally at $E$ and the side $AD$ internally at $F$. The line $EF$ meets the line $BC$ at $H$ and the line $BA$ at $K$.
Prove that the circumcircles of the triangles $CEH$ and $AFK$ both touch the circumcircle ... | [
"Because $DC \\parallel AB$ and $BEFA$ is cyclic, we have\n$\\angle EDF = \\angle KAF = 180^\\circ - \\angle FAB = \\angle FEB$ and\n$\\angle CEB = \\angle EBA = 180^\\circ - \\angle EFA = \\angle EFD = \\angle EHC,$\nthe last equality because $AD \\parallel BH$.\n\nThe first line shows tha... | Ireland | Irska 2014 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0erz | One of the numbers below is multiplied by $2016$ and the product is a perfect square. The number is
(A) $10$ (B) $14$ (C) $32$ (D) $100$ (E) $108$ | [
"Factorise $2016$. It is divisible by $2$, because it is an even number and it is also divisible by $9$ since the sum of the digits is $9$. The full factorisation is $2^5 \\times 3^2 \\times 7$. Each prime factor of a complete square must have an even number as the exponent (i.e. every prime factor must appear an e... | South Africa | South African Mathematics Olympiad First Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | B | |
0fa1 | Problem:
$ABC$ is an acute-angled triangle with circumcenter $O$. The circumcircle of $ABO$ intersects $AC$ and $BC$ at $M$ and $N$. Show that the circumradii of $ABO$ and $MNC$ are the same. | [
"Solution:\n\nIt is sufficient to show that $\\angle MBN = \\angle C$. But $\\angle MBN = \\angle MBO + \\angle OBN = \\angle MAO + \\angle OBN = \\angle MCO + \\angle OCN = \\angle C$."
] | Soviet Union | 25th ASU | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0exc | Problem:
a. A committee has met 40 times, with 10 members at every meeting. No two people have met more than once at committee meetings. Prove that there are more than 60 people on the committee.
b. Prove that you cannot make more than 30 subcommittees of 5 members from a committee of 25 members with no two subcommit... | [
"Solution:\n\na. Each meeting involves $\\frac{10 \\cdot 9}{2} = 45$ pairs. So after 40 meetings, there have been $40 \\times 45 = 1800$ pairs. We are told that these are all distinct. But if there are $N$ people on the committee, then there are only $\\frac{N(N-1)}{2}$ pairs available. For $N=60$, this is only $\\... | Soviet Union | 5th ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0gml | Let $ABCD$ be a convex quadrangle and $K$ be the point of intersection of its diagonals. If the points $L \in [AD]$, $M \in [AC]$, $N \in [BC]$ satisfy the conditions $KL \parallel AB$, $LM \parallel DC$, $MN \parallel AB$, show that
$$
\frac{\text{Area}(KLMN)}{\text{Area}(ABCD)} < \frac{8}{27}.
$$ | [] | Turkey | IMO TEAM SELECTION EXAM | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | proof only | null | |
03ph | Suppose a set $S$ satisfies the following conditions:
(1) every element in $S$ is a positive integer and not greater than $100$;
(2) for any two different elements $a$ and $b$ in $S$, there is an element $c$ in $S$ such that the greatest common divisor of $a$ and $c$ is equal to $1$, and the greatest common divisor of ... | [
"The maximum number of elements is $72$.\n\nA positive integer not greater than $100$ can be written as\n$$\nn = 2^{a_1} \\cdot 3^{a_2} \\cdot 5^{a_3} \\cdot 7^{a_4} \\cdot 11^{a_5} \\cdot q,\n$$\nwhere $q$ is a positive integer and not divisible by $2$, $3$, $5$, $7$ and $11$, and $a_1, a_2, a_3, a_4$ and $a_5$ ar... | China | China Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 72 | |
0551 | Solve the system of equations $3x + 7y + 14z = 252$, $xyz - u^2 = 2016$ for non-negative real numbers. | [
"$$\n\\begin{aligned}\n252 &= 3x + 7y + 14z \\ge 3\\sqrt[3]{3x \\cdot 7y \\cdot 14z} = 3\\sqrt[3]{3 \\cdot 7 \\cdot 14 \\cdot (2016 + u^2)} \\\\\n&\\ge 3\\sqrt[3]{3 \\cdot 7 \\cdot 14 \\cdot 2016} = 3\\sqrt[3]{2^6 \\cdot 3^3 \\cdot 7^3} = 2^2 \\cdot 3^2 \\cdot 7 = 252.\n\\end{aligned}\n$$\nTo not get a contradictio... | Estonia | Open Contests | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | (x, y, z, u) = (28, 12, 6, 0) | |
0drb | In an acute-angled triangle $ABC$, $M$ is a point on the side $BC$, the line $AM$ meets the circumcircle $\omega$ of $ABC$ at the point $Q$ distinct from $A$. The tangent to $\omega$ at $Q$ intersects the line through $M$ perpendicular to the diameter $AK$ of $\omega$ at the point $P$. Let $L$ be the point on $\omega$ ... | [
"\n\nLet $D$ be the foot of the perpendicular from $M$ onto the diameter $AK$. Since $AK$ is a diameter of $\\omega$, we have $\\angle MQK = 90^\\circ = \\angle MDK$ so that $M$, $Q$, $K$, $D$ are concyclic. Join $QK$. Then $\\angle PMQ = \\angle QKD = \\angle QKA = \\angle PQA = \\angle PQ... | Singapore | Singapur 2015 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0493 | Let $x$, $y$, $z$ be positive real numbers such that $x^3 + y^3 + z^3 = 1$. Prove that
$$
x^2 + y^2 + z^2 > x^5 + y^5 + z^5 + 2x^2y^2z^2(x + y + z).
$$ | [] | Croatia | CroatianCompetitions2011 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
039h | Find all pairs of functions $f, g : \mathbb{R} \to \mathbb{R}$ such that:
(1) $f(xg(y+1)) + y = x f(y) + f(x+g(y))$ for any two $x, y \in \mathbb{R}$;
(2) $f(0) + g(0) = 0$. | [] | Bulgaria | Second selection test for IMO 2007, Vietnam | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x and g(x) = x | |
0juv | Problem:
Let $S=\{a_{1}, \ldots, a_{n}\}$ be a finite set of positive integers of size $n \geq 1$, and let $T$ be the set of all positive integers that can be expressed as sums of perfect powers (including $1$) of distinct numbers in $S$, meaning
$$
T=\left\{\sum_{i=1}^{n} a_{i}^{e_{i}} \mid e_{1}, e_{2}, \ldots, e_{n}... | [
"Solution:\nIn general we can assume that each $a_{i}>1$, since replacing $a_{i}=1$ by some large integer $a$ creates a set $T$ containing the original $T$ as a subset (by setting $e_{i}=0$).\n\nWe proceed by induction on $n$. For the base case $n=1$, an arithmetic progression of length at least $3$ would give $a_{... | United States | HMMT Invitational Competition | [
"Number Theory > Algebraic Number Theory > Combinatorial number theory: van der Waerden's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
04o4 | Let $P(x), Q(x) \in \mathbb{R}[x]$ be polynomials with real coefficients such that
$$
P(P(x)) = (Q(x))^2
$$
holds for all real numbers $x$.
Does there necessarily exist a polynomial $R(x) \in \mathbb{R}[x]$ with real coefficients such that $P(x) = (R(x))^2$ holds for all real numbers $x$? | [
"Yes, there necessarily exists such polynomial $R(x)$.\n\nFirstly, notice that we are allowed to assume, without loss of generality, that polynomials $P(x)$ and $Q(x)$ have leading coefficient equal to $1$. If the polynomial $Q(x)$ is constant, then the polynomial $P(x)$ is constant as well, and we have $P(x) = (Q(... | Croatia | Croatian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | Yes | |
0d2d | Let $a_{1}, a_{2}, \ldots, a_{9}$ be integers. Prove that if $19$ divides $a_{1}^{9} + a_{2}^{9} + \cdots + a_{9}^{9}$ then $19$ divides the product $a_{1} a_{2} \cdots a_{9}$. | [
"Assume that $19$ does not divide the product $a_{1} a_{2} \\cdots a_{9}$. This means that $a_{1}, a_{2}, \\ldots, a_{9}$ are relatively prime with $19$. Using Fermat,\n$$\na_{1}^{18} \\equiv a_{2}^{18} \\equiv \\cdots \\equiv a_{9}^{18} \\equiv 1 \\pmod{19}.\n$$\nBut $a_{i}^{18} \\equiv 1 \\pmod{19}$ is equivalent... | Saudi Arabia | Preselection tests for the full-time training | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
05b1 | Find all positive integers $n$ for which you can replace a digit $4$ with the digits $22$ and obtain a number divisible by $n$. | [
"Let $n$ satisfy the condition and $n'$ be the number obtained by the replacement of digits. Denote by $x$ the number formed by the digits to the left of the replaced digit and denote by $y$ the number formed by the digits to the right of the replaced digit. Let $k$ be the number of digits in $y$. Then $n = x \\cdo... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | All n of the form 45*10^t or 2475*10^t for integer t >= 0. | |
08ku | Problem:
The nonnegative integer $n$ and $(2n+1) \times (2n+1)$ chessboard with squares colored alternatively black and white are given. For every natural number $m$ with $1 < m < 2n+1$, an $m \times m$ square of the given chessboard that has more than half of its area colored in black is called a $B$-square. If the g... | [
"Solution:\n\nEvery square with even side length will have an equal number of black and white $1 \\times 1$ squares, so it isn't a $B$-square. In a square with odd side length, there is one more $1 \\times 1$ black square than white squares, if it has black corner squares. So, a square with odd side length is a $B$... | JBMO | 2007 Shortlist JBMO | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | ((n+1)(2n^2 + 4n + 3))/3 | |
0io2 | Problem:
Show that for positive integers $n$,
$$
\sum_{d \mid n} \phi(d) = n
$$ | [
"Solution:\nBoth sides are multiplicative functions of $n$, the right side trivially and the left because for relatively prime positive integers $n$ and $n'$, \n$$\n\\left(\\sum_{d \\mid n} \\phi(d)\\right)\\left(\\sum_{d' \\mid n'} \\phi(d')\\right) = \\sum_{d \\mid n,\\ d' \\mid n'} \\phi(d) \\phi(d'),\n$$\nand $... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)"
] | null | proof only | null | |
0h6t | Number $2016$ is written on the board. Olesya and Andriy are playing game: they take turns (Olesya is the first to start) to reduce the number on the board to an integer, not exceeding the number of the move (on the first move Olesya must reduce the number to $1$, Andriy, in his turn, to $1$ or $2$, then Olesya to $1$,... | [
"Let us define the position for this game with a pair of numbers $(m, n)$, where $m$ – the number written on the board, and $n$ – the number of a move. We assume a position being winning, if after having got into it, a player still has a winning strategy. A losing one – when having got into it, an opponent will hav... | Ukraine | UkraineMO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | the second player wins | |
0ktf | Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$, distinct points $A_1, A_2, A_3, \dots, A_m$ lie on $\ell_A$, and distinct points $B_1, B_2, B_3, \dots, B_n$ lie on $\ell_B$. Additionally, when segments $A_iB_j$ are drawn for all $i = 1, 2, 3, \dots, m$ and $j = 1, 2, 3, \dot... | [] | United States | AIME II | [
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | final answer only | 96 | |
0j6w | Problem:
In preparation for a game of Fish, Carl must deal 48 cards to 6 players. For each card that he deals, he runs through the entirety of the following process:
1. He gives a card to a random player.
2. A player $Z$ is randomly chosen from the set of players who have at least as many cards as every other player (... | [
"Solution:\n\nAfter any number of cards are dealt, we see that the difference between the number of cards that any two players hold is at most one. Thus, after the first 47 cards have been dealt, there is only one possible distribution: there must be 5 players with 8 cards and 1 player with 7 cards. We have two cas... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 5/6 | |
0aky | Let $ABC$ be an acute triangle with $AB < AC$ and $AB < BC$. Let $P$ be a point on the segment $BC$ such that $\angle APB = \angle BAC$. The tangent to the circumcircle of $\triangle ABC$ at $A$ meets the circumcircle of $\triangle APB$ at a point $Q \neq A$. Let $Q'$ be the reflection of $Q$ relative to the midpoint o... | [] | North Macedonia | Team Selection Test for IMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof only | null | |
03kp | Problem:
Write the sum
$$
\sum_{k=0}^{n} \frac{(-1)^{k}\binom{n}{k}}{k^{3}+9 k^{2}+26 k+24}
$$
in the form $\frac{p(n)}{q(n)}$, where $p$ and $q$ are polynomials with integer coefficients. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 1/(2(n+3)(n+4)) | |
031b | Problem:
Let $M$ be an interior point of $\triangle ABC$. The lines $AM$, $BM$ and $CM$ meet the lines $BC$, $CA$ and $AB$ at points $A_1$, $B_1$ and $C_1$, respectively, such that $S_{CB_1M} = 2 S_{AC_1M}$. Prove that $A_1$ is the midpoint of the segment $BC$ if and only if $S_{BA_1M} = 3 S_{AC_1M}$. | [
"Solution:\n\nLet $A_1$ be the midpoint of the segment $BC$. Then Ceva's theorem implies that\n$$\n\\frac{AC_1}{C_1B} \\cdot \\frac{BA_1}{A_1C} \\cdot \\frac{CB_1}{B_1A} = 1\n$$\ni.e., $\\frac{AC_1}{C_1B} = \\frac{B_1A}{B_1C}$. Hence we have $B_1C_1 \\parallel BC$, i.e. $S_{BC_1M} = S_{CB_1M} = 2 S_{AC_1M}$ and we ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0hae | Is there a rectangle, that can be cut into 5 squares, among which there are two same ones, and all the other ones are distinct and differ from the same ones, and the same squares are
Fig. 33
a) the smallest of all;
b) the biggest of all? | [
"a) Let's show the corresponding example (Fig. 32).\n\nb) Consider a square of the smallest size. It can not be on the border of the rectangle (Fig. 33), as well as it can not be inside of the border of one of the other squares. This means, that only the following arrangement of the other four squares around the sm... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) yes; b) no | |
03d4 | Let $n \ge 2$ be a fixed positive integer. The set $M$ consists of $2n^2 - 3n + 2$ positive rational numbers. Prove that there exists an $n$-element subset $A$ of $M$ with the following property: for every $k \in \mathbb{N}$, $2 \le k \le n$, the sum of arbitrary $k$ (not necessarily distinct) elements of $A$ is not an... | [
"Without loss of generality we can consider the elements of $M$ as positive integers, as we can multiply them by the $\\gcd$ of their denominators, which does not affect the problem statement.\n\nThe arithmetic progression $\\{(2n-1)q+n\\}_{q=1}^{\\infty}$ contains infinitely many primes, due to the Dirichlet theor... | Bulgaria | Bulgaria 2022 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other"
] | null | proof only | null | |
06ap | For the various values of the positive integer $n$, determine all positive integers $N$ which are perfect squares and in their decimal representation have $n$ times the digit $2$ and one time the digit $5$. | [
"Answer: $5^2 = 25$ and $25^2 = 225$.\n\n**Proof.** Since a perfect square cannot have its last digit $2$, $N$ must be of the form:\n$$\nN = 22\\dots 225 = 22\\dots 200 + 25 = 100 \\cdot 2 \\cdot 11\\dots 1 + 25 = 100 \\cdot 2 \\cdot \\frac{10^{n-1} - 1}{9} + 25,\n$$\nwhere the digit $2$ there exists $n$ times, $n ... | Greece | 40th Hellenic Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 25 and 225 | |
0fkh | Problem:
En un cuadrilátero convexo se trazan las perpendiculares desde cada vértice a la diagonal que no pasa por él. Demostrar que los cuatro puntos de intersección de cada perpendicular con su correspondiente diagonal forman un cuadrilátero semejante al dado. | [
"Solution:\n\n\n\nVamos a probar que ambos cuadriláteros tienen todos sus ángulos iguales.\nPor construcción $AD A' D'$ es circunscriptible y de ahí que $\\angle C D' A' = \\angle A D A'$ por ser ambos suplementarios del mismo ángulo $\\angle A D' A'$.\nTambién el cuadrilátero $D C C' D'$ e... | Spain | FASE LOCAL DE LA XLIV OME | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hv5 | Problem:
Let $(x, y)$ be a point in the cartesian plane, $x, y > 0$. Find a formula in terms of $x$ and $y$ for the minimal area of a right triangle with hypotenuse passing through $(x, y)$ and legs contained in the $x$ and $y$ axes. | [
"Solution:\n$2 x y$."
] | United States | null | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 2xy | |
0k1c | Problem:
Let $\triangle ABC$ be an acute triangle, with $M$ being the midpoint of $\overline{BC}$, such that $AM = BC$. Let $D$ and $E$ be the intersection of the internal angle bisectors of $\angle AMB$ and $\angle AMC$ with $AB$ and $AC$, respectively. Find the ratio of the area of $\triangle DME$ to the area of $\t... | [
"Solution:\n\nLet $AM = \\ell$, let $DE = d$, and let the midpoint of $\\overline{DE}$ be $F$. Since $\\frac{AD}{AB} = \\frac{AE}{AC} = \\frac{2}{3}$ by the angle bisector theorem, $F$ lies on $\\overline{AM}$ and $\\triangle ADE$ is similar to $\\triangle ABC$. Note that $\\angle DME$ is formed by angle bisectors ... | United States | HMMT November 2018 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 2/9 | |
035z | Problem:
Find all values of the real parameters $a$ and $b$ such that the remainder in the division of the polynomial $x^{4}-3 a x^{3}+a x+b$ by the polynomial $x^{2}-1$ is equal to $(a^{2}+1)x+3 b^{2}$. | [
"Solution:\nIt follows from the condition that there is a polynomial $q(x)$ of degree $2$ such that\n$$\nx^{4}-3 a x^{3}+a x+b = q(x)(x^{2}-1) + (a^{2}+1)x + 3 b^{2}\n$$\nfor every $x$. Setting $x=1$ and $x=-1$ we obtain the system\n$$\n\\left\\lvert\\,\n\\begin{array}{cc}\na^{2}+3 b^{2}+2 a-b=0 \\\\\na^{2}-3 b^{2}... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | a = -1, b = (1 ± sqrt(13)) / 6 | |
0jo6 | Problem:
The fraction $\frac{1}{2015}$ has a unique "(restricted) partial fraction decomposition" of the form
$$
\frac{1}{2015}=\frac{a}{5}+\frac{b}{13}+\frac{c}{31}
$$
where $a, b, c$ are integers with $0 \leq a<5$ and $0 \leq b<13$. Find $a+b$. | [
"Solution:\n\nAnswer: 14 This is equivalent to $1=13 \\cdot 31 a+5 \\cdot 31 b+5 \\cdot 13 c$. Taking modulo $5$ gives $1 \\equiv 13 \\cdot 31 a \\pmod{5}$, and since $13 \\cdot 31 \\equiv 3 \\pmod{5}$, we have $1 \\equiv 3a \\pmod{5}$, so $a \\equiv 2 \\pmod{5}$. Taking modulo $13$ gives $1 \\equiv 5 \\cdot 31 b \... | United States | HMMT February 2015 | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 14 | |
0csz | Дан вписанный четырехугольник $ABCD$. Лучи $AB$ и $DC$ пересекаются в точке $K$. Оказалось, что точки $B$, $D$, а также середины отрезков $AC$ и $KC$ лежат на одной окружности. Какие значения может принимать угол $ADC$? | [
"Обозначим через $N$ и $M$ середины отрезков $KC$ и $AC$ соответственно. Тогда $MN$ — средняя линия в треугольнике $AKC$, поэтому $\\angle BAC = \\angle NMC$. Кроме того, $\\angle BAC = \\angle BDC$, так как четырехугольник $ABCD$ — вписанный.\n\nПусть точки $M$ и $N$ лежат с одной стороны от прямой $BD$. Тогда $M$... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 90 degrees | |
04om | The lengths of all sides of a right-angled triangle are positive integers. If the radius of its incircle is of length $4$, find all possible lengths of its legs. | [
"Let $a$ and $b$ be the lengths of legs, $c$ be the length of hypotenuse, and $r = 4$ be the radius of the incircle in the given triangle.\nExpressing its area in two different ways, we get\n$$\n r \\cdot \\frac{a + b + c}{2} = \\frac{ab}{2},\n$$\n$$\n ab - 4a - 4b = 4c.\n$$\nSince $c^2 = a^2 + b^2$, it follows tha... | Croatia | Croatian Mathematical Society Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | The leg pairs are (9, 40), (10, 24), (12, 16), (16, 12), (24, 10), (40, 9). | |
0kqx | Problem:
Let $ABC$ be a triangle with $AB = 8$, $AC = 12$, and $BC = 5$. Let $M$ be the second intersection of the internal angle bisector of $\angle BAC$ with the circumcircle of $ABC$. Let $\omega$ be the circle centered at $M$ tangent to $AB$ and $AC$. The tangents to $\omega$ from $B$ and $C$, other than $AB$ and $... | [
"Solution:\nRedefine $D$ as the reflection of $A$ across the perpendicular bisector $l$ of $BC$. We prove that $DB$ and $DC$ are both tangent to $\\omega$, and hence the two definitions of $D$ align. Indeed, this follows by symmetry; we have that $\\angle CBM = \\angle CAM = \\angle BAM = \\angle BCM$, so $BM = CM$... | United States | HMMT November 2022 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 16 | |
0ftz | Problem:
Sei $p$ eine ungerade Primzahl. Finde alle natürlichen Zahlen $k$, sodass
$$
\sqrt{k^{2}-p k}
$$
eine positive ganze Zahl ist. | [
"Solution:\nSei $\\sqrt{k^{2}-p k}=m$ mit $m>0$ ganz. Dann gilt $k(k-p)=m^{2}$. Wir unterscheiden nun zwei Fälle.\n\n$p \\mid k$ Dann ist auch $m$ durch $p$ teilbar. Schreibe $k=p \\cdot l$ und $m=p \\cdot n$, dann folgt $l(l-1)=n^{2}$. Wegen $l \\geq 2$ und\n$$\n(l-1)^{2}<l^{2}-l<l^{2}\n$$\nliegt $l(l-1)$ aber zwi... | Switzerland | SMO Finalrunde | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis... | null | proof and answer | k = ((p+1)/2)^2 | |
01hl | Determine all integers $C$ for which there exists a sequence $(a_1, a_2, ...)$ of positive integers satisfying
$$
a_{n+1}^2 = C + (n + 2021)a_n
$$
for all $n \ge 1$. | [
"Clearly for $C = 1$ we have the solution $(a_n)_{n=1}^\\infty = (n + 2019)_{n=1}^\\infty$. Let's prove that this is the only value for $C$ that works.\nAssume $(a_n)_{n=1}^\\infty$ is a solution and let $(b_n)_{n=1}^\\infty = (a_n - n)_{n=1}^\\infty$. We claim that for $n > |C| + 2021^2$:\n(i) If $b_n < 2019$, the... | Baltic Way | Baltic Way 2021 Shortlist | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1 | |
0jr7 | Problem:
For how many pairs of nonzero integers $(c, d)$ with $-2015 \leq c, d \leq 2015$ do the equations $c x = d$ and $d x = c$ both have an integer solution? | [
"Solution:\nWe need both $c / d$ and $d / c$ to be integers, which is equivalent to $|c| = |d|$, or $d = \\pm c$. So there are 4030 ways to pick $c$ and 2 ways to pick $d$, for a total of 8060 pairs."
] | United States | HMMT November 2015 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Divisibility / Factorization"
] | null | final answer only | 8060 | |
04gv | A circle of length $6N$ is divided by $3N$ marked points into $3N$ arcs: $N$ arcs of length $1$, $N$ arcs of length $2$ and $N$ arcs of length $3$.
Prove that among the marked points there exist two which are endpoints of some diameter of the circle. (V. Prasolov, Problems in plane and solid geometry) | [
"Let $3N$ marked points be called *outer points*. We divide each of $N$ arcs of length $2$ into two arcs of equal length using one point, and each of $N$ arcs of length $3$ into three arcs of equal length using two points. Let those $3N$ new points be called *inner points*. There are $6N$ outer and inner points in ... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0jst | Problem:
Let $ABC$ be a triangle with $AB = 13$, $BC = 14$, $CA = 15$. Let $H$ be the orthocenter of $ABC$. Find the distance between the circumcenters of triangles $AHB$ and $AHC$. | [
"Solution:\n\nLet $H_{B}$ be the reflection of $H$ over $AC$ and let $H_{C}$ be the reflection of $H$ over $AB$. The reflections of $H$ over $AB$, $AC$ lie on the circumcircle of triangle $ABC$. Since the circumcenters of triangles $AH_{C}B$, $AH_{B}C$ are both $O$, the circumcenters of $AHB$, $AHC$ are reflections... | United States | HMMT February 2016 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 14 | |
0b5z | Let $ABCD$ be a parallelogram of area $1$, and $AD \le AB \le BD < AC$. Prove that the parallelogram can be covered by a rectangle of area $\sqrt{3}$. | [] | Romania | Shortlisted Problems for the Romanian NMO | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
0ki8 | Each of 20 balls is tossed independently and at random into one of 5 bins. Let $p$ be the probability that some bin ends up with 3 balls, another with 5 balls, and the other three with 4 balls each. Let $q$ be the probability that every bin ends up with 4 balls. What is $\frac{p}{q}$?
(A) 1 (B) 4 (C) 8 (D) 12 (E) 16 | [
"Let us first compute $q$.\n\n$q$ is the probability that each bin gets exactly 4 balls when 20 balls are tossed into 5 bins.\n\nThe total number of ways to assign 20 balls to 5 bins is $5^{20}$.\n\nThe number of ways to assign 4 balls to each bin is:\n- First, choose which 4 balls go into bin 1: $\\binom{20}{4}$\n... | United States | AMC 12 A | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | MCQ | E | |
07k0 | For every natural number $n > 1$, find all subsets $\{r_0, r_1, \dots, r_k\}$ of distinct positive integers such that
$$
n^n + n^{n-1} + \dots + n + 1 \mid n^{r_k} + \dots + n^{r_1} + n^{r_0}
$$ | [
"First, note that since $S_n = n^n + n^{n-1} + \\dots + 1 = \\frac{n^{n+1}-1}{n-1}$, it divides $n^{n+1} - 1$. If $a \\equiv b \\pmod{n+1}$, then $n^a \\equiv n^b \\pmod{n^{n+1}-1}$, and thus $n^a \\equiv n^b \\pmod{S_n}$. Also, since $n$ and $S_n = n^n + n^{n-1} + \\dots + 1$ are coprime, we can, after a suitable ... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All and only those sets with exactly n+1 elements whose exponents form a complete residue system modulo n+1; equivalently, the residues {r_i mod (n+1)} are a permutation of {0, 1, ..., n}. | |
01bn | Triangle $ABC$ is given. Let $M$ be a midpoint of segment $AB$ and $T$ be a midpoint of arc $BC$ not containing $A$ of circumcircle of $ABC$. Let $K$ be a point inside triangle $ABC$ such that $MATK$ is an isosceles trapezoid with $AT||MK$. Show that $AK = KC$. | [
"Let $TK$ intersect the circumcircle of $ABC$ in points $T$, $S$. Then $\\angle ABS = \\angle ATS = \\angle BAT$ so $ASBT$ is a trapezoid. So $MK \\parallel AT \\parallel SB$ and $M$ is a midpoint of $AB$ thus $K$ is a midpoint of $TS$. But $\\angle TAC = \\angle BAT = \\angle ATS$ so $ACTS$ is an inscribed trapezo... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fd3 | Problem:
Dos semirrectas tienen su común origen en el punto $O$. Se considera una circunferencia $C_{1}$ tangente a ambas semirrectas, cuyo centro está situado a distancia $d_{1}$ de $O$, y cuyo radio es $r_{1}$. Se construyen sucesivamente las circunferencias $C_{n}$, de modo que $C_{n}$ es tangente a las semirrectas... | [
"Solution:\n\nEs claro de la figura que, por el Teorema de Thales, $\\frac{r_{n}}{d_{n}}=\\frac{r_{1}}{d_{1}}$ para todo $n$. Llamaremos\n\na este valor $\\alpha$. Además, se tiene que:\n$$\n\\begin{gathered}\n\\frac{r_{n}}{r_{n+1}}=\\frac{d_{n}}{d_{n+1}}=\\frac{d_{n+1}+r_{n+1}+r_{n}}{d_{n+... | Spain | XLVII Olimpiada Matemática Española Primera Fase | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | π r1 (d1 + r1)^2 / (4 d1) | |
07up | Show that $\frac{\sqrt{1+t} - \sqrt{|1-t|}}{1+\sqrt{t}} \le \frac{1}{\sqrt{2}}$, for all $t \ge 0$. | [
"$$\nf(t) = \\frac{\\sqrt{1+t} - \\sqrt{|1-t|}}{1+\\sqrt{t}}\n$$\nand observe that $f(1/t) = f(t)$, and that $f(0) = 0, f(1) = 1/\\sqrt{2}$. Therefore, it suffices to prove that $2(f(s^2))^2 \\le 1$ if $0 \\le s \\le 1$, i.e, that\n$$\n2(1 + s^2 + 1 - s^2 - 2\\sqrt{1 - s^4}) \\le (1 + s)^2.\n$$\nThis is equivalent ... | Ireland | IRL_ABooklet | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0e5m | Problem:
Za dani vrednosti $a=\sqrt{3}$ in $b=\sqrt{2}$ izračunaj natančno vrednost izraza
$$
\sqrt[4]{a^{-1} b^{3} \cdot \sqrt{6}} \cdot \sqrt[3]{\frac{6 a^{4} b^{-2}}{\sqrt{2}}}
$$ | [
"Solution:\n\nVstavimo vrednosti $a=\\sqrt{3}$ in $b=\\sqrt{2}$ v dani izraz. Dobimo $\\sqrt[4]{(\\sqrt{3})^{-1} \\cdot (\\sqrt{2})^{3} \\cdot \\sqrt{6}} \\cdot \\sqrt[3]{\\frac{6 \\cdot (\\sqrt{3})^{4} \\cdot (\\sqrt{2})^{-2}}{\\sqrt{2}}}$.\n\nPoenostavimo prvi faktor:\n$$(\\sqrt{3})^{-1} = 3^{-1/2}$$\n$$(\\sqrt{2... | Slovenia | Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Algebra > Intermediate Algebra > Other"
] | null | final answer only | 3 * 2^(1/3) | |
02xo | Problem:
Bronquinha consegue cortar a grama de seu quintal em 3 horas, mas se ele tomar suco de frutas Gummy, ele corta em 2 horas. Em determinado dia, Bronquinha começou a cortar a grama às 10 horas e, em certo momento, tomou o suco de frutas Gummy, terminando de cortar a grama às 12 horas e 30 minutos. Que horas Bro... | [
"Solution:\n\nSeja $A$ a área de grama a ser cortada. Assim, sem tomar o suco, a velocidade com a qual Bronquinha corta a grama é $\\frac{A}{3}$, enquanto que tomando o suco a velocidade é $\\frac{A}{2}$. Seja $t$ o tempo que Bronquinha corta a grama sem tomar o suco, na situação proposta, temos:\n$$\n\\begin{align... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 11:30 | |
06hr | Let $A$, $B$, $C$ be points on the same plane with $\angle ACB = 120^\circ$. There is a sequence of circles $\omega_0, \omega_1, \omega_2, \ldots$ on the same plane (with corresponding radii $r_0, r_1, r_2, \ldots$, where $r_0 > r_1 > r_2 > \ldots$) such that each circle is tangent to both segments $CA$ and $CB$. Furth... | [] | Hong Kong | HONG KONG PRELIMINARY SELECTION CONTEST | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English; Chinese | proof and answer | (3 + 2√3)/2 | |
06li | Let $ABCD$ be a cyclic quadrilateral inscribed in a circle $\Gamma$ such that $AB = AD$. Let $E$ be a point on the segment $CD$ such that $BC = DE$. The line $AE$ intersects $\Gamma$ again at $F$. The chords $AC$ and $BF$ meet at $M$. Let $P$ be the symmetric point of $C$ about $M$. Prove that $PE$ and $BF$ are paralle... | [
"Since $\\angle EFM = \\angle AFB = \\angle DCA = \\angle ECM$, the points $M$, $C$, $F$, $E$ are concyclic. Using the concyclic points, we obtain\n$$\n\\angle QME = \\angle FME = \\angle FCE = \\angle FCD = \\angle FBD = \\angle QBD.\n$$\nThis implies $EM \\parallel DB$, and hence $\\frac{DE}{EQ} = \\frac{BM}{MQ}$... | Hong Kong | CHKMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03kn | Problem:
If $\alpha, \beta, \gamma$ are the roots of $x^{3}-x-1=0$, compute
$$
\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma}
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | -7 | |
08ow | Problem:
The point $P$ is outside of the circle $\Omega$. Two tangent lines, passing from the point $P$, touch the circle $\Omega$ at the points $A$ and $B$. The median $A M$, $M \in (B P)$, intersects the circle $\Omega$ at the point $C$ and the line $P C$ intersects again the circle $\Omega$ at the point $D$. Prove ... | [
"Solution:\n\nSince $\\angle B A C = \\angle B A M = \\angle M B C$, we have $\\triangle M A B \\cong \\triangle M B C$.\n\n\n\nWe obtain $\\frac{M A}{M B} = \\frac{M B}{M C} = \\frac{A B}{B C}$. The equality $\\quad M B = M P$ implies $\\frac{M A}{M P} = \\frac{M P}{M C}$ and $\\angle P M ... | JBMO | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0if5 | Problem:
A true-false test has ten questions. If you answer five questions "true" and five "false," your score is guaranteed to be at least four. How many answer keys are there for which this is true? | [
"Solution:\nSuppose that either nine or ten of the questions have the same answer. Then no matter which five questions we pick to have this answer, we will be right at least four times. Conversely, suppose that there are at least two questions with each answer; we will show that we can get a score less than four. B... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 22 | |
0kde | Problem:
Let $A$ be the area of the largest semicircle that can be inscribed in a quarter-circle of radius $1$. Compute $\frac{120A}{\pi}$. | [
"Solution:\n\n\nThe optimal configuration is when the two ends $X$ and $Y$ of the semicircle lie on the arc of the quarter circle. Let $O$ and $P$ be the centers of the quarter circle and semicircle, respectively. Also, let $M$ and $N$ be the points where the semicircle is tangent to the ra... | United States | HMMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 20 | |
030p | Problem:
Fie $a \geqslant b \geqslant c \geqslant d$ numere reale cu proprietatea că
$$
(a-b)(b-c)(c-d)(d-a) = -3
$$
a) Dacă $a+b+c+d=6$, demonstrați că $d < 0,36$
b) Dacă $a^{2}+b^{2}+c^{2}+d^{2}=14$, arătați că $(a+c)(b+d) \leqslant 8$. Precizați cazurile de egalitate. | [] | Brazil | Al treilea baraj de selecție pentru OBMJ | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | a) d < 0.36.
b) (a + c)(b + d) ≤ 8, with equality when (a, b, c, d) = (3, 2, 1, 0) or (0, −1, −2, −3). | |
0dca | Let pairwise different positive integers $a$, $b$, $c$ with $\operatorname{gcd}(a, b, c) = 1$ are such that
$$
a \mid (b - c)^2, \quad b \mid (c - a)^2, \quad c \mid (a - b)^2.
$$
Prove that there is no non-degenerate triangle with side lengths $a$, $b$ and $c$. | [
"First, we will show that $a$, $b$, $c$ are pairwise coprime. Denote $d = \\operatorname{gcd}(b, c)$ and suppose that $d > 1$. Take $p$ as a prime divisor of $d$. We have\n$$\n\\left\\{\\begin{array}{l}\n p \\mid b \\\\\n p \\mid c\n\\end{array}\\right. \\Rightarrow \\left\\{\\begin{array}{l}\n p \\mid (c - a)^2 \\... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geo... | English | proof only | null | |
04gi | What is the smallest, and what is the largest value that the function $f: \mathbb{R} \to \mathbb{R}$, given by $f(x) = \cos^2 x + \sin x$, can take? (Ilko Brnetić) | [] | Croatia | Mathematica competitions in Croatia | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Smallest value: -1; Largest value: 5/4 | |
0bb5 | Let $f : [0, 1] \to (0, \infty)$ be a continuous function. For $n \in \mathbb{N}$, $n \ge 2$, consider $0 = t_0 < t_1 < \dots < t_n = 1$, such that
$$
\int_{t_0}^{t_1} f(t) \, dt = \int_{t_1}^{t_2} f(t) \, dt = \dots = \int_{t_{n-1}}^{t_n} f(t) \, dt.
$$
Compute
$$
\lim_{n \to \infty} \frac{n}{\frac{1}{f(t_1)} + \frac{... | [
"Denote by $(x_n)_{n \\ge 2}$ the given sequence. We have\n$$\n\\begin{aligned}\n\\frac{1}{x_n} &= \\frac{1}{n} \\sum_{k=1}^{n} \\frac{1}{f(t_k)} = \\frac{1}{n} \\sum_{k=1}^{n} \\frac{1}{f(F^{-1}(kI/n))} = \\frac{1}{n} \\sum_{k=1}^{n} \\frac{1}{F'(F^{-1}(kI/n))} \\\\\n&= \\frac{1}{n} \\sum_{k=1}^{n} (F^{-1})'(kI/n)... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Integral Calculus > Applications"
] | null | proof and answer | ∫_0^1 f(t) dt | |
0141 | Problem:
Is it possible to find $2005$ different positive square numbers such that their sum is also a square number? | [
"Solution:\n\nStart with a simple Pythagorean identity such as $3^{2} + 4^{2} = 5^{2}$. Multiply it by $5^{2}$\n$$\n3^{2} \\cdot 5^{2} + 4^{2} \\cdot 5^{2} = 5^{2} \\cdot 5^{2}\n$$\nand insert the identity for the first\n$$\n3^{2} \\cdot (3^{2} + 4^{2}) + 4^{2} \\cdot 5^{2} = 5^{2} \\cdot 5^{2}\n$$\nwhich gives\n$$... | Baltic Way | Baltic Way 2005 | [
"Number Theory > Diophantine Equations > Pythagorean triples"
] | null | proof only | null | |
0k71 | Problem:
In rectangle $A B C D$, points $E$ and $F$ lie on sides $A B$ and $C D$ respectively such that both $A F$ and $C E$ are perpendicular to diagonal $B D$. Given that $B F$ and $D E$ separate $A B C D$ into three polygons with equal area, and that $E F=1$, find the length of $B D$. | [
"Solution:\n\nObserve that $A E C F$ is a parallelogram. The equal area condition gives that $B E = D F = \\frac{1}{3} A B$. Let $C E \\cap B D = X$, then $\\frac{E X}{C X} = \\frac{B E}{C D} = \\frac{1}{3}$, so that $B X^2 = E X \\cdot C X = 3 E X^2 \\Rightarrow B X = \\sqrt{3} E X \\Rightarrow \\angle E B X = 30^... | United States | HMMT February 2019 February 16, 2019 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | √3 | |
094u | Problem:
A finite sequence $x_{1}, x_{2}, \ldots, x_{r}$ of positive integers is a palindrome if $x_{i}=x_{r+1-i}$ for all integers $1 \leq i \leq r$.
Let $a_{1}, a_{2}, \ldots$ be an infinite sequence of positive integers. For a positive integer $j \geq 2$, denote by $a[j]$ the finite subsequence $a_{1}, a_{2}, \ldot... | [
"Solution:\n\nDefine a break point to be a positive integer $k$ such that $a[k]$ is a palindrome. Let $c_{1}<c_{2}<\\ldots$ be a strictly increasing sequence of all break points. Then $c_{n+2} \\leq c_{n+1}+c_{n}$ also holds whenever $c_{n+2}>b_{2}$. Namely, if $b_{j-1}<c_{n+2} \\leq b_{j}$, then $c_{n+1} \\geq b_{... | Middle European Mathematical Olympiad (MEMO) | MEMO Szeged | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0aoe | Problem:
Solve for $c$ in the following system of equations:
$$
\begin{aligned}
& 16^{a+b} = \frac{\sqrt{2}}{2} \\
& 16^{b+c} = 4 \\
& 16^{a+c} = 2 \sqrt{2}
\end{aligned}
$$
(a) 0
(b) $\frac{3}{8}$
(c) $\frac{5}{8}$
(d) $\frac{1}{2}$ | [
"Solution:\n\nLet us write $16 = 2^4$.\n\nSo,\n$16^{a+b} = (2^4)^{a+b} = 2^{4(a+b)}$\n$16^{b+c} = 2^{4(b+c)}$\n$16^{a+c} = 2^{4(a+c)}$\n\nNow, rewrite the equations:\n\n1. $2^{4(a+b)} = \\frac{\\sqrt{2}}{2}$\n2. $2^{4(b+c)} = 4$\n3. $2^{4(a+c)} = 2\\sqrt{2}$\n\nExpress the right sides as powers of $2$:\n\n$\\frac{\... | Philippines | Qualifying Round | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | MCQ | d | |
0922 | Problem:
There are $n \geq 2$ houses on the northern side of a street. Going from the west to the east, the houses are numbered from $1$ to $n$. The number of each house is shown on a plate. One day the inhabitants of the street make fun of the postman by shuffling their number plates in the following way: for each pa... | [
"Solution:\n\nLet $f(n)$ denote the answer. We shall prove by induction that $f(n) = 2^{n-2}$. For $n=2$, the answer is clearly $2^{2-2} = 1$. We also define $f(1) = 1$. Now we consider arbitrary $n > 2$.\n\nLet $H_{i}$ denote the house with number $i$ at the start of the day, and let $(i \\rightleftarrows i+1)$ de... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2^{n-2} | |
0j45 | Problem:
Alice and Barbara play a game on a blackboard. At the start, zero is written on the board. Alice goes first, and the players alternate turns. On her turn, each player replaces $x$ – the number written on the board – with any real number $y$, subject to the constraint that $0 < y - x < 1$.
a. If the first pla... | [
"Solution:\n\nEach turn of the game is equivalent to adding a number between $0$ and $1$ to the number on the board.\n\na. Barbara has the winning strategy: whenever Alice adds $z$ to the number on the board, Barbara adds $1-z$. After Barbara's $i$th turn, the number on the board will be $i$. Therefore, after Barba... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a: Barbara; b: Alice | |
0krx | Problem:
Let $ABC$ be a triangle with centroid $G$, and let $E$ and $F$ be points on side $BC$ such that $BE = EF = FC$. Points $X$ and $Y$ lie on lines $AB$ and $AC$, respectively, so that $X$, $Y$, and $G$ are not collinear. If the line through $E$ parallel to $XG$ and the line through $F$ parallel to $YG$ intersect... | [
"Solution:\n\nLet $CG$ intersect $AB$ at $N$. Then $N$ is the midpoint of $AB$ and it is known that $\\frac{CG}{AB} = 2 = \\frac{CE}{EB}$, so $EG \\parallel AB$. Moreover, since $FE = EB$, we have $[EFG] = [EXG]$. Similarly, $[EFG] = [FYG]$. Now we have $[PXG] = [EXG] = [EFG] = [FYG] = [PYG]$, so $PG$ bisects $XY$,... | United States | HMMT February 2022 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
02hu | Problem:
Se eu der duas barras de chocolate para Tião, ele me empresta sua bicicleta por 3 horas. Se eu lhe der 12 bombons, ele me empresta a bicicleta por 2 horas. Amanhã, eu lhe darei uma barra de chocolate e 3 bombons. Por quantas horas ele me emprestará a bicicleta?
(A) $1 / 2$
(B) 1
(C) 2
(D) 3
(E) 4 | [
"Solution:\n\n$\\left\\{\\begin{array}{l}2 \\text{ barras } \\xrightarrow{\\text{ corresponde }} 3 \\text{ horas } \\\\ 12 \\text{ bombons } \\xrightarrow{\\text{ corresponde }} 2 \\text{ horas }\\end{array}\\right.$\n\nLogo,\n$\\left\\{\\begin{array}{l}1 \\text{ barra } \\xrightarrow{\\text{ corresponde }} 1,5 \\t... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | C | |
0i6u | Problem:
Our third and final item comes to us from Germany, I mean Geometry. It is known that a regular $n$-gon can be constructed with straightedge and compass if $n$ is a prime that is $1$ plus a power of $2$. It is also possible to construct a $2n$-gon whenever an $n$-gon is constructible, or a $p_{1} p_{2} \cdots p... | [
"Solution:\nThe known primes of this form (Fermat primes) are $3$, $5$, $17$, $257$, and $65537$, and the result is due to Gauss (German). If there are other such primes (unknown), then they are much bigger than $10^{10}$. So for each product of these primes, we can divide $4.3 \\cdot 10^{9}$ by that number and tak... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 4294967296 | |
0lbx | Let $n$ be a positive integer. Given $2n$ positive real numbers $a_1, a_2, a_3, \dots, a_n$ and $b_1, b_2, b_3, \dots, b_n$ satisfying the following condition:
$$
a_1 > a_2 > a_3 > \dots > a_n \text{ and } b_1 > b_2 > b_3 > \dots > b_n.
$$
A partition of these $2n$ numbers into $n$ disjoint pairs $(a_i, b_j), 1 \le i \... | [
"a. We call a pair $(a, b)$ regular if the difference between $a$ and $b$ is less than $2013$. Suppose that in the given regular partition, $a_1$ is paired with $b_i$, $1 \\le i \\le n$ and $b_1$ is paired with $a_j$, $1 \\le j \\le n$.\nWe have $-2013 < a_1 - b_i < 2013$ and $-2013 < a_j - b_1 < 2013$.\nTherefore,... | Vietnam | Vietnamese Mathematical Competitions | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2^{\lfloor (n+1)/2 \rfloor} | |
0j2s | Problem:
An ant starts out at $(0,0)$. Each second, if it is currently at the square $(x, y)$, it can move to $(x-1, y-1)$, $(x-1, y+1)$, $(x+1, y-1)$, or $(x+1, y+1)$. In how many ways can it end up at $(2010,2010)$ after $4020$ seconds? | [
"Solution:\n\nAnswer: $\\binom{4020}{1005}^{2}$\n\nNote that each of the coordinates either increases or decreases the $x$ and $y$ coordinates by $1$. In order to reach $2010$ after $4020$ steps, each of the coordinates must be increased $3015$ times and decreased $1005$ times. A permutation of $3015$ plusses and $... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | (4020 choose 1005)^2 | |
0iz6 | Problem:
Let $S=\{1,2,3,4,5,6,7,8,9,10\}$. How many (potentially empty) subsets $T$ of $S$ are there such that, for all $x$, if $x$ is in $T$ and $2x$ is in $S$ then $2x$ is also in $T$? | [
"Solution:\n\nWe partition the elements of $S$ into the following subsets: $\\{1,2,4,8\\}$, $\\{3,6\\}$, $\\{5,10\\}$, $\\{7\\}$, $\\{9\\}$.\n\nConsider the first subset, $\\{1,2,4,8\\}$. Say $2$ is an element of $T$. Because $2 \\cdot 2 = 4$ is in $S$, $4$ must also be in $T$. Furthermore, since $4 \\cdot 2 = 8$ i... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 180 | |
0k5m | Problem:
Find the sum of all real solutions to $x^{2} + \cos x = 2019$. | [
"Solution:\nThe left-hand side is an even function, hence for each $x$ that solves the equation, $-x$ will also be a solution. Pairing the solutions up in this way, we get that the sum must be $0$."
] | United States | HMMT February 2019 | [
"Precalculus > Functions",
"Precalculus > Trigonometric functions"
] | null | proof and answer | 0 | |
07b1 | In each unit cell of a $n \times m$ table an integer number is written. An **diagonal** is formed by those cells of the table for which the difference between their column number and their row number is a constant number. We want to make all the numbers in the table $0$ during a finite number of steps. In each step, we... | [
"Let $A_{i,j}$ denote the number in the cell located in the $i$-th row and $j$-th column of the table, and $S(i, j)$ the $3 \\times 3$ subtable whose rightmost upper number is $A_{i,j}$.\n\nFirst, assume that $m, n \\ge 3$ (If $m \\le 2$ or $n \\le 2$, there is no $3 \\times 3$ subtable). For $S(i, j)$, consider th... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0hf6 | There are four countries, each consists of several cities. The cities of any two countries are connected by at least $\frac{5}{6}$ of a number of all possible roads between these two countries. Prove that it is possible to choose one city from each country so that any two of them are connected. Any pair of cities can b... | [
"Let us denote countries as $X, Y, Z, T$. We randomly choose one city $x, y, z, t$ from each country respectively. In case there are all 6 roads between them, then the statement has been proved.\n\nIn case there is quadruple that is connected via less than 5 roads, then there has to exist a quadruple that is connec... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
02i7 | Problem:
Determine quais afirmações são verdadeiras:
(A) $|-108|>100$
(B) $|5-13|=|5|-|13|$
(C) $|2-9|=9-2$
(D) $|a^{2}+5|=a^{2}+5$
(E) $|-6 a|=6|a|$ | [
"Solution:\n\n(A) $|-108|=108>100$, verdadeira\n\n(B) $|5-13|=|-8|=8$ e $|5|-|13|=5-13=-8$, falsa.\n\n(C) $|2-9|=-(2-9)=9-2$ porque $2-9<0$, verdadeira.\n\n(D) $|a^{2}+5|=a^{2}+5$ porque $a^{2}+5>0$ para qualquer valor de $a$, verdadeira.\n\n$|x|=\\begin{cases}x \\text{ se } x \\geq 0 \\\\ -x \\text{ se } x<0\\end{... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | MCQ | A, C, D, E | |
0k3b | Problem:
The integers $1,2,3,4,5,6,7,8,9,10$ are written on a blackboard. Each day, a teacher chooses one of the integers uniformly at random and decreases it by $1$. Let $X$ be the expected value of the number of days which elapse before there are no longer positive integers on the board. Estimate $X$.
An estimate o... | [
"Solution:\n\nAnswer: $120.75280458176904$\n2967376198031036029043518754502337892568216567911415193776896659976433643147319195286268925688566342694"
] | United States | HMMT February 2018 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | 120.75280458176904 | |
0ilx | Problem:
We are given some similar triangles. Their areas are $1^{2}, 3^{2}, 5^{2}, \ldots$, and $49^{2}$. If the smallest triangle has a perimeter of $4$, what is the sum of all the triangles' perimeters? | [
"Solution:\n\nBecause the triangles are all similar, they all have the same ratio of perimeter squared to area, or, equivalently, the same ratio of perimeter to the square root of area. Because the latter ratio is $4$ for the smallest triangle, it is $4$ for all the triangles, and thus their perimeters are $4 \\cdo... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 2500 | |
06sq | Determine all functions $f: \mathbb{Q} \longrightarrow \mathbb{Z}$ satisfying
$$
f\left(\frac{f(x)+a}{b}\right)=f\left(\frac{x+a}{b}\right)
$$
for all $x \in \mathbb{Q}$, $a \in \mathbb{Z}$, and $b \in \mathbb{Z}_{>0}$. (Here, $\mathbb{Z}_{>0}$ denotes the set of positive integers.) | [
"Answer. There are three kinds of such functions, which are: all constant functions, the floor function, and the ceiling function.\n\nSolution 1.\nI. We start by verifying that these functions do indeed satisfy (1). This is clear for all constant functions. Now consider any triple $(x, a, b) \\in \\mathbb{Q} \\time... | IMO | International Mathematical Olympiad Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof and answer | All constant functions, the floor function, and the ceiling function. | |
0k84 | Problem:
Let $ABC$ be a triangle with $AB = 6$, $AC = 7$, $BC = 8$. Let $I$ be the incenter of $ABC$. Points $Z$ and $Y$ lie on the interior of segments $AB$ and $AC$ respectively such that $YZ$ is tangent to the incircle. Given point $P$ such that
$$
\angle ZPC = \angle YPB = 90^\circ
$$
find the length of $IP$. | [
"Solution:\nLet $PU$, $PV$ be tangents from $P$ to the incircle. We will invoke the dual of the Desargues Involution Theorem, which states the following:\nGiven a point $P$ in the plane and four lines $\\ell_1, \\ell_2, \\ell_3, \\ell_4$, consider the set of conics tangent to all four lines. Then we define a functi... | United States | HMMT February 2019 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadril... | null | proof and answer | √30/2 | |
0h2r | Let $C$, $E$ and $A$ be points on the sides $XY$, $YZ$ and $ZX$ of the triangle $XYZ$ respectively. Points $B$, $D$ and $F$ are chosen on the segments $AX$, $CY$ and $EZ$ respectively, so that $BC \perp AD$, $DE \perp CF$, $AF \perp BE$. Can it happen that the lines $XF$, $YB$ and $ZD$ are concurrent? | [
"**Answer:** No, it cannot.\n\nSuppose, it is possible (fig. 27). Let $O$ be the point of intersection of the lines $XF$, $YB$ and $ZD$. Then by the Ceva's theorem, we have\n$$\n\\frac{YD}{DX} \\cdot \\frac{XB}{BZ} \\cdot \\frac{ZF}{FY} = 1.\n$$\n\n\\begin{align*}\n\\frac{YD}{DX} &= \\frac{YD}{CD} \\cdot \\frac{CD}... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | No | |
0i9t | Problem:
For which $n$ is it possible to fill the $n$ by $n$ table with $0$'s, $1$'s and $2$'s so that the sums of numbers in rows and columns take all different values from $1$ to $2n$? | [
"Solution:\nFor odd $n$ it is impossible to create a table like that. Indeed, in such a table the sum of all column sums and row sums would be $1+2+\\ldots+2n = n(2n+1)$, and so would be odd. But it would also be twice the sum of all the numbers in the table (each number counted twice - once in the column sum, once... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | n is even | |
0dvc | Problem:
Trije Butalci, Bingo, Bunko in Balko, so oropali banko in odnesli 22 vreč z bankovci. Postavili so jih v vrsto tako, da je bilo v prvi vreči najmanj denarja, v vsaki naslednji pa en šop bankovcev več kot v tisti pred njo. Šef Bingo je ukradene vreče denarja delil po načelu: prva meni (Bingu), druga tebi (Bunk... | [
"Solution:\n\nŠopi šefa: $x + (x + 2) + \\cdots + (x + 20) = \\frac{6600000}{k}$\n\nŠopi Bunka in Balka: $(x + 1) + (x + 3) + \\cdots + (x + 21) = \\frac{6710000}{k}$\n\nUreditev obeh enačb:\n$$\n\\begin{aligned}\n& 11(x + 10) = \\frac{6600000}{k} \\quad \\text{in} \\\\\n& 11(x + 11) = \\frac{6710000}{k}\n\\end{ali... | Slovenia | 2. matematično tekmovanje dijakov srednjih tehniških in strokovnih šol | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 500000 SIT | |
098b | Problem:
Fie $ABC$ un triunghi ascuțitunghic, iar $H$ un punct din interiorul triunghiului, astfel încât $AB^{2} + CH^{2} = BC^{2} + AH^{2} = AC^{2} + BH^{2}$. Demonstrați că $H$ este ortocentrul triunghiului $ABC$. | [
"Solution:\n\n1. Fie $CC_{1} \\perp AB$, $C_{1} \\in AB$ și $HH_{1} \\perp AB$, $H_{1} \\in AB$.\n\n2. $AC^{2} - AC_{1}^{2} = BC^{2} - BC_{1}^{2}$,\n\n(1),\n$$\nAH^{2} - AH_{1}^{2} = BH^{2} - BH_{1}^{2}\n$$\n\n3. Din ipoteză obținem $AC^{2} - BC^{2} = AH^{2} - BH^{2}$, din relația (1) obținem $AC^{2} - BC^{2} = AC_... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
016c | Show that for each integer $n$,
$$
\frac{1}{1+n}\binom{4n+1}{2n}
$$
is an integer. | [
"It is easy to see that\n$$\n\\frac{1}{1+n}\\binom{4n+1}{2n} = \\frac{(4n)!(4n+1)}{((2n)!)^2(n+1)(2n+1)} \\\\ = \\frac{(4n)! \\cdot 2(4n+1)}{(2n)!(2n-2)!(2n-1)(2n)(2n+1)(2n+2)} \\quad (1)\n$$\n$$\n\\frac{1}{(2n-1) \\cdot (2n)} - \\frac{1}{(2n+1)(2n+2)} = \\frac{4n^2 + 6n + 2 - 4n^2 + 2n}{(2n-1)(2n)(2n+1)(2n+2)} \\\... | Baltic Way | Baltic Way SHL | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
0265 | Problem:
2. Potências de 2 - Calcule:
a) $1678^{2}-1677^{2}$
b) $1001^{2}+1000^{2}$
c) $19999^{2}$
d) $2001^{2}+2002^{2}+2003^{2}$ | [
"Solution:\n\n2. Potências de 2 - Fatorando temos:\n$a)$ $1678^{2}-1677^{2} = (1678+1677)(1678-1677) = 3355$.\n\nb) Como $(a+b)^{2} = a^{2} + 2ab + b^{2}$, temos:\n$$\n\\begin{aligned}\n1001^{2} + 1000^{2} & = (1000+1)^{2} + 1000^{2} = 1000^{2} + 2000 + 1 + 1000^{2} = \\\\\n& = 2 \\times 1000^{2} + 2001 = 2002001\n... | Brazil | null | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | a) 3355; b) 2002001; c) 399960001; d) 12024014 | |
0f6j | Problem:
A regular hexagon is divided into 24 equilateral triangles by lines parallel to its sides. 19 different numbers are assigned to the 19 vertices. Show that at least 7 of the 24 triangles have the property that the numbers assigned to its vertices increase counterclockwise. | [] | Soviet Union | 19th ASU | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Miscellaneous"
] | null | proof only | null | |
0dcz | Let $d$ be a positive divisor of a positive integer $m$ and $(a_{l}),(b_{l})$ two arithmetic sequences of positive integers. It is given that $\operatorname{gcd}(a_{i}, b_{j})=1$ and $\operatorname{gcd}(a_{k}, b_{n})= m$ for some positive integers $i, j, k$, and $n$. Prove that there exist positive integers $t$ and $s$... | [] | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | English | proof only | null | |
03lj | Problem:
Consider an equilateral triangle of side length $n$, which is divided into unit triangles, as shown. Let $f(n)$ be the number of paths from the triangle in the top row to the middle triangle in the bottom row, such that adjacent triangles in our path share a common edge and the path never travels up (from a l... | [
"Solution:\n\nWe shall show that $f(n) = (n-1)!$.\n\nLabel the horizontal line segments in the triangle $l_{1}, l_{2}, \\ldots$ as in the diagram below. Since the path goes from the top triangle to a triangle in the bottom row and never travels up, the path must cross each of $l_{1}, l_{2}, \\ldots, l_{n-1}$ exactl... | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2004! | |
099z | a, b, c are positive and $abc = 1$. Prove that
$$
\sqrt{\frac{a}{a+8}} + \sqrt{\frac{b}{b+8}} + \sqrt{\frac{c}{c+8}} \ge 1.
$$
(proposed by B. Ganbileg and U. Batzorig, inspired by Algebraic inequality book) | [
"First, we will prove the following inequality. If $a, b, c$ are positive real numbers then\n$$\n\\frac{a}{\\sqrt{a^2 + 8bc}} + \\frac{b}{\\sqrt{b^2 + 8ca}} + \\frac{c}{\\sqrt{c^2 + 8ab}} \\ge 1. \\quad (1)\n$$\nIndeed, we need to find $\\lambda \\in \\mathbb{R}$ such that\n$$\n\\frac{a}{\\sqrt{a^2 + bc}} \\ge \\fr... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof only | null | |
04f0 | In some country there are three cities $A$, $B$ and $C$. Between each of the two cities there are several roads (at least one) and all the roads are two-way. Apart from the direct road links between two cities, there are also indirect road links. An indirect road link between $X$ and $Y$ consists of a road that connect... | [
"Let $a$, $b$, and $c$ be the number of direct roads between $A$ and $B$, $B$ and $C$, and $A$ and $C$, respectively. Let $x$, $y$, and $z$ be the number of roads from $A$ to $B$, $B$ to $C$, and $A$ to $C$, respectively (so $a = x$, $b = y$, $c = z$).\n\nThe total number of road links between $A$ and $B$ is the su... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 23 or 97 |
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